id stringlengths 1 6 | url stringlengths 16 1.82k | content stringlengths 37 9.64M |
|---|---|---|
190700 | https://www.mathsisfun.com/geometry/steradian.html | Geometry Index
Show Ads
Hide Ads | About Ads
We may use Cookies
OK
Steradian
A steradian is used to measure "solid angles"
A steradian is related to the surface area of a sphere
in the same way a radian is related to the circumference of a circle:
| | | |
---
| A Radian "cuts out" a length of a circle's circumference equal to the radius | | |
| A Steradian "cuts out" an area of a sphere equal to (radius)2 | | |
The SI Unit abbreviation is sr
The name steradian is made up from the Greek stereos for "solid" and radian.
Sphere vs Steradian
The surface area of a sphere is 4πr2
The surface area of a steradian is r2
So a full sphere has a surface area of 4π steradians which is about 12.57 steradians. That means one steradian covers roughly 8% of the sphere.
And because we are measuring angles, it doesn't matter what size the sphere is it will always measure 4π steradians.
Example: The "unit sphere":
has a radius of 1
has a surface area of 4π,
a steradian "cuts out" an area of 1.
Radiant Intensity
Radiant intensity (how brightly something shines) can be measured in watts per steradian (W/sr).
Example: You measure the light coming from a powerful globe. You hold a 50 mm × 50 mm sensor 2 m away and it measures 0.1 Watts. What is the radiant intensity in W/sr (Watts per steradian)?
At 2 meters, one steradian covers an area of (2 m)2 = 4 m2
The sensor is small, so the area of sphere it occupies is about equal to its flat surface area: (0.05 m)2 = 0.0025 m2
Scale up the measured power: 0.1 W × 4 m20.0025 m2 = 160 W/sr
Square Degrees
Because we can convert from radians to degrees we can also convert from steradians to square degrees (deg2):
Since a radian is (180/π) degrees, we can square this to find that a steradian is (180/π)2 ≈ 3282.8 square degrees
Example: The Moon
The Moon's angular diameter is about 0.5°, so it covers:
π4 x (0.5°)2
≈ 0.2 deg2
That is a tiny fraction of the visible sky, which covers:
360° × 180°
≈ 64,800 deg2
Radian
Geometry Index
Copyright © 2025 Rod Pierce |
190701 | http://sharkphysics.weebly.com/uploads/8/5/9/5/8595301/retired_ap_optics.pdf | AP Optics Free Response Questions (1) AP® is a registered trademark of the College Board. The College Board was not involved in the production of and does not endorse this product. (2) Test Questions are Copyright © 1984-2008 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited. MIRRORS 1978 Q5 An object 6 centimeters high is placed 30 centimeters from a concave mirror of focal length 10 centimeters as shown above. (a) On the diagram above, locate the image by tracing two rays that begin at point P and pass through the focal point F. Is the image real or virtual? Is it located to the left or to the right of the mirror? (b) Calculate the position of the image. (c) Calculate the size of the image. (d) Indicate on the diagram above how the ray from point P to point Q is reflected, if aberrations are negligible. AP Optics Free Response Questions page 2 1983 Q5 The concave mirror shown above has a focal length of 20 centimeters. An object 3 centimeters high is placed 15 centimeters in front of the mirror. (a) Using at least two principal rays, locate the image on the diagram above. (b) Is the image real or virtual? Justify your answer. (c) Calculate the distance of the image from the mirror. (d) Calculate the height of the image. AP Optics Free Response Questions page 3 LENSES 1981 Q5 An object O is placed 18 centimeters from the center of a converging lens of focal length 6 centimeters as illustrated below: (a) On the illustration above, sketch a ray diagram to locate the image. (b) Is the image real or virtual? Explain your choice. (c) Using the lens equation, compute the distance of the image from the lens. A second converging lens, also of focal length 6 centimeters, is placed 6 centimeters to the right of the original lens as illustrated below. On the illustration above, sketch a ray diagram to locate the final image that now will be formed. Clearly indicate the final image. AP Optics Free Response Questions page 4 1982 Q6 An object is located a distance 3 2 f from a thin converging lens of focal length of as shown in the diagram below. (a) Calculate the position of the image. (b) Trace two of the principal rays to verify the position of the image. (c) Suppose the object remains fixed and the lens is removed. Another converging lens of focal length f2 is placed in exactly the same position as the first lens. A new real image larger than the first is now formed. Must the focal length of second lens be greater or less than f? Justify your answer. AP Optics Free Response Questions page 5 1986 Q6 An object is placed 3 centimeters to the left of a convex (converging) lens of focal length f = 2 centimeters, as shown below. (a) Sketch a ray diagram on the figure above to construct the image. It may be helpful to use a straightedge such as the edge of the green insert in your construction. (b) Determine the ratio of image size to object size. The converging lens is removed and a concave (diverging) lens of focal length f = −3 centimeters is placed as shown below. (c) Sketch a ray diagram on the figure above to construct the image. (d) Calculate the distance of this image from the lens. (e) State whether the image is real or virtual. The two lenses and the object are then placed as shown below. (f) Construct a complete ray diagram to show the final position of the image produced by the two-lens system. AP Optics Free Response Questions page 6 1989 Q5 The plano-convex lens shown above has a focal length f of 20 centimeters in air. An object is placed 50 centimeters (3f) from this lens. (a) State whether the image is real or virtual. (b) Determine the distance from the lens to the image. (c) Determine the magnification of this image (ratio of image size to object size). AP Optics Free Response Questions page 7 (d) The object, initially at a distance 3f from the lens, is moved toward the lens. On the axes below, sketch the image distance as the object distance varies from 3f to zero. (e) State whether the focal length of the lens would increase, decrease, or remain the same if the index of refraction of the lens were increased. Explain your reasoning. AP Optics Free Response Questions page 8 1992 Q6 A thin double convex lens of focal length f1 = +15 centimeters is located at the origin of the x-axis, as shown above. An object of height 8 centimeters is placed 45 centimeters to the left of the lens. (a) On the figure below, draw a ray diagram to show the formation of the image by the lens. Clearly show principal rays. (b) Calculate (do not measure) each of the following. i. The position of the image formed by the lens ii. The size of the image formed by the lens (c) Describe briefly what would happen to the image formed by the lens if the top half of the lens were blocked so that no light could pass through. AP Optics Free Response Questions page 9 A concave mirror with focal length f2 = +15 centimeters is placed at x = +30 centimeters. (d) On the figure below, indicate the position of the image formed by the lens, and draw a ray diagram to show the formation of the image by the mirror. Clearly show principal rays. AP Optics Free Response Questions page 10 1997 Q5 An object is placed 30 mm in front of a lens. An image of the object is located 90 mm behind the lens. (a) Is the lens converging or diverging? Explain your reasoning. (b) What is the focal length of the lens? (c) On the axis below, draw the lens at position x = 0. Draw at least two rays and locate the image to show the situation described above. Based on your diagram in (c), describe the image by answering the following questions in the blank spaces provided. i. Is the image real or virtual? ii. Is the image smaller than, larger than, or same size as the object? iii. Is the image inverted or upright compared to the object? (e) The lens is replaced by a concave mirror of focal length 20 mm. On the axis below, draw the mirror at position x = O so that a real image is formed. Draw at least two rays and locate the image to show this situation. AP Optics Free Response Questions page 11 2002 Q4 (15 points) A thin converging lens of focal length 10 cm is used as a simple magnifier to examine an object A that is held 6 cm from the lens. (a) On the figure below, draw a ray diagram showing the position and size of the image formed. (b) State whether the image is real or virtual. Explain your reasoning. (c) Calculate the distance of the image from the center of the lens. (d) Calculate the ratio of the image size to the object size. AP Optics Free Response Questions page 12 (e) The object A is now moved to the right from x = 6 cm to a position of x = 20 cm, as shown above. Describe the image position, size, and orientation when the object is at x = 20 cm. AP Optics Free Response Questions page 13 INDEX OF REFRACTION 1979 Q6 A light ray enters a block of plastic and travels along the path shown above. (a) By considering the behavior of the ray at point P, determine the speed of light in the plastic. (b) Determine what will happen to the light ray when it reaches point Q, using the diagram above to illustrate your conclusion. (c) There is an air bubble in the plastic block that happens to be shaped like a plano-convex lens as shown below. Sketch what happens to parallel rays of Light that strikes this air bubble. Explain your reasoning. AP Optics Free Response Questions page 14 1987 Q5 Light of frequency 6.0 × 1014 hertz strikes a glass/air boundary at an angle of incidence θ1. The ray is partially reflected and partially refracted at the boundary, as shown above. The index of refraction of this glass is 1.6 for light of this frequency. (a) Determine the value of θ3 if θ1 = 30°. (b) Determine the value of θ2 if θ1 = 30°. (c) Determine the speed of this light in the glass. (d) Determine the wavelength of this light in the glass. (e) What is the largest value of θ1 that will result in a refracted ray? AP Optics Free Response Questions page 15 1988 Q5 The triangular prism shown in Figure I above has index of refraction 1.5 and angles of 37°, 53°, and 90°. The shortest side of the prism is set on a horizontal table. A beam of light, initially horizontal, is incident on the prism from the left. (a) On Figure I above, sketch the path of the beam as it passes through and emerges from the prism. (b) Determine the angle with respect to the horizontal (angle of deviation) of the beam as it emerges from the prism. (c) The prism is replaced by a new prism of the same shape, which is set in the same position. The beam experiences total internal reflection at the right surface of this prism. What is the minimum possible index of refraction of this prism? The new prism having the index of refraction found in part (c) is then completely submerged in water (index of refraction = 1.33) as shown in Figure II below. A horizontal beam of light is again incident from the left. (d) On Figure II, sketch the path of the beam as it passes through and emerges from the prism. (e) Determine the angle with respect to the horizontal (angle of deviation) of the beam as it emerges from the prism. AP Optics Free Response Questions page 16 1990 Q6 A beam of light from a light source on the bottom of a swimming pool 3.0 meters deep strikes the surface of the water 2.0 meters to the left of the light source, as shown above. The index of refraction of the water in the pool is 1.33. (a) What angle does the reflected ray make with the normal to the surface? (b) What angle does the emerging ray make with the normal to the surface? (c) What is the minimum depth of water for which the light that strikes the surface of the water 2.0 meters to the left of the light source will be refracted into the air? AP Optics Free Response Questions page 17 In one section of the pool, there is a thin film of oil on the surface of the water. The thickness of the film is 1.0 × 10−7 meter and the index of refraction of the oil is 1.5. The light source is now held in the air and illuminates the film at normal incidence, as shown above. (d) At which of the interfaces (air-oil and oil-water), if either, does the light undergo a 180° phase change upon reflection? (e) For what wavelengths in the visible spectrum will the intensity be a maximum in the reflected beam? AP Optics Free Response Questions page 18 1993 Q4 The glass prism shown above has an index of refraction that depends on the wavelength of the light that enters it. The index of refraction is 1.50 for red light of wavelength 700 nanometers (700 × 10−9 meter) in vacuum and 1.60 for blue light of wavelength 480 nanometers in vacuum. A beam of white light is incident from the left, perpendicular to the first surface, as shown in the figure, and is dispersed by the prism into its spectral components. (a) Determine the speed of the blue light in the glass. (b) Determine the wavelength of the red light in the glass. (c) Determine the frequency of the red light in the glass. (d) On the figure above, sketch the approximate paths of both the red and the blue rays as they pass through the glass and back out into the vacuum. Ignore any reflected light. It is not necessary to calculate any angles, but do clearly show the change in direction of the rays, if any, at each surface and be sure to distinguish carefully any differences between the paths of the red and the blue beams. AP Optics Free Response Questions page 19 (e) The figure below represents a wedge-shaped hollow space in a large piece of the type of glass described above. On this figure, sketch the approximate path of the red and the blue rays as they pass through the hollow prism and back into the glass. Again, ignore any reflected light, clearly show changes in direction, if any, where refraction occurs, and carefully distinguish any differences in the two paths. AP Optics Free Response Questions page 20 1994 Q5 A point source S of monochromatic light is located on the bottom of a swimming pool filled with water to a depth of 1.0 meter, as shown above. The index of refraction of water is 1.33 for this light. Point P is located on the surface of the water directly above the light source. A person floats motionless on a raft so that the surface of the water is undisturbed. (a) Determine the velocity of the source's light in water. (b) On the diagram above, draw the approximate path of a ray of light from the source S to the eye of the person. It is not necessary to calculate any angles. (c) Determine the critical angle for the air-water interface. AP Optics Free Response Questions page 21 Suppose that a converging lens with focal length 30 centimeters in water is placed 20 centimeters above the light source, as shown in the diagram above. An image of the light source is formed by the lens. (d) Calculate the position of the image with respect to the bottom of the pool. (e) If, instead, the pool were filled with a material with a different index of refraction, describe the effect, if any, on the image and its position in each of the following cases. i. The index of refraction of the material is equal to that of the lens. ii. The index of refraction of the material is greater than that of water but less than that of the lens. AP Optics Free Response Questions page 22 2006 Q4 (15 points) A student performs an experiment to determine the index of refraction n of a rectangular glass slab in air. She is asked to use a laser beam to measure angles of incidence θi in air and corresponding angles of refraction θr in glass. The measurements of the angles for five trials are given in the table below. (a) Complete the last two columns in the table by calculating the quantities that need to be graphed to provide a linear relationship from which the index of refraction can be determined. Label the top of each column. (b) On the grid below, plot the quantities calculated in (a) and draw an appropriate graph from which the index of refraction can be determined. Label the axes. (c) Using the graph, calculate the index of refraction of the glass slab. AP Optics Free Response Questions page 23 The student is also asked to determine the thickness of a film of oil (n = 1.43) on the surface of water (n = 1.33). Light from a variable wavelength source is incident vertically onto the oil film as shown above. The student measures a maximum in the intensity of the reflected light when the incident light has a wavelength of 600 nm. (d) At which of the two interfaces does the light undergo a 180º phase change on reflection? The air-oil interface only The oil-water interface only Both interfaces Neither interface (e) Calculate the minimum possible thickness of the oil film. AP Optics Free Response Questions page 24 1999 Q6 (10 points) You are given the following equipment for use in the optics experiments in parts (a) and (b). A solid rectangular block made of transparent plastic A laser that produces a narrow, bright, monochromatic ray of light A protractor A meter stick A diffraction grating of known slit spacing A white opaque screen (a) Briefly describe the procedure you would use to determine the index of refraction of the plastic. Include a labeled diagram to show the experimental setup. Write down the corresponding equation you would use in your calculation and make sure all the variables in this equation are labeled on your diagram. AP Optics Free Response Questions page 25 (b) Since the index of refraction depends on wavelength, you decide you also want to determine the wavelength of your light source. Draw and label a diagram showing the experimental setup. Show the equation(s) you would use in your calculation and identify all the variables in the equation(s). State and justify any assumptions you make. AP Optics Free Response Questions page 26 2000 Q4 (15 points) A sheet of glass has an index of refraction ng = 1.50. Assume that the index of refraction for air is na = 1.00. (a) Monochromatic light is incident on the glass sheet, as shown in the figure below, at an angle of incidence of 60°. On the figure, sketch the path the light takes the first time it strikes each of the two parallel surfaces. Calculate and label the size of each angle (in degrees) on the figure, including angles of incidence, reflection, and refraction at each of the two parallel surfaces shown. (b) Next a thin film of material is to be tested on the glass sheet for use in making reflective coatings. The film has an index of refraction nf = 1.38. White light is incident normal to the surface of the film as shown below. It is observed that at a point where the light is incident on the film, light reflected from the surface appears green (λ = 525 nm). i. What is the frequency of the green light in air? ii. What is the frequency of the green light in the film? iii. What is the wavelength of the green light in the film? iv. Calculate the minimum thickness of film that would produce this green reflection. AP Optics Free Response Questions page 27 2001Q4 (15 points) In an experiment a beam of red light of wavelength 675 nm in air passes from glass into air, as shown above. The incident and refracted angles are θ1 and θ 2, respectively. In the experiment, angle θ 2 is measured for various angles of incidence θ 1, and the sines of the angles are used to obtain the line shown in the following graph. (a) Assuming an index of refraction of 1.00 for air, use the graph to determine a value for the index of refraction of the glass for the red light. Explain how you obtained this value. AP Optics Free Response Questions page 28 (b) For this red light, determine the following. i. The frequency in air ii. The speed in glass iii. The wavelength in glass (c) The index of refraction of this glass is 1.66 for violet light, which has wavelength 425 nm in air. i. Given the same incident angle θ1, show on the ray diagram on the previous page how the refracted ray for the violet light would vary from the refracted ray already drawn for the red light. ii. Sketch the graph of sin θ 2 versus sin θ 1 for the violet light on the figure on the previous page that shows the same graph already drawn for the red light. (d) Determine the critical angle of incidence θ c, for the violet light in the glass in order for total internal reflection to occur. AP Optics Free Response Questions page 29 DOUBLE SLIT 1991 Q6 Light consisting of two wavelengths, λa = 4.4 × 10−7 meter and λb 5.5 × 10−7 meter, is incident normally on a barrier with two slits separated by a distance d. The intensity distribution is measured along a plane that is a distance L = 0.85 meter from the slits, as shown above. The movable detector contains a photoelectric cell whose position y is measured from the central maximum. The first-order maximum for the longer wavelength λb occurs at λb 1.2 × 10−2 meter. (a) Determine the slit separation d. (b) At what position y, does the first-order maximum occur for the shorter wavelength λa? AP Optics Free Response Questions page 30 In a different experiment, light containing many wavelengths is incident on the slits. It is found that the photosensitive surface in the detector is insensitive to light with wavelengths longer than 6.0 × 10−7 m. (c) Determine the work function of the photosensitive surface. (d) Determine the maximum kinetic energy of electrons ejected from the photosensitive surface when exposed to light of wavelength λ = 4.4 10−7 m. AP Optics Free Response Questions page 31 1996 Q3 (15 points) Coherent monochromatic light of wavelength λ in air is incident on two narrow slits, the centers of which are 2.0 mm apart, as shown above. The interference pattern observed on a screen 5.0 m away is represented in the figure by the graph of light intensity I as a function of position x on the screen. (a) What property of light does this interference experiment demonstrate? (b) At point P in the diagram, there is a minimum in the interference pattern. Determine the path difference between the light arriving at this point from the two slits. (c) Determine the wavelength,λ of the light. AP Optics Free Response Questions page 32 (d) Briefly and qualitatively describe how the interference pattern would change under each of the following separate modifications and explain your reasoning. i. The experiment is performed in water, which has an index of refraction greater than 1. ii. One of the slits is covered. iii. The slits are moved farther apart. |
190702 | https://math.stackexchange.com/questions/1241854/converting-log-form-of-equation-into-linear-form | Skip to main content
Converting log form of equation into linear form
Ask Question
Asked
Modified
10 years, 4 months ago
Viewed
7k times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
I am trying to convert part of an equation from its log form into a linear form. Specifically, I am trying to convert 104log(x), into x4, but I'm really unsure of how to get from this first stage to the second. My experience with logarithms and exponents is limited, though I believe that 104log(x) can be re-written as 104+10log(x), but I'm not sure that this helps my plight! Any very basic guidance would be greatly appreciated.
algebra-precalculus
logarithms
Share
CC BY-SA 3.0
Follow this question to receive notifications
asked Apr 19, 2015 at 13:58
jasonjason
1311 silver badge33 bronze badges
1
Hint
104logx=10(logx)(4)=(10logx)4=…
– John Joy
Commented
Apr 19, 2015 at 16:17
Add a comment
|
2 Answers 2
Reset to default
This answer is useful
3
Save this answer.
Show activity on this post.
A cool rule about logarithmic functions is that
a⋅logb(x)=logb(xa).
From this,
104log(x)=10log(x4).
If we assume that you have a logarithm to the base 10, then
10log10(x4)=x4.
Share
CC BY-SA 3.0
Follow this answer to receive notifications
answered Apr 19, 2015 at 14:04
user230944user230944
Add a comment
|
This answer is useful
0
Save this answer.
Show activity on this post.
Taking logarithms,
log(104logx)=4logxlog10=logx(4log10)=log(x4log10),
so you can exponentiate both sides to get to
104logx=x4log10.
I shall spare you the rant about using log to mean log10.
Share
CC BY-SA 3.0
Follow this answer to receive notifications
answered Apr 19, 2015 at 14:03
ChappersChappers
69.2k1111 gold badges7474 silver badges153153 bronze badges
Add a comment
|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
algebra-precalculus
logarithms
See similar questions with these tags.
Featured on Meta
Community help needed to clean up goo.gl links (by August 25)
Related
0
How do you solve a equation by converting to logarithm form for the problem3ex−4+2=83?
3
Basic simultaneous equation with logarithms
1
Closed-form Solution of Log Sum
1
Convert (logn)n to the form nx for some x
2
What does the change of base rule for logarithms mean in terms of exponents?
1
re-writing non-linear into linear form, how to transform slope and intercept?
2
Converting logarithm to decimal form
0
How to make a variable in a log equation the subject?
0
Solve log equation 3log2(10)−log(x−9)=log(44)
2
Is this a special kind of exponential equation? How can I solve it?
Hot Network Questions
postgres-15 - WAL files being archived, but not recycled/removed from pg_wal
Do areas with high underground radon levels have uranium in it?
What would "Fruit Salad" mean in any other academic context than fruit salad?
Is it possible to identify this residual pattern as heteroscedastic or homoscedastic?
Are trills only for high-pitched notes?
How can I flatten a circular dependancy using C++20 modules?
It is widely reported that COGAT stopped aid starting from March 2nd. What official statements do we have from the Israeli government to that effect?
Is it mentioned that the 18-day Mahabharata war remained incomplete?
Buck LED driver inductor placement
Can I use super glue or hot glue as an insulator/solder mask for a small circuit?
Can this civilization land their biplanes in the air?
How is Zelensky able to challenge Putin to a face-to-face meeting if Ukrainian law prohibits negotiations with Putin?
Book series where a family living in the country is attacked by fae
Do I thank an ex-supervisor in my acknowledgements who was removed from my panel?
My Canadian employer is sending me to Germany to work on a project. Do I need a visa or a work permit?
Polynomial-Time Algorithms for Canonical Form of Ternary Matrices under Row/Column Permutations and Column Negations
Need help understanding the value of this infinite continued fraction intuitively
Michael Huemer's paper argues reincarnation is real: Is there any flaw in his reasoning?
How to top align panels
Why is there more than one model of electric guitar or bass
Why do word beginnings with X take a /z/ sound in English?
Will NASA send Juno to investigate 3i/Atlas?
Two Card Betting Game: Asymmetrical Problem
Koma-script 3.46 and caption
more hot questions
Question feed
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. |
190703 | https://www.youtube.com/watch?v=twm46yhp024 | Sensitivity Specificity and All That (tree diagrams)
Robert Cruikshank
4420 subscribers
16 likes
Description
591 views
Posted: 6 Feb 2021
This Two Minute Video offers a way to remember the meanings of various terms used in medical testing, by way of two tree diagrams.
As usual, if you have a specific idea for a video, please let me know in the comments and I will see what I can do!
4 comments
Transcript:
when you get a medical test there's a lot of terminology they throw around sensitivity specificity positive predictive value prevalence and so forth so here's a couple of diagrams tree diagrams to explain what's going on in the first place either you're sick or you're healthy and then say you go get a test and you either test positive or negative you're healthy and test positive or negative you can also view that the other way around and say first let's test positive or negative and then of those if you're tested positive what's the chance that you're sick and so forth turns out both of these trees are useful the first branch we run into is are you sick or healthy well if you're sick the probability of that happening is called the prevalence so if the chance of your being sick is p the chance of you being healthy would be one minus p if there's a 10 chance you're sick there's a 90 chance you're healthy now suppose you are sick chance that the test will come back positive correctly is called the sensitivity it's sensitive enough to detect that you're sick i'll label this s now on the other hand if you are healthy you want to test negative when the test is doing that correctly the chance of that happening is called the specificity if this is s then this is not s meaning 1 minus s if this is the specificity this is one minus the specificity now for this version you either test positive or negative if you tested positive the chance that you're sick is called the positive predictive value and they abbreviate that p p v if you test negative given that the chance that you're actually healthy is the negative predictive value so this would be 1 minus the positive predictive value and this would be 1 minus the negative predictive value |
190704 | https://kids.usmint.gov/ | Home | Coin Classroom
Skip to main content
An official website of the United States government
Here’s how you know
Here’s how you know
The .gov means it’s official.
Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site.
The site is secure.
The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely.
United States Mint Home
U.S. Mint Coin Classroom
U.S. Mint Coin Classroom
About the Mintexpand
Circulating Coins
U.S. Mint Locations
History Timeline
Life of a Coinexpand
Parts of a Coin
Designing Coins
Coin Composition
Collector's Cornerexpand
Get Started Collecting
Build Your Collection
Gamesexpand
Coin Stamper
Counting With Coins
Coin Memory Match
Gold Rush
Resourcesexpand
Videos
Coin Activities
Lesson Plans
Educator's Newsletter
Search
U.S. Mint Home
Games ----- How much gold can you catch? Play Gold Rush to find out!
Coin Activities --------------- Go on a quest for quarters to start your own coin collection!
Games ----- How much gold can you catch? Play Gold Rush to find out!
Coin Activities --------------- Go on a quest for quarters to start your own coin collection!
Games ----- How much gold can you catch? Play Gold Rush to find out!
1
2
Previous Next
GAMES
Go
VIDEOS
Play
Life of a Coin
Collector's Corner
Coloring Pages
About the Mint
Browse activities by grade
Browse the U.S. Mint Coin Classroom’s educational resources to find at-home activities, lesson plans, and more.
Learn More
Footer Utility
Coin Classroom Home
About the Mint
Life of a Coin
Collector's Corner
Games
Resources
U.S. Mint Home
About
Mint Tours
Coin & Medal Programs
History
News
Footer
FOIA
Accessibility
Privacy Policy
© 2025 United States Mint, U.S. Department of Treasury. All rights reserved. |
190705 | https://www.tutorchase.com/answers/a-level/physics/how-does-the-orientation-of-a-coil-affect-the-emf-induced-by | How does the orientation of a coil affect the EMF induced by electromagnetic induction?
The orientation of a coil affects the EMF induced by electromagnetic induction.
When a coil is moved through a magnetic field, an EMF is induced in the coil. The magnitude of the EMF depends on the rate of change of the magnetic flux through the coil. The magnetic flux is the product of the magnetic field strength and the area of the coil perpendicular to the field.
If the coil is oriented perpendicular to the magnetic field, the maximum amount of flux will pass through the coil and the induced EMF will be at its maximum. If the coil is oriented parallel to the magnetic field, the flux passing through the coil will be zero and no EMF will be induced.
If the coil is at an angle to the magnetic field, the flux passing through the coil will be less than the maximum, and the induced EMF will be less than the maximum. The angle between the coil and the magnetic field is known as the angle of incidence.
To understand this phenomenon more deeply, reviewing the basics of magnetic fields can provide additional insights into how changes in orientation impact the induced EMF. For further reading on the foundational concepts of magnetic fields, see Magnetic Field Basics.
Magnetic Field Basics
The orientation of the coil also affects the direction of the induced EMF. The direction of the EMF is given by Lenz's law, which states that the induced EMF will be in a direction that opposes the change in magnetic flux that produced it. This means that if the coil is moved in one direction through the magnetic field, the induced EMF will be in the opposite direction.
For more detailed exploration of how electromagnetic induction operates and its applications, consider reading about Electromagnetic Induction. Additionally, understanding how Magnetic Fields Due to Currents are generated can enhance comprehension of the interactions within electromagnetic systems.
Electromagnetic Induction
Magnetic Fields Due to Currents
Study and Practice for Free
Trusted by 100,000+ Students Worldwide
Achieve Top Grades in your Exams with our Free Resources.
Practice Questions, Study Notes, and Past Exam Papers for all Subjects!
Need help from an expert?
4.93/5 based on733 reviews in
The world’s top online tutoring provider trusted by students, parents, and schools globally.
Related Physics a-level Answers |
190706 | https://mathworld.wolfram.com/StandardForm.html | Standard Form -- from Wolfram MathWorld
TOPICS
AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld
Geometry
Line Geometry
Lines
MathWorld Contributors
Stover
Standard Form
"Standard form" is a general term used in mathematics to refer to a normalized form of an expression. The term "canonical form" is also used in certain contexts (for example, graphs).
The standard form of a line in the Cartesian plane is given by
(1)
for real numbers .
This form can be derived from any of the other forms (point-slope form, slope-intercept form, etc.), but can be seen most intuitively when starting from intercept form. Indeed, the intercept form of a line with x-intercept and y-intercept is given by
(2)
and so clearing denominators to obtain
(3)
gives the line in standard form with , , and .
See also
Intercept Form, Line, Linear Equation, Point-Slope Form, Slope-Intercept Form, Two-Point Form
Explore with Wolfram|Alpha
More things to try:
1->2, 2->3, 3->1, 3->4, 4->1
Conway constant to 200 digits
Laplace transform x^3
Cite this as:
Weisstein, Eric W. "Standard Form." From MathWorld--A Wolfram Resource.
Subject classifications
Geometry
Line Geometry
Lines
MathWorld Contributors
Stover
About MathWorld
MathWorld Classroom
Contribute
MathWorld Book
wolfram.com
13,278 Entries
Last Updated: Sun Sep 28 2025
©1999–2025 Wolfram Research, Inc.
Terms of Use
wolfram.com
Wolfram for Education
Created, developed and nurtured by Eric Weisstein at Wolfram Research
Created, developed and nurtured by Eric Weisstein at Wolfram Research |
190707 | https://www.gauthmath.com/solution/1833428526788657/Quest-Identify-the-constant-difference-for-a-hyperbola-with-foci-3-0-and-3-0-and | Solved: Quest Identify the constant difference for a hyperbola with foci (-3,0) and (3,0) and a po [Math]
Drag Image or
Click Here
to upload
Command+to paste
Upgrade
Sign in
Homework
Homework
Assignment Solver
Assignment
Calculator
Calculator
Resources
Resources
Blog
Blog
App
App
Gauth
Unlimited answers
Gauth AI Pro
Start Free Trial
Homework Helper
Study Resources
Math
Questions
Question
Quest Identify the constant difference for a hyperbola with foci (-3,0) and (3,0) and a point on the hyperbola (7,0). 10 14 2 6
Show transcript
Gauth AI Solution
100%(4 rated)
Answer
6
Explanation
Identify the foci and a point on the hyperbola. The foci are at (-3, 0) and (3, 0), and a point on the hyperbola is (7, 0).
Calculate the distance from the point (7,0) to each focus. The distance to (3,0) is |7-3| = 4. The distance to (-3,0) is |7-(-3)| = 10.
The constant difference is the difference between these two distances: 10 - 4 = 6
Helpful
Not Helpful
Explain
Simplify this solution
Gauth AI Pro
Back-to-School 3 Day Free Trial
Limited offer! Enjoy unlimited answers for free.
Join Gauth PLUS for $0
Previous questionNext question
Related
Identify the constant difference for a hyperbola with foci -6,0 and 6,0 and a point on the hyperbola 4,0. 8 14 12 10
100% (2 rated)
Identify the constant difference for a hyperbola with foci hyperbola 4,0. -6,0 and 6,0 and a point on the 10 12 8 14
94% (106 rated)
Identify the constant difference for a hyperbola with foci 0,-5 and 0,5 and a point on the hyperbola 0,3. 8 6 2 4
93% (533 rated)
Identify the constant difference for a hyperbola with foci -6,0 and 6,0 and a point on the hyperbola 4, 0. 12 8 10 14
98% (286 rated)
Identify the constant difference for a hyperbola with foci -7,0 and 7,0 and a point on the hyperbola 10,0 17 14 7 3
100% (2 rated)
Write an equation for the hyperbola with foci 12,0 and -12,0 and constant difference of 16.
100% (3 rated)
What is an equation for the hyperbola with foci 9,0 and -9,0 and constant difference of 14?
93% (589 rated)
Write an equation for the hyperbola. foci at 0,17 and 0,-17 and a constant difference of 16.
100% (4 rated)
Write an equation for the hyperbola. foci at 10,0 and -10,0 and a constant difference of 12.
99% (239 rated)
Write an equation for the hyperbola. foci at 50,0 and -50,0 and a constant difference of 28.
Gauth it, Ace it!
contact@gauthmath.com
Company
About UsExpertsWriting Examples
Legal
Honor CodePrivacy PolicyTerms of Service
Download App |
190708 | https://arxiv.org/pdf/2009.08817 | arXiv:2009.08817v1 [math.OC] 18 Sep 2020
Calmness of the solution mapping of Navier-Stokes problems modeled by hemivariational inequalities
D. Inoan a and J. Kolumb´ an b
a
Technical University of Cluj-Napoca, Cluj-Napoca, Romania; bBabe¸ s-Bolyai University, Cluj-Napoca, Romania
ARTICLE HISTORY
Compiled September 21, 2020
ABSTRACT
The main purpose of this paper is to find conditions for H¨ older calmness of the solution mapping, viewed as a function of the boundary data, of a hemivariational inequality governed by the Navier-Stokes operator. To this end, a more abstract model is studied first: a class of parametric equilibrium problems defined by trifunc-tions. The presence of trifunctions allows the extension of the monotonicity notions and of the duality principle in the theory of equilibrium problems.
KEYWORDS
Navier-Stokes equation; calmness; parametric equilibrium problems with trifunctions; hemi-variational inequalities
Introduction
In the papers and a class of hemivariational inequalities with the Navier-Stokes operator has been studied, where the nonslip boundary condition together with a Clarke subdifferential relation between the total pressure and the normal components of the velocity was assumed. The main feature of such a hemivariational inequality is that it is governed by a nonmonotone and nonlinear operator, and possibly by a multivalued boundary condition defined by the Clarke derivative of a locally Lipschitz superpotential. The problem under consideration comes from fluid flow control prob-lems and flow problems for semipermeable walls and membranes. It describes a model in which the boundary orifices in a channel are regulated to reduce the pressure of the fluid on the boundary when the normal velocity reaches a prescribed value. For the particular case when the superpotential is a lower semicontinuous convex functional the problem reduces to a variational inequality governed by a maximal monotone operator (see , , ). Existence results for nonconvex locally Lipschitz superpotentials were given in the above mentioned papers (stationary case), (evolution case) and , , (pe-riodic or antiperiodic case), for instance. The main purpose of this paper is to study the calmness (in the sense of and ) of the set of solutions of a hemivariatonal inequality governed by the Navier-Stokes operator, when the Clarke derivative is substituted by a more general control function,
CONTACT D. Inoan. Email: daniela.inoan@math.utcluj.ro
depending also on the state and time variables. Calmness is an important stability property since it gives a bound on the distance between perturbed solutions and unperturbed solutions. For real functions this prop-erty is weaker than the usual local Lipschitz continuity since one of the two points considered for comparison is required to be fixed, but it is stronger than the continuity at that point. A weaker property than calmness, which we could name lower calmness, was used for the first time by F.H. Clarke in the paper . Lower calmness is situated between calmness and lower semicontinuity. In optimization theory, lower calmness is frequently used even in the absence of calmness (see, for instance ). In the case of set-valued mappings, calmness is defined using the excess function defined by the Romanian mathematician D. Pompeiu (see ). In this case calmness is a generalization of the Aubin property, which in its turn is a generalization of the local Lipschitz property for set-valued mappings (see ). Calmness properties of solutions to parameterized equilibrium problems formulated with bifunctions have been studied widely. Most of this study has focused on particular models such as optimization problems (see , , and ) and variational or hemivariational inequalities (see , ). H¨ older calmness and H¨ older continuity of solution mappings of general parametric equilibrium problems have been studied, for instance, in the papers , , , , . The calmness property is strongly connected to the H¨ older metric subregularity of the inverse mapping (see , and the references therein). In the last years many papers about hemivariational inequalites similar to that of and , with important applications, were published. As an example we mention only the quite recently appeared article . Since the study of calmness of the solution sets for similar problems is undoubtely important, we will embed our problem in an abstract model of parametric equilibrium problem defined by trifunction and we will apply the abstract results obtained for the hemivariational inequalites governed by the Navier-Stokes operator, where the parameters are functions with some properties similar to those of Clarke generalized derivatives. Our main motivation to study equilibrium problems defined by trifunctions is the important role that the monotonicity has in existence and stability results for equilib-rium problems defined by bifunctions on one hand, and the existence of bifunctions that are not monotone, on the other hand. In we have shown that for trifunctions it is possible to define a monotonicity notion such that the monotone bifunctions that have value zero on the diagonal, generate monotone trifunctions and every bifunction is monotone as a trifunction. Therefore, for instance, the so-called mixed equilibrium problems can be formulated by monotone trifunctions. This makes possible to extend the duality principle to a large class of equilibrium problems, and to use it, for instance, by proving existence and stability of the solutions. This paper is structured in four sections, as follows: Section 2 contains several notions and results needed in the sequel. In Section 3 we prove our main calmness result (Theorem 3.1), in a general set-ting, for parametric equilibrium problems with trifunctions. It gives necessary and sufficient conditions for H¨ older calmness based on an apriori estimation for the dual problem, which is, in fact, a gradual uniform partial calmness property (see . To our best knowledge, calmness for such general problems is studied here for the first time. Existence theorems for equilibrium problems with trifunctions were given in . In the next section, we apply the previous abstract theorem to mixed equilibrium problems. Finally, returning to the main purpose of the paper, we focus on the Navier-Stokes problems modeled by hemivariational type inequalities with boundary control, obtaining necessary and sufficient conditions for H¨ older calmness of the solution map-2ping. Since calmness is stronger than continuity, our results can be seen as sharpening of other results obtained before on the behaviour of the solutions, when the data are perturbed (see, for example, Theorem 21 in and the convergence results in ).
Preliminaries
For a real number a we denote a− := d(a, R+) = max {− a, 0} and a+ := d(a, R−) = max {a, 0}. For a, b ∈ R the following properties hold: (a) ( −a)+ = a−, ( −a)− = a+;(b) If b ≥ 0 then a+ ≤ (a + b)+;(c) ( a + b)+ ≤ a+ + b+. The equality holds if and only if ab ≥ 0; (d) ( a + b)+ ≤ a+ + |b|;(e) If α > 0, then a+ ≤ α if and only if a ≤ α.In this paper, unless otherwise mentioned, M and X will be metric spaces, and -for convenience - both distances will be denoted by d. By B(u, r ) will be denoted the open ball centered at u, of radius r. The Euclidean norm on Rd (d = 1 , 2, . . . ) will be denoted by | · | , and the scalar product of u, v ∈ Rd by u · v.A function f : M → X is said to satisfy the H¨ older condition of rank k and exponent
ε if k ≥ 0, ε > 0, and
d(f (μ), f (μ′)) ≤ kd ε(μ, μ ′), (1) for all μ, μ ′ ∈ M . We say that f is a ( k, ε )-H¨ older function near ¯μ if there exists a neighbourhood U (¯ μ) of ¯ μ such that (1) is verified for all μ, μ ′ ∈ U (¯ μ). If ε = 1, we say that the function f is k-Lipschitz. A property between this and the continuity at ¯ μ is the calmness at ¯ μ.A function f : M → X is said to be H¨ older calm at ¯ μ if there exist k ≥ 0, ε > 0and a neighborhood U (¯ μ) of ¯ μ such that
d(f (μ), f (¯ μ)) ≤ kd ε(μ, ¯μ),
for all μ ∈ U (¯ μ). If ε = 1 we simply say that f is calm instead of ( k, 1)-H¨ older calm. For a function f : M → R, a property strictly between calmness at ¯ μ and lower semicontinuity at ¯ μ is the lower calmness at ¯ μ. The function f is said to be lower H¨ older calm at ¯ μ, with exponent ε > 0, if lim inf
μ→¯μ
f (μ) − f (¯ μ)
dε(μ, ¯μ) > −∞ . (2) The function f is said to be upper H¨ older calm at ¯ μ if the function −f is lower H¨ older calm at ¯ μ. It is clear that f is H¨ older calm at ¯ μ if and only if it is both lower and upper H¨ older calm. In optimization theory the lower calmness is frequently used even in the absence of upper calmness (see, for instance, ). (This could be the motivation why F.H. Clarke defined the notion of calmness by inequality (2) for M ⊂ R). Calmness can be generalized for set-valued functions too. For some sets A, B in the metric space ( X, d ) and a ∈ X, denote by
d(a, B ) := inf
b∈B
d(a, b )3the distance between the point a and the set B, and by
e(A, B ) := sup
a∈A
d(a, B )the Pompeiu excess of A with respect to B, where the convention
e(∅, B ) =
{ 0, when B 6 = ∅
+∞, otherwise is used and e(A, ∅) = + ∞, for any set A, including ∅.The Pompeiu-Hausdorff distance is defined by
h(A, B ) = max( e(A, B ), e (B, A )) .
As it is known, it does not furnish a metric on the space of all subsets, but it does on the space of nonempty closed and bounded subsets of X.A set-valued mapping S : M → 2X is said to be ( k, ε )-H¨ older continuous if k ≥ 0,
ε > 0, and
h(S(μ), S (μ′)) ≤ kd ε(μ, μ ′), for all μ, μ ′ ∈ M.
A mapping S : M → 2X is said to be outer (k, ε )-H¨ older at ¯ μ if k ≥ 0, ε > 0, and
e(S(μ), S (¯ μ)) ≤ kd ε(μ, ¯μ), for all μ ∈ M.
A mapping S : M → 2X is said to be ( k, ε )-H¨ older calm at (¯ μ, ¯u) if (¯ μ, ¯u) ∈ gph S,
k ≥ 0, ε > 0, and there exist neighbourhoods U (¯ μ) of ¯ μ and V (¯ u) of ¯ u such that
e(S(μ) ∩ V (¯ u), S (¯ μ)) ≤ kd ε(μ, ¯μ) for all μ ∈ U (¯ μ).
A mapping S : M → 2X is said to have the isolated H¨ older calmness property at (¯ μ, ¯u) ∈ gph S if there exist k ≥ 0, ε > 0, and the neighbourhoods U (¯ μ) of ¯ μ and V (¯ u)of ¯ u such that
d(u, ¯u) ≤ kd ε(μ, ¯μ) for all μ ∈ U (¯ μ) and u ∈ S(μ) ∩ V (¯ u).
If the above properties take place for ε = 1, then instead of ( k, 1)-H¨ older we say
k-Lipschitz. For more information on these notions we propose the book . Suppose Y is a normed space and D is an open subset of Y . If f : D → R is
k-Lipschitz near u ∈ D, then the Clarke generalized directional derivative f 0(u; v) at
u in the direction v ∈ Y is defined as follows:
f 0(u; v) = lim sup
w→u,t ց0
f (w + tv ) − f (w)
t .
Some of the basic properties of the function f 0 are summarised in the following
Proposition 2.1. (, Proposition 2.2.1 and 2.3.10) Let f be k-Lipschitz near u ∈
D. Then (a) The function v → f 0(u; v) is finite, sublinear, and k-Lipschitz on Y ;
4(b) The function (u, v ) → f 0(u; v) is upper semi-continuous; (c) f 0(u; −v) = ( −f )0(u; v);(d) If Z is a normed space, T : Y → Z is a surjective linear continuous operator and g : T (D) → R is k-Lipschitz near T u , then the function g ◦ T is k‖T ‖-Lipschitz near u and (g ◦ T )0(u; v) = g0(T u, T v ) for u, v ∈ Y .
Usually, a bifunction f : X × X → R is said to be monotone iff f (u, v ) + f (v, u ) ≤ 0, for all u, v ∈ X. It is called strongly monotone iff there exists m > 0 such that
md 2(u, v ) ≤ f (u, v ) + f (v, u ), for all u, v ∈ X. In the following definition, we extend these notions for trifunctions as follows:
Definition 2.2. The trifunction F : X × X × X → R is said to be monotone iff
F (u, v, u ) ≤ F (u, v, v ) for every u, v ∈ X.We say that F is H¨ older strongly monotone iff there exist m, β > 0 such that
md β (u, v ) ≤ F (u, v, v ) − F (u, v, u ), for every u, v ∈ X.
Remark 1. (a) If the trifunction F has the particular form F (u, v, w ) = f (w, v ) −
f (w, u ), with f (u, u ) = 0 for any u ∈ X, then F is monotone if and only if the bifunction f is monotone. (b) If F (u, v, w ) = g(u, v ), with g : X × X → R, obviously F is monotone as a trifunction. In consequence, if G : X × X × X → R is a monotone trifunction and
g : X × X → R is arbitrary, then the trifunction F (u, v, w ) = G(u, v, w ) + g(u, v )is also monotone. This fact simplifies, for instance, the theory of mixed equilibrium problems, and makes it more transparent (see ). (c) If X is a normed space with the dual X∗, then the operator T : X × X → X∗
is called semimonotone if it is monotone with respect to the second variable, that is
〈T (u, w 1) − T (u, w 2), w 1 − w2〉 ≥ 0, for every u, w 1, w 2 ∈ X.
Variational inequalities governed by such operators were studied, for instance, in for single-valued functions and in for set-valued mappings. In this last case, the function F : X ×X ×X → R defined by F (u, v, w ) = sup z∈T (u,w )〈z, v −u〉 is monotone, while f (u, v ) = sup z∈T (u,u )〈z, v − u〉 is not monotone, so, for the variational inequality governed by T (u, u ), the duality principle is not applicable.
An abstract model
In this section we consider a general equilibrium problem where the objective function is a trifunction that depends on a parameter μ. In the papers and we gave existence results for such problems, motivated by the fact that the classical theory for equilibrium problems with bifunctions can not be used for some problems that appear in applications. For a parameter μ ∈ M , the problem that we study is (P E )( μ) Find ¯ u ∈ K such that F (¯ u, z, ¯u; μ) ≥ 0, for every z ∈ K,
where X and M are metric spaces, F : X × X × X × M → R is a given function, and
K is a nonempty subset of X.5Denote by S(μ) the set of solutions of the problem that depends on the parameter
μ ∈ M . Throughout the paper we suppose that it is nonempty.
Theorem 3.1. Let ¯μ ∈ M be a nonisolated point and ¯u ∈ S(¯ μ) be fixed. Suppose that there exists a neighborhood U (¯ μ) of ¯μ, a neighborhood ˜V (¯ u) of ¯u and the numbers
a, c, θ ≥ 0, and b, m, α, β, ξ, θ > 0 such that (i) md β (¯ u, v ) ≤ F (¯ u, v, ¯u; ¯ μ)− + F (¯ u, v, v ; ¯ μ)+, for every v ∈ S(μ) ∩ ˜V (¯ u) and μ ∈
U (¯ μ);(ii) The estimation F (¯ u, v, v ; ¯ μ) ≤ cd β (¯ u, v ) + dθ(¯ u, v )[ ad α(¯ u, v ) + bd ξ (μ, ¯μ)] holds for every μ ∈ U (¯ μ) and v ∈ S(μ) ∩ ˜V (¯ u), with v 6 = ¯ u;(iii) θ < β and c < m .Then the mapping S : M → 2X is H¨ older calm at (¯ μ, ¯u) if and only if one of the following conditions is verified: 1) β < α + θ and a > 0;2) β = α + θ, a > 0 and a + c < m ;3) a = 0 .Moreover, in this case we have the isolated H¨ older calmness property at (¯ μ, ¯u). In the cases 2), 3) the solution ¯u is unique in the neighborhood ˜V (¯ u).The parameters from the definition of the calmness are: (α) V (¯ u) = ˜V (¯ u), δ = ξ
β − θ , k =
( b
m − c
) 1
β−θ
, for a = 0 ,(β) V (¯ u) = B(¯ u, r ) ∩ ˜V (¯ u), δ = ξ
β − θ , k = r
(
rβ−θ − a
m − c rα
) 1
θ−β
( b
m − c
) 1
β−θ
,where 0 < r <
( m − c
a
) 1
α+θ−β
, for a > 0.
For the proof, we will need the following:
Lemma 3.2. Let p > 0, q > 0, and l ≥ 0 be given real numbers. Then the following statements are equivalent: 1) p < q or p = q and l < 1, or l = 0 ;2) There exists ε0 > 0 such that for every ε ∈]0 , ε 0[, there exist δ > 0 and k > 0
such that, for all x ∈]0 , ε [ and y > 0 with xp − lx q ≤ y, the inequality x ≤ ky δ holds. Proof : Sufficiency. Let p < q , ε0 = l 1
p−q
and let ϕ : [0 , ∞) → R be the function defined by ϕ(ξ) = ξ−lξ q
p
.It is easy to see that on the interval ]0 , ε p[ this function has strictly positive values and is concave. From this, for ξ ∈]0 , ε p[, we have ϕ(εp )
εp
· ξ < ϕ (ξ) that is ξ < εp
ϕ(εp)
ϕ(ξ). Now consider x ∈]0 , ε [ with xp − lx q ≤ y and let ξ = xp. Then we get
xp < εp
ϕ(εp) ϕ(xp) ≤ εp
εp − lε q y.
and the conclusion is proved with
k = ε(εp − lε q )− 1
p
and δ = 1
p .
Let p = q and l < 1. For any x ∈]0 , +∞[ and y > 0, from xp − lx q ≤ y follows
x ≤ ky δ, with k = (1 − l)− 1
p
and δ = 1
p
.6Let l = 0. For any x ∈]0 , +∞[ and y > 0, from xp − lx q ≤ y follows the conclusion with k = 1 and δ = 1
p
.To prove the reverse implication, suppose that p > q . Then for ε ∈]0 , l 1
p−q
[ we have
ϕ(xp) < 0 for all x ∈]0 , ε [. If 2) holds, then we would have 0 < x ≤ ky δ , for every
y > 0, which is a contradiction.
Proof of Theorem 3.1 : Sufficiency. In the case β < α + θ and a > 0, let r be such that 0 < r < ( m−c
a
) 1
α+θ−β
, let the neighborhood of ¯ u be V (¯ u) = B(¯ u, r ) ∩ ˜V (¯ u), let
μ ∈ U (¯ μ) and v ∈ S(μ) ∩ V (¯ u) with v 6 = ¯ u.Since ¯ u ∈ S(¯ μ) and v ∈ K, we have
F (¯ u, v, ¯u; ¯ μ) ≥ 0. (3) From (i), (ii) and (3) follows that
md β (v, ¯u) ≤ F (¯ u, v, ¯u; ¯ μ)− + F (¯ u, v, v ; ¯ μ)+ = F (¯ u, v, v ; ¯ μ)+
= F (¯ u, v, v ; ¯ μ) ≤ cd β (v, ¯u) + dθ (v, ¯u)[ ad α(v, ¯u) + bd ξ (μ, ¯μ)] Therefore, (m − c)dβ−θ (v, ¯u) ≤ ad α(v, ¯u) + bd ξ (μ, ¯μ)and further on, since m − c > 0
dβ−θ(v, ¯u) ≤ a
m − c dα(v, ¯u) + b
m − c dξ (μ, ¯μ). (4) We apply Lemma 3.2, with x := d(v, ¯u), p := β −θ, q := α, y := b
m−c
dξ (μ, ¯μ), l := a
m−c
,
ε := r. Since v ∈ B(¯ u, r ) and v 6 = ¯ u, the conditions of the lemma are verified and we get
d(v, ¯u) ≤ kd δ (μ, ¯μ), (5) where δ = ξ
β − θ and k = r
(
rβ−θ − a
m − c rα
) 1
θ−β
( b
m − c
) 1
β−θ
.If v = ¯ u, (5) is obviously verified. In the case β = α + θ and a + c < m we have (m − c − a)dβ−θ (v, ¯u) ≤ bd ξ (μ, ¯μ)which implies
d(v, ¯u) ≤
( b
m − c − a
) 1
β−θ
d ξ
β−θ
(μ, ¯μ).
In the case a = 0, we can choose V (¯ u) = ˜ V (¯ u). Let μ ∈ U (¯ μ) and v ∈ S(μ) ∩ V (¯ u). In the same way as before, we get (m − c)dβ−θ (v, ¯u) ≤ bd ξ (μ, ¯μ),
7so directly
d(v, ¯u) ≤
( b
m − c
) 1
β−θ
d ξ
β−θ
(μ, ¯μ).
So, in all cases, there exist k and δ such that, for every v ∈ S(μ) ∩ V (¯ u),
d(v, S (¯ μ)) = inf
z∈S(¯ μ)
d(v, z ) ≤ d(v, ¯u) ≤ kd δ (μ, ¯μ).
This implies
e(S(μ) ∩ V (¯ u), S (¯ μ)) = sup
v∈S(μ)∩V(¯ u)
d(v, S (¯ μ)) ≤ kd δ (μ, ¯μ).
The necessity of at least one of the conditions 1), 2) or 3) follows from Lemma 3.2 and the fact that ¯ μ is nonisolated.
Remark 2. (a) If the function F is H¨ older strongly monotone, then condition (i) from Theorem 3.1 is verified. This follows directly from the fact that F (u, v, v )−F (u, v, u ) ≤
F (u, v, v )+ + F (u, v, u )−. The converse is not true (see for the case of bifunctions). (b) In Section 5 we will see how properties (i) and (ii) appear for hemivariational inequalities governed by the Navier-Stokes operator.
Parametric mixed equilibrium problems
Mixed equilibrium problems have an important role in applied mathematics. They were first studied in the paper . Consider the function F having the particular form
F (u, v, w ; μ) = f (w, v ; μ) − f (w, u ; μ) + g(u, v ; μ) (6) where f : X ×X ×M → R is such that the bifunction f (·, ·; μ) is monotone, f (u, u ; μ) = 0, for all u ∈ X, μ ∈ M and g : X × X × M → R is an arbitrary function. In this case we have
F (u, v, u ; μ) = f (u, v ; μ) + g(u, v ; μ) and F (u, v, v ; μ) = −f (v, u ; μ) + g(u, v ; μ). (7) The problem ( P E )( μ) becomes the mixed parametric equilibrium problem defined by f and g:(P M E )( μ) Find ¯ u ∈ K such that f (¯ u, z ; μ) + g(¯ u, z ; μ) ≥ 0, for every z ∈ K.
We denote by S(μ) the set of solutions of the problem ( P M E )( μ) and suppose that it is nonempty. The next result follows directly from Theorem 3.1.
Theorem 4.1. Let ¯μ ∈ M be nonisolated and ¯u ∈ S(¯ μ) be fixed. Suppose that there exist a neighborhood U (¯ μ) of ¯μ, a neighborhood ˜V (¯ u) of ¯u, and the numbers
a, b 1, b 2, c, θ ≥ 0, m, α, β, ξ, θ > 0 such that
8(i) md β (¯ u, v ) ≤ [f (¯ u, v ; ¯ μ) + g(¯ u, v ; ¯ μ)] − + [ f (v, ¯u; ¯ μ) − g(¯ u, v ; ¯ μ)] −, for every v ∈
S(μ) ∩ ˜V (¯ u) and μ ∈ U (¯ μ);(ii) f (v, ¯u; μ) − f (v, ¯u; ¯ μ) ≤ b1dθ(¯ u, v )dξ (μ, ¯μ), for every μ ∈ U (¯ μ), v ∈ S(μ) ∩ ˜V (¯ u),with v 6 = ¯ u;(iii) g(¯ u, v ; ¯ μ) + g(v, ¯u; μ) ≤ cd β (¯ u, v ) + dθ(¯ u, v )[ ad α(¯ u, v ) + b2dξ (μ, ¯μ)] , for every
μ ∈ U (¯ μ) and v ∈ S(μ) ∩ ˜V (¯ u), with v 6 = ¯ u;(iv) 0 < β − θ and c < m .Then the mapping S : M → 2X is H¨ older calm at (¯ μ, ¯u) if and only if one of the conditions 1), 2), 3) from Theorem 3.1 is verified. Moreover, in this case we have the isolated H¨ older calmness property at (¯ μ, ¯u). In the cases 2), 3) the solution ¯u is unique in the neighborhood ˜V (¯ u).Proof : We only have to check condition (ii) from Theorem 3.1. Let μ ∈ U (¯ μ) and
v ∈ S(μ) ∩ ˜V (¯ u). Then,
f (v, ¯u; μ) + g(v, ¯u; μ) ≥ 0.
We have, for b = b1 + b2,
F (¯ u, v, v ; ¯ μ) = −f (v, ¯u; ¯ μ) + g(¯ u, v ; ¯ μ)
≤ −f (v, ¯u; ¯ μ) + g(¯ u, v ; ¯ μ) + f (v, ¯u; μ) + g(v, ¯u; μ)
≤ cd β (¯ u, v ) + dθ(¯ u, v )[ ad α(¯ u, v ) + bd ξ (μ, ¯μ)] .
Therefore Theorem 3.1 can be applied.
Remark 3. Similar results on the H¨ older calmness of the solution mapping have been obtained, for instance, in , for g = 0. In this particular case, the set K was considered to depend also on a parameter λ. Theorem 4.1 can be extended in this sense, but it is not our aim in this paper. For stability results in the case of parametric mixed problems we mention also .
Navier-Stokes problems modeled by hemivariational inequalities
In the papers and , Mig´ orski and Ochal studied a class of hemivariational prob-lems for the Navier-Stokes operators, in the stationary and evolution case, respectively. When Ω is a bounded simply connected domain of Rd, d = 2 , 3, . . . , with boundary Γ of class C2, the Navier-Stokes equations that describe the flow of a viscous incompressible constant density fluid in the domain Ω are the following:
u′ − ν∆u + ( u · ∇ )u + ∇p = φ, (8)
∇ · u = 0 on Q = Ω ×]t0, t 1[ (9) Here u : Ω × [t0, t 1] → Rd is the velocity, ν is the kinematic viscosity of the fluid,
p : Ω × [t0, t 1] → R is the pressure, φ : Q → Rd is a vector field given by the external forces. To obtain a variational formulation of the previous equations, it is convenient to rewrite the problem in the equivalent Leray form (see ). 9For this let us consider the set
W = {w ∈ C∞(Ω , Rd) : div w = 0 on Ω }.
Denote by V and H the closure of W in the norms of W 12 (Ω , Rd) (the usual Sobolev space) and L2(Ω , Rd), respectively. We have V ⊂ H ≃ H∗ ⊂ V ∗ with the embeddings being dense, continuous, and compact. The space V is a Hilbert space with the associated scalar product (( u, v )) =
∑di=1 (Diu, D iv), where Di is the operator ∂
∂x i
. Consider the spaces
V = L2(t0, t 1; V ), H = L2(t0, t 1; H) and W = {w ∈ V : w′ ∈ V ∗},
where the time derivative w′ is understood in the sense of vector valued distributions. In this case W ⊂ V ⊂ H ⊂ V ∗ and the embeddings are continuous. The pairing between V and V ∗ will be denoted by 〈· , ·〉 , and the pairing between V
and V∗ will be denoted by 〈〈· , ·〉〉 . The space W is a separable reflexive Banach space with the norm ‖w‖W = ‖w‖V + ‖w′‖V∗ and is continuously embedded in C([ t0, t 1]; H)(see ). The norm on V will be denoted by ‖ · ‖ .To write the weak formulation of the problem (8)-(9), consider the operators A :
V → V ∗ and B : V × V → V ∗ defined by
〈Au, v 〉 = ν
∫
Ω
(( u, v )) d x, (10)
〈B(u, v ), w 〉 =
∫
Ω
d
∑
i,j =1
ui(Divj )wj dx, B[u] = B(u, u ), (11) and denote
〈Φ( t), v 〉 :=
∫
Ω
φ(x, t )v(x)d x,
for u, v, w ∈ V . It is well known (see , p. 162) that the operator B is well defined only if d ∈ { 2, 3, 4}; therefore in the following we suppose that this condition is fulfilled. For problem (8)-(9) to be well posed it is necessary to assign some boundary conditions. Let us consider, for instance, in the case d = 3, the Neumann condition u|Γ = h, where
h = pn − ν ∂γu
∂n
, n is the outward unit normal vector to ∂Ω, ∂
∂n
is the normal derivative operator, and γ : V → L2(Γ) is the trace operator. If we multiply the equation (8) by a test function v ∈ V , then using the Gauss formulae, we obtain the weak formulation of the Navier-Stokes equation with the Neumann boundary condition:
〈u′(t) + Au (t) + B[u(t)] , v 〉 +
∫
Γ
h · γv dσ(x) = 〈Φ( t), v 〉, a.e. t ∈]t0, t 1[, v ∈ V.
Similar to and , the Neumann boundary condition can be generalized by the subdifferential condition
h(x, t ) ∈ ∂j (x, t, γu (x, t )) on Γ × [t0, t 1],
10 where ∂j denotes the Clarke subdifferential of the locally Lipshitz function j : Γ ×
[t0, t 1] → R. In this case the problem becomes the following evolution hemivariational inequality: For a given K ⊆ W , find ¯ u ∈ K such that
〈¯u′(t)+ A¯u(t)+ B[¯ u(t)] −Φ( t), v − ¯u(t)〉+
∫
Γ
j0(x, t, γ ¯u(x, t ); γv (x)−γ ¯u(x, t ))d σ(x) ≥ 0,
(12) for all v ∈ V , a.e. t ∈ [t0, t 1], where j0 is the Clarke directional derivative of j. We have to note that, although K is a subset of W, in our results on hemivariational inequality (12), K will be endowed with the metric generated by the norm of V. This is motivated by the coercivity condition verified by the operator A.Using the notations (10) and (11), define the Navier-Stokes operator N : V → V ∗
by N u = Au + B[u], for u ∈ V . By , we have the following properties: I. A : V → V ∗ is linear, continuous, symmetric and 〈Au, u 〉 ≥ ν‖u‖2, for all u ∈ V ,II. B : V × V → V ∗ is bilinear, continuous and 〈B(u, v ), v 〉 = 0, for all u, v ∈ V ,III. The mapping B[·] : V → V ∗ is weakly continuous. From these follows that the function b, defined by b(u, v, z ) := 〈B(u, v ), z 〉 is trilinear and continuous. It follows that
〈B[u] − B[v], v − u〉 = 〈B(u − v, v ), u − v〉 ≤ c1 · ‖ u − v‖2 · ‖ v‖, (13) where c1 is a positive constant and u, v ∈ V . We can take c1 = sup ‖v‖,‖w‖=1 〈B(w, v ), w 〉.For u, v, z ∈ V we denote
〈〈A u, v 〉〉 =
∫ t1
t0
〈Au (t), v (t)〉dt, 〈〈B (u, v ), z 〉〉 =
∫ t1
t0
〈B(u(t), v (t)) , z (t)〉dt and
〈〈N u, v 〉〉 =
∫ t1
t0
〈N u (t), v (t)〉dt.
The generalized derivative Lu = u′ defines a linear operator L : W → V ∗ given by
〈〈L u, v 〉〉 =
∫ t1
t0
〈u′(t), v (t)〉dt, for all v ∈ V .
According to
〈〈L u, u 〉〉 =
∫ t1
t0
( 1
2 ‖u(t)‖2
H
)′
dt = 1
2
(‖u(t1)‖2
H
− ‖ u(t0)‖2
H
)
the monotonicity on K of L follows when for any u1, u 2 ∈ K the inequality
‖u2(t0) − u1(t0)‖H ≤ ‖u2(t1) − u1(t1)‖H holds. This happens, for instance, in the periodic case u(t0) = u(t1), in the anti-periodic case u(t0) = −u(t1), and for
K = {u ∈ W : u(t0) = u0} with a given u0 ∈ H.Let M be a nonempty set of functions μ : Γ × [t0, t 1] × Rd × Rd → R with the following properties: (M1) the function ( x, t ) 7 → μ(x, t, r ; s) is Lebesgue measurable for all ( r, s ) ∈ Rd×Rd;(M2) the function ( r, s ) 7 → μ(x, t, r ; s) is Borel measurable for all ( x, t ) ∈ Γ × [t0, t 1]; 11 (M3) there exists θ ∈ [0 , 1] such that for every μ, ¯μ ∈ M there exists a function
ϕμ, ¯μ ∈ L2(Γ × [t0, t 1]) for which
|μ(x, t, r ; s) − ¯μ(x, t, r ; s)| ≤ ϕμ, ¯μ(x, t )|s|θ
for all r, s ∈ Rd, s near 0, a.e. ( x, t ) ∈ Γ × [t0, t 1]. The conditions (M1) and (M2) guarantee that the function (x, t ) 7 →
μ(x, t, u (x, t ), v (x, t )) is Lebesque measurable for all u ∈ H (see , Proposition 6.34). For μ, ¯μ ∈ M we define the distance
d(μ, ¯μ) =
( ∫ t1
t0
∫
Γ
inf
0<ρ ≤1
sup
r,s ∈Rd,0<|s|≤ ρ
|s|−2θ · | μ(x, t, r ; s) − ¯μ(x, t, r ; s)|2dσ(x)d t
) 1
2
.
From (M3) it follows that d(μ, ¯μ) < +∞ for all μ, ¯μ ∈ M .For μ ∈ M and u, v ∈ V denote
Gμ(u; v) =
∫ t1
t0
∫
Γ
μ(x, t, γu (x, t ); γv (x, t ))d σ(x)d t.
Instead of (12) we consider a more general problem: (N S )( μ) Find u ∈ K such that, for all v ∈ K ,
〈〈L u + N u − Φ, v − u〉〉 + Gμ(u; v − u) ≥ 0. (14) We call such a problem hemivariational-like inequality with boundary control variable
μ. Denote by S(μ) the set of solutions of this problem and suppose it is nonempty for all μ ∈ M .Using Theorem 4.1 we are able to prove the following:
Theorem 5.1. Let K ⊂ W be such that the operator L is monotone on K. Let ¯μ ∈ M
be nonisolated and ¯u ∈ S(¯ μ). For ζ > 0, define
c(ζ) := lim sup
v→¯u,v ∈K
‖v − ¯u‖−ζ (G¯μ(¯ u; v − ¯u) + G¯μ(v; ¯ u − v)) (15)
and suppose that there exists τ > 0 such that 0 < c (τ ) < +∞. Suppose further that the conditions (M1)-(M3) are verified, and there exists ρ > 0 such that ‖¯u‖ < ρ and
ρc 1 < ν , where ν is the kinematic viscosity of the fluid, c1 is defined in (13), and the norms ‖ϕμ, ¯μ‖L2 are bounded for μ near ¯μ.Then the mapping μ 7 → S(μ) is H¨ older calm at (¯ μ, ¯u) if and only if one of the following conditions is verified: 1’) τ > 2;2’) τ = 2 and ρc 1 + c(2) < ν ;3’) G¯μ(¯ u, v − ¯u) + G¯μ(v, ¯u − v) ≤ 0 for v near ¯u.Moreover, if one of 1’), 2’) or 3’) is verified, then the solution set S has the isolated calmness property at (¯ μ, ¯u). In the cases 2’) and 3’) the solution ¯u is unique. Proof :12 Let the functions f, g : K × K × M → R be defined by
f (u, v ) := 〈〈A u, v − u〉〉 + 〈〈L u, v − u〉〉
and
g(u, v ; μ) = 〈〈B [u], v − u〉〉 + Gμ(u; v − u) − 〈〈 Φ, v − u〉〉 .
The problem is of the form (PME)( μ) studied in Section 4. The function f is strongly monotone, so the condition (i) of Theorem 4.1 is verified with β = 2 and m = ν. Indeed, for any v ∈ K we have
ν‖¯u − v‖2 ≤ − f (¯ u, v ) − f (v, ¯u) = −f (¯ u, v ) − g(¯ u, v ; ¯ μ) − f (v, ¯u) + g(¯ u, v ; ¯ μ)
≤ [f (¯ u, v ) + g(¯ u, v ; ¯ μ)] − + [ f (v, ¯u) − g(¯ u, v ; ¯ μ)] −.
Condition (ii) is trivially verified. For μ ∈ M and v ∈ K we have
g(¯ u, v ; ¯ μ) + g(v, ¯u; μ) = 〈〈B [¯ u] − B [v], v − ¯u〉〉 + G¯μ(¯ u; v − ¯u) + Gμ(v; ¯ u − v).
As it was mentioned before,
〈〈B [¯ u] − B [v], v − ¯u〉〉 = 〈〈B (¯ u − v, v ), ¯u − v〉〉 ≤ c1 · ‖ ¯u − v‖2 · ‖ v‖.
Further, by ‖¯u‖ ≤ ρ there exists δ > 0 such that, for ‖v − ¯u‖ < δ , we have ‖v‖ < ρ .On the other hand
|G μ(v, ¯u − v) − G ¯μ(v, ¯u − v)| ≤
∫ t1
t0
∫
Γ
|μ(x, t, γv (x, t ); γ ¯u(x, t ) − γv (x, t ))
− ¯μ(x, t, γv (x, t ); γ ¯u(x, t ) − γv (x, t )) |dσ(x)d t
≤
(∫ t1
t0
∫
Γ
|γ ¯u(x, t ) − γv (x, t )|2θ dσ(x)d t
) 1
2
· d(μ, ¯μ)
≤ c0‖v − ¯u‖θd(μ, ¯μ),
where c0 = ‖γ‖θ (see for the last inequality). Let us put
a =
0, if G¯μ(¯ u, v − ¯u) + G¯μ(v, ¯u − v) ≤ 0 for v near ¯ u,
1
2 (ν − ρc 1 + c(2)) if τ = 2 and 0 < c (2) < ν − ρc 1,
1 + c(τ ) if τ = 2 and c(2) ≥ ν − ρc 1 or τ 6 = 2 .
Then, for v ∈ K near ¯ u, we have
G¯μ(¯ u; v − ¯u) + Gμ(v; ¯ u − v)= G¯μ(¯ u; v − ¯u) + G¯μ(v, ¯u − v) − G ¯μ(v, ¯u − v) + Gμ(v; ¯ u − v)
≤ c0‖v − ¯u‖θ · d(μ, ¯μ) + a‖v − ¯u‖τ .
13 For ‖v − ¯u‖ < δ it follows that
g(¯ u, v ; ¯ μ) + g(v, ¯u; μ) ≤ ρc 1‖v − ¯u‖2 + a‖v − ¯u‖τ + c0‖v − ¯u‖θ · d(μ, ¯μ).
In this way, if ‖¯u‖ < ρ and ‖v − ¯u‖ < δ , conditions of Theorem 4.1 are fulfilled with
m = ν, b = c0, c = ρc 1, α = τ − θ, β = 2 and ξ = 1.
Remark 4. (a) Hypothesis ρc 1 < ν suggests that, if the viscosity coefficient ν is small, then the neighbourhood B(0 , ρ ) of 0, where the calmness property holds, is small too. If ν is very small, problems may arise concerning stability and the transition towards turbulent flows (see ). When fluctuations of flow velocity occur at very small spatial and temporal scales, one goes towards the so called turbulent models (see, for instance , , ). (b) Condition 3’) is obviously verified if the bifunction G¯μ is monotone near ¯ u. This is why the condition 0 < c (τ ) < +∞ from (15) is a generalization of the property which is sometimes named relaxed monotonicity (see for instance ). (c) If the function s 7 → μ(x, t, r, s ) is positively homogeneous then
d(μ, ¯μ) =
( ∫ t1
t0
∫
Γ
sup
r,s ∈Rd,0<|s|≤ 1
|s|−2θ · | μ(x, t, r ; s) − ¯μ(x, t, r ; s)|2dσ(x)d t
) 1
2
(16) (d) If the function s 7 → μ(x, t, r, s ) is linear and μ(x, t, r, s ) = μ0(x, t, r )( s) for all
r, s ∈ Rd, a.e. ( x, t ) ∈ Γ × [t0, t 1], then condition (M3), for θ = 1, can be substituted with the following: (M3’) For every μ, ¯μ ∈ M , there exists a function ϕμ, ¯μ ∈ V such that
|μ0(x, t, r ) − ¯μ0(x, t, r )| ≤ ϕμ, ¯μ(x, t ),
for all r ∈ Rd, a.e. ( x, t ) ∈ Γ × [t0, t 1]. In this case we have
d(μ, ¯μ) =
( ∫ t1
t0
∫
Γ
sup
r∈Rd
|μ0(x, t, r ) − ¯μ0(x, t, r )|)2dσ(x)d t
) 1
2
,
where ¯ μ0 is defined similarly to μ0.(e) If μ0 does not depend on ( x, t ), i.e. μ0(x, t, r ) = μ1(r) for every ( x, t, r ) ∈
Γ × [t0, t 1] × Rd then
d(μ, ¯μ) = b1 sup
r∈Rd
|μ1(r) − ¯μ1(r)|,
where b1 = mes(Γ × [t0, t 1]) and ¯ μ1 is defined by ¯ μ0 similar as μ1 by μ0.Finally, let us return to the particular case of evolution hemivariational inequalities governed by the Navier-Stokes operator. Let J be a family of functions j : Γ × [t0, t 1] ×
Rd → R for which the function ( x, t ) 7 → j(x, t, r ) is Lebesgue measurable for all r ∈ Rd
and r 7 → j(x, t, r ) is locally Lipschitz for a.e. ( x, t ). If we put μ(x, t, r ; s) = j0(x, t, r ; s), problem ( N S )( μ) reduces to (12). In this case, by Proposition 2.1 the bifunction
Gμ : V × V → R is well defined and, for θ ∈ [0 , 1], r, s ∈ Rd, and μ, ¯μ ∈ M , the equality (16) holds. As a consequence, for evolution hemivariational inequalities governed by the Navier-Stokes operator, Theorem 5.1 applies. 14 References
Mig´ orski S, Ochal A. Hemivariational inequalities for stationary Navier-Stokes equations. J Math Anal Appl. 2005;306:197–217. Mig´ orski S, Ochal A. Navier-Stokes problems modeled by evolution hemivariational inequalities. Discrete and Continuous Dynamical Systems. 2007;Supplement volume, 2007:731–740. Chebotarev A. Subdifferential boundary value problems for stationary navierstokes equa-tions. Differ Uravn (Differential Equations). 1992;28:1443–1450. Chebotarev A. Stationary variational inequalities in the model of inhomogeneous incom-pressible fluids. Sibirsk Math Zh (Siberian Math J). 1997;38:1184–1193. Konovalova DS. Subdifferential boundary value problems for evolution navierstokes equa-tions. Differ Uravn (Differential Equations). 2000;36:792–798. Chadli O, Ansari QH, Yao JC. Mixed equilibrium problems and anti-periodic solutions for nonlinear evolution equations. J Optim Theory Appl. 2016;168:410–440. Ben Aadi S, Chadli O, Koukkous A. Evolution hemivariational inequalities for non-stationary Navier-Stokes equations: Existence of periodic solutions by an equilibrium problem approach. Minimax Theory and its Applications. 2018;3:107–130. Chadli O, Kassay G, Saidi A. On the existence on antiperiodic solutions for hemi-variational inequalities: an equilibrium problem. Optimization Letters. 2019;:DOI: 10.1007/s11590–019–01490–1. Clarke FH. The generalized problem of Bolza. SIAM J Control and Optimization. 1976; 14:682–699. Dontchev AL, Rockafellar RT. Implicit functions and solution mappings, 2nd edn. Berlin: Springer; 2014. Clarke FH. Optimization and Nonsmooth Analysis. New York: John Wiley & Sons; 1983. Bonnans JF, Shapiro A. Optimization problems with perturbations: A guided tour. Siam Rev. 1998;40:228–264. Levy AB, Poliquin RA, Rockafellar RT. Stability of locally optimal solutions. SIAM Jour-nal on Optimization. 2000;10:580–604. Levy AB. Calm minima in parameterized finite-dimensional optimization. SIAM Journal on Optimization. 2000;11:160–178. Klatte D. On quantitative stability for non-isolated minima. Control and cybernetics. 1994;23:183–200. Yen ND. H¨ older continuity of solutions to a parametric variational inequality. Appl Math Optim. 1995;31:245–255. Mansour MA. Two remarks on the stability of generalized hemivariational inequalities. J Ineq Pure Appl Math. 2002;3:art. 28. Anh LQ, Khanh PQ. On the H¨older continuity of solutions to parametric multivalued vector equilibrium problems. J Math Anal Appl. 2006;321:308–315. Anh LQ, Kruger AY, Thao NH. On H¨older calmness of solution mappings in parametric equilibrium problems. TOP. 2014;22:331–342. Mansour MA, Riahi H. Sensitivity analysis for abstract equilibrium problems. J Math Anal Appl. 2005;306:684–691. Li SJ, Li XB. H¨ older continuity of solutions to parametric weak generalized Ky Fan inequality. J Optimiz Theory App. 2011;149:540–553. Bianchi M, Kassay G, Pini R. Stability of equilibria via regularity of the diagonal subd-ifferential operator. Set-Valued Var Anal. 2017;25:789805. Zhang B, Ng KF, Zheng XY, et al. H¨ older metric subregularity for multifunctions in type
Banach spaces. Optimization. 2016;65:1963–1982. Mig´ orski S, Gamorski P. A new class of quasistatic frictional contact problems governed by a variational-hemivariational inequality. Nonlinear Analysis: Real World Applications. 2019;50:583–602. Inoan D, Kolumb´ an J. Existence theorems via duality for equilibrium problems with
15 trifunctions. Optimization. 2018;67(5):537–547. Xiao Y, Sofonea M. On the optimal control of variational-hemivariational inequalities. J Math Anal Appl. 2019;475:364–384. Chen YQ. On semi-monotone operator theory and applications. J Math Anal Appl. 1999; 231:177–192. Kassay G, Kolumb´ an J. Variational inequalities given by semi-pseudomonotone mappings. Nonlinear Analysis Forum. 2000;5:35–50. Inoan D, Kolumb´ an J. On quasi-equilibrium problems with trifunctions. Minimax Theory and its Applications. 2018;3:161–172. Blum E, Oettli W. From optimization and variational inequalities to equilibrium prob-lems. Math Student. 1994;63:123–145. Temam R. Navier-Stokes equations. Amsterdam: North-Holland; 1979. Zeidler E. Nonlinear functional analysis and its applications II /B, Nonlinear monotone operators. New York: Springer; 1990. Clarke FH. Functional Analysis, Calculus of Variations and Optimal Control. London: Springer-Verlag; 2013. Berselli L, Iliescu T, Layton WJ. Mathematics of large eddy simulation of turbulent flows. Berlin-Heidelberg: Springer; 2006. Foias C, Manley O, Rosa R, et al. Navier-Stokes Equations and Turbulence. Cambridge: Cambridge University Press; 2004. De Lellis C, Szkelyhidi LJ. On turbulence and geometry from Nash to Onsager. Notices of the AMS. 2019;May2019:677–685.
16 |
190709 | https://pubmed.ncbi.nlm.nih.gov/39962011/ | Mullerian anomalies: revisiting imaging and classification - PubMed
Clipboard, Search History, and several other advanced features are temporarily unavailable.
Skip to main page content
An official website of the United States government
Here's how you know
The .gov means it’s official.
Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site.
The site is secure.
The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely.
Log inShow account info
Close
Account
Logged in as:
username
Dashboard
Publications
Account settings
Log out
Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation
Search: Search
AdvancedClipboard
User Guide
Save Email
Send to
Clipboard
My Bibliography
Collections
Citation manager
Display options
Display options
Format
Save citation to file
Format:
Create file Cancel
Email citation
Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page.
To:
Subject:
Body:
Format:
[x] MeSH and other data
Send email Cancel
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Add to My Bibliography
My Bibliography
Unable to load your delegates due to an error
Please try again
Add Cancel
Your saved search
Name of saved search:
Search terms:
Test search terms
Would you like email updates of new search results? Saved Search Alert Radio Buttons
Yes
No
Email: (change)
Frequency:
Which day?
Which day?
Report format:
Send at most:
[x] Send even when there aren't any new results
Optional text in email:
Save Cancel
Create a file for external citation management software
Create file Cancel
Your RSS Feed
Name of RSS Feed:
Number of items displayed:
Create RSS Cancel
RSS Link Copy
Full text links
SpringerFree PMC article
Full text links
Actions
Cite
Collections
Add to Collections
Create a new collection
Add to an existing collection
Name your collection:
Name must be less than 100 characters
Choose a collection:
Unable to load your collection due to an error
Please try again
Add Cancel
Permalink
Permalink
Copy
Display options
Display options
Format
Page navigation
Title & authors
Abstract
Conflict of interest statement
Figures
Similar articles
References
Related information
LinkOut - more resources
Insights Imaging
Actions
Search in PubMed
Search in NLM Catalog
Add to Search
. 2025 Feb 17;16(1):40.
doi: 10.1186/s13244-024-01879-2.
Mullerian anomalies: revisiting imaging and classification
Rashmi Dixit1,Chitty Suvarna Duggireddy2,Gaurav Shanker Pradhan1
Affiliations Expand
Affiliations
1 Department of Radiodiagnosis, Maulana Azad Medical College and associated Lok Nayak Hospital, New Delhi, India.
2 Department of Radiodiagnosis, Maulana Azad Medical College and associated Lok Nayak Hospital, New Delhi, India. duggireddychittysuvarna@gmail.com.
PMID: 39962011
PMCID: PMC11832971
DOI: 10.1186/s13244-024-01879-2
Item in Clipboard
Mullerian anomalies: revisiting imaging and classification
Rashmi Dixit et al. Insights Imaging.2025.
Show details
Display options
Display options
Format
Insights Imaging
Actions
Search in PubMed
Search in NLM Catalog
Add to Search
. 2025 Feb 17;16(1):40.
doi: 10.1186/s13244-024-01879-2.
Authors
Rashmi Dixit1,Chitty Suvarna Duggireddy2,Gaurav Shanker Pradhan1
Affiliations
1 Department of Radiodiagnosis, Maulana Azad Medical College and associated Lok Nayak Hospital, New Delhi, India.
2 Department of Radiodiagnosis, Maulana Azad Medical College and associated Lok Nayak Hospital, New Delhi, India. duggireddychittysuvarna@gmail.com.
PMID: 39962011
PMCID: PMC11832971
DOI: 10.1186/s13244-024-01879-2
Item in Clipboard
Full text links Cite
Display options
Display options
Format
Abstract
Mullerian duct anomalies (MDA) are a group of uncommon but treatable causes of infertility and pregnancy complications. This review describes the embryology, American Society of Reproductive Medicine (ASRM) classification 2021, and corresponding imaging features of MDA. The three phases of embryological development of Mullerian duct structures are described. The main emphasis is on the ASRM 2021 classification of MDA into nine descriptive categories, while the European Society of Human Reproduction and Embryology and the European Society for Gynecologic Endoscopy (ESHRE/ESGE) classification is also briefly described where necessary. MRI imaging features of MDA along with the acquisition techniques are discussed in detail, as MRI is the ideal imaging modality for MDA diagnosis. In addition, the current role of imaging modalities such as hysterosalpingography and ultrasound including 3D transvaginal ultrasound is also elucidated. The review aims to revisit the MRI imaging features of Mullerian anomalies and reiterates that an accurate description of each anomaly and precise communication with clinicians is the priority rather than rigidly fitting the anomaly into one particular category. CRITICAL RELEVANCE STATEMENT: The ASRM 2021 classification of Mullerian anomalies has re-defined the criteria for an arcuate uterus. Radiologists must know of the new classification and imaging features and try to describe each anomaly accurately rather than forcefully fitting an anomaly into a definite category. KEY POINTS: MDA has an important role in infertility and pregnancy complications. Knowledge of imaging features helps radiologists aid patient management; MRI is the preferred imaging modality for MDAs. An accurate MRI-based description of Mullerian anomalies is crucial, avoiding the pitfalls of rigid categorization.
Keywords: Embryology; Infertility; Magnetic resonance imaging; Mullerian duct anomalies; Reproductive medicine.
© 2025. The Author(s).
PubMed Disclaimer
Conflict of interest statement
Declarations. Ethics approval and consent to participate: Not applicable. Consent for publication: Not applicable. Competing interests: The authors declare that they have no competing interests.
Figures
Fig. 1
Development of female internal genitalia. …
Fig. 1
Development of female internal genitalia. a Indifferent gonads with undifferentiated genital ducts—Mullerian/paramesonephric ducts…
Fig. 1
Development of female internal genitalia. a Indifferent gonads with undifferentiated genital ducts—Mullerian/paramesonephric ducts lying lateral to mesonephric ducts at the cranial end and medial to them at the caudal end. b In female fetuses, mesonephric duct involution occurs gradually with Mullerian duct fusion and septal resorption eventually. c Uterus, cervix, and vagina with supports of the uterus and mesonephric remnants (epoöphoron, paroöphoron, and Gartner’s duct cyst)
Fig. 2
ESHRE/ESGE classification of Mullerian anomalies. …
Fig. 2
ESHRE/ESGE classification of Mullerian anomalies. # The interostial line connects the uterine ostia…
Fig. 2
ESHRE/ESGE classification of Mullerian anomalies. #The interostial line connects the uterine ostia of the fallopian tubes. Uterine fundal cleft or external indentation is used to differentiate fusion anomalies (didelphys and bicornuate) from resorption anomalies (septate and arcuate), which can be precisely demonstrated on a true coronal image . Fundal cleft > 1 cm or external indentation > 50% of uterine wall thickness suggests fusion anomalies, whereas < 1 cm or external indentation < 50% of uterine wall thickness represents resorption anomalies
Fig. 3
Diagrammatic representation of normal uterus
Fig. 3
Diagrammatic representation of normal uterus
Fig. 3
Diagrammatic representation of normal uterus
Fig. 4
Diagrammatic representation and imaging features…
Fig. 4
Diagrammatic representation and imaging features of Mullerian agenesis. 4.1: Diagrammatic representation of variants…
Fig. 4
Diagrammatic representation and imaging features of Mullerian agenesis. 4.1: Diagrammatic representation of variants of Mullerian agenesis. 4.2: MRKH syndrome type 1: (a) USG pelvis of a seventeen-year-old girl 46XX karyotype in the transverse plane reveals an ovoid hypoechoic structure posterior to the urinary bladder with poor zonal anatomy (arrow) suggesting the possibility of a hypoplastic uterus. Sagittal T2W MR image (b) reveals a small hypointense linear structure posterior to the bladder in the expected location of the uterus (arrow). The sagittal image obtained after partially emptying the bladder (c) better depicts the rudimentary zonal anatomy confirming the presence of a hypoplastic uterus(thick arrow). Also, note the vagina seen as a linear hyperintensity between the bladder and rectum (thin arrow). T2W coronal image (d) of the upper abdomen reveals both kidneys to be normal
Fig. 5
Imaging features of androgen insensitivity…
Fig. 5
Imaging features of androgen insensitivity syndrome. USG image in the transverse plane ( …
Fig. 5
Imaging features of androgen insensitivity syndrome. USG image in the transverse plane (a) of a seventeen-year-old phenotypic female with primary amenorrhea and 45XY karyotype reveals a linear echogenic structure posterior to the bladder likely vaginal canal with no evidence of any uterine tissue or ovaries. T2 weighted sagittal MR image (b) demonstrates a linear hyperintense vaginal canal (arrow) between the urinary bladder anteriorly, rectum, and anus posteriorly. Note the absence of a cervix or uterus. T2W axial image at the level of the inguinal canal (c) shows bilateral testes with epididymis seen in the inguinal region (arrowheads)
Fig. 6
Diagrammatic representation and imaging features…
Fig. 6
Diagrammatic representation and imaging features of cervical agenesis. 6.1: Diagrammatic representation of variants…
Fig. 6
Diagrammatic representation and imaging features of cervical agenesis. 6.1: Diagrammatic representation of variants of isolated cervical agenesis. Variants can have variable cervical length. 6.2 Isthmic agenesis: T2W coronal (a) and T1W axial (b) images of a nineteen-year-old girl with primary amenorrhea (a) reveal a normal-sized uterus with the endometrial cavity distended by T2 hypointense and T1 hyperintense contents suggesting blood products. Note the cervix (thin arrow) in a with no communication between the uterus and cervical canal. The uterus is pushed to the right by a hemorrhagic left tubo-ovarian mass due to retrograde menstruation. The normal H-shaped vagina (thick arrow) is well seen on a more inferior T2W axial image (c)
Fig. 7
Diagrammatic representation and imaging features…
Fig. 7
Diagrammatic representation and imaging features of unicornuate uterus. 7.1: Diagrammatic representation of variants…
Fig. 7
Diagrammatic representation and imaging features of unicornuate uterus. 7.1: Diagrammatic representation of variants of unicornuate uterus. 7.2. Unicornuate uterus. T2-FS coronal MR image (a) in a 31-year-old female with primary infertility reveals an off-midline left-sided banana-shaped uterine horn (black arrow) continuous with the cervix inferiorly s/o left-sided unicornuate uterus. 3D transvaginal USG image (b) in another patient shows a left-sided unicornuate uterus (white arrow) with a small non-cavitary non-communicating rudimentary horn on the right side (arrowhead). The right ovary is normal (asterisk)
Fig. 8
Diagrammatic representation and imaging features…
Fig. 8
Diagrammatic representation and imaging features of uterus didelphys. 8.1: Diagrammatic representation of variants…
Fig. 8
Diagrammatic representation and imaging features of uterus didelphys. 8.1: Diagrammatic representation of variants of uterus didelphys. 8.2. Uterus didelphys with double cervix and vaginal septum. Axial T2-weighted (a–c), and T2-FS coronal (d) MR images in a twenty-year-old female patient demonstrate two widely divergent uterine cavities showing normal zonal anatomy with no communication between them (a). Two cervical canals (arrows in b, d) are seen separated by an intervening myometrium suggesting a double cervix giving an owl’s eye appearance with increased intercervical distance. Two vaginal canals (arrowheads) are better appreciated in (c)
Fig. 9
Diagrammatic representation and imaging features…
Fig. 9
Diagrammatic representation and imaging features of bicornuate uterus. 9.1: Diagrammatic representation of variants…
Fig. 9
Diagrammatic representation and imaging features of bicornuate uterus. 9.1: Diagrammatic representation of variants of bicornuate uterus. 9.2. Bicornuate uterus with single cervix. T2 weighted coronal MR image demonstrates two widely separate endometrial cavities with external indentation greater than 1.5 cm (double-edged arrow) with a single cervix (thick arrow)
Fig. 10
Diagrammatic representation and imaging features…
Fig. 10
Diagrammatic representation and imaging features of septate uterus. 10.1: Diagrammatic representation of variants…
Fig. 10
Diagrammatic representation and imaging features of septate uterus. 10.1: Diagrammatic representation of variants of septate uterus. 10.2. Complete septate uterus with obstructed hemivagina and ipsilateral renal agenesis. T2 weighted coronal and T1 weighted axial MR images (a, b) in a twenty-one-year-old female patient demonstrate two endometrial cavities separated by a septum (arrow) extending into the cervix and vagina, the right hemivagina is distended with T2 hyperintense (black asterisk in a) and mild T1 hyperintense contents (white asterisk in b) suggesting blood products. To advantage, T2 weighted axial MR image (c) depicts the two vaginal canals separated by a longitudinal septum (short arrow) with layering of blood products as dependent hypointense and upper hyperintense contents (arrowhead) in distended right hemivagina. T2 weighted coronal MR image of the upper abdomen (d) reveals the absence of the right kidney. 10.3. Partial septate uterus. Axial T2 weighted MR image of a twenty-six-old year female patient with recurrent first-trimester pregnancy loss demonstrates a straight external fundal contour (black arrowheads) with internal indentation greater than 1.5 cm (double-headed arrow) from the interostial line (dashed line). The upper part of the septum is isointense to the adjacent myometrium suggesting the muscular septum does not reach up to the internal os. Note made of a tubo-ovarian hemorrhagic mass in the left adnexa
Fig. 11
Differences between arcuate, septate, and…
Fig. 11
Differences between arcuate, septate, and bicornuate uteri with imaging features of the arcuate…
Fig. 11
Differences between arcuate, septate, and bicornuate uteri with imaging features of the arcuate uterus. 11.1: Diagrammatic representation of differences between arcuate, septate, and bicornuate uteri. 11.2. Arcuate uterus: axial T2W MR image (a) of an arcuate uterus in a 26-year-old female patient with recurrent pregnancy loss shows a normal external fundal outline (white arrowheads) with an internal indentation lesser than 1.5 cm (double-headed arrow) from the interostial line. Note normal ovaries bilaterally (arrows). b 3D transvaginal USG image shows a broad saddle-shaped indentation of the arcuate uterus
Fig. 12
Diagrammatic representation and imaging features…
Fig. 12
Diagrammatic representation and imaging features of developmental vaginal anomalies. 12.1: Diagrammatic representation of…
Fig. 12
Diagrammatic representation and imaging features of developmental vaginal anomalies. 12.1: Diagrammatic representation of variants of the transverse vaginal septum and imperforate hymen. 12.2: Diagrammatic representation of longitudinal vaginal septum. 12.3: Transverse vaginal septum. T2W coronal image (a) in a nineteen-year-old female with primary amenorrhea shows a hypointense linear structure at the inferior aspect of vaginal cavity (white arrow) with upstream distension of vagina and endometrial cavity showing heterogeneously hypointense contents (white asterisks) within which appear hyperintense (black asterisks) on T1WI axial images (a) and (b) suggestive of hematometrocolpos
All figures (12)
See this image and copyright information in PMC
Similar articles
The accuracy of three-dimensional ultrasonography in the diagnosis of Müllerian duct anomalies and its concordance with magnetic resonance imaging.Cekdemir YE, Mutlu U, Acar D, Altay C, Secil M, Dogan OE.Cekdemir YE, et al.J Obstet Gynaecol. 2022 Jan;42(1):67-73. doi: 10.1080/01443615.2021.1877646. Epub 2021 May 3.J Obstet Gynaecol. 2022.PMID: 33938374
Comparison of the ESHRE-ESGE and ASRM classifications of Müllerian duct anomalies in everyday practice.Ludwin A, Ludwin I.Ludwin A, et al.Hum Reprod. 2015 Mar;30(3):569-80. doi: 10.1093/humrep/deu344. Epub 2014 Dec 22.Hum Reprod. 2015.PMID: 25534461 Free PMC article.
MRI Evaluation of Mullerian Duct Anomalies: Practical Classification by the New ASRM System.Al Najar MS, Al Ryalat NT, Sadaqah JS, Husami RY, Alzoubi KH.Al Najar MS, et al.J Multidiscip Healthc. 2022 Nov 9;15:2579-2589. doi: 10.2147/JMDH.S386936. eCollection 2022.J Multidiscip Healthc. 2022.PMID: 36388626 Free PMC article.
Mullerian duct anomalies: from diagnosis to intervention.Chandler TM, Machan LS, Cooperberg PL, Harris AC, Chang SD.Chandler TM, et al.Br J Radiol. 2009 Dec;82(984):1034-42. doi: 10.1259/bjr/99354802. Epub 2009 May 11.Br J Radiol. 2009.PMID: 19433480 Free PMC article.Review.
The spectrum of imaging appearances of müllerian duct anomalies: focus on MR imaging.Fukunaga T, Fujii S, Inoue C, Mukuda N, Murakami A, Tanabe Y, Harada T, Ogawa T.Fukunaga T, et al.Jpn J Radiol. 2017 Dec;35(12):697-706. doi: 10.1007/s11604-017-0681-4. Epub 2017 Sep 18.Jpn J Radiol. 2017.PMID: 28921452 Review.
See all similar articles
References
Behr SC, Courtier JL, Qayyum A (2012) Imaging of müllerian duct anomalies. Radiographics 32:E233–E250 - PubMed
Al Najar MS, Al Ryalat NT, Sadaqah JS, Husami RY, Alzoubi KH (2022) MRI evaluation of Mullerian duct anomalies: practical classification by the new ASRM system. J Multidiscip Healthc 15:2579–2589 - PMC - PubMed
Pfeifer SM, Attaran M, Goldstein J et al (2021) ASRM müllerian anomalies classification 2021. Fertil Steril 116:1238–1252 - PubMed
Ludwin A, Ludwin I (2015) Comparison of the ESHRE–ESGE and ASRM classifications of Müllerian duct anomalies in everyday practice. Hum Reprod 30:569–580 - PMC - PubMed
Robbins JB, Parry JP, Guite KM et al (2012) MRI of pregnancy-related issues: Mullerian duct anomalies. AJR Am J Roentgenol 198:302–310 - PubMed
Show all 31 references
Related information
MedGen
LinkOut - more resources
Full Text Sources
PubMed Central
Springer
Full text links[x]
SpringerFree PMC article
[x]
Cite
Copy Download .nbib.nbib
Format:
Send To
Clipboard
Email
Save
My Bibliography
Collections
Citation Manager
[x]
NCBI Literature Resources
MeSHPMCBookshelfDisclaimer
The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited.
Follow NCBI
Connect with NLM
National Library of Medicine
8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov |
190710 | https://www.britannica.com/science/Stefan-Boltzmann-law | SUBSCRIBE
SUBSCRIBE
Home
History & Society
Science & Tech
Biographies
Animals & Nature
Geography & Travel
Arts & Culture
ProCon
Money
Games & Quizzes
Videos
On This Day
One Good Fact
Dictionary
New Articles
History & Society
Lifestyles & Social Issues
Philosophy & Religion
Politics, Law & Government
World History
Science & Tech
Health & Medicine
Science
Technology
Biographies
Browse Biographies
Animals & Nature
Birds, Reptiles & Other Vertebrates
Bugs, Mollusks & Other Invertebrates
Environment
Fossils & Geologic Time
Mammals
Plants
Geography & Travel
Geography & Travel
Arts & Culture
Entertainment & Pop Culture
Literature
Sports & Recreation
Visual Arts
Image Galleries
Podcasts
Summaries
Top Questions
Britannica Kids
Ask the Chatbot Games & Quizzes History & Society Science & Tech Biographies Animals & Nature Geography & Travel Arts & Culture ProCon Money Videos
Stefan-Boltzmann law
Introduction References & Edit History Quick Facts & Related Topics
Quizzes
Physics and Natural Law
Stefan-Boltzmann law
physics
Print
verifiedCite
While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions.
Select Citation Style
Share
Share to social media
Facebook X
URL
Feedback
Thank you for your feedback
Our editors will review what you’ve submitted and determine whether to revise the article.
External Websites
American Meteorological Society Journal - The Role of the Nonlinearity of the Stefan–Boltzmann Law on the Structure of Radiatively Forced Temperature Change
Arizona State University - Stefan-Boltzmann law
Boston University Arts and Sciences - Physics - The Stefan Boltzmann Law
American Meteorological Society - Journal of Climate - The Role of the Nonlinearity of the Stefan?Boltzmann Law on the Structure of Radiatively Forced Temperature Change
EPJ Web of Conferences - On the effective Stefan-Boltzmann law and the thermodynamic origin of the initial radiation density in warm inflation
Joint Institute for Laboratory Astrophysics - The Stefan-Boltzmann Law
Florida International University - Stefan-Boltzmann Law
The College of Wooster - Physics Department - The Stefan-Boltzmann Law
Also known as: Stefan’s law
Written by
Written by
The Editors of Encyclopaedia Britannica
Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree....
The Editors of Encyclopaedia Britannica
Last Updated: •Article History
Key People:
: Ludwig Boltzmann
: Josef Stefan
Related Topics:
: heat
: thermal radiation
: blackbody
See all related content
Stefan-Boltzmann law, statement that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature. Formulated in 1879 by Austrian physicist Josef Stefan as a result of his experimental studies, the same law was derived in 1884 by Austrian physicist Ludwig Boltzmann from thermodynamic considerations: if E is the radiant heat energy emitted from a unit area in one second (that is, the power from a unit area) and T is the absolute temperature (in kelvins), then E = σT4, the Greek letter sigma (σ) representing the constant of proportionality, called the Stefan-Boltzmann constant. This constant has the value 5.670374419 × 10−8 watt per metre2 per K4. The law applies only to blackbodies, theoretical surfaces that absorb all incident heat radiation.
The Editors of Encyclopaedia Britannica This article was most recently revised and updated by Erik Gregersen. |
190711 | https://www.edn.com/squared-squares-who-knew/ | Squared Squares (Who Knew?) - EDN
Squared Squares (Who Knew?) - EDN
Aspencore networkNews & Analytics Products Design Tools About Us AspenCore Network News the global electronics community can trust
eetimes.com The trusted news source for power-conscious design engineers
powerelectronicsnews.com Supply chain news for the electronics industry
ebnonline.com The can't-miss forum engineers and hobbyists
elektroda.pl News the global electronics community can trust
eetimes.com The trusted news source for power-conscious design engineers
powerelectronicsnews.com Supply chain news for the electronics industry
ebnonline.com The can't-miss forum engineers and hobbyists
elektroda.pl News the global electronics community can trust
eetimes.com The trusted news source for power-conscious design engineers
powerelectronicsnews.com Supply chain news for the electronics industry
ebnonline.com The can't-miss forum engineers and hobbyists
elektroda.pl AspenCore Network Product news that empowers design decisions
electronicproducts.com Design engineer' search engine for electronic components
datasheets.com The electronic components resource for engineers and purchasers
eem.com Product news that empowers design decisions
electronicproducts.com Design engineer' search engine for electronic components
datasheets.com The electronic components resource for engineers and purchasers
eem.com Product news that empowers design decisions
electronicproducts.com Design engineer' search engine for electronic components
datasheets.com AspenCore Network The design site for hardware software, and firmware engineers
embedded.com Where makers and hobbyists share projects
electroschematics.com The design site for electronics engineers and engineering managers
edn.com The learning center for future and novice engineers
electronics-tutorials.ws The educational resource for the global engineering community
techonline.com The design site for hardware software, and firmware engineers
embedded.com Where makers and hobbyists share projects
electroschematics.com The design site for electronics engineers and engineering managers
edn.com The learning center for future and novice engineers
electronics-tutorials.ws The educational resource for the global engineering community
techonline.com The design site for hardware software, and firmware engineers
embedded.com Where makers and hobbyists share projects
electroschematics.com AspenCore Network Where electronics engineers discover the latest toolsThe design site for hardware software, and firmware engineers
eeweb.com Circuit simulation made easy
partsim.com Brings you all the tools to tackle projects big and small - combining real-world components with online collaboration
schematics.com Hardware design made easy
pcbweb.com A free online environment where users can create, edit, and share electrical schematics, or convert between popular file formats like Eagle, Altium, and OrCAD.
schematics.io Find the IoT board you’ve been searching for using this interactive solution space to help you visualize the product selection process and showcase important trade-off decisions.
transim.com/iot Transform your product pages with embeddable schematic, simulation, and 3D content modules while providing interactive user experiences for your customers.
transim.com/Products/Engage Where electronics engineers discover the latest toolsThe design site for hardware software, and firmware engineers
eeweb.com Circuit simulation made easy
partsim.com Brings you all the tools to tackle projects big and small - combining real-world components with online collaboration
schematics.com Hardware design made easy
pcbweb.com A free online environment where users can create, edit, and share electrical schematics, or convert between popular file formats like Eagle, Altium, and OrCAD.
schematics.io AspenCore Network A worldwide innovation hub servicing component manufacturers and distributors with unique marketing solutions
aspencore.com SiliconExpert provides engineers with the data and insight they need to remove risk from the supply chain.
siliconexpert.com Transim powers many of the tools engineers use every day on manufacturers' websites and can develop solutions for any company.
transim.com A worldwide innovation hub servicing component manufacturers and distributors with unique marketing solutions
aspencore.com SiliconExpert provides engineers with the data and insight they need to remove risk from the supply chain.
siliconexpert.com Transim powers many of the tools engineers use every day on manufacturers' websites and can develop solutions for any company.
transim.com A worldwide innovation hub servicing component manufacturers and distributors with unique marketing solutions
aspencore.com SiliconExpert provides engineers with the data and insight they need to remove risk from the supply chain.
siliconexpert.com
About Us
Advertise
Advertisement
EDN
Voice of the Engineer
Menu
Home
Design Centersexpand child menu
5G
Automotive
Communications
Components & Packaging
Consumer
DIY
IC Design
Medical
Power Management
Systems Design
Test & Measurement
Products & Teardownsexpand child menu
Products
Teardowns
Electroschematicsexpand child menu
Projects + Tutorials
Datasheets
Theory
Service Manuals
Gadgets
Tips
Blogs
Design Ideas
Subscribe
Login
Search
Search for:
Menu Search
Close
Search for:
Home
Design Centersexpand child menu
5G
Automotive
Communications
Components & Packaging
Consumer
DIY
IC Design
Medical
Power Management
Systems Design
Test & Measurement
Products & Teardownsexpand child menu
Products
Teardowns
Electroschematicsexpand child menu
Projects + Tutorials
Datasheets
Theory
Service Manuals
Gadgets
Tips
Blogs
Design Ideas
Subscribe
Login
News
Squared Squares (Who Knew?)
February 14, 2018
byMax Maxfield
Comments 0
Advertisement
Generating Squared Squares is not an easy task, until you realize that you can represent them as an electrical network and use math tools developed for engineering to solve the problem.
Generating Squared Squares is not an easy task, until you realize that you can represent them as an electrical network and use math tools developed for engineering to solve the problem.
Advertisement
My chum, Jay Dowling, is always sending me links to interesting stuff I never even knew I didn’t know (if you see what I mean). For example, I just learned that the number square shown in the photo below is the logo of the Math Society at Trinity College in Cambridge, England.
Advertisement
Partner Content
Leading the AI Evolution with Innovative 2mm Flash Memory Assembly Technology
09.26.2025
GUC Launches Next – Generation 2.5D/3D APT Platform Leveraging TSMC’s Latest 3DFabric® and Advanced Process Technologies
09.24.2025
How Huawei Reduces Design Time and Improves Confidence in Thermal Models
09.24.2025
(Source: Screenshot from YouTube)
All of the sub-squares are different sizes — for example, the big 50 in the upper left-hand corner indicates that its associated square is 50 units on each side — but they fit together to form the big square, which is 112 units on each side. The result is called a Squared Square. The reason the members of the Trinity College Math Society selected this as their logo — apart from the fact that they are all nerds, of course — is that this is the smallest possible Squared Square.
It turns out that generating Squared Squares is not an easy task. Before tackling Squared Squares, the Math Society started considering Squared Rectangles. One of the things they thought of was extending all the horizontal lines, and then representing each square as a vector connecting two of the lines. It then struck them that, if each vector represented a wire/current, then the result would be similar to an electrical network as illustrated below.
(Source: Screenshot from YouTube)
In turn, this meant that they could borrow all of the mathematical tools that had been created to address electrical networks and use those tools to solve the Squared Square problem. As you’ll see in this Numberphile video on Squared Squares, the resulting proofs are both clever and elegant.
I’m always amazed when I see something like this. It makes me realize how ingenious some people are to spot these underlying relationships in the first place, and to then to use them to perform the most amazing mental gymnastics. What about you? Were you already aware of this, or were you as surprised as I to see this video?
TagsEEWeb.com Archive
Max Maxfield
AboutMax Maxfield
When I was at university, we used analog computers and no one had conceived of anything like digital signal processing (DSP). Similarly, I pretty much predate electronic design automation (EDA) as we know and love it today. When I started my first career as a hardware design engineer, we captured our schematics at the gate-level using pencil and paper. Timing analysis involved identifying the critical paths and then adding the delays by hand (no one I knew could afford an electronic calculator). Functional verification involved the other engineers looking at your schematics and saying, "That looks like it will work." My first job out of college was as a member of a team designing CPUs for mainframe computers. Over the years, I've designed everything from silicon chips to circuit boards, and from brainwave amplifiers to steampunk "Display-O-Meters." Some of the things I built even worked. I also started writing magazine articles and books, all of which eventually led to my second career as a writer and editor (my mom still cannot believe I now make my living writing). So now, instead of actually doing engineering, I spend my days telling other people how to do it and how much harder things were when I was a lad.
Advertisement
0 comments on “Squared Squares (Who Knew?)”
Leave a Reply Cancel reply
You must Sign in or Register to post a comment.
Post navigation
Previous Post
Next Post
partner content
Leading the AI Evolution with Innovative 2mm Flash Memory Assembly Technology
09.26.2025
GUC Launches Next – Generation 2.5D/3D APT Platform Leveraging TSMC’s Latest 3DFabric® and Advanced Process Technologies
09.24.2025
How Huawei Reduces Design Time and Improves Confidence in Thermal Models
09.24.2025
Advertisement
Advertisement
Recent Posts
Hybrid system resolves edge AI’s on-chip memory conundrum
DC series motor caution
R&S expands VNA lineup to 54 GHz
2-in-1 SiC module raises power density
Redriver strengthens USB4v2 and DP 2.1a signals
Archives
Archives
Categories
Categories Search for:
Recent Comments
Robert Yankowitz on Unusual 2N3904 transistor circuit
mawyatt on Unusual 2N3904 transistor circuit
Robert Yankowitz on Unusual 2N3904 transistor circuit
asa70 on Unusual 2N3904 transistor circuit
mawyatt on Unusual 2N3904 transistor circuit
Advertisement
Aspencore Network
Products:
Electronic Products
TechOnline
News & Analysis:
EE Times
EE Times Europe
Power Electronics News
Elektroda.pl
Design:
EDN
Electronics-Tutorials
Planet Analog
Embedded
Electronics Know How
Tools:
EEWEB
Global Network
EE Times Asia
EE Times China
EE Times India
EE Times Taiwan
EE Times Japan
EDN Asia
EDN Taiwan
ESM China
EDN China
EDN Japan
For Advertisers
Contact Sales
Media Guide Request
Connect with us
Facebook
Twitter
Instagram
LinkedIn
YouTube
All contents are Copyright © 2025 by AspenCore, Inc. All Rights Reserved.
Contact UsAbout UsPrivacy PolicyTerms of UseSite MapNewslettersCalifornia Do Not Sell
[ninja_form id=2]
×
PowerUP is a free virtual eventthat will focus on critical topics such as power semiconductors, power management, sensors, and advanced power conversion technologies .
Register for Free>>
Close |
190712 | https://www.onemathematicalcat.org/Math/Precalculus_obj/ellipseDefn.htm | HomeAbout MeMain Table of ContentsTestimonialsContact
Definition of an Ellipse
The audio controller is at the bottom of the web page. (Look down!) This gives easy access while reading. The yellow highlighting follows my voice!
Depending on your connection speed, the audio may take some time to load. (It will display 0:00 until it's ready.) Of course I'm biased—but I think it's worth the wait!
(Click for cat book)
Ellipses were introduced in
Introduction to Conic Sections,
as one of
several different curves
(‘conic sections’)
that are formed by intersecting a plane with an
infinite double cone.
Identifying Conics by the Discriminant
introduced the general equation for any conic
section,
and gave conditions under which the graph
would be an ellipse.
In this current section,
we present and explore the standard
definition of an ellipse.
This definition facilitates
the derivation of standard equations for ellipses.
Recall that the notation
‘d(P,Q)’
denotes the distance between points P and Q.
DEFINITION
ellipse
An ellipse is the set of points
in a plane such that
the sum of the distances to
two fixed points is constant.
More precisely:
Let F1 and F2 be points;
they are called the foci
of the ellipse (pronounced FOE-sigh).
(The singular form of ‘foci’
is ‘focus’.)
Let k be a positive real number,
with k>d(F1,F2).
In this section, k
is referred to as the ellipse constant.
The ellipse determined by F1, F2
and k is the
set of all points P in a plane such that:
the sum of the distances to two fixed pointsd(P,F1)+d(P,F2)is=constantk
P is a general point on the ellipse.
d(P,F1)+d(P,F2)=constant
Note from Dr. Burns (the website creator)
Welcome—so glad you're here!
My sister and I have finished our house-clean-out adventure! Here it is on Zillow (30 Mt Laurel Way, Monterey, MA). Here are more photos (with captions).
Want to say hello? Sign my guestbook!
Old-Fashioned Playing with the Definition of an Ellipse
Got a piece of cardboard, paper, tape,
string/cord (not stretchy), and pen/pencil?
Then, you can create your own ellipse:
Tape the paper to the cardboard
(at the corners is sufficient).
Punch two small holes through
the paper/cardboard at the desired foci.
Put the cord/string through
the two holes from front to back,
and tie securely on the back.
The length of
the string that protrudes in the front,
when held taut by a pen (see photo),
is the ellipse constant.
Keeping the string taut,
trace the ellipse.
(The string gets a bit twisted
near the line through the two foci.
So, draw one continuous motion for
the upper half, then re-position and draw the lower half.)
(The sunflower in a vase
is optional. ☺
I grew my own sunflowers from seed in 2017,
when I was writing this section!)
More Playing with the Definition of an Ellipse
You can also play with ellipses using the dynamic
JSXGraph below:
0,0
F1
F2
The distance between the foci is: 10.0
12.00
The ellipse constant (slider value) is: 12.0
P
d(F1,P)+d(F2,P)=
12.0
F1 and F2 are the foci.
Move them around!
As you hover over each focus,
you can see the coordinates of the point.
The slider at the top sets
the ellipse constant.
The slider can be set to numbers
between 0 and 20 with increments
of 0.5.
The starting value is 12
(refresh the page as needed).
The current distance between
the foci is displayed near the top.
Watch this distance change
as you move the foci around.
The starting value for
d(F1,F2) is 10 (refresh the page as needed).
In order to see an ellipse,
the ellipse constant (slider value)
must be greater than the distance between the foci.
P is a general point on the ellipse.
Move it around!
When the ellipse constant equals
the distance between the foci,
the ‘ellipse’ degenerates
to a line segment.
Ignore the line segment
that you see when the ellipse constant
k is less than the distance
between the foci.
As discussed below, in this case
there are actually no points
P that satisfy d(F1,P)+d(F2,P)=k.
Notes
In the definition of ellipse,
the ellipse constant k is
required to be
strictly greater than
the distance between the two foci.
Why?
As shown below, other values of k don't give anything that
a reasonable person would want to call an ellipse!
A ‘Line Segment’ Ellipse: k=d(F1,F2)
Suppose the ellipse constant, k,
equals the distance between the foci:
that is, k=d(F1,F2).
In this case, the solution set to the equation
d(P,F1)+d(P,F2)=k
is the line segment between
F1 and F2 (including the endpoints).
Most people don't want to call a line segment
an ellipse!
This is why k is not allowed to
equal d(F1,F2) in the definition of ellipse.
d(P,F1) is the length of the
green segment
d(P,F2) is the length of the
red segment
Together, the green and red segments give d(F1,F2)
An ‘Empty’ Ellipse: k<d(F1,F2)
The shortest
distance between any two points
is a straight line.
In particular, the shortest distance from
F1 to F2 is the length of
the line segment between them,
and is denoted by d(F1,F2).
Thus, any path from F1
to F2 must have length greater than
or equal to d(F1,F2).
In particular (refer to sketch above),
the piecewise-linear path
from F1 to P
and then
from P to F2
always has length greater than or equal to d(F1,F2).
Therefore, if the ellipse constant
k is strictly less than d(F1,F2),
there are no points P
that make the following equation true:
always ≥d(F1,F2)d(P,F1)+d(P,F2)=<d(F1,F2)k
You might want to call this an empty ellipse, an
invisible ellipse, or an imaginary ellipse!
There's nothing there!
Master the ideas from this section by practicing below:
When you're done practicing, move on to:
Reflecting Property of an Ellipse
Concept Practice
Choose a specific problem type, or click ‘New problem’ for a random question.
Think about your answer.
Click ‘Check your answer’ to check!
PROBLEM TYPES:
1
2
3
4
5
6
7
8
9
AVAILABLE
MASTERED
IN PROGRESS |
190713 | https://www.youtube.com/watch?v=wf9hrC8sZPU | Derivative of Arcsine and Arccosine with the Chain Rule
Mathispower4u
330000 subscribers
13 likes
Description
17127 views
Posted: 9 Apr 2022
This video explains how to determine the derivative of inverse trigonometric functions.
Transcript:
we are given f of x equals inverse sine of 3x and asked to determine f prime of x first notice we do have a composite function where the inner function which we often refer to as u is 3x so when applying the chain rule if the inner function u is equal to 3x we know we have to find du dx or u prime which is a derivative of 3x with respect to x which is 3. and now if we take a look at our derivative formulas the derivative of inverse sine of u with respect to x is equal to one divided by the square root of the quantity one minus u squared times u prime which means in our case f prime of x is equal to one divided by the square root of the quantity one minus u squared which in our case is the square root of the quantity one minus the square of three x and then times u prime which is times three simplifying we have f prime of x equals three divided by the square root of the quantity one minus the square of three x is nine x squared looking at our second example we are given g of x equals inverse cosine of 4x cubed and asked to determine g prime of x once again notice how we do have a composite function where the inner function u is now 4x cubed so if u is equal to 4x cubed notice u prime is equal to 12x squared and now if we take a look at the derivative formula for inverse cosine and compare it to the derivative formula for inverse sine notice how the only difference is there's a negative sign included for the derivative of inverse cosine and therefore g prime of x is equal to negative and then we have one divided by the square root of the quantity one minus u squared which gives us the square root of the quantity one minus the square of four x cubed and then times u prime which is 12x squared simplifying we have g prime of x equals negative 12x squared divided by the square root of the quantity one minus the square of four x cubed is 16 x to the sixth i hope you found this helpful |
190714 | https://help.tableau.com/current/server/en-us/install_config_top.htm | Tableau Server on Windows Help
Install and Configure Tableau Server
The topics referenced at the bottom of this page describe the steps to install and configure Tableau Server. If you are installing a distributed deployment (cluster), use the steps in this topic to install the initial node, then, to install additional nodes, see Distributed and High Availability Tableau Server Installations.
After you run the installation, you must then continue setup by activating a license, registering Tableau Server, and configuring various settings including authentication.
What version are you installing or upgrading to?
Beginning with Tableau Server on Windows version 2019.4.0, a new Setup program is used to install and upgrade Tableau Server. If you are installing or upgrading to version 2019.3.x or earlier, see the 2019.3 Server Help for instructions.
Other installation methods
There are a few alternative methods that you can use to install Tableau Server.
If you want a quick start procedure to install Tableau Server in a non-production environment, see Jump-start Installation.
The topics included in this installation section describe how to install Tableau Server on Windows using the interactive installer. If you want to install Tableau Server using a command line, see Automated Installation of Tableau Server.
If you are installing Tableau Server in an environment without a connection to the internet, see Install Tableau Server in a Disconnected (Air-Gapped) Environment.
You can also install Tableau Server onto various cloud platforms. See Self-Host Tableau Server in a Public Cloud Service.
Validating your server deployment plan
Before you commit to installing a new Tableau Server deployment in your organization, be sure to carefully evaluate your options. For most organizations, Tableau Cloud will provide a more reliable, performant, and cost-effective analytics solution when compared to self-hosting Tableau Server. For information about the viability of Tableau Cloud for your organization, review this blog post, Should I move my analytics to the cloud?(Link opens in a new window)
Already running Tableau Server and want to migrate to Tableau Cloud? See Tableau Cloud Manual Migration Guide(Link opens in a new window).
Before you begin
To install Tableau Server you must have a computer that satisfies the hardware requirements. You will get an informational message if your computer meets the minimum requirements but does not satisfy the recommended minimum requirements. In this case, your computer hardware can handle a trial installation of Tableau but is not adequate for a production environment. For more information, see Before you install....
Installation steps
The following steps describe how to install Tableau Server on a single computer. Use the steps to install Tableau Server in a single server deployment. Use the steps to install the initial node in a multi-node Tableau Server deployment. Run the steps sequentially.
Install TSM
Activate and Register Tableau Server
Configure Initial Node Settings
Add an Administrator Account
Other articles in this section
Before you install...
Minimum Hardware Requirements and Recommendations
Install TSM
Activate Tableau
Configure Initial Node Settings
Add an Administrator Account
Initial Node Installation Defaults
Jump-start Installation
Install Switches and Properties for Tableau Server
Back to top
Cookie Consent Manager
General Information
Required Cookies
Functional Cookies
Advertising Cookies
General Information
We use three kinds of cookies on our websites: required, functional, and advertising. You can choose whether functional and advertising cookies apply. Click on the different cookie categories to find out more about each category and to change the default settings.
Privacy Statement
Required Cookies
Always Active
Required cookies are necessary for basic website functionality. Some examples include: session cookies needed to transmit the website, authentication cookies, and security cookies.
Functional Cookies
Functional cookies enhance functions, performance, and services on the website. Some examples include: cookies used to analyze site traffic, cookies used for market research, and cookies used to display advertising that is not directed to a particular individual.
Advertising Cookies
Advertising cookies track activity across websites in order to understand a viewer’s interests, and direct them specific marketing. Some examples include: cookies used for remarketing, or interest-based advertising.
Cookie List
Consent Leg.Interest
label
label
label |
190715 | https://www.youtube.com/watch?v=1ZuW53dC-vY | Find the Sum of the Series SUM((2^n + 1)/3^n)
The Math Sorcerer
1 subscribers
534 likes
Description
58963 views
Posted: 31 Mar 2021
Find the Sum of the Series SUM((2^n + 1)/3^n)
We write this as two infinite series which are both geometric in order to find the sum.
If you enjoyed this video please consider liking, sharing, and subscribing.
Udemy Courses Via My Website:
My FaceBook Page:
There are several ways that you can help support my channel:)
Consider becoming a member of the channel:
My GoFundMe Page:
My Patreon Page:
Donate via PayPal:
Udemy Courses(Please Use These Links If You Sign Up!)
Abstract Algebra Course
Advanced Calculus Course
Calculus 1 Course
Calculus 2 Course
Calculus 3 Course
Calculus Integration Insanity
Differential Equations Course
College Algebra Course
How to Write Proofs with Sets Course
How to Write Proofs with Functions Course
Statistics with StatCrunch Course
Math Graduate Programs, Applying, Advice, Motivation
Daily Devotionals for Motivation with The Math Sorcerer
Thank you:)
24 comments
Transcript:
find the sum of the series so the way we're going to do this is we are going to break it up so note that if you have 2 to the n plus 1 over 3 to the n we can write this as 2 to the n over 3 to the n plus 1 over 3 to the n then we can take a step further and use properties of exponents to write this as 2 over 3 and the whole thing is to the nth power and if we think of 1 as 1 to the n we can do the same thing here this is 1 over 3 and the whole thing is to the nth power so we can take this and write it as a sum of two infinite geometric series so this is the sum as n runs from 1 to infinity of 2 3 to the n plus and then we have another sum again running from one to infinity and this would be one over three to the n okay so both of these are geometric series and the number here you see is r it's called the common ratio and so whenever the absolute value of r is less than one the series will converge so both of these are convergent because here r is two thirds it's less than one an absolute value so we're good here r is one-third it's less than one an absolute value so we're good so to find the sum is a trick okay the trick tells us that basically you take whatever number is here in this case one and you plug it in for the n and the result goes up top so this is two thirds and then you always divide by one minus r so it's one minus two thirds again take whatever number is here plug it in for n put the result up top and then divide by 1 minus r same thing here you take this number you just put it here that goes up top and then you just divide by 1 minus r so 1 minus 1 3. and that always works okay if there's a 0 here you do the same thing you plug in the 0 put the result up top 1 minus r if there's a 7 here you take the 7 put it here and then just divide by 1 minus r so always works every time all right let's keep going this is equal to 2 over 3 over we can think of 1 as three over three so it's three over three minus two over three which is one over three plus one over three again think of one as three over three so three over three minus 1 over 3 is 2 over 3. really really nice so this is equal to well we're dividing so we multiply by the reciprocal in both cases so 2 two-thirds divided by one-third is two-thirds times three over one plus and then this is one-third divided by two-thirds so it's times the reciprocal which is three over two goes away goes away goes away goes away too much fun this is two plus one half if you think of two as four halves you can add these up and so you will get five over two and that would be the sum of the series i hope this video has been helpful to someone out there good luck |
190716 | https://www.analog.com/en/resources/technical-articles/frequently-asked-questions-thermoelectric-energy-harvesting-with-the-ltc3108-ltc3109.html | Frequently Asked Questions: Thermoelectric Energy Harvesting With the LTC3108 & LTC3109 | Analog Devices
Home
Resource Library
Technical Articles
Frequently Asked Questions: Thermoelectric Energy Harvesting With the LTC3108 & LTC3109
Back to Home
Frequently Asked Questions: Thermoelectric Energy Harvesting With the LTC3108 & LTC3109
by David Salerno
Oct 6 2014
×
Figure 1
Q. What are the target applications for the LTC3108 and LTC3109?
A. As discussed in the datasheet, these parts target applications that require a low average power (in the milliwatt range or less). This could be a micro-controller that is running continuously, or a wireless sensor/transmitter that is operating at a low duty cycle. The LTC3108 and LTC3109 were designed to be powered by any source that can provide a minimum input voltage of about 25mV, with a low source resistance (ideally 3Ω or less.). The LTC3109 can operate equally well from input voltages of either polarity (or a low frequency AC input).
Q. How much temperature differential do I need to start-up the LTC3108 & LTC3109?
A. With a 40mm square TEG and proper heatsinking, the LTC3108 & LTC3109 can startup and regulate VLDO & V OUT (no-load) with a dT as small as 1°K or less. Typically however, with smaller TEGs and heatsinks, a temperature differential of about 5°K is required.
Q. How much power can I get out of the LTC3108 or LTC3109?
A. This depends on the input voltage and the source resistance (i.e. which TEG you use, how much heatsinking is provided, and how much dT is available). There are curves in the datasheets showing the output current capability for different input voltages and different turns ratios. In general, the average output power capability is limited to about 15mW maximum, even at high V IN. At the other extreme, for very low V IN, the average output power may be less than 50µW.
Q. So my customer can’t even use the LTC3108 or LTC3109 to power an LED?
A. No – probably not continuously. The average output power required is too high (we must still obey the laws of physics). However it can pulse an LED very nicely!
Q. What if my customer needs more output power than the LTC3108 can provide?
A. The LTC3109 can be configured for uni-polar operation with a single transformer. In this configuration it will provide more than double the current of the LTC3108. The input resistance in this configuration is 1 Ohm, so it’s a better load match to low resistance voltage sources of 1 Ohm or less.
Q. What’s the difference between a TEC (thermoelectric cooler) and a TEG (thermoelectric generator)?
A. Usually nothing. In some cases, when making a thermoelectric device for power generation, manufacturers will use a higher temperature solder to allow for higher operating temperatures and higher output power. (When I use the term TEG in the datasheet, I am simply referring to the fact that we are using a Peltier module to generate power, not to cool.)
Q. How do I choose a TEG for a given application?
A. This will depend on the average power required by the application, and the dT available across the TEG. The output power of a typical TEG is around 90µW/°K – cm 2, assuming proper heatsinking. Taking into account conversion losses, figure about 25µW/°K – cm 2 maximum at the LTC3108 output. For example, a 40mm x 40mm TEG (with <2.5Ω source resistance) with a dT of 10°K across it will produce about 4mW of maximum output power from the LTC3108. Note that maintaining a dT of 10°K will require very good heat transfer with such a large TEG.
Q. TECs don’t seem to be specified for power generation purposes, so how do I know which ones will work best for harvesting?
A. Manufacturers specify the maximum voltage and maximum current rating of the TEC (when used for cooling purposes). For the highest output power in an energy harvesting application, choose the TEC with the highest voltage rating and the highest current rating. This will generally provide the highest output voltage for a given dT, and the lowest source resistance to allow for the most power transfer to the load. HOWEVER, the larger the TEG is, the lower its thermal resistance is, meaning that it will require more heatsinking on the cool side of the device to maintain a dT across it. Therefore, larger TEGs are only beneficial if you use a larger heatsink. Applications that don’t need as much power will perform well using a small TEG (say 15mm on a side) and a small heatsink.
Q. What kind of open-circuit output voltage does a TEG produce?
A. This depends on many factors (manufacturer, model number and absolute temperature), but as a rule of thumb, the open-circuit voltage is about N0.25mV per degree Kelvin across the device, where N is the number of couples (which is published in the TEG datasheet).
Q. What about the new thin-film TEGs from Micropelt and Nextreme? Do the LTC3108 & LTC3109 work well with them?
A. These new thin-film devices are very small and thin, and are generally designed to provide a higher output voltage (volts) at a low output current. In the case of the Micropelt device, the source resistance is very high (over 300 Ohms), and is therefore a very bad match to the LTC3108 or LTC3109 input. The Nextreme device has a much lower resistance and may work well, but requires a much higher dT (typically 50 – 70°K). The LTC3108 and LTC3109 were designed to work optimally with “traditional” TEGs (which usually have a source resistance between 0.5 and 5 Ohms) that are widely available from many manufacturers. The LTC3105, 250mV Boost converter with MPPC is a better match for higher resistance sources like the Micropelt device, as the input resistance of the converter can be programmed to match the source resistance.
Q. Isn’t the small size of these thin-film devices a big advantage?
A. Perhaps, but remember that any of these devices will require heatsinking to the ambient, and attachment to some source of heat (or cool), so in the end, the solution tends to be rather large, regardless of the size of the TEG.
Q. I understand that the load resistance should match the TEG source resistance for maximum power transfer. How do I do that?
A. You can approximate the TEG resistance by dividing the maximum voltage rating by the maximum current rating. Choose a module with a resistance of 2.5 Ohms or less for optimal power transfer. The input resistance of the LTC3108 harvester circuit is shown in the datasheet curves, but is generally >= 2.5 Ohms. The input resistance of the LTC3109 is typically 5.5Ω in autopolarity configuration. However, even if the matching is not perfect, it is preferable to use the lowest R S TEG available to maximize power capability. The caveat to this is that lower resistance TEGs are usually large and also have a lower thermal resistance. Therefore, they will also require a larger heatsink to maintain the desired dT across the device.
Q. You keep talking about the heatsinking. Why is this so important and how do I size the heatsink?
A. A TEG can only produce output power if there is a temperature drop across it, which requires heat flow through it. Therefore a heatsink is generally required on one side of the device, to maintain a dT across it. Otherwise the whole TEG will just reach the same temperature as the hot or cold side, and no power will be produced. (An exception to this would be if the TEG were sandwiched between two surfaces at different temperatures which each had a large thermal mass and low thermal resistance, such as a hot and cold water pipe. In this case of course, no added heatsink would be required.) For optimal power output, the heatsink should have a thermal resistance as low or lower than that of the TEG. Since the typical TEG has a thermal resistance of around 1-10°C/W (depending on surface area and thickness), this requires a reasonably good heatsink. Larger TEGs will have lower thermal resistance, and therefore require larger heatsinks. The heatsink directly impacts the dT, which directly affects the output power capability.
Q. How much do TEGs cost? I’ve heard they are expensive.
A. Getting TEC or TEG volume pricing seems to be a challenge. In single-piece quantities, the US manufacturer’s and distributor’s prices usually range from $15 to $30 apiece. However, don’t panic - you can buy a single 40mm TEG on Ebay for less than $5 including shipping from China, so volume pricing should be less than the cost of our IC. Some manufacturers are now offering small TEGs for about $1 in high volume.
Q. Where can my customers buy a TEC or TEG, and what part numbers are recommended?
A. Some of the big TEC/TEG manufacturers and distributors are: Tellurex, Marlow, CUI, Ferrotec, and Laird. The specific part number that will be required or work the best is application dependent.
Q. How do I size the C STORE and C OUT capacitors?
A. This is explained, with two detailed examples, in the LTC3108 and LTC3109 datasheets. Just remember that any pulse load current must come from the V OUT capacitor. The Store capacitor is only for providing a small current to recharge V OUT at times when the input voltage source is lost. For this reason, the storage element (battery or capacitor) does not need to be low ESR, and can in fact be hundreds of Ohms.
Q. Can I use Li-ion or the new thin-film lithium rechargeable batteries for the storage element?
A. Yes, but you will need other external circuitry (the LTC4070) to protect the battery from over-charge and over-discharge. Be very careful if using the new thin-film batteries from Infinite Power Solutions or Cymbet, as they have very little capacity. A special version of the protection circuit is required that draws NO current when the battery reaches UV. Otherwise the battery can become over-discharged and damaged if left uncharged for a day.
Q. What is the conversion efficiency of the LTC3108 & LTC3109? (and why is it so low?)
A. The datasheet has curves showing the efficiency for different input voltages and turns ratios. In general the efficiency from V IN to V OUT is only 20-40% (note that this does not include losses in the TEG itself due to source resistance). Remember – you are stepping the voltage up by a factor of 100! There are many sources of power loss (gate drive, transformer, rectifier), especially when the total power may only be 100’s of microwatts. However, the major source of power loss is usually the NMOS Rdson. At very low V IN, even the 6-7µA quiescent current of the IC becomes a significant loss.
Q. How do I choose the optimal step-up transformer ratio for an application?
A. The datasheet provides curves of I OUT vs V IN for different turns ratios to help in optimizing the transformer selection for a given application. However, you can use these simple rules to select the ratio:
If the application must startup at the lowest possible voltage (down to 20mV), use the 1:100 ratio
If the application must startup at the lowest possible voltage (down to 20mV), use the 1:100 ratio
If the minimum input voltage (under load) is at least 150mV, use the 1:20 ratio
Q. Are there other choices available for the step-up transformer besides those from Coilcraft?
A. Yes, Wurth now makes transformers that are essentially identical to the Coilcraft parts.
Q. Why is the maximum input voltage rating so low? (2V at SW, or V INRatio < 50)
A. There are a number of reasons. First of all, we are trying to avoid very high secondary winding voltages for safety reasons and to prevent exceeding the insulation rating of the transformer windings (conservatively rated for 100V), or the rating of the C1 and C2 capacitors. Secondly, we need to limit the maximum voltage seen from drain-gate of the NMOS devices when they are off (V GATE can equal -6V). Finally, if V IN is 300mV or more, you should use the LTC3105, and if it is 800mV or more, choose the LTC3525 or the LTC3526L.
Q. Do we have any patents on the LTC3108 or LTC3109?
A. We don’t on the LTC3108. The basic resonant architecture it uses is covered by numerous patents dating back over 20 years, and is now public domain. We do have a patent pending for the autopolarity architecture of the LTC3109.
Q. What about solar harvesting applications with the LTC3108?
A. In general, you should look at the LTC3105 first for these applications. The LTC3108 requires a minimum input current of a few milliamps (at the converter input) just to startup, and may not be a good load match for a PV cell. Therefore, small solar cells that have a short circuit current of less than a few milliamps will not work with the LTC3108 (or LTC3109).
About the Authors
David Salerno
David Salerno was a Design Section Leader for the Power by Linear group of Analog Device (former Linear Technology) for nearly 20 years. His main focus was the DC/DC Converter design on boost applications. David hold an B...
Add to myAnalog
Share
Copy Link
Send to Mail
Add to myAnalog
Add article to the Resources section of myAnalog, to an existing project or to a new project.
Add article to
Resources on myAnalog Resources on myAnalog
New Project
Create New Project
Project Name
Description
[x]
Cancel Confirm
Related to this Article
Products
LTC3108 RECOMMENDED FOR NEW DESIGNS Ultralow Voltage Step-Up Converter and Power ManagerLTC3109 RECOMMENDED FOR NEW DESIGNS Auto-Polarity, Ultralow Voltage Step-Up Converter and Power ManagerLTC3108-1 RECOMMENDED FOR NEW DESIGNS Ultralow Voltage Step-Up Converter and Power Manager
Product Categories
Power Management Switching Regulators
Step-Up (Boost) Regulators
Micropower Boost Regulators
Internal Power Switch Boost Regulators
Energy Harvesting
Latest Media 21
Video . Jul 24, 2025 02:29
Advanced Digital Power Management Learn More
Video . Jul 23, 2025 03:11
Maximizing OEE with AI Enabled Condition Monitoring Learn More
Webcast . Jul 23, 2025
Design Smarter: Integrated Precision Signal Chains for Scientific Instrumentation Learn More
Video . Jul 21, 2025 02:15
Low SWAP Radio Design with Nevis Learn More
Video . Jul 16, 2025 02:32
Precision Signal Chain Seminars with Hands-On Experience Learn More
Video . Jul 16, 2025 01:48
Signal Chain Designer Learn More
Video . Jul 16, 2025 02:24
Dude, where's my apple? Learn More
Video . Jul 16, 2025 02:43
LTspice 24.1: Fast, Free, Unlimited Learn More
Video . Jul 16, 2025 02:37
World's First Single Chip Multiturn Absolute Position Sensor - ADMT4000 Learn More
Video . Jul 16, 2025 03:54
Wideband RF Front-end Reference Design Learn More
Video . Jul 16, 2025 02:57
Synchronizing Systems with a High Number of ADCs/DACs Learn More
Video . Jul 11, 2025 02:00
Analog Devices at IMS 2025 Learn More
Video . Jul 9, 2025 02:05
Accelerate Your AI Development with AutoML for Embedded Learn More
Video . Jul 8, 2025 03:50
Silent Switcher 3 Learn More
Video . Jul 8, 2025 02:16
Apollo MxFE™: Agile Multi-Band Phased Array Radar Learn More
Video . Jul 8, 2025 02:34
6G FR3 RFFE for OTA Test Systems Learn More
Video . Jul 8, 2025 02:53
Apollo MxFE™ + Versal™: EW at the Edge Learn More
Video . Jul 7, 2025 03:57
0.1-20GHz 3UVPX Tuner+Digitizer+Processor Learn More
Video . Jul 7, 2025 03:37
Agility in Action: Frequency Hopping with DPD and AGC Learn More
Video . Jul 7, 2025 01:56
10GSPS digitizer for ToF mass spectroscopy Learn More
Video . Jul 2, 2025 01:29
Samana ADI DPD Demo on Titan Platform Learn More
Video . Jul 24, 2025 02:29
Advanced Digital Power Management Learn More
Video . Jul 23, 2025 03:11
Maximizing OEE with AI Enabled Condition Monitoring Learn More
Webcast . Jul 23, 2025
Design Smarter: Integrated Precision Signal Chains for Scientific Instrumentation Learn More
Video . Jul 21, 2025 02:15
Low SWAP Radio Design with Nevis Learn More
Video . Jul 16, 2025 02:32
Precision Signal Chain Seminars with Hands-On Experience Learn More
Video . Jul 16, 2025 01:48
Signal Chain Designer Learn More
Video . Jul 16, 2025 02:24
Dude, where's my apple? Learn More
Video . Jul 16, 2025 02:43
LTspice 24.1: Fast, Free, Unlimited Learn More
Video . Jul 16, 2025 02:37
World's First Single Chip Multiturn Absolute Position Sensor - ADMT4000 Learn More
Video . Jul 16, 2025 03:54
Wideband RF Front-end Reference Design Learn More
Video . Jul 16, 2025 02:57
Synchronizing Systems with a High Number of ADCs/DACs Learn More
Video . Jul 11, 2025 02:00
Analog Devices at IMS 2025 Learn More
Video . Jul 9, 2025 02:05
Accelerate Your AI Development with AutoML for Embedded Learn More
Video . Jul 8, 2025 03:50
Silent Switcher 3 Learn More
Video . Jul 8, 2025 02:16
Apollo MxFE™: Agile Multi-Band Phased Array Radar Learn More
Video . Jul 8, 2025 02:34
6G FR3 RFFE for OTA Test Systems Learn More
Video . Jul 8, 2025 02:53
Apollo MxFE™ + Versal™: EW at the Edge Learn More
Video . Jul 7, 2025 03:57
0.1-20GHz 3UVPX Tuner+Digitizer+Processor Learn More
Video . Jul 7, 2025 03:37
Agility in Action: Frequency Hopping with DPD and AGC Learn More
Video . Jul 7, 2025 01:56
10GSPS digitizer for ToF mass spectroscopy Learn More
Video . Jul 2, 2025 01:29
Samana ADI DPD Demo on Titan Platform Learn More
Subtitle
Learn More Add to myAnalog Share
Add to myAnalog
Add article to the Resources section of myAnalog, to an existing project or to a new project.
Add article to
Resources on myAnalog Resources on myAnalog
New Project
Create New Project
Project Name
Description
[x]
Cancel Confirm Copy Link
Send to Mail
Who We Are
Careers
Newsroom
What We Do (Signals+)
Investor Relations
Quality & Reliability
Sales and Distribution
What's New on Analog.com
Contact Us
Support
EngineerZone
Resources
Wiki
Analog Dialogue
ADI Developer Portal
Legal and Risk
Accessibility
Privacy Policy
Privacy Settings
Cookie Settings
Analog Devices Cookie Policy
By clicking “Accept All Cookies”, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts.
Cookie Settings Accept All Cookies
Manage Cookie Settings
When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences, or your device. It is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings below to learn more and to set your preferences. You cannot disable the Strictly Necessary Cookies as they are deployed to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, or to redirect you when you log out). For more information about the cookies click the following link.
Cookie Policy
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Targeting Cookies
[x] Targeting Cookies
These cookies may be set through our site by our advertising partners, to create a profile of the device owner’s interests in order to show relevant advertising on other sites. The technology does this by uniquely identifying your device. These cookies collect information to build a profile of your interests and to personalize your experience with targeted ads. If you do not allow these cookies, you will experience less targeted advertising (but will still see ads on other websites). If you have enabled privacy controls on your browser (such as the Do Not Track setting or Global Privacy Control signal), Targeting cookies will be disabled by default.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Confirm My Choices
Site Feedback |
190717 | https://link.springer.com/article/10.1007/s00234-025-03596-z | Pre-treatment and post-treatment nasopharyngeal carcinoma imaging: imaging updates, pearls and pitfalls | Neuroradiology
Your privacy, your choice
We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media.
By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection.
See our privacy policy for more information on the use of your personal data.
Manage preferences for further information and to change your choices.
Accept all cookies
Skip to main content
Advertisement
Log in
Menu
Find a journalPublish with usTrack your research
Search
Cart
Search
Search by keyword or author
Search
Navigation
Find a journal
Publish with us
Track your research
Home
Neuroradiology
Article
Pre-treatment and post-treatment nasopharyngeal carcinoma imaging: imaging updates, pearls and pitfalls
Review
Open access
Published: 11 April 2025
Volume 67,pages 1023–1047, (2025)
Cite this article
Download PDF
You have full access to this open access article
NeuroradiologyAims and scopeSubmit manuscript
Pre-treatment and post-treatment nasopharyngeal carcinoma imaging: imaging updates, pearls and pitfalls
Download PDF
Kwok Yan Li1,
Hoi Ming Kwok1,2,
Nin Yuan Pan1,
Lik Fai Cheng1&
…
Ka Fai Johnny Ma1
Show authors
2875 Accesses
Explore all metrics
Abstract
Purpose
Nasopharyngeal carcinoma (NPC) is endemic in Southeast Asia, requiring precise imaging for personalized treatment. This review highlights key imaging challenges and updates from recent literature, emphasizing findings that impact oncological management.
Methods
We discuss common and uncommon clinical entities, detailing salient imaging features and diagnostic distinctions to aid accurate interpretation.
Results
In the pre-treatment setting, leveraging the characteristic MR signals and spread patterns of NPC aids in defining the tumor volume for accurate staging and radiotherapy contouring. Key diagnostic challenges include differentiating tumor from benign hyperplasia, skull base osteomyelitis,and other skull base tumors. Perineural tumor spread, radiological extranodal extension and nodal necrosis further refine primary tumor and nodal assessment. In the post-treatment setting, the key question is whether tumor recurrence exists. Diagnostic challenges involve distinguishing tumor recurrence from scar tissue, post-radiation nasopharyngeal necrosis, or hypertrophied cervical ganglia. For recurrences, endoscopic nasopharyngectomy has emerged as the preferred approach over open surgery or re-irradiation. The text highlights characteristic post-treatment appearances and emphasizes recognizing these patterns to avoid misinterpretation and guide appropriate management.
Conclusion
Imaging plays a pivotal role in NPC precision oncology. Mastering imaging pearls and pitfalls empowers radiologists to provide clinicians with reliable, actionable guidance.
Similar content being viewed by others
Nasopharyngeal Neoplasms
Chapter© 2021
Enhancing Nasopharyngeal Carcinoma Survival Prediction: Integrating Pre- and Post-Treatment MRI Radiomics with Clinical Data
Article 30 April 2024
Diagnostic Imaging of Nasopharyngeal Carcinoma
Chapter© 2020
Explore related subjects
Discover the latest articles, books and news in related subjects, suggested using machine learning.
Cancer Imaging
Medical Imaging
Neuroradiology
Oncology
Radiation Oncology
Radiology
Use our pre-submission checklist
Avoid common mistakes on your manuscript.
Introduction
Nasopharyngeal carcinoma (NPC) is a prevalent tumor in Southeast Asia, especially in Southern China. It is related to Ebstein-Barr virus (EBV) infection. Its incidence is more than 20 per 100,000 person-years in endemic regions, while less than 1 per 100,000 person-years in most parts of the world [18
")]. Evolving evidence in pre- and post-treatment NPC urges increasingly personalized and precise oncological treatment, demanding expertise and close communication between radiologists and oncologists. This review provides insights into the key aspects and challenges in NPC imaging with updates from recently published literature, emphasizing those with oncological management impact (Table1). In the pre-treatment setting, the prime objective is to define the tumor volume accurately for staging and subsequent radiotherapy contouring. In the post-treatment setting, the key question is whether tumor recurrence exists.
Table 1 Summary of the pearls and pitfalls in NPC imaging that impact oncological decision making
Full size table
Pre-treatment imaging
1.
Interpretation of NPC MRI goes beyond local staging. Accurate primary tumor delineation forms the basis of precision radiotherapy in achieving the most dose-effective local or locoregional control. Radiologists need to recognize the characteristic tumor signal and have a tailored search pattern for the characteristic pattern of tumor spread.
Nowadays, NPC is primarily treated by radiotherapy via intensity-modulated radiotherapy (IMRT) with or without chemotherapy as the standard of care. Successful curative radiotherapy aims to maximize the primary tumor's loco-regional control and minimize treatment-related complications to the organ at risk (OAR). This relies on robust imaging techniques that accurately map the tumor, interpretation by subspecialized head and neck radiologists, and effective communication of the image findings with its management and prognostication information to the radiation oncologists.
MRI is the primary modality not only for local staging but also for radiotherapy planning due to its superior soft tissue visualization compared to 18F-fluorodeoxyglucose positron emission tomography/computed tomography (FDG-PET/CT) and computed tomography (CT) [25
")]. Fusion of MR images with planning CT images enables radiation oncologists to carry out precise radiotherapy planning. A state-of-the-art MR protocol for NPC imaging comprises both two-dimensional and three-dimensional sequences [2 MR imaging of nasopharyngeal carcinoma. Magn Reson Imaging Clin N Am 30(1):19–33.
")]. For two-dimensional sequences, axial fat-suppressed T2-weighted, T1-weighted pre- and post-contrast, and diffusion-weighted images (DWI) (with b-value at least 800 s/mm 2) should be performed with a scanned range from above the skull base to C3 [2 MR imaging of nasopharyngeal carcinoma. Magn Reson Imaging Clin N Am 30(1):19–33.
")]. Coronal T1-weighted pre- and post-contrast, as well as sagittal T1-weighted post-contrast sequences, could also be performed. Dedicated neck images of axial T1-weighted post-contrast sequence from C3 to the level below the suprasternal notch should be included for screening lower neck and supraclavicular nodes. The three-dimensional sequence can be obtained with a fat-suppressed T1-weighted sequence in the coronal plane with multiplanar reformat [2 MR imaging of nasopharyngeal carcinoma. Magn Reson Imaging Clin N Am 30(1):19–33.
")]. DWI with non-echoplanar imaging (non-EPI) or multishot echoplanar imaging (msEPI) techniques should be performed instead of single-shot echoplanar imaging (ssEPI) as they give less susceptibility-related distortion in the skull base. The examination should be performed on a 1.5-T or preferably a 3.0-T MRI with the use of a 64-channel head and neck coil, to achieve a better signal-to-noise ratio, spatial resolution, and a shorter scanning time.
Accurate target delineation for gross tumor volume (GTV) contouring is challenging given the complex locoregional anatomy and the intricate relation between the tumor and multiple OARs [33
")]. The clinical target volume (CTV) covers possible microscopic disease around the tumor. Its contouring relies on accurate and precise GTV contouring on both the primary tumor and nodal metastases. The traditional approach of "5 + 5" recommendation of GTV expansion was recommended from pathological evidence of head and neck squamous cell carcinoma, referring to a 5 mm volumetric expansion from the GTV as high-risk CTV, and another 5 mm expansion from the high risk CTV as low-risk CTV [3 Current radiotherapy considerations for nasopharyngeal carcinoma. Cancers (Basel) 14(23):5773.
")]. A tighter margin of 1–2 mm is recommended for primary tumor encroaching critical neural OARs (e.g., optic nerves, optic chiasm, brainstem, etc.) [3 Current radiotherapy considerations for nasopharyngeal carcinoma. Cancers (Basel) 14(23):5773.
")]. There has been ongoing research and advancement towards a more individualized approach to CTV contouring based on tumor extent and spread pattern, including the sparing of contralateral structures in unilateral disease and reduction in CTV upon response to induction chemotherapy [3 Current radiotherapy considerations for nasopharyngeal carcinoma. Cancers (Basel) 14(23):5773.
")]. Besides the primary tumor, CTV delineation of regional lymphatics moves towards the refinement of delineation boundaries in nodal level according to nodal spread pattern with the omission of lower neck irradiation in the uninvolved neck [3 Current radiotherapy considerations for nasopharyngeal carcinoma. Cancers (Basel) 14(23):5773.
")].
Radiologists need to have a tailored search pattern for nasopharyngeal tumor, given that the primary tumor spreads in a highly predictable and orderly fashion 45
"),[5 Extension of local disease in nasopharyngeal carcinoma detected by magnetic resonance imaging: improvement of clinical target volume delineation. Int J Radiat Oncol Biol Phys 75(3):742–750.
"),6 Extension patterns of nasopharyngeal carcinoma. Eur Radiol 17(10):2622–2630.
"),7 International guideline for the delineation of the clinical target volumes (CTV) for nasopharyngeal carcinoma. Radiother Oncol 126(1):25–36.
")]. The same principle also applies to the contouring of the CTV by radiation oncologists [7 International guideline for the delineation of the clinical target volumes (CTV) for nasopharyngeal carcinoma. Radiother Oncol 126(1):25–36.
")]. The primary tumor spreads via a stepwise fashion from the high-risk anatomical region around the nasopharynx (e.g., parapharyngeal space, prevertebral space, pterygoid process, and basisphenoid) to the more distal regions [4 Locoregional extension patterns of nasopharyngeal carcinoma and suggestions for clinical target volume delineation. Chin J Cancer 31(12):579–587.
"), 5 Extension of local disease in nasopharyngeal carcinoma detected by magnetic resonance imaging: improvement of clinical target volume delineation. Int J Radiat Oncol Biol Phys 75(3):742–750.
")]. Besides, the nodal metastases of NPC also follow an orderly fashion, and skip metastasis only occurs in 0.5–7.9% of cases [8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
")]. Both the level II and retropharyngeal nodes can be echelon nodes with involvement in 70% and 69% of the cases, respectively [3 Current radiotherapy considerations for nasopharyngeal carcinoma. Cancers (Basel) 14(23):5773.
"), 8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
")]. In interpreting tumoral involvement within the delicate head and neck structures, emphasis should be placed on the use of T1-weighted non-fat-suppressed images and the recognition of the characteristic tumoral signal on T2-weighted images. Careful inspection of structural asymmetry and soft tissue replacement of the fat signal in the skull base on a non-fat-suppressed T1-weighted image may be the first clue to guide radiologists to look for tumor involvement. NPC has a characteristic intermediate T2-weighted signal. It demonstrates intermediate enhancement after gadolinium injection [2 MR imaging of nasopharyngeal carcinoma. Magn Reson Imaging Clin N Am 30(1):19–33.
")]. This can serve as confirmation of tumoral involvement when both the structural asymmetry and the signal characteristic fit the pattern of tumor spread. Radiologists are adept in utilizing multisequence MRI, including DWI, and multiplanar reformats to navigate between the complex anatomy to map out the full extent of the tumor. NPC shows restricted diffusion on DWI due to hypercellularity, with a mean apparent diffusion coefficient (ADC) value of 871.7 ± 84.4 × 10⁻⁶ mm 2/s [9 Value of apparent diffusion coefficient histogram analysis in the differential diagnosis of nasopharyngeal lymphoma and nasopharyngeal carcinoma based on readout-segmented diffusion-weighted imaging. Front Oncol 12(11):632796.
")]. The distinctive high signal of the tumor on a high b-value DWI image and low signal in the ADC map relative to adjacent soft tissue structures can provide invaluable information on tumor margin for accurate tumor volume delineation [10 A review of the role of DWI in radiation therapy planning and treatment response assessment for NPC management. Curr Res Med Sci 2:84–90.
")]. A certain degree of peritumoral inflammation is commonly observed when NPC spreads to the parapharyngeal space adjacent to the masticator muscles, or the skull base. These can be differentiated from tumor involvement by a higher T2-weighted signal and a higher degree of enhancement, without associated restricted diffusion. This distinction is important in radiotherapy contouring.
2.Recognition of normal variants in the nasopharynx avoids misdiagnosis, over-investigation, and over-treatment. Radiologists should be aware of the coexistence of adenoidal hyperplasia and tumors.
The distinction of adenoidal hyperplasia from NPC on MRI is important, as the former is a benign condition that requires no further investigation. It is a common clinical encounter when patients are referred for MRI assessment of the non-specific nasopharyngeal soft tissue thickening in other imaging modalities, such as CT or PET/CT, or in MRI dedicated to the brain or other neck regions. With the advent of successful plasma Epstein-Barr virus DNA (EBV-DNA) screening for early-stage NPC, asymptomatic patients are referred for a screening MRI when they have a normal endoscopic exam or indeterminate soft tissue bulge in the nasopharynx 25
"), [11 Analysis of plasma epstein-barr virus DNA to screen for nasopharyngeal cancer. N Engl J Med 377(6):513–522.
"), 12 Complementary roles of MRI and endoscopic examination in the early detection of nasopharyngeal carcinoma. Ann Oncol 30(6):977–982.
")]. In these instances, the distinction between normal variants and tumors in the nasopharynx becomes crucial. Despite the international guideline recommendation for endoscopic biopsy of the nasopharynx to establish the definitive diagnosis of NPC, about 10%−30% of NPC are missed at initial endoscopy due to small size, submucosal location, coexistent hyperplasia, and anatomical difficulty in accessing the lateral pharyngeal recess [2 MR imaging of nasopharyngeal carcinoma. Magn Reson Imaging Clin N Am 30(1):19–33.
"), 12 Complementary roles of MRI and endoscopic examination in the early detection of nasopharyngeal carcinoma. Ann Oncol 30(6):977–982.
"),13 MRI detection of suspected nasopharyngeal carcinoma: a systematic review and meta-analysis. Neuroradiology 64(8):1471–1481.
"),14 SEOM clinical guideline in nasopharynx cancer (2017). Clin Transl Oncol 20(1):84–88.
")]. A meta-analysis with 7 out of 9 studies using contrast-enhanced MR protocols shows that MRI has a very high pooled sensitivity of 98.1% (95% CI 95.2–99.3%) and specificity of 91.7% (95% CI 88.3–94.2%) for the detection of NPC [13 MRI detection of suspected nasopharyngeal carcinoma: a systematic review and meta-analysis. Neuroradiology 64(8):1471–1481.
")]. MRI is also more sensitive compared with endoscopy (91.2% versus 76.5%) in the detection of early NPC in a prospective study in asymptomatic individuals with elevated plasma EBV-DNA [12 Complementary roles of MRI and endoscopic examination in the early detection of nasopharyngeal carcinoma. Ann Oncol 30(6):977–982.
")]. A recent prospective study using a quick and short contrast-free MRI screening protocol for NPC detection in EBV-positive individuals shows a high negative predictive value of 99.4%, complementing endoscopy and endoscopic-guided biopsy in the screening setting [15 Early detection of nasopharyngeal carcinoma: performance of a short contrast-free screening magnetic resonance imaging. J Natl Cancer Inst 116(5):665–672.
")]. Radiologists should be equipped with the essential knowledge for making this distinction and be aware of the possible challenge of a coexisting tumor and adenoidal hyperplasia.
Adenoidal hyperplasia is common and physiological in children between the ages of 6 and 10 years and then regresses by puberty. However, it is also possible to encounter nasopharyngeal tissue proliferation in adults, especially smokers 160
"), [17 Nasopharyngeal masses in adults—A retrospective analysis of 255 patients to evaluate symptoms, clinical findings, and histological results. World J Otorhinolaryngol Head Neck Surg 11(01):45–51.
")]. The typical MR appearance includes stripes in adenoid bulges on post-contrast T1-weighted images, preservation of deep white mucosal lines, and structural symmetry [18 Nasopharyngeal mucosa and adenoids: appearance at MR imaging. Radiology 263(2):437–443.
"), 19 MR imaging criteria for the detection of nasopharyngeal carcinoma: discrimination of early-stage primary tumors from benign hyperplasia. AJNR Am J Neuroradiol 39(3):515–523.
")]. Other findings such as cysts (41%) and Tornwaldt cysts (4%) can also be seen [18 Nasopharyngeal mucosa and adenoids: appearance at MR imaging. Radiology 263(2):437–443.
")]. Subsequent work by King et al. suggested several distinctive MRI imaging features to suggest NPC, including a significantly greater volume, size asymmetry, signal asymmetry, focal loss of the deep mucosal white line, and absence/ distortion of the adenoidal septa [19 MR imaging criteria for the detection of nasopharyngeal carcinoma: discrimination of early-stage primary tumors from benign hyperplasia. AJNR Am J Neuroradiol 39(3):515–523.
")] (Fig.1). The size asymmetry was the most accurate criterion (89.5%) for such distinction [19 MR imaging criteria for the detection of nasopharyngeal carcinoma: discrimination of early-stage primary tumors from benign hyperplasia. AJNR Am J Neuroradiol 39(3):515–523.
")].
Fig. 1
Distinction between nasopharyngeal tumor (a & b) and adenoidal hyperplasia (c & d). Axial T2-weighted, fat-suppressed image shows intermediate signal slightly asymmetrical soft tissue occupying the bilateral nasopharynx with absence of the adenoidal septa (asterisk) (a). Axial T1-weighted post-contrast image with fat suppression shows moderate enhancement of the bilateral nasopharyngeal soft tissue mass with focal loss of the deep white mucosal line in the left lateral and posterior wall (arrows) (b). Biopsy confirmed nasopharyngeal carcinoma. Axial T2-weighted, fat suppressed image shows the symmetrical soft tissue in the bilateral nasopharynx containing cystic focus (asterisk) and preservation of hypointense adenoidal septa (arrowhead) (c). Axial T1-weighted post-contrast image with fat suppression shows symmetrical hypo-enhancing soft tissue with a striated appearance and preservation of deep mucosal white line (d). Findings are consistent with adenoidal hyperplasia
Full size image
It can be challenging to distinguish a small T1 tumor or a fairly symmetric-looking tumor from adenoidal hyperplasia. In general, the above rules still apply in making the distinction. Identification of a small T1 tumor guides the site and depth of biopsy for endoscopic surgeons [25
")]. Furthermore, adenoidal hyperplasia and NPC can coexist [2 MR imaging of nasopharyngeal carcinoma. Magn Reson Imaging Clin N Am 30(1):19–33.
"), 20 Hypertrophic adenoids in patients with nasopharyngeal carcinoma: appearance at magnetic resonance imaging before and after treatment. Chin J Cancer 34(3):130–136.
")] (Fig.2), highlighting the danger of tumor underdetection from satisfaction of search. The identification of coexisting adenoidal hyperplasia allows precise delineation of GTV during radiotherapy planning and anticipation of a different post-treatment imaging appearance compared to the main tumor bulk on follow-up imaging, avoiding unnecessary repeated biopsy [20 Hypertrophic adenoids in patients with nasopharyngeal carcinoma: appearance at magnetic resonance imaging before and after treatment. Chin J Cancer 34(3):130–136.
")].
Fig. 2
Coexistence of nasopharyngeal tumor and lymphoid hyperplasia in a 70-year-old man referred for an enlarging right neck mass. Doppler ultrasound of the right neck shows a hypoechoic mass with increased peripheral vascular flow and a central anechoic cystic area with septations (a). Axial contrast CT of the neck shows an enlarged roundish right level II node with central hypoenhancing area suggestive of necrosis (b). Prominent soft tissue density in the nasopharynx was noted on the CT (not shown). Endoscopy found a right nasopharyngeal mass and biopsy was proven to be an undifferentiated carcinoma. Axial T2-weighted image shows soft tissue thickening of the bilateral nasopharynx. Small cysts are seen involving the bilateral nasopharynx, more on the left side. Minimal intermediate T2-weighted soft tissue thickening is seen at the lateral wall of right nasopharynx (asterisk) (c). The pharyngobasilar fascia is intact. Axial post-contrast T1-weighted, fat-suppressed image shows only non-specific moderately enhancing soft tissue thickening of the bilateral nasopharynx with intact deep mucosal white line (d). Coronal post-contrast, fat-suppressed T1-weighted image depicts a moderately enhancing mass in the right nasopharynx (asterisk) with focal loss of the deep mucosal white line at the roof (arrows) (e). Axial fusion FDG-PET/CT depicts the hyper-metabolic right nasopharyngeal mass and the hypermetabolic right level II node (not shown). Otherwise, no hypermetabolic distant metastasis (f). Radiological staging was T1N1M0 and the patient received radical radiotherapy. This case highlights three diagnostic pearls. First, the presence of benign lymphoid hyperplasia and tumor are not mutually exclusive. Second, the use of multiplanar reformat should be routinely performed for evaluation of the nasopharynx in order to avoid missing small tumors due to partial volume effect. Third, level II nodal metastasis can occur without involvement of the retropharyngeal node, especially for laterally located tumors in the nasopharynx
Full size image
3.
Perineural spread forms a tumor highway in NPC, potentially causing missing radiation targets and risk of residual tumor.
Perineural tumor growth is a well-recognized clinicopathological entity in the head and neck that is associated with poorer prognosis and an increased risk of locoregional recurrence [219
")]. It encompasses perineural invasion (PNI) and perineural spread (PNS). PNI is a histological diagnosis of tumor cell infiltration into nerves at the original tumor sites, which cannot be radiologically detected. PNS refers to macroscopic tumor involvement along a nerve extending away from the primary tumor, evading from the effective radiation field. This can be radiologically detected [21 Perineural invasion and spread in head and neck cancer. Expert Rev Anticancer Ther 12(3):359–371.
"),22 Perineural invasion in head and neck cancer. J Dent Res 97(7):742–750.
"),23 Pathology of perineural spread. J Neurol Surg B Skull Base 77(2):124–130.
")].
The incidence of PNS or PNI has been reported to be 27–82% in head and neck squamous cell carcinomas [219
")]. In NPC, Liu et al. evaluated MRI-detected PNI, together with PNS, giving a total incidence of approximately 39.9%. In this study, all tumors were locally advanced (T3–4) regardless of the presence of perineural involvement, with a particularly higher incidence of T4 tumor involvement. They are independent poor prognostic factors for distant metastasis-free survival and locoregional relapse-free survival [24 Prognostic value of magnetic resonance imaging-detected cranial nerve invasion in nasopharyngeal carcinoma. Br J Cancer 110(6):1465–1471.
")].
MRI has a sensitivity of 95–100% and a specificity of 85% in detecting PNS in head and neck cancers [252
")]. A recent meta-analysis showed FDG-PET/CT has a pooled sensitivity of 91.7%, pooled specificity of 92.35%, suggesting it is also an effective imaging modality [26 Does PET scan have any role in the diagnosis of perineural spread associated with the head and neck tumors? Adv Clin Exp Med 31(8):827–835.
")]. MRI is more sensitive than CT and FDG-PET/CT in the detection of PNS due to better soft tissue contrast [27 Comparison of diffusion-weighted MR imaging and 18F Fluorodeoxyglucose PET/CT in detection of residual or recurrent tumors and delineation of their local spread after (chemo) radiotherapy for head and neck squamous cell carcinoma. Eur J Radiol 130:109157.
")]. The excellent soft tissue contrast allows precise extent delineation for radiotherapy planning. However, the sensitivity for accurately depicting the entire disease extent is only about 63% compared to histology [21 Perineural invasion and spread in head and neck cancer. Expert Rev Anticancer Ther 12(3):359–371.
")]. Pathological evidence shows that PNS is usually continuous, and imaging may not be able to detect the segment of nerves with low tumor burden [21 Perineural invasion and spread in head and neck cancer. Expert Rev Anticancer Ther 12(3):359–371.
")]. There are also technical limitations from image acquisitions and image artifacts. From our experience, pre-contrast, non-fat-suppressed T1-weighted images in at least two planes and a volumetric T1-weighted post-contrast sequence with multiplanar reformatting play major roles in the imaging protocol to look for PNS. It is preferable to use a 3T magnet due to the ability to achieve a higher signal-to-noise ratio and thinner slices. Diligent search and scrutinization of the neural foramina and neural pathway for localizing the tumor spread are essential for accurate diagnosis and staging. Radiologists need to be skillful when scrutinizing skull base neuroforamina, as some of the foramina are better depicted on one plane than another [21 Perineural invasion and spread in head and neck cancer. Expert Rev Anticancer Ther 12(3):359–371.
")]. Both anterograde and retrograde spread may occur, and the whole pathway needs to be evaluated as “skipped lesion” has been reported on imaging [21 Perineural invasion and spread in head and neck cancer. Expert Rev Anticancer Ther 12(3):359–371.
")]. MRI signs for PNS include replacement of normal neural foramina fat by soft tissue signal, nerve enlargement, and nerve enhancement. Secondary imaging signs include denervation changes of involved musculature, with muscle edema in the acute phase and fatty atrophy in the chronic phase [23 Pathology of perineural spread. J Neurol Surg B Skull Base 77(2):124–130.
")].
From the authors’ experience, not just locally aggressive tumors but also small localized tumors in the nasopharynx are seen to present with PNS on MRI. Accurate identification of PNS may upstage the disease to T4, allowing intensification of treatment with concurrent chemoradiation and exact conformation of the radiation field to the PNS pathway by IMRT to minimize residual disease and risk of local recurrence (Fig.3).
Fig. 3
Perineural tumor spread in a 60-year-old man presented with left maxillary facial numbness with biopsy-proven undifferentiated carcinoma in left nasopharynx. Axial post-contrast T1-weighted image with fat suppression shows the left nasopharyngeal tumor at the fossa of Rosenmüller (asterisk) (a). Axial post-contrast T1-weighted image with fat suppression at a higher level reveals asymmetric enlargement and enhancement of the left vidian canal extending anterolaterally into the left pterygopalatine fossa (arrows) (b). Axial post-contrast T1-weighted MR image with fat suppression at a further higher level shows asymmetric enlargement and enhancement of the left foramen rotundum along the cavernous sinus (arrows) (c). Coronal T1-weighted image without fat suppression shows the asymmetric enlargement of the left foramen rotundum (black arrowhead) and vidian canal with fat effacement (white arrowhead). The nasopharyngeal tumor extends to the left vidian canal with perineural spread (asterisk) (d). The apparent small tumor in the nasopharynx with underdetection of perineural tumor spread may lead to understaging and subsequent undertreatment with missing irradiation target and over-estimation of survival
Full size image
4.
Nodal metastases of NPC follows a predictable and an orderly fashion. Nodal necrosis and radiological extranodal extension (ENE) carry significant prognostication and/ or treatment implications other than nodal staging itself.
Nodal metastasis is a common presentation for NPC, occurring in about 85% of cases [88
")]. The nodal metastases follow an orderly pattern of spread, with the most commonly involved regions being the retropharyngeal (69%) and level II lymph nodes (70%) as echelon nodes. Despite the common perception of nodal spread from the retropharyngeal station to level II, isolated level II node involvement can occur without involvement of the retropharyngeal nodes, particularly for lateralized tumors in the nasopharynx [3 Current radiotherapy considerations for nasopharyngeal carcinoma. Cancers (Basel) 14(23):5773.
"), 8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
")]. The level III (45%), IV (11%), and V (27%) nodal stations are less commonly involved [8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
")]. Skip metastases are rarely observed, occurring in 0.5–7.9% of cases [8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
")]. The presence of N2 or N3 disease, irrespective of any T stage according to the 8th American Joint Cancer Committee (AJCC) staging, should prompt the initiation of induction chemotherapy according to the latest National Comprehensive Cancer Network guidelines [28 NCCN clinical practice guidelines in oncology: head and neck cancers (version 4.2024) [PDF].
")]. The presence of N3 disease also serves as an indication for FDG-PET/CT to evaluate for distant metastases [2 MR imaging of nasopharyngeal carcinoma. Magn Reson Imaging Clin N Am 30(1):19–33.
")].
Other low-risk nodal groups are rarely involved, such as the supraclavicular nodes (3%), level IB nodes (3%), level IA and VI nodes (0%), and parotid nodes (1%). This unusual nodal involvement should prompt a search for involvement of the relevant drainage area or possible differential diagnosis. For example, level IB involvement can occur with tumors involving the oral cavity, anterior nasal cavity, or when there is bulky disease in level II nodes (short axis > 20mm) leading to retrograde spread [72
")]. Parotid node involvement is uncommon in NPC unless there is an advanced-stage disease or again a bulky node in level II leading to retrograde spread [29 Clinical treatment considerations in the intensity-modulated radiotherapy era for parotid lymph node metastasis in patients with nasopharyngeal carcinoma. Radiother Oncol 186:109802.
")]. Parotid node involvement is associated with a higher risk of treatment failure, distant metastasis, and regional recurrence, with similar disease-free survival (DFS) and distant metastasis-free survival (DMFS) as those with N3 disease [30 Prognosis and staging of parotid lymph node metastasis in nasopharyngeal carcinoma: An analysis in 10,126 patients. Oral Oncol 95:150–156.
")]. The involvement of these nodal stations should be communicated to the radiation oncologists to ensure appropriate field coverage in radiotherapy planning [8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
"), 29 Clinical treatment considerations in the intensity-modulated radiotherapy era for parotid lymph node metastasis in patients with nasopharyngeal carcinoma. Radiother Oncol 186:109802.
")].
Other important features that could be identified on CT or MRI but not routinely included in the 8th AJCC staging system include cervical nodal necrosis and radiological ENE. They should be identified, reported, and communicated to oncologists for better patient risk stratification and prognostication. Cervical nodal necrosis has an incidence of 44% among NPC cases. It can be diagnosed as a hypoenhancing area within the node on CT, or as a focal area of high signal intensity on T2-weighted images or a focal area of low signal intensity on contrast-enhanced T1-weighted images, with or without a surrounding rim of enhancement [311
")]. It is an independent negative prognostic factor with a significantly poorer survival and a considerably higher distant metastasis rate [30 Prognosis and staging of parotid lymph node metastasis in nasopharyngeal carcinoma: An analysis in 10,126 patients. Oral Oncol 95:150–156.
")]. A recent meta-analysis involving 4359 patients showed the presence of nodal necrosis predicted poorer DMFS, DFS, and overall survival (OS) [32 Prognostic value of cervical nodal necrosis on staging imaging of nasopharyngeal carcinoma in era of intensity-modulated radiotherapy: a systematic review and meta-analysis. Cancer Imaging 22(1):24.
")]. ENE is a histological diagnosis referring to tumor growth beyond the capsule of a lymph node [33 Assessment criteria and clinical implications of extranodal extension in head and neck cancer. Am Soc Clin Oncol Educ Book 41:265–278.
")]. It reflects the aggressiveness of a tumor and has an incidence of 33.6%—39.8% in NPC [34 Extranodal extension is a criterion for poor outcome in patients with metastatic nodes from cancer of the nasopharynx. Oral Oncol 88:124–130.
")]. In NPC, imaging serves as the primary modality for assessment of the presence of ENE, and radiological ENE is 90% specific to indicate the presence of pathological ENE [33 Assessment criteria and clinical implications of extranodal extension in head and neck cancer. Am Soc Clin Oncol Educ Book 41:265–278.
")]. It is important to recognize that radiological ENE appears as a spectrum of imaging abnormality, ranging from indistinct/ irregular nodal margin, extension to perinodal fat, conglomerate/ matted/ coalescent nodes and extension into adjacent structures such as muscle, skin and glands. Radiological ENE has an added value for risk stratification in future TNM but requires standardization in reporting [33 Assessment criteria and clinical implications of extranodal extension in head and neck cancer. Am Soc Clin Oncol Educ Book 41:265–278.
"), 35 Criteria for the diagnosis of extranodal extension detected on radiological imaging in head and neck cancer: Head and Neck Cancer International Group consensus recommendations. Lancet Oncol 25(7):e297–e307.
")]. The recently published Head and Neck Cancer International Group consensus recommendations have established a grading system for reporting (Fig.4) [35 Criteria for the diagnosis of extranodal extension detected on radiological imaging in head and neck cancer: Head and Neck Cancer International Group consensus recommendations. Lancet Oncol 25(7):e297–e307.
")]. It is worth noting that infiltration of muscle, skin and glands (grade 3) is associated with a significantly shorter regional relapse-free survival, DMFS, and OS than those with ENE to fat or without ENE [34 Extranodal extension is a criterion for poor outcome in patients with metastatic nodes from cancer of the nasopharynx. Oral Oncol 88:124–130.
")]. The addition of radiological ENE to N3 increases sensitivity to predict recurrence [36 Radiologic extranodal extension for nodal staging in nasopharyngeal carcinoma. Radiother Oncol 191:110050.
")]. The advanced ENE (grade 3) has been proposed as a criterion of N3 in the 9th AJCC staging [37 Ninth version of the AJCC and UICC nasopharyngeal cancer TNM staging classification. JAMA Oncol 10(12):1627–1635.
")]. Reporting radiological ENE is important not only because it is an independent predictor of poor prognosis, it also requires the expertise of subspecialized head and neck oncologists in radiotherapy contouring. The presence of radiological ENE requires an expansion of 10mm CTV instead of 5mm around the GTV [8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
")]. There should be prophylactic level 1B node coverage in low-risk CTV when a level II node with radiological ENE is present [8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
")]. Muscle infiltration indicates the inclusion of the muscle near the node with at least a 10 mm margin in all directions [8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
")].
Fig. 4
Grading of radiological extranodal extension from the Head and Neck Cancer International Group consensus recommendations [354
")]. Axial post-contrast T1-weighted image with fat suppression shows the left level II nodes demonstrating ill-defined nodal margin that extends into the perinodal fat (arrowhead). This is classified as grade 1 (a). Axial post-contrast T1-weighted image with fat suppression shows conglomerate left level II nodal mass (asterisk) with ill-defined nodal margin extending to the perinodal fat (arrowheads). This is classified as grade 2 (b). Axial post-contrast T1-weighted image with fat suppression shows conglomerate right level II nodal mass with extension to adjacent neck muscles (black arrows) (c). Axial post-contrast T1-weighted image with fat suppression shows conglomerate left level II nodal mass with extension to adjacent subcutaneous layer and skin (white arrows) (d). These are classified as grade 3
Full size image
5.
Various tumor mimics exist in the nasopharynx ranging from benign to malignant conditions. Accurate diagnosis avoids mismanagement.
There are various tumor mimics in the nasopharynx mimicking NPC, ranging from benign conditions such as skull base osteomyelitis to malignant conditions such as nasopharyngeal lymphoma, invasive pituitary macroadenoma, and sphenoid sinus squamous cell carcinoma.
Skull base osteomyelitis (SBO) is a rare and life-threatening complication of otogenic infection in elderly patients with diabetes mellitus 385
"),[39 Imaging spectrum and complications of otogenic infections: insights into delayed diagnosis. Br J Radiol 97(1156):726–733.
"),40 Skull base osteomyelitis: a comprehensive imaging review. AJNR Am J Neuroradiol 42(3):404–413.
")]. It is usually a sequela of untreated necrotizing external otitis caused by _Pseudomonas aeruginosa_ [38 Diagnostic dilemma between skull base osteomyelitis and nasopharyngeal carcinoma: a case series. Hong Kong Med J 27(4):300–302.
"), 39 Imaging spectrum and complications of otogenic infections: insights into delayed diagnosis. Br J Radiol 97(1156):726–733.
")]. It poses a diagnostic challenge to clinicians and radiologists and a delayed diagnosis of up to 6 months is common [38 Diagnostic dilemma between skull base osteomyelitis and nasopharyngeal carcinoma: a case series. Hong Kong Med J 27(4):300–302.
"), 39 Imaging spectrum and complications of otogenic infections: insights into delayed diagnosis. Br J Radiol 97(1156):726–733.
")]. MRI is often the investigation of choice for establishing the diagnosis and depicting the disease extent. NPC is the main differential diagnosis to consider [39 Imaging spectrum and complications of otogenic infections: insights into delayed diagnosis. Br J Radiol 97(1156):726–733.
"), 41 Diagnosis of skull base osteomyelitis. Radiographics 41(1):156–174.
")] and several helpful distinguishing imaging signs are illustrated (Fig.5) (Table2). Endoscopic biopsy or intraoperative image-guided biopsy is required in equivocal cases [38 Diagnostic dilemma between skull base osteomyelitis and nasopharyngeal carcinoma: a case series. Hong Kong Med J 27(4):300–302.
"), 41 Diagnosis of skull base osteomyelitis. Radiographics 41(1):156–174.
")].
Fig. 5
Illustrative comparative images of a 76-year-old female with skull base osteomyelitis (a) and a 53-year-old man with locally advanced NPC (clinical stage T4N3M0) (b). Axial post-contrast T1-weighted, fat-suppressed image shows intensely enhancing, infiltrative soft tissue signal involving the bilateral nasopharynx without significant mass effect. Note the preserved deep mucosal white line (arrowheads) and the preserved architecture of the nasopharynx. There is extension into the bilateral hypoglossal canals (black asterisks) and intracranial extension into the left posterior fossa. Lateral extension of the enhancing soft tissue signal to involve the left parotid gland is observed (white asterisk). Extensive enhancing marrow signal is noted in the bilateral basioccipital, left petrous and tympano-mastoid temporal bone. No retropharyngeal or cervical lymphadenopathy (a). Axial T2-weighted, fat-suppressed image shows the characteristic intermediate signal for NPC (asterisk). There is associated mass effect especially on the left nasopharynx and left prevertebral muscle. Disruption of the normal architecture of nasopharynx is evidenced by the partial disruption of the pharyngobasilar fascia (dotted line). Limited left lateral extension of the tumor is observed as the tumor is still bound by the tensor veli palatini muscle (arrows) (b). Extensive bilateral cervical lymphadenopathies (not shown)
Full size image
Table 2 Distinguishing MR features between skull base osteomyelitis and NPC
Full size table
Nasopharyngeal lymphoma and NPC differ in their imaging morphology. Lymphoma tends to appear as a midline mass with symmetry and involvement of Waldeyer’s ring while NPC tends to be an asymmetric mass of characteristic signals [427
")]. Lymphoma presents as an exophytic mass filling the airway with rare skull base invasion while NPC shows propensity for skull base invasion with or without intracranial spread [42 Multimodality imaging of extra-nodal lymphoma in the head and neck. Clin Radiol 77(8):e549–e559.
")]. The presence of parotid lymphadenopathy is more commonly observed in lymphoma than NPC [42 Multimodality imaging of extra-nodal lymphoma in the head and neck. Clin Radiol 77(8):e549–e559.
")]. Lymphoma tends to show significantly lower ADC values than NPC [42 Multimodality imaging of extra-nodal lymphoma in the head and neck. Clin Radiol 77(8):e549–e559.
"), 43 Differentiating nasopharyngeal carcinoma from lymphoma in the head and neck region using the apparent diffusion coefficient (ADC) value: a systematic review and meta-analysis. Pol J Radiol 17(88):e472–e482.
")]. Invasive pituitary macroadenoma, and sphenoid sinus squamous cell carcinoma tend to show different epicenters and signal characteristics as NPC (Fig.6).
Fig. 6
Sphenoid sinus squamous cell carcinoma in a 63-year-old man presented with headache and right 6th nerve palsy (a & b). Initial plain CT brain showed a hyperdense right sphenoid mass with clival destruction (not shown). Axial post-contrast T1-weighted image with fat-suppression shows an irregular and infiltrative mass centered at right sphenoid sinus with clival, bilateral petrous apex destruction, and intracranial extension (not shown). The lesion shows irregular peripheral enhancement with central areas of hypo-enhancement (asterisk) (a). Axial T2-weighted image with fat-suppression shows the corresponding mass with heterogenous, predominantly hypointense signal (asterisk) (b). Invasive pituitary macroadenoma in a 73-year-old woman presented with persistent headache (c & d). Axial T2-weighted image with fat-suppression shows an extensive tumor mass with intermediate signal containing multiple internal cystic foci involving the central skull base (arrowhead) (c). Coronal post-contrast T1-weighted image with fat-suppression shows the lesion centered at the sella turcica and clivus with bilateral cavernous sinus extension. The nasopharyngeal roof is unremarkable (asterisk) (d)
Full size image
Post-treatment imaging
#1 Recurrent tumors most commonly are in-field failure and a distinction has to be made with various post-treatment changes. Additional challenges in imaging interpretation are related to delayed tumor regression and time lag between histological and radiological regression. Identification of post-treatment changes from residual or recurrent tumors can avoid unnecessary biopsies.
Despite the advancement of IMRT techniques for the primary treatment of NPC, about 5–10% of patients will experience local or regional recurrence after treatment 449
"), [45 Patterns and prognosis of regional recurrence in nasopharyngeal carcinoma after intensity-modulated radiotherapy. Cancer Med 12(2):1399–1408.
")]. Direct visualization by endoscopy is challenging and may miss 27.8% of tumors due to submucosal or deep-seated locations [46 Imaging appearances for recurrent nasopharyngeal carcinoma and post-salvage nasopharyngectomy. Clin Radiol 68(11):e629–e638.
")]. Besides, endoscopic biopsy may have a risk of sampling errors when residual tumors are present only in small clusters [47 Utility of magnetic resonance imaging in determining treatment response and local recurrence in nasopharyngeal carcinoma treated curatively. BMC Cancer 20(1):193.
")]. Consequently, post-treatment imaging surveillance is a vital component of overall evaluation. The primary responsibility of radiologists in this phase is to identify the presence of any recurrent tumors. This is challenging given the complexity of various post-treatment anatomical distortions and changes. Knowledge of the characteristic signals of recurrent nasopharyngeal tumors and their failure patterns in the era of IMRT guides effective image interpretation. Key considerations to be addressed include selecting the appropriate imaging modality, determining the optimal timing for imaging, knowing what to expect on the post-treatment images, and finally deciding on the course of action when a recurrent tumor is detected.
The latency period for recurrent NPC is wide, with a median latency of 1.9 years (range 0.6–11.9) [480
")]. Most recurrences (81.7%) occur in the first three years after IMRT [45 Patterns and prognosis of regional recurrence in nasopharyngeal carcinoma after intensity-modulated radiotherapy. Cancer Med 12(2):1399–1408.
")]. However, there is still a small portion of recurrence that occurs more than 5 years post-treatment, and therefore continual surveillance is important. In-field recurrence (93.3%) is the most common pattern for loco-regional failure of NPC treatment [45 Patterns and prognosis of regional recurrence in nasopharyngeal carcinoma after intensity-modulated radiotherapy. Cancer Med 12(2):1399–1408.
")]. Distant metastases also occur in a certain portion of patients (17.5% of the cases) and remain the major cause of mortality [46 Imaging appearances for recurrent nasopharyngeal carcinoma and post-salvage nasopharyngectomy. Clin Radiol 68(11):e629–e638.
")].
Initial components for evaluation include clinical assessment, endoscopy, and plasma EBV DNA levels. Both MRI or FDG-PET/CT can be used for imaging response assessment. MRI is the modality of choice for local surveillance given its superior soft tissue contrast for assessment of local anatomical distortion, as well as disease extent in case of recurrent tumors. Accurate restaging and mapping of tumor extent forms an essential foundation for salvage therapy planning [480
")].
The optimal timing of follow-up imaging is a debated topic. International guidelines recommend that baseline follow-up imaging be done 3 to 12 months upon treatment completion 287
"), [49 Nasopharyngeal cancer: EHNS-ESMO-ESTRO Clinical Practice Guidelines for diagnosis, treatment and follow-up. Ann Oncol 21(Suppl 5):v187-9.
")]. The basis of scheduling post-treatment imaging originates from a study in 1999 showing that 93.2% of NPC patients treated with three-dimensional conformal radiotherapy can achieve histological remission by 12 weeks post-treatment [50 The time course of histologic remission after treatment of patients with nasopharyngeal carcinoma. Cancer 85(7):1446–1453")]. Another study in 2017 performed for post-treatment imaging after IMRT shows that only 83.3% of patients achieves complete clinical response by MRI and flexible endoscopy in 3–4 months, while this figure increases to 91.4% at 6–9 months. The prognosis of these two groups of patients showed no significant difference, therefore it is suggested that 6–9 months may be the optimal time point for maximal tumor response after IMRT [51 Delayed clinical complete response to intensity-modulated radiotherapy in nasopharyngeal carcinoma. Oral Oncol 75:120–126")]. Thus, delayed primary tumor regression and a time lag between histological and radiological regression can be potential pitfalls in follow-up imaging [47 Utility of magnetic resonance imaging in determining treatment response and local recurrence in nasopharyngeal carcinoma treated curatively. BMC Cancer 20(1):193.
")]. Anatomical regression also lags behind metabolic regression. Therefore, residual treated tissue is expected in the early phases of imaging surveillance and does not necessarily indicate the presence of viable tumor cells [46 Imaging appearances for recurrent nasopharyngeal carcinoma and post-salvage nasopharyngectomy. Clin Radiol 68(11):e629–e638.
"), 52 Posttreatment magnetic resonance imaging surveillance of head and neck cancers. Magn Reson Imaging Clin N Am 30(1):109–120.
")]. In a longitudinal cohort of 19 patients with "phantom tumor phenomenon" on MRI, 73.7% of them had histological correlation showing necrosis, inflammation, or reactive epithelial cells [53 Image-based diagnosis of residual or recurrent nasopharyngeal carcinoma may be a phantom tumor phenomenon. Medicine (Baltimore) 100(8):e24555.
")]. These false positive mimicking residual or recurrent tumors on MRI result in unnecessary biopsies, or even re-irradiation and surgery [53 Image-based diagnosis of residual or recurrent nasopharyngeal carcinoma may be a phantom tumor phenomenon. Medicine (Baltimore) 100(8):e24555.
")]. The optimal timing of the post-treatment MRI is a balance between early detection of residual or recurrent disease versus reduction of indeterminate rates in radiology reports. Most patients will require more than one post-treatment MRI to achieve a definitive disease status [47 Utility of magnetic resonance imaging in determining treatment response and local recurrence in nasopharyngeal carcinoma treated curatively. BMC Cancer 20(1):193.
")]. Only 33.2% cases will show complete remission on the first post-treatment MRI performed at a median time of 93 days (range 32–346 days), with the remaining cases either indeterminate (50.2%) or showing partial response (16.6%), requiring a second post-treatment MRI for follow-up [47 Utility of magnetic resonance imaging in determining treatment response and local recurrence in nasopharyngeal carcinoma treated curatively. BMC Cancer 20(1):193.
")] (Fig.7). Scheduling the first post-treatment MRI from 3 months to around 4 months upon treatment significantly increases the proportion of patients with radiological complete remission (32.8% vs 83.3%) [47 Utility of magnetic resonance imaging in determining treatment response and local recurrence in nasopharyngeal carcinoma treated curatively. BMC Cancer 20(1):193.
")].
Fig. 7
Delayed tumor regression and time lag between anatomical and metabolic regression in a 69-year-old man with locally extensive NPC (clinical stage T4N3M0) completed chemo-irradiation in 2023. Axial T1 post-contrast images with fat-suppression at pre-treatment stage show a large and infiltrative enhancing left nasopharyngeal tumor (arrow) and left level II lymphadenopathy (arrowhead) (a) & (b). There was evidence of pathological complete remission in the nasopharyngeal tumor at 10 weeks after completion of treatment. Axial T1 post-contrast images with fat suppression at 12 weeks show interval shrinkage in size of the known tumor with residual enhancing soft tissue in the left retropharyngeal and prevertebral space (arrow). The left level II node also shows interval shrinkage with residual heterogeneously enhancing soft tissue (arrowhead) (c) and (d). These were equivocal for residual tumors, which might warrant further palliative treatment. Axial fusion FDG PET/CT images at 12 weeks show the left nasopharyngeal lesion (arrow) as well as the left level II lesion (arrowhead) to be non-FDG avid, suggestive of metabolic resolution, despite the small residual soft tissue density. Apparent muscle activity along the bilateral longus colli muscles (asterisks) is a known diagnostic pitfall mimicking disease recurrence (e) and (f). FDG PET/ CT can be a helpful tool in this occasion due to its high negative predictive value.The patient subsequently underwent clinical observation instead of unnecessary biopsy or oncological therapy. Axial T1 post-contrast images at 24 weeks show almost complete resolution of the left nasopharyngeal lesion (asterisk) and the left level II node (arrowhead) (g) and (h)
Full size image
Radiologists must be aware of these pitfalls when interpreting the MRI. Conventional MRI is limited in the differentiation of the irradiated nasopharynx with anatomical distortion, granulation tissue, necrosis, and true viable tumor [54x
")]. Local anatomical distortion includes loss of symmetry of the nasopharynx and partial effacement of the lateral pharyngeal recess, posing challenges in assessment. Besides, a residual mass on follow-up MRI may not contain a viable tumor [54 Posttreatment imaging of the nasopharynx. Eur J Radiol 44(2):82–95.
")]. Non-enhancing tissue with a dark signal on T2-weighted imaging represents mature fibrosis. It may be difficult to distinguish between immature fibrosis, granulation tissue, and residual or recurrent tumors, as all of them enhance [55 Imaging recommendations for diagnosis, staging, and management of nasopharynx carcinoma. Indian J Med Paediatr Oncol 44:175–180.
")]. Correlation with endoscopic findings is important in imaging interpretation. Recurrent tumors typically show intermediate signal intensity on T2-weighted imaging with moderate contrast enhancement on T1-weighted contrast-enhanced images without fat saturation, and restricted diffusion on DWI [48 Recurrent nasopharyngeal carcinoma: the puzzles of long latency. Int J Radiat Oncol Biol Phys 44(1):149–56.
")] (Fig.8).
Fig. 8
Typical signal characteristics of the recurrent nasopharyngeal tumor in a 72-year-old man with history of NPC (clinical stage T1N1M0) and radical radiotherapy completed in 2013. Axial T2-weighted image with fat-suppression shows an intermediate signal mass in the right nasopharynx causing focal expansion of the pharyngobasilar fascia which appears ill-defined (arrows). There is also extension of signal abnormality to the right prevertebral muscle (asterisk) (a). Axial T1-weighted image with fat-suppression shows the moderately enhancing mass in the right nasopharynx (arrow) with extension to right prevertebral muscle (arrow) (b). On the ADC map there is hypointense signal corresponding to the right nasopharyngeal mass (arrow) (c). Axial fusion FDG-PET/CT confirms the presence of a hypermetabolic mass in the right nasopharynx (d). Biopsy confirmed recurrent NPC
Full size image
A practical solution includes the use of DWI and/ or FDG-PET/CT (Fig.9). DWI is helpful in addition to conventional MRI for the differentiation between benign post-treatment change and fibrosis versus tumor, as the latter will show restricted diffusion 562
. (ISSN 1043-1810)"), [57 Diagnostic value of diffusion-weighted magnetic resonance imaging for local and skull base recurrence of nasopharyngeal carcinoma after radiotherapy. Medicine (Baltimore) 97(34):e11929.
")]. Quantitative measurement of ADC value has been proven useful in the differentiation of benign post-treatment changes and fibrosis versus residual or recurrent tumors. The use of an ADC cutoff of 887 × 10–6 mm 2/s has a sensitivity of 87.2%, specificity of 94.1% and an area under curve of 0.967 in this regard [57 Diagnostic value of diffusion-weighted magnetic resonance imaging for local and skull base recurrence of nasopharyngeal carcinoma after radiotherapy. Medicine (Baltimore) 97(34):e11929.
")]. FDG-PET/CT has a very high negative predictive value of 95% for recurrent tumors, and a negative PET-CT at 6 months following completion of treatment indicates that residual or recurrent disease is highly unlikely [56 Essential imaging of the nasopharyngeal space with special focus on nasopharyngeal carcinoma. Oper Tech Otolaryngol Head Neck Surg 32(1):8–14.
. (ISSN 1043-1810)")]. Thus, it can be particularly useful in cases of indeterminate findings related to delayed tumor regression.
Fig. 9
Scar tissue in a 65-year-old man who had NPC (clinical stage T3N1M0) involving the right pterygopalatine fossa completed combined chemoirradiation in 2022. Post-treatment endoscopy revealed persistent tumor and therefore stereotactic radiotherapy was given. Axial T1-weighted post-contrast image with fat suppression shows the enhancing soft tissue at the right pterygopalatine fossa with local expansion (asterisk) suggestive of tumor (a). Axial T1-weighted post-contrast image with fat suppression performed at 18 months upon treatment completion shows interval shrinkage of the previous noted tumor at right pterygopalatine fossa but with persistent mildly enhancing soft tissue thickening (arrowhead) equivocal for residual tumor or scar (b). ADC map of the diffusion-weighted imaging performed at 18 months upon treatment completion showed no restricted diffusion in the right pterygopalatine fossa (arrowhead) (c). Axial fusion FDG-PET/CT at 18 months post-treatment depicts the right pterygopalatine fossa soft tissue to be non-FDG avid (d). The original tumor at the right nasopharynx has resolved (not shown). Findings are suggestive of scar tissue without viable tumor
Full size image
These limitations also support the fact that the combined use of both MRI and FDG-PET/CT is significantly more accurate than the use of either modality individually in detecting residual or recurrent NPC [583
")]. PET/MRI may be a useful one-stop investigation with both complementary anatomical and functional information, but there are technical challenges to overcome in combining PET and MRI such as MRI-based attenuation correction, clinical workflow integration, and motion artifacts owing to the long scanning time [59 PET/MRI: Technical challenges and recent advances. Nucl Med Mol Imaging 50(1):3–12.
")].
Identification of recurrent tumors should be followed by careful scrutiny to map out the full extent of the tumor and its nodal metastases, if any, for restaging. Attention should be paid to the relationship of the recurrent tumor and the internal carotid artery, as it determines the operability and the approach of salvage surgery 480
"), [60 Clinical advances in nasopharyngeal carcinoma surgery and a video demonstration. Vis Cancer Med 2:2")]. A close resection margin may limit the option of endoscopic salvage [60 Clinical advances in nasopharyngeal carcinoma surgery and a video demonstration. Vis Cancer Med 2:2")]. Nodal metastases should be evaluated by both MRI and FDG-PET/CT. Ultrasound-guided fine-needle aspiration is helpful for equivocal nodes. Selective neck dissection is the preferred treatment option due to better functional outcome [61 Comparing endoscopic surgeries with open surgeries in terms of effectiveness and safety in salvaging residual or recurrent nasopharyngeal cancer: Systematic review and meta-analysis. Head Neck 42(11):3415–3426.
")], therefore, accurate detection of nodal metastases is crucial. Systemic staging for distant metastases by FDG-PET/CT is vital, as the treatment strategy shifts from radical surgery to palliative chemotherapy if distant metastasis is present.
#2 Post-radiation nasopharyngeal necrosis is a rare but life-threatening delayed complication mimicking tumor recurrence to be recognized.
Post-radiation nasopharyngeal necrosis is a rare, delayed but life-threatening complication following radiation therapy for nasopharyngeal tumors [626
")]. It is a form of soft tissue necrosis involving the surrounding and affiliated tissues of the nasopharynx, such as the mucosa, longus capitis muscles, parapharyngeal tissues, and skull base [62 Postradiation nasopharyngeal necrosis in the patients with nasopharyngeal carcinoma. Head Neck 31(6):807–812.
")]. The incidence is about 1–2% in patients with NPC receiving primary radiotherapy, but increases to 30–40% after re-irradiation [63 Treating radiation-related nasopharyngeal necrosis with endostar in patient with nasopharyngeal carcinoma: A report of two cases and a literature review. Mol Clin Oncol 19(1):57.
")]. Predominant symptoms include foul nasal odor, persistent headache, and nasal hemorrhage [64 Clinical findings and imaging features of 67 nasopharyngeal carcinoma patients with postradiation nasopharyngeal necrosis. Chin J Cancer 32(10):533–538.
")]. Erosion of ICA may lead to potentially fatal hemorrhage [64 Clinical findings and imaging features of 67 nasopharyngeal carcinoma patients with postradiation nasopharyngeal necrosis. Chin J Cancer 32(10):533–538.
")].
A three-stage system for correlation of clinical and endoscopic findings has been described in the literature 626
"), [64 Clinical findings and imaging features of 67 nasopharyngeal carcinoma patients with postradiation nasopharyngeal necrosis. Chin J Cancer 32(10):533–538.
")]. The first stage mainly involves primary endoscopic findings of local mucosal structural alteration including ulceration. The second stage involves surrounding necrotic soft tissue on endoscopy, and the most severe stage involves osteonecrosis, with patients presenting with persistent headaches and foul odor [64 Clinical findings and imaging features of 67 nasopharyngeal carcinoma patients with postradiation nasopharyngeal necrosis. Chin J Cancer 32(10):533–538.
")].
Although endoscopic and imaging findings may be characteristic, the diagnosis remains challenging. Histopathological examination remains the gold standard to support the diagnosis and exclude the presence of recurrent tumors. Radiation-induced fibrosis is the main mechanism leading to the development of post-irradiation nasopharyngeal necrosis [633
")]. Distinctive radiation-induced or radiation-associated stromal changes are best recognized by the presence of fibrosis, fibrinous exudate, atypical fibroblasts, and paucity of cellular inflammatory exudate, which are characteristic findings [65 Demystifying the challenging diagnosis of post-radiation nasopharyngeal necrosis on multimodality imaging. J Med Imaging Radiat Oncol 68(7):805–807.
")].
Characteristic MRI features include local defects in the nasopharyngeal wall and rim-enhancing areas with central non-enhancement 626
"),[64 Clinical findings and imaging features of 67 nasopharyngeal carcinoma patients with postradiation nasopharyngeal necrosis. Chin J Cancer 32(10):533–538.
"),65 Demystifying the challenging diagnosis of post-radiation nasopharyngeal necrosis on multimodality imaging. J Med Imaging Radiat Oncol 68(7):805–807.
")] (Fig.10). It is important to scrutinize the internal carotid artery to look for pseudoaneurysm formation. CT may also depict gas densities presumed to be have tracked from the ulceration to the submucosal soft tissue. DWI may show restricted diffusion and FDG-PET/CT may show hypermetabolism [65 Demystifying the challenging diagnosis of post-radiation nasopharyngeal necrosis on multimodality imaging. J Med Imaging Radiat Oncol 68(7):805–807.
")]. These features overlap with those of recurrent tumors, potentially leading to misdiagnosis [65 Demystifying the challenging diagnosis of post-radiation nasopharyngeal necrosis on multimodality imaging. J Med Imaging Radiat Oncol 68(7):805–807.
")]. However, it is uncommon for recurrent nasopharyngeal tumors to show internal hypoenhancing areas related to necrosis. Understanding of this rare entity is important for early recognition and diagnosis. Imaging may also guide potential site for biopsy, which carries a higher risk if there is deep ulceration or if the internal carotid artery is exposed.
Fig. 10
Post-radiation nasopharyngeal necrosis in a 49-year-old man with NPC (clinical stage T4N2M0). He developed epistaxis and progressive headache 6 months after radiotherapy. Nasoendoscopy showed crusted and ulcerated mucosa without mass. Axial contrast-enhanced CT image shows mucosal irregularity in bilateral nasopharynx. Rim-enhancing lesions are seen in bilateral nasopharynx with extension to carotid spaces encasing the internal carotid arteries (arrowheads) (a). Axial T2-weighted image with fat suppression shows the lesions to be heterogeneous in signal intensity (arrowheads) (b). Axial and coronal post-contrast T1-weighted image with fat suppression show focal defects in the bilateral nasopharynx as discontinuity of the deep mucosal line (arrows). The rim-enhancing lesions encasing the bilateral internal carotid arteries are again seen (arrowheads) (c) and (d). Deep biopsy was suspended due to the close relationship with internal carotid arteries. The lesions gradually resolved on conservative management with restoration of the continuity of the deep mucosal white line on follow-up MRI (not shown)
Full size image
No specific treatment has been established other than conservative therapy, including nasal irrigation, systemic or topical antibiotics, intravenous nutritional supplements, hyperbaric oxygen, and debridement guided by nasal endoscopy [633
")]. Endoscopic or open surgery has also been suggested to remove necrotic tissue, followed by flap coverage; however these approaches are limited by available expertise [63 Treating radiation-related nasopharyngeal necrosis with endostar in patient with nasopharyngeal carcinoma: A report of two cases and a literature review. Mol Clin Oncol 19(1):57.
")]. The prognosis remains poor, with a reported 2-year overall survival of 51.6% [63 Treating radiation-related nasopharyngeal necrosis with endostar in patient with nasopharyngeal carcinoma: A report of two cases and a literature review. Mol Clin Oncol 19(1):57.
")].
#3 Not every enlarging retropharyngeal mass is a sign of malignancy: the hypertrophied superior cervical ganglia.
The superior cervical ganglion (SCG) can be mistaken for an enlarged retropharyngeal lymph node in patients with post-treatment NPC, leading to unnecessary surgery [661
")]. Recent case reports and studies have shown that the SCG has a propensity to become hypertrophic after radiotherapy, leading to their misidentification on imaging as metastatic lateral retropharyngeal lymph nodes [67 Hypertrophied superior cervical ganglia after radiotherapy for head and neck cancer. AJR Am J Roentgenol 221(5):695.
"), 68 Serial magnetic resonance imaging evaluations of irradiated superior cervical sympathetic ganglia: Not every retropharyngeal enlarging mass is a sign of malignancy. Eur J Radiol 98:126–129.
")].
Several key MRI features help in the discrimination. SCGs are usually medial to the internal carotid arteries and lateral to the prevertebral muscles, between the C2 and C4 vertebrae [688
")]. They typically exhibit a fusiform or elongated shape on the coronal plane, with a characteristic and consistent (> 90%) central dot sign best appreciated on T2-weighted and post-contrast T1-weighted images with fat-suppression, related to the underlying venule [66 Superior cervical ganglion mimicking retropharyngeal adenopathy in head and neck cancer patients: MRI features with anatomic, histologic, and surgical correlation. Neuroradiology 58(1):45–50.
"), 68 Serial magnetic resonance imaging evaluations of irradiated superior cervical sympathetic ganglia: Not every retropharyngeal enlarging mass is a sign of malignancy. Eur J Radiol 98:126–129.
"), 69 Superior cervical sympathetic ganglion: normal imaging appearance on 3T-MRI. Korean J Radiol. 17(5):657–63.
")]. SCGs show homogeneous and intense enhancement [69 Superior cervical sympathetic ganglion: normal imaging appearance on 3T-MRI. Korean J Radiol. 17(5):657–63.
")]. After radiotherapy, the size, T2-weighted signal, and the ADC value can increase up to about 1 year and then remain stable [69 Superior cervical sympathetic ganglion: normal imaging appearance on 3T-MRI. Korean J Radiol. 17(5):657–63.
"), 70 Magnetic resonance imaging features of the superior cervical ganglion and expected changes after radiation therapy to the head and neck in a long-term follow-up. Neuroradiology 62(4):519–524.
")]. The intraganglionic hypointensity, homogeneous enhancement, and well-defined margin should be maintained [69 Superior cervical sympathetic ganglion: normal imaging appearance on 3T-MRI. Korean J Radiol. 17(5):657–63.
")]. The enlargement is thought to be related to Schwann cell proliferation and/ or thickening of the perineurium and epineurium in response to radiation-induced damage [70 Magnetic resonance imaging features of the superior cervical ganglion and expected changes after radiation therapy to the head and neck in a long-term follow-up. Neuroradiology 62(4):519–524.
")].
To distinguish SCG from a retropharyngeal lymph node, the location, size, and signal characteristics on MRI are important considerations [714
")]. SCG is typically located more posterolaterally and inferiorly than a retropharyngeal node, which is most often (about 75%) located at the level of C1 [8 Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
")]. SCG tends to be larger in volume, has a higher degree of contrast enhancement, and a significantly higher ADC value than a retropharyngeal node (1.80 ± 0.28 × 10−3 mm 2/s versus 0.73 ± 0.10 × 10−3 mm 2/s, _P_< 0.001) [65 Demystifying the challenging diagnosis of post-radiation nasopharyngeal necrosis on multimodality imaging. J Med Imaging Radiat Oncol 68(7):805–807.
")]. FDG-PET/CT can be another problem-solving tool, as SCG usually (> 80%) does not show FDG avidity. (Fig.11).
Fig. 11
Illustrative comparative images of a 49-year-old female with a hypertrophied superior cervical ganglion with history of treated NPC in 2014 (clinical stage T4N1M0) (a) to (c) and a 58-year-old man with recurrent retropharyngeal node 2 years post-treatment of locally advanced NPC (clinical stage T4N3M0) (d) to (f). Axial T1-weighted post-contrast image with fat suppression at 6 months post-treatment shows a sub-centimeter enhancing lesion with central hypointense dot at left retropharyngeal space (arrowhead) located anteromedial to the ipsilateral internal carotid artery (a). The lesion showed interval enlargement and then stabilized at 6 and 7 years upon treatment completion (not shown). Axial post-contrast T1-weighted image with fat suppression at 9 years post-treatment shows an enlarged lesion in the left retropharyngeal space (arrowhead). Its T2-weighted hyperintense signal persists. The internal architecture with central hypointense dot is maintained (b). Otherwise, the nasopharynx showed no mass all along to indicate recurrence (not shown). Axial FDG-PET fusion image shows the lesion being non FDG-avid (c). Overall findings are in keeping with a hypertrophied superior cervical ganglion after radiotherapy. Axial T2-weighted image with fat-suppression at 2 years post-treatment shows a new sub-centimeter left retropharyngeal node with intermediate signal intensity (arrowhead) (d). Axial T1-weighted post-contrast image with fat suppression at 2 years post-treatment shows the corresponding lesion to be avidly enhancing (arrowhead) (e). Axial FDG-PET fusion image shows the lesion being hypermetabolic (f). No other local recurrence or distant metastases was depicted. The patient underwent salvage open nasopharyngectomy and confirmed recurrent tumor in the retropharyngeal node
Full size image
#4 Skull base marrow signal change after radiotherapy does not always equal recurrent or residual tumor.
In post-treatment MRI, surveillance of skull base marrow signal is important as it may be one of the first signs of tumor recurrence. Evaluation is challenging not only because of the slow recovery of the previously infiltrative bone marrow, but also due to the appearance of new bone marrow lesions related to radiation-induced injury (e.g. radiation osteitis, osteoradionecrosis (ORN)), or signal changes related to granulation tissues and fibrosis [72y
")]. In fact, skull base and clival marrow signal change can appear on post-treatment MRI with or without preceding tumoral invasion, and this can persist for many years after treatment completion [72 The evolution of bone marrow signal changes at the skull base in nasopharyngeal carcinoma patients treated with radiation therapy. Radiol Med 126(6):818–826.
")] (Fig.12). Moreover, these signal changes are also nonspecific for residual/recurrent tumor or benign post-treatment changes.
Fig. 12
Serial MRI demonstrating the signal change in the basisphenoid of a 56-year-old female with a history of localized NPC (clinical stage T1N0M0) in 2011 with pathological complete remission. Sagittal T1-weighted post-contrast MRI with fat suppression at baseline shows no evidence of signal abnormality at the clivus. The nasopharyngeal tumor is seen (asterisk) (a). There is progressive increase in the patchy enhancement in the clivus in 3-year and 9-year follow–up images (b) and (c), and the signal abnormality shows partial resolution in 12-year follow-up image (arrows) (d). Coronal T1 post-contrast MRI with fat suppression at baseline shows no evidence of signal abnormality at the basisphenoid (e). There is progressive increase in the patchy enhancement in the bilateral basisphenoid, more on left side, in 3-year follow-up image (f) and 9-year follow–up image (g), and the signal abnormality shows partial resolution in 12-year follow-up image (arrows) (h)
Full size image
A retrospective study reviewing the serial MRIs of 50 patients with NPC showed that clival bone marrow signal change persisted without any evidence of recurrence in 26 patients with clival infiltration at diagnosis, for a mean of 66.5 (maximum 137) months. There was accompanying contrast enhancement noted for up to 125 months [72y
")]. The presence of enhancing marrow signal in the post-irradiated skull base does not always indicate recurrent or residual tumor, and it is most often benign post-treatment changes.
There are several approaches to this diagnostic dilemma. Radiation osteitis is often asymptomatic and incidentally detected [731
")]. It appears as foci of low, intermediate and high signal intensity on T2-weighted images, with patchy areas of enhancement [73 Delayed complications of radiotherapy treatment for nasopharyngeal carcinoma: imaging findings. Clin Radiol 62(3):195–203.
")]. Patients with recurrent NPC may present with bloody nasal discharge, headache or painless enlargement of a neck node [74 Clinical characteristics of recurrent nasopharyngeal carcinoma in high-incidence area. ScientificWorldJournal 2012:719754.
")], and may be accompanied by a rising trend of plasma EBV DNA [75 Plasma epstein-barr virus DNA as an archetypal circulating tumour DNA marker. J Pathol 247(5):641–649.
")]. The time frame of occurrence and signal characteristics on conventional MRI of these two entities may overlap [73 Delayed complications of radiotherapy treatment for nasopharyngeal carcinoma: imaging findings. Clin Radiol 62(3):195–203.
"), 74 Clinical characteristics of recurrent nasopharyngeal carcinoma in high-incidence area. ScientificWorldJournal 2012:719754.
")]. ORN of the skull base occurs with a mean time of onset at 18 months upon completion of radiotherapy. Patients often present with pain, bleeding, foul odor and are typically found to have exposed and necrotic bone [76 Osteoradionecrosis of the skull base. J Neurooncol 150(3):477–482.
")]. It may show variable MRI signals with low signal on T1-weighted images, high signal on T2-weighted fat-suppressed images and avid enhancement on postcontrast T1-weighted fat-suppressed images [77 Diffusion-weighted and PET/MR imaging after radiation therapy for malignant head and neck tumors. Radiographics 35(5):1502–27.
")]. The identification of ORN on MRI alone is challenging and correlation with osseous findings on CT can be extremely helpful [78 Osteoradionecrosis after radiation therapy for head and neck cancer: differentiation from recurrent disease with CT and PET/CT imaging. AJNR Am J Neuroradiol 35(7):1405–1411.
")]. The presence of bone sclerosis, intraosseous gas and the absence of associated solid mass favors ORN rather than tumor recurrence [78 Osteoradionecrosis after radiation therapy for head and neck cancer: differentiation from recurrent disease with CT and PET/CT imaging. AJNR Am J Neuroradiol 35(7):1405–1411.
")]. The diagnosis of recurrent tumors is more straightforward when there are new areas of cortical destruction associated with an intermediately enhancing soft tissue mass [73 Delayed complications of radiotherapy treatment for nasopharyngeal carcinoma: imaging findings. Clin Radiol 62(3):195–203.
")]. Comparison of serial previous imaging is essential. Stable or decreasing size of bone marrow signal change or contrast enhancement favors post-treatment changes [71 MR Imaging of the superior cervical ganglion and inferior ganglion of the vagus nerve: structures that can mimic pathologic retropharyngeal lymph nodes. AJNR Am J Neuroradiol 39(1):170–176.
")]. Conversely, increasing size of bone marrow signal change or contrast enhancement is highly suspicious for recurrence [72 The evolution of bone marrow signal changes at the skull base in nasopharyngeal carcinoma patients treated with radiation therapy. Radiol Med 126(6):818–826.
")]. DWI can offer distinction between these entities. Prior work from Wang et al. has shown that the mean ADC values in clival recurrence are significantly lower than those without (0.780 ± 0.166 × 10−3 and 1.666 ± 0.342 × 10−3 mm 2/s, respectively (_P_ = 0.002)) [57 Diagnostic value of diffusion-weighted magnetic resonance imaging for local and skull base recurrence of nasopharyngeal carcinoma after radiotherapy. Medicine (Baltimore) 97(34):e11929.
")]. ORN generally exhibits high ADC value unless an abscess is present from superimposed infection [77 Diffusion-weighted and PET/MR imaging after radiation therapy for malignant head and neck tumors. Radiographics 35(5):1502–27.
")]. FDG-PET/CT may not be entirely reliable in this setting due to considerable overlap in SUV uptake between ORN and recurrent tumors leading to false positivity [77 Diffusion-weighted and PET/MR imaging after radiation therapy for malignant head and neck tumors. Radiographics 35(5):1502–27.
"), 78 Osteoradionecrosis after radiation therapy for head and neck cancer: differentiation from recurrent disease with CT and PET/CT imaging. AJNR Am J Neuroradiol 35(7):1405–1411.
")].
#5 Familiarity with the expected imaging appearance of nasopharyngectomy is essential for accurate image interpretation of post-operative changes versus recurrent tumors.
Salvage surgical resection in the form of nasopharyngectomy has been considered the primary treatment for residual or recurrent NPC at the primary site, as it can achieve reasonable long-term survival and better local control than reirradiation, while avoiding the toxicities of cumulative radiation injury to normal tissues [79, 802
")]. Nasopharyngectomy can be performed via open approaches (e.g. maxillary swing, midface degloving, transpalatal, transmaxillary, and trans-infratemporal fossa) or a transnasal endoscopic approach. The endoscopic approach has gained popularity in recent years as a minimally-invasive surgical option. In a recent meta-analysis, the endoscopic approach demonstrated higher survival rates compared to open approaches in rT1 (93% versus 87%), rT2 (77% versus 63%) and rT3 tumors (67% versus 53%), with fewer severe complications and a lower local recurrence rate (27% versus 32%) [61 Comparing endoscopic surgeries with open surgeries in terms of effectiveness and safety in salvaging residual or recurrent nasopharyngeal cancer: Systematic review and meta-analysis. Head Neck 42(11):3415–3426.
")]. The definition of resectability has significant variation across different institutions, and the indications for nasopharyngectomy have expanded. Recurrent tumors are considered inoperable only when there is substantial intracranial extension with cavernous sinus invasion or encasement of the petrosal internal carotid artery [79 Recent advances in the management of nasopharyngeal carcinoma. F1000Res 7:1829"), 81 MRI signal changes in the skull base bone after endoscopic nasopharyngectomy for recurrent NPC: a serial study of 9 patients. Eur J Radiol 82(2):309–315.
")].
The cornerstone of successful surveillance imaging relies on familiarity with the expected changes after surgery. It is crucial to review the pretreatment imaging, as recurrent tumors are expected to exhibit similar signal characteristics as the original tumor. Furthermore, knowledge of the specific operative procedures enhances the understanding of the expected imaging changes. The common imaging appearances of open nasopharyngectomy via the maxillary swing approach and transnasal endoscopic nasopharyngectomy will be discussed.
Open nasopharyngectomy via the maxillary swing approach involves mobilization of the maxilla to expose the nasopharynx and ipsilateral parapharyngeal space. This approach provides extensive access to the nasopharynx for tumor resection with a wide margin. With this approach, the maxilla, along with the anterior cheek tissue, is swung laterally after soft tissue dissection and osteotomies have been performed 480
"), [82 New approach to the nasopharynx: the maxillary swing approach. Head Neck 13(3):200–7.
")]. There can be expected signal alterations in the operative bed early on, which show gradual normalization over time. For example, drilling of the clivus to remove the tumor involving the longus colli muscle may lead to enhancing marrow signals in the early postoperative imaging [46 Imaging appearances for recurrent nasopharyngeal carcinoma and post-salvage nasopharyngectomy. Clin Radiol 68(11):e629–e638.
")]. A temporalis muscle flap or vastus lateralis flap can be employed to protect exposed arteries in the maxillary osteocutaneous flap, in order to prevent carotid artery blowout, postoperative maxillary osteonecrosis, and ascending infection [60 Clinical advances in nasopharyngeal carcinoma surgery and a video demonstration. Vis Cancer Med 2:2")]. The flap appears more swollen with hyperintense signals on T2-weighted images, which can be related to inflammatory changes in the early postoperative period. The fatty striated appearance should be preserved. It should not show restricted diffusion on DWI nor hypermetabolism on FDG-PET/CT (Figs.13 and 14). Over time, the T2-weighted signal and enhancement will decrease. Careful inspection of the interface between the flap and the surgical cavity (recipient bed) is important, as tumor recurrence is commonly seen at this site, where surgical or endoscopic access is difficult [83 Imaging of Surgical Free Flaps in Head and Neck Reconstruction. AJNR Am J Neuroradiol 40(1):5–13.
")]. Smooth, non-nodular and non-mass-like enhancement can be observed in the non-fatty portion of the flap [83 Imaging of Surgical Free Flaps in Head and Neck Reconstruction. AJNR Am J Neuroradiol 40(1):5–13.
")]. In contrast, nodularity, mass-like or focal enhancement with signal characteristics similar to those of the original tumor is a characteristic imaging appearance of recurrence [83 Imaging of Surgical Free Flaps in Head and Neck Reconstruction. AJNR Am J Neuroradiol 40(1):5–13.
")] (Fig.15).
Fig. 13
Same patient as Fig. 6. Normal expected early post-operative appearance of open nasopharyngectomy in a 72-year-old man with recurrent NPC (clinical restaging T2N0M0). He had open right nasopharyngectomy via maxillary swing approach with vastus lateralis flap followed by post-operative stereotactic radiotherapy. A follow-up MRI was performed at 2-month after the operation. Axial T2-weighted image with fat-suppression shows the flap at the right nasopharynx appearing diffusely swollen with markedly hyperintense signal (asterisk) with perifocal soft tissue edema (arrowheads) (a). Axial T1-weighted post-contrast image with fat suppression shows the flap with heterogenous enhancement (asterisk) and perifocal soft tissue enhancement in the operative bed (arrowheads) (b).The imaging appearance of the flap was confused with residual tumor and additional FDG-PET/CT as well as follow-up MRI with DWI were performed. Axial fusion FDG-PET/CT shows the flap being diffusely eumetabolic (asterisk) (c). Conventional sequences on follow-up MRI after a further 2 months showed no interval change in size and signal characteristics of the flap (not shown). ADC map shows that there is diffuse hyperintense signal within the flap suggestive of facilitated diffusion (asterisk) (d). No residual or recurrent tumor was evident
Full size image
Fig. 14
Normal expected post-operative appearance of open nasopharyngectomy in a 58-year-old man with NPC (clinical stage T4N3M0) completed chemo-irradiation in 2017. Two years after, he had a left retropharyngeal node recurrence on surveillance MRI (not shown) and subsequently an open nasopharyngectomy via maxillary swing approach and a flap reconstruction using temporalis muscle. Axial T1-weighted pre-contrast image without fat suppression one-year after the operation shows no recurrence, demonstrating normal fat striation of the temporalis muscle flap without soft tissue replacement (arrows) (a). Axial T1-weighted post-contrast image with fat-suppression shows non-mass like enhancement in the medial aspect of the flap (asterisk). The recipient bed of the flap shows a straight margin without discrete nodular enhancement (arrowheads) (b). Axial T2-weighted image with fat suppression shows mild non mass-like hyperintense signal (asterisk) (c). Coronal T1-weighted image without fat-suppression shows the smooth transition in the surgical bed and straight superior margin of the flap with respect to the sphenoid bone (arrowheads) (d)
Full size image
Fig. 15
Same patient as Fig.14. Recurrent tumor after open nasopharyngectomy in a 58-year-old man with NPC (clinical stage T4N3M0) completed chemo-irradiation in 2017 with retropharyngeal lymph node recurrence in 2019. He underwent open nasopharyngectomy via maxillary swing approach and a flap reconstruction using temporalis muscle in 2020. Axial T2-weighted image with fat suppression shows intermediate signal nodular lesion with irregular border at the recipient bed (arrowhead) (a). Axial T1-weighted image without fat-suppression shows the replacement of the loss of normal fatty streak of the flap with nodular soft tissue signal (arrowhead) (b). Axial post-contrast T1-weighted image with fat suppression shows the corresponding lesion to be intensely and homogenously enhancing (arrowhead). Encasement of ipsilateral internal carotid artery is noted indicating unresectable disease (asterisk) (c). Axial fusion FDG PET/CT image confirms hypermetabolism of the corresponding lesion (d). Overall findings are in keeping with recurrent tumor. He then received palliative chemotherapy for treatment
Full size image
Endoscopic nasopharyngectomy is a minimally invasive approach involving endoscopic access to the nasopharynx via the nasal cavities. Different techniques exist, and the extent of resection varies with the extent of the recurrent tumor 802
"), [84 Magnetic resonance imaging after nasopharyngeal endoscopic resection and skull base reconstruction. J Clin Med 13(9):2624.
")]. MRI signal alterations in the operative bed can be expected as early as 2 weeks, being more common at the site of the recurrent tumor and its adjacent area [80 Types of transnasal endoscopic nasopharyngectomy for recurrent nasopharyngeal carcinoma: Shanghai EENT hospital experience. Front Oncol 10:555862.
")]. Posterior septectomy and inferior turbinectomy can enhance exposure of the nasopharynx and surgical maneuverability [84 Magnetic resonance imaging after nasopharyngeal endoscopic resection and skull base reconstruction. J Clin Med 13(9):2624.
")]. Their corresponding defects should be expected on the postoperative imaging. Reconstruction via various flaps (e.g. nasoseptal flap or temporoparietal fascial flap) is needed to cover the bony defect, to avoid osteomyelitis and exposure of the internal carotid artery. Inflammatory signal change is commonly noted in the bony structures (e.g. clivus, pterygoid base) and soft tissue structures (e.g. retropharyngeal space, masticator space) in the early phase, and shows gradual resolution over time. The nasoseptal flap appears as a thin layer of tissue along the posterolateral border of the surgical field. Its mucosal layer shows homogeneous hyperintense signal on T2-weighted images and post-contrast enhancement. There should be minimal thickness reduction over time without significant signal alteration in one year [84 Magnetic resonance imaging after nasopharyngeal endoscopic resection and skull base reconstruction. J Clin Med 13(9):2624.
")] (Fig.16). The temporoparietal fascial flap assumes a tri-laminated appearance with mucosal, intermediate and deep layers. It typically displays a reduction in the T2 signal and associated enhancement of the intermediate layer over time, with an increase in T1 hyperintense signal suggesting regression of inflammatory change but progressive fatty change at about one year [84 Magnetic resonance imaging after nasopharyngeal endoscopic resection and skull base reconstruction. J Clin Med 13(9):2624.
")].
Fig. 16
Expected post-operative changes in a 51-year-old man who had NPC (clinical T3N2M0) and completed chemoradiation in 2022. He had recurrent NPC (rT1N0M0) one-year after completion of treatment and underwent transnasal endoscopic nasopharyngectomy. A surveillance MRI was performed at 6-month post-surgery. Axial T1-weighted image with fat suppression shows the posterior septectomy of the nasal septum (dashed line) and the absence of right inferior turbinate in keeping with inferior turbinectomy (white asterisk). The clivus shows diffusely increased T2-weighted signal suggestive of marrow edema (black asterisk) (a). Axial T2-weighted image with fat suppression shows the nasoseptal flap covering the bony defect of the bilateral posterior wall and right lateral wall of the nasopharynx with a deeper hypointense layer and a more superficial smooth thin T2-weighted hyperintense layer (arrows). The right torus tubarius has been resected. There is also enhancing marrow edema in the clivus related to bone drilling, in keeping with post-operative change (asterisk) (b)
Full size image
All these post-operative signal changes should be distinguished from the typical tumoral signal on conventional MRI. DWI and FDG-PET/CT are problem-solving tools in diagnostic dilemmas, and the presence of restricted diffusion on DWI and hypermetabolism on FDG-PET/CT will increase confidence in diagnosing recurrent tumors (Fig.17).
Fig. 17
Same patient as Fig.16. Recurrent tumor after endoscopic nasopharyngectomy in a 51-year-old man who had NPC (clinical T3N2M0) and completed chemoradiation in 2022. He had recurrent NPC (rT1N0M0) one-year after completion of treatment and underwent transnasal endoscopic nasopharyngectomy. A surveillance MRI was performed at 18-month post-surgery. Axial T2-weighted image with fat suppression at the superior resection margin shows infiltrative intermediate signal involving bilateral posterior and lateral walls of the nasopharynx (arrowheads). There is also perifocal patchy hyperintense signal involving the bilateral pterygoid bases and the clivus (a). ADC map shows the infiltrative soft tissue along bilateral nasopharyngeal walls to be hypointense (arrowheads) with hyperintensity on the high b-value DWI (not shown), in keeping with restricted diffusion (b). Axial post-contrast T1-weighted image with fat suppression shows the moderate enhancement of the soft tissue along bilateral nasopharyngeal walls (arrowheads), surrounded by marked enhancement in the adjacent soft tissue and the bone (c). Findings are suggestive of recurrent tumor along the nasopharyngeal walls on the background of post-operative change. Axial fusion FDG-PET/CT confirms the presence of hypermetabolic tumor along the nasopharyngeal walls. Otherwise no nodal or distant metastasis (d). Biopsy confirmed recurrent undifferentiated carcinoma
Full size image
Artificial intelligence (AI) in NPC imaging
There is growing interest in using AI for the clinical management of NPC, particularly through deep learning models that leverage imaging data for diagnosis, radiotherapy planning and prognosis prediction. It aims to enhance work efficiency, diagnostic accuracy and promote personalized treatment. Recent systematic reviews show most published researches are in areas of automatic segmentation for radiotherapy planning and prognosis prediction. Currently the use of AI is primarily limited to specific task performances in the clinical workflow. For radiotherapy planning, AI aids in automating the contouring of primary tumors and organ at risk on imaging, thereby enhancing the accuracy and work efficiency of radiation oncologists [852
")]. Additionally, AI models help predict patient outcomes (e.g. survival, response to treatment) based on imaging data. However, there are still conflicting results with regards to the diagnostic capacity of AI (e.g. tumor detection) compared to humans due to varying sizes, complexities of tumors, as well as differing reader experience affecting result comparison [86 Application of artificial intelligence to the diagnosis and therapy of nasopharyngeal carcinoma. J Clin Med 12(9):3077.
"), 87 Application of artificial intelligence for nasopharyngeal carcinoma management - a systematic review. Cancer Manag Res 26(14):339–366.
")]. For image acquisition, a recent study found that AI assisted compressed sensing technique not only reduced examination time, but also improved image quality [88 AI-assisted compressed sensing and parallel imaging sequences for MRI of patients with nasopharyngeal carcinoma: comparison of their capabilities in terms of examination time and image quality. Eur Radiol 33(11):7686–7696.
")]. Despite these advancements, barriers remain for the robust application of AI in clinical setting, including a lack of large-scale labelled dataset, standardization of scanning protocols, data privacy concerns and limited generalizability [87 Application of artificial intelligence for nasopharyngeal carcinoma management - a systematic review. Cancer Manag Res 26(14):339–366.
")].
Conclusion
Imaging is essential in precision oncological management in NPC. A good understanding of imaging pearls and pitfalls helps radiologists in stepping forward to provide reliable guidance for clinicians and improvement in patients’ health outcome.
Data availability
No datasets were generated or analysed during the current study.
References
Zhang Y, Rumgay H, Li M, Cao S, Chen W (2023) Nasopharyngeal cancer incidence and mortality in 185 countries in 2020 and the projected burden in 2040: population-based global epidemiological profiling. JMIR Public Health Surveill 20(9):e49968.
ArticleGoogle Scholar
King AD (2022) MR imaging of nasopharyngeal carcinoma. Magn Reson Imaging Clin N Am 30(1):19–33.
ArticlePubMedGoogle Scholar
Ng WT, Chow JCH, Beitler JJ, Corry J, Mendenhall W, Lee AWM, Robbins KT, Nuyts S, Saba NF, Smee R, Stokes WA, Strojan P, Ferlito A (2022) Current radiotherapy considerations for nasopharyngeal carcinoma. Cancers (Basel) 14(23):5773.
ArticlePubMedGoogle Scholar
Li WF, Sun Y, Chen M, Tang LL, Liu LZ, Mao YP, Chen L, Zhou GQ, Li L, Ma J (2012) Locoregional extension patterns of nasopharyngeal carcinoma and suggestions for clinical target volume delineation. Chin J Cancer 31(12):579–587.
ArticlePubMedPubMed CentralGoogle Scholar
Liang SB, Sun Y, Liu LZ, Chen Y, Chen L, Mao YP, Tang LL, Tian L, Lin AH, Liu MZ, Li L, Ma J (2009) Extension of local disease in nasopharyngeal carcinoma detected by magnetic resonance imaging: improvement of clinical target volume delineation. Int J Radiat Oncol Biol Phys 75(3):742–750.
ArticlePubMedGoogle Scholar
Dubrulle F, Souillard R, Hermans R (2007) Extension patterns of nasopharyngeal carcinoma. Eur Radiol 17(10):2622–2630.
ArticleCASPubMedGoogle Scholar
Lee AW, Ng WT, Pan JJ, Poh SS, Ahn YC, AlHussain H, Corry J, Grau C, Grégoire V, Harrington KJ, Hu CS, Kwong DL, Langendijk JA, Le QT, Lee NY, Lin JC, Lu TX, Mendenhall WM, O’Sullivan B, Ozyar E, Peters LJ, Rosenthal DI, Soong YL, Tao Y, Yom SS, Wee JT (2018) International guideline for the delineation of the clinical target volumes (CTV) for nasopharyngeal carcinoma. Radiother Oncol 126(1):25–36.
ArticlePubMedGoogle Scholar
Ho FC, Tham IW, Earnest A, Lee KM, Lu JJ (2012) Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
ArticleGoogle Scholar
Song C, Cheng P, Cheng J, Zhang Y, Xie S (2021) Value of apparent diffusion coefficient histogram analysis in the differential diagnosis of nasopharyngeal lymphoma and nasopharyngeal carcinoma based on readout-segmented diffusion-weighted imaging. Front Oncol 12(11):632796.
ArticleGoogle Scholar
Huang Z, Liao K (2023) A review of the role of DWI in radiation therapy planning and treatment response assessment for NPC management. Curr Res Med Sci 2:84–90.
ArticleGoogle Scholar
Chan KCA, Woo JKS, King A, Zee BCY, Lam WKJ, Chan SL, Chu SWI, Mak C, Tse IOL, Leung SYM, Chan G, Hui EP, Ma BBY, Chiu RWK, Leung SF, van Hasselt AC, Chan ATC, Lo YMD (2017) Analysis of plasma epstein-barr virus DNA to screen for nasopharyngeal cancer. N Engl J Med 377(6):513–522.
ArticleCASPubMedGoogle Scholar
King AD, Woo JKS, Ai QY, Chan JSM, Lam WKJ, Tse IOL, Bhatia KS, Zee BCY, Hui EP, Ma BBY, Chiu RWK, van Hasselt AC, Chan ATC, Lo YMD, Chan KCA (2019) Complementary roles of MRI and endoscopic examination in the early detection of nasopharyngeal carcinoma. Ann Oncol 30(6):977–982.
ArticleCASPubMedGoogle Scholar
Gorolay VV, Niles NN, Huo YR, Ahmadi N, Hanneman K, Thompson E, Chan MV (2022) MRI detection of suspected nasopharyngeal carcinoma: a systematic review and meta-analysis. Neuroradiology 64(8):1471–1481.
ArticlePubMedPubMed CentralGoogle Scholar
Pastor M, Lopez Pousa A, Del Barco E, Perez Segura P, Astorga BG, Castelo B, Bonfill T, Martinez Trufero J, Grau JJ, Mesia R (2018) SEOM clinical guideline in nasopharynx cancer (2017). Clin Transl Oncol 20(1):84–88.
ArticleCASPubMedGoogle Scholar
King AD, Ai QYH, Lam WKJ, Tse IOL, So TY, Wong LM, Tsang JYM, Leung HS, Zee BCY, Hui EP, Ma BBY, Vlantis AC, van Hasselt AC, Chan ATC, Woo JKS, Chan KCA (2024) Early detection of nasopharyngeal carcinoma: performance of a short contrast-free screening magnetic resonance imaging. J Natl Cancer Inst 116(5):665–672.
ArticleCASPubMedGoogle Scholar
Bage AM, Karthikeyan AD, Bage NN (2014) Adenoid hypertrophy existence in adulthood- a truth revealed. Natl J Clin Anat 3(1):12–16.
ArticleGoogle Scholar
Hoffmann TK, Hahn J (2025) Nasopharyngeal masses in adults—A retrospective analysis of 255 patients to evaluate symptoms, clinical findings, and histological results. World J Otorhinolaryngol Head Neck Surg 11(01):45–51.
ArticlePubMedGoogle Scholar
Bhatia KS, King AD, Vlantis AC, Ahuja AT, Tse GM (2012) Nasopharyngeal mucosa and adenoids: appearance at MR imaging. Radiology 263(2):437–443.
ArticlePubMedGoogle Scholar
King AD, Wong LYS, Law BKH, Bhatia KS, Woo JKS, Ai QY, Tan TY, Goh J, Chuah KL, Mo FKF, Chan KCA, Chan ATC, Vlantis AC (2018) MR imaging criteria for the detection of nasopharyngeal carcinoma: discrimination of early-stage primary tumors from benign hyperplasia. AJNR Am J Neuroradiol 39(3):515–523.
ArticleCASPubMedPubMed CentralGoogle Scholar
Wu YP, Cai PQ, Tian L, Xu JH, Mitteer RA Jr, Fan Y, Zhang Z (2015) Hypertrophic adenoids in patients with nasopharyngeal carcinoma: appearance at magnetic resonance imaging before and after treatment. Chin J Cancer 34(3):130–136.
ArticlePubMedGoogle Scholar
Johnston M, Yu E, Kim J (2012) Perineural invasion and spread in head and neck cancer. Expert Rev Anticancer Ther 12(3):359–371.
ArticleCASPubMedGoogle Scholar
Schmitd LB, Scanlon CS, D’Silva NJ (2018) Perineural invasion in head and neck cancer. J Dent Res 97(7):742–750.
ArticleCASPubMedPubMed CentralGoogle Scholar
Brown IS (2016) Pathology of perineural spread. J Neurol Surg B Skull Base 77(2):124–130.
ArticlePubMedPubMed CentralGoogle Scholar
Liu X, Liu LZ, Mao YP, Chen L, Tang LL, Zhou GQ, Sun Y, Yue D, Lin AH, Li L, Ma J (2014) Prognostic value of magnetic resonance imaging-detected cranial nerve invasion in nasopharyngeal carcinoma. Br J Cancer 110(6):1465–1471.
ArticleCASPubMedPubMed CentralGoogle Scholar
Lee H, Lazor JW, Assadsangabi R, Shah J (2019) An imager’s guide to perineural tumor spread in head and neck cancers: radiologic footprints on 18 F-FDG PET, with CT and MRI correlates. J Nucl Med 60(3):304–311.
ArticleCASPubMedGoogle Scholar
Nie X, Zhou J, Zeng J, Sun J, Chen W, Niu J (2022) Does PET scan have any role in the diagnosis of perineural spread associated with the head and neck tumors? Adv Clin Exp Med 31(8):827–835.
ArticlePubMedGoogle Scholar
Schroeder C, Lee JH, Tetzner U, Seidel S, Kim SY (2020) Comparison of diffusion-weighted MR imaging and 18 F Fluorodeoxyglucose PET/CT in detection of residual or recurrent tumors and delineation of their local spread after (chemo) radiotherapy for head and neck squamous cell carcinoma. Eur J Radiol 130:109157.
ArticlePubMedGoogle Scholar
National Comprehensive Cancer Network (2024) NCCN clinical practice guidelines in oncology: head and neck cancers (version 4.2024) [PDF].
Lin C, Lu N, Liang JL, Guo J, Gu LW, Sun R, Guo L, Yang Q (2023) Clinical treatment considerations in the intensity-modulated radiotherapy era for parotid lymph node metastasis in patients with nasopharyngeal carcinoma. Radiother Oncol 186:109802.
ArticlePubMedGoogle Scholar
Zhang Y, Zhang ZC, Li WF, Liu X, Liu Q, Ma J (2019) Prognosis and staging of parotid lymph node metastasis in nasopharyngeal carcinoma: An analysis in 10,126 patients. Oral Oncol 95:150–156.
ArticlePubMedGoogle Scholar
Lan M, Huang Y, Chen CY, Han F, Wu SX, Tian L, Zheng L, Lu TX (2015) Prognostic value of cervical nodal necrosis in nasopharyngeal carcinoma: analysis of 1800 patients with positive cervical nodal metastasis at MR imaging. Radiology 276(2):536–544.
ArticlePubMedGoogle Scholar
Ai QH, Hung KF, So TY, Mo FKF, Tsung Anthony Chin W, Hui EP, Ma BBY, Ying M, King AD (2022) Prognostic value of cervical nodal necrosis on staging imaging of nasopharyngeal carcinoma in era of intensity-modulated radiotherapy: a systematic review and meta-analysis. Cancer Imaging 22(1):24.
ArticlePubMedPubMed CentralGoogle Scholar
Huang SH, Chernock R, O’Sullivan B, Fakhry C (2021) Assessment criteria and clinical implications of extranodal extension in head and neck cancer. Am Soc Clin Oncol Educ Book 41:265–278.
ArticlePubMedGoogle Scholar
Ai QY, King AD, Poon DMC, Mo FKF, Hui EP, Tong M, Ahuja AT, Ma BBY, Chan ATC (2019) Extranodal extension is a criterion for poor outcome in patients with metastatic nodes from cancer of the nasopharynx. Oral Oncol 88:124–130.
ArticlePubMedGoogle Scholar
Henson C, Abou-Foul AK, Yu E, Glastonbury C, Huang SH, King AD, Lydiatt WM, McDowell L, Nagelschneider AA, Nankivell PC, O’Sullivan B, Rhys R, Xiao Y, Andrew D, Asmussen JT, Bidault F, Dankbaar JW, de Graaf P, Gebrim ES, Hu C, Ding J, Kanda T, Kim J, Kuno H, Medrano-Martorell S, Oikonomopoulos N, Goh JP, Santos-Armentia E, Schafigh DG, Subramaniam RM, Wu XC, Yom SS, Mehanna H (2024) Criteria for the diagnosis of extranodal extension detected on radiological imaging in head and neck cancer: Head and Neck Cancer International Group consensus recommendations. Lancet Oncol 25(7):e297–e307.
ArticlePubMedGoogle Scholar
Ai QYH, King AD, Yuan H, Vardhanabhuti V, Mo FKF, Hung KF, Hui EP, Kwong DL, Lee VH, Ma BBY (2024) Radiologic extranodal extension for nodal staging in nasopharyngeal carcinoma. Radiother Oncol 191:110050.
ArticleCASPubMedGoogle Scholar
Pan JJ, Mai HQ, Ng WT, Hu CS, Li JG, Chen XZ, Chow JCH, Wong E, Lee V, Ma LY, Guo QJ, Liu Q, Liu LZ, Xu TT, Gong XC, Qiang MY, Au KH, Liu TC, Chiang CL, Xiao YP, Lin SJ, Chen YB, Guo SS, Wong CHL, Tang LQ, Xu ZY, Jia YZ, Peng WS, Hu LP, Lu TZ, Jiang F, Cao CN, Xu W, Ma J, Blanchard P, Williams M, Glastonbury CM, King AD, Patel SG, Seethala RR, Colevas AD, Fan DM, Chua MLK, Huang SH, O’Sullivan B, Lydiatt W, Lee AWM (2024) Ninth version of the AJCC and UICC nasopharyngeal cancer TNM staging classification. JAMA Oncol 10(12):1627–1635.
ArticlePubMedPubMed CentralGoogle Scholar
Wong KY, Wong KC (2021) Diagnostic dilemma between skull base osteomyelitis and nasopharyngeal carcinoma: a case series. Hong Kong Med J 27(4):300–302.
ArticleCASPubMedGoogle Scholar
Lo ES, Kwok HM, Pan NY (2024) Imaging spectrum and complications of otogenic infections: insights into delayed diagnosis. Br J Radiol 97(1156):726–733.
ArticlePubMedPubMed CentralGoogle Scholar
Chapman PR, Choudhary G, Singhal A (2021) Skull base osteomyelitis: a comprehensive imaging review. AJNR Am J Neuroradiol 42(3):404–413.
ArticleCASPubMedPubMed CentralGoogle Scholar
Álvarez Jáñez F, Barriga LQ, Iñigo TR, Roldán Lora F (2021) Diagnosis of skull base osteomyelitis. Radiographics 41(1):156–174.
ArticlePubMedGoogle Scholar
Kwok HM, Ng FH, Chau CM, Lam SY, Ma JKF (2022) Multimodality imaging of extra-nodal lymphoma in the head and neck. Clin Radiol 77(8):e549–e559.
ArticleCASPubMedGoogle Scholar
Tabnak P, HajiEsmailPoor Z (2023) Differentiating nasopharyngeal carcinoma from lymphoma in the head and neck region using the apparent diffusion coefficient (ADC) value: a systematic review and meta-analysis. Pol J Radiol 17(88):e472–e482.
ArticleGoogle Scholar
Chen S, Yang D, Liao X, Lu Y, Yu B, Xu M, Bin Y, Zhou P, Yang Z, Liu K, Wang R, Zhao T, Kang M (2022) Failure patterns of recurrence and metastasis after intensity-modulated radiotherapy in patients with nasopharyngeal carcinoma: results of a multicentric clinical study. Front Oncol 11(11):693199.
ArticleCASPubMedPubMed CentralGoogle Scholar
Xiao XT, Wu YS, Chen YP, Liu X, Guo R, Tang LL, Ma J, Li WF (2023) Patterns and prognosis of regional recurrence in nasopharyngeal carcinoma after intensity-modulated radiotherapy. Cancer Med 12(2):1399–1408.
ArticleCASPubMedGoogle Scholar
Teo PT, Tan NC, Khoo JB (2013) Imaging appearances for recurrent nasopharyngeal carcinoma and post-salvage nasopharyngectomy. Clin Radiol 68(11):e629–e638.
ArticleCASPubMedGoogle Scholar
Meng K, Tey J, Ho FCH, Asim H, Cheo T (2020) Utility of magnetic resonance imaging in determining treatment response and local recurrence in nasopharyngeal carcinoma treated curatively. BMC Cancer 20(1):193.
ArticleCASPubMedPubMed CentralGoogle Scholar
Lee AW, Foo W, Law SC, Poon YF, Sze WM, Tung SY, Chappell R, Lau WH, Ho JH (1999) Recurrent nasopharyngeal carcinoma: the puzzles of long latency. Int J Radiat Oncol Biol Phys 44(1):149–56.
ArticleCASPubMedGoogle Scholar
Chan AT, Grégoire V, Lefebvre JL, Licitra L, Felip E, EHNS-ESMO-ESTRO Guidelines Working Group (2010) Nasopharyngeal cancer: EHNS-ESMO-ESTRO Clinical Practice Guidelines for diagnosis, treatment and follow-up. Ann Oncol 21(Suppl 5):v187-9.
ArticlePubMedGoogle Scholar
Kwong DL, Nicholls J, Wei WI, Chua DT, Sham JS, Yuen PW et al (1999) The time course of histologic remission after treatment of patients with nasopharyngeal carcinoma. Cancer 85(7):1446–1453
ArticleCASPubMedGoogle Scholar
Li WF, Zhang Y, Liu X, Tang LL, Tian L, Guo R et al (2017) Delayed clinical complete response to intensity-modulated radiotherapy in nasopharyngeal carcinoma. Oral Oncol 75:120–126
ArticlePubMedGoogle Scholar
Zuchowski C, Kemme J, Aiken AH, Baugnon KL, Abdel Razek AAK, Wu X (2022) Posttreatment magnetic resonance imaging surveillance of head and neck cancers. Magn Reson Imaging Clin N Am 30(1):109–120.
ArticlePubMedGoogle Scholar
Lee CC, Lee JC, Huang WY, Juan CJ, Jen YM, Lin LF (2021) Image-based diagnosis of residual or recurrent nasopharyngeal carcinoma may be a phantom tumor phenomenon. Medicine (Baltimore) 100(8):e24555.
ArticleCASPubMedGoogle Scholar
Ng SH, Liu HM, Ko SF, Hao SP, Chong VF (2002) Posttreatment imaging of the nasopharynx. Eur J Radiol 44(2):82–95.
ArticlePubMedGoogle Scholar
Ahuja CK, Agarwal V, Jain C, Vyas S, Kumar J, Singh P (2023) Imaging recommendations for diagnosis, staging, and management of nasopharynx carcinoma. Indian J Med Paediatr Oncol 44:175–180.
ArticleGoogle Scholar
So TY, Vardhanabhuti V (2021) Essential imaging of the nasopharyngeal space with special focus on nasopharyngeal carcinoma. Oper Tech Otolaryngol Head Neck Surg 32(1):8–14. (ISSN 1043-1810)
ArticleGoogle Scholar
Wang C, Liu L, Lai S, Su D, Liu Y, Jin G, Zhu X, Luo N (2018) Diagnostic value of diffusion-weighted magnetic resonance imaging for local and skull base recurrence of nasopharyngeal carcinoma after radiotherapy. Medicine (Baltimore) 97(34):e11929.
ArticlePubMedGoogle Scholar
Comoretto M, Balestreri L, Borsatti E, Cimitan M, Franchin G, Lise M (2008) Detection and restaging of residual and/or recurrent nasopharyngeal carcinoma after chemotherapy and radiation therapy: comparison of MR imaging and FDG PET/CT. Radiology 249(1):203–211.
ArticlePubMedGoogle Scholar
Jung JH, Choi Y, Im KC (2016) PET/MRI: Technical challenges and recent advances. Nucl Med Mol Imaging 50(1):3–12.
ArticlePubMedPubMed CentralGoogle Scholar
Ding XI, Liu Y-P, Hua Y-J, Zou X, Wang Z-Q, Xie Y-L, Chen M-Y (2021) Clinical advances in nasopharyngeal carcinoma surgery and a video demonstration. Vis Cancer Med 2:2
ArticleGoogle Scholar
Li G, Wang J, Tang H, Han R, Zhao Y, Wang X, Zhou H (2020) Comparing endoscopic surgeries with open surgeries in terms of effectiveness and safety in salvaging residual or recurrent nasopharyngeal cancer: Systematic review and meta-analysis. Head Neck 42(11):3415–3426.
ArticlePubMedGoogle Scholar
Hua YJ, Chen MY, Qian CN, Hong MH, Zhao C, Guo L, Guo X, Cao KJ (2009) Postradiation nasopharyngeal necrosis in the patients with nasopharyngeal carcinoma. Head Neck 31(6):807–812.
ArticlePubMedGoogle Scholar
Tang J, Li XW, Wu Y, Su Z, He Y, Sun XW, Cao XL, Li YH, Wang BC, Zou GR (2023) Treating radiation-related nasopharyngeal necrosis with endostar in patient with nasopharyngeal carcinoma: A report of two cases and a literature review. Mol Clin Oncol 19(1):57.
ArticlePubMedPubMed CentralGoogle Scholar
Chen MY, Mai HQ, Sun R, Guo X, Zhao C, Hong MH, Hua YJ (2013) Clinical findings and imaging features of 67 nasopharyngeal carcinoma patients with postradiation nasopharyngeal necrosis. Chin J Cancer 32(10):533–538.
ArticlePubMedPubMed CentralGoogle Scholar
Li KY, Kwok HM, Cheuk W, Ma KFJ (2024) Demystifying the challenging diagnosis of post-radiation nasopharyngeal necrosis on multimodality imaging. J Med Imaging Radiat Oncol 68(7):805–807.
ArticlePubMedGoogle Scholar
Loke SC, Karandikar A, Ravanelli M, Farina D, Goh JP, Ling EA, Maroldi R, Tan TY (2016) Superior cervical ganglion mimicking retropharyngeal adenopathy in head and neck cancer patients: MRI features with anatomic, histologic, and surgical correlation. Neuroradiology 58(1):45–50.
ArticleCASPubMedGoogle Scholar
Subodh A, Bhatt AA (2023) Hypertrophied superior cervical ganglia after radiotherapy for head and neck cancer. AJR Am J Roentgenol 221(5):695.
ArticlePubMedGoogle Scholar
Cho SJ, Lee JH, Park JE, Choi YJ, Kim JH, Kim HJ, Baek JH (2018) Serial magnetic resonance imaging evaluations of irradiated superior cervical sympathetic ganglia: Not every retropharyngeal enlarging mass is a sign of malignancy. Eur J Radiol 98:126–129.
ArticlePubMedGoogle Scholar
Lee JY, Lee JH, Song JS, Song MJ, Hwang SJ, Yoon RG, Jang SW, Park JE, Heo YJ, Choi YJ, Baek JH (2016) Superior cervical sympathetic ganglion: normal imaging appearance on 3T-MRI. Korean J Radiol. 17(5):657–63.
ArticlePubMedPubMed CentralGoogle Scholar
Ravanelli M, Tononcelli E, Leali M, Buffa I, Loke SC, Karandikar A, Chokkapan K, Yue GOC, Goh JPN, Tan TY, Farina D (2020) Magnetic resonance imaging features of the superior cervical ganglion and expected changes after radiation therapy to the head and neck in a long-term follow-up. Neuroradiology 62(4):519–524.
ArticlePubMedGoogle Scholar
Yokota H, Mukai H, Hattori S, Yamada K, Anzai Y, Uno T (2018) MR Imaging of the superior cervical ganglion and inferior ganglion of the vagus nerve: structures that can mimic pathologic retropharyngeal lymph nodes. AJNR Am J Neuroradiol 39(1):170–176.
ArticleCASPubMedPubMed CentralGoogle Scholar
Parlak S, Yazici G, Dolgun A, Ozgen B (2021) The evolution of bone marrow signal changes at the skull base in nasopharyngeal carcinoma patients treated with radiation therapy. Radiol Med 126(6):818–826.
ArticlePubMedGoogle Scholar
King AD, Ahuja AT, Yeung DK, Wong JK, Lee YY, Lam WW, Ho SS, Yu SC, Leung SF (2007) Delayed complications of radiotherapy treatment for nasopharyngeal carcinoma: imaging findings. Clin Radiol 62(3):195–203.
ArticleCASPubMedGoogle Scholar
Li JX, Lu TX, Huang Y, Han F (2012) Clinical characteristics of recurrent nasopharyngeal carcinoma in high-incidence area. ScientificWorldJournal 2012:719754.
ArticlePubMedPubMed CentralGoogle Scholar
Lam WKJ, Chan KCA, Lo YMD (2019) Plasma epstein-barr virus DNA as an archetypal circulating tumour DNA marker. J Pathol 247(5):641–649.
ArticlePubMedPubMed CentralGoogle Scholar
Leonetti JP, Weishaar JR, Gannon D, Harmon GA, Block A, Anderson DE (2020) Osteoradionecrosis of the skull base. J Neurooncol 150(3):477–482.
ArticlePubMedGoogle Scholar
Varoquaux A, Rager O, Dulguerov P, Burkhardt K, Ailianou A, Becker M (2015) Diffusion-weighted and PET/MR imaging after radiation therapy for malignant head and neck tumors. Radiographics 35(5):1502–27.
ArticlePubMedGoogle Scholar
Alhilali L, Reynolds AR, Fakhran S (2014) Osteoradionecrosis after radiation therapy for head and neck cancer: differentiation from recurrent disease with CT and PET/CT imaging. AJNR Am J Neuroradiol 35(7):1405–1411.
ArticleCASPubMedPubMed CentralGoogle Scholar
Lam WKJ, Chan JYK (2018) Recent advances in the management of nasopharyngeal carcinoma. F1000Res 7:1829
ArticleGoogle Scholar
Liu Q, Sun X, Li H, Zhou J, Gu Y, Zhao W, Li H, Yu H, Wang D (2021) Types of transnasal endoscopic nasopharyngectomy for recurrent nasopharyngeal carcinoma: Shanghai EENT hospital experience. Front Oncol 10:555862.
ArticlePubMedPubMed CentralGoogle Scholar
Li H, Wang DL, Liu XW, Chen MY, Mo YX, Geng ZJ, Xie CM (2013) MRI signal changes in the skull base bone after endoscopic nasopharyngectomy for recurrent NPC: a serial study of 9 patients. Eur J Radiol 82(2):309–315.
ArticlePubMedGoogle Scholar
Wei WI, Lam KH, Sham JS (1991) New approach to the nasopharynx: the maxillary swing approach. Head Neck 13(3):200–7.
ArticleCASPubMedGoogle Scholar
McCarty JL, Corey AS, El-Deiry MW, Baddour HM, Cavazuti BM, Hudgins PA (2019) Imaging of Surgical Free Flaps in Head and Neck Reconstruction. AJNR Am J Neuroradiol 40(1):5–13.
ArticleCASPubMedPubMed CentralGoogle Scholar
Rondi P, Ravanelli M, Rampinelli V, Hussain IZ, Ramanzin M, Di Meo N, Borghesi A, Tomasoni M, Schreiber A, Mattavelli D, Piazza C, Farina D (2024) Magnetic resonance imaging after nasopharyngeal endoscopic resection and skull base reconstruction. J Clin Med 13(9):2624.
ArticlePubMedPubMed CentralGoogle Scholar
Lin L, Dou Q, Jin YM, Zhou GQ, Tang YQ, Chen WL, Su BA, Liu F, Tao CJ, Jiang N, Li JY, Tang LL, Xie CM, Huang SM, Ma J, Heng PA, Wee JTS, Chua MLK, Chen H, Sun Y (2019) Deep learning for automated contouring of primary tumor volumes by MRI for nasopharyngeal carcinoma. Radiology 291(3):677–686.
ArticlePubMedGoogle Scholar
Yang X, Wu J, Chen X (2023) Application of artificial intelligence to the diagnosis and therapy of nasopharyngeal carcinoma. J Clin Med 12(9):3077.
ArticlePubMedPubMed CentralGoogle Scholar
Ng WT, But B, Choi HCW, de Bree R, Lee AWM, Lee VHF, López F, Mäkitie AA, Rodrigo JP, Saba NF, Tsang RKY, Ferlito A (2022) Application of artificial intelligence for nasopharyngeal carcinoma management - a systematic review. Cancer Manag Res 26(14):339–366.
ArticleGoogle Scholar
Liu H, Deng D, Zeng W, Huang Y, Zheng C, Li X, Li H, Xie C, He H, Xu G (2023) AI-assisted compressed sensing and parallel imaging sequences for MRI of patients with nasopharyngeal carcinoma: comparison of their capabilities in terms of examination time and image quality. Eur Radiol 33(11):7686–7696.
ArticlePubMedPubMed CentralGoogle Scholar
Download references
Funding
There is no funding received for the above study.
Author information
Authors and Affiliations
Department of Diagnostic and Interventional Radiology, Princess Margaret Hospital, Hong Kong SAR, China
Kwok Yan Li,Hoi Ming Kwok,Nin Yuan Pan,Lik Fai Cheng&Ka Fai Johnny Ma
Department of Imaging and Interventional Radiology, Chinese University of Hong Kong, Hong Kong SAR, China
Hoi Ming Kwok
Authors
1. Kwok Yan LiView author publications Search author on:PubMedGoogle Scholar
2. Hoi Ming KwokView author publications Search author on:PubMedGoogle Scholar
3. Nin Yuan PanView author publications Search author on:PubMedGoogle Scholar
4. Lik Fai ChengView author publications Search author on:PubMedGoogle Scholar
5. Ka Fai Johnny MaView author publications Search author on:PubMedGoogle Scholar
Contributions
H.M.K and K.Y.L conceptualised the study. N.Y.P., L.F.C, and K.F.J.M. supervised and supported the study. All authors collected the data. All authors contributed to drafting the initial manuscript. All authors interpreted the images. All authors substantially revised the manuscript. All authors reviewed and approved the final manuscript.
Corresponding author
Correspondence to Hoi Ming Kwok.
Ethics declarations
Ethical approval
The study was approved by the Central Institutional Review Board (CIRB-2024–441-2) and complied with the Declaration of Helsinki. Informed consent was waived, given the retrospective nature of the study.
Conflict of interest
The authors declare no competing interests.
Additional information
Publisher's Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Rights and permissions
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit
Reprints and permissions
About this article
Cite this article
Li, K.Y., Kwok, H.M., Pan, N.Y. et al. Pre-treatment and post-treatment nasopharyngeal carcinoma imaging: imaging updates, pearls and pitfalls. Neuroradiology67, 1023–1047 (2025).
Download citation
Received: 14 November 2024
Accepted: 18 March 2025
Published: 11 April 2025
Issue Date: April 2025
DOI:
Share this article
Anyone you share the following link with will be able to read this content:
Get shareable link
Sorry, a shareable link is not currently available for this article.
Copy to clipboard
Provided by the Springer Nature SharedIt content-sharing initiative
Keywords
Nasopharyngeal carcinoma
Imaging
Computed tomography
Positron emission tomography
Magnetic resonance imaging
Diffusion-weighted imaging
Profiles
Hoi Ming KwokView author profile
Use our pre-submission checklist
Avoid common mistakes on your manuscript.
Sections
Figures
References
Abstract
Introduction
Pre-treatment imaging
Post-treatment imaging
Artificial intelligence (AI) in NPC imaging
Conclusion
Data availability
References
Funding
Author information
Ethics declarations
Additional information
Rights and permissions
About this article
Advertisement
Fig. 1
View in articleFull size image
Fig. 2
View in articleFull size image
Fig. 3
View in articleFull size image
Fig. 4
View in articleFull size image
Fig. 5
View in articleFull size image
Fig. 6
View in articleFull size image
Fig. 7
View in articleFull size image
Fig. 8
View in articleFull size image
Fig. 9
View in articleFull size image
Fig. 10
View in articleFull size image
Fig. 11
View in articleFull size image
Fig. 12
View in articleFull size image
Fig. 13
View in articleFull size image
Fig. 14
View in articleFull size image
Fig. 15
View in articleFull size image
Fig. 16
View in articleFull size image
Fig. 17
View in articleFull size image
Zhang Y, Rumgay H, Li M, Cao S, Chen W (2023) Nasopharyngeal cancer incidence and mortality in 185 countries in 2020 and the projected burden in 2040: population-based global epidemiological profiling. JMIR Public Health Surveill 20(9):e49968.
ArticleGoogle Scholar
King AD (2022) MR imaging of nasopharyngeal carcinoma. Magn Reson Imaging Clin N Am 30(1):19–33.
ArticlePubMedGoogle Scholar
Ng WT, Chow JCH, Beitler JJ, Corry J, Mendenhall W, Lee AWM, Robbins KT, Nuyts S, Saba NF, Smee R, Stokes WA, Strojan P, Ferlito A (2022) Current radiotherapy considerations for nasopharyngeal carcinoma. Cancers (Basel) 14(23):5773.
ArticlePubMedGoogle Scholar
Li WF, Sun Y, Chen M, Tang LL, Liu LZ, Mao YP, Chen L, Zhou GQ, Li L, Ma J (2012) Locoregional extension patterns of nasopharyngeal carcinoma and suggestions for clinical target volume delineation. Chin J Cancer 31(12):579–587.
ArticlePubMedPubMed CentralGoogle Scholar
Liang SB, Sun Y, Liu LZ, Chen Y, Chen L, Mao YP, Tang LL, Tian L, Lin AH, Liu MZ, Li L, Ma J (2009) Extension of local disease in nasopharyngeal carcinoma detected by magnetic resonance imaging: improvement of clinical target volume delineation. Int J Radiat Oncol Biol Phys 75(3):742–750.
ArticlePubMedGoogle Scholar
Dubrulle F, Souillard R, Hermans R (2007) Extension patterns of nasopharyngeal carcinoma. Eur Radiol 17(10):2622–2630.
ArticleCASPubMedGoogle Scholar
Lee AW, Ng WT, Pan JJ, Poh SS, Ahn YC, AlHussain H, Corry J, Grau C, Grégoire V, Harrington KJ, Hu CS, Kwong DL, Langendijk JA, Le QT, Lee NY, Lin JC, Lu TX, Mendenhall WM, O’Sullivan B, Ozyar E, Peters LJ, Rosenthal DI, Soong YL, Tao Y, Yom SS, Wee JT (2018) International guideline for the delineation of the clinical target volumes (CTV) for nasopharyngeal carcinoma. Radiother Oncol 126(1):25–36.
ArticlePubMedGoogle Scholar
Ho FC, Tham IW, Earnest A, Lee KM, Lu JJ (2012) Patterns of regional lymph node metastasis of nasopharyngeal carcinoma: a meta-analysis of clinical evidence. BMC Cancer 21(12):98.
ArticleGoogle Scholar
Song C, Cheng P, Cheng J, Zhang Y, Xie S (2021) Value of apparent diffusion coefficient histogram analysis in the differential diagnosis of nasopharyngeal lymphoma and nasopharyngeal carcinoma based on readout-segmented diffusion-weighted imaging. Front Oncol 12(11):632796.
ArticleGoogle Scholar
Huang Z, Liao K (2023) A review of the role of DWI in radiation therapy planning and treatment response assessment for NPC management. Curr Res Med Sci 2:84–90.
ArticleGoogle Scholar
Chan KCA, Woo JKS, King A, Zee BCY, Lam WKJ, Chan SL, Chu SWI, Mak C, Tse IOL, Leung SYM, Chan G, Hui EP, Ma BBY, Chiu RWK, Leung SF, van Hasselt AC, Chan ATC, Lo YMD (2017) Analysis of plasma epstein-barr virus DNA to screen for nasopharyngeal cancer. N Engl J Med 377(6):513–522.
ArticleCASPubMedGoogle Scholar
King AD, Woo JKS, Ai QY, Chan JSM, Lam WKJ, Tse IOL, Bhatia KS, Zee BCY, Hui EP, Ma BBY, Chiu RWK, van Hasselt AC, Chan ATC, Lo YMD, Chan KCA (2019) Complementary roles of MRI and endoscopic examination in the early detection of nasopharyngeal carcinoma. Ann Oncol 30(6):977–982.
ArticleCASPubMedGoogle Scholar
Gorolay VV, Niles NN, Huo YR, Ahmadi N, Hanneman K, Thompson E, Chan MV (2022) MRI detection of suspected nasopharyngeal carcinoma: a systematic review and meta-analysis. Neuroradiology 64(8):1471–1481.
ArticlePubMedPubMed CentralGoogle Scholar
Pastor M, Lopez Pousa A, Del Barco E, Perez Segura P, Astorga BG, Castelo B, Bonfill T, Martinez Trufero J, Grau JJ, Mesia R (2018) SEOM clinical guideline in nasopharynx cancer (2017). Clin Transl Oncol 20(1):84–88.
ArticleCASPubMedGoogle Scholar
King AD, Ai QYH, Lam WKJ, Tse IOL, So TY, Wong LM, Tsang JYM, Leung HS, Zee BCY, Hui EP, Ma BBY, Vlantis AC, van Hasselt AC, Chan ATC, Woo JKS, Chan KCA (2024) Early detection of nasopharyngeal carcinoma: performance of a short contrast-free screening magnetic resonance imaging. J Natl Cancer Inst 116(5):665–672.
ArticleCASPubMedGoogle Scholar
Bage AM, Karthikeyan AD, Bage NN (2014) Adenoid hypertrophy existence in adulthood- a truth revealed. Natl J Clin Anat 3(1):12–16.
ArticleGoogle Scholar
Hoffmann TK, Hahn J (2025) Nasopharyngeal masses in adults—A retrospective analysis of 255 patients to evaluate symptoms, clinical findings, and histological results. World J Otorhinolaryngol Head Neck Surg 11(01):45–51.
ArticlePubMedGoogle Scholar
Bhatia KS, King AD, Vlantis AC, Ahuja AT, Tse GM (2012) Nasopharyngeal mucosa and adenoids: appearance at MR imaging. Radiology 263(2):437–443.
ArticlePubMedGoogle Scholar
King AD, Wong LYS, Law BKH, Bhatia KS, Woo JKS, Ai QY, Tan TY, Goh J, Chuah KL, Mo FKF, Chan KCA, Chan ATC, Vlantis AC (2018) MR imaging criteria for the detection of nasopharyngeal carcinoma: discrimination of early-stage primary tumors from benign hyperplasia. AJNR Am J Neuroradiol 39(3):515–523.
ArticleCASPubMedPubMed CentralGoogle Scholar
Wu YP, Cai PQ, Tian L, Xu JH, Mitteer RA Jr, Fan Y, Zhang Z (2015) Hypertrophic adenoids in patients with nasopharyngeal carcinoma: appearance at magnetic resonance imaging before and after treatment. Chin J Cancer 34(3):130–136.
ArticlePubMedGoogle Scholar
Johnston M, Yu E, Kim J (2012) Perineural invasion and spread in head and neck cancer. Expert Rev Anticancer Ther 12(3):359–371.
ArticleCASPubMedGoogle Scholar
Schmitd LB, Scanlon CS, D’Silva NJ (2018) Perineural invasion in head and neck cancer. J Dent Res 97(7):742–750.
ArticleCASPubMedPubMed CentralGoogle Scholar
Brown IS (2016) Pathology of perineural spread. J Neurol Surg B Skull Base 77(2):124–130.
ArticlePubMedPubMed CentralGoogle Scholar
Liu X, Liu LZ, Mao YP, Chen L, Tang LL, Zhou GQ, Sun Y, Yue D, Lin AH, Li L, Ma J (2014) Prognostic value of magnetic resonance imaging-detected cranial nerve invasion in nasopharyngeal carcinoma. Br J Cancer 110(6):1465–1471.
ArticleCASPubMedPubMed CentralGoogle Scholar
Lee H, Lazor JW, Assadsangabi R, Shah J (2019) An imager’s guide to perineural tumor spread in head and neck cancers: radiologic footprints on 18 F-FDG PET, with CT and MRI correlates. J Nucl Med 60(3):304–311.
ArticleCASPubMedGoogle Scholar
Nie X, Zhou J, Zeng J, Sun J, Chen W, Niu J (2022) Does PET scan have any role in the diagnosis of perineural spread associated with the head and neck tumors? Adv Clin Exp Med 31(8):827–835.
ArticlePubMedGoogle Scholar
Schroeder C, Lee JH, Tetzner U, Seidel S, Kim SY (2020) Comparison of diffusion-weighted MR imaging and 18 F Fluorodeoxyglucose PET/CT in detection of residual or recurrent tumors and delineation of their local spread after (chemo) radiotherapy for head and neck squamous cell carcinoma. Eur J Radiol 130:109157.
ArticlePubMedGoogle Scholar
National Comprehensive Cancer Network (2024) NCCN clinical practice guidelines in oncology: head and neck cancers (version 4.2024) [PDF].
Lin C, Lu N, Liang JL, Guo J, Gu LW, Sun R, Guo L, Yang Q (2023) Clinical treatment considerations in the intensity-modulated radiotherapy era for parotid lymph node metastasis in patients with nasopharyngeal carcinoma. Radiother Oncol 186:109802.
ArticlePubMedGoogle Scholar
Zhang Y, Zhang ZC, Li WF, Liu X, Liu Q, Ma J (2019) Prognosis and staging of parotid lymph node metastasis in nasopharyngeal carcinoma: An analysis in 10,126 patients. Oral Oncol 95:150–156.
ArticlePubMedGoogle Scholar
Lan M, Huang Y, Chen CY, Han F, Wu SX, Tian L, Zheng L, Lu TX (2015) Prognostic value of cervical nodal necrosis in nasopharyngeal carcinoma: analysis of 1800 patients with positive cervical nodal metastasis at MR imaging. Radiology 276(2):536–544.
ArticlePubMedGoogle Scholar
Ai QH, Hung KF, So TY, Mo FKF, Tsung Anthony Chin W, Hui EP, Ma BBY, Ying M, King AD (2022) Prognostic value of cervical nodal necrosis on staging imaging of nasopharyngeal carcinoma in era of intensity-modulated radiotherapy: a systematic review and meta-analysis. Cancer Imaging 22(1):24.
ArticlePubMedPubMed CentralGoogle Scholar
Huang SH, Chernock R, O’Sullivan B, Fakhry C (2021) Assessment criteria and clinical implications of extranodal extension in head and neck cancer. Am Soc Clin Oncol Educ Book 41:265–278.
ArticlePubMedGoogle Scholar
Ai QY, King AD, Poon DMC, Mo FKF, Hui EP, Tong M, Ahuja AT, Ma BBY, Chan ATC (2019) Extranodal extension is a criterion for poor outcome in patients with metastatic nodes from cancer of the nasopharynx. Oral Oncol 88:124–130.
ArticlePubMedGoogle Scholar
Henson C, Abou-Foul AK, Yu E, Glastonbury C, Huang SH, King AD, Lydiatt WM, McDowell L, Nagelschneider AA, Nankivell PC, O’Sullivan B, Rhys R, Xiao Y, Andrew D, Asmussen JT, Bidault F, Dankbaar JW, de Graaf P, Gebrim ES, Hu C, Ding J, Kanda T, Kim J, Kuno H, Medrano-Martorell S, Oikonomopoulos N, Goh JP, Santos-Armentia E, Schafigh DG, Subramaniam RM, Wu XC, Yom SS, Mehanna H (2024) Criteria for the diagnosis of extranodal extension detected on radiological imaging in head and neck cancer: Head and Neck Cancer International Group consensus recommendations. Lancet Oncol 25(7):e297–e307.
ArticlePubMedGoogle Scholar
Ai QYH, King AD, Yuan H, Vardhanabhuti V, Mo FKF, Hung KF, Hui EP, Kwong DL, Lee VH, Ma BBY (2024) Radiologic extranodal extension for nodal staging in nasopharyngeal carcinoma. Radiother Oncol 191:110050.
ArticleCASPubMedGoogle Scholar
Pan JJ, Mai HQ, Ng WT, Hu CS, Li JG, Chen XZ, Chow JCH, Wong E, Lee V, Ma LY, Guo QJ, Liu Q, Liu LZ, Xu TT, Gong XC, Qiang MY, Au KH, Liu TC, Chiang CL, Xiao YP, Lin SJ, Chen YB, Guo SS, Wong CHL, Tang LQ, Xu ZY, Jia YZ, Peng WS, Hu LP, Lu TZ, Jiang F, Cao CN, Xu W, Ma J, Blanchard P, Williams M, Glastonbury CM, King AD, Patel SG, Seethala RR, Colevas AD, Fan DM, Chua MLK, Huang SH, O’Sullivan B, Lydiatt W, Lee AWM (2024) Ninth version of the AJCC and UICC nasopharyngeal cancer TNM staging classification. JAMA Oncol 10(12):1627–1635.
ArticlePubMedPubMed CentralGoogle Scholar
Wong KY, Wong KC (2021) Diagnostic dilemma between skull base osteomyelitis and nasopharyngeal carcinoma: a case series. Hong Kong Med J 27(4):300–302.
ArticleCASPubMedGoogle Scholar
Lo ES, Kwok HM, Pan NY (2024) Imaging spectrum and complications of otogenic infections: insights into delayed diagnosis. Br J Radiol 97(1156):726–733.
ArticlePubMedPubMed CentralGoogle Scholar
Chapman PR, Choudhary G, Singhal A (2021) Skull base osteomyelitis: a comprehensive imaging review. AJNR Am J Neuroradiol 42(3):404–413.
ArticleCASPubMedPubMed CentralGoogle Scholar
Álvarez Jáñez F, Barriga LQ, Iñigo TR, Roldán Lora F (2021) Diagnosis of skull base osteomyelitis. Radiographics 41(1):156–174.
ArticlePubMedGoogle Scholar
Kwok HM, Ng FH, Chau CM, Lam SY, Ma JKF (2022) Multimodality imaging of extra-nodal lymphoma in the head and neck. Clin Radiol 77(8):e549–e559.
ArticleCASPubMedGoogle Scholar
Tabnak P, HajiEsmailPoor Z (2023) Differentiating nasopharyngeal carcinoma from lymphoma in the head and neck region using the apparent diffusion coefficient (ADC) value: a systematic review and meta-analysis. Pol J Radiol 17(88):e472–e482.
ArticleGoogle Scholar
Chen S, Yang D, Liao X, Lu Y, Yu B, Xu M, Bin Y, Zhou P, Yang Z, Liu K, Wang R, Zhao T, Kang M (2022) Failure patterns of recurrence and metastasis after intensity-modulated radiotherapy in patients with nasopharyngeal carcinoma: results of a multicentric clinical study. Front Oncol 11(11):693199.
ArticleCASPubMedPubMed CentralGoogle Scholar
Xiao XT, Wu YS, Chen YP, Liu X, Guo R, Tang LL, Ma J, Li WF (2023) Patterns and prognosis of regional recurrence in nasopharyngeal carcinoma after intensity-modulated radiotherapy. Cancer Med 12(2):1399–1408.
ArticleCASPubMedGoogle Scholar
Teo PT, Tan NC, Khoo JB (2013) Imaging appearances for recurrent nasopharyngeal carcinoma and post-salvage nasopharyngectomy. Clin Radiol 68(11):e629–e638.
ArticleCASPubMedGoogle Scholar
Meng K, Tey J, Ho FCH, Asim H, Cheo T (2020) Utility of magnetic resonance imaging in determining treatment response and local recurrence in nasopharyngeal carcinoma treated curatively. BMC Cancer 20(1):193.
ArticleCASPubMedPubMed CentralGoogle Scholar
Lee AW, Foo W, Law SC, Poon YF, Sze WM, Tung SY, Chappell R, Lau WH, Ho JH (1999) Recurrent nasopharyngeal carcinoma: the puzzles of long latency. Int J Radiat Oncol Biol Phys 44(1):149–56.
ArticleCASPubMedGoogle Scholar
Chan AT, Grégoire V, Lefebvre JL, Licitra L, Felip E, EHNS-ESMO-ESTRO Guidelines Working Group (2010) Nasopharyngeal cancer: EHNS-ESMO-ESTRO Clinical Practice Guidelines for diagnosis, treatment and follow-up. Ann Oncol 21(Suppl 5):v187-9.
ArticlePubMedGoogle Scholar
Kwong DL, Nicholls J, Wei WI, Chua DT, Sham JS, Yuen PW et al (1999) The time course of histologic remission after treatment of patients with nasopharyngeal carcinoma. Cancer 85(7):1446–1453
ArticleCASPubMedGoogle Scholar
Li WF, Zhang Y, Liu X, Tang LL, Tian L, Guo R et al (2017) Delayed clinical complete response to intensity-modulated radiotherapy in nasopharyngeal carcinoma. Oral Oncol 75:120–126
ArticlePubMedGoogle Scholar
Zuchowski C, Kemme J, Aiken AH, Baugnon KL, Abdel Razek AAK, Wu X (2022) Posttreatment magnetic resonance imaging surveillance of head and neck cancers. Magn Reson Imaging Clin N Am 30(1):109–120.
ArticlePubMedGoogle Scholar
Lee CC, Lee JC, Huang WY, Juan CJ, Jen YM, Lin LF (2021) Image-based diagnosis of residual or recurrent nasopharyngeal carcinoma may be a phantom tumor phenomenon. Medicine (Baltimore) 100(8):e24555.
ArticleCASPubMedGoogle Scholar
Ng SH, Liu HM, Ko SF, Hao SP, Chong VF (2002) Posttreatment imaging of the nasopharynx. Eur J Radiol 44(2):82–95.
ArticlePubMedGoogle Scholar
Ahuja CK, Agarwal V, Jain C, Vyas S, Kumar J, Singh P (2023) Imaging recommendations for diagnosis, staging, and management of nasopharynx carcinoma. Indian J Med Paediatr Oncol 44:175–180.
ArticleGoogle Scholar
So TY, Vardhanabhuti V (2021) Essential imaging of the nasopharyngeal space with special focus on nasopharyngeal carcinoma. Oper Tech Otolaryngol Head Neck Surg 32(1):8–14. (ISSN 1043-1810)
ArticleGoogle Scholar
Wang C, Liu L, Lai S, Su D, Liu Y, Jin G, Zhu X, Luo N (2018) Diagnostic value of diffusion-weighted magnetic resonance imaging for local and skull base recurrence of nasopharyngeal carcinoma after radiotherapy. Medicine (Baltimore) 97(34):e11929.
ArticlePubMedGoogle Scholar
Comoretto M, Balestreri L, Borsatti E, Cimitan M, Franchin G, Lise M (2008) Detection and restaging of residual and/or recurrent nasopharyngeal carcinoma after chemotherapy and radiation therapy: comparison of MR imaging and FDG PET/CT. Radiology 249(1):203–211.
ArticlePubMedGoogle Scholar
Jung JH, Choi Y, Im KC (2016) PET/MRI: Technical challenges and recent advances. Nucl Med Mol Imaging 50(1):3–12.
ArticlePubMedPubMed CentralGoogle Scholar
Ding XI, Liu Y-P, Hua Y-J, Zou X, Wang Z-Q, Xie Y-L, Chen M-Y (2021) Clinical advances in nasopharyngeal carcinoma surgery and a video demonstration. Vis Cancer Med 2:2
ArticleGoogle Scholar
Li G, Wang J, Tang H, Han R, Zhao Y, Wang X, Zhou H (2020) Comparing endoscopic surgeries with open surgeries in terms of effectiveness and safety in salvaging residual or recurrent nasopharyngeal cancer: Systematic review and meta-analysis. Head Neck 42(11):3415–3426.
ArticlePubMedGoogle Scholar
Hua YJ, Chen MY, Qian CN, Hong MH, Zhao C, Guo L, Guo X, Cao KJ (2009) Postradiation nasopharyngeal necrosis in the patients with nasopharyngeal carcinoma. Head Neck 31(6):807–812.
ArticlePubMedGoogle Scholar
Tang J, Li XW, Wu Y, Su Z, He Y, Sun XW, Cao XL, Li YH, Wang BC, Zou GR (2023) Treating radiation-related nasopharyngeal necrosis with endostar in patient with nasopharyngeal carcinoma: A report of two cases and a literature review. Mol Clin Oncol 19(1):57.
ArticlePubMedPubMed CentralGoogle Scholar
Chen MY, Mai HQ, Sun R, Guo X, Zhao C, Hong MH, Hua YJ (2013) Clinical findings and imaging features of 67 nasopharyngeal carcinoma patients with postradiation nasopharyngeal necrosis. Chin J Cancer 32(10):533–538.
ArticlePubMedPubMed CentralGoogle Scholar
Li KY, Kwok HM, Cheuk W, Ma KFJ (2024) Demystifying the challenging diagnosis of post-radiation nasopharyngeal necrosis on multimodality imaging. J Med Imaging Radiat Oncol 68(7):805–807.
ArticlePubMedGoogle Scholar
Loke SC, Karandikar A, Ravanelli M, Farina D, Goh JP, Ling EA, Maroldi R, Tan TY (2016) Superior cervical ganglion mimicking retropharyngeal adenopathy in head and neck cancer patients: MRI features with anatomic, histologic, and surgical correlation. Neuroradiology 58(1):45–50.
ArticleCASPubMedGoogle Scholar
Subodh A, Bhatt AA (2023) Hypertrophied superior cervical ganglia after radiotherapy for head and neck cancer. AJR Am J Roentgenol 221(5):695.
ArticlePubMedGoogle Scholar
Cho SJ, Lee JH, Park JE, Choi YJ, Kim JH, Kim HJ, Baek JH (2018) Serial magnetic resonance imaging evaluations of irradiated superior cervical sympathetic ganglia: Not every retropharyngeal enlarging mass is a sign of malignancy. Eur J Radiol 98:126–129.
ArticlePubMedGoogle Scholar
Lee JY, Lee JH, Song JS, Song MJ, Hwang SJ, Yoon RG, Jang SW, Park JE, Heo YJ, Choi YJ, Baek JH (2016) Superior cervical sympathetic ganglion: normal imaging appearance on 3T-MRI. Korean J Radiol. 17(5):657–63.
ArticlePubMedPubMed CentralGoogle Scholar
Ravanelli M, Tononcelli E, Leali M, Buffa I, Loke SC, Karandikar A, Chokkapan K, Yue GOC, Goh JPN, Tan TY, Farina D (2020) Magnetic resonance imaging features of the superior cervical ganglion and expected changes after radiation therapy to the head and neck in a long-term follow-up. Neuroradiology 62(4):519–524.
ArticlePubMedGoogle Scholar
Yokota H, Mukai H, Hattori S, Yamada K, Anzai Y, Uno T (2018) MR Imaging of the superior cervical ganglion and inferior ganglion of the vagus nerve: structures that can mimic pathologic retropharyngeal lymph nodes. AJNR Am J Neuroradiol 39(1):170–176.
ArticleCASPubMedPubMed CentralGoogle Scholar
Parlak S, Yazici G, Dolgun A, Ozgen B (2021) The evolution of bone marrow signal changes at the skull base in nasopharyngeal carcinoma patients treated with radiation therapy. Radiol Med 126(6):818–826.
ArticlePubMedGoogle Scholar
King AD, Ahuja AT, Yeung DK, Wong JK, Lee YY, Lam WW, Ho SS, Yu SC, Leung SF (2007) Delayed complications of radiotherapy treatment for nasopharyngeal carcinoma: imaging findings. Clin Radiol 62(3):195–203.
ArticleCASPubMedGoogle Scholar
Li JX, Lu TX, Huang Y, Han F (2012) Clinical characteristics of recurrent nasopharyngeal carcinoma in high-incidence area. ScientificWorldJournal 2012:719754.
ArticlePubMedPubMed CentralGoogle Scholar
Lam WKJ, Chan KCA, Lo YMD (2019) Plasma epstein-barr virus DNA as an archetypal circulating tumour DNA marker. J Pathol 247(5):641–649.
ArticlePubMedPubMed CentralGoogle Scholar
Leonetti JP, Weishaar JR, Gannon D, Harmon GA, Block A, Anderson DE (2020) Osteoradionecrosis of the skull base. J Neurooncol 150(3):477–482.
ArticlePubMedGoogle Scholar
Varoquaux A, Rager O, Dulguerov P, Burkhardt K, Ailianou A, Becker M (2015) Diffusion-weighted and PET/MR imaging after radiation therapy for malignant head and neck tumors. Radiographics 35(5):1502–27.
ArticlePubMedGoogle Scholar
Alhilali L, Reynolds AR, Fakhran S (2014) Osteoradionecrosis after radiation therapy for head and neck cancer: differentiation from recurrent disease with CT and PET/CT imaging. AJNR Am J Neuroradiol 35(7):1405–1411.
ArticleCASPubMedPubMed CentralGoogle Scholar
Lam WKJ, Chan JYK (2018) Recent advances in the management of nasopharyngeal carcinoma. F1000Res 7:1829
ArticleGoogle Scholar
Liu Q, Sun X, Li H, Zhou J, Gu Y, Zhao W, Li H, Yu H, Wang D (2021) Types of transnasal endoscopic nasopharyngectomy for recurrent nasopharyngeal carcinoma: Shanghai EENT hospital experience. Front Oncol 10:555862.
ArticlePubMedPubMed CentralGoogle Scholar
Li H, Wang DL, Liu XW, Chen MY, Mo YX, Geng ZJ, Xie CM (2013) MRI signal changes in the skull base bone after endoscopic nasopharyngectomy for recurrent NPC: a serial study of 9 patients. Eur J Radiol 82(2):309–315.
ArticlePubMedGoogle Scholar
Wei WI, Lam KH, Sham JS (1991) New approach to the nasopharynx: the maxillary swing approach. Head Neck 13(3):200–7.
ArticleCASPubMedGoogle Scholar
McCarty JL, Corey AS, El-Deiry MW, Baddour HM, Cavazuti BM, Hudgins PA (2019) Imaging of Surgical Free Flaps in Head and Neck Reconstruction. AJNR Am J Neuroradiol 40(1):5–13.
ArticleCASPubMedPubMed CentralGoogle Scholar
Rondi P, Ravanelli M, Rampinelli V, Hussain IZ, Ramanzin M, Di Meo N, Borghesi A, Tomasoni M, Schreiber A, Mattavelli D, Piazza C, Farina D (2024) Magnetic resonance imaging after nasopharyngeal endoscopic resection and skull base reconstruction. J Clin Med 13(9):2624.
ArticlePubMedPubMed CentralGoogle Scholar
Lin L, Dou Q, Jin YM, Zhou GQ, Tang YQ, Chen WL, Su BA, Liu F, Tao CJ, Jiang N, Li JY, Tang LL, Xie CM, Huang SM, Ma J, Heng PA, Wee JTS, Chua MLK, Chen H, Sun Y (2019) Deep learning for automated contouring of primary tumor volumes by MRI for nasopharyngeal carcinoma. Radiology 291(3):677–686.
ArticlePubMedGoogle Scholar
Yang X, Wu J, Chen X (2023) Application of artificial intelligence to the diagnosis and therapy of nasopharyngeal carcinoma. J Clin Med 12(9):3077.
ArticlePubMedPubMed CentralGoogle Scholar
Ng WT, But B, Choi HCW, de Bree R, Lee AWM, Lee VHF, López F, Mäkitie AA, Rodrigo JP, Saba NF, Tsang RKY, Ferlito A (2022) Application of artificial intelligence for nasopharyngeal carcinoma management - a systematic review. Cancer Manag Res 26(14):339–366.
ArticleGoogle Scholar
Liu H, Deng D, Zeng W, Huang Y, Zheng C, Li X, Li H, Xie C, He H, Xu G (2023) AI-assisted compressed sensing and parallel imaging sequences for MRI of patients with nasopharyngeal carcinoma: comparison of their capabilities in terms of examination time and image quality. Eur Radiol 33(11):7686–7696.
ArticlePubMedPubMed CentralGoogle Scholar
Discover content
Journals A-Z
Books A-Z
Publish with us
Journal finder
Publish your research
Language editing
Open access publishing
Products and services
Our products
Librarians
Societies
Partners and advertisers
Our brands
Springer
Nature Portfolio
BMC
Palgrave Macmillan
Apress
Discover
Your privacy choices/Manage cookies
Your US state privacy rights
Accessibility statement
Terms and conditions
Privacy policy
Help and support
Legal notice
Cancel contracts here
34.34.225.28
Not affiliated
© 2025 Springer Nature |
190718 | https://www.youtube.com/watch?v=yBtJAVXcNkw | How to Graph Quadratic Inequalities! (explanation, steps, and example)
You Can Learn Math with Alyssa
4590 subscribers
630 likes
Description
43470 views
Posted: 15 Sep 2022
Book 1-on-1 tutoring sessions at 🙂. And visit for full courses in Algebra and SAT/PSAT prep including videos, practice questions, study guides, and more.😊
How do you graph quadratic inequalities? These are inequalities where your quadratic is greater than or less than y, not greater than or less than 0. For quadratic inequalities greater or less than 0, those are in another video (listed below).
This video shows you step by step how to graph a quadratic inequality, from finding the vertex to charting points to determining where it should be a solid or dotted line. I also show you how to determine where to shade your graph after you've drawn your parabola.
0:58 - Graphing the parabola
8:45 - How to determine whether the parabola should be solid or dashed/dotted
11:02 - How to figure out where to shade your graph
Here are some additional related and helpful videos:
How to Solve Quadratic Inequalities! (explanation and examples):
How to Graph Inequalities on a Number Line!:
Graphing Systems of Linear Inequalities!:
How to Solve Quadratic Inequalities! (explanation and examples):
Solving Inequalities with Variables on Both Sides:
How to Factor Quadratic Equations! (leading coefficient is 1):
Factoring Negative Quadratic Equations:
Factoring Quadratic Equations with Coefficients:
How to Use the Quadratic Formula:
Thank you so much for watching! If you have any questions or requests for future videos, please leave a comment.
Keep an eye out for my new Math Merch - it's gonna be fun! :-)
If this was helpful in any way, please like, share, comment, or subscribe - it's the only way I can keep making more of these :-)
If you'd like more Algebra videos, go here:
inequality #inequalities #quadraticequation #quadraticequations #quadraticinequalities #actmathhelp #algebra #csecmathhelp #distancelearning #helpfulmathtips #helpmath #helpmewithmath #helpwithmath #homeschoolmathhelp #homeworkhelpmath #ilovemath #math #mathelp #mathematics #mathematicshelp #mathhelp #mathhomeworkhelp #mathproblems #mathshelp #mathstudent #mathteacher #mathtutor #onlinemathhelp #sscmathhelp #tutor
56 comments
Transcript:
hi everyone it's alyssa and welcome to you can learn math today we're talking about graphing quadratic inequalities not solving graphing this is really key solving a quadratic inequality is when you get something like x squared minus 3x plus 2 is greater than 0. this is the key over here 0. if it is one variable just an x and it's being said greater than less than a number usually zero that is solving a quadratic inequality not graphing graphing the one we're doing today is when it is greater than less than greater than or equal to less than or equal to y when you have an x and a y in this equation so if you're looking for solving quadratic inequalities that is a different video link is in the description below this is graphing okay so to graph a quadratic inequality we need a graph right there you go and i actually planned ahead and made a graph pat myself on the back for that one so you don't have to endure me awkwardly drawing a graph in front of you aren't you happy now we have to go through the main two phases of how we're going to solve these problems they are graph the parabola which is what a quadratic makes it makes a parabola in this instance it's going to be either opening up or down and then we're going to shade either we're either going to shade everywhere outside the parabola or everywhere inside the parabola seems simple enough right well if you remember how to graph a parabola it's probably been a while now in algebra then you are going to be one step ahead of this game if you don't and that's pretty common because a lot of times you know we learn things for tests and then we don't use them for a while and they can get a little bit rusty so i'm going to go through both of this i'm going to go through the process of graphing a parabola just a nice reminder and then we're going to talk about the shading so if you don't need the graphing the parabola part and you just need to know how to make sure it's a solid or dotted line and then shade it i'll put a link down below where you can jump forward to that point in the video but for everyone else who wants to remember how to graph the parabola this is where you're going to start okay so we are going to be graphing x squared minus 6x plus 12 is greater than y that's where we're starting now to graph this parabola we're going to start by doing the same things we would do if this was equal to y same basic process now some references i've seen will tell you to start with finding the zeros of this zeros are where a parabola crosses the x-axis i'm not a huge fan of starting with this because there are parabolas that don't cross the x-axis so if you spend a lot of time trying to find the zeros and it's one of these you've wasted your time now if your teacher specifically tells you start by trying to find the zeros please follow their lead we don't want you to lose points over this but if you are given the choice as to how to solve this i would recommend do not start that way i would start with finding the axis of symmetry which will lead you to the vertex vertex being the either the maximum or the minimum point of a parabola so the axis of symmetry the formula is x equals negative b over 2a again the axis of symmetry is if you have a parabola it's that straight line that goes up and down and splits the parabola perfectly into makes it into two symmetrical hey axis of symmetry two symmetrical parts and it's always x equals a number that's what we're looking for now the b and a hoping that looks a little familiar but if it doesn't here's just a quick reminder when you have a quadratic you have an x squared you always have an x squared you sometimes have a plain x and you sometimes have a plain number you sometimes have all three as in this case if you have all three a is the number in front of the x squared b is the number in front of the plane x and c is the number that's by itself if there is no number like in this case it is an understood imaginary number one so in this case our a and our b would be this our b there is negative six and our a is that imaginary i should say invisible is a better way of saying it not imaginary invisible number one so i'm going to plug those two numbers in since our b is negative 6 i plug that in for b and my a is that invisible or understood one so negative negative six on top that becomes positive so positive six two times one is two six divided by two is three now i know my axis of symmetry is x equals three so that means that there's this line x equals three and that somewhere on that is my parabola and i know it opens up because this is positive just a reminder in case that slipped your mind okay so it's a parabola it opens up and it's somewhere on this axis of symmetry its vertex is somewhere on this axis of symmetry how do i find that vertex well i now know that at the vertex x is 3. so to find the y i need to take 3 and plug it back into this equation so that's my next step i'm going to slide this over and i'm going to write it out over here so my next step i plug 3n so i have 3 squared minus six times three plus twelve i need to solve that three squared is nine minus six times three is eighteen plus twelve nine minus eight eight nine minus eighteen is negative nine plus 12 is positive 3. so when x is 3 y is 3 so my vertex is at 3 3 right that's my first big piece of information that i need to graph this parabola the second one i need is a chart with some points x and y i already know 3 3 is a point i'm going to go 2 to the left and 2 to the right some teachers may ask you to do 3 to the left 3 to the right follow their lead i'm just doing 2 for our purposes because it will show you the basic form of this parabola if i plug 1 into this equation because that's what i'm going to do i'm going to take these numbers and plug them in one at a time into this equation and see what y's what y values i get so when i plug 1 in 1 squared minus 6 times 1 plus 12 1 minus 6 plus 12. 1 minus 6 is negative 5 plus 12 is 7. i'll put a little seven right there and my other one let's change the color just try to make this a little more easy to see i'm gonna put two two squared minus six times two plus twelve four minus twelve plus twelve well four minus twelve plus twelve is just four now these are symmetrical parabolas are symmetrical on either side of the vertex so i know that this is four and seven on these other two sides the other two points i don't need to put four and five in there if you want to you can you will get these same numbers four and seven so now i have a chart of five points that will give me the basic idea of what this parabola looks like so i have my vertex at 3 3 i have the point at 2 4 the point at 4 4 the point at 1 7 and the point at 5 7. that gives me a generally pretty good idea of what this parabola is going to look like now if this had been a normal i say normal an example of one you'd done before where it was equals y you would at this point just draw a nice little solid line and you'd be done you'd have graphed this parabola well there's one more step to go in the graphing part before we get to the shading and the graphing part we need to decide whether this is going to be a solid or dotted line parabola and the way we determine that is by looking right here at this sign the rule is if it is greater than or less than you draw with a dotted line or dashed line if it is greater than or equal to or less than or equal to you draw it with a solid line so i'm going to do that and then i'm going to give you a quick little explanation as to why so this one over here is a greater than we just said that is a dashed or dotted line so i'm going to try to do that and i'm not going to do it very well because it's hard to do this with a stylus on a computer screen i admit it i'm not great at it but you can tell that is a parabola with dashes with breaks in it it's not solid now the reason we draw this with a dashed or dotted line is because we are trying to represent visually all the coordinates all the x and y pairs that make this true so we want all the x's where if we plug an x in there x and y pairs or if i plug an x in there i get a number that is greater than the y of that pair now if i plugged in any of the points that is actually on this parabola it would be where x squared minus 6x plus 12 is equal to y that's not part of the answer if i draw this as a solid line i'm saying the points on this parabola are part of the solution and they are not that's why if this were greater than or equal to we would draw it as a solid line because then the points on the parabola would be a part of the solution okay now here's we're at the shading part now for the shading you're going to pick a test point because we want to know all the points that make this original statement true my favorite test point is 0 0. you can only use a test point this is important if it is not part of the parabola it has to be somewhere outside or inside not on it if 0 0 is outside and not on the parabola it's great for this because it makes things very very simple so i'm going to zoom out just a little bit because we got a lot of writing going on here and scoot it up here and just here at the bottom i'm going to use my test point 0 0 and plug it into that original equation so 0 squared minus 6 times 0 plus 12. as you can see plugging in 0 for x just gets rid of those x's so i am left with 12 since 0 minus 6 times 0 0 plus 12 is 12. is 12 greater than and i apologize i did not write 0 in for y sorry about that is 12 greater than 0 because i plugged in x and y of 0 0. yes this is a true statement 12 is greater than 0. when you plug your test point in and you get a true answer you shade the area where the test point is in this case the test point is outside of this parabola so you are going to shade everywhere out here that is outside this parabola it doesn't have to be perfect you're just showing it's the outside portion of the parabola if i had done this test point and the answer i got at this point were false then i would have shaded the area here inside the parabola because it's the area you know then it would mean the area outside is false because that's where that test point was that that outside test point area was false so we need to shade where it would be true it would be the other part the part inside the parabola so for these again your steps are going to be first to graph the parabola you're going to my recommended way i should say to do that is to use the axis of symmetry to find the vertex and then chart a couple of points on either side put those points on your graph then look at your sign to say what it is if it is greater than or less than or greater than or equal to less than or equal to to determine whether you need it to be dashed or a solid line draw that parabola then use a test point i usually pick 0 0 plug that test point in for x and y and see if it is a true statement shade the area where the test point is it is a false statement shade the area where the test point isn't and you're done i hope you enjoyed that today um if it was helpful useful in any way please like share subscribe you know the drill thank you so much for stopping by hope you have a great day see you later bye |
190719 | https://www.khanacademy.org/math/multivariable-calculus/greens-theorem-and-stokes-theorem/formal-definitions-of-divergence-and-curl/a/defining-curl | Formal definition of curl in two dimensions (article) | Khan Academy
Skip to main content
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked.
Explore
Browse By Standards
Explore Khanmigo
Math: Pre-K - 8th grade
Math: Illustrative Math-aligned
Math: Eureka Math-aligned
Math: Get ready courses
Math: High school & college
Math: Multiple grades
Test prep
Science
Economics
Reading & language arts
Computing
Life skills
Social studies
Partner courses
Khan for educators
Select a category to view its courses
Search
AI for Teachers FreeDonateLog inSign up
Search for courses, skills, and videos
Help us do more
We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever.
Select gift frequency
One time
Recurring
Monthly
Yearly
Select amount
$10
$20
$30
$40
Other
Give now
By donating, you agree to our terms of service and privacy policy.
Skip to lesson content
Multivariable calculus
Course: Multivariable calculus>Unit 5
Lesson 1: Formal definitions of div and curl (optional reading)
Why care about the formal definitions of divergence and curl?
Formal definition of divergence in two dimensions
Formal definition of divergence in three dimensions
Formal definition of curl in two dimensions
Formal definition of curl in three dimensions
Math>
Multivariable calculus>
Green's, Stokes', and the divergence theorems>
Formal definitions of div and curl (optional reading)
© 2025 Khan Academy
Terms of usePrivacy PolicyCookie NoticeAccessibility Statement
Formal definition of curl in two dimensions
Google Classroom
Microsoft Teams
Learn how curl is really defined,which involves mathematically capturing the intuition of fluid rotation.This is good preparation for Green's theorem.
Background
Curl in two dimensions
Line integrals in a vector field
If you haven't already, you may also want to read "Why care about the formal definitions of divergence and curl" for motivation.
What we're building to
In two dimensions, curl is formally defined as the following limit of a line integral:
2d-curl F(x,y)=lim|A(x,y)|→0(1|A(x,y)|∮C F⋅d r)
Description of terms
F is a two-dimensional vector field.
(x,y) is some specific point in the plane.
A(x,y) represents some region around the point (x,y). For instance, it could be a circle centered at (x,y).
|A(x,y)| indicates the area of A(x,y).
lim|A(x,y)|→0 indicates we are considering the limit as the area of A(x,y) goes to zero, meaning the region is shrinking to the point (x,y).
C is the boundary of A(x,y), oriented counterclockwise.
∮C is the line integral around C, written as ∮ instead of ∫ to emphasize that C is a closed curve.
The line integral ∮C F⋅d r can be thought of as measuring the total fluid rotation around C.
This is complicated, but it will make sense as we build up to it one piece at a time.
Formalizing fluid rotation
Suppose you have a flowing fluid whose velocity is given by a vector field F(x,y), such as the one we looked at in the two-dimensional curl article.
Khan Academy video wrapper
See video transcript
If you didn't already know about curl, but you did just learn about line integrals through a vector field, how would you measure fluid rotation in a region?
To take a relatively simple example, consider the vector field
F(x,y)=[−y x]
This is the quintessential counterclockwise rotation vector field.
How can we make the idea of fluid rotation mathematical (before knowing about curl)? One way to do this is to imagine walking around the perimeter of some region, like a unit circle centered at the origin, and measuring if the fluid seems to flow with you or against you at each point.
Concept check: Let C represent the circumference of a unit circle centered at the origin, oriented counterclockwise. Given the picture of the vector field F above, consider the following line integral:
∮C F⋅d r
Without calculating it, what is the sign of this integral? (Recall that the symbol ∮ just emphasizes the fact that the line integral is being done over a closed loop, but it's computed the same way as any other line integral).
Choose 1 answer:
Choose 1 answer:
(Choice A) Positive A Positive
(Choice B) Negative B Negative
(Choice C) Zero C Zero
Check
Answer
Positive.
As you walk around the circle, the arrows of the vector field, F, always point in the same direction as your motion, d r. Therefore the dot product F⋅d r will always be positive, and hence the integral as a whole must be positive.
More generally, if a fluid tends to flow counterclockwise around a region, you would expect that the line integral of that fluid's velocity vector field around the perimeter of the region would be positive (when it's oriented counterclockwise).
You could also imagine a more complicated vector field, in which the fluid flows with you at some points on your counterclockwise walk around the circle, but against you at others.
The value F⋅d r will be positive while the flow is with you, and negative when it's against you. In a way, the integral ∮C F⋅d r is like a voting system that counts up how much these different directions cancel each other out and which one wins overall.
Letting the size of the region change
So, after mathematically expressing the idea of fluid rotation around a region, you might want to capture the more elusive idea of fluid rotation at a point. How might you go about that?
You could start by considering smaller and smaller regions around that point, such as circles of smaller and smaller radii, and seeing what the fluid flow around those regions looks like.
Concept check: Back to our vector field F=[−y x], rather than just looking at the unit circle, let C R represent a circle centered at the origin with radius R. This circle will still be oriented counterclockwise.
Compute the line integral of F around this circle as a function of R.
∮C R F⋅d r=
Check
Answer
First we need to parameterize this circle. We can do this with the function
r(t)=[R cos(t)R sin(t)]
For this parameterization to cover the circle exactly once, let t range from 0 to 2 π. With this, we expand the line integral as follows:
∮C R F⋅d r=∫0 2 π F(r(t))⋅r′(t)d t
Now plug the components of r(t) into the definition of F(x,y)
F(x,y)=[−y x]⇓F(r(t))=F(R cos(t),R sin(t))=[−R sin(t)R cos(t)]
Also, take the derivative of r(t)
r′(t)=[d d t R cos(t)d d t R sin(t)]=[−R sin(t)R cos(t)]
The fact that F(r(t))=r′(t) in this case is a coincidence, peculiar to this example. It makes the completion of the line integral pretty smooth, though.
∮C R F⋅d r=∫0 2 π F(r(t))⋅r′(t)d t=∫0 2 π[−R sin(t)R cos(t)]⋅[−R sin(t)R cos(t)]d t=∫0 2 π(R 2 sin 2(t)+R 2 cos 2(t))d t=∫0 2 π R 2(sin 2(t)+cos 2(t)⏟equals 1)d t=R 2∫0 2 π d t=2 π R 2
How does this value relate to the circle C R?
Choose 1 answer:
Choose 1 answer:
(Choice A) It equals the circumference of C R A It equals the circumference of C R
(Choice B) It is twice the area enclosed by the circle C R. B It is twice the area enclosed by the circle C R.
Check
Average rotation per unit area
The answer to this last question suggests something interesting. The rotation around a region seems to be proportional to the area of that region. Of course, you've only shown this for circles centered at the origin, not all possible regions, but it is nevertheless suggestive. This might give you an idea.
Key idea: Maybe if you take ∮C F⋅d r, which measures the fluid flow around a region, and divide it by the area of that region, it can give you some notion of the average rotation per unit area.
The idea of "average rotation per unit area" might feel a bit strange, but if you think back to the interpretation of curl, that's kind of what we want curl to represent. Rather than thinking about fluid rotation in a large region, curl is supposed to measure how fluid tends to rotate near a point.
Concept check: The vector field from the previous example is a little bit special in that the "rotation-per-unit-area" of circles around the origin is the same value for all circles. What is that value?
Check
Answer
When I say "rotation-per-unit-area", what I mean by "rotation" is the value of the integral
∮C F⋅d r
This is because this integral indicates how much the fluid is flowing counterclockwise around the curve C.
The "unit area" in question is the area enclosed by C.
We found in the previous question that, for our example, the integral equals 2 π R 2. Since the area of the circle is π R 2, the ratio is always 2.
Concept check: Recall that the formula for 2d-curl is
2d-curl F=∂F 2∂x−∂F 1∂y
where F 1 and F 2 are the components of F. Given the definition
F(x,y)=[−y x]
compute the curl of F.
2d-curl F=
Check
Answer
Applying the formula, plug in the following values:
F 1=−y
F 2=x,
Here's what we get:
2d-curl F=∂∂x(x)−∂∂y(−y)=1−(−1)=2
What makes this example atypical
I should point out that the fact that 2d-curl F is constant in this example and the fact that the average rotation per unit area was constant for all circles are both peculiar to this example. In general, the curl varies from point to point, and the average rotation per unit area depends on the size of the area.
Defining two-dimensional curl
Those last two questions show that the "average rotation per unit area" in circles centered at the origin happens to be the same as the curl of the function, at least for our specific example. This turns out to apply more broadly. In fact, the way we define the curl of a vector field F at a point (x,y) is to be the limit of this average rotation per unit area in smaller and smaller regions around the point (x,y).
Specifically, (drumroll please), Here's the formula defining two-dimensional curl:
2d-curl F(x,y)=lim|A(x,y)|→0(1|A(x,y)|∮C F⋅d r)⏟Average rotation per unit area
where
F is a two-dimensional vector field.
(x,y) is some specific point in the plane.
A(x,y) represents some region around the point (x,y). For instance, a circle centered at (x,y).
|A(x,y)| indicates the area of A(x,y).
lim|A(x,y)|→0 indicates we are considering the limit as the area of A(x,y) goes to 0, meaning this region is shrinking around (x,y).
C is the boundary of A(x,y), oriented counterclockwise.
∮C is the line integral around C, written as ∮ instead of ∫ to emphasize that C is a closed curve.
This formula is impractical for computation, but the connection between this and fluid rotation is very clear once you wrap your mind around it. It is a very beautiful fact that this definition gives the same thing as the formula used to compute two-dimensional curl.
2d-curl F=∂F 2∂x−∂F 1∂y
One more feature of conservative vector fields
Background: Conservative vector fields
If F(x,y) is a conservative vector field, all line integrals over closed loops are 0. Looking at the integral above, what does this imply?
Choose 1 answer:
Choose 1 answer:
(Choice A) Two-dimensional curl is always 0. A Two-dimensional curl is always 0.
(Choice B) Fluid rotation around region is not necessarily 0, but as the area of that region approaches 0, the integral ∮C F⋅d r approaches 0 faster than the area approaches 0. B Fluid rotation around region is not necessarily 0, but as the area of that region approaches 0, the integral ∮C F⋅d r approaches 0 faster than the area approaches 0.
Check
Answer
The expression ∮C F⋅d r is always 0, so the definition of two-dimensional curl reads
2d-curl F(x,y)=lim A(x,y)→0(1|A(x,y)|(0)⏞Line integral)=lim A(x,y)→0(0)=0
Hence, the 2d-curl of a conservative vector field is always zero.
This gives an important fact: If a vector field is conservative, it is irrotational, meaning the curl is zero everywhere.
In particular, since gradient fields are always conservative, the curl of the gradient is always zero. That is a fact you could find just by chugging through the formulas. However, I think it gives much more insight to understand it using the definition of curl together with the intuition for why gradient fields are conservative.
What about the converse? If a vector field has zero curl everywhere, does that mean it must be conservative?
Answer
It does, but unfortunately, you must wait until learning about Green's theorem to see why.
Summary
If a vector field represents fluid flow, you can quantify "fluid rotation in a region" by taking the line integral of that vector field along the border of that region.
To go from the idea of fluid rotation in a region to fluid flow around a point (which is what curl measures), we introduce the idea of "average rotation per unit area" in a region. Then consider what this value approaches as your region shrinks around a point.
In formulas, this gives us the definition of two-dimensional curl as follows:
2d-curl F(x,y)=lim A(x,y)→0(1|A(x,y)|∮C F⋅d r)⏟Average rotation per unit area
This relationship between curl and closed-loop line integrals implies that irrotational fields and conservative fields are one and the same.
Skip to end of discussions
Questions Tips & Thanks
Want to join the conversation?
Log in
Sort by:
Top Voted
Hexuan Sun 2 years ago Posted 2 years ago. Direct link to Hexuan Sun's post “I don't get how did the i...” more I don't get how did the integral of f dot dr go from representing fluid flow to fluid rotation
Answer Button navigates to signup page •Comment Button navigates to signup page
(3 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
Answer
Show preview Show formatting options
Post answer
bob 8 months ago Posted 8 months ago. Direct link to bob's post “it's the sum of how well ...” more it's the sum of how well aligned the counterclockwise contour is with the vector field which is close to what 2d-curl is. It's not a flux integral.
Comment Button navigates to signup page
(2 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
Mehmet Anil ASLIHAK 5 years ago Posted 5 years ago. Direct link to Mehmet Anil ASLIHAK's post “This definition of curl o...” more This definition of curl only gives a scalar, should not it be a vector? I mean what comes out from the integral in the definition is just a number, but mustn't curl be represented by a vector?
Answer Button navigates to signup page •Comment Button navigates to signup page
(1 vote)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
Answer
Show preview Show formatting options
Post answer
𝜏 Is Better Than 𝝅 5 years ago Posted 5 years ago. Direct link to 𝜏 Is Better Than 𝝅's post “Technically, curl should ...” more Technically, curl should be a vector quantity, but the vectorial aspect of curl only starts to matter in 3 dimensions, so when you're just looking at 2d-curl, the scalar quantity that you're mentioning is really the magnitude of the curl vector. The curl vector will always be perpendicular to the instantaneous plane of rotation, but in 2 dimensions it's implicit that the plane of rotation is the x-y plane so you don't really bother with the vectorial nature of curl until you get to 3 dimensional space. Then it starts to matter.
Comment Button navigates to signup page
(4 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
MustardManExtremeTurboEdition a year ago Posted a year ago. Direct link to MustardManExtremeTurboEdition's post “Is the shape of the regio...” more Is the shape of the region always a circle? In the breakdown of the formal definition at the top of the page, it feels like it's implied that the region could be something other than a circle. I don't really understand how that would work, so am I interpreting this wrong or did I just not pick up on something?
Answer Button navigates to signup page •1 comment Comment on MustardManExtremeTurboEdition's post “Is the shape of the regio...”
(2 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
Answer
Show preview Show formatting options
Post answer
Andrea Menozzi 8 years ago Posted 8 years ago. Direct link to Andrea Menozzi's post “where i can find the conn...” more where i can find the connection between the formal definition to the "practical" definition? do wo I go from one to the other?
Answer Button navigates to signup page •Comment Button navigates to signup page
(2 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
Answer
Show preview Show formatting options
Post answer
Up next: article
Use of cookies
Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy
Accept All Cookies Strictly Necessary Only
Cookies Settings
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session.
Functional Cookies
[x] Functional Cookies
These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences.
Targeting Cookies
[x] Targeting Cookies
These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service.
Performance Cookies
[x] Performance Cookies
These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
190720 | https://trace.tennessee.edu/cgi/viewcontent.cgi?article=3857&context=utk_graddiss | Published Time: Fri, 12 Mar 2021 02:19:16 GMT
University of Tennessee, Knoxville University of Tennessee, Knoxville
TRACE: Tennessee Research and Creative TRACE: Tennessee Research and Creative
Exchange Exchange
Doctoral Dissertations Graduate School
12-2005
Parity Theorems for Combinatorial Statistics Parity Theorems for Combinatorial Statistics
Mark A. Shattuck University of Tennessee - Knoxville
Follow this and additional works at:
Part of the Mathematics Commons
Recommended Citation Recommended Citation
Shattuck, Mark A., "Parity Theorems for Combinatorial Statistics. " PhD diss., University of Tennessee,
This Dissertation is brought to you for free and open access by the Graduate School at TRACE: Tennessee
Research and Creative Exchange. It has been accepted for inclusion in Doctoral Dissertations by an authorized
administrator of TRACE: Tennessee Research and Creative Exchange. For more information, please contact
trace@utk.edu .To the Graduate Council:
I am submitting herewith a dissertation written by Mark A. Shattuck entitled "Parity Theorems
for Combinatorial Statistics." I have examined the final electronic copy of this dissertation for
form and content and recommend that it be accepted in partial fulfillment of the requirements
for the degree of Doctor of Philosophy, with a major in Mathematics.
Carl G. Wagner, Major Professor
We have read this dissertation and recommend its acceptance:
John Nolt, Jan Rosinski, Pavlos Tzermias
Accepted for the Council:
Carolyn R. Hodges
Vice Provost and Dean of the Graduate School
(Original signatures are on file with official student records.) To the Graduate Council: I am submitting herewith a dissertation written by Mark A. Shattuck entitled “Parity Theorems for Combinatorial Statistics.”I have examined the final electronic copy of this dissertation for form and content and recommend that it be accepted in partial fulfillment of the requirements for the degree of Doctor of Philosophy, with a major in Mathematics. Carl G. Wagner
Major Professor We have read this dissertation and recommend its acceptance: John Nolt
Jan Rosinski
Pavlos Tzermias
Accepted for the Council: Anne Mayhew
Vice Chancellor and Dean of Graduate Studies (Original signatures are on file with official student records.) Parity Theorems for Combinatorial Statistics
A Dissertation Presented for the Doctor of Philosophy Degree The University of Tennessee, Knoxville
Mark A. Shattuck December 2005
ii Copyright c© 2005 by Mark A. Shattuck All rights reserved. iii
Acknowledgments
First, I must express my appreciation to my dissertation director, Dr. Carl Wagner. In addition to posing a very nice problem for my dissertation, he has assisted me academically, professionally, and personally in manifold ways over the past five years. I would also like to thank the other members of my doctoral committee for their time and input into this dissertation: Dr. John Nolt, Dr. Jan Rosinski, and Dr. Pavlos Tzermias. I would like to extend my appreciation to the following teachers who greatly enhanced by graduate school experience: Dr. G. Samuel Jordan, lin-ear analysis; Dr. Gretchen Matthews, algebra; Dr. Jan Rosinski, probability; Dr. Pavlos Tzermias, number theory; and Dr. Carl Wagner, combinatorics. In addition to spending extra time with me in exploring interesting prob-lems in their respective areas and thus further broadening my mathematical awareness, they offered me kind words of encouragement at various stages of my graduate school career. Finally, I wish to express my sincerest gratitude to my parents, Judge Clarence and Ruth Shattuck, whose encouragement and support for me throughout graduate school enabled me to complete the degree of Ph. D. iv
Abstract
A q-generalization Gn(q) of a combinatorial sequence Gn which reduces to that sequence when q = 1 is obtained by q-counting a statistic defined on a sequence of finite discrete structures enumerated by Gn. In what follows, we evaluate Gn(−1) for statistics on several classes of discrete structures, giving both algebraic and combinatorial proofs. For the latter, we define appropriate sign-reversing involutions on the associated structures. We shall call the actual algebraic result of such an evaluation at q = −1 a parity theorem (for the statistic on the associated class of discrete structures). Among the structures we study are permutations, binary sequences, La-guerre configurations, derangements, Catalan words, and finite set partitions. As a consequence of our results, we obtain bijective proofs of congruences in-volving Stirling, Catalan, and Bell numbers. In addition, we modify the ideas used to construct the aforementioned sign-reversing involutions to furnish bijective proofs of combinatorial identities involving sums with alternating signs. v
Contents Introduction 11 Parity Theorems for Statistics on Multiset Permutations and Laguerre Configurations 4
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Parity Theorems for Multiset Permutations . . . . . . . . . . . 51.3 Two Statistics on Laguerre Configurations . . . . . . . . . . . 91.3.1 The Statistic inv ρ . . . . . . . . . . . . . . . . . . . . . 10 1.3.2 The Statistic ˜ w . . . . . . . . . . . . . . . . . . . . . . 11 1.4 A Refinement of a Previous Result . . . . . . . . . . . . . . . 13
2Parity Theorems for Statistics on Permutations and Catalan Words 17
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.2 Permutation Statistics . . . . . . . . . . . . . . . . . . . . . . 18 2.2.1 Some Balanced Permutation Statistics . . . . . . . . . 18 2.2.2 An Unbalanced Permutation Statistic . . . . . . . . . . 19 2.3 Some Statistics for Derangements . . . . . . . . . . . . . . . . 22 2.4 Statistics for Catalan Words . . . . . . . . . . . . . . . . . . . 26
3 Parity Theorems for Partition Statistics 32
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 3.1.1 The Partition Statistics ˜ w, ˆ w, w∗, w . . . . . . . . . . 32 3.1.2 Algebraic Preliminaries . . . . . . . . . . . . . . . . . . 34 3.2 Parity Theorems for Partition Statistics . . . . . . . . . . . . . 35 3.3 A Notable Restriction . . . . . . . . . . . . . . . . . . . . . . 43 vi
Conclusion 45 References 47 Appendices 51
A A Combinatorial Proof of a q-Binomial Coefficient Identity . . 52 B Bijective Proofs of Alternating Sum Identities . . . . . . . . . 54 C Asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 58 2 Asymptotic Normality . . . . . . . . . . . . . . . . . . 59 3 The Ratio Gn(−1)
Gn(1)
. . . . . . . . . . . . . . . . . . . . . 62
Vita 65 vii
Notation
N nonnegative integers
P positive integers [n] the set {1, 2, . . . , n }, for n ∈ N (so = ∅)
⊆ containment (perhaps improper) of sets
⊂ proper containment of sets
|S| the cardinality (number of elements) of a finite set S
:= equals by definition 00 by convention, 0 0 = 1 throughout
x greatest integer x
x least integer xδi,j the Kronecker delta, equal to 1 if i = j and 0 otherwise
xn the product x(x − 1) · · · (x − n + 1), for n ∈ N (so x0 = 1)
nq the number qn−1 + qn−2 + · · · + 1, for n ∈ N (so 0 q = 0)
n!
q
the number nq (n − 1) q · · · 1q, for n ∈ N (so 0 !
q
= 1)
(nk
)
q
a q-binomial coefficient of order n
( nn1,...,n k
)
q
a q-multinomial coefficient of order nFq the finite field of q elements ( q implicitly a power of a prime)
F nq the finite n-dimensional vector space over Fq of qn elements 2s the power set (set of all subsets) of a set S
[k][n] the set of functions from [ n] to [ k]
Sn the symmetric group on n objects
Sn,k the subset of Sn whose members contain k cycles 1
Introduction
We’ll use the following notational conventions: N := {0, 1, 2, . . . }, P :=
{1, 2, . . . }, := ∅, and [ n] := {1, . . . , n } for n ∈ P. Empty sums take the value 0 and empty products the value 1, with 0 0 := 1. The letter q denotes an indeterminate, with 0 q := 0, nq := 1 + q + · · · + qn−1 for n ∈ P, 0 !
q
:= 1,
n!
q
:= 1 q2q · · · nq for n ∈ P, and
(nk
)
q
:=
n!
q
k!
q
(n − k)!
q
, if 0 k n;0, if k < 0 or 0 n < k.
(1) Let ∆ be a finite set of discrete structures and I : ∆ → N, with generating function
G (I, ∆; q) := ∑
δ∈∆
qI(δ) = ∑
k
|{ δ ∈ ∆ : I(δ) = k}| qk. (2) Of course, G(I, ∆; 1) = |∆|. If ∆ i := {δ ∈ ∆ : I(δ) ≡ i (mod 2) }, then
G (I, ∆; −1) = |∆0|−| ∆1|. Hence if G(I, ∆; −1) = 0, the set ∆ is “balanced” with respect to the parity of I. For example, setting q = −1 in the binomial theorem, (1 + q)n = ∑
S⊆[n]
q|S| =
n
∑
k=0
(nk
)
qk, (3) yields the familiar result that a finite nonempty set has as many subsets of odd cardinality as it has subsets of even cardinality. When G(I, ∆; −1) = 0 and hence |∆0| = |∆1|, it is instructive to identify an I-parity changing involution of ∆. (In what follows, we call the parity 2of I(δ) the I-parity of δ, and an involution δ → δ′ for which δ and δ′ have opposite I-parities an I-parity changing involution .) For the statistic |S| in (3), the map
S →
{
S ∪ { 1}, if 1 ∈ S;
S − { 1}, if 1 ∈ S,
furnishes such an involution. More generally, if G(I, ∆; −1) = |∆0|−| ∆1| = c,it suffices to identify a subset ∆ ∗ of ∆ of cardinality |c| contained completely within ∆ 0 or ∆ 1 (depending upon the sign of c) and then to define an I-parity changing involution on ∆ − ∆∗. The subset ∆ ∗ thus captures both the sign and magnitude of G(I, ∆; −1). Since each member of ∆ − ∆∗ is paired with another of opposite I-parity, we have |∆| ≡ | ∆∗| (mod 2). Thus, the I-parity changing involutions de-scribed above, in addition to conveying a visceral understanding of why
G(I, ∆; −1) takes a particular value, also supply combinatorial proofs of con-gruences of the form an ≡ bn (mod 2). Shattuck has, for example, given such a combinatorial proof of the congruence
S(n, k ) ≡
(n − k/ 2 − 1
n − k
)
(mod 2) (4) for Stirling numbers of the second kind, answering a question posed by Stan-ley [23, p. 46, Exercise 17b]. In what follows, we undergo a systematic study of the special case q = −1, giving both algebraic and combinatorial treatments. In Chapter 1, we estab-lish parity theorems for statistics on multiset permutations and Laguerre con-figurations, i.e., distributions of labeled balls to unlabeled, contents-ordered urns, algebraically by evaluating q-generating functions at q = −1 and combi-natorially by identifying appropriate parity changing involutions. In Chapter 2, we perform a similar task for statistics on permutations and Catalan words, i.e., binary sequences with an equal number of 1’s and 0’s in which no initial segment contains more 1’s and 0’s. In Chapter 3, we examine the parity of four closely related statistics on partitions of finite sets. As a consequence of our results, we obtain bijective proofs of congruences involving Stirling, Catalan, and Bell numbers. For a word w = w1w2 · · · wm in some alphabet consisting of integers, the inversion and major index statistics are given by
inv (w) := |{ (i, j ) : i < j and wi > w j }| 3and
maj (w) := ∑
i∈D(w)
i, where D(w) := {1 i m − 1 : wi > w i+1 } .
Most of the statistics of the first two chapters involve counting inversions or finding the major index of words used to encode various discrete structures. For the partition statistics studied in the third chapter, one first canonically orders the blocks of a finite partition and then computes various weighted sums involving block cardinalities. 4
Chapter 1 Parity Theorems for Statistics on Multiset Permutations and Laguerre Configurations 1.1 Introduction
We analyze the parity of two well known statistics on multiset permu-tations, thereby generalizing results found in for binary words. We also examine the parity of two statistics on what Garsia and Remmel term Laguerre configurations , i.e., distributions of labeled balls to unlabeled, contents-ordered urns. The generating functions for the statistics on multiset permutations involve q-multinomial coefficients, while those for the statistics on binary words and Laguerre configurations all involve q-binomial coeffi-cients. In 1.2, we evaluate q-multinomial coefficients and their sums, when q =
−1, giving both algebraic and bijective proofs. We also give a bijective proof of a recurrence for sums of q-binomial coefficients, known as Galois numbers ,furnishing an elementary alternative to Goldman and Rota’s proof by the method of linear functionals . In 1.3, we carry out a similar evaluation of the two types of q-Lah numbers that arise as generating functions for the aforementioned Laguerre configuration statistics. In 1.4, we refine a result of the second section by looking at the restriction of one of the statistics to binary words with a fixed number of descents. 5
1.2Parity Theorems for Multiset Permutations
We will use the notation {1n1 , 2n2 , . . . , k nk } for the multiset consisting of ni copies of i for all i ∈ [k]. A permutation of a multiset is just a way of listing all of its elements. The number of permutations of the multiset
{1n1 , 2n2 , . . . , k nk } is the multinomial coefficient
( nn1, n 2, . . . , n k
)
:= n!
n1! n2! · · · nk! ,
where n = n1 + n2 + · · · + nk.Given a permutation p = p1p2 · · · pn of a multiset, define the statistics inv
and maj by
inv (p) := |{ (i, j ) : i < j and pi > p j }|
and
maj (p) := ∑
i∈D(p)
i, where D(p) := {1 i n − 1 : pi > p i+1 } .
The statistics inv and maj record the number of inversions and the major index, respectively, of a multiset permutation p. The set D(p) is referred to as the down set or the set of descents of p.If ( n1, . . . , n k) is a sequence of nonnegative integers summing to n, then define the q-multinomial coefficient by
( nn1, . . . , n k
)
q
:= n!
q
n1!
q
· · · nk!
q
,
where j!
q
:= 1 q2q · · · jq and rq := 1 + q + · · · + qr−1 for j, r ∈ N.If K = {1n1 , 2n2 , . . . , k nk } and SK is the set of multiset permutations of
K, then
∑
p∈SK
qinv (p) =
( nn1, . . . , n k
)
q
= ∑
p∈SK
qmaj (p). (1.1) 6See and . Let
G(k)
q
(n) := ∑
n1+··· +nk=nni∈N
( nn1, . . . , n k
)
q
. (1.2) Then
∑
λ∈[k][n]
qinv (λ) = G(k)
q
(n) = ∑
λ∈[k][n]
qmaj (λ), (1.3) by (1.1), where members of [ k][n] are expressed as words.
Theorem 1.1. For all n ∈ N and sequences (n1, . . . , n k) in N with n =
n1 + · · · + nk,
( nn1, . . . , n k
)
−1
=
{( n/ 2
n1/2,..., nk/2
), if at most one ni is odd, i ∈ [k]; 0, otherwise, (1.4)
and
G(k)
−1
(n) := ∑
n1+··· +nk=nni∈N
( nn1, . . . , n k
)
−1
= k n/ 2 . (1.5)
Proof. Formula (1.5) follows from (1.4) and the multinomial theorem. To prove (1.4), first assume n is even and take k = 2 for simplicity. If n1 is even, then
( nn1, n 2
)
−1
= lim
q→− 1
( nn1, n 2
)
q
= lim
q→− 1
n1−1
∏
i=0
(n − i)q
(n1 − i)q
=
n1−2
∏
i=0
ieven
lim
q→− 1
( qn−i − 1
qn1−i − 1
)
=
n1−2
∏
i=0
ieven
n − i
n1 − i =
n1−2
∏
i=0
ieven
n/ 2 − i/ 2
n1/2 − i/ 2 =
( n/ 2
n1/2, n 2/2
)
.
If n1 is odd, then q = −1 is a zero of multiplicity one more in the numerator than in the denominator and hence ( nn1,n 2
)
−1
= 0 in that case. The case for 7
n odd is handled similarly or is gotten from the even case by taking limits as q → − 1 in the recurrence
( nn1, n 2
)
q
=
( n − 1
n1 − 1, n 2
)
q
qn1
( n − 1
n1, n 2 − 1
)
q
.
The preceding readily generalizes to the case k 3. We now give bijective proofs of formulas (1.5) and (1.4). By (1.3), formula (1.5) asserts that
|Λ0(n)| − | Λ1(n)| = k n/ 2 , (1.6) where Λ i(n) := {λ ∈ Λ( n) : inv (λ) ≡ i (mod 2) } and Λ( n) := [ k][n]. Our strategy for proving (1.6) is to identify a subset Λ +0 (n) of Λ 0(n) having cardi-nality k n/ 2 , along with an inv -parity changing involution of Λ( n) − Λ+0 (n). The set Λ +0 (n) comprises those λ = λ1λ2 · · · λn ∈ Λ( n) such that
λ2j−1 = λ2j , 1 j n/ 2 . (1.7) Clearly, Λ +0 (n) ⊆ Λ0(n) and |Λ+0 (n)| = k n/ 2 . If λ ∈ Λ( n) − Λ+0 (n), let j0
be the smallest j for which (1.7) fails to hold and let λ′ be the result of switching λ2j0−1 and λ2j0 in λ. The map λ → λ′ is clearly an involution of Λ( n) − Λ+0 (n), and the parity of the number of inversions in λ′ is opposite the parity of the number of inversions in λ. This establishes (1.6), and hence (1.5). Let Λ( n; K) denote the set of rearrangements of the multiset K = {1n1 ,
2n2 , . . . , k nk }, where n1 + n2 + · · · + nk = n. By (1.1), formula (1.4) asserts that
|Λ0(n;K)|−| Λ1(n;K)|=
{( n/ 2
n1/2,..., nk/2
), if at most one ni is odd, i ∈ [k]; 0, otherwise , (1.8) where Λ i(n; K) := Λ i(n) ∩ Λ( n; K). To show (1.8), let Λ +0 (n; K) := Λ +0 (n) ∩
Λ( n; K). The cardinality of Λ +0 (n; K) is given by the right-hand side of (1.8), and the restriction of the above map λ → λ′ to Λ( n; K) − Λ+0 (n; K) is again an involution and inherits the parity changing property. This establishes (1.8), and hence (1.4). 8Note that the preceding combinatorial arguments would have worked with the maj statistic in place of the inv statistic. Thus formulas (1.4) and (1.5) are parity theorems for both the inv and maj statistics. Letting k = n and ni = 1 for all i ∈ [k] in (1.1) yields the well known fact that inv and maj are equally distributed on the symmetric group Sn.Formula (1.4) then reveals that the inv and maj statistics are balanced on
Sn if n 2. The bijection for (1.4) in this case amounts to merely switching the first two positions of a permutation of [ n]. When k = 2, the q-multinomial coefficients are q-binomial coefficients and the numbers G(2)
q
(n) are the Galois numbers Gq(n) := n∑
i=0
(ni
)
q
of Goldman and Rota . We record the k = 2 case of Theorem 1.1.
Corollary 1.2. If 0 i n, then
(ni
)
−1
=
{0, if n is even and i is odd;
(n/ 2
i/ 2
), otherwise , (1.9)
and
G−1(n) :=
n
∑
i=0
(ni
)
−1
= 2 n/ 2 . (1.10) Note that (1.9) also follows upon substituting q = −1 into the well known identity [25, pp. 201–202]
∑
n>0
(ni
)
q
xn = xi
(1 − x)(1 − qx ) · · · (1 − qix) , i ∈ N, (1.11) and considering even and odd cases for i.Formula (1.9) is a parity theorem for both inv and maj on the set of sequential arrangements of the multiset {1i, 2n−i}. Each such sequential arrangement corresponds, geometrically, to a (minimal) lattice path from (0 , 0) to ( i, n −i), with 1 representing a horizontal and 2 a vertical step. Since the number of inversions of a sequential arrangement of {1i, 2n−i} equals the area subtended by the corresponding lattice path , one may also view (1.9) and (1.10) as parity theorems for area under lattice paths . For-mula (1.10) also follows by induction from the case q = −1 of the following recurrence for Gq(n): 9
Theorem 1.3. For all n 1,
Gq(n + 1) = 2 Gq(n) + ( qn − 1) Gq(n − 1) , (1.12)
with Gq(0) = 1 and Gq(1) = 2 .
Proof. Let a(n, i ) :=
∣∣∣{
λ ∈ [n] : inv (λ) = i
}∣ ∣∣ . By (1.3), showing (1.12) is equivalent to showing that
a(n + 1 , i ) = 2 a(n, i ) + a(n − 1, i − n) − a(n − 1, i )= a(n, i ) + ( a (n, i ) − a (n − 1, i )) + a (n − 1, i − n) , (1.13) for all i ∈ N, where a(m, j ) = 0 if m ∈ N and j < 0. The term a(n + 1 , i ) counts all λ = λ1λ2 · · · λn+1 ∈ [n+1] with i inver-sions. The term a(n, i ) counts the subclass of such words for which λn+1 = 2. The term a(n, i ) − a(n − 1, i ) counts the subclass of such words for which
λ1 = λn+1 = 1. For deletion of λ1 is a bijection from this subclass to the class of words u1u2 · · · un with i inversions and un = 1, and there are clearly
a(n, i ) − a(n − 1, i ) words of the latter type. Finally, the term a(n − 1, i − n)counts the subclass of words for which λ1 = 2 and λn+1 = 1. For deletion of
λ1 and λn+1 is a bijection from this subclass to the class of words v1v2 · · · vn−1
with i − n inversions (both classes being empty if i < n ).
The above proof provides an elementary alternative to Goldman and Rota’s proof of (1.12) using the method of linear functionals . It doesn’t appear though that the numbers G(k)
q
(n) satisfy a nice recurrence as in (1.12) in general for k 3.
1.3 Two Statistics on Laguerre Configurations
Let L(n, k ) denote the set of distributions of n balls, labeled 1 , 2, . . . , n ,among k unlabeled, contents-ordered urns with no urn left empty. Garsia and Remmel call such distributions Laguerre configurations . Members of L(n, k ) will be regarded as partitions of [ n] into k blocks, where members of each block are ordered. If L(n, k ) := |L (n, k )|, then L(n, 0) = δn, 0 ∀ n ∈ N, L(n, k ) = 0 if 0
n < k , and
L(n, k ) = n!
k!
(n − 1
k − 1
)
, 1 k n. (1.14) 10 The numbers L(n, k ), called Lah numbers , were introduced by Ivo Lah as connection constants in the polynomial identities
x(x+1) · · · (x+ n− 1) =
n
∑
k=0
L(n, k )x(x− 1) · · · (x− k+ 1) , ∀ n ∈ N. (1.15) In what follows, we analyze the parity of two statistics on Laguerre con-figurations. The main results of this section (namely Theorems 1.4 and 1.6) appear in . Recall the well known summation formula [23, p. 29],
(nk
)
q
= ∑
d0+d1+··· +dk=n−kdi∈N
q0d0+1 d1+··· +kd k , 0 k n. (1.16)
1.3.1 The Statistic inv ρ
Given δ ∈ L (n, k ), represent each ordered block by a word in [ n] and then arrange these words in a sequence W1, . . . , W k, by decreasing order of their least elements. Replace the commas in this sequence by zeros and count inversions in the resulting single word to obtain the value inv ρ(δ), i.e.,
inv ρ(δ) = inv ( W10W20 · · · 0Wk−10Wk) . (1.17) As an illustration, for δ = {3, 2, 5}, {7, 6, 8}, {1, 4} ∈ L (8 , 3), we have
inv ρ(δ) = 30, the number of inversions in the word 7680325014. The statistic inv ρ is due to Garsia and Remmel , who show that the generating function
Lq (n, k ) := ∑
δ∈L (n,k )
qinv ρ(δ) = qk(k−1) n!
q
k!
q
(n − 1
k − 1
)
q
, 1 k n, (1.18) which generalizes (1.14). Garsia and Remmel also show that
xq(x + 1) q · · · (x + n − 1) q =
n
∑
k=1
Lq (n, k )xq(x − 1) q · · · (x − k + 1) q, (1.19) where n ∈ P and xq := ( qx − 1) / (q − 1). Identity (1.19) has a polynomial version which doesn’t seem to have been previously noted: 11
x(qx + 1 q) · · · (qn−1x + ( n − 1) q)=
n
∑
k=1
Lq (n, k )x
(x − 1q
q
)
· · ·
(x − (k − 1) q
qk−1
)
, (1.20) which generalizes (1.15).
Theorem 1.4. If 1 k n, then
L−1(n, k ) = δn,k . (1.21)
Proof. It is obvious from (1.18) that L−1(n, n ) = 1. If 1 k < n with n
even or k odd, the factor n!
q
/k !
q
= nq · · · (k + 1) q in (1.18) is zero since j−1 = 0 if j is even. In the remaining case, where n is odd and k is even, the factor (n−1
k−1
)
−1
= 0 by (1.9). For a bijective proof of (1.21), let Li(n, k ) := {δ ∈ L (n, k ) : inv ρ(δ) ≡ i
(mod 2) } so that L−1(n, k ) = |L 0(n, k )| − |L 1(n, k )|. Then L−1(n, n ) = 1 since L(n, n ) consists of a single distribution whose inv ρ value is n(n − 1). If 1 k < n and δ ∈ L (n, k ) has the associated sequence W1, . . . , W k, locate the leftmost word Wi containing at least two letters and interchange its first two letters. The resulting map is a parity changing involution of L(n, k ), whence |L 0(n, k )| − |L 1(n, k )| = 0.
Note that L(n, 1) = Sn, the set of permutations of [ n], and so (1.18) generalizes the well known result that
∑
π∈Sn
qinv (π) = n!
q
.
Formula (1.21) then reflects the fact that among the permutations of [ n], if
n 2, there are as many with an odd number of inversions as there are with an even number of inversions. When n 2 and k = 1 in Theorem 1.4 above, the bijection then amounts to switching the first two letters of σ ∈ Sn, just as the bijection of Theorem 1.1 did when n 2 and ni = 1 for all i.
1.3.2The Statistic ˜ w
As before, we represent each ordered block of δ ∈ L (n, k ) by a word in [n]. Having now arranged these words in a sequence W1, . . . , W k by increasing order of their initial elements, we define ˜ w(δ) by the formula 12 ˜w(δ) =
k
∑
i=1
(i − 1) ( |Wi| − 1) , (1.22) where |Wi| denotes the length of the word Wi. As an illustration, for δ =
{3, 2, 5}, {7, 6, 8}, {1, 4} ∈ L (8 , 3), we have W1, W2, W3 = 14, 325, 768 and ˜w(δ) = 6. The statistic ˜w is an analogue (see ) of a now well known partition statistic first introduced by Carlitz .
Theorem 1.5. The generating function
˜Lq(n, k ) := ∑
δ∈L (n,k )
q ˜w(δ) = n!
k!
(n − 1
k − 1
)
q
, 1 k n. (1.23)
Proof. Given n1, . . . , n k in P with ∑ ni = n, there are (nk
)(n − k)! members of L(n, k ) whose corresponding words W1, . . . , W k, arranged by increasing order of initial elements, satisfy |Wi| = ni for all i ∈ [k], as there are (nk
)
ways to choose and place the initial elements and ( n − k)! ways to place the remaining elements. By (1.22) and (1.16), it follows that
∑
δ∈L (n,k )
q ˜w(δ) =
(nk
)
(n − k)! ∑
n1+··· +nk=nni∈P
q0( n1−1)+1( n2 −1)+ ··· +( k−1)( nk −1)
= n!
k!
(n − 1
k − 1
)
q
.
Theorem 1.6. If 1 k n, then
˜L−1(n, k ) =
{0, if n is odd and k is even;
n!
k!
((n−1) /2
(k−1) /2
), otherwise. (1.24)
Proof. Formula (1.24) is an immediate consequence of (1.23) and (1.9). Alternatively, with Li(n, k ) := {δ ∈ L (n, k ) : ˜w(δ) ≡ i (mod 2) }, we have ˜L−1(n, k ) = |L 0(n, k )| − |L 1(n, k )|. To prove (1.24), it thus suffices to iden-tify a subset L+0 (n, k ) of L0(n, k ) whose cardinality agrees with the right-hand side of (1.24), along with a ˜ w-parity changing involution of L(n, k )−L +0 (n, k ). 13 The set L+0 (n, k ) consists of those distributions whose associated se-quences W1, W 2, . . . , W k satisfy
|W2i−1| is odd and |W2i| = 1 , 1 i k/ 2 . (1.25) Clearly, L+0 (n, k ) = ∅ if n is odd and k is even. In the remaining cases, the factor n!/k ! arises just as in the proof of Theorem 1.5, and
( (n − 1) /2
(k − 1) /2
)
=
∣∣∣{
(n1, . . . , n k) : ∑ ni = n, n 2i−1 is odd, and n2i = 1 , 1 i k/ 2
}∣ ∣∣ . (1.26) Suppose now that δ ∈ L (n, k ) − L +0 (n, k ) has the associated sequence
W1, . . . , W k and that i0 is the smallest index for which (1.25) fails to hold. If |W2i0−1| is even, take the last member of W2i0−1 and place it at the end of
W2i0 . If |W2i0−1| is odd, whence |W2i0−1| 2, take the last member of W2i0
and place it at the end of W2i0−1. The resulting map is a parity changing involution of L(n, k ) − L +0 (n, k ).
1.4 A Refinement of a Previous Result
Recall that for a multiset permutation w = w1 · · · wn, the major index statistic is given by
maj (w) := ∑
i∈D(w)
i, where D(w) := {1 i n − 1 : wi > w i+1 } .
Let
S(a, b ; k) := {w ∈ S(a, b ) : |D (w)| = k} ,
where S(a, b ) is the set of binary words of length a + b with a 0’s. Given w ∈ S(a, b ), we shall call a (maximal) consecutive sequence of 0’s or 1’s in w a run (of 0’s or 1’s). So a member of S(a, b ; k) will have k + 1 or
k runs of 0’s depending on whether or not it starts with a 0 and will have
k + 1 or k runs of 1’s depending on whether or not it ends with a 1. Given w ∈ S(a, b ), associate the sequences wa = ( a1, a 2, . . . ) and wb =(b1, b 2, . . . ), where ai records the number of 0’s in the ith run of 0’s and bi14 records the number of 1’s in the ith run of 1’s. For members of S(a, b ; k), it will be convenient to think of the words wa and wb as sequences of length
k + 1, where the last entry is 0 if there is no ( k + 1) st run. The members of
S(a, b ; k) are then characterized by sequences wa and wb such that (i) a1 + · · · + ak+1 = a, ai ∈ P, 1 i k, ak+1 ∈ N;(ii) b1 + · · · + bk+1 = b, bi ∈ P, 1 i k, bk+1 ∈ N. (1.27) From (1.27), it follows that
|S (a, b ; k)| =
(ak
)( bk
)
. (1.28) The “problem of the runs”occurring in [8, p. 42] and [17, p. 47] is a closely related problem. F¨ urlinger and Hofbauer show that
Mq(a, b ; k) := ∑
w∈S(a,b ;k)
qmaj (w) = qk2
(ak
)
q
(bk
)
q
, (1.29) which is a q-generalization of (1.28).
Theorem 1.7. For all a, b, k ∈ N,
M−1(a, b ; k) =
{0, if a or b is even and k is odd ;(−1) k(a/ 2
k/ 2
)( b/ 2
k/ 2
), otherwise . (1.30)
Proof. Formula (1.30) is an immediate consequence of (1.29) and (1.9). Al-ternatively, let S±(a, b ; k) consist of those members of S(a, b ; k) with even or odd major index value, respectively. In each case, we shall identify a subset
S∗(a, b ; k) of S(a, b ; k) whose net weight matches the right-hand side of (1.30) as well as a maj -parity changing involution of S(a, b ; k) − S∗(a, b ; k). Given w ∈ S(a, b ; k), consider the two sequences wa and wb of length
k + 1 as described above (see (1.27)). First assume k is even. Let S∗(a, b ; k)consist of those words w whose associated sequences wa and wb satisfy (a) a2i−1 = 1 , a2i odd , 1 i k/ 2; (b) b2i−1 = 1 , b2i odd , 1 i k/ 2. (1.31) 15 Checking separately the four cases with regard to the parity of a and b shows that |S∗(a, b ; k)| = (a/ 2
k/ 2
)( b/ 2
k/ 2
) in each case with S∗(a, b ; k) ⊆ S+(a, b ; k). Suppose now that w ∈ S(a, b ; k) − S∗(a, b ; k) and that i0 is the smallest index i for which (a) or (b) fails to hold in (1.31). If both (a) and (b) fail when i = i0, consider only (a). In any event, we’ll refer to the appropriate pair of disqualifying runs of the same type simply as run 2 i0 − 1 and run 2 i0.If run 2 i0 is of even length, move a single character forward to run 2 i0 − 1. If run 2 i0 is of odd length, whence run 2 i0 − 1 has length of at least two, move a single character from run 2 i0 − 1 to run 2 i0.Next assume k is odd. Let S∗(a, b ; k) consist of those words w whose associated sequences wa and wb satisfy (a) a2i−1 = 1 , a2i odd , 1 i (k − 1) /2; (b) b2i−1 = 1 , b2i odd , 1 i (k − 1) /2; (c) bk odd , bk+1 = 0 with either ak+1 = 0 or
ak = 1 , ak+1 odd .
(1.32) If b is even, then S∗(a, b ; k) = ∅ as (b) and (c) in (1.32) cannot hold simulta-neously. If b is odd and a is even, then S∗(a, b ; k) contains ( a/ 2−1(k−1) /2
)( (b−1) /2(k−1) /2
)
positive and negative members which we pair as follows: (i) first switch the ith run of 1’s with the ith run of 0’s for 1 i k − 1, (ii) if ak = 1 , a k+1 odd, merge the two runs and place after the kth run of 1’s; if ak+1 = 0, whence ak is even, take one of the 0’s of the kth run and place it directly in front of the kth run of 1’s. If a and b are both odd, then S∗(a, b ; k) contains ((a−1) /2(k−1) /2
)( (b−1) /2(k−1) /2
) negative numbers. Suppose now w ∈ S(a, b ; k) − S∗(a, b ; k). First assume wa fails to satisfy (1.32)(a) or wb fails to satisfy (1.32)(b). Pair w with another member of
S(a, b ; k) − S∗(a, b ; k) of opposite parity exactly as described above when k
was even. So assume wa satisfies (1.32)(a) and wb satisfies (1.32)(b). Then (1.32)(c) fails to hold for wa or wb. We’ll consider two subcases: (I) wa fails (1.32)(c); (II) wa satisfies (1.32)(c), while wb fails (1.32)(c). Under (I), either
ak+1 2 even or ak+1 odd, ak 2, while under (II), either bk even or bk odd,
bk+1 1. For (I), move a single 0 forward from the ( k + 1) st run to the kth
run of 0’s if ak+1 2 even or move a single 0 from the kth run to the ( k + 1) st
run if ak+1 odd with ak 2. For (II), move a single 1 forward from the 16 (k + 1) st run to the kth run of 1’s if bk is odd or take a single 1 from the kth
run and place it at the end of the entire sequence if bk is even. In all cases, the resulting map is a parity changing involution of S(a, b ; k) − S∗(a, b ; k).
Remark. Note that Theorem 1.7 is a refinement of Corollary 1.2. Indeed, summing (1.29) over k 0 and using the q-Vandermonde identity yields the
q-binomial coefficient (a+ba
)
q
.17
Chapter 2 Parity Theorems for Statistics on Permutations and Catalan Words 2.1 Introduction
Recall that the inversion and major index statistics for a word w =
w1w2 · · · wm in some alphabet consisting of integers are given by
inv (w) := |{ (i, j ) : i < j and wi > w j }| ,
and
maj (w) := ∑
i∈D(w)
i, where D(w) := {1 i m − 1 : wi > w i+1 }.
We establish parity theorems for statistics on the symmetric group Sn, the derangements Dn, and the Catalan words Cn, giving both algebraic and bijective proofs. Most of the statistics involve counting inversions or finding the major index of various words. In 2.2, we establish parity theorems for several permutation statistics defined on all of Sn, algebraically by evaluating q-generating functions at
q = −1 and combinatorially by identifying appropriate parity changing in-volutions. In 2.3, we analyze the parity of some statistics on Dn, the set of derangements of [ n] (i.e., permutations of [ n] having no fixed points). 18 In Chapter 1, we derived parity theorems for the inversion and major index statistics on binary words of length n with k 1’s. In 2.4, we obtain comparable results for Cn, the set of binary words of length 2 n with n 1’s and with no initial segment containing more 1’s than 0’s (termed Catalan words ). Most of the material in this chapter appears in in revised form.
2.2 Permutation Statistics 2.2.1 Some Balanced Permutation Statistics
Let Sn be the set of permutations of [ n]. A function f : Sn → N is called a permutation statistic. Two important permutation statistics are
inv and maj , which record the number of inversions and the major index, respectively, of a permutation σ = σ1σ2 · · · σn, expressed as a word. The statistics inv and maj have the same q-generating function over Sn:
∑
σ∈Sn
qinv (σ) = n!
q
= ∑
σ∈Sn
qmaj (σ), (2.1) [23, Corollary 1.3.10] and [1, Corollary 3.8]. Substituting q = −1 into (2.1) reveals that n!
−1
= 0 if n 2, and hence
inv and maj are both balanced if n 2. Interchanging σ1 and σ2 in σ =
σ1σ2 · · · σn ∈ Sn changes both the inv and maj values by one and thus furnishes an appropriate involution. Note that switching the elements 1 and 2 in σ changes the inv -parity, but not necessarily the maj -parity. Now express σ ∈ Sn in the standard cycle form
σ = ( α1)( α2) · · · ,
where α1, α 2, . . . are the cycles of σ, ordered by increasing smallest elements with each cycle ( αi) written with its smallest element in the first position. Let
Sn,k denote the set of permutations of [ n] with k cycles and c(n, k ) := |Sn,k |,the signless Stirling number of the first kind. The c(n, k ) are connection constants in the polynomial identities
q(q + 1) · · · (q + n − 1) =
n
∑
k=0
c(n, k )qk. (2.2) 19 Setting q = −1 in (2.2) reveals that there are as many permutations of [ n]with an even number of cycles as there are with an odd number of cycles if
n 2. Alternatively, breaking apart or merging α1 and α2 as shown below, leaving the other cycles undisturbed, changes the parity of the number of cycles:
α1 = (1 · · · 2 · · · ), . . . ↔ α1 = (1 · · · ), α 2 = (2 · · · ), . . . .
This involution also shows that the statistic recording the number of cycles of σ with even cardinality is balanced if n 2. Given σ = ( α1)( α2) · · · , expressed in standard cycle form, let
w(σ) := ∑
i
(i − 1) |αi|.
Edelman, Simion, and White show that
∑
σ∈Sn
x|σ|qw(σ) =
n−1
∏
i=0
(xq i + i), (2.3) where |σ| denotes the number of cycles. Setting x = 1 in (2.3) yields
∑
σ∈Sn
qw(σ) =
n−1
∏
i=0
(qi + i), (2.4) another q-generalization of n!. Setting q = −1 in (2.4) shows that the w statistic is balanced if n 2. Alternatively, if the last cycle has cardinality greater than one, break off the last member and form a 1-cycle with it; if the last cycle contains a single member, place it at the end of the penultimate cycle.
2.2.2 An Unbalanced Permutation Statistic
Carlitz defines the statistic inv c on Sn as follows: express σ ∈ Sn in standard cycle form; then remove parentheses and count inversions in the resulting word to obtain inv c(σ). As an illustration, for the permutation
σ ∈ S7 given by 3241756, we have inv c(σ) = 3, the number of inversions in the word 1342576. 20 Let
cq(n, k ) := ∑
σ∈Sn,k
qinv c(σ), (2.5) where Sn,k is the set of permutations of [ n] with k cycles. Then cq(n, 0) =
δn, 0, c q(0 , k ) = δ0,k , and
cq(n, k ) = cq(n − 1, k − 1) + ( n − 1) qcq(n − 1, k ), ∀n, k ∈ P, (2.6) since n may go in a cycle by itself or come directly after any member of [ n−1] within a cycle. Using (2.6), it is easy to show that
x(x + 1 q) · · · (x + ( n − 1) q) =
n
∑
k=0
cq(n, k )xk. (2.7) Setting x = 1 in (2.7) gives
cq(n) :=
n
∑
k=0
cq(n, k ) = ∑
σ∈Sn
qinv c(σ) =
n−1
∏
j=0
(1 + jq). (2.8)
Theorem 2.1. For all n ∈ N,
c−1(n) := ∑
σ∈Sn
(−1) inv c(σ) = 2 n/ 2 . (2.9)
Proof. Put q = −1 in (2.8) and note that
jq |q=−1 =
{
0, if j is even; 1, if j is odd. Alternatively, with S+
n
, S −
n
denoting the members of Sn with even or odd inv c
values, respectively, we have c−1(n) = |S+
n
| − | S−
n
|. To prove (2.9), it thus suffices to identify a subset S∗
n
of S+
n
such that |S∗
n
| = 2 n/ 2 , along with an
inv c-parity changing involution of Sn − S∗
n
.First assume n is even. In this case, the set S∗
n
consists of those permu-tations expressible in standard cycle form as a product of 1-cycles and the transpositions (2 i − 1, 2i), 1 i n/ 2. Note that S∗
n
⊆ S+
n
with zero inv c
value for each of its 2 n/ 2 members. 21 Before giving the involution on Sn − S∗
n
, we make a definition: given
σ = ( α1)( α2) · · · ∈ Sm in standard cycle form and j, 1 j m, let σ[j]
be the permutation of [ j] (in standard cycle form) obtained by writing the members of [ j] in the order as they appear within the cycles of σ (e.g., if
σ = (163)(25)(4)(7) ∈ S7 and j = 4, then σ = (13)(2)(4) and σ = σ). Suppose now σ ∈ Sn − S∗
n
is expressed in standard cycle form and that
i0 is the smallest integer i, 1 i n/ 2, for which σ[2 i] ∈ S2i − S∗
2i
. Then it must be the case for σ that (i) neither 2 i0 − 1 nor 2 i0 starts a cycle, or (ii) exactly one of 2 i0 − 1, 2i0 starts a cycle with 2 i0 − 1 and 2 i0 not belonging to the same cycle. Switching 2 i0−1 and 2 i0 within σ, written in standard cycle form, changes the
inv c value by one, and the resulting map is thus a parity changing involution of Sn − S∗
n
.If n is odd, let S∗
n
⊆ S+
n
consist of those permutations expressible as a product of 1-cycles and the transpositions (2 i, 2i + 1), 1 i n−1
2
. Switch 2i0 and 2 i0 + 1 within σ ∈ Sn − S∗
n
, where i0 is the smallest i, 1 i n−1
2
,for which σ[2 i+1] ∈ S2i+1 − S∗
2i+1
.
The preceding parity theorem has the refinement
Theorem 2.2. For all n ∈ N,
c−1(n, k ) := ∑
σ∈Sn,k
(−1) inv c(σ) =
( n/ 2
n − k
)
, 0 k n. (2.10)
Proof. Set q = −1 in (2.7) to get
n
∑
k=0
c−1(n, k )xk = x n/ 2 (x + 1) n/ 2 =
n
∑
k=n/ 2
( n/ 2
n − k
)
xk.
Or let S±
n,k
:= Sn,k ∩ S±
n
and S∗
n,k
:= Sn,k ∩ S∗
n
. Then S∗
n,k
⊆ S+
n,k
and its cardinality agrees with the right-hand side of (2.10). The restriction of the map used for Theorem 2.1 to Sn,k − S∗
n,k
is again an involution and inherits the parity changing property. 22
Remark. The bijection of Theorem 2.2 also proves combinatorially that
c(n, k ) ≡
( n/ 2
n − k
)
(mod 2) , 0 k n, (2.11) since off of a set of cardinality (n/ 2
n−k
), each permutation σ ∈ Sn,k is paired with another of opposite inv c-parity. The congruences in (2.11) can also be obtained by taking mod 2 the polynomial identities in (2.2) (cf. [23, p. 46, Exercise 17c]).
2.3 Some Statistics for Derangements
A permutation σ of [ n] having no fixed points (i.e., i ∈ [n] such that
σ(i) = i) is called a derangement. Let Dn denote the set of derangements of [ n] and dn := |Dn|. A typical inclusion-exclusion argument gives the well known formula
dn = n!
n
∑
k=0
(−1) k
k! , ∀n ∈ N. (2.12) Given σ ∈ Dn, express it in the form
σ = ( α1)( α2) · · · ,
where α1, α 2, . . . are the cycles of σ arranged as follows: (i) the cycles α1, α 2, . . . are ordered by increasing second smallest ele-ments; (ii) each cycle ( αi) is written with the second smallest element in the last position. Garsia and Remmel term this the ordered cycle factorization (OCF for brief) of σ.Define the statistic inv o on Dn as follows: write out the cycles of σ ∈ Dn
in OCF form; then remove parentheses and count inversions in the resulting word to obtain inv o(σ). As an illustration, for the derangement σ ∈ D7
given by 4321756, we have inv o(σ) = 3, the number of inversions in the word 2314576. 23 The statistic inv o is due to Garsia and Remmel , who show that the generating function
Dq(n) := ∑
σ∈Dn
qinv o(σ) = n!
qn
∑
k=0
(−1) k
k!
q
, ∀n ∈ N, (2.13) which generalizes (2.12).
Theorem 2.3. For all n ∈ N,
D−1(n) =
{
1, if n is even;
0, if n is odd. (2.14)
Proof. Formula (2.14) is an immediate consequence of (2.13), for
n
∑
k=0
(−1) kn!
q
k!
q
∣∣∣∣∣q=−1
=
n
∑
k=0
(−1) kn∏
i=k+1
iq
∣∣∣∣∣q=−1
= ( −1) n−1n−1 + ( −1) n,
as
j−1 =
{
0, if j is even; 1, if j is odd. Alternatively, let σ = ( α1)( α2) · · · ∈ Dn be expressed in OCF form, first assuming n is odd. Locate the leftmost cycle of σ containing at least three members and interchange the first two members of this cycle. Now assume
n is even. If σ has a cycle of length greater than two, proceed as in the odd case. If all cycles of σ are transpositions and σ = (1 , 2)(3 , 4) · · · (n − 1, n ), let i0 be the smallest integer i for which the transposition (2 i − 1, 2i) fails to occur in σ. Switch 2 i0 − 1 and 2 i0 in σ, noting that 2 i0 − 1 and 2 i0 must both start cycles. Thus whenever n is even, every σ ∈ Dn is paired with another of opposite inv o-parity except for (1 , 2)(3 , 4) · · · (n − 1, n ), which has
inv o value zero.
Now consider the generating function dq(n) resulting when one restricts
inv to Dn, i.e.,
dq(n) := ∑
σ∈Dn
qinv (σ). (2.15) We have been unable to find a simple formula for dq(n) that generalizes (2.12) or a recurrence satisfied by dq(n) that generalizes one for dn. However, we do have the following parity result. 24
Theorem 2.4. For all n ∈ N,
d−1(n) = ( −1) n−1(n − 1) . (2.16)
Proof. Equivalently, we show that the numbers d−1(n) satisfy
d−1(n) = −d−1(n − 1) + ( −1) n−1, ∀n ∈ P, (2.17) with d−1(0) = 1. Let n 2, σ = σ1σ2 · · · σn ∈ Dn, and D∗
n
⊆ Dn consist of those derangements σ for which σ1 = 2 and σ2 3. Define an inv -parity changing involution f on Dn − D∗
n
− { n12 · · · n − 1} as follows: (i) if σ2 3, whence σ1 3, switch 1 and 2 in σ to obtain f (σ); (ii) if σ2 = 1, let k0 be the smallest integer k, 1 k ⌊n−1
2
⌋, such that
σ2kσ2k+1 = (2 k − 1)(2 k); switch 2 k0 and 2 k0 + 1 if σ2k0 = 2 k0 − 1 or switch 2 k0 − 1 and 2 k0 if σ2k0 2k0 + 1 to obtain f (σ). Thus,
d−1(n) := ∑
σ∈Dn
(−1) inv (σ) = ∑
σ∈D∗
n∪{ n12 ··· n−1}
(−1) inv (σ). (2.18) One can regard members σ of D∗
n
as 2 followed by a derangement of [ n − 1] since within the terminal segment σ′ := σ2σ3 · · · σn, we must have σ2 = 1 and σk = k for all k 3. Thus,
∑
σ′:σ∈D∗
n
(−1) inv (σ′ ) = d−1(n − 1) ,
from which ∑
σ∈D∗
n
(−1) inv (σ) = −d−1(n − 1) , (2.19) since the initial 2 adds an inversion. The recurrence (2.17) follows immedi-ately from (2.18) and (2.19) upon adding the contribution of ( −1) n−1 from the singleton {n12 · · · n − 1}.
Now consider the generating function rq(n) resulting when one restricts
maj to Dn, i.e.,
rq(n) := ∑
σ∈Dn
qmaj (σ). (2.20) We were unable to find a simple formula for rq(n) which generalizes (2.12). Yet when q = −1 we have 25
Theorem 2.5. For all n ∈ N,
r−1(n) =
{
(−1) n/ 2, if n is even;
0, if n is odd. (2.21)
Proof. First verify (2.21) for 0 n 3. Let n 4 and D∗
n
⊆ Dn consist of those derangements starting with 2143 when expressed as a word. We define a maj -parity changing involution of Dn − D∗
n
below. Note that for derangements of the form σ = 2143 σ5 · · · σn, the subword σ5 · · · σn is itself a derangement on n − 4 elements. Thus for n 4,
r−1(n) := ∑
σ∈Dn
(−1) maj (σ) = ∑
σ∈D∗
n
(−1) maj (σ) = r−1(n − 4) ,
which proves (2.21). We now define a maj -parity changing involution f of Dn −D∗
n
when n 4. Let σ = σ1σ2 · · · σn ∈ Dn − D∗
n
be expressed as a word. If possible, pair σ
with σ′ = f (σ) according to (I) and (II) below: (I) first, if both σ1 = 2 and σ2 = 1, then switch σ1 and σ2 within σ to obtain σ′;(II) if (I) cannot be implemented (i.e., σ1 = 2 or σ2 = 1) but σ3 = 4 and
σ4 = 3, then switch σ3 and σ4 within σ to obtain σ′.We now define f for the cases that remain. To do so, consider Sσ :=
σ1σ2σ3σ4 ∩ , where σ = σ1σ2 · · · σn ∈ Dn − D∗
n
is of a form not covered by rules (I) and (II) above. We consider cases depending upon |Sσ|. If |Sσ| = 2 or if |Sσ| = 4, first multiply σ by the transposition (34) and then exchange the letters in the third and fourth positions to obtain σ′. This corresponds to the pairings i) σ = a1b3 . . . 4 . . . ↔ σ′ = a14 b . . . 3 . . . ;ii) σ = 2 ab 3 . . . 4 . . . ↔ σ′ = 2 a4b . . . 3 . . . ;iii) σ = 2341 . . . ↔ σ′ = 2413 . . . ;iv) σ = 4123 . . . ↔ σ′ = 3142 . . . ,where a, b 5. 26 If |Sσ| = 3, then pair according to one of six cases shown below where
a 5, leaving the other letters undisturbed: i) σ = 314 a . . . ↔ σ′ = 41 a3 . . . ;ii) σ = 234 a . . . ↔ σ′ = 24 a3 . . . ;iii) σ = a123 . . . ↔ σ′ = 2 a13 . . . ;iv) σ = a142 . . . ↔ σ′ = 2 a41 . . . ;v) σ = 21 a3 . . . 4 . . . ↔ σ′ = 214 a . . . 3 . . . ;vi) σ = a143 . . . 2 . . . ↔ σ′ = 2 a43 . . . 1 . . . .It is easy to verify that σ and σ′ have opposite maj -parity in all cases.
2.4 Statistics for Catalan Words
The Catalan numbers cn are defined by the closed form
cn = 1
n + 1
(2nn
)
, n ∈ N, (2.22) as well as by the recurrence
cn+1 =
n
∑
j=0
cj cn−j , c0 = 1 . (2.23) If one defines the generating function
f (x) = ∑
n>0
cnxn, (2.24) then (2.23) is equivalent to
f (x) = 1 + xf (x)2. (2.25) Due to (2.23), the Catalan numbers enumerate many combinatorial struc-tures, among them the set Cn consisting of words w = w1w2 · · · w2n of n 1’s and n 0’s for which no initial segment contains more 1’s than 0’s (termed 27
Catalan words ). In this section, we’ll look at two q-analogues of the Catalan numbers, one of Carlitz which generalizes (2.25) and another of MacMahon which generalizes (2.22), when q = −1. These q-analogues arise as generating functions for statistics on Cn.If ˜Cq(n) := ∑
w∈Cn
qinv (w), (2.26) then ˜Cq(n + 1) =
n
∑
k=0
q(k+1)( n−k) ˜Cq(k) ˜Cq(n − k), ˜Cq(0) = 1 , (2.27) upon decomposing a Catalan word w ∈ Cn+1 into w = 0 w11w2 with w1 ∈ Ck,
w2 ∈ Cn−k for some k, 0 k n, and noting that the number of inversions of w is given by
inv (w) = inv (w1) + inv (w2) + ( k + 1)( n − k).
Taking reciprocal polynomials of both sides of (2.27) and writing
Cq(n) = q(n
2
) ˜ Cq−1 (n) (2.28) yields the recurrence
Cq(n + 1) =
n
∑
k=0
qkCq(k)Cq(n − k), Cq(0) = 1 . (2.29) If one defines the generating function
f (x) = ∑
n>0
Cq(n)xn, (2.30) then (2.29) is equivalent to the functional equation [3, 10]
f (x) = 1 + xf (x)f (qx ), (2.31) which generalizes (2.25). 28
Theorem 2.6. For all n ∈ N,
C−1(n) =
{δn, 0, if n is even ;(−1) n−1
2
cn−1
2
, if n is odd. (2.32)
Proof. Setting q = −1 in (2.31) gives
f (x) = 1 + xf (x)f (−x). (2.33) Putting −x for x in (2.33), solving the resulting system in f (x) and f (−x), and noting f (0) = 1 yields
f (x) = ∑
n>0
C−1(n)xn
= (2 x − 1) + √4x2 + 1
2x = 1 + ∑
n>1
(−1) n−1 1
n
(2n − 2
n − 1
)
x2n−1,
which implies (2.32). Alternatively, note that
C−1(n) = ( −1) (n
2
) ∑
w∈Cn
(−1) inv (w),
by (2.26) and (2.28). So (2.32) is equivalent to
∑
w∈Cn
(−1) inv (w) =
{δn, 0, if n is even;
cn−1
2
, if n is odd. (2.34) To prove (2.34), let C+
n
, C−
n
⊆ Cn consist of the Catalan words with even or odd inv values, respectively, and C∗
n
⊆ Cn consist of those words w =
w1w2 · · · w2n for which
w2iw2i+1 = 00 or 11 , 1 i n − 1. (2.35) Clearly, C∗
n
⊆ C+
n
with cardinality matching the right-hand side of (2.34). Suppose w ∈ Cn − C∗
n
and that i0 is the smallest index for which (2.35) fails to hold. Switch w2i0 and w2i0+1 in w. The resulting map is a parity changing involution of Cn − C∗
n
, which proves (2.34) and hence (2.32). 29 Another q-Catalan number arises as the generating function for the major index statistic on Cn . If ˜cq(n) := ∑
w∈Cn
qmaj (w), (2.36) then there is the closed form (see , [15, p. 215]) ˜cq(n) = 1
(n + 1) q
(2nn
)
q
, ∀n ∈ N, (2.37) which generalizes (2.22).
Theorem 2.7. For all n ∈ N,
˜c−1(n) =
( n
n/ 2
)
. (2.38)
Proof. If n is even, then by (2.37), ˜c−1(n) = lim
q→− 1
˜cq(n) = lim
q→− 1
1
(n + 1) qn−1∏
i=0
(2 n − i)q
(n − i)q
=
n−2
∏
i=0
ieven
lim
q→− 1
(q2n−i − 1
qn−i − 1
)
=
n−2
∏
i=0
ieven
2n − i
n − i =
n−2
∏
i=0
ieven
n − i/ 2
n/ 2 − i/ 2 =
( nn/ 2
)
,
with the odd case handled similarly. Alternatively, let C+
n
, C−
n
⊆ Cn consist of the Catalan words with even or odd major index value, respectively, and C∗
n
⊆ Cn consist of those words
w = w1w2 · · · w2n which satisfy the following two requirements: (i) one can express w as w = x1x2 · · · xn, where xi = 00, 11, or 01, 1
i n;(ii) for each i, xi = 01 only if the number of 00’s in the initial segment
x1x2 · · · xi−1 equals the number of 11’s. (A word in C∗
n
may start with either 01 or 00.) Clearly, C∗
n
⊆ C+
n
and below it is shown that |C∗
n
| = ( n
n/ 2
). Suppose w =
w1w2 · · · w2n ∈ Cn − C∗
n
and that i0 is the smallest integer i, 1 i n, such that one of the following two conditions holds: 30 (i) w2i−1w2i = 10, or (ii) w2i−1w2i = 01 and the number of 0’s in the initial segment w1w2 · · · w2i−2
is strictly greater than the number of 1’s. Switching w2i0−1 and w2i0 in w changes the major index by an odd amount and the resulting map is a parity changing involution of Cn − C∗
n
.We now show |C∗
n
| = ( n
n/ 2
) by defining a bijection between C∗
n
and the set Λ( n) of (minimal) lattice paths from (0 , 0) to ( n/ 2 , n − n/ 2 ). Given
w = x1x2 · · · xn ∈ C∗
n
as described in (i) and (ii) above, we construct a lattice path λw ∈ Λ( n) as follows. Let j1 < j 2 < . . . be the set of indices j, possibly empty and denoted S(w), for which xj = 01, with j0 := 0. For s 1, let step js in λw be a V (vertical step) if s is odd and an H (horizontal step) if
s is even. Suppose now i ∈ [n] − S(w) and that t, t 0, is the greatest integer such that jt < i . If t is even, put a V (resp., H) for the ith step of λw if xi = 11 (resp., 00). If t is odd, put a V (resp., H) for the ith step of λw if xi = 00 (resp., 11), which now specifies λw completely. The map w → λw is seen to be a bijection between C∗
n
and Λ( n); note that S(w) corresponds to the steps of λw in which it either rises above the line y = x or returns to y = x from above.
Note that the preceding supplies a combinatorial proof of the congruence
1
n+1
(2nn
) ≡ ( n
n/ 2
) (mod 2) for n ∈ N since off of a set of cardinality ( n
n/ 2
),each Catalan word w ∈ Cn is paired with another of opposite maj -parity. We also have for n ∈ P that cn is even if n is even and cn ≡ cn−1
2
(mod 2) if n is odd, by (2.34). Repeated use of this yields the well known fact that the nth Catalan number cn is odd if and only if n = 2 m − 1 for some
m ∈ N. By the first congruence noted, the same is true for the middle binomial coefficient ( n
n/ 2
).Let Pn ⊆ Sn consist of those permutations σ = σ1σ2 · · · σn avoiding the pattern 312, i.e., there are no indices i < j < k such that σj < σ k < σ i
(termed Catalan permutations ). Knuth [13, p. 238] describes a bijection g
between Pn and Cn in which
inv (σ) =
(n
2
)
− inv (g(σ)) , ∀σ ∈ Pn,31 and hence
Cq(n) := ∑
w∈Cn
q(n
2
)−inv (w) = ∑
σ∈Pn
qinv (σ). (2.39) By (2.32) and (2.39), we then have the parity result
∑
σ∈Pn
(−1) inv (σ) =
{δn, 0, if n is even; (−1) n−1
2
cn−1
2
, if n is odd. (2.40) The composite map g−1 ◦ h ◦ g, where h is the involution establishing (2.34), furnishes an appropriate involution for (2.40). 32
Chapter 3 Parity Theorems for Partition Statistics 3.1 Introduction 3.1.1 The Partition Statistics ˜w, ˆw, w∗, w
Let Π( n, k ) denote the set of all partitions of [ n] with k blocks and Π( n)the set of all partitions of [ n]. For all n, k ∈ N, let S(n, k ) := |Π( n, k )|
and B(n) := |Π( n)| = ∑
k
S(n, k ). The numbers S(n, k ) are called Stirling numbers of the second kind and the numbers B(n) are called Bell numbers .Then S(0 , 0) = 1, S(n, 0) = S(0 , k ) = 0 ∀n, k ∈ P, and
S(n, k ) = S(n − 1, k − 1) + kS (n − 1, k ), ∀n, k ∈ P, (3.1) upon considering whether or not n goes in a block by itself. Then B(0) = 1 and
B(n + 1) =
n
∑
k=0
(nk
)
B(k), ∀n ∈ N, (3.2) since (nk
)B(k) counts the partitions of [ n + 1] for which the size of the block containing n + 1 is n − k + 1. Associate to each π ∈ Π( n, k ) the unique ordered partition ( E1, . . . , E k)of [ n] comprising the same blocks as Π, arranged in increasing order of their smallest elements, and define statistics ˜ w, ˆ w, w∗, and w by 33 ˜w(π) :=
k
∑
i=1
(i − 1)( |Ei| − 1) , (3.3) ˆw(π) :=
k
∑
i=1
i(|Ei| − 1) = ˜ w(π) + n − k, (3.4)
w∗(π) :=
k
∑
i=1
i|Ei| = ˜ w(π) + n +
(k
2
)
, (3.5) and
w(π) :=
k
∑
i=1
(i − 1) |Ei| = ˜ w(π) +
(k
2
)
. (3.6) Consider the generating functions (see , , , and ) ˜Sq(n, k ) := ∑
π∈Π( n,k )
q ˜w(π), (3.7) ˆSq(n, k ) := ∑
π∈Π( n,k )
q ˆw(π) = qn−k ˜Sq(n, k ), (3.8)
S∗
q
(n, k ) := ∑
π∈Π( n,k )
qw∗(π) = q(k
2
)+n ˜Sq(n, k ), (3.9) and
Sq(n, k ) := ∑
π∈Π( n,k )
qw(π) = q(k
2
) ˜ Sq(n, k ). (3.10) Summing the q-Stirling numbers, ˜Sq(n, k ), ˆSq(n, k ), S∗
q
(n, k ), and Sq(n, k ), over k yields the respective q-Bell numbers, ˜Bq (n), ˆBq(n), B∗
q
(n), and Bq(n). These polynomials reduce to the classical Stirling and Bell numbers when
q = 1. In section 3.2, we evaluate these polynomials when q = −1, giving both algebraic and bijective proofs. Our algebraic arguments parallel the general scheme presented in , though several of our proofs are different. Our bi-jective proofs are those given in . In 3.3, we carry out a similar evaluation for the polynomials resulting when one restricts the w and w∗ statistics to the partitions of [ n] whose blocks have cardinality at most two. 34
3.1.2Algebraic Preliminaries
In this section, we establish some algebraic properties of the q-Stirling numbers which we’ll need in the next section. We first recall a theorem, due to Comtet , which greatly facilitates the analysis of many combinatorial arrays.
Theorem 3.1. Let D be an integral domain. If (un)n>0 is a sequence in D
and x is an indeterminate over D, then the following are equivalent charac-terizations of an array (U(n, k )) n,k >0:
U(n, k ) = U(n − 1, k − 1) + ukU(n − 1, k ), ∀n, k ∈ P, (3.11)
with U(n, 0) = un
0
and U(0 , k ) = δ0,k ∀n, k ∈ N,
U(n, k ) = ∑
d0+d1+··· +dk=n−kdi∈N
ud0
0
ud1
1
· · · udk
k
, ∀n, k ∈ N, (3.12)
∑
n>0
U(n, k )xn = xk
(1 − u0x)(1 − u1x) · · · (1 − ukx), ∀k ∈ N, (3.13)
and
xn =
n
∑
k=0
U(n, k )pk(x), ∀n ∈ N, (3.14)
where p0(x) := 1 and pk(x) := ( x − u0) · · · (x − uk−1) for k ∈ P.
Proof. Straightforward algebraic exercise.
We shall call the numbers U(n, k ) the Comtet numbers associated with the sequence (un)n>0, as in . By (3.1), the S(n, k ) are the Comtet numbers associated with the sequence (0 , 1, 2, . . . ).
Theorem 3.2. The q-Stirling numbers, ˜Sq(n, k ), are generated by the recur-rence relation
˜Sq(n, k ) = ˜Sq(n − 1, k − 1) + kq ˜Sq(n − 1, k ), ∀n, k ∈ P, (3.15)
with ˜Sq(0 , 0) = 1 and ˜Sq(n, 0) = ˜Sq(0 , k ) = 0 , ∀n, k ∈ P.35
Proof. The boundary conditions are obvious. To establish the recurrence (3.15), consider the contribution of the element n to the overall ˜w-weight of a member π = ( E1, . . . , E k) ∈ Π( n, k ). The sum of the weights of all members of Π( n, k ) for which n goes in a block by itself is ˜Sq (n − 1, k − 1), while the sum of the weights of all members of Π( n, k ) for which n goes in the ith block Ei, 1 i k, together with at least one member of [ n − 1], is
qi−1 ˜Sq(n−1, k ). Summing over i, 1 i k, and noting kq = 1+ q +· · · +qk−1
yields (3.15).
Recurrence (3.15) reveals that the numbers ˜Sq(n, k ) are the Comtet num-bers associated with the sequence ( nq)n>0. By Theorem 3.1, it follows imme-diately that ˜Sq(n, k ) = ∑
d1+··· +dk=n−kdi∈N
(1 q)d1 (2 q)d2 · · · (kq)dk , ∀n, k ∈ N, (3.16)
∑
n>0
˜Sq(n, k )xn = xk
(1 − 1qx)(1 − 2qx) · · · (1 − kqx) , ∀k ∈ N, (3.17) and
xn =
n
∑
k=0
˜Sq(n, k )φk(x), ∀n ∈ N, (3.18) where φ0(x) := 1 and φk(x) := x(x − 1q) · · · (x − (k − 1) q), ∀k ∈ P.Variants of (3.15)–(3.18) hold for the other q-Stirling numbers and follow from relations (3.7)–(3.10). For example, we have
S∗
q
(n, k ) = qkS∗
q
(n − 1, k − 1) + qk qS∗
q
(n − 1, k ), ∀n, k ∈ P, (3.19) and
∑
n>0
S∗
q
(n, k )xn = q(k+1 2 )xk
(1 − qx )(1 − qx − q2x) · · · (1 − qx − · · · − qkx), ∀k ∈ N.
(3.20)
3.2Parity Theorems for Partition Statistics
In this section, we derive simple expressions for the foregoing q-Stirling and q-Bell numbers when q = −1, giving both algebraic and bijective proofs. 36
Theorem 3.3. For all n ∈ N,
˜S−1(n, k ) =
(n − k/ 2 − 1
n − k
)
, 0 k n. (3.21)
Proof. Substituting q = −1 into (3.17) and noting
iq|q=−1 =
{
1, if i is odd; 0, if i is even, yields
∑
n>0
˜S−1(n, k )xn = xk
(1 − x) k/ 2
= ∑
n>k
(n − k/ 2 − 1
n − k
)
xn.
Note that (3.21) can also be obtained by letting q = −1 in (3.16).
Substituting (3.21) into (3.18) at q = −1 and applying the binomial theorem yields the orthogonality relation
min {n, 2m}
∑
k=m
(−1) k−m
(n − k/ 2 − 1
n − k
)( k/ 2
k − m
)
= δn,m , 0 m n. (3.22) Let F0 = F1 = 1 with Fn = Fn−1 + Fn−2 if n 2. As is well known,
Fn =
n/ 2
∑
i=0
(n − ii
)
, ∀n ∈ N. (3.23)
Theorem 3.4. For all n ∈ N,
˜B−1(n) :=
n
∑
k=0
˜S−1(n, k ) = Fn. (3.24) 37
Proof. Clearly, (3.24) holds for n = 0 , 1. If n 2, then by (3.21) and (3.23), ˜B−1(n) =
n
∑
k=0
(n − k/ 2 − 1
n − k
)
=
(n−1) /2
∑
i=0
(n − i − 1
i
)
+
n/ 2
∑
i=1
(n − i − 1
i − 1
)
=
(n−1) /2
∑
i=0
(n − 1 − ii
)
+
(n−2) /2
∑
i=0
(n − 2 − ii
)
= Fn−1 + Fn−2 = Fn,
proving the theorem.
Throughout, we’ll represent π ∈ Π( n) by ( E1, E 2, . . . ), the unique ordered partition of [ n] comprising the same blocks as π, arranged in increasing order of their smallest elements.
Combinatorial Proof of Theorem 3.4.
Let Π i(n) := {π ∈ Π( n) : ˜w(π) ≡ i (mod 2) } so that ˜B−1(n) = |Π0(n)| − |Π1(n)|. To prove (3.24), we’ll identify a subset ˜Π( n) of Π 0(n) such that
| ˜Π( n)| = Fn, along with a ˜ w-parity changing involution of Π( n) − ˜Π( n). The set ˜Π( n) consists of those partitions π = ( E1, E 2, . . . ) whose blocks satisfy the two conditions: each block of odd index comprises a set of consecutive integers; (3.25a) each block of even index is a singleton. (3.25b) Now | ˜Π( n)| = Fn, as | ˜Π( n)| is seen to satisfy the Fibonacci recurrence, upon considering whether or not {n} is a block. For if {n} is not a block and
n − 2 belongs to an odd-numbered (respectively, even-numbered) block of
π ∈ ˜Π( n), then {n − 1, n } constitutes a proper subset of (respectively, all of) the last block of π.Suppose now that π = ( E1, E 2, . . . ) belongs to Π( n) − ˜Π( n) and that i0
is the smallest of the integers i for which E2i−1 fails to satisfy (3.25a) or E2i
fails to satisfy (3.25b). Let M be the largest member of E2i0−1 ∪ E2i0 . If
M belongs to E2i0 −1, move it to E2i0 , while if M belongs to E2i0 , move it to
E2i0−1 (note that if |E2i0 | = 1, then necessarily M ∈ E2i0−1). The resulting map is a parity changing involution of Π( n) − ˜Π( n). 38 Below, we illustrate the fixed point set ˜Π( n) and the pairings of Π( n) −
˜Π( n) when n = 4, wherein the first two members of each row are paired. Π0(n) − ˜Π( n) Π1(n) ˜Π( n)
{1, 2, 4}, {3} {1, 2}, {3, 4} {1, 2, 3, 4}{1, 3, 4}, {2} {1, 3}, {2, 4} {1, 2, 3}, {4}{1}, {2, 3, 4} {1, 4}, {2, 3} {1}, {2}, {3, 4}{1, 3}, {2}, {4} {1}, {2, 3}, {4} {1, 2}, {3}, {4}{1, 4}, {2}, {3} {1}, {2, 4}, {3} {1}, {2}, {3}, {4}
Note that the above bijection preserves the number of blocks of π ∈ Π( n). We’ll use its restriction to Π( n, k ) to supply a
Combinatorial Proof of Theorem 3.3.
Let Π i(n, k ) := Π i(n) ∩ Π( n, k ) for i = 0 , 1, ˜Π( n, k ) := ˜Π( n) ∩ Π( n, k ), and π = ( E1, . . . , E k) ∈ ˜Π( n, k ). If k is even, identify each pair of blocks (E2i−1, E 2i), 1 i k/ 2, with summands xi in a composition x1 +· · · +xk/ 2 =
n, where each xi 2. If k is odd, identify ( E1, E 2), . . . , (Ek−2, E k−1), (Ek)with summands xi in x1 + · · · + x(k+1) /2 = n, where xi 2 for 1 i k−1
2
and x(k+1) /2 1. The cardinality of ˜Π( n, k ) is then given by the right-hand side of (3.21), and the restriction of the prior bijection to Π( n, k ) − ˜Π( n, k ) is again an involution, and inherits the parity changing property, which proves (3.21).
From (3.21) along with (3.8), (3.9), and (3.10), we have ˆS−1(n, k ) = ( −1) n−k
(n − k/ 2 − 1
n − k
)
, 0 k n, (3.26)
S∗−1(n, k ) = ( −1) (k
2
)+n
(n − k/ 2 − 1
n − k
)
, 0 k n, (3.27) and
S−1(n, k ) = ( −1) (k
2
)
(n − k/ 2 − 1
n − k
)
, 0 k n. (3.28) The bijection establishing (3.21) clearly applies to (3.26)–(3.28) as well. 39 The bijection of Theorem 3.3 also proves combinatorially that
S(n, k ) ≡
(n − k/ 2 − 1
n − k
)
(mod 2) , 0 k n, (3.29) since off of a set of cardinality (n− k/ 2 −1
n−k
), each partition π ∈ Π( n, k ) is paired with another of opposite ˜ w-parity. This furnishes an answer to a question raised by Stanley [23, p. 46, Exercise 17b].
Theorem 3.5. For all n ∈ N,
ˆB−1(n) :=
n
∑
k=0
ˆS−1(n, k ) = ( −1) n−1Fn−3, (3.30)
where F−3 = −1, F−2 = 1 , and F−1 = 0 .
Proof. Clearly, (3.30) holds for n = 0 , 1. If n 2, then by (3.23) and (3.26), ˆB−1(n) =
n
∑
k=0
(−1) n−k
(n − k/ 2 − 1
n − k
)
= ( −1) n−1
(n−1) /2
∑
i=0
((n − 1) − ii
)
−
(n−2) /2
∑
i=0
((n − 2) − ii
)
= ( −1) n−1(Fn−1 − Fn−2) = ( −1) n−1Fn−3.
Alternatively, let n 3, ˜Π( n) be as in the proof of Theorem 3.4, and ˆΠ( n) ⊆ ˜Π( n) consist of those partitions with an odd number of blocks and whose last block is a singleton. First, | ˆΠ( n)| = | ˜Π( n − 3) | = Fn−3 as the removal of n − 2, n − 1, and n from π ∈ ˆΠ( n) is seen to be a bijection between ˆΠ( n) and ˜Π( n − 3). Since ˆw(π) = ˜w(π) + n − k and since every
π ∈ ˆΠ( n) has an even ˜ w(π) value and an odd number of blocks, the ˆ w-parity of each π ∈ ˆΠ( n) is opposite the parity of n. Thus, ˆΠ( n) agrees with the right-hand side of (3.30) in both sign and magnitude. The ˜ w-parity changing involution of Theorem 3.4 defined on Π( n) − ˜Π( n)also changes the ˆw-parity. We now extend this involution to Π( n) − ˆΠ( n)as follows: if the last block of π ∈ ˜Π( n) − ˆΠ( n) is {n}, merge it with the penultimate block; if the last block is not a singleton, take n from this block and form the singleton {n}. The resulting extension is a ˆ w-parity changing involution of Π( n) − ˆΠ( n). 40 The Bell numbers B∗−1(n) are quite different from the numbers ˜B−1(n)and ˆB−1(n), as demonstrated by the following theorem.
Theorem 3.6. For all n ∈ N,
B∗−1(n) :=
n
∑
k=0
S∗−1(n, k ) =
1, if n ≡ 0 (mod 3);
−1, if n ≡ 1 (mod 3); 0, if n ≡ 2 (mod 3) .
(3.31)
Proof. Using the generating function method and (3.27), we have
∑
n>0
B∗−1(n)xn = ∑
n>0
( n∑
k=0
S∗−1(n, k )
)
xn
= ∑
n>0
( n∑
k=0
(−1) (k
2
)+n
(n − ⌊k
2
⌋ − 1
n − k
))
xn
= ∑
k>0
(−1) (k
2
) ∑
n>k
(−1) n
(n − ⌊k
2
⌋ − 1
n − k
)
xn
= 1 − ∑
keven
k>2
xk
2+1
∑
n>k
2−1
( n
k
2
− 1
)
(−x)n
− ∑
kodd
xk+1
2
∑
n>k−1
2
( n
k−1
2
)
(−x)n
= ∑
k>0
(−1) kx2k
(1 + x)k − ∑
k>0
(−1) kx2k+1
(1 + x)k+1 = 1
1 + x
∑
k>0
(−1) kx2k
(1 + x)k
= 1
1 + x + x2 = 1 − x
1 − x3 = ∑
m>0
(x3m − x3m+1 ),
which proves (3.31). Alternatively, let Π i(n) := {π ∈ Π( n) : w∗(π) ≡ i (mod 2) } and Π ∗(n)consist of those partitions π = ( E1, E 2, . . . ) whose blocks satisfy
E2i−1 = {3i − 2, 3i − 1}, E2i = {3i}, 1 i n/ 3 . (3.32) Then Π ∗(n) is a singleton contained in Π 0(n) if n ≡ 0 (mod 3) or contained in Π 1(n) if n ≡ 1 (mod 3). If n ≡ 2 (mod 3), Π ∗(n) is a doubleton containing two partitions of opposite w∗-parity, which we pair. 41 Suppose now that π = ( E1, E 2, . . . ) ∈ Π( n) − Π∗(n) and that i0 is the smallest index for which condition (3.32) fails to hold. Let n1 = 3 i0 − 2,
n2 = 3 i0 − 1, n3 = 3 i0 and V1 = E2i0−1, V2 = E2i0 , V3 = E2i0 +1 (the latter two if they occur). Consider the following four disjoint cases concerning the relative positions of the ni within the Vi:(I) n2 ∈ V2, n3 ∈ V3, and |V2 ∪ V3| 3; (II) Either (a) or (b) holds where (a) V2 = {n2} and V3 = {n3},(b) n2, n 3 ∈ V1;(III) n2 ∈ V2 and n3 ∈ V1 ∪ V2;(IV) n2 ∈ V1, n3 ∈ V2, and |V1 ∪ V2| 4. Within each case, we pair partitions of opposite parity as shown below, leav-ing the other blocks undisturbed: (i) V2 = {n2, . . . , M }, V3 = {n3, . . . } ↔ V2 = {n2, . . . }, V3 = {n3, . . . , M },where M is the largest member of V2 ∪ V3;(ii) V1 = {n1, . . . }, V2 = {n2}, V3 = {n3} ↔ V1 = {n1, n 2, n 3, . . . };(iii) V1 = {n1, n 3, . . . }, V2 = {n2, . . . } ↔ V1 = {n1, . . . }, V2 = {n2, n 3, . . . };(iv) V1 = {n1, n 2, . . . , N }, V2 = {n3, . . . } ↔ V1 = {n1, n 2, . . . },
V2 = {n3, . . . , N }, where N is the largest member of V1 ∪ V2.The resulting map is a parity changing involution of Π( n) − Π∗(n), which implies (3.31).
Below, we illustrate the fixed point set Π ∗(n) along with the pairings of Π( n) − Π∗(n) when n = 4. Π0(n) Π1(n) − Π∗(n) Π∗(n)
{1, 2, 3, 4} {1, 4}, {2}, {3} {1, 2}, {3}, {4}{1, 2}, {3, 4} {1, 2, 4}, {3}{1, 3}, {2, 4} {1}, {2, 3, 4}{1, 4}, {2, 3} {1, 3, 4}, {2}{1}, {2, 3}, {4} {1, 3}, {2}, {4}42
{1}, {2, 4}, {3} {1}, {2}, {3, 4}{1}, {2}, {3}, {4} {1, 2, 3}, {4}
Note that the bijection above, like the one used for Theorem 3.5, does not always preserve the number of blocks and hence has no meaningful restriction to Π( n, k ), unlike the bijection of Theorem 3.4.
Remark. In , Ehrlich evaluates σ(n) := − ∑
π∈Π( n)
(−1) α(π), where α(π) :=
∑
iodd
|Ei| for π = ( E1, E 2, . . . ) ∈ Π( n). The bijection of Theorem 3.6 estab-lishing B∗−1(n) also provides an alternative to Ehrlich’s iterative argument establishing his σ(n) since
σ(n) = − ∑
π=( E1,E 2,... )∈Π( n)
(−1) |E1|+|E3|+|E5|+···
= − ∑
π=( E1,E 2,... )∈Π( n)
(−1) |E1|+2 |E2|+3 |E3|+···
= −B∗−1(n).
It is easy to verify that one can write (3.31) more compactly as
B∗−1(n) = 1
1 − w wn − w
1 − w w2n,
where w is a primitive cube root of unity, which yields the pleasing exponen-tial generating function
∑
n>0
B∗−1(n)xn
n! = 1
1 − w ewx − w
1 − w ew2x. (3.33) Since Sq(n, k ) = q−nS∗
q
(n, k ),
B−1(n) :=
n
∑
k=0
S−1(n, k ) = ( −1) nB∗−1(n),
and so by (3.31),
B−1(n) =
(−1) n, if n ≡ 0 (mod 3); (−1) n+1 , if n ≡ 1 (mod 3); 0, if n ≡ 2 (mod 3) ,
(3.34) 43 with the above bijection clearly showing this. The preceding also supplies a combinatorial proof that B(n), the nth Bell number, is even if and only if
n ≡ 2 (mod 3) since every partition of [ n] is paired with another of opposite
w∗-parity when n ≡ 2 (mod 3) and since all partitions are so paired except for one otherwise (cf. Ehrlich [7, p. 512]).
3.3 A Notable Restriction
Let B(n, k ) := {π ∈ Π( n, k ) : all blocks of π have cardinality at most 2 },
B(n) := ∪kB(n, k ), A(n, k ) := |B (n, k )|, and A(n) := |B (n)|. The numbers
A(n, k ), known as Bessel numbers, are reparametrized coefficients of Bessel polynomials . By a fairly routine combinatorial argument,
A(n, k ) =
( n
2n − 2k
) (2 n − 2k)!
2n−k(n − k)! , n/ 2 k n. (3.35) Consider the q-generalizations of A(n, k ) and A(n) obtained by restricting the partition statistics w∗ and w to B(n, k ) and to B(n):
A∗
q
(n, k ) := ∑
π∈B (n,k )
qw∗(π), (3.36)
Aq(n, k ) := ∑
π∈B (n,k )
qw(π) = q−nA∗
q
(n, k ), (3.37)
A∗
q
(n) := ∑
π∈B (n)
qw∗(π), (3.38) and
Aq(n) := ∑
π∈B (n)
qw(π) = q−nA∗
q
(n). (3.39) The numbers A∗
q
(n, k ) satisfy the boundary conditions A∗
q
(n, 0) = δn, 0,
A∗
q
(0 , k ) = δ0,k , and A∗
q
(1 , 1) = q, along with the recurrence
A∗
q
(n, k ) = qnA∗
q
(n − 1, k − 1) + qn(n − 1) A∗
q
(n − 2, k − 1) , n 2, k 1, (3.40) upon considering whether the first block is the singleton {1} or a doubleton containing 1. 44 The recurrence in (3.40) is not of a form covered by Comtet’s Theo-rem. We were unable to find analogues of (3.16)–(3.18) for either A∗
q
(n, k )or Aq(n, k ), nor were we able to find simple expressions for A∗−1(n, k ) or
A−1(n, k ). However, we do have the following somewhat surprising parity theorem for B(n).
Theorem 3.7. For all n ∈ P,
A∗−1(4 n) = A∗−1(4 n − 1) =
n−1
∏
i=0
(4 i + 1)(4 i + 2) ,A∗−1(4 n − 2) = 0 , (3.41)
and
A∗−1(4 n − 3) = −
n−1
∏
i=1
(4 i + 1)(4 i − 2) .
Proof. Let am = A∗−1(m) and bm = A−1(m). Conditioning on whether or not the first block is a singleton yields
am = −bm−1 + ( m − 1) bm−2, m 2. (3.42) Since bm = ( −1) mam, we have, by (3.42), the second-order recurrence
am = ( −1) m[am−1 + ( m − 1) am−2], m 2, (3.43) with a0 = 1, a1 = −1. To solve (3.43), write out the first several terms of the sequence am and conjecture that a4m−2 = 0, along with a4m−1 = a4m if m 1. Assuming this conjecture to be true for the moment, let dm = a4m+1 and em = a4m+3 =
a4m+4 for m 0 with e−1 := 1. Then (3.43) implies em = −(4 m + 2) dm and
dm = −(4 m + 1) em−1 for m 0 so that
em = (4 m + 2)(4 m + 1) em−1, m 0,
whence
em =
m
∏
i=0
(4 i + 1)(4 i + 2) and dm = −
m
∏
i=1
(4 i + 1)(4 i − 2) , m 0.
The sequence am defined by a4n = a4n−1 = ∏n−1
i=0
(4 i + 1)(4 i + 2), a4n−2 = 0, and a4n−3 = − ∏n−1
i=1
(4 i + 1)(4 i − 2), if n 1 with a0 = 1, is easily shown to satisfy (3.43), which completes the proof. 45
Conclusion
The formulas of the first three chapters show that the case q = −1 is an interesting special case which differs in many respects from the general
q-case. In several instances, these formulas arise when either a q-recurrence or q-generating function assumes a particularly nice form for q = −1 (e.g., (3.24) in Ch. 3; (3.13) and (3.14) in ). The q = −1 case may have a nice closed form whereas there is no such closed form known for q general, as seen with (2.32). In a few instances, there seems to be no underlying algebraic development behind a particular parity result (e.g., (2.16), (2.21), and (3.41)). In addition, the author has encountered several instances in which the case q = −1 appears to have no simple closed form at all. One possible way to algebraically extend previous results for q = −1would be to allow q to be an arbitrary complex root of unity. Though in many instances such a generalization doesn’t seem possible, in a few, the generalization is straightforward. For example, when m 2 and q = e2πi/m ,then n!
q
= 0 if n m since mq = 1 + q + · · · + qm−1 = 0 if q = e2πi/m .To realize this combinatorially, it suffices to show, by (2.1), that the inv (or
maj ) statistic on Sn is balanced in the following sense whenever n m: the number of permutations of [ n] whose inv (or maj ) value is congruent to i
(mod m) is equal to the number whose value is congruent to j (mod m) for all i and j, where 0 i < j m − 1. If n m, then locate the largest member occurring in the first m po-sitions of σ ∈ Sn, expressed as a word, and cyclicly shift it through these positions, leaving the relative order of the other letters undisturbed. The m
members of Sn so obtained have different inv and maj values mod m. Since
Sn is partitioned into m-member equivalence classes by this procedure, it follows that inv and maj are both balanced as described. Note that for inv ,but not maj , this can also be realized by locating the positions occupied by the members of [ m] within σ ∈ Sn, expressed as a word, and shifting m in a cyclic fashion within these m positions, leaving the rest of σ undisturbed. 46 For another example, we let q = ρ = −1+ √3i
2
, a third root of unity, in the
q-binomial coefficient (nk
)
q
. By a calculation similar to the one in Theorem 1.1 or by substituting q = ρ into (1.11) and considering cases mod 3, we have the formula
(nk
)
ρ
=
−ρ2
( n/ 3
k/ 3
)
, if n ≡ k (mod 3) or k ≡ 0 (mod 3);
( n/ 3
k/ 3
)
, if n ≡ 2 (mod 3) and k ≡ 1 (mod 3); 0, otherwise .
(1) The combinatorial arguments of the first chapter used to evaluate (nk
)
−1
=
∑
w∈Λn,k
(−1) inv (w) can be modified to evaluate (nk
)
ρ
= ∑
w∈Λn,k
ρinv (w), where Λ n,k
denotes the set of binary words of length n with k 0’s. Instead of pairing members of Λ n,k of opposite inv -parity, we partition a portion of Λ n,k into tripletons each of whose members have different inv values mod 3. For each such tripleton {λ1, λ 2, λ 3}, we have that ρinv (λ1 ) + ρinv (λ2 ) + ρinv (λ3 ) = 0 since 1 + ρ + ρ2 = 0. Let Λ ′
n,k
consist of those words w = w1w2 · · · wn in Λ n,k satisfying
w3i−2 = w3i−1 = w3i, 1 i n/ 3 . (2) In all cases, the right-hand side of (1) above gives the net contribution of Λ ′
n,k
towards the sum (nk
)
ρ
= ∑
w∈Λn,k
ρinv (w); note that members of Λ ′
n,k
may end in either 01 or 10 if n ≡ 2 (mod 3) and k ≡ 1 (mod 3), hence the 1 + ρ = −ρ2
factor in this case. Suppose now that w = w1w2 · · · wn ∈ Λn,k − Λ′
n,k
, with i0 being the small-est i for which (2) fails to hold. Group the three members of Λ n,k −Λ′
n,k
gotten by circularly permuting w3i0−2, w3i0−1, and w3i0 within w = w1w2 · · · wn, leav-ing the rest of w undisturbed. Note that these three members of Λ n,k − Λ′
n,k
have different inv values mod 3. The preceding argument works equally well with maj in place of inv . See also formulas (4.1)–(4.6) in for similar behavior when q is a third root of unity. 47
References 48
References
G. Andrews, The theory of partitions, Encyclopedia of Mathematics and Applications, Vol. II , Addison-Wesley, Reading, MA (1976). L. Carlitz, Generalized Stirling numbers, Combinatorial Analysis Notes ,Duke University (1968), 1–15. L. Carlitz and R. Scoville, A note on weighted sequences, Fib. Quart.
13 (1975), 303–306. J. Choi and J. Smith, On the unimodality and combinatorics of Bessel numbers, Discrete Math. 264 (2003), 45–53. L. Comtet, Nombres de Stirling g´ en´ eraux et fonctions symmetriques, C. R. Acad. Sci. Paris S´ erie A 275 (1972), 747–750. P. Edelman, R. Simion, and D. White, Partition statistics on permuta-tions, Discrete Math. 99 (1992), 62–68. G. Ehrlich, Loopless algorithms for generating permutations, combina-tions, and other combinatorial configurations, J. ACM 20 (1973), 500– 513. W. Feller, An Introduction to Probability Theory and Its Applications ,John Wiley & Sons, Inc., New York, 1968. B. Fristedt and L. Gray, A Modern Approach to Probability Theory ,Birkh¨ auser, Boston, 1997. J. F¨ urlinger and J. Hofbauer, q-Catalan numbers, J. Combin. Theory, Ser. A , 40 (1985), 248–264. 49 A. Garsia and J. Remmel, A combinatorial interpretation of q-derangement and q-Laguerre numbers, Europ. J. Combinatorics 1(1980), 47–59. J. Goldman and G.-C. Rota, The number of subspaces of a vector space,
Recent Progress in Combinatorics , Academic Press, New York (1969), 75–83. D. Knuth, The Art of Computer Programming: Fundamental Algo-rithms, Vol. I , Addison-Wesley, Reading, MA, 1968. I. Lah, Eine neue Art von Zahlen, ihre Eigenschaften und Anwendung in der mathematischen Statistik, Mitteilungsbl. Math. Statist. 7 (1955), 203–212. P. MacMahon, Combinatorial Analysis, Vol. II , Cambridge Univ. Press, 1915–1916. Reprinted, Chelsea, New York, 1960. S. Milne, A q-analogue of restricted growth functions, Dobinski’s equal-ity, and Charlier polynomials, Trans. Amer. Math. Soc. 245 (1978), 89– 118. S. Ross, A First Course in Probability , Prentice-Hall, Inc., Englewood Cliffs, NJ, 1994. M. Shattuck, Bijective proofs of parity theorems for partition statistics,
J. Integer Seq. 8 (2005), Art. 5.1.5. M. Shattuck, Parity theorems for statistics on permutations and Catalan words, Integers 5 (2005), #A07. M. Shattuck and C. Wagner, Parity theorems for statistics on domino arrangements, Electron. J. Combin. 12 (2005), #N10. M. Shattuck and C. Wagner, Parity theorems for statistics on lattice paths and Laguerre configurations, J. Integer Seq. , in press (2005). R. Simion and D. Ullman, On the structure of the lattice of noncrossing partitions, Discrete Math . 98 (1991), 193–206. R. Stanley, Enumerative Combinatorics, Vol. I , Wadsworth and Brooks/ Cole, Monterey, 1986. 50 M. Wachs and D. White, p, q -Stirling numbers and partition statistics,
J. Combin. Theory, Ser. A , 56 (1991), 27–46. C. Wagner, Generalized Stirling and Lah numbers, Discrete Math. 160 (1996), 199–218. C. Wagner, Partition statistics and q-Bell numbers ( q = −1), J. Integer Seq. 7 (2004), Art. 4.1.1. C. Wagner, Enumerative Combinatorics Notes (Mathematics 521–522) ,University of Tennessee (2004–2005). 51
Appendices 52
Appendix A A Combinatorial Proof of a q-Binomial Coefficient Identity
Recall the q-binomial theorem (1 + x)(1 + qx ) · · · (1 + qn−1x) =
n
∑
k=0
q(k
2
)
(nk
)
q
xk, (1) originally due to Euler. Substituting − 1
x
for x in (1), multiplying both sides by xn, and reindexing yields the variant (x − 1)( x − q) · · · (x − qn−1) =
n
∑
k=0
(−1) n−kq(n−k
2
)
(nk
)
q
xk. (2) Our aim here is to give a direct combinatorial proof of (2). It suffices to prove equality whenever x = qm, m ∈ P, and q is a power of some prime. We’ll argue that, when x = qm, both sides of (2) count the injective linear transformations from Fnq to Fmq , the left-hand side clearly doing so, where Frq
denotes the finite r-dimensional vector space of qr elements over the finite field Fq of q elements. To show that the right-hand side of (2) also achieves this, we’ll use a weighted, sieve argument described below which starts with the set of all linear transformations. Let A denote the set of all ordered pairs α = ( U, T ), where T is a linear transformation from Fnq to Fmq and U is a subspace of the null space of T .Assign to each α ∈ A the weight ( −1) kq(k
2
), where k is the dimension of U.53 The right-hand side of (2) at x = qm, written in the form
n
∑
k=0
(−1) kq(k
2
)
(nk
)
q
qm(n−k), (3) then gives the total weight of all the members of A, according to the dimen-sion of U.Now let R be a linear transformation from Fnq to Fmq , with null space W
and dim( W) = j. Then the net weight of all the members of A with second coordinate R is given by
j
∑
i=0
∑
U ⊆W
dim( U)= i
(−1) iq(i
2
) =
j
∑
i=0
(−1) iq(i
2
)
(ji
)
q
= δj, 0, (4) the latter equality upon substituting x = 0 in the well known identity [25, pp. 201–202]
xn =
n
∑
k=0
(nk
)
q
(x − 1) · · · (x − qk−1), n 0. (5) Thus by (4), the net weight of all the members of A with second coordinate
R is zero when dim( W) is positive and one when dim( W) is zero. Therefore, the total weight of A equals the number of injective linear transformations, which completes the proof. 54
Appendix B Bijective Proofs of Alternating Sum Identities
The ideas used to construct the involutions of the first three chapters can be adapted to furnish bijective proofs of various identities involving sums with alternating signs, such as orthogonality relations, connection constant relations, and binomial coefficient identities, which are typically proven us-ing algebraic methods. In this section, we look at some specific identities illustrating how these combinatorial ideas can be applied. As far as pro-viding bijective proofs for these identities, constructing appropriate sign-reversing involutions seems to be more effective than other combinatorial proof techniques, such as direct argument or inclusion-exclusion. The author has used similar sign-reversing involutions in providing bijective proofs for several other combinatorial identities involving sums with alternating signs. The first identities we’ll look at are the well known orthogonality relations for Stirling numbers [23, Proposition 1.4.1],
n
∑
j=k
S(n, j )s(j, k ) = δn,k , 0 k n, (1) and
n
∑
j=k
s(n, j )S(j, k ) = δn,k , 0 k n, (2) where S(n, k ) is the Stirling number of the second kind, s(n, k ) = ( −1) n−k
c(n, k ) is the Stirling number of the first kind, and c(n, k ) is the signless 55 Stirling number of the first kind. We rewrite these relations in the more suggestive form
n
∑
j=k
(−1) j S(n, j )c(j, k ) = ( −1) kδn,k , 0 k n, (3) and
n
∑
j=k
(−1) j c(n, j )S(j, k ) = ( −1) nδn,k , 0 k n. (4) We first give a bijective proof of (3). Partition [ n] into j blocks, where
k j n, writing the members of [ n] within a block in ascending order. Now permute these j blocks, ordered lexicographically, according to a permutation on j elements with k cycles in standard cycle form. Let the resulting block arrangement have weight ( −1) j , where j is the number of individual blocks. The left-hand side of (3) then gives the total weight of all block arrangements with j individual blocks as j ranges from k to n.We now pair block arrangements of opposite sign as follows. Let m denote the number of the first cycle encountered, going from left to right within a block arrangement, possessing at least two members of [ n] in all and 8 denote the first, hence the smallest, member of [ n] encountered in cycle m. If {8} is not a block, split off 8 and start cycle m with the block {8}. If {8} is the first block of cycle m, hence not the last block, place 8 at the front of the second block of cycle m. All block arrangements are so paired except when k = n,in which case there is a single block arrangement of sign ( −1) k.Similarly for (4), take the j cycles, ordered lexicographically, of σ ∈ Sn,j ,expressed in standard cycle form, and arrange them according to a canonical ordered partition of [ j] with k blocks, writing the cycles occurring within a block in ascending order. Let the resulting cycle arrangement have weight (−1) j , where j is the number of individual cycles. Let c1, . . . , c r be the cycles in order which comprise the first block possessing at least two members of [n] in all. If cr is a 1-cycle, whence r 2, place the sole member of cr at the end of cr−1. If |cr| 2, break off the last member of cr and form a 1-cycle within the block. 56 The s(n, k ) (= ( −1) n−kc(n, k )) are the connection constants in the poly-nomial identities [23, p. 35]
xn =
n
∑
k=0
s(n, k )xk, n ∈ N, (5) where xn := x(x − 1) · · · (x − n + 1). Equivalently,
mn =
n
∑
k=0
(−1) n−kc(n, k )mk, n ∈ N, (6) for each m ∈ N.To prove (6), place the k cycles of σ ∈ Sn,k , expressed in standard cycle form, into m labeled urns in one of mk ways, ordering the cycles within an urn lexicographically. Let A denote the set of all possible arrangements as
k varies, 0 k n, and J ⊆ A those which consist of 1-cycles placed in distinct urns. Let sign( α) = ( −1) n−k, where k is the number of cycles in
α ∈ A . Note that |J | = mn, with each member of J having positive sign. Pair members of A−J of opposite sign by identifying the first urn possessing at least two members of [ n] in all and then either merging or breaking off the last member t of [ n] occurring within a cycle in this urn, depending upon whether or not t itself is a 1-cycle. Next, we generalize a well known orthogonality relation for binomial co-efficients (upon taking m = n in (7) below):
m
∑
k=r
(−1) k
( nm − k
)( kr
)
= ( −1) r
(n − r − 1
m − r
)
, 0 r m n. (7)
Proof of (7) . Note that both sides of (7) give the coefficient of xm in the convolution (1 + x)n · (−x)r
(1 + x)r+1 = ( −1) rxr(1 + x)n−r−1.
Alternatively, mark m − k members of [ n] with red in one of ( nm−k
) ways and consider the first k members of [ n] not marked red, where r k m.Choose r of these numbers to be marked blue in one of (kr
) ways and mark the remaining k − r numbers green. Any remaining members of [ n] will be 57 unmarked. Let such a coloring α of [ n] have sign ( −1) k. The left-hand side of (7) clearly gives the total weight of all possible colorings α.Now consider the smallest number marked either red or green. Switching to the other option changes the sign of α. This change can be effected provided that the first red or green number does not come later than the last blue number with at least one unmarked number in between. But this can occur only if the smallest red number is greater than r + 1 with k = r, of which there are (n−r−1
m−r
) possibilities each with sign ( −1) r.
We close with a bijective proof of the orthogonality relation (3.22):
min {n, 2m}
∑
k=m
(−1) k−m
(n − k/ 2 − 1
n − k
)( k/ 2
k − m
)
= δn,m , 0 m n. (8) Let An,m be the set of “marked compositions” x = ( x1, x 2, . . . ) of n
satisfying: (a) x itself is a composition of n with xi = 1 for every even index
i; (b) x has at least m parts and at most 2 m parts altogether; (c) if x has k
parts, where m k 2m, then k − m pairs of parts of the form ( x2i−1, x 2i),
i 1, are marked. We’ll assign the weight of ( −1) k−m to x ∈ A n,m possessing
k − m designated pairs. Note that the left-hand side of (8) then gives the net weight of all members of An,m .If n = m, then An,m is a singleton with positive sign. So suppose n > m
and x ∈ A n,m , with i0 the largest index i 1 such that one of the following holds: (i) (x2i−1, x 2i) is marked; (ii) x2i−1 2 with x2i−1 not part of marked pair. Note that xj = 1 for all j 2i0 + 1 and that the last part of a marked composition with an odd number of parts is, by definition, unmarked. Define a sign-reversing involution of An,m by: (I) if (i) holds, remove the designation and replace ( x2i0−1, x 2i0 ) with the single part ( x2i0 −1 + 1); (II) if (ii) holds, replace x2i0−1 with the two parts ( x2i0−1 − 1, 1) and desig-nate the pair ( x2i0−1 − 1, 1). 58
Appendix C Asymptotics 1 Introduction
Suppose we have a sequence of statistics In : ∆ n → N, with uniform probability measure on each finite discrete structure ∆ n. The random vari-able sequence ( In)n>1 is said to be asymptotically normal if, with E(In) = μn
and Var( In) = σ2
n
,lim
n→∞
P
(In − μn
σn
z
)
= Φ( z), (1) for all z, where Φ( z) = 1√2π
∫ z
−∞
e−t2/2 dt . For example, this is clearly true if ∆n = 2 [n] and In(A) = |A|, by the classical central limit theorem. In the next section, we’ll examine the asymptotic normality of some sta-tistics on discrete structures using a more generalized central limit theorem. The statistics we’ll look at all have probability generating functions (pgf’s) which factor completely into a product of simpler pgf’s. In the last section, we’ll briefly look at the asymptotics of the ratio rn := Gn(−1) /G n(1), where
Gn(q) := ∑
δ∈∆n
qIn(δ). (2) In most of the cases we look at, this ratio tends to zero exponentially fast (whenever Gn(1) itself increases exponentially). Thus, it appears that large imbalances for unbalanced statistics are usually rather short lived. 59
2Asymptotic Normality
For each n 1, let Xn, 1, X n, 2, . . . , X n,n be n mutually independent ran-dom variables, which we’ll refer to as X1, X 2, . . . , X n by a slight abuse of notation. We assume that the means and variances are finite and put
μi = E(Xi), σ2
i
= Var( Xi), 1 i n. (3) The sum Yn = X1 + · · · + Xn then has mean mn and variance s2
n
given by
mn = μ1 + · · · + μn, s2
n
= σ21 + · · · + σ2
n
. (4) The array (( Xn,i )ni=1 )n>1 is said to obey the central limit theorem if for all z,
P
(Yn − mn
sn
z
)
→ Φ( z). (5) The following theorem, a feat of early modern probability theory, gives nec-essary and sufficient conditions for the central limit theorem to hold (see [9, p. 322]):
Theorem 1 (Lindeberg-Feller). Suppose that lim
n→∞
max {σ2
k
: 1 k n}
s2
n
= 0 . (6)
Then the central limit theorem holds if and only if for every ε > 0, the truncated random variables Ui, 1 i n, defined by
Ui =
{
Xi − μi, if |Xi − μi| εs n;0, if |Xi − μi| > εs n, (7)
satisfy
1
s2
nn
∑
i=1
E(U2
i
) → 1. (8) Note that (8) easily implies (6). Feller uses the sufficiency in Theorem 1 to establish the asymptotic normality for the statistics recording the number of cycles and the number 60 of inversions of a randomly chosen member of Sn. We first outline Feller’s proof of asymptotic normality for the number of inversions. Let n 1 and Xi, 1 i n, be the random variable recording the number of inversions produced by i in σ ∈ Sn, expressed as a word. Then
Yn = X1 + · · · + Xn records the total number of inversions (i.e., the inv
value) of σ ∈ Sn. The number of inversions produced by i does not depend on the relative order of 1 , 2, . . . , i − 1 within σ ∈ Sn, which implies the Xi
are mutually independent. The Xi assume the values of 0 , 1, . . . , i − 1, each with probability 1
i
and therefore μi = i−1
2
and σ2
i
= i2−1
12
, 1 i n, from which one gets mn = n(n−1)
4
and s2
n
= n(n−1)(2 n+5)
72
.For large n, we have Ui = Xi −μi, 1 i n, in (7) as |Xi −μi| n εs n,which ensures that (8) is satisfied. Theorem 1 then gives the asymptotic normality for the inv statistic on Sn. Since mn ∼ n2
4
and s2
n
∼ n3
36
, one can conclude, for example, that the number, Nn, of members of Sn whose
inv value lies between the limits of n2
4
± βn 3/2
6
is asymptotically given by
n!(Φ( β) − Φ( −β)) for each β > 0. From the generating function (2.1),
∑
σ∈Sn
qinv (σ) =
n
∏
i=1
iq,
one sees that the probability generating function (pgf) for inv can be factored into a product of simpler pgf’s, 1
n!
∑
σ∈Sn
qinv (σ) = 1
1
( 1 + q
2
)
· · ·
( 1 + q + · · · + qn−1
n
)
, (9) hence the decomposition of inv into the simple independent components given above. We now apply similar reasoning to other combinatorial statistics. For ex-ample, recall that the Carlitz statistic, inv c, of Section 2.2 has the generating function
∑
σ∈Sn
qinv c(σ) =
n−1
∏
i=0
(1 + iq),
so that the pgf for inv c on Sn factors as 1
n!
∑
σ∈Sn
qinv c(σ) = 1
1 · 2
2 ·
(2 + q
3
)
· · ·
(2 + q + · · · + qn−2
n
)
. (10) 61 For n 2, let ( Xi)ni=1 be independent and given by P (Xi = 0) = 2
i
, P (Xi =
j) = 1
i
, 1 j i − 2, if i 2, with X1 = 0. If Yn records the inv c value of a randomly chosen member of Sn, then one can express Yn as Yn = X1+· · · +Xn,where the Xi are as given, by (10). Asymptotic normality for inv c then follows in much the same way as for inv .If m(σ) denotes the number of left-to-right maxima (i.e., record highs) of
σ ∈ Sn, expressed as a word, then there is the joint generating function [23, p. 49, Exercise 31]
∑
σ∈Sn
xm(σ)qinv (σ) =
n−1
∏
i=0
(x + qi q). (11) Setting x = q in (11) and dividing by n! gives the pgf for Yn recording the value of m(σ) + inv (σ) for σ ∈ Sn, chosen at random: 1
n!
∑
σ∈Sn
qm(σ)+ inv (σ) = q
1 · 2q
2 ·
(2q + q2
3
)
· · ·
(2q + q2 + · · · + qn−1
n
)
. (12) For n 2, define the independent sequence ( Xi)ni=1 by P (Xi = 1) = 2
i
,
P (Xi = j) = 1
i
, 2 j i−1, if i 2, with X1 = 1. Then Yn = X1 +· · · +Xn
records the m + inv value by (12), and the asymptotic normality follows much as before. Even though m and the number of cycles, π, are identically distributed on Sn, we cannot conclude from this that π+inv is asymptotically normal since m and π behave differently when considered jointly with inv .Next, we consider the statistic, denoted σ(S), recording the sum of the elements of S ⊂ N finite. If n 1, let Yn record the subset sum σ(S) for a randomly chosen subset S of {0, 1, . . . , n − 1}. If Xi records the contribution of i towards σ(S), 0 i n−1, then Yn = X0 +· · · +Xn−1, where the Xi are independent and given by P (Xi = 0) = 1
2
= P (Xi = i) if i 1, with X0 = 0. The asymptotic normality of σ(S) then follows from Theorem 1 since s2
n
is of order n3 with 0 Xi n.Setting x = 1 in the q-binomial theorem
n−1
∏
i=0
(1 + qix) =
n
∑
k=0
q(k
2
)
(nk
)
q
xk
gives
∑
S⊆{ 0,...,n −1}
qσ(S) =
n−1
∏
i=0
(1 + qi) =
n
∑
k=0
q(k
2
)
(nk
)
q
. (13) 62 Hence, the statistic on lattice paths given by α(λ)+ (k
2
), where α(λ) is the area subtended by a (minimal) lattice path λ of length n starting from the origin and k denotes the number of vertical steps, is asymptotically normal as σ(S)is. We were unable to determine whether or not α itself is asymptotically normal. Recall that the w statistic on permutations (Section 2.2) has the gener-ating function
∑
σ∈Sn
qw(σ) =
n−1
∏
i=0
(i + qi). (14) If Yn records the w value of a randomly chosen member of Sn, then, by (14), Yn = X1 + · · · + Xn, where the Xi are independent and given by
P (Xi = 0) = i−1
i
, P (Xi = i − 1) = 1
i
, if i 2, with X1 = 0. The Xi fail condition (8) as Var( Yn) has order only n2. As (6) clearly holds, the necessity of Theorem 1 implies that the w statistic on Sn fails to be asymptotically normal. This leaves open the questions of the existence and possible identity of a limiting distribution for w on Sn.
3 The Ratio Gn(−1)
Gn(1)In this section, we’ll briefly look at the ratio rn := Gn(−1)
Gn(1)
for n large, where
Gn(q) := ∑
δ∈∆n
qIn(δ) (15) and In = I is some statistic. Here, we’ll choose ∆ n to be the larger discrete structures (indexed by n) of Chapters 1–3 (e.g., Π( n), Sn, 2 [n], Cn, Dn). For the statistics I we’ve looked at in Chapters 1–3, the ratio rn tends to zero exponentially (i.e., |rn| rn for some r, 0 < r < 1, and all n sufficiently large) in almost all cases for these larger structures. In many cases, this is either obvious or follows from a short calculation. For instance, take ∆ n = Cn, the set of Catalan words of length 2 n, and
I = maj . By Theorem 2.7, we have for n ∈ N,1
2n = 2n/(n + 1)
22n/(n + 1) ( n
n/ 2
) / cn = rn63
2n
22n/(n + 1)(2 n + 1) = (n + 1)(2 n + 1)
2n ,
as 2 m/(m + 1) ( m
m/ 2
) 2m, m ∈ N, so that any r in the interval ( 1/2 , 1) will suffice. From Stirling’s formula for n! [8, p. 52], we have in fact rn ∼ √2n
2n
.For another example, we take ∆ n = Π( n), the set of partitions of [ n], and
I = ˜ w.
Proposition 1. ˜B−1(n)
˜B1(n)
tends to zero exponentially.
Proof. Note that B(n) > 2n−1 for n 3 since Π( n) contains a proper subset which is in 1-1 correspondence with the compositions of n in the obvious way whenever n 3. Since Fn ∼ c1
(1+ √5
2
)n
for large n, we then have, by Theorem 3.4, ˜B−1(n)
˜B1(n)
= Fn
B(n)
4c1
( 1+ √5
4
)n
for n sufficiently large, with
1+ √5
4
< 1 and c1 a positive constant.
Next, take I = w∗ and ∆ n = B(n), the partitions of [ n] whose blocks have cardinality at most 2. Recall that
A∗
q
(n) := ∑
π∈B (n)
qw∗(π), n 0. (16)
Proposition 2. A∗−1(n)
A∗
1(n)
tends to zero.
Proof. Let n 5 and first suppose 4 |n. Consider the class of partitions
B∗(n) ⊆ B (n) consisting of 4 singletons, with the rest of the elements of [ n]partitioned into doubletons such that the members of [ n] starting doubletons 2i − 1 and 2 i are the smallest numbers that haven’t been used in doubletons 1, 2, . . . , 2i − 2, 1 i n−4
4
, or in any of the singletons. Note that
|B ∗(n)| = (n
4
) n
4−2
∏
i=0
(4 i + 1)(4 i + 2) ,
so that by Theorem 3.7,
A∗−1(n)
A1(n) A∗−1(n)
|B ∗(n)| = (n − 2)( n − 3)
(n
4
) = 24
n(n − 1) .
If n ≡ 1 (mod 4) or if n ≡ 3 (mod 4), let B∗(n) be as above except that it now contains 5 singletons or 3 singletons, respectively. If n ≡ 2 (mod 4), then A∗−1(n) = 0. 64 By allowing the members of B∗(n) to contain arbitrarily many singletons, one can show that the ratio A∗−1(n)/ A∗
1
(n) tends to zero faster than the reciprocal of any polynomial. Yet it still isn’t clear, upon elementary consid-erations, whether or not this ratio tends to zero exponentially. For example, note that n− log n tends to zero faster than the reciprocal of any polynomial, yet fails to tend to zero exponentially. We were able to find an instance in which Gn(1) increases exponentially (i.e., Gn(1) rn for some r > 1 and all n sufficiently large), with rn failing to tend to zero exponentially. Letting x = q = −1 in (2.3) and noting
w∗(σ) = w(σ) + n for σ ∈ Sn reveals that
∑
σ∈Sn
(−1) w∗(σ)−| σ| =
{
(n − 1)! , if n is odd;
− (( n − 1)! + ( n − 2)!) , if n is even, (17) where n 1. Thus, rn ∼ 1
n
for large n when I = w∗ − | σ| and ∆ n = Sn.This imbalance for w∗ − | σ| is then numerically more significant and persists longer than the imbalances for the other statistics we’ve studied. We close with a bijective proof of (17). Let S±
n
consist of those per-mutations with even or odd w∗ − | σ| values, respectively, where members
σ = ( E1, E 2, . . . , E r) are expressed in the standard cycle form. First suppose
n is odd and let S∗
n
⊆ S+
n
consist of those permutations with a single cycle. Define a parity changing involution of Sn − S∗
n
as follows: (i) If |Er| is even, place the last member of Er after the last member of
Er−1.(ii) If |Er| is odd and |Er−1| 2 with the last member of Er−1 greater than the first member of Er, place the last member of Er−1 at the end of Er.(iii) If |Er| is odd and |Er−1| 2 with the last member of Er−1 less than the first member of Er , take the last member of Er−1 and form a 1-cycle with it. (iv) If |Er| is odd and |Er−1| = 1, place the singleton Er−1 at the end of
Er−2.Note that r 3 in (iv) since n is assumed odd. If n is even, let
S∗
n
⊆ S−
n
consist of those permutations with a single cycle or of the form (1)(2 i1 · · · in−2); note that |S∗
n
| = ( n − 1)! + ( n − 2)!. Use the same involution given by (i)–(iv) above, noting that r 3 in (iv) since (1)(2 i1 · · · in−2) is now disallowed. 65
Vita
Mark A. Shattuck was born in Chattanooga, Tennessee, where he at-tended McCallie School, graduating in 1995. Three years later, he earned the degree of Bachelor of Science in Mathematics from the University of Florida at Gainesville. After a short stint teaching at Chattanooga State Techni-cal Community College, he entered the University of Tennessee at Knoxville as a graduate student in the fall of 2000, earning the degrees of Master of Science and Doctor of Philosophy in Mathematics from that institution in 2001 and 2005. His chief mathematical interests lie in combinatorics, num-ber theory, Euclidean geometry, and elementary problem solving. His chief non-mathematical interests lie in entomology, the physical sciences, words, and the environment. |
190721 | https://dspace.mit.edu/bitstream/handle/1721.1/148363/7-340-fall-2006/contents/lecture-notes/index.htm | Lecture Notes | Avoiding Genomic Instability: DNA Replication, the Cell Cycle, and Cancer | Biology | MIT OpenCourseWare
This is an archived course. A more recent version may be available at ocw.mit.edu.
Subscribe to the OCW Newsletter
Help|Contact Us
FIND COURSES
Find courses by:
Topic
MIT Course Number
Department
Collections
New Courses
Most Visited Courses
OCW Scholar Courses
Audio/Video Lectures
Online Textbooks
Supplemental Resources
OCW Highlights for High School
MITx & Related OCW Courses
MIT Open Learning Library
Cross-Disciplinary Topic Lists
Energy
Entrepreneurship
Environment
Introductory Programming
Life Sciences
Transportation
Translated Courses
繁體字 / Traditional Chinese
Türkçe / Turkish
(비디오)한국 / Korean
For Educators
Chalk Radio Podcast
OCW Educator Portal
Instructor Insights by Department
Residential Digital Innovations
OCW Highlights for High School
Additional Resources
Give Now
Make a Donation
Why Give?
Our Supporters
Other Ways to Contribute
Become a Corporate Sponsor
About
About MIT OpenCourseWare
Site Statistics
OCW Stories
Newsletter
Chalk Radio Podcast
Open Matters Blog
Search Tips
XExclude words from your search
Put - in front of a word you want to leave out. For example, jaguar speed -car
Search for an exact match
Put a word or phrase inside quotes. For example, "tallest building".
Search for wildcards or unknown words
Put a in your word or phrase where you want to leave a placeholder. For example, "largest in the world".
Search within a range of numbers
Put .. between two numbers. For example, camera $50..$100.
Combine searches
Put "OR" between each search query. For example, marathon OR race.
Home » Courses » Biology » Avoiding Genomic Instability: DNA Replication, the Cell Cycle, and Cancer » Lecture Notes
Lecture Notes
Course Home
Syllabus
Calendar
Readings
Lecture Notes
Assignments
Related Resources
Lecture notes table.| SES# | TOPICS | LECTURE SUMMARIES |
---
| 1 | Introduction | The instructors and students will introduce themselves, their backgrounds and reasons for participating in the course. Instructors will discuss the structure of the course, including requirements and assignments. The instructors will then present a short overview of cell-cycle regulation of DNA replication and the steps of DNA replication to give the students background information. Finally, we will briefly discuss the topic for next week. |
| 2 | The Cell Cycle | For cells to properly divide, each stage of the cell cycle must be controlled so one stage does not begin before the completion of the previous stage. The two papers assigned for this week describe experiments that resulted in major breakthroughs in our understanding of how cellular division is controlled. The first paper, Evans et al., describes the identification of cyclins, which are now known to be specificity subunits of Cyclin-Dependent Kinases (Cdks). This paper was the first example of how proteolysis might be used to control the cell cycle. The second paper, Shirodkar et al., gives insight into how cyclins control the stages of the cell cycle by showing that cyclins interact with transcription factors in a stage-specific manner. In particular, this paper studies the transcription factor E2F, which interacts with the tumor-suppressor protein Rb. |
| 3 | G1 Phase and Initiation of DNA Replication | Although the DNA in a cell is not actively replicated until S phase, the process of initiating replication is begun earlier in the cell cycle. During the G1 phase of the cell cycle, a pre-Replicative Complex (pre-RC) forms at multiple sites on each chromosome. These sites are known as origins. This complex is required to mark sites that are capable of initiating replication and for recruiting components of the replication machinery. The first of the two assigned papers, Donovan et al., describes experiments that show that proteins of the pre-RC assemble at origins in a specific order. The second paper, Mailand and Diffley, provides data for how the cell regulates the formation of pre-RCs by stabilizing one of the pre-RC components, Cdc6. |
| 4 | S Phase and Replication Fork Regression | During the S phase of the cell cycle, replication is initiated at origins, which result in bi-directional replication forks. All origins do not initiate at the same time; rather there is a temporal program controlling when each origin initiates. The data in the paper by Dimitrova and Gilbert show that even though the temporal program is carried out during S phase, the temporal constraints are established during G1. In order for bi-directional forks to replicate DNA, a number of proteins are necessary to unwind the DNA for the polymerases to carry out replication. The second paper, Marinsek et al, provides insight into the function of a recently identified complex of proteins, GINS, at the replication fork. |
| 5 | Cell Cycle Control of DNA Replication | Once the cell replicates the DNA, it is imperative that a second round of replication does not begin before the cell divides. Failure to prevent a second round of replication can result in genomic instability, which can lead to tumorigenesis. The first paper this week, Hayles et al., is one of the many papers that show that Cdk activity directed by cyclins (CDK activity) is important to inhibit DNA replication. The second paper, Nguyen et al., shows the mechanisms of how CDK activity can prevent a second round of DNA replication. |
| 6 | Checkpoints | Each time a cell divides, not only does the genome need to be duplicated, but the genome must be free of damage. DNA damage can result in genomic instability and chromosomal rearrangements, both of which can lead to cancer. If damage occurs, a checkpoint is activated, resulting in a reversible cell-cycle arrest. During this arrest period, the cell can repair the DNA before continuing with the remainder of the cell cycle. The first paper this week, Kuerbitz et al., shows that p53 is required for the cell to activate a G1 checkpoint. This paper was one of the first articles that defined a physiological role for the tumor-suppressor protein p53. In the second paper, Katou et al. describe experiments that show that two proteins, Mrc1 and Tof1, are responsible for maintaining replication forks (i.e. ensuring that the forks do not collapse) during the repair. |
| 7 | DNA Repair | This week, we will discuss mechanisms for how the cell repairs different types of damage. During replication, polymerases could encounter altered nucleotides, or lesions, that cause the replication fork to stall. Depending on the type of lesion, this stalled fork might activate a checkpoint to repair the damage. However, the lesion can be bypassed by switching the replicative polymerase for a low-fidelity polymerase. The mechanism for recruiting these other polymerases is studied in this week’s first paper by Bienko et al. Sometimes the damage is too great to simply be replicated over, such as a double-strand break (DSB). In the second paper, Spangolo et al. describe the structure of a complex of proteins necessary to repair DSBs and the implications for the mechanism of these proteins. |
| 8 | Midterm Assignments | Field trip to the Oncology Department of Novartis. |
| 9 | Development and Endoreduplication | Not only is DNA replication regulated each cell cycle, but it is also regulated differently during different developmental stages. In some organisms, specific cells purposely induce multiple rounds of DNA replication without an intervening mitosis, which results in multiple copies of the genome within a single cell. This process is called endoreduplication. Verkest et al. demonstrate that the cell controls the initiation of endoreduplication via CDK activity. Initiation of origins can be affected by local transcription activity, which can change throughout the development of an organism. The results in the second paper show that changes in the transcriptional program of the HoxB domain during development affect the site of initiation at a nearby origin. |
| 10 | Recombination and Meiosis | Meiosis is the specialized cell cycle that produces four haploid cells from a single diploid cell. To accomplish this goal, the cell undergoes a single round of replication followed by two consecutive rounds of division. During these two divisions, chromosomes must segregate properly such that each haploid cell has the correct number and type of chromosomes. During chromosome segregation, genetic recombination occurs. This week we will discuss how the cell coordinates pre-meiotic replication with recombination and how chromosomes are monitored to ensure that they segregate properly. |
| 11 | DNA Replication and Cancer | One hallmark of cancer is the uncontrolled division of cells. Thus, many of the genes that are mutated in cancer cells encode products that regulate the cell cycle. As we have discussed previously, cancer arises through DNA damage and often genomic instability, both of which can occur due to errors in DNA replication. Recent studies on the mechanisms that lead to cancer have shown that there is a link between control of DNA replication and the onset of cancer. The two papers assigned for this week examine the roles of pre-RC components and pre-RC formation in causing a cancerous state. |
| 12 | Targeting Replication Components as Chemotherapy | As we discussed during the last class, regulation of DNA replication is often disrupted in cancerous cells. This disruption can lead to inappropriate DNA replication, which might lead to the amplification of oncogenes and the onset of cancer. It is becoming clear that one mechanism of the action of chemotherapeutic drugs is to target DNA replication. This week we will read two papers that attempt to understand the mechanism of two different anticancer drugs and how these mechanisms are related to DNA replication. |
| 13 | Viral Reprogramming of the Cell Cycle | Viruses must infect host cells to reproduce, because viruses do not contain all the necessary components to propagate themselves. Once inside of the host cell, the virus co-opts much of the cellular machinery to ensure its survival. The papers this week examine how viruses can control a host cell by corrupting the cell cycle. The first paper this week, Yew and Berk, describes how one of the adenovirus proteins blocks p53 function, which can result in cancer. The second paper, Wiebusch et al., describes the way that human cytomegalovirus prevents host genome replication to ensure the replication of the viral genome. |
| 14 | Final Assignments Presented | Each student will give a 10-12 minute oral presentation of a paper related to one of the topics we discussed during the semester. The student will be responsible for giving a brief statement of the relevant background, explaining key figures and tables, briefly discussing the conclusions of the paper and proposing subsequent experiments that could be done to answer questions raised in the paper. There will be time after each presentation for a short discussion, generated by questions from other students. |
Find Courses
Find by Topic
Find by Course Number
Find by Department
New Courses
Most Visited Courses
OCW Scholar Courses
Audio/Video Courses
Online Textbooks
Instructor Insights
Supplemental Resources
MITx & Related OCW Courses
MIT Open Learning Library
Translated Courses
For Educators
Chalk Radio Podcast
OCW Educator Portal
Instructor Insights by Department
Residential Digital Innovations
OCW Highlights for High School
Additional Resources
Give Now
Make a Donation
Why Give?
Our Supporters
Other Ways to Contribute
Become a Corporate Sponsor
About
About OpenCourseWare
Site Statistics
OCW Stories
Newsletter
Open Matters Blog
Tools
Help & FAQs
Contact Us
Accessibility
Site Map
Privacy & Terms of Use
RSS Feeds
Our Corporate Supporters
About MIT OpenCourseWare
MIT OpenCourseWare is an online publication of materials from over 2,500 MIT courses, freely sharing knowledge with learners and educators around the world. Learn more »
© 2001–2018
Massachusetts Institute of Technology
Your use of the MIT OpenCourseWare site and materials is subject to our Creative Commons License and other terms of use. |
190722 | https://dominilash.ru/sale/besplatnaya_dostavka_ot_5000_rubley/ | Бесплатная доставка от 5000 рублей
Каталог
Ресницы
Черные
Коричневые/Горький шоколад
Цветные
Мультиколор
Неон
Однотоновые
Омбре
Клей
Препараты
Пинцеты и Ножницы
Расходные материалы
Доп Оборудование
Доставка и оплата
Оплата
Доставка
Гарантия и возврат
Блог
Контакты
Напиши Нам
Напиши Нам Whats APP
Заказать звонок
E-mail
написать в WhatsApp
Адрес
Адрес склада: Москва, Ул.Каспийская 22к1 .
Вход через магазин Пятерочка, направо, 2 этаж. (м.Царицыно)
Режим работы
Пн - Пт: с 10:00 до 18:00
Элита Lash индустрии
Каталог
Акустика
Музыка на любой вкус
Ресницы
Черные
Коричневые/Горький шоколад
Цветные
Мультиколор
Неон
Однотоновые
Омбре
Клей
Препараты
Пинцеты и Ножницы
Расходные материалы
Доп Оборудование
Каталог
По всему сайту
По каталогу
Войти
0 Избранное
0 Корзина
ЛиквидацияРесницыКлейПрепаратыПинцеты и НожницыДополнительное оборудование
Каталог
Каталог
По всему сайту
По каталогу
Войти
0 Избранное
0 Корзина
Телефоны
Напиши Нам Whats APP
Заказать звонок
E-mail
написать в WhatsApp
Адрес
Адрес склада: Москва, Ул.Каспийская 22к1 .
Вход через магазин Пятерочка, направо, 2 этаж. (м.Царицыно)
Режим работы
Пн - Пт: с 10:00 до 18:00
0
0
Кабинет
0 Избранное
0 Корзина
Бренды
Акции
Блог
Контакты
Напиши Нам Whats APP
Назад
Телефоны
Напиши Нам Whats APP
Заказать звонок
написать в WhatsApp
Адрес склада: Москва, Ул.Каспийская 22к1 .
Вход через магазин Пятерочка, направо, 2 этаж. (м.Царицыно)
Пн - Пт: с 10:00 до 18:00
Темная тема Светлая тема
Интернет-магазин материалов для наращивания ресниц в Москве
Акции
Бесплатная доставка от 5000 рублей
Бесплатная доставка от 5000 рублей
ВКонтакте
Одноклассники
Мой Мир
Twitter
Viber
WhatsApp
Telegram
31 июля 2025
Наши специалисты ответят на любой интересующий вопрос по услуге
Написать сообщение
Мы хотим, чтобы ваши покупки в нашем магазине были еще более приятными и выгодными. Поэтому мы берем на себя все расходы по доставке, если сумма вашего заказа превышает 5 000 ₽.
Как это работает? Это очень просто!
Вам не нужно вводить промокоды или выполнять сложные действия. Наша система все сделает за вас.
Наполните корзину. Выбирайте все, что вам нравится, и добавляйте в корзину.
Достигните нужной суммы. Как только общая стоимость товаров в корзине превысит 5 000 ₽...
Получите бесплатную доставку! Стоимость доставки автоматически обнулится при оформлении заказа. Вы увидите это в строке "Доставка".
Условия акции
Мы сделали условия максимально прозрачными, чтобы у вас не осталось вопросов.
Сумма заказа: Бесплатная доставка применяется к заказам, итоговая сумма которых после применения всех скидок и промокодов составляет 5 000 ₽ или более.
География: Акция действует для всех заказов с доставкой по всей России.
Способ доставки: Бесплатной становится тип доставкиСДЭК (Самовывоз из пункта выдачи). На другие способы доставки (например, курьер, экспресс или авиа) акция не распространяется.
Сроки проведения: Акция действует на постоянной основе. Мы хотим, чтобы выгодные покупки были доступны вам всегда.
Часто задаваемые вопросы (FAQ)
Мой заказ на сумму 4 900 ₽. Могу ли я получить бесплатную доставку?
К сожалению, нет. Система автоматически применяет бесплатную доставку только к заказам на сумму ровно 5 000 ₽ и выше. Возможно, стоит добавить в корзину полезный аксессуар или сопутствующий товар?
Я применил скидку, и сумма стала меньше 5 000 ₽. Доставка будет бесплатной?
Нет, бесплатная доставка рассчитывается от финальной суммы к оплате. Если после применения скидочного купона итоговая сумма стала меньше 5 000 ₽, то доставка будет платной.
Как я пойму, что доставка стала бесплатной?
На странице оформления заказа, в блоке с расчетом итоговой стоимости, вы увидите, что цена доставки равна "0 ₽" или "Бесплатно".
Эта акция суммируется с другими скидками?
Да, вы можете использовать скидочные промокоды или покупать товары по распродаже. Главное, чтобы итоговая сумма к оплате была не менее 5 000 ₽.
Время для выгодных покупок!
Выберите свои идеальные товары прямо сейчас, а мы позаботимся о том, чтобы доставить их вам бесплатно.
Начать покупки
Товары по акции
Быстрый просмотр
Розничная цена
1 800 ₽
Оптовая от 3000 руб.
1 440 ₽
Оптовая от 7000 руб.
1 260 ₽
Пинцет DOMINI №7(зеркальный)"для выделения, с изгибом" профессиональная заточка
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
1 800 ₽
Оптовая от 3000 руб.
1 440 ₽
Оптовая от 7000 руб.
1 260 ₽
Пинцет DOMINI №6(зеркальный)"Прямой удлиненный" профессиональная заточка
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
1 800 ₽
Оптовая от 3000 руб.
1 440 ₽
Оптовая от 7000 руб.
1 260 ₽
Пинцет DOMINI №5(зеркальный)"Прямой с односторониим заужением" профессиональная заточка
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
1 800 ₽
Оптовая от 3000 руб.
1 440 ₽
Оптовая от 7000 руб.
1 260 ₽
Пинцет DOMINI №9(зеркальный)"Мини сапожок" профессиональная заточка
5
2
В наличии
В корзину
Быстрый просмотр
Розничная цена
650 ₽
Оптовая от 3000 руб.
520 ₽
Оптовая от 7000 руб.
455 ₽
Ремувер крем-гель DOMINI без аромата, 15 гр
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
2 500 ₽
Оптовая от 3000 руб.
2 000 ₽
Оптовая от 7000 руб.
1 750 ₽
Пинцет DOMINI №1 "Денежный" профессиональная заточка.
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
1 800 ₽
Оптовая от 3000 руб.
1 440 ₽
Оптовая от 7000 руб.
1 260 ₽
Пинцет DOMINI №4(зеркальный)"L 45°" профессиональная заточка
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
650 ₽
Оптовая от 3000 руб.
520 ₽
Оптовая от 7000 руб.
455 ₽
Обезжириватель DOMINI "Персик",15 мл
5
3
В наличии
В корзину
Быстрый просмотр
Розничная цена
80 ₽
Оптовая от 3000 руб.
64 ₽
Оптовая от 7000 руб.
56 ₽
Скотч бумажный
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
200 ₽
Оптовая от 3000 руб.
160 ₽
Оптовая от 7000 руб.
140 ₽
Щеточки нейлон (50 штук)
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
150 ₽
Оптовая от 3000 руб.
120 ₽
Оптовая от 7000 руб.
105 ₽
Лента пластиковая
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
15 ₽
Оптовая от 3000 руб.
12 ₽
Оптовая от 7000 руб.
10.50 ₽
Лунка для клея
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
150 ₽
Оптовая от 3000 руб.
120 ₽
Оптовая от 7000 руб.
105 ₽
Микробраши(100шт.)
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
2 500 ₽
Оптовая от 3000 руб.
2 000 ₽
Оптовая от 7000 руб.
1 750 ₽
Пинцет DOMINI №13 "Денежный" лазерная заточка.
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
2 500 ₽
Оптовая от 3000 руб.
2 000 ₽
Оптовая от 7000 руб.
1 750 ₽
Пинцет DOMINI №12 "Денежный" лазерная заточка.
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
2 500 ₽
Оптовая от 3000 руб.
2 000 ₽
Оптовая от 7000 руб.
1 750 ₽
Пинцет DOMINI №13 "Денежный" профессиональная заточка.
5
1
В наличии
В корзину
Быстрый просмотр
Розничная цена
2 500 ₽
Оптовая от 3000 руб.
2 000 ₽
Оптовая от 7000 руб.
1 750 ₽
Пинцет DOMINI №11 "Денежный" профессиональная заточка.
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
2 500 ₽
Оптовая от 3000 руб.
2 000 ₽
Оптовая от 7000 руб.
1 750 ₽
Пинцет DOMINI №10 "Денежный" профессиональная заточка.
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
2 500 ₽
Оптовая от 3000 руб.
2 000 ₽
Оптовая от 7000 руб.
1 750 ₽
Пинцет DOMINI №8 "Денежный" профессиональная заточка.
0
0
В наличии
В корзину
Быстрый просмотр
Розничная цена
2 500 ₽
Оптовая от 3000 руб.
2 000 ₽
Оптовая от 7000 руб.
1 750 ₽
Пинцет DOMINI №5 "Денежный" профессиональная заточка.
5
1
В наличии
В корзину
Назад к списку
О компании
Блог
Оплата и Доставка
ОПТ и сотрудничество
Каталог
Ресницы
Клей
Препараты
Пинцеты и Ножницы
Расходные материалы
Доп Оборудование
Связаться с нами
Напиши Нам
Напиши Нам Whats APP
Заказать звонок
E-mail
написать в WhatsApp
Адрес
Адрес склада: Москва, Ул.Каспийская 22к1 .
Вход через магазин Пятерочка, направо, 2 этаж. (м.Царицыно)
Режим работы
Пн - Пт: с 10:00 до 18:00
написать в WhatsApp
Адрес склада: Москва, Ул.Каспийская 22к1 .
Вход через магазин Пятерочка, направо, 2 этаж. (м.Царицыно)
Подписаться
на новости и акции
Продолжая, вы соглашаетесь с политикой конфиденциальности - [x]
© 2025 DOMINI LASH
Темная тема Светлая тема
Быстро с 1С-Битрикс
ГлавнаяКаталог0 Корзина0 ИзбранныеКабинетАкцииКонтактыУслугиБрендыОтзывыКомпанияЛицензииДокументыРеквизитыПоискБлогОбзоры
Файлы cookie
Мы используем файлы cookie, разработанные нашими специалистами и третьими лицами, для анализа событий на нашем веб-сайте, что позволяет нам улучшать взаимодействие с пользователями и обслуживание. Продолжая просмотр страниц нашего сайта, вы принимаете условия его использования. Более подробные сведения смотрите в нашей Политике в отношении файлов Cookie.
Принимаю |
190723 | https://tutorial.math.lamar.edu/problems/calciii/3DCoords.aspx | Paul's Online Notes
Go To
Notes
Practice Problems
Assignment Problems
Show/Hide
Show all Solutions/Steps/etc.
Hide all Solutions/Steps/etc.
Sections
3-Dimensional Space Introduction
Equations of Lines
Chapters
Partial Derivatives
Classes
Algebra
Calculus I
Calculus II
Calculus III
Differential Equations
Extras
Algebra & Trig Review
Common Math Errors
Complex Number Primer
How To Study Math
Cheat Sheets & Tables
Misc
Contact Me
MathJax Help and Configuration
Paul's Online Notes
Home
/
Calculus III
/
3-Dimensional Space
/ The 3-D Coordinate System
Show Mobile Notice
Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 12.1 : The 3-D Coordinate System
Give the projection of P=(3,−4,6) onto the three coordinate planes. Solution
Which of the points P=(4,−2,6) and Q=(−6,−3,2) is closest to the yz-plane? Solution
Which of the points P=(−1,4,−7) and Q=(6,−1,5) is closest to the z-axis? Solution
For problems 4 & 5 list all of the coordinates systems (R, R2, R3) that the given equation will have a graph in. Do not sketch the graph.
7x2−9y3=3x+1 Solution
x3+√y2+1−6z=2 Solution
| | |
--- |
| | | |
190724 | https://schoolnutrition.org/journal/fall-2020-a-qualitative-analysis-of-students-breakfast-habits-and-school-breakfast-participation-in-two-public-school-districts/ | Skip to content
FacebookTwitterLinkedinYoutubeInstagram
Abstract
Methods
A convenience sample of students in grades 1-12 from two public school districts in Connecticut participated in focus groups. Each of the 13 focus groups included students from two or three grade levels and was facilitated by a lead moderator and note-taker. Recorded discussions were transcribed with students’ comments coded to identify common themes. Within each theme, the grade levels of participants were identified to determine whether the theme represented comments by elementary school (grades 1-6), middle school (grades 7-8), and/or high school
(grades 9-12) participants.
Results
A total of 105 students (57% female, 43% male) participated. Across all grade levels, hunger, or a desire to avoid its adverse effects (e.g., grumpiness, headache), was identified as a consistently mentioned reason for eating breakfast. Other reasons included parental influence (elementary students), desire to improve energy (elementary students), and desire to improve school performance (elementary and middle school students). Breakfast skipping was often reported by middle and high school students, who cited inadequate time and lack of hunger as reasons. Across all grade levels, reasons for not participating in the SBP included distaste for, distrust of, and/or cost of school foods. Inconvenience of eating at school and perception of school foods as unhealthy or “junk” were cited by elementary and middle school students. Suggestions to improve school breakfast included more choices (all grades), improved freshness or quality of foods (all grades), and healthier options (elementary and middle school students).
Application To Child Nutrition Professionals
Emphasizing food quality and increasing convenience of school breakfast may help to increase SBP participation.
Full Article
Evidence from observational (Nicklas et al., 1993; Ptomey et al., 2016) and experimental studies (Wyon et al., 1995; Adolphus et al., 2013) supports the importance of breakfast for dietary intake and academic performance. Children who regularly eat breakfast may also have healthier body weights (Affenito et al., 2005; Fiore et al., 2006) and reduced insulin resistance (Donin et al., 2014; Karatzi et al., 2014). School Breakfast Programs (SBPs) can have positive effects on academic performance (Adolphus et al., 2013; Kleinman et al., 2002), diet quality (Ritchie et al., 2016), and school attendance (Anzman-Frasca et al., 2015). Despite these benefits of breakfast consumption, breakfast skipping is common among school-aged children (Dykstra et al., 2016; Hearst et al., 2016). Participation in the United States Department of Agriculture’s (USDA) National SBP is low, relative to participation in the National School Lunch Program (Food Research & Action Center, 2018). According to the School Nutrition Dietary Assessment Study-IV, factors associated with SBP participation included: lower grade level; eligibility for free or reduced-price meals; and lower meal price among students ineligible for free or reduced-price meals (USDA, 2012). However, few studies have examined determinants of breakfast consumption in general (Olsta, 2013; Reddan et al., 2002; Sweeney & Horishita, 2005), or participation in SBPs in particular (Sabol et al., 2011; Bailey-Davis et al., 2013) among U.S. children and adolescents. This qualitative study was designed to identify the factors influencing breakfast consumption and SBP participation among students in grades 1-12 in two Connecticut school districts.
Methods
Sample
Four school districts were invited to participate. Superintendents of two districts (referred to as Districts A and B) granted approval. The study was approved by the Griffin Hospital Institutional Review Board. In 2017, 12 focus groups were held with 11 in District A (nine in elementary schools, one in a middle school, and one in a high school) and one focus group in District B’s middle school. Characteristics of school districts and study participants are presented in Table 1. Students were included if they: (1) were enrolled in a participating school; (2) were invited by a teacher or administrator; (3) provided signed parental consent and their written or verbal assent, as appropriate for age; and (4) were present in school on the day of the focus group.
Data Collection
School administrators asked teachers to identify students who were willing to share their opinions and diverse in race/ethnicity, gender, and academic performance. Each focus group included 5 to 11 students from two or three grade levels and ranged in length from 30 to 60 minutes. Two researchers served as facilitator and note-taker, respectively. They jointly developed an interview guide based on a literature review and input from the districts’ school foodservice directors. Key questions included:
If you eat/don’t eat breakfast every day or almost every day, why/why not?
If you eat breakfast on days that you go to school, do you usually eat breakfast at home or at school? Why?
How could your school make school breakfast better?
Follow-up questions varied depending on students’ responses. Sessions were audio recorded and transcribed. Each participant’s grade level and gender were noted, but not race or ethnicity, to minimize the collection of personal information.
Data Analysis
NVivo 11 software was used to code transcripts. Two independent coders, including the focus group facilitator, assigned codes to participant statements based on their content. Codes generated by each coder were compared and discussed until agreement was reached on a final set of codes. Each coder then re-coded the transcripts using the established codebook. Codes were compared again, with any discrepancies reviewed and resolved, and collapsed into themes based on how they related to our research questions. Responses in each theme were reviewed to determine whether some categories were populated more by comments from students in elementary (grades 1-6), middle (grades 7-8), or high school (grades 9-12).
Results and Discussion
Themes
A total of 105 students participated. Distributions by grade level and gender are presented in Table 1.
| | | |
---
| Table 1. School District and Participant Characteristics (N=105) | | |
| | District A, n (%) | District B, n (%) |
| District Characteristics (2016-2017) | | |
| Number of Schools | 8 | 6 |
| Total Enrollment | 4,815 | 2,364 |
| White | 3,507 (72.8) | 951 (40.2) |
| Black or African American | 243 (5.0) | 428 (18.1) |
| Hispanic or Latino | 604 (12.5) | 828 (35.0) |
| Eligible for Free or Reduced-Price Meals | 1,37 (21.5) | 1,597 (67.6) |
| Overall SBP Participation Rate (%) | 14.0 | 76.4 |
| Free/Reduced SBP Participation Rate (%) | 35.8 | 72.6 |
| School Breakfast Program Features | | |
| Chef involved in meal program | No | No |
| Breakfast in the Classroom | No | No |
| Grab-and-Go Breakfast Carts | Elementary schools only | No |
| Second-Chance Breakfast/Extended Service Hours | High school only | No |
| Focus Group Participant Characteristics | | |
| Total participants | 99 | 6 |
| Grades 1-4 | 67 (67.7) | 0 (0.0) |
| Female, n (%) within grades 1-4 | 34 (50.7) | 0 (0.0) |
| Grades 5-8 | 11 (11.1) | 6 (100.0) |
| Female, n (%) within grades 5-8 | 7 (63.6) | 4 (66.7) |
| Grades 9-12 | 21 (21.2) | 0 (0) |
| Female, n (%) within grades 9-12 | 15 (71.4) | 0 (0) |
Sources: EndHunger CT, 2018; Connecticut Department of Education, 2018
Students’ comments were organized into four themes: (1) reasons for eating breakfast, (2) reasons for skipping breakfast, (3) reasons for not participating in the SBP, and (4) opportunities to improve school breakfast. Within each theme, subthemes were identified (14 total). The themes and subthemes summarized below are outlined in Table 2, which includes examples of quotes related to each theme.
| | | | |
--- --- |
| Table 2. Representative Quotations Regarding Breakfast Consumption and School Breakfast | | | |
| Themes and Subthemes | REPRESENTATIVE QUOTATIONS | | |
| | Elementary School (Grades 1-6) | Middle School (Grades 7 & 8) | High School (Grades 9-12) |
| Reasons for Eating Breakfast | | | |
| Parental Influence | I always eat breakfast because my dad says breakfast is the most important meal of the day. (female, grade 2) | — | — |
| Hunger/To Avoid Physical Symptoms | Like for me, whenever I don’t eat breakfast, my stomach gets upset and I can’t focus as well. (female, grade 6) | I always make sure to have something in the morning because around like, 4th period, my stomach starts to growl if I don’t. It’s really embarrassing. (female, grade 8) | ‘cus then [if I’m hungry and don’t eat], I like, get grumpy. It’s never good for me. (female, grade 10) |
| Desire to Improve Energy Level | The reason I get breakfast every day is ‘cause I feel like it just energizes me…and I feel like it gets me ready for the day. (male, grade 3) | — | — |
| Desire to Improve School Performance | Whenever I don’t eat breakfast, it’s the only thing I’m thinking about. I can’t think about what I’m supposed to be learning in class and yeah. It becomes a struggle so I feel that breakfast should be an important part of your morning. (male, grade 5) | ‘Cause [being hungry in class] distracts me from my work. I keep thinking about food and stuff, wanting to eat something, so I make sure I’m always fed. (male, grade 8) | — |
| | | | |
--- --- |
| Table 2. Representative Quotations Regarding Breakfast Consumption and School Breakfast (cont.) | | | |
| Themes and Subthemes | REPRESENTATIVE QUOTATIONS | | |
| | Elementary School (Grades 1-6) | Middle School (Grades 7 & 8) | High School (Grades 9-12) |
| Reasons for Skipping Breakfast | | | |
| Lack of Hunger | — | I feel like it’s the time I wake up, also, because on the weekends and if I were usually to just wake up on my own time, I wouldn’t wake up ‘til probably like, an hour from now. So I feel that when its 6:30 in the morning I’m so focused on actually trying to wake myself up that I’m just not hungry or think I’m not hungry. I’m just not focused on that. (female, grade 8) | It’s the time thing. Just getting up so early, I’m half asleep. I don’t even want to eat. (female, grade 12) |
| Inadequate Time | — | I do it when I have time, because most likely I’m running late when I’m going to school. (female, grade 8) | I just don’t have the time. (male, grade 12) |
| Reasons for Not Participating in School Breakfast | | | |
| Distaste for School Foods | And, like, sometimes, the food that’s at the school, I don’t like it so, and the food that’s at my house, I like it, so that why I eat at my house. (female, grade 1) | I find [school breakfast] kind of disgusting. (male, grade 7) | The bagels here are soggy. They keep them in a bag and put them in the microwave. They get really moist. (male, grade 12) |
| Cost of School Foods | If there’s cereal that you don’t like and you have cereal that you do like at home, you’re gonna waste money for nothing. (male, grade 3) Once my brother got a lot of breakfasts and it costs way more money, and like, my parents had to bring in, like $30 every week, ‘cus he got breakfast and it was really bad. Now we just if we, like, don’t have any more time to eat breakfast [at home]. (female, grade 1) | Yeah, I found disgusting things [in my school lunch] and plus school lunch costs money too, you know. (female, grade 8) | So, like, the smoothies here, I think they charge $3. (female, grade 10) |
Table 2. Representative Quotations Regarding Breakfast Consumption and School Breakfast (cont.)
| | | | |
--- --- |
| Themes and Subthemes | REPRESENTATIVE QUOTATIONS | | |
| | Elementary School (Grades 1-6) | Middle School (Grades 7 & 8) | High School (Grades 9-12) |
| School Foods are Unhealthy or Junk | ‘Cus they have really bad stuff here. Like Poptarts, they have these.. I mean I’m sure they could have cereal balls.. or I mean bars, but they could have cinnamon buns. All of these things you don’t, like… apple strudel.. like, they’re all desserts. They’re not really breakfast. (male, grade 4) | I feel like I’m looking for something that will actually benefit my body. I obviously want to feed my body good foods. I ‘m not into feed it junk that’s not gonna benefit it whatsoever and it’s just gonna tear it down in a sense. Like, I want something that is fresh and is appealing and it’s also gonna benefit me throughout my day and like, throughout my lifetime. I want to have something healthy, I feel like, and I feel that, although it is easier to get the packaged food that you could heat up in the microwave, there’s so many bad things in them, I feel like, and I don’t think I want to put that in my body. And that’s why most kids don’t want to eat breakfast here. (female, grade 8) | — |
| Distrust of School Foods | So once, this person told me that once he found that his chicken nugget was raw and the other one was overdone. (female, grade 3) | We’ve had expired, we’ve had expired milk there and it smells disgusting we don’t drink it. Cause that could get us sick. And then on top of that eating food that is from a freezer that could get us sick too. (female, grade 8) We don’t trust it much. We don’t trust the school lunches. (female, grade 8) | Yeah I trust the food at my house like 100% more than the food here. Like, I won’t mess with the food here. (female, grade 12) I’m pretty sure prison food is safer than school food. (male, grade 12) |
| Inconvenient to Eat at School | I have breakfast at home because it’s a lot quicker than here, because there’s normally a line. (female, grade 4) | It’s like your morning rush hour, even though it’s like a small 15 minute slot, that’s basically when you have to prepare for your day and you have to get everything done and completed. Like you have to talk to your teachers and I hate to say it’s one of the most social points of the day but you get to see your friends in the morning and obviously everyone wants to talk to them, because you’re not gonna see | — |
Table 2. Representative Quotations Regarding Breakfast Consumption and School Breakfast (cont.)
| | | | |
--- --- |
| Themes and Subthemes | REPRESENTATIVE QUOTATIONS | | |
| | Elementary School (Grades 1-6) | Middle School (Grades 7 & 8) | High School (Grades 9-12) |
| | | them all day. And I feel that most kids are more distracted with the things they prioritize above than breakfast than actually making their way down to get breakfast. (female, grade 8) | |
| Opportunities for Improvement in School Breakfast | | | |
| More Choices | They kind of just have the same things, like carrots, oranges and apples and bananas. Like, some people don’t like those so I think they should have like, strawberries and olives and grapes. Like, fruits and vegetables. Like, more variety of those two, so I’d just like a bigger variety of options. (female, grade 4) | Plus I think they should give us a variety of bagels instead of just one kind, you have to eat this kind, and varieties of cream cheese.(female, grade 8) | They’re always saying that we don’t have enough money. There’s not enough in the budget to pay for things, but if we had the variety of food that we want, then we would put our money towards that and actually get that. (female, grade 10) |
| Fresher, Real, Better Foods | k they should have eggs but not like… they are all soggy ‘cus they’re shipped so like, if they cooked them there it would taste better. (male, grade 4) | [The school should serve] Real meat, not fake meat; not plastic cheap cheese, real cheese. (male, grade 7) We were learning about processed foods and stuff and it gives us a perfect example about school foods, about how it sits in boxes inside of a freezer for months. And they only cook 6 out of 36 meals for us and they have all these different meals here and they could make something else good besides their everyday thing instead. (female, grade 8) | They could have the same menu, but just have better quality stuff. Like get fresher food and higher quality ingredients and like, this trash that they give us. (male, grade 12) I think, part of it is that they make stuff that’s kind of real food, but like, like they want you to be able to take it to class, kind of, because we don’t have time to like, sit and eat real food. But like, we just came from study hall so we would have time, if they had good breakfast food, we could actually like, sit and actually eat a real breakfast. (male, grade 12) |
| Healthier Options | I think they should take away all the junk food and put in more healthy stuff. (male, grade 1) | I know that they serve, like, brownie bars and if they were looking for something packaged that was easy to give a lot of kids, I think a better | — |
Table 2. Representative Quotations Regarding Breakfast Consumption and School Breakfast (cont.)
| | | |
---
| Themes and Subthemes | REPRESENTATIVE QUOTATIONS | |
| Elementary School (Grades 1-6) | Middle School (Grades 7 & 8) | High School (Grades 9-12) |
| alternative would be like a granola bar instead of a brownie, ‘cus it’s more nutritious even though it’s not the most nutritious. (female, grade 8) | | |
| [My friend] gets breakfast, and he gets, like, junk food all the time, like brownies and cinnamon rolls and I think they should change it to, like, healthy stuff. (male, grade 2) | | |
Note. Cells are left blank when a theme does not apply to the grade level.
Reasons For Eating Breakfast
Most elementary school students reported eating breakfast every day or almost every day. Many middle and high school students reported not eating breakfast regularly. Across grade levels, the most commonly mentioned reasons for eating breakfast were: parental influence, hunger, desire to avoid physical symptoms (e.g., fatigue, upset stomach, headache), and desire to improve school performance. Parental influence and/or desire to improve energy levels were important among elementary students.
Reasons For Skipping Breakfast
Main reasons for skipping breakfast, as reported by middle and high school students, included lack of hunger and inadequate time. Younger children generally reported feeling hungry and wanting to eat in the morning, whereas older children more often said they were not hungry. Many middle and high school students who skipped breakfast described feeling rushed and having little time to eat breakfast on school days.
Reasons For Not Participating In Sbps
More students preferred eating breakfast at home rather than at school. The top reasons were distaste for and/or distrust of school foods, followed by cost. Many also mentioned inadequate time to eat at school, and perceived school food as unhealthy or “junk.”
Dislike of school foods was frequently noted in generic terms (e.g., “I don’t really like the food”). Specific reasons for disliking school foods included a perception that they were low in quality, less good than foods at home, or “fake.” Students at all grade levels were concerned about the cost of school meals. Many students in elementary schools, and some in middle schools, felt that school breakfast foods were unhealthy. In contrast, high school students expressed less concern about the healthfulness of school breakfast. Some said that efforts to provide healthful foods such as whole grain products were “a turn off” and that students should be able to decide whether or not to eat healthfully at school.
Middle and high school students frequently expressed a distrust of school foods. Paradoxically, although some complained about schools’ heavy reliance on packaged foods, they also often chose these foods at school because they seemed “safer.” Students at all grade levels shared stories of discovering non-food items in in their foods (e.g. hair, plastic, a bug, “pink fuzz” resembling insulation) and discovering that items were expired or spoiled. Among the food quality complaints were incidents of curdled milk, undercooked cheeseburgers, and discolored salami.
For elementary and middle school students, inconvenience was also a barrier to SBP participation. Some students reported feeling too rushed to eat at school, while others said they had more important things to do, such as talking to teachers or friends or getting organized for class. Students also mentioned late buses, cafeteria lines, long walking distances to the foodservice area, and a perception that eating in the classroom could be messy.
While some students said they would eat school breakfast when their preferred foods were served (e.g. breakfast sandwiches or “brownie bars”), more students reported eating breakfast at school when hungry and lacking other options due to time constraints at home. However, some students made positive comments about school breakfast. For example, a female student in grade 4 said: “Everyone thinks [the apple strudel] disgusting, but I think it’s really good….And then, ometimes, they’ll have new things, so I mean, whatever there is, I usually just have it.”
Opportunities To Improve School Breakfast
A common suggestion for improvement was greater variety of food choices. Students expressed a desire for more options (e.g., varieties of fruits, bagels, and beverages) and opportunities to customize meals (e.g., by selecting breakfast sandwich or bagel toppings). Many students, especially in middle and high school, prioritized “freshly made,” “cooked,” “good quality,” “real” foods over those that were “packaged,” “frozen,” or “wrapped.” Several elementary and middle school students wanted healthier options, such as fresh fruits not already offered (e.g., melon, berries, mango), Cheerios, and eggs.
Discussion Of Findings
Prior qualitative studies of students in grades 4 and 5 in Alabama (Sabol et al., 2011) and grades 6 to 8 in Philadelphia (Bailey-Davis et al., 2013) suggested that SBP participation was influenced by: belief that breakfast is important; dislike of foods offered; cost; stigma; and time. Sabol et al. and Bailey-Davis et al. reported that students ate breakfast to improve concentration or prevent feeling sick, and that distrust of school food was prevalent. These findings are similar to ours, with the exceptions of themes concerning cost and stigma. Sabol et al. reported cost as a concern for parents, but not students. In our study, students themselves cited cost as a barrier to SBP participation. Sabol et al. and Bailey-Davis et al. found that stigma associated with SBP participation was important, but this issue was absent from all focus group discussions in our study. Our findings also generally agreed with those of previous quantitative surveys that reported lack of time and lack of hunger as reasons for skipping breakfast (Olsta, 2012; Reddan et al., 2002; Shaw, 1998; Sweeny & Horishita, 2005) and belief that breakfast provides energy and improves ability to pay attention in school (Reddan et al., 2002).
Conclusions And Application
While our results are consistent with prior research on breakfast consumption and SBP participation among school-aged children, they also provide new information about potential commonalities and differences between age groups.
Across grade levels, our study participants perceived that school meals were often made from processed foods and were low in quality. They wanted more foods to be cooked from scratch, a recommendation similar to those made in previous focus group studies with high school students (Asada et al., 2017; James, Rienzo, & Frazee, 1996) Prior studies designed to improve staff culinary training and increase scratch cooking have shown associations with improved nutrient profiles of school meals served (Cohen et al., 2012; Schober et al., 2016), Although scratch cooking requires more labor, one study suggested that total food and labor costs of scratch-cooked entrees may not exceed those of entrees requiring no scratch cooking (Woodward-Lopez et al., 2014).
Another consistent theme across age groups was lack of time or lack of hunger as a reason for skipping breakfast at the middle and high school level. Breakfast was served at “grab-and-go” carts in elementary schools in District A, but the cafeteria was the only place where middle and high school students could get breakfast. School foodservice administrators could consider adopting other school breakfast service models, such as longer service hours and/or breakfast in the classroom. Among high school students, “grab-and-go” carts (Larson et al., 2018; Olsta, 2013) and expanded cafeteria hours (Olsta, 2013) have increased SBP participation rates. Notably, while high school students in our study did have access to breakfast over an extended time, many reported not taking advantage of it, suggesting that addressing the timing alone may be insufficient to increase SBP participation. When planning “grab-and-go” meals, it is important to recognize that students may still expect these meals to be made onsite using fresh foods.
Lack of time and hunger in the morning among middle and high school students may be related to early school start times and changes in circadian rhythms that occur during puberty
(Hagenauer et al., 2009). A delay of 1-3 hours in the circadian phase and timing of sleep in adolescents has been observed, peaking between the ages of 15 and 21 (Hagenauer et al., 2009). For many teens, just as they naturally shift their sleep later in the evening, they are required to be at school earlier in the morning. The high school attended by our participants in District A started at 7:40am, whereas elementary schools started at 9:00am. The impact of later school start times on SBP participation has not yet been investigated and could be explored in future research.
School nutrition professionals may use the findings of our study to inform their SBP marketing efforts. For example, messaging to elementary school students and their parents should emphasize the importance of breakfast to “start the day” and the nutritional value of school breakfasts. On the other hand, when targeting middle and high school students, it may be more effective to de-emphasize nutrition and to instead highlight the freshness and quality of foods. When cooking foods to order is not feasible, school nutrition professionals may consider assembling items at the time of purchase or offering students the opportunity to choose their own sandwich or salad toppings.
Due to the nature of focus group research, this study had inherent limitations. Although school administrators and teachers were asked to invite a representative group of students to participate, participants may have differed in important ways from other students. Additionally, some participants were more engaged than others in the discussions; therefore, their opinions are overrepresented in the analysis.
While the results of our study cannot be generalized to other settings and populations, when considered in conjunction with the strikingly similar findings of other studies, they can be used to inform practices in SBPs. Attention to student perceptions of food quality and freshness should be paramount at every grade level. While participants’ comments in this study were not always factually correct—for example, claims that the food was “fake” or that meat was undercooked—they reflect a pervasive perception among students about school food that influences their likelihood of participating in school meal programs. Making school breakfast more convenient and ensuring adequate time for students to eat when hungry (which may be later for older students) are also strategies that could reduce barriers, but these strategies alone will be insufficient without also addressing students’ negative attitudes towards school meals.
References
Adolphus, K., Lawton, C. L., & Dye, L. (2013). The effects of breakfast on behavior and academic performance in children and adolescents. Frontiers in Human Neuroscience, 7, 425. doi: 10.3389/fnhum.2013.00425
Affenito, S. G., Thompson, D. R., Barton, B. A., Franko, D. L., Daniels, S. R., Obarzanek,
E.,Schreiber, G. B., & Striegel-Moore, R. H. (2005). Breakfast consumption by AfricanAmerican and white adolescent girls correlates positively with calcium and fiber intake
and negatively with body mass index. Journal of the American Dietetic Association,
105(6), 938-945. doi: 10.1016/j.jada.2005.03.003
Anzman-Frasca, S., Djang, H. C., Halmo, M. M., Dolan, P. R., & Economos, C. D. (2015).
Estimating impacts of a breakfast in the classroom program on school outcomes. JAMA
Pediatrics, 169(1), 71-77. doi: 10.1001/jamapediatrics.2014.2042
Asada, Y., Hughes, A. G., Read, M., Schwartz, M. B., & Chriqui, J. F. (2017). High school
students’ recommendations to improve school food environments: Insights from a critical
stakeholder group. Journal of School Health, 87(11), 842-849. doi: 10.1111/josh.12562
Bailey-Davis, L., Virus, A., McCoy, T. A., Wojtanowski, A., Vander Veur, S. S., & Foster, G. D.
(2013). Middle school student and parent perceptions of government-sponsored free
school breakfast and consumption: A qualitative inquiry in an urban setting. Journal of
the Academy of Nutrition and Dietetics, 113(2), 251-257.
Cohen, J. F., Smit, L. A., Parker, E., Austin, S. B., Frazier, A. L., Economos, C. D., & Rimm, E.
B. (2012). Long-term impact of a chef on school lunch consumption: Findings from a 2-
year pilot study in Boston middle schools. Journal of the Academy of Nutrition and
Dietetics, 112(6), 927-933. doi: 10.1016/j.jand.2012.01.015
Connecticut Department of Education. (2018). District Profile and Performance Reports for
Academic Year 2016-2017. Retrieved May 13, 2020 from
Donin, A. S., Nightingale, C. M., Owen, C. G., Rudnicka, A. R., Perkin, M. R., Jebb, S. A.,
Stephen, A., M., Sattar, N., Cook, D. G., & Whincup, P. H. (2014). Regular breakfast
consumption and type 2 diabetes risk markers in 9- to 10-year-old children in the Child
Heart and Health Study in England (CHASE): A cross-sectional analysis. PLoS
Medicine, 11(9), e1001703. doi: 10.1371/journal.pmed.1001703
Dykstra, H., Davey, A., Fisher, J. O., Polonsky, H., Sherman, S., Abel, M. L., Dale, L., C.,
Foster, G., D., & Bauer, K. W. (2016). Breakfast-skipping and selecting low-nutritionalquality foods for breakfast are common among low-income urban children, regardless of
food security status. Journal of Nutrition, 146(3), 630-636. doi: 10.3945/jn.115.225516
End Hunger Connecticut. (2018). 2018 Connecticut School Breakfast Report Card. Retrieved
May 13, 2020, from
2017-School-Breakfast-Report-Card-EHC.pdf
Fiore, H., Travis, S., Whalen, A., Auinger, P., & Ryan, S. (2006). Potentially protective factors
associated with healthful body mass index in adolescents with obese and nonobese
parents: A secondary data analysis of the third National Health and Nutrition
Examination Survey, 1988-1994. Journal of the American Dietetic Association, 106(1),
55-64; quiz 76-59. doi: 10.1016/j.jada.2005.09.046
Hagenauer, M. H., Perryman, J. I., Lee, T. M., & Carskadon, M. A. (2009). Adolescent changes
in the homeostatic and circadian regulation of sleep. Developmental Neuroscience, 31(4),
276-284. doi: 10.1159/000216538
Hearst, M. O., Shanafelt, A., Wang, Q., Leduc, R., & Nanney, M. S. (2016). Barriers, Benefits,
and behaviors related to breakfast consumption among rural adolescents. Journal of
School Health, 86(3), 187-194. doi: 10.1111/josh.12367
James, D. C., Rienzo, B. A., & Frazee, C. (1996). Using focus group interviews to understand
school meal choices. Journal of School Health, 66(4), 128-131.
Karatzi, K., Moschonis, G., Barouti, A. A., Lionis, C., Chrousos, G. P., Manios, Y., & Healthy
Growth Study, G. (2014). Dietary patterns and breakfast consumption in relation to
insulin resistance in children. The Healthy Growth Study. Public Health Nutrition,
17(12), 2790-2797. doi: 10.1017/S1368980013003327
Kleinman, R. E., Hall, S., Green, H., Korzec-Ramirez, D., Patton, K., Pagano, M. E., & Murphy,
J. M. (2002). Diet, breakfast, and academic performance in children. Annals of Nutrition
and Metabolism, 46(Suppl. 1), 24-30.
Larson, N., Wang, Q., Grannon, K., Wei, S., Nanney, M. S., & Caspi, C. (2018). A low-cost,
grab-and-go breakfast intervention for rural high school students: Changes in school
breakfast program participation among at-risk students in Minnesota. Journal of Nutrition
Education and Behavior, 50(2), 125-132 e121. doi: 10.1016/j.jneb.2017.08.001
Lawman, H. G., Polonsky, H. M., Vander Veur, S. S., Abel, M. L., Sherman, S., Bauer, K. W.,
Sanders, T. D., Fisher, J. O., Bailey-Davis, L., Ng, J., Van Wye, G., & Foster, G. D.
(2014). Breakfast patterns among low-income, ethnically-diverse 4th-6th grade children
in an urban area. BMC Public Health, 14, 604. doi: 10.1186/1471-2458-14-604
Neumark-Sztainer, D., Story, M., Perry, C., & Casey, M. A. (1999). Factors influencing food
choices of adolescents: Findings from focus-group discussions with adolescents. Journal
of the American Dietetic Association, 99(8), 929-937. doi: 10.1016/S0002-
8223(99)00222-9
Nicklas, T. A., Bao, W., Webber, L. S., & Berenson, G. S. (1993). Breakfast consumption affects
adequacy of total daily intake in children. Journal of the American Dietetic Association,
93(8), 886-891.
O’Dea J, A. (2003). Why do kids eat healthful food? Perceived benefits of and barriers to
healthful eating and physical activity among children and adolescents. Journal of the
American Dietetic Association, 103(4), 497-501. doi: 10.1053/jada.2003.50064
Olsta, J. (2013). Bringing breakfast to our students: A program to increase school breakfast
participation. Journal of School Nursing, 29(4), 263-270. doi:
10.1177/1059840513476094
Ptomey, L. T., Steger, F. L., Schubert, M. M., Lee, J., Willis, E. A., Sullivan, D. K., Szabo-Reed,
A., N., Washburn, R. A., & Donnelly, J. E. (2016). Breakfast intake and composition is
associated with superior academic achievement in elementary schoolchildren. Journal of
the American College of Nutrition, 35(4), 326-333. doi: 10.1080/07315724.2015.1048381
Reddan, J., Wahlstrom, K., & Reicks, M. (2002). Children’s perceived benefits and barriers in
relation to eating breakfast in schools with or without Universal School Breakfast.
Journal of Nutrition Education and Behavior, 34(1), 47-52.
Ritchie, L. D., Rosen, N. J., Fenton, K., Au, L. E., Goldstein, L. H., & Shimada, T. (2016).
School breakfast policy is associated with dietary intake of fourth- and fifth-grade
students. Journal of the Academy of Nutrition and Dietetics, 116(3), 449-457. doi:
10.1016/j.jand.2015.08.020
Sabol, A., Struempler, B., & Zizza, C. (2011). Student and parent perceptions of barriers to and
benefits of the School Breakfast Program in elementary schools in southeast Alabama.
Journal of Child Nutrition and Management, 35(2).
Schober, D. J., Carpenter, L., Currie, V., & Yaroch, A. L. (2016). Evaluation of the
LiveWell@School Food Initiative shows increases in scratch cooking and improvement
in nutritional content. Journal of School Health, 86(8), 604-611. doi: 10.1111/josh.12413
Shaw, M. E. (1998). Adolescent breakfast skipping: An Australian study. Adolescence, 33(132),
851-861.
Sweeney, N. M., & Horishita, N. (2005). The breakfast-eating habits of inner city high school
students. Journal of School Nursing, 21(2), 100-105. doi:
10.1177/10598405050210020701
U.S. Department of Agriculture, Food and Nutrition Service, Office of Research and Analysis.
(2012). School Nutrition Dietary Assessment Study IV, Vol. I: School Foodservice
Operations, School Environments, and Meals Offered and Served. Alexandria, VA.
Woodward-Lopez, G., Kao, J., Kiesel, K., Lewis Miller, M., Boyle, M., Drago-Ferguson, S.,
Braff-Guajardo, E., & Crawford, P. (2014). Is scratch-cooking a cost-effective way to
prepare healthy school meals with US Department of Agriculture foods? Journal of the
Academy of Nutrition and Dietetics, 114(9), 1349-1358. doi: 10.1016/j.jand.2014.05.002
Wyon, D., Abrahamsson, L., Jartelius, M., & Fletcher, R. J. (1995). Energy intake at breakfast
improves school performance of 10 year old Swedish children. Journal of the American
Dietetic Association, 95(9), A92.
Zellner, D. A., & Cobuzzi, J. L. (2017). Eat your veggies: A chef-prepared, family style school
lunch increases vegetable liking and consumption in elementary school students. Food
Quality and Preference, 55, 8-15.
Biography
Kimberly N. Doughty, MPH, PhD is currently with the Egan School of Nursing and Health Studies at Fairfield University in Fairfield Connecticut. At the time this research was completed, she was Research Scientist at Yale-Griffin Prevention Research Center in Derby, Connecticut. Judith A. Treu, MS, RD is a Research Associate and Kerstin Eckner, BS a Student Intern at the Yale-Griffin Prevention Research Center.
Jump to Full Article
Download PDF
Purpose / Objectives
The purpose of this study was to identify key drivers of students’ breakfast habits, including eating breakfast at home or at school, to inform strategies that school nutrition professionals might use to increase participation in School Breakfast Programs (SBPs).
Join SNA
Get started.
Make a difference in the lives of America’s students while growing your professional career.
Become a Member
Scroll To Top
By continuing to browse the site you are agreeing to our use of cookies and similar tracking technologies described in our privacy policy. |
190725 | https://www.lenovo.com/us/en/glossary/gigabyte/?srsltid=AfmBOooiqXGcID2pQRI5pn1X0PDlBiiIhlJMP8EdfJ4OUHhu4OlD2Pbo | What is a Gigabyte (GB) & How Much Data Does it Hold? | Lenovo US
Please note: This website includes an accessibility system. Press Control-F11 to adjust the website to people with visual disabilities who are using a screen reader; Press Control-F10 to open an accessibility menu.
Accessibility
Popup heading
Press enter for Accessibility for blind people who use screen readers
Press enter for Keyboard Navigation
Press enter for Accessibility menu
What is gigabyte (GB)?
Lenovo ProBusiness StoreLog In / Sign Up
Learn More
Community
Accessibility
Sales:
Home:
1-855-253-6686 Business:
1-866-426-0911
Chat Now >
Visit Sales Support Page > Order Support:
Order Lookup >
Visit Order Support Page >
Technical Support >
skip to main content
All
All
Laptops
Desktops
Workstations
Monitors
PC Accessories & Electronics
Tablets
Servers & Storage
Servers Accessories
Sign In / Create Account My Lenovo Account Keep track of your wishlist, orders, and rewards all in one place Sign In / Create Account Welcome Back! Access your order, subscriptions, saved carts, rewards balance, and profile View My Account Orders View & track your ordersRewards Earn & redeem RewardsProfile Edit name, password, and account settingsWishlist Manage a wishlist of your favorite productsProducts Manage your devices, accessoriesProduct Registration Register your product and/or update your warranty datesLenovo Family Hub Manage your family's orders all in one place Sign Out
My Lenovo Rewards
Cart
Products
Solutions
Services
Support
About Lenovo
Deals
Business
Student
Gaming
Laptops
Desktops
Workstations
Accessories
Software
Monitors
Tablets
Servers & Storage
Home & Office
AI
Deals
TEMPORARILY UNAVAILABLE
DISCONTINUED
Temporary Unavailable
Cooming Soon!
. Additional units will be charged at the non-eCoupon price. Purchase additional now
We're sorry, the maximum quantity you are able to buy at this amazing eCoupon price is
Sign in or Create an Account to Save Your Cart!
Sign in or Create an Account to Join Rewards
Temporarilyunavailable
Discontinued
comingsoon
View Cart
Remove
minicart_error_please_view
Your cart is empty! Don’t miss out on the latest products and savings — find your next favorite laptop, PC, or accessory today.
item(s) in cart
Some items in your cart are no longer available. Please visit cart for more details.
has been deleted
Please review your cart as items have changed.
of
Contains Add-ons
Subtotal
Proceed to Checkout
Yes
No
Popular Searches
What are you looking for today ?
Trending
Recent Searches
Items
All
Cancel
Top Suggestions
View All >
Starting at
Fall Tech Fest!Power up your season with these unbeatable deals on PCs and tech
Buy online, pick up select products at Best Buy.Shop Pick Up >
My Lenovo Rewards!Earn 3%-9% in rewards and get free expedited delivery on select products.Join for Free >
Lease-to-own today with Katapult. Get started with an initial lease payment as low as $1! Learn More >
Shopping for a business?New Lenovo Pro members get $100 off first order of $1,000+, exclusive savings & 1:1 tech support.Learn More >
Home>Glossary> What is gigabyte (GB)?
Learn More
Annual Sale
Laptop Deals
Desktop Deals
Workstation Deals
Gaming PC Deals
PC Accessories Deals
Monitor Deals
Tablets & Phones Deals
Server & Storage Deals
Clearance Sale
Knowledgebase
AI-Glossary
SMB-Glossary
ISG Glossary
Monitors Glossary
What is gigabyte (GB)?
GB is a unit of digital information storage capacity. It is commonly used in technology, computing, programming, and communications. GB is a term you often hear in the tech world. It's a unit of measurement used to describe the amount of digital information that can be stored or transmitted.
How much data can a gigabyte hold?
A gigabyte can hold approximately 1 billion bytes of data. When you have a gigabyte of storage, it means you have the capacity to store around 1 billion characters or bytes of data.
What is the equivalent of a gigabyte in other units?
In terms of smaller units, a gigabyte is equal to 1,024 megabytes (MB). If you divide a gigabyte into smaller units, you will have around 1,024 megabytes. However, a gigabyte is equal to 0.001 terabytes (TB). If you want to express a gigabyte in larger units, it would be around 0.001 terabytes.
What is a megabyte?
In the realm of digital storage, understanding the various units is key. A megabyte (MB) is one of these units, often used to measure data size in computers and electronic devices.
Data storage involves saving files, applications, photos, and videos. Computers and smartphones typically measure storage in gigabytes (GB) and terabytes (TB). However, beneath this hierarchy lies the megabyte.
It’s important not to confuse storage with memory. Storage—measured in megabytes, gigabytes, or terabytes—is where long-term data resides, like apps and photos. Memory, or RAM (Random Access Memory), is temporary storage for data that a device is actively using. This explains why your phone might house 128GB for storage but only 4GB for RAM.
Let's break down the megabyte in relation to other units:
1 Megabyte (MB) = 1,024 Kilobytes (KB)
1 Kilobyte (KB) = 1,024 Bytes
1 Byte = 8 bits
1 Bit = smallest data unit, either a 0 or 1
Each step in this hierarchy builds on the previous one, multiplying by 1,024—a system based on powers of two, which is standard in computing.
In summary, a megabyte serves as a crucial building block of digital storage, fitting into a broader system necessary for organizing and utilizing data effectively. Understanding this helps demystify how our devices store and access the information we rely on every day.
What is the difference between megabytes and megabits?
When exploring digital storage and data transfer, understanding the difference between megabytes (MB) and megabits (Mb) is crucial.
Megabytes vs. Megabits
Megabyte (MB):A megabyte is a unit of digital information commonly used to describe the size of a file. One megabyte is equivalent to 1,024 kilobytes (KB) or 8,388,608 bits. This measurement is often used to denote the size of documents, images, and videos on your devices.
Megabit (Mb):A megabit is primarily used to measure data transfer rates and internet speed. One megabit corresponds to 1,024 kilobits (Kb). Internet speed, such as 100 Mb/s (megabits per second), tells you how fast data can be downloaded or uploaded over your internet connection.
Key Differences
Purpose:Megabytes measure file size, while megabits evaluate speed and bandwidth.
Conversion:1 Megabyte = 8 Megabits. Thus, a file size of 1 MB will take 8 seconds to download at a speed of 1 Mb/s.
Usage Context:When purchasing storage, you'll deal in MB. For internet service plans, you'll likely encounter Mb.
Quick Reference
1 MB = 8 Mb
Storage size:
Data transfer speed:Megabits
By grasping these differences, you can better navigate data storage needs and understand what to expect from your internet connection.
What can I do with a gigabyte of storage?
With a gigabyte of storage, you can save thousands of text documents, hundreds of photos, or a few hours of high-quality video. It provides you with enough space to store a significant amount of data, depending on the type of files you're dealing with.
To give you a clearer picture:
Text Documents:A typical Word document without images takes up about 20KB. This means you can store a massive number of these files without worrying about space.
Photos and Songs: An average MP3 song or a photo generally takes up around 5MB. This allows you to comfortably store hundreds of your favorite tracks or cherished memories.
Videos:Video files vary widely in size. An hour-long video can range from 500MB to 10GB, depending on the quality. Therefore, storage capacity can quickly be consumed by high-resolution videos.
Video files vary widely in size. An hour-long video can range from 500MB to 10GB, depending on the quality. Therefore, storage capacity can quickly be consumed by high-resolution videos.
How much storage does an hour-long video take up?
The storage required for an hour-long video can vary significantly, ranging from around 500MB to over 10GB, depending on several key factors.
Resolution:Higher resolutions like 1080p or 4K result in larger file sizes compared to lower resolutions like 720p. For example, a 720p video might need 1-2GB per hour, while a 4K video could exceed 15GB.
Frame Rate:Videos with higher frame rates (e.g., 60fps versus 30fps) are smoother but require more storage due to the increased amount of data being captured.
Compression:The level of video compression plays a huge role. A highly compressed format like MP4 (H.264 codec) reduces the file size significantly while maintaining decent quality.
Bitrate:A high bitrate delivers better quality but increases storage requirements. Lower bitrates, on the other hand, save storage by sacrificing some quality.
File Format:Different formats store data differently. MP4 is more space-efficient, which may retain higher quality but consume more storage.
Example Scenarios:
A 1080p video at 30fps with moderate compression may require about 4GB per hour.
The same video at 4K resolution and 60fps without heavy compression could easily exceed 12GB per hour.
If stored in a highly compressed format, even a 1080p video might shrink to under 1GB.
Understanding these factors can help you estimate storage needs and choose the right balance between quality and file size based on your requirements.
What is the typical siz of a photo file?
The size of a photo file can vary widely, usually ranging from 1MB to over 50MB, depending on several important factors.
Resolution:Photos with higher resolutions, like 4K (3840 x 2160 pixels), contain more pixels and therefore require more storage compared to lower-resolution images such as 1080p (1920 x 1080 pixels). For example, a standard 12-megapixel photo taken on a smartphone typically takes up 2-6MB when saved as a JPEG.
File Format:Compressed formats like JPEG are smaller, often around 5MB, and suitable for everyday use. Formats like PNG, which offer lossless compression, tend to have larger file sizes due to higher data retention. RAW files, used in professional photography, can easily exceed 20-50MB per photo as they retain all image data without compression.
Compression Levels:The amount of compression applied impacts file size significantly. A JPEG with high compression might shrink to 1-2MB but at the cost of noticeable quality loss, while the same image with minimal compression will appear sharper but take up more space.
Color Depth:Images with higher color depth (e.g., 16-bit versus 8-bit) store more color information, resulting in larger file sizes. This is especially true for professional or high-dynamic-range (HDR) photos.
How much storage does a typical MP3 song require?
The storage required for an MP3 song can vary depending on factors like song length, bitrate, and compression. For a standard three-minute song, the file size is usually around 3MB to 5MB at a bitrate of 128kbps, which is considered standard quality for most listening.
Key Factors Affecting MP3 File Size:
Bitrate: higher bitrate provides better audio quality but increases file size. For example:
128kbps (standard quality): ~1MB per minute of audio.
256kbps (high quality): ~2MB per minute.
Song Length:Longer tracks naturally require more storage. A 10-minute song at 128kbps can take up about 10MB.
Compression:MP3 uses lossy compression, balancing file size and quality. More aggressive compression reduces size but sacrifices some audio fidelity.Understanding these factors lets you manage storage while enjoying your preferred audio quality.
What is the approximate size of a Word document without images?
A typical Word document without images generally uses around 20KB to 50KB, but the actual file size can vary based on several factors.
Factors Influencing Word Document File Size:
Amount of Text:Documents with just a few pages of plain text have smaller file sizes, usually around 20KB. However, as the word count increases, so does the file size, with longer documents potentially reaching 100KB or more.
Formatting:Fancy text styling, such as different fonts, sizes, colors, or intricate layouts, can slightly increase the file size. For instance, a 5-page document with detailed formatting might take closer to 40KB versus a plain-text version at 20KB.
Embedded Elements:Even without images, including tables, hyperlinks, or footnotes can add complexity and contribute to a larger file size. For example, a document with several tables could be twice the size of one with only text.
Revisions and Metadata:Word files keep metadata like tracked changes, comments, or author details. These can add up, especially in heavily edited or collaborative documents.
Can I install a large software program in a gigabyte of storage?
It depends on the size of the software program. Some large software programs, such as video editing software or computer games, can take up several gigabytes of storage space. If the software program you want to install is relatively small, you can typically fit it within a gigabyte of storage. However, for larger software programs, you may need more storage space than just a gigabyte.
How long can I record video with a gigabyte of storage?
The recording time for video in a gigabyte of storage depends on several factors. These include video resolution, frame rate, bitrate, and compression settings. On average, you can record approximately 5 to 10 minutes of high-definition video (1080p) in a gigabyte of storage. If you are recording in standard definition (480p), you can store more video, around 20 to 30 minutes, in a gigabyte of storage. However, keep in mind that these numbers are approximate and can vary based on the specific video settings.
Factors Influencing Video Storage Needs
Resolution:Higher resolutions, like 4K, require more storage compared to 1080p or 480p. The clarity and detail of the video increase with resolution, but so does the file size.
Format:The format of the video file, such as MP4 or AVI, determines how data is compressed and stored. MP4 is a common format known for efficient compression without significant quality loss.
Bitrate:This refers to the amount of data processed per second in a video file. A higher bitrate generally means better video quality but also larger file sizes.
Compression Settings:Different compression settings can significantly alter how much storage a video requires. Efficient compression reduces file size while maintaining quality.
Understanding these factors can help you better estimate your storage needs and make informed decisions about your video recording settings.
How much storage space does an average smartphone have?
The storage capacity of smartphones can vary, but most modern smartphones come with a minimum of 32 gigabytes of storage. Some high-end smartphones can have storage capacities of 128 gigabytes or even more. Typically, a modern smartphone will have at least 32 gigabytes of storage, which allows you to store a significant amount of apps, photos, videos, and other data.
Understanding Your Storage Needs
When determining how much storage you need, consider the types of files you plan to store. Here's a quick breakdown:
Documents:A standard Word document without images takes up about 20KB, which is negligible in terms of storage.
Music:A typical three-minute song in MP3 format might use approximately 5MB.
Photos:Similar to music files, a photo can also take up around 5MB, depending on resolution.
Videos:These require the most space, ranging from 500MB for an hour-long video at lower resolutions to as much as 10GB for high-definition formats.
Factors to Consider
Video storage is particularly variable due to resolution and format differences. Common formats like MP4 can differ greatly in size depending on these factors. Therefore, if you frequently capture or store video, consider opting for more storage.
General Storage Recommendations
Whether using a smartphone, cloud service, or external hard drive, the more storage capacity you have, the better. This ensures you have ample room for apps, music, videos, and photos without constantly managing storage limits. Additionally, future-proofing your device with extra storage can save you from potential regrets as your data needs grow.
How is storage capacity affected by the device's operating system?
When buying a hard drive, you might notice discrepancies between the advertised storage capacity and what your device actually displays. This difference often boils down to how each operating system interprets data measurements.
Binary vs. Decimal Measurement
Hard drive manufacturers commonly utilize the power-of-ten system, measuring 1 kilobyte (KB) as 1,000 bytes. In contrast, many operating systems, to various extents, stick with the power-of-two system that defines 1 KB as 1,024 bytes. This shift in standards is where the confusion begins.
Decimal Measurement (Manufacturer):
1 KB = 1,000 Bytes
1 MB = 1,000 KB
1 GB = 1,000 MB
Binary Measurement (Traditional OS View):
1 KB = 1,024 Bytes
1 MB = 1,024 KB
1 GB = 1,024 MB
Interpretation by Operating Systems
Different operating systems handle these conversions uniquely:
Windows OS:Sticks with the binary measurement. For instance, when you buy a 1TB hard drive, Windows displays it roughly as 931GB because it calculates storage using the binary method.
Linux OS:Often aligns with the decimal system, which might result in displaying storage closer to what manufacturers advertise.
Example of Discrepancies
Take a 250GB hard drive. In binary terms, Windows will show about 232GB. On the other hand, the storage appearing in environments that use decimal, like certain Linux distributions, will be closer to the intended 250GB.
Understanding these differences is essential for consumers. It sets realistic expectations when managing data storage, particularly when dealing with large files or purchasing additional storage solutions.
Keep in mind that while storage capacities may seem inconsistent across different devices, the number of bytes available remains the same. It’s purely a matter of how each system chooses to calculate and display those bytes.
Is a gigabyte the same as a gibibyte?
No, a gigabyte (GB) and a gibibyte (GiB) are not the same. A gigabyte is based on the decimal system and represents 1 billion bytes, while a gibibyte is based on the binary system and represents 1,073,741,824 bytes. The confusion arises because computer systems often use binary values, but hard drives use decimal values to define storage capacity. Therefore, when you see storage capacities in operating systems, they are usually expressed in gibibytes (GiB) rather than gigabytes (GB).
To understand the difference between the power-of-two and power-of-ten systems in quantifying storage, consider how they interpret units:
Decimal System (Power of Ten):This system is straightforward—1,000 bytes equal 1 kilobyte (kB), 1,000 kilobytes equal 1 megabyte (MB), and so forth. This is the system typically used by hard drive manufacturers to define storage capacity.
Binary System (Power of Two):Here, 1,024 bytes make up a kilobyte (kB), 1,024 kilobytes make up a megabyte (MB), and so on. This system aligns with how computers process and store data.
When you purchase a 1TB hard drive, it’s marketed as having 1,000,000,000,000 bytes. However, when your computer, which uses the binary system, reads this drive, it calculates the capacity as approximately 931GB. Why? Because the computer divides by 1,024 at each stage of conversion:
1. Convert bytes to kilobytes:Divide by 1,024, resulting in 976,562,500 kB.
2. Convert kilobytes to megabytes:Again, divide by 1,024, yielding 953,674.3 MB.
3. Convert megabytes to gigabytes:One more division by 1,024 gives you 931.32 GB.
This discrepancy is why storage appears smaller on your computer than what's advertised. Understanding this can help manage expectations when evaluating storage needs and capacities.
How do different operating systems interpret hard drive capacity?
Different operating systems interpret hard drive capacity based on the way they calculate storage, which can lead to discrepancies between the advertised capacity and what users see on their devices. This comes down to two measurement systems – binary (base-2) and decimal (base-10).
1. Binary vs. Decimal Systems
1. Hard drive manufacturers typically use the decimal system (base-10), where 1 kilobyte (KB) equals 1,000 bytes, 1 megabyte (MB) equals 1,000 KB, and so on. Under this system, a 1 terabyte (TB) drive is advertised as 1,000 gigabytes or 1 trillion bytes.
2. Operating systems like Windows and Linux, however, often use the binary system (base-2), where 1 KB equals 1,024 bytes. This difference accumulates, so a 1TB drive marketed by the manufacturer will appear as approximately 931GB when interpreted by the operating system.
2. Formatted Storage and File Systems
Beyond measurement systems, formatting the drive and adding a file system (like NTFS or APFS) also reduces available space. This space is used for metadata, partitioning, and organizing files. For example, a 500GB drive may provide only 465GB of usable space after formatting and applying a file system.
3. Examples of OS ReportingOn Windows, right-clicking the hard drive and viewing "Properties" will show the capacity in both bytes and GB (using the binary system).
Why These Differences Matter
Understanding these discrepancies helps manage storage expectations. For instance, knowing that a 2TB drive will show as roughly 1.81TB on many systems can help users estimate real-world capacity more accurately. Additionally, to maximize storage, it’s helpful to consider drive formatting and file system setup during installation.
What makes some SSD storage capacities unique compared to standard sizes?
Some SSDs have non-standard storage capacities due to a combination of design factors and technological trade-offs. Here's why this happens:
1. Overprovisioning for Performance and Longevity
Many SSDs include extra storage beyond what's shown to the user, known as overprovisioning. This reserved storage allows for:
1. Wear leveling, which helps evenly distribute write and erase cycles across memory cells, extending the SSD's lifespan.
2. Error correction and replacement of faulty memory cells, ensuring better reliability.
For example, an SSD advertised as 1TB may actually have 1,024GB of raw capacity, but a portion is reserved for these maintenance tasks, leaving 977GB as usable space.
2. Manufacturing and Memory Configuration
SSDs use NAND flash storage, which is manufactured in specific block and die sizes. Creating exact decimal-based capacities like 500GB or 1TB can result in non-standard sizes when aligning with hardware design and production yields. This is why you may find SSDs with capacities like 480GB or 960GB instead.
3. Impact on Usable Space
On top of overprovisioning, the usable capacity shown in an operating system can differ further due to formatting and how different systems calculate storage (binary vs. decimal). For instance:
1. A 1TB SSD (technically 1,000GB as marketed by manufacturers) may appear as approximately 931GB in a system using binary calculations.By understanding these factors, users can better interpret the capacity of SSDs and make informed decisions based on their performance and storage needs. When purchasing, always check the advertised vs. usable capacity to ensure it meets your requirements.
What are examples of how storage capacities are displayed on different devices?
A: Storage capacities on devices like hard drives, SSDs, and USB drives often appear differently from what’s advertised. These discrepancies are due to varying measurement systems and the way operating systems interpret usable space. Here are some examples:
1. Hard Drives
1. An advertised 1TB (1,000GB) hard drive will typically display as 931GB on a Windows system.
2. This occurs because manufacturers use the decimal system, where 1GB equals 1,000MB. However, operating systems like Windows operate on the binary system, where 1GB equals 1,024MB, leading to the smaller displayed capacity.
2. SSDs (Solid-State Drives)
1. A 512GB SSD will commonly show 477GB of usable space. This accounts for two factors:
3. The binary interpretation of storage reduces the total shown capacity.
4. Overprovisioning reserves a portion of the SSD’s space for system maintenance, wear leveling, and optimizing performance.
5. USB Drives
1. A USB drive marketed as 32GB often displays as approximately 28.8GB.
2. This difference is due to the same binary vs. decimal system discrepancy, along with formatting overhead that uses part of the storage for file system structure details.
What is the process for calculating the binary interpretation of a hard drive's capacity?
To calculate the binary interpretation of a hard drive's capacity, start with the total advertised bytes (e.g., 1,000,000,000,000 bytes for a 1TB drive). Divide this number by 1,024 to convert it into kilobytes (KB). Repeat the process to further convert kilobytes into megabytes (MB), and then megabytes into gigabytes (GB). For a 1TB drive, this will result in approximately 931.32GB. This method reflects the base-2 (binary) system that operating systems like Windows use, where 1KB equals 1,024 bytes, ensuring consistency in technical storage representation.
How do different operating systems interpret hard drive capacity?
Operating systems interpret hard drive capacity differently due to the measurement systems they use, which can create noticeable discrepancies. Here's how it works:
1. Binary vs. Decimal Systems
1. Hard drive manufacturers use the decimal system (base-10), where 1 kilobyte (KB) equals 1,000 bytes, 1 megabyte (MB) equals 1,000 KB, and so on. For example, a 1TB drive contains 1,000,000,000,000 bytes according to this system.
2. Operating systems like Windows rely on the binary system (base-2), where 1KB equals 1,024 bytes, 1MB equals 1,024KB, etc. When a drive’s capacity is converted using binary calculations, the reported size appears smaller than the advertised size because a binary "GB" requires more bytes than a decimal one. For instance, the same 1TB drive translates to approximately 931.32GB in binary terms.
2. How Operating Systems Interpret Capacity
1. Windows:Windows systems (Windows 7, 10, 11) display storage capacity using the binary system, leading to the discrepancy noted above. This is why a 500GB hard drive might show as around 465GB.
2. Linux:Modern Linux distributions may also use binary measurements for file systems, but tools like df or partition managers can display drive capacity in either base-2 or base-10 depending on configuration.
3. Impact on Reported Capacity
The measurement system and formatting overhead (file system setup, metadata storage, etc.) combine to produce the usable space reported by the OS. Users might notice that their advertised storage capacity is lower once the drive is installed in their computer or formatted for use.
By understanding these interpretation differences, users can better estimate the real amount of usable storage space and make informed decisions when selecting storage devices.
What are Kibibytes and Mibibytes, and why are they used?
Kibibytes (KiB) and Mibibytes (MiB) are terms used to precisely define digital information storage sizes. Unlike the commonly used kilobytes (KB) and megabytes (MB), which often refer to 1,000 bytes and 1,000,000 bytes respectively, kibibytes and mibibytes adhere to a binary system of measurement.
Binary-Based Measurements
1 Kibibyte (KiB) = 1,024 bytes
1 Mebibyte (MiB) = 1,024 kibibytes (KiB) = 1,048,576 bytes
These binary-based metrics were standardized by the International Electrotechnical Commission (IEC) to eliminate ambiguity.
Why Are They Used?
The primary purpose of using kibibytes and mibibytes is to provide clarity and precision in digital data measurement. Here's why they matter:
1. Clarity in Communication:The traditional use of kilobytes and megabytes can create confusion because they are often assumed to mean 1,024 bytes and 1,024 kilobytes respectively. Kibibytes and mibibytes make it clear that the count is based on a binary system.
2. Standards and Accuracy:In technical and scientific contexts, precision is crucial. Using kibibytes and mibibytes ensures that everyone speaks the same "language," especially in fields where accurate data storage metrics are essential.
3. Aligned with File Sizes:Most computer systems, including file system architectures, operate on powers of two. Using binary-based units like kibibytes and mibibytes aligns with this natural framework, making calculations more straightforward.
By using these precise terms, industries involving data storage and processing can avoid misunderstandings and achieve greater accuracy in their operations.
Why do computers use the power-of-two system?
Computers and most electronic devices operate using the power-of-two system because they fundamentally rely on binary code. Unlike humans, who use a decimal system (base-10), computers work with a binary system (base-2). This is because computers are composed of countless tiny switches called transistors that exist in one of two states: on or off.
Binary
Binary is simple yet effective for processing data. Each switch corresponds to a binary digit (or "bit")—either a 0 or a 1. By combining these bits in sequences, computers can perform complex calculations and store vast amounts of data.
Logical Efficiency
The binary system aligns perfectly with the physical structure of electronic circuits. It allows operations to be performed quickly and reliably, making it the most logical choice for computing. Numbers in binary are expressed in powers of two, such as 2, 4, 8, 16, and so on. This is the backbone for various computer processes, including data storage and memory allocation.
Universal Application
From laptops to smartphones and tablets, every digital device uses this essential binary framework. Developers and manufacturers optimize hardware and software to function efficiently within this system, leading to faster processing and more reliable technology.
In summary, computers depend on the power-of-two system due to their binary nature, which is integral to the operation of all modern electronics. This system ensures speed, efficiency, and operational reliability across all digital platforms.
Why does a hard drive often have less usable capacity than advertised?
Hard drives frequently appear to have less storage than what’s advertised due to differences in how data capacity is calculated and the space reserved for system files. Hard drive manufacturers typically use the decimal system (power of ten) for measuring storage capacity. In this system:
1,000 Bytes = 1 kilobyte (KB)
1,000 kilobytes (KB) = 1 Megabyte (MB)
1,000 Megabytes (MB) = 1 Gigabyte (GB)
1,000 Gigabytes (GB) = 1 Terabyte (TB)
1,000 Gigabytes (GB) = 1 Terabyte (TB)
1,024 Bytes = 1 kibibyte (KiB)
1,024 kibibytes (KiB) = 1 Mebibyte (MiB)
1,024 Mebibytes (MiB) = 1 Gibibyte (GiB)
1,024 Gibibytes (GiB) = 1 Tebibyte (TiB)
This discrepancy means a hard drive advertised as 1TB (1,000GB) is typically seen by a computer as around 931GB because the computer calculates based on powers of two.
Moreover, devices such as PCs, phones, and tablets require some of the storage capacity for system files and operating systems. This means if a phone advertises 64GB of space, the actual available space for apps, photos, and other files will be less. This reduction accounts for pre-installed software and system updates.
In summary, the difference between marketing terms and computing reality, combined with necessary system allocations, leads to hard drives showing lower usable capacity than advertised.
Can I use a gigabyte to measure internet data usage?
Yes, a gigabyte is commonly used to measure internet data usage. Internet service providers often provide data plans with specific limits, such as 100 gigabytes per month. This limit refers to the amount of data you can transfer over the internet within that time frame. If you have a data plan with a limit of 100 gigabytes per month, it means you can use up to 100 gigabytes of data for activities like browsing the web, streaming videos, downloading files, and more.
Is a gigabyte the largest unit of storage?
No, a gigabyte is not the largest unit of storage. There are larger units such as terabytes (TB), petabytes (PB), exabytes (EB), and beyond. While a gigabyte provides a substantial amount of storage, there are larger units available for storing and managing massive amounts of data.
Can I convert a gigabyte to a different unit of storage?
Yes, you can convert a gigabyte to a different unit of storage. For example, to convert gigabytes to megabytes, you multiply the number of gigabytes by 1,024. To convert gigabytes to terabytes, you divide the number of gigabytes by 1,024. Converting between different units of storage is a straightforward process using simple multiplication or division by the appropriate conversion factor.
Is a gigabyte the same as a gigabit?
No, a gigabyte (GB) and a gigabit (Gb) are not the same. A gigabyte is a unit of storage capacity, while a gigabit is a unit of data transfer rate. A gigabyte represents 8 gigabits, so when you see internet speeds advertised in gigabits per second (Gbps), it means the data transfer rate, not the storage capacity. It's important to differentiate between the two, as they measure different aspects of data.
Can I measure the size of files and folders on my computer in gigabytes?
Yes, you can measure the size of files and folders on your computer in gigabytes. Operating systems like Windows provide tools that display the size of files and folders in the appropriate units, including gigabytes. By checking the properties or information of a file or folder, you can see its size in gigabytes and understand how much storage space it occupies on your computer.
What is the largest file size that can be stored in a gigabyte?
The largest file size that can be stored in a gigabyte is 1 gigabyte itself. Since a gigabyte is a unit of storage, it can accommodate a file size equal to its own capacity. So, the maximum file size that can fit in a gigabyte is exactly 1 gigabyte.
Can I use a gigabyte flash drive to transfer files between computers?
Yes, you can use a gigabyte flash drive (also known as a universal serial bus (USB) drive or thumb drive) to transfer files between computers. Flash drives come in various storage capacities, including gigabyte sizes. You can copy files from one computer onto the flash drive, and then plug it into another computer to access those files. Using a gigabyte flash drive allows you to carry and transfer a significant amount of data between different devices.
How is memory (RAM) different from storage space?
Understanding the distinction between memory (RAM) and storage space is crucial, especially when evaluating your device’s capabilities.
Memory vs. Storage: A Breakdown
Memory (RAM):This is your device's short-term storage. RAM, or Random Access Memory, holds data that your device actively uses so that it can be quickly accessed by the processor. Think of it as your device's workspace. The more RAM available, the more tasks or apps your device can handle simultaneously. However, RAM is limited, often ranging from 4GB to 16GB in many devices, because it only holds data temporarily and is cleared or lost when the device is turned off.
Storage Space:This refers to your device's long-term storage capacity, where all your files, apps, photos, and videos are stored. Unlike RAM, storage space isn't wiped clean when you power down your device. Storage space is significantly larger — typically measured in gigabytes (GB) or terabytes (TB) — with common devices offering between 128GB and 1TB. It functions like a library, keeping all your information safe and ready for use even after your device is turned off.
In essence, while both components are essential for your device's performance, RAM provides quick access for currently used data, whereas storage space retains information long term, even beyond shutdowns. Understanding their different purposes helps you make informed decisions when purchasing phones, tablets, or computers, ensuring the device meets both your immediate and sustained data handling needs.
How long does it take to transfer a gigabyte of data over a fast internet connection?
The time it takes to transfer a gigabyte of data over a fast internet connection depends on several factors, such as your internet speed, network congestion, and the quality of your connection. With a fast and stable internet connection, it can take anywhere from a few seconds to a few minutes to transfer a gigabyte of data. However, keep in mind that real-world transfer speeds can vary, and it's influenced by multiple factors beyond just the size of the data being transferred.
Looking for a Great Deal?
Shop Lenovo.com for great deals on A+ Education PCs, Accessories, Bundles and more.
Shop Deals Now
Recommended Education Products
Laptop Starting at $1,359.00 Learn More Starting at $1,999.99 Learn More Laptop Starting at $1,439.99 Learn More Laptop Starting at $1,219.99 Learn More Mobile Workstation Starting at $2,419.00 Learn More
Shop
Student Deals
K-12 Student Laptops
Student Accessories
Laptops by Major
Explore
What is STEM?
Best Laptops for College
Student & Teacher Discounts
Lenovo Sustainability Programs
See All STEM Articles
Online STEM Education
STEM Career Tips & Resources
Women in STEM
STEM Educator Training
STEM Grants & Funding
STEM Education: Coding for Kids
STEM Education: Robotics
STEM Education: Biotechnology
STEM Education: Sustainability
STEM Education: AI & ML
While every effort has been made to ensure accuracy, this glossary is provided for reference purposes only and may contain errors or inaccuracies. It serves as a general resource for understanding commonly used terms and concepts. For precise information or assistance regarding our products, we recommend visiting our dedicated support site, where our team is readily available to address any questions or concerns you may have.
Why Lenovo?
Go Greener with Leno...
Lenovo is committed to Smarter Climate Action with lower energy laptops, use of sustainable materials and packaging, and available CO2 Offset Services. Learn More
Get It Now, Pay For ...
Lenovo has multiple financing option: the Lenovo Credit Card, installment plans, and lease-to-own financing op... Learn More
Productivity & Peace of Mind
Stay productive with Premium Care Plus - 24/7 priority support and coverage against accidental damage. Power through your day with our Smart Performance and extended battery solutions. Learn More
Fast & Secure
Get the most from your laptop with Lenovo’s state-of-the-art Smart Performance, delivering powerful, all-in-on... Learn More
Assistance and Suppo...
Chat with a trained professional who can help you find the right products, place or check on an order, or setup your Education laptop. Contact Us
Enter email to receive Lenovo marketing and promotional emails. Review our Privacy Statement for more details.
Email
Please enter the correct email address!
Email address is required
×
Select Country / Region:
About Lenovo
Our Company
News
Investors Relations
Compliance
ESG
Product Recycling
Product Recalls
Executive Briefing Center
Lenovo Cares
Careers
Formula 1 Partnership
Products & Services
Laptops & Ultrabooks
Smarter AI for You
Desktop Computers
Workstations
Gaming
Tablets
Servers, Storage, & Networking
Accessories & Software
Services & Warranty
Product FAQs
Outlet
Deals
Lenovo Coupons
Cloud Security Software
Windows 11 Upgrade
Shop By Industry
Small Business Solutions
Large Enterprise Solutions
Government Solutions
Healthcare Solutions
Higher Education Solutions
Education Discounts
Discount Programs
Resources
Legion Gaming Community
Lenovo Creator Community
Lenovo Pro Community
Lenovo Pro for Business
My Lenovo Rewards
Lenovo Financing
Customer Discounts
Affiliate Program
Affinity Program
Employee Purchase Program
Lenovo Partner Hub
Laptop Buying Guide
Where to Buy
Glossary
Knowledgebase
Server Buying Guide
Customer Support
Contact Us
Policy FAQs
Return Policy
Shipping Information
Order Lookup
Register a Product
Replacement Parts
Technical Support
Forums
Provide Feedback;%20window.lenovoSurvey('p3094292724','feedback');%20return%20false)
© 2025 Lenovo. All rights reserved.
PrivacyCookie Consent ToolDo Not Sell or Share My Personal InformationU.S. Privacy NoticeSite MapTerms of UseExternal Submission PolicySales terms and conditionsAnti-Slavery and Human Trafficking Statement
This will open a new window☺Feedback
Compare ()
removeAll x
Call
Need Help?Call:
1-855-253-6686
Select Your Store
Add items to your cart?
We've noticed that you've entered a different store. Do you want to add these items to your cart in this Store?
No Yes.Add in Public Store
Add items to your cart in the Lenovo Pro Store?
We've noticed that you've entered a different store. Do you want to add these items to your cart in the Lenovo Pro Store?
No Yes. Add items in Lenovo Pro.
Add items to your Affinity Store?
We've noticed that you've entered a different store. Do you want to add these items to your cart in Affinity Store?
No Yes.Add items in Affinity store.
Add items to your Education Store?
We've noticed that you've entered a different store. Do you want to add these items to your cart in Education Store?
No Yes.Add items in Education store.
You will be automatically signed in and redirected to your shopping experience after
seconds or click the “Continue” button to go there immediately.
Continue
Success!
flash.idme.banner.success.dialog.success
You now have a verified Lenovo Education account, and your discounts will automatically be applied to eligible products. Enjoy your benefits for a full year!
flash.idme.banner.success.dialog.close
Sign In
Success!
You will be automatically signed in and redirected to your shopping experience after seconds or click the “Continue” button to go there immediately.
Continue
flash.idme.banner.success.dialog.success
You now have a verified Lenovo Education account, and your discounts will automatically be applied to eligible products. Enjoy your benefits for a full year!
flash.idme.banner.success.dialog.close
We use cookies and similar technologies to personalize content and ads and analyze our traffic. By browsing our site or by selecting "X", you agree to our use of cookies and our sharing of information with our partners. View the Lenovo Website Privacy Statement. Visit our Cookie Consent Tool to manage your preferences. Click Here to Reject All non-essential cookies.
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now
Buy Now |
190726 | https://online.stat.psu.edu/stat504/lesson/4/4.5 | Skip to main content
Keyboard Shortcuts
Help
: F1 or
?
Previous Page
: ← +
CTRL (Windows)
: ← +
⌘ (Mac)
Next Page
: → +
CTRL (Windows)
: → +
⌘ (Mac)
Search Site
: CTRL +
SHIFT +
F (Windows)
: ⌘ +
⇧ +
F (Mac)
Close Message
: ESC
4.5 - Fisher's Exact Test
The tests discussed so far that use the chi-square approximation, including the Pearson and LRT for nominal data as well as the Mantel-Haenszel test for ordinal data, perform well when the contingency tables have a reasonable number of observations in each cell, as already discussed in Lesson 1. When samples are small, the distributions of , , and (and other large-sample based statistics) are not well approximated by the chi-squared distribution; thus their -values are not to be trusted. In such situations, we can perform inference using an exact distribution (or estimates of exact distributions), but we should keep in mind that -values based on exact tests can be conservative (i.e, measured to be larger than they really are).
We may use anexact testif:
the row totals and the column totals are both fixed by design of the study.
we have a small sample size ,
more than 20% of cells have expected cell counts less than 5, and no expected cell count is less than 1.
Example: Lady tea tasting Section
Here we consider the famous tea tasting example! In a summer tea-part in Cambridge, England, a lady claimed to be able to discern, by taste alone, whether a cup of tea with milk had the tea poured first or the milk poured first. An experiment was performed by Sir R.A. Fisher himself, then and there, to see if her claim was valid. Eight cups of tea were prepared and presented to her in random order. Four had the milk poured first, and other four had the tea poured first. The lady tasted each one and rendered her opinion. The results are summarized in the following table:
| Actually poured first | Lady says poured first | |
---
| tea | milk |
| tea | 3 | 1 |
| milk | 1 | 3 |
The row totals are fixed by the experimenter. The column totals are fixed by the lady, who knows that four of the cups are "tea first" and four are "milk first." Under , the lady has no discerning ability, which is to say the four cups she calls "tea first" are a random sample from the eight. If she selects four at random, the probability that three of these four are actually "tea first" comes from the hypergeometric distribution, :
A -value is the probability of getting a result as extreme or more extreme than the event actually observed, assuming is true. In this example, the -value would be , where is the observed value of , which in this case is 3. The only result more extreme would be the lady's (correct) selection of all four the cups that are truly "tea first," which has probability
As it turns out, the -value is , which is only weak evidence against the null. In other words, there is not enough evidence to reject the null hypothesis that the lady is just purely guessing. To be fair, experiments with small amounts of data are generally not very powerful, to begin with, given the limited information.
Here is how we can do this computation in SAS and R. Further below we describe in a bit more detail the underlying idea behind these calculations.
Code
/----------------------------
| Example: Fisher's Tea Lady
-----------------------------/
data tea;
input poured $ lady $ count;
datalines;
tea tea 3
tea milk 1
milk tea 1
milk milk 3
;
run;
proc freq data=tea order=data;
weight count;
tables pouredlady/ chisq relrisk riskdiff expected;
exact fisher chisq or;
run;
Output
The SAS System
The FREQ Procedure
| | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- |
| | Frequency Expected Percent Row Pct Col Pct | | | Table of poured by lady | | | | | poured | lady | | | | tea | milk | Total | | tea | 3 2 37.50 75.00 75.00 | 1 2 12.50 25.00 25.00 | 4 50.00 | | milk | 1 2 12.50 25.00 25.00 | 3 2 37.50 75.00 75.00 | 4 50.00 | | Total | 4 50.00 | 4 50.00 | 8 100.00 | |
Statistics for Table of poured by lady
| Statistic | DF | Value | Prob |
| WARNING: 100% of the cells have expected counts less than 5. (Asymptotic) Chi-Square may not be a valid test. | | | |
| Chi-Square | 1 | 2.0000 | 0.1573 |
| Likelihood Ratio Chi-Square | 1 | 2.0930 | 0.1480 |
| Continuity Adj. Chi-Square | 1 | 0.5000 | 0.4795 |
| Mantel-Haenszel Chi-Square | 1 | 1.7500 | 0.1859 |
| Phi Coefficient | | 0.5000 | |
| Contingency Coefficient | | 0.4472 | |
| Cramer's V | | 0.5000 | |
| Pearson Chi-Square Test | |
| Chi-Square | 2.0000 |
| DF | 1 |
| Asymptotic Pr > ChiSq | 0.1573 |
| Exact Pr >= ChiSq | 0.4857 |
| Likelihood Ratio Chi-Square Test | |
| Chi-Square | 2.0930 |
| DF | 1 |
| Asymptotic Pr > ChiSq | 0.1480 |
| Exact Pr >= ChiSq | 0.4857 |
| Mantel-Haenszel Chi-Square Test | |
| Chi-Square | 1.7500 |
| DF | 1 |
| Asymptotic Pr > ChiSq | 0.1859 |
| Exact Pr >= ChiSq | 0.4857 |
| Fisher's Exact Test | |
| Cell (1,1) Frequency (F) | 3 |
| Left-sided Pr <= F | 0.9857 |
| Right-sided Pr >= F | 0.2429 |
| | |
| Table Probability (P) | 0.2286 |
| Two-sided Pr <= P | 0.4857 |
| Column 1 Risk Estimates | | | | | | |
| | Risk | ASE | 95% Confidence Limits | | Exact 95% Confidence Limits | |
| Difference is (Row 1 - Row 2) | | | | | | |
| Row 1 | 0.7500 | 0.2165 | 0.3257 | 1.0000 | 0.1941 | 0.9937 |
| Row 2 | 0.2500 | 0.2165 | 0.0000 | 0.6743 | 0.0063 | 0.8059 |
| Total | 0.5000 | 0.1768 | 0.1535 | 0.8465 | 0.1570 | 0.8430 |
| Difference | 0.5000 | 0.3062 | -0.1001 | 1.0000 | | |
| Column 2 Risk Estimates | | | | | | |
| | Risk | ASE | 95% Confidence Limits | | Exact 95% Confidence Limits | |
| Difference is (Row 1 - Row 2) | | | | | | |
| Row 1 | 0.2500 | 0.2165 | 0.0000 | 0.6743 | 0.0063 | 0.8059 |
| Row 2 | 0.7500 | 0.2165 | 0.3257 | 1.0000 | 0.1941 | 0.9937 |
| Total | 0.5000 | 0.1768 | 0.1535 | 0.8465 | 0.1570 | 0.8430 |
| Difference | -0.5000 | 0.3062 | -1.0000 | 0.1001 | | |
| Odds Ratio and Relative Risks | | | |
| Statistic | Value | 95% Confidence Limits | |
| Odds Ratio | 9.0000 | 0.3666 | 220.9270 |
| Relative Risk (Column 1) | 3.0000 | 0.5013 | 17.9539 |
| Relative Risk (Column 2) | 0.3333 | 0.0557 | 1.9949 |
| Odds Ratio | |
| Odds Ratio | 9.0000 |
| | |
| Asymptotic Conf Limits | |
| 95% Lower Conf Limit | 0.3666 |
| 95% Upper Conf Limit | 220.9270 |
| | |
| Exact Conf Limits | |
| 95% Lower Conf Limit | 0.2117 |
| 95% Upper Conf Limit | 626.2435 |
Sample Size = 8
OPTION EXACT in SAS indicates that we are doing exact tests which consider ALL tables with the exact same margins as the observed table. This option will work for any table. OPTION FISHER, more specifically performs Fisher's exact test which is an exact test only for a table in SAS.
For R, see TeaLady.R where you can see we used the fisher.test() function to perform Fisher's exact test for the table in question.
```
one-sided Fisher's exact test
fisher.test(tea, alternative = "greater")
two-sided Fisher's exact test
fisher.test(tea)
```
The same could be done using chisq.test() with option, simulate.p.value=TRUE. By reading the help file on fisher. test() function, you will see that certain options in this function only work for tables. For the output, see TeaLady.out
The basic idea behind exact tests is that they enumerate all possible tables that have the same margins, e.g., row sums and column sums. Then to compute the relevant statistics, e.g., , , odds-ratios, we look for all tables where the values are more extreme than the one we have observed. The key here is that in the set of tables with the same margins, once we know the value in one cell, we know the rest of the cells. Therefore, to find a probability of observing a table, we need to find the probability of only one cell in the table (rather than the probabilities of four cells). Typically we use the value of cell (1,1).
Under the null hypothesis of independence, more specifically when odds-ratio , the probability distribution of that one cell is hypergeometric, as discussed in the Tea lady example.
Stop and Think!
In the Lady tasting tea example, there are 5 possible tables that have the same observed margins. Can you figure out which are those?
Extension of Fisher's test Section
For problems where the number of possible tables is too large, Monte Carlo methods are used to approximate "exact" statistics (e.g., option MC in SAS FREQ EXACT and in R under chisq.test() you need to specify simulate.p.value = TRUE and indicate how many runs you want MC simulation to do; for more see the help files). This extension, and thus these options in SAS and R, of the Fisher's exact test for a table, in effect, takes samples from a large number of possibilities in order to simulate the exact test.
This test is "exact" because no large-sample approximations are used. The -value is valid regardless of the sample size. Asymptotic results may be unreliable when the distribution of the data is sparse, or skewed. Exact computations are based on the statistical theory of exact conditional inference for contingency tables.
Fisher's exact test is definitely appropriate when the row totals and column totals are both fixed by design. Some have argued that it may also be used when only one set of margins is truly fixed. This idea arises because the marginal totals provide little information about the odds ratio .
Exact non-null inference for Section
When , this distribution is hypergeometric, which we used in Fisher's exact test. More generally, Fisher (1935) gave this distribution for any value of . Using this distribution, it is easy to compute Fisher's exact -value for testing the null hypothesis for any . The set of all values that cannot be rejected at the level test forms an exact 95% confidence region for .
Let's look at a part of the SAS output a bit closer, we get the same CIs in the R ouput. First, notice the sample estimate of the odds ratio equal to 9, which we can compute from the cross-product ratio as we have discussed earlier. Note also that fisher.test() in R for tables will give so-called "conditional estimate" of the odds-ratio so the value will be different (in this case, approximately 6.408).
Notice the difference between theexact and asymptotic CIs for the odds ratio for the two-sided alternative (e.g., ). The exact version is larger. Recalling the interpretation of the odds ratio, what do these CIs tell us about true unknown odds-ratio? This is a simple example of how inference may vary if you have small samples or sparseness.
| Odds Ratio | |
| Odds Ratio | 9.0000 |
| | |
| Asymptotic Conf Limits | |
| 95% Lower Conf Limit | 0.3666 |
| 95% Upper Conf Limit | 220.9270 |
| | |
| Exact Conf Limits | |
| 95% Lower Conf Limit | 0.2117 |
| 95% Upper Conf Limit | 626.2435 |
Sample Size = 8
Bias correction for estimating Section
Earlier, we learned that the natural estimate of is
and that is approximately normally distributed about with estimated variance
Advanced note: this is the estimated variance of the limiting distribution, not an estimate of the variance of itself. Because there is a nonzero probability that the numerator or the denominator of may be zero, the moments of and do not actually exist. If the estimate is modified by adding to each , we have
with estimated variance
In smaller samples, may be slightly less biased than . |
190727 | https://math.stackexchange.com/questions/1587735/prove-the-identity-cosh2x-cosh2x-sinh2x-using-the-cauchy-product | sequences and series - Prove the identity $\cosh(2x)=\cosh^2(x)+\sinh^2(x)$ using the Cauchy product. - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Prove the identity cosh(2 x)=cosh 2(x)+sinh 2(x)cosh(2 x)=cosh 2(x)+sinh 2(x) using the Cauchy product. [closed]
Ask Question
Asked 9 years, 9 months ago
Modified9 years, 9 months ago
Viewed 2k times
This question shows research effort; it is useful and clear
5
Save this question.
Show activity on this post.
Closed. This question is off-topic. It is not currently accepting answers.
This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.
Closed 9 years ago.
Improve this question
Prove the identity
cosh(2 x)=cosh 2(x)+sinh 2(x)cosh(2 x)=cosh 2(x)+sinh 2(x)
using the Cauchy product and the Taylor series expansions of cosh(x)cosh(x) and sinh(x)sinh(x). The relations involving the exponential function are not to be used.
sequences-and-series
power-series
taylor-expansion
hyperbolic-functions
cauchy-product
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Dec 24, 2015 at 13:40
wythagoras
25.8k 6 6 gold badges 65 65 silver badges 120 120 bronze badges
asked Dec 24, 2015 at 13:38
SteepleSteeple
111 5 5 bronze badges
3
2 You may want to include your work into your post.C. Falcon –C. Falcon 2015-12-24 13:50:54 +00:00 Commented Dec 24, 2015 at 13:50
Do you know what the Cauchy product is? It's pretty much a very direct proof if you know how to use taylor series expansions and cauchy products...Eleven-Eleven –Eleven-Eleven 2015-12-24 14:25:11 +00:00 Commented Dec 24, 2015 at 14:25
@Eleven-Eleven I get how the Cauchy product works by definition, but after writing the RHS in the expanded form, I have no clue how to proceed further...Steeple –Steeple 2015-12-24 14:29:49 +00:00 Commented Dec 24, 2015 at 14:29
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
10
Save this answer.
Show activity on this post.
Given two power series
f(x)=∑n=0∞a n x n,f(x)=∑n=0∞a n x n,
g(x)=∑n=0∞b n x n g(x)=∑n=0∞b n x n
The Cauchy product is just their product
f(x)g(x)=∑n=0∞a n x n∑n=0∞b n x n=∑n=0∞(∑k=0 n a k b n−k)x n f(x)g(x)=∑n=0∞a n x n∑n=0∞b n x n=∑n=0∞(∑k=0 n a k b n−k)x n
So consider f(x)=cosh x=∑∞n=0 x 2 n(2 n)!,g(x)=sinh x=∑∞n=0 x 2 n+1(2 n+1)!f(x)=coshx=∑n=0∞x 2 n(2 n)!,g(x)=sinhx=∑n=0∞x 2 n+1(2 n+1)!. Now f(x)2=cosh 2 x f(x)2=cosh 2x is
∑n=0∞x 2 n(2 n)!∑n=0∞x 2 n(2 n)!=∑n=0∞∑k=0 n 1(2 k)!1(2 n−2 k)!x 2 n=∑n=0∞∑k=0 n(2 n 2 k)x 2 n(2 n)!∑n=0∞x 2 n(2 n)!∑n=0∞x 2 n(2 n)!=∑n=0∞∑k=0 n 1(2 k)!1(2 n−2 k)!x 2 n=∑n=0∞∑k=0 n(2 n 2 k)x 2 n(2 n)!
For g(x)2=sinh 2 x,g(x)2=sinh 2x, rewrite the sum as
∑n=0∞x 2 n+1(2 n+1)!=x∑n=0∞x 2 n(2 n+1)!∑n=0∞x 2 n+1(2 n+1)!=x∑n=0∞x 2 n(2 n+1)!
It is easier to take the x x out and put it back, as it will stay in line with the Cauchy Product definition (won't change the power of x x in the expansion). Thus
x∑n=0∞x 2 n(2 n+1)!x∑n=0∞x 2 n(2 n+1)!=x 2∑n=0∞∑k=0 n 1(2 k+1)!1(2 n−2 k+1)!x 2 n x∑n=0∞x 2 n(2 n+1)!x∑n=0∞x 2 n(2 n+1)!=x 2∑n=0∞∑k=0 n 1(2 k+1)!1(2 n−2 k+1)!x 2 n
=∑n=0∞∑k=0 n 1(2 k+1)!1(2 n−2 k+1)!x 2 n+2=∑n=1∞∑k=0 n−1 1(2 k+1)!1(2 n−2−2 k+1)!x 2 n=∑n=0∞∑k=0 n 1(2 k+1)!1(2 n−2 k+1)!x 2 n+2=∑n=1∞∑k=0 n−1 1(2 k+1)!1(2 n−2−2 k+1)!x 2 n
=∑n=1∞∑k=0 n−1 1(2 k+1)!1(2 n−2 k−1)!x 2 n=∑n=1∞∑k=0 n−1(2 n 2 k+1)x 2 n(2 n)!=∑n=1∞∑k=0 n−1 1(2 k+1)!1(2 n−2 k−1)!x 2 n=∑n=1∞∑k=0 n−1(2 n 2 k+1)x 2 n(2 n)!
Now you can add the two sums together, and notice, you are getting a sum of binomial coefficients up to 2 n 2 n! So
∑k=0 n(2 n 2 k)+∑k=0 n−1(2 n 2 k+1)=∑k=0 2 n(2 n k)=2 2 n∑k=0 n(2 n 2 k)+∑k=0 n−1(2 n 2 k+1)=∑k=0 2 n(2 n k)=2 2 n
So
cosh 2 x+sinh 2 x=∑n=0∞2 2 n x 2 n(2 n)!=∑n=0∞(2 x)2 n(2 n)!=cosh 2 x cosh 2x+sinh 2x=∑n=0∞2 2 n x 2 n(2 n)!=∑n=0∞(2 x)2 n(2 n)!=cosh2 x
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Dec 31, 2015 at 18:05
answered Dec 24, 2015 at 15:13
Eleven-ElevenEleven-Eleven
8,989 9 9 gold badges 44 44 silver badges 82 82 bronze badges
3
2 Wow, very clear and well presented! I feel the crucial steps that an amateur like me could get stuck on are the index transformation from n n to n−1 n−1 in the expansion of sinh 2(x)sinh 2(x), as well as the observation that the two sums added together combine to give a simple result. Thank you very much!Steeple –Steeple 2015-12-24 15:25:39 +00:00 Commented Dec 24, 2015 at 15:25
1 very nice ...........+1 Bhaskara-III –Bhaskara-III 2015-12-24 15:27:22 +00:00 Commented Dec 24, 2015 at 15:27
1 The index transformation is crucial. Glad I could be of help Eleven-Eleven –Eleven-Eleven 2015-12-24 15:44:34 +00:00 Commented Dec 24, 2015 at 15:44
Add a comment|
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
sequences-and-series
power-series
taylor-expansion
hyperbolic-functions
cauchy-product
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
2Partials sums of cosh(x) and sinh(x)
2cosh 4 x−sinh 4 x=cosh 2 x cosh 4x−sinh 4x=cosh2 x
0Find the Taylor series of argtanh(x) using sinh(x) and cosh(x)
2Sinh, cosh, and parabolas.
1Show cosh x coshx and sinh x sinhx are continious using ε−δ ε−δ proof
6Simplifying cosh x+sinh x coshx+sinhx, cosh 2 x+sinh 2 x cosh 2x+sinh 2x, cosh 2 x−sinh 2 x cosh 2x−sinh 2x using only the Taylor Series of cosh,sinh cosh,sinh
4How to prove d d x sinh x=cosh x d d x sinhx=coshx when sinh sinh and cosh cosh are defined by an integral?
4If 1 cosh x+1 cosh y+1 cosh z=1 1 coshx+1 coshy+1 coshz=1, prove that sinh x⋅sinh y⋅sinh z≥16 2–√sinhx⋅sinhy⋅sinhz≥16 2.
Hot Network Questions
How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done?
How can the problem of a warlock with two spell slots be solved?
Repetition is the mother of learning
With with auto-generated local variables
How to home-make rubber feet stoppers for table legs?
Why include unadjusted estimates in a study when reporting adjusted estimates?
Bypassing C64's PETSCII to screen code mapping
Calculating the node voltage
Is direct sum of finite spectra cancellative?
Non-degeneracy of wedge product in cohomology
In Dwarf Fortress, why can't I farm any crops?
Making sense of perturbation theory in many-body physics
How to convert this extremely large group in GAP into a permutation group.
Riffle a list of binary functions into list of arguments to produce a result
Another way to draw RegionDifference of a cylinder and Cuboid
Cannot build the font table of Miama via nfssfont.tex
With line sustain pedal markings, do I release the pedal at the beginning or end of the last note?
Do we need the author's permission for reference
Why are LDS temple garments secret?
Overfilled my oil
The rule of necessitation seems utterly unreasonable
Is there a way to defend from Spot kick?
What’s the usual way to apply for a Saudi business visa from the UAE?
Is it safe to route top layer traces under header pins, SMD IC?
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
190728 | https://www.youtube.com/watch?v=RadBGUSomV4 | Charming Yet Devious Time Travel Puzzle Game! - Meet Me At Noon
Aliensrock
720000 subscribers
5185 likes
Description
143929 views
Posted: 22 Jun 2022
Meet Me At Noon Part 1- incredibly digestible and clever puzzles, this game's got my number! Control two halves as an hourglass through the day/night cycle, collecting stars and making sure each one hits its goal. Starts simple but becomes more and more mind-bending!
Edited by: Tyler
aliensrock #puzzle #meetmeatnoon
New to the channel? These links will help:
VOD Channel:
Discord:
Twitch:
Twitter:
159 comments
Transcript:
Intro hey guys tyler here today i'm trying out meet me at noon this is a cute and clever time travel puzzle game it's like kind of puzzle cut mostly kind of programming sort of you'll see it's neat you have two characters the sun hourglass half and the moon hourglass half both of their goals i'm going to skip the tips because i read it in my own time and i can explain it decently the goal is to get each of their characters to their respective symbols this is the sun hourglass character so i must get to the sun symbol same goes for the moon one now the stars that i pass over i think are optional the game doesn't specify but that's you know get the stars important i'm gonna be getting the stars and here's my moon character you can jump up to get the stars and you gotta make it to the moon symbol now you may have noticed that all the moves are being tracked in the bottom middle hence the programming eventually i'll sort of program the characters to make specific moves that i want and probably have them work together now notice i had two sun moves and now two moon moves when i move the moon the sun goes backwards and i think it stops once it reaches its special symbol so as long as it passes 3 NEWBORNS REUNION over the special symbol it's fine so gotta control the two characters i have five sun moves then five moon moves so i will go over the symbol and then go and fetch that star and now it's night time so i would go and fetch the moon symbol at the same time my previous character fetches the sun symbol so it's neat uh sun is just in a backwards little time loop so four sun moves four moon moves i'm gonna run this guy off a cliff just trust me it'll work out because when he goes back in time he gets the star and they're happy together in perfect harmony yin and yang here we go statues i only see a moon symbol oh maybe the sun symbol is actually oh yeah i'm on the sun symbol right now you can see there's a little icon above my guy's head so what i can do is i just press down three times so it doesn't go anywhere and that way you can stay perfectly safe don't run off the edge [Music] PAVE THE WAY pave the way of course so i go over twice i'm gonna rest once by staying down and then go left three times to make it to a symbol and then oh you know i think that's a little backwards i was hoping to stand on the other character's head let me try again so maybe i'll get the sun symbol first then go over and then wait twice so when i move over i could use it as a stepping stool and get to my own symbol missing step four steps for each of them well so what's stopping me from doing a similar solution to the previous puzzle go over there and then just wait oh i get it i get it instantly go over here jump and then wait so hear me out i'll go over here it jumps so i can get off and get the start myself incredible this requires a little pushing i see so i run off the edge and i make a move and that doesn't get me to the sun but you know what if i go here and push it over it'll make it on its own because it's programmed to move in those ways and i've now completed a chapter one nine down onto two nine possibly introducing new mechanics new ways to manipulate time perhaps oh but i didn't get the star let me just take a quick look so what if i try something like this i go left left down left and then get the star now and push it again and then go back so i'm not fully sure why that character 1-9 NEWBORNS A LITTLE PUSH moves to the right twice no i do get why because it goes backwards and it moved left in its programming so it moves right in actuality it's basically just the inverse of the inputs okay on to the next chapter so i imagine i've got to create another bridge here yeah yeah yeah get you a jump up and then collect the star stay jump to keep going so again i think i'm gonna have to become the method of transportation well if i want to go backwards i have to think about it in reverse so right then two left means that i can go with it crouch again and then jump to my goal yeah think about the sun character going 2 TWIN SPRITS BRIDGE backwards during the night time and that's all i need to remember so four sun daytime moves interesting so i could go left up down down so then you move with me and then you jump nice [Music] ELEVATOR and i must make an elevator huh how the hell am i getting moon man up that cliff well let's start with something simple why just move this way send here oh sick yeah it automatically teleports up there i've seen that happen before oh no but you moved the wrong way so i gotta stay down so you go back and i can jump off now i gotta make an escalator at seven moves apiece this does not look easy so i mean i should start by moving all the way over here i should definitely jump and then just wait for the moon character to come i got the moon star i don't think i got the sun star i should wait now and then wait to be carried up i think i missed the sun star though 5 TWIN SPRITS ESCALATOR no i got it next level the conductor i see on the timeline there's maybe only specific times where the star is active so during the third phase of the moon cycle is when i must collect the star so one of the last things i should do is make sure the moon can get over and then do a jump here and then wait i'm thinking in backwards time so now i wait jumps go over i don't collect the star oh go back move over oh that's wrong too okay let me restart then yeah you go it jump why does it happen right away maybe if i jump twice it'll be better you jump once yeah maybe it's just better to be faster and then i could jump down here collecting a star i could jump again to get back up and the only one of us have made it to the goal okay hear me out then sun character want to start with a down move to wait then go over jump go over again jump and then make it here so now it's just a lot faster as a whole collect the star jump up and make it to the goal let's go brainy power farther away looks like another pushy level well if i move twice to the left and twice twice to the right i stop the movement back to the left push you over and then you'll do like a super move oh but you're just a little short too far away okay back it up let me see some how many moves total is this that's five moves to the right so in order to get five right moves i mean basically it's a combination of the right moves with the sun character and the right word pushes from the moon character so maybe three moves right and two pushes but i also need to get the star well i could collect the star right away something like that and because the sun's got the star the moon doesn't have to worry about it oh back it up back it up i have an idea do another push here and then go over yes two pushes three rightward moves A NEW DIRECTION and a new direction so what if my moves just start like this and then i make sure this falls down and moves over on the bottom floor that one was simple next level tortoise something must move slowly well the final two moves of the sun character should be two moves to the right so i should start with two moves to the left and then should come a jump and that's why that notch is there and then i'm just gonna stand still three times so that way i can push except i miss the star so i solve the puzzle miss this star i'm going to go back and get that star so still left left jump but this time what about right left right that gets the star and i get to jump and finish the puzzle awesome 2-9 TWIN SPRITS TORTOISE dude this game's so cute can i do one more chapter let's do one more chapter yeah i really like how this game feels on the brain not too hard not too easy all very intuitive and now we got day the night of the day well it seems simple maybe i could just start by getting the star and during night time just collect the moon oh and now we get to see the moon playback it seems that the moon plays back in the same order that you make its moves NEW HORIZON A NEW DAWN so the day reverses but the moon stays the same four four four does this push the moon oh interesting starting with the day i can't actually move much then i guess i can do some i can't even jump actually all i can do is run off a cliff which is illegal so i must press down four times so then night time must just rest down here seems simple enough oh no i should probably do a jump i see so instead of staying down i could jump and then rest because i have to start by pressing down and now it pushes me up and i can move over to the goal without running out of moves [Music] [Applause] BRANCHING OUT branching out wait what is that star and that sun well first off kind of collect the star i should be able to so then with the moon i actually want to go this way first and then go over and then moon has to make it to its goal so sun can jump oops wait wait again but that's not right i won't have time to get to my goal then i i won't have the ability to get my goal whoa whoa oh no no no don't shatter the hourglass back it up a little probably to the moon cycle because something's wrong here instead i think the play is just to move over wait jump and then move right twice probably easier because again i have to start by waiting i tried moving right and it wouldn't let me but it's no matter because i make it my goal beautiful music too got the way fairer so because the sun character moves backwards why don't i try moving twice and then once so that the moon character can come here and prevent the backwards movement so that way it's three easy moves to make it a star and to my goal i really like the extra bonus the star provides you always know when the star is going to show up which just creates an extra level of planning speaking of extra level planning this is three whole three and a half day moon cycles i mean how hard could be can i just go down and then right no that's automatically wrong let me try going left then left beginning the night cycle [Music] so you can't go back but now you can go back once now it's the day cycle so i wanted to go this way oh but the night cycle starts back up here again so i can beat the level but getting the 5 NEW HORIZON FOUNTAIN star is the hard part i see who wants to get the star probably the sun so maybe move right twice block your backwards movement and jump just in case that matters it doesn't but i got the star at least but now now how do i get the sun all the way back oh i know instead of jumping up with the moon i should instead move to left to get a nice little push right from my moon friend oh i didn't really get a bush if i could go here and now push okay that works too it seems like a forgiving level PILGRIMAGE pilgrimage all right it looks like i gotta get that star really early on okay right now and then somehow we've got to make it very far over let's at least block that maybe again block it and then move towards my goal oh and actually i could just walk four times alright fairly simple WALK ON AIR walk on air again i have to get that star early well what happens if i move left then right then right and then say if i move my moon character here you're just stuck i get pushed over though and it's literally walking on air it's always so cute when the the solution to the puzzle is like in the title but it still isn't immediately clear and this one's called airplane well here's what i know the moon's moves have to be three to the right otherwise it's not making it's my goal so the stars i guess doesn't have to be hit no the star can be hit by the moon character actually because if i jump and then go left left i think this will still work okay on stop moving on your own and hit the star and then go over i guess it didn't have to be a right right right because the airplane moves it oh [Music] hold on the moon still needs help to get to the edge [Music] like that but now the sun's not at its goal so you know what okay let me try my original idea but make a small amendment where i go here what if you jump okay don't jump wait again and then go over so now it's got a repeat but what if i jump oh i can't make you jump that worked because i left my son character in its 8 NEW HORIZON AIRPLANE spot it was a lot to do that i'm going to be honest that one i'm gonna have to look back and recording to fully understand i'll carry on to the next level now i'm sure i'll understand it once looking back so it's called stairway to heaven and the star can only be hit in the final day cycle well six moves isn't that just six moves to the right okay so it seems like maybe most the action hap has to excuse me has to happen during the night time [Music] so if i jump here this doesn't move over and i'm at a slightly different spot oh but i still gotta wait okay i can't make it to that star easily all the final levels are easy to beat hard to get the star for how do we do this i still maintain the first six moves have to be going down and move up here oh interesting i can jump and keep moving through this and then block you from moving over further [Music] what if i jump oh i get pushed back up i like that okay that's not right then either jump jump move here oh that's how it's done yeah this should give me enough time i think because you jump over me and now i can make this star and back and that's the world first three worlds taken down really cute game i think the word programming was misleading but you kind of programmed two characters it's just very basic programming anyways that was meet me for noon out on steam now if you want to check it out thank you guys all for watching hope you enjoyed if you want to see more let me know and i'll see y'all in the next video have a wonderful day and peace |
190729 | https://math.stackexchange.com/questions/2469501/calculating-the-position-of-the-median | statistics - Calculating the position of the median - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Calculating the position of the median
Ask Question
Asked 7 years, 11 months ago
Modified4 years, 9 months ago
Viewed 6k times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
I know this is a simple question, but I cannot find a straight answer anywhere.
When calculating the medium of listed data, the formula is (n+1)/2. My Statistics teacher said for grouped data the position of the medium is n/2. However, this seems contradictory as for discrete grouped data, the data could be written in a list if the original values were known. Therefore two different values for the median are found.
I know this question is probably going to get flagged as it has already been asked, however, it has never been answered. I find it frustrating that such a fundamental concept in Statistics, what is supposed to be precise and never subjective has a wishy-washy answer.
Rant over, I think the position should be (n+1)/2 for discrete data and n/2 for continuous data. Moreover, this then leads into the obvious question of how we should calculate quartiles as well.
It would be nice if everyone could agree on a certain method for calculating the medium as all my textbooks are saying different things.
Someone please dispell the confusion and this statistical mess.
statistics
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
asked Oct 12, 2017 at 18:31
InquirerInquirer
875 2 2 gold badges 9 9 silver badges 22 22 bronze badges
0
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
1
Save this answer.
Show activity on this post.
Answer 1: The n+1 2 n+1 2 formula you provided is, at best, a bit sloppy unless interpreted properly.
Example: If your data is {3,5,9,10}{3,5,9,10}, then your median can be found in position 2.5. Since that's not a whole number, we usually interpret it as "the average of entries 2 and 3," or in this case, 7 7. This is a convention; there's no theoretically deep reason for why it's the "correct thing to do."
Answer 2: You're hoping for too much from the formula for the grouped median. It is not possible to actually extract a median with any accuracy, because the process of grouping destroys so much information in the dataset.
Example: The above data could be presented as
Bin 1−3 4−6 7−9 10−12 Count 1 1 1 1 Bin Count 1−3 1 4−6 1 7−9 1 10−12 1
There are many different ways to compute the median from grouped data presented in this way. (What number do you use as the "class representative?" For the 1−3 1−3 class, is the representative 2 2 as the midpoint of 1 1 and 3 3, or is it 2.5 2.5 as the midpoint of the adjacent lower class limits 1 1 and 4 4?) Most of the calculations you could do won't reproduce the original answer of 7 7. And just in case any of them would, notice that I could switch the 5 5 in my original dataset for 4 4, which would change the median calculated in example 1 to 6.5 6.5 but possess the exact same frequency distribution above.
Example 3: You're hoping for too much in general with consistency of medians and quartiles at all. If you play around with some different statistical software, you might notice that medians, quartiles, etc. of the exact same datasets could be calculated differently. There is no "true correct" algorithm for what these should be; there are various conventions that are all fairly reasonable for how one could compute them. In many introductory statistics experiences, one averages two adjacent numbers when needed; however, some statistical software packages like R will do some linear interpolation of numbers and get slightly different results.
TL;DR: There's no right way to do this, so go with what your teacher says in each context.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Oct 12, 2017 at 19:18
answered Oct 12, 2017 at 19:09
Aaron MontgomeryAaron Montgomery
11k 1 1 gold badge 19 19 silver badges 29 29 bronze badges
1
Thanks, I think I'll just have to accept that that this is a case where there is no right answer and just a series of best practices.Inquirer –Inquirer 2017-10-12 19:17:25 +00:00 Commented Oct 12, 2017 at 19:17
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
statistics
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
2Can I calculate a percentile given only quartile values?
0Getting the upper and lower quartiles in data with an even number of observations, or where the quartile lands on a decimal number
0Calculating the Median from a Withington Census
1Can grouped data be discrete?
1Why is finding the median and quartiles different on a cumulative frequency graph? (Or is it?)
Hot Network Questions
Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations?
Does "An Annotated Asimov Biography" exist?
ConTeXt: Unnecessary space in \setupheadertext
Do we need the author's permission for reference
How to solve generalization of inequality problem using substitution?
Is existence always locational?
Is direct sum of finite spectra cancellative?
Gluteus medius inactivity while riding
Discussing strategy reduces winning chances of everyone!
Storing a session token in localstorage
What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left?
Stress in "agentic"
Xubuntu 24.04 - Libreoffice
Overfilled my oil
How can the problem of a warlock with two spell slots be solved?
Why do universities push for high impact journal publications?
Is encrypting the login keyring necessary if you have full disk encryption?
How to locate a leak in an irrigation system?
The rule of necessitation seems utterly unreasonable
Alternatives to Test-Driven Grading in an LLM world
Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation
Childhood book with a girl obsessed with homonyms who adopts a stray dog but gives it back to its owners
What's the expectation around asking to be invited to invitation-only workshops?
Do sum of natural numbers and sum of their squares represent uniquely the summands?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
190730 | https://math.stackexchange.com/questions/1500547/evaluate-p1-p-1 | real analysis - Evaluate $P(1)/P(-1)$ - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Evaluate P(1)/P(−1)P(1)/P(−1)
Ask Question
Asked 9 years, 11 months ago
Modified5 years ago
Viewed 2k times
This question shows research effort; it is useful and clear
3
Save this question.
Show activity on this post.
The polynomial f(x)=x 2007+17 x 2006+1 f(x)=x 2007+17 x 2006+1 has distinct zeroes r 1,…,r 2007 r 1,…,r 2007. A polynomial P P of degree 2007 2007 has the property that P(r j+1 r j)=0 P(r j+1 r j)=0 for j=1,…,2007 j=1,…,2007. Determine the value of P(1)/P(−1)P(1)/P(−1).
Let P(x)=a 2007 x 2007+...+a 0 P(x)=a 2007 x 2007+...+a 0.
r j+1 r j=r 2 j+1 r j r j+1 r j=r j 2+1 r j
So we need the map: x+1 x→x x+1 x→x.
We need a u→x u→x. Where u(x+1/x)=x u(x+1/x)=x.
The inverse function is: u=1 2⋅(x±x 2−4−−−−−√)u=1 2⋅(x±x 2−4)[USED WOLFRAM ALPHA]
So,
P(x)=(1 2⋅(x±x 2−4−−−−−√))2007+17(1 2⋅(x±x 2−4−−−−−√))2006+1 P(x)=(1 2⋅(x±x 2−4))2007+17(1 2⋅(x±x 2−4))2006+1
But letting x=1 x=1 or −1−1 makes it so that the answer is imaginary!
real-analysis
complex-analysis
algebra-precalculus
analysis
polynomials
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
asked Oct 27, 2015 at 18:44
Amad27Amad27
11.6k 5 5 gold badges 49 49 silver badges 119 119 bronze badges
2
Not only that, your P(x)P(x) isn't even a polynomial!Akiva Weinberger –Akiva Weinberger 2015-10-27 19:46:26 +00:00 Commented Oct 27, 2015 at 19:46
Let's write P+(x)P+(x) for the version with ++ instead of ±±, and P−(x)P−(x) for the version with −− instead of ±±. Neither of those are polynomials. However, you can use these to find something that is.Akiva Weinberger –Akiva Weinberger 2015-10-27 19:48:09 +00:00 Commented Oct 27, 2015 at 19:48
Add a comment|
2 Answers 2
Sorted by: Reset to default
This answer is useful
4
Save this answer.
Show activity on this post.
Continuing after step (2)(2) in @Jack D'Aurizio's answer.
P(1)P(−1)=∏i=1 2007 r 2 i−r i+1 r 2 i+r i+1=∏i=1 2007(r i+ω)(r i+ω 2)(r i−ω)(r i−ω 2)=∏i=1 2007(−ω−r i)(−ω 2−r i)(ω−r i)(ω 2−r i)=∏i=1 2007 f(−ω)f(−ω 2)f(ω)f(ω 2)=(17 ω 2)(17 ω)(2+17 ω 2)(2+17 ω)=289 259 P(1)P(−1)=∏i=1 2007 r i 2−r i+1 r i 2+r i+1=∏i=1 2007(r i+ω)(r i+ω 2)(r i−ω)(r i−ω 2)=∏i=1 2007(−ω−r i)(−ω 2−r i)(ω−r i)(ω 2−r i)=∏i=1 2007 f(−ω)f(−ω 2)f(ω)f(ω 2)=(17 ω 2)(17 ω)(2+17 ω 2)(2+17 ω)=289 259
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
answered Sep 14, 2020 at 7:46
SarGeSarGe
3,081 2 2 gold badges 10 10 silver badges 29 29 bronze badges
Add a comment|
This answer is useful
3
Save this answer.
Show activity on this post.
For first, 1 r i+r i=1 r j+r j 1 r i+r i=1 r j+r j is equivalent to (r i−r j)(r i r j−1)=0(r i−r j)(r i r j−1)=0. Since f(x)f(x) is not a palyndromic polynomial, the set of numbers r i+1 r i r i+1 r i is exactly the set of roots of P P.
Assuming that P P is a monic polynomial, Vieta's formulas give:
P(1)=∏i=1 2007(1−r i−1 r i),P(−1)=−∏i=1 2007(1+r i+1 r i)(1)(1)P(1)=∏i=1 2007(1−r i−1 r i),P(−1)=−∏i=1 2007(1+r i+1 r i)
hence it follows that:
P(1)P(−1)=∏i=1 2007 r 2 i−r i+1 r 2 i+r i+1=∏i=1 2007 r i−1 r i+1⋅∏i=1 2007 r 3 i−1 r 3 i+1(2)(2)P(1)P(−1)=∏i=1 2007 r i 2−r i+1 r i 2+r i+1=∏i=1 2007 r i−1 r i+1⋅∏i=1 2007 r i 3−1 r i 3+1
where the first term of the RHS can be easily computed from f(1)f(1) and f(−1)f(−1), and the whole expression just depends on the values of f f over the sixth roots of unity, due to r 2 i+r i+1=(r i−ω 2)(r i−ω 4)r i 2+r i+1=(r i−ω 2)(r i−ω 4), where ω=exp(2 π i 6)ω=exp(2 π i 6). Given (2)(2), it follows that:
P(1)P(−1)=f(ω)⋅f(ω 5)f(ω 2)⋅f(ω 4)=17 ω 2⋅(−17 ω)(2−17 ω)⋅(2+17 ω 2)=1 297 578−i 3√2.(3)(3)P(1)P(−1)=f(ω)⋅f(ω 5)f(ω 2)⋅f(ω 4)=17 ω 2⋅(−17 ω)(2−17 ω)⋅(2+17 ω 2)=1 297 578−i 3 2.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited Oct 27, 2015 at 20:59
answered Oct 27, 2015 at 20:07
Jack D'AurizioJack D'Aurizio
372k 42 42 gold badges 419 419 silver badges 886 886 bronze badges
3
The answer is 289/259. But why the sixth root of unity?Amad27 –Amad27 2015-10-28 08:00:55 +00:00 Commented Oct 28, 2015 at 8:00
I mean, did you just solve x 2+x+1=0 x 2+x+1=0?Amad27 –Amad27 2015-10-28 08:02:11 +00:00 Commented Oct 28, 2015 at 8:02
Old post, but Vieta's formulas have nothing to do with this argument. It's just factoring the polynomial and plugging in values.Joshua P. Swanson –Joshua P. Swanson 2020-09-17 19:58:37 +00:00 Commented Sep 17, 2020 at 19:58
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
real-analysis
complex-analysis
algebra-precalculus
analysis
polynomials
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Linked
2To find the equation whose roots are some function of roots of another given equation.
Related
7Prove that any cubic polynomial with one real root has Galois group isomorphic to S 3 S 3
1Splitting field of x 5−1 x 5−1
0Do roots of a monic polynomial contain an open cube?
8The Galois group of polynomial p(x)∈K[x]p(x)∈K[x] is cyclic and is generated by q(x)∈K[x]q(x)∈K[x].
0Sum and product of roots of a polynomial
6Polynomial Big List: Find the polynomial whose roots are given by some functions of the roots of given polynomials.
5Why do we need to include the factor a 2 n−2 n a n 2 n−2 in the discriminant of a polynomial?
3Polynomial problem with vieta formulas
0Proof of Unique Factorisation of Polynomials over C C by Identity Principle
Hot Network Questions
Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator
The rule of necessitation seems utterly unreasonable
For every second-order formula, is there a first-order formula equivalent to it by reification?
What were "milk bars" in 1920s Japan?
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
Why include unadjusted estimates in a study when reporting adjusted estimates?
Cannot build the font table of Miama via nfssfont.tex
Determine which are P-cores/E-cores (Intel CPU)
Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward"
в ответе meaning in context
An odd question
alignment in a table with custom separator
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left?
Implications of using a stream cipher as KDF
With line sustain pedal markings, do I release the pedal at the beginning or end of the last note?
Who is the target audience of Netanyahu's speech at the United Nations?
What's the expectation around asking to be invited to invitation-only workshops?
Bypassing C64's PETSCII to screen code mapping
My dissertation is wrong, but I already defended. How to remedy?
Childhood book with a girl obsessessed with homonyms who adopts a stray dog but gives it back to its owners
How can the problem of a warlock with two spell slots be solved?
I have a lot of PTO to take, which will make the deadline impossible
Xubuntu 24.04 - Libreoffice
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
190731 | https://people.chem.ucsb.edu/neuman/robert/orgchembyneuman.book/04%20Stereochemistry/04FullChapt.pdf | (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 0 Chapter 4 Stereochemistry from Organic Chemistry by Robert C. Neuman, Jr. Professor of Chemistry, emeritus University of California, Riverside orgchembyneuman@yahoo.com Chapter Outline of the Book I. Foundations 1. Organic Molecules and Chemical Bonding 2. Alkanes and Cycloalkanes 3. Haloalkanes, Alcohols, Ethers, and Amines 4. Stereochemistry 5. Organic Spectrometry II. Reactions, Mechanisms, Multiple Bonds 6. Organic Reactions (Not yet Posted) 7. Reactions of Haloalkanes, Alcohols, and Amines. Nucleophilic Substitution 8. Alkenes and Alkynes 9. Formation of Alkenes and Alkynes. Elimination Reactions 10. Alkenes and Alkynes. Addition Reactions 11. Free Radical Addition and Substitution Reactions III. Conjugation, Electronic Effects, Carbonyl Groups 12. Conjugated and Aromatic Molecules 13. Carbonyl Compounds. Ketones, Aldehydes, and Carboxylic Acids 14. Substituent Effects 15. Carbonyl Compounds. Esters, Amides, and Related Molecules IV. Carbonyl and Pericyclic Reactions and Mechanisms 16. Carbonyl Compounds. Addition and Substitution Reactions 17. Oxidation and Reduction Reactions 18. Reactions of Enolate Ions and Enols 19. Cyclization and Pericyclic Reactions (Not yet Posted) V. Bioorganic Compounds 20. Carbohydrates 21. Lipids 22. Peptides, Proteins, and α−Amino Acids 23. Nucleic Acids Note: Chapters marked with an () are not yet posted. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 1 4: Stereochemistry Preview 4-3 4.1 Tetrahedral Carbon Configurations 4-3 Two Configurations at Tetrahedral Carbon (4.1A) 4-3 Non-Superimposable Mirror Images Handedness and Chirality Chiral Atoms (4.1B) 4-4 Chiral Carbon Atoms Other Chiral Atoms Molecular Chirality 4.2 Stereoisomers and R,S Assignments 4-6 R and S Nomenclature (4.2A) 4-6 Clockwise and Counterclockwise Isomers The Assignments of R and S R and S Assignment Rules (4.2B) 4-8 Case 1. Each Atom Directly Bonded to a Chiral C is Different Case 2. Two or More Atoms Bonded to a Chiral C are the Same Case 3. Groups with Double and Triple Bonds More Complex Molecules 4.3 The Number and Types of Stereoisomers 4-13 Compounds Can Have 2n Stereoisomers (4.3A) 4-13 2-Bromo-3-chlorobutane Configuration at C2 in the (2R,3R) Isomer Configuration at C2 in the other Stereoisomers Relationships Between Stereoisomers (4.3B) 4-15 Enantiomers Diastereomers Compounds with Fewer than 2n Stereoisomers (4.3C) 4-17 2,3-Dibromobutane Meso Form 4.4 Drawing Structures of Stereoisomers 4-21 3-D Conformations of Stereoisomers (4.4A) 4-21 Many Ways to Draw the Same Stereoisomer 3-D Structures for Comparing Stereoisomers Fischer Projections (4.4B) 4-24 Definition of Fischer Projections Manipulations of Fischer Projections Using Fischer Projections to Draw Stereoisomers (continued next page) (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 2 4.5 Cyclic Molecules 4-33 Cyclic Stereoisomers (4.5A) 4-33 Chiral Centers in 1-Bromo-3-methylcyclohexane Stereoisomers of 1-Bromo-3-methylcyclohexane Stereochemical Relationships between cis and trans Isomers Isomeric Bromomethylcyclohexanes Drawings of Cyclic Stereoisomers (4.5B) 4-37 Wedge-Bond Structures Chair Forms Haworth Projections 4.6 Optical Activity 4-39 Rotation of Plane Polarized Light and the Polarimeter (4.6A) 4-39 Polarimeter Light Rotation by the Sample Magnitude and Sign of Light Rotation (4.6B) 4-41 Observed versus Specific Rotation Specific Rotations of Enantiomers Relative and Absolute Configurations Specific Rotations of Diastereomers d and l Isomers Racemic Mixture Appendix A: Resolution of Stereoisomers 4-43 Resolution of Diastereomers Resolution of Enantiomers Appendix B: Optical Purity 4-46 %Optical Purity Enantiomeric Excess (%ee) Appendix C: Absolute Configuration 4-47 Chapter Review 4-49 Feature: What a Difference a Configuration Makes 4-51 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 3 4: Stereochemistry •Tetrahedral Carbon Configurations •Stereoisomers and R,S Assignments •The Number and Types of Stereoisomers •Drawing Structures of Stereoisomers •Cyclic Molecules •Optical Activity Preview Nomenclature rules for organic compounds allow us to draw their chemical bonds and show specific positions of atoms and groups on their carbon skeletons. We can draw 3-dimensional structures for these molecules based on the tetrahedral structure of their C atoms and we know that they have many different conformations due to rotation about their chemical bonds. In this chapter, we will learn that there is a property of tetrahedral carbon atoms that causes some chemical names that we have learned to inadequately describe a unique molecule. For example, there are two different molecules with the name 2-bromobutane because there are two different ways to bond a set of four atoms or groups to a tetrahedral atom. This stereochemical property of tetrahedral C is present in all molecules, but only leads to different structures in some of them. This chapter will vigorously exercise your ability to picture objects in three dimensions. Your molecular model kit will be a very important aid to learning the material in this chapter. 4.1 Tetrahedral Carbon Configurations There are two different ways to bond four different atoms or groups to a tetrahedral carbon. Two Configurations at Tetrahedral Carbon (4.1A) We use bromochlorofluromethane (CHBrClF) to illustrate the two ways of bonding four different atoms to a tetrahedral C. [graphic 4.1] Non-Superimposable Mirror Images. The two structures of CHBrClF labelled (A) and (B) are different from each other because no matter how they are each oriented in space, they can never be superimposed on each other. If you correctly superimpose each of the halogens (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 4 atoms of (A) and (B) on each other, you will find that their carbon atoms, and also their hydrogen atoms, are far away from each other as we show in Figure [grapic 4.2]. [graphic 4.2] Alternatively if we superimpose the C atoms and H atoms of (A) and (B) on each other, the halogen atoms do not correctly overlap with each other. As a result, we say that (A) and (B) are non-superimposable and that their C atoms have different configurations. We illustrate that (A) and (B) are mirror images of each other by showing in Figure [graphic 4.3] that the mirror image of one of them is identical to the other. [graphic 4.3] If you rotate the mirror image of (A) around the axis shown, it is completely superimposable on (B). The mirror image of (A) is (B), and the mirror image of (B) is (A). Handedness and Chirality. (A) and (B) differ from each other like a right hand differs from a left hand. Right and left hands have the same component parts attached in the same way to each other, but they cannot be superimposed on each other. Like right and left hands, (A) and (B) are mirror images of each other. Because of this analogy with hands, chemists say that the two different configurations of C in CHBrClF ((A) and (B)) have the property of handedness. Chemists use the term chirality to refer to the property of handedness when it applies to molecules. A molecule is chiral if it cannot be superimposed on its mirror image. As a result, the (A) and (B) structures of CHBrClF are chiral molecules. Chiral Atoms (4.1B) A molecule is usually chiral because it contains one or more chiral atoms. However we will see below that specialized molecules can be chiral even when they have no chiral atoms. Chiral Carbon Atoms. A carbon atom must have four different atoms or groups bonded to it in order to be chiral. If two or more of the groups or atoms on a tetrahedral C are identical, the C cannot be chiral and we describe it as achiral. While CHBrClF has a chiral C, the compound CH2BrCl is achiral because it has a tetrahedral C on which two of the bonded atoms are the same (the two H's). We confirm that CH2BrCl is not a chiral molecule by showing in Figure [graphic 4.4] that its mirror image is superimposable on the original molecule. [graphic 4.4] Other Chiral Atoms. Chiral molecules can also result from the presence of chiral atoms other than C such as the chiral N in a tetraalkylaminium ion. [graphic 4.5] The N is chiral because it is an atom with tetrahedral bond angles like C and it has four different alkyl groups bonded to it. As a result, the mirror image of this molecule is non-superimposable on the original structure so it is a chiral molecule. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 5 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 6 Amines. The nitrogen atom of an amine (R3N:) can also be chiral since amines have a tetrahedral (pyramidal) structure. If the three R groups are different from each other as shown in Figure [graphic 4.6], the mirror image of an amine is not superimposable on the original amine. [graphic 4.6] The unshared electron pair in the sp3 orbital is like a fourth "group". Even when they possess chiral N atoms, amines are not considered chiral compounds because they undergo amine inversion (Chapter 3) at a very rapid rate (about 1011 times per second for NH3) as we show in Figure [graphic 4.7]. [graphic 4.7] This inversion allows an amine to rapidly change into its non-superimposable mirror image. As a result, unlike aminium ions and compounds with chiral C, it is not possible to individually isolate just one of the chiral forms of an amine. Molecular Chirality Without Chiral Atoms. An example of a chiral molecule without a chiral atom is (A) in Figure [graphic 4.8]. [graphic 4.8] While it has no chiral atoms, this molecule is chiral because it is not superimposable on its mirror image. The mirror image of (A) cannot be superimposed on (A) no matter how it is oriented in space so it is a different compound that we label as (B). You can demonstrate this by making models of (A) and its mirror image (B) using a molecular model set. There are relatively few chiral molecules that have no chiral atoms. 4.2 Stereoisomers and R,S Assignments A chiral molecule and the molecule that is its non-superimposable mirror image are stereoisomers of each other. Based on the nomenclature rules that we have learned so far, stereoisomers have the same chemical name such as the pair of stereoisomers (A) and (B) of CHBrClF that both have the name bromochlorofluoromethane. In order to distinguish (A) and (B), we use additional nomenclature that we describe here. R and S Nomenclature (4.2A) We distinguish the two different stereoisomers of CHBrClF with the prefixes R or S so that their complete names are (R)-bromochlorofluoromethane and (S)-bromochlorofluoromethane. R and S describe the two different configurations at the chiral C and we will show below how we assign them to the two stereoisomers using a set of rules applicable to any chiral atom. Clockwise and Counterclockwise Isomers. In order to assign R and S to a chiral C, we will learn a set of rules that allows us to uniquely give the priority numbers "1", "2", "3", and "4" to each atom or group on a chiral C. For the moment, let's not worry about these rules. We first need to recognize that once the priority numbers are correctly assigned to the (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 7 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 8 four atoms or groups on the chiral C, there are two different ways that these priority numbers can appear on the tetrahedral C. [graphic 4.10] When we orient each of these two structures so that "4" is behind the chiral C, our views of these structures when we look at the chiral C's show "1", "2", and "3" progressing "clockwise" in one structure and "counterclockwise" in the other. The Assignments of R and S. Chemists use a set of rules called the Cahn-Ingold-Prelog system for assigning the priority numbers "1" through "4" to the atoms or groups on a chiral C or other chiral atom. After we use these rules to assign the priority numbers to the specific atoms or groups, we refer to the clockwise isomer in Figure [graphic 4.10] as the R isomer, and the counterclockwise isomer as the S isomer. R comes from the latin word "rectus" which means the direction "right". When the numbers "1", "2", and "3" progress in a clockwise direction we think of them as progressing toward the "right" as we show in Figure [graphic 4.11]. [graphic 4.11] The letter S comes from the latin word "sinister" which means the direction "left". When the numbers "1", "2", and "3" progress in a counterclockwise direction we think of them as progressing toward the "left". We show in the next section how we use the Cahn-Ingold-Prelog rules to assign the numbers "1", "2", "3", and "4" to atoms or groups bonded to a tetrahedral atom. Remembering R and S. If you have trouble remembering that a "clockwise" progression of the priority numbers is progression to the "right" with respect to direction, you can think of the "clockwise" progression as being "right" with respect to "correctness". That does not mean that "counterclockwise" progression is "wrong", it still is "left" R and S Assignment Rules (4.2B) The Cahn-Ingold-Prelog method uses the atomic numbers of the atoms bonded directly or indirectly to the chiral atom. We illustrate each rule with an example before stating the rule. Case 1. Each Atom Directly Bonded to a Chiral C is Different. Our example is CHBrClF that we first used to illustrate a chiral molecule. Br has the highest atomic number (35) so we assign it priority number "1", Cl has the next highest atomic number (17) so we assign it priority number "2", we assign priority number "3" to F since it has the third highest atomic number (9), while we assign priority number "4" to H because it has the lowest atomic number (1). (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 9 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 10 When we orient the molecule so that the priority "4" atom (H in this case) points away from us, and then view each stereoisomer along the C-"4" (C-H) bond, we see the two stereoisomers labelled with their respective priority numbers as the views that we show in Figure [graphic 4.12]. [graphic 4.12] The first structure with the clockwise progression of the atoms labelled "1", "2", and "3" has the R configuration at C. The other structure, with the counterclockwise progression has the S configuration at C. Chiral C's directly bonded to four different atoms are easy to designate R or S even when these atoms are part of a larger group. We show two compounds where we have assigned priority numbers to the groups based only on the atomic numbers of the atoms directly bonded to the chiral C. [graphic 4.13] All of these examples fit the first "R,S assignment rule" that states: Rule 1. Substituents on a chiral atom are given the priority numbers "1", "2", "3", "4" in order of decreasing atomic number of the atom directly connected to the chiral atom. Unfortunately, there are not many molecules that fit Rule 1. While molecules with chiral C's must have four different groups on each chiral C, two or more of the atoms directly bonded to the chiral C are frequently the same. Case 2. Two or More Atoms Bonded to a Chiral C are the Same. When a compound has four different groups bonded to the chiral C (labelled as C), and two or more of the directly bonded atoms are the same (as in 2-bromobutane shown in Figure [graphic 4.14]), you must examine atoms beyond those directly bonded to C to assign the correct priority numbers. [graphic 4.14] The four different atoms or groups bonded to the C are Br, H, CH3, and CH2CH3 so the atoms directly bonded to C are Br, H, C and C. We immediately assign priority number "1" to Br since it has the highest atomic number, and priority number "4" to H since it has the lowest atomic number. However we need another rule to assign priority numbers "2" and "3" to CH3, and CH2CH3 since they both have C bonded to C. When we cannot initially distinguish two groups such as CH3 and CH2CH3 because their "first level" atoms directly connected to C are the same (both are C in this case), we look at their "second level" atoms. The second level of atoms in each group are those bonded to the C directly bonded to C. For CH3, they are the three H's that are shaded in Figure [graphic 4.15]. For CH2CH3, they are the two shaded H's and the shaded C of its CH3 group. [graphic 4.15] (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 11 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 12 For each of these groups we list these "second level" shaded atoms in decreasing order of atomic number and this gives the sequence (H, H, H) for CH3, and (C, H, H) for CH2CH3. We then compare the atomic numbers of the atoms in these sequences in the order that they are written until we find the first point of difference. In (C, H, H), the first atom (C) has atomic number 6, while in (H, H, H), the first atom (H) has atomic number 1. Since atomic number 6 is higher than 1, we immediately assign the higher priority number "2" to CH2CH3 (with the sequence (C,H,H)) and the lower priority number "3" to CH3 (with the sequence (H,H,H)) without comparing any other atoms in the sequences. [graphic 4.16] In another example, C is bonded to Br, CH2OH, (CH3)3C, and H. [graphic 4.17] We immediately assign priority numbers "1" and "4" to Br and H, respectively, because they have the highest and lowest atomic numbers of all of the atoms directly bonded to C, but once again the directly bonded (first level) atoms for the other two groups are C. The sequence of "second level" atoms bonded to those C's in order of decreasing atomic number is (C, C, C) for C(CH3)3 and (O, H, H) for CH2OH as we show in Figure [graphic 4.18]. [graphic 4.18] When we begin our comparison of the sequences with the first atom in each sequence, O has a higher atomic number than C so we immediately give CH2OH a higher priority number than C(CH3)3. Although there are 2 additional C's in the sequence (C, C, C) compared to 2 H's in the sequence (O, H, H), we ignore these additional atoms once we have identified the first point of difference. The assigned priority numbers, and the resulting R and S isomers, are shown below. [graphic 4.19] These compounds require the use of a second "R,S Assignment Rule" that states: Rule 2. When two or more groups have the same type of atom directly bonded to the chiral atom, their priority numbers depend on the atomic numbers of their "second level" atoms. If the second level atoms in one group are X, Y, and Z, whose order of atomic numbers is X>Y>Z, we group them as (X, Y, Z) and compare them in that order (one at a time) with "second level" atoms arranged in the same way for the other group or groups bonded to the chiral atom. When we find a difference in atomic number, we assign the higher priority number to the group with the higher atomic number atom in the comparison. Case 3. Groups with Double and Triple Bonds. There are special rules for molecules with double or triple bonds (eg. C=C, C≡C, C=O, or C≡N ) in groups bonded to C. [graphic 4.20] In the example with C=O, we treat C=O as if it is O-C-O. As a result, the sequence of (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 13 "second level" atoms bonded to C in that example is (O, O, H) and these are circled in the structure. Similarly, we treat the C of C≡N as if it has three bonded N atoms so the sequence of "second level" atoms in that example is (N, N, N) and we have circled them. The rule that applies to double and triple bonds in groups bonded to chiral atoms is: Rule 3. A double bond to an atom is replaced by two single bonds to the same type of atom and a triple bond to an atom is replaced by three single bonds to the same type of atom. More Complex Molecules. In more complex molecules, you will need to go beyond the "second level" of atoms in groups bonded to C to find the first point of difference. In each case you must compare sequences of atoms at the same level with each other until you find the point of difference as we have described for comparisons of "second level" atoms. 4.3 The Number and Types of Stereoisomers The maximum number of stereoisomers for a molecule increases exponentially as the number of chiral atoms in the molecule increases. Compounds Can Have 2n Stereoisomers (4.3A) We have seen that a molecule with 1 chiral C has 2 stereomers. If a molecule contains 2 chiral C's, it is possible for it to have 4 stereoisomers, while a molecule with 3 chiral carbons can have 8 stereoisomers. A molecule with n chiral atoms may have up to 2n stereoisomers. 2-Bromo-3-chlorobutane. A molecule with 2 chiral C's and 4 stereoisomers is 2-bromo-3-chlorobutane. [graphic 4.22] C1 and C4 are achiral because they each have 3 H's, but C2 and C3 are chiral because they each have 4 different groups. Those on C2 are CH3, H, Br, and CHClCH3, while those on C3 are CH3, H, Cl, and CHBrCH3. [graphic 4.23] We treat C2 and C3 as separate chiral centers, so the fact that some groups on C2 are the same as some on C3 does not affect the individual chirality of C2 or C3. These separate chiral centers (C2 and C3) can each be R or S, so the names of the four possible stereoisomers are (2R,3R)-2-bromo-3-chlorobutane, (2S,3S)-2-bromo-3-chlorobutane, (2R,3S)-2-bromo-3-chlorobutane, and (2S,3R)-2-bromo-3-chlorobutane. [graphic 4.24] For easy comparison, we have shown each stereoisomer in an orientation where C2 and C3 lie on a vertical line, their H's project back into the paper, and their other two groups project out toward you. (We will see later in Section 4.4 that there are other ways to draw these stereoisomers). (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 14 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 15 Before we discuss the relationships between these four stereoisomers, we need to be sure that we have made the correct assignments of R and S at C2 and C3 in Figure [graphic 4.24]. We verify this below for C2 of the (2R, 3R) isomer using the rules that we presented earlier. Configuration at C2 in the (2R,3R) Isomer. We have taken (2R,3R)-2-bromo-3-chlorobutane from Figure [graphic 4.24] and redrawn it as structure (1) in Figure [graphic 4.25]. [graphic 4.25] This figure is a stepwise illustration verifying that C2 has the R configuration. The assignment of configuration to C2 does not depend on the configuration of C3 so in the first step we remove the wedges and dashes from C3, and write it and its groups simply as CHClCH3 as shown in structure (2). The atoms directly bonded to C2 in order of decreasing atomic number are Br, C, C, and H so in the second step we can immediately assign the priority numbers "1" and "4" to Br and H, respectively. However to assign priority numbers "2" and "3", we must analyze the "second level" atoms in the other two groups on C2. These are (H, H, H) for CH3, and (Cl, C, H) for CHClCH3, so the first atom in each sequence leads us to the assignment of "2" to CHClCH3 and "3" to CH3 as we show on structure (3). We orient the molecule so that the priority "4" atom (H) is directed away from us by "lifting up" the CHClCH3 group from the plane of the paper in the third step. This moves the H further away from us as we show in structure (4). Finally in the fourth step, we connect priority numbers "1", "2", and "3" with arrows that show the direction of their progression in structure (5). The "clockwise" progression verifies the assignment of R to C2 in this (2R,3R) stereoisomer. Configuration at C2 in the other Stereoisomers. At this point we can go back and look at C2 in all four stereoisomers of 2-bromo-3-chlorobutane (Figure [graphic 4.24]). By inspecting these structures you should be able to see that the configuration at C2 in the (2R,3S) isomer is the same as that at C2 in the (2R,3R) isomer that we just analyzed. You should also be able to see that the configurations at C2 in either (2S,3S) or (2S,3R) are th same as each other and are mirror images of the C2 configuration in either (2R,3R) or (2R,3S). We ignored the stereochemistry of the C3 group when we analyzed C2, but you can now separately verify the configuration at C3 in the same way that we just described for C2. Relationships Between Stereoisomers (4.3B) The four stereoisomers of 2-bromo-3-chlorobutane in Figure [graphic 4.24] differ from each other because of the differences in their configurations at C2 and C3. While we can uniquely (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 16 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 17 refer to each of them by their designations (2R,3R) etc., we also use the general terms enantiomer and diastereomer to describe their relationships to each other. Enantiomers. Enantiomers are stereoisomers that are mirror images of each other, but non-superimposable. We have seen that (R) and (S)-bromochlorofluoromethane are non-superimposable mirror images, so they are enantiomers of each other as we illustrate in Figure [graphic 4.26]. [graphic 4.26] Among the four stereoisomers of 2-bromo-3-chlorobutane (Figure [graphic 4.24]), the (2R,3R) and (2S,3S) isomers are mirror images of each other, and so are the (2R,3S) and (2S,3R) isomers. Since they are also non-superimposable, (2R,3R) and (2S,3S) are enantiomers of each other, and (2R,3S) and (2S,3R) are also enantiomers of each other (Figure [graphic 4.26]). Diastereomers. Any pair of stereoisomers of a compound that is not a pair of enantiomers is a pair of diastereoisomers. For example, the (2R,3R) and (2R,3S) stereoisomers of 2-bromo-3-chlorobutane are not enantiomers of each other since they are not mirror images. As a result, these two stereoisomers are diastereomers of each other. Based on these definitions, a pair of stereoisomers of a compound is either a pair of enantiomers or a pair of diastereomers. We see this in Table 4.1 that summarizes all of the pair-wise relationships between the stereoisomers of 2-bromo-3-chlorobutane. Table 4.1. Relationships Between the Stereoisomers of 2-Bromo-3-Chlorobutane Isomer Pair Mirror Super- Relationship Images? imposable? (2R,3R),(2S,3S) Yes No Enantiomers (2R,3S),(2S,3R) Yes No Enantiomers (2R,3R),(2S,3R) No No Diastereomers (2R,3R),(2R,3S) No No Diastereomers (2S,3S),(2R,3S) No No Diastereomers (2S,3S),(2S,3R) No No Diastereomers Compounds with Fewer than 2n Stereoisomers (4.3C) Some compounds with n chiral centers have fewer than 2n stereoisomers. 2,3-Dibromobutane. Both C2 and C3 in 2,3-dibromobutane are chiral, and each has an R and an S configuration. [graphic 4.27] While we can show 4 possible stereoisomers in this figure, we will see that the compound has only 3 different stereoisomers. The (2R,3R) and (2S,3S) isomers are mirror images of each other and non-superimposable so they are a pair of (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 18 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 19 enantiomers. In contrast, while the (2R,3S) and (2S,3R) structures are also mirror images of each other, we will see that they are superimposable so they are identical to each other. [graphic 4.28] If you lift the (2S,3R) structure off of the page and rotate it by 180°, you can superimpose it exactly on top of the (2R,3S) structure (Figure [graphic 4.28]). These two structures are superimposable because each has a plane of symmetry (Figure [graphic 4.29]) in which the groups on C2 (Br, CH3, H) are chemically identical to, and the exact mirror images of, the groups on C3 (Br, CH3, H). [graphic 4.29] Meso Form. Since (2R,3S) and (2S,3R)-2,3-dibromobutane (Figures [graphic 4.27] and [graphic 4.28]) are identical to each other (superimposable on each other), they are a single stereoisomer that we call a meso form or meso isomer. A meso form is a stereoisomer of a compound with two or more chiral centers that is superimposable on its own mirror image. You may wonder why we can refer to this single meso isomer as either (2R,3S) or (2S,3R). This results from the mirror plane of symmetry that allows us to number the molecule beginning at either terminal CH3 group. If we reverse the numbers on C2 and C3 of the (2R,3S) stereoisomer, without changing the configurations at the two C's, then (2R,3S) becomes (3R,2S). This new designation (3R,2S) is completely equivalent to the designation (2S,3R) so we see that the designations (2R,3S) or (2S,3R) are completely interchangeable. Since meso forms are stereoisomers with mirror planes of symmetry, you can identify a meso form by identifying its mirror plane. Another way to predict the presence of a mirror plane in a stereoisomer, and its identity as a meso form, is to recognize that there are two identical ways to number the carbon atoms when you name the molecule. Because it has a meso form, 2,3-dibromobutane has only 3 unique stereoisomers. We summarize the relationships between pairs of its stereoisomers in Table 4.2 [next page]. Table 4.2. Relationships Between the Stereoisomers of 2,3-Dibromobutane Isomer Pair Mirror Super- Relationship Images? imposable? (2R,3R),(2S,3S) Yes No Enantiomers (2R,3S),(2S,3R) Yes Yes Identical (Meso form) (2R,3R),(2S,3R) No No Diastereomers (2R,3R),(2R,3S) No No Diastereomers (2S,3S),(2R,3S) No No Diastereomers (2S,3S),(2S,3R) No No Diastereomers (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 20 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 21 4.4 Drawing Structures of Stereoisomers In this section we consider different ways of drawing stereoisomers and methods for interconverting these various types of drawings. This will help us compare stereoisomers and determine their stereochemical relationships to each other. 3-D Conformations of Stereoisomers (4.4A) There are a variety of ways to draw 3-D structures for stereoisomers since each stereoisomer can be drawn in many different orientations in space, and in different conformations resulting from rotation about C-C and other single bonds in the molecule. Many Ways to Draw the Same Stereoisomer. All of the structures that we show in Figure [graphic 4.32] are the same stereoisomer (2R,3S)-2-bromo-3-chlorobutane. [graphic 4.32] While they appear different from each other, we can use the the R,S assignment rules to confirm in each structure that the configuration at C2 is R and that at C3 is S. Configurations at a chiral atom do not change when we rotate about C-C bonds, or when we rotate a molecule in space. In order to change a configuration at a chiral C, we must exchange the positions of two atoms and/or groups bonded to that C. This means that we must break the chemical bonds between these two atoms or groups and the chiral center, and then form new bonds to the atoms or groups in the opposite spatial orientation. 3-D Structures for Comparing Stereoisomers. Because it is visually difficult to relate structures of stereoisomers with different conformations (C-C rotation) or different spatial orientations such as those in Figure [graphic 4.32], the best way to draw a complete set of stereoisomers of a compound is to arbitrarily choose one structure for a stereoisomer and model the rest of the stereoisomers on it. We illustrate this for 3-bromo-2-butanol in Figure [graphic 4.33] starting with four different arbitrary structures for its (2S,3R) stereoisomer on lines (A) through (D). [graphic 4.33] We draw the second structure in each group as the mirror image of the first structure. The configuration at each chiral C in a mirror image is opposite that in the original structure, so the mirror image second structure in each case is (2R,3S)-3-bromo-2-butanol. We draw the third stereoisomer in groups (A) through (D) by arbitrarily changing the configuration at one chiral C in each second structure. Because we have arbitrarily changed the configuration at C2 from R to S, the third structure in each group is the new stereoisomer (2S,3S)-3-bromo-2-butanol and it is different from both the first structure (2S,3R) and second structure (2R,3S) in each group. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 22 Figure 4.33 (See Next Page) (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 23 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 24 We draw the fourth structure in each group as the mirror image of the third structure. The configuration at each chiral C changes in a mirror image, so all of these fourth stereoisomers are (2R,3R)-3-bromo-2-butanol. Since 3-bromo-2-butanol has no mirror plane, its four stereoisomers (shown in different orientations on lines (A)-(D)) are different from each other and there are no meso forms. Each group of stereoisomers on lines (A)-(D) are valid representations of the four stereoisomers of 3-bromo-2-butanol. However you will frequently see 3-D stereoisomers drawn as wedge-bond structures such as those in groups (A) or (B) in Figure [graphic 4.33] with their chiral C's on a vertical line and their horizontal bonds pointing out from the paper. Organic chemists usually draw groups of stereoisomers with the maximum number of C's on the vertical line as in group (A), while the conformations in group (B) are useful for assigning R and S configurations to the chiral carbons since they have priority "4" groups on C2 and on C3 at the top and bottom of the structures . Configuration of C2 in (2R,3R)-3-bromo-2-butanol. Let's verify the assignment of the R configuration to C2 in (2R,3S)-3-bromo-2-butanol using the structure from group (B) of Figure [graphic 4.33] that we show again in Figure [graphic 4.34]. [graphic 4.34] Our first step is to assign priority numbers to the atoms and groups on C2. We then want to view this C2 chiral carbon down its C2-"priority '4'-group" bond so we lift the C2-C3 bond from the plane of the paper so that C3 is pointing towards us. As a result, the priority "4" H atom moves further away from us as shown in the second structure of Figure [graphic 4.34]. When we now connect the groups numbered "1" through "3" by arrows, we see that the configuration at C2 is R because the arrows rotate in a clockwise or "right" direction. We can carry out the same process at C3 to verify that it has the S configuration. In more complex molecules, you may find it more challenging to orient a particular chiral C so that you can determine its R,S assignment as we describe above. This takes practice and is usually made easier by the use of molecular models. However, molecular models may not always be available, so you need to practice making these configurational assignments both with and without them. Fischer Projections (4.4B) Chemists sometimes use Fischer projections, rather than 3-D wedge-bond drawings, to represent structures of stereoisomers. We will make extensive use of Fischer projections to draw structures of sugars (carbohydrates) in Chapter 20. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 25 Definition of Fischer Projections. We illustrate the definition of a Fischer projection with the structures in Figure [graphic 4.35]. [graphic 4.35] A Fischer projection symbolizes a 3-D wedge-bond structure whose horizontal bonds at each chiral C are solid wedges projecting out of the plane of the paper, and whose vertical bonds at each chiral C are dashed wedges projecting back into the plane of the paper (or lying in the plane of the paper as is the case for the central C-C bond in the structure with two chiral C's). Two stereoisomers of molecules that we have already discussed are shown in Figure [graphic 4.36] using both 3-D wedge-bond structures and their Fischer projections. [graphic 4.36] You can see that we have drawn the horizontal bonds in the 3-D wedge-bond structures as solid wedges, and the top and bottom vertical bonds as dashed-wedges. When we draw the 3-D wedge-bond drawings in this way, we are able to immediately draw their adjacent Fischer projections. Alternatively, had we first drawn these Fischer projections, they would have immediately defined their adjacent 3-D wedge-bond structures. It is important to realize that we can draw each of these stereoisomers in a variety of different conformations and orientations in space. However, only when we draw wedge-bond structures with horizontal bonds projecting toward us, and vertical bonds projecting away from us, can we immediately convert them into the Fischer projections. Note in Fischer projections that (1) all of the bonds are solid lines, (2) all of the atoms or groups are in the same relative positions as in the wedge-bond structures, and (3) the chiral C's are not labelled with the letter "C". It is customary not to label the chiral C's with the letter "C" in a Fischer projection in order to distinguish it from line-bond structures such as those shown early in Chapter 1 (see also Figure [graphic 4.1]. Fischer projections specify the configuration and stereochemistry at chiral C's, but line-bond structures do not! Manipulations of Fischer Projections. Fischer projections eliminate the need to draw solid and dashed wedge-bonds required in 3-D structures, but without those solid and dashed-wedge bonds we must be very careful how we reorient Fischer projections on a piece of paper. We illustrate in Figure [graphic 4.37] a problem that occurs when we reorient a Fischer projection. [graphic 4.37] When we simply rotate the Fischer projection of (R)-CHBrClF in the plane of the paper by 90° (step (A)), it is now (S)-CHBrClF! We can verify this by converting that rotated Fischer projection into its wedge-bond structure and assigning a configuration to the chiral C using the R,S assignment rules. This inversion of configuration where R becomes S at the chiral C upon 90° rotation of a Fischer projection results from the different meaning of its horizontal and vertical bonds. If (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 26 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 27 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 28 we rotate that middle Fischer projection by another 90° in the plane of the paper (step (B)), we can show by converting it into a wedge-bond structure that it has once again become (R)-CHBrClF. This demonstrates that rotation of a Fischer projection by a full 180° in the plane of a piece of paper (the result of two separate 90° rotations), does not change the configuration at its chiral carbons. Using Fischer Projections to Draw Stereoisomers. We draw Fischer projections in Figure [graphic 4.41] of the possible stereoisomers of 2-bromo-3-chlorobutane and of 2,3-dibromobutane that we discussed earlier using wedge-bond drawings. [graphic 4.41] For each compound, we arbitrarily draw the first Fischer projection so that the chiral C's and CH3 groups are on the imaginary vertical line mentioned earlier. The second Fischer projections are "mirror images" of the first ones. Since they are also non-superimposable on these first structures, they are enantiomers of these first structures. You can verify this by rotating the second structure for each compound by 180° in the plane of the paper and noting that the atoms do not match up with those on the first structures. Remember that a 180° rotation of a Fischer projection does not change the configuration at any chiral C in the molecule. We draw the third Fischer projections by changing the configuration at just one C in the second structure. As a result, each is a diastereomer of the first and second structures of each compound. The fourth Fischer projections are mirror images of the third Fischer projections. For 2-bromo-3-chlorobutane, this fourth Fischer projection cannot be superimposed on its mirror image (the third Fischer projection) so it is an enantiomer of the third structure. The situation is different for 2,3-dibromobutane. If we rotate its fourth Fischer projection by 180° in the plane of the paper, we can superimpose it exactly on the third Fischer projection that is its mirror image. Since these third and fourth Fischer projections of 2,3-dibromo-butane are identical mirror images, they are each a representation of the single meso form of 2,3-dibromobutane. R,S Assignments Using Fischer Projections. We have shown Fischer projections of the 4 stereoisomers of 2-bromo-3-chlorobutane and the 3 stereoisomers of 2,3-dibromobutane in Figure [graphic 4.41], but have not labelled their chiral C's R or S. You can provide these assignments by converting each Fischer projection into a wedge-bond structure, and then analyzing each chiral C as we did earlier in this chapter. However, you can also assign R or S to chiral C's in Fischer projections without converting them to wedge-bond structures. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 29 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 30 If the priority "4" group of each chiral C is on the top or bottom position of the Fischer projection, all you need to do is to connect the remaining priority numbers with the curved arrows as shown in Figure [graphic 4.42] because a group on the top or bottom of a Fischer projection is defined as pointing back into the page away from you. [graphic 4.42] However, if the chiral C on which you want to do R,S assignments in the Fischer projection does not have the lowest priority group at the top or bottom position, you can "move" the priority "4" group to a top or bottom position by exchanging it with one of the other groups bonded to the chiral C using the Exchange Rule that we describe here. The Fischer Projection Exchange Rule. Exchanging the positions of any two groups bonded to the same chiral carbon in a Fischer projection changes the configuration at that chiral carbon. We demonstrate that this is correct below, but first we present a step-by-step outline of how you can use this exchange rule to assign the R or S configuration to a chiral carbon atom for a stereoisomer of 2,3-dibromobutane, and we illustrate it with the 6 steps in Figure [graphic 4.43]: [graphic 4.43] (Step 1) Identify the chiral C in a Fischer projection whose R,S configuration you wish to assign. (Step 2) Assign priority numbers to the groups on that chiral C. (Step 3) Put the priority 4 group at the top (or bottom) of the Fischer projection of that chiral C, by exchanging its position with the group already in the top (or bottom) position. (Step 4) Exchange positions of any two groups on that chiral C other than the priority 4 group. (Step 5) Connect the priority 1, 2, and 3 groups with arrows and assign R or S to that chiral C as appropriate. (Step 6) This R or S assignment will be the same as that for the original chiral C. The configuration at the chiral C at the end of this sequence of steps is the same as it was in the original Fischer projection because you made two exchanges. The first (done in step (3)) inverts the configuration at the chiral C according to the basic exchange rule, and the second (done in step (4)) returns the inverted configuration to the configuration in the original projection. Confirmation of the Exchange Rule. A way to prove that the exchange of positions of two groups on a chiral C changes configuration at that C is illustrated by the following steps: (1) Draw a Fischer projection for a chiral C and then convert that into a wedge-bond structure. (2) Determine the R,S configuration at the chiral C using R,S assignment rules described earlier. (3) Exchange two groups on the Fischer projection that you drew in (1) and convert this new Fischer projection into a wedge-bond structure. (4) Determine the R,S configuration at the chiral C in the wedge-bond structure created in (3). If you do all of this correctly you will find out that the two wedge-bond structures have opposite configurations. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 31 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 32 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 33 4.5 Cyclic Molecules Cyclic molecules can have chiral centers in the ring or in a group bonded to the ring. [graphic 4.45] When the chiral center is not in the ring (as in (A)), we treat the molecule just like those we have already described. This section describes stereoisomers of molecules that have chiral centers in the ring (as in (B)). Cyclic Stereoisomers (4.5A) To draw stereoisomers for cyclic systems, we first identify the chiral atom or atoms in the ring just as we did for acyclic systems. As with acyclic molecules, a chiral C in a ring must have four different groups bonded to it. Chiral Centers in 1-bromo-3-methylcyclohexane. Both C1 and C3 of 1-bromo-3-methylcyclohexane (structure (B) in Figure [graphic 4.45]) are chiral although at first glance you may think that neither of them has four different groups. For example, C1 is bonded to a Br, H, and two ring CH2's, while C3 is bonded to a CH3, H, and two ring CH2's. You may initially think that the two ring CH2's on C1 or on C3 do not qualify as different groups, but a closer look at the two ring CH2's on C1 shows that one is CH2-CH(CH3)-, while the other is CH2-CH2-. These two groups differ at the next C bonded to each CH2. (There is also one more CH2 group in the ring between the CH2CH2- and CH2CH(CH3)- groups, but we ignore it in our analysis because we found a point of difference before we reached that group). In a similar way, the four groups on C3 are CH3, H, CH2CH2-, and CH2CH(Br)- ([graphic 4.46]). A Method to Identify Chiral Centers in Rings. If you have difficulty identifying chiral C's in rings, imagine a pair of twins standing on the C in the ring that you are testing for chirality. Allow each of the twins to simultaneously walk in opposite directions on the ring one step at a time using the atoms in the ring as stepping stones. If the two twins encounter a difference in substitution after the same number of steps, the C that they were both originally standing on is chiral. If they meet again without encountering any differences, the original C is achiral. Stereoisomers of 1-bromo-3-methylcyclohexane. Since 1-bromo-3-methylcyclohexane has 2 chiral centers, it can have no more than the four stereoisomers (2n = 22 = 4) whose structures we show in Figure [graphic 4.49]. To help determine the relationships between these possible stereoisomers, we have reoriented (B) and (D) in Figure [graphic 4.50] so that you can see more easily that (B) is the mirror image of (A), and (D) is the mirror image of (C). [graphic 4.50] Since (A) and (B) are non-superimposable mirror images (see Figure [graphic 4.49]), they are enantiomers of each other, and the same is true for (C) and (D). As a result, (A) and (C) (or (D)) are diastereomers like (B) and (C) (or (D)). (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 34 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 35 Stereochemical Relationships between cis and trans Isomers. You can also see in Figure [graphic 4.50] that (A) and (B) are both cis isomers (Chapter 2) of 1-bromo-3-methylcyclohexane, while (C) and (D) are both trans isomers. This means that cis-1-bromo-3-methylcyclohexane is the pair of enantiomers (A) and (B), while trans-1-bromo-3-methylcyclohexane is the pair of enantiomers (C) and (D). Since (A) or (B) are each diastereomers of (C) or (D), we can think of cis and trans-1-bromo-3-methylcyclohexane as diastereomers of each other. R,S Assignments at Ring Carbons. You can assign R or S configurations to the chiral carbons in cyclic systems as we described for acyclic systems. One method is to isolate a specific chiral ring C from a wedge drawing of a stereoisomer and make the assignment on that isolated chiral center. We illustrate this in Figure [graphic 4.51] for the C1 carbon in a stereoisomer of 1-bromo-3-methylcyclohexane. [graphic 4.51] We first identify the four different groups on C1 and give them priority numbers as we illustrate in the middle structure of Figure [graphic 4.51]. We then extract the C1 carbon from the molecule and identify some of its groups using just priority numbers to make viewing easier as we show in the third structure. Since these priority numbers progress in a clockwise direction using the method outlined for acyclic systems, we have identified the configuration at C1 as R. Isomeric Bromomethylcyclohexanes. There are 3 other bromomethylcyclohexanes that are structural isomers of 1-bromo-3-methylcyclohexane. Of these, only 1-bromo-2-methylcyclohexane has chiral carbon atoms. [graphic 4.52] 1-Bromo-1-methylcyclohexane is a single compound with two different chair conformations and no chiral C's. Five of the six ring carbons are CH2 groups that cannot be chiral because they each have two H's. While the C1 center has a Br and a CH3, its other "two groups" are identical to each other so it is achiral. C1 and C4 of 1-bromo-4-methylcyclohexane are also achiral as are the ring CH2's. However, 1-bromo-4-methylcyclohexane does have a cis and a trans isomer that are stereoisomers because they differ only in the spatial arrangement of their Br and CH3 groups in space. [graphic 4.53] They are diastereomers because they are not mirror images, and since each has a mirror plane, they are meso forms. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 36 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 37 In contrast, 1-bromo-2-methylcyclohexane has two chiral C's and four stereoisomers. [graphic 4.54] You can do a stereochemical analysis of these stereoisomers in a way that is completely analogous to that we have illustrated for 1-bromo-3-methylcyclohexane. Drawings of Cyclic Stereoisomers (4.5B) There are several ways to draw cyclic stereoisomers. Wedge-Bond Structures. Wedge-bond drawings (Figures [graphic 4.49] and [graphic 4.50]) are the easiest way to depict stereoisomers of cyclic systems. They are simple to draw and we can use them to depict rings of any size as we illustrated in Chapter 2. However, wedge-bond structures do not show 3-dimensional aspects of these stereoisomers. Chair Forms. Chair forms are the most accurate way to depict the 3-D structures of cyclohexane stereoisomers (Figure [graphic 4.55]). [graphic 4.55] Each wedge-bond structure in Figure [graphic 4.50] corresponds to the view we "see" when we look at the top of each chair form (A) through (D) in Figure [graphic 4.55]. The chair forms remind us that each stereoisomer is an equilibrium mixture of two conformations. This is no different than the situation for acyclic systems, since every acyclic stereoisomer that we have seen is a mixture of different conformations. Haworth Projections. Organic chemists sometimes depict stereoisomers of cyclic systems with Haworth projections such as those we show for 1-bromo-3-methylcyclohexane in Figure [graphic 4.56]. [graphic 4.56] Haworth projections of rings are flat, so they clearly show the cis or trans relationships between groups on a ring. They also have the advantage that they can be used for rings of any size. You can see that the Haworth projections of (A) and (B) in Figure [graphic 4.56] look very much like the top two chair conformations of (A) and (B) in Figure [graphic 4.55] if we flatten the rings in the chair forms. It's harder to see that the Haworth projections for (C) and (D) arise by flattening the rings of the chair conformations of (C) and (D). However you can see this clearly if you make a molecular model of chair (C) or (D) and then carefully flatten its ring. We will use Haworth projections to depict sugar molecules (carbohydrates) (Chapter 20) that often have six-membered rings such as that we show for the sugar α-D-glucose in Figure [graphic 4.57]. [graphic 4.57] (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 38 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 39 4.6 Optical Activity Stereoisomers with non-superimposable mirror images are optically active. Optically active compounds in solution or in the form of pure liquids rotate plane polarized light. Organic chemists use optical activity to identify stereoisomers, to assess their purity, and to relate stereoisomers to each other. Rotation of Plane Polarized Light and the Polarimeter (4.6A) Organic chemists use polarimeters to measure optical activity. Polarimeter. We illustrate the general features of a polarimeter, and the rotation of plane polarized light by an optically active compound in the polarimeter, in the schematic drawing in Figure [graphic 4.58]. [graphic 4.58] The beam of light from the light source oscillates in an infinite series of planes that intersect each other like the intersecting arrows at the center of the light source . The polarizer allows light in only one of these planes to pass through it, so we say that the light (represented by the single arrow on the right side of the polarizer) is plane polarized. Light Rotation by the Sample. As plane polarized light passes through the polarimeter tube containing the solution or liquid sample of the optically active compound, the compound rotates the plane of the light. We illustrate this with the sequence of tilting arrows in the polarimeter tube. The analyzer measures the amount of the rotation and displays it as a positive (+) or negative (-) rotation between 0° and 180° compared to the plane of the light before it encounters the optically active sample. Chemists use the (+) or (-) sign of the observed rotation as part of the name of the stereoisomer. For example, (+)-2-bromobutane is the name of the stereoisomer of 2-bromobutane that rotates light in the (+) direction. Some Cautionary Words. You may see references to (+) light rotations as "clockwise" rotations and (-) rotations as "counterclockwise" rotations. However when used to describe light rotation, clockwise and counterclockwise have no connection with their use in R and S assignments. In addition, although you can assign R or S configurations using rules we have presented, you usually cannot predict direction or magnitude of light rotation by an optically active molecule just from its structure. Both the direction and magnitude of light rotation by an optically compound are physical properties of a compound like its boiling point or melting point. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 40 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 41 Magnitude and Sign of Light Rotation (4.6B) The number of degrees (°) that an optically active compound rotates light depends on several factors. Observed versus Specific Rotation. We use the symbol α for the observed rotation of plane polarized light that we measure in the polarimeter. α depends on the wavelength of the light, any solvent that we use, the temperature of the sample, the concentration of the stereoisomer in a solvent (or the density of a pure liquid), and the length of the polarimeter tube (the cell length ). The specific rotation [α] for the optically active compound is independent of the concentration (or density) of the compound and the cell pathlength and we can calculate it from the observed rotation α using equation (1)[next page]. [α] = α/(c x l) (1) α is the observed rotation in degrees (°) c is the concentration (g/mL) of the chiral compound (or its density (g/mL) if a pure liquid) l is the pathlength of the cell (decimeters, dm)(1 dm =10 cm) Chemists usually measure α values at room temperature (20 to 25°C) using a specific wavelength of light from a sodium vapor lamp called the sodium D line. It is customary to specify the solvent, temperature, wavelength of light, and concentration when reporting a value of [α]. For example, if you calculated a specific rotation of -37° from an observed rotation measured at a temperature of 23°C using the sodium D line and a solution of 0.03 g of the compound in 1.00 mL of CHCl3, you should write it as: [α]23D = -37° (c, 0.03 in CHCl3). Specific Rotations of Enantiomers. While we cannot predict the actual value of a specific rotation [α] for an optically active compound, we do know that the specific rotation values for a stereoisomer and its enantiomer must have exactly the same magnitude withopposite signs since the two enantiomers are mirror images. For this reason, chemists historically describe one enantiomer of a pair of enantiomers as the (+) enantiomer and the other as the (-) enantiomer. Relative and Absolute Configurations. Since the two members of an enantiomeric pair are mirror images, each chiral center in one enantiomer must have a configuration opposite to that of the corresponding chiral center in the other enantiomer. The corresponding chiral centers in the two enantiomers have opposite relative configurations. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 42 When we know whether the configuration of a chiral center is R or S in a stereoisomer for which we also know the sign of its specific (or observed) rotation, then we know the absolute configuration of that center. If we know absolute configurations for chiral centers in one stereoisomer, then we know them for the corresponding centers in its enantiomer since they have opposite relative configurations at each chiral center. We discuss determination of absolute configurations in Appendix C at the end of this chapter. Specific Rotations of Diastereomers. We can predict the specific rotation [α] of a stereoisomer if we know [α] for its enantiomer, but this information usually does not allow us to predict [α] values for its diastereomers. Meso forms are an exception since they are not optically active. The plane of symmetry of the meso form not only causes the meso form to be superimposable on its mirror image, but makes its optical rotation 0°. You can imagine that the two mirror image parts of the meso form (see Figure [graphic 4.29]) rotate light in equal and opposite directions so that they cancel each other giving a net rotation of 0° for the whole molecule. d and l Isomers. Chemists have historically referred to (+)-enantiomers as d-enantiomers, and (-)-enantiomers as l-enantiomers. These lower case letters d and l are derived from the Latin words dextro meaning "right" and levo meaning "left" and refer to the direction that the stereoisomer rotates plane polarized light. We will refer to optically active stereoisomers exclusively by the designations (+) and (-), however you will encounter d and l in older books and literature and chemists still use these terms. (In order to have the meaning that we have just described, d and l must be lower case letters. We will learn in a later Chapter that the upper case letters (capital letters) D and L have completely different meanings! Optical Isomers. Historically, chemists have also referred to enantiomers as optical isomers because they rotate light in equal but opposite directions. However, this term is confusing because chemists sometimes also use it to describe diastereomers of each other. For this reason, authorities in stereochemistry recommend against the use of the term optical isomer. Nonetheless, you will encounter it in textbooks and literature, and it continues to be used informally by chemists in their conversations. Racemic Mixture. Chemists refer to an equimolar mixture of the two enantiomers of an enantiomeric pair as a racemic mixture or racemate and often label them as such by placing (±) or "d,l" in front of their chemical names. Racemic mixtures show no light rotation in a polarimeter. Since the two enantiomers in the racemic mixture rotate light in equal and (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 43 opposite directions, and are present in the same concentrations in the racemic mixture, the net rotation of the mixture as observed in a polarimeter is 0°. Appendix A: Resolution of Stereoisomers Chemists often wish to resolve (isolate) individual stereoisomers of a compound. One reason is because molecules of biological importance are often single stereoisomers as we describe in the Feature at the end of this chapter. Resolution of stereoisomers can be more difficult than separation of unrelated organic compounds since most separation methods depend on differences in boiling points, or solubilities in solvents of the components of a mixture. These differences are small or non-existent for stereoisomers because they have the same mass, functional groups, and chemical structures except for configurations at chiral centers. Resolution of Diastereomers. When there is a difference in physical properties between two diastereomers, it is conceivable that they can be separated by fractional crystallization, fractional distillation, or various types of chromatography. Fractional crystallization makes use of differences in solubilities of compounds in solvents to aid in their separation while fractional distillation uses differences in the boiling points of compounds for the same purpose. Chromatography is a general term for a number of different techniques used to separate mixtures of compounds. All chromatographic methods involve passing a mixture of compounds through a column containing a stationary phase that interacts differentially with individual compounds or stereoisomers in the mixture leading to its separation into individual components that can be individually isolated. Resolution of Enantiomers. Individual enantiomers of an enantiomeric pair have identical physical properties, so they cannot be resolved by fractional crystallization or fractional distillation. However, if we chemically convert each enantiomer into a new compound with an additional chiral atom of the same configuration, the pair of enantiomers becomes a pair of diastereomers with different physical properties. We illustrate this schematically in Figure [graphic 4.59] and describe it in more detail below. [graphic 4.59] (+)A and (-)A are enantiomers of each other so each has identical physical properties (b.p., m.p., etc.) except for the direction that they rotate plane polarized light. However, if we react each of them with the same stereoisomeric reagent (+)B and the chiral centers in A and B are not changed, the reaction products are the diastereomers (+)A-(+)B and (-)A-(+)B with (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 44 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 45 different physical properties. If these different properties allow us to separate (+)A-(+)B from (-)A-(+)B, we can theoretically obtain separate samples of pure (+)A and pure (-)A after we remove the (+)B groups. Alternatively, we might separate a pair of enantiomers by chromatography using some specific chiral stereoisomer as the stationary phase in the chromatography column. Enantiomers have identical properties in an achiral environment, but we would expect two enantiomers to interact differently with a single stereoisomer in a chromatography column since they will appear as different compounds to the single stereoisomer in the stationary phase. Some Reported Physical Properties for Stereoisomers. We compare in Table 4.3 some experimentally determined physical properties reported in the chemical literature for stereoisomers of several different compounds. Table 4.3. Physical Properties of Some Stereoisomers. Compound Relationship [α]D m.p.(°C) 2,3-butanediol (2R,3R) enantiomeric -13.0° 19.7 (2S,3S) pair +12.4° 25 meso form diastereomer (0°) 34.4 3-amino-2-butanol (2R,3R) enantiomeric -15.84° 15-16 (2S,3S) pair +15.69° 7-11 (2R,3S) enantiomeric +0.80° 49 (2S,3R) pair Data not available bromochlorofluoromethane (+) enantiomeric +0.20° (-) pair -0.13° 1,2-cyclohexanediol (1R,2R), trans enantiomeric -46.5° 113-4 (1S,2S), trans pair +36.7° 108-9 meso form, cis diastereomer (0°) 98 1,2-cyclohexanediamine (1R,2R), trans enantiomeric -36° (1S,2S), trans pair +34° meso form, cis diastereomer (0°) The enantiomers in the enantiomeric pairs should have values of [α]D with equal magnitudes and opposite signs, but the experimental values are not identical. This probably means that each enantiomer is not absolutely pure, and it may also reflect experimental error in their measurement. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 46 Nonetheless, you can see that the data for each enantiomer in a pair are very similar. The (2R,3R) and (2S,3S) enantiomers of 3-amino-2-butanol have [α]D values that are close to each other, but significantly different from that of their (2R,3S) diastereomer. In addition, the m.p.'s of (2R,3R) and (2S,3S) are similar, but very different than that of the (2R,3S) diastereomer. You can also see that m.p.'s of meso forms differ more from those of the members of enantiomeric pairs than do the m.p.'s of the individual enantiomers from each other. Appendix B: Optical Purity When specific rotations of individual enantiomers of an enantiomeric pair do not have the same magnitude (eg. see Table 4.3) when calculated from α values measured under identical conditions, one or both enantiomers may be impure. While the impurity may be a diastereomer or some other compound, if repeated purification does not lead to the same specific rotation for each enantiomer the impurity is probably the other enantiomer. %Optical Purity. Two enantiomers of an enantiomeric pair always have equal but opposite rotations, so contamination by the other enantiomer always leads to an observed rotation that is smaller than that of the pure enantiomer. As a result, you can expect that as the purity of an enantiomer increases, so will the magnitude of its apparent specific rotation. We describe the purity of a stereoisomer contaminated with its enantiomer as %Optical Purity define it using equations (2) or (3). %Optical Purity = ([α]expt/[α]o) x 100 (2) %Optical Purity = ⎢%(+) - %(-) ⎢ = %Enantiomeric Excess = % ee (3) Equation (2) states that the %Optical Purity of a sample contaminated by its enantiomer is proportional to the experimentally determined specific rotation ([α]expt) divided by the true specific rotation ([α]o). If a sample of an enantiomer has an [α]expt value of +36.7° while ([α]o) is +46.5°, the %Optical Purity is [(36.7)/(46.5)]x100 % or 78.9%. Since %Optical Purity is also equal to the difference in the % of each enantiomer in the sample (%(+) and %(-) in equation (3)), then the absolute value of the difference ⎢%(+) - %(-) ⎢ also equals 78.9%. If the sample contains only the two enantiomers, %(+) - %(-) must equal 100% so one enantiomer is about 89.4% of the mixture while the other is about 10.6%. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 47 Enantiomeric Excess (%ee). In order to obtain %Optical Purity from equation (2), we must know [α]o and this requires that we have a pure sample of enantiomer. This presents a dilemma since we may not know when we have an absolutely pure enantiomer. Equation (3) provides a solution to this problem since %Optical Purity is also equal to the difference in the percentage amounts of the two enantiomers in the mixture (the %Enantiomeric Excess or %ee) if no other impurities are present. If we can determine the relative amounts of the two enantiomers (%ee) by a method other than optical activity, then we can calculate %Optical Purity from equation (3) and use it in equation (2) along with [α]expt to calculate a value for [α]o . Fortunately, there are instrumental methods widely used today in chemical research that frequently allow us to independently determine %ee values. One of these is Nuclear Magnetic Resonance (NMR) that we describe in Chapter 5. Appendix C: Absolute Configuration At the beginning of this chapter we saw that bromochlorofluromethane (CHBrClF) has two stereoisomers that are its R and S enantiomers. If you are given each of these enantiomers in a separate unlabelled bottle with no additional information, you would not know which one is R and which one is S because they have identical properties except for the direction that they rotate plane polarized light. Because of this difference in light rotation, you can determine which is (+) and which is (-). However without knowing the answer to begin with, you cannot say which is R and which is S because there is no connection between R,S configuration and direction of light rotation. This means that you do not know the absolute configurations of these two enantiomers. In order to know the absolute configuration of a stereoisomer with a known specific rotation ([α]) you must be able to specify the R or S configuration at each of its chiral centers. Absolute configurations for chiral centers in compounds were unknown until 1951. In that year a Dutch chemist J. M. Bijvoet (1892-?) reported his use of X-ray diffraction to determine that (+)-tartaric acid and (-)-isoleucine were the (R,R) stereoisomers that we show in Figure [graphic 4.60]. [graphic 4.60] These X-ray diffraction experiments on (+)-tartaric acid and (-)-isoleucine also provided absolute configurations for chiral centers in other molecules whose relative configurations were known with respect to the chiral centers in (+)-tartaric acid or (-)-isoleucine such as those related to (+)-tartaric acid that we show here: (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 48 (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 49 (+)-tartaric acid HO2C-CH(OH)-CH(OH)-CO2H ↓ ↓ (+)-malic acid HO2C-CH(OH)-CH2-CO2H ↓ ↓ (+)-isoserine HO2C-CH(OH)-CH2-NH2 ↓ ↓ (-)-glyceric acid HO2C-CH(OH)-CH2-OH ↑ ↑ (+)-glyceraldehyde H(O=)C-CH(OH)-CH2-OH The configuration at C is the same in each of these compounds and is not changed by the chemical reactions (indicated by the arrows) that interconvert these compounds. As a result, if C in (+)-tartaric acid has the configuration shown in Figure [graphic 4.61], then the configuration at C in (+)-glyceraldehyde is that shown in this same figure. [graphic 4.61] Since Bijovet showed that this Fischer projection for (+)-tartaric acid has the correct absolute configuration at each chiral C, the Fischer projection for (+)-glyceraldehyde also has the correct configuration. Two Chiral Centers in (+)-Tartaric Acid. Note that (+)-tartaric acid has two chiral C's while there is only one in (+)-glyceraldehyde. Which chiral C in (+)-tartaric acid becomes the chiral C in (+)-glyceraldehyde? It turns out that it makes no difference since the two chiral C's in (+)-tartaric acid are chemically and configurationally identical (they are both R). No matter which CH(OH)-CO2H group of tartaric acid becomes CH2-OH in glyceraldehyde, the stereochemical result for glyceraldehyde is the same. Long before Bijvoet carried out his experiments, Emil Fischer (1852-1919) arbitrarily assigned the structure in Figure [graphic 4.61] to (+)-glyceraldehyde. [graphic 4.61] He knew that the odds were 50/50 that it was correct, so chemists were pleased when Bijvoet's structure determination of (+)-tartaric acid showed that Fischer's guess was correct. Chapter Review Tetrahedral Carbon Configurations. (1) Four different atoms or groups can bond in two different ways to tetrahedral C atoms. (2) These two different configurations are non-superimposable mirror images and the C is chiral. (3) Chiral C's can cause molecules containing them to be chiral molecules. (4) Some molecules have structural features causing them to be chiral without chiral atoms. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 50 Stereoisomers and R,S Assignments. (1) R and S uniquely identify the two possible configurations at a chiral C. (2) R,S assignment rules use priority numbers ("1" through "4") for each atom or group on the chiral C based on the relative atomic numbers of atoms in the groups. (3) Chiral C's with R configurations have a clockwise progression of priority numbers from "1" to "3" when viewed down the C-"4" bond with "4" in back, while C's with S configurations have a counterclockwise progression. The Number and Types of Stereoisomers. (1) Molecules with n chiral centers can have up to 2n stereoisomers. (2) Enantiomers are stereoisomers that are non-superimposable mirror images of each other. (3) Diastereomers are stereoisomers of the same compound that are not enantiomers. (4) A meso form is a stereoisomer that has a superimposable mirror image because it has a plane of symmetry. (5) Compounds with meso forms have fewer than 2n stereoisomers. Drawing Structures of Stereoisomers. (1) Solid and dashed wedge-bond drawings are the clearest way to show R or S configurations at chiral centers. (2) A set of stereoisomers is best drawn starting with one arbitrary conformation for a stereoisomer and interchanging atoms and groups at each chiral center on this conformation to obtain the other stereoisomers. (3) Fischer projections use line bonds to represent wedge-bonds following a set of specific rules. (4) The exchange of any two groups bonded to a single chiral center in a Fischer projection changes the configuration at that chiral center from R to S (or S to R). Cyclic Molecules. (1) Atoms in rings can be chiral, and cyclic molecules with chiral atoms have enantiomers, diastereomers, and meso forms like acyclic molecules. (2) cis and trans isomers of cyclic molecules are diastereomers of each other. (3) Haworth projections of cyclic stereoiomers are useful for comparing configurations at chiral atoms in rings. (4) Conformational changes in rings (eg. cyclohexane ring-flipping) do not interconvert stereoisomers nor change configurations at chiral atoms. Optical Activity. (1) Chiral molecules are optically active and rotate plane polarized light. (2) The magnitude of light rotation is measured in degrees (°) using a polarimeter and its sign is (+) or (-). (3) [α] = α/(c x l) where [α] is the specific rotation, α is the observed rotation, c is concentration (or density), and l is pathlength. (4) [α] values of pure enantiomers are equal in magnitude but opposite in sign. (5) The absolute configuration of a chiral atom is known when it can be assigned as R or S in a pure stereoisomer. (6) Relative configurations of chiral centers in two stereoisomers are known when they can be identified as the same or opposite to each other without knowing which is R and S. (7) Meso forms are optically inactive because they possess a plane of symmetry. (8) An equimolar mixture of enantiomers (a racemate or a racemic mixture) is optically inactive because it contains equal numbers of molecules that rotate light in equal but opposite directions. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 51 What a Difference a Configuration Makes! The R,S configuration at a chiral carbon can have enormous consequences with respect to the biological activity of organic compounds. Frequently, one enantiomer of a pair of enantiomers is biologically active while the other enantiomer is biologically inactive. Sometimes, one enantiomer may actually inhibit the desired function of the other enantiomer. It is also possible that one enantiomer is beneficial while the other enantiomer is hamful to an organism. Examples of these effects of configuration on biological activity are found among amino acids, sugars, sex pheremones, and pharmaceuticals. Amino Acids and Sugars The configurations at chiral carbons in amino acids and sugars have dramatic effects on their biological activity. Amino acids, the building blocks of proteins, have the general line-bond structure shown below where the R group has a variety of different structures that we describe in Chapter 22. [graphic 4.62] In all of these amino acids the configuration of the chiral C must be that shown in the wedge-bond structure (A). [graphic 4.63] The enantiomeric form (B) of each of these amino acids is not incorporated into protein molecules and is therefore not produced by the conversion of protein molecules into their component amino acids. In a similar way, sugar molecules such as α-D-glucose are biologically active and can serve as metabolic energy sources for a variety of organisms including humans. [graphic 4.64] This is not the case for its enantiomer (α-L-glucose) that we and other organisms cannot metabolize. Unlike amino acids that generally have only one chiral center, sugar molecules have several chiral centers. For example α-D-glucose shown above has 5 chiral centers and the configuration at each of these is the mirror image of those in α-L-glucose. The importance of all of these chiral centers in determining the characteristics of sugars will be explored in the carbohydrate chapter (Chapter 20). (D and L designate enantiomers, but we will see in Chapters 20 and 22 that they are defined differently than d and l ). Sex Pheromones Pheromones are organic compounds that serve as a means of chemical communication between organisms. Sex pheromones provide a way for a member of one sex of a species to find or attact a member of the opposite sex of the same species. A number of years ago, the compound shown in Figure [graphic 4.65] was isolated from females of the organism Popillia japonica (commonly known as Japanese beetles because of the country of their origin). [graphic 4.65] Extracts of this compound taken from female beetles were shown to be a powerful attractant to male members of this same species. However, when this compound was synthesized in the laboratory it provided no attraction to the discriminating males of Popillia japonica. Chemists and entomologists (scientists who study insects) determined that the feature responsible for these effects is the chiral carbon C in the wedge-bond drawings of the two enantiomers of this pheromone shown in Figure [graphic 4.66]. [graphic 4.66] They were able to show that only the R enantiomer was a sex attractant. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 52 Of even greater interest was the fact that the presence of the S enantiomer actually prevents the R enantiomer from serving as an attractant! As a result, a racemic mixture of the R and S forms is essentially biologically inactive. Pharmaceuticals Pharmaceutical chemists have found that a wide variety of compounds with pharmaceutical properties have one or more chiral atoms. For example the compound ibuprofen, which is the active ingredient in such brand name drugs as Advil™, Motrin™, and Nuprin™, has the chemical structure shown here with its chiral carbon designated as C. [graphic 4.67] It is sold in these preparations as a racemic mixture of the R and S enantiomers, but only the S enantiomer has the desired biological activity. While the R enantiomer is biologically inactive, it isomerizes in vivo to the biologically active S stereoisomer after it is ingested. While ibuprofen appears so far to be a harmless example of the dramatic difference in biological activity of two enantiomers, the case of thalidomide is a tragic demonstration of what a difference a configuration can make. [graphic 4.68] This compound with one chiral C was extensively marketed for a number of years as a sedative, but it was subsequently found to have dreadful side effects on fetal development when taken by women during pregnancy. It is now known that these "teratogenic" side effects are caused by only one of the enantiomers while the other enantiomer possesses only the desired sedative effects for which thalidomide was originally marketed. As a result of these unintended side effects, the drug was removed from the market for a number of years. However as of this writing it has been reintroduced in a carefully controlled manner designed to preclude its use by women who may become pregnant because it has been shown to have significant positive benefits in treatment of conditions for which there are few pharmaceutical options. The thalidomide experience served as a major impetus to pharmaceutical companies to take great care in testing drugs with chiral atoms and to avoid racemic mixtures whenever possible. In addition, these companies try to develop pharmaceuticals without chiral atoms wherever possible not only to avoid problems like that described above, but also to reduce costs associated with synthesis of enantiomerically pure compounds and/or large scale resolutions of optical isomers. However because of the chiral character of living systems that we will explore in Chapters 20 ot 23 of this text, it seems certain that many new and effective pharmaceuticals will continue to have chiral atoms. (2/94)(6,7,9/95)(8,9/97)(12/99)(1/00) Neuman Chapter 4 53 |
190732 | https://zoologicalbulletin.de/BzB_Volumes/Volume_57_2/119_126_BzB57_2_K%C3%BChnel_Susanne_et_al.PDF | INTRODUCTION Reproduction is a crucial trait in the life history of any or-ganism and scientists have been intrigued and challenged by this event, and the structures associated with it, ever since the days of Darwin (1871). Both aquatic and terres-trial vertebrates have evolved an extraordinary diversity of reproductive strategies and morphologies, including va-rieties of oviparity and viviparity (Meisenheimer 1921; Lombardi 1998). Among tetrapods, the diversity of repro-ductive modes is exceptionally high in amphibians. In this group, we also see multiple trends towards terrestrialisa-tion and internal fertilisation (e.g. Haddad & Prado 2005).
Reproductive modes such as viviparity have evolved in-dependently in all three lineages of modern amphibians (e.g. Noble 1927; Wake & Dickie 1998; Wells 2007).
Herein we interpret a reproductive mode as a combina-tion of several reproductive traits, including oviposition site, clutch characteristics, stage and size of hatchling, and type of parental care (sensu Salthe 1969). Internal fertilization is a precondition for viviparity (Wourms & Lombardi 1992; Böhme & Ziegler 2008). It is associated with different strategies of sperm transfer, which have evolved within all three amphibian orders, ranging from cloacal apposition in anurans to a true cop-ula via a male intromittent organ in caecilians (Sever et al. 2001; Kupfer et al. 2006). Like most other tetrapods, amphibians have a cloaca, a chamber that receives prod-ucts from the kidneys, the intestine and the gonads, and opens to the outside through a cloacal opening or vent (Kardong 2006). Below we review the diversity of amphibian cloacal mor-phologies involved in ensuring a secure direct sperm trans-fer and internal fertilization among anurans, urodeles and caecilians. REPRODUCTIVE MORPHOLOGY OF AMPHIBIANS Anura The majority of anurans, currently including almost 6000 species (AmphibiaWeb 2010), practice external fertiliza-tion, and thus have no special male cloacal arrangements facilitating direct sperm transfer (recently summarised by Wells 2007). During copulation, males grasp females firm-ly with their forearms (termed amplexus). In most cases, sperm is directly released on the eggs protruding from the female cloaca, but in some cases fertilisation takes place without amplexus (e.g. Crump 1974; Kunte 2004). Inter-nal fertilisation is rare among anurans, and mostly con-nected to viviparity or other complex parental care mech-anisms (e.g. Wake 1993; Beck 1998). Exceptionally, the phylogenetically basal tailed frogs As-caphus truei and A. montanus are the only anurans known to have evolved a true intromittent organ in males (see Figs 1A–B). During courtship they practise a combination of amplexus and copulation called “copulexus” (see Sever et al. 2001; Stephenson & Verrell 2003). The so-called Evolutionary reproductive morphology of amphibians: an overview Susanne Kühnel1, Sandy Reinhard1 & Alexander Kupfer1,2 1 Institut für Spezielle Zoologie und Evolutionsbiologie mit Phyletischem Museum, Friedrich-Schiller-Universität Jena, Erbertstr. 1, D-07743 Jena, Germany 2 Biologie und Didaktik, Universität Siegen, Adolf-Reichwein-Str. 2, D-57068 Siegen Correspondence; E-mail: alexander.kupfer@uni-jena.de Bonn zoological Bulletin Volume 57 Issue 2 pp. 119–126 Bonn, November 2010 Abstract. Reproduction is a crucial trait in the life history of any organism, and vertebrates, whether aquatic or terres-trial, have evolved an extraordinary diversity of reproductive strategies and morphologies. Among tetrapods, the diver-sity of reproductive modes is exceptionally high in amphibians, who also show multiple trends towards terrestrialisation and internal fertilisation. Herein we give a brief overview of the diversity of amphibian reproductive morphology, with a special emphasis on the cloaca, for all three major lineages, i.e., anurans, urodeles and caecilians. Key words. Reproduction, genital morphology, Amphibia.
Bonn zoological Bulletin 57 (2): 119–126 ©ZFMK 120 Susanne Kühnel et al.
Bonn zoological Bulletin 57 (2): 119–126 ©ZFMK Fig. 1.
Reproductive morphology of anurans and salamanders. (A) Inguinal amplexus (“copulexus”) of Ascaphus truei. (B) ma-le Ascaphus truei. The “tail”, a cloacal extension, can be inserted into the cloaca of the female during amplexus, ventrolateral view.
(C) male Mertensophryne micranotis (Anura: Bufonidae), left, dorsal view and its cloaca, right, caudal view (after Grandison 1980).
(D) cephalic amplexus of Notophthalmus viridescens (Urodela: Salamandridae). The male grasps the females’s neck whilst fan-ning pheromones towards her nostrils. (E) cloacal region of lentic breeding Cynops pyrrhogaster (Urodela: Salamandridae). The male’s cloaca (left) is heavily swollen compared to that of the female (right). (F) cloacal region of lotic breeding Euproctus mon-tanus (Urodela: Salamandridae, after Brizzi et al. 1995). Males (left) possess a cloacal protuberance (cp) which bears a protusible pseudopenis (pp), whereas the female cloaca is slightly conical shaped and its opening is located ventrally (right). (G) amplecting pair of Calotriton arnoldi (Urodela: Salamandridae). The male grasps the female’s trunk with his tail.
“tail” resembles the posteriorly extended cloaca, proxi-mately attended by Nobelian rods and strengthened by vascularized tissue that is engorged with blood during cop-ulation. This gives the ventral cloacal surface a pinkish colour (Noble & Putnam 1931; Duellman & Trueb 1994).
To insert the posterior pointing “tail” into the female vent, the male first flexes his pelvis at a right angle to the ver-tebral column. Contraction of the paired Musculi compres-sores cloacae (Duellman & Trueb 1994) bend the intro-mittent organ ventrally, with the male vent pointing ante-riorly (Slater 1931). Keratinised spines are present with-in the male cloacal orifice, but whether they function to enhance the attachment of the male to the female remains unclear (Noble & Putnam 1931; Metter 1964).
Additionally, internal fertilisation including an amplexus and cloacal apposition occurs in a few anurans, such as several species of viviparous African dwarf toads Nec-tophrynoides (Wake 1980; Wake & Dickie 1998) and Nim-baphrynoides (Sandberger et al. 2010), and in two species of Caribbean Eleutherodactylus, the viviparous E. jasperi (Dewry & Kirkland 1976; Wake 1978) and the oviparous-direct developing E. coqui (Townsend et al. 1981). Mat-ing has only been observed in couples of E. coqui in a spe-cial amplectic position called reverse hind leg clasp, that is initiated by the female (Townsend & Stewart 1986).
Males do not clasp, and the female rests her hind legs on top of the male´s legs. This behaviour might be correlat-ed with terrestrial reproduction and internal fertilization.
It is also thought to be present in the viviparous E. jasperi but has not yet been observed (Wake 1978). Within the African Bufonidae, all species of Nectophrynoides (and also Altiphrynoides malcolmi and Nimbaphrynoides oc-cidentalis, former members of Nectophrynoides, see Frost et al. 2006) practice internal fertilization. Altiphrynoides and Nimbaphrynoides both show a dimorphism of the male and female vent, and an inguinal amplexus in a unique belly-to-belly position has been reported as well (Grandison 1978).
As in the internally fertilising Ascaphus ssp., males of the East African toad Mertensophryne micranotis (Bufonidae) exhibit modifications of the cloacal region (Duellman & Trueb 1994). They have small conical spines around the rim of the vent and at the entrance to the cloacal tube re-stricted to the ridges of the puckered vent (Grandison 1980, see also Fig 1C). Males and females keep a very tight cloacal contact during mating. Although the cloacal spines play a role to ensure a close apposition of the vents, to secure internal fertilisation, there is no evidence for a direct interlocking mechanism in the furrows of the female vent (Grandison & Ashe 1983).
Another potential record of internal fertilisation is provid-ed for the neotropical Pumpkin Toadlet Brachycephalus ephippium (Pombal et al. 1994). During mating, males shift from an inguinal to an axillary amplexus to optimal-ly allow positioning of the vents, and thus maximize fer-tilization of the relatively large eggs (5.1 to 5.3 mm). A further record of viviparity in fanged frogs (Limnonectes spec.) from Sulawesi probably also involves internal fer-tilisation(Emerson 2001).
It can be hypothesised that (1) many terrestrially breed-ing species with large direct-developing clutches are in-ternal fertilizers and (2) if additional viviparous species are encountered they will also show internal fertilisation.
Thus, internal fertilisation and viviparity in anurans might be more widespread than currently recognized (see also Wake 1978).
Data on the reproductive biology, including the mating be-haviour, of many species is still lacking (Duellman & Trueb, 1994; Wells 2007). Life history data from around 23 % of the currently known species is missing, as listed in the data deficient category of the IUCN (Stuart et al.
2008).
Urodela The majority of the 590 species of urodeles exhibit inter-nal fertilization, whilst only males of the basal families Hynobiidae, Cryptobranchidae, and presumably Sirenidae, fertilise eggs externally (summarised in Duellmann & Trueb 1994; Wells 2007). The complex and elaborate courtship behavior of most salamanders includes the dep-osition of a spermatophore by the male, which is subse-quently received by the female. A true intromittent organ in salamanders is lacking, although direct sperm transfer can be found in one species – the Corsican brook newt Euproctus montanus, a lotic breeding endemic of the is-land of Corsica. The cloaca of the male brook newt re-sembles a conical protuberance (Fig 1F). The cloacal chamber hosts a “pseudopenis”, a broad conspicuous papilla, which can be evaginated during mating (Brizzi et al. 1995; Carranza & Amat 2005). The male grasps the female during amplexus, holding her tail with his jaws and wrapping his tail around her trunk, whilst his backward projecting cloaca is positioned close to that of the female.
A deep groove along the ventral surface of the pseudope-nis, which is aligned with the cloacal tube, guarantees a guided, unidirectional flux of cloacal products. Thus, sperm mixed with secretory products is transferred direct-ly into the female’s cloaca. The Salamandroidea that prac-tice internal fertilization possess a distinct set of male cloa-cal glands necessary for spermatophore production (Sev-er 2002). The glands are hormonally controlled and hy-pertrophied during the breeding season, often causing a sexual dimorphism in cloacal shape. However, in Euproc-121 Amphibian reproductive morphology Bonn zoological Bulletin 57 (2): 119–126 ©ZFMK tus montanus, cloacal glands are reduced or partly lack-ing (Brizzi et al. 1995; Sever 2002). Males of six salaman-drid genera possess a so-called “pseudopenis”, a projec-tion of the dorsal roof which nearly fills the entire ante-rior chamber of the cloaca. It is involved in shaping and expulsion of the spermatophore (Halliday 1998), but can-not be everted as in the Corsican brook newt (Brizzi et al.
1995; Carranza & Amat 2005).
Sexes of most species, regardless of the mode of fertili-sation, show a sexual dimorphism in cloacal shape (Figs 1E–F). Usually, the male cloaca appears larger and more swollen than the female one. This is caused by the activ-ity of the glands mentioned above (see also Sever 2002).
Species that breed in the water and show elaborate courtship dances or walks, such as some members of the family Salamandridae, produce courtship pheromones, which are fanned towards the female using the tail. Sala-manders that mate terrestrially also use courtship pheromones secreted from specialised glands to attract fe-males. Pheromone-producing cloacal glands are therefore highly influenced by sexual selection (e.g. Sever 2002; Houck et al. 2008). Usually, female cloacae are less promi-nent, but they may also possess up to three types of cloa-cal glands in Salamandroidea, mainly accounting for sperm storage (spermathecae), a unique feature among vertebrates (Sever 1994). Females may retain and mix vi-able spermatozoa from multiple matings in the spermath-ecae for longer periods (e.g. Steinfartz et al. 2006). Fe-male Eurycea fertilise eggs from stored sperm up to eight months after insemination, female Notophthalmus viri-descens effectively store sperm for up to six months, and female Salamandra salamandra are reported to store sperm for up to two years (Sever et al. 1996; Stebbins & Cohen 1997; Sever & Brizzi 1998). Additionally, the shape of female cloacae can be adapted to a specific substrate for oviposition and type of water body. Females of stream-breeding species, such as Calotriton asper, sometimes have a conically shaped cloa-ca for egg deposition and safe attachment between stones and in crevices (e.g. Carranza & Amat 2005). Lotic breeders such as Calotriton ssp. often engage in an amplexus directly transferring the spermatophore into the female cloaca (Fig 1G). It ensures direct and rapid sper-matophore uptake, and thus reduces energy wasting, which can hardly be avoided during aquatic breeding where the male and the female often have no physical contact. Breed-ing patterns including an amplexus are common in sala-mandrids. Multiple ways of female capture are known, such as the cephalic capture of Notophthalmus ssp. (see Fig 1D), the dorsal capture of Taricha, or the ventral cap-ture performed by fire salamanders (Stebbins & Cohen 1997). The mating amplexus may last up to several hours, depending on the species. Salamanders of the family Am-bystomatidae mate in the water, and the males guide fe-males to spermatophore-uptake using a “tail-nudging-walk”, except in Ambystoma gracile, A. laterale, A. jef-fersonarium and A. macrodactylum, which capture fe-males in an amplexus (Duellmann & Trueb 1994; Verrell & Krenz 1998). In contrast, some plethodontids perform a unique “tail-straddling-walk” behaviour (e.g. Arnold 1977). Gymnophiona In contrast to all salamanders (with the exception of Eu-proctus montanus) and frogs (with the exception of As-caphus ssp.), the male caecilian cloaca is evertible through the vent and operates as an intromittent organ or phallus, a unique structure among tetrapods (Tonutti 1931, see al-so Fig 2A). Presumably all ca. 190 caecilian species (oviparous and viviparous) practice internal fertilisation with the help of the phallodeum (Tonutti 1931; Wake 1972; Gower & Wilkinson 2002), which is inserted into the female vent during copulation (e.g. Kupfer et al.
2006a).
The caecilian vent is simply surrounded by several folds, which are variably arranged among the groups and dis-play sexual dimorphism in some species, such as mem-bers of the Typhlonectidae (e.g. Taylor 1968; Kupfer 2007). In contrast, the cloaca is highly complex and di-verse. The male caecilian cloaca is an elongated tube di-vided into two distinct chambers. The cranial urodeum is rather simply built, bearing longitudinal ridges, and con-nects to the intestine and the urogenital ducts, which en-ter after performing a U-bend (Sawaya 1942; Gower & Wilkinson 2002). An extraordinary feature is the presence of Müllerian ducts, which become glandular during repro-ductive activity, and secrete a fluid containing lipids and sugars necessary for sperm motility (e.g. Wake 1981). The caudal phallodeum is more broadly built and the inner structure is very different. The ridges are arranged in a more complex pattern (running transversely). In adults it is often equipped with tuberosities or crests, which give the phallodeum a characteristic morphology and gives rise to an extraordinary variation in shape (Wiedersheim 1879; Tonutti 1931, 1933; Wake 1972; Exbrayat 1991; Gower & Wilkinson 2002, see also Fig 2), that is impotant for caecilian systematics (Müller et al. 2005). East-African scolecomorphid caecilians even have cartilaginous spicules (Wake 1998). In many species, pouchy dorsolat-eral appendixes – so called “blind sacs” – extend anteri-or to the phallodeum. During eversion, the luminal sur-face of the phallodeum represents the outer structure of the phallus, with the urodeum lying in-between (see Tonut-ti 1931; Gower & Wilkinson 2002, see also Fig. C right).
122 Susanne Kühnel et al.
Bonn zoological Bulletin 57 (2): 119–126 ©ZFMK 123 Amphibian reproductive morphology Bonn zoological Bulletin 57 (2): 119–126 ©ZFMK Fig. 2.
Genital morphology of caecilian amphibians. (A) male Chthonerpeton indistincum (Gymnophiona: Typhlonectidae) show-ing an everted phallus, MHNM 09323, right – detail. (B) Geotrypetes seraphini (Gymnophiona: Caeciliidae), lateral (left), dorsal (central) and ventral (right) view of the everted phallus, AK 01149. (C) Typhlonectes natans (Gymnophiona: Typhlonectidae), SRµCT-Scan of the everted phallus. Right – virtual clipping, frontal view. (D) SRµCT-Scan of female cloaca (Ichthyophis cf. kohtaoen-sis). Dorsolateral view, virtual cut of cloacal sheath, cranial part and blind sacs, green - cloaca, violet – oviducts, yellow – blad-der. Abbreviations: MHNM = Museo Nacional de Historia Natural Montevideo Uruguay, AK = Alexander Kupfer collection.
To retract the cloaca within the body after copulation, cae-cilians possess a specific muscle (musculus retractor cloa-cae), which is also found in some females (Wilkinson 1990). The female cloaca of caecilians has received little atten-tion (e.g. Hypogeophis rostratus Tonutti 1931; Ty-phlonectes compressicauda, Exbrayat 2006), the only ded-icated morphological study was presented by Wake (1972), proposing a functional association between the specific male and female morphologies. The female cloa-ca is supposed to be non-eversible (Wilkinson 1990), therefore displaying a different morphology. Generally it is shorter than in males, and the urogenital ducts lack a copulatory loop (see Fig. 2D). There is also evidence for a bisection of the female cloaca (Exbrayat 1991; Kühnel et al. submitted). The cranial chamber is homologous to the male urodeum. The caudal chamber is marked by a different arrangement of longitudinal cloacal folds most-ly lacking tuberosities, and therefore easily recognised.
Nothing is at present known about how far the male phal-lus inserts into the female cloaca, and if special structures corresponding to the male ornamentation are present, help-ing in fixation during copulation. Copulations in caecilians have rarely been observed. Da-ta are available for two aquatic/semiaquatic species, the typhlonectids Typhlonectes compressicauda and Chthon-erpeton indistinctum. Pairs of C. indistinctum copulated for betwen 30 minutes and 5 hours (Barrio 1969) and those of T. compressicauda for between 75 minutes and 3 hours (Murphy et al. 1977; Billo et al. 1985). Observations on copulations in terrestrial caecilians have, to the best of our knowledge, only been presented for the Indian ichthyophi-id Ichthyophis beddomei (Bhatta 1999) and Ichthyophis cf. kohtaoensis (Kupfer et al. 2006a). Bhatta reports on a copulation lasting for about 40 or 45 minutes, an obser-vation fitting well with the duration time of about 45 min-utes that was observed in Ichthyophis cf. kohtaoensis (Kupfer et al. 2006a).
Caecilians show a remarkable diversity of reproductive modes associated with parental care (e.g. Wake 1977; Himstedt 1996; Wilkinson & Nussbaum 1998). Oviparous caecilians guarding egg clutches in terrestrial chambers (e.g. Sarasin & Sarasin 1887–1890) either have the pre-sumed ancestral amphibian life cycle with aquatic larvae, or show direct development of juveniles with no aquatic larval stage (e.g. Brauer 1897). Females of viviparous species retain fertilised eggs. Embryogenesis is complet-ed within the oviducts, and after hatching the foetuses feed mainly intrauterinely on the hypertrophied oviductal lin-ing (e.g. Parker 1956; Welsch et al. 1977). After a long gestation period, the females give birth to fully metamor-phosed, precocial young with the adult-type morphology (e.g. Billo et al. 1985; Exbrayat & Delsol 1985). Recent-ly, a novel form of parental investment, maternal derma-totrophy, a.k.a. skin feeding, where altricial young feed externally on the mother´s hyperthrophied skin, has been described (Kupfer et al. 2006b; Wilkinson et al. 2008).
SUMMARY AND PERSPECTIVES In addition to their remarkable diversity of reproductive modes, amphibians also show large variation in their re-productive morphology. Many morphological peculiari-ties are related to the evolution of internal fertilisation, and ultimately to viviparity. In relation to fertilisation and sperm transfer, different strategies have evolved within the three amphibian orders, ranging from cloacal apposition in anurans to a true copula via a highly complex male in-tromittent organ in caecilians. Amphibians offer a prime system for comparative studies of evolutionary reproduc-tive biology. Research on the reproductive or genital mor-phology should include modern methodology, such as 3D reconstruction and soft tissue synchrotron radiation based X-ray microtomography (SRµCT, see Fig. 2 C–D). Be-cause amphibian diversity is steadily increasing (although at the same time many species are declining or even go-ing extinct) we envisage that many more unexpected re-productive strategies and morphologies remain to be dis-covered.
Acknowledgements. Phillipp Wagner is congratulated for or-ganising this magnificent Festschrift volume. Many thanks for giving us the opportunity to contribute. We would like to dedi-cate our contribution to Wolfgang Böhme for his continuous ef-fort in studying vertebrate, especially squamate genitalia and their evolution. Synchrotron radiation based x-ray microtomog-raphy (SRµCT) of caecilian genitalia was carried out at beam-line BW2 at the Deutschen Elektronen Synchroton (DESY, Ham-burg) under experimental projects I-20080054 and I-20090089.
Travel of SK and AK has been generously supported by DESY.
Felix Beckmann (DESY) provided initial 3D reconstructions and Frank Friedrich (University of Hamburg) and Thomas Kleinte-ich (University of Washington) aided in the successful process-ing of 3D models. Alexander Haas (University of Hamburg) gen-erously provided laboratory space and technical support for SK during her collections-based research at the Museum für Zoolo-gie Hamburg (ZMH). SK is financially supported by the Volk-swagen Stiftung (grant initiative “evolutionary biology”). An anonymous referee, Lennart Olsson and Philipp Wagner gave valuable comments on earlier versions of the manuscript.
REFERENCES Arnold SJ (1977) The courtship behavior of North American salamanders with some comments on Old World salamandrids.
Pp. 141–183 in: Taylor D & Guttman S (eds.) The Reproduc-tive Biology of Amphibians. Plenum Press, New York Barrio A (1969) Observaciones sobre Chthonerpeton indistinc-tum (Gymnophiona, Caecilidae) y su reproduccion. Physis 28: 499–503 124 Susanne Kühnel et al.
Bonn zoological Bulletin 57 (2): 119–126 ©ZFMK Beck CW (1998) Mode of fertilization and parental care in anu-rans. Animal Behaviour 55: 439–449 Bhatta G (1999) Some aspects of general activity, foraging and breeding in Ichthyophis beddomei (Peters) and Ichthyophis malabarensis (Taylor) (Apoda: Ichthyophiidae) in captivity.
Zoos’ Print Journal 14: 23–36 Billo RR, Straub JO, Senn DG (1985) Vivipare Apoda (Amphib-ia: Gymnophiona), Typhlonectes compressicaudus (Duméril & Bibron, 1841): Kopulation, Tragzeit und Geburt. Amphib-ia-Reptilia 6: 1–9 Böhme W, Ziegler T (2009) A review of iguanian and an-guimorph lizard genitalia (Squamata: Chamaeleonidae; Vara-noidea, Shinisauridae, Xenosauridae, Anguidae) and their phy-logenetic significance: comparisons with molecular data sets.
Journal of Zoological Systematics and Evolutionary Research 47: 189–202 Brauer A (1897) Beiträge zur Kenntnis der Entwicklungs-geschichte und der Anatomie der Gymnophionen. Zoologis-che Jahrbücher, Abteilung für Anatomie und Ontogenie der Thiere. 10: 389–472 Brizzi R, Calloni C, Delfino G, Tanteri G (1995) Notes on the Male Cloacal Anatomy and Reproductive Biology of Euproc-tus montanus (Amphibia: Salamandridae). Herpetologica 51: 8–18 Carranza S, Amat F (2005) Taxonomy, biogeography and evo-lution of Euproctus (Amphibia : Salamandridae), with the res-urrection of the genus Calotriton and the description of a new endemic species from the Iberian Peninsula. Zoological Jour-nal of the Linnean Society 145: 555–582 Crump ML (1972) Territoriality and Mating Behavior in Den-drobates granuliferus (Anura: Dendrobatidae). Herpetologi-ca 28: 195–198 Darwin CR (1871) The descent of man, and selection in rela-tion to sex. John Murray Press, London Drewry GE, Kirkland LJ (1976) A New Ovoviviparous Frog, Eleutherodactylus jasperi (Amphibia, Anura, Leptodactyli-dae), from Puerto Rico. Journal of Herpetology 10: 161–165 Duellman WE & Trueb L (1994) Biology of Amphibians. John Hopkins, Baltimore Emerson SB (2001) Amacroevolutionary study of historical con-tingency in the fanged frogs of Southeast Asia. Biological Journal of the Linnean Society 73: 139–151 Exbrayat J-M & Delsol M (1985) Reproduction and Growth of Typhlonectes compressicaudus: A Viviparous Gymnophione.
Copeia 1985: 950–955 Exbrayat J-M, Estabel J (2006) Anatomy with particular refer-ence to the reproductive system. Pp. 79–156 in: Exbrayat J-M (ed.) Reproductive Biology and Phylogeny of Gymnophiona (Caecilians). Science Publishers Inc., Enfield, NH Exbrayat JM (1991) Anatomie du cloaque chez quelques Gymnophiones. Bulletin de la Société Herpétologique de France 58: 30–42 Frost DR, Grant T, Faivovich J, Bain RH, Haas A, Haddad CFB, De Sa RO, Channing A, Wilkinson M, Donnellan SC, Rax-worthy CJ, Campbell JA, Blotto BL, Moler P, Drewes RC, Nussbaum RA, Lynch JD, Green DM, Wheeler WC (2006) The amphibian tree of life. Bulletin of the American Muse-um of Natural History 297: 1–370 Gower DJ, Wilkinson M (2002) Phallus morphology in caecil-ians and its systematic utility. Bulletin of the Natural Histo-ry Museum, Zoology Series 68: 143–154 Grandison AGC (1978) The occurrence of Nectophrynoides (Anura: Bufonidae) in Ethiopia. A new concept of the genus with a description of a new species. Monitore zoologico ital-iano (NS) supplemento 11: 119–172 Grandison AGC (1980) Aspects of breeding morphology in Mertensophryne micranotis (Anura: Bufonidae): secondary sexual characters, eggs and tadpole. Bulletin of the British Mu-seum (Natural History) Zoology 39: 299–304 Grandison AGC, Ashe S (1983) The distribution, behavioural ecology and breeding strategy of the pygmy toad, Merten-sophryne micranotis (Lov.). Bulletin of the British Museum (Natural History) Zoology 45: 85–93 Haddad CFB, Prado CPA (2005) Reproductive modes in frogs and their unexpected diversity in the Atlantic forest of Brazil.
Bioscience 55: 207–217 Halliday T (1998) Sperm competition in amphibians. Pp.
465–502 in: Birkhead TR & Møller AP (eds.) Sperm Compe-tition and Sexual Selection. Academic Press, London Himstedt W (1996) Die Blindwühlen. Westarp Wissenschaften, Magdeburg Houck LD, Watts RA, Arnold SJ, Bowen KE, Kiemnec KM, Godwin HA, Feldhoff PW, Feldhoff RC (2008) A Recombi-nant Courtship Pheromone Affects Sexual Receptivity in a Plethodontid Salamander. Chemical Senses 33: 623–631 Kardong KV (2006) Vertebrates: comparative anatomy, function, evolution. Fourth edition. Mc Graw Hill, New York Kühnel S, Herzen J, Kleinteich T, Beckmann F, Kupfer A (ac-cepted pending revision) The female cloaca of an oviparous caecilian amphibian (Gymnophiona): functional and season-al aspects. Acta Zoologica Kunte K (2004) Natural History and Reproductive Behavior of Nyctibatrachus cf. humayuni (Anura: Ranidae). Herpetolog-ical Review 35: 137–140 Kupfer A (2007) Sexual size dimorphism in amphibians: an overview. in: Fairbairn DJ, Blanckenhorn WU & Szekely T (eds.) Sex, Size and Gender Roles: Evolutionary Studies of Sexual Size Dimorphism. Oxford University Press, Oxford Kupfer A, Kramer A, Himstedt W, Greven H (2006a) Copula-tion and egg retention in an oviparous Caecilian (Amphibia: Gymnophiona). Zoologischer Anzeiger 244: 223–228 Kupfer A, Müller H, Antoniazzi MM, Jared C, Greven H, Nuss-baum RA, Wilkinson M (2006b) Parental investment by skin feeding in a caecilian amphibian. Nature 440: 926–929 Lombardi J (1998) Comparative vertebrate reproduction. Kluw-er Academic Publishers, Norwell, Massachusetts USA Meisenheimer J (1921) Geschlecht und Geschlechter im Tier-reiche. I. Die natürlichen Beziehungen. Gustav Fischer, Jena Metter DE (1964) A Morphological and Ecological Comparison of Two Populations of the Tailed Frog, Ascaphus truei Stej-neger. Copeia 1964: 181–195 Murphy JB, Quinn H, Campbell J (1977) Observations on the breeding habits of the aquatic caecilian Typhlonectes compres-sicaudus. Copeia 1: 66–69 Noble GK (1927) The value of life history data in the study of evolution. Annals of the New York Academy of Sciences 30: 31–128 Noble GK, Putnam PG (1931) Observations on the Life Histo-ry of Ascaphus truei Stejneger. Copeia 1931: 97–101 Parker HW (1956) Viviparous caecilians and amphibian phyloge-ny. Nature 178: 250–252 Pombal JP, Jr., Sazima I, Haddad ClFB (1994) Breeding Behav-ior of the Pumpkin Toadlet, Brachycephalus ephippium (Brachycephalidae). Journal of Herpetology 28: 516–519 Salthe SN (1969) Reproductive modes and the numbers and size of ova in the urodeles. American Midland Naturalist 81: 467–490 125 Amphibian reproductive morphology Bonn zoological Bulletin 57 (2): 119–126 ©ZFMK Sandberger L, Hillers A, Doumbia J, Loua N-S, Brede C, Rödel M-O (2010) Rediscovery of the Liberian Nimba toad, Nim-baphrynoides liberiensis (Xavier, 1978) (Amphibia: Anura: Bufonidae), and reassessment of its taxonomic status. Zootaxa 2355: 56–68 Sarasin P, Sarasin F (1887–1890) Ergebnisse naturwis-senschaftlicher Forschungen auf Ceylon in den Jahren 1884–1886. Zur Entwicklunggeschichte und Anatomie der Ceylonesischen Blindwühle Ichthyophis glutinosus. C. W.
Kreidel, Wiesbaden Sawaya MP (1942) Sobre la cloaca dos Siphonops. Boletins da Faculdade de Filosofia, Ciencias e Letras, Zoologia, Sao Pao-lo 15: 3–43 Sever DM (1994) Observations on regionalization of secretory activity in the spermathecae of salamanders and comments on phylogeny of sperm storage in female amphibians. Herpeto-logica 50: 383–397 Sever DM (2002) Female sperm storage in amphibians. Journal of Experimental Zoology 292: 165–179 Sever DM, Brizzi R (1998) Comparative biology of sperm stor-age in female salamanders. Journal of Experimental Zoology 282: 460–476 Sever DM, Moriarty EC, Rania LC, Hamlett WC (2001) Sperm storage in the oviduct of the internal fertilizing frog Ascaphus truei. Journal of Morphology 248: 1–21 Sever DM, Rania LC, Krenz JD (1996) Annual cycle of sperm storage in spermathecae of the red-spotted newt, Notophthal-mus viridescens (Amphibia: Salamandridae). Journal of Mor-phology 227: 155–170 Slater JR (1931) The Mating of Ascaphus truei Stejneger. Copeia 1931: 62–63 Stebbins RC, Cohen NW (1997) A Natural History of Amphib-ians. Princeton University Press, Princeton, NJ Steinfartz S, Stemshorn K, Kuesters D, Tautz D (2006) Patterns of multiple paternity within and between annual reproduction cycles of the fire salamander (Salamandra salamandra) under natural conditions. Journal of Zoology 268: 1–8 Stephenson B, Verrell P (2003) Courtship and mating of the tailed frog (Ascaphus truei). Journal of Zoology 259: 15–22 Stuart S, Hoffmann M, Chanson J, Cox N, Berridge R, Ramani P, Young B (2008) Threatened Amphibians of the World. Lynx Edicions, Barcelona Taylor EH (1968) The Caecilians of the World – A taxonomic review. University of Kansas Press, Lawrence Tonutti E (1931) Beitrag zur Kenntnis der Gymnophionen XV.
Das Genitalsystem. Morphologisches Jahrbuch 68: 151–293 Tonutti E (1933) Beitrag zur Kenntnis der Gymnophionen XIX.
Untersuchungen der Kopulationsorgane bei weiteren Gymnophionenarten. Morphologisches Jahrbuch 72: 156–211 Townsend DS, Stewart MM (1986) Courtship and Mating Be-havior of a Puerto Rican Frog, Eleutherodactylus coqui. Her-petologica 42: 165–170 Townsend DS, Stewart MM, Pough FH, Brussard PF (1981) In-ternal Fertilization in an Oviparous Frog. Science 212: 469–471 Verrell PA, Krenz JD (1998) Competition for Mates in the Mole Salamander, Ambystoma talpoideum: Tactics That May Max-imize Male Mating Success. Behaviour 135: 121–138 Wake MH (1972) Evolutionary morphology of the caecilian uro-genital system. IV. The cloaca. Journal of Morphology 136: 353–365 Wake MH (1977) The reproductive biology of caecilians: an evo-lutionary perspective. Pp. 73–101 in: Taylor DH & Guttman SI (eds.) The Reproductive Biology of Amphibians. Plenum Publishing Corporation, New York Wake MH (1978) The Reproductive Biology of Eleutherodacty-lus jasperi (Amphibia, Anura, Leptodactylidae), with Com-ments on the Evolution of Live-Bearing Systems. Journal of Herpetology 12: 121–133 Wake MH (1980) The Reproductive Biology of Nectophrynoides malcolmi (Amphibia: Bufonidae), with Comments on the Evo-lution of Reproductive Modes in the Genus Nectophrynoides.
Copeia 1980: 193–209 Wake MH (1981) Structure and function of the male Mueller-ian gland in caecilians, with comments on Its evolutionary sig-nificance. Journal of Herpetology 15: 17–22 Wake MH (1993) Evolution of oviductal gestation in amphib-ians. Journal of Experimental Zoology 266: 394–413 Wake MH (1998) Cartilage in the cloaca: Phallodeal spicules in caecilians (Amphibia : Gymnophiona). Journal of Morphol-ogy 237: 177–186 Wake MH, Dickie R (1998) Oviduct structure and function and reproductive modes in amphibians. Journal of Experimental Zoology 282: 477–506 Wells KD (2007) The ecology and behaviour of amphibians. Uni-versity of Chicago, Chicago and London Welsch U, Müller M, Schubert C (1977) Elektronenmikroskopis-che und histochemische Beobachtungen zur Fortpflanzungs-biologie viviparer Gymnophionen (Chthonerpeton indistinc-tum). Zoologische Jahrbücher für Anatomie 97: 532–549 Wilkinson M (1990) The presence of a Musculus Retractor Cloa-cae in female caecilians (Amphibia: Gymnophiona). Amphib-ia-Reptilia 11: 300–304 Wilkinson M, Kupfer A, Marques-Porto R, Jeffkins H, Antoni-azzi MM, Jared C (2008) One hundred million years of skin feeding? Extended parental care in a Neotropical caecilian (Amphibia : Gymnophiona). Biology Letters 4: 358–361 Wilkinson M, Nussbaum RA(1998) Caecilian viviparity and am-niote origins. Journal of Natural History 32: 1403–1409 Wourms JP, Lombardi J (1992) Reflections on the evolution of piscine viviparity. American Zoologist 32: 276–293 Received: 25.IX.2010 Accepted: 16.X.2010 126 Susanne Kühnel et al.
Bonn zoological Bulletin 57 (2): 119–126 ©ZFMK |
190733 | https://www.neuroskills.com/education-and-resources/drs-scoring-instructions/ | DRS Scoring Instructions | Centre for Neuro Skills
Press Option+1 for screen-reader mode, Option+0 to cancelAccessibility Screen-Reader Guide, Feedback, and Issue Reporting | New window
[x] EN
ES
Contact US | Refer a Patient888.811.3065
Search for:
Search for:
About Us Mission StatementHistoryFounder’s StoryCompany History LeadershipClinical LeadershipMedical LeadershipResidential Leadership CareersWho We AreBenefitsStudent ProgramCareer GrowthCareer FAQs Why Choose CNSPatient OutcomesEnvironmental Validity FAQs News September 26, 2025 ##### After unimaginable loss, a North Texas family struggles to rebuild their lives September 22, 2025 ##### BACK TO LIFE: Ex-firefighter recounts near-death experience September 15, 2025 ##### E-scooter injuries: Rising trends and ways to be safe
Recovery Journey Clinical Evaluation AdmissionsFunding ProgramsResidential InpatientDay TreatmentAdolescent ProgramContinued CareTelerehabilitation Therapies Services TreatmentTraumatic Brain Injury (TBI)StrokeMild Traumatic Brain Injury (MTBI)Post COVID-19 Neuro Injury Discharge Community Reintegration
Locations Bakersfield Los Angeles San Francisco Austin Dallas Fort Worth Houston
Patients & Caregivers Patient Experiences Patient Videos Patient Outcomes FinancesMake a PaymentFunding & Payor OverviewSocial Security Benefits Patient SupportAdvocacy ToolkitAssistive TechnologyBrain Injury Support GroupsCaregiving & TBICoping With FearLegal IssuesStress ManagementSelecting a Rehab Program Family ResourcesKey Points for FamiliesConsequences of Brain Injury on FamiliesStress Management for FamiliesParents of Adults With TBITherapeutic ActivitiesFamily Resource Links Children & Brain InjuryHelping Children CopeSiblings & PeersStudents With TBIParenting After Brain Injury
Professionals Continued Education CNS Outcomes & Value Return to Work Payor Relations Workers’ Compensation Sport Concussion Assessment Tool 5 CNS IN THE NEWS September 26, 2025 ##### After unimaginable loss, a North Texas family struggles to rebuild their lives September 22, 2025 ##### BACK TO LIFE: Ex-firefighter recounts near-death experience
News & Media News Events Videos Press Room NewsSeptember 26, 2025 ##### After unimaginable loss, a North Texas family struggles to rebuild their lives September 22, 2025 ##### BACK TO LIFE: Ex-firefighter recounts near-death experience September 15, 2025 ##### E-scooter injuries: Rising trends and ways to be safe EventsOctober 24 - 25 Austin TX ##### Team Luke Hope for Minds – Making Connections Conference October 08 San Diego, CA ##### MacroPro and Friends-Detecting Malingering Following Mild Traumatic Brain Injury: Diagnostic Complexities and Practical Treatment Considerations February 03 - 06 New Orleans, Louisiana ##### International Stroke Conference 2026
About Brain Injury Brain InjuryBrain FunctionBrain Injury OverviewMTBI & ConcussionStrokeNeuroplasticityVeterans & TBIVision & Brain InjuryBrain Injury StatisticsCNS Pinterest Bookstore Vision & Brain InjuryBrain Injury StatisticsCNS Pinterest Bookstore Rating ScalesCNS Independent Living ScaleRancho Los Amigos – RevisedMayo-Portland Adaptability Inventory (MPAI)Glasgow Coma Scale (GCS)Disability Rating Scale (DRS)DRS Scoring Instructions
Research Who We Are Research Projects Publications Articles & Abstracts RECENT POSTS##### Neuroendocrine dysfunction after TBI: A guide for the clinical neuropsychologist ##### Postacute Rehabilitation May Reduce Rehospitalization, Emergency Room Visits ##### Centre for Neuro Skills and University of Texas Medical Branch Study ##### Sleep architecture following stroke is distinct from TBI
Search for:
DRS Scoring Instructions
Item Definitions
Eye opening
0 - SPONTANEOUS: eyes open with sleep/wake rhythms, indicating active arousal mechanisms; does not assume awareness.
1 - TO SPEECH AND/OR SENSORY STIMULATION: a response to any verbal approach, whether spoken or shouted, not necessarily the command to open the eyes. Also, response to touch, mild pressure.
2 - TO PAIN: tested by a painful stimulus.
3 - NONE: no eye opening even to painful stimulation.
Best communication ability
If patient cannot use voice because of tracheostomy, is aphasic or dysarthric, or has vocal cord paralysis or voice dysfunction, then estimate patient’s best response and enter note under comments.
0 - ORIENTED: implies awareness of self and the environment. Patient is able to tell you a) who they are; b) where they are; C) why they are there; d) year; e) season; f) month; g) day; h) time of day.
1 - CONFUSED: attention can be held and patient responds to questions but responses are delayed and/or indicate varying degrees of disorientation and confusion.
2 - INAPPROPRIATE: intelligible articulation but speech is used only in an exclamatory or random way (such as shouting and swearing); no sustained communication exchange is possible.
3 - INCOMPREHENSIBLE: moaning, groaning, or sounds without recognizable words; no consistent communication signs.
4 - NONE: no sounds or communications signs from patient.
Best motor response
0 - OBEYING: obeying command to move finger on best side. If no response or not suitable, try another command such as move lips, blink eyes, etc. Do not include grasp or other reflex responses.
1 - LOCALIZING: a painful stimulus at more than one site causes a limb to move (even slightly) in an attempt to remove it. It is a deliberate motor act to move away from or remove the source of noxious stimulation. If there is doubt as to whether withdrawal or localization has occurred after 3 or 4 painful stimulations, rate as localization.
2 - WITHDRAWING: any generalized movement away from a noxious stimulus that is more than a simple reflex response.
3 - FLEXING: painful stimulation results in either flexion at the elbow, rapid withdrawal with abduction of the shoulder, or a slow withdrawal with adduction of the shoulder. If there is confusion between flexing and withdrawing, use pin prick on hands, then face.
4 - EXTENDING: painful stimulation results in extension of the limb.
5 - NONE: no response can be elicited. Usually associated with hypotonia. Exclude spinal transection as an explanation of lack of response; be satisfied that an adequate stimulus has been applied.
Cognitive ability for feeding, toileting, and grooming
Rate each of the 3 functions separately. For each function, answer the question, “Does the patient show awareness of how and when to perform each specified activity?” Ignore motor disabilities that interfere with carrying out a function. (This is rated under Level of Functioning described below.) Rate best response for toileting based on bowel and bladder behavior. Grooming refers to bathing, washing, brushing of teeth, shaving, combing or brushing of hair, and dressing.
0 - COMPLETE: continuously shows awareness that they know how to feed, toilet, or groom self and can convey unambiguous information that they know when this activity should occur.
1 - PARTIAL: intermittently shows awareness that they know how to feed, toilet, or groom self and/or can intermittently convey reasonably clear information that they know when the activity should occur.
2 - MINIMAL: shows in a primitive way how to feed, toilet, or groom self and/or shows infrequently by certain signs, sounds, or activities that they are vaguely aware of when the activity should occur.
3 - NONE: shows virtually no awareness at any time that they know how to feed, toilet, or groom self and cannot convey information by signs, sounds, or activity that they know when the activity should occur.
Level of functioning
0 - COMPLETELY INDEPENDENT: able to live as they wish, requiring no restriction due to physical, mental, emotional, or social problems.
1 - INDEPENDENT IN SPECIAL ENVIRONMENT: capable of functioning independently when needed requirements are met (mechanical aids).
2 - MILDLY DEPENDENT: able to care for most of own needs but requires limited assistance due to physical, cognitive, and/or emotional problems (eg, needs nonresident helper).
3 - MODERATELY DEPENDENT: able to care for self partially but needs another person at all times.
4 - MARKEDLY DEPENDENT: needs help with all major activities and the assistance of another person at all times.
5 - TOTALLY DEPENDENT: not able to assist in own care and requires 24-hour nursing care.
Employability
The psychosocial adaptability or employability item takes into account overall cognitive and physical ability to be an employee, homemaker, or student This determination should take into account considerations such as the following:
Able to understand, remember, and follow instructions;
Can plan and carry out tasks at least at the level of an office clerk or ri simple routine, repetitive industrial situations, or can do school assignments;
Ability to remain oriented, relevant, and appropriate in work and other psychosocial situations;
Ability to get to and from work or shopping centers using private or public transportation effectively;
Ability to deal with number concepts;
Ability to make purchases and handle simple money exchange problems;
Ability to keep track of time schedules and appointments.
0 - NOT RESTRICTED: can compete in the open market for a relatively wide range of jobs commensurate with existing skills; or can initiate, plan, execute, and assume responsibilities associated with homemaking; or can understand and carry out most age-relevant school assignments.
1 - SELECTED JOBS, COMPETITIVE: can compete in a limited job market for a relatively narrow range of jobs because of limitations of the type described above and/or because of some physical limitations; or can initiate, plan, execute, and assume many but not all responsibilities associated with homemaking; or can understand and carry out many but not al school assignments.
2 - SHELTERED WORKSHOP, NON-COMPETITIVE: cannot compete successfully in job market because of limitations described above and/or because of moderate or severe physical limitations; or cannot without major assistance initiate, plan, execute, and assume responsibilities for homemaking; or cannot understand and carry out even relatively simple school assignments without assistance.
3 - NOT EMPLOYABLE: completely unemployable because of extreme psychosocial limitations of the type described above; or completely unable to initiate, plan, execute, and assume any responsibilities associated with homemaking; or cannot understand or carry out any school assignments.
Instructions
Place date of rating at top of column. Place appropriate rating next to each of the eight items listed. Add eight ratings together to obtain total DR score.
Quick Links
Home
About Us
Careers
Recovery Journey
Locations
Patients & Caregivers
Professionals
News & Media
About Brain Injury
Research
Locations
Austin, TX
Bakersfield, CA
Dallas, TX
Fort Worth, TX
Houston, TX
Los Angeles, CA
Plano, TX
San Francisco, CA
Stay Connected ==============
CNS Monthly Newsletter
The latest CNS updates, including events, company information, and patient care developments
The Inside View
Quarterly magazine focused on brain injury research, rehabilitation, and advancements shaping the field
Subscribe Now
Sign-up for one or both to stay connected with brain injury news and recover
888.811.3065
Notice of Information PracticesPrivacy Policy & Practices
© Copyright 2025 Centre for Neuro Skills. All Rights Reserved. |
190734 | https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Multivariable_Calculus/1%3A_Vectors_in_Space/Intersection_of_a_Line_and_a_Plane | Published Time: 2019-10-01T15:11:41Z
Intersection of a Line and a Plane - Mathematics LibreTexts
Skip to main content
Table of Contents menu
search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode
school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Search
Search this book
Submit Search
x
Text Color
Reset
Bright Blues Gray Inverted
Text Size
Reset
+-
Margin Size
Reset
+-
Font Type
Enable Dyslexic Font - [x]
Downloads expand_more
Download Page (PDF)
Download Full Book (PDF)
Resources expand_more
Periodic Table
Physics Constants
Scientific Calculator
Reference expand_more
Reference & Cite
Tools expand_more
Help expand_more
Get Help
Feedback
Readability
x
selected template will load here
Error
This action is not available.
chrome_reader_mode Enter Reader Mode
1: Vectors in Space
Multivariable Calculus
{ }
{ "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Finding_the_Angle_a_Given_Vector_Makes_with_the_Positive_x-axis" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Intersection_of_a_Line_and_a_Plane : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
{ "1:_Vectors_in_Space" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "2:_Topics_in_Vector-Valued_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3:_Topics_in_Partial_Derivatives" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
Tue, 17 Nov 2020 15:35:09 GMT
Intersection of a Line and a Plane
24528
24528
admin
{ }
Anonymous
Anonymous
2
false
false
[ "article:topic", "authorname:pseeburger", "license:ccby" ]
[ "article:topic", "authorname:pseeburger", "license:ccby" ]
Search site Search Search Go back to previous article
Sign in
Username Password Sign in
Sign in
Sign in
Forgot password
Contents
1. Home
2. Bookshelves
3. Calculus
4. Supplemental Modules (Calculus)
5. Multivariable Calculus
6. 1: Vectors in Space
7. Intersection of a Line and a Plane
Expand/collapse global location
Home
Campus Bookshelves
Bookshelves
Arithmetic and Basic Math
Pre-Algebra
Algebra
Geometry
Precalculus & Trigonometry
Calculus
Supplemental Modules (Calculus)
Differential Calculus
Integral Calculus
Vector Calculus
Multivariable Calculus
1: Vectors in Space
2: Topics in Vector-Valued Functions
3: Topics in Partial Derivatives
Method of Lagrange Multipliers (Trench)/Method_of_Lagrange_Multipliers_(Trench))
Calculus (OpenStax))
Calculus (Guichard))
Calculus 3e (Apex))
Map: Calculus - Early Transcendentals (Stewart))
Map: University Calculus (Hass et al.))
Yet Another Calculus Text - A Short Introduction with Infinitesimals (Sloughter))
Applied Calculus (Calaway, Hoffman and Lippman))
Informal Calculus with Applications to Biological and Environmental Sciences (Seacrest))
Book: Active Calculus (Boelkins et al.))
CLP-1 Differential Calculus (Feldman, Rechnitzer, and Yeager))
CLP-2 Integral Calculus (Feldman, Rechnitzer, and Yeager))
CLP-3 Multivariable Calculus (Feldman, Rechnitzer, and Yeager))
CLP-4 Vector Calculus (Feldman, Rechnitzer, and Yeager))
Vector Calculus (Corral))
The Calculus of Functions of Several Variables (Sloughter))
Differential Calculus for the Life Sciences (Edelstein-Keshet))
Elementary Calculus 2e (Corral))
Calculus by David Guichard (Improved))
Differential Equations
Analysis
Linear Algebra
Abstract and Geometric Algebra
Combinatorics and Discrete Mathematics
Mathematical Logic and Proofs
Applied Mathematics
Scientific Computing, Simulations, and Modeling
Learning Objects
Intersection of a Line and a Plane
Last updated Nov 17, 2020
Save as PDF
Finding the Angle a Given Vector Makes with the Positive x-axis
Back Matter
Page ID 24528
Paul Seeburger
Monroe Community College
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents No headers
A given line and a given plane may or may not intersect. If the line does intersect with the plane, it's possible that the line is completely contained in the plane as well. How can we differentiate between these three possibilities?
Example 8 8: Finding the intersection of a Line and a plane
Determine whether the following line intersects with the given plane. If they do intersect, determine whether the line is contained in the plane or intersects it in a single point. Finally, if the line intersects the plane in a single point, determine this point of intersection.
Line:x y z=2−t=1+t=3 t Plane:3 x−2 y+z=10 Line:x=2−t Plane:3 x−2 y+z=10 y=1+t z=3 t
Solution
Notice that we can substitute the expressions of t t given in the parametric equations of the line into the plane equation for x x, y y, and z z.
3(2−t)−2(1+t)+3 t=10 3(2−t)−2(1+t)+3 t=10
Solving this equation for t t:
6−3 t−2−2 t+3 t=10 6−3 t−2−2 t+3 t=10
4−2 t=10 4−2 t=10
−2 t=6−2 t=6
t=−3 t=−3
Since we found a single value of t t from this process, we know that the line should intersect the plane in a single point, here where t=−3 t=−3. So the point of intersection can be determined by plugging this value in for t t in the parametric equations of the line.
Here: x=2−(−3)=5,y=1+(−3)=−2,and z=3(−3)=−9 x=2−(−3)=5,y=1+(−3)=−2,and z=3(−3)=−9.
So the point of intersection of this line with this plane is (5,−2,−9)(5,−2,−9). We can verify this by putting the coordinates of this point into the plane equation and checking to see that it is satisfied.
Check: 3(5)−2(−2)+(−9)=15+4−9=10✓3(5)−2(−2)+(−9)=15+4−9=10✓
Now that we have examined what happens when there is a single point of intersection between a line and a point, let's consider how we know if the line either does not intersect the plane at all or if it lies on the plane (i.e., every point on the line is also on the plane).
Example 9 9: Other relationships between a line and a plane
Determine whether the following line intersects with the given plane. If they do intersect, determine whether the line is contained in the plane or intersects it in a single point. Finally, if the line intersects the plane in a single point, determine this point of intersection.
Line:x y z=1+2 t=−2+3 t=−1+4 t Plane:x+2 y−2 z=5 Line:x=1+2 t Plane:x+2 y−2 z=5 y=−2+3 t z=−1+4 t
Solution
Substituting the expressions of t t given in the parametric equations of the line into the plane equation gives us:
(1+2 t)+2(−2+3 t)−2(−1+4 t)=5(1+2 t)+2(−2+3 t)−2(−1+4 t)=5
Simplifying the left side gives us:
1+2 t−4+6 t+2−8 t=5 1+2 t−4+6 t+2−8 t=5
Collecting like terms on the left side causes the variable t t to cancel out and leaves us with a contradiction:
−1=5−1=5
Since this is not true, we know that there is no value of t t that makes this equation true, and thus there is no value of t t that will give us a point on the line that is also on the plane. This means that this line does not intersect with this plane and there will be no point of intersection.
How can we tell if a line is contained in the plane?
What if we keep the same line, but modify the plane equation to be x+2 y−2 z=−1 x+2 y−2 z=−1? In this case, repeating the steps above would again cause the variable t t to be eliminated from the equation, but it would leave us with an identity, −1=−1−1=−1, rather than a contradiction. This means that every value of t t will produce a point on the line that is also on the plane, telling us that the line is contained in the plane whose equation is x+2 y−2 z=−1 x+2 y−2 z=−1.
This page titled Intersection of a Line and a Plane is shared under a CC BY license and was authored, remixed, and/or curated by Paul Seeburger.
Back to top
Finding the Angle a Given Vector Makes with the Positive x-axis
Back Matter
Was this article helpful?
Yes
No
Recommended articles
Front Matter
Back Matter
Finding the Angle a Given Vector Makes with the Positive x-axis
TitlePage
InfoPage
Article typeSection or PageAuthorPaul SeeburgerLicenseCC BY
Tags
This page has no tags.
© Copyright 2025 Mathematics LibreTexts
Powered by CXone Expert ®
?
The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org.
Support Center
How can we help?
Contact Support Search the Insight Knowledge Base Check System Status×
contents readability resources tools
☰
Finding the Angle a Given Vector Makes with the Positive x-axis
Back Matter |
190735 | https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.norm.html | GitHub
Scientific Python Forum
GitHub
Scientific Python Forum
scipy.stats.norm#
scipy.stats.norm = object>[source]#
: A normal continuous random variable.
The location (`loc`) keyword specifies the mean. The scale (`scale`) keyword specifies the standard deviation.
As an instance of the `rv_continuous` class, `norm` object inherits from it a collection of generic methods (see below for the full list), and completes them with details specific for this particular distribution.
Methods
| | |
--- |
| rvs(loc=0, scale=1, size=1, random\_state=None) | Random variates. |
| pdf(x, loc=0, scale=1) | Probability density function. |
| logpdf(x, loc=0, scale=1) | Log of the probability density function. |
| cdf(x, loc=0, scale=1) | Cumulative distribution function. |
| logcdf(x, loc=0, scale=1) | Log of the cumulative distribution function. |
| sf(x, loc=0, scale=1) | Survival function (also defined as `1 - cdf`, but sf is sometimes more accurate). |
| logsf(x, loc=0, scale=1) | Log of the survival function. |
| ppf(q, loc=0, scale=1) | Percent point function (inverse of `cdf` — percentiles). |
| isf(q, loc=0, scale=1) | Inverse survival function (inverse of `sf`). |
| moment(order, loc=0, scale=1) | Non-central moment of the specified order. |
| stats(loc=0, scale=1, moments=’mv’) | Mean(‘m’), variance(‘v’), skew(‘s’), and/or kurtosis(‘k’). |
| entropy(loc=0, scale=1) | (Differential) entropy of the RV. |
| fit(data) | Parameter estimates for generic data. See scipy.stats.rv\_continuous.fit for detailed documentation of the keyword arguments. |
| expect(func, args=(), loc=0, scale=1, lb=None, ub=None, conditional=False, \\kwds) | Expected value of a function (of one argument) with respect to the distribution. |
| median(loc=0, scale=1) | Median of the distribution. |
| mean(loc=0, scale=1) | Mean of the distribution. |
| var(loc=0, scale=1) | Variance of the distribution. |
| std(loc=0, scale=1) | Standard deviation of the distribution. |
| interval(confidence, loc=0, scale=1) | Confidence interval with equal areas around the median. |
Notes
The probability density function for `norm` is:
\[f(x) = \frac{\exp(-x^2/2)}{\sqrt{2\pi}}\]
for a real number \(x\).
The probability density above is defined in the “standardized” form. To shift and/or scale the distribution use the `loc` and `scale` parameters. Specifically, `norm.pdf(x,loc,scale)` is identically equivalent to `norm.pdf(y)/ scale` with `y =(x -loc)/ scale`. Note that shifting the location of a distribution does not make it a “noncentral” distribution; noncentral generalizations of some distributions are available in separate classes.
Examples
```
>>> import numpy as np>>> from scipy.stats import norm>>> import matplotlib.pyplot as plt>>> fig, ax = plt. subplots(1, 1)
```
Get the support:
```
>>> lb, ub = norm. support()
```
Calculate the first four moments:
```
>>> mean, var, skew, kurt = norm. stats(moments = 'mvsk')
```
Display the probability density function (`pdf`):
```
>>> x = np. linspace(norm. ppf(0.01),... norm. ppf(0.99), 100)>>> ax. plot(x, norm. pdf(x),... 'r-', lw = 5, alpha =0.6, label = 'norm pdf')
```
Alternatively, the distribution object can be called (as a function) to fix the shape, location and scale parameters. This returns a “frozen” RV object holding the given parameters fixed.
Freeze the distribution and display the frozen `pdf`:
```
>>> rv = norm()>>> ax. plot(x, rv. pdf(x),'k-', lw = 2, label = 'frozen pdf')
```
Check accuracy of `cdf` and `ppf`:
```
>>> vals = norm. ppf([0.001,0.5,0.999])>>> np. allclose([0.001,0.5,0.999], norm. cdf(vals)) True
```
Generate random numbers:
```
>>> r = norm. rvs(size = 1000)
```
And compare the histogram:
```
>>> ax. hist(r, density = True, bins = 'auto', histtype = 'stepfilled', alpha =0.2)>>> ax. set_xlim([x, x[- 1]])>>> ax. legend(loc = 'best', frameon = False)>>> plt. show()
```
On this page |
190736 | https://www.clarke-energy.com/us/heating-value/ | Load More Search Results
English (US)
English
Français (French)
Ελληνικά (Greek)
Romnă (Romanian)
Heating Value
Home → Heating Value
The lower heating value (LHV) or higher heating value (HHV) of a gas is an important consideration when selecting a gas engine or CHP plant. Gas engines efficiency is typically quoted based upon the LHV of the gas.
Whenever a hydrocarbon fuel is burned one product of combustion is water. The quantity of water produced is dependent upon the amount of hydrogen in the fuel. Due to high combustion temperatures, this water takes the form of steam which stores a small fraction of the energy released during combustion as the latent heat of vaporization; in simple terms, as heat energy stored in the vaporized ‘state’ of water.
The total amount of heat liberated during the combustion of a unit of fuel, the HHV or HCV, includes the latent heat stored in the vaporized water. In some applications it is possible to condense this vapor back to its liquid state and ‘recover’ a proportion of this energy. However, engine exhaust temperatures are above that at which the water vapor would condense, and hence the steam ‘escapes’ with the exhaust gases carrying with it the stored energy.
The amount of heat available from a fuel after the latent heat of vaporization, the LHV or LCV, is deducted from the HHV, and it is this, that is available when the fuel is burned in an engine. The energy input into a gas engine should be defined using the LHV of the fuel.
Fuel suppliers will usually quote the HHV and it will be this measure that will be used when kWh unit charges are applied for the fuel. In the case of natural gas the ratio of HHV to LHV is approximately 1.108:1. Hence, when performing a cost benefit analysis for a CHP application, it is the HHV figure which should be used.
The LHV of a fuel determines the fuel flow rate required when going into the engine because the total quantity of energy input necessary for the engine to produce a specific output power is defined and fixed. Hence the gas flow rate has to be such in order to provide the required energy input.
Fuel LHV is normally quoted using units of kWh/Nm3 and therefore, if the energy input to the engine is known, the gas flow rate in m3/hr can easily be calculated.
See also
Laminar Flame Speed Methane Number
Service Offerings
Service
Commissioning
Maintenance Agreements
Technical Training
Remote Monitoring & Diagnostics
Field Service
Spare Parts
Upgrades, Repair & Overhaul
Conversions, Modifications and Upgrades
Can We Help? Contact Us Now >
Any Further Questions?
If you have any technical questions that need answering, would like to arrange to speak to a sales advisor or book a feasibility study.
Contact Us |
190737 | https://www.brown.edu/Departments/Engineering/Courses/En4/notes_old/straightlline/Straightlineex.htm | EN4: Dynamics and Vibrations
Division of Engineering
Brown University
Examples of straight line motion problems
Review: Straight Line Motion with Constant Acceleration.
A particle is at position and has speed at time t=0. If the particle
accelerates at a constant rate ,
find expressions for the speed and position of the particle as functions of time.
Start by integrating the acceleration to find the velocity
Now integrate the velocity to find the position
You will probably remember these results from your carefree youth… You can use the
results as formulae for any problem where the acceleration is constant. In general,
however, you will have to do the calculus from scratch.
Example 1: An accelerometer is attached to a structure to measure its vibration
characteristics. The structure is set in motion by releasing it from rest at an initial
position =3cm. The resulting acceleration may be approximated by , where A= and rad/sec. Find expressions for the velocity and position
of the structure as a function of time.
As always we integrate the acceleration to get the velocity and then integrate the
velocity to get the position
where we have noted that because
the system was released from rest.
Now do the calculus again to get the position
We can now substitute numerical values to see that
Example 2. In many problems, you will need to calculate the position and
velocity of an object as a function of time, and then use the results to answer some
problem. Here is an example. Suppose you want to throw some keys to a friend standing on a
second floor balcony. If you release the keys at m above the ground, what initial vertical velocity is
needed for them to reach your friend’s hand a height h=6m above the ground?
We clearly will need to be able to calculate the position of the keys as a function of
time to be able to answer this problem, so that’s where we start. We first need to
get the acceleration of the keys: draw a free body diagram showing the force acting on the
keys
Hence, Newton II shows that
so that a=-g.
This is therefore a case of constant acceleration. We can merely write down the
position and velocity from the results we calculated earlier:
Now let’s review the data we are given. We know that the keys are released from . This leaves us with 4 more
unknown quantities in our equations: s,t,v,and . We need to find some physical conditions
that will give us these values.
Two observations:
(i) If you throw something vertically up in the air, it slows down, comes to a stop at
its maximum height, and falls back down again;
(ii) If you throw the keys too fast, they will go higher than 6m; if you throw them too
slowly, they will not reach 6m.
Thus, the critical vertical velocity is such that when the keys come to a stop, they
just reach 6m.
Suppose the keys reach their maximum height at a time . Since the velocity is zero at this time,
we have
We need to find such that s=6mat time . Thus
Example 3: An aircraft carrier of mass M is traveling with speed when its engines are shut down.
Find an expression that may be used to determine how far the carrier will travel and how
long it will take to stop.
First, describe the problem. Idealize the ship as a particle, let s denote the
distance traveled after engine shutdown, and let v denote the ship’s speed.
We will assume that the carrier is slowed by hydrodynamic drag. As a first
approximation, we will assume that the drag force may be expressed as
where v is the speed of the ship and is a constant.
As always, we start with a free body diagram showing the forces acting on the ship.
There are three forces: weight, buoyancy and drag
The acceleration of the ship may be written in terms of either its speed or distance
traveled as
Write down Newton’s second law
so, comparing I and j coefficients we learn that
Now we can do some calculus. The second equation will give us an expression for
velocity as a function of time, as follows
We could integrate this to find the position of the ship as a function of time. Recall
that v=ds/dt, so
The third equation will give us an expression for the speed of the carrier as a
function of its distance traveled
Now, let’s see what the math is telling us. We have an expression for velocity as
a function of time
so we can try to find the time at which v=0, but the answer is ! This is
not good.
Let’s see if we can do better finding the distance required to stop. We have that
and again, when v=0, .
Bummer…
What has gone wrong here? Well, we should ask three questions:
(i) Is the answer physically reasonable? Not really: we wouldn’t expect a ship to
go on traveling indefinitely (a possible exception is the `Flying Dutchman,’ a famous
ghost ship…)
(ii) Did we do the math right? A good way to check is to look at the physical
dimensions of all our equations. Everything seems to check out.
(iii) Did we make any incorrect assumptions? Well, what assumptions did we make. First,
we idealized the tanker as a particle. That may be a stretch, but the tanker doesn’t
deform or rotate significantly while it’s slowing down so this approximation should
be OK. We assumed that Newton’s laws will work. That’s probably true as well. We
also assumed that drag force is proportional to the square of the velocity. This is the
assumption that is most likely to be incorrect.
Indeed, it turns out that is valid only for high velocities, when the drag force
is primarily associated with the inertia of the fluid – as the ship pushes through
the water, it must exert a force on the fluid to move it aside. An equal and opposite
force acts on the ship. At lower velocities, the accelerations in the fluid are much
smaller so the hydrodynamic drag force is less significant. Instead, the ship’s
motion is resisted by viscous drag in the water. Viscous drag is caused by the
(rather weak) attractive forces that act between water molecules. It varies linearly with
velocity
You can verify for yourself that with viscous drag only, the speed and distance
traveled by the ship are given as functions of time by
while the distance traveled is related to the speed by
Now, the ship still takes an infinite time to stop, but the total distance traveled is
finite and is given by
Finally, we could combine our results to estimate the total distance traveled by the
ship. To do this, we could split the motion into two parts. If the hydrodynamic force is
greater than viscous drag, we will assume that motion is resisted by hydrodynamic drag
only. If the viscous drag is greater, we will assume that motion is resisted by viscous
drag only.
The velocity at which the two drag forces are equal is given by
Assuming that , the distance required to slow from initial speed to speed under hydrodynamic drag is given by
The additional distance required to slow from speed to zero under viscous drag is
so the total distance to stop is
The time required to stop is still infinite.
Let’s substitute some numbers to get an estimate. Specifications for aircraft
carriers are easy to find on the web. For example, for the Nimitz class:
Length: 317m (1040 feet – that’s a mighty short runway to land a jet on!)
Beam (that’s width to landlubbers): 40.8m (134 feet)
Displacement (that’s mass ): 87,998 metric tons (a ton is 1000 kg)
Speed: 30 knots (that’s sea miles per hour, or 55km/hr or 15 m/s)
It’s harder to estimate the constants C and B. There are a few
computer simulations of hull drag published on the web. Expressions for drag force are
normally expressed in the form
where is the density of the fluid, A is the immersed area of the hull, and
is the drag coefficient –
a number which is either computed or measured for a particular hull. The drag coefficient
appears to be a number around according
to simulations.
Our constant C therefore follows as
The density of sea water is of the order of 1000 , and we can estimate the immersed area by
approximating the hull as a rectangle length L, width W and height H,
and using the fact that the weight of water displaced by the hull must be equal to the
weight of the ship (Archimedes principle) Thus
This gives . A draft of about 20 feet seems about right.
The immersed area is then
Therefore 9000 is a reasonable number for C.
The number B is even harder to find, because viscous drag kicks in at such low
velocities that it is of little consequence to ship design. We’ll make a wild guess
and say that viscous drag kicks in at about 1 knot (0.5m/s). This gives
Luckily our estimate turns out not to be particularly sensitive to the value used for B.
Finally, plug all our estimated numbers into the expression we calculated for the
stopping distance to conclude that the stopping distance is around 30km – that’s
about 20 miles!
Controlling the drag forces acting on a ship is a matter of considerable practical
importance. The drag forces are actually rather more complex than the simple forms
we’ve used here – there are several different contributions to drag, including
skin friction drag (the viscous term), form drag (primarily due to the inertia of the
fluid) and wave drag (a large ship pushes a big wave of water along with it – lifting
the water in this wave contributes to drag and is very significant at high velocities).
There’s an interesting non-technical article on the subject in Scientific American.
Example 5: A spring-mass oscillator consists of a mass and a spring connected as
shown. The coordinate s measures the displacement of the mass relative to its
position when the spring is unstretched. If the spring is linear, the mass is subjected to
a deceleration proportional to s. Suppose that and that you give the mass a velocity v=1 m/s in
the position s=0.
(a) How far will the mass move to the right before the spring brings it to a
stop?
To answer this, we need an expression for distance moved as a function of velocity. We
can get one by integrating the alternative expression for acceleration:
We are given that the initial velocity and the initial position . Consequently, when the mass stops (v=0), we have
.
Why do we get two solutions? Well, the mass vibrates back and forth: first it moves 1m
to the right (s=+1/2m) and then 1m to the left (s=-1/2m). Consequently,
there are two positions where v=0. We are told to find how far the mass moves to
the right, which is
.
(b) What will be the velocity of the mass when it has returned to the position s=0?
We already know the relationship between s and v from the preceding part
of the question:
We can simply set in the
result to see that
Not enough information is given in the question to decide whether the velocity is
positive or negative. The first time the mass returns to s=0, the velocity is
negative, the second time it is positive, and so on.
Although we are not asked to do so in the question, we may calculate s as a
function of time. We have that
The velocity and acceleration follow as
Observe that a=-4s, as required.
We now have all the techniques we need to compute position and velocity from
acceleration, at least for the case of straight line motion. One final remark is in order.
The procedure we described here only works if we are given , or , that is, acceleration is
either: (i) only a function of time; (ii) only a function of velocity or (iii) only a
function of position.
It sometimes happens that acceleration is a function of all three! For example, when we
analyze forced vibrations of a spring-mass-damper system later in the course, we will find
that the equation for acceleration is
Here, ,
, and are constants.
The good news is that this equation can be solved. The bad news is that the solution is
much more difficult than the problems we’ve discussed here – the worst news is
that you will have to learn how to solve it. |
190738 | https://rosalind.info/glossary/chargaffs-rules/ | Glossary
Chargaff's rules
Chargaff's rules is a two main rules of nucleotide distribution in DNA strings, discovered by Austrian chemist
Erwin Chargaff in early 1950s in Columbia University.
First Chargaff's rule (or first parity rule) holds that in double-stranded DNA molecule
observed percentage base pair equality: %A = %T and %G = %C.
This finding, with the results of x-ray diffraction analysis by Rosalind Franklin, served as one of the
grounds for the Watson-Crick double helix model.
Second Chargaff's rule (second parity rule) holds that for each of the DNA strands
observed following approximate equality: %A ~ %T and %G ~ %C.
It was an empirical observation, and the basis for this rule is still under investigation.
It was shown that it does not apply to organellar genomes (mitochondria and plastids) smaller
than ~20-30 kbp, single stranded DNA (viral) genomes or any type of RNA genome.
Wikipedia
Report a typo
Page:
Context: |
190739 | https://link.springer.com/article/10.1007/s00029-021-00714-6 | Promotion of Kreweras words
You have full access to this open access article
2272 Accesses
6 Citations
6
Altmetric
Explore all metrics
Abstract
Kreweras words are words consisting of n (\mathrm {A})’s, n (\mathrm {B})’s, and n (\mathrm {C})’s in which every prefix has at least as many (\mathrm {A})’s as (\mathrm {B})’s and at least as many (\mathrm {A})’s as (\mathrm {C})’s. Equivalently, a Kreweras word is a linear extension of the poset (\mathsf{V}\times [n]). Kreweras words were introduced in 1965 by Kreweras, who gave a remarkable product formula for their enumeration. Subsequently they became a fundamental example in the theory of lattice walks in the quarter plane. We study Schützenberger’s promotion operator on the set of Kreweras words. In particular, we show that 3n applications of promotion on a Kreweras word merely swaps the (\mathrm {B})’s and (\mathrm {C})’s. Doing so, we provide the first answer to a question of Stanley from 2009, asking for posets with ‘good’ behavior under promotion, other than the four families of shapes classified by Haiman in 1992. We also uncover a strikingly simple description of Kreweras words in terms of Kuperberg’s (\mathfrak {sl}_3)-webs, and Postnikov’s trip permutation associated with any plabic graph. In this description, Schützenberger’s promotion corresponds to rotation of the web.
Similar content being viewed by others
Weyl Group ({\varvec{q}})-Kreweras Numbers and Cyclic Sieving
A Proof of a Conjecture of Gukov–Pei–Putrov–Vafa
On the Support of Grothendieck Polynomials
Explore related subjects
Avoid common mistakes on your manuscript.
1 Introduction
The famous ballot problem, whose history stretches back to the 19th century, asks in how many ways we can order the ballots of an election between two candidates Alice and Bob, who each receive n votes, so that during the counting of ballots Alice never trails Bob. These ballot orderings co length 2n in the letters (\mathrm {A}) and (\mathrm {B}), with as many (\mathrm {A})’s as (\mathrm {B})’s, for which every prefix has at least as many (\mathrm {A})’s as (\mathrm {B})’s. Such words are called Dyck words, and they are counted by the ubiquitous Catalan numbers
In 1965, Kreweras considered the following version of a 3-candidate ballot problem: in how many ways can we order the ballots of an election between three candidates Alice, Bob, and Charlie, who each receive n votes, so that during the counting Alice never trails Bob and Alice never trails Charlie—although the relative position of Bob and Charlie may change during the counting? These ballot orderings correspond to words of length 3n in the letters (\mathrm {A}), (\mathrm {B}), and (\mathrm {C}), with equally many (\mathrm {A})’s, (\mathrm {B})’s, and (\mathrm {C})’s, for which every prefix has at least as many (\mathrm {A})’s as (\mathrm {B})’s and also at least as many (\mathrm {A})’s as (\mathrm {C})’s. We call such words Kreweras words. Kreweras proved that they are counted by the formula
For many years Kreweras’s formula seemed like an isolated enumerative curiosity, although simplified proofs were presented by Niederhausen [19, 20] and Kreweras–Niederhausen in the 1980s. Gessel gave yet another proof which demonstrated that the generating function (\sum _{n=0}^{\infty }K_n \, x^n) for this sequence of numbers is algebraic. Interest in Kreweras’s result was revived decades later in the context of lattice walk enumeration. Kreweras words evidently correspond to walks in ({\mathbb {Z}}^2) with steps of the form (\mathrm {A}=(1,1)), (\mathrm {B}=(-1,0)), and (\mathrm {C}=(0,-1)) from the origin to itself which always remain in the nonnegative orthant. Such walks are called Kreweras walks. Bousquet-Mélou gave another proof of Kreweras’s product formula counting Kreweras walks using the kernel method from analytic combinatorics. Indeed, the Kreweras walks are nowadays a fundamental example in the study of “walks with small step sizes in the quarter plane,” a program successfully carried out over a number of years in the 2000s by Bousquet-Mélou and others (see, e.g., ). Finally, we note that Bernardi gave a purely combinatorial proof of the product formula for the number of Kreweras walks via a bijection with (decorated) cubic maps.
In this paper we study a cyclic group action on Kreweras words.
Let (w=(w_1,w_2,\ldots ,w_{3n})) be a Kreweras word of length 3n. The promotion of w, denoted ({{\,\mathrm{{Pro}}\,}}(w)), is obtained from w as follows. Let (\iota (w)) be the smallest index (\iota \ge 1) for which the prefix ((w_1,w_2,\ldots ,w_{\iota })) has either the same number of (\mathrm {A})’s as (\mathrm {B})’s or the same number of (\mathrm {A})’s as (\mathrm {C})’s. Then
It is easy to verify that ({{\,\mathrm{{Pro}}\,}}(w)) is also a Kreweras word, and that promotion is an invertible action on the set of Kreweras words.
Example 1.1
Let . Here we circled the letter (w_{\iota (w)}), and hence ({{\,\mathrm{{Pro}}\,}}(w) = \mathrm {A}\mathrm {B}\mathrm {A}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}\mathrm {B}). We can further compute that the first several iterates of promotion applied to w are
Note that ({{\,\mathrm{{Pro}}\,}}^9(w)) is obtained from w by swapping all (\mathrm {B})’s for (\mathrm {C})’s and vice-versa.
Our first result predicts the order of promotion on Kreweras words:
Theorem 1.2
Let w be a Kreweras word of length 3n. Then ({{\,\mathrm{{Pro}}\,}}^{3n}(w)) is obtained from w by swapping all (\mathrm {B})’s for (\mathrm {C})’s and vice-versa. In particular, ({{\,\mathrm{{Pro}}\,}}^{6n}(w)=w).
Promotion of Kreweras words comes from the theory of partially ordered sets. In a series of papers from the 60s and 70s, Schützenberger [28,29,30] introduced and developed the theory of a cyclic action called promotion, as well as a closely related involutive action called evacuation, on the linear extensions of any poset. Let V(n) denote the Cartesian product of the 3-element “V”-shaped poset and the n-element chain [n]. Then, as observed by Kreweras–Niederhausen , the linear extensions of V(n) are in obvious bijection with the Kreweras words of length 3n. And promotion of Kreweras words as described above is the same as Schützenberger’s promotion on the linear extensions of V(n).
Previously there were only four known (non-trivial) families of posets for which the order of promotion can be predicted; see Fig. 4. These were classified by Haiman in the 1990s [10, 11]. In a survey on promotion and evacuation, Stanley [33, §4, Question 3] asked whether there were any other families of posets for which the order of promotion is given by a simple formula. Our work shows that V(n) is such an example.
Promotion of Dyck words as rotation of noncrossing matchings
Promotion of Kreweras words as rotation of webs
Dyck words of length 2n correspond to linear extensions of (\times [n]), and hence carry an action of promotion. Figure 1 depicts a well-known bijection between Dyck words of length 2n and noncrossing matchings of ([2n] {:}{=}{1,2,\ldots ,2n}), and shows how under this bijection promotion of Dyck words corresponds to rotation of noncrossing matchings (this was first observed by Dennis White: see [26, §8]). This observation immediately implies that ({{\,\mathrm{{Pro}}\,}}^{2n}(w)=w) for w a Dyck word of length 2n.
Our proof of Theorem 1.2 is also essentially based on a diagrammatic representation of Kreweras words for which promotion corresponds to rotation; see Fig. 2. However, these diagrams are not coming from Bernardi’s cubic map bijection. Instead, they are related to Kuperberg’s webs.
Webs are certain trivalent bipartite planar graphs which Kuperberg [17, 109–151 (1996)")] introduced in order to study the invariant theory of Lie algebras. Khovanov and Kuperberg [14$$
sl
(
3
)
are not dual canonical. Pac. J. Math. 188(1), 129–153 (1999)")] showed that a particular class of webs (namely, irreducible \(\mathfrak {sl}\_3\)-webs with 3n white boundary vertices) are in bijection with linear extensions of \(\times [n]\). Petersen, Pylyavskyy, and Rhoades [22, 19–41 (2009)")] (see also Tymoczko [37, 611–632 (2012)")]) showed that, via the Khovanov-Kuperberg bijection, rotation of webs corresponds to promotion of linear extensions.
We say that ({\mathcal {W}}) is a Kreweras web if ({\mathcal {W}}) is an irreducible (\mathfrak {sl}_3)-web with all boundary vertices white and having no internal face with a multiple of four sides. We define a surjective map (w \mapsto {\mathcal {W}}_w) from Kreweras words to Kreweras webs. This map behaves well with respect to Schützenberger’s operators:
Theorem 1.3
Let w be a Kreweras word. Then,
({\mathcal {W}}_{{{\,\mathrm{{Pro}}\,}}(w)} = {{\,\mathrm{{Rot}}\,}}({\mathcal {W}}_{w}));
({\mathcal {W}}_{{{\,\mathrm{{Evac}}\,}}(w)} = {{\,\mathrm{{Flip}}\,}}({\mathcal {W}}_{w})),
where ({{\,\mathrm{{Rot}}\,}}) denotes the rotation of a web and ({{\,\mathrm{{Flip}}\,}}) its reflection across a diameter.
The map between Kreweras words and Kreweras webs can be made bijective by keeping track of a certain 3-edge-coloring of the web. We then obtain the following enumerative corollaries.
Theorem 1.4
We have
where the sum is over all Kreweras webs ({\mathcal {W}}) with 3n boundary vertices, and (\kappa ({\mathcal {W}})) is the number of connected components of ({\mathcal {W}}). Moreover, the number of connected Kreweras webs ({\mathcal {W}}) with 3n boundary vertices is
A curious feature of our work not present in any previous work we are aware of is the use of trip permutations, in the sense of Postnikov’s theory of plabic graphs [24")], to study webs.
The rest of the paper is organized as follows. In Sect. 2 we review promotion and evacuation of linear extensions. In Sect. 3 we prove Theorem 1.2 by using what we call the Kreweras bump diagram of a Kreweras word w, and extracting from this diagram a permutation (\sigma _w) which rotates under promotion. This permutation (\sigma _w) is in fact a trip permutation. In Sect. 4 we discuss evacuation of Kreweras words, using the theory of growth diagrams. In Sect. 5 we reinterpret our results in the language of webs and plabic graphs. Finally, in Sect. 6 we briefly discuss some possible future directions.
2 Promotion and evacuation of linear extensions
In this section we quickly review background on promotion of linear extensions of posets, and explain precisely how promotion of Kreweras words as defined in Sect. 1 fits into that framework. We assume that the reader is familiar with basic notions from poset theory as laid out for instance in [34, Chapter 3]. Throughout, all posets are assumed to be finite.
Let P be a poset with (\ell ) elements. For us, a linear extension of P will be a list ((p_1,p_2,\ldots ,p_{\ell })) of all of the elements of P, each appearing once, for which (p_i \le p_j) implies that (i \le j). We use ({\mathcal {L}}(P)) denote the set of linear extensions of P.
For each (1 \le i \le \ell -1), we define the involution (\tau _i:{\mathcal {L}}(P) \rightarrow {\mathcal {L}}(P)) by
In other words, (\tau _i) swaps the ith and ((i+1))st entries of a linear extension, if possible.
Definition 2.1
Promotion, ({{\,\mathrm{{Pro}}\,}}:{\mathcal {L}}(P)\rightarrow {\mathcal {L}}(P)), is the following composition of (\tau _i):
Being a composition of involutions, promotion is evidently invertible. There is another definition of promotion in terms of jeu de taquin slides, but we find the definition as composition of the (\tau _i) more convenient; see Stanley’s survey [33, §2] for the details.
The poset V(n) whose linear extensions are Kreweras words
Recall that in Sect. 1 we defined to be the poset which is the Cartesian product of the 3-element “V”-shaped poset and the n-element chain. We label the 3n elements of V(n) by the symbols (\mathrm {A}_1,\mathrm {A}_2,\ldots ,\mathrm {A}_n), (\mathrm {B}_1,\mathrm {B}_2,\ldots ,\mathrm {B}_n), and (\mathrm {C}_1,\mathrm {C}_2,\ldots ,\mathrm {C}_n) as depicted in Fig. 3.
Now let us show that promotion of Kreweras words is the same as promotion of linear extensions of V(n):
Proposition 2.2
“Removing the subscripts” is a bijection from linear extensions of V(n) to Kreweras words of length 3n, and under this bijection, promotion of linear extensions corresponds to promotion of Kreweras words as was defined in Sect. 1.
Proof
The claim about removing the subscripts being a bijection is straightforward. The comparison of the two promotions is also essentially straightforward. Let w be a Kreweras word of length 3n, and consider the (\tau _i) as acting on w via the aforementioned bijection. Then the effect of (\tau _i) on w is
Consider the application of (\tau _{3n-1} \circ \cdots \circ \tau _1) to w. Recall the definition of (\iota (w)) from Sect. 1. When applying (\tau _i) for (i=1,\ldots ,\iota (w)-2) to w, we will never be in the “otherwise” case above; so the effect of (\tau _{\iota (w)-2} \circ \cdots \circ \tau _{1}) will be to cyclically rotate the substring ((w_1,\ldots ,w_{\iota (w)-1})). In other words, we “push” the initial (\mathrm {A}) in w as far to the right as we can. Then, when applying (\tau _{\iota (w)-1}) we will for the first time be in the “otherwise” case, so (\tau _{\iota (w)-1}) will act as the identity. Finally, when applying (\tau _i) for (i=\iota (w)+1,\ldots ,3n-1) to w, we will again never be in the “otherwise” case; so the effect of (\tau _{3n-1} \circ \cdots \circ \tau _{\iota (w)}) will be to cyclically rotate the substring ((w_{\iota (w)},\ldots ,w_{3n})). In other words, we “push” the (\mathrm {B}) or (\mathrm {C}) in position (\iota (w)) all the way to the end. Thus (\tau _{3n-1} \circ \cdots \circ \tau _1) indeed acts on w in the same way as ({{\,\mathrm{{Pro}}\,}}), as was defined in Sect. 1. (\square )
Next, in order to motivate the study of (powers of) promotion, let us recall the basic results concerning promotion and its close cousin evacuation which were established by Schützenberger.
Definition 2.3
Evacuation, ({{\,\mathrm{{Evac}}\,}}:{\mathcal {L}}(P)\rightarrow {\mathcal {L}}(P)), is the following composition of (\tau _i):
Again, there is an alternative definition of evacuation in terms of jeu de taquin. Historically, interest in evacuation arose because of its close connection with the Robinson-Schensted correspondence.
There is an obvious duality between the linear extensions of P and of its dual poset (P^): for (L = (p_1,\ldots ,p_{\ell }) \in {\mathcal {L}}(P)), we use (L^) to denote the linear extension (L^ {:}{=}(p_{\ell },p_{\ell -1},\ldots ,p_1) \in {\mathcal {L}}(P^)) of the dual poset.
Dual evacuation, ({{\,\mathrm{{Evac}}\,}}^{}:{\mathcal {L}}(P)\rightarrow {\mathcal {L}}(P)), is defined by ({{\,\mathrm{{Evac}}\,}}^{}(L) {:}{=}({{\,\mathrm{{Evac}}\,}}(L^))^). In some sense, dual evacuation is “just as natural” as evacuation. In terms of the involutions (\tau _i), we have
Note that (L \mapsto ({{\,\mathrm{{Pro}}\,}}(L^))^) is just ({{\,\mathrm{{Pro}}\,}}^{-1}), so we do not give it a different name.
The basic results of Schützenberger [28,29,30] on promotion and evacuation of an arbitrary poset P are summarized in the following proposition; again, see Stanley [33, §2] for a modern presentation.
Proposition 2.4
For any poset P, we have that:
({{\,\mathrm{{Evac}}\,}}) and ({{\,\mathrm{{Evac}}\,}}^{}) are both involutions;
({{\,\mathrm{{Evac}}\,}}\circ {{\,\mathrm{{Pro}}\,}}= {{\,\mathrm{{Pro}}\,}}^{-1} \circ {{\,\mathrm{{Evac}}\,}});
({{\,\mathrm{{Pro}}\,}}^{#P} = {{\,\mathrm{{Evac}}\,}}^{} \circ {{\,\mathrm{{Evac}}\,}}).
Because ({{\,\mathrm{{Pro}}\,}}^{#P}) is the composition of the “natural” involutions ({{\,\mathrm{{Evac}}\,}}) and ({{\,\mathrm{{Evac}}\,}}^{}), this power of ({{\,\mathrm{{Pro}}\,}}) is somehow the “right” power to consider. (Sometimes older sources refer to ({{\,\mathrm{{Pro}}\,}}^{#P}) as total promotion, but we will eschew this terminology as it tends to confuse.) Following [33, §4], we now review the posets P for which the behavior of ({{\,\mathrm{{Pro}}\,}}^{#P}) is understood.
The previously known posets with good behavior of promotion
The fundamental properties of jeu de taquin which Schützenberger established in imply that for (P=[k] \times [n]) a product of two chains, a.k.a. a rectangle, ({{\,\mathrm{{Pro}}\,}}^{#P}) is the identity. While studying reduced decompositions in the symmetric group, Edelman and Greene showed that for P a staircase (depicted in Fig. 4b by way of example), ({{\,\mathrm{{Pro}}\,}}^{#P}) is transposition (i.e., the poset automorphism that is the reflection across the vertical axis of symmetry). Finally, Haiman showed that for P a shifted double staircase (Fig. 4c) or a shifted trapezoid (Fig. 4d), ({{\,\mathrm{{Pro}}\,}}^{#P}) is the identity; and he did this using a general method (based on his notion of dual equivalence) which recaptures the rectangle and staircase results as well. In a follow-up paper, Haiman and Kim proved moreover that among Young diagram shapes and shifted shapes, the posets in Fig. 4 are the only families for which ({{\,\mathrm{{Pro}}\,}}^{#P}) is the identity or transposition.
In [33, §4, Question 3], Stanley asked whether there are any other (non-trivial) families of posets, beyond those depicted in Fig. 4, for which ({{\,\mathrm{{Pro}}\,}}^{#P}) can be described. Our main result, Theorem 1.2, shows that V(n) is such an example: for (P=V(n)), ({{\,\mathrm{{Pro}}\,}}^{#P}) is “transposition” (i.e., reflection across the vertical axis of symmetry).
3 The order of promotion
In this section we prove Theorem 1.2. Throughout we fix a nonnegative integer n as in the statement of that theorem. Also, from now on we adopt the notational convention (-\mathrm {B}: = \mathrm {C}) and (-\mathrm {C}{:}{=}\mathrm {B}).
Our strategy is to associate to each Kreweras word a permutation, such that promotion of the Kreweras word corresponds to rotation of the permutation.
Definition 3.1
We use ({\mathfrak {S}}_{m}) to denote the symmetric group on m letters. We represent permutations (\sigma \in {\mathfrak {S}}_{m}) either via cycle notation, or via one-line notation as (\sigma = [\sigma (1),\ldots ,\sigma (m)]). For (\sigma \in {\mathfrak {S}}_{m}), the rotation of (\sigma ), denoted ({{\,\mathrm{{Rot}}\,}}(\sigma )), is the (right) conjugation of (\sigma ) by the long cycle ((1,2,\ldots ,m) \in {\mathfrak {S}}_{m}); i.e.,
Rotation of a permutation as we have just defined it corresponds to rotation of its functional digraph representation.
We first associate a diagram to a Kreweras word, which we will then use to obtain the desired permutation. This diagram will be built out of arcs, and the crossings of the arcs in the diagram will play a significant role. So let us review notions related to arcs and crossings.
Definition 3.2
An arc is a pair (i, j) of positive integers with (i < j). A crossing is a set ({(i,j),(k,\ell )}) of two arcs such that (i\le k< j < \ell ).
Note that this definition slightly deviates from the usual notion, in that the pairs (i, j) and ((i,\ell )) form a crossing. However, this modification is only relevant when considering Kreweras bump diagrams, defined below.
Definition 3.3
Let ({\mathcal {A}}) be a collection of arcs. For a set of positive integers S, we say that ({\mathcal {A}}) is a noncrossing matching of S if for every ((i,j)\in {\mathcal {A}}) we have (i,j \in S), every (i\in S) belongs to a unique arc in ({\mathcal {A}}), and no two arcs in ({\mathcal {A}}) form a crossing. The set of openers of ({\mathcal {A}}) is ({i:(i,j)\in {\mathcal {A}}}) and the set the set of closers of ({\mathcal {A}}) is ({j:(i,j)\in {\mathcal {A}}}).
The bijection between Dyck words and noncrossing matchings suggested by Fig. 1 enters into the definition of the diagram we associate to a Kreweras word, so let us now formalize this bijection. Let w be a Dyck word of length 2n. We associate to w the unique noncrossing matching of [2n] whose set of openers is ({i\in [2n]:w_i=\mathrm {A}}) and whose set of closers is ({i\in [2n]:w_i=B}). This sets up a (well-known) bijection between the Dyck words of length 2n and the noncrossing matchings of [2n].
A Kreweras word can be thought of as two overlapping Dyck words, and hence has two noncrossing matchings naturally associated to it. As we now explain, the diagram for our Kreweras word will be the union of these two noncrossing matchings.
Definition 3.4
Let w be a Kreweras word of length 3n. Let (\varepsilon \in {\mathrm {B},\mathrm {C}}). We use ({\mathcal {M}}^\varepsilon _w) to denote the noncrossing matching of ({i\in [3n]:w_i\ne -\varepsilon }) whose set of openers is ({i\in [3n]:w_i= \mathrm {A}}) and whose set of closers is ({i\in [3n]:w_i=\varepsilon }).
The Kreweras bump diagram ({\mathcal {D}}_w) of w is obtained by placing the numbers (1,\ldots ,3n) in this order on a line, and drawing a semicircle above the line connecting i and j for each arc ((i, j)\in {\mathcal {M}}^{B}_w \cup {\mathcal {M}}^{C}_w). The arc is solid blue if ((i,j)\in {\mathcal {M}}^{B}_w) and dashed crimson (i.e., red) if ((i,j) \in {\mathcal {M}}^{C}_w). The arcs are drawn in such a fashion that only pairs of arcs which form a crossing intersect, and any two arcs intersect at most once.
Example 3.5
As in Example 1.1, let (w = \mathrm {A}\mathrm {A}\mathrm {B}\mathrm {B}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}). The two noncrossing matchings ({\mathcal {M}}^B_w) and ({\mathcal {M}}^C_w), drawn as arc diagrams, are
The Kreweras bump diagram ({\mathcal {D}}_w) is obtained by placing these two arc diagrams on top of one another:
Now we extract from ({\mathcal {D}}_w) a permutation (\sigma _w \in {\mathfrak {S}}_{3n}).
The rules of the road when taking a trip in a Kreweras bump diagram: a depicts what happens at an internal crossing, and b depicts what happens at a boundary crossing. Note that the color of the arcs in the crossing is irrelevant
Definition 3.6
Let w be a Kreweras word of length 3n and let ({\mathcal {D}}_w) be its associated Kreweras bump diagram. We define the trip permutation (\sigma _w \in {\mathfrak {S}}_{3n}) of w as follows. For each (i\in {1,\ldots ,3n}), we define (\sigma _w(i)) by taking a trip in ({\mathcal {D}}_w) starting at i, as we now describe. If i is a closer of ({\mathcal {M}}^B_w\cup {\mathcal {M}}^C_w), then we start our trip by walking from i towards (i') along the unique arc ((i',i)) incident to i; if i is an opener of ({\mathcal {M}}^B_w\cup {\mathcal {M}}^C_w), then we start our trip by walking from i towards (i') along the arc ((i,i')) incident to i with the smallest value of (i'). We continue walking until we encounter a crossing. Whenever we encounter a crossing of arcs (a, b) and (c, d) with (a\le c< b < d), we follow the rules of the road: we continue towards
b, if coming from a;
a, if coming from c – this is a left turn;
d, if coming from b – this is a right turn; and
c, if coming from d.
These rules of the road are depicted in Fig. 5. Finally, if j is the terminal vertex of the trip, we set (\sigma _w(i) {:}{=}j).
Example 3.7
As in Example 3.5, let (w = \mathrm {A}\mathrm {A}\mathrm {B}\mathrm {B}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}).
First let us compute (\sigma _w(1)). We start by walking along the arc (1, 4) from 1 towards 4. We encounter the crossing ({(1,4),(2,5)}), but following the first rule we continue towards 4, where we terminate. Thus, (\sigma _w(1)=4). This trip looks pictorially as follows:
Next, let us compute (\sigma _w(4)). We start by walking along the arc (1, 4) from 4 towards 1. As we encounter the crossing ({(1,4),(2,5)}) we turn right, and continue along the arc (2, 5) from 2 towards 5, where we terminate. Thus, (\sigma _w(4)=5):
Next, let us compute (\sigma _w(3)). We start by walking along the arc (2, 3) from 3 towards 2. As we encounter the boundary crossing ({(2,3),(2,5)}) we turn right, and continue along the arc (2, 5) from 2 towards 5. Then we encounter the crossing ({(1,4),(2,5)}) turn left, and continue along the arc (1, 4) from 4 towards 1. At the boundary crossing ({(1,4),(1,8)}) we turn right, and continue along the arc (1, 8) from 1 towards 8. Next we encounter the crossing ({(1,8),(6,9)}), but continue straight along the arc (1, 8) from 1 towards 8, where we terminate. Thus, (\sigma _w(3) = 8).
Finally, let us compute (\sigma _w(7)). We start by walking along the arc (6, 7) from 7 towards 6. There we encounter the boundary crossing ({(6,7),(6,9)}) and turn right, so that we continue along the arc (6, 9) from 6 towards 9. As we encounter the crossing ({(1,8),(6,9)}) we turn left, and continue along the arc (1, 8) from 8 towards 1. We encounter the boundary crossing ({(1,4),(1,8)}), and terminate at 1. Thus, (\sigma _w(7)=1).
We could further compute ([\sigma _w(1),\ldots ,\sigma _w(9)] = [4,3,8,5,2,7,1,9,6]).
Next we establish some basic properties of (\sigma _w).
Proposition 3.8
Let w be a Kreweras word of length 3n.
Definition 3.6 yields a permutation (\sigma _w \in {\mathfrak {S}}_{3n}).
Let (1 \le i \le 3n) with (w_i=\mathrm {A}). Then (w_{\sigma (i)} \in {\mathrm {B},\mathrm {C}}).
Let (1 \le i \le 3n) with (w_i \in {\mathrm {B},\mathrm {C}}). Then either (w_{\sigma _w(i)}=-w_i), or (w_{\sigma _w(i)}=\mathrm {A}) and (w_{\sigma _w(\sigma _w(i))}=-w_i).
We have
In particular, (\sigma _w) has no fixed points.
Proof
For (a): this follows from the fact that the rules of the road permute the incoming and outgoing directions locally at each crossing.
For (b): if (w_i=\mathrm {A}), then the trip in ({\mathcal {D}}_w) starting at i will never turn at a crossing. Hence (\sigma _w(i)) will be the index of the nearer (\mathrm {B}) or (\mathrm {C}) that is matched with the (\mathrm {A}) at i.
For (c): observe that if (w_i\ne \mathrm {A}), then the trip in ({\mathcal {D}}_w) starting at i starts heading towards an index j with (w_j = \mathrm {A}). Moreover, the sequence of turns such a trip makes is a right turn, followed by a left turn, followed by a right turn, et cetera. Whenever the trip turns right, it heads towards an index j with (w_j \ne \mathrm {A}), and whenever it turns left, it heads towards a j with (w_j = \mathrm {A}). If the trip turns an odd number of times in total, it terminates at a j with (w_j\ne \mathrm {A}), and because only oppositely colored arcs of ({\mathcal {D}}_w) cross, this means in fact (w_j = -w_i). If the trip turns an even number of times it terminates at a j with (w_j = \mathrm {A}), and again because only oppositely colored arcs of ({\mathcal {D}}_w) cross, we see from the proof of (b) above that (w_{\sigma _w(j)}=-w_i).
For (d): we will show that (\sigma _w(j) < j) if and only if (\sigma _w(j)=\mathrm {A}), for any (1 \le j \le 3n). If (w_j=\mathrm {A}) then this is clear from the proof of (b) above. So suppose (w_j \ne \mathrm {A}). If the trip in ({\mathcal {D}}_w) starting at j never turns, the claim is also clear. So suppose further the trip starting at j does turn, and suppose the arcs we traverse along the trip are, in order, ((i_0,j_0=j), (i_1,j_1),\ldots , (i_{\ell },j_{\ell })). Then, first note that (i_0< j < j_1). Furthermore, we claim that for all (2\le k\le \ell ), the arc ((i_k,j_k)) nests ((i_{k-2},j_{k-2})): i.e., (i_k< i_{k-2}< j_{k-2} < j_k). Indeed, otherwise either we would not encounter a crossing with ((i_k,j_k)) when traveling along ((i_{k-1},j_{k-1})) from the crossing with ((i_{k-2},j_{k-2})), or we would not turn at such a crossing. To finish the proof of the claim, note that if (\ell ) is even then (\sigma _w(j)=i_{\ell }) and (w_{i_{\ell }}=\mathrm {A}) [see the proof of (c) above], and then the nesting property implies (i_{\ell }< i_0 < j). Similarly, if (\ell ) is odd then (\sigma _w(j)=j_{\ell }) and (w_{j_{\ell }}\ne \mathrm {A}), and then the nesting property implies (j< j_1 < j_{\ell }). (\square )
The permutation (\sigma _w) does not quite determine the Kreweras word w. For example, if (w') is obtained from w by swapping all (\mathrm {B})’s for (\mathrm {C})’s and vice-versa, then clearly we have (\sigma _w = \sigma _{w'}). So we need to keep track of a little extra data along with (\sigma _w). To that end, we define the map (\varepsilon _w:{1,\ldots ,3n} \rightarrow {\mathrm {B}, \mathrm {C}}) by setting
for all (1 \le i \le 3n). Proposition 3.8 (b) guarantees that (\varepsilon _w(i)\in {\mathrm {B}, \mathrm {C}}). As a shorthand we write (\varepsilon _w=[\varepsilon _w(1),\ldots ,\varepsilon _w(3n)]). Thanks to Proposition 3.8 (d), the pair ((\sigma _w,\varepsilon _w)) determines w:
Corollary 3.9
For any Kreweras word w of length 3n,
for all (1 \le i \le 3n).
We now come to the key lemma in the proof of our main result, which says that (\sigma _w) and (\varepsilon _w) evolve in a simple way under promotion.
Lemma 3.10
Let w be a Kreweras word of length 3n. Then,
(\sigma _{{{\,\mathrm{{Pro}}\,}}(w)} = {{\,\mathrm{{Rot}}\,}}(\sigma _w));
(\varepsilon _{{{\,\mathrm{{Pro}}\,}}(w)} = [\varepsilon _w(2),\varepsilon _w(3),\ldots ,\varepsilon _w(3n), -\varepsilon _w(1)]).
Before we prove Lemma 3.10, let’s see an example.
Example 3.11
As in Example 3.7, let (w = \mathrm {A}\mathrm {A}\mathrm {B}\mathrm {B}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}). We saw above that (\sigma _w = [4,3,8,5,2,7,1,9,6]). We also have (\varepsilon _w=[\mathrm {B}, \mathrm {B}, \mathrm {C}, \mathrm {C}, \mathrm {B}, \mathrm {C}, \mathrm {B}, \mathrm {B}, \mathrm {C}]).
As we saw in Example 1.1, ({{\,\mathrm{{Pro}}\,}}(w) = \mathrm {A}\mathrm {B}\mathrm {A}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}\mathrm {B}). Its associated bump diagram is
From the diagram ({\mathcal {D}}_{{{\,\mathrm{{Pro}}\,}}(w)}) one could compute that (\sigma _{{{\,\mathrm{{Pro}}\,}}(w)}=[2, 7, 4, 1, 6, 9, 8, 5, 3]) and (\varepsilon _{{{\,\mathrm{{Pro}}\,}}(w)}=[\mathrm {B}, \mathrm {C}, \mathrm {C}, \mathrm {B}, \mathrm {C}, \mathrm {B}, \mathrm {B}, \mathrm {C}, \mathrm {C}]), in agreement with Lemma 3.10.
We proceed to prove Lemma 3.10.
Proof of Lemma 3.10
The key to the proof is the following procedure to obtain the Kreweras bump diagram ({\mathcal {D}}_{{{\,\mathrm{{Pro}}\,}}(w)}) from ({\mathcal {D}}_w).
Figure illustrating the proof of Lemma 3.10
Let ((1,b)\in {\mathcal {M}}^\mathrm {B}_w) and ((1,c)\in {\mathcal {M}}^\mathrm {C}_w), and suppose that (b < c) without loss of generality. Let ({\widetilde{{\mathcal {N}}}}^\mathrm {B}{:}{=}({\mathcal {M}}^\mathrm {B}_w\setminus {(1,b)} )\cup {(b, 3n+1)}). This is a noncrossing matching, because (i< b< j < 3n+1) would imply (1< i< b < j). Furthermore, let
where ({(i_1,j_1),\dots ,(i_m,j_m)}) is the set of arcs in ({\mathcal {M}}^\mathrm {C}_w) with
that is, the set of arcs which cross (1, b). ({\widetilde{{\mathcal {N}}}}^\mathrm {C}) is a noncrossing matching: for example, suppose that an arc (i, j) satisfies (i< i_\ell< j < j_{\ell -1}); then we have in fact (j < b) because (i, j) cannot cross (1, b). The other cases are dealt with similarly.
Let (\varepsilon \in {\mathrm {B}, \mathrm {C}}) and let ({\mathcal {N}}^\varepsilon ) be obtained from ({\widetilde{{\mathcal {N}}}}^\varepsilon ) by replacing every arc (i, j) with ((i-1, j-1)). Then ({\mathcal {N}}^\varepsilon ) is a noncrossing matching of ({i\in [3n]:{{\,\mathrm{{Pro}}\,}}(w)_i \ne \varepsilon }) with set of openers ({i\in [3n]:{{\,\mathrm{{Pro}}\,}}(w)_i= \mathrm {A}}) and set of closers ({i\in [3n]:{{\,\mathrm{{Pro}}\,}}(w)_i=\varepsilon }). Since the set of openers and the set of closers uniquely determine a noncrossing matching, ({\mathcal {D}}_{{{\,\mathrm{{Pro}}\,}}(w)} = {\mathcal {N}}^\mathrm {B}\cup {\mathcal {N}}^\mathrm {C}).
We define the trip permutation of ({\widetilde{{\mathcal {N}}}}^\mathrm {B}\cup {\widetilde{{\mathcal {N}}}}^\mathrm {C}) in the obvious way, by taking trips starting at (2 \le i \le 3n+1) and using the rules of the road. We now show that (\sigma _w) coincides with the trip permutation of ({\widetilde{{\mathcal {N}}}}^\mathrm {B}\cup {\widetilde{{\mathcal {N}}}}^\mathrm {C}), provided we identify 1 and (3n+1). To do so, we subdivide every arc of ({\mathcal {D}}_w) crossing (1, b) into an initial, a middle and a final part, such that the middle part contains the crossing with (1, b) and no other crossings. Additionally, slightly abusing language, we say that the arc (1, b) only has a middle part and (1, c) consists only of a middle part (containing only the crossing with (1, b)) and a final part.
Similarly, we subdivide every arc of ({\widetilde{{\mathcal {N}}}}^\mathrm {B}\cup {\widetilde{{\mathcal {N}}}}^\mathrm {C}) in ({(i_1,c),(i_2,j_1),\dots ,(i_m,j_{m-1})}) into an initial, a middle and a final part, such that the middle part contains the crossing with ((b,3n+1)) and no other crossings. Additionally, we say that ((b,j_m)) consists only of a middle part (containing only the crossing with ((b,3n+1))) and a final part, and ((b,3n+1)) only has a middle part.
We now note that the initial and the final parts of the arcs in ({\mathcal {D}}_w) and ({\widetilde{{\mathcal {N}}}}^\mathrm {B}\cup {\widetilde{{\mathcal {N}}}}^\mathrm {C}) are identical. It therefore suffices to check that the portions of a trip proceeding in a middle part also begin and end at the same arcs in ({\mathcal {D}}_w) and ({\widetilde{{\mathcal {N}}}}^\mathrm {B}\cup {\widetilde{{\mathcal {N}}}}^\mathrm {C}) (provided we identify 1 and (3n+1)). Labeling the beginnings of the middle parts in ({\mathcal {D}}_w) from left to right with (1, s_1,\dots , s_m) and their endings with (b, t_m,\dots , t_1, t_0) and the beginnings of the middle parts in ({\widetilde{{\mathcal {N}}}}^\mathrm {B}\cup {\widetilde{{\mathcal {N}}}}^\mathrm {C}) from left to right with (s_1,\dots , s_m, b) and their endings with (t_m,\dots , t_1, t_0, 1=3n+1 ), we find that in both cases the trip connects these as follows:
This is depicted in Fig. 6.
This proves (a). For (b): define (\varepsilon '_w = [\varepsilon '_w(1),\ldots ,\varepsilon '_w(3n)]) by
Recall [see the proof of Proposition 3.8(b)] that for i with (w_i=\mathrm {A}), (\sigma _w(i)) is just the index of the nearer (\mathrm {B}) or (\mathrm {C}) that is matched with the (\mathrm {A}) at i. Hence
Together with (a), this observation proves (b). (\square )
Theorem 1.2 follows easily from Lemma 3.10:
Proof of Theorem 1.2
Lemma 3.10 says (\varepsilon _{{{\,\mathrm{{Pro}}\,}}^{3n}(w)} = [-\varepsilon _{w}(1),-\varepsilon _{w}(2),\ldots ,-\varepsilon _{w}(3n)]) and (\sigma _{{{\,\mathrm{{Pro}}\,}}^{3n}(w)} = \sigma _w). Thus Corollary 3.9 implies that ({{\,\mathrm{{Pro}}\,}}^{3n}(w)) is obtained from w by swapping all (\mathrm {B})’s for (\mathrm {C})’s and vice-versa, as claimed. (\square )
4 Evacuation of Kreweras words
Via the bijection between Kreweras words of length 3n and linear extensions of V(n) described in Proposition 2.2, we can also view evacuation as acting on the set of Kreweras words. In this section we will describe the evacuation of a Kreweras word, using some of the machinery from Sect. 3. As we will see, just as with promotion, evacuation has a very simple effect on (\sigma _w) and (\varepsilon _w).
In order to study evacuation of Kreweras words, we will employ another formulation of promotion and evacuation of linear extensions in terms of growth diagrams. This approach is discussed in [33, §5]; it is essentially due to Fomin (see, e.g., [32, Chapter 7: Appendix 1]).
Let P be a poset with (\ell ) elements. An order ideal of P is a subset (I\subseteq P) that is downwards-closed, i.e., for which (q \in I) and (p \le q \in P) implies that (p \in I). The set of order ideals of P is denoted ({\mathcal {J}}(P)). A linear extension ((p_1,p_2,\ldots ,p_{\ell }) \in {\mathcal {L}}(P)) corresponds to a chain (I_0 \subset I_1 \subset \cdots \subset I_{\ell } \in {\mathcal {J}}(P)) of order ideals of length (\ell ), where we set (I_{i} {:}{=}{p_1,p_2,\ldots ,p_i}) for (0 \le i \le \ell ). This sets up a (well-known) bijection between linear extensions of P and maximal chains of order ideals of P.
Definition 4.1
Let (L \in {\mathcal {L}}(P)) be a linear extension. The growth diagram of L is a labeling of the subset (D {:}{=}{(x,y) \in {\mathbb {Z}}^2:-y-\ell \le x \le -y}) of the two-dimensional grid ({\mathbb {Z}}^2) by order ideals (I_{(x,y)} \in {\mathcal {J}}(P)), ((x,y)\in D), subject to the following conditions:
(I_{(-\ell +k,-k)}=\varnothing ) and (I_{(k,-k)}=P) for all (k\in {\mathbb {Z}});
(I_{(-\ell ,0)} \subset I_{(-\ell +1,0)} \subset \cdots \subset I_{(0,0)}) is the chain corresponding to L;
for any four points ((x,y), (x,y+1), (x+1,y), (x+1,y+1) \in D), the labeling obeys the following local rule: if (I_{(x,y)} = I), (I_{(x,y+1)} = I\cup {p}), and (I_{(x+1,y+1)} = I\cup {p,q}), then
This rule is depicted in Fig. 7.
The local rule for growth diagrams of linear extensions
In the following proposition we summarize the basic results about growth diagrams of linear extensions. The essential idea is that all paths of length (\ell ) from a point of the form ((-\ell +k,-k)) to a point of the form ((j,-j)) correspond to linear extensions, and the local rule in Fig. 7 reflects the behavior of the involutions (\tau _i) from Sect. 2. See [33, §5] for the details and references.
Proposition 4.2
For any (L \in {\mathcal {L}}(P)),
The growth diagram of L is well-defined, i.e., there is a unique order ideal labeling (I_{(x,y)}\in {\mathcal {J}}(P)), ((x,y)\in D) satisfying the conditions in Definition 4.1;
For any (k \in {\mathbb {Z}}), the chain (I_{(-\ell +k,-k)} \subset I_{(-\ell +k+1,-k)} \subset \cdots \subset I_{(k,-k)}) corresponds to ({{\,\mathrm{{Pro}}\,}}^{k}(L));
The chain (I_{(0,-\ell )} \subset I_{(0,-\ell +1)} \subset \cdots \subset I_{(0,0)}) corresponds to ({{\,\mathrm{{Evac}}\,}}(L)).
Example 4.3
Let P be the following three-element poset:
Consider the linear extension (L=(p,q,r)\in {\mathcal {L}}(P)). Then, writing subsets as strings for shorthand, the portion of the growth diagram of L with y coordinate between 0 and (-3) is
From this diagram we can read off ({{\,\mathrm{{Pro}}\,}}^3(L)=L) and ({{\,\mathrm{{Evac}}\,}}(L) = (r,p,q)).
Proposition 4.2 implies that simple geometric operations on growth diagrams have combinatorial meaning:
Corollary 4.4
For any (L \in {\mathcal {L}}(P)),
For any (k \in {\mathbb {Z}}), the growth diagram for ({{\,\mathrm{{Pro}}\,}}^{k}(L)) is obtained from the growth diagram for L by translating by the vector ((-k,k));
The growth diagram for ({{\,\mathrm{{Evac}}\,}}(L)) is obtained from the growth diagram for L by reflecting across the line (y=x).
Proof
The first bulleted item is an immediate consequence of Proposition 4.2. The second is also an immediate consequence of Proposition 4.2, as soon as one observes that the local rule in Fig. 7 is symmetric under swapping x- and y-coordinates. (\square )
Via the geometric operations described in Corollary 4.4, the basic properties concerning promotion and evacuation summarized in Proposition 2.4 are easily obtained via this growth diagram approach.
Now let’s think about what growth diagrams look like for our poset of interest, V(n). There is an obvious bijection
which sends I to (a, b, c) where (a= \max {i \in {\mathbb {N}}:{\mathrm {A}_1,\mathrm {A}_2,\ldots ,\mathrm {A}_i}\subseteq I}) and similarly for b and c. We consider growth diagrams for linear extensions of V(n) labeled by elements of J(n) via this bijection. The local rule then becomes: if (I_{(x,y)} = (a,b,c)), (I_{(x,y+1)} = (a,b,c)+e_i), and (I_{(x+1,y+1)} = (a,b,c)+e_i+e_j), then
Here we have (i,j \in {\mathrm {A},\mathrm {B},\mathrm {C}}), and we use the conventions that (e_{\mathrm {A}} {:}{=}(1,0,0)), (e_{\mathrm {B}} {:}{=}(0,1,0)), and (e_{\mathrm {C}} {:}{=}(0,0,1)). This is depicted in Fig. 8.
The local rule for growth diagrams of Kreweras words
Let us further decorate the growth diagram of a linear extension of V(n) in the following way. We refer to ((x,y), (x,y+1), (x+1,y), (x+1,y+1)) as the square in position (x, y). If in a growth diagram of a linear extension of V(n) these four points constitute an “otherwise” case in Fig. 8, then we fill this square with (j \in {\mathrm {B},\mathrm {C}}), where (e_j) is as in that figure.
Via the bijection of Proposition 2.2, linear extensions of V(n) are the same as Kreweras words of length 3n. In this way, for such a Kreweras word w we obtain a labeling of ({(x,y) \in {\mathbb {Z}}^2:-y-3n \le x \le -y}) by (J(n) = {(a,b,c)\in {\mathbb {N}}^3:b,c\le a \le n}), which furthermore has some of its squares filled with (\mathrm {B})’s and (\mathrm {C})’s. We call this whole object the decorated growth diagram of w.
From now on in this section we will work with decorated growth diagrams of Kreweras words of length 3n.
Example 4.5
As in Example 3.11, let (w = \mathrm {A}\mathrm {A}\mathrm {B}\mathrm {B}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}). Then Fig. 9 depicts the portion of the decorated growth diagram for w with y coordinate between 0 and (-9). In this figure we use the string abc as shorthand for ((a,b,c) \in J(n)).
A decorated growth diagram for a Kreweras word
For (i \in {\mathbb {Z}}), let us refer to the set of squares in positions of the form ((x,-i)) as the ith row of a diagram. Similarly, for (j\in {\mathbb {Z}}), let us refer to the set of squares in positions of the form ((-3n-1+j,y)) as the jth column of a diagram. Example 4.5 may have suggested that in every row of the decorated growth diagram of a Kreweras word there is a unique filled square. This is true:
Proposition 4.6
Let w be a Kreweras word of length 3n and consider its decorated growth diagram. Recall the definition of (\iota (w)) from Sect. 1. Then for any (i \in {\mathbb {Z}}), the square in the ith row and ((\iota (w')+i-1))th column is filled with the letter (w'_{\iota (w')}), where (w' {:}{=}{{\,\mathrm{{Pro}}\,}}^{i-1}(w)). This is the unique filled square in the ith row.
Proof
By the translation symmetry of growth diagrams, Corollary 4.4 (a), it is enough to prove this for (i=1). As mentioned above, the local rule defining growth diagrams of linear extensions reflects the behavior of the involutions (\tau _i). In particular, a square in the 1st row and jth column corresponds to an application of (\tau _{j-1}) when carrying out the product (\tau _{\ell } \circ \cdots \circ \tau _2 \circ \tau _1) to perform promotion. We can view these (\tau _i) as acting directly on the Kreweras word w as in the proof of Proposition 2.2, and we will be in an “otherwise” for a square in Fig. 8 exactly when we are in an “otherwise” case for corresponding (\tau _i). As the proof of that proposition explains, the unique (\tau _i) for which an “otherwise” case occurs is (i=\iota (w)-1). (\square )
Thanks to the x/y symmetry of growth diagrams in Corollary 4.4 (b), Proposition 4.6 also implies that every column of a decorated growth diagram contains a unique filled square.
Now we will demonstrate how (\sigma _w) and (\varepsilon _w) from Sect. 3 can easily be read off from decorated growth diagrams.
Lemma 4.7
Let w be a Kreweras word of length 3n and consider its decorated growth diagram. Let (1 \le i \le 3n), and suppose the unique filled square in the ith row is in the jth column and is filled with (\varepsilon \in {\mathrm {B},\mathrm {C}}). Then we have (\varepsilon _w(i) = \varepsilon ) and (\sigma _w(i) = \langle j \rangle _{3n}), where (\langle k \rangle _{3n}) denotes the unique element of ({1,\ldots ,3n}) congruent to k modulo 3n.
Proof
By Proposition 4.6, we need to show that (\sigma _w(i) = \langle \iota ({{\,\mathrm{{Pro}}\,}}^{i-1}(w))+i-1 \rangle _{3n}) and (\varepsilon _w(i) = {{\,\mathrm{{Pro}}\,}}^{i-1}(w)_{\iota ({{\,\mathrm{{Pro}}\,}}^{i-1}(w))}) for all (1 \le i \le 3n).
First let us explain why this holds for (i=1). Note that ((1,\iota (w))) is the “near” arc emanating from 1 in the Kreweras bump diagram ({\mathcal {D}}_w). Moreover, the rules of the road are such that in the trip starting at 1 we will never make any turns on our way from 1 towards (\iota (w)). So indeed (\sigma _w(1) = \iota (w)), and (\varepsilon _w(1) = w_{\iota (w)}).
Then we have thanks to Lemma 3.10 that (\sigma _w(i) = \langle \sigma _{{{\,\mathrm{{Pro}}\,}}^{i-1}(w)}(1) +i-1\rangle _{3n}) and (\varepsilon _w(i) = \varepsilon _{{{\,\mathrm{{Pro}}\,}}^{i-1}(w)}(1)) for any (1\le i \le n). Thus, by the result in the previous paragraph applied to ({{\,\mathrm{{Pro}}\,}}^{i-1}(w)), we are done. (\square )
Example 4.8
As in Example 4.5, let (w = \mathrm {A}\mathrm {A}\mathrm {B}\mathrm {B}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}). We saw in Example 3.11 that (\sigma _w = [4,3,8,5,2,7,1,9,6]) and (\varepsilon _w=[\mathrm {B}, \mathrm {B}, \mathrm {C}, \mathrm {C}, \mathrm {B}, \mathrm {C}, \mathrm {B}, \mathrm {B}, \mathrm {C}]). In agreement with Lemma 4.7, we can also easily read off this (\sigma _w) and (\varepsilon _w) from w’s decorated growth diagram, which is depicted in Fig. 9.
We could have defined (\sigma _w) by setting (\sigma _w(i) {:}{=}\langle \iota ({{\,\mathrm{{Pro}}\,}}^{i-1}(w))+i-1 \rangle _{3n}) in light of Lemma 4.7. However, if we did so, it would not be at all clear that (\sigma _w) is a permutation. This is why we defined (\sigma _w) in terms of trips in Kreweras bump diagrams.
Let us record just a few more basic properties of decorated growth diagrams in the following proposition.
Proposition 4.9
Let w be a Kreweras word of length 3n and consider its decorated growth diagram. Then,
For any ((x,y) \in {\mathbb {Z}}^2), if the square in position (x, y) is filled with (\varepsilon ), then the square in position ((x+3n,y-3n)) is filled with (-\varepsilon );
For any (i \in {\mathbb {Z}}), if the unique filled square in the ith row is filled with (\varepsilon ), then the unique filled square in the ith column is filled with (-\varepsilon ).
Proof
For (a): this is an immediate consequence of the translation symmetry of growth diagrams in Corollary 4.4 (a), and our main result, Theorem 1.2.
For (b): by the translation symmetry of Corollary 4.4 (a) it is enough to prove this statement for a single row/column pair. Let us prove it for the (\sigma _w(1))th row. For the (\sigma _w(1))th row, this statement is a consequence of the interpretation of (\varepsilon _w) in Lemma 4.7 and Proposition 3.8 (c). (\square )
We now state our main results about evacuation of Kreweras words. For these we need the notion of reverse-complementation of permutations.
Definition 4.10
For a permutation (\sigma \in {\mathfrak {S}}_m), the reverse-complement of (\sigma ), denoted ({{\,\mathrm{{RevComp}}\,}}(\sigma )), is the conjugation of (\sigma ) by the longest element of the symmetric group (w_0 {:}{=}[m,m-1,\ldots ,1] \in {\mathfrak {S}}_{m}); i.e.,
Note that reverse-complementation commutes with inversion because (w_0) is an involution.
Lemma 4.11
Let w be a Kreweras word of length 3n. Then,
(\sigma _{{{\,\mathrm{{Evac}}\,}}(w)} = {{\,\mathrm{{RevComp}}\,}}(\sigma ^{-1}_w));
(\varepsilon _{{{\,\mathrm{{Evac}}\,}}(w)} = [-\varepsilon _w(3n), -\varepsilon _w(3n-1), \ldots , -\varepsilon _w(1)]).
Theorem 4.12
Let w be a Kreweras word of length 3n. Then
One nice property of the operations in Lemma 4.11 is that they are evidently involutive. It is also easy to see that they have the “right” interaction (in the sense of Proposition 2.4) with the operations in Lemma 3.10.
On the other hand, from the definition of (\sigma _w) in terms of trips in Kreweras bump diagrams it is far from clear why the word ((w_{\sigma _w(3n)}, w_{\sigma _w(3n-1)}, \ldots , w_{\sigma _w(1)})) appearing in Theorem 4.12 is a Kreweras word.
Before we prove these results, let us do an example.
Example 4.13
As in Example 4.8, let (w = \mathrm {A}\mathrm {A}\mathrm {B}\mathrm {B}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}). We saw above that (\sigma _w = [4,3,8,5,2,7,1,9,6]) and (\varepsilon _w=[\mathrm {B}, \mathrm {B}, \mathrm {C}, \mathrm {C}, \mathrm {B}, \mathrm {C}, \mathrm {B}, \mathrm {B}, \mathrm {C}]).
Thanks to Proposition 4.2, we can read off ({{\,\mathrm{{Evac}}\,}}w) from w’s growth diagram, which is depicted in Fig. 9: we have ({{\,\mathrm{{Evac}}\,}}w = \mathrm {A}\mathrm {B}\mathrm {A}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}\mathrm {B}). This agrees with Theorem 4.12.
The Kreweras bump diagram of ({{\,\mathrm{{Evac}}\,}}(w)) is
From the diagram ({\mathcal {D}}_{{{\,\mathrm{{Evac}}\,}}(w)}) one could compute that (\sigma _{{{\,\mathrm{{Evac}}\,}}(w)}=[2, 7, 4, 1, 6, 9, 8, 5, 3]) and (\varepsilon _{{{\,\mathrm{{Evac}}\,}}(w)}=[\mathrm {B}, \mathrm {C}, \mathrm {C}, \mathrm {B}, \mathrm {C}, \mathrm {B}, \mathrm {B}, \mathrm {C}, \mathrm {C}]), in agreement with Lemma 4.11.
By comparing this example with Example 3.11, we see that ({{\,\mathrm{{Pro}}\,}}(w)={{\,\mathrm{{Evac}}\,}}(w)) in this case, but that’s a coincidence for this particular Kreweras word w which does not always happen.
We proceed to prove Lemma 4.11 and Theorem 4.12.
Proof of Lemma 4.11
Thanks to the x/y symmetry of growth diagrams in Corollary 4.4 (b), Lemma 4.7 says that for (1 \le j \le 3n), if the unique filled square in the (3n+1-j)th column of the decorated growth diagram of w is in the (3n+1-i)th row, then (\sigma _{{{\,\mathrm{{Evac}}\,}}(w)}(j) {:}{=}\langle i \rangle _{3n}); and if this square is filled with (\varepsilon \in {\mathrm {B}, \mathrm {C}}) then (\varepsilon _{{{\,\mathrm{{Evac}}\,}}(w)}(j) {:}{=}\varepsilon ).
Then, the periodicity property of decorated growth diagrams in Proposition 4.9 (a) (along with the interpretation of (\sigma _w) in Lemma 4.7) gives (\sigma _{{{\,\mathrm{{Evac}}\,}}(w)} = {{\,\mathrm{{RevComp}}\,}}(\sigma ^{-1}_w)).
Meanwhile, Proposition 4.9 (b) (along with the interpretation of (\varepsilon _w) in Lemma 4.7) gives (\varepsilon _{{{\,\mathrm{{Evac}}\,}}(w)} = [-\varepsilon _w(3n), -\varepsilon _w(3n-1), \ldots , -\varepsilon _w(1)]). (\square )
Proof of Theorem 4.12
This is easy enough to see from the decorated growth diagram of w directly, but we can also deduce it from Lemma 4.11.
For any permutation (\sigma \in {\mathfrak {S}}^m), a straightforward unraveling of the definitions shows that
Hence Corollary 3.9 and Lemma 4.11 (a) imply that at least the positions of the (\mathrm {A})’s are the same in ({{\,\mathrm{{Evac}}\,}}(w)) and ((w_{\sigma _w(3n)}, w_{\sigma _w(3n-1)}, \ldots , w_{\sigma _w(1)})).
Now let (1\le i \le 3n) be such that ({{\,\mathrm{{Evac}}\,}}(w)_i \in {\mathrm {B},\mathrm {C}}). Then Corollary 3.9 and Lemma 4.11 imply that ({{\,\mathrm{{Evac}}\,}}(w)_i = -\varepsilon _w(\sigma _w(3n+1-i))). Since ({{\,\mathrm{{Evac}}\,}}(w)_i \ne \mathrm {A}), the previous paragraph tells us that (w_{\sigma _w(3n+1-i)}\in {\mathrm {B},\mathrm {C}}), and hence Proposition 3.8 (c) tells us that (-\varepsilon _w(\sigma _w(3n+1-i)) = w_{\sigma _w(3n+1-i)}). Thus ({{\,\mathrm{{Evac}}\,}}(w)_i=w_{\sigma _w(3n+1-i)}) in this case as well. (\square )
Of course, it is also reasonable to ask how dual evacuation acts on Kreweras words. But Proposition 2.4 says that ({{\,\mathrm{{Evac}}\,}}^{}(w)={{\,\mathrm{{Evac}}\,}}({{\,\mathrm{{Pro}}\,}}^{3n}(w))) for any Kreweras word w of length 3n, and thus our main result, Theorem 1.2, says that ({{\,\mathrm{{Evac}}\,}}^{}(w)) is obtained from ({{\,\mathrm{{Evac}}\,}}(w)) by swapping all (\mathrm {B})’s for (\mathrm {C})’s and vice-versa. Similarly, we can see that (\sigma _{{{\,\mathrm{{Evac}}\,}}^{}(w)} = {{\,\mathrm{{RevComp}}\,}}(\sigma ^{-1}_w)) and (\varepsilon _{{{\,\mathrm{{Evac}}\,}}^{}(w)} = [\varepsilon _w(3n), \varepsilon _w(3n-1), \ldots , \varepsilon _w(1)]) thanks to Lemmas 3.10 and 4.11.
5 Webs
In this section we reinterpret our results from the previous sections in the language of webs. We recall the notion of an (\mathfrak {sl}_3)-web, which is due to Kuperberg [17, 109–151 (1996)")]:
Definition 5.1
An (\mathfrak {sl}_3)-web ({\mathcal {W}}) is a planar graph, embedded in a disk, with boundary vertices labeled (1,2,\ldots ,m) arranged on the rim of the disk in counterclockwise order, and any number of (unlabeled) internal vertices such that
({\mathcal {W}}) is trivalent: all the boundary vertices have degree one, while all the internal vertices have degree three;
({\mathcal {W}}) is bipartite: the vertices (both boundary and internal) are colored white and black, with edges only between oppositely colored vertices.
We call the face of ({\mathcal {W}}) containing the boundary vertices the outer face, and all other faces internal. We say that ({\mathcal {W}}) is irreducible (or non-elliptic) if it has no internal faces with fewer than 6 sides.
Among all the (\mathfrak {sl}_3)-webs, the irreducible ones play a distinguished role. For instance, there are only finitely many irreducible webs with a fixed number of boundary vertices.
We will now explain how to convert a Kreweras bump diagram of a Kreweras word into a web by “breaking apart” its crossings.
Breaking apart the crossings in a Kreweras bump diagram to obtain a web. In a we show what happens at an internal crossing, and in b we show what happens at a boundary crossing
Construction 1
Let w be a Kreweras word and ({\mathcal {D}}_w) its associated Kreweras bump diagram. We obtain a planar graph ({\mathcal {W}}_w), embedded into a disk, together with a 3-coloring (c_w) of its edges as follows.
We replace each crossing of two arcs in ({\mathcal {D}}_w) with a pair of a vertices, one white and one black, joined by a wavy avocado (i.e. green) edge, as in Fig. 10. The white vertex in this pair is “to the left” of the black vertex, that is, closer to the openers of ({\mathcal {D}}_w). We color all vertices of degree one in the resulting graph, corresponding to the openers and closers of ({\mathcal {D}}_w), white, and keep the labels of these vertices. Finally, the color of the non-avocado edges of ({\mathcal {W}}_w) is inherited from ({\mathcal {D}}_w).
Example 5.2
As in Example 4.13, let (w = \mathrm {A}\mathrm {A}\mathrm {B}\mathrm {B}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}). Recall that the Kreweras bump diagram ({\mathcal {D}}_w) of w is:
Breaking apart the crossings of ({\mathcal {D}}_w) gives the following 3-edge-colored web:
Forgetting the 3-edge-coloring, and drawing the graph embedded in a disk, we obtain the web ({\mathcal {W}}_w):
Proposition 5.3
Let w be a Kreweras word and let ((W_w, c_w)) be the 3-edge-colored graph obtained by Construction 1.
Then (W_w) is an irreducible (\mathfrak {sl}_3)-web with 3n boundary vertices, all of which are white. Moreover, (W_w) has no internal face having a multiple of four sides.
The 3-coloring (c_w) of the edges of (W_w) is proper, i.e., each vertex is incident to at most one edge in each color class.
Finally, the construction is injective, that is, given (({\mathcal {W}}_w,c_w)) we can recover w: the boundary vertices incident to an avocado edge correspond to the (\mathrm {A})’s in w, those incident to a blue edge correspond to (\mathrm {B})’s, and those incident to a crimson edge correspond to (\mathrm {C})’s.
For the proof of Proposition 5.3: how faces of ({\mathcal {D}}_w) correspond to faces of ({\mathcal {W}}_w)
Proof
The only non-trivial claim is that the number of sides of any face cannot be a multiple of 4. As depicted in Fig. 11, internal faces of ({\mathcal {D}}_w) with k sides correspond to internal faces of ({\mathcal {W}}_w) with (2k-2) sides. Since ({\mathcal {D}}_w) only has crossings between arcs of different colors, the number of sides of any internal face of ({\mathcal {D}}_w) is even. (\square )
The web ({\mathcal {W}}_w) without its 3-edge-coloring is not quite enough to recover w. However, as we now explain, it gives information equivalent to the permutation (\sigma _w). In fact, we can associate a permutation to any (\mathfrak {sl}_3)-web by taking trips in the web, similar to what we did in Sect. 3 for Kreweras bump diagrams.
The rules of the road when taking a trip in a web
Definition 5.4
Let ({\mathcal {W}}) be an (\mathfrak {sl}_3)-web with m boundary vertices. The trip permutation of ({\mathcal {W}}), denoted (\mathrm {trip}_{{\mathcal {W}}} \in {\mathfrak {S}}_m), is obtained as follows. For (1\le i \le m) we take a trip in ({\mathcal {W}}) starting at i. To do this, we start by walking from boundary vertex i along the unique edge incident to it. When we come to any internal vertex in ({\mathcal {W}}), we continue our trip by following the rules of the road:
If the vertex is black, we turn right, i.e., we walk out along the next edge counterclockwise from where we came in;
If the vertex is white, we turn left, i.e., we walk out along the next edge clockwise from where we came in.
These rules of the road are depicted in Fig. 12. We stop our trip when we reach a boundary vertex. If j is the boundary vertex we reach from the trip starting at i, then we set (\mathrm {trip}_{{\mathcal {W}}}(i) {:}{=}j).
That (\mathrm {trip}_{{\mathcal {W}}}) is genuinely a permutation again follows from the fact that the rules of the road around any vertex locally permute the entry and exit points.
Our reason for considering trip permutations is the follow proposition:
Proposition 5.5
Let w be a Kreweras word. Then (\sigma _w= \mathrm {trip}_{{\mathcal {W}}}).
Proof
This is simply a matter of checking that locally at a crossing of arcs, the rules of the road for trips in the Kreweras bump diagram ({\mathcal {D}}_w) agree with the rules of the road for the trips in the web ({\mathcal {W}}_w). And to do that, we just need to look at Figs. 5, 10 and 12.
(\square )
The notion of trip permutations is due to Postnikov [24")], and comes from his theory of plabic graphs. A plabic (“planar bicolored”) graph is a planar graph, embedded in a disk, whose internal vertices are colored black or white, and whose boundary vertices have degree one. There are some differences between plabic graphs and \(\mathfrak {sl}\_3\)-webs:
The boundary vertices of a plabic graph are not colored;
The internal vertices of a plabic graph need not be trivalent;
The coloring of internal vertices of a plabic graph does not have to be proper, i.e., vertices of the same color may be adjacent.
Except for the small technicality about boundary vertices being colored, an (\mathfrak {sl}_3)-web is a special case of a plabic graph. Postnikov [24"), §13] defined trip permutations for plabic graphs in exactly the same way as we have done for webs in Definition 5.4 above: turn right at black vertices and left at white vertices.Footnote 1
If ({\mathcal {W}}) and ({\mathcal {W}}') are two (\mathfrak {sl}_3)-webs with m boundary vertices, and they differ only in the way their boundary vertices are colored, then (\mathrm {trip}_{{\mathcal {W}}}=\mathrm {trip}_{{\mathcal {W}}'}), since the color of boundary vertices does not enter into the definition of trip permutations in any way. However, note that the color of any boundary vertex which is adjacent to an internal vertex has its color determined by the bipartiteness condition. Hence, if ({\mathcal {W}}) and ({\mathcal {W}}') differ only in the way their boundary vertices are colored, then ({\mathcal {W}}') is obtained from ({\mathcal {W}}) by swapping the colors of pairs of oppositely colored, adjacent boundary vertices. In particular, if ({\mathcal {W}}) has all its boundary vertices the same color, then there is no web that differs from ({\mathcal {W}}') only in the way its boundary vertices are colored.
We now explain how Postnikov’s work implies that for irreducible webs, the situation discussed in the previous paragraph is the only way that trip permutations can coincide.
Lemma 5.6
Let ({\mathcal {W}}) and ({\mathcal {W}}') be irreducible (\mathfrak {sl}_3)-webs with m boundary vertices. Suppose that (\mathrm {trip}_{{\mathcal {W}}} = \mathrm {trip}_{{\mathcal {W}}'}). Then ({\mathcal {W}}) and ({\mathcal {W}}') differ at most in the way their boundary vertices are colored. In particular, if all the boundary vertices of ({\mathcal {W}}) are the same color, then ({\mathcal {W}}={\mathcal {W}}').
Proof
Postnikov [24"), §12] defined certain transformations of plabic graphs he called moves and reductions. If \({\mathcal {W}}\) is an irreducible \(\mathfrak {sl}\_3\)-web (viewed as a plabic graph), then the only moves or reductions we can apply to it are “trivial” moves which add 2-valent vertices by subdividing an edge, or remove such 2-valent vertices by un-subdividing edges. (Crucially, the fact that all internal faces have at least 6 sides means we will never be able to carry out a square move, which is the fundamental, nontrivial move in the theory.) In particular, we will never be able to apply a reduction to \({\mathcal {W}}\), so \({\mathcal {W}}\) is reduced. Then [24"), Theorem 13.2(4)] says that \(\mathrm {trip}\_{{\mathcal {W}}}\) has no fixed points, so we don’t have to worry about the issue of decorated fixed points. Finally, a key result [24"), Theorem 13.4] from Postnikov’s paper says that two reduced plabic graphs have the same trip permutation if and only if they are related via a series of moves. Since, as mentioned, the only moves we can apply either add or remove 2-valent vertices, we will not be able to reach any other web than \({\mathcal {W}}\) via these moves. Hence, Postnikov’s result tells us that any other web with the same trip permutation as \({\mathcal {W}}\) is equal to \({\mathcal {W}}\)—except in the way the boundary vertices are colored, which the plabic graph story does not see. \(\square \)
Lemma 5.6 lets us apply our knowledge about how ({{\,\mathrm{{Pro}}\,}}) and ({{\,\mathrm{{Evac}}\,}}) affect (\sigma _w) to understand how they affect ({\mathcal {W}}_w) (Theorem 1.3 from Sect. 1). We just need to define the corresponding web operations.
Definition 5.7
Let ({\mathcal {W}}) be an (\mathfrak {sl}_3)-web with m boundary vertices. The rotation of ({\mathcal {W}}), denote ({{\,\mathrm{{Rot}}\,}}({\mathcal {W}})), is obtained from ({\mathcal {W}}) be relabeling its vertices according to the inverse long cycle ((m,m-1,\ldots ,2,1) \in {\mathfrak {S}}_m). The flip of ({\mathcal {W}}), denoted ({{\,\mathrm{{Flip}}\,}}({\mathcal {W}})), is obtained from ({\mathcal {W}}) by drawing a chord in the disk separating 1 and m, reflecting ({\mathcal {W}}) across this chord, and then relabeling its vertices according to the longest element ([m,m-1,\ldots ,1] \in {\mathfrak {S}}_m).
Theorem 5.8
Let w be a Kreweras word. Then,
({\mathcal {W}}_{{{\,\mathrm{{Pro}}\,}}(w)} = {{\,\mathrm{{Rot}}\,}}({\mathcal {W}}_{w}));
({\mathcal {W}}_{{{\,\mathrm{{Evac}}\,}}(w)} = {{\,\mathrm{{Flip}}\,}}({\mathcal {W}}_{w})).
Proof
We have (\mathrm {trip}_{{\mathcal {W}}_w} = \sigma _w) because of Proposition 5.5. Thus Lemmas 3.10 and 4.11 imply (\mathrm {trip}_{{\mathcal {W}}_{{{\,\mathrm{{Pro}}\,}}(w)}}={{\,\mathrm{{Rot}}\,}}(\mathrm {trip}_{{\mathcal {W}}_w})) and (\mathrm {trip}_{{\mathcal {W}}_{{{\,\mathrm{{Evac}}\,}}(w)}}={{\,\mathrm{{RevComp}}\,}}(\mathrm {trip}_{{\mathcal {W}}_w}^{-1})). For any (\mathfrak {sl}_3)-web ({\mathcal {W}}), it is straightforward to verify that (\mathrm {trip}_{{{\,\mathrm{{Rot}}\,}}({\mathcal {W}})}={{\,\mathrm{{Rot}}\,}}(\mathrm {trip}_{{\mathcal {W}}})) and (\mathrm {trip}_{{{\,\mathrm{{Flip}}\,}}({\mathcal {W}})}={{\,\mathrm{{RevComp}}\,}}(\mathrm {trip}_{{\mathcal {W}}}^{-1})). But then thanks to Lemma 5.6, we know that ({{\,\mathrm{{Rot}}\,}}({\mathcal {W}}_w)) and ({{\,\mathrm{{Flip}}\,}}({\mathcal {W}}_w)) are the only irreducible (\mathfrak {sl}_3)-webs with trip permutations equal to (\mathrm {trip}_{{\mathcal {W}}_{{{\,\mathrm{{Pro}}\,}}(w)}}) and (\mathrm {trip}_{{\mathcal {W}}_{{{\,\mathrm{{Evac}}\,}}(w)}}). Therefore, we must have ({\mathcal {W}}_{{{\,\mathrm{{Pro}}\,}}(w)}={{\,\mathrm{{Rot}}\,}}({\mathcal {W}}_w)) and ({\mathcal {W}}_{{{\,\mathrm{{Evac}}\,}}(w)}={{\,\mathrm{{Flip}}\,}}({\mathcal {W}}_w)), as claimed. (\square )
Example 5.9
As in Example 5.2, let (w = \mathrm {A}\mathrm {A}\mathrm {B}\mathrm {B}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}). We saw in Example 4.13 that ({{\,\mathrm{{Pro}}\,}}(w) = {{\,\mathrm{{Evac}}\,}}(w) = w') where (w' = \mathrm {A}\mathrm {B}\mathrm {A}\mathrm {C}\mathrm {A}\mathrm {C}\mathrm {C}\mathrm {B}\mathrm {B}). Recall that the Kreweras bump diagram ({\mathcal {D}}_{w'}) of (w') is:
Breaking apart the crossings of ({\mathcal {D}}_{w'}) gives the following 3-edge-colored web:
Forgetting the 3-edge-coloring, and drawing the graph embedded in a disk, we see that the web ({\mathcal {W}}_w) is:
Comparing with Example 5.2, we can see that ({\mathcal {W}}_{w'} = {{\,\mathrm{{Rot}}\,}}({\mathcal {W}}_w) = {{\,\mathrm{{Flip}}\,}}({\mathcal {W}}_w)), in agreement with Theorem 5.8.
It is also possible to describe how promotion and evacuation affect the 3-edge-coloring (c_w) (briefly: we “swap” colors of edges along trips), but we will not go into details about that here.
However, a question we will answer in the following subsection is: which webs ({\mathcal {W}}) are equal to ({\mathcal {W}}_w) for some Kreweras word w? As we will see, the restriction coming from Proposition 5.3 is the only restriction.
5.1 Kreweras webs
Definition 5.10
A Kreweras web is an irreducible (\mathfrak {sl}_3)-web such that all boundary vertices are white and there are no internal faces with a multiple of 4 sides.
We note that a simple counting argument shows that any (\mathfrak {sl}_3)-web with all white boundary vertices has a multiple of 3 boundary vertices.
We have already seen from Proposition 5.3 that any web ({\mathcal {W}}_w) corresponding to a Kreweras word w must be a Kreweras web. Our goal in this subsection is to show that all Kreweras webs arise this way.
Theorem 5.11
Let ({\mathcal {W}}) be an (\mathfrak {sl}_3)-web. Then there is a Kreweras word w for which ({\mathcal {W}}={\mathcal {W}}_w) if and only ({\mathcal {W}}) is a Kreweras web. Moreover, if ({\mathcal {W}}) is a Kreweras web, then the number of Kreweras words w for which ({\mathcal {W}}={\mathcal {W}}_w) is (2^{\kappa ({\mathcal {W}})}), where (\kappa ({\mathcal {W}})) is the number of connected components of ({\mathcal {W}}).
Let us first note an enumerative consequence (which is stated as part of Theorem 1.4 in Sect. 1):
Corollary 5.12
We have
where the sum is over all Kreweras webs ({\mathcal {W}}) with 3n boundary vertices.
Proof
This follows from combining Theorem 5.11 with Kreweras’s product formula enumerating Kreweras words . (\square )
For more discussion of enumeration of webs (including an explanation of the rest of Theorem 1.4), see Sect. 6.2.
We prove Theorem 5.11 by demonstrating that we can appropriately edge-color any Kreweras web ({\mathcal {W}}). This is achieved via the following construction:
Construction 2
Let ({\mathcal {W}}) be Kreweras web with boundary vertices labeled counterclockwise 1 to 3n, and let (c_1,\dots ,c_{\kappa ({\mathcal {W}})}) be a choice of color, either blue or crimson, for each connected component of ({\mathcal {W}}). We create a proper 3-edge-coloring of ({\mathcal {W}}) (with colors avocado, blue, and crimson), and a system of 2n colored directed paths (1_L, 1_R, \dots , n_L, n_R) in ({\mathcal {W}}), with the following properties:
Paths (i_L) and (i_R) begin at the same boundary vertex, and (i_L) turns left when leaving the unique edge e incident to this vertex, while (i_R) turns right when leaving e;
The first and every other edge of a path is colored avocado, and all the other edges of the path have the same color (either blue or crimson) – which we call the color of the path;
Every avocado edge is traversed by precisely two paths, and every other edge is traversed by precisely one path;
Any two paths share at most one (necessarily avocado) edge, and if they do, they are of different color;
If two paths (i_X) and (j_Y) with (i < j) share an (avocado) edge e, then the path (i_X) turns to the left when visiting e, and continues the the right when leaving it.
The system of paths is created inductively. Once paths (1_L, 1_R,\dots , i{-}1_L, i{-}1_R) are determined, the paths (i_L) and (i_R) start at the boundary vertex with smallest label incident to an uncolored edge. If (i_L) is in a connected component of ({\mathcal {W}}) different from the connected components containing (1_L,\dots ,i{-}1_L), the color of (i_L) is (c_k), where k is the number of connected components containing (1_L,\dots ,i_L). The 3-edge-coloring is then inherited from the colors of the paths.
We say that Construction 2succeeds on ({\mathcal {W}}) if the requested properties can be satisfied when creating the paths.
Lemma 5.13
Construction 2 succeeds on ({\mathcal {W}}) if and only if ({\mathcal {W}}={\mathcal {W}}_w) for some Kreweras word w. And in this case, the (2^{\kappa ({\mathcal {W}})}) 3-edge-colorings produced by applying Construction 2 to ({\mathcal {W}}) with different choices of (c_1,\ldots ,c_{\kappa ({\mathcal {W}})}) are exactly all the (c_w) for such Kreweras words w.
Proof
Let w be a Kreweras word. Proposition 5.3 says that ({\mathcal {W}}_w) must be a Kreweras web. Moreover, it is easy to see that the arcs of ({\mathcal {D}}_w) determine a system of colored paths in ({\mathcal {W}}_w) satisfying the properties required in Construction 2.
Conversely, let ({\mathcal {W}}) be a Kreweras web on which Construction 2 succeeds. Then, by the properties of the construction, the paths of the same color form two noncrossing perfect matchings, with the same set of openers. Thus, they yield a Kreweras bump diagram of a Kreweras word. (\square )
Corollary 5.14
The set of Kreweras webs ({\mathcal {W}}) on which Construction 2 succeeds is closed under ({{\,\mathrm{{Rot}}\,}}) and ({{\,\mathrm{{Flip}}\,}}).
Lemma 5.15
Suppose that the boundary vertices 1 and 2 are in the same connected component of a Kreweras web ({\mathcal {W}}), and suppose that the shortest path from 1 to 2 (i.e., the one that turns right at every vertex) consists of (4k+2) edges, for (k\ge 1). Then, if it succeeds, the coloring produced by Construction 2 colors the edges incident to 1 and 2 avocado. Moreover, the path (1_R) and the path (2_L) have the same color.
Proof
The edge incident to 2 will be colored a non-avocado color if and only if the distance between vertices 1 and 2 is two: if not, the path (1_R) turns left after the second edge and therefore does not visit vertex 2. Thus, after the paths (1_L) and (1_R) are created, the edge incident to vertex 2 is uncolored, and is therefore chosen as the initial edge of paths (2_L) and (2_R).
So now let us focus on the claim about the colors of (1_R) and (2_L). Let x be the first white non-boundary vertex on the colored path (1_R). It suffices to show that every other edge of the shortest path from 2 to x is colored avocado, and the colors of the remaining edges alternate.
The situation in Lemma 5.15
Suppose that a non-avocado edge on this path belongs to the colored path (i_X) and the two following edges, (e_\ell ) to the left and (e_r) to the right, belong to the colored path (j_Y). Then, since colored paths share at most one (avocado) edge, we have (2< i < j). This situation is depicted in Fig. 13.
It follows that the colored path (i_X) continues on edge (e_\ell ), which is therefore colored avocado. Furthermore, the left edge after (e_\ell ) belongs to path (j_Y), whose color is therefore different from the color of (i_X). (\square )
Lemma 5.16
A planar cubic bipartite simple graph has at least six 4-cycles.
Proof
Let G be a planar cubic bipartite simple graph. Without loss of generality, we can assume that G is connected.
By the handshaking lemma, a cubic graph has an even number of vertices, say 2n, and 3n edges. For (k\ge 2), let (f_{2k}) be the number of faces bounded by 2k edges of G. By Euler’s formula, the total number of faces of G equals (3n-2n+2 = n+2).
Since G is cubic, every vertex is contained in three faces. Thus
which implies that (f_4 \ge 6). (\square )
Lemma 5.17
Let ({\mathcal {W}}) be a Kreweras web with at least one internal face. Then there is an internal face of ({\mathcal {W}}) which has at least three consecutive sides on its boundary with the outer face.
Proof
Let G be the graph obtained from ({\mathcal {W}}) by removing all vertices not contained in any internal face. Let (v_1,\ldots ,v_k) be the list of vertices of degree 2 in G. The lemma’s claim is equivalent to the assertion that there are two vertices among (v_1,\ldots ,v_k) which are adjacent, which we now show.
Let (G') be a copy of G, and let (v'_1, \ldots , v'_k) be the vertices of (G') corresponding to (v_1,\ldots ,v_k). We construct a planar bipartite cubic graph H by adding edges ({v_i, v'_i}) to (G\cup G') for (i\in {1,\dots ,k}), as depicted in Fig. 14.
For the proof of Lemma 5.17: how to obtain a bipartite cubic graph from a web
Suppose that G has no pair of adjacent vertices of degree 2. Then H contains no 4-cycles, which is impossible by Lemma 5.16. (\square )
Lemma 5.18
Construction 2 succeeds on any Kreweras web ({\mathcal {W}}).
Proof
We use induction on the number of internal faces and the number of vertices of ({\mathcal {W}}). We also freely relabel the boundary vertices via Corollary 5.14.
If ({\mathcal {W}}) has a white vertex which is not contained in an internal face, we replace this vertex by three independent boundary vertices and obtain three graphs ({\mathcal {W}}_1), ({\mathcal {W}}_2), and ({\mathcal {W}}_3), ordered counterclockwise. We label the boundary vertices of ({\mathcal {W}}_1) and ({\mathcal {W}}_2) such that the split vertex is the last boundary vertex and those of ({\mathcal {W}}_3) such that the split vertex has label 1.
By induction, Construction 2 succeeds on all three graphs. Moreover, we can color ({\mathcal {W}}_1) and ({\mathcal {W}}_2) such that the colors of the edges incident to their last boundary vertices are distinct. Finally, we choose the coloring of ({\mathcal {W}}_3) so that the edge to the left of the first boundary edge has the same color as the edge incident to the last boundary vertex of ({\mathcal {W}}_2), and, accordingly, the edge to the right of the first boundary edge has the same color as the edge incident to the last boundary vertex of ({\mathcal {W}}_1). It is now clear that we obtain a coloring which coincides with the coloring produced by Construction 2.
For the proof of Lemma 5.18: how to break apart and reattach faces for the coloring in Construction2
Therefore, we can assume that all white vertices of ({\mathcal {W}}) are contained in internal faces. By Lemma 5.17, there is an edge separating an internal face from the outer face, such that one of its vertices is black and adjacent to a boundary vertex, and the other vertex is white and adjacent to a black internal vertex which in turn is adjacent to two boundary vertices. Via Corollary 5.14, we may assume that these latter two boundary vertices are labeled 1 and 2, and the former boundary vertex is labeled 3.
We construct ({\mathcal {W}}') by removing the edge incident to the white internal vertex and the attached boundary vertices, and then splitting this white vertex into two independent boundary vertices. By construction, the number of internal faces of ({\mathcal {W}}') is one less than the number of faces of ({\mathcal {W}}). We label its boundary vertices so that the two split vertices have labels 1 and 2. By induction, Construction 2 succeeds on ({\mathcal {W}}').
Using the properties of this coloring guaranteed by Lemma 5.15, we can obtain a coloring of ({\mathcal {W}}) which coincides with the coloring produced by Construction 2, as depicted in Fig. 15. (\square )
Lemmas 5.13 and 5.18 implies Theorem 5.11.
6 Future directions
In this section we discuss some potential connections and possible threads of future research.
6.1 Relation of our work to previous work on webs and promotion
Webs were introduced by Kuperberg [17, 109–151 (1996)")] to study invariant tensors of representations of simple Lie algebras (and their relatives like quantum groups). In [14$$
sl
(
3
)
are not dual canonical. Pac. J. Math. 188(1), 129–153 (1999)")], Khovanov and Kuperberg described a bijection between the set of linear extensions of \(\times [n]\) (i.e., standard Young tableaux (SYTs) of \(3 \times n\) rectangular shape) and irreducible \(\mathfrak {sl}\_3\)-webs with 3n white boundary vertices. In [22, 19–41 (2009)")] Petersen, Pylyavskyy, and Rhoades showed that, under the bijection of Khovanov–Kuperberg, promotion of linear extensions of \(\times [n]\) corresponds to rotation of webs. This should be seen as directly analogous to the fact, mentioned in Sect. 1, that promotion of linear extensions of \( \times [n]\) (i.e., promotion of Dyck words) corresponds to rotation of noncrossing matchings: indeed, noncrossing matchings can be seen as “\(\mathfrak {sl}\_2\)-webs.” Later, Tymoczko [37, 611–632 (2012)")] gave a different, simpler description of the Khovanov–Kuperberg bijection and used this description to reprove the results of Petersen–Pylyavskyy–Rhoades as well. Building on Tymoczko’s work, Russell [27, 851–862 (2013)")] (see also Patrias [21")]) related rotation of irreducible \(\mathfrak {sl}\_3\)-webs with arbitrarily colored boundary vertices to promotion of 3-rowed semistandard (as opposed to standard) tableaux.
At its core, the proof of our main results boils down to showing that for a Kreweras word w, the web ({\mathcal {W}}_{{{\,\mathrm{{Pro}}\,}}(w)}) is the rotation of the web ({\mathcal {W}}_w). Hence, our work would seem to be closely related to the aforementioned work relating webs and promotion. And indeed, our procedure of obtaining a web from a Kreweras bump diagram by breaking apart its crossings is very similar to Tymoczko’s procedure of converting a so-called “m-diagram” into a web. However, the exact relation between our work and prior work is not clear to us. Let us emphasize some points of contrast.
In the same way that linear extensions of (\times [n]) naturally correspond to Dyck words, linear extensions of (\times [n]) naturally correspond to words of length 3n in the letters (\mathrm {A}), (\mathrm {B}), and (\mathrm {C}), with equally many (\mathrm {A})’s, (\mathrm {B})’s, and (\mathrm {C})’s, for which every prefix has at least as many (\mathrm {A})’s as (\mathrm {B})’s and at least as many (\mathrm {B})’s as (\mathrm {C})’s (this representation is usually called the lattice word or Yamanouchi word of the tableau). In this way, the linear extensions of (\times [n]) can be viewed as a subset of the Kreweras words of length 3n. However, promotion of a linear extension of (\times [n]) is not the same as promotion of its corresponding Kreweras word. Furthermore, the web obtained from a linear extension of (\times [n]) via the Khovanov–Kuperberg/Tymoczko bijection is not the same as the web ({\mathcal {W}}_w) for its corresponding Kreweras word w. Indeed, as we have already seen with Theorem 5.11, only a subset of irreducible (\mathfrak {sl}_3)-webs with 3n white boundary vertices arise as ({\mathcal {W}}_w) for some Kreweras word w of length 3n. And, on the other hand, unlike the situation with SYTs, we also need extra decoration (the 3-edge-coloring) to recover w from ({\mathcal {W}}_w).
Another way that our work differs from the work mentioned above is that for us, the trip permutation (\sigma _w) associated to the web ({\mathcal {W}}_w) plays a central role, in contrast to previous work on webs and promotion. In fact, it seems that viewing a web as a plabic graph in order to extract a trip permutation is a new idea, although Lam and Fraser, Lam and Le discuss some relationships between webs and plabic graphs. We also note that one could adapt the argument in Lemma 4.7 to show that for a linear extension of (\times [n]), the trip permutation of its Khovanov–Kuperberg/Tymoczko web can similarly be read off from its growth diagram.
At any rate, it would certainly be interesting to understand more precisely the connection between our work and the previous work on webs and promotion, if there is some precise connection.
6.2 More enumeration, and connection with planar maps
As we just saw in Sect. 6.1, the total number of irreducible (\mathfrak {sl}_3)-webs with 3n white boundary vertices is the same as the number of standard Young tableaux of (3 \times n) rectangular shape, for which there is a well-known product formula (2 \, \frac{(3n)!}{n!(n+1)!(n+2)!}) (sometimes these numbers are called “three-dimensional Catalan numbers”). Corollary 5.12 gives a product formula for a weighted enumeration of Kreweras webs. Let us now explain how one can enumerate Kreweras webs, without this weighting. As we will see, certain famous sequences of numbers counting planar maps naturally arise.
We will employ a small amount of generatingfunctionology for this task. When dealing with combinatorial generating functions it is often useful to reduce to “connected” objects. We say a (non-empty) Kreweras word w is connected if it contains no proper consecutive substring which is a Kreweras word. Evidently,
Let us form the generating functions
where
As we have seen in Sect. 1, (K_n =\frac{4^n}{(n+1)(2n+1)}\left( {\begin{array}{c}3n\ n\end{array}}\right) ) ( Note that every Kreweras word is obtained, in a unique way, from a connected Kreweras word w by inserting an arbitrary (possibly empty) Kreweras word after each letter of w. This yields the generating function equation
From the above equation, and the formula for (K_n), it is possible to use Lagrange inversion to deduce that (K^c_n = 2^{n+1} \, \frac{(4n-3)!}{(3n-1)!n!}).
Next, we do the same but with Kreweras webs instead of Kreweras words. That is, we form the generating functions
where
It follows from Theorem 5.11 that (W^c_n = \frac{1}{2}K^c_n). Moreover, the same reasoning as in the case of Kreweras words implies the generating function equation
From the above equation, and the formula for (W^c_n), it is possible to use Lagrange inversion to obtain the coefficients (W_n), although the answer one obtains is not as nice as for (K_n).
Finally, let us explain the connection with planar maps. Recall that a planar map is a topological equivalence class of embeddings of a connected planar graph in the sphere. The number of rooted, bridgeless, cubic planar maps with 2n vertices is
Bernardi defined a bijection from Kreweras words of length 3n to rooted, bridgeless, cubic planar maps with 2n vertices decorated with a depth tree, which is a certain kind of spanning tree. He also explained why every such map has exactly (2^{n}) depth trees, and thus combinatorially explained the above equality.
Meanwhile, the number of rooted, 3-connected, cubic planar maps with 2n vertices is
We believe that under Bernardi’s bijection, a Kreweras word is connected if and only if its corresponding cubic planar map is 3-connected. Thus Bernardi’s bijection also explains, combinatorially, the above equality. However, it would be desirable to directly explain why the enumeration of connected Kreweras webs is related to the enumeration of rooted, 3-connected, cubic planar maps, without going through Kreweras words. There is some reason to hope this is possible because webs seem at least superficially similar to cubic planar maps.
6.3 An algebraic model
Is there an algebraic model which explains the behavior of promotion on Kreweras words?
The Henriques–Kamnitzer cactus group action on the tensor product of crystals of representations of a (simple, finite-dimensional) Lie algebra gives rise to a notion of promotion acting on the highest weight words of weight zero for such a tensor product: see [6, 23, 38]. For example, letting V be the vector representation of (\mathfrak {sl}_k), this cactus group promotion action on the weight zero highest weight words of (\otimes ^{kn} V) corresponds to promotion of linear extensions of ([k] \times [n]) (i.e., promotion of SYTs of (k \times n) rectangular shape). Moreover, there are general results (see the references above) which imply that cactus group promotion of weight zero highest weight words always has good behavior.
As mentioned in the previous subsection, Kuperberg first defined webs in order to study invariant tensors in tensor products of representations of Lie algebras. So it is not so unreasonable to think that promotion of Kreweras words could be connected to invariant tensors and the Henriques–Kamnitzer cactus group action in some way. Perhaps the proper algebraic model for Kreweras words will come from representations of some variant of a simple, finite-dimensional Lie algebra, like a Lie superalgebra or a Kac–Moody algebra. A better understanding of the algebraic significance of the “no 4k-sided internal faces” condition for the Kreweras webs could be the key to uncovering the proper algebraic model for Kreweras words.
We note that the cactus group promotion of highest weight words can be described via local rules: see [23, §4.2]. As we saw in Sect. 4, promotion of Kreweras words can also be described via local rules. However, local rules alone are not enough to guarantee good behavior of promotion: again, as we saw in Sect. 4, promotion of the linear extensions of any poset can be described by local rules, but most posets have bad behavior of promotion.
6.4 Cyclic sieving
Diagrammatic and/or algebraic models are often useful for establishing cyclic sieving results. Let us recall this notion from Reiner–Stanton–White :
Definition 6.1
Let X be a finite set. Let (C=\langle c \rangle ) be a cyclic group of order (\ell ) acting on X, generated by element (c\in C). Let (f(q) \in {\mathbb {N}}[q]) be a polynomial in q with nonnegative integer coefficients. Then we say the triple ((X,\Phi ,f(q))) exhibits cyclic sieving if for all k we have
where (\omega {:}{=}e^{2\pi i/\ell }) is a primitive (\ell )th root of unity.
Cyclic sieving phenomena (CSPs) involving polynomials which have an expression as a ratio of products of q-numbers are especially valued, because they imply that every symmetry class has a product formula.
In the case of promotion of SYTs of (k\times n) rectangular shape, Rhoades obtained such a CSP. He showed (({\text {SYTs of shape }k\times n},\langle {{\,\mathrm{{Pro}}\,}}\rangle ,f(q))) exhibits cyclic sieving, where f(q) is the major index generating function for the SYTs of shape (k \times n), which has the product formula
assuming by symmetry that (k\le n). Note that this product formula is the well-known q-hook length formula [32, Cor. 7.21.5].
We conjecture the following CSP for promotion of Kreweras words:
Conjecture 6.2
For all (n\ge 1), the rational expression
is a polynomial in q with nonnegative integer coefficients, and the triple
exhibits cyclic sieving.
Conjecture 6.2 strongly suggests that some good algebraic model for Kreweras word promotion should exist, although we do not know of the precise algebraic or combinatorial significance of the polynomial f(q) appearing in the conjecture.
We also conjecture similarly that there is a product formula for the number of Kreweras words fixed by ({{\,\mathrm{{Evac}}\,}}) and ({{\,\mathrm{{Evac}}\,}}^{}):
Conjecture 6.3
For all (n\ge 1), the number of Kreweras words of length 3n with ({{\,\mathrm{{Evac}}\,}}^{}(w)=w) is
The number of Kreweras words of length 3n with ({{\,\mathrm{{Evac}}\,}}(w)=w) is this same number if n is even, and is 0 if n is odd.
It is possible that Conjecture 6.3 could be phrased as a “(q=-1)” result for a polynomial which has a product formula as a rational expression, although we do not have a candidate for such a polynomial. Note that every poset has a “(q=-1)” result for counting self-evacuating linear extensions, where the polynomial is essentially the major index generating function: see [33, §3].
6.5 Order polynomial product formulas
It is reasonable to ask “where is this Kreweras word promotion really coming from?,” or in other words, “what is it about the poset V(n) that would lead one to suspect that it has good promotion behavior?” Let us attempt to answer this question.
Let P be a poset. A P-partition of height m is a weakly order-preserving map (\pi :P \rightarrow {0,1,\ldots ,m}). It is well-known that the number of P-partitions of height m is given by a polynomial (\Omega _P(m)) in m, called the order polynomial of P, whose degree is (#P) and whose leading coefficient is (1/#P!) times the number of linear extensions of P (see, e.g., [34, §3.12]).
In , Kreweras–Niederhausen obtained the following product formula for the entire order polynomial of the poset V(n):
They deduced the product formula counting Kreweras words (i.e., linear extensions of V(n)) as a corollary.
In , the first author (S.H.) presented the following heuristic: “posets with good dynamical behavior = posets with order polynomial product formulas.” Here “good dynamical behavior” includes good behavior of promotion of linear extensions. It is via this heuristic that promotion for Kreweras words was discovered: S.H. asked a question on MathOverflow about posets with order polynomial product formulas, and was lead to the paper [15, 55–60 (1981)")] and the V(n) poset by an answer of Ira Gessel [8")].
6.6 Rowmotion
Rowmotion is a certain invertible action on the order ideals of any poset P which has been studied by many authors over a number of years, with a renewed interest especially in the last 10 or so years. Rowmotion and promotion are “similar” in many respects. For an overview and history of rowmotion see, e.g., or . Einstein and Propp introduced a piecewise-linear extension of rowmotion, which in particular yields a (piecewise-linear) action of rowmotion on the set of P-partitions of height m.
In the “posets with good dynamical behavior = posets with order polynomial product formulas” heuristic just mentioned, “good dynamical behavior” also includes good behavior of rowmotion of order ideals, and more generally good behavior of (piecewise-linear) rowmotion of P-partitions.
In agreement with this heuristic, it (experimentally) appears that the poset V(n) has good behavior of rowmotion of order ideals and P-partitions. Conjectures concerning rowmotion for V(n) appeared in the aforementioned paper .
Notes
Technically Postnikov considered decorated permutations, which have their fixed points colored either black or white. None of the trip permutations we obtain from irreducible webs will have fixed points (see the proof of Lemma 5.6), so this issue of fixed point decoration will not concern us.
References
Bernardi, O.: Bijective counting of Kreweras walks and loopless triangulations. J. Combin. Theory Ser. A 114(5), 931–956 (2007)
Article
MathSciNet
Google Scholar
Bousquet-Mélou, M.: Walks in the quarter plane: Kreweras’ algebraic model. Ann. Appl. Probab. 15(2), 1451–1491 (2005)
Article
MathSciNet
Google Scholar
Bousquet-Mélou, M., and Mishna, M.: Walks with small steps in the quarter plane. In: Algorithmic Probability and Combinatorics, volume 520 of Contemp. Math., pp 1–39. Am. Math. Soc., Providence, RI (2010)
Edelman, P., Greene, C.: Balanced tableaux. Adv. Math. 63(1), 42–99 (1987)
Article
MathSciNet
Google Scholar
Einstein, D., Propp, J.: Combinatorial, piecewise-linear, and birational homomesy for products of two chains. Algebr. Comb. 4(2), 201–224 (2021)
MathSciNet
MATH
Google Scholar
Fontaine, B., Kamnitzer, J.: Cyclic sieving, rotation, and geometric representation theory. Selecta Math. (N.S.) 20(2), 609–625 (2014)
Article
MathSciNet
Google Scholar
Fraser, C., Lam, T., Le, I.: From dimers to webs. Trans. Am. Math. Soc. 371(9), 6087–6124 (2019)
Article
MathSciNet
Google Scholar
Gessel, I.: Unified Framework for Posets with Order Polynomial Product Formulas. MathOverflow. (version: 2019-10-06)
Gessel, I.M.: A probabilistic method for lattice path enumeration. J. Stat. Plann. Inference 14(1), 49–58 (1986)
Article
MathSciNet
Google Scholar
Haiman, M.D.: Dual equivalence with applications, including a conjecture of Proctor. Discrete Math. 99(1–3), 79–113 (1992)
Article
MathSciNet
Google Scholar
Haiman, M.D., Kim, D.: A characterization of generalized staircases. Discrete Math. 99(1–3), 115–122 (1992)
Article
MathSciNet
Google Scholar
Henriques, A., Kamnitzer, J.: Crystals and coboundary categories. Duke Math. J. 132(2), 191–216 (2006)
Article
MathSciNet
Google Scholar
Hopkins, S.: Order polynomial product formulas and poset dynamics. 2020. For the “Open Problems in Algebraic Combinatorics" AMS volume to accompany the OPAC 2022 conference at U. Minnesota. Preprint, arXiv:2006.01568
Khovanov, M., Kuperberg, G.: Web bases for ({\rm sl}(3)) are not dual canonical. Pac. J. Math. 188(1), 129–153 (1999)
Article
MathSciNet
Google Scholar
Kreweras, G., Niederhausen, H.: Solution of an enumerative problem connected with lattice paths. Eur. J. Combin. 2(1), 55–60 (1981)
Article
MathSciNet
Google Scholar
Kreweras, G.: Sur une classe de problemes de dénombrement liés au treillis des partitions des entiers. Cahiers du B.U.R.O. 6, 5–105 (1965)
Google Scholar
Kuperberg, G.: Spiders for rank (2) Lie algebras. Commun. Math. Phys. 180(1), 109–151 (1996)
Article
MathSciNet
Google Scholar
Lam, T.: Dimers, webs, and positroids. J. Lond. Math. Soc. (2) 92(3), 633–656 (2015)
Article
MathSciNet
Google Scholar
Niederhausen, H.: Sheffer polynomials in path enumeration. In Proceedings of the West Coast Conference on Combinatorics, Graph Theory and Computing (Humboldt State Univ., Arcata, Calif., 1979), Congress. Numer., XXVI, pp 281–294. Utilitas Math., Winnipeg, Man. (1980)
Niederhausen, H.: The ballot problem with three candidates. Eur. J. Combin. 4(2), 161–167 (1983)
Article
MathSciNet
Google Scholar
Patrias, R.: Promotion on generalized oscillating tableaux and web rotation. J. Combin. Theory Ser. A 161, 1–28 (2019)
Article
MathSciNet
Google Scholar
Petersen, T.K., Pylyavskyy, P., Rhoades, B.: Promotion and cyclic sieving via webs. J. Algebr. Combin. 30(1), 19–41 (2009)
Article
MathSciNet
Google Scholar
Pfannerer, S., Rubey, M., Westbury, B.W.: Promotion on oscillating and alternating tableaux and rotation of matchings and permutations. Algebr. Comb. 3(1), 107–141 (2020)
MathSciNet
MATH
Google Scholar
Postnikov, A.: Total Positivity, Grassmannians, and Networks. Preprint, arXiv:math/0609764 (2006)
Reiner, V., Stanton, D., White, D.: The cyclic sieving phenomenon. J. Comb. Theory Ser. A 108(1), 17–50 (2004)
Article
MathSciNet
Google Scholar
Rhoades, B.: Cyclic sieving, promotion, and representation theory. J. Comb. Theory Ser. A 117(1), 38–76 (2010)
Article
MathSciNet
Google Scholar
Russell, H.M.: An explicit bijection between semistandard tableaux and non-elliptic (sl_3) webs. J. Algebr. Comb. 38(4), 851–862 (2013)
Article
Google Scholar
Schützenberger, M.P.: Quelques remarques sur une construction de Schensted. Math. Scand. 12, 117–128 (1963)
Article
MathSciNet
Google Scholar
Schützenberger, M.P.: Promotion des morphismes d’ensembles ordonnés. Discrete Math. 2, 73–94 (1972)
Article
MathSciNet
Google Scholar
Schützenberger, M.P.: Evacuations. In Colloquio Internazionale sulle Teorie Combinatorie (Rome, 1973), Tomo I, pages 257–264. Atti dei Convegni Lincei, No. 17. (1976)
Schützenberger, M.-P.: La correspondance de Robinson. In Combinatoire et représentation du groupe symétrique (Actes Table Ronde CNRS, Univ. Louis-Pasteur Strasbourg, Strasbourg, 1976), pp 59–113. Lecture Notes in Math., Vol. 579 (1977)
Stanley, R.P.: Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999. With a foreword by Gian-Carlo Rota and appendix 1 by Sergey Fomin
Stanley, R.P.: Promotion and evacuation. Electron. J. Combin., 16(2, Special volume in honor of Anders Björner):Research Paper 9, 24 (2009)
Stanley, R.P.: Enumerative Combinatorics. Volume 1, volume 49 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, second edition (2012)
Striker, J., Williams, N.: Promotion and rowmotion. Eur. J. Comb. 33(8), 1919–1942 (2012)
Article
MathSciNet
Google Scholar
Thomas, H., Williams, N.: Rowmotion in slow motion. Proc. Lond. Math. Soc. (3) 119(5), 1149–1178 (2019)
Article
MathSciNet
Google Scholar
Tymoczko, Julianna: A simple bijection between standard (3\times n) tableaux and irreducible webs for (\mathfrak{sl}_3). J. Algebr. Comb. 35(4), 611–632 (2012)
Article
MathSciNet
Google Scholar
Westbury, B.W.: Invariant tensors and the cyclic sieving phenomenon. Electron. J. Comb. 23(4): Paper 4.25, 40 (2016)
Download references
Acknowledgements
S.H. thanks Ira Gessel, whose answer to a MathOverflow question of his [8")] made him aware of the paper [15, 55–60 (1981)")] and the poset V(n), and thus initiated this research.
Funding
Open access funding provided by TU Wien (TUW).
Author information
Authors and Affiliations
Department of Mathematics, Howard University, Washington, DC, USA
Sam Hopkins
Fakultät für Mathematik und Geoinformation, TU Wien, Vienna, Austria
Martin Rubey
Search author on:PubMed Google Scholar
Search author on:PubMed Google Scholar
Corresponding author
Correspondence to
Martin Rubey.
Additional information
Publisher's Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Sam Hopkins was supported by NSF Grant #1802920. Martin Rubey was supported by the the Austrian Science Fund (FWF): P 29275.
Rights and permissions
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit
Reprints and permissions
About this article
Cite this article
Hopkins, S., Rubey, M. Promotion of Kreweras words.
Sel. Math. New Ser. 28, 10 (2022).
Download citation
Accepted: 17 September 2021
Published: 26 November 2021
DOI:
Share this article
Anyone you share the following link with will be able to read this content:
Sorry, a shareable link is not currently available for this article.
Provided by the Springer Nature SharedIt content-sharing initiative
Keywords
Mathematics Subject Classification
Profiles
Avoid common mistakes on your manuscript.
Advertisement
Search
Navigation
Discover content
Publish with us
Products and services
Our brands
3.84.242.119
Not affiliated
© 2025 Springer Nature |
190740 | https://math.answers.com/basic-math/What_is_negative_19_as_a_decimal | What is negative 19 as a decimal? - Answers
Create
0
Log in
Subjects>Math>Basic Math
What is negative 19 as a decimal?
Anonymous ∙ 7 y ago
Updated: 9/27/2023
A decimal number is simply a way of representing a number in such a way that the place value of each digit is ten times that of the digit to its right. A decimal representation does not require a decimal point. So the required decimal representation is -19, exactly as in the question.
Wiki User ∙ 7 y ago
Copy
Add Your Answer
What else can I help you with?
Search
Loading...
Loading...
Trending Questions
What do people notice about the factors of a square number?Why all the multiple of 6 are even numbers?What is 0.55 rounded to the nearest hundredth?What is complex math?What is 30 percent off of 25.00 equal?What is the percentage of an eighth?What is what is 402 rounded to the nearest hundred?What percentage is 85 of 250?What percent is three quarters?What is your grade with 26 out of 50?What is the actual sum and the estimated sum of 3874 1845?What is 3.25 percent of 125000?What fraction is smaller than 1 12?How many percent of 50 is 10?How many us dollars equals 1.00 us?What is the minimum daily requiremtns of potassism?What is 510 percent as a decimal?What are the common factors of 35 and 36?What is the HCF of 14 and 20?How do you write 9.68 in word form?
Resources
LeaderboardAll TagsUnanswered
Top Categories
AlgebraChemistryBiologyWorld HistoryEnglish Language ArtsPsychologyComputer ScienceEconomics
Product
Community GuidelinesHonor CodeFlashcard MakerStudy GuidesMath SolverFAQ
Company
About UsContact UsTerms of ServicePrivacy PolicyDisclaimerCookie PolicyIP Issues
Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers. |
190741 | https://www.ck12.org/book/ck-12-middle-school-math-concepts-grade-8/r57/section/12.1/ | Welcome to CK-12 Foundation | CK-12 Foundation
AI Teacher Tools – Save Hours on Planning & Prep.
Try it out!
What are you looking for? Search
Math
Grade 6
Grade 7
Grade 8
Algebra 1
Geometry
Algebra 2
PreCalculus
Science
Earth Science
Life Science
Physical Science
Biology
Chemistry
Physics
Social Studies
Economics
Geography
Government
Philosophy
Sociology
Subject
Math
Elementary Math
Grade 1
Grade 2
Grade 3
Grade 4
Grade 5
Interactive
Math 6
Math 7
Math 8
Algebra I
Geometry
Algebra II
Conventional
Math 6
Math 7
Math 8
Algebra I
Geometry
Algebra II
Probability & Statistics
Trigonometry
Math Analysis
Precalculus
Calculus
What's the difference?
Science
Grade K to 5
Earth Science
Life Science
Physical Science
Biology
Chemistry
Physics
Advanced Biology
FlexLets
Math FlexLets
Science FlexLets
English
Writing
Spelling
Social Studies
Economics
Geography
Government
History
World History
Philosophy
Sociology
More
Astronomy
Engineering
Health
Photography
Technology
College
College Algebra
College Precalculus
Linear Algebra
College Human Biology
The Universe
Adult Education
Basic Education
High School Diploma
High School Equivalency
Career Technical Ed
English as 2nd Language
Country
Bhutan
Brasil
Chile
Georgia
India
Translations
Spanish
Korean
Deutsch
Chinese
Greek
Polski
Explore
EXPLORE
Flexi A FREE Digital Tutor for Every Student
FlexBooks 2.0 Customizable, digital textbooks in a new, interactive platform
FlexBooks Customizable, digital textbooks
Schools FlexBooks from schools and districts near you
Study Guides Quick review with key information for each concept
Adaptive Practice Building knowledge at each student’s skill level
Simulations Interactive Physics & Chemistry Simulations
PLIX Play. Learn. Interact. eXplore.
CCSS Math Concepts and FlexBooks aligned to Common Core
NGSS Concepts aligned to Next Generation Science Standards
Certified Educator Stand out as an educator. Become CK-12 Certified.
Webinars Live and archived sessions to learn about CK-12
Other Resources
CK-12 Resources
Concept Map
Testimonials
CK-12 Mission
Meet the Team
CK-12 Helpdesk
FlexLets
Know the essentials.
Pick a Subject
Donate
Sign InSign Up
Difficulty Level: | Created by:
Last Modified:
Read Resources Details
Loading...
Notes/Highlights
| Color | Highlighted Text | Notes | |
--- --- |
| | Please Sign In to create your own Highlights / Notes |
Currently there are no resources to be displayed.
Description
No description available here...
Difficulty Level
Basic
Standards Correlations
-
Date Created
Invalid Date
Last Modified
Invalid Date
.
Show Details ▼
Back to the top of the page ↑
Support
CK-12 Foundation is a non-profit organization that provides free educational materials and resources.
FLEXIAPPS
ABOUT
Our missionMeet the teamPartnersPressCareersSecurityBlogCK-12 usage mapTestimonials
SUPPORT
Certified Educator ProgramCK-12 trainersWebinarsCK-12 resourcesHelpContact us
BYCK-12
Common Core MathK-12 FlexBooksCollege FlexBooksTools and apps
CONNECT
TikTokInstagramYouTubeTwitterMediumFacebookLinkedIn
v2.11.10.20250923073248-b88c97d744
© CK-12 Foundation 2025 | FlexBook Platform®, FlexBook®, FlexLet® and FlexCard™ are registered trademarks of CK-12 Foundation.
Terms of usePrivacyAttribution guide
Curriculum Materials License
Oops, looks like cookies are disabled on your browser. Click on this link to see how to enable them.
X
Student Sign Up
Are you a teacher? Sign up here
Sign in with Google
Having issues? Click here
Sign in with Microsoft Sign in with Apple
or
Sign up using email
By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy
Already have an account? Sign In
No Results Found
Your search did not match anything in .
Got It |
190742 | https://www.youtube.com/watch?v=Ifb4Jwywefg | How To Identify The Relationship In Analogies? - Law School Prep Hub
Law School Prep Hub
465 subscribers
Description
10 views
Posted: 6 Aug 2025
How To Identify The Relationship In Analogies? Understanding analogies is a key skill for anyone preparing for the Law School Admission Test. In this informative video, we will guide you through the process of identifying relationships in analogies, a common question type on the exam. We’ll start by breaking down what an analogy is and how it compares two pairs of items, revealing the connections between them.
You’ll learn how to analyze the first pair of items to determine their relationship, whether it’s based on similarity, contrast, causation, or other connections. We’ll also emphasize the importance of articulating these relationships clearly, ensuring you can apply the same reasoning to the second pair.
We’ll discuss common pitfalls in analogy arguments, helping you to recognize flaws that may weaken your analysis. Mastering this skill not only enhances your performance on the Law School Admission Test but also proves beneficial in legal practice, where recognizing analogies can clarify case law and strengthen arguments.
Join us as we break down this essential skill for future law students, and don’t forget to subscribe for more helpful content related to law school preparation and legal reasoning.
⬇️ Subscribe to our channel for more valuable insights.
🔗Subscribe:
LawSchool #Analogies #LSATPrep #LegalReasoning #StudyTips #LawStudents #TestPreparation #AnalyticalSkills #CriticalThinking #Logic #LawSchoolAdmissionTest #LegalEducation #ArgumentAnalysis #ReasoningSkills #LegalPractice #ExamSuccess
About Us: Welcome to Law School Prep Hub! Our channel is dedicated to guiding you through the journey of law school admissions and beyond. Whether you're preparing for the LSAT, researching law school rankings, or seeking tips for writing a compelling personal statement, we provide practical advice and strategies to help you succeed.
Transcript:
How to identify the relationship in analogies. Have you ever found yourself puzzled by analogy questions while preparing for the law school admission test? Understanding how to identify relationships in analogies can make a significant difference in your performance. Let's break it down step by step to make it easier for you. First, let's clarify what an analogy is. An analogy compares to pairs of items, showing how the first pair relates to each other in a way that is similar to how the second pair relates. The law school admission test often includes these types of questions to assess your reasoning skills. To start, focus on the first pair of items in the analogy. Take a moment to analyze how these two elements connect. There are several common relationships you might encounter. For instance, they could share a similarity, contrast in some way, or one could cause the other. You might also see a part to whole relationship where one item is a component of the other. Another possibility is a functional relationship where one item serves a purpose related to the other. Next, it's essential to articulate the relationship clearly. This means putting it into words. For example, you might say item A causes item B or item A is a type of item B. This step helps you clarify your understanding of the relationship before moving on. Once you have defined the first relationship, turn your attention to the second pair of items. Check if the same relationship applies here. The goal is to see if the connection you identified in the first pair holds true for the second pair as well. Consistency is key. Make sure that the relationship is logical and remains intact in both pairs. If you find that the connection breaks down or only partially holds true in the second pair, then the analogy may not be strong. It's also helpful to be aware of common flaws in analogy arguments. Often these arguments assume that the two items being compared are similar in all relevant aspects without sufficient evidence. A typical flaw is overlooking important differences that could weaken the analogy. For law students, mastering this skill is particularly useful. When you read legal cases, recognizing analogies can help you understand how past cases relate to current ones. This skill also aids in evaluating legal reasoning, allowing you to spot strengths or weaknesses in arguments. Furthermore, it can assist you in crafting persuasive arguments by drawing parallels between facts or legal principles. In summary, to identify the relationship in analogies, carefully analyze the first pair to understand their connection. Then apply that same connection to the second pair, ensuring the relationship is logical and consistent. This skill is essential for anyone preparing for the law school admission test and for effective legal reasoning in practice. |
190743 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4543740/ | Caffey disease in neonatal period: the importance of the family! - PMC
Skip to main content
An official website of the United States government
Here's how you know
Here's how you know
Official websites use .gov
A .gov website belongs to an official government organization in the United States.
Secure .gov websites use HTTPS
A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites.
Search
Log in
Dashboard
Publications
Account settings
Log out
Search… Search NCBI
Primary site navigation
Search
Logged in as:
Dashboard
Publications
Account settings
Log in
Search PMC Full-Text Archive
Search in PMC
Journal List
User Guide
View on publisher site
Download PDF
Add to Collections
Cite
Permalink PERMALINK
Copy
As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health.
Learn more: PMC Disclaimer | PMC Copyright Notice
BMJ Case Rep
. 2012 Oct 9;2012:bcr2012006996. doi: 10.1136/bcr-2012-006996
Search in PMC
Search in PubMed
View in NLM Catalog
Add to search
Caffey disease in neonatal period: the importance of the family!
Ana Rita Prior
Ana Rita Prior
1 Department of Paediatrics, Hospital Santa Maria, Centro Hospitalar Lisboa Norte, Lisbon, Portugal
Find articles by Ana Rita Prior
1, Oana Moldovan
Oana Moldovan
2 Department of Paediatrics, Genetics, Hospital Santa Maria, Centro Hospitalar Lisboa Norte, Lisbon, Portugal
Find articles by Oana Moldovan
2, António Azevedo
António Azevedo
1 Department of Paediatrics, Hospital Santa Maria, Centro Hospitalar Lisboa Norte, Lisbon, Portugal
Find articles by António Azevedo
1, Carlos Moniz
Carlos Moniz
1 Department of Paediatrics, Hospital Santa Maria, Centro Hospitalar Lisboa Norte, Lisbon, Portugal
Find articles by Carlos Moniz
1
Author information
Article notes
Copyright and License information
1 Department of Paediatrics, Hospital Santa Maria, Centro Hospitalar Lisboa Norte, Lisbon, Portugal
2 Department of Paediatrics, Genetics, Hospital Santa Maria, Centro Hospitalar Lisboa Norte, Lisbon, Portugal
Series information
Rare disease
Collection date 2012.
2012 BMJ Publishing Group Ltd
PMC Copyright notice
PMCID: PMC4543740 PMID: 23047998
Abstract
A male newborn was apparently well until his second day of life, when increased irritability and a swelling in his right leg were noted. He was rooming-in with his mother since birth. On examination, a mass on the anterior surface of the right leg was noticed. The mass was firm, elongated, ill-defined, unmovable and painful at palpation. No overlying skin changes were seen. The newborn had a family history of neonatal bone swelling with resolution before the age of 2. Subsequent images showed hyperostosis in the diaphysis of the right tibia. After exclusion of other conditions such as trauma, osteomyelitis and congenital syphilis, the involvement of the tibial diaphysis, sparing the epiphyses and the benign course of the disease in family history, were indicative of Caffey disease. The genetic study confirmed this diagnosis. Caffey disease, although rare, should not be overlooked in the diagnostic approach to childhood bone swelling.
Background
Caffey disease is a rare cause of bone swelling in a newborn. The manifestations of this disorder can sometimes resemble those of child's physical abuse, as well as other diseases. For this reason, a correct diagnosis of the cause of unexplained trauma, fractures or swellings in children requires that clinicians be familiar with such diseases, showing the importance of well-interpreted imaging studies in these diagnoses.12
Case presentation
A male newborn was apparently well until 28 h of life when his mother identified a swelling in his right leg (figure 1). He was irritable since then. Apart from the painful swollen leg, the newborn had no other systemic symptoms such as fever or poor feeding.
Figure 1.
Open in a new tab
Tumefaction in the middle third of the anterior surface of the right leg.
On examination, there was a firm tumefaction in the middle third of the anterior surface of the right leg, painful on palpation, with no changes in colour or temperature of skin over the affected limb. Apart from this, the physical examination was normal. There were no external wounds or bruises.
The mother was 33-year-old, ARh+ and gravida 3 para 2. The current pregnancy was planned and uneventful. Biochemical screening was negative. Three routine fetal ultrasounds were normal. In the third trimester (34 weeks of gestational age), serological maternal investigation was irrelevant (negative for hepatitis B virus, hepatitis C virus, toxoplasma, HIV and Venereal Disease Research Laboratory, VDRL; immune to Rubella). Cytomegalovirus serology was not studied. Group B Streptococcus screening test was positive.
The child was born at term (39 weeks) by normal vaginal delivery, non-traumatic with cephalic presentation. No history of prolonged rupture of membranes or maternal fever during labour. Intrapartum antibiotic prophylaxis was performed. Apgar scores were 8 and 9 at 1 and 5 min, respectively. At birth, his weight was 3490 g (25th–50th percentile), length 50 cm (50th percentile) and head circumference 35 cm (25th percentile). From birth, he was in hospital rooming-in with mother, the primary caregiver who denied any history of trauma or injury. He was exclusively breastfed since birth and not received any supplements including vitamin syrup.
He is the second child of non-consanguineous parents, with a previously healthy daughter. Tracing back his family history (figure 2), mother and two maternal uncles reported a ‘bone disease’ manifested as a swelling-bone in the neonatal period, which resolved without sequelae in the first 2 years of life (24 , 12 and 5 months, respectively). In all cases, tibia was affected and one of the maternal uncles had involvement of other bones (skull, legs and arms). No family history of multiple fractures was reported.
Figure 2.
Open in a new tab
Family pedigree where the mother's and two maternal uncles involvement can be seen.
Investigations
Biochemical and haematological investigations were normal (haemoglobin 17 g/dl, total white blood cell count 12 780×10 9/l, platelet 335×10 9/l, calcium (total) 9.7 mg/dl, phosphorous 7.4 mg/dl, urea 16 mg/dl, creatinine 0.6 mg/dl and C reactive protein 0.3 mg/dl).
x-Rays of the affected leg showed thickening of the periosteum at tibial diaphysis without epiphysial involvement and no signs of fracture (figure 3).
Figure 3.
Open in a new tab
Anteroposterior (ap) and lateral (perfil) x-rays of right lower limb at 2 days of life, showing cortical new bone formation associated with subperiosteal thickening of tibial dyaphysis.
Maternal serology for syphilis (VDRL) was not performed after birth.
A germinal heterozygous missense mutation c.3040C>T (p.Arg1014Cys) in exon 41 of COL1A1 gene, associated with Caffey disease, was identified in the newborn, as well as in his mother and in his uncle (figure 2).
Differential diagnosis
The initial diagnosis hypothesis was trauma. In a newborn with non-traumatic vaginal delivery, cephalic presentation, with normal medical examination in first day of life and who seems well until 28 h of life when increased irritability and a swelling in his right leg were noted, trauma (inflicted or otherwise) must be excluded. There was a single region involved and periosteal reaction of a single bone that except for the mandible, could be suggestive of trauma.3 However, the absence of external wounds or bruises and the fact that he shared the hospital room not only with his mother but also with other people (who denied any trauma) under the surveillance of hospital staff, suggested otherwise.
Osteomyelitis was also regarded as a possible diagnosis, due to the swollen and tender extremity. Nevertheless, a bone infection was thought to be unlikely because the newborn showed good physical condition with normal vital signs and no fever. In addition, bone changes are not usually present in early acute osteomyelitis and children with osteomyelitis usually have increased acute-phase reactants’ levels (C reactive protein) and possibly increased or left-shifted white blood cell count, which did not occur in this case.
Despite a negative VDRL in the third trimester of pregnancy, this test was performed more than 4 weeks before delivery, not excluding a recent infection. As abnormal long-bone radiographs with diaphyseal periostitis are a common manifestation of early congenital syphilis (60–80%) and may even be the only manifestation in infants born to mothers with untreated syphilis, we had to consider congenital syphilis as a possible diagnosis too. However, in congenital syphilis the skeletal changes usually involve not only diaphysis but also metaphysis of the long bones and the lesions are usually symmetric. In this case, maternal serology for syphilis (VDRL) to exclude or confirm this diagnosis was not performed after birth.
There are other causes of periosteum thickening, including scurvy and hypervitaminosis A, but these were not considered in differential diagnosis as they usually do not affect such small infants.4
After the exclusion of the conditions aforementioned, the characteristic triad of irritability, swelling and bone lesions with periosteal reaction involving the tibia diaphysis sparing the epiphyses, the age at presentation and benign family history, pointed to the Caffey disease. Subsequently, the genetic study confirmed this diagnosis.
Treatment
Symptomatic treatment with acetaminophen was given, with good clinical response. The newborn was discharged on the third day of life, keeping surveillance on an outpatient basis and with recommendation for analgesic therapy during periods of increased irritability.
Outcome and follow-up
With 1 month of age, the swelling in the middle third of the right leg persisted, without other inflammatory signs, rarely requiring medication. Radiography of the right leg showed worsening of hyperostosis with thickening of the cortex in the diaphysis of the right tibia (figure 4).
Figure 4.
Open in a new tab
Anteroposterior (ap) and lateral (lat) x-rays of right lower limb at 1 month of life, showing worsening of periosteal reaction of tibial dyaphysis.
Discussion
Caffey disease or infantile cortical hyperostosis is a proliferative bone disease, resulting in new bone formation at the periosteum, sometimes exuberant, becoming compact and with a pronounced cortical thickening.2
Skeletal growth is a strictly controlled phenomenon. While bone growth in length is performed by adding endochondral bone formation in metaphyseal spongy bone, the growth in diameter is achieved by subperiosteal new bone apposition. The periosteum is a fibrous connective tissue capsule that covers the external surface of the bone, except articular surfaces that are covered by hyaline cartilage. Periosteum has an outer layer which resembles other dense connective tissues and an inner layer, more cellular that contains the osteoprogenitor cells. The size of the marrow cavity is controlled by a combination of bone apposition versus resorption at the endocortical surface. In Caffey disease, there is an exacerbated subperiosteal intramembranous bone formation (hyperostosis), triggered by local inflammation (periostitis). This excess of bone tissue could be resorbed either at the endocortical surface with an expansion of the marrow cavity and a more persistent deformity, or at the exocortical surface, with no effect on the size of the marrow cavity.5
Two forms of cortical hyperostosis or Caffey disease have been described: prenatal and infantile. The prenatal form is characterised by extensive hyperostotic bone involvement, angulations and shortening of long bones, polyhydramnios and fetal hydrops, being a rare disorder with a high death rate due to prematurity and lung hypoplasia. The diagnosis of this form is based on ultrasound findings (as early as 14 weeks of gestation, average 27 weeks) of short and angulated bones with irregular diaphyses. Conversely, in the reported case, the pregnancy was uneventful, with normal ultrasounds and a term delivery of a healthy newborn. Consequently, the prenatal form of the Caffey disease should be excluded.6 The prenatal form of Caffey disease appears to be an autosomal recessive disorder.78
In our case, the classical mild infantile form seems more likely. It was first described in 1930 by Roske (Europe), but was further characterised by Caffey and Silverman in 1945.6910 The classical presentation of this disease has an onset within the first 6 months of life and includes a triad of irritability, swelling and bone lesions with underlying cortical bone thickening.26910 The swelling appears suddenly, is deep and firm, and may be tender, as occurred in our case. Fever may also occur. It has been described in many bones, most often involving the mandible (95% of cases), as well as ribs, shoulder blades and long bones (affect only the diaphyses, sparing the epiphyses and metaphyses). It is often multifocal and asymmetric,2 although cases of monostotic involvement have also been reported.10
Although the aetiology of Caffey disease is not completely understood, familial and sporadic forms appear to exist. Some of these sporadic cases have been attributed to administration of prostaglandins E1 and E2 for treatment of ductal-dependent cardiac lesions, with radiographs changes occurring in 62% of all infants receiving prostaglandins infusion for more than 60 days. Review of familial form, with autosomal-dominant pattern of inheritance, showed evidence of variable penetrance, as there were obligate carriers who lacked episodes of cortical hyperostosis. However, the signs and symptoms can be subtle and even missed in some of these unaffected carriers.8
Affected members of some families with Caffey disease, including both infantile and prenatal form, were found to be heterozygous for a missense mutation (c.3040C>T) in exon 41 of the gene encoding the α1 chain of type I collagen (COL1A1, 17q21), altering residue 836 (R836C) in the triple-helical domain of this chain.26–9 This mutation was identified in our newborn, as well as in his mother and in his uncle.
The COL1A1 gene is responsible for other collagen diseases such as osteogenesis imperfecta and Ehlers-Danlos syndrome, and for this reason, studies place the Caffey disease in the family of type I collagen-related diseases.211
Caffey disease is rare, being reported to affect 3 per 1000 children under 6 months, with no sex or racial predilection.26 It has a low prevalence, probably due to underdiagnosis.4
Diagnosis may be delayed as this disorder mimics a wide range of diseases, including osteomyelitis, chronic hypervitaminosis A, bone tumour, scurvy and prolonged PGE1 infusion.7 The effects of Caffey disease can also resemble those of child physical abuse. However, a misdiagnosis of child abuse can also pose a significant burden to the child, the parents and the suspected aggressor, being of crucial importance that clinics be aware of other conditions that may mimic a non-accidental injury, such as infantile hyperostosis.712
While no laboratory tests are specific for diagnosis of infantile cortical hyperostosis,7 radiography is the most valuable diagnostic study in infantile cortical hyperostosis, showing layers of periosteal new bone formation, with cortical thickening. Soft-tissue swelling is evident as well. Other diagnostic imaging studies have not proved to be helpful.2
Awareness of the existence of this rare condition and its typical clinicoradiological profile will avoid the patient being subjected to unnecessary investigation and treatment.7
No specific treatment exists for Caffey disease. Management is mainly palliative, aimed at pain relief. Some authors described a good response to immunoglobulin, corticosteroids and non-steroidal anti-inflammatory, but generally these agents do not have any effect on the bone lesions. For this reason, there is no consensus on its use.
Caffey disease is a self-limited disease, sometimes with periods of exacerbation/remission and usually resolves without sequelae by 2 years.23 However, when paired bones such as tibia and fibula, or radius and ulna, are affected, cross-fusion may be a long-term complication.10 Although some authors refer that the bone changes persist into adolescence and adulthood, others as Gensure et al report that studied patients with COL1A1 mutations had no skeletal deformity after childhood. In the study of four generations of a family affected with Caffey disease (Cerruti-Mainardi et al13), COL1A1 mutation was identified in nine members of this family, reaching all target height and showing no bone changes sequelae in adulthood.
Learning points.
Caffey disease or infantile cortical hyperostosis is an inflammatory proliferative bone disease, which has an onset within the first 6 months of life and usually resolves without sequelae by 2 years.
It is characterised by a triad of systemic symptoms (irritability and/or fever), soft-tissue swelling and underlying cortical bone thickening; most often involving the mandible (95% of cases), as well as ribs, shoulder blades and long bones (affect only the diaphyses, sparing the epiphyses and metaphyses).
In most cases, a good clinical history, basic laboratory evaluation and plain radiographs demonstrating periosteal involvement are sufficient to confirm this diagnosis.
Caffey disease, although rare, should not be forgotten in the diagnostic approach to early childhood bone swellings.
A high index of suspicion is necessary to establish this diagnosis, avoiding protracted investigations and unnecessary treatment for this otherwise self-limiting illness.
Footnotes
Competing interests: None.
Patient consent: Obtained.
References
1.Kaissy A, Petje G, Brauwer V, et al. Professional awareness is needed to distinguish between child physical abuse from other disorders that can mimic signs of abuse (skull base sclerosis in infant manifesting features of infantile cortical hyperostosis): a case report and review of the literature. Case J 2009;2:133. [DOI] [PMC free article] [PubMed] [Google Scholar]
2.Novick A, Gellman H. Infantile cortical hyperostosis. eMedicine (accessed 29 Jun 2012).
3.Scherl SA. Differential diagnosis of the orthopedic manifestations of child abuse. UpToDate 2011. (accessed 10 Oct 2011)
4.Ludman A, Bravo M, Moguillansky S. Hiperostosis cortical infantil. Enfermedad de Caffey. Arch Argent Pediatr 2010;108:360–2. [DOI] [PubMed] [Google Scholar]
5.Horton WA. Caffey disease is a type I collagenopathy. GGH 2005;21:40–1. [Google Scholar]
6.Hochwald O, Osiovich H. Prenatal Caffey disease. Isr Med Assoc J 2011;13:113–14. [PubMed] [Google Scholar]
7.Kutty N, Thomas D, George L, et al. Caffey disease or infantile cortical hyperostosis: a case report. Oman Med J 2010;25:134–6. [DOI] [PMC free article] [PubMed] [Google Scholar]
8.Gensure RC, Makitie O, Barclay C, et al. A novel COL1A1 mutation in infantile cortical hyperostosis (Caffey disease) expands the spectrum of collagen-related disorders. J Clin Invest 2005;115:1250–7. [DOI] [PMC free article] [PubMed] [Google Scholar]
9.Maclachlan AK, Gerrard JW, Houston CS, et al. Familial infantile cortical hyperostosis in a large Canadian family. Can Med Assoc J 1984;130:1172–4. [PMC free article] [PubMed] [Google Scholar]
10.Hall C. Caffey disease. Orphanet encyclopedia. 2005, (accessed 20 Jun 2012).
11.Glorieux FH. Caffey disease: an unlikely collagenopathy. J Clin Invest 2005;115:1142–3. [DOI] [PMC free article] [PubMed] [Google Scholar]
12.Lo HPW, Lau HY, Li CH, et al. Infantile cortical hyperostosis (Caffey disease): a possible misdiagnosis as physical abuse. Hong Kong Med J 2010;16: 397. [PubMed] [Google Scholar]
13.Cerruti-Mainardi P, Venturi G, Spunton M, et al. Infantile cortical hyperostosis and COL1A1 mutation in four generations. Eur J Pediatr 2011;170:1385–90. [DOI] [PMC free article] [PubMed] [Google Scholar]
Articles from BMJ Case Reports are provided here courtesy of BMJ Publishing Group
ACTIONS
View on publisher site
PDF (229.2 KB)
Cite
Collections
Permalink PERMALINK
Copy
RESOURCES
Similar articles
Cited by other articles
Links to NCBI Databases
On this page
Abstract
Background
Case presentation
Investigations
Differential diagnosis
Treatment
Outcome and follow-up
Discussion
Footnotes
References
Cite
Copy
Download .nbib.nbib
Format:
Add to Collections
Create a new collection
Add to an existing collection
Name your collection
Choose a collection
Unable to load your collection due to an error
Please try again
Add Cancel
Follow NCBI
NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed
Connect with NLM
NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube
National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894
Web Policies
FOIA
HHS Vulnerability Disclosure
Help
Accessibility
Careers
NLM
NIH
HHS
USA.gov
Back to Top |
190744 | https://math.washington.edu/~burke/crs/408/notes/nlp/unoc.pdf | OPTIMALITY CONDITIONS 1. Unconstrained Optimization 1.1. Existence. Consider the problem of minimizing the function f : Rn →R where f is continuous on all of Rn: P min x∈Rn f(x).
As we have seen, there is no guarantee that f has a minimum value, or if it does, it may not be attained. To clarify this situation, we examine conditions under which a solution is guaranteed to exist. Recall that we already have at our disposal a rudimentary existence result for constrained problems. This is the Weierstrass Extreme Value Theorem.
Theorem 1.1. (Weierstrass Extreme Value Theorem) Every continuous function on a compact set attains its extreme values on that set.
We now build a basic existence result for unconstrained problems based on this theorem.
For this we make use of the notion of a coercive function.
Definition 1.1. A function f : Rn →R is said to be coercive if for every sequence {xν} ⊂Rn for which ∥xν∥→∞it must be the case that f(xν) →+∞as well.
Continuous coercive functions can be characterized by an underlying compactness property on their lower level sets.
Theorem 1.2. (Coercivity and Compactness) Let f : Rn →R be continuous on all of Rn.
The function f is coercive if and only if for every α ∈R the set {x |f(x) ≤α} is compact.
Proof. We first show that the coercivity of f implies the compactness of the sets {x |f(x) ≤α}.
We begin by noting that the continuity of f implies the closedness of the sets {x |f(x) ≤α}.
Thus, it remains only to show that any set of the form {x |f(x) ≤α} is bounded. We show this by contradiction.
Suppose to the contrary that there is an α ∈Rn such that the set S = {x |f(x) ≤α} is unbounded. Then there must exist a sequence {xν} ⊂S with ∥xν∥→∞. But then, by the coercivity of f, we must also have f(xν) →∞. This contra-dicts the fact that f(xν) ≤α for all ν = 1, 2, . . . . Therefore the set S must be bounded.
Let us now assume that each of the sets {x |f(x) ≤α} is bounded and let {xν} ⊂Rn be such that ∥xν∥→∞. Let us suppose that there exists a subsequence of the integers J ⊂N such that the set {f(xν)}J is bounded above. Then there exists α ∈Rn such that {xν}J ⊂{x |f(x) ≤α}. But this cannot be the case since each of the sets {x |f(x) ≤α} is bounded while every subsequence of the sequence {xν} is unbounded by definition. Therefore, the set {f(xν)}J cannot be bounded, and so the sequence {f(xν)} contains no bounded subsequence, i.e. f(xν) →∞.
□ This result in conjunction with Weierstrass’s Theorem immediately yields the following existence result for the problem P.
Theorem 1.3. (Coercivity implies existence) Let f : Rn →R be continuous on all of Rn. If f is coercive, then f has at least one global minimizer.
1 2 Proof. Let α ∈R be chosen so that the set S = {x |f(x) ≤α} is non-empty. By coercivity, this set is compact. By Weierstrass’s Theorem, the problem min {f(x) |x ∈S } has at least one global solution. Obviously, the set of global solutions to the problem min {f(x) |x ∈S } is a global solution to P which proves the result.
□ Remark: It should be noted that we only need to know that the coercivity hypothesis is stronger than is strictly required in order to establish the existence of a solution. Indeed, a global minimizer must exist if there exist one non-empty compact lower level set. We do not need all of them to be compact. However, in practice, coercivity is easy to check.
1.2. First-Order Optimality Conditions. This existence result can be quite useful, but unfortunately it does not give us a constructive test for optimality. That is, we may know a solution exists, but we still do not have a method for determining whether any given point may or may not be a solution. We now present such a test using the derivatives of the objective function f. For this we will assume that f is twice continuously differentiable on Rn and develop constructible first- and second-order necessary and sufficient conditions for optimality.
The optimality conditions we consider are built up from those developed in first term calculus for functions mapping from R to R. The reduction to the one dimensional case comes about by considering the functions φ : R →R given by φ(t) = f(x + td) for some choice of x and d in Rn. The key variational object in this context is the directional derivative of f at a point x in the direction d given by f ′(x; d) = lim t↓0 f(x + td) −f(x) t .
When f is differentiable at the point x ∈Rn, then f ′(x; d) = ∇f(x)Td = φ′(0).
Note that if f ′(x; d) < 0, then there must be a ¯ t > 0 such that f(x + td) −f(x) t < 0 whenever 0 < t < ¯ t .
In this case, we must have f(x + td) < f(x) whenever 0 < t < ¯ t .
That is, we can always reduce the function value at x by moving in the direction d an arbitrarily small amount. In particular, if there is a direction d such that f ′(x; d) exists with f ′(x; d) < 0, then x cannot be a local solution to the problem minx∈Rn f(x). Or equivalently, if x is a local to the problem minx∈Rn f(x), then f ′(x; d) ≥0 whenever f ′(x; d) exists. We state this elementary result in the following lemma.
Lemma 1.1 (Basic First-Order Optimality Result). Let f : Rn →R and let ¯ x ∈Rn be a local solution to the problem minx∈Rn f(x). Then f ′(x; d) ≥0 3 for every direction d ∈Rn for which f ′(x; d) exists.
We now apply this result to the case in which f : Rn →R is differentiable.
Theorem 1.4. Let f : Rn →R be differentiable at a point x ∈Rn. If x is a local minimum of f, then ∇f(x) = 0.
Proof. By Lemma 1.1 we have 0 ≤f ′(¯ x; d) = ∇f(¯ x)Td for all d ∈Rn .
Taking d = −∇f(¯ x) we get 0 ≤−∇f(¯ x)T∇f(¯ x) = −∥∇f(¯ x)∥2 ≤0.
Therefore, ∇f(¯ x) = 0.
□ When f : Rn →R is differentiable, any point x ∈Rn satisfying ∇f(x) = 0 is said to be a stationary (or, equivalently, a critical) point of f. In our next result we link the notions of coercivity and stationarity.
Theorem 1.5. Let f : Rn →R be differentiable on all of Rn. If f is coercive, then f has at least one global minimizer these global minimizers can be found from among the set of critical points of f.
Proof. Since differentiability implies continuity, we already know that f has at least one global minimizer. Differentiabilty implies that this global minimizer is critical.
□ This result indicates that one way to find a global minimizer of a coercive differentiable function is to first find all critical points and then from among these determine those yielding the smallest function value.
1.3. Second-Order Optimality Conditions. To obtain second-order conditions for opti-mality we must first recall a few properties of the Hessian matrix ∇2f(x). The calculus tells us that if f is twice continuously differentiable at a point x ∈Rn, then the hessian ∇2f(x) is a symmetric matrix. Symmetric matrices are orthogonally diagonalizable. That is, there exists and orthonormal basis of eigenvectors of ∇2f(x) , v1, v2, . . . , vn ∈Rn such that ∇2f(x) = V λ1 0 0 . . .
0 0 λ2 0 . . .
0 .
.
.
...
.
.
.
0 0 . . .
. . .
λn V T where λ1, λ2, . . . , λn are the eigenvalues of ∇2f(x) and V is the matrix whose columns are given by their corresponding vectors v1, v2, . . . , vn: V = v1, v2, . . . , vn .
It can be shown that ∇2f(x) is positive semi-definite if and only if λi ≥0, i = 1, 2, . . . , n, and it is positive definite if and only if λi > 0, i = 1, 2, . . . , n. Thus, in particular, if ∇2f(x) is positive definite, then dT∇2f(x)d ≥λmin ∥d∥2 for all d ∈Rn, 4 where λmin is the smallest eigenvalue of ∇2f(x).
We now give our main result on second-order necessary and sufficient conditions for opti-mality in the problem minx∈Rn f(x). The key tools in the proof are the notions of positive semi-definiteness and definiteness along with the second-order Taylor series expansion for f at a given point x ∈Rn: (1.1) f(x) = f(x) + ∇f(x)T(x −x) + 1 2(x −x)T∇2f(x)(x −x) + o(∥x −x∥2) where lim x→x o(∥x −x∥2) ∥x −x∥2 = 0.
Theorem 1.6. Let f : Rn →R be twice continuously differentiable at the point x ∈Rn.
(1) (Necessity) If x is a local minimum of f, then ∇f(x) = 0 and ∇2f(x) is positive semi-definite.
(2) (Sufficiency) If ∇f(x) = 0 and ∇2f(x) is positive definite, then there is an α > 0 such that f(x) ≥f(x) + α∥x −x∥2 for all x near x.
Proof.
(1) We make use of the second-order Taylor series expansion (1.1) and the fact that ∇f(¯ x) = 0 by Theorem 1.4. Given d ∈Rn and t > 0 set x := x + td, plugging this into (1.1) we find that 0 ≤f(¯ x + td) −f(¯ x) t2 = 1 2dT∇2f(x)d + o(t2) t2 since ∇f(x) = 0 by Theorem 1.4. Taking the limit as t →0 we get that 0 ≤dT∇2f(x)d.
Since d was chosen arbitrarily, ∇2f(x) is positive semi-definite.
(2) The Taylor expansion (1.1) and the hypothesis that ∇f(¯ x) = 0 imply that (1.2) f(x) −f(x) ∥x −x∥2 = 1 2 (x −x)T ∥x −x∥∇2f(x) (x −x) ∥x −x∥+ o(∥x −x∥2) ∥x −x∥2 .
If λmin > 0 is the smallest eigenvalue of ∇2f(x), choose ϵ > 0 so that (1.3) o(∥x −x∥2) ∥x −x∥2 ≤λmin 4 whenever ∥x −x∥< ϵ. Then, for all ∥x −x∥< ϵ, we have from (1.2) and (1.3) that f(x) −f(x) ∥x −x∥2 ≥ 1 2λmin + o(∥x −x∥2) ∥x −x∥2 ≥ 1 4λmin.
Consequently, if we set α = 1 4λmin, then f(x) ≥f(x) + α∥x −x∥2 whenever ∥x −x∥< ϵ.
□ 5 In order to apply the second-order sufficient condition one must be able to check that a symmetric matrix is positive definite. As we have seen, this can be done by computing the eigenvalues of the matrix and checking that they are all positive. But there is another approach that is often easier to implement using the principal minors of the matrix.
Theorem 1.7. Let H ∈Rn×n be symmetric.
We define the kth principal minor of H, denoted ∆k(H), to be the determinant of the upper-left k × k submatrix of H. Then (1) H is positive definite if and only if ∆k(H) > 0, k = 1, 2, . . . , n.
(2) H is negative definite if and only if (−1)k∆k(H) > 0, k = 1, 2, . . . , n.
Definition 1.2. Let f : Rn →R be continuously differentiable at ¯ x. If ∇f(¯ x) = 0, but ¯ x is neither a local maximum or a local minimum, we call ¯ x a saddle point for f.
Theorem 1.8. Let f : Rn →R be twice continuously differentiable at ¯ x. If ∇f(¯ x) = 0 and ∇2f(¯ x) has both positive and negative eigenvalues, then ¯ x is a saddle point of f.
Theorem 1.9. Let H ∈Rn×n be symmetric. If H is niether positive definite or negative definite and all of its principal minors are non-zero, then H has both positive and negative eigenvalues. In this case we say that H is indefinite.
Example: Consider the matrix H = 1 1 −1 1 5 1 −1 1 4 .
We have ∆1(H) = 1, ∆2(H) = 1 1 1 5 = 4, and ∆3(H) = det(H) = 8.
Therefore, H is positive definite.
1.4. Convexity. In the previous section we established first- and second-order optimality conditions. These conditions we based on only local information and so only refer to prop-erties of local extrema. In this section we study the notion of convexity which allows us to provide optimality conditions for global solutions.
Definition 1.3.
(1) A set C ⊂Rn is said to be convex if for every x, y ∈C and λ ∈[0, 1] one has (1 −λ)x + λy ∈C .
(2) A function f : Rn →R is said to be convex if for every two points x1, x2 ∈Rn and λ ∈[0, 1] we have (1.4) f(λx1 + (1 −λ)x2) ≤λf(x1) + (1 −λ)f(x2).
The function f is said to be strictly convex if for every two distinct points x1, x2 ∈Rn and λ ∈(0, 1) we have (1.5) f(λx1 + (1 −λ)x2) < λf(x1) + (1 −λ)f(x2).
6 The inequality (1.4) is equivalent to the statement that the secant line connecting (x1, f(x1)) and (x2, f(x2)) lies above the graph of f on the line segment λx1 + (1 −λ)x2, λ ∈[0, 1].
x x + (1 - )x 2 x 2 1 (x , f (x )) 1 2 (x , f (x )) λ 1 λ 2 1 That is, the set epi (f) = {(x, µ) : f(x) ≤µ}, called the epi-graph of f is a convex set. Indeed, it can be shown that the convexity of the set epi (f) is equivalent to the convexity of the function f. This observation allows us to extend the definition of the convexity of a function to functions taking potentially infinite values.
Definition 1.4. A function f : Rn →R ∪{+∞} = ¯ R is said to be convex if the set epi (f) = {(x, µ) : f(x) ≤µ} is a convex set. We also define the essential domain of f to be the set dom (f) = {x : f(x) < +∞} .
We say that f is strictly convex if the strict inequality (1.5) holds whenever x1, x2 ∈dom (f) are distinct.
Example: cTx, ∥x∥, ex, x2 The role of convexity in linking the global and the local in optimization theory is illustrated by the following result.
Theorem 1.10. Let f : Rn →¯ R be convex. If x ∈Rn is a local minimum for f, then x is a global minimum for f.
Proof. Suppose to the contrary that there is a b x ∈Rn with f(b x) < f(x). Since ¯ x is a local solution, there is an ϵ > 0 such that f(x) ≤f(x) whenever ∥x −x∥≤ϵ .
Taking ϵ smaller if necessary, we may assume that ϵ < 2∥x −b x∥.
Set λ := ϵ(2∥x −b x∥)−1 < 1 and xλ := x + λ(b x −x). Then ∥xλ −x∥≤ϵ/2 and f(xλ) ≤ (1 −λ)f(x) + λf(b x) < f(x). This contradicts the choice of ϵ and so no such b x exists.
□ Strict convexity implies the uniqueness of solutions.
Theorem 1.11. Let f : Rn →¯ R be strictly convex. If f has a global minimizer, then it is unique.
7 Proof. Let x1 and x2 be distinct global minimizers of f. Then, for λ ∈(0, 1), f((1 −λ)x1 + λx2) < (1 −λ)f(x1) + λf(x2) = f(x1) , which contradicts the assumption that x1 is a global minimizer.
□ If f is a differentiable convex function, much more can be said. We begin with the following lemma.
Lemma 1.2. Let f : Rn →¯ R be convex (not necessarilty differentiable).
(1) Given x, d ∈Rn the difference quotient (1.6) f(x + td) −f(x) t is a non-decreasing function of t on (0, +∞).
(2) For every x, d ∈Rn the directional derivative f ′(x; d) always exists and is given by (1.7) f ′(x; d) := inf t>0 f(x + td) −f(x) t .
Proof. We first assume (1) is true and show (2). Recall that (1.8) f ′(x; d) := lim t↓0 f(x + td) −f(x) t .
Now if the difference quotient (1.6) is non-decreasing in t on (0, +∞), then the limit in (1.8) is necessarily given by the infimum in (1.7). This infimum always exists and so f ′(x; d) always exists and is given by (1.7).
We now prove (1). Let x, d ∈Rn and let 0 < t1 < t2. Then f(x + t1d) = f x + t1 t2 t2d = f h 1 − t1 t2 x + t1 t2 (x + t2d) i ≤ 1 −t1 t2 f(x) + t1 t2 f(x + t2d).
Hence f(x + t1d) −f(x) t1 ≤f(x + t2d) −f(x) t2 .
□ A very important consequence of Lemma 1.2 is the subdifferential inequality. This inequal-ity is obtained by plugging t = 1 and d = y −x into the right hand side of (1.7) where y is any other point in Rn. This substitution gives the inequality (1.9) f(y) ≥f(x) + f ′(x; y −x) for all y ∈Rn and x ∈dom (f) .
The subdifferential inequality immediately yields the following result.
Theorem 1.12 (Convexity and Optimality). Let f : Rn →¯ R be convex (not necessarilty differentiable) and let ¯ x ∈dom (f). Then the following three statements are equivalent.
(i) ¯ x is a local solution to minx∈Rn f(x).
(ii) f ′(¯ x; d) ≥0 for all d ∈Rn.
8 (iii) ¯ x is a global solution to minx∈Rn f(x).
Proof. Lemma 1.1 gives the implication (i)⇒(ii). To see the implication (ii)⇒(iii) we use the subdifferential inequality and the fact that f ′(¯ x; y −¯ x) exists for all y ∈Rn to obtain f(y) ≥f(¯ x) + f ′(¯ x; y −¯ x) ≥f(¯ x) for all y ∈Rn.
The implication (iii)⇒(i) is obvious.
□ If it is further assumed that f is differentiable, then we obtain the following elementary consequence of Theorem 1.12.
Theorem 1.13. Let f : Rn →R be convex and suppose that x ∈Rn is a point at which f is differentiable. Then x is a global minimum of f if and only if ∇f(x) = 0.
As Theorems 1.12 and 1.13 demonstrate, convex functions are well suited to optimization theory. Thus, it is important that we be able to recognize when a function is convex. For this reason we give the following result.
Theorem 1.14. Let f : Rn →¯ R.
(1) If f is differentiable on Rn, then the following statements are equivalent: (a) f is convex, (b) f(y) ≥f(x) + ∇f(x)T(y −x) for all x, y ∈Rn (c) (∇f(x) −∇f(y))T(x −y) ≥0 for all x, y ∈Rn.
(2) If f is twice differentiable then f is convex if and only if ∇2f(x) is positive semi-definite for all x ∈Rn.
Remark: The condition in Part (c) is called monotonicity.
Proof. (a) ⇒(b) If f is convex, then 1.14 holds. By setting t := 1 and d := y −x we obtain (b).
(b) ⇒(c) Let x, y ∈Rn. From (b) we have f(y) ≥f(x) + ∇f(x)T(y −x) and f(x) ≥f(y) + ∇f(y)T(x −y).
By adding these two inequalities we obtain (c).
(c) ⇒(b) Let x, y ∈Rn. By the Mean Value Theorem there exists 0 < λ < 1 such that f(y) −f(x) = ∇f(xλ)T(y −x) where xλ := λy + (1 −λ)x. By hypothesis, 0 ≤ [∇f(xλ) −∇f(x)]T(xλ −x) = λ[∇f(xλ) −∇f(x)]T(y −x) = λ[f(y) −f(x) −∇f(x)T(y −x)].
Hence f(y) ≥f(x) + ∇f(x)T(y −x).
9 (b) ⇒(a) Let x, y ∈Rn and set α := max λ∈[0,1] ϕ(λ) := [f(λy + (1 −λ)x) −(λf(y) + (1 −λ)f(x))].
We need to show that α ≤0. Since [0, 1] is compact and ϕ is continuous, there is a λ ∈[0, 1] such that ϕ(λ) = α. If λ equals zero or one, we are done. Hence we may as well assume that 0 < λ < 1 in which case 0 = ϕ′(λ) = ∇f(xλ)T(y −x) + f(x) −f(y) where xλ = x + λ(y −x), or equivalently λf(y) = λf(x) −∇f(xλ)T(x −xλ).
But then α = f(xλ) −(f(x) + λ(f(y) −f(x))) = f(xλ) + ∇f(xλ)T(x −xλ) −f(x) ≤ 0 by (b).
2) Suppose f is convex and let x, d ∈Rn, then by (b) of Part (1), f(x + td) ≥f(x) + t∇f(x)Td for all t ∈R. Replacing the left hand side of this inequality with its second-order Taylor expansion yields the inequality f(x) + t∇f(x)Td + t2 2 dT∇2f(x)d + o(t2) ≥f(x) + t∇f(x)Td, or equivalently, 1 2dt∇2f(x)d + o(t2) t2 ≥0.
Letting t →0 yields the inequality dT∇2f(x)d ≥0.
Since d was arbitrary, ∇2f(x) is positive semi-definite.
Conversely, if x, y ∈Rn, then by the Mean Value Theorem there is a λ ∈(0, 1) such that f(y) = f(x) + ∇f(x)T(y −x) + 1 2(y −x)T∇2f(xλ)(y −x) where xλ = λy + (1 −λ)x. Hence f(y) ≥f(x) + ∇f(x)T(y −x) since ∇2f(xλ) is positive semi-definite. Therefore, f is convex by (b) of Part (1).
Convexity is also preserved by certain operations on convex functions. A few of these are given below.
Theorem 1.15. Let fi : Rn →¯ R be convex functions for i = 1, 2, . . . , m, and let λi ≥0, i = 1, . . . , m. Then the following functions are also convex.
(1) f(x) := φ(f1(x)), where φ : R →R is any non-decreasing function on R.
10 (2) f(x) := Pm i=1 λifi(x) (Non-negative linear combinations) (3) f(x) := max{f1(x), f2(x), . . . , fm(x)} (pointwise max) (4) f(x) := sup {Pm i=1 fi(xi) |x = Pm i=1 xi} (infimal convolution) (5) f ∗ 1(y) := supx∈Rn[yTx −f1(x)] (convex conjugation) 1.4.1. More on the Directional Derivative. It is a powerful fact that convex function are directionally differentiable at every point of their domain in every direction. But this is just the beginning of the story. The directional derivative of a convex function possess several other important and surprising properties. We now develop a few of these.
Definition 1.5. Let h : Rn →R ∪{+∞}. We say that h is positively homogeneous if h(λx) = λh(x) for all x ∈R and λ > 0.
We say that h is subadditive if h(x + y) ≤h(x) + h(y) for all x, y ∈R.
Finally, we say that h is sublinear if it is both positively homogeneous and subadditive.
There are numerous important examples of sublinear functions (as we shall soon see), but perhaps the most familiar of these is the norm ∥x∥. Positive homogeneity is obvious and subadditivity is simply the triangle inequality. In a certain sense the class of sublinear function is a generalization of norms. It is also important to note that sublinear functions are always convex functions. Indeed, given x, y ∈dom (h) and 0 ≤λ ≤1, h(λx + (1 −λ)y) ≤ h(λx) + h(1 −λ)y) = λh(x) + (1 −λ)h(y).
Theorem 1.16. Let f : Rn →R ∪{+∞} be a convex function.
Then at every point x ∈dom (f) the directional derivative f ′(x; d) is a sublinear function of the d argument, that is, the function f ′(x; ·) : Rn →R ∪{+∞} is sublinear. Thus, in particular, the function f ′(x; ·) is a convex function.
Remark: Since f is convex and x ∈dom (f), f ′(x; d) exists for all d ∈Rn.
Proof. Let x ∈dom (f), d ∈Rn, and λ > 0. Then f ′(x; λd) = lim t↓0 f(x + tλd) −f(x) t = lim t↓0 λf(x + tλd) −f(x) λt = λ lim (λt)↓0 f(x + (tλ)d) −f(x) (λt) = λf ′(x; d), showing that f ′(x; ·) is positively homogeneous.
11 Next let d1, d2 ∈Rn, Then f ′(x; d1 + d2) = lim t↓0 f(x + t(d1 + d2)) −f(x) t = lim t↓0 f( 1 2(x + 2td1) + 1 2(x + 2td2)) −f(x) t ≤ lim t↓0 1 2f(x + 2td1) + 1 2f(x + 2td2) −f(x) t ≤ lim t↓0 1 2(f(x + 2td1) −f(x)) + 1 2(f(x + 2td2) −f(x)) t = lim t↓0 f(x + 2td1) −f(x) 2t + lim t↓0 f(x + 2td2) −f(x) 2t = f ′(x; d1) + f ′(x; d2), showing that f ′(x; ·) is subadditive and completing the proof.
□ 12 Exercises (1) Show that the functions f(x1, x2) = x2 1 + x3 2, and g(x1, x2) = x2 1 + x4 2 both have a critical point at (x1, x2) = (0, 0) and that their associated hessians are positive semi-definite. Then show that (0, 0) is a local (global) minimizer for g and not for f.
(2) Find the local minimizers and maximizers for the following functions if they exist: (a) f(x) = x2 + cos x (b) f(x1, x2) = x2 1 −4x1 + 2x2 2 + 7 (c) f(x1, x2) = e−(x2 1+x2 2) (d) f(x1, x2, x3) = (2x1 −x2)2 + (x2 −x3)2 + (x3 −1)2 (3) Which of the functions in problem 2 above are convex and why?
(4) If f : Rn →¯ R = R ∪{+∞} is convex, show that the sets levf(α) = {x : f(x) ≤α} are convex sets for every α ∈R. Let h(x) = x3. Show that the sets levh(α) are convex for all α, but the function h is not itself a convex function.
(5) Show that each of the following functions is convex.
(a) f(x) = e−x (b) f(x1, x2, . . . , xn) = e−(x1+x2+···+xn) (c) f(x) = ∥x∥ (6) Consider the linear equation Ax = b, where A ∈Rm×n and b ∈Rm. When n < m it is often the case that this equation is over-determined in the sense that no solution x exists. In such cases one often attempts to locate a ‘best’ solution in a least squares sense. That is one solves the linear least squares problem (lls) : minimize 1 2 ∥Ax −b∥2 2 for x. Define f : Rn →R by f(x) := 1 2 ∥Ax −b∥2 2 .
(a) Show that f can be written as a quadratic function, i.e. a function of the form f(x) := 1 2x TQx −a Tx + α .
(b) What are ∇f(x) and ∇2f(x)?
(c) Show that ∇2f(x) is positive semi-definite.
(d) ∗Show that a solution to (lls) must always exist.
(e) ∗Provide a necessary and sufficient condition on the matrix A (not on the matrix ATA) under which (lls) has a unique solution and then display this solution in terms of the data A and b.
13 (7) Consider the functions f(x) = 1 2x TQx −c Tx and ft(x) = 1 2x TQx −c Tx + tφ(x), where t > 0, Q ∈Rn×n is positive semi-definite, c ∈Rn, and φ : Rn →R ∪{+∞} is given by φ(x) = −Pn i=1 ln xi , if xi > 0, i = 1, 2, . . . , n, +∞ , otherwise.
(a) Show that φ is a convex function.
(b) Show that both f and ft are convex functions.
(c) Show that the solution to the problem min ft(x) always exists and is unique.
(8) Classify each of the following functions as either coercive or non-coercive showing why you classification is correct.
(a) f(x, y, z) = x3 + y3 + z3 −xyz (b) f(x, y, z) = x4 + y4 + z2 −3xy −z (c) f(x, y, z) = x4 + y4 + z2 −7xyz2 (d) f(x, y) = x4 + y4 −2xy2 (e) f(x, y, z) = log(x2y2z2) −x −y −z (f) f(x, y, z) = x2 + y2 + z2 −sin(xyz) (9) Show that each of the following functions is convex or strictly convex.
(a) f(x, y) = 5x2 + 2xy + y2 −x + 2y + 3 (b) f(x, y) = (x + 2y + 1)8 −log((xy)2), if 0 < x, 0 < y, +∞, otherwise.
(c) f(x, y) = 4e3x−y + 5ex2+y2 (d) f(x, y) = x + 2 x + 2y + 4 y, if 0 < x, 0 < y, +∞, otherwise.
(10) Compute the global minimizers of the functions given in the previous problem if they exist.
□ |
190745 | https://www.euramet.org/download?tx_eurametfiles_download%5Baction%5D=download&tx_eurametfiles_download%5Bcontroller%5D=File&tx_eurametfiles_download%5Bfiles%5D=37681&tx_eurametfiles_download%5Bidentifier%5D=%252Fdocs%252FPublications%252Fcalguides%252FI-CAL-GUI-017_Calibration_Guideline_No._17_web.pdf&cHash=9bc897cc2a061f86b88891d775fba0e5 | Published Time: Tue, 21 Feb 2023 09:21:29 GMT
Guidelines on the Calibration of Electromechanical and Mechanical Manometers
EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )EURAMET Calibration Guide No. 17
Version 4.1(09 /20 22 )
I-CAL -GUI -017 /v 4.1/20 22 /09
Authorship and imprint
This document was developed by the EURAMET e.V., Technical Committee for Mass and Related
Quantit ies.
Authors: Wladimir Sabuga (PTB, Germany ), Oliver Ott (PTB, Germany), Yasin Durgut (UME, T urkey ),
Janez Šetina (IMT, Slovenia ), Pierre Otal (LNE, F rance ), Nieves Medina (CEM, Spain ), Aykurt Altintas
(FORCE Technology, Denmark ).
EURAMET e.V.
Bundesallee 100
38116 Braunschweig
Germany
E-mail: secretariat@euramet.org
Phone: +49 531 592 1960
Versions
Version 4.1 September 2022
Version 4.0 April 2019
Version 3.0 April 2017
Version 2.0 March 2011
Vers ion 1.0 July 2007
The changes since the previous version of the Guide concern the correction of errors on page 21 (a text
typo and an error in equation (20)) and page 23 (incorrect specification of measurement series below
equation (23b)).
Official language
The English language version of this document is the definitive version. The EURAMET Secretariat can
give permission to translate this text into other languages, subject to certain conditions available on
application. In case of any inconsistency between the terms of the translation and the t erms of this
document, this document shall prevail.
Copyright
The copyright of this document (EURAMET Calibration Guide No. 1 7, version 4.1 – English version) is held
by © EURAMET e.V. 20 22 . The English language version of this publication is the definiti ve version. The
text may not be copied for resale and may not be reproduced other than in full. Extracts may be taken only
with the permission of the EURAMET Secretariat.
ISBN 978 -3-942992 -72 -5
Image on cover page by PTB.
Guidance for Users
This document gives guidance on measurement practices in the specified fields of measurements. By
applying the recommendations presented in this document, laboratories can produce calibration results
that can be recogni sed and accepted throughout Europe. The ap proaches taken are not mandatory and
are for the guidance of calibration laboratories. The document has been produced as a means of promoting
a consistent approach to good measurement practice leading to and supporting laboratory accreditation.
The guide may be used by third parties, e.g. National Accreditation Bodies, peer reviewers, witnesses to
measurements, etc., as a reference only. Should the guide be adopted as part of a requirement of any such
party, this shall be for that application only and the EURAMET Secretariat should be informed of any such
adoption.
On request EURAMET may involve third parties in stakeholder consultations when a review of the guide is
planned. If you are interested, p lease contact the EURAMET Sec retariat. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
I-CAL -GUI -017/v4. 1/20 22 /0 9
No representation is made nor warranty given that this document or the information contained in it will be
suitable for any particular purpose. In no event shall EURAMET, the authors or anyone else involved in the
creation of the document be liable for any da mages whatsoever arising out of the use of the information
contained herein. The parties using the guide shall indemnify EURAMET accordingly.
Further information
For further information about this document, please contact your national contact person in t he EURAMET
Technical Committee for Mass and Related Quantities (see www.euramet.org ).EURAMET Calibration Guide No. 17
Version 4.1(09 /20 22 )
Conditions for the Use and Translation of EURAMET Publications
To stimulate international harmonisation of technical procedures in metrology, EURAMET e.V. welcomes
the use of its Calibration Guides and Technical Guides by other organisations, e.g. National Metrology
Institutes, Regional Metrology Organisations, National Accreditation Bodies, or Regional Accreditation
Organisations beyond Europe.
General Information
EURAMET e.V. holds the copyright on all documents developed withi n its committees.
EURAMET documents may be translated and / or published by other organisations with the permission of
EURAMET under the following conditions:
1) the use of the documents is for non -commercial purposes only ,
2) resulting d ocuments , in electroni c or printed form , shall be distributed free of charge ,
3) addenda, e.g. foreword or annexes, may be added, but must be clearly recognisable as such,
4) if a printed version is prepared, four copies shall be sent to the EURAMET Secretariat.
For national dissemination, EURAMET Members or Associates may add the name and/or logo of the
National Metrology Institute (NMI) or Designated Institute (DI) to the docum ent. The EURAMET Secretariat
must be informed in advance.
Permission to translate or distribute reformatted EURAMET documents, including Calibration Guides and
Technical Guides must be obtained from the EURAMET Secretariat in written form by e-mail to:
secretariat@euramet.org .
Publication of Calibration Guides and Technical Guides by other Organisations
If an organisation intends to publish a Guide in its own version,
1) the document shall not be modified , except for refer ence s to specific European standar ds,
organisations or situations . In this case , modifications must be recognisable as such, e.g. in a
footn ote,
2) the document may have the organisation’s own cover page and registration number. A reference
to the original document must be made directly under the registration number as follows: ‘This
document is identical to [title of the document, version number, publication year]. The copyright of
the orig inal version is held by © EURAMET e.V.’
Add itionally, the following rules apply if a document is translated into another language.
Translation of EURAMET Publications
If an organisation intends to translate a EURAMET publication,
1) the document shall not be modified and shall be clearly recognisable as a translation of the
corresponding EURAMET document,
2) reference must be made to the original document as follows: ‘This document is a translation of [title
of the document, version number, publication year]. The copyright of the original version is held by
© EURAMET e.V.’.
In case of any inconsistency between the terms of the translation and the terms of the original document
the original document shall prevail. EURAMET Calibration Guide No. 17
Version 4.1 (09 /20 22 )
EURAMET Calibration Guide No. 17
Version 4.1(09 /20 22 )
Guidelines on the Calibration of
Electromechanical and Mechanical
Manometers
Purpose
This document has been produced to enhance the equivalence and mutual recognition of
calibration results obtained by laboratories performing calibrations of electromechanical and
mechanical manometers. EURAMET Calibration Guide No. 17
Version 4 .1(0 9/20 22 )
Content
1 INTRODUCTION ................................ ................................ ................................ ................. 2
2 REFERENCE DOCUMENTS AND LITERATURE ................................ ............................... 2
3 DEFINITIONS ................................ ................................ ................................ ...................... 2
4 PRINCIPLES OF THE ELECTROMECHANICAL AND MECHANICAL MANOMETERS ...... 3
4.1 Pressure transducers ................................ ................................ ................................ .... 3
4.2 Pressure transmitters ................................ ................................ ................................ .... 3
4.3 Manometers with digital or analogue indication ................................ ............................. 3
4.4 Bourdon tube manometers ................................ ................................ ........................... 4
5 LABORATORY CALIBRATION PROCEDURES ................................ ................................ . 4
5.1 Installation of the equipment ................................ ................................ ......................... 4
5.2 Methods of calibration ................................ ................................ ................................ ... 5
5.2.1 Basic Calibration Procedure ................................ ................................ .............. 5
5.2.2 Standard Calibration Procedure ................................ ................................ ........ 5
5.2.3 Comprehensive Calibration Procedure ................................ .............................. 6
5.3 Equipment set -up ................................ ................................ ................................ .......... 6
5.3.1 Reference instrument ................................ ................................ ........................ 6
5.3.2 Mechanical set -up ................................ ................................ ............................. 6
5.3.3 Electrical set -up ................................ ................................ ................................ . 8
5.4 Calibration sequences ................................ ................................ ................................ 10
5.4.1 Preparatory work ................................ ................................ ............................. 10
5.4.2 Calibration procedures ................................ ................................ .................... 11
5.4.3 Presentation of results ................................ ................................ ..................... 13
6 DETERMINATION OF THE UNCERTAINTY OF MEASUREMENT ................................ ... 14
6.1 Common aspects of determining the uncertainty of measurement .............................. 14
6.2 Guidance on uncertainty calculation for selected practical cases ................................ 18
6.2.1 Calibration of a digital or analogue manometer ................................ ............... 18
6.2.2 Calibration of a pressure transducer with electrical output ............................... 20
6.3 Determination of the characteristic values significant for the uncertainty ..................... 22
7 EXAMPLES ................................ ................................ ................................ ....................... 24
7.1 Example 1 – Ca libration of an indicating digital pressure gauge ................................ . 24
7.1.1 Example 1a – Basic Calibration Procedure ................................ ...................... 26
7.1.2 Example 1b – Standard Calibration Procedure ................................ ................ 28
7.1.3 Example 1c – Comprehensive Calibration Procedure ................................ ...... 31
7.2 Example 2 – Calibration of a pressure transducer ................................ ....................... 33
7.2.1 Example 2a – Modelling the output si gnal by using a linear characteristic ....... 35
7.2.2 Example 2b – Determination of the transmission coefficient ............................ 37 EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-2-
1 INTRODUCTION
This document deals with the calibration of electromechanical and mechanical
manometers.
The Guidelines provide the users of electromechanical and mechanical manometers with
the fundamentals necessary for establishing and applying calibration procedures.
These Guidelines apply to all electromechanical manometers for measuring absolute,
posit ive and negative gauge and differential pressures, as well as to bourdon tube
manometers.
Notes: The Guidelines refer to the "measurement" function of a measuring pressure
controller in particular.
The Guidelines do not refer to piezoelectric pressure tra nsducers.
2 REFERENCE DOCUMENTS AND LITERATURE
DIN EN 472, Pressure gauges – Vocabulary, 1994
DIN EN 837 -1, Pressure gauges – Part 1: Bourdon tube pressure gauges – Dimensions,
metrology, requirements and testing, 1997
DIN EN 837 -2, Pressure gauges – Part 2: Selection and installation recommendations for
pressure gauges, 1997
DIN EN 837 -3, Pressure gauges – Part 3: Diaphragm and capsule pressure gauges –
Dimensions, metrology, requirements and testing, 1997
DIN EN 12645, Tyre pressure measuring instrum ents – Devices for inspection of pressure
and/or inflation/deflation of tyres for motor vehicles – Metrology, requirements and testing,
2015
DIN EN 61298 -2, Process measurement and control devices – General methods and
procedures for evaluating performance – Part 2: Tests under reference conditions, 2009
EA -4/02 M (rev 01), Evaluation of the uncertainty of measurement in calibration, 2013
EURAMET cg -3 (Version 1.0), Calibration of pressure balances, 2011
IEC 60770, Transmitters for use in industrial -process control; Part 1: Methods for
performance evaluation, 2010; Part 2: Methods for inspection and routine testing, 2010
ILAC P10:01/2013 ILAC policy on the traceability of measurement results, 2013
JCGM 100, GUM 1995 with minor corrections, Evaluation of meas urement data - Guide
to the expression of uncertainty in measurements, issued by BIPM, IEC, IFCC, ISO,
IUPAC, IUPAP and OIML, 2008 (revised in 2010)
JCGM 200 (3rd edition), International vocabulary of metrology - Basic and general
concepts and associated t erms (VIM), issued by BIPM, IEC, IFCC, ISO, IUPAC, IUPAP
and OIML, 2012
RM Aero 802 41, Calibration and check of electromechanical manometers, Bureau de
Normalisation de l'Aéronautique et de l'Espace, BNAE, 1993 (in French)
3 DEFINITIONS
In order to avoid ambiguity, the terms mentioned below have the following meanings:
Line pressure : Static pressure used as a reference for differential pressures.
Reference level : Level at which the value of the applied pressure is quantified. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-3-
Note: The m anufacturer of the instrument specifies this level. If this is not the
case, the calibration laboratory shall specify it.
4 PRINCIPLES OF THE ELECTROMECHANICAL AND MECHANICAL MANOMETERS
The Guidelines deal with three types of electromechanical manometers:
• pr essure transducers,
• pressure transmitters,
• manometers with digital or analogue indication
and with mechanical bourdon tube manometers.
4.1 Pressure transducers
Pressure transducers convert the measured pressure into an analogue electrical signal
that is proportional to the applied pressure.
According to the model, the output signal can be:
• a voltage,
• a current,
• a frequency.
To ensure that they function, the pressure transducers need a continuous power supply
stabilised to a level in relation to the expected uncertainty of the pressure measurement.
4.2 Pressure transmitters
A pressure transmitter is generally a unit that consists of a pressure transducer and a
module for conditioning and amplifying the transducer signal.
Depending on the model, the output information of a pressure transmitter may be:
• a voltage (5 V; 10 V; ...),
• a current (4 mA – 20 mA; ...),
• a relative resistance change (1 mV/V; ...),
• a frequency, or
• a digital format (RS 232; ...).
For operation, pressure transmitters need a continuous electrical supply, which need not
be specifically stabilised.
4.3 Manometers with digital or analogue indication
This type of manometer is a complete measuring instrument that indicates units of
pressure. According to the pattern, it may consist of the following units:
(a) Manometer with a digital indication:
• pressure transducer,
• analogue conditioning module,
• analogue -to -digital converter,
• digital processing module, EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-4-
• digital indication (in the unit(s) specified by the manufacturer),
• electrical power supply unit (generally incorporated).
(b) Manometer with an analogue indication:
• pressure transducer,
• analogue conditioning module,
• analogue indicating module,
• electrical power supply unit (generally incorporated).
These elements may be a ccommodated in one housing (internal transducer) or constitute
separate devices, one of which is the transducer (external transducer).
The manometers may also be equipped with analogue or digital output ports.
Note: Calibration of such an instrument shoul d be performed for each output of interest.
4.4 Bourdon tube manometers
This type of manometer is a complete measuring instrument that indicates units of
pressure. It consists of the following units:
• pressure -responsive element (Bourdon tube),
• movement,
• pointe r,
• dial.
These parts shall be installed inside an enclosing case.
5 LABORATORY CALIBRATION PROCEDURES
5.1 Installation of the equipment
The following instructions should be followed when installing the equipment:
• Protect the equipment from direct sunlight.
• Place the instrument to be calibrated as close as possible to the reference
standard.
• Ensure that the pressure reference levels of both instruments are as close as
possible and account for the difference in the pressure reference level when
calcu lating corrections and uncertainties.
• The equipment should be switched on in the calibration environment before
starting the calibration in order to reach thermal equilibrium in the whole system.
• Take into account the manufacturer's specification for mount ing position, torque,
warm -up, for example.
The calibration is to be carried out after temperature equalization between calibration item
and environment. A period for warming up the calibration item or potential warming -up of
the calibration item due to t he supply voltage is to be taken into account. The calibration
is to be performed at an ambient temperature stable to within ±1 °C; this temperature must
lie between 18 °C and 28 °C and is to be recorded. If the air density has an effect on the
calibration result, not only the ambient temperature but also the atmospheric pressure and
the relative humidity are to be recorded. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-5-
5.2 Methods of calibration
If appropriate, the procedure of calibration should allow – according to the client's
requirements – the evaluation of the hysteresis, the linearity and the repeatability of the
instrument to be calibrated.
The applied procedure depends on the target uncertainty of the instrument according to
the client's requirements and can limit the lowest achievable unce rtainty.
Generally, the calibration is to be carried out in pressure points uniformly distributed over
the calibration range. In the Sections 5.2.1, 5.2.2 and 5.2.3, there are exemplary
recommendations for a calibration range from 0 % FS (full scale) to 1 00 % FS including
the zero point as one point to illustrate the procedures.
The comparison between the measurement values of the instrument to be calibrated and
the reference standard can be performed by two different methods:
• adjustment of the pressure according to the indication of the instrument to be
calibrated,
• adjustment of the pressure according to the indication of the reference standard.
5.2.1 Basic Calibration Procedure
The Basic Calibration Procedure should be used for instruments where the expanded
measurement target uncertainty ( k = 2) is U ≥ 0.2 % FS. Calibration is performed in one
measuring cycle at six different pressure points consisting of a series of increasing
pressure and a series of decreasing pressure.
Repeatability is estimate d from calibration in a measuring series of increasing pressure at
two pressure points (preferably 0 % FS and another one around the centre of the
calibrated pressure range, e.g. 40 % FS or 60 % FS) that is measured three times. The
repeatability value is then applied to all other pressure points.
Using the Basic Calibration Procedure the expanded measurement uncertainty ( k = 2) of
the calibrated instrument may not be reported smaller than 0.2 % FS.
5.2.2 Standard C alibration Procedure
The Standard Calibration P rocedure should be used for instruments where the expanded
measurement target uncertainty ( k = 2) is 0.05 % FS ≤ U < 0.2 % FS. Calibration is
performed in one measuring cycle at 11 pressure points consisting of a series of
increasing pressure and a series of decreasing pressure.
Repeatability is estimated from calibration in a measuring series of increasing pressure at
four pressure points (preferably 0 % FS, 20 % FS, 50 % FS and 80 % FS) measured three
times. At the remaining pressure points, the biggest of the repeatability values as
measured in the neighbouring points is taken :
• At 10 % FS, the repeatability at 0 % FS or at 20 % FS is taken, whichever is bigger.
• At 30 % FS and 40 % FS, the repeatability at 20 % FS or at 50 % FS is taken,
whichever is bigg er.
• At 60 % FS and 70 % FS, the repeatability at 50 % FS or at 80 % FS is taken,
whichever is bigger.
• At 90 % FS and 100 % FS, the repeatability at 80 % FS is taken.
Using the Standard Calibration P rocedure the expanded measurement uncertainty ( k = 2)
of the calibrated instrument may not be reported smaller than 0.05 % FS. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-6-
5.2.3 Comprehensive Calibration Procedure
The Comprehensive Calibration Procedure should be used for instruments where the
expanded measurement target uncertainty ( k = 2) is U < 0.0 5 % FS. Calibration is
performed at 11 pressure points in three measuring cycles, each consisting of a series of
increasing pressure and a series of decreasing pressure.
5.3 Equipment set -up
5.3.1 Reference instrument
The reference instrument shall comply with the following requirements:
• It shall be traceable to national or international standards.
• Its uncertainty shall be better (if practicable) than that of the instrument to be
calibrated, the ratio being in general equal to or greater than 2.
5.3.2 Mechanical set -up
For calibrations in gaseous media it is strongly recommended to use a pressurised
container with dry and clean gas as the pressure source. The container must be equipped
with a pressure -reducing valve or connected to a pressure control valve if required by the
measurement range of the instrument to be calibrated.
In order to ensure the quality of the gas, the vacuum pump – if required – shall be equipped
with accessories such as traps and isolating valves.
5.3.2.1 Positive gauge pressure in gaseous media
The typical set -up may be as follows (see Figure 1):
1reference standard
2instrument to be calibrated, mounted as normally used
3fine -regulated inlet valve
4fine -regulated pressure relief valve
5volume regulator
6pressure source
Figure 1 – Set -up in positive gauge pressure, gaseous media
The required pressure is roughly established using inlet or outlet valves depending on
whether the pressure is supposed to be set up from low pressure or from high pressure.
The final pressure adjustment is performed using a volume regulator. 56
3
4
21EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
7 -
5.3.2.2 Absolute and negative gauge pressure in gaseous media
The typical set -up may be as follows (see Figure 2):
1 reference standard
2 instrument to be calibrated, mounted as normally used
3 fine -regulated inlet valves
4 fine -regulated pressure relief valve
5 volume regulator
6 pressure source
7 vacuum pump (for pressures below atmospheric pressure)
Figure 2 – Set -up in absolute and negative gauge pressure, gaseous media
The required pressure is roug hly established using inlet or outlet valves depending on
whether the pressure is supposed to be set up from low pressure or from high pressure.
The final pressure adjustment is performed using a volume regulator.
In the case of absolute pressures signifi cantly higher than the atmospheric pressure, the
use of a gauge pressure reference standard and a barometric pressure -measuring
reference standard is acceptable. The set -up recommended for positive gauge pressures
is applicable. The value of the absolute p ressure is obtained by the summation of the
values of the pressures measured with the two reference standards.
5.3.2.3 Differential pressure in gaseous media
The set -up depicted in Figure 3 is recommended.
The required line pressure is roughly established using i nlet or outlet valves depending on
whether the pressure is supposed to be set up from low pressure or from high pressure.
The final pressure adjustment is performed using a volume regulator. During this
procedure the bypass valve is open.
The required diff erential pressure is set up using one of the volume regulators.
Instead of using two reference standards, a differential pressure standard or a twin
pressure balance may be used.
3
6
7
3
5
21
4EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
8 -
A vacuum pump arranged downstream of the inlet valve allows the line pressure to be
lower than the atmospheric pressure.
1 two reference standards (or a differential pressure standard 1 ')
2 instrument to be calibrated, mounted as normally used
3 bypass valve
4 fine -regulated i nlet valves
5 fine -regulated pressure relief valves
6 two volume regulators
7 vacuum pump (for line pressures below atmospheric pressure)
8 pressure source
Figure 3 – Set -up in differential pressure, gaseous media
5.3.2.4 Hydraulic pressure
The set -up for hydraulic gauge pressure and hydraulic differential pressure is basically the
same as that for gaseous media (cf. Sections 5.3.2.1 and 5.3.2.3) with the following
options:
• the relief valves are replaced with discharge valves connected to a reservoir of
pre ssure transmitting fluid,
• the pressure sources are replaced by screw press and/or priming pump,
• the vacuum pump (cf. Figure 3) is not required.
For absolute liquid pressures, refer additionally to the last paragraph of Section 5.3.2.2.
5.3.3 Electrical set -up
Th is section refers only to transducers and transmitters with an analogue output signal.
If the transducer being calibrated is equipped with a signal conditioner, then follow the
manufacturer's instructions concerning the electrical set -up.
If no signal c onditioner is available, a relevant data sheet with the manufacturer's
specifications should be available.
If applicable, the voltmeter and the reference standard resistor shall be calibrated and
traceable to the corresponding national/international stand ard.
535
66
4
8
7
4
112
1’ EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-9-
In every case, it is important to follow the recommendations concerning the electrical
shielding, to ensure the quality of the connections (of the "low -level" transducers in
particular), to meet the safety requirements. Some instruments may be suppli ed with a
power supply system or are supposed to be connected to such a system.
According to the instrument type, various set -ups are possible. This guide deals only with
the three most typical set -ups which are presented below.
5.3.3.1 Two -wire transmitters
Gene rally, this is the case of instruments with DC loop (4 – 20) mA. However, some other
output signals (0 to 10 mA, 0 to 20 mA or 0 to 50 mA) are applicable.
The typical set -up may be as follows (see Figure 4):
1 transmitter
2power supply
3measurement (reading unit)
Figure 4 – Electrical set -up, two -wire transmitters
The current I can be determined by measuring the voltage V at the terminals of a
calibrated standard resistor R in the circuit via RVI = .
It is recommended to follow the manufact urer's instructions concerning the values of the
power supply voltage and the resistor or the client's specifications when appropriate.
5.3.3.2 Three -wire transmit ters or transducers
These are generally instruments with a Wheatstone bridge. The typical set -up may be as
follows (see Figure 5):
1 transmitter or transducer
2power supply
3measurement (reading unit)
Figure 5 – Electrical set -up, three -wire transmitters or transducers EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-10 -
For the selection of the power supply and the voltage -measuring instrument, it is
recommended to follow the manufacturer's specifications. The resistor of this instrument
shall, however, be sufficiently high (at least 10 times) compared with the internal
resistance of the transmitter or transducer.
5.3.3.3 Four -wire transmitte rs or transducers
These are generally instruments with a Wheatstone bridge.
The typical set -up is as follows (see Figure 6):
1 transmitter or transducer
2power supply
3measurement (reading unit)
Figure 6 – Electrical set -up, four -wire transmitters or transducers
As the output signal is a low -level signal, it is important to ensure the appropriate quality
of the earth connections and of the shielding.
Variants:
• The output signal is an amplified signal from the amplifier (high -level ou tputs)
incorporated in the transmitter.
• Some instruments may include a probe for temperature compensation; the output
of this probe may consist of one or two supplementary wires.
5.4 Calibration sequences
5.4.1 Preparatory work
Prior to the calibration itself, the good working condition of the instrument shall be
visually confirmed, especially regarding:
• the inscriptions,
• the instrument being undamaged, the cleanliness of the instrument (free of
contamination),
• the good quality of the electrical contacts ,
• the readability of indications, the setting elements adjusted in defined positions
• the documents necessary for calibration (technical data, operating instructions). EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-11 -
It is recommended to perform the following operations:
• identify the reference levels
o of the reference,
o of the instrument to be calibrated (at the level of the connection or at the
reference level defined by the manufacturer),
• minimise the difference between the reference levels,
• for torque sensitive devices, follow the manufacturer' s instructions,
• make sure that there is no leakage by checking the set -up at the upper pressure
limit, the pressure must be stable .
5.4.2 Calibration procedures
In the case of instruments with several outputs, it is sufficient to perform the calibration for
the output(s) specified by the user.
Irrespective of the instrument to be calibrated and of the procedure to be used (refer to
Section 5.2), the operations are performed in three successive steps:
• check of a limited number of pressure points of the measurement range to
determine the initial metrological condition of the instrument,
• adjustment of the instrument according to the manufacturer's specification,
• calibration appropriate to the instrument over its whole measurement range or
span.
Each of these operatio ns, especially the adjustment of the instrument, shall be performed
only with the agreement of the client and shall be reported in the calibration certificate.
The general proceeding for calibration measurements is as follows:
• Prior to the calibration procedure itself, bring the instrument at least twice to its
upper pressure limit and keep the pressure for at least one minute; the time
between the preloadings should be at least one minute. The preloadings should
be repeated after a change of the mounting for the determination of the
reproducibility.
• After preloading and after reaching steady -state conditions the instrument is set
to zero, if possible. The zero reading is carried out immediately afterwards.
• For the variation of the pressure points in a measurement series, the time
between two successive pressure points should be the same and should not be
shorter than 30 seconds. The reading should be made 30 seconds after the end
of the pressure change at the earliest.
• The measurement value for the upper limit of the calibration range is to be
recorded prior to and after the waiting time of at least two minutes (five minutes
for bourdon tube manometers).
• The zero reading at the end of a measuring cycle is made 30 seconds after
complete relief at the earliest.
• The adjustment of the pressure during the repeated measuring series should be
as close as possible to the values of the individual pressures of the first
measuring series. The difference between the reference pressures in the
repeated mea suring series must not be bigger than 1 % of the calibration range.
Note: The reading of a bourdon tube manometer shall be obtained after the gauge has EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-12 -
been lightly tapped in order to minimise the error caused by the friction of
movement components.
5.4.2.1 Initia l check
To determine the long -term drift of the instrument, it is necessary to provide the user with
some information on its condition prior to any potential adjustment.
If the user does not apply for a complete calibration to be carried out prior to the
adjustment, it is recommended to perform the following operations:
• operate the instrument and preload it as described above,
• during the first pressure rise, check the indication obtained for conformity with the
specifications,
• read the indications of the i nstrument at 0 %, 50 % and 100 % of its measurement
span.
5.4.2.2 Adjustment
If the response of the instrument does not conform to the conventional response during
the initial check in any of the check points (5.4.2.1), e.g.:
• for a digital manometer with direct re ading, if there is a difference between the
indicated pressure and the applied pressure,
• for a transmitter with electrical output, if there is a deviation from the conventional
signal of, for example, 4 mA to 20 mA,
• for a bourdon tube manometer with direct reading, if there is a difference between
the indicated pressure and the applied pressure based on manometer accuracy
class,
then perform an adjustment of the instrument according to the client’s requirements.
Depending on the capabilities of the c alibration laboratory such a procedure shall be
performed:
• with the aid of the means normally accessible to the user (potentiometers for zero
and full scale, sometimes with mid -scale),
• with the internal adjustment facilities of the instrument (potentiomete rs, storage
of a calibration curve, etc.), in conformity with the information contained in the
technical description, after agreement of the client.
Note: This operation obviously presumes a detailed knowledge of the adjustment
procedures and requires spec ialised operators and calibration means that are more
powerful than the instrument to be calibrated.
If the instrument provides scale marks which are useful to the user (calibration notches,
restitution of a calibration curve for example), it is recommende d to determine these
elements in order to report them in the calibration certificate.
5.4.2.3 Main calibration
The calibration procedure to be used (cf. Section 5.2) is selected according to the target
uncertainty of measurement for the instrument to be calibrated .
At each calibration point at least the following data shall be recorded:
• the pressure indicated by the reference instrument or the elements necessary for
calculating the pressure actually measured (values of masses and temperature EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
13 -
for a pressure balance , for example),
• the indication of the instrument to be calibrated.
The following data shall also be recorded:
• the identification parameters of the reference instrument,
• the values of the influence quantities such as temperature, atmospheric pressure
and re lative humidity,
• the identification parameters of the instrument to be calibrated,
• the identification of the instruments included in the measuring system and/or the
instrument used for measuring the output signal.
5.4.3 Presentation of results
In general, it is recommended to present the results of a calibration in a form that can be
easily evaluated by the users of the measuring instrument under calibration. It is essential
to present clearly the results of the calibration and the methods of model ling or
interpolation (if applicable).
In order to take into account a specific method of measurement uncertainty evaluation and
calculation, the results are presented differently depending on whether the measuring
instrument under calibration provides:
• an output signal in an electric unit (pressure transducers and transmitters),
• an indication in a pressure unit (manometers with digital or analogue indication).
Table 1
Calibration results Model
Applied
pressure
pref
Applied
pressure
pref
Mean of
output
signal
Standard
deviation
of output
signal
Modelled
indicated
pressure
pind,mod
Deviation
pind,mod –
pref
Expanded
uncertainty of
deviation
1) 2) 3) 3), 4) 5) 5) 5), 6)
Increasing
pressure
Decreasing
pressure
1) The pressure measured by the reference instrument at the reference level of the instrument
to be calibrated, expressed in pascals or multiples. Instead of this column, the conversion
coefficient of the instrument pressure unit to the pascal can be given.
2) The pressure measured by the reference instrument at the reference level of the instrument
to be calibrated, expressed in the unit of the output signal of the instrument to be calibrated.
3) Value expressed in the unit of the output signal of the instrument to be calibrated.
4) Calculated at every measurement point if at least three values are available.
5) Value expressed in the pressure unit of the instrument to be calibrated. Reporting the model
in the calibration certificate is o ptional.
6) The uncertainty determined according to Section 6. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
14 -
Table 2
Applied
pressure
pref
Applied
pressure
pref
Indicated
pressure
pind
Standard
deviation of
Measurement
Deviation
pind -pref
Expanded
uncertainty of
deviation
1) 2) 3) 3) 3) 3), 4)
Increasing
pressures
Decreasing
pressures
1) The pressure measured by the reference instrument at the reference level of the instrument
to be calibrated, expressed in pascals or multiples. Instead of this column, the conversion
coefficient of the instrument pressure unit to the pascal can be given.
2) Pressure measured by the reference instrument at the reference level of the instrument to be
calibrated, expressed in the pressure unit of the instrument to b e calibrated.
3) Value expressed in the pressure unit of the instrument to be calibrated.
4) Evaluated according to Section 6.
5.4.3.1 Case of pressure transducers and transmitters
Whatever the modelling is, calibration results may be presented in the form of Table 1.
It should be noted that the standard deviation of the input signal (generally very small) is
not presented in this table because the deviation is taken into account in the uncertainty
of the measurements performed with the reference instrument.
5.4.3.2 Ca se of manometers with digital or analogue indication
Calibration results may be presented in the form of Table 2.
6 DETERMINATION OF THE UNCERTAINTY OF MEASUREMENT
6.1 Common aspects of determining the uncertainty of measurement
The principal elements to be taken into account for the evaluation of the uncertainty of the
calibration result for an electromechanical or mechanical manometer are,
for a pressure transducer or transmitter:
• uncertainty of the reference instrument in the conditions of use (cf. calibra tion
certificate, long -term stability, environmental conditions, for example),
• uncertainty due to zero error,
• uncertainty due to repeatability,
• uncertainty due to reproducibility of the instrument under calibration where
applicable,
• uncertainty due to reve rsibility (hysteresis) of the instrument under calibration,
• uncertainty of the measuring instruments used during the calibration (voltage,
current, frequency, etc.) including their resolution,
• uncertainty due to influence quantities,
• uncertainty due to pow er supply for the low -level transducers (in the case where EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-15 -
the output signal is proportional to the supply voltage the uncertainty of
measurement and the short -term stability of the supply voltage have to be taken
into account),
• uncertainty due to modelli ng (standard deviation of measurement results from the
model equation),
• uncertainty due to estimation of the head correction between the instrument to be
calibrated and the reference instrument;
for a manometer with digital or analogue indication:
• uncerta inty of the reference instrument in the conditions of use (cf. calibration
certificate, long -term stability, environmental conditions, for example),
• uncertainty due to zero error,
• uncertainty due to repeatability,
• uncertainty due to reproducibility of the instrument under calibration where
applicable,
• uncertainty due to the resolution of the instrument to be calibrated,
• uncertainty due to reversibility (hysteresis) of the instrument under calibration,
• uncertainty due to estimation of the head correction between the instrument to be
calibrated and the reference instrument.
Procedure
The uncertainty of the calibration results shall be evaluated following the principles
published in the EA document 4/02 and JCGM 100.
When analysing the u ncertainty budget, the terms and rules of calculation stated in
Table 3 are used assuming that no correlation between the input quantities must be taken
into consideration .
Relative uncertainties are stated by the variables w, W to distinguish them from th e
absolute uncertainties u, U.
In addition to this general rule of calculating uncertainties, there are two special cases
which lead to a simple quadratic addition of either the absolute uncertainties u, U or the
relative uncertainties w, W. In the case of the sum/difference model, the sensitivity
coefficients take the value 1=ic ; in the case of the product/quotient model the sensitivity
coefficients result in 1−
= xyci .EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
16 -
Table 3
Model function ),... ,f( Nxxxy 21=
Standard
uncertainty of
measurement ( )ixu
the standard uncertainty
associated with the input quantity
xiic
sensitivity coefficient ii xc f( )yui
contribution to the standard
uncertainty associated with the
result, caused by the standard
uncertainty ( )ixu of the input
quantity xi)()( iii xucyu ( )yu
standard uncertainty associated
with the result =
= Nii yuyu 122 )()(=
= Nii yuyu 12 )()(
Expanded
uncertainty of
measurement ( )yU
expanded uncertainty of
measurement )()( yukyU =
k coverage factor 2=k 1
Sum/difference model =
+= NiiXXY 1
(1)
Y output quantity
X input quantity/quantities on which the measurand depends
Xi unknown uncorrected error(s) 0=iXE
expected values (no contributions to the output quantity but
to the uncertainty of measurement)
This model is suited to determine, for example, the deviations of indicating pressure
gauges (cf. Section 6.2 for the meaning of the indices): =
+−= Niipppp 1ref ind
(2)
1
The expanded uncertainty of measurement ( )yU shall encompass the shortest possible interval with a
coverage probability of approximately 95 %. The coverage factor k is implicitly defined by ( ) ( ) yukyU = .
If, as is usually the case in practice, the probability distribution associated with t he measurand is normal
(Gaussian), then ( )yU shall be taken as ( ) yu2 , i.e. k = 2. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
17 -
Product/ quotient model =
= NiiKXY 1
(3)
Y output quantity
X input quantity/quantities on which the measurand depends ( )jii XXK
+= 1
correction factor(s)
Xi unknown uncorrected error(s)
Xj quantity/quantities to which
Xi is related 0=iXE
; 1=iKE expected values (no contributions to the output quantity but
to the uncertainty of measurement)
This model is suited to determine, for example, the transmission coefficient S of a pressure
transducer with electrical output using preferably relative uncertainties of measurem ent
for the uncertainty analysis: =
== NiiKpVGVXXS 1ref PS ind input output )/(
(4)
Input quantities
The uncertainties of measurement associated with the input quantities are grouped into
two categories according to the way in which they have been determined:
Type A: The value and the associated standard uncertainty are determined by methods
of statistical analysis for measurement series carried out under repeatability
conditions.
Type B: The value and the associated standard uncertainty are determined on the basis
of other information, for example:
• previous measurement data (for example, from type approvals),
• general knowledge of and experience with the properties and the behaviour of
measuring instruments and materials,
• manufacturer’s specifications,
• calibration cer tificates or other certificates,
• reference data taken from handbooks.
In many cases, only the upper and lower limi ts a+ and a– can be stated for the value xi of
a quantity, and a probability distribution with constant probability density between these
lim its can be assumed. This situation is described by a rectangular probability distribution.
With aaa 2=− −+
(5)
the estimate of the input quantity ( )−+ += aaxi 21
(6)
and the attributed standard uncertainty EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
18 -( ) 3
axu i =
(7)
are obtained.
6.2 Guidance on uncertainty calculation for selected prac tical cases
6.2.1 Calibration of a digital or analogue manometer
Choice of the model
The sum/difference model is used to determine the indication deviation and its uncertainty
separately for values measured at increasing (index up) and decreasing (index dn, for
down) pressure: up/dn repeat, zero up/dn ref, up/dn ind, 21up/dn ref, up/dn ind, up/dn pppppppp ii
++−=+−= =
(8)
The symbols are explained in Table 4.
Table 4
Y = pup/dn/mean measurand; deviation of the indication
X1 = pind,up/dn/mean indication of the pressure gauge
X2 = pref,up/dn/mean pressure generated by reference standard 2
X1 =
pzero unknown uncorrected measurement error due to zero error
X2 =
prepeat,up/dn/mean unknown uncorrected measurement error due to repeatability
The limited resolution of the indication (in Table 5 given by the variability interval 2
a =
r
for a device with digital indication) has to be taken into account as part of the uncertainty
analysis.
The mean values of the indication and the reference pressure are obtained by :2
dn ind, up ind, mean ind,
ppp +=
,2
dn ref, up ref, mean ref,
ppp += (9)
To calculate the deviation Δpmean of the mean indication, the contribution of the hysteresis
effect has to be taken into account:
X3 =
physteresis unknown uncorrected measurement error due to hysteresis hysteresis mean repeat, zero mean ref, mean ind, 31mean ref, mean ind, mean
ppppppppp ii
+++−=+−= =
(10)
In addition, a contribution of the reproducibility after reinstallation has to be taken into
2
The pressure generated by the reference standard in the reference level of the calibration object must be
corrected for the influence of the conditions of use. In consequence, the uncertainty analysis also covers
uncertainty components which take the difference between reference and calibration conditions into
account. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
19 -
account where applicable:
X4 =
preprod,up/dn/mean unknown uncorrected measurement error due to reproducibility
Uncertainty calculation
When the series at increasing and decreasing pressures are analysed separately, the
expanded uncertainty of measurement ( k = 2) becomes 2up/dn repeat, 2zero 2up/dn ind, 2ref up/dn uuuukpU +++= )(
(11)
For the usage of symbols refer also to Table 5.
In the applications of the calibration object it is often useful to combine the expa nded
uncertainty U with the deviation p. This provides information about the maximum
deviation of one single measurement result from the correct value (as issued from the
value that would have been measured with the standard instrument).
For this purpose , a so -called error span 3U can be defined: up/dn up/dn up/dn ppUpU += )()(
(12)
To calculate the uncertainty of the mean values of the increasing and decreasing pressure
series, the contribution of the hysteresis effect must be included: 2hysteresis 2mean repeat, 2zero 2mean ind, 2ref mean uuuuukpU ++++= )(
(13)
The largest uncertainty of the repeatabilities urepeat,up/dn at each calibration pressure is used
as the value urepeat, mean . The error span )( mean
pU is obtained accordingly :mean mean mean ppUpU += )()(
(14)
Information available about the input quantities
The knowledge about the input q uantities can be summarised in a table (Table 5
exemplifies a device with digital indication).
Statement of a single value
In addition to the error span for each calibration pressure, the maximum error span in the
range covered by the calibration (in pre ssure units or related to the measured value or the
measurement span) may be stated. Compliance with specified maximum permissible
errors can also be confirmed (statement of compliance).
3
The error span is the maximum difference to be expected between the m easured value and the
conventional true value of the measurand. The error span can be used in technical specifications to
characterise the accuracy of the calibrated instrument. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
20 -
Table 5
No. Quantity Estimate Unit 4 Varia -
bility
interval
Proba -
bility
distri -
bution
Divi -
sor
Standard
uncer -
tainty
Sensi -
tivity
coeffi -
cient
Contribution to
uncertainty
Xi xi 2a( )i
xP( )i
xu ci( ) yui
1 Pind,up/dn/mean pind,up/dn/mean, j Pa
r
(reso -
lution)
rectan -
gular 32
231 r 1 uind ,up/dn/mean
2 pref,up/dn/mean pref,up/dn/mean, j Pa normal 2)( j
pu ref, -1 uref
3
pzero 0 Pa f0
rectan -
gular 320
231 f 1 uzero
4
prepeat,up/dn/mean 0 Pa b’ up/dn/mean
rectan -
gular 32
231 b 1 urepeat,up/dn/mean
5
physteresis 0 Pa h rectan -
gular 32
231 h 1 uhysteresis
Y pup/dn/mean Pa ( ) yu
Notes: The formulae recommended to determine the quantities f0, b’ and h from a limited set of
measured data are defined by equations (21) to (28) in the section Determination of the
characteristic values significant for the uncertainty .
If sufficient data are available, the repeatability should be expressed by the empirical
standard deviation.
For a device with analogue indication, the variability interval 2 a for No. 1 is equal to 2 r (refer
to Section 6.3).
6.2.2 Calibration of a pressure transducer with electrical output
Choice of the model
Usually the dependence of the output qu antity of a pressure transducer (any electrical
quantity) on the input quantity (the pressure) is described by a so -called characteristic
Y = f(p), generally a straight line passing through Y = 0 or some defined point Y = Y0 and
having a slope adjusted by the manufacturer to meet a specified value within certain limits .
The calibration of the pressure transducer can now be based on the model equation +−= iYpYY
)f( ref ind
(15)
where the function f( p) is regarded as defined in the mathematical se nse, i.e. in the case
of a polynomial by coefficients without uncertainties, and the output quantity Yind has
values yind measured at the calibration pressures pref obtained from the standard.
Equation (15) corresponds to equation (8) and the sum/differen ce model can be used to
determine the deviation Y and its uncertainty separately for values measured at
increasing and decreasing pressure or for the mean values. However, contributions ( )ind Yu
must be included to account for the measurement uncertainty of the instruments used to
measure the output signal of the transducer.
4
It is recommended to state the unit of the uncertainty contributions (unit of the physical quantity, unit of
indication, related (dimensionless) quantity, etc.). EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
21 -
A formally different approach is to determine the transmission coefficient S, using the
product/quotient model – again separately for values measured at increasing and
decreasing pressures: up/dn repeat, zero up/dn ref, PS up/dn ind, 21up/dn ref, PS up/dn ind, input up/dn output, up/dn
KKpGV VKpGV VXXS ii
)/( )/(
=== =
(16)
Table 6
Y = Sup/dn/mean measurand; transmission coefficient
X1 = Vind,up/dn/mean indication of the output device (voltmeter)
X2 = G transmission coefficient of amplifier
X3 = VPS power supply voltage (auxiliary device)
X4 = pref,up/dn/mean pressure generated by the reference standard
K1 = Kzero correction factor for zero error
K2 = Krepeat,up/dn/mean correction factor for repeatability
K3 = Khysteresis correction factor for hysteresis
K4 = Kreprod,up/dn/mean if appropriate, correction factor for reproducibility (for example, when
the effect of torque is estimated during the calibration)
The corresponding result for the mean values of the transmission coefficient is obtained
by including the correction factor for hysteresis: hysteresis repeat zero mean ref, PS mean ind, 31mean ref, PS mean ind, input mean output, mean
KKKpGV VKpGV VXXS ii
)/( )/(
=== =
(17)
Uncertainty calculation
When the increasing and decreasing pressure series are analysed separately, the relative
expanded uncertainty ( k = 2) of the transmission coefficient is obtained as 2up/dn repeat, 2zero 2PS 22up/dn ind, 2ref up/dn wwwwwwkSW G +++++=)(
(18)
For the usage of symbols refer also to Table 7.
When the mean value of the increasing and decreasing pressure series is used, 2hysteresis 2mean repeat, 2zero 2PS 2G2mean ind, 2ref mean wwwwwwwkSW ++++++=)(
(19)
the largest uncertainty of the repeatabilities wrepeat,up/dn at each calibration pressure is used
as the value wrepeat,mean .
The relative error span is ( ) = + mean mean mean mean
( ) SW S W S S
(20) EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
22 -
with 0mean mean SSS −=
The single transmission coefficient S0 is preferably the slope of the straight line fitted
through all measured values of the output signal.
Information available about the input quantities
The knowledge about the input quantities can be summarised in a table:
Table 7
No. Quantity Estimate Unit Variability
interval
Proba -
bility
distri -
bution
Divi -
sor
Relative standard
uncertainty
Expo -
nent
asso -
ciated
with Xi
Contribution to
uncertainty
Xi xi 2a( )i
xP( )i
xw1
ii
c x y −( ) ywi
1 Vind,up/dn/mean Vind,up/dn/mean, j V normal 2)( j
Vw ,up/dn/mean ind, 1 wind,up/dn/mean
2 G G G normal 2)(Gw -1 wG
3 VPS VPS V normal 2)( PS
Vw -1 wPS
4 pref,up/dn/mean pref,up/dn/mean, j Pa normal 2)( j
pw ref, -1 wref
5 Kzero 1 1 f0,rel.
rectan -
gular 32r e l. 0 ,
231
f 1 wzero
6 Krepeat,up/dn/mean 1 1 b’ up/dn/mean,rel.
rectan -
gular 32r e l.
231
b 1 wrepeat,up/dn/mean
7 Kreprod ,up/dn/mean 1 1 bup/dn/mean,rel.
rectan -
gular 32r e l.
231
b 1 wreprod ,up/dn/mean
8 Khysteresis 1 1 hrel.
rectan -
gular 32r e l.
231
h 1 whysteresis
Y Sup/dn/mean S( ) yw
Notes : The characteristic quantities f0,rel. , b' rel. , brel. and hrel. here are relative quantities, i.e.
quantities related to the measured value (the indication).
In the determination of the transmission factor, the zero point is not a calibration point.
Despite this, the zero shift observed enters into the uncertainty of the measured value s of
the output signal and thus influences the uncertainty of the calibration result for the output
quantity S.
See also notes to Table 5
6.3 Determination of the characteristic values significant for the uncertainty
Preliminary remark: According to page 17 , the type A contributions to the uncertainty
should be stated in the form of empirical standard deviations. In the case of measuring
instruments affected by hysteresis, where the measurements in the direction of increasing
and decre asing pressures must be evaluated separately, a maximum of only three
measured values is available at each calibration point and the assumption that these
values are normally distributed is often not justified. Some simple formulae are, therefore,
given in the following. They are not based on statistical assumptions and, according to
experience, they furnish useful substitutes for the standard deviations. Their application
is, however, optional. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
23 -
Resolution r
For devices with digital indication the resolution corresponds to the digit step, provided the
indicated value does not vary by more than one digit step when the pressure -measuring
device is unloaded. A variability interval 2 a = r of a rectangular distribution is to be
estimated.
If, with the pr essure -measuring device unloaded, the indicated value varies by more than
the value of the resolution determined before, the resolution r is to be taken as the span
of the fluctuation, additionally added with a digital step.
For devices with analogue indi cation the indicated value is estimated by visual
interpolation between any two successive scale marks. Thus, the smallest assessable
division of the scale interval is called the interpolated division r, differentiating between
different values indicated. For the uncertainty analysis a variability interval 2 a = 2r of a
rectangular distribution is to be estimated.
Zero error f0
The zero point may be set prior to each measurement cycle comprising one measurement
series each at increasing and decreasing pre ssures, and it must be recorded prior to and
after each measurement cycle. The reading in units of pressure – when indicated after
conversion – must be taken after the complete removal of the load. The zero error is
calculated as follows:
f0 = max {| pind ,2,0 – p ind ,1,0 – (pref,2,0 – pref1,0 )|,| p ind ,4,0 – p ind ,3,0 – (pref, 4,0 – pref 3,0 )|,
|p ind ,6,0 –pind ,5,0 – (pref, 6,0 – pref5 ,0 )|} (21)
The indices number the measured values pind, i,j read at the zero points j = 0 of
measurement series M1 to M6 with i = 1 to 6.
Repeatability b'
The repeatability, with the mounting unchanged, is determined from the difference
between the deviations measured in the corresponding measurement series, corrected by
the zero signal (the index j numbers the nominal pressure values):
b' up, j = max {|( pind,3, j – pind,3,0 ) – (pind,1, j – pind,1,0 ) – (pref,3, j – pref,3,0 ) + (pref,1, j – pref,1,0 )|,
|( pind,5, j – pind,5,0 ) – (pind,1, j – pind,1,0 ) – (pref,5, j – pref,5,0 ) + (pref,1, j – pref,1,0 )|,
|( pind,5, j – pind,5,0 ) – (pind,3, j – pind,3,0 ) – (pref,5, j – pref,5,0 ) + (pref,3, j – pref,3,0 )|} (22a)
b' dn, j = max {|( pind,4, j – pind,3,0 ) – (pind,2, j – pind,1,0 ) – (pref,4, j – pref,3,0 ) + (pref,2, j – pref,1,0 )|,
|( pind,6, j – pind,5,0 ) – (pind,2, j – pind,1,0 ) – (pref,6, j – pref,5,0 ) + (pref,2, j – pref,1,0 )|,
|( pind,6, j – pind,5,0 ) – (pind,4, j – pind,3,0 ) – (pref,6, j – pref,5,0 ) + ( pref,4, j – pref,3,0 )|} (23a)
b' mean, j = max {b' up, j, b' dn, j} (24)
The equations (22a) and (23a) are replaced with (22b) and (23b)
b' up, j = |( pind,3, j – pind,3,0 ) – (pind,1, j – pind,1,0 ) – (pref,3, j – pref,3,0 ) + (pref,1, j – pref,1,0 )| (22b)
b' dn, j = |( pind,4, j – pind,3,0 ) – (pind,2, j – pind,1,0 ) – (pref,4, j – pref,3,0 ) + (pref,2, j – pref,1,0 )| (23b)
if measurement series M5 and M6 are performed after reinstallation to check
reproducibility.
Reproducibility b
The reproducibility, with the mounting changed by reinstallation, is determined from the EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
24 -
difference between the values measured in the cor responding measurement series,
corrected by the zero signal:
bup, j = max {|( pind ,5, j – pind ,5,0 ) – (pind ,1, j – pind ,1,0 ) – (pref, 5, j – pref, 5,0 ) + (pref, 1, j –pref, 1,0 )|,
|( pind ,5, j – pind ,5,0 ) – (pind ,3, j – pind ,3,0 ) – (pref, 5, j – pref, 5,0 ) + (pref, 3, j – pref, 3,0 )|} (25)
bdn, j = max {|( pind ,6, j – pind ,5 ,0 ) – (pind ,2, j – pind ,1 ,0 ) – (pref, 6, j – pref,5 ,0 ) + (pref, 2, j –pref,1 ,0 )|,
|( pind ,6, j – pind ,5 ,0 ) – (pind ,4, j – pind ,3 ,0 ) – (pref, 6, j – pref,5 ,0 ) + (pref, 4, j – pref,3 ,0 )|} (26)
bmean, j = max {bup, j, bdn, j} (27)
Hysteresis h (Reversibility)
The hysteresis is determined from the difference between the corresponding deviations of
the output values measured at increasing and decreasing pressures, corrected by the zero
signal: ( ) ( )( ) ( )012ref, 12ref, 012ref, 2ref, 1012ind, 12ind, 012ind, 2ind,
1
,,,,,,,,
−−−=−−−
−+−−−−−=
mjmmjmnmmjmmjmj
ppppppppnh
(28)
The variable n refers to the number of the complete measuring cycles m.
7 EXAMPLES
General remarks
Two examples have been chosen:
Example 1: Calibration of an indicating digital pressure gauge with oil as the pressure
medium. The three different calibration procedures referring to Section 5.2
are presented:
(a) Basic Calibration Procedure (Example 1a),
(b) Standard Calibration Procedure (Example 1b),
(c) Comprehensive Calibration Procedure (Example 1c).
Example 2: Calibration of a pressure transducer with oil as the pressure medium.
Example 2 is evaluated in two different ways:
(a) by using a linear characteristic to model the output signal
(Example 2a),
(b) by determining the transmission coefficient (Example 2b).
In these examples it is assumed that the pressures g enerated by the reference standard,
pref , were exactly the same in the series measured at increasing and decreasing pressures
as well as in repeated measurement series. If it is not the case, individual reference
pressure values must be taken for each measurement series.
7.1 Example 1 – Calibration of an indicating digital pressure gauge
Calibration object
The calibration object was an indicating digital pressure gauge with the following
parameters:
Range: 0 MPa to 25 MPa (gauge pressure)
Resolution 0.1 kPa EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )-25 -
Reference temperature: 20 °C
Note : At pressures below some small critical value, the zero reading appears in the
display. The zero reading does not correspond exactly to
pref = 0.
Reference standard
The reference standard was an oil -operated pressure balance operated at piston -cylinder
temperature
tref , at ambient pressure
pamb and ambient temperature
tamb , i.e. at an ambient
air density
amb (
pamb ,
tamb , 60 % relative humidity).
The expanded uncertainty of the pressures measured a t calibration conditions in the
reference level of the calibration object is ( ) ref 5ref 10 0.8kPa 02 .0 ppU += −
for pref > 1 MPa
Calibration procedure
Before calibration the instrument was twice brought to its nominal pressure and kept at
this pressure for one minute.
The difference h in height between the pressure reference levels of the calibration object
and the standard instrument was adjusted to zero.
The calibration temperature was equal to the reference temperature within ±0.5 °C .
Depending on the chosen calibration procedure, one or three complete cycles of
comparison measurements were carried out. In the case of the Basic and Standard
Calibration Procedures, additional measurements were carried out for determining the
repeatability as described. Tables E1a.1, E1b.1 and E1c.1 show the raw data. The
evaluation of the data is given in Tables E1a.2, E1b.2 and E1c.2. Finally, the uncertainty
analyses for a selected reference pressure are presented in Tables E1a.3, E1b.3 and
E1c.3. The numerical results for the Comprehensive Calibra tion Procedure are depicted
in Figure 7.
Evaluation of the uncertainty of measurement
The uncertainty of the observed difference between the indicated pressure and the
reference pressure as obtained from the reference standard is calculated from the
sum /difference model separately for pressures measured at increasing and decreasing
pressures. The uncertainty of the mean values of the indicated pressure is calculated by
adding the uncertainty contribution due to reversibility (hysteresis). If no correctio ns are
applied to the readings, the accuracy of the pressures measured with the calibrated
instrument is given by its error span. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
26 -
7.1.1 Example 1a – Basic Calibration Procedure
Table E1a.1: Raw data
Applied
pressure
Expanded
uncertainty
Reading
pind
pref ( )ref
pU M1 (up) M2 (dn) M3 (up) M5 (up)
bar bar bar bar bar bar
0.000 0.0000 -0.079 -0.076 -0.084 -0.071
50.029 0.0042 49.934 49.974
100.057 0.0082 99.972 100.005
150.085 0.0122 150.038 150.046 150.006 150.018
200.113 0.0162 200.088 200.075
250.140 0.0202 250.098 250.108
Table E1a.2: Calculation and results
Expanded
uncertainty
Applied
pressure
Applied
pressure Reading Repeatability
interval
Devia -
tion
Expanded
uncertainty of
measurement 5( )ref
pU
pref pref pind,up/dn b' up/dn Δpup/dn ( )up/dn
pU
kPa MPa bar bar bar bar bar
0.00 0.0000 0.000 -0.079 0.028 -0.079 0.016
0.42 5.0029 50.029 49.934 0.028 -0.095 0.017
0.82 10.0057 100.057 99.972 0.028 -0.085 0.018
1.22 15.0085 150.085 150.038 0.028 -0.047 0.020
1.62 20.0113 200.113 200.088 0.028 -0.025 0.023
2.02 25.0140 250.140 250.098 0.028 -0.042 0.026
2.02 25.0140 250.140 250.108 0.028 -0.032 0.026
1.62 20.0113 200.113 200.075 0.028 -0.038 0.023
1.22 15.0085 150.085 150.046 0.028 -0.039 0.020
0.82 10.0057 100.057 100.005 0.028 -0.052 0.018
0.42 5.0029 50.029 49.974 0.028 -0.055 0.017
0.00 0.0000 0.000 -0.076 0.028 -0.076 0.016
5
The stated values for ( )up/dn
pU are only calculative results from the uncertainty analysis and not
applicable for reporting. Due to th e chosen Basic Calibration Procedure, the expanded uncertainties are
not smaller than 0.2 % FS. Thus, for the whole calibration range ( )up/dn pU = 0.50 bar is valid. EURAMET Calibration Guide No. 17
Version 4. 1 (09 /20 22 ) - 27 -
Expanded
uncertainty
Applied
pressure
Applied
pressure
Mean
reading
Repeatability
interval
Hyster -
esis
Devia -
tion
Expanded
uncertainty of
measurement 6 Error span ( )ref
pU
pref pref pind,mean b' mean h Δpmean ( )mean
pU ( )mean
pU
kPa MPa bar bar bar bar bar bar bar
0.00 0.0000 0.000 -0.078 0.028 0.003 -0.078 0.016 0.578
0.42 5.0029 50.029 49.954 0.028 0.040 -0.075 0.029 0.575
0.82 10.0057 100.057 99.989 0.028 0.033 -0.069 0.026 0.569
1.22 15.0085 150.085 150.042 0.028 0.008 -0.043 0.021 0.543
1.62 20.0113 200.113 200.082 0.028 0.013 -0.031 0.024 0.531
2.02 25.0140 250.140 250.103 0.028 0.010 -0.037 0.027 0.537
Table E1a.3: Uncertainty budget at calibration pressure 100 bar
Quantity Estimate Variability
interval
Probability
distribution Divisor Standard
uncertainty
Sensitivity
coefficient
Contribution
to
uncertainty
Variance
Xi xi 2a( )i
xP( )i
xu ci( ) yui( ) yui
2
pref
100.057
bar
0.0164
bar normal 2 0.0041
bar -1 0.0041 bar 1.68·10 -5
bar 2
pind.mean
99.989
bar
0.0010
bar rectangular 3 2.89·10 -4
bar 1 2.89·10 -4
bar
8.33·10 -8
bar 2
δpzero
0.000
bar
0.0030
bar rectangular 3 8.66·10 -4
bar 1 8.66·10 -4
bar
7.50·10 -7
bar 2
δprepeat.mean
0.000
bar
0.0280
bar rectangular 3 0.0081
bar 1 0.0081 bar 6.53·10 -5
bar 2
δphysteresis
0.000
bar
0.0330
bar rectangular 3 0.0095
bar 1 0.0095 bar 9.08·10 -5
bar 2
Δpmean
-0.069
bar 0.0132 bar 1.74·10 -4
bar 2
Δpmean = -0.07 bar Due to the chosen Basic Calibration Procedure, the
expanded uncertainty is not smaller than 0.2 % FS. ( )mean
pU
= 0.50 bar
at k = 2
6 The stated values for ( )mean pU are only calculative results from the uncertainty analysis and not
applicable for reporting. Due to the chosen Basic Calibration Procedure, the expanded uncertainties are
not smaller than 0.2 % FS. Thus, for the whole calibration range ( )mean pU = 0.50 bar is valid. The error
span ( )mean pU is calculated on the basis of the valid uncertainty. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
28 -
7.1.2 Example 1b – Standard Calibration Procedure
Table E1b.1: Raw data
Applied
pressure
Expanded
uncertainty
Reading
pind
pref ( )ref
pU M1 (up) M2 (dn) M3 (up) M5 (up)
bar bar bar bar bar bar
0.000 0.0000 -0.079 -0.076 -0.084 -0.071
25.015 0.0022 24.943 24.931
50.029 0.0042 49.934 49.974 49.967 49.956
75.043 0.0062 74.940 74.960
100.057 0.0082 99.972 100.005
125.071 0.0102 125.019 125.031 124.994 124.974
150.085 0.0122 150.038 150.046
175.099 0.0142 175.004 175.051
200.113 0.0162 200.088 200.075 200.033 200.078
225.127 0.0182 225.079 225.064
250.140 0.0202 250.098 250.108 EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
29 -
Table E1b.2: Calculation and results
Expanded
uncertainty
Applied
pressure
Applied
pressure Reading Repeatability
interval
Devia -
tion
Expanded
uncertainty of
measurement 7( )ref
pU
pref pref pind,up/dn b' up/dn Δpup/dn ( )up/dn
pU
kPa MPa bar bar bar bar bar
0.00 0.0000 0.000 -0.079 0.000 -0.079 0.002
0.22 2.5015 25.015 24.943 0.038 -0.072 0.022
0.42 5.0029 50.029 49.934 0.038 -0.095 0.022
0.62 7.5043 75.043 74.940 0.053 -0.103 0.031
0.82 10.0057 100.057 99.972 0.053 -0.085 0.032
1.02 12.5071 125.071 125.019 0.053 -0.052 0.032
1.22 15.0085 150.085 150.038 0.053 -0.047 0.033
1.42 17.5099 175.099 175.004 0.053 -0.095 0.034
1.62 20.0113 200.113 200.088 0.050 -0.025 0.033
1.82 22.5127 225.127 225.079 0.050 -0.048 0.034
2.02 25.0140 250.140 250.098 0.050 -0.042 0.035
2.02 25.0140 250.140 250.108 0.050 -0.032 0.035
1.82 22.5127 225.127 225.064 0.050 -0.063 0.034
1.62 20.0113 200.113 200.075 0.050 -0.038 0.033
1.42 17.5099 175.099 175.051 0.053 -0.048 0.034
1.22 15.0085 150.085 150.046 0.053 -0.039 0.033
1.02 12.5071 125.071 125.031 0.053 -0.040 0.032
0.82 10.0057 100.057 100.005 0.053 -0.052 0.032
0.62 7.5043 75.043 74.960 0.053 -0.083 0.031
0.42 5.0029 50.029 49.974 0.038 -0.055 0.022
0.22 2.5015 25.015 24.931 0.038 -0.084 0.022
0.00 0.0000 0.000 -0.076 0.000 -0.076 0.002
7
The stated values for ( )up/dn
pU are only calculative results from the uncertainty analysis and not
applicable for reporting. Due to the chosen Standard Calibration Procedure, the expanded uncertainties
are not to be smaller than 0.05 % FS. Thus, for the whole calibration range ( )up/dn
pU = 0.13 bar is valid. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
30 -
Expanded
uncertainty
Applied
pressure
Applied
pressure
Mean
reading
Repeatability
interval
Hyster -
esis
Devia -
tion
Expanded
uncertainty of
measurement 8 Error span ( )ref
pU
pref pref pind,mean b' mean h Δpmean ( )mean
pU ( )mean
pU
kPa MPa bar bar bar bar bar bar bar
0.00 0.0000 0.000 -0.078 0.000 0.003 -0.078 0.003 0.203
0.22 2.5015 25.015 24.937 0.038 0.012 -0.078 0.023 0.203
0.42 5.0029 50.029 49.954 0.038 0.040 -0.075 0.032 0.200
0.62 7.5043 75.043 74.950 0.053 0.020 -0.093 0.033 0.218
0.82 10.0057 100.057 99.989 0.053 0.033 -0.069 0.037 0.194
1.02 12.5071 125.071 125.025 0.053 0.012 -0.046 0.033 0.171
1.22 15.0085 150.085 150.042 0.053 0.008 -0.043 0.033 0.168
1.42 17.5099 175.099 175.028 0.053 0.047 -0.072 0.043 0.197
1.62 20.0113 200.113 200.082 0.050 0.013 -0.031 0.034 0.156
1.82 22.5127 225.127 225.072 0.050 0.015 -0.055 0.035 0.180
2.02 25.0140 250.140 250.103 0.050 0.010 -0.037 0.036 0.162
Table E1b.3: Uncertainty budget at calibration pressure 100 bar
Quantity Estimate Variability
interval
Probability
distribution Divisor Standard
uncertainty
Sensitivity
coefficient
Contribution
to
uncertainty
Variance
Xi xi 2a( )i
xP( )i
xu ci( ) yui( ) yui
2
pref
100.057
bar
0.0164
bar normal 2 0.0041
bar -1 0.0041
bar
1.68·10 -5
bar 2
pind.mean
99.989
bar
0.0010
bar rectangular 3 2.89·10 -4
bar 1 2.89·10 -4
bar
8.33·10 -8
bar 2
δpzero
0.000
bar
0.0030
bar rectangular 3 8.66·10 -4
bar 1 8.66·10 -4
bar
7.50·10 -7
bar 2
δprepeat.mean
0.000
bar
0.0530
bar rectangular 3 0.0153
bar 1 0.0153
bar
2.34·10 -4
bar 2
δphysteresis
0.000
bar
0.0330
bar rectangular 3 0.0095
bar 1 0.0095
bar
9.08·10 -5
bar 2
Δpmean
-0.069
bar
0.0185
bar
3.42·10 -4
bar 2
Δpmean = -0.07 bar Due to the chosen Standard Calibration Procedure, the
expanded uncertainty is not smaller than 0.05 % FS. ( )mean
pU
= 0.13 bar
at k = 2
8
The stated values for ( )mean pU are only calculative results from the uncertainty analysis and not
applicable for reporting. Due to the chosen Standard Calibration Procedure, the expanded uncertainties
are not to be smaller than 0.05 % FS. Thus, for the whole calibration range ( )mean pU = 0.13 bar is valid.
The error span ( )mean pU is calculated on the basis of the valid uncertainty. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
31 -
7.1.3 Example 1c – Comprehensive Calibration Procedure
Table E1c.1: Raw data
Applied
pressure
Expanded
uncertainty
Reading
pind
pref ( )ref
pU M1 (up) M2 (dn) M3 (up) M4 (dn) M5 (up) M6 (dn)
bar bar bar bar bar bar bar bar
0.000 0.0000 -0.079 -0.076 -0.084 -0.067 -0.071 -0.078
25.015 0.0022 24.943 24.931 24.898 24.958 24.952 24.946
50.029 0.0042 49.934 49.974 49.967 49.997 49.956 49.975
75.043 0.0062 74.940 74.960 74.957 74.994 74.971 74.983
100.057 0.0082 99.972 100.005 99.985 99.981 99.991 100.016
125.071 0.0102 125.019 125.031 124.994 125.018 124.974 125.001
150.085 0.0122 150.038 150.046 150.006 150.061 150.018 150.025
175.099 0.0142 175.004 175.051 175.038 175.062 175.045 175.061
200.113 0.0162 200.088 200.075 200.033 200.061 200.078 200.027
225.127 0.0182 225.079 225.064 225.044 225.102 225.069 225.079
250.140 0.0202 250.098 250.108 250.084 250.072 250.052 250.077
Table E1c.2: Calculation and results
Expanded
uncertainty
Applied
pressure
Applied
pressure
Mean
reading
Repeatability
interval
Devia -
tion
Expanded
uncertainty of
measurement ( )ref
pU
pref pref pind,up/dn b' up/dn Δpup/dn ( )up/dn
pU
kPa MPa bar bar bar bar bar
0.00 0.0000 0.000 -0.078 0.000 -0.078 0.010
0.22 2.5015 25.015 24.931 0.041 -0.084 0.026
0.42 5.0029 50.029 49.952 0.038 -0.077 0.024
0.62 7.5043 75.043 74.956 0.023 -0.087 0.018
0.82 10.0057 100.057 99.983 0.018 -0.074 0.016
1.02 12.5071 125.071 124.996 0.053 -0.075 0.034
1.22 15.0085 150.085 150.021 0.028 -0.064 0.023
1.42 17.5099 175.099 175.029 0.039 -0.070 0.028
1.62 20.0113 200.113 200.066 0.050 -0.047 0.035
1.82 22.5127 225.127 225.064 0.030 -0.063 0.027
2.02 25.0140 250.140 250.078 0.054 -0.062 0.038
2.02 25.0140 250.140 250.086 0.039 -0.054 0.032
1.82 22.5127 225.127 225.082 0.043 -0.045 0.032
1.62 20.0113 200.113 200.054 0.056 -0.059 0.037
1.42 17.5099 175.099 175.058 0.016 -0.041 0.020
1.22 15.0085 150.085 150.044 0.049 -0.041 0.032
1.02 12.5071 125.071 125.017 0.038 -0.054 0.026
0.82 10.0057 100.057 100.001 0.022 -0.056 0.018
0.62 7.5043 75.043 74.979 0.039 -0.064 0.025
0.42 5.0029 50.029 49.982 0.035 -0.047 0.023
0.22 2.5015 25.015 24.945 0.032 -0.070 0.021
0.00 0.0000 0.000 -0.074 0.024 -0.074 0.017 EURAMET Calibration Guide No. 17
Version 4. 1 (09 /20 22 ) - 32 -
Expanded
uncertainty
Applied
pressure
Applied
pressure
Mean
reading
Repeatability
interval
Hyster -
esis
Devia -
tion
Expanded
uncertainty of
measurement
Error span ( )ref
pU
pref pref pind,mean b' mean h Δpmean ( )mean
pU ( )mean
pU
kPa MPa bar bar bar bar bar bar bar
0.00 0.0000 0.000 -0.076 0.024 0.009 -0.076 0.018 0.094
0.22 2.5015 25.015 24.938 0.041 0.026 -0.077 0.030 0.107
0.42 5.0029 50.029 49.967 0.038 0.030 -0.062 0.030 0.092
0.62 7.5043 75.043 74.968 0.039 0.023 -0.076 0.029 0.104
0.82 10.0057 100.057 99.992 0.022 0.021 -0.065 0.022 0.087
1.02 12.5071 125.071 125.006 0.053 0.021 -0.065 0.036 0.101
1.22 15.0085 150.085 150.032 0.049 0.023 -0.053 0.035 0.088
1.42 17.5099 175.099 175.044 0.039 0.029 -0.055 0.033 0.088
1.62 20.0113 200.113 200.060 0.056 0.031 -0.053 0.041 0.094
1.82 22.5127 225.127 225.073 0.043 0.028 -0.054 0.036 0.090
2.02 25.0140 250.140 250.082 0.054 0.016 -0.058 0.039 0.098
Table E1c.3: Uncertainty budget at calibration pressure 100 bar
Quantity Estimate Variability
interval
Probability
distribution
Divi -
sor
Standard
uncertainty
Sensitivity
coefficient
Contribution
to
uncertainty
Variance
Xi xi 2a( )i
xP( )i
xu ci( ) yui( ) yui
2
pref
100.057
bar
0.0164
bar normal 2 0.0041
bar -1 0.0041
bar
1.68·10 -5
bar 2
pind,mean
99.992
bar
0.0010
bar rectangular 3 2.89·10 -4
bar 1 2.89·10 -4
bar
8.33·10 -8
bar 2
δpzero
0.000
bar
0.0170
bar rectangular 3 0.0049
bar 1 0.0049
bar
2.41·10 -5
bar 2
δprepeat,mean
0.000
bar
0.0220
bar rectangular 3 0.0064
bar 1 0.0064
bar
4.03·10 -5
bar 2
δphysteresis
0.000
bar
0.0207
bar rectangular 3 0.0060
bar 1 0.0060
bar
3.56·10 -5
bar 2
Δpmean
-0.065
bar
0.0108
bar
1.17·10 -4
bar 2
Δpmean = -0.065 bar ( ) ( )mean mean pukpU = (k = 2) ( )mean
pU = 0.022 bar EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
33 -
The error bars indicate the expanded uncertainty of mean values.
Figure 7 – Calibration of an indicating pressure gauge using the Comprehensive
Calibration Procedure – Example 1c
7.2 Example 2 – Calibration of a pressure transducer
Calibration object
The calibration object was a pressure transducer with a Wheatstone bridge on a metal
diaphragm as the sensing element. It had the following parameters:
Range: 0 MPa to 20 MPa (gauge pressure)
Reference temperature: 20 °C.
Reference standard
The reference standard was an oil -operated pressure balance operated at piston -cylinder
temperature
tref , at ambient pressure
pamb and ambient temperature
tamb , i.e. at an ambient
air density
amb (
pamb ,
tamb , 60 % rel. humidity).
The expanded uncertainty of the pressures measured at calibration conditions in the
reference level of the calibration object is ( ) ref 4ref 10 01 ppU = −
.
for pref > 1 MPa
Calibration procedure
The output signal of the pressure transducer was measured as ( )PS ind ind VGVI in units
mV/V using a digital compensator, the expanded measurement uncertainty of which was
0.00005 mV/V.
Before calibration the instrument was twice brought to its maxi mum pressure and kept at
this pressure for one minute.
-0,12
-0,10
-0,08
-0,06
-0,04
-0,02
0,00
0 20 40 60 80 100 120 140 160 180 200 220 240 260
deviation Δ
p
up/dn/mean / bar
pressure
pref / bar
increasing pressure decreasing pressure mean value EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
34 -
The difference h in height between the pressure reference levels of the calibration object
and the standard instrument was adjusted to zero.
The calibration temperature was equal to the re ference te mperature within ±0.5 °C .
Three complete cycles of comparison measurements were carried out (Comprehensive
Calibration Procedure). To check the reproducibility the pressure transducer was
reinstalled before starting the third measuring cycle. Table E2 show s the raw data.
Table E2: Raw data
Applied
pressure
Relative
expanded
uncertainty
Indication of the digital compensator
Iind
pref
( )ref
pWM1 (up) M2 (dn) M3 (up) M4 (dn) M5 (up) M6 (dn)
bar mV/V mV/V mV/V mV/V mV/V mV/V
0.000 0.00000 -0.00003 0.00000 0.00002 0.00000 -0.00002
20.010 1.0 ·10 -40.20009 0.20026 0.20019 0.20033 0.20021 0.20032
40.022 1.0 ·10 -40.40026 0.40063 0.40032 0.40067 0.40033 0.40064
60.033 1.0 ·10 -40.60041 0.60094 0.60049 0.60097 0.60049 0.60092
80.045 1.0 ·10 -40.80053 0.80118 0.80062 0.80120 0.80062 0.80110
100.056 1.0 ·10 -41.00063 1.00139 1.00072 1.00135 1.00075 1.00125
120.068 1.0 ·10 -41.20074 1.20149 1.20080 1.20141 1.20082 1.20132
140.079 1.0 ·10 -41.40080 1.40158 1.40089 1.40150 1.40090 1.40133
160.091 1.0 ·10 -41.60082 1.60157 1.60091 1.60148 1.60091 1.60126
180.102 1.0 ·10 -41.80084 1.80148 1.80097 1.80135 1.80091 1.80111
200.113 1.0 ·10 -42.00079 2.00100 2.00088 2.00114 2.00086 2.00087
The evaluation of Example 2a is based on the defined linear characteristic of the
instrument. The pressures calculated from the measured output signals using this
characteristic are compared with the pressures obtained from the standard instrument.
The sum /difference model is applied to calculate the uncertainty of measurement using
procedures described in Section 6.3 "Determination of the characteristic values significant
for the uncertainty" (page 22 ). The n umerical results are presented in Table E2a.1 and
are depicted in Figure 8.
In Example 2b, the transmission coefficient of the same instrument is determined at the
same calibration points. Zero error, repeatability, reproducibility and hysteresis are
calc ulated using the formulae presented on page 23 each divided by the corresponding
indication Iind to obtain the relative values. The numerical results are presented in Tables
E2b.1 and E2b.2 and are shown in Figure 9.
Figure 8 demonstrates that the calibration methods 2a and 2b are equivalent.
The error spans ( )mean pU plotted in Figure 8 can be calculated f rom the error spans ( )mean SU
of the values of the transmission coefficient Smean as ( ) ( ) ( ) ( ) ref mean 0ref mean mean 1 pSWSpSUpU ==
Evaluation of the uncertainty of measurement
The uncertainty of the observed difference p between the pressure calculated from the
characteristic straight line and the reference pressure as obtained from the reference
standard is calculated from the sum/difference model separately for pressures measured
at increasing and decreasing pressures. The uncertainty of the mean values of p is
calculated by adding the uncertainty contribution due to reversibility (hysteresis). If no EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
35 -
corrections are applied to the readings, the accuracy of the pressures measured with the
calibrated instrument is given b y its error span.
Note: The slope of the linear characteristic, the single transmission coefficient S0, is
obtained from a straight line fitted to the calibration data. It replaces the nominal
value 1.000000·10 -2 (mV/V)/bar (corresponding to an output signal of 2 mV/V FS)
as defined by the manufacturer and – like the nominal value – has to be regarded
as a defined value without uncertainty.
7.2.1 Example 2a – Modelling the output signal by using a linear characteristic
Table E2a.1: Calculation and results
Expanded
uncer -
tainty
Applied
pressure
Applied
pressure
Mean
indication
Repeat -
ability
interval
Reprodu -
cibility
interval
Expanded
uncertainty of
digital
compensator
Modelled
indicated
pressure
Devia -
tion
Expanded
unce rtainty
of measure -
ment ( )ref
pU
pref pref Iind,up/dn b' up/dn bup/dn ( )up/dn ind,
VU pind,mod,up/dn Δpup/dn ( )up/dn
pU
kPa MPa bar mV/V mV/V mV/V mV/V bar bar bar
0.00 0.0000 0.000 0.000000 0.000000 0.000000 0.000050 0.000 0.000 0.005
0.20 2.0010 20.010 0.200163 0.000100 0.000120 0.000050 20.013 0.003 0.011
0.40 4.0022 40.022 0.400303 0.000060 0.000070 0.000050 40.024 0.002 0.009
0.60 6.0033 60.033 0.600463 0.000080 0.000080 0.000050 60.037 0.004 0.010
0.80 8.0045 80.045 0.800590 0.000090 0.000090 0.000050 80.047 0.002 0.012
1.00 10.0056 100.056 1.000700 0.000090 0.000120 0.000050 100.055 -0.001 0.014
1.20 12.0068 120.068 1.200787 0.000060 0.000080 0.000050 120.061 -0.007 0.014
1.40 14.0079 140.079 1.400863 0.000090 0.000100 0.000050 140.065 -0.014 0.017
1.60 16.0091 160.091 1.600880 0.000090 0.000090 0.000050 160.064 -0.027 0.018
1.80 18.0102 180.102 1.800907 0.000130 0.000070 0.000050 180.063 -0.039 0.021
2.00 20.0113 200.113 2.000843 0.000090 0.000070 0.000050 200.054 -0.059 0.022
2.00 20.0113 200.113 2.001003 0.000140 0.000270 0.000050 200.070 -0.043 0.027
1.80 18.0102 180.102 1.801313 0.000130 0.000370 0.000050 180.104 0.002 0.029
1.60 16.0091 160.091 1.601437 0.000090 0.000310 0.000050 160.119 0.028 0.025
1.40 14.0079 140.079 1.401470 0.000080 0.000250 0.000050 140.126 0.047 0.021
1.20 12.0068 120.068 1.201407 0.000080 0.000170 0.000050 120.123 0.055 0.017
1.00 10.0056 100.056 1.001330 0.000040 0.000140 0.000050 100.118 0.062 0.014
0.80 8.0045 80.045 0.801160 0.000020 0.000100 0.000050 80.104 0.059 0.011
0.60 6.0033 60.033 0.600943 0.000030 0.000050 0.000050 60.085 0.052 0.009
0.40 4.0022 40.022 0.400647 0.000040 0.000030 0.000050 40.059 0.037 0.007
0.20 2.0010 20.010 0.200303 0.000070 0.000060 0.000050 20.027 0.017 0.008
0.00 0.0000 0.000 -0.000010 0.000050 0.000040 0.000050 -0.001 -0.001 0.006
Modelled pressure: pind,mod = Iind /S0 with 1/ S0 = 1/0.01000151 bar/(mV/V) = 99.9849 bar/(mV/V) EURAMET Calibration Guide No. 17
Version 4. 1 (09 /20 22 ) - 36 -
Applied
pressure
Mean
indication
Variability intervals of
corresponding series Hysteresis
Modelled
indicated
pressure
Devia -
tion
Expanded
uncertainty
of mea -
surement
Error span Error span
from E2b 9
pref Iind,mean b' mean bmean h pind,mod,mean Δpmean ( )mean
pU ( )mean
pU ( )mean
pU
bar mV/V mV/V mV/V mV/V bar bar bar bar bar
0.000 -0.000005 0.000050 0.000040 0.000023 0.000 0.000 0.007 0.007 n/a
20.010 0.200233 0.000100 0.000120 0.000140 20.020 0.010 0.013 0.024 0.024
40.022 0.400475 0.000060 0.000070 0.000343 40.041 0.019 0.022 0.041 0.041
60.033 0.600703 0.000080 0.000080 0.000480 60.061 0.028 0.030 0.058 0.058
80.045 0.800875 0.000090 0.000100 0.000570 80.075 0.030 0.035 0.066 0. 06 6
100.056 1.001015 0.000090 0.000140 0.000630 100.086 0.030 0.039 0.070 0.070
120.068 1.201097 0.000080 0.000170 0.000620 120.092 0.024 0.040 0.063 0. 06 3
140.079 1.401167 0.000090 0.000250 0.000607 140.096 0.017 0.041 0.058 0.058
160.091 1.601158 0.000090 0.000310 0.000557 160.092 0.001 0.041 0.041 0.042
180.102 1.801110 0.000130 0.000370 0.000407 180.084 -0.018 0.038 0.056 0. 05 6
200.113 2.000923 0.000140 0.000270 0.000160 200.062 -0.051 0.029 0.080 0. 07 9
Table E2a.2: Uncertainty budget at calibration pressure 100 bar
Quantity Estimate Variability
interval
Probability
distribution Divisor Standard
uncertainty
Sensitivity
coefficient
Contribution
to
uncertainty
Variance
Xi xi 2a( )i
xP( )i
xu ci( ) yui( ) yui
2
pref
100.056
bar
0.020
bar normal 2 0.005
bar -1 0.0050
bar
2.50·10 -5
bar 2
Iind,mean
1.001015
mV/V
0.000100
mV/V normal 2 0.000025
mV/V
99.9849
bar/(mV/V)
0.0025
bar
6.25·10 -6
bar 2
δIzero
0.000000
mV/V
0.000030
mV/V rectangular 3 0.000009
mV/V
99.9849
bar/(mV/V)
0.0009
bar
7.50·10 -7
bar 2
δIrepeat,mean
0.000000
mV/V
0.000090
mV/V rectangular 3 0.000026
mV/V
99.9849
bar/(mV/V)
0.0026
bar
6.75·10 -6
bar 2
δIreprod,mean
0.000000
mV/V
0.000140
mV/V rectangular 3 0.000040
mV/V
99.9849
bar/(mV/V)
0.0040
bar
1.63·10 -5
bar 2
δIhysteresis
0.000000
mV/V
0.000630
mV/V rectangular 3 0.000182
mV/V
99.9849
bar/(mV/V)
0.0182
bar
3.31·10 -4
bar 2
Δpmean
0.030
bar
0.0196
bar
3.86·10 -4
bar 2
Δpmean = 0.030 bar ( ) ( )mean mean pukpU = (k = 2) ( )mean
pU = 0.039 bar
9 See Table E2b.1 for comparison with the other way of estimating the error span. EURAMET Calibration Guide No. 17
Version 4. 1 (09 /20 22 ) - 37 -
7.2.2 Example 2b – Determination of the transmission coefficient
Table E2b.1 Evaluation
Applied
pressure
Relative
expanded
uncertainty
Mean output
signal Zero error Repeatability Reproducibility Hysteresis
pref ( )mean ind,
IW Iind,mean f0,rel. b' mean,rel. b mean,rel. hrel.
bar mV/V
0.000 n/a -0.000005 n/a n/a n/a n/a
20.010 2.50·10 -4 0.200233 1.5 0·10 -4 4.99 ·10 -4 5.99 ·10 -4 6.99 ·10 -4
40.022 1.25·10 -4 0.400475 7. 49 ·10 -5 1.5 0·10 -4 1.7 5·10 -4 8.57 ·10 -4
60.033 8.32 ·10 -5 0.600703 4.99 ·10 -5 1.3 3·10 -4 1.3 3·10 -4 7.99 ·10 -4
80.045 6.24 ·10 -5 0.800875 3.7 5·10 -5 1.1 2·10 -4 1. 25 ·10 -4 7.1 2·10 -4
100.056 4.99 ·10 -5 1.001015 3.0 0·10 -5 8.99 ·10 -5 1. 40 ·10 -4 6. 29 ·10 -4
120.068 4.16 ·10 -5 1.201097 2.5 0·10 -5 6.66 ·10 -5 1. 42 ·10 -4 5. 16 ·10 -4
140.079 3.57 ·10 -5 1.401167 2.1 4·10 -5 6.42 ·10 -5 1. 78 ·10 -4 4.3 3·10 -4
160.091 3.12 ·10 -5 1.601158 1. 87 ·10 -5 5.62 ·10 -5 1.94 ·10 -4 3. 48 ·10 -4
180.102 2.78 ·10 -5 1.801110 1. 67·10 -5 7.22 ·10 -5 2. 05 ·10 -4 2. 26 ·10 -4
200.113 2.50 ·10 -5 2.000923 1.5 0·10 -5 7.00 ·10 -5 1.35 ·10 -4 8.0 0·10 -5
Table E2b.2 Results
Applied
pressure
Transmission
coefficient Deviation
Relative expanded
uncertainty of
measurement
Expanded uncertainty
of measurement Error span
pref Smean Smean ( )mean
SW( )mean
SU( )mean
SU
Iind, mean /pref 0mean SS −( ) mean 2
2 Sw i( ) mean mean SSW ( ) mean mean SSU +
bar (mV/V)/bar (mV/V)/bar (mV/V)/bar (mV/V)/bar
0.000 n/a n/a n/a n/a n/a
20.010 0.01000666 0.00000515 6. 68 ·10 -4 6.68 ·10 -6 1.18 ·10 -5
40.022 0.01000637 0.00000486 5. 39 ·10 -4 5.39 ·10 -6 1.03 ·10 -5
60.033 0.01000622 0.00000471 4.9 2·10 -4 4.93 ·10 -6 9.64 ·10 -6
80.045 0.01000531 0.00000380 4. 39 ·10 -4 4.39 ·10 -6 8.19 ·10 -6
100.056 0.01000455 0.00000304 3.9 2·10 -4 3.93 ·10 -6 6.97 ·10 -6
120.068 0.01000347 0.00000196 3.3 0·10 -4 3.30 ·10 -6 5.27 ·10 -6
140.079 0.01000269 0.00000118 2.93 ·10 -4 2.93 ·10 -6 4.11 ·10 -6
160.091 0.01000155 0.00000004 2. 55 ·10 -4 2.55 ·10 -6 2.59 ·10 -6
180.102 0.01000050 -0.00000101 2. 09 ·10 -4 2.09 ·10 -6 3.10 ·10 -6
200.113 0.00999897 -0.00000254 1. 43 ·10 -4 1.43 ·10 -6 3.97 ·10 -6
Single value S0: 0.01000151
(mV/V)/bar EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
38 -
Table E2b. 3 Uncertainty budget at the calibration pressure p = 100 bar
Quantity Estimate Variability
interval
Probability
distribution
Divi -
sor
Relative
standard
uncertainty
Exponent
associate
d with Xi
Contribution
to
uncertainty
Variance
Xi xi 2a( )i
xP( )i
xw1
ii
c x y −( ) ywi( ) ywi
2
pref
100.056
bar
0.020
bar normal 2 5.00·10 -5 -1 5.00·10 -5 2.50·10 -9
Iind,mean
1.001015
mV/V
0.00010
mV/V normal 2 2.50·10 -5 1 2.50·10 -5 6. 24·10 -10
Kzero 1 3.0 0·10 -5 rectangular 3 8. 65·10 -6 1 8. 65·10 -6 7. 48 ·10 -11
Krepeat,mean 1 8.99 ·10 -5 rectangular 3 2.60·10 -5 1 2.60·10 -5 6. 74·10 -10
Kreprod,mean 1 1. 40 ·10 -4 rectangular 3 4. 04 ·10 -5 1 4. 04 ·10 -5 1. 63 ·10 -9
Khysteresis 1 6. 29 ·10 -4 rectangular 3 1.82·10 -4 1 1.82·10 -4 3. 30·10 -8
Smean
0.01000455
(mV/V)/bar 1. 96·10 -4 3. 85·10 -8
Smean = 0.0100045
(mV/V)/bar) 10 ( ) ( )mean mean SwkSW = (k = 2) ( )mean
SW
= 3.9·10 -4
At the calibration pressure
pref = 100 bar the expanded uncertainty ( ) bar 100 mean SU of the
value bar 100 mean
S of the transmission factor is calculated as ( ) ( ) bar 100 mean mean bar 100 mean SSWSU =
= 3.9·10 -4·0.0100045 (mV/V)/bar
= 3.9·10 -6 (mV/V)/bar.
Statement of a single value of the transmission coefficient
The general use of a pressure transducer does not imply the application of different
transmission coefficients for the individual load steps (i.e. calibration pressures) but of only
a single transmission coefficient for the total range covered by the calibration. This is
preferably the slope of the straight line fitted through all measured values of the output
signal .
When this characteristic quantity of the pressure transducer is used, a statement of
compliance replaces the uncertainties associated with the individual values measured for
the transmission co efficient.
This requires that the limits of permissible error be fixed, which can be done on the basis
of the calibration results by the calculation of the error spans, i.e. by adding :
• the uncertainties associated with the individual values measured for the
transmission coefficient, and
• the deviations of these values from the single value stated for the transmission
coefficient.
Normally, error span values decrease with increasing pressure (see Figure 9). Two
methods for fixing the limits of permissible er ror are possible:
• one may choose the largest calculated error span as the limiting value, or
10
The transmission coefficient is valid for the c alibration pressure pref = 100.056 bar. It differs from the single
transmission coefficient calculated from all calibration pressures. EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
39 -
• limiting values of the errors are described by suitable curves, e.g. polynomials.
Note: The use of pressure -dependent limiting values of the errors is not common p ractice.
However, it allows smaller uncertainties to be stated when pressure measurements
are carried out with the calibrated instrument in the upper part of the measuring
range.
In the case of objects to be calibrated whose transmission coefficient has be en specified
by the manufacturer, the limits of permissible error may alternatively be identified with the
tolerance assigned to the specified value. In this case it must, however, always be checked
whether the values of the transmission coefficient determ ined upon calibration, including
their associated uncertainties and their systematic deviations from the specified single
value, do not exceed the limits of permissible error.
The error bars indicate the expanded uncertainties of the mean values whereas the blue solid lines
indicate the error spans ( )mean pU as obtained in Example 2a. For comparison, the corresponding
error spans obtained from the evaluation method of Example 2b are shown as black dashed lines.
Ideally, the solid and dashed lines should c oincide. Differences reflect the different ways of evaluation,
including an interpolation step by modelling the output signal using a single transmission coefficient
for Example 2a. Obviously, the overall result does not depend very much on such difference s as was
demonstrated. In Example 2a, the single transmission coefficient was chosen as
S0 = 0.01000151 (mV/V)/bar.
Figure 8 – Calibration of a pressure transducer EURAMET Calibration Guide No. 17
Version 4. 1(09 /20 22 )
40 -
The solid line indicates the value of the single transmission coeff icient. The error bars indicate the
error spans corresponding to transmission coefficients obtained in Example 2b.
Fi gure 9 - Measured values and error spans of the transmission coefficient
0,009990
0,009995
0,010000
0,010005
0,010010
0,010015
0,010020
0 20 40 60 80 100 120 140 160 180 200
transmission coefficient
S
/ (mV/V)/bar
pressure
pref / bar
mean values single transmission coefficient EURAMET e.V.
Bundesallee 100
38116 Braunschweig
Germany
EURAMET e.V. is a non-profit association under German law.
Phone: +49 531 592 1960
Fax: +49 531 592 1969
E-mail: secretariat@euramet.org |
190746 | https://gcsetime.com/wp-content/uploads/2022/03/Vector-Mechanics-For-Engineers-Statics-By-Ferdinand-P.-Beer-E.-Russell-Johnston-Jr-David-F.-Mazurek-.pdf | Vector Mechanics For Engineers Statics This page intentionally left blank Eleventh Edition Vector Mechanics For Engineers Ferdinand P. Beer Late of Lehigh University E. Russell Johnston, Jr.
Late of University of Connecticut David F. Mazurek U.S. Coast Guard Academy Statics VECTOR MECHANICS FOR ENGINEERS: STATICS, ELEVENTH EDITION Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2016 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2013, 2010, and 2007. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.
Some ancillaries, including electronic and print components, may not be available to customers outside the United States.
This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5 ISBN 978-0-07-768730-4 MHID 0-07-768730-2 Managing Director: Thomas Timp Global Brand Manager: Raghothaman Srinivasan Director of Development: Rose Koos Product Developer: Robin Reed Brand Manager: Thomas Scaife, Ph.D.
Digital Product Analyst: Dan Wallace Editorial Coordinator: Samantha Donisi-Hamm Marketing Manager: Nick McFadden LearnSmart Product Developer: Joan Weber Content Production Manager: Linda Avenarius Content Project Managers: Jolynn Kilburg and Lora Neyens Buyer: Laura Fuller Designer: Matthew Backhaus Content Licensing Specialist (Image): Carrie Burger Media Project Manager: TK Typeface: 10/12 Times LT Std Printer: R. R. Donnelley All credits appearing on page or at the end of the book are considered to be an extension of the copyright page.
The photo on the cover shows One World Trade Center in New York City, the tallest skyscraper in the Western Hemisphere. From its foundation to its structural components and mechanical systems, the design and operation of the tower is based on the fundamentals of engineering mechanics.
Library of Congress Cataloging-in-Publication Data On File The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the information presented at these sites.
www.mhhe.com v About the Authors Ferdinand P. Beer. Born in France and educated in France and Switzer-land, Ferd received an M.S. degree from the Sorbonne and an Sc.D. degree in theoretical mechanics from the University of Geneva. He came to the United States after serving in the French army during the early part of World War II and taught for four years at Williams College in the Williams-MIT joint arts and engineering program. Following his service at Williams College, Ferd joined the faculty of Lehigh University where he taught for thirty-seven years. He held several positions, includ-ing University Distinguished Professor and chairman of the Department of Mechanical Engineering and Mechanics, and in 1995 Ferd was awarded an honorary Doctor of Engineering degree by Lehigh University.
E. Russell Johnston, Jr. Born in Philadelphia, Russ received a B.S. degree in civil engineering from the University of Delaware and an Sc.D. degree in the field of structural engineering from the Massachusetts Institute of Tech-nology. He taught at Lehigh University and Worcester Polytechnic Institute before joining the faculty of the University of Connecticut where he held the position of chairman of the Department of Civil Engineering and taught for twenty-six years. In 1991 Russ received the Outstanding Civil Engineer Award from the Connecticut Section of the American Society of Civil Engineers.
David F. Mazurek. David holds a B.S. degree in ocean engineering and an M.S. degree in civil engineering from the Florida Institute of Technol-ogy and a Ph.D. degree in civil engineering from the University of Con-necticut. He was employed by the Electric Boat Division of General Dynamics Corporation and taught at Lafayette College prior to joining the U.S. Coast Guard Academy, where he has been since 1990. He is a reg-istered Professional Engineer in Connecticut and Pennsylvania, and has served on the American Railway Engineering & Maintenance-of-Way Association’s Committee 15—Steel Structures since 1991. He is a Fellow of the American Society of Civil Engineers, and was elected to the Con-necticut Academy of Science and Engineering in 2013. He was the 2014 recipient of both the Coast Guard Academy’s Distinguished Faculty Award and its Center for Advanced Studies Excellence in Scholarship Award. Professional interests include bridge engineering, structural forensics, and blast-resistant design.
vi Brief Contents 1 Introduction 1 2 Statics of Particles 15 3 Rigid Bodies: Equivalent Systems of Forces 82 4 Equilibrium of Rigid Bodies 169 5 Distributed Forces: Centroids and Centers of Gravity 230 6 Analysis of Structures 297 7 Internal Forces and Moments 367 8 Friction 429 9 Distributed Forces: Moments of Inertia 485 10 Method of Virtual Work 573 Appendix: Fundamentals of Engineering Examination A1 Answers to Problems AN1 Photo Credits C1 Index I1 vii Contents Preface x Guided Tour xiv Digital Resources xvi Acknowledgments xviii List of Symbols xix 1 Introduction 1 1.1 What is Mechanics? 2 1.2 Fundamental Concepts and Principles 3 1.3 Systems of Units 5 1.4 Converting between Two Systems of Units 10 1.5 Method of Solving Problems 12 1.6 Numerical Accuracy 14 2 Statics of Particles 15 2.1 Addition of Planar Forces 16 2.2 Adding Forces by Components 29 2.3 Forces and Equilibrium in a Plane 39 2.4 Adding Forces in Space 52 2.5 Forces and Equilibrium in Space 66 Review and Summary 75 Review Problems 79 3 Rigid Bodies: Equivalent Systems of Forces 82 3.1 Forces and Moments 84 3.2 Moment of a Force about an Axis 105 3.3 Couples and Force-Couple Systems 120 3.4 Simplifying Systems of Forces 136 Review and Summary 161 Review Problems 166 viii Contents 4 Equilibrium of Rigid Bodies 169 4.1 Equilibrium in Two Dimensions 172 4.2 Two Special Cases 195 4.3 Equilibrium in Three Dimensions 204 Review and Summary 225 Review Problems 227 5 Distributed Forces: Centroids and Centers of Gravity 230 5.1 Planar Centers of Gravity and Centroids 232 5.2 Further Considerations of Centroids 249 5.3 Additional Applications of Centroids 262 5.4 Centers of Gravity and Centroids of Volumes 273 Review and Summary 291 Review Problems 295 6 Analysis of Structures 297 6.1 Analysis of Trusses 299 6.2 Other Truss Analyses 317 6.3 Frames 330 6.4 Machines 348 Review and Summary 361 Review Problems 364 7 Internal Forces and Moments 367 7.1 Internal Forces in Members 368 7.2 Beams 378 7.3 Relations among Load, Shear, and Bending Moment 391 7.4 Cables 403 7.5 Catenary Cables 416 Review and Summary 424 Review Problems 427 ix Contents 8 Friction 429 8.1 The Laws of Dry Friction 431 8.2 Wedges and Screws 450 8.3 Friction on Axles, Disks, and Wheels 459 8.4 Belt Friction 469 Review and Summary 479 Review Problems 482 9 Distributed Forces: Moments of Inertia 485 9.1 Moments of Inertia of Areas 487 9.2 Parallel-Axis Theorem and Composite Areas 498 9.3 Transformation of Moments of Inertia 513 9.4 Mohr’s Circle for Moments of Inertia 523 9.5 Mass Moments of Inertia 529 9.6 Additional Concepts of Mass Moments of Inertia 549 Review and Summary 564 Review Problems 570 10 Method of Virtual Work 573 10.1 The Basic Method 574 10.2 Work, Potential Energy, and Stability 595 Review and Summary 609 Review Problems 612 Appendix: Fundamentals of Engineering Examination A1 Answers to Problems AN1 Photo Credits C1 Index I1 Advanced or specialty topics x Preface Objectives A primary objective in a first course in mechanics is to help develop a student’s ability first to analyze problems in a simple and logical manner, and then to apply basic principles to its solution. A strong conceptual understanding of these basic mechanics principles is essential for success-fully solving mechanics problems. We hope that this text, as well as the proceeding volume, Vector Mechanics for Engineers: Dynamics, will help instructors achieve these goals.† General Approach Vector analysis is introduced early in the text and is used in the presentation and discussion of the fundamental principles of mechanics. Vector methods are also used to solve many problems, particularly three-dimensional prob-lems where these techniques result in a simpler and more concise solution. The emphasis in this text, however, remains on the correct understanding of the principles of mechanics and on their application to the solution of engi-neering problems, and vector analysis is presented chiefly as a convenient tool.‡ Practical Applications Are Introduced Early. One of the characteristics of the approach used in this book is that mechanics of particles is clearly separated from the mechanics of rigid bodies. This approach makes it possible to consider simple practical applications at an early stage and to postpone the introduction of the more difficult concepts. For example: • In Statics, statics of particles is treated first (Chap. 2); after the rules of addition and subtraction of vectors are introduced, the principle of equilibrium of a particle is immediately applied to practical situations involving only concurrent forces. The statics of rigid bodies is consid-ered in Chaps. 3 and 4. In Chap. 3, the vector and scalar products of two vectors are introduced and used to define the moment of a force about a point and about an axis. The presentation of these new con-cepts is followed by a thorough and rigorous discussion of equivalent systems of forces leading, in Chap. 4, to many practical applications involving the equilibrium of rigid bodies under general force systems.
• In Dynamics, the same division is observed. The basic concepts of force, mass, and acceleration, of work and energy, and of impulse and momentum are introduced and first applied to problems involv-ing only particles. Thus, students can familiarize themselves with †Both texts also are available in a single volume, Vector Mechanics for Engineers: Statics and Dynamics, eleventh edition.
‡In a parallel text, Mechanics for Engineers: Statics, fifth edition, the use of vector algebra is limited to the addition and subtraction of vectors.
2.2 ADDING FORCES BY COMPONENTS In Sec. 2.1E, we described how to resolve a force into components. Here we discuss how to add forces by using their components, especially rectangular components. This method is often the most convenient way to add forces and, in practice, is the most common approach. (Note that we can readily extend the properties of vectors established in this section to the rectangular components of any vector quantity, such as velocity or momentum.) 2.2A Rectangular Components of a Force: Unit Vectors In many problems, it is useful to resolve a force into two components that are perpendicular to each other. Figure 2.14 shows a force F resolved into a component Fx along the x axis and a component Fy along the y axis. The parallelogram drawn to obtain the two components is a rectangle, and Fx and Fy are called rectangular components.
The x and y axes are usually chosen to be horizontal and vertical, respectively, as in Fig. 2.14; they may, however, be chosen in any two perpendicular directions, as shown in Fig. 2.15. In determining the O F Fy Fx x y Fig. 2.14 Rectangular components of a force F.
Fy Fx F x y O Fig. 2.15 Rectangular components of a force F for axes rotated away from horizontal and vertical. The 11th edition has undergone a complete rewrite to modernize and streamline the language throughout the text.
NEW!
xi Preface the three basic methods used in dynamics and learn their respective advantages before facing the difficulties associated with the motion of rigid bodies.
New Concepts Are Introduced in Simple Terms. Since this text is designed for the first course in statics, new concepts are presented in simple terms and every step is explained in detail. On the other hand, by discussing the broader aspects of the problems considered, and by stressing methods of general applicability, a definite maturity of approach is achieved. For example, the concepts of partial constraints and statical indeterminacy are introduced early and are used throughout.
Fundamental Principles Are Placed in the Context of Simple Applications. The fact that mechanics is essentially a deduc-tive science based on a few fundamental principles is stressed. Derivations have been presented in their logical sequence and with all the rigor war-ranted at this level. However, the learning process being largely inductive, simple applications are considered first. For example: • The statics of particles precedes the statics of rigid bodies, and prob-lems involving internal forces are postponed until Chap. 6.
• In Chap. 4, equilibrium problems involving only coplanar forces are considered first and solved by ordinary algebra, while problems involving three-dimensional forces and requiring the full use of vector algebra are discussed in the second part of the chapter.
Systematic Problem-Solving Approach. New to this edition of the text, all the sample problems are solved using the steps of Strategy, Modeling, Analysis, and Reflect & Think, or the “SMART” approach. This methodology is intended to give students confidence when approach-ing new problems, and students are encouraged to apply this approach in the solution of all assigned problems.
Free-Body Diagrams Are Used Both to Solve Equilibrium Problems and to Express the Equivalence of Force Systems. Free-body diagrams are introduced early, and their impor-tance is emphasized throughout the text. They are used not only to solve equilibrium problems but also to express the equivalence of two systems of forces or, more generally, of two systems of vectors. The advantage of this approach becomes apparent in the study of the dynamics of rigid bodies, where it is used to solve three-dimensional as well as two-dimensional problems. By placing the emphasis on “free-body-diagram equations” rather than on the standard algebraic equations of motion, a more intuitive and more complete understanding of the fundamental prin-ciples of dynamics can be achieved. This approach, which was first intro-duced in 1962 in the first edition of Vector Mechanics for Engineers, has now gained wide acceptance among mechanics teachers in this country. It is, therefore, used in preference to the method of dynamic equilibrium and to the equations of motion in the solution of all sample problems in this book. 4.1 EQUILIBRIUM IN TWO DIMENSIONS In the first part of this chapter, we consider the equilibrium of two-dimensional structures; i.e., we assume that the structure being analyzed and the forces applied to it are contained in the same plane. Clearly, the reactions needed to maintain the structure in the same position are also contained in this plane.
4.1A Reactions for a Two-Dimensional Structure The reactions exerted on a two-dimensional structure fall into three cat-egories that correspond to three types of supports or connections. 1. Reactions Equivalent to a Force with a Known Line of Action. Sup-ports and connections causing reactions of this type include rollers, rockers, frictionless surfaces, short links and cables, collars on friction-less rods, and frictionless pins in slots. Each of these supports and connections can prevent motion in one direction only. Figure 4.1 shows these supports and connections together with the reactions they produce. Each reaction involves one unknown––specifically, the magnitude of the reaction. In problem solving, you should denote this magnitude by an appropriate letter. The line of action of the reaction is known and should be indicated clearly in the free-body diagram. The sense of the reaction must be as shown in Fig. 4.1 for cases of a frictionless surface (toward the free body) or a cable (away from the free body). The reaction can be directed either way in the cases of double-track rollers, links, collars on rods, or pins in slots. Generally, we NEW!
xii Preface A Four-Color Presentation Uses Color to Distinguish Vectors. Color has been used, not only to enhance the quality of the illustrations, but also to help students distinguish among the various types of vectors they will encounter. While there was no intention to “color code” this text, the same color is used in any given chapter to represent vectors of the same type. Throughout Statics, for example, red is used exclusively to represent forces and couples, while position vectors are shown in blue and dimensions in black. This makes it easier for the students to identify the forces acting on a given particle or rigid body and to follow the discussion of sample problems and other examples given in the text.
A Careful Balance Between SI and U.S. Customary Units Is Consistently Maintained. Because of the current trend in the American government and industry to adopt the international system of units (SI metric units), the SI units most frequently used in mechanics are introduced in Chap. 1 and are used throughout the text. Approximately half of the sample problems and 60 percent of the homework problems are stated in these units, while the remainder are in U.S. customary units. The authors believe that this approach will best serve the need of students, who, as engineers, will have to be conversant with both systems of units. It also should be recognized that using both SI and U.S. customary units entails more than the use of conversion factors. Since the SI system of units is an absolute system based on the units of time, length, and mass, whereas the U.S. customary system is a gravitational system based on the units of time, length, and force, different approaches are required for the solution of many problems. For example, when SI units are used, a body is generally specified by its mass expressed in kilograms; in most prob-lems of statics it will be necessary to determine the weight of the body in newtons, and an additional calculation will be required for this purpose. On the other hand, when U.S. customary units are used, a body is speci-fied by its weight in pounds and, in dynamics problems, an additional calculation will be required to determine its mass in slugs (or lb?s2/ft). The authors, therefore, believe that problem assignments should include both systems of units. The Instructor’s and Solutions Manual provides six different lists of assignments so that an equal number of problems stated in SI units and in U.S. customary units can be selected. If so desired, two complete lists of assignments can also be selected with up to 75 percent of the problems stated in SI units.
Optional Sections Offer Advanced or Specialty Topics. A large number of optional sections have been included. These sections are indicated by asterisks and thus are easily distinguished from those which form the core of the basic statics course. They can be omitted without prejudice to the understanding of the rest of the text. Among the topics covered in these additional sections are the reduc-tion of a system of forces to a wrench, applications to hydrostatics, equi-librium of cables, products of inertia and Mohr’s circle, the determination of the principal axes and the mass moments of inertia of a body of arbi-trary shape, and the method of virtual work. The sections on the inertia Sample Problem 3.10 Three cables are attached to a bracket as shown. Replace the forces exerted by the cables with an equivalent force-couple system at A.
STRATEGY: First determine the relative position vectors drawn from point A to the points of application of the various forces and resolve the forces into rectangular components. Then sum the forces and moments.
MODELING and ANALYSIS: Note that FB 5 (700 N)lBE where lBE 5 BE BE 5 75i 2 150j 1 50k 175 Using meters and newtons, the position and force vectors are rB/A 5 AB 5 0.075i 1 0.050k FB 5 300i 2 600j 1 200k rC/A 5 AC 5 0.075i 2 0.050k FC 5 707i 2 707k rD/A 5 AD 5 0.100i 2 0.100j FD 5 600i 1 1039j The force-couple system at A equivalent to the given forces con-sists of a force R 5 oF and a couple MR A 5 o(r 3 F). Obtain the force R by adding respectively the x, y, and z components of the forces: R 5 oF 5 (1607 N)i 1 (439 N)j 2 (507 N)k b (continued) 50 mm 50 mm 100 mm 100 mm 75 mm 1000 N 1200 N 700 N x y z O A B C D 45º 45º 30º 60º E(150 mm, –50 mm, 100 mm) Remark: Since all the forces are contained in the plane of the figure, you would expect the sum of their moments to be perpendicular to that plane. Note that you could obtain the moment of each force component directly from the diagram by first forming the product of its magnitude and perpendicular distance to O and then assigning to this product a posi-tive or a negative sign, depending upon the sense of the moment.
b. Single Tugboat. The force exerted by a single tugboat must be equal to R, and its point of application A must be such that the moment of R about O is equal to MR O (Fig. 3). Observing that the position vector of A is r 5 xi 1 70j you have r 3 R 5 MR O (xi 1 70j) 3 (9.04i 2 9.79j) 5 21035k 2x(9.79)k 2 633k 5 21035k x 5 41.1 ft b REFLECT and THINK: Reducing the given situation to that of a single force makes it easier to visualize the overall effect of the tugboats in maneuvering the ocean liner. But in practical terms, having four boats applying force allows for greater control in slowing and turning a large ship in a crowded harbor.
Fig. 3 Point of application of single tugboat to create same effect as given force system.
70 ft x 9.04i – 9.79j R A O xiii Preface properties of three-dimensional bodies are primarily intended for students who will later study in dynamics the three-dimensional motion of rigid bodies. The material presented in the text and most of the problems require no previous mathematical knowledge beyond algebra, trigonometry, and elementary calculus; all the elements of vector algebra necessary to the understanding of the text are carefully presented in Chaps. 2 and 3. In general, a greater emphasis is placed on the correct understanding of the basic mathematical concepts involved than on the nimble manipulation of mathematical formulas. In this connection, it should be mentioned that the determination of the centroids of composite areas precedes the calculation of centroids by integration, thus making it possible to establish the concept of the moment of an area firmly before introducing the use of integration.
xiv Guided Tour Chapter Introduction. Each chapter begins with a list of learning objectives and an outline that previews chapter topics. An introductory section describes the material to be covered in simple terms, and how it will be applied to the solution of engineering problems. Chapter Lessons. The body of the text is divided into sections, each consisting of one or more sub-sections, several sample problems, and a large number of end-of-section problems for students to solve. Each sec-tion corresponds to a well-defined topic and generally can be covered in one lesson. In a number of cases, however, the instructor will find it desir-able to devote more than one lesson to a given topic. The Instructor’s and Solutions Manual contains suggestions on the coverage of each lesson.
Concept Applications. Concept Applications are used within selected theory sections to amplify certain topics, and they are designed to reinforce the specific material being presented and facilitate its understanding.
Sample Problems. The Sample Problems are set up in much the same form that students will use when solving assigned problems, and they employ the SMART problem-solving methodology that students are encouraged to use in the solution of their assigned problems. They thus serve the double purpose of amplifying the text and demonstrating the type of neat and orderly work that students should cultivate in their own solutions. In addition, in-problem references and captions have been added to the sample problem figures for contextual linkage to the step-by-step solution.
Solving Problems on Your Own. A section entitled Solving Prob-lems on Your Own is included for each lesson, between the sample problems and the problems to be assigned. The purpose of these sections is to help students organize in their own minds the preceding theory of the text and the solution methods of the sample problems so that they can more success-fully solve the homework problems. Also included in these sections are spe-cific suggestions and strategies that will enable the students to more efficiently attack any assigned problems.
Homework Problem Sets. Most of the problems are of a practical nature and should appeal to engineering students. They are primarily designed, however, to illustrate the material presented in the text and to help students understand the principles of mechanics. The problems are grouped according to the portions of material they illustrate and, in general, are arranged in order of increasing difficulty. Problems requiring special attention are indi-cated by asterisks. Answers to 70 percent of the problems are given at the end of the book. Problems for which the answers are given are set in straight type in the text, while problems for which no answer is given are set in italic and red font color. Sample Problem 4.10 A 450-lb load hangs from the corner C of a rigid piece of pipe ABCD that has been bent as shown. The pipe is supported by ball-and-socket joints A and D, which are fastened, respectively, to the floor and to a vertical wall, and by a cable attached at the midpoint E of the portion BC of the pipe and at a point G on the wall. Determine (a) where G should be located if the tension in the cable is to be minimum, (b) the correspond-ing minimum value of the tension.
12 ft 12 ft 450 lb A B C D E G 6 ft 6 ft 6 ft STRATEGY: Draw the free-body diagram of the pipe showing the reac-tions at A and D. Isolate the unknown tension T and the known weight W by summing moments about the diagonal line AD, and compute values from the equilibrium equations.
MODELING and ANALYSIS: Free-Body Diagram. The free-body diagram of the pipe includes the load W 5 (2450 lb)j, the reactions at A and D, and the force T exerted by the cable (Fig. 1). To eliminate the reactions at A and D from the computations, take the sum of the moments of the forces about the line AD and set it equal to zero. Denote the unit vector along AD by λ, which enables you to write oMAD 5 0: λ ? (AE 3 T) 1 λ ? (AC 3 W) 5 0 (1) Fig. 1 Free-body diagram of pipe.
A B C D E x y z T Dxi Dyj Dzk A xi Ay j Azk W = –450 j 6 ft 6 ft 12 ft 12 ft 12 ft NEW! Over 300 of the homework problems in the text are new or revised.
NEW!
xv Guided Tour Chapter Review and Summary. Each chapter ends with a review and summary of the material covered in that chapter. Marginal notes are used to help students organize their review work, and cross-references have been included to help them find the portions of material requiring their special attention.
Review Problems. A set of review problems is included at the end of each chapter. These problems provide students further opportunity to apply the most important concepts introduced in the chapter.
Computer Problems. Accessible through Connect are problem sets for each chapter that are designed to be solved with computational software. Many of these problems are relevant to the design process; they may involve the analysis of a structure for various configurations and loadings of the structure, or the determination of the equilibrium positions of a given mech-anism that may require an iterative method of solution. Developing the algorithm required to solve a given mechanics problem will benefit the students in two different ways: (1) it will help them gain a better understand-ing of the mechanics principles involved; (2) it will provide them with an opportunity to apply their computer skills to the solution of a meaningful engineering problem.
75 In this chapter, we have studied the effect of forces on particles, i.e., on bodies of such shape and size that we may assume all forces acting on them apply at the same point.
Resultant of Two Forces Forces are vector quantities; they are characterized by a point of application, a magnitude, and a direction, and they add according to the parallelogram law (Fig. 2.30). We can determine the magnitude and direction of the resultant R of two forces P and Q either graphically or by trigonometry using the law of cosines and the law of sines [Sample Prob. 2.1].
Components of a Force Any given force acting on a particle can be resolved into two or more com-ponents, i.e., it can be replaced by two or more forces that have the same effect on the particle. A force F can be resolved into two components P and Q by drawing a parallelogram with F for its diagonal; the components P and Q are then represented by the two adjacent sides of the parallelogram (Fig. 2.31). Again, we can determine the components either graphically or by trigonom-etry [Sec. 2.1E].
Review and Summary Q R P A Fig. 2.30 Q F P A Fig. 2.31 F x y Fy = Fy j Fx = Fxi j i Fig. 2.32 Rectangular Components; Unit Vectors A force F is resolved into two rectangular components if its components Fx and Fy are perpendicular to each other and are directed along the coordinate axes (Fig. 2.32). Introducing the unit vectors i and j along the x and y axes, respectively, we can write the components and the vector as [Sec. 2.2A] Fx 5 Fxi Fy 5 Fy j (2.6) and F 5 Fxi 1 Fyj (2.7) where Fx and Fy are the scalar components of F. These components, which can be positive or negative, are defined by the relations Fx 5 F cos θ Fy 5 F sin θ (2.8) 79 2.127 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigo-nometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B. 2.128 Determine the x and y components of each of the forces shown.
Review Problems A B 40° 20° Fig. P2.127 106 lb 102 lb 200 lb x y 24 in.
28 in.
45 in.
40 in.
30 in.
O Fig. P2.128 a a 200 lb 400 lb P Fig. P2.129 30° 20° α 300 lb A B C Fig. P2.130 2.129 A hoist trolley is subjected to the three forces shown. Knowing that α 5 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant. 2.130 Knowing that α 5 55° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.
xvi Connect® Engineering provides online presen-tation, assignment, and assessment solutions. It connects your students with the tools and resources they’ll need to achieve success. With Connect Engineering you can deliver assignments, quizzes, and tests online. A robust set of ques-tions and activities are presented and aligned with the textbook’s learning outcomes. As an instructor, you can edit existing questions and author entirely new problems. Integrate grade reports easily with Learning Man-agement Systems (LMS), such as WebCT and Blackboard—and much more. Connect Engineering also provides students with 24/7 online access to a media-rich eBook, allowing seamless integration of text, media, and assessments. To learn more, visit connect.mheducation.com Find the following instructor resources available through Connect: • Instructor’s and Solutions Manual. The Instructor’s and Solutions Manual that accompanies the eleventh edition features solutions to all end of chapter problems. This manual also features a number of tables designed to assist instructors in creating a schedule of assignments for their course. The various topics covered in the text have been listed in Table I and a suggested number of periods to be spent on each topic has been indicated. Table II prepares a brief description of all groups of problems and a classification of the problems in each group accord-ing to the units used. Sample lesson schedules are shown in Tables III, IV, and V, together with various alternative lists of assigned homework problems.
• Lecture PowerPoint Slides for each chapter that can be modified. These generally have an introductory application slide, animated worked-out problems that you can do in class with your students, concept questions, and “what-if?” questions at the end of the units.
• Textbook images • Computer Problem sets for each chapter that are designed to be solved with computational software.
• C.O.S.M.O.S., the Complete Online Solutions Manual Organization System that allows instructors to create custom homework, quizzes, and tests using end-of-chapter problems from the text. LearnSmart is available as an integrated feature of McGraw-Hill Connect. It is an adaptive learning system designed to help students learn faster, study more efficiently, and retain more knowledge for greater success. LearnSmart assesses a student’s knowledge of course content through a series of adaptive questions. It pinpoints concepts the student does not understand and maps out a personalized study plan for success. This innovative study tool also has features that allow instructors to see exactly what students have accomplished and a built-in assessment tool for graded assignments. Digital Resources NEW!
xvii NEW!
Digital Resources SmartBook™ is the first and only adaptive reading experience available for the higher education market. Powered by an intelligent diagnostic and adaptive engine, SmartBook facilitates the reading process by identifying what content a student knows and doesn’t know through adaptive assessments. As the student reads, the reading material constantly adapts to ensure the student is focused on the content he or she needs the most to close any knowledge gaps.
Visit the following site for a demonstration of LearnSmart or SmartBook: www.learnsmartadvantage.com CourseSmart. This text is offered through CourseSmart for both instructors and students. CourseSmart is an online browser where students can purchase access to this and other McGraw-Hill textbooks in a digital format. Through their browser, students can access the complete text online at almost half the cost of a traditional text. Purchasing the eText-book also allows students to take advantage of CourseSmart’s web tools for learning, which include full text search, notes and highlighting, and e-mail tools for sharing notes among classmates. To learn more about CourseSmart options, contact your sale s representative or visit www.coursesmart.com.
xviii A special thanks to Amy Mazurek who thoroughly checked the solutions and answers of all problems in this edition and then prepared the solutions for the Instructor’s and Solutions Manual. The authors thank the many companies that provided photographs for this edition. We are pleased to acknowledge David Chelton, who carefully reviewed the entire text and provided many helpful suggestions for revising this edition. The authors also thank the members of the staff at McGraw-Hill for their support and dedication during the preparation of this new edition. We particularly wish to acknowledge the contributions of Global Brand Manager Raghu Srinivasan, Brand Manager Thomas Scaife, Product Developers Robin Reed & Joan Weber, Content Project Manager Jolynn Kilburg, and Program Manager Lora Neyens.
David F. Mazurek The authors gratefully acknowledge the many helpful comments and suggestions offered by focus group attendees and by users of the previous editions of Vector Mechanics for Engineers: Acknowledgments George Adams Northeastern University William Altenhof University of Windsor Sean B. Anderson Boston University Manohar Arora Colorado School of Mines Gilbert Baladi Michigan State University Brock E. Barry United States Military Francois Barthelat McGill University Oscar Barton, Jr U.S. Naval Academy M. Asghar Bhatti University of Iowa Shaohong Cheng University of Windsor Philip Datseris University of Rhode Island Daniel Dickrell, III University of Florida Timothy A. Doughty University of Portland Howard Epstein University of Conneticut Asad Esmaeily Kansas State University, Civil Engineering Department David Fleming Florida Institute of Technology Ali Gordon University of Central Florida, Orlando Jeff Hanson Texas Tech University David A. Jenkins University of Florida Shaofan Li University of California, Berkeley Tom Mase California Polytechnic State University Gregory Miller University of Washington William R. Murray Cal Poly State University Eric Musslman University of Minnesota, Duluth Masoud Olia Wentworth Institute of Technology Mark Olles Renssalaer Polytechnic Institute Renee K. B. Petersen Washington State University Carisa Ramming Oklohoma State University Amir G Rezaei California State Polytechnic University, Pomona Martin Sadd University of Rhode Island Stefan Seelecke North Carolina State University Yixin Shao McGill University Muhammad Sharif The University of Alabama Anthony Sinclair University of Toronto Lizhi Sun University of California, lrvine Jeffrey Thomas Northwestern University Robert J. Witt University of Wisconsin, Madison Jiashi Yang University of Nebraska Xiangwa Zeng Case Western Reserve University xix a Constant; radius; distance A, B, C, . . . Reactions at supports and connections A, B, C, . . . Points A Area b Width; distance c Constant C Centroid d Distance e Base of natural logarithms F Force; friction force g Acceleration of gravity G Center of gravity; constant of gravitation h Height; sag of cable i, j, k Unit vectors along coordinate axes I, Ix, . . . Moments of inertia I Centroidal moment of inertia Ixy, . . . Products of inertia J Polar moment of inertia k Spring constant kx, ky, kO Radii of gyration k Centroidal radius of gyration l Length L Length; span m Mass M Couple; moment MO Moment about point O MR O Moment resultant about point O M Magnitude of couple or moment; mass of earth MOL Moment about axis OL N Normal component of reaction O Origin of coordinates p Pressure P Force; vector Q Force; vector r Position vector r Radius; distance; polar coordinate R Resultant force; resultant vector; reaction R Radius of earth s Position vector s Length of arc; length of cable S Force; vector t Thickness T Force T Tension U Work V Vector product; shearing force V Volume; potential energy; shear w Load per unit length W, W Weight; load x, y, z Rectangular coordinates; distances x, y, z Rectangular coordinates of centroid or center of gravity α, β, g Angles g Specific weight δ Elongation δr Virtual displacement δU Virtual work l Unit vector along a line η Efficiency θ Angular coordinate; angle; polar coordinate μ Coefficient of friction ρ Density f Angle of friction; angle List of Symbols This page intentionally left blank The tallest skyscraper in the Western Hemisphere, One World Trade Center is a prominent feature of the New York City skyline. From its foundation to its structural components and mechanical systems, the design and operation of the tower is based on the fundamentals of engineering mechanics.
Introduction 1 2 Introduction Introduction 1.1 WHAT IS MECHANICS? 1.2 FUNDAMENTAL CONCEPTS AND PRINCIPLES 1.3 SYSTEMS OF UNITS 1.4 CONVERTING BETWEEN TWO SYSTEMS OF UNITS 1.5 METHOD OF SOLVING PROBLEMS 1.6 NUMERICAL ACCURACY Objectives • Define the science of mechanics and examine its fundamental principles.
• Discuss and compare the International System of Units and U.S. Customary Units.
• Discuss how to approach the solution of mechanics problems, and introduce the SMART problem-solving methodology.
• Examine factors that govern numerical accuracy in the solution of a mechanics problem.
1.1 What is Mechanics?
Mechanics is defined as the science that describes and predicts the condi-tions of rest or motion of bodies under the action of forces. It consists of the mechanics of rigid bodies, mechanics of deformable bodies, and mechanics of fluids.
The mechanics of rigid bodies is subdivided into statics and dynamics. Statics deals with bodies at rest; dynamics deals with bodies in motion. In this text, we assume bodies are perfectly rigid. In fact, actual structures and machines are never absolutely rigid; they deform under the loads to which they are subjected. However, because these deformations are usu-ally small, they do not appreciably affect the conditions of equilibrium or the motion of the structure under consideration. They are important, though, as far as the resistance of the structure to failure is concerned. Deformations are studied in a course in mechanics of materials, which is part of the mechanics of deformable bodies. The third division of mechan-ics, the mechanics of fluids, is subdivided into the study of incompressible fluids and of compressible fluids. An important subdivision of the study of incompressible fluids is hydraulics, which deals with applications involving water.
Mechanics is a physical science, since it deals with the study of physical phenomena. However, some teachers associate mechanics with mathematics, whereas many others consider it as an engineering subject. Both these views are justified in part. Mechanics is the foundation of most engineering sciences and is an indispensable prerequisite to their study. However, it does not have the empiricism found in some engineering sci-ences, i.e., it does not rely on experience or observation alone. The rigor of mechanics and the emphasis it places on deductive reasoning makes it resemble mathematics. However, mechanics is not an abstract or even a pure science; it is an applied science. The purpose of mechanics is to explain and predict physical phe-nomena and thus to lay the foundations for engineering applications. You need to know statics to determine how much force will be exerted on a point in a bridge design and whether the structure can withstand that force. Determining the force a dam needs to withstand from the water in a river requires statics. You need statics to calculate how much weight a crane can lift, how much force a locomotive needs to pull a freight train, or how 1.2 Fundamental Concepts and Principles 3 much force a circuit board in a computer can withstand. The concepts of dynamics enable you to analyze the flight characteristics of a jet, design a building to resist earthquakes, and mitigate shock and vibration to pas-sengers inside a vehicle. The concepts of dynamics enable you to calculate how much force you need to send a satellite into orbit, accelerate a 200,000-ton cruise ship, or design a toy truck that doesn’t break. You will not learn how to do these things in this course, but the ideas and methods you learn here will be the underlying basis for the engineering applications you will learn in your work.
1.2 Fundamental Concepts and Principles Although the study of mechanics goes back to the time of Aristotle (384– 322 b.c.) and Archimedes (287–212 b.c.), not until Newton (1642–1727) did anyone develop a satisfactory formulation of its fundamental princi-ples. These principles were later modified by d’Alembert, Lagrange, and Hamilton. Their validity remained unchallenged until Einstein formulated his theory of relativity (1905). Although its limitations have now been recognized, newtonian mechanics still remains the basis of today’s engi-neering sciences.
The basic concepts used in mechanics are space, time, mass, and force. These concepts cannot be truly defined; they should be accepted on the basis of our intuition and experience and used as a mental frame of reference for our study of mechanics.
The concept of space is associated with the position of a point P. We can define the position of P by providing three lengths measured from a certain reference point, or origin, in three given directions. These lengths are known as the coordinates of P.
To define an event, it is not sufficient to indicate its position in space. We also need to specify the time of the event.
We use the concept of mass to characterize and compare bodies on the basis of certain fundamental mechanical experiments. Two bodies of the same mass, for example, are attracted by the earth in the same manner; they also offer the same resistance to a change in translational motion.
A force represents the action of one body on another. A force can be exerted by actual contact, like a push or a pull, or at a distance, as in the case of gravitational or magnetic forces. A force is characterized by its point of application, its magnitude, and its direction; a force is repre-sented by a vector (Sec. 2.1B).
In newtonian mechanics, space, time, and mass are absolute con-cepts that are independent of each other. (This is not true in relativistic mechanics, where the duration of an event depends upon its position and the mass of a body varies with its velocity.) On the other hand, the concept of force is not independent of the other three. Indeed, one of the funda-mental principles of newtonian mechanics listed below is that the resultant force acting on a body is related to the mass of the body and to the manner in which its velocity varies with time.
In this text, you will study the conditions of rest or motion of par-ticles and rigid bodies in terms of the four basic concepts we have intro-duced. By particle, we mean a very small amount of matter, which we 4 Introduction assume occupies a single point in space. A rigid body consists of a large number of particles occupying fixed positions with respect to one another. The study of the mechanics of particles is clearly a prerequisite to that of rigid bodies. Besides, we can use the results obtained for a particle directly in a large number of problems dealing with the conditions of rest or motion of actual bodies.
The study of elementary mechanics rests on six fundamental prin-ciples, based on experimental evidence.
• The Parallelogram Law for the Addition of Forces. Two forces acting on a particle may be replaced by a single force, called their resultant, obtained by drawing the diagonal of the parallelogram with sides equal to the given forces (Sec. 2.1A).
• The Principle of Transmissibility. The conditions of equilibrium or of motion of a rigid body remain unchanged if a force acting at a given point of the rigid body is replaced by a force of the same magnitude and same direction, but acting at a different point, pro-vided that the two forces have the same line of action (Sec. 3.1B).
• Newton’s Three Laws of Motion. Formulated by Sir Isaac Newton in the late seventeenth century, these laws can be stated as follows: FIRST LAW. If the resultant force acting on a particle is zero, the particle remains at rest (if originally at rest) or moves with constant speed in a straight line (if originally in motion) (Sec. 2.3B). SECOND LAW. If the resultant force acting on a particle is not zero, the particle has an acceleration proportional to the magnitude of the resultant and in the direction of this resultant force.
As you will see in Sec. 12.1, this law can be stated as F 5 ma (1.1) where F, m, and a represent, respectively, the resultant force acting on the particle, the mass of the particle, and the acceleration of the particle expressed in a consistent system of units. THIRD LAW. The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and oppo-site sense (Ch. 6, Introduction).
• Newton’s Law of Gravitation. Two particles of mass M and m are mutually attracted with equal and opposite forces F and 2F of magnitude F (Fig. 1.1), given by the formula F 5 G Mm r2 (1.2) where r 5 the distance between the two particles and G 5 a uni-versal constant called the constant of gravitation. Newton’s law of gravitation introduces the idea of an action exerted at a distance and extends the range of application of Newton’s third law: the action F and the reaction 2F in Fig. 1.1 are equal and opposite, and they have the same line of action.
A particular case of great importance is that of the attraction of the earth on a particle located on its surface. The force F exerted by the earth on the particle is defined as the weight W of the particle. Suppose we set M –F F m r Fig. 1.1 From Newton‘s law of gravitation, two particles of masses M and m exert forces upon each other of equal magnitude, opposite direction, and the same line of action. This also illustrates Newton‘s third law of motion.
1.3 Systems of Units 5 M equal to the mass of the earth, m equal to the mass of the particle, and r equal to the earth’s radius R. Then introducing the constant g 5 GM R2 (1.3) we can express the magnitude W of the weight of a particle of mass m as† W 5 mg (1.4) The value of R in formula (1.3) depends upon the elevation of the point considered; it also depends upon its latitude, since the earth is not truly spherical. The value of g therefore varies with the position of the point considered. However, as long as the point actually remains on the earth’s surface, it is sufficiently accurate in most engineering computations to assume that g equals 9.81 m/s2 or 32.2 ft/s2.
The principles we have just listed will be introduced in the course of our study of mechanics as they are needed. The statics of particles carried out in Chap. 2 will be based on the parallelogram law of addition and on Newton’s first law alone. We introduce the principle of transmis-sibility in Chap. 3 as we begin the study of the statics of rigid bodies, and we bring in Newton’s third law in Chap. 6 as we analyze the forces exerted on each other by the various members forming a structure. We introduce Newton’s second law and Newton’s law of gravitation in dynamics. We will then show that Newton’s first law is a particular case of Newton’s second law (Sec. 12.1) and that the principle of transmissibility could be derived from the other principles and thus eliminated (Sec. 16.1D). In the meantime, however, Newton’s first and third laws, the parallelogram law of addition, and the principle of transmissibility will provide us with the necessary and sufficient foundation for the entire study of the statics of particles, rigid bodies, and systems of rigid bodies.
As noted earlier, the six fundamental principles listed previously are based on experimental evidence. Except for Newton’s first law and the prin-ciple of transmissibility, they are independent principles that cannot be derived mathematically from each other or from any other elementary physical prin-ciple. On these principles rests most of the intricate structure of newtonian mechanics. For more than two centuries, engineers have solved a tremendous number of problems dealing with the conditions of rest and motion of rigid bodies, deformable bodies, and fluids by applying these fundamental prin-ciples. Many of the solutions obtained could be checked experimentally, thus providing a further verification of the principles from which they were derived. Only in the twentieth century has Newton’s mechanics found to be at fault, in the study of the motion of atoms and the motion of the planets, where it must be supplemented by the theory of relativity. On the human or engineering scale, however, where velocities are small compared with the speed of light, Newton’s mechanics have yet to be disproved.
1.3 Systems of Units Associated with the four fundamental concepts just discussed are the so-called kinetic units, i.e., the units of length, time, mass, and force. These units cannot be chosen independently if Eq. (1.1) is to be satisfied. †A more accurate definition of the weight W should take into account the earth’s rotation.
Photo 1.1 When in orbit of the earth, people and objects are said to be weightless even though the gravitational force acting is approximately 90% of that experienced on the surface of the earth. This apparent contradiction will be resolved in Chapter 12 when we apply Newton’s second law to the motion of particles.
6 Introduction Three of the units may be defined arbitrarily; we refer to them as basic units. The fourth unit, however, must be chosen in accordance with Eq. (1.1) and is referred to as a derived unit. Kinetic units selected in this way are said to form a consistent system of units.
International System of Units (SI Units).† In this system, which will be in universal use after the United States has completed its conver-sion to SI units, the base units are the units of length, mass, and time, and they are called, respectively, the meter (m), the kilogram (kg), and the second (s). All three are arbitrarily defined. The second was originally chosen to represent 1/86 400 of the mean solar day, but it is now defined as the duration of 9 192 631 770 cycles of the radiation corresponding to the transition between two levels of the fundamental state of the cesium-133 atom. The meter, originally defined as one ten-millionth of the distance from the equator to either pole, is now defined as 1 650 763.73 wave-lengths of the orange-red light corresponding to a certain transition in an atom of krypton-86. (The newer definitions are much more precise and with today’s modern instrumentation, are easier to verify as a standard.) The kilogram, which is approximately equal to the mass of 0.001 m3 of water, is defined as the mass of a platinum-iridium standard kept at the International Bureau of Weights and Measures at Sèvres, near Paris, France. The unit of force is a derived unit. It is called the newton (N) and is defined as the force that gives an acceleration of 1 m/s2 to a body of mass 1 kg (Fig. 1.2). From Eq. (1.1), we have 1 N 5 (1 kg)(1 m/s2) 5 1 kg?m/s2 (1.5) The SI units are said to form an absolute system of units. This means that the three base units chosen are independent of the location where measure-ments are made. The meter, the kilogram, and the second may be used anywhere on the earth; they may even be used on another planet and still have the same significance.
The weight of a body, or the force of gravity exerted on that body, like any other force, should be expressed in newtons. From Eq. (1.4), it follows that the weight of a body of mass 1 kg (Fig. 1.3) is W 5 mg 5 (1 kg)(9.81 m/s2) 5 9.81 N Multiples and submultiples of the fundamental SI units are denoted through the use of the prefixes defined in Table 1.1. The multiples and submultiples of the units of length, mass, and force most frequently used in engineering are, respectively, the kilometer (km) and the millimeter (mm); the megagram‡ (Mg) and the gram (g); and the kilonewton (kN). Accord-ing to Table 1.1, we have 1 km 5 1000 m 1 mm 5 0.001 m 1 Mg 5 1000 kg 1 g 5 0.001 kg 1 kN 5 1000 N The conversion of these units into meters, kilograms, and newtons, respec-tively, can be effected by simply moving the decimal point three places †SI stands for Système International d’Unités (French) ‡Also known as a metric ton.
a = 1 m/s2 m = 1 kg F = 1 N Fig. 1.2 A force of 1 newton applied to a body of mass 1 kg provides an acceleration of 1 m/s2.
a = 9.81 m/s2 m = 1 kg W = 9.81 N Fig. 1.3 A body of mass 1 kg experiencing an acceleration due to gravity of 9.81 m/s2 has a weight of 9.81 N.
1.3 Systems of Units 7 to the right or to the left. For example, to convert 3.82 km into meters, move the decimal point three places to the right: 3.82 km 5 3820 m Similarly, to convert 47.2 mm into meters, move the decimal point three places to the left: 47.2 mm 5 0.0472 m Using engineering notation, you can also write 3.82 km 5 3.82 3 103 m 47.2 mm 5 47.2 3 1023 m The multiples of the unit of time are the minute (min) and the hour (h). Since 1 min 5 60 s and 1 h 5 60 min 5 3600 s, these multiples cannot be converted as readily as the others.
By using the appropriate multiple or submultiple of a given unit, you can avoid writing very large or very small numbers. For example, it is usually simpler to write 427.2 km rather than 427 200 m and 2.16 mm rather than 0.002 16 m.† Units of Area and Volume. The unit of area is the square meter (m2), which represents the area of a square of side 1 m; the unit of volume is the cubic meter (m3), which is equal to the volume of a cube of side 1 m. In order to avoid exceedingly small or large numerical values when com-puting areas and volumes, we use systems of subunits obtained by respec-tively squaring and cubing not only the millimeter, but also two intermediate †Note that when more than four digits appear on either side of the decimal point to express a quantity in SI units––as in 427 000 m or 0.002 16 m––use spaces, never commas, to sepa-rate the digits into groups of three. This practice avoids confusion with the comma used in place of a decimal point, which is the convention in many countries.
Table 1.1 Sl Prefixes Multiplication Factor Prefix† Symbol 1 000 000 000 000 5 1012 tera T 1 000 000 000 5 109 giga G 1 000 000 5 106 mega M 1 000 5 103 kilo k 100 5 102 hecto‡ h 10 5 101 deka‡ da 0.1 5 1021 deci‡ d 0.01 5 1022 centi‡ c 0.001 5 1023 milli m 0.000 001 5 1026 micro µ 0.000 000 001 5 1029 nano n 0.000 000 000 001 5 10212 pico p 0.000 000 000 000 001 5 10215 femto f 0.000 000 000 000 000 001 5 10218 atto a †The first syllable of every prefix is accented, so that the prefix retains its identity. Thus, the preferred pronunciation of kilometer places the accent on the first syllable, not the second.
‡The use of these prefixes should be avoided, except for the measurement of areas and volumes and for the nontechnical use of centimeter, as for body and clothing measurements.
8 Introduction submultiples of the meter: the decimeter (dm) and the centimeter (cm). By definition, 1 dm 5 0.1 m 5 1021 m 1 cm 5 0.01 m 5 1022 m 1 mm 5 0.001 m 5 1023 m Therefore, the submultiples of the unit of area are 1 dm2 5 (1 dm)2 5 (1021 m)2 5 1022 m2 1 cm2 5 (1 cm)2 5 (1022 m)2 5 1024 m2 1 mm2 5 (1 mm)2 5 (1023 m)2 5 1026 m2 Similarly, the submultiples of the unit of volume are 1 dm3 5 (1 dm)3 5 (1021 m)3 5 1023 m3 1 cm3 5 (1 cm)3 5 (1022 m)3 5 1026 m3 1 mm3 5 (1 mm)3 5 (1023 m)3 5 1029 m3 Note that when measuring the volume of a liquid, the cubic decimeter (dm3) is usually referred to as a liter (L).
Table 1.2 shows other derived SI units used to measure the moment of a force, the work of a force, etc. Although we will introduce these units in later chapters as they are needed, we should note an important rule at Table 1.2 Principal SI Units Used in Mechanics Quantity Unit Symbol Formula Acceleration Meter per second squared . . . m/s2 Angle Radian rad † Angular acceleration Radian per second squared . . . rad/s2 Angular velocity Radian per second . . . rad/s Area Square meter . . . m2 Density Kilogram per cubic meter . . . kg/m3 Energy Joule J N?m Force Newton N kg?m/s2 Frequency Hertz Hz s–1 Impulse Newton-second . . . kg?m/s Length Meter m ‡ Mass Kilogram kg ‡ Moment of a force Newton-meter . . . N?m Power Watt W J/s Pressure Pascal Pa N/m2 Stress Pascal Pa N/m2 Time Second s ‡ Velocity Meter per second . . . m/s Volume Solids Cubic meter . . . m3 Liquids Liter L 10–3 m3 Work Joule J N?m †Supplementary unit (1 revolution 5 2π rad 5 360°).
‡Base unit.
1.3 Systems of Units 9 this time: When a derived unit is obtained by dividing a base unit by another base unit, you may use a prefix in the numerator of the derived unit, but not in its denominator. For example, the constant k of a spring that stretches 20 mm under a load of 100 N is expressed as k 5 100 N 20 mm 5 100 N 0.020 m 5 5000 N/m or k 5 5 kN/m but never as k 5 5 N/mm.
U.S. Customary Units. Most practicing American engineers still commonly use a system in which the base units are those of length, force, and time. These units are, respectively, the foot (ft), the pound (lb), and the second (s). The second is the same as the corresponding SI unit. The foot is defined as 0.3048 m. The pound is defined as the weight of a platinum standard, called the standard pound, which is kept at the National Institute of Standards and Technology outside Washington D.C., the mass of which is 0.453 592 43 kg. Since the weight of a body depends upon the earth’s gravitational attraction, which varies with location, the standard pound should be placed at sea level and at a latitude of 45° to properly define a force of 1 lb. Clearly the U.S. customary units do not form an absolute system of units. Because they depend upon the gravitational attraction of the earth, they form a gravitational system of units.
Although the standard pound also serves as the unit of mass in com-mercial transactions in the United States, it cannot be used that way in engineering computations, because such a unit would not be consistent with the base units defined in the preceding paragraph. Indeed, when acted upon by a force of 1 lb––that is, when subjected to the force of gravity–– the standard pound has the acceleration due to gravity, g 5 32.2 ft/s2 (Fig. 1.4), not the unit acceleration required by Eq. (1.1). The unit of mass consistent with the foot, the pound, and the second is the mass that receives an acceleration of 1 ft/s2 when a force of 1 lb is applied to it (Fig. 1.5). This unit, sometimes called a slug, can be derived from the equation F 5 ma after substituting 1 lb for F and 1 ft/s2 for a. We have F 5 ma 1 lb 5 (1 slug)(1 ft/s2) This gives us 1 slug 5 1 lb 1 ft/s2 5 1 lb?s2/ft (1.6) Comparing Figs. 1.4 and 1.5, we conclude that the slug is a mass 32.2 times larger than the mass of the standard pound.
The fact that, in the U.S. customary system of units, bodies are characterized by their weight in pounds rather than by their mass in slugs is convenient in the study of statics, where we constantly deal with weights and other forces and only seldom deal directly with masses. However, in the study of dynamics, where forces, masses, and accelerations are involved, the mass m of a body is expressed in slugs when its weight W is given in pounds. Recalling Eq. (1.4), we write m 5 W g (1.7) where g is the acceleration due to gravity (g 5 32.2 ft/s2).
a = 32.2 ft /s2 m = 1 lb mass F = 1 lb Fig. 1.4 A body of 1 pound mass acted upon by a force of 1 pound has an acceleration of 32.2 ft/s2.
a = 1 ft /s2 m = 1 slug (= 1 lb • s2/ft) F = 1 lb Fig. 1.5 A force of 1 pound applied to a body of mass 1 slug produces an acceleration of 1 ft/s2.
10 Introduction Other U.S. customary units frequently encountered in engineering problems are the mile (mi), equal to 5280 ft; the inch (in.), equal to (1/12) ft; and the kilopound (kip), equal to 1000 lb. The ton is often used to represent a mass of 2000 lb but, like the pound, must be converted into slugs in engineering computations.
The conversion into feet, pounds, and seconds of quantities expressed in other U.S. customary units is generally more involved and requires greater attention than the corresponding operation in SI units. For exam-ple, suppose we are given the magnitude of a velocity v 5 30 mi/h and want to convert it to ft/s. First we write v 5 30 mi h Since we want to get rid of the unit miles and introduce instead the unit feet, we should multiply the right-hand member of the equation by an expression containing miles in the denominator and feet in the numerator. However, since we do not want to change the value of the right-hand side of the equation, the expression used should have a value equal to unity. The quotient (5280 ft)/(1 mi) is such an expression. Operating in a similar way to transform the unit hour into seconds, we have v 5 a30mi h b a5280 ft 1 mi b a 1 h 3600 sb Carrying out the numerical computations and canceling out units that appear in both the numerator and the denominator, we obtain v 5 44 ft s 5 44 ft/s 1.4 Converting Between Two Systems of Units In many situations, an engineer might need to convert into SI units a numerical result obtained in U.S. customary units or vice versa. Because the unit of time is the same in both systems, only two kinetic base units need be converted. Thus, since all other kinetic units can be derived from these base units, only two conversion factors need be remembered.
Units of Length. By definition, the U.S. customary unit of length is 1 ft 5 0.3048 m (1.8) It follows that 1 mi 5 5280 ft 5 5280(0.3048 m) 5 1609 m or 1 mi 5 1.609 km (1.9) Also, 1 in. 5 1 12 ft 5 1 12(0.3048 m) 5 0.0254 m 1.4 Converting Between Two Systems of Units 11 or 1 in. 5 25.4 mm (1.10) Units of Force. Recall that the U.S. customary unit of force (pound) is defined as the weight of the standard pound (of mass 0.4536 kg) at sea level and at a latitude of 45° (where g 5 9.807 m/s2). Then, using Eq. (1.4), we write W 5 mg 1 lb 5 (0.4536 kg)(9.807 m/s2) 5 4.448 kg?m/s2 From Eq. (1.5), this reduces to 1 lb 5 4.448 N (1.11) Units of Mass. The U.S. customary unit of mass (slug) is a derived unit. Thus, using Eqs. (1.6), (1.8), and (1.11), we have 1 slug 5 1 lb?s2/ft 5 1 lb 1 ft/s2 5 4.448 N 0.3048 m/s2 5 14.59 N?s2/m Again, from Eq. (1.5), 1 slug 5 1 lb?s2/ft 5 14.59 kg (1.12) Although it cannot be used as a consistent unit of mass, recall that the mass of the standard pound is, by definition, 1 pound mass 5 0.4536 kg (1.13) We can use this constant to determine the mass in SI units (kilograms) of a body that has been characterized by its weight in U.S. customary units (pounds).
To convert a derived U.S. customary unit into SI units, simply multiply or divide by the appropriate conversion factors. For example, to convert the moment of a force that is measured as M 5 47 lb?in. into SI units, use formulas (1.10) and (1.11) and write M 5 47 lb?in. 5 47(4.448 N)(25.4 mm) 5 5310 N?mm 5 5.31 N?m You can also use conversion factors to convert a numerical result obtained in SI units into U.S. customary units. For example, if the moment of a force is measured as M 5 40 N?m, follow the procedure at the end of Sec. 1.3 to write M 5 40 N?m 5 (40 N?m)a 1 ÿ lb 4.448 Nb a 1 ft 0.3048 mb Carrying out the numerical computations and canceling out units that appear in both the numerator and the denominator, you obtain M 5 29.5 lb?ft The U.S. customary units most frequently used in mechanics are listed in Table 1.3 with their SI equivalents.
Photo 1.2 In 1999, The Mars Climate Orbiter entered orbit around Mars at too low an altitude and disintegrated. Investigation showed that the software on board the probe interpreted force instructions in newtons, but the software at mission control on the earth was generating those instructions in terms of pounds.
12 Introduction 1.5 Method of Solving Problems You should approach a problem in mechanics as you would approach an actual engineering situation. By drawing on your own experience and intuition about physical behavior, you will find it easier to understand and formulate the problem. Once you have clearly stated and understood the problem, however, there is no place in its solution for arbitrary methodologies. The solution must be based on the six fundamental principles stated in Sec. 1.2 or on theorems derived from them. Every step you take in the solution must be justified on this basis. Strict rules must be followed, which lead to the solution in an almost automatic fashion, leaving no room for your intuition or “feeling.” After you have obtained an answer, you should check it. Here again, you may call upon Table 1.3 U.S. Customary Units and Their SI Equivalents Quantity U.S. Customary Unit SI Equivalent Acceleration ft/s2 0.3048 m/s2 in./s2 0.0254 m/s2 Area ft2 0.0929 m2 in2 645.2 mm2 Energy ft?lb 1.356 J Force kip 4.448 kN lb 4.448 N oz 0.2780 N Impulse lb?s 4.448 N?s Length ft 0.3048 m in. 25.40 mm mi 1.609 km Mass oz mass 28.35 g lb mass 0.4536 kg slug 14.59 kg ton 907.2 kg Moment of a force lb?ft 1.356 N?m lb?in. 0.1130 N?m Moment of inertia Of an area in4 0.4162 3 106 mm4 Of a mass lb?ft?s2 1.356 kg?m2 Momentum lb?s 4.448 kg?m/s Power ft?lb/s 1.356 W hp 745.7 W Pressure or stress lb/ft2 47.88 Pa lb/in2 (psi) 6.895 kPa Velocity ft/s 0.3048 m/s in./s 0.0254 m/s mi/h (mph) 0.4470 m/s mi/h (mph) 1.609 km/h Volume ft3 0.02832 m3 in3 16.39 cm3 Liquids gal 3.785 L qt 0.9464 L Work ft?lb 1.356 J 1.5 Method of Solving Problems 13 your common sense and personal experience. If you are not completely satisfied with the result, you should carefully check your formulation of the problem, the validity of the methods used for its solution, and the accuracy of your computations.
In general, you can usually solve problems in several different ways; there is no one approach that works best for everybody. However, we have found that students often find it helpful to have a general set of guidelines to use for framing problems and planning solutions. In the Sample Prob-lems throughout this text, we use a four-step method for approaching prob-lems, which we refer to as the SMART methodology: Strategy, Modeling, Analysis, and Reflect and Think. 1. Strategy. The statement of a problem should be clear and precise, and it should contain the given data and indicate what information is required. The first step in solving the problem is to decide what con-cepts you have learned that apply to the given situation and to connect the data to the required information. It is often useful to work backward from the information you are trying to find: Ask yourself what quanti-ties you need to know to obtain the answer, and if some of these quanti-ties are unknown, how can you find them from the given data. 2. Modeling. The first step in modeling is to define the system; that is, clearly define what you are setting aside for analysis. After you have selected a system, draw a neat sketch showing all quantities involved with a separate diagram for each body in the problem. For equilibrium problems, indicate clearly the forces acting on each body along with any relevant geometrical data, such as lengths and angles. (These diagrams are known as free-body diagrams and are described in detail in Sec. 2.3C and the beginning of Ch. 4.) 3. Analysis. After you have drawn the appropriate diagrams, use the fundamental principles of mechanics listed in Sec. 1.2 to write equations expressing the conditions of rest or motion of the bodies considered. Each equation should be clearly related to one of the free-body dia-grams and should be numbered. If you do not have enough equations to solve for the unknowns, try selecting another system, or reexamine your strategy to see if you can apply other principles to the problem. Once you have obtained enough equations, you can find a numerical solution by following the usual rules of algebra, neatly recording each step and the intermediate results. Alternatively, you can solve the resulting equations with your calculator or a computer. (For multipart problems, it is sometimes convenient to present the Modeling and Analysis steps together, but they are both essential parts of the overall process.) 4. Reflect and Think. After you have obtained the answer, check it carefully. Does it make sense in the context of the original problem? For instance, the problem may ask for the force at a given point of a structure. If your answer is negative, what does that mean for the force at the point? You can often detect mistakes in reasoning by checking the units. For example, to determine the moment of a force of 50 N about a point 0.60 m from its line of action, we write (Sec. 3.3A) M 5 Fd 5 (30 N)(0.60 m) 5 30 N?m 14 Introduction The unit N?m obtained by multiplying newtons by meters is the correct unit for the moment of a force; if you had obtained another unit, you would know that some mistake had been made.
You can often detect errors in computation by substituting the numerical answer into an equation that was not used in the solution and verifying that the equation is satisfied. The importance of correct computa-tions in engineering cannot be overemphasized. 1.6 Numerical Accuracy The accuracy of the solution to a problem depends upon two items: (1) the accuracy of the given data and (2) the accuracy of the computations per-formed. The solution cannot be more accurate than the less accurate of these two items. For example, suppose the loading of a bridge is known to be 75 000 lb with a possible error of 100 lb either way. The relative error that measures the degree of accuracy of the data is 100 lb 75 000 lb 5 0.0013 5 0.13% In computing the reaction at one of the bridge supports, it would be mean-ingless to record it as 14 322 lb. The accuracy of the solution cannot be greater than 0.13%, no matter how precise the computations are, and the possible error in the answer may be as large as (0.13/100)(14 322 lb) ≈ 20 lb. The answer should be properly recorded as 14 320 6 20 lb.
In engineering problems, the data are seldom known with an accu-racy greater than 0.2%. It is therefore seldom justified to write answers with an accuracy greater than 0.2%. A practical rule is to use four figures to record numbers beginning with a “1” and three figures in all other cases. Unless otherwise indicated, you should assume the data given in a problem are known with a comparable degree of accuracy. A force of 40 lb, for example, should be read as 40.0 lb, and a force of 15 lb should be read as 15.00 lb.
Electronic calculators are widely used by practicing engineers and engineering students. The speed and accuracy of these calculators facili-tate the numerical computations in the solution of many problems. How-ever, you should not record more significant figures than can be justified merely because you can obtain them easily. As noted previously, an accu-racy greater than 0.2% is seldom necessary or meaningful in the solution of practical engineering problems.
Many engineering problems can be solved by considering the equilibrium of a “particle.” In the case of this beam that is being hoisted into position, a relation between the tensions in the various cables involved can be obtained by considering the equilibrium of the hook to which the cables are attached.
Statics of Particles 2 16 Statics of Particles Introduction 2.1 ADDITION OF PLANAR FORCES 2.1A Force on a Particle: Resultant of Two Forces 2.1B Vectors 2.1C Addition of Vectors 2.1D Resultant of Several Concurrent Forces 2.1E Resolution of a Force into Components 2.2 ADDING FORCES BY COMPONENTS 2.2A Rectangular Components of a Force: Unit Vectors 2.2B Addition of Forces by Summing X and Y Components 2.3 FORCES AND EQUILIBRIUM IN A PLANE 2.3A Equilibrium of a Particle 2.3B Newton’s First Law of Motion 2.3C Problems Involving the Equilibrium of a Particle: Free-Body Diagrams 2.4 ADDING FORCES IN SPACE 2.4A Rectangular Components of a Force in Space 2.4B Force Defined by Its Magnitude and Two Points on Its Line of Action 2.4C Addition of Concurrent Forces in Space 2.5 FORCES AND EQUILIBRIUM IN SPACE Objectives • Describe force as a vector quantity.
• Examine vector operations useful for the analysis of forces.
• Determine the resultant of multiple forces acting on a particle.
• Resolve forces into components.
• Add forces that have been resolved into rectangular components.
• Introduce the concept of the free-body diagram.
• Use free-body diagrams to assist in the analysis of planar and spatial particle equilibrium problems.
Introduction In this chapter, you will study the effect of forces acting on particles. By the word “particle” we do not mean only tiny bits of matter, like an atom or an electron. Instead, we mean that the sizes and shapes of the bodies under consideration do not significantly affect the solutions of the problems. Another way of saying this is that we assume all forces acting on a given body act at the same point. This does not mean the object must be tiny—if you were modeling the mechanics of the Milky Way galaxy, for example, you could treat the Sun and the entire Solar System as just a particle.
Our first step is to explain how to replace two or more forces acting on a given particle by a single force having the same effect as the original forces. This single equivalent force is called the resultant of the original forces. After this step, we will derive the relations among the various forces acting on a particle in a state of equilibrium. We will use these relations to determine some of the forces acting on the particle.
The first part of this chapter deals with forces contained in a single plane. Because two lines determine a plane, this situation arises any time we can reduce the problem to one of a particle subjected to two forces that support a third force, such as a crate suspended from two chains or a traffic light held in place by two cables. In the second part of this chap-ter, we examine the more general case of forces in three-dimensional space.
2.1 ADDITION OF PLANAR FORCES Many important practical situations in engineering involve forces in the same plane. These include forces acting on a pulley, projectile motion, and an object in equilibrium on a flat surface. We will examine this situ-ation first before looking at the added complications of forces acting in three-dimensional space.
2.1 Addition of Planar Forces 17 2.1A Force on a Particle: Resultant of Two Forces A force represents the action of one body on another. It is generally char-acterized by its point of application, its magnitude, and its direction. Forces acting on a given particle, however, have the same point of applica-tion. Thus, each force considered in this chapter is completely defined by its magnitude and direction.
The magnitude of a force is characterized by a certain number of units. As indicated in Chap. 1, the SI units used by engineers to measure the mag-nitude of a force are the newton (N) and its multiple the kilonewton (kN), which is equal to 1000 N. The U.S. customary units used for the same purpose are the pound (lb) and its multiple the kilopound (kip), which is equal to 1000 lb. We saw in Chapter 1 that a force of 445 N is equivalent to a force of 100 lb or that a force of 100 N equals a force of about 22.5 lb.
We define the direction of a force by its line of action and the sense of the force. The line of action is the infinite straight line along which the force acts; it is characterized by the angle it forms with some fixed axis (Fig. 2.1). The force itself is represented by a segment of that line; through the use of an appropriate scale, we can choose the length of this segment to represent the magnitude of the force. We indicate the sense of the force by an arrowhead. It is important in defining a force to indicate its sense. Two forces having the same magnitude and the same line of action but a different sense, such as the forces shown in Fig. 2.1a and b, have directly opposite effects on a particle.
(a) A 30° Fixed axis Fixed axis 10 lb (b) A 30° 10 lb Fig. 2.1 The line of action of a force makes an angle with a given fixed axis. (a) The sense of the 10-lb force is away from particle A; (b) the sense of the 10-lb force is toward particle A.
Experimental evidence shows that two forces P and Q acting on a particle A (Fig. 2.2a) can be replaced by a single force R that has the same effect on the particle (Fig. 2.2c). This force is called the resultant of the forces P and Q. We can obtain R, as shown in Fig. 2.2b, by con-structing a parallelogram, using P and Q as two adjacent sides. The diago-nal that passes through A represents the resultant. This method for finding the resultant is known as the parallelogram law for the addition of two forces. This law is based on experimental evidence; it cannot be proved or derived mathematically.
2.1B Vectors We have just seen that forces do not obey the rules of addition defined in ordinary arithmetic or algebra. For example, two forces acting at a right angle to each other, one of 4 lb and the other of 3 lb, add up to a force of Fig. 2.2 (a) Two forces P and Q act on particle A. (b) Draw a parallelogram with P and Q as the adjacent sides and label the diagonal that passes through A as R. (c) R is the resultant of the two forces P and Q and is equivalent to their sum.
A P Q (a) A P R Resultant Parallelogram Q (b) A R (c) 18 Statics of Particles 5 lb acting at an angle between them, not to a force of 7 lb. Forces are not the only quantities that follow the parallelogram law of addition. As you will see later, displacements, velocities, accelerations, and momenta are other physical quantities possessing magnitude and direction that add according to the parallelogram law. All of these quantities can be repre-sented mathematically by vectors. Those physical quantities that have mag-nitude but not direction, such as volume, mass, or energy, are represented by plain numbers often called scalars to distinguish them from vectors.
Vectors are defined as mathematical expressions possessing magnitude and direction, which add according to the parallelogram law. Vectors are represented by arrows in diagrams and are distinguished from scalar quantities in this text through the use of boldface type (P). In longhand writing, a vector may be denoted by drawing a short arrow above the letter used to represent it ( P →). The magnitude of a vector defines the length of the arrow used to represent it. In this text, we use italic type to denote the magnitude of a vector. Thus, the magnitude of the vector P is denoted by P.
A vector used to represent a force acting on a given particle has a well-defined point of application––namely, the particle itself. Such a vec-tor is said to be a fixed, or bound, vector and cannot be moved without modifying the conditions of the problem. Other physical quantities, how-ever, such as couples (see Chap. 3), are represented by vectors that may be freely moved in space; these vectors are called free vectors. Still other physical quantities, such as forces acting on a rigid body (see Chap. 3), are represented by vectors that can be moved along their lines of action; they are known as sliding vectors.
Two vectors that have the same magnitude and the same direction are said to be equal, whether or not they also have the same point of application (Fig. 2.3); equal vectors may be denoted by the same letter.
The negative vector of a given vector P is defined as a vector having the same magnitude as P and a direction opposite to that of P (Fig. 2.4); the negative of the vector P is denoted by 2P. The vectors P and 2P are commonly referred to as equal and opposite vectors. Clearly, we have P 1 (2P) 5 0 2.1C Addition of Vectors By definition, vectors add according to the parallelogram law. Thus, we obtain the sum of two vectors P and Q by attaching the two vectors to the same point A and constructing a parallelogram, using P and Q as two adjacent sides (Fig. 2.5). The diagonal that passes through A represents the sum of the vectors P and Q, denoted by P 1 Q. The fact that the sign 1 is used for both vector and scalar addition should not cause any confusion if vector and scalar quantities are always carefully distinguished. Note that the magnitude of the vector P 1 Q is not, in general, equal to the sum P 1 Q of the magnitudes of the vectors P and Q.
Since the parallelogram constructed on the vectors P and Q does not depend upon the order in which P and Q are selected, we conclude that the addition of two vectors is commutative, and we write P 1 Q 5 Q 1 P (2.1) P 1 Q 5 Q 1 P Fig. 2.3 Equal vectors have the same magnitude and the same direction, even if they have different points of application.
P P Fig. 2.4 The negative vector of a given vector has the same magnitude but the opposite direction of the given vector.
P –P Fig. 2.5 Using the parallelogram law to add two vectors.
A P P + Q Q Photo 2.1 In its purest form, a tug-of-war pits two opposite and almost-equal forces against each other. Whichever team can generate the larger force, wins. As you can see, a competitive tug-of-war can be quite intense.
2.1 Addition of Planar Forces 19 From the parallelogram law, we can derive an alternative method for determining the sum of two vectors, known as the triangle rule. Consider Fig. 2.5, where the sum of the vectors P and Q has been deter-mined by the parallelogram law. Since the side of the parallelogram oppo-site Q is equal to Q in magnitude and direction, we could draw only half of the parallelogram (Fig. 2.6a). The sum of the two vectors thus can be found by arranging P and Q in tip-to-tail fashion and then connecting the tail of P with the tip of Q. If we draw the other half of the parallelo-gram, as in Fig. 2.6b, we obtain the same result, confirming that vector addition is commutative.
We define subtraction of a vector as the addition of the correspond-ing negative vector. Thus, we determine the vector P 2 Q, representing the difference between the vectors P and Q, by adding to P the negative vector 2Q (Fig. 2.7). We write P 2 Q 5 P 1 (2Q) (2.2) Fig. 2.6 The triangle rule of vector addition. (a) Adding vector Q to vector P equals (b) adding vector P to vector Q.
A A P P Q Q P + Q P + Q (a) (b) Fig. 2.7 Vector subtraction: Subtracting vector Q from vector P is the same as adding vector –Q to vector P.
P – Q P P Q –Q (a) (b) A P Q S P + Q P + Q + S P Q S P + Q + S A A P Q S Q + S P + Q + S P P Q Q S S P + Q + S = S + Q + P A (a) (b) (c) (d) Fig. 2.8 Graphical addition of vectors. (a) Applying the triangle rule twice to add three vectors; (b) the vectors can be added in one step by the polygon rule; (c) vector addition is associative; (d) the order of addition is immaterial.
Here again we should observe that, although we use the same sign to denote both vector and scalar subtraction, we avoid confusion by taking care to distinguish between vector and scalar quantities.
We now consider the sum of three or more vectors. The sum of three vectors P, Q, and S is, by definition, obtained by first adding the vectors P and Q and then adding the vector S to the vector P 1 Q. We write P 1 Q 1 S 5 (P 1 Q) 1 S (2.3) Similarly, we obtain the sum of four vectors by adding the fourth vector to the sum of the first three. It follows that we can obtain the sum of any number of vectors by applying the parallelogram law repeatedly to suc-cessive pairs of vectors until all of the given vectors are replaced by a single vector.
If the given vectors are coplanar, i.e., if they are contained in the same plane, we can obtain their sum graphically. For this case, repeated application of the triangle rule is simpler than applying the parallelogram law. In Fig. 2.8a, we find the sum of three vectors P, Q, and S in this manner. The triangle rule is first applied to obtain the sum P 1 Q of the 20 Statics of Particles vectors P and Q; we apply it again to obtain the sum of the vectors P 1 Q and S. However, we could have omitted determining the vector P 1 Q and obtain the sum of the three vectors directly, as shown in Fig. 2.8b, by arranging the given vectors in tip-to-tail fashion and connecting the tail of the first vector with the tip of the last one. This is known as the polygon rule for the addition of vectors.
The result would be unchanged if, as shown in Fig. 2.8c, we had replaced the vectors Q and S by their sum Q 1 S. We may thus write P 1 Q 1 S 5 (P 1 Q) 1 S 5 P 1 (Q 1 S) (2.4) which expresses the fact that vector addition is associative. Recalling that vector addition also has been shown to be commutative in the case of two vectors, we can write P 1 Q 1 S 5 (P 1 Q) 1 S 5 S 1 (P 1 Q) (2.5) 5 S 1 (Q 1 P) 5 S 1 Q 1 P This expression, as well as others we can obtain in the same way, shows that the order in which several vectors are added together is immaterial (Fig. 2.8d ).
Product of a Scalar and a Vector. It is convenient to denote the sum P 1 P by 2P, the sum P 1 P 1 P by 3P, and, in general, the sum of n equal vectors P by the product nP. Therefore, we define the product nP of a positive integer n and a vector P as a vector having the same direction as P and the magnitude nP. Extending this definition to include all scalars and recalling the definition of a negative vector given earlier, we define the product kP of a scalar k and a vector P as a vector having the same direction as P (if k is positive) or a direction opposite to that of P (if k is negative) and a magnitude equal to the product of P and the absolute value of k (Fig. 2.9).
2.1D Resultant of Several Concurrent Forces Consider a particle A acted upon by several coplanar forces, i.e., by several forces contained in the same plane (Fig. 2.10a). Since the forces all pass through A, they are also said to be concurrent. We can add the vectors representing the forces acting on A by the polygon rule (Fig. 2.10b). Since the use of the polygon rule is equivalent to the repeated application of the parallelogram law, the vector R obtained in this way represents the resul-tant of the given concurrent forces. That is, the single force R has the same effect on the particle A as the given forces. As before, the order in which we add the vectors P, Q, and S representing the given forces is immaterial.
2.1E Resolution of a Force into Components We have seen that two or more forces acting on a particle may be replaced by a single force that has the same effect on the particle. Conversely, a single P 1 Q 1 S 5 (P 1 Q) 1 S 5 P 1 (Q 1 S) P 1.5 P –2 P Fig. 2.9 Multiplying a vector by a scalar changes the vector’s magnitude, but not its direction (unless the scalar is negative, in which case the direction is reversed).
A A P P Q Q S S (a) R (b) Fig. 2.10 Concurrent forces can be added by the polygon rule.
2.1 Addition of Planar Forces 21 force F acting on a particle may be replaced by two or more forces that, together, have the same effect on the particle. These forces are called components of the original force F, and the process of substituting them for F is called resolving the force F into components.
Clearly, each force F can be resolved into an infinite number of possible sets of components. Sets of two components P and Q are the most important as far as practical applications are concerned. However, even then, the number of ways in which a given force F may be resolved into two components is unlimited (Fig. 2.11). A A A P P P Q Q Q F F F (a) (b) (c) Fig. 2.11 Three possible sets of components for a given force vector F.
A P Q F Fig. 2.12 When component P is known, use the triangle rule to find component Q.
Fig. 2.13 When the lines of action are known, use the parallelogram rule to determine components P and Q.
A P Q F In many practical problems, we start with a given vector F and want to determine a useful set of components. Two cases are of particular interest: 1. One of the Two Components, P, Is Known. We obtain the second component, Q, by applying the triangle rule and joining the tip of P to the tip of F (Fig. 2.12). We can determine the magnitude and direction of Q graphically or by trigonometry. Once we have determined Q, both components P and Q should be applied at A. 2. The Line of Action of Each Component Is Known. We obtain the magnitude and sense of the components by applying the parallelogram law and drawing lines through the tip of F that are parallel to the given lines of action (Fig. 2.13). This process leads to two well-defined com-ponents, P and Q, which can be determined graphically or computed trigonometrically by applying the law of sines.
You will encounter many similar cases; for example, you might know the direction of one component while the magnitude of the other component is to be as small as possible (see Sample Prob. 2.2). In all cases, you need to draw the appropriate triangle or parallelogram that satisfies the given conditions.
22 Statics of Particles Sample Problem 2.1 Two forces P and Q act on a bolt A. Determine their resultant.
STRATEGY: Two lines determine a plane, so this is a problem of two coplanar forces. You can solve the problem graphically or by trigonometry.
MODELING: For a graphical solution, you can use the parallelogram rule or the triangle rule for addition of vectors. For a trigonometric solu-tion, you can use the law of cosines and law of sines or use a right-triangle approach.
ANALYSIS: Graphical Solution. Draw to scale a parallelogram with sides equal to P and Q (Fig. 1). Measure the magnitude and direction of the resultant. They are R 5 98 N α 5 35° R 5 98 N a 35° b You can also use the triangle rule. Draw forces P and Q in tip-to-tail fashion (Fig. 2). Again measure the magnitude and direction of the resul-tant. The answers should be the same. R 5 98 N α 5 35° R 5 98 N a 35° b Trigonometric Solution. Using the triangle rule again, you know two sides and the included angle (Fig. 3). Apply the law of cosines. R2 5 P2 1 Q2 2 2PQ cos B R2 5 (40 N)2 1 (60 N)2 2 2(40 N)(60 N) cos 155° R 5 97.73 N Now apply the law of sines: sin A Q 5 sin B R sin A 60 N 5 sin 1558 97.73 N (1) Solving Eq. (1) for sin A, you obtain sin A 5 (60 N) sin1558 97.73 N Using a calculator, compute this quotient, and then obtain its arc sine: A 5 15.04° α 5 20° 1 A 5 35.04° Use three significant figures to record the answer (cf. Sec. 1.6): R 5 97.7 N a 35.0° b Alternative Trigonometric Solution. Construct the right triangle BCD (Fig. 4) and compute CD 5 (60 N) sin 25° 5 25.36 N BD 5 (60 N) cos 25° 5 54.38 N 25° 20° A Q = 60 N P = 40 N A P Q R a Fig. 1 Parallelogram law applied to add forces P and Q.
A P Q R Fig. 2 Triangle rule applied to add forces P and Q.
155º 25° 20° R B C P = 40 N Q = 60 N a A Fig. 3 Geometry of triangle rule applied to add forces P and Q.
25° 20° = 60 N Q R B C D 40 25.36 54.38 94.38 a A Fig. 4 Alternative geometry of triangle rule applied to add forces P and Q.
2.1 Addition of Planar Forces 23 Sample Problem 2.2 Two tugboats are pulling a barge. If the resultant of the forces exerted by the tugboats is a 5000-lb force directed along the axis of the barge, deter-mine (a) the tension in each of the ropes, given that α 5 45°, (b) the value of α for which the tension in rope 2 is minimum.
STRATEGY: This is a problem of two coplanar forces. You can solve the first part either graphically or analytically. In the second part, a graphi-cal approach readily shows the necessary direction for rope 2, and you can use an analytical approach to complete the solution.
MODELING: You can use the parallelogram law or the triangle rule to solve part (a). For part (b), use a variation of the triangle rule.
ANALYSIS: a. Tension for α 5 45°. Graphical Solution. Use the parallelogram law. The resultant (the diagonal of the parallelogram) is equal to 5000 lb and is directed to the right. Draw the sides parallel to the ropes (Fig. 1). If the drawing is done to scale, you should measure T1 5 3700 lb T2 5 2600 lb b (continued) Then, using triangle ACD, you have tan A 5 25.36 N 94.38 N A 5 15.048 R 5 25.36 sin A R 5 97.73 N Again, α 5 20° 1 A 5 35.04° R 5 97.7 N a 35.0° b REFLECT and THINK: An analytical solution using trigonometry pro-vides for greater accuracy. However, it is helpful to use a graphical solu-tion as a check.
30° 45° 30° 45° 5000 lb T1 T2 B Fig. 1 Parallelogram law applied to add forces T1 and T2.
30° 1 2 a A C B 24 Statics of Particles Trigonometric Solution. Use the triangle rule. Note that the triangle in Fig. 2 represents half of the parallelogram shown in Fig. 1. Using the law of sines, T1 sin 458 5 T2 sin 308 5 5000 lb sin 1058 With a calculator, compute and store the value of the last quotient. Mul-tiply this value successively by sin 45° and sin 30°, obtaining T1 5 3660 lb T2 5 2590 lb b b. Value of α for Minimum T2. To determine the value of α for which the tension in rope 2 is minimum, use the triangle rule again. In Fig. 3, line 1-19 is the known direction of T1. Several possible directions of T2 are shown by the lines 2-29. The minimum value of T2 occurs when T1 and T2 are perpendicular (Fig. 4). Thus, the minimum value of T2 is T2 5 (5000 lb) sin 30° 5 2500 lb Corresponding values of T1 and α are T1 5 (5000 lb) cos 30° 5 4330 lb α 5 90° 2 30° α 5 60° b REFLECT and THINK: Part (a) is a straightforward application of resolving a vector into components. The key to part (b) is recognizing that the minimum value of T2 occurs when T1 and T2 are perpendicular.
45° 30° 5000 lb 105° T1 T2 B Fig. 2 Triangle rule applied to add forces T1 and T2.
1 2 2 2 5000 lb 1' 2' 2' 2' Fig. 3 Determination of direction of minimum T2.
30° 5000 lb T1 T2 90° a B Fig. 4 Triangle rule applied for minimum T2.
25 25 T he preceding sections were devoted to adding vectors by using the parallelogram law, triangle rule, and polygon rule with application to forces.
We presented two sample problems. In Sample Prob. 2.1, we used the parallelogram law to determine the resultant of two forces of known magnitude and direction. In Sample Prob. 2.2, we used it to resolve a given force into two components of known direction.
You will now be asked to solve problems on your own. Some may resemble one of the sample problems; others may not. What all problems and sample problems in this section have in common is that they can be solved by direct application of the paral-lelogram law.
Your solution of a given problem should consist of the following steps: 1. Identify which forces are the applied forces and which is the resultant. It is often helpful to write the vector equation that shows how the forces are related. For example, in Sample Prob. 2.1 you could write R 5 P 1 Q You may want to keep this relation in mind as you formulate the next part of the solution.
2. Draw a parallelogram with the applied forces as two adjacent sides and the resultant as the included diagonal (Fig. 2.2). Alternatively, you can use the triangle rule with the applied forces drawn in tip-to-tail fashion and the resultant extending from the tail of the first vector to the tip of the second (Fig. 2.6).
3. Indicate all dimensions. Using one of the triangles of the parallelogram or the triangle constructed according to the triangle rule, indicate all dimensions—whether sides or angles—and determine the unknown dimensions either graphically or by trigonometry.
4. Recall the laws of trigonometry. If you use trigonometry, remember that the law of cosines should be applied first if two sides and the included angle are known [Sample Prob. 2.1], and the law of sines should be applied first if one side and all angles are known [Sample Prob. 2.2].
If you have had prior exposure to mechanics, you might be tempted to ignore the solution techniques of this lesson in favor of resolving the forces into rectangular components. The component method is important and is considered in the next sec-tion, but use of the parallelogram law simplifies the solution of many problems and should be mastered first.
SOLVING PROBLEMS ON YOUR OWN 26 Problems 2.1 Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of their resultant using (a) the parallelo-gram law, (b) the triangle rule. 2.2 Two forces are applied as shown to a bracket support. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. 2.3 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P 5 10 kN and Q 5 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P 5 6 kips and Q 5 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.
45° 30° 900 N 600 N Fig. P2.1 500 lb 800 lb 35° 60° Fig. P2.2 C A B 25° 50° P Q Fig. P2.3 and P2.4 27 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α 5 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. 120 N P α 25° Fig. P2.5 A B 25° 15° T1 T2 Fig. P2.6 and P2.7 30° B C A α Fig. P2.8 and P2.9 2.6 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 5 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.
2.7 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 5 1000 lb, deter-mine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. 2.8 A disabled automobile is pulled by means of two ropes as shown. The tension in rope AB is 2.2 kN, and the angle α is 25°. Knowing that the resultant of the two forces applied at A is directed along the axis of the automobile, determine by trigonometry (a) the tension in rope AC, (b) the magnitude of the resultant of the two forces applied at A. 2.9 A disabled automobile is pulled by means of two ropes as shown. Knowing that the tension in rope AB is 3 kN, determine by trigo-nometry the tension in rope AC and the value of α so that the resul-tant force exerted at A is a 4.8-kN force directed along the axis of the automobile.
28 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.
Fig. P2.19 and P2.20 A 55° 25° 85° P Q 50 N 25° P a Fig. P2.10 425 lb A P 30° a Fig. P2.11, P2.12 and P2.13 2.11 A steel tank is to be positioned in an excavation. Knowing that α 5 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. 2.12 A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. 2.13 A steel tank is to be positioned in an excavation. Determine by trigo-nometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R. 2.14 For the hook support of Prob. 2.10, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R. 2.15 For the hook support shown, determine by trigonometry the magni-tude and direction of the resultant of the two forces applied to the support. 2.16 Solve Prob. 2.1 by trigonometry. 2.17 Solve Prob. 2.4 by trigonometry.
2.18 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N. 2.19 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P 5 48 N and Q 5 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.
2.20 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P 5 60 N and Q 5 48 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.
25° 45° 200 lb 300 lb Fig. P2.15 2.2 Adding Forces by Components 29 2.2 ADDING FORCES BY COMPONENTS In Sec. 2.1E, we described how to resolve a force into components. Here we discuss how to add forces by using their components, especially rectangular components. This method is often the most convenient way to add forces and, in practice, is the most common approach. (Note that we can readily extend the properties of vectors established in this section to the rectangular components of any vector quantity, such as velocity or momentum.) 2.2A Rectangular Components of a Force: Unit Vectors In many problems, it is useful to resolve a force into two components that are perpendicular to each other. Figure 2.14 shows a force F resolved into a component Fx along the x axis and a component Fy along the y axis. The parallelogram drawn to obtain the two components is a rectangle, and Fx and Fy are called rectangular components.
The x and y axes are usually chosen to be horizontal and vertical, respectively, as in Fig. 2.14; they may, however, be chosen in any two perpendicular directions, as shown in Fig. 2.15. In determining the O F Fy Fx x y Fig. 2.14 Rectangular components of a force F.
Fy Fx F x y O Fig. 2.15 Rectangular components of a force F for axes rotated away from horizontal and vertical.
x y Magnitude = 1 j i Fig. 2.16 Unit vectors along the x and y axes.
rectangular components of a force, you should think of the construction lines shown in Figs. 2.14 and 2.15 as being parallel to the x and y axes, rather than perpendicular to these axes. This practice will help avoid mis-takes in determining oblique components, as in Sec. 2.1E.
Force in Terms of Unit Vectors. To simplify working with rect-angular components, we introduce two vectors of unit magnitude, directed respectively along the positive x and y axes. These vectors are called unit vectors and are denoted by i and j, respectively (Fig. 2.16). Recalling the 30 Statics of Particles definition of the product of a scalar and a vector given in Sec. 2.1C, note that we can obtain the rectangular components Fx and Fy of a force F by multiplying respectively the unit vectors i and j by appropriate scalars (Fig. 2.17). We have Fx 5 Fxi Fy 5 Fyj (2.6) and F 5 Fx i 1 F y j (2.7) The scalars Fx and Fy may be positive or negative, depending upon the sense of Fx and of Fy, but their absolute values are equal to the magnitudes of the component forces Fx and Fy, respectively. The scalars Fx and Fy are called the scalar components of the force F, whereas the actual component forces Fx and Fy should be referred to as the vector components of F. However, when there exists no possibility of confusion, we may refer to the vector as well as the scalar components of F as simply the components of F. Note that the scalar component Fx is positive when the vector com-ponent Fx has the same sense as the unit vector i (i.e., the same sense as the positive x axis) and is negative when Fx has the opposite sense. A simi-lar conclusion holds for the sign of the scalar component Fy.
Scalar Components. Denoting by F the magnitude of the force F and by θ the angle between F and the x axis, which is measured counter-clockwise from the positive x axis (Fig. 2.17), we may express the scalar components of F as Fx 5 F cos θ Fy 5 F sin θ (2.8) These relations hold for any value of the angle θ from 0° to 360°, and they define the signs as well as the absolute values of the scalar compo-nents Fx and Fy.
F 5 Fx F i 1 F y j Fx F 5 F cos θ θ Fy F 5 F sin F θ Concept Application 2.1 A force of 800 N is exerted on a bolt A as shown in Fig. 2.18a. Determine the horizontal and vertical components of the force.
Solution In order to obtain the correct sign for the scalar components Fx and Fy, we could substitute the value 180° 2 35° 5 145° for θ in Eqs. (2.8). However, it is often more practical to determine by inspection the signs of Fx and Fy (Fig. 2.18b) and then use the trigonometric functions of the angle α 5 35°. Therefore, Fx 5 2F cos α 5 2(800 N) cos 35° 5 2655 N Fy 5 1F sin α 5 1(800 N) sin 35° 5 1459 N The vector components of F are thus Fx 5 2(655 N)i Fy 5 1(459 N)j and we may write F in the form F 5 2(655 N)i 1 (459 N)j b Fig. 2.18 (a) Force F exerted on a bolt; (b) rectangular components of F.
F = 800 N F = 800 N 35º A A (a) (b) x y Fy Fx = 35º = 145º F x y Fy = Fy j = F sin j Fx = Fxi = F cos i j i Fig. 2.17 Expressing the components of F in terms of unit vectors with scalar multipliers.
2.2 Adding Forces by Components 31 Concept Application 2.2 A man pulls with a force of 300 N on a rope attached to the top of a building, as shown in Fig. 2.19a. What are the horizontal and vertical components of the force exerted by the rope at point A?
Solution You can see from Fig. 2.19b that Fx 5 1(300 N) cos α Fy 5 2(300 N) sin α Observing that AB 5 10 m, we find from Fig. 2.19a cos α 5 8 m AB 5 8 m 10 m 5 4 5 sin α 5 6 m AB 5 6 m 10 m 5 3 5 We thus obtain Fx 5 1(300 N)4 5 5 1240 N Fy 5 2(300 N)3 5 5 2180 N This gives us a total force of F 5 (240 N)i 2 (180 N)j b Fig. 2.19 (a) A man pulls on a rope attached to a building; (b) components of the rope’s force F.
(a) 6 m 8 m A B (b) F = 300 N A Fy Fx x y Direction of a Force. When a force F is defined by its rectangular components Fx and Fy (see Fig. 2.17), we can find the angle θ defining its direction from tan θ 5 Fy Fx (2.9) We can obtain the magnitude F of the force by applying the Pythagorean theorem, F 5 2F x 2 1 F y 2 (2.10) or by solving for F from one of the Eqs. (2.8).
32 Statics of Particles Concept Application 2.3 A force F 5 (700 lb)i 1 (1500 lb)j is applied to a bolt A. Determine the magnitude of the force and the angle θ it forms with the horizontal.
Solution First draw a diagram showing the two rectangular components of the force and the angle θ (Fig. 2.20). From Eq. (2.9), you obtain tan θ 5 Fy Fx 5 1500 lb 700 lb Using a calculator, enter 1500 lb and divide by 700 lb; computing the arc tangent of the quotient gives you θ 5 65.0°. Solve the second of Eqs. (2.8) for F to get F 5 Fy sin θ 5 1500 lb sin 65.08 5 1655 lb The last calculation is easier if you store the value of Fy when originally entered; you may then recall it and divide it by sin θ.
Fig. 2.20 Components of a force F exerted on a bolt.
A x y F Fx = (700 lb) i Fy = (1500 lb) j 2.2B Addition of Forces by Summing X and Y Components We described in Sec. 2.1A how to add forces according to the parallelo-gram law. From this law, we derived two other methods that are more readily applicable to the graphical solution of problems: the triangle rule for the addition of two forces and the polygon rule for the addition of three or more forces. We also explained that the force triangle used to define the resultant of two forces could be used to obtain a trigonometric solution.
However, when we need to add three or more forces, we cannot obtain any practical trigonometric solution from the force polygon that defines the resultant of the forces. In this case, the best approach is to obtain an analytic solution of the problem by resolving each force into two rectangular components. Consider, for instance, three forces P, Q, and S acting on a particle A (Fig. 2.21a). Their resultant R is defined by the relation R 5 P 1 Q 1 S (2.11) Resolving each force into its rectangular components, we have Rx i 1 R y j 5 Px i 1 Py j 1 Qx i 1 Qy j 1 Sx i 1 Sy j 5 (Px 1 Qx 1 Sx )i 1 (Py 1 Qy 1 Sy)j S P Q A (a) Fig. 2.21 (a) Three forces acting on a particle.
2.2 Adding Forces by Components 33 From this equation, we can see that Rx 5 Px 1 Qx 1 Sx Ry 5 Py 1 Qy 1 Sy (2.12) or for short, Rx 5 oFx Ry 5 oFy (2.13) We thus conclude that when several forces are acting on a particle, we obtain the scalar components Rx and Ry of the resultant R by adding algebraically the corresponding scalar components of the given forces. (Clearly, this result also applies to the addition of other vector quantities, such as velocities, accelerations, or momenta.) In practice, determining the resultant R is carried out in three steps, as illustrated in Fig. 2.21. 1. Resolve the given forces (Fig. 2.21a) into their x and y components (Fig. 2.21b). (d) A R q Fig. 2.21 (d) Determining the resultant from its components.
(b) A Pyj Sy j Sxi Qyj Qxi Pxi Fig. 2.21 (b) Rectangular components of each force. 2. Add these components to obtain the x and y components of R (Fig. 2.21c). (c) A Ry j R xi Fig. 2.21 (c) Summation of the components. 3. Apply the parallelogram law to determine the resultant R 5 Rx i 1 Ry j (Fig. 2.21d ). The procedure just described is most efficiently carried out if you arrange the computations in a table (see Sample Problem 2.3). Although this is the only practical analytic method for adding three or more forces, it is also often preferred to the trigonometric solution in the case of adding two forces.
34 Statics of Particles Sample Problem 2.3 Four forces act on bolt A as shown. Determine the resultant of the forces on the bolt.
STRATEGY: The simplest way to approach a problem of adding four forces is to resolve the forces into components.
MODELING: As we mentioned, solving this kind of problem is usually easier if you arrange the components of each force in a table. In the table below, we entered the x and y components of each force as determined by trigonometry (Fig. 1). According to the convention adopted in this section, the scalar number representing a force component is positive if the force component has the same sense as the corresponding coordinate axis. Thus, x components acting to the right and y components acting upward are represented by positive numbers.
ANALYSIS: Force Magnitude, N x Component, N y Component, N F1 150 1129.9 175.0 F2 80 227.4 175.2 F3 110 0 2110.0 F4 100 196.6 225.9 Rx 5 1199.1 Ry 5 114.3 Thus, the resultant R of the four forces is R 5 Rx i 1 R y j R 5 (199.1 N)i 1 (14.3 N)j b You can now determine the magnitude and direction of the resultant. From the triangle shown in Fig. 2, you have tan α 5 Ry Rx 5 14.3 N 199.1 N α 5 4.18 R 5 14.3 N sinα 5 199.6 N R 5 199.6 N a 4.1° b F2 = 80 N F1 = 150 N F3 = 110 N F4 = 100 N 20° 30° 15° x y A (F2 cos 20°)j (F1 sin 30°)j (F1 cos 30°)i –(F2 sin 20°)i (F4 cos 15°)i –(F4 sin 15°)j –F3 j Fig. 1 Rectangular components of each force.
R Ry = (14.3 N)j Rx = (199.1 N)i a Fig. 2 Resultant of the given force system.
REFLECT and THINK: Arranging data in a table not only helps you keep track of the calculations, but also makes things simpler for using a calculator on similar computations.
35 35 Y ou saw in the preceding lesson that we can determine the resultant of two forces either graphically or from the trigonometry of an oblique triangle.
A. When three or more forces are involved, the best way to determine their resultant R is by first resolving each force into rectangular components. You may encounter either of two cases, depending upon the way in which each of the given forces is defined.
Case 1. The force F is defined by its magnitude F and the angle α it forms with the x axis. Obtain the x and y components of the force by multiplying F by cos α and sin α, respectively [Concept Application 2.1].
Case 2. The force F is defined by its magnitude F and the coordinates of two points A and B on its line of action (Fig. 2.19). Find the angle α that F forms with the x axis by trigonometry, and then use the process of Case 1. However, you can also find the components of F directly from proportions among the various dimensions involved without actually determining α [Concept Application 2.2].
B. Rectangular components of the resultant. Obtain the components Rx and Ry of the resultant by adding the corresponding components of the given forces algebraically [Sample Prob. 2.3].
You can express the resultant in vectorial form using the unit vectors i and j, which are directed along the x and y axes, respectively: R 5 R x i 1 R y j Alternatively, you can determine the magnitude and direction of the resultant by solv-ing the right triangle of sides Rx and Ry for R and for the angle that R forms with the x axis.
SOLVING PROBLEMS ON YOUR OWN 36 Problems 2.21 and 2.22 Determine the x and y components of each of the forces shown.
A C B 720 mm 650 mm Fig. P2.25 29 lb 51 lb O x y 90 in.
96 in.
28 in.
84 in.
80 in.
48 in.
50 lb Fig. P2.21 O Dimensions in mm 424 N 408 N 800 N x y 900 800 600 560 480 Fig. P2.22 60 lb 50 lb 40 lb 25° y x 60° 50° Fig. P2.24 80 N 120 N 150 N 30° 35° 40° y x Fig. P2.23 2.23 and 2.24 Determine the x and y components of each of the forces shown.
2.25 Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.
37 2.26 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, deter-mine (a) the magnitude of the force P, (b) its vertical component. 2.27 The hydraulic cylinder BC exerts on member AB a force P directed along line BC. Knowing that P must have a 600-N component per-pendicular to member AB, determine (a) the magnitude of the force P, (b) its component along line AB.
A B C D 35° Q Fig. P2.26 45° 30° B A M C Fig. P2.27 A B C 55° Q Fig. P2.28 60° 50° B C D A Q Fig. P2.29 2.28 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. 2.29 The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component per-pendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC.
2.30 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpen-dicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC. 2.31 Determine the resultant of the three forces of Prob. 2.21. 2.32 Determine the resultant of the three forces of Prob. 2.23.
2.33 Determine the resultant of the three forces of Prob. 2.24. 2.34 Determine the resultant of the three forces of Prob. 2.22.
A B C D 7 m 2.4 m Fig. P2.30 38 2.35 Knowing that α 5 35°, determine the resultant of the three forces shown.
200 N 150 N 100 N 30° a a Fig. P2.35 500 N 200 N 7 25 24 5 3 4 A B C L = 1460 mm 1100 mm 960 mm Fig. P2.36 120 lb 80 lb 60 lb a a' α α 20° Fig. P2.37 and P2.38 75 lb 50 lb 25° 65° 35° A B C Fig. P2.41 2.36 Knowing that the tension in rope AC is 365 N, determine the resul-tant of the three forces exerted at point C of post BC. 2.37 Knowing that α 5 40°, determine the resultant of the three forces shown. 2.38 Knowing that α 5 75°, determine the resultant of the three forces shown. 2.39 For the collar of Prob. 2.35, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant. 2.40 For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant. 2.41 Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant. 2.42 For the block of Probs. 2.37 and 2.38, determine (a) the required value of α if the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant.
2.3 Forces and Equilibrium in a Plane 39 2.3 FORCES AND EQUILIBRIUM IN A PLANE Now that we have seen how to add forces, we can proceed to one of the key concepts in this course: the equilibrium of a particle. The connection between equilibrium and the sum of forces is very direct: a particle can be in equilibrium only when the sum of the forces acting on it is zero.
2.3A Equilibrium of a Particle In the preceding sections, we discussed methods for determining the resul-tant of several forces acting on a particle. Although it has not occurred in any of the problems considered so far, it is quite possible for the resultant to be zero. In such a case, the net effect of the given forces is zero, and the particle is said to be in equilibrium. We thus have the definition: When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium.
A particle acted upon by two forces is in equilibrium if the two forces have the same magnitude and the same line of action but opposite sense. The resultant of the two forces is then zero, as shown in Fig. 2.22.
Another case of equilibrium of a particle is represented in Fig. 2.23a, where four forces are shown acting on particle A. In Fig. 2.23b, we use the polygon rule to determine the resultant of the given forces. Starting from point O with F1 and arranging the forces in tip-to-tail fashion, we find that the tip of F4 coincides with the starting point O. Thus, the resultant R of the given system of forces is zero, and the particle is in equilibrium.
A 100 lb 100 lb Fig. 2.22 When a particle is in equilibrium, the resultant of all forces acting on the particle is zero.
Fig. 2.23 (a) Four forces acting on particle A; (b) using the polygon law to find the resultant of the forces in (a), which is zero because the particle is in equilibrium.
A F1 = 300 lb F2 = 173.2 lb F4 = 400 lb F3 = 200 lb 30º 30º (a) F4 = 400 lb F1 = 300 lb F3 = 200 lb F2 = 173.2 lb O (b) Photo 2.2 Forces acting on the carabiner include the weight of the girl and her harness, and the force exerted by the pulley attachment. Treating the carabiner as a particle, it is in equilibrium because the resultant of all forces acting on it is zero.
The closed polygon drawn in Fig. 2.23b provides a graphical expres-sion of the equilibrium of A. To express algebraically the conditions for the equilibrium of a particle, we write Equilibrium of a particle R 5 oF 5 0 (2.14) R 5 oF 5 0 40 Statics of Particles Resolving each force F into rectangular components, we have o (F x i 1 Fy j) 5 0 or (o F x )i 1 (o F y ) j 5 0 We conclude that the necessary and sufficient conditions for the equilib-rium of a particle are Equilibrium of a particle (scalar equations) o Fx 5 0 o Fy 5 0 (2.15) Returning to the particle shown in Fig. 2.23, we can check that the equi-librium conditions are satisfied. We have o Fx 5 300 lb 2 (200 lb) sin 30° 2 (400 lb) sin 30° 5 300 lb 2 100 lb 2 200 lb 5 0 o Fy 5 2173.2 lb 2 (200 lb) cos 30° 1 (400 lb) cos 30° 5 2173.2 lb 2 173.2 lb 1 346.4 lb 5 0 2.3B Newton’s First Law of Motion As we discussed in Section 1.2, Sir Isaac Newton formulated three fun-damental laws upon which the science of mechanics is based. The first of these laws can be stated as: If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in motion).
From this law and from the definition of equilibrium just presented, we can see that a particle in equilibrium is either at rest or moving in a straight line with constant speed. If a particle does not behave in either of these ways, it is not in equilibrium, and the resultant force on it is not zero. In the following section, we consider various problems concerning the equilibrium of a particle. Note that most of statics involves using Newton’s first law to ana-lyze an equilibrium situation. In practice, this means designing a bridge or a building that remains stable and does not fall over. It also means understanding the forces that might act to disturb equilibrium, such as a strong wind or a flood of water. The basic idea is pretty simple, but the applications can be quite complicated.
2.3C Free-Body Diagrams and Problem Solving In practice, a problem in engineering mechanics is derived from an actual physical situation. A sketch showing the physical conditions of the problem is known as a space diagram.
The methods of analysis discussed in the preceding sections apply to a system of forces acting on a particle. A large number of problems involving actual structures, however, can be reduced to problems concern-ing the equilibrium of a particle. The method is to choose a significant particle and draw a separate diagram showing this particle and all the oFx F 5 0 oFy F 5 0 2.3 Forces and Equilibrium in a Plane 41 forces acting on it. Such a diagram is called a free-body diagram. (The name derives from the fact that when drawing the chosen body, or particle, it is “free” from all other bodies in the actual situation.) As an example, consider the 75-kg crate shown in the space diagram of Fig. 2.24a. This crate was lying between two buildings, and is now being lifted onto a truck, which will remove it. The crate is supported by a vertical cable that is joined at A to two ropes, which pass over pulleys attached to the buildings at B and C. We want to determine the tension in each of the ropes AB and AC.
In order to solve this problem, we first draw a free-body diagram showing a particle in equilibrium. Since we are interested in the rope ten-sions, the free-body diagram should include at least one of these tensions or, if possible, both tensions. You can see that point A is a good free body for this problem. The free-body diagram of point A is shown in Fig. 2.24b. It shows point A and the forces exerted on A by the vertical cable and the two ropes. The force exerted by the cable is directed downward, and its magni-tude is equal to the weight W of the crate. Recalling Eq. (1.4), we write W 5 mg 5 (75 kg)(9.81 m/s2) 5 736 N and indicate this value in the free-body diagram. The forces exerted by the two ropes are not known. Since they are respectively equal in magni-tude to the tensions in rope AB and rope AC, we denote them by TAB and TAC and draw them away from A in the directions shown in the space diagram. No other detail is included in the free-body diagram.
Since point A is in equilibrium, the three forces acting on it must form a closed triangle when drawn in tip-to-tail fashion. We have drawn this force triangle in Fig. 2.24c. The values TAB and TAC of the tensions in the ropes may be found graphically if the triangle is drawn to scale, or they may be found by trigonometry. If we choose trigonometry, we use the law of sines: TAB sin 608 5 TAC sin 408 5 736 N sin 808 TAB 5 647 N TAC 5 480 N When a particle is in equilibrium under three forces, you can solve the problem by drawing a force triangle. When a particle is in equilibrium under more than three forces, you can solve the problem graphically by drawing a force polygon. If you need an analytic solution, you should solve the equations of equilibrium given in Sec. 2.3A: oFx 5 0 oFy 5 0 (2.15) These equations can be solved for no more than two unknowns. Similarly, the force triangle used in the case of equilibrium under three forces can be solved for only two unknowns.
The most common types of problems are those in which the two unknowns represent (1) the two components (or the magnitude and direc-tion) of a single force or (2) the magnitudes of two forces, each of known direction. Problems involving the determination of the maximum or mini-mum value of the magnitude of a force are also encountered (see Probs. 2.57 through 2.61).
TAB TAC A A B C 50º 30º 50º 30º (a) Space diagram (b) Free-body diagram (c) Force triangle 736 N TAB TAC 736 N 40º 60º 80º Fig. 2.24 (a) The space diagram shows the physical situation of the problem; (b) the free-body diagram shows one central particle and the forces acting on it; (c) the force triangle can be solved with the law of sines. Note that the forces form a closed triangle because the particle is in equilibrium and the resultant force is zero.
Photo 2.3 As illustrated in Fig. 2.24, it is possible to determine the tensions in the cables supporting the shaft shown by treating the hook as a particle and then applying the equations of equilibrium to the forces acting on the hook.
42 Statics of Particles Sample Problem 2.4 In a ship-unloading operation, a 3500-lb automobile is supported by a cable. A worker ties a rope to the cable at A and pulls on it in order to center the automobile over its intended position on the dock. At the moment illustrated, the automobile is stationary, the angle between the cable and the vertical is 2°, and the angle between the rope and the hori-zontal is 30°. What are the tensions in the rope and cable?
STRATEGY: This is a problem of equilibrium under three coplanar forces. You can treat point A as a particle and solve the problem using a force triangle.
MODELING and ANALYSIS: Free-Body Diagram. Choose point A as the particle and draw the complete free-body diagram (Fig. 1). TAB is the tension in the cable AB, and TAC is the tension in the rope.
Equilibrium Condition. Since only three forces act on point A, draw a force triangle to express that it is in equilibrium (Fig. 2). Using the law of sines, TAB sin 1208 5 TAC sin 28 5 3500 lb sin 588 With a calculator, compute and store the value of the last quotient. Multiplying this value successively by sin 120° and sin 2°, you obtain TAB 5 3570 lb TAC 5 144 lb b REFLECT and THINK: This is a common problem of knowing one force in a three-force equilibrium problem and calculating the other forces from the given geometry. This basic type of problem will occur often as part of more complicated situations in this text.
2° 30° A C B TAB TAC 2° 30° A 3500 lb Fig. 1 Free-body diagram of particle A.
TAB TAC 2° 3500 lb 120° 58° Fig. 2 Force triangle of the forces acting on particle A.
2.3 Forces and Equilibrium in a Plane 43 Sample Problem 2.5 Determine the magnitude and direction of the smallest force F that main-tains the 30-kg package shown in equilibrium. Note that the force exerted by the rollers on the package is perpendicular to the incline.
STRATEGY: This is an equilibrium problem with three coplanar forces that you can solve with a force triangle. The new wrinkle is to determine a minimum force. You can approach this part of the solution in a way similar to Sample Problem 2.2.
MODELING and ANALYSIS: Free-Body Diagram. Choose the package as a free body, assuming that it can be treated as a particle. Then draw the corresponding free-body diagram (Fig. 1).
Equilibrium Condition. Since only three forces act on the free body, draw a force triangle to express that it is in equilibrium (Fig. 2). Line 1-19 represents the known direction of P. In order to obtain the minimum value of the force F, choose the direction of F to be perpendicular to that of P. From the geometry of this triangle, F 5 (294 N) sin 15° 5 76.1 N α 5 15° F 5 76.1 N b15° b REFLECT and THINK: Determining maximum and minimum forces to maintain equilibrium is a common practical problem. Here the force needed is about 25% of the weight of the package, which seems reason-able for an incline of 15°.
Sample Problem 2.6 For a new sailboat, a designer wants to determine the drag force that may be expected at a given speed. To do so, she places a model of the proposed hull in a test channel and uses three cables to keep its bow on the center-line of the channel. Dynamometer readings indicate that for a given speed, the tension is 40 lb in cable AB and 60 lb in cable AE. Determine the drag force exerted on the hull and the tension in cable AC.
STRATEGY: The cables all connect at point A, so you can treat that as a particle in equilibrium. Because four forces act at A (tensions in three cables and the drag force), you should use the equilibrium conditions and sum forces by components to solve for the unknown forces.
15° 30 kg F 15° F P W = (30 kg)(9.81 m/s2) = 294 N Fig. 1 Free-body diagram of package, treated as a particle.
F P 15° 1 1' 294 N Fig. 2 Force triangle of the forces acting on package.
Flow A B C E 4 ft 4 ft 7 ft 1.5 ft a b (continued) 44 Statics of Particles MODELING and ANALYSIS: Determining the Angles. First, determine the angles α and β defin-ing the direction of cables AB and AC: tan α 5 7 ft 4 ft 5 1.75 tan β 5 1.5 ft 4 ft 5 0.375 α 5 60.268 β 5 20.568 Free-Body Diagram. Choosing point A as a free body, draw the free-body diagram (Fig. 1). It includes the forces exerted by the three cables on the hull, as well as the drag force FD exerted by the flow.
Equilibrium Condition. Because point A is in equilibrium, the resul-tant of all forces is zero: R 5 TAB 1 TAC 1 TAE 1 FD 5 0 (1) Because more than three forces are involved, resolve the forces into x and y components (Fig. 2): TAB 5 2(40 lb) sin 60.26°i 1 (40 lb) cos 60.26°j 5 2(34.73 lb)i 1 (19.84 lb)j TAC 5 TAC sin 20.56°i 1 TAC cos 20.56°j 5 0.3512TAC i 1 0.9363TAC j TAE 5 2(60 lb)j FD 5 FDi Substituting these expressions into Eq. (1) and factoring the unit vectors i and j, you have (234.73 lb 1 0.3512TAC 1 FD)i 1 (19.84 lb 1 0.9363TAC 2 60 lb)j 5 0 This equation is satisfied if, and only if, the coefficients of i and j are each equal to zero. You obtain the following two equilibrium equations, which express, respectively, that the sum of the x components and the sum of the y components of the given forces must be zero. (oFx 5 0:) 234.73 lb 1 0.3512TAC 1 FD 5 0 (2) (oFy 5 0:) 19.84 lb 1 0.9363TAC 2 60 lb 5 0 (3) From Eq. (3), you find TAC 5 142.9 lb b Substituting this value into Eq. (2) yields FD 5 119.66 lb b REFLECT and THINK: In drawing the free-body diagram, you assumed a sense for each unknown force. A positive sign in the answer indicates that the assumed sense is correct. You can draw the complete force poly-gon (Fig. 3) to check the results.
TAC FD TAB = 40 lb TAE = 60 lb a = 60.26° b = 20.56° A Fig. 1 Free-body diagram of particle A.
FDi TACsin 20.56°i TACcos 20.56°j 20.56° 60.26° (40 lb) cos 60.26°j –(40 lb) sin 60.26°i –(60 lb)j y x A Fig. 2 Rectangular components of forces acting on particle A.
TAC = 42.9 lb TAE = 60 lb TAB = 40 lb FD = 19.66 lb b = 20.56° a = 60.26° Fig. 3 Force polygon of forces acting on particle A.
45 45 W hen a particle is in equilibrium, the resultant of the forces acting on the particle must be zero. Expressing this fact in the case of a particle under coplanar forces provides you with two relations among these forces. As in the preceding sample problems, you can use these relations to determine two unknowns—such as the mag-nitude and direction of one force or the magnitudes of two forces.
Drawing a clear and accurate free-body diagram is a must in the solution of any equilibrium problem. This diagram shows the particle and all of the forces acting on it. Indicate in your free-body diagram the magnitudes of known forces, as well as any angle or dimensions that define the direction of a force. Any unknown magnitude or angle should be denoted by an appropriate symbol. Nothing else should be included in the free-body diagram. Skipping this step might save you pencil and paper, but it is very likely to lead you to a wrong solution.
Case 1. If the free-body diagram involves only three forces, the rest of the solution is best carried out by drawing these forces in tip-to-tail fashion to form a force triangle. You can solve this triangle graphically or by trigonometry for no more than two unknowns [Sample Probs. 2.4 and 2.5].
Case 2. If the free-body diagram indicates more than three forces, it is most practi-cal to use an analytic solution. Select x and y axes and resolve each of the forces into x and y components. Setting the sum of the x components and the sum of the y components of all the forces to zero, you obtain two equations that you can solve for no more than two unknowns [Sample Prob. 2.6].
We strongly recommend that, when using an analytic solution, you write the equations of equilibrium in the same form as Eqs. (2) and (3) of Sample Prob. 2.6. The practice adopted by some students of initially placing the unknowns on the left side of the equation and the known quantities on the right side may lead to confusion in assigning the appropriate sign to each term.
Regardless of the method used to solve a two-dimensional equilibrium problem, you can determine at most two unknowns. If a two-dimensional problem involves more than two unknowns, you must obtain one or more additional relations from the infor-mation contained in the problem statement.
SOLVING PROBLEMS ON YOUR OWN 46 FREE-BODY PRACTICE PROBLEMS 2.F1 Two cables are tied together at C and loaded as shown. Draw the free-body diagram needed to determine the tension in AC and BC. 2.F2 Two forces of magnitude TA 5 8 kips and TB 5 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, draw the free-body diagram needed to determine the magnitudes of the forces TC and TD. 2.F3 The 60-lb collar A can slide on a frictionless vertical rod and is connected as shown to a 65-lb counterweight C. Draw the free-body diagram needed to determine the value of h for which the system is in equilibrium. 2.F4 A chairlift has been stopped in the position shown. Knowing that each chair weighs 250 N and that the skier in chair E weighs 765 N, draw the free-body diagrams needed to determine the weight of the skier in chair F.
Problems A B C 1600 kg 960 mm 1100 mm 400 mm Fig. P2.F1 14 m 24 m 6 m 8.25 m 10 m 1.10 m A B C D F E Fig. P2.F4 40° TB TA TC TD Fig. P2.F2 65 lb 60 lb C A B h 15 in.
Fig. P2.F3 47 END-OF-SECTION PROBLEMS 2.43 Two cables are tied together at C and are loaded as shown. Deter-mine the tension (a) in cable AC, (b) in cable BC. 2.44 Two cables are tied together at C and are loaded as shown. Knowing that α 5 30°, determine the tension (a) in cable AC, (b) in cable BC.
2.45 Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. 2.46 Two cables are tied together at C and are loaded as shown. Knowing that P 5 500 N and α 5 60°, determine the tension in (a) in cable AC, (b) in cable BC.
2.47 Two cables are tied together at C and are loaded as shown. Deter-mine the tension (a) in cable AC, (b) in cable BC.
Fig. P2.43 A B C 400 lb 50° 30° Fig. P2.44 A B C 6 kN 55° α Fig. P2.45 3.4 m 2 m 4.8 m 3 m 1.98 kN A B C 3.6 m Fig. P2.46 45º A B C P 25º a Fig. P2.47 75° 75° 200 kg C A B 48 2.48 Knowing that α 5 20°, determine the tension (a) in cable AC, (b) in rope BC. 2.49 Two cables are tied together at C and are loaded as shown. Knowing that P 5 300 N, determine the tension in cables AC and BC. 2.50 Two cables are tied together at C and are loaded as shown. Deter-mine the range of values of P for which both cables remain taut. 2.51 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P 5 500 lb and Q 5 650 lb, determine the magnitudes of the forces exerted on rods A and B.
Fig. P2.48 5° A C B α 1200 lb Fig. P2.49 and P2.50 A B C 45° 30° 30° 200 N P 50° 40° A B P Q FA FB Fig. P2.51 and P2.52 FD FC FA FB B A D C 3 4 Fig. P2.53 and P2.54 2.52 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magni-tudes of the forces exerted on rods A and B are FA 5 750 lb and FB 5 400 lb, determine the magnitudes of P and Q. 2.53 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA 5 8 kN and FB 5 16 kN, determine the magnitudes of the other two forces. 2.54 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA 5 5 kN and FD 5 6 kN, determine the magnitudes of the other two forces.
49 2.55 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α 5 30° and β 5 10° and that the combined weight of the boatswain’s chair and the sailor is 200 lb, determine the tension (a) in the support cable ACB, (b) in the traction cable CD. 2.56 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α 5 25° and β 5 15° and that the tension in cable CD is 20 lb, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) the ten-sion in the support cable ACB. 2.57 For the cables of Prob. 2.44, find the value of α for which the ten-sion is as small as possible (a) in cable BC, (b) in both cables simul-taneously. In each case determine the tension in each cable. 2.58 For the cables of Prob. 2.46, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corre-sponding value of α. 2.59 For the situation described in Fig. P2.48, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. 2.60 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable.
A B C a b D Fig. P2.55 and P2.56 A B C P = 75 lb 30º 30º 60º Q Fig. P2.60 2.61 A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN.
A C 0.7 m B 1.2 m Fig. P2.61 50 2.62 For W 5 800 N, P 5 200 N, and d 5 600 mm, determine the value of h consistent with equilibrium. 500 N 150 N 150 N 50° 30° A α Fig. P2.65 P W d h d Fig. P2.62 50 lb x C B A P 20 in.
Fig. P2.63 and P2.64 2.64 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P 5 48 lb. 2.65 Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the mag-nitude of the resultant of the forces acting at A is less than 600 N. 2.63 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x 5 4.5 in., (b) x 5 15 in.
51 2.66 A 200-kg crate is to be supported by the rope-and-pulley arrange-ment shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilib-rium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chap. 4.) 2.4 m P A α 200 kg 0.75 m B Fig. P2.66 T T T T T (a) (b) (c) (d) (e) Fig. P2.67 A D B C P 25° 55° Q Fig. P2.69 and P2.70 2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Prob. 2.66.) 2.68 Solve parts b and d of Prob. 2.67, assuming that the free end of the rope is attached to the crate. 2.69 A load Q is applied to pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P 5 750 N, determine (a) the tension in cable ACB, (b) the magni-tude of load Q. 2.70 An 1800-N load Q is applied to pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Deter-mine (a) the tension in cable ACB, (b) the magnitude of load P.
52 Statics of Particles 2.4 ADDING FORCES IN SPACE The problems considered in the first part of this chapter involved only two dimensions; they were formulated and solved in a single plane. In the last part of this chapter, we discuss problems involving the three dimensions of space.
2.4A Rectangular Components of a Force in Space Consider a force F acting at the origin O of the system of rectangular coordinates x, y, and z. To define the direction of F, we draw the vertical plane OBAC containing F (Fig. 2.25a). This plane passes through the vertical y axis; its orientation is defined by the angle ϕ it forms with the xy plane. The direction of F within the plane is defined by the angle θy that F forms with the y axis. We can resolve the force F into a vertical component Fy and a horizontal component Fh; this operation, shown in Fig. 2.25b, is carried out in plane OBAC according to the rules developed earlier. The corresponding scalar components are Fy 5 F cos θy Fh 5 F sin θy (2.16) However, we can also resolve Fh into two rectangular components Fx and Fz along the x and z axes, respectively. This operation, shown in Fig. 2.25c, is carried out in the xz plane. We obtain the following expressions for the corresponding scalar components: Fx 5 Fh cos ϕ 5 F sin θy cos ϕ Fz 5 Fh sin ϕ 5 F sin θy sin ϕ (2.17) The given force F thus has been resolved into three rectangular vector components Fx , Fy , Fz , which are directed along the three coordinate axes.
We can now apply the Pythagorean theorem to the triangles OAB and OCD of Fig. 2.25: F 2 5 (OA)2 5 (OB)2 1 (BA)2 5 F 2 y 1 F 2 h F 2 h 5 (OC)2 5 (OD)2 1 (DC)2 5 F 2 x 1 F 2 z Eliminating F 2 h from these two equations and solving for F, we obtain the following relation between the magnitude of F and its rectangular scalar components: Magnitude of a force in space F 5 2F 2 x 1 F 2 y 1 F 2 z (2.18) The relationship between the force F and its three components Fx , Fy , and Fz is more easily visualized if we draw a “box” having Fx , Fy , and Fz for edges, as shown in Fig. 2.26. The force F is then represented by the main diagonal OA of this box. Figure 2.26b shows the right triangle F 5 2F 2 2 x 1 F 2 y 1 F 2 z 2 (a) A B C z y x O F y (b) Fh Fy A B C z y x O F y (c) Fh Fy Fx Fz E D B C z y x O Fig. 2.25 (a) A force F in an xyz coordinate system; (b) components of F along the y axis and in the xz plane; (c) components of F along the three rectangular axes.
2.4 Adding Forces in Space 53 OAB used to derive the first of the formulas (2.16): Fy 5 F cos θy . In Fig. 2.26a and c, two other right triangles have also been drawn: OAD and OAE. These triangles occupy positions in the box comparable with that of triangle OAB. Denoting by θx and θz , respectively, the angles that F forms with the x and z axes, we can derive two formulas similar to Fy 5 F cos θy . We thus write Scalar components of a force F Fx 5 F cos θx Fy 5 F cos θy Fz 5 F cos θz (2.19) The three angles θx , θy, and θz define the direction of the force F; they are more commonly used for this purpose than the angles θy and ϕ intro-duced at the beginning of this section. The cosines of θx, θy, and θz are known as the direction cosines of the force F.
Introducing the unit vectors i, j, and k, which are directed respec-tively along the x, y, and z axes (Fig. 2.27), we can express F in the form Vector expression of a force F F 5 F x i 1 F y j 1 F z k (2.20) where the scalar components Fx, Fy, and Fz are defined by the relations in Eq. (2.19).
Fx F 5 F cos F θx θ x Fy F 5 F cos F θy θ Fz F 5 F cos F θz θ F 5 F x i 1 F y j 1 F z k Fig. 2.26 (a) Force F in a three-dimensional box, showing its angle with the x axis; (b) force F and its angle with the y axis; (c) force F and its angle with the z axis.
Fx Fy Fz F x x y A D E O B C z (a) Fx Fy Fz F x y A D E O B C z (b) y Fx Fy Fz F z x y A D E O B C z (c) Fig. 2.27 The three unit vectors i, j, k lie along the three coordinate axes x, y, z, respectively.
y x z i k j 54 Statics of Particles Concept Application 2.4 A force of 500 N forms angles of 60°, 45°, and 120°, respectively, with the x, y, and z axes. Find the components Fx, Fy, and Fz of the force and express the force in terms of unit vectors.
Solution Substitute F 5 500 N, θx 5 60°, θy 5 45°, and θz 5 120° into formulas (2.19). The scalar components of F are then Fx 5 (500 N) cos 60° 5 1250 N Fy 5 (500 N) cos 45° 5 1354 N Fz 5 (500 N) cos 120° 5 2250 N Carrying these values into Eq. (2.20), you have F 5 (250 N)i 1 (354 N)j 2 (250 N)k As in the case of two-dimensional problems, a plus sign indicates that the component has the same sense as the corresponding axis, and a minus sign indicates that it has the opposite sense.
The angle a force F forms with an axis should be measured from the positive side of the axis and is always between 0 and 180°. An angle θx smaller than 90° (acute) indicates that F (assumed attached to O) is on the same side of the yz plane as the positive x axis; cos θx and Fx are then positive. An angle θx larger than 90° (obtuse) indicates that F is on the other side of the yz plane; cos θx and Fx are then negative. In Concept Application 2.4, the angles θx and θy are acute and θz is obtuse; conse-quently, Fx and Fy are positive and Fz is negative.
Substituting into Eq. (2.20) the expressions obtained for Fx, Fy, and Fz in Eq. (2.19), we have F 5 F (cos θx i 1 cos θy j 1 cos θz k) (2.21) This equation shows that the force F can be expressed as the product of the scalar F and the vector l 5 cos θx i 1 cos θy j 1 cos θz k (2.22) Clearly, the vector l is a vector whose magnitude is equal to 1 and whose direction is the same as that of F (Fig. 2.33). The vector l is referred to as the unit vector along the line of action of F. It follows from Eq. (2.22) that the components of the unit vector l are respectively equal to the direction cosines of the line of action of F: lx 5 cos θx ly 5 cos θy lz 5 cos θz (2.23) Fig. 2.28 Force F can be expressed as the product of its magnitude F and a unit vector l in the direction of F. Also shown are the components of F and its unit vector.
x y z λ λ (Magnitude = 1) F = Fλ λ Fyj Fxi Fzk cos yj cos zk cos xi 2.4 Adding Forces in Space 55 Note that the values of the three angles θx, θy, and θz are not inde-pendent. Recalling that the sum of the squares of the components of a vector is equal to the square of its magnitude, we can write l2 x 1 l2 y 1 l2 z 5 1 Substituting for lx, ly, and lz from Eq. (2.23), we obtain Relationship among direction cosines cos 2θx 1 cos 2θy 1 cos 2θz 5 1 (2.24) In Concept Application 2.4, for instance, once the values θx 5 60° and θy 5 45° have been selected, the value of θz must be equal to 60° or 120° in order to satisfy the identity in Eq. (2.24).
When the components Fx, Fy, and Fz of a force F are given, we can obtain the magnitude F of the force from Eq. (2.18). We can then solve relations in Eq. (2.19) for the direction cosines as cos θx 5 Fx F cos θy 5 Fy F cos θz 5 Fz F (2.25) From the direction cosines, we can find the angles θx, θy, and θz character-izing the direction of F.
cos 2θx θ 1 cos 2θy θ 1 cos 2θz θ 5 1 Concept Application 2.5 A force F has the components Fx 5 20 lb, Fy 5 230 lb, and Fz 5 60 lb. Determine its magnitude F and the angles θx, θy, and θz it forms with the coordinate axes.
Solution You can obtain the magnitude of F from formula (2.18): F 5 2F 2 x 1 F 2 y 1 F 2 z 5 2(20 lb)2 1 (230 lb)2 1 (60 lb)2 5 24900 lb 5 70 lb Substituting the values of the components and magnitude of F into Eqs. (2.25), the direction cosines are cos θx 5 Fx F 5 20 lb 70 lb cos θy 5 Fy F 5 230 lb 70 lb cos θz 5 Fz F 5 60 lb 70 lb Calculating each quotient and its arc cosine gives you θx 5 73.4° θy 5 115.4° θz 5 31.0° These computations can be carried out easily with a calculator.
56 Statics of Particles 2.4B Force Defined by its Magnitude and Two Points on its Line of Action In many applications, the direction of a force F is defined by the coordi-nates of two points, M(x1, y1, z1) and N(x2, y2, z2), located on its line of action (Fig. 2.29). Consider the vector MN joining M and N and of the same Fig. 2.29 A case where the line of action of force F is determined by the two points M and N. We can calculate the components of F and its direction cosines from the vector MN .
y x z O M(x1, y1, z1) N(x2, y2, z2) dy = y2 – y1 dz = z2 – z1 < 0 d x = x2 – x1 F λ sense as a force F. Denoting its scalar components by dx, dy, and dz, respectively, we write MN 5 dxi 1 dy j 1 dzk (2.26) We can obtain a unit vector l along the line of action of F (i.e., along the line MN) by dividing the vector MN by its magnitude MN. Substituting for MN from Eq. (2.26) and observing that MN is equal to the distance d from M to N, we have l 5 MN ¡ MN 5 1 d1dxi 1 dy j 1 dzk2 (2.27) Recalling that F is equal to the product of F and l, we have F 5 Fl 5 F d 1d x i 1 d y j 1 d z k2 (2.28) It follows that the scalar components of F are, respectively, Scalar components of force F Fx 5 Fdx d Fy 5 Fdy d Fz 5 Fdz d (2.29) The relations in Eq. (2.29) considerably simplify the determination of the components of a force F of given magnitude F when the line of action of F is defined by two points M and N. The calculation consists of Fx F 5 Fd F x d d Fy F 5 Fd F y d d Fz F 5 Fd F z d d 2.4 Adding Forces in Space 57 first subtracting the coordinates of M from those of N, then determining the components of the vector MN and the distance d from M to N. Thus, dx 5 x 2 2 x1 dy 5 y2 2 y1 dz 5 z 2 2 z1 d 5 2d 2 x 1 d 2 y 1 d 2 z Substituting for F and for dx, dy, dz, and d into the relations in Eq. (2.29), we obtain the components Fx, Fy, and Fz of the force.
We can then obtain the angles θx, θy, and θz that F forms with the coordinate axes from Eqs. (2.25). Comparing Eqs. (2.22) and (2.27), we can write Direction cosines of force F cos θx 5 dx d cos θy 5 dy d cos θz 5 dz d (2.30) In other words, we can determine the angles θx, θy, and θz directly from the components and the magnitude of the vector MN .
2.4C Addition of Concurrent Forces in Space We can determine the resultant R of two or more forces in space by sum-ming their rectangular components. Graphical or trigonometric methods are generally not practical in the case of forces in space.
The method followed here is similar to that used in Sec. 2.2B with coplanar forces. Setting R 5 oF we resolve each force into its rectangular components: Rxi 1 Ry j 1 Rzk 5 o (Fxi 1 Fy j 1 Fzk) 5 (oFx)i 1 (oFy)j 1 (oFz)k From this equation, it follows that Rectangular components of the resultant Rx 5 oFx R y 5 oFy Rz 5 oFz (2.31) The magnitude of the resultant and the angles θx, θy, and θz that the resul-tant forms with the coordinate axes are obtained using the method dis-cussed earlier in this section. We end up with Resultant of concurrent forces in space R 5 2R 2 x 1 R 2 y 1 R 2 z (2.32) cos θx 5 Rx R cos θy 5 Ry R cos θz 5 Rz R (2.33) cosθx θ 5 dx d d cosθy θ 5 dy d d cosθz θ 5 dz d d Rx 5 oFx F x R y 5 oFy F Rz 5 oFz F R 5 2R 2 2 x 1 R 2 y 1 R 2 z 2 cosθx θ 5 Rx R cosθy θ 5 Ry R cosθz θ 5 Rz R 58 Statics of Particles Sample Problem 2.7 A tower guy wire is anchored by means of a bolt at A. The tension in the wire is 2500 N. Determine (a) the components Fx, Fy, and Fz of the force acting on the bolt and (b) the angles θx, θy, and θz defining the direction of the force.
STRATEGY: From the given distances, we can determine the length of the wire and the direction of a unit vector along it. From that, we can find the components of the tension and the angles defining its direction.
MODELING and ANALYSIS: a. Components of the Force. The line of action of the force acting on the bolt passes through points A and B, and the force is directed from A to B. The components of the vector AB , which has the same direction as the force, are dx 5 240 m dy 5 180 m dz 5 130 m The total distance from A to B is AB 5 d 5 2d 2 x 1 d 2 y 1 d 2 z 5 94.3 m Denoting the unit vectors along the coordinate axes by i, j, and k, you have AB 5 2(40 m)i 1 (80 m)j 1 (30 m)k Introducing the unit vector λ 5 AB /AB (Fig. 1), you can express F in terms of AB → as F 5 Fλ 5 F AB AB 5 2500 N 94.3 m AB Substituting the expression for AB gives you F 5 2500 N 94.3 m 32(40 m)i 1 (80 m)j 1 (30 m)k4 5 2(1060 N)i 1 (2120 N)j 1 (795 N)k The components of F, therefore, are Fx 5 21060 N Fy 5 12120 N Fz 5 1795 N b b. Direction of the Force. Using Eqs. (2.25), you can write the direction cosines directly (Fig. 2): cos θx 5 Fx F 5 21060 N 2500 N cos θy 5 Fy F 5 12120 N 2500 N cos θz 5 Fz F 5 1795 N 2500 N A B 80 m 40 m 30 m A B F y z x k j i 80 m 40 m 30 m λ Fig. 1 Cable force acting on bolt at A, and its unit vector.
A B y z x qy qx qz Fig. 2 Direction angles for cable AB.
2.4 Adding Forces in Space 59 Calculating each quotient and its arc cosine, you obtain θx 5 115.1° θy 5 32.0° θz 5 71.5° b (Note. You could have obtained this same result by using the components and magnitude of the vector AB rather than those of the force F.) REFLECT and THINK: It makes sense that, for a given geometry, only a certain set of components and angles characterize a given resultant force. The methods in this section allow you to translate back and forth between forces and geometry.
Sample Problem 2.8 A wall section of precast concrete is temporarily held in place by the cables shown. If the tension is 840 lb in cable AB and 1200 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted by cables AB and AC on stake A.
STRATEGY: This is a problem in adding concurrent forces in space. The simplest approach is to first resolve the forces into components and to then sum the components and find the resultant.
MODELING and ANALYSIS: Components of the Forces. First resolve the force exerted by each cable on stake A into x, y, and z components. To do this, determine the components and magnitude of the vectors AB and AC , measuring them from A toward the wall section (Fig. 1). Denoting the unit vectors along the coordinate axes by i, j, k, these vectors are AB 5 2(16 ft)i 1 (8 ft)j 1 (11 ft)k AB 5 21 ft AC 5 2(16 ft)i 1 (8 ft)j 2 (16 ft)k AC 5 24 ft 27 ft C D A B 8 ft 16 ft 11 ft Fig. 1 Cable forces acting on stake at A, and their unit vectors.
C B A 16 ft 16 ft 8 ft 11 ft y z x i k j TAB = (840 lb) λ λAB TAC = (1200 lb) λ λAC λAB λ λAC 60 Statics of Particles Denoting by lAB the unit vector along AB, the tension in AB is TAB 5 TABlAB 5 TAB AB ¡ AB 5 840 lb 21 ft AB ¡ Substituting the expression found for AB , the tension becomes TAB 5 840 lb 21 ft 32(16 ft)i 1 (8 ft)j 1 (11 ft)k4 TAB 5 2(640 lb)i 1 (320 lb)j 1 (440 lb)k Similarly, denoting by lAC the unit vector along AC, the tension in AC is TAC 5 TAClAC 5 TAC AC ¡ AC 5 1200 lb 24 ft AC ¡ TAC 5 2(800 lb)i 1 (400 lb)j 2 (800 lb)k Resultant of the Forces. The resultant R of the forces exerted by the two cables is R 5 TAB 1 TAC 5 2(1440 lb)i 1 (720 lb)j 2 (360 lb)k You can now determine the magnitude and direction of the resultant as R 5 2R2 x 1 R2 y 1 R2 z 5 2(21440)2 1 (720)2 1 (2300)2 R 5 1650 lb b The direction cosines come from Eqs. (2.33): cos θx 5 Rx R 5 21440 lb 1650 lb cos θy 5 Ry R 5 1720 lb 1650 lb cos θz 5 Rz R 5 2360 lb 1650 lb Calculating each quotient and its arc cosine, the angles are θx 5 150.8° θy 5 64.1° θz 5 102.6° b REFLECT and THINK: Based on visual examination of the cable forces, you might have anticipated that θx for the resultant should be obtuse and θy should be acute. The outcome of θz was not as apparent.
61 61 SOLVING PROBLEMS ON YOUR OWN I n this section, we saw that we can define a force in space by its magnitude and direction or by the three rectangular components Fx, Fy, and Fz.
A. When a force is defined by its magnitude and direction, you can find its rect-angular components Fx, Fy, and Fz as follows.
Case 1. If the direction of the force F is defined by the angles θy and f shown in Fig. 2.25, projections of F through these angles or their complements will yield the components of F [Eqs. (2.17)]. Note that to find the x and z components of F, first project F onto the horizontal plane; the projection Fh obtained in this way is then resolved into the components Fx and Fz (Fig. 2.25c).
Case 2. If the direction of the force F is defined by the angles θx, θy, and θz that F forms with the coordinate axes, you can obtain each component by multiplying the magnitude F of the force by the cosine of the corresponding angle [Concept Application 2.4]: Fx 5 F cos θx Fy 5 F cos θy Fz 5 F cos θz Case 3. If the direction of the force F is defined by two points M and N located on its line of action (Fig. 2.29), first express the vector MN drawn from M to N in terms of its components dx, dy, and dz and the unit vectors i, j, and k: MN 5 dxi 1 dy j 1 dzk Then determine the unit vector l along the line of action of F by dividing the vector MN by its magnitude MN. Multiplying l by the magnitude of F gives you the desired expression for F in terms of its rectangular components [Sample Prob. 2.7]: F 5 Fl 5 F d 1d x i 1 d y j 1 d z k2 It is helpful to use a consistent and meaningful system of notation when determining the rectangular components of a force. The method used in this text is illustrated in Sample Prob. 2.8, where the force TAB acts from stake A toward point B. Note that the subscripts have been ordered to agree with the direction of the force. We recom-mend that you adopt the same notation, as it will help you identify point 1 (the first subscript) and point 2 (the second subscript).
When calculating the vector defining the line of action of a force, you might think of its scalar components as the number of steps you must take in each coordinate direc-tion to go from point 1 to point 2. It is essential that you always remember to assign the correct sign to each of the components.
(continued) 62 B. When a force is defined by its rectangular components Fx, Fy, and Fz, you can obtain its magnitude F from F 5 2F 2 x 1 F 2 y 1 F 2 z You can determine the direction cosines of the line of action of F by dividing the components of the force by F: cos θx 5 Fx F cos θy 5 Fy F cos θz 5 Fz F From the direction cosines, you can obtain the angles θx, θy, and θz that F forms with the coordinate axes [Concept Application 2.5].
C. To determine the resultant R of two or more forces in three-dimensional space, first determine the rectangular components of each force by one of the procedures described previously. Adding these components will yield the components Rx, Ry, and Rz of the resultant. You can then obtain the magnitude and direction of the resultant as indicated previously for a force F [Sample Prob. 2.8].
63 Problems END-OF-SECTION PROBLEMS 2.71 Determine (a) the x, y, and z components of the 600-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. 2.72 Determine (a) the x, y, and z components of the 450-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) 2.74 Solve Prob. 2.73 assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. 2.75 The angle between spring AB and the post DA is 30°. Knowing that the tension in the spring is 50 lb, determine (a) the x, y, and z com-ponents of the force exerted on the circular plate at B, (b) the angles θx, θy, and θz defining the direction of the force at B. 2.76 The angle between spring AC and the post DA is 30°. Knowing that the tension in the spring is 40 lb, determine (a) the x, y, and z com-ponents of the force exerted on the circular plate at C, (b) the angles θx, θy, and θz defining the direction of the force at C. 2.77 Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles θx, θy, and θz defining the direc-tion of that force. 2.78 Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Deter-mine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles θx, θy, and θz defining the direction of that force. 2.79 Determine the magnitude and direction of the force F 5 (240 N)i – (270 N)j 1 (680 N)k. 2.80 Determine the magnitude and direction of the force F 5 (320 N)i 1 (400 N)j – (250 N)k. 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles θx 5 69.3° and θz 5 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle θy, (b) the other components and the magnitude of the force.
y x z 600 N 450 N 25º 30º 40º 35º Fig. P2.71 and P2.72 z x 35° 35° D B C y A Fig. P2.75 and P2.76 x D A y 56 ft α O 50° 20° B C z Fig. P2.77 and P2.78 64 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles θx 5 70.9° and θy 5 144.9°. Knowing that the z component of the force is 252.0 lb, determine (a) the angle θz, (b) the other components and the magnitude of the force. 2.83 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx 5 80 N, θz 5 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy. 2.84 A force F of magnitude 1200 N acts at the origin of a coordinate system. Knowing that θx 5 65°, θy 5 40°, and Fz > 0, determine (a) the components of the force, (b) the angle θz. 2.85 A frame ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. 2.86 For the frame and cable of Prob. 2.85, determine the components of the force exerted by the cable on the support at E. 2.87 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AB is 2 kips, determine the components of the force exerted at A by the cable. 2.88 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable. 2.89 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AB is 408 N, determine the components of the force exerted on the plate at B.
y x z A B E D C O 600 mm 400 mm 480 mm 510 mm 280 mm 210 mm Fig. P2.85 36 ft 28.8 ft 18 ft 45 ft 54 ft 30° A B C Fig. P2.87 and P2.88 x y z A B C D O 250 130 360 360 320 450 480 Dimensions in mm Fig. P2.89 and P2.90 2.90 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 429 N, determine the components of the force exerted on the plate at D.
65 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P 5 300 N and Q 5 400 N. 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P 5 400 N and Q 5 300 N. 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.
z x y 30° 20° 15° 50° P Q Fig. P2.91 and P2.92 y x z A B C D O 40 in.
60 in.
60 in.
45 in.
Fig. P2.93 and P2.94 z 24 in.
29 in.
25 in.
48 in.
A C B O y 36 in.
x P Fig. P2.97 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. 2.95 For the frame of Prob. 2.85, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N. 2.96 For the plate of Prob. 2.89, determine the tensions in cables AB and AD knowing that the tension in cable AC is 54 N and that the resul-tant of the forces exerted by the three cables at A must be vertical. 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. 2.98 For the boom and loading of Prob. 2.97, determine the magnitude of the load P.
66 Statics of Particles 2.5 FORCES AND EQUILIBRIUM IN SPACE According to the definition given in Sec. 2.3, a particle A is in equilibrium if the resultant of all the forces acting on A is zero. The components Rx, Ry, and Rz of the resultant of forces in space are given by equations (2.31); when the components of the resultant are zero, we have oFx 5 0 oFy 5 0 oFz 5 0 (2.34) Equations (2.34) represent the necessary and sufficient conditions for the equilibrium of a particle in space. We can use them to solve problems dealing with the equilibrium of a particle involving no more than three unknowns.
The first step in solving three-dimensional equilibrium problems is to draw a free-body diagram showing the particle in equilibrium and all of the forces acting on it. You can then write the equations of equilibrium (2.34) and solve them for three unknowns. In the more common types of problems, these unknowns will represent (1) the three components of a single force or (2) the magnitude of three forces, each of known direction.
Photo 2.4 Although we cannot determine the tension in the four cables supporting the car by using the three equations (2.34), we can obtain a relation among the tensions by analyzing the equilibrium of the hook.
2.5 Forces and Equilibrium in Space 67 Sample Problem 2.9 A 200-kg cylinder is hung by means of two cables AB and AC that are attached to the top of a vertical wall. A horizontal force P perpendicular to the wall holds the cylinder in the position shown. Determine the mag-nitude of P and the tension in each cable.
STRATEGY: Connection point A is acted upon by four forces, including the weight of the cylinder. You can use the given geometry to express the force components of the cables and then apply equilibrium conditions to calculate the tensions.
MODELING and ANALYSIS: Free-Body Diagram. Choose point A as a free body; this point is subjected to four forces, three of which are of unknown magnitude. Introducing the unit vectors i, j, and k, resolve each force into rectangular components (Fig. 1): P 5 Pi W 5 2mgj 5 2(200 kg)(9.81 m/s2)j 5 2(1962 N)j (1) A B C P 8 m 10 m 1.2 m 2 m 200kg 12 m W 12 m C B z y x A O P 8 m 10 m 1.2 m 2 m TAB j TAC k i AB AC Fig. 1 Free-body diagram of particle A.
68 Statics of Particles For TAB and TAC, it is first necessary to determine the components and magnitudes of the vectors AB and AC . Denoting the unit vector along AB by lAB, you can write TAB as AB ¡ 5 211.2 m2i 1 110 m2j 1 18 m2k AB 5 12.862 m lAB 5 AB ¡ 12.862 m 5 20.09330i 1 0.7775j 1 0.6220k TAB 5 TABlAB 5 20.09330TABi 1 0.7775TAB j 1 0.6220TABk (2) Similarly, denoting the unit vector along AC by lAC, you have for TAC AC ¡ 5 211.2 m2i 1 110 m2j 2 110 m2k AC 5 14.193 m lAC 5 AC ¡ 14.193 m 5 20.08455i 1 0.7046j 2 0.7046k TAC 5 TAClAC 5 20.08455TACi 1 0.7046TAC j 2 0.7046TACk (3) Equilibrium Condition. Since A is in equilibrium, you must have oF 5 0: TAB 1 TAC 1 P 1 W 5 0 or substituting from Eqs. (1), (2), and (3) for the forces and factoring i, j, and k, you have (20.09330TAB 2 0.08455TAC 1 P)i 1 (0.7775TAB 1 0.7046TAC 2 1962 N)j 1 (0.6220TAB 2 0.7046TAC)k 5 0 Setting the coefficients of i, j, and k equal to zero, you can write three scalar equations, which express that the sums of the x, y, and z components of the forces are respectively equal to zero.
(o Fx 5 0:) 20.09330TAB 2 0.08455TAC 1 P 5 0 (o Fy 5 0:) 10.7775TAB 1 0.7046TAC 2 1962 N 5 0 (o Fz 5 0:) 10.6220TAB 2 0.7046TAC 5 0 Solving these equations, you obtain P 5 235 N TAB 5 1402 N TAC 5 1238 N b REFLECT and THINK: The solution of the three unknown forces yielded positive results, which is completely consistent with the physical situation of this problem. Conversely, if one of the cable force results had been negative, thereby reflecting compression instead of tension, you should recognize that the solution is in error.
69 69 W e saw earlier that when a particle is in equilibrium, the resultant of the forces acting on the particle must be zero. In the case of the equilibrium of a particle in three-dimensional space, this equilibrium condition provides you with three rela-tions among the forces acting on the particle. These relations may be used to deter-mine three unknowns—usually the magnitudes of three forces.
The solution usually consists of the following steps: 1. Draw a free-body diagram of the particle. This diagram shows the particle and all the forces acting on it. Indicate on the diagram the magnitudes of known forces, as well as any angles or dimensions that define the direction of a force. Any unknown magnitude or angle should be denoted by an appropriate symbol. Nothing else should be included in the free-body diagram.
2. Resolve each force into rectangular components. Following the method used earlier, determine for each force F the unit vector l defining the direction of that force, and express F as the product of its magnitude F and l. You will obtain an expression of the form F 5 Fλ 5 F d (dxi 1 dy j 1 dzk) where d, dx, dy, and dz are dimensions obtained from the free-body diagram of the particle. If you know the magnitude as well as the direction of the force, then F is known and the expression obtained for F is well defined; otherwise F is one of the three unknowns that should be determined.
3. Set the resultant, or sum, of the forces exerted on the particle equal to zero. You will obtain a vector equation consisting of terms containing the unit vectors i, j, or k. Group the terms containing the same unit vector and factor that vector. For the vector equation to be satisfied, you must set the coefficient of each of the unit vectors equal to zero. This yields three scalar equations that you can solve for no more than three unknowns [Sample Prob. 2.9].
SOLVING PROBLEMS ON YOUR OWN 70 Problems FREE-BODY PRACTICE PROBLEMS 2.F5 Three cables are used to tether a balloon as shown. Knowing that the tension in cable AC is 444 N, draw the free-body diagram needed to determine the vertical force P exerted by the balloon at A. 2.F6 A container of mass m 5 120 kg is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each cable. 2.F7 A 150-lb cylinder is supported by two cables AC and BC that are attached to the top of vertical posts. A horizontal force P, which is perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to deter-mine the magnitude of P and the force in each cable. 2.F8 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. Knowing that the tension in wire AB is 630 lb, draw the free-body diagram needed to determine the vertical force P exerted by the tower on the pin at A. A B C D O 4.20 m 4.20 m 3.30 m 5.60 m 2.40 m x y z Fig. P2.F5 x y z A B D C O 600 mm 320 mm 360 mm 500 mm 450 mm Fig. P2.F6 y A 90 ft 30 ft O B 30 ft 20 ft 45 ft z D C 60 ft 65 ft x Fig. P2.F8 O B C A y x z 15 ft 7.2 ft 3.6 ft 10.8 ft 10.8 ft P Fig. P2.F7 71 END-OF-SECTION PROBLEMS 2.99 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container knowing that the tension in cable AB is 6 kN. 2.100 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN. 2.101 Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AD is 481 N.
Fig. P2.99 and P2.100 y x z 450 mm 500 mm 360 mm 320 mm 600 mm A C D B Fig. P2.101 and P2.102 A B C D O 4.20 m 4.20 m 3.30 m 5.60 m 2.40 m x y z Fig. P2.105 and P2.106 x y z A B C D O 36 in.
27 in.
60 in.
32 in.
40 in.
Fig. P2.103 y 16 in.
8 in.
a a 24 in.
A C D B x z 2.102 Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable. 2.103 A 36-lb triangular plate is supported by three wires as shown. Deter-mine the tension in each wire, knowing that a 5 6 in. 2.104 Solve Prob. 2.103, assuming that a 5 8 in. 2.105 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AC is 544 lb. 2.106 A 1600-lb crate is supported by three cables as shown. Determine the tension in each cable.
72 2.107 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q 5 0, find the value of P for which the tension in cable AD is 305 N.
Fig. P2.107 and P2.108 y x z 220 mm 240 mm 960 mm Q P A B C D O 380 mm 320 mm 960 mm x y z A B C D O 250 130 360 360 320 450 480 Dimensions in mm Fig. P2.109 and P2.110 y A 100 ft B C O D 60 ft z x 74 ft 18 ft 20 ft 25 ft 20 ft Fig. P2.111 and P2.112 2.108 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that P 5 1200 N, determine the values of Q for which cable AD is taut. 2.109 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate. 2.110 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate. 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 840 lb, determine the vertical force P exerted by the tower on the pin at A. 2.112 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 590 lb, determine the vertical force P exerted by the tower on the pin at A.
73 2.113 In trying to move across a slippery icy surface, a 175-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope.
z 16 ft 8 ft B A C O x y 4 ft 30 ft 32 ft 12 ft Fig. P2.113 180 lb D A B C 18 in.
16 in.
22 in.
24 in.
24 in.
Fig. P2.120 2.114 Solve Prob. 2.113 assuming that a friend is helping the man at A by pulling on him with a force P 5 2(45 lb)k. 2.115 For the rectangular plate of Probs. 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. 2.116 For the cable system of Probs. 2.107 and 2.108, determine the ten-sion in each cable knowing that P 5 2880 N and Q 5 0. 2.117 For the cable system of Probs. 2.107 and 2.108, determine the ten-sion in each cable knowing that P 5 2880 N and Q 5 576 N. 2.118 For the cable system of Probs. 2.107 and 2.108, determine the ten-sion in each cable knowing that P 5 2880 N and Q 5 –576 N (Q is directed downward). 2.119 For the transmission tower of Probs. 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 1800 lb. 2.120 Three wires are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. Determine the tension in each wire when a 180-lb cylinder is suspended from point D as shown.
74 2.121 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W 5 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) 200 mm x y y z z B Q P A O Fig. P2.125 D x E O B 25 in.
17.5 in.
45 in.
60 in.
80 in.
y C A z P Fig. P2.123 2.122 Knowing that the tension in cable AC of the system described in Prob. 2.121 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container. 2.123 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Know-ing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. 2.124 Knowing that the tension in cable AE of Prob. 2.123 is 75 lb, deter-mine (a) the magnitude of the load P, (b) the tension in cables BAC and AD. 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P 5 (341 N)j is applied to collar A, determine (a) the tension in the wire when y 5 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system. 2.126 Solve Prob. 2.125 assuming that y 5 275 mm.
y x z 0.78 m 0.40 m 0.40 m P O B F E C W A D 1.60 m 0.86 m 1.20 m 1.30 m Fig. P2.121 75 In this chapter, we have studied the effect of forces on particles, i.e., on bodies of such shape and size that we may assume all forces acting on them apply at the same point.
Resultant of Two Forces Forces are vector quantities; they are characterized by a point of application, a magnitude, and a direction, and they add according to the parallelogram law (Fig. 2.30). We can determine the magnitude and direction of the resultant R of two forces P and Q either graphically or by trigonometry using the law of cosines and the law of sines [Sample Prob. 2.1].
Components of a Force Any given force acting on a particle can be resolved into two or more com-ponents, i.e., it can be replaced by two or more forces that have the same effect on the particle. A force F can be resolved into two components P and Q by drawing a parallelogram with F for its diagonal; the components P and Q are then represented by the two adjacent sides of the parallelogram (Fig. 2.31). Again, we can determine the components either graphically or by trigonom-etry [Sec. 2.1E].
Review and Summary Q R P A Fig. 2.30 Q F P A Fig. 2.31 F x y Fy = Fy j Fx = Fxi j i Fig. 2.32 Rectangular Components; Unit Vectors A force F is resolved into two rectangular components if its components Fx and Fy are perpendicular to each other and are directed along the coordinate axes (Fig. 2.32). Introducing the unit vectors i and j along the x and y axes, respectively, we can write the components and the vector as [Sec. 2.2A] Fx 5 Fxi Fy 5 Fy j (2.6) and F 5 Fxi 1 Fyj (2.7) where Fx and Fy are the scalar components of F. These components, which can be positive or negative, are defined by the relations Fx 5 F cos θ Fy 5 F sin θ (2.8) 76 When the rectangular components Fx and Fy of a force F are given, we can obtain the angle θ defining the direction of the force from tan θ 5 Fy Fx (2.9) We can obtain the magnitude F of the force by solving one of the equations (2.8) for F or by applying the Pythagorean theorem: F 5 2F x 2 1 F y 2 (2.10) Resultant of Several Coplanar Forces When three or more coplanar forces act on a particle, we can obtain the rectangular components of their resultant R by adding the corresponding com-ponents of the given forces algebraically [Sec. 2.2B]: Rx 5 oFx Ry 5 oFy (2.13) The magnitude and direction of R then can be determined from relations similar to Eqs. (2.9) and (2.10) [Sample Prob. 2.3].
Forces in Space A force F in three-dimensional space can be resolved into rectangular com-ponents Fx, Fy, and Fz [Sec. 2.4A]. Denoting by θx, θy, and θz, respectively, the angles that F forms with the x, y, and z axes (Fig. 2.33), we have Fx 5 F cos θx Fy 5 F cos θy Fz 5 F cos θz (2.19) Direction Cosines The cosines of θx, θy, and θz are known as the direction cosines of the force F. Introducing the unit vectors i, j, and k along the coordinate axes, we can write F as F 5 Fxi 1 Fy j 1 Fzk (2.20) or F 5 F(cos θxi 1 cos θy j 1 cos θzk) (2.21) Fig. 2.33 x y z A B C D E F Fx Fy Fz x y z (a) x y z A B C D E F Fx Fy Fz x y z A B C D E F Fx Fy Fz (b) (c) O O O 77 This last equation shows (Fig. 2.34) that F is the product of its magnitude F and the unit vector expressed by l 5 cos θxi 1 cos θy j 1 cos θzk Since the magnitude of l is equal to unity, we must have cos2 θx 1 cos2 θy 1 cos2 θz 5 1 (2.24) When we are given the rectangular components Fx, Fy, and Fz of a force F, we can find the magnitude F of the force by F 5 2F 2 x 1 F 2 y 1 F 2 z (2.18) and the direction cosines of F are obtained from Eqs. (2.19). We have cos θx 5 Fx F cos θy 5 Fy F cos θz 5 Fz F (2.25) When a force F is defined in three-dimensional space by its magnitude F and two points M and N on its line of action [Sec. 2.4B], we can obtain its rectangular components by first expressing the vector MN joining points M and N in terms of its components dx, dy, and dz (Fig. 2.35): MN 5 dxi 1 dyj 1 dzk (2.26) We next determine the unit vector l along the line of action of F by dividing MN by its magnitude MN 5 d: l 5 MN y MN 5 1 d 1dxi 1 dy j 1 dzk2 (2.27) Recalling that F is equal to the product of F and l, we have F 5 Fl 5 F d 1dxi 1 dy j 1 dzk2 (2.28) x y z λ λ (Magnitude = 1) F = Fλ λ Fyj Fxi Fzk cos yj cos zk cos xi Fig. 2.34 x y z F O M(x1, y1, z1) N(x2, y2, z2) dy = y2 – y1 dx = x2 – x1 dz = z2 – z1 < 0 λ Fig. 2.35 78 From this equation it follows [Sample Probs. 2.7 and 2.8] that the scalar components of F are, respectively, Fx 5 Fdx d Fy 5 Fdy d Fz 5 Fdz d (2.29) Resultant of Forces in Space When two or more forces act on a particle in three-dimensional space, we can obtain the rectangular components of their resultant R by adding the corre-sponding components of the given forces algebraically [Sec. 2.4C]. We have Rx 5 oFx Ry 5 oFy Rz 5 oFz (2.31) We can then determine the magnitude and direction of R from relations simi-lar to Eqs. (2.18) and (2.25) [Sample Prob. 2.8].
Equilibrium of a Particle A particle is said to be in equilibrium when the resultant of all the forces acting on it is zero [Sec. 2.3A]. The particle remains at rest (if originally at rest) or moves with constant speed in a straight line (if originally in motion) [Sec. 2.3B].
Free-Body Diagram To solve a problem involving a particle in equilibrium, first draw a free-body diagram of the particle showing all of the forces acting on it [Sec. 2.3C]. If only three coplanar forces act on the particle, you can draw a force triangle to express that the particle is in equilibrium. Using graphical methods of trigonometry, you can solve this triangle for no more than two unknowns [Sample Prob. 2.4]. If more than three coplanar forces are involved, you should use the equations of equilibrium: oFx 5 0 oFy 5 0 (2.15) These equations can be solved for no more than two unknowns [Sample Prob. 2.6].
Equilibrium in Space When a particle is in equilibrium in three-dimensional space [Sec. 2.5], use the three equations of equilibrium: oFx 5 0 oFy 5 0 oFz 5 0 (2.34) These equations can be solved for no more than three unknowns [Sample Prob. 2.9].
79 2.127 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigo-nometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.
2.128 Determine the x and y components of each of the forces shown.
Review Problems A B 40° 20° Fig. P2.127 106 lb 102 lb 200 lb x y 24 in.
28 in.
45 in.
40 in.
30 in.
O Fig. P2.128 a a 200 lb 400 lb P Fig. P2.129 30° 20° α 300 lb A B C Fig. P2.130 2.129 A hoist trolley is subjected to the three forces shown. Knowing that α 5 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant.
2.130 Knowing that α 5 55° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.
80 2.131 Two cables are tied together at C and loaded as shown. Knowing that P 5 360 N, determine the tension (a) in cable AC, (b) in cable BC.
A B P Q = 480 N C 3 4 600 mm 250 mm Fig. P2.131 35º A B C P 50º a Fig. P2.132 36° 60° 48° 20° x y z A B C E D Fig. P2.133 x y z A B D C O 600 mm 920 mm 360 mm 900 mm Fig. P2.134 2.132 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 800 N, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. 2.133 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. 2.134 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C.
81 2.135 Find the magnitude and direction of the resultant of the two forces shown knowing that P 5 600 N and Q 5 450 N. 2.136 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P 5 Pi and Q 5 Qk are applied to the ring to main-tain the container in the position shown. Knowing that W 5 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) z x O y 30° 25° 40° 55° P Q Fig. P2.135 Q P O A C B y x z W 160 mm 400 mm 130 mm 150 mm 240 mm Fig. P2.136 20 in.
x x y z z B Q P A O Fig. P2.137 and P2.138 2.137 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when x 5 9 in., (b) the corresponding magnitude of the force P required to maintain the equilibrium of the system. 2.138 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when P 5 120 lb and Q 5 60 lb.
Four tugboats work together to free the oil tanker Coastal Eagle Point that ran aground while attempting to navigate a channel in Tampa Bay. It will be shown in this chapter that the forces exerted on the ship by the tugboats could be replaced by an equivalent force exerted by a single, more powerful, tugboat.
t b t k t th t f th il t k C t l E l Rigid Bodies: Equivalent Systems of Forces 3 Introduction 83 Introduction In Chapter 2, we assumed that each of the bodies considered could be treated as a single particle. Such a view, however, is not always possible. In general, a body should be treated as a combination of a large number of particles. In this case, we need to consider the size of the body as well as the fact that forces act on different parts of the body and thus have different points of application.
Most of the bodies considered in elementary mechanics are assumed to be rigid. We define a rigid body as one that does not deform. Actual structures and machines are never absolutely rigid and deform under the loads to which they are subjected. However, these deformations are usu-ally small and do not appreciably affect the conditions of equilibrium or the motion of the structure under consideration. They are important, though, as far as the resistance of the structure to failure is concerned and are considered in the study of mechanics of materials.
In this chapter, you will study the effect of forces exerted on a rigid body, and you will learn how to replace a given system of forces by a simpler equivalent system. This analysis rests on the fundamental assump-tion that the effect of a given force on a rigid body remains unchanged if that force is moved along its line of action (principle of transmissibility). It follows that forces acting on a rigid body can be represented by sliding vectors, as indicated earlier in Sec. 2.1B.
Two important concepts associated with the effect of a force on a rigid body are the moment of a force about a point (Sec. 3.1E) and the Introduction 3.1 FORCES AND MOMENTS 3.1A External and Internal Forces 3.1B Principle of Transmissibility: Equivalent Forces 3.1C Vector Products 3.1D Rectangular Components of Vector Products 3.1E Moment of a Force about a Point 3.1F Rectangular Components of the Moment of a Force 3.2 MOMENT OF A FORCE ABOUT AN AXIS 3.2A Scalar Products 3.2B Mixed Triple Products 3.2C Moment of a Force about a Given Axis 3.3 COUPLES AND FORCE-COUPLE SYSTEMS 3.3A Moment of a Couple 3.3B Equivalent Couples 3.3C Addition of Couples 3.3D Couple Vectors 3.3E Resolution of a Given Force into a Force at O and a Couple 3.4 SIMPLIFYING SYSTEMS OF FORCES 3.4A Reducing a System of Forces to a Force-Couple System 3.4B Equivalent and Equipollent Systems of Forces 3.4C Further Reduction of a System of Forces 3.4D Reduction of a System of Forces to a Wrench Review and Summary Objectives • Discuss the principle of transmissibility that enables a force to be treated as a sliding vector.
• Define the moment of a force about a point.
• Examine vector and scalar products, useful in analysis involving moments.
• Apply Varignon’s Theorem to simplify certain moment analyses.
• Define the mixed triple product and use it to determine the moment of a force about an axis.
• Define the moment of a couple, and consider the particular properties of couples.
• Resolve a given force into an equivalent force-couple system at another point.
• Reduce a system of forces into an equivalent force-couple system.
• Examine circumstances where a system of forces can be reduced to a single force.
• Define a wrench and consider how any general system of forces can be reduced to a wrench. 84 Rigid Bodies: Equivalent Systems of Forces moment of a force about an axis (Sec. 3.2C). The determination of these quantities involves computing vector products and scalar products of two vectors, so in this chapter, we introduce the fundamentals of vector algebra and apply them to the solution of problems involving forces acting on rigid bodies.
Another concept introduced in this chapter is that of a couple, i.e., the combination of two forces that have the same magnitude, parallel lines of action, and opposite sense (Sec. 3.3A). As you will see, we can replace any system of forces acting on a rigid body by an equivalent system con-sisting of one force acting at a given point and one couple. This basic combination is called a force-couple system. In the case of concurrent, coplanar, or parallel forces, we can further reduce the equivalent force-couple system to a single force, called the resultant of the system, or to a single couple, called the resultant couple of the system.
3.1 FORCES AND MOMENTS The basic definition of a force does not change if the force acts on a point or on a rigid body. However, the effects of the force can be very different, depending on factors such as the point of application or line of action of that force. As a result, calculations involving forces acting on a rigid body are generally more complicated than situations involving forces acting on a point. We begin by examining some general classifications of forces acting on rigid bodies.
3.1A External and Internal Forces Forces acting on rigid bodies can be separated into two groups: (1) external forces and (2) internal forces. 1. External forces are exerted by other bodies on the rigid body under consideration. They are entirely responsible for the external behavior of the rigid body, either causing it to move or ensuring that it remains at rest. We shall be concerned only with external forces in this chapter and in Chaps. 4 and 5. 2. Internal forces hold together the particles forming the rigid body. If the rigid body is structurally composed of several parts, the forces hold-ing the component parts together are also defined as internal forces. We will consider internal forces in Chaps. 6 and 7.
As an example of external forces, consider the forces acting on a disabled truck that three people are pulling forward by means of a rope attached to the front bumper (Fig. 3.1a). The external forces acting on the truck are shown in a free-body diagram (Fig. 3.1b). Note that this free-body diagram shows the entire object, not just a particle representing the object. Let us first consider the weight of the truck. Although it embodies the effect of the earth’s pull on each of the particles forming the truck, the weight can be represented by the single force W. The point of application of this force––that is, the point at which the force acts––is defined as the center of gravity of the truck. (In Chap. 5, we will show how to determine the location of centers of gravity.) The weight W tends to make the truck move vertically downward. In fact, it would actually cause the truck to W F R1 R2 (a) (b) Fig. 3.1 (a) Three people pulling on a truck with a rope; (b) free-body diagram of the truck, shown as a rigid body instead of a particle.
3.1 Forces and Moments 85 move downward, i.e., to fall, if it were not for the presence of the ground. The ground opposes the downward motion of the truck by means of the reactions R1 and R2. These forces are exerted by the ground on the truck and must therefore be included among the external forces acting on the truck.
The people pulling on the rope exert the force F. The point of appli-cation of F is on the front bumper. The force F tends to make the truck move forward in a straight line and does actually make it move, since no external force opposes this motion. (We are ignoring rolling resistance here for simplicity.) This forward motion of the truck, during which each straight line keeps its original orientation (the floor of the truck remains horizontal, and the walls remain vertical), is known as a translation. Other forces might cause the truck to move differently. For example, the force exerted by a jack placed under the front axle would cause the truck to pivot about its rear axle. Such a motion is a rotation. We conclude, therefore, that each external force acting on a rigid body can, if unop-posed, impart to the rigid body a motion of translation or rotation, or both.
3.1B Principle of Transmissibility: Equivalent Forces The principle of transmissibility states that the conditions of equilibrium or motion of a rigid body remain unchanged if a force F acting at a given point of the rigid body is replaced by a force F9 of the same magnitude and same direction, but acting at a different point, provided that the two forces have the same line of action (Fig. 3.2). The two forces F and F9 have the same effect on the rigid body and are said to be equivalent forces. This principle, which states that the action of a force may be transmitted along its line of action, is based on experimental evidence. It cannot be derived from the properties established so far in this text and therefore must be accepted as an experimental law. (You will see in Sec. 16.1D that we can derive the principle of transmissibility from the study of the dynamics of rigid bodies, but this study requires the use of Newton’s second and third laws and of several other concepts as well.) Therefore, our study of the statics of rigid bodies is based on the three principles introduced so far: the parallelogram law of vector addition, Newton’s first law, and the principle of transmissibility.
We indicated in Chap. 2 that we could represent the forces acting on a particle by vectors. These vectors had a well-defined point of application––namely, the particle itself––and were therefore fixed, or bound, vectors. In the case of forces acting on a rigid body, however, the point of application of the force does not matter, as long as the line of action remains unchanged. Thus, forces acting on a rigid body must be represented by a different kind of vector, known as a sliding vector, since forces are allowed to slide along their lines of action. Note that all of the properties we derive in the following sections for the forces acting on a rigid body are valid more generally for any system of sliding vectors. In order to keep our presentation more intuitive, however, we will carry it out in terms of physical forces rather than in terms of mathematical sliding vectors.
Returning to the example of the truck, we first observe that the line of action of the force F is a horizontal line passing through both the front = F F' Fig. 3.2 Two forces F and F9 are equivalent if they have the same magnitude and direction and the same line of action, even if they act at different points.
86 Rigid Bodies: Equivalent Systems of Forces and rear bumpers of the truck (Fig. 3.3). Using the principle of transmis-sibility, we can therefore replace F by an equivalent force F9 acting on the rear bumper. In other words, the conditions of motion are unaffected, and all of the other external forces acting on the truck (W, R1, R2) remain unchanged if the people push on the rear bumper instead of pulling on the front bumper.
The principle of transmissibility and the concept of equivalent forces have limitations. Consider, for example, a short bar AB acted upon by equal and opposite axial forces P1 and P2, as shown in Fig. 3.4a. According W F F′ Equivalent forces R1 R2 W R1 R2 = Fig. 3.3 Force F9 is equivalent to force F, so the motion of the truck is the same whether you pull it or push it.
= P1 P2 A B (a) = P1 P' 2 A B (b) A B (c) = P1 P2 A B (d) = P1 P' 2 A B (e) A B ( f ) Fig. 3.4 (a–c) A set of equivalent forces acting on bar AB; (d–f) another set of equivalent forces acting on bar AB. Both sets produce the same external effect (equilibrium in this case) but different internal forces and deformations.
to the principle of transmissibility, we can replace force P2 by a force P9 2 having the same magnitude, the same direction, and the same line of action but acting at A instead of B (Fig. 3.4b). The forces P1 and P9 2 acting on the same particle can be added according to the rules of Chap. 2, and since these forces are equal and opposite, their sum is equal to zero. Thus, in terms of the external behavior of the bar, the original system of forces shown in Fig. 3.4a is equivalent to no force at all (Fig. 3.4c).
Consider now the two equal and opposite forces P1 and P2 acting on the bar AB as shown in Fig. 3.4d. We can replace the force P2 by a force P9 2 having the same magnitude, the same direction, and the same line of action but acting at B instead of at A (Fig. 3.4e). We can add forces P1 and P9 2, and their sum is again zero (Fig. 3.4f ). From the point of view of the mechanics of rigid bodies, the systems shown in Fig. 3.4a and d are thus equivalent. However, the internal forces and deformations produced by the two systems are clearly different. The bar of Fig. 3.4a is in tension and, if not absolutely rigid, increases in length slightly; the bar of Fig. 3.4d is in compression and, if not absolutely rigid, decreases in length slightly. Thus, although we can use the principle of transmissibility to determine the 3.1 Forces and Moments 87 conditions of motion or equilibrium of rigid bodies and to compute the external forces acting on these bodies, it should be avoided, or at least used with care, in determining internal forces and deformations.
3.1C Vector Products In order to gain a better understanding of the effect of a force on a rigid body, we need to introduce a new concept, the moment of a force about a point. However, this concept is more clearly understood and is applied more effectively if we first add to the mathematical tools at our disposal the vector product of two vectors.
The vector product of two vectors P and Q is defined as the vector V that satisfies the following conditions. 1. The line of action of V is perpendicular to the plane containing P and Q (Fig. 3.5a). 2. The magnitude of V is the product of the magnitudes of P and Q and of the sine of the angle θ formed by P and Q (the measure of which is always 180° or less). We thus have Magnitude of a vector product V 5 PQ sin θ (3.1) 3. The direction of V is obtained from the right-hand rule. Close your right hand and hold it so that your fingers are curled in the same sense as the rotation through θ that brings the vector P in line with the vector Q. Your thumb then indicates the direction of the vector V (Fig. 3.5b). Note that if P and Q do not have a common point of appli-cation, you should first redraw them from the same point. The three vectors P, Q, and V—taken in that order—are said to form a right-handed triad.† As stated previously, the vector V satisfying these three conditions (which define it uniquely) is referred to as the vector product of P and Q. It is represented by the mathematical expression Vector product V 5 P 3 Q (3.2) Because of this notation, the vector product of two vectors P and Q is also referred to as the cross product of P and Q.
It follows from Eq. (3.1) that if the vectors P and Q have either the same direction or opposite directions, their vector product is zero. In the general case when the angle θ formed by the two vectors is neither 0° nor 180°, Eq. (3.1) has a simple geometric interpretation: The magnitude V of the vector product of P and Q is equal to the area of the parallelogram that has P and Q for sides (Fig. 3.6). The vector product P 3 Q is V 5 PQ sin θ †Note that the x, y, and z axes used in Chap. 2 form a right-handed system of orthogonal axes and that the unit vectors i, j, and k defined in Sec. 2.4A form a right-handed orthogonal triad. V 5 P 3 Q Q P V = P × Q q (a) V V points in the direction of the thumb Fingers curl in the direction from P to Q (b) Fig. 3.5 (a) The vector product V has the magnitude PQ sin θ and is perpendicular to the plane of P and Q; (b) you can determine the direction of V by using the right-hand rule.
Q Q' P V Fig. 3.6 The magnitude of the vector product V equals the area of the parallelogram formed by P and Q. If you change Q to Q9 in such a way that the parallelogram changes shape but P and the area are still the same, then the magnitude of V remains the same.
88 Rigid Bodies: Equivalent Systems of Forces therefore unchanged if we replace Q by a vector Q9 that is coplanar with P and Q such that the line joining the tips of Q and Q9 is parallel to P: V 5 P 3 Q 5 P 3 Q9 (3.3) From the third condition used to define the vector product V of P and Q––namely, that P, Q, and V must form a right-handed triad––it fol-lows that vector products are not commutative; i.e., Q 3 P is not equal to P 3 Q. Indeed, we can easily check that Q 3 P is represented by the vector 2V, which is equal and opposite to V: Q 3 P 5 2(P 3 Q) (3.4) Concept Application 3.1 Let us compute the vector product V 5 P 3 Q, where the vector P is of magnitude 6 and lies in the zx plane at an angle of 30° with the x axis, and where the vector Q is of magnitude 4 and lies along the x axis (Fig. 3.7).
Solution It follows immediately from the definition of the vector product that the vector V must lie along the y axis, directed upward, with the magnitude V 5 PQ sin θ 5 (6)(4) sin 30° 5 12 ■ y x z Q P 60° 30° Fig. 3.7 Two vectors P and Q with angle between them.
We saw that the commutative property does not apply to vector products. However, it can be demonstrated that the distributive property P 3 (Q1 1 Q2) 5 P 3 Q1 1 P 3 Q2 (3.5) does hold.
A third property, the associative property, does not apply to vector products; we have in general (P 3 Q) 3 S Þ P 3 (Q 3 S) (3.6) 3.1D Rectangular Components of Vector Products Before we turn back to forces acting on rigid bodes, let’s look at a more convenient way to express vector products using rectangular components. To do this, we use the unit vectors i, j, and k that were defined in Chap. 2.
Consider first the vector product i 3 j (Fig. 3.8a). Since both vectors have a magnitude equal to 1 and since they are at a right angle to each other, their vector product is also a unit vector. This unit vector must be k, since the vectors i, j, and k are mutually perpendicular and form a y x z i j i × j = k (a) y x z i j (b) j × i = –k Fig. 3.8 (a) The vector product of the i and j unit vectors is the k unit vector; (b) the vector product of the j and i unit vectors is the 2k unit vector.
3.1 Forces and Moments 89 right-handed triad. Similarly, it follows from the right-hand rule given in Sec. 3.1C that the product j 3 i is equal to 2k (Fig. 3.8b). Finally, note that the vector product of a unit vector with itself, such as i 3 i, is equal to zero, since both vectors have the same direction. Thus, we can list the vector products of all the various possible pairs of unit vectors: i 3 i 5 0 j 3 i 5 2k k 3 i 5 j i 3 j 5 k j 3 j 5 0 k 3 j 5 2i (3.7) i 3 k 5 2j j 3 k 5 i k 3 k 5 0 We can determine the sign of the vector product of two unit vectors simply by arranging them in a circle and reading them in the order of the multi-plication (Fig. 3.9). The product is positive if they follow each other in counterclockwise order and is negative if they follow each other in clock-wise order.
j Unit vector products read in this direction are positive Unit vector products read in this direction are negative i k Fig. 3.9 Arrange the three letters i, j, k in a counterclockwise circle. You can use the order of letters for the three unit vectors in a vector product to determine its sign.
We can now easily express the vector product V of two given vectors P and Q in terms of the rectangular components of these vectors. Resolving P and Q into components, we first write V 5 P 3 Q 5 (Pxi 1 Py j 1 Pz k) 3 (Qxi 1 Qy j 1 Qz k) Making use of the distributive property, we express V as the sum of vector products, such as Px i 3 Q y j. We find that each of the expressions obtained is equal to the vector product of two unit vectors, such as i 3 j, multiplied by the product of two scalars, such as Px Qy. Recalling the identities of Eq. (3.7) and factoring out i, j, and k, we obtain V 5 (PyQz 2 PzQy)i 1 (PzQx 2 PxQz)j 1 (PxQy 2 PyQx)k (3.8) Thus, the rectangular components of the vector product V are Rectangular components of a vector product Vx 5 PyQz 2 PzQy Vy 5 PzQx 2 PxQz (3.9) Vz 5 PxQy 2 PyQx Vx V 5 Py P Qz 2 PzQy Vy V 5 PzQx 2 Px P Qz Vz V 5 Px P Qy 2 Py P Qx 90 Rigid Bodies: Equivalent Systems of Forces Returning to Eq. (3.8), notice that the right-hand side represents the expan-sion of a determinant. Thus, we can express the vector product V in the following form, which is more easily memorized:† Rectangular components of a vector product (determinant form) V 5 † i j k Px Py Pz Qx Qy Qz † (3.10) 3.1E Moment of a Force about a Point We are now ready to consider a force F acting on a rigid body (Fig. 3.10a). As we know, the force F is represented by a vector that defines its mag-nitude and direction. However, the effect of the force on the rigid body depends also upon its point of application A. The position of A can be conveniently defined by the vector r that joins the fixed reference point O with A; this vector is known as the position vector of A. The position vector r and the force F define the plane shown in Fig. 3.10a.
We define the moment of F about O as the vector product of r and F: Moment of a force about a point O MO 5 r 3 F (3.11) According to the definition of the vector product given in Sec. 3.1C, the moment MO must be perpendicular to the plane containing O and force F. The sense of MO is defined by the sense of the rotation that will bring vector r in line with vector F; this rotation is observed as counterclockwise by an observer located at the tip of MO. Another way of defining the sense of MO is furnished by a variation of the right-hand rule: Close your right hand and hold it so that your fingers curl in the sense of the rotation that F would impart to the rigid body about a fixed axis directed along the line of action of MO. Then your thumb indicates the sense of the moment MO (Fig. 3.10b).
Finally, denoting by θ the angle between the lines of action of the position vector r and the force F, we find that the magnitude of the moment of F about O is Magnitude of the moment of a force MO 5 rF sin θ 5 Fd (3.12) V 5 † i j k Px P Py P Pz Qx Qy Qz † MO 5 r 3 F MO M 5 rF sin θ 5 Fd †Any determinant consisting of three rows and three columns can be evaluated by repeating the first and second columns and forming products along each diagonal line. The sum of the products obtained along the red lines is then subtracted from the sum of the products obtained along the black lines.
i j k i j Px Py Pz Px Py Qx Qy Qz Qx Qy MO d A F r q O (a) MO Vector MO points in the direction of the thumb Fingers curl in the direction from r to F (b) Fig. 3.10 Moment of a force about a point. (a) The moment MO is the vector product of the position vector r and the force F; (b) a right-hand rule indicates the sense of MO.
3.1 Forces and Moments 91 where d represents the perpendicular distance from O to the line of action of F (see Fig. 3.10). Experimentally, the tendency of a force F to make a rigid body rotate about a fixed axis perpendicular to the force depends upon the distance of F from that axis, as well as upon the magnitude of F. For example, a child’s breath can exert enough force to make a toy propeller spin (Fig. 3.11a), but a wind turbine requires the force of a substantial wind to rotate the blades and generate electrical power (Fig. 3.11b). How-ever, the perpendicular distance between the rotation point and the line of action of the force (often called the moment arm) is just as important. If you want to apply a small moment to turn a nut on a pipe without break-ing it, you might use a small pipe wrench that gives you a small moment Fig. 3.11 (a, b) The moment of a force depends on the magnitude of the force; (c, d) it also depends on the length of the moment arm.
(c) Small moment arm (d) Large moment arm (b) Large force (a) Small force 92 Rigid Bodies: Equivalent Systems of Forces arm (Fig. 3.11c). But if you need a larger moment, you could use a large wrench with a long moment arm (Fig. 3.11d). Therefore, The magnitude of MO measures the tendency of the force F to make the rigid body rotate about a fixed axis directed along MO.
In the SI system of units, where a force is expressed in newtons (N) and a distance in meters (m), the moment of a force is expressed in newton-meters (N?m). In the U.S. customary system of units, where a force is expressed in pounds and a distance in feet or inches, the moment of a force is expressed in lb?ft or lb?in.
Note that although the moment MO of a force about a point depends upon the magnitude, the line of action, and the sense of the force, it does not depend upon the actual position of the point of application of the force along its line of action. Conversely, the moment MO of a force F does not characterize the position of the point of application of F.
However, as we will see shortly, the moment MO of a force F of a given magnitude and direction completely defines the line of action of F. Indeed, the line of action of F must lie in a plane through O perpendicular to the moment MO; its distance d from O must be equal to the quotient MO /F of the magnitudes of MO and F; and the sense of MO determines whether the line of action of F occurs on one side or the other of the point O.
Recall from Sec. 3.1B that the principle of transmissibility states that two forces F and F9 are equivalent (i.e., have the same effect on a rigid body) if they have the same magnitude, same direction, and same line of action. We can now restate this principle: Two forces F and F9 are equivalent if, and only if, they are equal (i.e., have the same magnitude and same direction) and have equal moments about a given point O. The necessary and sufficient conditions for two forces F and F9 to be equivalent are thus F 5 F9 and MO 5 M9 O (3.13) We should observe that if the relations of Eqs. (3.13) hold for a given point O, they hold for any other point.
Two-Dimensional Problems. Many applications in statics deal with two-dimensional structures. Such structures have length and breadth but only negligible depth. Often, they are subjected to forces contained in the plane of the structure. We can easily represent two-dimensional struc-tures and the forces acting on them on a sheet of paper or on a blackboard. Their analysis is therefore considerably simpler than that of three-dimensional structures and forces.
Consider, for example, a rigid slab acted upon by a force F in the plane of the slab (Fig. 3.12). The moment of F about a point O, which is chosen in the plane of the figure, is represented by a vector MO perpen-dicular to that plane and of magnitude Fd. In the case of Fig. 3.12a, the vector MO points out of the page, whereas in the case of Fig. 3.12b, it points into the page. As we look at the figure, we observe in the first case F 5 F9 and MO 5 M9 O MO F d O (a) MO = + Fd F (b) MO = – Fd MO d O Fig. 3.12 (a) A moment that tends to produce a counterclockwise rotation is positive; (b) a moment that tends to produce a clockwise rotation is negative.
3.1 Forces and Moments 93 that F tends to rotate the slab counterclockwise and in the second case that it tends to rotate the slab clockwise. Therefore, it is natural to refer to the sense of the moment of F about O in Fig. 3.12a as counterclockwise l, and in Fig. 3.12b as clockwise i.
Since the moment of a force F acting in the plane of the figure must be perpendicular to that plane, we need only specify the magnitude and the sense of the moment of F about O. We do this by assigning to the magnitude MO of the moment a positive or negative sign according to whether the vector MO points out of or into the page.
3.1F Rectangular Components of the Moment of a Force We can use the distributive property of vector products to determine the moment of the resultant of several concurrent forces. If several forces F1, F2, . . . are applied at the same point A (Fig. 3.13) and if we denote by r the position vector of A, it follows immediately from Eq. (3.5) that r 3 (F1 1 F2 1 . . .) 5 r 3 F1 1 r 3 F2 1 . . . (3.14) In words, The moment about a given point O of the resultant of several concurrent forces is equal to the sum of the moments of the various forces about the same point O.
This property, which was originally established by the French mathemati-cian Pierre Varignon (1654–1722) long before the introduction of vector algebra, is known as Varignon’s theorem.
The relation in Eq. (3.14) makes it possible to replace the direct deter mination of the moment of a force F by determining the moments of two or more component forces. As you will see shortly, F is generally resolved into components parallel to the coordinate axes. However, it may be more expeditious in some instances to resolve F into components that are not parallel to the coordinate axes (see Sample Prob. 3.3).
In general, determining the moment of a force in space is consider-ably simplified if the force and the position vector of its point of application are resolved into rectangular x, y, and z components. Consider, for example, the moment MO about O of a force F whose components are Fx, Fy, and Fz and that is applied at a point A with coordinates x, y, and z (Fig. 3.14). Since the components of the position vector r are respectively equal to the coordinates x, y, and z of the point A, we can write r and F as r 5 xi 1 yj 1 zk (3.15) F 5 Fxi 1 Fyj 1 Fzk (3.16) Substituting for r and F from Eqs. (3.15) and (3.16) into MO 5 r 3 F (3.11) and recalling Eqs. (3.8) and (3.9), we can write the moment MO of F about O in the form MO 5 Mxi 1 Myj 1 Mzk (3.17) where the components Mx, My, and Mz are defined by the relations r 3 (F1 1 F2 1 . . .) 5 r 3 F1 1 r 3 F2 1 . . .
y x z O A r F1 F2 F3 F4 Fig. 3.13 Varignon’s theorem says that the moment about point O of the resultant of these four forces equals the sum of the moments about point O of the individual forces.
Fz k x y z O zk yj xi r A (x, y, z) Fy j Fxi Fig. 3.14 The moment MO about point O of a force F applied at point A is the vector product of the position vector r and the force F, which can both be expressed in rectangular components.
94 Rigid Bodies: Equivalent Systems of Forces Rectangular components of a moment Mx 5 yFz 2 zFy My 5 zFx 2 xFz Mz 5 xFy 2 yFx (3.18) As you will see in Sec. 3.2C, the scalar components Mx, My, and Mz of the moment MO measure the tendency of the force F to impart to a rigid body a rotation about the x, y, and z axes, respectively. Substituting from Eq. (3.18) into Eq. (3.17), we can also write MO in the form of the deter-minant, as MO 5 † i j k x y z Fx Fy Fz † (3.19) To compute the moment MB about an arbitrary point B of a force F applied at A (Fig. 3.15), we must replace the position vector r in Eq. (3.11) by a vector drawn from B to A. This vector is the position vector of A relative to B, denoted by rA/B. Observing that rA/B can be obtained by subtracting rB from rA, we write MB 5 rA/B 3 F 5 (rA 2 rB) 3 F (3.20) or using the determinant form, MB 5 † i j k xA/B yA/B zA/B Fx Fy Fz † (3.21) where xA/B, yA/B, and zA/B denote the components of the vector rA/B: xA/B 5 xA 2 xB yA/B 5 yA 2 yB zA/B 5 zA 2 zB In the case of two-dimensional problems, we can assume without loss of generality that the force F lies in the xy plane (Fig. 3.16). Setting z 5 0 and Fz 5 0 in Eq. (3.19), we obtain MO 5 (xFy 2 yFx)k We can verify that the moment of F about O is perpendicular to the plane of the figure and that it is completely defined by the scalar MO 5 Mz 5 xFy 2 yFx (3.22) As noted earlier, a positive value for MO indicates that the vector MO points out of the paper (the force F tends to rotate the body counter clockwise about O), and a negative value indicates that the vector MO points into the paper (the force F tends to rotate the body clockwise about O).
To compute the moment about B(xB, yB) of a force lying in the xy plane and applied at A(xA, yA) (Fig. 3.17), we set zA/B 5 0 and Fz 5 0 in Eq. (3.21) and note that the vector MB is perpendicular to the xy plane and is defined in magnitude and sense by the scalar MB 5 (xA 2 xB)Fy 2 (yA 2 yB)Fx (3.23) Mx M 5 yFz F 2 zFy F My M 5 zFx F 2 xFz F Mz M 5 xFy F 2 yFx F MO 5 † i j k x y z Fx F Fy F Fz F † MB 5 † i j k xA x /B / yA y /B / zA z /B / Fx F Fy F Fz F † Fz k x y z B O A rA/B (xA – xB)i (zA – zB)k (yA – yB)j Fy j Fxi Fig. 3.15 The moment MB about the point B of a force F applied at point A is the vector product of the position vector rA/B and force F.
y x z O Fy j Fx i F xi yj r MO = Mzk A (x, y,0) Fig. 3.16 In a two-dimensional problem, the moment MO of a force F applied at A in the xy plane reduces to the z component of the vector product of r with F.
y x z O B Fy j Fx i F A (yA – yB)j (xA – xB)i rA/B MB = MBk Fig. 3.17 In a two-dimensional problem, the moment MB about a point B of a force F applied at A in the xy plane reduces to the z component of the vector product of rA/B with F.
3.1 Forces and Moments 95 Sample Problem 3.1 A 100-lb vertical force is applied to the end of a lever, which is attached to a shaft at O. Determine (a) the moment of the 100-lb force about O; (b) the horizontal force applied at A that creates the same moment about O; (c) the smallest force applied at A that creates the same moment about O; (d) how far from the shaft a 240-lb vertical force must act to create the same moment about O; (e) whether any one of the forces obtained in parts b, c, or d is equivalent to the original force.
100 lb 60° A O 24 in.
STRATEGY: The calculations asked for all involve variations on the basic defining equation of a moment, MO 5 Fd.
MODELING and ANALYSIS: a. Moment about O. The perpendicular distance from O to the line of action of the 100-lb force (Fig. 1) is d 5 (24 in.) cos 60° 5 12 in.
The magnitude of the moment about O of the 100-lb force is MO 5 Fd 5 (100 lb)(12 in.) 5 1200 lb?in.
Since the force tends to rotate the lever clockwise about O, represent the moment by a vector MO perpendicular to the plane of the figure and point-ing into the paper. You can express this fact with the notation MO 5 1200 lb?in. i b b. Horizontal Force. In this case, you have (Fig. 2) d 5 (24 in.) sin 60° 5 20.8 in.
Since the moment about O must be 1200 lb?in., you obtain MO 5 Fd 1200 lb?in. 5 F(20.8 in.) F 5 57.7 lb F 5 57.7 lb y b c. Smallest Force. Since MO 5 Fd, the smallest value of F occurs when d is maximum. Choose the force perpendicular to OA and note that d 5 24 in. (Fig. 3); thus MO 5 Fd 1200 lb?in. 5 F(24 in.) F 5 50 lb F 5 50 lb c30° b (continued) Fig. 2 Determination of horizontal force at A that creates same moment about O.
F 60° MO A O 24 in.
d Fig. 3 Determination of smallest force at A that creates same moment about O.
F MO 60° A O 24 in.
60° MO 100 lb A O 24 in.
d Fig. 1 Determination of the moment of the 100-lb force about O using perpendicular distance d.
96 Rigid Bodies: Equivalent Systems of Forces d. 240-lb Vertical Force. In this case (Fig. 4), MO 5 Fd yields 1200 lb?in. 5 (240 lb)d d 5 5 in.
but OB cos 60° 5 d so OB 5 10 in. b e. None of the forces considered in parts b, c, or d is equivalent to the original 100-lb force. Although they have the same moment about O, they have different x and y components. In other words, although each force tends to rotate the shaft in the same direction, each causes the lever to pull on the shaft in a different way.
REFLECT and THINK: Various combinations of force and lever arm can produce equivalent moments, but the system of force and moment pro-duces a different overall effect in each case. Sample Problem 3.2 A force of 800 N acts on a bracket as shown. Determine the moment of the force about B.
STRATEGY: You can resolve both the force and the position vector from B to A into rectangular components and then use a vector approach to complete the solution.
MODELING and ANALYSIS: Obtain the moment MB of the force F about B by forming the vector product MB 5 rA/B 3 F where rA/B is the vector drawn from B to A (Fig. 1). Resolving rA/B and F into rectangular components, you have rA/B 5 2(0.2 m)i 1 (0.16 m)j F 5 (800 N) cos 60°i 1 (800 N) sin 60°j 5 (400 N)i 1 (693 N)j Recalling the relations in Eq. (3.7) for the cross products of unit vectors (Sec. 3.5), you obtain MB 5 rA/B 3 F 5 [2(0.2 m)i 1 (0.16 m)j] 3 [(400 N)i 1 (693 N)j] 5 2(138.6 N?m)k 2 (64.0 N?m)k 5 2(202.6 N?m)k MB 5 203 N?m i b The moment MB is a vector perpendicular to the plane of the figure and pointing into the page.
(continued) Fig. 4 Position of vertical 240-lb force that creates same moment about O.
240 lb MO 60° A B O d 800 N 60° B A 160 mm 200 mm 60° Fy = (693 N)j Fx = (400 N) i rA/B MB F = 800 N + (0.16 m) j – (0.2 m) i A B Fig. 1 The moment MB is determined from the vector product of position vector rA/B and force vector F.
3.1 Forces and Moments 97 REFLECT and THINK: We can also use a scalar approach to solve this problem using the components for the force F and the position vector rA/B. Following the right-hand rule for assigning signs, we have 1lMB 5 oMB 5 oFd 5 2(400 N)(0.16 m) 2 (693 N)(0.2 m) 5 2202.6 N?m MB 5 203 N?m i b Sample Problem 3.3 A 30-lb force acts on the end of the 3-ft lever as shown. Determine the moment of the force about O.
STRATEGY: Resolving the force into components that are perpendicu-lar and parallel to the axis of the lever greatly simplifies the moment calculation.
MODELING and ANALYSIS: Replace the force by two components: one component P in the direction of OA and one component Q perpendicu-lar to OA (Fig. 1). Since O is on the line of action of P, the moment of P about O is zero. Thus, the moment of the 30-lb force reduces to the moment of Q, which is clockwise and can be represented by a negative scalar.
Q 5 (30 lb) sin 20° 5 10.26 lb MO 5 2Q(3 ft) 5 2(10.26 lb)(3 ft) 5 230.8 lb?ft Since the value obtained for the scalar MO is negative, the moment MO points into the page. You can write it as MO 5 30.8 lb?ft i b REFLECT and THINK: Always be alert for simplifications that can reduce the amount of computation.
A O 20° 50° 30 lb 3 ft Fig. 1 30-lb force at A resolved into components P and Q to simplify the determination of the moment MO.
MO P Q A O 20° 30 lb 3 ft Sample Problem 3.4 A rectangular plate is supported by brackets at A and B and by a wire CD. If the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire on point C.
STRATEGY: The solution requires resolving the tension in the wire and the position vector from A to C into rectangular components. You will need a unit vector approach to determine the force components.
MODELING and ANALYSIS: Obtain the moment MA about A of the force F exerted by the wire on point C by forming the vector product MA 5 rC/A 3 F (1) (continued) 80 mm 80 mm A B C D 240 mm 240 mm 300 mm 98 Rigid Bodies: Equivalent Systems of Forces where rC/A is the vector from A to C rC/A 5 AC 5 (0.3 m)i 1 (0.08 m)k (2) and F is the 200-N force directed along CD (Fig. 1). Introducing the unit vector l 5 CD /CD, you can express F as F 5 Fλ 5 (200 N) CD CD (3) Resolving the vector CD into rectangular components, you have CD 5 2(0.3 m)i 1 (0.24 m)j 2 (0.32 m)k CD 5 0.50 m Substituting into (3) gives you F 5 200 N 0.50m [2(0.3 m)i 1 (0.24 m)j 2 (0.32 m)k] 5 2(120 N)i 1 (96 N)j 2 (128 N)k (4) Substituting for rC/A and F from (2) and (4) into (1) and recalling the relations in Eq. (3.7) of Sec. 3.1D, you obtain (Fig. 2) MA 5 rC/A 3 F 5 (0.3i 1 0.08k) 3 (2120i 1 96j 2 128k) 5 (0.3)(96)k 1 (0.3)(2128)(2j) 1 (0.08)(2120)j 1 (0.08)(96)(2i) MA 5 2(7.68 N?m)i 1 (28.8 N?m)j 1 (28.8 N?m)k b Fig. 2 Components of moment MA applied at A.
A C D (28.8 N•m) j (28.8 N•m) k – (7.68 N•m) i F = (200 N) Alternative Solution. As indicated in Sec. 3.1F, you can also express the moment MA in the form of a determinant: MA 5 † i j k xC 2 xA yC 2 yA zC 2 zA Fx Fy Fz † 5 † i j k 0.3 0 0.08 2120 96 2128 † MA 5 2(7.68 N?m)i 1 (28.8 N?m)j 1 (28.8 N?m)k b REFLECT and THINK: Two-dimensional problems often are solved eas-ily using a scalar approach, but the versatility of a vector analysis is quite apparent in a three-dimensional problem such as this.
Fig. 1 The moment MA is determined from position vector rC/A and force vector F.
rC/A A B C D x y z O 0.08 m 0.08 m 0.3 m 200 N 0.24 m 0.24 m 99 99 I n this section, we introduced the vector product or cross product of two vectors. In the following problems, you will use the vector product to compute the moment of a force about a point and also to determine the perpendicular distance from a point to a line.
We defined the moment of the force F about the point O of a rigid body as MO 5 r 3 F (3.11) where r is the position vector from O to any point on the line of action of F. Since the vector product is not commutative, it is absolutely necessary when computing such a product that you place the vectors in the proper order and that each vector have the correct sense. The moment MO is important because its magnitude is a measure of the tendency of the force F to cause the rigid body to rotate about an axis directed along MO.
1. Computing the moment MO of a force in two dimensions. You can use one of the following procedures: a. Use Eq. (3.12), MO 5 Fd, which expresses the magnitude of the moment as the product of the magnitude of F and the perpendicular distance d from O to the line of action of F [Sample Prob. 3.1].
b. Express r and F in component form and formally evaluate the vector product MO 5 r 3 F [Sample Prob. 3.2]. c. Resolve F into components respectively parallel and perpendicular to the position vector r. Only the perpendicular component contributes to the moment of F [Sample Prob. 3.3].
d. Use Eq. (3.22), MO 5 Mz 5 xFy 2 yFx. When applying this method, the simplest approach is to treat the scalar components of r and F as positive and then to assign, by observation, the proper sign to the moment produced by each force component [Sample Prob. 3.2].
2. Computing the moment MO of a force F in three dimensions. Following the method of Sample Prob. 3.4, the first step in the calculation is to select the most convenient (simplest) position vector r. You should next express F in terms of its rectangular components. The final step is to evaluate the vector product r 3 F to determine the moment. In most three-dimensional problems you will find it easiest to calculate the vector product using a determinant.
3. Determining the perpendicular distance d from a point A to a given line. First assume that a force F of known magnitude F lies along the given line. Next determine its moment about A by forming the vector product MA 5 r 3 F, and calculate this product as indicated above. Then compute its magnitude MA. Finally, substitute the values of F and MA into the equation MA 5 Fd and solve for d.
SOLVING PROBLEMS ON YOUR OWN 100 Problems 3.1 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E. 3.2 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E. 3.3 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if α 5 10°, and (c) the smallest force P that creates the same moment about B.
4 in.
A B P 18 in.
C a 70° Fig. P3.3 3.4 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D. 3.5 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D.
A B C E W 0.85 m 0.5 m 0.6 m 0.6 m D Fig. P3.1 and P3.2 300 N A B D C 25° 100 mm 200 mm 200 mm 125 mm Fig. P3.4 and P3.5 101 3.6 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α 5 25°, determine the moment of the force about point B by resolving the force into hori-zontal and vertical components. 3.7 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α 5 25°, determine the moment of the force about point B by resolving the force into com-ponents along AB and in a direction perpendicular to AB.
3.8 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that the moment of the force about B is 120 lb∙in. clockwise, determine the value of α. 3.9 Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied (a) at point A, (b) at point C. 3.10 Rod AB is held in place by the cord AC. Knowing that c 5 840 mm and that the moment about B of the force exerted by the cord at point A is 756 N∙m, determine the tension in the cord. 3.11 and 3.12 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A.
A B C 15.3 in.
12.0 in.
12.0 in.
2.33 in.
Fig. P3.11 17.2 in.
4.38 in.
7.62 in.
20.5 in.
A B C Fig. P3.12 A B 20 lb 65° a Fig. P3.6 through P3.8 A B C 240 mm c 450 mm Fig. P3.9 and P3.10 102 3.13 and 3.14 It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C.
B C A 42 mm 144 mm 56 mm Fig. P3.13 A B C 88 mm 56 mm 42 mm Fig. P3.14 3.15 Form the vector products B 3 C and B9 3 C, where B 5 B9, and use the results obtained to prove the identity sin α cos β 5 1 2 sin (α 1 β) 1 1 2 sin (α 2 β).
y x C B B' a b b Fig. P3.15 3.16 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P 5 28i 1 4j 2 4k and Q 5 3i 1 3j 1 6k, (b) P 5 7i 2 6j 2 3k and Q 5 23i 1 6j 2 2k. 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) 2i 1 3j 2 6k and 5i 2 8j 2 6k, (b) 4i 2 4j 1 3k and 23i 1 7j 2 5k. 3.18 A line passes through the points (12 m, 8 m) and (23 m, 25 m). Determine the perpendicular distance d from the line to the origin O of the system of coordinates.
3.19 Determine the moment about the origin O of the force F 5 4i 2 3j 1 5k that acts at a point A. Assume that the position vector of A is (a) r 5 2i 1 3j 2 4k, (b) r 5 28i 1 6j 2 10k, (c) r 5 8i 2 6j 1 5k.
103 3.20 Determine the moment about the origin O of the force F 5 2i 1 3j 2 4k that acts at a point A. Assume that the position vector of A is (a) r 5 3i 2 6j 1 5k, (b) r 5 i 2 4j 2 2k, (c) r 5 4i 1 6j 2 8k. 3.21 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B. 3.22 The 12-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 380 lb, determine the moment about A of the force exerted by the cable at B.
B C A x y z 4.8 ft 12 ft 8 ft Fig. P3.22 3.23 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A.
B A x y z 50 mm 60 mm 25 mm 200 N 30° 60° C Fig. P3.23 3.24 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E.
y B A C 1 m 4.25 m 7 m 6 m 0.75 m x z Fig. P3.21 y z x B D O E C A 160 mm 90 mm 120 mm 120 mm Fig. P3.24 104 3.25 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Deter-mine the moment about A of the force exerted by the line at B.
x y z A B D C 45° 30° 8° Fig. P3.25 3.26 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about point O of the force exerted by the cable at B. 3.27 In Prob. 3.22, determine the perpendicular distance from point A to cable BC. 3.28 In Prob. 3.24, determine the perpendicular distance from point O to wire AE. 3.29 In Prob. 3.24, determine the perpendicular distance from point B to wire AE. 3.30 In Prob. 3.25, determine the perpendicular distance from point A to a line drawn through points B and C. 3.31 In Prob. 3.25, determine the perpendicular distance from point D to a line drawn through points B and C. 3.32 In Prob. 3.26, determine the perpendicular distance from point O to cable BD. 3.33 In Prob. 3.26, determine the perpendicular distance from point C to cable BD. 3.34 Determine the value of a that minimizes the perpendicular distance from point C to a section of pipeline that passes through points A and B.
A O B E C D 2 m 2.5 m 2 m 1 m y z x Fig. P3.26 x y z A B C 8 ft 3 ft 2 ft 10 ft a 24 ft 18 ft 16 ft Fig. P3.34 3.2 Moment of a Force about an Axis 105 3.2 MOMENT OF A FORCE ABOUT AN AXIS We want to extend the idea of the moment about a point to the often use-ful concept of the moment about an axis. However, first we need to intro-duce another tool of vector mathematics. We have seen that the vector product multiplies two vectors together and produces a new vector. Here we examine the scalar product, which multiplies two vectors together and produces a scalar quantity.
3.2A Scalar Products The scalar product of two vectors P and Q is defined as the product of the magnitudes of P and Q and of the cosine of the angle θ formed between them (Fig. 3.18). The scalar product of P and Q is denoted by P ? Q. Scalar product P ? Q 5 PQ cos θ (3.24) Note that this expression is not a vector but a scalar, which explains the name scalar product. Because of the notation used, P ? Q is also referred to as the dot product of the vectors P and Q.
It follows from its very definition that the scalar product of two vectors is commutative, i.e., that P ? Q 5 Q ? P (3.25) It can also be proven that the scalar product is distributive, as shown by P ? (Q1 1 Q2) 5 P ? Q1 1 P ? Q2 (3.26) As far as the associative property is concerned, this property cannot apply to scalar products. Indeed, (P ? Q) ? S has no meaning, because P ? Q is not a vector but a scalar.
We can also express the scalar product of two vectors P and Q in terms of their rectangular components. Resolving P and Q into compo-nents, we first write P ? Q 5 (Pxi 1 Pyj 1 Pzk) ? (Qxi 1 Qyj 1 Qzk) Making use of the distributive property, we express P ? Q as the sum of scalar products, such as Pxi ? Qxi and Pxi ? Qyj. However, from the defini-tion of the scalar product, it follows that the scalar products of the unit vectors are either zero or one. i ? i 5 1 j ? j 5 1 k ? k 5 1 i ? j 5 0 j ? k 5 0 k ? i 5 0 (3.27) Thus, the expression for P ? Q reduces to Scalar product P ? Q 5 PxQx 1 PyQy 1 PzQz (3.28) P ? Q 5 PQcosθ P ? Q 5 Px P Qx 1 Py P Qy 1 PzQz Q P q Fig. 3.18 Two vectors P and Q and the angle θ between them.
106 Rigid Bodies: Equivalent Systems of Forces In the particular case when P and Q are equal, we note that P ? P 5 P2 x 1 P2 y 1 P2 z 5 P2 (3.29) Applications of the Scalar Product 1. Angle formed by two given vectors. Let two vectors be given in terms of their components: P 5 Pxi 1 Py j 1 Pzk Q 5 Qxi 1 Qyj 1 Qzk To determine the angle formed by the two vectors, we equate the expressions obtained in Eqs. (3.24) and (3.28) for their scalar product, PQ cos θ 5 PxQx 1 PyQy 1 PzQz Solving for cos θ, we have cos θ 5 PxQx 1 PyQy 1 PzQz PQ (3.30) 2. Projection of a vector on a given axis. Consider a vector P forming an angle θ with an axis, or directed line, OL (Fig. 3.19a). We define the projection of P on the axis OL as the scalar POL 5 P cos θ (3.31) The projection POL is equal in absolute value to the length of the seg-ment OA. It is positive if OA has the same sense as the axis OL––that is, if θ is acute––and negative otherwise. If P and OL are at a right angle, the projection of P on OL is zero. Now consider a vector Q directed along OL and of the same sense as OL (Fig. 3.19b). We can express the scalar product of P and Q as P ? Q 5 PQ cos θ 5 POLQ (3.32) from which it follows that POL 5 P ? Q Q 5 PxQx 1 PyQy 1 PzQz Q (3.33) In the particular case when the vector selected along OL is the unit vector l (Fig. 3.19c), we have POL 5 P ? l (3.34) Recall from Sec. 2.4A that the components of l along the coordinate axes are respectively equal to the direction cosines of OL. Resolving P and l into rectangular components, we can express the projection of P on OL as POL 5 Px cos θx 1 Py cos θy 1 Pz cos θz (3.35) where θx, θy, and θz denote the angles that the axis OL forms with the coordinate axes.
POL 5 P ? l y x z O A P L q y x z A P L q Q O y x z O A P L qx qy qz (a) (b) (c) Fig. 3.19 (a) The projection of vector P at an angle θ to a line OL; (b) the projection of P and a vector Q along OL; (c) the projection of P, a unit vector λ along OL, and the angles of OL with the coordinate axes.
3.2 Moment of a Force about an Axis 107 3.2B Mixed Triple Products We have now seen both forms of multiplying two vectors together: the vector product and the scalar product. Here we define the mixed triple product of the three vectors S, P, and Q as the scalar expression Mixed triple product S ? (P 3 Q) (3.36) This is obtained by forming the scalar product of S with the vector product of P and Q. [In Chapter 15, we will introduce another kind of triple product, called the vector triple product, S 3 (P 3 Q).] The mixed triple product of S, P, and Q has a simple geometrical interpretation (Fig. 3.20a). Recall from Sec. 3.4 that the vector P 3 Q is perpendicular to the plane containing P and Q and that its magnitude is equal to the area of the parallelogram that has P and Q for sides. Also, Eq. (3.32) indicates that we can obtain the scalar product of S and P 3 Q by multiplying the magnitude of P 3 Q (i.e., the area of the parallelogram defined by P and Q) by the projection of S on the vector P 3 Q (i.e., by the projection of S on the normal to the plane containing the parallelo-gram). The mixed triple product is thus equal, in absolute value, to the volume of the parallelepiped having the vectors S, P, and Q for sides (Fig. 3.20b). The sign of the mixed triple product is positive if S, P, and Q form a right-handed triad and negative if they form a left-handed triad. [That is, S ? (P 3 Q) is negative if the rotation that brings P into line with Q is observed as clockwise from the tip of S]. The mixed triple product is zero if S, P, and Q are coplanar.
Since the parallelepiped defined in this way is independent of the order in which the three vectors are taken, the six mixed triple products that can be formed with S, P, and Q all have the same absolute value, although not the same sign. It is easily shown that S ? (P 3 Q) 5 P ? (Q 3 S) 5 Q ? (S 3 P) 5 2S ? (Q 3 P) 5 2P ? (S 3 Q) 5 2Q ? (P 3 S) (3.37) Arranging the letters representing the three vectors counterclockwise in a circle (Fig. 3.21), we observe that the sign of the mixed triple product remains unchanged if the vectors are permuted in such a way that they still read in counterclockwise order. Such a permutation is said to be a circular permutation. It also follows from Eq. (3.37) and from the com-mutative property of scalar products that the mixed triple product of S, P, and Q can be defined equally well as S ? (P 3 Q) or (S 3 P) ? Q.
We can also express the mixed triple product of the vectors S, P, and Q in terms of the rectangular components of these vectors. Denoting P 3 Q by V and using formula (3.28) to express the scalar product of S and V, we have S ? (P 3 Q) 5 S ? V 5 SxVx 1 SyVy 1 SzVz Substituting from the relations in Eq. (3.9) for the components of V, we obtain S ? (P 3 Q) 5 Sx(PyQz 2 PzQy) 1 Sy(PzQx 2 PxQz) 1 Sz(PxQy 2 PyQx) (3.38) S?(P 3 Q) S P Q P × Q S P Q (a) (b) Fig. 3.20 (a) The mixed triple product is equal to the magnitude of the cross product of two vectors multiplied by the projection of the third vector onto that cross product; (b) the result equals the volume of the parallelepiped formed by the three vectors. S P Q Fig. 3.21 Counterclockwise arrangement for determining the sign of the mixed triple product of three vectors P, Q, and S.
108 Rigid Bodies: Equivalent Systems of Forces We can write this expression in a more compact form if we observe that it represents the expansion of a determinant: Mixed triple product, determinant form S ? (P 3 Q) 5 † Sx Sy Sz Px Py Pz Qx Qy Qz † (3.39) By applying the rules governing the permutation of rows in a determinant, we could easily verify the relations in Eq. (3.37), which we derived earlier from geometrical considerations.
3.2C Moment of a Force about a Given Axis Now that we have the necessary mathematical tools, we can introduce the concept of moment of a force about an axis. Consider again a force F acting on a rigid body and the moment MO of that force about O (Fig. 3.22). Let OL be an axis through O. We define the moment MOL of F about OL as the projection OC of the moment MO onto the axis OL. y x z r L A C O MO F Fig. 3.22 The moment MOL of a force F about the axis OL is the projection on OL of the moment MO. The calculation involves the unit vector l along OL and the position vector r from O to A, the point upon which the force F acts.
Suppose we denote the unit vector along OL by l and recall the expres-sions (3.34) and (3.11) for the projection of a vector on a given axis and for the moment MO of a force F. Then we can express MOL as Moment about an axis through the origin MOL 5 λ ? MO 5 λ ? (r 3 F) (3.40) S?(P 3 Q) 5 † Sx S Sy S Sz S Px P Py P Pz Qx Qy Qz † MOL 5 λ ? MO 5 λ ? (r 3 F) 3.2 Moment of a Force about an Axis 109 This shows that the moment MOL of F about the axis OL is the scalar obtained by forming the mixed triple product of l, r, and F. We can also express MOL in the form of a determinant, MOL 5 † λx λy λz x y z Fx Fy Fz † (3.41) where lx, ly, lz 5 direction cosines of axis OL x, y, z 5 coordinates of point of application of F Fx, Fy, Fz 5 components of force F The physical significance of the moment MOL of a force F about a fixed axis OL becomes more apparent if we resolve F into two rectangular components F1 and F2, with F1 parallel to OL and F2 lying in a plane P perpendicular to OL (Fig. 3.23). Resolving r similarly into two compo-nents r1 and r2 and substituting for F and r into Eq. (3.40), we get MOL 5 l ? [(r1 1 r2) 3 (F1 1 F2)] 5 l ? (r1 3 F1) 1 l ? (r1 3 F2) 1 l ? (r2 3 F1) 1 l ? (r2 3 F2) Note that all of the mixed triple products except the last one are equal to zero because they involve vectors that are coplanar when drawn from a common origin (Sec. 3.2B). Therefore, this expression reduces to MOL 5 l ? (r2 3 F2) (3.42) The vector product r2 3 F2 is perpendicular to the plane P and represents the moment of the component F2 of F about the point Q where OL inter-sects P. Therefore, the scalar MOL, which is positive if r2 3 F2 and OL have the same sense and is negative otherwise, measures the tendency of F2 to make the rigid body rotate about the fixed axis OL. The other com-ponent F1 of F does not tend to make the body rotate about OL, because F1 and OL are parallel. Therefore, we conclude that The moment MOL of F about OL measures the tendency of the force F to impart to the rigid body a rotation about the fixed axis OL.
From the definition of the moment of a force about an axis, it follows that the moment of F about a coordinate axis is equal to the component of MO along that axis. If we substitute each of the unit vectors i, j, and k for l in Eq. (3.40), we obtain expressions for the moments of F about the coordinate axes. These expressions are respectively equal to those obtained earlier for the components of the moment MO of F about O: Mx 5 yFz 2 zFy My 5 zFx 2 xFz Mz 5 xFy 2 yFx (3.18) Just as the components Fx, Fy, and Fz of a force F acting on a rigid body measure, respectively, the tendency of F to move the rigid body in the x, y, and z directions, the moments Mx, My, and Mz of F about the coor-dinate axes measure the tendency of F to impart to the rigid body a rota-tion about the x, y, and z axes, respectively.
MOL 5 † λx λy λz x y z Fx F Fy F Fz F † Mx M 5 yFz F 2 zFy F My M 5 zFx F 2 xFz F Mz M 5 xFy F 2 yFx F r r1 r2 F1 F2 P Q L A O F Fig. 3.23 By resolving the force F into components parallel to the axis OL and in a plane perpendicular to the axis, we can show that the moment MOL of F about OL measures the tendency of F to rotate the rigid body about the axis.
110 Rigid Bodies: Equivalent Systems of Forces y x z L A B O F C rA/B = rA – rB Fig. 3.24 The moment of a force about an axis or line L can be found by evaluating the mixed triple product at a point B on the line. The choice of B is arbitrary, since using any other point on the line, such as C, yields the same result.
More generally, we can obtain the moment of a force F applied at A about an axis that does not pass through the origin by choosing an arbitrary point B on the axis (Fig. 3.24) and determining the projection on the axis BL of the moment MB of F about B. The equation for this projection is given here.
Moment about an arbitrary axis MBL 5 l ? MB 5 l ? (rA/B 3 F) (3.43) where rA/B 5 rA 2 rB represents the vector drawn from B to A. Expressing MBL in the form of a determinant, we have MBL 5 † lx ly lz xA/B yA/B zA/B Fx Fy Fz † (3.44) where lx, ly, lz 5 direction cosines of axis BL xA/B 5 xA 2 xB yA/B 5 yA 2 yB zA/B 5 zA 2 zB Fx, Fy, Fz 5 components of force F Note that this result is independent of the choice of the point B on the given axis. Indeed, denoting by MCL the moment obtained with a different point C, we have MCL 5 l ? [(rA 2 rC) 3 F] 5 l ? [(rA 2 rB) 3 F] 1 l ? [(rB 2 rC) 3 F] However, since the vectors l and rB 2 rC lie along the same line, the volume of the parallelepiped having the vectors l, rB 2 rC, and F for sides is zero, as is the mixed triple product of these three vectors (Sec. 3.2B). The expression obtained for MCL thus reduces to its first term, which is the expression used earlier to define MBL. In addition, it follows from Sec. 3.1E that, when computing the moment of F about the given axis, A can be any point on the line of action of F.
MBL M 5 l ? MB 5 l ? (rA/B r 3 F) MB M L 5 † lx ly lz xA x /B / yA y /B / zA z /B / Fx F Fy F Fz F † 3.2 Moment of a Force about an Axis 111 Sample Problem 3.5 A cube of side a is acted upon by a force P along the diagonal of a face, as shown. Determine the moment of P (a) about A, (b) about the edge AB, (c) about the diagonal AG of the cube. (d) Using the result of part c, determine the perpendicular distance between AG and FC.
A B C D E F G a P STRATEGY: Use the equations presented in this section to compute the moments asked for. You can find the distance between AG and FC from the expression for the moment MAG.
MODELING and ANALYSIS: a. Moment about A. Choosing x, y, and z axes as shown (Fig. 1), resolve into rectangular components the force P and the vector rF/A 5 AF drawn from A to the point of application F of P. rF/A 5 ai 2 aj 5 a(i 2 j) P 5 (P/12)j 2 (P/12)k 5 (P/12)(j 2 k) Fig. 1 Position vector rF/A and force vector P relative to chosen coordinate system.
i k j A B C D E F G x y z a a a P rF/A O The moment of P about A is the vector product of these two vectors: MA 5 rF/A 3 P 5 a(i 2 j) 3 (P/12)(j 2 k) MA 5 (aP/12)(i 1 j 1 k) b b. Moment about AB. You want the projection of MA on AB: MAB 5 i ? MA 5 i ? (aP/12)(i 1 j 1 k) MAB 5 aP/12 b (continued) 112 Rigid Bodies: Equivalent Systems of Forces You can verify that since AB is parallel to the x axis, MAB is also the x component of the moment MA.
c. Moment about diagonal AG. You obtain the moment of P about AG by projecting MA on AG. If you denote the unit vector along AG by l (Fig. 2), the calculation looks like this: λ 5 AG AG 5 ai 2 aj 2 ak a13 5 (1/13)(i 2 j 2 k) MAG 5 l ? MA 5 (1/13)(i 2 j 2 k)?(aP/12)(i 1 j 1 k) MAG 5 (aP/16)(1 2 1 2 1) MAG 5 2aP/16 b Alternative Method. You can also calculate the moment of P about AG from the determinant form: MAG 5 † lx ly lz xF/A yF/A zF/A Fx Fy Fz † 5 † 1/13 21/13 21/13 a 2a 0 0 P/12 2P/12 † 5 2aP/16 d. Perpendicular Distance between AG and FC. First note that P is perpendicular to the diagonal AG. You can check this by forming the scalar product P ? l and verifying that it is zero: P ? λ 5 (P/12)(j 2 k) ? (1/13)(i 2 j 2 k) 5 (P16)(0 2 1 1 1) 5 0 You can then express the moment MAG as 2Pd, where d is the perpen-dicular distance from AG to FC (Fig. 3). (The negative sign is needed because the rotation imparted to the cube by P appears as clockwise to an observer at G.) Using the value found for MAG in part c, MAG 5 2Pd 5 2aP/16 d 5 a/16 b Fig. 3 Perpendicular distance d from AG to FC.
O A B C D E F G d P REFLECT and THINK: In a problem like this, it is important to visual-ize the forces and moments in three dimensions so you can choose the appropriate equations for finding them and also recognize the geometric relationships between them.
Fig. 2 Unit vector l used to determine moment of P about AG.
A B C D E F G x y z O P 113 113 SOLVING PROBLEMS ON YOUR OWN I n the problems for this section, you will apply the scalar product (or dot product) of two vectors to determine the angle formed by two given vectors and the projec-tion of a force on a given axis. You will also use the mixed triple product of three vectors to find the moment of a force about a given axis and the perpendicular dis-tance between two lines.
1. Calculating the angle formed by two given vectors. First express the vectors in terms of their components and determine the magnitudes of the two vectors. Then find the cosine of the desired angle by dividing the scalar product of the two vectors by the product of their magnitudes [Eq. (3.30)].
2. Computing the projection of a vector P on a given axis OL. In general, begin by expressing P and the unit vector l, which defines the direction of the axis, in component form. Take care that l has the correct sense (that is, l is directed from O to L). The required projection is then equal to the scalar product P ? l. However, if you know the angle θ formed by P and l, the projection is also given by P cos θ.
3. Determining the moment MOL of a force about a given axis OL. We defined MOL as MOL 5 l ? MO 5 l ? (r 3 F) (3.40) where l is the unit vector along OL and r is a position vector from any point on the line OL to any point on the line of action of F. As was the case for the moment of a force about a point, choosing the most convenient position vector will simplify your calculations. Also, recall the warning of the preceding section: The vectors r and F must have the correct sense, and they must be placed in the proper order. The proce-dure you should follow when computing the moment of a force about an axis is illustrated in part c of Sample Prob. 3.5. The two essential steps in this procedure are (1) express l, r, and F in terms of their rectangular components and (2) evaluate the mixed triple product l ? (r 3 F) to determine the moment about the axis. In most three-dimensional problems, the most convenient way to compute the mixed triple product is by using a determinant.
As noted in the text, when l is directed along one of the coordinate axes, MOL is equal to the scalar component of MO along that axis.
4. Determining the perpendicular distance between two lines. Remember that it is the perpendicular component F2 of the force F that tends to make a body rotate about a given axis OL (Fig. 3.23). It then follows that MOL 5 F2 d (continued) 114 where MOL is the moment of F about axis OL and d is the perpendicular distance between OL and the line of action of F. This last equation provides a simple technique for determining d. First assume that a force F of known magnitude F lies along one of the given lines and that the unit vector l lies along the other line. Next compute the moment MOL of the force F about the second line using the method discussed above. The magnitude of the parallel component, F1, of F is obtained using the scalar product: F1 5 F ? l The value of F2 is then determined from F2 5 2F2 2 F2 1 Finally, substitute the values of MOL and F2 into the equation MOL 5 F2 d and solve for d.
You should now realize that the calculation of the perpendicular distance in part d of Sample Prob. 3.5 was simplified by P being perpendicular to the diagonal AG. In general, the two given lines will not be perpendicular, so you will have to use the technique just outlined when determining the perpendicular distance between them.
115 Problems 3.35 Given the vectors P 5 2i 1 3j 2 k, Q 5 5i 2 4j 1 3k, and S 5 23i 1 2j 2 5k, compute the scalar products P ? Q, P ? S, and Q ? S. 3.36 Form the scalar product B ? C and use the result obtained to prove the identity cos (α – β) 5 cos α cos β 1 sin α sin β B C y x a b Fig. P3.36 3.37 Three cables are used to support a container as shown. Determine the angle formed by cables AB and AD. 3.38 Three cables are used to support a container as shown. Determine the angle formed by cables AC and AD. 3.39 Knowing that the tension in cable AC is 280 lb, determine (a) the angle between cable AC and the boom AB, (b) the projection on AB of the force exerted by cable AC at point A.
3 ft 6 ft 7.5 ft A P 4.5 ft 6 ft B C D y x z 6.5 ft Fig. P3.39 and P3.40 3.40 Knowing that the tension in cable AD is 180 lb, determine (a) the angle between cable AD and the boom AB, (b) the projection on AB of the force exerted by cable AD at point A.
x y z A B D C O 600 mm 320 mm 360 mm 500 mm 450 mm Fig. P3.37 and P3.38 116 3.41 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at point B.
x z 3 m 3 m 1 m 0.38 m 0.08 m 0.16 m Detail of the stake at B 1.5 m A B C D B y Fig. P3.41 and P3.42 3.42 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope BC is 490 N, determine (a) the angle between rope BC and the stake, (b) the projection on the stake of the force exerted by rope BC at point B. 3.43 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x 5 11 in., determine the angle formed by the two cords, (a) using Eq. (3.30), (b) applying the law of cosines to triangle ABC. 3.44 Solve Prob. 3.43 for the position corresponding to x 5 4 in. 3.45 Determine the volume of the parallelepiped of Fig. 3.20b when (a) P 5 4i 2 3j 1 2k, Q 5 22i 2 5j 1 k, and S 5 7i 1 j 2 k, (b) P 5 5i 2 j 1 6k, Q 5 2i 1 3j 1 k, and S 5 23i 2 2j 1 4k. 3.46 Given the vectors P 5 3i 2 j 1 k, Q 5 4i 1 Qyj 2 2k, and S 5 2i 2 2j 1 2k, determine the value of Qy for which the three vectors are coplanar. 3.47 A crane is oriented so that the end of the 25-m boom AO lies in the yz plane. At the instant shown, the tension in cable AB is 4 kN. Determine the moment about each of the coordinate axes of the force exerted on A by cable AB. 3.48 The 25-m crane boom AO lies in the yz plane. Determine the maxi-mum permissible tension in cable AB if the absolute value of moments about the coordinate axes of the force exerted on A by cable AB must be |Mx| # 60 kN?m, |My| # 12 kN?m, |Mz| # 8 kN?m y z x x C O A B 12 in.
24 in.
20 in.
Fig. P3.43 Fig. P3.47 and P3.48 A C B y 2.5 m 15 m O x z 117 3.49 To loosen a frozen valve, a force F with a magnitude of 70 lb is applied to the handle of the valve. Knowing that θ 5 25°, Mx 5 261 lb?ft, and Mz 5 243 lb?ft, determine ϕ and d.
x y d z B A q F 4 in.
11 in.
f Fig. P3.49 and P3.50 3.50 When a force F is applied to the handle of the valve shown, its moments about the x and z axes are Mx 5 277 lb?ft and Mz 5 281 lb?ft, respectively. For d 5 27 in., determine the moment My of F about the y axis.
3.51 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N?m and 2460 N?m, determine the distance a.
x y z A B C D O a 1.6 m 2.2 m 4.8 m Fig. P3.51 and P3.52 3.52 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the man applies a 195-N force to end A of the rope and that the moment of that force about the y axis is 132 N?m, determine the distance a.
118 3.53 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x axis of the forces exerted by the cables on the barn at points A and D is equal to 4728 lb?ft, determine the magnitude of TDE when TAB 5 255 lb.
3.54 Solve Prob. 3.53 when the tension in cable AB is 306 lb. 3.55 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P.
z y x O H D Q P C A B 24 in.
21 in.
18 in.
12 in.
16 in.
17 in.
30 in.
32 in.
G Fig. P3.55 and P3.56 3.56 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. 3.57 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.
x y z A B C D G O P H 0.35 m 0.875 m 0.75 m 0.75 m 0.925 m 0.5 m 0.5 m Fig. P3.57 E B z y x C D A F 12 ft 12 ft 1.5 ft 1 ft 14 ft Fig. P3.53 119 3.58 In Prob. 3.57, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. 3.59 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining points D and B. 3.60 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining points D and B. 3.61 A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA.
x y z O A B C P Fig. P3.61 and P3.62 3.62 A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Prob. 3.61 to deter-mine the perpendicular distance between edges OA and BC. 3.63 Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1. 3.64 In Prob. 3.55, determine the perpendicular distance between rod AB and the line of action of P. 3.65 In Prob. 3.56, determine the perpendicular distance between rod AB and the line of action of Q. 3.66 In Prob. 3.57, determine the perpendicular distance between portion BH of the cable and the diagonal AD. 3.67 In Prob. 3.58, determine the perpendicular distance between portion BG of the cable and the diagonal AD. 3.68 In Prob. 3.59, determine the perpendicular distance between cable AE and the line joining points D and B. 3.69 In Prob. 3.60, determine the perpendicular distance between cable CF and the line joining points D and B.
y z x A E B D C 0.6 m 0.6 m 0.6 m 0.9 m 0.9 m 0.3 m 0.4 m 0.4 m 0.7 m 0.2 m 0.35 m F Fig. P3.59 and P3.60 120 Rigid Bodies: Equivalent Systems of Forces 3.3 COUPLES AND FORCE-COUPLE SYSTEMS Now that we have studied the effects of forces and moments on a rigid body, we can ask if it is possible to simplify a system of forces and moments without changing these effects. It turns out that we can replace a system of forces and moments with a simpler and equivalent system. One of the key ideas used in such a transformation is called a couple. 3.3A Moment of a Couple Two forces F and 2F, having the same magnitude, parallel lines of action, and opposite sense, are said to form a couple (Fig. 3.25). Clearly, the sum of the components of the two forces in any direction is zero. The sum of the moments of the two forces about a given point, however, is not zero. The two forces do not cause the body on which they act to move along a line (translation), but they do tend to make it rotate.
Let us denote the position vectors of the points of application of F and 2F by rA and rB, respectively (Fig. 3.26). The sum of the moments of the two forces about O is rA 3 F 1 rB 3 (2F) 5 (rA 2 rB) 3 F Setting rA 2 rB 5 r, where r is the vector joining the points of application of the two forces, we conclude that the sum of the moments of F and 2F about O is represented by the vector M 5 r 3 F (3.45) The vector M is called the moment of the couple. It is perpendicular to the plane containing the two forces, and its magnitude is M 5 rF sin θ 5 Fd (3.46) where d is the perpendicular distance between the lines of action of F and 2F and θ is the angle between F (or 2F) and r. The sense of M is defined by the right-hand rule.
Note that the vector r in Eq. (3.45) is independent of the choice of the origin O of the coordinate axes. Therefore, we would obtain the same result if the moments of F and 2F had been computed about a different point O9. Thus, the moment M of a couple is a free vector (Sec. 2.1B), which can be applied at any point (Fig. 3.27).
M 5 r 3 F M 5 rF sin F θ 5 Fd –F F Fig. 3.25 A couple consists of two forces with equal magnitude, parallel lines of action, and opposite sense.
–F F y x z A O d M r rB rA B q Fig. 3.26 The moment M of the couple about O is the sum of the moments of F and of 2F about O.
–F F d M Fig. 3.27 The moment M of a couple equals the product of F and d, is perpendicular to the plane of the couple, and may be applied at any point of that plane.
–F F Photo 3.1 The parallel upward and downward forces of equal magnitude exerted on the arms of the lug nut wrench are an example of a couple.
3.3 Couples and Force-Couple Systems 121 From the definition of the moment of a couple, it also follows that two couples––one consisting of the forces F1 and 2F1, the other of the forces F2 and 2F2 (Fig. 3.28)––have equal moments if F1d1 5 F2d2 (3.47) provided that the two couples lie in parallel planes (or in the same plane) and have the same sense (i.e., clockwise or counterclockwise).
3.3B Equivalent Couples Imagine that three couples act successively on the same rectangular box (Fig. 3.29). As we have just seen, the only motion a couple can impart to a rigid body is a rotation. Since each of the three couples shown has the same moment M (same direction and same magnitude M 5 120 lb?in.), we can expect each couple to have the same effect on the box.
– F1 F1 d1 – F2 F2 d2 Fig. 3.28 Two couples have the same moment if they lie in parallel planes, have the same sense, and if F1d1 5 F2d2.
y x z 20 lb 20 lb (a) (b) (c) 6 in.
4 in.
4 in.
M y x z 30 lb 30 lb 4 in.
M y x z 30 lb 30 lb 4 in.
M Fig. 3.29 Three equivalent couples. (a) A couple acting on the bottom of the box, acting counterclockwise viewed from above; (b) a couple in the same plane and with the same sense but larger forces than in (a); (c) a couple acting in a different plane but same sense.
As reasonable as this conclusion appears, we should not accept it hastily. Although intuition is of great help in the study of mechanics, it should not be accepted as a substitute for logical reasoning. Before stating that two systems (or groups) of forces have the same effect on a rigid body, we should prove that fact on the basis of the experimental evidence introduced so far. This evidence consists of the parallelogram law for the addition of two forces (Sec. 2.1A) and the principle of transmissibility (Sec. 3.1B). Therefore, we state that two systems of forces are equivalent (i.e., they have the same effect on a rigid body) if we can transform one of them into the other by means of one or several of the following operations: (1) replacing two forces acting on the same particle by their resultant; (2) resolving a force into two components; (3) canceling two equal and opposite forces acting on the same particle; (4) attaching to the same particle two equal and opposite forces; and (5) moving a force along its line of action. Each of these operations is easily justified on the basis of the parallelogram law or the principle of transmissibility.
Let us now prove that two couples having the same moment M are equivalent. First consider two couples contained in the same plane, and 122 Rigid Bodies: Equivalent Systems of Forces assume that this plane coincides with the plane of the figure (Fig. 3.30). The first couple consists of the forces F1 and 2F1 of magnitude F1, located at a distance d1 from each other (Fig. 3.30a). The second couple consists of the forces F2 and 2F2 of magnitude F2, located at a distance d2 from each other (Fig. 3.30d). Since the two couples have the same moment M, which is perpendicular to the plane of the figure, they must have the same sense (assumed here to be counterclockwise), and the relation F1d1 5 F2d2 (3.47) must be satisfied. To prove that they are equivalent, we shall show that the first couple can be transformed into the second by means of the opera-tions listed previously.
Let us denote by A, B, C, and D the points of intersection of the lines of action of the two couples. We first slide the forces F1 and 2F1 until they are attached, respectively, at A and B, as shown in Fig. 3.30b. We then resolve force F1 into a component P along line AB and a component Q along AC (Fig. 3.30c). Similarly, we resolve force 2F1 into 2P along AB and 2Q along BD. The forces P and 2P have the same magnitude, the same line of action, and opposite sense; we can move them along their common line of action until they are applied at the same point and may then be canceled. Thus, the couple formed by F1 and 2F1 reduces to a couple consisting of Q and 2Q.
We now show that the forces Q and 2Q are respectively equal to the forces 2F2 and F2. We obtain the moment of the couple formed by Q and 2Q by computing the moment of Q about B. Similarly, the moment of the couple formed by F1 and 2F1 is the moment of F1 about B. However, by Varignon’s theorem, the moment of F1 is equal to the sum of the moments of its components P and Q. Since the moment of P about B is zero, the moment of the couple formed by Q and 2Q must be equal to the moment of the couple formed by F1 and 2F1. Recalling Eq. (3.47), we have Qd2 5 F1d1 5 F2d2 and Q 5 F2 Thus, the forces Q and 2Q are respectively equal to the forces 2F2 and F2, and the couple of Fig. 3.30a is equivalent to the couple of Fig. 3.30d.
Now consider two couples contained in parallel planes P1 and P2. We prove that they are equivalent if they have the same moment. In view of the preceding discussion, we can assume that the couples consist of forces of the same magnitude F acting along parallel lines (Fig. 3.31a and d). We propose to show that the couple contained in plane P1 can be trans-formed into the couple contained in plane P2 by means of the standard operations listed previously.
–F1 F1 d1 –F1 F1 –F1 F1 –F2 F2 d2 (a) (b) (c) (d) A B C D Q – Q P A B C D – P = = = Fig. 3.30 Four steps in transforming one couple to another couple in the same plane by using simple operations. (a) Starting couple; (b) label points of intersection of lines of action of the two couples; (c) resolve forces from first couple into components; (d) final couple.
–F1 F1 P1 P2 (a) –F2 F2 P2 P1 (d) (b) (c) –F3 F3 –F1 F1 –F2 F2 –F3 F3 Fig. 3.31 Four steps in transforming one couple to another couple in a parallel plane by using simple operations. (a) Initial couple; (b) add a force pair along the line of intersection of two diagonal planes; (c) replace two couples with equivalent couples in the same planes; (d) final couple.
3.3 Couples and Force-Couple Systems 123 Let us consider the two diagonal planes defined respectively by the lines of action of F1 and 2F2 and by those of 2F1 and F2 (Fig. 3.31b). At a point on their line of intersection, we attach two forces F3 and 2F3, which are respectively equal to F1 and 2F1. The couple formed by F1 and 2F3 can be replaced by a couple consisting of F3 and 2F2 (Fig. 3.31c), because both couples clearly have the same moment and are contained in the same diagonal plane. Similarly, the couple formed by 2F1 and F3 can be replaced by a couple consisting of 2F3 and F2. Canceling the two equal and opposite forces F3 and 2F3, we obtain the desired couple in plane P2 (Fig. 3.31d). Thus, we conclude that two couples having the same moment M are equiva-lent, whether they are contained in the same plane or in parallel planes.
The property we have just established is very important for the correct understanding of the mechanics of rigid bodies. It indicates that when a couple acts on a rigid body, it does not matter where the two forces forming the couple act or what magnitude and direction they have. The only thing that counts is the moment of the couple (magnitude and direction). Couples with the same moment have the same effect on the rigid body.
3.3C Addition of Couples Consider two intersecting planes P1 and P2 and two couples acting respec-tively in P1 and P2. Recall that each couple is a free vector in its respective plane and can be represented within this plane by any combination of equal, opposite, and parallel forces and of perpendicular distance of separation that provides the same sense and magnitude for this couple. Thus, we can assume, without any loss of generality, that the couple in P1 consists of two forces F1 and 2F1 perpendicular to the line of intersection of the two planes and acting respectively at A and B (Fig. 3.32a). Similarly, we can assume that the couple in P2 consists of two forces F2 and 2F2 perpendicular to AB and acting respectively at A and B. It is clear that the resultant R of F1 and F2 and the resultant 2R of 2F1 and 2F2 form a couple. Denoting the vector joining B to A by r and recalling the definition of the moment of a couple (Sec. 3.3A), we express the moment M of the resulting couple as M 5 r 3 R 5 r 3 (F1 1 F2) By Varignon’s theorem, we can expand this expression as M 5 r 3 F1 1 r 3 F2 The first term in this expression represents the moment M1 of the couple in P1, and the second term represents the moment M2 of the couple in P2. Therefore, we have M 5 M1 1 M2 (3.48) We conclude that the sum of two couples of moments M1 and M2 is a couple of moment M equal to the vector sum of M1 and M2 (Fig. 3.32b). We can extend this conclusion to state that any number of couples can be added to produce one resultant couple, as M 5 oM 5 o(r 3 F) 3.3D Couple Vectors We have seen that couples with the same moment, whether they act in the same plane or in parallel planes, are equivalent. Therefore, we have no need to draw the actual forces forming a given couple in order to define its effect –F1 –F2 –R F1 F2 M1 M2 P1 P2 (a) (b) r A B O R M Fig. 3.32 (a) We can add two couples, each acting in one of two intersecting planes, to form a new couple. (b) The moment of the resultant couple is the vector sum of the moments of the component couples.
124 Rigid Bodies: Equivalent Systems of Forces y x z –F F (a) (b) (c) (d) d O y x z O y x z O x O M M My Mx Mz (M = Fd) y z = = = Fig. 3.33 (a) A couple formed by two forces can be represented by (b) a couple vector, oriented perpendicular to the plane of the couple. (c) The couple vector is a free vector and can be moved to other points of application, such as the origin. (d) A couple vector can be resolved into components along the coordinate axes.
on a rigid body (Fig. 3.33a). It is sufficient to draw an arrow equal in mag-nitude and direction to the moment M of the couple (Fig. 3.33b). We have also seen that the sum of two couples is itself a couple and that we can obtain the moment M of the resultant couple by forming the vector sum of the moments M1 and M2 of the given couples. Thus, couples obey the law of addition of vectors, so the arrow used in Fig. 3.33b to represent the couple defined in Fig. 3.33a truly can be considered a vector.
The vector representing a couple is called a couple vector. Note that, in Fig. 3.33, we use a red arrow to distinguish the couple vector, which represents the couple itself, from the moment of the couple, which was represented by a green arrow in earlier figures. Also note that we added the symbol l to this red arrow to avoid any confusion with vectors representing forces. A couple vector, like the moment of a couple, is a free vector. Therefore, we can choose its point of application at the origin of the system of coordinates, if so desired (Fig. 3.33c). Furthermore, we can resolve the couple vector M into component vectors Mx, My, and Mz that are directed along the coordinate axes (Fig. 3.33d). These component vectors represent couples acting, respectively, in the yz, zx, and xy planes.
3.3E Resolution of a Given Force into a Force at O and a Couple Consider a force F acting on a rigid body at a point A defined by the position vector r (Fig. 3.34a). Suppose that for some reason it would simplify the analysis to have the force act at point O instead. Although we can move F along its line of action (principle of transmissibility), we cannot move it to a point O that does not lie on the original line of action without modifying the action of F on the rigid body.
–F F (a) (b) (c) = = O MO r F F O r A A F O A Fig. 3.34 Replacing a force with a force and a couple. (a) Initial force F acting at point A; (b) attaching equal and opposite forces at O; (c) force F acting at point O and a couple.
3.3 Couples and Force-Couple Systems 125 We can, however, attach two forces at point O, one equal to F and the other equal to 2F, without modifying the action of the original force on the rigid body (Fig. 3.34b). As a result of this transforma tion, we now have a force F applied at O; the other two forces form a couple of moment MO 5 r 3 F. Thus, Any force F acting on a rigid body can be moved to an arbitrary point O provided that we add a couple whose moment is equal to the moment of F about O. The couple tends to impart to the rigid body the same rotational motion about O that force F tended to produce before it was transferred to O. We represent the couple by a couple vector MO that is perpendicular to the plane containing r and F. Since MO is a free vector, it may be applied anywhere; for convenience, however, the couple vector is usually attached at O together with F. This combination is referred to as a force-couple system (Fig. 3.34c).
O O r A O' s r' F (a) MO r A F (b) (c) MO' O' s r' = = O r A O' s r' F Fig. 3.35 Moving a force to different points. (a) Initial force F acting at A; (b) force F acting at O and a couple; (c) force F acting at O9 and a different couple.
If we move force F from A to a different point O9 (Fig. 3.35a and c), we have to compute the moment MO9 5 r9 3 F of F about O9 and add a new force-couple system consisting of F and the couple vector MO9 at O9. We can obtain the relation between the moments of F about O and O9 as MO9 5 r9 3 F 5 (r 1 s) 3 F 5 r 3 F 1 s 3 F MO9 5 MO 1 s 3 F (3.49) where s is the vector joining O9 to O. Thus, we obtain the moment MO9 of F about O9 by adding to the moment MO of F about O the vector product s 3 F, representing the moment about O9 of the force F applied at O.
We also could have established this result by observing that, in order to transfer to O9 the force-couple system attached at O (Fig. 3.35b and c), we could freely move the couple vector MO to O9. However, to move force F from O to O9, we need to add to F a couple vector whose moment is equal to the moment about O9 of force F applied at O. Thus, the couple vector MO9 must be the sum of MO and the vector s 3 F.
As noted here, the force-couple system obtained by transferring a force F from a point A to a point O consists of F and a couple vector MO perpendicular to F. Conversely, any force-couple system consisting of a force F and a couple vector MO that are mutually perpendicular can be replaced by a single equivalent force. This is done by moving force F in the plane perpendicular to MO until its moment about O is equal to the moment of the couple being replaced.
MO9 5 MO 1 s 3 F Photo 3.2 The force exerted by each hand on the wrench could be replaced with an equivalent force-couple system acting on the nut.
126 Rigid Bodies: Equivalent Systems of Forces Sample Problem 3.6 Determine the components of the single couple equivalent to the two couples shown.
STRATEGY: Look for ways to add equal and opposite forces to the diagram that, along with already known perpendicular distances, will pro-duce new couples with moments along the coordinate axes. These can be combined into a single equivalent couple.
MODELING: You can simplify the computations by attaching two equal and opposite 20-lb forces at A (Fig. 1). This enables you to replace the original 20-lb-force couple by two new 20-lb-force couples: one lying in the zx plane and the other in a plane parallel to the xy plane. ANALYSIS: You can represent these three couples by three couple vec-tors Mx, My, and Mz directed along the coordinate axes (Fig. 2). The corresponding moments are Mx 5 2(30 lb)(18 in.) 5 2540 lb?in.
My 5 1(20 lb)(12 in.) 5 1240 lb?in.
Mz 5 1(20 lb)(9 in.) 5 1180 lb?in.
These three moments represent the components of the single couple M equivalent to the two given couples. You can write M as M 5 2(540 lb?in.)i 1 (240 lb?in.)j 1 (180 lb?in.)k b REFLECT and THINK: You can also obtain the components of the equiva-lent single couple M by computing the sum of the moments of the four given forces about an arbitrary point. Selecting point D, the moment is (Fig. 3) M 5 MD 5 (18 in.)j 3 (230 lb)k 1 [(9 in.)j 2 (12 in.)k] 3 (220 lb)i After computing the various cross products, you get the same result, as M 5 2(540 lb?in.)i 1 (240 lb?in.)j 1 (180 lb?in.)k b y x A B C D E 30 lb 30 lb 12 in.
7 in.
20 lb z 9 in.
9 in.
20 lb Fig. 1 Placing two equal and opposite 20-lb forces at A to simplify calculations.
y x A B C D E 30 lb 30 lb 12 in.
7 in.
20 lb 20 lb 20 lb 20 lb z 9 in.
9 in.
Fig. 2 The three couples represented as couple vectors.
y x z My = +(240 lb•in.)j Mx = –(540 lb•in.)i Mz = +(180 lb•in.)k Fig. 3 Using the given force system, the equivalent single couple can also be determined from the sum of moments of the forces about any point, such as point D.
z y x A B C D E 30 lb 30 lb 12 in.
7 in.
20 lb 20 lb 9 in.
9 in.
3.3 Couples and Force-Couple Systems 127 Sample Problem 3.7 Replace the couple and force shown by an equivalent single force applied to the lever. Determine the distance from the shaft to the point of applica-tion of this equivalent force.
STRATEGY: First replace the given force and couple by an equivalent force-couple system at O. By moving the force of this force-couple system a distance that creates the same moment as the couple, you can then replace the system with one equivalent force.
MODELING and ANALYSIS: To replace the given force and couple, move the force F 5 2(400 N)j to O, and at the same time, add a couple of moment MO that is equal to the moment about O of the force in its original position (Fig. 1). Thus, MO 5 OB 3 F 5 [(0.150 m)i 1 (0.260 m)j] 3 (2400 N)j 5 2(60 N?m)k Fig. 1 Replacing given force and couple with an equivalent force-couple at O.
= B 150 mm O F = – (400 N)j – (400 N)j – (24 N•m)k – (24 N•m)k – (60 N•m)k O 260 mm When you add this new couple to the couple of moment 2(24 N?m)k formed by the two 200-N forces, you obtain a couple of moment 2(84 N?m)k (Fig. 2). You can replace this last couple by applying F at a point C chosen in such a way that 2(84 N ? m)k 5 OC 3 F 5 [(OC) cos 608 i 1 (OC) sin 608 j] 3 (2400 N)j 5 2(OC)cos 608(400 N)k The result is (OC) cos 60° 5 0.210 m 5 210 mm OC 5 420 mm b REFLECT and THINK: Since the effect of a couple does not depend on its location, you can move the couple of moment 2(24 N?m)k to B, obtaining a force-couple system at B (Fig. 3). Now you can eliminate this couple by applying F at a point C chosen in such a way that 2(24 N?m)k 5 BC 3 F 5 2(BC) cos 608(400 N)k The conclusion is (BC) cos 60° 5 0.060 m 5 60 mm BC 5 120 mm OC 5 OB 1 BC 5 300 mm 1 120 mm OC 5 420 mm b B 400 N 200 N 200 N 150 mm 60 mm O 60° 300 mm Fig. 2 Resultant couple eliminated by moving force F.
= C – (400 N)j – (400 N)j – (84 N•m)k O O 60° Fig. 3 Couple can be moved to B with no change in effect. This couple can then be eliminated by moving force F.
= –(400 N)j –(400 N)j –(24 N•m)k –(24 N•m)k B 150 mm O B O = C –(400 N)j –(400 N)j –(24 N•m)k B O O B 60° 128 128 SOLVING PROBLEMS ON YOUR OWN I n this section, we discussed the properties of couples. To solve the following problems, remember that the net effect of a couple is to produce a moment M. Since this moment is independent of the point about which it is computed, M is a free vector and remains unchanged if you move it from point to point. Also, two couples are equivalent (that is, they have the same effect on a given rigid body) if they produce the same moment.
When determining the moment of a couple, all previous techniques for computing moments apply. Also, since the moment of a couple is a free vector, you should compute its value relative to the most convenient point.
Because the only effect of a couple is to produce a moment, it is possible to represent a couple with a vector, called the couple vector, that is equal to the moment of the couple. The couple vector is a free vector and is represented by a special symbol, , to distinguish it from force vectors. In solving the problems in this section, you will be called upon to perform the following operations: 1. Adding two or more couples. This results in a new couple, the moment of which is obtained by adding vectorially the moments of the given couples [Sample Prob. 3.6].
2. Replacing a force with an equivalent force-couple system at a specified point. As explained in Sec. 3.3E, the force of a force-couple system is equal to the original force, whereas the required couple vector is equal to the moment of the original force about the given point. In addition, it is important to note that the force and the couple vector are perpendicular to each other. Conversely, it follows that a force-couple system can be reduced to a single force only if the force and couple vector are mutually perpendicular (see the next paragraph).
3. Replacing a force-couple system (with F perpendicular to M) with a single equivalent force. The requirement that F and M be mutually perpendicular is satis-fied in all two-dimensional problems. The single equivalent force is equal to F and is applied in such a way that its moment about the original point of application is equal to M [Sample Prob. 3.7].
129 Problems 3.70 Two 80-N forces are applied as shown to the corners B and D of a rectangular plate. (a) Determine the moment of the couple formed by the two forces by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples. (b) Use the result obtained to determine the perpendicular distance between lines BE and DF. 3.71 Two parallel 40-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about point A.
A B C 40 N 40 N 20° 55° 270 mm 390 mm Fig. P3.71 3.72 Four 11 2-in.-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indi-cated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that minimum tension? 3.73 Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resul-tant couple applied to the board is 1132.5 lb?in. counterclockwise.
D A F E C B 80 N 80 N 500 mm 300 mm 50° 50° Fig. P3.70 A B C D 60 lb 60 lb 40 lb 40 lb 9 in.
12 in.
Fig. P3.72 and P3.73 130 3.74 A piece of plywood in which several holes are being drilled succes-sively has been secured to a workbench by means of two nails. Knowing that the drill exerts a 12-N∙m couple on the piece of plywood, determine the magnitude of the resulting forces applied to the nails if they are located (a) at A and B, (b) at B and C, (c) at A and C.
A B C 450 mm 240 mm Fig. P3.74 3.75 The two shafts of a speed-reducer unit are subjected to couples of magnitude M1 5 15 lb∙ft and M2 5 3 lb∙ft, respectively. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis. 3.76 If P 5 0 in the figure, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.
x y z B C D A E –P P 16 lb 16 lb 40 lb 40 lb 15 in.
15 in.
10 in.
10 in.
10 in.
Fig. P3.76 and P3.77 3.77 If P 5 20 lb in the figure, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.
3.78 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis. 3.79 Solve Prob. 3.78, assuming that two 10-N vertical forces have been added, one acting upward at C and the other downward at B.
M2 M1 y z x Fig. P3.75 144 mm 160 mm 192 mm 120 mm y x z 120 mm 50 N 50 N 12.5 N 12.5 N A B E C F D Fig. P3.78 131 3.80 Shafts A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts by a single equivalent couple, specifying its magnitude and the direction of its axis.
x z y C B A 20° 20° 900 lb·ft 840 lb·ft 1200 lb·ft Fig. P3.80 3.81 A 500-N force is applied to a bent plate as shown. Determine (a) an equivalent force-couple system at B, (b) an equivalent system formed by a vertical force at A and a force at B.
B A 500 N 30° 300 mm 125 mm 175 mm 75 mm Fig. P3.81 3.82 The tension in the cable attached to the end C of an adjustable boom ABC is 560 lb. Replace the force exerted by the cable at C with an equivalent force-couple system (a) at A, (b) at B.
10 ft 20° 30° A B C T 8 ft Fig. P3.82 132 3.83 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C.
A B C D 60° 6.7 m 4 m Fig. P3.83 3.84 A 30-lb vertical force P is applied at A to the bracket shown, which is held by screws at B and C. (a) Replace P with an equivalent force-couple system at B. (b) Find the two horizontal forces at B and C that are equivalent to the couple obtained in part a. 3.85 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of part a.
A B C D 360 N 0.4 m 0.35 m 2.4 m 0.3 m 40° 30° Fig. P3.85 and P3.86 3.86 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. If two workers attempt to move the same rock by applying a force at A and a parallel force at C, determine these two forces so that they will be equivalent to the single 360-N force shown in the figure. 3.87 The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N hori-zontal forces as shown. Replace this force and couple with a single force F applied at point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.) P 5 in.
2 in.
A B C 3 in.
Fig. P3.84 A D B C E 900 N 250 N 250 N 120 mm 90 mm 90 mm x Fig. P3.87 133 3.88 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at point C, and determine the distance d from C to a line drawn through points D and E. (b) Solve part a if the directions of the two 360-N forces are reversed. 3.89 Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in part a, and specify its point of appli-cation on the lever.
20 lb 20 lb 30º 20º 20º 48 lb A C B 40 in.
30 in.
55º Fig. P3.89 3.90 A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. (a) For α 5 40°, specify the magnitude and line of action of the equivalent force. (b) Specify the value of α if the line of action of the equivalent force is to intersect line CD 300 mm to the right of D. 3.91 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle. 3.92 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.
a A B C D E F 300 N 300 N P 0.2 m Fig. P3.92 450 mm 150 mm 360 N 360 N B d D 600 N E C A y x z Fig. P3.88 a a A B C D 15 N 15 N 48 N 240 mm 400 mm Fig. P3.90 A B C 3.2 in.
2.8 in.
2.9 lb 2.65 lb 25° 25° x y z D Fig. P3.91 134 3.93 Replace the 250-kN force P with an equivalent force-couple system at G. 3.94 A 2.6-kip force is applied at point D of the cast-iron post shown. Replace that force with an equivalent force-couple system at the center A of the base section.
6 in.
5 in.
12 in.
x z y B A D E 2.6 kips Fig. P3.94 3.95 Replace the 150-N force with an equivalent force-couple system at A. 3.96 To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C.
z 990 mm 594 mm 100 mm O A B C y 750 mm 67 mm x 1850 mm Fig. P3.96 P y A G z x 60 mm 30 mm Fig. P3.93 x y z A C 120 mm 40 mm 60 mm 20 mm 35° 150 N D B 200 mm Fig. P3.95 135 3.97 A 46-lb force F and a 2120-lb?in. couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H.
M F 3 in.
x y z A B C D E F H J 14 in.
18 in.
25 in.
45 in.
Fig. P3.97 3.98 A 110-N force acting in a vertical plane parallel to the yz plane is applied to the 220-mm-long horizontal handle AB of a socket wrench. Replace the force with an equivalent force-couple system at the origin O of the coordinate system. 3.99 An antenna is guyed by three cables as shown. Knowing that the tension in cable AB is 288 lb, replace the force exerted at A by cable AB with an equivalent force-couple system at the center O of the base of the antenna.
16 ft x y z O A B C D 128 ft 96 ft 128 ft 64 ft Fig. P3.99 and P3.100 3.100 An antenna is guyed by three cables as shown. Knowing that the tension in cable AD is 270 lb, replace the force exerted at A by cable AD with an equivalent force-couple system at the center O of the base of the antenna.
150 mm 110 N A B x y z O 35° 15° Fig. P3.98 136 Rigid Bodies: Equivalent Systems of Forces 3.4 SIMPLIFYING SYSTEMS OF FORCES We saw in the preceding section that we can replace a force acting on a rigid body with a force-couple system that may be easier to analyze. However, the true value of a force-couple system is that we can use it to replace not just one force but a system of forces to simplify analysis and calculations.
3.4A Reducing a System of Forces to a Force-Couple System Consider a system of forces F1, F2, F3, . . . , acting on a rigid body at the points A1, A2, A3, . . . , defined by the position vectors r1, r2, r3, etc. (Fig. 3.36a). As seen in the preceding section, we can move F1 from A1 to a given point O if we add a couple of moment M1 equal to the moment r1 3 F1 of F1 about O. Repeating this procedure with F2, F3, . . . , we obtain the system shown in Fig. 3.36b, which consists of the original forces, now acting at O, and the added couple vectors. Since the forces are now concurrent, they can be added vectorially and replaced by their resultant R. Similarly, the couple vectors M1, M2, M3, . . . , can be added vectorially and replaced by a single couple vector MR O. Thus, We can reduce any system of forces, however complex, to an equivalent force-couple system acting at a given point O. Note that, although each of the couple vectors M1, M2, M3, . . . in Fig. 3.36b is perpendicular to its corresponding force, the resultant force R and the resultant couple vector MR O shown in Fig. 3.36c are not, in general, perpendicular to each other.
(a) F1 F2 F3 r2 r3 A2 A3 = O r1 A1 (b) F1 F2 M1 M2 M3 = O F3 (c) R MO R O Fig. 3.36 Reducing a system of forces to a force-couple system. (a) Initial system of forces; (b) all the forces moved to act at point O, with couple vectors added; (c) all the forces reduced to a resultant force vector and all the couple vectors reduced to a resultant couple vector.
The equivalent force-couple system is defined by Force-couple system R 5 oF MR O 5 oMO 5 o(r 3 F) (3.50) These equations state that we obtain force R by adding all of the forces of the system, whereas we obtain the moment of the resultant couple vector MR O, called the moment resultant of the system, by adding the moments about O of all the forces of the system.
R 5 oF MR O 5 oMO 5 o(r 3 F) 3.4 Simplifying Systems of Forces 137 Once we have reduced a given system of forces to a force and a couple at a point O, we can replace it with a force and a couple at another point O9. The resultant force R will remain unchanged, whereas the new moment resultant MR O9 will be equal to the sum of MR O and the moment about O9 of force R attached at O (Fig. 3.37). We have MR O9 5 MR O 1 s 3 R (3.51) In practice, the reduction of a given system of forces to a single force R at O and a couple vector MR O is carried out in terms of compo-nents. Resolving each position vector r and each force F of the system into rectangular components, we have r 5 xi 1 yj 1 zk (3.52) F 5 Fxi 1 Fy j 1 Fzk (3.53) Substituting for r and F in Eq. (3.50) and factoring out the unit vectors i, j, and k, we obtain R and MR O in the form R 5 Rxi 1 Ryj 1 Rzk MR O 5 Mx Ri 1 My Rj 1 Mz Rk (3.54) The components Rx, Ry, and Rz represent, respectively, the sums of the x, y, and z components of the given forces and measure the tendency of the system to impart to the rigid body a translation in the x, y, or z direction. Similarly, the components MR x, MR y, and MR z represent, respectively, the sum of the moments of the given forces about the x, y, and z axes and measure the tendency of the system to impart to the rigid body a rotation about the x, y, or z axis.
If we need to know the magnitude and direction of force R, we can obtain them from the components Rx, Ry, and Rz by means of the relations in Eqs. (2.18) and (2.19) of Sec. 2.4A. Similar computations yield the magnitude and direction of the couple vector MR O.
3.4B Equivalent and Equipollent Systems of Forces We have just seen that any system of forces acting on a rigid body can be reduced to a force-couple system at a given point O. This equivalent force-couple system characterizes completely the effect of the given force system on the rigid body. Two systems of forces are equivalent if they can be reduced to the same force-couple system at a given point O. Recall that the force-couple system at O is defined by the relations in Eq. (3.50). Therefore, we can state that Two systems of forces, F1, F2, F3, . . . , and F9 1, F9 2, F9 3, . . . , that act on the same rigid body are equivalent if, and only if, the sums of the forces and the sums of the moments about a given point O of the forces of the two systems are, respectively, equal. Mathematically, the necessary and sufficient conditions for the two sys-tems of forces to be equivalent are Conditions for equivalent systems of forces oF 5 oF9 and oMO 5 oM9 O (3.55) MR O9 5 MR O 1 s 3 R oF 5 oF9 and oMO 5 oM9 O O O' s O O' s R R MO R MO' R = Fig. 3.37 Once a system of forces has been reduced to a force-couple system at one point, we can replace it with an equivalent force-couple system at another point. The force resultant stays the same, but we have to add the moment of the resultant force about the new point to the resultant couple vector.
138 Rigid Bodies: Equivalent Systems of Forces Note that to prove that two systems of forces are equivalent, we must establish the second of the relations in Eq. (3.55) with respect to only one point O. It will hold, however, with respect to any point if the two systems are equivalent.
Resolving the forces and moments in Eqs. (3.55) into their rectan-gular components, we can express the necessary and sufficient conditions for the equivalence of two systems of forces acting on a rigid body as oFx 5 oF9 x oFy 5 oF9 y oFz 5 oF9 z oMx 5 oM9 x oMy 5 oM9 y oMz 5 oM9 z (3.56) These equations have a simple physical significance. They express that Two systems of forces are equivalent if they tend to impart to the rigid body (1) the same translation in the x, y, and z directions, respectively, and (2) the same rotation about the x, y, and z axes, respectively.
In general, when two systems of vectors satisfy Eqs. (3.55) or (3.56), i.e., when their resultants and their moment resultants about an arbitrary point O are respectively equal, the two systems are said to be equipollent. The result just established can thus be restated as If two systems of forces acting on a rigid body are equipollent, they are also equivalent.
It is important to note that this statement does not apply to any system of vectors. Consider, for example, a system of forces acting on a set of independent particles that do not form a rigid body. A different system of forces acting on the same particles may happen to be equipollent to the first one; i.e., it may have the same resultant and the same moment resul-tant. Yet, since different forces now act on the various particles, their effects on these particles are different; the two systems of forces, while equipollent, are not equivalent.
3.4C Further Reduction of a System of Forces We have now seen that any given system of forces acting on a rigid body can be reduced to an equivalent force-couple system at O, consisting of a force R equal to the sum of the forces of the system, and a couple vector MR O of moment equal to the moment resultant of the system.
When R 5 0, the force-couple system reduces to the couple vector MR O. The given system of forces then can be reduced to a single couple called the resultant couple of the system.
What are the conditions under which a given system of forces can be reduced to a single force? It follows from the preceding section that we can replace the force-couple system at O by a single force R acting along a new line of action if R and MR O are mutually perpendicular. The systems of forces that can be reduced to a single force, or resultant, are therefore the systems for which force R and the couple vector MR O are mutually perpendicular. This condition is generally not satisfied by systems of forces in space, but it is satisfied by systems consisting of (1) concurrent forces, (2) coplanar forces, or (3) parallel forces. Let’s look at each case separately. 1. Concurrent forces act at the same point; therefore, we can add them directly to obtain their resultant R. Thus, they always reduce to a single force. Concurrent forces were discussed in detail in Chap. 2.
oFx F 5 oF9 x F x oFy F 5 oF9 y F oFz F 5 oF9 z F oMx M 5 oM9 x M x oMy M 5 oM9 y M oMz M 5 oM9 z M Fpush Force-couple Fpull Photo 3.3 The forces exerted by the children upon the wagon can be replaced with an equivalent force-couple system when analyzing the motion of the wagon.
3.4 Simplifying Systems of Forces 139 2. Coplanar forces act in the same plane, which we assume to be the plane of the figure (Fig. 3.38a). The sum R of the forces of the system also lies in the plane of the figure, whereas the moment of each force about O and thus the moment resultant MR O are perpendicular to that plane. The force-couple system at O consists, therefore, of a force R and a couple vector MR O that are mutually perpendicular (Fig. 3.38b).† We can reduce them to a single force R by moving R in the plane of the figure until its moment about O becomes equal to MR O. The distance from O to the line of action of R is d 5 MR O/R (Fig. 3.38c).
F1 F2 F3 x y O (a) = x y O (b) MO R R = x y O (c) R A d = MO/R R Fig. 3.38 Reducing a system of coplanar forces. (a) Initial system of forces; (b) equivalent force-couple system at O; (c) moving the resultant force to a point A such that the moment of R about O equals the couple vector.
†Because the couple vector MR O is perpendicular to the plane of the figure, we represent it by the symbol l . A counterclockwise couple l represents a vector pointing out of the page and a clockwise couple i represents a vector pointing into the page.
x y O (a) MO R Rx Ry R = x y O (b) Rx Ry R = B x = MO/Ry R x y O (c) Rx Ry R y = –MO/Rx R C Fig. 3.39 Reducing a system of coplanar forces by using rectangular components. (a) From Fig. 3.38(b), resolve the resultant into components along the x and y axes; (b) determining the x intercept of the final line of action of the resultant; (c) determining the y intercept of the final line of action of the resultant. As noted earlier, the reduction of a system of forces is consider-ably simplified if we resolve the forces into rectangular components. The force-couple system at O is then characterized by the components (Fig. 3.39a) Rx 5 oFx Ry 5 oFy Mz R 5 MO R 5 oMO (3.57) 140 Rigid Bodies: Equivalent Systems of Forces To reduce the system to a single force R, the moment of R about O must be equal to MR O. If we denote the coordinates of the point of application of the resultant by x and y and apply equation (3.22) of Sec. 3.1F, we have xRy 2 yRx 5 MR O This represents the equation of the line of action of R. We can also determine the x and y intercepts of the line of action of the resultant directly by noting that MR O must be equal to the moment about O of the y component of R when R is attached at B (Fig. 3.39b) and to the moment of its x component when R is attached at C (Fig. 3.39c). 3. Parallel forces have parallel lines of action and may or may not have the same sense. Assuming here that the forces are parallel to the y axis (Fig. 3.40a), we note that their sum R is also parallel to the y axis. y x z F1 F2 F3 O (a) = y x z O (b) MO R Mz R k Mx Ri R = y x z O (c) r A x z R Fig. 3.40 Reducing a system of parallel forces. (a) Initial system of forces; (b) equivalent force-couple system at O, resolved into components; (c) moving R to point A, chosen so that the moment of R about O equals the resultant moment about O.
On the other hand, since the moment of a given force must be perpen-dicular to that force, the moment about O of each force of the system and thus the moment resultant MR O lie in the zx plane. The force-couple system at O consists, therefore, of a force R and a couple vector MR O that are mutually perpendicular (Fig. 3.40b). We can reduce them to a single force R (Fig. 3.40c) or, if R 5 0, to a single couple of moment MR O.
In practice, the force-couple system at O is characterized by the components Ry 5 oFy MR x 5 oMx MR z 5 oMz (3.58) The reduction of the system to a single force can be carried out by moving R to a new point of application A(x, 0, z), which is chosen so that the moment of R about O is equal to MR O.
r 3 R 5 MR O (xi 1 zk) 3 Ry j 5 Mx Ri 1 Mz Rk Photo 3.4 The parallel wind forces acting on the highway signs can be reduced to a single equivalent force. Determining this force can simplify the calculation of the forces acting on the supports of the frame to which the signs are attached.
3.4 Simplifying Systems of Forces 141 By computing the vector products and equating the coefficients of the corresponding unit vectors in both sides of the equation, we obtain two scalar equations that define the coordinates of A: 2zRy 5 MR x and xRy 5 MR z These equations express the fact that the moments of R about the x and z axes must be equal, respectively, to MR x and MR z.
3.4D Reduction of a System of Forces to a Wrench In the general case of a system of forces in space, the equivalent force-couple system at O consists of a force R and a couple vector MR O that are not perpendicular and where neither is zero (Fig. 3.41a). This system of forces cannot be reduced to a single force or to a single couple. However, we still have a way of simplifying this system further.
The simplification method consists of first replacing the couple vec-tor by two other couple vectors that are obtained by resolving MR O into a component M1 along R and a component M2 in a plane perpendicular to R (Fig. 3.41b). Then we can replace the couple vector M2 and force R by a single force R acting along a new line of action. The original system of forces thus reduces to R and to the couple vector M1 (Fig. 3.41c), i.e., to R and a couple acting in the plane perpendicular to R. (a) O MO R = R M2 (b) O = M1 R (c) O M1 R A Fig. 3.41 Reducing a system of forces to a wrench. (a) General force system reduced to a single force and a couple vector, not perpendicular to each other; (b) resolving the couple vector into components along the line of action of the force and perpendicular to it; (c) moving the force and collinear couple vector (the wrench) to eliminate the couple vector perpendicular to the force.
This particular force-couple system is called a wrench because the resulting combination of push and twist is the same as that caused by an actual wrench. The line of action of R is known as the axis of the wrench, and the ratio p 5 M1/R is called the pitch of the wrench. A wrench there-fore consists of two collinear vectors: a force R and a couple vector M1 5 pR (3.59) M1 5 pR 142 Rigid Bodies: Equivalent Systems of Forces Recall the expression in Eq. (3.33) for the projection of a vector on the line of action of another vector. Using this equation, we note that the projection of MR O on the line of action of R is M1 5 R ? MR O R Thus, we can express the pitch of the wrench as† p 5 M1 R 5 R ? MR O R2 (3.60) To define the axis of the wrench, we can write a relation involving the position vector r of an arbitrary point P located on that axis. We first attach the resultant force R and couple vector M1 at P (Fig. 3.42). Then, since the moment about O of this force-couple system must be equal to the moment resultant MR O of the original force system, we have M1 1 r 3 R 5 MR O (3.61) Alternatively, using Eq. (3.59), we have pR 1 r 3 R 5 MR O (3.62) p 5 M1 R 5 R ? MR O R2 †The expressions obtained for the projection of the couple vector on the line of action of R and for the pitch of the wrench are independent of the choice of point O. Using the relation (3.51) of Sec. 3.4A, we note that if a different point O 9 had been used, the numerator in (3.60) would have been R ? MR O9 5 R ? (MR O 1 s 3 R) 5 R ? MR O 1 R ? (s 3 R) Since the mixed triple product R ? (s 3 R ) is identically equal to zero, we have R ? MR O9 5 R ? MR O Thus, the scalar product R ? MR O is independent of the choice of point O. O MO R R M1 O = R Axis of wrench P r Fig. 3.42 By finding the position vector r that locates any arbitrary point on the axis of the wrench, you can define the axis.
R M1 Photo 3.5 The pushing-turning action associated with the tightening of a screw illustrates the collinear lines of action of the force and couple vector that constitute a wrench.
3.4 Simplifying Systems of Forces 143 Sample Problem 3.8 A 4.80-m-long beam is subjected to the forces shown. Reduce the given system of forces to (a) an equivalent force-couple system at A, (b) an equivalent force-couple system at B, (c) a single force or resultant. Note: Since the reactions at the supports are not included in the given system of forces, the given system will not maintain the beam in equilibrium.
STRATEGY: The force part of an equivalent force-couple system is simply the sum of the forces involved. The couple part is the sum of the moments caused by each force relative to the point of interest. Once you find the equivalent force-couple at one point, you can transfer it to any other point by a moment calculation.
MODELING and ANALYSIS: a. Force-Couple System at A. The force-couple system at A equiv-alent to the given system of forces consists of a force R and a couple MR A defined as (Fig. 1): R 5 oF 5 (150 N)j 2 (600 N)j 1 (100 N)j 2 (250 N)j 5 2(600 N)j MR A 5 o(r 3 F) 5 (1.6i) 3 (2600j) 1 (2.8i) 3 (100j) 1 (4.8i) 3 (2250j) 5 2(1880 N?m)k The equivalent force-couple system at A is thus R 5 600 Nw MR A 5 1880 N?m i b b. Force-Couple System at B. You want to find a force-couple system at B equivalent to the force-couple system at A determined in part a. The force R is unchanged, but you must determine a new couple MR B, the moment of which is equal to the moment about B of the force-couple system determined in part a (Fig. 2). You have MR B 5 MR A 1 BA 3 R 5 2(1880 N?m)k 1(24.8 m)i 3(2600 N)j 5 2(1880 N?m)k 1 (2880 N?m)k 5 1(1000 N?m)k The equivalent force-couple system at B is thus R 5 600 Nw MR B 5 1000 N?m l b c. Single Force or Resultant. The resultant of the given system of forces is equal to R, and its point of application must be such that the moment of R about A is equal to MR A (Fig. 3). This equality of moments leads to r 3 R 5 MR A xi 3 (2600 N)j 5 2(1880 N?m)k 2x(600 N)k 5 2(1880 N?m)k 150 N 600 N 100 N 250 N A B 1.6 m 1.2 m 2 m Fig. 1 Force-couple system at A that is equivalent to given system of forces.
A B 150 j – 600 j 100 j – 250 j 1.6 i 2.8 i 4.8 i A B – (600 N) j – (1880 N•m) k Fig. 2 Finding force-couple system at B equivalent to that determined in part a.
A B – (600 N) j – (1880 N•m) k (2880 N•m) k 4.8 m A – (600 N) j (1000 N•m) k B Fig. 3 Single force that is equivalent to given system of forces.
A B – (600 N) j x (continued) 144 Rigid Bodies: Equivalent Systems of Forces Solving for x, you get x 5 3.13 m. Thus, the single force equivalent to the given system is defined as R 5 600 Nw x 5 3.13 m b REFLECT and THINK: This reduction of a given system of forces to a single equivalent force uses the same principles that you will use later for finding centers of gravity and centers of mass, which are important parameters in engineering mechanics.
Sample Problem 3.9 Four tugboats are bringing an ocean liner to its pier. Each tugboat exerts a 5000-lb force in the direction shown. Determine (a) the equivalent force-couple system at the foremast O, (b) the point on the hull where a single, more powerful tugboat should push to produce the same effect as the original four tugboats.
STRATEGY: The equivalent force-couple system is defined by the sum of the given forces and the sum of the moments of those forces at a par-ticular point. A single tugboat could produce this system by exerting the resultant force at a point of application that produces an equivalent moment.
MODELING and ANALYSIS: a. Force-Couple System at O. Resolve each of the given forces into components, as in Fig. 1 (kip units are used). The force-couple system at O equivalent to the given system of forces consists of a force R and a couple MR O defined as R 5 oF 5 (2.50i 2 4.33j) 1 (3.00i 2 4.00j) 1 (25.00j) 1 (3.54i 1 3.54j) 5 9.04i 2 9.79j MR O 5 o(r 3 F) 5 (290i 1 50j) 3 (2.50i 2 4.33j) 1 (100i 1 70j) 3 (3.00i 2 4.00j) 1 (400i 1 70j) 3 (25.00j) 1 (300i 2 70j) 3 (3.54i 1 3.54j) 5 (390 2 125 2 400 2 210 2 2000 1 1062 1 248)k 5 21035k The equivalent force-couple system at O is thus (Fig. 2) R 5 (9.04 kips)i 2 (9.79 kips)j MR O 5 2(1035 kip?ft)k or R 5 13.33 kips c47.3° MR O 5 1035 kip?ft i b 3 2 3 4 1 4 60° 50 ft 90 ft 110 ft 200 ft O 70 ft 45° 100 ft 100 ft 100 ft Fig. 1 Given forces resolved into components.
– 4.33j – 4 j – 5 j F1 F2 F3 F4 3i 3.54 j 3.54 i 2.5i 50 ft 110 ft 200 ft O 70 ft 90 ft 100 ft 100 ft 100 ft Fig. 2 Equivalent force-couple system at O.
MO = –1035 k R 9.04i –9.79j 47.3° R O (continued) 3.4 Simplifying Systems of Forces 145 Remark: Since all the forces are contained in the plane of the figure, you would expect the sum of their moments to be perpendicular to that plane. Note that you could obtain the moment of each force component directly from the diagram by first forming the product of its magnitude and perpendicular distance to O and then assigning to this product a posi-tive or a negative sign, depending upon the sense of the moment.
b. Single Tugboat. The force exerted by a single tugboat must be equal to R, and its point of application A must be such that the moment of R about O is equal to MR O (Fig. 3). Observing that the position vector of A is r 5 xi 1 70j you have r 3 R 5 MR O (xi 1 70j) 3 (9.04i 2 9.79j) 5 21035k 2x(9.79)k 2 633k 5 21035k x 5 41.1 ft b REFLECT and THINK: Reducing the given situation to that of a single force makes it easier to visualize the overall effect of the tugboats in maneuvering the ocean liner. But in practical terms, having four boats applying force allows for greater control in slowing and turning a large ship in a crowded harbor.
Fig. 3 Point of application of single tugboat to create same effect as given force system.
70 ft x 9.04i – 9.79j R A O Sample Problem 3.10 Three cables are attached to a bracket as shown. Replace the forces exerted by the cables with an equivalent force-couple system at A.
STRATEGY: First determine the relative position vectors drawn from point A to the points of application of the various forces and resolve the forces into rectangular components. Then sum the forces and moments.
MODELING and ANALYSIS: Note that FB 5 (700 N)lBE where lBE 5 BE BE 5 75i 2 150j 1 50k 175 Using meters and newtons, the position and force vectors are rB/A 5 AB 5 0.075i 1 0.050k FB 5 300i 2 600j 1 200k rC/A 5 AC 5 0.075i 2 0.050k FC 5 707i 2 707k rD/A 5 AD 5 0.100i 2 0.100j FD 5 600i 1 1039j The force-couple system at A equivalent to the given forces con-sists of a force R 5 oF and a couple MR A 5 o(r 3 F). Obtain the force R by adding respectively the x, y, and z components of the forces: R 5 oF 5 (1607 N)i 1 (439 N)j 2 (507 N)k b (continued) 50 mm 50 mm 100 mm 100 mm 75 mm 1000 N 1200 N 700 N x y z O A B C D 45º 45º 30º 60º E(150 mm, –50 mm, 100 mm) 146 Rigid Bodies: Equivalent Systems of Forces The computation of MR A is facilitated by expressing the moments of the forces in the form of determinants (Sec. 3.1F). Thus, rByA 3 FB 5 † i j k 0.075 0 0.050 300 2600 200 † 5 30i 245k rCyA 3 FC 5 † i j k 0.075 0 20.050 707 0 2707 † 5 17.68j rDyA 3 FD 5 † i j k 0.100 20.100 0 600 1039 0 † 5 163.9k Adding these expressions, you have MA R 5 o(r 3 F) 5 (30 N?m)i 1 (17.68 N?m)j 1 (118.9 N?m)k b Figure 1 shows the rectangular components of the force R and the couple MR A.
REFLECT and THINK: The determinant approach to calculating moments shows its advantages in a general three-dimensional problem such as this.
x y z O (17.68 N•m)j (439 N)j –(507 N)k (1607 N)i (118.9 N•m)k (30 N•m)i Fig. 1 Rectangular components of equivalent force-couple system at A.
Sample Problem 3.11 A square foundation mat supports the four columns shown. Determine the magnitude and point of application of the resultant of the four loads.
A B C 4 ft 5 ft 5 ft 6 ft 40 kips 20 kips 12 kips x z O 8 kips y STRATEGY: Start by reducing the given system of forces to a force-couple system at the origin O of the coordinate system. Then reduce the system further to a single force applied at a point with coordinates x, z.
MODELING: The force-couple system consists of a force R and a couple vector MR O defined as R 5 oF MR O 5 o(r 3 F) (continued) 3.4 Simplifying Systems of Forces 147 ANALYSIS: After determining the position vectors of the points of application of the various forces, you may find it convenient to arrange the computations in tabular form. The results are shown in Fig. 1.
r, ft F, kips r 3 F, kip?ft 0 240j 0 10i 212j 2 120k 10i 1 5k 28j 40i 2 80k 4i 1 10k 220j 200i 2 80k R 5 280j MR O 5 240i 2 280k The force R and the couple vector MR O are mutually perpendicular, so you can reduce the force-couple system further to a single force R. Select the new point of application of R in the plane of the mat and in such a way that the moment of R about O is equal to MR O. Denote the position vector of the desired point of application by r and its coordinates by x and z (Fig. 2). Then r 3 R 5 MR O (xi 1 zk) 3 (280j) 5 240i 2 280k 280xk 1 80zi 5 240i 2 280k It follows that 280x 5 2280 80z 5 240 x 5 3.50 ft z 5 3.00 ft The resultant of the given system of forces is R 5 80 kipsw at x 5 3.50 ft, z 5 3.00 ft b REFLECT and THINK: The fact that the given forces are all parallel simplifies the calculations, so the final step becomes just a two-dimensional analysis.
Fig. 1 Force-couple system at O that is equivalent to given force system.
x z O –(280 kip•ft)k –(80 kips)j (240 kip•ft)i y Fig. 2 Single force that is equivalent to given force system.
x y z O –(80 kips)j xi zk y x z A B C D E O F1 = Pi F2 = Pj a a a Sample Problem 3.12 Two forces of the same magnitude P act on a cube of side a as shown. Replace the two forces by an equivalent wrench, and determine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz plane.
STRATEGY: The first step is to determine the equivalent force-couple system at the origin O. Then you can reduce this system to a wrench and determine its properties.
(continued) 148 Rigid Bodies: Equivalent Systems of Forces MODELING and ANALYSIS: Equivalent Force-Couple System at O. The position vectors of the points of application E and D of the two given forces are rE 5 ai 1 aj and rD 5 aj 1 ak. The resultant R of the two forces and their moment resultant MR O about O are (Fig. 1) R 5 F1 1 F2 5 Pi 1 Pj 5 P(i 1 j) (1) MR O 5 rE 3 F1 1 rD 3 F2 5 (ai 1 aj) 3 Pi 1 (aj 1 ak) 3 Pj 5 2Pak 2 Pai 5 2Pa(i 1 k) (2) a. Resultant Force R. It follows from Eq. (1) and Fig. 1 that the resul-tant force R has a magnitude of R 5 P22, lies in the xy plane, and forms angles of 45° with the x and y axes. Thus R 5 P12 θx 5 θy 5 458 θz 5 908 b b. Pitch of the Wrench. Using equation (3.60) of Sec. 3.4D and Eqs. (1) and (2) above, the pitch p of the wrench is p 5 R ? MR O R2 5 P(i 1 j) ? (2Pa)(i 1 k) (P22)2 5 2P2a(1 1 0 1 0) 2P2 p 5 2a 2 b c. Axis of the Wrench. From the pitch and from Eq. (3.59), the wrench consists of the force R found in Eq. (1) and the couple vector M1 5 pR 5 2 a 2 P(i 1 j) 5 2 Pa 2 (i 1 j) (3) To find the point where the axis of the wrench intersects the yz plane, set the moment of the wrench about O equal to the moment resultant MR O of the original system: M1 1 r 3 R 5 MR O Alternatively, noting that r 5 yj 1 zk (Fig. 2) and substituting for R, MR O, and M1 from Eqs. (1), (2), and (3), we have 2 Pa 2 (i 1 j) 1 (yj 1 zk) 3 P(i 1 j) 5 2Pa(i 1 k) 2 Pa 2 i 2 Pa 2 j 2 Pyk 1 Pzj 2 Pzi 5 2Pai 2 Pak Equating the coefficients of k and then the coefficients of j, the final result is y 5 a z 5 a/2 b REFLECT and THINK: Conceptually, reducing a system of forces to a wrench is simply an additional application of finding an equivalent force-couple system.
y x z MO R R O Pi Pj – Pai – Pak Fig. 1 Force-couple system at O that is equivalent to the given force system.
y x z R O r M1 = pR yj zk Fig. 2 Wrench that is equivalent to the given force system.
149 149 SOLVING PROBLEMS ON YOUR OWN I n this section you studied the reduction and simplification of force systems. In solving the problems that follow, you will be asked to perform the following operations.
1. Reducing a force system to a force and a couple at a given point A.
The force is the resultant R of the system that is obtained by adding the various forces. The moment of the couple is the moment resultant of the system that is obtained by adding the moments about A of the various forces. We have R 5 oF MR A 5 o(r 3 F) where the position vector r is drawn from A to any point on the line of action of F.
2. Moving a force-couple system from point A to point B. If you wish to reduce a given force system to a force-couple system at point B, you need not recompute the moments of the forces about B after you have reduced it to a force-couple system at point A. The resultant R remains unchanged, and you can obtain the new moment resultant MR B by adding the moment about B of the force R applied at A to MR A [Sample Prob. 3.8]. Denoting the vector drawn from B to A as s, you have MR B 5 MR A 1 s 3 R 3. Checking whether two force systems are equivalent. First reduce each force system to a force-couple system at the same, but arbitrary, point A (as explained in the first operation). The two force systems are equivalent (that is, they have the same effect on the given rigid body) if the two reduced force-couple systems are identical; that is, if oF 5 oF9 and oMA 5 oM9A You should recognize that if the first of these equations is not satisfied––that is, if the two systems do not have the same resultant R––the two systems cannot be equivalent, and there is no need to check whether or not the second equation is satisfied.
4. Reducing a given force system to a single force. First reduce the given system to a force-couple system consisting of the resultant R and the couple vector MR A at (continued) 150 some convenient point A (as explained in the first operation). Recall from Section 3.4 that further reduction to a single force is possible only if the force R and the couple vector MR A are mutually perpendicular. This will certainly be the case for systems of forces that are either concurrent, coplanar, or parallel. You can then obtain the required single force by moving R until its moment about A is equal to MR A, as you did in several problems in Section 3.4. More formally, the position vector r drawn from A to any point on the line of action of the single force R must satisfy the equation r 3 R 5 MR A This procedure was illustrated in Sample Probs. 3.8, 3.9, and 3.11.
5. Reducing a given force system to a wrench. If the given system includes forces that are not concurrent, coplanar, or parallel, the equivalent force-couple system at a point A will consist of a force R and a couple vector MR A that, in general, are not mutually perpendicular. (To check whether R and MR A are mutually perpendicular, form their scalar product. If this product is zero, they are mutually perpendicular; otherwise, they are not.) If R and MR A are not mutually perpendicular, the force-couple system (and thus the given system of forces) cannot be reduced to a single force. However, the system can be reduced to a wrench—the combination of a force R and a couple vector M1 directed along a common line of action called the axis of the wrench (Fig. 3.42). The ratio p 5 M1/R is called the pitch of the wrench.
To reduce a given force system to a wrench, you should follow these steps, a. Reduce the given system to an equivalent force-couple system (R, MR O), typically located at the origin O. b. Determine the pitch p from Eq. (3.60), p 5 M1 R 5 R ? MR O R2 and the couple vector from M1 5 pR. c. Set the moment about O of the wrench equal to the moment resultant MR O of the force-couple system at O: M1 1 r 3 R 5 MR O (3.61) This equation allows you to determine the point where the line of action of the wrench intersects a specified plane, since the position vector r is directed from O to that point. These steps are illustrated in Sample Prob. 3.12. Although determining a wrench and the point where its axis intersects a plane may appear difficult, the process is simply the application of several of the ideas and techniques developed in this chapter. Once you have mastered the wrench, you can feel confident that you understand much of Chap. 3.
151 Problems 3.101 A 3-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent force-couple system at end A of the beam. (b) Which of the loadings are equivalent? 3.102 A 3-m-long beam is loaded as shown. Determine the loading of Prob. 3.101 that is equivalent to this loading.
3.103 Determine the single equivalent force and the distance from point A to its line of action for the beam and loading of (a) Prob. 3.101a, (b) Prob. 3.101b, (c) Prob. 3.102. 3.104 Five separate force-couple systems act at the corners of a piece of sheet metal that has been bent into the shape shown. Determine which of these systems is equivalent to a force F 5 (10 lb)i and a couple of moment M 5 (15 lb?ft)j 1 (15 lb?ft)k located at the origin.
5 lb•ft 5 lb•ft 15 lb•ft 5 lb•ft 15 lb•ft 15 lb•ft 15 lb•ft 15 lb•ft 80 lb•ft 25 lb•ft 10 lb 10 lb 10 lb 10 lb 10 lb y z O H A C J I B D G x F E 2 ft 2 ft 2 ft 1 ft 2.5 ft Fig. P3.104 A B 200 N 400 N•m 3 m 300 N 300 N 400 N•m 200 N (a) 300 N 400 N•m 200 N (c) 500 N 400 N•m (d) (b) 800 N 400 N•m 1000 N•m 1000 N•m 300 N 200 N 400 N•m 300 N (e) 300 N 400 N•m 800 N (g) 250 N 400 N•m (h) ( f ) 1000 N•m 250 N Fig. P3.101 A B 300 N 500 N•m 200 N•m 3 m 200 N Fig. P3.102 152 3.105 The weights of two children sitting at ends A and B of a seesaw are 84 lb and 64 lb, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 60 lb, (b) 52 lb? 3.106 Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 4.1 lb, while the one at C weighs 3.5 lb. (a) If d 5 25 in., determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe.
D B C E d 34 in.
10 in.
84 in.
A Fig. P3.106 3.107 A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position. If b 5 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the equivalent force and its point of application on the beam.
b a A B 1300 N 400 N 600 N a 2 a b 400 N 9 m Fig. P3.107 3.108 A 6 3 12-in. plate is subjected to four loads as shown. Find the resultant of the four loads and the two points at which the line of action of the resultant intersects the edge of the plate.
A B C 6 ft 6 ft Fig. P3.105 A B C F E D 6 in.
6 in.
6 in.
50 lb 50 lb 20 lb 40 lb Fig. P3.108 153 3.109 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor. 3.110 To test the strength of a 625 3 500-mm suitcase, forces are applied as shown. If P 5 88 N, (a) determine the resultant of the applied forces, (b) locate the two points where the line of action of the resultant intersects the edge of the suitcase.
450 mm 100 mm 80 mm 280 mm 180 N 212 N 100 N P D C A B Fig. P3.110 3.111 Solve Prob. 3.110, assuming that P 5 138 N. 3.112 Pulleys A and B are mounted on bracket CDEF. The tension on each side of the two belts is as shown. Replace the four forces with a single equivalent force, and determine where its line of action inter-sects the bottom edge of the bracket. 3.113 A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line drawn through points A and G.
C A B D F E G 240 lb 160 lb 300 lb 40° 180 lb 70° x y 4 ft 8 ft 8 ft 8 ft 8 ft 8 ft 6 ft Fig. P3.113 3.114 A couple of magnitude M 5 80 lb?in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resul-tant intersects line AB and line BC. 3.115 A couple M and the three forces shown are applied to an angle bracket. Find the moment of the couple if the line of action of the resultant of the force system is to pass through (a) point A, (b) point B, (c) point C.
140 lb 30° 60 lb O W 2 in.
2 in.
Fig. P3.109 2 in.
1 in.
210 lb 150 lb 25° 25° 4 in.
120 lb 160 lb A B C D E F r = 2 in.
r = 1 in.
1 2 6 in.
6 in.
Fig. P3.112 A B C 10 lb 25 lb 60° 12 in.
40 lb M 8 in.
Fig. P3.114 and P3.115 154 3.116 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P 5 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH. 3.117 Solve Prob. 3.116, assuming that P 5 60 N. 3.118 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at the point D obtained by drawing the perpendicular from the point of contact to the x axis. (b) For a 5 1 m and b 5 2 m, determine the value of x for which the moment of the equivalent force-couple system at D is maximum.
y b a C B A F D x y = b(1 – ) x2 a2 Fig. P3.118 3.119 A machine component is subjected to the forces shown, each of which is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at A. 3.120 Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz plane. Replace the belt forces shown with an equivalent force-couple system at A.
A B C D x y z 20º 10º 10º 155 N 240 N 145 N 215 N 180 mm 225 mm 225 mm Fig. P3.120 C A B D F E G H P 200 N 240 mm 120 N 70° 15° 50 mm 50 mm 50 mm 80 N 42 N•m 40 N•m 180 mm 640 mm 520 mm Fig. P3.116 Fig. P3.119 30 mm 90 mm 50 mm 60 mm 75 mm x x y C D O B A 150 N 300 N 125 N 240 N 155 3.121 As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A if R 5 21.2 lb and M 5 13.25 lb?ft.
A B R M C x y z 64 in.
96 in.
42 in.
48 in.
Fig. P3.121 3.122 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench at a distance of 10 in. from the axis of the pipe and in a direction per-pendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus he avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equiva-lent to the two 40-lb forces when this condition is satisfied.
40 lb 40 lb y x z 25 in.
18 in.
7.5 in.
10 in.
10 in.
A B C D E F 7.5 in.
Fig. P3.122 3.123 Assuming θ 5 60° in Prob. 3.122, replace the two 40-lb forces with an equivalent force-couple system at D and determine whether the plumber’s action tends to tighten or loosen the joint between (a) pipe CD and elbow D, (b) elbow D and pipe DE. Assume all threads to be right-handed.
156 3.124 Four forces are applied to the machine component ABDE as shown. Replace these forces with an equivalent force-couple system at A.
z 200 mm 40 mm 160 mm 100 mm 20 mm x 50 N 250 N 120 N 300 N y B E D A Fig. P3.124 3.125 A blade held in a brace is used to tighten a screw at A. (a) Determine the forces exerted at B and C, knowing that these forces are equiva-lent to a force-couple system at A consisting of R 5 2(25 N)i 1 Ry j 1 Rzk and MR A 5 2(13.5 N?m)i. (b) Find the corresponding values of Ry and Rz. (c) What is the orientation of the slot in the head of the screw for which the blade is least likely to slip when the brace is in the position shown?
x B Czk Cxi Cyj –B k 240 mm 240 mm 180 mm y z A C Fig. P3.125 3.126 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at points A and B and applies forces at these points. Knowing that these forces are equiva-lent to a force-couple system at C consisting of the force C 5 2(8 lb)i 1 (4 lb)k and the couple MC 5 (360 lb?in.)i, determine the forces applied at A and at B when Az 5 2 lb.
2 in.
8 in.
10 in.
Ax Ay Az Bx By Bz A B x y C z Fig. P3.126 157 3.127 Three children are standing on a 5 3 5-m raft. If the weights of the children at points A, B, and C are 375 N, 260 N, and 400 N, respec-tively, determine the magnitude and the point of application of the resultant of the three weights. 3.128 Three children are standing on a 5 3 5-m raft. The weights of the children at points A, B, and C are 375 N, 260 N, and 400 N, respec-tively. If a fourth child weighing 425 N climbs onto the raft, deter-mine where she should stand if the other children remain in the positions shown and if the line of action of the resultant of the four weights is to pass through the center of the raft. 3.129 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine the magnitude and the point of application of the resultant of the four wind forces when a 5 1 ft and b 5 12 ft.
D A B C x y z E F G H a 2.5 ft 90 lb 160 lb 50 lb 105 lb 9 ft 5.5 ft b 5 ft 8 ft 3 ft Fig. P3.129 and P3.130 3.130 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine a and b so that the point of application of the resultant of the four forces is at G.
3.131 A concrete foundation mat of 5-m radius supports four equally spaced columns, each of which is located 4 m from the center of the mat. Determine the magnitude and the point of application of the resultant of the four loads.
3.132 Determine the magnitude and the point of application of the smallest additional load that must be applied to the foundation mat of Prob. 3.131 if the resultant of the five loads is to pass through the center of the mat.
A B C x y z E F G O 0.5 m 0.25 m 0.25 m 1.5 m 1 m 2 m Fig. P3.127 and P3.128 z x y 125 kN 25 kN 75 kN 100 kN 5 m O Fig. P3.131 158 3.133 Three forces of the same magnitude P act on a cube of side a as shown. Replace the three forces with an equivalent wrench and deter-mine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench. 3.134 A piece of sheet metal is bent into the shape shown and is acted upon by three forces. If the forces have the same magnitude P, replace them with an equivalent wrench and determine (a) the mag-nitude and the direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench.
A B C D E F G H O x y z F1 F2 F3 a a a 3 2 a a Fig. P3.134 3.135 and 3.136 The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.
11 lb 6 lb•in.
6 lb•in.
10 lb 15 in.
x z y A B O 20 in.
Fig. P3.136 y x z A B C D O F1 F2 F3 a a a Fig. P3.133 1 N•m 4 N•m 15 N 20 N 100 mm x z y A O Fig. P3.135 159 3.137 and 3.138 Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz plane.
0.1 m 0.3 m 0.6 m x z y A B 0.4 m 30 N•m 84 N 32 N•m 80 N Fig. P3.137 3.139 Two ropes attached at A and B are used to move the trunk of a fallen tree. Replace the forces exerted by the ropes with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz plane. 3.140 A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane.
A B x y z C D E O 4a 20a 12a 18a 15a 9a Fig. P3.140 10 in.
10 in.
10 in.
30 in.
16 in.
x z y A B 238 lb•in.
17 lb 26.4 lb 220 lb•in.
Fig. P3.138 5 m 2 m 2 m 6 m x y z O A B C D 1650 N 1500 N 14 m 9 m 12 m 9 m Fig. P3.139 160 3.141 and 3.142 Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz plane. If it cannot be reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz plane.
x y z A B C D E F G H I 160 lb•in.
34 lb 30 lb K 12 in.
6 in.
6 in.
6 in.
3 in.
8 in.
18 in.
18 in.
Fig. P3.142 3.143 Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y axis and applied respectively at A and B.
x y z B A O M R b a Fig. P3.143 3.144 Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane. 3.145 Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point. 3.146 Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.
160 mm 120 mm 120 mm 60 mm 60 mm 40 mm 40 mm 50 N 70 N B C H D I y A E z F G x 10 N•m 14 N•m Fig. P3.141 161 Review and Summary Principle of Transmissibility In this chapter, we presented the effects of forces exerted on a rigid body. We began by distinguishing between external and internal forces [Sec. 3.1A]. We then explained that, according to the principle of transmissibility, the effect of an external force on a rigid body remains unchanged if we move that force along its line of action [Sec. 3.1B]. In other words, two forces F and F9 acting on a rigid body at two different points have the same effect on that body if they have the same magnitude, same direction, and same line of action (Fig. 3.43). Two such forces are said to be equivalent.
Vector Product Before proceeding with the discussion of equivalent systems of forces, we introduced the concept of the vector product of two vectors [Sec. 3.1C]. We defined the vector product V 5 P 3 Q of the vectors P and Q as a vector perpendicular to the plane containing P and Q (Fig. 3.44) with a magnitude of V 5 PQ sin θ (3.1) and directed in such a way that a person located at the tip of V will observe the rotation to be counterclockwise through θ, bringing the vector P in line with the vector Q. The three vectors P, Q, and V—taken in that order—are said to form a right-handed triad. It follows that the vector products Q 3 P and P 3 Q are represented by equal and opposite vectors: Q 3 P 5 2(P 3 Q) (3.4) It also follows from the definition of the vector product of two vectors that the vector products of the unit vectors i, j, and k are i 3 i 5 0 i 3 j 5 k j 3 i 5 2k and so on. You can determine the sign of the vector product of two unit vec-tors by arranging in a circle and in counterclockwise order the three letters representing the unit vectors (Fig. 3.45): The vector product of two unit vec-tors is positive if they follow each other in counterclockwise order and nega-tive if they follow each other in clockwise order.
Rectangular Components of Vector Product The rectangular components of the vector product V of two vectors P and Q are expressed [Sec. 3.1D] as Vx 5 PyQz 2 PzQy Vy 5 PzQx 2 PxQz (3.9) Vz 5 PxQy 2 PyQx F F' = Fig. 3.43 Q P V = P × Q q (a) V (b) Fig. 3.44 i j k Fig. 3.45 162 We can also express the components of a vector product as a determinant: V 5 † i j k Px Py Pz Qx Qy Qz † (3.10) Moment of a Force about a Point We defined the moment of a force F about a point O [Sec. 3.1E] as the vector product MO 5 r 3 F (3.11) where r is the position vector drawn from O to the point of application A of the force F (Fig. 3.46). Denoting the angle between the lines of action of r and F as θ, we found that the magnitude of the moment of F about O is MO 5 rF sin θ 5 Fd (3.12) where d represents the perpendicular distance from O to the line of action of F.
Rectangular Components of Moment The rectangular components of the moment MO of a force F [Sec. 3.1F] are Mx 5 yFz 2 zFy My 5 zFx 2 xFz (3.18) Mz 5 xFy 2 yFx where x, y, and z are the components of the position vector r (Fig. 3.47). Using a determinant form, we also wrote MO 5 † i j k x y z Fx Fy Fz † (3.19) In the more general case of the moment about an arbitrary point B of a force F applied at A, we had MB 5 † i j k xA/B yA/B zA/B Fx Fy Fz † (3.21) where xA/B, yA/B, and zA/B denote the components of the vector rA/B: xA/B 5 xA 2 xB yA/B 5 yA 2 yB zA/B 5 zA 2 zB In the case of problems involving only two dimensions, we can assume the force F lies in the xy plane. Its moment MB about a point B in the same plane is perpendicular to that plane (Fig. 3.48) and is completely defined by the scalar MB 5 (xA 2 xB)Fy 2 (yA 2 yB)Fx (3.23) Various methods for computing the moment of a force about a point were illustrated in Sample Probs. 3.1 through 3.4.
Scalar Product of Two Vectors The scalar product of two vectors P and Q [Sec. 3.2A], denoted by P ? Q, is defined as the scalar quantity P ? Q 5 PQ cos θ (3.24) MO d A F r θ O Fig. 3.46 Fy j Fx i Fz k x y z O zk yj xi r A (x, y, z) Fig. 3.47 y x z O B Fy j Fx i F A (yA – yB)j (xA – xB)i rA/B MB = MB k Fig. 3.48 163 where θ is the angle between P and Q (Fig. 3.49). By expressing the scalar product of P and Q in terms of the rectangular components of the two vectors, we determined that P ? Q 5 PxQx 1 PyQy 1 PzQz (3.28) Projection of a Vector on an Axis We obtain the projection of a vector P on an axis OL (Fig. 3.50) by forming the scalar product of P and the unit vector l along OL. We have POL 5 P ? l (3.34) Using rectangular components, this becomes POL 5 Px cos θx 1 Py cos θy 1 Pz cos θz (3.35) where θx, θy, and θz denote the angles that the axis OL forms with the coor-dinate axes.
Mixed Triple Product of Three Vectors We defined the mixed triple product of the three vectors S, P, and Q as the scalar expression S ? (P 3 Q) (3.36) obtained by forming the scalar product of S with the vector product of P and Q [Sec. 3.2B]. We showed that S ? (P 3 Q) 5† Sx Sy Sz Px Py Pz Qx Qy Qz † (3.39) where the elements of the determinant are the rectangular components of the three vectors.
Moment of a Force about an Axis We defined the moment of a force F about an axis OL [Sec. 3.2C] as the projection OC on OL of the moment MO of the force F (Fig. 3.51), i.e., as the mixed triple product of the unit vector l, the position vector r, and the force F: MOL 5 l ? MO 5 l ? (r 3 F) (3.40) y x z r L A C O MO F Fig. 3.51 Q P q Fig. 3.49 y x z O A P L qx qy qz Fig. 3.50 164 Force-Couple System Any force F acting at a point A of a rigid body can be replaced by a force-couple system at an arbitrary point O consisting of the force F applied at O The determinant form for the mixed triple product is MOL 5 † lx ly lz x y z Fx Fy Fz † (3.41) where lx, ly, lz 5 direction cosines of axis OL x, y, z 5 components of r Fx, Fy, Fz 5 components of F An example of determining the moment of a force about a skew axis appears in Sample Prob. 3.5.
Couples Two forces F and 2F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple [Sec. 3.3A]. The moment of a couple is independent of the point about which it is computed; it is a vector M perpendicular to the plane of the couple and equal in magnitude to the product of the common magnitude F of the forces and the perpendicular distance d between their lines of action (Fig. 3.52).
–F F d M Fig. 3.52 Two couples having the same moment M are equivalent, i.e., they have the same effect on a given rigid body [Sec. 3.3B]. The sum of two couples is itself a couple [Sec. 3.3C], and we can obtain the moment M of the resultant couple by adding vectorially the moments M1 and M2 of the original couples [Sample Prob. 3.6]. It follows that we can represent a couple by a vector, called a couple vector, equal in magnitude and direction to the moment M of the couple [Sec. 3.3D]. A couple vector is a free vector that can be attached to the origin O if so desired and resolved into components (Fig. 3.53).
y x z –F F (a) d O = (b) y x z O M (M = Fd) = (c) y x z O M = (d) x O My Mx Mz y z Fig. 3.53 165 and a couple of moment MO, which is equal to the moment about O of the force F in its original position [Sec. 3.3E]. Note that the force F and the couple vector MO are always perpendicular to each other (Fig. 3.54).
O MO r A A F F O = Fig. 3.54 Reduction of a System of Forces to a Force-Couple System It follows [Sec. 3.4A] that any system of forces can be reduced to a force-couple system at a given point O by first replacing each of the forces of the system by an equivalent force-couple system at O (Fig. 3.55) and then adding all of the forces and all of the couples to obtain a resultant force R and a resultant couple vector MR O [Sample Probs. 3.8 through 3.11]. In general, the resultant R and the couple vector MR O will not be perpendicular to each other.
(a) F1 F2 F3 r2 r3 A2 A3 = O r1 A1 (b) F1 F2 M1 M2 M3 = O F3 (c) R MO R O Fig. 3.55 Equivalent Systems of Forces We concluded [Sec. 3.4B] that, as far as rigid bodies are concerned, two systems of forces, F1, F2, F3, . . . and F9 1, F9 2, F9 3, . . . , are equivalent if, and only if, oF 5 oF9 and oMO 5 oM9 O (3.55) Further Reduction of a System of Forces If the resultant force R and the resultant couple vector MR O are perpendicular to each other, we can further reduce the force-couple system at O to a single resultant force [Sec. 3.4C]. This is the case for systems consisting of (a) concurrent forces (cf. Chap. 2), (b) coplanar forces [Sample Probs. 3.8 and 3.9], or (c) parallel forces [Sample Prob. 3.11]. If the resultant R and the couple vector MR O are not perpendicular to each other, the system cannot be reduced to a single force. We can, however, reduce it to a special type of force-couple system called a wrench, consisting of the resultant R and a cou-ple vector M1 directed along R [Sec. 3.4D and Sample Prob. 3.12].
166 Review Problems 3.147 A 300-N force P is applied at point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part a, determine the perpendicular distance from O to the line of action of P.
3.148 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolv-ing that force into horizontal and vertical components applied (a) at point C, (b) at point E.
A B C D E d 0.875 m 0.2 m Fig. P3.148 3.149 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A. 3.150 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC.
x y z A B C D 2 ft 1 ft 8 ft 6.5 ft 4 ft 6 ft Fig. P3.150 A P 30° B O 40° 120 mm 48° 200 mm Fig. P3.147 3 ft x y z A C D 7.75 ft 6 ft B Fig. P3.149 167 3.151 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Determine the moment Mx of P about the x axis when θ 5 65°, knowing that My 5 215 N?m and Mz 5 236 N?m.
3.152 A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z axis of the resul-tant force RA exerted on the davit at A must not exceed 279 lb?ft in absolute value. Determine the largest allowable tension in line ABAD when x 5 6 ft.
3 ft x y z A C D 7.75 ft x B Fig. P3.152 3.153 In a manufacturing operation, three holes are drilled simultaneously in a workpiece. If the holes are perpendicular to the surfaces of the workpiece, replace the couples applied to the drills with a single equivalent couple, specifying its magnitude and the direction of its axis.
x y z 20° 25° 1.75 N •m 1.5 N •m 1.5 N •m Fig. P3.153 3.154 A 260-lb force is applied at A to the rolled-steel section shown. Replace that force with an equivalent force-couple system at the center C of the section. 3.155 The force and couple shown are to be replaced by an equivalent single force. Knowing that P 5 2Q, determine the required value of α if the line of action of the single equivalent force is to pass through (a) point A, (b) point C.
O y z 150 mm C B A f q x P 200 mm 100 mm Fig. P3.151 A B D C a a –Q Q P Fig. P3.155 A B C 260 lb 2 in.
2.5 in.
4 in.
4 in.
Fig. P3.154 168 3.156 A 77-N force F1 and a 31-N?m couple M1 are applied to corner E of the bent plate shown. If F1 and M1 are to be replaced with an equiva-lent force-couple system (F2, M2) at corner B and if (M2)z 5 0, determine (a) the distance d, (b) F2 and M2.
x z y B A C E D G H J F1 70 mm 30 mm 30 mm d 60 mm 83.3 mm 250 mm M1 Fig. P3.156 3.157 Three horizontal forces are applied as shown to a vertical cast-iron arm. Determine the resultant of the forces and the distance from the ground to its line of action when (a) P 5 200 N, (b) P 5 2400 N, (c) P 5 1000 N. 3.158 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C know-ing that these forces and the couple are equivalent to a force-couple system at A consisting of the force R 5 (2.6 lb)i 1 Ry j 2 (0.7 lb)k and the couple MR A 5 Mxi 1 (1.0 lb?ft)j 2 (0.72 lb?ft)k. (b) Find the corresponding values of Ry and Mx.
B B 1 lb•ft 3.5 in.
1.75 in.
2 in.
x y z A C Cxi –Czk –Cyj Fig. P3.158 600 N 400 N A B C D P 150 mm 150 mm 150 mm Fig. P3.157 The Tianjin Eye is a Ferris wheel that straddles a bridge over the Hai River in China. The structure is designed so that the support reactions at the wheel bearings as well as those at the base of the frame maintain equilibrium under the effects of vertical gravity and horizontal wind forces.
Equilibrium of Rigid Bodies 4 170 Equilibrium of Rigid Bodies Introduction We saw in Chapter 3 how to reduce the external forces acting on a rigid body to a force-couple system at some arbitrary point O. When the force and the couple are both equal to zero, the external forces form a system equivalent to zero, and the rigid body is said to be in equilibrium. We can obtain the necessary and sufficient conditions for the equi-librium of a rigid body by setting R and MR O equal to zero in the relations of Eq. (3.50) of Sec. 3.4A: oF 5 0 oMO 5 o (r 3 F) 5 0 (4.1) Resolving each force and each moment into its rectangular components, we can replace these vector equations for the equilibrium of a rigid body with the following six scalar equations: oFx 5 0 oFy 5 0 oFz 5 0 (4.2) oMx 5 0 oMy 5 0 oMz 5 0 (4.3) We can use these equations to determine unknown forces applied to the rigid body or unknown reactions exerted on it by its supports. Note that Eqs. (4.2) express the fact that the components of the external forces in the x, y, and z directions are balanced; Eqs. (4.3) express the fact that the moments of the external forces about the x, y, and z axes are balanced. Therefore, for a rigid body in equilibrium, the system of external forces imparts no translational or rotational motion to the body.
In order to write the equations of equilibrium for a rigid body, we must first identify all of the forces acting on that body and then draw the corresponding free-body diagram. In this chapter, we first consider the equilibrium of two-dimensional structures subjected to forces contained in their planes and study how to draw their free-body diagrams. In addition to the forces applied to a structure, we must also consider the reactions exerted on the structure by its supports. A specific reaction is associated with each type of support. You will see how to determine whether the structure is properly supported, so that you can know in advance whether you can solve the equations of equilibrium for the unknown forces and reactions.
oF 5 0 oMO 5 o (r 3 F) 5 0 oFx F 5 0 oFy F 5 0 oFz F 5 0 oMx M 5 0 oMy M 5 0 oMz M 5 0 Introduction Free-Body Diagrams 4.1 EQUILIBRIUM IN TWO DIMENSIONS 4.1A Reactions for a Two-Dimensional Structure 4.1B Rigid-Body Equilibrium in Two Dimensions 4.1C Statically Indeterminate Reactions and Partial Constraints 4.2 TWO SPECIAL CASES 4.2A Equilibrium of a Two-Force Body 4.2B Equilibrium of a Three-Force Body 4.3 EQUILIBRIUM IN THREE DIMENSIONS 4.3A Rigid-Body Equilibrium in Three Dimensions 4.3B Reactions for a Three-Dimensional Structure Objectives • Analyze the static equilibrium of rigid bodies in two and three dimensions.
• Consider the attributes of a properly drawn free-body diagram, an essential tool for the equilibrium analysis of rigid bodies.
• Examine rigid bodies supported by statically indeter-minate reactions and partial constraints.
• Study two cases of particular interest: the equilibrium of two-force and three-force bodies.
Later in this chapter, we consider the equilibrium of three-dimensional structures, and we provide the same kind of analysis to these structures and their supports.
Free-Body Diagrams In solving a problem concerning a rigid body in equilibrium, it is essential to consider all of the forces acting on the body. It is equally important to exclude any force that is not directly applied to the body. Omitting a force or adding an extraneous one would destroy the conditions of equilibrium. Therefore, the first step in solving the problem is to draw a free-body diagram of the rigid body under consideration. We have already used free-body diagrams on many occasions in Chap. 2. However, in view of their importance to the solution of equilib-rium problems, we summarize here the steps you must follow in drawing a correct free-body diagram. 1. Start with a clear decision regarding the choice of the free body to be analyzed. Mentally, you need to detach this body from the ground and separate it from all other bodies. Then you can sketch the contour of this isolated body. 2. Indicate all external forces on the free-body diagram. These forces rep-resent the actions exerted on the free body by the ground and by the bodies that have been detached. In the diagram, apply these forces at the various points where the free body was supported by the ground or Photo 4.1 A tractor supporting a bucket load. As shown, its free-body diagram should include all external forces acting on the tractor.
Load Boom weight Axes Body y Body weight Reactions Front wheel reaction Rear wheel reaction, vertical Rear wheel reaction, horizontal x Bucket load Tractor weight Photo 4.2 Tractor bucket and boom. In Chap. 6, we will see how to determine the internal forces associated with interconnected members such as these using free-body diagrams like the one shown.
Load Boom weight Piston reaction Body Body weight Reactions Bucket load Boom reaction, vertical Boom reaction, horizontal Free-Body Diagrams 171 172 Equilibrium of Rigid Bodies was connected to the other bodies. Generally, you should include the weight of the free body among the external forces, since it represents the attraction exerted by the earth on the various particles forming the free body. You will see in Chapter 5 that you should draw the weight so it acts at the center of gravity of the body. If the free body is made of several parts, do not include the forces the various parts exert on each other among the external forces. These forces are internal forces as far as the free body is concerned. 3. Clearly mark the magnitudes and directions of the known external forces on the free-body diagram. Recall that when indicating the directions of these forces, the forces are those exerted on, and not by, the free body. Known external forces generally include the weight of the free body and forces applied for a given purpose. 4. Unknown external forces usually consist of the reactions through which the ground and other bodies oppose a possible motion of the free body. The reactions constrain the free body to remain in the same position; for that reason, they are sometimes called constraining forces. Reactions are exerted at the points where the free body is supported by or con-nected to other bodies; you should clearly indicate these points. Reac-tions are discussed in detail in Secs. 4.1 and 4.3. 5. The free-body diagram should also include dimensions, since these may be needed for computing moments of forces. Any other detail, however, should be omitted.
4.1 EQUILIBRIUM IN TWO DIMENSIONS In the first part of this chapter, we consider the equilibrium of two-dimensional structures; i.e., we assume that the structure being analyzed and the forces applied to it are contained in the same plane. Clearly, the reactions needed to maintain the structure in the same position are also contained in this plane.
4.1A Reactions for a Two-Dimensional Structure The reactions exerted on a two-dimensional structure fall into three cat-egories that correspond to three types of supports or connections. 1. Reactions Equivalent to a Force with a Known Line of Action. Sup-ports and connections causing reactions of this type include rollers, rockers, frictionless surfaces, short links and cables, collars on friction-less rods, and frictionless pins in slots. Each of these supports and connections can prevent motion in one direction only. Figure 4.1 shows these supports and connections together with the reactions they produce. Each reaction involves one unknown––specifically, the magnitude of the reaction. In problem solving, you should denote this magnitude by an appropriate letter. The line of action of the reaction is known and should be indicated clearly in the free-body diagram. The sense of the reaction must be as shown in Fig. 4.1 for cases of a frictionless surface (toward the free body) or a cable (away from the free body). The reaction can be directed either way in the cases of double-track rollers, links, collars on rods, or pins in slots. Generally, we 4.1 Equilibrium in Two Dimensions 173 assume that single-track rollers and rockers are reversible, so the cor-responding reactions can be directed either way. 2. Reactions Equivalent to a Force of Unknown Direction and Magni-tude. Supports and connections causing reactions of this type include frictionless pins in fitted holes, hinges, and rough surfaces. They can prevent translation of the free body in all directions, but they cannot prevent the body from rotating about the connection. Reactions of this group involve two unknowns and are usually represented by their x and Fig. 4.1 Reactions of supports and connections in two dimensions.
Support or Connection Reaction Number of Unknowns Rollers Rocker Frictionless surface Force with known line of action perpendicular to surface Force with known line of action along cable or link Force with known line of action perpendicular to rod or slot 1 1 1 Short cable Short link Collar on frictionless rod Frictionless pin in slot 90º Frictionless pin or hinge Rough surface Force of unknown direction or or 2 Fixed support Force and couple This rocker bearing supports the weight of a bridge. The convex surface of the rocker allows the bridge to move slightly horizontally.
Links are often used to support suspended spans of highway bridges.
Force applied to the slider exerts a normal force on the rod, causing the window to open.
Pin supports are common on bridges and overpasses.
This cantilever support is fxed at one end and extends out into space at the other end.
3 a a 174 Equilibrium of Rigid Bodies y components. In the case of a rough surface, the component normal to the surface must be directed away from the surface. 3. Reactions Equivalent to a Force and a Couple. These reactions are caused by fixed supports that oppose any motion of the free body and thus constrain it completely. Fixed supports actually produce forces over the entire surface of contact; these forces, however, form a system that can be reduced to a force and a couple. Reactions of this group involve three unknowns usually consisting of the two components of the force and the moment of the couple.
When the sense of an unknown force or couple is not readily appar-ent, do not attempt to determine it. Instead, arbitrarily assume the sense of the force or couple; the sign of the answer will indicate whether the assumption is correct or not. (A positive answer means the assumption is correct, while a negative answer means the assumption is incorrect.) 4.1B Rigid-Body Equilibrium in Two Dimensions The conditions stated in Sec. 4.1A for the equilibrium of a rigid body become considerably simpler for the case of a two-dimensional structure. Choosing the x and y axes to be in the plane of the structure, we have Fz 5 0 Mx 5 My 5 0 Mz 5 MO for each of the forces applied to the structure. Thus, the six equations of equilibrium stated in Sec. 4.1 reduce to three equations: oFx 5 0 oFy 5 0 oMO 5 0 (4.4) Since oMO 5 0 must be satisfied regardless of the choice of the origin O, we can write the equations of equilibrium for a two-dimensional structure in the more general form Equations of equilibrium in two dimensions oFx 5 0 oFy 5 0 oMA 5 0 (4.5) where A is any point in the plane of the structure. These three equations can be solved for no more than three unknowns.
You have just seen that unknown forces include reactions and that the number of unknowns corresponding to a given reaction depends upon the type of support or connection causing that reaction. Referring to Fig. 4.1, note that you can use the equilibrium equations (4.5) to determine the reactions associated with two rollers and one cable, or one fixed support, or one roller and one pin in a fitted hole, etc.
For example, consider Fig. 4.2a, in which the truss shown is in equi-librium and is subjected to the given forces P, Q, and S. The truss is held in place by a pin at A and a roller at B. The pin prevents point A from moving by exerting a force on the truss that can be resolved into the components Ax and Ay. The roller keeps the truss from rotating about A by exerting the vertical force B. The free-body diagram of the truss is shown in Fig. 4.2b; it includes the reactions Ax, Ay, and B as well as the applied forces P, Q, and S (in x and y component form) and the weight W of the truss. Since the truss is in equilibrium, the sum of the moments about A of all of the forces shown in Fig. 4.2b is zero, or oMA 5 0. We can use oFx F 5 0 oFy F 5 0 oMA M 5 0 C A B D P Q S (a) C A B D (b) Py Qy Qx Sy Sx W Px B Ax Ay Fig. 4.2 (a) A truss supported by a pin and a roller; (b) free-body diagram of the truss.
4.1 Equilibrium in Two Dimensions 175 this equation to determine the magnitude B because the equation does not contain Ax or Ay. Then, since the sum of the x components and the sum of the y components of the forces are zero, we write the equations oFx 5 0 and oFy 5 0. From these equations, we can obtain the components Ax and Ay, respectively.
We could obtain an additional equation by noting that the sum of the moments of the external forces about a point other than A is zero. We could write, for instance, oMB 5 0. This equation, however, does not contain any new information, because we have already established that the system of forces shown in Fig. 4.2b is equivalent to zero. The additional equation is not independent and cannot be used to determine a fourth unknown. It can be useful, however, for checking the solution obtained from the original three equations of equilibrium.
Although the three equations of equilibrium cannot be augmented by additional equations, any of them can be replaced by another equation. Properly chosen, the new system of equations still describes the equilib-rium conditions but may be easier to work with. For example, an alterna-tive system of equations for equilibrium is oFx 5 0 oMA 5 0 oMB 5 0 (4.6) Here the second point about which the moments are summed (in this case, point B) cannot lie on the line parallel to the y axis that passes through point A (Fig. 4.2b). These equations are sufficient conditions for the equi-librium of the truss. The first two equations indicate that the external forces must reduce to a single vertical force at A. Since the third equation requires that the moment of this force be zero about a point B that is not on its line of action, the force must be zero, and the rigid body is in equilibrium.
A third possible set of equilibrium equations is oMA 5 0 oMB 5 0 oMC 5 0 (4.7) where the points A, B, and C do not lie in a straight line (Fig. 4.2b). The first equation requires that the external forces reduce to a single force at A; the second equation requires that this force pass through B; and the third equation requires that it pass through C. Since the points A, B, C do not lie in a straight line, the force must be zero, and the rigid body is in equilibrium.
Notice that the equation oMA 5 0, stating that the sum of the moments of the forces about pin A is zero, possesses a more definite physical meaning than either of the other two equations (4.7). These two equations express a similar idea of balance but with respect to points about which the rigid body is not actually hinged. They are, however, as useful as the first equation. The choice of equilibrium equations should not be unduly influenced by their physical meaning. Indeed, in practice, it is desirable to choose equations of equilibrium containing only one unknown, since this eliminates the necessity of solving simulta neous equations. You can obtain equations containing only one unknown by summing moments about the point of intersection of the lines of action of two unknown forces or, if these forces are parallel, by summing force components in a direction perpendicular to their common direction. For example, in Fig. 4.3, in which the truss shown is held by rollers at A and B and a short link at D, we can eliminate the reactions at A and B by summing x components. We can eliminate the reactions at A and D oFx F 5 0 oMA M 5 0 oMB M 5 0 oMA M 5 0 oMB M 5 0 oMC 5 0 C A B D D P Q S (a) C A B D (b) Py Qy Qx Sy Sx A W Px B Fig. 4.3 (a) A truss supported by two rollers and a short link; (b) free-body diagram of the truss.
176 Equilibrium of Rigid Bodies by summing moments about C and the reactions at B and D by summing moments about D. The resulting equations are oFx 5 0 oMC 5 0 oMD 5 0 Each of these equations contains only one unknown. 4.1C Statically Indeterminate Reactions and Partial Constraints In the two examples considered in Figs. 4.2 and 4.3, the types of supports used were such that the rigid body could not possibly move under the given loads or under any other loading conditions. In such cases, the rigid body is said to be completely constrained. Recall that the reactions cor-responding to these supports involved three unknowns and could be deter-mined by solving the three equations of equilibrium. When such a situation exists, the reactions are said to be statically determinate.
Consider Fig. 4.4a, in which the truss shown is held by pins at A and B. These supports provide more constraints than are necessary to keep the truss from moving under the given loads or under any other loading conditions. Note from the free-body diagram of Fig. 4.4b that the corre-sponding reactions involve four unknowns. We pointed out in Sec. 4.1D that only three independent equilibrium equations are available; therefore, in this case, we have more unknowns than equations. As a result, we cannot determine all of the unknowns. The equations oMA 5 0 and oMB 5 0 yield the vertical components By and Ay, respectively, but the equation oFx 5 0 gives only the sum Ax 1 Bx of the horizontal components of the reactions at A and B. The components Ax and Bx are statically indeterminate. We could determine their magnitudes by considering the deformations pro-duced in the truss by the given loading, but this method is beyond the scope of statics and belongs to the study of mechanics of materials.
Let’s consider the opposite situation. The supports holding the truss shown in Fig. 4.5a consist of rollers at A and B. Clearly, the constraints provided by these supports are not sufficient to keep the truss from mov-ing. Although they prevent any vertical motion, the truss is free to move horizontally. The truss is said to be partially constrained.† From the free-body diagram in Fig. 4.5b, note that the reactions at A and B involve only two unknowns. Since three equations of equilibrium must still be satisfied, we have fewer unknowns than equations. In such a case, one of the equilibrium equations will not be satisfied in general. The equations oMA 5 0 and oMB 5 0 can be satisfied by a proper choice of reactions at A and B, but the equation oFx 5 0 is not satisfied unless the sum of the horizontal components of the applied forces happens to be zero. We thus observe that the equilibrium of the truss of Fig. 4.5 cannot be main-tained under general loading conditions.
From these examples, it would appear that, if a rigid body is to be completely constrained and if the reactions at its supports are to be statically determinate, there must be as many unknowns as there are equations of equilibrium. When this condition is not satisfied, we can be certain that either the rigid body is not completely constrained or that the reactions at its supports †Partially constrained bodies are often referred to as unstable. However, to avoid confusion between this type of instability, due to insufficient constraints, and the type of instability considered in Chap. 10, which relates to the behavior of a rigid body when its equilibrium is disturbed, we shall restrict the use of the words stable and unstable to the latter case.
C A B D P Q S (a) C A B D (b) Py Qy Qx Sy Sx Bx By Ax Ay W Px Fig. 4.4 (a) Truss with statically indeterminate reactions; (b) free-body diagram.
C A B D P Q S (a) C A B D (b) Py Qy Qx Sy Sx A W Px B Fig. 4.5 (a) Truss with partial constraints; (b) free-body diagram.
4.1 Equilibrium in Two Dimensions 177 are not statically determinate. It is also possible that the rigid body is not completely constrained and that the reactions are statically indeterminate.
You should note, however, that, although this condition is necessary, it is not sufficient. In other words, the fact that the number of unknowns is equal to the number of equations is no guarantee that a body is completely con-strained or that the reactions at its supports are statically determinate. Consider Fig. 4.6a, which shows a truss held by rollers at A, B, and E. We have three unknown reactions of A, B, and E (Fig. 4.6b), but the equation oFx 5 0 is not satisfied unless the sum of the horizontal components of the applied forces happens to be zero. Although there are a sufficient number of constraints, these constraints are not properly arranged, so the truss is free to move horizontally. We say that the truss is improperly constrained. Since only two equilibrium equations are left for determining three unknowns, the reactions are statically indeterminate. Thus, improper constraints also produce static indeterminacy.
The truss shown in Fig. 4.7 is another example of improper constraints— and of static indeterminacy. This truss is held by a pin at A and by rollers at B and C, which altogether involve four unknowns. Since only three inde-pendent equilibrium equations are available, the reactions at the supports are statically indeterminate. On the other hand, we note that the equation oMA 5 0 cannot be satisfied under general loading conditions, since the lines of action of the reactions B and C pass through A. We conclude that the truss can rotate about A and that it is improperly constrained.† The examples of Figs. 4.6 and 4.7 lead us to conclude that A rigid body is improperly constrained whenever the supports (even though they may provide a sufficient number of reactions) are arranged in such a way that the reactions must be either concurrent or parallel.‡ In summary, to be sure that a two-dimensional rigid body is com-pletely constrained and that the reactions at its supports are statically determinate, you should verify that the reactions involve three—and only three—unknowns and that the supports are arranged in such a way that they do not require the reactions to be either concurrent or parallel.
Supports involving statically indeterminate reactions should be used with care in the design of structures and only with a full knowledge of the problems they may cause. On the other hand, the analysis of structures possessing statically indeterminate reactions often can be partially carried out by the methods of statics. In the case of the truss of Fig. 4.4, for example, we can determine the vertical components of the reactions at A and B from the equilibrium equations.
For obvious reasons, supports producing partial or improper constraints should be avoided in the design of stationary structures. However, a partially or improperly constrained structure will not necessarily collapse; under par-ticular loading conditions, equilibrium can be maintained. For example, the trusses of Figs. 4.5 and 4.6 will be in equilibrium if the applied forces P, Q, and S are vertical. Besides, structures designed to move should be only partially constrained. A railroad car, for instance, would be of little use if it were completely constrained by having its brakes applied permanently.
†Rotation of the truss about A requires some “play” in the supports at B and C. In practice such play will always exist. In addition, we note that if the play is kept small, the displacements of the rollers B and C and, thus, the distances from A to the lines of action of the reactions B and C will also be small. The equation oMA 5 0 then requires that B and C be very large, a situation which can result in the failure of the supports at B and C.
‡Because this situation arises from an inadequate arrangement or geometry of the supports, it is often referred to as geometric instability.
C A B E E D P Q S (a) C A B D (b) Py Qy Qx Sy Sx A W Px B E Fig. 4.6 (a) Truss with improper constraints; (b) free-body diagram.
B C Ax C D P Q S (a) C A B A B D (b) Py Qy Qx Sy Sx Ay W Px Fig. 4.7 (a) Truss with improper constraints; (b) free-body diagram.
178 Equilibrium of Rigid Bodies Sample Problem 4.1 A fixed crane has a mass of 1000 kg and is used to lift a 2400-kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B.
STRATEGY: Draw a free-body diagram to show all of the forces acting on the crane, then use the equilibrium equations to calculate the values of the unknown forces.
MODELING: Free-Body Diagram. By multiplying the masses of the crane and of the crate by g 5 9.81 m/s2, you obtain the corresponding weights––that is, 9810 N or 9.81 kN, and 23 500 N or 23.5 kN (Fig. 1). The reaction at pin A is a force of unknown direction; you can represent it by components Ax and Ay. The reaction at the rocker B is perpendicular to the rocker surface; thus, it is horizontal. Assume that Ax, Ay, and B act in the directions shown.
ANALYSIS: Determination of B. The sum of the moments of all external forces about point A is zero. The equation for this sum contains neither Ax nor Ay, since the moments of Ax and Ay about A are zero. Multiplying the magnitude of each force by its perpendicular distance from A, you have 1loMA 5 0: 1B(1.5 m) 2 (9.81 kN)(2 m) 2 (23.5 kN)(6 m) 5 0 B 5 1107.1 kN B 5 107.1 kN y b Since the result is positive, the reaction is directed as assumed.
Determination of Ax. Determine the magnitude of Ax by setting the sum of the horizontal components of all external forces to zero.
y 1 oFx 5 0: Ax 1 B 5 0 Ax 1 107.1 kN 5 0 Ax 5 2107.1 kN Ax 5 107.1 kN z b Since the result is negative, the sense of Ax is opposite to that assumed originally.
Determination of Ay. The sum of the vertical components must also equal zero. Therefore, 1 x oFy 5 0: Ay 2 9.81 kN 2 23.5 kN 5 0 Ay 5 133.3 kN Ay 5 33.3 kNx b Adding the components Ax and Ay vectorially, you can find that the reaction at A is 112.2 kN b17.3°.
REFLECT and THINK: You can check the values obtained for the reactions by recalling that the sum of the moments of all the external forces about any point must be zero. For example, considering point B (Fig. 2), you can show 1loMB 5 2(9.81 kN)(2 m) 2 (23.5 kN)(6 m) 1 (107.1 kN)(1.5 m) 5 0 2400 kg A B G 4 m 2 m 1.5 m Fig. 1 Free-body diagram of crane.
A B B 23.5 kN Ay Ax 9.81 kN 1.5 m 4 m 2 m Fig. 2 Free-body diagram of crane with solved reactions.
33.3 kN 107.1 kN 107.1 kN A B 23.5 kN 9.81 kN 4 m 2 m 1.5 m 4.1 Equilibrium in Two Dimensions 179 Sample Problem 4.2 Three loads are applied to a beam as shown. The beam is supported by a roller at A and by a pin at B. Neglecting the weight of the beam, determine the reactions at A and B when P 5 15 kips.
STRATEGY: Draw a free-body diagram of the beam, then write the equilibrium equations, first summing forces in the x direction and then summing moments at A and at B.
MODELING: Free-Body Diagram. The reaction at A is vertical and is denoted by A (Fig. 1). Represent the reaction at B by components Bx and By. Assume that each component acts in the direction shown.
ANALYSIS: Equilibrium Equations. Write the three equilibrium equations and solve for the reactions indicated: y 1 oFx 5 0: Bx 5 0 Bx 5 0 b 1loMA 5 0: 2(15 kips)(3 ft) 1 By(9 ft) 2 (6 kips)(11 ft) 2 (6 kips)(13 ft) 5 0 By 5 121.0 kips By 5 21.0 kipsx b 1loMB 5 0: 2A(9 ft) 1 (15 kips)(6 ft) 2 (6 kips)(2 ft) 2 (6 kips)(4 ft) 5 0 A 5 16.00 kips A 5 6.00 kipsx b REFLECT and THINK: Check the results by adding the vertical com-ponents of all of the external forces: 1 xoFy 5 16.00 kips 2 15 kips 1 21.0 kips 2 6 kips 2 6 kips 5 0 Remark. In this problem, the reactions at both A and B are vertical; however, these reactions are vertical for different reasons. At A, the beam is supported by a roller; hence, the reaction cannot have any horizontal component. At B, the horizontal component of the reaction is zero because it must satisfy the equilibrium equation oFx 5 0 and none of the other forces acting on the beam has a horizontal component.
You might have noticed at first glance that the reaction at B was vertical and dispensed with the horizontal component Bx. This, however, is bad practice. In following it, you run the risk of forgetting the compo-nent Bx when the loading conditions require such a component (i.e., when a horizontal load is included). Also, you found the component Bx to be zero by using and solving an equilibrium equation, oFx 5 0. By setting Bx equal to zero immediately, you might not realize that you actually made use of this equation. Thus, you might lose track of the number of equa-tions available for solving the problem.
3 ft 2 ft 2 ft 6 kips 6 kips P 6 ft A B Fig.1 Free-body diagram of beam.
3 ft 2 ft 2 ft 6 kips 15 kips 6 kips 6 ft By Bx A A B 180 Equilibrium of Rigid Bodies Sample Problem 4.3 A loading car is at rest on a track forming an angle of 25° with the verti-cal. The gross weight of the car and its load is 5500 lb, and it acts at a point 30 in. from the track, halfway between the two axles. The car is held by a cable attached 24 in. from the track. Determine the tension in the cable and the reaction at each pair of wheels.
STRATEGY: Draw a free-body diagram of the car to determine the unknown forces, and write equilibrium equations to find their values, sum-ming moments at A and B and then summing forces.
MODELING: Free-Body Diagram. The reaction at each wheel is perpendicular to the track, and the tension force T is parallel to the track. Therefore, for convenience, choose the x axis parallel to the track and the y axis perpen-dicular to the track (Fig. 1). Then resolve the 5500-lb weight into x and y components.
Wx 5 1(5500 lb) cos 25° 5 14980 lb Wy 5 2(5500 lb) sin 25° 5 22320 lb ANALYSIS: Equilibrium Equations. Take moments about A to eliminate T and R1 from the computation.
1loMA 5 0: 2(2320 lb)(25 in.) 2 (4980 lb)(6 in.) 1 R2(50 in.) 5 0 R2 5 11758 lb R2 5 1758 lb w b Then take moments about B to eliminate T and R2 from the computation. 1loMB 5 0: (2320 lb)(25 in.) 2 (4980 lb)(6 in.) 2 R1(50 in.) 5 0 R1 5 1562 lb R1 5 1562 lb w b Determine the value of T by summing forces in the x direction.
w 1oFx 5 0: 14980 lb 2 T 5 0 T 5 14980 lb T 5 4980 lb w b Figure 2 shows the computed values of the reactions.
REFLECT and THINK: You can verify the computations by summing forces in the y direction.
w 1oFy 5 1562 lb 1 1758 lb 2 2320 lb 5 0 You could also check the solution by computing moments about any point other than A or B.
24 in.
25º G 25 in.
25 in.
30 in.
y x R1 R2 2320 lb 6 in.
A T B G 25 in.
25 in.
4980 lb Fig. 1 Free-body diagram of car.
562 lb 1758 lb y x 4980 lb 25 in.
25 in.
2320 lb 6 in.
A B G 4980 lb Fig. 2 Free-body diagram of car with solved reactions.
4.1 Equilibrium in Two Dimensions 181 Sample Problem 4.4 The frame shown supports part of the roof of a small building. Knowing that the tension in the cable is 150 kN, determine the reaction at the fixed end E.
STRATEGY: Draw a free-body diagram of the frame and of the cable BDF. The support at E is fixed, so the reactions here include a moment; to determine its value, sum moments about point E.
MODELING: Free-Body Diagram. Represent the reaction at the fixed end E by the force components Ex and Ey and the couple ME (Fig. 1). The other forces acting on the free body are the four 20-kN loads and the 150-kN force exerted at end F of the cable.
Fig. 1 Free-body diagram of frame.
6 m 150 kN Ey Ex ME 20 kN 20 kN 20 kN 20 kN A B C D E F 4.5 m 1.8 m 1.8 m 1.8 m 1.8 m ANALYSIS: Equilibrium Equations. First note that DF 5 2(4.5 m)2 1 (6 m)2 5 7.5 m Then you can write the three equilibrium equations and solve for the reactions at E.
y 1 oFx 5 0: Ex 1 4.5 7.5(150 kN) 5 0 Ex 5 290.0 kN Ex 5 90.0 kN z b 1 x oFy 5 0: Ey 2 4(20 kN) 2 6 7.5(150 kN) 5 0 Ey 5 1200 kN Ey 5 200 kNx b 1loME 5 0: (20 kN)(7.2 m) 1 (20 kN)(5.4 m) 1 (20 kN)(3.6 m) 1(20 kN)(1.8 m) 2 6 7.5 (150 kN)(4.5 m) 1 ME 5 0 ME 5 1180.0 kN?m ME 5 180.0 kN?m l b REFLECT and THINK: The cable provides a fourth constraint, making this situation statically indeterminate. This problem therefore gave us the value of the cable tension, which would have been determined by means other than statics. We could then use the three available independent static equilibrium equations to solve for the remaining three reactions.
20 kN 20 kN 20 kN 20 kN A B C D E F 1.8 m 1.8 m 1.8 m 1.8 m 2.25 m 3.75 m 4.5 m 182 Equilibrium of Rigid Bodies Sample Problem 4.5 A 400-lb weight is attached at A to the lever shown. The constant of the spring BC is k 5 250 lb/in., and the spring is unstretched when θ 5 0. Determine the position of equilibrium.
A B C O k = 250 lb/in.
r = 3 in.
l = 8 in.
W = 400 lb q STRATEGY: Draw a free-body diagram of the lever and cylinder to show all forces acting on the body (Fig. 1), then sum moments about O. Your final answer should be the angle θ.
MODELING: Free-Body Diagram. Denote by s the deflection of the spring from its unstretched position and note that s 5 rθ. Then F 5 ks 5 krθ.
ANALYSIS: Equilibrium Equation. Sum the moments of W and F about O to eliminate the reactions supporting the cylinder. The result is 1loMO 5 0: Wl sin θ 2 r(krθ) 5 0 sin θ 5 kr 2 Wl θ Substituting the given data yields sin θ 5 (250 lb/in.)(3 in.)2 (400 lb)(8 in.) θ sin θ 5 0.703 θ Solving by trial and error, the angle is θ 5 0 θ 5 80.3˚ b REFLECT and THINK: The weight could represent any vertical force acting on the lever. The key to the problem is to express the spring force as a function of the angle θ. Fig. 1 Free-body diagram of the lever and cylinder.
A s O W F = ks Ry Rx Unstretched position q r l sin q 183 183 Y ou saw that, for a rigid body in equilibrium, the system of external forces is equivalent to zero. To solve an equilibrium problem, your first task is to draw a neat, reasonably large free-body diagram on which you show all external forces. You should include both known and unknown forces.
For a two-dimensional rigid body, the reactions at the supports can involve one, two, or three unknowns, depending on the type of support (Fig. 4.1). A correct free-body diagram is essential for the successful solution of a problem. Never proceed with the solution of a problem until you are sure that your free-body diagram includes all loads, all reactions, and the weight of the body (if appropriate). 1. You can write three equilibrium equations and solve them for three unknowns.
The three equations might be oFx 5 0 oFy 5 0 oMO 5 0 However, usually several alternative sets of equations are possible, such as oFx 5 0 oMA 5 0 oMB 5 0 where point B is chosen in such a way that the line AB is not parallel to the y axis, or oMA 5 0 oMB 5 0 oMC 5 0 where the points A, B, and C do not lie along a straight line.
2. To simplify your solution, it may be helpful to use one of the following solution techniques. a. By summing moments about the point of intersection of the lines of action of two unknown forces, you obtain an equation in a single unknown. b. By summing components in a direction perpendicular to two unknown parallel forces, you also obtain an equation in a single unknown.
3. After drawing your free-body diagram, you may find that one of the following special situations arises. a. The reactions involve fewer than three unknowns. The body is said to be partially constrained and motion of the body is possible. b. The reactions involve more than three unknowns. The reactions are said to be statically indeterminate. Although you may be able to calculate one or two reac-tions, you cannot determine all of them. c. The reactions pass through a single point or are parallel. The body is said to be improperly constrained and motion can occur under a general loading condition.
SOLVING PROBLEMS ON YOUR OWN 184 Problems FREE-BODY PRACTICE PROBLEMS 4.F1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pick-up truck. Draw the free-body diagram needed to determine the reactions at each of the two rear wheels A and front wheels B. 4.F2 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, draw the free-body diagram needed to determine the tension in the cable and the reaction at C.
A B D 12 in.
20° 75 lb C 10 in.
15 in.
Fig. P4.F2 4.F3 A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120-lb force is applied at D. Draw the free-body diagram needed to determine the reactions at A, B, and C. 4.F4 A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, draw the free-body diagram needed to determine the reac-tion at C.
C 20 N 20 N 75 mm 45 mm A B 75 mm Fig. P4.F4 C D G 1.7 m 2.8 m A B 1.8 m 1.2 m 0.75 m Fig. P4.F1 120 lb 30° A B C D 8 in.
8 in.
8 in.
Fig. P4.F3 185 END-OF-SECTION PROBLEMS 4.1 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle? 4.2 The gardener of Prob. 4.1 wishes to transport a second 250-N bag of fertilizer at the same time as the first one. Determine the maxi-mum allowable horizontal distance from the axle A of the wheelbar-row to the center of gravity of the second bag if she can hold only 75 N with each arm. 4.3 A 2100-lb tractor is used to lift 900 lb of gravel. Determine the reaction at each of the two (a) rear wheels A, (b) front wheels B.
20 in.
40 in.
50 in.
900 lb A B G Fig. P4.3 4.4 For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC. 4.5 A load of lumber of weight W 5 25 kN is being raised by a mobile crane. The weight of boom ABC and the combined weight of the truck and driver are as shown. Determine the reaction at each of the two (a) front wheels H, (b) rear wheels K. 4.6 A load of lumber of weight W 5 25 kN is being raised by a mobile crane. Knowing that the tension is 25 kN in all portions of cable AEF and that the weight of boom ABC is 3 kN, determine (a) the tension in rod CD, (b) the reaction at pin B. 4.7 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a 5 10 in., (b) if a 5 7 in.
6 in.
6 in.
8 in.
10 lb 30 lb 50 lb 40 lb A B a Fig. P4.7 4.8 For the bracket and loading of Prob. 4.7, determine the smallest distance a if the bracket is not to move.
0.15 m 0.15 m 60 N 250 N A 0.7 m Fig. P4.1 A C B 15 lb 20 lb 35 lb 15 lb 20 lb 6 in.
8 in.
8 in.
6 in.
Fig. P4.4 2.0 m 2.0 m 2.0 m 0.5 m 0.3 m 0.4 m 0.6 m 0.9 m A B C D E F 50 kN 3 kN W K H Fig. P4.5 and P4.6 186 4.9 Three loads are applied as shown to a light beam supported by cables attached at B and D. Neglecting the weight of the beam, determine the range of values of Q for which neither cable becomes slack when P 5 0. 4.10 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 12 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P 5 0. 4.11 For the beam of Prob. 4.10, determine the range of values of Q for which the loading is safe when P 5 5 kN. 4.12 For the beam of Sample Prob. 4.2, determine the range of values of P for which the beam will be safe, knowing that the maximum allow-able value of each of the reactions is 25 kips and that the reaction at A must be directed upward. 4.13 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the dis-tance d for which the beam is safe. 4.14 For the beam and loading shown, determine the range of the distance a for which the reaction at B does not exceed 100 lb downward or 200 lb upward.
a A D C B 6 in.
300 lb 300 lb 50 lb 8 in.
4 in.
12 in.
Fig. P4.14 4.15 Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C.
A B E D C 90° 60 mm 90 mm 80 mm 120 mm Fig. P4.15 and P4.16 4.16 Two links AB and DE are connected by a bell crank as shown. Determine the maximum force that can be safely exerted by link AB on the bell crank if the maximum allowable value for the reaction at C is 1600 N.
0.5 m 0.75 m 0.75 m 1.5 m 7.5 kN P Q A B D C E Fig. P4.9 and P4.10 50 N 100 N 150 N 450 mm d A B 450 mm Fig. P4.13 187 4.17 The required tension in cable AB is 200 lb. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C. 4.18 Determine the maximum tension that can be developed in cable AB if the maximum allowable value of the reaction at C is 250 lb. 4.19 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C.
240 N 240 N 0.24 m 0.4 m 0.4 m A B C D a = 0.18 m Fig. P4.19 4.20 Solve Prob. 4.19, assuming that a 5 0.32 m. 4.21 The 40-ft boom AB weighs 2 kips; the distance from the axle A to the center of gravity G of the boom is 20 ft. For the position shown, determine (a) the tension T in the cable, (b) the reaction at A.
C 5 kips 2 kips 30° 10° T B A G Fig. P4.21 4.22 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C.
B D 30° 500 N C 200 mm 250 mm 250 mm A Fig. P4.22 P D A B C 15 in.
7 in.
60° Fig. P4.17 and P4.18 188 4.23 and 4.24 For each of the plates and loadings shown, determine the reactions at A and B.
40 lb 40 lb 50 lb 50 lb A B (a) 30° 10 in.
A B (b) 20 in.
10 in.
4 in.
4 in.
20 in.
Fig. P4.23 40 lb 40 lb 50 lb 50 lb A B (a) 30º 20 in.
10 in.
A B (b) 20 in.
10 in.
4 in.
4 in.
Fig. P4.24 4.25 A rod AB, hinged at A and attached at B to cable BD, supports the loads shown. Knowing that d 5 200 mm, determine (a) the tension in cable BD, (b) the reaction at A. 4.26 A rod AB, hinged at A and attached at B to cable BD, supports the loads shown. Knowing that d 5 150 mm, determine (a) the tension in cable BD, (b) the reaction at A. 4.27 Determine the reactions at A and B when (a) α 5 0, (b) α 5 90°, (c) α 5 30°.
10 in.
10 in.
12 in.
a A B 75 lb Fig. P4.27 90 N 100 mm 100 mm 100 mm 100 mm A B d D 90 N Fig. P4.25 and P4.26 189 4.28 Determine the reactions at A and C when (a) α 5 0, (b) α 5 30°. 4.29 Rod ABC is bent in the shape of an arc of circle of radius R. Know-ing that θ 5 30°, determine the reaction (a) at B, (b) at C.
A B C R P q Fig. P4.29 and P4.30 4.30 Rod ABC is bent in the shape of an arc of circle of radius R. Know-ing that θ 5 60°, determine the reaction (a) at B, (b) at C. 4.31 Neglecting friction, determine the tension in cable ABD and the reac-tion at C when θ 5 60°.
A B D C 90° P q a a 2a Fig. P4.31 and P4.32 4.32 Neglecting friction, determine the tension in cable ABD and the reac-tion at C when θ 5 45°. 4.33 A force P of magnitude 90 lb is applied to member ACDE that is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a 5 3 in., determine (a) the tension in the cable, (b) the reac-tion at D. 4.34 Solve Prob. 4.33 for a 5 6 in.
800 mm 200 mm 300 N 200 mm 300 N a A B C Fig. P4.28 C E A D P a 7 in.
5 in.
12 in.
B Fig. P4.33 190 4.35 Bar AC supports two 400-N loads as shown. Rollers at A and C rest against frictionless surfaces and a cable BD is attached at B. Deter-mine (a) the tension in cable BD, (b) the reaction at A, (c) the reac-tion at C.
400 N 400 N 100 mm 150 mm 100 mm 300 mm 500 mm A B C D 250 mm Fig. P4.35 4.36 A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with friction-less vertical walls. Determine the tension in cable BE and the reac-tions at A and D.
A 20 kg B C D E 125 mm 200 mm 175 mm 75 mm Fig. P4.36 4.37 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when θ 5 30°. 4.38 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine (a) the smallest value of θ for which the equilibrium of the bracket is main-tained, (b) the corresponding reactions at C, D, and E. 4.39 A movable bracket is held at rest by a cable attached at C and by frictionless rollers at A and B. For the loading shown, determine (a) the tension in the cable, (b) the reactions at A and B.
A B C 600 N 475 mm 75 mm 50 mm 90 mm Fig. P4.39 A B C D E 3 in.
3 in.
2 in.
20 lb 40 lb q 4 in.
4 in.
Fig. P4.37 and P4.38 191 4.40 A light bar AB supports a 15-kg block at its midpoint C. Rollers at A and B rest against frictionless surfaces, and a horizontal cable AD is attached at A. Determine (a) the tension in cable AD, (b) the reac-tions at A and B.
250 mm 250 mm B C D A 350 mm 15 kg Fig. P4.40 4.41 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P 5 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F.
P A B D E F 4 in.
4 in.
7 in.
2 in.
30º 30 lb 3 in.
Fig. P4.41 4.42 For the plate of Prob. 4.41, the reaction at F must be directed down-ward, and its maximum value is 20 lb. Neglecting friction at the pins, determine the required range of values of P. 4.43 The rig shown consists of a 1200-lb horizontal member ABC and a vertical member DBE welded together at B. The rig is being used to raise a 3600-lb crate at a distance x 5 12 ft from the vertical member DBE. If the tension in the cable is 4 kips, determine the reaction at E, assuming that the cable is (a) anchored at F as shown in the figure, (b) attached to the vertical member at a point located 1 ft above E. 4.44 For the rig and crate of Prob. 4.43 and assuming that cable is anchored at F as shown, determine (a) the required tension in cable ADCF if the maximum value of the couple at E as x varies from 1.5 to 17.5 ft is to be as small as possible, (b) the corresponding maxi-mum value of the couple. 4.45 A 175-kg utility pole is used to support at C the end of an electric wire. The tension in the wire is 600 N, and the wire forms an angle of 15° with the horizontal at C. Determine the largest and smallest allowable tensions in the guy cable BD if the magnitude of the cou-ple at A may not exceed 500 N?m.
A C B F E x D 5 ft 10 ft 17.5 ft 6.5 ft 3.75 ft W = 1200 lb 3600 lb Fig. P4.43 A D C B 3.6 m 1.5 m 4.5 m 15° Fig. P4.45 192 4.46 Knowing that the tension in wire BD is 1300 N, determine the reac-tion at the fixed support C of the frame shown. 4.47 Determine the range of allowable values of the tension in wire BD if the magnitude of the couple at the fixed support C is not to exceed 100 N?m. 4.48 Beam AD carries the two 40-lb loads shown. The beam is held by a fixed support at D and by the cable BE that is attached to the counterweight W. Determine the reaction at D when (a) W 5 100 lb, (b) W 5 90 lb.
A B C D 40 lb 40 lb E 5 ft 4 ft 4 ft W Fig. P4.48 and P4.49 4.49 For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 40 lb?ft. 4.50 An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm radius, determine the reac-tion at A in each case.
B A A A B B 8 kg 8 kg 8 kg (a) (b) (c) 1.6 m 1.6 m 1.6 m Fig. P4.50 4.51 A uniform rod AB with a length of l and weight of W is suspended from two cords AC and BC of equal length. Determine the angle θ corresponding to the equilibrium position when a couple M is applied to the rod.
A B C a a q W M Fig. P4.51 750 N 500 mm 150 mm 250 mm 600 mm 450 N A B C D 400 mm Fig. P4.46 and P4.47 193 4.52 Rod AD is acted upon by a vertical force P at end A and by two equal and opposite horizontal forces of magnitude Q at points B and C. Neglecting the weight of the rod, express the angle θ cor-responding to the equilibrium position in terms of P and Q.
B C A D q P −Q Q a a a Fig. P4.52 4.53 A slender rod AB with a weight of W is attached to blocks A and B that move freely in the guides shown. The blocks are connected by an elastic cord that passes over a pulley at C. (a) Express the tension in the cord in terms of W and θ. (b) Determine the value of θ for which the tension in the cord is equal to 3W.
A B C W q l Fig. P4.53 4.54 and 4.55 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P 5 2W.
B A C q W P l l Fig. P4.54 P B C l l q W A Fig. P4.55 194 4.56 A collar B with a weight of W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when θ 5 0. (a) Derive an equation in θ, W, k, and l that must be satisfied when the collar is in equilibrium. (b) Knowing that W 5 300 N, l 5 500 mm, and k 5 800 N/m, determine the value of θ corresponding to equilibrium. 4.57 Solve Sample Prob. 4.5, assuming that the spring is unstretched when θ 5 90°. 4.58 A vertical load P is applied at end B of rod BC. The constant of the spring is k, and the spring is unstretched when θ 5 60°. (a) Neglect-ing the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, k, and l. (b) Determine the value of θ corresponding to equilibrium if P 5 1 4 kl. 4.59 Eight identical 500 3 750-mm rectangular plates, each of mass m 5 40 kg, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. In each case, deter-mine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indetermi-nate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions.
A B C D 1 2 3 4 5 6 7 8 Fig. P4.59 4.60 The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. For each case, answer the questions listed in Prob. 4.59, and, wherever possible, compute the reactions, assuming that the magnitude of the force P is 100 lb. B A C 1 3 ft 2 ft 2 ft 2 3 4 5 6 7 8 P P P P P P P P Fig. P4.60 A B q l Fig. P4.56 B C l A q l P Fig. P4.58 4.2 Two Special Cases 195 4.2 TWO SPECIAL CASES In practice, some simple cases of equilibrium occur quite often, either as part of a more complicated analysis or as the complete models of a situ-ation. By understanding the characteristics of these cases, you can often simplify the overall analysis.
4.2A Equilibrium of a Two-Force Body A particular case of equilibrium of considerable interest in practical appli-cations is that of a rigid body subjected to two forces. Such a body is commonly called a two-force body. We show here that, if a two-force body is in equilibrium, the two forces must have the same magnitude, the same line of action, and opposite sense.
Consider a corner plate subjected to two forces F1 and F2 acting at A and B, respectively (Fig. 4.8a). If the plate is in equilibrium, the sum of the moments of F1 and F2 about any axis must be zero. First, we sum moments about A. Since the moment of F1 is obviously zero, the moment of F2 also must be zero and the line of action of F2 must pass through A (Fig. 4.8b). Similarly, summing moments about B, we can show that the line of action of F1 must pass through B (Fig. 4.8c). Therefore, both forces have the same line of action (line AB). You can see from either of the equations oFx 5 0 and oFy 5 0 that they must also have the same mag-nitude but opposite sense.
(c) A B F1 F2 (b) A B F2 (a) A B F1 F2 F1 Fig. 4.8 A two-force body in equilibrium. (a) Forces act at two points of the body; (b) summing moments about point A shows that the line of action of F2 must pass through A; (c) summing moments about point B shows that the line of action of F1 must pass through B.
If several forces act at two points A and B, the forces acting at A can be replaced by their resultant F1, and those acting at B can be replaced by their resultant F2. Thus, a two-force body can be more generally defined as a rigid body subjected to forces acting at only two points.
The resultants F1 and F2 then must have the same line of action, the same magnitude, and opposite sense (Fig. 4.8).
Later, in the study of structures, frames, and machines, you will see how the recognition of two-force bodies simplifies the solution of certain problems.
196 Equilibrium of Rigid Bodies 4.2B Equilibrium of a Three-Force Body Another case of equilibrium that is of great practical interest is that of a three-force body, i.e., a rigid body subjected to three forces or, more generally, a rigid body subjected to forces acting at only three points. Consider a rigid body subjected to a system of forces that can be reduced to three forces F1, F2, and F3 acting at A, B, and C, respectively (Fig. 4.9a). We show that if the body is in equilibrium, the lines of action of the three forces must be either concurrent or parallel.
F2 F3 F1 B C D A (a) (b) (c) F2 F3 F1 B C D A F2 F3 F1 B C A Fig. 4.9 A three-force body in equilibrium. (a–c) Demonstration that the lines of action of the three forces must be either concurrent or parallel.
Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about any axis must be zero. Assuming that the lines of action of F1 and F2 intersect and denoting their point of intersection by D, we sum moments about D (Fig. 4.9b). Because the moments of F1 and F2 about D are zero, the moment of F3 about D also must be zero, and the line of action of F3 must pass through D (Fig. 4.9c). Therefore, the three lines of action are concurrent. The only exception occurs when none of the lines intersect; in this case, the lines of action are parallel.
Although problems concerning three-force bodies can be solved by the general methods of Sec. 4.1, we can use the property just established to solve these problems either graphically or mathematically using simple trigonometric or geometric relations (see Sample Problem 4.6).
4.2 Two Special Cases 197 Sample Problem 4.6 A man raises a 10-kg joist with a length of 4 m by pulling on a rope. Find the tension T in the rope and the reaction at A.
STRATEGY: The joist is acted upon by three forces: its weight W, the force T exerted by the rope, and the reaction R of the ground at A. Therefore, it is a three-force body, and you can compute the forces by using a force triangle.
MODELING: First note that W 5 mg 5 (10 kg)(9.81 m/s2) 5 98.1 N Since the joist is a three-force body, the forces acting on it must be con-current. The reaction R therefore must pass through the point of intersec-tion C of the lines of action of the weight W and the tension force T, as shown in the free-body diagram (Fig. 1). You can use this fact to determine the angle α that R forms with the horizontal.
ANALYSIS: Draw the vertical line BF through B and the horizontal line CD through C (Fig. 2). Then AF 5 BF 5 (AB) cos 458 5 (4 m) cos 458 5 2.828 m CD 5 EF 5 AE 5 1 2(AF) 5 1.414 m BD 5 (CD) cot (458 1 258) 5 (1.414 m) tan 208 5 0.515 m CE 5 DF 5 BF 2 BD 5 2.828 m 2 0.515 m 5 2.313 m Fig. 2 Geometry analysis of the lines of action for the three forces acting on joist, concurrent at point C.
45° 45° 4 m A B C G D E F 25° a From these calculations, you can determine the angle α as tan α 5 CE AE 5 2.313 m 1.414 m 5 1.636 α 5 58.68 b You now know the directions of all the forces acting on the joist.
Force Triangle. Draw a force triangle as shown (Fig. 3) with its inte-rior angles computed from the known directions of the forces. You can then use the law of sines to find the unknown forces.
T sin 31.48 5 R sin 1108 5 98.1 N sin 38.68 T 5 81.9 N b R 5 147.8 N a58.68 b REFLECT and THINK: In practice, three-force members occur often, so learning this method of analysis is useful in many situations.
Fig. 1 Free-body diagram of joist.
45° 25° 4 m B A A B C G T R W = 98.1 N a Fig. 3 Force triangle.
T R 98.1 N 110° 38.6° 20° 31.4° a = 58.6° 198 198 SOLVING PROBLEMS ON YOUR OWN T his section covered two particular cases of equilibrium of a rigid body.
1. A two-force body is subjected to forces at only two points. The resultants of the forces acting at each of these points must have the same magnitude, the same line of action, and opposite sense. This property allows you to simplify the solutions of some problems by replacing the two unknown components of a reaction by a single force of unknown magnitude but of known direction.
2. A three-force body is subjected to forces at only three points. The resultants of the forces acting at each of these points must be concurrent or parallel. To solve a problem involving a three-force body with concurrent forces, draw the free-body dia-gram showing that these three forces pass through the same point. You may be able to complete the solution by using simple geometry, such as a force triangle and the law of sines [see Sample Prob. 4.6].
This method for solving problems involving three-force bodies is not difficult to understand, but in practice, it can be difficult to sketch the necessary geometric con-structions. If you encounter difficulty, first draw a reasonably large free-body diagram and then seek a relation between known or easily calculated lengths and a dimension that involves an unknown. Sample Prob. 4.6 illustrates this technique, where we used the easily calculated dimensions AE and CE to determine the angle α.
199 Problems 4.61 A 500-lb cylindrical tank, 8 ft in diameter, is to be raised over a 2-ft obstruction. A cable is wrapped around the tank and pulled horizon-tally as shown. Knowing that the corner of the obstruction at A is rough, find the required tension in the cable and the reaction at A.
A B G 2 ft 8 ft T Fig. P4.61 4.62 Determine the reactions at A and B when a 5 180 mm. 4.63 For the bracket and loading shown, determine the range of values of the distance a for which the magnitude of the reaction at B does not exceed 600 N.
4.64 The spanner shown is used to rotate a shaft. A pin fits in a hole at A, while a flat, frictionless surface rests against the shaft at B. If a 60-lb force P is exerted on the spanner at D, find the reactions at A and B.
15 in.
3 in.
P A B C D 50º Fig. P4.64 4.65 Determine the reactions at B and C when a 5 30 mm.
100 mm 40 mm 60 mm 60 mm 250 N A C B D a Fig. P4.65 A B C 240 mm 300 N a Fig. P4.62 and P4.63 200 4.66 A 12-ft wooden beam weighing 80 lb is supported by a pin and bracket at A and by cable BC. Find the reaction at A and the tension in the cable. 4.67 Determine the reactions at B and D when b 5 60 mm.
A B C D 75 mm 80 N 90 mm b 250 mm Fig. P4.67 4.68 For the frame and loading shown, determine the reactions at C and D. 4.69 A 50-kg crate is attached to the trolley-beam system shown. Know-ing that a 5 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B.
A B C D 55° 1.4 m 0.4 m a W Fig. P4.69 4.70 One end of rod AB rests in the corner A and the other end is attached to cord BD. If the rod supports a 150-N load at its midpoint C, find the reaction at A and the tension in the cord. 4.71 For the boom and loading shown, determine (a) the tension in cord BD, (b) the reaction at C.
D C B A 3 kips 32 in.
16 in.
12 in.
32 in.
Fig. P4.71 C A B 80 lb 8 ft 6 ft 6 ft 6 ft Fig. P4.66 150 lb 3 ft 3 ft 1.5 ft 1.5 ft D B A C Fig. P4.68 150 N A B C D 240 mm 240 mm 360 mm 200 mm Fig. P4.70 201 4.72 A 40-lb roller of 8-in. diameter, which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 0.3 in., determine the force P required to move the roller onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right. 4.73 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α 5 45°.
A C 300 N B 300 mm 250 mm 150 mm α Fig. P4.73 and P4.74 4.74 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α 5 60°. 4.75 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B.
4.76 Solve Prob. 4.75, assuming that the 170-N force applied at B is horizontal and directed to the left. 4.77 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B.
A B D C 72 lb a = 12 in.
7 in.
24 in.
Fig. P4.77 4.78 Using the method of Sec. 4.2B, solve Prob. 4.22. 4.79 Knowing that θ 5 30°, determine the reaction (a) at B, (b) at C. 4.80 Knowing that θ 5 60°, determine the reaction (a) at B, (b) at C.
Fig. P4.72 30° P A B C 170 N 150 mm 150 mm 160 mm Fig. P4.75 A B C R P q Fig. P4.79 and P4.80 202 4.81 Determine the reactions at A and B when β 5 50°.
A B C 100 N 250 mm 150 mm 25° b Fig. P4.81 and P4.82 4.82 Determine the reactions at A and B when β 5 80°. 4.83 Rod AB is bent into the shape of an arc of circle and is lodged between two pegs D and E. It supports a load P at end B. Neglecting friction and the weight of the rod, determine the distance c corre-sponding to equilibrium when a 5 20 mm and R 5 100 mm.
P A R C D E a a c B Fig. P4.83 4.84 A slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium, derive an expression for angle θ in terms of angle β.
A B q b L Fig. P4.84 and P4.85 4.85 An 8-kg slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium and that β 5 30°, determine (a) the angle θ that the rod forms with the vertical, (b) the reactions at A and B.
203 4.86 A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle θ cor-responding to equilibrium.
4.87 A slender rod BC with a length of L and weight W is held by two cables as shown. Knowing that cable AB is horizontal and that the rod forms an angle of 40° with the horizontal, determine (a) the angle θ that cable CD forms with the horizontal, (b) the tension in each cable.
40° C B D L q A Fig. P4.87 4.88 A thin ring with a mass of 2 kg and radius r 5 140 mm is held against a frictionless wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in the string, (c) the reaction at C.
4.89 A slender rod with a length of L and weight W is attached to a collar at A and is fitted with a small wheel at B. Knowing that the wheel rolls freely along a cylindrical surface of radius R, and neglecting friction, derive an equation in θ, L, and R that must be satisfied when the rod is in equilibrium.
R L A B C q Fig. P4.89 4.90 Knowing that for the rod of Prob. 4.89, L 5 15 in., R 5 20 in., and W 5 10 lb, determine (a) the angle θ corresponding to equilibrium, (b) the reactions at A and B.
A B q 2R Fig. P4.86 140 mm 125 mm d A B C Fig. P4.88 204 Equilibrium of Rigid Bodies 4.3 EQUILIBRIUM IN THREE DIMENSIONS The most general situation of rigid-body equilibrium occurs in three dimensions. The approach to modeling and analyzing these situations is the same as in two dimensions: Draw a free-body diagram and then write and solve the equilibrium equations. However, you now have more equa-tions and more variables to deal with. In addition, reactions at supports and connections can be more varied, having as many as three force com-ponents and three couples acting at one support. As you will see in the Sample Problems, you need to visualize clearly in three dimensions and recall the vector analysis from Chapters 2 and 3.
4.3A Rigid-Body Equilibrium in Three Dimensions We saw in Sec. 4.1 that six scalar equations are required to express the condi-tions for the equilibrium of a rigid body in the general three-dimensional case: oFx 5 0 oFy 5 0 oFz 5 0 (4.2) oMx 5 0 oMy 5 0 oMz 5 0 (4.3) We can solve these equations for no more than six unknowns, which gen-erally represent reactions at supports or connections.
In most problems, we can obtain the scalar equations (4.2) and (4.3) more conveniently if we first write the conditions for the equilibrium of the rigid body considered in vector form: oF 5 0 oMO 5 o(r 3 F) 5 0 (4.1) Then we can express the forces F and position vectors r in terms of scalar components and unit vectors. This enables us to compute all vector prod-ucts either by direct calculation or by means of determinants (see Sec. 3.1F). Note that we can eliminate as many as three unknown reaction components from these computations through a judicious choice of the point O. By equating to zero the coefficients of the unit vectors in each of the two relations in Eq. (4.1), we obtain the desired scalar equations.† Some equilibrium problems and their associated free-body diagrams might involve individual couples Mi either as applied loads or as support reactions. In such situations, you can accommodate these couples by expressing the second part of Eq. (4.1) as oMO 5 o(r 3 F) 1 oMi 5 0 (4.19) 4.3B Reactions for a Three-Dimensional Structure The reactions on a three-dimensional structure range from a single force of known direction exerted by a frictionless surface to a force-couple system oFx F 5 0 oFy F 5 0 oFz F 5 0 oMx M 5 0 oMy M 5 0 oMz M 5 0 oF 5 0 oMO 5 o(r 3 F) 5 0 oMO 5 o(r 3 F) 1 oMi 5 0 †In some problems, it may be convenient to eliminate from the solution the reactions at two points A and B by writing the equilibrium equation oMAB 5 0. This involves determining the moments of the forces about the axis AB joining points A and B (see Sample Prob. 4.10). 4.3 Equilibrium in Three Dimensions 205 exerted by a fixed support. Consequently, in problems involving the equi-librium of a three-dimensional structure, between one and six unknowns may be associated with the reaction at each support or connection. Figure 4.10 shows various types of supports and connections with their corresponding reactions. A simple way of determining the type of reaction corresponding to a given support or connection and the number of unknowns involved is to find which of the six fundamental motions (translation in the x, y, and z directions and rotation about the x, y, and z axes) are allowed and which motions are prevented. The number of motions prevented equals the number of reactions.
Ball supports, frictionless surfaces, and cables, for example, prevent translation in one direction only and thus exert a single force whose line of action is known. Therefore, each of these supports involves one unknown–– namely, the magnitude of the reaction. Rollers on rough surfaces and wheels on rails prevent translation in two directions; the corresponding reactions consist of two unknown force components. Rough surfaces in direct contact and ball-and-socket supports prevent translation in three directions while still allowing rotation; these supports involve three unknown force components.
Some supports and connections can prevent rotation as well as trans-lation; the corresponding reactions include couples as well as forces. For example, the reaction at a fixed support, which prevents any motion (rota-tion as well as translation) consists of three unknown forces and three unknown couples. A universal joint, which is designed to allow rotation about two axes, exerts a reaction consisting of three unknown force com-ponents and one unknown couple.
Other supports and connections are primarily intended to prevent trans-lation; their design, however, is such that they also prevent some rotations. The corresponding reactions consist essentially of force components but may also include couples. One group of supports of this type includes hinges and bearings designed to support radial loads only (for example, journal bearings or roller bearings). The corresponding reactions consist of two force com-ponents but may also include two couples. Another group includes pin-and-bracket supports, hinges, and bearings designed to support an axial thrust as well as a radial load (for example, ball bearings). The corresponding reactions consist of three force components but may include two couples. However, these supports do not exert any appreciable couples under normal conditions of use. Therefore, only force components should be included in their analysis unless it is clear that couples are necessary to maintain the equilibrium of the rigid body or unless the support is known to have been specifically designed to exert a couple (see Probs. 4.119 through 4.122).
If the reactions involve more than six unknowns, you have more unknowns than equations, and some of the reactions are statically indeterminate. If the reactions involve fewer than six unknowns, you have more equations than unknowns, and some of the equations of equilibrium cannot be satisfied under general loading conditions. In this case, the rigid body is only partially constrained. Under the particular loading conditions corresponding to a given problem, however, the extra equations often reduce to trivial identities, such as 0 5 0, and can be disregarded; although only partially constrained, the rigid body remains in equilibrium (see Sample Probs. 4.7 and 4.8). Even with six or more unknowns, it is possible that some equations of equilibrium are not satisfied. This can occur when the reactions associated with the given supports either are parallel or inter-sect the same line; the rigid body is then improperly constrained.
Photo 4.3 Universal joints, seen on the drive shafts of rear-wheel-drive cars and trucks, allow rotational motion to be transferred between two noncollinear shafts.
Photo 4.4 This pillow block bearing supports the shaft of a fan used in an industrial facility.
206 Equilibrium of Rigid Bodies Ball Frictionless surface Force with known line of action, perpendicular to surface (one unknown) Force with known line of action, along cable (one unknown) Cable F F Roller on rough surface Rough surface Universal joint Hinge and bearing supporting radial load only Wheel on rail Two force components, one perpendicular to surface and one parallel to axis of wheel Three force components, mutually perpendicular at point of contact Three force components, one couple Three force components, three couples (no translation, no rotation) Three force components and up to two couples Two force components and up to two couples Fx Fx Mx Fy Fz Fx Fy Fz Fy Fz Fy Fz My (Mz) (My) (Mz) (My) Mz Ball and socket Fixed support Hinge and bearing supporting axial thrust and radial load Pin and bracket Fy Fz Fx Mx Fy Fz Fig. 4.10 Reactions at supports and connections in three dimensions.
4.3 Equilibrium in Three Dimensions 207 Sample Problem 4.7 A 20-kg ladder used to reach high shelves in a storeroom is supported by two flanged wheels A and B mounted on a rail and by a flangeless wheel C resting against a rail fixed to the wall. An 80-kg man stands on the ladder and leans to the right. The line of action of the combined weight W of the man and ladder intersects the floor at point D. Determine the reactions at A, B, and C.
STRATEGY: Draw a free-body diagram of the ladder, then write and solve the equilibrium equations in three dimensions.
MODELING: Free-Body Diagram. The combined weight of the man and ladder is W 5 2mgj 5 2(80 kg 1 20 kg)(9.81 m/s2)j 5 2(981 N)j You have five unknown reaction components: two at each flanged wheel and one at the flangeless wheel (Fig. 1). The ladder is thus only partially constrained; it is free to roll along the rails. It is, however, in equilibrium under the given load because the equation oFx 5 0 is satisfied.
ANALYSIS: Equilibrium Equations. The forces acting on the ladder form a sys-tem equivalent to zero: oF 5 0: Ayj 1 Azk 1 Byj 1 Bzk 2 (981 N)j 1 Ck 5 0 (Ay 1 By 2 981 N)j 1 (Az 1 Bz 1 C)k 5 0 (1) oMA 5 o(r 3 F) 5 0: 1.2i 3 (Byj 1 Bzk) 1 (0.9i 2 0.6k) 3 (2981j) 1 (0.6i 1 3j 2 1.2k) 3 Ck 5 0 Computing the vector products gives you† 1.2Byk 2 1.2Bzj 2 882.9k 2 588.6i 2 0.6Cj 1 3Ci 5 0 (3C 2 588.6)i 2 (1.2Bz 1 0.6C)j 1 (1.2By 2 882.9)k 5 0 (2) Setting the coefficients of i, j, and k equal to zero in Eq. (2) produces the following three scalar equations, which state that the sum of the moments about each coordinate axis must be zero: 3C 2 588.6 5 0 C 5 1196.2 N 1.2Bz 1 0.6C 5 0 Bz 5 298.1 N 1.2By 2 882.9 5 0 By 5 1736 N The reactions at B and C are therefore B 5 1(736 N)j 2 (98.1 N)k C 5 1(196.2 N)k b †The moments in this sample problem, as well as in Sample Probs. 4.8 and 4.9, also can be expressed as determinants (see Sample Prob. 3.10).
A B C D 0.6 m 0.6 m 0.9 m 0.3 m W 3 m Fig. 1 Free-body diagram of ladder.
A 0.6 m 0.6 m 0.9 m 0.3 m x y z Ck –(981 N)j Ayj Azk Bzk Byj 3 m 208 Equilibrium of Rigid Bodies Setting the coefficients of j and k equal to zero in Eq. (1), you obtain two scalar equations stating that the sums of the components in the y and z directions are zero. Substitute the values above for By, Bz, and C to get Ay 1 By 2 981 5 0 Ay 1 736 2 981 5 0 Ay 5 1245 N Az 1 Bz 1 C 5 0 Az 2 98.1 1 196.2 5 0 Az 5 298.1 N Therefore, the reaction at A is A 5 1(245 N)j 2 (98.1 N)k b REFLECT and THINK: You summed moments about A as part of the analysis. As a check, you could now use these results and demonstrate that the sum of moments about any other point, such as point B, is also zero.
Sample Problem 4.8 A 5 3 8-ft sign of uniform density weighs 270 lb and is supported by a ball-and-socket joint at A and by two cables. Determine the tension in each cable and the reaction at A.
STRATEGY: Draw a free-body diagram of the sign, and express the unknown cable tensions as Cartesian vectors. Then determine the cable tensions and the reaction at A by writing and solving the equilibrium equations.
MODELING: Free-Body Diagram. The forces acting on the sign are its weight W 5 2(270 lb)j and the reactions at A, B, and E (Fig. 1). The reaction at A is a force of unknown direction represented by three unknown components. Since the directions of the forces exerted by the cables are known, these forces involve only one unknown each: specifically, the magnitudes TBD and TEC. The total of five unknowns means that the sign is partially con-strained. It can rotate freely about the x axis; it is, however, in equilibrium under the given loading, since the equation oMx 5 0 is satisfied.
ANALYSIS: You can express the components of the forces TBD and TEC in terms of the unknown magnitudes TBD and TEC as follows: BD 5 2(8 ft)i 1 (4 ft)j 2 (8 ft)k BD 5 12 ft EC 5 2(6 ft)i 1 (3 ft)j 1 (2 ft)k EC 5 7 ft TBD 5 TBDaBD BDb 5 TBD(22 3i 1 1 3j 2 2 3k) TEC 5 TEC aEC ECb 5 TEC(26 7i 1 3 7j 2 2 7k) A B C D E x y z 6 ft 2 ft 2 ft 5 ft 4 ft 8 ft 3 ft Fig. 1 Free-body diagram of sign.
W = – (270 lb)j A xi Azk A yj TEC TBD A B C D E x y z 6 ft 2 ft 2 ft 4 ft 4 ft 4 ft 8 ft 3 ft 4.3 Equilibrium in Three Dimensions 209 Equilibrium Equations. The forces acting on the sign form a system equivalent to zero: oF 5 0: Axi 1 Ayj 1 Azk 1 TBD 1 TEC 2 (270 lb)j 5 0 (Ax 2 2 3 TBD 2 6 7 TEC)i 1 (Ay 1 1 3 TBD 1 3 7 TEC 2 270 lb)j 1 (Az 2 2 3 TBD 1 2 7 TEC)k 5 0 (1) oMA 5 o(r 3 F) 5 0: (8 ft)i 3 TBD(22 3 i 1 1 3 j 2 2 3 k) 1 (6 ft)i 3 TEC(26 7 i 1 3 7 j 1 2 7 k) 1 (4 ft)i 3 (2270 lb)j 5 0 (2.667TBD 1 2.571TEC 2 1080 lb)k 1 (5.333TBD 2 1.714TEC)j 5 0 (2) Setting the coefficients of j and k equal to zero in Eq. (2) yields two scalar equations that can be solved for TBD and TEC: TBD 5 101.3 lb TEC 5 315 lb b Setting the coefficients of i, j, and k equal to zero in Eq. (1) produces three more equations, which yield the components of A. A 5 1(338 lb)i 1 (101.2 lb)j 2 (22.5 lb)k b REFLECT and THINK: Cables can only act in tension, and the free-body diagram and Cartesian vector expressions for the cables were con-sistent with this. The solution yielded positive results for the cable forces, which confirms that they are in tension and validates the analysis. Sample Problem 4.9 A uniform pipe cover of radius r 5 240 mm and mass 30 kg is held in a horizontal position by the cable CD. Assuming that the bearing at B does not exert any axial thrust, determine the tension in the cable and the reactions at A and B.
r = 240 mm A B C D 160 mm 240 mm 240 mm 240 mm 210 Equilibrium of Rigid Bodies STRATEGY: Draw a free-body diagram with the coordinate axes shown (Fig. 1) and express the unknown cable tension as a Cartesian vector. Then apply the equilibrium equations to determine this tension and the support reactions.
MODELING: Free-Body Diagram. The forces acting on the free body include its weight, which is W 5 2mgj 5 2(30 kg)(9.81 m/s2)j 5 2(294 N)j The reactions involve six unknowns: the magnitude of the force T exerted by the cable, three force components at hinge A, and two at hinge B. Express the components of T in terms of the unknown magnitude T by resolving the vector DC into rectangular components: DC 5 2(480 mm)i 1 (240 mm)j 2 (160 mm)k DC 5 560 mm T 5 T DC DC 5 26 7 T i 1 3 7 T j 2 2 7 T k ANALYSIS: Equilibrium Equations. The forces acting on the pipe cover form a system equivalent to zero. Thus, oF 5 0: Axi 1 Ayj 1 Azk 1 Bxi 1 Byj 1 T 2 (294 N)j 5 0 (Ax 1 Bx 2 6 7T )i 1 (Ay 1 By 1 3 7T 2 294 N)j 1 (Az 2 2 7T )k 5 0 (1) oMB 5 o(r 3 F) 5 0: 2rk 3 (Axi 1 Ayj 1 Azk) 1 (2ri 1 rk) 3 (2 6 7T i 1 3 7T j 2 2 7T k) 1 (ri 1 rk) 3 (2294 N)j 5 0 (22Ay 2 3 7T 1 294 N)ri 1 (2Ax 2 2 7T )rj 1 ( 6 7 T 2 294 N)rk 5 0 (2) Setting the coefficients of the unit vectors equal to zero in Eq. (2) gives three scalar equations, which yield Ax 5 149.0 N Ay 5 173.5 N T 5 343 N b Setting the coefficients of the unit vectors equal to zero in Eq. (1) produces three more scalar equations. After substituting the values of T, Ax, and Ay into these equations, you obtain Az 5 198.0 N Bx 5 1245 N By 5 173.5 N The reactions at A and B are therefore A 5 1(49.0 N)i 1 (73.5 N)j 1 (98.0 N)k b B 5 1(245 N)i 1 (73.5 N)j b REFLECT and THINK: As a check, you can determine the tension in the cable using a scalar analysis. Assigning signs by the right-hand rule (rhr), we have (1rhr) oMz 5 0: 3 7T(0.48 m) 2 (294 N)(0.24 m) 5 0 T 5 343 N b r = 240 mm A B C D W = – (294 N)j Bxi Byj A xi Ayj Azk 160 mm 80 mm T r = 240 mm r = 240 mm x y z 240 mm Fig. 1 Free-body diagram of pipe cover.
4.3 Equilibrium in Three Dimensions 211 Sample Problem 4.10 A 450-lb load hangs from the corner C of a rigid piece of pipe ABCD that has been bent as shown. The pipe is supported by ball-and-socket joints A and D, which are fastened, respectively, to the floor and to a vertical wall, and by a cable attached at the midpoint E of the portion BC of the pipe and at a point G on the wall. Determine (a) where G should be located if the tension in the cable is to be minimum, (b) the correspond-ing minimum value of the tension.
12 ft 12 ft 450 lb A B C D E G 6 ft 6 ft 6 ft STRATEGY: Draw the free-body diagram of the pipe showing the reac-tions at A and D. Isolate the unknown tension T and the known weight W by summing moments about the diagonal line AD, and compute values from the equilibrium equations.
MODELING and ANALYSIS: Free-Body Diagram. The free-body diagram of the pipe includes the load W 5 (2450 lb)j, the reactions at A and D, and the force T exerted by the cable (Fig. 1). To eliminate the reactions at A and D from the computations, take the sum of the moments of the forces about the line AD and set it equal to zero. Denote the unit vector along AD by λ, which enables you to write oMAD 5 0: l ? (AE 3 T) 1 l ? (AC 3 W) 5 0 (1) Fig. 1 Free-body diagram of pipe.
A B C D E x y z T Dxi Dyj Dzk A xi Ay j Azk W = –450 j 6 ft 6 ft 12 ft 12 ft 12 ft 212 Equilibrium of Rigid Bodies You can compute the second term in Eq. (1) as follows: AC 3 W 5 (12i 1 12j) 3 (2450j) 5 25400k l 5 AD AD 5 12i 1 12j 2 6k 18 5 2 3 i 1 2 3 j 2 1 3 k l ? (AC 3 W) 5 (2 3 i 1 2 3 j 2 1 3 k) ? (25400k) 5 11800 Substituting this value into Eq. (1) gives l ? (AE 3 T) 5 21800 lb?ft (2) Minimum Value of Tension. Recalling the commutative property for mixed triple products, you can rewrite Eq. (2) in the form T ? (l 3 AE ) 5 21800 lb?ft (3) This shows that the projection of T on the vector λ 3 AE is a constant. It follows that T is minimum when it is parallel to the vector l 3 AE 5 (2 3 i 1 2 3 j 2 1 3 k) 3 (6i 1 12j) 5 4i 2 2j 1 4k The corresponding unit vector is 2 3 i 2 1 3 j 1 2 3 k, which gives Tmin 5 T(2 3 i 2 1 3 j 1 2 3 k) (4) Substituting for T and l 3 AE in Eq. (3) and computing the dot products yields 6T 5 21800 and, thus, T 5 2300. Carrying this value into Eq. (4) gives you Tmin 5 2200i 1 100j 2 200k Tmin 5 300 lb b Location of G. Since the vector EG and the force Tmin have the same direction, their components must be proportional. Denoting the coordi-nates of G by x, y, and 0 (Fig. 2), you get x 2 6 2200 5 y 2 12 1100 5 0 2 6 2200 x 5 0 y 5 15 ft b Fig. 2 Location of point G for minimum tension in cable.
A B C D G(x, y, 0) E(6, 12, 6) x y z W Tmin REFLECT and THINK: Sometimes you have to rely on the vector analysis presented in Chapters 2 and 3 as much as on the conditions for equilibrium described in this chapter.
213 213 SOLVING PROBLEMS ON YOUR OWN I n this section, you considered the equilibrium of a three-dimensional body. It is again most important that you draw a complete free-body diagram as the first step of your solution.
1. Pay particular attention to the reactions at the supports as you draw the free-body diagram. The number of unknowns at a support can range from one to six (Fig. 4.10). To decide whether an unknown reaction or reaction component exists at a support, ask yourself whether the support prevents motion of the body in a certain direction or about a certain axis.
a. If motion is prevented in a certain direction, include in your free-body diagram an unknown reaction or reaction component that acts in the same direction. b. If a support prevents rotation about a certain axis, include in your free-body diagram a couple of unknown magnitude that acts about the same axis.
2. The external forces acting on a three-dimensional body form a system equiva-lent to zero. Writing oF 5 0 and oMA 5 0 about an appropriate point A and setting the coefficients of i, j, k in both equations equal to zero provides you with six scalar equations. In general, these equations contain six unknowns and may be solved for these unknowns.
3. After completing your free-body diagram, you may want to seek equations involving as few unknowns as possible. The following strategies may help you. a. By summing moments about a ball-and-socket support or a hinge, you obtain equations from which three unknown reaction components have been eliminated [Sample Probs. 4.8 and 4.9]. b. If you can draw an axis through the points of application of all but one of the unknown reactions, summing moments about that axis will yield an equation in a single unknown [Sample Prob. 4.10].
4. After drawing your free-body diagram, you may find that one of the following situations exists. a. The reactions involve fewer than six unknowns. The body is partially con-strained and motion of the body is possible. However, you may be able to determine the reactions for a given loading condition [Sample Prob. 4.7]. b. The reactions involve more than six unknowns. The reactions are statically indeterminate. Although you may be able to calculate one or two reactions, you cannot determine all of them [Sample Prob. 4.10]. c. The reactions are parallel or intersect the same line. The body is improperly constrained, and motion can occur under a general loading condition.
214 Problems FREE-BODY PRACTICE PROBLEMS 4.F5 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 1.5 in. and the radius of spool C is 2 in. Knowing that TB 5 20 lb and that the system rotates at a constant rate, draw the free-body diagram needed to determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle.
TB x D C B A z y 4.5 in.
4.5 in.
6 in.
TC Fig. P4.F5 4.F6 A 12-m pole supports a horizontal cable CD and is held by a ball and socket at A and two cables BE and BF. Knowing that the ten-sion in cable CD is 14 kN and assuming that CD is parallel to the x axis (ϕ 5 0), draw the free-body diagram needed to determine the tension in cables BE and BF and the reaction at A. 4.F7 A 20-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 30° with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Draw the free-body diagram needed to determine the magnitude of the force exerted by the brace and the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust.
E C D z A y x B 0.9 m 0.9 m 0.6 m 30° Fig. P4.F7 C B F E A D x z 6 m 6 m 8 m 12 m 7.5 m y f Fig. P4.F6 215 END-OF-SECTION PROBLEMS 4.91 Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Know-ing that when the system is at rest, the tension is 90 N in both portions of belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust. 4.92 Solve Prob. 4.91, assuming that the pulley rotates at a constant rate and that TB 5 104 N, T 9 B 5 84 N, and TC 5 175 N. 4.93 A small winch is used to raise a 120-lb load. Find (a) the magnitude of the vertical force P that should be applied at C to maintain equi-librium in the position shown, (b) the reactions at A and B, assuming that the bearing at B does not exert any axial thrust.
Fig. P4.93 120 lb P A B C x y z 10 in.
10 in.
9 in.
10 in.
8 in.
3 in.
30° 4.94 A 4 3 8-ft sheet of plywood weighing 34 lb has been temporarily placed among three pipe supports. The lower edge of the sheet rests on small collars at A and B and its upper edge leans against pipe C. Neglecting friction at all surfaces, determine the reactions at A, B, and C.
Fig. P4.94 4 ft y z B A x 1 ft 3.75 ft 3 ft 5 ft 3 ft 4 ft C 4.95 A 250 3 400-mm plate of mass 12 kg and a 300-mm-diameter pulley are welded to axle AC that is supported by bearings at A and B. For β 5 30°, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the bearing at B does not exert any axial thrust. 4.96 Solve Prob. 4.95 for β 5 60°.
Fig. P4.91 z A T9B T9C x y B 150 mm 200 mm 100 mm TC TB C D T B A z y x D C 150 mm 160 mm 160 mm 250 mm 200 mm 400 mm b Fig. P4.95 216 4.97 The 20 3 20-in. square plate shown weighs 56 lb and is supported by three vertical wires. Determine the tension in each wire.
4.98 The 20 3 20-in. square plate shown weighs 56 lb and is supported by three vertical wires. Determine the weight and location of the lightest block that should be placed on the plate if the tensions in the three wires are to be equal. 4.99 An opening in a floor is covered by a 1 3 1.2-m sheet of plywood with a mass of 18 kg. The sheet is hinged at A and B and is main-tained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C.
y z x A B E C 0.15 m 0.2 m 0.2 m 0.6 m 1.2 m D Fig. P4.99 4.100 Solve Prob. 4.99, assuming that the small block C is moved and placed under edge DE at a point 0.15 m from corner E. 4.101 Two steel pipes AB and BC, each having a mass per unit length of 8 kg/m, are welded together at B and supported by three vertical wires. Knowing that a 5 0.4 m, determine the tension in each wire.
B A C D y x z a 1.2 m 0.6 m Fig. P4.101 x y z B C A 10 in.
10 in.
16 in.
4 in.
16 in.
Fig. P4.97 and P4.98 217 4.102 For the pipe assembly of Prob. 4.101, determine (a) the largest per-missible value of a if the assembly is not to tip, (b) the corresponding tension in each wire. 4.103 The 24-lb square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when a 5 10 in., (b) the value of a for which the tension in each wire is 8 lb. 4.104 The table shown weighs 30 lb and has a diameter of 4 ft. It is sup-ported by three legs equally spaced around the edge. A vertical load P with a magnitude of 100 lb is applied to the top of the table at D. Determine the maximum value of a if the table is not to tip over. Show, on a sketch, the area of the table over which P can act without tipping the table.
A B C D a P Fig. P4.104 4.105 A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A. 4.106 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a 5 3 m, determine the tension in each cable and the reaction at A.
A B C F x y z D E 455 N 1.5 m 1.5 m a 2 m 3 m 3 m 3 m 3 m Fig. P4.106 4.107 Solve Prob. 4.106 for a 5 1.5 m. 4.108 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. 4.109 Solve Prob. 4.108, assuming that the 3.6-kN load is applied at point A.
y x B C A a 30 in.
a 30 in.
z Fig. P4.103 840 lb x y z E A B C D 4 ft 6 ft 7 ft 6 ft 6 ft Fig. P4.105 A B C x y z D E 3.6 kN 1.2 m 1.2 m 1.2 m 0.6 m 0.8 m 0.8 m Fig. P4.108 218 4.110 The 10-ft flagpole AC forms an angle of 30° with the z axis. It is held by a ball-and-socket joint at C and by two thin braces BD and BE. Knowing that the distance BC is 3 ft, determine the tension in each brace and the reaction at C. 4.111 A 48-in. boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C.
Fig. P4.111 A B C F x y z D E 20 in.
16 in.
320 lb 30 in.
20 in.
48 in. 4.112 Solve Prob. 4.111, assuming that the 320-lb load is applied at A. 4.113 A 10-kg storm window measuring 900 3 1500 mm is held by hinges at A and B. In the position shown, it is held away from the side of the house by a 600-mm stick CD. Assuming that the hinge at A does not exert any axial thrust, determine the magnitude of the force exerted by the stick and the components of the reactions at A and B.
Fig. P4.113 x z y A C D E B 1500 mm 1500 mm 900 mm 4.114 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.
z x y A 3 ft 3 ft 30° 3 ft 75 lb B C D E Fig. P4.110 250 mm 50 mm 300 mm 400 N C D E F x z 50 mm 250 mm A B H y 30° Fig. P4.114 219 4.115 The horizontal platform ABCD weighs 60 lb and supports a 240-lb load at its center. The platform is normally held in position by hinges at A and B and by braces CE and DE. If brace DE is removed, determine the reactions at the hinges and the force exerted by the remaining brace CE. The hinge at A does not exert any axial thrust. 4.116 The lid of a roof scuttle weighs 75 lb. It is hinged at corners A and B and maintained in the desired position by a rod CD pivoted at C. A pin at end D of the rod fits into one of several holes drilled in the edge of the lid. For α 5 50°, determine (a) the magnitude of the force exerted by rod CD, (b) the reactions at the hinges. Assume that the hinge at B does not exert any axial thrust.
x y B A D z 26 in.
C 15 in.
7 in.
32 in.
α Fig. P4.116 4.117 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust.
690 mm 960 mm x y z E D A B C 675 mm 90 mm 450 mm 270 mm 90 mm Fig. P4.117 4.118 Solve Prob. 4.117, assuming that cable DCE is replaced by a cable attached to point E and hook C.
D z x y B A 4 ft 2 ft 2 ft 3 ft 300 lb E C Fig. P4.115 220 4.119 Solve Prob. 4.113, assuming that the hinge at A has been removed and that the hinge at B can exert couples about axes parallel to the x and y axes. 4.120 Solve Prob. 4.115, assuming that the hinge at B has been removed and that the hinge at A can exert an axial thrust, as well as couples about axes parallel to the x and y axes. 4.121 The assembly shown is used to control the tension T in a tape that passes around a frictionless spool at E. Collar C is welded to rods ABC and CDE. It can rotate about shaft FG but its motion along the shaft is prevented by a washer S. For the loading shown, determine (a) the tension T in the tape, (b) the reaction at C. 4.122 The assembly shown is welded to collar A that fits on the vertical pin shown. The pin can exert couples about the x and z axes but does not prevent motion about or along the y axis. For the loading shown, determine the tension in each cable and the reaction at A.
Fig. P4.122 480 N A C D E F x y z 60 mm 45 mm 90 mm 120 mm 80 mm 4.123 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 1.8-kN load is applied at F, determine the tension in each cable.
Fig. P4.123 x z y A B C D E F 1.8 kN 240 mm 320 mm 210 mm 210 mm 420 mm 420 mm T A B C F D E S G x y z 6 lb 2 in.
1.6 in.
4.2 in.
2.4 in.
T Fig. P4.121 221 4.124 Solve Prob. 4.123, assuming that the 1.8-kN load is applied at C.
4.125 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A.
Fig. P4.125 x y z A B C D E F G J H 24 lb 24 lb 9 in.
16 in.
16 in.
8 in.
12 in.
16 in.
8 in.
8 in.
8 in.
O 4.126 Solve Prob. 4.125, assuming that the load at C has been removed. 4.127 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P 5 240 lb, a 5 12 in., b 5 8 in., and c 5 10 in.
Fig. P4.127 x y z b c A B C P a 4.128 Solve Prob. 4.127, assuming that the force P is removed and is replaced by a couple M 5 1 (600 lb?in.)j acting at B.
222 4.129 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a 5 150 mm, determine the tension in each cable and the reaction at A.
Fig. P4.129 and P4.130 A B C H D E F G x y z 140 mm 350 N 300 mm 140 mm 200 mm a 480 mm 4.130 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. Knowing that the 350-N load is applied at D (a 5 300 mm), determine the tension in each cable and the reaction at A. 4.131 The assembly shown consists of an 80-mm rod AF that is welded to a cross frame consisting of four 200-mm arms. The assembly is supported by a ball-and-socket joint at F and by three short links, each of which forms an angle of 45° with the vertical. For the load-ing shown, determine (a) the tension in each link, (b) the reaction at F.
x y z E F A B P C D 45º 45º 45º 200 mm 200 mm 200 mm 200 mm 80 mm Fig. P4.131 4.132 The uniform 10-kg rod AB is supported by a ball-and-socket joint at A and by the cord CG that is attached to the midpoint G of the rod. Knowing that the rod leans against a frictionless vertical wall at B, determine (a) the tension in the cord, (b) the reactions at A and B.
x y z G O A B C 150 mm 150 mm 400 mm 600 mm Fig. P4.132 223 4.133 The frame ACD is supported by ball-and-socket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the frame supports at point C a load of magnitude P 5 268 N, determine the tension in the cable.
x y z A B C D G O P H 0.35 m 0.875 m 0.75 m 0.75 m 0.925 m 0.5 m 0.5 m Fig. P4.133 4.134 Solve Prob. 4.133, assuming that cable GBH is replaced by a cable GB attached at G and B. 4.135 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 60-lb load is applied at C as shown, determine the tension in the cable.
x y z A B C D E F 8 in.
7 in.
9 in.
60 lb 11 in.
16 in.
10 in.
14 in.
Fig. P4.135 4.136 Solve Prob. 4.135, assuming that cable DF is replaced by a cable connecting B and F. 4.137 Two rectangular plates are welded together to form the assembly shown. The assembly is supported by ball-and-socket joints at B and D and by a ball on a horizontal surface at C. For the loading shown, determine the reaction at C.
B A y z D C x 6 in.
12 in.
8 in.
9 in.
80 lb Fig. P4.137 224 4.138 The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown.
640 N x y z A B C D E F 240 mm 160 mm 480 mm 200 mm 490 mm Fig. P4.138 4.139 Solve Prob. 4.138, assuming that wire DF is replaced by a wire connecting C and F. 4.140 Two 2 3 4-ft plywood panels, each with a weight of 12 lb, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension.
z 2 ft 2 ft 2 ft 2 ft 2 ft x x y y A B C O H D E F 12 lb 12 lb Fig. P4.140 4.141 Solve Prob. 4.140, subject to the restriction that H must lie on the y axis.
225 Review and Summary Equilibrium Equations This chapter was devoted to the study of the equilibrium of rigid bodies, i.e., to the situation when the external forces acting on a rigid body form a system equivalent to zero [Introduction]. We then have oF 5 0 oMO 5 o(r 3 F) 5 0 (4.1) Resolving each force and each moment into its rectangular components, we can express the necessary and sufficient conditions for the equilibrium of a rigid body with the following six scalar equations: oFx 5 0 oFy 5 0 oFz 5 0 (4.2) oMx 5 0 oMy 5 0 oMz 5 0 (4.3) We can use these equations to determine unknown forces applied to the rigid body or unknown reactions exerted by its supports.
Free-Body Diagram When solving a problem involving the equilibrium of a rigid body, it is essential to consider all of the forces acting on the body. Therefore, the first step in the solution of the problem should be to draw a free-body diagram showing the body under consideration and all of the unknown as well as known forces acting on it.
Equilibrium of a Two-Dimensional Structure In the first part of this chapter, we considered the equilibrium of a two-dimensional structure; i.e., we assumed that the structure considered and the forces applied to it were contained in the same plane. We saw that each of the reactions exerted on the structure by its supports could involve one, two, or three unknowns, depending upon the type of support [Sec. 4.1A]. In the case of a two-dimensional structure, the equations given previ-ously reduce to three equilibrium equations: oFx 5 0 oFy 5 0 oMA 5 0 (4.5) where A is an arbitrary point in the plane of the structure [Sec. 4.1B]. We can use these equations to solve for three unknowns. Although the three equilib-rium equations (4.5) cannot be augmented with additional equations, any of them can be replaced by another equation. Therefore, we can write alternative sets of equilibrium equations, such as oFx 5 0 oMA 5 0 oMB 5 0 (4.6) where point B is chosen in such a way that the line AB is not parallel to the y axis, or oMA 5 0 oMB 5 0 oMC 5 0 (4.7) where the points A, B, and C do not lie in a straight line.
Static Indeterminacy, Partial Constraints, Improper Constraints Since any set of equilibrium equations can be solved for only three unknowns, the reactions at the supports of a rigid two-dimensional structure cannot be 226 completely determined if they involve more than three unknowns; they are said to be statically indeterminate [Sec. 4.1C]. On the other hand, if the reac-tions involve fewer than three unknowns, equilibrium is not maintained under general loading conditions; the structure is said to be partially constrained.
The fact that the reactions involve exactly three unknowns is no guarantee that you can solve the equilibrium equations for all three unknowns. If the supports are arranged in such a way that the reactions are either concurrent or parallel, the reactions are statically indeterminate, and the structure is said to be improperly constrained.
Two-Force Body, Three-Force Body We gave special attention in Sec. 4.2 to two particular cases of equilibrium of a rigid body. We defined a two-force body as a rigid body subjected to forces at only two points, and we showed that the resultants F1 and F2 of these forces must have the same magnitude, the same line of action, and opposite sense (Fig. 4.11), which is a property that simplifies the solution of certain problems in later chapters. We defined a three-force body as a rigid body subjected to forces at only three points, and we demonstrated that the resultants F1, F2, and F3 of these forces must be either concurrent (Fig. 4.12) or parallel. This property provides us with an alternative approach to the solution of problems involving a three-force body [Sample Prob. 4.6].
Equilibrium of a Three-Dimensional Body In the second part of this chapter, we considered the equilibrium of a three-dimensional body. We saw that each of the reactions exerted on the body by its supports could involve between one and six unknowns, depending upon the type of support [Sec. 4.3A]. In the general case of the equilibrium of a three-dimensional body, all six of the scalar equilibrium equations (4.2) and (4.3) should be used and solved for six unknowns [Sec. 4.3B]. In most problems, however, we can obtain these equations more conveniently if we start from oF 5 0 oMO 5 o (r 3 F) 5 0 (4.1) and then express the forces F and position vectors r in terms of scalar com-ponents and unit vectors. We can compute the vector products either directly or by means of determinants, and obtain the desired scalar equations by equat-ing to zero the coefficients of the unit vectors [Sample Probs. 4.7 through 4.9]. We noted that we may eliminate as many as three unknown reaction components from the computation of oMO in the second of the relations (4.1) through a judicious choice of point O. Also, we can eliminate the reactions at two points A and B from the solution of some problems by writing the equation oMAB 5 0, which involves the computation of the moments of the forces about an axis AB joining points A and B [Sample Prob. 4.10]. We observed that when a body is subjected to individual couples Mi, either as applied loads or as support reactions, we can include these couples by expressing the second part of Eq. (4.1) as oMO 5 o(r 3 F) 1 oMi 5 0 (4.19) If the reactions involve more than six unknowns, some of the reactions are statically indeterminate; if they involve fewer than six unknowns, the rigid body is only partially constrained. Even with six or more unknowns, the rigid body is improperly constrained if the reactions associated with the given sup-ports are either parallel or intersect the same line.
Fig. 4.11 A B F1 F2 Fig. 4.12 F2 F3 F1 B C D A 227 4.142 A 3200-lb forklift truck is used to lift a 1700-lb crate. Determine the reaction at each of the two (a) front wheels A, (b) rear wheels B.
4.143 The lever BCD is hinged at C and attached to a control rod at B. If P 5 100 lb, determine (a) the tension in rod AB, (b) the reaction at C.
P A B C D 7.5 in.
3 in.
4 in.
90° Fig. P4.143 4.144 Determine the reactions at A and B when (a) h 5 0, (b) h 5 200 mm.
60° 300 mm 250 mm 250 mm 150 N G B A h Fig. P4.144 4.145 Neglecting friction and the radius of the pulley, determine (a) the tension in cable ADB, (b) the reaction at C.
A B C 150 mm 200 mm 80 mm 80 mm 120 N D Fig. P4.145 Review Problems B A 12 in.
G' G 16 in.
24 in.
Fig. P4.142 228 4.146 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α 5 0), determine the tension in the cord and the reactions at A and C. 4.147 A slender rod AB, with a weight of W, is attached to blocks A and B that move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when θ 5 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ when W 5 75 lb, l 5 30 in., and k 5 3 lb/in.
A B W q l Fig. P4.147 4.148 Determine the reactions at A and B when a 5 150 mm. 4.149 For the frame and loading shown, determine the reactions at A and C.
A B C D 30 lb 4 in.
6 in.
3 in.
Fig. P4.149 4.150 A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N verti-cal load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.
T 720 N y 80 mm 120 mm 120 mm 200 mm A E B C D x z 40 mm Fig. P4.150 A B E C D 30° 80 N a 0.2 m 0.2 m 30° 0.2 m Fig. P4.146 A B 240 mm 80 mm 320 N a Fig. P4.148 229 4.151 The 45-lb square plate shown is supported by three vertical wires. Determine the tension in each wire.
4.152 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.
x y z D H F E A B C 25 in.
20 in.
4 in.
12 in.
8 in.
4 in.
30 in.
Fig. P4.152 4.153 A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case, if possible, determine the reactions at the supports.
A A B B C C P A B C P P A B C P 45° 45° (a) (b) (c) (d) a = 30° 30° a a a a a a a a a a a a Fig. P4.153 A C B z x y 20 in.
20 in.
15 in.
5 in.
Fig. P4.151 Loads on dams include three types of distributed forces: the weights of its constituent elements, the pressure forces exerted by the water on its submerged face, and the pressure forces exerted by the ground on its base.
Distributed Forces: Centroids and Centers of Gravity 5 Introduction We have assumed so far that we could represent the attraction exerted by the earth on a rigid body by a single force W. This force, called the force due to gravity or the weight of the body, is applied at the center of gravity of the body (Sec. 3.1A). Actually, the earth exerts a force on each of the particles forming the body, so we should represent the attraction of the earth on a rigid body by a large number of small forces distributed over the entire body. You will see in this chapter, however, that all of these small forces can be replaced by a single equivalent force W. You will also see how to determine the center of gravity—i.e., the point of application of the resultant W—for bodies of various shapes.
In the first part of this chapter, we study two-dimensional bodies, such as flat plates and wires contained in a given plane. We introduce two concepts closely associated with determining the center of gravity of a plate or a wire: the centroid of an area or a line and the first moment of an area or a line with respect to a given axis. Computing the area of a surface of revolution or the volume of a body of revolution is directly related to determining the centroid of the line or area used to generate that surface or body of revolution (theorems of Pappus-Guldinus). Also, as we show in Sec. 5.3, the determination of the centroid of an area sim-plifies the analysis of beams subjected to distributed loads and the com-putation of the forces exerted on submerged rectangular surfaces, such as hydraulic gates and portions of dams.
In the last part of this chapter, you will see how to determine the center of gravity of a three-dimensional body as well as how to calculate the centroid of a volume and the first moments of that volume with respect to the coordinate planes.
Introduction 5.1 PLANAR CENTERS OF GRAVITY AND CENTROIDS 5.1A Center of Gravity of a Two-Dimensional Body 5.1B Centroids of Areas and Lines 5.1C First Moments of Areas and Lines 5.1D Composite Plates and Wires 5.2 FURTHER CONSIDERATIONS OF CENTROIDS 5.2A Determination of Centroids by Integration 5.2B Theorems of Pappus-Guldinus 5.3 ADDITIONAL APPLICATIONS OF CENTROIDS 5.3A Distributed Loads on Beams 5.3B Forces on Submerged Surfaces 5.4 CENTERS OF GRAVITY AND CENTROIDS OF VOLUMES 5.4A Three-Dimensional Centers of Gravity and Centroids 5.4B Composite Bodies 5.4C Determination of Centroids of Volumes by Integration Objectives • Describe the centers of gravity of two and three-dimensional bodies.
• Define the centroids of lines, areas, and volumes.
• Consider the fi rst moments of lines and areas, and examine their properties.
• Determine centroids of composite lines, areas, and volumes by summation methods.
• Determine centroids of composite lines, areas, and volumes by integration.
• Apply the theorems of Pappus-Guldinus to analyze surfaces and bodies of revolution.
• Analyze distributed loads on beams and forces on submerged surfaces.
Introduction 231 Photo 5.1 The precise balancing of the components of a mobile requires an understanding of centers of gravity and centroids, the main topics of this chapter.
232 Distributed Forces: Centroids and Centers of Gravity 5.1 PLANAR CENTERS OF GRAVITY AND CENTROIDS In Chapter 4, we showed how the locations of the lines of action of forces affects the replacement of a system of forces with an equivalent system of forces and couples. In this section, we extend this idea to show how a distributed system of forces (in particular, the elements of an object’s weight) can be replaced by a single resultant force acting at a specific point on an object. The specific point is called the object’s center of gravity.
5.1A Center of Gravity of a Two-Dimensional Body Let us first consider a flat horizontal plate (Fig. 5.1). We can divide the plate into n small elements. We denote the coordinates of the first element by x1 and y1, those of the second element by x2 and y2, etc. The forces exerted by the earth on the elements of the plate are denoted, respectively, by DW1, DW2, . . . , DWn. These forces or weights are directed toward the center of the earth; however, for all practical purposes, we can assume them to be parallel. Their resultant is therefore a single force in the same direction. The magnitude W of this force is obtained by adding the mag-nitudes of the elemental weights.
oFz: W 5 DW1 1 DW2 1 ? ? ? 1 DWn Fig. 5.1 The center of gravity of a plate is the point where the resultant weight of the plate acts. It is the weighted average of all the elements of weight that make up the plate.
O ∆W1 x y z y1 x1 O ∆W1 x y z (x2, y2) (x1, y1) (xm, yn) ∆W2 ∆Wn x y z G O x y W (a) Single element of the plate (b) Multiple elements of the plate (c) Center of gravity x 5 #x dW W y 5 #y dW W To obtain the coordinates x and y of point G where the resultant W should be applied, we note that the moments of W about the y and x axes are equal to the sum of the corresponding moments of the elemental weights: oMy: xW 5 x1DW1 1 x2DW2 1 ? ? ? 1 xnDWn oMx: yW 5 y1DW1 1 y2DW2 1 ? ? ? 1 ynDWn (5.1) Solving these equations for x and y gives us x 5 x1DW1 1 x2DW2 1 ? ? ? 1 xnDWn W y 5 y1DW1 1 y2DW2 1 ? ? ? 1 ynDWn W 5.1 Planar Centers of Gravity and Centroids 233 We could use these equations in this form to find the center of gravity of a collection of n objects, each with a weight of Wi.
If we now increase the number of elements into which we divide the plate and simultaneously decrease the size of each element, in the limit of infinitely many elements of infinitesimal size, we obtain the expressions Weight, center of gravity of a flat plate W 5 #dW x W 5 #x dW y W 5 #y dW (5.2) Or, solving for x and y, we have W 5 #dW x 5 #x dW W y 5 #y dW W (5.29) These equations define the weight W and the coordinates x and y of the center of gravity G of a flat plate. The same equations can be derived for a wire lying in the xy plane (Fig. 5.2). Note that the center of gravity G of a wire is usually not located on the wire.
W 5 #dW xW 5 #x dW yW 5 #y dW (a) Single element of the wire (b) Multiple elements of the wire (c) Center of gravity x 5 #x dW W y 5 #y dW W ∆W1 x y z O ∆Wn ∆W1 x y y1 z O x1 x y z O y W x G ∆W2 (x2, y2) (x1, y1) (xm, yn) Fig. 5.2 The center of gravity of a wire is the point where the resultant weight of the wire acts. The center of gravity may not actually be located on the wire.
5.1B Centroids of Areas and Lines In the case of a flat homogeneous plate of uniform thickness, we can express the magnitude DW of the weight of an element of the plate as DW 5 γ t DA where γ 5 specific weight (weight per unit volume) of the material t 5 thickness of the plate DA 5 area of the element Similarly, we can express the magnitude W of the weight of the entire plate as W 5 γ tA where A is the total area of the plate.
G Photo 5.2 The center of gravity of a boomerang is not located on the object itself.
234 Distributed Forces: Centroids and Centers of Gravity If U.S. customary units are used, the specific weight γ should be expressed in lb/ft3, the thickness t in feet, and the areas DA and A in square feet. Then DW and W are expressed in pounds. If SI units are used, γ should be expressed in N/m3, t in meters, and the areas DA and A in square meters; the weights DW and W are then expressed in newtons.† Substituting for DW and W in the moment equations (5.1) and divid-ing throughout by γ t, we obtain oMy: xA 5 x1 DA1 1 x2 DA2 1 ? ? ? 1 xn DAn oMx: yA 5 y1 DA1 1 y2 DA2 1 ? ? ? 1 yn DAn If we increase the number of elements into which the area A is divided and simultaneously decrease the size of each element, in the limit we obtain Centroid of an area A xA 5#x dA y A 5 #y dA (5.3) Or, solving for x and y, we obtain x 5 #x dA A y 5 #y dA A (5.39) These equations define the coordinates x and y of the center of gravity of a homogeneous plate. The point whose coordinates are x and y is also known as the centroid C of the area A of the plate (Fig. 5.3). If the plate is not homogeneous, you cannot use these equations to determine the center of gravity of the plate; they still define, however, the centroid of the area.
xA xA 5#x dA d y A 5 #y dA d O x x y y O x y A ∆A C y x O x y (a) Divide area into elements (b) Element DA at point x, y (c) Centroid located at x 5 #x dA A y 5 #y dA A Fig. 5.3 The centroid of an area is the point where a homogeneous plate of uniform thickness would balance.
†We should note that in the SI system of units, a given material is generally characterized by its density ρ (mass per unit volume) rather than by its specific weight γ. You can obtain the specific weight of the material from the relation γ 5 ρg where g 5 9.81 m/s2. Note that since ρ is expressed in kg/m3, the units of γ are (kg/m3)(m/s2), or N/m3.
5.1 Planar Centers of Gravity and Centroids 235 Fig. 5.4 The centroid of a line is the point where a homogeneous wire of uniform cross section would balance.
(a) Divide line into elements (b) Element DL at point x, y (c) Centroid located at x 5 #x dL L y 5 #y dL L x x y y O x y O y x ∆ L L C x y O In the case of a homogeneous wire of uniform cross section, we can express the magnitude DW of the weight of an element of wire as DW 5 γa DL where γ 5 specific weight of the material a 5 cross-sectional area of the wire DL 5 length of the element The center of gravity of the wire then coincides with the centroid C of the line L defining the shape of the wire (Fig. 5.4). We can obtain the coordinates x and y of the centroid of line L from the equations Centroid of a line L xL 5#x dL yL 5#y dL (5.4) Solving for x and y gives us x 5 #x dL L y 5 #y dL L (5.49) xL 5#x dL d yL 5#y dL d 5.1C First Moments of Areas and Lines The integral ∫ x dA in Eqs. (5.3) is known as the first moment of the area A with respect to the y axis and is denoted by Qy. Similarly, the integral ∫ y dA defines the first moment of A with respect to the x axis and is denoted by Qx. That is, First moments of area A Qy 5 #x dA Qx 5 #y dA (5.5) Qy 5 #xdA d Qx 5 #ydA d 236 Distributed Forces: Centroids and Centers of Gravity Comparing Eqs. (5.3) with Eqs. (5.5), we note that we can express the first moments of the area A as the products of the area and the coordinates of its centroid: Qy 5 xA Qx 5 yA (5.6) It follows from Eqs. (5.6) that we can obtain the coordinates of the centroid of an area by dividing the first moments of that area by the area itself. The first moments of the area are also useful in mechanics of mate-rials for determining the shearing stresses in beams under transverse load-ings. Finally, we observe from Eqs. (5.6) that, if the centroid of an area is located on a coordinate axis, the first moment of the area with respect to that axis is zero. Conversely, if the first moment of an area with respect to a coordinate axis is zero, the centroid of the area is located on that axis.
We can use equations similar to Eqs. (5.5) and (5.6) to define the first moments of a line with respect to the coordinate axes and to express these moments as the products of the length L of the line and the coordi-nates x and y of its centroid.
An area A is said to be symmetric with respect to an axis BB9 if for every point P of the area there exists a point P9 of the same area such that the line PP9 is perpendicular to BB9 and is divided into two equal parts by that axis (Fig. 5.5a). The axis BB9 is called an axis of symmetry. A line L is said to be symmetric with respect to an axis BB9 if it satisfies similar conditions. When an area A or a line L possesses an axis of sym-metry BB9, its first moment with respect to BB9 is zero, and its centroid is located on that axis. For example, note that, for the area A of Fig. 5.5b, which is symmetric with respect to the y axis, every element of area dA Qy 5 xA xA Qx 5 yA yA x x y O C A – x dA dA' P P' B' (a) (b) B Fig. 5.5 Symmetry about an axis. (a) The area is symmetric about the axis BB9. (b) The centroid of the area is located on the axis of symmetry.
5.1 Planar Centers of Gravity and Centroids 237 with abscissa x corresponds to an element dA9 of equal area and with abscissa 2x. It follows that the integral in the first of Eqs. (5.5) is zero and, thus, that Qy 5 0. It also follows from the first of the relations in Eq. (5.3) that x 5 0. Thus, if an area A or a line L possesses an axis of symmetry, its centroid C is located on that axis.
We further note that if an area or line possesses two axes of sym-metry, its centroid C must be located at the intersection of the two axes (Fig. 5.6). This property enables us to determine immediately the centroids of areas such as circles, ellipses, squares, rectangles, equilateral triangles, or other symmetric figures, as well as the centroids of lines in the shape of the circumference of a circle, the perimeter of a square, etc.
C C B B' B B' D D' D' D (a) (b) Fig. 5.6 If an area has two axes of symmetry, the centroid is located at their intersection. (a) An area with two axes of symmetry but no center of symmetry; (b) an area with two axes of symmetry and a center of symmetry.
We say that an area A is symmetric with respect to a center O if, for every element of area dA of coordinates x and y, there exists an ele-ment dA9 of equal area with coordinates 2x and 2y (Fig. 5.7). It then follows that the integrals in Eqs. (5.5) are both zero and that Qx 5 Qy 5 0. It also follows from Eqs. (5.3) that x 5 y 5 0; that is, that the centroid of the area coincides with its center of symmetry O. Similarly, if a line possesses a center of symmetry O, the centroid of the line coincides with the center O.
Note that a figure possessing a center of symmetry does not neces-sarily possess an axis of symmetry (Fig. 5.7), whereas a figure possessing two axes of symmetry does not necessarily possess a center of symmetry (Fig. 5.6a). However, if a figure possesses two axes of symmetry at right angles to each other, the point of intersection of these axes is a center of symmetry (Fig. 5.6b).
Determining the centroids of unsymmetrical areas and lines and of areas and lines possessing only one axis of symmetry will be discussed in the next section. Centroids of common shapes of areas and lines are shown in Fig. 5.8A and B.
5.1D Composite Plates and Wires In many instances, we can divide a flat plate into rectangles, triangles, or the other common shapes shown in Fig. 5.8A. We can determine the abscissa X of the plate’s center of gravity G from the abscissas x1, x2, . . . , xn of the centers of gravity of the various parts. To do this, we equate the moment of the weight of the whole plate about the y axis to the sum of x y O A dA dA' x y – y – x Fig. 5.7 An area may have a center of symmetry but no axis of symmetry.
238 Distributed Forces: Centroids and Centers of Gravity ⎯x O C r Shape Area ⎯x ⎯x ⎯x ⎯y ⎯y Triangular area Quarter-circular area Semicircular area Quarter-elliptical area Semielliptical area Semiparabolic area Parabolic area Parabolic spandrel General spandrel Circular sector C C C ⎯y ⎯y C C ⎯x ⎯x C C ⎯y ⎯y h 3 bh 2 b 2 b 2 4r 4r 3a 4 3a 8 3h 5 3h 10 2ah 3 3h 5 4ah 3 ah 3 4b 4 4r 2 2 4b 4 4a O O O O O O O O r h 0 a a a a a b y = kx2 h h n + 1 n + 2 a n + 1 4n + 2h ah n + 1 C 0 0 0 ⎯x ⎯y y = kxn h C 3α 2r sin α r2 α 3 3 3 3 3 3 r2 r2 ab ab Fig. 5.8A Centroids of common shapes of areas.
5.1 Planar Centers of Gravity and Centroids 239 the moments of the weights of the various parts about the same axis (Fig. 5.9). We can obtain the ordinate Y of the center of gravity of the plate in a similar way by equating moments about the x axis. Mathemati-cally, we have oMy: X(W1 1 W2 1 . . . 1 Wn) 5 x1W1 1 x2W2 1 . . . 1 xnWn oMx: Y(W1 1 W2 1 . . . 1 Wn) 5 y1W1 1 y2W2 1 . . . 1 ynWn x y r sin a a 2r 2r 2r 2 r r Shape Quarter-circular arc Semicircular arc Arc of circle Length 0 2a r 0 O O O C C r r C x y x a a Fig. 5.8B Centroids of common shapes of lines.
= x y z x y z O G ⎯X ⎯Y W1 W2 W3 G1 G2 G3 ΣW ΣMy : ⎯X Σ W = Σ ⎯x W ΣMx : ⎯Y Σ W = Σ ⎯y W O Fig. 5.9. We can determine the location of the center of gravity G of a composite plate from the centers of gravity G1, G2, . . . of the component plates.
In more condensed notation, this is Center of gravity of a composite plate X 5 o xW W Y 5 o yW W (5.7) X 5 oxW W Y 5 oyW W 240 Distributed Forces: Centroids and Centers of Gravity We can use these equations to find the coordinates X and Y of the center of gravity of the plate from the centers of gravity of its component parts.
If the plate is homogeneous and of uniform thickness, the center of gravity coincides with the centroid C of its area. We can determine the abscissa X of the centroid of the area by noting that we can express the first moment Qy of the composite area with respect to the y axis as (1) the product of X and the total area and (2) as the sum of the first moments of the elementary areas with respect to the y axis (Fig. 5.10). We obtain the = x y O C ⎯X ⎯Y A1 A3 A2 C1 C2 C3 ΣA Qy = ⎯X Σ A = Σ ⎯x A Qx = ⎯Y Σ A = Σ ⎯y A x y O Fig. 5.10 We can find the location of the centroid of a composite area from the centroids of the component areas.
x y z x y ⎯x1 ⎯x2 ⎯xA ⎯x W1 W2 W3 A1 A1 Semicircle A2 Full rectangle A3 Circular hole A2 A3 + – A ⎯x3 ⎯x1 ⎯x3 ⎯x2 + + – + + – – Fig. 5.11 When calculating the centroid of a composite area, note that if the centroid of a component area has a negative coordinate distance relative to the origin, or if the area represents a hole, then the first moment is negative.
ordinate Y of the centroid in a similar way by considering the first moment Qx of the composite area. We have Qy 5 X(A1 1 A2 1 . . . 1 An) 5 x1A1 1 x2 A2 1 . . . 1 xn An Qx 5 Y(A1 1 A2 1 . . . 1 An) 5 y1A1 1 y2A2 1 . . . 1 yn An Again, in shorter form, Centroid of a composite area Qy 5 X oA 5 oxA Qx 5 Y oA 5 oyA (5.8) These equations yield the first moments of the composite area, or we can use them to obtain the coordinates X and Y of its centroid.
First moments of areas, like moments of forces, can be positive or negative. Thus, you need to take care to assign the appropriate sign to the moment of each area. For example, an area whose centroid is located to the left of the y axis has a negative first moment with respect to that axis. Also, the area of a hole should be assigned a negative sign (Fig. 5.11).
Similarly, it is possible in many cases to determine the center of gravity of a composite wire or the centroid of a composite line by dividing the wire or line into simpler elements (see Sample Prob. 5.2).
Qy 5 XoA o 5 oxA xA Qx 5 YoA o 5 oyA yA 5.1 Planar Centers of Gravity and Centroids 241 Sample Problem 5.1 For the plane area shown, determine (a) the first moments with respect to the x and y axes; (b) the location of the centroid.
STRATEGY: Break up the given area into simple components, find the centroid of each component, and then find the overall first moments and centroid.
MODELING: As shown in Fig. 1, you obtain the given area by adding a rectangle, a triangle, and a semicircle and then subtracting a circle. Using the coordinate axes shown, find the area and the coordinates of the centroid of each of the component areas. To keep track of the data, enter them in a table. The area of the circle is indicated as negative because it is sub-tracted from the other areas. The coordinate y of the centroid of the triangle is negative for the axes shown. Compute the first moments of the compo-nent areas with respect to the coordinate axes and enter them in your table.
y x 80 mm 60 mm 60 mm 40 mm 120 mm Fig. 1 Given area modeled as the combination of simple geometric shapes.
y y x 80 mm 60 mm r1 = 60 mm r2 = 40 mm 120 mm x x x x y y y = + + _ 40 mm 40 mm –20 mm = 25.46 mm 4r1 3 r1 = 60 mm r2 = 40 mm 60 mm 60 mm 60 mm 80 mm 105.46 mm 80 mm Component A, mm2 x, mm y, mm x A, mm3 y A, mm3 Rectangle (120)(80) 5 9.6 3 103 60 40 1576 3 103 1384 3 103 Triangle 1 2(120)(60) 5 3.6 3 103 40 220 1144 3 103 272 3 103 Semicircle 1 2π(60)2 5 5.655 3 103 60 105.46 1339.3 3 103 1596.4 3 103 Circle 2π(40)2 5 25.027 3 103 60 80 2301.6 3 103 2402.2 3 103 oA 5 13.828 3 103 oxA 5 1757.7 3 103 oyA 5 1506.2 3 103 ANALYSIS: a. First Moments of the Area. Using Eqs. (5.8), you obtain Qx 5 oyA 5 506.2 3 103 mm3 Qx 5 506 3 103 mm3 b Qy 5 oxA 5 757.7 3 103 mm3 Qy 5 758 3 103 mm3 b b. Location of Centroid. Substituting the values given in the table into the equations defining the centroid of a composite area yields (Fig. 2) X oA 5 oxA: X(13.828 3 103 mm2) 5 757.7 3 103 mm3 X 5 54.8 mm b Y oA 5 oyA: Y(13.828 3 103 mm2) 5 506.2 3 103 mm3 Y 5 36.6 mm b y x C X = 54.8 mm Y = 36.6 mm Fig. 2 Centroid of composite area.
242 Distributed Forces: Centroids and Centers of Gravity REFLECT and THINK: Given that the lower portion of the shape has more area to the left and that the upper portion has a hole, the location of the centroid seems reasonable upon visual inspection.
Sample Problem 5.2 The figure shown is made from a piece of thin, homogeneous wire. Deter-mine the location of its center of gravity.
STRATEGY: Since the figure is formed of homogeneous wire, its center of gravity coincides with the centroid of the corresponding line. Therefore, you can simply determine that centroid.
10 in.
12 in.
5 in.
24 in.
C y x B A 26 in.
Fig. 1 Location of each line segment’s centroid.
MODELING: Choosing the coordinate axes shown in Fig. 1 with the origin at A, determine the coordinates of the centroid of each line segment and compute the first moments with respect to the coordinate axes. You may find it convenient to list the data in a table.
Segment L, in. x, in. y, in. x L, in2 y L, in2 AB 24 12 0 288 0 BC 26 12 5 312 130 CA 10 0 5 0 50 oL 5 60 ox L 5 600 oy L 5 180 ANALYSIS: Substituting the values obtained from the table into the equations defining the centroid of a composite line gives X oL 5 ox L: X(60 in.) 5 600 in2 X 5 10 in. b Y oL 5 oy L: Y(60 in.) 5 180 in2 Y 5 3 in. b REFLECT and THINK: The centroid is not on the wire itself, but it is within the area enclosed by the wire.
26 in.
10 in.
24 in.
C B A 5.1 Planar Centers of Gravity and Centroids 243 Sample Problem 5.3 A uniform semicircular rod of weight W and radius r is attached to a pin at A and rests against a frictionless surface at B. Determine the reactions at A and B.
STRATEGY: The key to solving the problem is finding where the weight W of the rod acts. Since the rod is a simple geometrical shape, you can look in Fig. 5.8 for the location of the wire’s centroid.
MODELING: Draw a free-body diagram of the rod (Fig. 1). The forces acting on the rod are its weight W, which is applied at the center of grav-ity G (whose position is obtained from Fig. 5.8B); a reaction at A, repre-sented by its components Ax and Ay; and a horizontal reaction at B.
G B Ax A Ay W B 2r 2r Fig. 1 Free-body diagram of rod.
ANALYSIS: 1l oMA 5 0: B(2r) 2 W a2r π b 5 0 B 5 1W π B 5 W π y b y 1 oFx 5 0: Ax 1 B 5 0 Ax 5 2B 5 2W π Ax 5 W π z 1 xoFy 5 0: Ay 2 W 5 0 Ay 5 W x Adding the two components of the reaction at A (Fig. 2), we have A 5 c W 2 1 aW π b 2 d 1/2 A 5 W a1 1 1 π 2b 1/2 b tan α 5 W W/π 5 π α 5 tan21π b The answers can also be expressed as A 5 1.049W b72.3° B 5 0.318Wy b REFLECT and THINK: Once you know the location of the rod’s center of gravity, the problem is a straightforward application of the concepts in Chapter 4.
A B O r Ay = W a Ax = W A Fig. 2 Reaction at A.
244 244 SOLVING PROBLEMS ON YOUR OWN I n this section, we developed the general equations for locating the centers of gravity of two-dimensional bodies and wires [Eqs. (5.2)] and the centroids of plane areas [Eqs. (5.3)] and lines [Eqs. (5.4)]. In the following problems, you will have to locate the centroids of composite areas and lines or determine the first moments of the area for composite plates [Eqs. (5.8)].
1. Locating the centroids of composite areas and lines. Sample Problems 5.1 and 5.2 illustrate the procedure you should follow when solving problems of this type. However, several points are worth emphasizing. a. The first step in your solution should be to decide how to construct the given area or line from the common shapes of Fig. 5.8. You should recognize that for plane areas it is often possible to construct a particular shape in more than one way. Also, showing the different components (as is done in Sample Prob. 5.1) can help you cor-rectly establish their centroids and areas or lengths. Do not forget that you can subtract areas as well as add them to obtain a desired shape. b. We strongly recommend that for each problem you construct a table listing the areas or lengths and the respective coordinates of the centroids. Remember, any areas that are “removed” (such as holes) are treated as negative. Also, the sign of negative coordinates must be included. Therefore, you should always carefully note the location of the origin of the coordinate axes. c. When possible, use symmetry [Sec. 5.1C] to help you determine the location of a centroid. d. In the formulas for the circular sector and for the arc of a circle in Fig. 5.8, the angle α must always be expressed in radians.
2. Calculating the first moments of an area. The procedures for locating the cen-troid of an area and for determining the first moments of an area are similar; however, it is not necessary to compute the total area for finding first moments. Also, as noted in Sec. 5.1C, you should recognize that the first moment of an area relative to a centroidal axis is zero.
3. Solving problems involving the center of gravity. The bodies considered in the following problems are homogeneous; thus, their centers of gravity and centroids coincide. In addition, when a body that is suspended from a single pin is in equilib-rium, the pin and the body’s center of gravity must lie on the same vertical line.
It may appear that many of the problems in this section have little to do with the study of mechanics. However, being able to locate the centroid of composite shapes will be essential in several topics that you will study later in this course.
245 Problems 5.1 through 5.9 Locate the centroid of the plane area shown.
x y 45 mm 45 mm 27 mm Fig. P5.1 x y 60 mm 60 mm 60 mm 75 mm 75 mm 75 mm Fig. P5.4 x y 24 mm 18 mm 32 mm 12 mm 12 mm Fig. P5.3 1 in.
1 in.
2 in.
5 in.
4 in.
x y Fig. P5.2 x y 6 in.
4 in.
3 in.
Fig. P5.5 Fig. P5.6 x y 60 mm 60 mm Fig. P5.7 x y 8 in.
8 in.
5 in.
8 in.
Fig. P5.8 r = 38 in.
x y 16 in.
20 in.
Fig. P5.9 x y 75 mm 75 mm 75 mm 246 5.10 through 5.15 Locate the centroid of the plane area shown. 5.16 Determine the y coordinate of the centroid of the shaded area in terms of r1, r2, and α. 5.17 Show that as r1 approaches r2, the location of the centroid approaches that for an arc of circle of radius (r1 1 r2)/2. 5.18 Determine the x coordinate of the centroid of the trapezoid shown in terms of h1, h2, and a.
x h2 a h1 y Fig. P5.18 5.19 For the semiannular area of Prob. 5.12, determine the ratio r1 to r2 so that the centroid of the area is located at x 5 21 2 r2 and y 5 0.
Parabola Vertex 10 in.
3 in.
16 in.
y x Fig. P5.10 x y a = 8 in.
x = ky2 b = 4 in.
Fig. P5.13 x y Vertex Parabola 75 mm 60 mm 60 mm Fig. P5.11 3 m 4.5 m 4.5 m r = 1.8 m Vertex Parabola x y Fig. P5.14 x y r1 = 72 mm r2 = 120 mm Fig. P5.12 x y r 90 mm 60 mm 60 mm 60 mm Fig. P5.15 x y α α r1 r2 Fig. P5.16 and Fig. P5.17 247 5.20 A composite beam is constructed by bolting four plates to four 60 3 60 3 12-mm angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the cen-troidal x axis of the red shaded areas shown, respectively, in parts a and b of the figure. Knowing that the force exerted on the bolt at A is 280 N, determine the force exerted on the bolt at B.
300 mm 12 mm 12 mm 12 mm 12 mm 60 mm 60 mm A C C x x B (a) (b) 450 mm Fig. P5.20 5.21 and 5.22 The horizontal x axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained. 0.60 in.
0.84 in.
0.24 in. 0.24 in.
0.72 in.
x y 0.72 in.
A2 A1 C Fig. P5.21 65 20 40 20 Dimensions in mm x y 15 40 A2 A1 C Fig. P5.22 5.23 The first moment of the shaded area with respect to the x axis is denoted by Qx. (a) Express Qx in terms of b, c, and the distance y from the base of the shaded area to the x axis. (b) For what value of y is Qx maximum, and what is that maximum value?
x y b c y c C Fig. P5.23 248 5.24 through 5.27 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed.
5.24 Fig. P5.1.
5.25 Fig. P5.3.
5.26 Fig. P5.5.
5.27 Fig. P5.8. 5.28 The homogeneous wire ABC is bent into a semicircular arc and a straight section as shown and is attached to a hinge at A. Determine the value of θ for which the wire is in equilibrium for the indicated position. 5.29 The frame for a sign is fabricated from thin, flat steel bar stock of mass per unit length 4.73 kg/m. The frame is supported by a pin at C and by a cable AB. Determine (a) the tension in the cable, (b) the reaction at C.
A C B R 0.75 m 0.8 m 0.2 m 1.35 m 0.6 m Fig. P5.29 5.30 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which portion BCD of the wire is horizontal.
80 mm B L C A D 60 mm Fig. P5.30 and P5.31 5.31 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which portion AB of the wire is horizontal. 5.32 Determine the distance h for which the centroid of the shaded area is as far above line BB9 as possible when (a) k 5 0.10, (b) k 5 0.80. 5.33 Knowing that the distance h has been selected to maximize the dis-tance y from line BB9 to the centroid of the shaded area, show that y 5 2h/3.
A B C r q r Fig. P5.28 B B' b kb a h Fig. P5.32 and P5.33 5.2 Further Considerations of Centroids 249 5.2 FURTHER CONSIDERATIONS OF CENTROIDS The objects we analyzed in Sec. 5.1 were composites of basic geometric shapes like rectangles, triangles, and circles. The same idea of locating a center of gravity or centroid applies for an object with a more complicated shape, but the mathematical techniques for finding the location are a little more difficult.
5.2A Determination of Centroids by Integration For an area bounded by analytical curves (i.e., curves defined by algebraic equations), we usually determine the centroid by evaluating the integrals in Eqs. (5.39): x 5 #x dA A y 5 #y dA A (5.39) If the element of area dA is a small rectangle of sides dx and dy, evaluat-ing each of these integrals requires a double integration with respect to x and y. A double integration is also necessary if we use polar coordinates for which dA is a small element with sides dr and r dθ.
In most cases, however, it is possible to determine the coordinates of the centroid of an area by performing a single integration. We can achieve this by choosing dA to be a thin rectangle or strip, or it can be a thin sector or pie-shaped element (Fig. 5.12). The centroid of the thin rectangle is located at its center, and the centroid of the thin sector is located at a distance (2/3)r from its vertex (as it is for a triangle). Then we obtain the coordinates of the centroid of the area under consideration xel = x yel = y/2 dA = ydx (c) yel = y dA = (a – x) dy (b) xel = a + x 2 (a) xel = 2r 3 yel = 2r 3 dA = 1 2 cosθ sinθ r2 dθ xel yel xel xel yel yel x a y x y x x x y y y O O O dx dy P(x, y) P(x, y) r θ 2r 3 P( , r) θ Fig. 5.12 Centroids and areas of differential elements. (a) Vertical rectangular strip; (b) horizontal rectangular strip; (c) triangular sector.
250 Distributed Forces: Centroids and Centers of Gravity by setting the first moment of the entire area with respect to each of the coordinate axes equal to the sum (or integral) of the corresponding moments of the elements of the area. Denoting the coordinates of the centroid of the element dA by xel and yel, we have First moments of area Qy 5 xA 5#xel dA Qx 5 yA 5#yel dA (5.9) If we do not already know the area A, we can also compute it from these elements.
In order to carry out the integration, we need to express the coordi-nates xel and yel of the centroid of the element of area dA in terms of the coordinates of a point located on the curve bounding the area under con-sideration. Also, we should express the area of the element dA in terms of the coordinates of that point and the appropriate differentials. This has been done in Fig. 5.12 for three common types of elements; the pie-shaped ele-ment of part (c) should be used when the equation of the curve bounding the area is given in polar coordinates. You can substitute the appropriate expressions into formulas (5.9), and then use the equation of the bounding curve to express one of the coordinates in terms of the other. This process reduces the double integration to a single integration. Once you have deter-mined the area and evaluated the integrals in Eqs. (5.9), you can solve these equations for the coordinates x and y of the centroid of the area.
When a line is defined by an algebraic equation, you can determine its centroid by evaluating the integrals in Eqs. (5.49): x 5 #x dL L y 5 #y dL L (5.49) You can replace the differential length dL with one of the following expressions, depending upon which coordinate, x, y, or θ, is chosen as the independent variable in the equation used to define the line (these expres-sions can be derived using the Pythagorean theorem): dL 5 B1 1 ady dxb 2 dx dL 5 B1 1 adx dyb 2 dy dL 5 Br2 1 a dr dθb 2 dθ After you have used the equation of the line to express one of the coor-dinates in terms of the other, you can perform the integration and solve Eqs. (5.4) for the coordinates x and y of the centroid of the line.
5.2B Theorems Of Pappus-Guldinus These two theorems, which were first formulated by the Greek geometer Pappus during the third century C.E. and later restated by the Swiss math-ematician Guldinus or Guldin (1577–1643), deal with surfaces and bodies Qy 5 xA xA 5#xel dA d Qx 5 yA yA 5#yel dA d 5.2 Further Considerations of Centroids 251 of revolution. A surface of revolution is a surface that can be generated by rotating a plane curve about a fixed axis. For example, we can obtain the surface of a sphere by rotating a semicircular arc ABC about the diameter AC (Fig. 5.13). Similarly, rotating a straight line AB about an axis AC produces the surface of a cone, and rotating the circumference of a circle about a nonintersecting axis generates the surface of a torus or ring. A body of revolution is a body that can be generated by rotating a plane area about a fixed axis. As shown in Fig. 5.14, we can generate a sphere, a cone, and a torus by rotating the appropriate shape about the indicated axis.
Sphere Cone Torus Fig. 5.14. Rotating plane areas about an axis generates volumes of revolution.
Theorem I. The area of a surface of revolution is equal to the length of the generating curve times the distance traveled by the centroid of the curve while the surface is being generated.
x x dL dA C L ⎯y y 2 ⎯y Fig. 5.15 An element of length dL rotated about the x axis generates a circular strip of area dA. The area of the entire surface of revolution equals the length of the line L multiplied by the distance traveled by the centroid C of the line during one revolution.
Proof. Consider an element dL of the line L (Fig. 5.15) that is revolved about the x axis. The circular strip generated by the element dL has an area A B C A C B Sphere Cone A C Torus Fig. 5.13. Rotating plane curves about an axis generates surfaces of revolution.
Photo 5.3 The storage tanks shown are bodies of revolution. Thus, their surface areas and volumes can be determined using the theorems of Pappus-Guldinus.
252 Distributed Forces: Centroids and Centers of Gravity dA equal to 2πy dL. Thus, the entire area generated by L is A 5 ∫ 2πy dL. Recall our earlier result that the integral ∫ y dL is equal to yL. Therefore, we have A 5 2πyL (5.10) Here 2πy is the distance traveled by the centroid C of L (Fig. 5.15). ✷ Note that the generating curve must not cross the axis about which it is rotated; if it did, the two sections on either side of the axis would generate areas having opposite signs, and the theorem would not apply.
Theorem II. The volume of a body of revolution is equal to the generating area times the distance traveled by the centroid of the area while the body is being generated.
Proof. Consider an element dA of the area A that is revolved about the x axis (Fig. 5.16). The circular ring generated by the element dA has a volume dV equal to 2πy dA. Thus, the entire volume generated by A is V 5 ∫2πy dA, and since we showed earlier that the integral ∫ y dA is equal to yA, we have V 5 2πyA (5.11) Here 2πy is the distance traveled by the centroid of A. ✷ y x dV dA y x A C 2 y Fig. 5.16 An element of area dA rotated about the x axis generates a circular ring of volume dV. The volume of the entire body of revolution equals the area of the region A multiplied by the distance traveled by the centroid C of the region during one revolution.
Again, note that the theorem does not apply if the axis of rotation intersects the generating area. The theorems of Pappus-Guldinus offer a simple way to compute the areas of surfaces of revolution and the volumes of bodies of revolution. Conversely, they also can be used to determine the centroid of a plane curve if you know the area of the surface generated by the curve or to determine the centroid of a plane area if you know the volume of the body generated by the area (see Sample Prob. 5.8).
A 5 2πy π L V 5 2πy π A yA 5.2 Further Considerations of Centroids 253 Sample Problem 5.4 Determine the location of the centroid of a parabolic spandrel by direct integration.
STRATEGY: First express the parabolic curve using the parameters a and b. Then choose a differential element of area and express its area in terms of a, b, x, and y. We illustrate the solution first with a vertical ele-ment and then a horizontal element.
MODELING: Determination of the Constant k. Determine the value of k by substituting x 5 a and y 5 b into the given equation. We have b 5 ka2 or k 5 b/a2. The equation of the curve is thus y 5 b a2 x2 or x 5 a b1/2 y1/2 ANALYSIS: Vertical Differential Element. Choosing the differential element shown in Fig. 1, the total area of the region is A 5# dA 5# y dx 5# a 0 b a2 x2 dx 5 c b a2 x3 3 d a 0 5 ab 3 a x y y dA = y dx yel = y 2 xel = x Fig. 1 Vertical differential element used to determine centroid.
The first moment of the differential element with respect to the y axis is xel dA; hence, the first moment of the entire area with respect to this axis is Qy 5 # xel dA 5 # xy dx 5 # a 0 x a b a2 x2b dx 5 c b a2 x4 4 d a 0 5 a2b 4 Since Qy 5 xA, you have xA 5#xel dA x ab 3 5 a2b 4 x 5 3 4 a b Likewise, the first moment of the differential element with respect to the x axis is yel dA, so the first moment of the entire area about the x axis is Qx 5# yel dA 5# y 2 y dx 5# a 0 1 2 a b a2 x2b 2 dx 5 c b2 2a4 x5 5 d a 0 5 ab2 10 Since Qx 5 yA, you get yA 5# yel dA y ab 3 5 ab2 10 y 5 3 10 b b a x y = kx2 y b 254 Distributed Forces: Centroids and Centers of Gravity Horizontal Differential Element. You obtain the same results by considering a horizontal element (Fig. 2). The first moments of the area are Qy 5# xel dA 5# a 1 x 2 (a 2 x) dy 5# b 0 a2 2 x2 2 dy 5 1 2 # b 0 aa2 2 a2 b yb dy 5 a2b 4 Qx 5# yel dA 5# y(a 2 x) dy 5# y aa 2 a b1/2 y1/2b dy 5# b 0 aay 2 a b1/2 y3/2b dy 5 ab2 10 To determine x and y, again substitute these expressions into the equations defining the centroid of the area.
REFLECT and THINK: You obtain the same results whether you choose a vertical or a horizontal element of area, as you should. You can use both methods as a check against making a mistake in your calculations.
x b yel = y xel = a + x 2 dA = (a – x) dy a y x Fig. 2 Horizontal differential element used to determine centroid.
Sample Problem 5.5 Determine the location of the centroid of the circular arc shown.
STRATEGY: For a simple figure with circular geometry, you should use polar coordinates.
MODELING: The arc is symmetrical with respect to the x axis, so y 5 0. Choose a differential element, as shown in Fig. 1. ANALYSIS: Determine the length of the arc by integration.
L 5# dL 5# α 2α r dθ 5 r # α 2α dθ 5 2rα The first moment of the arc with respect to the y axis is Qy 5# x dL 5# α 2α (r cos θ)(r dθ) 5 r2 # α 2α cos θ dθ 5 r2[sin θ]α 2α 5 2r2 sin α Since Qy 5 xL, you obtain x(2rα) 5 2r2 sin α x 5 r sin α α b REFLECT and THINK: Observe that this result matches that given for this case in Fig. 5.8B.
O α α r Fig. 1 Differential element used to determine centroid.
x y θ O r = θ α dθ dL = r dθ x = r cosθ = – θ α 5.2 Further Considerations of Centroids 255 Sample Problem 5.6 Determine the area of the surface of revolution shown that is obtained by rotating a quarter-circular arc about a vertical axis.
STRATEGY: According to the first Pappus-Guldinus theorem, the area of the surface of revolution is equal to the product of the length of the arc and the distance traveled by its centroid. MODELING and ANALYSIS: Referring to Fig. 5.8B and Fig. 1, you have x 5 2r 2 2r π 5 2r a1 2 1 πb A 5 2πxL 5 2πc 2r a1 2 1 πbd aπr 2 b A 5 2πr2(π 2 1) b r 2r Fig. 1 Centroid location of arc.
y x x 2r C 2r 20 mm 20 mm 20 mm 60 mm 30 mm 400 mm 100 mm Sample Problem 5.7 The outside diameter of a pulley is 0.8 m, and the cross section of its rim is as shown. Knowing that the pulley is made of steel and that the density of steel is ρ 5 7.85 3 103 kg/m3, determine the mass and weight of the rim.
STRATEGY: You can determine the volume of the rim by applying the second Pappus-Guldinus theorem, which states that the volume equals the product of the given cross-sectional area and the distance traveled by its centroid in one complete revolution. However, you can find the volume more easily by observing that the cross section can be formed from rect-angle I with a positive area and from rectangle II with a negative area (Fig. 1).
MODELING: Use a table to keep track of the data, as you did in Sec. 5.1. Distance Traveled Area, mm2 y, mm by C, mm Volume, mm3 I 15000 375 2π(375) 5 2356 (5000)(2356) 5 11.78 3 106 II 21800 365 2π(365) 5 2293 (21800)(2293) 5 24.13 3 106 Volume of rim 5 7.65 3 106 ANALYSIS: Since 1 mm 5 1023 m, you have 1 mm23 5 (1023 m)3 5 1029 m3. Thus you obtain V 5 7.65 3 106 mm3 5 (7.65 3 106)(1029 m3) 5 7.65 3 1023 m3.
m 5 ρV 5 (7.85 3 103 kg/m3)(7.65 3 1023 m3) m 5 60.0 kg b W 5 mg 5 (60.0 kg)(9.81 m/s2) 5 589 kg?m/s2 W 5 589 N b _ 100 mm 60 mm 50 mm 30 mm CII CI II I 375 mm 365 mm Fig. 1 Modeling the given area by subtracting area II from area I.
256 Distributed Forces: Centroids and Centers of Gravity Sample Problem 5.8 Using the theorems of Pappus-Guldinus, determine (a) the centroid of a semicircular area and (b) the centroid of a semicircular arc. Recall that the volume and the surface area of a sphere are 4 3πr3 and 4πr2, respectively.
STRATEGY: The volume of a sphere is equal to the product of the area of a semicircle and the distance traveled by the centroid of the semicircle in one revolution about the x axis. Given the volume, you can determine the distance traveled by the centroid and thus the distance of the centroid from the axis. Similarly, the area of a sphere is equal to the product of the length of the generating semicircle and the distance traveled by its centroid in one revolution. You can use this to find the location of the centroid of the arc.
MODELING: Draw diagrams of the semicircular area and the semicir-cular arc (Fig. 1) and label the important geometries.
x x r r2 A = 2 L = ⎯y ⎯y r r Fig. 1 Semicircular area and semicircular arc.
ANALYSIS: Set up the equalities described in the theorems of Pappus-Guldinus and solve for the location of the centroid.
V 5 2π yA 4 3 π r3 5 2π y a1 2 πr2b y 5 4r 3π b A 5 2πyL 4πr2 5 2πy(πr) y 5 2r π b REFLECT and THINK: Observe that this result matches those given for these cases in Fig. 5.8.
REFLECT and THINK: When a cross section can be broken down into multiple common shapes, you can apply Theorem II of Pappus–Guldinus in a manner that involves finding the products of the centroid (y) and area (A), or the first moments of area (yA), for each shape. Thus, it was not necessary to find the centroid or the area of the overall cross section.
257 257 SOLVING PROBLEMS ON YOUR OWN I n the problems for this section, you will use the equations x 5 #x dA A y 5 #y dA A (5.39) x 5 #x dL L y 5 #y dL L (5.49) to locate the centroids of plane areas and lines, respectively. You will also apply the theorems of Pappus-Guldinus to determine the areas of surfaces of revolution and the volumes of bodies of revolution.
1. Determining the centroids of areas and lines by direct integration. When solving problems of this type, you should follow the method of solution shown in Sample Probs. 5.4 and 5.5. To compute A or L, determine the first moments of the area or the line, and solve Eqs. (5.3) or (5.4) for the coordinates of the centroid. In addition, you should pay particular attention to the following points. a. Begin your solution by carefully defining or determining each term in the appli-cable integral formulas. We strongly encourage you to show on your sketch of the given area or line your choice for dA or dL and the distances to its centroid.
b. As explained in Sec. 5.2A, x and y in Eqs. (5.3) and (5.4) represent the coor-dinates of the centroid of the differential elements dA and dL. It is important to recognize that the coordinates of the centroid of dA are not equal to the coordinates of a point located on the curve bounding the area under consideration. You should carefully study Fig. 5.12 until you fully understand this important point. c. To possibly simplify or minimize your computations, always examine the shape of the given area or line before defining the differential element that you will use. For example, sometimes it may be preferable to use horizontal rectangular elements instead of vertical ones. Also, it is usually advantageous to use polar coordinates when a line or an area has circular symmetry. d. Although most of the integrations in this section are straightforward, at times it may be necessary to use more advanced techniques, such as trigonometric substitu-tion or integration by parts. Using a table of integrals is often the fastest method to evaluate difficult integrals.
2. Applying the theorems of Pappus-Guldinus. As shown in Sample Probs. 5.6 through 5.8, these simple, yet very useful theorems allow you to apply your knowl-edge of centroids to the computation of areas and volumes. Although the theorems refer to the distance traveled by the centroid and to the length of the generating curve or to the generating area, the resulting equations [Eqs. (5.10) and (5.11)] contain the products of these quantities, which are simply the first moments of a line (yL) and an area (yA), respectively. Thus, for those problems for which the generating line or area consists of more than one common shape, you need only determine yL or yA; you do not have to calculate the length of the generating curve or the generating area.
258 5.34 through 5.36 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.
Problems x y h a Fig. P5.34 x y y = mx y = kx2 h a Fig. P5.35 x y a a h y = kx2 Fig. P5.36 x y a a 2 a 2 a a Fig. P5.37 x y r1 r2 Fig. P5.38 x y b a x2 a2 y2 b2 + = 1 Fig. P5.39 5.37 through 5.39 Determine by direct integration the centroid of the area shown. 5.40 and 5.41 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.
x y b a y = k(x – a)2 Fig. P5.40 x y a b y1 = k1x2 y2 = k2x3 Fig. P5.41 259 5.42 Determine by direct integration the centroid of the area shown. 5.43 and 5.44 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b.
x y b 2 b 2 a 2 a 2 x = ky2 Fig. P5.43 x y y = kx2 a a b b Fig. P5.44 5.45 and 5.46 A homogeneous wire is bent into the shape shown. Deter-mine by direct integration the x coordinate of its centroid. x y a a π 2 0 ≤ ≤ θ x = a cos3 y = a sin3 θ θ Fig. P5.45 y x r 45° 45° Fig. P5.46 5.47 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. Express your answer in terms of a. 5.48 and 5.49 Determine by direct integration the centroid of the area shown.
x y q r = a eq Fig. P5.48 y x L a y 5 a sin p x L L 2 Fig. P5.49 5.50 Determine the centroid of the area shown in terms of a. 5.51 Determine the centroid of the area shown when a 5 4 in. 5.52 Determine the volume and the surface area of the solid obtained by rotating the area of Prob. 5.1 about (a) the x axis, (b) the line x 5 72 mm. 5.53 Determine the volume and the surface area of the solid obtained by rotating the area of Prob. 5.2 about (a) the x axis, (b) the y axis.
y x L L a y = a 1 – x L x2 L2 + ) ( Fig. P5.42 x y a a y = kx 3 2 Fig. P5.47 1 x x y a a y = Fig. P5.50 and P5.51 260 5.54 Determine the volume and the surface area of the solid obtained by rotating the area of Prob. 5.6 about (a) the line x 5 –60 mm, (b) the line y 5 120 mm. 5.55 Determine the volume and the surface area of the chain link shown, which is made from a 6-mm-diameter bar, if R 5 10 mm and L 5 30 mm.
L R R Fig. P5.55 5.56 Determine the volume of the solid generated by rotating the para-bolic area shown about (a) the x axis, (b) the axis AA9.
x y h a a a A A' Fig. P5.56 5.57 Verify that the expressions for the volumes of the first four shapes in Fig. 5.21 are correct. 5.58 Knowing that two equal caps have been removed from a 10-in.-diameter wooden sphere, determine the total surface area of the remaining portion.
5.59 Three different drive belt profiles are to be studied. If at any given time each belt makes contact with one-half of the circumference of its pulley, determine the contact area between the belt and the pulley for each design.
0.625 in.
(a) (b) (c) 0.08 in.
r = 0.25 in.
40° 40° 0.375 in.
0.125 in.
3 in.
3 in.
3 in.
Fig. P5.59 4 in.
4 in.
10 in.
Fig. P5.58 261 5.60 Determine the capacity, in liters, of the punch bowl shown if R 5 250 mm.
R R Fig. P5.60 5.61 Determine the volume and total surface area of the bushing shown. 5.62 Determine the volume and weight of the solid brass knob shown, knowing that the specific weight of brass is 0.306 lb/in3. 5.63 Determine the total surface area of the solid brass knob shown. 5.64 The aluminum shade for the small high-intensity lamp shown has a uniform thickness of 1 mm. Knowing that the density of aluminum is 2800 kg/m3, determine the mass of the shade.
32 mm 26 mm 32 mm 56 mm 28 mm 66 mm 8 mm Fig. P5.64 5.65 The shade for a wall-mounted light is formed from a thin sheet of translucent plastic. Determine the surface area of the outside of the shade, knowing that it has the parabolic cross section shown.
100 mm y x y = kx2 250 mm Fig. P5.65 60 mm 42 mm 52 mm 20 mm Fig. P5.61 1.25 in.
r = 0.75 in.
r = 0.75 in.
Fig. P5.62 and P5.63 262 Distributed Forces: Centroids and Centers of Gravity 5.3 ADDITIONAL APPLICATIONS OF CENTROIDS We can use the concept of the center of gravity or the centroid of an area to solve other problems besides those dealing with the weights of flat plates. The same techniques allow us to deal with other kinds of distrib-uted loads on objects, such as the forces on a straight beam (a bridge girder or the main carrying beam of a house floor) or a flat plate under water (the side of a dam or a window in an aquarium tank).
5.3A Distributed Loads on Beams Consider a beam supporting a distributed load; this load may consist of the weight of materials supported directly or indirectly by the beam, or it may be caused by wind or hydrostatic pressure. We can represent the distributed load by plotting the load w supported per unit length (Fig. 5.17); this load is expressed in N/m or in lb/ft. The magnitude of the force exerted on an element of the beam with length dx is dW 5 w dx, and the total load supported by the beam is W 5 # L 0 w dx Note that the product w dx is equal in magnitude to the element of area dA shown in Fig. 5.17a. The load W is thus equal in magnitude to the total area A under the load curve, as W 5 #dA 5 A We now want to determine where a single concentrated load W, of the same magnitude W as the total distributed load, should be applied on the beam if it is to produce the same reactions at the supports (Fig. 5.17b). However, this concentrated load W, which represents the resultant of the given distrib-uted loading, is equivalent to the loading only when considering the free-body diagram of the entire beam. We obtain the point of application P of the equivalent concentrated load W by setting the moment of W about point O equal to the sum of the moments of the elemental loads dW about O. Thus, (OP)W 5 #x dW Then, since dW 5 w dx 5 dA and W 5 A, we have (OP)A 5 # L 0 x dA (5.12) Since this integral represents the first moment with respect to the w axis of the area under the load curve, we can replace it with the product xA. We therefore have OP 5 x, where x is the distance from the w axis to the centroid C of the area A (this is not the centroid of the beam).
We can summarize this result: We can replace a distributed load on a beam by a concentrated load; the magnitude of this single load is equal to the area under the load curve, and its line of action passes through the centroid of that area. (a) (b) w O w dx x L B dW = dA x d W w O B x L P W = A W C x = Fig. 5.17 (a) A load curve representing the distribution of load forces along a horizontal beam, with an element of length dx; (b) the resultant load W has magnitude equal to the area A under the load curve and acts through the centroid of the area.
Photo 5.4 The roof of a building shown must be able to support not only the total weight of the snow but also the nonsymmetric distributed loads resulting from drifting of the snow.
5.3 Additional Applications of Centroids 263 Note, however, that the concentrated load is equivalent to the given loading only so far as external forces are concerned. It can be used to determine reac-tions, but should not be used to compute internal forces and deflections.
5.3B Forces on Submerged Surfaces The approach used for distributed loads on beams works in other applica-tions as well. Here, we use it to determine the resultant of the hydrostatic pressure forces exerted on a rectangular surface submerged in a liquid. We can use these methods to determine the resultant of the hydrostatic forces exerted on the surfaces of dams, rectangular gates, and vanes. In Chap. 9, we discuss the resultants of forces on submerged surfaces of variable width.
Consider a rectangular plate with a length of L and width of b, where b is measured perpendicular to the plane of the figure (Fig. 5.18). As for the case of distributed loads on a beam, the load exerted on an element of the plate with a length of dx is w dx, where w is the load per unit length and x is the distance along the length. However, this load also can be expressed as p dA 5 pb dx, where p is the gage pressure in the liquid† and b is the width of the plate; thus, w 5 bp. Since the gage pres-sure in a liquid is p 5 γh, where γ is the specific weight of the liquid and h is the vertical distance from the free surface, it follows that w 5 bp 5 bγh (5.13) This equation shows that the load per unit length w is proportional to h and, thus, varies linearly with x.
From the results of Sec. 5.3A, the resultant R of the hydrostatic forces exerted on one side of the plate is equal in magnitude to the trap-ezoidal area under the load curve, and its line of action passes through the centroid C of that area. The point P of the plate where R is applied is known as the center of pressure.‡ Now consider the forces exerted by a liquid on a curved surface of constant width (Fig. 5.19a). Since determining the resultant R of these forces by direct integration would not be easy, we consider the free body obtained by detaching the volume of liquid ABD bounded by the curved surface AB and by the two plane surfaces AD and DB shown in Fig. 5.19b. The forces acting on the free body ABD are the weight W of the detached volume of liquid, the resultant R1 of the forces exerted on AD, the resul-tant R2 of the forces exerted on BD, and the resultant 2R of the forces exerted by the curved surface on the liquid. The resultant 2R is both equal and opposite to and has the same line of action as the resultant R of the forces exerted by the liquid on the curved surface. We can determine the forces W, R1, and R2 by standard methods. After their values have been found, we obtain the force 2R by solving the equations of equilibrium for the free body of Fig. 5.19b. The resultant R of the hydrostatic forces exerted on the curved surface is just the reverse of 2R.
†The pressure p, which represents a load per unit area, is measured in N/m2 or in lb/ft2. The derived SI unit N/m2 is called a pascal (Pa).
‡The area under the load curve is equal to wE L, where wE is the load per unit length at the center E of the plate. Then from Eq. (5.13), we have R 5 wE L 5 (bpE)L 5 pE(bL) 5 pE A where A denotes the area of the plate. Thus, we can obtain the magnitude of R by multiply-ing the area of the plate by the pressure at its center E. Note, however, that the resultant R should be applied at P, not at E.
Fig. 5.18 The waterside face of a hydroelectric dam can be modeled as a rectangular plate submerged under water. Shown is a side view of the plate.
C R w L E P A Surface of water B x dx (a) (b) A B A D B R R1 R2 –R W Fig. 5.19 (a) Force R exerted by a liquid on a submerged curved surface of constant width; (b) free-body diagram of the volume of liquid ABD.
264 Distributed Forces: Centroids and Centers of Gravity Sample Problem 5.9 A beam supports a distributed load as shown. (a) Determine the equivalent concentrated load. (b) Determine the reactions at the supports.
STRATEGY: The magnitude of the resultant of the load is equal to the area under the load curve, and the line of action of the resultant passes through the centroid of the same area. Break down the area into pieces for easier calculation, and determine the resultant load. Then, use the calculated forces or their resultant to determine the reactions.
MODELING and ANALYSIS: a. Equivalent Concentrated Load. Divide the area under the load curve into two triangles (Fig. 1), and construct the table below. To simplify the computations and tabulation, the given loads per unit length have been converted into kN/m.
Component A, kN x, m x A, kN?m Triangle I 4.5 2 9 Triangle II 13.5 4 54 oA 5 18.0 oxA 5 63 Thus, XoA 5 oxA: X(18 kN) 5 63 kN?m X 5 3.5 m The equivalent concentrated load (Fig. 2) is W 5 18 kN w b Its line of action is located at a distance X 5 3.5 m to the right of A b b. Reactions.
The reaction at A is vertical and is denoted by A. Rep-resent the reaction at B by its components Bx and By. Consider the given load to be the sum of two triangular loads (see the free-body diagram, Fig. 3). The resultant of each triangular load is equal to the area of the triangle and acts at its centroid. Write the following equilibrium equations from the free-body diagram: H 1 oFx 5 0: Bx 5 0 b 1l oMA 5 0: 2(4.5 kN)(2 m) 2 (13.5 kN)(4 m) 1 By(6 m) 5 0 By 5 10.5 kNx b 1l oMB 5 0: 1(4.5 kN)(4 m) 1 (13.5 kN)(2 m) 2 A(6 m) 5 0 A 5 7.5 kNx b REFLECT and THINK: You can replace the given distributed load by its resultant, which you found in part a. Then you can determine the reac-tions from the equilibrium equations oFx 5 0, oMA 5 0, and oMB 5 0. Again the results are Bx 5 0 By 5 10.5 kNx A 5 7.5 kNx b A B wA = 1500 N/m wB = 4500 N/m L = 6 m I II 4.5 kN/m 1.5 kN/m 6 m x = 2 m x = 4 m x Fig. 1 The load modeled as two triangular areas.
A B 18 kN X = 3.5 m Fig. 2 Equivalent concentrated load.
A Bx By 4.5 kN 13.5 kN 2 m 4 m 6 m Fig. 3 Free-body diagram of beam.
5.3 Additional Applications of Centroids 265 Sample Problem 5.10 The cross section of a concrete dam is shown. Consider a 1-ft-thick section of the dam, and determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the resultant of the pressure forces exerted by the water on the face BC of the dam. The specific weights of concrete and water are 150 lb/ft3 and 62.4 lb/ft3, respectively.
5 ft Vertex Parabola 18 ft A B C 22 ft 9 ft 10 ft STRATEGY: Draw a free-body diagram of the section of the dam, breaking it into parts to simplify the calculations. Model the resultant of the reactions as a force-couple system at A. Use the method described in Sec. 5.3B to find the force exerted by the dam on the water and reverse it to find the force exerted by the water on face BC.
MODELING and ANALYSIS: a. Ground Reaction. Choose as a free body the 1-ft-thick section AEFCDB of the dam and water (Fig. 1). The reaction forces exerted by the ground on the base AB are represented by an equivalent force-couple system at A. Other forces acting on the free body are the weight of the dam represented by the weights of its components W1, W2, and W3; the weight of the water W4; and the resultant P of the pressure forces exerted on section BD by the water to the right of section BD. x y 2.5 ft 4 ft E F C D A B 3 ft H M V P W1 W3 W4 W2 w = bp = (1 ft)(18 ft)(62.4 lb/ft3) 9 ft 22 ft 14 ft 6 ft 18 ft 6 ft 6 ft Fig. 1 Free-body diagram of dam and water.
266 Distributed Forces: Centroids and Centers of Gravity Calculate each of the forces that appear in the free-body diagram, Fig. 1: W1 5 1 2(9 ft)(22 ft)(1 ft)(150 lb/ft3) 5 14,850 lb W2 5 (5 ft)(22 ft)(1 ft)(150 lb/ft3) 5 16,500 lb W3 5 1 3(10 ft)(18 ft)(1 ft)(150 lb/ft3) 5 9000 lb W4 5 2 3(10 ft)(18 ft)(1 ft)(62.4 lb/ft3) 5 7488 lb P 5 1 2(18 ft)(1 ft)(18 ft)(62.4 lb/ft3) 5 10,109 lb Equilibrium Equations. Write the equilibrium equations for the sec-tion of the dam, and calculate the forces and moment labeled at A in Fig. 1.
H 1 oFx 5 0: H 2 10,109 lb 5 0 H 5 10,110 lb y b 1 xoFy 5 0: V 2 14,850 lb 2 16,500 lb 2 9000 lb 2 7488 lb 5 0 V 5 47,840 lbx b 1l oMA 5 0: 2(14,850 lb)(6 ft) 2 (16,500 lb)(11.5 ft) 2 (9000 lb)(17 ft) 2 (7488 lb)(20 ft) 1 (10,109 lb)(6 ft) 1 M 5 0 M 5 520,960 lb?ft l b You can replace the force-couple system by a single force acting at a distance d to the right of A, where d 5 520,960 lb?ft 47,840 lb 5 10.89 ft b. Resultant R of Water Forces. Draw a free-body diagram for the parabolic section of water BCD (Fig. 2). The forces involved are the resultant 2R of the forces exerted by the dam on the water, the weight W4, and the force P. Since these forces must be concurrent, 2R passes through the point of intersection G of W4 and P. Draw a force triangle to determine the magnitude and direction of 2R. The resultant R of the forces exerted by the water on the face BC is equal and opposite. Hence, R 5 12,580 lb d36.5° b x y 4 ft C D B G P W4 = 7488 lb W4 –R –R P = 10,109 lb α α = 36.5° R = 12,580 lb 6 ft Fig. 2 Free-body diagram of parabolic section of water BCD.
REFLECT and THINK: Note that if you found the distance d to be negative—that is, if the moment reaction at A had been acting in the opposite direction—this would have indicated an instability condition of the dam. In this situation, the effects of the water pressure would over-come the weight of the dam, causing it to tip about A.
267 267 T he problems in this section involve two common and very important types of loading: distributed loads on beams and forces on submerged surfaces of constant width. As we discussed in Sec. 5.3 and illustrated in Sample Probs. 5.9 and 5.10, determining the single equivalent force for each of these loadings requires a knowl-edge of centroids.
1. Analyzing beams subjected to distributed loads. In Sec. 5.3A, we showed that a distributed load on a beam can be replaced by a single equivalent force. The mag-nitude of this force is equal to the area under the distributed load curve, and its line of action passes through the centroid of that area. Thus, you should begin solving this kind of problem by replacing the various distributed loads on a given beam by their respective single equivalent forces. You can then determine the reactions at the sup-ports of the beam by using the methods of Chap. 4.
When possible, divide complex distributed loads into the common-shape areas shown in Fig. 5.8A (Sample Prob. 5.9). You can replace each of these areas under the loading curve by a single equivalent force. If required, you can further reduce the system of equivalent forces to a single equivalent force. As you study Sample Prob. 5.9, note how we used the analogy between force and area under the loading curve and applied the techniques for locating the centroid of a composite area to analyze a beam sub-jected to a distributed load.
2. Solving problems involving forces on submerged bodies. Remember the follow-ing points and techniques when solving problems of this type.
a. The pressure p at a depth h below the free surface of a liquid is equal to γh or ρgh, where γ and ρ are the specific weight and the density of the liquid, respectively. The load per unit length w acting on a submerged surface of constant width b is then w 5 bp 5 bγh 5 bρgh b. The line of action of the resultant force R acting on a submerged plane surface is perpendicular to the surface. c. For a vertical or inclined plane rectangular surface with a width of b, you can represent the loading on the surface using a linearly distributed load that is trapezoidal in shape (Fig. 5.18). The magnitude of the resultant R is given by R 5 γhE A where hE is the vertical distance to the center of the surface and A is the area of the surface.
SOLVING PROBLEMS ON YOUR OWN 268 d. The load curve is triangular (rather than trapezoidal) when the top edge of a plane rectangular surface coincides with the free surface of the liquid, since the pres-sure of the liquid at the free surface is zero. For this case, it is straightforward to determine the line of action of R, because it passes through the centroid of a trian-gular distributed load.
e. For the general case, rather than analyzing a trapezoid, we suggest that you use the method indicated in part b of Sample Prob. 5.9. First divide the trapezoidal dis-tributed load into two triangles, and then compute the magnitude of the resultant of each triangular load. (The magnitude is equal to the area of the triangle times the width of the plate.) Note that the line of action of each resultant force passes through the centroid of the corresponding triangle and that the sum of these forces is equivalent to R. Thus, rather than using R, you can use the two equivalent resultant forces whose points of application are easily calculated. You should use the equation given for R here in paragraph c when you need only the magnitude of R.
f. When the submerged surface of a constant width is curved, you can obtain the resultant force acting on the surface by considering the equilibrium of the volume of liquid bounded by the curved surface and by using horizontal and vertical planes (Fig. 5.19). Observe that the force R1 of Fig. 5.19 is equal to the weight of the liquid lying above the plane AD. The method of solution for problems involving curved surfaces is shown in part b of Sample Prob. 5.10.
In subsequent mechanics courses (in particular, mechanics of materials and fluid mechanics), you will have ample opportunity to use the ideas introduced in this section.
269 5.66 and 5.67 For the beam and loading shown, determine (a) the mag-nitude and location of the resultant of the distributed load, (b) the reactions at the beam supports.
Problems 1600 N/m 400 N/m A B 6 m Fig. P5.66 900 N/m 2000 N/m A B 6 m Parabola Vertex Fig. P5.67 A B 4 ft 3 ft 150 lb/ft 200 lb/ft Fig. P5.68 50 lb/in.
A B 12 in.
6 in.
20 in.
400 lb Fig. P5.69 600 lb/ft 480 lb/ft A D B C 2 ft 6 ft 3 ft Fig. P5.70 5.68 through 5.73 Determine the reactions at the beam supports for the given loading.
400 N/m 900 N/m A B 0.6 m 0.4 m 1.5 m Fig. P5.71 100 lb/ft 200 lb/ft A B 6 ft 12 ft Parabolas Fig. P5.72 A B 6 m 900 N/m 300 N/m Parabola Vertex Fig. P5.73 270 5.74 Determine (a) the distance a so that the vertical reactions at supports A and B are equal, (b) the corresponding reactions at the supports. 5.75 Determine (a) the distance a so that the reaction at support B is minimum, (b) the corresponding reactions at the supports. 5.76 Determine the reactions at the beam supports for the given loading when w0 5 400 lb/ft.
300 lb/ft w0 A B C 5 ft 7 ft Fig. P5.76 and P5.77 5.77 Determine (a) the distributed load w0 at the end A of the beam ABC for which the reaction at C is zero, (b) the corresponding reaction at B. 5.78 The beam AB supports two concentrated loads and rests on soil that exerts a linearly distributed upward load as shown. Determine the values of wA and wB corresponding to equilibrium. 5.79 For the beam and loading of Prob. 5.78, determine (a) the distance a for which wA 5 20 kN/m, (b) the corresponding value of wB. In the following problems, use γ 5 62.4 lb/ft3 for the specific weight of fresh water and γc 5 150 lb/ft3 for the specific weight of concrete if U.S. customary units are used. With SI units, use ρ 5 103 kg/m3 for the density of fresh water and ρc 5 2.40 3 103 kg/m3 for the density of concrete. (See the footnote on page 234 for how to determine the specific weight of a material given its density.) 5.80 The cross section of a concrete dam is as shown. For a 1-ft-wide dam section determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.
15 ft 18 ft 9 ft 6 ft 6 ft A B C Fig. P5.80 5.81 The cross section of a concrete dam is as shown. For a 1-m-wide dam section determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam.
A B wA wB 24 kN 30 kN 0.3 m 1.8 m a = 0.6 m Fig. P5.78 4 m 3 m 1.5 m 2 m Parabola Vertex A B C Fig. P5.81 A B 4 m 600 N/m a 1800 N/m Fig. P5.74 and P5.75 271 5.82 The dam for a lake is designed to withstand the additional force caused by silt that has settled on the lake bottom. Assuming that silt is equivalent to a liquid of density ρs 5 1.76 3 103 kg/m3 and con-sidering a 1-m-wide section of dam, determine the percentage increase in the force acting on the dam face for a silt accumulation of depth 2 m.
5.83 The base of a dam for a lake is designed to resist up to 120 percent of the horizontal force of the water. After construction, it is found that silt (that is equivalent to a liquid of density ρs 5 1.76 3 103 kg/m3) is settling on the lake bottom at the rate of 12 mm/year. Considering a 1-m-wide section of dam, determine the number of years of service until the dam becomes unsafe. 5.84 An automatic valve consists of a 9 3 9-in. square plate that is piv-oted about a horizontal axis through A located at a distance h 5 3.6 in. above the lower edge. Determine the depth of water d for which the valve will open. 5.85 An automatic valve consists of a 9 3 9-in. square plate that is piv-oted about a horizontal axis through A. If the valve is to open when the depth of water is d 5 18 in., determine the distance h from the bottom of the valve to the pivot A. 5.86 The 3 3 4-m side AB of a tank is hinged at its bottom A and is held in place by a thin rod BC. The maximum tensile force the rod can withstand without breaking is 200 kN, and the design specifications require the force in the rod not to exceed 20 percent of this value. If the tank is slowly filled with water, determine the maximum allow-able depth of water d in the tank.
A B C T 3 m d Fig. P5.86 and P5.87 5.87 The 3 3 4-m side of an open tank is hinged at its bottom A and is held in place by a thin rod BC. The tank is to be filled with glycerine with a density of 1263 kg/m3. Determine the force T in the rod and the reaction at the hinge after the tank is filled to a depth of 2.9 m. 5.88 A 0.5 3 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a friction-less stop at B. Determine the reactions at A and B when cable BCD is slack. 5.89 A 0.5 3 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a friction-less stop at B. Determine the minimum tension required in cable BCD to open the gate.
Water Silt 6.6 m Fig. P5.82 and P5.83 h d A B 9 in.
Fig. P5.84 and P5.85 A B C D T 0.27 m 0.45 m 0.48 m 0.64 m Fig. P5.88 and P5.89 272 5.90 A 4 3 2-ft gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 828 lb/ft. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor.
5.91 Solve Prob. 5.90 if the gate weighs 1000 lb. 5.92 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. The pin is located at a distance h 5 0.10 m below the center of gravity C of the gate. Determine the depth of water d for which the gate will open.
B C h 0.75 m 0.40 m d A Fig. P5.92 and P5.93 5.93 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. The pin is located at a distance h below the center of gravity C of the gate. Determine the distance h if the gate is to open when d 5 0.75 m. 5.94 A long trough is supported by a continuous hinge along its lower edge and by a series of horizontal cables attached to its upper edge. Determine the tension in each of the cables at a time when the trough is completely full of water.
A r = 24 in.
20 in.
20 in.
20 in.
Fig. P5.94 5.95 The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d 5 3.5 ft, determine the force exerted on the gate by the shear pin.
A B C D 2 ft 3 ft d 4 ft Fig. P5.90 30° A B 1.8 ft d Fig. P5.95 5.4 Centers of Gravity and Centroids of Volumes 273 5.4 CENTERS OF GRAVITY AND CENTROIDS OF VOLUMES So far in this chapter, we have dealt with finding centers of gravity and centroids of two-dimensional areas and objects such as flat plates and plane surfaces. However, the same ideas apply to three-dimensional objects as well. The most general situations require the use of multiple integration for analysis, but we can often use symmetry considerations to simplify the calculations. In this section, we show how to do this.
5.4A Three-Dimensional Centers of Gravity and Centroids For a three-dimensional body, we obtain the center of gravity G by divid-ing the body into small elements. The weight W of the body acting at G is equivalent to the system of distributed forces DW representing the weights of the small elements. Choosing the y axis to be vertical with positive sense upward (Fig. 5.20) and denoting the position vector of G to be r, we set W equal to the sum of the elemental weights DW and set its moment about O equal to the sum of the moments about O of the elemental weights. Thus, oF: 2Wj 5 o(2DWj) (5.14) oMO: r 3 (2Wj) 5 o[r 3 (2DWj)] G y O = ∆W y x x z z O r r W = –Wj ∆W = –∆Wj Fig. 5.20 For a three-dimensional body, the weight W acting through the center of gravity G and its moment about O is equivalent to the system of distributed weights acting on all the elements of the body and the sum of their moments about O.
We can rewrite the last equation in the form rW 3 (2j) 5 (or DW) 3 (2j) (5.15) From these equations, we can see that the weight W of the body is equiva-lent to the system of the elemental weights DW if the following conditions are satisfied: W 5 o DW rW 5 or DW Photo 5.5 To predict the flight characteristics of the modified Boeing 747 when used to transport a space shuttle, engineers had to determine the center of gravity of each craft.
274 Distributed Forces: Centroids and Centers of Gravity Increasing the number of elements and simultaneously decreasing the size of each element, we obtain in the limit as Weight, center of gravity of a three-dimensional body W 5# dW r W 5# r dW (5.16) Note that these relations are independent of the orientation of the body. For example, if the body and the coordinate axes were rotated so that the z axis pointed upward, the unit vector 2j would be replaced by 2k in Eqs. (5.14) and (5.15), but the relations in Eqs. (5.16) would remain unchanged. Resolving the vectors r and r into rectangular components, we note that the second of the relations in Eqs. (5.16) is equivalent to the three scalar equations x W 5# x dW y W 5# y dW z W 5# z dW (5.17) or x 5 #x dW W y 5 #y dW W z 5 #z dW W (5.179) If the body is made of a homogeneous material of specific weight γ, we can express the magnitude dW of the weight of an infinitesimal ele-ment in terms of the volume dV of the element and express the magnitude W of the total weight in terms of the total volume V. We obtain dW 5 γ dV W 5 γV Substituting for dW and W in the second of the relations in Eqs. (5.16), we have r V 5# r dV (5.18) In scalar form, this becomes Centroid of a volume V x V 5# x dV y V 5# y dV z V 5# z dV (5.19) or x 5 #x dV V y 5 #y dV V z 5 #z dV V (5.199) The center of gravity of a homogeneous body whose coordinates are x, y, z is also known as the centroid C of the volume V of the body. If the body is not homogeneous, we cannot use Eqs. (5.19) to determine the center of gravity of the body; however, Eqs. (5.19) still define the centroid of the volume.
The integral ∫ x dV is known as the first moment of the volume with respect to the yz plane. Similarly, the integrals ∫ y dV and ∫ z dV define the first moments of the volume with respect to the zx plane and W 5# dW rW 5# r dW xW 5# x dW yW 5# y dW zW 5# z dW xV 5# x dV yV 5# y dV zV 5# z dV 5.4 Centers of Gravity and Centroids of Volumes 275 the xy plane, respectively. You can see from Eqs. (5.19) that if the centroid of a volume is located in a coordinate plane, the first moment of the volume with respect to that plane is zero.
A volume is said to be symmetrical with respect to a given plane if, for every point P of the volume, there exists a point P9 of the same volume such that the line PP9 is perpendicular to the given plane and is bisected by that plane. We say the plane is a plane of symmetry for the given volume. When a volume V possesses a plane of symmetry, the first moment of V with respect to that plane is zero, and the centroid of the volume is located in the plane of symmetry. If a volume possesses two planes of symmetry, the centroid of the volume is located on the line of intersection of the two planes. Finally, if a volume possesses three planes of symmetry that intersect at a well-defined point (i.e., not along a com-mon line), the point of intersection of the three planes coincides with the centroid of the volume. This property enables us to determine immediately the locations of the centroids of spheres, ellipsoids, cubes, rectangular parallelepipeds, etc.
For unsymmetrical volumes or volumes possessing only one or two planes of symmetry, we can determine the location of the centroid by integration (Sec. 5.4C). The centroids of several common volumes are shown in Fig. 5.21. Note that, in general, the centroid of a volume of revolution does not coincide with the centroid of its cross section. Thus, the centroid of a hemisphere is different from that of a semicircular area, and the centroid of a cone is different from that of a triangle.
5.4B Composite Bodies If a body can be divided into several of the common shapes shown in Fig. 5.21, we can determine its center of gravity G by setting the moment about O of its total weight equal to the sum of the moments about O of the weights of the various component parts. Proceeding in this way, we obtain the following equations defining the coordinates X, Y, Z of the cen-ter of gravity G as Center of gravity of a body with weight W X oW 5 ox W Y oW 5 oy W Z oW 5 oz W (5.20) or X 5 o xW o W Y 5 o yW o W Z 5 o zW o W (5.209) If the body is made of a homogeneous material, its center of gravity coincides with the centroid of its volume, and we obtain Centroid of a volume V X oV 5 ox V Y oV 5 oy V Z oV 5 oz V (5.21) or X 5 o xV o V Y 5 o yV o V Z 5 o zV o V (5.219) XoW 5 oxW YoW 5 oyW ZoW 5 ozW XoV 5 oxV YoV 5 oyV ZoV 5 ozV 276 Distributed Forces: Centroids and Centers of Gravity Fig. 5.21 Centroids of common shapes and volumes.
Shape Semiellipsoid of revolution Paraboloid of revolution Cone Pyramid Hemisphere C Volume 3a 8 3h 8 h 3 h 4 h 4 1 3 abh ⎯x a a a a a b C C C C h h h h ⎯x ⎯x ⎯x ⎯x ⎯x 2 3 a3 2 3 a2h 1 2 a2h 1 3 a2h 5.4 Centers of Gravity and Centroids of Volumes 277 5.4C Determination of Centroids of Volumes by Integration We can determine the centroid of a volume bounded by analytical surfaces by evaluating the integrals given earlier in this section: x V 5# x dV y V 5# y dV z V 5# z dV (5.22) If we choose the element of volume dV to be equal to a small cube with sides dx, dy, and dz, the evaluation of each of these integrals requires a triple integration. However, it is possible to determine the coordinates of the centroid of most volumes by double integration if we choose dV to be equal to the volume of a thin filament (Fig. 5.22). We then obtain the coordinates of the centroid of the volume by rewriting Eqs. (5.22) as x V 5# xel dV y V 5# yel dV z V 5# zel dV (5.23) Then we substitute the expressions given in Fig. 5.22 for the volume dV and the coordinates xel, yel, zel. By using the equation of the surface to express z in terms of x and y, we reduce the integration to a double inte-gration in x and y.
If the volume under consideration possesses two planes of symmetry, its centroid must be located on the line of intersection of the two planes. Choosing the x axis to lie along this line, we have y 5 z 5 0 and the only coordinate to determine is x. This can be done with a single integration by dividing the given volume into thin slabs parallel to the yz plane and expressing dV in terms of x and dx in the equation x V 5# xel dV (5.24) For a body of revolution, the slabs are circular, and their volume is given in Fig. 5.23.
dx r xel z y x xel = x dV = r2 dx Fig. 5.23 Determining the centroid of a body of revolution.
xV 5# xel dV yV 5# yel dV zV 5# zel dV xV 5# xel dV P(x,y,z) z y x z zel xel yel xel = x, yel = y, zel = dV = z dx dy z 2 dx dy Fig. 5.22 Determining the centroid of a volume by double integration.
278 Distributed Forces: Centroids and Centers of Gravity Sample Problem 5.11 Determine the location of the center of gravity of the homogeneous body of revolution shown that was obtained by joining a hemisphere and a cylinder and carving out a cone.
STRATEGY: The body is homogeneous, so the center of gravity coin-cides with the centroid. Since the body was formed from a composite of three simple shapes, you can find the centroid of each shape and combine them using Eq. (5.21).
MODELING: Because of symmetry, the center of gravity lies on the x axis. As shown in Fig. 1, the body is formed by adding a hemisphere to a cylinder and then subtracting a cone. Find the volume and the abscissa of the centroid of each of these components from Fig. 5.21 and enter them in a table (below). Then you can determine the total volume of the body and the first moment of its volume with respect to the yz plane.
Component Volume, mm3 x, mm x V, mm4 Hemisphere 1 2 4π 3 (60)3 5 0.4524 3 106 222.5 210.18 3 106 Cylinder π(60)2(100) 5 1.1310 3 106 150 156.55 3 106 Cone 2π 3 (60)2(100) 5 20.3770 3 106 175 228.28 3 106 oV 5 1.206 3 106 oxV 5 118.09 3 106 Thus, X oV 5 oxV: X(1.206 3 106 mm3) 5 18.09 3 106 mm4 X 5 15 mm b 100 mm x z 60 mm 60 mm y O Fig. 1 The given body modeled as the combination of simple geometric shapes.
50 mm x x x y y y O O O 60 mm 3 8 (60 mm) = 22.5 mm 3 4 (100 mm) = 75 mm + – ANALYSIS: Note that the location of the centroid of the hemisphere is negative because it lies to the left of the origin.
REFLECT and THINK: Adding the hemisphere and subtracting the cone have the effect of shifting the centroid of the composite shape to the left of that for the cylinder (50 mm). However, because the first moment of volume 5.4 Centers of Gravity and Centroids of Volumes 279 for the cylinder is larger than for the hemisphere and cone combined, you should expect the centroid for the composite to still be in the positive x domain. Thus, as a rough visual check, the result of 115 mm is reasonable.
Sample Problem 5.12 Locate the center of gravity of the steel machine part shown. The diame ter of each hole is 1 in.
0.5 in.
0.5 in.
1 in.
1 in.
1 in.
x z y 4.5 in.
2.5 in.
2 in.
2 in.
STRATEGY: This part can be broken down into the sum of two volumes minus two smaller volumes (holes). Find the volume and centroid of each volume and combine them using Eq. (5.21) to find the overall centroid.
MODELING: As shown in Fig. 1, the machine part can be obtained by adding a rectangular parallelepiped (I) to a quarter cylinder (II) and then subtracting two 1-in.-diameter cylinders (III and IV). Determine the volume and the coordinates of the centroid of each component and enter them in a table (below). Using the data in the table, determine the total volume and the moments of the volume with respect to each of the coordinate planes.
ANALYSIS: You can treat each component volume as a planar shape using Fig. 5.8A to find the volumes and centroids, but the right-angle joining of components I and II requires calculations in three dimensions. You may find it helpful to draw more detailed sketches of components with the centroids carefully labeled (Fig. 2).
0.5 in.
0.5 in.
CII CII CI CIII CIV CI, CIII, CIV 1 in.
1 in.
2 in.
1.5 in.
2.25 in.
0.25 in.
0.25 in.
4r 3 = = 0.8488 in.
4 (2) x z y y 8 in.
p 3p 3p Fig. 2 Centroids of components.
4.5 in.
2 in.
I II III IV 2 in.
1 in. diam.
+ _ _ Fig. 1 The given body modeled as the combination of simple geometric shapes.
280 Distributed Forces: Centroids and Centers of Gravity Sample Problem 5.13 Determine the location of the centroid of the half right circular cone shown.
y z x h a STRATEGY: This is not one of the shapes in Fig. 5.21, so you have to determine the centroid by using integration.
MODELING: Since the xy plane is a plane of symmetry, the centroid lies in this plane, and z 5 0. Choose a slab of thickness dx as a differential element. The volume of this element is dV 5 1 2πr2 dx V, in3 x, in. y, in. z, in. x V, in4 y V, in4 z V, in4 I (4.5)(2)(0.5) 5 4.5 0.25 21 2.25 1.125 24.5 10.125 II 1 4π(2)2(0.5) 5 1.571 1.3488 20.8488 0.25 2.119 21.333 0.393 III 2π(0.5)2(0.5) 5 20.3927 0.25 21 3.5 20.098 0.393 21.374 IV 2π(0.5)2(0.5) 5 20.3927 0.25 21 1.5 20.098 0.393 20.589 oV 5 5.286 oxV 5 3.048 oyV 5 25.047 ozV 5 8.555 Thus, XoV 5 oxV: X(5.286 in3) 5 3.048 in4 X 5 0.577 in. b YoV 5 oyV: Y(5.286 in3) 5 25.047 in4 Y 5 20.955 in. b ZoV 5 ozV: Z(5.286 in3) 5 8.555 in4 Z 5 1.618 in. b REFLECT and THINK: By inspection, you should expect X and Z to be considerably less than (1/2)(2.5 in.) and (1/2)(4.5 in.), respectively, and Y to be slightly less in magnitude than (1/2)(2 in.). Thus, as a rough visual check, the results obtained are as expected.
5.4 Centers of Gravity and Centroids of Volumes 281 Obtain the coordinates xel and yel of the centroid of the element from Fig. 5.8 (semicircular area): xel 5 x yel 5 4r 3π Noting that r is proportional to x, use similar triangles (Fig. 1) to write r x 5 a h r 5 a h x y z x h ⎯yel a r ⎯xel = x Fig. 1 Geometry of the differential element.
ANALYSIS: The volume of the body is V 5# dV 5# h 0 1 2 πr2 dx 5# h 0 1 2 π aa h xb 2 dx 5 πa2h 6 The moment of the differential element with respect to the yz plane is xel dV; the total moment of the body with respect to this plane is #xel dV 5# h 0 x(1 2 πr2) dx 5# h 0 x(1 2 π) aa h xb 2 dx 5 πa2h2 8 Thus, xV 5# xel dV x πa2h 6 5 πa2h2 8 x 5 3 4h b Similarly, the moment of the differential element with respect to the zx plane is yel dV; the total moment is # yel dV 5# h 0 4r 3π (1 2πr2)dx 5 2 3# h 0 aa h xb 3 dx 5 a3h 6 Thus, yV 5# yel dV y πa2h 6 5 a3h 6 y 5 a π b REFLECT and THINK: Since a full right circular cone is a body of revolution, its x is unchanged for any portion of the cone bounded by planes intersecting along the x axis. The same centroid location in the x direction was therefore obtained for the half cone that Fig. 5.21 shows for the full cone. Similarly, the same x result would be obtained for a quarter cone.
282 282 I n the problems for this section, you will be asked to locate the centers of gravity of three-dimensional bodies or the centroids of their volumes. All of the techniques we previously discussed for two-dimensional bodies—using symmetry, dividing the body into common shapes, choosing the most efficient differential element, etc.— also may be applied to the general three-dimensional case. 1. Locating the centers of gravity of composite bodies. In general, you must use Eqs. (5.20): XoW 5 oxW YoW 5 oyW ZoW 5 ozW (5.20) However, for the case of a homogeneous body, the center of gravity of the body coincides with the centroid of its volume. Therefore, for this special case, you can also use Eqs. (5.21) to locate the center of gravity of the body: X oV 5 oxV YoV 5 oyV ZoV 5 ozV (5.21) You should realize that these equations are simply an extension of the equations used for the two-dimensional problems considered earlier in the chapter. As the solutions of Sample Probs. 5.11 and 5.12 illustrate, the methods of solution for two- and three-dimensional problems are identical. Thus, we once again strongly encourage you to construct appropriate diagrams and tables when analyzing composite bodies. Also, as you study Sample Prob. 5.12, observe how we obtained the x and y coordinates of the centroid of the quarter cylinder using the equations for the centroid of a quarter circle.
Two special cases of interest occur when the given body consists of either uniform wires or uniform plates made of the same material. a. For a body made of several wire elements of the same uniform cross section, the cross-sectional area A of the wire elements factors out of Eqs. (5.21) when V is replaced with the product AL, where L is the length of a given element. Equations (5.21) thus reduce in this case to XoL 5 oxL YoL 5 oyL ZoL 5 ozL b. For a body made of several plates of the same uniform thickness, the thickness t of the plates factors out of Eqs. (5.21) when V is replaced with the product tA, where A is the area of a given plate. Equations (5.21) thus reduce in this case to XoA 5 oxA YoA 5 oyA ZoA 5 ozA 2. Locating the centroids of volumes by direct integration. As explained in Sec. 5.4C, you can simplify evaluating the integrals of Eqs. (5.22) by choosing either a thin filament (Fig. 5.22) or a thin slab (Fig. 5.23) for the element of volume d V. Thus, you should begin your solution by identifying, if possible, the d V that produces the single or double integrals that are easiest to compute. For bodies of revolution, this may be a thin slab (as in Sample Prob. 5.13) or a thin cylindrical shell. However, it is important to remember that the relationship that you establish among the variables (like the relationship between r and x in Sample Prob. 5.13) directly affects the com-plexity of the integrals you have to compute. Finally, we again remind you that xel, yel, and zel in Eqs. (5.23) are the coordinates of the centroid of dV.
SOLVING PROBLEMS ON YOUR OWN 283 5.96 Consider the composite body shown. Determine (a) the value of x when h 5 L/2, (b) the ratio h/L for which x 5 L.
y a z x b 2 L h b Fig. P5.96 5.97 Determine the location of the centroid of the composite body shown when (a) h 5 2b, (b) h 5 2.5b.
a C B A h b Fig. P5.97 5.98 The composite body shown is formed by removing a semiellipsoid of revolution of semimajor axis h and semiminor axis a/2 from a hemisphere of radius a. Determine (a) the y coordinate of the cen-troid when h 5 a/2, (b) the ratio h/a for which y 5 20.4a. 5.99 Locate the centroid of the frustum of a right circular cone when r1 5 40 mm, r2 5 50 mm, and h 5 60 mm.
h r1 r2 Fig. P5.99 Problems y x z h a a 2 Fig. P5.98 284 5.100 For the machine element shown, locate the x coordinate of the center of gravity.
5.101 For the machine element shown, locate the z coordinate of the center of gravity. 5.102 For the machine element shown, locate the y coordinate of the center of gravity.
1.5 in.
1.5 in.
1.5 in.
2.25 in.
0.75 in.
0.5 in.
x y z r = 0.95 in.
r = 0.95 in.
1.5 in.
1.5 in.
Fig. P5.102 and P5.103 5.103 For the machine element shown, locate the z coordinate of the center of gravity. 5.104 For the machine element shown, locate the y coordinate of the center of gravity.
50 mm 50 mm 60 mm z x O r = 30 mm r = 40 mm 60 mm 60 mm 10 mm 10 mm 10 mm y Fig. P5.104 and P5.105 5.105 For the machine element shown, locate the x coordinate of the center of gravity.
Dimensions in mm y x 19 40 24 10 19 10 90 20 z O r = 12 Fig. P5.100 and P5.101 285 5.106 and 5.107 Locate the center of gravity of the sheet-metal form shown.
y x z 80 mm 125 mm 150 mm 250 mm Fig. P5.106 x y z 1.5 m r = 1.8 m 1.2 m 0.8 m Fig. P5.107 5.108 A corner reflector for tracking by radar has two sides in the shape of a quarter circle with a radius of 15 in. and one side in the shape of a triangle. Locate the center of gravity of the reflector, knowing that it is made of sheet metal with a uniform thickness.
y z x 15 in.
15 in.
Fig. P5.108 5.109 A wastebasket, designed to fit in the corner of a room, is 16 in. high and has a base in the shape of a quarter circle with a radius of 10 in. Locate the center of gravity of the wastebasket, knowing that it is made of sheet metal with a uniform thickness. 5.110 An elbow for the duct of a ventilating system is made of sheet metal with a uniform thickness. Locate the center of gravity of the elbow.
x z y 76 mm 100 mm r = 200 mm r = 400 mm Fig. P5.110 x y z 16 in.
10 in.
10 in.
Fig. P5.109 286 5.111 A window awning is fabricated from sheet metal with a uniform thickness. Locate the center of gravity of the awning.
5.112 A mounting bracket for electronic components is formed from sheet metal with a uniform thickness. Locate the center of gravity of the bracket.
x y z r = 0.625 in.
3 in.
1.25 in.
0.75 in.
0.75 in.
1 in.
2.5 in.
6 in.
Fig. P5.112 5.113 A thin sheet of plastic with a uniform thickness is bent to form a desk organizer. Locate the center of gravity of the organizer.
x z r = 6 mm r = 6 mm r = 6 mm y 60 mm 74 mm 30 mm r = 5 mm 69 mm 75 mm Fig. P5.113 5.114 A thin steel wire with a uniform cross section is bent into the shape shown. Locate its center of gravity.
1.0 m x y z A B C O 2.4 m 2.4 m Fig. P5.114 x y z 4 in.
34 in.
r = 25 in.
Fig. P5.111 287 5.115 The frame of a greenhouse is constructed from uniform aluminum channels. Locate the center of gravity of the portion of the frame shown.
5 ft 3 ft 2 ft x r Fig. P5.115 5.116 and 5.117 Locate the center of gravity of the figure shown, know-ing that it is made of thin brass rods with a uniform diameter.
x y z A B E D O 30 in.
r = 16 in.
Fig. P5.117 x y z A B D O 1.5 m 0.6 m 1 m Fig. P5.116 5.118 A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is 1030 kg/m3 and of steel is 7860 kg/m3, locate the center of gravity of the awl.
10 mm 3.5 mm r 90 mm 25 mm 80 mm 50 mm Fig. P5.118 5.119 A bronze bushing is mounted inside a steel sleeve. Knowing that the specific weight of bronze is 0.318 lb/in3 and of steel is 0.284 lb/in3, determine the location of the center of gravity of the assembly.
0.40 in.
1.00 in.
1.80 in.
1.125 in.
0.5 in.
0.75 in.
Fig. P5.119 288 5.120 A brass collar with a length of 2.5 in. is mounted on an aluminum rod with a length of 4 in. Locate the center of gravity of the com-posite body. (Specific weights: brass 5 0.306 lb/in3, aluminum 5 0.101 lb/in3.) 4 in.
1.6 in.
2.5 in.
3 in.
Fig. P5.120 5.121 The three legs of a small glass-topped table are equally spaced and are made of steel tubing that has an outside diameter of 24 mm and a cross-sectional area of 150 mm2. The diameter and the thickness of the table top are 600 mm and 10 mm, respectively. Knowing that the density of steel is 7860 kg/m3 and of glass is 2190 kg/m3, locate the center of gravity of the table. 5.122 through 5.124 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. 5.122 A hemisphere 5.123 A semiellipsoid of revolution 5.124 A paraboloid of revolution. 5.125 and 5.126 Locate the centroid of the volume obtained by rotating the shaded area about the x axis.
Fig. P5.126 y x a h y = k(x – h)2 y x 1 m 3 m y = (1 – ) 1 x Fig. P5.125 5.127 Locate the centroid of the volume obtained by rotating the shaded area about the line x 5 h.
r = 280 mm r = 180 mm Fig. P5.121 y x x2 h2 y2 a2 + = 1 h a Fig. P5.127 289 5.128 Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the x axis.
y x b a a y5b sin x 2a Fig. P5.128 and P5.129 5.129 Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the y axis. (Hint: Use a thin cylindrical shell of radius r and thickness dr as the element of volume.) 5.130 Show that for a regular pyramid of height h and n sides (n 5 3, 4, . . .) the centroid of the volume of the pyramid is located at a distance h/4 above the base. 5.131 Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R. 5.132 The sides and the base of a punch bowl are of uniform thickness t. If t ,, R and R 5 250 mm, determine the location of the center of gravity of (a) the bowl, (b) the punch.
R R Fig. P5.132 5.133 Locate the centroid of the section shown, which was cut from a thin circular pipe by two oblique planes.
y x z h h 3 a a Fig. P5.133 x y z R R Fig. P5.131 290 5.134 Locate the centroid of the section shown, which was cut from an elliptical cylinder by an oblique plane.
x z h y a a b b Fig. P5.134 5.135 Determine by direct integration the location of the centroid of the volume between the xz plane and the portion shown of the surface y 5 16h(ax – x2)(bz – z2)/a2b2.
y x z b a Fig. P5.135 5.136 After grading a lot, a builder places four stakes to designate the corners of the slab for a house. To provide a firm, level base for the slab, the builder places a minimum of 3 in. of gravel beneath the slab. Determine the volume of gravel needed and the x coordinate of the centroid of the volume of the gravel. (Hint: The bottom surface of the gravel is an oblique plane, which can be represented by the equa-tion y 5 a 1 bx 1 cz.) y x z 5 in.
3 in.
8 in.
6 in.
30 ft 50 ft Fig. P5.136 291 This chapter was devoted chiefly to determining the center of gravity of a rigid body, i.e., to determining the point G where we can apply a single force W—the weight of the body—to represent the effect of Earth’s attraction on the body.
Center of Gravity of a Two-Dimensional Body In the first part of this chapter, we considered two-dimensional bodies, such as flat plates and wires contained in the xy plane. By adding force components in the vertical z direction and moments about the horizontal y and x axes [Sec. 5.1A], we derived the relations W 5# dW xW 5# x dW yW 5# y dW (5.2) These equations define the weight of the body and the coordinates x and y of its center of gravity.
Centroid of an Area or Line In the case of a homogeneous flat plate of uniform thickness [Sec. 5.1B], the center of gravity G of the plate coincides with the centroid C of the area A of the plate. The coordinates are defined by the relations xA 5# x dA yA 5# y dA (5.3) Similarly, determining the center of gravity of a homogeneous wire of uniform cross section contained in a plane reduces to determining the centroid C of the line L representing the wire; we have xL 5#x dL yL 5#y dL (5.4) First Moments The integrals in Eqs. (5.3) are referred to as the first moments of the area A with respect to the y and x axes and are denoted by Qy and Qx, respectively [Sec. 5.1C]. We have Qy 5 xA Qx 5 yA (5.6) The first moments of a line can be defined in a similar way.
Properties of Symmetry Determining the centroid C of an area or line is simplified when the area or line possesses certain properties of symmetry. If the area or line is symmetric with respect to an axis, its centroid C lies on that axis; if it is symmetric with respect to two axes, C is located at the intersection of the two axes; if it is symmetric with respect to a center O, C coincides with O.
Review and Summary 292 Center of Gravity of a Composite Body The areas and the centroids of various common shapes are tabulated in Fig. 5.8. When a flat plate can be divided into several of these shapes, the coordinates X and Y of its center of gravity G can be determined from the coordinates x1, x2, . . . and y1, y2, . . . of the centers of gravity G1, G2, . . . of the various parts [Sec. 5.1D]. Equating moments about the y and x axes, respectively (Fig. 5.24), we have XoW 5 oxW YoW 5 oyW (5.7) x y z x y z O O G ⎯X ⎯Y ΣW = G1 G2 G3 W1 W2 W3 Fig. 5.24 If the plate is homogeneous and of uniform thickness, its center of gravity coincides with the centroid C of the area of the plate, and Eqs. (5.7) reduce to Qy 5 XoA 5 oxA Qx 5 YoA 5 oyA (5.8) These equations yield the first moments of the composite area, or they can be solved for the coordinates X and Y of its centroid [Sample Prob. 5.1]. Deter-mining the center of gravity of a composite wire is carried out in a similar fashion [Sample Prob. 5.2].
Determining a Centroid by Integration When an area is bounded by analytical curves, you can determine the coor-dinates of its centroid by integration [Sec. 5.2A]. This can be done by evaluat-ing either the double integrals in Eqs. (5.3) or a single integral that uses one of the thin rectangular or pie-shaped elements of area shown in Fig. 5.12. Denoting by xel and yel the coordinates of the centroid of the element dA, we have Qy 5 xA 5# xel dA Qx 5 yA 5# yel dA (5.9) It is advantageous to use the same element of area to compute both of the first moments Qy and Qx; we can also use the same element to determine the area A [Sample Prob. 5.4].
Theorems of Pappus–Guldinus The theorems of Pappus-Guldinus relate the area of a surface of revolution or the volume of a body of revolution to the centroid of the generating curve or area [Sec. 5.2B]. The area A of the surface generated by rotating a curve of length L about a fixed axis (Fig. 5.25a) is A 5 2πyL (5.10) where y represents the distance from the centroid C of the curve to the fixed axis. Similarly, the volume V of the body generated by rotating an area A (a) (b) x C L y y x A C 2 y 2 y Fig. 5.25 293 about a fixed axis (Fig. 5.25b) is V 5 2πyA (5.11) where y represents the distance from the centroid C of the area to the fixed axis.
Distributed Loads The concept of centroid of an area also can be used to solve problems other than those dealing with the weight of flat plates. For example, to determine the reactions at the supports of a beam [Sec. 5.3A], we can replace a distributed load w by a concentrated load W equal in magnitude to the area A under the load curve and passing through the centroid C of that area (Fig. 5.26). We can use this same approach to determine the resultant of the hydrostatic forces exerted on a rectangular plate submerged in a liquid [Sec. 5.3B].
w w O O w dx x L B B dW = dA x x L P x W = A W d W C = Fig. 5.26 Center of Gravity of a Three-Dimensional Body The last part of this chapter was devoted to determining the center of gravity G of a three-dimensional body. We defined the coordinates x, y, z of G by the relations xW 5# x dW yW 5# y dW z W 5# z dW (5.17) Centroid of a Volume In the case of a homogeneous body, the center of gravity G coincides with the centroid C of the volume V of the body. The coordinates of C are defined by the relations xV 5# x dV yV 5# y dV zV 5# z dV (5.19) If the volume possesses a plane of symmetry, its centroid C lies in that plane; if it possesses two planes of symmetry, C is located on the line of intersection of the two planes; if it possesses three planes of symmetry that intersect at only one point, C coincides with that point [Sec. 5.4A].
Center of Gravity of a Composite Body The volumes and centroids of various common three-dimensional shapes are tabulated in Fig. 5.21. When a body can be divided into several of these shapes, we can determine the coordinates X, Y, Z of its center of gravity G from the corresponding coordinates of the centers of gravity of its various parts [Sec. 5.4B]. We have XoW 5 oxW YoW 5 oyW ZoW 5 ozW (5.20) 294 If the body is made of a homogeneous material, its center of gravity coincides with the centroid C of its volume, and we have [Sample Probs. 5.11 and 5.12] XoV 5 oxV YoV 5 oyV ZoV 5 ozV (5.21) Determining a Centroid by Integration When a volume is bounded by analytical surfaces, we can find the coordinates of its centroid by integration [Sec. 5.4C]. To avoid the computation of triple integrals in Eqs. (5.19), we can use elements of volume in the shape of thin filaments, as shown in Fig. 5.27. Denoting the coordinates of the centroid of the element dV as xel, yel, zel, we rewrite Eqs. (5.19) as xV 5# xel dV yV 5# yel dV zV 5# zel dV (5.23) P(x,y,z) z y x z zel xel yel xel = x, yel = y, zel = dV = z dx dy z 2 dx dy Fig. 5.27 that involve only double integrals. If the volume possesses two planes of sym-metry, its centroid C is located on their line of intersection. Choosing the x axis to lie along that line and dividing the volume into thin slabs parallel to the yz plane, we can determine C from the relation xV 5# xel dV (5.24) with a single integration [Sample Prob. 5.13]. For a body of revolution, these slabs are circular and their volume is given in Fig. 5.28. dx r xel z y x xel = x dV = r2 dx Fig. 5.28 295 5.137 and 5.138 Locate the centroid of the plane area shown.
Review Problems y 6 in.
6 in.
6 in.
6 in.
3 in.
x Fig. P5.137 x y 120 mm r = 75 mm Fig. P5.138 5.139 A uniform circular rod with a weight of 8 lb and radius of 10 in. is attached to a pin at C and to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C.
5.140 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h.
x y y = h(1 – kx3) h a Fig. P5.140 5.141 Determine by direct integration the centroid of the area shown.
x y h L y = h 1 + – 2 x2 L2 x L ( ( Fig. P5.141 5.142 The escutcheon (a decorative plate placed on a pipe where the pipe exits from a wall) shown is cast from brass. Knowing that the den-sity of brass is 8470 kg/m3, determine the mass of the escutcheon.
75 mm 25 mm 75 mm 26° 26° Fig. P5.142 B r C A Fig. P5.139 296 5.143 Determine the reactions at the beam supports for the given loading.
5.144 A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE that exert uniformly distrib-uted upward loads as shown. Determine the values of wBC and wDE corresponding to equilibrium when wA 5 600 N/m.
A wA wBC wDE B C D E F 6 m 3.1 m 0.6 m 1.0 m 0.8 m 1200 N/m Fig. P5.144 5.145 A tank is divided into two sections by a 1 3 1-m square gate that is hinged at A. A couple with a magnitude of 490 N∙m is required for the gate to rotate. If one side of the tank is filled with water at the rate of 0.1 m3/min and the other side is filled simultaneously with methyl alcohol (density ρma 5 789 kg/m3) at the rate of 0.2 m3/min, determine at what time and in which direction the gate will rotate. 5.146 Determine the y coordinate of the centroid of the body shown.
x y z 12 in.
12 in.
4 in.
8 in.
Fig. P5.147 192 mm 64 mm 96 mm 120° 120° Fig. P5.148 A Water 0.4 m 0.2 m Methyl Alcohol 0.6 m Fig. P5.145 y x z h a b a 2 Fig. P5.146 5.147 An 8-in.-diameter cylindrical duct and a 4 3 8-in. rectangular duct are to be joined as indicated. Knowing that the ducts were fabricated from the same sheet metal, which is of uniform thickness, locate the center of gravity of the assembly.
5.148 Three brass plates are brazed to a steel pipe to form the flagpole base shown. Knowing that the pipe has a wall thickness of 8 mm and that each plate is 6 mm thick, determine the location of the center of gravity of the base. (Densities: brass 5 8470 kg/m3; steel 5 7860 kg/m3.) 9 ft A B 200 lb/ft 6 ft 6 ft Fig. P5.143 Trusses, such as this cantilever arch bridge over Deception Pass in Washington State, provide both a practical and an economical solution to many engineering problems.
Analysis of Structures 6 298 Analysis of Structures Introduction In the preceding chapters, we studied the equilibrium of a single rigid body, where all forces involved were external to the rigid body. We now consider the equilibrium of structures made of several connected parts. This situation calls for determining not only the external forces acting on the structure, but also the forces that hold together the various parts of the structure. From the point of view of the structure as a whole, these forces are internal forces.
Consider, for example, the crane shown in Fig. 6.1a that supports a load W. The crane consists of three beams AD, CF, and BE connected by frictionless pins; it is supported by a pin at A and by a cable DG. The free-body diagram of the crane is drawn in Fig. 6.1b. The external forces shown in the diagram include the weight W, the two components Ax and Ay of the reaction at A, and the force T exerted by the cable at D. The internal forces holding the various parts of the crane together do not appear in the free-body diagram. If, however, we dismember the crane and draw a free-body diagram for each of its component parts, we can see the forces holding the three beams together, since these forces are external forces from the point of view of each component part (Fig. 6.1c). Introduction 6.1 ANALYSIS OF TRUSSES 6.1A Simple Trusses 6.1B The Method of Joints 6.1C Joints Under Special Loading Conditions 6.1D Space Trusses 6.2 OTHER TRUSS ANALYSES 6.2A The Method of Sections 6.2B Trusses Made of Several Simple Trusses 6.3 FRAMES 6.3A Analysis of a Frame 6.3B Frames That Collapse Without Supports 6.4 MACHINES Objectives • Define an ideal truss, and consider the attributes of simple trusses.
• Analyze plane and space trusses by the method of joints.
• Simplify certain truss analyses by recognizing special loading and geometry conditions.
• Analyze trusses by the method of sections.
• Consider the characteristics of compound trusses.
• Analyze structures containing multiforce members, such as frames and machines.
Fig. 6.1 A structure in equilibrium. (a) Diagram of a crane supporting a load; (b) free-body diagram of the crane; (c) free-body diagrams of the components of the crane.
T T A B C D E F W B C D E E F E F W W G (a) B B C C D (b) (c) Ay Ax A Ay Ax A 6.1 Analysis of Trusses 299 Note that we represent the force exerted at B by member BE on member AD as equal and opposite to the force exerted at the same point by member AD on member BE. Similarly, the force exerted at E by BE on CF is shown equal and opposite to the force exerted by CF on BE, and the components of the force exerted at C by CF on AD are shown equal and opposite to the components of the force exerted by AD on CF. These representations agree with Newton’s third law, which states that The forces of action and reaction between two bodies in contact have the same magnitude, same line of action, and opposite sense.
We pointed out in Chap. 1 that this law, which is based on experimental evidence, is one of the six fundamental principles of elementary mechan-ics. Its application is essential for solving problems involving connected bodies.
In this chapter, we consider three broad categories of engineering structures: 1. Trusses, which are designed to support loads and are usually stationary, fully constrained structures. Trusses consist exclusively of straight mem-bers connected at joints located at the ends of each member. Members of a truss, therefore, are two-force members, i.e., members acted upon by two equal and opposite forces directed along the member. 2. Frames, which are also designed to support loads and are also usually stationary, fully constrained structures. However, like the crane of Fig. 6.1, frames always contain at least one multi-force member, i.e., a member acted upon by three or more forces that, in general, are not directed along the member. 3. Machines, which are designed to transmit and modify forces and are structures containing moving parts. Machines, like frames, always con-tain at least one multi-force member.
6.1 ANALYSIS OF TRUSSES The truss is one of the major types of engineering structures. It provides a practical and economical solution to many engineering situations, espe-cially in the design of bridges and buildings. In this section, we describe the basic elements of a truss and study a common method for analyzing the forces acting in a truss.
Photo 6.1 The structures you see around you to support loads or transmit forces are generally trusses, frames, or machines.
Two-force member (a) A truss bridge Multi-force member (b) A bicycle frame Multi-force member (c) A hydraulic machine arm 300 Analysis of Structures 6.1A Simple Trusses A truss consists of straight members connected at joints, as shown in Fig. 6.2a. Truss members are connected at their extremities only; no mem-ber is continuous through a joint. In Fig. 6.2a, for example, there is no member AB; instead we have two distinct members AD and DB. Most actual structures are made of several trusses joined together to form a space framework. Each truss is designed to carry those loads that act in its plane and thus may be treated as a two-dimensional structure.
In general, the members of a truss are slender and can support little lateral load; all loads, therefore, must be applied at the various joints and not to the members themselves. When a concentrated load is to be applied between two joints or when the truss must support a distributed load, as in the case of a bridge truss, a floor system must be provided. The floor transmits the load to the joints through the use of stringers and floor beams (Fig. 6.3).
Photo 6.2 Shown is a pin-jointed connection on the approach span to the San Francisco–Oakland Bay Bridge.
Fig. 6.2 (a) A typical truss consists of straight members connected at joints; (b) we can model a truss as two-force members connected by pins.
A B C D (a) (b) P A B C D P Fig. 6.4 A two-force member of a truss can be in tension or compression.
(a) Tension (b) Compression We assume that the weights of the truss members can be applied to the joints, with half of the weight of each member applied to each of the two joints the member connects. Although the members are actually joined together by means of welded, bolted, or riveted connections, it is custom-ary to assume that the members are pinned together; therefore, the forces acting at each end of a member reduce to a single force and no couple. This enables us to model the forces applied to a truss member as a single force at each end of the member. We can then treat each member as a two-force member, and we can consider the entire truss as a group of pins and two-force members (Fig. 6.2b). An individual member can be acted upon as shown in either of the two sketches of Fig. 6.4. In Fig. 6.4a, the forces tend to pull the member apart, and the member is in tension; in Fig. 6.4b, the forces tend to push the member together, and the member is in compression. Some typical trusses are shown in Fig. 6.5.
Consider the truss of Fig. 6.6a, which is made of four members connected by pins at A, B, C, and D. If we apply a load at B, the truss will greatly deform, completely losing its original shape. In contrast, the truss of Fig. 6.6b, which is made of three members connected by pins at A, B, and C, will deform only slightly under a load applied at B. The only possible Floor beam Stringer Joints Fig. 6.3 A floor system of a truss uses stringers and floor beams to transmit an applied load to the joints of the truss.
6.1 Analysis of Trusses 301 deformation for this truss is one involving small changes in the length of its members. The truss of Fig. 6.6b is said to be a rigid truss, the term ‘rigid’ being used here to indicate that the truss will not collapse.
As shown in Fig. 6.6c, we can obtain a larger rigid truss by adding two members BD and CD to the basic triangular truss of Fig. 6.6b. We can repeat this procedure as many times as we like, and the resulting truss will be rigid if each time we add two new members they are attached to two existing joints and connected at a new joint. (The three joints must not be in a straight line.) A truss that can be constructed in this manner is called a simple truss.
Note that a simple truss is not necessarily made only of triangles. The truss of Fig. 6.6d, for example, is a simple truss that we constructed from triangle ABC by adding successively the joints D, E, F, and G. Fig. 6.5 You can often see trusses in the design of a building roof, a bridge, or other other larger structures.
Pratt Pratt Howe Howe Fink Typical Roof Trusses Typical Bridge Trusses Baltimore Warren K truss Stadium Cantilever portion of a truss Bascule Other Types of Trusses Fig. 6.6 (a) A poorly designed truss that cannot support a load; (b) the most elementary rigid truss consists of a simple triangle; (c) a larger rigid truss built up from the triangle in (b); (d) a rigid truss not made up of triangles alone.
A B B' C A B C A B C C' D D A B C D E F G (a) (b) (c) (d) 302 Analysis of Structures On the other hand, rigid trusses are not always simple trusses, even when they appear to be made of triangles. The Fink and Baltimore trusses shown in Fig. 6.5, for instance, are not simple trusses, because they cannot be constructed from a single triangle in the manner just described. All of the other trusses shown in Fig. 6.5 are simple trusses, as you may easily check. (For the K truss, start with one of the central triangles.) Also note that the basic triangular truss of Fig. 6.6b has three mem-bers and three joints. The truss of Fig. 6.6c has two more members and one more joint; i.e., five members and four joints altogether. Observing that every time we add two new members, we increase the number of joints by one, we find that in a simple truss the total number of members is m 5 2n 2 3, where n is the total number of joints.
6.1B The Method of Joints We have just seen that a truss can be considered as a group of pins and two-force members. Therefore, we can dismember the truss of Fig. 6.2, whose free-body diagram is shown in Fig. 6.7a, and draw a free-body diagram for each pin and each member (Fig. 6.7b). Each member is acted upon by two forces, one at each end; these forces have the same magnitude, same line of action, and opposite sense (Sec. 4.2A). Furthermore, Newton’s third law states that the forces of action and reaction between a member and a pin are equal and opposite. Therefore, the forces exerted by a mem-ber on the two pins it connects must be directed along that member and be equal and opposite. The common magnitude of the forces exerted by a member on the two pins it connects is commonly referred to as the force in the member, even though this quantity is actually a scalar. Since we know the lines of action of all the internal forces in a truss, the analysis of a truss reduces to computing the forces in its various members and determining whether each of its members is in tension or compression.
Since the entire truss is in equilibrium, each pin must be in equilib-rium. We can use the fact that a pin is in equilibrium to draw its free-body diagram and write two equilibrium equations (Sec. 2.3A). Thus, if the truss contains n pins, we have 2n equations available, which can be solved for 2n unknowns. In the case of a simple truss, we have m 5 2n 2 3; that is, 2n 5 m 1 3, and the number of unknowns that we can determine from the free-body diagrams of the pins is m 1 3. This means that we can find the forces in all the members, the two components of the reaction RA, and the reaction RB by considering the free-body diagrams of the pins.
We can also use the fact that the entire truss is a rigid body in equi-librium to write three more equations involving the forces shown in the free-body diagram of Fig. 6.7a. Since these equations do not contain any new information, they are not independent of the equations associated with the free-body diagrams of the pins. Nevertheless, we can use them to deter-mine the components of the reactions at the supports. The arrangement of pins and members in a simple truss is such that it is always possible to find a joint involving only two unknown forces. We can determine these forces by using the methods of Sec. 2.3C and then transferring their values to the adjacent joints, treating them as known quantities at these joints. We repeat this procedure until we have determined all unknown forces.
As an example, let’s analyze the truss of Fig. 6.7 by considering the equilibrium of each pin successively, starting with a joint at which only Photo 6.3 Two K trusses were used as the main components of the movable bridge shown, which moved above a large stockpile of ore. The bucket below the trusses picked up ore and redeposited it until the ore was thoroughly mixed. The ore was then sent to the mill for processing into steel.
Fig. 6.7 (a) Free-body diagram of the truss as a rigid body; (b) free-body diagrams of the five members and four pins that make up the truss.
D A B C C B P P (a) (b) RB RB D RA A RA 6.1 Analysis of Trusses 303 two forces are unknown. In this truss, all pins are subjected to at least three unknown forces. Therefore, we must first determine the reactions at the supports by considering the entire truss as a free body and using the equations of equilibrium of a rigid body. In this way we find that RA is vertical, and we determine the magnitudes of RA and RB.
This reduces the number of unknown forces at joint A to two, and we can determine these forces by considering the equilibrium of pin A. The reaction RA and the forces FAC and FAD exerted on pin A by members AC and AD, respectively, must form a force triangle. First we draw RA (Fig. 6.8); noting that FAC and FAD are directed along AC and AD, respec-tively, we complete the triangle and determine the magnitude and sense of FAC and FAD. The magnitudes FAC and FAD represent the forces in members AC and AD. Since FAC is directed down and to the left––that is, toward joint A––member AC pushes on pin A and is in compression. (From Newton’s third law, pin A pushes on member AC.) Since FAD is directed away from joint A, member AD pulls on pin A and is in tension. (From Newton’s third law, pin A pulls away from member AD.) Photo 6.4 Because roof trusses, such as those shown, require support only at their ends, it is possible to construct buildings with large, unobstructed interiors.
Fig. 6.8 Free-body diagrams and force polygons used to determine the forces on the pins and in the members of the truss in Fig. 6.7.
Free-body diagram Joint A Joint D Joint C Joint B B Force polygon FAC FAC FAD FDA FCA FCB FCD FCD FCA FCB RB RB FBD FBD FBC FBC FDA FDC FDC FDB FDB P P FAD RA RA A D C 304 Analysis of Structures We can now proceed to joint D, where only two forces, FDC and FDB, are still unknown. The other forces are the load P, which is given, and the force FDA exerted on the pin by member AD. As indicated previously, this force is equal and opposite to the force FAD exerted by the same member on pin A. We can draw the force polygon corresponding to joint D, as shown in Fig. 6.8, and determine the forces FDC and FDB from that polygon. However, when more than three forces are involved, it is usually more convenient to solve the equations of equilibrium oFx 5 0 and oFy 5 0 for the two unknown forces. Since both of these forces are directed away from joint D, members DC and DB pull on the pin and are in tension.
Next, we consider joint C; its free-body diagram is shown in Fig. 6.8. Both FCD and FCA are known from the analysis of the preceding joints, so only FCB is unknown. Since the equilibrium of each pin provides sufficient information to determine two unknowns, we can check our analysis at this joint. We draw the force triangle and determine the magnitude and sense of FCB. Since FCB is directed toward joint C, member CB pushes on pin C and is in compression. The check is obtained by verifying that the force FCB and member CB are parallel.
Finally, at joint B, we know all of the forces. Since the correspond-ing pin is in equilibrium, the force triangle must close, giving us an addi-tional check of the analysis. Note that the force polygons shown in Fig. 6.8 are not unique; we could replace each of them by an alternative configuration. For example, the force triangle corresponding to joint A could be drawn as shown in Fig. 6.9. We obtained the triangle actually shown in Fig. 6.8 by drawing the three forces RA, FAC, and FAD in tip-to-tail fashion in the order in which we cross their lines of action when moving clockwise around joint A.
6.1C Joints Under Special Loading Conditions Some geometric arrangements of members in a truss are particularly simple to analyze by observation. For example, Fig. 6.10a shows a joint connecting four members lying along two intersecting straight lines. The free-body dia-gram of Fig. 6.10b shows that pin A is subjected to two pairs of directly opposite forces. The corresponding force polygon, therefore, must be a paral-lelogram (Fig. 6.10c), and the forces in opposite members must be equal.
Fig. 6.9 Alternative force polygon for joint A in Fig. 6.8.
RA FAD FAC Fig. 6.10 (a) A joint A connecting four members of a truss in two straight lines; (b) free-body diagram of pin A; (c) force polygon (parallelogram) for pin A. Forces in opposite members are equal.
(a) A A D C B E (b) (c) FAD FAB FAE FAC FAD FAB FAE FAC 6.1 Analysis of Trusses 305 Consider next Fig. 6.11a, in which a joint connects three members and supports a load P. Two members lie along the same line, and load P acts along the third member. The free-body diagram of pin A and the corresponding force polygon are the same as in Fig. 6.10b and c, with FAE replaced by load P. Thus, the forces in the two opposite members must be equal, and the force in the other member must equal P. Figure 6.11b shows a particular case of special interest. Since, in this case, no external load is applied to the joint, we have P 5 0, and the force in member AC is zero. Member AC is said to be a zero-force member.
Now consider a joint connecting two members only. From Sec. 2.3A, we know that a particle acted upon by two forces is in equilibrium if the two forces have the same magnitude, same line of action, and opposite sense. In the case of the joint of Fig. 6.12a, which connects two members AB and AD lying along the same line, the forces in the two members must be equal for pin A to be in equilibrium. In the case of the joint of Fig. 6.12b, pin A cannot be in equilibrium unless the forces in both members are zero. Members connected as shown in Fig. 6.12b, therefore, must be zero-force members.
Fig. 6.12 (a) A joint in a truss connecting two members in a straight line. Forces in the members are equal. (b) If the two members are not in a straight line, they must be zero-force members.
(a) A D B (b) A D B Spotting joints that are under the special loading conditions just described will expedite the analysis of a truss. Consider, for example, a Howe truss loaded as shown in Fig. 6.13. We can recognize all of the members represented by green lines as zero-force members. Joint C con-nects three members, two of which lie in the same line, and is not sub-jected to any external load; member BC is thus a zero-force member. Applying the same reasoning to joint K, we find that member JK is also a zero-force member. But joint J is now in the same situation as joints C and K, so member IJ also must be a zero-force member. Examining joints C, J, and K also shows that the forces in members AC and CE are equal, that the forces in members HJ and JL are equal, and that the forces in members IK and KL are equal. Turning our attention to joint I, where the 20-kN load and member HI are collinear, we note that the force in member HI is 20 kN (tension) and that the forces in members GI and IK are equal. Hence, the forces in members GI, IK, and KL are equal.
Note that the conditions described here do not apply to joints B and D in Fig. 6.13, so it is wrong to assume that the force in member DE is 25 kN or that the forces in members AB and BD are equal. To determine the forces in these members and in all remaining members, you need to Fig. 6.11 (a) Joint A in a truss connects three members, two in a straight line and the third along the line of a load. Force in the third member equals the load. (b) If the load is zero, the third member is a zero-force member.
(a) A D C B (b) A P D C B Fig. 6.13 An example of loading on a Howe truss; identifying special loading conditions.
A B C D E F G H 25 kN 25 kN 50 kN 20 kN I J K L 306 Analysis of Structures carry out the analysis of joints A, B, D, E, F, G, H, and L in the usual manner. Thus, until you have become thoroughly familiar with the condi-tions under which you can apply the rules described in this section, you should draw the free-body diagrams of all pins and write the correspond-ing equilibrium equations (or draw the corresponding force polygons) whether or not the joints being considered are under one of these special loading conditions.
A final remark concerning zero-force members: These members are not useless. For example, although the zero-force members of Fig. 6.13 do not carry any loads under the loading conditions shown, the same members would probably carry loads if the loading conditions were changed. Besides, even in the case considered, these members are needed to support the weight of the truss and to maintain the truss in the desired shape.
6.1D Space Trusses When several straight members of a truss are joined together at their extremities to form a three-dimensional configuration, the resulting struc-ture is called a space truss. Recall from Sec. 6.1A that the most elemen-tary two-dimensional rigid truss consists of three members joined at their extremities to form the sides of a triangle. By adding two members at a time to this basic configuration and connecting them at a new joint, we could obtain a larger rigid structure that we defined as a simple truss. Similarly, the most elementary rigid space truss consists of six members joined at their extremities to form the edges of a tetrahedron ABCD (Fig. 6.14a). By adding three members at a time to this basic configuration, such as AE, BE, and CE (Fig. 6.14b), attaching them to three existing joints, and connecting them at a new joint, we can obtain a larger rigid structure that we define as a simple space truss. (The four joints must not lie in a plane.) Note that the basic tetrahedron has six members and four joints, and every time we add three members, the number of joints increases by one. Therefore, we conclude that in a simple space truss the total num-ber of members is m 5 3n 2 6, where n is the total number of joints.
If a space truss is to be completely constrained and if the reactions at its supports are to be statically determinate, the supports should consist of a combination of balls, rollers, and balls and sockets, providing six unknown reactions (see Sec. 4.3B). We can determine these unknown reactions by solving the six equations expressing that the three-dimen-sional truss is in equilibrium.
Although the members of a space truss are actually joined together by means of bolted or welded connections, we assume for analysis purposes that each joint consists of a ball-and-socket connection. Thus, no couple is applied to the members of the truss, and we can treat each member as a two-force member. The conditions of equilibrium for each joint are expressed by the three equations oFx 5 0, oFy 5 0, and oFz 5 0. Thus, in the case of a simple space truss containing n joints, writing the conditions of equilibrium for each joint yields 3n equations. Since m 5 3n 2 6, these equations suffice to determine all unknown forces (forces in m members and six reactions at the supports). However, to avoid the necessity of solving simultaneous equations, you should take care to select joints in such an order that no selected joint involves more than three unknown forces.
Photo 6.5 Three-dimensional or space trusses are used for broadcast and power transmission line towers, roof framing, and spacecraft applications, such as components of the International Space Station.
Fig. 6.14 (a) The most elementary space truss consists of six members joined at their ends to form a tetrahedron. (b) We can add three members at a time to three joints of an existing space truss, connecting the new members at a new joint, to build a larger simple space truss.
A B C D A B C D E (a) (b) 6.1 Analysis of Trusses 307 Sample Problem 6.1 Using the method of joints, determine the force in each member of the truss shown.
STRATEGY: To use the method of joints, you start with an analysis of the free-body diagram of the entire truss. Then look for a joint connecting only two members as a starting point for the calculations. In this example, we start at joint A and proceed through joints D, B, E, and C, but you could also start at joint C and proceed through joints E, B, D, and A.
MODELING and ANALYSIS: You can combine these steps for each joint of the truss in turn. Draw a free-body diagram; draw a force polygon or write the equilibrium equations; and solve for the unknown forces.
Entire Truss. Draw a free-body diagram of the entire truss (Fig. 1); external forces acting on this free body are the applied loads and the reactions at C and E. Write the equilibrium equations, taking moments about C.
1loMC 5 0: (2000 lb)(24 ft) 1 (1000 lb)(12 ft) 2 E(6 ft) 5 0 E 5 110,000 lb E 5 10,000 lbx y 1 oFx 5 0: Cx 5 0 1 xoFy 5 0: 22000 lb 2 1000 lb 1 10,000 lb 1 Cy 5 0 Cy 5 27000 lb Cy 5 7000 lb w Joint A. This joint is subject to only two unknown forces: the forces exerted by AB and those by AD. Use a force triangle to determine FAB and FAD (Fig. 2). Note that member AB pulls on the joint so AB is in tension, and member AD pushes on the joint so AD is in compression. Obtain the magnitudes of the two forces from the proportion 2000 lb 4 5 FAB 3 5 FAD 5 FAB 5 1500 lb T b FAD 5 2500 lb C b Joint D. Since you have already determined the force exerted by mem-ber AD, only two unknown forces are now involved at this joint. Again, use a force triangle to determine the unknown forces in members DB and DE (Fig. 3).
(continued) 12 ft 12 ft 12 ft 6 ft 6 ft 8 ft A B C D E 2000 lb 1000 lb Fig. 1 Free-body diagram of entire truss.
12 ft 12 ft 12 ft 6 ft 6 ft 8 ft A B C D E E 2000 lb 1000 lb Cy Cx FAD FAD FAB FAB A 2000 lb 2000 lb 3 3 4 4 5 5 Fig. 2 Free-body diagram of joint A.
FDA = 2500 lb D FDB FDB FDE FDE FDA 3 3 4 4 5 5 Fig. 3 Free-body diagram of joint D.
308 Analysis of Structures FDB 5 FDA FDB 5 2500 lb T b FDE 5 2A3 5BFDA FDE 5 3000 lb C b Joint B. Since more than three forces act at this joint (Fig. 4), determine the two unknown forces FBC and FBE by solving the equilib-rium equations oFx 5 0 and oFy 5 0. Suppose you arbitrarily assume that both unknown forces act away from the joint, i.e., that the members are in tension. The positive value obtained for FBC indicates that this assumption is correct; member BC is in tension. The negative value of FBE indicates that the second assumption is wrong; member BE is in compression.
FBA = 1500 lb FBD = 2500 lb FBE B FBC 1000 lb 3 3 4 4 Fig. 4 Free-body diagram of joint B.
1 xoFy 5 0: 21000 2 4 5(2500) 2 4 5FBE 5 0 FBE 5 23750 lb FBE 5 3750 lb C b y 1 oFx 5 0: FBC 2 1500 2 3 5(2500) 2 3 5(3750) 5 0 FBC 5 15250 lb FBC 5 5250 lb T b Joint E. Assume the unknown force FEC acts away from the joint (Fig. 5). Summing x components, you obtain y 1 oFx 5 0: 3 5FEC 1 3000 1 3 5(3750) 5 0 FEC 5 28750 lb FEC 5 8750 lb C b Summing y components, you obtain a check of your computations: 1 xoFy 5 10,000 2 4 5(3750) 2 4 5(8750) 5 10,000 2 3000 2 7000 5 0 (checks) REFLECT and THINK: Using the computed values of FCB and FCE, you can determine the reactions Cx and Cy by considering the equilibrium of Joint C (Fig. 6). Since these reactions have already been determined from the equilibrium of the entire truss, this provides two checks of your com putations. You can also simply use the computed values of all forces acting on the joint (forces in members and reactions) and check that the joint is in equilibrium: y 1 oFx 5 25250 1 3 5(8750) 5 25250 1 5250 5 0 (checks) 1 xoFy 5 27000 1 4 5(8750) 5 27000 1 7000 5 0 (checks) FEB = 3750 lb FEC FED = 3000 lb E = 10,000 lb E 3 3 4 4 Fig. 5 Free-body diagram of joint E.
FCB = 5250 lb FCE = 8750 lb Cy = 7000 lb Cx = 0 C 3 4 Fig. 6 Free-body diagram of joint C.
309 309 SOLVING PROBLEMS ON YOUR OWN I n this section, you learned to use the method of joints to determine the forces in the members of a simple truss; that is, a truss that can be constructed from a basic triangular truss by adding to it two new members at a time and connecting them at a new joint.
The method consists of the following steps: 1. Draw a free-body diagram of the entire truss, and use this diagram to deter-mine the reactions at the supports.
2. Locate a joint connecting only two members, and draw the free-body diagram of its pin. Use this free-body diagram to determine the unknown force in each of the two members. If only three forces are involved (the two unknown forces and a known one), you will probably find it more convenient to draw and solve the corresponding force triangle. If more than three forces are involved, you should write and solve the equilibrium equations for the pin, oFx 5 0 and oFy 5 0, assuming that the members are in tension. A positive answer means that the member is in tension, a negative answer means that the member is in compression. Once you have found the forces, enter their values on a sketch of the truss with T for tension and C for compression.
3. Next, locate a joint where the forces in only two of the connected members are still unknown. Draw the free-body diagram of the pin and use it as indicated in Step 2 to determine the two unknown forces.
4. Repeat this procedure until you have found the forces in all the members of the truss. Since you previously used the three equilibrium equations associated with the free-body diagram of the entire truss to determine the reactions at the supports, you will end up with three extra equations. These equations can be used to check your computations.
5. Note that the choice of the first joint is not unique. Once you have determined the reactions at the supports of the truss, you can choose either of two joints as a starting point for your analysis. In Sample Prob. 6.1, we started at joint A and pro-ceeded through joints D, B, E, and C, but we could also have started at joint C and proceeded through joints E, B, D, and A. On the other hand, having selected a first joint, you may in some cases reach a point in your analysis beyond which you cannot proceed. You must then start again from another joint to complete your solution. Keep in mind that the analysis of a simple truss always can be carried out by the method of joints. Also remember that it is helpful to outline your solution before starting any computations.
310 6.1 through 6.8 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.
PROBLEMS Fig. P6.4 A B C D E F 24 kN 8 kN 7 kN 7 kN 0.8 m 1.5 m 1.5 m Fig. P6.5 B A C D 10 kips 10 kips 5 ft 10 ft 10 ft Fig. P6.2 C A B 2.8 kN 1.4 m 0.4 m 0.75 m Fig. P6.3 300 lb 15 in.
48 in.
20 in.
A C B Fig. P6.1 B C A 240 lb 20 in.
16 in.
15 in.
Fig. P6.6 B C D A E 24 kN 4.5 m 3.2 m 6 m 6 m Fig. P6.8 A B C D 5 kN 5 kN 4 m 4 m 2 m Fig. P6.7 A B C D E 12 ft 693 lb 5 ft 5 ft 11 ft 311 6.9 and 6.10 Determine the force in each member of the truss shown. State whether each member is in tension or compression.
Fig. P6.9 E H D 30° 30° G C F B A 4 kN 4 kN a a a a Fig. P6.10 E H G B C F D A 5 kN a a a 6.11 Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression. 6.12 Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression.
Fig. P6.12 A B C D E F G H 600 lb 600 lb 300 lb 600 lb 300 lb 8 ft 8 ft 8 ft 8 ft 6 ft 6 ft 6.13 Determine the force in each member of the roof truss shown. State whether each member is in tension or compression. 6.14 Determine the force in each member of the Fink roof truss shown. State whether each member is in tension or compression.
Fig. P6.14 C D E F G 1.5 kN 1.5 kN 2.25 m 2.25 m 3 kN 3 kN 3 kN 1 m 1 m A B 3 m 3 m 3 m Fig. P6.11 C D E F G H A B 6 ft 8 ft 8 ft 8 ft 8 ft 300 lb 300 lb 600 lb 600 lb 600 lb 2 ft 4 in.
Fig. P6.13 C D E F A B 1.2 kN 2.4 kN 9 m 9 m 1.2 kN 2.4 kN 6 m 6 m 6 m 7.5 m 312 Fig. P6.22 and P6.23 C D E F G H A B 400 lb 400 lb 5.76 ft 5.76 ft 5.76 ft 5.76 ft 800 lb 800 lb 800 lb 10.54 ft 12.5 ft 6.72 ft 6.15 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression.
6.16 Solve Prob. 6.15 assuming that the load applied at E has been removed. 6.17 Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression.
Fig. P6.17 5.7 kN 10.5 kN 5.7 kN 10.5 kN 9.6 kN A B C D E F G H 2.4 m 3.8 m 3.2 m 3.2 m 3.8 m 6.18 The truss shown is one of several supporting an advertising panel. Determine the force in each member of the truss for a wind load equivalent to the two forces shown. State whether each member is in tension or compression. 6.19 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression.
6.20 Solve Prob. 6.19 assuming that the load applied at G has been removed. 6.21 Determine the force in each of the members located to the left of FG for the scissors roof truss shown. State whether each member is in tension or compression.
Fig. P6.21 C D E F G H I J K L A B 1 kN 1 kN 2 kN 2 kN 2 m 2 m 2 m 2 m 2 m 2 m 1 m 1 m 1 m 1 m 1 m 6.22 Determine the force in member DE and in each of the members located to the left of DE for the inverted Howe roof truss shown. State whether each member is in tension or compression. 6.23 Determine the force in each of the members located to the right of DE for the inverted Howe roof truss shown. State whether each member is in tension or compression.
Fig. P6.15 C D E F G A B 12 ft 9 ft 9 ft 6 kips 6 kips 18 ft 18 ft 18 ft 18 ft 18 ft Fig. P6.18 A B C D E F 800 N 800 N 2 m 2 m 3.75 m 3.75 m Fig. P6.19 C D E F G H A B 4 kips 4 kips 4 kips 12 ft 9 ft 9 ft 9 ft 9 ft 313 6.24 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each mem-ber is in tension or compression.
Fig. P6.24 C D E F G H I J K L M N O P Q R A B 1.60 m 1.2 kN 1.2 kN 1.2 kN 1.2 kN 1.2 kN 1.2 kN 0.60 m 0.60 m 0.60 m 0.60 m 0.60 m 0.60 m S T 2.21 m 2.21 m 1.20 m 1.20 m 2.97 m 6.25 For the tower and loading of Prob. 6.24 and knowing that FCH 5 FEJ 5 1.2 kN C and FEH 5 0, determine the force in member HJ and in each of the members located between HJ and NO. State whether each member is in tension or compression.
6.26 Solve Prob. 6.24 assuming that the cables hanging from the right side of the tower have fallen to the ground. 6.27 and 6.28 Determine the force in each member of the truss shown. State whether each member is in tension or compression.
Fig. P6.27 12 ft 12 ft 12 ft 15 ft 15 kips 12 ft A B C E D G F H Fig. P6.28 A B C D E F G H 48 kN 4 m 4 m 4 m 4 m 4.5 m 314 6.29 Determine whether the trusses of Probs. 6.31a, 6.32a, and 6.33a are simple trusses. 6.30 Determine whether the trusses of Probs. 6.31b, 6.32b, and 6.33b are simple trusses.
6.31 For the given loading, determine the zero-force members in each of the two trusses shown. 6.32 For the given loading, determine the zero-force members in each of the two trusses shown.
Fig. P6.32 A B C D E F G H I J K L P (a) A B C D E F G H I J K L M P Q N O P (b) a 2 a 2 a a a a 6.33 For the given loading, determine the zero-force members in each of the two trusses shown.
Fig. P6.33 P A F G H I J D E B C C F A D E H G B (a) (b) P P P Q 6.34 Determine the zero-force members in the truss of (a) Prob. 6.21, (b) Prob. 6.27.
Fig. P6.31 A B C D E F G H I J K L M P Q N O (a) A B C D E F G H I J K L M P Q N O (b) 315 6.35 The truss shown consists of six members and is supported by a short link at A, two short links at B, and a ball-and-socket at D. Determine the force in each of the members for the given loading. 6.36 The truss shown consists of six members and is supported by a ball-and-socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P 5 (22184 N)j and Q 5 0.
Fig. P6.36 and P6.37 z 2.1 m 2.1 m A B C D P Q O x y 0.8 m 4.8 m 2 m 6.37 The truss shown consists of six members and is supported by a ball-and-socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P 5 0 and Q 5 (2968 N)i. 6.38 The truss shown consists of nine members and is supported by a ball-and-socket at A, two short links at B, and a short link at C. Determine the force in each of the members for the given loading.
Fig. P6.38 6 ft 6 ft 6 ft 1600 lb 7.5 ft x y z A B C D E 8 ft 6 ft Fig. P6.35 A B C D O x y z 7 ft 7 ft 10 ft 24 ft 400 lb 316 6.39 The truss shown consists of nine members and is supported by a ball-and-socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Deter-mine the force in each member for P 5 (21200 N)j and Q 5 0.
Fig. P6.39 y A B C D E O P Q z 1.2 m 0.6 m 0.6 m x 0.75 m 2.25 m 3 m 6.40 Solve Prob. 6.39 for P 5 0 and Q 5 (2900 N)k. 6.41 The truss shown consists of 18 members and is supported by a ball-and-socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely con-strained, and that the reactions at its supports are statically determi-nate. (b) For the given loading, determine the force in each of the six members joined at E.
Fig. P6.41 and P6.42 A E H G C F D B x y z 10.08 ft 9.60 ft 11.00 ft (275 lb) i (240 lb) k 6.42 The truss shown consists of 18 members and is supported by a ball-and-socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at G.
6.2 Other Truss Analyses 317 6.2 OTHER TRUSS ANALYSES The method of joints is most effective when we want to determine the forces in all the members of a truss. If, however, we need to determine the force in only one member or in a very few members, the method of sections is more efficient.
6.2A The Method of Sections Assume, for example, that we want to determine the force in member BD of the truss shown in Fig. 6.15a. To do this, we must determine the force with which member BD acts on either joint B or joint D. If we were to use the method of joints, we would choose either joint B or joint D as a free body. However, we can also choose a larger portion of the truss that is composed of several joints and members, provided that the force we want to find is one of the external forces acting on that portion. If, in addition, we choose the portion of the truss as a free body where a total of only three unknown forces act upon it, we can obtain the desired force by solving the equations of equilibrium for this portion of the truss. In practice, we isolate a portion of the truss by passing a section through three members of the truss, one of which is the desired member. That is, we draw a line that divides the truss into two completely separate parts but does not intersect more than three members. We can then use as a free body either of the two portions of the truss obtained after the intersected members have been removed.† In Fig. 6.15a, we have passed the section nn through members BD, BE, and CE, and we have chosen the portion ABC of the truss as the free body (Fig. 6.15b). The forces acting on this free body are the loads P1 and P2 at points A and B and the three unknown forces FBD, FBE, and FCE. Since we do not know whether the members removed are in tension or compression, we have arbitrarily drawn the three forces away from the free body as if the members are in tension.
We use the fact that the rigid body ABC is in equilibrium to write three equations that we can solve for the three unknown forces. If we want to determine only force FBD, say, we need write only one equation, pro-vided that the equation does not contain the other unknowns. Thus, the equation oME 5 0 yields the value of the magnitude FBD (Fig. 6.15b). A positive sign in the answer will indicate that our original assumption regarding the sense of FBD was correct and that member BD is in tension; a negative sign will indicate that our assumption was incorrect and that BD is in compression.
On the other hand, if we want to determine only force FCE, we need to write an equation that does not involve FBD or FBE; the appropriate equation is oMB 5 0. Again, a positive sign for the magnitude FCE of the desired force indicates a correct assumption, that is, tension; and a nega-tive sign indicates an incorrect assumption, that is, compression.
If we want to determine only force FBE, the appropriate equation is oFy 5 0. Whether the member is in tension or compression is again determined from the sign of the answer.
Fig. 6.15 (a) We can pass a section nn through the truss, dividing the three members BD, BE, and CE. (b) Free-body diagram of portion ABC of the truss. We assume that members BD, BE, and CE are in tension.
A B C A B C D E E G (a) (b) n n P1 P2 P1 P2 P3 FCE FBD FBE †In the analysis of some trusses, we can pass sections through more than three members, provided we can write equilibrium equations involving only one unknown that we can use to determine the forces in one, or possibly two, of the intersected members. See Probs. 6.61 through 6.64. 318 Analysis of Structures If we determine the force in only one member, no independent check of the computation is available. However, if we calculate all of the unknown forces acting on the free body, we can check the computations by writing an additional equation. For instance, if we determine FBD, FBE, and FCE as indicated previously, we can check the work by verifying that oFx 5 0.
6.2B Trusses Made of Several Simple Trusses Consider two simple trusses ABC and DEF. If we connect them by three bars BD, BE, and CE as shown in Fig. 6.16a, together they form a rigid truss ABDF. We can also combine trusses ABC and DEF into a single rigid truss by joining joints B and D at a single joint B and connecting joints C and E by a bar CE (Fig. 6.16b). This is known as a Fink truss. The trusses of Fig. 6.16a and b are not simple trusses; you cannot con-struct them from a triangular truss by adding successive pairs of members as described in Sec. 6.1A. They are rigid trusses, however, as you can check by comparing the systems of connections used to hold the simple trusses ABC and DEF together (three bars in Fig. 6.16a, one pin and one bar in Fig. 6.16b) with the systems of supports discussed in Sec. 4.1. Trusses made of several simple trusses rigidly connected are known as compound trusses.
Fig. 6.16 Compound trusses. (a) Two simple trusses ABC and DEF connected by three bars. (b) Two simple trusses ABC and DEF connected by one joint and one bar (a Fink truss).
A B C D E F (a) A B C E F (b) In a compound truss, the number of members m and the number of joints n are still related by the formula m 5 2n 2 3. You can verify this by observing that if a compound truss is supported by a frictionless pin and a roller (involving three unknown reactions), the total number of unknowns is m 1 3, and this number must be equal to the number 2n of equations obtained by expressing that the n pins are in equilibrium. It follows that m 5 2n 2 3. Compound trusses supported by a pin and a roller or by an equiva-lent system of supports are statically determinate, rigid, and completely constrained. This means that we can determine all of the unknown reac-tions and the forces in all of the members by using the methods of statics, and the truss will neither collapse nor move. However, the only way to determine all of the forces in the members using the method of joints requires solving a large number of simultaneous equations. In the case of the compound truss of Fig. 6.16a, for example, it is more efficient to pass 6.2 Other Truss Analyses 319 a section through members BD, BE, and CE to determine the forces in these members.
Suppose, now, that the simple trusses ABC and DEF are connected by four bars; BD, BE, CD, and CE (Fig. 6.17). The number of members m is now larger than 2n 2 3. This truss is said to be overrigid, and one of the four members BD, BE, CD, or CE is redundant. If the truss is supported by a pin at A and a roller at F, the total number of unknowns is m 1 3. Since m . 2n 2 3, the number m 1 3 of unknowns is now larger than the number 2n of available independent equations; the truss is statically indeterminate.
Fig. 6.17 A statically indeterminate, overrigid compound truss, due to a redundant member.
A B C D E F Finally, let us assume that the two simple trusses ABC and DEF are joined by a single pin, as shown in Fig. 6.18a. The number of mem bers, m, is now smaller than 2n 2 3. If the truss is supported by a pin at A and a roller at F, the total number of unknowns is m 1 3. Since m , 2n 2 3, the number m 1 3 of unknowns is now smaller than the number 2n of equilibrium equations that need to be satisfied. This truss is nonrigid and will collapse under its own weight. However, if two pins are used to support it, the truss becomes rigid and will not collapse (Fig. 6.18b). Note that the total number of unknowns is now m 1 4 and is equal to the number 2n of equations. More generally, if the reactions at the supports involve r unknowns, the condition for a compound truss to be statically determinate, rigid, and completely constrained is m 1 r 5 2n. However, although this condition is necessary, it is not sufficient for the equilibrium of a structure that ceases to be rigid when detached from its supports (see Sec. 6.3B).
Fig. 6.18 Two simple trusses joined by a pin. (a) Supported by a pin and a roller, the truss will collapse under its own weight. (b) Supported by two pins, the truss becomes rigid and does not collapse.
A B C E F (a) (b) A B C E F 320 Analysis of Structures Sample Problem 6.2 Determine the forces in members EF and GI of the truss shown.
A B C D E F G H I J K 28 kips 28 kips 16 kips 10 ft 8 ft 8 ft 8 ft 8 ft 8 ft STRATEGY: You are asked to determine the forces in only two of the members in this truss, so the method of sections is more appropriate than the method of joints. You can use a free-body diagram of the entire truss to help determine the reactions, and then pass sections through the truss to isolate parts of it for calculating the desired forces.
MODELING and ANALYSIS: You can go through the steps that follow for the determination of the support reactions, and then for the analysis of portions of the truss.
Free-Body: Entire Truss. Draw a free-body diagram of the entire truss. External forces acting on this free body consist of the applied loads and the reactions at B and J (Fig. 1). Write and solve the following equi-librium equations.
1loMB 5 0: 2(28 kips)(8 ft) 2 (28 kips)(24 ft) 2 (16 kips)(10 ft) 1 J(32 ft) 5 0 J 5 133 kips J 5 33 kips x y 1 oFx 5 0: Bx 1 16 kips 5 0 Bx 5 216 kips Bx 5 16 kipsz 1loMJ 5 0: (28 kips)(24 ft) 1 (28 kips)(8 ft) 2 (16 kips)(10 ft) 2 By(32 ft) 5 0 By 5 123 kips By 5 23 kipsx A B C D E F G H I J K 28 kips 28 kips 16 kips 10 ft 8 ft 8 ft 8 ft 8 ft 8 ft J By Bx Fig. 1 Free-body diagram of entire truss.
Force in Member EF. Pass section nn through the truss diagonally so that it intersects member EF and only two additional members (Fig. 2). 6.2 Other Truss Analyses 321 Remove the intersected members and choose the left-hand portion of the truss as a free body (Fig. 3). Three unknowns are involved; to eliminate the two horizontal forces, we write 1 xoFy 5 0: 123 kips 2 28 kips 2 FEF 5 0 FEF 5 25 kips A B C D E F G H I J K 28 kips 28 kips 16 kips 16 kips n n m m 23 kips 33 kips Fig. 2 Sections nn and mm that will be used to analyze members EF and GI.
FEG FEF FDF D 28 kips 16 kips 23 kips A B C E Fig. 3 Free-body diagram to analyze member EF.
The sense of FEF was chosen assuming member EF to be in tension; the negative sign indicates that the member is in compression.
FEF 5 5 kips C b Force in Member GI. Pass section mm through the truss vertically so that it intersects member GI and only two additional members (Fig. 2). Remove the intersected members and choose the right-hand portion of the truss as a free body (Fig. 4). Again, three unknown forces are involved; to eliminate the two forces passing through point H, sum the moments about that point. 1loMH 5 0: (33 kips)(8 ft) 2 (16 kips)(10 ft) 1 FGI(10 ft) 5 0 FGI 5 210.4 kips FGI 5 10.4 kips C b REFLECT and THINK: Note that a section passed through a truss does not have to be vertical or horizontal; it can be diagonal as well. Choose the orientation that cuts through no more than three members of unknown force and also gives you the simplest part of the truss for which you can write equilibrium equations and determine the unknowns.
FGI FHI FHJ 10 ft 8 ft H I J K 16 kips 33 kips Fig. 4 Free-body diagram to analyze member GI.
322 Analysis of Structures Sample Problem 6.3 Determine the forces in members FH, GH, and GI of the roof truss shown.
STRATEGY: You are asked to determine the forces in only three members of the truss, so use the method of sections. Determine the reactions by treating the entire truss as a free body and then isolate part of it for analysis. In this case, you can use the same smaller part of the truss to determine all three desired forces.
MODELING and ANALYSIS: Your reasoning and computation should go something like the sequence given here.
Free Body: Entire Truss. From the free-body diagram of the entire truss (Fig. 1), find the reactions at A and L: A 5 12.50 kN x L 5 7.50 kNx Note that tan α 5 FG GL 5 8 m 15 m 5 0.5333 α 5 28.078 Force in Member GI. Pass section nn vertically through the truss (Fig. 1). Using the portion HLI of the truss as a free body (Fig. 2), obtain the value of FGI : 1loMH 5 0: (7.50 kN)(10 m) 2 (1 kN)(5 m) 2 FGI(5.33 m) 5 0 FGI 5 113.13 kN FGI 5 13.13 kN T b Force in Member FH. Determine the value of FFH from the equation oMG 5 0. To do this, move FFH along its line of action until it acts at point F, where you can resolve it into its x and y components (Fig. 3). The moment of FFH with respect to point G is now (FFH cos α)(8 m).
1loMG 5 0: (7.50 kN)(15 m) 2 (1 kN)(10 m) 2 (1 kN)(5 m) 1 (FFH cos α)(8 m) 5 0 FFH 5 213.81 kN FFH 5 13.81 kN C b Force in Member GH. First note that tan β 5 GI HI 5 5 m 2 3(8 m) 5 0.9375 β 5 43.158 Then determine the value of FGH by resolving the force FGH into x and y components at point G (Fig. 4) and solving the equation oML 5 0.
1loML 5 0: (1 kN)(10 m) 1 (1 kN)(5 m) 1 (FGH cos β)(15 m) 5 0 FGH 5 21.371 kN FGH 5 1.371 kN C b REFLECT and THINK: Sometimes you should resolve a force into components to include it in the equilibrium equations. By first sliding this force along its line of action to a more strategic point, you might eliminate one of its components from a moment equilibrium equation.
h = 8 m A B C D F G H I J K L E 1 kN 1 kN 1 kN 1 kN 1 kN 5 kN 5 kN 5 kN 6 panels @ 5 m = 30 m A B C D F G H I J K L E 1 kN 1 kN 1 kN 1 kN 1 kN 5 kN 5 kN 5 kN n n 12.50 kN 7.50 kN a = 28.07° Fig. 1 Free-body diagram of entire truss.
Fig. 2 Free-body diagram to analyze member GI.
H I J K L FGI FFH FGH 1 kN 1 kN 7.50 kN (8 m) = 5.33 m 2 3 5 m 5 m Fig. 3 Simplifying the analysis of member FH by first sliding its force to point F.
F G H I J K L FGI FGH FFH sin a FFH cos a 1 kN 1 kN 7.50 kN a = 28.07° 5 m 5 m 8 m 5 m Fig. 4 Simplifying the analysis of member GH by first sliding its force to point G.
G H I J K L FGI FFH FGH sin b b = 43.15° FGH cos b 1 kN 1 kN 7.50 kN 5 m 5 m 5 m 323 323 SOLVING PROBLEMS ON YOUR OWN T he method of joints that you studied in Sec. 6.1 is usually the best method to use when you need to find the forces in all of the members of a simple truss. However, the method of sections, which was covered in this section, is more efficient when you need to find the force in only one member or the forces in a very few members of a simple truss. The method of sections also must be used when the truss is not a simple truss.
A. To determine the force in a given truss member by the method of sections, follow these steps: 1. Draw a free-body diagram of the entire truss, and use this diagram to determine the reactions at the supports.
2. Pass a section through three members of the truss, one of which is the member whose force you want to find. After you cut through these members, you will have two separate portions of truss.
3. Select one of these two portions of truss and draw its free-body diagram. This diagram should include the external forces applied to the selected portion as well as the forces exerted on it by the intersected members that were removed.
4. You can now write three equilibrium equations that can be solved for the forces in the three intersected members.
5. An alternative approach is to write a single equation that can be solved for the force in the desired member. To do so, first observe whether the forces exerted by the other two members on the free body are parallel or whether their lines of action intersect.
a. If these forces are parallel, you can eliminate them by writing an equilib-rium equation involving components in a direction perpendicular to these two forces. b. If their lines of action intersect at a point H, you can eliminate them by writing an equilibrium equation involving moments about H.
6. Keep in mind that the section you use must intersect three members only. The reason is that the equilibrium equations in Step 4 can be solved for only three unknowns. However, you can pass a section through more than three members to find the force in one of those members if you can write an equilibrium equation containing only that force as an unknown. Such special situations are found in Probs. 6.61 through 6.64.
324 B. About completely constrained and determinate trusses: 1. Any simple truss that is simply supported is a completely constrained and deter-minate truss.
2. To determine whether any other truss is or is not completely constrained and determinate, count the number m of its members, the number n of its joints, and the number r of the reaction components at its supports. Compare the sum m 1 r repre-senting the number of unknowns and the product 2n representing the number of available independent equilibrium equations.
a. If m 1 r < 2n, there are fewer unknowns than equations. Thus, some of the equations cannot be satisfied, and the truss is only partially constrained. b. If m 1 r > 2n, there are more unknowns than equations. Thus, some of the unknowns cannot be determined, and the truss is indeterminate. c. If m 1 r 5 2n, there are as many unknowns as there are equations. This, however, does not mean that all of the unknowns can be determined and that all of the equations can be satisfied. To find out whether the truss is completely or improp-erly constrained, try to determine the reactions at its supports and the forces in its members. If you can find all of them, the truss is completely constrained and determinate.
325 Problems 6.43 A Mansard roof truss is loaded as shown. Determine the force in members DF, DG, and EG.
1.2 kN 1.2 kN 1.2 kN 1.2 kN 1.2 kN A C E G I K L B D F H J 4 m 3 m 4 m 4 m 4 m 2.25 m 2.25 m Fig. P6.43 and P6.44 6.44 A Mansard roof truss is loaded as shown. Determine the force in members GI, HI, and HJ. 6.45 Determine the force in members BD and CD of the truss shown.
C E G F H D A B 7.5 ft 36 kips 36 kips 4 panels at 10 ft = 40 ft Fig. P6.45 and P6.46 6.46 Determine the force in members DF and DG of the truss shown. 6.47 Determine the force in members CD and DF of the truss shown. 6.48 Determine the force in members FG and FH of the truss shown. 6.49 Determine the force in members CD and DF of the truss shown.
5 m 10 kN 10 kN 10 kN 10 kN 3 m 3 m 3 m 3 m A C E G I H F D B Fig. P6.49 and P6.50 6.50 Determine the force in members CE and EF of the truss shown.
1.8 m 4 panels @ 2.4 m = 9.6 m 12 kN 12 kN A C E G J I D F H B Fig. P6.47 and P6.48 326 6.51 Determine the force in members DE and DF of the truss shown when P 5 20 kips. 6.52 Determine the force in members EG and EF of the truss shown when P 5 20 kips. 6.53 Determine the force in members DF and DE of the truss shown.
20 kN 30 kN G E C A B D F 1.5 m 2 m 2 m 2 m 2 m Fig. P6.53 and P6.54 6.54 Determine the force in members CD and CE of the truss shown. 6.55 A monosloped roof truss is loaded as shown. Determine the force in members CE, DE, and DF. 1 kN 1 kN 2 kN 2 kN 2 kN 0.46 m 2.4 m 2.4 m 2.4 m 2.4 m A B C D E F G H I J 2.62 m Fig. P6.55 and P6.56 6.56 A monosloped roof truss is loaded as shown. Determine the force in members EG, GH, and HJ. 6.57 A Howe scissors roof truss is loaded as shown. Determine the force in members DF, DG, and EG.
A B C D E F G H I J K L 0.8 kip 0.8 kip 1.6 kips 1.6 kips 1.6 kips 1.6 kips 1.6 kips 6 ft 4.5 ft 8 ft 8 ft 8 ft 8 ft 8 ft 8 ft Fig. P6.57 and P6.58 6.58 A Howe scissors roof truss is loaded as shown. Determine the force in members GI, HI, and HJ.
A B C D E F G H I J 7.5 ft 6 panels @ 6 ft = 36 ft K L P P P P P Fig. P6.51 and P6.52 327 6.59 Determine the force in members AD, CD, and CE of the truss shown.
A B C D E F G H I K 9 kips 5 kips 5 kips 15 ft 15 ft 15 ft 8 ft J Fig. P6.59 and P6.60 6.60 Determine the force in members DG, FG, and FH of the truss shown. 6.61 Determine the force in members DG and FI of the truss shown. (Hint: Use section aa.) 5 kN 5 kN 5 kN 2 m 2 m 3 m 3 m 3 m A D G J F I K B E H C a a b b Fig. P6.61 and P6.62 6.62 Determine the force in members GJ and IK of the truss shown. (Hint: Use section bb.) 6.63 Determine the force in members EH and GI of the truss shown. (Hint: Use section aa.) B D a G I b L O N P M K H F a b J E A 15 ft 15 ft 15 ft 15 ft 15 ft 15 ft C 12 kips 12 kips 12 kips 8 ft 8 ft Fig. P6.63 and P6.64 6.64 Determine the force in members HJ and IL of the truss shown. (Hint: Use section bb.) 328 6.65 and 6.66 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter.
6.65 Counters CJ and HE. 6.66 Counters IO and KN.
C D E F G H I J K L M N O P Q R A B 1.60 m 1.2 kN 1.2 kN 1.2 kN 1.2 kN 1.2 kN 1.2 kN 0.60 m 0.60 m 0.60 m 0.60 m 0.60 m 20° 20° 20° 20° 20° 20° 0.60 m S T 2.21 m 2.21 m 2.21 m 1.20 m 1.20 m 2.97 m Fig. P6.65 and P6.66 6.67 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the force in member DE and in the counters that are acting under the given loading. 6.68 Solve Prob. 6.67 assuming that the 9-kip load has been removed. 6.69 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) A B D F C E G H 6 kips 9 kips 12 kips 8 ft 8 ft 8 ft 8 ft 6 ft Counters Fig. P6.67 P P (a) P P (b) (c) P P P P Fig. P6.69 329 6.70 through 6.74 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) P (a) (b) (c) P P P (a) (b) (c) P P P P P (a) (b) (c) P P P P P P P P P (a) (b) (c) P P P P P P P P (a) (b) (c) P P P P Fig. P6.70 Fig. P6.71 Fig. P6.72 Fig. P6.73 Fig. P6.74 330 Analysis of Structures 6.3 FRAMES When we study trusses, we are looking at structures consisting entirely of pins and straight two-force members. The forces acting on the two-force members are directed along the members themselves. We now consider struc-tures in which at least one of the members is a multi-force member, i.e., a member acted upon by three or more forces. These forces are generally not directed along the members on which they act; their directions are unknown; therefore, we need to represent them by two unknown components.
Frames and machines are structures containing multi-force members. Frames are designed to support loads and are usually stationary, fully constrained structures. Machines are designed to transmit and modify forces; they may or may not be stationary and always contain moving parts.
Photo 6.6 Frames and machines contain multi-force members. Frames are fully constrained structures, whereas machines like this prosthetic hand are movable and designed to transmit or modify forces.
6.3A Analysis of a Frame As the first example of analysis of a frame, we consider again the crane described in Sec. 6.1 that carries a given load W (Fig. 6.19a). The free-body diagram of the entire frame is shown in Fig. 6.19b. We can use this diagram to determine the external forces acting on the frame. Summing moments about A, we first determine the force T exerted by the cable; summing x and y components, we then determine the components Ax and Ay of the reaction at the pin A.
Fig. 6.19 A frame in equilibrium. (a) Diagram of a crane supporting a load; (b) free-body diagram of the crane; (c) free-body diagrams of the components of the crane.
A B C D E F W G (a) F W T B C D E (b) Ay Ax A (c) A B B C C D E E F W FBE FBE –FBE – FBE T Ay Ax Cy Cx –Cy –Cx 6.3 Frames 331 In order to determine the internal forces holding the various parts of a frame together, we must dismember it and draw a free-body diagram for each of its component parts (Fig. 6.19c). First, we examine the two-force members. In this frame, member BE is the only two-force member. The forces acting at each end of this member must have the same mag-nitude, same line of action, and opposite sense (Sec. 4.2A). They are therefore directed along BE and are denoted, respectively, by FBE and 2FBE. We arbitrarily assume their sense as shown in Fig. 6.19c; the sign obtained for the common magnitude FBE of the two forces will confirm or deny this assumption.
Next, we consider the multi-force members, i.e., the members that are acted upon by three or more forces. According to Newton’s third law, the force exerted at B by member BE on member AD must be equal and opposite to the force FBE exerted by AD on BE. Similarly, the force exerted at E by member BE on member CF must be equal and opposite to the force 2FBE exerted by CF on BE. Thus, the forces that the two-force member BE exerts on AD and CF are, respectively, equal to 2FBE and FBE; they have the same magnitude FBE, opposite sense, and should be directed as shown in Fig. 6.19c.
Joint C connects two multi-force members. Since neither the direc-tion nor the magnitude of the forces acting at C are known, we represent these forces by their x and y components. The components Cx and Cy of the force acting on member AD are arbitrarily directed to the right and upward. Since, according to Newton’s third law, the forces exerted by member CF on AD and by member AD on CF are equal and opposite, the components of the force acting on member CF must be directed to the left and downward; we denote them, respectively, by 2Cx and 2Cy. Whether the force Cx is actually directed to the right and the force 2Cx is actually directed to the left will be determined later from the sign of their common magnitude Cx with a plus sign indicating that the assump-tion was correct and a minus sign that it was wrong. We complete the free-body diagrams of the multi-force members by showing the external forces acting at A, D, and F.† We can now determine the internal forces by considering the free-body diagram of either of the two multi-force members. Choosing the free-body diagram of CF, for example, we write the equations oMC 5 0, oME 5 0, and oFx 5 0, which yield the values of the magnitudes FBE, Cy, and Cx, respectively. We can check these values by verifying that member AD is also in equilibrium.
Note that we assume the pins in Fig. 6.19 form an integral part of one of the two members they connected, so it is not necessary to show their free-body diagrams. We can always use this assumption to simplify the analysis of frames and machines. However, when a pin connects three †It is not strictly necessary to use a minus sign to distinguish the force exerted by one member on another from the equal and opposite force exerted by the second member on the first, since the two forces belong to different free-body diagrams and thus are not easily confused. In the Sample Problems, we use the same symbol to represent equal and opposite forces that are applied to different free bodies. Note that, under these conditions, the sign obtained for a given force component does not directly relate the sense of that component to the sense of the corresponding coordinate axis. Rather, a positive sign indicates that the sense assumed for that component in the free-body diagram is correct, and a negative sign indicates that it is wrong. 332 Analysis of Structures or more members, connects a support and two or more members, or when a load is applied to a pin, we must make a clear decision in choosing the member to which we assume the pin belongs. (If multi-force members are involved, the pin should be attached to one of these members.) We then need to identify clearly the various forces exerted on the pin. This is illustrated in Sample Prob. 6.6.
6.3B Frames That Collapse Without Supports The crane we just analyzed was constructed so that it could keep the same shape without the help of its supports; we therefore considered it to be a rigid body. Many frames, however, will collapse if detached from their supports; such frames cannot be considered rigid bodies. Consider, for example, the frame shown in Fig. 6.20a that consists of two members AC and CB carrying loads P and Q at their midpoints. The members are supported by pins at A and B and are connected by a pin at C. If we detach this frame from its supports, it will not maintain its shape. There-fore, we should consider it to be made of two distinct rigid parts AC and CB.
The equations oFx 5 0, oFy 5 0, and oM 5 0 (about any given point) express the conditions for the equilibrium of a rigid body (Chap. 4); we should use them, therefore, in connection with the free-body diagrams of members AC and CB (Fig. 6.20b). Since these members are multi- force members and since pins are used at the supports and at the connection, we represent each of the reactions at A and B and the forces at C by two components. In accordance with Newton’s third law, we represent the components of the force exerted by CB on AC and the components of the force exerted by AC on CB by vectors of the same magnitude and opposite sense. Thus, if the first pair of components consists of Cx and Cy, the second pair is represented by 2Cx and 2Cy.
Note that four unknown force components act on free body AC, whereas we need only three independent equations to express that the body is in equilibrium. Similarly, four unknowns, but only three equations, are associated with CB. However, only six different unknowns are involved in the analysis of the two members, and altogether, six equations Fig. 6.20 (a) A frame of two members supported by two pins and joined together by a third pin. Without the supports, the frame would collapse and is therefore not a rigid body. (b) Free-body diagrams of the two members. (c) Free-body diagram of the whole frame.
A B C (a) Q P A B C C (b) Ay Ax By Bx Cy Cx –Cy –Cx Q P A B C (c) Ay Ax By Bx Q P 6.3 Frames 333 are available to express that the members are in equilibrium. Setting oMA 5 0 for free body AC and oMB 5 0 for CB, we obtain two simul-taneous equations that we can solve for the common magnitude Cx of the components Cx and 2Cx and for the common magnitude Cy of the components Cy and 2Cy. We then have oFx 5 0 and oFy 5 0 for each of the two free bodies, successively obtaining the magnitudes Ax , Ay, Bx, and By.
Observe that, since the equations of equilibrium oFx 5 0, oFy 5 0, and oM 5 0 (about any given point) are satisfied by the forces acting on free body AC and since they are also satisfied by the forces acting on free body CB, they must be satisfied when the forces acting on the two free bodies are considered simultaneously. Since the internal forces at C cancel each other, we find that the equations of equilibrium must be satisfied by the external forces shown on the free-body diagram of the frame ACB itself (Fig. 6.20c), even though the frame is not a rigid body. We can use these equations to determine some of the components of the reactions at A and B. We will find, however, that the reactions cannot be completely determined from the free-body diagram of the whole frame. It is thus necessary to dismember the frame and consider the free-body diagrams of its component parts (Fig. 6.20b), even when we are interested in determining external reactions only. The reason is that the equilibrium equations obtained for free body ACB are necessary conditions for the equilibrium of a nonrigid structure, but these are not sufficient conditions.
The method of solution outlined here involved simultaneous equations. We now present a more efficient method that utilizes the free body ACB as well as the free bodies AC and CB. Writing oMA 5 0 and oMB 5 0 for free body ACB, we obtain By and Ay. From oMC 5 0, oFx 5 0, and oFy 5 0 for free body AC, we successively obtain Ax, Cx, and Cy. Finally, setting oFx 5 0 for ACB gives us Bx.
We noted previously that the analysis of the frame in Fig. 6.20 involves six unknown force components and six independent equilibrium equations. (The equilibrium equations for the whole frame were obtained from the original six equations and, therefore, are not independent.) More-over, we checked that all unknowns could be actually determined and that all equations could be satisfied. This frame is statically determinate and rigid. (We use the word “rigid” here to indicate that the frame maintains its shape as long as it remains attached to its supports.) In general, to determine whether a structure is statically determinate and rigid, you should draw a free-body diagram for each of its component parts and count the reactions and internal forces involved. You should then deter-mine the number of independent equilibrium equations (excluding equa-tions expressing the equilibrium of the whole structure or of groups of component parts already analyzed). If you have more unknowns than equations, the structure is statically indeterminate. If you have fewer unknowns than equations, the structure is nonrigid. If you have as many unknowns as equations and if all unknowns can be determined and all equations satisfied under general loading conditions, the structure is stati-cally determinate and rigid. If, however, due to an improper arrangement of members and supports, all unknowns cannot be determined and all equations cannot be satisfied, the structure is statically indeterminate and nonrigid.
334 Analysis of Structures Sample Problem 6.4 In the frame shown, members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force in link DE and the components of the force exerted at C on member BCD.
STRATEGY: Follow the general procedure discussed in this section. First treat the entire frame as a free body, which will enable you to find the reac-tions at A and B. Then dismember the frame and treat each member as a free body, which will give you the equations needed to find the force at C.
MODELING and ANALYSIS: Since the external reactions involve only three unknowns, compute the reactions by considering the free-body dia-gram of the entire frame (Fig. 1). 1 xoFy 5 0: Ay 2 480 N 5 0 Ay 5 1480 N Ay 5 480 N x 1loMA 5 0: 2(480 N)(100 mm) 1 B(160 mm) 5 0 B 5 1300 N B 5 300 Ny y 1 oFx 5 0: B 1 Ax 5 0 300 N 1 Ax 5 0 Ax 5 2300 N Ax 5 300 Nz Now dismember the frame (Figs. 2 and 3). Since only two members are connected at C, the components of the unknown forces acting on ACE and BCD are, respectively, equal and opposite. Assume that link DE is in tension (Fig. 3) and exerts equal and opposite forces at D and E, directed as shown.
Fig. 2 Free-body diagram of member BCD.
B C D 60 mm 60 mm 480 N 100 mm 150 mm a Cy Cx FDE 300 N Free Body: Member BCD. Using the free body BCD (Fig. 2), you can write and solve three equilibrium equations: 1 ioMC 5 0: (FDE sin α)(250 mm) 1 (300 N)(80 mm) 1 (480 N)(100 mm) 5 0 FDE 5 2561 N FDE 5 561 N C b y 1 oFx 5 0: Cx 2 FDE cos α 1 300 N 5 0 Cx 2 (2561 N) cos 28.07° 1 300 N 5 0 Cx 5 2795 N 1 x oFy 5 0: Cy 2 FDE sin α 2 480 N 5 0 Cy 2 (2561 N) sin 28.07° 2 480 N 5 0 Cy 5 1216 N From the signs obtained for Cx and Cy, the force components Cx and Cy exerted on member BCD are directed to the left and up, respectively. Thus, you have Cx 5 795 Nz, Cy 5 216 Nx b REFLECT and THINK: Check the computations by considering the free body ACE (Fig. 3). For example, 1loMA 5 (FDE cos α)(300 mm) 1 (FDE sin α)(100 mm) 2 Cx(220 mm) 5 (2561 cos α)(300) 1 (2561 sin α)(100) 2 (2795)(220) 5 0 A B C D E 60 mm 60 mm 80 mm 480 N 100 mm 150 mm 160 mm Fig. 1 Free-body diagram of entire frame.
A B C D E 160 mm 80 mm 480 N 100 mm 150 mm Ay B Ax a a = tan–1 = 28.07° 80 150 Fig. 3 Free-body diagrams of member ACE and DE.
C A E D E 80 mm 480 N 100 mm a Cy Cx FDE FDE FDE 300 N 220 mm 6.3 Frames 335 Sample Problem 6.5 Determine the components of the forces acting on each member of the frame shown.
STRATEGY: The approach to this analysis is to consider the entire frame as a free body to determine the reactions, and then consider separate members. However, in this case, you will not be able to determine forces on one member without analyzing a second member at the same time.
MODELING and ANALYSIS: The external reactions involve only three unknowns, so compute the reactions by considering the free-body diagram of the entire frame (Fig. 1).
1loME 5 0: 2(2400 N)(3.6 m) 1 F(4.8 m) 5 0 F 5 11800 N F 5 1800 Nx b 1 x oFy 5 0: 22400 N 1 1800 N 1 Ey 5 0 Ey 5 1600 N Ey 5 600 N x b y 1 oFx 5 0: Ex 5 0 b Now dismember the frame. Since only two members are connected at each joint, force components are equal and opposite on each member at each joint (Fig. 2).
Free Body: Member BCD.
1loMB 5 0: 2(2400 N)(3.6 m) 1 Cy(2.4 m) 5 0 Cy 5 13600 N b 1loMC 5 0: 2(2400 N)(1.2 m) 1 By(2.4 m) 5 0 By 5 11200 N b y 1 oFx 5 0: 2Bx 1 Cx 5 0 Neither Bx nor Cx can be obtained by considering only member BCD; you need to look at member ABE. The positive values obtained for By and Cy indicate that the force components By and Cy are directed as assumed.
Free Body: Member ABE.
1loMA 5 0: Bx(2.7 m) 5 0 Bx 5 0 b y 1 oFx 5 0: 1Bx 2 Ax 5 0 Ax 5 0 b 1 x oFy 5 0: 2Ay 1 By 1 600 N 5 0 2Ay 1 1200 N 1 600 N 5 0 Ay 5 11800 N b Free Body: Member BCD. Returning now to member BCD, you have y 1 oFx 5 0: 2Bx 1 Cx 5 0 0 1 Cx 5 0 Cx 5 0 b REFLECT and THINK: All unknown components have now been found. To check the results, you can verify that member ACF is in equilibrium.
1loMC 5 (1800 N)(2.4 m) 2 Ay(2.4 m) 2 Ax(2.7 m) 5 (1800 N)(2.4 m) 2 (1800 N)(2.4 m) 2 0 5 0 (checks) 2400 N A B C D E F 2.7 m 3.6 m 4.8 m 2.7 m 2400 N A C D E F 3.6 m 4.8 m Ey F Ex B Fig. 1 Free-body diagram of entire frame.
600 N 1800 N 2.7 m 2.7 m By Cy Bx By Ay Ay Ax Ax Bx Cx Cy Cx A A B B C E F 2400 N C D 2.4 m 2.4 m 1.2 m Fig. 2 Free-body diagrams of individual members.
336 Analysis of Structures Sample Problem 6.6 A 600-lb horizontal force is applied to pin A of the frame shown. Determine the forces acting on the two vertical members of the frame.
STRATEGY: Begin as usual with a free-body diagram of the entire frame, but this time you will not be able to determine all of the reactions. You will have to analyze a separate member and then return to the entire frame analysis in order to determine the remaining reaction forces.
MODELING and ANALYSIS: Choosing the entire frame as a free body (Fig. 1), you can write equilibrium equations to determine the two force components Ey and Fy. However, these equations are not sufficient to determine Ex and Fx.
1loME 5 0: 2(600 lb)(10 ft) 1 Fy(6 ft) 5 0 Fy 5 11000 lb Fy 5 1000 lb x b 1 x oFy 5 0: Ey 1 Fy 5 0 Ey 5 21000 lb Ey 5 1000 lb w b To proceed with the solution, now consider the free-body diagrams of the various members (Fig. 2). In dismembering the frame, assume that pin A is attached to the multi-force member ACE so that the 600-lb force is applied to that member. Note that AB and CD are two-force members.
Free Body: Member ACE 1 xoFy 5 0: 2 5 13FAB 1 5 13FCD 2 1000 lb 5 0 1loME 5 0: 2(600 lb)(10 ft) 2 (12 13FAB)(10 ft) 2 (12 13FCD)(2.5 ft) 5 0 Solving these equations simultaneously gives you FAB 5 21040 lb FCD 5 11560 lb b The signs indicate that the sense assumed for FCD was correct and the sense for FAB was incorrect. Now summing x components, you have y 1 oFx 5 0: 600 lb 1 12 13(21040 lb) 1 12 13(11560 lb) 1 Ex 5 0 Ex 5 21080 lb Ex 5 1080 lbz b Free Body: Entire Frame. Now that Ex is determined, you can return to the free-body diagram of the entire frame.
y 1 oFx 5 0: 600 lb 2 1080 lb 1 Fx 5 0 Fx 5 1480 lb Fx 5 480 lby b REFLECT and THINK: Check your computations by verifying that the equation oMB 5 0 is satisfied by the forces acting on member BDF. 1loMB 5 2( 12 13FCD)(2.5 ft) 1 (Fx)(7.5 ft) 5 212 13(1560 lb)(2.5 ft) 1 (480 lb)(7.5 ft) 5 23600 lb?ft 1 3600 lb?ft 5 0 (checks) 600 lb A B C D E F Ey Ex Fy Fx 6 ft 10 ft Fig. 1 Free-body diagram of entire frame.
600 lb A B C D E F 2.5 ft 2.5 ft 2.5 ft 2.5 ft 6 ft A B C D FAB FAB FCD FCD Fig. 2 Free-body diagrams of individual members.
600 lb A B C D E F FAB FAB FCD FCD Ey = 1000 lb Fy = 1000 lb Ex Fx 12 12 13 13 5 5 2.5 ft 5 ft 7.5 ft 2.5 ft 337 337 SOLVING PROBLEMS ON YOUR OWN I n this section, we analyzed frames containing one or more multi-force members. In the problems that follow, you will be asked to determine the external reactions exerted on the frame and the internal forces that hold together the members of the frame.
In solving problems involving frames containing one or more multi-force members, follow these steps.
1. Draw a free-body diagram of the entire frame. To the greatest extent possible, use this free-body diagram to calculate the reactions at the supports. (In Sample Prob. 6.6 only two of the four reaction components could be found from the free body of the entire frame.) 2. Dismember the frame, and draw a free-body diagram of each member.
3. First consider the two-force members. Equal and opposite forces apply to each two-force member at the points where it is connected to another member. If the two-force member is straight, these forces are directed along the axis of the member. If you cannot tell at this point whether the member is in tension or compression, assume that the member is in tension and direct both of the forces away from the member. Since these forces have the same unknown magnitude, give them both the same name and, to avoid any confusion later, do not use a plus sign or a minus sign.
4. Next consider the multi-force members. For each of these members, show all of the forces acting on the member, including applied loads, reactions, and internal forces at connections. Clearly indicate the magnitude and direction of any reaction or reaction component found earlier from the free-body diagram of the entire frame. a. Where a multi-force member is connected to a two-force member, apply a force to the multi-force member that is equal and opposite to the force drawn on the free-body diagram of the two-force member, giving it the same name.
b. Where a multi-force member is connected to another multi-force member, use horizontal and vertical components to represent the internal forces at that point, since the directions and magnitudes of these forces are unknown. The direction you choose for each of the two force components exerted on the first multi-force member is arbitrary, but you must apply equal and opposite force components of the same name to the other multi-force member. Again, do not use a plus sign or a minus sign.
(continued) 338 5. Now determine the internal forces as well as any reactions that you have not already found. a. The free-body diagram of each multi-force member can provide you with three equilibrium equations. b. To simplify your solution, seek a way to write an equation involving a single unknown. If you can locate a point where all but one of the unknown force components intersect, you can obtain an equation in a single unknown by summing moments about that point. If all unknown forces except one are parallel, you can obtain an equation in a single unknown by summing force components in a direction perpendicular to the parallel forces. c. Since you arbitrarily chose the direction of each of the unknown forces, you cannot determine whether your guess was correct until the solution is complete. To do that, consider the sign of the value found for each of the unknowns: a positive sign means that the direction you selected was correct; a negative sign means that the direction is opposite to the direction you assumed.
6. To be more effective and efficient as you proceed through your solution, observe the following rules. a. If you can find an equation involving only one unknown, write that equa-tion and solve it for that unknown. Immediately replace that unknown wherever it appears on other free-body diagrams by the value you have found. Repeat this process by seeking equilibrium equations involving only one unknown until you have found all of the internal forces and unknown reactions. b. If you cannot find an equation involving only one unknown, you may have to solve a pair of simultaneous equations. Before doing so, check that you have included the values of all of the reactions you obtained from the free-body diagram of the entire frame. c. The total number of equations of equilibrium for the entire frame and for the individual members will be larger than the number of unknown forces and reactions. After you have found all of the reactions and all of the internal forces, you can use the remaining equations to check the accuracy of your computations.
339 Problems FREE-BODY PRACTICE PROBLEMS 6.F1 For the frame and loading shown, draw the free-body diagram(s) needed to determine the force in member BD and the components of the reaction at C. 6.F2 For the frame and loading shown, draw the free-body diagram(s) needed to determine the components of all forces acting on member ABC.
A B C D E 4 ft 20 kips 5 ft 5 ft Fig. P6.F2 6.F3 Draw the free-body diagram(s) needed to determine all the forces exerted on member AI if the frame is loaded by a clockwise couple of magnitude 1200 lb?in. applied at point D. 6.F4 Knowing that the pulley has a radius of 0.5 m, draw the free-body diagram(s) needed to determine the components of the reactions at A and E.
1 m 1 m 3 m 3 m 2 m 700 N C B D A E Fig. P6.F4 A B C D 510 mm 240 mm 135 mm 120 mm 400 N 450 mm Fig. P6.F1 C D E F H I G B A 20 in.
10 in.
20 in.
10 in.
10 in.
20 in.
10 in.
48 in.
Fig. P6.F3 340 END-OF-SECTION PROBLEMS 6.75 and 6.76 Determine the force in member BD and the components of the reaction at C.
160 lb 24 in.
14 in.
8 in.
10 in.
8 in.
A B C D J Fig. P6.76 6.77 For the frame and loading shown, determine the force acting on member ABC (a) at B, (b) at C.
A B C D 200 N 120 mm 90 mm 120 mm 120 mm Fig. P6.77 6.78 Determine the components of all forces acting on member ABCD of the assembly shown. 6.79 For the frame and loading shown, determine the components of all forces acting on member ABC. 6.80 Solve Prob. 6.79 assuming that the 18-kN load is replaced by a clockwise couple with a magnitude of 72 kN?m applied to member CDEF at point D. 6.81 Determine the components of all forces acting on member ABCD when θ 5 0.
A B C D J E F 8 in.
12 in.
4 in.
4 in.
2 in.
q 60 lb 6 in.
Fig. P6.81 and P6.82 6.82 Determine the components of all forces acting on member ABCD when θ 5 90°.
r = 1.4 m 310 N A B D C 30° 1.92 m 0.56 m Fig. P6.75 D C E B J A 120 lb 4 in.
2 in.
2 in.
2 in.
4 in.
4 in.
Fig. P6.78 C D E F B A 3.6 m 18 kN 2 m 2 m 2 m Fig. P6.79 341 6.83 Determine the components of the reactions at A and E, (a) if the 800-N load is applied as shown, (b) if the 800-N load is moved along its line of action and is applied at point D.
800 N C B D E 200 mm A 300 mm 600 mm 300 mm Fig. P6.83 6.84 Determine the components of the reactions at D and E if the frame is loaded by a clockwise couple of magnitude 150 N?m applied (a) at A, (b) at B.
A C B D E 0.4 m 0.4 m 0.6 m 0.6 m 0.6 m Fig. P6.84 6.85 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D. 6.86 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple with a magnitude of 36 N∙m applied (a) at B, (b) at D. 6.87 Determine the components of the reactions at A and B, (a) if the 100-lb load is applied as shown, (b) if the 100-lb load is moved along its line of action and is applied at point F.
A B C D F E 100 lb 4 in.
5 in.
5 in.
10 in.
Fig. P6.87 A B C D E 80 mm 170 mm 75 mm 125 mm Fig. P6.85 and P6.86 342 6.88 The 48-lb load can be moved along the line of action shown and applied at A, D, or E. Determine the components of the reactions at B and F if the 48-lb load is applied (a) at A, (b) at D, (c) at E. 6.89 The 48-lb load is removed and a 288-lb?in. clockwise couple is applied successively at A, D, and E. Determine the components of the reac-tions at B and F if the couple is applied (a) at A, (b) at D, (c) at E. 6.90 (a) Show that, when a frame supports a pulley at A, an equivalent loading of the frame and of each of its component parts can be obtained by removing the pulley and applying at A two forces equal and parallel to the forces that the cable exerted on the pulley. (b) Show that, if one end of the cable is attached to the frame at a point B, a force of mag-nitude equal to the tension in the cable should also be applied at B.
T T T T T T T T A A A B A B = = (a) (b) Fig. P6.90 6.91 Knowing that each pulley has a radius of 250 mm, determine the components of the reactions at D and E.
2 m 1.5 m 2 m 4.8 kN C B D A E Fig. P6.91 6.92 Knowing that the pulley has a radius of 75 mm, determine the com-ponents of the reactions at A and B.
240 N C D A B E 75 mm 300 mm 300 mm 125 mm Fig. P6.92 A D B C E F 5 in.
7 in.
48 lb 8 in.
8 in.
Fig. P6.88 and P6.89 343 6.93 A 3-ft-diameter pipe is supported every 16 ft by a small frame like that shown. Knowing that the combined weight of the pipe and its contents is 500 lb/ft and assuming frictionless surfaces, determine the components (a) of the reaction at E, (b) of the force exerted at C on member CDE. 6.94 Solve Prob. 6.93 for a frame where h 5 6 ft. 6.95 A trailer weighing 2400 lb is attached to a 2900-lb pickup truck by a ball-and-socket truck hitch at D. Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional load on each of the truck wheels due to the trailer.
8 ft 6 ft h 5 9 ft r 5 1.5 ft A B C D E Fig. P6.93 6.96 In order to obtain a better weight distribution over the four wheels of the pickup truck of Prob. 6.95, a compensating hitch of the type shown is used to attach the trailer to the truck. The hitch consists of two bar springs (only one is shown in the figure) that fit into bearings inside a support rigidly attached to the truck. The springs are also connected by chains to the trailer frame, and specially designed hooks make it possible to place both chains in tension. (a) Determine the tension T required in each of the two chains if the additional load due to the trailer is to be evenly distributed over the four wheels of the truck. (b) What are the resulting reactions at each of the six wheels of the trailer-truck combination?
D E F Bar spring Chain under tension T 1.7 ft Fig. P6.96 A B C D 2400 lb 2900 lb 2 ft 9 ft 3 ft 5 ft 4 ft Fig. P6.95 344 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reac-tions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.
A C D B Gc Gm 3.2 m 0.8 m 1.2 m 2.8 m 2 m 75 kN 100 kN 300 kN Gl Fig. P6.97 6.98 Solve Prob. 6.97 assuming that the 75-kN load has been removed. 6.99 Knowing that P 5 90 lb and Q 5 60 lb, determine the components of all forces acting on member BCDE of the assembly shown. 6.100 Knowing that P 5 60 lb and Q 5 90 lb, determine the components of all forces acting on member BCDE of the assembly shown. 6.101 and 6.102 For the frame and loading shown, determine the compo-nents of all forces acting on member ABE.
A B C D E F 0.3 m 12 kN 0.9 m 0.9 m 1.2 m 0.6 m Fig. P6.101 2.7 m 2.7 m 3.6 m 2400 N 1.5 m 4.8 m A B C D J E F Fig. P6.102 6.103 For the frame and loading shown, determine the components of all forces acting on member ABD. 6.104 Solve Prob. 6.103 assuming that the 360-lb load has been removed.
A B C E D 6 in.
6 in.
4 in.
8 in.
4 in.
P Q Fig. P6.99 and P6.100 6 in.
9 in.
9 in.
9 in.
12 in.
12 in.
360 lb 240 lb A B D E C Fig. P6.103 345 6.105 For the frame and loading shown, determine the components of the forces acting on member DABC at B and D.
6.106 Solve Prob. 6.105 assuming that the 6-kN load has been removed. 6.107 The axis of the three-hinge arch ABC is a parabola with vertex at B. Knowing that P 5 112 kN and Q 5 140 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB.
A B C P Q 1.8 m 1.4 m 3 m 3 m 8 m 6 m Fig. P6.107 and P6.108 6.108 The axis of the three-hinge arch ABC is a parabola with vertex at B. Knowing that P 5 140 kN and Q 5 112 kN, determine (a) the components of the reaction at A, (b) the components of the force exerted at B on segment AB. 6.109 and 6.110 Neglecting the effect of friction at the horizontal and vertical surfaces, determine the forces exerted at B and C on member BCE.
A B C D E 6 in.
6 in.
50 lb 4 in.
2 in.
6 in.
12 in.
Fig. P6.109 6.111, 6.112, and 6.113 Members ABC and CDE are pin-connected at C and supported by four links. For the loading shown, determine the force in each link.
C B A D G H E F 12 kN 6 kN 0.5 m 0.5 m 0.6 m 0.2 m 0.4 m Fig. P6.105 A B C D E F G H P a a a a a Fig. P6.111 A B C D E F G H P a a a a a Fig. P6.112 I A B C D E F G H P a a a a a Fig. P6.113 A B C E 6 in.
6 in.
50 lb 6 in.
6 in.
12 in.
Fig. P6.110 346 6.114 Members ABC and CDE are pin-connected at C and supported by the four links AF, BG, DG, and EH. For the loading shown, deter-mine the force in each link. 6.115 Solve Prob. 6.112 assuming that the force P is replaced by a clock-wise couple of moment M0 applied to member CDE at D. 6.116 Solve Prob. 6.114 assuming that the force P is replaced by a clock-wise couple of moment M0 applied at the same point. 6.117 Four beams, each with a length of 2a, are nailed together at their midpoints to form the support system shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at A, D, E, and H.
A B C D E F G H P Fig. P6.117 6.118 Four beams, each with a length of 3a, are held together by single nails at A, B, C, and D. Each beam is attached to a support located at a distance a from an end of the beam as shown. Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at E, F, G, and H.
B C D E F G H a a 2a 2a A P Fig. P6.118 P 2a 2a a a a a A B G C D E H F Fig. P6.114 347 6.119 through 6.121 Each of the frames shown consists of two L-shaped members connected by two rigid links. For each frame, determine the reactions at the supports and indicate whether the frame is rigid.
B A P (a) 2a 2a a a (c) B A P 2a 2a a a A C B P (b) 2a 2a a a Fig. P6.119 (a) B P 2a 2a a A a 1 4 a (c) B P 2a 2a a A a 1 4 a (b) P 2a 2a a A B a 1 4 a Fig. P6.120 (a) A B P 2a 2a a a A C B P 2a 2a a a (c) A B P 2a 2a a a (b) Fig. P6.121 348 Analysis of Structures 6.4 MACHINES Machines are structures designed to transmit and modify forces. Whether they are simple tools or include complicated mechanisms, their main pur-pose is to transform input forces into output forces. Consider, for exam-ple, a pair of cutting pliers used to cut a wire (Fig. 6.21a). If we apply two equal and opposite forces P and 2P on the handles, the pliers will exert two equal and opposite forces Q and 2Q on the wire (Fig. 6.21b).
Fig. 6.21 (a) Input forces on the handles of a pair of cutting pliers; (b) output forces cut a wire.
A (a) (b) P –P Q –Q b a Fig. 6.23 Free-body diagrams of the members of the pliers, showing components of the internal forces at joint A.
–Ax A A (a) (b) Ay –Ay Ax P Q –P –Q a b To determine the magnitude Q of the output forces when we know the magnitude P of the input forces (or, conversely, to determine P when Q is known), we draw a free-body diagram of the pliers alone (i.e., without the wire), showing the input forces P and 2P and the reactions 2Q and Q that the wire exerts on the pliers (Fig. 6.22). However, since a pair of pliers forms a nonrigid structure, we must treat one of the component parts as a free body in order to determine the unknown forces. Consider Fig. 6.23a, for example. Taking moments about A, we obtain the relation Pa 5 Qb, which defines the magnitude Q in terms of P (or P in terms of Q). We can use the same free-body diagram to determine the components of the internal force at A; we find Ax 5 0 and Ay 5 P 1 Q.
Fig. 6.22 To show a free-body diagram of the pliers in equilibrium, we include the input forces and the reactions to the output forces.
Q –Q A P –P In the case of more complicated machines, it is generally necessary to use several free-body diagrams and, possibly, to solve simultaneous equa-tions involving various internal forces. You should choose the free bodies to include the input forces and the reactions to the output forces, and the total number of unknown force components involved should not exceed the num-ber of available independent equations. It is advisable, before attempting to solve a problem, to determine whether the structure considered is determi-nate. There is no point, however, in discussing the rigidity of a machine, since a machine includes moving parts and thus must be nonrigid.
Photo 6.7 This lamp can be placed in many different positions. To determine the forces in the springs and the internal forces at the joints, we need to consider the components of the lamp as free bodies.
6.4 Machines 349 Sample Problem 6.7 A hydraulic-lift table is used to raise a 1000-kg crate. The table consists of a platform and two identical linkages on which hydraulic cylinders exert equal forces. (Only one linkage and one cylinder are shown.) Members EDB and CG are each of length 2a, and member AD is pinned to the mid-point of EDB. If the crate is placed on the table so that half of its weight is supported by the system shown, determine the force exerted by each cylinder in raising the crate for θ 5 60°, a 5 0.70 m, and L 5 3.20 m. Show that the result is inde-pendent of the distance d.
STRATEGY: The free-body diagram of the entire frame will involve more than three unknowns, so it alone can not be used to solve this problem. Instead, draw free-body diagrams of each component of the machine and work from them.
MODELING: The machine consists of the platform and the linkage. Its free-body diagram (Fig. 1) includes an input force FDH exerted by the cylinder; the weight W/2, which is equal and opposite to the output force; and reactions at E and G, which are assumed to be directed as shown. Dismember the mechanism and draw a free-body diagram for each of its com-ponent parts (Fig. 2). Note that AD, BC, and CG are two-force members. Member CG has already been assumed to be in compression; now assume that AD and BC are in tension and direct the forces exerted on them as shown. Use equal and opposite vectors to represent the forces exerted by the two-force members on the platform, on member BDE, and on roller C.
(continued) A B C D E G H 2a W 1 2 q L 2 L 2 d FDH FCG Ey Ex E G A B C D W 1 2 Fig. 1 Free-body diagram of machine.
FAD A B B C C W 1 2 q d A D FAD FAD FAD FDH FBC Ey Ex a a f B B D E q B C FBC FBC FCG FCG G C FCG FBC C C q Fig. 2 Free-body diagram of each component part.
350 Analysis of Structures ANALYSIS: Free Body: Platform ABC (Fig. 3). y 1 oFx 5 0: FAD cos θ 5 0 FAD 5 0 1 xoFy 5 0: B 1 C 2 1 2W 5 0 B 1 C 5 1 2W (1) Free-Body Roller C (Fig. 4). Draw a force triangle and obtain FBC 5 C cot θ.
Free Body: Member BDE (Fig. 5). Recalling that FAD 5 0, you have 1loME 5 0: FDH cos (ϕ 2 90°)a 2 B(2a cos θ) 2 FBC(2a sin θ) 5 0 FDHa sin ϕ 2 B(2a cos θ) 2 (C cot θ)(2a sin θ) 5 0 FDH sin ϕ 2 2(B 1 C) cos θ 5 0 From Eq. (1), you obtain FDH 5 W cos θ sin ϕ (2) Note that the result obtained is independent of d. b Applying first the law of sines to triangle EDH (Fig. 6), you have sin ϕ EH 5 sin θ DH sin ϕ 5 EH DH sin θ (3) Now using the law of cosines, you get (DH)2 5 a2 1 L2 2 2aL cos θ 5 (0.70)2 1 (3.20)2 2 2(0.70)(3.20) cos 60° (DH)2 5 8.49 DH 5 2.91 m Also note that W 5 mg 5 (1000 kg)(9.81 m/s2) 5 9810 N 5 9.81 kN Substituting for sin ϕ from Eq. (3) into Eq. (2) and using the numerical data, your result is FDH 5 W DH EH cot θ 5 (9.81 kN) 2.91 m 3.20 m cot 608 FDH 5 5.15 kN b REFLECT and THINK: Note that link AD ends up having zero force in this situation. However, this member still serves an important func-tion, as it is necessary to enable the machine to support any horizontal load that might be exerted on the platform.
FAD A B B C C W 1 2 q d Fig. 3 Free-body diagram of platform ABC.
FAD FDH FBC Ey Ex a a f B B D E q Fig. 5 Free-body diagram of member BDE.
Fig. 4 Free-body diagram of roller C and its force triangle.
FCG FBC C C q FCG FBC C q a f D H E q L Fig. 6 Geometry of triangle EDH.
351 351 SOLVING PROBLEMS ON YOUR OWN T his section dealt with the analysis of machines. Since machines are designed to transmit or modify forces, they always contain moving parts. However, the machines considered here are always at rest, and you will be working with the set of forces required to maintain the equilibrium of the machine.
Known forces that act on a machine are called input forces. A machine transforms the input forces into output forces, such as the cutting forces applied by the pliers of Fig. 6.21. You will determine the output forces by finding the equal and opposite forces that should be applied to the machine to maintain its equilibrium.
In Sec. 6.3, you analyzed frames; you will use almost the same procedure to analyze machines by following these steps.
1. Draw a free-body diagram of the whole machine, and use it to determine as many as possible of the unknown forces exerted on the machine.
2. Dismember the machine and draw a free-body diagram of each member.
3. First consider the two-force members. Apply equal and opposite forces to each two-force member at the points where it is connected to another member. If you can-not tell at this point whether the member is in tension or in compression, assume that the member is in tension and direct both of the forces away from the member. Since these forces have the same unknown magnitude, give them both the same name.
4. Next consider the multi-force members. For each of these members, show all of the forces acting on it, including applied loads and forces, reactions, and internal forces at connections. a. Where a multi-force member is connected to a two-force member, apply to the multi-force member a force that is equal and opposite to the force drawn on the free-body diagram of the two-force member, giving it the same name.
b. Where a multi-force member is connected to another multi-force member, use horizontal and vertical components to represent the internal forces at that point. The directions you choose for each of the two force components exerted on the first multi-force member are arbitrary, but you must apply equal and opposite force components of the same name to the other multi-force member.
5. Write equilibrium equations after you have completed the various free-body diagrams.
a. To simplify your solution, you should, whenever possible, write and solve equilibrium equations involving single unknowns. b. Since you arbitrarily chose the direction of each of the unknown forces, you must determine at the end of the solution whether your guess was correct. To that effect, consider the sign of the value found for each of the unknowns. A positive sign indicates that your guess was correct, and a negative sign indicates that it was not.
6. Finally, check your solution by substituting the results obtained into an equilib-rium equation that you have not previously used.
352 FREE-BODY PRACTICE PROBLEMS 6.F5 An 84-lb force is applied to the toggle vise at C. Knowing that θ 5 90°, draw the free-body diagram(s) needed to determine the vertical force exerted on the block at D. 6.F6 For the system and loading shown, draw the free-body diagram(s) needed to determine the force P required for equilibrium.
200 mm 100 N 50 N A B C D E 30° P 75 mm Fig. P6.F6 6.F7 A small barrel weighing 60 lb is lifted by a pair of tongs as shown. Knowing that a 5 5 in., draw the free-body diagram(s) needed to determine the forces exerted at B and D on tong ABD. 6.F8 The position of member ABC is controlled by the hydraulic cylinder CD. Knowing that θ 5 30°, draw the free-body diagram(s) needed to determine the force exerted by the hydraulic cylinder on pin C, and the reaction at B.
C B A D q 10 kN 1.5 m 0.5 m 0.8 m 90° Fig. P6.F8 Problems A B C D q 84 lb 7 in.
24 in.
24 in.
9 in.
40 in.
Fig. P6.F5 A B C D P a a 6 in.
9 in.
18 in.
Fig. P6.F7 353 END-OF-SECTION PROBLEMS 6.122 The shear shown is used to cut and trim electronic-circuit-board laminates. For the position shown, determine (a) the vertical component of the force exerted on the shearing blade at D, (b) the reaction at C. 6.123 A 100-lb force directed vertically downward is applied to the toggle vise at C. Knowing that link BD is 6 in. long and that a 5 4 in., determine the horizontal force exerted on block E.
D A B a C E 100 lb 6 in.
15° Fig. P6.123 and P6.124 6.124 A 100-lb force directed vertically downward is applied to the toggle vise at C. Knowing that link BD is 6 in. long and that a 5 8 in., determine the horizontal force exerted on block E. 6.125 The control rod CE passes through a horizontal hole in the body of the toggle system shown. Knowing that link BD is 250 mm long, determine the force Q required to hold the system in equilibrium when β 5 20°.
A B C D E 100 N 35 mm β Q 200 mm 150 mm Fig. P6.125 6.126 Solve Prob. 6.125 when (a) β 5 0, (b) β 5 6°. 6.127 The press shown is used to emboss a small seal at E. Knowing that P 5 250 N, determine (a) the vertical component of the force exerted on the seal, (b) the reaction at A. 6.128 The press shown is used to emboss a small seal at E. Knowing that the vertical component of the force exerted on the seal must be 900 N, determine (a) the required vertical force P, (b) the corre-sponding reaction at A.
400 N 300 mm 60 mm 45 mm 30° 30° A C E B 25 mm 30 mm D Fig. P6.122 A B C D E 20° 60° 15° P 400 mm 200 mm Fig. P6.127 and P6.128 354 6.129 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ 5 30°. 6.130 The pin at B is attached to member ABC and can slide freely along the slot cut in the fixed plate. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ 5 60°. 6.131 Arm ABC is connected by pins to a collar at B and to crank CD at C. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ 5 0.
A B C D θ M 160 mm 90 mm 240 N 180 mm 320 mm 300 mm 125 mm Fig. P6.131 and P6.132 6.132 Arm ABC is connected by pins to a collar at B and to crank CD at C. Neglecting the effect of friction, determine the couple M required to hold the system in equilibrium when θ 5 90°. 6.133 The Whitworth mechanism shown is used to produce a quick-return motion of point D. The block at B is pinned to the crank AB and is free to slide in a slot cut in member CD. Determine the couple M that must be applied to the crank AB to hold the mechanism in equilibrium when (a) α 5 0, (b) α 5 30°. 6.134 Solve Prob. 6.133 when (a) α 5 60°, (b) α 5 90°. 6.135 and 6.136 Two rods are connected by a frictionless collar B. Knowing that the magnitude of the couple MA is 500 lb?in., deter-mine (a) the couple MC required for equilibrium, (b) the correspond-ing components of the reaction at C.
A B C MA MC 8 in.
6 in.
14 in.
Fig. P6.135 A B C MA MC 8 in.
6 in.
14 in.
Fig. P6.136 A B C D θ M 25 lb 10 in.
6 in.
8 in.
Fig. P6.129 and P6.130 700 mm 100 mm 400 mm 1200 N D B A C M a Fig. P6.133 355 6.137 and 6.138 Rod CD is attached to the collar D and passes through a collar welded to end B of lever AB. Neglecting the effect of fric-tion, determine the couple M required to hold the system in equilib-rium when θ 5 30°. 6.139 Two hydraulic cylinders control the position of the robotic arm ABC. Knowing that in the position shown the cylinders are parallel, determine the force exerted by each cylinder when P 5 160 N and Q 5 80 N.
A B C D E F G 150 mm 150 mm 200 mm P Q 600 mm 300 mm 400 mm Fig. P6.139 and P6.140 6.140 Two hydraulic cylinders control the position of the robotic arm ABC. In the position shown, the cylinders are parallel and both are in ten-sion. Knowing that FAE 5 600 N and FDG 5 50 N, determine the forces P and Q applied at C to arm ABC. 6.141 A 39-ft length of railroad rail of weight 44 lb/ft is lifted by the tongs shown. Determine the forces exerted at D and F on tong BDF.
D A C B 9.6 in.
9.6 in.
F E 6 in.
8 in.
12 in.
0.8 in.
0.8 in.
Fig. P6.141 A B C D M q 150 N 100 mm 80 mm Fig. P6.137 q A B C D M 300 N 200 mm Fig. P6.138 356 6.142 A log weighing 800 lb is lifted by a pair of tongs as shown. Deter-mine the forces exerted at E and F on tong DEF.
A B C D E F G 3 in.
3 in.
1.5 in.
800 lb 1.5 in.
12 in.
2.5 in.
3.5 in.
12 in.
Fig. P6.142 6.143 The tongs shown are used to apply a total upward force of 45 kN on a pipe cap. Determine the forces exerted at D and F on tong ADF. 6.144 If the toggle shown is added to the tongs of Prob. 6.143 and a single vertical force is applied at G, determine the forces exerted at D and F on tong ADF. 6.145 The pliers shown are used to grip a 0.3-in.-diameter rod. Knowing that two 60-lb forces are applied to the handles, determine (a) the magnitude of the forces exerted on the rod, (b) the force exerted by the pin at A on portion AB of the pliers.
B A C 1.2 in.
60 lb 60 lb 30° 9.5 in.
Fig. P6.145 6.146 Determine the magnitude of the gripping forces exerted along line aa on the nut when two 50-lb forces are applied to the handles as shown. Assume that pins A and D slide freely in slots cut in the jaws.
A B C D E F 25 mm 60 mm 75 mm 85 mm 90 mm Fig. P6.143 55 mm 55 mm 45 kN 22 mm G A B Fig. P6.144 A B a a C D E 50 lb 50 lb 4.5 in.
0.75 in.
0.5 in.
Fig. P6.146 357 6.147 In using the bolt cutter shown, a worker applies two 300-N forces to the handles. Determine the magnitude of the forces exerted by the cutter on the bolt. 6.148 Determine the magnitude of the gripping forces produced when two 300-N forces are applied as shown. 6.149 and 6.150 Determine the force P that must be applied to the toggle CDE to maintain bracket ABC in the position shown.
150 mm 150 mm 150 mm 30 mm 910 N P A B C D E 150 mm 150 mm Fig. P6.149 30 mm 910 N P A B C D E 150 mm 150 mm 150 mm 150 mm 150 mm Fig. P6.150 6.151 Since the brace shown must remain in position even when the mag-nitude of P is very small, a single safety spring is attached at D and E. The spring DE has a constant of 50 lb/in. and an unstretched length of 7 in. Knowing that l 5 10 in. and that the magnitude of P is 800 lb, determine the force Q required to release the brace.
l A D B E C Q P 15 in.
20 in.
2 in.
1 in.
Fig. P6.151 6.152 The specialized plumbing wrench shown is used in confined areas (e.g., under a basin or sink). It consists essentially of a jaw BC pinned at B to a long rod. Knowing that the forces exerted on the nut are equivalent to a clockwise (when viewed from above) couple with a magnitude of 135 lb?in., determine (a) the magnitude of the force exerted by pin B on jaw BC, (b) the couple M0 that is applied to the wrench.
12 mm 24 mm 24 mm 24 mm 300 N 300 N 460 mm 96 mm A B C D E Fig. P6.147 A B C D 300 N 300 N 12 mm 120 mm 36 mm 30 mm 30 mm 6 mm 42 mm 96 mm Fig. P6.148 A B C M0 5 8 in.
3 8 in.
1 8 in.
1 Fig. P6.152 358 6.153 The motion of the bucket of the front-end loader shown is controlled by two arms and a linkage that are pin-connected at D. The arms are located symmetrically with respect to the central, vertical, and longi-tudinal plane of the loader; one arm AFJ and its control cylinder EF are shown. The single linkage GHDB and its control cylinder BC are located in the plane of symmetry. For the position and loading shown, determine the force exerted (a) by cylinder BC, (b) by cylinder EF.
A B C D E F G H 12 in.
12 in.
12 in.
20 in.
20 in.
24 in.
22 in.
28 in.
75 in.
4500 lb 10 in.
18 in.
J Fig. P6.153 6.154 The bucket of the front-end loader shown carries a 3200-lb load. The motion of the bucket is controlled by two identical mechanisms, only one of which is shown. Knowing that the mechanism shown supports one-half of the 3200-lb load, determine the force exerted (a) by cylinder CD, (b) by cylinder FH.
A B C D E F 3200 lb Dimensions in inches G H 8 15 15 16 12 6 24 15 20 16 24 6 Fig. P6.154 6.155 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together have a mass of 200 kg and have a combined center of gravity located directly above C. For the position when θ 5 20°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A. 6.156 The telescoping arm ABC of Prob. 6.155 can be lowered until end C is close to the ground, so that workers can easily board the platform. For the position when θ 5 220°, determine (a) the force exerted at B by the single hydraulic cylinder BD, (b) the force exerted on the supporting carriage at A.
A B C D 5 m 2.4 m 0.9 m 0.5 m θ Fig. P6.155 359 6.157 The motion of the backhoe bucket shown is controlled by the hydrau-lic cylinders AD, CG, and EF. As a result of an attempt to dislodge a portion of a slab, a 2-kip force P is exerted on the bucket teeth at J. Knowing that θ 5 45°, determine the force exerted by each cylinder.
A B C G H J D E P F I 12 in.
10 in.
16 in.
60 in.
20 in.
48 in.
10 in.
15 in.
35 in.
36 in.
40 in.
8 in.
16 in.
16 in.
18 in.
10 in.
θ Fig. P6.157 6.158 Solve Prob. 6.157 assuming that the 2-kip force P acts horizontally to the right (θ 5 0). 6.159 The gears D and G are rigidly attached to shafts that are held by frictionless bearings. If rD 5 90 mm and rG 5 30 mm, determine (a) the couple M0 that must be applied for equilibrium, (b) the reac-tions at A and B.
x y z H E A B D G rG rD M0 30 N.m C F 180 mm 120 mm 200 mm 120 mm Fig. P6.159 360 6.160 In the planetary gear system shown, the radius of the central gear A is a 5 18 mm, the radius of each planetary gear is b, and the radius of the outer gear E is (a 1 2b). A clockwise couple with a magnitude of MA 5 10 N?m is applied to the central gear A and a counterclock-wise couple with a magnitude of MS 5 50 N?m is applied to the spider BCD. If the system is to be in equilibrium, determine (a) the required radius b of the planetary gears, (b) the magnitude ME of the couple that must be applied to the outer gear E. 6.161 Two shafts AC and CF, which lie in the vertical xy plane, are con-nected by a universal joint at C. The bearings at B and D do not exert any axial force. A couple with a magnitude of 500 lb?in. (clock-wise when viewed from the positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the cross-piece must be zero.) A B C D E x y z 4 in.
6 in.
5 in.
30° 500 lb-in.
F Fig. P6.161 6.162 Solve Prob. 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical. 6.163 The large mechanical tongs shown are used to grab and lift a thick 7500-kg steel slab HJ. Knowing that slipping does not occur between the tong grips and the slab at H and J, determine the components of all forces acting on member EFH. (Hint: Consider the symmetry of the tongs to establish relationships between the components of the force acting at E on EFH and the components of the force acting at D on DGJ.) A B C D E Fig. P6.160 A B C D E F G H J W 0.3 m 0.5 m 0.5 m 1.8 m 0.9 m 1.8 m 1.3 m 1 m Fig. P6.163 361 Review and Summary In this chapter, you studied ways to determine the internal forces holding together the various parts of a structure.
Analysis of Trusses The first half of the chapter presented the analysis of trusses, i.e., structures consisting of straight members connected at their extremities only. Because the members are slender and unable to support lateral loads, all of the loads must be applied at the joints; thus, we can assume that a truss consists of pins and two-force members [Sec. 6.1A].
Simple Trusses A truss is rigid if it is designed in such a way that it does not greatly deform or collapse under a small load. A triangular truss consisting of three members connected at three joints is clearly a rigid truss (Fig. 6.24a). The truss obtained by adding two new members to the first one and connecting them at a new joint (Fig. 6.24b) is also rigid. Trusses obtained by repeating this procedure are called simple trusses. We may check that, in a simple truss, the total number of members is m 5 2n 2 3, where n is the total number of joints [Sec. 6.1A].
(a) (b) A B C A B C D Fig. 6.24 Method of Joints We can determine the forces in the various members of a simple truss by using the method of joints [Sec. 6.1B]. First, we obtain the reactions at the supports by considering the entire truss as a free body. Then we draw the free-body diagram of each pin, showing the forces exerted on the pin by the members or supports it connects. Since the members are straight two-force members, the force exerted by a member on the pin is directed along that member, and only the magnitude of the force is unknown. In the case of a simple truss, it is always possible to draw the free-body diagrams of the pins in such an order that only two unknown forces are included in each diagram. We obtain these forces from the corresponding two equilibrium equations or—if only three forces are involved—from the corresponding force triangle. If the force exerted by a member on a pin is directed toward 362 362 that pin, the member is in compression; if it is directed away from the pin, the member is in tension [Sample Prob. 6.1]. The analysis of a truss is sometimes expedited by first recognizing joints under special loading conditions [Sec. 6.1C]. The method of joints also can be extended for the analysis of three-dimensional or space trusses [Sec. 6.1D].
Method of Sections The method of sections is usually preferable to the method of joints when we want to determine the force in only one member—or very few members— of a truss [Sec. 6.2A]. To determine the force in member BD of the truss of Fig. 6.25a, for example, we pass a section through members BD, BE, and CE; remove these members; and use the portion ABC of the truss as a free body (Fig. 6.25b). Setting oME 5 0, we determine the magnitude of force FBD that represents the force in member BD. A positive sign indicates that the member is in tension; a negative sign indicates that it is in compression [Sample Probs. 6.2 and 6.3].
A B C A B C D E E G (a) (b) n n P1 P2 P1 P2 P3 FCE FBD FBE Fig. 6.25 Compound Trusses The method of sections is particularly useful in the analysis of compound trusses, i.e., trusses that cannot be constructed from the basic triangular truss of Fig. 6.24a but are built by rigidly connecting several simple trusses [Sec. 6.2B]. If the component trusses are properly connected (e.g., one pin and one link, or three non-concurrent and unparallel links) and if the resulting structure is properly supported (e.g., one pin and one roller), the compound truss is statically determinate, rigid, and completely constrained. The following necessary—but not sufficient—condition is then satisfied: m 1 r 5 2n, where m is the number of members, r is the number of unknowns representing the reactions at the supports, and n is the number of joints.
363 Frames and Machines In the second part of the chapter, we analyzed frames and machines. These structures contain multi-force members, i.e., members acted upon by three or more forces. Frames are designed to support loads and are usually stationary, fully constrained structures. Machines are designed to transmit or modify forces and always contain moving parts [Sec. 6.3].
Analysis of a Frame To analyze a frame, we first consider the entire frame to be a free body and write three equilibrium equations [Sec. 6.3A]. If the frame remains rigid when detached from its supports, the reactions involve only three unknowns and may be determined from these equations [Sample Probs. 6.4 and 6.5]. On the other hand, if the frame ceases to be rigid when detached from its supports, the reactions involve more than three unknowns, and we cannot determine them completely from the equilibrium equations of the frame [Sec. 6.3B; Sample Prob. 6.6].
Multi-force Members We then dismember the frame and identify the various members as either two-force members or multi-force members; we assume pins form an integral part of one of the members they connect. We draw the free-body diagram of each of the multi-force members, noting that, when two multi-force members are connected to the same two-force member, they are acted upon by that member with equal and opposite forces of unknown magnitude but known direction. When two multi-force members are connected by a pin, they exert on each other equal and opposite forces of unknown direction that should be represented by two unknown components. We can then solve the equilibrium equations obtained from the free-body diagrams of the multi-force members for the various internal forces [Sample Probs. 6.4 and 6.5]. We also can use the equilibrium equations to complete the determination of the reactions at the supports [Sample Prob. 6.6]. Actually, if the frame is statically determinate and rigid, the free-body diagrams of the multi-force members could provide as many equations as there are unknown forces (including the reactions) [Sec. 6.3B]. However, as suggested previously, it is advisable to first consider the free-body diagram of the entire frame to minimize the number of equations that must be solved simultaneously.
Analysis of a Machine To analyze a machine, we dismember it and, following the same procedure as for a frame, draw the free-body diagram of each multi-force member. The corresponding equilibrium equations yield the output forces exerted by the machine in terms of the input forces applied to it as well as the internal forces at the various connections [Sec. 6.4; Sample Prob. 6.7].
364 Review Problems 6.164 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression.
6.165 Using the method of joints, determine the force in each member of the double-pitch roof truss shown. State whether each member is in tension or compression.
C D E F G H A B 4 m 4 m 4 m 3 m 3 m 1 kN 2 kN 2 kN 1.75 kN 1.5 kN 0.75 kN 6 m 6 m 6 m 6 m Fig. P6.165 6.166 A stadium roof truss is loaded as shown. Determine the force in members AB, AG, and FG.
A B C D E F G H I J K L 0.9 kips 0.9 kips 1.8 kips 1.8 kips 8 ft 8 ft 31.5 ft 9 ft 12 ft 14 ft 14 ft Fig. P6.166 and P6.167 6.167 A stadium roof truss is loaded as shown. Determine the force in members AE, EF, and FJ. 6.168 Determine the components of all forces acting on member ABD of the frame shown.
1.2 m 0.9 m 1.2 m A C D E B 6 kN 3 kN Fig. P6.164 6 ft 4 ft 4 ft 4 ft 4 ft A B J D E C 300 lb 450 lb Fig. P6.168 365 6.169 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N?m applied (a) at B, (b) at D.
6.170 Knowing that the pulley has a radius of 50 mm, determine the com-ponents of the reactions at B and E.
A B C D E 300 N 180 mm 120 mm 150 mm Fig. P6.170 6.171 For the frame and loading shown, determine the components of the forces acting on member CFE at C and F.
40 lb A B C D E F 6 in.
4 in.
5 in.
4 in.
4 in.
Fig. P6.171 6.172 For the frame and loading shown, determine the reactions at A, B, D, and E. Assume that the surface at each support is frictionless.
8 in.
8 in.
6 in.
6 in.
30° 1000 lb A B C D E Fig. P6.172 A B C D E 240 mm 240 mm 240 mm 160 mm Fig. P6.169 366 6.173 Water pressure in the supply system exerts a downward force of 135 N on the vertical plug at A. Determine the tension in the fusible link DE and the force exerted on member BCE at B.
24 mm A D B E C 24 mm 6 mm 16 mm Fig. P6.173 6.174 A couple M with a magnitude of 1.5 kN?m is applied to the crank of the engine system shown. For each of the two positions shown, determine the force P required to hold the system in equilibrium.
M A B P (a) (b) C 50 mm 75 mm 175 mm A B M P C 75 mm 100 mm 50 mm Fig. P6.174 6.175 The compound-lever pruning shears shown can be adjusted by plac-ing pin A at various ratchet positions on blade ACE. Knowing that 300-lb vertical forces are required to complete the pruning of a small branch, determine the magnitude P of the forces that must be applied to the handles when the shears are adjusted as shown.
A B C D E 3.5 in.
1.6 in.
0.5 in.
0.55 in.
0.25 in.
0.65 in.
0.75 in.
P –P Fig. P6.175 The Assut de l’Or Bridge in the City of Arts and Science in Valencia, Spain, is cable-stayed, where the bridge deck is supported by cables attached to the curved tower. The tower itself is partially supported by four anchor cables. The deck of the bridge consists of a system of beams that support the roadway.
Internal Forces and Moments 7 368 Internal Forces and Moments Introduction In previous chapters, we considered two basic problems involving struc-tures: (1) determining the external forces acting on a structure (Chap. 4) and (2) determining the internal forces that hold together the various mem-bers forming a structure (Chap. 6). Now we consider the problem of deter-mining the internal forces that hold together the parts of a given individual member.
We will first analyze the internal forces in the members of a frame, such as the crane considered in Fig. 6.1. Note that, whereas the internal forces in a straight two-force member can produce only tension or com-pression in that member, the internal forces in any other type of member usually produce shear and bending as well.
Most of this chapter is devoted to the analysis of the internal forces in two important types of engineering elements: 1. Beams, which are usually long, straight prismatic members designed to support loads applied at various points along it. 2. Cables, which are flexible members capable of withstanding only ten-sion and are designed to support either concentrated or distributed loads. Cables are used in many engineering applications, such as suspension bridges and power transmission lines.
7.1 INTERNAL FORCES IN MEMBERS Consider a straight two-force member AB (Fig. 7.1a). From Sec. 4.2A, we know that the forces F and 2F acting at A and B, respectively, must be directed along AB in opposite sense and have the same magnitude F. Suppose we cut the member at C. To maintain equilibrium of the resulting free bodies AC and CB, we must apply to AC a force 2F equal and Introduction 7.1 INTERNAL FORCES IN MEMBERS 7.2 BEAMS 7.2A Various Types of Loading and Support 7.2B Shear and Bending Moment in a Beam 7.2C Shear and Bending-Moment Diagrams 7.3 RELATIONS AMONG LOAD, SHEAR, AND BENDING MOMENT 7.4 CABLES 7.4A Cables with Concentrated Loads 7.4B Cables with Distributed Loads 7.4C Parabolic Cables 7.5 CATENARY CABLES Objectives • Consider the general state of internal member forces, which includes axial force, shearing force, and bending moment.
• Apply equilibrium analysis methods to obtain specifi c values, general expressions, and diagrams for shear and bending-moment in beams.
• Examine relations among load, shear, and bending-moment, and use these to obtain shear and bending-moment diagrams for beams.
• Analyze the tension forces in cables subjected to concentrated loads, loads uniformly distributed along the horizontal, and loads uniformly distributed along the cable itself.
(a) (b) C A B F –F –F C A F –F B C F Fig. 7.1 A straight two-force member in tension. (a) External forces act at the ends of the member; (b) internal axial forces do not depend on the location of section C.
7.1 Internal Forces in Members 369 opposite to F and to CB a force F equal and opposite to 2F (Fig. 7.1b). These new forces are directed along AB in opposite sense and have the same magnitude F. Since the two parts AC and CB were in equilibrium before the member was cut, internal forces equivalent to these new forces must have existed in the member itself. We conclude that, in the case of a straight two-force member, the internal forces that the two portions of the member exert on each other are equivalent to axial forces. The com-mon magnitude F of these forces does not depend upon the location of the section C and is referred to as the force in member AB. In the case shown in Fig. 7.1, the member is in tension and elongates under the action of the internal forces. In the case represented in Fig. 7.2, the member is in com-pression and decreases in length under the action of the internal forces.
(a) (b) C A B F –F –F C A F –F B C F Fig. 7.2 A straight two-force member in compression. (a) External forces act at the ends; (b) internal axial forces are independent of the location of section C.
T FBE Cx Ay Ax Cy T A B C D E F W G (a) A B C D J (b) FBE Ay Ax Cy Cx –F –M –V A B C J (d) Ay Ax A (e) FBE Cy Cx B C V M F T D D J (c) Fig. 7.3 (a) Crane from Chapter 6; (b) free-body diagram of multi-force member AD; (c,d) free-body diagrams of sections of member AD showing internal force-couple systems; (e) deformation of member AD.
Next, consider a multi-force member. Take, for instance, member AD of the crane analyzed in Sec. 6.3A. This crane is shown again in Fig. 7.3a, and we drew the free-body diagram of member AD in Fig. 7.3b. Suppose we cut member AD at J and draw a free-body diagram for each of the portions JD and AJ (Fig. 7.3c and d). Considering the free body JD, we find that, to maintain its equilibrium, we need to apply at J a force F to balance the vertical component of T; a force V to balance the 370 Internal Forces and Moments horizontal component of T; and a couple M to balance the moment of T about J. Again, we conclude that internal forces must have existed at J before member AD was cut, which is equivalent to the force-couple system shown in Fig. 7.3c. According to Newton’s third law, the internal forces acting on AJ must be equivalent to an equal and opposite force-couple system, as shown in Fig. 7.3d. Clearly, the action of the internal forces in member AD is not limited to producing tension or compression, as in the case of straight two-force members; the internal forces also produce shear and bending. The force F is an axial force; the force V is called a shearing force; and the moment M of the couple is known as the bending moment at J. Note that, when determining internal forces in a member, you should clearly indicate on which portion of the member the forces are supposed to act. The deformation that occurs in member AD is sketched in Fig. 7.3e. The actual analysis of such a deformation is part of the study of mechanics of materials.
Also note that, in a two-force member that is not straight, the internal forces are also equivalent to a force-couple system. This is shown in Fig. 7.4, where the two-force member ABC has been cut at D.
(a) D B C A P –P (b) D A P M V F (c) D B C – P –F –M –V Fig. 7.4 (a) Free-body diagram of a two-force member that is not straight; (b, c) free-body diagrams of sections of member ABC showing internal force-couple systems.
Photo 7.1 The design of the shaft of a circular saw must account for the internal forces resulting from the forces applied to the teeth of the blade. At a given point in the shaft, these internal forces are equivalent to a force-couple system consisting of axial and shearing forces and couples representing the bending and torsional moments.
7.1 Internal Forces in Members 371 Sample Problem 7.1 In the frame shown, determine the internal forces (a) in member ACF at point J, (b) in member BCD at point K. This frame was previously ana-lyzed in Sample Prob. 6.5.
STRATEGY: After isolating each member, you can cut it at the given point and treat the resulting parts as objects in equilibrium. Analysis of the equilibrium equations, as we did before in Sample Problem 6.5, will determine the internal force-couple system.
MODELING: The reactions and the connection forces acting on each member of the frame were determined previously in Sample Prob. 6.5. The results are repeated in Fig. 1.
1.2 m 1.5 m 2400 N a A B C D K J F E 3.6 m 2.7 m 2.7 m 4.8 m 2400 N A A B B C C D K J F E 1200 N 1200 N 3600 N 3600 N 1800 N 1800 N 1800 N 600 N Fig. 1 Reactions and connection forces acting on each member of the frame.
372 Internal Forces and Moments ANALYSIS: a. Internal Forces at J. Cut member ACF at point J, obtaining the two parts shown in Fig. 2. Represent the internal forces at J by an equivalent force-couple system, which can be determined by considering the equilibrium of either part. Considering the free body AJ, you have 1l oMJ 5 0: 2(1800 N)(1.2 m) 1 M 5 0 M 5 12160 N?m M 5 2160 N?m l b 1 oFx 5 0: F 2 (1800 N) cos 41.7° 5 0 F 5 11344 N F 5 1344 N b 1 oFy 5 0: 2V 1 (1800 N) sin 41.7° 5 0 V 5 11197 N V 5 1197 N b The internal forces at J are therefore equivalent to a couple M, an axial force F, and a shearing force V. The internal force-couple system acting on part JCF is equal and opposite. b. Internal Forces at K. Cut member BCD at K, obtaining the two parts shown in Fig. 3. Considering the free body BK, you obtain 1l oMK 5 0: (1200 N)(1.5 m) 1 M 5 0 M 5 21800 N?m M 5 1800 N?m b yoFx 5 0: F 5 0 F 5 0 b 1xoFy 5 0: 21200 N 2 V 5 0 V 5 21200 N V 5 1200 Nx b 1 REFLECT and THINK: The mathematical techniques involved in solv-ing a problem of this type are not new; they are simply applications of concepts presented in earlier chapters. However, the physical interpretation is new: we are now determining the internal forces and moments within a structural member. These are of central importance in the study of mechanics of materials.
2400 N 1200 N 3600 N V M F –M –V –F B C D K K y x 1.5 m Fig. 3 Free-body diagrams of portion BK and DK of member BCD.
A C J J F 3600 N 1800 N 1800 N a = 41.7° y x V M F 1.2 m –F –M –V Fig. 2 Free-body diagrams of portion AJ and FJ of member ACF.
373 373 I n this section, we discussed how to determine the internal forces in the member of a frame. The internal forces at a given point in a straight two-force member reduce to an axial force, but in all other cases, they are equivalent to a force-couple system consisting of an axial force F, a shearing force V, and a couple M representing the bending moment at that point.
To determine the internal forces at a given point J of the member of a frame, you should take the following steps.
1. Draw a free-body diagram of the entire frame, and use it to determine as many of the reactions at the supports as you can.
2. Dismember the frame and draw a free-body diagram of each of its members.
Write as many equilibrium equations as are necessary to find all of the forces acting on the member on which point J is located.
3. Cut the member at point J and draw a free-body diagram of each resulting portion. Apply to each portion at point J the force components and couple represent-ing the internal forces exerted by the other portion. These force components and couples are equal in magnitude and opposite in sense.
4. Select one of the two free-body diagrams you have drawn and use it to write three equilibrium equations for the corresponding portion of the member. a. Summing moments about J and equating them to zero yields the bending moment at point J. b. Summing components in directions parallel and perpendicular to the mem-ber at J and equating them to zero yields, respectively, the axial and shearing forces.
5. When recording your answers, be sure to specify the portion of the member you have used, since the forces and couples acting on the two portions have opposite senses.
The solutions of the problems in this section require you to determine the forces exerted on each other by the various members of a frame, so be sure to review the methods used in Chap. 6 to solve this type of problem. When frames involve pulleys and cables, for instance, remember that the forces exerted by a pulley on the member of the frame to which it is attached have the same magnitude and direction as the forces exerted by the cable on the pulley [Prob. 6.90].
SOLVING PROBLEMS ON YOUR OWN 374 7.1 and 7.2 Determine the internal forces (axial force, shearing force, and bending moment) at point J of the structure indicated. 7.1 Frame and loading of Prob. 6.76 7.2 Frame and loading of Prob. 6.78 7.3 Determine the internal forces at point J when α 5 90°.
Problems 300 mm 300 mm 480 mm 240 mm J A B D a C 780 N Fig. P7.3 and P7.4 80 mm 80 mm 40 mm 20 mm 20 mm 20 mm A K J B 250 N 250 N C Fig. P7.5 and P7.6 B A J C D 24 in.
8 in.
16 in.
16 in.
32 in.
Fig. P7.7 7.4 Determine the internal forces at point J when α 5 0. 7.5 and 7.6 For the frame and loading shown, determine the internal forces at the point indicated: 7.5 Point J 7.6 Point K 7.7 An archer aiming at a target is pulling with a 45-lb force on the bowstring. Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at point J. 7.8 For the bow of Prob. 7.7, determine the magnitude and location of the maximum (a) axial force, (b) shearing force, (c) bending moment.
375 7.9 A semicircular rod is loaded as shown. Determine the internal forces at point J.
A B J K 120 N 60° 30° 180 mm 180 mm Fig. P7.9 and P7.10 A B J C D 280 N q 160 mm 120 mm 160 mm 160 mm Fig. P7.11 and P7.12 P a A B J L h Fig. P7.13 and P7.14 7.10 A semicircular rod is loaded as shown. Determine the internal forces at point K. 7.11 A semicircular rod is loaded as shown. Determine the internal forces at point J knowing that θ 5 30°. 7.12 A semicircular rod is loaded as shown. Determine the magnitude and location of the maximum bending moment in the rod.
7.13 The axis of the curved member AB is a parabola with vertex at A. If a vertical load P of magnitude 450 lb is applied at A, determine the internal forces at J when h 5 12 in., L 5 40 in., and a 5 24 in.
7.14 Knowing that the axis of the curved member AB is a parabola with vertex at A, determine the magnitude and location of the maximum bending moment. 7.15 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at point J of the frame shown.
0.6 m 0.2 m 0.2 m 0.8 m 0.8 m A B C D K J E F 360 N 1 m 1.8 m Fig. P7.15 and P7.16 7.16 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at point K of the frame shown.
376 7.17 A 5-in.-diameter pipe is supported every 9 ft by a small frame con-sisting of two members as shown. Knowing that the combined weight of the pipe and its contents is 10 lb/ft and neglecting the effect of friction, determine the magnitude and location of the maxi-mum bending moment in member AC.
A B C D E r = 2.5 in.
9 in.
6.75 in.
12 in.
Fig. P7.17 0.2 m 0.8 m 0.8 m 0.8 m A B C D K J E 360 N 1 m 1.8 m Fig. P7.19 and P7.20 7.18 For the frame of Prob. 7.17, determine the magnitude and location of the maximum bending moment in member BC. 7.19 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at point J of the frame shown. 7.20 Knowing that the radius of each pulley is 200 mm and neglecting friction, determine the internal forces at point K of the frame shown. 7.21 and 7.22 A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at point J.
3 4 3 4 B C D J A P (a) (b) (c) a a a a B C D J A P 3 4 a a a a B C D J A P a a a a Fig. P7.21 B C D J P (a) (b) (c) a a a a B C D J P 3 4 3 4 a a a a B C D J A A A P a a a a Fig. P7.22 377 7.23 A quarter-circular rod of weight W and uniform cross section is sup-ported as shown. Determine the bending moment at point J when θ 5 30°.
B J A q r Fig. P7.23 7.24 For the rod of Prob. 7.23, determine the magnitude and location of the maximum bending moment. 7.25 A semicircular rod of weight W and uniform cross section is sup-ported as shown. Determine the bending moment at point J when θ 5 60°.
q B J A r r Fig. P7.25 and P7.26 r q B J C A O Fig. P7.27 r q A B O C J Fig. P7.28 7.26 A semicircular rod of weight W and uniform cross section is sup-ported as shown. Determine the bending moment at point J when θ 5 150°. 7.27 and 7.28 A half section of pipe rests on a frictionless horizontal surface as shown. If the half section of pipe has a mass of 9 kg and a diameter of 300 mm, determine the bending moment at point J when θ 5 90°.
378 Internal Forces and Moments 7.2 BEAMS A structural member designed to support loads applied at various points along the member is known as a beam. In most cases, the loads are per-pendicular to the axis of the beam and cause only shear and bending in the beam. When the loads are not at a right angle to the beam, they also produce axial forces in the beam.
Beams are usually long, straight prismatic bars. Designing a beam for the most effective support of the applied loads is a two-part process: (1) determine the shearing forces and bending moments produced by the loads and (2) select the cross section best suited to resist these shearing forces and bending moments. Here we are concerned with the first part of the problem of beam design. The second part belongs to the study of mechanics of materials.
7.2A Various Types of Loading and Support A beam can be subjected to concentrated loads P1, P2, . . . that are expressed in newtons, pounds, or their multiples, kilonewtons and kips (Fig. 7.5a). We can also subject a beam to a distributed load w, expressed in N/m, kN/m, lb/ft, or kips/ft (Fig. 7.5b). In many cases, a beam is subjected to a combina-tion of both types of load. When the load w per unit length has a constant value over part of the beam (as between A and B in Fig. 7.5b), the load is said to be uniformly distributed over that part of the beam. Determining the reactions at the supports is considerably simplified if we replace distrib-uted loads by equivalent concentrated loads, as explained in Sec. 5.3A. However, you should not do this substitution, or at least perform it with care, when calculating internal forces (see Sample Prob. 7.3).
Beams are classified according to the way in which they are sup-ported. Figure 7.6 shows several types of beams used frequently. (a) Concentrated loads (b) Distributed load A B C A B C D P1 P2 w Fig. 7.5 A beam may be subjected to (a) concentrated loads or (b) distributed loads, or a combination of both.
Fig. 7.6 Some common types of beams and their supports.
(a) Simply supported beam (d) Continuous beam (b) Overhanging beam (e) Beam fixed at one end and simply supported at the other end ( f ) Fixed beam (c) Cantilever beam Statically Determinate Beams Statically Indeterminate Beams L L L L L L1 L2 Roof support (Continuous beam) Roof extension (Overhanging beam) Viewing platform (Cantilever beam) Examples 7.2 Beams 379 The distance L between supports is called the span. Note that the reactions are determinate if the supports involve only three unknowns. If more unknowns are involved, the reactions are statically indeterminate, and the methods of statics are not sufficient to determine the reactions. In such a case, we must take into account the properties of the beam with regard to its resistance to bending. Beams supported by only two rollers are not shown here; they are partially constrained and move under certain types of loadings.
Sometimes two or more beams are connected by hinges to form a single continuous structure. Two examples of beams hinged at a point H are shown in Fig. 7.7. Here the reactions at the supports involve four unknowns and cannot be determined from the free-body diagram of the two-beam system. However, we can determine the reactions by consider-ing the free-body diagram of each beam separately. Analysis of this situ-ation involves six unknowns (including two force components at the hinge), and six equations are available.
7.2B Shear and Bending Moment in a Beam Consider a beam AB subjected to various concentrated and distributed loads (Fig. 7.8a). We propose to determine the shearing force and bending moment at any point of the beam. In the example considered here, the beam is simply supported, but the method used could be applied to any type of statically determinate beam.
First we determine the reactions at A and B by choosing the entire beam as a free body (Fig. 7.8b). Setting oMA 5 0 and oMB 5 0, we obtain, respectively, RB and RA.
A A B B C H H (a) (b) Fig. 7.7 Examples of two-beam systems connected by a hinge. In both cases, free-body diagrams of each individual beam enable you to determine the support reactions.
Photo 7.2 As a truck crosses a highway overpass, the internal forces vary in the beams of the overpass.
A B C A B C A B C C w1 w2 (a) (b) (c) P1 P2 P3 w1 w2 P1 w1 P1 P2 P3 w2 P2 P3 M M' V V' RA RB RA RB Fig. 7.8 (a) A simply supported beam AB; (b) free-body diagram of the beam; (c) free-body diagrams of portions AC and CB of the beam, showing internal shearing forces and couples.
380 Internal Forces and Moments To determine the internal forces at an arbitrary point C, we cut the beam at C and draw the free-body diagrams of the portions AC and CB (Fig. 7.8c). Using the free-body diagram of AC, we can determine the shearing force V at C by equating the sum of the vertical components of all forces acting on AC to zero. Similarly, we can find the bending moment M at C by equating the sum of the moments about C of all forces and couples acting on AC to zero. Alternatively, we could use the free-body diagram of CB† and determine the shearing force V9 and the bending moment M9 by equating the sum of the vertical components and the sum of the moments about C of all forces and couples acting on CB to zero. Although this choice of free bodies may make the computation of the numerical values of the shearing force and bending moment easier, it requires us to indicate on which portion of the beam the internal forces considered are acting. If we want to calculate and efficiently record the shearing force and bending moment at every point of the beam, we must devise a way to avoid having to specify which portion of the beam is used as a free body every time. Therefore, we shall adopt the following conventions.
In determining the shearing force in a beam, we always assume that the internal forces V and V9 are directed as shown in Fig. 7.8c. A positive value obtained for their common magnitude V indicates that this assumption is correct and that the shearing forces are actually directed as shown. A negative value obtained for V indicates that the assumption is wrong and the shearing forces are directed in the opposite way. Thus, to define completely the shearing forces at a given point of the beam, we only need to record the magnitude V, together with a plus or minus sign. The scalar V is commonly referred to as the shear at the given point of the beam.
Similarly, we always assume that the internal couples M and M9 are directed as shown in Fig. 7.8c. A positive value obtained for their magnitude M, commonly referred to as the bending moment, indicates that this assumption is correct, whereas a negative value indicates that it is wrong. Summarizing these sign conventions, we state: The shear V and the bending moment M at a given point of a beam are said to be positive when the internal forces and couples acting on each portion of the beam are directed as shown in Fig. 7.9a.
You may be able to remember these conventions more easily by noting that: 1. The shear at C is positive when the external forces (loads and reac-tions) acting on the beam tend to shear off the beam at C as indicated in Fig. 7.9b. 2. The bending moment at C is positive when the external forces acting on the beam tend to bend the beam at C in a concave-up fashion as indicated in Fig. 7.9c.
† We now designate the force and couple representing the internal forces acting on CB by V9 and M9, rather than by 2V and 2M as done earlier. The reason is to avoid confusion when applying the sign convention we are about to introduce. C (b) Effect of external forces (positive shear) (a) Internal forces at section (positive shear and positive bending moment) (c) Effect of external forces (positive bending moment) C M V M' V' Fig. 7.9 Figure for remembering the signs of shear and bending moment.
7.2 Beams 381 It may also help to note that the situation described in Fig. 7.9, in which the values of both the shear and the bending moment are positive, is precisely the situation that occurs in the left half of a simply supported beam carrying a single concentrated load at its midpoint. This particular example is fully discussed in the following section.
7.2C Shear and Bending-Moment Diagrams Now that we have clearly defined shear and bending moment in sense as well as in magnitude, we can easily record their values at any point along a beam by plotting these values against the distance x measured from one end of the beam. The graphs obtained in this way are called, respectively, the shear diagram and the bending-moment diagram. As an example, consider a simply supported beam AB of span L subjected to a single concentrated load P applied at its midpoint D (Fig. 7.10a). We first determine the reactions at the supports from the free-body diagram of the entire beam (Fig. 7.10b); we find that the mag-nitude of each reaction is equal to P/2.
Next we cut the beam at a point C between A and D and draw the free-body diagrams of AC and CB (Fig. 7.10c). Assuming that shear and bending moment are positive, we direct the internal forces V and V9 and the internal couples M and M9 as indicated in Fig. 7.9a. Considering the free body AC, we set the sum of the vertical components and the sum of the moments about C of the forces acting on the free body to zero. From this, we find V 5 1P/2 and M 5 1Px/2. Therefore, both shear and bending moment are positive. (You can check this by observing that the reaction at A tends to shear off and to bend the beam at C as indi-cated in Fig. 7.9b and c.) Now let’s plot V and M between A and D (Fig. 7.10e and f ). The shear has a constant value V 5 P/2, whereas the bending moment increases linearly from M 5 0 at x 5 0 to M 5 PL/4 at x 5 L/2.
Proceeding along the beam, we cut it at a point E between D and B and consider the free body EB (Fig. 7.10d). As before, the sum of the vertical components and the sum of the moments about E of the forces acting on the free body are zero. We obtain V 5 2P/2 and M 5 P(L 2 x)/2. The shear is therefore negative and the bending moment is positive. (Again, you can check this by observing that the reaction at B bends the beam at E as indicated in Fig. 7.9c but tends to shear it off in a manner opposite to that shown in Fig. 7.9b.) We can now complete the shear and bending-moment diagrams of Fig. 7.10e and f. The shear has a constant value V 5 2P/2 between D and B, whereas the bending moment decreases linearly from M 5 PL/4 at x 5 L/2 to M 5 0 at x 5 L.
Note that when a beam is subjected to concentrated loads only, the shear is of constant value between loads and the bending moment varies linearly between loads. However, when a beam is subjected to distributed loads, the shear and bending moment vary quite differently (see Sample Prob. 7.3).
x L L M x V D D D E B A B E L 2 M' M M P V' (c) (b) (a) (d) (e) ( f ) P 2 RA = P 2 RB = P 2 RB = P 2 RA = P 2 RA = P 2 – PL 4 P 2 A x A B A B D P 2 RB = P P P V V C C M' V' x L – x L 2 L 2 L 2 Fig. 7.10 (a) A beam supporting a single concentrated load at its midpoint; (b) free-body diagram of the beam; (c) free-body diagrams of parts of the beam after a cut at C ; (d) free-body diagrams of parts of the beam after a cut at E; (e) shear diagram of the beam; (f) bending-moment diagram of the beam.
382 Internal Forces and Moments Sample Problem 7.2 Draw the shear and bending-moment diagrams for the beam and loading shown.
STRATEGY: Treat the entire beam as a free body to determine the reac-tions, then cut the beam just before and just after each external concen-trated force (Fig. 1) to see how the shear and bending moment change along the length of the beam.
MODELING and ANALYSIS: Free-Body, Entire Beam. From the free-body diagram of the entire beam, find the reactions at B and D: RB 5 46 kNx RD 5 14 kNx Shear and Bending Moment. First, determine the internal forces just to the right of the 20-kN load at A. Consider the stub of beam to the left of point 1 as a free body, and assume V and M are positive (according to the standard convention). Then you have 1 xoFy 5 0: 220 kN 2 V1 5 0 V1 5 220 kN 1 l oM1 5 0: (20 kN)(0 m) 1 M1 5 0 M1 5 0 Next, consider the portion of the beam to the left of point 2 as a free body: 1 xoFy 5 0: 220 kN 2 V2 5 0 V2 5 220 kN 1 l oM2 5 0: (20 kN)(2.5 m) 1 M2 5 0 M2 5 250 kN?m Determine the shear and bending moment at sections 3, 4, 5, and 6 in a similar way from the free-body diagrams. The results are V3 5 126 kN M3 5 250 kN?m V4 5 126 kN M4 5 128 kN?m V5 5 214 kN M5 5 128 kN?m V6 5 214 kN M6 5 0 For several of the later cuts, the results are easier to obtain by considering as a free body the portion of the beam to the right of the cut. For example, consider the portion of the beam to the right of point 4. You have 1 xoFy 5 0: V4 2 40 kN 1 14 kN 5 0 V4 5 126 kN 1 l oM4 5 0: 2M4 1 (14 kN)(2 m) 5 0 M4 5 128 kN?m Shear and Bending-Moment Diagrams. Now plot the six points shown on the shear and bending-moment diagrams. As indicated in Sec. 7.2C, the shear is of constant value between concentrated loads, and the bending moment varies linearly. You therefore obtain the shear and bending-moment diagrams shown in Fig. 1.
REFLECT and THINK: The calculations are pretty similar for each new choice of free body. However, moving along the beam, the shear changes magnitude whenever you pass a transverse force and the graph of the bending moment changes slope at these points.
A D B C 20 kN 40 kN 2.5 m 3 m 2 m A D B C V M M1 V1 M2 V2 M3 V3 M4 V4 M5 V5 M6 V6 M' 4 V'4 20 kN 20 kN 20 kN 20 kN 20 kN 20 kN 20 kN 40 kN 40 kN 40 kN 40 kN 46 kN 46 kN 46 kN 46 kN 46 kN 14 kN 1 2 3 4 5 6 –14 kN –20 kN +28 kN·m –50 kN·m +26 kN 2.5 m 3 m 2 m 2.5 m 3 m 2 m x x 14 kN Fig. 1 Free-body diagrams of beam sections, and the resulting shear and bending-moment diagrams.
7.2 Beams 383 Sample Problem 7.3 Draw the shear and bending-moment diagrams for the beam AB. The distributed load of 40 lb/in. extends over 12 in. of the beam from A to C, and the 400-lb load is applied at E.
STRATEGY: Again, consider the entire beam as a free body to find the reactions. Then cut the beam within each region of continuous load. This will enable you to determine continuous functions for shear and bending moment, which you can then plot on a graph.
MODELING and ANALYSIS: Free-Body, Entire Beam. Determine the reactions by considering the entire beam as a free body (Fig. 1).
1l oMA 5 0: By(32 in.) 2 (480 lb)(6 in.) 2 (400 lb)(22 in.) 5 0 By 5 1365 lb By 5 365 lbx 1l oMB 5 0: (480 lb)(26 in.) 1 (400 lb)(10 in.) 2 A(32 in.) 5 0 A 5 1515 lb A 5 515 lbx y 1 oFx 5 0: Bx 5 0 Bx 5 0 Now, replace the 400-lb load by an equivalent force-couple system acting on the beam at point D and cut the beam at several points (Fig. 2).
A B C D E 40 lb/in.
400 lb 6 in. 4 in.
32 in.
12 in.
10 in.
A A B C D E 400 lb 6 in.
10 in.
480 lb Bx By 16 in.
Fig. 1 Free-body diagram of entire beam.
A B M V M V M M V V C D 40 lb/in.
400 lb 12 in.
6 in.
40x 480 lb 480 lb 400 lb 515 lb 515 lb 515 lb 515 lb 515 lb 365 lb 1600 lb·in.
3300 lb·in.
3510 lb·in.
x 2 x – 6 x x x – 18 32 in.
12 in.
18 in.
–365 lb 35 lb 1 2 3 1600 lb·in.
5110 lb·in.
x x x x – 6 14 in.
Fig. 2 Free-body diagrams of beam sections, and the resulting shear and bending-moment diagrams.
384 Internal Forces and Moments Shear and Bending Moment. From A to C. Determine the internal forces at a distance x from point A by considering the portion of the beam to the left of point 1. Replace that part of the distributed load acting on the free body by its resultant. You get 1 xoFy 5 0: 515 2 40 x 2 V 5 0 V 5 515 2 40x 1 l oM1 5 0: 2515x 1 40x(1 2 x) 1 M 5 0 M 5 515x 2 20x 2 Note that V and M are not numerical values, but they are expressed as functions of x. The free-body diagram shown can be used for all values of x smaller than 12 in., so the expressions obtained for V and M are valid throughout the region 0 , x , 12 in.
From C to D. Consider the portion of the beam to the left of point 2. Again replacing the distributed load by its resultant, you have 1 x oFy 5 0: 515 2 480 2 V 5 0 V 5 35 lb 1 l oM2 5 0: 2515x 1 480(x 2 6) 1 M 5 0 M 5 (2880 1 35x) lb?in.
These expressions are valid in the region 12 in. , x , 18 in.
From D to B. Use the portion of the beam to the left of point 3 for the region 18 in. , x , 32 in. Thus, 1 x oFy 5 0: 515 2 480 2 400 2 V 5 0 V 5 2365 lb 1 l oM3 5 0: 2515x 1 480(x 2 6) 2 1600 1 400(x 2 18) 1 M 5 0 M 5 (11,680 2 365x) lb?in.
Shear and Bending-Moment Diagrams. Plot the shear and bend-ing-moment diagrams for the entire beam. Note that the couple of moment 1600 lb?in. applied at point D introduces a discontinuity into the bending-moment diagram. Also note that the bending-moment diagram under the distributed load is not straight but is slightly curved.
REFLECT and THINK: Shear and bending-moment diagrams typically feature various kinds of curves and discontinuities. In such cases, it is often useful to express V and M as functions of location x as well as to determine certain numerical values.
385 385 SOLVING PROBLEMS ON YOUR OWN I n this section, you saw how to determine the shear V and the bending moment M at any point in a beam. You also learned to draw the shear diagram and the bending-moment diagram for the beam by plotting, respectively, V and M against the distance x measured along the beam.
A. Determining the shear and bending moment in a beam. To determine the shear V and the bending moment M at a given point C of a beam, take the following steps.
1. Draw a free-body diagram of the entire beam, and use it to determine the reac-tions at the beam supports.
2. Cut the beam at point C, and using the original loading, select one of the two resulting portions of the beam.
3. Draw the free-body diagram of the portion of the beam you have selected. Show: a. The loads and the reactions exerted on that portion of the beam, replacing each distributed load by an equivalent concentrated load, as explained in Sec. 5.3A. b. The shearing force and the bending moment representing the internal forces at C. To facilitate recording the shear V and the bending moment M after determining them, follow the convention indicated in Figs. 7.8 and 7.9. Thus, if you are using the portion of the beam located to the left of C, apply at C a shearing force V directed downward and a bending moment M directed counterclockwise. If you are using the portion of the beam located to the right of C, apply at C a shearing force V9 directed upward and a bending moment M9 directed clockwise [Sample Prob. 7.2].
4. Write the equilibrium equations for the portion of the beam you have selected. Solve the equation oFy 5 0 for V and the equation oMC 5 0 for M.
5. Record the values of V and M with the sign obtained for each of them. A positive sign for V means that the shearing forces exerted at C on each of the two portions of the beam are directed as shown in Figs. 7.8 and 7.9; a negative sign means they have the opposite sense. Similarly, a positive sign for M means that the bending couples at C are directed as shown in these figures, and a negative sign means that they have the opposite sense. In addition, a positive sign for M means that the concav-ity of the beam at C is directed upward, and a negative sign means that it is directed downward.
386 B. Drawing the shear and bending-moment diagrams for a beam. Obtain these diagrams by plotting, respectively, V and M against the distance x measured along the beam. However, in most cases, you need to compute the values of V and M at only a few points.
1. For a beam supporting only concentrated loads, note [Sample Prob. 7.2] that a. The shear diagram consists of segments of horizontal lines. Thus, to draw the shear diagram of the beam, you need to compute V only just to the left or just to the right of the points where the loads or reactions are applied. b. The bending-moment diagram consists of segments of oblique straight lines. Thus, to draw the bending-moment diagram of the beam, you need to compute M only at the points where the loads or reactions are applied.
2. For a beam supporting uniformly distributed loads, note [Sample Prob. 7.3] that under each of the distributed loads: a. The shear diagram consists of a segment of an oblique straight line. Thus, you need to compute V only where the distributed load begins and where it ends. b. The bending-moment diagram consists of an arc of parabola. In most cases, you need to compute M only where the distributed load begins and where it ends.
3. For a beam with a more complicated loading, you need to consider the free-body diagram of a portion of the beam of arbitrary length x and determine V and M as functions of x. This procedure may have to be repeated several times, since V and M are often represented by different functions in various parts of the beam [Sample Prob. 7.3].
4. When a couple is applied to a beam, the shear has the same value on both sides of the point of application of the couple, but the bending-moment diagram shows a discontinuity at that point, rising or falling by an amount equal to the magnitude of the couple. Note that a couple can either be applied directly to the beam or result from the application of a load on a member rigidly attached to the beam [Sample Prob. 7.3].
387 7.29 through 7.32 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
7.33 and 7.34 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment. 7.35 and 7.36 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
7.37 and 7.38 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
Problems w A B C D L 4 L 2 L 4 Fig. P7.29 w0 L A B Fig. P7.30 A B C P L 3 2L 3 Fig. P7.31 P P A B C a a Fig. P7.32 M0 A B C L 2 L 2 Fig. P7.33 M0 = PL P A B L Fig. P7.34 15 kN 25 kN 20 kN 30 kN A D B E C 1 m 0.8 m 0.4 m 0.3 m Fig. P7.35 0.6 m 0.9 m 0.2 m A B C D E 40 kN 32 kN 16 kN 1.5 m Fig. P7.36 6 kips 12 kips 4.5 kips 2 ft 2 ft 2 ft 2 ft A B C D E Fig. P7.37 120 lb 120 lb 300 lb 10 in.
20 in.
15 in.
A B C D E 25 in.
Fig. P7.38 388 7.39 through 7.42 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
60 kN 25 kN/m 1 m A B C D 2 m 2 m Fig. P7.39 50 kN 20 kN/m A B C D 2 m 2 m 2 m Fig. P7.40 8 kips 8 kips 2 ft 2 ft C D 5 ft 4 kips/ft A B Fig. P7.41 2.5 kips/ft 12 kips A B C 6 ft 4 ft Fig. P7.42 A B C E D w w P a a a a Fig. P7.43 7.43 Assuming the upward reaction of the ground on beam AB to be uniformly distributed and knowing that P 5 wa, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
7.44 Solve Prob. 7.43 knowing that P 5 3wa. 7.45 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment dia-grams, (b) determine the maximum absolute values of the shear and bending moment. 7.46 Solve Prob. 7.45 assuming that the 12-kip load has been removed. 7.47 and 7.48 Assuming the upward reaction of the ground on beam AB to be uniformly distributed, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
A B C 2 ft 2 ft 12 kips 4 ft 4 ft 6 kips 6 kips Fig. P7.45 A B C D 3 m 1.5 m 1.5 m 8 kN/m Fig. P7.47 A B C D 3 m 1.5 m 1.5 m 8 kN/m 8 kN/m Fig. P7.48 120 N 120 N 200 mm 200 mm A B C Fig. P7.49 A B C D E 300 mm 300 mm 300 mm 300 mm 300 mm 400 N 400 N 400 N 150 mm 150 mm Fig. P7.50 7.49 and 7.50 Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.
389 389 7.51 and 7.52 Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.
6 in.
6 in.
4 in.
10 in.
10 in.
8 in.
10 in.
50 lb 50 lb 100 lb A E F G H D C B Fig. P7.51 A B C G D E F 4 in.
5 in.
9 in.
6 in.
45 lb 120 lb 6 in.
Fig. P7.52 7.53 Two small channel sections DF and EH have been welded to the uniform beam AB of weight W 5 3 kN to form the rigid structural member shown. This member is being lifted by two cables attached at D and E. Knowing that θ 5 30° and neglecting the weight of the channel sections, (a) draw the shear and bending-moment diagrams for beam AB, (b) determine the maximum absolute values of the shear and bending moment in the beam. 7.54 Solve Prob. 7.53 when θ 5 60°. 7.55 For the structural member of Prob. 7.53, determine (a) the angle θ for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M|max. (Hint: Draw the bending-moment diagram and then equate the absolute values of the largest positive and negative bending moments obtained.) 7.56 For the beam of Prob. 7.43, determine (a) the ratio k 5 P/wa for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of |M|max. (See hint for Prob. 7.55.) 7.57 Determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M|max. (See hint for Prob. 7.55.) A D F G H C q q 0.5 m B E 1.5 m 1.5 m 1 m 1 m Fig. P7.53 80 N a a 100 40 50 100 A B C Dimensions in mm E D F Fig. P7.57 390 7.58 For the beam and loading shown, determine (a) the distance a for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of |M|max. (See hint for Prob. 7.55.) E A B C D 3 kN 3 kN 2 kN 0.6 m a 1 m 0.8 m Fig. P7.58 B A L a a Fig. P7.59 P Q A B C D 30 in.
30 in.
a Fig. P7.60 w A B L W Fig. P7.62 7.59 A uniform beam is to be picked up by crane cables attached at A and B. Determine the distance a from the ends of the beam to the points where the cables should be attached if the maximum absolute value of the bending moment in the beam is to be as small as pos-sible. (Hint: Draw the bending-moment diagram in terms of a, L, and the weight per unit length w, and then equate the absolute values of the largest positive and negative bending moments obtained.) 7.60 Knowing that P 5 Q 5 150 lb, determine (a) the distance a for which the maximum absolute value of the bending moment in beam AB is as small as possible, (b) the corresponding value of |M|max. (See hint for Prob. 7.55.) 7.61 Solve Prob. 7.60 assuming that P 5 300 lb and Q 5 150 lb. 7.62 In order to reduce the bending moment in the cantilever beam AB, a cable and counterweight are permanently attached at end B. Deter-mine the magnitude of the counterweight for which the maximum absolute value of the bending moment in the beam is as small as possible and the corresponding value of |M|max. Consider (a) the case when the distributed load is permanently applied to the beam, (b) the more general case when the distributed load may either be applied or removed.
7.3 Relations Among Load, Shear, and Bending Moment 391 7.3 RELATIONS AMONG LOAD, SHEAR, AND BENDING MOMENT If a beam carries more than two or three concentrated loads or if it carries a distributed load, the method outlined in Sec. 7.2 for plotting shear and bending-moment diagrams is likely to be quite cumbersome. However, constructing a shear diagram and, especially, a bending-moment diagram, are much easier if we take into consideration some relations among load, shear, and bending moment.
Consider a simply supported beam AB carrying a distributed load w per unit length (Fig. 7.11a). Let C and C9 be two points of the beam at a distance Dx from each other. We denote the shear and bending moment at C by V and M, respectively, and we assume they are positive. We denote the shear and bending moment at C9 by V 1 DV and M 1 DM.
Let us now detach the portion of beam CC9 and draw its free-body diagram (Fig. 7.11b). The forces exerted on the free body include a load with a magnitude of w Dx (indicated by a dashed arrow to distinguish it from the original distributed load from which it is derived) and internal forces and couples at C and C9. Since we assumed both shear and bending moment are positive, the forces and couples are directed as shown in the figure.
Relations Between Load and Shear. Because the free body CC9 is in equilibrium, we set the sum of the vertical components of the forces acting on it to zero: V 2 (V 1 DV) 2 w D x 5 0 DV 5 2w D x Dividing both sides of this equation by D x and then letting D x approach zero, we obtain dV dx 5 2w (7.1) Equation (7.1) indicates that, for a beam loaded as shown in Fig. 7.11a, the slope d V/dx of the shear curve is negative and the numerical value of the slope at any point is equal to the load per unit length at that point.
Integrating (7.1) between arbitrary points C and D, we have VD 2 VC 5 2# xD xC w dx (7.2) or VD 2 VC 5 2(area under load curve between C and D) (7.29) Note that we could also obtain this result by considering the equilibrium of the portion of beam CD, since the area under the load curve represents the total load applied between C and D.
Equation (7.1) is not valid at a point where a concentrated load is applied; the shear curve is discontinuous at such a point, as we saw in dV dx d 5 2w VD V 2 VC V 5 2(area under load curve between C and C D) (b) w A B C C' C C' D w∆x (a) M V M + ∆M V + ∆V ∆x 2 ∆x x ∆x w Fig. 7.11 (a) A simply supported beam carrying a distributed load; (b) free-body diagram of a portion CC 9 of the beam.
392 Internal Forces and Moments Sec. 7.2. Similarly, formulas (7.2) and (7.29) cease to be valid when con-centrated loads are applied between C and D, since they do not take into account the sudden change in shear caused by a concentrated load. For-mulas (7.2) and (7.29), therefore, should be applied only between succes-sive concentrated loads.
Relations Between Shear and Bending Moment. Returning to the free-body diagram of Fig. 7.11b, we can set the sum of the moments about C9 to be zero, obtaining (M 1 DM) 2 M 2 V Dx 1 wDx Dx 2 5 0 DM 5 V Dx 2 1 2 w(Dx)2 Dividing both sides of this equation by Dx and then letting Dx approach zero, we have dM dx 5 V (7.3) Equation (7.3) indicates that the slope dM/dx of the bending-moment curve is equal to the value of the shear. This is true at any point where the shear has a well-defined value, i.e., at any point where no concentrated load is applied. Formula (7.3) also shows that the shear is zero at points where the bending moment is maximum. This property simplifies the determina-tion of points where the beam is likely to fail under bending.
Integrating Eq. (7.3) between arbitrary points C and D, we obtain MD 2 MC 5 # xD xC V dx (7.4) MD 2 MC 5 area under shear curve between C and D (7.49) Note that the area under the shear curve should be considered positive where the shear is positive and negative where the shear is negative. For-mulas (7.4) and (7.49) are valid even when concentrated loads are applied between C and D, as long as the shear curve has been drawn correctly. The formulas cease to be valid, however, if a couple is applied at a point between C and D, since they do not take into account the sudden change in bending moment caused by a couple (see Sample Prob. 7.7).
In most engineering applications, you need to know the value of the bending moment at only a few specific points. Once you have drawn the shear diagram and determined M at one end of the beam, you can obtain the value of the bending moment at any given point by computing the area under the shear curve and using formula (7.49). For instance, since MA 5 0 for the beam of Fig. 7.12, you can determine the maximum value of the bending moment for that beam simply by measuring the area of the shaded triangle in the shear diagram as Mmax 5 1 2 L 2 wL 2 5 wL2 8 In this example, the load curve is a horizontal straight line, the shear curve is an oblique straight line, and the bending-moment curve dM d dx d 5 V MD M 2 MC 5 area under shear curve between C and C D (b) w A B C C' C C' D w∆x (a) M V M + ∆M V + ∆V ∆x 2 ∆x x ∆x w Fig. 7.11 (repeated) 7.3 Relations Among Load, Shear, and Bending Moment 393 is a parabola. If the load curve had been an oblique straight line (first degree), the shear curve would have been a parabola (second degree), and the bending-moment curve would have been a cubic (third degree). The equations of the shear and bending-moment curves are always, respec-tively, one and two degrees higher than the equation of the load curve. Thus, once you have computed a few values of the shear and bending moment, you should be able to sketch the shear and bending-moment diagrams without actually determining the functions V(x) and M(x). The sketches will be more accurate if you make use of the fact that, at any point where the curves are continuous, the slope of the shear curve is equal to 2w and the slope of the bending-moment curve is equal to V.
Concept Application 7.1 Consider a simply supported beam AB with a span of L carrying a uni-formly distributed load w (Fig. 7.12a). From the free-body diagram of the entire beam, we determine the magnitude of the reactions at the supports: RA 5 RB 5 wL/2 (Fig. 7.12b). Then we draw the shear diagram. Close to end A of the beam, the shear is equal to RA; that is, to wL/2, as we can check by considering a very small portion of the beam as a free body. Using formula (7.2), we can then determine the shear V at any distance x from A as V 2 VA 5 2# x 0 w dx 5 2wx V 5 VA 2 wx 5 wL 2 2 wx 5 w aL 2 2 xb The shear curve is thus an oblique straight line that crosses the x axis at x 5 L/2 (Fig. 7.12c). Now consider the bending moment. We first observe that MA 5 0. The value M of the bending moment at any distance x from A then can be obtained from Eq. (7.4), as M 2 MA 5 # x 0 V dx M 5 # x 0 w aL 2 2 xb dx 5 w 2 (Lx 2 x2) The bending-moment curve is a parabola. The maximum value of the bending moment occurs when x 5 L/2, since V (and thus dM/dx) is zero for that value of x. Substituting x 5 L/2 in the last equation, we obtain Mmax 5 wL2/8.
(b) (c) (d) A A B B L (a) wL 2 RA = wL 2 RB = wL 2 wL2 8 L 2 L 2 wL 2 – V L L M x x w w Fig. 7.12 (a) A simply supported beam carrying a uniformly distributed load; (b) free-body diagram of the beam to determine the reactions at the supports; (c) the shear curve is an oblique straight line; (d) the bending-moment diagram is a parabola.
394 Internal Forces and Moments Sample Problem 7.4 Draw the shear and bending-moment diagrams for the beam and loading shown.
STRATEGY: The beam supports two concentrated loads and one dis-tributed load. You can use the equations in this section between these loads and under the distributed load, but you should expect certain changes in the diagrams at the load points.
MODELING and ANALYSIS: Free-Body, Entire Beam. Consider the entire beam as a free body and determine the reactions (Fig. 1): 1l oMA 5 0: D(24 ft) 2 (20 kips)(6 ft) 2 (12 kips)(14 ft) 2 (12 kips)(28 ft) 5 0 D 5 126 kips D 5 26 kipsx 1xoFy 5 0: Ay 2 20 kips 2 12 kips 1 26 kips 2 12 kips 5 0 Ay 5 118 kips Ay 5 18 kipsx y 1 oFx 5 0: Ax 5 0 Ax 5 0 Note that the bending moment is zero at both A and E; thus, you know two points (indicated by small circles) on the bending-moment diagram.
Shear Diagram. Since dV/dx 5 2w, the slope of the shear diagram is zero (i.e., the shear is constant between concentrated loads and reactions). To find the shear at any point, divide the beam into two parts and consider either part as a free body. For example, using the portion of the beam to the left of point 1 (Fig. 1), you can obtain the shear between B and C: 1 x oFy 5 0: 118 kips 2 20 kips 2 V 5 0 V 5 22 kips You can also find that the shear is 112 kips just to the right of D and zero at end E. Since the slope dV/dx 5 2w is constant between D and E, the shear diagram between these two points is a straight line.
Bending-Moment Diagram. Recall that the area under the shear curve between two points is equal to the change in bending moment between the same two points. For convenience, compute the area of each portion of the shear diagram and indicate it on the diagram (Fig. 1). Since you know the bending moment MA at the left end is zero, you have MB 2 MA 5 1108 MB 5 1108 kip?ft MC 2 MB 5 216 MC 5 192 kip?ft MD 2 MC 5 2140 MD 5 248 kip?ft ME 2 MD 5 148 ME 5 0 Since you know ME is zero, this gives you a check of the calculations. Between the concentrated loads and reactions, the shear is constant; thus, the slope dM/dx is constant. Therefore, you can draw the bending-moment diagram by connecting the known points with straight lines. A B C D E 20 kips 12 kips 1.5 kips/ft 6 ft 8 ft 8 ft 10 ft Ax Ay 12 kips 1.5 kips/ft 4 ft M V V(kips) +18 M(kip·ft) (+108) (–16) +12 (+48) –14 2 +108 +92 48 D 6 ft 8 ft 8 ft 10 ft A B C D E 20 kips 12 kips B 1 C D E 20 kips 18 kips 18 kips 20 kips 26 kips 12 kips A x x (140) Fig. 1 Free-body diagrams of beam, free-body diagram of section to left of cut, shear diagram, bending-moment diagram.
7.3 Relations Among Load, Shear, and Bending Moment 395 Between D and E, where the shear diagram is an oblique straight line, the bending-moment diagram is a parabola. From the V and M diagrams, note that Vmax 5 18 kips and Mmax 5 108 kip?ft.
REFLECT and THINK: As expected, the values of shear and slopes of the bending-moment curves show abrupt changes at the points where con-centrated loads act. Useful for design, these diagrams make it easier to determine the maximum values of shear and bending moment for a beam and its loading.
Sample Problem 7.5 Draw the shear and bending-moment diagrams for the beam and loading shown and determine the location and magnitude of the maximum bending moment.
STRATEGY: The load is a distributed load over part of the beam with no concentrated loads. You can use the equations in this section in two parts: for the load and no load regions. From the discussion in this section, you can expect the shear diagram will show an oblique line under the load, followed by a horizontal line. The bending-moment diagram should show a parabola under the load and an oblique line under the rest of the beam.
MODELING and ANALYSIS: Free-Body, Entire Beam. Consider the entire beam as a free body (Fig. 1) to obtain the reactions RA 5 80 kN x RC 5 40 kN x Shear Diagram. The shear just to the right of A is VA 5 180 kN. Because the change in shear between two points is equal to minus the area under the load curve between these points, you can obtain VB by writing VB 2 VA 5 2(20 kN/m)(6 m) 5 2120 kN VB 5 2120 1 VA 5 2120 1 80 5 240 kN Since the slope dV/dx 5 2w is constant between A and B, the shear diagram between these two points is represented by a straight line. Between B and C, the area under the load curve is zero; therefore, VC 2 VB 5 0 VC 5 VB 5 240 kN and the shear is constant between B and C (Fig. 1).
Bending-Moment Diagram. The bending moment at each end of the beam is zero. In order to determine the maximum bending moment, you need to locate the section D of the beam where V 5 0. You have VD 2 VA 5 2wx 0 2 80 kN 5 2(20 kN/m)x A B C 20 kN/m 6 m 3 m (40) A A A B B D C C 20 kN/m 6 m w V M x x x 80 kN 80 kN 40 kN 40 kN (+160) x = 4 m (120) 160 kN·m 120 kN·m Fig. 1 Free-body diagram of beam, shear diagram, bending-moment diagram.
396 Internal Forces and Moments Solving for x: x 5 4 m b The maximum bending moment occurs at point D, where we have dM/dx 5 V 5 0. Calculate the areas of the various portions of the shear diagram and mark them (in parentheses) on the diagram (Fig. 1). Since the area of the shear diagram between two points is equal to the change in bending moment between those points, you can write MD 2 MA 5 1160 kN?m MD 5 1160 kN?m MB 2 MD 5 240 kN?m MB 5 1120 kN?m MC 2 MB 5 2120 kN?m MC 5 0 The bending-moment diagram consists of an arc of parabola followed by a segment of straight line; the slope of the parabola at A is equal to the value of V at that point. The maximum bending moment is Mmax 5 MD 5 1160 kN?m b REFLECT and THINK: The analysis conforms to our initial expecta-tions. It is often useful to predict what the results of analysis will be as a way of checking against large-scale errors. However, final results can only depend on detailed modeling and analysis.
Sample Problem 7.6 Sketch the shear and bending-moment diagrams for the cantilever beam shown.
STRATEGY: Because no support reactions appear until the right end of the beam, you can rely on the equations from this section without needing to use free-body diagrams and equilibrium equations. Due to the non-uniform load, you should expect the results to involve equations of higher degree with a parabolic curve in the shear diagram and a cubic curve in the bending-moment diagram.
MODELING and ANALYSIS: Shear Diagram. At the free end of the beam, VA 5 0. Between A and B, the area under the load curve is 1 2 w 0 a; we find VB by writing VB 2 VA 5 21 2 w0 a VB 5 2 1 2 w0 a Between B and C, the beam is not loaded; thus, VC 5 VB. At A, we have w 5 w0, and according to Eq. (7.1), the slope of the shear curve is dV/dx 5 2w0. At B, the slope is dV/dx 5 0. Between A and B, the loading decreases linearly, and the shear diagram is parabolic (Fig. 1). Between B and C, w 5 0 and the shear diagram is a horizontal line.
A B C x M V w0 1 2 w0a 1 2 w0a 1 2 [ w0a(L a)] 1 3 [ w0a2] 1 6 w0a(3L a) 1 3 w0a2 a L x Fig. 1 Beam with load, shear diagram, bending-moment diagram.
7.3 Relations Among Load, Shear, and Bending Moment 397 Bending-Moment Diagram. Note that MA 5 0 at the free end of the beam. You can compute the area under the shear curve, obtaining MB 2 MA 5 21 3 w0 a2 MB 5 21 3 w0 a2 MC 2 MB 5 21 2 w0 a(L 2 a) MC 5 21 6 w0a(3L 2 a) You can complete the sketch of the bending-moment diagram by recalling that dM/dx 5 V. The result is that between A and B the diagram is rep-resented by a cubic curve with zero slope at A and between B and C the diagram is represented by a straight line.
REFLECT and THINK: Although not strictly required for the solution of this problem, determining the support reactions would serve as an excellent check of the final values of the shear and bending-moment diagrams.
Sample Problem 7.7 The simple beam AC is loaded by a couple of magnitude T applied at point B. Draw the shear and bending-moment diagrams for the beam.
STRATEGY: The load supported by the beam is a concentrated couple. Since the only vertical forces are those associated with the support reac-tions, you should expect the shear diagram to be of constant value. How-ever, the bending-moment diagram will have a discontinuity at B due to the couple.
MODELING and ANALYSIS: Free-Body, Entire Beam. Consider the entire beam as a free body and determine the reactions: RA 5 T L↑ RB 5 T Lw Shear and Bending-Moment Diagrams (Fig. 1). The shear at any section is constant and equal to T/L. Since a couple is applied at B, the bending-moment diagram is discontinuous at B; because the couple is counterclockwise, the bending moment decreases suddenly by an amount equal to T. You can demonstrate this by taking a section to the immediate right of B and applying equilibrium to solve for the bending moment at this location.
REFLECT and THINK: You can generalize the effect of a couple applied to a beam. At the point where the couple is applied, the bending-moment diagram increases by the value of the couple if it is clockwise and decreases by the value of the couple if it is counterclockwise.
x V M a T L A C B T L x a L T –T(1 – ) a L Fig. 1 Beam with load, shear diagram, bending-moment diagram.
398 398 I n this section, we described how to use the relations among load, shear, and bending moment to simplify the drawing of shear and bending-moment diagrams. These relations are dV dx 5 2w dM dx 5 V (7.1) (7.3) VD 2 VC 5 2(area under load curve between C and D) (7.29) MD 2 MC 5 (area under shear curve between C and D) (7.49) Taking these relations into account, you can use the following procedure to draw the shear and bending-moment diagrams for a beam.
1. Draw a free-body diagram of the entire beam, and use it to determine the reac-tions at the beam supports.
2. Draw the shear diagram. This can be done as in the preceding section by cutting the beam at various points and considering the free-body diagram of one of the two resulting portions of the beam [Sample Prob. 7.3]. You can, however, consider one of the following alternative procedures. a. The shear V at any point of the beam is the sum of the reactions and loads to the left of that point; an upward force is counted as positive, and a downward force is counted as negative. b. For a beam carrying a distributed load, you can start from a point where you know V and use Eq. (7.29) repeatedly to find V at all other points of interest.
3. Draw the bending-moment diagram, using the following procedure. a. Compute the area under each portion of the shear curve, assigning a posi-tive sign to areas above the x axis and a negative sign to areas below the x axis. b. Apply Eq. (7.49) repeatedly [Sample Probs. 7.4 and 7.5], starting from the left end of the beam, where M 5 0 (except if a couple is applied at that end, or if the beam is a cantilever beam with a fixed left end). c. Where a couple is applied to the beam, be careful to show a discontinuity in the bending-moment diagram by increasing the value of M at that point by an amount equal to the magnitude of the couple if the couple is clockwise, or decreasing the value of M by that amount if the couple is counterclockwise [Sample Prob. 7.7].
SOLVING PROBLEMS ON YOUR OWN 399 399 4. Determine the location and magnitude of |M|max. The maximum absolute value of the bending moment occurs at one of the points where dM/dx 5 0 [according to Eq. (7.3), that is at a point where V is equal to zero or changes sign]. You should a. Determine from the shear diagram the value of |M| where V changes sign; this will occur under a concentrated load [Sample Prob. 7.4]. b. Determine the points where V 5 0 and the corresponding values of |M|; this will occur under a distributed load. To find the distance x between point C where the distributed load starts and point D where the shear is zero, use Eq. (7.29). For VC, use the known value of the shear at point C; for VD, use zero and express the area under the load curve as a function of x [Sample Prob. 7.5].
5. You can improve the quality of your drawings by keeping in mind that, at any given point according to Eqs. (7.1) and (7.3), the slope of the V curve is equal to 2w and the slope of the M curve is equal to V.
6. Finally, for beams supporting a distributed load expressed as a function w(x), remember that you can obtain the shear V by integrating the function 2w(x), and you can obtain the bending moment M by integrating V(x) [Eqs. (7.2) and (7.4)].
400 Problems 7.63 Using the method of Sec. 7.3, solve Prob. 7.29. 7.64 Using the method of Sec. 7.3, solve Prob. 7.30. 7.65 Using the method of Sec. 7.3, solve Prob. 7.31. 7.66 Using the method of Sec. 7.3, solve Prob. 7.32.
7.67 Using the method of Sec. 7.3, solve Prob. 7.33.
7.68 Using the method of Sec. 7.3, solve Prob. 7.34. 7.69 and 7.70 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment. 7.71 Using the method of Sec. 7.3, solve Prob. 7.39. 7.72 Using the method of Sec. 7.3, solve Prob. 7.40. 7.73 Using the method of Sec. 7.3, solve Prob. 7.41. 7.74 Using the method of Sec. 7.3, solve Prob. 7.42.
7.75 and 7.76 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
8 kN 10 kN 3 m 3 m 3 m 3 m 24 kN⋅m A B C 8 kN D E Fig. P7.69 9 kN 18 kN 1 m 1.5 m 2 m 3 kN⋅m 12 kN⋅m A B C D Fig. P7.70 A E 3 ft 3 ft 5 ft 4 ft 16 kips B C D 45 kips 8 kips Fig. P7.75 A B 10 in.
6 in.
6 in.
6 in.
6 in.
10 in.
100 lb 16 lb/in.
D C E F G 100 lb 150 lb 16 lb/in.
Fig. P7.76 A B C 15 kN/m 1.5 m 6 m 45 kN⋅m Fig. P7.77 2.5 kN/m A C B 2.5 m 1 m Fig. P7.78 7.77 and 7.78 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.
401 7.79 and 7.80 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.
20 kN/m A B C D 1.25 m 0.5 m 2 m Fig. P7.79 3.2 m 0.8 m 2 kN/m 4 kN A B C Fig. P7.80 A C 9 ft 6 ft 800 lb/ft B 600 lb Fig. P7.81 A C 2.5 ft 20 ft 400 lb/ft B 3200 lb Fig. P7.82 A B C D 300 lb/ft 300 lb 2 ft 2 ft 4 ft Fig. P7.83 7.81 and 7.82 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.
7.83 (a) Draw the shear and bending-moment diagrams for beam AB, (b) determine the magnitude and location of the maximum absolute value of the bending moment.
A B x w w = w0 cos p x 2L L Fig. P7.85 A B x w L w0 Fig. P7.86 7.84 Solve Prob. 7.83 assuming that the 300-lb force applied at D is directed upward.
7.85 and 7.86 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magni-tude and location of the maximum bending moment.
402 7.87 and 7.88 For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magni-tude and location of the maximum bending moment.
w0 1 2 B L x w0 w A Fig. P7.87 A B w L x w = w0 sin px L Fig. P7.88 0.3 m 0.3 m 0.3 m 0.3 m A B C D E 20 kN/m P Q Fig. P7.89 7.89 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is 1800 N∙m at D and 11300 N∙m at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam.
2 ft 2 ft 4 ft 2 ft 2 ft A B C D E F P Q 250 lb/ft Fig. P7.91 7.90 Solve Prob. 7.89 assuming that the bending moment was found to be 1650 N∙m at D and 11450 N∙m at E. 7.91 The beam AB is subjected to the uniformly distributed load shown and to two unknown forces P and Q. Knowing that it has been experimentally determined that the bending moment is 16.10 kip∙ft at D and 15.50 kip∙ft at E, (a) determine P and Q, (b) draw the shear and bending-moment diagrams for the beam. 7.92 Solve Prob. 7.91 assuming that the bending moment was found to be 15.96 kip∙ft at D and 16.84 kip∙ft at E.
7.4 Cables 403 7.4 CABLES Cables are used in many engineering applications, such as suspension bridges, transmission lines, aerial tramways, guy wires for high towers, etc. Cables may be divided into two categories, according to their loading: (1) supporting concentrated loads and (2) supporting distributed loads. 7.4A Cables with Concentrated Loads Consider a cable attached to two fixed points A and B and supporting n vertical concentrated loads P1, P2, . . . , Pn (Fig. 7.13a). We assume that the cable is flexible, i.e., that its resistance to bending is small and can be neglected. We further assume that the weight of the cable is negligible compared with the loads supported by the cable. We can therefore approxi-mate any portion of cable between successive loads as a two-force mem-ber. Thus, the internal forces at any point in the cable reduce to a force of tension directed along the cable.
We assume that each of the loads lies in a given vertical line, i.e., that the horizontal distance from support A to each of the loads is known. We also assume that we know the horizontal and vertical distances between the supports. With these assumptions, we want to determine the shape of the cable (i.e., the vertical distance from support A to each of the points C1, C2, . . . , Cn ) and also the tension T in each portion of the cable.
We first draw the free-body diagram of the entire cable (Fig. 7.13b). Since we do not know the slopes of the portions of cable attached at A and B, we represent the reactions at A and B by two components each. Thus, four unknowns are involved, and the three equations of equilibrium are not sufficient to determine the reactions. (Clearly, a cable is not a rigid body; thus, the equilibrium equations represent necessary but not sufficient conditions. See Sec. 6.3B.) We must therefore obtain an additional equa-tion by considering the equilibrium of a portion of the cable. This is possible if we know the coordinates x and y of a point D of the cable. We draw the free-body diagram of the portion of cable AD (Fig. 7.14a). From the equilibrium condition oMD 5 0, we obtain an addi-tional relation between the scalar components Ax and Ay and can determine Photo 7.3 The weight of the chairlift cables is negligible compared to the weights of the chairs and skiers, so we can use the methods of this section to determine the force at any point in the cable.
(b) D Ay Ax By Bx A B C1 C2 C3 P1 P2 P3 L d x1 x2 x3 (a) A B C1 C2 C3 P1 P2 P3 y3 y2 y1 L d x1 x2 x3 Fig. 7.13 (a) A cable supporting vertical concentrated loads; (b) free-body diagram of the entire cable.
x1 x (a) D Ay Ax C1 P1 T A y P2 q x1 x2 (b) Ay Ax C1 C2 P1 T A y2 Fig. 7.14 (a) Free-body diagram of the portion of cable AD; (b) free-body diagram of the portion of cable AC2.
404 Internal Forces and Moments the reactions at A and B. However, the problem remains indeterminate if we do not know the coordinates of D unless we are given some other rela-tion between Ax and Ay (or between Bx and By). The cable might hang in any of various possible ways, as indicated by the dashed lines in Fig. 7.13b.
Once we have determined Ax and Ay, we can find the vertical distance from A to any point of the cable. Considering point C2, for example, we draw the free-body diagram of the portion of cable AC2 (Fig. 7.14b). From oFC2 5 0, we obtain an equation that we can solve for y2. From oFx 5 0 and oFy 5 0, we obtain the components of force T representing the tension in the portion of cable to the right of C2. Note that T cos θ 5 2Ax; that is, the horizontal component of the tension force is the same at any point of the cable. It follows that the tension T is maximum when cos θ is minimum, i.e., in the portion of cable that has the largest angle of inclination θ. Clearly, this portion of cable must be adjacent to one of the two supports of the cable.
7.4B Cables with Distributed Loads Consider a cable attached to two fixed points A and B and carrying a dis-tributed load (Fig. 7.15a). We just saw that for a cable supporting concen-trated loads, the internal force at any point is a force of tension directed along the cable. By contrast, in the case of a cable carrying a distributed load, the cable hangs in the shape of a curve, and the internal force at a point D is a force of tension T directed along the tangent to the curve. Here we examine how to determine the tension at any point of a cable supporting a given distributed load. In the following sections, we will deter-mine the shape of the cable for two common types of distributed loads.
Considering the most general case of distributed load, we draw the free-body diagram of the portion of cable extending from the lowest point C to a given point D of the cable (Fig. 7.15b). The three forces acting on the free body are the tension force T0 at C, which is horizontal; the ten-sion force T at D, which is directed along the tangent to the cable at D; and the resultant W of the distributed load supported by the portion of cable CD. Drawing the corresponding force triangle (Fig. 7.15c), we obtain the relations T cos θ 5 T0 T sin θ 5 W (7.5) T 5 2T 2 0 1 W 2 tan θ 5 W T0 (7.6) T cos θ 5 T0 T T sin θ 5 W T 5 2T 2 0 1 W 2 2 tan θ 5 W T0 T A B C (b) (c) (a) D T T W T0 T0 q q W C D Fig. 7.15 (a) A cable carrying a distributed load; (b) free-body diagram of the portion of the cable CD; (c) force triangle for the free-body diagram in part (b).
x1 x (a) D Ay Ax C1 P1 T P2 A q y x1 x2 (b) Ay Ax C1 C2 P1 T A y2 Fig. 7.14 (repeated) 7.4 Cables 405 From the relations in Eqs. (7.5), we see that the horizontal component of the tension force T is the same at any point. Furthermore, the vertical component of T at any point is equal to the magnitude W of the load when measured from the lowest point (C) to the point in question (D). Relations in Eq. (7.6) show that the tension T is minimum at the lowest point and maximum at one of the two support points.
7.4C Parabolic Cables Now suppose that cable AB carries a load uniformly distributed along the horizontal (Fig. 7.16a). We can approximate the load on the cables of a suspension bridge in this way, since the weight of the cables is small compared with the uniform weight of the roadway. We denote the load per unit length by w (measured horizontally) and express it in N/m or lb/ft. Choosing coordinate axes with the origin at the lowest point C of the cable, we find that the magnitude W of the total load carried by the portion of cable extending from C to the point D with coordinates x and y is W 5 wx. The relations in Eqs. (7.6) defining the magnitude and direc-tion of the tension force at D become T 5 2T 2 0 1 w2x2 tan θ 5 wx T0 (7.7) Moreover, the distance from D to the line of action of the resultant W is equal to half of the horizontal distance from C to D (Fig. 7.16b). Summing moments about D, we have 1l oMD 5 0: wx x 2 2 T0 y 5 0 Solving for y, we have Equation of parabolic cable y 5 wx2 2T0 (7.8) This is the equation of a parabola with a vertical axis and its vertex at the origin of coordinates. Thus, the curve formed by cables loaded uni-formly along the horizontal is a parabola.‡ When the supports A and B of the cable have the same elevation, the distance L between the supports is called the span of the cable and the vertical distance h from the supports to the lowest point is called the sag of the cable (Fig. 7.17a). If you know the span and sag of a cable and if the load w per unit horizontal length is given, you can find the minimum tension T0 by substituting x 5 L /2 and y 5 h in Eq. (7.8). Equations (7.7) then yield the tension and the slope at any point of the cable and Eq. (7.8) defines the shape of the cable.
‡ Cables hanging under their own weight are not loaded uniformly along the horizontal and do not form parabolas. However, the error introduced by assuming a parabolic shape for cables hanging under their own weight is small when the cable is sufficiently taut. In the next section, we give a complete discussion of cables hanging under their own weight.
y 5 wx w 2 2T0 T (b) (a) A B C y y T T0 q D(x,y) x x w W = wx x 2 D C y x 2 Fig. 7.16 (a) A cable carrying a uniformly distributed load along the horizontal; (b) free-body diagram of the portion of cable CD.
Photo 7.4 The main cables of suspension bridges, like the Golden Gate Bridge above, may be assumed to carry a loading uniformly distributed along the horizontal.
406 Internal Forces and Moments When the supports have different elevations, the position of the low-est point of the cable is not known, and we must determine the coordinates xA, yA and xB, yB of the supports. To do this, we note that the coordinates of A and B satisfy Eq. (7.8) and that xB 2 xA 5 L and yB 2 yA 5 d where L and d denote, respectively, the horizontal and vertical distances between the two supports (Fig. 7.17b and c).
We can obtain the length of the cable from its lowest point C to its support B from the formula sB 5 # xB 0 B1 1 ady dxb 2 dx (7.9) Differentiating Eq. (7.8), we obtain the derivative dy/dx 5 wx/T0. Substi-tuting this into Eq. (7.9) and using the binomial theorem to expand the radical in an infinite series, we have sB 5# xB 0 B1 1 w2x2 T 2 0 dx 5# xB 0 a1 1 w2x2 2T 2 0 2 w4x4 8T 4 0 1 . . .b dx sB 5 xB a1 1 w2x 2 B 6T 2 0 2 w4x4 B 40T 4 0 1 . . .b Then, since wx 2 B/2T0 5 yB, we obtain sB 5 xB c 1 1 2 3 ayB xB b 2 2 2 5 ayB xB b 4 1 . . .d (7.10) This series converges for values of the ratio yB /xB less than 0.5. In most cases, this ratio is much smaller, and only the first two terms of the series need be computed.
(a) A B C y x L h (b) yB yA xB xA y x L d A B C xA < 0 A B C y x L xB yB yA d (c) Fig. 7.17 (a) The shape of a parabolic cable is determined by its span L and sag h; (b, c) span and vertical distance between supports for cables with supports at different elevations.
7.4 Cables 407 Sample Problem 7.8 The cable AE supports three vertical loads from the points indicated. If point C is 5 ft below the left support, determine (a) the elevation of points B and D, (b) the maximum slope and the maximum tension in the cable.
STRATEGY: To solve for the support reactions at A, consider a free-body diagram of the entire cable as well as one that takes a section at C, since you know the coordinates of this point. Taking subsequent sections at B and D will then enable you to determine their elevations. The result-ing cable geometry establishes the maximum slope, which is where the maximum tension in the cable occurs.
MODELING and ANALYSIS: Free Body, Entire Cable. Determine the reaction components Ax and Ay as 1loME 5 0: Ax(20 ft) 2 Ay(60 ft) 1 (6 kips)(40 ft) 1 (12 kips)(30 ft) 1 (4 kips)(15 ft) 5 0 20Ax 2 60Ay 1 660 5 0 Free Body, ABC. Consider the portion ABC of the cable as a free body (Fig. 1). Then you have 1loMC 5 0: 2Ax(5 ft) 2 Ay(30 ft) 1 (6 kips)(10 ft) 5 0 25Ax 2 30Ay 1 60 5 0 Solving the two equations simultaneously, you obtain Ax 5 218 kips Ax 5 18 kips z Ay 5 15 kips Ay 5 5 kipsx a. Elevation of Points B and D: Free Body, AB. Considering the portion of cable AB as a free body, you obtain 1l oMB 5 0: (18 kips)yB 2 (5 kips)(20 ft) 5 0 yB 5 5.56 ft below A b Free Body, ABCD. Using the portion of cable ABCD as a free body gives you 1loMD 5 0: 2(18 kips)yD 2 (5 kips)(45 ft) 1 (6 kips)(25 ft) 1 (12 kips)(15 ft) 5 0 yD 5 5.83 ft above A b b. Maximum Slope and Maximum Tension. Note that the maximum slope occurs in portion DE. Since the horizontal component of the tension is constant and equal to 18 kips, you have tan θ 5 14.17 15 ft θ 5 43.4° b Tmax 5 18 kips cos θ Tmax 5 24.8 kips b D B A C E 6 kips 12 kips 4 kips 20 ft 5 ft 20 ft 15 ft 15 ft 10 ft D B A C E 6 kips 12 kips 4 kips 20 ft 5 ft 20 ft 15 ft 15 ft 10 ft yB yD Ax Ay Ex Ey D B A C E 6 kips 18 kips 5 kips 12 kips 4 kips 14.17 ft 5.83 ft 15 ft Ex =18 kips Ey B A C 6 kips 12 kips 5 ft 20 ft 10 ft Ax Ay q B A 6 kips 5 kips 20 ft 20 ft 15 ft 10 ft D B A C 6 kips 18 kips 18 kips 5 kips 12 kips 4 kips Fig. 1 Free-body diagrams of cable system.
408 Internal Forces and Moments Sample Problem 7.9 A light cable is attached to a support at A, passes over a small frictionless pulley at B, and supports a load P. The sag of the cable is 0.5 m and the mass per unit length of the cable is 0.75 kg/m. Determine (a) the magnitude of the load P, (b) the slope of the cable at B, (c) the total length of the cable from A to B. Since the ratio of the sag to the span is small, assume the cable is para-bolic. Also, neglect the weight of the portion of cable from B to D.
STRATEGY: Because the pulley is frictionless, the load P is equal in magnitude to the tension in the cable at B. You can determine the tension using the methods of this section and then use that value to determine the slope and length of the cable.
MODELING and ANALYSIS: a. Load P. Denote the lowest point of the cable by C and draw the free-body diagram of the portion CB of cable (Fig. 1). Assuming the load is uniformly distributed along the horizontal, you have w 5 (0.75 kg/m)(9.81 m/s2) 5 7.36 N/m The total load for the portion CB of cable is W 5 wxB 5 (7.36 N/m)(20 m) 5 147.2 N This load acts halfway between C and B. Summing moments about B gives you 1loMB 5 0: (147.2 N)(10 m) 2 T0(0.5 m) 5 0 T0 5 2944 N From the force triangle (Fig. 2), you obtain TB 5 2T2 0 1 W2 5 2(2944 N)2 1 (147.2 N)2 5 2948 N Since the tension on each side of the pulley is the same, you end up with P 5 TB 5 2948 N b b. Slope of Cable at B. The force triangle also tells us that tan θ 5 W T0 5 147.2 N 2944 N 5 0.05 θ 5 2.9° b c. Length of Cable. Applying Eq. (7.10) between C and B (Fig. 3) gives you sB 5 xB c 1 1 2 3 ayB xB b 2 1 d 5 (20 m) c 1 1 2 3 a0.5 m 20 m b 2 1 d 5 20.00833 m The total length of the cable between A and B is twice this value. Thus, Length 5 2sB 5 40.0167 m b REFLECT and THINK: Notice that the length of the cable is only very slightly more than the length of the span between A and B. This means that the cable must be very taut, which is consistent with the relatively large value of load P (compared to the weight of the cable).
A B D 0.5 m P 40 m C B y x yB = 0.5 m xB = 20 m Fig. 3 Dimensions used to determine length of cable.
C B W = 147.2 N T0 TB q y 10 m 10 m 0.5 m x Fig. 1 Free-body diagram of cable portion CB.
W = 147.2 N T0 TB q Fig. 2 Force triangle for cable portion CB.
409 409 I n the problems of this section, you will apply the equations of equilibrium to cables that lie in a vertical plane. We assume that a cable cannot resist bending, so the force of tension in the cable is always directed along the cable.
A. In the first part of this lesson, we considered cables subjected to concentrated loads. Since we assume the weight of the cable is negligible, the cable is straight between loads. Your solution will consist of the following steps.
1. Draw a free-body diagram of the entire cable showing the loads and the hori-zontal and vertical components of the reaction at each support. Use this free-body diagram to write the corresponding equilibrium equations.
2. You will have four unknown components and only three equations of equilib-rium (see Fig. 7.13). You must therefore find an additional piece of information, such as the position of a point on the cable or the slope of the cable at a given point.
3. After you have identified the point of the cable where the additional informa-tion exists, cut the cable at that point, and draw a free-body diagram of one of the two resulting portions of the cable. a. If you know the position of the point where you have cut the cable, set oM 5 0 about that point for the new free body. This will yield the additional equation required to solve for the four unknown components of the reactions [Sample Prob. 7.8]. b. If you know the slope of the portion of the cable you have cut, set oFx 5 0 and oFy 5 0 for the new free body. This will yield two equilibrium equations that, together with the original three, you can solve for the four reaction components and for the tension in the cable where it has been cut.
4. To find the elevation of a given point of the cable and the slope and tension at that point once you have found the reactions at the supports, you should cut the cable at that point and draw a free-body diagram of one of the two resulting portions of the cable. Setting oM 5 0 about the given point yields its elevation. Writing oFx 5 0 and oFy 5 0 yields the components of the tension force from which you can find its magnitude and direction.
5. For a cable supporting vertical loads only, the horizontal component of the ten-sion force is the same at any point. It follows that, for such a cable, the maximum tension occurs in the steepest portion of the cable.
SOLVING PROBLEMS ON YOUR OWN (continued ) 410 B. In the second portion of this section, we considered cables carrying a load that is uniformly distributed along the horizontal. The shape of the cable is then parabolic.
Your solution will use one or more of the following concepts.
1. Place the origin of coordinates at the lowest point of the cable and direct the x and y axes to the right and upward, respectively. Then the equation of the parabola is y 5 wx 2 2T0 (7.8) The minimum cable tension occurs at the origin, where the cable is horizontal. The maximum tension is at the support where the slope is maximum.
2. If the supports of the cable have the same elevation, the sag h of the cable is the vertical distance from the lowest point of the cable to the horizontal line joining the supports. To solve a problem involving such a parabolic cable, use Eq. (7.8) for one of the supports; this equation can be solved for one unknown.
3. If the supports of the cable have different elevations, you will have to write Eq. (7.8) for each of the supports (see Fig. 7.17).
4. To find the length of the cable from the lowest point to one of the supports, you can use Eq. (7.10). In most cases, you will need to compute only the first two terms of the series.
411 7.93 Three loads are suspended as shown from the cable ABCDE. Know-ing that dC 5 4 m, determine (a) the components of the reaction at E, (b) the maximum tension in the cable.
Problems 2 kN 6 kN 4 kN A 5 m B E D C dC dD dB 5 m 5 m 5 m Fig. P7.93 and P7.94 A B C D E 300 lb 300 lb 200 lb 8 ft 8 ft 8 ft 8 ft 6 ft d C Fig. P7.95 and P7.96 7.94 Knowing that the maximum tension in cable ABCDE is 25 kN, deter-mine the distance dC. 7.95 If dC 5 8 ft, determine (a) the reaction at A, (b) the reaction at E. 7.96 If dC 5 4.5 ft, determine (a) the reaction at A, (b) the reaction at E. 7.97 Knowing that dC 5 3 m, determine (a) the distances dB and dD, (b) the reaction at E. 7.98 Determine (a) distance dC for which portion DE of the cable is hori-zontal, (b) the corresponding reactions at A and E. 7.99 Knowing that dC 5 15 ft, determine (a) the distances dB and dD, (b) the maximum tension in the cable. 7.100 Determine (a) the distance dC for which portion BC of the cable is horizontal, (b) the corresponding components of the reaction at E.
A B C D E d B 2 m 2 m 5 kN 5 kN 10 kN d C d D 4 m 3 m 3 m Fig. P7.97 and P7.98 A B C 7.5 ft 6 ft 9 ft 6 ft 9 ft d B D d D d C E 2 kips 2 kips 2 kips Fig. P7.99 and P7.100 412 7.101 Knowing that mB 5 70 kg and mC 5 25 kg, determine the magnitude of the force P required to maintain equilibrium.
P A B C D 4 m 4 m 6 m 3 m 5 m mB mC Fig. P7.101 and P7.102 7.102 Knowing that mB 5 18 kg and mC 5 10 kg, determine the magnitude of the force P required to maintain equilibrium. 7.103 Cable ABC supports two loads as shown. Knowing that b 5 21 ft, determine (a) the required magnitude of the horizontal force P, (b) the corresponding distance a. 7.104 Cable ABC supports two loads as shown. Determine the distances a and b when a horizontal force P of magnitude 200 lb is applied at C.
7.105 If a 5 3 m, determine the magnitudes of P and Q required to main-tain the cable in the shape shown.
180 lb 140 lb P 12 ft 9 ft a b A B C Fig. P7.103 and P7.104 P A B C D E 2 m 2 m 120 kN Q 4 m 4 m 4 m 4 m a Fig. P7.105 and P7.106 7.106 If a 5 4 m, determine the magnitudes of P and Q required to main-tain the cable in the shape shown. 7.107 An electric wire having a mass per unit length of 0.6 kg/m is strung between two insulators at the same elevation that are 60 m apart. Knowing that the sag of the wire is 1.5 m, determine (a) the maxi-mum tension in the wire, (b) the length of the wire.
7.108 The total mass of cable ACB is 20 kg. Assuming that the mass of the cable is distributed uniformly along the horizontal, determine (a) the sag h, (b) the slope of the cable at A.
A B C h E D 150 kg 8 m 4.5 m 6 m Fig. P7.108 413 7.109 The center span of the George Washington Bridge, as originally con-structed, consisted of a uniform roadway suspended from four cables. The uniform load supported by each cable was w 5 9.75 kips/ft along the horizontal. Knowing that the span L is 3500 ft and that the sag h is 316 ft, determine for the original configuration (a) the maxi-mum tension in each cable, (b) the length of each cable. 7.110 The center span of the Verrazano-Narrows Bridge consists of two uniform roadways suspended from four cables. The design of the bridge allows for the effect of extreme temperature changes that cause the sag of the center span to vary from hw 5 386 ft in winter to hs 5 394 ft in summer. Knowing that the span is L 5 4260 ft, determine the change in length of the cables due to extreme tem-perature changes. 7.111 Each cable of the Golden Gate Bridge supports a load w 5 11.1 kips/ft along the horizontal. Knowing that the span L is 4150 ft and that the sag h is 464 ft, determine (a) the maximum tension in each cable, (b) the length of each cable. 7.112 Two cables of the same gauge are attached to a transmission tower at B. Since the tower is slender, the horizontal component of the resultant of the forces exerted by the cables at B is to be zero. Know-ing that the mass per unit length of the cables is 0.4 kg/m, determine (a) the required sag h, (b) the maximum tension in each cable. 7.113 A 76-m length of wire having a mass per unit length of 2.2 kg/m is used to span a horizontal distance of 75 m. Determine (a) the approximate sag of the wire, (b) the maximum tension in the wire. [Hint: Use only the first two terms of Eq. (7.10).] 7.114 A cable of length L 1 D is suspended between two points that are at the same elevation and a distance L apart. (a) Assuming that D is small compared to L and that the cable is parabolic, determine the approximate sag in terms of L and D. (b) If L 5 100 ft and D 5 4 ft, determine the approximate sag. [Hint: Use only the first two terms of Eq. (7.10).] 7.115 The total mass of cable AC is 25 kg. Assuming that the mass of the cable is distributed uniformly along the horizontal, determine the sag h and the slope of the cable at A and C.
A B C h 3 m 90 m 60 m Fig. P7.112 h 450 kg A B C 2.5 m 2.5 m 3 m 5 m Fig. P7.115 414 7.116 Cable ACB supports a load uniformly distributed along the horizontal as shown. The lowest point C is located 9 m to the right of A. Deter-mine (a) the vertical distance a, (b) the length of the cable, (c) the components of the reaction at A.
A C B 2.25 m 60 kg/m 6 m 9 m a Fig. P7.116 A B 1100 ft 496 ft 10.2 kips/ft C h = 30 ft Fig. P7.117 7.117 Each cable of the side spans of the Golden Gate Bridge supports a load w 5 10.2 kips/ft along the horizontal. Knowing that for the side spans the maximum vertical distance h from each cable to the chord AB is 30 ft and occurs at midspan, determine (a) the maximum tension in each cable, (b) the slope at B. 7.118 A steam pipe weighing 45 lb/ft that passes between two buildings 40 ft apart is supported by a system of cables as shown. Assuming that the weight of the cable system is equivalent to a uniformly distributed loading of 5 lb/ft, determine (a) the location of the lowest point C of the cable, (b) the maximum tension in the cable.
40 ft A B C 5 ft 4 ft 4 ft Fig. P7.118 415 7.119 A cable AB of span L and a simple beam A9B9 of the same span are subjected to identical vertical loadings as shown. Show that the mag-nitude of the bending moment at a point C9 in the beam is equal to the product T0h, where T0 is the magnitude of the horizontal compo-nent of the tension force in the cable and h is the vertical distance between point C and the chord joining the points of support A and B.
B C A' A B' C' P1 P1 P2 P2 P3 P3 Pn Pn a h L Fig. P7.119 A B C D x y a a q Fig. P7.126 7.120 through 7.123 Making use of the property established in Prob. 7.119, solve the problem indicated by first solving the corresponding beam problem. 7.120 Prob. 7.94. 7.121 Prob. 7.97a. 7.122 Prob. 7.99a. 7.123 Prob. 7.100a. 7.124 Show that the curve assumed by a cable that carries a distributed load w(x) is defined by the differential equation d2y/dx2 5 w(x)/T0, where T0 is the tension at the lowest point. 7.125 Using the property indicated in Prob. 7.124, determine the curve assumed by a cable of span L and sag h carrying a distributed load w 5 w0 cos (πx/L), where x is measured from midspan. Also deter-mine the maximum and minimum values of the tension in the cable. 7.126 If the weight per unit length of the cable AB is w0 / cos2 θ, prove that the curve formed by the cable is a circular arc. (Hint: Use the prop-erty indicated in Prob. 7.124.) 416 Internal Forces and Moments 7.5 CATENARY CABLES Let us now consider a cable AB carrying a load that is uniformly distrib-uted along the cable itself (Fig. 7.18a). Cables hanging under their own weight are loaded in this way. We denote the load per unit length by w (measured along the cable) and express it in N/m or lb/ft. The magnitude W of the total load carried by a portion of cable with a length of s, extending A B y C O c x D(x,y) (a) ds dx q q T T D C W = ws W = ws (b) (c) T0 T0 s s dy Fig. 7.18 (a) A cable carrying a load uniformly distributed along the cable; (b) free-body diagram of a portion of the cable CD; (c) force triangle for part (b).
from the lowest point C to some point D, is W 5 ws. Substituting this value for W in formula (7.6), we obtain the tension at D, as T 5 2T 2 0 1 w 2s 2 In order to simplify the subsequent computations, we introduce the con-stant c 5 T0 /w. This gives us T0 5 wc W 5 ws T 5 w2c2 1 s2 (7.11) The free-body diagram of the portion of cable CD is shown in Fig. 7.18b. However, we cannot use this diagram directly to obtain the equation of the curve assumed by the cable, because we do not know the horizontal distance from D to the line of action of the resultant W of the load. To obtain this equation, we note that the horizontal projection (a) High-voltage power lines (b) A spider’s web (c) The Gateway Arch Photo 7.5 Catenary cables occur in nature as well as in engineered structures. (a) High-voltage power lines, common all across the country and in much of the world, support only their own weight. (b) Catenary cables can be as delicate as the silk threads of a spider’s web. (c) The Gateway to the West Arch in St. Louis is an inverted catenary arch cast in concrete (which is in compression instead of tension).
7.5 Catenary Cables 417 of a small element of cable of length ds is dx 5 ds cos θ. Observing from Fig. 7.18c that cos θ 5 T0 /T and using Eq. (7.11), we have dx 5 ds cos θ 5 T0 T ds 5 wc ds w2c2 1 s2 5 ds 21 1 s2/c2 Selecting the origin O of the coordinates at a distance c directly below C (Fig. 7.18a) and integrating from C(0, c) to D(x, y), we obtain† x 5# s 0 ds 21 1 s2/c2 5 c c sinh21 s c d s 0 5 c sinh21 s c This equation, which relates the length s of the portion of cable CD and the horizontal distance x, can be written in the form Length of catenary cable s 5 c sinh x c (7.15) We can now obtain the relation between the coordinates x and y by writing dy 5 dx tan θ. Observing from Fig. 7.18c that tan θ 5 W/T0 and using (7.11) and (7.15), we have dy 5 dx tan θ 5 W T0 dx 5 s c dx 5 sinh x c dx Integrating from C(0, c) to D(x, y) and using Eqs. (7.12) and (7.13), we obtain y 2 c 5 # x 0 sinh x c dx 5 c c cosh x c d x 0 5 c acosh x c 2 1b y 2 c 5 c cosh x c 2 c † This integral appears in all standard integral tables. The function z 5 sinh21u (read “arc hyperbolic sine u”)is the inverse of the function u 5 sinh z (read “hyperbolic sine z”). This function and the function v 5 cosh z (read “hyperbolic cosine z”) are defined as u 5 sinh z 5 1 2(ez 2 e2z) v 5 cosh z 5 1 2(ez 1 e2z) Numerical values of the functions sinh z and cosh z are listed in tables of hyperbolic func-tions and also may be computed on most calculators, either directly or from the definitions. Refer to any calculus text for a complete description of the properties of these functions. In this section, we use only the following properties, which are easy to derive from the definitions: d sinh z dz 5 cosh z d cosh z dz 5 sinh z (7.12) sinh 0 5 0 cosh 0 5 1 (7.13) cosh2 z 2 sinh2 z 5 1 (7.14) s 5 c sinh x c A B y C O c x D(x,y) (a) q T W = ws (c) T0 s ds dx q T D C W = ws (b) T0 s dy Fig. 7.18 (continued).
418 Internal Forces and Moments which reduces to Equation of catenary cable y 5 c cosh x c (7.16) This is the equation of a catenary with vertical axis. The ordinate c of the lowest point C is called the parameter of the catenary. By squaring both sides of Eqs. (7.15) and (7.16), subtracting, and taking Eq. (7.14) into account, we obtain the following relation between y and s: y2 2 s2 5 c2 (7.17) Solving Eq. (7.17) for s2 and carrying into the last of the relations in Eqs. (7.11), we write these relations as T0 5 wc W 5 ws T 5 wy (7.18) The last relation indicates that the tension at any point D of the cable is proportional to the vertical distance from D to the horizontal line repre-senting the x axis.
When the supports A and B of the cable have the same elevation, the distance L between the supports is called the span of the cable and the vertical distance h from the supports to the lowest point C is called the sag of the cable. These definitions are the same as those given for para-bolic cables; note that, because of our choice of coordinate axes, the sag h is now h 5 yA 2 c (7.19) Also note that some catenary problems involve transcendental equations, which must be solved by successive approximations (see Sample Prob. 7.10). When the cable is fairly taut, however, we can assume that the load is uniformly distributed along the horizontal and replace the catenary by a parabola. This greatly simplifies the solution of the problem, and the error introduced is small.
When the supports A and B have different elevations, the position of the lowest point of the cable is not known. We can then solve the problem in a manner similar to that indicated for parabolic cables by not-ing that the cable must pass through the supports and that xB 2 xA 5 L and yB 2 yA 5 d, where L and d denote, respectively, the horizontal and vertical distances between the two supports.
y 5 c cosh x c y2 2 s2 5 c2 T0 T 5 wc W 5 ws T 5 wy h 5 yA y 2 c 7.5 Catenary Cables 419 Sample Problem 7.10 A uniform cable weighing 3 lb/ft is suspended between two points A and B as shown. Determine (a) the maximum and minimum values of the tension in the cable, (b) the length of the cable.
STRATEGY: This is a cable carrying only its own weight that is sup-ported by its ends at the same elevation. You can use the analysis in this section to solve the problem.
MODELING and ANALYSIS: Equation of Cable. Place the origin of coordinates at a distance c below the lowest point of the cable (Fig. 1). The equation of the cable is given by Eq. (7.16), as y 5 c cosh x c The coordinates of point B are xB 5 250 ft yB 5 100 1 c Substituting these coordinates into the equation of the cable, you obtain 100 1 c 5 c cosh 250 c 100 c 1 1 5 cosh 250 c Determine the value of c by substituting successive trial values, as shown in the following table.
c 250 c 100 c 100 c 1 1 cosh 250 c 300 0.833 0.333 1.333 1.367 350 0.714 0.286 1.286 1.266 330 0.758 0.303 1.303 1.301 328 0.762 0.305 1.305 1.305 Taking c 5 328, you have yB 5 100 1 c 5 428 ft a. Maximum and Minimum Values of the Tension. Using Eqs. (7.18), you obtain Tmin 5 T0 5 wc 5 (3 lb/ft)(328 ft) Tmin 5 984 lb b Tmax 5 TB 5 wyB 5 (3 lb/ft)(428 ft) Tmax 5 1284 lb b b. Length of Cable. You can find one-half of the length of the cable by solving Eq. (7.17). Hence, y2 B 2 s2 CB 5 c2 s2 CB 5 y2 B 2 c2 5 (428)2 2 (328)2 sCB 5 275 ft The total length of the cable is therefore sAB 5 2sCB 5 2(275 ft) sAB 5 550 ft b REFLECT and THINK: The sag in the cable is one-fifth of the cable’s span, so it is not very taut. The weight of the cable is ws 5 (3 lb/ft)(550 ft) 5 1650 lb, while its maximum tension is only 1284 lb. This demonstrates that the total weight of a cable can exceed its maximum tension. A B 100 ft 500 ft xB yB y x O c A C B Fig. 1 Cable geometry.
420 420 SOLVING PROBLEMS ON YOUR OWN I n the last section of this chapter, we described how to solve problems involving a cable carrying a load uniformly distributed along the cable. The shape assumed by the cable is a catenary and is defined by y 5 c cosh x c (7.16) 1. Keep in mind that the origin of coordinates for a catenary is located at a distance c directly below its lowest point. The length of the cable from the origin to any point is expressed as s 5 c sinh x c (7.15) 2. You should first identify all of the known and unknown quantities. Then con-sider each of the equations listed in the text (Eqs. 7.15 through 7.19) and solve an equation that contains only one unknown. Substitute the value found into another equation, and solve that equation for another unknown.
3. If the sag h is given, use Eq. (7.19) to replace y by h 1 c in Eq. (7.16) if x is known [Sample Prob. 7.10] or in Eq. (7.17) if s is known, and solve the resulting equation for the constant c.
4. Many of the problems you will encounter will involve the solution by trial and error of an equation involving a hyperbolic sine or cosine. You can make your work easier by keeping track of your calculations in a table, as in Sample Prob. 7.10, or by applying a numerical methods approach using a computer or calculator.
421 7.127 A 25-ft chain with a weight of 30 lb is suspended between two points at the same elevation. Knowing that the sag is 10 ft, determine (a) the distance between the supports, (b) the maximum tension in the chain. 7.128 A 500-ft-long aerial tramway cable having a weight per unit length of 2.8 lb/ft is suspended between two points at the same elevation. Knowing that the sag is 125 ft, find (a) the horizontal distance between the supports, (b) the maximum tension in the cable. 7.129 A 40-m cable is strung as shown between two buildings. The maxi-mum tension is found to be 350 N, and the lowest point of the cable is observed to be 6 m above the ground. Determine (a) the horizontal distance between the buildings, (b) the total mass of the cable.
Problems A B C 14 m L 6 m Fig. P7.129 A B C P h L Fig. P7.131, P7.132, and P7.133 7.130 A 50-m steel surveying tape has a mass of 1.6 kg. If the tape is stretched between two points at the same elevation and pulled until the tension at each end is 60 N, determine the horizontal distance between the ends of the tape. Neglect the elongation of the tape due to the tension.
7.131 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the force P for which h 5 8 m, (b) the corresponding span L. 7.132 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Knowing that the magnitude of the horizontal force applied to the collar is P 5 20 N, determine (a) the sag h, (b) the span L. 7.133 A 20-m length of wire having a mass per unit length of 0.2 kg/m is attached to a fixed support at A and to a collar at B. Neglecting the effect of friction, determine (a) the sag h for which L 5 15 m, (b) the corresponding force P. 7.134 Determine the sag of a 30-ft chain that is attached to two points at the same elevation that are 20 ft apart.
422 7.135 A counterweight D is attached to a cable that passes over a small pulley at A and is attached to a support at B. Knowing that L 5 45 ft and h 5 15 ft, determine (a) the length of the cable from A to B, (b) the weight per unit length of the cable. Neglect the weight of the cable from A to D.
A B C D h 80 lb L Fig. P7.135 A B h L Fig. P7.138 A B C M h 10 m Fig. P7.139 and P7.140 A B C a 12 m 1.8 m Fig. P7.141 and P7.142 7.136 A 90-m wire is suspended between two points at the same elevation that are 60 m apart. Knowing that the maximum tension is 300 N, determine (a) the sag of the wire, (b) the total mass of the wire.
7.137 A cable weighing 2 lb/ft is suspended between two points at the same elevation that are 160 ft apart. Determine the smallest allow-able sag of the cable if the maximum tension is not to exceed 400 lb.
7.138 A uniform cord 50 in. long passes over a pulley at B and is attached to a pin support at A. Knowing that L 5 20 in. and neglecting the effect of friction, determine the smaller of the two values of h for which the cord is in equilibrium. 7.139 A motor M is used to slowly reel in the cable shown. Knowing that the mass per unit length of the cable is 0.4 kg/m, determine the maximum tension in the cable when h 5 5 m. 7.140 A motor M is used to slowly reel in the cable shown. Knowing that the mass per unit length of the cable is 0.4 kg/m, determine the maximum tension in the cable when h 5 3 m. 7.141 The cable ACB has a mass per unit length of 0.45 kg/m. Knowing that the lowest point of the cable is located at a distance a 5 0.6 m below the support A, determine (a) the location of the lowest point C, (b) the maximum tension in the cable.
7.142 The cable ACB has a mass per unit length of 0.45 kg/m. Knowing that the lowest point of the cable is located at a distance a 5 2 m below the support A, determine (a) the location of the lowest point C, (b) the maximum tension in the cable.
423 7.143 A uniform cable weighing 3 lb/ft is held in the position shown by a horizontal force P applied at B. Knowing that P 5 180 lb and θA 5 60°, determine (a) the location of point B, (b) the length of the cable. 7.144 A uniform cable weighing 3 lb/ft is held in the position shown by a horizontal force P applied at B. Knowing that P 5 150 lb and θA 5 60°, determine (a) the location of point B, (b) the length of the cable. 7.145 To the left of point B, the long cable ABDE rests on the rough hori-zontal surface shown. Knowing that the mass per unit length of the cable is 2 kg/m, determine the force F when a 5 3.6 m.
A B P qA a b Fig. P7.143 and P7.144 D A B E F h = 4 m a Fig. P7.145 and P7.146 A B a q = 30° Fig. P7.147 qA qB A B h L Fig. P7.151, P7.152 and P7.153 7.146 To the left of point B, the long cable ABDE rests on the rough hori-zontal surface shown. Knowing that the mass per unit length of the cable is 2 kg/m, determine the force F when a 5 6 m. 7.147 The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the rod; a stop attached to the rod prevents the collar at B from moving on the rod. Neglecting the effect of friction and the weight of the collars, determine the distance a. 7.148 Solve Prob. 7.147 assuming that the angle θ formed by the rod and the horizontal is 45°. 7.149 Denoting the angle formed by a uniform cable and the horizontal by θ, show that at any point (a) s 5 c tan θ, (b) y 5 c sec θ. 7.150 (a) Determine the maximum allowable horizontal span for a uniform cable with a weight per unit length of w if the tension in the cable is not to exceed a given value Tm. (b) Using the result of part a, determine the maximum span of a steel wire for which w 5 0.25 lb/ft and Tm 5 8000 lb. 7.151 A cable has a mass per unit length of 3 kg/m and is supported as shown. Knowing that the span L is 6 m, determine the two values of the sag h for which the maximum tension is 350 N. 7.152 Determine the sag-to-span ratio for which the maximum tension in the cable is equal to the total weight of the entire cable AB. 7.153 A cable with a weight per unit length of w is suspended between two points at the same elevation that are a distance L apart. Deter-mine (a) the sag-to-span ratio for which the maximum tension is as small as possible, (b) the corresponding values of θB and Tm.
424 In this chapter, you learned to determine the internal forces that hold together the various parts of a given member in a structure.
Forces in Straight Two-Force Members Considering first a straight two-force member AB [Sec. 7.1], recall that such a member is subjected at A and B to equal and opposite forces F and 2F directed along AB (Fig. 7.19a). Cutting member AB at C and drawing the free-body diagram of portion AC, we concluded that the internal forces exist-ing at C in member AB are equivalent to an axial force 2F equal and opposite to F (Fig. 7.19b). Note that, in the case of a two-force member that is not straight, the internal forces reduce to a force-couple system and not to a single force.
Forces in Multi-Force Members Consider next a multi-force member AD (Fig. 7.20a). Cutting it at J and drawing the free-body diagram of portion JD, we concluded that the internal forces at J are equivalent to a force-couple system consisting of the axial force F, the shearing force V, and a couple M (Fig. 7.20b). The magnitude of the shearing force measures the shear at point J, and the moment of the couple is referred to as the bending moment at J. Since an equal and opposite force-couple system is obtained by considering the free-body diagram of por-tion AJ, it is necessary to specify which portion of member AD is used when recording the answers [Sample Prob. 7.1].
FBE Cx Ay Ax Cy T A B C D J (a) V M F T D J (b) Fig. 7.20 Forces in Beams Most of the chapter was devoted to the analysis of the internal forces in two important types of engineering structures: beams and cables. Beams are usu-ally long, straight prismatic members designed to support loads applied at various points along the member. In general, the loads are perpendicular to the axis of the beam and produce only shear and bending in the beam. The loads may be either concentrated at specific points or distributed along the entire length or a portion of the beam. The beam itself may be supported in various ways; since only statically determinate beams are considered in this Review and Summary (a) (b) C A B F – F – F C A F Fig. 7.19 425 text, we limited our analysis to that of simply supported beams, overhanging beams, and cantilever beams [Sec. 7.2].
Shear and Bending Moment in a Beam To obtain the shear V and bending moment M at a given point C of a beam, we first determine the reactions at the supports by considering the entire beam as a free body. We then cut the beam at C and use the free-body diagram of one of the two resulting portions to determine V and M. In order to avoid any confusion regarding the sense of the shearing force V and couple M (which act in opposite directions on the two portions of the beam), we adopted the sign convention illustrated in Fig. 7.21 [Sec. 7.2B]. Once we have determined the values of the shear and bending moment at a few selected points of the beam, it is usually possible to draw a shear diagram and a bending-moment diagram representing, respectively, the shear and bending moment at any point of the beam [Sec. 7.2C]. When a beam is subjected to concentrated loads only, the shear is of constant value between loads, and the bending moment varies linearly between loads [Sample Prob. 7.2]. When a beam is subjected to distributed loads, the shear and bending moment vary quite differently [Sample Prob. 7.3].
Internal forces at section (positive shear and positive bending moment) M V M' V' Fig. 7.21 Relations among Load, Shear, and Bending Moment Construction of the shear and bending-moment diagrams is simplified by tak-ing into account the following relations. Denoting the distributed load per unit length by w (assumed positive if directed downward), we have [Sec. 7.3]: dV dx 5 2w (7.1) dM dx 5 V (7.3) In integrated form, these equations become VD 2 VC 5 2(area under load curve between C and D) (7.29) MD 2 MC 5 area under shear curve between C and D (7.49) Equation (7.29) makes it possible to draw the shear diagram of a beam from the curve representing the distributed load on that beam and the value of V at one end of the beam. Similarly, Eq. (7.49) makes it possible to draw the bending-moment diagram from the shear diagram and the value of M at one end of the beam. However, discontinuities are introduced in the shear diagram by concentrated loads and in the bending-moment diagram by concentrated couples, none of which are accounted for in these equations [Sample Probs. 7.4 and 7.7]. Finally, we note from Eq. (7.3) that the points of the beam where the bending moment is maximum or minimum are also the points where the shear is zero [Sample Prob. 7.5].
426 Cables with Concentrated Loads The second half of the chapter was devoted to the analysis of flexible cables. We first considered a cable of negligible weight supporting concentrated loads [Sec. 7.4A]. Using the entire cable AB as a free body (Fig. 7.22), we noted that the three available equilibrium equations were not sufficient to determine the four unknowns representing the reactions at supports A and B. However, if the coordinates of a point D of the cable are known, we can obtain an additional equation by considering the free-body diagram of portion AD or DB of the cable. Once we have determined the reactions at the supports, we can find the elevation of any point of the cable and the tension in any portion of the cable from the appropriate free-body diagram [Sample Prob. 7.8]. We noted that the horizontal component of the force T representing the ten-sion is the same at any point of the cable.
Cables with Distributed Loads We next considered cables carrying distributed loads [Sec. 7.4B]. Using as a free body a portion of cable CD extending from the lowest point C to an arbi-trary point D of the cable (Fig. 7.23), we observed that the horizontal component of the tension force T at D is constant and equal to the tension T0 at C, whereas its vertical component is equal to the weight W of the portion of cable CD. The magnitude and direction of T were obtained from the force triangle: T 5 2T 2 0 1 W 2 tan θ 5 W T0 (7.6) Parabolic Cable In the case of a load uniformly distributed along the horizontal—as in a sus-pension bridge (Fig. 7.24)—the load supported by portion CD is W 5 wx, where w is the constant load per unit horizontal length [Sec. 7.4C]. We also found that the curve formed by the cable is a parabola with equation y 5 wx2 2T0 (7.8) and that the length of the cable can be found by using the expansion in series given in Eq. (7.10) [Sample Prob. 7.9].
Catenary In the case of a load uniformly distributed along the cable itself—e.g., a cable hanging under its own weight (Fig. 7.25)—the load supported by portion CD is W 5 ws, where s is the length measured along the cable and w is the con-stant load per unit length [Sec. 7.5]. Choosing the origin O of the coordinate axes at a distance c 5 T0 /w below C, we derived the relations s 5 c sinh x c (7.15) y 5 c cosh x c (7.16) y2 2 s2 5 c2 (7.17) T0 5 wc W 5 ws T 5 wy (7.18) These equations can be used to solve problems involving cables hanging under their own weight [Sample Prob. 7.10]. Equation (7.16), which defines the shape of the cable, is the equation of a catenary.
A B C y D(x,y) x w Fig. 7.24 A B y C O c x D(x,y) s Fig. 7.25 Ax Ay A Bx By C1 C2 C3 D P1 P2 P3 B L d x1 x2 x3 Fig. 7.22 D C T T W T0 q q W T0 Fig. 7.23 427 Review Problems 7.154 and 7.155 Knowing that the turnbuckle has been tightened until the tension in wire AD is 850 N, determine the internal forces at the point indicated: 7.154 Point J 7.155 Point K 7.156 Two members, each consisting of a straight and a quarter-circular portion of rod, are connected as shown and support a 75-lb load at A. Determine the internal forces at point J.
A B E F J C D 75 lb K 6 in.
3 in.
3 in.
6 in.
3 in.
3 in.
3 in.
Fig. P7.156 7.157 Knowing that the radius of each pulley is 150 mm, that α 5 20°, and neglecting friction, determine the internal forces at (a) point J, (b) point K.
A B C K D J 500 N 0.6 m a 0.6 m 0.9 m 0.9 m Fig. P7.157 7.158 For the beam shown, determine (a) the magnitude P of the two upward forces for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding value of |M|max.
2 ft 2 ft 2 ft 2 ft 2 ft A B C D E F P P 60 kips 60 kips Fig. P7.158 120 mm A E F D J K C B 100 mm 100 mm 100 mm 280 mm Fig. P7.154 and P7.155 428 7.159 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the magnitude and location of the maximum absolute value of the bending moment.
7.160 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.
7.161 For the beam shown, draw the shear and bending-moment diagrams, and determine the magnitude and location of the maximum absolute value of the bending moment, knowing that (a) M 5 0, (b) M 5 24 kip∙ft.
7.162 The beam AB, which lies on the ground, supports the parabolic load shown. Assuming the upward reaction of the ground to be uniformly distributed, (a) write the equations of the shear and bending-moment curves, (b) determine the maximum bending moment.
A B w L x w0 w = L x – x2 4w0 L2 ( ) Fig. P7.162 7.163 Two loads are suspended as shown from the cable ABCD. Knowing that dB 5 1.8 m, determine (a) the distance dC, (b) the components of the reaction at D, (c) the maximum tension in the cable. 7.164 A wire having a mass per unit length of 0.65 kg/m is suspended from two supports at the same elevation that are 120 m apart. If the sag is 30 m, determine (a) the total length of the wire, (b) the maximum tension in the wire. 7.165 A 10-ft rope is attached to two supports A and B as shown. Deter-mine (a) the span of the rope for which the span is equal to the sag, (b) the corresponding angle θB.
A B C qB h L Fig. P7.165 4 m 25 kN/m 20 kN⋅m A B Fig. P7.159 A B C 4 kips/ft 4 ft 4 ft M Fig. P7.161 B C D A 6 kN 10 kN 3 m 3 m 4 m dB dC Fig. P7.163 A E B C D 9 in.
12 in.
12 in.
12 in.
20 lb/in.
125 lb 125 lb Fig. P7.160 The tractive force that a railroad locomotive can develop depends upon the frictional resistance between the drive wheels and the rails. When the potential exists for wheel slip to occur, such as when a train travels upgrade over wet rails, sand is deposited on top of the railhead to increase this friction.
Friction 8 430 Friction Introduction 8.1 THE LAWS OF DRY FRICTION 8.1A Coefficients of Friction 8.1B Angles of Friction 8.1C Problems Involving Dry Friction 8.2 WEDGES AND SCREWS 8.2A Wedges 8.2B Square-Threaded Screws 8.3 FRICTION ON AXLES, DISKS, AND WHEELS 8.3A Journal Bearings and Axle Friction 8.3B Thrust Bearings and Disk Friction 8.3C Wheel Friction and Rolling Resistance 8.4 BELT FRICTION Objectives • Examine the laws of dry friction and the associated coeffi cients and angles of friction.
• Consider the equilibrium of rigid bodies where dry friction at contact surfaces is modeled.
• Apply the laws of friction to analyze problems involv-ing wedges and square-threaded screws.
• Study engineering applications of the laws of friction, such as in modeling axle, disk, wheel, and belt friction.
Introduction In the previous chapters, we assumed that surfaces in contact are either frictionless or rough. If they are frictionless, the force each surface exerts on the other is normal to the surfaces, and the two surfaces can move freely with respect to each other. If they are rough, tangential forces can develop that prevent the motion of one surface with respect to the other.
This view is a simplified one. Actually, no perfectly frictionless sur-face exists. When two surfaces are in contact, tangential forces, called friction forces, always develop if you attempt to move one surface with respect to the other. However, these friction forces are limited in magnitude and do not prevent motion if you apply sufficiently large forces. Thus, the distinction between frictionless and rough surfaces is a matter of degree. You will see this more clearly in this chapter, which is devoted to the study of friction and its applications to common engineering situations.
There are two types of friction: dry friction, sometimes called Coulomb friction, and fluid friction or viscosity. Fluid friction develops Photo 8.1 Examples of friction in an automobile. Depending upon the application, the degree of friction is controlled by design engineers.
Low friction— pistons in engine cylinders Low friction— journal bearings on front axle High friction— disk brakes Moderate friction— shock absorbers High friction— tire treads Low friction— air bag release High friction— drive belt from engine 8.1 The Laws of Dry Friction 431 between layers of fluid moving at different velocities. This is of great importance in analyzing problems involving the flow of fluids through pipes and orifices or dealing with bodies immersed in moving fluids. It is also basic for the analysis of the motion of lubricated mechanisms. Such problems are considered in texts on fluid mechanics. The present study is limited to dry friction, i.e., to situations involving rigid bodies that are in contact along unlubricated surfaces.
In the first section of this chapter, we examine the equilibrium of various rigid bodies and structures, assuming dry friction at the surfaces of contact. Afterward, we consider several specific engineering applications where dry friction plays an important role: wedges, square-threaded screws, journal bearings, thrust bearings, rolling resistance, and belt friction.
8.1 THE LAWS OF DRY FRICTION We can illustrate the laws of dry friction by the following experiment. Place a block of weight W on a horizontal plane surface (Fig. 8.1a). The forces acting on the block are its weight W and the reaction of the surface. Since the weight has no horizontal component, the reaction of the surface also has no horizontal component; the reaction is therefore normal to the surface and is represented by N in Fig. 8.1a. Now suppose that you apply a hori-zontal force P to the block (Fig. 8.1b). If P is small, the block does not move; some other horizontal force must therefore exist, which balances P. This other force is the static-friction force F, which is actually the resultant of a great number of forces acting over the entire surface of contact between the block and the plane. The nature of these forces is not known exactly, but we generally assume that these forces are due to the irregularities of the surfaces in contact and, to a certain extent, to molecular attraction.
W N P (a) F P Fm Fk Equilibrium Motion Impending motion A B W N (b) (c) A B F Fig. 8.1 (a) Block on a horizontal plane, friction force is zero; (b) a horizontally applied force P produces an opposing friction force F; (c) graph of F with increasing P.
If you increase the force P, the friction force F also increases, continu-ing to oppose P, until its magnitude reaches a certain maximum value Fm (Fig. 8.1c). If P is further increased, the friction force cannot balance it any-more, and the block starts sliding. As soon as the block has started in motion, the magnitude of F drops from Fm to a lower value Fk. This happens because less interpenetration occurs between the irregularities of the surfaces in contact when these surfaces move with respect to each other. From then on, the block keeps sliding with increasing velocity while the friction force, denoted by Fk and called the kinetic-friction force , remains approximately constant.
432 Friction Note that, as the magnitude F of the friction force increases from 0 to Fm, the point of application A of the resultant N of the normal forces of contact moves to the right. In this way, the couples formed by P and F and by W and N, respectively, remain balanced. If N reaches B before F reaches its maximum value Fm, the block starts to tip about B before it can start sliding (see Sample Prob. 8.4).
8.1A Coefficients of Friction Experimental evidence shows that the maximum value Fm of the static-friction force is proportional to the normal component N of the reaction of the surface. We have Static friction Fm 5 µsN (8.1) where µs is a constant called the coefficient of static friction. Similarly, we can express the magnitude Fk of the kinetic-friction force in the form Kinetic friction Fk 5 µkN (8.2) where µk is a constant called the coefficient of kinetic friction. The coef-ficients of friction µs and µk do not depend upon the area of the surfaces in contact. Both coefficients, however, depend strongly on the nature of the surfaces in contact. Since they also depend upon the exact condition of the surfaces, their value is seldom known with an accuracy greater than 5%. Approximate values of coefficients of static friction for various com-binations of dry surfaces are given in Table 8.1. The corresponding values of the coefficient of kinetic friction are about 25% smaller. Since coeffi-cients of friction are dimensionless quantities, the values given in Table 8.1 can be used with both SI units and U.S. customary units.
Table 8.1 Approximate Values of Coefficient of Static Friction for Dry Surfaces Metal on metal 0.15–0.60 Metal on wood 0.20–0.60 Metal on stone 0.30–0.70 Metal on leather 0.30–0.60 Wood on wood 0.25–0.50 Wood on leather 0.25–0.50 Stone on stone 0.40–0.70 Earth on earth 0.20–1.00 Rubber on concrete 0.60–0.90 From this discussion, it appears that four different situations can occur when a rigid body is in contact with a horizontal surface: 1. The forces applied to the body do not tend to move it along the surface of contact; there is no friction force (Fig. 8.2a). 2. The applied forces tend to move the body along the surface of contact but are not large enough to set it in motion. We can find the static-friction force F that has developed by solving the equations of equilibrium for the body. Since there is no evidence that F has reached its maximum Fm F 5 µsN s Fk F 5 µkN k W P N F = 0 Py Px F = Px N = Py + W F < sN N = P + W (a) No friction (Px = 0) W P N F (b) No motion (Px < Fm) Py Px Fm = Px N = Py + W Fm = sN W P N Fm (c) Motion impending (Px = Fm) Py Px Fk < Px N = Py + W Fk = kN W P N Fk (d) Motion (Px > Fk) μ μ μ Fig. 8.2 (a) Applied force is vertical, friction force is zero; (b) horizontal component of applied force is less than Fm, no motion occurs; (c) horizontal component of applied force equals Fm, motion is impending; (d) horizontal component of applied force is greater than Fk, forces are unbalanced and motion continues.
8.1 The Laws of Dry Friction 433 value, the equation Fm 5 µsN cannot be used to determine the friction force (Fig. 8.2b). 3. The applied forces are such that the body is just about to slide. We say that motion is impending. The friction force F has reached its maximum value Fm and, together with the normal force N, balances the applied forces. Both the equations of equilibrium and the equation Fm 5 µsN can be used. Note that the friction force has a sense opposite to the sense of impending motion (Fig. 8.2c). 4. The body is sliding under the action of the applied forces, and the equa-tions of equilibrium no longer apply. However, F is now equal to Fk, and we can use the equation Fk 5 µkN. The sense of Fk is opposite to the sense of motion (Fig. 8.2d).
8.1B Angles of Friction It is sometimes convenient to replace the normal force N and the friction force F by their resultant R. Let’s see what happens when we do that.
Consider again a block of weight W resting on a horizontal plane surface. If no horizontal force is applied to the block, the resultant R reduces to the normal force N (Fig. 8.3a). However, if the applied force P has a horizontal component Px that tends to move the block, force R has a horizontal component F and, thus, forms an angle f with the normal to the surface (Fig. 8.3b). If you increase Px until motion becomes impend-ing, the angle between R and the vertical grows and reaches a maximum value (Fig. 8.3c). This value is called the angle of static friction and is denoted by fs. From the geometry of Fig. 8.3c, we note that Angle of static friction tan fs 5 Fm N 5 µs N N tan fs 5 µs (8.3) If motion actually takes place, the magnitude of the friction force drops to Fk; similarly, the angle between R and N drops to a lower value fk, which is called the angle of kinetic friction (Fig. 8.3d). From the geo-metry of Fig. 8.3d, we have Angle of kinetic friction tan fk 5 Fk N 5 µk N N tan fk 5 µk (8.4) Another example shows how the angle of friction can be used to advan-tage in the analysis of certain types of problems. Consider a block resting on a board and subjected to no other force than its weight W and the reaction R of the board. The board can be given any desired inclination. If the board is horizontal, the force R exerted by the board on the block is perpendicular to the board and balances the weight W (Fig. 8.4a). If the board is given a small angle of inclination θ, force R deviates from the perpendicular to the board by angle θ and continues to balance W (Fig. 8.4b). The reaction R now has a normal component N with a magnitude of N 5 W cos θ and a tangential component F with a magnitude of F 5 W sin θ.
tan fs 5 µs tan fk 5 µk R = N P P (a) No friction (b) No motion (c) Motion impending (d) Motion f < fs P R N Fk < Px R N Fm = Px R N F = Px Px Py Px Py Py Px P W W W W f = fs f = fk Fig. 8.3 (a) Applied force is vertical, friction force is zero; (b) applied force is at an angle, its horizontal component balanced by the horizontal component of the surface resultant; (c) impending motion, the horizontal component of the applied force equals the maximum horizontal component of the resultant; (d) motion, the horizontal component of the resultant is less than the horizontal component of the applied force.
434 Friction If we keep increasing the angle of inclination, motion soon becomes impending. At that time, the angle between R and the normal reaches its maximum value θ 5 fs (Fig. 8.4c). The value of the angle of inclination corresponding to impending motion is called the angle of repose. Clearly, the angle of repose is equal to the angle of static friction fs. If we further increase the angle of inclination θ, motion starts and the angle between R and the normal drops to the lower value fk (Fig. 8.4d). The reaction R is not vertical anymore, and the forces acting on the block are unbalanced.
8.1C Problems Involving Dry Friction Many engineering applications involve dry friction. Some are simple situ-ations, such as variations on the block sliding on a plane just described. Others involve more complicated situations, as in Sample Prob. 8.3. Many problems deal with the stability of rigid bodies in accelerated motion and will be studied in dynamics. Also, several common machines and mecha-nisms can be analyzed by applying the laws of dry friction, including wedges, screws, journal and thrust bearings, and belt transmissions. We will study these applications in the following sections.
The methods used to solve problems involving dry friction are the same that we used in the preceding chapters. If a problem involves only a motion of translation with no possible rotation, we can usually treat the body under consideration as a particle and use the methods of Chap. 2. If the problem involves a possible rotation, we must treat the body as a rigid body and use the methods of Chap. 4. If the structure considered is made of several parts, we must apply the principle of action and reaction, as we did in Chap. 6.
If the body being considered is acted upon by more than three forces (including the reactions at the surfaces of contact), the reaction at each surface is represented by its components N and F, and we solve the prob-lem using the equations of equilibrium. If only three forces act on the body under consideration, it may be more convenient to represent each reaction by the single force R and solve the problem by using a force triangle.
Most problems involving friction fall into one of the following three groups. 1. All applied forces are given, and we know the coefficients of friction; we are to determine whether the body being considered remains at rest or slides. The friction force F required to maintain equilibrium is W R W W (a) No friction (b) No motion q = 0 q < fs R R R W q q (c) Motion impending (d) Motion q = fs = angle of repose W sin q W cos q F = W sin q N = W cos q N = W cos q q Fm = W sin q Fk < W sin q q > fs N = W cos q q q = fs fk Fig. 8.4 (a) Block on horizontal board, friction force is zero; (b) board's angle of inclination is less than angle of static friction, no motion; (c) board's angle of inclination equals angle of friction, motion is impending; (d) angle of inclination is greater than angle of friction, forces are unbalanced and motion occurs.
Photo 8.2 The coefficient of static friction between a crate and the inclined conveyer belt must be sufficiently large to enable the crate to be transported without slipping.
8.1 The Laws of Dry Friction 435 unknown (its magnitude is not equal to µsN) and needs to be deter-mined, together with the normal force N, by drawing a free-body dia-gram and solving the equations of equilibrium (Fig. 8.5a). We then compare the value found for the magnitude F of the friction force with the maximum value Fm 5 µsN. If F is smaller than or equal to Fm, the body remains at rest. If the value found for F is larger than Fm, equi-librium cannot be maintained and motion takes place; the actual mag-nitude of the friction force is then Fk 5 µkN. 2. All applied forces are given, and we know the motion is impending; we are to determine the value of the coefficient of static friction. Here again, we determine the friction force and the normal force by drawing a free-body diagram and solving the equations of equilibrium (Fig. 8.5b). Since we know that the value found for F is the maximum value Fm, we determine the coefficient of friction by solving the equation Fm 5 µsN. 3. The coefficient of static friction is given, and we know that the motion is impending in a given direction; we are to determine the magnitude or the direction of one of the applied forces. The friction force should be shown in the free-body diagram with a sense opposite to that of the impending motion and with a magnitude Fm 5 µsN (Fig. 8.5c). We can then write the equations of equilibrium and determine the desired force.
As noted previously, when only three forces are involved, it may be more convenient to represent the reaction of the surface by a single force R and to solve the problem by drawing a force triangle. Such a solution is used in Sample Prob. 8.2.
When two bodies A and B are in contact (Fig. 8.6a), the forces of friction exerted, respectively, by A on B and by B on A are equal and opposite (Newton’s third law). In drawing the free-body diagram of one of these bodies, it is important to include the appropriate friction force with its correct sense. Observe the following rule: The sense of the friction force acting on A is opposite to that of the motion (or impending motion) of A as observed from B (Fig. 8.6b). (It is therefore the same as the motion of B as observed from A.) The sense of the friction force acting on B is determined in a similar way (Fig. 8.6c). Note that the motion of A as observed from B is a relative motion. For example, if body A is fixed and body B moves, body A has a relative motion with respect to B. Also, if both B and A are moving down but B is moving faster than A, then body A is observed, from B, to be moving up.
Fm = msN W P N Frequired (a) W P N (b) Fm = msN W P N (c) Sense of impending motion Fig. 8.5 Three types of friction problems: (a) given the forces and coefficient of friction, will the block slide or stay? (b) given the forces and that motion is pending, determine the coefficient of friction; (c) given the coefficient of friction and that motion is impending, determine the applied force.
–F –N –P –Q –Q Motion of A with respect to B Motion of B with respect to A –P P P F N Q Q (c) (b) (a) A A B B Fig. 8.6 (a) Two blocks held in contact by forces; (b) free-body diagram for block A, including direction of friction force; (c) free-body diagram for block B, including direction of friction force.
436 Friction Sample Problem 8.1 A 100-lb force acts as shown on a 300-lb crate placed on an inclined plane. The coefficients of friction between the crate and the plane are µs 5 0.25 and µk 5 0.20. Determine whether the crate is in equilibrium, and find the value of the friction force.
STRATEGY: This is a friction problem of the first type: You know the forces and the friction coefficients and want to determine if the crate moves. You also want to find the friction force.
MODELING and ANALYSIS Force Required for Equilibrium. First determine the value of the friction force required to maintain equilibrium. Assuming that F is directed down and to the left, draw the free-body diagram of the crate (Fig. 1) and solve the equilibrium equations: 1 oFx 5 0: 100 lb 2 3 5(300 lb) 2 F 5 0 F 5 280 lb F 5 80 lb 1 oFy 5 0: N 2 4 5(300 lb) 5 0 N 5 1240 lb N 5 240 lb The force F required to maintain equilibrium is an 80-lb force directed up and to the right; the tendency of the crate is thus to move down the plane.
Maximum Friction Force. The magnitude of the maximum friction force that may be developed between the crate and the plane is Fm 5 µsN Fm 5 0.25(240 lb) 5 60 lb Since the value of the force required to maintain equilibrium (80 lb) is larger than the maximum value that may be obtained (60 lb), equilibrium is not maintained and the crate will slide down the plane.
Actual Value of Friction Force. The magnitude of the actual friction force is Factual 5 Fk 5 µkN 5 0.20(240 lb) 5 48 lb The sense of this force is opposite to the sense of motion; the force is thus directed up and to the right (Fig. 2): Factual 5 48 lb b Note that the forces acting on the crate are not balanced. Their resultant is 3 5(300 lb) 2 100 lb 2 48 lb 5 32 lb REFLECT and THINK: This is a typical friction problem of the first type. Note that you used the coefficient of static friction to determine if the crate moves, but once you found that it does move, you needed the coefficient of kinetic friction to determine the friction force.
x x x x x x 100 lb 3 4 5 300 lb 100 lb F N x 300 lb 3 4 5 y Fig. 1 Free-body diagram of crate showing assumed direction of friction force.
Motion F = 48 lb N = 240 lb 100 lb 300 lb Fig. 2 Free-body diagram of crate showing actual friction force.
8.1 The Laws of Dry Friction 437 Sample Problem 8.2 A support block is acted upon by two forces as shown. Know-ing that the coefficients of friction between the block and the incline are µs 5 0.35 and µk 5 0.25, determine the force P required to (a) start the block moving up the incline; (b) keep it moving up; (c) prevent it from sliding down.
STRATEGY: This problem involves practical variations of the third type of friction problem. You can approach the solu-tions through the concept of the angles of friction.
MODELING: Free-Body Diagram. For each part of the problem, draw a free-body diagram of the block and a force triangle including the 800-N vertical force, the horizontal force P, and the force R exerted on the block by the incline. You must determine the direction of R in each separate case. Note that, since P is per-pendicular to the 800-N force, the force triangle is a right tri-angle, which easily can be solved for P. In most other problems, however, the force triangle will be an oblique triangle and should be solved by applying the law of sines.
ANALYSIS: a. Force P to Start Block Moving Up. In this case, motion is impending up the incline, so the resultant is directed at the angle of static friction (Fig. 1). Note that the resultant is oriented to the left of the normal such that its friction compo-nent (not shown) is directed opposite the direction of impending motion.
P 5 (800 N) tan 44.29° P 5 780 Nz b b. Force P to Keep Block Moving Up. Motion is continuing, so the resultant is directed at the angle of kinetic friction (Fig. 2). Again, the resultant is oriented to the left of the normal such that its friction component is directed opposite the direction of motion.
P 5 (800 N) tan 39.04° P 5 649 Nz b c. Force P to Prevent Block from Sliding Down. Here, motion is impending down the incline, so the resultant is directed at the angle of static friction (Fig. 3). Note that the resultant is oriented to the right of the normal such that its fric-tion component is directed opposite the direction of impending motion. P 5 (800 N) tan 5.71° P 5 80.0 Nz b REFLECT and THINK: As expected, considerably more force is required to begin moving the block up the slope than is necessary to restrain it from sliding down the slope.
800 N 25° P fs tan fs = ms 25° + 19.29° = 44.29° fs = 19.29° = 0.35 800 N 800 N 25° P R P R Fig. 1 Free-body diagram of block and its force triangle—motion impending up incline.
tan fk = mk fk 25° + 14.04° = 39.04° fk = 14.04° = 0.25 P R 800 N 800 N 25° P R Fig. 2 Free-body diagram of block and its force triangle—motion continuing up incline.
25° – 19.29° = 5.71° fs = 19.29° P R fs 800 N 800 N 25° P R Fig. 3 Free-body diagram of block and its force triangle—motion prevented down the slope.
438 Friction Sample Problem 8.3 The movable bracket shown may be placed at any height on the 3-in.- diameter pipe. If the coefficient of static friction between the pipe and bracket is 0.25, determine the minimum distance x at which the load W can be supported. Neglect the weight of the bracket.
STRATEGY: In this variation of the third type of friction problem, you know the coefficient of static friction and that motion is impending. Since the problem involves consideration of resistance to rotation, you should apply both moment equilibrium and force equilibrium.
MODELING: Free-Body Diagram. Draw the free-body diagram of the bracket (Fig. 1). When W is placed at the minimum distance x from the axis of the pipe, the bracket is just about to slip, and the forces of friction at A and B have reached their maximum values: FA 5 µsNA 5 0.25 NA FB 5 µsNB 5 0.25 NB ANALYSIS: Equilibrium Equations.
y 1 oFx 5 0: NB 2 NA 5 0 NB 5 NA 1 xoFy 5 0: FA 1 FB 2 W 5 0 0.25NA 1 0.25NB 5 W Since NB is equal to NA, 0.50NA 5 W NA 5 2W 1loMB 5 0: NA(6 in.) 2 FA(3 in.) 2 W(x 2 1.5 in.) 5 0 6NA 2 3(0.25NA) 2 Wx 1 1.5W 5 0 6(2W) 2 0.75(2W) 2 Wx 1 1.5W 5 0 Dividing through by W and solving for x, you have x 5 12 in. b REFLECT and THINK: In a problem like this, you may not figure out how to approach the solution until you draw the free-body diagram and examine what information you are given and what you need to find. In this case, since you are asked to find a distance, the need to evaluate moment equilibrium should be clear.
W 6 in.
3 in.
x NA NB FA FB W A B 3 in.
x – 1.5 in.
x 6 in.
Fig. 1 Free-body diagram of bracket.
8.1 The Laws of Dry Friction 439 Sample Problem 8.4 An 8400-kg truck is traveling on a level horizontal curve, resulting in an effective lateral force H (applied at the center of gravity G of the truck). Treating the truck as a rigid system with the center of gravity shown, and knowing that the distance between the outer edges of the tires is 1.8 m, determine (a) the maximum force H before tipping of the truck occurs, (b) the minimum coefficient of static friction between the tires and road-way such that slipping does not occur before tipping.
STRATEGY: For the direction of H shown, the truck would tip about the outer edge of the right tire. At the verge of tip, the normal force and friction force are zero at the left tire, and the normal force at the right tire is at the outer edge. You can apply equilibrium to determine the value of H neces-sary for tip and the required friction force such that slipping does not occur.
MODELING: Draw the free-body diagram of the truck (Fig. 1), which reflects impending tip about point B. Obtain the weight of the truck by multiplying its mass of 8400 kg by g 5 9.81 m/s2; that is, W 5 82 400 N or 82.4 kN.
ANALYSIS: Free Body: Truck (Fig. 1).
1loMB 5 0: (82.4 kN)(0.8 m) 2 H(1.4 m) 5 0 H 5 147.1 kN H 5 47.1 kN y b y 1 oFx 5 0: 47.1 kN 2 FB 5 0 FB 5 147.1 kN 1 xoFy 5 0: NB 2 82.4 kN 5 0 NB 5 182.4 kN Minimum Coefficient of Static Friction. The magnitude of the maximum friction force that can be developed is Fm 5 µsNB 5 µs (82.4 kN) Setting this equal to the friction force required, FB 5 47.1 kN, gives µs (82.4 kN) 5 47.1 kN µs 5 0.572 b REFLECT and THINK: Recall from physics that H represents the force due to the centripetal acceleration of the truck (of mass m), and its mag-nitude is H 5 m(v2/ρ) where v 5 velocity of the truck ρ 5 radius of curvature In this problem, if the truck were traveling around a curve of 100-m radius (measured to G), the velocity at which it would begin to tip would be 23.7 m/s (or 85.2 km/h). You will learn more about this aspect in your study of dynamics.
0.8 m G H FB NB 1.4 m B A 82.4 kN 1.8 m G H 1.0 m 1.4 m Fig. 1 Free-body diagram of truck.
440 440 SOLVING PROBLEMS ON YOUR OWN I n this section, you studied and applied the laws of dry friction. Previously, you had encountered only (a) frictionless surfaces that could move freely with respect to each other or (b) rough surfaces that allowed no motion relative to each other.
A. In solving problems involving dry friction, keep the following ideas in mind.
1. The reaction R exerted by a surface on a free body can be resolved into a normal component N and a tangential component F. The tangential component is known as the friction force. When a body is in contact with a fixed surface, the direction of the friction force F is opposite to that of the actual or impending motion of the body. a. No motion will occur as long as F does not exceed the maximum value Fm 5 µsN, where µs is the coefficient of static friction. b. Motion will occur if a value of F larger than Fm is required to maintain equi-librium. As motion takes place, the actual value of F drops to Fk 5 µkN, where µk is the coefficient of kinetic friction [Sample Prob. 8.1]. c. Motion may also occur at a value of F smaller than Fm if tipping of the rigid body is a possibility [Sample Prob. 8.4} 2. When only three forces are involved, you might prefer an alternative approach to the analysis of friction [Sample Prob. 8.2]. The reaction R is defined by its magnitude R and the angle f it forms with the normal to the surface. No motion occurs as long as f does not exceed the maximum value fs, where tan fs 5 µs. Motion does occur if a value of f larger than fs is required to maintain equilibrium, and the actual value of f drops to fk, where tan fk 5 µk.
3. When two bodies are in contact, you must determine the sense of the actual or impending relative motion at the point of contact. On each of the two bodies, a friction force F is in a direction opposite to that of the actual or impending motion of the body as seen from the other body (see Fig. 8.6).
B. Methods of solution. The first step in your solution is to draw a free-body diagram of the body under consideration, resolving the force exerted on each surface where friction exists into a normal component N and a friction force F. If several bodies are involved, draw a free-body diagram for each of them, labeling and directing the forces at each surface of contact, as described for analyzing frames in Chap. 6.
441 441 The problem you have to solve may fall in one of the following three categories.
1. You know all the applied forces and the coefficients of friction, and you must determine whether equilibrium is maintained. In this situation, the friction force is unknown and cannot be assumed to be equal to µsN. a. Write the equations of equilibrium to determine N and F.
b. Calculate the maximum allowable friction force, Fm 5 µsN. If F # Fm, equilibrium is maintained. If F $ Fm, motion occurs, and the magnitude of the friction force is Fk 5 µkN [Sample Prob. 8.1].
2. You know all the applied forces, and you must find the smallest allowable value of µs for which equilibrium is maintained. Assume that motion is impending, and determine the corresponding value of µs. a. Write the equations of equilibrium to determine N and F. b. Since motion is impending, F 5 Fm. Substitute the values found for N and F into the equation Fm 5 µsN and solve for µs [Sample Prob. 8.4].
3. The motion of the body is impending and µs is known; you must find some unknown quantity, such as a distance, an angle, the magnitude of a force, or the direc-tion of a force. a. Assume a possible motion of the body and, on the free-body diagram, draw the friction force in a direction opposite to that of the assumed motion. b. Since motion is impending, F 5 Fm 5 µsN. Substituting the known value for µs, you can express F in terms of N on the free-body diagram, thus eliminating one unknown. c. Write and solve the equilibrium equations for the unknown you seek [Sample Prob. 8.3].
442 Problems FREE-BODY PRACTICE PROBLEMS 8.F1 Knowing that the coefficient of friction between the 25-kg block and the incline is µs 5 0.25, draw the free-body diagram needed to determine both the smallest value of P required to start the block moving up the incline and the corresponding value of β. 8.F2 Two blocks A and B are connected by a cable as shown. Knowing that the coefficient of static friction at all surfaces of contact is 0.30 and neglecting the friction of the pulleys, draw the free-body diagrams needed to determine the smallest force P required to move the blocks.
P A B 60 lb 40 lb Fig. P8.F2 8.F3 A cord is attached to and partially wound around a cylinder with a weight of W and radius r that rests on an incline as shown. Know-ing that θ 5 30°, draw the free-body diagram needed to determine both the tension in the cord and the smallest allowable value of the coefficient of static friction between the cylinder and the incline for which equilibrium is maintained.
25° θ D B A C T Fig. P8.F3 8.F4 A 40-kg packing crate must be moved to the left along the floor without tipping. Knowing that the coefficient of static friction between the crate and the floor is 0.35, draw the free-body diagram needed to determine both the largest allowable value of α and the corresponding magnitude of the force P. 0.5 m 0.8 m B A D C α P Fig. P8.F4 P 30° β 25 kg Fig. P8.F1 443 END-OF-SECTION PROBLEMS 8.1 Determine whether the block shown is in equilibrium, and find the magnitude and direction of the friction force when P 5 150 N.
Fig. P8.1 and P8.2 P ms = 0.30 mk = 0.25 20° 500 N 8.2 Determine whether the block shown is in equilibrium, and find the magnitude and direction of the friction force when P 5 400 N. 8.3 Determine whether the block shown is in equilibrium, and find the magnitude and direction of the friction force when P 5 120 lb. 8.4 Determine whether the block shown is in equilibrium, and find the magnitude and direction of the friction force when P 5 80 lb. 8.5 Determine the smallest value of P required to (a) start the block up the incline, (b) keep it moving up. 8.6 The 20-lb block A hangs from a cable as shown. Pulley C is con-nected by a short link to block E, which rests on a horizontal rail. Knowing that the coefficient of static friction between block E and the rail is µs 5 0.35 and neglecting the weight of block E and the friction in the pulleys, determine the maximum allowable value of θ if the system is to remain in equilibrium. 8.7 The 10-kg block is attached to link AB and rests on a moving belt. Knowing that µs 5 0.30 and µk 5 0.25 and neglecting the weight of the link, determine the magnitude of the horizontal force P that should be applied to the belt to maintain its motion (a) to the left as shown, (b) to the right.
Fig. P8.7 P 10 kg 35° B A 8.8 Considering only values of θ less than 90°, determine the smallest value of θ required to start the block moving to the right when (a) W 5 75 lb, (b) W 5 100 lb.
Fig. P8.3, P8.4, and P8.5 50 lb 30° 40° P ms = 0.40 mk = 0.30 Fig. P8.6 q T 20 lb D C B A E Fig. P8.8 W q mk = 0.20 ms = 0.25 30 lb 444 8.9 Knowing that θ 5 40°, determine the smallest force P for which equilibrium of the 7.5-kg block is maintained. 8.10 Knowing that P 5 100 N, determine the range of values of θ for which equilibrium of the 7.5-kg block is maintained. 8.11 The 50-lb block A and the 25-lb block B are supported by an incline that is held in the position shown. Knowing that the coefficient of static friction is 0.15 between the two blocks and zero between block B and the incline, determine the value of θ for which motion is impending. 8.12 The 50-lb block A and the 25-lb block B are supported by an incline that is held in the position shown. Knowing that the coefficient of static friction is 0.15 between all surfaces of contact, determine the value of θ for which motion is impending. 8.13 Three 4-kg packages A, B, and C are placed on a conveyor belt that is at rest. Between the belt and both packages A and C, the coeffi-cients of friction are µs 5 0.30 and µk 5 0.20; between package B and the belt, the coefficients are µs 5 0.10 and µk 5 0.08. The packages are placed on the belt so that they are in contact with each other and at rest. Determine which, if any, of the packages will move and the friction force acting on each package.
Fig. P8.13 A B C 15° 4 kg 4 kg 4 kg 8.14 Solve Prob. 8.13 assuming that package B is placed to the right of both packages A and C.
8.15 A uniform crate with a mass of 30 kg must be moved up along the 15° incline without tipping. Knowing that force P is horizontal, determine (a) the largest allowable coefficient of static friction between the crate and the incline, (b) the corresponding magnitude of force P. 8.16 A worker slowly moves a 50-kg crate to the left along a loading dock by applying a force P at corner B as shown. Knowing that the crate starts to tip about edge E of the loading dock when a 5 200 mm, determine (a) the coefficient of kinetic friction between the crate and the loading dock, (b) the corresponding magnitude P of the force.
Fig. P8.16 P A C B D 1.2 m 0.9 m 15° a E Fig. P8.11 and P8.12 B q A Fig. P8.15 A B C D 15° P L L Fig. P8.9 and P8.10 μk = 0.35 μs = 0.45 P 7.5 kg θ 445 8.17 A half-section of pipe weighing 200 lb is pulled by a cable as shown. The coefficient of static friction between the pipe and the floor is 0.40. If α 5 30°, determine (a) the tension T required to move the pipe, (b) whether the pipe will slide or tip.
Fig. P8.17 a T B A 8.18 A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. Assuming that the casters at both A and B are locked, determine (a) the force P required to move the cabinet to the right, (b) the largest allowable value of h if the cabinet is not to tip over. 8.19 Wire is being drawn at a constant rate from a spool by applying a vertical force P to the wire as shown. The spool and the wire wrapped on the spool have a combined weight of 20 lb. Knowing that the coefficients of friction at both A and B are µs 5 0.40 and µk 5 0.30, determine the required magnitude of force P. 8.20 Solve Prob. 8.19 assuming that the coefficients of friction at B are zero. 8.21 The cylinder shown has a weight W and radius r. Express in terms of W and r the magnitude of the largest couple M that can be applied to the cylinder if it is not to rotate, assuming the coefficient of static friction to be (a) zero at A and 0.30 at B, (b) 0.25 at A and 0.30 at B.
A B M Fig. P8.21 and P.22 8.22 The cylinder shown has a weight W and radius r, and the coefficient of static friction µs is the same at A and B. Determine the magnitude of the largest couple M that can be applied to the cylinder if it is not to rotate. 8.23 and 8.24 End A of a slender, uniform rod with a length of L and weight W bears on a surface as shown, while end B is supported by a cord BC. Knowing that the coefficients of friction are µs 5 0.40 and µk 5 0.30, determine (a) the largest value of θ for which motion is impending, (b) the corresponding value of the tension in the cord.
A B P 3 in.
3 in.
Fig. P8.19 C A B P h 24 in.
Fig. P8.18 L L B C A q Fig. P8.23 A C B L L q Fig. P8.24 446 8.25 A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient of static friction µs is zero at B, determine the smallest value of µs at A for which equilibrium is maintained. 8.26 A 6.5-m ladder AB leans against a wall as shown. Assuming that the coefficient of static friction µs is the same at A and B, determine the smallest value of µs for which equilibrium is maintained. 8.27 The press shown is used to emboss a small seal at E. Knowing that the coefficient of static friction between the vertical guide and the embossing die D is 0.30, determine the force exerted by the die on the seal.
A B C D E 20° 60° 15° 250 N 400 mm 200 mm Fig. P8.27 8.28 The machine base shown has a mass of 75 kg and is fitted with skids at A and B. The coefficient of static friction between the skids and the floor is 0.30. If a force P with a magnitude of 500 N is applied at corner C, determine the range of values of θ for which the base will not move. 8.29 The 50-lb plate ABCD is attached at A and D to collars that can slide on the vertical rod. Knowing that the coefficient of static friction is 0.40 between both collars and the rod, determine whether the plate is in equilibrium in the position shown when the magnitude of the vertical force applied at E is (a) P 5 0, (b) P 5 20 lb.
A B C D E G P 50 lb 5 ft 2 ft 3 ft Fig. P8.29 and P8.30 8.30 In Prob. 8.29, determine the range of values of the magnitude P of the vertical force applied at E for which the plate will move downward.
A B 6 m 2.5 m Fig. P8.25 and P8.26 A B C G q 0.4 m P 0.8 m 0.5 m 0.6 m Fig. P8.28 447 8.31 A window sash weighing 10 lb is normally supported by two 5-lb sash weights. Knowing that the window remains open after one sash cord has broken, determine the smallest possible value of the coef-ficient of static friction. (Assume that the sash is slightly smaller than the frame and will bind only at points A and D.) A B C D 27 in.
36 in.
Fig. P8.31 8.32 A 500-N concrete block is to be lifted by the pair of tongs shown. Determine the smallest allowable value of the coefficient of static friction between the block and the tongs at F and G. 8.33 A pipe with a diameter of 60 mm is gripped by the stillson wrench shown. Portions AB and DE of the wrench are rigidly attached to each other, and portion CF is connected by a pin at D. If the wrench is to grip the pipe and be self-locking, determine the required mini-mum coefficients of friction at A and C.
A B C D E 60 mm 15 mm 50 mm F P 500 mm Fig. P8.33 8.34 A safety device used by workers climbing ladders fixed to high struc-tures consists of a rail attached to the ladder and a sleeve that can slide on the flange of the rail. A chain connects the worker’s belt to the end of an eccentric cam that can be rotated about an axle attached to the sleeve at C. Determine the smallest allowable common value of the coefficient of static friction between the flange of the rail, the pins at A and B, and the eccentric cam if the sleeve is not to slide down when the chain is pulled vertically downward.
A B C D E F G 90 mm 90 mm 45 mm 500 N 45 mm 75 mm 105 mm 360 mm 500 N 315 mm Fig. P8.32 P 0.8 in.
B D C A E 4 in.
3 in.
4 in.
6 in.
Fig. P8.34 448 8.35 To be of practical use, the safety sleeve described in Prob. 8.34 must be free to slide along the rail when pulled upward. Determine the largest allowable value of the coefficient of static friction between the flange of the rail and the pins at A and B if the sleeve is to be free to slide when pulled as shown in the figure. Assume (a) θ 5 60°, (b) θ 5 50°, (c) θ 5 40°. 8.36 Two 10-lb blocks A and B are connected by a slender rod of negligible weight. The coefficient of static friction is 0.30 between all surfaces of contact, and the rod forms an angle θ 5 30° with the vertical. (a) Show that the system is in equilibrium when P 5 0. (b) Deter-mine the largest value of P for which equilibrium is maintained.
A B W = 10 lb W = 10 lb P q Fig. P8.36 8.37 A 1.2-m plank with a mass of 3 kg rests on two joists. Knowing that the coefficient of static friction between the plank and the joists is 0.30, determine the magnitude of the horizontal force required to move the plank when (a) a 5 750 mm, (b) a 5 900 mm. 8.38 Two identical uniform boards, each with a weight of 40 lb, are temporarily leaned against each other as shown. Knowing that the coefficient of static friction between all surfaces is 0.40, determine (a) the largest magnitude of the force P for which equilibrium will be maintained, (b) the surface at which motion will impend.
4 ft A C D B P 6 ft 6 ft 8 ft Fig. P8.38 8.39 Two rods are connected by a collar at B. A couple MA with a mag-nitude of 15 N∙m is applied to rod AB. Knowing that the coefficient of static friction between the collar and the rod is 0.30, determine the largest couple MC for which equilibrium will be maintained. 8.40 In Prob. 8.39, determine the smallest couple MC for which equilib-rium will be maintained.
P B D A E 4 in.
4 in.
3 in.
C θ Fig. P8.35 P C a b B A L = 1.2 m Fig. P8.37 C B A 325 mm 100 mm 200 mm MA MC Fig. P8.39 449 8.41 A 10-ft beam, weighing 1200 lb, is to be moved to the left onto the platform as shown. A horizontal force P is applied to the dolly, which is mounted on frictionless wheels. The coefficients of friction between all surfaces are µs 5 0.30 and µk 5 0.25, and initially, x 5 2 ft. Knowing that the top surface of the dolly is slightly higher than the platform, determine the force P required to start moving the beam. (Hint: The beam is supported at A and D.) 8.42 (a) Show that the beam of Prob. 8.41 cannot be moved if the top surface of the dolly is slightly lower than the platform. (b) Show that the beam can be moved if two 175-lb workers stand on the beam at B, and determine how far to the left the beam can be moved. 8.43 Two 8-kg blocks A and B resting on shelves are connected by a rod of negligible mass. Knowing that the magnitude of a horizontal force P applied at C is slowly increased from zero, determine the value of P for which motion occurs and what that motion is when the coefficient of static friction between all surfaces is (a) µs 5 0.40, (b) µs 5 0.50. 8.44 A slender steel rod with a length of 225 mm is placed inside a pipe as shown. Knowing that the coefficient of static friction between the rod and the pipe is 0.20, determine the largest value of θ for which the rod will not fall into the pipe.
A B q 75 mm Fig. P8.44 8.45 In Prob. 8.44, determine the smallest value of θ for which the rod will not fall out of the pipe. 8.46 Two slender rods of negligible weight are pin-connected at C and attached to blocks A and B, each with a weight W. Knowing that θ 5 80° and that the coefficient of static friction between the blocks and the horizontal surface is 0.30, determine the largest value of P for which equilibrium is maintained. 8.47 Two slender rods of negligible weight are pin-connected at C and attached to blocks A and B, each with a weight W. Knowing that P 5 1.260W and that the coefficient of static friction between the blocks and the horizontal surface is 0.30, determine the range of values of θ between 0 and 180° for which equilibrium is maintained.
100 mm 200 mm 25° 8 kg 8 kg C A B P Fig. P8.43 A C B W W 30° P 60° q Fig. P8.46 and P8.47 x 2 ft C D P 10 ft A B Fig. P8.41 450 Friction 8.2 WEDGES AND SCREWS Friction is a key element in analyzing the function and operation of several types of simple machines. Here we examine the wedge and the screw, which are both extensions of the inclined plane we analyzed in Sect. 8.1.
8.2A Wedges Wedges are simple machines used to raise large stone blocks and other heavy loads. These loads are raised by applying to the wedge a force usually considerably smaller than the weight of the load. In addition, because of the friction between the surfaces in contact, a properly shaped wedge remains in place after being forced under the load. In this way, you can use a wedge advantageously to make small adjustments in the position of heavy pieces of machinery.
Consider the block A shown in Fig. 8.7a. This block rests against a vertical wall B, and we want to raise it slightly by forcing a wedge C between block A and a second wedge D. We want to find the minimum value of the force P that we must apply to wedge C to move the block. We assume that we know the weight W of the block, which is either given in pounds or determined in newtons from the mass of the block expressed in kilograms.
We have drawn the free-body diagrams of block A and wedge C in Fig. 8.7b and c. The forces acting on the block include its weight and the normal and friction forces at the surfaces of contact with wall B and wedge C. The magnitudes of the friction forces F1 and F2 are equal, respectively, to µsN1 and µsN2, because the motion of the block must be started. It is important to show the friction forces with their correct sense. Since the block will move upward, the force F1 exerted by the wall on the block must be directed downward. On the other hand, since wedge C moves to the right, the relative motion of A with respect to C is to the left, and the force F2 exerted by C on A must be directed to the right.
Now consider the free body C in Fig. 8.7c. The forces acting on C include the applied force P and the normal and friction forces at the surfaces of contact with A and D. The weight of the wedge is small compared with the other forces involved and can be neglected. The forces exerted by A on C are equal and opposite to the forces N2 and F2 exerted by C on A, so we denote them, respectively, by 2N2 and 2F2; the friction force 2F2 therefore must be directed to the left. We check that the force F3 exerted by D is also directed to the left.
We can reduce the total number of unknowns involved in the two free-body diagrams to four if we express the friction forces in terms of the normal forces. Then, since block A and wedge C are in equilibrium, we obtain four equations that we can solve to obtain the magnitude of P. Note that, in the example considered here, it is more convenient to replace each pair of normal and friction forces by their resultant. Each free body is then subjected to only three forces, and we can solve the problem by drawing the corresponding force triangles (see Sample Prob. 8.5).
8.2B Square-Threaded Screws Square-threaded screws are frequently part of jacks, presses, and other mechanisms. Their analysis is similar to the analysis of a block sliding along an inclined plane. (Screws are also commonly used as fasteners, but the threads on these screws are shaped differently.) Photo 8.3 Wedges are used as shown to split tree trunks because the normal forces exerted by a wedge on the wood are much larger than the force required to insert the wedge.
Fig. 8.7 (a) A wedge C used to raise a block A; (b) free-body diagram of block A; (c) free-body diagram of wedge C. Note the directions of the friction forces.
W P N2 (b) (a) (c) A A B C P C D N1 N3 –N2 –F2 F1 = msN1 F2 = msN2 F3 = msN3 6° 6° 8.2 Wedges and Screws 451 Consider the jack shown in Fig. 8.8. The screw carries a load W and is supported by the base of the jack. Contact between the screw and the base takes place along a portion of their threads. By applying a force P on the handle, the screw can be made to turn and to raise the load W.
In Fig. 8.9a, we have unwrapped the thread of the base and shown it as a straight line. We obtained the correct slope by horizontally drawing the product 2πr, where r is the mean radius of the thread, and vertically drawing the lead L of the screw, i.e., the distance through which the screw advances in one turn. The angle θ this line forms with the horizontal is the lead angle. Since the force of friction between two surfaces in contact does not depend upon the area of contact, we can assume a much smaller than actual area of contact between the two threads, which allows us to represent the screw as the block shown in Fig. 8.9a. Note that, in this analysis of the jack, we neglect the small friction force between cap and screw.
The free-body diagram of the block includes the load W, the reac-tion R of the base thread, and a horizontal force Q, which has the same effect as the force P exerted on the handle. The force Q should have the same moment as P about the axis of the screw, so its magnitude should be Q 5 Pa/r. We can obtain the value of force Q, and thus that of force P required to raise load W, from the free-body diagram shown in Fig. 8.9a. The friction angle is taken to be equal to fs , since presumably the load is raised through a succession of short strokes. In mechanisms providing for the continuous rotation of a screw, it may be desirable to distinguish between the force required to start motion (using fs) and that required to maintain motion (using fk).
If the friction angle fs is larger than the lead angle θ, the screw is said to be self-locking; it will remain in place under the load. To lower the load, we must then apply the force shown in Fig. 8.9b. If fs is smaller than θ, the screw will unwind under the load; it is then necessary to apply the force shown in Fig. 8.9c to maintain equilibrium.
The lead of a screw should not be confused with its pitch. The lead is defined as the distance through which the screw advances in one turn; the pitch is the distance measured between two consecutive threads. Lead and pitch are equal in the case of single-threaded screws, but they are different in the case of multiple-threaded screws, i.e., screws having several independent threads. It is easily verified that for double-threaded screws the lead is twice as large as the pitch; for triple-threaded screws, it is three times as large as the pitch; etc.
Fig. 8.8 A screw as part of a jack carrying a load W.
Cap Screw Base P W r a Fig. 8.9 Block-and-incline analysis of a screw. We can represent the screw as a block, because the force of friction does not depend on the area of contact between two surfaces.
(c) Impending motion downward with fs < q (b) Impending motion downward with fs > q (a) Impending motion upward fs Q W R q q fs Q W R q q fs Q W R q q L 2 r Photo 8.4 An example of a square-threaded screw, fitted to a sleeve, as might be used in an industrial application.
Pitch 452 Friction 19.3° 19.3° 11.3° 11.3° P P R3 R3 R1 = 381 lb 381 lb 90° – 19.3° = 70.7° 19.3° + 11.3° = 30.6° A Fig. 4 19.3° 19.3° 11.3° 11.3° P P R3 R3 R1 = 381 lb 381 lb 90° – 19.3° = 70.7° 19.3° + 11.3° = 30.6° A Fig. 4 Sample Problem 8.5 The position of the machine block B is adjusted by moving the wedge A. Knowing that the coefficient of static friction is 0.35 between all surfaces of contact, determine the force P required to (a) raise block B, (b) lower block B.
STRATEGY: For both parts of the problem, normal forces and friction forces act between the wedge and the block. In part (a), you also have normal and friction forces at the left surface of the block; for part (b), they are on the right surface of the block. If you combine the normal and friction forces at each surface into resultants, you have a total of three forces acting on each body and can use force triangles to solve.
400 lb P B A 8° φs = 19.3° φs = 19.3° R2 R2 R1 R1 400 lb 400 lb 8° 8° + 19.3° = 27.3° 27.3° 90° + 19.3° = 109.3° 180° – 27.3° – 109.3° = 43.4° B Fig. 1 Free-body diagram of block and its force triangle—block being raised.
19.3° 19.3° R3 R3 P P R1 = 549 lb 549 lb 27.3° 27.3° 90° – 19.3° = 70.7° 27.3° + 19.3° = 46.6° A Fig. 2 Free-body diagram of wedge and its force triangle—block being raised.
φs = 19.3° φs = 19.3° 11.3° R2 R2 R1 R1 400 lb 400 lb 8° 90° – 19.3° = 70.7° 19.3° – 8° = 11.3° 180° – 70.7° – 11.3° = 98.0° B Fig. 3 Free-body diagram of block and its force triangle—block being lowered.
MODELING: For each part, draw the free-body diagrams of block B and wedge A together with the corresponding force triangles. Then use the law of sines to find the desired forces. Note that, since µs 5 0.35, the angle of friction is fs 5 tan21 0.35 5 19.3° ANALYSIS: a. Force P to raise block Free Body: Block B (Fig. 1). The friction force on block B due to wedge A is to the left, so the resul-tant R1 is at an angle equal to the slope of the wedge plus the angle of friction. R1 sin 109.38 5 400 lb sin 43.48 R1 5 549 lb Free Body: Wedge A (Fig. 2). The friction forces on wedge A are to the right.
P sin 46.68 5 549 lb sin 70.78 P 5 423 lb z b b. Force P to lower block Free Body: Block B (Fig. 3). Now the friction force on block B due to wedge A is to the right, so the resultant R1 is at an angle equal to the angle of friction minus the slope of the wedge. R1 sin 70.78 5 400 lb sin 98.08 R1 5 381 lb Free Body: Wedge A (Fig. 4). The friction forces on wedge A are to the left. P sin 30.68 5 381 lb sin 70.78 P 5 206 lb y b REFLECT and THINK: The force needed to lower the block is much less than the force needed to raise the block, which makes sense.
19.3° 19.3° 11.3° 11.3° P P R3 R3 R1 = 381 lb 381 lb 90° – 19.3° = 70.7° 19.3° + 11.3° = 30.6° A Fig. 4 Free-body diagram of wedge and its force triangle—block being lowered.
8.2 Wedges and Screws 453 Sample Problem 8.6 A clamp is used to hold two pieces of wood together as shown. The clamp has a double square thread with a mean diameter of 10 mm and a pitch of 2 mm. The coefficient of friction between threads is µs 5 0.30. If a maximum couple of 40 N?m is applied in tightening the clamp, determine (a) the force exerted on the pieces of wood, (b) the couple required to loosen the clamp.
STRATEGY: If you represent the screw by a block, as in the analysis of this section, you can determine the incline of the screw from the geo-metry given in the problem, and you can find the force applied to the block by setting the moment of that force equal to the applied couple.
MODELING and ANALYSIS: a. Force Exerted by Clamp. The mean radius of the screw is r 5 5 mm. Since the screw is double-threaded, the lead L is equal to twice the pitch: L 5 2(2 mm) 5 4 mm. Obtain the lead angle θ and the friction angle fs from tan θ 5 L 2πr 5 4 mm 10π mm 5 0.1273 θ 5 7.38 tan fs 5 µs 5 0.30 fs 5 16.7° You can find the force Q that should be applied to the block representing the screw by setting its moment Qr about the axis of the screw equal to the applied couple. Q(5 mm)5 40 N?m Q 5 40 N?m 5 mm 5 40 N?m 5 3 1023 m 5 8000 N 5 8 kN Now you can draw the free-body diagram and the corresponding force triangle for the block (Fig. 1). Solve the triangle to find the magnitude of the force W exerted on the pieces of wood.
W 5 Q tan(θ 1 ϕs) 5 8 kN tan 24.08 W 5 17.97 kN b b. Couple Required to Loosen Clamp. You can obtain the force Q required to loosen the clamp and the corresponding couple from the free-body diagram and force triangle shown in Fig. 2. Q 5 W tan (fs 2 θ) 5 (17.97 kN) tan 9.4° 5 2.975 kN Couple 5 Qr 5 (2.975 kN)(5 mm) 5 (2.975 3 103 N)(5 3 1023 m) 5 14.87 N?m Couple 5 14.87 N?m b REFLECT and THINK: In practice, you often have to determine the force effectively acting on a screw by setting the moment of that force about the axis of the screw equal to an applied couple. However, the rest of the analy-sis is mostly an application of dry friction. Also note that the couple required to loosen a screw is not the same as the couple required to tighten it.
Fig. 1 Free-body diagram of block and its force triangle—clamp being tightened.
fs = 16.7° Q = 8 kN q = 7.3° q = 7.3° R W L = 4 mm 2 r = 10 mm q + fs = 24.0° Q = 8 kN R W Fig. 2 Free-body diagram of block and its force triangle—clamp being loosened.
W = 17.97 kN Q fs = 16.7° q = 7.3° q = 7.3° R L = 4 mm 2 r = 10 mm fs – q = 9.4° Q R W = 17.97 kN 454 454 SOLVING PROBLEMS ON YOUR OWN I n this section, you saw how to apply the laws of friction to the solution of problems involving wedges and square-threaded screws.
1. Wedges. Keep the following steps in mind when solving a problem involving a wedge. a. First draw a free-body diagram of the wedge and of each of the other bodies involved. Carefully note the sense of the relative motion of all surfaces of contact and show each friction force acting in a direction opposite to the direction of that relative motion. b. Show the maximum static friction force Fm at each surface if the wedge is to be inserted or removed, since motion will be impending in each of these cases. c. The reaction R and the angle of friction, rather than the normal force and the friction force, are most useful in many applications. You can then draw one or more force triangles and determine the unknown quantities either graphically or by trigonometry [Sample Prob. 8.5].
2. Square-Threaded Screws. The analysis of a square-threaded screw is equivalent to the analysis of a block sliding on an incline. To draw the appropriate incline, you need to unwrap the thread of the screw and represent it as a straight line [Sample Prob. 8.6]. When solving a problem involving a square-threaded screw, keep the following steps in mind. a. Do not confuse the pitch of a screw with the lead of a screw. The pitch of a screw is the distance between two consecutive threads, whereas the lead of a screw is the distance the screw advances in one full turn. The lead and the pitch are equal only in single-threaded screws. In a double-threaded screw, the lead is twice the pitch. b. The couple required to tighten a screw is different from the couple required to loosen it. Also, screws used in jacks and clamps are usually self-locking; that is, the screw will remain stationary as long as no couple is applied to it, and a couple must be applied to the screw to loosen it [Sample Prob. 8.6].
455 Problems 8.48 The machine part ABC is supported by a frictionless hinge at B and a 10° wedge at C. Knowing that the coefficient of static friction is 0.20 at both surfaces of the wedge, determine (a) the force P required to move the wedge to the left, (b) the components of the correspond-ing reaction at B. 8.49 Solve Prob. 8.48 assuming that the wedge is moved to the right. 8.50 and 8.51 Two 8° wedges of negligible weight are used to move and position the 800-kg block. Knowing that the coefficient of static friction is 0.30 at all surfaces of contact, determine the smallest force P that should be applied as shown to one of the wedges.
8° 8° 800 kg P Fig. P8.50 8° 8° 800 kg P Fig. P8.51 8.52 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The base plate CD has been welded to the lower flange of the beam, and the end reaction of the beam is known to be 18 kips. The coefficient of static friction is 0.30 between two steel surfaces and 0.60 between steel and concrete. If horizontal motion of the beam is prevented by the force Q, determine (a) the force P required to raise the beam, (b) the corresponding force Q. 8.53 Solve Prob. 8.52 assuming that the end of the beam is to be lowered. 8.54 Block A supports a pipe column and rests as shown on wedge B. Knowing that the coefficient of static friction at all surfaces of contact is 0.25 and that θ 5 45°, determine the smallest force P required to raise block A. 8.55 Block A supports a pipe column and rests as shown on wedge B. Knowing that the coefficient of static friction at all surfaces of contact is 0.25 and that θ 5 45°, determine the smallest force P for which equilibrium is maintained. 8.56 Block A supports a pipe column and rests as shown on wedge B. The coefficient of static friction at all surfaces of contact is 0.25. If P 5 0, determine (a) the angle θ for which sliding is impending, (b) the corresponding force exerted on the block by the vertical wall.
600 N P 200 mm 250 mm 10° C D B A Fig. P8.48 and P8.49 18 kips 12° Q P A B C D E F Fig. P8.52 3 kN A B P q Fig. P8.54, P8.55, and P8.56 456 8.57 A wedge A of negligible weight is to be driven between two 100-lb blocks B and C resting on a horizontal surface. Knowing that the coefficient of static friction between all surfaces of contact is 0.35, determine the smallest force P required to start moving the wedge (a) if the blocks are equally free to move, (b) if block C is securely bolted to the horizontal surface.
C B 4 in.
0.75 in.
0.75 in.
A P Fig. P8.57 8.58 A 15° wedge is forced into a saw cut to prevent binding of the circular saw. The coefficient of static friction between the wedge and the wood is 0.25. Knowing that a horizontal force P with a magni-tude of 30 lb was required to insert the wedge, determine the mag-nitude of the forces exerted on the board by the wedge after insertion. 8.59 A 12° wedge is used to spread a split ring. The coefficient of static friction between the wedge and the ring is 0.30. Knowing that a force P with a magnitude of 120 N was required to insert the wedge, determine the magnitude of the forces exerted on the ring by the wedge after insertion. 8.60 The spring of the door latch has a constant of 1.8 lb/in. and in the position shown exerts a 0.6-lb force on the bolt. The coefficient of static friction between the bolt and the strike plate is 0.40; all other surfaces are well lubricated and may be assumed frictionless. Deter-mine the magnitude of the force P required to start closing the door.
A B C 3 8 in.
1 2 in.
P 45° Fig. P8.60 8.61 In Prob. 8.60, determine the angle that the face of the bolt near B should form with line BC if the force P required to close the door is to be the same for both the position shown and the position when B is almost at the strike plate.
7.5° P Fig. P8.58 12° P Fig. P8.59 457 8.62 A 5° wedge is to be forced under a 1400-lb machine base at A. Knowing that the coefficient of static friction at all surfaces is 0.20, (a) determine the force P required to move the wedge, (b) indicate whether the machine base will move. 8.63 Solve Prob. 8.62 assuming that the wedge is to be forced under the machine base at B instead of A. 8.64 A 15° wedge is forced under a 50-kg pipe as shown. The coefficient of static friction at all surfaces is 0.20. (a) Show that slipping will occur between the pipe and the vertical wall. (b) Determine the force P required to move the wedge. 8.65 A 15° wedge is forced under a 50-kg pipe as shown. Knowing that the coefficient of static friction at both surfaces of the wedge is 0.20, determine the largest coefficient of static friction between the pipe and the vertical wall for which slipping will occur at A. 8.66 A 200-N block rests as shown on a wedge of negligible weight. The coefficient of static friction µs is the same at both surfaces of the wedge, and friction between the block and the vertical wall may be neglected. For P 5 100 N, determine the value of µs for which motion is impending. (Hint: Solve the equation obtained by trial and error.) P 200 N 15° A B Fig. P8.66 8.67 Solve Prob. 8.66 assuming that the rollers are removed and that µs is the coefficient of friction at all surfaces of contact. 8.68 Derive the following formulas relating the load W and the force P exerted on the handle of the jack discussed in Sec. 8.2B. (a) P 5 (Wr/a) tan (θ 1 fs) to raise the load; (b) P 5 (Wr/a) tan (fs 2 θ) to lower the load if the screw is self-locking; (c) P 5 (Wr/a) tan (θ 2 fs) to hold the load if the screw is not self-locking. 8.69 The square-threaded worm gear shown has a mean radius of 2 in. and a lead of 0.5 in. The large gear is subjected to a constant clock-wise couple of 9.6 kip∙in. Knowing that the coefficient of static friction between the two gears is 0.12, determine the couple that must be applied to shaft AB in order to rotate the large gear coun-terclockwise. Neglect friction in the bearings at A, B, and C. 8.70 In Prob. 8.69, determine the couple that must be applied to shaft AB in order to rotate the large gear clockwise.
G P 50 in.
20 in.
B A 1400 lb Fig. P8.62 P A B G 15° Fig. P8.64 and P8.65 16 in.
9.6 kip⋅in.
A B C Fig. P8.69 458 8.71 High-strength bolts are used in the construction of many steel structures. For a 1-in.-nominal-diameter bolt, the required minimum bolt tension is 51 kips. Assuming the coefficient of friction to be 0.30, determine the required couple that should be applied to the bolt and nut. The mean diameter of the thread is 0.94 in., and the lead is 0.125 in. Neglect friction between the nut and washer, and assume the bolt to be square-threaded. 8.72 The position of the automobile jack shown is controlled by a screw ABC that is single-threaded at each end (right-handed thread at A, left-handed thread at C). Each thread has a pitch of 2.5 mm and a mean diameter of 9 mm. If the coefficient of static friction is 0.15, determine the magnitude of the couple M that must be applied to raise the automobile.
A B C D E M 20° 20° 6000 N Fig. P8.72 8.73 For the jack of Prob. 8.72, determine the magnitude of the couple M that must be applied to lower the automobile. 8.74 The vise shown consists of two members connected by two double-threaded screws with a mean radius of 0.25 in. and pitch of 0.08 in. The lower member is threaded at A and B (µs 5 0.35), but the upper member is not threaded. It is desired to apply two equal and opposite forces of 120 lb on the blocks held between the jaws. (a) What screw should be adjusted first? (b) What is the maximum couple applied in tightening the second screw?
8.75 The ends of two fixed rods A and B are each made in the form of a single-threaded screw with a mean radius of 6 mm and pitch of 2 mm. Rod A has a right-handed thread, and rod B has a left-handed thread. The coefficient of static friction between the rods and the threaded sleeve is 0.12. Determine the magnitude of the couple that must be applied to the sleeve in order to draw the rods closer together.
2 kN 2 kN A B Fig. P8.75 8.76 Assuming that in Prob. 8.75 a right-handed thread is used on both rods A and B, determine the magnitude of the couple that must be applied to the sleeve in order to rotate it.
Fig. P8.71 5 in.
A B 5 in.
Fig. P8.74 8.3 Friction on Axles, Disks, and Wheels 459 8.3 FRICTION ON AXLES, DISKS, AND WHEELS Journal bearings are used to provide lateral support to rotating shafts and axles. Thrust bearings are used to provide axial support to shafts and axles. If the journal bearing is fully lubricated, the frictional resis-tance depends upon the speed of rotation, the clearance between axle and bearing, and the viscosity of the lubricant. As indicated in Sec. 8.1, such problems are studied in fluid mechanics. However, we can apply the methods of this chapter to the study of axle friction when the bear-ing is not lubricated or only partially lubricated. In this case, we can assume that the axle and the bearing are in direct contact along a single straight line.
8.3A Journal Bearings and Axle Friction Consider two wheels, each with a weight of W, rigidly mounted on an axle supported symmetrically by two journal bearings (Fig. 8.10a). If the wheels rotate, we find that, to keep them rotating at constant speed, it is Fig. 8.10 (a) Two wheels supported by two journal bearings; (b) point of contact when the axle is rotating; (c) free-body diagram of one wheel and corresponding half axle; (d) frictional resistance produces a couple that opposes the couple maintaining the axle in motion; (e) graphical analysis with circle of friction.
φ –M W R R M W r rf (b) Wheels Journal bearings Axle (a) (c) (d) (e) N F B R r B O O O N F R B A O M W M k φk 460 Friction necessary to apply a couple M to each of them. The free-body diagram in Fig. 8.10c represents one of the wheels and the corresponding half axle in projection on a plane perpendicular to the axle. The forces acting on the free body include the weight W of the wheel, the couple M required to maintain its motion, and a force R representing the reaction of the bearing. This force is vertical, equal, and opposite to W, but it does not pass through the center O of the axle; R is located to the right of O at a distance such that its moment about O balances the moment M of the couple. Therefore, when the axle rotates, contact between the axle and bearing does not take place at the lowest point A. Instead, contact takes place at point B (Fig. 8.10b) or, rather, along a straight line intersecting the plane of the figure at B. Physically, the location of contact is explained by the fact that, when the wheels are set in motion, the axle “climbs” in the bearings until slip-page occurs. After sliding back slightly, the axle settles more or less in the position shown. This position is such that the angle between the reac-tion R and the normal to the surface of the bearing is equal to the angle of kinetic friction fk. The distance from O to the line of action of R is thus r sin fk, where r is the radius of the axle. Setting oMO 5 0 for the forces acting on the free body (the wheel), we obtain the magnitude of the couple M required to overcome the frictional resistance of one of the bearings: M 5 Rr sin fk (8.5) For small values of the angle of friction, we can replace sin fk by tan fk; that is, by µk. This gives us the approximate formula M < R r µ k (8.6) In the solution of certain problems, it may be more convenient to let the line of action of R pass through O, as it does when the axle does not rotate. In such a case, you need to add a couple 2M, with the same magnitude as the couple M but of opposite sense, to the reaction R (Fig. 8.10d). This couple represents the frictional resistance of the bearing.
If a graphical solution is preferred, you can readily draw the line of action of R (Fig. 8.10e) if you note that it must be tangent to a circle centered at O and with a radius rf 5 r sin fk < r µ k (8.7) This circle is called the circle of friction of the axle and bearing, and it is independent of the loading conditions of the axle.
8.3B Thrust Bearings and Disk Friction Two types of thrust bearings are commonly used to provide axial support to rotating shafts and axles: (1) end bearings and (2) collar bearings (Fig. 8.11). In the case of collar bearings, friction forces develop between 8.3 Friction on Axles, Disks, and Wheels 461 Fig. 8.11 In thrust bearings, an axial force keeps the rotating axle in contact with the support bearing.
M M P P (a) End bearing (b) Collar bearing Fig. 8.12 Geometry of the frictional contact surface in a thrust bearing.
M M P R1 R2 ∆N ∆F ∆A r q the two ring-shaped areas in contact. In the case of end bearings, friction takes place over full circular areas or over ring-shaped areas when the end of the shaft is hollow. Friction between circular areas, called disk friction, also occurs in other mechanisms, such as disk clutches.
To obtain a formula for the most general case of disk friction, let us consider a rotating hollow shaft. A couple M keeps the shaft rotating at constant speed, while an axial force P maintains it in contact with a fixed bearing (Fig. 8.12). Contact between the shaft and the bearing takes place over a ring-shaped area with an inner radius of R1 and an outer radius of R2. Assuming that the pressure between the two surfaces in contact is uniform, we find that the magnitude of the normal force DN exerted on an element of area DA is DN 5 P DA/A, where A 5 π (R2 2 2 R2 1) and that the magnitude of the friction force DF acting on DA is DF 5 µk DN. Let’s use r to denote the distance from the axis of the shaft to the element of area DA. Then the magnitude DM of the moment of DF about the axis of the shaft is DM 5 r DF 5 rmkP DA π(R2 2 2 R2 1) Equilibrium of the shaft requires that the moment M of the couple applied to the shaft be equal in magnitude to the sum of the moments of the 462 Friction friction forces DF opposing the motion of the shaft. Replacing DA by the infinitesimal element dA 5 r dθ dr used with polar coordinates and integrating over the area of contact, the expression for the magnitude of the couple M required to overcome the frictional resistance of the bearing is M 5 µkP π(R2 2 2 R2 1)# 2π 0 # R2 R1 r2 dr dθ 5 mkP π(R2 2 2 R2 1)# 2π 0 1 3(R3 2 2 R3 1)dθ M 5 2 3 mkP R3 2 2 R3 1 R2 2 2 R2 1 (8.8) When contact takes place over a full circle with a radius of R, formula (8.8) reduces to M 5 2 3mkPR (8.9) This value of M is the same value we would obtain if contact between shaft and bearing took place at a single point located at a distance 2R/3 from the axis of the shaft.
The largest couple that can be transmitted by a disk clutch without causing slippage is given by a formula similar to Eq. (8.9), where µk has been replaced by the coefficient of static friction µs.
8.3C Wheel Friction and Rolling Resistance The wheel is one of the most important inventions of our civilization. Among many other uses, with a wheel we can move heavy loads with relatively little effort. Because the point where the wheel is in contact with the ground at any given instant has no relative motion with respect to the ground, use of the wheel avoids the large friction forces that would arise if the load were in direct contact with the ground. However, some resis-tance to the wheel’s motion does occur. This resistance has two distinct causes. It is due to (1) a combined effect of axle friction and friction at the rim and (2) the fact that the wheel and the ground deform, causing contact between wheel and ground to take place over an area rather than at a single point.
To understand better the first cause of resistance to the motion of a wheel, consider a railroad car supported by eight wheels mounted on 8.3 Friction on Axles, Disks, and Wheels 463 axles and bearings. We assume the car is moving to the right at constant speed along a straight horizontal track. The free-body diagram of one of the wheels is shown in Fig. 8.13a. The forces acting on the free body include the load W supported by the wheel and the normal reaction N of the track. Since W passes through the center O of the axle, we rep-resent the frictional resistance of the bearing by a counterclockwise couple M (see Sec. 8.3A). Then to keep the free body in equilibrium, we must add two equal and opposite forces P and F, forming a clock-wise couple of moment 2M. The force F is the friction force exerted by the track on the wheel, and P represents the force that should be applied to the wheel to keep it rolling at constant speed. Note that the forces P and F would not exist if there were no friction between the wheel and the track. The couple M representing the axle friction would then be zero; the wheel would slide on the track without turning in its bearing.
The couple M and the forces P and F also reduce to zero when there is no axle friction. For example, a wheel that is not held in bearings but rolls freely and at constant speed on horizontal ground (Fig. 8.13b) is subjected to only two forces: its own weight W and the normal reaction N of the ground. No friction force acts on the wheel regardless of the value of the coefficient of friction between wheel and ground. Thus, a wheel rolling freely on horizontal ground should keep rolling indefinitely.
Experience, however, indicates that a free wheel does slow down and eventually come to rest. This is due to the second type of resistance mentioned at the beginning of this section, known as rolling resistance. Under the load W, both the wheel and the ground deform slightly, caus-ing the contact between wheel and ground to take place over a certain area. Experimental evidence shows that the resultant of the forces exerted by the ground on the wheel over this area is a force R applied at a point B, which is not located directly under the center O of the wheel but slightly in front of it (Fig. 8.13c). To balance the moment of W about B and to keep the wheel rolling at constant speed, it is necessary to apply a horizontal force P at the center of the wheel. Setting oMB 5 0, we obtain Pr 5 Wb (8.10) where r 5 radius of wheel b 5 horizontal distance between O and B The distance b is commonly called the coefficient of rolling resistance. Note that b is not a dimensionless coefficient, since it represents a length; b is usually expressed in inches or in millimeters. The value of b depends upon several parameters in a manner that has not yet been clearly established. Values of the coefficient of rolling resistance vary from about 0.01 in. or 0.25 mm for a steel wheel on a steel rail to 5.0 in. or 125 mm for the same wheel on soft ground.
Fig. 8.13 (a) Free-body diagram of a rolling wheel, showing the effect of axle friction; (b) free-body diagram of a free wheel, not connected to an axle; (c) free-body diagram of a rolling wheel, showing the effect of rolling resistance.
W N O A (b) Free wheel P W O B b R (c) Rolling resistance r M P W F N O A (a) Effect of axle friction 464 Friction Sample Problem 8.7 A pulley with a diameter of 4 in. can rotate about a fixed shaft with a diameter of 2 in. The coefficient of static friction between the pulley and shaft is 0.20. Determine (a) the smallest vertical force P required to start raising a 500-lb load, (b) the smallest vertical force P required to hold the load, (c) the smallest horizontal force P required to start raising the same load.
STRATEGY: You can use the radius of the circle of friction to position the reaction of the pulley in each scenario and then apply the principles of equilibrium.
MODELING and ANALYSIS: a. Vertical Force P Required to Start Raising the Load. When the forces in both parts of the rope are equal, contact between the pulley and shaft takes place at A (Fig. 1). When P is increased, the pulley rolls around the shaft slightly and contact takes place at B. Draw the free-body diagram of the pulley when motion is impending. The perpendicular dis-tance from the center O of the pulley to the line of action of R is rf 5 r sin fs < r µ s rf < (1 in.)0.20 5 0.20 in.
Summing moments about B, you obtain 1l oMB 5 0: (2.20 in.)(500 lb) 2 (1.80 in.)P 5 0 P 5 611 lb P 5 611 lbw b b. Vertical Force P to Hold the Load. As the force P is decreased, the pulley rolls around the shaft, and contact takes place at C (Fig. 2). Considering the pulley as a free body and summing moments about C, you find 1l oMC 5 0: (1.80 in.)(500 lb) 2 (2.20 in.)P 5 0 P 5 409 lb P 5 409 lbw b c. Horizontal Force P to Start Raising the Load. Since the three forces W, P, and R are not parallel, they must be concurrent (Fig. 3). The direction of R is thus determined from the fact that its line of action must pass through the point of intersection D of W and P and must be tangent to the circle of friction. Recall that the radius of the circle of friction is rf 5 0.20 in., so you can calculate the angle marked θ in Fig. 3 as sin θ 5 OE OD 5 0.20 in.
(2 in.)12 5 0.0707 θ 5 4.18 From the force triangle, you can determine P 5 W cot (45° 2 θ) 5 (500 lb) cot 40.9° 5 577 lb P 5 577 lb y b REFLECT and THINK: Many elementary physics problems treat pul-leys as frictionless, but when you do take friction into account, the results can be quite different, depending on the direction of motion, the directions of the forces involved, and especially the coefficient of friction.
Fig. 1 Free-body diagram of pulley—smallest vertical force to raise the load.
W = 500 lb A B O R P fs 2.20 in.
1.80 in.
Fig. 2 Free-body diagram of pulley—smallest vertical force to hold load.
W = 500 lb C A R P 1.80 in.
2.20 in.
O fs Fig. 3 Free-body diagram of pulley and force triangle— smallest horizontal force to raise load.
W = 500 lb D O E R P q rf 45° – q W = 500 lb P R 465 465 SOLVING PROBLEMS ON YOUR OWN I n this section, we described several additional engineering applications of the laws of friction.
1. Journal bearings and axle friction. In journal bearings, the reaction does not pass through the center of the shaft or axle that is being supported. The distance from the center of the shaft or axle to the line of action of the reaction (Fig. 8.10) is defined by rf 5 r sin fk < r µ k if motion is actually taking place. It is defined by rf 5 r sin fs < r µ s if motion is impending. Once you have determined the line of action of the reaction, you can draw a free-body diagram and use the corresponding equations of equilibrium to complete the solution [Sample Prob. 8.7]. In some problems, it is useful to observe that the line of action of the reaction must be tangent to a circle with a radius of rf < r µ k or rf < r µ s, which is known as the circle of friction [Sample Prob. 8.7, part c].
2. Thrust bearings and disk friction. In a thrust bearing, the magnitude of the couple required to overcome frictional resistance is equal to the sum of the moments of the kinetic friction forces exerted on the end of the shaft [Eqs. (8.8) and (8.9)]. An example of disk friction is the disk clutch. It is analyzed in the same way as a thrust bearing, except that to determine the largest couple that can be transmitted you must compute the sum of the moments of the maximum static friction forces exerted on the disk.
3. Wheel friction and rolling resistance. The rolling resistance of a wheel is caused by deformations of both the wheel and the ground. The line of action of the reaction R of the ground on the wheel intersects the ground at a horizontal distance b from the center of the wheel. The distance b is known as the coefficient of rolling resistance and is expressed in inches or millimeters.
4. In problems involving both rolling resistance and axle friction, the free-body dia-gram should show that the line of action of the reaction R of the ground on the wheel is tangent to the friction circle of the axle and intersects the ground at a horizontal distance from the center of the wheel equal to the coefficient of rolling resistance.
466 Problems 8.77 A lever of negligible weight is loosely fitted onto a 75-mm-diameter fixed shaft. It is observed that the lever will just start rotating if a 3-kg mass is added at C. Determine the coefficient of static friction between the shaft and the lever.
A B O D C 150 mm 100 mm 30 kg 20 kg 75 mm O Fig. P8.77 8.78 A hot-metal ladle and its contents weigh 130 kips. Knowing that the coefficient of static friction between the hooks and the pinion is 0.30, determine the tension in cable AB required to start tipping the ladle. 8.79 and 8.80 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the force P required to start raising the load.
T 64 in.
16 in.
B A Fig. P8.78 20 kg P 90 mm 45 mm Fig. P8.79 and P8.81 20 kg P 90 mm 45 mm Fig. P8.80 and P8.82 8.81 and 8.82 The double pulley shown is attached to a 10-mm-radius shaft that fits loosely in a fixed bearing. Knowing that the coefficient of static friction between the shaft and the poorly lubricated bearing is 0.40, determine the magnitude of the smallest force P required to maintain equilibrium.
467 8.83 The block and tackle shown are used to raise a 150-lb load. Each of the 3-in.-diameter pulleys rotates on a 0.5-in.-diameter axle. Know-ing that the coefficient of static friction is 0.20, determine the tension in each portion of the rope as the load is slowly raised. 8.84 The block and tackle shown are used to lower a 150-lb load. Each of the 3-in.-diameter pulleys rotates on a 0.5-in.-diameter axle. Knowing that the coefficient of static friction is 0.20, determine the tension in each portion of the rope as the load is slowly lowered.
8.85 A scooter is to be designed to roll down a 2 percent slope at a con-stant speed. Assuming that the coefficient of kinetic friction between the 25-mm-diameter axles and the bearings is 0.10, determine the required diameter of the wheels. Neglect the rolling resistance between the wheels and the ground. 8.86 The link arrangement shown is frequently used in highway bridge construction to allow for expansion due to changes in temperature. At each of the 60-mm-diameter pins A and B, the coefficient of static friction is 0.20. Knowing that the vertical component of the force exerted by BC on the link is 200 kN, determine (a) the horizontal force that should be exerted on beam BC to just move the link, (b) the angle that the resulting force exerted by beam BC on the link will form with the vertical. 8.87 and 8.88 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating counterclockwise.
2.5 in.
5 in.
B A 50 lb P 2 in.
Fig. P8.87 and P8.89 8.89 and 8.90 A lever AB of negligible weight is loosely fitted onto a 2.5-in.-diameter fixed shaft. Knowing that the coefficient of static friction between the fixed shaft and the lever is 0.15, determine the force P required to start the lever rotating clockwise. 8.91 A loaded railroad car has a mass of 30 Mg and is supported by eight 800-mm-diameter wheels with 125-mm-diameter axles. Knowing that the coefficients of friction are µs 5 0.020 and µk 5 0.015, determine the horizontal force required (a) to start the car moving, (b) to keep the car moving at a constant speed. Neglect rolling resis-tance between the wheels and the rails.
A B C D E F 150 lb TEF Fig. P8.83 and P8.84 500 mm A B C Fig. P8.86 2.5 in.
2 in.
5 in.
50 lb P B A Fig. P8.88 and P8.90 468 8.92 Knowing that a couple of magnitude 30 N∙m is required to start the vertical shaft rotating, determine the coefficient of static friction between the annular surfaces of contact. 8.93 A 50-lb electric floor polisher is operated on a surface for which the coefficient of kinetic friction is 0.25. Assuming that the normal force per unit area between the disk and the floor is uniformly distributed, determine the magnitude Q of the horizontal forces required to pre-vent motion of the machine. 8.94 The frictional resistance of a thrust bearing decreases as the shaft and bearing surfaces wear out. It is generally assumed that the wear is directly proportional to the distance traveled by any given point of the shaft and thus to the distance r from the point to the axis of the shaft. Assuming then that the normal force per unit area is inversely proportional to r, show that the magnitude M of the couple required to overcome the frictional resistance of a worn-out end bearing (with contact over the full circular area) is equal to 75 percent of the value given by Eq. (8.9) for a new bearing. 8.95 Assuming that bearings wear out as indicated in Prob. 8.94, show that the magnitude M of the couple required to overcome the fric-tional resistance of a worn-out collar bearing is M 5 1 2 µk P(R1 1 R2) where P 5 magnitude of the total axial force R1, R2 5 inner and outer radii of collar 8.96 Assuming that the pressure between the surfaces of contact is uni-form, show that the magnitude M of the couple required to overcome frictional resistance for the conical bearing shown is M 5 2 3 mk P sin θ R3 2 2 R3 1 R2 2 2 R2 1 8.97 Solve Prob. 8.93 assuming that the normal force per unit area between the disk and the floor varies linearly from a maximum at the center to zero at the circumference of the disk. 8.98 Determine the horizontal force required to move a 2500-lb automo-bile with 23-in.-diameter tires along a horizontal road at a constant speed. Neglect all forms of friction except rolling resistance, and assume the coefficient of rolling resistance to be 0.05 in. 8.99 Knowing that a 6-in.-diameter disk rolls at a constant velocity down a 2 percent incline, determine the coefficient of rolling resistance between the disk and the incline. 8.100 A 900-kg machine base is rolled along a concrete floor using a series of steel pipes with outside diameters of 100 mm. Knowing that the coefficient of rolling resistance is 0.5 mm between the pipes and the base and 1.25 mm between the pipes and the concrete floor, deter-mine the magnitude of the force P required to slowly move the base along the floor. 8.101 Solve Prob. 8.85 including the effect of a coefficient of rolling resis-tance of 1.75 mm. 8.102 Solve Prob. 8.91 including the effect of a coefficient of rolling resis-tance of 0.5 mm.
M 4 kN 50 mm 120 mm Fig. P8.92 P θ θ R1 R2 M Fig. P8.96 P Fig. P8.100 20 in.
18 in.
Q –Q Fig. P8.93 8.4 Belt Friction 469 8.4 BELT FRICTION Another common application of dry friction concerns belts, which serve many different purposes in engineering, such as transmitting a torque from a lawn mower engine to its wheels. Some of the same analysis affects the design of band brakes and the operation of ropes and pulleys. Consider a flat belt passing over a fixed cylindrical drum (Fig. 8.14a). We want to determine the relation between the values T1 and T2 of the tension in the two parts of the belt when the belt is just about to slide toward the right.
Fig. 8.14 (a) Tensions at the ends of a belt passing over a drum; (b) free-body diagram of an element of the belt, indicating the condition that the belt is about to slip to the right.
P O P' θ β P1 P2 T1 T T2 (a) P O P' (b) x y ∆N ∆F = s∆N T'= T + ∆T 2 2 θ ∆ θ ∆ θ ∆ θ ∆ µ First we detach from the belt a small element PP9 subtending an angle Dθ. Denoting the tension at P by T and the tension at P9 by T 1 DT, we draw the free-body diagram of the element of the belt (Fig. 8.14b). Besides the two forces of tension, the forces acting on the free body are the normal component DN of the reaction of the drum and the friction force DF. Since we assume motion is impending, we have DF 5 µs DN. Note that if Dθ approaches zero, the magnitudes DN and DF and the difference DT between the tension at P and the tension at P9 also approach zero; the value T of the tension at P, however, remains unchanged. This observation helps in understanding our choice of notation.
Choosing the coordinate axes shown in Fig. 8.14b, we can write the equations of equilibrium for the element PP9 as oFx 5 0: (T 1 DT) cos Dθ 2 2 T cos Dθ 2 2 msDN 5 0 (8.11) 470 Friction oFy 5 0: DN 2 (T 1 DT) sin Dθ 2 2 T sin Dθ 2 5 0 (8.12) Solving Eq. (8.12) for DN and substituting into Eq. (8.11), we obtain after reductions DT cos Dθ 2 2 µs(2T 1 DT) sin Dθ 2 5 0 Now we divide both terms by Dθ. For the first term, we do this simply by dividing DT by Dθ. We carry out the division of the second term by dividing the terms in parentheses by 2 and the sine by Dθ/2. The result is DT Dθ cos Dθ 2 2 µs aT 1 DT 2 b sin(Dθ/2) Dθ/2 5 0 If we now let Dθ approach zero, the cosine approaches one and DT/2 approaches zero, as noted above. The quotient of sin (Dθ/2) over Dθ/2 approaches one, according to a lemma derived in all calculus textbooks. Since the limit as Dθ approaches 0 of DT/Dθ is equal to the derivative dT/dθ by definition, we get dT dθ 2 µsT 5 0 dT T 5 µsdθ Now we integrate both members of the last equation from P1 to P2 (see Fig. 8.14a). At P1, we have θ 5 0 and T 5 T1; at P2, we have θ 5 β and T 5 T2. Integrating between these limits, we have # T2 T1 dT T 5 # β 0 ms dθ ln T2 2 ln T1 5 µsβ Noting that the left-hand side is equal to the natural logarithm of the quotient of T2 and T1, this reduces to ln T2 T1 5 µsβ (8.13) We can also write this relation in the form Belt friction, impending slip T2 T1 5 eµsβ (8.14) The formulas we have derived apply equally well to problems involving flat belts passing over fixed cylindrical drums and to problems ln T2 T T1 T 5 µsβ T2 T T1 T 5 eµsβ Fig. 8.14a (repeated) P O P' θ β P1 P2 T1 T2 (a) θ ∆ Photo 8.5 A sailor wraps a rope around the smooth post (called a bollard) in order to control the rope using much less force than the tension in the taut part of the rope.
8.4 Belt Friction 471 involving ropes wrapped around a post or capstan. They also can be used to solve problems involving band brakes. (In this situation, the drum is about to rotate, but the band remains fixed.) The formulas also can be applied to problems involving belt drives. In these problems, both the pulley and the belt rotate; our concern is then to find whether the belt will slip; i.e., whether it will move with respect to the pulley.
Formulas (8.13) and (8.14) should be used only if the belt, rope, or brake is about to slip. Generally, it is easier to use Eq. (8.14) if you need to find T1 or T2; it is preferable to use Eq. (8.13) if you need to find either µs or the angle of contact β. Note that T2 is always larger than T1. T2 therefore represents the tension in that part of the belt or rope that pulls, whereas T1 is the tension in the part that resists. Also observe that the angle of contact β must be expressed in radians. The angle of contact β may be larger than 2π ; for example, if a rope is wrapped n times around a post, β is equal to 2π n.
If the belt, rope, or brake is actually slipping, you should use formulas similar to Eqs. (8.13) and (8.14) involving the coefficient of kinetic friction µk to find the difference in forces. If the belt, rope, or brake is not slipping and is not about to slip, none of these formulas can be used.
The belts used in belt drives are often V-shaped. In the V belt shown in Fig. 8.15a, contact between belt and pulley takes place along the sides of the groove. Again, we can obtain the relation between the values T1 and T2 of the tension in the two parts of the belt when the belt is just about to slip by drawing the free-body diagram of an element of the belt (Fig. 8.15b and c). Formulas similar to Eqs. (8.11) and (8.12) are derived, but the magnitude of the total friction force acting on the element is now 2 DF, and the sum of the y components of the normal forces is 2 DN sin (α/2). Proceeding as previoulsy, we obtain ln T2 T1 5 µsβ sin (α/2) (8.15) or T2 T1 5 eµs β/sin (α/2) (8.16) Fig. 8.15 (a) A V belt lying in the groove of a pulley; (b) free-body diagram of a cross-sectional element of the belt; (c) free-body diagram of a short length of belt.
(a) (b) (c) x y y z α 2 α 2 ∆N ∆N 2 T sin 2 (T + ∆T) sin 2 2 T + ∆T T 2∆F α 2 2∆N sin α ∆ α θ ∆θ ∆θ ∆θ ∆θ 472 Friction Sample Problem 8.8 A hawser (a thick docking rope) thrown from a ship to a pier is wrapped two full turns around a bollard. The tension in the hawser is 7500 N; by exerting a force of 150 N on its free end, a dockworker can just keep the hawser from slipping. (a) Determine the coefficient of friction between the hawser and the bollard. (b) Determine the tension in the hawser that could be resisted by the 150-N force if the hawser were wrapped three full turns around the bollard.
STRATEGY: You are given the difference in forces and the angle of con-tact through which the friction acts. You can insert these data in the equations of belt friction to determine the coefficient of friction, and then you can use the result to determine the ratio of forces in the second situation. MODELING and ANALYSIS: a. Coefficient of Friction. Since slipping of the hawser is impend-ing, we use Eq. (8.13): ln T2 T1 5 ms β Since the hawser is wrapped two full turns around the bollard, you have β 5 2(2π rad) 5 12.57 rad T1 5 150 N T2 5 7500 N Therefore, µsβ 5 ln T2 T1 ms(12.57 rad) 5 ln 7500 N 150 N 5 ln 50 5 3.91 µs 5 0.311 µs 5 0.311 b b. Hawser Wrapped Three Turns Around Bollard. Using the value of µs obtained in part a, you now have (Fig. 1) β 5 3(2π rad) 5 18.85 rad T1 5 150 N µs 5 0.311 Substituting these values into Eq. (8.14), you obtain T2 T1 5 eµs β T2 150 N 5 e(0.311)(18.85) 5 e5.862 5 351.5 T2 5 52 725 N T2 5 52.7 kN b REFLECT and THINK: You can see how the use of a simple post or pulley can have an enormous effect of the magnitude of a force. This is why such systems are commonly used to control, load, and unload container ships in a harbor.
150 N 7500 N T1 = 150 N T2 Fig. 1 Hawser wrapped three turns around bollard.
Photo 8.6 Dockworker mooring a ship using a hawser wrapped around a bollard.
8.4 Belt Friction 473 Sample Problem 8.9 A flat belt connects pulley A, which drives a machine tool, to pulley B, which is attached to the shaft of an electric motor. The coefficients of friction are µs 5 0.25 and µk 5 0.20 between both pulleys and the belt. Knowing that the maximum allowable tension in the belt is 600 lb, deter-mine the largest torque that the belt can exert on pulley A.
STRATEGY: The key to solving this problem is to identify the pulley where slippage would first occur, and then find the corresponding belt tensions when slippage is impending. The resistance to slippage depends upon the angle of contact β between pulley and belt, as well as upon the coefficient of static friction µs. Since µs is the same for both pulleys, slip-page occurs first on pulley B, for which β is smaller (Fig. 1).
MODELING and ANALYSIS: Pulley B. Using Eq. (8.14) with T2 5 600 lb, µs 5 0.25, and β 5 120° 5 2π/3 rad (Fig. 2), you obtain T2 T1 5 eµs β 600 lb T1 5 e0.25(2π/3) 5 1.688 T1 5 600 lb 1.688 5 355.4 lb Pulley A. Draw the free-body diagram of pulley A (Fig. 3). The couple MA is applied to the pulley using the machine tool to which it is attached and is equal and opposite to the torque exerted by the belt. Setting the sum of the moments equal to zero gives 1l oMA 5 0: MA 2 (600 lb)(8 in.) 1 (355.4 lb)(8 in.) 5 0 MA 5 1957 lb?in. MA 5 163.1 lb?ft b REFLECT and THINK: You may check that the belt does not slip on pulley A by computing the value of µs required to prevent slipping at A and verify that it is smaller than the actual value of µs. From Eq. (8.13), you have µsβ 5 ln T2 T1 5 ln 600 lb 355.4 lb 5 0.524 Since β 5 240° 5 4π/3 rad, 4π 3 µs 5 0.524 µs 5 0.125 , 0.25 60° A B r = 1 in.
8 in.
Fig. 1 Angles of contact for the pulleys.
60° 30° b = 240° b = 120° A B T2 = 600 lb T1 b = 120° B Fig. 2 Belt tensions at pulley B.
T1 = 355.4 lb A x A y MA T2 = 600 lb A 8 in.
Fig. 3 Free-body diagram of pulley A.
474 474 SOLVING PROBLEMS ON YOUR OWN I n the preceding section, you studied belt friction. The problems you will solve include belts passing over fixed drums, band brakes in which the drum rotates when the band remains fixed, and belt drives.
1. Problems involving belt friction fall into one of the following two categories. a. Problems in which slipping is impending. You can use one of the following formulas involving the coefficient of static friction µs. ln T2 T1 5 µsβ (8.13) or T2 T1 5 eµsβ (8.14) b. Problems in which slipping is occurring. You can obtain the formulas to be used from Eqs. (8.13) and (8.14) by replacing µs with the coefficient of kinetic friction µk.
2. As you start solving a belt-friction problem, remember these conventions: a. The angle β must be expressed in radians. In a belt-and-drum problem, this is the angle subtending the arc of the drum on which the belt is wrapped. b. The larger tension is always denoted by T2 and the smaller tension is denoted by T1. c. The larger tension occurs at the end of the belt which is in the direction of the motion, or impending motion, of the belt relative to the drum.
3. In each of the problems you will be asked to solve, three of the four quantities T1, T2, β, and µs (or µk) will either be given or readily found, and you will then solve the appropriate equation for the fourth quantity. You will encounter two kinds of problems. a. Find µs between belt and drum, knowing that slipping is impending. From the given data, determine T1, T2, and β; substitute these values into Eq. (8.13) and solve for µs [Sample Prob. 8.8, part a]. Follow the same procedure to find the smallest value of µs for which slipping will not occur. b. Find the magnitude of a force or couple applied to the belt or drum, know-ing that slipping is impending. The given data should include µs and β. If it also includes T1 or T2, use Eq. (8.14) to find the other tension. If neither T1 nor T2 is known but some other data is given, use the free-body diagram of the belt-drum system to write an equi-librium equation that you can solve simultaneously with Eq. (8.14) for T1 and T2. You then will be able to find the magnitude of the specified force or couple from the free-body diagram of the system. Follow the same procedure to determine the largest value of a force or couple that can be applied to the belt or drum if no slipping is to occur [Sample Prob. 8.9].
475 Problems 8.103 A rope having a weight per unit length of 0.4 lb/ft is wound 21 2 times around a horizontal rod. Knowing that the coefficient of static fric-tion between the rope and the rod is 0.30, determine the minimum length x of rope that should be left hanging if a 100-lb load is to be supported. 8.104 A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can resist a force of 5000 lb on the other end of the hawser. Determine (a) the coefficient of static friction between the hawser and the bollard, (b) the number of times the hawser should be wrapped around the bollard if a 20,000-lb force is to be resisted by the same 80-lb force. 8.105 Two cylinders are connected by a rope that passes over two fixed rods as shown. Knowing that the coefficient of static friction between the rope and the rods is 0.40, determine the range of the mass m of cylinder D for which equilibrium is maintained. 8.106 Two cylinders are connected by a rope that passes over two fixed rods as shown. Knowing that for cylinder D upward motion impends when m 5 20 kg, determine (a) the coefficient of static friction between the rope and the rods, (b) the corresponding tension in portion BC of the rope. 8.107 Knowing that the coefficient of static friction is 0.25 between the rope and the horizontal pipe and 0.20 between the rope and the vertical pipe, determine the range of values of P for which equilib-rium is maintained.
P 400 N Fig. P8.107 and P8.108 8.108 Knowing that the coefficient of static friction is 0.30 between the rope and the horizontal pipe and that the smallest value of P for which equilibrium is maintained is 80 N, determine (a) the largest value of P for which equilibrium is maintained, (b) the coefficient of static friction between the rope and the vertical pipe.
x 10 ft 100 lb Fig. P8.103 D A B 50 kg m C Fig. P8.105 and P8.106 476 8.109 A band belt is used to control the speed of a flywheel as shown. Determine the magnitude of the couple being applied to the fly-wheel, knowing that the coefficient of kinetic friction between the belt and the flywheel is 0.25 and that the flywheel is rotating clock-wise at a constant speed. Show that the same result is obtained if the flywheel rotates counterclockwise. 8.110 The setup shown is used to measure the output of a small turbine. When the flywheel is at rest, the reading of each spring scale is 14 lb. If a 105-lb∙in. couple must be applied to the flywheel to keep it rotating clockwise at a constant speed, determine (a) the reading of each scale at that time, (b) the coefficient of kinetic friction. Assume that the length of the belt does not change.
B A 18.75 in.
Fig. P8.110 and P8.111 8.111 The setup shown is used to measure the output of a small turbine. The coefficient of kinetic friction is 0.20, and the reading of each spring scale is 16 lb when the flywheel is at rest. Determine (a) the reading of each scale when the flywheel is rotating clockwise at a constant speed, (b) the couple that must be applied to the flywheel. Assume that the length of the belt does not change. 8.112 A flat belt is used to transmit a couple from drum B to drum A. Knowing that the coefficient of static friction is 0.40 and that the allowable belt tension is 450 N, determine the largest couple that can be exerted on drum A. 8.113 A flat belt is used to transmit a couple from pulley A to pulley B. The radius of each pulley is 60 mm, and a force of magnitude P 5 900 N is applied as shown to the axle of pulley A. Knowing that the coefficient of static friction is 0.35, determine (a) the largest couple that can be transmitted, (b) the corresponding maximum value of the tension in the belt.
A B P 240 mm Fig. P8.113 8.114 Solve Prob. 8.113 assuming that the belt is looped around the pulleys in a figure eight.
80 mm B C A D P = 100 N 320 mm 150 mm 80 mm E Fig. P8.109 B A 15° 15° rA = 120 mm rB = 50 mm Fig. P8.112 477 8.115 The speed of the brake drum shown is controlled by a belt attached to the control bar AD. A force P with a magnitude of 25 lb is applied to the control bar at A. Determine the magnitude of the couple being applied to the drum knowing that the coefficient of kinetic friction between the belt and the drum is 0.25, that a 5 4 in., and that the drum is rotating at a constant speed (a) counterclockwise, (b) clockwise. 8.116 The speed of the brake drum shown is controlled by a belt attached to the control bar AD. Knowing that a 5 4 in., determine the maxi-mum value of the coefficient of static friction for which the brake is not self-locking when the drum rotates counterclockwise. 8.117 The speed of the brake drum shown is controlled by a belt attached to the control bar AD. Knowing that the coefficient of static friction is 0.30 and that the brake drum is rotating counterclockwise, determine the minimum value of a for which the brake is not self-locking. 8.118 Bucket A and block C are connected by a cable that passes over drum B. Knowing that drum B rotates slowly counterclockwise and that the coefficients of friction at all surfaces are µs 5 0.35 and µk 5 0.25, determine the smallest combined mass m of the bucket and its contents for which block C will (a) remain at rest, (b) start moving up the incline, (c) continue moving up the incline at a constant speed. 8.119 Solve Prob. 8.118 assuming that drum B is frozen and cannot rotate. 8.120 and 8.122 A cable is placed around three parallel pipes. Knowing that the coefficients of friction are µs 5 0.25 and µk 5 0.20, deter-mine (a) the smallest weight W for which equilibrium is maintained, (b) the largest weight W that can be raised if pipe B is slowly rotated counterclockwise while pipes A and C remain fixed.
50 lb W A C B Fig. P8.120 and P8.121 8.121 and 8.123 A cable is placed around three parallel pipes. Two of the pipes are fixed and do not rotate; the third pipe is slowly rotated. Knowing that the coefficients of friction are µs 5 0.25 and µk 5 0.20, determine the largest weight W that can be raised (a) if only pipe A is rotated counterclockwise, (b) if only pipe C is rotated clockwise.
A B C D a P 24 in.
r 8 in.
E Fig. P8.115, P8.116, and P8.117 A B C m 100 kg 30° Fig. P8.118 A C B W 50 lb Fig. P8.122 and P8.123 478 8.124 A recording tape passes over the 20-mm-radius drive drum B and under the idler drum C. Knowing that the coefficients of friction between the tape and the drums are µs 5 0.40 and µk 5 0.30 and that drum C is free to rotate, determine the smallest allowable value of P if slipping of the tape on drum B is not to occur. 8.125 Solve Prob. 8.124 assuming that the idler drum C is frozen and cannot rotate. 8.126 The strap wrench shown is used to grip the pipe firmly without marring the external surface of the pipe. Knowing that the coefficient of static friction is the same for all surfaces of contact, determine the smallest value of µs for which the wrench will be self-locking when a 5 200 mm, r 5 30 mm, and θ 5 65°.
θ r a D P C B A Fig. P8.126 8.127 Solve Prob. 8.126 assuming that θ 5 75°. 8.128 The 10-lb bar AE is suspended by a cable that passes over a 5-in.-radius drum. Vertical motion of end E of the bar is prevented by the two stops shown. Knowing that µs 5 0.30 between the cable and the drum, determine (a) the largest counterclockwise couple M0 that can be applied to the drum if slipping is not to occur, (b) the corresponding force exerted on end E of the bar. 8.129 Solve Prob. 8.128 assuming that a clockwise couple M0 is applied to the drum. 8.130 Prove that Eqs. (8.13) and (8.14) are valid for any shape of surface provided that the coefficient of friction is the same at all points of contact. 8.131 Complete the derivation of Eq. (8.15), which relates the tension in both parts of a V belt. 8.132 Solve Prob. 8.112 assuming that the flat belt and drums are replaced by a V belt and V pulleys with α 5 36°. (The angle α is as shown in Fig. 8.15a.) 8.133 Solve Prob. 8.113 assuming that the flat belt and pulleys are replaced by a V belt and V pulleys with α 5 36°. (The angle α is as shown in Fig. 8.15a.) A C E B D 5 in.
5 in. 3 in.
10 lb M0 Fig. P8.128 A B C D P TA 0.3 N⋅m Fig. P8.124 T2 T1 β T2 T1 β Fig. P8.130 479 This chapter was devoted to the study of dry friction, i.e., to problems involv-ing rigid bodies in contact along unlubricated surfaces.
Fig. 8.16 N F W P P F Equilibrium Motion Fm Fk Static and Kinetic Friction If we apply a horizontal force P to a block resting on a horizontal surface [Sec. 8.1], we note that at first the block does not move. This shows that a friction force F must have developed to balance P (Fig. 8.16). As the magnitude of P increases, the magnitude of F also increases until it reaches a maximum value Fm. If P is further increased, the block starts sliding, and the magnitude of F drops from Fm to a lower value Fk. Experimental evidence shows that Fm and Fk are proportional to the normal component N of the reaction of the surface. We have Fm 5 µsN Fk 5 µkN (8.1, 8.2) where µs and µk are called, respectively, the coefficient of static friction and the coefficient of kinetic friction. These coefficients depend on the nature and the condition of the surfaces in contact. Approximate values of the coef-ficients of static friction are given in Table 8.1.
Angles of Friction It is sometimes convenient to replace the normal force N and the friction force F by their resultant R (Fig. 8.17). As the friction force increases and reaches its maximum value Fm 5 µsN, the angle f that R forms with the normal to the surface increases and reaches a maximum value fs, which is called the angle of static friction. If motion actually takes place, the magnitude of F drops to Fk; similarly, the angle f drops to a lower value fk, which is called the angle of kinetic friction. As shown in Sec. 8.1B, we have tan fs 5 µs tan fk 5 µk (8.3, 8.4) Problems Involving Friction When solving equilibrium problems involving friction, you should keep in mind that the magnitude F of the friction force is equal to Fm 5 µsN only if the body is about to slide [Sec. 8.1C]. If motion is not impending, you should Review and Summary Fig. 8.17 R W P φ N F 480 480 treat F and N as independent unknowns to be determined from the equilibrium equations (Fig. 8.18a). You should also check that the value of F required to maintain equilibrium is not larger than Fm; if it were, the body would move, and the magnitude of the friction force would be Fk 5 µkN [Sample Prob. 8.1]. On the other hand, if motion is known to be impending, F has reached its maximum value Fm 5 µsN (Fig. 8.18b), and you should substitute this expres-sion for F in the equilibrium equations [Sample Prob. 8.3]. When only three forces are involved in a free-body diagram, including the reaction R of the surface in contact with the body, it is usually more convenient to solve the problem by drawing a force triangle [Sample Prob. 8.2]. In some problems, impending motion can be due to tipping instead of slipping; the assessment of this condition requires a moment equilibrium analysis of the body [Sample Prob. 8.4].
Fig. 8.18 W P N W P N Frequired Fm = msN (a) (b) When a problem involves the analysis of the forces exerted on each other by two bodies A and B, it is important to show the friction forces with their correct sense. The correct sense for the friction force exerted by B on A, for instance, is opposite to that of the relative motion (or impending motion) of A with respect to B [Fig. 8.6].
Wedges and Screws In the later sections of this chapter, we considered several specific engineering applications where dry friction plays an important role. In the case of wedges, which are simple machines used to raise heavy loads [Sec. 8.2A], we must draw two or more free-body diagrams, taking care to show each friction force with its correct sense [Sample Prob. 8.5]. The analysis of square-threaded screws, which are frequently used in jacks, presses, and other mechanisms, is reduced to the analysis of a block sliding on an incline by unwrapping the thread of the screw and showing it as a straight line [Sec. 8.2B]. This is shown again in Fig. 8.19, where r denotes the mean radius of the thread, L is the lead of the screw (i.e., the distance through which the screw advances in one turn), W is the load, and Qr is equal to the couple exerted on the screw. We noted in the case of multiple-threaded screws that the lead L of the screw is not equal to its pitch, which is the distance measured between two consecutive threads. Other engineering applications considered in this chapter were journal bearings and axle friction [Sec. 8.3A], thrust bearings and disk friction [Sec. 8.3B], wheel friction and rolling resistance [Sec. 8.3C], and belt friction [Sec. 8.4].
Fig. 8.19 φs Q W R θ θ L 2 r p 481 Belt Friction In solving a problem involving a flat belt passing over a fixed cylinder, it is important to first determine the direction in which the belt slips or is about to slip. If the drum is rotating, the motion or impending motion of the belt should be determined relative to the rotating drum. For instance, if the belt shown in Fig. 8.20 is about to slip to the right relative to the drum, the friction Fig. 8.20 P O P' q b ∆q P1 P2 T1 T2 forces exerted by the drum on the belt are directed to the left, and the tension is larger in the right-hand portion of the belt than in the left-hand portion. Denoting the larger tension by T2, the smaller tension by T1, the coefficient of static friction by µs, and the angle (in radians) subtended by the belt by β, we derived in Sec. 8.4 the formulas ln T2 T1 5 µsβ (8.13) T2 T1 5 eµsβ (8.14) that we used in solving Sample Probs. 8.8 and 8.9. If the belt actually slips on the drum, the coefficient of static friction µs should be replaced by the coefficient of kinetic friction µk in both of these formulas.
482 8.134 and 8.135 The coefficients of friction are µs 5 0.40 and µk 5 0.30 between all surfaces of contact. Determine the smallest force P required to start the 30-kg block moving if cable AB (a) is attached as shown, (b) is removed.
Fig. P8.135 P A B 20 kg 30 kg 8.136 A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. If h 5 32 in., determine the magnitude of the force P required to move the cabinet to the right (a) if all casters are locked, (b) if the casters at B are locked and the casters at A are free to rotate, (c) if the casters at A are locked and the casters at B are free to rotate. Fig. P8.136 C A B P h 24 in. 8.137 A slender rod with a length of L is lodged between peg C and the vertical wall, and supports a load P at end A. Knowing that the coefficient of static friction between the peg and the rod is 0.15 and neglecting friction at the roller, determine the range of values of the ratio L/a for which equilibrium is maintained.
Fig. P8.137 A B C L a 30° P Review Problems Fig. P8.134 P A B 20 kg 30 kg 483 8.138 The hydraulic cylinder shown exerts a force of 3 kN directed to the right on point B and to the left on point E. Determine the magnitude of the couple M required to rotate the drum clockwise at a constant speed.
Fig. P8.138 D E 150 mm 300 mm A B 150 mm 300 mm 250 mm ms = 0.40 mk = 0.30 M C 150 mm 150 mm 8.139 A rod DE and a small cylinder are placed between two guides as shown. The rod is not to slip downward, however large the force P may be; i.e., the arrangement is said to be self-locking. Neglecting the weight of the cylinder, determine the minimum allowable coef-ficients of static friction at A, B, and C.
8.140 Bar AB is attached to collars that can slide on the inclined rods shown. A force P is applied at point D located at a distance a from end A. Knowing that the coefficient of static friction µs between each collar and the rod upon which it slides is 0.30 and neglecting the weights of the bar and of the collars, determine the smallest value of the ratio a/L for which equilibrium is maintained.
Fig. P8.140 45° 45° A D B P a L 8.141 Two 10° wedges of negligible weight are used to move and position the 400-lb block. Knowing that the coefficient of static friction is 0.25 at all surfaces of contact, determine the smallest force P that should be applied as shown to one of the wedges.
Fig. P8.141 400 lb P 10° Fig. P8.139 B q A D E C P 484 8.142 A 10° wedge is used to split a section of a log. The coefficient of static friction between the wedge and the log is 0.35. Knowing that a force P with a magnitude of 600 lb was required to insert the wedge, determine the magnitude of the forces exerted on the wood by the wedge after insertion.
Fig. P8.142 P 10° 8.143 In the gear-pulling assembly shown, the square-threaded screw AB has a mean radius of 15 mm and a lead of 4 mm. Knowing that the coefficient of static friction is 0.10, determine the couple that must be applied to the screw in order to produce a force of 3 kN on the gear. Neglect friction at end A of the screw.
8.144 A lever of negligible weight is loosely fitted onto a 30-mm-radius fixed shaft as shown. Knowing that a force P of magnitude 275 N will just start the lever rotating clockwise, determine (a) the coefficient of static friction between the shaft and the lever, (b) the smallest force P for which the lever does not start rotating counterclockwise.
Fig. P8.144 100 mm 160 mm P 30 mm 40 kg A B C 8.145 In the pivoted motor mount shown, the weight W of the 175-lb motor is used to maintain tension in the drive belt. Knowing that the coefficient of static friction between the flat belt and drums A and B is 0.40 and neglecting the weight of platform CD, determine the largest couple that can be transmitted to drum B when the drive drum A is rotating clockwise.
Fig. P8.143 A B Fig. P8.145 A B C D W 10 in.
12 in.
6 in.
The strength of structural members used in the construction of buildings depends to a large extent on the properties of their cross sections. This includes the second moments of area, or moments of inertia, of these cross sections.
Distributed Forces: Moments of Inertia 9 486 Distributed Forces: Moments of Inertia Introduction 9.1 MOMENTS OF INERTIA OF AREAS 9.1A Second Moment, or Moment of Inertia, of an Area 9.1B Determining the Moment of Inertia of an Area by Integration 9.1C Polar Moment of Inertia 9.1D Radius of Gyration of an Area 9.2 PARALLEL-AXIS THEOREM AND COMPOSITE AREAS 9.2A The Parallel-Axis Theorem 9.2B Moments of Inertia of Composite Areas 9.3 TRANSFORMATION OF MOMENTS OF INERTIA 9.3A Product of Inertia 9.3B Principal Axes and Principal Moments of Inertia 9.4 MOHR’S CIRCLE FOR MOMENTS OF INERTIA 9.5 MASS MOMENTS OF INERTIA 9.5A Moment of Inertia of a Simple Mass 9.5B Parallel-Axis Theorem for Mass Moments of Inertia 9.5C Moments of Inertia of Thin Plates 9.5D Determining the Moment of Inertia of a Three-Dimensional Body by Integration 9.5E Moments of Inertia of Composite Bodies 9.6 ADDITIONAL CONCEPTS OF MASS MOMENTS OF INERTIA 9.6A Mass Products of Inertia 9.6B Principal Axes and Principal Moments of Inertia 9.6C Principal Axes and Moments of Inertia for a Body of Arbitrary Shape Objectives • Describe the second moment, or moment of inertia, of an area.
• Determine the rectangular and polar moments of inertia of areas and their corresponding radii of gyration by integration.
• Develop the parallel-axis theorem and apply it to determine the moments of inertia of composite areas.
• Introduce the product of inertia and apply it to analyze the transformation of moments of inertia when coordinate axes are rotated.
• Describe the moment of inertia of a mass with respect to an axis.
• Apply the parallel-axis theorem to facilitate mass moment of inertia computations.
• Analyze the transformation of mass moments of inertia when coordinate axes are rotated.
Introduction In Chap. 5, we analyzed various systems of forces distributed over an area or volume. The three main types of forces considered were (1) weights of homogeneous plates of uniform thickness (Secs. 5.1 and 5.2); (2) distrib-uted loads on beams and submerged surfaces (Sec. 5.3); and (3) weights of homogeneous three-dimensional bodies (Sec. 5.4). In all of these cases, the distributed forces were proportional to the elemental areas or volumes associated with them. Therefore, we could obtain the resultant of these forces by summing the corresponding areas or volumes, and we deter-mined the moment of the resultant about any given axis by computing the first moments of the areas or volumes about that axis.
In the first part of this chapter, we consider distributed forces DF where the magnitudes depend not only upon the elements of area DA on which these forces act but also upon the distance from DA to some given axis. More precisely, we assume the magnitude of the force per unit area DF/DA varies linearly with the distance to the axis. Forces of this type arise in the study of the bending of beams and in problems involving submerged nonrectangular surfaces. Starting with the assumption that the elemental forces involved are distributed over an area A and vary linearly with the distance y to the x axis, we will show that the magnitude of their resultant R depends upon the first moment Qx of the area A. However, the location of the point where R is applied depends upon the second moment, or moment of inertia, Ix of the same area with respect to the x axis. You will see how to compute the moments of inertia of various areas with respect to given x and y axes. We also introduce the polar moment of inertia JO of an area. To facilitate these computations, we establish a relation between the moment of inertia Ix 9.1 Moments of Inertia of Areas 487 of an area A with respect to a given x axis and the moment of inertia Ix9 of the same area with respect to the parallel centroidal x9 axis (a relation known as the parallel-axis theorem). You will also study the transforma-tion of the moments of inertia of a given area when the coordinate axes are rotated.
In the second part of this chapter, we will explain how to determine the moments of inertia of various masses with respect to a given axis. Moments of inertia of masses are common in dynamics problems involv-ing the rotation of a rigid body about an axis. To facilitate the computation of mass moments of inertia, we introduce another version of the parallel-axis theorem. Finally, we will analyze the transformation of moments of inertia of masses when the coordinate axes are rotated.
9.1 MOMENTS OF INERTIA OF AREAS In the first part of this chapter, we consider distributed forces DF whose magnitudes DF are proportional to the elements of area DA on which the forces act and, at the same time, vary linearly with the distance from DA to a given axis.
9.1A Second Moment, or Moment of Inertia, of an Area Consider a beam with a uniform cross section that is subjected to two equal and opposite couples: one applied at each end of the beam. Such a beam is said to be in pure bending. The internal forces in any section of the beam are distributed forces whose magnitudes DF 5 ky DA vary lin-early with the distance y between the element of area DA and an axis passing through the centroid of the section. (This statement can be derived in a course on mechanics of materials.) This axis, represented by the x axis in Fig. 9.1, is known as the neutral axis of the section. The forces on one side of the neutral axis are forces of compression, whereas those on the other side are forces of tension. On the neutral axis itself, the forces are zero.
The magnitude of the resultant R of the elemental forces DF that act over the entire section is R 5 #ky dA 5 k#y dA You might recognize this last integral as the first moment Qx of the section about the x axis; it is equal to yA and is thus equal to zero, since the centroid of the section is located on the x axis. The system of forces DF thus reduces to a couple. The magnitude M of this couple (bending moment) must be equal to the sum of the moments DMx 5 y DF 5 ky2 DA of the elemental forces. Integrating over the entire section, we obtain M 5 #ky2 dA 5 k#y2 dA Fig. 9.1 Representative forces on a cross section of a beam subjected to equal and opposite couples at each end.
y x y ∆F = ky ∆A ∆A 488 Distributed Forces: Moments of Inertia This last integral is known as the second moment, or moment of inertia,† of the beam section with respect to the x axis and is denoted by Ix. We obtain it by multiplying each element of area dA by the square of its distance from the x axis and integrating over the beam section. Since each product y2 dA is positive, regardless of the sign of y, or zero (if y is zero), the integral Ix is always positive.
Another example of a second moment, or moment of inertia, of an area is provided by the following problem from hydrostatics. A vertical circular gate used to close the outlet of a large reservoir is submerged under water as shown in Fig. 9.2. What is the resultant of the forces exerted by the water on the gate, and what is the moment of the resultant about the line of intersection of the plane of the gate and the water surface (x axis)?
If the gate were rectangular, we could determine the resultant of the forces due to water pressure from the pressure curve, as we did in Sec. 5.3B. Since the gate is circular, however, we need to use a more general method. Denoting the depth of an element of area DA by y and the specific weight of water by γ, the pressure at an element is p 5 γ y, and the magnitude of the elemental force exerted on DA is DF 5 p DA 5 γ y DA. The magnitude of the resultant of the elemental forces is thus R 5 #γ y dA 5 γ#y dA We can obtain this by computing the first moment Qx 5 e y dA of the area of the gate with respect to the x axis. The moment Mx of the resultant must be equal to the sum of the moments DMx 5 y DF 5 γy2 DA of the elemental forces. Integrating over the area of the gate, we have Mx 5 #γ y2 dA 5 γ#y2 dA Here again, the last integral represents the second moment, or moment of inertia, Ix of the area with respect to the x axis.
9.1B Determining the Moment of Inertia of an Area by Integration We just defined the second moment, or moment of inertia, Ix of an area A with respect to the x axis. In a similar way, we can also define the moment of inertia Iy of the area A with respect to the y axis (Fig. 9.3a): Moments of inertia of an area Ix 5 #y2 dA Iy 5 #x2 dA (9.1) Ix I 5 #y2dA d Iy I 5 #x2dA d Fig. 9.2 Vertical circular gate, submerged under water, used to close the outlet of a reservoir.
y x y C ∆A ∆F = gy ∆A †The term second moment is more proper than the term moment of inertia, which logically should be used only to denote integrals of mass (see Sec. 9.5). In engineering practice, however, moment of inertia is used in connection with areas as well as masses. 9.1 Moments of Inertia of Areas 489 We can evaluate these integrals, which are known as the rectangular moments of inertia of the area A, more easily if we choose dA to be a thin strip parallel to one of the coordinate axes. To compute Ix, we choose the strip parallel to the x axis, so that all points of the strip are at the same distance y from the x axis (Fig. 9.3b). We obtain the moment of inertia dIx of the strip by multiplying the area dA of the strip by y2. To compute Iy, we choose the strip parallel to the y axis, so that all points of the strip are at the same distance x from the y axis (Fig. 9.3c). Then the moment of inertia dIy of the strip is x2 dA.
Moment of Inertia of a Rectangular Area. As an example, let us determine the moment of inertia of a rectangle with respect to its base (Fig. 9.4). Dividing the rectangle into strips parallel to the x axis, we have dA 5 b dy dIx 5 y2b dy Ix 5 # h 0 by2 dy 5 1 3 bh3 (9.2) Computing lx and ly Using the Same Elemental Strips. We can use Eq. (9.2) to determine the moment of inertia dIx with respect to the x axis of a rectangular strip that is parallel to the y axis, such as the strip shown in Fig. 9.3c. Setting b 5 dx and h 5 y in formula (9.2), we obtain dIx 5 1 3 y3 dx We also have dIy 5 x2 dA 5 x2y dx Thus, we can use the same element to compute the moments of inertia Ix and Iy of a given area (Fig. 9.5).
Fig. 9.3 (a) Rectangular moments of inertia dIx and dIy of an area dA; (b) calculating Ix with a horizontal strip; (c) calculating Iy with a vertical strip.
x y y x (a) dA = dx dy dx dy dIx = y2 dA dIy = x2 dA x y y x (b) a dA = ( a – x ) dy dy dIx = y2 dA y x y x (c) dA = y dx dx dIy = x2 dA Fig. 9.4 Calculating the moment of inertia of a rectangular area with respect to its base.
h y y b dy x dA = b dy Fig. 9.5 Using the same strip element of a given area to calculate Ix and Iy.
y x y x dx dIx = y3 dx 1 3 dIy = x2 y dx 490 Distributed Forces: Moments of Inertia 9.1C Polar Moment of Inertia An integral of great importance in problems concerning the torsion of cylindrical shafts and in problems dealing with the rotation of slabs is Polar moment of inertia JO 5 #r2 dA (9.3) where r is the distance from O to the element of area dA (Fig. 9.6). This integral is called the polar moment of inertia of the area A with respect to the “pole” O.
We can compute the polar moment of inertia of a given area from the rectangular moments of inertia Ix and Iy of the area if these quantities are already known. Indeed, noting that r2 5 x2 1 y2, we have JO 5 #r2d A 5 # (x2 1 y2)d A 5 #y2d A 1 #x 2d A that is, JO 5 Ix 1 Iy (9.4) 9.1D Radius of Gyration of an Area Consider an area A that has a moment of inertia Ix with respect to the x axis (Fig. 9.7a). Imagine that we concentrate this area into a thin strip parallel to the x axis (Fig. 9.7b). If the concentrated area A is to have the same moment of inertia with respect to the x axis, the strip should be placed at a distance kx from the x axis, where kx is defined by the relation Ix 5 kx 2 A Solving for kx, we have Radius of gyration kx 5 B Ix A (9.5) The distance kx is referred to as the radius of gyration of the area with respect to the x axis. In a similar way, we can define the radii of gyration ky and kO (Fig. 9.7c and d); we have Iy 5 ky 2 A ky 5 B Iy A (9.6) JO 5 k2 O A kO 5 B JO A (9.7) If we rewrite Eq. (9.4) in terms of the radii of gyration, we find that kO 2 5 kx 2 1 ky 2 (9.8) Fig. 9.7 (a) Area A with given moment of inertia Ix; (b) compressing the area to a horizontal strip with radius of gyration kx; (c) compressing the area to a vertical strip with radius of gyration ky; (d) compressing the area to a circular ring with polar radius of gyration kO.
kx y x A O (a) y x A O (b) ky y x A O (c) kO y x A O (d) Fig. 9.6 Distance r used to evaluate the polar moment of inertia of area A.
y y x dA A x r O 9.1 Moments of Inertia of Areas 491 Sample Problem 9.1 Determine the moment of inertia of a triangle with respect to its base.
STRATEGY: To find the moment of inertia with respect to the base, it is expedient to use a differential strip of area parallel to the base. Use the geometry of the situation to carry out the integration.
MODELING: Draw a triangle with a base b and height h, choosing the x axis to coincide with the base (Fig. 1). Choose a differential strip parallel to the x axis to be dA. Since all portions of the strip are at the same dis-tance from the x axis, you have dIx 5 y2 dA dA 5 l dy ANALYSIS: Using similar triangles, you have l b 5 h 2 y h l 5 b h 2 y h d A 5 b h 2 y h d y Integrating dIx from y 5 0 to y 5 h, you obtain Ix 5# y2 dA 5# h 0 y2b h 2 y h dy 5 b h# h 0 (hy2 2 y3) dy 5 b h c h y3 3 2 y4 4 d h 0 Ix 5 bh3 12 b REFLECT and THINK: This problem also could have been solved using a differential strip perpendicular to the base by applying Eq. (9.2) to express the moment of inertia of this strip. However, because of the geometry of this triangle, you would need two integrals to complete the solution.
x y y dy b h h – y l Fig. 1 Triangle with differential strip element parallel to its base.
Concept Application 9.1 For the rectangle shown in Fig. 9.8, compute the radius of gyration kx with respect to its base. Using formulas (9.5) and (9.2), you have k2 x 5 Ix A 5 1 3 bh3 bh 5 h2 3 kx 5 h 13 The radius of gyration kx of the rectangle is shown in Fig. 9.8. Do not confuse it with the ordinate y 5 h/2 of the centroid of the area. The radius of gyration kx depends upon the second moment of the area, whereas the ordinate y is related to the first moment of the area.
Fig. 9.8 Radius of gyration of a rectangle with respect to its base.
h b kx y C 492 Distributed Forces: Moments of Inertia Sample Problem 9.2 (a) Determine the centroidal polar moment of inertia of a circular area by direct integration. (b) Using the result of part (a), determine the moment of inertia of a circular area with respect to a diameter.
STRATEGY: Since the area is circular, you can evaluate part (a) by using an annular differential area. For part (b), you can use symmetry and Eq. (9.4) to solve for the moment of inertia with respect to a diameter.
MODELING and ANALYSIS: a. Polar Moment of Inertia. Choose an annular differential element of area to be dA (Fig. 1). Since all portions of the differential area are at the same distance from the origin, you have dJO 5 u2 dA dA 5 2πu du JO 5# dJO 5# r 0 u2(2πu du) 5 2π# r 0 u3 du JO 5 π 2 r4 b b. Moment of Inertia with Respect to a Diameter. Because of the symmetry of the circular area, Ix 5 Iy. Then from Eq. (9.4), you have JO 5 Ix 1 Iy 5 2Ix π 2 r4 5 2Ix Idiameter 5 Ix 5 π 4 r4 b REFLECT and THINK: Always look for ways to simplify a problem by the use of symmetry. This is especially true for situations involving circles or spheres.
x y r du u O Fig. 1 Circular area with an annular differential element.
Sample Problem 9.3 (a) Determine the moment of inertia of the shaded region shown with respect to each of the coordinate axes. (Properties of this region were considered in Sample Prob. 5.4.) (b) Using the results of part (a), determine the radius of gyration of the shaded area with respect to each of the coordinate axes.
STRATEGY: You can determine the moments of inertia by using a single differential strip of area; a vertical strip will be more convenient. You can calculate the radii of gyration from the moments of inertia and the area of the region.
x y b y = kx2 a 9.1 Moments of Inertia of Areas 493 MODELING: Referring to Sample Prob. 5.4, you can find the equation of the curve and the total area using y 5 b a2 x2 A 5 1 3ab ANALYSIS: a. Moments of Inertia.
Moment of Inertia Ix. Choose a vertical differential element of area for dA (Fig. 1). Since all portions of this element are not at the same distance from the x axis, you must treat the element as a thin rectangle. The moment of inertia of the element with respect to the x axis is then dIx 5 1 3 y3 dx 5 1 3 a b a2 x2b 3 dx 5 1 3 b3 a6 x6 dx Ix 5# dIx 5# a 0 1 3 b3 a6 x6 dx 5 c 1 3 b3 a6 x7 7 d a 0 Ix 5 ab3 21 b Moment of Inertia Iy. Use the same vertical differential element of area. Since all portions of the element are at the same distance from the y axis, you have dIy 5 x2 dA 5 x2(y dx) 5 x2 a b a2 x2b dx 5 b a2 x4 dx Iy 5# dIy 5# a 0 b a2 x4 dx 5 c b a2 x5 5 d a 0 Iy 5 a3b 5 b b. Radii of Gyration kx and ky. From the definition of radius of gyration, you have k2 x 5 Ix A 5 ab3/21 ab/3 5 b2 7 kx 5 21 7 b b and k2 y 5 Iy A 5 a3b/5 ab/3 5 3 5a2 ky 5 23 5a b REFLECT and THINK: This problem demonstrates how you can calculate Ix and Iy using the same strip element. However, the general mathematical approach in each case is distinctly different.
dx x x y a y Fig. 1 Subject area with vertical differential strip element.
494 494 SOLVING PROBLEMS ON YOUR OWN I n this section, we introduced the rectangular and polar moments of inertia of areas and the corresponding radii of gyration. Although the problems you are about to solve may appear more appropriate for a calculus class than for one in mechanics, we hope that our introductory comments have convinced you of the relevance of moments of inertia to your study of a variety of engineering topics.
1. Calculating the rectangular moments of inertia Ix and Iy. We defined these quantities as Ix 5 #y2 dA Iy 5 #x2 dA (9.1) where dA is a differential element of area dx dy. The moments of inertia are the second moments of the area; it is for that reason that Ix, for example, depends on the perpendicular distance y to the area dA. As you study Sec. 9.1, you should recognize the importance of carefully defining the shape and the orientation of dA. Furthermore, you should note the following points.
a. You can obtain the moments of inertia of most areas by means of a single integration. You can use the expressions given in Figs. 9.3b and c and Fig. 9.5 to calculate Ix and Iy. Regardless of whether you use a single or a double integration, be sure to show the element dA that you have chosen on your sketch. b. The moment of inertia of an area is always positive, regardless of the location of the area with respect to the coordinate axes. The reason is that the moment of inertia is obtained by integrating the product of dA and the square of distance. (Note how this differs from the first moment of the area.) Only when an area is removed (as in the case for a hole) does its moment of inertia enter in your computations with a minus sign. c. As a partial check of your work, observe that the moments of inertia are equal to an area times the square of a length. Thus, every term in an expression for a moment of inertia must be a length to the fourth power.
2. Computing the polar moment of inertia JO. We defined JO as JO 5 #r2 dA (9.3) where r2 5 x2 1 y2. If the given area has circular symmetry (as in Sample Prob. 9.2), it is possible to express dA as a function of r and to compute JO with a single integration. When the area lacks circular symmetry, it is usually easier first to calculate Ix and Iy and then to determine JO from JO 5 Ix 1 Iy (9.4) Lastly, if the equation of the curve that bounds the given area is expressed in polar coor-dinates, then dA 5 r dr dθ, and you need to perform a double integration to compute the integral for JO [see Prob. 9.27].
3. Determining the radii of gyration kx and ky and the polar radius of gyration kO. These quantities are defined in Sec. 9.1D. You should realize that they can be determined only after you have computed the area and the appropriate moments of inertia. It is impor-tant to remember that kx is measured in the y direction, whereas ky is measured in the x direction; you should carefully study Sec. 9.1D until you understand this point.
495 Problems 9.1 through 9.4 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.
9.5 through 9.8 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.
y b x a y 5 k(x 2 a)2 Fig. P9.1 and P9.5 x y b y 5 kx1/3 a Fig. P9.2 and P9.6 h b y x Fig. P9.3 and P9.7 x y y 5 kx3 b a Fig. P9.4 and P9.8 9.9 through 9.11 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis. 9.12 through 9.14 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis.
b y x a 2a y = k(x − a)3 Fig. P9.9 and P9.12 b b y x a y = c(1 − kx1/2) y = −c(1 − kx1/2) Fig. P9.10 and P9.13 b y x a y = kex/a Fig. P9.11 and P9.14 496 9.15 and 9.16 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.
y h x a a y 5 kx2 Fig. P9.16 and P9.18 9.17 and 9.18 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis. 9.19 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.
h h y x a a y = mx + b y = c sin kx Fig. P9.19 and P9.20 9.20 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis. 9.21 and 9.22 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to point P.
a a a a a a P Fig. P9.21 b b 3b a a P Fig. P9.22 Fig. P9.15 and P9.17 b y x a y2 = k2x1/2 y1 = k1x2 497 9.23 and 9.24 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to point P.
a a 2a 2a P y x y = c + k2x2 y = k1x2 Fig. P9.23 9.25 (a) Determine by direct integration the polar moment of inertia of the annular area shown with respect to point O. (b) Using the result of part a, determine the moment of inertia of the given area with respect to the x axis.
R1 R2 y x O Fig. P9.25 and P9.26 9.26 (a) Show that the polar radius of gyration kO of the annular area shown is approximately equal to the mean radius Rm 5 (R1 1 R2)/2 for small values of the thickness t 5 R2 2 R1. (b) Determine the percentage error introduced by using Rm in place of kO for the following values of t/Rm: 1, 1 2, and 1 10. 9.27 Determine the polar moment of inertia and the polar radius of gyration of the shaded area shown with respect to point O. 9.28 Determine the polar moment of inertia and the polar radius of gyration of the isosceles triangle shown with respect to point O. 9.29 Using the polar moment of inertia of the isosceles triangle of Prob. 9.28, show that the centroidal polar moment of inertia of a circular area of radius r is πr4/2. (Hint: As a circular area is divided into an increasing number of equal circular sectors, what is the approximate shape of each circular sector?) 9.30 Prove that the centroidal polar moment of inertia of a given area A cannot be smaller than A2/2π. (Hint: Compare the moment of inertia of the given area with the moment of inertia of a circle that has the same area and the same centroid.) O y x a 2a q R = a + kq Fig. P9.27 y x O b 2 b 2 h Fig. P9.28 P r r 2 Fig. P9.24 498 Distributed Forces: Moments of Inertia 9.2 PARALLEL-AXIS THEOREM AND COMPOSITE AREAS In practice, we often need to determine the moment of inertia of a com-plicated area that can be broken down into a sum of simple areas. How-ever, in doing these calculations, we have to determine the moment of inertia of each simple area with respect to the same axis. In this section, we first derive a formula for computing the moment of inertia of an area with respect to a centroidal axis parallel to a given axis. Then we show how you can use this formula for finding the moment of inertia of a composite area.
9.2A The Parallel-Axis Theorem Consider the moment of inertia I of an area A with respect to an axis AA9 (Fig. 9.9). We denote the distance from an element of area dA to AA9 by y. This gives us I 5 #y2 dA Let us now draw through the centroid C of the area an axis BB9 parallel to AA9; this axis is called a centroidal axis. Denoting the distance from the element dA to BB9 by y9, we have y 5 y9 1 d, where d is the distance between the axes AA9 and BB9. Substituting for y in the previous integral, we obtain I 5# y2 dA 5# (y¿ 1 d)2 dA 5# y92 dA 1 2d# y9 dA 1 d2# dA The first integral represents the moment of inertia I of the area with respect to the centroidal axis BB9. The second integral represents the first moment of the area with respect to BB9, but since the centroid C of the area is located on this axis, the second integral must be zero. The last integral is equal to the total area A. Therefore, we have Parallel-axis theorem I 5 I 1 Ad2 (9.9) This formula states that the moment of inertia I of an area with respect to any given axis AA9 is equal to the moment of inertia I of the area with respect to a centroidal axis BB9 parallel to AA9 plus the product of the area A and the square of the distance d between the two axes. This theorem is known as the parallel-axis theorem. Substituting k 2A for I and k 2A for I , we can also express this theorem as k 2 5 k 2 1 d 2 (9.10) A similar theorem relates the polar moment of inertia JO of an area about a point O to the polar moment of inertia JC of the same area about its centroid C. Denoting the distance between O and C by d, we have JO 5 JC 1 Ad2 or k2 O 5 k 2 C 1 d2 (9.11) I 5 I 1 Ad2 Fig. 9.9 The moment of inertia of an area A with respect to an axis AA9 can be determined from its moment of inertia with respect to the centroidal axis BB9 by a calculation involving the distance d between the axes.
A' A B' B C y y' d dA 9.2 Parallel-Axis Theorem and Composite Areas 499 9.2B Moments of Inertia of Composite Areas Consider a composite area A made of several component areas A1, A2, A3, . . . . The integral representing the moment of inertia of A can be subdivided into integrals evaluated over A1, A2, A3, . . . . Therefore, we can obtain the moment of inertia of A with respect to a given axis by adding the moments of inertia of the areas A1, A2, A3, . . . with respect to the same axis. Concept Application 9.2 As an application of the parallel-axis theorem, let us determine the moment of inertia IT of a circular area with respect to a line tangent to the circle (Fig. 9.10). We found in Sample Prob. 9.2 that the moment of inertia of a circular area about a centroidal axis is I 5 1 4πr4. Therefore, we have IT 5 I I 1 Ad2 5 1 4πr4 1 (πr2)r2 5 5 4πr4 r T C d = r Fig. 9.10 Finding the moment of inertia of a circle with respect to a line tangent to it.
Concept Application 9.3 We can also use the parallel-axis theorem to determine the centroidal moment of inertia of an area when we know the moment of inertia of the area with respect to a parallel axis. Consider, for instance, a triangular area (Fig. 9.11). We found in Sample Prob. 9.1 that the moment of inertia of a triangle with respect to its base AA9 is equal to 1 12 bh3. Using the parallel-axis theorem, we have IAA¿ 5 I BB¿ 1 Ad2 I BB¿ 5 IAA¿ 2 Ad2 5 1 12bh3 2 1 2bh(1 3h)2 5 1 36bh3 Note that we subtracted the product Ad2 from the given moment of inertia in order to obtain the centroidal moment of inertia of the triangle. That is, this product is added when transferring from a centroidal axis to a parallel axis, but it is subtracted when transfer-ring to a centroidal axis. In other words, the moment of inertia of an area is always smaller with respect to a centroidal axis than with respect to any parallel axis. Returning to Fig. 9.11, we can obtain the moment of inertia of the triangle with respect to the line DD9 (which is drawn through a vertex) by writing IDD¿ 5 I BB¿ 1 Ad¿2 5 1 36bh3 1 1 2bh(2 3h)2 5 1 4bh3 Note that we could not have obtained IDD9 directly from IAA9. We can apply the parallel -axis theorem only if one of the two parallel axes passes through the centroid of the area.
Fig. 9.11 Finding the centroidal moment of inertia of a triangle from the moment of inertia about a parallel axis.
b A' A C B' B D' D h d' = h 2 3 d = h 1 3 500 Distributed Forces: Moments of Inertia Fig. 9.12 Moments of inertia of common geometric shapes.
Quarter circle C Rectangle Triangle Circle Semicircle Ellipse b y y' x' x 1 12 Ix' = bh3 1 12 Iy' = b3h 1 8 Ix = Iy = r4 1 4 JO = r4 1 4 Ix = Iy = r 4 1 2 JO = r4 1 36 Ix' = bh3 1 12 Ix = bh3 1 3 Iy = b3h 1 12 JC = bh(b2 + h2) 1 3 Ix = bh3 h b x' x x r O y h C h 3 x O C y r x O C y r x b y a 1 16 Ix = Iy = r4 1 8 JO = r4 1 4 Ix = ab3 1 4 Iy = a3b 1 4 JO = ab(a2 + b2) O Figure 9.12 shows several common geometric shapes along with formulas for the moments of inertia of each one. Before adding the moments of inertia of the component areas, however, you may have to use the parallel-axis theorem to transfer each moment of inertia to the desired axis. Sample Probs. 9.4 and 9.5 illustrate the technique.
Properties of the cross sections of various structural shapes are given in Fig. 9.13. As we noted in Sec. 9.1A, the moment of inertia of a beam Photo 9.1 Figure 9.13 tabulates data for a small sample of the rolled-steel shapes that are readily available. Shown above are examples of wide-flange shapes that are commonly used in the construction of buildings.
9.2 Parallel-Axis Theorem and Composite Areas 501 Fig. 9.13A Properties of rolled-steel shapes (U.S. customary units). Designation Area in2 Depth in.
Width in.
Axis X–X X X X X X X X X Y Y Y Y Y Y Y Y Axis Y–Y W Shapes (Wide-Flange Shapes) S Shapes (American Standard Shapes) C Shapes (American Standard Channels) Angles ∗Courtesy of the American Institute of Steel Construction, Chicago, Illinois †Nominal depth in inches and weight in pounds per foot ‡Depth, width, and thickness in inches W18 × 76† W16 × 57 W14 × 38 W8 × 31 22.3 16.8 11.2 9.12 18.2 16.4 14.1 8.00 11.0 7.12 6.77 8.00 1330 758 385 110 7.73 6.72 5.87 3.47 152 43.1 26.7 37.1 2.61 1.60 1.55 2.02 S18 × 54.7† S12 × 31.8 S10 × 25.4 S6 × 12.5 16.0 9.31 7.45 3.66 18.0 12.0 10.0 6.00 801 217 123 22.0 6.00 5.00 4.66 3.33 7.07 4.83 4.07 2.45 4.61 3.87 3.11 2.34 20.7 9.33 6.73 1.80 1.14 1.00 0.950 0.702 C12 × 20.7† C10 × 15.3 C8 × 11.5 C6 × 8.2 6.08 4.48 3.37 2.39 12.0 10.0 8.00 6.00 2.94 2.60 2.26 1.92 129 67.3 32.5 13.1 35.4 5.52 1.23 17.3 9.43 1.09 3.86 2.27 1.31 0.687 0.797 0.711 0.623 0.536 0.698 0.634 0.572 0.512 11.0 3.75 1.44 4.75 3.75 1.19 1.79 1.21 0.926 1.91 1.58 0.953 1.86 1.18 0.836 1.98 1.74 0.980 35.4 5.52 1.23 6.22 2.55 0.390 1.79 1.21 0.926 1.14 0.824 0.569 1.86 1.18 0.836 0.981 0.746 0.487 L6 × 6 × 1‡ L4 × 4 × L3 × 3 × L6 × 4 × L5 × 3 × L3 × 2 × ⎯Ix, in4 ⎯kx, in.
⎯y, in.
⎯Iy, in4 ⎯ky, in.
⎯x, in.
4 1 2 1 2 1 2 1 4 1 y x x section about its neutral axis is closely related to the computation of the bending moment in that section of the beam. Thus, determining moments of inertia is a prerequisite to the analysis and design of structural members.
Note that the radius of gyration of a composite area is not equal to the sum of the radii of gyration of the component areas. In order to deter-mine the radius of gyration of a composite area, you must first compute the moment of inertia of the area.
502 Distributed Forces: Moments of Inertia Fig. 9.13B Properties of rolled-steel shapes (SI units).
⎯Ix 106 mm4 ⎯Iy 106 mm4 ⎯y mm ⎯x mm ⎯x Designation Area mm2 Depth mm Width mm Axis X–X X X X X X X Y Y Y Y Y Y ⎯kx mm ⎯ky mm Axis Y–Y W Shapes (Wide-Flange Shapes) S Shapes (American Standard Shapes) C Shapes (American Standard Channels) †Nominal depth in millimeters and mass in kilograms per meter ‡Depth, width, and thickness in millimeters W460 × 113† W410 × 85 W360 × 57.8 W200 × 46.1 14 400 10 800 7230 5880 462 417 358 203 279 181 172 203 554 316 160 45.8 196 171 149 88.1 63.3 17.9 11.1 15.4 66.3 40.6 39.4 51.3 S460 × 81.4† S310 × 47.3 S250 × 37.8 S150 × 18.6 10 300 6010 4810 2360 457 305 254 152 333 90.3 51.2 9.16 152 127 118 84.6 180 123 103 62.2 8.62 3.88 2.80 0.749 29.0 25.4 24.1 17.8 C310 × 30.8† C250 × 22.8 C200 × 17.1 C150 × 12.2 3920 2890 2170 1540 305 254 203 152 74.7 66.0 57.4 48.8 53.7 28.0 13.5 5.45 117 98.3 79.0 59.4 1.61 0.945 0.545 0.286 20.2 18.1 15.8 13.6 17.7 16.1 14.5 13.0 7100 2420 929 3060 2420 768 14.7 2.30 0.512 7.20 3.93 0.454 45.5 30.7 23.5 48.5 40.1 24.2 47.2 30.0 21.2 50.3 44.2 24.9 14.7 2.30 0.512 2.59 1.06 0.162 45.5 30.7 23.5 29.0 20.9 14.5 47.2 30.0 21.2 24.9 18.9 12.4 L152 × 152 × 25.4‡ L102 × 102 × 12.7 L76 × 76 × 6.4 L152 × 102 × 12.7 L127 × 76 × 12.7 L76 × 51 × 6.4 X X Y Y Angles y x 9.2 Parallel-Axis Theorem and Composite Areas 503 Sample Problem 9.4 The strength of a W14 3 38 rolled-steel beam is increased by attaching a 9 3 3/4-in. plate to its upper flange as shown. Determine the moment of inertia and the radius of gyration of the composite section with respect to an axis that is parallel to the plate and passes through the centroid C of the section.
STRATEGY: This problem involves finding the moment of inertia of a composite area with respect to its centroid. You should first determine the location of this centroid. Then, using the parallel-axis theorem, you can determine the moment of inertia relative to this centroid for the overall section from the centroidal moment of inertia for each component part.
MODELING and ANALYSIS: Place the origin O of coordinates at the centroid of the wide-flange shape, and compute the distance Y to the centroid of the composite section by using the methods of Chap. 5 (Fig. 1). Refer to Fig. 9.13A for the area of the wide-flange shape. The area and the y coordinate of the centroid of the plate are A 5 (9 in.)(0.75 in.) 5 6.75 in2 y 5 1 2(14.1 in.) 1 1 2(0.75 in.) 5 7.425 in.
Section Area, in2 y, in. yA, in3 Plate 6.75 7.425 50.12 Wide-fl ange shape 11.2 0 0 oA 5 17.95 oyA 5 50.12 YoA 5 oyA Y(17.95) 5 50.12 Y 5 2.792 in.
Moment of Inertia. Use the parallel-axis theorem to determine the moments of inertia of the wide-flange shape and the plate with respect to the x9 axis. This axis is a centroidal axis for the composite section but not for either of the elements considered separately. You can obtain the value of I x for the wide-flange shape from Fig. 9.13A. For the wide-flange shape, Ix9 5 I x 1 AY 2 5 385 1 (11.2)(2.792)2 5 472.3 in4 For the plate, Ix9 5 I x 1 Ad2 5 ( 1 12)(9)(3 4)3 1 (6.75)(7.425 2 2.792)2 5 145.2 in4 For the composite area, Ix9 5 472.3 1 145.2 5 617.5 in4 Ix9 5 618 in4 b Radius of Gyration. From the moment of inertia and area just calculated, you obtain k2 x¿ 5 Ix¿ A 5 617.5 in4 17.95 in2 kx¿ 5 5.87 in. b REFLECT and THINK: This is a common type of calculation for many different situations. It is often helpful to list data in a table to keep track of the numbers and identify which data you need.
9 in.
14.1 in.
6.77 in.
C 3 4 in.
x y d C O 7.425 in.
x' Y Fig. 1 Origin of coordinates placed at centroid of wide-flange shape.
504 Distributed Forces: Moments of Inertia Sample Problem 9.5 Determine the moment of inertia of the shaded area with respect to the x axis.
STRATEGY: You can obtain the given area by subtracting a half circle from a rectangle (Fig. 1). Then compute the moments of inertia of the rectangle and the half circle separately.
240 mm 120 mm y x r = 90 mm A C a 240 mm 120 mm y y y x x x A' x'= b − Fig. 1 Modeling given area by subtracting a half circle from a rectangle.
A' A C a = 38.2 mm x' 120 mm y x b = 81.8 mm Fig. 2 Centroid location of the half circle.
MODELING and ANALYSIS: Moment of Inertia of Rectangle. Referring to Fig. 9.12, you have Ix 5 1 3 bh3 5 1 3 (240 mm)(120 mm)3 5 138.2 3 106 mm4 Moment of Inertia of Half Circle. Refer to Fig. 5.8 and determine the location of the centroid C of the half circle with respect to diameter AA9. As shown in Fig. 2, you have a 5 4r 3π 5 (4)(90 mm) 3π 5 38.2 mm The distance b from the centroid C to the x axis is b 5 120 mm 2 a 5 120 mm 2 38.2 mm 5 81.8 mm Referring now to Fig. 9.12, compute the moment of inertia of the half circle with respect to diameter AA9 and then compute the area of the half circle. IAA¿ 5 1 8 πr4 5 1 8 π(90 mm)4 5 25.76 3 106 mm4 A 5 1 2 πr2 5 1 2 π(90 mm)2 5 12.72 3 103 mm2 Next, using the parallel-axis theorem, obtain the value of Ix9 as IAA¿ 5 I x¿ 1 Aa2 25.76 3 106 mm4 5 I x9 1 (12.72 3 103 mm2)(38.2 mm)2 I x¿ 5 7.20 3 106 mm4 Again using the parallel-axis theorem, obtain the value of Ix as Ix 5 I x¿ 1 Ab2 5 7.20 3 106 mm4 1 (12.72 3 103 mm2)(81.8 mm)2 5 92.3 3 106 mm4 Moment of Inertia of Given Area. Subtracting the moment of inertia of the half circle from that of the rectangle, you obtain Ix 5 138.2 3 106 mm4 2 92.3 3 106 mm4 Ix 5 45.9 3 106 mm4 b REFLECT and THINK: Figures 5.8 and 9.12 are useful references for locating centroids and moments of inertia of common areas; don’t forget to use them.
505 505 SOLVING PROBLEMS ON YOUR OWN I n this section, we introduced the parallel-axis theorem and showed how to use it to simplify the computation of moments and polar moments of inertia of composite areas. The areas that you will consider in the following problems will consist of common shapes and rolled-steel shapes. You will also use the parallel-axis theorem to locate the point of application (the center of pressure) of the resultant of the hydrostatic forces acting on a submerged plane area.
1. Applying the parallel-axis theorem. In Sec. 9.2, we derived the parallel-axis theorem I 5 I 1 Ad2 (9.9) which states that the moment of inertia I of an area A with respect to a given axis is equal to the sum of the moment of inertia I of that area with respect to a parallel centroidal axis and the product Ad 2, where d is the distance between the two axes. It is important that you remember the following points as you use the parallel-axis theorem.
a. You can obtain the centroidal moment of inertia I _ of an area A by subtracting the product Ad 2 from the moment of inertia I of the area with respect to a parallel axis. It follows that the moment of inertia I is smaller than the moment of inertia I of the same area with respect to any parallel axis.
b. You can apply the parallel-axis theorem only if one of the two axes involved is a centroidal axis. Therefore, as we noted in Concept Application 9.3, to compute the moment of inertia of an area with respect to a noncentroidal axis when the moment of inertia of the area is known with respect to another noncentroidal axis, it is necessary to first compute the moment of inertia of the area with respect to a centroidal axis parallel to the two given axes.
2. Computing the moments and polar moments of inertia of composite areas. Sample Probs. 9.4 and 9.5 illustrate the steps you should follow to solve problems of this type. As with all composite-area problems, you should show on your sketch the common shapes or rolled-steel shapes that constitute the various elements of the given area, as well as the distances between the centroidal axes of the elements and the axes about which the moments of inertia are to be computed. In addition, it is important to note the following points. a. The moment of inertia of an area is always positive, regardless of the location of the axis with respect to which it is computed. As pointed out in the comments for the preceding section, only when an area is removed (as in the case of a hole) should you enter its moment of inertia in your computations with a minus sign.
(continued) 506 b. The moments of inertia of a semiellipse and a quarter ellipse can be determined by dividing the moment of inertia of an ellipse by 2 and 4, respectively. Note, however, that the moments of inertia obtained in this manner are with respect to the axes of symmetry of the ellipse. To obtain the centroidal moments of inertia of these shapes, use the parallel-axis theorem. This remark also applies to a semicircle and to a quarter circle. Also note that the expressions given for these shapes in Fig. 9.12 are not centroidal moments of inertia.
c. To calculate the polar moment of inertia of a composite area, you can use either the expressions given in Fig. 9.12 for JO or the relationship JO 5 Ix 1 Iy (9.4) depending on the shape of the given area. d. Before computing the centroidal moments of inertia of a given area, you may find it necessary to first locate the centroid of the area using the methods of Chap. 5.
3. Locating the point of application of the resultant of a system of hydrostatic forces. In Sec. 9.1, we found that R 5 γ # y dA 5 γ y A Mx 5 γ # y2 dA 5 γIx where y is the distance from the x axis to the centroid of the submerged plane area. Since R is equivalent to the system of elemental hydrostatic forces, it follows that oMx: yPR 5 Mx where yP is the depth of the point of application of R. Then yP(γ y A) 5 γIx or yP 5 Ix yA In closing, we encourage you to carefully study the notation used in Fig. 9.13 for the rolled-steel shapes, as you will likely encounter it again in subsequent engineering courses.
507 Problems 9.31 and 9.32 Determine the moment of inertia and the radius of gyra-tion of the shaded area with respect to the x axis.
y x O 12 mm 12 mm 8 mm 24 mm 24 mm 24 mm 6 mm 24 mm 6 mm Fig. P9.31 and P9.33 9.33 and 9.34 Determine the moment of inertia and the radius of gyra-tion of the shaded area with respect to the y axis. 9.35 and 9.36 Determine the moments of inertia of the shaded area shown with respect to the x and y axes.
y x a a a a O Fig. P9.35 y 2a 2a a a a a 3 2 a 3 2 x O Fig. P9.36 9.37 The centroidal polar moment of inertia JC of the 24-in2 shaded area is 600 in4. Determine the polar moments of inertia JB and JD of the shaded area knowing that JD 5 2JB and d 5 5 in. 9.38 Determine the centroidal polar moment of inertia JC of the 25-in2 shaded area knowing that the polar moments of inertia of the area with respect to points A, B, and D are, respectively, JA 5 281 in4, JB 5 810 in4, and JD 5 1578 in4.
Fig. P9.37 and P9.38 y x C A D 2a B a d y x O 2 in.
in.
2 in.
2 in.
1 in.
1 in.
1 in.
1 in.
1 2 in.
1 2 in.
1 2 in.
1 2 Fig. P9.32 and P9.34 508 9.39 Determine the shaded area and its moment of inertia with respect to the centroidal axis parallel to AA9 knowing that d1 5 25 mm and d2 5 10 mm and that its moments of inertia with respect to AA9 and BB9 are 2.2 3 106 mm4 and 4 3 106 mm4, respectively.
C A' d1 d2 A B' B Fig. P9.39 and P9.40 9.40 Knowing that the shaded area is equal to 6000 mm2 and that its moment of inertia with respect to AA9 is 18 3 106 mm4, determine its moment of inertia with respect to BB9 for d1 5 50 mm and d2 5 10 mm. 9.41 through 9.44 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB. 1.2 in.
A B 1.8 in.
5.0 in.
0.9 in.
2.0 in.
2.1 in.
Fig. P9.43 A B 1.3 in.
1.0 in.
0.5 in.
3.8 in.
0.5 in.
3.6 in.
Fig. P9.44 9.45 and 9.46 Determine the polar moment of inertia of the area shown with respect to (a) point O, (b) the centroid of the area.
O 6 in.
6 in.
4.5 in.
Semicircle Fig. P9.45 4 in.
4 in.
4 in.
4 in.
O Fig. P9.46 180 A B 60 60 60 Dimensions in mm 80 40 Fig. P9.41 42 mm 28 mm 36 mm A B Fig. P9.42 509 9.47 and 9.48 Determine the polar moment of inertia of the area shown with respect to (a) point O, (b) the centroid of the area.
O 40 Dimensions in mm 40 40 40 60 80 Fig. P9.47 9.49 Two channels and two plates are used to form the column section shown. For b 5 200 mm, determine the moments of inertia and the radii of gyration of the combined section with respect to the centroi-dal x and y axes.
10 mm C250 3 22.8 C b y x 375 mm Fig. P9.49 9.50 Two L6 3 4 3 1 2-in. angles are welded together to form the section shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes. 9.51 Four L3 3 3 3 1 4-in. angles are welded to a rolled W section as shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes. 9.52 Two 20-mm steel plates are welded to a rolled S section as shown. Determine the moments of inertia and the radii of gyration of the combined section with respect to the centroidal x and y axes.
S310 × 47.3 C x 80 mm 80 mm 20 mm y Fig. P9.52 O 84 mm 54 mm 27 mm 42 mm Semiellipses Fig. P9.48 C y x 5 in.
5 in.
1 4 W 8 3 31 L 3 3 3 3 Fig. P9.51 C y x 6 in.
4 in.
in.
1 2 Fig. P9.50 510 9.53 A channel and a plate are welded together as shown to form a section that is symmetrical with respect to the y axis. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes.
12 in.
0.5 in.
y x C8 × 11.5 C Fig. P9.53 9.54 The strength of the rolled W section shown is increased by welding a channel to its upper flange. Determine the moments of inertia of the combined section with respect to its centroidal x and y axes. 9.55 Two L76 3 76 3 6.4-mm angles are welded to a C250 3 22.8 channel. Determine the moments of inertia of the combined section with respect to centroidal axes respectively parallel and perpendicu-lar to the web of the channel.
C250 × 22.8 L76 × 76 × 6.4 Fig. P9.55 9.56 Two steel plates are welded to a rolled W section as indicated. Knowing that the centroidal moments of inertia Ix and Iy of the com-bined section are equal, determine (a) the distance a, (b) the moments of inertia with respect to the centroidal x and y axes. 9.57 and 9.58 The panel shown forms the end of a trough that is filled with water to the line AA9. Referring to Sec. 9.1A, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).
A a h A' Fig. P9.57 A A' h h b b Fig. P9.58 C W460 × 113 C250 × 22.8 y x Fig. P9.54 C y x 13 in.
1.0 in.
a 13 in.
W 14 × 38 Fig. P9.56 511 9.59 and 9.60 The panel shown forms the end of a trough that is filled with water to the line AA9. Referring to Sec. 9.1A, determine the depth of the point of application of the resultant of the hydrostatic forces acting on the panel (the center of pressure).
r A′ A Fig. P9.59 9.61 A vertical trapezoidal gate that is used as an automatic valve is held shut by two springs attached to hinges located along edge AB. Know-ing that each spring exerts a couple of magnitude 1470 N?m, deter-mine the depth d of water for which the gate will open. 9.62 The cover for a 0.5-m-diameter access hole in a water storage tank is attached to the tank with four equally spaced bolts as shown. Determine the additional force on each bolt due to the water pressure when the center of the cover is located 1.4 m below the water surface.
0.25 m 0.32 m C D A B Fig. P9.62 9.63 Determine the x coordinate of the centroid of the volume shown. (Hint: The height y of the volume is proportional to the x coordinate; consider an analogy between this height and the water pressure on a submerged surface.) 9.64 Determine the x coordinate of the centroid of the volume shown; this volume was obtained by intersecting an elliptic cylinder with an oblique plane. (See hint of Prob. 9.63.) x y z 64 mm 64 mm 39 mm 39 mm Fig. P9.64 a h a A' A Parabola Fig. P9.60 1.2 m 0.84 m 0.51 m 0.28 m d A B D E Fig. P9.61 z a b h 2b y x Fig. P9.63 512 9.65 Show that the system of hydrostatic forces acting on a submerged plane area A can be reduced to a force P at the centroid C of the area and two couples. The force P is perpendicular to the area and has a magnitude of P 5 γAy sin θ, where γ is the specific weight of the liquid. The couples are Mx9 5 (γIx9 sin θ)i and My9 5 (γI x9y9 sin θ)j, where Ix9y9 5 ex9y9dA (see Sec. 9.3). Note that the couples are independent of the depth at which the area is submerged.
x x' y y' C A q ⎯y Mx' My' ⎯x P Fig. P9.65 9.66 Show that the resultant of the hydrostatic forces acting on a sub-merged plane area A is a force P perpendicular to the area and of magnitude P 5 γAy sin θ 5 pA, where γ is the specific weight of the liquid and p is the pressure at the centroid C of the area. Show that P is applied at a point CP, called the center of pressure, whose coordinates are xp 5 Ixy /Ay and yp 5 Ix /Ay, where Ixy 5 exy dA (see Sec. 9.3). Show also that the difference of ordinates yp 2 y is equal to k 2 x¿/ y and thus depends upon the depth at which the area is submerged.
P x y x' y' A C CP q ⎯y ⎯x yP xP Fig. P9.66 9.3 Transformation of Moments of Inertia 513 9.3 TRANSFORMATION OF MOMENTS OF INERTIA The moments of inertia of an area can have different values depending on what axes we use to calculate them. It turns out that it is often important to determine the maximum and minimum values of the moments of inertia, which means finding the particular orientation of axes that produce these values. The first step in calculating moments of inertia with regard to rotated axes is to determine a new kind of second moment, called the product of inertia. In this section, we illustrate the procedures for this.
9.3A Product of Inertia The product of inertia of an area A with respect to the x and y axes is defined by the integral Product of inertia Ixy 5 #xy dA (9.12) We calculate it by multiplying each element dA of an area A by its coor-dinates x and y and integrating over the area (Fig. 9.14). Unlike the moments of inertia Ix and Iy, the product of inertia Ixy can be positive, negative, or zero. We will see shortly that the product of inertia is neces-sary for transforming moments of inertia with respect to a different set of axes; in a course on mechanics of materials, you will find other applica-tions of this quantity.
When one or both of the x and y axes are axes of symmetry for the area A, the product of inertia Ixy is zero. Consider, for example, the channel section shown in Fig. 9.15. Since this section is symmetrical with respect to the x axis, we can associate with each element dA of coordinates x and y an element dA9 of coordinates x and 2y. Clearly, the contributions to Ixy of any pair of elements chosen in this way cancel out, and the integral of Eq. (9.12) reduces to zero.
We can derive a parallel-axis theorem for products of inertia similar to the one established in Sec. 9.2 for moments of inertia. Consider an area A and a system of rectangular coordinates x and y (Fig. 9.16). Through the centroid C of the area, with coordinates x and y, we draw two centroidal axes x9 and y9 that are parallel, respectively, to the x and y axes. We denote the coordinates of an element of area dA with respect to the original axes by x and y, and the coordinates of the same element with respect to the centroidal axes by x9 and y9. This gives us x 5 x¿ 1 x and y 5 y¿ 1 y Substituting into Eq. (9.12), we obtain the expression for the product of inertia Ixy as Ixy 5# xy dA 5# (x¿ 1 x)(y¿ 1 y) dA 5# x¿y¿ dA 1 y # x¿ dA 1 x # y¿ dA 1 x y # dA Fig. 9.14 An element of area dA with coordinates x and y.
dA x y A O x y Fig. 9.15 If an area has an axis of symmetry, its product of inertia is zero.
dA' dA x y O –y y x Fig. 9.16 An element of area dA with respect to x and y axes and the centroidal axes x9 and y9 for area A.
x y O C y x dA x y y' y' x' x' 514 Distributed Forces: Moments of Inertia The first integral represents the product of inertia Ixy of the area A with respect to the centroidal axes x9 and y9. The next two integrals represent first moments of the area with respect to the centroidal axes; they reduce to zero, since the centroid C is located on these axes. The last integral is equal to the total area A. Therefore, we have Parallel-axis theorem for products of inertia Ixy 5 Ix9y9 1 xyA (9.13) 9.3B Principal Axes and Principal Moments of Inertia Consider an area A with coordinate axes x and y (Fig. 9.17) and assume that we know the moments and product of inertia of the area A. We have Ix 5# y2 dA Iy 5# x2 dA Ixy 5# xy dA (9.14) We propose to determine the moments and product of inertia Ix9, Iy9, and Ix9y9 of A with respect to new axes x9 and y9 that we obtain by rotating the original axes about the origin through an angle θ.
Fig. 9.17 An element of area dA with respect to x and y axes and a set of x9 and y9 axes rotated about the origin through an angle θ.
dA x x y y O y' y' x' x' q x cos q y sin q We first note that the relations between the coordinates x9, y9 and x, y of an element of area dA are x9 5 x cos θ 1 y sin θ y9 5 y cos θ 2 x sin θ Substituting for y9 in the expression for Ix9, we obtain Ix¿ 5# (y¿)2 dA 5# (y cos θ 2 x sin θ)2 dA 5 cos2 θ # y2 dA 2 2 sin θ cos θ # xy dA 1 sin2 θ # x2 dA Ix I y 5 Ix I 9y9 1 xyA yA 9.3 Transformation of Moments of Inertia 515 Fig. 9.18 Plots of Ix9y9 versus (a) Ix9 and (b) Iy9 for different values of the parameter θ are identical circles. The circle in part (a) indicates the average, maximum, and minimum values of the moment of inertia.
O M C A R B Ix'y' Ix' Ix' Ix'y' Imin Iave Imax (a) Iy' O C N R Ix'y' (b) Iave Iy' –Ix'y' Using the relations in Eq. (9.14), we have Ix9 5 Ix cos2 θ 2 2Ixy sin θ cos θ 1 Iy sin2 θ (9.15) Similarly, we obtain for Iy9 and Ix9y9 the expressions Iy9 5 Ix sin2 θ 1 2Ixy sin θ cos θ 1 Iy cos2 θ (9.16) Ix9y9 5 (Ix 2 Iy) sin θ cos θ 1 Ixy(cos2 θ 2 sin2 θ) (9.17) Recalling the trigonometric relations sin 2θ 5 2 sin θ cos θ cos 2θ 5 cos2 θ 2 sin2 θ and cos2 θ 5 1 1 cos 2θ 2 sin2 θ 5 1 2 cos 2θ 2 we can write Eqs. (9.15), (9.16), and (9.17) as Ix¿ 5 Ix 1 Iy 2 1 Ix 2 Iy 2 cos 2θ 2 Ixy sin 2θ (9.18) Iy¿ 5 Ix 1 Iy 2 2 Ix 2 Iy 2 cos 2θ 1 Ixy sin 2θ (9.19) Ix¿y¿ 5 Ix 2 Iy 2 sin 2θ 1 Ixy cos 2θ (9.20) Now, adding Eqs. (9.18) and (9.19), we observe that Ix9 1 Iy9 5 Ix 1 Iy (9.21) We could have anticipated this result, since both members of Eq. (9.21) are equal to the polar moment of inertia JO.
Equations (9.18) and (9.20) are the parametric equations of a circle. This means that, if we choose a set of rectangular axes and plot a point M of abscissa Ix9 and ordinate Ix9y9 for any given value of the parameter θ, all of the points will lie on a circle. To establish this property algebraically, we can eliminate θ from Eqs. (9.18) and (9.20) by transposing (Ix 1 Iy)/2 in Eq. (9.18), squaring both sides of Eqs. (9.18) and (9.20), and adding. The result is aIx9 2 Ix 1 Iy 2 b 2 1 I2 x9y9 5 a Ix 2 Iy 2 b 2 1 I2 xy (9.22) Setting Iave 5 Ix 1 Iy 2 and R 5 Ba Ix 2 Iy 2 b 2 1 I2 xy (9.23) we can write the identity equation (9.22) in the form (Ix9 2 Iave)2 1 I2 x9y9 5 R2 (9.24) This is the equation of a circle of radius R centered at the point C whose x and y coordinates are Iave and 0, respectively (Fig. 9.18a). Note that Eqs. (9.19) and (9.20) are parametric equations of the same circle. Furthermore, because of the symmetry of the circle about the 516 Distributed Forces: Moments of Inertia horizontal axis, we would obtain the same result if we plot a point N of coordinates Iy9 and 2 Ix9y9 (Fig. 9.18b) instead of plotting M. We will use this property in Sec. 9.4.
The two points A and B where this circle intersects the horizontal axis (Fig. 9.18a) are of special interest: Point A corresponds to the maximum value of the moment of inertia Ix9, whereas point B corresponds to its minimum value. In addition, both points correspond to a zero value of the product of inertia Ix9y9. Thus, we can obtain the values θm of the parameter θ corresponding to the points A and B by setting Ix9y9 5 0 in Eq. (9.20). The result is† tan 2θm 5 2Ixy Ix 2 Iy (9.25) This equation defines two values (2θm) that are 180° apart and thus two values (θm) that are 90° apart. One of these values corresponds to point A in Fig. 9.18a and to an axis through O in Fig. 9.17 with respect to which the moment of inertia of the given area is maximum. The other value corresponds to point B and to an axis through O with respect to which the moment of inertia of the area is minimum. These two perpendicular axes are called the principal axes of the area about O. The correspond-ing values Imax and Imin of the moment of inertia are called the principal moments of inertia of the area about O. Since we obtained the two values θm defined by Eq. (9.25) by setting Ix9y9 5 0 in Eq. (9.20), it is clear that the product of inertia of the given area with respect to its prin-cipal axes is zero.
Note from Fig. 9.18a that Imax 5 Iave 1 R Imin 5 Iave 2 R (9.26) Using the values for Iave and R from formulas (9.23), we obtain Imax, min 5 Ix 1 Iy 2 6 Ba Ix 2 Iy 2 b 2 1 I2 xy (9.27) Unless you can tell by inspection which of the two principal axes corre-sponds to Imax and which corresponds to Imin, you must substitute one of the values of θm into Eq. (9.18) in order to determine which of the two corresponds to the maximum value of the moment of inertia of the area about O.
Referring to Sec. 9.3A, note that, if an area possesses an axis of symmetry through a point O, this axis must be a principal axis of the area about O. On the other hand, a principal axis does not need to be an axis of symmetry; whether or not an area possesses any axes of symmetry, it will always have two principal axes of inertia about any point O.
The properties we have established hold for any point O located inside or outside the given area. If we choose the point O to coincide with the centroid of the area, any axis through O is a centroidal axis; the two principal axes of the area about its centroid are referred to as the principal centroidal axes of the area.
†We can also obtain this relation by differentiating Ix9 in Eq. (9.18) and setting dIx9/dθ 5 0.
9.3 Transformation of Moments of Inertia 517 Sample Problem 9.6 Determine the product of inertia of the right triangle shown (a) with respect to the x and y axes and (b) with respect to centroidal axes parallel to the x and y axes.
STRATEGY: You can approach this problem by using a vertical differential strip element. Because each point of the strip is at a different distance from the x axis, it is necessary to describe this strip mathemati-cally using the parallel-axis theorem. Once you have completed the solu-tion for the product of inertia with respect to the x and y axes, a second application of the parallel-axis theorem yields the product of inertia with respect to the centroidal axes.
MODELING and ANALYSIS: a. Product of Inertia Ixy. Choose a vertical rectangular strip as the differential element of area (Fig. 1). Using a differential version of the parallel-axis theorem, you have dIxy 5 dIx¿y¿ 1 xel yel dA The element is symmetrical with respect to the x9 and y9 axes, so dIx9y9 5 0. From the geometry of the triangle, you can express the variables in terms of x and y. y 5 h a1 2 x bb dA 5 y dx 5 h a1 2 x bb dx xel 5 x yel 5 1 2y 5 1 2h a1 2 x bb Integrating dIxy from x 5 0 to x 5 b gives you Ixy: Ixy 5# dIxy 5# xel yel dA 5# b 0 x(1 2)h2 a1 2 x bb 2 dx 5 h2# b 0 ax 2 2 x2 b 1 x3 2b2b dx 5 h2 c x2 4 2 x3 3b 1 x4 8b2 d b 0 Ixy 5 1 24 b2h2 b b. Product of Inertia I _ x0y0 . The coordinates of the centroid of the triangle relative to the x and y axes are (Fig. 2 and Fig. 5.8A) x 5 1 3 b y 5 1 3 h Using the expression for Ixy obtained in part a, apply the parallel-axis theorem again: Ixy 5 I x–y– 1 x yA 1 24b2h2 5 I x–y– 1 (1 3b)(1 3h)(1 2bh) I x–y– 5 1 24b2h2 2 1 18b2h2 Ix–y– 5 2 1 72b2h2 b REFLECT and THINK: An equally effective alternative strategy would be to use a horizontal strip element. Again, you would need to use the parallel-axis theorem to describe this strip, since each point in the strip would be a different distance from the y axis.
y x h b y y' x x' y h x dx b xel yel Fig. 1 Using a vertical rectangular strip as the differential element.
y y x x C b y x h Fig. 2 Centroid of the triangular area.
518 Distributed Forces: Moments of Inertia Sample Problem 9.7 For the section shown, the moments of inertia with respect to the x and y axes have been computed and are known to be Ix 5 10.38 in4 Iy 5 6.97 in4 Determine (a) the orientation of the principal axes of the section about O, (b) the values of the principal moments of inertia of the section about O.
STRATEGY: The first step is to compute the product of inertia with respect to the x and y axes, treating the section as a composite area of three rectangles. Then you can use Eq. (9.25) to find the principal axes and Eq. (9.27) to find the principal moments of inertia.
MODELING and ANALYSIS: Divide the area into three rectangles as shown (Fig. 1). Note that the product of inertia Ix9y9 with respect to centroidal axes parallel to the x and y axes is zero for each rectangle. Thus, using the parallel-axis theorem Ixy 5 Ix¿y¿ 1 xÊ yÊA you find that Ixy reduces to x y A for each rectangle.
Rectangle Area, in2 x , in. y , in. x _y _A, in4 I 1.5 21.25 11.75 23.28 II 1.5 0 0 0 III 1.5 11.25 21.75 23.28 oxyA 5 26.56 Ixy 5 ox yA 5 26.56 in4 a. Principal Axes. Since you know the magnitudes of Ix, Iy, and Ixy, you can use Eq. (9.25) to determine the values of θm (Fig. 2): tan 2θm 5 2 2Ixy Ix 2 Iy 5 2 2(26.56) 10.38 2 6.97 5 13.85 2θm 5 75.48 and 255.48 θm 5 37.78 and θm 5 127.78 b b. Principal Moments of Inertia. Using Eq. (9.27), you have Imax,min 5 Ix 1 Iy 2 6 Ba Ix 2 Iy 2 b 2 1 I2 xy 5 10.38 1 6.97 2 6 Ba10.38 2 6.97 2 b 2 1 (26.56)2 Imax 5 15.45 in4 Imin 5 1.897 in4 b REFLECT and THINK: Note that the elements of the area of the section are more closely distributed about the b axis than about the a axis. Therefore, you can conclude that Ia 5 Imax 5 15.45 in4 and Ib 5 Imin 5 1.897 in4. You can verify this conclusion by substituting θ 5 37.7° into Eqs. (9.18) and (9.19).
O 3 in.
3 in.
in.
1 2 4 in.
y x in.
1 2 in.
1 2 1.25 in.
1.25 in.
1.75 in.
1.75 in.
I II III O y x Fig. 1 Modeling the given area as three rectangles.
b a qm = 127.7° qm = 37.7° O y x Fig. 2 Orientation of principal axes.
519 519 SOLVING PROBLEMS ON YOUR OWN I n the problems for this section, you will continue your work with moments of inertia and use various techniques for computing products of inertia. Although the problems are generally straightforward, several items are worth noting.
1. Calculating the product of inertia Ixy by integration. We defined this quantity as Ixy 5 #xy dA (9.12) and stated that its value can be positive, negative, or zero. You can compute the product of inertia directly from this equation using double integration, or you can find it by using single integration as shown in Sample Prob. 9.6. When applying single integration and using the parallel-axis theorem, it is important to remember that in the equation dIxy 5 dIx9y9 1 xel yel dA x el and y el are the coordinates of the centroid of the element of area dA. Thus, if dA is not in the first quadrant, one or both of these coordinates is negative.
2. Calculating the products of inertia of composite areas. You can easily compute these quantities from the products of inertia of their component parts by using the parallel-axis theorem, as Ixy 5 I x9y9 1 x yA (9.13) The proper technique to use for problems of this type is illustrated in Sample Probs. 9.6 and 9.7. In addition to the usual rules for composite-area problems, it is essential that you remember the following points. a. If either of the centroidal axes of a component area is an axis of symmetry for that area, the product of inertia I _ x9y9 for that area is zero. Thus, I x9y9 is zero for com-ponent areas such as circles, semicircles, rectangles, and isosceles triangles, which possess an axis of symmetry parallel to one of the coordinate axes.
b. Pay careful attention to the signs of the coordinates x _ and y _ of each component area when you use the parallel-axis theorem [Sample Prob. 9.7].
3. Determining the moments of inertia and the product of inertia for rotated coor-dinate axes. In Sec. 9.3B, we derived Eqs. (9.18), (9.19), and (9.20) from which you can compute the moments of inertia and the product of inertia for coordinate axes that have been rotated about the origin O. To apply these equations, you must know a set of values Ix, Iy, and Ixy for a given orientation of the axes, and you must remember that θ is positive for counterclockwise rotations of the axes and negative for clockwise rotations of the axes.
4. Computing the principal moments of inertia. We showed in Sec. 9.3B that a particular orientation of the coordinate axes exists for which the moments of inertia attain their maxi-mum and minimum values, Imax and Imin, and for which the product of inertia is zero. Equa-tion (9.27) can be used to compute these values that are known as the principal moments of inertia of the area about O. The corresponding axes are referred to as the principal axes of the area about O, and their orientation is defined by Eq. (9.25). To determine which of the principal axes corresponds to Imax and which corresponds to Imin, you can either follow the procedure outlined in the text after Eq. (9.27) or observe about which of the two principal axes the area is more closely distributed; that axis corresponds to Imin [Sample Prob. 9.7].
520 520 Problems 9.67 through 9.70 Determine by direct integration the product of inertia of the given area with respect to the x and y axes.
x y h b Fig. P9.68 x y a a a y 5 k x Fig. P9.70 9.71 through 9.74 Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
x y 40 mm 40 mm 60 mm 60 mm C Fig. P9.72 0.25 in.
3 in.
2 in.
0.25 in.
0.487 in.
0.980 in.
y x C L3 × 2 × 1 4 Fig. P9.74 x y O x2 4a2 y2 a2 + = 1 a 2a Fig. P9.67 x y b y 5 kx1/2 a Fig. P9.69 20 mm 20 mm 60 mm 10 mm 10 mm 10 mm 100 mm 60 mm y x C Fig. P9.71 6 in.
6 in.
y x C Fig. P9.73 521 19 in.
15 in.
9 in.
3 in.
9 in.
3 in.
2 in.
2 in.
y x C Fig. P9.76 y C x 50.3 mm 152 mm L 152 3 102 3 12.7 24.9 mm 12.7 mm 102 mm 12.7 mm Fig. P9.78 9.75 through 9.78 Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
40 mm 8 mm 100 mm 8 mm 8 mm 40 mm y x C Fig. P9.75 1.3 in.
5.3 in.
1.0 in.
0.412 in.
0.5 in.
0.5 in.
2.25 in.
3.6 in.
y C x Fig. P9.77 9.79 Determine for the quarter ellipse of Prob. 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45° counterclockwise, (b) through 30° clockwise. 9.80 Determine the moments of inertia and the product of inertia of the area of Prob. 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 30° counterclockwise. 9.81 Determine the moments of inertia and the product of inertia of the area of Prob. 9.73 with respect to new centroidal axes obtained by rotating the x and y axes 60° counterclockwise. 9.82 Determine the moments of inertia and the product of inertia of the area of Prob. 9.75 with respect to new centroidal axes obtained by rotating the x and y axes 45° clockwise. 9.83 Determine the moments of inertia and the product of inertia of the L3 3 2 3 1 4-in. angle cross section of Prob. 9.74 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise.
522 9.84 Determine the moments of inertia and the product of inertia of the L152 3 102 3 12.7-mm angle cross section of Prob. 9.78 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise. 9.85 For the quarter ellipse of Prob. 9.67, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. 9.86 through 9.88 For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. 9.86 Area of Prob. 9.72 9.87 Area of Prob. 9.73 9.88 Area of Prob. 9.75 9.89 and 9.90 For the angle cross section indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia. 9.89 The L3 3 2 3 1 4-in. angle cross section of Prob. 9.74 9.90 The L152 3 102 3 12.7-mm angle cross section of Prob. 9.78 9.4 Mohr’s Circle for Moments of Inertia 523 9.4 MOHR’S CIRCLE FOR MOMENTS OF INERTIA The circle introduced in the preceding section to illustrate the relations between the moments and products of inertia of a given area with respect to axes passing through a fixed point O was first introduced by the German engineer Otto Mohr (1835–1918) and is known as Mohr’s circle. Here, we show that, if we know the moments and product of inertia of an area A with respect to two rectangular x and y axes that pass through a point O, we can use Mohr’s circle to graphically determine (a) the principal axes and principal moments of inertia of the area about O and (b) the moments and product of inertia of the area with respect to any other pair of rect-angular axes x9 and y9 through O.
Consider a given area A and two rectangular coordinate axes x and y (Fig. 9.19a). Assuming that we know the moments of inertia Ix and Iy and the product of inertia Ixy, we can represent them on a diagram by plotting a point X with coordinates Ix and Ixy and a point Y with coordinates Iy and 2Ixy (Fig. 9.19b). If Ixy is positive, as assumed in Fig. 9.19a, then point X is located above the horizontal axis and point Y is located below, as shown in Fig. 9.19b. If Ixy is negative, X is located below the horizontal axis and Y is located above. Joining X and Y with a straight line, we denote the point of intersection of line XY with the horizontal axis by C. Then we draw the circle of center C and diameter XY. Noting that the abscissa of C and the radius of the circle are respectively equal to the quantities Iave and R defined by formula (9.23), we conclude that the circle obtained is Mohr’s circle for the given area about point O. Thus, the abscissas of the points A and B where the circle intersects the horizontal axis represent, respectively, the principal moments of inertia Imax and Imin of the area.
Also note that, since tan (XCA) 5 2Ixy/(Ix 2 Iy), the angle XCA is equal in magnitude to one of the angles 2θm that satisfy Eq. (9.25). Thus, the angle θm, which in Fig. 9.19a defines the principal axis Oa corresponding Fig. 9.19 (a) An area A with principal axes Oa and Ob and axes Ox9 and Oy9 obtained by rotation through an angle θ; (b) Mohr’s circle used to calculate angles and moments of inertia.
x' x y' q qm O b a y (a) 2q A B C Y 2qm O Imin Imax Ixy Ix' Ix'y' Ix'y' Ixy Ixy Ix, Iy Iy X X' Ix Iy' Y' (b) 524 Distributed Forces: Moments of Inertia to point A in Fig. 9.19b, is equal to half of the angle XCA of Mohr’s circle. In addition, if Ix . Iy and Ixy . 0, as in the case considered here, the rota-tion that brings CX into CA is clockwise. Also, under these conditions, the angle θm obtained from Eq. (9.25) is negative; thus, the rotation that brings Ox into Oa is also clockwise. We conclude that the senses of rotation in both parts of Fig. 9.19 are the same. If a clockwise rotation through 2θm is required to bring CX into CA on Mohr’s circle, a clockwise rotation through θm will bring Ox into the corresponding principal axis Oa in Fig. 9.19a.
Since Mohr’s circle is uniquely defined, we can obtain the same circle by considering the moments and product of inertia of the area A with respect to the rectangular axes x9 and y9 (Fig. 9.19a). The point X9 with coordinates Ix9 and Ix9y9 and the point Y9 with coordinates Iy9 and 2Ix9y9 are thus located on Mohr’s circle, and the angle X9CA in Fig. 9.19b must be equal to twice the angle x9Oa in Fig. 9.19a. Since, as noted previously, the angle XCA is twice the angle xOa, it follows that the angle XCX9 in Fig. 9.19b is twice the angle xOx9 in Fig. 9.19a. The diameter X9Y9, which defines the moments and product of inertia Ix9, Iy9, and Ix9y9 of the given area with respect to rectangular axes x9 and y9 forming an angle θ with the x and y axes, can be obtained by rotating through an angle 2θ the diameter XY, which corresponds to the moments and product of inertia Ix , Iy, and Ixy. Note that the rotation that brings the diameter XY into the diameter X9Y9 in Fig. 9.19b has the same sense as the rotation that brings the x and y axes into the x9 and y9 axes in Fig. 9.19a.
Finally, also note that the use of Mohr’s circle is not limited to graphical solutions, i.e., to solutions based on the careful drawing and measuring of the various parameters involved. By merely sketching Mohr’s circle and using trigonometry, you can easily derive the various relations required for a numerical solution of a given problem (see Sample Prob. 9.8).
Sample Problem 9.8 For the section shown, the moments and product of inertia with respect to the x and y axes are Ix 5 7.20 3 106 mm4 Iy 5 2.59 3 106 mm4 Ixy 5 22.54 3 106 mm4 Using Mohr’s circle, determine (a) the principal axes of the section about O, (b) the values of the principal moments of inertia of the section about O, and (c) the moments and product of inertia of the section with respect to the x9 and y9 axes that form an angle of 60° with the x and y axes.
STRATEGY: You should carefully draw Mohr’s circle and use the geometry of the circle to determine the orientation of the principal axes. Then complete the analysis for the requested moments of inertia.
MODELING: Drawing Mohr’s Circle. First plot point X with coordinates Ix 5 7.20, Ixy 5 22.54, and plot point Y with coordinates Iy 5 2.59, 2Ixy 5 12.54. Join X and Y with a straight line to define the center C y x y x O q = 60° L152 × 102 × 12.7 9.4 Mohr’s Circle for Moments of Inertia 525 of Mohr’s circle (Fig. 1). You can measure the abscissa of C, which represents Iave, and the radius R of the circle either directly or using Iave 5 OC 5 1 2(Ix1 Iy) 5 1 2(7.20 3106 1 2.59 3106) 54.895 3106 mm4 CD 5 1 2(Ix 2 Iy) 5 1 2(7.20 3 106 2 2.59 3 106) 5 2.305 3 106 mm4 R 5 2(CD)2 1 (DX)2 5 2(2.305 3 106)2 1 (2.54 3 106)2 5 3.430 3 106 mm4 ANALYSIS: a. Principal Axes. The principal axes of the section correspond to points A and B on Mohr’s circle, and the angle through which you should rotate CX to bring it into CA defines 2θm. You obtain tan 2θm 5 DX CD 5 2.54 2.305 5 1.102 2θm 5 47.88 l θm 5 23.98 l b Thus, the principal axis Oa corresponding to the maximum value of the moment of inertia is obtained by rotating the x axis through 23.9° counterclockwise; the principal axis Ob corresponding to the minimum value of the moment of inertia can be obtained by rotating the y axis through the same angle (Fig. 2).
b. Principal Moments of Inertia. The principal moments of inertia are represented by the abscissas of A and B. The results are Imax 5 OA 5 OC 1 CA 5 Iave 1 R 5 (4.895 1 3.430)106 mm4 Imax 5 8.33 3 106 mm4 b Imin 5 OB 5 OC 2 BC 5 Iave 2 R 5 (4.895 2 3.430)106 mm4 Imin 5 1.47 3 106 mm4 b c. Moments and Product of Inertia with Respect to the x9 and y9 Axes. On Mohr’s circle, you obtain the points X9 and Y9, which correspond to the x9 and y9 axes, by rotating CX and CY through an angle 2θ 5 2(60°) 5 120° counterclockwise (Fig. 3). The coordinates of X9 and Y9 yield the desired moments and product of inertia. Noting that the angle that CX9 forms with the horizontal axis is ϕ 5 120° 2 47.8° 5 72.2°, you have Ix9 5 OF 5 OC 1 CF 5 4.895 3 106 mm4 1 (3.430 3 106 mm4) cos 72.2° Ix9 5 5.94 3 106 mm4 b Iy9 5 OG 5 OC 2 GC 5 4.895 3 106 mm4 2 (3.430 3 106 mm4) cos 72.2° Iy9 5 3.85 3 106 mm4 b Ix9y9 5 FX9 5 (3.430 3 106 mm4) sin 72.2° Ix9y9 5 3.27 3 106 mm4 b REFLECT and THINK: This problem illustrates typical calculations with Mohr’s circle. The technique is a useful one to learn and remember.
2qm = 47.8° 2q = 120° 4.895 × 106 mm4 3.430 × 106 mm4 Ixy X Y F G X' Y' f O C Ix, Iy Fig. 3 Using Mohr’s circle to determine the moments and product of inertia with respect to x9 and y9 axes.
O B E C D A Ixy(106 mm4) Y(2.59, +2.54) X(7.20, –2.54) 2qm Ix, Iy (106 mm4) Fig. 1 Mohr’s circle.
y x b a O qm = 23.9° Fig. 2 Orientation of the principal axes.
526 526 SOLVING PROBLEMS ON YOUR OWN I n the problems for this section, you will use Mohr’s circle to determine the moments and products of inertia of a given area for different orientations of the coordinate axes. Although in some cases using Mohr’s circle may not be as direct as substituting into the appropriate equations [Eqs. (9.18) through (9.20)], this method of solution has the advan-tage of providing a visual representation of the relationships among the variables involved. Also, Mohr’s circle shows all of the values of the moments and products of inertia that are possible for a given problem.
Using Mohr’s circle. We presented the underlying theory in Sec. 9.3B, and we discussed the application of this method in Sec. 9.4 and in Sample Prob. 9.8. In the same problem, we presented the steps you should follow to determine the principal axes, the principal moments of inertia, and the moments and product of inertia with respect to a specified orientation of the coordinates axes. When you use Mohr’s circle to solve problems, it is important that you remember the following points. a. Mohr’s circle is completely defined by the quantities R and Iave, which represent, respectively, the radius of the circle and the distance from the origin O to the center C of the circle. You can obtain these quantities from Eqs. (9.23) if you know the moments and product of inertia for a given orientation of the axes. However, Mohr’s circle can be defined by other combinations of known values [Probs. 9.103, 9.106, and 9.107]. For these cases, it may be necessary to first make one or more assumptions, such as choosing an arbitrary location for the center when Iave is unknown, assigning relative magnitudes to the moments of inertia (for example, Ix . Iy), or selecting the sign of the product of inertia. b. Point X of coordinates (Ix, Ixy) and point Y of coordinates (Iy, 2Ixy) are both located on Mohr’s circle and are diametrically opposite. c. Since moments of inertia must be positive, all of Mohr’s circle must lie to the right of the Ixy axis; it follows that Iave . R for all cases. d. As the coordinate axes are rotated through an angle θ, the associated rotation of the diameter of Mohr’s circle is equal to 2θ and is in the same sense (clockwise or coun-terclockwise). We strongly suggest that you label the known points on the circumference of the circle with the appropriate capital letter, as was done in Fig. 9.19b and for the Mohr circles of Sample Prob. 9.8. This will enable you to determine the sign of the correspond-ing product of inertia for each value of θ and which moment of inertia is associated with each of the coordinate axes [Sample Prob. 9.8, parts a and c].
Although we have introduced Mohr’s circle within the specific context of the study of moments and products of inertia, the Mohr circle technique also applies to the solution of analogous but physically different problems in mechanics of materials. This multiple use of a specific technique is not unique, and as you pursue your engineering studies, you will encounter several methods of solution that can be applied to a variety of problems.
527 Problems 9.91 Using Mohr’s circle, determine for the quarter ellipse of Prob. 9.67 the moments of inertia and the product of inertia with respect to new axes obtained by rotating the x and y axes about O (a) through 45° counterclockwise, (b) through 30° clockwise. 9.92 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Prob. 9.72 with respect to new centroidal axes obtained by rotating the x and y axes 30° counterclockwise. 9.93 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Prob. 9.73 with respect to new centroidal axes obtained by rotating the x and y axes 60° counterclockwise. 9.94 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area of Prob. 9.75 with respect to new centroidal axes obtained by rotating the x and y axes 45° clockwise. 9.95 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the L3 3 2 3 1 4-in. angle cross section of Prob. 9.74 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise. 9.96 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the L152 3 102 3 12.7-mm angle cross section of Prob. 9.78 with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise. 9.97 For the quarter ellipse of Prob. 9.67, use Mohr’s circle to determine the orientation of the principal axes at the origin and the correspond-ing values of the moments of inertia. 9.98 through 9.102 Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes and the corresponding values of the moments of inertia. 9.98 Area of Prob. 9.72 9.99 Area of Prob. 9.76 9.100 Area of Prob. 9.73 9.101 Area of Prob. 9.74 9.102 Area of Prob. 9.77 (The moments of inertia Ix and Iy of the area of Prob. 9.102 were determined in Prob. 9.44) 9.103 The moments and product of inertia of an L4 3 3 3 1 4-in. angle cross section with respect to two rectangular axes x and y through C are, respectively, Ix 5 1.33 in4, Iy 5 2.75 in4, and Ixy , 0, with the mini-mum value of the moment of inertia of the area with respect to any axis through C being Imin5 0.692 in4. Using Mohr’s circle, determine (a) the product of inertia Ixy of the area, (b) the orientation of the principal axes, (c) the value of Imax.
528 9.104 and 9.105 Using Mohr’s circle, determine the orientation of the principal centroidal axes and the corresponding values of the moments of inertia for the cross section of the rolled-steel angle shown. (Properties of the cross sections are given in Fig. 9.13.) C x y 18.9 mm L127 × 76 × 12.7 12.7 mm 12.7 mm 127 mm 44.2 mm 76 mm Fig. P9.105 9.106 For a given area, the moments of inertia with respect to two rectangular centroidal x and y axes are Ix 5 1200 in4 and Iy 5 300 in4, respectively. Knowing that, after rotating the x and y axes about the centroid 30° counterclockwise, the moment of inertia relative to the rotated x axis is 1450 in4, use Mohr’s circle to determine (a) the orientation of the principal axes, (b) the principal centroidal moments of inertia. 9.107 It is known that for a given area Iy 5 48 3 106 mm4 and Ixy 5 220 3 106 mm4, where the x and y axes are rectangular centroidal axes. If the axis corresponding to the maximum product of inertia is obtained by rotating the x axis 67.5° counterclockwise about C, use Mohr’s circle to determine (a) the moment of inertia Ix of the area, (b) the principal centroidal moments of inertia. 9.108 Using Mohr’s circle, show that for any regular polygon (such as a pentagon) (a) the moment of inertia with respect to every axis through the centroid is the same, (b) the product of inertia with respect to every pair of rectangular axes through the centroid is zero. 9.109 Using Mohr’s circle, prove that the expression Ix¿Iy¿ 2 I2 x¿y¿ is independent of the orientation of the x9 and y9 axes, where Ix9, Iy9, and Ix9y9 represent the moments and product of inertia, respectively, of a given area with respect to a pair of rectangular axes x9 and y9 through a given point O. Also show that the given expression is equal to the square of the length of the tangent drawn from the origin of the coordinate system to Mohr’s circle. 9.110 Using the invariance property established in the preceding problem, express the product of inertia Ixy of an area A with respect to a pair of rectangular axes through O in terms of the moments of inertia Ix and Iy of A and the principal moments of inertia Imin and Imax of A about O. Use the formula obtained to calculate the product of inertia Ixy of the L3 3 2 3 1 4-in. angle cross section shown in Fig. 9.13A, knowing that its maximum moment of inertia is 1.257 in4.
6.4 mm 76 mm 51 mm 6.4 mm 12.4 mm 24.9 mm y x C L76 × 51 × 6.4 Fig. P9.104 9.5 Mass Moments of Inertia 529 9.5 MASS MOMENTS OF INERTIA So far in this chapter, we have examined moments of inertia of areas. In the rest of this chapter, we consider moments of inertia associated with the masses of bodies. This will be an important concept in dynamics when studying the rotational motion of a rigid body about an axis.
9.5A Moment of Inertia of a Simple Mass Consider a small mass Dm mounted on a rod of negligible mass that can rotate freely about an axis AA9 (Fig. 9.20a). If we apply a couple to the system, the rod and mass (assumed to be initially at rest) start rotating about AA9. We will study the details of this motion later in dynamics. At present, we wish to indicate only that the time required for the system to reach a given speed of rotation is proportional to the mass Dm and to the square of the distance r. The product r 2 Dm thus provides a measure of the inertia of the system; i.e., a measure of the resistance the system offers when we try to set it in motion. For this reason, the product r 2 Dm is called the moment of inertia of the mass Dm with respect to axis AA9.
Now suppose a body of mass m is to be rotated about an axis AA9 (Fig. 9.20b). Dividing the body into elements of mass Dm1, Dm2, etc., we find that the body’s resistance to being rotated is measured by the sum r 2 1 Dm1 1 r 2 2 Dm2 1 . . . . This sum defines the moment of inertia of the body with respect to axis AA9. Increasing the number of elements, we find that the moment of inertia is equal, in the limit, to the integral Moment of inertia of a mass I 5 #r 2 dm (9.28) I 5 #r 2dm d Fig. 9.20 (a) An element of mass Dm at a distance r from an axis AA9; (b) the moment of inertia of a rigid body is the sum of the moments of inertia of many small masses; (c) the moment of inertia is unchanged if all the mass is concentrated at a point at a distance from the axis equal to the radius of gyration.
A' A r1 r2 r3 ∆m1 ∆m2 ∆m ∆m3 A' A m A' A r k (a) (b) (c) 530 Distributed Forces: Moments of Inertia We define the radius of gyration k of the body with respect to axis AA9 by the relation Radius of gyration of a mass I 5 k2m or k 5 B I m (9.29) The radius of gyration k represents the distance at which the entire mass of the body should be concentrated if its moment of inertia with respect to AA9 is to remain unchanged (Fig. 9.20c). Whether it stays in its original shape (Fig. 9.20b) or is concentrated as shown in Fig. 9.20c, the mass m reacts in the same way to a rotation (or gyration) about AA9.
If SI units are used, the radius of gyration k is expressed in meters and the mass m in kilograms, so the unit for the moment of inertia of a mass is kg?m2. If U.S. customary units are used, the radius of gyration is expressed in feet and the mass in slugs (i.e., in lb?s2/ft), so the derived unit for the moment of inertia of a mass is lb?ft?s2.† We can express the moment of inertia of a body with respect to a coordinate axis in terms of the coordinates x, y, z of the element of mass dm (Fig. 9.21). Noting, for example, that the square of the distance r from the element dm to the y axis is z2 1 x2, the moment of inertia of the body with respect to the y axis is Iy 5 #r2 dm 5 # (z2 1 x2) dm We obtain similar expressions for the moments of inertia with respect to the x and z axes. Moments of inertia with respect to coordinate axes Ix 5 # (y2 1 z2) dm Iy 5 # (z2 1 x2) dm (9.30) Iz 5 # (x2 1 y2) dm 9.5B Parallel-Axis Theorem for Mass Moments of Inertia Consider again a body of mass m and let Oxyz be a system of rectangular coordinates whose origin is at the arbitrary point O. Let Gx9y9z9 be a system of parallel centroidal axes; i.e., a system whose origin is at the I 5 k2m or k 5 B I m B Ix I 5 # (y2 1 z2)dm d Iy I 5 # (z2 1 x2)dm d Iz I 5 # (x2 1 y2)dm d Fig. 9.21 An element of mass dm in an x, y, z coordinate system.
dm x y y O z r z x †When converting the moment of inertia of a mass from U.S. customary units to SI units, keep in mind that the base unit (pound) used in the derived unit (lb?ft?s2) is a unit of force (not of mass). Therefore, it should be converted into newtons. We have 1 lb ?ft?s2 5 (4.45 N)(0.3048 m)(1 s)2 5 1.356 N?m?s2 or since 1 N 5 1 kg?m/s2 1 lb?ft?s2 5 1.356 kg?m2 Photo 9.2 The rotational behavior of this crankshaft depends upon its mass moment of inertia with respect to its axis of rotation, as you will see in a dynamics course.
9.5 Mass Moments of Inertia 531 center of gravity G of the body and whose axes x9, y9, z9 are parallel to the x, y, and z axes, respectively (Fig. 9.22). (Note that we use the term centroidal here to define axes passing through the center of gravity G of the body, regardless of whether or not G coincides with the centroid of the volume of the body.) We denote by x, y, z the coordinates of G with respect to Oxyz. Then we have the following relations between the coor-dinates x, y, z of the element dm with respect to Oxyz and its coordinates x9, y9, z9 with respect to the centroidal axes Gx9y9z9: x 5 x9 1 x y 5 y9 1 y z 5 z9 1 z (9.31) Referring to Eqs. (9.30), we can express the moment of inertia of the body with respect to the x axis as Ix 5# (y2 1 z2) dm 5# [(y¿ 1 y )2 1 (z¿ 1 z )2] dm 5# (y¿2 1 z¿2) dm 1 2 y # y¿ dm 1 2z # z¿ dm 1 (y 2 1 z 2)# dm The first integral in this expression represents the moment of inertia I x¿ of the body with respect to the centroidal axis x9. The second and third integrals represent the first moment of the body with respect to the z9x9 and x9y9 planes, respectively, and since both planes contain G, these two integrals are zero. The last integral is equal to the total mass m of the body. Therefore, we have Ix 5 Ix¿ 1 m(y 2 1 z 2) (9.32) Similarly, Iy 5 Iy9 1 m(z 2 1 x 2) Iz 5 Iz9 1 m(x 2 1 y 2) (9.329) We easily verify from Fig. 9.22 that the sum z 2 1 x 2 represents the square of the distance OB between the y and y9 axes. Similarly, y 2 1 z 2 and x 2 1 y 2 represent the squares of the distance between the x and x9 axes and the z and z9 axes, respectively. We denote the distance between an arbitrary axis AA9 and a parallel centroidal axis BB9 by d (Fig. 9.23). Then the general relation between the moment of inertia I of the body with respect to AA9 and its moment of inertia I with respect to BB9, known as the parallel-axis theorem for mass moments of inertia, is Parallel-axis theorem for mass moments of inertia I 5 I 1 md 2 (9.33) Expressing the moments of inertia in terms of the corresponding radii of gyration, we can also write k 2 5 k 2 1 d 2 (9.34) where k and k represent the radii of gyration of the body about AA9 and BB9, respectively.
Ix I 5 Ix I ¿ 1 m(y 2 1 z 2) Iy I 5 Iy I 9 1 m(z 2 1 x 2) Iz I 5 Iz I 9 1 m(x 2 1 y 2) I 5 I 1 md 2 Fig. 9.23 We use d to denote the distance between an arbitrary axis AA9 and a parallel centroidal axis BB9.
A' B' A B G d Fig. 9.22 A body of mass m with an arbitrary rectangular coordinate system at O and a parallel centroidal coordinate system at G. Also shown is an element of mass dm.
dm y O G B z x y' x' z' z y x 532 Distributed Forces: Moments of Inertia 9.5C Moments of Inertia of Thin Plates Now imagine a thin plate of uniform thickness t, made of a homogeneous material of density ρ (density 5 mass per unit volume). The mass moment of inertia of the plate with respect to an axis AA9 contained in the plane of the plate (Fig. 9.24a) is IAA9, mass 5 #r2 dm Fig. 9.24 (a) A thin plate with an axis AA9 in the plane of the plate; (b) an axis BB9 in the plane of the plate and perpendicular to AA9; (c) an axis CC9 perpendicular to the plate and passing through the intersection of AA9 and BB9.
t B' C' r r A' A dA C (a) (b) (c) A' A B B' B dA dA r r t t Since dm 5 ρt dA, we have IAA9, mass 5 ρt#r2 d A However, r represents the distance of the element of area dA to the axis AA9. Therefore, the integral is equal to the moment of inertia of the area of the plate with respect to AA9. IAA9, mass 5 ρtIAA9, area (9.35) Similarly, for an axis BB9 that is contained in the plane of the plate and is perpendicular to AA9 (Fig. 9.24b), we have IBB9, mass 5 ρtIBB9, area (9.36) Consider now the axis CC9, which is perpendicular to the plate and passes through the point of intersection C of AA9 and BB9 (Fig. 9.24c). This time we have ICC9, mass 5 ρtJC, area (9.37) where JC is the polar moment of inertia of the area of the plate with respect to point C.
Recall the relation JC 5 IAA9 1 IBB9 between the polar and rectan-gular moments of inertia of an area. We can use this to write the relation between the mass moments of inertia of a thin plate as ICC9 5 IAA9 1 IBB9 (9.38) 9.5 Mass Moments of Inertia 533 Rectangular Plate. In the case of a rectangular plate of sides a and b (Fig. 9.25), we obtain the mass moments of inertia with respect to axes through the center of gravity of the plate as IAA¿, mass 5 rtIAA¿, area 5 rt( 1 12 a3b) IBB¿, mass 5 rtIBB¿, area 5 rt( 1 12 ab3) Since the product ρabt is equal to the mass m of the plate, we can also write the mass moments of inertia of a thin rectangular plate as IAA9 5 1 12 ma2 IBB9 5 1 12 mb2 (9.39) ICC¿ 5 IAA¿ 1 IBB¿ 5 1 12 m(a2 1 b2) (9.40) Circular Plate. In the case of a circular plate, or disk, of radius r (Fig. 9.26), Eq. (9.35) becomes IAA¿, mass 5 rtIAA¿, area 5 rt(1 4 pr4) In this case, the product ρπr2t is equal to the mass m of the plate, and IAA9 5 IBB9. Therefore, we can write the mass moments of inertia of a circular plate as IAA¿ 5 IBB¿ 5 1 4 mr2 (9.41) ICC¿ 5 IAA¿ 1 IBB¿ 5 1 2 mr2 (9.42) 9.5D Determining the Moment of Inertia of a Three-Dimensional Body by Integration We obtain the moment of inertia of a three-dimensional body by evaluating the integral I 5 er2 dm. If the body is made of a homogeneous material with a density ρ, the element of mass dm is equal to ρ dV, and we have I 5 ρer2 dV. This integral depends only upon the shape of the body. Thus, in order to compute the moment of inertia of a three-dimensional body, it is generally necessary to perform a triple, or at least a double, integration.
However, if the body possesses two planes of symmetry, it is usually possible to determine the body’s moment of inertia with a single integra-tion. We do this by choosing as the element of mass dm a thin slab that is perpendicular to the planes of symmetry. In the case of bodies of revo-lution, for example, the element of mass is a thin disk (Fig. 9.27). Using formula (9.42), we can express the moment of inertia of the disk with respect to the axis of revolution as indicated in Fig. 9.27. Its moment of inertia with respect to each of the other two coordinate axes is obtained by using formula (9.41) and the parallel-axis theorem. Integration of these expressions yields the desired moment of inertia of the body.
9.5E Moments of Inertia of Composite Bodies Figure 9.28 lists the moments of inertia of a few common shapes. For a body consisting of several of these simple shapes in combination, you can obtain the moment of inertia of the body with respect to a given axis by first computing the moments of inertia of its component parts about the desired axis and then adding them together. As was the case for areas, the radius of gyration of a composite body cannot be obtained by adding the radii of gyration of its component parts.
Fig. 9.27 Using a thin disk to determine the moment of inertia of a body of revolution.
O y' y z dx r z' x x dm = r r2 dx dIx = r2 dm 1 2 dIy = dIy' + x2 dm = ( r2 + x2)dm dIz = dIz' + x2 dm = ( r2 + x2)dm 1 4 1 4 Fig. 9.25 A thin rectangular plate of sides a and b.
t C' B' A B b a A' C Fig. 9.26 A thin circular plate of radius r.
C' C B' A B A' t r 534 Distributed Forces: Moments of Inertia Fig. 9.28 Mass moments of inertia of common geometric shapes.
Slender rod Thin rectangular plate Rectangular prism Thin disk Circular cylinder Circular cone Sphere G L Iy = Iz = mL2 1 12 Iy = Iz = mr2 1 4 Ix = m(b2 + c2) 1 12 Ix = m(b2 + c2) 1 12 Ix = mr2 1 2 Ix = ma2 3 10 Iy = m(c2 + a2) 1 12 Iy = Iz = m( a2 + h2) 3 5 1 4 Ix = ma2 1 2 Iy = Iz = m(3a2 + L2) 1 12 Ix = Iy = Iz = ma2 2 5 Iz = m(a2 + b2) 1 12 Iy = mc2 1 12 Iz = mb2 1 12 a x x x x a x x x r z z z z z z z y y y y y y h L a c c b b G r y 9.5 Mass Moments of Inertia 535 Sample Problem 9.9 Determine the moment of inertia of a slender rod of length L and mass m with respect to an axis that is perpendicular to the rod and passes through one end.
STRATEGY: Approximating the rod as a one-dimensional body enables you to solve the problem by a single integration.
MODELING and ANALYSIS: Choose the differential element of mass shown in Fig. 1 and express it as a mass per unit length.
dm 5 m L dx Iy 5# x2 dm 5# L 0 x2 m L dx 5 c m L x3 3 d L 0 Iy 5 1 3 mL2 b REFLECT and THINK: This problem could also have been solved by starting with the moment of inertia for a slender rod with respect to its centroid, as given in Fig. 9.28, and using the parallel-axis theorem to obtain the moment of inertia with respect to an end of the rod.
x y z L x x y z L dx Fig. 1 Differential element of mass.
Sample Problem 9.10 For the homogeneous rectangular prism shown, determine the moment of inertia with respect to the z axis.
STRATEGY: You can approach this problem by choosing a differential element of mass perpendicular to the long axis of the prism; find its moment of inertia with respect to a centroidal axis parallel to the z axis; and then apply the parallel-axis theorem.
MODELING and ANALYSIS: Choose as the differential element of mass the thin slab shown in Fig. 1. Then dm 5 ρbc dx Referring to Sec. 9.5C, the moment of inertia of the element with respect to the z9 axis is dIz9 5 1 12 b2 dm Applying the parallel-axis theorem, you can obtain the mass moment of inertia of the slab with respect to the z axis.
dIz 5 dIz9 1 x2 dm 5 1 12 b2 dm 1 x2 dm 5 ( 1 12 b2 1 x2)rbc dx Integrating from x 5 0 to x 5 a gives you Iz 5# dIz 5# a 0 ( 1 12 b2 1 x2)ρbc dx 5 ρabc( 1 12 b2 1 1 3a2) Since the total mass of the prism is m 5 ρabc, you can write Iz 5 m( 1 12 b2 1 1 3 a2) Iz 5 1 12 m(4a2 1 b2) b REFLECT and THINK: Note that if the prism is thin, b is small com-pared to a, and the expression for Iz reduces to 1 3 ma2, which is the result obtained in Sample Prob. 9.9 when L 5 a.
x y a c b z dx y z' x x z Fig. 1 Differential element of mass.
536 Distributed Forces: Moments of Inertia Sample Problem 9.11 Determine the moment of inertia of a right circular cone with respect to (a) its longitudinal axis, (b) an axis through the apex of the cone and perpendicular to its longitudinal axis, (c) an axis through the centroid of the cone and perpendicular to its longitudinal axis.
STRATEGY: For parts (a) and (b), choose a differential element of mass in the form of a thin circular disk perpendicular to the longitudinal axis of the cone. You can solve part (c) by an application of the parallel-axis theorem.
MODELING and ANALYSIS: Choose the differential element of mass shown in Fig. 1. Express the radius and mass of this disk as r 5 a x h dm 5 ρπr2 dx 5 ρπ a2 h2 x2dx a. Moment of Inertia Ix. Using the expression derived in Sec. 9.5C for a thin disk, compute the mass moment of inertia of the differential element with respect to the x axis.
dIx 5 1 2 r2 dm 5 1 2 aa x hb 2 arπ a2 h2 x2 dxb 5 1 2 rπ a4 h4 x4 dx Integrating from x 5 0 to x 5 h gives you Ix 5# dIx 5# h 0 1 2 rπ a4 h4 x4 dx 5 1 2 rπ a4 h4 h5 5 5 1 10 rπa4h Since the total mass of the cone is m 5 1 3 rπa2h, you can write this as Ix 5 1 10rπa4h 5 3 10a2(1 3rπa2h) 5 3 10ma2 Ix 5 3 10ma2 b b. Moment of Inertia Iy. Use the same differential element. Apply-ing the parallel-axis theorem and using the expression derived in Sec. 9.5C for a thin disk, you have dIy 5 dIy9 1 x2dm 5 1 4r2dm 1 x2dm 5 (1 4 r2 1 x2)dm Substituting the expressions for r and dm into this equation yields dIy 5 a1 4 a2 h2 x2 1 x2b arπ a2 h2 x2 dxb 5 rπ a2 h2 a a2 4h2 1 1b x4 dx Iy 5# dIy 5# h 0 rπa2 h2 a a2 4h2 1 1b x4 dx 5 rπa2 h2 a a2 4h2 1 1b h5 5 Introducing the total mass of the cone m, you can rewrite Iy as Iy 5 3 5(1 4 a2 1 h2)1 3rπa2h Iy 5 3 5 m(1 4 a2 1 h2) b c. Moment of Inertia I _ y0. Apply the parallel-axis theorem to obtain Iy 5 Iy– 1 mx 2 Solve for Iy0 and recall from Fig. 5.21 that x 5 3 4 h (Fig. 2). The result is Iy– 5 Iy 2 mx 2 5 3 5 m(1 4 a2 1 h2) 2 m(3 4 h)2 Iy– 5 3 20 m(a2 1 1 4 h2) b REFLECT and THINK: The parallel-axis theorem for masses can be just as useful as the version for areas. Don’t forget to use the reference figures for centroids of volumes when needed.
z y x h a z y x dx y' x h r a Fig. 1 Differential element of mass.
z y x h y" x = h 3 4 Fig. 2 Centroid of a right circular cone.
9.5 Mass Moments of Inertia 537 Sample Problem 9.12 A steel forging consists of a 6 3 2 3 2-in. rectangular prism and two cylinders with a diameter of 2 in. and length of 3 in. as shown. Deter-mine the moments of inertia of the forging with respect to the coordi-nate axes. The specific weight of steel is 490 lb/ft3.
STRATEGY: Compute the moments of inertia of each component from Fig. 9.28 using the parallel-axis theorem when necessary. Note that all lengths should be expressed in feet to be consistent with the units for the given specific weight.
MODELING and ANALYSIS: Computation of Masses.
Prism V 5 (2 in.)(2 in.)(6 in.) 5 24 in3 W 5 (24 in3)(490 lb/ft3) 1728 in3/ft3 5 6.81 lb m 5 6.81 lb 32.2 ft/s2 5 0.211 lb?s2/ft Each Cylinder V 5 π(1 in.)2(3 in.) 5 9.42 in3 W 5 (9.42 in3)(490 lb/ft3) 1728 in3/ft3 5 2.67 lb m 5 2.67 lb 32.2 ft/s2 5 0.0829 lb?s2/ft Moments of Inertia (Fig. 1). Prism Ix 5 Iz 5 1 12 (0.211 lb?s2/ft)[( 6 12 ft)2 1 ( 2 12 ft)2] 5 4.88 3 1023 lb?ft?s2 Iy 5 1 12 (0.211 lb?s2/ft)[( 2 12 ft)2 1 ( 2 12 ft)2] 5 0.977 3 1023 lb?ft?s2 Each Cylinder Ix 5 1 2 ma2 1 my 2 5 1 2(0.0829 lb?s2/ft)( 1 12 ft)2 1 (0.0829 lb?s2/ft)( 2 12 ft)2 5 2.59 3 1023 lb?ft?s2 Iy 5 1 12 m(3a2 1 L2) 5 mx 2 5 1 12(0.0829 lb?s2/ft)[3( 1 12 ft)2 1 ( 3 12 ft)2] 1 (0.0829 lb?s2/ft)(2.5 12 ft)2 5 4.17 3 1023 lb?ft?s2 Iz 5 1 12 m(3a21 L2) 1 m(x 2 1 y 2) 5 1 12(0.0829 lb?s2/ft)[3( 1 12 ft)21 ( 3 12 ft)2] 1 (0.0829 lb?s2/ft)[(2.5 12 ft)2 1 ( 2 12 ft)2] 5 6.48 3 1023 lb?ft?s2 Entire Body. Adding the values obtained for the prism and two cylinders, you have Ix 5 4.88 3 1023 1 2(2.59 3 1023) Ix 5 10.06 3 1023 lb?ft?s2 b Iy 5 0.977 3 1023 1 2(4.17 3 1023) Iy 5 9.32 3 1023 lb?ft?s2 b Iz 5 4.88 3 1023 1 2(6.48 3 1023) Iz 5 17.84 3 1023 lb?ft?s2 b REFLECT and THINK: The results indicate this forging has more resistance to rotation about the z axis (largest moment of inertia) than about the x or y axes. This makes intuitive sense, because more of the mass is farther from the z axis than from the x or y axes.
2 in.
2 in.
1 in.
A B y x 3 in.
2 in.
2 in.
z 2 in.
6 in.
1 in.
A B y z x 3 in.
2.5 in.
2 in.
2 in.
2 in.
Fig. 1 Geometry of each component.
538 Distributed Forces: Moments of Inertia Sample Problem 9.13 A thin steel plate that is 4 mm thick is cut and bent to form the machine part shown. The density of the steel is 7850 kg/m3. Determine the moments of inertia of the machine part with respect to the coordinate axes.
STRATEGY: The machine part consists of a semicircular plate and a rectangular plate from which a circular plate has been removed (Fig. 1). After calculating the moments of inertia for each part, add those of the semicircular plate and the rectangular plate, then subtract those of the circular plate to determine the moments of inertia for the entire machine part.
MODELING and ANALYSIS: Computation of Masses. Semicircular Plate V1 5 1 2 pr2t 5 1 2 p(0.08 m)2(0.004 m) 5 40.21 3 1026 m3 m1 5 ρV1 5 (7.85 3 103 kg/m3)(40.21 3 1026 m3) 5 0.3156 kg Rectangular Plate V2 5 (0.200 m)(0.160 m)(0.004 m) 5 128 3 1026 m3 m2 5 ρV2 5 (7.85 3 103 kg/m3)(128 3 1026 m3) 5 1.005 kg Circular Plate V3 5 πa2t 5 π(0.050 m)2(0.004 m) 5 31.42 3 1026 m3 m3 5 ρV3 5 (7.85 3 103 kg/m3)(31.42 3 1026 m3) 5 0.2466 kg Moments of Inertia. Compute the moments of inertia of each com-ponent, using the method presented in Sec. 9.5C.
Semicircular Plate. Observe from Fig. 9.28 that, for a circular plate of mass m and radius r, Ix 5 1 2 mr2 Iy 5 Iz 5 1 4 mr2 Because of symmetry, halve these values for a semicircular plate. Thus, Ix 5 1 2(1 2 mr2) Iy 5 Iz 5 1 2(1 4 mr2) Since the mass of the semicircular plate is m1 5 1 2 m, you have Ix5 1 2 m1r2 5 1 2(0.3156 kg)(0.08 m)2 5 1.010 31023 kg? m2 Iy5Iz5 1 4(1 2 mr2)5 1 4 m1r25 1 4(0.3156 kg)(0.08 m)25 0.505 3 1023 kg? m2 Rectangular Plate Ix 5 1 12 m2c2 5 1 12(1.005 kg)(0.16 m)2 5 2.144 3 1023 kg?m2 Iz 5 1 3 m2b2 5 1 3(1.005 kg)(0.2 m)2 5 13.400 3 1023 kg?m2 Iy 5 Ix 1 Iz 5 (2.144 1 13.400)(1023) 5 15.544 3 1023 kg?m2 Circular Plate Ix 5 1 4 m3a2 5 1 4(0.2466 kg)(0.05 m)2 5 0.154 3 1023 kg? m2 Iy 5 1 2 m3a2 1 m3d2 5 1 2(0.2466 kg)(0.05 m)2 1(0.2466 kg)(0.1 m)2 5 2.774 3 1023 kg ? m2 Iz 5 1 4 m3a2 1 m3d 2 5 1 4(0.2466 kg)(0.05 m)2 1 (0.2466 kg)(0.1 m)2 5 2.620 3 1023 kg? m2 Entire Machine Part Ix 5 (1.010 1 2.144 2 0.154)(1023) kg?m2 Ix 5 3.00 3 1023 kg?m2 b Iy 5 (0.505 1 15.544 2 2.774)(1023) kg?m2 Iy 5 13.28 3 1023 kg?m2 b Iz 5 (0.505 1 13.400 2 2.620)(1023) kg?m2 Iz 5 11.29 3 1023 kg?m2 b y z 80 x Dimensions in mm 80 80 50 100 100 z z x x r = 0.08 m d = 0.1 m c = 0.16 m a = 0.05 m + + _ y y y z x b = 0.2 m Fig. 1 Modeling the machine part as a combination of simple geometric shapes.
539 539 SOLVING PROBLEMS ON YOUR OWN I n this section, we introduced the mass moment of inertia and the radius of gyration of a three-dimensional body with respect to a given axis [Eqs. (9.28) and (9.29)]. We also derived a parallel-axis theorem for use with mass moments of inertia and discussed the computation of the mass moments of inertia of thin plates and three-dimensional bodies.
1. Computing mass moments of inertia. You can calculate the mass moment of inertia I of a body with respect to a given axis directly from the definition given in Eq. (9.28) for simple shapes [Sample Prob. 9.9]. In most cases, however, it is necessary to divide the body into thin slabs, compute the moment of inertia of a typical slab with respect to the given axis—using the parallel-axis theorem if necessary—and integrate the resulting expression.
2. Applying the parallel-axis theorem. In Sec. 9.5B, we derived the parallel-axis theo-rem for mass moments of inertia as I 5 I 1 md2 (9.33) This theorem states that the moment of inertia I of a body of mass m with respect to a given axis is equal to the sum of the moment of inertia I of that body with respect to a parallel centroidal axis and the product md 2, where d is the distance between the two axes. When you calculate the moment of inertia of a three-dimensional body with respect to one of the coordinate axes, you can replace d 2 by the sum of the squares of distances measured along the other two coordinate axes [Eqs. (9.32) and (9.329)].
3. Avoiding unit-related errors. To avoid errors, you must be consistent in your use of units. Thus, all lengths should be expressed in meters or feet, as appropriate, and for problems using U.S. customary units, masses should be given in lb?s2/ft. In addition, we strongly recommend that you include units as you perform your calculations [Sample Probs. 9.12 and 9.13].
4. Calculating the mass moment of inertia of thin plates. We showed in Sec. 9.5C that you can obtain the mass moment of inertia of a thin plate with respect to a given axis by multiplying the corresponding moment of inertia of the area of the plate by the density ρ and the thickness t of the plate [Eqs. (9.35) through (9.37)]. Note that, since the axis CC9 in Fig. 9.24c is perpendicular to the plate, ICC9, mass is associated with the polar moment of inertia JC, area. Instead of calculating the moment of inertia of a thin plate with respect to a specified axis directly, you may sometimes find it convenient to first compute its moment of inertia with respect to an axis parallel to the specified axis and to then apply the parallel-axis theorem. Furthermore, to determine the moment of inertia of a thin plate with respect to an axis perpendicular to the plate, you may wish to first determine its moments of inertia with respect to two perpendicular in-plane axes and to then use Eq. (9.38). Finally, remem-ber that the mass of a plate consists of area A, thickness t, and density ρ, as m 5 ρtA.
(continued) 540 5. Determining the moment of inertia of a body by direct single integration. We discussed in Sec. 9.5D and illustrated in Sample Probs. 9.10 and 9.11 how you can use single integration to compute the moment of inertia of a body that can be divided into a series of thin, parallel slabs. For such cases, you will often need to express the mass of the body in terms of the body’s density and dimensions. Assuming that the body has been divided, as in the sample problems, into thin slabs perpendicular to the x axis, you will need to express the dimensions of each slab as functions of the variable x.
a. In the special case of a body of revolution, the elemental slab is a thin disk, and you can use the equations given in Fig. 9.27 to determine the moments of inertia of the body [Sample Prob. 9.11]. b. In the general case, when the body is not a solid of revolution, the differential element is not a disk but a thin slab of a different shape. You cannot use the equations of Fig. 9.27 in this case. See, for example, Sample Prob. 9.10, where the element was a thin, rectangular slab. For more complex configurations, you may want to use one or more of the following equations, which are based on Eqs. (9.32) and (9.329) of Sec. 9.5B. dIx 5 dIx9 1 (y 2 el 1 z 2 el) dm dIy 5 dIy9 1 (z 2 el 1 x 2 el) dm dIz 5 dIz9 1 (x 2 el 1 y 2 el) dm Here, the primes denote the centroidal axes of each elemental slab and xel, yel, and zel represent the coordinates of its centroid. Determine the centroidal moments of inertia of the slab in the manner described earlier for a thin plate: Refer to Fig. 9.12, calculate the corresponding moments of inertia of the area of the slab, and multiply the result by the density ρ and the thickness t of the slab. Also, assuming that the body has been divided into thin slabs perpendicular to the x axis, remember that you can obtain dIx9 by adding dIy9 and dIz9 instead of computing it directly. Finally, using the geometry of the body, express the result obtained in terms of the single variable x, and integrate in x.
6. Computing the moment of inertia of a composite body. As stated in Sec. 9.5E, the moment of inertia of a composite body with respect to a specified axis is equal to the sum of the moments of its components with respect to that axis. Sample Probs. 9.12 and 9.13 illustrate the appropriate method of solution. Also remember that the moment of inertia of a component is negative only if the component is removed (as in the case of a hole).
Although the composite-body problems in this section are relatively straightforward, you will have to work carefully to avoid computational errors. In addition, if some of the moments of inertia that you need are not given in Fig. 9.28, you will have to derive your own formulas, using the techniques described in this section.
541 Problems 9.111 A thin plate with a mass m is cut in the shape of an equilateral triangle of side a. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axes AA9 and BB9, (b) the centroidal axis CC9 that is perpendicular to the plate.
B A' A B' C' C Fig. P9.111 A B A C B C r1 r2 Fig. P9.112 9.112 A ring with a mass m is cut from a thin uniform plate. Determine the mass moment of inertia of the ring with respect to (a) the axis AA9, (b) the centroidal axis CC9 that is perpendicular to the plane of the ring. 9.113 A thin, semielliptical plate has a mass m. Determine the mass moment of inertia of the plate with respect to (a) the centroidal axis BB9, (b) the centroidal axis CC9 that is perpendicular to the plate. 9.114 The parabolic spandrel shown was cut from a thin, uniform plate. Denoting the mass of the spandrel by m, determine its mass moment of inertia with respect to (a) the axis BB9, (b) the axis DD9 that is perpendicular to the spandrel. (Hint: See Sample Prob. 9.3.) A A' B B' D' D a b y = kx2 Fig. P9.114 C a A B A9 B9 C9 b Fig. P9.113 542 9.115 A piece of thin, uniform sheet metal is cut to form the machine com-ponent shown. Denoting the mass of the component by m, determine its mass moment of inertia with respect to (a) the x axis, (b) the y axis.
9.116 A piece of thin, uniform sheet metal is cut to form the machine com-ponent shown. Denoting the mass of the component by m, determine its mass moment of inertia with respect to (a) the axis AA9, (b) the axis BB9, where the AA9 and BB9 axes are parallel to the x axis and lie in a plane parallel to and at a distance a above the xz plane. 9.117 A thin plate with a mass m has the trapezoidal shape shown. Deter-mine the mass moment of inertia of the plate with respect to (a) the x axis, (b) the y axis.
C' C x y z a 1.5a 2a 2a A' A Fig. P9.117 and P9.118 9.118 A thin plate with a mass m has the trapezoidal shape shown. Determine the mass moment of inertia of the plate with respect to (a) the cen-troidal axis CC9 that is perpendicular to the plate, (b) the axis AA9 that is parallel to the x axis and is located at a distance 1.5a from the plate. 9.119 Determine by direct integration the mass moment of inertia with respect to the z axis of the right circular cylinder shown, assuming that it has a uniform density and a mass m. 9.120 The area shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Using direct integration, express the mass moment of inertia of the solid with respect to the x axis in terms of m and h. 9.121 The area shown is revolved about the x axis to form a homogeneous solid of revolution of mass m. Determine by direct integration the mass moment of inertia of the solid with respect to (a) the x axis, (b) the y axis. Express your answers in terms of m and the dimen-sions of the solid.
a h 2a y x y = k x Fig. P9.121 A B A' B' x y z a 2 a a a a Fig. P9.115 and P9.116 h 2h a x y Fig. P9.120 Fig. P9.119 y x a z L 543 9.122 Determine by direct integration the mass moment of inertia with respect to the x axis of the tetrahedron shown, assuming that it has a uniform density and a mass m.
y z x b a h Fig. P9.122 and P9.123 9.123 Determine by direct integration the mass moment of inertia with respect to the y axis of the tetrahedron shown, assuming that it has a uniform density and a mass m. 9.124 Determine by direct integration the mass moment of inertia and the radius of gyration with respect to the x axis of the paraboloid shown, assuming that it has a uniform density and a mass m. 9.125 A thin, rectangular plate with a mass m is welded to a vertical shaft AB as shown. Knowing that the plate forms an angle θ with the y axis, determine by direct integration the mass moment of inertia of the plate with respect to (a) the y axis, (b) the z axis.
x y z A 1 b a 2 1 b 2 B q Fig. P9.125 9.126 A thin steel wire is bent into the shape shown. Denoting the mass per unit length of the wire by m9, determine by direct integration the mass moment of inertia of the wire with respect to each of the coordinate axes.
z x y kx y2 z2 + = a h Fig. P9.124 a a z x y y = (a2/3 – x2/3)3/2 Fig. P9.126 544 9.127 Shown is the cross section of an idler roller. Determine its mass moment of inertia and its radius of gyration with respect to the axis AA9. (The specific weight of bronze is 0.310 lb/in3; of aluminum, 0.100 lb/in3; and of neoprene, 0.0452 lb/in3.) 9.128 Shown is the cross section of a molded flat-belt pulley. Determine its mass moment of inertia and its radius of gyration with respect to the axis AA9. (The density of brass is 8650 kg/m3, and the density of the fiber-reinforced polycarbonate used is 1250 kg/m3.) Brass Polycarbonate 2 mm A' A 11 mm 22 mm 9.5 mm 17.5 mm 5 mm 17 mm 28 mm Fig. P9.128 9.129 The machine part shown is formed by machining a conical surface into a circular cylinder. For b 5 1 2h, determine the mass moment of inertia and the radius of gyration of the machine part with respect to the y axis.
x z a a h b y Fig. P9.129 9.130 Knowing that the thin hemispherical shell shown has a mass m and thickness t, determine the mass moment of inertia and the radius of gyration of the shell with respect to the x axis. (Hint: Consider the shell as formed by removing a hemisphere of radius r from a hemi-sphere of radius r 1 t; then neglect the terms containing t2 and t3 and keep those terms containing t.) A A' Neoprene Aluminum Bronze in.
11 16 in.
13 16 in.
1 1 8 in.
3 8 in.
1 2 in.
1 4 Fig. P9.127 z x y r Fig. P9.130 545 9.131 A square hole is centered in and extends through the aluminum machine component shown. Determine (a) the value of a for which the mass moment of inertia of the component is maximum with respect to the axis AA9 that bisects the top surface of the hole, (b) the corresponding values of the mass moment of inertia and the radius of gyration with respect to the axis AA9. (The specific weight of aluminum is 0.100 lb/in3.) 9.132 The cups and the arms of an anemometer are fabricated from a mate-rial of density ρ. Knowing that the mass moment of inertia of a thin, hemispherical shell with a mass m and thickness t with respect to its centroidal axis GG9 is 5ma2/12, determine (a) the mass moment of inertia of the anemometer with respect to the axis AA9, (b) the ratio of a to l for which the centroidal moment of inertia of the cups is equal to 1 percent of the moment of inertia of the cups with respect to the axis AA9.
l d a A A' G G' a 2 Fig. P9.132 9.133 After a period of use, one of the blades of a shredder has been worn to the shape shown and is of mass 0.18 kg. Knowing that the mass moments of inertia of the blade with respect to the AA9 and BB9 axes are 0.320 g?m2 and 0.680 g?m2, respectively, determine (a) the loca-tion of the centroidal axis GG9, (b) the radius of gyration with respect to axis GG9.
A A' B B' 80 mm G' G Fig. P9.133 9.134 Determine the mass moment of inertia of the 0.9-lb machine com-ponent shown with respect to the axis AA9.
A A' 4.2 in.
a a a 2 4.2 in.
15 in.
Fig. P9.131 A A' 0.4 in.
1.2 in.
1.6 in.
2.4 in.
Fig. P9.134 546 9.135 and 9.136 A 2-mm thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to each of the coordinate axes.
z x y 350 mm 150 mm 195 mm Fig. P9.136 9.137 A subassembly for a model airplane is fabricated from three pieces of 1.5-mm plywood. Neglecting the mass of the adhesive used to assemble the three pieces, determine the mass moment of inertia of the subassembly with respect to each of the coordinate axes. (The density of the plywood is 780 kg/m3.) 9.138 A section of sheet steel 0.03 in. thick is cut and bent into the sheet metal machine component shown. Determine the mass moment of inertia of the component with respect to each of the coordinate axes. (The specific weight of steel is 490 lb/ft3.) y z x 6 in.
4.5 in.
4.5 in.
6 in.
Fig. P9.138 9.139 A framing anchor is formed of 0.05-in.-thick galvanized steel. Deter-mine the mass moment of inertia of the anchor with respect to each of the coordinate axes. (The specific weight of galvanized steel is 470 lb/ft3.) 200 mm 100 mm 120 mm y z x Fig. P9.135 y x z 300 mm 120 mm Fig. P9.137 x y z 1 in.
1.25 in.
2 in.
2.25 in.
3.5 in.
Fig. P9.139 547 9.140 A farmer constructs a trough by welding a rectangular piece of 2-mm-thick sheet steel to half of a steel drum. Knowing that the density of steel is 7850 kg/m3 and that the thickness of the walls of the drum is 1.8 mm, determine the mass moment of inertia of the trough with respect to each of the coordinate axes. Neglect the mass of the welds.
y x z 285 mm 840 mm 210 mm Fig. P9.140 9.141 The machine element shown is fabricated from steel. Determine the mass moment of inertia of the assembly with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m3.) 40 mm 40 mm 40 mm 20 mm 60 mm 20 mm 80 mm x y z Fig. P9.141 9.142 Determine the mass moments of inertia and the radii of gyration of the steel machine element shown with respect to the x and y axes. (The density of steel is 7850 kg/m3.) 40 44 44 20 20 y z x 120 120 70 70 Dimensions in mm Fig. P9.142 548 9.143 Determine the mass moment of inertia of the steel machine element shown with respect to the x axis. (The specific weight of steel is 490 lb/ft3.) 2.4 in.
1.8 in.
0.6 in.
0.6 in.
1.8 in.
2 in.
1.6 in.
1.8 in.
1.5 in.
y x z Fig. P9.143 and P.9.144 9.144 Determine the mass moment of inertia of the steel machine element shown with respect to the y axis. (The specific weight of steel is 490 lb/ft3.) 9.145 Determine the mass moment of inertia of the steel fixture shown with respect to (a) the x axis, (b) the y axis, (c) the z axis. (The density of steel is 7850 kg/m3.) 9.146 Aluminum wire with a weight per unit length of 0.033 lb/ft is used to form the circle and the straight members of the figure shown. Determine the mass moment of inertia of the assembly with respect to each of the coordinate axes. 9.147 The figure shown is formed of 1 8-in.-diameter steel wire. Knowing that the specific weight of the steel is 490 lb/ft3, determine the mass moment of inertia of the wire with respect to each of the coordinate axes.
x y z 18 in.
18 in.
18 in.
Fig. P9.147 9.148 A homogeneous wire with a mass per unit length of 0.056 kg/m is used to form the figure shown. Determine the mass moment of inertia of the wire with respect to each of the coordinate axes.
50 mm 70 mm 40 mm 16 mm 80 mm y x z 50 mm 38 mm 24 mm Fig. P9.145 x z 8 in.
8 in.
8 in.
y 16 in.
8 in.
Fig. P9.146 1.2 m 1.2 m 1.2 m z x y Fig. P9.148 9.6 Additional Concepts of Mass Moments of Inertia 549 9.6 ADDITIONAL CONCEPTS OF MASS MOMENTS OF INERTIA In this final section of the chapter, we present several concepts involving mass moments of inertia that are analogous to material presented in Sec. 9.4 involving moments of inertia of areas. These ideas include mass products of inertia, principal axes of inertia, and principal moments of inertia for masses, which are necessary for the study of the dynamics of rigid bodies in three dimensions.
9.6A Mass Products of Inertia In this section, you will see how to determine the moment of inertia of a body with respect to an arbitrary axis OL through the origin (Fig. 9.29) if its moments of inertia with respect to the three coordinate axes, as well as certain other quantities defined here, have already been determined.
The moment of inertia IOL of the body with respect to OL is equal to ep2 dm, where p denotes the perpendicular distance from the element of mass dm to the axis OL. If we denote the unit vector along OL by l and the position vector of the element dm by r, the perpendicular distance p is equal to r sin θ, which is the magnitude of the vector product l 3 r. We therefore have IOL 5 #p2 dm 5 #Zl 3 rZ2 dm (9.43) Expressing Zl 3 rZ2 in terms of the rectangular components of the vector product, we have IOL 5# [(lx y 2 ly x)2 1 (ly z 2 lz y)2 1 (lz x 2 lx z)2] dm Here, the components lx, ly, lz of the unit vector l represent the direction cosines of the axis OL, and the components x, y, z of r represent the coordinates of the element of mass dm. Expanding the squares and rear-ranging the terms, we obtain IOL 5 l2 x # (y2 1 z2) dm 1 l2 y # (z2 1 x2) dm 1 l2 z # (x2 1 y2) dm 2 2lx ly # xy dm 2 2ly lz # yz dm 2 2lzlx # zx dm (9.44) Referring to Eqs. (9.30), note that the first three integrals in Eq. (9.44) represent, respectively, the moments of inertia Ix, Iy, and Iz of the body with respect to the coordinate axes. The last three integrals in Eq. (9.44), which involve products of coordinates, are called the products of inertia of the body with respect to the x and y axes, the y and z axes, and the z and x axes, respectively. Mass products of inertia Ixy 5# xy dm Iyz 5# yz dm Izx 5# zx dm (9.45) Ix I y x 5# xy x dm d Iy I z 5# yz dm d Iz I x 5# zx dm d Fig. 9.29 An element of mass dm of a body and its perpendicular distance to an arbitrary axis OL through the origin.
y dm z x p L O q r 550 Distributed Forces: Moments of Inertia Rewriting Eq. (9.44) in terms of the integrals defined in Eqs. (9.30) and (9.45), we have IOL 5 Ixlx 2 1 Iyl2 y 1 Izlz 2 2 2Ixylxly 2 2Iyzlylz 2 2Izxlzlx (9.46) The definition of the products of inertia of a mass given in Eqs. (9.45) is an extension of the definition of the product of inertia of an area (Sec. 9.3). Mass products of inertia reduce to zero under the same conditions of sym-metry as do products of inertia of areas, and the parallel-axis theorem for mass products of inertia is expressed by relations similar to the formula derived for the product of inertia of an area. Substituting the expressions for x, y, and z given in Eqs. (9.31) into Eqs. (9.45), we find that Parallel-axis theorem for mass products of inertia Ixy 5 Ix¿y¿ 1 mx y Iyz 5 Iy¿z¿ 1 my z (9.47) Izx 5 Iz¿x¿ 1 mz x Here x, y, z are the coordinates of the center of gravity G of the body and Ix¿y¿, Iy¿z¿, Iz¿x¿ denote the products of inertia of the body with respect to the centroidal axes x9, y9, and z9 (see Fig. 9.22).
9.6B Principal Axes and Principal Moments of Inertia Let us assume that we have determined the moment of inertia of the body considered in the preceding section with respect to a large number of axes OL through the fixed point O. Suppose that we plot a point Q on each axis OL at a distance OQ 5 1/2IOL from O. The locus of the points Q forms a surface (Fig. 9.30). We can obtain the equation of that surface by sub-stituting 1/(OQ)2 for IOL in Eq. (9.46) and then multiplying both sides of the equation by (OQ)2. Observing that (OQ) lx 5 x (OQ) ly 5 y (OQ) lz 5 z where x, y, z denote the rectangular coordinates of Q, we have Ix x2 1 Iyy2 1 Izz2 2 2Ix y xy 2 2Iy z yz 2 2Iz xzx 5 1 (9.48) This is the equation of a quadric surface. Since the moment of inertia IOL is different from zero for every axis OL, no point Q can be at an infinite distance from O. Thus, the quadric surface obtained is an ellipsoid. This ellipsoid, which defines the moment of inertia of the body with respect to any axis through O, is known as the ellipsoid of inertia of the body at O.
Observe that, if we rotate the axes in Fig. 9.30, the coefficients of the equation defining the ellipsoid change, since they are equal to the moments and products of inertia of the body with respect to the rotated coordinate axes. However, the ellipsoid itself remains unaffected, since its IOL I 5 Ix I lx l2 l 1 Iy I l2 y 1 Iz I lz 2 2 2Ixy I lx l ly 2 2Iyz I lylz 2 2Izx I lzlx l Ix I y x 5 Ix I ¿y¿ 1 mx m y Iy I z 5 Iy I ¿z¿ 1 my m z Iz I x z 5 Iz I ¿x¿ 1 mzx Fig. 9.30 The ellipsoid of inertia defines the moment of inertia of a body with respect to any axis through O.
x L y z O 1/√IOL Q(x, y, z) 9.6 Additional Concepts of Mass Moments of Inertia 551 shape depends only upon the distribution of mass in the given body. Sup-pose that we choose as coordinate axes the principal axes x9, y9, and z9 of the ellipsoid of inertia (Fig. 9.31). The equation of the ellipsoid with respect to these coordinate axes is known to be of the form Ix9x92 1 Iy9y92 1 Iz9z92 5 1 (9.49) which does not contain any products of the coordinates. Comparing Eqs. (9.48) and (9.49), we observe that the products of inertia of the body with respect to the x9, y9, and z9 axes must be zero. The x9, y9, and z9 axes are known as the principal axes of inertia of the body at O, and the coefficients Ix9, Iy9, and Iz9 are referred to as the principal moments of inertia of the body at O. Note that, given a body of arbitrary shape and a point O, it is always possible to find principal axes of inertia of the body at O; that is, axes with respect to which the products of inertia of the body are zero. Indeed, whatever the shape of the body, the moments and prod-ucts of inertia of the body with respect to the x, y, and z axes through O define an ellipsoid, and this ellipsoid has principal axes that, by definition, are the principal axes of inertia of the body at O.
If the principal axes of inertia x9, y9, and z9 are used as coordinate axes, the expression in Eq. (9.46) for the moment of inertia of a body with respect to an arbitrary axis through O reduces to IOL 5 Ix9l2 x9 1 Iy9l2 y9 1 Iz9l2 z9 (9.50) The determination of the principal axes of inertia of a body of arbi-trary shape is somewhat involved and is discussed in the next section. In many cases, however, these axes can be spotted immediately. Consider, for instance, the homogeneous cone of elliptical base shown in Fig. 9.32; this cone possesses two mutually perpendicular planes of symmetry OAA9 and OBB9. From the definition of Eq. (9.45), we observe that if we choose the x9y9 and y9z9 planes to coincide with the two planes of symmetry, all of the products of inertia are zero. The x9, y9, and z9 axes selected in this way are therefore the principal axes of inertia of the cone at O. In the IOL I 5 Ix I 9l2 x9 1 Iy I 9l2 y9 1 Iz I 9l2 z9 Fig. 9.32 A homogeneous cone with elliptical base has two mutually perpendicular planes of symmetry.
z' A' B' A B O x' y' Fig. 9.31 Principal axes of inertia x9, y9, z9 of the body at O.
x z' y' x' y z O 552 Distributed Forces: Moments of Inertia case of the homogeneous regular tetrahedron OABC shown in Fig. 9.33, the line joining the corner O to the center D of the opposite face is a principal axis of inertia at O, and any line through O perpendicular to OD is also a principal axis of inertia at O. This property is apparent if we observe that rotating the tetrahedron through 120° about OD leaves its shape and mass distribution unchanged. It follows that the ellipsoid of inertia at O also remains unchanged under this rotation. The ellipsoid, therefore, is a body of revolution whose axis of revolution is OD, and the line OD, as well as any perpendicular line through O, must be a principal axis of the ellipsoid.
9.6C Principal Axes and Moments of Inertia for a Body of Arbitrary Shape The method of analysis described in this section extends the analysis in the preceding section. However, generally speaking, you should use it only when the body under consideration has no obvious property of symmetry.
Consider the ellipsoid of inertia of a body at a given point O (Fig. 9.34). Let r be the radius vector of a point P on the surface of the ellipsoid, and let n be the unit vector along the normal to that surface at P. We observe that the only points where r and n are collinear are points P1, P2, and P3, where the principal axes intersect the visible portion of the surface of the ellipsoid (along with the corresponding points on the other side of the ellipsoid).
Recall from calculus that the direction of the normal to a surface of equation f(x, y, z) 5 0 at a point P(x, y, z) is defined by the gradient =f of the function f at that point. To obtain the points where the principal axes intersect the surface of the ellipsoid of inertia, we must therefore express that r and =f are collinear, =f 5 (2K)r (9.51) where K is a constant, r 5 xi 1 yj 1 zk, and §f 5 0f 0x i 1 0f 0y j 1 0f 0z k Recalling Eq. (9.48), we note that the function f(x, y, z) corresponding to the ellipsoid of inertia is f(x, y, z) 5 Ixx2 1 Iyy2 1 Izz2 2 2Ixyxy 2 2Iyzyz 2 2Izxzx 2 1 Substituting for r and =f into Eq. (9.51) and equating the coefficients of the unit vectors, we obtain Ixx 2 Ixyy 2 Izxz 5 Kx 2Ixyx 1 Iyy 2 Iyzz 5 Ky (9.52) 2Izxx 2 Iyzy 1 Izz 5 Kz Fig. 9.33 A line drawn from a corner to the center of the opposite face of a homogeneous regular tetrahedron is a principal axis, since each 120° rotation of the body about this axis leaves its shape and mass distribution unchanged.
B D A C O Fig. 9.34 The principal axes intersect an ellipsoid of inertia at points where the radius vectors are collinear with the unit normal vectors to the surface.
x P1 P P3 P2 r n z' y' x' y z O 9.6 Additional Concepts of Mass Moments of Inertia 553 Dividing each term by the distance r from O to P, we obtain similar equa-tions involving the direction cosines lx, ly, and lz: Ixlx 2 Ixyly 2 Iz xlz 5 Klx 2Ix ylx 1 Iyly 2 Iyzlz 5 Kly (9.53) 2Iz xlx 2 Iyzly 1 Izlz 5 Klz Transposing the right-hand members leads to the homogeneous linear equations, as (Ix 2 K)lx 2 Ix yly 2 Iz xlz 5 0 2Ix ylx 1 (Iy 2 K)ly 2 Iyzlz 5 0 (9.54) 2Iz xlx 2 Iyzly 1 (Iz 2 K)lz 5 0 For this system of equations to have a solution different from lx 5 ly 5 lz 5 0, its discriminant must be zero. Thus, † Ix 2 K 2Ix y 2Iz x 2Ix y Iy 2 K 2Iyz 2Iz x 2Iyz Iz 2 K † 5 0 (9.55) Expanding this determinant and changing signs, we have K3 2 (Ix 1 Iy 1 Iz)K2 1 (IxIy 1 Iy Iz 1 IzIx 2 I 2 x y 2 I 2 y z 2 I 2 z x)K 2 (Ix Iy Iz 2 Ix I 2 yz 2 Iy I 2 z x 2 IzI 2 x y 2 2Ix yIyzIz x) 5 0 (9.56) This is a cubic equation in K, which yields three real, positive roots: K1, K2, and K3.
To obtain the direction cosines of the principal axis corresponding to the root K1, we substitute K1 for K in Eqs. (9.54). Since these equations are now linearly dependent, only two of them may be used to determine lx, ly, and lz. We can obtain an additional equation, however, by recalling from Sec. 2.4A that the direction cosines must satisfy the relation lx 2 1 l2 y 1 lz 2 5 1 (9.57) Repeating this procedure with K2 and K3, we obtain the direction cosines of the other two principal axes.
We now show that the roots K1, K2, and K3 of Eq. (9.56) are the principal moments of inertia of the given body. Let us substitute for K in Eqs. (9.53) the root K1, and for lx, ly, and lz the corresponding values (lx)1, (ly)1, and (lz)1 of the direction cosines; the three equations are satis-fied. We now multiply by (lx)1, (ly)1, and (lz)1, respectively, each term in the first, second, and third equation and add the equations obtained in this way. The result is I x 2(lx)2 1 1 I 2 y(ly)2 1 1 Iz 2(lz)2 1 2 2Ix y(lx)1(ly)1 2 2Iyz(ly)1(lz)1 2 2Iz x(lz)1(lx)1 5 K1[(lx)2 1 1 (ly)2 1 1 (lz)2 1] Recalling Eq. (9.46), we observe that the left-hand side of this equation represents the moment of inertia of the body with respect to the principal axis corresponding to K1; it is thus the principal moment of inertia cor-responding to that root. On the other hand, recalling Eq. (9.57), we note that the right-hand member reduces to K1. Thus, K1 itself is the principal moment of inertia. In the same fashion, we can show that K2 and K3 are the other two principal moments of inertia of the body.
554 Distributed Forces: Moments of Inertia Sample Problem 9.14 Consider a rectangular prism with a mass of m and sides a, b, and c. Determine (a) the moments and products of inertia of the prism with respect to the coordinate axes shown, (b) its moment of inertia with respect to the diagonal OB.
STRATEGY: For part (a), you can introduce centroidal axes and apply the parallel-axis theorem. For part (b), determine the direction cosines of line OB from the given geometry and use either Eq. (9.46) or (9.50).
MODELING and ANALYSIS: a. Moments and Products of Inertia with Respect to the Coordinate Axes.
Moments of Inertia. Introduce the centroidal axes x9, y9, and z9 with respect to which the moments of inertia are given in Fig. 9.28, and then apply the parallel-axis theorem (Fig. 1). Thus, Ix 5 Ix¿ 1 m(y2 1 z2) 5 1 12m(b2 1 c2) 1 m(1 4b2 1 1 4c2) Ix 5 1 3 m(b2 1 c2) b Similarly, Iy 5 1 3m(c2 1 a2) Iz 5 1 3m(a2 1 b2) b Products of Inertia. Because of symmetry, the products of inertia with respect to the centroidal axes x9, y9, and z9 are zero, and these axes are principal axes of inertia. Using the parallel-axis theorem, you have Ixy 5 I x9y9 1 mxy 5 0 1 m(1 2a)(1 2b) Ixy 5 1 4mab b Similarly, Iyz 5 1 4mbc Izx 5 1 4mca b b. Moment of Inertia with Respect to OB. Recall Eq. (9.46): IOB 5 Ixlx 2 1 Iyl2 y 1 Izlz 2 2 2Ixylxly 2 2Iyzlylz 2 2Izxlzlx where the direction cosines of OB are (Fig. 2) lx 5 cos θx 5 OH OB 5 a (a2 1 b2 1 c2)1/2 ly 5 b (a2 1 b2 1 c2)1/2 lz 5 c (a2 1 b2 1 c2)1/2 Substituting the values obtained in part (a) for the moments and products of inertia and for the direction cosines into the equation for IOB, you obtain IOB 5 1 a2 1 b2 1 c2 [1 3m(b2 1 c2)a2 1 1 3m(c2 1 a2)b2 1 1 3m(a2 1 b2)c2 21 2ma2b2 2 1 2mb2c2 2 1 2mc2a2] IOB 5 m 6 a2b2 1 b2c2 1 c2a2 a2 1 b2 1 c2 b REFLECT and THINK: You can also obtain the moment of inertia IOB directly from the principal moments of inertia Ix9, Iy9, and Iz9, since the line OB passes through the centroid O9. Since the x9, y9, and z9 axes are principal axes of inertia (Fig. 3), use Eq. (9.50) to write IOB 5 Ix9l2 x 1 Iy9l2 y 1 Iz9l2 z 5 1 a2 1 b2 1 c2 c m 12(b2 1 c2)a2 1 m 12 (c2 1 a2)b2 1 m 12 (a2 1 b2)c2 d IOB 5 m 6 a2b2 1 b2c2 1 c2a2 a2 1 b2 1 c2 b O O' y' z' x' z x y E A B F H D C a b c Fig. 1 Centroidal axes for the rectangular prism.
O z x y E A B F H D C a b c qz qx qy O E B H D a b c z x y Fig. 2 Direction angles for OB.
qz qx qy O O B y z x Fig. 3 Line OB passes through the centroid O9.
9.6 Additional Concepts of Mass Moments of Inertia 555 Sample Problem 9.15 If a 5 3c and b 5 2c for the rectangular prism of Sample Prob. 9.14, determine (a) the principal moments of inertia at the origin O, (b) the principal axes of inertia at O.
STRATEGY: Substituting the data into the results from Sample Prob. 9.14 gives you values you can use with Eq. (9.56) to determine the principal moments of inertia. You can then use these values to set up a system of equations for finding the direction cosines of the principal axes.
MODELING and ANALYSIS: a. Principal Moments of Inertia at the Origin O. Substituting a 5 3c and b 5 2c into the solution to Sample Prob. 9.14 gives you Ix 5 5 3mc2 Iy 5 10 3 mc2 Iz 5 13 3 mc2 Ixy 5 3 2mc2 Iyz 5 1 2mc2 Izx 5 3 4mc2 Substituting the values of the moments and products of inertia into Eq. (9.56) and collecting terms yields K3 2 (28 3 mc2)K2 1 (3479 144 m2c4)K 2 589 54 m3c6 5 0 Now solve for the roots of this equation; from the discussion in Sec. 9.6C, it follows that these roots are the principal moments of inertia of the body at the origin.
K1 5 0.568867mc2 K2 5 4.20885mc2 K3 5 4.55562mc2 K1 5 0.569mc2 K2 5 4.21mc2 K3 5 4.56mc2 b b. Principal Axes of Inertia at O. To determine the direction of a principal axis of inertia, first substitute the corresponding value of K into two of the equations (9.54). The resulting equations, together with Eq. (9.57), constitute a system of three equations from which you can determine the direction cosines of the corresponding principal axis. Thus, for the first principal moment of inertia K1, you have (5 3 2 0.568867)mc2(lx)1 2 3 2mc2(ly)1 2 3 4mc2(lz)1 5 0 23 2mc2(lx)1 1 (10 3 2 0.568867) mc2(ly)1 2 1 2mc2(lz)1 5 0 (lx)2 1 1 (ly)2 1 1 (lz)2 1 5 1 Solving yields (lx)1 5 0.836600 (ly)1 5 0.496001 (lz)1 5 0.232557 The angles that the first principal axis of inertia forms with the coordinate axes are then (θx)1 5 33.2° (θy)1 5 60.3° (θz)1 5 76.6° b Using the same set of equations successively with K2 and K3, you can find that the angles associated with the second and third principal moments of inertia at the origin are, respectively, (θx)2 5 57.8° (θy)2 5 146.6° (θz)2 5 98.0° b and (θx)3 5 82.8° (θy)3 5 76.1° (θz)3 5 164.3° b 556 556 SOLVING PROBLEMS ON YOUR OWN I n this section, we defined the mass products of inertia Ixy, Iyz, and Izx of a body and showed you how to determine the moments of inertia of that body with respect to an arbitrary axis passing through the origin O. You also saw how to determine at the origin O the principal axes of inertia of a body and the corresponding principal moments of inertia.
1. Determining the mass products of inertia of a composite body. You can express the mass products of inertia of a composite body with respect to the coordinate axes as the sums of the products of inertia of its component parts with respect to those axes. For each component part, use the parallel-axis theorem to write Eqs. (9.47) Ixy 5 Ix9y9 1 mx y Iyz 5 Iy9z9 1 my z Izx 5 Iz9x9 1 mz x Here the primes denote the centroidal axes of each component part, and x, y, and z represent the coordinates of its center of gravity. Keep in mind that the mass products of inertia can be positive, negative, or zero, and be sure to take into account the signs of x, y, and z. a. From the properties of symmetry of a component part, you can deduce that two or all three of its centroidal mass products of inertia are zero. For instance, you can verify for a thin plate parallel to the xy plane, a wire lying in a plane parallel to the xy plane, a body with a plane of symmetry parallel to the xy plane, and a body with an axis of sym-metry parallel to the z axis that the products of inertia Iy¿z¿ and Iz¿x¿ are zero. For rectangular, circular, or semicircular plates with axes of symmetry parallel to the coordinate axes, straight wires parallel to a coordinate axis, circular and semicircular wires with axes of symmetry parallel to the coordinate axes, and rectangular prisms with axes of symmetry parallel to the coordinate axes, the products of inertia I x9y9, I y9z9, and I z9x9 are all zero. b. Mass products of inertia that are different from zero can be computed from Eqs. (9.45). Although, in general, you need a triple integration to determine a mass product of inertia, you can use a single integration if you can divide the given body into a series of thin, parallel slabs. The computations are then similar to those discussed in the preced-ing section for moments of inertia.
2. Computing the moment of inertia of a body with respect to an arbitrary axis OL. In Sec. 9.6A, we derived an expression for the moment of inertia IOL that was given in Eq. (9.46). Before computing IOL, you must first determine the mass moments and products of inertia of the body with respect to the given coordinate axes, as well as the direction cosines of the unit vector l along OL.
557 557 3. Calculating the principal moments of inertia of a body and determining its principal axes of inertia. You saw in Sec. 9.6B that it is always possible to find an orientation of the coordinate axes for which the mass products of inertia are zero. These axes are referred to as the principal axes of inertia, and the corresponding moments of inertia are known as the principal moments of inertia of the body. In many cases, you can determine the principal axes of inertia of a body from its properties of symmetry. The procedure required to determine the principal moments and principal axes of inertia of a body with no obvious property of symmetry was discussed in Sec. 9.6C and was illustrated in Sample Prob. 9.15. It consists of the following steps. a. Expand the determinant in Eq. (9.55) and solve the resulting cubic equation. You can obtain the solution by trial and error or (preferably) with an advanced scientific cal-culator or appropriate computer software. The roots K1, K2, and K3 of this equation are the principal moments of inertia of the body. b. To determine the direction of the principal axis corresponding to K1, substitute this value for K in two of the equations (9.54) and solve these equations, together with Eq. (9.57), for the direction cosines of the principal axis corresponding to K1. c. Repeat this procedure with K2 and K3 to determine the directions of the other two principal axes. As a check of your computations, you may wish to verify that the scalar product of any two of the unit vectors along the three axes you have obtained is zero and, thus, that these axes are perpendicular to each other.
558 Problems 9.149 Determine the mass products of inertia Ixy, Iyz, and Izx of the steel fixture shown. (The density of steel is 7850 kg/m3.) 50 mm 70 mm 40 mm 16 mm 80 mm y x z 50 mm 38 mm 24 mm Fig. P9.149 9.150 Determine the mass products of inertia Ixy, Iyz, and Izx of the steel machine element shown. (The density of steel is 7850 kg/m3.) Dimensions in mm y x 35 35 60 60 20 10 10 22 22 z Fig. P9.150 9.151 and 9.152 Determine the mass products of inertia Ixy, Iyz, and Izx of the cast aluminum machine component shown. (The specific weight of aluminum is 0.100 lb/in3.) r = 0.55 in.
0.6 in.
5.4 in.
2.4 in.
3.6 in.
0.8 in.
r = 0.8 in.
y x z Fig. P9.152 x y z 1.4 in.
1.1 in.
1.1 in.
1.2 in.
0.3 in.
1.8 in.
0.7 in.
4.5 in.
Fig. P9.151 559 9.153 through 9.156 A section of sheet steel 2 mm thick is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass products of inertia Ixy, Iyz, and Izx of the component.
225 mm z x y 225 mm 400 mm 180 mm Fig. P9.153 z x y 350 mm 150 mm 195 mm Fig. P9.155 9.157 The figure shown is formed of 1.5-mm-diameter aluminum wire. Knowing that the density of aluminum is 2800 kg/m3, determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.
180 mm 250 mm 300 mm z x y Fig. P9.157 200 mm 100 mm 120 mm y z x Fig. P9.154 y z x 225 mm r = 135 mm Fig. P9.156 560 9.158 Thin aluminum wire of uniform diameter is used to form the figure shown. Denoting the mass per unit length of the wire by m9, determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure. 9.159 and 9.160 Brass wire with a weight per unit length w is used to form the figure shown. Determine the mass products of inertia Ixy, Iyz, and Izx of the wire figure.
x y z 2a 2a a a a 3 2 Fig. P9.159 x y z 2a a a a 3 2 Fig. P9.160 9.161 Complete the derivation of Eqs. (9.47) that expresses the parallel-axis theorem for mass products of inertia. 9.162 For the homogeneous tetrahedron of mass m shown, (a) determine by direct integration the mass product of inertia Izx, (b) deduce Iyz and Ixy from the result obtained in part a. 9.163 The homogeneous circular cone shown has a mass m. Determine the mass moment of inertia of the cone with respect to the line joining the origin O and point A.
3a 3a a O A x y z a 3 2 Fig. P9.163 9.164 The homogeneous circular cylinder shown has a mass m. Determine the mass moment of inertia of the cylinder with respect to the line joining the origin O and point A that is located on the perimeter of the top surface of the cylinder.
z x R1 R2 y Fig. P9.158 a b c x y z Fig. P9.162 h a O y x z A Fig. P9.164 561 9.165 Shown is the machine element of Prob. 9.141. Determine its mass moment of inertia with respect to the line joining the origin O and point A.
40 mm 40 mm 40 mm 20 mm 60 mm 20 mm 80 mm x y z A O Fig. P9.165 9.166 Determine the mass moment of inertia of the steel fixture of Probs. 9.145 and 9.149 with respect to the axis through the origin that forms equal angles with the x, y, and z axes. 9.167 The thin, bent plate shown is of uniform density and weight W. Determine its mass moment of inertia with respect to the line joining the origin O and point A.
x y z O A a a a Fig. P9.167 9.168 A piece of sheet steel with thickness t and specific weight γ is cut and bent into the machine component shown. Determine the mass moment of inertia of the component with respect to the line joining the origin O and point A. 9.169 Determine the mass moment of inertia of the machine component of Probs. 9.136 and 9.155 with respect to the axis through the origin characterized by the unit vector l 5 (24i 1 8j 1 k)/9. 9.170 through 9.172 For the wire figure of the problem indicated, determine the mass moment of inertia of the figure with respect to the axis through the origin characterized by the unit vector l 5 (23i 2 6j 1 2k)/7. 9.170 Prob. 9.148 9.171 Prob. 9.147 9.172 Prob. 9.146 y z x 2a a A O Fig. P9.168 562 9.173 For the homogeneous circular cylinder shown with radius a and length L, determine the value of the ratio a/L for which the ellipsoid of inertia of the cylinder is a sphere when computed (a) at the cen-troid of the cylinder, (b) at point A. 9.174 For the rectangular prism shown, determine the values of the ratios b/a and c/a so that the ellipsoid of inertia of the prism is a sphere when computed (a) at point A, (b) at point B.
x y z b 2 b 2 A B c 2 c 2 a 2 a 2 Fig. P9.174 9.175 For the right circular cone of Sample Prob. 9.11, determine the value of the ratio a/h for which the ellipsoid of inertia of the cone is a sphere when computed (a) at the apex of the cone, (b) at the center of the base of the cone. 9.176 Given an arbitrary body and three rectangular axes x, y, and z, prove that the mass moment of inertia of the body with respect to any one of the three axes cannot be larger than the sum of the mass moments of inertia of the body with respect to the other two axes. That is, prove that the inequality Ix # Iy 1 Iz and the two similar inequalities are satisfied. Furthermore, prove that Iy $ 1 2 Ix if the body is a homo-geneous solid of revolution, where x is the axis of revolution and y is a transverse axis. 9.177 Consider a cube with mass m and side a. (a) Show that the ellipsoid of inertia at the center of the cube is a sphere, and use this prop-erty to determine the moment of inertia of the cube with respect to one of its diagonals. (b) Show that the ellipsoid of inertia at one of the corners of the cube is an ellipsoid of revolution, and determine the principal moments of inertia of the cube at that point. 9.178 Given a homogeneous body of mass m and of arbitrary shape and three rectangular axes x, y, and z with origin at O, prove that the sum Ix 1 Iy 1 Iz of the mass moments of inertia of the body cannot be smaller than the similar sum computed for a sphere of the same mass and the same material centered at O. Furthermore, using the result of Prob. 9.176, prove that, if the body is a solid of revolution where x is the axis of revolution, its mass moment of inertia Iy about a transverse axis y cannot be smaller than 3ma2/10, where a is the radius of the sphere of the same mass and the same material.
x y z A a L 2 L 4 L 4 Fig. P9.173 563 9.179 The homogeneous circular cylinder shown has a mass m, and the diameter OB of its top surface forms 45° angles with the x and z axes. (a) Determine the principal mass moments of inertia of the cylinder at the origin O. (b) Compute the angles that the principal axes of inertia at O form with the coordinate axes. (c) Sketch the cylinder, and show the orientation of the principal axes of inertia relative to the x, y, and z axes.
a O x z y a B Fig. P9.179 9.180 through 9.184 For the component described in the problem indi-cated, determine (a) the principal mass moments of inertia at the origin, (b) the principal axes of inertia at the origin. Sketch the body and show the orientation of the principal axes of inertia relative to the x, y, and z axes. 9.180 Prob. 9.165 9.181 Probs. 9.145 and 9.149 9.182 Prob. 9.167 9.183 Prob. 9.168 9.184 Probs. 9.148 and 9.170 564 In the first half of this chapter, we discussed how to determine the resultant R of forces DF distributed over a plane area A when the magnitudes of these forces are proportional to both the areas DA of the elements on which they act and the distances y from these elements to a given x axis; we thus had DF 5 ky DA. We found that the magnitude of the resultant R is proportional to the first moment Qx 5 ey dA of area A, whereas the moment of R about the x axis is proportional to the second moment, or moment of inertia, Ix 5 ey2 dA of A with respect to the same axis [Sec. 9.1A].
Rectangular Moments of Inertia The rectangular moments of inertia Ix and Iy of an area [Sec. 9.1B] are obtained by evaluating the integrals Ix 5 #y 2 d A Iy 5 #x 2 dA (9.1) We can reduce these computations to single integrations by choosing dA to be a thin strip parallel to one of the coordinate axes. We also recall that it is possible to compute Ix and Iy from the same elemental strip (Fig. 9.35) using the formula for the moment of inertia of a rectangular area [Sample Prob. 9.3].
y x y x dx dIx = y3 dx 3 1 dIy = x2y dx Fig. 9.35 Polar Moment of Inertia We defined the polar moment of inertia of an area A with respect to the pole O [Sec. 9.1C] as JO 5 #r2 dA (9.3) where r is the distance from O to the element of area dA (Fig. 9.36). Observ-ing that r 2 5 x 2 1 y 2, we established the relation JO 5 Ix 1 Iy (9.4) Review and Summary y y x dA A x r O Fig. 9.36 565 Radius of Gyration We defined the radius of gyration of an area A with respect to the x axis [Sec. 9.1D] as the distance kx, where Ix 5 k 2 x A. With similar definitions for the radii of gyration of A with respect to the y axis and with respect to O, we have kx 5 B Ix A ky 5 B Iy A kO 5 B JO A (9.5–9.7) Parallel-Axis Theorem The parallel-axis theorem, presented in Sec. 9.2A, states that the moment of inertia I of an area with respect to any given axis AA9 (Fig. 9.37) is equal to the moment of inertia I of the area with respect to the centroidal axis BB9 that is parallel to AA9 plus the product of the area A and the square of the distance d between the two axes: I 5 I 1 Ad2 (9.9) You can use this formula to determine the moment of inertia I of an area with respect to a centroidal axis BB9 if you know its moment of inertia I with respect to a parallel axis AA9. In this case, however, the product Ad2 should be subtracted from the known moment of inertia I. A similar relation holds between the polar moment of inertia JO of an area about a point O and the polar moment of inertia JC of the same area about its centroid C. Letting d be the distance between O and C, we have JC 5 JC 1 Ad2 (9.11) Composite Areas The parallel-axis theorem can be used very effectively to compute the moment of inertia of a composite area with respect to a given axis [Sec. 9.2B]. Considering each component area separately, we first compute the moment of inertia of each area with respect to its centroidal axis, using the data provided in Figs. 9.12 and 9.13 whenever possible. Then apply the parallel-axis theorem to determine the moment of inertia of each component area with respect to the desired axis, and add the values [Sample Probs. 9.4 and 9.5].
Product of Inertia Section 9.3 was devoted to the transformation of the moments of inertia of an area under a rotation of the coordinate axes. First, we defined the product of inertia of an area A as Ixy 5 #xy dA (9.12) and showed that Ixy 5 0 if the area A is symmetrical with respect to either or both of the coordinate axes. We also derived the parallel-axis theorem for products of inertia as Ixy 5 Ix¿y¿ 1 xÊyÊA (9.13) where Ix¿y¿ is the product of inertia of the area with respect to the centroidal axes x9 and y9 that are parallel to the x and y axes and x and y are the coordinates of the centroid of the area [Sec. 9.3A].
A' B' B A C d Fig. 9.37 566 Rotation of Axes In Sec. 9.3B, we determined the moments and product of inertia Ix9, Iy9, and Ix9y9 of an area with respect to x9 and y9 axes obtained by rotating the original x and y coordinate axes counterclockwise through an angle θ (Fig. 9.38). We expressed Ix9, Iy9, and Ix9y9 in terms of the moments and product of inertia Ix, Iy, and Ixy computed with respect to the original x and y axes. Ix¿ 5 Ix 1 Iy 2 1 Ix 2 Iy 2 cos 2θ 2 Ixy sin 2θ (9.18) Iy¿ 5 Ix 1 Iy 2 2 Ix 2 Iy 2 cos 2θ 1 Ixy sin 2θ (9.19) Ix¿y¿ 5 Ix 2 Iy 2 sin 2θ 1 Ixy cos 2θ (9.20) Principal Axes We defined the principal axes of the area about O as the two axes perpen-dicular to each other with respect to which the moments of inertia of the area are maximum and minimum. The corresponding values of θ, denoted by θm, were obtained from tan 2θm 5 2 2Ixy Ix 2 Iy (9.25) Principal Moments of Inertia The corresponding maximum and minimum values of I are called the principal moments of inertia of the area about O: Imax,min 5 Ix 1 Iy 2 ; Ba Ix 2 Iy 2 b 2 1 I2 xy (9.27) We also noted that the corresponding value of the product of inertia is zero.
Mohr’s Circle The transformation of the moments and product of inertia of an area under a rotation of axes can be represented graphically by drawing Mohr’s circle [Sec. 9.4]. Given the moments and product of inertia Ix, Iy, and Ixy of the area with respect to the x and y coordinate axes, we plot points X (Ix, Ixy) and Y (Iy, –Ixy) and draw the line joining these two points (Fig. 9.39). This line is a diameter of Mohr’s circle and thus defines this circle. As the coordinate axes are rotated through θ, the diameter rotates through twice that angle, and the coordinates of X9 and Y9 yield the new values Ix9, Iy9, and Ix9y9 of the moments and product of inertia of the area. Also, the angle θm and the coordinates of points A and B define the principal axes a and b and the principal moments of inertia of the area [Sample Prob. 9.8].
y y' x' x O q Fig. 9.38 567 567 Moments of Inertia of Masses The second half of the chapter was devoted to determining moments of inertia of masses, which are encountered in dynamics problems involving the rotation of a rigid body about an axis. We defined the mass moment of inertia of a body with respect to an axis AA9 (Fig. 9.40) as I 5# r2 dm (9.28) where r is the distance from AA9 to the element of mass [Sec. 9.5A]. We defined the radius of gyration of the body as k 5 B I m (9.29) The moments of inertia of a body with respect to the coordinate axes were expressed as Ix 5# (y2 1 z2) dm Iy 5# (z2 1 x2) dm (9.30) Iz 5# (x2 1 y2) dm Parallel-Axis Theorem We saw that the parallel-axis theorem also applies to mass moments of inertia [Sec. 9.5B]. Thus, the moment of inertia I of a body with respect to an arbitrary axis AA9 (Fig. 9.41) can be expressed as I 5 I 1 md2 (9.33) where I is the moment of inertia of the body with respect to the centroidal axis BB9 that is parallel to the axis AA9, m is the mass of the body, and d is the distance between the two axes.
x' x y' q qm 2q O b a y A B C Y 2qm O Imin Imax Ixy Ix' Ix'y' –Ix'y' –Ixy Ixy Ix, Iy Iy X X' Ix Iy' Y' Fig. 9.39 A' A r1 r2 r3 Δm1 Δm2 Δm3 Fig. 9.40 A' B' A B G d Fig. 9.41 568 Moments of Inertia of Thin Plates We can readily obtain the moments of inertia of thin plates from the moments of inertia of their areas [Sec. 9.5C]. We found that for a rectangular plate the moments of inertia with respect to the axes shown (Fig. 9.42) are IAA9 5 1 12ma2 IBB9 5 1 12mb2 (9.39) ICC9 5 IAA9 1 IBB9 5 1 12m(a2 1 b2) (9.40) whereas for a circular plate (Fig. 9.43), they are IAA9 5 IBB9 5 1 4mr 2 (9.41) ICC9 5 IAA9 1 IBB9 5 1 2mr 2 (9.42) t C' B' A B b a A' C Fig. 9.42 C' C B' A B A' t r Fig. 9.43 Composite Bodies When a body possesses two planes of symmetry, it is usually possible to use a single integration to determine its moment of inertia with respect to a given axis by selecting the element of mass dm to be a thin plate [Sample Probs. 9.10 and 9.11]. On the other hand, when a body consists of several common geo-metric shapes, we can obtain its moment of inertia with respect to a given axis by using the formulas given in Fig. 9.28 together with the parallel-axis theorem [Sample Probs. 9.12 and 9.13].
Moment of Inertia with Respect to an Arbitrary Axis In the last section of the chapter, we described how to determine the moment of inertia of a body with respect to an arbitrary axis OL that is drawn through the origin O [Sec. 9.6A]. We denoted the components of the unit vector l along OL by lx, ly, and lz (Fig. 9.44) and introduced the products of inertia as Ixy 5# xy dm Iyz 5# yz dm Izx 5# zx dm (9.45) We found that the moment of inertia of the body with respect to OL could be expressed as IOL 5 Ixl2 x 1 Iyl2 y 1 Izlz 2 2 2Ixylxly 2 2Iyzlylz 2 2Izxlzlx (9.46) y dm z x p L O q r Fig. 9.44 569 Ellipsoid of Inertia By plotting a point Q along each axis OL at a distance OQ 5 1/2IOL from O [Sec. 9.6B], we obtained the surface of an ellipsoid, known as the ellipsoid of inertia of the body at point O.
x z' y' x' y z O Fig. 9.45 Principal Axes and Principal Moments of Inertia The principal axes x9, y9, and z9 of this ellipsoid (Fig. 9.45) are the principal axes of inertia of the body; that is, the products of inertia Ix9y9, Iy9z9, and Iz9x9 of the body with respect to these axes are all zero. In many situations, you can deduce the principal axes of inertia of a body from its properties of symmetry. Choosing these axes to be the coordinate axes, we can then express IOL as IOL 5 Ix9lx 2 9 1 Iy9l2 y9 1 Iz9lz 2 9 (9.50) where Ix9, Iy9, and Iz9 are the principal moments of inertia of the body at O. When the principal axes of inertia cannot be obtained by observation [Sec. 9.6B], it is necessary to solve the cubic equation K 3 2 (Ix 1 Iy 1 Iz)K 2 1 (Ix Iy 1 Iy Iz 1 IzIx 2 I 2 xy 2 I 2 yz 2 I 2 z x)K 2 (Ix Iy Iz 2 Ix I 2 yz 2 Iy I 2 z x 2 IzI 2 x y 2 2Ix yIyzIz x) 5 0 (9.56) We found [Sec. 9.6C] that the roots K1, K2, and K3 of this equation are the principal moments of inertia of the given body. The direction cosines (lx)1, (ly)1, and (lz)1 of the principal axis corresponding to the principal moment of inertia K1 are then determined by substituting K1 into Eqs. (9.54) and by solv-ing two of these equations and Eq. (9.57) simultaneously. The same procedure is then repeated using K2 and K3 to determine the direction cosines of the other two principal axes [Sample Prob. 9.15].
570 9.185 Determine by direct integration the moments of inertia of the shaded area with respect to the x and y axes.
h y x a y = 4h( ) x a x2 a2 − Fig. P9.185 9.186 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the y axis. 9.187 Determine the moment of inertia and the radius of gyration of the shaded area shown with respect to the x axis.
2b b a y x y = kx2 y = 2b − cx2 Fig. P9.187 9.188 Determine the moments of inertia I x and I y of the area shown with respect to centroidal axes respectively parallel and perpendicular to side AB.
12 mm 18 mm 18 mm 12 mm 22 mm 72 mm 14 mm A B Fig. P9.188 Review Problems x b y a = 1 y2 b2 x2 a2 + Fig. P9.186 571 9.189 Determine the polar moment of inertia of the area shown with respect to (a) point O, (b) the centroid of the area.
9.190 Two L4 3 4 3 1 2-in. angles are welded to a steel plate as shown. Determine the moments of inertia of the combined section with respect to centroidal axes respectively parallel and perpendicular to the plate.
L4 × 4 × 1 2 in.
1 2 10 in.
Fig. P9.190 9.191 Using the parallel-axis theorem, determine the product of inertia of the L5 3 3 3 1 2-in. angle cross section shown with respect to the centroidal x and y axes. 9.192 For the L5 3 3 3 1 2-in. angle cross section shown, use Mohr’s circle to determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30° clockwise, (b) the orientation of the principal axes through the centroid and the corresponding values of the moments of inertia. 9.193 A thin plate with a mass m was cut in the shape of a parallelogram as shown. Determine the mass moment of inertia of the plate with respect to (a) the x axis, (b) the axis BB9 that is perpendicular to the plate.
B B' y A A' x z a a a Fig. P9.193 and P9.194 9.194 A thin plate with mass m was cut in the shape of a parallelogram as shown. Determine the mass moment of inertia of the plate with respect to (a) the y axis, (b) the axis AA9 that is perpendicular to the plate.
54 mm 54 mm Semiellipse 36 mm 18 mm O Fig. P9.189 L5 × 3 × 0.746 in.
1.74 in.
5 in.
3 in.
y x C 1 2 in.
1 2 in.
1 2 Fig. P9.191 and P9.192 572 9.195 A 2-mm-thick piece of sheet steel is cut and bent into the machine component shown. Knowing that the density of steel is 7850 kg/m3, determine the mass moment of inertia of the component with respect to each of the coordinate axes.
x y z 0.48 m 0.76 m Fig. P9.195 9.196 Determine the mass moment of inertia of the steel machine element shown with respect to the z axis. (The specific weight of steel is 490 lb/ft3.) x y z 3.7 in.
0.9 in.
1.35 in.
1.2 in.
1.4 in.
0.6 in.
0.6 in.
0.9 in.
0.9 in.
3.1 in.
9 in.
Fig. P9.196 The method of virtual work is particularly effective when a simple relation can be found among the displacements of the points of application of the various forces involved. This is the case for the scissor lift platform being used by workers to gain access to a highway bridge under construction.
Method of Virtual Work 10 574 Method of Virtual Work Introduction In the preceding chapters, we solved problems involving the equilibrium of rigid bodies by expressing the balance of external forces acting on the bodies. We wrote the equations of equilibrium oFx 5 0, oFy 5 0, and oMA 5 0 and solved them for the desired unknowns. We now consider a different method, which turns out to be more effective for solving certain types of equilibrium problems. This method, based on the principle of virtual work, was first formally used by the Swiss mathematician Jean Bernoulli in the eighteenth century.
As you will see in Sec. 10.1B, the principle of virtual work consid-ers a particle or rigid body (or more generally, a system of connected rigid bodies) that is in equilibrium under various external forces. The principle states that, if the body is given an arbitrary displacement from that position of equilibrium, the total work done by the external forces during the dis-placement is zero. This principle is particularly effective when applied to the solution of problems involving the equilibrium of machines or mecha-nisms consisting of several connected members.
In the second part of this chapter, we apply the method of virtual work in an alternative form based on the concept of potential energy. We will show in Sec. 10.2 that, if a particle, rigid body, or system of rigid bodies is in equilibrium, the derivative of its potential energy with respect to a variable defining its position must be zero.
You will also learn in this chapter to evaluate the mechanical efficiency of a machine (Sec. 10.1D) and to determine whether a given position of equilibrium is stable, unstable, or neutral (Sec. 10.2D).
10.1 THE BASIC METHOD The first step in explaining the method of virtual work is to define the terms displacement and work as they are used in mechanics. Then we can state the principle of virtual work and show how to apply it in practical situations. We also take the opportunity to define mechanical efficiency, which is a useful and important parameter for the design of real machines.
Objectives • Define the work of a force, and consider the circum-stances when a force does no work.
• Examine the principle of virtual work, and apply it to analyze the equilibrium of machines and mechanisms.
• Apply the concept of potential energy to determine the equilibrium position of a rigid body or a system of rigid bodies.
• Evaluate the mechanical effi ciency of machines, and consider the stability of equilibrium. Introduction 10.1 THE BASIC METHOD 10.1A Work of a Force 10.1B Principle of Virtual Work 10.1C Applying the Principle of Virtual Work 10.1D Mechanical Efficiency of Real Machines 10.2 WORK, POTENTIAL ENERGY, AND STABILITY 10.2A Work of a Force During a Finite Displacement 10.2B Potential Energy 10.2C Potential Energy and Equilibrium 10.2D Stability of Equilibrium 10.1 The Basic Method 575 10.1A Work of a Force Consider a particle that moves from a point A to a neighboring point A9 (Fig. 10.1). If r denotes the position vector corresponding to point A, we denote the small vector joining A and A9 by the differential dr; we call the vector dr the displacement of the particle. Now let us assume that a force F is acting on the particle. The work dU of force F corresponding to the displacement dr is defined as the quantity Definition of work dU 5 F ? dr (10.1) That is, dU is the scalar product of the force F and the displacement dr. Suppose we denote the magnitudes of the force by F, the displacement by ds, and the angle formed by F and dr by α. Then, recalling the defini-tion of the scalar product of two vectors (Sec. 3.2A), we have dU 5 F ds cos α (10.19) Work is a scalar quantity, so it has a magnitude and a sign, but no direc-tion. Note that work should be expressed in units obtained by multiplying units of length by units of force. Thus, if we use U.S. customary units, we should express work in ft?lb or in?lb. If we use SI units, we express work in N ? m. This unit of work is called a joule (J).† It follows from (10.19) that work dU is positive if the angle α is acute and negative if α is obtuse. Three particular cases are of special interest. • If the force F has the same direction as d r, the work dU reduces to F ds. • If F has a direction opposite to that of d r, the work is dU 5 2F ds. • Finally, if F is perpendicular to d r, the work dU is zero.
We can also consider the work dU of a force F during a displace-ment d r to be the product of F and the component ds cos α of the displacement dr along F (Fig. 10.2a). This view is particularly useful in computing the work done by the weight W of a body (Fig. 10.2b). The work of W is equal to the product of W and the vertical displacement dy of the center of gravity G of the body. If the displacement is downward, the work is positive; if the displacement is upward, the work is negative.
Some forces frequently encountered in statics do no work, such as forces applied to fixed points (ds 5 0) or acting in a direction perpendicu-lar to the displacement (cos α 5 0). Among these forces are the reaction at a frictionless pin when the body supported rotates about the pin; the reaction at a frictionless surface when the body in contact moves along the surface; the reaction at a roller moving along its track; the weight of a body when its center of gravity moves horizontally; and the friction force dU 5 F ? dr Fig. 10.1 The work of a force acting on a particle is the scalar product of the force and the particle's displacement.
a dr A A' O r F r + dr †The joule is the SI unit of energy, whether in mechanical form (work, potential energy, kinetic energy) or in chemical, electrical, or thermal form. Note that even though 1 N ? m 5 1 J, we must express the moment of a force in N ? m, and not in joules, since the moment of a force is not a form of energy.
Fig. 10.2 (a) You can think of work as the product of a force and the component of displacement in the direction of the force. (b) This is useful for computing the work done by an object's weight.
ds cos a a a dr dr A dy G W F (a) (b) A' G' 576 Method of Virtual Work acting on a wheel rolling without slipping (since at any instant the point of contact does not move). Examples of forces that do work are the weight of a body (except in the case considered previously), the friction force acting on a body sliding on a rough surface, and most forces applied on a moving body.
In certain cases, the sum of the work done by several forces is zero. Consider, for example, two rigid bodies AC and BC that are connected at C by a frictionless pin (Fig. 10.3a). Among the forces acting on AC is the force F exerted at C by BC. In general, the work of this force is not zero, but it is equal in magnitude and opposite in sign to the work of the force 2F exerted by AC on BC, since these forces are equal and opposite and are applied to the same particle. Thus, when the total work done by all the forces acting on AB and BC is considered, the work of the two internal forces at C cancels out. We obtain a similar result if we consider a system consisting of two blocks connected by a cord AB that is not extensible (Fig. 10.3b). The work of the tension force T at A is equal in magnitude to the work of the tension force T9 at B, since these forces have the same magnitude and the points A and B move through the same distance; but in one case, the work is positive, and in the other, it is negative. Thus, the work of the internal forces again cancels out.
Photo 10.1 (a) In analyzing a crane, we might consider displacements associated with vertical movement of a container. (b) A force does work if it has a component in the direction of a displacement. (c) A force does no work if there is no displacement or if the force is perpendicular to a displacement.
Force in hydraulic cylinder Tension in cable Boom reactions Weight of boom if not moving vertically Weight of load Fig. 10.3 (a) For a frictionless pin or (b) a cord that is not extensible, the total work done by the pairs of internal forces is zero.
A C B B –F F (a) (b) T' T A We can show that the total work of the internal forces holding together the particles of a rigid body is zero. Consider two particles A and B of a rigid body and the two equal and opposite forces F and 2F they exert on each other (Fig. 10.4). Although, in general, small displacements dr and dr9 of the two particles are different, the components of these displacements along AB must be equal; otherwise, the particles would not Fig. 10.4 As demonstrated here for an arbitrary pair of particles, the total work of the internal forces holding a rigid body together is zero.
B B' A' –F F dr A dr' 10.1 The Basic Method 577 remain at the same distance from each other, so the body would not be rigid. Therefore, the work of F is equal in magnitude and opposite in sign to the work of 2F, and their sum is zero.
In computing the work of the external forces acting on a rigid body, it is often convenient to determine the work of a couple without consider-ing separately the work of each of the two forces forming the couple. Consider the two forces F and 2F forming a couple of moment M and acting on a rigid body (Fig. 10.5). Any small displacement of the rigid body bringing A and B, respectively, into A9 and B0 can be divided into two parts: one in which points A and B undergo equal displacements dr1, the other in which A9 remains fixed while B9 moves into B0 through a displacement dr2 with a magnitude of ds2 5 r dθ. In the first part of the motion, the work of F is equal in magnitude and opposite in sign to the work of 2F, and their sum is zero. In the second part of the motion, only force F works, and its work is dU 5 F ds2 5 Fr dθ. But the product Fr is equal to the magnitude M of the moment of the couple. Thus, the work of a couple of moment M acting on a rigid body is Work of a couple dU 5 M dθ (10.2) where dθ is the small angle (expressed in radians) through which the body rotates. We again note that work should be expressed in units obtained by multiplying units of force by units of length.
10.1B Principle of Virtual Work Consider a particle acted upon by several forces F1, F2, . . . , Fn (Fig. 10.6). We can imagine that the particle undergoes a small displacement from A to A9. This displacement is possible, but it does not necessarily take place. The forces may be balanced and the particle remains at rest, or the particle may move under the action of the given forces in a direction different from that of AA9. Since the considered displacement does not actually occur, it is called a virtual displacement, which is denoted by δ r. The symbol δ r represents a differential of the first order; it is used to distin-guish the virtual displacement from the displacement d r that would take place under actual motion. As you will see, we can use virtual displace-ments to determine whether the conditions of equilibrium of a particle are satisfied.
The work of each of the forces F1, F2, . . . , Fn during the virtual displacement δ r is called virtual work. The virtual work of all the forces acting on the particle of Fig. 10.6 is δU 5 F1 ? δr 1 F2 ? δr 1 . . . 1 Fn ? δr 5 (F1 1 F2 1 . . . 1 Fn) ? δr or δU 5 R ? δr (10.3) where R is the resultant of the given forces. Thus, the total virtual work of the forces F1, F2, . . . , Fn is equal to the virtual work of their resultant R.
dU 5 M dθ Fig. 10.5 The work of a couple acting on a rigid body is the moment of the couple times the angular rotation.
B' B'' B A –F F dr1 dr1 dr2 A' r dq Fig. 10.6 Forces acting on a particle that goes through a virtual displacement.
F2 F1 Fn A A' dr 578 Method of Virtual Work The principle of virtual work for a particle states: If a particle is in equilibrium, the total virtual work of the forces acting on the particle is zero for any virtual displacement of the particle.
This condition is necessary: if the particle is in equilibrium, the resultant R of the forces is zero, and it follows from Eq. (10.3) that the total virtual work δU is zero. The condition is also sufficient: if the total virtual work δU is zero for any virtual displacement, the scalar product R ? δr is zero for any δr, and the resultant R must be zero.
In the case of a rigid body, the principle of virtual work states: If a rigid body is in equilibrium, the total virtual work of the external forces acting on the rigid body is zero for any virtual displacement of the body.
The condition is necessary: if the body is in equilibrium, all the particles forming the body are in equilibrium and the total virtual work of the forces acting on all the particles must be zero. However, we have seen in the preceding section that the total work of the internal forces is zero; there-fore, the total work of the external forces also must be zero. The condition can also be proven to be sufficient.
The principle of virtual work can be extended to the case of a system of connected rigid bodies. If the system remains connected during the virtual displacement, only the work of the forces external to the system need be considered, since the total work of the internal forces at the various connections is zero.
10.1C Applying the Principle of Virtual Work The principle of virtual work is particularly effective when applied to the solution of problems involving machines or mechanisms consisting of several connected rigid bodies. Consider, for instance, the toggle vise ACB of Fig. 10.7a used to compress a wooden block. Suppose we wish to deter-mine the force exerted by the vise on the block when a given force P is applied at C, assuming there is no friction. Denoting the reaction of the block on the vise by Q, we draw the free-body diagram of the vise and Fig. 10.7 (a) A toggle vise used to compress a wooden block, assuming no friction; (b) a virtual displacement of the vise.
l l B A q q P N B A x P (a) (b) C' xB yC q dxB –dyC dq C y Q C B' Ax Ay 10.1 The Basic Method 579 consider the virtual displacement obtained by giving a positive increment δθ to angle θ (Fig. 10.7b). Choosing a system of coordinate axes with origin at A, we note that xB increases as yC decreases. This is indicated in the figure, where we indicate a positive increment δxB and a negative increment 2δyC. The reactions Ax, Ay, and N do no work during the virtual displacement considered, so we need only compute the work done by P and Q. Since Q and δxB have opposite senses, the virtual work of Q is δUQ 5 2Q δxB. Since P and the increment shown (2δyC) have the same sense, the virtual work of P is δUP 5 1P(2δyC) 5 2P δyC. (We could have predicted the minus signs by simply noting that the forces Q and P are directed opposite to the positive x and y axes, respectively.) Expressing the coordinates xB and yC in terms of the angle θ and differentiating, we obtain xB 5 2l sin θ yC 5 l cos θ δxB 5 2l cos θ δθ δyC 5 2l sin θ δθ (10.4) The total virtual work of the forces Q and P is thus δU 5 δUQ 1 δUP 5 2Q δxB 2 P δyC 5 22Ql cos θ δθ 1 Pl sin θ δθ Setting δU 5 0, we obtain 2Ql cos θ δθ 5 Pl sin θ δθ (10.5) and Q 5 1 2 P tan θ (10.6) The superiority of the method of virtual work over the conventional equilibrium equations in the problem considered here is clear: by using the method of virtual work, we were able to eliminate all unknown reac-tions, whereas the equation oMA 5 0 would have eliminated only two of the unknown reactions. This property of the method of virtual work can be used in solving many problems involving machines and mechanisms. If the virtual displacement considered is consistent with the constraints imposed by the supports and connections, all reactions and internal forces are eliminated and only the work of the loads, applied forces, and friction forces need be considered.
We can also use the method of virtual work to solve problems involv-ing completely constrained structures, although the virtual displacements considered never actually take place. Consider, for example, the frame ACB shown in Fig. 10.8a. If point A is kept fixed while point B is given a hori-zontal virtual displacement (Fig. 10.8b), we need consider only the work of P and Bx. We can thus determine the reaction component Bx in the same way as the force Q of the preceding example (Fig. 10.7b); we have Bx 5 1 2 P tan θ By keeping B fixed and giving a horizontal virtual displacement to A, we can similarly determine the reaction component Ax. Then we can determine the components Ay and By by rotating the frame ACB as a rigid body about B and A, respectively.
Photo 10.2 The method of virtual work is useful for determining the forces exerted by the hydraulic cylinders positioning the bucket lift. The reason is that a simple relation exists among the displacements of the points of application of the forces acting on the members of the lift.
580 Method of Virtual Work We can also use the method of virtual work to determine the configuration of a system in equilibrium under given forces. For example, we can obtain the value of the angle θ for which the linkage of Fig. 10.7 is in equilibrium under two given forces P and Q by solving Eq. (10.6) for tan θ.
Note, however, that the attractiveness of the method of virtual work depends to a large extent upon the existence of simple geometric relations between the various virtual displacements involved in the solution of a given problem. When no such simple relations exist, it is usually advisable to revert to the conventional method of Chap. 6.
10.1D Mechanical Efficiency of Real Machines In analyzing the toggle vise of Fig. 10.7, we assumed that no friction forces were involved. Thus, the virtual work consisted only of the work of the applied force P and of the reaction Q. However, the work of reac-tion Q is equal in magnitude and opposite in sign to the work of the force exerted by the vise on the block. Therefore, Equation (10.5) states that the output work 2Ql cos θ δθ is equal to the input work Pl sin θ δθ. A machine in which input and output work are equal is said to be an “ideal” machine. In a “real” machine, friction forces always do some work, and the output work is smaller than the input work.
Consider again the toggle vise of Fig. 10.7a. and now assume that a friction force F develops between the sliding block B and the horizontal plane (Fig. 10.9). Using the conventional methods of statics and summing moments about A, we find N 5 P/2. Denoting the coefficient of friction between block B and the horizontal plane by μ, we have F 5 μN 5 μP/2. Fig. 10.8 (a) A completely constrained frame ACB; (b) a virtual displacement of the frame in order to determine Bx, keeping A fixed.
q q x q C y C A B P P B' C' Bx xB Ay A B By Ax l l yC dq dxB –dyC (a) (b) Photo 10.3 The clamping force of the toggle clamp shown can be expressed as a function of the force applied to the handle by first establishing the geometric relations among the members of the clamp and then applying the method of virtual work.
10.1 The Basic Method 581 Recalling formulas (10.4), we find that the total virtual work of the forces Q, P, and F during the virtual displacement shown in Fig. 10.9 is δU 5 2Q δxB 2 P δyC 2 F δxB 5 22Ql cos θ δθ 1 Pl sin θ δθ 2 μPl cos θ δθ Setting δU 5 0, we obtain 2Ql cos θ δθ 5 Pl sin θ δθ 2 μPl cos θ δθ (10.7) This equation states that the output work is equal to the input work minus the work of the friction force. Solving for Q, we have Q 5 1 2 P (tan θ 2 μ) (10.8) Note that Q 5 0 when tan θ 5 μ, that is, when θ is equal to the angle of friction ϕ, and that Q , 0 when θ , ϕ. Thus, we can use the toggle vise only for values of θ larger than the angle of friction.
We define the mechanical efficiency η of a machine as the ratio Mechanical efficiency h 5 output work input work (10.9) Clearly, the mechanical efficiency of an ideal machine is η 5 1 when input and output work are equal, whereas the mechanical efficiency of a real machine is always less than 1.
In the case of the toggle vise we have just analyzed, we have h 5 output work input work 5 2Ql cos θ δθ Pl sin θ δθ (10.10) We can check that, in the absence of friction forces, we would have μ 5 0 and η 5 1. In the general case when μ is different from zero, the efficiency η becomes zero for μ cot θ 5 1, that is, for tan θ 5 μ or θ 5 tan21 μ 5 ϕ. We note again that the toggle vise can be used only for values of θ larger than the angle of friction ϕ.
h 5 output work input work Fig. 10.9 A virtual displacement of the toggle vise with friction.
N B A x P C' xB yC q dxB –dyC dq C y Q B' Ax Ay F = mN 582 Method of Virtual Work Sample Problem 10.1 Using the method of virtual work, determine the magnitude of the couple M required to maintain the equilibrium of the mechanism shown.
STRATEGY: For a virtual displacement consistent with the constraints, the reactions do no work, so you can focus solely on the force P and the moment M. You can solve for M in terms of P and the geometric parameters.
MODELING: Choose a coordinate system with origin at E (Fig. 1). Then xD 5 3l cos θ δxD 5 23l sin θ δθ ANALYSIS: Principle of Virtual Work. Since the reactions A, Ex, and Ey do no work during the virtual displacement, the total virtual work done by M and P must be zero. Notice that P acts in the positive x direc-tion and M acts in the positive θ direction. You obtain δU 5 0: 1M δθ 1 P δxD 5 0 1M δθ 1 P(23l sin θ δθ) 5 0 M 5 3Pl sin θ b REFLECT and THINK: This problem illustrates that the principle of virtual work can help determine a moment as well as a force in a straight-forward computation.
P M A B D l q q l l C F E P A Ex Ey M A B D q C F x y E dq – dxD xD Fig. 1 Free-body diagram of mechanism showing a virtual displacement.
Sample Problem 10.2 Determine the expressions for θ and for the tension in the spring that correspond to the equilibrium position of the mechanism. The unstretched length of the spring is h, and the spring constant is k. Neglect the weight of the mechanism.
STRATEGY: The tension in the spring is a force F exerted at C. Apply-ing the principle of virtual work, you can obtain a relationship between F and the applied force P.
MODELING: With the coordinate system shown in Fig. 1, yB 5 l sin θ yC 5 2l sin θ δyB 5 l cos θ δθ δyC 5 2l cos θ δθ The elongation of the spring is s 5 yC 2 h 5 2l sin θ 2 h. The magnitude of the force exerted at C by the spring is F 5 ks 5 k(2l sin θ 2 h) (1) ANALYSIS: Principle of Virtual Work. Since the reactions Ax, Ay, and C do no work, the total virtual work done by P and F must be zero.
δU 5 0: P δyB 2 F δyC 5 0 P(l cos θ δθ) 2 k(2l sin θ 2 h)(2l cos θ δθ) 5 0 sin θ 5 P 1 2kh 4kl b Substituting this expression into Eq. (1), you obtain F 5 1 2P b REFLECT and THINK: You can verify these results by applying the appropriate equations of equilibrium.
A B C l l q q P A x A y yC yB dyB dyC A B C q dq P F C h x y s Fig. 1 Free-body diagram of mechanism showing a virtual displacement.
10.1 The Basic Method 583 Sample Problem 10.3 A hydraulic-lift table is used to raise a 1000-kg crate. The table consists of a platform and two identical linkages on which hydraulic cylinders exert equal forces. (Only one link-age and one cylinder are shown.) Members EDB and CG are each of length 2a, and member AD is pinned to the midpoint of EDB. If the crate is placed on the table so that half of its weight is supported by the system shown, determine the force exerted by each cylinder in raising the crate for θ 5 60°, a 5 0.70 m, and L 5 3.20 m. (This mechanism was previously considered in Sample Prob. 6.7.) STRATEGY: The principle of virtual work allows you to find a relationship between the force applied by the cylinder and the weight without involving the reactions. However, you need a relationship between the virtual displacement and the change in angle θ, which is found from the law of cosines applied to the given geometry.
MODELING: The free body consists of the platform and the linkage (Fig. 1), with an input force FDH exerted by the cylinder and an output force equal and opposite to 1 2 W.
FDH FCG Ey Ex E G A B C D W 1 2 Fig. 1 Free-body diagram of the platform and linkage.
ANALYSIS: Principle of Virtual Work. First observe that the reactions at E and G do no work. Denoting the eleva-tion of the platform above the base by y and the length DH of the cylinder-and-piston assembly by s (Fig. 2), you have δU 5 0: 21 2W δy 1 FDH δs 5 0 (1) You can express the vertical displacement δy of the platform in terms of the angular displacement δθ of EDB as y 5 (EB) sin θ 5 2a sin θ δy 5 2a cos θ δθ To express δs similarly in terms of δθ, first note that by the law of cosines (Fig. 3), s2 5 a2 1 L2 2 2aL cos θ A B C D E G H 2a W 1 2 θ L 2 L 2 d H s y FDH E B D W 1 2 q B' dy ds dq D' Fig. 2 Virtual displacement of the machine.
H E D a q L s Fig. 3 Geometry associated with the cylinder-and-piston assembly.
584 Method of Virtual Work Differentiating, 2s δs 5 22aL(2sin θ) δθ ds 5 aL sin u s du Substituting for δy and δs into Eq. (1), you have (21 2 W )2a cos u du 1 FDH aL sin u s du 5 0 FDH 5 W s L cot u With the given numerical data, you obtain W 5 mg 5 (1000 kg)(9.81 m/s2) 5 9810 N 5 9.81 kN s2 5 a2 1 L2 2 2aL cos θ 5 (0.70)2 1 (3.20)2 2 2(0.70)(3.20) cos 608 5 8.49 s 5 2.91 m FDH 5 W s L cot u 5 (9.81 kN)2.91 m 3.20 m cot 60° FDH 5 5.15 kN b REFLECT and THINK: The principle of virtual work gives you a rela-tionship between forces, but sometimes you need to review the geometry carefully to find a relationship between the displacements.
585 585 SOLVING PROBLEMS ON YOUR OWN I n this section, we described how to use the method of virtual work, which is a different way of solving problems involving the equilibrium of rigid bodies.
The work done by a force during a displacement of its point of application or by a couple during a rotation is found, respectively, by using: dU 5 F ds cos α (10.1) dU 5 M dθ (10.2) Principle of virtual work. In its more general and more useful form, this principle can be stated as: If a system of connected rigid bodies is in equilibrium, the total virtual work of the external forces applied to the system is zero for any virtual displacement of the system.
As you apply the principle of virtual work, keep in mind the following points.
1. Virtual displacement. A machine or mechanism in equilibrium has no tendency to move. However, we can cause—or imagine—a small displacement. Since it does not actually occur, such a displacement is called a virtual displacement.
2. Virtual work. The work done by a force or couple during a virtual displacement is called virtual work.
3. You need consider only the forces that do work during the virtual displacement.
4. Forces that do no work during a virtual displacement that are consistent with the constraints imposed on the system are a. Reactions at supports b. Internal forces at connections c. Forces exerted by inextensible cords and cables None of these forces need be considered when you use the method of virtual work.
5. Be sure to express the various virtual displacements involved in your computa-tions in terms of a single virtual displacement. This is done in each of the three preceding sample problems, where the virtual displacements are all expressed in terms of δθ.
6. Remember that the method of virtual work is effective only in those cases where the geometry of the system makes it relatively easy to relate the displacements involved.
586 Problems 10.1 Determine the vertical force P that must be applied at C to maintain the equilibrium of the linkage. 10.2 Determine the horizontal force P that must be applied at A to main-tain the equilibrium of the linkage.
10 in.
5 in.
4 in.
6 in.
9 in.
6 in.
A B C D E F G 30 lb 80 lb 240 lb·in.
60 lb Fig. P10.2 and P10.4 10.3 and 10.4 Determine the couple M that must be applied to member ABC to maintain the equilibrium of the linkage. 10.5 A spring of constant 15 kN/m connects points C and F of the linkage shown. Neglecting the weight of the spring and linkage, determine the force in the spring and the vertical motion of point G when a vertical downward 120-N force is applied (a) at point C, (b) at points C and H.
H C F G D E B A Fig. P10.5 and P10.6 10.6 A spring of constant 15 kN/m connects points C and F of the linkage shown. Neglecting the weight of the spring and linkage, determine the force in the spring and the vertical motion of point G when a vertical downward 120-N force is applied (a) at point E, (b) at points E and F.
60 N 80 N 20 N 40 N 0.3 m 0.3 m 0.3 m A D B C G E F Fig. P10.1 and P10.3 587 10.7 The two-bar linkage shown is supported by a pin and bracket at B and a collar at D that slides freely on a vertical rod. Determine the force P required to maintain the equilibrium of the linkage.
100 lb A B C D E F P 8 in.
8 in.
8 in.
9 in.
150 lb Fig. P10.7 10.8 Knowing that the maximum friction force exerted by the bottle on the cork is 60 lb, determine (a) the force P that must be applied to the corkscrew to open the bottle, (b) the maximum force exerted by the base of the corkscrew on the top of the bottle. 10.9 Rod AD is acted upon by a vertical force P at end A and by two equal and opposite horizontal forces of magnitude Q at points B and C. Derive an expression for the magnitude Q of the horizontal forces required for equilibrium. 10.10 and 10.11 The slender rod AB is attached to a collar A and rests on a small wheel at C. Neglecting the radius of the wheel and the effect of friction, derive an expression for the magnitude of the force Q required to maintain the equilibrium of the rod.
Q A B C q a P l Fig. P10.10 Q A B C P q a l Fig. P10.11 10.12 Knowing that the line of action of the force Q passes through point C, derive an expression for the magnitude of Q required to maintain equilibrium.
10.13 Solve Prob. 10.12 assuming that the force P applied at point A acts horizontally to the left.
P Fig. P10.8 P D C A B a Q −Q q a a Fig. P10.9 P D E C A B l l l Q q q Fig. P10.12 588 10.14 The mechanism shown is acted upon by the force P; derive an expression for the magnitude of the force Q required to maintain equilibrium. 10.15 and 10.16 Derive an expression for the magnitude of the couple M required to maintain the equilibrium of the linkage shown.
P q A B C l l 1 2 M Fig. P10.15 P P A B C l l q M Fig. P10.16 10.17 A uniform rod AB with length l and weight W is suspended from two cords AC and BC of equal length. Derive an expression for the magnitude of the couple M required to maintain equilibrium of the rod in the position shown. 10.18 The pin at C is attached to member BCD and can slide along a slot cut in the fixed plate shown. Neglecting the effect of friction, derive an expression for the magnitude of the couple M required to maintain equilibrium when the force P that acts at D is directed (a) as shown, (b) vertically downward, (c) horizontally to the right.
P A B C D q q q M l l l Fig. P10.18 10.19 For the linkage shown, determine the couple M required for equilibrium when l 5 1.8 ft, Q 5 40 lb, and θ 5 65°. 10.20 For the linkage shown, determine the force Q required for equilib-rium when l 5 18 in., M 5 600 lb?in., and θ 5 70°.
E F l l l Q P D C B A q q Fig. P10.14 W A B q a a M C Fig. P10.17 Q q A B C l l 1 2 M Fig. P10.19 and P10.20 589 10.21 A 4-kN force P is applied as shown to the piston of the engine system. Knowing that AB 5 50 mm and BC 5 200 mm, determine the couple M required to maintain the equilibrium of the system when (a) θ 5 30°, (b) θ 5 150°. 10.22 A couple M with a magnitude of 100 N?m is applied as shown to the crank of the engine system. Knowing that AB 5 50 mm and BC 5 200 mm, determine the force P required to maintain the equilibrium of the system when (a) θ 5 60°, (b) θ 5 120°. 10.23 Rod AB is attached to a block at A that can slide freely in the vertical slot shown. Neglecting the effect of friction and the weights of the rods, determine the value of θ corresponding to equilibrium.
800 N 160 N q D B C A 200 mm 200 mm 100 mm Fig. P10.23 10.24 Solve Prob. 10.23 assuming that the 800-N force is replaced by a 24-N?m clockwise couple applied at D. 10.25 Determine the value of θ corresponding to the equilibrium position of the rod of Prob. 10.10 when l 5 30 in., a 5 5 in., P 5 25 lb, and Q 5 40 lb. 10.26 Determine the values of θ corresponding to the equilibrium position of the rod of Prob. 10.11 when l 5 24 in., a 5 4 in., P 5 10 lb, and Q 5 18 lb. 10.27 Determine the value of θ corresponding to the equilibrium position of the mechanism of Prob. 10.12 when P 5 80 N and Q 5 100 N. 10.28 Determine the value of θ corresponding to the equilibrium position of the mechanism of Prob. 10.14 when P 5 270 N and Q 5 960 N. 10.29 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is k, and the spring is unstretched when θ 5 30°. For the loading shown, derive an equation in P, θ, l, and k that must be satisfied when the system is in equilibrium. 10.30 Two rods AC and CE are connected by a pin at C and by a spring AE. The constant of the spring is 1.5 lb/in., and the spring is unstretched when θ 5 30°. Knowing that l 5 10 in. and neglecting the weight of the rods, determine the value of θ corresponding to equilibrium when P 5 40 lb. 10.31 Solve Prob. 10.30 assuming that force P is moved to C and acts vertically downward.
q P M A B C Fig. P10.21 and P10.22 A B C D E q q P l l Fig. P10.29 and P10.30 590 10.32 Two bars AD and DG are connected by a pin at D and by a spring AG. Knowing that the spring is 300 mm long when unstretched and that the constant of the spring is 5 kN/m, determine the value of x corresponding to equilibrium when a 900-N load is applied at E as shown. 10.33 Solve Prob. 10.32 assuming that the 900-N vertical force is applied at C instead of E.
10.34 Two 5-kg bars AB and BC are connected by a pin at B and by a spring DE. Knowing that the spring is 150 mm long when unstretched and that the constant of the spring is 1 kN/m, determine the value of x corresponding to equilibrium.
A 200 mm 200 mm 400 mm 400 mm B E C D x Fig. P10.34 10.35 A vertical force P with a magnitude of 150 N is applied to end E of cable CDE that passes over a small pulley D and is attached to the mechanism at C. The constant of the spring is k 5 4 kN/m, and the spring is unstretched when θ 5 0. Neglecting the weight of the mechanism and the radius of the pulley, determine the value of θ corresponding to equilibrium.
200 mm 100 mm 200 mm P q B A C D E Fig. P10.35 A 200 mm G B F E C D x 900 N 200 mm 200 mm 200 mm 200 mm 200 mm Fig. P10.32 591 10.36 A load W with a magnitude of 72 lb is applied to the mechanism at C. Neglecting the weight of the mechanism, determine the value of θ corresponding to equilibrium. The constant of the spring is k 5 20 lb/in., and the spring is unstretched when θ 5 0. 10.37 and P10.38 Knowing that the constant of spring CD is k and that the spring is unstretched when rod ABC is horizontal, determine the value of θ corresponding to equilibrium for the data indicated. 10.37 P 5 300 N, l 5 400 mm, and k 5 5 kN/m 10.38 P 5 75 lb, l 5 15 in., and k 5 20 lb/in. 10.39 The lever AB is attached to the horizontal shaft BC that passes through a bearing and is welded to a fixed support at C. The torsional spring constant of the shaft BC is K; that is, a couple of magnitude K is required to rotate end B through 1 rad. Knowing that the shaft is untwisted when AB is horizontal, determine the value of θ corresponding to the position of equilibrium when P 5 100 N, l 5 250 mm, and K 5 12.5 N?m/rad.
P A B C l q Fig. P10.39 10.40 Solve Prob. 10.39 assuming that P 5 350 N, l 5 250 mm, and K 5 12.5 N?m/rad. Obtain answers in each of the following quad-rants: 0 , θ , 90°, 270° , θ , 360°, and 360° , θ , 450°. 10.41 The position of boom ABC is controlled by the hydraulic cylinder BD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin B when θ 5 70°.
C D 2 ft q 8 kips 1.5 ft 3 ft B A Fig. P10.41 and P10.42 10.42 The position of boom ABC is controlled by the hydraulic cylinder BD. For the loading shown, determine the largest allowable value of the angle θ if the maximum force that the cylinder can exert on pin B is 25 kips.
W q A B C 6 in.
15 in.
Fig. P10.36 P A B C D l l l q Fig. P10.37 and P10.38 592 10.43 The position of member ABC is controlled by the hydraulic cylinder CD. For the loading shown, determine the force exerted by the hydraulic cylinder on pin C when θ 5 55°.
B C D A 0.5 m 0.8 m 90° 10 kN 1.5 m θ Fig. P10.43 and P10.44 10.44 The position of member ABC is controlled by the hydraulic cylinder CD. Determine angle θ, knowing that the hydraulic cylinder exerts a 15-kN force on pin C. 10.45 The telescoping arm ABC is used to provide an elevated platform for construction workers. The workers and the platform together weigh 500 lb, and their combined center of gravity is located directly above C. For the position when θ 5 20°, determine the force exerted on pin B by the single hydraulic cylinder BD. 10.46 Solve Prob. 10.45 assuming that the workers are lowered to a point near the ground so that θ 5 220°.
10.47 Denoting the coefficient of static friction between collar C and the vertical rod by μs, derive an expression for the magnitude of the largest couple M for which equilibrium is maintained in the position shown. Explain what happens if μs $ tan θ.
P q A B C l l 1 2 M Fig. P10.47 and P10.48 10.48 Knowing that the coefficient of static friction between collar C and the vertical rod is 0.40, determine the magnitude of the largest and smallest couple M for which equilibrium is maintained in the position shown, when θ 5 35°, l 5 600 mm, and P 5 300 N.
A B C D 15 ft 7.2 ft 2.7 ft 1.5 ft q Fig. P10.45 593 10.49 A block with weight W is pulled up a plane forming an angle α with the horizontal by a force P directed along the plane. If μ is the coefficient of friction between the block and the plane, derive an expression for the mechanical efficiency of the system. Show that the mechanical efficiency cannot exceed 1 2 if the block is to remain in place when the force P is removed. 10.50 Derive an expression for the mechanical efficiency of the jack discussed in Sec. 8.2B. Show that if the jack is to be self-locking, the mechanical efficiency cannot exceed 1 2.
10.51 Denoting the coefficient of static friction between the block attached to rod ACE and the horizontal surface by μs, derive expressions in terms of P, μs, and θ for the largest and smallest magnitude of the force Q for which equilibrium is maintained.
Q E F l l l P D C B q q A Fig. P10.51 and P10.52 10.52 Knowing that the coefficient of static friction between the block attached to rod ACE and the horizontal surface is 0.15, determine the magnitude of the largest and smallest force Q for which equilib-rium is maintained when θ 5 30°, l 5 0.2 m, and P 5 40 N. 10.53 Using the method of virtual work, determine separately the force and couple representing the reaction at A.
800 N G F C E D B 1.5 m 2.4 m 1.2 m 1.8 m 600 N 1.5 m 1.8 m A Fig. P10.53 and P10.54 10.54 Using the method of virtual work, determine the reaction at D.
594 10.55 Referring to Prob. 10.43 and using the value found for the force exerted by the hydraulic cylinder CD, determine the change in the length of CD required to raise the 10-kN load by 15 mm.
10.56 Referring to Prob. 10.45 and using the value found for the force exerted by the hydraulic cylinder BD, determine the change in the length of BD required to raise the platform attached at C by 2.5 in. 10.57 Determine the vertical movement of joint D if the length of member BF is increased by 1.5 in. (Hint: Apply a vertical load at joint D, and using the methods of Chap. 6, compute the force exerted by member BF on joints B and F. Then apply the method of virtual work for a virtual displacement resulting in the specified increase in length of member BF. This method should be used only for small changes in the lengths of members.) A B C D E F G H 30 ft 40 ft 40 ft 40 ft 40 ft Fig. P10.57 and P10.58 10.58 Determine the horizontal movement of joint D if the length of member BF is increased by 1.5 in. (See the hint for Prob. 10.57.) 10.2 Work, Potential Energy, and Stability 595 10.2 WORK, POTENTIAL ENERGY, AND STABILITY The concept of virtual work has another important connection with equilibrium, leading to criteria for conditions of stable, unstable, and neutral equilibrium. However, to explain this connection, we first need to introduce expressions for the work of a force during a finite displacement and then to define the concept of potential energy.
10.2A Work of a Force During a Finite Displacement Consider a force F acting on a particle. In Sec. 10.1A, we defined the work of F corresponding to an infinitesimal displacement dr of the particle as dU 5 F ? dr (10.1) We obtain the work of F corresponding to a finite displacement of the particle from A1 to A2 (Fig. 10.10a) that is denoted by U1y2 by integrating Eq. (10.1) along the curve described by the particle. Thus, Work during a finite displacement U1y2 5# A2 A1 F ? dr (10.11) Using the alternative expression dU 5 F ds cos α (10.19) given in Sec. 10.1 for the elementary work dU, we can also express the work U1y2 as U1y2 5# s2 s1 (F cos α) ds (10.119) Here, the variable of integration s measures the distance along the path traveled by the particle. We can represent the work U1y2 by the area under the curve obtained by plotting F cos α against s (Fig. 10.10b). In the case of a force F of constant magnitude acting in the direction of motion, formula (10.119) yields U1y2 5 F(s2 2 s1).
U1y2 5# A2 A # 1 F ? dr Fig. 10.10 (a) A force acting on a particle moving along a path from A1 to A2; (b) the work done by the force in (a) equals the area under the graph of F cos α versus s.
s (b) O s1 s2 F cos a (a) O ds A A1 s1 s2 A2 a F s 596 Method of Virtual Work Recall from Sec. 10.1 that the work of a couple of moment M during an infinitesimal rotation dθ of a rigid body is dU 5 M dθ (10.2) Therefore, we can express the work of the couple during a finite rotation of the body as Work during a finite rotation U1y2 5# θ2 θ1 M dθ (10.12) In the case of a constant couple, formula (10.12) yields U1y2 5 M(θ2 2 θ1) Work of a Weight. We stated in Sec. 10.1 that the work of a body’s weight W during an infinitesimal displacement of the body is equal to the product of W and the vertical displacement of the body’s center of gravity. With the y axis pointing upward, we obtain the work of W during a finite displacement of the body (Fig. 10.11) from dU 5 2W dy Integrating from A1 to A2, we have U1y2 5 2# y2 y1 W dy 5 Wy1 2 Wy2 (10.13) or U1y2 5 2W(y2 2 y1) 5 2W Dy (10.139) where Dy is the vertical displacement from A1 to A2. The work of the weight W is thus equal to the product of W and the vertical displace-ment of the center of gravity of the body. The work is positive when Dy , 0, that is, when the body moves down.
Work of the Force Exerted by a Spring. Consider a body A attached to a fixed point B by a spring. We assume that the spring is undeformed when the body is at A0 (Fig. 10.12a). Experimental evidence shows that the mag-nitude of the force F exerted by the spring on a body A is proportional to the deflection x of the spring measured from position A0. We have F 5 kx (10.14) where k is the spring constant expressed in SI units of N/m or U.S. customary units of lb/ft or lb/in. The work of force F exerted by the spring during a finite displacement of the body from A1 (x 5 x1) to A2 (x 5 x2) is obtained from dU 5 2F dx 5 2kx dx U1y2 5 2# x2 x1 kx dx 5 1 2kx2 1 2 1 2kx2 2 (10.15) You should take care to express k and x in consistent units. For example, if you use U.S. customary units, k should be expressed in lb/ft and x expressed in feet, or k is given in lb/in. and x in inches. In the first case, the work is obtained in ft?lb; in the second case, it is in in?lb. We note that the work of U1y2 5# θ2 θ θ # 1 Mdθ Fig. 10.11 The work done by the weight of a body equals the magnitude of the weight times the vertical displacement of its center of gravity.
A A1 A2 y1 y2 dy y W Fig. 10.12 (a) When a body is attached to a fixed point by a spring, the force on it is the product of the spring constant and the displacement from the undeformed position; (b) the work of the force equals the area under the graph of F versus x between x1 and x2.
Spring undeformed A0 A B B x1 x1 x2 x2 x F (a) (b) F = kx ∆x F F1 F2 A2 B A1 x 10.2 Work, Potential Energy, and Stability 597 the force F exerted by the spring on the body is positive when x2 , x1, that is, when the spring is returning to its undeformed position.
Since Eq. (10.14) is the equation of a straight line of slope k passing through the origin, we can obtain the work U1y2 of F during the displace-ment from A1 to A2 by evaluating the area of the trapezoid shown in Fig. 10.12b. This is done by computing the values F1 and F2 and multiply-ing the base Dx of the trapezoid by its mean height as 1 2(F1 1 F2). Since the work of the force F exerted by the spring is positive for a negative value of Dx, we have U1y2 5 21 2(F1 1 F2) Dx (10.16) Equation (10.16) is usually more convenient to use than Eq. (10.15) and affords fewer chances of confusing the units involved.
10.2B Potential Energy Let’s consider again the body of Fig. 10.11. Using Eq. (10.13), we obtain the work of weight W during a finite displacement by subtracting the value of the function Wy corresponding to the second position of the body from its value corresponding to the first position. Thus, the work of W is independent of the actual path followed; it depends only upon the initial and final values of the function Wy. This function is called the potential energy of the body with respect to the force due to gravity W and is denoted by Vg. Thus, U1y2 5 (Vg)1 2 (Vg)2 with Vg 5 Wy (10.17) Note that if (Vg)2 . (Vg)1, that is, if the potential energy increases during the displacement (as in the case considered here), the work U1y2 is negative. If, on the other hand, the work of W is positive, the potential energy decreases. Therefore, the potential energy Vg of the body provides a measure of the work that can be done by its weight W. Since only the change in potential energy—not the actual value of Vg—is involved in formula (10.17), we can add an arbitrary constant to the expression obtained for Vg. In other words, the level from which the elevation y is measured can be chosen arbitrarily. Note that potential energy is expressed in the same units as work, i.e., in joules (J) if SI units are used† and in ft?lb or in?lb if U.S. customary units are used.
Now consider the body of Fig. 10.12a. Using Eq. (10.15), we obtain the work of the elastic force F by subtracting the value of the function 1 2 kx2 corresponding to the second position of the body from its value corresponding to the first position. This function, denoted by Ve, is called the potential energy of the body with respect to the elastic force F. We have U1y2 5 (Ve)1 2 (Ve)2 with Ve 5 1 2 kx2 (10.18) Note that during the displacement considered, the work of force F exerted by the spring on the body is negative and the potential energy Ve increases. Also note that the expression obtained for Ve is valid only if the deflection of the spring is measured from its undeformed position.
We can use the concept of potential energy when forces other than gravity forces and elastic forces are involved. It remains valid as long as †See footnote, p. 575.
598 Method of Virtual Work the elementary work dU of the force considered is an exact differential.
It is then possible to find a function V, called potential energy, such that dU 5 2dV (10.19) Integrating Eq. (10.19) over a finite displacement, we obtain Potential energy, general formulation U1y2 5 V1 2 V2 (10.20) This equation says that the work of the force is independent of the path followed and is equal to minus the change in potential energy. A force that satisfies Eq. (10.20) is called a conservative force.† 10.2C Potential Energy and Equilibrium Applying the principle of virtual work is considerably simplified if we know the potential energy of a system. In the case of a virtual displace-ment, formula (10.19) becomes δU 5 2δV. Moreover, if the position of the system is defined by a single independent variable θ, we can write δV 5 (dV/dθ) δθ. Since δθ must be different from zero, the condition δU 5 0 for the equilibrium of the system becomes Equilibrium condition dV dθ 5 0 (10.21) In terms of potential energy, therefore, the principle of virtual work states: If a system is in equilibrium, the derivative of its total potential energy is zero. If the position of the system depends upon several independent variables (the system is then said to possess several degrees of freedom), the partial deriva-tives of V with respect to each of the independent variables must be zero.
Consider, for example, a structure made of two members AC and CB and carrying a load W at C. The structure is supported by a pin at A and a roller at B, and a spring BD connects B to a fixed point D (Fig. 10.13a). The constant of the spring is k, and we assume that the natural length of the spring is equal to AD, so that the spring is undeformed when B coincides with A. Neglecting friction forces and the weights of the members, we find that the only forces that do work during a virtual displacement of the structure are the weight W and the force F exerted by the spring at point B (Fig. 10.13b). Therefore, we can obtain the total potential energy of the system by adding the potential energy Vg corresponding to the gravity force W and the potential energy Ve corresponding to the elastic force F.
Choosing a coordinate system with the origin at A and noting that the deflection of the spring measured from its undeformed position is AB 5 xB, we have Ve 5 1 2 kx2 B and Vg 5 WyC dU 5 2dV U1y2 5 V1 V 2 V2 V dV dθ 5 0 †A detailed discussion of conservative forces is given in Sec. 13.2B of Dynamics.
Fig. 10.13 (a) Structure carrying a load at C with a spring from B to D; (b) free-body diagram of the structure, and a virtual displacement.
q q C A B C W A D B xB Ay B Ax l l yC W (a) (b) F = kxB 10.2 Work, Potential Energy, and Stability 599 Expressing the coordinates xB and yC in terms of the angle θ, we have xB 5 2l sin θ yC 5 l cos θ Ve 5 1 2k(2l sin θ)2 Vg 5 W(l cos θ) V 5 Ve 1 Vg 5 2kl2 sin2 θ 1 Wl cos θ (10.22) We obtain the positions of equilibrium of the system by setting the derivative of the potential energy V to zero, as dV dθ 5 4kl2 sin θ cos θ 2 Wl sin θ 5 0 or, factoring out l sin θ, as dV dθ 5 l sin θ(4kl cos θ 2 W) 5 0 There are therefore two positions of equilibrium corresponding to the values θ 5 0 and θ 5 cos21 (W/4kl), respectively.† 10.2D Stability of Equilibrium Consider the three uniform rods with a length of 2a and weight W shown in Fig. 10.14. Although each rod is in equilibrium, there is an important difference between the three cases considered. Suppose that each rod is slightly disturbed from its position of equilibrium and then released. Rod a moves back toward its original position; rod b keeps moving away from its original position; and rod c remains in its new position. In case a, the equilibrium of the rod is said to be stable; in case b, it is unstable; and in case c, it is neutral.
Recall from Sec. 10.2B that the potential energy Vg with respect to gravity is equal to Wy, where y is the elevation of the point of application of W measured from an arbitrary level. We observe that the potential energy of rod a is minimum in the position of equilibrium considered, that the potential energy of rod b is maximum, and that the potential energy of rod c is constant. Equilibrium is thus stable, unstable, or neutral according to whether the potential energy is minimum, maximum, or constant (Fig. 10.15).
Fig. 10.14 (a) Rod supported from above, stable equilibrium; (b) rod supported from below, unstable equilibrium; (c) rod supported at its midpoint, neutral equilibrium.
q (a) Stable equilibrium A B W 2a y q q (b) Unstable equilibrium A W 2a a y (c) Neutral equilibrium A B B C y = a †The second position does not exist if W . 4kl.
600 Method of Virtual Work This result is quite general, as we now show. We first observe that a force always tends to do positive work and thus to decrease the potential energy of the system on which it is applied. Therefore, when a system is disturbed from its position of equilibrium, the forces acting on the system tend to bring it back to its original position if V is minimum (Fig. 10.15a) and to move it farther away if V is maximum (Fig. 10.15b). If V is constant (Fig. 10.15c), the forces do not tend to move the system either way.
Recall from calculus that a function is minimum or maximum according to whether its second derivative is positive or negative. There-fore, we can summarize the conditions for the equilibrium of a system with one degree of freedom (i.e., a system for which the position is defined by a single independent variable θ) as dV dθ 5 0 d2V dθ 2 . 0: stable equilibrium dV dθ 5 0 d2V dθ 2 , 0: unstable equilibrium (10.23) dV dθ 5 0 d2 d V dθ2 . 0:stable equilibrium dV dθ 5 0 d2 d V dθ2 , 0:unstable equilibrium Fig. 10.15 Stable, unstable, and neutral equilibria correspond to potential energy values that are minimum, maximum, or constant, respectively.
(a) Stable equilibrium q (b) Unstable equilibrium V (c) Neutral equilibrium q V q V If both the first and the second derivatives of V are zero, it is necessary to examine derivatives of a higher order to determine whether the equilibrium is stable, unstable, or neutral. The equilibrium is neutral if all derivatives are zero, since the potential energy V is then a constant. The equilibrium is stable if the first derivative found to be different from zero is of even order and positive. In all other cases, the equilibrium is unstable.
If the system of interest possesses several degrees of freedom, the potential energy V depends upon several variables. Thus, it becomes necessary to apply the theory of functions of several variables to determine whether V is minimum. It can be verified that a system with two degrees of freedom is stable, and the corresponding potential energy V(θ1, θ2) is minimum, if the following relations are satisfied simultaneously: 0V 0θ1 5 0V 0θ2 5 0 a 02V 0θ1 0θ2 b 2 2 02V 0θ 2 1 02V 0θ 2 2 , 0 (10.24) 02V 0θ 2 1 . 0 or 02V 0θ 2 2 . 0 10.2 Work, Potential Energy, and Stability 601 Sample Problem 10.4 A 10-kg block is attached to the rim of a 300-mm-radius disk as shown. Knowing that spring BC is unstretched when θ 5 0, determine the position or positions of equilibrium, and state in each case whether the equilibrium is stable, unstable, or neutral.
STRATEGY: The first step is to determine a potential energy function V for the system. You can find the positions of equilibrium by determining where the derivative of V is zero. You can find the types of stability by finding where V is maximum or minimum.
MODELING and ANALYSIS: Potential Energy. Denote the deflection of the spring from its undeformed position by s, and place the origin of coordinates at O (Fig. 1). You obtain Ve 5 1 2 ks2 Vg 5 Wy 5 mgy Measuring θ in radians, you have s 5 aθ y 5 b cos θ Substituting for s and y in the expressions for Ve and Vg gives you Ve 5 1 2 ka2θ 2 Vg 5 mgb cos θ V 5 Ve 1 Vg 5 1 2 ka2θ 2 1 mgb cos θ Positions of Equilibrium. Setting dV/dθ 5 0, you obtain dV dθ 5 ka2θ 2 mgb sin θ 5 0 sin θ 5 ka2 mgb θ Now substitute a 5 0.08 m, b 5 0.3 m, k 5 4 kN/m, and m 5 10 kg. The result is sin θ 5 (4 kN/m)(0.08 m)2 (10 kg)(9.81 m/s2)(0.3 m) θ sin θ 5 0.8699 θ where θ is expressed in radians. Solving by trial and error for θ, you find θ 5 0 and θ 5 0.902 rad θ 5 0 and θ 5 51.7° b Stability of Equilibrium. The second derivative of the potential energy V with respect to θ is d2V dθ2 5 ka2 2 mgb cos θ 5 (4 kN/m)(0.08 m)2 2 (10 kg)(9.81 m/s2)(0.3 m) cos θ 5 25.6 2 29.43 cos θ For θ 5 0, d2V dθ2 5 25.6 2 29.43 cos 08 5 23.83 , 0 The equilibrium is unstable for θ 5 0 b For θ 5 51.7°, d2V dθ2 5 25.6 2 29.43 cos 51.78 5 17.36 . 0 The equilibrium is stable for θ 5 51.7° b REFLECT and THINK: If you just let the block-and-disk system fall on its own, it will come to rest at θ 5 51.7°. If you balance the system at θ 5 0, the slightest touch will put it in motion.
q 10 kg A B O C a = 80 mm b = 300 mm k = 4 kN/m s q A O y y x b a Undeformed position W = mg F = ks Fig. 1 Free-body diagram of rotated disk, showing only those forces that do work.
602 602 SOLVING PROBLEMS ON YOUR OWN I n this section, we defined the work of a force during a finite displacement and the potential energy of a rigid body or a system of rigid bodies. You saw how to use the concept of potential energy to determine the equilibrium position of a rigid body or a system of rigid bodies.
1. The potential energy V of a system is the sum of the potential energies associated with the various forces acting on the system that do work as the system moves. In the problems of this section, you will determine the following energies. a. Potential energy of a weight. This is the potential energy due to gravity, Vg 5 Wy, where y is the elevation of the weight W measured from some arbitrary reference level. You can use the potential energy Vg with any vertical force P of constant magnitude directed downward; we write Vg 5 Py. b. Potential energy of a spring. This is the potential energy due to the elastic force exerted by a spring, Ve 5 1 2k x2, where k is the constant of the spring and x is the deformation of the spring measured from its unstretched position.
Reactions at fixed supports, internal forces at connections, forces exerted by inexten-sible cords and cables, and other forces that do no work do not contribute to the potential energy of the system.
2. Express all distances and angles in terms of a single variable, such as an angle θ, when computing the potential energy V of a system. This is necessary because determining the equilibrium position of the system requires computing the derivative dV/dθ.
3. When a system is in equilibrium, the first derivative of its potential energy is zero. Therefore, a. To determine a position of equilibrium of a system, first express its potential energy V in terms of the single variable θ, and then compute its derivative and solve the equation dV/dθ 5 0 for θ. b. To determine the force or couple required to maintain a system in a given position of equilibrium, substitute the known value of θ in the equation dV/dθ 5 0, and solve this equation for the desired force or couple.
4. Stability of equilibrium. The following rules generally apply: a. Stable equilibrium occurs when the potential energy of the system is minimum, that is, when dV/dθ 5 0 and d2V/dθ2 . 0 (Figs. 10.14a and 10.15a). b. Unstable equilibrium occurs when the potential energy of the system is maximum, that is, when dV/dθ 5 0 and d2V/dθ2 , 0 (Figs. 10.14b and 10.15b). c. Neutral equilibrium occurs when the potential energy of the system is constant; dV/dθ, d2V/dθ2, and all the successive derivatives of V are then equal to zero (Figs. 10.14c and 10.15c).
See page 600 for a discussion of the case when dV/dθ, d2V/dθ2, but not all of the successive derivatives of V are equal to zero.
603 Problems 10.59 Using the method of Sec. 10.2C, solve Prob. 10.29. 10.60 Using the method of Sec. 10.2C, solve Prob. 10.30. 10.61 Using the method of Sec. 10.2C, solve Prob. 10.31. 10.62 Using the method of Sec. 10.2C, solve Prob. 10.32. 10.63 Using the method of Sec. 10.2C, solve Prob. 10.34. 10.64 Using the method of Sec. 10.2C, solve Prob. 10.35. 10.65 Using the method of Sec. 10.2C, solve Prob. 10.37. 10.66 Using the method of Sec. 10.2C, solve Prob. 10.38. 10.67 Show that equilibrium is neutral in Prob. 10.1. 10.68 Show that equilibrium is neutral in Prob. 10.7. 10.69 Two uniform rods, each with a mass m, are attached to gears of equal radii as shown. Determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral. 10.70 Two uniform rods, AB and CD, are attached to gears of equal radii as shown. Knowing that WAB 5 8 lb and WCD 5 4 lb, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral. 10.71 Two uniform rods AB and CD, of the same length l, are attached to gears as shown. Knowing that rod AB weighs 3 lb and that rod CD weighs 2 lb, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral.
A C D B 2a a q 2q Fig. P10.71 q q A B D C l l Fig. P10.69 and P10.70 604 10.72 Two uniform rods, each of mass m and length l, are attached to drums that are connected by a belt as shown. Assuming that no slip-ping occurs between the belt and the drums, determine the positions of equilibrium of the system and state in each case whether the equilibrium is stable, unstable, or neutral. 10.73 Using the method of Sec. 10.2C, solve Prob. 10.39. Determine whether the equilibrium is stable, unstable, or neutral. (Hint: The potential energy corresponding to the couple exerted by a torsion spring is 1 2 Kθ 2, where K is the torsional spring constant and θ is the angle of twist.) 10.74 In Prob. 10.40, determine whether each of the positions of equilibrium is stable, unstable, or neutral. (See hint for Prob. 10.73.) 10.75 A load W with a magnitude of 100 lb is applied to the mechanism at C. Knowing that the spring is unstretched when θ 5 15°, determine that value of θ corresponding to equilibrium and check that the equilibrium is stable.
W q A B C l = 20 in.
r = 5 in.
k = 50 lb/in.
Fig. P10.75 and P10.76 10.76 A load W with a magnitude of 100 lb is applied to the mechanism at C. Knowing that the spring is unstretched when θ 5 30°, determine that value of θ corresponding to equilibrium and check that the equilibrium is stable. 10.77 A slender rod AB with a weight W is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when y 5 0, determine the value of y corresponding to equilibrium when W 5 80 N, l 5 500 mm, and k 5 600 N/m.
y l C B W l A Fig. P10.77 D A B C a 2a 2q q Fig. P10.72 605 10.78 A slender rod AB with a weight W is attached to two blocks A and B that can move freely in the guides shown. Knowing that both springs are unstretched when y 5 0, determine the value of y corresponding to equilibrium when W 5 80 N, l 5 500 mm, and k 5 600 N/m.
C B A l l W y Fig. P10.78 10.79 A slender rod AB with a weight W is attached to two blocks A and B that can move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when AB is horizontal. Neglecting the weight of the blocks, derive an equation in θ, W, l, and k that must be satisfied when the rod is in equilibrium.
C A B q l W Fig. P10.79 and P10.80 10.80 A slender rod AB with a weight W is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when AB is horizontal, determine three values of θ corresponding to equilibrium when W 5 300 lb, l 5 16 in., and k 5 75 lb/in. State in each case whether the equilibrium is stable, unstable, or neutral. 10.81 A spring AB of constant k is attached to two identical gears as shown. Knowing that the spring is undeformed when θ 5 0, deter-mine two values of the angle θ corresponding to equilibrium when P 5 30 lb, a 5 4 in., b 5 3 in., r 5 6 in., and k 5 5 lb/in. State in each case whether the equilibrium is stable, unstable, or neutral. 10.82 A spring AB of constant k is attached to two identical gears as shown. Knowing that the spring is undeformed when θ 5 0, and given that a 5 60 mm, b 5 45 mm, r 5 90 mm, and k 5 6 kN/m, determine (a) the range of values of P for which a position of equi-librium exists, (b) two values of θ corresponding to equilibrium if the value of P is equal to half the upper limit of the range found in part a.
A q q B a a b r r b P Fig. P10.81 and P10.82 606 10.83 A slender rod AB is attached to two collars A and B that can move freely along the guide rods shown. Knowing that β 5 30° and P 5 Q 5 400 N, determine the value of the angle θ corresponding to equilibrium.
10.84 A slender rod AB is attached to two collars A and B that can move freely along the guide rods shown. Knowing that β 5 30°, P 5 100 N, and Q 5 25 N, determine the value of the angle θ corresponding to equilibrium.
10.85 and 10.86 Cart B, which weighs 75 kN, rolls along a sloping track that forms an angle β with the horizontal. The spring constant is 5 kN/m, and the spring is unstretched when x 5 0. Determine the distance x corresponding to equilibrium for the angle β indicated. 10.85 Angle β 5 30° 10.86 Angle β 5 60° 4 m x A B b Fig. P10.85 and P10.86 10.87 and 10.88 Collar A can slide freely on the semicircular rod shown. Knowing that the constant of the spring is k and that the unstretched length of the spring is equal to the radius r, determine the value of θ corresponding to equilibrium when W 5 50 lb, r 5 9 in., and k 5 15 lb/in.
A B C q r W Fig. P10.87 B A C q r W Fig. P10.88 P A B L q b Q Fig. P10.83 and P10.84 607 10.89 Two bars AB and BC of negligible weight are attached to a single spring of constant k that is unstretched when the bars are horizontal. Determine the range of values of the magnitude P of two equal and opposite forces P and 2P for which the equilibrium of the system is stable in the position shown.
l l −P P A B C Fig. P10.89 10.90 A vertical bar AD is attached to two springs of constant k and is in equilibrium in the position shown. Determine the range of values of the magnitude P of two equal and opposite vertical forces P and 2P for which the equilibrium position is stable if (a) AB 5 CD, (b) AB 5 2CD. 10.91 Rod AB is attached to a hinge at A and to two springs, each of constant k. If h 5 25 in., d 5 12 in., and W 5 80 lb, determine the range of values of k for which the equilibrium of the rod is stable in the position shown. Each spring can act in either tension or compression. 10.92 Rod AB is attached to a hinge at A and to two springs, each of constant k. If h 5 45 in., k 5 6 lb/in., and W 5 60 lb, determine the smallest distance d for which the equilibrium of the rod is stable in the position shown. Each spring can act in either tension or compression. 10.93 and 10.94 Two bars are attached to a single spring of constant k that is unstretched when the bars are vertical. Determine the range of values of P for which the equilibrium of the system is stable in the position shown.
P A B D P A B C D L 3 L 3 L 3 Fig. P10.93 Fig. P10.94 A D C B l a P –P Fig. P10.90 A B W d h Fig. P10.91 and P10.92 608 10.95 The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of Q for which the equilibrium of the system is stable in the position shown when a 5 24 in., b 5 20 in., and P 5 150 lb. 10.96 The horizontal bar BEH is connected to three vertical bars. The collar at E can slide freely on bar DF. Determine the range of values of P for which the equilibrium of the system is stable in the position shown when a 5 150 mm, b 5 200 mm, and Q 5 45 N. 10.97 Bars AB and BC, each with a length l and of negligible weight, are attached to two springs, each of constant k. The springs are undeformed, and the system is in equilibrium when θ1 5 θ2 5 0. Determine the range of values of P for which the equilibrium position is stable.
P A B C q1 q2 Fig. P10.97 10.98 Solve Prob. 10.97 knowing that l 5 800 mm and k 5 2.5 kN/m. 10.99 Two rods of negligible weight are attached to drums of radius r that are connected by a belt and spring of constant k. Knowing that the spring is undeformed when the rods are vertical, determine the range of values of P for which the equilibrium position θ1 5 θ2 5 0 is stable. 10.100 Solve Prob. 10.99 knowing that k 5 20 lb/in., r 5 3 in., l 5 6 in., and (a) W 5 15 lb, (b) W 5 60 lb.
P Q Q D A B C F I H G E a b Fig. P10.95 and P10.96 W 2 q 1 q A B D C r r l l P Fig. P10.99 609 Review and Summary Work of a Force The first section of this chapter was devoted to the principle of virtual work and to its direct application to the solution of equilibrium problems. We first defined the work of a force F corresponding to the small displacement dr [Sec. 10.1A] as the quantity dU 5 F?dr (10.1) obtained by forming the scalar product of the force F and the displacement dr (Fig. 10.16). Denoting the magnitudes of the force and of the displace-ment by F and ds, respectively, and the angle formed by F and dr by α, we have dU 5 F ds cos α (10.19) The work dU is positive if α , 90°, zero if α 5 90°, and negative if α . 90°. We also found that the work of a couple of moment M acting on a rigid body is dU 5 M dθ (10.2) where dθ is the small angle expressed in radians through which the body rotates.
Virtual Displacement Considering a particle located at A and acted upon by several forces F1, F2, . . . , Fn [Sec. 10.1B], we imagined that the particle moved to a new position A9 (Fig. 10.17). Since this displacement does not actually take place, we refer to it to as a virtual displacement denoted by δr. The corresponding work of the forces is called virtual work and is denoted by δU. We have δU 5 F1?δr 1 F2?δr 1 . . . 1 Fn?δr Principle of Virtual Work The principle of virtual work states that if a particle is in equilibrium, the total virtual work δU of the forces acting on the particle is zero for any virtual displacement of the particle. The principle of virtual work can be extended to the case of rigid bodies and systems of rigid bodies. Since it involves only forces that do work, its application provides a useful alternative to the use of the equilibrium equations in the solution of many engineering problems. It is particularly effective in the case of machines and mechanisms consisting of connected rigid bodies, since the work of the reactions at the supports is zero and the work of the internal forces at the pin connections cancels out [Sec. 10.1C; Sample Probs. 10.1, 10.2, and 10.3].
a dr A A' F Fig. 10.16 F2 F1 Fn A A' dr Fig. 10.17 610 Mechanical Efficiency In the case of real machines, however [Sec. 10.1D], the work of the friction forces should be taken into account with the result that the output work is less than the input work. We defined the mechanical efficiency of a machine as the ratio h 5 output work input work (10.9) We noted that, for an ideal machine (no friction), η 5 1, whereas for a real machine, η , 1.
Work of a Force over a Finite Displacement In the second section of this chapter, we considered the work of forces corresponding to finite displacements of their points of application. We obtained the work U1y2 of the force F corresponding to a displacement of the particle A from A1 to A2 (Fig. 10.18) by integrating the right-hand side of Eqs. (10.1) or (10.19) along the curve described by the particle [Sec. 10.2A]. Thus, U1y2 5# A2 A1 F?dr (10.11) or U1y2 5# s2 s1 (F cos α) ds (10.119) Similarly, we expressed the work of a couple of moment M corresponding to a finite rotation from θ1 to θ2 of a rigid body as U1y2 5# θ2 θ1 M dθ (10.12) Work of a Weight We obtained the work of the weight W of a body as its center of gravity moves from the elevation y1 to y2 (Fig. 10.19) by setting F 5 W and α 5 180° in Eq. (10.119) as U1y2 5 2# y2 y1 W dy 5 Wy1 2 Wy2 (10.13) The work of W is therefore positive when the elevation y decreases.
A A1 A2 y1 y2 dy y W Fig. 10.19 O ds A A1 s1 s2 A2 a F s Fig. 10.18 611 Work of the Force Exerted by a Spring The work of the force F exerted by a spring on a body A as the spring is stretched from x1 to x2 (Fig. 10.20) can be obtained by setting F 5 kx, where k is the constant of the spring, and α 5 180° in Eq. (10.119). Hence, U1y2 5 2# x2 x1 k x dx 5 1 2 k x2 1 2 1 2 k x2 2 (10.15) The work of F is therefore positive when the spring is returning to its unde-formed position.
Potential Energy When the work of a force F is independent of the path actually followed between A1 and A2, the force is said to be a conservative force, and we can express its work as U1y2 5 V1 2 V2 (10.20) Here V is the potential energy associated with F, and V1 and V2 represent the values of V at A1 and A2, respectively [Sec. 10.2B]. We found the potential energies associated, respectively, with the force of gravity W and the elastic force F exerted by a spring to be Vg 5 Wy and Ve 5 1 2 kx2 (10.17, 10.18) Alternative Expression for the Principle of Virtual Work When the position of a mechanical system depends upon a single independent variable θ, the potential energy of the system is a function V(θ) of that variable, and it follows from Eq. (10.20) that δU 5 2δV 5 2(dV/dδ ) δθ. The condition δU 5 0 required by the principle of virtual work for the equilibrium of the system thus can be replaced by the condition dV dθ 5 0 (10.21) When all the forces involved are conservative, it may be preferable to use Eq. (10.21) rather than apply the principle of virtual work directly [Sec. 10.2C; Sample Prob. 10.4].
Stability of Equilibrium This alternative approach presents another advantage, since it is possible to determine from the sign of the second derivative of V whether the equilibrium of the system is stable, unstable, or neutral [Sec. 10.2D]. If d2V/dθ 2 . 0, V is minimum and the equilibrium is stable; if d2V/dθ 2 , 0, V is maximum and the equilibrium is unstable; if d2V/dθ 2 5 0, it is necessary to examine derivatives of a higher order.
Spring undeformed A0 A B B x1 x2 x F A2 B A1 Fig. 10.20 612 Review Problems 10.101 Determine the vertical force P that must be applied at G to maintain the equilibrium of the linkage.
300 lb 100 lb 6 in.
A B C D E F G 10 in.
12 in.
8 in.
Fig. P10.101 and P10.102 10.102 Determine the couple M that must be applied to member DEFG to maintain the equilibrium of the linkage. 10.103 Determine the force P required to maintain the equilibrium of the linkage shown. All members are of the same length, and the wheels at A and B roll freely on the horizontal rod. 10.104 Derive an expression for the magnitude of the force Q required to maintain the equilibrium of the mechanism shown. 10.105 Derive an expression for the magnitude of the couple M required to maintain the equilibrium of the linkage shown.
q D C A B E a a a a F M P P a Fig. P10.105 10.106 A vertical load W is applied to the linkage at B. The constant of the spring is k, and the spring is unstretched when AB and BC are horizontal. Neglecting the weight of the linkage, derive an equa-tion in θ, W, l, and k that must be satisfied when the linkage is in equilibrium.
400 N 100 N C F D E G H B A P 150 N 75 N Fig. P10.103 90° A B C P P q q Q D 90° l l l Fig. P10.104 A B C D l l W q Fig. P10.106 613 10.107 A force P with a magnitude of 240 N is applied to end E of cable CDE, which passes under pulley D and is attached to the mechanism at C. Neglecting the weight of the mechanism and the radius of the pulley, determine the value of θ corresponding to equilibrium. The constant of the spring is k 5 4 kN/m, and the spring is unstretched when θ 5 90°.
A B C 120 mm 300 mm 300 mm q D E P Fig. P10.107 10.108 Two identical rods ABC and DBE are connected by a pin at B and by a spring CE. Knowing that the spring is 4 in. long when unstretched and that the constant of the spring is 8 lb/in., determine the distance x corresponding to equilibrium when a 24-lb load is applied at E as shown.
10.109 Solve Prob. 10.108 assuming that the 24-lb load is applied at C instead of E. 10.110 Two uniform rods each with a mass m and length l are attached to gears as shown. For the range 0 # θ # 180°, determine the positions of equilibrium of the system, and state in each case whether the equilibrium is stable, unstable, or neutral.
10.111 A homogeneous hemisphere with a radius r is placed on an incline as shown. Assuming that friction is sufficient to prevent slipping between the hemisphere and the incline, determine the angle θ corresponding to equilibrium when β 5 10°.
q b G C Fig. P10.111 and P10.112 10.112 A homogeneous hemisphere with a radius r is placed on an incline as shown. Assuming that friction is sufficient to prevent slipping between the hemisphere and the incline, determine (a) the largest angle β for which a position of equilibrium exists, (b) the angle θ corresponding to equilibrium when the angle β is equal to half the value found in part a.
9 in.
6 in.
A x D B E C 24 lb Fig. P10.108 q 1.5q A B 3a D 2a C Fig. P10.110 This page intentionally left blank A1 Fundamentals of Engineering Examination Engineers are required to be licensed when their work directly affects the public health, safety, and welfare. The intent is to ensure that engineers have met minimum qualifications involving competence, ability, experi-ence, and character. The licensing process involves an initial exam, called the Fundamentals of Engineering Examination; professional experience; and a second exam, called the Principles and Practice of Engineering.
Those who successfully complete these requirements are licensed as a Professional Engineer. The exams are developed under the auspices of the National Council of Examiners for Engineering and Surveying. The first exam, the Fundamentals of Engineering Examination, can be taken just before or after graduation from a four-year accredited engineer-ing program. The exam stresses subject material in a typical undergraduate engineering program, including statics. The topics included in the exam cover much of the material in this book. The following is a list of the main topic areas, with references to the appropriate sections in this book. Also included are problems that can be solved to review this material.
Concurrent Force Systems (2.1–2.2; 2.4) Problems: 2.31, 2.35, 2.36, 2.37, 2.77, 2.83, 2.92, 2.94, 2.97 Vector Forces (3.1–3.2) Problems: 3.17, 3.18, 3.26, 3.33, 3.37, 3.39 Equilibrium in Two Dimensions (2.3; 4.1–4.2) Problems: 4.1, 4.13, 4.14, 4.17, 4.31, 4.33, 4.67, 4.77 Equilibrium in Three Dimensions (2.5; 4.3) Problems: 4.99, 4.101, 4.103, 4.108, 4.115, 4.117, 4.127, 4.129, 4.135 Centroids of Areas and Volumes (5.1–5.2; 5.4) Problems: 5.9, 5.16, 5.30, 5.35, 5.41, 5.55, 5.62, 5.96, 5.102, 5.103, 5.125 Analysis of Trusses (6.1–6.2) Problems: 6.3, 6.4, 6.32, 6.43, 6.44, 6.53 Equilibrium of Two-Dimensional Frames (6.3) Problems: 6.75, 6.81, 6.85, 6.93, 6.94 Shear and Bending Moment (7.1–7.3) Problems: 7.22, 7.30, 7.36, 7.41, 7.45, 7.49, 7.71, 7.79 Friction (8.1–8.2; 8.4) Problems: 8.11, 8.18, 8.19, 8.30, 8.49, 8.52, 8.103, 8.104, 8.105 Moments of Inertia (9.1–9.4) Problems: 9.6, 9.31, 9.32, 9.33, 9.72, 9.74, 9.80, 9.83, 9.98, 9.103 APPENDIX This page intentionally left blank AN1 Answers to Problems CHAPTER 2 2.1 1391 N a 47.8°. 2.2 906 lb a 26.6°. 2.4 8.03 kips d 3.88. 2.5 (a) 101.4 N. (b) 196.6 N. 2.6 (a) 853 lb. (b) 567 lb. 2.8 (a) TAC 5 2.60 kN. (b) R 5 4.26 kN. 2.9 (a) TAC 5 2.66 kN c 34.3°. 2.10 (a) 37.18. (b) 73.2 N. 2.11 (a) 392 lb. (b) 346 lb. 2.13 (a) 368 lb y. (b) 213 lb. 2.14 (a) 21.1 Nw. (b) 45.3 N. 2.15 414 lb c 72.0°. 2.16 1391 N a 47.8°. 2.17 8.03 kips d 3.8°. 2.19 104.4 N b 86.7°. 2.21 (29 lb) 21.0 lb, 20.0 lb; (50 lb)214.00 lb, 48.0 lb; (51 lb) 24.0 lb, 245.0 lb. 2.23 (80 N) 61.3 N, 51.4 N; (120 N) 41.0 N, 112.8 N; (150 N)2122.9 N, 86.0 N. 2.24 (40 lb) 20.0 lb, 234.6 lb; (50 lb) 238.3 lb, 232.1 lb; (60 lb) 54.4 lb, 25.4 lb. 2.26 (a) 523 lb. (b) 428 lb. 2.27 (a) 621 N. (b) 160.8 N. 2.28 (a) 610 lb. (b) 500 lb. 2.29 (a) 2190 N. (b) 2060 N. 2.31 38.6 lb a 36.6°. 2.32 251 N b 85.3°. 2.34 654 N c 21.5°. 2.35 309 N d 86.6°. 2.36 474 N c 32.5°. 2.37 203 lb a 8.46°. 2.39 (a) 21.78. (b) 229 N. 2.40 (a) 26.5 N. (b) 623 N. 2.42 (a) 56.38. (b) 204 lb. 2.43 (a) 352 lb. (b) 261 lb. 2.44 (a) 5.22 kN. (b) 3.45 kN 2.46 (a) 305 N. (b) 514 N. 2.48 (a) 1244 lb. (b) 115.4 lb. 2.49 TCA 5 134.6 N; TCB 5 110.4 N. 2.50 179.3 N < P < 669 N. 2.51 FA 5 1303 lb; FB 5 420 lb. 2.53 FC 5 6.40 kN; FD 5 4.80 kN. 2.54 FB 5 15.00 kN; FC 5 8.00 kN. 2.55 (a) TACB 5 269 lb. (b) TCD 5 37.0 lb. 2.57 (a) α 5 35.08; TAC 5 4.91 kN; TBC 5 3.44 kN, (b) α 5 55.08; TAC 5 TBC 5 3.66 kN. 2.58 (a) 784 N. (b) α 5 71.08. 2.59 (a) α 5 5.008. (b) 104.6 lb. 2.61 1.250 m. 2.62 75.6 mm. 2.63 (a) 10.98 lb. (b) 30.0 lb. 2.65 27.48 # α # 222.68. 2.67 (a) 300 lb. (b) 300 lb. (c) 200 lb. (d) 200 lb. (e) 150.0 lb. 2.68 (a) 200 lb. (b) 150.0 lb. 2.69 (a) 1293 N. (b) 2220 N. 2.71 (a) 220 N, 544 N, 126.8 N. (b) 68.58, 25.08, 77.88. 2.72 (a) 2237 N, 258 N, 282 N. (b) 121.88, 55.08, 51.1°. 2.73 (a) 2175.8 N, 2257 N, 251 N. (b) 116.18, 130.08, 51.18. 2.74 (a) 350 N, 2169.0 N, 93.8 N. (b) 28.98, 115.08, 76.48. 2.75 (a) 220.5 lb, 43.3 lb, 214.33 lb. (b) 114.28, 30.08, 106.78. 2.77 (a) 21861 lb, 3360 lb, 677 lb. (b) 118.58, 30.58, 80.08. 2.79 (a) 770 N; 71.88; 110.58; 28.08. 2.81 (a) 140.38. (b) Fx 5 79.9 lb, Fz 5 120.1 lb; F 5 226 lb. 2.82 (a) 118.28. (b) Fx 5 36.0 lb, Fy 5 290.0 lb; F 5 110.0 lb. 2.84 (a) Fx 5 507 N, Fy 5 919 N, Fz 5 582 N. (b) 61.08. 2.85 240 N; 2255 N; 160.0 N. 2.87 21.260 kips; 1.213 kips; 0.970 kips. 2.88 20.820 kips; 0.978 kips; 20.789 kips. 2.89 192.0 N; 288 N; 2216 N. 2.91 515 N; θx 5 70.28; θy 5 27.68; θz 5 71.58 2.92 515 N; θx 5 79.88; θy 5 33.48; θz 5 58.68. 2.94 913 lb; θx 5 50.68; θy 5 117.68; θz 551.88. 2.95 748 N; θx 5120.18; θy 5 52.58; θz 5128.08. 2.96 TAB 5 490 N; TAD 5 515 N. 2.97 130.0 lb. 2.99 13.98 kN. 2.101 926 N ↑. 2.103 TDA 5 14.42 lb; TDB 5 TDC 5 13.00 lb. 2.104 TDA 5 14.42 lb; TDB 5 TDC 5 13.27 lb. 2.106 TAB 5 571 lb; TAC 5 830 lb; TAD 5 528 lb. 2.107 960 N. 2.108 0 # Q , 300 N. 2.109 845 N. 2.110 768 N. 2.112 2000 lb. 2.113 TAB 5 30.8 lb; TAC 5 62.5 lb. 2.115 TAB 5 510 N; TAC 5 56.2 N; TAD 5 536 N. 2.116 TAB 5 1340 N; TAC 5 1025 N; TAD 5 915 N. 2.117 TAB 5 1431 N; TAC 5 1560 N; TAD 5 183.0 N. 2.118 TAB 5 1249 N; TAC 5 490 N; TAD 5 1647 N. 2.119 TAB 5 974 lb; TAC 5 531 lb; TAD 5 533 lb. 2.121 378 N. 2.123 TBAC 5 76.7 lb; TAD 5 26.9 lb; TAE 5 49.2 lb. 2.124 (a) 305 lb. (b) TBAC 5 117.0 lb; TAD 5 40.9 lb. 2.125 (a) 1155 N. (b) 1012 N. 2.127 21.8 kN c 73.4°. 2.128 (102 lb) 248.0 lb, 90.0 lb; (106 lb) 56.0 lb, 90.0 lb; (200 lb) 2160.0 lb, 2120.0 lb 2.130 (a) 172.7 lb. (b) 231 lb. 2.131 (a) 312 N. (b) 144.0 N. 2.133 (a) 56.4 lb; 2103.9 lb; 220.5 lb. (b) 62.08, 150.08, 99.88. 2.135 940 N; 65.78, 28.28, 76.48. 2.136 P 5 131.2 N; Q 5 29.6 N. 2.137 (a) 125.0 lb. (b) 45.0 lb.
AN2 Answers to Problems 3.84 (a) F 5 30.0 lbw; M 5 150.0 lb.in. l. (b) B 5 50.0 lb ←; C 5 50.0 lb y. 3.86 FA 5 168.0 N d 50.0°; FC 5 192.0 N d50.0°. 3.87 F 5 900 Nw; x 5 50.0 mm. 3.89 (a) F 5 48.0 lb a 65.0°; M 5 490 lb.in. i. (b) F 5 48.0 lb a 65.0° applied 17.78 in. to the left of B. 3.90 (a) 48.0 N intersecting line AB 144.0 mm to the right of A. (b) 77.78 or 215.728. 3.91 (0.227 lb)i 1 (0.1057 lb)k; 63.6 in. to the right of B. 3.93 F 5 2(250 kN)j; M 5 (15.00 kN?m)i 1 (7.50 kN?m)k. 3.95 F 5 2(122.9 N)j 2 (86.0 N)k; M 5 (22.6 N?m)i 1 (15.49 N?m)j 2 (22.1 N?m)k. 3.96 F 5 (5.00 N)i 1 (150.0 N)j 2 (90.0 N)k; M 5 (77.4 N?m)i 1 (61.5 N?m)j 1 (106.8 N?m)k. 3.97 F 5 (36.0 lb)i 2 (28.0 lb)j 2 (6.00 lb)k; M 5 2(157.0 lb?ft)i 1 (22.5 lb?ft)j 2 (240 lb?ft)k. 3.98 F 5 2(28.5 N)i 1 (106.3 N)k; M 5 (12.35 N?m)i 2 (19.16 N?m)j 2 (5.13 N?m)k. 3.99 F 5 2(128.0 lb)i 2 (256 lb)j 1 (32.0 lb)k; M 5 (4.10 kip?ft)i 1 (16.38 kip?ft)k. 3.101 (a) Loading a: 500 Nw; 1000 N?m i. Loading b: 500 Nx; 500 N?m l. Loading c: 500 Nw; 500 N?m i. Loading d: 500 Nw; 1100 N?m i. Loading e: 500 Nw; 1000 N?m i. Loading f: 500 Nw; 200 N?m i. Loading g: 500 Nw; 2300 N?m l. Loading h: 500 Nw; 600 N?m l. (b) Loadings a and e are equivalent. 3.102 Equivalent to case f of problem 3.101. 3.104 Equivalent force-couple system at D. 3.105 (a) 2.00 ft to the right of C. (b) 2.31 ft to the right of C. 3.106 (a) 39.6 in. to the right of D. (b) 33.1 in. 3.108 44.7 lb b 26.6°; 10.61 in. to the left of C and 5.30 in. below C. 3.110 (a) 224 N c 63.4°. (b) 130.0 mm to the left of B and 260 mm below B. 3.111 (a) 269 N c 68.2°. (b) 120.0 mm to the left of B and 300 mm below B. 3.113 773 lb d 79.0°; 9.54 ft to the right of A. 3.114 (a) 29.9 lb b 23.0°. (b) AB: 10.30 in. to the left of B; BC: 4.36 in. below B. 3.115 (a) 60.2 lb?in. l. (b) 200 lb?in.l. (C) 20.0 lb?in. i. 3.116 (a) 0.365 m above G. (b) 0.227 m to the right of G. 3.117 (a) 0.299 m above G. (b) 0.259 m to the right of G. 3.118 (a) R 5 F d tan 211a2/2bx2; M 5 2Fb21x 2 x3/a22/ 2a4 1 4b2x2 l. (b) 0.369 m. 3.119 R 5 2(300 N)i 2 (240 N)j 1 (25.0 N)k; M 5 2(3.00 N?m)i 1 (13.50 N?m)j 1 (9.00 N?m)k. 3.120 R 5 (420 N)j 2 (339 N)k; M 5 (1.125 N?m)i 1 (163.9 N?m)j 2 (109.9 N?m)k. 3.122 (a) 60.08. (b) (20.0 lb)i 2 (34.6 lb)j; (520 lb?in.)i. 3.124 R 5 2(420 N)i 2 (50.0 N)j 2 (250 N)k; M 5 (30.8 N?m)j 2 (22.0 N?m)k. 3.125 (a) B 5 2(75.0 N)k, C 5 2(25.0 N)i 1 (37.5 N)k. (b) Ry 5 0, Rz 5 237.5 N. (c) when the slot is vertical. 3.126 A5(1.600 lb)i 2(36.0 lb)j 1 (2.00 lb)k, B 5 2(9.60 lb)i 1 (36.0 lb)j 1 (2.00 lb)k. 3.127 1035 N; 2.57 m from OG and 3.05 m from OE. 3.128 2.32 m from OG and 1.165 m from OE. 3.129 405 lb; 12.60 ft to the right of AB and 2.94 ft below BC. 3.130 a 5 0.722 ft; b 5 20.6 ft. 3.133 (a) P23; θx 5 θy 5 θz 5 54.78. (b) 2a (c) Axis of the wrench is diagonal OA.
CHAPTER 3 3.1 (a) 196.2 N?m i. (b) 199.0 N b 59.5°. 3.2 (a) 196.2 N?m i. (b) 321 N d 35.0°. (c) 231 Nx at point D. 3.4 (a) 41.7 N?m l. (b) 147.4 N a 45.0°. 3.5 (a) 41.7 N?m l. (b) 334 N. (c) 176.8 N a 58.0°. 3.6 115.7 lb?in. 3.7 115.7 lb?in. 3.9 (a) 292 N?m i. (b) 292 N?m i. 3.11 116.2 lb?ft l. 3.12 128.2 lb?ft l. 3.13 140.0 N?m l. 3.17 (a) l 5 20.677i 2 0.369j 2 0.636k. (b) l 5 20.0514i 1 0.566j 1 0.823k. 3.18 1.184 m. 3.20 (a) 9i 1 22j 1 21k. (b) 22i 1 11k. (c) 0. 3.22 (2400 lb?ft)j 1 (1440 lb?ft)k. 3.23 (7.50 N?m)i 2 (6.00 N?m)j 2(10.39 N?m)k. 3.25 (225.4 lb?ft)i 2 (12.60 lb?ft)j 2(12.60 lb?ft)k. 3.26 (1200 N?m)i 2 (1500 N?m)j 2 (900 N?m)k. 3.27 7.37 ft. 3.28 100.8 mm. 3.29 144.8 mm. 3.30 5.17 ft. 3.32 2.36 m. 3.33 1.491 m. 3.35 P?Q 5 25; P?S 5 15; Q?S 5 238. 3.37 77.98. 3.39 (a) 59.08. (b) 144.0 lb. 3.40 (a) 70.58. (b) 60.0 lb 3.41 (a) 52.98. (b) 326 N. 3.43 26.88. 3.44 33.38. 3.45 (a) 67.0. (b) 111.0. 3.46 2. 3.47 Mx 5 78.9 kN?m, My 5 13.15 kN?m, Mz 5 29.86 kN?m. 3.48 3.04 kN. 3.49 f 5 24.68; d 5 34.6 in. 3.51 1.252 m. 3.52 1.256 m. 3.53 283 lb. 3.55 1207 lb?ft. 3.57 290.0 N?m. 3.58 2111.0 N?m. 3.59 12.28 N?m. 3.60 29.50 N?m. 3.61 a P/22. 3.64 13.06 in. 3.65 12.69 in. 3.67 0.249 m. 3.68 0.1198 m. 3.70 (a) 7.33 N?m l. (b) 91.6 mm. 3.71 6.19 N?m i. 3.73 1.125 in. 3.74 (a) 26.7 N. (b) 50.0 N. (c) 23.5 N. 3.76 M 5 604 lb?in.; θx 5 72.88, θy 5 27.38, θz 5 110.58. 3.77 M 5 1170 lb?in.; θx 5 81.28, θy 5 13.708, θz 5 100.48. 3.78 M 5 3.22 N?m; θx 5 90.08, θy 5 53.18, θz 5 36.98. 3.79 M 5 2.72 N?m; θx 5 134.9°, θy 5 58.0°; θz 5 61.9°. 3.80 M 5 2150 lb?ft; θx 5 113.08, θy 5 92.78, θz 5 23.28. 3.82 (a) FA 5 560 lb c 20.0°; MA 5 7720 lb?ft i. (b) FB5 560 lb c 20.0°; MB 5 4290 lb?ft i. 3.83 FA 5 389 N c 60.0°; FC 5 651 N c 60.0°.
Answers to Problems AN3 4.48 (a) D 5 20.0 lbw; MD 5 20.0 lb?ft l. (b) D 5 10.00 lbw; MD 5 30.0 lb?ft i. 4.50 (a) A 5 78.5 Nx; MA 5 125.6 N?m l. (b) A 5 111.0 Nx; MA 5 125.6 N?m l. (c) A 5 157.0 Nx; MA 5 251 N?m l. 4.51 θ 5 sin 2112M cot α/W l2. 4.52 θ 5 tan 21(Q/3P). 4.53 (a) T 5 (W/2)/(1 2 tan θ). (b) 39.88. 4.54 (a) θ 5 2 cos 21C1 41W P 6 2W 2 P 2 1 82D. (b) 65.18. 4.55 (a) θ 5 2 sin 211W/2P2. (b) 29.08. 4.57 141.18. 4.59 (1) completely constrained; determinate; A 5 C 5 196.2 Nx . (2) completely constrained; determinate; B 5 0, C 5 D 5 196.2 Nx . (3) completely constrained; indeterminate; Ax5 294 N y; Dx5 294 ←. (4) improperly constrained; indeterminate; no equilibrium. (5) partially constrained; determinate; equilibrium; C 5 D 5 196.2 Nx . (6) completely constrained; determinate; B 5 294 N y, D 5 491 N b 53.1°. (7) partially constrained; no equilibrium. (8) completely constrained; indeterminate; B 5 196.2 Nx , Dy5 196.2 Nx . 4.61 T 5 289 lb; A 5 577 lb a 60.08. 4.62 A 5 400 Nx; B 5 500 N c 53.1°. 4.63 a $ 138.6 mm. 4.65 B 5 501 N b 56.3°; C 5 324 N c 31.0°. 4.66 A 5 82.5 lb a 14.048; T 5 100.0 lb. 4.67 B 5 888 N c 41.3°; D 5 943 N b 45.0°. 4.69 (a) 499 N. (b) 457 N b 26.6°. 4.71 (a) 5.63 kips. (b) 4.52 kips d4.76°. 4.72 (a) 24.9 lb d30.0°. (b) 15.34 lb a 30.08. 4.73 A 5 778 Nw; C 5 1012 N b 77.9°. 4.75 A 5 170.0 N b 33.9°; C 5 160.0 N a 28.18. 4.77 T 5 100.0 lb; B 5 111.1 lb c 30.3°. 4.78 (a) 400 N. (b) 458 N a 49.18. 4.79 (a) 2P b 60.0°. (b) 1.239P c 36.2°. 4.80 (a) 1.55 P b 30.0°. (b) 1.086 P a 22.98. 4.81 A 5 163.1 N c 74.1°; B 5 258 N b 65.0°. 4.83 60.0 mm. 4.84 tan θ 5 2 tan β. 4.85 (a) 49.18. (b) A 5 45.3 N z; B 5 90.6 N a 60.08. 4.86 32.58. 4.88 (a) 225 mm. (b) 23.1 N. (c) 12.21 N y. 4.90 (a) 59.48. (b) A 5 8.45 lb y; B 5 13.09 lb b 49.8°. 4.91 A 5 (120.0 N)j 1 (133.3 N)k; D 5 (60.0 N)j 1 (166.7 N)k. 4.93 (a) 96.0 lb. (b) A 5 (2.40 lb)j; B 5 (214 lb)j. 4.94 A 5 (22.9 lb)i 1 (8.50 lb)j; B 5 (22.9 lb)i 1 (25.5 lb)j; C 5 2(45.8 lb)i. 4.95 (a) 78.5 N. (b) A 5 2(27.5 N)i 1 (58.9 N)j; B 5 (106.0 N)i 1 (58.9 N)j. 4.97 TA 5 21.0 lb; TB 5 TC 5 17.50 lb. 4.99 (a) 121.9 N. (b) 246.2 N. (c) 100.9 N. 4.100 (a) 95.6 N. (b) 27.36 N. (c) 88.3 N. 4.101 TA 5 23.5 N; TC 5 11.77 N; TD 5 105.9 N. 4.102 (a) 0.480 in. (b) TA 5 23.5 N; TC 5 0; TD 5 117.7 N. 4.103 (a) TA 5 6.00 lb; TB 5 TC 5 9.00 lb. (b) 15.00 in. 4.105 TBD 5 1100 lb; TBE 5 1100 lb; A 5 (1200 lb)i 2 (560 lb)j. 4.106 TBD 5 780 N; TBE 5 390 N; A 5 2(195.0 N)i 1 (1170 N)j 1 (130.0 N)k. 4.107 TBD 5 525 N; TBE 5 105.0 N; A 5 2(105.0 N)i 1 (840 N)j 1 (140.0 N)k. 3.134 (a) P; θx 5 90.08, θy 5 90.08, θz 5 0. (b) 5a/2. (c) Axis of the wrench is parallel to the z-axis at x 5 a, y 5 2a. 3.136 (a) 2(21.0 lb) j. (b) 0.571 in. (c) At x 5 0, z 5 1.667 in; and is parallel to the y axis. 3.137 (a) 2(84.0 N)j 2 (80.0 N)k. (b) 0.477 m. (c) x 5 0.526 m, y 5 0, z 5 20.1857 m. 3.140 (a) 3P (2i 2 20j 2 k)/25. (b) 20.0988a. (c) x 5 2.00a, y 5 0, z 5 21.990a. 3.141 R 5 (20.0 N)i 1 (30.0 N)j 2 (10.00 N)k; y 5 20.540 m, z 5 20.420 m. 3.143 FA 5 (M/b)i 1 R [1 1 (a/b)]k; FB 5 2(M/b)i 2 (a R/b)k. 3.147 (a) 20.5 N?m l. (b) 68.4 mm. 3.148 760 N?m l. 3.150 43.68. 3.151 23.0 N?m. 3.153 M 5 4.50 N?m; θx 5 90.08, θy 5 177.18, θz 5 87.18. 3.154 F 5 260 lb d67.4°; Mc 5 200 lb.in i. 3.156 (a) 135.0 mm. (b) F2 5 (42.0 N)i 1 (42.0 N)j 2 (49.0 N)k; M2 5 2(25.9 N?m)i 1 (21.2 N?m)j 3.158 (a) B 5 (2.50 lb)i, C 5 (0.1000 lb)i 2 (2.47 lb)j 2(0.700 lb)k. (b) Ry5 22.47 lb; Mx5 1.360 lb?ft. CHAPTER 4 4.1 42.0 Nx . 4.2 0.264 m. 4.4 (a) 245 lb x . (b) 140.0 lb. 4.5 (a) 34.0 kNx . (b) 4.96 kN ↑. 4.6 (a) 81.1 kN. (b) 134.1 kN ↑. 4.7 (a) A 5 20.0 lbw; B 5 150.0 lb ↑. (b) A 5 10.00 lbw; B 5 140.0 lb x . 4.9 1.250 kN # Q # 27.5 kN. 4.12 6.00 kips # P # 27.0 kips. 4.13 150.0 mm # d # 400 mm. 4.14 2.00 in. # a # 10.00 in. 4.15 (a) 600 N. (b) 1253 N a 69.88. 4.17 (a) 80.8 lbw. (b) 216 lb a 22.08. 4.18 232 lb. 4.19 (a) 2.00 kN. (b) 2.32 kN a 46.48. 4.22 (a) 400 N. (b) 458 N a 49.18. 4.23 (a) A 5 44.7 lb b 26.6°; B 5 30.0 lb ↑. (b) A 5 30.2 lb b 41.4°; B 5 34.6 lb b 60.0°. 4.24 (a) A 5 20.0 lb x; B 5 50.0 lb b 36.9°. (b) A 5 23.1 lb a 60.08; B 5 59.6 lb b 30.2°. 4.25 (a) 190.9 N. (b) 142.3 N a 18.438. 4.26 (a) 324 N. (b) 270 N y. 4.28 (a) A 5 225 Nx; C 5 641 N d 20.6°. (b) A 5 365 N a 60.08; C 5 884 N d 22.0°. 4.31 T 5 2P/3; C 5 0.577P y. 4.32 T 5 0.586P; C 5 0.414P y. 4.33 (a) 117.0 lb. (b) 129.8 lb c 56.3°. 4.34 (a) 195.0 lb. (b) 225 lb c 45.0°. 4.35 (a) 1432 N. (b) 1100 Nx . (c) 1400 N ←. 4.36 TBE 5 196.2 N; A 5 73.6 N y; D 5 73.6 N ←. 4.39 (a) 600 N. (b) A 5 4.00 kN ←; B 5 4.00 kN y. 4.40 (a) 105.1 N. (b) A 5 147.2 N↑; B 5 105.1 N ←. 4.41 (a) A 5 20.2 lb x; B 5 30.0 lb b 60.0°. (b) 16.21 lbw. 4.42 5.44 lb # P # 17.23 lb. 4.43 (a) E 5 8.80 kipsx; ME 5 36.0 kip?ft i. (b) E 5 4.80 kipsx; ME 5 51.0 kip?ft i. 4.45 Tmax 5 2240 N; Tmin 5 1522 N. 4.46 C 5 1951 N b88.5°; MC 5 75.0 N?m i. 4.47 1.232 kN # T # 1.774 kN.
AN4 Answers to Problems 5.17 Y 5 (r1 1 r2)(cos α)/(π 2 2α). 5.19 0.520. 5.20 459 N. 5.21 0.235 in3 for A1, 20.235 in3 for A2. 5.23 (a) b(c2 2 y2)/2. (b) y 5 0; Qx 5 bc2/2. 5.24 X 5 40.9 mm, Y 5 25.3 mm. 5.26 X 5 3.38 in., Y 5 2.93 in. 5.29 (a) 125.3 N. (b) 137.0 N a 56.7°. 5.30 120.0 mm. 5.31 99.5 mm. 5.32 (a) 0.513a. (b) 0.691a. 5.34 x 5 2 3 a, y 5 2 3 h. 5.35 x 5 a/2, y 5 2h/5. 5.37 x 5 a(3 2 4 sin α)/6 (1 2 α), y 5 0. 5.39 x 5 2a/3(4 2 π), y 5 2b/3(4 2 π). 5.40 x 5 a/4, y 5 3b/10. 5.41 x 5 3a/5, y 5 12b/35. 5.43 x 5 17a/130, y 5 11b/26. 5.44 x 5 a, y 5 17b/35. 5.45 2a /5. 5.46 2222r/3π. 5.48 x 5 29.27a, y 5 3.09a. 5.49 x 5 L/π, y 5 πa/8. 5.51 x 5 y 5 1.027 in. 5.52 (a) V 5 401 3 103 mm3; A 5 34.1 3 103 mm2. (b) V 5 492 3 103 mm3; A 5 41.9 3 103 mm2. 5.53 (a) V 5 248 in3; A 5 547 in2. (b) V 5 72.3 in3; A 5 169.6 in2. 5.54 (a) V 5 2.26 3 106 mm3; A 5 116.3 3 103 mm2. (b) V 5 1.471 3 106 mm3; A 5 116.3 3 103 mm2. 5.55 V 5 3470 mm3; A 5 2320 mm2. 5.58 308 in2. 5.60 31.9 liters. 5.62 V 5 3.96 in3, W 5 1.211 lb. 5.63 14.52 in2. 5.64 0.0305 kg. 5.66 (a) R 5 6000 Nw, x 5 3.60 m. (b) A 5 6000 Nx , MA 5 21.6 kN?m l. 5.67 (a) R 5 7.60 kNw, x 5 2.57 m. (b) A 5 4.35 kNx , B 5 3.25 kNx . 5.69 A 5 900 lb x; MA 5 9200 lb?in. l. 5.70 B 5 1360 lb x; C 5 2360 lb x . 5.71 A 5 105.0 Nx; B 5 270 Nx . 5.73 A 5 3.00 kNx; MA 5 12.60 kN?m l. 5.74 (a) 0.536 m. (b) A 5 B 5 761 Nx . 5.76 B 5 3770 lb x; C 5 429 lb x . 5.77 (a) 900 lb/ft. (b) 7200 lb x . 5.78 wA 5 10.00 kN/m; wB 5 50 kN/m. 5.80 (a) H 5 10.11 kips y, V 5 37.8 kipsx . (b) 10.48 ft to the right of A. (c) R 5 10.66 kips d 18.43°. 5.81 (a) H 5 44.1 kN y, V 5 228 kNx . (b) 1.159 m to the right of A. (c) R 5 59.1 kN d 41.6°. 5.82 6.98%. 5.84 12.00 in. 5.85 4.00 in. 5.87 T 5 6.72 kN z ; A 5 141.2 kN z. 5.88 A 5 1197 N b 53.1°; B 5 1511 N b 53.1°. 5.89 3570 N. 5.90 6.00 ft. 5.92 0.683 m. 4.108 TAD 5 2.60 kN; TAE 5 2.80 kN; C 5 (1.800 kN)j 1 (4.80 kN)k. 4.109 TAD 5 5.20 kN; TAE 5 5.60 kN; C 5 (9.60 kN)k. 4.110 TBD 5 TBE 5 176.8 lb; C 5 2(50.0 lb)j 1 (216.5 lb)k. 4.113 FCD 5 19.62 N; A 5 2(19.22 N)i 1 (45.1 N)j; B 5 (49.1 N)j. 4.115 A 5 2(56.3 lb)i; B 5 2(56.2 lb)i 1 (150.0 lb)j 2 (75.0 lb)k; FCE 5 202 lb. 4.116 (a) 116.6 lb. (b) A 5 2 (72.7 lb)j 2 (38.1 lb)k; B 5 (37.5 lb)j. 4.117 (a) 345 N. (b) A 5 (114.4 N)i 1 (377 N)j 1 (141.5 N)k; B 5 (113.2 N)j 1 (185.5 N)k. 4.119 FCD 5 19.62 N; B 5 2(19.22 N)i 1 (94.2 N)j; MB 5 2(40.6 N?m)i 2 (17.30 N?m)j. 4.120 A 5 2(112.5 lb)i 1 (150.0 lb)j 2 (75.0 lb)k; MA 5 (600 lb?ft)i 1 (225 lb?ft)j; FCE 5 202 lb. 4.121 (a) 5.00 lb. (b) C 5 2(5.00 lb)i 1 (6.00 lb)j 2 (5.00 lb)k; MC 5 (8.00 lb?in.)j 2 (12.00 lb?in)k. 4.122 TCF 5 200 N; TDE 5 450 N; A 5 (160.0 N)i 1 (270 N)k; MA 5 2(16.20 N?m)i. 4.123 TBD 5 2.18 kN; TBE 5 3.96 kN; TCD 5 1.500 kN. 4.124 TBD 5 0; TBE 5 3.96 kN; TCD 5 3.00 kN. 4.127 A 5 (120.0 lb)j 2 (150.0 lb)k; B 5 (180.0 lb)i 1 (150.0 lb)k; C 5 2(180.0 lb)i 1 (120.0 lb)j. 4.128 A 5 (20.0 lb)j 1 (25.0 lb)k; B 5 (30.0 lb)i 2 (25.0 lb)k; C 5 2(30.0 lb)i 2 (20.0 lb)j. 4.129 TBE 5 975 N; TCF 5 600 N; TDG 5 625 N; A 5 (2100 N)i 1 (175.0 N)j 2 (375 N)k. 4.131 TB 5 20.366 P; TC 5 1.219 P; TD 5 20.853 P; F 5 20.345 Pi 1 Pj 2 0.862 Pk. 4.133 360 N. 4.135 85.3 lb. 4.136 181.7 lb. 4.137 (45.0 lb)j. 4.138 343 N. 4.140 (a) x 5 4.00 ft, y 5 8.00 ft. (b) 10.73 lb. 4.141 (a) x 5 0, y 5 16.00 ft. (b) 11.31 lb. 4.142 (a) 1761 lbx . (b) 689 lb x . 4.143 (a) 150.0 lb. (b) 225 lb d 32.3°. 4.145 (a) 130.0 N. (b) 224 d 2.05°. 4.146 T 5 80.0 N; A 5 160.0 N c 30.0°; C 5 160.0 N b 30.0°. 4.148 A 5 680 N a 28.1°; B 5 600 N z. 4.149 A 5 63.6 lb c 45.0°; C 5 87.5 lb b 59.0°. 4.151 TA 5 5.63 lb; TB 5 16.88 lb; TC 5 22.5 lb. 4.153 (a) A 5 0.745 P a 63.4°; C 5 0.471 P b 45.0°. (b) A 5 0.812 P a 60.0°; C 5 0.503 P d 36.2°. (c) A 5 0.448 P b 60.0°; C 5 0.652 P a 69.9°. (d) improperly constrained; no equilibrium. CHAPTER 5 5.1 X 5 42.2 mm, Y 5 24.2 mm. 5.2 X 5 1.045 in., Y 5 3.59 in. 5.3 X 5 2.84 mm, Y 5 24.8 mm. 5.4 X 5 52.0 mm, Y 5 65.0 mm. 5.5 X 5 3.27 in., Y 5 2.82 in. 5.6 X 5 210.00 mm, Y 5 87.5 mm. 5.9 X 5 Y 5 16.75 mm. 5.10 X 5 10.11 in., Y 5 3.88 in. 5.11 X 5 30.0 mm, Y 5 64.8 mm. 5.13 X 5 3.20 in., Y 5 2.00 in. 5.14 X 5 0, Y 5 1.372 m. 5.16 Y 5 a2 3bar3 2 2 r3 1 r2 2 2 r2 1 ba 2 cos α π 2 2αb.
Answers to Problems AN5 6.12 FAB 5 FFH 5 1500 lb C; FAC 5 FCE 5 FEG 5 FGH 5 1200 lb T; FBC 5 FFG 5 0; FBD 5 FDF 5 1000 lb C; FBE 5 FEF 5 500 lb C; FDE 5 600 lb T. 6.13 FAB 5 6.24 kN C; FAC 5 2.76 kN T; FBC 5 2.50 kN C; FBD 5 4.16 kN C; FCD 5 1.867 kN T; FCE 5 2.88 kN T; FD 5 3.75 kN C; FDF 5 0; FEF 5 1.200 kN C. 6.15 FAB 5 FFG 5 7.50 kips C; FAC 5 FEG 5 4.50 kips T; FBC 5 FEF 5 7.50 kips T; FBD 5 FDF 5 9.00 kips C; FCD 5 FDE 5 0; FCE 5 9.00 kips T. 6.17 FAB 5 47.2 kN C; FAC 5 44.6 kN T; FBC 5 10.50 kN C; FBD 5 47.2 kN C; FCD 5 17.50 kN T; FCE 5 30.6 kN T; FDE 5 0. 6.18 FAB 5 2250 N C; FAC 5 1200 N T; FBC 5 750 N T; FBD 5 1700 N C; FBE 5 400 N C; FCE 5 850 N C; FCF 5 1600 N T; FDE 5 1500 N T; FEF 5 2250 N T. 6.19 FAB 5 FFH 5 7.50 kips C; FAC 5 FGH 5 4.50 kips T; FBC 5 FFG 5 4.00 kips T; FBD 5 FDF 5 6.00 kips C; FBE 5 FEF 5 2.50 kips T; FCE 5 FEG 5 4.50 kips T; FDE 5 0. 6.21 FAB 5 9.90 kN C; FAC 5 7.83 kN T; FBC 5 0; FBD 5 7.07 kN C; FBE 5 2.00 kN C; FCE 5 7.83 kN T; FDE 5 1.000; kN T; FDF 5 5.03 kN C; FDG 5 0.559 kN C; FEG 5 5.59 kN T. 6.22 FAB 5 3610 lb C; FAC 5 4110 lb T; FBC 5 768 lb C; FBD 5 3840 lb C; FCD 5 1371 lb T; FCE 5 2740 lb T; FDE 5 1536 lb C. 6.23 FDF 5 4060 lb C; FDG 5 1371 lb T; FEG 5 2740 lb T; FFG 5 768 lb C; FFH 5 4290 lb C; FGH 5 4110 lb T. 6.24 FAB 5 FDF 5 2.29 kN T; FAC 5 FEF 5 2.29 kN C; FBC 5 FDE 5 0.600 kN C; FBD 5 2.21 kN T; FBE 5 FEH 5 0; FCE 5 2.21 kN C; FCH 5 FEJ 5 1.200 kN C. 6.27 FAB 5 FBC 5 FCD 5 36.0 kips T; FAE 5 57.6 kips T; FAF 5 45.0 kips C; FBF 5 FBG 5 FCG 5 FCH 5 0; FDH 5 FFG 5 FGH 5 39.0 kips C; FEF 5 36.0 kips C. 6.28 FAB 5 128.0 kN T; FAC 5 136.7 kN C; FBD 5 FDF 5 FFH 5 128.0 kN T; FCE 5 FEG 5 136.7 kN C; FGH 5 192.7 kN C; FBC 5 FBE 5 FDE 5 FDG 5 FFG 5 0. 6.29 Truss of Prob. 6.33a is the only simple truss. 6.30 Trusses of Prob. 6.32b and Prob. 6.33b are simple trusses. 6.32 (a) AI, BJ, CK, DI, EI, FK, GK. (b) FK, IO. 6.34 (a) BC, HI, IJ, JK. (b) BF, BG, CG, CH. 6.35 FAB 5 FAD 5 244 lb C; FAC 5 1040 lb T; FBC 5 FCD 5 500 lb C; FBD 5 280 lb T. 6.36 FAB 5 FAD 5 861 N C; FAC 5 676 N C; FBC 5 FCD 5 162.5 N T; FBD 5 244 N T. 6.37 FAB 5 FAD 5 2810 N T; FAC 5 5510 N C; FBC 5 FCD 5 1325 N T; FBD 5 1908 N C. 6.38 FAB 5 FAC 5 1061 lb C; FAD 5 2500 lb T; FBC 5 2100 lb T; FBD 5 FCD 5 1250 lb C; FBE 5 FCE 5 1250 lb C; FDE 5 1500 lb T. 6.39 FAB 5 840 N C; FAC 5 110.6 N C; FAD 5 394 N C; FAE 5 0; FBC 5 160.0 N T; FBE 5 200 N T; FCD 5 225 N T; FCE 5 233 N C; FDE 5 120.0 N T. 6.40 FAB 5 FAE 5 FBC 5 0; FAC 5 995 N T; FAD 5 1181 N C; FBE 5 600 N T; FCD 5 375 N T; FCE 5 700 N C; FDE 5 360 N T. 6.43 FDF 5 5.45 kN C; FDG 5 1.000 kN T; FEG 5 4.65 kN T. 6.44 FGI 5 4.65 kN T; FHI 5 1.800 kN C; FHJ 5 4.65 kN C. 6.45 FBD 5 36.0 kips C; FCD 5 45.0 kips C. 6.46 FDF 5 60.0 kips C; FDG 5 15.00 kips C. 6.49 FCD 5 20.0 kN C; FDF 5 52.0 kN C. 6.50 FCE 5 36.0 kN T; FEF 5 15.00 kN C. 6.51 FDE 5 25.0 kips T; FDF 5 13.00 kips C. 5.93 0.0711 m. 5.94 208 lb. 5.96 (a) 0.548 L. (b) 223. 5.97 (a) b/10 to the left of base of cone. (b) 0.1136b to the right of base of cone. 5.98 (a) 20.402 a. (b) h/a 5 2/5 or 2/3. 5.99 27.8 mm above base of cone. 5.100 18.28 mm. 5.102 20.0656 in. 5.103 2.57 in. 5.104 219.02 mm. 5.106 X 5 125.0 mm, Y 5 167.0 mm, Z 5 33.5 mm. 5.107 X 5 0.295 m, Y 5 0.423 m, Z 5 1.703 m. 5.109 X 5 Z 5 4.21 in., Y 5 7.03 in. 5.110 X 5 180.2 mm, Y 5 38.0 mm, Z 5 193.5 mm. 5.111 X 5 17.00 in., Y 5 15.68 in., Z 5 14.16 in. 5.113 X 5 46.5 mm, Y 5 27.2 mm, Z 5 30.0 mm. 5.114 X 5 0.909 m, Y 5 0.1842 m, Z 5 0.884 m. 5.116 X 5 0.410 m, Y 5 0.510 m, Z 5 0.1500 m. 5.117 X 5 0, Y 5 10.05 in., Z 5 5.15 in. 5.118 X 5 61.1 mm from the end of the handle. 5.119 Y 5 0.526 in. above the base. 5.121 Y 5 421 mm above the floor. 5.122 (x1) 5 21a/88; (x2) 5 27a/40. 5.123 (x1) 5 21h/88; (x2) 5 27h/40. 5.124 (x1) 5 2h/9; (x2) 5 2 h/3. 5.125 x 5 2.34 m; y 5 z 5 0. 5.128 x 5 1.297a; y 5 z 5 0. 5.129 x 5 z 5 0; y 5 0.374b. 5.132 (a) x 5 z 5 0, y 5 2121.9 mm. (b) x 5 z 5 0, y 5 290.2 mm. 5.134 x 5 0, y 5 5h/16, z 5 2b/4. 5.135 x 5 a/2, y 5 8h/25, z 5 b/2. 5.136 V 5 688 ft3; x 5 15.91 ft. 5.137 X 5 5.67 in., Y 5 5.17 in. 5.138 X 5 92.0 mm, Y 5 23.3 mm. 5.139 (a) 5.09 lb. (b) 9.48 lb b 57.5°. 5.141 x 5 2L/5, y 5 12h/25. 5.143 A 5 2860 lb x; B 5 740 lb x . 5.144 wBC 5 2810 N/m; wDE 5 3150 N/m. 5.146 2(2h2 2 3b2)/2 (4h 2 3b). 5.148 X 5 Z 5 0, Y 5 83.3 mm above the base.
CHAPTER 6 6.1 FAB 5 900 lb T; FAC 5 780 lb C; FBC 5 720 lb T. 6.2 FAB 5 1.700 kN T; FAC 5 2.00 kN T; FBC 5 2.50 kN T. 6.3 FAB 5 720 lb T; FAC 5 1200 lb C; FBC 5 780 lb C. 6.4 FAB 5 FBC 5 0; FAD 5 FCF 5 7.00 kN C; FBD 5 FBF 5 34.0 kN C; FBE 5 8.00 kN T; FDE 5 FEF 5 30.0 kN T. 6.6 FAC 5 80.0 kN T; FCE 5 45.0 kN T; FDE 5 51.0 kN C; FBD 5 51.0 kN C; FCD 5 48.0 kN T; FBC 19.00 kN C. 6.8 FAB 5 20.0 kN T; FAD 5 20.6 kN C; FBC 5 30.0 kN T; FBD 5 11.18 kN C; FCD 5 10.00 kN T. 6.9 FAB 5 FDE 5 8.00 kN C; FAF 5 FFG 5 FGH 5 FEH 5 6.93 kN T; FBC 5 FCD 5 FBG 5 FDG 5 4.00 kN C; FBF 5 FDH 5 FCG 5 4.00 kN T. 6.11 FAB 5 FFH 5 1500 lb C; FAC 5 FCE 5 FEG 5 FGH 5 1200 lb T; FBC 5 FFG 5 0; FBD 5 FDF 5 1200 lb C; FBE 5 FEF 5 60.0 lb C; FDE 5 72.0 lb T.
AN6 Answers to Problems 6.101 Ax 5 13.00 kN z, Ay 5 4.00 kNw; Bx 5 36.0 kN y. By 5 6.00 kNx; Ex 5 23.0 kN z, Ey 5 2.00 kNw. 6.102 Ax 5 2025 N z, Ay 5 1800 kNw; Bx 5 4050 N y, By 5 1200 N x; Ex 5 2025 N z, Ey 5 600 N x . 6.103 Ax 5 1110 lb z, Ay 5 600 lb x; Bx 5 1110 lb z, By 5 800 lbw; Dx 5 2220 lb y, Dy 5 200 lb x . 6.104 Ax 5 660 lb z, Ay 5 240 lb x ; Bx 5 660 lb z, By 5 320 lbw ; Dx 5 1320 lb y, Dy 5 80.0 lb x . 6.107 (a) Ax 5 200 kN y, Ay 5 122.0 kNx . (b) Bx 5 200 kN z, By 5 10.00 kNw. 6.108 (a) Ax 5 205 kN y, Ay 5 134.5 kNx .
(b) Bx 5 205 kN z. By 5 5.50 kNx . 6.109 B 5 98.5 lb a 24.0°; C 5 90.6 lb b 6.34°. 6.110 B 5 25.0 lb x; C 5 79.1 lb b 18.43°. 6.112 FAF 5 P/4 C; FBG 5 FDG 5 P/22 C; FEH 5 P/4 T. 6.113 FAG 5 22P/6 C; FBF 5 222P/3 C; FDI 5 22P/3 C; FEH 5 22P/6 T. 6.115 FAF 5 M0/4a C; FBG 5 FDG 5 M0/22a T; FEH 5 3M0/4a C. 6.116 FAF 5 M0/6a T; FBG 5 22M0/6a T; FDG 5 22M0/3a T; FEH 5 M0/6a C. 6.117 A 5 P/15x; D 5 2P/15x; E 5 8P/15x; H 5 4P/15x . 6.118 E 5 P/5w; F 5 8P/5 x; G 5 4P/5w; H 5 2P/5x . 6.120 (a) A 5 2.06P a 14.04°; B 5 2.06 b 14.04°; frame is rigid. (b) Frame is not rigid. (c) A 5 1.25P b 36.9°. B 5 1.031P a 14.04°; frame is rigid. 6.122 (a) 2860 Nw. (b) 2700 N d 68.5°. 6.123 564 lb y. 6.124 275 lb y. 6.125 764 N z. 6.127 (a) 746 Nw. (b) 565 N c 61.3°. 6.129 832 lb?in. l. 6.130 360 lb?in. l. 6.131 195.0 kN?m i. 6.132 40.5 kN?m l. 6.133 (a) 160.8 N?m l. (b) 155.9 N?m l. 6.134 (a) 117.8 N?m l. (b) 47.9 N?m l. 6.137 18.43 N?m i. 6.138 208 N?m i. 6.139 FAE 5 800 N T; FDG 5 100.0 N C. 6.140 P 5 120.0 Nw; Q 5 110.0 N z. 6.141 F 5 3290 lb c 15.12°; D 5 4550 lb z. 6.143 D 5 30.0 kN z; F 5 37.5 kN c 36.9°. 6.144 D 5 150.0 kN z; F 5 96.4 kN c 13.50°. 6.145 (a) 475 lb. (b) 528 lb b 63.3°. 6.147 44.8 kN. 6.148 8.45 kN. 6.149 140.0 N. 6.151 315 lb. 6.152 (a) 312 lb. (b) 135.0 lb?in.i. 6.153 (a) 4.91 kips C. (b) 10.69 kips C. 6.154 (a) 2.86 kips C. (b) 9.43 kips C. 6.155 (a) 9.29 kN b 44.4°. (b) 8.04 kN c 34.4°. 6.159 (a) (90.0 N?m)i. (b) A 5 0; MA 5 2(48.0 N?m)i, B 5 0; MB 5 2(72.0 N?m)i. 6.160 (a) 27.0 mm. (b) 40.0 N?m i. 6.163 Ex 5 100.0 kN y, Ey 5 154.9 kNx; Fx 5 26.5 kN y. Fy 5 118.1 kNw; Hx 5 126.5 kN z, Hy 5 36.8 kNw. 6.164 FAB 5 4.00 kN T; FAD 5 15.00 kN T; FBD 5 9.00 kN C; FBE 5 5.00 kN T; FCD 5 16.00 kN C; FDE 5 4.00 kN C. 6.52 FEG 5 16.00 kips T; FEF 5 6.40 kips C. 6.53 FDF 5 91.4 kN T; FDE 5 38.6 kN C. 6.54 FCD 5 64.2 kN T; FCE 5 92.1 kN C. 6.55 FCE 5 7.20 kN T; FDE 5 1.047 kN C; FDF 5 6.39 kN C. 6.56 FEG 5 3.46 kN T; FGH 5 3.78 kN C; FHJ 5 3.55 kN C. 6.59 FAD 5 3.38 kips C; FCD 5 0; FCE 5 14.03 kips T. 6.60 FDG 5 18.75 kips C; FFG 5 14.03 kips T; FFH 5 17.43 kips T. 6.61 FDG 5 3.75 kN T; FFI 5 3.75 kN C. 6.62 FGJ 5 11.25 kN T; FIK 5 11.25 kN C. 6.65 (a) CJ. (b) 1.026 kN T. 6.66 (a) IO. (b) 2.05 kN T. 6.67 FBE 5 10.00 kips T; FDE 5 0; FEF 5 5.00 kips T. 6.68 FBE 5 2.50 kips T; FDE 5 1.500 kips C; FDG 5 2.50 kips T. 6.69 (a) improperly constrained. (b) completely constrained, determinate. (c) completely constrained, indeterminate. 6.70 (a) completely constrained, determinate. (b) partially constrained. (c) improperly constrained. 6.71 (a) completely constrained, determinate. (b) completely constrained, indeterminate. (c) improperly constrained. 6.72 (a) partially constrained. (b) completely constrained. determinate. (c) completely constrained, indeterminate. 6.75 FBD 5 375 N C; Cx 5 205 Nz; Cy 5 360 Nw. 6.76 FBD 5 780 lb T; Cx 5 720 lb z, Cy 5 140.0 lbw. 6.77 (a) 125.0 N b 36.9°. (b) 125.0 N d 36.9°. 6.78 Ax 5 120.0 lb y, Ay 5 30.0 lb x; Bx 5 120.0 lb z, By 5 80.0 lbw; C 5 30.0 lbw; D 5 80.0 lb x . 6.79 Ax 5 18.00 kN z, Ay 5 20.0 kNw; B 5 9.00 kN y; Cx 5 9.00 kN y, Cy 5 20.0 kNx . 6.80 A 5 20.0 kNw, B 5 18.00 kN z; Cx 5 18.00 kN y, Cy 5 20.0 kNx . 6.81 A 5 150.0 lb y; Bx 5 150.0 lb z, By 5 60.0 lb x; C 5 20.0 lb x; D 5 80.0 lbw. 6.83 (a) Ax 5 2700 N y, Ay 5 200 Nx; Ex 5 2700 N z, Ey 5 600 Nx . (b) Ax 5 300 N y, Ay 5 200 Nx; Ex 5 300 N z, Ey 5 600 Nx . 6.85 (a) Ax 5 300 N z, Ay 5 660 Nx; Ex 5 300 N y, Ey 5 90.0 Nx . (b) Ax 5 300 N z, Ay 5 150.0 Nx; Ex 5 300 N y, Ey 5 600 N x . 6.87 (a) Ax 5 80.0 lb z, Ay 5 40.0 lb x; Bx 5 80.0 lb y, By 5 60.0 lb x . (b) Ax 5 0, Ay 5 40.0 lb x ; Bx 5 0, By 5 60.0 lb x . 6.88 (a) and (c) Bx 5 32.0 lb y, By 5 10.00 lb x; Fx 5 32.0 lb z, Fy 5 38.0 lb x . (b) Bx 5 32.0 lb z, By 5 34.0 lb x; Fx 5 32.0 lb y, Fy 5 14.00 lb x . 6.89 (a) and (c) Bx 5 24.0 lb z, By 5 7.50 lbw; Fx 5 24.0 lb y, Fy 5 7.50 lb x . (b) Bx 5 24.0 lb z, By 5 10.50 lb x; Fx 5 24.0 lb y, Fy 5 10.50 lbw. 6.91 Dx 5 13.60 kN y, Dy 5 7.50 kNx; Ex 5 13.60 kN z, Ey 5 2.70 kNw. 6.92 Ax 5 45.0 N z, Ay 5 30.0 Nw; Bx 5 45.0 N y, By 5 270 Nx . 6.93 (a) Ex 5 2.00 kips z, Ey 5 2.25 kipsx . (b) Cx 5 4.00 kips z, Cy 5 5.75 kipsx . 6.94 (a) Ex 5 3.00 kips z, Ey 5 1.500 kipsx . (b) Cx 5 3.00 kips z, Cy 5 6.50 kipsx . 6.95 (a) A 5 982 lb x; B 5 935 lb x; C 5 733 lb x . (b) D B 5 1291 lb; D C 5 272.7 lb. 6.96 (a) 572 lb. (b) A 5 1070 lb x; B 5 709 lb x; C 5 870 lb x . 6.99 B 5 152.0 lbw; Cx 5 60.0 lb z, Cy 5 200 lb x; Dx 5 60.0 lb y, 42.0 lb x . 6.100 B 5 108.0 lbw; Cx 5 90.0 lb z, Cy 5 150.0 lb x; Dx 5 90.0 lb y, Dy 5 18.00 lb x .
Answers to Problems AN7 7.55 (a) 54.5°. (b) 675 N?m. 7.56 (a) 1.236. (b) 0.1180 wa2. 7.57 (a) 40.0 mm. (b) 1.600 N?m. 7.58 (a) 0.840 m. (b) 1.680 N?m. 7.59 0.207 L. 7.62 (a) 0.414 wL; 0.0858 wL2. (b) 0.250 wL; 0.250 wL2. 7.69 (a) ZVZmax 5 15.00 kN; ZMZmax 5 42.0 kN?m. 7.70 (b) ZVZmax 5 17.00 kN; ZMZmax 5 17.00 kN?m. 7.77 (b) 75.0 kN?m, 4.00 m from A. 7.78 (b) 1.378 kN?m, 1.050 m from A. 7.79 (b) 26.4 kN?m, 2.05 m from A. 7.80 (b) 5.76 kN?m, 2.40 m from A. 7.81 (b) 14.40 kip?ft, 6.00 ft from A. 7.82 (b) 16.20 kip?ft, 13.50 ft from A. 7.86 (a) V 5 (w0/6L)(L2 2 3x2); M 5 (w0/6L)(L2x 2 x3). (b) 0.0642 w0 L2A a 5 0.577L. 7.87 (a) V 5 (w0 L/4)[3(x/L)2 2 4(x/L) 1 1]; M 5 (w0L2/4) [(x/L)3 2 2(x/L)2 1 (x/L)]. (b) w0L2/27, at x 5 L/3. 7.89 (a) P 5 4.00 kNw; Q 5 6.00 kNw. (b) MC 5 2900 N?m. 7.90 (a) P 5 2.50 kNw; Q 5 7.50 kNw. (b) MC 5 2900 N?m. 7.91 (a) P 5 1.350 kipsw; Q 5 0.450 kipsw. (b) Vmax 5 2.70 kips at A; Mmax 5 6.345 kip?ft, 5.40 ft from A. 7.92 (a) P 5 0.540 kipsw; Q 5 1.860 kipsw. (b) Vmax 5 3.14 kips at B; Mmax 5 7.00 kip?ft, 6.88 ft from A. 7.93 (a) Ex 5 10.00 kN y, Ey 5 7.00 kNx . (b) 12.21 kN. 7.94 1.667 m. 7.95 (a) 838 lb b 17.35°. (b) 971 lb a 34.5°. 7.96 (a) 2670 lb d 2.10°. (b) 2810 lb a 18.65°. 7.97 (a) dB 5 1.733 m; dD 5 4.20 m. (b) 21.5 kN a 3.81°. 7.98 (a) 2.80 m. (b) A 5 32.0 kN b 38.7°; E 5 25.0 kN y. 7.101 196.2 N. 7.102 157.0 N. 7.103 (a) 240 lb. (b) 9.00 ft. 7.104 a 5 7.50 ft; b 5 17.50 ft 7.107 (a) 1775 N. (b) 60.1 m. 7.109 (a) 50,200 kips. (b) 3580 ft. 7.110 3.75 ft. 7.111 (a) 56,400 kips. (b) 4284 ft. 7.112 (a) 6.75 m. (b) TAB 5 615 N; TBC 5 600 N. 7.114 (a) 23LD/8. (b) 12.25 ft. 7.115 h 5 27.6 mm; θA 5 25.5°; θC 5 27.6°. 7.116 (a) 4.05 m. (b) 16.41 m. (c) Ax 5 5890 N z, Ay 5 5300 Nx . 7.117 (a) 58,900 kips, (b) 29.2°. 7.118 (a) 16.00 ft to the left of B. (b) 2000 lb. 7.125 Y 5 h[1 2 cos(πx/L)]; Tmin 5 w0L2/hπ2; Tmax 5 (w0L/π)21L2/h2π 22 1 1 7.127 (a) 12.36 ft. (b) 15.38 lb. 7.128 (a) 412 ft. (b) 875 lb. 7.129 (a) 35.6 m. (b) 49.2 kg. 7.130 49.86 ft. 7.133 (a) 5.89 m. (b) 10.89 N y. 7.134 10.05 ft. 7.135 (a) 56.3 ft. (b) 2.36 lb/ft. 7.136 (a) 30.2 m. (b) 56.6 kg. 7.139 31.8 N. 7.140 29.8 N. 7.143 (a) a 5 79.0 ft; b 5 60.0 ft. (b) 103.9 ft. 7.144 (a) a 5 65.8 ft; b 5 50.0 ft. (b) 86.6 ft. 7.145 119.1 N y. 7.146 177.6 N y. 7.147 3.50 ft. 6.165 FAB 5 7.83 kN C; FAC 5 7.00 kN T; FBC 5 1.886 kN C; FBD 5 6.34 kN C; FCD 5 1.491 kN T; FCE 5 5.00 kN T; FDE 5 2.83 kN C; FDF 5 3.35 kN C; FEF 5 2.75 kN T; FEG 5 1.061 kN C; FEH 5 3.75 kN T; FFG 5 4.24 kN C; FGH 5 5.30 kN C. 6.166 FAB 5 8.20 kips T; FAG 5 4.50 kips T; FFG 5 11.60 kips C. 6.168 Ax 5 900 lb z; Ay 5 75.0 lb x; B 5 825 lbw; Dx 5 900 lb y; Dy 5 750 lb x . 6.170 Bx 5 700 N z, By 5 200 Nw; Ex 5 700 N y, Ey 5 500 Nx . 6.171 Cx 5 78.0 lb y, Cy 5 28.0 lb x; Fx 5 78.0 lb z, Fy 5 12.00 lb x . 6.172 A 5 327 lb y; B 5 827 lb z; D 5 621 lb x; E 5 246 lb x . 6.174 (a) 21.0 kN z. (b) 5 52.5 kN z.
CHAPTER 7 7.1 F 5 720 lb y; V 5 140.0 lb x; M 5 1120 lb?in. l (On JC). 7.2 F 5 120.0 lb z; V 5 30.0 lbw; M 5 120.0 lb?in. l. 7.3 F 5 125.0 N a 67.4°; V 5 300 N c 22.6°; M 5 156.0 N?m. i. 7.4 F 5 2330 N a 67.4°; V 5 720 N c 22.6°; M 5 374 N?m. i. 7.7 F 5 23.6 lb a 76.0°; V 5 29.1 lb a 14.04°; M 5 540 lb?in. i. 7.8 (a) 30.0 lb at C. (b) 33.5 lb at B and D. (c) 960 lb?in. at C. 7.9 F 5 103.9 N b 60.0°; V 5 60.0 N a 30.0°; M 5 18.71 N?m i (On AJ). 7.10 F 5 60.0 N d 30.0°; V 5 103.9 c 60.0°; M 5 10.80 N?m l (On BK). 7.11 F 5 194.6 N c 60.0°; V 5 257 N a 30.0°; M 5 24.7 N?m i (On AJ). 7.12 45.2 N?m for θ 5 82.9°. 7.15 F 5 250 N c 36.9°; V 5 120.0 N a 53.1; M 5 120.0 N?m l (On BJ). 7.16 F 5 560 N z; V 5 90.0 Nw; M 5 72.0 N?m i (On AK). 7.17 150.0 lb?in. at D. 7.18 105.0 lb?in. at E. 7.19 F 5 200 N c 36.9°; V 5 120.0 N a 53.1°; M 5 120.0 N?m l (On BJ). 7.20 F 5 520 N z; V 5 120.0 Nw; M 5 96.0 N?mi (On AK). 7.23 0.0557 Wr (On AJ). 7.24 0.1009 Wr for θ 5 57.3°. 7.25 0.289 Wr (On BJ). 7.26 0.417 Wr (On BJ). 7.29 (b) ZVZmax 5 wL/4; ZMZmax 5 3wL2/32. 7.30 (b) ZVZmax 5 w0L/2; ZMZmax 5 w0L2/6. 7.31 (b) ZVZmax 5 2P/3; ZMZmax 5 2PL/9. 7.32 (b) ZVZmax 5 2P; ZMZmax 5 3Pa. 7.35 (b) ZVZmax 5 40.0 kN; ZMZmax 5 55.0 kN?m. 7.36 (b) ZVZmax 5 50.5 kN; ZMZmax 5 39.8 kN?m. 7.39 (b) ZVZmax 5 64.0 kN; ZMZmax 5 92.0 kN?m. 7.40 (b) ZVZmax 5 40.0 kN; ZMZmax 5 40.0 kN?m. 7.41 (b) ZVZmax 5 18.00 kips; ZMZmax 5 48.5 kip?ft. 7.42 (b) ZVZmax 5 15.30 kips; ZMZmax 5 46.8 kip?ft. 7.45 (b) ZVZmax 5 6.00 kips; ZMZmax 5 12.00 kip?ft. 7.46 (b) ZVZmax 5 4.00 kips; ZMZmax 5 6.00 kip?ft. 7.47 (b) ZVZmax 5 6.00 kN; ZMZmax 5 9.00 kN?m. 7.48 (b) ZVZmax 5 6.00 kN; ZMZmax 5 9.00 kN?m. 7.49 ZVZmax 5 180.0 N; ZMZmax 5 36.0 N?m. 7.50 ZVZmax 5 800 N; ZMZmax 5 180.0 N?m. 7.51 ZVZmax 5 90.0 lb; ZMZmax 5 1400 lb?in. 7.52 ZVZmax 5 165.0 lb; ZMZmax 5 1625 lb?in.
AN8 Answers to Problems 8.55 9.13 N z. 8.56 (a) 28.1°. (b) 728 N a 14.04°. 8.57 (a) 50.4 lbw. (b) 50.4 lbw. 8.59 143.4 N. 8.60 1.400 lb. 8.62 (a) 197.0 lb y. (b) Base will not move. 8.63 (a) 280 lb z. (b) Base moves. 8.64 (b) 283 N z. 8.65 0.442. 8.66 0.1103. 8.67 0.1013. 8.71 693 lb?ft. 8.72 35.8 N?m. 8.73 9.02 N?m. 8.74 (a) Screw A. (b) 14.06 lb?in. 8.77 0.226. 8.78 4.70 kips. 8.79 450 N. 8.80 412 N. 8.81 334 N. 8.82 376 N. 8.84 TAB 5 77.5 lb; TCD 5 72.5 lb. TEF 5 67.8 lb. 8.86 (a) 4.80 kN. (b) 1.375°. 8.88 22.0 lb z. 8.89 1.948 lbw. 8.90 18.01 lb z. 8.92 0.1670. 8.93 3.75 lb. 8.98 10.87 lb. 8.99 0.0600 in. 8.100 154.4 N. 8.101 300 mm. 8.102 (a) 1.288 kN. (b) 1.058 kN. 8.103 2.34 ft. 8.104 (a) 0.329. (b) 2.67 turns. 8.105 14.23 kg # m # 175.7 kg. 8.106 (a) 0.292. (b) 310 N. 8.109 31.8 N?m l. 8.110 (a) TA 5 8.40 lb; TB 5 19.60 lb. (b) 0.270. 8.111 (a) TA 5 11.13 lb; TB 5 20.9 lb. (b) 91.3 lb?in. i. 8.112 35.1 N?m. 8.113 (a) 27.0 N?m. (b) 675 N. 8.114 (a) 39.0 N?m. (b) 844 N. 8.117 4.49 in. 8.118 (a) 11.66 kg. (b) 38.6 kg. (c) 34.4 kg. 8.119 (a) 9.46 kg. (b) 167.2 kg. (c) 121.0 kg. 8.120 (a) 10.39 lb. (b) 58.5 lb. 8.121 (a) 28.9 lb. (b) 28.9 lb. 8.124 5.97 N. 8.125 9.56 N. 8.126 0.350. 8.128 (a) 30.3 lb?in. l. (b) 3.78 lbw. 8.129 (a) 17.23 lb?in. i. (b) 2.15 lb x . 8.133 (a) 51.0 N?m. (b) 875 N. 8.134 (a) 353 N z. (b) 196.2 N z. 8.136 (a) 136.0 lb y. (b) 30.0 lb y. (c) 12.86 lb y. 8.137 6.35 # L/a # 10.81. 8.138 151.5 N?m. 8.140 0.225. 8.141 313 lb y. 8.143 6.44 N?m. 8.144 (a) 0.238. (b) 218 Nw. 7.148 5.71 ft. 7.151 0.394 m and 10.97 m. 7.152 0.1408. 7.153 (a) 0.338. (b) 56.5°; 0.755 wL. 7.154 (On AJ) F 5 750 Nx; V 5 400 N z; M 5 130.0 N?m l. 7.156 (On BJ) F 5 12.50 lb a 30.0°; V 5 21.7 lb b 60.0°; M 5 75.0 lb?in.i. 7.157 (a) (On AJ) F 5 500 N z; V 5 500 Nx; M 5 300 N?m i. (b) (On AK) F 5 970 Nx; V 5 171.0 Nz; M 5 446 N?m i. 7.158 (a) 40.0 kips. (b) 40.0 kip·ft. 7.161 (a) 18.00 kip·ft, 3.00 ft from A. (b) 34.1 kip·ft, 2.25 ft from A. 7.163 (a) 2.28 m. (b) Dx 5 13.67 kN y; Dy 5 7.80 kNx . (c) 15.94 kN. 7.164 (a) 138.1 m. (b) 602 N. 7.165 (a) 4.22 ft. (b) 80.3°.
CHAPTER 8 8.1 Block is in equilibrium, F 5 30.1 N b 20.0°. 8.2 Block moves up, F 5 151.7 N c 20.0°. 8.3 Block moves, F 5 36.1 lb c 30.0°. 8.4 Block is in equilibrium, F 5 36.3 lb c 30.0°. 8.5 (a) 83.2 lb. (b) 66.3 lb. 8.7 (a) 29.7 N z. (b) 20.9 N y. 8.9 74.5 N. 8.10 17.91° # θ # 66.4°. 8.11 31.0°. 8.12 46.4°. 8.13 Package C does not move; FC 5 10.16 N Q . Package A and B move; FA 5 7.58 N Q; FB 5 3.03 N Q . 8.14 All packages move; FA 5 FC 5 7.58 N Q; FB 5 3.03 N Q . 8.17 (a) 75.0 lb. (b) Pipe will slide. 8.18 (a) P 5 36.0 lb y. (b) hmax 5 40.0 in. 8.19 P 5 8.34 lb. 8.20 P 5 7.50 lb. 8.21 (a) 0.300 Wr. (b) 0.349 Wr. 8.22 M 5 Wrμs (1 1 μs)/(1 1 μs 2). 8.23 (a) 136.4°. (b) 0.928 W. 8.25 0.208. 8.27 664 Nw. 8.29 (a) Plate in equilibrium. (b) Plate moves downward. 8.30 10.00 lb , P , 36.7 lb. 8.32 0.860. 8.34 0.0533. 8.35 (a) 1.333. (b) 1.192. (c) 0.839. 8.36 (b) 2.69 lb. 8.37 (a) 2.94 N. (b) 4.41 N. 8.39 30.6 N?m l. 8.40 18.90 N?m l. 8.41 135.0 lb. 8.43 (a) System slides; P 5 62.8 N. (b) System rotates about B; P 5 73.2 N. 8.44 35.8°. 8.45 20.5°. 8.46 1.225 W. 8.47 46.4° # θ # 52.4° and 67.6° # θ # 79.4°. 8.48 (a) 283 N z. (b) Bx 5 413 N z; By 5 480 Nw. 8.49 (a) 107.0 N z. (b) Bx 5 611 N z; By 5 480 Nw. 8.52 (a) 15.26 kips. (b) 5.40 kips. 8.53 (a) 6.88 kips. (b) 5.40 kips. 8.54 9.86 kN z.
Answers to Problems AN9 9.81 Ix9 5 1033 in4; Iy9 5 2020 in4; Ix9y9 5 2873 in4. 9.83 Ix9 5 0.236 in4; Iy9 5 1.244 in4; Ix9y9 5 0.1132 in4. 9.85 20.2° and 110.2°; 1.754a4; 0.209a4. 9.86 25.1° and 115.1°; Imax 5 8.32 3 106 mm4; Imin 5 2.08 3 106 mm4. 9.87 29.7° and 119.7°; 2530 in4; 524 in4. 9.89 223.7° and 66.3°; 1.257 in4; 0.224 in4. 9.91 (a) Ix9 5 0.482a4; Iy9 5 1.482a4; Ix9y9 5 20.589a4. (b) Ix9 5 1.120a4; Iy9 5 0.843a4; 0.760a4. 9.92 Ix9 5 2.12 3 106 mm4; Iy9 5 8.28 3 106 mm4; Ix9y9 5 20.532 3 106 mm4. 9.93 Ix9 5 1033 in4; Iy9 5 2020 in4; Ix9y9 5 2873 in4. 9.95 Ix9 5 0.236 in4; Iy9 5 1.244 in4; Ix9y9 5 0.1132 in4. 9.97 20.2°; 1.754a4; 0.209a4. 9.98 23.9°; 8.33 3 106 mm4; 1.465 3 106 mm4. 9.99 33.4°; 22.1 3 103 in4; 2490 in4. 9.100 29.7°; 2530 in4; 524 in4. 9.103 (a) 21.146 in4. (b) 29.1° clockwise. (c) 3.39 in4. 9.104 23.8° clockwise; 0.524 3 106 mm4; 0.0917 3 106 mm4. 9.105 19.54° counterclockwise; 4.34 3 106 mm4; 0.647 3 106 mm4. 9.106 (a) 25.3°. (b) 1459 in4; 40.5 in4. 9.107 (a) 88.0 3 106 mm4. (b) 96.3 3 106 mm4; 39.7 3 106 mm4. 9.111 (a) IAA9 5 IBB9 5 ma2/24. (b) ma2/12. 9.112 (a) m (r1 2 1 r2 2)/4. (b) m (r1 2 1 r2 2)/2. 9.113 (a) 0.0699 mb2. (b) m(a2 1 0.279 b2)/4. 9.114 (a) mb2/7. (b) m(7a2 1 10b2)/70. 9.117 (a) 5ma2/18. (b) 3.61 ma2. 9.118 (a) 0.994 ma2. (b) 2.33 ma2. 9.119 m(3a2 1 4L2)/12. 9.120 1.329 mh2. 9.121 (a) 0.241 mh2. (b) m(3a2 1 0.1204 h2). 9.122 m(b2 1 h2)/10. 9.124 ma2/3; a/23. 9.126 Ix 5 Iy 5 ma2/4; Iz 5 ma2/2. 9.127 1.160 3 1026 lb?ft?s2; 0.341 in. 9.128 837 3 1029 kg?m2; 6.92 mm. 9.130 2 mr2/3; 0.816r. 9.131 (a) 2.30 in. (b) 20.6 3 1023 lb?ft?s2; 2.27 in. 9.132 (a) πpl2 [6a2t(5a2/3l2 1 2a/l 1 1) 1 d2l/4]. (b) 0.1851. 9.133 (a) 27.5 mm to the right of A. (b) 32.0 mm. 9.135 Ix 5 7.11 3 1023 kg?m2; Iy 5 16.96 3 1023 kg?m2; Iz 5 15.27 3 1023 kg?m2. 9.136 Ix 5 175.5 3 1023 kg?m2; Iy 5 309.1023 kg?m2; Iz 5 154.4 3 1023 kg?m2. 9.138 Ix 5 334 3 1026 lb?ft?s2; Iy 5 Iz 5 1.356 3 1023 lb?ft?s2. 9.139 Ix 5 344 3 1026 lb?ft?s2; Iy 5 132.1 3 1026 lb·ft?s2; Iz 5 453 3 1026 lb·ft?s2. 9.141 (a) 13.99 3 1023 kg?m2. (b) 20.6 3 1023 kg?m2. (c) 14.30 3 1023 kg?m2. 9.142 Ix 5 28.3 3 1023 kg?m2; Iy 5 183.8 3 1023 kg?m2; kx 5 42.9 mm; ky 5 109.3 mm. 9.143 30.5 3 1023 lb?ft?s2. 9.145 (a) 26.4 3 1023 kg?m2. (b) 31.2 3 1023 kg?m2. (c) 8.58 3 1023 kg?m2. 9.147 Ix 5 0.0392 lb?ft?s2; Iy 5 0.0363 lb?ft?s2; Iz 5 0.0304 lb?ft?s2. 9.148 Ix 5 0.323 kg?m2; Iy 5 Iz 5 0.419 kg?m2. 9.149 Ixy 5 2.50 3 1023 kg?m2; Iyz 5 4.06 3 1023 kg?m2; Izx 5 8.81 3 1023 kg?m2. 9.150 Ixy 5 286 3 1026 kg?m2; Iyz 5 Izx 5 0. 9.151 Ixy 5 21.726 3 1023 lb?ft?s2; Iyz 5 0.507 3 1023 lb?ft?s2; Izx 5 22.12 3 1023 lb?ft?s2. 9.152 Ixy 5 2538 3 1026 lb?ft?s2; Iyz 5 2171.4 3 1026 lb?ft?s2; Izx 5 1120 3 1026 lb?ft?s2.
CHAPTER 9 9.1 a3b/30. 9.2 3a3b/10. 9.3 b3h/12. 9.4 a3b/6. 9.6 ab3/6. 9.8 3ab3/10. 9.9 ab3/15. 9.10 ab3/15. 9.11 0.1056 ab3. 9.12 3.43 a3b. 9.15 3a3/35; b29/35. 9.16 0.0945ah3; 0.402h. 9.17 3a3b/35; a29/35. 9.18 31a3h/20; a293/35. 9.21 20a4; 1.826a. 9.22 4ab(a2 1 4b2)/3; 21a2 1 4b22/3. 9.23 64a4/15; 1.265a. 9.25 (π/2)(R2 4 2 R1 4); (π/4)(R2 4 2 R1 4). 9.26 (b) for t/Rm 51, 210.56 %; for t/Rm 5 1/2, 22.99%; for t/Rm 51/10; 20.1250 %. 9.28 bh (12h2 1 b2)/48; 2112h2 1 b22/24. 9.31 390 3 103 mm4; 21.9 mm. 9.32 46.0 in4; 1.599 in. 9.33 64.3 3 103 mm4; 8.87 mm. 9.34 46.5 in4; 1.607 in. 9.37 JB 5 1800 in4; JD 5 3600 in4. 9.39 3000 mm2; 325 3 103 mm4. 9.40 24.6 3 106 mm4. 9.41 Ix 5 13.89 × 106 mm4; Iy 5 20.9 3 106 mm4. 9.42 Ix 5 479 × 103 mm4; Iy 5 149.7 3 103 mm4. 9.43 Ix 5 191.3 in4; Iy 5 75.2 in4. 9.44 Ix 5 18.13 in4; Iy 5 4.51 in4. 9.47 (a) 11.57 3 106 mm4. (b) 7.81 3 106 mm4. 9.48 (a) 12.16 3 106 mm4. (b) 9.73 3 106 mm4. 9.49 Ix 5 186.7 3 106 mm4; kx 5 118.6 mm; Iy 5 167.7 3 106 mm4. ky 5 112.4 mm. 9.50 Ix 5 44.5 in4; kx 5 2.16 in.; Iy 5 27.7 in4; ky 5 1.709 in. 9.51 Ix 5 250 in4; kx 5 4.10 in.; Iy 5 141.9 in4; ky 5 3.09 in. 9.52 Ix 5 260 3 106 mm4; kx 5 144.6 mm; Iy 5 17.53 mm4; ky 5 37.6 mm. 9.54 Ix 5 745 3 106 mm4; Iy 5 91.3 3 106 mm4. 9.55 Ix 5 3.55 3 106 mm4; Iy 5 49.8 3 106 mm4. 9.57 h/2. 9.58 15h/14. 9.59 3πr/16. 9.60 4h/7. 9.63 5a/8. 9.64 80.0 mm. 9.67 a4/2. 9.68 b2h2/4. 9.69 a2b2/6. 9.71 21.760 3 106 mm4. 9.72 2.40 3 106 mm4. 9.74 20.380 in4. 9.75 471 3 103 mm4. 9.76 29010 in4. 9.78 2.54 3 106 mm4. 9.79 (a) Ix9 5 0.482a4; Iy9 5 1.482a4; Ix9y9 5 20.589a4. (b) Ix9 5 1.120a4; Iy9 5 0.843a4; Ix9y9 5 0.760a4. 9.80 Ix9 5 2.12 3 106 mm4; Iy9 5 8.28 3 106 mm4; Ix9y9 5 20.532 3 106 mm4.
AN10 Answers to Problems 10.19 85.2 lb·ft i. 10.20 22.8 lb d 70.0°. 10.23 38.7°. 10.24 68.0°. 10.27 36.4°. 10.28 67.1°. 10.30 25.0°. 10.31 39.7° and 69.0°. 10.32 390 mm. 10.33 330 mm. 10.35 38.7°. 10.36 52.4°. 10.37 22.6°. 10.38 51.1°. 10.39 59.0°. 10.40 78.7°, 324°, 379°. 10.43 12.03 kN R. 10.44 20.4°. 10.45 2370 lb a. 10.46 2550 lb a. 10.48 300 N?m, 81.8 N?m. 10.49 η 5 1/(1 1 μ cot α). 10.50 η 5 tan θ/tan (θ 1 fs). 10.52 37.6 N, 31.6 N. 10.53 A 5 250 Nx; MA 5 450 N?m l. 10.54 1050 Nx . 10.57 0.833 in.w. 10.58 0.625 in. y. 10.60 25.0°. 10.61 39.7° and 69.0°. 10.62 390 mm. 10.69 θ 5 245.0°, unstable; θ 5 135.0°, stable. 10.70 θ 5 263.4°, unstable; θ 5 116.6°, stable. 10.71 θ 5 90.0° and θ 5 270°, unstable; θ 5 22.0° and θ 5 158.0°, stable. 10.72 θ 5 0 and θ 5 180.0°, unstable; θ 5 75.5° and θ 5 284°, stable. 10.73 59.0°, stable. 10.74 78.7°, stable; 324°, unstable; 379°, stable. 10.77 357 mm. 10.78 252 mm. 10.80 9.39° and 90.0°, stable; 34.2°, unstable. 10.81 17.11°, stable; 72.9°, unstable. 10.83 49.1°. 10.86 16.88 m. 10.87 54.8°. 10.88 37.4°. 10.89 P , kl/2. 10.91 k . 6.94 lb/in. 10.92 15.00 in. 10.93 P , 2kL/9. 10.94 P , kL/18. 10.96 P , 160.0 N. 10.98 P , 764 N. 10.100 (a) P , 10.00 lb. (b) P , 20.0 lb. 10.101 60.0 lbw. 10.102 600 lb?in. i. 10.103 500 Nx . 10.105 M 5 7Pa cos θ 10.107 19.40°. 10.108 7.13 in. 10.110 θ 5 0, unstable; θ 5 137.8°, stable. 10.112 (a) 22.0°. (b) 30.6°. 9.155 Ixy 5 28.04 3 1023 kg?m2; Iyz 5 12.90 3 1023 kg?m2; Izx 5 94.0 3 1023 kg?m2. 9.156 Ixy 5 0; Iyz 5 48.3 × 1026 kg?m2; Izx 5 24.43 3 1023 kg?m2. 9.157 Ixy 5 47.9 3 1026 kg?m2; Iyz 5 102.1 3 1026 kg?m2; Izx 5 64.1 3 1026 kg?m2. 9.158 Ixy 5 2m9 R1 3/2; Iyz 5 m9 R1 3/2; Izx 5 2m9 R2 3/2. 9.159 Ixy 5 wa3(1 2 5π)g; Iyz 5 211π wa3/g; Izx 5 4wa3(1 1 2π)/g. 9.160 Ixy 5211wa3/g; Iyz 5 wa3(π 1 6)/2g; Izx 5 2wa3/4g. 9.162 (a) mac/20. (b) Ixy 5 mab/20; Iyz 5 mbc/20. 9.165 18.17 3 1023 kg?m2. 9.166 11.81 3 1023 kg?m2. 9.167 5 Wa2/18g. 9.168 4.41 γta4/g. 9.169 281 3 1023 kg?m2. 9.170 0.354 kg?m2. 9.173 (a) 1/ 23. (b) 27/12. 9.174 (a) b/a 5 2; c/a 5 2. (b) b/a 5 1; c/a 5 0.5. 9.175 (a) 2. (b) 22/3. 9.179 (a) K1 5 0.363ma2; K2 5 1.583ma2; K3 5 1.720ma2. (b) (θx)1 5 (θz)1 5 49.7°, (θy)1 5 113.7°; (θx)2 5 45.0° (θy)2 5 90.0°, (θz)2 5 135.0°; (θx)3 5 (θz)3 5 73.5°, (θy)3 5 23.7°. 9.180 (a) K1 5 14.30 3 1023 kg?m2; K2 5 13.96 3 1023 kg?m2; K3 5 20.6 3 1023 kg?m2. (b) (θx)1 5 (θy)1 5 90.0°, (θz)1 5 0°; (θx)2 5 3.42°, (θy)2 5 86.6°. (θz)2 5 90.0°; (θx)3 5 93.4°, (θy)3 5 3.43°, (θz)3 5 90.0° 9.182 (a) K1 5 0.1639Wa2/g; K2 5 1.054Wa2/g; K3 5 1.115Wa2/g. (b) (θx)1 5 36.7°, (θy)1 5 71.6°; (θz)1 5 59.5°; (θx)2 5 74.9°, (θy)2 5 54.5°, (θz)2 5 140.5°; (θx)3 5 57.5°, (θy)3 5 138.8°, (θz)3 5 112.4° 9.183 (a) K1 5 2.26γta4/g; K2 5 17.27γta4/g; K3 5 19.08γta4/g. (b) (θx)1 5 85.0°, (θy)1 5 36.8°, (θz)1 5 53.7°; (θx)2 5 81.7°, (θy)2 5 54.7°; (θz)2 5 143.4°; (θx)3 5 9.70°, (θy)3 5 99.0°, (θz)3 5 86.3°. 9.185 Ix 5 16ah3/105; Iy 5 ha3/5. 9.186 πa3b/8; a/2. 9.188 Ix 5 1.874 3 106 mm4; Iy 5 5.82 3 106 mm4. 9.189 (a) 3.13 3 106 mm4. (b) 2.41 3 106 mm4. 9.191 22.81 in4. 9.193 (a) ma2/3. (b) 3ma2/2. 9.195 Ix 5 0.877 kg?m2; Iy 5 1.982 kg?m2; Iz 5 1.652 kg?m2. 9.196 0.0442 lb?ft?s2.
CHAPTER 10 10.1 65.0 Nw. 10.2 132.0 lb y. 10.3 39.0 N·m i. 10.4 1320 lb?in. l. 10.5 (a) 60.0 N C, 8.00 mmw. (b) 300 N C, 40.0 mmw. 10.6 (a) 120.0 N C, 16.00 mmw. (b) 300 N C, 40.0 mmw. 10.9 Q 5 3 P tan θ. 10.10 Q 5 P[(l/a)] cos3 θ 2 1]. 10.12 Q 5 2 P sin θ/cos (θ/2). 10.14 Q 5 (3P/2) tan θ. 10.15 M 5 Pl/2 tan θ. 10.16 M 5 Pl(sin θ 1 cos θ). 10.17 M 5 1 2Wl tan α sin θ. 10.18 (a) M 5 Pl sin 2θ. (b) M 5 3Pl cos θ. (c) M 5 Pl sin θ.
Photo Credits CHAPTER 1 Opener: ©Renato Bordoni/Alamy; Photo 1.1: NASA; Photo 1.2: NASA/JPL-Caltech.
CHAPTER 2 Opener: ©Getty Images RF; Photo 2.1: ©DGB/Alamy; Photo 2.2: ©Michael Doolittle/Alamy; Photo 2.3: ©Flirt/ SuperStock; Photo 2.4: ©WIN-Initiative/Neleman/Getty Images.
CHAPTER 3 Opener: ©St Petersburg Times/ZumaPress/Newscom; Fig. 3.11a: ©Gavela Montes Productions/Getty Images RF; Fig. 3.11b: ©Image Source/Getty Images RF; Fig. 3.11c: ©Valery Voennyy/Alamy RF; Fig. 3.11d: ©Monty Rakusen/Getty Images RF; Photo 3.1: ©McGraw-Hill Education/Photo by Lucinda Dowell; Photo 3.2: ©Steve Hix; Photo 3.3: ©Jose Luis Pelaez/Getty Images; Photo 3.4: ©Images-USA/Alamy RF; Photo 3.5: ©Dana White/PhotoEdit.
CHAPTER 4 Opener: ©View Stock/Getty Images RF; Photos 4.1, 4.2: ©McGraw-Hill Education/Photos by Lucinda Dowell; Fig. 4.1(Rocker bearing): Courtesy Godden Collection. National Information Service for Earthquake Engineering, University of California, Berkeley; Fig. 4.1(Links): Courtesy Michigan Department of Transportation; Fig. 4.1(Slider and rod): ©McGraw-Hill Education/Photo by Lucinda Dowell; Fig. 4.1(Pin support): Courtesy Michigan Department of Transportation; Fig. 4.1(Cantilever support): ©Richard Ellis/Alamy; Photo 4.3: ©McGraw-Hill Education/Photo by Lucinda Dowell; Photo 4.4: Courtesy of SKF, Limited.
CHAPTER 5 Opener: ©Akira Kaede/Getty Images RF; Photo 5.1: ©Christie’s Images Ltd./SuperStock; Photo 5.2: ©C Squared Studios/Getty Images RF; Photo 5.3: ©Michel de Leeuw/Getty Images RF; Photo 5.4: ©maurice joseph/Alamy; Fig. 5.18: ©North Light Images/agefotostock; Photo 5.5: NASA/Carla Thomas.
CHAPTER 6 Opener: ©Lee Rentz/Photoshot; Photo 6.1a: ©Datacraft Co Ltd/Getty Images RF; Photo 6.1b: ©Fuse/Getty Images RF; Photo 6.1c: ©Design Pics/Ken Welsh RF; Photo 6.2: Courtesy Godden Collection. National Information Service for Earthquake Engineering, University of California, Berkeley; Photo 6.3: Courtesy of Ferdinand Beer; Photo 6.4: ©McGraw-Hill Education/ Photo by Sabina Dowell; Photo 6.5: ©James Hardy/PhotoAlto RF; Photo 6.6: ©Mark Thiessen/National Geographic Society/Corbis; Photo 6.7: ©Getty Images RF.
CHAPTER 7 Opener: ©Jonatan Martin/Getty Images RF; Photo 7.1: ©McGraw-Hill Education/Photo by Sabina Dowell; Fig. 7.6(Continuous beam): ©Ross Chandler/Getty Images RF; Fig. 7.6(Overhanging beam): ©Goodshoot/Getty Images RF; Fig. 7.6(Cantilever beam): ©Ange/ Alamy; Photo 7.2: ©Alan Thornton/Getty Images; Photo 7.3: ©Bob Edme/AP Photo; Photo 7.4: ©Thinkstock/Getty Images RF; Photo 7.5a: ©Ingram Publishing/Newscom; Photo 7.5b: ©Karl Weatherly/Getty Images RF; Photo 7.5c: ©Eric Nathan/Alamy.
CHAPTER 8 Opener: ©Bicot (Marc Caya); Photo 8.2: ©Tomohiro Ohsumi/ Bloomberg/Getty Images; Photo 8.3: ©Leslie Miller/agefotostock; Photo 8.4: Courtesy of REMPCO Inc.; Photo 8.5: ©Bart Sadowski/Getty Images RF; Photo 8.6: ©Fuse/Getty Images RF.
CHAPTER 9 Opener: ©ConstructionPhotography.com/Photoshot; Photo 9.1: ©Barry Willis/Getty Images; Photo 9.2: ©loraks/Getty Images RF.
CHAPTER 10 Opener: ©Tom Brakefield/SuperStock; Photo 10.1: Courtesy Brian Miller; Photo 10.2: Courtesy of Altec, Inc.; Photo 10.3: Courtesy of DE-STA-CO.
C1 This page intentionally left blank I1 A Addition of couples, 123–124 of forces, 4, 32–33 of vectors, 18–20 parallelogram law for, 4, 18 polygon rule for, 19–20 summing x and y components, 32–33 triangle rule for, 19 Analysis, see Structural analysis Angles formed by two vectors, 106, 113 lead, 451 of friction, 433–434 of repose, 434 Arbitrary shaped bodies, moments of inertia of, 552–553, 556 Area centroid of common, 238 composite, 240 first moment of, 231, 235–237, 244, 250 integration centroids determined by, 249–250 moments of inertia determined by, 488–489, 494 moment of inertia, 487–494, 498–506, 513–519 for a rectangular area, 488–489 for composite areas, 499–506 for hydrostatic force system, 488, 506 of common geometric shapes, 500 polar moments, 490, 494 principal axis and moments, 514–516, 519 product of inertia, 513–514, 519 second moments, 487–494 transformation of, 513–519 using same strip elements, 489 radius of gyration, 490–491 theorems of Pappus-Guldinus, 250–252 two-dimensional bodies, 231, 235–237, 244 units of, 7–9 Axle friction, 459–460, 465 Axis moments of a force about, 84, 105–114 arbitrary point for, 109–110 given origin for, 108–109 mixed triple products, 107–108 scalar products, 105–106 neutral, 487 principal axis and moments of inertia about the centroid, 516 ellipsoid of inertia, 550–552 for a body of arbitrary shape, 552–553, 556 of a mass, 551–553, 557 of an area, 514–516, 519 B Ball and socket supports, 206 Basic units, 6 Beams bending moment in, 379–381, 385, 391–399 centroids of, 262–264, 267 classification of, 378 loading conditions concentrated, 262–263, 378 distributed, 262–264, 267, 378 internal forces and, 368, 378–386 uniformly distributed, 378 pure bending, 487 shear and bending moment diagrams for, 381, 386 shearing forces, 379–381, 385, 391–399 span, 379 Belt friction, 469–474 Bending moments, 370–373 beams, 379–381, 385, 391–399 diagrams for, 381, 386 external forces and, 380 internal forces as, 368, 370–373 shearing force relations with, 392–393 Bracket supports, 206 C Cable supports, 173, 206 Cables, 368, 403–410, 416–420 catenary, 416–420 internal forces of, 368, 403–410 parabolic, 405–406, 410 sag, 405 solutions for reactions, 409–410, 420 span, 405 supporting concentrated loads, 403–404, 409 supporting distributed loads, 404–405, 410, 416–420 supporting vertical loads, 403–404, 409 Catenary cables, 416–420 Center of gravity, 84 composite area, 240 composite bodies, 275 composite plate, 239–240 location of, 232–233 problem solving with, 239–244 three-dimensional bodies, 273–275, 282 two-dimensional bodies, 231–233 Center of pressure, 263 Centroid, 231, 233–235 distributed load problems using, 262–268 integration for determination of, 249–250, 277 location of, 233–235, 244 of areas, 233–235, 238, 244, 249–250, 262–268 of common shapes, 238–239, 276 of lines, 233–235, 239, 244 of volume, 274–277 theorems of Pappus-Guldinus, 250–252 three-dimensional bodies, 274–277 two-dimensional bodies, 233–235, 238–239 Circle of friction, 460, 465 Coefficients of friction, 432–433 kinetic, 432 rolling resistance, 463, 465 static, 432 Collar bearings, 460–461 Commutative property, 88, 105 Composite bodies, 275–276 center of gravity of, 275 centroid of, 275–276 mass moment of inertia for, 533–540, 556 Composite plates and wires, 237–240 Compound truss, 318–319 Compressibility of fluids, 2 Compression, deformation from, 86 Concentrated loads beams, 262–263, 378 cables supporting, 403–404, 409 Concurrent forces resultants, 20, 57 system reduction of, 138 Connections, 172–174. See also Support reactions Conservative force, 597 Constraining forces, 172, 176–177, 205 completely constrained, 176 free-body diagram reactions, 172 improperly constrained, 177, 205 partially constrained, 176, 205 three-dimensional rigid bodies, 205 two-dimensional rigid bodies, 176–177 Conversion of units, 10–12 Coplanar forces resultants, 19–20 system reduction of, 139–140 Coplanar vectors, 19–20 Coulomb friction, see Dry friction Couples, 120–128 addition of, 123–124 equivalent, 121–123 force-couple system resolution, 124–125 moment of, 120–121 work of, 577 Cross product, see Vector product D Deformable bodies, mechanics of, 2 Deformation, 86–87 internal forces and, 86 principle of transmissibility for prevention of, 86–87 Degrees of freedom, 600 Derived units, 6 Dick clutch, 465 Direction cosines, 53, 55 Direction of a force, 17, 31. See also Line of action Disk friction, 460–462, 465 Index I2 Index Displacement finite, 595–597 of a particle, 575 virtual, 577–578, 585 work and, 577–578, 585, 595–597 Distributed forces, 230–296. See also Centroids beam loads, 262–264, 267, 378 cables supporting loads, 404–405, 410, 416–120 concentrated load and, 262–263 integration methods for centroid location, 249–257, 277, 282 moments of inertia, 485–572 of areas, 487–494, 498–506 of masses, 487 polar, 486, 490, 494 transformation of, 513–519 submerged surfaces, 263, 265–268 theorems of Pappus-Guldinus, 250–252 three-dimensional bodies, 273–282 center of gravity, 273–275, 282 centroid of volume location, 274–277, 282 composite bodies, 275–276 two-dimensional bodies, 232–244 center of gravity, 232–233, 244 centroid of area and line location, 233–235, 238–239, 244 composite plates and wires, 237–240 first moment of an area or line, 235–237, 244 planar elements, 232–244 Distributive property, 88 Dot product, 105. See also Scalar products Double integration, 249, 277 Dry friction, 430–441 angles of, 433–434 coefficients of, 432–433 kinetic friction force, 431–432 laws of, 431 problems involving, 434–441 static friction force, 431–432 E Efficiency, mechanical, 580–581 Elastic force, 596–597, 602 Ellipsoid of inertia, 550–552 End bearings, 460–461 Equal and opposite vectors, 18 Equilibrium, 39 equations of, 39–40 force relations and, 16, 39–45 frame determinacy and, 332–333 free-body diagrams for, 40–41, 170–172 Newton’s first law of motion and, 40 neutral, 599 of a particle, 39–40, 66–74 three-dimensional (space) problems, 66–74 two-dimensional (planar) problems, 39–40 of rigid bodies, 169–229 statically determinate reactions, 176 statically indeterminate reactions, 176–177, 205 support reactions, 172–174, 204–206 three-dimensional structures, 204–213 three-force body, 196–198 two-dimensional structures, 172–183 two-force body, 195, 198 principle of transmissibility and, 4 stable, 599–600 unstable, 599–600 virtual work conditions, 598–602 potential energy and, 598–599, 602 stability and, 599–602 Equipollent systems, 138 Equivalent couples, 121–123 Equivalent systems of forces, 82–168 deformation and, 86–87 external, 84–85 internal, 84, 86–87 point of application, 84–85 principle of transmissibility and, 83, 85–87 reduction to force-couple system, 136–137 rigid bodies, 82–168 simplifying, 136–150 weight and, 84–85 External forces, 84–85 equivalent systems and, 84–85 shear and bending moment conventions, 380 F First moment of an area or line, 231, 235–237, 244 of volume, 274 Fixed (bound) vector, 18 Fixed supports, 174, 206 Fluid friction, 430–431 Fluids, 2 Force. See also Distributed forces; Equivalent forces; Force systems concept of, 3 concurrent, 20, 138 conservative, 597 constraining, 172, 176–177, 205 conversion of units of, 11 coplanar, 19–20, 139–140 direction, 17, 31 elastic, 597 equilibrium and, 16, 39–45 equivalent, 85 friction, 430–431 gravity, 597 input (machines), 348 kinetic friction, 431–432 line of action (direction), 17, 56–57 magnitude, 17, 52, 56–57 output (machines), 348 parallel, 140–141 parallelogram law for addition of, 4 particle equilibrium and, 15–81 planar (two-dimensional), 16–51 concurrent force resultants, 20 parallelogram law for, 17 rectangular components, 29–32 resolution into components, 20–21 resultant of two forces, 17 scalar components, 30 summing x and y components, 32–33 unit vectors for, 29–32 point of application, 17 scalar representation, 18, 20 sense of, 17 static friction, 431–432 three dimensions of space, 52–74 concurrent force resultants, 57 rectangular components, 52–55 scalar components, 53 unit vectors for, 54–55 vector representation, 17–20 weight, 4–5 work of, 575–577, 595–597 Force-couple systems, 124–125 conditions for, 137–138 equipollent, 138 equivalent systems reduced to, 137–138 reactions equivalent to, 174 reducing a system of forces into, 136–137 resolution of a given force into, 124–125, 128 resultant couples, 138–141 wrench, 141–142 Force systems, 82–168 center of gravity, 84 concurrent, 138 coplanar, 139–140 couples, 120–128 equipollent, 138 equivalent, 84–87, 136–150 external forces, 84–85 force-couple, 124–125 internal forces, 84–87, 298–299 moment about a point, 83 moment about an axis, 84 parallel, 140–141 point of application, 84–85 position vectors defining, 90, 136 reducing into force-couple system, 136–137 resolution of a given force into force-couple system, 124–125, 128 simplifying, 136–150 virtual work application to connected rigid bodies, 578–580 weight and, 84–85 Force triangle, 41 Frames, 299, 330–338 collapse of without supports, 332–333 equilibrium of forces, 332–333 free-body diagrams of force members, 330–332 multi-force members, 299, 330 statically determinate and rigid, 333 statically indeterminate and nonrigid, 333 Free-body diagrams, 13 equilibrium particle force, 40–41 rigid-body force, 170–172 frame analysis using, 330–332 machine analysis of members, 348, 351 truss analysis of joint forces, 303 two-dimensional problems, 40–41, 170–172 Free vector, 18 Friction, 429–484 angles of, 433–434 axle, 459–460, 465 belt, 469–474 circle of, 460 coefficients of, 432–433 disk, 460–462, 465 dry, 430–441 fluid, 430–431 forces of, 430–431 I3 Index journal bearings, 459–460, 465 lubrication and, 431, 459 rolling resistance, 462–463, 465 screws, 450–451, 453–454 slipping, 470–471, 474 thrust bearings, 459, 460–462, 465 wedges, 450, 452, 454 wheel, 462–463, 465 Frictionless pins, 173–174 Frictionless surface supports, 173, 206 Fundamentals of Engineering Exam, A1 G Gravitational units, 9 Gravity (weight) potential energy with respect to, 597, 602 work of, 596 H Hydraulics, 2 Hydrostatic force system, 488, 506 I Impending motion, 432–433, 441 Impending slip, 470 Input forces, 348, 580 Integration centroids determined by of an area, 249–250 of volume, 277 double, 249, 277 moments of inertia determined by for a body of revolution, 533, 540 for a rectangular area, 488–489 for a three-dimensional body, 533 of a mass, 533, 540 of an area, 488–489, 494 using the same elemental strips, 489 theorems of Pappus-Guldinus applied to, 250–252 triple, 277 Internal forces, 84, 367–428 axial forces as, 369, 370 beams, 368, 378–386 bending moments, 368, 370, 379–381, 385, 391–399 cables, 368, 403–410, 416–420 deformation and, 86–87 equivalent systems and, 84, 86–87 in compression, 86, 368 in members, 368–373 in tension, 86, 368 loadings, 378–379, 391–399, 403–405 principle of transmissibility for equilibrium of, 86–87 relations among load, shear, and bending moments, 391–399 rigid bodies, 84, 86–87 shear and bending moment diagrams for, 381, 386 shearing forces as, 368, 370, 379–381, 385, 391–399 structural analysis and, 298–299 International System of Units (SI), 6–9 J Joints under special loading conditions, 304–306 Journal bearings, axle friction of, 459–460, 465 K Kinetic friction force, 431–432 Kinetic units, 5–6 L Lead and lead angle, 451, 454 Length, conversion of units of, 10–11 Line of action, 17 force direction representation, 4, 17, 56–57 magnitude and, 3–4, 17 moment of a force, 91–92 particles, 17, 54–57 planar (two dimensional) force, 17 reactions with known, 172–173 rigid bodies, 91–92, 172–173 three-dimensional (space) force, 54–57 unit vector along, 54–55 Lines centroid of common, 239 first moment of, 231, 235–237, 244 two-dimensional bodies, 231, 235–237, 244 Loading conditions beams, 262–264, 267, 368, 378–386 cables, 368, 403–410, 416–420 center of pressure, 263 centroid of the area, 262–268 concentrated, 262–263, 378, 403–404, 409 distributed, 262–268, 378, 404–405, 410, 416–120 relations with shear and bending moments, 391–399 submerged surfaces, 263, 265–268 uniformly distributed, 378 Lubrication, friction and, 431, 459 M Machines free-body diagrams of members, 348, 351 input forces, 348 mechanical efficiency of, 580–581 multi-force members, 299, 330 output forces, 348 structural analysis of, 299, 330, 348–351 Magnitude of a force force characteristics, 3–4, 17 line of action and, 4, 17, 56–57 moments of a force, 90–92 particles, 17, 52, 56–57 reactions with unknown direction and, 173–174 rigid bodies, 90–92 units of, 71 vector characteristics, 52 Mass concept of, 3 conversion of units of, 11 moments of inertia, 487, 529–540 integration used to determine, 533, 540 of common geometric shapes, 500, 534 of composite bodies, 533–540, 556 of simple mass, 529–530 of thin plates, 532–533 parallel-axis theorem for, 530–531, 539 principal axis and moments, 551–553, 557 product of inertia, 549–550, 556 Mechanical efficiency of machines, 580–581 Mechanics conversion of units, 10–12 fundamental concepts and principles, 3–5 method of solving problems, 12–14 newtonian, 3 numerical accuracy, 14 of deformable bodies, 2 of fluids, 2 of particles, 3–4 of rigid bodies, 2, 4 relativistic, 3 role of statics and dynamics in, 2 study of, 2–3 systems of units, 5–10 Members axial forces in, 369, 370 free-body diagrams of, 330–332 internal forces in, 368–373 machine analysis of, 348, 351 multi-force, 299, 330, 369–370 redundant, 319 shearing force in, 370 two-force, 299, 300, 370 zero-force, 305 Method of joints, 302–309 Method of sections, 317–323 Mixed triple products of vectors, 107–108 Mohr’s circle for moments of inertia, 523–526 Moment arm, 91. See also Line of Action Moments, see Moments of a force; Moments of inertia Moments of a force about a point, 83, 90–99 line of action (moment arm), 91–92 magnitude of, 90–92 position vector of, 90 rectangular components of, 93–94 right-hand rule for, 90 three-dimensional problems, 93–94, 99 two-dimensional problems, 92–93, 94, 99 Varignon’s theorem for, 93 vector products, 90 about an axis, 84, 105–114 angles formed by two vectors, 106, 113 arbitrary point for, 109–110 given origin for, 108–109 mixed triple products, 107–108 perpendicular distance between lines, 109, 113–114 projection of a vector for, 106, 113 scalar products, 105–106 of a couple, 120–121 Moments of inertia, 485–572 integration used to determine, 488–489, 494, 533, 540 Mohr’s circle for, 523–526 neutral axis, 487 of arbitrary shaped bodies, 552–553, 556 of areas, 487–494, 498–506 for composite areas, 499–506 for hydrostatic force system, 488, 506 of common geometric shapes, 500 I4 Index of masses, 487, 529–540 for composite bodies, 533–540, 556 for thin plates, 532–533, 539 of common geometric shapes, 500, 534 of simple mass, 529–530 parallel-axis theorem for, 498–506, 514, 530–531, 539, 550 polar, 486, 490, 494 radius of gyration, 490–491, 494, 530 transformation of, 513–519, 549–556 ellipsoid of inertia, 550–552 mass product of inertia, 549–550, 556 principal axis and moments, 514–516, 519, 551–553, 557 product of inertia, 513–514, 519 second moment as, 486–487, 494 unit-related errors, 539 Motion external forces and, 84–85 impending, 432–433, 441, 470 Newton’s first law of, 4, 40 particles, 40 relative, 435 rigid bodies, 84–85 rotation, 85 translation, 85 slipping, 470–471, 474 weight and, 84–85 Multi-force members, 299, 330, 369–370. See also Frames; Machines N Neutral axis, 487 Neutral equilibrium, 599, 602 Newtonian mechanics, 3 Newton’s laws, 4–5 first law of motion, 4, 40 gravitation, 4–5 motion, 4, 40 particles in equilibrium and, 40 second law of motion, 4 third law of motion, 4, 299 Nonrigid truss, 319 Numerical accuracy, 14 O Output forces, 348, 580 Overrigid truss, 319 P Pappus-Guldinus, theorems of, 250–252 Parabolic cables, 405–406, 410 Parallel-axis theorem composite area application of, 499–506 for mass moments of inertia, 530–531, 539 for mass product of inertia, 550 for moments of inertia of an area, 498–506 for product of inertia, 514 Parallel forces, reduction of system of, 140–141 Parallelogram law, 4 addition of two vectors, 18 addition of forces, 4 resultant of two forces, 17 Particles, 3–4 direction of a force, 17, 31 displacement of, 575 line of action (direction), 17, 56–57 magnitude of force, 17, 52, 56–57 mechanics of, 3–4 resultant of forces, 16–17, 20 scalars for force representation, 18, 20, 30 statics of, 15–81 three-dimensional (space) problems, 52–74 adding forces in space, 52–62 concurrent force resultants, 57 direction cosines for, 53, 55 equilibrium of, 39–45 force defined by magnitude and two points, 56–57 rectangular components, 52–55 two-dimensional (planar) problems, 16–50 adding forces by components, 29–35 concurrent force resultants, 20 equilibrium of, 39–45 free-body diagrams, 40–41 Newton’s first law of motion for, 40 planar forces in, 16–25 rectangular components, 29–32 resolving several forces into two components, 32–33 unit vectors for, 29–32, 54–55 vectors for force representation, 17–20, 29–30 Perpendicular distance between lines, 109, 113–114 Pin supports, 173–174, 206 Pitch, 451, 454 Planar forces, 16–51 equilibrium of, 39–45 line of action, 17 magnitude of, 17 parallelogram law for, 17 rectangular components, 29–32 resolution into components, 20–21 resultant of several concurrent forces, 20 resultant of two forces, 17 scalar components, 30 scalar representation of, 18, 20 summing x and y components, 32–33 unit vectors for, 29–32 vector representation of, 17–20 Plates center of gravity for, 239–240 circular, 533 composite, 239–240 mass moment of inertia for, 532–533 rectangular, 533 thin, 532–533, 539 Point of application, 17, 84–85 Polar moment of inertia, 486, 490, 494 Position vector, 90, 136 Potential energy equations of, 597–598 equilibrium and, 598–599, 602 virtual work and, 574, 597–599, 602 with respect to elastic force (a spring), 597, 602 with respect to gravity (weight), 597, 602 Principal axis and moments of inertia about the centroid, 516 ellipsoid of inertia, 550–552 for a body of arbitrary shape, 552–553, 556 of a mass, 551–553, 557 of an area, 514–516, 519 Principle of transmissibility, 4, 83, 85–87 equivalent forces of, 85–87 rigid-body applications, 83, 85–87 sliding vectors from, 83, 85 Principle of virtual work, 574, 577–580, 585 application of, 578–580 virtual displacement, 577–578, 585 Problems, 12–14 error detection, 13–14 force triangle, 41 free-body diagrams for, 13, 40–41 methods for solving, 12–14 SMART method for solving, 13 solution basis, 12–13 space diagram for, 40 Product of inertia, 513–514, 519 Projection of a vector, 106, 113 Pure bending, 487 Q Quadratic surface equation, 550 R Radius of gyration, 490–491, 494, 530 Reactions, 172 constraining forces, 172, 176–177 equilibrium of rigid bodies and, 172–174, 204–206 equivalent to force and couple, 174 free-body diagrams showing, 172 support, 172–174, 204–206 three-dimensional structures, 204–206 two-dimensional structures, 172–176 with known line of action, 172–173 with unknown direction and magnitude, 173–174 Rectangular components moments of a force, 93–94 particles, 29–32, 52–55 planar (two-dimensional) forces, 29–32 rigid bodies, 88–90, 93–94 space (three-dimensional) forces, 52–55 unit vectors for, 29–32, 54–55 vector products, 88–90 Redundant members, 319 Relative motion, 435 Relativistic mechanics, 3 Resultant couples, 138–141 Resultant of forces, 16–17 concurrent, 20, 57 of several concurrent forces, 20 of two forces, 17 parallelogram law for, 17 particle statics, 17, 20 planar forces, 17, 20 statics and, 16 three-dimensional (space), 57 Revolution, mass moment of inertia for a body of, 533, 540 Right-hand rule, 87, 90 Right-handed triad, 87–88 Rigid bodies, 83, 82–168, 169–229 constraining forces, 172, 176–177, 205 Moments of inertia (Cont.) I5 Index couples, 120–128 equilibrium of, 169–229 statically determinate reactions, 176 statically indeterminate reactions, 176–177 support reactions for, 172–174, 204–206 three-dimensional structures, 204–213 three-force body, 196–198 two-dimensional structures, 172–183 two-force body, 195, 198 equivalent systems of forces center of gravity, 84 deformation and, 86–87 external forces, 84–85 internal forces, 84, 86–87 point of application, 84–85 reduction to force-couple system, 136–137 simplifying, 136–150 weight and, 84–85 force-couple systems equipollent, 138 equivalent systems reduced to, 137–138 reducing a system of forces into, 136–137 resolution of force into, 124–125, 128 resultant couples, 138–141 wrench, 141–142 free-body diagrams for, 170–172 mechanics of, 2, 4 moments of a couple, 120–128 of a force about a point, 83, 90–99 of a force about an axis, 84, 105–114 principle of transmissibility and, 4, 83, 85–87 reactions, 172–174 rectangular components, 88–90, 93–94 scalar (dot) products, 105–106 sliding vector representation, 18, 83 vector products, 87–90, 105–108 virtual work application to systems of connected, 578–580 Rigid truss, 301 Rocker supports, 172–173 Roller supports, 172–173 Rolling resistance, 462–463, 465 Rotation, 85 Rough surface supports, 173, 206 S Sag, 405 Scalar products of vectors, 105–106 Scalars, 18 particle force representation, 18 product of vector and, 20 rectangular force components, 30, 53 Screws, 450–451, 453–454 friction and, 450–451, 453–454 lead and lead angle, 451, 454 pitch, 451, 454 self-locking, 451 square-threaded, 450–451, 454 Self-locking screws, 451 Shearing forces, 370–373 beams, 368, 370, 379–381, 385, 391–399 bending moment relations with, 392–393 diagrams for, 381, 386 external forces and, 380 internal forces as, 368, 370–373 load relations with, 391–392 Simple truss, 300–302, 306 Sliding vectors, 18, 83, 85 Slipping, belt friction and, 470–471, 474 SMART method for solving problems, 13 Space, concept of, 3. See also Three-dimensional problems Space diagram, 40 Space truss, 306 Span, 379, 405 Springs (elastic force), work of, 596–597, 602 Square-threaded screws, 450–451 Stable equilibrium, 599–600, 602 Static friction force, 431–432 Statically determinate reactions, 176, 333 Statically indeterminate reactions, 176–177, 205, 333 Statics of particles, 15–81 resultant of forces, 16–17, 20 role of in mechanics, 2 state of equilibrium, 16 Structural analysis, 297–366 frames, 299, 330–338 internal force reactions, 298–299 machines, 299, 330, 348–351 multi-force members, 299, 330 Newton’s third law for, 299 trusses, 299–309, 317–324 two-force members, 299, 300 virtual work applications, 578–580 zero-force members, 305 Structures analysis of, 297–366 equilibrium of, 172–183, 204–213 statically determinate reactions, 176, 333 statically indeterminate reactions, 176–177, 205, 333 three-dimensional, 204–213 two-dimensional, 172–183 Submerged surfaces, distributed forces on, 263, 265–268 Support reactions, 172–174, 204–206 fixed, 174 frictionless pins, 173–174 of one unknown and one direction, 172–173 rollers and rockers, 172–173 static determinacy and, 333 three-dimensional structures, 204–206 two-dimensional structures, 172–174 Symmetry, planes of, 277 Systems of units, 5–12 converting between, 10–12 International System of Units (SI), 6–9 U.S. customary units, 9–10, 12 T Tension, deformation from internal forces of, 86 Theorems Pappus-Guldinus, 250–252 parallel-axis, 498–506, 514, 530–531, 539 Varignon’s, 93 Theory of relativity, 3 Thin plates, mass moment of inertia for, 532–533, 539 Three-dimensional bodies, 273–282 center of gravity, 273–275, 282 centroid of volume location, 274–277, 282 composite bodies, 275–276 Three-dimensional (space) problems, 52–74 adding forces in, 52–65 concurrent force resultants, 57 direction cosines for, 53, 55 equilibrium in, 66–74, 204–213 forces in, 52–74 line of action, 54–57 magnitude of force, 56–57 moments of a force about a point, 93–94, 99 particles, 52–74 rectangular components, 52–55 rigid bodies, 93–94, 99, 204–213 support reactions, 204–206 unit vector for, 54–55 Three-force body, equilibrium of, 196–198 Thrust bearings, disk friction of, 459, 460–462, 465 Time, concept of, 3 Translation, 85 Transmissibility, see Principle of Transmissibility Triangle rule for addition of vectors, 19 Triple integration, 277 Trusses, 299–309, 317–324 analysis of, 299–309, 317–324 compound, 318–319 free-body joint diagrams, 303 joints under special loading conditions, 304–306 method of joints, 302–309 method of sections, 317–323 nonrigid, 319 overrigid, 319 redundant members, 319 rigid, 301 simple, 300–302, 306 space, 306 two-force members, 299, 300 zero-force members, 305 Two-dimensional bodies, 232–244 center of gravity, 232–233, 244 centroid of area and line location, 233–235, 238–239, 244 composite plates and wires, 237–240 first moment of an area or line, 235–237, 244 planar elements, 232–244 Two-dimensional (planar) problems equilibrium in, 172–183 moments of a force, 92–93, 94, 99 rigid-body structures, 172–183 statically determinate reactions, 176 statically indeterminate reactions, 176–177 support reactions, 172–174 Two-force body, equilibrium of, 195, 198 Two-force members, 299, 300, 370. See also Trusses U Uniformly distributed loads, 378 Unit vectors, 29–32, 54–55 Units, 5–12 basic, 6 converting between systems, 10–12 derived, 6 I6 Index energy, 575 force, conversion of, 11 gravitational, 9 International System of Units (SI), 6–9 kinetic, 5–6 length, conversion of, 10–11 mass, conversion of, 11 of area and volume, 7–9 quantity equivalents of SI and U.S. customary, 12 SI abbreviations (formulas) of, 8 SI prefixes, 7 systems of, 5–10 U.S. customary, 9–10, 12 Universal joint supports, 206 Unstable equilibrium, 599–600, 602 U.S. customary units, 9–10, 12 V Varignon’s theorem, 93 Vector products, 87–90, 105–108 commutative property and, 88 distributive property and, 88 mixed triple products, 107–108 moment of force about a given axis, 105–108 moment of force about a point, 87–90 rectangular components of, 88–90 right-hand rule for, 87, 90 scalar products, 105–106 Vectors, 17–20 addition of, 18–20 parallelogram law for, 18 polygon rule for, 19–20 triangle rule for, 19 angle formed by, 106 coplanar, 19–20 equal and opposite, 18 fixed (bound), 18 force addition using, 17–20, 52–55 free, 18 mixed triple products, 107–108 moments of a force, 90, 105–114 about a given axis, 105–114 about a point, 90 negative, 18–19 particle force representation, 17–20 planar forces, 17–20 position, 90, 136 product of scalar and, 20 projection of, 106 rectangular force components, 30, 29–32 rigid-body representation, 83, 85 sliding, 18, 83, 85 three-dimensional forces, 53–55 subtraction of, 19 unit, 29–32, 54–55 Virtual work, 573–613 displacement of a particle, 575 equilibrium conditions, 598–602 mechanical efficiency of machines, 580–581 method of, 574–585 potential energy and, 574, 597–599, 602 principle of, 574, 577–580, 585 application to systems of connected rigid bodies, 578–580 virtual displacement, 577–578, 585 work during finite displacement, 595–597 input, 580 of a couple, 577 of a force, 575–577, 595–597 output, 580 virtual, 577–578, 585 Viscosity, see Fluid friction Volume, units of, 7–9 W Wedges, 450, 452, 454 Weight, 4–5 as a force, 4–5 center of gravity, 84 external force as, 84–85 gravity and, 596–597, 602 potential energy effected by, 597, 602 point of application, 84 rigid-body motion and, 84–85 work of, 596 Wheel friction, 462–463, 465 Work during finite displacement, 595–597 input, 580 of a couple, 577 of a force, 575–577, 595–597 of a spring (elastic force), 596–597 of a weight (gravity), 596 output, 580 virtual, 577–578, 585 Wrench, reduction of force-couple forces into, 141–142 Z Zero-force members, 305 Units (Cont.) |
190747 | https://home.cc.umanitoba.ca/~harland/Research/Carmichael.pdf | THE ITERATED CARMICHAEL LAMBDA FUNCTION NICK HARLAND Abstract. The Carmichael lambda function λ(n) is defined to be the smallest positive integer m such that am is congruent to 1 modulo n, for all a and n relatively prime. The function λk(n) is defined to be the kth iterate of λ(n). Previous results show a normal order for n/λk(n) where k = 1, 2. We will show a normal order for all k.
1. Introduction The Carmichael lambda function λ(n) is defined to be the order of the largest cyclic subgroup of the multiplicative subgroup (Z/nZ)×. It can be computed using the identity λ(lcm{a, b}) = lcm{λ(a), λ(b)} and its values at prime powers which are λ(pk) = φ(pk) = pk −pk−1 for odd primes p and λ(2) = 1, λ(4) = 2, and λ(2k) = φ(2k)/2 = 2k−2 for k ≥3.
Several properties of λ(n) were studied by Erd˝ os, Pomerance, and Schmutz in . In partic-ular they showed that λ(n) = n exp(−(1+o(1)) log log n log log log n) as n →∞for almost all n. Martin and Pomerance showed in that λ(λ(n)) = n exp(−(1+o(1))(log log n)2 log log log n) as n →∞for almost all n. The k–fold iterated Carmichael lambda function is defined re-cursively to be λ1(n) = λ(n), λk(n) = λ(λk−1(n)).
We define φk(n) similarly. In it is conjectured that λk(n) = n exp − 1 (k −1)!(1 + ok(1))(log log n)k log log log n for almost all n. In this paper we prove that conjecture.
Theorem 1. For fixed k, the normal order of log n λk(n) is 1 (k−1)!(log log n)k log log log n.
We’ll actually prove the theorem in the following slightly stronger form. Given any function ψ(x) = o(log log log x) and ψ(x) →∞as x →∞we have log n λk(n) = 1 (k −1)!(log log n)k log log log n + Ok ψ(n) for all but O(x/ψ(x)) integers up to x.
We will also turn our attention to finding an asymptotic formula involving iterates involv-ing λ and φ. Banks, Luca, Saidak and St˘ anic˘ a in showed that for almost all n, λ(φ(n)) = n exp(−(1 + o(1))(log log n)2 log log log n) and φ(λ(n)) = n exp(−(1 + o(1))(log log n) log log log n).
2010 Mathematics Subject Classification. 11N56 (11N37).
Key words and phrases. arithmetic functions, normal order, iterated Carmichael function.
1 As a corollary to Theorem 1 we will obtain asymptotic formulas for higher iterates involving λ and φ. Specifically we prove the following.
Theorem 2. For l ≥0 and k ≥1, let g(n) = φl(λ(f(n))), where f(n) is a (k −1) iterated arithmetic function consisting of iterates of φ and λ. Then the normal order of log(n/g(n)) is 1 (k−1)!(log log n)k log log log n.
An example of the use of this theorem is for φφλφφλλφ(n). Since l = 2, k = 5, we get that the normal order of log n φφλφφλλφ(n) is 1 4!(log log n)5 log log log n.
The proof of Theorem 1 involves breaking down n λk(n) in terms of the iterated Euler φ function by using (1) n λk(n) = n φ(n) φ(n) φ2(n) . . .
φk−1(n) φk(n) φk(n) λk(n) of which estimates for all but the last term are known. Hence log n λk(n) can be written as a sum of the logarithms on the right side of (1) and so we’ll analyze the term log(φk(n)/λk(n)).
The following notations and conventions will be used throughout the paper. The letters p, q, r will always denote primes and k ≥2 will be a fixed integer. Note that the theorem has already been proven for k = 1. Let vp(n) be the largest power of p which divides n, so that n = Y p pvp(n).
Let the set Pn be {p : p ≡1 (mod n)}. Throughout the paper we will assume x > eee and y = y(x) = log log x. Also let ψ(x) be any function going to ∞such that ψ(x) = o(log y) = o(log log log x). Whenever we use the phrase “for almost all n ≤x” in a result, we mean that the result is true for all n ≤x except a set of size O(x/ψ(x)). Lastly we note that any implicit constant may depend on k.
2. Required Estimates The following estimates will be used throughout the paper. We use the Chebeshev bound (2) X n≤x Λ(n) = X p≤x log p ≪x where Λ(n) is the von–Mangoldt function. We also require a formula of Mertens (See [7, Theorem 2.7(b)]) (3) X q≤x log q q = log x + O(1).
Using partial summation on (2) we can obtain the tail estimates (4) X q>x log q q2 ≪1 x 2 and (5) X q>x 1 q2 ≪ 1 x log x.
Given m, x, let A be the smallest a for which ma > x. We can then manipulate the sums X a∈N P(a) ma = 1 m ∞ X a=0 P(a) ma and X a∈N ma>x P(a) ma ≪1 x ∞ X a=0 P(a) ma−A = 1 x ∞ X a=A Q(a) ma for Q(x) = P(x + A). Then by noting that P∞ a=A P(a) ma ≪P 1 uniformly for m ≥2 and A ≥0 we obtain the estimates (6) X a∈N P(a) ma ≪P 1 m, X a∈N ma>x P(a) ma ≪P 1 x.
From [7, Corollary 1.15] we get (7) X s≤x 1 s = log x + O(1) from which it easily follows that (8) X D≤s≤x s≡a (mod C) 1 s ≪1 D + log x C .
We will also make frequent use of the Brun-Titchmarsh inequality [7, Theorem 3.9] (9) π(t; n, a) ≪ t φ(n) log(t/n).
By partial summation on (9) we can obtain (10) X p≤t p∈Pn 1 p ≪log log t φ(n) .
Whenever n/φ(n) is bounded, as it will be whenever n is a prime, prime power or a product of two prime powers, we can replace this bound with (11) X p≤t p∈Pn 1 p ≤c log log t n for some absolute constant c. We include the c because occasionally we require an inequality as opposed to an estimate. We will also require the following asymptotic from [8, Theorem 1] (12) X p∈Pn p≤t 1 p = log log t φ(n) + O log n φ(n) , 3 which easily implies that (13) X p∈Pn p≤t 1 p −1 = log log t φ(n) + O log n φ(n) , since the difference is X p∈Pn p≤t 1 p(p −1) ≤ ∞ X m=1 1 mn(mn + 1) < 1 n2 ∞ X m=1 1 m2 ≪1 n2.
3. Required Propositions and Proof of Theorem 1 As mentioned previously, the main contribution to log(n/λk(n)) will come from log(φk(n)/λk(n)).
Finding this term will involve a summation over prime powers which divide each of φk(n) and λk(n). It turns out that the largest contribution to this term will come from small primes which divide φk(n). By small, we mean primes q ≤(log log x)k = yk. Hence we will split the sum into small primes and large primes q > yk. Therefore to prove Theorem 1 we will require the following propositions. The first summations deal with the large primes which divide φk(n) and the second involves the large primes whose prime powers divide φk(n). We will show that the contribution of these primes to the main sum is small and hence it will end up as part of the error term.
Proposition 3.
X q>yk νq(φk(n))=1 (νq(φk(n)) −νq(λk(n))) log q ≪ykψ(x) for almost all n ≤x.
Proposition 4.
X q>yk νq(φk(n))≥2 νq(φk(n)) log q ≪ykψ(x) for almost all n ≤x.
Since the main contribution will come from small primes dividing φk(n), the next propos-tion will show that the contribution of small primes dividing λk(n) to the main sum can also be merged into the error term.
Proposition 5.
X q≤yk νq(λk(n)) log q ≪ykψ(x) for almost all n ≤x.
That will leave us with the contribution of small primes dividing φk(n). We will use an additive function to approximate this sum. Let hk(n) be the additive function defined by hk(n) = X p1|n X p2|p1−1 · · · X pk|pk−1−1 X q≤yk νq(pk −1) log q.
4 The following propostion shows that the difference between the sum involving the small primes dividing φk(n) and the term hk(n) is small.
Proposition 6.
X q≤yk νq(φk(n)) log q = hk(n) + O(yk−1 log y · ψ(x)) for almost all n ≤x, That leaves us with log(φk(n)/λk(n)) being approximated by hk(n). The last proposition will obtain an asymptotic formula for hk(n). From there we will have enough armoury to tackle Theorem 1.
Proposition 7.
hk(n) = 1 (k −1)!yk log y + O(yk) for almost all n ≤x.
Proof of Theorem 1. We start by breaking down the function log(n/λk(n)).
log n λk(n) = log n φ(n) + log φ(n) φ2(n) + · · · + log φk−1(n) φk(n) + log φk(n) λk(n) .
Using the lower bound φ(m) ≫m/ log log m, see [7, Theorem 2.3] we have that log n φ(n) + log φ(n) φ2(n) + · · · + log φk−1(n) φk(n) ≪log log log n and so log n λk(n) = log φk(n) λk(n) + O(log log log n).
In fact we could have used a more precise estimate for φi(n)/φi+1(n) for i ≥1 which can be found in but the one we used is more than good enough. Next we break down the remaining term into summations. We will break it up into small primes and large primes.
log φk(n) λk(n) = X q>yk (νq(φk(n)) −νq(λk(n))) log q + X q≤yk (νq(φk(n)) −νq(λk(n))) log q = X q>yk νq(φk(n))=1 (νq(φk(n)) −νq(λk(n))) log q + X q>yk νq(φk(n))≥2 (νq(φk(n)) −νq(λk(n))) log q + X q≤yk νq(φk(n)) log q − X q≤yk νq(λk(n)) log q.
Note that if a | b, then λ(a) | φ(b) since λ(a) | φ(a) | φ(ma) for any m. This quickly implies that λk(n) always divides φk(n) for all k and so we get 0 ≤ X q>yk νq(φk(n))≥2 (νq(φk(n)) −νq(λk(n))) log q ≤ X q>yk νq(φk(n))≥2 (νq(φk(n)) log q.
5 Using Propositions 3,4,5 and 6 we get log n λk(n) = hk(n) + O ykψ(x) for almost all n ≤x. Finally by using Proposition 7 we get log n λk(n) = 1 (k −1)!yk log y + O ykψ(x) for almost all n ≤x, finishing the proof of Theorem 1.
□ 4. Prime Power Divisors of φk(n) For various reasons thoughout this paper, we are concerned with the number of n ≤x such that qa can divide φk(n). We will analyze a few of those situations here: Case 1: q2 | n. Clearly the number of such n is at most x q2.
Case 2: There exists p1 ∈Pq2, p2 ∈Pp1, p3 ∈Pp2, ..., pl ∈Ppl−1 where pl | n. By using (11) repeatedly we get that the number of such n is X n≤x X p1∈Pq2 X p2∈Pp1 ...
X pl∈Ppl−1 pl|n 1 = X p1∈Pq2 X p2∈Pp1 ...
X pl∈Ppl−1 n≤x pl|n 1 ≪ X p1∈Pq2 X p2∈Pp1 ...
X pl∈Ppl−1 x pl ≪ X p1∈Pq2 X p2∈Pp1 ...
X pl−1∈Ppl−2 xy pl−1 ≪ X p1∈Pq2 X p2∈Pp1 xyl−2 p2 ≪ X p1∈Pq2 xyl−1 p1 ≪xyl q2 Now that we’ve taken care of any case where p ∈Pq2, we are just left with the possibilities not containing any powers of q. Unfortunately these cases still allow for many possibilities which we will display in an array. There are lots of ways for a prime power qa to arise in φk(n) we now define various sets of primes that are involved in generating these powers of q, and we will eventually sum over all possibilities for these sets of primes. The set Lh,i will denote a finite set of primes. To begin, the set L1,2 will be an arbitrary finite set of primes in Pq and let L1,1 be empty. That is: Case 3: Level (1,2) L1,2 ⊆Pq.
Level (2,1) (Obtaining the primes in the previous level) 6 L2,1 is any set of primes with the property that for all p ∈L1,1 ∪L1,2, there exists a unique prime r ∈L2,1 such that r ∈Pp. In other words p will divide φ(r) and hence the primes in L2,1 will create the primes in L1,1 ∪L1,2.
Level (2,2) (New primes in Pq) L2,2 ⊆Pq.
In general for all 1 < h ≤k we have for all p ∈Lh−1,1 ∪Lh−1,2 there exists a unique prime r ∈Lh,1 such that r ∈Pp, Lh,2 is an arbitrary subset of Pq, and r ∈Lk,1 ∪Lk,2 ⇒r | n.
Some description of the terms are in order including some helpful definitions.
Definition 8. An incarnation I of Case 3 is some specified description of how the primes in a lower level create the primes in the level directly above.
For example, for k = 3, an incarnation I for which q4 | φ3(n) would be s1, s2, s3, r3, r4 ∈Pq where r1 ∈Ps1, r2 ∈Ps2s3, p1 ∈Pr1r2, p2 ∈Pr3r4, with p1p2 | n.
Definition 9. An subincarnation of I is an incarnation with added conditions. In other words if J is a subincarnation of I and an integer n satisfies incarnation J, then it will also satisfy incarnation I.
For example, I is a subincarnation of the incarnation s1, s3, r3, r4 ∈Pq where r1 ∈Ps1, r2 ∈ Ps3, p1 ∈Pr1r2, p2 ∈Pr3r4, with p1p2 | n.
Let p be a prime in Lh,i which we need to divide φk−h+1(n). The definition of Lh,i ensures that there is a unique prime dividing φk−h(n) for which p | r −1. The primes in levels (k, 1), (k, 2) dividing n are for the base case of the recursion, so that each prime divides φ0(n) = n. When i = 2 we are introducing new primes to get greater powers of q in φk(n).
Note that it’s not necessary to have any primes on the levels (i, 2). In fact the “worst case scenario” that we will see has no primes on these except Level (1,2).
Now that we’ve described the way to get qa | φk(n), what is our exponent a? Let mh,i = #Lh,i. From the recursion above we can see that qmk,2 | φ(n) and so do the primes in Lk−1,1.
For the second iteration of φ, qmk,2−1+mk−1,2 | φ2(n) and so do the primes in Lk−2,1. Hence the power of q which divides φk(n) is (14) max 1≤j≤k(m1,1 + X 2≤h≤j (mh,2 −1)) where the sum can be empty if there are no primes in the second level (j, 2) or there are not enough to survive, i.e. mj,2 < j −1 and hence q ∤φj(Q Lj,2 p). Without loss of generality, we can assume the former, since the later is a subincarnation of the former.
Now we’ll introduce some notation to be used in future propositions.
For any single incarnation of Case 3, let M be the total number of primes, N be the total new primes introduced at the levels (h, 2) and H be the maximum necessary level (h, 2). Specifically M = X h (mh,1 + mh,2) N = X h≤H mh,2 and H yields the maximum value in (14). Note that under this notation, qN−H+1 | φk(n).
For example, in the incarnation I above, L1,2 = {s1, s2, s3}, L2,1 = {r1, r2}, L2,2 = {r3, r4}, L3,1 = {p1, p2}, L3,2 = ∅ 7 as well as m1,2 = 3, m2,1 = 2, m2,2 = 2, m3,1 = 2, m3,2 = 0.
Hence M = 9, N = 5, H = 2 and so the power of q which divides φ3(n) is 5 −2 + 1 = 4 as expected.
Now that we’ve described Case 3, how many possible n are in that case?
Lemma 10. The number of n ≤x satisfying any incarnation of Case 3 is O cM xyM qN where c is the constant from equation (11).
Proof. Let Lh = Lh,1 ∪Lh,2. We use Brun-Titchmarsh (11) for all the primes at each level of Case 3, so the number of n is X n≤x X p1∈L1 X p2∈L2 · · · X pk∈Lk 1 = X p1∈L1 X p2∈L2 · · · X pk∈Lk X pk|n n≤x 1 ≪ X p1∈L1 X p2∈L2 · · · X pk∈Lk x Q pk∈Lk pk .
Note that we have repeatedly counted the same primes in the sum as we can reorder the primes in each level. It won’t be important here, but will need to be more carefully addressed later. Since the primes in level (k, 1) gave us some pk ∈Ppk−1 for all the primes in Lk−1, and for p ∈Lk,k we have p ∈Pq. By Brun–Titchmarsh (11) we get that the above sum is ≪ X p1∈L1 X p2∈L2 · · · X pk−1∈Lk−1 x(cy)mk,1+mk,2 Q pk−1∈Lk−1 pk−1qmk,2 .
Once again we get mk−1,1 + mk−1,2 new applications of Brun-Titchmarsh giving the new primes in level k −2 as well as mk−1,2 new powers of q. Continuing along in this manner we get: ≪ X p1∈L1 x(cy) P 2≤i≤k(mi,1+mi,2) Q p1∈L1 p1q P 2≤i≤k mi,2 ≪x(cy) P 1≤i≤k(mi,1+mi,2) q P 1≤i≤k mi,2 = x(cy)M qN .
□ The last thing we’ll consider in this section about the ways to obtain φk(n) is to determine the number of possible incarnations of Case 3. We note that there are lots of incarnations which are subincarnations of others. We will develop a concept of minimality.
Definition 11. An incarnation of Case 3 is minimal if it does not contain any strings of p1 ∈Pp2, p2 ∈Pp3 . . . pk−1 ∈Ppk where pk | n.
Note that any incarnation of Case 3 is a subincarnation of a minimal one. We now use this concept to show the number of necessary incarnations of Case 3 is small.
8 5. Large Primes Dividing φk(n) In this section we will prove the two propostions dealing with q being large. We’ll start with the propostion where νq(φk(n)) = 1.
Proof of Proposition 3. It suffices to show X n≤x X q>yk νq(φk(n))=1 (νq(φk(n)) −νq(λk(n))) log q ≪xyk as then there are at most O xyk ykψ(x) = O x ψ(x) such n where the bound for the sum in Proposition 3 fails to hold. We examine the cases where νq(φk(n)) = 1. Using the notation in Lemma 10 we have two subcases for Case 3, whether N = 1 or N > 1.
Suppose N = 1, then H = 1, m1,2 = 1 and mh,2 = 0 for 1 < h ≤k. Since mh,1 ≤ mh−1,1 + mh−1,2 we get mh,1 ≤1 for all 1 ≤h ≤k. Hence mh,1 = 1 for all h ≤k and so we get the case: p1 ∈Pq, p2 ∈Pp1, p3 ∈Pp2, . . . , pk ∈Ppk−1 where pk | n. However in this case we also get νq(λk(n))) = 1 giving us no additions to our sum.
Suppose N > 1, then M = P h(mh,1 + mh,2) ≤k P h mh,2 = kN so the number of cases we get are O cM xyM qN ≪cMxykN qN ≪cMxy2k q2 since y > qk. Since vq(φk(n)) = N −H + 1 and H ≤k, N ≤k implying that M ≤k2.Hence cM is bounded as a function of k. Also since M is bounded in terms of k, there are Ok(1) possible incarnations of Case 3, and the bound already absorbs the possiblities from Cases 1 and 2. Hence we have X q>yk X n≤x νq(φk(n))=1 (νq(φk(n)) −νq(λk(n))) log q ≤ X q>yk X n≤x νq(φk(n))=1 N>1 log q ≪ X q>yk xy2k log q q2 ≪xyk by (4).
□ We turn our attention to vq(φk(n)) > 1. We have to be more careful here since we can’t guarantee that the number of incarnations of Case 3 is Ok(1). We’ll start by proving a lemma which can eliminate a lot of those cases.
9 Lemma 12. Let q > yk and Sq = Sq(x) consist of all n ≤x such that Case 1,2 or Case 3 where M ≤k(N −1) occurs. Then #Sq ≪xyk q2 Proof. There are clearly Ok(1) incarnations of Cases 1 and 2 and each yield at most O(xyk/q2) such n. By Lemma 10 for each incarnation of Case 3, we get at most O cMyM qN ≪cMyk q2 such n since M ≤k(N−1) and q > yk. It remains to show we only require Ok(1) such incarna-tions. Suppose n satisfies an incarnation with M ≤k(N −1). Then it also satisfies a minimal incarnation with M ≤k(N −1) since removing a string of p1 ∈Pp2, p2 ∈Pp3 . . . pk−1 ∈Ppk, would decrease N by 1 and M by k leaving the inequality unchanged. Secondly we can as-sume that n also satisfies an incarnation where k(N −2) < M ≤k(N −1) since we can keep eliminating primes in the Li,2, which decrease N by 1, but M by at most k. This must eventu-ally produce an incarnation where k(N −2) < M ≤k(N −1) since if we eliminate all primes in the Li,2 but 1, then M > k(N −1). Also note that the condition mh,1 ≤mh−1,1 + mh−1,2 forces M ≤kN. If M is bounded between k(N −2) and kN and the incarnation is minimal, we get that N is bounded by 2k since eliminating a prime in Li,2 can only shrink M by at most k −1 since our incarnation is minimal.
Therefore n satisifies an incarnation where N and hence M are bounded functions of k. Since there are only Ok(1) such incarnations, we get our result, noting that cM can be absorbed into the constant as well.
□ Proof of Proposition 4. Let S = S(x) = S q>yk Sq. Using Lemma 12 we have #S ≤ X q>yk #Sq ≪ X q>yk xyk q2 ≪xyk X q>yk 1 q2 ≪ xyk log(yk)yk ≪ x ψ(x) by (5). As for the n with n / ∈S and a = νq(φk(n)) > 1, the only remaining case is that M > k(N −1). Recall that a = N +H −1. If H = 1, then N = m1,2 = a, and so m2,1 = a−1 or a. Otherwise for k ≥2, M = X h mh,1 ≤a + (k −1)m2,1 ≤a + (k −1)(a −2) = k(a −1) −k + 2 ≤(k −1)N leading to a contradiction. If H > 1, then we again wish to show that m2,1 ≥a −k.
M = X h (mh,1 + mh,2) ≤km1,2 + (k −1) X h>1 mh,2 = m1,2 + (k −1)N = k(N −1) −N + k + m1,2 10 which implies m1,2 > N −k and so P h>1 mh,2 = N −m1,1 < k. Therefore if m2,1 < a −k, then M = X h (mh,1 + mh,2) ≤m1,2 + (k −1)m2,1 + (k −1) X h>1 mh,2 ≤a + (k −1)(a −k −1) + (k −1)(k −1) = ak −2k ≤k(N −1) as N > a again leading to a contradiction. Hence m2,1 ≥a −k and so we can get X n/ ∈S n≤x X q>yk νq(φk(n))>1 (νq(φk(n)) log q ≤2 X n/ ∈S n≤x X q>yk νq(φk(n))>1 (νq(φk(n)) −1) log q ≪ X q>yk log q X a≥2 a X n≤x n/ ∈S νq(φk(n))=a 1.
Unfortunately, just blindly using the Brun-Titchmarsh inequality in (11) won’t be good enough as we must sum over all a. Let g(a, k) = (a −k)! if a ≥k or 1 otherwise and note that since we have m1,2 ≥a −k, we have at least g(a, k) permutations of the same primes.
Then by using Lemma 10 we get a X q>yk log q X n≤x n/ ∈S νq(φk(n))=a 1 ≪a x(cy)M qNg(a, k) ≪ack(a+k−1)xy2k q2g(a, k) using the assumption that q > yk and M ≤kN ≤k(a + k −1). Hence we get our sum is X n/ ∈S n≤x X q>yk νq(φk(n))>1 (νq(φk(n)) log q ≪ X q>yk log q X a≥2 ack(a+k−1)xy2k q2g(a, k) = xy2k X q>yk log q q2 X a≥2 ack(a+k−1) g(a, k) However the latter sum converges to some function depending on k, and so we get ≪xy2k X q>yk log q q2 ≪xyk by (4).
□ 11 6. Small Primes Dividing λk(n) We now turn our attention to the bound involving λk(n) in the summand. Just like when we were dealing with the number of cases where qa | φk(n), we will need a lemma to deal with the number of cases where qa | λk(n). Fortunately this case is much simpler as the only two ways for qa | λ(n) is for qa+1 | n or for there to exist p | n with p ∈Pqa. Note that these conditions aren’t sufficient, but are necessary when q = 2.
Lemma 13. The number of positive integers n ≤x for which qa | λk(n) is O( xyk qa ).
Proof. We’ll proceed by induction on k. If k = 1, then qa | λ(n) if qa+1 | n or p ∈Pqa with p | n. The number of such n is at most X n≤x qa+1|n 1 + X n≤x p∈Pqa p|n 1 ≪ x qa+1 + X p∈Pqa x p ≪ x qa+1 + xy qa ≪xy qa .
using (11). Suppose the number of n ≤x for which qa | λk−1(n) is O( xyk−1 qa ). If qa | λk(n), then either qa+1 | λk−1(n) or p ∈Pqa with p | λk−1(n). Hence the number of such n is bounded by X n≤x qa+1|λk−1(n) 1 + X n≤x p∈Pqa p|λk−1(n) 1 ≪xyk−1 qa+1 + X p∈Pqa xyk−1 p ≪xyk−1 qa+1 + xyk qa ≪xyk qa as needed.
□ Proof of Proposition 5. Like in the proof of previous propositions, we’ll show X n≤x X q≤yk νq(λk(n)) log q ≪xyk.
The left hand side is equal to X n≤x X q≤yk νq(λk(n)) log q = X n≤x X q≤yk log q X a∈N qa|λk(n) 1 ≤ X n≤x X q≤yk log q X a∈N qa≤yk 1 + X n≤x X q≤yk log q X a∈N qa|λk(n) qa>yk 1.
The first sum is X n≤x X q≤yk log q X a∈N qa≤yk 1 = X n≤x X m≤yk Λ(m) ≪ X n≤x yk ≪xyk, and by Lemma 13 and using the geometric estimate in (6) the second sum becomes X n≤x X q≤yk log q X a∈N qa|λk(n) qa>yk 1 ≪ X q≤yk log q X a∈N qa>yk xyk qa ≪ X q≤yk log qxyk yk ≪xyk.
12 □ 7. Reduction To hk(n) For Small Primes The small primes dividing φk(n) are what contributes to the asymptotic term of log(n/λk(n)).
In this section we show that the important case is the supersquarefree case of p dividing φk(n) which is when p ∈Pp1, p1 ∈Pp2 . . . pk−1 ∈Ppk, pk | n.
For this reason we will approximate the sum P q≤yk vq(φk(n)) log q with (15) hk(n) = X p1|n X p2|p1−1 · · · X pk|pk−1−1 X q≤yk νq(pk −1) log q.
Proof of Proposition 6. For any fixed prime q, we know that vq(φ(m)) = max{0, vq(m) −1} + X p|m vq(p −1), which implies X p|m vq(p −1) ≤vq(φ(m)) ≤vq(m) + X p|m vq(p −1).
Repeated use of this inequality for m = φl(n) where l ranges from k −1 to 0 yields X p|φk−1(n) vq(p −1) ≤vq(φk(n)) ≤ X p|φk−1(n) vq(p −1) + X p|φk−2(n) vq(p −1) + · · · + X p|φ(n) vq(p −1) + vq(n).
(16) A prime p divides φk−1(n) either in the supersquarefree case (ssf), or not in the supersquare-free case (nssf), yielding X ssf vq(p −1) ≤ X p|φk−1(n) vq(p −1) ≤ X ssf vq(p −1) + X nssf vq(p −1).
Combining this inequality with (16) yields X ssf vq(p −1) ≤vq(φk(n)) ≤ X ssf vq(p −1) + X nssf vq(p −1) + X p|φk−2(n) vq(p −1) + · · · + X p|φ(n) vq(p −1) + vq(n).
Subtracting the sum over the supersquarefree case, multiplying through by log q and sum-ming over q ≤yk we get 13 0 ≤ X q≤yk νq(φk(n)) log q −hk(n) ≤ X q≤yk X nssf vq(p −1) log q + X q≤yk X p|φk−2(n) vq(p −1) log q + · · · + X q≤yk X p|n vq(p −1) log q where we get hk(n) from (15). Hence it suffices to show that the sum on the right side becomes our error term. For the sum X n≤x X q≤yk X p|φm(n) vq(p −1) log q = X n≤x X q≤yk X p|φm(n) X a∈N qa|p−1 log q = X n≤x X q≤yk log q X a∈N X p∈Pqa p|φm(n) 1, we’ll split the sum over values of p ≤yk−1 and p > yk−1. For p ≤yk−1 we uniformly get for all n that X q≤yk log q X a∈N X p∈Pqa p≤yk−1 p|φm(n) 1 ≤ X q≤yk log q X a∈N π(yk−1; qa, 1) ≪ X q≤yk log q X a∈N yk−1 φ(qa) ≪yk−1 X q≤yk log q q ≪yk−1 log y using the geometric estimate (6) and the prime number theorem for arithmetic progressions.
As for p > yk−1 we fix an M and N from case 3 for which p | φm(n), of which there are at most Ok(1) such M, N since vp(φ(m)) = 1. Therefore 14 X n≤x X q≤yk log q X a∈N X p>yk−1 p∈Pqa p|φm(n) 1 ≪ X q≤yk log q X a∈N X p∈Pqa p>yk−1 xyM pN ≤ X q≤yk log q X a∈N X p∈Pqa xyM−(k−1)(N−1) p ≪ X q≤yk log q X a∈N xyM−(k−1)(N−1)+1 qa ≪ X q≤yk xyM−(k−1)(N−1)+1 log q q ≪xyM−(k−1)(N−1)+1 log yk ≪xyM−(k−1)(N−1)+1 log y.
Since the M, N were chosen for φm(n) we know that M ≤mN where equality holds if and only if we are in the supersquarefree case. Now either m ≤k −2 or m = k −1 and we are not in the supersquarefreecase. In the former case we have an error of O(xy(k−2)N−(k−1)(N−1)+1 log y) = O(xyk−N log y) = O(xyk−1 log y) since N ≥1, or in the latter case O(xy(k−1)N−1−(k−1)(N−1)+1 log y) = O(xyk−1 log y).
Thus we get X n≤x X q≤yk X nssf vq(p −1) log q + X q≤yk X p|φk−2(n) vq(p −1) log q + . . .
+ X q≤yk X p|n vq(p −1) log q ≪xyk−1 log y and so X q≤yk X nssf vq(p −1) log q + X q≤yk X p|φk−2(n) vq(p −1) log q + . . .
+ X q≤yk X p|n vq(p −1) log q ≪yk−1 log y · ψ(x) as required.
□ 15 8. Reduction to the First and Second Moments The Tur´ an-Kubilius inequality [5, Lemma 3.1] asserts that if f(n) is a complex additive function, then there exists an absolute constant C such that (17) X n≤x |f(n) −M1(x)|2 ≤CxM2(x) where M1(x) = P p≤x|f(p)|/p and M2(x) = P p≤x|f(p)|2/p. Since hk(n) is additive we can apply this inequality where M1(x) = P p≤x hk(p)/p, M2(x) = P p≤x hk(p)2/p. We will need to find bounds on M1 and M2 therefore it’s our goal to prove the following two propositions: Proposition 14. For all x > eee, M1(x) = 1 (k −1)!yk log y + O(yk) Proposition 15. For all x > eee, M2(x) ≪y2k−1 logk−1 y.
These will lead to a proof of Proposition 7.
Proof of Proposition 7. Let N denote the number of n ≤x for which |hk(n) −M1(x)| > yk.
The contribution of such n to the sum in (17) is at least Ny2k. Thus Propostion 15 implies N ≪x logk−1 y/y and so Proposition 14 implies that hk(n) = 1 (k−1)!yk log y + O(yk) except for a set of size O(x(log y)k−1/y).
□ 9. Lots of Summations In our proofs of Propositions 14 and 15 we will see that M1(x) and M2(x) will reduce to summations involving π(x; p, 1). We will be using some sieve techniques to bound these sums and those will require some bounds on sums on multiplicative functions involving φ(m). This section will involve the estimation of the latter sums.
Lemma 16. For any non-negative integer L we have (18) X m≤t mL φ(m)L+1 ≪L log t.
Proof. If f(n) is a non-negative multiplicative function, we know that (19) X n≤t f(n) ≤ Y p≤t ∞ X r=0 f(pr).
16 Applying (19) with mL φ(m)L+1 yields X m≤t mL φ(m)L+1 ≤ Y p≤t 1 + ∞ X r=1 prL (pr −pr−1)L+1 = Y p≤t 1 + ∞ X r=1 pL−r+1 (p −1)L+1 = Y p≤t 1 + 1 (p −1)L+1 pL 1 −1 p = Y p≤t 1 + pL+1 (p −1)L+2 ≤exp X p≤t log 1 + pL+1 (p −1)L+2 = exp X p≤t pL+1 (p −1)L+2 + OL 1 p2 = exp X p≤t 1 p + OL 1 p2 ≪L log t using (3).
□ Lemma 17. Given a positive integer C ≤tγ and non-negative integer L we have (20) X m≤t (Cm + 1)L φ(Cm + 1)Lφ(m) ≪L,γ log t.
Proof. It will suffice to show X m≤t (Cm + 1)2L−1 φ(Cm + 1)2L ≪L log t C as then by Cauchy–Schwarz we can get that X m≤t (Cm + 1)L φ(Cm + 1)Lφ(m) 2 ≤ X m≤t (Cm + 1)2L−1 φ(Cm + 1)2L X m≤t (Cm + 1) φ(m)2 ≪L log t C C log t ≪L log2 t by using (18). Using Mobius inversion, let s(n) be the multiplicative function defined by n2L φ(n)2L = 1 ∗s = X d|n s(d).
17 Testing at prime powers, we can easily see that s(1) = 1, s(p) = 1 −1 p −2L −1 and s(pk) = 0 for all k ≥2.
Hence X m≤t (Cm + 1)2L−1 φ(Cm + 1)2L = X C<n≤Ct+1 n≡1 (mod C) n2L−1 φ(n)2L = X C<n≤Ct+1 n≡1 (mod C) 1 n n2L φ(n)2L = X C<n≤Ct+1 n≡1 (mod C) 1 n X d|n s(d) = X d≤Ct+1 s(d) X C<n≤Ct+1 d|n n≡1 (mod C) 1 n.
By (8) and noticing that C and d are relatively prime we get X C ee be a real number and let constants α, α1, α2 satisfy 0 < α < 1/2 and 0 < α1 < α2 < 1/2.
(a) If b > tα, then (23) X pk∈Pb X pk−1∈Ppk · · · X p2∈Pp3 π(t; p2, 1) ≪t log t(log log t)k−2 b .
(b) If b ≤tα1, then (24) X pl∈Pb pl>tα2 X pl−1∈Ppl · · · X p2∈Pp3 π(t; p2, 1) ≪ bl−1t φ(b)l log t.
(c) If b ≤tα1, then (25) X pl∈Pb X pl−1∈Ppl · · · X p2∈Pp3 π(t; p2, 1) ≪t(log log t)l−1 φ(b) log t .
The implicit constants in (a) −(c) depend on the choices of the α.
20 Proof. For (23) we just use the trivial estimate π(t; p2, 1) ≤t/p2 and several uses of Brun-Titchmarsh (11) to get X pk∈Pb X pk−1∈Pk · · · X p2∈P3 π(t; p2, 1) ≤ X pk∈Pb X pk−1∈Pk · · · X p2∈P3 t p2 ≪t X pk∈Pb X pk−1∈Pk · · · X p3∈P4 log log t p3 ≪t X pk∈Pb (log log t)k−2 pk ≤t X m≡1 (mod b) tα≤m≤t (log log t)k−2 m ≤t log t(log log t)k−2 b where m > 1 and m ≡1 (mod b) imply that m > b and by using (7). As for (24) we get X pl∈Pb l>tα2 X pl−1∈Pl · · · X p2∈P3 π(t; p2, 1) = X pl∈Pb l>tα2 X pl−1∈Pl · · · X p3∈P4 #{(m1, p2) : p2 = 1 (mod p3), p2 > tα2, m1p2 + 1 ≤t, p2, m1p2 + 1 prime} = X pl∈Pb l>tα2 X pl−1∈Pl · · · X p4∈P5 #{(m1, m2, p3) : p3 = 1 (mod p4), p3 > tα2, m1(m2p3 + 1) + 1 ≤t, {p3, m2p3 + 1, m1(m2p3 + 1) + 1} prime} = #{(m1, m2, . . . , ml−1, pl) : pl = 1 (mod b), pl > tα2, m1(m2 . . . (ml−2(ml−1pl + 1) + 1) + . . .
+ 1 ≤t, {pl, ml−1pl + 1, ml−2(ml−1pl + 1) + 1, . . . , m1(m2 . . . (ml−2(ml−1pl + 1) + 1) + · · · + 1} prime} ≤ X m1···l−1≤t1−α2 #{pl < t/m1 . . . ml−1 : pl = 1 (mod b), {pl, ml−1pl + 1, ml−2(ml−1pl + 1) + 1, . . . , m1(m2...(ml−2(ml−1pl + 1) + 1) + · · · + 1} prime}.
From here will need to use Brun’s Sieve method (see [4, Theorem 2.4]) to get that #{pl <t/m1 . . . ml−1 : pl = 1 (mod b), {pl, ml−1pl + 1, ml−2(ml−1pl + 1) + 1, . . . , m1(m2 . . . (ml−2(ml−1pl + 1) + 1) + · · · + 1} prime} ≪ El−1 φ(E)l−1 bl−1 φ(b)l−1 bc1 . . . cl−1 φ(bc1 . . . cl−1) t/m1 . . . ml−1b (log t/m1 . . . ml−1b)l 21 where the ci and E are E = l−1 Y i=1 mi(i+1)/2 i (1 + m1 + m1m2 + · · · + m1 . . . ml−3)(1 + m2 + m2m3 + · · · + m2 . . . ml−3) . . . (1 + ml−3)(1 + m1 + m1m2 + · · · + m1 . . . ml−4)(1 + m2 + m2m3 + · · · + m2 . . . ml−4) . . . (1 + ml−4) . . . (1 + m1) and for 1 ≤i ≤l −1, ci = 1 + mi + mimi+1 + · · · + mi . . . ml−2, cl−1 = 1.
Now using φ(mn) ≤φ(m)φ(n) and m1 . . . ml−1b ≤t1+α1−α2 where 1 + α1 −α2 < 1 we get ≪ El−1 φ(E)l−1 bl−1 φ(b)l c1 φ(c1) . . .
cl−1 φ(cl−1) t m1 . . . ml−1(log t)l.
Using mL φ(mL) = m φ(m), we get the sum X m1...ml−1≤t1−α2 El−1 φ(E)l−1 c1 φ(c1) . . .
cl−1 φ(cl−1) 1 m1 . . . ml−1 = X m1...ml−1≤t1−α2 (E∗)l−1 φ(E∗)l−1 c1 φ(c1) . . .
cl−1 φ(cl−1) 1 m1 . . . ml−1 where E∗=(1 + m1 + m1m2 + · · · + m1 . . . ml−3)(1 + m2 + m2m3 + · · · + m2 . . . ml−3) . . . (1 + ml−3)(1 + m1 + m1m2 + · · · + m1 . . . ml−4)(1 + m2 + m2m3 + · · · + m2 . . . ml−4) . . . (1 + ml−4) . . . (1 + m1).
We have that every factor in E∗as well as the ci are of the form 1+Cmi for some i or of the form mL i . Hence using l −1 applications of Lemmas 16, 18 or 19 we can pick offthe factors of the form (1 + Cmi) one at a time.
X m1...ml−1≤t1−α2 El−1 φ(E)l−1 c1 φ(c1)... cl−1 φ(cl−1) 1 m1 . . . ml−1 ≪ X m2...ml−1≤t1−α2 (E′)l−1 φ(E′)l−1 c′ 1 φ(c′ 1) . . .
c′ l−1 φ(c′ l−1) 1 m2 . . . ml−1 (log t) ≪ X m3...ml−1≤t1−α2 (E′′)l−1 φ(E′′)l−1 c′′ 1 φ(c′′ 1) . . .
c′′ l−1 φ(c′′ l−1) 1 m3 . . . ml−1 (log2 t) ≪· · · ≪(log t)l−1.
where the E(e), c(e) i denote the E∗and ci terms with the factors of the form 1+Cm1 through 1 + Cme removed. Note that the C are at most 1 + t + t2 + · · · + tk−3 ≤tk−2 and l ≤k so 22 the implied constant only depends on k. Therefore X pl∈Pb l>tα2 X pl−1∈Pl · · · X p2∈P3 π(t; p2, 1) ≪ tbl−1 φ(b)l(log t)l(log t)l−1 = tbl−1 φ(b)l log t.
As for part (c), first note that b/φ(b) ≪log log b, so for pl > tα2, we get that part (b) implies our bound. As for pl ≤tα2 we’ll split it into cases where p3 is less than or greater than tα2.
If p3 ≤tα2, then X pl∈Pb pl≤tα2 X pl−1∈Pl · · · X p2∈P3 p2≤tα2 π(t; p2, 1) ≪ X pl∈Pb pl≤tα2 X pl−1∈Pl · · · X p2∈P3 p2≤tα2 t φ(p2) log t/p2 ≪ X pl∈Pb pl≤tα2 X pl−1∈Pl · · · X p2∈P3 p2≤tα2 t p2 log t ≪ X pl∈Pb t(log log t)l−2 pl log t ≪t(log log t)l−1 φ(b) log t If p3 > tα2, then since b ≤tα2 there is a minimum m such that pm ≤tα2. So using part (b) with l = m we get X pl∈Pb pl≤tα2 X pl−1∈Pl · · · X p2∈P3 p2>tα2 π(t; p2, 1) ≪ X pl∈Pb pl≤tα2 X pl−1∈Pl · · · X pm+1∈Pm+2 (pm−1)m−1t φ(pm−1)m log t ≪ X pl∈Pb pl≤tα2 X pl−1∈Pl · · · X pm+1∈Pm+2 t pm−1 log t ≪t(log log t)l−m φ(b) log t ≪t(log log t)l−1 φ(b) log t since m ≥2 and by using Brun-Titchmarsh (11) which finishes part (c) and the lemma.
□ As for the summations requires for the second moment, we’ll note that we need twice as many sums due to hk(p)2. However the techniques required are similar.
Lemma 21. Let t > ee and 0 < 2α1 < α2 < 1/2. Then (a) If b1 > tα1 or b2 > tα1 then (26) X p2∈Pb1 r2∈Pb2 π(t; p2r2, 1) ≪t log2 t b1b2 .
23 (b) If neither b1 nor b2 exceeds tα1, then (27) X pk∈Pb1 rk∈Pb2 pkrk>tα2 ...
X p2∈Pp3 r2∈Pr3 π(t; p2r2, 1) ≪t(log log t)k−1bk−1 2 φ(b1)φ(b2)k log t + t(log log t)k−1bk−1 1 φ(b2)φ(b1)k log t .
(c) If neither b1 nor b2 exceeds tα1, then (28) X pk∈Pb1 rk∈Pb2 ...
X p2∈Pp3 r2∈Pr3 π(t; p2r2, 1) ≪t(log log t)2k−2 φ(b1)φ(b2) log t.
(d) If neither b1 nor b2 exceeds tα1, then (29) X pk∈Pb1 rk∈Pb2 ...
X p3∈Pp4 r3∈Pr4 X s∈Pp3∩Pr3 π(t; s, 1) ≪t(log log t)2k−2 φ(b1)φ(b2) log t.
Again the implicit constants depend on our choice of the α.
Proof. (a) is similar to part (a) of Lemma 20. For part (b) we first assume that pk ≤rk, then X pk∈Pb1 rk∈Pb2 pk≤rk pkrk>tα2 ...
X p2∈Pp3 r2∈Pr3 π(t; p2r2, 1) = X pk∈Pb1 rk∈Pb2 pkrk>tα2 · · · X p3∈Pp4 r3∈Pr4 #{(m1, p2, r2) : p2 = 1 (mod p3), r2 = 1 (mod r3), r2p2 > tα2, m1r2p2 + 1 ≤t, p2, m1r2p2 + 1 prime} = X pk∈Pb1 pk≤rk X pk−1∈Pk · · · X p2∈P3 X rk∈Pb2 pkrk>tα2 X rk−1∈Prk · · · X r4∈Pr5 #{(m1, m2, r3) : r3 = 1 (mod r4), r3p2 > tα2, m1p2(m2r3 + 1) + 1 ≤t, {r3, m2r3 + 1, m1p2(m2r3 + 1) + 1} prime} = X pk∈Pb1 pk≤rk X pk−1∈Pk · · · X p2∈P3 #{(m1, m2, . . . , ml−1, rl) : rl = 1 (mod b2), p2rk > tα2, m1p2(m2 . . . (mk−2(mk−1rk + 1) + 1) + · · · + 1 ≤t, {rk, mk−1rk + 1, mk−2(mk−1rk + 1) + 1, . . . , m1p2(m2 . . . (mk−2(mk−1rk + 1) + 1) + · · · + 1} prime} ≤ X m1...ml−1≤t1−α2 X pk∈Pb1 pk≤rk X pk−1∈Pk · · · X p2∈P3 #{rk < t/p2m1...mk−1 : rk = 1 (mod b2), {rk, mk−1rk + 1, mk−2(mk−1rk + 1) + 1, . . . , p2m1(m2 . . . (mk−2(mk−1rk + 1) + 1) + · · · + 1} prime} 24 Just like in Lemma 20 we use Brun’s Sieve. However, notice that we have almost the same set, except with m1 replaced with m1p2. Hence we have #{rk < t/p2m1 · · ·k−1 : rk = 1 (mod b1), {rk, mk−1rk + 1, mk−2(mk−1rk + 1) + 1, . . . , p2m1(m2 . . . (mk−2(mk−1rk + 1) + 1) + · · · + 1} prime} ≪ Ek−1 φ(E)k−1 bk−1 2 φ(b2)k−1 b2c1 . . . ck−1 φ(b2c1 . . . ck−1) t/p2m1 . . . mk−1b2 (log t/p2m1 . . . mk−1b2)k where the ci and E are E =p2 l−1 Y i=1 mi(i+1)/2 i (1 + p2m1 + p2m1m2 + ... + p2m1 . . . mk−3)(1 + m2 + m2m3 + . . .
+ m2 . . . mk−3) . . . (1 + mk−3)(1 + p2m1 + p2m1m2 + · · · + p2m1 . . . mk−4) (1 + m2 + m2m3 + · · · + m2 . . . mk−4) . . . (1 + mk−4) . . . (1 + p2m1) and for 2 ≤i ≤k −1, c1 = 1 + p2m1 + p2m1m2 + · · · + p2m1 . . . mk−2, ci = 1 + mi + mimi+1 + · · · + mi . . . mk−2, ck−1 = 1.
By the same methods as Lemma 20 and using that p2/φ(p2) is bounded and noting that t p2m1...mk−1b2 > rk b1 > tα2/2−α1 = tϵ for some ϵ > 0 since α2 > 2α1, we get that X pk∈Pb1 rk∈Pb2 pkrk>tα2 · · · X p2∈Pp3 r2∈Pr3 π(t; p2r2, 1) ≪ tbk−1 2 φ(b2)k log t X pk∈Pb1 X pk−1∈Pk · · · X p2∈P3 1 p2 ≪ tbk−1 2 φ(b2)k log t X pk∈Pb1 (log log t)k−2 pk ≪t(log log t)k−1bk−1 2 φ(b1)φ(b2)k log t .
25 The case for rk ≤pk is similar. As for part (c), first note that bi/φ(bi) ≪log log bi for i ∈{1, 2}. taking care of the case where pkrk > tα2. As for pkrk ≤tα2 we get X pk∈Pb1 rk∈Pb2 pkrk≤tα2 · · · X p2∈Pp3 r2∈Pr3 π(t; p2r2, 1) ≪ X pk∈Pb1 rk∈Pb2 pkrk≤tα2 · · · X p2∈Pp3 r2∈Pr3 t φ(p2r2) log t/p2r2 ≪ X pk∈Pb1 rk∈Pb2 pkrk≤tα2 · · · X p2∈Pp3 r2∈Pr3 t p2r2 log t ≪ X pk∈Pb1 rk∈Pb2 pkrk≤tα2 t(log log t)2k−4 pkrk log t ≪t(log log t)2k−2 φ(b1)φ(b2) log t using Brun-Titchmarsh, (11) finishing part (c). As for part (d) we note that X p3∈Pp4 r3∈Pr4 X s∈Pp3∩Pr3 π(t; s, 1) = X p3∈Pp4 r3∈Pr4 #{(m1, s) : s = 1 (mod p3r3), m1s + 1 ≤t, s, m1s + 1 prime} = X p3∈Pp4 #{(m1, m2, r3) : r3 = 1 (mod r4), m1(m2p3r3 + 1) + 1 ≤t, {m2p3r3 + 1, m1(m2p3r3 + 1) + 1 prime} and so on, yielding a similar sieve as part (b).
□ 11. Reduction of P hk(p) to small values of pk We will be using Euler Summation on the sum P p≤t hk(p) in our efforts to find our estimate for M1(x). It will turn out that the large primes do not contribute much to the some. The sum will involve estimating π(t; p, 1) by li(t)/p−1. The following lemma will deal with those errors and will involve the Bombieri–Vinogradov Theorem.
26 Lemma 22. For all 2 ≤l ≤k, x > eee and v > ee, X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 π(v, pk−l+2, 1) −li(v) pk−l+2 ≪v log y log v + li(v)(log log v)l−2.
Proof. Let E(t; r, 1) = π(t; r, 1) −li(t) r−1. Then we have X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 π(v, pk−l+2, 1) − li(v) pk−l+2 −1 = X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 E(v; pk−l+2, 1) ≪ X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 |E(v; pk−l+2, 1)|.
Let Ω(m) denote the number of divisors of m which are primes or prime powers. We use the estimate Ω(m) ≪log m to get X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 |E(v; pk−l+2, 1)| ≤log(yk) X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 |E(v; pk−l+2, 1)| X pk−l+3|pk−l+2−1 p3≤v1/9 X pk−l+4|pk−l+3−1 pk−l+4≤v1/27 · · · X q≤yk X a∈N qa|pk−1 1 ≤log(yk) X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 |E(v; pk−l+2, 1)| X pk−l+3|pk−l+2−1 p3≤v1/9 X pk−l+4|pk−l+3−1 pk−l+4≤v1/27 · · · X pk≤v1/3k−1 pk|pk−1−1 Ω(pk −1) ≪log y X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 |E(v; pk−l+2, 1)| X pk−l+3|pk−l+2−1 p3≤v1/9 X pk−l+4|pk−l+3−1 pk−l+4≤v1/27 · · · X pk≤v1/3k−1 pk|pk−1−1 log t.
Continuing in this manner we obtain X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 |E(v; pk−l+2, 1)| ≪log y(log v)l−1 X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 |E(v; pk−l+2, 1)| ≪v log y log t 27 using Bombieri–Vinogradov. As for the difference between X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 li(v) pk−l+2 −1 and (30) X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 li(v) pk−l+2 we get that it is X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 li(v) pk−l+2(pk−l+2 −1) ≤ X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · ∞ X i=1 li(v) (ipk−l+3 + 1)(ipk−l+3) ≪ X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+3∈Ppk−l+4 pk−l+3≤v1/9 li(v) p2 k−l+3 ≪ X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+3∈Ppk−l+4 pk−l+3≤v1/9 li(v) pk−l+3qa ≪ X q≤yk log q X a∈N li(v)(log log v)l−2 q2a ≪ X q≤yk li(v)(log log v)l−2 log q q2 ≪li(v)(log log v)l−2 using the Brun–Titchmarsh inequality (11), the inequality pk−l+3 ≥qa and noting that the sum over q converges.
□ Lemma 23. For all x > eee and t > ee, X p≤t hk(p) = X q≤yk log q X a∈N X pk∈Pqa pk≤t1/3k−1 X pk−1∈Ppk pk−1≤t1/3k−2 · · · X p2∈Pp3 p2≤t1/3 π(t; p2, 1) + O t1−1/3k log t(log log t)k−2yk + t(log log t)k−2 log y log t .
28 Proof. For a prime p, hk(p) = X p1|p X p2|p1−1 · · · X pk|pk−1−1 X q≤yk νq(pk −1) log q = X p2|p−1 · · · X pk|pk−1−1 X q≤yk νq(pk −1) log q since the only prime which can divide p is p itself. Hence X p≤t hk(p) = X p≤t X p2|p−1 · · · X pk|pk−1−1 X q≤yk νq(pk −1) log q = X p≤t X p2|p1−1 · · · X pk|pk−1−1 X q≤yk X pk∈Pqa a∈N log q = X q≤yk log q X a∈N X pk∈Pqa X pk−1∈Ppk · · · X p2∈Pp3 X p≤t p∈Pp2 1 = X q≤yk log q X a∈N X pk∈Pqa X pk−1∈Ppk · · · X p2∈Pp3 π(t; p2, 1).
We wish to approximate π(t; p2, 1) by li(t) p2−1 and use the Bombieri-Vinogradov Theorem to deal with the error. However this approximation only allows primes up to say t1/3. So we use the estimations in Lemma 20 to bound these errors. We will see that the main contribution comes from pi ≤t1/3i−1 and qa ≤t1/3k.
Using Lemma 20, we get for large qa X q≤yk log q X a∈N qa>t1/3k X pk∈Pqa X pk−1∈Ppk · · · X p3∈Pp2 π(t; p2, 1) ≪ X q≤yk log q X a∈N qa>t1/3k t log t(log log t)k−2 qa .
By geometric estimates, if a∗is the smallest a where qa > t1/3k, then we get that the above is ≪t log t(log log t)k−2 X q≤yk log q qa∗ ≤t1−1/3k log t(log log t)k−2 X q≤yk log q ≪t1−1/3k log t(log log t)k−2yk.
Now suppose qa ≤t1/3k. Let l be the last index (supposing one exists) where pi > t1/3i−1 By using (24) where l ranges from 2 to k, we can bound the large values of the pi.
29 X q≤yk log q X a∈N qa≤t1/3k X pk∈Pqa pk≤t1/3k−1 · · · X pl+1∈Ppl+2 pl+1≤t1/3l X pl∈Ppl+1 pl>t1/3l−1 X pl−1∈Ppl · · · X p2∈Pp3 π(t; p2, 1) ≪ X q≤yk log q X a∈N qa≤t1/3k X pk∈Pqa pk≤t1/3k−1 · · · X pl+2∈Ppl+3 pl+2≤t1/3l+1 X pl+1∈Ppl+2 pl+1>t1/3l (pl)l−1t φ(pl)l log t ≪ X q≤yk log q X a∈N qa≤t1/3k X pk∈Pqa pk≤t1/3k−1 · · · X pl+2∈Ppl+3 pl+2≤t1/3l+1 X pl+1∈Ppl+2 pl+1>t1/3l t pl+1 log t since pl is prime and l ≤k. By Brun-Titchmarsh (11) we get ≪ X q≤yk log q X a∈N qa≤t1/3k t(log log t)k−l qa log t ≪ X q≤yk t(log log t)k−l log q q log t ≪t(log log t)k−2 log y log t by (3) and since l ≥2. Hence we get X p≤t hk(p) = X q≤yk log q X a∈N qa≤t1/3k X pk∈Pqa pk≤t1/3k−1 X pk−1∈Ppk pk−1≤t1/3k−2 · · · X p2∈Pp3 p2≤t1/3 π(t, p2, 1) + O t1−1/3k log t(log log t)k−2yk + t(log log t)k−2 log y log t finishing the lemma.
□ 12. Evaluation of the Main Term Now we’ll deal with the main term from Lemma 23. We will deal with estimating the individual sums recursively. Hence we wish to make the following definition.
Definition 24. Let 2 ≤l ≤k and 2 ≤u ≤t. Then define gk,l(u) = X q≤yk log q X a∈N X pk∈Pqa pk≤u1/3l−1 X pk−1∈Ppk pk−1≤u1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤u1/3 π(u; pk−l+2, 1).
Note that gk,k(t) is the summation in Lemma 23. Next we’ll exhibit the recursive formula satisfied by the gk,l.
30 Lemma 25. Let 3 ≤l ≤k, then (31) gk,l(v) = li(v) Z v1/3 2 1 u2gk,l−1(u)du + O v(log log v)l−2 log y log v .
Proof. We’ll proceed by approximating π by li and then use partial summation to recover π.
Using Lemma 22 we get gk,l(v) = X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 π(v; pk−l+2, 1) ≪ X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 li(v) pk−l+2 + O v log y log v + li(v)(log log v)l−2 .
We use Euler summation on the inner sum to get X pk−l+2∈Ppk−l+3 pk−l+2≤v1/3 1 pk−l+2 = π(v1/3; pk−l+3, 1) v1/3 + Z v1/3 2 π(u; pk−l+3, 1) u2 du and so we get that gk,l(v) = li(v) X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+3∈Ppk−l+4 pk−l+3≤v1/3 π(v1/3; pk−l+3, 1) v1/3 + Z v1/3 2 π(u; pk−l+3, 1) u2 du + O v log y log v + li(v)(log log v)l−2 .
Inside the sum by trivially estimating π(x; q, 1) by x/q inside the sum and using Brun– Titchmarsh (11) we get X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+3∈Ppk−l+4 pk−l+3≤v1/3 π(v1/3; pk−l+3, 1) v1/3 ≪ X q≤yk log q X a∈N X pk∈Pqa pk≤v1/3l−1 X pk−1∈Ppk pk−1≤v1/3l−2 · · · X pk−l+3∈Ppk−l+4 pk−l+3≤v1/3 1 pk−l+3 ≪ X q≤yk log q X a∈N (log log v)l−2 qa ≪ X q≤yk log q(log log v)l−2 q ≪(log log v)l−2 log y.
Multiplying through by li(v) finishes the lemma.
□ 31 We now require a lemma to find the asymptotic formula for hk using the previous recur-rence relation Lemma 26. Let 2 ≤l ≤k.
gk,l(u) = ku(log log u)l−1 log y (l −1)! log u + O u(log log u)l−1 log u + u(log log u)l−2 log2 y log u which implies X p≤t hk(p) = kt(log log t)k−1 log y (k −1)! log t + O t(log log t)k−1 log t + t(log log t)k−2 log2 y log t + t1−1/3k log t(log log t)k−2yk .
Proof. The second formula is derived from the first by setting l = k, u = t and using Lemma 23. We’ll proceed with the first formula by induction on l. Using the estimates we obtained via Bombieri–Vinogradov in Lemma 22, we have for l = 2 gk,2(u) = X q≤yk log q X a∈N X pk∈Pqa pk≤u1/3 π(u; pk, 1) = li(u) X q≤yk log q X a∈N X pk∈Pqa pk≤u1/3 1 pk + O li(u) + u log y log u .
We then use (13) and log log(u1/3) = log log u + O(1) to get gk,2(u) = li(u) X q≤yk log q X a∈N log log u1/3 φ(qa) + O log(qa) φ(qa) + O u log y log u = li(u)(log log u + O(1)) X q≤yk log q X a∈N 1 qa + O 1 qa+1 + O li(u) X q≤yk log2 q X a∈N a qa + O u log y log u = li(u)(log log u + O(1)) X q≤yk log q q + O log q q2 + O li(u) X q≤yk log2 q q + u log y log u = li(u) log log u log(yk) + O li(u)(log y + log log u + log2 y) + u log y log u = ku log log u log y log u + O u log log u log u + u log2 y log u , 32 completing the base case. Now using Lemma 25 we get gk,l(v) = li(v) Z v1/3 2 1 u2gk,l−1(u)du + O v(log log v)l−2 log y log v = li(v) Z v1/3 2 1 u2 ku(log log u)l−2 log y (l −2)! log u + O u(log log u)l−2 log u + u(log log u)l−3 log2 y log u du + O v(log log v)l−2 log y log v = li(v) Z v1/3 2 k(log log u)l−2 log y (l −2)!u log u + O (log log u)l−2 u log u + (log log u)l−3 log2 y u log u du + O v(log log v)l−2 log y log v = k li(v)(log log v1/3)l−1 log y (l −1)!
+ O li(v)(log log v1/3)l−1 + li(v)(log log v1/3)l−2 log2 y + v(log log v)l−2 log y log v .
Once again by using log log v1/3 = log log v + O(1) we get kv(log log v)l−1 log y (l −1)! log v + O v(log log v)l−1 log v + v(log log v)l−2 log2 y log v + v(log log v)l−2 log y log v = kv(log log v)l−1 log y (l −1)! log v + O v(log log v)l−1 log v + v(log log v)l−2 log2 y log v , completing the induction.
□ 13. The Proof of the First Moment We now are in a position to prove the propostion for the first moment.
Proof of Proposition 14.
M1(x) = X p≤x hk(p) p = X p≤ee hk(p) p + X ee 0.
Proof. Our sum is X q1,q2≤yk log q1 log q2 X a1,a2∈N X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X p3∈Pp4 r3∈Pr4 X s∈Pp3∩Pr3 X p≤t p∈Ps 1 = X q1,q2≤yk log q1 log q2 X a1,a2∈N X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X p3∈Pp4 r3∈Pr4 X s∈Pp3r3 π(t; s, 1).
We split up into two cases.
If qa1 1 qa2 2 > tα, then suppose qa1 1 > tα/2. (the other case is analogous) We get from the trivial bound on π(t; s, 1) that X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 >t α 2 X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X p3∈Pp4 r3∈Pr4 X s∈Pp3r3 π(t; s, 1) = X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 >t α 2 X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X p3∈Pp4 r3∈Pr4 X s∈Pp3r3 t log t s = X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 >t α 2 X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X p3∈Pp4 r3∈Pr4 t log t log log t p3r3 = X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 >t α 2 t log t(log log t)2k−3 qα1 1 qα2 2 .
By letting A = min{a|qa1 1 > t α 2 } we get ≪ X q1,q2≤yk log q1 log q2 t log t(log log t)k−1 qA 1 q2 ≤t1−α 2 log t(log log t)2k−3 X q1≤yk log q1 X q2≤yk log q2 q ≪t1−ϵyk log y.
35 If qa1 1 qa2 2 > tα, then by Lemma 21 part (d) we get X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 qa2 2 ≤tα X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X p3∈Pp4 r3∈Pr4 X s∈Pp3r3 π(t; s, 1) ≪ X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 qa2 2 ≤tα t(log log t)2k−2 qa1 1 qa2 2 log t ≪ X q1,q2≤yk log q1 log q2 t(log log t)2k−2 q1q2 log t = t(log log t)2k−2 log t X q≤yk log q q 2 ≪t(log log t)2k−2 log t log2 y by (3), completing the lemma.
□ We now have enough to finish the second moment which is the final piece of the puzzle.
Proof of Proposition 15.
X p≤t hk(p)2 = X p≤x X p1|p X p2|p1−1 · · · X pk|pk−1−1 X q≤yk νq(pk −1) log q 2 = X q1,q2≤yk log q1 log q2 X a1,a2∈N X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X p2∈Pp3 r2∈Pr3 X p≤t p∈Pp2 p∈Pr2 1 since the condition p1 | p only occurs if p1 = p. We then split up the sum according to whether or not p2 = r2. Lemma 27 deals with the part where s = p2 = r2 leaving us with X q1,q2≤yk log q1 log q2 X a1,a2∈N X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk ...
X p2∈Pp3 r2∈Pr3 p2̸=r2 X p≤t p∈Pp2 p∈Pr2 1 + O t1−ϵyk log y + t(log log t)2k−2 log t log2 y .
The sum becomes X q1,q2≤yk log q1 log q2 X a1,a2∈N X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X p2∈Pp3 r2∈Pr3 π(t; p2r2, 1).
If qa1 1 > tα1, then so is p2, and hence by (26) we get 36 X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 >tα1 X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X p3∈Pp4 r3∈Pr4 t log2 t p3r3 ≪ X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 >tα1 t log2 t(log log t)2k−4 qa1 1 qa2 2 ≪t1−α1 log2 t(log log t)2k−4 X q1,q2≤yk log q1 log q2 X a2∈N 1 qa2 2 ≪t1−α1 log2 t(log log t)2k−4 X q1,q2≤yk log q1 log q2 q2 ≪t1−α1 log2 t(log log t)2k−4(yk log y).
We similarly get the same bound if qa2 2 > tα1. If neither of qa1 1 , qa2 2 exceed tα1, then by (28) and using that for bi = qai i bi φ(bi) ≪1, 1 φ(bi) ≪1 bi , we get X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 ,qa2 2 ≤tα1 X pk∈Pqa1 1 rk∈Pqa2 2 X pk−1∈Ppk rk−1∈Prk · · · X pi∈Ppi+1 ri∈Pri+1 X pi−1∈Ppi ri−1∈Pri · · · X p2∈Pp3 r2∈Pr3 π(t; p2r2, 1) ≪ X q1,q2≤yk log q1 log q2 X a1,a2∈N qa1 1 ,qa2 2 ≤tα1 t(log log t)2k−2 qa1 1 qa2 2 log t ≪t(log log t)2k−2 log t X q1,q2≤yk log q1 log q2 q1q2 ≪t(log log t)2k−2 log2 y log t .
Hence the above gives us that X p≤t hk(p)2 ≪t1−ϵyk log y + t(log log t)2k−2 log2 y log t .
37 Using partial summation we have M2(x) = X p≤x hk(p)2 p = X p≤ee hk(p)2 p + 1 x X ee≤p≤x hk(p)2 + Z x ee dt t2 X ee≤p≤t hk(p)2 ≪1 + 1 x x1−ϵyk log y + x(log log x)2k−2 log2 y log x + Z x ee t−1−ϵyk log y + (log log t)2k−2 log2 y t log t dt ≪y2k−2 log2 y log x + x−ϵyk log y + (log log x)2k−1 log2 y ≪y2k−1 log2 y completing the proof of Proposition 15 and hence Theorem 1.
□ 15. Theorem 2 We now turn our attention to the proof of Theorem 2. It will be necessary to use the following upper bound for the Carmichael function of a product.
Lemma 28. Let a, b be natural numbers, then (32) λ(ab) ≤bλ(a).
Proof. We first note that it suffices to show the inequality whenever b is prime, because if b = p1 . . . pk where the pi are not necessarily distinct, then repeated use of the theorem where b is prime yields λ(ab) = λ(ap1 . . . pk) ≤p1λ(ap2 . . . pk) ≤· · · ≤p1 . . . pkλ(a) = bλ(a).
If b is a prime which divides a, then a = bepe1 1 . . . pek k and ab = be+1pe1 1 . . . pek k .
Therefore λ(ab) = lcm λ(be+1), λ(pe1 1 ), . . . , λ(pek k ) ≤lcm bλ(be), λ(pe1 1 ), . . . , λ(pek k ) ≤b ∗lcm λ(be), λ(pe1 1 ), . . . , λ(pek k ) = bλ(a) where the first inequality is in fact an equality if be = 4. Also note that in this case, it would not be hard to show that λ(ab) | bλ(a). If (a, b) = 1, then 38 λ(ab) = lcm b −1, λ(pe1 1 ), . . . , λ(pek k ) ≤(b −1) lcm λ(pe1 1 ), . . . , λ(pek k ) < bλ(a), ending the proposition.
□ Suppose that g(n) is an arithmetic function of the form φ(h(n)) where h(n) is a (k−1)–fold iterate involving φ and λ. Then we can use equation (32) to get λl+k(n) ≤λl(g(n)) ≤λl g(n) λk(n)λk(n) ≤λl+k(n) g(n) λk(n).
Since g(n) ≤n we have that g(n) λk(n) ≤ n λk(n) = exp 1 (k −1)!(1 + ok(1))(log log n)k log log log n by Theorem 1 and hence λl+k(n) ≤λl(g(n)) ≤λl g(n) λk(n)λk(n) ≤λl+k(n) exp 1 (k −1)!(log log n)k(1+ok(1)) log log log n .
From the fact that λl+k(n) = n exp − 1 (k + l −1)!(1 + ol,k(1))(log log n)k+l log log log n we get λl(g(n)) = n exp − 1 (k + l −1)!(1 + ol,k(1))(log log n)k+l log log log n .
As for φ(g(n)) we note that unless g(n) = φk(n), g(n) can be writen as φl(h(n)) where h(n) is a (k −l)–fold iterate beginning with a λ. From above we can see that h(n) = n exp − 1 (k −l −1)!(1 + ok(1))(log log n)k−l log log log n and so φ(h(n)) is bounded above by h(n) and below by h(n) eγ log log h(n) + 3 log log h(n) = h(n) eγ log log n − 1 (k−l−1)!(1 + ok(1))(log log n)k−l log log log n = h(n) eγ log log n −O 1 (k−l−1)! log n(1 + ok(1))(log log n)k−l log log log n = h(n) exp O(log log log n) which is within the error of h(n).
Hence any string of φ will not change our estimate.
Therefore if g(n) is a k–fold iteration of φ and λ which is not φk(n), but which begins with l copies of φ, then 39 g(n) = n exp − 1 (k −l −1)!(1 + ok(1))(log log n)k−l log log log n yielding our theorem.
Acknowledgements The author would like to thank Greg Martin for his assistance and guidance.
References W.D. Banks, F. Luca, F. Saidak and P. St˘ anic˘ a. Compositions with the Euler and Carmichael Functions, Abh. Math. Sem. Univ. Hamburg., 75 (2005), 215–244.
P. Erd˝ os, A. Granville, C. Pomerance, and C. Spiro, On the Normal Behavior of the Iterates of some Arithmetic Functions, in Analytic number theory (Allerton Park, IL, 1989), Progr. Math., 85 Birkh¨ auser Boston, Boston, MA, (1990), 165–204.
P. Erd˝ os, C. Pomerance, and E. Schmutz, Carmichael’s Lambda Function, Acta Arith., 58 (1991), 363–385.
H. Halberstam and H. Richert Sieve Methods, Academic Press [A subsidiary of Harcourt Brace Jo-vanovich, Publishers], London-New York (2004) London Mathematical Society Monographs, No.4.
J.P. Kubilius, Probabilistic Methods in the Theory of Numbers, Translations of Mathetical Monographs, Vol. 11, American Math. Soc., Providence (1964) G.Martin and C. Pomerance, The Iterated Carmichael λ–Function and the Number of Cycles of the Power Generator, Acta Arith., 118 (2005), no. 4, 305–335.
H. Montgomery and R. Vaughan, Multiplicative Number Theory I. Classical Theory, Cambridge Uni-versity Press, New York (2007) C. Pomerance, On the Distribution of Amicable Numbers, J. Reine Angew. Math., 293/294 (1977), 217–222.
Department of Mathematics, University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, BC, V6T 1Z2, Canada E-mail address: harlandn@math.ubc.ca 40 |
190748 | https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/15%3A_Benzene_and_Aromaticity/15.S%3A_Benzene_and_Aromaticity_(Summary) | 15.S: Benzene and Aromaticity (Summary) - Chemistry LibreTexts
Skip to main content
Table of Contents menu
search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode
school Campus Bookshelves
menu_book Bookshelves
perm_media Learning Objects
login Login
how_to_reg Request Instructor Account
hub Instructor Commons
Search
Search this book
Submit Search
x
Text Color
Reset
Bright Blues Gray Inverted
Text Size
Reset
+-
Margin Size
Reset
+-
Font Type
Enable Dyslexic Font - [x]
Downloads expand_more
Download Page (PDF)
Download Full Book (PDF)
Resources expand_more
Periodic Table
Physics Constants
Scientific Calculator
Reference expand_more
Reference & Cite
Tools expand_more
Help expand_more
Get Help
Feedback
Readability
x
selected template will load here
Error
This action is not available.
chrome_reader_mode Enter Reader Mode
15: Benzene and Aromaticity
Organic Chemistry (Morsch et al.)
{ }
{ "15.00:Introduction" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15.01:_Naming_Aromatic_Compounds" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15.02:_Structure_and_Stability_of_Benzene" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15.03:_Aromaticity_and_the_Huckel_4n__2_Rule" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15.04:_Aromatic_Ions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15.05:_Aromatic_Heterocycles-Pyridine_and_Pyrrole" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15.06:_Polycyclic_Aromatic_Compounds" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15.07:_Spectroscopy_of_Aromatic_Compounds" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15.S:_Benzene_and_Aromaticity(Summary)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
{ "00:Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Structure_and_Bonding" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Polar_Covalent_Bonds_Acids_and_Bases" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Organic_Compounds-_Alkanes_and_Their_Stereochemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Organic_Compounds-Cycloalkanes_and_their_Stereochemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Stereochemistry_at_Tetrahedral_Centers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_An_Overview_of_Organic_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Alkenes-_Structure_and_Reactivity" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Alkenes-_Reactions_and_Synthesis" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Alkynes-An_Introduction_to_Organic_Synthesis" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Organohalides" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Structure_Determination-Mass_Spectrometry_and_Infrared_Spectroscopy" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Structure_Determination-Nuclear_Magnetic_Resonance_Spectroscopy" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "14:_Conjugated_Compounds_and_Ultraviolet_Spectroscopy" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15:_Benzene_and_Aromaticity" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "16:_Chemistry_of_Benzene-Electrophilic_Aromatic_Substitution" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "17:_Alcohols_and_Phenols" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "18:_Ethers_and_Epoxides_Thiols_and_Sulfides" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "19:_Aldehydes_and_Ketones-_Nucleophilic_Addition_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "20:_Carboxylic_Acids_and_Nitriles" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "21:_Carboxylic_Acid_Derivatives-_Nucleophilic_Acyl_Substitution_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "22:_Carbonyl_Alpha-Substitution_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "23:_Carbonyl_Condensation_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "24:_Amines_and_Heterocycles" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "25:_Biomolecules-_Carbohydrates" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "26:_Biomolecules-_Amino_Acids_Peptides_and_Proteins" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "27:_Biomolecules-Lipids" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "28:_Biomolecules-Nucleic_Acids" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "29:_Orbitals_and_Organic_Chemistry-_Pericyclic_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "30:_Synthetic_Polymers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
Mon, 18 Mar 2024 05:32:25 GMT
15.S: Benzene and Aromaticity (Summary)
214191
214191
Delmar Larsen
{ }
Anonymous
Anonymous User
2
false
false
[ "showtoc:no", "cssprint:dense", "author@Layne Morsch", "authorname:libretexts" ]
[ "showtoc:no", "cssprint:dense", "author@Layne Morsch", "authorname:libretexts" ]
Search site Search Search Go back to previous article
Sign in
Username Password Sign in
Sign in
Sign in
Forgot password
Contents
1. Home
2. Bookshelves
3. Organic Chemistry
4. Organic Chemistry (Morsch et al.)
5. 15: Benzene and Aromaticity
6. 15.S: Benzene and Aromaticity (Summary)
Expand/collapse global location
Organic Chemistry (Morsch et al.)
Front Matter
1: Structure and Bonding
2: Polar Covalent Bonds; Acids and Bases
3: Organic Compounds- Alkanes and Their Stereochemistry
4: Organic Compounds - Cycloalkanes and their Stereochemistry
5: Stereochemistry at Tetrahedral Centers
6: An Overview of Organic Reactions
7: Alkenes- Structure and Reactivity
8: Alkenes - Reactions and Synthesis
9: Alkynes - An Introduction to Organic Synthesis
10: Organohalides
11: Reactions of Alkyl Halides- Nucleophilic Substitutions and Eliminations
12: Structure Determination - Mass Spectrometry and Infrared Spectroscopy
13: Structure Determination - Nuclear Magnetic Resonance Spectroscopy
14: Conjugated Compounds and Ultraviolet Spectroscopy
15: Benzene and Aromaticity
16: Chemistry of Benzene - Electrophilic Aromatic Substitution
17: Alcohols and Phenols
18: Ethers and Epoxides; Thiols and Sulfides
19: Aldehydes and Ketones- Nucleophilic Addition Reactions
20: Carboxylic Acids and Nitriles
21: Carboxylic Acid Derivatives- Nucleophilic Acyl Substitution Reactions
22: Carbonyl Alpha-Substitution Reactions
23: Carbonyl Condensation Reactions
24: Amines and Heterocycles
25: Biomolecules- Carbohydrates
26: Biomolecules- Amino Acids, Peptides, and Proteins
27: Biomolecules - Lipids
28: Biomolecules - Nucleic Acids
29: Orbitals and Organic Chemistry - Pericyclic Reactions
30: Synthetic Polymers
Back Matter
15.S: Benzene and Aromaticity (Summary)
Last updated Mar 18, 2024
Save as PDF
15.7: Spectroscopy of Aromatic Compounds
16: Chemistry of Benzene - Electrophilic Aromatic Substitution
Page ID 214191
Layne Morsch
LibreTexts
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Concepts & Vocabulary
2. Skills to Master
Concepts & Vocabulary
15.0 Introduction
Aromatic compounds contain ring structures with a special type of resonance delocalization.
Aromatic compounds can be drawn with alternating single and double bonds, each atom in the ring must have a p-orbital available.
15.1 Naming Aromatic Compounds
Disubstituted benzene derivatives are often named using ortho (1,2), meta (1,3) and para (1,4).
There are common benzene derivative names that are used by IUPAC such as toluene, phenol, benzoic acid and benzaldehyde.
A benzene group that is named as a substituent is called phenyl.
A benzene with a CH 2 as a substituent group is called benzyl.
15.2 Structure and Stability of Benzene
Benzene does not undergo the same reactions that alkenes do, due to its aromatic stability.
Aromatic molecules must have all ring atoms in the same plane to allow delocalization of the pi electrons.
Heats of hydrogenation can be used to show the special stability of benzene compared to what would be expected for a theoretical cyclohexatriene molecule.
15.3 Aromaticity and the Hückel 4n + 2 Rule
The four criteria for aromaticity are that the molecule must:
be cyclic
be planar
be fully conjugated
have 4n+2 π Electrons
Ionic molecules and heterocyclic molecules can also be aromatic if they meet the four criteria.
15.4 Aromatic Ions
Carbanions and carbocations that meet the rules for aromaticity are also aromatic.
15.5 Aromatic Heterocycles: Pyridine and Pyrrole
Heterocycles that meet the rules for aromaticty are also aromatic.
If a lone pair of electrons on a ring atom can result in 4n+2 π Electrons, they will be in a p-orbital. If not, they will remain in hybrid orbitals.
15.6 Polycyclic Aromatic Compounds
Benzene rings can be fused together to give larger aromatic compounds with mutliple rings called polycyclic aromatic compounds (or polycyclic aromatic hydrocarbons).
15.7 Spectroscopy of Aromatic Compounds
Aromatic compounds can be identified by common infrared absorptions in the 3000-3100 cm-1 and 1500-1600 cm-1.
In 1 H NMR, aromatic hydrogens appear in the 6.5-8 ppm region.
Skills to Master
Skill 15.1 Using IUPAC rules to name substituted benzene molecules.
Skill 15.2 Use heats of hydrogenation to explain aromatic stabilization.
Skill 15.3 Draw molecular orbital diagram for benzene (all 6 MO's).
Skill 15.4 Use the criteria for aromaticity to determine if a molecule is aromatic or not.
Skill 15.5 Determine whether lone pairs of electrons for ions and heterocycles will be in p orbitals or hybrid orbitals.
Skill 15.6 Identify aromatic absorbances in infrared spectroscopy.
Skill 15.7 Identify aromatic resonances in 1 H NMR spectroscopy.
This page titled 15.S: Benzene and Aromaticity (Summary) is shared under a not declared license and was authored, remixed, and/or curated by Layne Morsch.
Back to top
15.7: Spectroscopy of Aromatic Compounds
16: Chemistry of Benzene - Electrophilic Aromatic Substitution
Was this article helpful?
Yes
No
Recommended articles
15.0: IntroductionAromatic hydrocarbons contain ring structures with delocalized pi electron systems and are more stable than their saturated analogs.
15.1: Naming Aromatic Compounds
15.2: Structure and Stability of Benzene
15.3: Aromaticity and the Hückel 4n + 2 RuleIn 1931, German chemist and physicist Erich Hückel proposed a theory to help determine if a planar ring molecule would have aromatic properties. His r...
15.4: Aromatic IonsCarbanions and carbocations may also show aromatic stabilization.
AuthorLibreTextsPrint CSSDenseShow Page TOCno on page
Tags
author@Layne Morsch
© Copyright 2025 Chemistry LibreTexts
Powered by CXone Expert ®
?
The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org.
Support Center
How can we help?
Contact Support Search the Insight Knowledge Base Check System Status×
contents readability resources tools
☰ |
190749 | https://www.biorxiv.org/content/10.1101/2021.06.10.445299v1.full-text | A deep learning approach to capture the essence of Candida albicans morphologies | bioRxiv
Skip to main content
Home
Submit
FAQ
Blog
ALERTS / RSS
Resources
About
Channels
4D Nucleome
Academia Sinica
Advances in Genome Biology and Technology (AGBT) General Meeting 2016 #AGBT16
Albert Einstein College of Medicine
Aligning Science Across Parkinson's (ASAP)
Allen Institute for Cell Science
Arc Institute
Babraham Institute
BioImaging North America
Biology of Genomes 2016 #BOG16
Brown University
California Institute of Technology
Carnegie Mellon University
Case Western Reserve University
Central Oxford Structural Microscopy and Imaging
Centre for Microbiology and Environmental Systems Science
Chan Zuckerberg Biohub
Columbia University
Donders Institute for Brain, Cognition and Behaviour
DREAM
Drug Development and Clinical Therapeutics
ENCODE
Ernst Strüngmann Institute (ESI) for Neuroscience
European Molecular Biology Laboratory (EMBL)
Francis Crick Institute
Fred Hutchinson Cancer Center
Georgia Institute of Technology
Harvard Program in Therapeutic Sciences
Human Cell Atlas
Human Pangenome Reference Consortium (HPRC)
IMO Workshop
Impact of Genomic Variation on Function (IGVF)
Institute of Science and Technology Austria
International Human Epigenome Consortium (IHEC)
International Mouse Phenotyping Consortium (IMPC)
Iowa State University
Johns Hopkins University
Mathematical Oncology
Micron Oxford
Michigan State University
National Taiwan University
NCI Cancer Systems Biology Consortium
NF Open Science Initiative
NCI Human Tumor Atlas Network
Neuromatch Conference
NeurotechEU
North Carolina State University
Northeastern University
Oregon Health & Sciences University
Rosetta Commons
Rutgers University
SeroNet
Simons Foundation Autism Research Initiative (SFARI)
Society for Molecular Biology and Evolution #SMBE2016
Somatic Cell Genome Editing Program
SPARC
Stowers Institute for Medical Research
Stockholm University
Tel Aviv University
The Michael J. Fox Foundation
The Rockefeller University
The Sainsbury Laboratory
The Whitehead Institute
University of Connecticut Health Center
University of California, Berkeley
University of California, San Diego
University of California, San Francisco
University of Chicago
University of Geneva
University of Guelph
University of Hong Kong
University of Illinois Chicago
University of Iowa
University of Kansas
University of Massachusetts Chan Medical School
University of New South Wales
University of Ottawa
University of Sydney
Vanderbilt University
Vienna BioCenter
Washington University in St. Louis
Weizmann Institute of Science
Yale University
Search for this keyword
Advanced Search
New Results Follow this preprint
A deep learning approach to capture the essence of Candida albicans morphologies
V Bettauer, ACBP Costa, RP Omran, S Massahi, E Kirbizakis, S Simpson, V Dumeaux, C Law, M Whiteway, MT Hallett
doi:
Now published in Microbiology Spectrum doi: 10.1128/spectrum.01472-22
V Bettauer
1 Centre for Applied Synthetic Biology
3 Department of Computer Science and Software Engineering
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
ACBP Costa
1 Centre for Applied Synthetic Biology
2 Department of Biology
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
RP Omran
1 Centre for Applied Synthetic Biology
2 Department of Biology
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
S Massahi
1 Centre for Applied Synthetic Biology
2 Department of Biology
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
E Kirbizakis
1 Centre for Applied Synthetic Biology
2 Department of Biology
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
S Simpson
1 Centre for Applied Synthetic Biology
3 Department of Computer Science and Software Engineering
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
V Dumeaux
2 Department of Biology
4 PERFORM Centre
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
C Law
5 Centre for Microscopy and Cellular Imaging
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
M Whiteway
1 Centre for Applied Synthetic Biology
2 Department of Biology
6 Centre for Structural and Functional Genomics Concordia University, Montreal, Quebec, Canada
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
MT Hallett
1 Centre for Applied Synthetic Biology
2 Department of Biology
6 Centre for Structural and Functional Genomics Concordia University, Montreal, Quebec, Canada
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
For correspondence: michael.hallett@concordia.ca
Abstract
Full Text
Info/History
Metrics
Supplementary material
Data/Code
Preview PDF
Abstract
We present deep learning-based approaches for exploring the complex array of morphologies exhibited by the opportunistic human pathogen C. albicans. Our system entitled Candescence automatically detects C. albicans cells from Differential Image Contrast microscopy, and labels each detected cell with one of nine vegetative, mating-competent or filamentous morphologies. The software is based upon a fully convolutional one-stage object detector and exploits a novel cumulative curriculum-based learning strategy that stratifies our images by difficulty from simple vegetative forms to more complex filamentous architectures. Candescence achieves very good performance on this difficult learning set which has substantial intermixing between the predicted classes. To capture the essence of each C. albicans morphology, we develop models using generative adversarial networks and identify subcomponents of the latent space which control technical variables, developmental trajectories or morphological switches. We envision Candescence as a community meeting point for quantitative explorations of C. albicans morphology.
Introduction
Fungal infections represent an urgent and significant threat to human health affecting 1.2 billion people yearly1–3. They kill approximately the same number of people (1.6 million) as malaria4, and are implicated in cancer progression5. Candida albicans is one of the most important of these human pathogens6 and carries a significant socio-economic burden2,7–9. As such, it has become an important system for studying fungal pathogenicity. C. albicans is morphologically classified as a pleomorphic yeast-like fungus and systematically classified as an ascomycete. It is well adapted to its role as an opportunistic pathogen with a diverse range of morphologies that predominate in different niches. The morphologies can be broadly partitioned into vegetative and mating-competent forms10,11. We provide a brief review of the morphologies central to our effort here (Figure 1A-E).
Download figure
Open in new tab
Figure 1.
A. An example of a typical DIC image in Labelbox, a learning-set creation tool for group annotation. The color codes for bounding boxes described in panel B. The artifact class is used to bound imperfections and technical artifacts in the images, whereas the unknown class is used to bound cells for which we could not judge morphology. C. Labelling of an image that is enriched for pseudohyphae. Note that the overall intensity of our images are not required to be the same. D. Given the size and complexity of the filamentous forms, we annotated each hyphae or pseudohyphae using three classes. D, E. H-start and P-start are intended to label only the “start” of the hyphae or pseudohyphae. Since junctions are different between hyphae and pseudohyphae, we labelled them with the H- and P-junction classes. Finally we bounded the entire filamentous object with a rectangle; note that such bounding boxes overlap with surrounding cells. F. Our system is built upon the FCOS software with structure unaltered from Tian et al. (2019). G. The Varasana learning set is first partitioned into three components (training, validation, and testing). Then both validation and testing are split into six grades. The grades are ordered from white to hyphae. When an image appears first in grade i, it is also included in all grades j > i. The test set contains examples of both wildtype SC5314 and SN148a cells, but also a collection of genetic and environmental perturbations that induced abnormal C. albicans morphologies. H. Using the Varasana learning set, we performed a grid search across several hyperparameters (n=∼30K combinations). Bold face indicates the combination of hyperparameters that had the best performance.
Vegetative forms differ in cell and colony morphologies, their molecular state and their behaviors as infectious agents. These yeast forms are commonly found on mucosal and skin surfaces where they grow benignly and are tolerated by the host immune system12. The white form has a round unicellular morphology (Figure 1A,B). If engulfed by macrophages, the white yeast will switch to the hyphae morphology as a means of immune escape13. The opaque form is larger and has a more rectangular shape than white cells (Figure 1A,B). C. albicans can mate only when both cells are in the opaque morphology14. Switching between the two morphologies is rare (∼104 cell divisions), stochastic, and strongly influenced by environmental cues and homozygosity of mating type. White cells are believed to be better suited to internal infections, while opaque cells thrive in skin infections15. The gray morphology is an alternative, stable form between the white and opaque morphologies (Figure 1A,B). Gray cells are “shiny” and small like white cells yet elongated like opaque cells in solid media16.
C. albicans can also assume mating-competent morphologies including the budding yeast white cell (Figure 1B) and two distinct filamentous forms (Figure 1C-E). The hyphal form is characterized by long tube-like filaments without constrictions at the site of septation. Hyphae are able to invade epithelial and endothelial cells, and damage host tissue in mucosal infections in order to gain access to the bloodstream17. The yeast to hyphal switch is initiated under a variety of environmental conditions including presence of serum, neutral pH, 5% CO2, N-acetyl-D-glucosamine, amongst others18. In the first cell cycle, the germ tube morphology is observed, manifesting as tube projecting from the round yeast cell (Figure 1D). The second filamentous growth form, pseudohyphae, are significantly different from hyphae at the molecular level in terms of cellular regulation and growth19–21 but also in terms of their phenotypic presentation. Pseudohyphae tend to have more branch points, because mother-daughter attachments are more easily disrupted than in hyphae (Figure 1E). Unlike hyphae, pseudohyphal cells exhibit synchronous cell divisions and septation at the mother-bud neck22. Weak filament-inducing conditions (e.g. high-phosphate medium) favour a pseudohyphae over a hyphae morphology. Like hyphal cells, they interact with the mouth, vagina and bloodstream of the host. The shmoo morphology arises from opaque cells which have formed a projection produced by the a and ɑ types when preparing to mate, leading to tetraploid zygote formation.
Some less frequent C. albicans morphologies are not considered in our effort here. This includes chlamydospores which are formed at the end of pseudohyphae and hyphae filaments (suspensor cells). They have a thicker cell wall and are larger than blastospores. The final chlamydospore has an elaborate septin-derived substructure. We also do not consider Gastrointestinally-IndUced Transition (GUT) cells, which are derived from the white morphology when passaging through the gut of the host. The morphology confers survival in the digestive tract via metabolic adaptations to available nutrients in this environment23. The trimera morphology is sometimes observed after unequal chromosome segregation under, for example, stress from antifungal exposure. Finally, we do not consider the goliath morphology which arises in response to mammalian host nutritional immunity strategies including zinc sequestration24. When C. albicans is unable to scavenge host zinc during endothelial invasion, the fungus transforms to a giant yeast cell morphology with advanced adhesion, a property shared with the hyphal morphology.
There have been several earlier efforts in fungal image analysis25 including approaches to computationally differentiate between types of fungi and allergenic fungal spores26, and to characterize the macro-structure of mycelium or filamentous growth27–34. Tleis and Verbeek experimented with a suite of machine learning techniques to segment S. cerevisiae cells and measured a range of features and textures from two channel images acquired by a laser scanning confocal microscope35. Wang et al. provided a segmentation method with an efficient edge-tracing algorithm for bright-field images of fission yeast, which have a consistent oblong morphology36. Another effort employed microfluidics to capture individual S. cerevisiae cells; non-fluorescent images were used to train a classifier of cell cycle state for each cell37.
Deep learning was first applied to biomedical imaging in 2012 with the approach from Ciresan and colleagues38 to automatically segment neuronal structures depicted in stacks of electron microscopy images. Shortly thereafter, Rosenberger et al.39 developed a generic convolution network (termed U-net) for image segmentation that has since been used in many biomedical-related image recognition challenges. Our work here builds upon a different but well-established deep learning architecture entitled Res-Net40. These and other deep learning-based approaches have been extended to cell counting, detection and morphology. To date, these methods have been primarily used in the context of human tissue analyses41–44.
Deep learning has opened up many new avenues of investigation in microscopy through the introduction of new techniques for image transformation, object localization in super resolution computing, and cross-modality imaging45. One challenge is to establish a model of the morphology of a single cell, and then detect when cellular perturbagens change that morphology46–48. Approaches use both generative adversarial training49 and variational autoencoder-based methods50. Recently, a very powerful form of deep learning-based image analysis entitled Faster Region-based Convolutional Neural Network (Faster R-CNN)51 was applied to images of blood smears from individuals infected with malaria52. The goal was to identify different human cell types concomitantly with the protist Plasmodium vivax in images.
With respect to fungi, there have been significant advances to integrate robotic-controlled hardware with high-content image analysis tools to rapidly screen and analyze hundreds of thousands of images of S. cerevisiae to, for example, identify essential genes and genetic interactions53,54. These works were the first to bring large-scale image analysis at the subcellular level; the first deep learning approach (DeepLoc) expanded on this subcellular microscopy work in S. cerevisiae55.
Our effort investigates two novel inter-related image recognition challenges with C. albicans using methodology from deep learning. We first develop a system that can automatically detect C. albicans cells from microscopy images and label each detected object with its morphology. The model is trained using our large compendium of Differential Interference Contrast (DIC) images containing individual C. albicans cells in different morphological states. We treat this problem as a multi-class, multi-object object detection problem where 15 distinct classes are used to describe nine morphologies. Some images contain as many as one hundred tightly packed cells. Our method builds upon a fully convolutional one-stage object detector (FCOS)56 and a novel cumulative curriculum-based learning strategy that stratifies images by difficulty. Then, building upon our ability to accurately identify and classify cells, we develop a deep learning-based model that captures the essence of each C. albicans morphology. These models, which are based upon generative adversarial networks (GANs), can be interrogated to identify components of its latent space that control various features of the images. This includes technical variables but also biologically-relevant processes such as developmental trajectories or transitions between morphologies. These models provide the first dynamic and continuous approach to capturing the essence of the different morphological states and transitions between them. We show how these models can be used to automatically identify subtle changes from wild-type phenotype when given images from genetically perturbed C. albicans populations.
Results
A deep learning approach to recognizing C. albicans morphologies
Our first goal is to develop a fully automated tool that identifies the location and morphology of cells from microscopy images of C. albicans populations. We cast this challenge as a multi-object, multi-class detection problem: (i) the system identifies the location of all C. albicans cells in each image, and then (ii) predicts the morphology of each hypothesized cell. Images may contain an arbitrary number of individual objects (cells or artifacts). Our system classifies nine of the 12 reported C. albicans morphologies: yeast white, budding yeast, opaque, budding opaque, shmoo, gray-like, budding gray-like, hyphae, and pseudohyphae. Each detected object is assigned the morphology with the highest likelihood, provided the probability of the prediction exceeds a user-defined threshold τ.
Our software (entitled Candescence) is based on what is termed a fully-convolutional one-stage object detector (FCOS)56, a new approach that has several benefits over other deep learning algorithms for multi-object, multi-class problems (Figure 1F). One of its key advantages resides from how it flags areas of an image likely to contain an object. Rather than requiring many parameters, which collectively control the size, location and total number of bounding boxes, the FCOS considers each individual pixel in an image as potentially centering an object, abating the need to optimize many hyperparameters simultaneously. FCOSs are able to handle objects of variable size, an important property given the difference between, for example, yeast white and hyphal cells.
FCOSs can make use of transfer learning, a technique where a neural network is first trained in a distinct but similar context compared to the problem at hand. In our case, we begin with a FCOS trained with the ResNet-101 dataset, a convolutional neural network trained on more than a million images partitioned into 1000 classes of common household objects and animals40. Intuitively, transfer learning guarantees that our neural network comes equipped with the basic circuitry to recognize simple shapes, shades, edges and textures. Candescence allows this set of rich feature representations to be retrained for our morphologies, requiring only a tractable number of C. albicans images. Methods 1 describes the architecture and technical parameters of the FCOS.
An image compendium of C. albicans morphologies
Supervised machine learning problems including this image recognition problem exploit a learning set, which is a curated collection of images where bounding boxes have been drawn and labelled for every object. We constructed such a learning set by growing C. albicans SC5314 or SN148A colonies under conditions necessary to induce each of the nine morphologies (Table 1A, Methods 2): yeast white, budding white, opaque, budding opaque, gray-like, budding gray-like, shmoo, hyphae, and pseudohyphae. The number of observed goliath, zygote, chlamydospores and trimera cells was too small for computational training and were excluded from the analysis. This first version of our software also did not consider C. albican s GUT cells as they require special culturing conditions. Each population was stained with calcofluor white and prepared for DIC and fluorescent microscopy at various magnifications (40x-100x; Methods 3). We also cultured and imaged C. albicans colonies that had specific genetic perturbations known to affect morphology. In total, 1214 images were generated (Supplemental Table 1).
View this table:
View inline
View popup
Download powerpoint
Table 1.
Summary of the Varasana learning set from Supplemental Table 1. An A indicates that some images from this type of colony were used in the training and validation datasets, and B indicates that some images were used in the test set. All genetic perturbations were performed in SC5314 cells. In the Serum column, a No indicates only YPD media was used, otherwise the percent serum added to YPD is recorded. RM indicates room temperature. The significance column indicates the results of applying the proportionality test to determine if there was a difference in performance between the validation and test set. Here and indicate p-values below 0.05 and 0.01 respectively and -- indicate a p-values > 0.05. In cases where a p-value is provided for a type A colony, only images that were omitted from the training and validation datasets were used.
In each of the resultant DIC images, a bounding box was manually drawn around each C. albicans cell and labelled with its morphological class following the guidelines of Sudbery et al.21, Whiteway and Bachewich11, Tao et al.16 and Noble et al.23 (Methods 4). More specifically, cells with a round-to-oval morphology (4.9μm × 6.8μm) were labelled white unless they were attached to a second smaller white cell with concomitant evidence of a bud site (as sometime assisted by manual inspection of the calcofluor white signal in the matched fluorescent image); in this case they were labelled budding white (Figure 1A,B). Ellipsoidal cells approximately twice the size of white cells (9.5μm × 11.8μm) were labelled opaque. Like white cells, they were labelled budding opaque if there was evidence of a smaller opaque cell with a new bud site. Cells were labelled shmoo if they had an irregular (often boomerang) shape and large vacuoles. Many of our images contain a significant number of vegetative-like cells that are visibly distinct from both white and opaque morphologies. These cells are ellipsoidal in shape, with a size in the upper quantile of white cells, but below the size of opaque cells. These properties are close in spirit to the gray morphology of Tao and colleagues16, although we did not grow C. albicans colonies under conditions that specifically induce this morphology. To date, confirmation of the gray cell morphology requires the use of specific molecular markers. However, given the distinctiveness and ubiquitousness of these cells in our data, we decided to explicitly model this structure and use the term gray-like.
The diversity and complexity of filamentous cells necessitated the development of an approach that uses several markers concomitantly to reliably identify them. Hyphae are thin tube-shaped cells with a width of ∼2μm. Pseudohyphae are multicellular entities which tend to be elongated and ellipsoidal in structure. The minimum width is 2.8μm. Our markup scheme first bounds the entire hyphae or pseudohyphae. Since filamentous cells are often long and irregular in shape, the bounding box almost always overlaps, or completely subsumes, the bounding box of other objects. Since this overlap represents a significant challenge for learning algorithms, we also labelled the location of the original germ bud of each hyphae, or estimated the start cell of the pseudohyphae; these bounding boxes were labelled H- and P-start respectively. Compared to hyphae/pseudohyphae bounding boxes, H- and P-start are much smaller, and therefore disjoint from other objects in the image (Figure 1C-E). As is the case for the start sites, the septal junctions in hyphae and pseudohyphae are visually distinct from one another. Only pseudohyphae have constrictions at the mother-bud neck and subsequent septal junctions. We placed bounding boxes at these junctions and labelled them as H- or P-junction. Those cells for which we could not reach agreement on morphology were labelled as unknown; non-cellular events in the images were labelled as artifacts.
We stress here that our strategy attempted to label each cell using only its visual appearance and independent of other factors including the predominant morphology of cells in the image. For example, in an image of a colony grown under conditions which will enrich for yeast white cells, we still expect one in 104 cells to stochastically assume the opaque morphology; cells on the border between these morphologies may have been biased towards the background yeast white form. In some cases, a consensus was difficult to achieve across labellers.
Varasana: a cumulative curriculum-based C. albicans learning set for Candescence
Learning sets are typically tri-partitioned (Figure 1G). The training set contains a large set of images which are presented to the neural network during training and used to fit the model (i.e. update the weights of each arc in the neural network). The validation set is typically a smaller sample of data which is used to generate an unbiased estimation of the model fit during training. It provides a means to tune the hyper-parameters of the model. The test dataset is used only once after all training is complete to assess the final fit of the model. We partitioned the images from Table 1A so that cells from each morphology are assigned to the training and validation set at a 7:3 ratio. This procedure was confounded by the fact that there is significant variability in both the number of cells per image and the composition of morphologies per image. From the collection of 1214 images, the training and validation dataset contains 216 and 94 images respectively. These images in turn contain 4880 and 1958 objects respectively. The independent test contains the remaining 904 images.
FCOSs have several hyper-parameters that affect the overall performance of the system including the learning rate (the amount weights are updated during gradient descent), momentum (a parameter that stipulates how many previous steps can be used to determine the direction of a weight update), decay (a regularization parameter restraining the complexity of our model), epoch number (the number of times the learning phase cycles through the complete training and validation sets), the IoU (the intersection of union statistic controls how accurate the regression of the bounding box must be), the threshold τ (the lower bound for the probability an object is assigned a class), and others (Figure 1H). We performed a grid search across a range of values for these parameters and measured convergence, model complexity and performance after each trial (Methods 4-5). Although the resultant classifiers had good performance for some morphologies (e.g, F1 ∼78% for white, budding white, opaque, gray and shmoo), several classes remained poorly predicted (e.g. F1 ∼50% for pseudohyphae and hyphae related classes) and budding classes were poorly distinguished from their parent (e.g. F1 ∼60% for budding white, gray and opaque).
The steep increase in difficulty between yeast white and the filamentous classes highlighted the need for a more structured learning set. We opted for a curriculum approach57,58, a well-established concept in psychology which has re-emerged in the deep learning community. The fundamental idea is to structure the learning set so that the neural network is exposed to concepts according to their difficulty, with the easiest concepts presented first. Using the results from our original grid search to judge easy and difficult objects, we re-designed our learning set into grades one through six. Each grade was enriched for a specific subset of classes, although at least one example of all classes was present at each grade. Simple examples of yeast white for example were placed in grade 1, and more complicated examples (e.g. crowded images, poor image quality) were placed in later grades. In general, grade 1 is enriched for yeast white and budding white cells, grade 2 introduces opaque and budding opaque, grade 3 presents gray and budding gray, grade 4 focuses on the shmoo form, grade 5 on pseudohyphae and grade 6 concludes with hyphae. Once an image appears at a certain grade, it appears in all subsequent grades. We hypothesize that this cumulative strategy ensures that lessons learnt early with populous morphologies such as yeast white and opaque are retained when the complicated filamentous morphologies are presented to the learner. All grades had many examples of artifacts and unknowns. To the best of our knowledge this is the first use of a cumulative curriculum learning approach.
A new grid search was conducted but this time the hyper-parameters were allowed to vary across the different grades. This grid search also considered different levels of freezing in our four layer FCOS. Freezing refers to the process of disallowing layers of the neural network to change. For example, the most restrictive freezing regimen disallows any of the four layers to change in response to new examples, implying that our classification is solely based on the original transferred ResNet-101. The most permissive freezing strategy allows all layers to be updated during training when presented with our images. Supplemental Figure 1 breaks down the learning set by grade and classes, and provides the distribution of the number of cells per image.
Searching for classifiers of C. albicans morphologies
A second grid search was performed with the validation dataset across the hyperparameter space. The search required ∼60 days of continuous computation on a system with 10 GPUs. We judged performance initially using the mean average precision (mAP)59, a standard approach for measuring the performance of multi-object/multi-class problems, across all trials with a final value of 0.407 (Methods 5). This best classifier has a learning rate of 0.01, momentum of 0.97, and decay of 0.001 (Figure 1H). Analysis of the loss curves established that 1,000 epochs at each grade sufficed for convergence for each of the three component loss curves (Methods 5, Supplemental Figure 2). An initial performance assessment suggested that an IoU of 0.5 and a τ of 0.25 maximized the recall, precision and F1 which were estimated to be 82.4%, 66.5% and 73.7% respectively (Table 2).
View this table:
View inline
View popup
Download powerpoint
Table 2.
Exploration of the performance of the FCOS in the grid search. Panel A provides the relevant summary statistics for blindspots, hallucinations and classifications. With a posterior analysis and correction of the hallucinations, performance was recalculated and reported in the Adj Sens (adjusted sensitivity) column. True positives correspond to a correctly identified object that is also correctly classified. In an FCOS, all pixels that are not part of a bounding box and which are not predicted to be part of a bounding box are true negatives. False positives are hallucinations and incorrect classifications. False negatives are blindspots only. Panel B reports the blindspots per class for the best performing classifier at τ=0.25.
Somewhat surprisingly, the performance of classifiers with more liberal freezing strategies was better than strategies that froze layers. We hypothesize that the cumulative nature of our learning set guaranteed that the network retained learnt rules from early grades without the need for freezing. To the best of our knowledge, we have not seen any exploration of cumulative curriculum learning nor the observation that a cumulative curriculum learning approach may suffice in lieu of freezing. Figure 2 depicts the ground truth labels (left) and predicted objects and their classifications (right) across three typical images containing many of the morphologies. The images help to show that Candescence is able to cope with overlapping objects in dense images.
Download figure
Open in new tab
Figure 2.
Representative images of the ground truth in the Varasana validation set (left) each of which consists of a bounding box around the object and a class label, and the predictions made by the FCOS Candescence (right) which also have a score between 0 and 1 from the softmax layer of the FCOS representing the strength of belief in the classification. Panel A depicts an image we found difficult to label as the objects have both pseudohyphal and hyphal properties. Nevertheless, Candescence recapitulates our classifications. Note that if an object is identified as hyphae, the junctions and start are also labelled as H; this is also true for pseudohyphae. This suggests that Candescence is learning to classify not only on the image but also as a function of the predicted labels of the same object. The image in Panel B contains a diverse collection of classes. Note that we labelled gray-like only by their size and “texture” (smaller but rectangular like opaque, and more gaunt than white cells). Overall, these categories witnessed the highest number of classification errors. Candescence predicts well even for highly dense and diverse images such as Panel C.
Candescence almost never hallucinates but has some blind spots
The performance of our classifier can be decomposed into its object detection and object classification components. For object detection, false positives refer to cases where Candescence predicts a bounding box in a location of the image that does not have a ground truth bounding box. Using the 94 images of the validation dataset with an IoU of 0.5 and a range of thresholds for τ, we manually examined all false positive predictions. Supplemental Figure 3 depicts all 358 such hallucinations for τ = 0.25. It is difficult in some dark images (n=13) to detect an object and these instances were confirmed as true hallucinations by Candescence. In approximately 30 of the remaining cases, there is in fact an object in the bounding buy but the predicted classification is incorrect. Essentially the labellers (missed bounding box) and Candescence (incorrect class) made a mistake. In all remaining 315 cases, Candescence was correct. After adjusting the performance measures by removing the effects of these human errors, the recall, precision and F1 rose to 85.1%, 80.7% and 83.2% respectively (Table 2A, bold red, Methods 5). Figure 3A provides a small sample of correct and incorrect hallucinations.
Download figure
Open in new tab
Figure 3.
A. Representative images chosen from the complete collection in Supplemental Figure 3 of all false positive object identifications from Candescence. This set of hallucinations is subdivided by the class label returned by Candescence. X indicates that we truly are not able to see an object (a true hallucination) and M indicates that Candescence was indeed correct to predict a bounding box at that location (missed by the human annotators) but its subsequent classification disagrees with our criteria. Subimages lacking an annotation indicate that Candescence correctly identified and classified the object, which represents a human error. Panel B depicts three partial images with false negatives (“blindspots”) labelled with red boxes. Unannotated cells in (i-iii) were correctly handled by Candescence and their bounding boxes have been removed for clarity. (i) This is an example of a recurrent problem where the probability from the softmax is distributed over two or more labels (e.g. white and budding white) presumably because it has difficulty to guess whether they are only touching versus still attached. The shared probability causes both to fall below our threshold τ. (ii) Although Candescence performs well, blindspots arise presumably due to the dense packing of cells. (iii) Our strategy during labelling was to leave cells that were partially outside of the field of view unlabelled. Such edge effects cause some problems and are often predicted as Unknown. Lastly, bounding boxes for large filamentous C. albicans are sometimes missed, especially if other cells are within the vicinity. Panel C depicts the confusion matrix for Candescence with columns corresponding to ground truth classifications and rows corresponding to Candescence predictions.
Candescence however does miss several bounding boxes in the ground truth dataset (n=301 false negatives). Approximately one-quarter of such blindspots correspond to artifacts. Artifact is a heterogeneous class that was used to label all defects in the microscopy images regardless of their individual visual qualities. As such, it is perhaps understandable that Candescence does not learn to predict this class well as there is no consistent set of attributes. Approximately one-quarter of the blindspots are related to objects labelled as yeast white in the ground truth dataset (Table 2B). We observed a recurring pattern across these cases: often the white cell was physically adjacent to a second cell (Figure 3Bi). Forensic investigation of the neural network suggests that the softmax function of the FCOS distributes the probability between the yeast white and budding yeast white uniformly. This causes the score for both categories to fall below our chosen threshold τ = 0.25; hence Candescence fails to identify a bounding box in that location of the image. This also occurred between opaque and budding opaque. In fact, more than one third of the images of the validation dataset had at least one problematic prediction of this form.
Candescence failed to identify several white cells in dense images (Figure 3Bii), and failed to label several germ tubes cells as H-start. As well, P-junctions in pseudohyphae enriched images were often hard to identify (∼10% of all blindspots). A P-junction looks like two adjacent cells with a bright region in the fluorescent image. This pattern is not unlike countless other locations in the images that capture two adjacent cells, or budding cells. Our hope was that Candescence would learn to associate the presence of P-junctions with the larger bounding box of the pseudohyphae itself along with the “nearby” P-start. There is some indication that Candescence has learnt this calculus, although there are many P-junctions per instance of pseudohyphae and they can be quite distal to the P-start. Lastly, some cells near the edges of the image were overlooked such as the example in Figure 3B iii. Overall, many of the blindspots occur in dense, crowded images similar to Figure 3Bii-iii.
Candescence exhibits high classification accuracy
Here we restrict attention to only those objects which have been correctly located in the images (n=2104) and investigate the classification performance of Candescence. In total there are 382 errors (accuracy of 81.8%; Table 2). Figure 3C depicts the confusion matrix across the 15 classes in the validation dataset with diagonal entries corresponding to correct classifications. Comparing our labels to predictions, we observe pronounced confusion between opaque and gray-like with some 51 opaque cells classified as gray-like, and 19 gray-like as opaque. Significant but slightly reduced confusion exists between white and gray-like. In fact, misclassifications between white, gray-like and opaque account for almost one-third of all classification errors. Although there is considerable confusion between these three classes, we note that the statistical performance is still far better than random, suggesting that our visual inspection and labelling was at least partially consistent. A smaller degree of confusion exists between hyphae and H-start. The softmax appears to diffuse probabilities across H-start, white and budding white, for some bounding boxes. This is perhaps another manifestation of the difficulties we observed with blindspots depicted in Figure 3B.
Candescence retains its capacity to classify genetically perturbed C. albicans in the test set
Our test set of 904 images serves as an independent measure of performance. Some images of SC5314 or SN148a strains without genetic modifications had been left out of both the training and validation set but were grown under the same conditions needed to induce different morphologies. We compared Candescence predictions to manual curation for images referenced in Table 1A but differences in performance between the validation and test set were insignificant (comparison of proportions χ 2 test, Methods 5).
The test also contains C. albicans colonies grown with a variety of conditions, preparation protocols, and genetic perturbations (Table 1B, Supplemental Table 1). The four perturbed genes all have well-established and central roles in filamentation processes and the regulation of morphology. Modifications of these genes generated cells that can differ visually from wildtype morphologies, creating an interesting challenge to the ability of Candescence to identify and classify cells.
Our panel includes a transformed SC5314 strain with a CRISPR/Cas9-based homozygous deletion of Unscheduled Meiotic gene Expression (UME6), a Zn(II)2Cys6 (zinc cluster) transcription factor that controls transition to true hyphae by maintaining expression of filament-specific genes in response to inducing conditions. Although cells lacking UME6 are able to form germ tubes, hyphal extension is limited60. Our panel also includes Biofilm ReGulatory 1 (BRG1) encoding a transcription factor that recruits the histone deacetylase Hda1 to hyphal-specific promoters and removes Nrg1 inhibition to promote filamentation. Filamentation is decreased in brg1Δ/brg1Δ cells61. We did not detect a statistical significant difference in performance of Candescence for both UME6 null and BRG 1 null cells under conditions inducing both the white morphology (Supplemental Figure 4A) and the pseudohyphae morphology (Supplemental Figure 4B).
Regulator of Hyphal Activity 1 (RHA1) encodes a zinc cluster transcription factor that serves as a regulator of the Nrg1/Brg1 switch. Hyperactivation of Rha1 can trigger filamentous growth in the absence of external signals or in the presence of serum can bypass the need for Brg162. Loss of Rha1 function leads to reduced ability to generate hyphal growth in the presence of external signatures. Loss of both Rha1 and Ume6 ablates filamentation completely. Although Rha1 null cells generate yeast white cells with standard appearance when grown at 30℃ in YPD, they produce smallish pseudohyphae with reduced branching when grown at 30℃ with serum added to the YPD. Candescence does successfully label pseudohyphal substructures including P-start, P-junction, but there is some increased confusion with hyphae as exemplified in Supplemental Figure 4C. A minor depreciation in accuracy was observed and this was statistically significant (p < 0.05, type 20 Table 1B).
We cultured RHA1 gain of function (GOF) mutants constructed using the zinc cluster hyper-activation technique from Schillig and Morschhaeuser63 and observed that cells grown at 30°C in YPD formed pseudohyphae that tended to look like wildtype (Supplemental Figure 4D). Some of these images were used in the training and validation datasets and performance remains the same on the omitted test images. However, the performance of Candescence is statistically worse in RHA1 GOF/UME6 null cells. Although these cells present a small pseudohyphae morphology, the images tend to contain multiple tightly packed clusters, which we hypothesize contributes to an increase in the number of blindspots (Supplemental Figure 4E).
RHA1 GOF/BRG1 null cells have a morphology distinct from wildtype hyphae and pseudohyphae, comprising elongated chains without branching and with less pinching at junctions (Supplemental Figure 4F). Here Candescence most often labels these cells as hyphae. Interestingly, if an object has been labelled as hyphae, the junctions are labeled as H-junctions and not P-junctions, although often the characteristic pinching of pseudohyphae is clearly present. This may suggest that the classification rules discovered by the deep learner go beyond appearance and use information regarding the labels of nearby objects.
Candescence again experiences a loss in performance with RHA1 GOF/BCR1 null cells grown at 30°C in YPD only. Bcr1 is a C2H2 zinc finger transcription factor which regulates a/α biofilm formation and cell-surface-associated genes. The homozygous null variant exhibited decreased adhesion, biofilm formation, and cell size. There is conflicting data regarding whether it promotes or inhibits filamentous growth, however cells which do transit to a filamentous morphology appear abnormal64,65. In the Varasana image set, this mutant strain generates cells that appear to branch similarly to pseudohyphae but the individual cells are yeast white or budding yeast in appearance. They tend to form thick clusters in the image (Supplemental Figure 4G). The performance decrease is highly significant (p < 0.001) with many cells in dense clumps labelled as white or budding white. Candescence does however identify the location of the vast majority of cells and often correctly labels isolated objects as pseudohyphae. When RHA1 GOF/BCR1 null cells are grown at 37°C with serum added to the medium (types 13, 17, 20, Table 1B), we observed large pseudohyphae and Candescence classifies correctly at the same rate as the validation set (Supplemental Figure 4H).
The space of C. albicans morphologies is complex and continuous
We observed considerable cell-to-cell morphological variability across the images. Heterogeneity arises due to technical variations (e.g. light intensity, focus), natural biological programs (e.g. cell cycle affecting size/shape), and transitions between morphologies (e.g. growth of a hyphal cell from germ tube). These sources of heterogeneity complicate both the manual labeling procedure during construction of the learning set and the downstream ability of an FCOS to correctly assign morphology. Our goal here is to quantitatively explore the complexity of the Varasana compendium in an unbiased manner.
Towards this end, we developed an unsupervised approach based on a variational autoencoder (VAE)66. A VAE is a generative artificial neural network that uses a game theoretic approach involving two players: the encoder and the decoder. Instances of each object (individual cells from our ground truth annotations) are provided to the encoder. The encoder re-represents these images in the latent (hidden) layers of its network. Typically the latent space is much smaller than the size of the original input, forcing the encoder to build succinct models capturing the most salient features of each image. The goal of the decoder is to reconstruct the original image from only the encoder’s latent representation. The encoder and decoder together are penalized according to how much the reconstructed image differs from the original. This cycle is repeated for many epochs across the training and validation sets. Note that this encoding is built in an unsupervised manner as it does not make use of the class labels (morphological assignments). Our particular VAE uses several convolutional layers to encode each image in a 2D latent space across all wildtype (SC5314, SN148a) cells detected by the FCOC-classifier and β parameter that down-weights the Kullbach-Leibler component of the loss function (Methods 6, Supplemental Figure 5).
After training, the resultant encoder was applied to all cells from the test and validation sets and the two-dimensional latent space visualised (Figure 4). Rather than islands of distinct isolated cells, we observe an unbroken continuum in both the V1 and V2 axes. Although some dimensions of the latent space capture specific morphologies (e.g. shmoo, hyphae and pseudohyphae on the left end of the first V1 dimension), there is a ubiquitous imperfect separation between all of the morphologies. This suggests that almost all morphologies have instances that are difficult to differentiate from one another. Yeast white cells span almost the entire first latent dimension, overlapping in some regions heavily with gray-like and opaque. Several technical artifacts including light intensity drive the scatterplot especially in the first V1 dimension. The second V2 dimension primarily captures differences in the size and texture of the cells.
Download figure
Open in new tab
Figure 4
depicts a scatterplot of the two dimensional latent space of the VAE on the training and validation datasets. The junction, hyphae, pseudohyphae, and unknown class have been removed from the training and visualization of the VAE. The V1 dimension of the VAE is strongly associated with light intensity of the image. The vertical V2 dimension captures other variability primarily related to size and texture of the cells.
Capturing the canonical forms of C. albicans morphologies: generative adversarial networks
A sufficiently large collection of images will likely capture snapshots of cells in all technical (e.g. different light intensities), developmental (e.g. across each step of the cell cycle) and morphological states (e.g. along the transition from germ tube to hyphae). Our goal here is to build pseudotime models from these images that capture the progression of these technical and biological variables. Our approach is based on generative adversarial networks (GANs). Intuitively, this deep learning technique re-represents the images in a latent space in a manner that captures this step-by-step progression. Then computational techniques can search for trajectories through the latent space that correspond to a specific effect of interest (e.g. a specific transformation between two morphologies). This allows us to derive continuous “movie-like” models of cells morphing along this trajectory.
GANs are trained using a game-theoretic adversarial approach that pits two deep networks - the generator and the discriminator - against each other67. The purpose of the generator model is to create “fake” examples of C. albicans cells with different morphologies. These fake images are created to deceive its adversary, the discriminator model. The generator is allowed to feed both fake and real images to the discriminator model, whose goal is to differentiate between the two. The generator is then told which images the discriminator got right or wrong; this information is used to update the generative model (which corresponds to updating parameters of the neural network). This creates an “arms race” between the deep networks. After a sufficient number of epochs, the generator will ideally produce fake images of C. albicans morphologies which cannot be distinguished from true images by the discriminator. Our GAN is based on the computationally accessible method from Liu and colleagues68 and learnt using the training and validation components of the Varasana dataset (Methods 7).
After training, we exploit the generator to explore trajectories in our latent space. We start with two images representing the end-points of a process of interest. For example, the left end-point might be a real image of a yeast white cell while the right end-point might correspond to an opaque cell. We then used a process called inversion69 to find a representation of these two images in the latent space of the generator (Methods 7). Figure 5A depicts examples of target images and their nearest neighbour in the latent space. Next, the system finds a linear path between these two points in the latent space, so that the nearest neighbour of the final “fake” image at t 7 is the real image at the right endpoint. Lastly, the intermediate “fake” images are reconstructed from the latent space to provide a visualization of the trajectory using the generator function. Figure 5B (i) depicts the results of applying this procedure to find yeast white to opaque morphological switch. Interestingly, the system seems to arrive at a decision point at t 7. At this point, the trajectory could bifurcate towards budding white, although in this case it continues towards opaque. As a comparison, when we asked for a trajectory from yeast white to budding white (Figure 5B ii), the trajectory stays true to the yeast form and does not appear to wander towards the opaque morphology. The hypothesized bud is perhaps somewhat disproportionately large compared to its mother, and larger than the bud of the real image. Panel 5B (iii) presents a trajectory that starts with a budding opaque cell. Both cells appear to bud a second time from t 4 through t 6. The progeny of the daughter cell (left) disappears from the trajectory. Although imperfect, the system appears to have learnt a reasonable model of pseudohyphal development from relatively few (∼200) images. Continuous movies for each of these trajectories are available in the supplementary material.
Download figure
Open in new tab
Figure 5
A. For each pair of images, the left is a synthesized image produced by our model, and the right image is the image found from the real training data that is its nearest neighbour under the LPIPS metric. B. Three separate trajectories. Here the left and right most images correspond to real images from the training data. The sequence of images at points s 1 to s 8 are produced by a linear interpolation between the real images in the generator’s latent space in the GAN.
Detecting deviations from standard C. albicans morphologies: anomaly detection with GANs
Microscopy is routinely used to judge whether a specific genetic or environmental perturbation has led to an observable phenotype. This could manifest as a visual change in the composition of cells in an image (e.g. an increase in opaque cells versus control), a difference in the spatial distribution of cells in the image (e.g. clumping of cells), or a change in appearance of the cell (e.g. shortened hyphae, large vacuoles). In this section, we build upon our GAN-based morphology models to address the third issue, namely a system capable of deciding whether the cells in an image deviate significantly from the space of wildtype C. albicans morphologies. Our hope is that the deep learner is more sensitive than “eyeballing” microscopy images when manually attempting to investigate if an experimental strain is abnormal.
The algorithm starts by mapping a target bounding box (representing a single cell, hyphae or pseudohyphae) into the latent space of the GAN’s generator (Methods 8). The nearest neighbour in the latent space is found; this corresponds to the object from the training and validation dataset that is visually most similar to the target. The distance between the target cell and nearest neighbour is computed using a specialized similarity metric developed from the Learned Perceptual Image Patch Similarity70 measure. The intuition is that distances between a target with a normal morphology will have a nearest neighbour that is closer in the latent space than a target with a very abnormal morphology.
Figure 6 represents an example of using this test with a RHA1 GOF/BCR1 null colony. Candescence is first used to identify the objects and their morphology in the image, and the anomaly score is computed for each such object in the latent space built from the Varsana learning set. When we compare the distribution of anomaly scores between all images from this mutant colony (type 13) and compare them against a collection of “wildtype” pseudohyphal cells (type 65), we observe a statistical enrichment of outliers with abnormal morphology (Kolmogorov-Smirnoff test, p < 0.01). We generally do not observe differences at the low end of anomaly scores, as almost all images from genetically perturbed colonies still contain many examples of cells with normal morphology.
Download figure
Open in new tab
Figure 6
A. Example of anomaly detection using a RHA1 GOF/BCR1 null colony (UID 230 of type 13). Panel B enlarges a series of bounding boxes predicted by Candescence with their associated anomaly score. The histogram of panel C compares the distribution of anomaly scores between all images of type 13 versus type 30, a collection of normal appearing pseudohyphal and hyphal cells. There is a small but statistically significant enrichment of cells from the RHA1 GOF/BCR1 null colony with elevated anomaly score differences (Kolmogorov-Smirnoff test, p < 0.01).
Conclusions
Candescence includes a multi-object detection algorithm capable of accurately classifying nine C. albicans morphologies. It is based on a fully convolutional one-stage (FCOS) architecture which both locates objects and classifies them with high accuracy. The system is trained with the Varasana learning set consisting of ∼1,200 total images. The training and validation datasets consist of 310 images which have been manually annotated with bounding boxes and class assignments. Using transfer learning, the starting point for training is ResNet-101, a network capable of locating and classifying common household items and pets. It is possible that the image building blocks (textures, edges, colours) encoded in ResNet-101 are not optimal for (DIC) microscopy. As our collection of C. albicans images grows, it may be feasible to build a microscopy-specific analog of ResNet-101 and improve performance.
Since a flat-structured learning set led to suboptimal performance, we developed a six grade curriculum learning set and ordered examples by increasing difficulty. To the best of our knowledge we are the first to use a cumulative strategy where images appear at some grade and re-appear in all subsequent grades. We hypothesize that this approach removes the need for layered freezing strategies. This is advantageous, since optimization of hyper-parameters is computationally expensive and the space of possible freezing strategies grows exponentially in the number of grades and network layers.
Candescence has very good performance. Given that an object has been correctly regressed in the image, it will be assigned the correct class label with ∼82% probability. Misclassifications tend to occur between morphologies that ‘overlap’ as highlighted by the VAE plots of Figure 4. Confusion tends to exist between vegetative morphologies and their budding/mating forms (e.g. from white yeast to budding white). The difficulty here for Candescence is to judge cases where the daughter cell is large but still attached versus detached but still physically adjacent to its mother. We hypothesize that a larger set of training examples will remove this confusion. Inclusion of time lapse images of specific biological processes would also improve performance. Moreover, integration of the matched fluorescent images with the current grayscale DIC image would perhaps provide the learning procedure with the necessary information to judge whether separation has occurred. It would be straightforward to encode the fluorescent image as an added dimension beyond the gray scale 800×800 input currently used. We did not pursue this avenue in the first version due to complications in the training procedure, since some fluorescent images were not available.
The construction of any learning set requires consistent labelling rules. In our setting, there were visible differences in sizable subpopulations that stretch from yeast white to opaque. Figure 4 reinforces that this heterogeneity is continuous with several areas enriched for cells with a common but perhaps non-canonical appearance. This includes a large number of small rectangular “gaunt” cells we labelled as gray-like. Candescence is at times confused between the yeast white, gray-like and opaque morphologies, an expected observation given that there were many images for which we were unable to form consensus as a group. Absolute assessment of true gray versus our gray-like is out of reach for this study as we lack the necessary molecular markers as per the original findings of Tao and colleagues16. Although false classifications are enriched between white, opaque and gray-like, the success rate is still very good, suggesting that this dichotomy does exist in the images. If these cells had instead been randomly assigned the three class labels without regards to physical appearance, it is highly unlikely a classifier could learn to distinguish them with statistically significant performance. Although the relationship to the Tian et al. gray cells remains unsettled, it does suggest that there is interesting substructure across the vegetative forms that is perhaps not captured by our current dichotomy; other cryptic morphologies could exist. These physically distinct subcolonies could represent simply canonical vegetative forms at specific moments of their development, or could represent phenotype diversity that arises in response to an environmental or communal cue71. Building upon this primeval version, Candescence may eventually allow for the interrogation of community structure and interaction.
Imagery of stained or fluorescent reporter molecules would help resolve issues of cryptic morphologies and would extend the capacity of the system to classify using subcellular features. This technique has been used successfully to explore changes to S. cerevisiae in the presence of genetic perturbations55, although Baker’s yeast does not have as large a range of morphologies as C. albicans.
Our system initially appeared to have significant challenges during the object detection step. However, careful analysis of the false positive detections suggest that many such events are in fact not “hallucinations” but “false false positives”. That is, they correspond to true events in the image files that we missed during the manual annotation procedure. A large portion of these events correspond to either subtle technical artifacts or small cells in crowded images.
Candescence appears to have some blind spots, missing cells that are annotated in the ground truth dataset. The tradeoff between false negatives and positives is controlled by underlying IoU and threshold τ parameters. We observed that accurate bounding boxes were indeed regressed for the majority of these false negative objects but such objects required an abatement of either the IoU or τ parameters before they were reported as positive predictions. Our forensic analysis of the neural network suggests that this is due to confusion between, for example, yeast white and yeast budding white; the probability is amortized over two or more classes and therefore drops below τ.
Furthermore with respect to blindspots, we hypothesize that the differences in size and shape between the morphologies (e.g. yeast white versus pseudohyphae) induce different distributions of IoU and τ scores. Therefore, in images containing diverse cell types, the single universal τ is essentially too conservative for some morphologies but too liberal for others. It is an interesting future challenge to modify FCOS-based classifiers to adjust for this heterogeneity in a statistically sound manner.
We explored in the independent test set a range of genetically altered C. albicans populations involving genes RHA1, UME6, BCR1 and BGR1 with established roles controlling filamentation. The observed changes in classification accuracy is consistent with the fact that these morphologies represent shifts away from wildtype forms. The Candescence response to these perturbations is intuitive and largely retains its performance.
Variational autoencoders (VAEs) provide a convenient tool for modelling the diversity of cells caused by natural cellular programs, morphologies and technical artifacts. We observe a nearly unbroken, continuous distribution of points in our two-dimensional embedding, suggesting that the underlying space of C. albicans morphologies are also continuous and overlapping, as perhaps expected. We certainly cannot rule out the possibility that a more advanced architecture for the VAE would better separate the morphologies. However we stress that this specific VAE easily separates the vast majority of points in other deep learning sets including MNIST72 and Omniglot73, suggesting the C. albicans morphology is at least as difficult as these well-studied learning challenges.
To the best of our knowledge, this is the first attempt to capture the space of C. albicans morphology in a continuous manner that respects technical variation, developmental processes and morphological transitions. Our intention is that the GAN models can be used to automatically detect new morphologies that are perhaps subtly different from our current dichotomy. Using tools for detecting anomalies via the generator’s latent space, we show how cells displaying non-canonical morphological forms can be flagged and quantified. This should find practical value in microscopy based studies: the tool will provide the community with a central resource that not only houses all C. albicans images but also unbiased models developed from those images that extended to non-canonical morphologies. Future studies will benefit from the ability to compare their images across this synthesized sum of knowledge. The technique should be straightforward to transfer to other fungi.
We have shown that the deep learning-based approaches are able to recapitulate the classification rules that are encoded by our choice of labelling strategy. It is unlikely that our labelling strategy is correct and other labellers with more or different expertise may have chosen a different way to partition the learning set and assign labels to individual examples. Although we attempted to be as consistent as possible when assigning class labels in Varasana version 1.0, our labels are certainly imperfect and open to debate. The classifier does however function far beyond random guesses, suggesting that our scheme has some value.
There is evidence that Candescence is able to “overcome” errors and inconsistencies between labellers. This is a well documented problem in image recognition research including computational pathology74. We argue that the computational techniques introduced to computational pathology are largely applicable to fungal systems. For example, machine learning based analysis of medical images has been shown to be more sensitive than trained pathologists identifying small events and complex patterns beyond perhaps human capacities75. As Candescence evolves to include stainings, fluorescent markers and a greater spectrum of microscopy imaging techniques, it may reveal cryptic cell or subcellular structure or community organization. Computational approaches in imaging provide a means to combine different modes of data, and to provide downstream analyses that integrate this information in a statistically sound manner76–78. For C. albicans, this might entail the integration of information concerning strain, growth conditions and genetic challenges together with images to better understand the composition and dynamics of colonies. Imaging standards analogous to the Digital Imaging and Communication (DICOM)79 for microbial systems including imaging of host tissue would enable better data sharing across the fungal community and allow for “hive analysis”80.
Last and perhaps most importantly, a surprising degree of disagreement has been observed (and quantified via Cohen’s k score) between expert pathologists when challenged with the same images81. Our limited experience with C. albicans morphology suggests that there may be similar disagreements across experts in this field. Such differences may hint at important alternative classification schemes. These differences may be important as Candescence is extended into clinical samples where the presence of C. albicans and its morphology are considered concomitantly within their host tissue. Our effort here represents an opportunity for the community to kernelize their knowledge of the dynamics of morphologies in a quantitative objective manner.
Materials and Methods
1. A fully convolutional one-stage object detector for morphology classification
There are fundamentally two computational problems underlying multi-class, multi-object detection. The first problem is to detect the locations in an image where objects exist; this is a regression to determine the four coordinates corresponding to the corners of the bounding box. In our case, the number of objects per image ranges up to ∼100 (Supplemental Figure 1) and objects may overlap in these images especially with respect to the filamentous morphology. Most object detectors rely on pre-defined anchor boxes. An anchor box is a rectangle that bounds an object in an image. Approaches that use anchor boxes make educated guesses where objects might exist in the image in addition to guesses regarding the size, aspect ratio and number of such boxes. The fact that there are exponentially many potential anchor boxes in any image makes this a computationally demanding exercise. Moreover, there are many parameters (size, aspect ratio and number of boxes) that require optimization and re-design on new datasets. The second problem is then to correctly label each object by its class (morphology, start or junction attribute, unknown or artifact).
Here we have opted to use a fully convolutional one-stage object detector (FCOS) for classifying C.albicans morphology56. FCOSs represent an anchor box-free reformulation of object detection. This is achieved by predicting, for each point in each feature map, the offset position to the top-left and bottom-right coordinates of a bounding box. Five feature maps are produced and each such map is limited to predicting bounding boxes of a predetermined size. For example, the first feature map predicts bounding boxes with maximum area of 30×30 pixels, while the final feature map is used to predict bounding boxes with maximum area of 128×128 pixels. A standard convolutional neural network is used with a softmax head for the classification component. We developed Candescence on top of the open-source implementation of FCOS provided by the machine vision platform MMDetection82 (Figure 1F).
2. Strains and Media
Images of yeast white, opaque and shmoo morphologies were acquired from C. albicans SN148a cells grown on YPD agar (1% yeast extract, 2% bacto-peptone, 2% D-glucose, 2% agar and 50 µg/mL uridine). Opaque switching was induced by growing cells on SC Glucosamine media (0.67% yeast nitrogen base lacking amino acids, 0.15% complete amino acid mixture, 2% agar, 1.25% N-acetylglucosamine (GlcNAc), 100 µg/mL uridine). We used 5 µg/mL phloxine B to stain opaque colonies. The shmoo morphology was induced by treating opaque cells with 10 µg/mL α-pheromone for 24 hrs in room temperature shaking at 220rpm.
To Induce filamentation in wild type SC5314, two colonies of cells was grown separately in 5ml of glucose-phosphate-proline (GPP) media83, (2.5 mM KH2PO4 (pH 6.5), 10.2 mM L-proline, 2.6 mM N-acetyl-D-glucosamine and 3 mM MgSO47H2O, 20% glucose) for 12 to 16 hours in 30 and 37℃ shaker incubator. The next day, 1 ml of cells from each colony were washed twice with 1ml 1X PBS. Different genetic variants of the SC5314 strain were also used to generate colonies enriched for filamentous morphologies. After growth, cells were washed with 1X PBS and diluted to different OD600 values in fresh liquid Spider, YPD or combinations of YPD and Fetal Bovine serum with or without centrifugation as per Supplemental Table 1, which provides a complete list of strains, conditions and protocols.
3. Microscopy
C. albicans colonies were mounted on slides and stained with a concentration of 2 μg/ml calcofluor white for 20 minutes before imaging. Images of C. albicans were captured using a Leica DM6000 upright microscope equipped with 100x (NA 1.3), 60x (NA 1.4) and 40x (NA 0.75) lenses and a Hamamatsu Orca ER camera. For DIC images, samples were captured using DIC optics and the built-in transmitted illuminator of the microscope. For cells labelled with fluorescent probes, samples were illuminated with a 100W mercury bulb (Osram) and passed through filter cubes optimised for illumination of calcofluor white-labelled samples (ex 377/50, em 447/60). We used the fluorescent images during the manual labelling procedure to help decide the best morphological assignment. This was particularly relevant to distinguish between white/opaque/gray-like and their budding variants, and also between junction types for the filamentous morphologies as bud scars are clearly visible. However the fluorescent images were not submitted to the FCOS (or other deep learning tool) during training. They are available as part of the Varasana learning set.
4. Image annotation and development of the learning set
In general, all computations were done using Python version 3.7 and R version 3.6.3. Our learning set was prepared using Labelbox ( software that facilitates the distributed annotation of image files. As a group, we labelled images following the guidelines from Sudbery et al.21, Whiteway and Bachewich11, Noble et al.23 and Tao et al.16. Labelbox assigns images in a manner that guarantees the same image is scored by multiple labellers.
One labeller (MH) modified assignments after the first round of labelling to ensure consistency as best possible across the labellers. A second round of quality control was performed by MH after the grid search was completed and our best classifier identified. Here all false positives and negatives were examined and a decision was made as to whether the instance was a labelling mistake, or a mistake made by the classifier.
5. Measures of system performance
Throughout the following TP, TN, FP, FN denote true positives, true negatives, false positives and false negatives respectively. Recall (a.k.a. sensitivity) measures the rate of false negatives whereas precision measures the rate of false positives:
The F 1 measure is convenient as it combines both the recall and precision into a single summary statistic:
In multi-object/multi-class problems there are two fundamental parameters. The first parameter is related to the object detection component of the FCOS and is termed the Intersection over Union (IoU) value:
Throughout this manuscript, we use a threshold of 0.5 for the IOU. The IoU controls how closely the predicted bounding box must overlap with the ground truth bounding box to be considered a positive. More stringent IOUs tend to decrease the recall of the system significantly with recall dropping due to a rapid increase in the number of false negatives84.
The second parameter is related to the classification component of the FCOS. Here τ represents the minimum value from the softmax of the classification head of the FCOS that must be exceeded if an object is to be assigned a class. More precisely, the score that an object belongs to each of the 15 classes is computed. It is assigned a class if and only if (i) the class has the highest score and (ii) the score exceeds τ.
The dual nature of object detection/object classification problems requires refinement of these fundamental concepts. We say that an object in an image is a TP if and only if the bounding box is predicted correctly (i.e. the IoU > 0.5) and the object is then classified correctly:
The concept of a true negative in this setting is tricky, since any pixel, which does not belong to a bounding box in an image, is in essence a TN. An object is a FP if and only if either (i) the predicted bounding box does not overlap sufficiently with a ground truth bounding box, or (ii) the predicted bounding box does overlap sufficiently with a ground truth bounding box but the classification is incorrect. Type (i) is termed as hallucination. Type (ii) is a misclassification. An object is a FN if and only if we fail to predict a bounding box with sufficient overlap with a ground truth bounding box. We term these FNs blindspots.
The mean average precision (mAP) is the standard and preferred approach for measuring the performance of multi-object/multi-class problems59. The mAP computes the average of the average precision over τ. Here the average precision corresponds to the area under the precision recall curve induce by a specific threshold for the IoU and varying τ. The maximum value for the mAP is 1.
Within an FCOS, total loss is computed as the sum of three individual losses: center-ness, bounding box and classification loss (Supplemental Figure 2). The concept of center-ness loss is specific to the FCOS and represents a mechanism to avoid the identification of multiple, spurious bounding boxes for a single object. It converges to a value of 0.5756,85. Bounding box loss measures disagreement between the ground truth location of bounding boxes with the regression produced by the deep learner. Finally, classification loss measures how well correctly identified objects are assigned classes.
We searched the hyperparameter space defined by the cross-product of different settings for the learning rate, momentum, decay, number of epochs per grade, IoU, threshold τ and various freezing rates (Figure 1H) where boldface font denotes the choice of parameters for the final FCOS. Throughout these experiments, the 800 × 800 input images were subjected to augmentation (random multi-scale flipping). Other parameters included 1000 warmup iterations, a warmup ratio of ⅓ and an SGD optimizer with grad clipping. The mAP and three loss functions described above were used to initially judge the quality of the classifier. The code from running the FCOS and interpreting results is depicted in Supplemental Figure 7.
When investigating differences in performance of a classifier between the test and validation sets, we manually curated a subset of test images for each entry in Table 1A and B in a manner to ensure that at least 100 cells were labelled (with the exception of the first two entries of Table 1A where there were too few cells). Then we built a contingency table where rows correspond to correct and incorrect predictions, and columns correspond to the test and validation dataset. A 2 test was used with the null hypothesis of no difference between the overall number of correct predictions in the test and validation sets.
6. Variational autoencoder (VAE) for unsupervised analyses
The variational autoencoder was constructed using the Keras for R (version 2.4) package86. Supplemental Figure 5 depicts the architecture of the model. Briefly here, the input to the network is a 128 × 128 pixel image. The image is subjected to a series of convolutional layers with a 5 × 5 kernels, 64 filters and strides that successively transformation the representation, followed by a flattening operation that reshapes the representation into a 262, 144 real vector, before a final reduction to a two dimensional latent space. Layers for z mean, the z log score and for decoding all follow standard VAE procedures (see code for details). Training used the wildtype Varasana training and validation sets in batch sizes of 100 across 20 epochs. The test set objects, which include the genetically modified C. albicans variants from Varasana, were not used in training. In particular, we extracted each (ground truth) bounding box across these files and reshaped the images to size 128 × 128. We used a β=0.4 parameter to down-weight the Kullbach-Leibler portion of the loss function87. It is straightforward to visualize the resultant two dimensional latent space with a scatterplot.
7. Generative Adversarial Networks
We follow the approach from Liu et al.68 to build unconditional GAN images. This approach requires computational resources that are accessible to most labs and requires few training samples, a feature important for our setting here. The model is particularly amenable to the disentanglement procedures utilized below. Supplemental Figure 6A, B depict the structure of the Generator and Discriminator respectively. We extended the Pytorch-based code available from the authors88. The input used for training corresponds to the ground truth bounding boxes of the training and validation datasets, reshaped as 128 × 128 image as was done for the VAE described in Methods 6. Here we used a shift predictor and deformator learning rate of 0.0001, with 2000 steps in batches of size 4.
We follow the framework of Creswell and Bharath69 to build trajectories between two (real) target images. Here however we opted to use the Learned Perceptual Image Patch Similarity (LPIPS) metric as the loss function70. The core idea is to find both “real” target images in the Generator’s latent space. This requires a so-called inversion which we perform using Algorithm INFER of Creswell and Bharath. Then we use linear interpolation between these two points in the latent space and reconstruct visual representations at user defined points along this path.
8. Anomaly detection
Anomaly detection proceeds using the framework from Schlegl et al.89 but replacing their residual and discriminator loss with the Learned Perceptual Image Patch Similarity (LPIPS) metric for estimating the similarity between two images70. Our approach proceeds as follows.
a. Each object in a image file is bounded manually, or;
b. Candescence is used to automatically regress bounding boxes for the objects in the target image file.
For each target object (that is, for each bounding box or patch) T, we find its optimal inversion z in the latent space of the generator G using the INFER(T, G) algorithm from Creswell and Bharath modified to use the LPIPS similarity metric. Here G(z) is a synthetic image.
Find the nearest real neighbour d of G(z) across all images d in the training set D under the LPIPS metric.
The anomaly score is then defined as follows: where D is the set of training images and ℒ is the LPIPS function between two images.
With respect to 1b, we remark that the performance of Candescence was not severely reduced when given an image with non-canonical morphologies and in this setting the classification returned by the FCOS is not used. Therefore, this automated approach should suffice in most scenarios, unless the change in morphology is very large. However, when the change in morphology is very obvious, we will not need a sensitive algorithm to detect it.
Funding Information
This work was supported by Canadian Research Chair Tier I awards to MW and MTH, and an NSERC Discovery award to MTH.
Author Contributions
VB developed the deep learning algorithm, performed experiments and helped prepare the manuscript. ACBPC, RPO and SM constructed the library of microscopy images used for training. EK and SS developed the Varasana learning set from these images. CL provided expertise in microscopy and analysis of the images. VD, MW and MTH prepared the manuscript. MTH designed the study and obtained funding for the project.
Conflict of Interest
The authors declare no conflict of interest.
Data Availability and Reproducibility
The Varasana dataset, code, trained models, supplemental movies and a Jupyter notebook to run Candescence is available from the Open Science Framework at
Download figure
Open in new tab
Supplemental Figure 1.
Description of the cumulative curriculum learning set. Panel A depicts the frequency of the different classes across the six grades in both training and validation. An image is part of all subsequent grades once it appears for the first time. Images are included in the training and validation sets at a ratio of 7:3. Panel B provides histograms of the number of objects (cells, junctions, unknown, artifacts) per image across all six grades. The maximum number of objects in any image was 97, although we note that several test set images exceeded this bound (not shown here).
Download figure
Open in new tab
Supplemental Figure 2.
The curves for the three FCOS notions of loss across the six grades for the chosen Candescence classifier (τ=0.25) from Table 2. Although the number of iterations (x-axis) varies across the grades because of differences in the size of the learning set for each grade, in all cases this translates to a total of 5,000 epochs. From this, an epoch number of 1,000 was chosen, as it appears that convergence has been reached after one-fifth of the epochs. It is well-established that the center-ness loss converges to ∼0.57. All other losses are negligibly above 0.
Download figure
Open in new tab
Download figure
Open in new tab
Supplemental Figure 3.
A compendium of all hallucinations (false positive object predictions) produced by Candescence. As in Figure 3(i), X indicates that we truly are not able to see an object (a true hallucination) and M indicates that Candescence was indeed correct to predict a bounding box at that location (missed by the human annotators) but its subsequent classification disagrees with our criteria. Subimages lacking an annotation indicate that Candescence correctly identified and classified the object, which represents a human error.
Download figure
Open in new tab
Download figure
Open in new tab
Supplemental Figure 4.
Exploration of the performance of Candescence on the test set. Each panel refers to one of ∼1,000 test set images with the UID available in Supplemental Table 1.
Download figure
Open in new tab
Supplemental Figure 5
sketches the general design of our Keras R-based VAE. Input is a 128×128 subimage from the original images of the training and validation datasets. We opted here for a two-dimensional latent space for ease of visualization.
Download figure
Open in new tab
Supplemental Figure 6
sketches the general design of our Generative Adversarial Network (GAN) trained on subimages as per the VAE. This design remains unchanged from the original presentation in Liu et al. (2021), although hyperparameters were fit using Varasana.
Download figure
Open in new tab
Supplemental Figure 7.
A description of the source code for (src/4-fcos-perf). The routines start with a classifier that has been learnt via experiment src/3-curriculum, and use this to measure its performance in different ways on the validation and test set. The scripts are written in Python or R, and make use of the FCOS implementation provided by MMDETECTION.
Supplemental Table 1. The Varasana learning set. UID is a distinct integer for each file. Type corresponds to a single colony that was photographed under the microscope and repetition refers to individual images taken of that colony. Gene target 1-3 describe the specific genetic modifications. ON under the Time column indicates overnight for 24 hours total. Columns S-AM provide the (manually assigned) number of cells per class per image across the training and validation datasets.
Footnotes
References
1.↵Perea, S.&Patterson, T. F. Antifungal resistance in pathogenic fungi. Clin. Infect. Dis. Off. Publ. Infect. Dis. Soc. Am. 35, 1073–1080 (2002). OpenUrlGoogle Scholar
2.↵Pfaller, M. A., Diekema, D. J., Turnidge, J. D., Castanheira, M.&Jones, R. N.Twenty Years of the SENTRY Antifungal Surveillance Program: Results for Candida Species From 1997-2016. Open Forum Infect. Dis. 6, S79–S94 (2019). OpenUrlCrossRefGoogle Scholar
3.↵Vandeputte, P., Ferrari, S.&Coste, A. T. Antifungal resistance and new strategies to control fungal infections. Int. J. Microbiol. 2012, 713687 (2012). OpenUrlCrossRefPubMedGoogle Scholar
4.↵Brown, G. D.et al.Hidden Killers: Human Fungal Infections. Sci. Transl. Med. 4, 165rv13–165rv13 (2012). OpenUrlFREE Full TextGoogle Scholar
5.↵Dambuza, I. M.&Brown, G. D. Fungi accelerate pancreatic cancer. Nature 574, 184–185 (2019). OpenUrlGoogle Scholar
6.↵Kullberg, B. J.&Arendrup, M. C.Invasive Candidiasis. N. Engl. J. Med. 373, 1445–1456 (2015). OpenUrlCrossRefPubMedGoogle Scholar
7.↵Benedict, K., Jackson, B. R., Chiller, T.&Beer, K. D. Estimation of Direct Healthcare Costs of Fungal Diseases in the United States. Clin. Infect. Dis. Off. Publ. Infect. Dis. Soc. Am. 68, 1791–1797 (2019). OpenUrlGoogle Scholar
8.Benjamin, D. K.et al.Neonatal candidiasis: epidemiology, risk factors, and clinical judgment. Pediatrics 126, e865–873 (2010). OpenUrlAbstract/FREE Full TextGoogle Scholar
9.↵Brown, G. D.&Netea, M. G. Exciting developments in the immunology of fungal infections. Cell Host Microbe 11, 422–424 (2012). OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar
10.↵Gow, N. A. R.&Yadav, B. Microbe Profile: Candida albicans: a shape-changing, opportunistic pathogenic fungus of humans. Microbiology 163, 1145–1147 (2017). OpenUrlCrossRefPubMedGoogle Scholar
11.↵Whiteway, M.&Bachewich, C. Morphogenesis in Candida albicans. Annu. Rev. Microbiol. 61, 529–553 (2007). OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar
12.↵Lohse, M. B.&Johnson, A. D. White-opaque switching in Candida albicans. Curr. Opin. Microbiol. 12, 650–654 (2009). OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar
13.↵Lorenz, M. C., Bender, J. A.&Fink, G. R. Transcriptional response of Candida albicans upon internalization by macrophages. Eukaryot. Cell 3, 1076–1087 (2004). OpenUrlAbstract/FREE Full TextGoogle Scholar
14.↵Lockhart, S. R.et al.In Candida albicans, white-opaque switchers are homozygous for mating type. Genetics 162, 737–745 (2002). OpenUrlAbstract/FREE Full TextGoogle Scholar
15.↵Lachke, S. A., Lockhart, S. R., Daniels, K. J.&Soll, D. R. Skin facilitates Candida albicans mating. Infect. Immun. 71, 4970–4976 (2003). OpenUrlAbstract/FREE Full TextGoogle Scholar
16.↵Tao, L.et al.Discovery of a “White-Gray-Opaque” Tristable Phenotypic Switching System in Candida albicans: Roles of Non-genetic Diversity in Host Adaptation. PLoS Biol. 12, e1001830 (2014). OpenUrlCrossRefPubMedGoogle Scholar
17.↵Basmaciyan, L., Bon, F., Paradis, T., Lapaquette, P.&Dalle, F. Candida Albicans Interactions With The Host: Crossing The Intestinal Epithelial Barrier. Tissue Barriers 7, 1612661 (2019). OpenUrlGoogle Scholar
18.↵Sudbery, P. E. Growth of Candida albicans hyphae. Nat. Rev. Microbiol. 9, 737–748 (2011). OpenUrlCrossRefPubMedGoogle Scholar
19.↵Crampin, H.et al.Candida albicans hyphae have a Spitzenkörper that is distinct from the polarisome found in yeast and pseudohyphae. J. Cell Sci. 118, 2935–2947 (2005). OpenUrlAbstract/FREE Full TextGoogle Scholar
20.Sudbery, P. E. The germ tubes of Candida albicans hyphae and pseudohyphae show different patterns of septin ring localization. Mol. Microbiol. 41, 19–31 (2001). OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar
21.↵Sudbery, P., Gow, N.&Berman, J. The distinct morphogenic states of Candida albicans. Trends Microbiol. 12, 317–324 (2004). OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar
22.↵Veses, V.&Gow, N. A. R. Pseudohypha budding patterns of Candida albicans. Med. Mycol. 47, 268–275 (2009). OpenUrlCrossRefPubMedGoogle Scholar
23.↵Noble, S. M., Gianetti, B. A.&Witchley, J. N. Candida albicans cell-type switching and functional plasticity in the mammalian host. Nat. Rev. Microbiol. 15, 96–108 (2017). OpenUrlCrossRefPubMedGoogle Scholar
24.↵Kehl-Fie, T. E.&Skaar, E. P. Nutritional immunity beyond iron: a role for manganese and zinc. Curr. Opin. Chem. Biol. 14, 218–224 (2010). OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar
25.↵Papagianni, M. Characterization of Fungal Morphology using Digital Image Analysis Techniques. J. Microb. Biochem. Technol. 06, (2014). Google Scholar
26.↵Benyon, F. H. L., Jones, A. S., Tovey, E. R.&Stone, G. Differentiation of allergenic fungal spores by image analysis, with application to aerobiological counts. Aerobiologia 15, 211–223 (1999). OpenUrlCrossRefGoogle Scholar
27.↵Obara, B., Grau, V.&Fricker, M. D. A bioimage informatics approach to automatically extract complex fungal networks. Bioinforma. Oxf. Engl. 28, 2374–2381 (2012). OpenUrlGoogle Scholar
28.Hardy, N.et al.Advanced digital image analysis method dedicated to the characterization of the morphology of filamentous fungus: ADVANCED DIGITAL IMAGE ANALYSIS. J. Microsc. 266, 126–140 (2017). OpenUrlGoogle Scholar
29.Lecault, V., Patel, N.&Thibault, J. Morphological Characterization and Viability Assessment of Trichoderma reesei by Image Analysis. Biotechnol. Prog. 23, 734–740 (2008). OpenUrlGoogle Scholar
30.Posch, A. E., Spadiut, O.&Herwig, C. A novel method for fast and statistically verified morphological characterization of filamentous fungi. Fungal Genet. Biol. 49, 499–510 (2012). OpenUrlPubMedGoogle Scholar
31.Sachs, C. C., Koepff, J., Wiechert, W., Grünberger, A.&Nöh, K.mycelyso - high-throughput analysis of Streptomyces mycelium live cell imaging data. BMC Bioinformatics 20, 452 (2019). OpenUrlGoogle Scholar
32.Tronnolone, H.et al.TAMMiCol: Tool for analysis of the morphology of microbial colonies. PLoS Comput. Biol. 14, e1006629 (2018). OpenUrlGoogle Scholar
33.Wurster, S.et al.Live Monitoring and Analysis of Fungal Growth, Viability, and Mycelial Morphology Using the IncuCyte NeuroTrack Processing Module. mBio 10, e00673–19, /mbio/10/3/mBio.00673-19.atom (2019). OpenUrlGoogle Scholar
34.↵Kritikos, G.et al.A tool named Iris for versatile high-throughput phenotyping in microorganisms. Nat. Microbiol. 2, 17014 (2017). OpenUrlGoogle Scholar
35.↵Tleis, M. S.&Verbeek, F. J. Machine Learning approach to discriminate Saccharomyces cerevisiae yeast cells using sophisticated image features. J. Integr. Bioinforma. 12, 276 (2015). OpenUrlGoogle Scholar
36.↵Wang, L.et al.Segmentation of yeast cell’s bright-field image with an edge-tracing algorithm. J. Biomed. Opt. 23, 1–7 (2018). OpenUrlCrossRefGoogle Scholar
37.↵Yang Yu, B., Elbuken, C., Ren, C. L.&Huissoon, J. P.Image processing and classification algorithm for yeast cell morphology in a microfluidic chip. J. Biomed. Opt. 16, 066008 (2011). OpenUrlPubMedGoogle Scholar
38.↵Cireşan, D. C., Giusti, A., Gambardella, L. M.&Schmidhuber, J. Deep neural networks segment neuronal membranes in electron microscopy images. in Proceedings of the 25th International Conference on Neural Information Processing Systems - Volume 2 2843–2851 (Curran Associates Inc., 2012). Google Scholar
39.↵Ronneberger, O., Fischer, P.&Brox, T.U-Net: Convolutional Networks for Biomedical Image Segmentation. ArXiv arxiv:150504597 Cs (2015). Google Scholar
40.↵He, K., Zhang, X., Ren, S.&Sun, J.Deep Residual Learning for Image Recognition. ArXiv arxiv:151203385 Cs (2015). Google Scholar
41.↵Falk, T.et al.U-Net: deep learning for cell counting, detection, and morphometry. Nat. Methods 16, 67–70 (2019). OpenUrlCrossRefPubMedGoogle Scholar
42.McQuin, C.et al.CellProfiler 3.0: Next-generation image processing for biology. PLOS Biol. 16, e2005970 (2018). OpenUrlCrossRefGoogle Scholar
43.Greenwald, N. F.et al.Whole-cell segmentation of tissue images with human-level performance using large-scale data annotation and deep learning. (2021)doi:10.1101/2021.03.01.431313. OpenUrlAbstract/FREE Full TextGoogle Scholar
44.↵Berg, S.et al.ilastik: interactive machine learning for (bio)image analysis. Nat. Methods 16, 1226–1232 (2019). OpenUrlCrossRefPubMedGoogle Scholar
45.↵Strack, R. Deep learning in imaging. Nat. Methods 16, 17–17 (2019). OpenUrlGoogle Scholar
46.↵Ando, D. M., McLean, C. Y.&Berndl, M. Improving Phenotypic Measurements in High-Content Imaging Screens. (2017)doi:10.1101/161422. OpenUrlAbstract/FREE Full TextGoogle Scholar
47.Pawlowski, N., Caicedo, J. C., Singh, S., Carpenter, A. E.&Storkey, A. Automating Morphological Profiling with Generic Deep Convolutional Networks. (2016)doi:10.1101/085118. OpenUrlAbstract/FREE Full TextGoogle Scholar
48.↵Godinez, W. J., Hossain, I., Lazic, S. E., Davies, J. W.&Zhang, X. A multi-scale convolutional neural network for phenotyping high-content cellular images. Bioinformatics 33, 2010–2019 (2017). OpenUrlCrossRefGoogle Scholar
49.↵Goldsborough, P., Pawlowski, N., Caicedo, J. C., Singh, S.&Carpenter, A. E. CytoGAN: Generative Modeling of Cell Images. (2017)doi:10.1101/227645. OpenUrlAbstract/FREE Full TextGoogle Scholar
50.↵Lafarge, M. W.et al.Capturing Single-Cell Phenotypic Variation via Unsupervised Representation Learning. in International Conference on Medical Imaging with Deep Learning 315–325 (2019). Google Scholar
51.↵Girshick, R., Donahue, J., Darrell, T.&Malik, J. Region-Based Convolutional Networks for Accurate Object Detection and Segmentation. IEEE Trans. Pattern Anal. Mach. Intell. 38, 142–158 (2016). OpenUrlCrossRefGoogle Scholar
52.↵Hung, J.et al.Applying Faster R-CNN for Object Detection on Malaria Images. ArXiv arxiv:180409548 Cs (2019). Google Scholar
53.↵Mattiazzi Usaj, M.et al.High-Content Screening for Quantitative Cell Biology. Trends Cell Biol. 26, 598–611 (2016). OpenUrlCrossRefPubMedGoogle Scholar
54.↵Styles, E. B., Friesen, H., Boone, C.&Andrews, B. J. High-Throughput Microscopy-Based Screening in Saccharomyces cerevisiae. Cold Spring Harb. Protoc. 2016, pdb.top087593 (2016). Google Scholar
55.↵Kraus, O. Z.et al.Automated analysis of high-content microscopy data with deep learning. Mol. Syst. Biol. 13, 924 (2017). OpenUrlAbstract/FREE Full TextGoogle Scholar
56.↵Tian, Z., Shen, C., Chen, H.&He, T.FCOS: Fully Convolutional One-Stage Object Detection. ArXiv arxiv:190401355 Cs (2019). Google Scholar
57.↵Elman, J. L. Learning and development in neural networks: the importance of starting small. Cognition 48, 71–99 (1993). OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar
58.↵Bengio, Y., Louradour, J., Collobert, R.&Weston, J.Curriculum learning. in Proceedings of the 26th Annual International Conference on Machine Learning - ICML ’09 1–8 (ACM Press, 2009).doi:10.1145/1553374.1553380. OpenUrlCrossRefGoogle Scholar
59.↵Salton, G.&McGill, M. Introduction to modern information retrieval. (McGraw-Hill, 1983). Google Scholar
60.↵Zeidler, U.et al.UME6 is a crucial downstream target of other transcriptional regulators of true hyphal development in Candida albicans. FEMS Yeast Res. 9, 126–142 (2009). OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar
61.↵Su, C., Yu, J.&Lu, Y. Hyphal development in Candida albicans from different cell states. Curr. Genet. 64, 1239–1243 (2018). OpenUrlGoogle Scholar
62.↵Omran, R. P., Law, C., Dumeaux, V., Morschhäuser, J.&Whiteway, M. The zinc cluster transcription factor Rha1 is a positive filamentation regulator in Candida albicans. (2020)doi:10.1101/2020.01.21.901744. OpenUrlAbstract/FREE Full TextGoogle Scholar
63.↵Schillig, R.&Morschhäuser, J. Analysis of a fungus-specific transcription factor family, the Candida albicans zinc cluster proteins, by artificial activation: Artificial activation of a transcription factor family. Mol. Microbiol. 89, 1003–1017 (2013). OpenUrlCrossRefPubMedGoogle Scholar
64.↵Guan, G.et al.Bcr1 plays a central role in the regulation of opaque cell filamentation in Candida albicans. Mol. Microbiol. 89, 732–750 (2013). OpenUrlCrossRefPubMedGoogle Scholar
65.↵Villa, S.et al.Transcriptional control of hyphal morphogenesis in Candida albicans. FEMS Yeast Res. 20, (2020). Google Scholar
66.↵Kingma, D. P.&Welling, M. Auto-Encoding Variational Bayes. ArXiv arxiv:13126114 Cs Stat (2013). Google Scholar
67.↵Goodfellow, I. J.et al.Generative Adversarial Networks. ArXiv arxiv:14062661 Cs Stat (2014). Google Scholar
68.↵Liu, B., Zhu, Y., Song, K.&Elgammal, A.Towards Faster and Stabilized GAN Training for High-fidelity Few-shot Image Synthesis. ArXiv arxiv:210104775 Cs (2021). Google Scholar
69.↵Creswell, A.&Bharath, A. A.Inverting The Generator Of A Generative Adversarial Network. ArXiv arxiv:161105644 Cs (2016). Google Scholar
70.↵Zhang, R., Isola, P., Efros, A. A., Shechtman, E.&Wang, O.The Unreasonable Effectiveness of Deep Features as a Perceptual Metric. ArXiv arxiv:180103924 Cs (2018). Google Scholar
71.↵Ackermann, M. A functional perspective on phenotypic heterogeneity in microorganisms. Nat. Rev. Microbiol. 13, 497–508 (2015). OpenUrlCrossRefPubMedGoogle Scholar
72.↵LeCun, Y.The mnist database of handwritten digits. (1989). Google Scholar
73.↵Lake, B. M., Salakhutdinov, R.&Tenenbaum, J. B. Human-level concept learning through probabilistic program induction. Science 350, 1332–1338 (2015). OpenUrlAbstract/FREE Full TextGoogle Scholar
74.↵Cui, M.&Zhang, D. Y.Artificial intelligence and computational pathology. Lab. Invest. 101, 412–422 (2021). OpenUrlGoogle Scholar
75.↵Liu, Y.et al.Detecting Cancer Metastases on Gigapixel Pathology Images. ArXiv arxiv:170302442 Cs (2017). Google Scholar
76.↵Saltz, J.et al.Spatial Organization and Molecular Correlation of Tumor-Infiltrating Lymphocytes Using Deep Learning on Pathology Images. Cell Rep. 23, 181–193.e7 (2018). OpenUrlPubMedGoogle Scholar
77.Harmon, S. A., Tuncer, S., Sanford, T., Choyke, P. L.&Türkbey, B. Artificial intelligence at the intersection of pathology and radiology in prostate cancer. Diagn. Interv. Radiol. Ank. Turk. 25, 183–188 (2019). OpenUrlGoogle Scholar
78.↵Moscatelli, M.et al.An infrastructure for precision medicine through analysis of big data. BMC Bioinformatics 19, 351 (2018). OpenUrlGoogle Scholar
79.↵Digital Imaging and Communication (DICOM). The Medical Imaging Technology Association (MITA). (2021). Google Scholar
80.↵Mazzanti, M., Shirka, E., Gjergo, H.&Hasimi, E.Imaging, Health Record, and Artificial Intelligence: Hype or Hope?Curr. Cardiol. Rep. 20, 48 (2018). Google Scholar
81.↵Warth, A.et al.Interobserver variability in the application of the novel IASLC/ATS/ERS classification for pulmonary adenocarcinomas. Eur. Respir. J. 40, 1221–1227 (2012). OpenUrlAbstract/FREE Full TextGoogle Scholar
82.↵Chen, K.et al.MMDetection: Open MMLab Detection Toolbox and Benchmark. ArXiv arxiv:190607155 Cs Eess (2019). Google Scholar
83.↵Hornby, J. M., Dumitru, R.&Nickerson, K. W. High phosphate (up to 600 mM) induces pseudohyphal development in five wild type Candida albicans. J. Microbiol. Methods 56, 119–124 (2004). OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar
84.↵Arlen, T.Understanding the mAP Evaluation Metric for Object Detection. Google Scholar
85.↵Zhi, T.FCOS centerness loss value for convergence. Google Scholar
86.↵Chollet, F.&Allaire, J.R interface to Keras. (2017). Google Scholar
87.↵Higgins, I.et al.Unsupervised deep learning identifies semantic disentanglement in single inferotemporal neurons. ArXiv arxiv:200614304 Q-Bio (2020). Google Scholar
88.↵Liu, B.FastGAN-pytorch. (2021). Google Scholar
89.↵Schlegl, T., Seeböck, P., Waldstein, S. M., Schmidt-Erfurth, U.&Langs, G.Unsupervised Anomaly Detection with Generative Adversarial Networks to Guide Marker Discovery. ArXiv arxiv:170305921 Cs (2017). Google Scholar
Back to top
PreviousNext
Posted June 10, 2021.
Download PDF
Print/Save Options
Download PDFFull Text & In-line FiguresXML
More Info
Supplementary Material
Data/Code
Email
Share
A deep learning approach to capture the essence of Candida albicans morphologies
V Bettauer, ACBP Costa, RP Omran, S Massahi, E Kirbizakis, S Simpson, V Dumeaux, C Law, M Whiteway, MT Hallett
bioRxiv 2021.06.10.445299; doi:
Now published in Microbiology Spectrum doi: 10.1128/spectrum.01472-22
Share This Article: Copy
Citation Tools
Get QR code
Subject Area
Microbiology
Reviews and Context
0
Comment
0
TRIP Peer Reviews
0
Community Reviews
0
Automated Services
0
Blogs/Media
0
Author Videos
Subject Areas
All Articles
Animal Behavior and Cognition(6862)
Biochemistry(15684)
Bioengineering(12074)
Bioinformatics(37392)
Biophysics(19268)
Cancer Biology(16409)
Cell Biology(22772)
Clinical Trials(138)
Developmental Biology(12176)
Ecology(17925)
Epidemiology(2067)
Evolutionary Biology(22314)
Genetics(14528)
Genomics(20574)
Immunology(15718)
Microbiology(36255)
Molecular Biology(15269)
Neuroscience(79337)
Paleontology(595)
Pathology(2541)
Pharmacology and Toxicology(4292)
Physiology(6778)
Plant Biology(13591)
Scientific Communication and Education(1934)
Synthetic Biology(3820)
Systems Biology(8999)
Zoology(2074)
The Clinical Trials and Epidemiology subject categories are now closed to new submissions following the completion of bioRxiv's clinical research pilot project and launch of the dedicated health sciences server medRxiv (submit.medrxiv.org). New papers that report results of Clinical Trials must now be submitted to medRxiv. Most new Epidemiology papers also should be submitted to medRxiv, but if a paper contains no health-related information, authors may choose to submit it to another bioRxiv subject category (e.g., Genetics or Microbiology).
Reviews and Context x
Comments 0 TRiP 0 Community 0 Automated 0 Blogs/Media 0 Video 0
Comments
bioRxiv aims to provide a venue for anyone to comment on a bioRxiv preprint. Comments are moderated for offensive or irrelevant content (this can take ~24 h). Please avoid duplicate submissions and read our Comment Policy before commenting. The content of a comment is not endorsed by bioRxiv.
Share this comments tab (click to copy link)Copied!
TRiP
bioRxiv partners with journals and review services to enable posting of peer reviews and editorial decisions related to preprints they are evaluating. Reviews are posted with the consent of the authors.
TRiP reviews are part of a peer review pilot project in which participating organizations post peer reviews of manuscripts they are evaluating
There is no TRiP material for this paper.
Community Reviews
bioRxiv aims to inform readers about online discussion of this preprint occurring elsewhere. The content at the links below is not endorsed by either bioRxiv or the preprint's authors.
Community reviews for this article:
There are no community reviews for this paper.
Automated Services
A variety of services now perform automated analyses of papers. Outputs from automated tools that summarize and extract information from bioRxiv preprints using AI and other technologies are displayed below. Note these tools can generate errors and the information has not been verified by bioRxiv or the authors.
Automated Services:
There are no automated services for this paper.
Blog/Media Links
bioRxiv aims to inform readers about online discussion of this preprint occurring elsewhere. The content at the links below is not endorsed by either bioRxiv or the preprint's authors.
Video
bioRxiv partners with conferences and institutions to display recordings of talks and seminars related to preprints. These are posted with the consent of the authors.
Video:
There are no videos for this paper.
Powered by
Powered by
Follow this preprint
X
You can now receive automatic notifications when a preprint is revised, withdrawn, commented on, peer reviewed, or published in a journal. Select the events you would like to follow below and click "Submit". To see all of the preprints you are currently following, please go to the bioRxiv Alerts Page.
Sign In to Follow this Preprint
Email
Email this Article close
Thank you for your interest in spreading the word about bioRxiv.
NOTE: Your email address is requested solely to identify you as the sender of this article.
Your Email
Your Name
Send To
Enter multiple addresses on separate lines or separate them with commas.
You are going to email the following A deep learning approach to capture the essence of Candida albicans morphologies
Message Subject (Your Name) has forwarded a page to you from bioRxiv
Message Body (Your Name) thought you would like to see this page from the bioRxiv website.
Your Personal Message
CAPTCHA
This question is for testing whether or not you are a human visitor and to prevent automated spam submissions.
Citation Tools close
A deep learning approach to capture the essence of Candida albicans morphologies
V Bettauer, ACBP Costa, RP Omran, S Massahi, E Kirbizakis, S Simpson, V Dumeaux, C Law, M Whiteway, MT Hallett
bioRxiv 2021.06.10.445299; doi:
Now published in Microbiology Spectrum doi: 10.1128/spectrum.01472-22
Citation Manager Formats
BibTeX
Bookends
EasyBib
EndNote (tagged)
EndNote 8 (xml)
Medlars
Mendeley
Papers
RefWorks Tagged
Ref Manager
RIS
Zotero
We use cookies on this site to enhance your user experience. By clicking any link on this page you are giving your consent for us to set cookies.
Continue Find out more |
190750 | https://surlalunefairytales.com/oldsite/authors/halliwell/riddlerhymes.html | SurLaLune Fairy Tales: Popular Rhymes and Nursery Tales by James Orchard Halliwell
Introduction | Annotated Tales | eBooks | Bookstore | Illustration Gallery | Discussion Board | Blog Canadian Wonder Tales by Cyrus MacMillan Celtic Fairy Tales by Joseph Jacobs More Celtic Fairy Tales by Joseph JacobsCinderella: 345 Variants by Marian Roalfe CoxCzecholovak Fairy Tales by Parker FillmoreDutch Fairy Tales For Young Folks by William E. GriffisEast O' the Sun and West O' the Moon by Gudrun Thorne-ThomsenEnglish Fairy Tales by Joseph JacobsMore English Fairy Tales by Joseph JacobsEuropean Folk and Fairy Tales by Joseph JacobsFacetious Nights of Straparola by Giovanni Francesco StraparolaFairy Tales; Their Origin and Meaning by John Thackray Bunce The Fairy Tales of Charles Perrault The Fairy Tales of Hans Christian Andersen The Fairy Tales of Madame D'Aulnoy The Fairy Tales of Oscar WildeFairy Tales of Old Japan by William E. GriffisFolk Tales from the Russian by Verra Xenophontovna Kalamatiano de BlumenthalGrimm Tales Made Gay by Guy Wetmore Carryl Household Tales by Jacob and Wilhelm Grimm (including all notes)Il Pentamerone by Giambattista Basile Indian Fairy Tales by Joseph JacobsIrish Fairy Tales by James StephensItalian Popular Tales (tales only) by Thomas CraneJapanese Fairy Tales by Yei Theodora OzakiJataka Tales by Ellen C. BabbittMore Jataka Tales by Ellen C. BabbittKaffir Folk-Lore by Georg McCall ThealMyths and Folklore of Ireland by Jeremiah CurtinOld Hungarian Fairy Tales by Baroness Emmuska OrczyOld Peter's Russian Tales by Arthur RansomePhilippine Folk Tales by Mabel Cook ColePopular Rhymes and Nursery Tales by James Orchard HalliwellPopular Tales of the Norse by Asbjornsen and MoePortuguese Folk-Tales by Consiglieri PedrosoRussian Wonder Tales by Post WheelerSixty Folk-Tales From Exclusively Slavonic Sources by A. H. WratislawThe Sleeping Beauty and Other Fairy Tales From the Old French by Sir Arthur Quiller-CouchSouth African Folk-Tales by James A. HoneyA Study of Fairy Tales by Laura F. KreadyTales of Old Japan by Algernon Bertram Freeman-MitfordTom Tit Tot: An Essay on Savage Philosophy in Folk-Tale by Edward CloddThe True Story of My Life: A Sketch by Hans Christian AndersenWest African Folk-Tales by William H. Barker and Cecilia Sinclair Riddle-Rhymes Return to Popular Rhymes and Nursery Tales by James Orchard Halliwell A VERY favorite class of rhymes with children, though the solutions are often most difficult to guess. Nursery riddle-rhymes are extremely numerous, and a volume might be filled with them without much difficulty. Many of the most common ones are found in manuscript collections of the sixteenth and seventeenth centuries. I'm in every one's way, But no one I stop; My four horns every day In every way play, And my head is nailed on at the top! —A turnstile. There was a king met a king In a straight lane; Says the king to the king, Where have you been? I've been in the wood, Hunting the doe: Pray lend me your dog, That I may do so. Call him, call him! What must I call him? Call him as you and I, We've done both. —The dog's name was Been, and the name of the persons who met each other was King. This riddle was obtained recently from oral tradition. I observe, however, a version of it in MS. Harl. 1962, of the seventeenth century. The cuckoo and the gowk, The laverock and the lark, The twire-snipe, the weather-bleak; How many birds is that? —Three, for the second name in each line is a synonyme. The cuckoo is called a gowk in the North of England; the lark, a laverock; and the twire-snipe and weather-bleak, or weather-bleater, are the same birds. Hoddy-doddy, With a round black body! Three feet and a wooden hat; What's that? —An iron pot. In the country, an iron pot with three legs, and a wooden cover, the latter raised or put on by means of a peg at the top, is used for suspending over a fire, or to place on the hearth with a wood fire. Riddle me, riddle me, what is that Over the head and under the hat? —Hair. From Kent. The fiddler and his wife, The piper and his mother, Ate three half-cakes, three whole cakes, And three quarters of another. How much did each get? —The fiddler's wife was the piper's mother. Each one there got ½ + 1 + ¼ or 1¾. There was a little green house, And in the little green house There was a little brown house, And in the little brown house There was a little yellow house, And in the little yellow house There was a little white house, And in the little white house There was a little heart. —A walnut. A flock of white sheep On a red hill; Here they go, there they go, Now they stand still! —The teeth and gums. Old Father Greybeard, Without tooth or tongue, If you'll give me your finger, I'll give you my thumb. —Greybeard, says Moor, Suffolk Words, p. 155, was the appropriate name for a fine large handsome stone bottle, holding perhaps three or four, or more gallons, having its handle terminating in a venerable Druidic face. This riddle appears to be alluded to in MS. Harl. 7316, p. 61: I'm a dull senseless blockhead, 'tis true, when I'm young, And like old grandsire Greyberd without tooth or tongue, But by the kind help and assistance of arts I sometimes attain to politeness of parts: What God never sees, What the king seldom sees; What we see every day: Read my riddle,—I pray. —An equal. This riddle is well known in Sweden. The following version was given me by Mr. Stephens: Jag ser det dagligen; Kungen ser det sällan; Gud ser det aldrig. "I see it daily; The king sees it seldom; God sees it never." As white a milk, And not milk; As green as grass, And not grass; As red as blood, And not blood; As black as soot, And not soot! —A bramble-blossom. The land was white, The seed was black; It'll take a good scholar To riddle me that. —Paper and writing. As high as a castle, As weak as a wastle; And all the king's horses Cannot pull it down. —Smoke. A wastle is a North country term for a twig or withy, possibly connected with A. S. wædl. I've seen you where you never was, And where you ne'er will be; And yet you in that very same place May still be seen by me. —The reflection of a face in a looking-glass. Banks full, braes full, Though ye gather all day, Ye'll not gather your hands full. —The mist. From Northumberland. Sometimes thus: A hill full, a hole full, Ye cannot catch a bowl full. A young man and a young woman quarrelled, and the former, in his anger, exclaimed,— Three words I know to be true, All which begin with W. The young woman immediately guessed the enigma, and replied in a similar strain,— I too know them, And eke three which begin with M. —Woman wants wit. Man much more. The calf, the goose, the bee, The world is ruled by these three. —Parchment, pens, and wax. A house full, a yard full, And ye can't catch a bowl full. —Smoke. As I was going o'er London bridge, I heard something crack; Not a man in all England Can mend that! —Ice. I had a little sister, They called her Pretty Peep; She wades in the waters, Deep, deep, deep! She climbs up the mountains, High, high, high; My poor little sister, She has but one eye. —A star. This charming little riddle is always a great favorite with children. As I was going o'er yon moor of moss, I met a man on a gray horse; He whipp'd and he wail'd, I ask'd him what he ail'd; He said he was going to his father's funeral, Who died seven years before he was born! — His father was a dyer. As I look'd out o' my chamber window, I heard something fall; I sent my maid to pick it up, But she couldn't pick it all. —Snuff. From Yorkshire. Black within, and red without, Four corners round about. —A chimney. From Yorkshire. As I was going o'er London bridge, I met a drove of guinea pigs; They were nick'd and they were nack'd, And they were all yellow back'd. —A swarm of bees; not a very likely family to meet in that neighbourhood, at least nowadays, but some of the authors of these poems seem to have been continually traversing London bridge. Higher than a house, higher than a tree; Oh! whatever can that be? —A star. From Yorkshire. Which weighs heavier— A stone of lead Or a stone of feather? —They both weigh alike. Lilly low, lilly low, set up on an end, See little baby go out at town end. —A candle. Lillylow is a North country term for the flame of a candle. Low, A.-S. lig, is universal. At the end of my yard there is a vat, Four-and-twenty ladies dancing in that: Some in green gowns, and some with blue hat: He is a wise man who can tell me that. —A field of flax. Jackatawad ran over the moor, Never behind, but always before! —The ignis fatuus, or Will o' the Wisp. Jackatawad is a provincial term for this phenomenon. Black'm, saut'm, rough'm, glower'm, saw, Click'm, gatt'm, flaug'm into girnigaw. —Eating a sloe. A North country riddle, given by Brockett. Girnigaw is the cavity of the mouth. There was a man rode through our town, Gray Grizzle was his name; His saddle-bow was gilt with gold; Three time I've named his name. —Gaffer Was. From Yorkshire. There was a man went over the Wash, Grizzle grey was his horse; Bent was his saddle-bow: I've told you his name three times, And yet you don't know! —The same as the last. From Norfolk. I am become of flesh and blood, As other creatures be; Yet there's neither flesh nor blood Doth remain in me. I make kings that they fall out, I make them agree; And yet there's neither flesh nor blood Doth remain in me. —A pen. Riddles similar to this are current in most languages. Mr.Stephens has kindly furnished me with the following one obtained in Sweden: Af kött och blod är jag upprunnen, Men ingen blod är i mig funnen; Många herrar de mig bära, Med hvassa knifvar de mig skära. Mången har jag gifvit ära, Mången har jag tagit af, Mången har jag lagt i graf. Of flesh and blood sprung am I ever; But blood in me that find ye never. Many great lords bear me proudly, With sharp knives cutting me loudly. Many I've graced right honorably: Rich ones many I've humble made; Many within their grave I've laid! The pen has been a fertile subject for the modern riddle-writer. The best production of the kind was printed a few months ago in the Times newspaper, contributed by Miss Agnes Strickland. Into my house came neighbour John, With three legs and a wooden one; If one be taken from the same, Then just five there will remain. —He had a IV legged stool with him, and taking away the left-hand numeral, there remains V. Link Lank, on a bank, Ten against four. —A milkmaid. Two legs sat upon three legs, With four legs standing by; Four then were drawn by ten: Read my riddle ye can't, However much ye try. —An amplification of the above, the milkmaid of course sitting on a three-legged stool. Over the water, And under the water, And always with its head down! —A nail in the bottom of a ship. As straight as a maypole, As little as a pin, As bent as a bucker, And as round as a ring. I do not know the solution of this riddle. A bucker is a bent piece of wood by which slaughtered sheep are hung up by their expanded hind legs, before being cut out. Hitty Pitty within the wall, Hitty Pitty without the wall: If you touch Hitty Pitty, Hitty Pitty will bite you. —A nettle. MS. Harl. 1962, xvii. cent. The first letter of our fore-fadyr, A worker of wax, An I and an N; The colour of an ass: And what have you then? —Abindon, or Abingdon, in Berks. An ancient rebus given in Lelandi Itin. ed. 1744, ii. 136. I saw a fight the other day; A damsel did begin the fray. She with her daily friend did meet, Then standing in the open street; She gave such hard and sturdy blows, He bled ten gallons at the nose; Yet neither seem to faint nor fall, Nor gave her any abuse at all. —A pump. MS. Harl. 1962, xvij. cent. A water there is I must pass, A broader water never was; And yet of all waters I ever did see, To pass over with less jeopardy. —The dew. From the same MS. There is a bird of great renown, Useful in city and in town; None work like unto him can do; He's yellow, black, red, and green, A very pretty bird I mean; Yet he's both fierce and fell: I count him wise that can this tell —A bee. From the same MS. As I went over Hottery Tottery, I looked into Harbora Lilly; I spied a cutterell Playing with her cambril. I cryed, Ho, neighbour, ho! Lend me your cue and your goe, To shoot at yonder cutterell Playing with her cambril, And you shall have the curle of her loe. —A man calling to his neighbour for a gun to shoot a deer, and he should have her humbles. MS. ibid. As I went through my houter touter, Houter trouter, verly; I see one Mr. Higamgige Come over the hill of Parley. But if I had my early verly, Carly verly verly; I would have bine met with Mr. Higamgige Come over the hill of Parley —A man going over a hill, and a fly lighting on his head. MS. ibid. THE FOUR SISTERS. I have four sisters beyond the sea, Para-mara, dictum, domine. And they did send four presents to me, Partum, quartum, paradise, tempum, Para-mara, dictum, domine! The first it was a bird without e'er a bone; Para-mara, dictum, domine. The second was a cherry without e'er a stone; Partum, quartum, &c. The third it was a blanket without e'er a thread, Para-mara, dictum, &c. The fourth it was a book which no man could read, Partum, quartum, &c. How can there be a bird without e'er a bone? Para-mara, dictum, &c. How can there be a cherry without e'er a stone? Partum, quartum, &c. How can there be a blanket without e'er a thread? Para-mara, dictum, &c. How can there be a book which no man can read? Partum, quartum, &c. When the bird's in the shell, there is no bone; Para-mara, dictum, &c. When the cherry's in the bud, there is no stone; Partum, quartum, &c. When the blanket's in the fleece, there is no thread; Para-mara, dictum, &c. When the book's in the press, no man can read; Partum, quartum, &c. Several versions of this metrical riddle are common in the North of England, and an ingenious antiquary has suggested that it is a parody on the old monkish songs! It will remind the reader of the Scottish ballad of Captain Wedderburn's Courtship, O hold away from me, kind sir, I pray you let me be; For I will not go to your bed, Till you dress me dishes three: Dishes three you must dress to me, And I must have them a', Before that I lie in your bed, Either at stock or wa'. O I must have to my supper A cherry without a stone; And I must have to my supper A chicken without a bone: And I must have to my supper A bird without a ga', Before I lie into your bed, Either at stock or wa'. When the cherry is in the bloom, I'm sure it hath no stone; And when the chicken is in its shell, I'm sure it hath no bone: The dove it is a gentle bird, It flies without a ga', And we shall both lie in ae bed, And thou's lie next the wa'. THE DEMANDS JOYOUS. It is not generally known that many of our popular riddles are centuries old. Yet such is the fact, and those whose course of reading has made them acquainted with ancient collections are not unfrequently startled by observing a quibble of the fifteenth or sixteenth century go the round of modern newspapers as a new invention, or perhaps as an importation from America! A few months ago, an instance of this species of resuscitation took place in the publication of the question, "Which were made first, elbows or knees?" This was an enigma current in England in the time of Queen Elizabeth, and is found in a manuscript in the British Museum written before the close of the sixteenth century. The earliest collection of riddles printed in this country came from the press of Wynkyn de Worde in the year 1511, in black letter, under the title of the "Demaundes Joyous." Only one copy of this tract, which was "imprynted at London, in Flete Strete, at the sygne of the Sonne," is known to exist, and it is now preserved in the public library at Cambridge. It is chiefly a compilation from an early French tract under a similar title, but which is far more remarkable for its grossness. The reader may be amused with the following specimens, and perhaps recognise some of them as old favorites: "Demand. Who bore the best burden that ever was borne?—R. The ass on which our Lady rode when she fled with our Lord into Egypt. D. What became of that ass?—R. Adam's mother did eat her. D. Who is Adam's mother?—R. The earth. Demand. What space is from the surface of the sea to its greatest depth? — R. A stone's cast. Demand. How many calves' tails behoveth to reach from the earth to the sky? — R. No more but one, an' it be long enough. Demand. Which is the most profitable beast, and that which men eat least of? — R. Bees. Demand. Which is the broadest water, and the least jeopardy to pass over? — R. The dew. Demand. What thing is that which never was nor never will be? — R. A mouse making her nest in a cat's ear. Demand. Why doth a dog turn himself thrice round before he layeth down? — Because he knoweth not the bed's head from its foot. Demand. Why do men make an oven in the town? — R. For because they cannot make the town in the oven. Demand. How may a man know or perceive a cow in a flock of sheep? — R. By sight. Demand. What alms are worst bestowed that men give. — R. Alms to a blind man, for he would willingly see him hanged by the neck that gave it him. Demand. What thing is that which hath no end? — R. A bowl. Demand. What people be they that never go a-procession? — R. Those that ring the bells in the mean time. Demand. What is that that freezeth never? — R. Hot water. Demand. What thing is that that is most likest unto a horse? — R. That is a mare. Demand. What thing is that which is more frightful the smaller it is? — R. A bridge. Demand. Why doth an ox lie down? — R. Because he cannot sit. Demand. How many straws go to a goose's nest? — R. None, for lack of feet. Demand. Who slew the fourth part of the world? — Cain, when he killed his brother Abel. Demand. What man is he that getteth his living backwards? — R. A ropemaker. The reader will please to recollect the antiquity of these, and their curiousity, before he condemns their triviality. Let the worst be said of them, they are certainly as good as some of Shakespeare's jokes, which no doubt elicited peals of laughter from an Elizabethan audience. This may be said to be only a negative kind of recommendation, and, indeed, when we reflect on the apparent poverty of verbal humour in those days, the wonder is that it could have been so well relished. The fact must be that we often do not understand the greater part of the meaning intended to be conveyed. To revert to the lengthened transmission of jokes, I may mention my discovery of the following in MS. Addit. 5008, in the British Museum, a journal of the time of Queen Elizabeth. The anecdote, by some means, went the round of the provincial press in 1843, as of modern composition. "On a very rainy day, a man, entering his house, was accosted by his wife in the following manner: 'Now, my dear, while you are wet, go and fetch me a bucket of water.' He obeyed, brought the water and threw it all over her, saying at the same time, 'Now, my dear, while you are wet, go and fetch another!'" Halliwell, James Orchard. Popular Rhymes and Nursery Tales. London: John Russell Smith, 1849. Return to Popular Rhymes and Nursery Tales by James Orchard Halliwell Support SurLaLune Available on CafePress.com ##### Available on Amazon.com Available on Art.com ©Heidi Anne Heiner, SurLaLune Fairy Tales E-mail: heidi@surlalunefairytales.com Page created 10/25/07; Last updated 10/25/07 www.surlalunefairytales.com |
190751 | https://economics.ucsd.edu/~jsobel/172aw02/notes1.pdf | Linear Programming Notes I: Introduction and Problem Formulation 1 Introduction to Operations Research Economics 172 is a two quarter sequence in Operations Research. Management Science majors are required to take the course. I do not know what Management Science is. Most of you picked the major. I assume that you either know what it is or do not care. You may not know what Operations Research is. I am going to tell you, but it will leave you disappointed.
Operations Research is research into operations.
The field began during the Second World War. The military needed to solve a lot of different kinds of resource allocation problems.
A prototypical problem was a form of the transportation problem that we’ll study later in the course. In this problem the military had supplies available in several different locations (ammunition factories), it had several different locations that needed the supplies (battle fronts), it knew how much it cost to ship supplies from any factory to any front.
It knew how much was produced at each factory and how much was needed in each front. It wanted to figure out how to minimize the cost of shipping the supplies to the various locations while meeting two types of resource availability constraints (that you do not send more ammunition from a factory than is available and that you send as much as necessary to each battle front). Many other resource allocation problems arose in the planning of military operations.
Operations Research was a field of study that tried to come up with practical solutions to these problems.
People need to allocate resources even in peacetime. Economics is a discipline devoted to the study of methods to allocate scarce resources.
It is natural to study the methods of Operations Research in an economics class.
Some of the methods developed have direct relevance to decision making. Courses in Operations Research are therefore traditional parts of undergraduate and graduate business programs.
Operations Research as a discipline involves several different things. First, there is the identification of real world situations that lend themselves to for-mulation as mathematical optimization problems. Second, there is a process of translating these problems into mathematical language. Third, there is the de-velopment of mathematics that explains the general structure of the mathemat-ical problems that arise in the second stage. Fourth, there is the development of methods for solving these problems.
The Operations Research sequence introduces some of the basic mathemat-ical techniques for describing and solving problems (steps 3 and 4 above). It provides practice in the formulation of problems (steps 1 and 2 above).
A mathematical programming problem is an optimization problem subject to constraints. In the general problem, you are given a function f and a set S.
You are asked to find a solution to the problem: 1 max f(x) subject to x ∈S.
(1) A linear programming problem is a mathematical programming problem in which the function f is linear and the set S is described using linear inequalities or equations. It turns out that lots of interesting problems can be described as linear programming problems. It turns out that there is an efficient algorithm that solves linear programming problems efficiently and exactly. It turns out that the solutions to linear programming problems provide interesting economic information. Economics 172A concentrates on these problems.
Economics 172B primarily studies non-linear programming. That is, prob-lems in which the function f is non-linear and the set S is described using non-linear inequalities or equations. This theory uses calculus techniques.
In the Economics 172 sequence, the word “programming” has nothing to do with computer programming (although it is true that there are computer programs that can be used to solve mathematical programming problems). This terminology is confusing, but it is standard.
2 Introduction to Linear Programming Economics 172A studies linear programming. So you need to know what a linear function is. The function f of n variables x = (x1, . . . , xn) is linear if there are constants a1, . . . , an such that f(x) = a1x1 + . . . + anxn.
(2) This expression is also written: f(x) = n X i=1 aixi = a · x, (3) where a = (a1, . . . , an).
Two properties characterize linear functions: additivity and constant returns to scale. Additivity means that f(x + y) = f(x) + f(y). Constant returns to scale means that f(cx) = cf(x) for any constant c. These properties make sense sometimes. Other times they are silly. Suppose f(z) is how much it costs you to buy z. Remember, z has n components, so you can think of z as a list. The ith entry in the list, zi, tells you the amount of good i that you are buying.
Additivity says that if you first buy x and then buy y, it costs the same amount if you bought x + y at one time. Constant returns to scale says that buying half as much costs half as much and buying twice as much costs twice as much.
Provided that there are no specials (“buy two, get one free”), most of what you buy at the grocery store satisfies these properties. It certainly describes how you pay for gasoline at the pump. On the other hand, linearity does not hold in 2 milk prices: a gallon container costs less than two half gallon containers. Still, one of the most basic linear functions that we deal with is the one that assigns value to lists of goods. If pi is the price per unit of good i, then the linear function f(x) = p · x gives the cost of buying the ‘bundle’ x consisting of x1 units of good 1, x2 units of good 2, and so on.
Linearity is usually not a very good assumption for utility functions.
If f(z) represents the utility (loosely, happiness) you get from having z, then both additivity and constant returns to scale are likely to fail. For example, if x represents having a CD player (and nothing else), while y represents having a CD of your favorite music, then presumably f(x + y) > f(x) + f(y) and also 2f(x+y) > f(2(x+y)). The first inequality says that having both the CD player and the CD is better than the sum of utilities available from having exactly one.
(You could argue that having only the CD player or only the CD is worthless.) The second inequality says that having twice the utility from both CD player and CD is better than having the utility of two CD players and two copies of the CD. (You could argue that the second copy of the disk and the second player is useless.) In economics it is typical to assume diminishing marginal utility.
In our context that is just a fancy way of saying that doubling what you have does less than double your utility (the second $100,000,000 does not generate as much additional utility as the first $100,000,000.) The linearity assumption does not apply to production processes that have fixed costs (the first unit costs much more than subsequent units) or capacity constraints. It does not apply to situations in which ‘units’ are not perfectly di-visible (that is, the components of x theoretically measure continuous quantities not numbers of people). Divisibility is a standard simplifying assumption.
The point is that linearity is an assumption. You should reflect on whether it is a reasonable assumption in the applications that arise during the quarter.
Now return to (1).
It is time to get a better understanding of what a mathematical programming problem is. The next few paragraphs will contain several really important definitions. You’ll hear them over and over again.
S is your constraint set or feasible set. Maybe it is the different combina-tions of things that you can afford to buy. Maybe it is the different combination of things that you have the available raw materials to manufacture.
In any event, it is what keeps you from doing whatever you want.
The function f is your objective function. It is what you are trying to optimize (optimize means either minimize or maximize). It is possible that the set S is empty.
If this is true, then your problem is infeasible. You can’t solve it. This is a perfectly reasonable mathematical possibility. Economically, it means that your constraints are inconsistent. You will see examples soon enough. If S is not empty, then the problem is feasible.
What does it mean to solve a mathematical programming problem? A so-lution to (1) is a special value x∗that has two properties: 1. Feasibility. x∗∈S.
2. Optimality. If x ∈S, then f(x∗) ≥f(x).
3 That is, a solution must satisfy the constraints of the problem and, among all things that satisfy the constaints, yield the highest objective function value. If x∗is a solution to (1), then f(x∗) is called the optimal value (or sometimes just value) of the problem. Not all problems have solutions (for example, infeasible problems have no solution). Problems may have more than one solution. (There may be two different ways to solve the problem.) If a problem has a solution, then the value must be unique (otherwise the lower number can’t be the value).
Our problems will turn out to fall into one of three categories. They will either be infeasible or they will have a solution or they will be unbounded. A problem is unbounded it if is possible to make the objective function arbitrarily large. In symbols, this means that for any M, there exists an x ∈S such that f(x) > M. In words, a problem is unbounded if for any target value of the objective function (M) it is possible to find a way to make f even bigger than M using a feasible point x.
These definitions apply to any problem like (1). The course restricts at-tention to linear programming problems. A linear programming problem is a mathematical programming problem is which f is linear and the set S is de-scribed by linear inequalities or equations. There is a standard form for writing linear programming problems (LPs).
max c · x subject to Ax ≤b and x ≥0.
(4) In this formulation, c = (c1, . . . , cn), b = (b1, . . . , bm), 0 denotes an n dimen-sional list of zeros, and A is a matrix with m rows and n columns (an m × n matrix); the entry of A in row i and column j is aij. In this basic problem, the given data are c (the coefficients of the variables in the objective function), b (the resources constraints), and A (the technology). In order to formulate the problem, you must know these things. The problem that I have described has n variables (the components of x) and m + n constraints. The first m constraints come from the set of inequalities summarized by Ax ≤b. The remaining n constraints are the non-negativity constraints on the components of x. The notation Ax ≤b is short hand for the system of m inequalities. A representative inequality (the ith inequality) takes the form n X j=1 aijxj ≤bi.
The objective function and the constraints in the problem are all linear. In principle, the objective in a linear programming problem can be to maximize or to minimize; the constraints can be written in the form of equations or inequalities of either direction, and inequality constraints need be present for some (or none) of the variables.
It turns out that any linear programming problem can be written in the standard form above. I’ll say more about that later. At this point, note only that (4) describes the set of problems we will study.
4 Now I can comment on the contents of the course outline. The first topic is problem formulation. This is the process of taking a situation described in words and translating it into a mathematical problem in the form (4). This process probably represents the most likely application you might make of the techniques of the class in the “real world.” The classroom is not the real world.
You will see rather contrived examples. During the first week of the class, I will describe possible linear programming problem and formulate a couple slowly.
There will be formulations throughout the class. My experience is that students have trouble formulating problems. You might find that the first topic is the most challenging part of the course.
When there are only two variables, it is possible to solve linear programming problems graphically. The second topic shows you how to do this. Graphical solution is easy and illustrates most of the basic ideas about solutions of linear programming problems. The problem is that most problems involve more than two variables and graphical methods do not apply.
Algorithms exist that can solve any linear programming problem. These algorithms are widely used in industry. The oldest and still most widely used algorithm is the simplex algorithm. Versions of the algorithms are available as part of common spreadsheet programs. Since your computer already knows the algorithm and can do computations more easily than you can, it makes no sense to teach you the entire procedure. Still, the essentials of the simplex algorithm are straightforward and instructive. Knowing how the algorithm works is useful on its own and also helps you interpret solutions provided by computers. I will spend some time teaching you a bit about the algorithm in Topic 3.
The fourth topic is the heart of the course. It turns out that when you solve a linear programming problem you automatically solve another linear program-ming problem (called the dual of the original problem). The theory of duality is beautiful and interesting (to the mathematically inclined). It also provides truly important economic information about solutions to linear programming prob-lems. Sensitivity Analysis refers to the study of what happens to the solution to a linear programming problem when one changes the problem (by varying the objective function or the resource constraints). There is a lot to say here. We will say some things theoretically. Other things we will illustrate using solutions to problems obtained by the computer.
Game Theory is a big topic (there is an entire undergraduate course devoted to it). It is a mathematical theory of strategic interaction. Zero-sum games are a special class of game that includes most of the things called games by normal people (chess, poker, tic-tac-toe) and generally situations where players have completely opposed interests. It turns out that there is an intimate relationship between zero-sum games and linear programming. I will tell you about it. (I should warn you that game theory rarely provides practical advice on how to play a game.) The final topic covers a special class of linear programming problem. This problem has special structure. It provides a useful way to introduce integer linear programming (that is, linear programming problems with the additional restriction that all variables must be whole numbers).
5 3 Introduction to Problem Formulations Problem formulation is the most important part of a Operations Research course for a Management Science major. When you are the boss, you’ll hire a geeky engineer to do some basic math and write software. You’ll earn big bucks by identifying the important problems and translating them from a verbal identifi-cation to a mathematical form. The engineer will then solve the mathematical problem. You will interpret the solution and put it into practice. It is impor-tant for you to know enough about the basic mathematics for you to be able to frame questions that the engineer might be able to answer and to be able to judge whether the answers provided are sensible. Formulation, however, is key.
Unfortunately, I have little useful to say on the topic. In order to formulate problems, you need to be able to understand symbols, you need common sense, and you need practice. I am not aware of a mechanical series of steps you can take in order to complete a formulation.
Now I will got through a particular (and standard) linear programming problem and formulate it.
The problem is called the Diet Problem. Here is the story.
You run a small institution (prison, junior high school, third world country).
People work in your institution and you must feed them. Your job is to meet their basic requirements for nutrients at minimum cost. In order to do this, you need to know several things. You must know what foods are available and the cost of each food. You must also know which nutrients are necessary and the nutritional content of each of the foods. With this information you can figure out how much any combination of food costs and you can figure out the nutritional content of any combination of food. You can decide which combinations of food are sufficient to meet the nutritional requirements and then pick the cheapest combination that meets the nutritional requirement. (Perhaps the story makes more sense if you imagine that your job is to feed the animals on your farm.) So that is the basic verbal story. It has a surface plausibility. That is, you can imagine someone wanting to find cheap ways to feed people. It is a bit bizarre because it contains no mention of what the foods taste like. The story does not place restrictions on food (for example, not too much salt, sugar, or fat; or no meat), although these restrictions can be included without much trouble.
In order to formulate the problem as a linear programming problem, we need notation to describe the given data. This information typically is given to you in the statement of a formulation problem. Assume that there are n different kinds of food. The price per unit of the jth food is pj. Assume that there are m different nutrients. The nutritional requirement of nutrient i is ci. Finally, let aij be the amount of the ith nutrient in one unit of the jth food.
Let me repeat this information less abstractly. The n different foods could be things like lettuce, hamburger, potatoes, oranges, pizza, and so on. When I talk about the jth food, I mean one of these (maybe I list all available foods in alphabetical order and number them 1 through n). pj is the unit price of food j.
So, p1 might be the price of a head of lettuce; p2 might be the price of a pound of hamburger; and so on. The m different nutrients could be things like vitamin 6 C, iron, niacin, and so on. ci is the daily minimum requirement of nutrient i. The units of these things are weird (I think that Vitamin E is measured in “International Units,” other nutrients are measured in grams). This does not matter, as long as you can figure out how much of each nutrient you can find in each food. That is where the aij comes in. Suppose that there are .5 grams of niacin in a head of lettuce. If lettuce is food 1 and niacin is nutrient 5, then this means that a51 = .5.
Now we have a description of the problem in words and a description of the basic data of the problem. Notice that you (as the manager of the institution) could find out the data. You look up food prices at the grocery store. You consult a nutritionist to figure out the entries in the matrix A.
You check government standards to figure out the nutritional requirements. Your problem is to figure out what to buy.
In order to formulate this as a mathematical problem, you need to invent a name for what you are looking for.
Step 1: Identify Variables.
You are looking for amounts of food. Therefore, your variables are quantities of each of the n foods. These are unknowns and need names. Let xj be the number of units of food j purchased. You want to find x = (x1, . . . , xn).
Now you need to use this notation to figure out the objective function and the constraints of the problem.
Step 2: Write Down the Objective Function.
The objective is to minimize the cost of the food that you buy. If you buy x how much will it cost? Break it down. Buying x means that you buy x1 units of the first food, x2 units of the second food, and so on. How much do you spend on the first food? It costs p1 per unit. Therefore you spend p1x1 on the first food. How much do you spend in total? You just add up what you spend on each of the foods. This quantity is: p1x1 + · · · + pjxj + · · · + pnxn = n X j=1 pjxj = p · x.
(5) (5) is the objective function. That is, you want to find x to min p · x.
I invoked linearity assumptions to write the objective function. I assumed constant returns to scale when I asserted that if pj is the price of one unit, then pjxj is the price of xj units. This is an assumption. Maybe it is impossible to buy goods in tiny quantities. Maybe it is possible to get large purchases at lower costs per unit. If so, then the linearity assumption is not appropriate (although it may be a reasonable approximation). I also invoked additivity when I claimed that the cost of the entire purchase is just the sum of the amount spent on each food. This assumption is reasonable, but you can imagine settings where people get discounts for buying large quantities.
If the problem was simply to minimize costs, then the answer would be easy.
Buy no food. After all, that costs you nothing. The problem with that is that the people in your institution will die. You want to minimize expenditures, but 7 only after you have met the nutritional requirements. You need a way to decide whether the food you buy actually satisfies nutritional requirements.
Step 3: Write Down the Constraints.
The constraints are that you satisfy nutritional requirements. You need to buy enough food to supply all nutrients in (at least) the recommended amounts.
How much nutrient i do you need? ci. How much of this nutrient is supplied when you have x? Again, take it one food at a time. You have x1 units of the first food. This means that you obtain ai1x1 units of the ith nutrient coming from the first food. (The units of x1 might be pounds (of hamburger); the units of aij might be grams (of iron) per pound (of hamburger)). Hence the product gives you a quantity of grams (of iron). How much nutrient i do you get from x? Add up the amount of nutrient i you get from each food.
ai1x1 + · · · + aijxj + · · · + ainxn = n X j=1 aijxj.
(6) Therefore, to supply enough of nutrient i you must satisfy the constraint that (6) be greater than or equal to ci. The constraint: ai1x1 + · · · + aijxj + · · · + ainxn = n X j=1 aijxj ≥ci (7) (7) describes the ith nutritional constraint. The entire problem imposes such a constraint for each nutrient. That is, (7) must hold for i = 1, . . . , m.
I can lump the constraints together using matrix notation: the m constraints described by (7) are: Ax ≥c. Once again notice that I made linearity assump-tions to formulate the constraints.
If the Vitamin C you get from oranges detracts from the Vitamin C you get from kiwis, then additivity fails. If your body cannot process more than 3 potatoes in a day (causing them to pass from your system without supplying nutrients), then the constant returns to scale assumption fails.
It is also natural to add the restriction that you cannot buy negative quan-tities of food. In symbols: x ≥0.
Step 4: Write Down the Entire Problem.
The work is over. Now just summarize it. The problem is to find x to solve: min p · x subject to Ax ≥c and x ≥0.
In practice, you will be given values for the parameters of the problem (A, p, and c) and then would go ahead and try to find a numerical solution. In fact, you can amuse yourself by going to: and finding out the costs of sample diets (I am not sure where the prices or requirements came from). This program allows you to select the foods that you 8 are willing to eat. I selected about thirty foods and was told that I should limit my diet to carrots, peanut butter, potatoes, and skim milk. With these, I could meet my nutritional requirements for 99 cents per day. This diet was heavy on the peanut butter. I decided that maybe I didn’t want to survive by spreading it on carrots so I ruled out peanut butter. When I did, my optimal diet cost $4.32 and involved nine different foods. I am grateful that my enormous university salary provides me the luxury of spending even more than this on food every day.
During the formulation problem, it is useful to think about what the solu-tion of the problem might look like. Would you expect the problem to have a solution? In theory two things can go wrong. Maybe the problem is infeasible.
That would mean that it is impossible to find any foods that would satisfy the nutritional requirements. This could happen if the government decided that everyone needed to consume positive quantities of Vitamin X, but there was no food that contained Vitamin X. (My son eats chicken soup, pasta, corn bread, chicken nuggets, and chocolate desserts. It is possible that this would not be enough to satisfy nutritional requirements without vitamin supplements.) On the other hand, if you could find every nutrient in some food, then (by buying enough) you could guarantee that you satisfy all of the requirements. That is, it is sensible to assume that the problem is feasible. Could it be unbounded?
For a minimization problem, this would mean that you could make the cost of the optimal diet arbitrarily small - not close to zero, but smaller than any num-ber. It does not make sense that the diet would cost less than, say, −$100. (I would interpret this as meaning that the store paid you $100 to take the food.) Indeed, if you assume that prices are all non-negative (true everywhere but in Mom’s kitchen), then any bundle of food you purchase will cost a non-negative amount, so the cost of diets are bounded below. In summary, it is sensible to assume that the diet problem has a solution.
Before leaving the diet problem, I want to describe another problem that is based on precisely the same data as the diet problem. Here is the story. You still run the institution. Someone approaches you and says: “Why bother with food? All you care about is that your animals get nutrients. I sell pills (one kind of pill for each nutrient). I have set my prices so that you can get nutrients more cheaply from me than through food. I will sell you exactly the nutrients you need and you will be better off.” You think about this and decide that it sounds reasonable. The new problem is to figure out how the pill seller should behave. Her problem is to set prices of pills that maximize the amount she can get selling you the necessary nutrients subject to the constraint that the pills provide nutrients more cheaply than food.
Step 1: Variables.
The pill seller wants to find prices for each nutrient pill. That is, she is looking for y = (y1, . . . , yi, . . . , ym), where yi is the price charged for a pill that supplies one unit of nutrient i.
Step 2: Objective.
The pill seller wants to maximize her profit. She sells c. If she can charge the prices y, then she earns c · y.
9 Step 3: Constraints.
What does it mean for the pills to be cheaper than food? Consider the first food. You don’t care what it looks like or what it tastes like. You only care what nutrients it provides. Suppose you try to replace food one with pills. What kinds of pill would you need? Food one supplies (in theory) amounts of all m nutrients. If you wanted to replace the nutrient i found in one unit of food one with nutrient i pills, you would need ai1 pills. This means that replacing the nutrient i in food one with pills would cost ai1yi. The total amount you would need to replace the nutrients in food one with pills is therefore Pm i=1 ai1yi.
In order for the (nutrients in the) pills to be cheaper than (the nutrients in) food one, it must be that m X i=1 ai1yi ≤p1.
I want to impose this kind of constraint for each food. That is, each food is at least as expensive as the cost of its nutrients.
This leads to, for each j = 1, . . . , n, m X i=1 aijyi ≤pj.
In concise notation, this becomes Aty ≤p (where At is the transpose of A: the matrix you get when you interchange rows and columns). Throughout this course I will write this expression yA ≤p. Those comfortable with linear algebra will know that this notation confuses row vectors with column vectors, but it is convenient and should lead to no confusion.
Also, I add a non-negativity constraint (that states that the pill seller does not give people money to take her pills): y ≥0.
Step 4: Conclusion.
Put the constraints together and we have the pill seller’s problem: Find y = (y1, . . . , ym) to solve: max c · y subject to yA ≤p and y ≥0.
On one hand, the pill seller’s problem is just a contrived way to practice problem formulation. It turns out, however, that it illustrates an important idea that will appear later in the course. At this stage, I want to point out several things.
Both the diet problem and the pill seller’s problem use the same basic data (A, c, and p). I constructed the pill seller’s problem so that there will be a relationship between its value and the value of the diet problem. Specifically, the optimal value of the diet problem (the minimum cost) will be greater than or equal to the optimal value of the pill problem (the maximum earnings of the seller).
Why?
The constraints in the pill problem guarantee that pills are cheaper than food. What the pill seller earns is what you would need to pay to buy all of the necessary nutrients. Since these nutrients cost less when 10 purchased in pill form (by the construction of the prices) than when purchased in food form, it must be that the cost of the pills is cheaper than the cost of the food. You can prove this. Suppose that x is feasible for the diet problem and y is feasible for the pill problem. That means that x satisfies Ax ≥c and x ≥0 and y satisfies yA ≤p and y ≥0. It follows that yAx ≥y · c (This follows because Ax ≥c and y ≥0. All you are doing is multiplying m separate inequalities by a non-negative numbers and then adding them up.
Note that yAx and y · c are both numbers.) and also yAx ≤p · x.
Combining these two inequalities yields y · c ≤yAx ≤p · x and, in particu-lar, y · c ≤p · x. This inequality says in symbols what I said in words earlier: The cost of the pills (priced so that pills are cheaper than food) is no greater than the cost of a feasible diet.
The general property of linear programming problems that you’ll learn is that when you actually solve these problems, the values are equal. That is, when you find the minimum cost diet, it will cost exactly the same amount as you would pay a profit-maximizing pill seller for the pills. This relationship (and consequences of it) allow us to interpret the prices obtained when you solve the pill seller’s problems as interesting economic quantities. It turns out that they actually provide the economic value of nutrients as seen by you (in your role as institutional menu planner). These prices give simple ways to answer questions of the form: how much extra would it cost to satisfy the diet problem if the nutritional requirement of the first nutrient went up by one unit.
11 |
190752 | https://journals.lww.com/amjforensicmedicine/Fulltext/2012/12000/An_Autopsy_Case_of_Acute_Carbon_Monoxide_Poisoning.16.aspx?generateEpub=Article%7Camjforensicmedicine:2012:12000:00016%7C10.1097/paf.0b013e318273b823%7C | The American Journal of Forensic Medicine and Pathology
Log in
or
Register
Subscribe to journal Subscribe
Get new issue alerts Get alerts;;)
Subscribe to eTOC;;)
### Secondary Logo
Enter your Email address:
Privacy Policy
### Journal Logo
Articles
Advanced Search
Toggle navigation
SubscribeRegisterLogin
Browsing History
Articles & Issues
Current Issue
Previous Issues
Published Ahead-of-Print
Collections
Journal Club Articles
The Scalpel is Passed
NAME Eckert Award Winners
Firearm News
TOX NEWS
Video Abstracts
For Authors
Submit a Manuscript
Information for Authors
Language Editing Services
Author Permissions
Journal Info
About the Journal
Editorial Board
Advertising
Open Access
Subscription Services
Reprints
Rights and Permissions
Articles
Advanced Search
December 2012 - Volume 33 - Issue 4
Previous Abstract
Next Abstract
Cite
Copy
Export to RIS
Export to EndNote
Share
Email
Facebook
X
LinkedIn
Favorites
Permissions
More
Cite
Permissions
Article as EPUB
Export All Images to PowerPoint FileAdd to My Favorites
Email to Colleague
Colleague's E-mail is Invalid
Your Name:
Colleague's Email:
Separate multiple e-mails with a (;).
Message:
Your message has been successfully sent to your colleague.
Some error has occurred while processing your request. Please try after some time.
Export to
End Note
Procite
Reference Manager
[x] Save my selection
Case Reports
An Autopsy Case of Acute Carbon Monoxide Poisoning After a Long-Term Vegetative State
Sato, Hiroaki MD, PhD; Tanaka, Toshiko PhD; Kasai, Kentaro MPSc; Tanaka, Noriyuki MD, PhD
Author Information
From the Department of Forensic Medicine, School of Medicine, University of Occupational and Environmental Health, Kitakyushu, Japan.
Manuscript received December 20, 2009; accepted April 6, 2010.
The authors report no conflicts of interest.
Reprints: Hiroaki Sato, MD, PhD, Department of Forensic Medicine, School of Medicine, University of Occupational and Environmental Health, Japan, Iseigaoka1-1, Yahata-Nishi, Kitakyushu, 807-8555, Japan. E-mail: h-sato@med.uoeh-u.ac.jp.
The American Journal of Forensic Medicine and Pathology 33(4):p 341-343, December 2012. | DOI: 10.1097/PAF.0b013e318273b823
Buy
Abstract
A 23-year-old woman was rescued from an accidental fire in a state of cardiopulmonary arrest. Based on the diagnosis of carbon monoxide (CO) poisoning, she received hyperbaric oxygen therapy and survived in a vegetative state. After 1 and a half years, she died without recovering from the vegetative state. At autopsy, the brain was observed to be moderately softened with a severely atrophied appearance and ventricular enlargement. In addition, a characteristic damage of hypoxic-ischemic leukoencephalopathy was also observed clearly in both the bilateral globus pallidus and cerebral white matter, which are typical findings of past acute CO poisoning. A long-term vegetative state causes the brain to soften and liquefy because of reactive gliosis and autolytic change. The cause of death becomes difficult to diagnose only from the autopsy findings in general. This case is rare in that the past acute CO poisoning could be diagnosed from the remaining typical cerebral findings even after a long-term vegetative state.
© 2012 Lippincott Williams & Wilkins, Inc.
Full Text Access for Subscribers:
##### Individual Subscribers Log in for access
##### Institutional Users Access through Ovid®
Not a Subscriber?
Buy
Subscribe
Request Permissions
You can read the full text of this article if you:
Log InAccess through Ovid
Source
An Autopsy Case of Acute Carbon Monoxide Poisoning After a Long-Term Vegetative State
The American Journal of Forensic Medicine and Pathology33(4):341-343, December 2012.
Full-Size
Email
Favorites
Export
View in Gallery
Email to Colleague
Colleague's E-mail is Invalid
Your Name:
Colleague's Email:
Separate multiple e-mails with a (;).
Message:
Your message has been successfully sent to your colleague.
Some error has occurred while processing your request. Please try after some time.
Readers Of this Article Also Read
Accidental Carbon Monoxide Poisoning While Driving: A Case Report With Review...
Unusual Presentation of Death Due to Carbon Monoxide Poisoning: A Report of Two ...
Peanut Butter and Fatal Fluoride Poisoning: A Case of Mistaken Identity
Delayed Death Following Carbon Monoxide Poisoning: A Case Report
Most Popular
Asphyxial Suicide by Inhalation of Helium Inside a Plastic Bag
Asphyxial Suicide with Helium and a Plastic Bag
Observation of Two Suicides by Helium Inhalation in a Prefilled Environment
Suicide by Asphyxiation Due to Helium Inhalation
Nitrogen–Plastic Bag Suicide: A Case Report
Back to Top
Never Miss an Issue
Get new journal Tables of Contents sent right to your email inbox Get New Issue Alerts
Browse Journal Content
Most Popular
For Authors
About the Journal
Past Issues
Current Issue
Register on the website
Subscribe
Get eTOC Alerts;;)
For Journal Authors
Submit an article
How to publish with us
Customer Service
Live Chat
Chat Offline
Activate your journal subscription
Activate Journal Subscription
Browse the help center
Help
Contact us at:
Support: Submit a Service Request
TEL: (USA): TEL: (Int’l): 800-638-3030 (within USA) 301-223-2300 (international)
Manage Cookie Preferences
Privacy Policy
Legal Disclaimer
Terms of Use
Open Access Policy
Feedback
Sitemap
RSS Feeds
LWW Journals
Your California Privacy Choices
Copyright©2025
Wolters Kluwer Health, Inc. and/or its subsidiaries. All rights reserved. | Content use for text and data mining and artificial intelligence training is not permitted.
Your Privacy
To give you the best possible experience we use cookies and similar technologies. We use data collected through these technologies for various purposes, including to enhance website functionality, remember your preferences, show the most relevant content, and show the most useful ads. You can select your preferences by clicking the link. For more information, please review ourPrivacy & Cookie Notice
Accept All Cookies
Manage Cookie Preferences
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device. Because we respect your right to privacy, you can choose not to allow certain types of cookies on our website. Click on the different category headings to find out more and manage your cookie preferences. However, blocking some types of cookies may impact your experience on the site and the services we are able to offer.
Privacy & Cookie Notice
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function. They are usually set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, this may have an effect on the proper functioning of (parts of) the site.
View Vendor Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality, user experience and personalization, and may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies, then some or all of these services may not function properly.
View Vendor Details
Performance Cookies
[x] Performance Cookies
These cookies support analytic services that measure and improve the performance of our site. They help us know which pages are the most and least popular and see how visitors move around the site.
View Vendor Details
Advertising Cookies
[x] Advertising Cookies
These cookies may collect insights to issue personalized content and advertising on our own and other websites, and may be set through our site by third party providers. If you do not allow these cookies, you may still see basic advertising on your browser that is generic and not based on your interests.
View Vendor Details
Vendors List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
190753 | https://learn.concord.org/resources/779/mutations | Mutations | STEM Resource Finder
Skip to main content.
Learn about the Concord Consortium
Home
Collections
About
Help
Register
Log In
Menu Close
« Find more educational resources
Mutations
Explore how changing the DNA sequence can change the amino acid sequence of a protein. Proteins are composed of long strings of amino acids. Proteins are coded for in the DNA. DNA is composed of four different types of nucleotides. Converting the information in DNA into protein is a two-step process, involving transcription and translation. In transcription each mRNA nucleotide pairs with the complementary DNA nucleotide. In translation, each tRNA nucleotide pairs with the complementary mRNA nucleotide. Thus, a change in the DNA sequence can change the amino acid sequence of the protein. There are three basic types of mutations: insertion, deletion and substitution. Some mutations are silent, meaning that there is no change in the protein, while others can cause major changes in the protein.
PreviewAssign or Share
Requirements
This activity runs entirely in a Web browser. Preferred browsers are: Google Chrome (versions 30 and above) Safari (versions 7 and above), Firefox (version 30 and above), Internet Explorer (version 10 or higher), and Microsoft Edge.
Standards
Performance Expectations
MS-LS3-1
Heredity: Inheritance and Variation of Traits. Students who demonstrate understanding can: Develop and use a model to describe why structural changes to genes (mutations) located on chromosomes may affect proteins and may result in harmful, beneficial, or neutral effects to the structure and function of the organism.
Subject Areas
Life Science
Grade Levels
7-8
9-12
Higher Ed
Learn More
This resource is part of the Concord Consortium's Next-Generation Molecular Workbench project.
FacebookTwitterEmail
© 2025 The Concord Consortium. All Rights Reserved.
The Concord Consortium is a 501(c)(3) nonprofit charity registered in the U.S. under EIN 04-3254131.
Privacy Policy · Questions/Feedback: Send us an email
Version: v2.25.5 (React 18.3.1) |
190754 | https://www.michigan.gov/mdhhs/adult-child-serv/childrenfamilies/hereditary/newborn-screening----list-of-disorders | Newborn Screening - List of Disorders
Skip to main content
MDHHS
Health & Human Services
Assistance Programs Assistance Programs collapsed link
Adult & Children's Services Adult & Children's Services collapsed link
Safety & Injury Prevention Safety & Injury Prevention collapsed link
Keeping Michigan Healthy Keeping Michigan Healthy collapsed link
Doing Business with MDHHS Doing Business with MDHHS collapsed link
Inside MDHHS Inside MDHHS collapsed link
Assistance Programs
Go to Assistance Programs
Benefit Updates
Women, Infants & Children
Cash Assistance
Child Care Assistance
Children's Special Health Care Services
Emergency Relief: Home, Utilities & Burial
Food Assistance
Health Care Coverage
Help from Other Programs
Medicaid
Migrant Services
Housing and Homeless Services
Water Assistance Program
Refugee Assistance
Universal Caseload Action Plan
Benefit Updates
Go to Benefit Updates
Learn about Federal changes to Medicaid and Food Assistance Programs..
Women, Infants & Children
Go to Women, Infants & Children
Michigan's Women, Infants & Children program, providing supplemental nutrition, breastfeeding information, and other resources for healthy mothers & babies.
Formula Information for WIC Clients
About WIC
WIC EBT Transition Information
Breastfeeding for WIC Clients
WIC Project FRESH
Fraud & Abuse
Frequently Asked Questions
Local WIC Agencies
Health Care Providers
Non-Discrimination Statement
Nutrition
BF Resources for Families
CIAO
Client Resources
WIC Data & System Management
WIC Eligibility
Michigan WIC Client Advisory Council
Share Your WIC Joy
Submitting a Michigan WIC Complaint
Welcome To WIC
WIC Reminders
WIC Staff
WIC Vendors/Grocers
Vaccine Information
Cash Assistance
Go to Cash Assistance
Information on the Family Independence Program, State Disability Assistance, SSI, Refugee, and other cash assistance.
Application Process
Disability Assistance
Eligibility
Forms & Publications
Payment - Bridge Card
Refugee Assistance
Child Care Assistance
Go to Child Care Assistance
If you cannot afford child care, payment assistance is available.
Children's Special Health Care Services
Go to Children's Special Health Care Services
Children's Special Health Care Services information and FAQ's.
Bullying Prevention Initiative
Children with Medical Complexity Targeted Case Management
Children’s Special Health Care Services program eligibility expanded to age 26
General Information For Families About CSHCS
Home Care Childrens HCC TEFRA Program
Michigan Pediatric Epilepsy Project
Emergency Relief: Home, Utilities & Burial
Go to Emergency Relief: Home, Utilities & Burial
Information on assistance with home repairs, heat and utility bills, relocation, home ownership, burials, home energy, and eligibility requirements.
Burial
Earned Income Tax Credit (EITC)
Energy & Weatherization
Forms & Publications
Heat & Utilities
Home Ownership
Home Repairs
Relocation
Food Assistance
Go to Food Assistance
Information on the Food Assistance Program, eligibility requirements, and other food resources.
Application Process
Eligibility
Forms and Publications
Payment
Supplemental Nutrition Assistance Program Education
Buy Food Online
Restaurant Meal Program
Summer EBT
Disaster Food Assistance Program
Health Care Coverage
Go to Health Care Coverage
Health Care Coverage information and resources.
Adults
Children & Teens
Healthy Michigan Plan
Help Finding Health Care
Persons With Disabilities
Pregnant Women
Protected Health Information (HIPAA)
Services for Seniors
Medical Debt Relief
Help from Other Programs
Go to Help from Other Programs
Help from Other Programs
Medicaid
Go to Medicaid
Information about the health care programs available through Medicaid and how to qualify.
1115 Reentry Services Demonstration
Community Health Workers
Rural Health Transformation Program
Migrant Services
Go to Migrant Services
Our migrant program works with a number of organizations to provide services for Michigan’s migrant and seasonal farmworkers.
Servicios para Campesinos
Housing and Homeless Services
Go to Housing and Homeless Services
Housing and Homeless Services
About Us
Training & Events
Funding
Resources for the Public
Resources for Providers
Partners
Glossary
Water Assistance Program
Go to Water Assistance Program
Water Assistance Program
Refugee Assistance
Go to Refugee Assistance
A federal program which helps persons admitted into the U.S. as refugees to become self-sufficient after their arrival.
Universal Caseload Action Plan
Go to Universal Caseload Action Plan
Universal caseload, or task-based processing, is a different way of handling public assistance cases.
Adult & Children's Services
Go to Adult & Children's Services
Abuse & Neglect
Adoption
Adults & Seniors
Child Fatality Registry
Child Support
Children & Families
Foster Care
Hope For A Home
Informed Consent for Abortion
Juvenile Justice
Child Welfare Medical and Behavioral Health Resources
Prevention and Family Preservation Services
Abuse & Neglect
Go to Abuse & Neglect
Information on the Children's Protective Services Program, child abuse reporting procedures, and help for parents in caring for their children.
Adult Protective Services
Children's Protective Services
Children's Trust Fund
Forms and Publications
Human Trafficking
Adoption
Go to Adoption
Information on adoption programs, adoption resources, locating birth parents and obtaining information from adoption records.
Forms and Publications
Post-Adoption Parent Resources
Records
Support
Adults & Seniors
Go to Adults & Seniors
Information on Adult Protective Services, Independent Living Services, Adult Community Placement Services, and HIV/AIDS Support Services.
Aging, Community Living and Supports Bureau (ACLS Bureau)
Adult Community Placement
Forms and Publications
Independent Living (Home Help)
GetSetUp - Michigan
Services for Seniors
Child Fatality Registry
Go to Child Fatality Registry
Child Fatality Registry
Child Support
Go to Child Support
Information on child support services for participants and partners.
Child Support Community Advisory Council
How Do I?
Structured Child Support
Resources
Our Partners
Children & Families
Go to Children & Families
Programs for healthy children & families, including immunization, lead poisoning prevention, prenatal smoking cessation, and many others.
Developmental Delays - Early On
Early Hearing Detection and Intervention
Healthy Children & Healthy Families
Hereditary Disorders
Division of Immunization
Maternal & Child Health Epidemiology
Pregnancy Risk Assessment Monitoring
Trauma & Toxic Stress
Child & Adolescent Health
Michigan Maternal Mortality Surveillance Program
Foster Care
Go to Foster Care
Information on the Children's Foster Care program and becoming a Foster Parent.
Forms and Publications
Fostering Mental Health
Learn More at Events Near You
Support for Foster Youth
Support for Parents
Respite Care
Hope For A Home
Go to Hope For A Home
Hope For A Home
Faces of Foster Care
Informed Consent for Abortion
Go to Informed Consent for Abortion
Information about the Michigan law that requires certain information be made available to a woman who is seeking an abortion at least 24 hours prior to the abortion procedure.
Informed Consent for Abortion for Providers
Juvenile Justice
Go to Juvenile Justice
Information on treatment and services for juvenile offenders, success stories, and more.
Bed Management Tool
Pura Strong OpEd
Facilities
Grants
Interstate Compact
Policy & Compliance
Youth Resources
Juvenile Competency
Regional Detention Support Services
Prison Rape Elimination Act
Child Welfare Medical and Behavioral Health Resources
Go to Child Welfare Medical and Behavioral Health Resources
Child Welfare Medical and Behavioral Health Resources
Serious Emotional Disturbance Waiver
Contacts
Education and Training
Emergency Services
Grants and Current Projects
Guides
Policies and Forms
Prevention and Family Preservation Services
Go to Prevention and Family Preservation Services
MDHHS and partners support a continuum of prevention services. The continuum is inclusive of primary, secondary, and tertiary prevention activities. Visit this page to learn more.
Prevention and Family Preservation Data
Safety & Injury Prevention
Go to Safety & Injury Prevention
Children's & Adult Protective Services
Children's Trust Fund - Abuse Prevention
Injury & Violence Prevention
Public Safety
Safe Delivery
Infant Safe Sleep
Environmental Health
Firearm Injury Prevention
Children's & Adult Protective Services
Go to Children's & Adult Protective Services
Children's Trust Fund - Abuse Prevention
Go to Children's Trust Fund - Abuse Prevention
Injury & Violence Prevention
Go to Injury & Violence Prevention
Information about injury and violence prevention programs in Michigan
Public Safety
Go to Public Safety
information about the Department's public safety programs.
Bureau of Emergency Preparedness, EMS, and Systems of Care
Division of Victim Services
Division of Emergency Preparedness & Response
Great Lakes Border Health Initiative
Michigan Sports Concussion Law
Safe Delivery
Go to Safe Delivery
Information on the Safe Delivery Program, laws, and publications.
Safe Delivery Publications
Infant Safe Sleep
Go to Infant Safe Sleep
Information on Safe Sleep for your baby, how to protect your baby's life.
Information for Professionals
Safe Sleep for Your Baby
Infant Safe Sleep for EMS Agencies and Fire Departments
Safe Sleep Resources by County
Stories on Infant Safe Sleep
Environmental Health
Go to Environmental Health
The Michigan Department of Health and Human Services' (MDHHS) Environmental Health Bureau (EHB) uses the best available science to reduce, eliminate, or prevent harm from environmental, chemical, and physical hazards.
Menominee Warehouse Fire Response
Environmental Health Home Page
Your Health and Wildfire Smoke
How Can I Be Exposed?
Our Programs
Data and Research
For Health Care Providers
Firearm Injury Prevention
Go to Firearm Injury Prevention
Information, sources of data and resources to support Firearm Injury Prevention efforts across the state of Michigan.
Secure Storage Law
Firearm Safety
Keeping Michigan Healthy
Go to Keeping Michigan Healthy
Medical Resources for Child Protection
Infectious Disease
Public Health
Local Health Services
Adult Behavioral Health & Developmental Disability
Neighborhood Wellness Centers
Plan of Safe Care
Virtual Baby Fair
Office of Transformation, Engagement and Community Health
Chronic Diseases
Communicable & Chronic Diseases
Health Statistics & Reports
Maternal & Infant Health
Medical Resources for Child Protection
Go to Medical Resources for Child Protection
Medical Resources for Child Protection
Infectious Disease
Go to Infectious Disease
Public Health
Go to Public Health
Local Health Services
Go to Local Health Services
Serves as liaison and provides funding to Michigan’s 45 local health departments.
What Is Local Public Health
Accreditation
Block Grant
Adult Behavioral Health & Developmental Disability
Go to Adult Behavioral Health & Developmental Disability
Adult Behavioral Health & Developmental Disability Services
Developmental Disability
CCBHC Demonstration
Michigan Crisis and Access Line (MiCAL)
Behavioral Health Information Sharing & Privacy
Marihuana
Mental Health Diversion Council
PIHPs
Stay Well Program Resources
Syringe Service Program (SSP)
Veteran Navigators
WHODAS
Substance Use Gambling and Epidemiology
Problem Gambling
Integrated Treatment for Co-occurring Disorders
Mental Health
MI Psychiatric Care Improvement Project
Reporting Requirements
Neighborhood Wellness Centers
Go to Neighborhood Wellness Centers
NWC Hours
Plan of Safe Care
Go to Plan of Safe Care
Information and resources for development of Plans of Safe Care.
POSC for Parents
POSC for Professionals
Resources
Templates
Virtual Baby Fair
Go to Virtual Baby Fair
Virtual Baby Fair OnDemand
Office of Transformation, Engagement and Community Health
Go to Office of Transformation, Engagement and Community Health
Multicultural Health
State Legal Authorities
About OEMH
Resources, Videos, & Other
TEaCH Data and Epidemiology
Health Equity Advisory Councils
TTA
Capacity Building
Addressing COVID 19 Disparities
Race and Ethnicity Standards
Behavioral Health Grant
Vaccine Equity
COVID-19
Relevant Legislation
Toolkits
Annual Reports/Documents
Programs
Trainings
Heritage Months and Special Observances
Chronic Diseases
Go to Chronic Diseases
Prevention of diseases & conditions such as heart disease, cancer, diabetes and many others.
Long COVID
Seasonal Respiratory Viruses
Michigan Arthritis Program
Asthma
Cancer
Cardiovascular Health, Nutrition & Physical Activity
Dementia
Diabetes
Disability Health
Hepatitis
HIV/STI
Lead
Tobacco
Communicable & Chronic Diseases
Go to Communicable & Chronic Diseases
Information on communicable & chronic diseases.
Coronavirus
Communicable Disease Information and Resources
Healthcare-Associated Infections
Michigan Disease Surveillance System
Health Statistics & Reports
Go to Health Statistics & Reports
Maternal & Infant Health
Go to Maternal & Infant Health
Home to an array of public health programs, initiatives and interventions aimed at improving the health and well-being of women, infants, families and communities.
Advancing Healthy Births
Clinical Quality Initiatives
Breastfeeding
Doula Initiative
Michigan Perinatal Quality Collaborative (MI PQC)
Our Programs and Initiatives
Partners and Resources
Get Involved
Mother Infant Health & Equity Collaborative (MIHEC) Meetings
Doing Business with MDHHS
Go to Doing Business with MDHHS
Birth, Death, Marriage and Divorce Records
Boards and Commissions
Bridge Card Participation
Child & Adult Provider Payments
Child Care Fund
Child Welfare
Contractor and Subrecipient Resources
Community & Faith-Based Programs
Farmworker Outreach Services Division
Forms & Applications
MIBridges Partners
State Health Improvement Plan
State Health Assessment
Health Care Providers
Licensing
Michigan Statewide Systems of Care
State Innovation Model
MiSACWIS Training
Pathways to Potential
Protect MiFamily -Title IV-E Waiver
Weatherization Assistance Program
Birth, Death, Marriage and Divorce Records
Go to Birth, Death, Marriage and Divorce Records
Birth, Death, Marriage and Divorce Records
Infectious Agent Reporting
Order A Record Online
Correct A Birth Record
Schedule Appointment
Order A Record by Mail
Eligibility Requirements
Additional Information
Correct A Death Record
GENDIS
Heirloom Birth Certificates
Boards and Commissions
Go to Boards and Commissions
Boards and Commissions
Michigan Gun Violence Prevention Task Force
Children Trust Michigan
Community Action & Economic Opportunity
Health Information Technology Commission
Certificate of Need
Child Lead Exposure Elimination Commission
Coronavirus Task Force on Racial Disparities
Domestic Violence
Human Trafficking Health Advisory Board
Institutional Review Board
Michigan Commission on Services to the Aging
Nursing Home Workforce Stabilization Council
Guy Thompson Parent Advisory Council (GTPAC)
Prescription Drug Task Force
Protect Michigan Commission
Michigan Suicide Prevention Commission
Bridge Card Participation
Go to Bridge Card Participation
Bridge Card Participation Information on Electronic Benefits for clients and businesses, lists of participating retailers and ATMs, and QUEST.
Child & Adult Provider Payments
Go to Child & Adult Provider Payments
Provider Payments Information on the direct deposit of State of Michigan payments into a provider's bank account.
Child Care Fund
Go to Child Care Fund
Child Care Fund
Contact Information
Child Welfare
Go to Child Welfare
Child Welfare Resources
Child Welfare Law Manual
Child Welfare Training
MCI Delegation of Authority
Every Student Succeeds Act
MiTEAM Practice Model
Strengthening Our Focus on Children & Families
Contractor and Subrecipient Resources
Go to Contractor and Subrecipient Resources
Information on How to Bid, Requests for Proposals, forms and publications, contractor rates, and manuals.
Community & Faith-Based Programs
Go to Community & Faith-Based Programs
Information & resources for Community and Faith-Based partners
Community and Faith Based Initiative
Workshops
Farmworker Outreach Services Division
Go to Farmworker Outreach Services Division
Photo Galleries
About the Farmworker Outreach Services Division
Farmworker Committees and Councils
Farmworker Resources
Farmworker Research in Michigan
COVID
División de Servicios de Extensión para Campesinos
MI Interagency Migrant Services Committee
Migrant Resource Councils
Forms & Applications
Go to Forms & Applications
Information on DHS Applications and Forms grouped by category.
MIBridges Partners
Go to MIBridges Partners
MIBridges Partners
Features
Frequently Asked Questions
Become a Partner
Community Partner Training
Tools and Resources
State Health Improvement Plan
Go to State Health Improvement Plan
State Health Improvement Plan (SHIP)
State Health Assessment
Go to State Health Assessment
State Health Assessment
Michigan 2019 SHA-Call to Action
Health Care Providers
Go to Health Care Providers
Health Care Providers
Certificate of Need
Highly Integrated Dual Eligible Special Needs Plan
MI Coordinated Health
Nurse Loan Repayment Program
Civil Monetary Penalty (CMP) Grant Program
Community Mental Health Services
Departmental Forms
Health Professional Shortage Area
High Utilizers
HIPAA
Institutional Review Board
International Medical Graduate Programs
Community Transition Services
Lab Services
Nurse Aide Training and Testing Reimbursement Forms and Instructions
MI Health Link
Office of Nursing Programs
State Loan Repayment Program
Behavioral Health Loan Repayment Program (BHLRP)
Substance Abuse Providers
Michigan Opioid Treatment Access Loan Repayment Program
Licensing
Go to Licensing
Licensing information for Adult Foster Care and Homes for the Aged, Child Day Care Facilities, Child Caring Institutions, Children's Foster Care Homes, Child Placing Agencies, Juvenile Court Operated Facilities and Children's or Adult Foster Care Camps.
Body Art Licensing
Child Welfare Licensing
Michigan Statewide Systems of Care
Go to Michigan Statewide Systems of Care
Michigan Statewide Systems of Care
Trauma
Stroke STEMI
Committees
Contacts
State Innovation Model
Go to State Innovation Model
Information on the grant awarded for the State Innovation Model Proposal
MiSACWIS Training
Go to MiSACWIS Training
MiSACWIS Training
Pathways to Potential
Go to Pathways to Potential
Pathways to Potential site
About
Strategic Priorities
Success Coaches in Action
Why Pathways to Potential
Get Involved
Resources
Success Stories
Contact Us
Protect MiFamily -Title IV-E Waiver
Go to Protect MiFamily -Title IV-E Waiver
Protect MiFamily -Title IV-E Waiver
Assessment Tools
Contractors
Evaluation
Expected Outcomes
Forms - Publications
Protect MI Family Contacts
Understanding Michigan's Waiver project
Weatherization Assistance Program
Go to Weatherization Assistance Program
Offers resources for agencies who operate the Weatherization Assistance Program in the state of Michigan
CSPM Manual
Prospective Clients
Prospective Contractors
Prospective Workforce
State Map of Weatherization Operators
Students in Energy Efficiency-Related Field
Resources for Weatherization Operators
Photo Gallery
Inside MDHHS
Go to Inside MDHHS
Newsroom
How To Find Resources
SEC 238
Tribal Government Services and Policy
Contact MDHHS
County Offices
Executive Staff Bios
Careers With MDHHS
Budget & Finance
Complaints
Office of Inspector General
Community & Volunteer Opportunities
Reports & Statistics - Health Services
Hotlines
Legal
MDHHS Audit
Policy and Planning
Reports & Statistics - Human Services
Newsroom
Go to Newsroom
MDHHS News, Press Releases, Media toolkit, and Media Inquiries.
How To Find Resources
Go to How To Find Resources
SEC 238
Go to SEC 238
Tribal Government Services and Policy
Go to Tribal Government Services and Policy
Contact MDHHS
Go to Contact MDHHS
County Offices
Go to County Offices
Locate your County Office
East Michigan
Genesee County
Local Health Department Maps
Macomb County
Northern Mid-Michigan
Oakland County
U.P. and Northern Michigan
Urban Counties
Wayne County
West Michigan
Executive Staff Bios
Go to Executive Staff Bios
Careers With MDHHS
Go to Careers With MDHHS
Careers With MDHHS
New Hire Forms
Budget & Finance
Go to Budget & Finance
Boilerplate Reports
Complaints
Go to Complaints
Complaints
Office of Inspector General
Go to Office of Inspector General
Office of Inspector General
About Us
Sample of Investigations
Michigan Law
Annual Reports
Media Resources
Contact Us
Community & Volunteer Opportunities
Go to Community & Volunteer Opportunities
Information on resources in your community and volunteer recruitment and training, and services provided at local DHS offices.
Reports & Statistics - Health Services
Go to Reports & Statistics - Health Services
Information is collected to monitor the general health and well-being of Michigan citizens.
Birth Defects
Cancer Statistics
Communicable Diseases
Community Health Information
Chronic Disease and Health Indicators
Health Disparities
Health Care Statistics
Maternal & Child Health Data
Michigan Public Health Data Dashboards
Public Health Advisory Council
Services Statistics for Flint
Population Trends
Other Chronic Disease & Injury Control Data
Other Published Reports and Surveys
Vital Statistics
Hotlines
Go to Hotlines
Helpful phone numbers to MDHHS services.
Legal
Go to Legal
Equal Opportunity, Legal Base, Laws and Reporting Welfare Fraud information.
Child Welfare Reform
Equal Opportunity
Federal Child Welfare Reviews
Freedom of Information Act
Laws
Legal Base
Medical Records Access Act
Nondiscrimination Statement (No discriminacion)
Subpoena
Records from Closed State Facilities
Flint Water Settlement
Hepatitis C Lawsuit
Severance Pay FY21
Severance Pay FY22
Severance Pay FY23
Text Campaign Terms & Conditions
MDHHS Audit
Go to MDHHS Audit
Information about audits conducted by the Office of Audit.
Audit Reporting
Long Term Care Audit
Policy and Planning
Go to Policy and Planning
Legislation policy and planning information.
2022-2024 Social Determinants of Health Strategy
EMS
Future of Behavioral Health
Workforce/Access & Grants Management Section
Reports & Statistics - Human Services
Go to Reports & Statistics - Human Services
A variety of reports & statistics for programs and services.
Annual Report
Capital Improvement Plan
Green Book Filtered Search Page
Reports, Evaluations & Studies
State Plans & Federal Regulations
Trend Report
Assistance Programs
Adult & Children's Services
Safety & Injury Prevention
Keeping Michigan Healthy
Doing Business with MDHHS
Inside MDHHS
BackAssistance Programs
Benefit Updates
Women, Infants & Children
Cash Assistance
Child Care Assistance
Children's Special Health Care Services
Emergency Relief: Home, Utilities & Burial
Food Assistance
Health Care Coverage
Help from Other Programs
Medicaid
Migrant Services
Housing and Homeless Services
Water Assistance Program
Refugee Assistance
Universal Caseload Action Plan
BackBenefit Updates
BackWomen, Infants & Children
Formula Information for WIC Clients
About WIC
WIC EBT Transition Information
Breastfeeding for WIC Clients
WIC Project FRESH
Fraud & Abuse
Frequently Asked Questions
Local WIC Agencies
Health Care Providers
Non-Discrimination Statement
Nutrition
BF Resources for Families
CIAO
Client Resources
WIC Data & System Management
WIC Eligibility
Michigan WIC Client Advisory Council
Share Your WIC Joy
Submitting a Michigan WIC Complaint
Welcome To WIC
WIC Reminders
WIC Staff
WIC Vendors/Grocers
Vaccine Information
BackCash Assistance
Application Process
Disability Assistance
Eligibility
Forms & Publications
Payment - Bridge Card
Refugee Assistance
BackChild Care Assistance
BackChildren's Special Health Care Services
Bullying Prevention Initiative
Children with Medical Complexity Targeted Case Management
Children’s Special Health Care Services program eligibility expanded to age 26
General Information For Families About CSHCS
Home Care Childrens HCC TEFRA Program
Michigan Pediatric Epilepsy Project
BackEmergency Relief: Home, Utilities & Burial
Burial
Earned Income Tax Credit (EITC)
Energy & Weatherization
Forms & Publications
Heat & Utilities
Home Ownership
Home Repairs
Relocation
BackFood Assistance
Application Process
Eligibility
Forms and Publications
Payment
Supplemental Nutrition Assistance Program Education
Buy Food Online
Restaurant Meal Program
Summer EBT
Disaster Food Assistance Program
BackHealth Care Coverage
Adults
Children & Teens
Healthy Michigan Plan
Help Finding Health Care
Persons With Disabilities
Pregnant Women
Protected Health Information (HIPAA)
Services for Seniors
Medical Debt Relief
BackHelp from Other Programs
BackMedicaid
1115 Reentry Services Demonstration
Community Health Workers
Rural Health Transformation Program
BackMigrant Services
Servicios para Campesinos
BackHousing and Homeless Services
About Us
Training & Events
Funding
Resources for the Public
Resources for Providers
Partners
Glossary
BackWater Assistance Program
BackRefugee Assistance
BackUniversal Caseload Action Plan
BackAdult & Children's Services
Abuse & Neglect
Adoption
Adults & Seniors
Child Fatality Registry
Child Support
Children & Families
Foster Care
Hope For A Home
Informed Consent for Abortion
Juvenile Justice
Child Welfare Medical and Behavioral Health Resources
Prevention and Family Preservation Services
BackAbuse & Neglect
Adult Protective Services
Children's Protective Services
Children's Trust Fund
Forms and Publications
Human Trafficking
BackAdoption
Forms and Publications
Post-Adoption Parent Resources
Records
Support
BackAdults & Seniors
Aging, Community Living and Supports Bureau (ACLS Bureau)
Adult Community Placement
Forms and Publications
Independent Living (Home Help)
GetSetUp - Michigan
Services for Seniors
BackChild Fatality Registry
BackChild Support
Child Support Community Advisory Council
How Do I?
Structured Child Support
Resources
Our Partners
BackChildren & Families
Developmental Delays - Early On
Early Hearing Detection and Intervention
Healthy Children & Healthy Families
Hereditary Disorders
Division of Immunization
Maternal & Child Health Epidemiology
Pregnancy Risk Assessment Monitoring
Trauma & Toxic Stress
Child & Adolescent Health
Michigan Maternal Mortality Surveillance Program
BackFoster Care
Forms and Publications
Fostering Mental Health
Learn More at Events Near You
Support for Foster Youth
Support for Parents
Respite Care
BackHope For A Home
Faces of Foster Care
BackInformed Consent for Abortion
Informed Consent for Abortion for Providers
BackJuvenile Justice
Bed Management Tool
Pura Strong OpEd
Facilities
Grants
Interstate Compact
Policy & Compliance
Youth Resources
Juvenile Competency
Regional Detention Support Services
Prison Rape Elimination Act
BackChild Welfare Medical and Behavioral Health Resources
Serious Emotional Disturbance Waiver
Contacts
Education and Training
Emergency Services
Grants and Current Projects
Guides
Policies and Forms
BackPrevention and Family Preservation Services
Prevention and Family Preservation Data
BackSafety & Injury Prevention
Children's & Adult Protective Services
Children's Trust Fund - Abuse Prevention
Injury & Violence Prevention
Public Safety
Safe Delivery
Infant Safe Sleep
Environmental Health
Firearm Injury Prevention
BackChildren's & Adult Protective Services
BackChildren's Trust Fund - Abuse Prevention
BackInjury & Violence Prevention
BackPublic Safety
Bureau of Emergency Preparedness, EMS, and Systems of Care
Division of Victim Services
Division of Emergency Preparedness & Response
Great Lakes Border Health Initiative
Michigan Sports Concussion Law
BackSafe Delivery
Safe Delivery Publications
BackInfant Safe Sleep
Information for Professionals
Safe Sleep for Your Baby
Infant Safe Sleep for EMS Agencies and Fire Departments
Safe Sleep Resources by County
Stories on Infant Safe Sleep
BackEnvironmental Health
Menominee Warehouse Fire Response
Environmental Health Home Page
Your Health and Wildfire Smoke
How Can I Be Exposed?
Our Programs
Data and Research
For Health Care Providers
BackFirearm Injury Prevention
Secure Storage Law
Firearm Safety
BackKeeping Michigan Healthy
Medical Resources for Child Protection
Infectious Disease
Public Health
Local Health Services
Adult Behavioral Health & Developmental Disability
Neighborhood Wellness Centers
Plan of Safe Care
Virtual Baby Fair
Office of Transformation, Engagement and Community Health
Chronic Diseases
Communicable & Chronic Diseases
Health Statistics & Reports
Maternal & Infant Health
BackMedical Resources for Child Protection
BackInfectious Disease
BackPublic Health
BackLocal Health Services
What Is Local Public Health
Accreditation
Block Grant
BackAdult Behavioral Health & Developmental Disability
Developmental Disability
CCBHC Demonstration
Michigan Crisis and Access Line (MiCAL)
Behavioral Health Information Sharing & Privacy
Marihuana
Mental Health Diversion Council
PIHPs
Stay Well Program Resources
Syringe Service Program (SSP)
Veteran Navigators
WHODAS
Substance Use Gambling and Epidemiology
Problem Gambling
Integrated Treatment for Co-occurring Disorders
Mental Health
MI Psychiatric Care Improvement Project
Reporting Requirements
BackNeighborhood Wellness Centers
NWC Hours
BackPlan of Safe Care
POSC for Parents
POSC for Professionals
Resources
Templates
BackVirtual Baby Fair
Virtual Baby Fair OnDemand
BackOffice of Transformation, Engagement and Community Health
State Legal Authorities
About OEMH
Resources, Videos, & Other
TEaCH Data and Epidemiology
Health Equity Advisory Councils
TTA
Capacity Building
Addressing COVID 19 Disparities
Race and Ethnicity Standards
Behavioral Health Grant
Vaccine Equity
COVID-19
Relevant Legislation
Toolkits
Annual Reports/Documents
Programs
Trainings
Heritage Months and Special Observances
BackChronic Diseases
Long COVID
Seasonal Respiratory Viruses
Michigan Arthritis Program
Asthma
Cancer
Cardiovascular Health, Nutrition & Physical Activity
Dementia
Diabetes
Disability Health
Hepatitis
HIV/STI
Lead
Tobacco
BackCommunicable & Chronic Diseases
Coronavirus
Communicable Disease Information and Resources
Healthcare-Associated Infections
Michigan Disease Surveillance System
BackHealth Statistics & Reports
BackMaternal & Infant Health
Advancing Healthy Births
Clinical Quality Initiatives
Breastfeeding
Doula Initiative
Michigan Perinatal Quality Collaborative (MI PQC)
Our Programs and Initiatives
Partners and Resources
Get Involved
Mother Infant Health & Equity Collaborative (MIHEC) Meetings
BackDoing Business with MDHHS
Birth, Death, Marriage and Divorce Records
Boards and Commissions
Bridge Card Participation
Child & Adult Provider Payments
Child Care Fund
Child Welfare
Contractor and Subrecipient Resources
Community & Faith-Based Programs
Farmworker Outreach Services Division
Forms & Applications
MIBridges Partners
State Health Improvement Plan
State Health Assessment
Health Care Providers
Licensing
Michigan Statewide Systems of Care
State Innovation Model
MiSACWIS Training
Pathways to Potential
Protect MiFamily -Title IV-E Waiver
Weatherization Assistance Program
BackBirth, Death, Marriage and Divorce Records
Infectious Agent Reporting
Order A Record Online
Correct A Birth Record
Schedule Appointment
Order A Record by Mail
Eligibility Requirements
Additional Information
Correct A Death Record
GENDIS
Heirloom Birth Certificates
BackBoards and Commissions
Michigan Gun Violence Prevention Task Force
Children Trust Michigan
Community Action & Economic Opportunity
Health Information Technology Commission
Certificate of Need
Child Lead Exposure Elimination Commission
Coronavirus Task Force on Racial Disparities
Domestic Violence
Human Trafficking Health Advisory Board
Institutional Review Board
Michigan Commission on Services to the Aging
Nursing Home Workforce Stabilization Council
Guy Thompson Parent Advisory Council (GTPAC)
Prescription Drug Task Force
Protect Michigan Commission
Michigan Suicide Prevention Commission
BackBridge Card Participation
BackChild & Adult Provider Payments
BackChild Care Fund
Contact Information
BackChild Welfare
Child Welfare Law Manual
Child Welfare Training
MCI Delegation of Authority
Every Student Succeeds Act
MiTEAM Practice Model
Strengthening Our Focus on Children & Families
BackContractor and Subrecipient Resources
BackCommunity & Faith-Based Programs
Community and Faith Based Initiative
Workshops
BackFarmworker Outreach Services Division
Photo Galleries
About the Farmworker Outreach Services Division
Farmworker Committees and Councils
Farmworker Resources
Farmworker Research in Michigan
COVID
División de Servicios de Extensión para Campesinos
MI Interagency Migrant Services Committee
Migrant Resource Councils
BackForms & Applications
BackMIBridges Partners
Features
Frequently Asked Questions
Become a Partner
Community Partner Training
Tools and Resources
BackState Health Improvement Plan
BackState Health Assessment
Michigan 2019 SHA-Call to Action
BackHealth Care Providers
Certificate of Need
Highly Integrated Dual Eligible Special Needs Plan
MI Coordinated Health
Nurse Loan Repayment Program
Civil Monetary Penalty (CMP) Grant Program
Community Mental Health Services
Departmental Forms
Health Professional Shortage Area
High Utilizers
HIPAA
Institutional Review Board
International Medical Graduate Programs
Community Transition Services
Lab Services
Nurse Aide Training and Testing Reimbursement Forms and Instructions
MI Health Link
Office of Nursing Programs
State Loan Repayment Program
Behavioral Health Loan Repayment Program (BHLRP)
Substance Abuse Providers
Michigan Opioid Treatment Access Loan Repayment Program
BackLicensing
Body Art Licensing
Child Welfare Licensing
BackMichigan Statewide Systems of Care
Trauma
Stroke STEMI
Committees
Contacts
BackState Innovation Model
BackMiSACWIS Training
BackPathways to Potential
About
Strategic Priorities
Success Coaches in Action
Why Pathways to Potential
Get Involved
Resources
Success Stories
Contact Us
BackProtect MiFamily -Title IV-E Waiver
Assessment Tools
Contractors
Evaluation
Expected Outcomes
Forms - Publications
Protect MI Family Contacts
Understanding Michigan's Waiver project
BackWeatherization Assistance Program
CSPM Manual
Prospective Clients
Prospective Contractors
Prospective Workforce
State Map of Weatherization Operators
Students in Energy Efficiency-Related Field
Resources for Weatherization Operators
Photo Gallery
BackInside MDHHS
Newsroom
How To Find Resources
SEC 238
Tribal Government Services and Policy
Contact MDHHS
County Offices
Executive Staff Bios
Careers With MDHHS
Budget & Finance
Complaints
Office of Inspector General
Community & Volunteer Opportunities
Reports & Statistics - Health Services
Hotlines
Legal
MDHHS Audit
Policy and Planning
Reports & Statistics - Human Services
BackNewsroom
BackHow To Find Resources
BackSEC 238
BackTribal Government Services and Policy
BackContact MDHHS
BackCounty Offices
East Michigan
Genesee County
Local Health Department Maps
Macomb County
Northern Mid-Michigan
Oakland County
U.P. and Northern Michigan
Urban Counties
Wayne County
West Michigan
BackExecutive Staff Bios
BackCareers With MDHHS
New Hire Forms
BackBudget & Finance
Boilerplate Reports
BackComplaints
BackOffice of Inspector General
About Us
Sample of Investigations
Michigan Law
Annual Reports
Media Resources
Contact Us
BackCommunity & Volunteer Opportunities
BackReports & Statistics - Health Services
Birth Defects
Cancer Statistics
Communicable Diseases
Community Health Information
Chronic Disease and Health Indicators
Health Disparities
Health Care Statistics
Maternal & Child Health Data
Michigan Public Health Data Dashboards
Public Health Advisory Council
Services Statistics for Flint
Population Trends
Other Chronic Disease & Injury Control Data
Other Published Reports and Surveys
Vital Statistics
BackHotlines
BackLegal
Child Welfare Reform
Equal Opportunity
Federal Child Welfare Reviews
Freedom of Information Act
Laws
Legal Base
Medical Records Access Act
Nondiscrimination Statement (No discriminacion)
Subpoena
Records from Closed State Facilities
Flint Water Settlement
Hepatitis C Lawsuit
Severance Pay FY21
Severance Pay FY22
Severance Pay FY23
Text Campaign Terms & Conditions
BackMDHHS Audit
Audit Reporting
Long Term Care Audit
BackPolicy and Planning
2022-2024 Social Determinants of Health Strategy
EMS
Future of Behavioral Health
Workforce/Access & Grants Management Section
BackReports & Statistics - Human Services
Annual Report
Capital Improvement Plan
Green Book Filtered Search Page
Reports, Evaluations & Studies
State Plans & Federal Regulations
Trend Report
View Result for:
/mdhhs/adult-child-serv/childrenfamilies/hereditary/newborn-screening----list-of-disorders
Back to Department of Health & Human Services
Search is currently unavailable. Please try again later.
Popular on michigan.gov
Agriculture and Rural Development
Civil Rights
Environment
Health and Human Services
Natural Resources
Secretary of State
How Do I...
Register to Vote
Renew My License Plate
View assistance programs
Unsupported Browser Detected
The web Browser you are currently using is unsupported, and some features of this site may not work as intended. Please update to a modern browser such as Chrome, Firefox or Edge to experience all features Michigan.gov has to offer.
Supported Browsers
Google Chrome
Safari
Microsoft Edge
Firefox
Newborn Screening - List of Disorders
Newborn Screening - List of Disorders
Adult & Children's Services
Children & Families
Hereditary Disorders
Newborn Screening
All infants born in Michigan are screened for50+ disorders and hearing.Shown below are the disorders currently included on the screening panel. (Fact sheets are available for some disorders.)
Argininemia (ARG)
Argininosuccinic acidemia (ASA)
Citrullinemia Type I (CIT-I)
Citrullinemia Type II (CIT-II)
Homocystinuria (HCY)
Hypermethioninemia (MET)
Maple syrup urine disease (MSUD)
Phenylketonuria (PKU)
Benign hyperphenylalaninemia defect (H-PHE)
Biopterin cofactor biosynthesis defect (BIOPT-BS)
11.Biopterin cofactor regeneration defect (BIOPT-REG)
Tyrosinemia (TYR-I)
13.Tyrosinemia Type II(TYR-II)
Tyrosinemia Type III(TYR-III)
Fatty Acid Oxidation Disorders:
Carnitine acylcarnitine translocase deficiency (CACT)
Carnitine palmitoyltransferase I deficiency (CPT-IA)
Carnitine palmitoyltransferase II deficiency (CPT-II)
Carnitine uptake deficiency (CUD)
Dienoyl-CoA reductase deficiency (DERED)
Glutaric acidemia type II (GA-2)
Long-chain L-3-hydroxy acyl-CoA dehydrogenase deficiency (LCHAD)
Medium/short-chain L-3-hydroxy acyl-CoA dehydrogenase deficiency (M/SCHAD)
Medium-chain acyl-CoA dehydrogenase deficiency (MCAD)
Medium-chain ketoacyl-CoA thiolase deficiency (MCKAT)
Trifunctional protein deficiency (TFP)
Very long-chain acyl-CoA dehydrogenase deficiency (VLCAD)
Organic Acid Disorders:
2-Methyl-3-hydroxy butyric aciduria (2M3HBA)
28.2-Methylbutyryrl-CoA dehydrogenase deficiency (2MBG)
3-hydroxy 3-methylglutaric aciduria (HMG)
3-Methylcrotonyl-CoA carboxylase deficiency (3-MCC)
3-Methylglutaconic aciduria (3MGA)
Beta-ketothiolase deficiency (BKT)
Glutaric acidemia type I (GA1)
Isovaleric acidemia (IVA)
Malonic acidemia (MAL)
Methylmalonic acidemia cobalamin disorders (Cbl A,B)
Methylmalonic aciduria with homocystinuria (Cbl C,D)
Methylmalonic acidemia methylmalonyl-CoA mutase (MUT)
Multiple carboxylase deficiency (MCD)
Propionic acidemia (PROP)
Hemoglobinopathies:
Interpretation of Newborn Hemoglobin Screening results
S/Beta thalassemia
S/C disease
Sickle cell anemia
Variant hemoglobinopathies
Hemoglobin H disease
Endocrine Disorders:
Congenital adrenal hyperplasia (CAH)
Congenital hypothyroidism (CH)
Lysosomal Storage Disorders:
Glycogen Storage Disease Type II (Pompe)
Mucopolysaccharidosis Type I (MPS I)
Other Disorders:
Biotinidase deficiency (BIOT)
Galactosemia (GALT)
Cystic Fibrosis (CF)
Severe combined immunodeficiency (SCID)
T-cell related lymphocyte deficiencies
55.X-linked Adrenoleukodystrophy (X-ALD)
Spinal muscular atrophy (SMA)
Guanidinoacetate methyltransferase (GAMT) deficiency
Early Hearing Detection and Intervention (EHDI) Program
Critical Congenital Heart Disease (CCHD)
Disorders labeled with one asterisk were added to the NBS panel in 2005 using the tandem mass spectrometry (MS/MS) platform.
Over 1800 mutations have been identified in the gene associated with Cystic Fibrosis, the Cystic Fibrosis Transmembrane Regulator (CFTR) gene. Most newborn screening programs only screen for the most common disease-causing CFTR gene mutations.
Provided is a list of CFTR mutations currently included in Michigan's Cystic Fibrosis Newborn Screening Panel
NBS Main Page
Follow us
Newborn Screening - List of Disorders
Copyright State of Michigan
Sign up for email updates
Submit
MDHHS Is Hiring
Apply Here
MDHHS FOIA
Submit a FOIA Request
Get personalized voter information on early voting and other topics.
Michigan.gov/Vote
Copyright 2025 State of MichiganMi.gov HomePoliciesAccessibilityDisability ResourcesStatewide FOIA DirectoryDepartments |
190755 | https://www.youtube.com/watch?v=21PvoMvwqGY | Unit's Digit of Large Numbers (Concept and Practice Questions for ALL EXAMS)
IPM Leap
1490 subscribers
34 likes
Description
1567 views
Posted: 21 Sep 2021
Finding the Unit's place Digit of a given Exponential is a popular question type in all Competitive Exams. This video discusses the tips and tricks for solving all questions related to Unit's Digit.
The following practice questions have been discussed to build the concept fundamentals and give students ample practice to ace the topic:
Question. Find the remainder when 47^3 is divided by 10?
Question. Find the unit’s digit of:
71"×" 72"×" 73"×" 74"×" 75"×" 76"×" 77"×" 78"×" 79
Question. If you add up the first 55 whole numbers, what will be the unit’s digit of the Sum?
Question. Find the unit’s digit of: 21^201-35^305 + 46^406
Question. Find the unit’s digit of the following Exponents:
Unit’s Digit of 74^1536=
Unit’s Digit of 139^237= ?
Unit’s Digit of 52^43= ?
Unit’s Digit of 192^284= ?
Unit’s Digit of 2022^2021= ?
Unit’s Digit of 3^100= ?
Unit’s Digit of 263^262= ?
Unit’s Digit of 3^75-3^57= ?
Unit’s Digit of 37^40= ?
Unit’s Digit of 387^387= ?
Unit’s Digit of 57^75+37^73= ?
Unit’s Digit of 18^18= ?
Unit’s Digit of 488^273= ?
Unit’s Digit of 8^(8^8 )
Question. If X = 527^725+984^367 - 283^823
Find the unit’s digit of X
(1) 7 (2) 3 (3) 4 (4) 6
Question. Find the unit’s digit of 34^10!× 33^9!× 32^8!
(1) 8 (2) 4 (3) 2 (4) 6
Question. Find the last non-zero digit of 70^2021
(1) 7 (2) 3 (3) 9 (4) 1
Question. Find the digit in the thousand’s position of:
5^3× 2^2021
(1) 8 (2) 4 (3) 2 (4) 6
Question. Find the unit’s digit of: 〖"(" 123^123)〗^41
(1) 7 (2) 3 (3) 9 (4) 1
Question. Find the unit’s digit of:〖 28〗^(29^30 )
(1) 8 (2) 4 (3) 2 (4) 6
5 comments
Transcript:
hi friends today's video is about finding the units digit of large expressions like the ones you can see on your screen this is a very very popular topic in all competitive exams and this single video will help you ace this mass concept and score full mass in this topic so this is the complete list of all the questions we are going to discuss in this video and they might look scary to you at first but believe me by the end of this video you will be able to solve all of them with confidence just keep a paper and pen ready to take down the notes and let's begin so let us suppose we have to find the unit's digit of 53 into 72 we already know that in our decimal system of numbers the last digit of a number is called the units digit the second last is called the tens digit the third blast is called the hundreds digit and so on when we multiply 53 and 72 and this is how we do it pay attention we first multiply 2 the units digit of 72 with the units digit of 53 and jot down the product of these two digits which is 6 and later we multiply it with the rest of the number once we are done with 2 the units digit we move on to the next line and put a cross here before multiplying 53 with 7 which is the tens digit the point is when you add these two lines to find out the product at the units place we will simply copy this number 6 which we got by multiplying the last digits of these two numbers so the idea is very simple when we talk about the units digit of a product of numbers only their units digits matter and we can safely ignore the rest of the digits in those numbers same holds true for addition or subtraction if we have to find the units digit of 153 plus 2376 minus 207 while solving this sum we will align the units digit of all the three numbers like this and therefore the units digit of our answer will depend only and only on the units digit of these three numbers now the maximum application of this concept of units digit is when we have numbers with powers so let's say we have to calculate 83 square we know it is 83 into 83 when we calculated 53 into 72 we learned that the units digit of a product depends only on the units digits of the numbers involved that's why we can safely ignore the tens digit 8 and simply calculate 3 square 3 into 3 is 9. therefore the units digit of 83 squared will be the same as that of 3 squared which is 9. in the next example we have to find the remainder when 47 cube is divided by 10. you might be wondering what's a remainders question doing in the concept of unit digit but hang on you will soon understand so someone who is doing this question for the first time will likely calculate 47 cube 47 into 47 into 47 47 square is 2 2 0 9 that into 47 is 1 0 3 8 2 3. this when divided by 10 the remainder will be nothing else but the units digit did you realize this question is just a different way of asking the units digit of 47 cube because any number when divided by 10 leaves the units digit as remainder we know the better way of doing this was just multiplying the units digit of 47 with itself three times seven into seven into seven is seven sevens are 49 and nine sevens are 63. the units digit has to be three and you will save a lot of time by doing this question like this time management is a very very important aspect in all entrance exams so the questions we have done so far teach us a very important tip and that is to find the units digit of an expression when we add subtract multiply only the units digits of the numbers matter let's apply what we have learned so far to the question on your screen find the units digit of 71 into 72 into 73 up to 79 units digit only depends on the units digit of the numbers involved so when we multiply the first three numbers 1 into 2 into 3 gives us 6 as the units digit of these 3 numbers multiplying 6 with the units digit of the fourth number gives us 6 into 4 24 that is 4 as the units digit of first four numbers now 4 multiplied with the units digit of the fifth number which is 5 gives us 20 therefore the units digit of the first five numbers multiplied together is 0. can you guess what will happen next since 0 multiplied with anything else is also 0 0 into 6 into 7 into 8 into 9 is also 0. so our answer for the units digit of this expression is 0. there's a very important take away in this question 5 multiplied with any even digit results in 0. there are many more questions based on this and if you have a slight hit it helps you solve such questions easily with confidence isn't it let's move on to our next question if you add up the first 55 whole numbers what will be the units digit of the sum now whole numbers are all numbers starting 0 which you can count on your fingertips natural numbers start from 1. sum of first 55 whole numbers will be equal to the sum of first 54 natural numbers now obviously you won't add these numbers one by one we will use the formula we know that the sum of first and natural numbers is n into n plus 1 by 2. there is a very interesting story behind this formula which you can learn through our video on the same topic so sum of first 54 natural numbers is 54 into 54 plus 1 by 2. this simplifies to 27 into 55 the units digit will be same as that of 7 into 5 7 5s are 35 so our answer for this question is 5 now with this basic background let's start the most important and tricky part of this chapter finding the units digit of a number raised to a power say x raised power y when we have numbers like these x is called the base and y is called the power index or exponent we know that the number in the base has to end with one of the digits from 0 to 9 or in other words the units digit will vary from 0 to 9 right these 10 digits from 0 to 9 can further be divided into three categories out of which the simplest category is number ending in 0 or 1 or 5 or 6. why so because you may raise these numbers to an arbitrary power but their units digit will remain the same for example take ten ten raise per one ends with zero so does ten raise per two digit for ten raise per three and so on the zeros keep on increasing if the number has a units digit of zero no matter what power it is raised to it will always end in zero same is the case for numbers within this digit one one into one is always 1. so if the given number has a units digit of 1 no matter what power it is raised to it will always end in 1. again 5 into 5 is 25 which also ends with 5 that means if the given number has a units digit of 5 it will always end in 5 no matter what power it is raised to and the last digit in this category is 6. 6 into 6 is 36 so a number with a units digit of 6 will always end in 6 regardless of its power or exponent simple so far right let's apply what we have learned to the following question find the units digit of 21 raised bar 201 minus 35 raised power 305 plus 46 raised power 406 so we know that the units digit remains the same in the case of 0 1 5 or 6. our answer will be 1 minus 5 plus 6 which is 2 our answer for this long expression will be 2. category 2 is when units digit of the base is 4 or 9. let's look at powers of 4 to get an idea of how these questions are to be attempted 4 a power 1 is 4 4 s bar 2 16 4 a spa 3 64 4 s per 4 2 56 4 5 1 0 2 4 and so on look at the units digit of these powers to see how they behave 4 6 4 6 4. there is an alternating pattern of 4 and 6. whenever the power of 4 is odd the unit's digit is 4 and whenever it is even the units digit is 6 and indeed if i need to find out the units digit of 4 raised per 6 i just need to multiply the units digit of 4 raised power 5 which is equal to 1 0 2 4 with 4 4 x 4 is 16 so next units digit will be 6 and 6 multiplied by 4 is 24 so the units digit of the number after that which is 4 s plus 7 will again be 4. so we say that the powers of 4 have a cyclicity of 2 4 raised power 1 or 4 is per any odd number ends in 4 whereas 4 s bar 2 or 4 raised bar any even number ends in a 6. for example if we have to find out the unit's digit of 74 is per 1536 how do we go about solving this question first of all just keep the units digit of the base which is 4 and ignore all other ridges before that so essentially we have to find out the units digit of 4 race power per 1536 and since 4 spa any even number ends in 6 our answer is 6. similar pattern is observed in powers of 9 9 raised power 1 is 9 9 raised per 2 81 9 raised by 3 7 29 9 raise power 4 6 5 6 1 and so on nine raised power five will have a unit digit of one into nine that is 9 and 9 raised per 6 will again have a units digit which is same as that of 9 into 9 81 which ends in 1. 919191 is the alternating pattern for the units digit of 9 again two different possibilities for units digit so cyclicity is 2 so just remember that 9 raised power 1 and also any odd power of 9 will have a units digit of 9 whereas 9 raised by 2 or any even power of 9 will have a units digit of 1. let's try out a question what's the unit's digit of 139 raised power of 237 so first of all we ignore all other digits in the base except the units digit that means we have to find out the units digit of 9 raised power of 237 and 9 raised bar of any odd number has a units digit which is same as that of 9 raised power of 1 that is 9. so our answer is 9. and the last category category number three is when the units digit of the base is two three seven or eight let's discuss the base two looking at its powers to raise for one is two two race path two four to race path three eight to raise power four sixteen and two rays per five is thirty two there is again a recurring pattern isn't it unit digits are 2 4 8 6 and again 2. next is 2 raised past 6 units digit will be 2 raised by 2 that is 4. for 2 days past seven units digit will be four into two that is eight for two days for eight units digit will be same as that of eight into two that is sixteen so we keep six since we are interested only in the units digit for 2 raised per 9 units it will be same as that of 6 into 2 that is 12 so we keep 2 so units digit are repeating themselves in a pattern of four the cyclicity is four two four eight six two four eight six and so on so for powers of two just remember that there are four steps in the cycle at the first step the unit's digit will be same as that of 2 raised power 1 which is 2. at the second step the units digit will be same as that of 2 raised bar 2 which is 4. at the third step the units digit will be same as that of to race part three which is eight and at the fourth step the units digit will be same as that of two race power four which is six so pay attention the fourth step of this cycle is when you have to raise power four or two days per eight or two days per 12 or two days past 16 or two days for 20 and so on when powers are multiples of four we are at the fourth step of the cycle when the units digit will be six next important question is how do we find out which step of the cycle a given power lies yeah so let's try out a question to understand this challenge what's the unit's digit of 52 raised per 43 just ignore the five in the base and keep only two find the units digit of 2 raised power 43 which step of the cycle would this be okay so if the power were 40 a multiple of 4 that would be at the fourth step right 43 minus 40 is 3 so the power of 43 means we are at the third step of the cycle units digit will be the same as that of 2 raised power of 3 that is it will be eight our answer is eight so what exactly do we need to do we must extract the maximum multiple of four from the given power which means we must divide the power given in the question by 4 and find out the remainder if the remainder is 0 that means the given power is a multiple of 4 and we are at the fourth step of the cycle else if the remainder is 1 2 or 3 then we are at the same step of the cycle as the remainder we obtained like in this case the power was 43 43 divided by 4 the remainder was 3 so we were at the third step of the cycle next question is to find the unit's digit of 192 raised power of 284 just keep two in the base and ignore the digits before that to raise bar 284 now 284 1 divided by 4 and you can use the divisibility rule of 4 which means just divide the last 2 digits by 4 and c so 84 it turns out is the multiple of four the remainder is zero we are at the fourth step of the cycle so units digit will be same as that of two raised power four sixteen so our answer is simply six and the last practice question for base 2 is to find the units digit of 2022 raised bar of 2021. ignore these digits in base power is 2021. which step of the cycle divide 2021 by 4 21 divided by 4 remainder is 1 so we are at the first step of the cycle and units digit will be the same as that of 2 raised power 1 which is 2 our answer is 2 let's move on to base 3 looking at its powers three raised power one is three three is part two nine three is per three twenty seven three raised per four eighty one three is per five is 243 and so on there is a cyclicity of 4 units digits are repeating in a pattern of 3 9 7 1 and again 3 9 7 1 and this pattern continues for higher powers of 3 as well so for powers of three the four steps in the cycle would be at the first step the units digit will be the same as that of three raised power one which is three at the second step the unit digit will be the same as that of three days per two which is nine at the third step the units digit will be the same as that of three days per three which is seven and at the fourth step the units digit will be the same as that of three days per 4 which is 1 and just remember the 4th step of this cycle is when the power is a multiple of 4 when it is exactly divisible by 4. let's try out some questions now find the units digit of 3 raised power of 100 the power 100 is clearly divisible by 4 so we are at the fourth step of the cycle units digit will be the same as that of 3 raised power 4 which is 81 so our answer is 1. next we have to find the units digit of 263 raised power of 262. in the base just keep the units digit 3 and ignore the digits before that so what's the universe digit of 3 days per 262 divide the power by 4 to see which step of the cycle you are at now 262 when divided by 4 it leaves the remainder 2. so that means we are at the second step of the cycle so units digit will be same as that of 3 raised bar 2. so our answer is 9. next question is a little different find the units digital 3 raised past 75 minus 3 raised per 57 so we just try to find the respective units digit of these two numbers 3s by 75 3 3s per 75 75 when divided by 4 leaves a remainder 3 third step of the cycle so units digit will be same as that of 3 days but 3 which is 27 so we jot down 7 3 is per 57 the power 57 when divided by 4 leaves the remainder 1 first step of the cycle so units digit will be same as that of 3 raised power one we jot down three our answer is simply seven minus three that is four units digit of this expression is four then we have base 7 looking at his powers 7 raised power 1 is 7 7 raised by 2 is 49 severe spot 3 is 343 so embrace per 4 is 2 one for seven raised power five units digit will be one into seven so eunice digits have started repeating in a cyclicity of four they are repeating in a pattern of seven nine one and again 7931 and this pattern will continue for higher powers of seven as well for powers of seven the four steps in the cycle would be at the first step the unit digit will be same as that of 7 raised power 1 which is 7. at the second step the units digit will be same as that of 7 raised bar 2 which is 49 so 9 at the third step the units digit will be same as that of seven raised part three which is the same as that of nine into seven so three and at the fourth step the units digit will be same as that of seven raised power 4 which is the same as that of 3 into 7 so 1 and again we must remember that the 4th step of this cycle is when the power is a multiple of 4 and is exactly divisible by 4. let's try out some questions now find the unit's digit of 37 raised power 40 first ignore 3 in the base next the power 40 is clearly divisible by 4 so we are at the fourth step of the cycle units digit will be the same as that of 7 raised power 4 which is 1 so our answer is 1. next we have to find the units digit of 387 race power of 387 now don't get confused if the base and power are the same a procedure to solve this question doesn't change because of that in the base just keep the unit digit 7 and ignore the digits before that so what's the units digit of 7 race path 387 divide the power by 4 to see which step of the cycle you are at now 387 when divided by 4 leaves the remainder 3 we are at the third step of the cycle so units digit will be same as that of 7 raised part 3 so our answer is 3 in the next question we have to find the units digit of 57 raised to 75 plus 37 raised per 73 again ignore the extra strategies in the basis of these two numbers and just keep their units digit which is 7 for both 7 days per 75 75 when divided by 4 leaves the remainder 3 third step of the cycle so units digit will be same as that of seven days per three which is three seven raised past 73 the power 73 when divided by four leaves the remainder one first step of the cycle so unit digit will be same as that of 7 raised power 1. we jot down 7. our answer is simply 3 plus 7 which is equal to 10 and the last digit of 10 is 0. so i'll answer for the units digit of this expression is also zero finally the base of eight looking at its powers a trace for one is eight eight race per two is sixty four eight race path three is five hundred and twelve forty threes power four units digit will be two into eight which is sixteen so we write 6 for 8 raised power 5 units digit will be 6 into 8 which is 48 so we write 8. are you able to see that the unit digits have started repeating and there is a cyclicity of four the unit's digits of the powers of eight are repeating in a pattern of eight four two six and again eight four two six and this pattern will continue for higher powers of 8 as well for powers of 8 the 4 steps in the cycle would be at the first step the units digit will be same as that of 8 raised per 1 which is 8 obviously at the second step the earnest digit will be same as that of 8 raised part 2 which is 64. so 4. at the third step the units digit will be same as that of a trace part 3 which is the same as that of 4 into 8 the unit's digit is 2 and at the fourth step the units digit will be same as that of 8 raised power 4 which is the same as that of 2 into 8 so it will be six a reminder the fourth step of this cycle is when the power is a multiple of four and is exactly divisible by four let's try out practice questions specific to base 8. find the units digit of 18 raised bar 18. first ignore 1 in the base next the bar 18 when divided by 4 leaves the remainder 2 so we are at the second step of the cycle units digit will be the same as that of 8 raised part 2 which is 4 so our answer is 4. next we have to find the units digit of 488 raised bar of 273 in the base just keep the units digit 8 and ignore the resist before that so what's the units digit of 8 raised bar 273 next divide the power by 4 to see which step of the cycle you are at 273 when divided by 4 leaves the remainder 1 we are at the first step of the cycle so unit's digit will be same as that of 8 raised power 1 so our answer is 8. and finally we have a unique question where we have power of a power find units digit of 8 raised by decrease bar rate wow focus on its power which is 8 race per 8 and divided by 4 since 8 is divisible by 4 8 raised by 8 will also be exactly divisible by 4 isn't it that means we are at the fourth step of the cycle and our units digit will be same as that of 8 raised power 4 which is 6. so our answer for the units digit of this expression is 6. now let's mix up all the three categories and whatever else we have learned so far and let's try out some awesome questions from this topic so the question on your screen is if x is equal to 527 raised power of 725 plus 984 raised power of 367 minus 283 raised power of 823 find the units digit of x a quick look at this this expression tells us that 7 4 and 3 are the units digits of these numbers so we can just ignore the rest of the bridges and reduce the clutter that will free up our mind so the reduced bases are seven four and three seven and three have a cyclicity of four whereas the base four has a cyclicity of two 4 raised power of an odd number gives us 4 as a units digit of the middle term for the terms containing 7 and 3 we will need to know which step of the cycle we are at so we divide the powers by 4. 725 divided by 4 gives us 1 as remainder so the unit's digit of 7 raised by 725 is same as that of 7 raised power of 1 and 823 divided by 4 gives us 3s remainder the units digit of this term will be same as that of 3 raised part 3 so our units digit of the expression will be 7 plus 4 minus 7 the result is 4 next we have to find the units digit of 34 days per of 10 factorial into 33 raised per of nine factorial into 32 race per of eight factorial you may pause the video if you want to do this question at your own pace let's discuss first step is to reduce the clutter and drop the tenth digit which is three for all these numbers now we look at the powers if you want a quick overview of factorials you may watch our lesson on that topic it's a 10 minute lesson which will give you a good enough introduction and practice of factorials for now we just need to know that n factorial is all natural numbers from 1 to n multiplied together clearly all powers that is 10 factorial 9 factorial and 8 factorial are even also and divisible by 4 also because to evaluate these factorials we have to first multiply numbers from 1 to 4 and then the further numbers they're all multiples of 4. so the unit's digit of 4 is part 10 factorial will be the same as that of 4 raised bar of any even number such as 4 raised power of 2 4 is per 2 is 16 so we write 6 as this unit's digit uns digit of 3 raised by 10 factorial will be the same as that of 3 raised power 4 because 10 factorial is a multiple of 4. since 3 raised power 4 is 81 we write 1 as a units digit and 2 raised per 8 factorial units digit will be the same as that of 2 days per 4 today's path 4 being 16 we write 6 again multiplying 6 into 1 into 6 is 36 the units digit of which is 6. so our answer for the unit digit of this expression is also 6. find the last non-zero digit of 70 raised power of 2021 again you might pause the video if you want to try it at your own comfortable pace and if you have tried let's discuss so 70 days per of 2021 it's clear will end in a lots of zeros do you know exactly how many well 78 7 into 10 so 70 days per 2021 can be written as 7 raised by 2021 into 10 raised per of 2021. all zeroes come from 10 race per of 2021. so 2021 zeros and more importantly we have the value of 7 raised per of 2021 written before all these zeros so if they are asking the last non-zero digit of 70 race part 2021 that means they are asking the units digit of 7 raised part of 2021 which we know how to find isn't it so 7 days per 2021 7 has a cyclicity of 4 so which step of the cycle for that divide the power 2021 by 4 we get 1 as remainder the power lies at the first step of the cycle so units digit of 7 raised by 2021 will be the same as that of 7 raised power 1 which is 7. our answer is 7. find the digit in the thousandths position of 5 raised by 3 into 2 raised power of 2021 take your time pause the video and try out the question okay so if you have tried let's discuss so this expression has three fives multiplied with 2021 twos and we know that 5 into 2 is equal to 10. the three fives in this expression will combine with three twos to add triple zero at the end of this number we can see this happening from the following equation 5 raised by 3 into 2 2 raised power of 2021 is equal to pi raised power 3 into 2 raised by 3 into 2 days by 2021 minus 3 which is equal to 2 raised power of 2018 into 1000 clearly the thousands digit of this expression will be the units digit of 2 raised per of 2018 because we'll have 3 digits 0 0 0 written after that what's the units digit of 2 days per 18 the base 2 has a cyclicity of 4 for its powers so divide 2018 by 4 the remainder will be 2. that means second step of the cycle our answer for the units digit will be 2 raised by 2 that is 4. and finally let's try the last two questions on your screen at this stage for these two questions and only for these two last questions we see an overlapping between the concept of uranus digit and the topic remainders i will try to explain these in the best possible way that i can but to have a complete understanding of remainders do watch our next lesson which will be on remainders so find the unit's digit of 123 raised power 123 whole raised power of 41 now we know that a raised power m whole raised power of n is equal to a raised bar n into n so 123 raised power of 123 4 raised power 41 can be written as 123 raised power of 123 into 41. we need the units digit of this number so just keep three and ignore the digits before that the base three has a cyclicity of 4 so we need to know which step of the cycle does our power lie and the power is 123 into 41. let's divide it by 4 and find the remainder the trick for remainder of this is very simple we don't need to multiply 123 and 41. just divide 123 and 41 individually by 4 and multiply the respective remainders to find our net remainder 123 divided by 4 leaves remainder 3 and 41 divided by 4 leaves remainder 1. so our net remainder for 123 into 41 will be 3 into 1 that is 3 that means the power lies at the third step of the cycle and units digit will be the same as that of three rays per three which is 27 and ends in seven our answer is seven next question is find the unit's digit of 28 raised by 29 race bar of 30 ignore the 10th digit in the base which is 2 now the base is 8. again it has a cyclicity of 4. so we need to divide the power by 4 and the power is 29 raised bar of 30. again for finding remainder we don't need to calculate the value of 29 raised power of 30 rather what we need to do is divide 29 by 4. 29 divided by 4 leaves the remainder 1. so remainder will be 1 raised power of 30 which we know is equal to 1 so the remainder is 1. hence the power lies at the first step of the cycle our answer for units digit will be 8 raised bar 1 our answer is 8. that brings us to the end of this video i hope the concept of unis digit is crystal clear to you now and you can attempt all questions from this topic with full confidence do press the like button and share the lesson with your friends as well we'll be glad to take up any queries or doubts that you may have and you can post them in the comments box below happy learning and take care |
190756 | https://www2.atmos.umd.edu/~hmdaley/EMU_OCch23_blanks.pdf | 1 Topic/ Objective: Name: Class: Date: Essential Questions and Themes: Subheadings -> Questions Notes: Prof. Hannah Daley Organic Chemistry II 02/16/2024 Regiochemistry (defining Ortho/Para and Meta directors), Inductive and resonance effects from substitu-ents, Deactivating Vs Activating Groups , and impacts of substituent on the outcome of EAS reactions Chapter 23: Aromatic Substitution 2 Reactions of substituted Benzenes E E Review the general reaction of electrophilic aromatic substitution to get a monosubstituted benzene + + A monosubstituted benzene has three chemically distinct hydrogens that can lead to three different possible products: ___disubstituted benzene. In order to determine the regiochemistry (or where disubstitution will occur), we need to consider the __and __ effects of the substituent on the benzene ring Sub Sub +E Sub Sub + + E Disubstituted benzene regiochemistry: 2 Important Takeaways! The inductive effect is negative (-I) when the substituent is an electron withdrawing group ( halogen or adjacent positive charge). The inductive effect is positive (+I) when the substituent is an electron donating group ( anion or adjacent negative charge). The inductive effect is distance dependent, decreasing rapidly with distance, so it has a stronger influence on the Ortho and Meta positions. Inductive effect on benzene substituent groups What is the inductive effect? the ___across a sigma bond that arises due to a ___ (EN) of surrounding atoms Electronegativity by Pauling Scale Positive inductive effect (+ I) examples: O– has a ______ ____ Anions _ electrons across the sigma bond away from it, inducing a _ charge on the benzene ring and _ the electron density of the system. “Electron __ Group (EDG)” Also, if the substituent bond connecting to the benzene ring is fully or partially _, the substituent will __electrons to the ring. Ex: alkyl(-R), alkoxy(-OR), and amine(-CNHR) groups H C S I Br N Cl O F 2.20 2.58 2.66 2.96 3.04 3.16 3.44 Negative inductive effect (- I) examples: Chlorine is _____ ____ Halogens __ electrons across the sigma bond towards it, inducing a _ charge on the benzene ring and_ the electron density of the system. “Electron __ Group (EWG)” Also, if the substituent bond connecting to the benzene ring is fully or partially _, the substituent will __ from the ring. Ex: nitro(-NO2), carbonyl (-CO), and cyano (-CN) groups 3 Questions: Topic: Resonance stability from the lone pair of electrons on the hydroxyl group make phenol groups ortho and para directing 2-nitrophenol __ _ m-Nitrophenol 4-nitrophenol _ + + Ortho Intermediate: ___ resonance structures Meta Intermediate: ___ resonance structures Para Intermediate: ___ resonance structures HNO3 CH3CO2H, 60C Ortho: Meta:0% Para:50% All non– hydrogen atoms have octets, making these res-onance structures __. Important Take away! If the first substituent is attached by an atom with at least one lone pair of electrons (ex. –OH, -O-, -OR, halogens ) the disubstituted atom/group will have one additional stable resonance structure in the ortho and para position and favor ortho/para disubstituent. This substituent groups have a positive resonance (+ R) effect. + + + + + + + + “Positive __ effect” 4 Questions: Topic: Show how the methyl substituent on benzene is Orth-/Para– directing using resonance structures __ o-Nitrotoluene 3-nitrotoluene _ _ p-Nitrotoluene Important Take away! In addition to considering the number of resonance structures formed in the intermediate step, you need to consider the stability of structure and the substituents inductive effect on stability. EDGs will help stabilize an adjacent positive charge and act as ortho/para directors, because of the methyl groups positive inductive (+ I) effect. Ortho:63% Meta: Para: Ortho Intermediate: __ resonance structures Meta Intermediate: __ resonance structures Especially _structure because CH3 group can act as an electron donating group (EDG) and stabilizes the adjacent _ charge + + Para Intermediate: 3 resonance structures similar to ortho resonance + + + + + + “Positive __ effect” 5 Questions: Topic: Show how the nitro group (-NO2) on benzene is Meta– directing using resonance structures _ _ Important Take away! EWGs will destabilize an adjacent positive charge and act as meta directors due to their negative inductive effect. Nitro groups are strongly electron withdrawing Ortho:7% Meta: Para:2% Meta Intermediate: __ resonance structures NO2 group is an __ and significantly __the adjacent positive charge + + + + + + Para Intermediate: ___ resonance structures similar to ortho resonance ___ effect __ _ Ortho Intermediate: __ resonance structures ___ _ 6 Questions: Topic: Simple guide to determining disubstitution regiochemistry 1. Substituents attached by an atom with at least one ___ are _ directors, because the _ intermediate will have one additional stable resonance structures. 2. Substituents attached by an atom with _ lone pair of electrons are: • _ directors if they are ___. There is not an extra resonance structure, but the alkyl group is _ and stabilizes the adjacent __ • Mets directors if the atom at the point of attachment is electronegative or if it is bonded to highly ____. There is no extra resonance structure AND the electron withdrawing group attached makes the resonance structure with the ______ Key takeaway! The major product of EAS is the one with the most stable (lower in energy) arenium intermediate state. To determine this, one most consider: (1) The substituents ___. (2) Resonance structure _ and __ (3) how _ the reaction is formed. Activating and Deactivating groups determine the rate of EAS reaction Activating groups are substituent groups that ____ towards electrophilic aromatic substitution (EAS). Activating groups make EAS occurs _ relative to unsubstituted benzene. Deactivating groups are substituent groups that ___ towards electrophilic aromatic substitution (EAS), so EAS occurs __relative to unsubstituted benzene. Substituent Resonance (+/-R) effect Inductive (+/- I) effect Activating or Deactivating Ortho/Para or Meta -O- -OH -Cl -NO2 CO-OR (ester) 7 Questions: Topic: Important Take aways! 1) Activating /Deactivating groups determine speed of reaction. Activating = _. Deactivating = ___. 2) Activating Groups are generally ___directors and Deactivating groups are generally ___directors (except for halogens) 3) If resonance and inductive effect are completing: __ tends to have a greater impact. Strongly Activating Groups: -O-, -NH2, -NR2, -OH, -OR Moderately Strong Activating Groups: amide (-NHCOR/-NHCOH) Weakly Activating Group: Alkyl groups (-R) —————— Reference Benzene (-H)———————- Weakly Deactivating: halogens (-Cl, -Br, -I) Moderately Deactivating: Esters(-COOR), carboxylic acid (-COOH) , Ketones (-COR), aldehydes (-CHO) Strongly Deactivating: nitrile (-CN), nitro (-NO2), sulfonic acid (-SO2OH), amine (-NH3 +), trimethyl amine(-N(CH3)3 +) Textbook table references for activating and deactivating group relative reaction rates 8 Questions: Topic: If substituent group is deactivating, disubstitution is slower and more difficult Substituent impacts on Friedel-Crafts reactions If the substituent group is a deactivating group, the second substitution will be __ difficult requiring a __to increase the electrophile. Ex. 1. Friedel-Crafts reactions do not readily take place ______ groups. Why? Deactivation groups slow the reaction and the cation electrophile from Friedel crafts reactions will ___ before the ring attacks it. 2. Friedel-Crafts alkylations are subject to polyalkylation Why? Because alkyl groups __and each alkyl group added will __ the overall reaction rate and make subsequent alkylations _. Solution! First add an acyl group (moderately deactivating) to the ring, then reduce C=O to CH2 with an acid. 9 Questions: Topic: Impact of reaction conditions on substituent effects The substituents effect on regiochemistry (ie. orth-, meta-, and para-) and reaction rate is not always absolute. The reaction conditions impact the regiochemistry There are two ways to slow this reaction down and induce single bromination 1. Add a __ to make the substituent less activating Acetic acid will ___ of the solution, resulting in an increased concentration of H+ substantially _ the concentration of phenoxide anion (a very powerful _group). The only route available is through phenol which is slower and __ after a single bromination 2. Decreasing the _ and adding a nonpolar solvent instead of water A nonpolar solvent, such as CS2, will not aid in the separation of Br2 and will ______available for the reaction, making it difficult to form two bromine additions. Similarly, if the substituent is a ___ and is likely to bond with a H and gain a __ (ex. Amino group) . Increasing the pH can lead to a _, ___ group that avoids Friedel– Crafts reaction. Br2 H2O, 10 Questions: Topic: Consider regiochemistry with multi-substitutions If you have a disubstituted benzene and you want to determine where the next substitution will occur you need to consider 3 things: 1. 2. 3. Ideal scenario! Both groups direct the same way p-Nitrotoluene Directing group disagreement p-Nitrotoluene 2,4-Dinitrotoluene 1. Where each existing group is directing? 2. The Activating/Deactivating strength of each group? 3. Steric hindrance? m-Nitrotoluene 3,4-Dinitrotoluene 2,5-Dinitrotoluene 11 Questions: Topic: High activation groups offer more resonance stability What if directing groups disagree and activating groups have the same ability? p-Ethyltoluene 4-Ethyl-2-Nitrotoluene % 4-Ethyl-3-Nitrotoluene % How does multiple substitutions impact overall rate of EAS? Substituent effects on overall rate are ___ meaning, each activating group __reaction rate and each ____ group decreases the reaction rate by how __activating or deactivating the group is. ___ have one destabilizing and one stabilizing resonance intermediate structure. Its tempting to think these will __ __, however the contribution to the resonance hybrid increases with ____. Thus, ______ is favored 12 Questions: Topic: A B C D Now you try: rank these groups from fastest to slowest reaction Summary: 24-72 hours after taking these notes, reread them and summarize what you learned and what you are still confused on. (~3-5 sentences) Link to blank Notes: ~hmdaley/EMU_OCch23_blanks.pdf Link to completed Notes: ~hmdaley/EMU_OCch23_blanks.pdf |
190757 | https://brainly.com/question/38777091 | [FREE] Which expression is equivalent to m - 4m - 4m ? - brainly.com
Search
Learning Mode
Cancel
Log in / Join for free
Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions
Log in
Join for free
Tutoring Session
+74,6k
Smart guidance, rooted in what you’re studying
Get Guidance
Test Prep
+43,1k
Ace exams faster, with practice that adapts to you
Practice
Worksheets
+5,6k
Guided help for every grade, topic or textbook
Complete
See more
/
Mathematics
Expert-Verified
Expert-Verified
Which expression is equivalent to m−4 m−4 m?
1
See answer Explain with Learning Companion
NEW
Asked by rkbsv8v8g4 • 09/29/2023
0:03
/
0:15
Read More
Community
by Students
Brainly
by Experts
ChatGPT
by OpenAI
Gemini
Google AI
Community Answer
This answer helped 14684050 people
14M
5.0
0
Upload your school material for a more relevant answer
The expression 'm - 4m - 4m' simplifies to '-7m' in algebra. This is done by combining like terms, which in this case are the terms with 'm', to get the final simplified expression.
Explanation
The expression you've mentioned in your question is m - 4m - 4m.
This is a basic algebraic expression. First, let's understand what this expression represents:
Here, we're subtracting 4m from m two times in succession. What matters here is the coefficients of 'm', a variable.
When we simplify this, we'll treat the 'm' as a common part of each term.
If we sum these together, the equation becomes:
m - 4m - 4m = 1m - 4m - 4m = -7m.
So, the equivalent expression for m - 4m - 4m in simple form is -7m.
Learn more about Algebraic Expression here:
brainly.com/question/34192827
SPJ1
Answered by AntonHugh •56.5K answers•14.7M people helped
Thanks 0
5.0
(1 vote)
Expert-Verified⬈(opens in a new tab)
This answer helped 14684050 people
14M
5.0
0
Upload your school material for a more relevant answer
The expression m−4 m−4 m simplifies to −7 m by combining like terms. We combine the coefficients of m to arrive at this result. This process involves adding the numbers in front of m to simplify the expression accurately.
Explanation
To simplify the expression m−4 m−4 m, we will follow the steps to combine like terms.
Start with the original expression: m−4 m−4 m.
Notice that we can combine the terms that have the variable m.
First, rewrite the expression to show the coefficients:
1 m−4 m−4 m
4. Add the coefficients together:
1−4−4=−7
5. Substitute this back to get the final simplified expression:
−7 m
Thus, the expression m−4 m−4 m simplifies to −7 m.
Examples & Evidence
An example of combining like terms can be seen in the expression 3 x+4 x−2 x, which simplifies to 5 x (by adding 3+4−2 to get 5). Another example is 2 y−3 y+y, which simplifies to 0 y or simply 0 because the coefficients add up to zero. These steps show how combining coefficients works, similar to what we did with m−4 m−4 m.
Combining like terms is a fundamental concept in algebra. When you have numbers in front of the same variable, they can be treated as coefficients and added or subtracted just like regular numbers. This method ensures the accuracy of results and is commonly used in algebraic expressions.
Thanks 0
5.0
(1 vote)
Advertisement
rkbsv8v8g4 has a question! Can you help?
Add your answer See Expert-Verified Answer
### Free Mathematics solutions and answers
Community Answer 4 Which expressions are equivalent to 4m - 2 + (-8m)
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish?
Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit?
Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25
Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points.
New questions in Mathematics
A business based in South Africa pays a consultant in the U.S. 1,420,000 rand per year. Assume that the conversion is 1 rand to 0.074 dollars. What is the consultant's salary in dollars per month? First, fill in the two blanks on the left side of the equation using two of the provided ratios. Then, write your answer rounded to the nearest hundredth on the right side of the equation. Provided Ratios: 0.074 dollars 1 rand1 rand 0.074 dollars1 year 12 months12 months 1 year Equation to complete: 1 year 1,420,000 rand×[blank]×[blank]=[answer in dollars per month]
The total time in minutes that the operators at a call center are on the phone with customers t days after the call center goes live is modeled by f(t) = 6t^3 + 16t + 15. The total number of customers calling t days after the call center goes live is modeled by c(t) = 7t^2 + 7t + 10. The average time spent on the phone per customer is g(t) = f(t)/c(t). How fast is the average time spent on the phone per customer changing on the tenth day? (Round to two decimal places.)
Solve. −3 w(w−5)(w+2)=0
Which of the following method or rule can be used to multiply (1+2 i)(1+5 i)? A. Square of a binomial formula B. FOIL method C. Product of a sum and difference formula D. Product rule
A population model is given by P(t)=t 2+2 t+13 l n(t 2+6),t≥0. What is the rate of change of the population at t=1? (Round your answer to 2 decimal places.)
Previous questionNext question
Learn
Practice
Test
Open in Learning Companion
Company
Copyright Policy
Privacy Policy
Cookie Preferences
Insights: The Brainly Blog
Advertise with us
Careers
Homework Questions & Answers
Help
Terms of Use
Help Center
Safety Center
Responsible Disclosure Agreement
Connect with us
(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)
Brainly.com
Dismiss
Materials from your teacher, like lecture notes or study guides,
help Brainly adjust this answer to fit your needs.
Dismiss |
190758 | https://zhuanlan.zhihu.com/p/656639527 | 高一数学必修 一元二次函数、方程和不等式 - 知乎
首页
知乎直答 焕新
知乎知学堂
等你来答
切换模式
登录/注册
高一数学必修 一元二次函数、方程和不等式
首发于高中数学
切换模式
登录/注册
高一数学必修 一元二次函数、方程和不等式
hybase
音视频 flvAnalyser | hysAnalyser
收录于 · 高中数学
14 人赞同了该文章
目录
学习目标
了解等式的性质，理解不等式的概念，掌握不等式基本性质，能运用这些性质解决有关问题
了解基本不等式的证明过程
熟练掌握基本不等式 a+b 2≥a b(a,b>0)\frac{a+b}{2} \geq \sqrt{ab} (a,b>0) 及变形的应用,能解决简单最大值、最小值问题
会结合一元二次函数的图像，判断一元二次方程实根的存在性及实根个数，了解函数的零点与方程根的关系
了解一元二次不等式的意义，掌握一元二次不等式和相应函数，方程的联系
能抽象出一元二次不等式和对应函数，并通过构建一元二次函数模型解决实际问题
知识结构
图1 知识结构说明
一、等式与不等式
基本知识
两个实数a，b，其大小关系有三种可能，(可转化为比较它们的差与0的大小)，即有
a>b⇔a−b>0 a>b \Leftrightarrow a-b>0
a=b⇔a−b=0 a=b \Leftrightarrow a-b=0
a<b \Leftrightarrow a-b<0
重要不等式
\forall 实数a, b\in R，有 a^{2} + b^{2} \geq 2ab ，其中当且仅当 a=b 时，等号成立。
等式的基本性质
（1）如果 a＝b，那么 b＝a.
（2）如果 a＝b，b＝c，那么 a＝c.
（3）如果 a＝b，那么 a±c＝b±c.
（4）如果 a＝b，那么 ac＝bc.
（5）如果 a＝b，c≠0，那么 \frac{a}{c} = \frac{b}{c}
不等式性质
性质 别名 性质内容 注意
1 对称性 a>b⇔bb，b>c⇒a>c 不可逆
3 可加性 a>b⇔a＋c>b＋c 可逆
4 可乘性 a>b，c>0⇒ac>bc
a>b，c<0⇒acb，c>d⇒a＋c>b＋d 同向
6 同向同正可乘性 a>b>0，c>d>0⇒ac>bd 同向
7 乘方法则 a>b>0⇒a^n>b^n(n∈N，n≥2)同正
8 开方法则 a>b>0⇒a^n>b^n(n∈R，0<n<1)开方和乘方雷同
不等式性质的应用
注意事项：（解题时必须利用不等式性质，步步有据）
1）能运用作差法，用不等式解决解决现实生活中的问题（包括证明不等式的成立）：解题思路是将解决的问题转化成不等关系，利用作差法比较大小，进而解决实际问题．
作差法步骤为
2）能运行不等式性质，求不等式的解集（未知数的取值范围）
转为同向相加；注意不等式性质的传递性；
注意事项，需要特别小心谨慎，同向不等式相加不是等价变换，防止多次同向不等式相加造成范围被扩大（样例很多）。
举例说明
图2 多次同向相加造成范围的变化
3）能正确运行不等式性质判断命题的真假；（需特别留意带负数的乘法和除法，求倒数等）
二、基本不等式
1) \sqrt{ab} \leq \frac{a+b}{2}，(a>0,b>0)
2) a^{2}+b^{2} \geq 2ab(a,b\in R) （重要不等式）
3) \frac{b}{a} + \frac{a}{b} \geq 2 (a，b同号)
以上不等式等号成立条件均为 a=b .
算术平均数和几何平均数
如果 a>0，b>0，则 \sqrt{ab} \leq \frac{a+b}{2}，当 a=b 时，等号成立．
其中 \frac{a+b}{2} 叫做正数 a，b 的算术平均数， \sqrt{ab} 叫做正数 a，b 的几何平均数．
基本不等式的变形
1） ab\leq (\frac{a+b}{2})^{2}，a，b∈R，当且仅当 a=b 时，等号成立．
2）a+b\geq 2\sqrt{ab} ，a，b 都是正数，当且仅当 a=b 时，等号成立．
3） \frac{a^{2} + b^{2}}{2} \geq \left( \frac{a+b}{2} \right)^{2} (a,b\in R)
基本不等式的应用
1、用基本不等式求最值
用基本不等式 \frac{a+b}{2} \geq \sqrt{ab} 求最值应注意三点：
(1) a，b必须是正数．
(2) ①如果 ab 等于定值 P，那么当a＝b时，和 a＋b 有最小值 2\sqrt{P} ；（积定值，和最小）
②如果 a＋b等于定值 S，那么当a＝b时，积 ab 有最大值 \frac{S^{2}}{4} .（和定值，积最大）
(3)其中“=”是否成立，是最值的重要条件，因此求解时务必要判断等号成立条件是否满足．
补充：
函数法求最值：若利用基本不等式时等号取不到，则无法利用基本不等式求最值，则可将要求的式子看成一个函数，利用函数的单调性求最值.
2、利用基本不等式解决实际问题的步骤
解实际问题时，首先审清题意，然后将实际问题转化为数学问题，再利用数学知识(函数及不等式性质等)解决问题.用基本不等式解决此类问题时，应按如下步骤进行：
先理解题意，设变量，设变量时一般把要求最大值或最小值的变量定为函数.
建立相应的函数关系式，把实际问题抽象为函数的最大值或最小值问题.
在定义域内，求出函数的最大值或最小值.
正确写出答案.
3、利用基本不等式证明不等式
1)要注意基本不等式的要求 a>0, b>0
三、一元二次函数与一元二次方程的对应关系
一元二次函数： y = ax^{2} + bx+c (其中a>0)
一元二次方程：ax^{2} + bx+c=0 (其中a>0)
函数和方程对应关系
判别式\Delta = b^{2} -4ac，根据判别式的值有三种对应关系：
1） \Delta > 0
二次函数在直角坐标系的图象如下所示，
二次函数曲线与x轴有两个交点x1,x2
对应方程有两个不相等实数根 x1、x2，其中x1<x2
x1=\frac{-b-\sqrt{b^{2}-4ac}}{2a}
x2=\frac{-b+\sqrt{b^{2}-4ac}}{2a}
2） \Delta = 0
二次函数在直角坐标系的图象如下所示
二次函数曲线和x轴有一个交点
对应方程有两个相等实数根 x1=x2=-\frac{b}{2a} ；
3） \Delta <0
二次函数在直角坐标系的图象如下所示
二次函数曲线和x轴没有交点
对应方程，则没有实数根
一元二次函数的零点
一般对于二次函数： y = ax^{2} + bx+c (其中a\ne 0)
我们把使得 ax^{2} + bx+c=0 的实数叫做二次函数 y = ax^{2} + bx+c (其中a\ne 0) 的零点
二次函数的零点不是指点，而是二次函数曲线与x轴较大的横坐标
四、一元二次不等式
不等式的解和函数，方程对应关系
一元二次函数： y = ax^{2} + bx+c (其中a>0)
一元二次方程：ax^{2} + bx+c=0 (其中a>0)
判别式 \Delta = b^{2} -4ac，根据判别式的值，求解不等式的结果：
1） \Delta > 0
ax^{2} + bx+c>0 (其中a>0) 解集: x < x1 或 x > x2 （在x轴上方曲线区域）
ax^{2} + bx+c<0 (其中a>0) 解集：x1<x<x2 (在x轴下方曲线区域)
2） \Delta = 0
ax^{2} + bx+c>0 (其中a>0) 解集: x\ne -\frac{b}{2a} 所有实数
ax^{2} + bx+c<0 (其中a>0) 解集： \phi 无解 (曲线全部在x轴上或上方)
3） \Delta < 0
ax^{2} + bx+c>0 (其中a>0) 解集: 全体实数R
ax^{2} + bx+c<0 (其中a>0) 解集：\phi 无解
函数，方程，不等式解的对应关系
如图所示
不等式，二次方程、二次函数的对应关系图
一元二次不等式的概念
只含有一个未知数，并且未知数的最高次数是2的不等式，叫做一元二次不等式
常见形式，其中a,b,c均为常数，且 a\ne0
ax^{2} + bx+c>0 (其中a\ne 0)
ax^{2} + bx+c<0 (其中a\ne 0)
ax^{2} + bx+c\geq0 (其中a\ne0)
ax^{2} + bx+c\leq0 (其中a\ne0)
一元二次不等式的应用
1）分式不等式的解法
分式不等式转换为一元二次不等式
2）一元二次不等式恒成立问题
1．转化为一元二次不等式解集为R的情况，即
ax^{2} + bx+c>0 (a\ne0) 恒成立 ⇔ \left{ a>0 并且 \Delta <0 \right}
ax^{2} + bx+c<0 (a\ne0) 恒成立 ⇔ \left{ a<0 并且 \Delta <0 \right}
3）利用不等式解决实际问题的一般步骤
选取合适的字母表示题目中的未知数．
由题目中给出的不等关系，列出关于未知数的不等式(组)．
求解所列出的不等式(组)．
结合题目的实际意义确定答案．
求解形如ax^{2} + bx+c>0 或 ax^{2} + bx+c<0 (a\ne0) 的一元二次不等式解集的一般步骤
如果 a<0 ，通过对不等式变形，使二次项系数大于零；
计算对应方程的判别式，求出相应的一元二次方程的根，或根据判别式说明方程没有实根；
画出对应二次函数 y=ax^{2} + bx+c 的图像
由图像得出不等式的解集．
复习和巩固
选择题
1．已知实数 a、b、c 满足 c<b<a，且 ac<0，那么下列不等式一定成立的是（ ）
A． ac(a-c)>0 B． c(b-a)<0
C． cb^{2} < ab^{2} D． ab>ac
2．若 b<a<0 ，给出下列不等式：
① \frac{1}{a+b} < \frac{1}{ab} ；② \left| a \right| + b>0 ；③ a - \frac{1}{a} > b- \frac{1}{b} ；④ ln a^{2} > ln b^{2}
其中正确的不等式是（ ）
A．①④ B．②③
C．①③ D．②④
3．下列命题中，正确的是（ ）
A．若 \frac{a}{c^{2}} < \frac{b}{c^{2}}，则 a<b
B．若 ac>bc ，则 a<b
C．若 a>b,c>d ，则 a-c>b-d
D．若 a>b,c>d ，则 ac>bd
4．已知 x>0，y>0 ，且 2x+8y-xy=0 ，则 x+y 的最小值是（ ）
A．10 B．15 C．18 D．23
解答题
已知 α，β 满足 －1≤α＋β≤1，1≤α＋2β≤3 ，求 α＋3β 的取值范围.
6．设 f(x)＝(4a－3)x＋b－2a，x∈[0，1] ，若 f(0)≤2，f(1)≤2 ，求 a＋b 的取值范围．
7．（1）已知 x>2 ，求 y=x+\frac{1}{x-2} 的最小值．
（2）已知 a,b,c 是不全相等的实数，求证： a^{2}+b^{2}+c^{2} > ab+bc+ac ．
8．设 a>0, b>0 且 a+2b=3 ．
（1）求 ab 的最大值；
（2）求 \frac{2}{a}+\frac{6}{b} 的最小值．
（2023.09.24定稿 hybase@qq.com）
下一章
编辑于 2023-12-02 16:38・广东
一元二次函数
一元二次方程
一元二次不等式
赞同 14添加评论
分享
喜欢收藏申请转载
写下你的评论...
还没有评论，发表第一个评论吧
关于作者
hybase
音视频 flvAnalyser | hysAnalyser
回答 33文章 77关注者 297
关注他发私信
推荐阅读
开眼看世界之教材分析实录——一元二次函数、方程和不等式（一） ============================== Aladd...发表于开眼看世界...【高中数学基础知识】（八）二次函数、一元二次方程和不等式 ============================ 在初中阶段，我们学习过一次函数、一元一次方程和一元一次不等式之间的关系。本节内容将沿用类似的思路探讨二次函数、一元二次方程和一元二次不等式之间的关系。但下一节课才会出现的“函数… 门酱胡安 开眼看世界之教材分析实录——一元二次函数、方程和不等式（二） ============================== Aladd...发表于开眼看世界...实数（等式和不等式） ========== 在高中数学的第二章，A版教材称其为一元二次函数、方程和不等式，而B版教材仅称其为等式和不等式，这不是说A版教材的内容丰富，反而是说B版教材更突出数学的一般性。 不论你学哪一版本，都… 杨树森 发表于做以数学为...
想来知乎工作？请发送邮件到 jobs@zhihu.com
打开知乎App
在「我的页」右上角打开扫一扫
其他扫码方式：微信
下载知乎App
无障碍模式
验证码登录
密码登录
开通机构号
中国 +86
获取短信验证码
获取语音验证码
登录/注册
其他方式登录
未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》
扫码下载知乎 App
关闭二维码
打开知乎App
在「我的页」右上角打开扫一扫
其他扫码方式：微信
下载知乎App
无障碍模式
验证码登录
密码登录
开通机构号
中国 +86
获取短信验证码
获取语音验证码
登录/注册
其他方式登录
未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》
扫码下载知乎 App
关闭二维码 |
190759 | https://en.wikipedia.org/wiki/Law_of_sines | Jump to content
Search
Contents
1 Proof
2 The ambiguous case of triangle solution
3 Examples
3.1 Example 1
3.2 Example 2
4 Relation to the circumcircle
4.1 Proof
4.2 Relationship to the area of the triangle
5 Spherical law of sines
5.1 Vector proof
5.2 Geometric proof
5.3 Other proofs
6 Hyperbolic case
7 The case of surfaces of constant curvature
8 Higher dimensions
9 History
10 See also
11 References
12 External links
Law of sines
Alemannisch
العربية
Azərbaycanca
বাংলা
Беларуская
Български
Bosanski
Català
Чӑвашла
Čeština
Dansk
Deutsch
Eesti
Ελληνικά
Español
Esperanto
Euskara
فارسی
Français
Galego
한국어
Հայերեն
हिन्दी
Hrvatski
Bahasa Indonesia
Íslenska
Italiano
עברית
ქართული
Қазақша
Latviešu
Lietuvių
Magyar
Македонски
Bahasa Melayu
Монгол
Nederlands
日本語
Norsk bokmål
Oʻzbekcha / ўзбекча
ភាសាខ្មែរ
Piemontèis
Polski
Português
Romnă
Русский
Shqip
සිංහල
Simple English
Slovenčina
Slovenščina
کوردی
Српски / srpski
Srpskohrvatski / српскохрватски
Suomi
Svenska
தமிழ்
Татарча / tatarça
ไทย
Türkçe
Українська
اردو
Tiếng Việt
文言
吴语
粵語
中文
Edit links
Article
Talk
Read
Edit
View history
Tools
Actions
Read
Edit
View history
General
What links here
Related changes
Upload file
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Print/export
Download as PDF
Printable version
In other projects
Wikimedia Commons
Wikibooks
Wikidata item
Appearance
From Wikipedia, the free encyclopedia
Property of all triangles on a Euclidean plane
This article is about the law of sines in trigonometry. For the law of sines in physics, see Snell's law.
Law of sines
Figure 1, with circumcircle
Figure 2, without circumcircle
Two triangles labelled with components of the law of sines. The angles α, β and γ are associated with the respective vertices A, B, and C; the respective sides of lengths a, b, and c are opposite these (e.g., side a is opposite vertex A with angle α).
| Trigonometry |
| Outline History Usage Functions (sin, cos, tan, inverse) Generalized trigonometry |
| Reference |
| Identities Exact constants Tables Unit circle |
| Laws and theorems |
| Sines Cosines Tangents Cotangents Pythagorean theorem |
| Calculus |
| Trigonometric substitution Integrals (inverse functions) Derivatives Trigonometric series |
| Mathematicians |
| Hipparchus Ptolemy Brahmagupta al-Hasib al-Battani Regiomontanus Viète de Moivre Euler Fourier |
| v t e |
In trigonometry, the law of sines (sometimes called the sine formula or sine rule) is a mathematical equation relating the lengths of the sides of any triangle to the sines of its angles. According to the law, where a, b, and c are the lengths of the sides of a triangle, and α, β, and γ are the opposite angles (see figure 2), while R is the radius of the triangle's circumcircle. When the last part of the equation is not used, the law is sometimes stated using the reciprocals; The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known—a technique known as triangulation. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the triangle is not uniquely determined by this data (called the ambiguous case) and the technique gives two possible values for the enclosed angle.
The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in scalene triangles, with the other being the law of cosines.
The law of sines can be generalized to higher dimensions on surfaces with constant curvature.
Proof
[edit]
With the side of length a as the base, the triangle's altitude can be computed as b sin γ or as c sin β. Equating these two expressions gives and similar equations arise by choosing the side of length b or the side of length c as the base of the triangle.
The ambiguous case of triangle solution
[edit]
When using the law of sines to find a side of a triangle, an ambiguous case occurs when two separate triangles can be constructed from the data provided (i.e., there are two different possible solutions to the triangle). In the case shown below they are triangles ABC and ABC′.
Given a general triangle, the following conditions would need to be fulfilled for the case to be ambiguous:
The only information known about the triangle is the angle α and the sides a and c.
The angle α is acute (i.e., α < 90°).
The side a is shorter than the side c (i.e., a < c).
The side a is longer than the altitude h from angle β, where h = c sin α (i.e., a > h).
If all the above conditions are true, then each of angles β and β′ produces a valid triangle, meaning that both of the following are true:
From there we can find the corresponding β and b or β′ and b′ if required, where b is the side bounded by vertices A and C and b′ is bounded by A and C′.
Examples
[edit]
The following are examples of how to solve a problem using the law of sines.
Example 1
[edit]
Given: side a = 20, side c = 24, and angle γ = 40°. Angle α is desired.
Using the law of sines, we conclude that
Note that the potential solution α = 147.61° is excluded because that would necessarily give α + β + γ > 180°.
Example 2
[edit]
If the lengths of two sides of the triangle a and b are equal to x, the third side has length c, and the angles opposite the sides of lengths a, b, and c are α, β, and γ respectively then
Relation to the circumcircle
[edit]
In the identity the common value of the three fractions is actually the diameter of the triangle's circumcircle. This result dates back to Ptolemy.
Proof
[edit]
As shown in the figure, let there be a circle with inscribed and another inscribed that passes through the circle's center O. The has a central angle of and thus , by Thales's theorem. Since is a right triangle, where is the radius of the circumscribing circle of the triangle. Angles and lie on the same circle and subtend the same chord c; thus, by the inscribed angle theorem, . Therefore,
Rearranging yields
Repeating the process of creating with other points gives
Relationship to the area of the triangle
[edit]
The area of a triangle is given by , where is the angle enclosed by the sides of lengths a and b. Substituting the sine law into this equation gives
Taking as the circumscribing radius,
It can also be shown that this equality implies where T is the area of the triangle and s is the semiperimeter
The second equality above readily simplifies to Heron's formula for the area.
The sine rule can also be used in deriving the following formula for the triangle's area: denoting the semi-sum of the angles' sines as , we have
where is the radius of the circumcircle: .
Spherical law of sines
[edit]
The spherical law of sines deals with triangles on a sphere, whose sides are arcs of great circles.
Suppose the radius of the sphere is 1. Let a, b, and c be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere, a, b, and c are the angles at the center of the sphere subtended by those arcs, in radians. Let A, B, and C be the angles opposite those respective sides. These are dihedral angles between the planes of the three great circles.
Then the spherical law of sines says:
Vector proof
[edit]
Consider a unit sphere with three unit vectors OA, OB and OC drawn from the origin to the vertices of the triangle. Thus the angles α, β, and γ are the angles a, b, and c, respectively. The arc BC subtends an angle of magnitude a at the centre. Introduce a Cartesian basis with OA along the z-axis and OB in the xz-plane making an angle c with the z-axis. The vector OC projects to ON in the xy-plane and the angle between ON and the x-axis is A. Therefore, the three vectors have components:
The scalar triple product, OA ⋅ (OB × OC) is the volume of the parallelepiped formed by the position vectors of the vertices of the spherical triangle OA, OB and OC. This volume is invariant to the specific coordinate system used to represent OA, OB and OC. The value of the scalar triple product OA ⋅ (OB × OC) is the 3 × 3 determinant with OA, OB and OC as its rows. With the z-axis along OA the square of this determinant is Repeating this calculation with the z-axis along OB gives (sin c sin a sin B)2, while with the z-axis along OC it is (sin a sin b sin C)2. Equating these expressions and dividing throughout by (sin a sin b sin c)2 gives where V is the volume of the parallelepiped formed by the position vector of the vertices of the spherical triangle. Consequently, the result follows.
It is easy to see how for small spherical triangles, when the radius of the sphere is much greater than the sides of the triangle, this formula becomes the planar formula at the limit, since and the same for sin b and sin c.
Geometric proof
[edit]
Consider a unit sphere with:
Construct point and point such that
Construct point such that
It can therefore be seen that and
Notice that is the projection of on plane . Therefore
By basic trigonometry, we have:
But
Combining them we have:
By applying similar reasoning, we obtain the spherical law of sines:
See also: Spherical trigonometry, Spherical law of cosines, and Half-side formula
Other proofs
[edit]
A purely algebraic proof can be constructed from the spherical law of cosines. From the identity and the explicit expression for from the spherical law of cosines Since the right hand side is invariant under a cyclic permutation of the spherical sine rule follows immediately.
The figure used in the Geometric proof above is used by and also provided in Banerjee (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices.
Hyperbolic case
[edit]
In hyperbolic geometry when the curvature is −1, the law of sines becomes
In the special case when B is a right angle, one gets
which is the analog of the formula in Euclidean geometry expressing the sine of an angle as the opposite side divided by the hypotenuse.
See also: Hyperbolic triangle
The case of surfaces of constant curvature
[edit]
Define a generalized sine function, depending also on a real parameter :
The law of sines in constant curvature reads as
By substituting , , and , one obtains respectively , , and , that is, the Euclidean, spherical, and hyperbolic cases of the law of sines described above.
Let indicate the circumference of a circle of radius in a space of constant curvature . Then . Therefore, the law of sines can also be expressed as:
This formulation was discovered by János Bolyai.
Higher dimensions
[edit]
A tetrahedron has four triangular facets. The absolute value of the polar sine (psin) of the normal vectors to the three facets that share a vertex of the tetrahedron, divided by the area of the fourth facet will not depend upon the choice of the vertex:
More generally, for an n-dimensional simplex (i.e., triangle (n = 2), tetrahedron (n = 3), pentatope (n = 4), etc.) in n-dimensional Euclidean space, the absolute value of the polar sine of the normal vectors of the facets that meet at a vertex, divided by the hyperarea of the facet opposite the vertex is independent of the choice of the vertex. Writing V for the hypervolume of the n-dimensional simplex and P for the product of the hyperareas of its (n − 1)-dimensional facets, the common ratio is
Note that when the vectors v1, ..., vn, from a selected vertex to each of the other vertices, are the columns of a matrix V then the columns of the matrix are outward-facing normal vectors of those facets that meet at the selected vertex. This formula also works when the vectors are in a m-dimensional space having m > n. In the m = n case that V is square, the formula simplifies to
History
[edit]
An equivalent of the law of sines, that the sides of a triangle are proportional to the chords of double the opposite angles, was known to the 2nd century Hellenistic astronomer Ptolemy and used occasionally in his Almagest.
Statements related to the law of sines appear in the astronomical and trigonometric work of 7th century Indian mathematician Brahmagupta. In his Brāhmasphuṭasiddhānta, Brahmagupta expresses the circumradius of a triangle as the product of two sides divided by twice the altitude; the law of sines can be derived by alternately expressing the altitude as the sine of one or the other base angle times its opposite side, then equating the two resulting variants. An equation even closer to the modern law of sines appears in Brahmagupta's Khaṇḍakhādyaka, in a method for finding the distance between the Earth and a planet following an epicycle; however, Brahmagupta never treated the law of sines as an independent subject or used it systematically for solving triangles.
The spherical law of sines is sometimes credited to 10th century scholars Abu-Mahmud Khujandi or Abū al-Wafāʾ (it appears in his Almagest), but it is given prominence in Abū Naṣr Manṣūr's Treatise on the Determination of Spherical Arcs, and was credited to Abū Naṣr Manṣūr by his student al-Bīrūnī in his Keys to Astronomy. Ibn Muʿādh al-Jayyānī's 11th-century Book of Unknown Arcs of a Sphere also contains the spherical law of sines.
The 13th-century Persian mathematician Naṣīr al-Dīn al-Ṭūsī stated and proved the planar law of sines:
In any plane triangle, the ratio of the sides is equal to the ratio of the sines of the angles opposite to those sides. That is, in triangle ABC, we have AB : AC = Sin(∠ACB) : Sin(∠ABC)
By employing the law of sines, al-Tusi could solve triangles where either two angles and a side were known or two sides and an angle opposite one of them were given. For triangles with two sides and the included angle, he divided them into right triangles that he could then solve. When three sides were given, he dropped a perpendicular line and then used Proposition II-13 of Euclid's Elements (a geometric version of the law of cosines). Al-Tusi established the important result that if the sum or difference of two arcs is provided along with the ratio of their sines, then the arcs can be calculated.
According to Glen Van Brummelen, "The Law of Sines is really Regiomontanus's foundation for his solutions of right-angled triangles in Book IV, and these solutions are in turn the bases for his solutions of general triangles." Regiomontanus was a 15th-century German mathematician.
See also
[edit]
Gersonides – Medieval Jewish philosopher
Half-side formula – for solving spherical triangles
Law of cosines
Law of tangents
Law of cotangents
Mollweide's formula – for checking solutions of triangles
Solution of triangles
Surveying
References
[edit]
^ a b c "Generalized law of sines". mathworld.
^ Coxeter, H. S. M. and Greitzer, S. L. Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp. 1–3, 1967
^ a b "Law of Sines". www.pballew.net. Archived from the original on December 29, 2002. Retrieved 2018-09-18.
^ Mr. T's Math Videos (2015-06-10), Area of a Triangle and Radius of its Circumscribed Circle, archived from the original on 2021-12-11, retrieved 2018-09-18
^ Mitchell, Douglas W., "A Heron-type area formula in terms of sines," Mathematical Gazette 93, March 2009, 108–109.
^ Banerjee, Sudipto (2004), "Revisiting Spherical Trigonometry with Orthogonal Projectors" (PDF), The College Mathematics Journal, 35 (5), Mathematical Association of America: 375–381, doi:10.1080/07468342.2004.11922099, archived from the original (PDF) on 2004-10-29, retrieved 2024-02-08
^ Katok, Svetlana (1992). Fuchsian groups. Chicago: University of Chicago Press. p. 22. ISBN 0-226-42583-5.
^ Eriksson, Folke (1978). "The law of sines for tetrahedra and n-simplices". Geometriae Dedicata. 7 (1): 71–80. doi:10.1007/bf00181352.
^ Toomer, Gerald J., ed. (1998). Ptolemy's Almagest. Princeton University Press. pp. 7, fn. 10, 462, fn. 96.
^ Winter, Henry James Jacques (1952). Eastern Science. John Murray. p. 46. Colebrooke, Henry Thomas (1817). Algebra, with Arithmetic and Mensuration from the Sanscrit of Brahmegupta and Bhascara. London: John Murray. pp. 299–300.
^ Van Brummelen, Glen (2009). The Mathematics of the Heavens and the Earth. Princeton University Press. pp. 109–111. ISBN 978-0-691-12973-0. Brahmagupta (1934). The Khandakhadyaka: An Astronomical Treatise of Brahmagupta. Translated by Sengupta, Prabodh Chandra. University of Calcutta.
^ Sesiano, Jacques (2000). "Islamic mathematics". In Selin, Helaine; D'Ambrosio, Ubiratan (eds.). Mathematics Across Cultures: The History of Non-western Mathematics. Springer. pp. 137–157. ISBN 1-4020-0260-2. Van Brummelen, Glen (2009). The Mathematics of the Heavens and the Earth. Princeton University Press. pp. 183–185. ISBN 978-0-691-12973-0.
^ O'Connor, John J.; Robertson, Edmund F., "Abu Abd Allah Muhammad ibn Muadh Al-Jayyani", MacTutor History of Mathematics Archive, University of St Andrews
^ "Nasir al-Din al-Tusi - Biography". Maths History. Retrieved 2025-03-10.
^ Katz, Victor J. (2017-03-21). A History of Mathematics: An Introduction. Pearson. p. 315. ISBN 978-0-13-468952-4.
^ Van Brummelen, Glen (2009). The Mathematics of the Heavens and the Earth: The Early History of Trigonometry. Princeton University Press. p. 259. ISBN 978-0-691-12973-0.
External links
[edit]
Wikimedia Commons has media related to Law of sines.
"Sine theorem", Encyclopedia of Mathematics, EMS Press, 2001
The Law of Sines at cut-the-knot
Degree of Curvature
Finding the Sine of 1 Degree
Generalized law of sines to higher dimensions
| v t e Ancient Greek mathematics |
| Mathematicians(timeline) | Anaxagoras Anthemius Apollonius Archimedes Archytas Aristaeus the Elder Aristarchus Autolycus Bion Bryson Callippus Carpus Chrysippus Cleomedes Conon Ctesibius Democritus Dicaearchus Dinostratus Diocles Dionysodorus of Caunus Dionysodorus of Amisene Diophantus Domninus Eratosthenes Euclid Eudemus Eudoxus Eutocius Geminus Heliodorus Heron Hipparchus Hippasus Hippias Hippocrates Hypatia Hypsicles Isidore of Miletus Leon Marinus Menaechmus Menelaus Metrodorus Nicomachus Nicomedes Nicoteles Oenopides Pandrosion Pappus Perseus Philolaus Philon Philonides Porphyry of Tyre Posidonius Proclus Ptolemy Pythagoras Serenus Sosigenes Sporus Thales Theaetetus Theodorus Theodosius Theon of Alexandria Theon of Smyrna Thymaridas Xenocrates Zeno of Elea Zeno of Sidon Zenodorus |
| Treatises | Almagest Arithmetica Conics (Apollonius) Catoptrics Data (Euclid) Elements (Euclid) Little Astronomy Measurement of a Circle On Conoids and Spheroids On the Sizes and Distances (Aristarchus) On Sizes and Distances (Hipparchus) On the Moving Sphere (Autolycus) Optics (Euclid) On Spirals On the Sphere and Cylinder Ostomachion Phaenomena (Euclid) Planisphaerium Spherics (Theodosius) Spherics (Menelaus) The Quadrature of the Parabola The Sand Reckoner |
| Conceptsand definitions | Chord Circles of Apollonius + Apollonian circles + Apollonian gasket + Problem of Apollonius Commensurability Diophantine equation Euclidean geometry Golden ratio Lune of Hippocrates Method of exhaustion Parallel postulate Platonic solid Regular polygon Straightedge and compass construction + Angle trisection + Doubling the cube + Squaring the circle + Quadratrix of Hippias + Neusis construction |
| Results | | | | --- | | In Elements | Angle bisector theorem Exterior angle theorem Euclidean algorithm Euclid's theorem Geometric mean theorem Hinge theorem Inscribed angle theorem Intercept theorem Intersecting chords theorem Intersecting secants theorem Law of cosines Pons asinorum Pythagorean theorem Tangent-secant theorem Thales's theorem Theorem of the gnomon | Apollonius's theorem Aristarchus's inequality Heron's formula Law of sines Menelaus's theorem Pappus's area theorem Problem II.8 of Arithmetica Ptolemy's inequality Ptolemy's table of chords Ptolemy's theorem Spiral of Theodorus |
| Centers/Schools | - Cyrene - Platonic Academy - Pythagoreanism - School of Chios |
| Related | | | | Ancient Greek astronomy Attic numerals Greek numerals | | History of | A History of Greek Mathematics + by Thomas Heath Archimedes Palimpsest algebra + timeline arithmetic + timeline calculus + timeline geometry + timeline logic + timeline mathematics + timeline numbers + prehistoric counting numeral systems + list | | Other cultures | Arabian/Islamic Babylonian Chinese Egyptian Incan Indian Japanese | |
| Ancient Greece portal • Mathematics portal |
Retrieved from "
Categories:
Trigonometry
Angle
Theorems about triangles
Hidden categories:
CS1: unfit URL
Articles with short description
Short description matches Wikidata
Pages using multiple image with auto scaled images
Commons category link from Wikidata
Articles containing proofs
Law of sines
Add topic |
190760 | https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-negative-number-topic/negative-symbol-as-opposite/v/opposite-of-a-number | Use of cookies
Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account.
For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session.
Functional Cookies
These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences.
Targeting Cookies
These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission.
We do not use cookies to serve third party ads on our Service.
Performance Cookies
These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates.
For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users.
We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
Cookie List
Consent Leg.Interest
label
label
label |
190761 | https://www.texastech.edu/offices/treasury/currency-conversion.php | Office of Treasury
Comprehensive Cash and Investment Pool
Debt Information
Staff
Texas Tech University System
System Administration Building
1508 Knoxville Ave., Ste. 315
Box 41098
Lubbock, TX 79409-1098
Phone: 806.742.1700
8 a.m. - 5:00 p.m. Central, Monday - Friday
Home /
Offices /
Office of Treasury /
International Currency Conversion
International Currency Conversion
Exchange Rate Estimation
International currency exchange rates are to be used only as a forecasting tool. Rates
change constantly, and are greatly affected by international events. A link to Reuters
is provided below for current exchange rates. The exchange rate quotes provided by
Reuters are “mid-market” rates which are not available for individuals, but rather
are the published rates for currency purchases over $1 million U.S. dollars. Therefore,
the exchange rate from Reuters will not match the exchange rate the System obtains
through its global payments provider.
International Currency Exchange Rates
There are a few things to consider when working with exchange rates:
Exchange rates change continuously.
Rates are greatly affected by world events, more specifically events that may directly
affect a specific country, meaning that rates can vary greatly from day-to-day, should
unusual events transpire.
The TTU System negotiates the best possible exchange rate. However, there are fees
associated with conducting international business. Our global payments provider builds
those fees into the rate of exchange. The rates provided online do not have these
costs built in.
Looking online to obtain a currency conversion is a smart business decision, and helps
you estimate your costs. Just remember that the figures are estimates, and will differ
at the time of currency conversion.
International Currency Format
U.S. currency is formatted with a decimal point (.) as a separator between dollars
and cents. Some countries use a comma (,) instead of a decimal point to indicate the
separation. In addition, while the U.S. and a number of other countries use a comma
to separate thousands, some countries use a decimal point for this purpose.
For these reasons, it is critically important to recognize the differences in the
placement of decimal points and commas in international currency. Failure to do so
could result in the loss of significant amounts of money for the System.
To help you identify the formatting for currency, below is a list of countries and
their respective currency formats. It is also important to note that most currencies
utilize the format of threes and twos for dollars and cents (see the examples below),
which means that the decimal or comma should be irrelevant, especially when the cents
are shown. This is what you should look for when converting currencies.
Examples:
500 or 500,00 or 500.00 = five hundred dollars and no cents
500,15 or 500.15 = five hundred dollars and fifteen cents
500,150 or 500.150 or 500,150.00 or 500.150,00 = five hundred thousand, one hundred
fifty dollars and no cents
International Currency Format
| | |
--- |
| Australia | Nepal |
| Botswana | New Zealand |
| British West Indies | Nicaragua |
| Brunei | Nigeria |
| Canada (English-speaking) | Pakistan |
| Dominican Republic | People's Republic of China |
| Guatemala | Philippines |
| Hong Kong | Singapore |
| India | Sri Lanka |
| Ireland | Switzerland (only when the amount is in Swiss francs) |
| Israel | Taiwan |
| Japan | Tanzania |
| Kenya | Thailand |
| Korea (both North and South) | Uganda |
| Lebanon | United Kingdom |
| Malaysia | United States (including insular areas) |
| Malta | Zimbabwe |
| Mexico | |
Countries Using a Comma to Separate Dollars & Cents
| | |
--- |
| Albania | Italy |
| Andorra | Kazakhstan |
| Argentina | Kirghistan |
| Armenia | Latvia |
| Austria | Lebanon |
| Azerbaijan | Lithuania |
| Belarus | Luxembourg (uses both marks officially) |
| Belgium | Macau (in Portuguese text) |
| Bolivia | Macedonia |
| Bosnia and Herzegovina | Moldova |
| Brazil | Mongolia |
| Bulgaria | Morocco |
| Cameroon | Netherlands |
| Canada (French-speaking) | Norway |
| Chile | Panama |
| Colombia | Paraguay |
| Costa Rica | Peru |
| Croatia | Poland |
| Cuba | Portugal |
| Cyprus | Romania |
| Czech Republic | Russia |
| Denmark | Serbia |
| Dominican Republic | Slovakia |
| Estonia | Slovenia |
| Faroes | South Africa (officially) |
| Finland | Spain |
| France | Sweden |
| Germany | Tunisia |
| Georgia | Turkey |
| Greece | Ukraine |
| Greenland | Uruguay |
| Honduras | Uzbekistan |
| Hungary | Venezuela |
| Iceland | Vietnam |
| Indonesia | | |
190762 | https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Hydrides | Skip to main content
Hydrides
Last updated
: Jun 30, 2023
Save as PDF
Carbonates
Oxides
Page ID
: 543
( \newcommand{\kernel}{\mathrm{null}\,})
The term hydride is commonly named after binary compounds that hydrogen forms with other elements of the periodic table. Hydride compounds in general form with almost any element, except a few noble gases. The trends and properties vary according to the type of intermolecular force that bonds the elements together, the temperature, its molecular masses, and other components. Hydrides are classified into three major groups, depending on what elements the hydrogen bonds to. The three major groups are covalent, ionic, and metallic hydrides. Formally, hydride is known as the negative ion of a hydrogen, H-, also called a hydride ion. Because of this negative charge, hydrides have reducing, or basic properties. Its special characteristics will be further discussed.
Covalent Hydrides
The first major group is covalent hydrides, which is when a hydrogen atom and one or more non-metals form compounds. This occurs when hydrogen covalently bonds to a more electropositive element by sharing electron pairs. These hydrides can be volatile or non-volatile. Volatile simply means being readily able to be vaporized at low temperatures. One such example of a covalent hydride is when hydrogen bonds with chlorine and forms hydrochloric acid (HCl). Examples are listed below:
H2(g)+Cl2(g)⟶2HCl(g)(1)
3H2(g)+N2(g)⟶2NH3(g)(2)
The hydrides of nonmetals on the periodic table become more electronegative as you move from group 13 to 17. This means that they are less capable of donating an electron, and want to keep them because their electron orbital becomes fuller. Instead of donating a H−, they would instead donate a H+ because they are more acidic.
Example 1: Boron Hydrides
Boron can form many different types of hydrides; one of them is borane (BH3), which reacts violently with air and is easily oxidized. Borane occurs as a gaseous substance, and can form B2H6 by two borane molecules combined with each other. Borane is not a stable compound because it does not follow a complete octet rule since it has only six valence electrons.
Example 2: Nitrogen Hydrides
Ammonia is an important nitrogen hydride that is possible due to the synthesis of nitrogen and water which is called the Haber-Bosch process. The chemical equation for this reaction is:
N2(g)+3H2(g)↽−−⇀2NH3(g)
To yield ammonia, there needs to be a catalyst to speed up the reaction, a high temperature and a high pressure. Ammonia is a reagent used in many chemistry experiments and is used as fertilizer. Ammonia can react with sulfuric acid to produce ammonium sulfate, which is also an important fertilizer. In this reaction, ammonia acts as a base since it receives electrons while sulfuric acid gives off electrons.
2NH3(aq)+H2SO4(aq)↽−−⇀(NH4)2SO4(aq)
Other hydrides of nitrogen include ammonium chloride, hydrazine and hydroxylamine. Ammonium chloride is widely used in dry-cell batteries and clean metals.
Ionic Hydrides
The second category of hydrides are ionic hydrides (also known as saline hydrides or pseudohalides). These compounds form between hydrogen and the most active metals, especially with the alkali and alkaline-earth metals of group one and two elements. In this group, the hydrogen acts as the hydride ion (H−). They bond with more electropositive metal atoms. Ionic hydrides are usually binary compounds (i.e., only two elements in the compound) and are also insoluble in solutions.
A(s)+H2(g)⟶2AH(s)(3)
with A as any group 1 metal.
A(s)+H2(g)⟶AH2(s)(4)
with A as any group 2 metal.
Ionic hydrides combine vigorously with water to produce hydrogen gas.
Example 3: Alkali Metal Hydrides
As ionic hydrides, alkali metal hydrides contain the hydride ion H− as well. They are all very reactive and readily react with various compounds. For example, when an alkali metal reacts with hydrogen gas under heat, an ionic hydride is produced. Alkali metal hydrides also react with water to produce hydrogen gas and a hydroxide salt:
MH(s)+H2O(l)⟶MOH(aq)+H2(g)
Metallic Hydrides
The third category of hydrides are metallic hydrides, also known as interstitial hydrides. Hydrogen bonds with transition metals. One interesting and unique characteristic of these hydrides are that they can be nonstoichiometric, meaning basically that the fraction of H atoms to the metals are not fixed. Nonstoichiometric compounds have a variable composition. The idea and basis for this is that with metal and hydrogen bonding there is a crystal lattice that H atoms can and may fill in between the lattice while some might, and is not a definite ordered filling. Thus it is not a fixed ratio of H atoms to the metals. Even so, metallic hydrides consist of more basic stoichiometric compounds as well.
Intermolecular Interactions
You may think that hydrides are all intact through hydrogen bonding because of the presence of at least a hydrogen atom, but that is false. Only some hydrides are connected with hydrogen bonding. Hydrogen bonds have energies of the order of 15-40 kJ/mol, which are fairly strong but in comparison with covalent bonds at energies greater than 150 kJ/mol, they are still much weaker. Some hydrogen bonding can be weak if they are mildly encountered with neighboring molecules. Specifically fluorine, oxygen, and nitrogen are more vulnerable to hydrogen bonding.
In hydrides, hydrogen is bonded with a highly electronegative atom so their properties are more distinguished. Such that in the chart below comparing boiling points of groups 14-17 hydrides, the values of ammonia (NH3), water (H2O), and hydrogen fluoride (HF) break the increasing boiling point trend.
Supposedly, as the molecular mass increases, the boiling points increase as well. Due to the hydrogen bonds of the three following hydrides, they distinctly have high boiling points instead of the initial assumption of having the lowest boiling points. What occur in these hydrogen bonds are strong dipole-dipole attractions because of the high ionic character of the compounds.
References
General Chemistry:Principles and Modern Applications.-9th ed./Ralph H.Petrucci...
General Chemistry.7th ed./Kenneth W. Whitten.
Peruzzini, Maurizio, and Rinaldo Poli. Recent Advances in Hydride Chemistry. St. Louis: Elsevier, 2001.
General Chemistry 9e. Boston: Houghton Mifflin, 2008.
Contributors
Tandis Arani (UCD) |
190763 | https://www.mdpi.com/2077-0383/13/13/3782 | Next Article in Journal
Analysis of Postoperative Complication and Revision Rates and Mid- to Long-Term Implant Survival in Primary Short-Stem Total Hip Arthroplasty
Previous Article in Journal
Impact of Periodontitis on Endothelial Risk Dysfunction and Oxidative Stress Improvement in Patients with Cardiovascular Disease
Previous Article in Special Issue
Cardiological Challenges Related to Long-Term Mechanical Circulatory Support for Advanced Heart Failure in Patients with Chronic Non-Ischemic Cardiomyopathy
Active Journals Find a Journal Journal Proposal Proceedings Series
## Topics
For Authors For Reviewers For Editors For Librarians For Publishers For Societies For Conference Organizers
Open Access Policy Institutional Open Access Program Special Issues Guidelines Editorial Process Research and Publication Ethics Article Processing Charges Awards Testimonials
## Author Services
Sciforum MDPI Books Preprints.org Scilit SciProfiles Encyclopedia JAMS Proceedings Series
Overview Contact Careers News Press Blog
Sign In / Sign Up
Notice
clear
Notice
You are accessing a machine-readable page. In order to be human-readable, please install an RSS reader.
Continue Cancel
clear
All articles published by MDPI are made immediately available worldwide under an open access license. No special permission is required to reuse all or part of the article published by MDPI, including figures and tables. For articles published under an open access Creative Common CC BY license, any part of the article may be reused without permission provided that the original article is clearly cited. For more information, please refer to
Feature papers represent the most advanced research with significant potential for high impact in the field. A Feature Paper should be a substantial original Article that involves several techniques or approaches, provides an outlook for future research directions and describes possible research applications.
Feature papers are submitted upon individual invitation or recommendation by the scientific editors and must receive positive feedback from the reviewers.
Editor’s Choice articles are based on recommendations by the scientific editors of MDPI journals from around the world. Editors select a small number of articles recently published in the journal that they believe will be particularly interesting to readers, or important in the respective research area. The aim is to provide a snapshot of some of the most exciting work published in the various research areas of the journal.
Original Submission Date Received: .
Journals
Active Journals
Find a Journal
Journal Proposal
Proceedings Series
Topics
Information
For Authors
For Reviewers
For Editors
For Librarians
For Publishers
For Societies
For Conference Organizers
Open Access Policy
Institutional Open Access Program
Special Issues Guidelines
Editorial Process
Research and Publication Ethics
Article Processing Charges
Awards
Testimonials
Author Services
Initiatives
Sciforum
MDPI Books
Preprints.org
Scilit
SciProfiles
Encyclopedia
JAMS
Proceedings Series
About
Overview
Contact
Careers
News
Press
Blog
Sign In / Sign Up Submit
Journals
JCM
Volume 13
Issue 13
10.3390/jcm13133782
Submit to this Journal Review for this Journal Propose a Special Issue
► ▼ Article Menu
Article Menu
Academic Editor
Michael Dandel
Subscribe SciFeed
Recommended Articles
Author Biographies
Related Info Links
PubMed/Medline
Google Scholar
More by Authors Links
on DOAJ
Ahmad, T.
Manohar, S. A.
Stencel, J. D.
Le Jemtel, T. H.
on Google Scholar
Ahmad, T.
Manohar, S. A.
Stencel, J. D.
Le Jemtel, T. H.
on PubMed
Ahmad, T.
Manohar, S. A.
Stencel, J. D.
Le Jemtel, T. H.
/ajax/scifeed/subscribe
Article Views
Citations -
Table of Contents
Altmetric share Share announcement Help format_quote Cite question_answer Discuss in SciProfiles
Need Help?
Support
Find support for a specific problem in the support section of our website.
Get Support
Feedback
Please let us know what you think of our products and services.
Give Feedback
Information
Visit our dedicated information section to learn more about MDPI.
Get Information
clear
JSmol Viewer
clear
first_page
Download PDF
settings
Order Article Reprints
Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Open AccessReview
Dobutamine in the Management of Advanced Heart Failure
by
Tanjeev Ahmad
Tanjeev Ahmad
SciProfiles Scilit Preprints.org Google Scholar
Tajnjeev Ahmad MD is a second year internal medicine resident at Tulane University School of with a [...]
Tajnjeev Ahmad MD is a second year internal medicine resident at Tulane University School of Medicine with a strong interest in cardiovascular disease. He graduated from college at University of Michigan- Ann Arbor with a bachelor's degree in biochemistry and neuroscience and medical school from Wright State University School of Medicine.
Read moreRead less
,
Shamitha A. Manohar
Shamitha A. Manohar
SciProfiles Scilit Preprints.org Google Scholar
,
Jason D. Stencel
Jason D. Stencel
SciProfiles Scilit Preprints.org Google Scholar
and
Thierry H. Le Jemtel
Thierry H. Le Jemtel
SciProfiles Scilit Preprints.org Google Scholar
John W. Deming Department of Medicine Tulane University, 131 S., New Orleans, LA 70112, USA
Author to whom correspondence should be addressed.
J. Clin. Med. 2024, 13(13), 3782;
Submission received: 14 May 2024 / Revised: 16 June 2024 / Accepted: 20 June 2024 / Published: 27 June 2024
(This article belongs to the Special Issue New Insights into the Management of Advanced (Stage D) Heart Failure)
Download keyboard_arrow_down
Download PDF Download PDF with Cover Download XML Download Epub
Browse Figures
Versions Notes
Abstract
Background: The potential harm and clinical benefits of inotropic therapy in patients with decompensated heart failure with reduced ejection fraction or advanced heart failure were debated for three decades. Nonetheless, confronted with a dismal quality of life in the last months to years of life, continuous home inotropic therapy has recently gained traction for palliative therapy in patients who are not candidates for left ventricular mechanical circulatory support or heart transplantation. Methods: As continuous inotropic therapy is only considered for patients who experience symptomatic relief and display objective evidence of improvement, clinical equipoise is no longer present, and randomized controlled trials are hard to conduct. Results: We first outline the transient use of inotropic therapy in patients with decompensated heart failure with reduced ejection fraction and emphasize the hemodynamic requisite for inotropic therapy, which is a demonstration of a low cardiac output through a low mixed venous oxygen saturation. Lastly, we review the current experience with the use of home inotropic therapy in patients who are not candidates or are awaiting mechanical circulatory support or heart transplantation. Conclusions: Evidence-based clinical data are needed to guide inotropic therapy for refractory decompensated heart failure with reduced ejection fraction in patients who are ineligible or awaiting mechanical circulatory support or heart transplantation.
Keywords:
decompensated heart failure with reduced ejection fraction; cardiovascular disease; advanced heart failure; inotrope; dobutamine; milrinone; levosimendan decompensated reduced ejection fraction
1. Intravenous Inotropic Therapy
Positive inotropic therapy has been and remains the subject of controversy in the treatment of heart failure (HF) . Thirty years ago, the detrimental effects of oral milrinone on mortality and morbidity in the Prospective Randomized Milrinone Evaluation Survival Evaluation (PROMISE) trial and intravenous (IV) milrinone in the Outcomes of a Prospective Trial of Intravenous Milrinone for Exacerbations of Chronic Heart Failure (OPTIME-CHF) study tempered the initial enthusiasm for positive inotropic therapy in the management of decompensated heart failure with reduced ejection fraction (HFrEF) [2,3]. Subsequently, three post hoc analyses of landmark randomized controlled trials (RTCs) questioned the safety of positive inotropic therapy in decompensated HFrEF [4,5,6].
In the Flolan International Randomized Survival Trial (FIRST), decompensated HFrEF patients who were receiving dobutamine had a worse outcome than that of patients who did not after adjustment for age, sex baseline functional status/left ventricular ejection fraction, randomization to epoprostenol, and ischemic cause of HF . However, no specific adjustment addressed the decision to add dobutamine to the management of decompensated HFrEF before the randomization to epoprostenol or to standard care. The decision was likely related to a markedly depressed cardiac output (CO).
Next, in the Randomized Evaluation of Mechanical Assistance in treatment of Chronic heart Failure (REMATCH) trial, where 71% of patients receiving IV inotropic therapy at randomization and 29% of patients were not receiving IV inotropic therapy, two or more inotropic agents were administered at randomization in 43% of patients; 75% of patients received dobutamine, 43% received milrinone, and 30% received dopamine . Patients not receiving IV inotropic therapy at randomization had higher survival rates independent of Left Ventricular Assist Device (LVAD) implantation. As in the post hoc analysis of the FIRST trial, no specific adjustment addressed the rationale for inotropic therapy before the randomization to LVAD or to continued medical therapy.
Although positive inotropic therapy may hasten the progression of the underlying myocardial disease in patients with HF, it alleviates symptoms and returns decompensated HFrEF to a compensated state when optimal neuro-hormonal modulation fails or cannot be attempted due to systemic hypotension . Nowadays, intravenous (IV) positive inotropic therapy is commonly considered in three clinical situations. The first is the escalation to inotropic therapy in patients hospitalized for decompensated HFrEF who do not respond to or do not tolerate optimal doses of guideline-directed medical therapy (GDMT). The second situation is palliative inotropic therapy in hospitalized patients for decompensated HFrEF who, not candidates for LVAD or heart transplantation (HT), experience marked symptomatic deterioration after any attempt to taper or discontinue IV inotropic therapy [8,9]. The third situation is patients approved for heart transplantation who are awaiting a donor heart.
2. Aims and Safety of Intravenous Inotropic Therapy
The cardinal hemodynamic aim of positive inotropic therapy is to restore a normal or near-normal CO. A sine qua non requisite of positive inotropic therapy is the demonstration of a low CO . Regrettably, and likely due to spiraling costs, Medicare no longer requires demonstration of hemodynamic changes when initiating home inotropic therapy .
The measurement of CO by thermodilution requires the insertion of a pulmonary artery catheter, which, although routinely performed in the intensive care unit, is expensive. Alternatively, the insertion of a peripherally inserted central catheter (PICC line) is a less expensive and safer procedure than pulmonary artery catheterization, which provides an estimate of CO through measurement of central vein oxygen saturation (ScvO2%).
Half a century ago, ScvO2 was found to reflect myocardial function in patients admitted to an intensive care unit for acute myocardial infarction . Vein O2 saturation was >60% in AMI patients without evidence of myocardial dysfunction, ≤60% in AMI patients with clinical evidence of HF, and ≤45% in AMI patients in shock, as evidenced by a blood pressure ≤90 mmHg and decreased peripheral perfusion. Subsequently, several studies underlined the poor correlation between ScvO2 and mixed venous O2 saturation (SvO2) after cardiac surgery or in critically ill patients [13,14,15,16,17]. Nonetheless, many still view ScvO2 as a useful estimate of Svo2, although ScvO2 and SvO2 may differ [18,19,20,21].
In healthy individuals, O2 saturation is greater in the inferior vena cava (IVC) than the superior vena cava (SVC) due to the contribution of highly O2 saturated blood from the renal veins to the IVC . In patients with HF or shock, O2 saturation is lower in IVC than in SVC due to a marked reduction in renal perfusion and preferential cerebral perfusion . Hence, O2 saturation in the SVC (ScvO2) is usually 5% higher than in the IVC in patients with HF or shock. Further, contrary to SvO2, ScvO2 does not reckon the contribution of low O2 saturated (40–50%) blood from the coronary sinus [14,24]. Central vein O2 saturation (ScvO2) can underestimate SvO2 by 10–20% in patients with cardiogenic shock . The right atrium (RA) drains blood from the SVC, IVC, and coronary sinus. Hence, advancing the central vein catheter into the right atrium (RA) obliterates the difference between ScvO2 and SvO2 [13,21,25]. Last, femoral vein O2 saturation poorly correlates with ScvO2 in critically ill patients . Sepsis, a left-to-right shunt, or a systemic inflammatory syndrome should be excluded in patients with decompensated HFrEF and ScvO2 or SvO2 >60% [27,28]. When ScvO2 is <50%, patients who are hospitalized for decompensated HFrEF should undergo right heart catheterization to confirm or exclude a low CO state. In summary, while ScvO2 is an imperfect surrogate for SvO2, it should be obtained when the insertion of a pulmonary artery catheter is not feasible.
The limited energy supplies of a depressed myocardium hinder the use of IV-positive inotropes, especially dobutamine, in patients with advanced HF. By increasing myocardial contractility and heart rate, IV-positive inotropes tend to increase myocardial oxygen consumption (MVO2), as the left ventricular (LV) wall stress may not decrease sufficiently to offset the metabolic cost of increasing myocardial contractility and heart rate . The metabolic cost of IV dobutamine is clearly of concern in patients with ischemic cardiomyopathy. It also concerns patients with dilated cardiomyopathy, as blood flow in the LV subendocardial layer is as reduced in dilated cardiomyopathy as in ischemic cardiomyopathy . In a pig model of LV systolic dysfunction, MVO2 increases out of proportion to hemodynamic changes when dobutamine is administered at high doses . At low doses, dobutamine does not waste oxygen in the same model . At a rate of infusion of 10 mcg/kg/min, dobutamine increases MVO2 by 54% . At a rate of infusion of 5.0 mcg/kg/min, dobutamine does not alter heart rate or blood pressure, while it decreases pulmonary capillary wedge pressure and increases CO . Coronary blood flow increases proportionally to the rise in CO, and the coronary arterial–venous difference does not change, resulting in a 20% increase in MVO2 . Hence, dobutamine needs to be infused at the lowest rate of infusion that provides the needed CO improvement. We seldom use dobutamine at rates of infusion >5.0 mcg/kg/min to prevent large increases in MVO2, ventricular arrhythmias, and hemodynamic tolerance [34,35].
Last, positive inotropic agents like dobutamine and milrinone worsen the defective intracellular [Ca2+] handling of patients with HF. By increasing intracellular diastolic [Ca2+] positive inotropic agents, especially dobutamine, this increases the risk of fatal ventricular arrhythmias at higher rates of infusion. The addition of oral amiodarone to long-term intermittent dobutamine infusion lowers the risk of death from any cause (hazard ratio, 0.403; 95% confidence interval [CI] (0.164–0.048) over a follow-up of 2 years in a randomized controlled double-blind trial of 30 patients with advanced HF. The median survival time was 574 days with the addition of oral amiodarone and 144 days with placebo . Besides oral amiodarone and close monitoring of the serum potassium concentration, the only effective intervention for the treatment of dobutamine-induced ventricular tachycardia/fibrillation is an implantable cardioverter defibrillator (ICD).
3. Escalation to Dobutamine in Hospitalized Patients with Advanced Heart Failure
The choice of a positive inotropic agent between dobutamine, milrinone, and levosimendan depends on clinical situations and physician preferences. When long-term beta-adrenergic blockade (BAB) cannot be tapered and discontinued due to coexisting conditions like atrial fibrillation (AF) or portal hypertension, milrinone and levosimendan are the preferred inotropic agents. Further, investigators prefer milrinone to dobutamine in patients with biventricular failure and pulmonary hypertension. Milrinone exerts a positive inotropic effect at low doses (intravenous [IV] boluses of 25.0 mcg/kg) . At higher doses (IV boluses of 75 mcg/kg), milrinone exerts potent vasodilator effects . Milrinone increases myocardial contractility at low doses that do not affect heart rate or blood pressure. The positive inotropic effect of milrinone is modest on the right ventricle (RV) . In a subset of patients with isolated RV failure, milrinone was not superior to dobutamine in the Dobutamine Compared with Milrinone (DOREMI) trial . The DOREMI trial compared milrinone to dobutamine in patients with cardiogenic shock (CS) stages B, C, D, or E with a combined endpoint of in-hospital death, cardiac arrest, heart transplantation, need for circulatory support, transient ischemic attack/stroke, or renal replacement therapy . Aside from the infrequent use of pulmonary artery catheters, the limited use of mechanical circulatory support in stages B and C of CS (12% in patients randomized to milrinone and 15% in patients randomized to dobutamine) does not reflect contemporary practice in the United States [40,41]. Dobutamine had no significant advantage over milrinone or vice versa in the DOREMI trial . However, due to the dismal prognosis of patients in CS, the DOREMI trial was not powered to detect a small treatment effect . A meta-analysis of nine studies and two randomized controlled trials of milrinone versus dobutamine in patients in low CO or in CS revealed that dobutamine may be associated with a shorter length of hospitalization and a greater all-cause mortality .
Levosimendan exerts a positive inotropic action through calcium sensitization of troponin C, a vasodilator action through the activation of ATP-sensitive potassium channels in smooth vascular muscle cells, and a protective action through the activation of ATP-sensitive mitochondrial potassium channels (Table 1) .
From 2007 to 2021, six major randomized controlled trials of levosimendan were reported [44,45,46,47,48,49]. Three trials were positive [44,45,46]. Compared to dobutamine, levosimendan reduced mortality in advanced HF patients receiving beta-adrenergic blockade, while intermittent use of levosimendan reduced functional decline and hospitalizations for HF, along with the incidence of acute HF decompensation [44,45,46]. Three trials were neutral or negative [47,48,49]. Levosimendan did not reduce all-cause mortality compared to dobutamine, while intermittent use of levosimendan did not improve a composite endpoint of death/LVAD/HT nor improved functional capacity and quality of life compared to a placebo. In contrast to randomized controlled trials, systematic reviews, meta-analyses, and single-center experiences uniformly reported that the intermittent use of levosimendan is safe and improves morbidity and mortality in patients with advanced HF, providing great attention is given to patient dosing, dosing intervals, and patient monitoring [50,51,52,53,54]. When intravenously administered at 0.2 mcg/kg/min for a total dose of 6.25 mg every 2 weeks, levosimendan facilitates medical optimization in advanced HF patients who are intolerant to GDMT . Last, compared to dobutamine, the use of levosimendan appears to have increased from 2012 to 2021 in patients with cardiogenic shock . Levosimendan is marketed worldwide but is not currently approved in the United States by the Food and Drug Administration (FDA). A randomized, placebo-controlled trial of oral levosimendan is currently underway in the United States for the treatment of pulmonary hypertension in patients with HF with preserved ejection fraction. Currently, dobutamine and milrinone are the only two available positive inotropic agents for the treatment of decompensated HFrEF, or long-term inotropic support, in the United States.
A short half-life is the overwhelming advantage of dobutamine over other inotropic agents. The short half-life of dobutamine allows rapid titration to an effective and safe rate of infusion as well as the elimination of dobutamine within 5–10 min of an adverse effect. Dobutamine-mediated symptomatic relief needs to be corroborated by consonant hemodynamic changes. The positive inotropic action of dobutamine aims to increase CO through an increase in stroke volume rather than heart rate. Dobutamine commonly increases the heart rate in patients with advanced HF and atrial fibrillation. In patients with ischemic cardiomyopathy, a 10–15% increase in heart rate may require reducing the rate of infusion of dobutamine, while a similar increase in heart rate may be tolerated in patients with dilated cardiomyopathy. Dobutamine at a rate of infusion of 10 mcg/kg/min recently demonstrated a large increase in cardiac output in young patients after heart transplantation .
The initiation and continuation of dobutamine therapy vary based on clinical scenarios. In patients with a first hospitalization for decompensated HFrEF and renal insufficiency, dobutamine therapy may not be considered for several days as satisfactory decongestion necessitates prolonged bed rest and close titration of the loop diuretic regimen (Figure 1). In contrast to a first hospitalization for advanced HF decompensation, dobutamine therapy may be initiated early after admission in patients with recurrent hospitalizations for advanced HF and continued until full decongestion and a return to baseline renal function are achieved (Figure 2). Slow tapering of dobutamine therapy at a rate of 1 mcg/kg/min per 6, 12, or 24 h may result in a shorter hospitalization than accelerated dobutamine discontinuation . Last, BAB may not be resumed for 1–2 weeks until complete recovery and regain of baseline status.
4. Palliative Dobutamine Therapy
Despite the above-mentioned safety concerns, home IV inotropic therapy has increasingly been considered an integral part of palliation over the past 15 years [58,59]. Of the 98 patients who received IV-positive inotropes as part of palliative management from January 2007 to March 2013 at the University of Alabama in Birmingham, 84.8% and 15.2% received milrinone and dobutamine, respectively . Notably, 80.7% of patients who received positive inotropes remained on BAB. The possibility to continue BAB underlies the preferential use of milrinone over dobutamine. Unfortunately, continuing BAB thwarts any further analysis, as patients who tolerate BAB are likely to be less sick than patients who cannot no longer tolerate BAB.
Continuous IV administration of milrinone or dobutamine had similar effects on clinical outcome in 121 patients hospitalized with advanced HF at the Cleveland Clinic . However, prior to the initiation of continuous IV inotrope therapy, 5% of patients who received dobutamine were kept on BAB, and 34% of patients who received milrinone were kept on BAB. Hence, patients who received milrinone or dobutamine had a similar clinical outcome, although patients who received dobutamine were likely to be sicker than those who received milrinone. Among the 49 patients who received palliative IV-positive inotropic therapy from January 2011 to January 2017 at Columbia University Irving Medical Center, 6.1% received dobutamine and 77.6% received milrinone . Besides the preference to continue BAB in the framework of palliative therapy, the preferential use of milrinone over dobutamine is unexpected for financial reasons . Medicare spent an average of USD 64,152 per patient for home milrinone therapy, compared to USD 848 for home dobutamine therapy .
Two recent single-center observational studies investigated the safety and effectiveness of dobutamine for palliative therapy in patients with advanced HF. In the East Limburg Hospital, continuous IV administration of dobutamine alleviated symptoms and reduced HF hospitalizations at 3, 6, and 12 months, as well as health-care-related costs, in 21 consecutive patients with advanced HF . The rate of dobutamine infusion was 4.0 mcg/kg/min. The mortality rate was 48% at 1 year, with 75% of patients dying at home. Nineteen patients with advanced HF were discharged home from the Hospices Civils de Lyon while receiving dobutamine at a mean rate of infusion of 2.6 mcg/kg/min. The median follow-up was 203 days (79–434). At 3, 6, and 12 months, 74, 53, and 32% of patients were alive, respectively.
In a systematic literature search of 10 studies of patients with acutely decompensated HF and CS, including 1 randomized controlled trial, milrinone lowered the risk of in-hospital mortality compared to dobutamine in all patients with acutely decompensated HF with a relative risk [RR] of 0.86, 95% CI: 0.79–0.95; p < 0.05) . The benefit of milrinone over dobutamine was marginal in patients with acutely decompensated HF with destination therapy (RR of 0.76 95%, CI: 0.60–0.96, p < 0.05). In-hospital mortality was similar with milrinone and dobutamine in patients with acutely decompensated HF and CS, as well as in patients with acutely decompensated HF awaiting HT .
Retrospective analyses report higher survival rates with milrinone than with dobutamine in patients who are hospitalized for advanced HF and are discharged home in the framework of palliative therapy [42,43]. However, all these retrospective analyses have the same shortcomings as the post hoc analyses of the FIRST, REMATCH, and ESCAPE landmark RCTs. The clinical and laboratory parameters that led to the selection of milrinone or dobutamine cannot be retrospectively addressed.
In brief, in the absence of evidence-based data, the experience of the practitioner remains a critical resource. The successful management of positive inotropic therapy for patients with advanced HF hinges on continuity of care and close patient monitoring. Despite the challenges of conducting randomized controlled trials in sick and unstable patients, further evidence-based data are unquestionably needed. Due to a lack of standardization surrounding the measurement of patient-centered outcomes in studies of inotropes for advanced HF, the effect of inotropes on patient-reported health status remains unknown .
Author Contributions
Conceptualization, T.A. and S.A.M., Writing—original draft preparation, T.A., S.A.M. and J.D.S., writing—reviewing and editing, T.H.L.J. and S.A.M. All authors have read and agreed to the published version of the manuscript.
Funding
This research received no external funding.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
The original contributions presented in the study are included in the article, further inquiries can be directed to the corresponding author.
Conflicts of Interest
The authors declare no conflicts of interest.
References
Rao, A.; Singh, M.; Maini, M.; Anderson, K.M.; Crowell, N.A.; Henderson, P.R.; Gholami, S.S.; Sheikh, F.H.; Najjar, S.S.; Groninger, H. Bridge to Nowhere: A Retrospective Single-Center Study on Patients Using Chronic Intravenous Inotropic Support as Bridge Therapy Who Do Not Receive Surgical Therapy. Front. Cardiovasc. Med. 2022, 9, 918146. [Google Scholar] [CrossRef] [PubMed]
Packer, M.; Carver, J.R.; Rodeheffer, R.J.; Ivanhoe, R.J.; DiBianco, R.; Zeldis, S.M.; Hendrix, G.H.; Bommer, W.J.; Elkayam, U.; Kukin, M.L.; et al. Effect of Oral Milrinone on Mortality in Severe Chronic Heart Failure. N. Engl. J. Med. 1991, 325, 1468–1475. [Google Scholar] [CrossRef] [PubMed]
Cuffe, M.S.; Califf, R.M.; Adams, K.F.J.; Benza, R.; Bourge, R.; Colucci, W.S.; Massie, B.M.; O’Connor, C.M.; Pina, I.; Quigg, R.; et al. Short-Term Intravenous Milrinone for Acute Exacerbation of Chronic Heart Failure: A Randomized Controlled Trial. JAMA 2002, 287, 1541–1547. [Google Scholar] [CrossRef] [PubMed]
O’Connor, C.M.; Gattis, W.A.; Uretsky, B.F.; Adams, K.F.J.; McNulty, S.E.; Grossman, S.H.; McKenna, W.J.; Zannad, F.; Swedberg, K.; Gheorghiade, M.; et al. Continuous Intravenous Dobutamine Is Associated with an Increased Risk of Death in Patients with Advanced Heart Failure: Insights from the Flolan International Randomized Survival Trial (FIRST). Am. Heart J. 1999, 138, 78–86. [Google Scholar] [CrossRef] [PubMed]
Stevenson, L.W.; Miller, L.W.; Desvigne-Nickens, P.; Ascheim, D.D.; Parides, M.K.; Renlund, D.G.; Oren, R.M.; Krueger, S.K.; Costanzo, M.R.; Wann, L.S.; et al. Left Ventricular Assist Device as Destination for Patients Undergoing Intravenous Inotropic Therapy: A Subset Analysis from REMATCH (Randomized Evaluation of Mechanical Assistance in Treatment of Chronic Heart Failure). Circulation 2004, 110, 975–981. [Google Scholar] [CrossRef] [PubMed]
Elkayam, U.; Tasissa, G.; Binanay, C.; Stevenson, L.W.; Gheorghiade, M.; Warnica, J.W.; Young, J.B.; Rayburn, B.K.; Rogers, J.G.; DeMarco, T.; et al. Use and Impact of Inotropes and Vasodilator Therapy in Hospitalized Patients with Severe Heart Failure. Am. Heart J. 2007, 153, 98–104. [Google Scholar] [CrossRef]
LeJemtel, T.H.; Sonnenblick, E.H. Should the Failing Heart Be Stimulated? N. Engl. J. Med. 1984, 310, 1384–1385. [Google Scholar] [CrossRef] [PubMed]
Malotte, K.; Saguros, A.; Groninger, H. Continuous Cardiac Inotropes in Patients with End-Stage Heart Failure: An Evolving Experience. J. Pain Symptom Manag. 2018, 55, 159–163. [Google Scholar] [CrossRef]
Gilstrap, L.G.; DeFilippis, E.M.; Stevenson, L.W. An Unintended Consequence of the 21st-Century Cures Act for Patients with Heart Failure. Circulation 2017, 136, 123–125. [Google Scholar] [CrossRef]
Guglin, M. Inotropes and Mortality. JACC Heart Fail. 2018, 6, 1050–1051. [Google Scholar] [CrossRef]
Chuzi, S.; Allen, L.A.; Dunlay, S.M.; Warraich, H.J. Palliative Inotrope Therapy: A Narrative Review. JAMA Cardiol. 2019, 4, 815–822. [Google Scholar] [CrossRef] [PubMed]
Goldman, R.H.; Klughaupt, M.; Metcalf, T.; Spivack, A.P.; Harrison, D.C. Measurement of Central Venous Oxygen Saturation in Patients with Myocardial Infarction. Circulation 1968, 38, 941–946. [Google Scholar] [CrossRef] [PubMed]
Scheinman, M.M.; Brown, M.A.; Rapaport, E. Critical Assessment of Use of Central Venous Oxygen Saturation as a Mirror of Mixed Venous Oxygen in Severely Ill Cardiac Patients. Circulation 1969, 40, 165–172. [Google Scholar] [CrossRef] [PubMed]
Chawla, L.S.; Zia, H.; Gutierrez, G.; Katz, N.M.; Seneff, M.G.; Shah, M. Lack of Equivalence between Central and Mixed Venous Oxygen Saturation. Chest 2004, 126, 1891–1896. [Google Scholar] [CrossRef] [PubMed]
Lorentzen, A.-G.; Lindskov, C.; Sloth, E.; Jakobsen, C.-J. Central Venous Oxygen Saturation Cannot Replace Mixed Venous Saturation in Patients Undergoing Cardiac Surgery. J. Cardiothorac. Vasc. Anesth. 2008, 22, 853–857. [Google Scholar] [CrossRef] [PubMed]
Yazigi, A.; El Khoury, C.; Jebara, S.; Haddad, F.; Hayeck, G.; Sleilaty, G. Comparison of Central Venous to Mixed Venous Oxygen Saturation in Patients with Low Cardiac Index and Filling Pressures After Coronary Artery Surgery. J. Cardiothorac. Vasc. Anesth. 2008, 22, 77–83. [Google Scholar] [CrossRef] [PubMed]
Riva, J.A.; Bouchacourt, J.P.; Kohn, W.E.; Hurtado, F.J. The changes in the oxygen saturations in the superior vena cava and the pulmonary artery are not the same during cardiac surgery. Rev. Esp. Anestesiol. Reanim. 2015, 62, 140–144. [Google Scholar] [CrossRef]
Dueck, M.H.; Klimek, M.; Appenrodt, S.; Weigand, C.; Boerner, U. Trends but Not Individual Values of Central Venous Oxygen Saturation Agree with Mixed Venous Oxygen Saturation during Varying Hemodynamic Conditions. Anesthesiology 2005, 103, 249–257. [Google Scholar] [CrossRef] [PubMed]
Hassan, E.; Roffman, D.S.; Applefeld, M.M. The Value of Mixed Venous Oxygen Saturation as a Therapeutic Indicator in the Treatment of Advanced Congestive Heart Failure. Am. Heart J. 1987, 113, 743–749. [Google Scholar] [CrossRef]
Berridge, J.C. Influence of Cardiac Output on the Correlation between Mixed Venous and Central Venous Oxygen Saturation. Br. J. Anaesth. 1992, 69, 409–410. [Google Scholar] [CrossRef]
Alkhouli, M.; Yusuf Solaiman, A.; Zhao, H.; Morad, A.; Escarcega, R.O.; O’Murchu, B.; Bove, A.A. Clinical Utility of Central Venous Saturation for the Calculation of Cardiac Index in Cardiac Patients. J. Card. Fail. 2014, 20, 716–722. [Google Scholar] [CrossRef] [PubMed]
Cargill, W.H.; Hickam, J.B. The oxygen consumption of the normal and the diseased human kidney. J. Clin. Investig. 1949, 28, 526–532. [Google Scholar] [CrossRef] [PubMed]
Glamann, D.B.; Lange, R.A.; Hillis, L.D. Incidence and Significance of a “Step-down” in Oxygen Saturation from Superior Vena Cava to Pulmonary Artery. Am. J. Cardiol. 1991, 68, 695–697. [Google Scholar] [CrossRef] [PubMed]
Kopterides, P.; Bonovas, S.; Mavrou, I.; Kostadima, E.; Zakynthinos, E.; Armaganidis, A. Venous Oxygen Saturation and Lactate Gradient from Superior Vena Cava to Pulmonary Artery in Patients with Septic Shock. Shock 2009, 31, 561–567. [Google Scholar] [CrossRef] [PubMed]
Cavaliere, F.; Zamparelli, R.; Martinelli, L.; Scapigliati, A.; De Paulis, S.; Caricato, A.; Gargaruti, R.; Cina, A. Blood from the Right Atrium May Provide Closer Estimates of Mixed Venous Saturation than Blood from the Superior Vena Cava. A Pilot Study. Minerva Anestesiol. 2014, 80, 11–18. [Google Scholar] [PubMed]
Davison, D.L.; Chawla, L.S.; Selassie, L.; Jones, E.M.; McHone, K.C.; Vota, A.R.; Junker, C.; Sateri, S.; Seneff, M.G. Femoral-Based Central Venous Oxygen Saturation Is Not a Reliable Substitute for Subclavian/Internal Jugular-Based Central Venous Oxygen Saturation in Patients Who Are Critically Ill. Chest 2010, 138, 76–83. [Google Scholar] [CrossRef] [PubMed]
Roswell, R.O.; Wong, M.P.; Stefanescu Schmidt, A.C.; Petranovic, M.; Zern, E.K.; Burkhoff, D.; Sundt, T.M.; O’Gara, P.T.; Harris, C.K. Case 8-2024: A 55-Year-Old Man with Cardiac Arrest, Cardiogenic Shock, and Hypoxemia. N. Engl. J. Med. 2024, 390, 1030–1043. [Google Scholar] [CrossRef] [PubMed]
Urina Jassir, D.; Chaanine, A.H.; Desai, S.; Rajapreyar, I.; Le Jemtel, T.H. Therapeutic Dilemmas in Mixed Septic-Cardiogenic Shock. Am. J. Med. 2023, 136, 27–32. [Google Scholar] [CrossRef] [PubMed]
Grose, R.; Strain, J.; Greenberg, M.; LeJemtel, T.H. Systemic and Coronary Effects of Intravenous Milrinone and Dobutamine in Congestive Heart Failure. J. Am. Coll. Cardiol. 1986, 7, 1107–1113. [Google Scholar] [CrossRef]
Parodi, O.; De Maria, R.; Oltrona, L.; Testa, R.; Sambuceti, G.; Roghi, A.; Merli, M.; Belingheri, L.; Accinni, R.; Spinelli, F. Myocardial Blood Flow Distribution in Patients with Ischemic Heart Disease or Dilated Cardiomyopathy Undergoing Heart Transplantation. Circulation 1993, 88, 509–522. [Google Scholar] [CrossRef]
Müller, S.; How, O.-J.; Jakobsen, Ø.; Hermansen, S.E.; Røsner, A.; Stenberg, T.A.; Myrmel, T. Oxygen-Wasting Effect of Inotropy: Is There a Need for a New Evaluation? An Experimental Large-Animal Study Using Dobutamine and Levosimendan. Circ. Heart Fail. 2010, 3, 277–285. [Google Scholar] [CrossRef] [PubMed]
Bendersky, R.; Chatterjee, K.; Parmley, W.W.; Brundage, B.H.; Ports, T.A. Dobutamine in Chronic Ischemic Heart Failure: Alterations in Left Ventricular Function and Coronary Hemodynamics. Am. J. Cardiol. 1981, 48, 554–558. [Google Scholar] [CrossRef] [PubMed]
Magorien, R.D.; Unverferth, D.V.; Brown, G.P.; Leier, C.V. Dobutamine and Hydralazine: Comparative Influences of Positive Inotropy and Vasodilation on Coronary Blood Flow and Myocardial Energetics in Nonischemic Congestive Heart Failure. J. Am. Coll. Cardiol. 1983, 1, 499–505. [Google Scholar] [CrossRef] [PubMed]
Weinstein, J.S.; Baim, D.S. The Effects of Acute Dobutamine Administration on Myocardial Metabolism and Energetics. Heart Fail. 1986, 3, 110–116. [Google Scholar]
Rao, A.; Anderson, K.M.; Mohammed, S.; Hofmeyer, M.; Gholami, S.S.; Sheikh, F.H.; Rodrigo, M.E.; Crowell, N.A.; Javed, H.; Gupta, S.; et al. Chronic Intravenous Inotropic Support as Palliative Therapy and Bridge Therapy for Patients with Advanced Heart Failure: A Single-Center Experience. J. Card. Fail. 2021, 27, 974–980. [Google Scholar] [CrossRef] [PubMed]
Nanas, J.N.; Tsagalou, E.P.; Kanakakis, J.; Nanas, S.N.; Terrovitis, J.V.; Moon, T.; Anastasiou-Nana, M.I. Long-Term Intermittent Dobutamine Infusion, Combined with Oral Amiodarone for End-Stage Heart Failure: A Randomized Double-Blind Study. Chest 2004, 125, 1198–1204. [Google Scholar] [CrossRef] [PubMed]
Jaski, B.E.; Fifer, M.A.; Wright, R.F.; Braunwald, E.; Colucci, W.S. Positive Inotropic and Vasodilator Actions of Milrinone in Patients with Severe Congestive Heart Failure. Dose-Response Relationships and Comparison to Nitroprusside. J. Clin. Investig. 1985, 75, 643–649. [Google Scholar] [CrossRef]
Rettig, G.F.; Schieffer, H.J. Acute Effects of Intravenous Milrinone in Heart Failure. Eur. Heart J. 1989, 10 (Suppl. C), 39–43. [Google Scholar] [CrossRef] [PubMed]
Mathew, R.; Di Santo, P.; Jung, R.G.; Marbach, J.A.; Hutson, J.; Simard, T.; Ramirez, F.D.; Harnett, D.T.; Merdad, A.; Almufleh, A.; et al. Milrinone as Compared with Dobutamine in the Treatment of Cardiogenic Shock. N. Engl. J. Med. 2021, 385, 516–525. [Google Scholar] [CrossRef]
Garan, A.R.; Kanwar, M.; Thayer, K.L.; Whitehead, E.; Zweck, E.; Hernandez-Montfort, J.; Mahr, C.; Haywood, J.L.; Harwani, N.M.; Wencker, D.; et al. Complete Hemodynamic Profiling with Pulmonary Artery Catheters in Cardiogenic Shock Is Associated with Lower In-Hospital Mortality. JACC Heart Fail. 2020, 8, 903–913. [Google Scholar] [CrossRef]
Wagaman, B. The Efficacy of Pulmonary Artery Catheters in Reducing Mortality in Acute Heart Failure Cardiogenic Shock: A Systematic Review. Heart Lung 2024, 66, 123–128. [Google Scholar] [CrossRef] [PubMed]
Abdel-Razek, O.; Di Santo, P.; Jung, R.G.; Parlow, S.; Motazedian, P.; Prosperi-Porta, G.; Visintini, S.; Marbach, J.A.; Ramirez, F.D.; Simard, T.; et al. Efficacy of Milrinone and Dobutamine in Cardiogenic Shock: An Updated Systematic Review and Meta-Analysis. Crit. Care Explor. 2023, 5, e0962. [Google Scholar] [CrossRef] [PubMed]
Masarone, D.; Melillo, E.; Gravino, R.; Errigo, V.; Martucci, M.L.; Caiazzo, A.; Petraio, A.; Pölzl, G.; Pacileo, G. Inotropes in Patients with Advanced Heart Failure: Not Only Palliative Care. Heart Fail. Clin. 2021, 17, 587–598. [Google Scholar] [CrossRef] [PubMed]
Mebazaa, A.; Nieminen, M.S.; Filippatos, G.S.; Cleland, J.G.; Salon, J.E.; Thakkar, R.; Padley, R.J.; Huang, B.; Cohen-Solal, A. Levosimendan vs. Dobutamine: Outcomes for Acute Heart Failure Patients on Beta-Blockers in SURVIVE. Eur. J. Heart Fail. 2009, 11, 304–311. [Google Scholar] [CrossRef]
Comín-Colet, J.; Manito, N.; Segovia-Cubero, J.; Delgado, J.; García Pinilla, J.M.; Almenar, L.; Crespo-Leiro, M.G.; Sionis, A.; Blasco, T.; Pascual-Figal, D.; et al. Efficacy and Safety of Intermittent Intravenous Outpatient Administration of Levosimendan in Patients with Advanced Heart Failure: The LION-HEART Multicentre Randomised Trial. Eur. J. Heart Fail. 2018, 20, 1128–1136. [Google Scholar] [CrossRef] [PubMed]
García-González, M.J.; Aldea Perona, A.; Lara Padron, A.; Morales Rull, J.L.; Martínez-Sellés, M.; de Mora Martin, M.; López Díaz, J.; López Fernandez, S.; Ortiz Oficialdegui, P.; Jiménez Sosa, A. Efficacy and Safety of Intermittent Repeated Levosimendan Infusions in Advanced Heart Failure Patients: The LAICA Study. ESC Heart Fail. 2021, 8, 4820–4831. [Google Scholar] [CrossRef] [PubMed]
Mebazaa, A.; Nieminen, M.S.; Packer, M.; Cohen-Solal, A.; Kleber, F.X.; Pocock, S.J.; Thakkar, R.; Padley, R.J.; Põder, P.; Kivikko, M. Levosimendan vs Dobutamine for Patients with Acute Decompensated Heart Failure: The SURVIVE Randomized Trial. JAMA 2007, 297, 1883–1891. [Google Scholar] [CrossRef] [PubMed]
Altenberger, J.; Parissis, J.T.; Costard-Jaeckle, A.; Winter, A.; Ebner, C.; Karavidas, A.; Sihorsch, K.; Avgeropoulou, E.; Weber, T.; Dimopoulos, L.; et al. Efficacy and Safety of the Pulsed Infusions of Levosimendan in Outpatients with Advanced Heart Failure (LevoRep) Study: A Multicentre Randomized Trial. Eur. J. Heart Fail. 2014, 16, 898–906. [Google Scholar] [CrossRef]
Pölzl, G.; Altenberger, J.; Comín-Colet, J.; Delgado, J.F.; Fedele, F.; García-González, M.J.; Gustafsson, F.; Masip, J.; Papp, Z.; Störk, S.; et al. Repetitive Levosimendan Infusions for Patients with Advanced Chronic Heart Failure in the Vulnerable Post-Discharge Period: The Multinational Randomized LeoDOR Trial. Eur. J. Heart Fail. 2023, 25, 2007–2017. [Google Scholar] [CrossRef]
Oliva, F.; Perna, E.; Marini, M.; Nassiacos, D.; Cirò, A.; Malfatto, G.; Morandi, F.; Caico, I.; Perna, G.; Meloni, S.; et al. Scheduled Intermittent Inotropes for Ambulatory Advanced Heart Failure. The RELEVANT-HF Multicentre Collaboration. Int. J. Cardiol. 2018, 272, 255–259. [Google Scholar] [CrossRef]
Elsherbini, H.; Soliman, O.; Zijderhand, C.; Lenzen, M.; Hoeks, S.E.; Kaddoura, R.; Izham, M.; Alkhulaifi, A.; Omar, A.S.; Caliskan, K. Intermittent Levosimendan Infusion in Ambulatory Patients with End-Stage Heart Failure: A Systematic Review and Meta-Analysis of 984 Patients. Heart Fail. Rev. 2022, 27, 493–505. [Google Scholar] [CrossRef] [PubMed]
Cholley, B.; Bojan, M.; Guillon, B.; Besnier, E.; Mattei, M.; Levy, B.; Ouattara, A.; Tafer, N.; Delmas, C.; Tonon, D.; et al. Overview of the Current Use of Levosimendan in France: A Prospective Observational Cohort Study. Ann. Intensive Care 2023, 13, 69. [Google Scholar] [CrossRef] [PubMed]
Dobarro, D.; Donoso-Trenado, V.; Solé-González, E.; Moliner-Abós, C.; Garcia-Pinilla, J.M.; Lopez-Fernandez, S.; Ruiz-Bustillo, S.; Diez-Lopez, C.; Castrodeza, J.; Méndez-Fernández, A.B.; et al. Intermittent Inotropic Support with Levosimendan in Advanced Heart Failure as Destination Therapy: The LEVO-D Registry. ESC Heart Fail. 2023, 10, 1193–1204. [Google Scholar] [CrossRef] [PubMed]
Silvetti, S.; Pollesello, P.; Belletti, A. Repeated Levosimendan Infusions in the Management of Advanced Heart Failure: Review of the Evidence and Meta-Analysis of the Effect on Mortality. J. Cardiovasc. Pharmacol. 2024, 83, 144–157. [Google Scholar] [CrossRef]
Masarone, D.; Kittleson, M.M.; Martucci, M.L.; Valente, F.; Gravino, R.; Verrengia, M.; Ammendola, E.; Contaldi, C.; Di Palma, V.; Caiazzo, A.; et al. Levosimendan as a “Bridge to Optimization” in Patients with Advanced Heart Failure with Reduced Ejection-A Single-Center Study. J. Clin. Med. 2022, 11, 4227. [Google Scholar] [CrossRef] [PubMed]
Belfioretti, L.; Francioni, M.; Battistoni, I.; Angelini, L.; Matassini, M.V.; Pongetti, G.; Shkoza, M.; Piangerelli, L.; Piva, T.; Nicolini, E.; et al. Evolution of Cardiogenic Shock Management and Development of a Multidisciplinary Team-Based Approach: Ten Years Experience of a Single Center. J. Clin. Med. 2024, 13, 2101. [Google Scholar] [CrossRef]
Latus, H.; Raap, R.; Klingel, K.; Happel, C.; Moysich, A.; Khalil, M.; Kerst, G.; Milla, J.; Skrzypek, S.; Thul, J.; et al. Left Ventricular Physiology and Ventricular-Vascular Interactions in Young Patients After Heart Transplantation. J. Am. Heart Assoc. 2023, 12, e028766. [Google Scholar] [CrossRef] [PubMed]
Rao, A.; Maini, M.; Anderson, K.M.; Crowell, N.A.; Gholami, S.S.; Foley Lgsw, C.; Violanti, D.; Singh, M.; Sheikh, F.H.; Najjar, S.S.; et al. Benefits and Harms of Continuous Intravenous Inotropic Support as Palliative Therapy: A Single-Institution, Retrospective Analysis. Am. J. Hosp. Palliat. Care 2024, 41, 50–55. [Google Scholar] [CrossRef]
Neicheril, R.; Snipelisky, D. A Review of the Contemporary Use of Inotropes in Patients with Heart Failure. Curr. Opin. Cardiol. 2024, 39, 104–109. [Google Scholar] [CrossRef]
Martens, P.; Vercammen, J.; Ceyssens, W.; Jacobs, L.; Luwel, E.; Van Aerde, H.; Potargent, P.; Renaers, M.; Dupont, M.; Mullens, W. Effects of Intravenous Home Dobutamine in Palliative End-Stage Heart Failure on Quality of Life, Heart Failure Hospitalization, and Cost Expenditure. ESC Heart Fail. 2018, 5, 562–569. [Google Scholar] [CrossRef]
Biswas, S.; Malik, A.H.; Bandyopadhyay, D.; Gupta, R.; Goel, A.; Briasoulis, A.; Fonarow, G.C.; Lanier, G.M.; Naidu, S.S. Meta-Analysis Comparing the Efficacy of Dobutamine Versus Milrinone in Acute Decompensated Heart Failure and Cardiogenic Shock. Curr. Probl. Cardiol. 2023, 48, 101245. [Google Scholar] [CrossRef] [PubMed]
Clarke, J.-R.D.; Riello, R.; Allen, L.A.; Psotka, M.A.; Teerlink, J.R.; Lindenfeld, J.; Desai, N.R.; Ahmad, T. Effect of Inotropes on Patient-Reported Health Status in End-Stage Heart Failure: A Review of Published Clinical Trials. Circ. Heart Fail. 2021, 14, e007759. [Google Scholar] [CrossRef] [PubMed]
Figure 1. Flowchart of sequence of events for initial hospitalization of advanced HF.
Figure 1. Flowchart of sequence of events for initial hospitalization of advanced HF.
Figure 2. Flowchart of sequence of events for recurrent hospitalization of advanced HF.
Figure 2. Flowchart of sequence of events for recurrent hospitalization of advanced HF.
Table 1. Mechanism of action of levosimendan, dobutamine, and milrinone.
Table 1. Mechanism of action of levosimendan, dobutamine, and milrinone.
| | Phosphodiesterase III Inhibitor | β1 and β2 Agonist | Troponin C Synthesizer | ATP-Sensitive K Channel |
--- ---
| Levosimendan | x | | x | x |
| Dobutamine | | x | | |
| Milrinone | x | | | |
| |
| Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. |
© 2024 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
Share and Cite
MDPI and ACS Style
Ahmad, T.; Manohar, S.A.; Stencel, J.D.; Le Jemtel, T.H. Dobutamine in the Management of Advanced Heart Failure. J. Clin. Med. 2024, 13, 3782.
AMA Style
Ahmad T, Manohar SA, Stencel JD, Le Jemtel TH. Dobutamine in the Management of Advanced Heart Failure. Journal of Clinical Medicine. 2024; 13(13):3782.
Chicago/Turabian Style
Ahmad, Tanjeev, Shamitha A. Manohar, Jason D. Stencel, and Thierry H. Le Jemtel. 2024. "Dobutamine in the Management of Advanced Heart Failure" Journal of Clinical Medicine 13, no. 13: 3782.
APA Style
Ahmad, T., Manohar, S. A., Stencel, J. D., & Le Jemtel, T. H. (2024). Dobutamine in the Management of Advanced Heart Failure. Journal of Clinical Medicine, 13(13), 3782.
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.
Article Metrics
No
Article Access Statistics
For more information on the journal statistics, click here.
Multiple requests from the same IP address are counted as one view.
Zoom | Orient | As Lines | As Sticks | As Cartoon | As Surface | Previous Scene | Next Scene
Export citation file: BibTeX)
MDPI and ACS Style
Ahmad, T.; Manohar, S.A.; Stencel, J.D.; Le Jemtel, T.H. Dobutamine in the Management of Advanced Heart Failure. J. Clin. Med. 2024, 13, 3782.
AMA Style
Ahmad T, Manohar SA, Stencel JD, Le Jemtel TH. Dobutamine in the Management of Advanced Heart Failure. Journal of Clinical Medicine. 2024; 13(13):3782.
Chicago/Turabian Style
Ahmad, Tanjeev, Shamitha A. Manohar, Jason D. Stencel, and Thierry H. Le Jemtel. 2024. "Dobutamine in the Management of Advanced Heart Failure" Journal of Clinical Medicine 13, no. 13: 3782.
APA Style
Ahmad, T., Manohar, S. A., Stencel, J. D., & Le Jemtel, T. H. (2024). Dobutamine in the Management of Advanced Heart Failure. Journal of Clinical Medicine, 13(13), 3782.
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.
clear
J. Clin. Med., EISSN 2077-0383, Published by MDPI
RSS Content Alert
Further Information
Article Processing Charges Pay an Invoice Open Access Policy Contact MDPI Jobs at MDPI
Guidelines
For Authors For Reviewers For Editors For Librarians For Publishers For Societies For Conference Organizers
MDPI Initiatives
Sciforum MDPI Books Preprints.org Scilit SciProfiles Encyclopedia JAMS Proceedings Series
Follow MDPI
LinkedIn Facebook X
© 1996-2025 MDPI (Basel, Switzerland) unless otherwise stated
Disclaimer
Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.
Terms and Conditions Privacy Policy
We use cookies on our website to ensure you get the best experience.Read more about our cookies here.
Share Link
clear
clear
Back to TopTop |
190764 | https://blog.csdn.net/qq_52489160/article/details/120423395 | 数学定理证明解析-CSDN博客
博客
下载
学习
社区
GitCode
InsCodeAI
会议
搜索
AI 搜索
登录
登录后您可以：
复制代码和一键运行
与博主大V深度互动
解锁海量精选资源
获取前沿技术资讯
立即登录
会员·新人礼包
消息
历史
创作中心
创作
证明公理1.1
最新推荐文章于 2025-09-28 20:44:01 发布
原创 于 2021-09-22 21:25:09 发布·197 阅读
· 0
· 0·
CC 4.0 BY-SA版权
版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。
文章标签：
#c++
本文通过反证法探讨了一个关于整数除法的数学定理，假设r大于等于b，并由此推导出矛盾，从而证明了r的唯一性。
假设r>=b，所以有r=b+k(k>0)
有a=qb+r=qb+b+k=(q+1)b+k
因为q不等于q+1，所以不符合定理
所以假设不成立
存在唯一性，r必须为0<=r
确定要放弃本次机会？
福利倒计时
: :
立减 ¥
普通VIP年卡可用
立即使用
Hi起来！！！
关注关注
0点赞
踩
0 收藏 觉得还不错? 一键收藏
0评论
分享复制链接 分享到 QQ 分享到新浪微博 扫一扫
举报 举报
集合论导引：选择 公理 之独立性
AI天才研究院
07-10 830
集合论导引：选择 公理 之独立性 1 . 背景介绍 1 . 1 问题的由来 集合论作为数学的基础之一，对于理解现代数学、计算机科学以及逻辑学具有至关重要的作用。集合论中的基本概念，如集合、元素、空集、并集、交集等，为构建更高级的概念和理论体系提供了基础框架。
参与评论 您还未登录，请先 登录 后发表或查看评论
如何理解与学习数学分析——第一部分——数学分析概观
9-16
读过<<如何学习数学>>( How to Study for/as a Mathematics, Degree/Major)的人会认识到本章中的一些思想;对于数学分析而言,这里的讨论更简短但更具体。 2 . 2 公理(Axioms) 公理 是数学家们一致同意视为正确的表述;公理 构成了我们发展理论的基础。(译注:公理(也称公设)(postulate):没有经过 证明,
塔斯基的真理定义解析 | 语义构想与形式构建_塔斯基真理理论
9-13
There is a consequence for the foundations of mathematics . First-order Zermelo-Fraenkel set theory is widely regarded as the standard of mathematical correctness, in the sense that a proof is correct if and only if it can be formalised as a formal proof in set theory . We would like to be . . .
7、数学系统、直接 证明 与反例
metal的博客
06-21 22
本博客详细介绍了数学系统的构成，包括 公理、定义和未定义术语，并探讨了定理、推论和引理的概念。通过直接 证明 的方法，展示了如何 证明 数学命题，并使用反例来反驳全称量化陈述。博客还包含大量关于整数性质、有理数运算以及集合运算的 证明 示例与反例，旨在帮助读者掌握数学 证明 的基本方法和逻辑推理技巧。
RSA的 证明
Lion Chen i 咖啡
04-11 8315
RSA是一种非对称加密算法，能够将大家都能看懂的明文加密成旁人无法读懂的密文。相比之下，最早的加密，如凯撒移位加密，其加密和解密用的是相同的钥匙（密钥），被称作对称加密算法。这种算法，一旦知道了一个密钥，便可以获取所有加密通讯的明文。但是，如果加密用一把钥匙，而解密只能用另一把钥匙，那么，即便窃听消息的人获得了加密密钥，也无法还原出明文。这样便极大的提高了加密通讯的安全性。这种加密与解密密钥不同的方
ES: 机器学习、专家系统、控制系统的数学映射_专家系统 映射
9-4
“机器学习是用数据或以往的经验,以此优化计算机程序的性能标准。” 一种经常引用的英文定义是:A computer program is said to learn from experience E with respect to some cl as s of t as ks T and performance me as ure P, if its performance at t as ks in T, as me as ured by P, improves with experience E . . . .
人工机器:Neural Turing Machines(NTM)
9-22
We considered the set of all possible 6-Gram distributions over binary sequences . Each 6-Gram distribution can be expressed as a table of 2^5 = 32 numbers, specifying the probability that the next bit will be one, given all possible length five binary histories . For each training example, . . .
5、自动定理 证明 中的关键概念与技术
ll5678的博客
05-31 365
本文详细探讨了自动定理 证明 中的关键概念与技术，包括抽象一致性属性的定义与验证、数学 公理 的应用、推理规则与逻辑性质的定义、推理系统的构建与验证、实际应用与优化策略，以及通过案例分析展示了推理系统的工作原理。最后对未来的多模态推理、深度学习和跨领域应用进行了展望。
1 6、定理 证明 与常识推理
f9g0h的博客
06-21 67
本博客深入探讨了定理 证明 与常识推理的基本原理、方法及其结合应用。从经典逻辑和非经典逻辑的定理 证明 技术，到默认推理、非单调逻辑、因果推理等常识推理方式，文章详细解析了各类逻辑推理的核心概念与实际案例。同时，还介绍了定理 证明 与常识推理在自然语言处理、推荐系统、机器人导航及智能家居等领域的应用，并讨论了其面临的挑战与未来发展方向，旨在提升复杂环境下的智能推理与决策能力。
离散数学-逻辑与 证明 基础 1 . 8(证明 方法和策略)_完美幂
9-26
解答:在本定理的 证明 中,我们利用∣a∣=a|a| = a∣a∣=a当a≥0a \geq 0a≥0且∣a∣=−a|a| = -a∣a∣=−a当a≤0a \leq 0a≤0来去除绝对值。由于公式中包含了∣x∣|x|∣x∣和∣y∣|y|∣y∣,我们将需要四种情况:(i)xxx和yyy都为非负,(ii)xxx为非负且yyy为负,(iii)xxx为负且yyy . . .
提示工程架构师,如何选对自动化部署工具_自动化部署架构
9-21
提示即代码(Prompt-as-Code):将提示模板、参数和逻辑实现为版本化代码的实践,使提示开发遵循软件工程最佳实践。 推理服务:提供AI模型预测能力的生产服务,负责接收输入、处理请求、调用模型并返回结果,通常需要处理高并发、低延迟和动态扩展。 模型注册表:集中存储和管理模型版本、元数据和性能指标的系统,支持模型发现、. . .
【LaTeX】IEEE模板中定理、公理 和 证明 语句的使用 热门推荐
TangPlusHPC的博客
07-05 1万+
正如题目所言，笔者第一次使用IEEE的会议LaTex模板，涉及到定理和 证明 语句时之前的写法不可行，于是笔者研究了模板中自带的文件《How to Use the IEEEtran LATEX Cl as s》文件中一节，将学习成果记录如下。 用户必须首先通过以下命令声明结构名 其中，是用户为结构选择的标识符，是指结构的标题，是计数器的可选名称，其编号将与结构编号一起显示，其更新将重置结构计数器。 大多数IEEE论文在整个工作中都使用序列化定理编号，因此通常不指定。 在定义该结构之后，可以通过以下语句使用该结构
在精益(Lean)中对定理进行 证明(第一章)
cmmsdwj的博客
12-24 3351
1 . 介绍 1 . 1 计算机与定理 证明 形式验证包括使用逻辑和计算方法， 以此建立以精确数学术语表达的论述。这些论述可以包括普通的数学定理，以及硬件或软件，网络协议以及机械和混合系统满足其规范的要求的论述。实际上，在验证一个数学定理和验证系统的正确性之间没有明显的区别：形式验证需要用数学术语描述硬件和软件系统，此时建立关于它们正确性的声明成为一种定理 证明 形式。相反，数学定理的 证明 可能需要冗长的计 . . .
【面向计算机的数理逻辑/软件理论基础笔记】命题逻辑系统的推理机制_蕴 . . .
9-24
第一类:AiA_{i}Ai本身就是 公理; 第二类:存在j,k<ij,k<ij,k<i,使得AiA_{i}Ai是由AjA_{j}Aj和AkA_{k}Ak使用MPMPMP规则推导出来的定理; 第三类:最后一项公式AnA_{n}An是推导出来的最终定理,通常称为结论,记作⊢An\vd as h A_{n}⊢An,nnn叫做 证明 的长度。
R 深度学习入门指南(一)
9-16
也就是说,计算概率并不像看起来那么简单,因为必须经常评估依赖性和独立性。举个简单的例子,假设我们正在评估两个事件 A 和 B 的概率,我们还假设事件 B 发生的概率依赖于 A 的发生。因此,如果 A 不发生,B 发生的概率为 0。数学上,我们将两个事件 A 和 B 的依赖性与独立性定义如下: . . .
几何学小课堂：几何学 公理 化体系(逻辑是从一个结论通向另一个结论唯一的通道)
专注于计算机研发知识和资讯分享
03-01 1032
L 1（即AB）和L2（即CD）是两条直线，它们相交于O点，∠1 和∠2被称为对顶角。当一条直线L和另一条直线M相交后，左右两边的夹角相等，则称M和L垂直。如果直线L和M垂直，那么夹角就是直角。从直角和垂直的定义可以得到的结论：一条直线自身的角度，等于左右两个直角相加。
minimax定理 证明
蒜法
03-29 747
minimax 定理 证明
从初中到大学:数学公式与定理完全指南
9-16
roots = [(-b / (2a),)] else: # 没有实数根 roots = [] # 输出结果 print("根为:", roots) AI写代码python 运行 通过这段代码,我们可以看到如何利用符号计算库sympy来求解二次方程的根。这对于处理复杂数学问题非常有帮助。二次函数的应用不仅限于数学领域,在物理、工程等自然科学中也有广泛应用。
罗素“类型论”与计算机程序设计艺术 / By 陈光剑&AI天才研究院-CSDN博 . . .
9-28
集合论:通过 公理 化方法(如ZFC集合论)来避免悖论。 类型论:通过引入类型层次和严格的类型规则来预防悖论的产生。 在数学基础中的角色: 集合论:提供了一种统一的语言来描述几乎所有的数学结构。 类型论:提供了一种更细致的方法来构造数学对象,特别适合于构造主义数学。
高三数学 1 三个 公理 试题
08-19
1 . 公理 1 . 1 - 共面问题：这是关于如何确定平面的基本规则。例如，如果空间中有四个点，其中任意三个点都不在同一条直线上，根据 公理，如果这些点共面，那么只可能有一个平面；如果不共面，那么可以确定四个平面 . . .
C语言速成秘籍——循环结构(while、do while、for)和跳转语句(break，continue) 最新发布
2402_89218457的博客
09-28 431
循环结构（for循环，while循环，do while循环）的语法学习和使用 循环结构中的跳转语句（break，continue），break永久跳离循环结构；continue在while/do while循环及for循环中的的用法是不一样的，一个是跳到循环条件判断，一个是跳到变量调整、再进行条件判断
C++ —— F / string
我会等
09-24 981
（一）C 语言字符串的局限存储形式：以 '\0' 结尾的字符集合操作问题：依赖 str 系列库函数，与字符串分离，不符合 OOP 思想管理风险：底层空间需用户自行管理，易出现越界访问（二）实际应用需求OJ 题目：多以 string 类形式出题工作场景：为追求简单、方便、快捷，优先使用 string 类，极少用 C 库字符串操作函数（三）相关面试题预览字符串转整形数字字符串相加。
C语言中的浮点数与整数的存储方式
jackispy的博客
09-27 766
特性整数（int, long等）浮点数（float, double）表示内容整数值（正、负或零）带小数部分的实数（可能有精度损失）存储格式二进制原码或补码IEEE 754 标准（符号、指数、尾数）是否有精度问题无有（特别是小数，如 0 . 1 无法精确表示）比较操作直接比较是否相等不可直接比较是否相等，应考虑误差范围典型用途计数、索引、离散值科学计算、测量值、图形坐标等连续值常见类型（32位系统）int（32位）、long（32/64位）
C语言核心知识点总表（1）
sqym1的博客
09-27 847
本文整理了C语言入门核心知识点，适合初学者和复习使用。主要内容包括：1）程序基础结构（Hello World组成、main函数形式、头文件规则、编译四阶段）；2）数据类型（基本类型占用字节和取值范围、有符号与无符号区别、补码机制）；3）变量（命名规则、初始化方式、生命周期与作用域）。重点提示了常见易错点，如变量未初始化、数据类型溢出、编译错误处理等。文中通过表格对比和代码示例直观展示知识点，建议收藏备用。
C++ 输入输出优化：关闭流同步cin、cout详解
小天狼星_布莱克 的博客
09-24 473
C++ 输入输出优化摘要：通过ios::sync_with_stdio(false)关闭 C++ 与C标准流的同步，能显著提升I/O性能，配合cin . tie(0)解除cin与cout绑定可进一步优化。适用于算法竞赛等需处理大量数据的场景，但要避免混合使用 C++ 和C的I/O函数。优化后cin读取速度可从 1 200ms降至300ms（百万数据级）。完整优化代码示例包含这两项关键设置，并强调了注意事项。
3 . 2 组帧 (答案见原书 P57)(注:HDLC 协议已从最新大纲中删除)
qq_46118239的博客
09-27 411
(1) 【20 1 3 统考真题】 HDLC 协议对 0 1 1 1 1 1 000 1 1 1 1 1 0 组帧后, 对应的比特串为 ( ) (注:HDLC 协议已从最新大纲中删除)的三种主要方法：字符计数法、字符（字节）填充法和比特填充法。本题全面考察了数据链路层。
C++ 中的模板是个啥机制？好用吗？
Skrrapper的博客
09-27 675
C++ 的模板（Template）可以看作是一种“泛型”编程机制，它让我们编写与的代码。简单来说，使用模板时，编译器会在编译期，从而避免了为不同类型重复编写相同逻辑的代码。模板解决了代码复用性和类型安全的问题：相同的算法或数据结构可以应用于多种类型，而编译器会在实例化时保证类型一致，不会出现像宏那样的盲目替换导致的类型混乱。与 C++ 中的宏相比，宏在预处理阶段只做简单的文本替换，不做类型检查，易引入难以察觉的错误；而模板是语言层面支持的泛型工具，在才展开代码，保证了类型安全和可维护性。
高通平台WiFi学习--深入解析 WCN39xx/PMIC GPIO/LDO 状态读取与调试
weixin_47456647的博客
09-28 13
在嵌入式系统尤其是涉及无线局域网（WLAN）模块的设备中，电源管理集成电路（PMIC）以及通用输入输出（GPIO）、低压差线性稳压器（LDO）的状态对于模块能否正常工作至关重要。本文将详细介绍如何读取 WCN39xx/PMIC 的 GPIO 和 LDO 状态，以及相关的调试方法。
CMake学习篇---CMake入门demo+从0开始+含项目文件
Edwinwzy的博客
09-24 549
本文介绍了CMake的基本概念和使用方法，重点讲解了CMakeLists . txt的编写和常用命令。内容包括：1) CMake的跨平台特性，通过生成本地Makefile实现编译；2) 项目结构组织，展示了一个简单的CMake项目示例；3) 详细解析了三个核心命令：cmake_minimum_required指定最低版本要求，project定义项目名称和属性，add_executable生成可执行文件。文章还提到CMake注释的两种语法形式。该教程适合CMake初学者快速入门，为后续深入学习打下基础。
关于我们
招贤纳士
商务合作
寻求报道
400-660-0108
kefu@csdn.net
在线客服
工作时间 8:30-22:00
公安备案号11010502030143
京ICP备19004658号
京网文〔2020〕1039-165号
经营性网站备案信息
北京互联网违法和不良信息举报中心
家长监护
网络110报警服务
中国互联网举报中心
Chrome商店下载
账号管理规范
版权与免责声明
版权申诉
出版物许可证
营业执照
©1999-2025北京创新乐知网络技术有限公司
Hi起来！！！
博客等级
码龄5年
11 原创2 点赞 1 收藏 1 粉丝
关注
私信
🛒 B2B2C商家入驻平台系统java版 Java+vue+uniapp 功能强大 稳定 支持diy 方便二开广告
热门文章
CINTA作业八：CRT 817
CINTA四：群、子群 727
CINTA作业七：同态 698
CINTA作业五：循环群 696
CINTA作业九：QR 367
上一篇： 用 C 语言编程实现一种迭代版本的简单乘法 下一篇： 2021-09-22 证明命题1.1
最新评论
CINTA作业七：同态
Bintou:第一题我都讲过答案了，怎么还答成那样？
CINTA作业三：同余、模指数、费尔马小定理、欧拉定理
Bintou:有很大的改进空间。
CINTA作业二：GCD与EGCD
Bintou:你的程序有调试的吗？
大家在看
嵌入式开发具体介绍及未来工作展望 655
GEO营销怎么做？要从向AI数据投毒，转向借AI生成的答案，占据用户心智！ 266
SEO与GEO实操底层逻辑：互联网到AI时代获客手段变迁史！
微信机器人-通过ntchat获取聊天记录自动发送聊天信息 1
Mysql杂志（二十七）——数据库状态
最新文章
CINTA作业九：QR
CINTA作业八：CRT
CINTA作业七：同态
2021年 11篇
🛒 B2B2C商家入驻平台系统java版 Java+vue+uniapp 功能强大 稳定 支持diy 方便二开广告
上一篇： 用 C 语言编程实现一种迭代版本的简单乘法 下一篇： 2021-09-22 证明命题1.1
最新文章
CINTA作业九：QR
CINTA作业八：CRT
CINTA作业七：同态
2021年 11篇
登录后您可以享受以下权益：
免费复制代码
和博主大V互动
下载海量资源
发动态/写文章/加入社区
×立即登录
评论
被折叠的 条评论 为什么被折叠?到【灌水乐园】发言
查看更多评论
添加红包
祝福语
请填写红包祝福语或标题
红包数量
个
红包个数最小为10个
红包总金额
元
红包金额最低5元
余额支付
当前余额 3.43 元 前往充值 >
需支付：10.00 元
取消 确定
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付 元
使用余额支付
点击重新获取
扫码支付
钱包余额 0
抵扣说明：
1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。
余额充值
确定 取消
举报
选择你想要举报的内容（必选）
内容涉黄
政治相关
内容抄袭
涉嫌广告
内容侵权
侮辱谩骂
样式问题
其他
原文链接（必填）
请选择具体原因（必选）
包含不实信息
涉及个人隐私
请选择具体原因（必选）
侮辱谩骂
诽谤
请选择具体原因（必选）
搬家样式
博文样式
补充说明（选填）
取消
确定
AI助手
AI 搜索 智能体 AI 编程 AI 作业助手
下载APP 程序员都在用的中文IT技术交流社区 公众号 专业的中文 IT 技术社区，与千万技术人共成长 视频号 关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部 |
190765 | https://stackoverflow.com/questions/28613436/if-statements-without-brackets | c - If statements without brackets - Stack Overflow
Join Stack Overflow
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
Sign up with GitHub
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Overflow
1. About
2. Products
3. For Teams
Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers
Advertising Reach devs & technologists worldwide about your product, service or employer brand
Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models
Labs The future of collective knowledge sharing
About the companyVisit the blog
Loading…
current community
Stack Overflow helpchat
Meta Stack Overflow
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Let's set up your homepage Select a few topics you're interested in:
python javascript c#reactjs java android html flutter c++node.js typescript css r php angular next.js spring-boot machine-learning sql excel ios azure docker
Or search from our full list:
javascript
python
java
c#
php
android
html
jquery
c++
css
ios
sql
mysql
r
reactjs
node.js
arrays
c
asp.net
json
python-3.x
.net
ruby-on-rails
sql-server
swift
django
angular
objective-c
excel
pandas
angularjs
regex
typescript
ruby
linux
ajax
iphone
vba
xml
laravel
spring
asp.net-mvc
database
wordpress
string
flutter
postgresql
mongodb
wpf
windows
xcode
amazon-web-services
bash
git
oracle-database
spring-boot
dataframe
azure
firebase
list
multithreading
docker
vb.net
react-native
eclipse
algorithm
powershell
macos
visual-studio
numpy
image
forms
scala
function
vue.js
performance
twitter-bootstrap
selenium
winforms
kotlin
loops
express
dart
hibernate
sqlite
matlab
python-2.7
shell
rest
apache
entity-framework
android-studio
csv
maven
linq
qt
dictionary
unit-testing
asp.net-core
facebook
apache-spark
tensorflow
file
swing
class
unity-game-engine
sorting
date
authentication
go
symfony
t-sql
opencv
matplotlib
.htaccess
google-chrome
for-loop
datetime
codeigniter
perl
http
validation
sockets
google-maps
object
uitableview
xaml
oop
visual-studio-code
if-statement
cordova
ubuntu
web-services
email
android-layout
github
spring-mvc
elasticsearch
kubernetes
selenium-webdriver
ms-access
ggplot2
user-interface
parsing
pointers
c++11
google-sheets
security
machine-learning
google-apps-script
ruby-on-rails-3
templates
flask
nginx
variables
exception
sql-server-2008
gradle
debugging
tkinter
delphi
listview
jpa
asynchronous
web-scraping
haskell
pdf
jsp
ssl
amazon-s3
google-cloud-platform
jenkins
testing
xamarin
wcf
batch-file
generics
npm
ionic-framework
network-programming
unix
recursion
google-app-engine
mongoose
visual-studio-2010
.net-core
android-fragments
assembly
animation
math
svg
session
intellij-idea
hadoop
rust
next.js
curl
join
winapi
django-models
laravel-5
url
heroku
http-redirect
tomcat
google-cloud-firestore
inheritance
webpack
image-processing
gcc
keras
swiftui
asp.net-mvc-4
logging
dom
matrix
pyspark
actionscript-3
button
post
optimization
firebase-realtime-database
web
jquery-ui
cocoa
xpath
iis
d3.js
javafx
firefox
xslt
internet-explorer
caching
select
asp.net-mvc-3
opengl
events
asp.net-web-api
plot
dplyr
encryption
magento
stored-procedures
search
amazon-ec2
ruby-on-rails-4
memory
canvas
audio
multidimensional-array
random
jsf
vector
redux
cookies
input
facebook-graph-api
flash
indexing
xamarin.forms
arraylist
ipad
cocoa-touch
data-structures
video
azure-devops
model-view-controller
apache-kafka
serialization
jdbc
woocommerce
razor
routes
awk
servlets
mod-rewrite
excel-formula
beautifulsoup
filter
docker-compose
iframe
aws-lambda
design-patterns
text
visual-c++
django-rest-framework
cakephp
mobile
android-intent
struct
react-hooks
methods
groovy
mvvm
ssh
lambda
checkbox
time
ecmascript-6
grails
google-chrome-extension
installation
cmake
sharepoint
shiny
spring-security
jakarta-ee
plsql
android-recyclerview
core-data
types
sed
meteor
android-activity
activerecord
bootstrap-4
websocket
graph
replace
scikit-learn
group-by
vim
file-upload
junit
boost
memory-management
sass
import
async-await
deep-learning
error-handling
eloquent
dynamic
soap
dependency-injection
silverlight
layout
apache-spark-sql
charts
deployment
browser
gridview
svn
while-loop
google-bigquery
vuejs2
dll
highcharts
ffmpeg
view
foreach
makefile
plugins
redis
c#-4.0
reporting-services
jupyter-notebook
unicode
merge
reflection
https
server
google-maps-api-3
twitter
oauth-2.0
extjs
terminal
axios
pip
split
cmd
pytorch
encoding
django-views
collections
database-design
hash
netbeans
automation
data-binding
ember.js
build
tcp
pdo
sqlalchemy
apache-flex
mysqli
entity-framework-core
concurrency
command-line
spring-data-jpa
printing
react-redux
java-8
lua
html-table
ansible
jestjs
neo4j
service
parameters
enums
material-ui
flexbox
module
promise
visual-studio-2012
outlook
firebase-authentication
web-applications
webview
uwp
jquery-mobile
utf-8
datatable
python-requests
parallel-processing
colors
drop-down-menu
scipy
scroll
tfs
hive
count
syntax
ms-word
twitter-bootstrap-3
ssis
fonts
rxjs
constructor
google-analytics
file-io
three.js
paypal
powerbi
graphql
cassandra
discord
graphics
compiler-errors
gwt
socket.io
react-router
solr
backbone.js
memory-leaks
url-rewriting
datatables
nlp
oauth
terraform
datagridview
drupal
oracle11g
zend-framework
knockout.js
triggers
neural-network
interface
django-forms
angular-material
casting
jmeter
google-api
linked-list
path
timer
django-templates
arduino
proxy
orm
directory
windows-phone-7
parse-platform
visual-studio-2015
cron
conditional-statements
push-notification
functional-programming
primefaces
pagination
model
jar
xamarin.android
hyperlink
uiview
visual-studio-2013
vbscript
google-cloud-functions
gitlab
azure-active-directory
jwt
download
swift3
sql-server-2005
configuration
process
rspec
pygame
properties
combobox
callback
windows-phone-8
linux-kernel
safari
scrapy
permissions
emacs
scripting
raspberry-pi
clojure
x86
scope
io
expo
azure-functions
compilation
responsive-design
mongodb-query
nhibernate
angularjs-directive
request
bluetooth
reference
binding
dns
architecture
3d
playframework
pyqt
version-control
discord.js
doctrine-orm
package
f#
rubygems
get
sql-server-2012
autocomplete
tree
openssl
datepicker
kendo-ui
jackson
yii
controller
grep
nested
xamarin.ios
static
null
statistics
transactions
active-directory
datagrid
dockerfile
uiviewcontroller
webforms
discord.py
phpmyadmin
sas
computer-vision
notifications
duplicates
mocking
youtube
pycharm
nullpointerexception
yaml
menu
blazor
sum
plotly
bitmap
asp.net-mvc-5
visual-studio-2008
yii2
electron
floating-point
css-selectors
stl
jsf-2
android-listview
time-series
cryptography
ant
hashmap
character-encoding
stream
msbuild
asp.net-core-mvc
sdk
google-drive-api
jboss
selenium-chromedriver
joomla
devise
cors
navigation
anaconda
cuda
background
frontend
binary
multiprocessing
pyqt5
camera
iterator
linq-to-sql
mariadb
onclick
android-jetpack-compose
ios7
microsoft-graph-api
rabbitmq
android-asynctask
tabs
laravel-4
environment-variables
amazon-dynamodb
insert
uicollectionview
linker
xsd
coldfusion
console
continuous-integration
upload
textview
ftp
opengl-es
macros
operating-system
mockito
localization
formatting
xml-parsing
vuejs3
json.net
type-conversion
data.table
kivy
timestamp
integer
calendar
segmentation-fault
android-ndk
prolog
drag-and-drop
char
crash
jasmine
dependencies
automated-tests
geometry
azure-pipelines
android-gradle-plugin
itext
fortran
sprite-kit
header
mfc
firebase-cloud-messaging
attributes
nosql
format
nuxt.js
odoo
db2
jquery-plugins
event-handling
jenkins-pipeline
nestjs
leaflet
julia
annotations
flutter-layout
keyboard
postman
textbox
arm
visual-studio-2017
gulp
stripe-payments
libgdx
synchronization
timezone
uikit
azure-web-app-service
dom-events
xampp
wso2
crystal-reports
namespaces
swagger
android-emulator
aggregation-framework
uiscrollview
jvm
google-sheets-formula
sequelize.js
com
chart.js
snowflake-cloud-data-platform
subprocess
geolocation
webdriver
html5-canvas
centos
garbage-collection
dialog
sql-update
widget
numbers
concatenation
qml
tuples
set
java-stream
smtp
mapreduce
ionic2
windows-10
rotation
android-edittext
modal-dialog
spring-data
nuget
doctrine
radio-button
http-headers
grid
sonarqube
lucene
xmlhttprequest
listbox
switch-statement
initialization
internationalization
components
apache-camel
boolean
google-play
serial-port
gdb
ios5
ldap
youtube-api
return
eclipse-plugin
pivot
latex
frameworks
tags
containers
github-actions
c++17
subquery
dataset
asp-classic
foreign-keys
label
embedded
uinavigationcontroller
copy
delegates
struts2
google-cloud-storage
migration
protractor
base64
queue
find
uibutton
sql-server-2008-r2
arguments
composer-php
append
jaxb
zip
stack
tailwind-css
cucumber
autolayout
ide
entity-framework-6
iteration
popup
r-markdown
windows-7
airflow
vb6
g++
ssl-certificate
hover
clang
jqgrid
range
gmail
Next You’ll be prompted to create an account to view your personalized homepage.
Home
Questions
AI Assist Labs
Tags
Challenges
Chat
Articles
Users
Jobs
Companies
Collectives
Communities for your favorite technologies. Explore all Collectives
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Collectives™ on Stack Overflow
Find centralized, trusted content and collaborate around the technologies you use most.
Learn more about Collectives
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
If statements without brackets
Ask Question
Asked 10 years, 7 months ago
Modified2 years, 4 months ago
Viewed 50k times
This question shows research effort; it is useful and clear
14
Save this question.
Show activity on this post.
I'm hoping to get some clarification on if and if else statements that do not have brackets and how to read them. I can read if else, and else if statements with brackets easily and they make sense to me but these have always confused me, here is a sample question.
c
if (x > 10)
if (x > 20)
printf("YAY\n");
else printf("TEST\n");
c
if-statement
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Improve this question
Follow
Follow this question to receive notifications
edited Feb 19, 2015 at 21:22
Mureinik
315k 54 54 gold badges 391 391 silver badges 403 403 bronze badges
asked Feb 19, 2015 at 17:53
Phlash6Phlash6
169 1 1 gold badge 2 2 silver badges 12 12 bronze badges
5
2 this is highly language dependent. Please specify the language this code is in (I'd guess C, but could be otherwise).aruisdante –aruisdante 2015-02-19 17:56:54 +00:00 Commented Feb 19, 2015 at 17:56
Having said that, the general rule in most curly-brace languages is that a flow-control statement need not have bracketed scope if the contents of the statement is a single expression. In this case, it's leveraging the fact that if(x<20) printf("YAY\n") collapses into a single expression, so passes muster to be contained under if(x<10) without the need for braces.aruisdante –aruisdante 2015-02-19 18:00:18 +00:00 Commented Feb 19, 2015 at 18:00
2 an else always refer to the if conditioning the previous expression or block.njzk2 –njzk2 2015-02-19 18:07:12 +00:00 Commented Feb 19, 2015 at 18:07
4 as a side note an personal view, i think actions should be taken against people who use ifs without brackets in languages that allow both behaviors.njzk2 –njzk2 2015-02-19 18:08:12 +00:00 Commented Feb 19, 2015 at 18:08
3 @njzk2 Indeed. Apple learned this the hard way. Unbracketed flow-control statements are usually outlawed by most style guides (sometimes with an exception for if (condition) return statements) because they are an endless source of bugs.aruisdante –aruisdante 2015-02-19 18:09:14 +00:00 Commented Feb 19, 2015 at 18:09
Add a comment|
3 Answers 3
Sorted by: Reset to default
This answer is useful
18
Save this answer.
Show activity on this post.
If there are no brackets on an if/else, the first statement after the if will get executed.
If statement:
c
if (condition)
printf("this line will get executed if condition is true");
printf("this line will always get executed");
If/else:
c
if (condition)
printf("this line will get executed if condition is true");
else
printf("this line will get executed if condition is false");
printf("this line will always get executed");
Note: Your code will break if there are multiple commands between an if and its matching else.
c
if (condition)
printf("this line will get executed if condition is true");
printf("this line will always get executed");
else
printf("this else will break since there is no longer a matching if statement");
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Improve this answer
Follow
Follow this answer to receive notifications
answered Feb 19, 2015 at 18:02
Victor JohnsonVictor Johnson
465 2 2 silver badges 8 8 bronze badges
Comments
Add a comment
This answer is useful
10
Save this answer.
Show activity on this post.
Without the brackets, the else will relate to the if it's immediately after. So, to properly indent your example:
c
if (x > 10)
if (x > 20)
printf("YAY\n");
else // i.e., x > 10 but not x > 20
printf("TEST\n");
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Improve this answer
Follow
Follow this answer to receive notifications
edited May 22, 2023 at 5:22
InSync
12.1k 5 5 gold badges 21 21 silver badges 59 59 bronze badges
answered Feb 19, 2015 at 17:58
MureinikMureinik
315k 54 54 gold badges 391 391 silver badges 403 403 bronze badges
Comments
Add a comment
This answer is useful
3
Save this answer.
Show activity on this post.
An if statement without brackets will only take into account the next expression after the if:
c
if(foo > 5)
foo = 10;
bar = 5;
Here, foo will only be set to 10 if foo is bigger than 5 but bar will always be set to 5 because it's not inside the scope of the if statement, that would require brackets.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Improve this answer
Follow
Follow this answer to receive notifications
edited Feb 19, 2015 at 18:02
answered Feb 19, 2015 at 18:00
Hatted RoosterHatted Rooster
36.6k 7 7 gold badges 67 67 silver badges 124 124 bronze badges
1 Comment
Add a comment
aruisdante
aruisdanteOver a year ago
It actually doesn't have to be on a new line. It's the next expression/statement after it. if (foo > 5) foo = 10; would be perfectly valid as well.
2015-02-19T18:01:45.2Z+00:00
3
Reply
Copy link
Your Answer
Thanks for contributing an answer to Stack Overflow!
Please be sure to answer the question. Provide details and share your research!
But avoid …
Asking for help, clarification, or responding to other answers.
Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Draft saved
Draft discarded
Sign up or log in
Sign up using Google
Sign up using Email and Password
Submit
Post as a guest
Name
Email
Required, but never shown
Post Your Answer Discard
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
c
if-statement
See similar questions with these tags.
The Overflow Blog
The history and future of software development (part 1)
Getting Backstage in front of a shifting dev experience
Featured on Meta
Spevacus has joined us as a Community Manager
Introducing a new proactive anti-spam measure
New and improved coding challenges
New comment UI experiment graduation
Policy: Generative AI (e.g., ChatGPT) is banned
Report this ad
Report this ad
Linked
1Is 'If Else' statement indentation important or not in C++?
0I am trying to write a two sum function but I keep getting an extra value on my output
Related
2Advanced IF statement in C
7Function call in if statement without parentheses
4Why does this if statement require brackets?
3If statement without condition
0else if & curly brackets
0Working of Nested IF-Else Without the Braces
0Why must this 'if' block with a single statement require a bracket to work properly?
1if statement in C
0Why no brackets needed among conditions in if sentence?
3If statement without curly braces
Hot Network Questions
My dissertation is wrong, but I already defended. How to remedy?
Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation
Does the curvature engine's wake really last forever?
Matthew 24:5 Many will come in my name!
Passengers on a flight vote on the destination, "It's democracy!"
RTC battery and VCC switching circuit
Does the Mishna or Gemara ever explicitly mention the second day of Shavuot?
Alternatives to Test-Driven Grading in an LLM world
ConTeXt: Unnecessary space in \setupheadertext
Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term?
Can you formalize the definition of infinitely divisible in FOL?
Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish?
What is a "non-reversible filter"?
Childhood book with a girl obsessessed with homonyms who adopts a stray dog but gives it back to its owners
A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man
Can a cleric gain the intended benefit from the Extra Spell feat?
Is direct sum of finite spectra cancellative?
What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel?
Cannot build the font table of Miama via nfssfont.tex
Is it ok to place components "inside" the PCB
Interpret G-code
Who is the target audience of Netanyahu's speech at the United Nations?
What meal can come next?
Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
lang-c
Why are you flagging this comment?
Probable spam.
This comment promotes a product, service or website while failing to disclose the author's affiliation.
Unfriendly or contains harassment/bigotry/abuse.
This comment is unkind, insulting or attacks another person or group. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Stack Overflow
Questions
Help
Chat
Products
Teams
Advertising
Talent
Company
About
Press
Work Here
Legal
Privacy Policy
Terms of Service
Contact Us
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings |
190766 | https://www.effortlessmath.com/wp-content/uploads/2020/09/One-Step-Inequalities.pdf?srsltid=AfmBOopAZ2DLoSR9L5FAdM8ruFoHq4xj8HylVHMBjV4WqUFKpLb8M9EK | Math Worksheets Name: ____
Date: _____
… So Much More Online! Please visit: www.EffortlessMath.com
One –Step Ine qualities
Solve inequalities.
1) 𝑥 + 8 ≥ 18
2) 𝑣 − 1 < 3
3) 𝑟 + 13 < 9
4) 𝑛 − 2 ≤ 4
5) 𝑝 + 8 > −4
6) 17 + 𝑘 ≤ 10
7) 𝑥 + 13 ≥ 5
8) 𝑥 − 12 < 11
9) 𝑥 − 5 ≤ 8
10 ) 𝑥 − 1 ≥ 2
11 ) 9𝑥 > −54
12 ) 4𝑥 ≤ 12
13 ) 𝑥 + 4 < 2
14 ) 𝑥 + 2 > 7
15 ) −11 < −9 + 𝑦
16 ) 7 ≥ 𝑥 − 5
17 ) − 4 + 𝑘 < 19
18 ) −9 + 𝑎 > 8
19 ) 12 > 𝑐 + 5
20 ) 1 + 𝑞 > −10
Graph each inequality.
21 ) 6𝑥 > − 24
22 ) − 2 < 𝑥
4
23 ) 𝑥
3 > 2
24 ) 4𝑥 < 8
25 ) 12 < 3𝑥
26 ) 𝑥
3 < 1
-4-1-2-3-5-6-7-8-9-10 12345678910 0
-4-1-2-3-5-6-7-8-9-10 12345678910 0
-4-1-2-3-5-6-7-8-9-10 12345678910 0
-4-1-2-3-5-6-7-8-9-10 12345678910 0
-4-1-2-3-5-6-7-8-9-10 12345678910 0
-4-1-2-3-5-6-7-8-9-10 12345678910 0
Math Worksheets Name: ____
Date: _____
… So Much More Online! Please visit: www.EffortlessMath.com
Answers
One –Step Inequalities
1) 𝑥 ≥ 10
2) 𝑣 < 4
3) 𝑟 < −4
4) 𝑛 ≤ 6
5) 𝑝 > −12
6) 𝑘 ≤ −7
7) 𝑥 ≥ −8
8) 𝑥 23
9) 𝑥 ≤ 13
10 ) 𝑥 ≥ 3
11 ) 𝑥 > −6
12 ) 𝑥 ≤ 3
13 ) 𝑥 < −2
14 ) 𝑥 > 5
15 ) – 2 < 𝑦
16 ) 12 ≥ 𝑥
17 ) 𝑘 < 23
18 ) 𝑎 > 17
19 ) 7 > 𝑐
20 ) 𝑞 > −11
21 )
22 )
23 )
24 )
25 )
26 ) |
190767 | http://www.360doc.com/content/24/0318/14/35070365_1117573149.shtml | 题目解答
搜索
我的图书馆
查看信箱
系统消息
官方通知
设置
开始对话
有 11 人和你对话，查看 忽略
历史对话记录
通知设置
发文章
发文工具
撰写网文摘手文档视频思维导图随笔相册原创同步助手
其他工具
图片转文字文件清理AI助手
留言交流
搜索
分享
QQ空间QQ好友新浪微博微信
生成长图转Word打印朗读全屏修改转藏+1
×
微信扫一扫关注
查看更多精彩文章
×
微信扫一扫
将文章发送给好友
题目解答
lhyfsxb8kc6ks9 2024-03-18 发布于河南来源|17阅读|1 转藏
大 中 小
转藏全屏朗读打印转Word生成长图分享
QQ空间QQ好友新浪微博微信
展开全文
已知函数f（x）=x 2 -4x+a+3，g（x）=mx+5-2m．（Ⅰ）若y=f（x）在[-1，1]上存在零点，求实数a的取值范围；（Ⅱ）当a=0时，若对任意的x 1 ∈[1，4]，总存在x 2 ∈[1，4]，使f（x 1 ）=g（x 2 ）成立，求实数m的取值范围．
（Ⅰ）：因为函数f（x）=x 2 -4x+a+3的对称轴是x=2，所以f（x）在区间[-1，1]上是减函数，因为函数在区间[-1，1]上存在零点，则必有： f(1)≤0 f(-1)≥0 即 a≤0 a+8≥0 ，解得-8≤a≤0，故所求实数a的取值范围为[-8，0]．（Ⅱ）若对任意的x 1 ∈[1，4]，总存在x 2 ∈[1，4]，使f（x 1 ）=g（x 2 ）成立，只需函数y=f（x）的值域为函数y=g（x）的值域的子集．f（x）=x 2 -4x+3，x∈[1，4]的值域为[-1，3]，下求g（x）=mx+5-2m的值域．①当m=0时，g（x）=5-2m为常数，不符合题意舍去；②当m＞0时，g（x）的值域为[5-m，5+2m]，要使[-1，3]⊆[5-m，5+2m]，需 5-m≤-1 5+2m≥3 ，解得m≥6；③当m＜0时，g（x）的值域为[5+2m，5-m]，要使[-1，3]⊆[5+2m，5-m]，需 5+2m≤-1 5-m≥3 ，解得m≤-3；综上，m的取值范围为（-∞，-3]∪[6，+∞）．
817509
本站是提供个人知识管理的网络存储空间，所有内容均由用户发布，不代表本站观点。请注意甄别内容中的联系方式、诱导购买等信息，谨防诈骗。如发现有害或侵权内容，请点击 一键举报。
转藏分享
QQ空间QQ好友新浪微博微信
献花（0） +1
来自： lhyfsxb8kc6ks9>《高中数学》
举报/认领
上一篇：
裂项放缩，思考逻辑，王者段位，独步天下 #高中数学
下一篇：
题目解答
猜你喜欢
0 条评论
发表
请遵守用户评论公约
查看更多评论
类似文章更多
设f（x）定义域为D，若满足（1）f（x）在D内是单调函数（2）存在[a，b]?D使f（x）在x∈[a，b]值域为[a，b]，则称f（x）为D上的闭函数．当f(x)=k+x+2为闭函数时，k的范...
设f（x）定义域为D，若满足（1）f（x）在D内是单调函数（2）存在[a，b]?D使f（x）在x∈[a，b]值域为[a，b]，则称f（x）为D上的闭函数．当f(x)=k+x+2为闭函数时，k的范围是题文。设f（x）定义域为D，若...
高中数学求函数定义域和值域专题训练含答案
[高中数学求函数定义域和值域专题训练含答案。1、?已知函数的定义域为，值域是，则的值域是，的定义域是??．.（2）若函数的定义域为，求函数的定义域；1、(1)函数的定义域为{-1,0,1,2,3},则f(-1)=(-1)-...
高中数学：函数学习中几对易混问题
高中数学：函数学习中几对易混问题。而值域为R应转化为函数的值域包含，即函数值取遍所有的正数。（2）令，则函数，的值域为取遍一切正实数值是值域的子集。而值域为，是指自变量x在定义域内取一切值时...
第2练 不等式（解析版）-2023年高考一轮复习精讲精练必备
对于D，由，得，，，而函数在R上单调递增，因此，，D不正确.故选：AC12．已知正数a，b满足， 则（?）A．的最大值是B．的最大值是C．的最小值是D．的最小值为【答案】ABD【详解】由得，当且仅当时取等，...
12月26日:函数的概念
【查看答案】【答案】C【解析】 A．这两个函数的定义域不同,所以不是同一函数; B．这两个函数的定义域不同,所以不是同一函数; C．这两个函数的定义域、值域与对应关系均相同,所以是同一函数;...
新课标高中数学（必修1）第二章：基本初等函数1（提高训练）题
第二章基本初等函数（1）1．函数上的最大值和最小值之和为，2．已知在上是的减函数，的取值范围是()4．设函数，则的值为（）5．定义在上的任意函数都可以表示成一个奇函数与一个.1．若函数的定义域为，...
用判别式求取值范围合理吗？
求函数的值域，以及更一般代数式的取值范围的问题，很多人常常会用到一元二次方程的判别式.一个简单的例子是函数的值域，如果用判别式来...
高中数学
(1)、二次函数在定轴定区间的取值范围。【例题4】函数 y = x^2 - 2x + 3 在闭区间 [0,2] 上取值范围为 .【例题5】已知函数 y = x^2 - 2x...
1.3.1函数的单调性
1.3.1函数的单调性。以上是通过观察图象的方法来说明函数在某一区间的单调性，是一种比较粗略的方法，那么，对于任给函数，我们怎样根据增减函数的定义来证明它的单调性呢？变式训练3：在经济学中，函...
个图VIP年卡，限时优惠价168元>>x
lhyfsxb8kc6ks9
关注对话
TA的最新馆藏
如何快速记住泰勒展开式？ 三角函数，图像法求解交点个数 导数高手必修课：极值点偏移问题 导数21 利用恒成立求最值 #利用恒成立求最值 #导数 #高考数学 高一数学培优——基本不等式7大拓展方法总结 高中老师不讲 但是考试一定会考的两个函数
喜欢该文的人也喜欢更多
学会这几个公式，夸人就会变得超级简单阅73
人生的本质意义是经历、体验、试错、感悟阅124
摄影：游南京栖霞山景区【4】阅173
聚焦伊能静三伏天的瘦身方案，顺应天时，兼顾健康与身材管理阅142
黄君璧《仿古山水册》：古韵新声中的时空对话阅80
热门阅读换一换
商业银行关联方到底该如何认定？（图解+案例）阅11778
新人教版八年级物理下册全册教案阅14639
飞花令“春”的诗句100首，含“春”字的诗句大全阅40916
小学英语新课程标准（2011版）阅81758
《易经》全文及详解阅117270
复制
打印文章
发送到手机微信扫码，在手机上查看选中内容
全屏阅读
朗读全文
分享文章QQ空间QQ好友新浪微博微信
AI解释
复制
打印文章
adsbygoogle.js 发送到手机微信扫码，在手机上查看选中内容
全屏阅读
朗读全文
AI助手
阅读时有疑惑？点击向AI助手提问吧
联系客服
在线客服：
360doc小助手2
客服QQ：
1732698931
联系电话：4000-999-276
客服工作时间9:00-18:00，晚上非工作时间，请在QQ留言，第二天客服上班后会立即联系您。 |
190768 | https://ida.mtholyoke.edu/bitstream/10166/695/1/181.pdf | Seidel Switching Michelle A. Lastrina A thesis presented to the faculty of Mount Holyoke College in partial fulfillment of the requirements for the degree of Bachelor of Arts with Honors.
Department of Mathematics South Hadley, Massachusetts May 28, 2006 Acknowledgements I would like to thank Profs. Giuliana Davidoffand Harriet Pollatsek for all of their guidance throughout this project, it has been greatly appreciated. Special thanks to Prof. Robert Perlis of Louisiana State University for introducing me to this topic while participating in a Mathematics REU during the summer of 2005. Thanks to Prof. Gregory Quenell of the State Unviersity of New York, Plattsburgh for providing a wonderful starting point with . Thanks to the Mount Holyoke College Mathematics Department for all of their support and the wonderful education they have provided me with. I would also like to thank my friends and family for their support and encouragement.
Abstract Seidel switching is a technique for generating pairs of graphs that are cospectral 1but not necessarily isomorphic.
We will discuss and prove some important properties related to this graph construction.
Cospectral pairs of regular graphs are rarer than cospectral pairs of non-regular graphs. As a result, after looking at how to construct graphs via the Seidel technique, we will look specifically at generating regular pairs of graphs via the Seidel swtiching technique.
1In the term isospectral is used, while other sources use cospectral. Since we will be referring to pairs of graphs, the term cospectral will be used for consistency.
Contents 1 Introduction 5 2 Definitions 6 3 Eigenvalues and k-walks 9 4 An introduction to Seidel switching 13 5 Seidel switching generates cospectral graphs 14 6 Seidel switching and regularity 23 7 Seidel switching and the Ihara zeta function 30 8 Small, regular Seidel pairs 31 9 Further questions 32 A All 3-walks in ΓA and ΓB (Combinatorial Proof) 33 B Symmetric matrices have real eigenvalues 47 B.1 Schur’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . .
47 B.2 Spectral Theorem for Hermitian Matrices . . . . . . . . . . . .
50 B.3 Fundamental Theorem of Real Symmetric Matrices . . . . . .
51 5 1 Introduction We will be exploring a topic in graph theory known as Seidel switching. It was first introduced in as part of a discussion on equilateral point sets in elliptic space and later explored by Robert Brooks and Gregory Quenell. In , Gregory Quenell explored Seidel switching in order to answer the question “can you hear the shape of a graph?” In other words, does the spectrum of a graph determine the structure (or shape) of a graph? In proving that Seidel switching generates cospectral graphs that are not necessarily isomorphic, we can see that the answer to this question is no, graph structure is not determined by its spectrum. This question is directly related to the question posed by Marc Kac in 1966, “can you hear the shape of a drum?” This question explores whether or not a manifold can be determined uniquely by the spectrum. We will be taking the work of Quenell in and expanding on it. Providing proofs of facts and theorems stated, as well as specific examples of facts mentioned throughout.
We will begin by introducing background information to be used throughout.
This includes definitions, theorems and proofs that will be important to our results. We denote graphs as G or Γ. These graphs are undirected, may contain more than one component and we allow for loops and multiple edges. We will 6 then describe the Seidel switching construction and work through a proof of the fact that graphs constructed via the Seidel technique are cospectral by providing a nontrivial example illustrating the ideas presented. We will also look at the construction with a restriction on it to generate non-isomorphic pairs of regular graphs that are cospectral. We will discuss a conjecture that arises naturally from certain examples and will consider a function, called the Ihara zeta function, that gives us information about the graphs. Throughout we will discuss various properties that arise from Seidel switching and provide specific examples to illustrate these.
7 2 Definitions Before beginning our discussion of Seidel switching, we introduce some defin-itions that will be helpful.
Definition 2.1. A graph G = (V, E) is a collection of vertices vi ∈V and edges ϵi ∈E such that each edge has a vertex at each of its endpoints for edges ϵi = {vj, vk} where vj, vk ∈V .
Definition 2.2. The degree, or valency, of vertex v in a graph G is the number of edges incident to v, where each loop at v is counted twice.
Definition 2.3. G is regular if each vertex is of the same degree, and q-regular if the common degree is q.
Definition 2.4. For a graph G, a walk from vertex v to vertex w is a sequence of edges ϵ1ϵ2ϵ3...ϵn such that the initial vertex of ϵ1 is v and the terminal vertex of ϵn is w where the terminal vertex of ϵj is the initial vertex of ϵj+1. A closed walk occurs when the initial vertex is the same as the terminal vertex of the walk. We refer to a walk such that n=k as a k-walk.
Definition 2.5. A prime walk C is a closed walk without backtracking or a 8 tail (backtracking that occurs at the last step) and there does not exist a walk B such that C = Bk for k > 1.
Definition 2.6. The adjacency matrix A of G is the symmetric n × n matrix in which entry aij = aji is the number of edges in G with endpoints (vi, vj). (Note that the diagonal entry aii is twice the number of loops at vertex i. This is because for an undirected graph, a loop can be traversed in either of two directions.) Definition 2.7. The set of eigenvalues of A is the graph spectrum of G.
Definition 2.8. The length spectrum of G is the sequence l0(G), l1(G), l2(G)...
where lk(G) is the number of closed k-walks in G for integers k ≥0.
Definition 2.9. Isospectral graphs are graphs with the same graph spectrum, or their adjacency matrices have the same eigenvalues.
Definition 2.10. If a pair of graphs are isospectral, they are called cospectral graphs .
Definition 2.11. Graphs are length isospectral if they have the same length spectrum.
9 Definition 2.12. A complete graph Kn is a graph where each pair of ver-tices is connected by an edge. Kn will have n vertices, n(n+1) 2 edges and is regular of degree n −1.
Definition 2.13. Two graphs G1 and G2 are said to be isomorphic if there is a 1-1 and onto mapping f : V1 →V2 such that {vi, vj} ∈E1 if and only if {f(vi), f(vj)} ∈E2.
Definition 2.14. The automorphism group, Aut(G), of a graph G is the group of isomorphisms from G to G.
Definition 2.15. The symmetric group Sn of degree n is the group of all permutations on n elements. Sn is therefore a permutation group of order n!.
We note that every group of order n is isomorphic to a subgroup of Sn.
Definition 2.16. The dihedral group D4 is the group generated by the per-mutations (1234) and (13) and corresponds to the symmetries of the square.
Definition 2.17. A graph Γ is md2, or minimal degree 2, provided the degree of each vertex is at least 2.
Hence such a Γ is not a tree and does not contain any vertices of degree zero.
Definition 2.18. The Ihara zeta function of a regular graph Γ is given by 10 the following infinite product ZΓ(u) = Y primes C∈Γ (1 −udegC)−1 for prime walks C of Γ. Where deg C is the number of edges in C.
We also have the following: Theorem 2.19. Ihara’s Theorem: If Γ is a connected (q +1)-regular graph with adjacency matrix A, n vertices and e edges, then the Ihara zeta function is defined as the following rational function ZΓ(u) = (1 −u2)n−e det(I −Au + Qu2).
Here Q = qI is the diagonal matrix where d is the degree of the vertices.2 For a proof of Ihara’s Theorem see pp.417-418.
2We use Q = qI because Γ is regular. For a non-regular graph, Q is the diagonal matrix such that each aii along the diagonal is equal to the degree q + 1i minus 1 at each vertex vi of Γ.
11 3 Eigenvalues and k-walks The theorems and corresponding proofs of this section are an important part of to show that two graphs have the same length spectrum if and only if they have the same spectrum. This will then be used in the proof that Seidel switching produces cospectral graphs. One conclusion that will we be using in this section is the fact that because A is symmetric, its eigenvalues will be real. For a proof of this, we refer you to section B of the appendix.
Theorem 3.1. Let A = (aij) be the adjacency matrix of an undirected graph G. For each integer k ≥0, the number of k-walks from vi to vj in G is equal to [Ak]ij, the ijth entry of the kth power of A.
Proof. We will prove this theorem using induction.
Let G be an undirected graph with vertices v1, v2, ..., vn. Let A be the adja-cency matrix of G.
We will show that for all integers k ≥0, [Ak]ij = the number of k-walks from vi to vj for all vertices of G.
12 n = 1: A1 = A ⇒[A1]ij = [A]ij The entry aij equals the number of edges between vi and vj (by the definition of adjacency matrix), but this is the same as the number of 1-walks from vi to vj for i ̸= j.
We recall from earlier that for a loop at vi, we count two 1-walks from vi to vi since a loop contributes to the valency twice.
n = k −1: Assume [Ak−1]ij =the number of k −1-walks from vi to vj. (This is our induction hypothesis.) n = k: Assuming the induction hypothesis, we want to show that [Ak]ij = the number of k-walks from vi to vj.
Let A = (aij) and Ak−1 = (bij). So Ak = AAk−1 = Ak−1A and the ij entry of Ak can be found by multiplying the ith row of A by the jth column of Ak−1. Hence, [Ak]ij = ai1b1j + ai2b2j + ... + ainbnj for all i, j ∈{1, 2, ..., n}.
13 We will now look at each term of the above sum: ai1 = the number of edges between vi and v1 b1j = the number of k −1-walks between v1 and vj (by the induction hypothesis).
Clearly, any edge between vi and v1 can connect with any k −1-walk from v1 to vj and form a k-walk from vi to vj with v1 as its second vertex.
By multiplication, we now have ai1b1j = the number of k-walks from vi to vj in G with v1 as the second vertex of the walk.
We may now generalize this for all m ∈{1, 2, ..., n} to obtain the following: aimbmj = the number of k-walks from vi to vj in G such that vm is the second vertex of the walk.
For every k-walk from vi to vj there is a vertex of G that is the second vertex of the walk. Thus, the total number of k-walks from vi to vj is ai1b1j + ai2b2j + ... + ainbnj.
14 From earlier, we know that this is [Ak]ij. Thus, [Ak]ij = the number of k-walks from vi to vj in G.
We can now use this conclusion to prove the following theorem that will in-dicate the direct relationship between the spectrum and length spectrum of a graph G.
Theorem 3.2. Let G be a graph with adjacency matrix A and let λ1, λ2, ..., λN be the eigenvalues of A. For each integer k ≥0, the total number of closed k-walks in G is equal to λk 1 + λk 2 + ... + λk N.
Proof. The total number of closed k-walks in G equals the sum of the number of closed k-walks at each vertex vi in G. A closed k-walk at a vertex vi = [Ak]ii.
Let us recognize the following lemma and corresponding proof.
Lemma 3.3. For a diagonalizable matrix A and its corresponding diagonal matrix D = QAQ−1, tr(A)=tr(D).
Proof. We first show tr(AB) = tr(BA).
X i [AB]ii = X i X k aikbki 15 = X k X i bkiaik = X k [BA]kk We now apply tr(AB) = tr(BA) to get tr(QAQ−1) = tr(Q AQ−1) = tr(AQ−1 Q) = tr(A) This tells us, since the entries along the diagonal of a diagonal matrix are its eigenvalues, tr(D)=the sum of the eigenvalues= tr(A).
We now have the total number of closed k-walks in G = N X i=1 [Ak]ii = tr(Ak).
Since λi are eigenvalues of A, λk i are eigenvalues of Ak.
Thus, we have tr(Ak) = λk 1 + λk 2 + ... + λk N.
16 4 An introduction to Seidel switching Let us begin by explaining how to construct a pair of cospectral graphs using the Seidel switching technique. This can also be found in pp. 5-6.
Given the graphs G1 = (V1, E1) and G2 = (V2, E2) such that G2 is regular and has an even number of vertices, we create a set of edges E. The set E joins each vertex in V1 to half of the vertices in V2. There is no predetermined way to choose E. We also form the set of edges EC such that there is an edge in EC between a vertex in V1 and a vertex in V2 if and only if there is no edge between these vertices in E.
These two edge sets form two different graphs ΓA and ΓB such that ΓA = (VA, EA) = (V1 ∪V2, E1 ∪E2 ∪E) and ΓB = (VB, EB) = (V1 ∪V2, E1 ∪E2 ∪EC) We provide illustrations to show this construction more clearly. We begin with G1 and G2 as shown in Figure 4.1 on p.16a. Note that G2 has an even number of vertices and is 2-regular. We then introduce the sets E and EC in Figure 4.2 on p.16a. We can now construct the graphs ΓA and ΓB as shown in Figure 4.3 on p.16a. We see that in this example, ΓA and ΓB are non-isomorphic. To see 17 this, we can compare the location of the loops in ΓA and ΓB.
Definition 4.1. We can refer to the construction involving G1, G2, E, EC, ΓA, and ΓB as the Seidel pair (G1, G2, E).
Definition 4.2. (ΓA, ΓB) is a Seidel switch of (G1, G2).
The dependence on E is important because a different choice of the edge set E combined with G1 and G2 can create a different pair of ΓA and ΓB as the following example shows. While alludes to this fact by noting that E uniquely identifies a Seidel pair, it does not directly state the following: Example 4.3. Given graphs G1 and G2, there may exist more than one choice for E.
Figures 4.4-4.8 on pp.16b-c illustrate this example. We can see that we begin with the same G1 and G2 such that they both have four vertices and G1 and G2 are both 2-regular. However, we choose two different edge sets E, as shown, that generate two different Seidel switches of (G1, G2). In this case we produce two pairs of isomorphic graphs.
18 5 Seidel switching generates cospectral graphs Theorem 5.1. Graphs constructed via the Seidel technique are cospectral.
Section 4 of provides a combinatorial proof that shows graphs constructed via the Seidel technique are length isospectral and thus cospectral. We now outline this proof and illustrate it with the following nontrivial example.
We begin with the Seidel pair (G1, G2, E) as shown in Figures 5.1-5.3 on pp.18a-b. We then provide (v, w, k) A and B-patches as defined by .
For vertices v and w in V1 and a nonnegative integer k (for our example we will let v = vertex 1, w = vertex 2, and k = 3), we define a (v, w, k) A-patch to be the sequence of edges ϵ0ϵ1ϵ2...ϵkϵk+1 in which 1. ϵ0 ∈E and ϵk+1 ∈E 2. ϵ0 begins at vertex v and ϵk+1 ends at w 3. ϵi ∈E2 for i = 0, ...k 4. If ϵi ends at vertex u, then ϵi+1 begins at vertex u for i = 0, ..., k.
19 Thus, a (v, w, k) A-patch is a walk in ΓA from v to w where only the first and last edges are in E. We also have a (v, w, k) B-patch that is a walk from v to w in ΓB where only the first and last edges are in EC. We define this formally as a sequence of directed edges ϵ0ϵ1ϵ2...ϵkϵk+1 such that 1. ϵ0 ∈EC and ϵk+1 ∈EC 2. ϵ0 begins at vertex v and ϵk+1 ends at w 3. ϵi ∈E2 for i = 0, ...k 4. If ϵi ends at vertex u, then ϵi+1 begins at vertex u for i = 0, ..., k.
We provide examples of both an A-patch and B-patch that can be seen in Figures 5.2 and 5.3 on pp.18a-b. For our example, we can see that the walk 1 →8 →6 →7 →8 →4 in ΓA is an example of a (1, 4, 3) A-patch. The walk 1 →11 →(cw)11 →(cw)11 →(cw)11 →4 in ΓB is an example of a (1, 4, 3) B-patch. Here (cw) denotes clockwise move-ment around the loop at vertex 11.
20 We are then presented with the following lemma that is a significant part of the proof of the validity of Seidel switching: Lemma 5.2. Given a Seidel pair (G1, G2, E), for each pair v and w of vertices in V1 and each non-negative integer k, the number of (v, w, k) A-patches is equal to the number of (v, w, k) B-patches.
To show this we are first required to show the following result.
Lemma 5.3. Let G be an r-regular graph. Given a vertex v in G and a non-negative integer k, the number of k-walks in G which begin at v is rk. Also, the number of k-walks in G which end at v is rk.
We can see this is true by examining our example, let k = 3. We thus move on to prove Lemma 5.2. We begin by considering vertices v and w in G1 and partitioning the vertices of V2 as follows: V2 = V(v,A) ∪V(v,B) where V(v,A) is the set of vertices in V2 adjacent to v in ΓA and V(v,B) is the set of vertices in V2 that are adjacent to v in ΓB. We then create another partition in the same sense: V2 = V(w,A) ∪V(w,B) where the vertices in V(w,A) are adjacent to w in ΓA and the vertices in V(w,B) are adjacent to w in ΓB. For our example, the partitions are as follows: 21 • V(v,A) = vertices 6, 7, and 8 • V(v,B) = vertices 9, 10 and 11 • V(w,A) = vertices 8, 9, and 10 • V(w,B) = vertices 6, 7, and 11 We can also note that the following holds for our example, and is equal to 3: |V(v,A)| = |V(v,B)| = |V(w,A)| = |V(w,B)| = |V2| 2 .
We then partition all of the k-walks in G2 into four sets, WAA ∪WAB ∪WBA ∪WBB according to where their beginning and ending vertices lie. This is done as follows: • WAA contains k-walks beginning in V(v,A) and ending in V(w,A).
• WAB contains k-walks beginning in V(v,B) and ending in V(w,A).
• WBA contains k-walks beginning in V(v,A) and ending in V(w,B).
• WBB contains k-walks beginning in V(v,B) and ending in V(w,B).
For our example, we can now partition all 3-walks in G2.
22 WAA: We recall that these are the 3-walks beginning in V(v,A) and ending in V(w,A). Here we will have 3-walks beginning at vertices 6, 7, and 8 and ending at vertex 8, as listed below: • 6 →8 →6 →8 • 6 →8 →7 →8 • 6 →7 →6 →8 • 7 →8 →6 →8 • 7 →8 →7 →8 • 7 →6 →7 →8 • 8 →6 →7 →8 • 8 →7 →6 →8 Here we can see that |WAA| = 8.
WAB: All 3-walks in G2 beginning in V(v,B) and ending in V(w,A). The beginning vertices will be 9 and 10, and the ending vertices will be 9 and 10. Note that here we denote 9 →(a)10 and 9 →(b)10 to indicate the difference between the two edges connecting vertices 9 and 10.
23 • 9 →(a)10 →(b)9 →(b)10 • 9 →(a)10 →(b)9 →(a)10 • 9 →(a)10 →(a)9 →(b)10 • 9 →(a)10 →(a)9 →(a)10 • 9 →(b)10 →(b)9 →(b)10 • 9 →(b)10 →(b)9 →(a)10 • 9 →(b)10 →(a)9 →(b)10 • 9 →(b)10 →(a)9 →(a)10 • 10 →(a)9 →(b)10 →(b)9 • 10 →(a)9 →(b)10 →(a)9 • 10 →(a)9 →(a)10 →(b)9 • 10 →(a)9 →(a)10 →(a)9 • 10 →(b)9 →(b)10 →(b)9 • 10 →(b)9 →(b)10 →(a)9 • 10 →(b)9 →(a)10 →(b)9 24 • 10 →(b)9 →(a)10 →(a)9 |WAB| = 16 WBA: We have all 3-walks in G2 beginning in V(v,A) and ending in V(w,B). The beginning vertices will be 6, 7 and 8, and the ending vertices will be 6 and 7.
• 6 →8 →6 →7 • 6 →8 →7 →6 • 6 →7 →8 →6 • 6 →7 →8 →7 • 6 →7 →6 →7 • 7 →8 →6 →7 • 7 →8 →7 →6 • 7 →6 →7 →6 • 7 →6 →8 →6 • 7 →6 →8 →7 • 8 →6 →7 →6 25 • 8 →6 →8 →6 • 8 →6 →8 →7 • 8 →7 →6 →7 • 8 →7 →8 →6 • 8 →7 →8 →7 |WBA| = 16 WBB: All 3-walks from V(v,B) to V(w,B) beginning and ending at vertex 11.
Here we use 11 →(cw)11 and 11 →(ccw)11 to denote the difference between clockwise and counterclockwise movement around the loop at vertex 11.
• 11 →(cw)11 →(cw)11 →(cw)11 • 11 →(cw)11 →(cw)11 →(ccw)11 • 11 →(cw)11 →(ccw)11 →(cw)11 • 11 →(cw)11 →(ccw)11 →(ccw)11 • 11 →(ccw)11 →(cw)11 →(cw)11 • 11 →(ccw)11 →(cw)11 →(ccw)11 • 11 →(ccw)11 →(ccw)11 →(cw)11 26 • 11 →(ccw)11 →(ccw)11 →(ccw)11 |WBB| = 8 We can see that WAA ∪WAB contains all k-walks in G2 that begin at a vertex in V(v,A) and end in V2.
Since G2 is r-regular, in the case of our example 2-regular, and there are exactly |V2| 2 , or 3, vertices in V(v,A), we apply Lemma 5.3 to get |WAA| + |WAB| = |V(v,A)|rk = |V2| 2 rk, (5.1) For our example, 8 + 16 = 3 ∗23 = 6 2 ∗23 = 32.
(5.2) We also have that WAB ∪WBB contains all of the k-walks, or 3-walks, in G2 beginning in V2 and ending at a vertex in V(w,B). This gives us |WAB| + |WBB| = |V(w,B)|rk = |V2| 2 rk.
(5.3) Thus, we have, 16 + 8 = 3 ∗23 = 6 2 ∗23 = 32.
(5.4) From (5.1) and (5.3), we have |WAA| + |WAB| = |WAB| + |WBB|, (5.5) and it follows that |WAA| = |WBB|.
27 We can see that these equalities hold for our example.
To finish the proof of Lemma 5.2, we can see that each 3-walk in WAA corre-sponds to exactly one (v, w, 3) A-patch, and each 3-walk in WBB corresponds to exactly one (v, w, 3) B-patch.
The validity of the following corollary is a result of WAA and WBB having the same cardinality.
Corollary 5.4. Let v and w be vertices in V1 and k a non-negative integer.
Let PA (v,w,k) be the set of (v, w, k) A-patches from v to w and PB (v,w,k) be the set of (v, w, k) B-patches from v to w. Then there is a one-to-one correspondence PA (v,w,k) ↔PB (v,w,k).
We can now show that ΓA and ΓB of a Seidel pair are length isospectral.
Theorem 5.5. Let k be a non-negative integer and (G1, G2, E) a Seidel pair.
There is a one-to-one correspondence between the set of all closed k-walks in ΓA and the set of all closed k-walks in ΓB.
We divide every closed k-walk W in ΓA, recall Figure 6.5, into the following three types: • Type I: W is contained in G1 28 • Type II: W is contained in G2 • Type III: W contains some edges in E.
We do the same for all closed k-walks Z in ΓB, recall Figure 6.6: • Type I′: Z is contained in G1 • Type II′: Z is contained in G2 • Type III′: Z contains some edges in EC.
For a comprehensive list of all such walks in our example, see Section A of the Appendix.
For a closed k-walk, or 3-walk, W in ΓA, if W is Type I or Type II, then it is also a closed k-walk, or 3-walk, of Type I′ or Type II′ in ΓB. Thus we have a correspondence for these walks via the identity mapping.
The completion of the proof requires the construction of a bijection from Type III closed k-walks in ΓA to Type III′ closed k-walks in ΓB. Let W = ϵ1ϵ2...ϵk be a Type III closed k-walk in ΓA. For our example, we have W = ϵ1ϵ2ϵ3.
29 Consider W = {3, 1}, {1, 7}, {7, 3}.
Since W is closed and contains some edge in E, it contains some edge ϵm in E that goes from a vertex in V1 to a vertex in V2. By permuting the edges in W, we obtain W = ϵmϵm+1ϵm+2...ϵm+k−1 where we read the subscripts modulo k. This is a closed k-walk beginning and ending at some vertex v in V1. For our example, we have W = ϵmϵm+1ϵm+2 and W = {1, 7}, {7, 3}, {3, 1}, a closed 3-walk beginning and ending at vertex 1.
We can view W as the following sequence: W = P1W1P2W2...PjWj where each Pi is a (vi, wi, li) A-patch and each Wi is a walk in G1 from wi to vi+1, with subscripts modulo j. For our example, we have W = P1W1 30 such that P1 = (1, 3, 2) A-patch passing through vertex 7 in ΓA and W1 = a walk of length 1 from vertex 3 to 1.
By Corollary 5.4, we have a (vi, wi, li) B-patch P ′ i that corresponds to each Pi patch. Thus, we can replace each Pi in W with its corresponding P ′ i to get a new walk W ′ = P ′ 1W1P ′ 2W2...P ′ jWj.
For our example, we have P ′ 1 = (1, 3, 2) B-patch passing through vertex 11 in ΓB and W ′ = P ′ 1W1.
We now have a closed k-walk, or 3-walk, beginning and ending at v = vertex 1.
However, each ϵh of W contained in E has been replaced by ϵ′ h in EC. Therefore, W is a closed k-walk in ΓB. To complete this mapping, we take W ′ = ϵ′ mϵ′ m+1ϵ′ m+2...ϵ′ m+k−1 such that each ϵ′ h = ϵh or its corresponding edge in some P ′ i. Thus, we have W ′ = ϵ′ mϵ′ m+1ϵ′ m+2 31 such that ϵ′ m = {1, 11}, ϵ′ m+1 = {11, 3}, and ϵ′ m+2 = {3, 1} = ϵm+2.
We obtain a Type III′ closed 3-walk in ΓB by applying another cyclic permu-tation to the edges in W W ′ = ϵ′ 1ϵ′ 2...ϵ′ k W ′ = ϵ′ 1ϵ′ 2ϵ′ 3 W ′ = {3, 1}, {1, 11}, {11, 3} We may invert this mapping W 7→W ′ as follows: a closed 3-walk W ′ in ΓB contains some first edge ϵ′ m in E from V1 to V2.
W ′ = {3, 1}, {1, 11}, {11, 3} such that ϵ′ m = {1, 11} in EC from vertex 1 of G1 to vertex 11 of G2. We then apply a cyclic permutation to the edges in W ′ such that ϵ′ m is first W ′ = {1, 11}, {11, 3}, {3, 1} and replace each (vi, wi, li) B-patch with its corresponding A-patch which is known to exist by Corollary 5.4.
P ′ 1 = (1, 3, 2) A-patch through vertex 11 ↓ P ′ 1 = (1, 3, 2) B-patch through vertex 7.
32 Undoing the cyclic permutation so that the image of ϵ′ 1 is first is our last step W = {3, 1}, {1, 7}, {7, 3}.
We may verify this by observing the following of our example: • The first edge in W from V1 to V2 is in the same position as the first edge in W ′ from V1 to V2.
• Replacing a B-patch with an A-patch is the inverse of replacing an A-patch with a B-patch.
33 6 Seidel switching and regularity We shall now look at Seidel switching with a further restriction on the con-struction.
This is done because, although one might intuitively think the opposite is true, cospectral non-isomorphic pairs of regular graphs appear to be rarer than cospectral pairs of non-regular graphs. Thus we look at Seidel switching under the following restrictions for a regular Seidel pair provided in p.14.
Theorem 6.1. Suppose (G1, G2, E) is a Seidel pair, and that ΓA is q−regular.
Then 1. |V1| is even.
2. G1 is regular.
3.
|V1| 2 + r = |V2| 2 + s = q, where r is the valency of G2 and s is the valency of G1.
4. ΓB is q−regular.
We begin by providing an example of a regular construction. Figures 6.1-6.3 on p.33a show G1, G2, and a pair of isomorphic ΓA and ΓB that fit the conditions of Theorem 6.1. One can refer back to this example while working through 34 the following proof. A proof of Theorem 6.1 similar to the following can be found in .
Proof. As in all cases of Seidel switching, G2 is r-regular, so in ΓA each vertex from V2 will be an endpoint of the same number of edges in E. This number will be |E| |V2|. This is because in E edges will be distributed evenly for all vi ∈V2.
Using the previously described method of Seidel construction, we know that |E| = |V1| × |V2| 2 = |V1| × |V2| 2 .
So, the number of edges in E at each vertex in G2 will be |V1| 2 . This indicates that |V1| must be even and q = r + |V1| 2 .
(6.1) We now consider a vertex v of G1 with a valency of s. After constructing ΓA there are |V2| 2 new edges adjacent to v. This follows from the reasoning used above for G2. Thus, the valency of v as a vertex of ΓA is s + |V2| 2 .
We know that ΓA is q−regular. So, we can see that s = q −|V2| 2 , for each choice of v. Thus, we have that G1 is regular and q = s + |V2| 2 .
(6.2) 35 With this we have proved the third part of Theorem 6.1 in (6.1) and (6.2).
We now use a similar argument to prove that ΓB is q-regular.
Using the fact that EC is made up of edges that join each vertex of the graph G1 to exactly half of the vertices of the graph G2, it is clear that the valency of each vertex in the set V1 in the graph ΓB is s + |V2| 2 = q.
Now let v be a vertex of G2. We have |V1| 2 vertices in V1 that are adjacent to v via edges in E. Thus, there are also |V1| 2 vertices in V1 that are not adjacent to v via edges in E. These vertices make up the set of vertices that will be connected to v by edges in the set EC. Therefore, the valency of a vertex v in the graph ΓB is r + |V1| 2 = q.
Definition 6.2. When all of the conditions of Theorem 6.1 are met, we will refer to (G1, G2, E) as a regular Seidel pair.
In , Quenell indicates without example that conditions 1, 2 and 3 of Theorem 6.1 are not sufficient for the regularity of ΓA and ΓB. Here we provide an example illustrating this insufficiency. Figure 6.4 on p.35a gives G1 and G2 36 such that G1 has an even number of vertices, 6, and is 1-regular. We also see that |V1| 2 +r = 3+2 = 5 and |V2| 2 +s = 4+1 = 5, thus satisfying conditions 1, 2, and 3 of Theorem 6.1. However, once we add E as shown in Figure 6.5 on p.35a, we generate ΓA and ΓB that are non-regular. To see this, compare vertices 1, 7, and 10. We note that there do exist E that would induce regularity, but it is not guaranteed that such a E will be chosen.
Corollary 6.3. When the conditions of Theorem 6.1 are met, |V1| = |V2| if and only if r = s.
Proof. Consider |V1| 2 + r = |V2| 2 + s Then we have |V1| = |V2| ⇔ |V1| 2 + r = |V1| 2 + s ⇔ r = s Corollary 6.4. If (G1, G2, E) is a Seidel pair in which either G1 or G2 contains only two vertices, then ΓA is isomorphic to ΓB.
37 Quenell indicates in , without proof, that using the above and “a little more work”, the following can be shown using the idea that G1 must be the union of two K2s and the automorphism group of G1 is large enough to induce an isomorphism between ΓA and ΓB for any choice of E.
Corollary 6.5. If |V1| = |V2| = 4 and G1 is 1-regular, then the graphs ΓA and ΓB in any Seidel pair (G1, G2, E) are isomorphic.
We are able to give a proof as follows: Proof. Corollary 5.2 implies that G2 must also be 1-regular.
We now look to give a concrete proof.
We first notice that Theorem 5.1, Corollary 5.2 and our hypothesis imply that G2 must also be the union of two K2s. See Figure 6.6 on p.37a.
We provide four different choices of E and consider the mapping of vertices in VA to vertices in VB. Figures 6.7-6.18 on pp.37a-d show us G1 and G2 as well as these four different examples of E that generate pairs of isomorphic graphs.
Note the differences in location of vertices 1, 2, 3, and 4 in each ΓA and ΓB.
We determine the automorphism group of G1 as follows: • σ0 : 1 2 3 4 (1) • σ1 : 1 2 4 3 (34) 38 • σ2 : 2 1 3 4 (12) • σ3 : 2 1 4 3 (12)(34) • σ4 : 3 4 2 1 (13)(24) • σ5 : 3 4 1 2 (1324) • σ6 : 4 3 1 2 (1423) • σ7 : 4 3 2 1 (14)(23) Thus, Aut(G1) = {1, (12), (34), (12)(34), (13)(24), (1324), (1423), (14)(23)}.
This is D4, a subgroup of S4, corresponding to the symmetries of the square as shown in Figure 6.20 on p.38a.
We represent the vertices of G1 as the vertices of a square, as shown in Figure 6.19 on p.38a. We have two adjacent pairs corresponding to two diagonals of the square and four non-adjacent pairs corresponding to the four sides of the square.
We note that Aut(G1) is transitive on the two adjacent pairs and the four nonadjacent pairs. In fact, the following is clear: Proposition 6.6. For vertices {a, b, c, d} of G1, 39 1. If a and d are adjacent and b and c are adjacent in G1, then (ad)(bc) ∈ Aut(G1).
2. If a and d are nonadjacent and b and c are nonadjacent in G1, then (ad)(bc) ∈Aut(G1).
Given G1 and G2 as shown and labeled earlier, we let vertices of G1 = {1, 2, 3, 4} = {a, b, c, d} vertices of G2 = {5, 6, 7, 8} = {r, s, t, u} and think of the choice of E as a map φ : {1, 2, 3, 4} →pairs{56, 57, 58, 67, 68, 78} and EC ↔φC.
We can divide these mappings into two cases, |φ−1{r, s}| = 2 for some pair {r, s} and |φ−1{r, s}| ≤1 for all pairs {r, s}.
Case 1:|φ−1{r, s}| = 2 for some pair {r, s}. This forces the following: φ ΓA ΓB φ(a) = {r, s} ar,as∈E at,au∈EC φ(b) = {r, s} br,bs∈E bt,bu∈EC φ(c) = {t, u} ct,cu∈E cr,cs∈EC φ(d) = {t, u} dt,du∈E dr,ds∈EC 40 If a and b are adjacent, then c and d are also adjacent and if a and b are nonadjacent, then c and d are also nonadjacent. We can say that there exists σ ∈Aut(G1) such that σ{a, b} = {c, d}. This σ induces ΓA ≃ΓB.
Case 2: |φ−1{r, s}| ≤1 for all pairs {r, s}.
Suppose φ(a) = {r, s}.
Then, φ(b) = {r, t} φ(c) = {s, u} φ(d) = {t, u} because r and s cannot occur together again, so t and u must occur in combi-nation with r and s in some order. Thus, we have the following: ΓA ΓB ar,as∈E at,au∈EC br,bt∈E bs,bu∈EC cs,cu∈E cr,ct∈EC dt,du∈E dr,ds∈EC We look at Figures 6.21 and 6.22 on p.40a to see the differences in E and EC.
Claim: σ = (ad)(bc) ∈Aut(G1) induces ΓA ≃ΓB.
Case A: If a and d are adjacent then b and c are adjacent. This can be seen in Figure 6.23 on p.40b.
41 Case B: If a and d are nonadjacent then b and c are nonadjacent. There are two possibilities for this case, as shown in Figures 6.24 and 6.25 on p.40b.
After looking at various examples of Seidel pairs for different combinations of G1 and G2, the following conjecture emerged.
Conjecture: If G1, G2 are both regular and |V1| and |V2| are both even, then ΓA and ΓB will both be regular.
We look to Figures 6.26-6.28 on p.41a for an example illustrating the ideas of this conjecture.
There are, however, a few questions that arise from this conjecture: 1. Do G1 and G2 both need to have the same number of vertices?
2. Do G1 and G2 need to have the same valency?
We now attempt to answer these questions in an effort to support the conjec-ture and determine if it is indeed true.
1. The example shown in Figures 6.29 and 6.30 on p.41b shows that when G1 and G2 have a different number of vertices, our conjecture does not hold. This is clear from comparing the degrees of vertices 4 and 10. This counterexample shows us that the original conjecture is false, indicating a need for further 42 restriction on the hypothesis. Thus, we now provide a revised version of the conjecture to account for what we have just learned: Conjecture (Revised): If G1, G2 are both regular and |V1| = |V2| = 2N, then ΓA, ΓB will both be regular.
2. Figures 6.31-6.33 on p.42a give us a pair of G1 and G2 with different va-lencies. This indicates to us that the second condition we are questioning is not necessary. We have been able to prove a special case of the conjecture and also show that the resulting ΓA and ΓB are cospectral. The following theo-rem illustrates an example of the conjecture with the two mentioned possible restrictions.
Theorem 6.7. Given q-regular G1 and G2 such that |V1| = |V2| = 2N.
Then 1. ΓA and ΓB are (q + N)-regular.
2. ΓA and ΓB are cospectral.
Proof. Here we provide the proof of (1).
We are given the following: 43 |V1| = |V2| = 2N ΓA = (VA, EA) = (V1 ∪V2, E1 ∪E2 ∪E) ΓB = (VB, EB) = (V1 ∪V2, E1 ∪E2 ∪EC) |VA| = |VB| = 4N |E| = |EC| Since E is a set of edges connecting half of the vertices in G1 to every vertex in G2, we are adding N edges to each vertex upon construction of ΓA. The same occurs with the construction of ΓB using EC. Given that G1 and G2 are q-regular, we can now see that ΓA and ΓB are (q + N)-regular.
We refer you to section 5 for an illustrated example that works through the proof provided in of the fact that ΓA and ΓB will be cospectral for any Seidel switch; this is a special case.
This is also a special case of the regular Seidel pair described in Theorem 6.1.
44 7 Seidel switching and the Ihara zeta function We now return to the Ihara zeta function defined earlier and discuss a connec-tion between Seidel switching and two graphs with equal Ihara zeta functions.
Ihara defined the zeta function as a product over p-adic group elements. It was not until Serre looked at the product that the Ihara zeta function had a graph theoretic interpretation. Sunada, Hashimoto, Bass and others extended the theory.
Through the following theorem, Aubi Mellien was able to show a direct con-nection between the equality of Ihara zeta functions of two graphs and their cospectrality. Mellien, a student participating in the mathematics REU at LSU during the summer of 2001, developed this theorem while working with Robert Perlis.
Theorem 7.1. Aubi’s Theorem: Let Γ, Γ′ be regular md2 graphs. Then ZΓ(u) = ZΓ′(u) if and only if Γ and Γ′ are cospectral.
A proof of Aubi’s Theorem can be found on pp.22-24 of and is referred to as Theorem 3.1.2.
The next theorem follows from Aubi’s Theorem, as stated above, and the proven fact that ΓA and ΓB are cospectral (Theorem 5.1).
45 Theorem 7.2. For q-regular graphs G1 and G2 such that q ≥1, ZΓA(u) = ZΓB(u) where (ΓA, ΓB) is any Seidel switch of G1, G2.
46 8 Small, regular Seidel pairs One aspect of interest in regards to Seidel switching is the construction of regular Seidel pairs such that ΓA and ΓB are non-isomorphic. It is of particular interest to look at small examples. From , Lemma 8.1. There exist at least three regular Seidel pairs with ΓA not iso-morphic to ΓB in which |V1| = |V2| = 4 and G1 and G2 are both 2-regular.
In (p.17), Quenell provides one example of such a Seidel pair. Here we provide two more examples of Seidel pairs of this type.
Figures 8.1-8.3 on p.45a show us the Seidel pair given in . Figures 8.4-8.9 on p.45b-c then provide the two more non-isomorphic Seidel pairs we have found such that the described conditions have been met.
Thus we can now see that there are at least three regular Seidel pairs of this type.
47 9 Further questions While exploring the preceding results, many other questions have come up that would be of interest to pursue. We provide some of these questions below: • Is there a way to compute the number of possible edge sets that exist when given G1 and G2? (Excluding redundancy of isomorphisms) • Do there exist more generalizations of examples we have referred to?
• Are there any consistent results when G1 = G2?
• Can you determine when ΓA and ΓB will be isomorphic?
• Are there any general results when G1 has an odd number of vertices and is not regular?
• What happens when G1 and G2 both have an even number or vertices, but G1 is not regular?
• What happens when G1 is regular and has an odd number of vertices?
• Are there any general results when G1 and G2 have the same regularity versus different regularity?
• How often do symmetric Seidel pairs arise?
48 • What affects do the presence of loops in G1 and/or G2 have on ΓA and ΓB?
• Why is it that cospectral pairs of regular graphs are rarer than cospectral pairs of non-regular graphs?
49 A All 3-walks in ΓA and ΓB (Combinatorial Proof) We provide a comprehensive list of all closed 3-walks W and Z in ΓA and ΓB respectively such that they are divided into the three types as described.
Type I: all closed 3-walks such that W is contained in G1. There will be none containing vertices 4 or 5.
• 1 →2 →(a)3 →1 • 1 →2 →(b)3 →1 • 1 →3 →(a)2 →1 • 1 →3 →(b)2 →1 • 2 →1 →3 →(a)2 • 2 →1 →3 →(b)2 • 2 →(a)3 →1 →2 • 2 →(b)3 →1 →2 • 3 →1 →2 →(a)3 • 3 →1 →2 →(b)3 50 • 3 →(a)2 →1 →3 • 3 →(b)2 →1 →3 |Type I W| = 12 Type II: all closed 3-walks W in G2. These will contain vertices 6, 7, and 8 or vertex 11, but not vertices 9 or 10.
• 6 →7 →8 →6 • 6 →8 →7 →6 • 7 →6 →8 →7 • 7 →8 →6 →7 • 8 →6 →7 →8 • 8 →7 →6 →8 • 11 →(cw)11 →(cw)11 →(cw)11 • 11 →(cw)11 →(cw)11 →(ccw)11 • 11 →(cw)11 →(ccw)11 →(cw)11 • 11 →(cw)11 →(ccw)11 →(ccw)11 • 11 →(ccw)11 →(cw)11 →(cw)11 51 • 11 →(ccw)11 →(cw)11 →(ccw)11 • 11 →(ccw)11 →(ccw)11 →(cw)11 • 11 →(ccw)11 →(ccw)11 →(ccw)11 |Type II W| = 14 Type III: W contains some edges in E Let us divide these up by the vertex at which the 3-walk begins and end.
Vertex 1: • 1 →3 →7 →1 • 1 →3 →8 →1 • 1 →6 →7 →1 • 1 →6 →8 →1 • 1 →7 →3 →1 • 1 →7 →6 →1 • 1 →7 →8 →1 • 1 →8 →3 →1 • 1 →8 →6 →1 52 • 1 →8 →7 →1 Vertex 2: • 2 →(a)3 →9 →2 • 2 →(b)3 →9 →2 • 2 →9 →3 →(a)2 • 2 →9 →3 →(b)2 • 2 →9 →(a)10 →2 • 2 →9 →(b)10 →2 • 2 →10 →(a)9 →2 • 2 →10 →(b)9 →2 • 2 →11 →(cw)11 →2 • 2 →11 →(ccw)11 →2 Vertex 3: • 3 →1 →7 →3 • 3 →1 →8 →3 53 • 3 →(a)2 →9 →3 • 3 →(b)2 →9 →3 • 3 →7 →1 →3 • 3 →7 →8 →3 • 3 →8 →1 →3 • 3 →8 →7 →3 • 3 →9 →2 →(a)3 • 3 →9 →2 →(b)3 Vertex 4: • 4 →(a)5 →9 →4 • 4 →(a)5 →10 →4 • 4 →(b)5 →9 →4 • 4 →(b)5 →10 →4 • 4 →9 →5 →(a)4 • 4 →9 →5 →(b)4 54 • 4 →9 →(a)10 →4 • 4 →9 →(b)10 →4 • 4 →10 →5 →(a)4 • 4 →10 →5 →(b)4 • 4 →10 →(a)9 →4 • 4 →10 →(b)9 →4 Vertex 5: • 5 →(a)4 →9 →5 • 5 →(a)4 →10 →5 • 5 →(b)4 →9 →5 • 5 →(b)4 →10 →5 • 5 →9 →4(a) →5 • 5 →9 →4 →(b)5 • 5 →9 →(a)10 →5 • 5 →9 →(b)10 →5 55 • 5 →10 →4 →(a)5 • 5 →10 →4 →(b)5 • 5 →10 →(a)9 →5 • 5 →10 →(b)9 →5 • 5 →11 →(cw)11 →5 • 5 →11 →(ccw)11 →5 Vertex 6: • 6 →1 →7 →6 • 6 →1 →8 →6 • 6 →7 →1 →6 • 6 →8 →1 →6 Vertex 7: • 7 →1 →3 →7 • 7 →1 →6 →7 • 7 →1 →8 →7 56 • 7 →3 →1 →7 • 7 →3 →8 →7 • 7 →6 →1 →7 • 7 →8 →1 →7 • 7 →8 →3 →7 Vertex 8: • 8 →1 →3 →8 • 8 →1 →6 →8 • 8 →1 →7 →8 • 8 →3 →1 →8 • 8 →3 →7 →8 • 8 →6 →1 →8 • 8 →7 →1 →8 • 8 →7 →3 →8 Vertex 9: 57 • 9 →2 →(a)3 →9 • 9 →2 →(b)3 →9 • 9 →2 →10 →(a)9 • 9 →2 →10 →(b)9 • 9 →3 →(a)2 →9 • 9 →3 →(b)2 →9 • 9 →4 →(a)5 →9 • 9 →4 →(b)5 →9 • 9 →4 →10 →(a)9 • 9 →4 →10 →(b)9 • 9 →5 →(a)4 →9 • 9 →5 →(b)4 →9 • 9 →5 →10 →(a)9 • 9 →5 →10 →(b)9 • 9 →(a)10 →2 →9 58 • 9 →(a)10 →4 →9 • 9 →(a)10 →5 →9 • 9 →(b)10 →2 →9 • 9 →(b)10 →4 →9 • 9 →(b)10 →5 →9 Vertex 10: • 10 →2 →9 →(a)10 • 10 →2 →9 →(b)10 • 10 →4 →(a)5 →10 • 10 →4 →(b)5 →10 • 10 →4 →9 →(a)10 • 10 →4 →9 →(b)10 • 10 →5 →(a)4 →10 • 10 →5 →(b)4 →10 • 10 →5 →9 →(a)10 59 • 10 →5 →9 →(b)10 • 10 →(a)9 →2 →10 • 10 →(a)9 →4 →10 • 10 →(a)9 →5 →10 • 10 →(b)9 →2 →10 • 10 →(b)9 →4 →10 • 10 →(b)9 →5 →10 Vertex 11: • 11 →2 →11 →(cw)11 • 11 →2 →11 →(ccw)11 • 11 →5 →11 →(cw)11 • 11 →5 →11 →(ccw)11 • 11 →(cw)11 →2 →11 • 11 →(cw)11 →5 →11 • 11 →(cw)11 →(cw)11 →(cw)11 60 • 11 →(cw)11 →(cw)11 →(ccw)11 • 11 →(cw)11 →(ccw)11 →(cw)11 • 11 →(cw)11 →(ccw)11 →(ccw)11 • 11 →(ccw)11 →2 →11 • 11 →(ccw)11 →5 →11 • 11 →(ccw)11 →(cw)11 →(cw)11 • 11 →(ccw)11 →(cw)11 →(ccw)11 • 11 →(ccw)11 →(ccw)11 →(cw)11 • 11 →(ccw)11 →(ccw)11 →(ccw)11 |Type III W| = 128 Type I′ for 3-walks Z contained in G1 and Type II′ 3-walks Z in G2 will be the same as Type I and II W walks respectively. Thus, we now list all of the Type III′ 3-walks Z.
Type III′: Z is a 3-walk containing some edges in EC.
Vertex 1: • 1 →3 →10 →1 61 • 1 →3 →11 →1 • 1 →9 →(a)10 →1 • 1 →9 →(b)10 →1 • 1 →10 →3 →1 • 1 →10 →(a)9 →1 • 1 →10 →(b)9 →1 • 1 →11 →3 →1 • 1 →11 →(cw)11 →1 • 1 →11 →(ccw)11 →1 Vertex 2: • 2 →(a)3 →6 →2 • 2 →(b)3 →6 →2 • 2 →6 →3 →(a)2 • 2 →6 →3 →(b)2 • 2 →6 →7 →2 62 • 2 →6 →8 →2 • 2 →7 →6 →2 • 2 →7 →8 →2 • 2 →8 →6 →2 • 2 →8 →7 →2 Vertex 3: • 3 →1 →10 →3 • 3 →1 →11 →3 • 3 →(a)2 →6 →3 • 3 →(b)2 →6 →3 • 3 →6 →2 →(a)3 • 3 →6 →2 →(b)3 • 3 →10 →1 →3 • 3 →11 →1 →3 • 3 →11 →(cw)11 →3 63 • 3 →11 →(ccw)11 →3 Vertex 4: • 4 →(a)5 →6 →4 • 4 →(a)5 →7 →4 • 4 →(b)5 →6 →4 • 4 →(b)5 →7 →4 • 4 →6 →5 →(a)4 • 4 →6 →5 →(b)4 • 4 →6 →7 →4 • 4 →7 →5 →(a)4 • 4 →7 →5 →(b)4 • 4 →7 →6 →4 • 4 →11 →(cw)11 →4 • 4 →11 →(ccw)11 →4 Vertex 5: 64 • 5 →(a)4 →6 →5 • 5 →(a)4 →7 →5 • 5 →(b)4 →6 →5 • 5 →(b)4 →7 →5 • 5 →6 →4 →(a)5 • 5 →6 →4 →(b)5 • 5 →6 →7 →5 • 5 →6 →8 →5 • 5 →7 →4 →(a)5 • 5 →7 →4 →(b)5 • 5 →7 →6 →5 • 5 →7 →8 →5 • 5 →8 →6 →(a)5 • 5 →8 →6 →(b)5 • 5 →8 →7 →(a)5 65 • 5 →8 →7 →(b)5 Vertex 6: • 6 →2 →(a)3 →6 • 6 →2 →(b)3 →6 • 6 →2 →7 →6 • 6 →2 →8 →6 • 6 →3 →(a)2 →6 • 6 →3 →(b)2 →6 • 6 →4 →(a)5 →6 • 6 →4 →(b)5 →6 • 6 →4 →7 →6 • 6 →5 →(a)4 →6 • 6 →5 →(b)4 →6 • 6 →5 →7 →6 • 6 →5 →8 →6 66 • 6 →7 →2 →6 • 6 →7 →4 →6 • 6 →7 →5 →6 • 6 →8 →2 →6 • 6 →8 →5 →6 Vertex 7: • 7 →2 →6 →7 • 7 →2 →8 →7 • 7 →4 →(a)5 →7 • 7 →4 →(b)5 →7 • 7 →4 →6 →7 • 7 →5 →(a)4 →7 • 7 →5 →(b)4 →7 • 7 →5 →6 →7 • 7 →5 →8 →7 67 • 7 →6 →2 →7 • 7 →6 →4 →7 • 7 →6 →5 →7 • 7 →8 →2 →7 • 7 →8 →5 →7 Vertex 8: • 8 →2 →6 →8 • 8 →2 →7 →8 • 8 →5 →6 →8 • 8 →5 →7 →8 • 8 →6 →2 →8 • 8 →6 →5 →8 • 8 →7 →2 →8 • 8 →7 →5 →8 Vertex 9: 68 • 9 →1 →10 →(a)9 • 9 →1 →10 →(b)9 • 9 →(a)10 →1 →9 • 9 →(b)10 →1 →9 Vertex 10: • 10 →1 →3 →10 • 10 →3 →1 →10 • 10 →(a)9 →1 →10 • 10 →(b)9 →1 →10 Vertex 11: • 11 →1 →3 →11 • 11 →1 →11 →(cw)11 • 11 →1 →11 →(ccw)11 • 11 →3 →1 →11 • 11 →3 →11 →(cw)11 69 • 11 →3 →11 →(ccw)11 • 11 →4 →11 →(cw)11 • 11 →4 →11 →(ccw)11 • 11 →(cw)11 →1 →11 • 11 →(cw)11 →3 →11 • 11 →(cw)11 →4 →11 • 11 →(cw)11 →(cw)11 →(cw)11 • 11 →(cw)11 →(cw)11 →(ccw)11 • 11 →(cw)11 →(ccw)11 →(cw)11 • 11 →(cw)11 →(ccw)11 →(ccw)11 • 11 →(ccw)11 →1 →11 • 11 →(ccw)11 →3 →11 • 11 →(ccw)11 →4 →11 • 11 →(ccw)11 →(cw)11 →(cw)11 • 11 →(ccw)11 →(cw)11 →(ccw)11 70 • 11 →(ccw)11 →(ccw)11 →(cw)11 • 11 →(ccw)11 →(ccw)11 →(ccw)11 |Type III′ Z| = 128 71 B Symmetric matrices have real eigenvalues Our ultimate goal is to present the proof of the Fundamental Theorem of Real Symmetric Matrices. In order to do so, we must first present and provide the proofs of two other theorems that will be needed in this proof.
B.1 Schur’s Lemma We begin with the theorem known as Schur’s Lemma. This will be used in the proof of the next preparatory theorem whose proof we will be working through.
Theorem B.1. Let A be an n×n (complex) matrix. There is a unitary matrix U such that U −1AU is upper triangular.
Proof. We will prove Schur’s Lemma by induction.
If n = 1, then Schur’s Lemma is trivially true.
Let us assume this theorem holds for all m × m matrices such that 1 ≤m ≤ (n−1).
We will now show that it holds for an n × n matrix A.
Let λ1 be an eigenvalue of A, and let v1 be a corresponding unit eigenvector.
72 We know that A, and every complex matrix, will have at least 1 eigenvalue by the Fundamental Theorem of Algebra (Section 9.1 of ).
We now expand v1 so that it becomes a basis for Cn by choosing v2,...,vn so v1,...,vn is an orthonormal basis.
Using the Gram-Schmidt process we can then transform it into an orthonormal basis v1, v2, ..., vn.
Let U1 be the unitary matrix with jth column vector vj. The first column vector of AU1 will be Av1 = λ1v1.
We know the ith row vector of U ∗ 1 is v∗ i . Since the vectors vj are orthogonal, v∗ i vj = 0 for i ̸= j.
So the first column vector of U ∗ 1AU1 will be U ∗ 1(λ1v1) = λ1 0 : : 0 .
73 Using this, we can write U ∗ 1AU1 = λ1 ∗ ∗ ..
..
∗ 0 0 : A1 : 0 , (B.1) where A1 is an (n −1) × (n −1) submatrix.
Our induction hypothesis indicates there is an (n−1)×(n−1) unitary matrix C such that C∗A1C = B for an upper triangular matrix B.
Let U2 = 1 0 0 ..
..
0 0 0 : C : 0 .
(B.2) Since C is unitary, we can show that U2 is also a unitary matrix.
U ∗ 2U2 = 1 0 0 ..
..
0 0 0 : C∗ : 0 1 0 0 ..
..
0 0 0 : C : 0 = 1 0 0 ..
..
0 0 0 : C∗C : 0 74 = 1 0 0 ..
..
0 0 0 : I : 0 = I and thus U2 is unitary.
We now consider U = U1U2 and calculate U ∗U = U ∗ 2(U ∗ 1U1)U2 = U ∗ 2IU2 = U ∗ 2U2 = I.
This indicates that U is also unitary.
Using U ∗AU = U ∗ 2U ∗ 1AU1U2 (B.3) we can combine (2.3) with (2.1) and (2.2): U ∗AU = 1 0 ..
..
0 0 : C∗ : 0 λ1 ∗ ..
..
∗ 0 : A1 : 0 1 0 ..
..
0 0 : C : 0 = λ1 ∗ ..
..
∗ 0 : C∗A1 : 0 1 0 ..
..
0 0 : C : 0 75 = λ1 ∗ ..
..
∗ 0 : C∗A1C : 0 We obtain U ∗AU = λ1 ∗ ..
..
∗ 0 : B : 0 .
Since B is upper triangular, this tells us U ∗AU is upper triangular as well.
Now that we have completed this proof, we can use Schur’s Lemma in the proof of the Spectral Theorem for Hermitian Matrices. This next theorem will then be used in the proof of the Fundamental Theorem of Real Symmetric Matrices.
B.2 Spectral Theorem for Hermitian Matrices Theorem B.2. If A is a Hermitian matrix, there exists a unitary matrix U such that U −1AU is a diagonal matrix. Also, all eigenvalues of A are real.
Proof. We will begin by proving the first part of the theorem.
From Schur’s Lemma, we have that there is a unitary matrix U such that U −1AU is upper triangular.
76 Given that U is unitary, we see that U −1 = U ∗and U ∗U = I. Because A is Hermitian, A∗= A.
So, (U −1AU)∗= (U ∗AU)∗= U ∗A∗(U ∗)∗= U ∗AU = U −1AU.
Therefore, U −1AU is Hermitian as well.
The conjugate transpose of an upper triangular matrix will be a lower triangu-lar matrix. Knowing that U −1AU is Hermitian indicates it is both upper tri-angular and lower triangular, so it must be a diagonal matrix D: U −1AU = D.
Therefore, we now know that the matrix A is unitarily diagonalizable, U −1AU = D.
We can now show that the eigenvalues of A are real. Since D is a diagonal matrix, we know that the entries along it’s diagonal will be the eigenvalues of A.
We also know that D is Hermitian, so D = D∗ D = λ1 0 0 .
.
0 λ2 .
.
.
.
.
.
.
.
.
.
0 0 .
.
.
0 λn 77 D∗= λ∗ 1 0 0 .
.
0 λ∗ 2 .
.
.
.
.
.
.
.
.
.
0 0 .
.
.
0 λ∗ n Since D = D∗, λj = λ∗ j for j = 1, ..., n.
Therefore, the eigenvalues λ1, ..., λn must be real.
We are now able to use this important conclusion in the proof of the Fun-damental Theorem of Real Symmetric Matrices.
B.3 Fundamental Theorem of Real Symmetric Matrices Theorem B.3. Every n × n real symmetric matrix has n real eigenvalues (counted with algebraic multiplicity) and is diagonalizable by a real orthogonal matrix.
Proof. We know every real n × n symmetric matrix A is also Hermitian.
By the Spectral Theorem for Hermitian Matrices, A is diagonalizable with real eigenvalues λ1, ..., λn.
Therefore, A has n real eigenvalues counted with algebraic multiplicity.
We must count the eigenvalues with multiplicity as a result of the following theorem found on p.313 of .
78 Theorem B.4. A Criterion for Diagonalization: An n × n matrix A is diagonalizable if and only if the algebraic multiplicity of each (possibly complex) eigenvalue is equal to its geometric multiplicity.
From the Spectral Theorem for Hermitian Matrices we also know that A is unitarily diagonalizable. We will let it be diagonalized by the unitary matrix U.
By Theorem 5.2 of , Matrix Summary of Eigenvalues of A, and its proof (p.306), the column vectors of U will be eigenvectors of A.
We can determine these eigenvectors by row-reducing A −λiI for i = 1, ..., n.
We have already proved the λi are real. We will now row-reduce over the real numbers.
This tells us the row-reduced echelon form of A −λiI will have only real elements.
This will be best illustrated through an example.
Example: Let A = 1 0 1 0 2 0 1 0 1 We will first determine the eigenvalues of A and their corresponding row-79 reduced echelon forms of A −λiI.
The characteristic polynomial of A is: det(A −λI) = det 1 −λ 0 1 0 2 −λ 0 1 0 1 −λ = (2 −λ)[(1 −λ)2 −1] = −λ(λ −2)2 λ1 = 0, λ2 = λ3 = 2 Now that we have found the eigenvalues of A, we can determine the row-reduced echelon forms of the A−λiI and show all their entries are real numbers.
λ1 = 0 A −λ1I = 1 0 1 0 2 0 1 0 1 ∼ 1 0 1 0 1 0 0 0 0 λ2 = λ3 = 2 A −2 = −1 0 1 0 0 0 1 0 −1 ∼ 1 0 −1 0 0 0 0 0 0 80 Thus showing the row-reduced echelon forms of A −λiI contain only real entries.
Since A is diagonalizable, A −λiI in row-reduced echelon form will have dim(nullspace) = geometric multiplicity of λi geometric multiplicity of λi = algebraic multiplicity of λi.
If, for M = A−λiI, M is an n×n real matrix with rank M = k ≤n, then the nullspace of M is a subspace of Rn of dimension n−k. Therefore the nullspace of M has a basis consisting of vectors in Rn.
The Gram-Schmidt process lets us assume the basis of each eigenspace is orthonormal.
If we let vectors of these orthonormal bases of the eigenspaces form the columns of the matrices, C1, C2, ..., Cr, we obtain a real orthogonal matrix C that di-agonalizes A.
C = C1 C2 .
.
.
Cr 81 References Allan Clark. Elements of Abstract Algebra. Dover Publications, Mineola, NY, 1984.
Debra Czarneski. Zeta Functions of Finite Graphs. PhD Dissertation, Louisiana State University. August 2005.
Susanna S. Epp. Discrete Mathematics with Applications. PWS Publishing Company, Boston, second edition, 1995.
John B. Fraleigh and Raymond A. Bauregard Linear Algebra 3rd Edition.
1995: Addison-Wesley Publsihing Company, Reading, MA.
C.D. Godsill and B.D. McKay. Constructing Cospectral Graphs. Aequa-tiones Mathematicae, 25:257-268, 1982.
Robert Perlis. Finite Graphs and their Zeta Functions. Lecture at Louisiana State University, transcribed by Christopher Belford. Baton Rouge, 12 June 2003.
Gregory Quenell. The Combinatorics of Seidel Switching. Preprint, Decem-ber 16, 1997.
82 Audrey Terras. Fourier Analysis on Finite Groups and Applications. Lon-don Mathematical Society Student Texts 43. Cambridge University Press, New York, NY, 1999.
J.H. van Lint and J.J. Seidel. Equilateral point sets in elliptic geometry.
Proceedings. Koninklijke Nederlanse Academie van Wetenschppen, Series A, 69:335-348, 1966.
Eric W. Weisstein. ”Isospectral Graphs.” From MathWorld–A Wolfram Web Resource. Douglas B. West. Introduction to Graph Theory. Prentice Hall, Upper Saddle River, NJ, second edition, 2001. |
190769 | https://ocw.mit.edu/courses/18-226-probabilistic-methods-in-combinatorics-fall-2022/mit18_226_f22_lec_full.pdf | MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao Lecture notes (MIT 18.226) Probabilistic Methods in Combinatorics Yufei Zhao Massachusetts Institute of Technology MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao Last updated: June 18, 2024 These are lecture notes for a graduate class that I taught at MIT. The main textbook reference is Alon and Spencer, The Probabilistic Method, Wiley, 4ed.
Please report errors via the Google Form
Asymptotic notation convention We adopt the following asymptotic notation. Each line below has the same meaning for positive functions 𝑓and 𝑔as some parameter, usually 𝑛, tends to infinity: • 𝑓≲𝑔, 𝑓= 𝑂(𝑔), 𝑔= Ω( 𝑓), 𝑓≤𝐶𝑔(for some constant 𝐶> 0) • 𝑓/𝑔→0, 𝑓≪𝑔, 𝑓= 𝑜(𝑔) (and sometimes 𝑔= 𝜔( 𝑓)) • 𝑓= Θ(𝑔), 𝑓≍𝑔, 𝑔≲𝑓≲𝑔 • 𝑓∼𝑔, 𝑓= (1 + 𝑜(1))𝑔 • whp (= with high probability) means with probability 1 −𝑜(1) Warning. Analytic number theorists use ≪to mean 𝑂(·) (Vinogradov notation), differently from how the symbol is used in these notes.
Subscripts (e.g., 𝑂𝑠( ), ≲𝑠) are used to emphasize that the hidden constants may depend on the subscripted parameters. For example, 𝑓(𝑠, 𝑥) ≲𝑠𝑔(𝑠, 𝑥) means that for every 𝑠 there is some constant 𝐶𝑠so that 𝑓(𝑠, 𝑥) ≤𝐶𝑠𝑔(𝑠, 𝑥) for all 𝑥.
We write [𝑁] := {1, . . . , 𝑁}.
MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao Contents Contents 1 Introduction 1 1.1 Lower bounds to Ramsey numbers . . . . . . . . . . . . . . . . . . .
1 1.2 Set systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 1.3 2-colorable hypergraphs . . . . . . . . . . . . . . . . . . . . . . . .
10 1.4 List chromatic number of 𝐾𝑛,𝑛 . . . . . . . . . . . . . . . . . . . . .
12 2 Linearity of Expectations 17 2.1 Hamiltonian paths in tournaments . . . . . . . . . . . . . . . . . . .
17 2.2 Sum-free subset . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 2.3 Turán’s theorem and independent sets . . . . . . . . . . . . . . . . .
19 2.4 Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 2.5 Unbalancing lights . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 2.6 Crossing number inequality . . . . . . . . . . . . . . . . . . . . . . .
25 3 Alterations 29 3.1 Dominating set in graphs . . . . . . . . . . . . . . . . . . . . . . . .
29 3.2 Heilbronn triangle problem . . . . . . . . . . . . . . . . . . . . . . .
30 3.3 Markov’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 3.4 High girth and high chromatic number . . . . . . . . . . . . . . . . .
32 3.5 Random greedy coloring . . . . . . . . . . . . . . . . . . . . . . . .
33 4 Second Moment 37 4.1 Does a typical random graph contain a triangle? . . . . . . . . . . . .
37 4.2 Thresholds for fixed subgraphs . . . . . . . . . . . . . . . . . . . . .
42 4.3 Thresholds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46 4.4 Clique number of a random graph . . . . . . . . . . . . . . . . . . .
55 4.5 Hardy–Ramanujan theorem on the number of prime divisors . . . . .
57 4.6 Distinct sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61 4.7 Weierstrass approximation theorem . . . . . . . . . . . . . . . . . . .
63 5 Chernoff Bound 69 5.1 Discrepancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao Contents 5.2 Nearly equiangular vectors . . . . . . . . . . . . . . . . . . . . . . .
73 5.3 Hajós conjecture counterexample . . . . . . . . . . . . . . . . . . . .
75 6 Lovász Local Lemma 79 6.1 Statement and proof . . . . . . . . . . . . . . . . . . . . . . . . . . .
79 6.2 Coloring hypergraphs . . . . . . . . . . . . . . . . . . . . . . . . . .
83 6.3 Independent transversal . . . . . . . . . . . . . . . . . . . . . . . . .
89 6.4 Directed cycles of length divisible by 𝑘 . . . . . . . . . . . . . . . .
90 6.5 Lopsided local lemma . . . . . . . . . . . . . . . . . . . . . . . . . .
92 6.6 Algorithmic local lemma . . . . . . . . . . . . . . . . . . . . . . . .
97 7 Correlation Inequalities 107 7.1 Harris–FKG inequality . . . . . . . . . . . . . . . . . . . . . . . . . 107 7.2 Applications to random graphs . . . . . . . . . . . . . . . . . . . . . 110 8 Janson Inequalities 115 8.1 Probability of non-existence . . . . . . . . . . . . . . . . . . . . . . 115 8.2 Lower tails . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 8.3 Chromatic number of a random graph . . . . . . . . . . . . . . . . . 124 9 Concentration of Measure 129 9.1 Bounded differences inequality . . . . . . . . . . . . . . . . . . . . . 129 9.2 Martingales concentration inequalities . . . . . . . . . . . . . . . . . 130 9.3 Chromatic number of random graphs . . . . . . . . . . . . . . . . . . 135 9.4 Isoperimetric inequalities: a geometric perspective . . . . . . . . . . 139 9.5 Talagrand’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 152 9.6 Euclidean traveling salesman problem . . . . . . . . . . . . . . . . . 162 10 Entropy 173 10.1 Basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 10.2 Permanent, perfect matchings, and Steiner triple systems . . . . . . . 178 10.3 Sidorenko’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . 185 10.4 Shearer’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 11 Containers 201 11.1 Containers for triangle-free graphs . . . . . . . . . . . . . . . . . . . 203 11.2 Graph containers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 11.3 Hypergraph container theorem . . . . . . . . . . . . . . . . . . . . . 208 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1 Introduction The probabilistic method is an important technique in combinatorics. In a typical application, we wish to prove the existence of something with certain desirable prop-erties. To do so, we devise a random construction and show that it works with positive probability.
Let us begin with a simple example of this method.
Theorem 1.0.1 (Large bipartite subgraph) Every graph with 𝑚edges has a bipartite subgraph with at least 𝑚/2 edges.
Proof. Let the graph by 𝐺= (𝑉, 𝐸). Assign every vertex a color, randomly either black or white, uniformly and independently at random.
Let 𝐸′ be the set of edges with one black endpoint and one white endpoint. Then (𝑉, 𝐸′) is a bipartite subgraph of 𝐺.
Every edge belongs to 𝐸′ with probability 1/2. So by the linearity of expectation, the expected size of 𝐸′ is E[|𝐸′|] = 1 2 |𝐸| .
Thus there is some coloring with |𝐸′| ≥1 2 |𝐸|. Then (𝑉, 𝐸′) is the desired subgraph.
□ 1.1 Lower bounds to Ramsey numbers Ramsey number 𝑹(𝒌, ℓ) = smallest 𝑛such that in every red-blue edge coloring of 𝐾𝑛, there exists a red 𝐾𝑘or a blue 𝐾ℓ.
For example, 𝑅(3, 3) = 6 (every red/blue edge-coloring of 𝐾6 has a monochromatic triangle, but one can color 𝐾5 without any monochromatic triangle).
Ramsey (1929) proved that 𝑅(𝑘, ℓ) exists (i.e., is finite). This is known as Ramsey’s theorem.
1 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1 Introduction Paul Erdős (1913–1996) is considered the father of the probabilistic method.
He published around 1,500 papers during his lifetime, and had more than 500 collaborators. To learn more about Erdős, see his biography The man who loved only numbers by Hoffman and the documentary N is a number (You may be able to watch this movie for free on Kanopy using your local public library account).
Frank Ramsey (1903–1930) wrote seminal papers in philosophy, economics, and mathematical logic, before his untimely death at the age of 26 from liver problems.
See a recent profile of him in the New Yorker.
Finding quantitative estimates of Ramsey numbers (and its generalizations) is generally a difficult and often fundamental problem in Ramsey theory.
Remark 1.1.1 (Hungarian names). Many Hungarian mathematicians, notable due to Erdős’ influence, made foundational contributions to this field. So we will encounter many Hungarian names in this subject.
How to type “Erdős” in L AT EX: Erd\H{o}s (incorrect: Erd\"os, which produces “Erdös”) How to pronounce Hungarian names: Hungarian spelling Sounds like Examples s sh Erdős, Simonovits sz s Szemerédi, Lovász Erdős’ original proof One of the earliest application of the probabilistic method in combinatorics given by Erdős in his seminal paper: P. Erdős, Some remarks on the theory of graphs, Bull. Amer. Math. Soc, 1947.
2 Image courtesy of Kmhkmh. Source: Wikimedia Commons. License CC BY.
© Cambridge Wittgenstein archive. All rights reserved. This content is excluded from our Creative Commons license. For more information, see
MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1.1 Lower bounds to Ramsey numbers Here is the main result of this paper.
Theorem 1.1.2 (Lower bound to Ramsey numbers; Erdős 1947) If 𝑛 𝑘 21−(𝑘 2) < 1, then 𝑅(𝑘, 𝑘) > 𝑛.
In other words, there exists a red-blue edge-coloring of 𝐾𝑛with no monochromatic 𝐾𝑘.
In the proof below, we will apply the union bound: for events 𝐸1, . . . , 𝐸𝑚, P(𝐸1 ∪· · · ∪𝐸𝑚) ≤P(𝐸1) + · · · + P(𝐸𝑚).
We usually think of each 𝐸𝑖as a “bad event” that we are trying to avoid.
Proof. Color edges of 𝐾𝑛with red or blue independently and uniformly at random.
For every fixed subset 𝑆of 𝑘vertices, let 𝐴𝑆denote the event that 𝑆induces a monochromatic 𝐾𝑘, so that P(𝐴𝑆) = 21−(𝑘 2). Then, by the union bound, P(there is a monochromatic 𝐾𝑘) = P © « Ø 𝑆∈([𝑛] 𝑘) 𝐴𝑆 ª ® ® ¬ ≤ ∑︁ 𝑆∈([𝑛] 𝑘) P(𝐴𝑆) = 𝑛 𝑘 21−(𝑘 2) < 1.
Thus, with positive probability, the random coloring gives no monochromatic 𝐾𝑘. So there exists some coloring with no monochromatic 𝐾𝑘.
□ Remark 1.1.3 (Quantitative bound). By optimizing 𝑛as a function of 𝑘in the theorem above (using Stirling’s formula), we obtain 𝑅(𝑘, 𝑘) > 1 𝑒 √ 2 + 𝑜(1) 𝑘2𝑘/2.
Erdős’ 1947 paper actually was phrased in terms of counting: of all 2(𝑛 2) possible colorings, the total number of bad colors is strictly less than 2(𝑛 2).
In this course, we mostly consider finite probability spaces. While in principle the finite probability arguments can be rephrased as counting, some of the later more involved arguments are impractical without a probabilistic perspective.
Remark 1.1.4 (Constructive lower bounds). The above proof only gives the existence of a red-blue edge-coloring of 𝐾𝑛without monochromatic cliques. Is there a way to find algorithmically find one? With an appropriate 𝑛, even though a random coloring achieves the goal with very high probability, there is no efficient method (in polynomial running time) to certify that any specific edge-coloring avoids monochromatic cliques.
3 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1 Introduction So even though there are lots of Ramsey colorings, it is hard to find and certify an actual one. This difficulty has been described as finding hay in a haystack.
Finding constructive lower bounds is a major open problem. There was major progress on this problem stemming from connections to randomness extractors in computer science (e.g., Barak et al. 2012, Chattopadhyay & Zuckerman 2016, Cohen 2017) Remark 1.1.5 (Ramsey number upper bounds). Although Ramsey proved that Ram-sey numbers are finite, his upper bounds are quite large. Erdős–Szekeres (1935) used a simple and nice inductive argument to show 𝑅(𝑘+ 1, ℓ+ 1) ≤ 𝑘+ ℓ 𝑘 .
For diagonal Ramsey numbers 𝑅(𝑘, 𝑘), this bound has the form 𝑅(𝑘, 𝑘) ≤(4−𝑜(1))𝑘.
Recently, in a major and surprising breakthrough, Campos, Griffiths, Morris, and Sahasrabudhe (2023+) show that there is some constant 𝑐> 0 so that for all sufficiently large 𝑘, 𝑅(𝑘, 𝑘) ≤(4 −𝑐)𝑘.
This is the first exponential improvement over the Erdős–Szekeres bound.
Alteration method Let us give another argument that slightly improves the earlier lower bound on Ramsey numbers.
Instead of just taking a random coloring and analyzing it, we first randomly color, and then fix some undesirable features. This is called the alteration method (sometimes also the deletion method).
Theorem 1.1.6 (Ramsey lower bound via alteration) For any 𝑘, 𝑛, we have 𝑅(𝑘, 𝑘) > 𝑛− 𝑛 𝑘 21−(𝑘 2).
Proof. We construct an edge-coloring of a clique in two steps: (1) Randomly color each edge of 𝐾𝑛with red or blue (independently and uniformly at random); (2) Delete a vertex from every monochromatic 𝐾𝑘.
The process yields a 2-edge-colored clique with no monochromatic 𝐾𝑘(since the second step destroyed all monochromatic cliques).
4 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1.1 Lower bounds to Ramsey numbers Let us now analyze how many vertices we get at the end.
Let 𝑋be the number of monochromatic 𝐾𝑘’s in the first step.
Since each 𝐾𝑘is monochromatic with probability 21−(𝑘 2), by the linearity of expectations, E𝑋= 𝑛 𝑘 21−(𝑘 2).
In the second step, we delete at most |𝑋| vertices (since we delete one vertex from every clique). Thus final graph has size ≥𝑛−|𝑋|, which has expectation 𝑛− 𝑛 𝑘 21−(𝑘 2).
Thus with positive probability, the remaining graph has ≥𝑛− 𝑛 𝑘 21−(𝑘 2) vertices (and no monochromatic 𝐾𝑘by construction).
□ Remark 1.1.7 (Quantitative bound). By optimizing the choice of 𝑛in the theorem, we obtain 𝑅(𝑘, 𝑘) > 1 𝑒+ 𝑜(1) 𝑘2𝑘/2, which improves the previous bound by a constant factor of √ 2.
Lovász local lemma Often we wish to avoid a set of “bad events” 𝐸1, . . . , 𝐸𝑛. Here are two easy extremes: • (Union bound) If Í 𝑖P(𝐸𝑖) < 1, then union bound tells us that we can avoid all bad events.
• (Independence) If all bad events are independent, then the probability that none of 𝐸𝑖occurs is Î𝑛 𝑖=1(1 −P(𝐸𝑖)) > 0 (provided that all P(𝐸𝑖) < 1).
What if we are in some intermediate situation, where the union bound is not good enough, and the bad events are not independent, but there are only few dependencies?
The Lovász local lemma provides us a solution when each event is only independent with all but a small number of other events.
Here is a version of the Lovász local lemma, which we will prove later in Chapter 6.
5 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1 Introduction Theorem 1.1.8 (Lovász local lemma — random variable model) Let 𝑥1, . . . , 𝑥𝑁be independent random variables. Let 𝐵1, . . . , 𝐵𝑚⊆[𝑁]. For each 𝑖, let 𝐸𝑖be an event that depends only on the variables indexed by 𝐵𝑖(i.e., 𝐸𝑖is allowed to depend only on {𝑥𝑗: 𝑗∈𝐵𝑖}).
Suppose, for every 𝑖∈[𝑚], 𝐵𝑖has non-empty intersections with at most 𝑑other 𝐵𝑗’s, and P[𝐸𝑖] ≤ 1 (𝑑+ 1)𝑒.
Then with positive probability, none of the events 𝐸𝑖occur.
Here 𝑒= 2.71 · · · is the base of the natural logarithm. This constant turns out to be optimal in the above theorem.
Using the Lovász local lemma, let us give one more improvement to the Ramsey number lower bounds.
Theorem 1.1.9 (Ramsey lower bound via local lemma; Spencer 1977) If 𝑘 2 𝑛 𝑘−2 + 1 21−(𝑘 2) < 1/𝑒, then 𝑅(𝑘, 𝑘) > 𝑛.
Proof. Color the edges of 𝐾𝑛with red/blue uniformly and independently at random.
For each 𝑘-vertex subset 𝑆, let 𝐸𝑆be the event that 𝑆induces a monochromatic 𝐾𝑘. So P[𝐸𝑆] = 21−(𝑘 2).
In the setup of the local lemma, we have one independent random variable correspond-ing to each edge. Each event 𝐸𝑆depends only on the variables corresponding to the edges in 𝑆.
If 𝑆and 𝑆′ are both 𝑘-vertex subsets, their cliques share an edge if and only if |𝑆∩𝑆′| ≥2. So for each 𝑆, there are at most 𝑘 2 𝑛 𝑘−2 choices 𝑘-vertex sets 𝑆′ with |𝑆∩𝑆′| ≥2. So the local lemma applies provided that 21−(𝑘 2) < 1 𝑒 1 𝑘 2 𝑛 𝑘−2 + 1 .
So with positive probability none of the events 𝐸𝑆occur, which means an edge-coloring with no monochromatic 𝐾𝑘’s.
□ Remark 1.1.10 (Quantitative lower bounds). By optimizing the choice of 𝑛, we obtain 𝑅(𝑘, 𝑘) > √ 2 𝑒+ 𝑜(1) !
𝑘2𝑘/2 6 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1.2 Set systems once again improving the previous bound by a constant factor of √ 2. This is the best known lower bound to 𝑅(𝑘, 𝑘) to date.
1.2 Set systems In extremal set theory, we often wish to understand the maximum size of a set system with some given property. A set system F is a collection of subsets of some ground set, usually [𝑛]. That is, each element of F is a subset of [𝑛]. We will see some classic results from extremal set theory each with a clever probabilistic proof.
Sperner’s theorem We say that a set family F is an antichain if no element of F is a subset of another element of F (i.e., the elements of F are pairwise incomparable by containment).
Question 1.2.1 What is the maximum number of sets in an antichain of subsets of [𝑛]?
The set F = [𝑛] 𝑘 (i.e., all 𝑘-element subsets of [𝑛]) has size 𝑛 𝑘 . It is an antichain (why?). The size 𝑛 𝑘 is maximized when 𝑘= 𝑛 2 or 𝑛 2 . The next result shows that this is indeed the best we can do.
Theorem 1.2.2 (Sperner’s theorem, 1928) If F is an antichain of subsets of {1, 2, . . . , 𝑛}, then |F | ≤ 𝑛 ⌊𝑛/2⌋ .
In fact, we will show an even stronger result: Theorem 1.2.3 (LYM inequality; Bollobás 1965, Lubell 1966, Meshalkin 1963, and Yamamoto 1954) If F is an antichain of subsets of [𝑛], then ∑︁ 𝐴∈F 𝑛 |𝐴| −1 ≤1.
Sperner’s theorem follows since 𝑛 |𝐴| ≤ 𝑛 ⌊𝑛/2⌋ for all 𝐴.
Proof. Let 𝜎(1), . . . , 𝜎(𝑛) be a permutation of 1, . . . , 𝑛chosen uniformly at random.
Consider the chain: ∅, {𝜎(1)} , {𝜎(1), 𝜎(2)} , {𝜎(1), 𝜎(2), 𝜎(3)} , . . . , {𝜎(1), . . . , 𝜎(𝑛)} .
7 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1 Introduction For each 𝐴⊆{1, 2, . . . , 𝑛}, let 𝐸𝐴denote the event that 𝐴appears in the above chain.
Then 𝐸𝐴occurs if and only if all the elements of 𝐴appears first in the permutation 𝜎, followed by all the elements of [𝑛] \ 𝐴.
The number of such permutations is |𝐴|!(𝑛−|𝐴|)!. Hence P(𝐸𝐴) = |𝐴|!(𝑛−|𝐴|)!
𝑛!
= 𝑛 |𝐴| −1 .
Since F is an antichain, if 𝐴, 𝐵∈F are distinct, then 𝐸𝐴and 𝐸𝐵cannot both occur (as otherwise 𝐴and 𝐵both appear in the above chain and so one of them contains the other, violating the antichain hypothesis). So {𝐸𝐴: 𝐴∈F } is a set of disjoint events, and thus their probabilities sum to at most 1. So ∑︁ 𝐴∈F 𝑛 |𝐴| −1 ≤ ∑︁ 𝐴∈F P(𝐸𝐴) ≤1.
□ Bollobás’ two families theorem Theorem 1.2.4 (Bollobás’ two families theorem 1965) Let 𝐴1, . . . , 𝐴𝑚be 𝑟-element sets and 𝐵1, . . . , 𝐵𝑚be 𝑠-element sets such that 𝐴𝑖∩𝐵𝑖= ∅for all 𝑖and 𝐴𝑖∩𝐵𝑗≠∅for all 𝑖≠𝑗. Then 𝑚≤ 𝑟+𝑠 𝑟 .
This bound is sharp: let 𝐴𝑖range over all 𝑟-element subsets of [𝑟+ 𝑠] and set 𝐵𝑖= [𝑟+ 𝑠] \ 𝐴𝑖.
Let us give an application/motivation for Bollobás’ two families theorem in terms of transversals. Given a set family F , say that 𝑇is a transversal for F if 𝑇∩𝑆≠∅for all 𝑆∈F (i.e., 𝑇hits every element of F ). Let 𝝉(F), the transversal number of F , be the size of the smallest transversal of F . Say that F is 𝝉-critical if 𝜏(F ′) < 𝜏(F ) whenever F ′ is a proper subset of F .
Question 1.2.5 What is the maximum size of a 𝜏-critical 𝑟-uniform F with 𝜏(F ) = 𝑠+ 1?
We claim that the answer is 𝑟+𝑠 𝑟 . Indeed, let F = {𝐴1, . . . , 𝐴𝑚}, and 𝐵𝑖an 𝑠-element transversal of F \ {𝐴𝑖} for each 𝑖. Then the hypothesis of Bollobás’ two families theorem is satisfied. Thus 𝑚≤ 𝑟+𝑠 𝑟 .
Conversely, F = [𝑟+𝑠] 𝑟 is 𝜏-critcal 𝑟-uniform with 𝜏(F ) = 𝑠+ 1 (why?).
Here is a more general statement of the Bollobás’ two-family theorem.
8 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1.2 Set systems Theorem 1.2.6 (Bollobás’ two families theorem 1965) Let 𝐴1, . . . , 𝐴𝑚and 𝐵1, . . . , 𝐵𝑚be finite sets such that 𝐴𝑖∩𝐵𝑖= ∅for all 𝑖and 𝐴𝑖∩𝐵𝑗≠∅for all 𝑖≠𝑗. Then 𝑚 ∑︁ 𝑖=1 |𝐴𝑖| + |𝐵𝑖| |𝐴𝑖| −1 ≤1.
Note that Sperner’s theorem and LYM inequality are also special cases, since if {𝐴1, . . . , 𝐴𝑚} is an antichain, then setting 𝐵𝑖= [𝑛] \ 𝐴𝑖for all 𝑖satisfies the hypothesis.
Proof. The proof is a modification of the proof of the LYM inequality earlier.
Consider a uniform random ordering of 𝐴1 ∪· · · ∪𝐴𝑚∪𝐵1 ∪· · · ∪𝐵𝑚.
Let 𝐸𝑖be the event that all elements of 𝐴𝑖appear before 𝐵𝑖. Then P(𝐸𝑖) = |𝐴𝑖| + |𝐵𝑖| |𝐴𝑖| −1 .
Note that the events 𝐸𝑖are disjoint, since 𝐸𝑖and 𝐸𝑗both occurring would contradict the hypothesis for 𝐴𝑖, 𝐵𝑖, 𝐴𝑗, 𝐵𝑗(why?). Thus Í 𝑖P(𝐸𝑖) ≤1. This yields the claimed inequality.
□ Bollobas’ two families theorem has many interesting generalizations that we will not discuss here (e.g., see Gil Kalai’s blog post). There are also beautiful linear algebraic proofs of this theorem and its extensions.
Erdős–Ko–Rado theorem on intersecting families We say that a family F is intersecting if 𝐴∩𝐵≠∅for all 𝐴, 𝐵∈F . That is, no two sets in F are disjoint.
Here is an easy warm up.
Question 1.2.7 (Intersecting family—unrestricted sizes) What is the largest intersecting family of subsets of [𝑛]?
One way to generate a large intersecting family is to include all sets that contain a fixed element (say, the element 1). This family has size 2𝑛−1 and is clearly intersecting.
(This isn’t the only example with size 2𝑛−1; can you think of other intersecting families with the same size?) 9 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1 Introduction It turns out that one cannot do better than 2𝑛−1. Since we can pair up each subset of [𝑛] with its complement. At most one of 𝐴and [𝑛] \ 𝐴can be in an intersecting family.
And so at most half of all sets can be in an intersecting family.
The question becomes much more interesting if we restrict to 𝑘-uniform families.
Question 1.2.8 (𝑘-uniform intersecting family) What is the largest intersecting family of 𝑘-element subsets of [𝑛]?
Example: F = all subsets containing the element 1.
Then F is intersecting and |F | = 𝑛−1 𝑘−1 .
Theorem 1.2.9 (Erdős–Ko–Rado 1961; proved in 1938) If 𝑛≥2𝑘, then every intersecting family of 𝑘-element subsets of [𝑛] has size at most 𝑛−1 𝑘−1 .
Remark 1.2.10. The assumption 𝑛≥2𝑘is necessary since if 𝑛< 2𝑘, then the family of all 𝑘-element subsets of [𝑛] is automatically intersecting by pigeonhole.
Proof. Consider a uniform random circular permutation of 1, 2, . . . , 𝑛(arrange them randomly around a circle) For each 𝑘-element subset 𝐴of [𝑛], we say that 𝐴is contiguous if all the elements of 𝐴lie in a contiguous block on the circle.
The probability that 𝐴forms a contiguous set on the circle is exactly 𝑛/ 𝑛 𝑘 .
So the expected number of contiguous sets in F is exactly 𝑛|F | / 𝑛 𝑘 .
Since F is intersecting, there are at most 𝑘contiguous sets in F (under every circular ordering of [𝑛]). Indeed, suppose that 𝐴∈F is contiguous. Then there are 2(𝑘−1) other contingous sets (not necessarily in F ) that intersect 𝐴, but they can be paired off into disjoint pairs (check! Here we use the hypothesis that 𝑛≥2𝑘). Since F is intersecting, it follows that it contains at most 𝑘contiguous sets.
Combining with result from the previous paragraph, we see that 𝑛|F | / 𝑛 𝑘 ≤𝑘, and hence |F | ≤𝑘 𝑛 𝑛 𝑘 = 𝑛−1 𝑘−1 .
□ 1.3 2-colorable hypergraphs An 𝒌-uniform hypergraph (or 𝒌-graph) is a pair 𝐻= (𝑉, 𝐸), where 𝑉(vertices) is a finite set and 𝐸(edges) is a set of 𝑘-element subsets of 𝐸, i.e., 𝐸⊆ 𝑉 𝑘 . (So hypergraphs and set families are the same concept, just different names.) 10 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1.3 2-colorable hypergraphs We say that 𝐻is 𝒓-colorable if the vertices can be colored using 𝑟colors so that no edge is monochromatic.
Let 𝒎(𝒌) denote the minimum number of edges in a 𝑘-uniform hypergraph that is not 2-colorable (elsewhere in the literature, “2-colorable” = “property B”, named after Bernstein who introduced the concept in 1908). Some small values: • 𝑚(2) = 3 • 𝑚(3) = 7. Example: Fano plane (below) is not 2-colorable (the fact there are no 6-edge non-2-colorable 3-graphs can be proved by exhaustive search).
• 𝑚(4) = 23, proved via exhaustive computer search (Östergård 2014) Exact value of 𝑚(𝑘) is unknown for all 𝑘≥5. However, we can get some asymptotic lower and upper bounds using the probability method.
Theorem 1.3.1 (Erdős 1964) 𝑚(𝑘) ≥2𝑘−1 for every 𝑘≥2.
(In other words, every 𝑘-uniform hypergraph with fewer than 2𝑘−1 edges is 2-colorable.) Proof. Let there be 𝑚< 2𝑘−1 edges. In a random 2-coloring, the probability that there is a monochromatic edge is ≤2−𝑘+1𝑚< 1.
□ Remark 1.3.2. Later in Section 3.5 we will prove an better lower bound 𝑚(𝑘) ≳ 2𝑘√︁ 𝑘/log 𝑘, which is the best known to date.
Perhaps somewhat surprisingly, the state of the art upper bound is also proved using probabilistic method (random construction).
Theorem 1.3.3 (Erdős 1964) 𝑚(𝑘) = 𝑂(𝑘22𝑘).
(In other words, there exists a 𝑘-uniform hypergraph with 𝑂(𝑘22𝑘) edges that is not 2-colorable.) Proof. Let |𝑉| = 𝑛= 𝑘2 (this choice is motivated by the displayed inequality be-low). Let 𝐻be the 𝑘-uniform hypergraph obtained by choosing 𝑚edges 𝑆1, . . . , 𝑆𝑚 independently and uniformly at random (i.e., with replacement) among 𝑉 𝑛 .
11 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1 Introduction Given a coloring 𝜒: 𝑉→, if 𝜒colors 𝑎vertices with one color and 𝑏vertices with the other color, then the probability that the (random) edge 𝑆1 is monochromatic under the (non-random) coloring 𝜒is 𝑎 𝑘 + 𝑏 𝑘 𝑛 𝑘 ≥ 2 𝑛/2 𝑘 𝑛 𝑘 = 2(𝑛/2)(𝑛/2 −1) · · · (𝑛/2 −𝑘+ 1) 𝑛(𝑛−1) · · · (𝑛−𝑘+ 1) ≥2 𝑛/2 −𝑘+ 1 𝑛−𝑘+ 1 𝑘 = 2−𝑘+1 1 − 𝑘−1 𝑛−𝑘+ 1 𝑘 = 2−𝑘+1 1 − 𝑘−1 𝑘2 −𝑘+ 1 𝑘 ≥𝑐2−𝑘 for some constant 𝑐> 0.
Since the edges are chosen independently at random, for any coloring 𝜒, P(𝜒is a proper coloring) ≤(1 −𝑐2−𝑘)𝑚≤𝑒−𝑐2−𝑘𝑚 (using 1 + 𝑥≤𝑒𝑥for all real 𝑥). By the union bound, P(the random hypergraph has a proper 2-coloring) ≤ ∑︁ 𝜒 P(𝜒is a proper coloring) ≤2𝑛𝑒−𝑐2−𝑘𝑚< 1 for some 𝑚= 𝑂(𝑘22𝑘) (recall 𝑛= 𝑘2). Thus there exists a non-2-colorable 𝑘-uniform hypergraph with 𝑚edges.
□ 1.4 List chromatic number of 𝐾𝑛,𝑛 Given a graph 𝐺, its chromatic number 𝝌(𝑮) is the minimum number of colors required to proper color its vertices.
In list coloring, each vertex of 𝐺is assigned a list of allowable colors. We say that 𝐺 is 𝒌-choosable (also called 𝒌-list colorable) if it has a proper coloring no matter how one assigns a list of 𝑘colors to each vertex.
We write ch(𝑮), called the list chromatic number (also called: choice number, choosability, list colorability) of 𝐺, to be the smallest 𝑘so that 𝐺is 𝑘-choosable.
We have 𝜒(𝐺) ≤ch(𝐺) by assigning the same list of colors to each vertex. The inequality may be strict, as we will see below.
For example, while every bipartite graph is 2-colorable, 𝐾3,3 is not 2-choosable.
Indeed, no list coloring of 𝐾3,3 is possible with color lists (check!): 12 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1.4 List chromatic number of 𝐾𝑛,𝑛 {2, 3} {1, 3} {1, 2} {2, 3} {1, 3} {1, 2} Exercise: check that ch(𝐾3,3) = 3.
Question 1.4.1 What is the asymptotic behavior of ch(𝐾𝑛,𝑛)?
First we prove an upper bound on ch(𝐾𝑛,𝑛).
Theorem 1.4.2 If 𝑛< 2𝑘−1, then 𝐾𝑛,𝑛is 𝑘-choosable.
In other words, ch(𝐾𝑛,𝑛) ≤ log2(2𝑛) + 1.
Proof. For each color, mark it either L or R independently and uniformly at random.
For any vertex of 𝐾𝑛,𝑛on the left part, remove all its colors marked R.
For any vertex of 𝐾𝑛,𝑛on the right part, remove all its colors marked L.
The probability that some vertex has no colors remaining is at most 2𝑛2−𝑘< 1 by the union bound. So with positive probability, every vertex has some color remaining.
Assign the colors arbitrarily for a valid coloring.
□ The lower bound on ch(𝐾𝑛,𝑛) turns out to follow from the existence of non-2-colorable 𝑘-uniform hypergraph with many edges.
Theorem 1.4.3 If there exists a non-2-colorable 𝑘-uniform hypergraph with 𝑛edges, then 𝐾𝑛,𝑛is not 𝑘-choosable.
Proof. Let 𝐻= (𝑉, 𝐸) be a non-2-colorable 𝑘-uniform hypergraph |𝐸| = 𝑛edges.
Now, view 𝑉as colors and assign to the 𝑖-th vertex of 𝐾𝑛on both the left and right bipartitions a list of colors given by the 𝑖-th edge of 𝐻. We leave it as an exercise to check that this 𝐾𝑛,𝑛is not list colorable.
□ Recall from Theorem 1.3.3 that there exists a non-2-colorable 𝑘-uniform hypergraph with 𝑂(𝑘22𝑘) edges. Thus ch(𝐾𝑛,𝑛) > (1 −𝑜(1)) log2 𝑛.
Putting these bounds together: 13 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1 Introduction Corollary 1.4.4 (List chromatic number of a complete bipartite graph) ch(𝐾𝑛,𝑛) = (1 + 𝑜(1)) log2 𝑛 It turns out that, unlike the chromatic number, the list chromatic number always grows with the average degree. The following result was proved using the method of hypergraph containers, a very important modern development in combinatorics that we will see a glimpse of in Chapter 11. It provides the optimal asymptotic dependence (the example of 𝐾𝑛,𝑛shows optimality).
Theorem 1.4.5 (Saxton and Thomason 2015) If a graph 𝐺has average degree 𝑑, then, as 𝑑→∞, ch(𝐺) > (1 + 𝑜(1)) log2 𝑑.
They also proved similar results for the list chromatic number of hypergraphs. For graphs, a slightly weaker result, off by a factor of 2, was proved earlier by Alon (2000).
Exercises 1. Verify the following asymptotic calculations used in Ramsey number lower bounds: a) For each 𝑘, the largest 𝑛satisfying 𝑛 𝑘 21−(𝑘 2) < 1 has 𝑛= 1 𝑒 √ 2 + 𝑜(1) 𝑘2𝑘/2.
b) For each 𝑘, the maximum value of 𝑛− 𝑛 𝑘 21−(𝑘 2) as 𝑛ranges over positive integers is 1 𝑒+ 𝑜(1) 𝑘2𝑘/2.
c) For each 𝑘, the largest 𝑛satisfying 𝑒 𝑘 2 𝑛 𝑘−2 + 1 21−(𝑘 2) < 1 satisfies 𝑛= √ 2 𝑒+ 𝑜(1) 𝑘2𝑘/2.
2. Prove that, if there is a real 𝑝∈[0, 1] such that 𝑛 𝑘 𝑝(𝑘 2) + 𝑛 𝑡 (1 −𝑝)(𝑡 2) < 1 then the Ramsey number 𝑅(𝑘, 𝑡) satisfies 𝑅(𝑘, 𝑡) > 𝑛. Using this show that 𝑅(4, 𝑡) ≥𝑐 𝑡 log 𝑡 3/2 for some constant 𝑐> 0.
14 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 1.4 List chromatic number of 𝐾𝑛,𝑛 3. Let 𝐺be a graph with 𝑛vertices and 𝑚edges. Prove that 𝐾𝑛can be written as a union of 𝑂(𝑛2(log 𝑛)/𝑚) isomorphic copies of 𝐺(not necessarily edge-disjoint).
4. Prove that there is an absolute constant 𝐶> 0 so that for every 𝑛× 𝑛matrix with distinct real entries, one can permute its rows so that no column in the permuted matrix contains an increasing subsequence of length at least 𝐶√𝑛. (A subsequence does not need to be selected from consecutive terms. For example, (1, 2, 3) is an increasing subsequence of (1, 5, 2, 4, 3).) 5. Generalization of Sperner’s theorem. Let F be a collection of subset of [𝑛] that does not contain 𝑘+ 1 elements forming a chain: 𝐴1 ⊊· · · ⊊𝐴𝑘+1. Prove that F is no larger than taking the union of the 𝑘levels of the Boolean lattice closest to the middle layer.
6. Let 𝐺be a graph on 𝑛≥10 vertices. Suppose that adding any new edge to 𝐺 would create a new clique on 10 vertices. Prove that 𝐺has at least 8𝑛−36 edges.
Hint: Apply Bollobás’ two families theorem 7. Let 𝑘≥4 and 𝐻a 𝑘-uniform hypergraph with at most 4𝑘−1/3𝑘edges. Prove that there is a coloring of the vertices of 𝐻by four colors so that in every edge all four colors are represented.
8. Given a set F of subsets of [𝑛] and 𝐴⊆[𝑛], write F |𝐴:= {𝑆∩𝐴: 𝑆∈F } (its projection onto 𝐴). Prove that for every 𝑛and 𝑘, there exists a set F of subsets of [𝑛] with |F | = 𝑂(𝑘2𝑘log 𝑛) such that for every 𝑘-element subset 𝐴of [𝑛], F |𝐴contains all 2𝑘subsets of 𝐴.
9. Let 𝐴1, . . . , 𝐴𝑚be 𝑟-element sets and 𝐵1, . . . , 𝐵𝑚be 𝑠-element sets. Suppose 𝐴𝑖∩𝐵𝑖= ∅for each 𝑖, and for each 𝑖≠𝑗, either 𝐴𝑖∩𝐵𝑗≠∅or 𝐴𝑗∩𝐵𝑖≠∅.
Prove that 𝑚≤(𝑟+ 𝑠)𝑟+𝑠/(𝑟𝑟𝑠𝑠).
10. ★Show that in every non-2-colorable 𝑛-uniform hypergraph, one can find at least 𝑛 2 2𝑛−1 𝑛 unordered pairs of edges with each pair intersecting in exactly one vertex.
15 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2 Linearity of Expectations Linearity of expectations refers to the following basic fact about the expectation: given random variables 𝑋1, . . . , 𝑋𝑛and constants 𝑐1, . . . , 𝑐𝑛, E[𝑐1𝑋1 + · · · + 𝑐𝑛𝑋𝑛] = 𝑐1E[𝑋1] + · · · + 𝑐𝑛E[𝑋𝑛].
This identity does not require any assumption of independence. On the other hand, generally E[𝑋𝑌] ≠E[𝑋]E[𝑌] unless 𝑋and 𝑌are uncorrelated (independent random variables are always uncorrelated).
Here is a simple application (there are also much more involved solutions via enumer-ation methods).
Question 2.0.1 (Expected number of fixed points) What is the average number of fixed points of a uniform random permutation of an 𝑛 element set?
Solution. Let 𝑋𝑖be the event that element 𝑖∈[𝑛] is fixed. Then E[𝑋𝑖] = 1/𝑛. The expected number of fixed points is E[𝑋1 + · · · + 𝑋𝑛] = E[𝑋1] + · · · + E[𝑋𝑛] = 1.
□ 2.1 Hamiltonian paths in tournaments We frequently use the following fact: with positive probability, 𝑋≥E[𝑋] (likewise for 𝑋≤E[𝑋]).
A tournament is a directed complete graph. A Hamilton path in a directed graph is a directed path that contains every vertex exactly once.
Question 2.1.1 (Number of Hamilton paths in a tournament) What is the maximum (and minimum) number of Hamilton paths in an 𝑛-vertex tournament?
17 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2 Linearity of Expectations The minimization problem is easier. The transitive tournament (i.e., respecting a fixed linear ordering of vertices) has exactly one Hamilton path. On the other hand, every tournament has at least one Hamilton path (Exercise: prove this! Hint: consider a longest directed path).
The maximization problem is more difficult and interesting.
Here we have some asymptotic results.
Theorem 2.1.2 (Tournaments wit many Hamilton paths; Szele 1943) There is a tournament on 𝑛vertices with at least 𝑛!2−(𝑛−1) Hamilton paths Proof. Consider a random tournament where every edge is given a random orientation chosen uniformly and independently. Each of the 𝑛! permutations of vertices forms a directed path with probability 2−𝑛+1. So that expected number of Hamilton paths is 𝑛!2−𝑛+1. Thus, there exists a tournament with at least this many Hamilton paths.
□ This was considered the first use of the probabilistic method. Szele conjectured that the maximum number of Hamilton paths in a tournament on 𝑛players is 𝑛!/(2 −𝑜(1))𝑛.
This was proved by Alon (1990) using the Minc–Brégman theorem on permanents (we will see this later in Chapter 10 on the entropy method).
2.2 Sum-free subset A subset 𝐴in an abelian group is sum-free if there do not exist 𝑎, 𝑏, 𝑐∈𝐴with 𝑎+ 𝑏= 𝑐.
Does every 𝑛-element set contain a large sum-free set?
Theorem 2.2.1 (Large sum-free subsets; Erdős 1965) Every set of 𝑛nonzero integers contains a sum-free subset of size ≥𝑛/3.
Proof. Let 𝐴⊆Z \ {0} with |𝐴| = 𝑛. For 𝜃∈[0, 1], let 𝐴𝜃:= {𝑎∈𝐴: {𝑎𝜃} ∈(1/3, 2/3)} where {·} denotes fractional part. Then 𝐴𝜃is sum-free since (1/3, 2/3) is sum-free in R/Z.
For 𝜃uniformly chosen at random, {𝑎𝜃} is also uniformly random in [0, 1], so P(𝑎∈ 𝐴𝜃) = 1/3. By linearity of expectations, E|𝐴𝜃| = 𝑛/3.
□ 18 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2.3 Turán’s theorem and independent sets Remark 2.2.2 (Additional results). Alon and Kleitman (1990) noted that one can improve the bound to ≥(𝑛+ 1)/3 by noting that |𝐴𝜃| = 0 for 𝜃close to zero (say, |𝜃| < (3 max 𝑎∈𝐴|𝑎|)−1), so that |𝐴𝜃| < 𝑛/3 with positive probability, and hence |𝐴𝜃| > 𝑛/3 with positive probability. Note that since |𝐴𝜃| is an integer, being > 𝑛/3 is the same as being ≥(𝑛+ 1)/3.
Bourgain (1997) improved it to ≥(𝑛+ 2)/3 via a difficult Fourier analytic argument.
This is currently the best lower bound known.
It remains an open problem to prove ≥(𝑛+ 𝑓(𝑛))/3 for some function 𝑓(𝑛) →∞.
In the other direction, Eberhard, Green, and Manners (2014) showed that there exist 𝑛-element sets of integers whose largest sum-free subset has size ≤(1/3 + 𝑜(1))𝑛.
2.3 Turán’s theorem and independent sets Question 2.3.1 (Turán problem) What is the maximum number of edges in an 𝑛-vertex 𝐾𝑘-free graph?
Taking the complement of a graph changes its independent sets to cliques and vice versa. So the problem is equivalent to one about graphs without large independent sets.
The following result, due to Caro (1979) and Wei (1981), shows that a graph with small degrees much contain large independent sets. The probabilistic method proof shown here is due to Alon and Spencer.
Theorem 2.3.2 (Caro 1979, Wei 1981) Every graph 𝐺contains an independent set of size at least ∑︁ 𝑣∈𝑉(𝐺) 1 𝑑𝑣+ 1, where 𝑑𝑣is the degree of vertex 𝑣.
Proof. Consider a random ordering (permutation) of the vertices. Let 𝐼be the set of vertices that appear before all of its neighbors. Then 𝐼is an independent set.
For each 𝑣∈𝑉, P(𝑣∈𝐼) = 1 1+𝑑𝑣(this is the probability that 𝑣appears first among {𝑣} ∪𝑁(𝑣)). Thus E|𝐼| = Í 𝑣∈𝑉(𝐺) 1 𝑑𝑣+1. Thus with positive probability, |𝐼| is at least this expectation.
□ Remark 2.3.3. Equality occurs if 𝐺is a disjoint union of cliques.
19 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2 Linearity of Expectations Remark 2.3.4 (Derandomization). Here is an alternative “greedy algorithm” proof of the Caro–Wei inequality. At each step, take a vertex of smallest degree, and remove it and all its neighbors. If each vertex 𝑣is assigned weight 1/(𝑑𝑣+ 1), then the total weight removed at each step is at most 1. Thus there must be at least Í 𝑣1/(𝑑𝑣+ 1) steps.
Some probabilistic proofs, especially those involving linearity of expectations, can be derandomized this way into an efficient deterministic algorithm. However, for many other proofs (such as Ramsey lower bounds from Section 1.1), it is not known how to derandomize the proof.
By taking the complement of the graph, independent sets become cliques, and so we obtain the following corollary.
Corollary 2.3.5 Every 𝑛-vertex graph 𝐺contains a clique of size at least ∑︁ 𝑣∈𝑉(𝐺) 1 𝑛−𝑑𝑣 .
Note that equality is attained when 𝐺is multipartite.
Now let us answer the earlier question about maximizing the number of edges in a 𝐾𝑟+1-free graph.
The Turán graph 𝑻𝒏,𝒓is the complete multipartite graph formed by partitioning 𝑛 vertices into 𝑟parts with sizes as equal as possible (differing by at most 1).
Example: 𝑇10,3 = 𝐾3,3,4 = It is easy to see that 𝑇𝑛,𝑟is 𝐾𝑟+1-free.
Turán’s theorem (1941) tells us that 𝑇𝑛,𝑟indeed maximizes the number of edges among 𝑛-vertex 𝐾𝑟+1-free graphs. We will prove a slightly weaker statement, below, which is tight when 𝑛is divisible by 𝑟.
20 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2.4 Sampling Theorem 2.3.6 (Turán’s theorem 1941) The number of edges in an 𝑛-vertex 𝐾𝑟+1-free graph is at most 1 −1 𝑟 𝑛2 2 .
Proof. Let 𝑚be the number of edges. Since 𝐺is 𝐾𝑟+1-free, by Corollary 2.3.5, the size 𝜔(𝐺) of the largest clique of 𝐺satisfies 𝑟≥𝜔(𝐺) ≥ ∑︁ 𝑣∈𝑉 1 𝑛−𝑑𝑣 ≥ 𝑛 𝑛−1 𝑛 Í 𝑣𝑑𝑣 = 𝑛 𝑛−2𝑚 𝑛 .
Rearranging gives 𝑚≤ 1 −1 𝑟 𝑛2 2 .
□ Remark 2.3.7. By a careful refinement of the above argument, we can deduce Turán’s theorem that 𝑇𝑛,𝑟maximizes the number of edges in a 𝑛-vertex 𝐾𝑟+1-free graph, by noting that Í 𝑣∈𝑉 1 𝑛−𝑑𝑣is minimized over fixed Í 𝑣𝑑𝑣when the degrees are nearly equal.
Also, Theorem 2.3.6 is asymptotically tight in the sense that the Turán graph 𝑇𝑛,𝑟, for fixed 𝑟and 𝑛→∞, as (1 −1/𝑟−𝑜(1))𝑛2/2 edges.
For more on this topic, see Chapter 1 of my textbook Graph Theory and Additive Combinatorics and the class with the same title.
2.4 Sampling By Turán’s theorem (actually Mantel’s theorem, in this case for triangles, the maximum number of edges in an 𝑛-vertex triangle-free graph is 𝑛2/4 .
How about the problem for hypergraphs? A tetrahedron, denoted 𝐾(3) 4 , is a complete 3-uniform hypergraph (3-graph) on 4 vertices (think of the faces of a usual 3-dimensional tetrahedron).
Question 2.4.1 (Hypergraph Turán problem for tetrahra) What is the maximum number of edges in an 𝑛-vertex 3-uniform hypergraph not containing any tetrahedra?
This turns out to be a notorious open problem. Turán conjectured that the answer is 5 9 + 𝑜(1) 𝑛 3 , 21 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2 Linearity of Expectations which can be achieved using the 3-graph illustrated below: 𝑉1 𝑉2 𝑉3 Above, the vertices are partitioned into three nearly equal sets 𝑉1,𝑉2,𝑉3, and all the edges come in two types: (i) with one vertex in each of the three parts, and (ii) two vertices in 𝑉𝑖and one vertex in 𝑉𝑖+1, with the indices considered mod 3.
Let us give some easy upper bounds, in order to illustrate a simple yet important technique of bounding by sampling.
Proposition 2.4.2 (A cheap sampling bound) Every tetrahedron-free 3-graph on 𝑛≥4 vertices has at most 3 4 𝑛 3 edges.
Proof. Let 𝑆be a 4-vertex subset chosen uniformly at random.
If the graph has 𝑝 𝑛 3 edges, then the expected number of edges induced by 𝑆is 4𝑝by linearity of expectations (why?).
Since the 3-graph is tetrahedron-free, 𝑆induces at most 3 edges. Therefore, 4𝑝≤3.
Thus the total number of edges is 𝑝 𝑛 3 ≤3 4 𝑛 3 .
□ Why stop at sampling four vertices? Can we do better by sampling five vertices? To run the above argument, we will know how many edges can there be in a 5-vertex tetrahedron-free graph.
Lemma 2.4.3 A 5-vertex tetrahedron-free 3-graph has at most 7 edges.
Proof. We can convert a 5-vertex 3-graph 𝐻to a 5-vertex graph 𝐺, by replacing each triple by its complement. Then 𝐻being tetrahedron-free is equivalent to 𝐺not having a vertex of degree 4. The maximum number of edges in a 5-vertex graph with maximum degree at most 3 is ⌊3 · 5/2⌋= 7 (check this can be achieved).
□ We can improve Proposition 2.4.2 by sampling 5 vertices instead of 4 in its proof. This yields (check): 22 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2.5 Unbalancing lights Proposition 2.4.4 Every tetrahedron-free 3-graph on 𝑛≥4 vertices has at most 7 10 𝑛 3 edges.
By sampling 𝑠vertices and using brute-force search to solve the 𝑠-vertex problem, we can improve the upper bound by taking larger values of 𝑠. In fact in principle, if we had unlimited computational power, we can arbitrarily close to optimum by taking sufficiently large 𝑠(why?). However, this is not a practical method due to the cost of the brute-force search. There are more clever ways to get better bounds (also with the help of a computer). The best known upper bound notably via a method known as flag algebras (using sums of squares) due to Razborov, which can give ≤(0.561 · · · ) 𝑛 3 ).
For more on the Hypergraph Turán problem, see the survey by Keevash (2011).
2.5 Unbalancing lights Consider an 𝑛× 𝑛array of light bulbs. Initially some arbitrary subset of the light bulbs are turned on. We are allowed up toggle the lights (on/off) for an entire row or column at a time. How many lights can be guarantee to turn on?
If we flip each row/column independently with probability 1/2, then on expectation, we get exactly half of the lights to turn on. Can we do better?
In the probabilistic method, not every step has to be random. A better strategy is to first flip all the columns randomly, and then decide what to do with each row greedily based on what has happened so far. This is captured in the following theorem, where the left-hand side represents # {bulbs on} −# {bulbs off} .
Theorem 2.5.1 Let 𝑎𝑖𝑗∈{−1, 1} for all 𝑖, 𝑗∈[𝑛]. There exists 𝑥𝑖, 𝑦𝑗∈{−1, 1} for all 𝑖, 𝑗∈[𝑛] such that 𝑛 ∑︁ 𝑖,𝑗=1 𝑎𝑖𝑗𝑥𝑖𝑦𝑗≥ √︂ 2 𝜋+ 𝑜(1) !
𝑛3/2.
Proof. Choose 𝑦1, . . . , 𝑦𝑛∈{−1, 1} independently and uniformly at random. For each 𝑖, let 𝑅𝑖= 𝑛 ∑︁ 𝑗=1 𝑎𝑖𝑗𝑦𝑗 23 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2 Linearity of Expectations and set 𝑥𝑖∈{−1, 1} to be the sign of 𝑅𝑖(arbitrarily choose 𝑥𝑖if 𝑅𝑖= 0. Then the LHS sum is 𝑛 ∑︁ 𝑖=1 𝑅𝑖𝑥𝑖= 𝑛 ∑︁ 𝑖=1 |𝑅𝑖| .
For each 𝑖, 𝑅𝑖has the same distribution as a sum of 𝑛i.i.d. uniform {−1, 1}: 𝑆𝑛= 𝜀1 + · · · + 𝜀𝑛(note that 𝑅𝑖’s are not independent for different 𝑖’s). Thus, for each 𝑖 E[|𝑅𝑖|] = E[|𝑆𝑛|] = √︂ 2 𝜋+ 𝑜(1) !
√𝑛, since by the central limit theorem lim 𝑛→∞E |𝑆𝑛| √𝑛 = E[|𝑋|] where 𝑋∼Normal(0, 1) = 1 √ 2𝜋 ∫ R |𝑥|𝑒−𝑥2/2 𝑑𝑥= √︂ 2 𝜋 (one canalsousebinomialsumidentitiestocomputeexactly: E[|𝑆𝑛|] = 𝑛21−𝑛 𝑛−1 ⌊(𝑛−1)/2⌋ , though it is rather unnecessary to do so.) Thus E 𝑛 ∑︁ 𝑖=1 |𝑅𝑖| = √︂ 2 𝜋+ 𝑜(1) !
𝑛3/2.
Thus with positive probability, the sum is ≥ √︃ 2 𝜋+ 𝑜(1) 𝑛3/2.
□ The next example is tricky. The proof will set up a probabilistic process where the parameters are not given explicitly. A compactness argument will show that a good choice of parameters exists.
Theorem 2.5.2 Let 𝑘≥2. Let 𝑉= 𝑉1 ∪· · · ∪𝑉𝑘, where 𝑉1, . . . ,𝑉𝑘are disjoint sets of size 𝑛. The edges of the complete 𝑘-uniform hypergraph on 𝑉are colored with red/blue. Suppose that every edge formed by taking one vertex from each 𝑉1, . . . ,𝑉𝑘is colored blue.
Then there exists 𝑆⊆𝑉such that the number of red edges and blue edges in 𝑆differ by more than 𝑐𝑘𝑛𝑘, where 𝑐𝑘> 0 is a constant.
Proof. We will write this proof for 𝑘= 3 for notational simplicity. The same proof works for any 𝑘.
Let 𝑝1, 𝑝2, 𝑝3 be real numbers to be decided. We are going to pick 𝑆randomly by 24 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2.6 Crossing number inequality including each vertex in 𝑉𝑖with probability 𝑝𝑖, independently. Let 𝑎𝑖,𝑗,𝑘= #{blue edges in 𝑉𝑖× 𝑉𝑗× 𝑉𝑘} −#{red edges in 𝑉𝑖× 𝑉𝑗× 𝑉𝑘}.
Then E[#{blue edges in 𝑆} −#{red edges in 𝑆}] equals to some polynomial 𝑓(𝑝1, 𝑝2, 𝑝3) = ∑︁ 𝑖≤𝑗≤𝑘 𝑎𝑖,𝑗,𝑘𝑝𝑖𝑝𝑗𝑝𝑘= 𝑛3𝑝1𝑝2𝑝3 + 𝑎1,1,1𝑝3 1 + 𝑎1,1,2𝑝2 1𝑝2 + · · · .
(note that 𝑎1,2,3 = 𝑛3 by hypothesis). We would be done if we can find 𝑝1, 𝑝2, 𝑝3 ∈ [0, 1] such that | 𝑓(𝑝1, 𝑝2, 𝑝3)| > 𝑐for some constant 𝑐> 0 (not depending on the 𝑎𝑖,𝑗,𝑘’s). Note that |𝑎𝑖,𝑗,𝑘| ≤𝑛3. We are done after the following lemma Lemma 2.5.3 Let 𝑃𝑘denote the set of polynomials 𝑔(𝑝1, . . . , 𝑝𝑘) of degree 𝑘, whose coefficients have absolute value ≤1, and the coefficient of 𝑝1𝑝2 · · · 𝑝𝑘is 1.
Then there is a constant 𝑐𝑘> 0 such that for all 𝑔∈𝑃𝑘, there is some 𝑝1, . . . , 𝑝𝑘∈[0, 1] with |𝑔(𝑝1, . . . , 𝑝𝑘)| ≥𝑐.
Proof of Lemma. Set 𝑀(𝑔) = sup𝑝1,...,𝑝𝑘∈[0,1] |𝑔(𝑝1, . . . , 𝑝𝑘)| (note that sup is achieved as max due to compactness).
For 𝑔∈𝑃𝑘, since 𝑔is nonzero (its coefficient of 𝑝1𝑝2 · · · 𝑝𝑘is 1), we have 𝑀(𝑔) > 0. As 𝑃𝑘is compact and 𝑀: 𝑃𝑘→R is continuous, 𝑀attains a minimum value 𝑐= 𝑀(𝑔) > 0 for some 𝑔∈𝑃𝑘.
■ □ 2.6 Crossing number inequality Consider drawings of graphs on a plane using continuous curves as edges.
The crossing number cr(𝑮) is the minimum number of crossings in a drawing of 𝐺.
A graph is planar if cr(𝐺) = 0.
The graphs 𝐾3,3 and 𝐾5 are non-planar. Furthermore, the following theorem charac-terizes these two graphs as the only obstructions to planarity: Kuratowski’s theorem (1930). Every non-planar graph contains a subgraph that is topologically homeomorphic to 𝐾3,3 or 𝐾5.
Wagner’s theorem (1937). A graph is planar if and only if it does not have 𝐾3,3 or 𝐾5 as a minor.
25 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2 Linearity of Expectations (It is not too hard to show that Wagner’s theorem and Kuratowski’s theorem are equivalent) If a graph has a lot of edges, is it guaranteed to have a lot of crossings no matter how it is drawn in the plane?
Question 2.6.1 What is the minimum possible number of crossings that a drawing of: • 𝐾𝑛? (Hill’s conjecture) • 𝐾𝑛,𝑛? (Zarankiewicz conjecture; Turán’s brick factory problem) • a graph on 𝑛vertices and 𝑛2/100 edges?
The following result, due to Ajtai–Chvátal–Newborn–Szemerédi (1982) and Leighton (1984), lower bounds the number of crossings for graphs with many edges.
Theorem 2.6.2 (Crossing number inequality) In a graph 𝐺= (𝑉, 𝐸), if |𝐸| ≥4|𝑉|, then cr(𝐺) ≳|𝐸|3 |𝑉|2 .
Remark 2.6.3. The constant 4 in |𝐸| ≥4 |𝑉| can be replaced by any constant greater than 3 (at the cost of changing the constant in the conclusion). On the other hand, by considering a large triangular grid, we get a planar graph with average degree arbitrarily close to 6.
Corollary 2.6.4 In a graph 𝐺= (𝑉, 𝐸), if |𝐸| ≳|𝑉|2, then cr(𝐺) ≳|𝑉|4.
Proof. The proof has three steps, starting with some basic facts on planar graphs.
Step 1: From zero to one.
Recall Euler’s formula: 𝑣−𝑒+ 𝑓= 2 for every connected planar drawing of graph.
Here 𝑣is the number of vertices, 𝑒the number of edges, and 𝑓the number of faces (connected components of the complement of the drawing, including the outer infinite region).
For every connected planar graph with at least one cycle, 3|𝐹| ≤2|𝐸| since every face is adjacent to ≥3 edges, whereas every edge is adjacent to exactly 2 faces. Plugging into Euler’s formula, |𝐸| ≤3|𝑉| −6.
26 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2.6 Crossing number inequality Thus |𝐸| ≤3|𝑉| for all planar graphs. Hence cr(𝐺) > 0 whenever |𝐸| > 3|𝑉|.
Step 2: From one to many.
The above argument gives us one crossing.
Next, we will use it to obtain many crossings.
By deleting one edge for each crossing, we get a planar graph, so |𝐸| −cr(𝐺) ≤3|𝑉|, that is cr(𝐺) ≥|𝐸| −3|𝑉|.
This is a “cheap bound.” For graphs with |𝐸| = Θ(𝑛2), this gives cr(𝐺) ≳𝑛2. This is not a great bound. We next will use the probabilistic method to boost this bound.
Step 3: Bootstrapping.
Let 𝑝∈[0, 1] to be decided. Let 𝐺′ = (𝑉′, 𝐸′) be obtained from 𝐺by randomly keeping each vertex with probability 𝑝. Then cr(𝐺′) ≥|𝐸′| −3|𝑉′|.
So E cr(𝐺′) ≥E|𝐸′| −3E|𝑉′| We have E cr(𝐺′) ≤𝑝4 cr(𝐺), E|𝐸′| = 𝑝2|𝐸| and E|𝑉′| = 𝑝E|𝑉|. So 𝑝4 cr(𝐺) ≥𝑝2|𝐸| −3𝑝|𝑉|.
Thus cr(𝐺) ≥𝑝−2|𝐸| −3𝑝−3|𝑉|.
Setting 𝑝= 4 |𝑉| /|𝐸| ∈[0, 1] (here is where we use the hypothesis that |𝐸| ≥4 |𝑉|) so that 4𝑝−3|𝑉| = 𝑝−2|𝐸|, we obtain cr(𝐺) ≳|𝐸|3 /|𝑉|2.
□ Remark 2.6.5. The above idea of boosting a cheap bound to a better bound is an important one. We saw a version of this idea in Section 2.4 where we sampled a constant number of vertices to deduce upper bounds on the hypergraph Turán num-ber. In the above crossing number inequality application, we are also applying some preliminary cheap bound to some sampled induced subgraph, though this time the sampled subgraph has super-constant size.
It is tempting to modify the proof by sampling edges instead of vertices, but this does not work.
27 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 2 Linearity of Expectations Exercises 1. Let 𝐴be a measurable subset of the unit sphere in R3 (centered at the origin) containing no pair of orthogonal points.
a) Prove that 𝐴occupies at most 1/3 of the sphere in terms of surface area.
b) ★Prove an upper bound smaller than 1/3 (give your best bound).
2. ★Prove that every set of 10 points in the plane can be covered by a union of disjoint unit disks.
3. Let r = (𝑟1, . . . , 𝑟𝑘) be a vector of nonzero integers whose sum is nonzero.
Prove that there exists a real 𝑐> 0 (depending on r only) such that the following holds: for every finite set 𝐴of nonzero reals, there exists a subset 𝐵⊆𝐴with |𝐵| ≥𝑐|𝐴| such that there do not exist 𝑏1, . . . , 𝑏𝑘∈𝐵with 𝑟1𝑏1 +· · ·+𝑟𝑘𝑏𝑘= 0.
4. Prove that every set 𝐴of 𝑛nonzero integers contains two disjoint subsets 𝐵1 and 𝐵2, such that both 𝐵1 and 𝐵2 are sum-free, and |𝐵1| + |𝐵2| > 2𝑛/3.
5. Let 𝐺be an 𝑛-vertex graph with 𝑝𝑛2 edges, with 𝑛≥10 and 𝑝≥10/𝑛.
Prove that 𝐺contains a pair of vertex-disjoint and isomorphic subgraphs (not necessarily induced) each with at least 𝑐𝑝2𝑛2 edges, where 𝑐> 0 is a constant.
6. ★Prove that for every positive integer 𝑟, there exists an integer 𝐾such that the following holds. Let 𝑆be a set of 𝑟𝑘points evenly spaced on a circle. If we partition 𝑆= 𝑆1 ∪· · · ∪𝑆𝑟so that |𝑆𝑖| = 𝑘for each 𝑖, then, provided 𝑘≥𝐾, there exist 𝑟congruent triangles where the vertices of the 𝑖-th triangle lie in 𝑆𝑖, for each 1 ≤𝑖≤𝑟.
7. ★Prove that [𝑛]𝑑cannot be partitioned into fewer than 2𝑑sets each of the form 𝐴1 × · · · × 𝐴𝑑where 𝐴𝑖⊊[𝑛].
28 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 3 Alterations We saw the alterations method in Section 1.1 to give lower bounds to Ramsey numbers.
The basic idea is to first make a random construction, and then fix the blemishes.
3.1 Dominating set in graphs In a graph 𝐺= (𝑉, 𝐸), we say that 𝑈⊆𝑉is dominating if every vertex in 𝑉\𝑈has a neighbor in 𝑈.
Theorem 3.1.1 Every graph on 𝑛vertices with minimum degree 𝛿> 1 has a dominating set of size ≤ log(𝛿+ 1) + 1 𝛿+ 1 𝑛.
Naive attempt: take out vertices greedily. The first vertex eliminates 1 + 𝛿vertices, but subsequent vertices eliminate possibly fewer vertices.
Proof. Two-step process (alteration method): 1. Choose a random subset 2. Add enough vertices to make it dominating Let 𝑝∈[0, 1] to be decided later. Let 𝑋be a random subset of 𝑉where every vertex is included with probability 𝑝independently.
Let 𝑌= 𝑉\ (𝑋∪𝑁(𝑋)). Each 𝑣∈𝑉lies in 𝑌with probability ≤(1 −𝑝)1+𝛿.
Then 𝑋∪𝑌is dominating, and E[|𝑋∪𝑌|] = E[|𝑋|] + E[|𝑌|] ≤𝑝𝑛+ (1 −𝑝)1+𝛿𝑛≤(𝑝+ 𝑒−𝑝(1+𝛿))𝑛 using 1 + 𝑥≤𝑒𝑥for all 𝑥∈R. Finally, setting 𝑝= log(𝛿+1) 𝛿+1 to minimize 𝑝+ 𝑒−𝑝(1+𝛿), we bound the above expression by ≤ 1 + log(𝛿+ 1) 𝛿+ 1 .
□ 29 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 3 Alterations 3.2 Heilbronn triangle problem Question 3.2.1 How can one place 𝑛points in the unit square so that no three points forms a triangle with small area?
Let Δ(𝑛) = sup 𝑆⊆[0,1]2 |𝑆|=𝑛 min 𝑝,𝑞,𝑟∈𝑆 distinct area(𝑝𝑞𝑟)/ Naive constructions fair poorly. E.g., 𝑛points around a circle has a triangle of area Θ(1/𝑛3) (the triangle formed by three consectutive points has side lengths ≍1/𝑛and angle 𝜃= (1 −1/𝑛)2𝜋). Even worse is arranging points on a grid, as you would get triangles of zero area.
Heilbronn conjectured that Δ(𝑛) = 𝑂(𝑛−2).
Komlós, Pintz, and Szemerédi (1982) disproved the conjecture, showing Δ(𝑛) ≳ 𝑛−2 log 𝑛. They used an elaborate probabilistic construction. Here we show a much simpler version probabilistic construction that gives a weaker bound Δ(𝑛) ≳𝑛−2.
Remark 3.2.2 (Upper bounds). For a long time, the best upper bound known was Δ(𝑛) ≤𝑛−8/7+𝑜(1) due to Komlós, Pintz, and Szemerédi (1981). This was recently improved to Δ(𝑛) ≤𝑛−8/7−𝑐by Cohen, Pohoata, and Zakharov (2023+).
Theorem 3.2.3 (Many points without small area triangles) For every positive integer 𝑛, there exists a set of 𝑛points in [0, 1]2 such that every triple spans a triangle of area ≥𝑐𝑛−2, for some absolute constant 𝑐> 0.
Proof. Choose 2𝑛points at random. For every three random points 𝑝, 𝑞, 𝑟, let us estimate P𝑝,𝑞,𝑟(area(𝑝, 𝑞, 𝑟) ≤𝜀).
By considering the area of a circular annulus around 𝑝, with inner and outer radii 𝑥 and 𝑥+ Δ𝑥, we find P𝑝,𝑞(|𝑝𝑞| ∈[𝑥, 𝑥+ Δ𝑥]) ≤𝜋((𝑥+ Δ𝑥)2 −𝑥2).
30 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 3.3 Markov’s inequality So the probability density function satisfies P𝑝,𝑞(|𝑝𝑞| ∈[𝑥, 𝑥+ 𝑑𝑥]) ≤2𝜋𝑥𝑑𝑥.
For fixed 𝑝, 𝑞 P𝑟(area(𝑝𝑞𝑟) ≤𝜀) = P𝑟 dist(𝑝𝑞, 𝑟) ≤2𝜀 |𝑝𝑞| ≲ 𝜀 |𝑝𝑞| .
Thus, with 𝑝, 𝑞, 𝑟at random P𝑝,𝑞,𝑟(area(𝑝𝑞𝑟) ≤𝜀) ≲ ∫√ 2 0 2𝜋𝑥𝜀 𝑥𝑑𝑥≍𝜀.
Given these 2𝑛random points, let 𝑋be the number of triangles with area ≤𝜀. Then E𝑋= 𝑂(𝜀𝑛3).
Choose 𝜀= 𝑐/𝑛2 with 𝑐> 0 small enough so that E𝑋≤𝑛.
Delete a point from each triangle with area ≤𝜀.
The expected number of remaining points is E[2𝑛−𝑋] ≥𝑛, and no triangles with area ≤𝜀= 𝑐/𝑛2.
Thus with positive probability, we end up with ≥𝑛points and no triangle with area ≤𝑐/𝑛2.
□ Algebraic construction. Here is another construction due to Erdős (in appendix of Roth (1951)) also giving Δ(𝑛) ≳𝑛−2: Let 𝑝be a prime. The set {(𝑥, 𝑥2) ∈F2 𝑝: 𝑥∈F𝑝} has no 3 points collinear (a parabola meets every line in ≤2 points). Take the corresponding set of 𝑝points in [𝑝]2 ⊆Z2.
Then every triangle has area ≥1/2 due to Pick’s theorem. Scale back down to a unit square. (If 𝑛is not a prime, then use that there is a prime between 𝑛and 2𝑛.) 3.3 Markov’s inequality We note an important tool that will be used next.
Theorem 3.3.1 (Markov’s inequality) Let 𝑋≥0 be random variable. Then for every 𝑎> 0, P(𝑋≥𝑎) ≤E[𝑋] 𝑎 .
31 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 3 Alterations Proof. E[𝑋] ≥E[𝑋1𝑋≥𝑎] ≥E[𝑎1𝑋≥𝑎] = 𝑎P(𝑋≥𝑎) □ Take-home message: for r.v. 𝑋≥0, if E𝑋is very small, then typically 𝑋is small.
3.4 High girth and high chromatic number If a graph has a 𝑘-clique, then you know that its chromatic number is at least 𝑘.
Conversely, if a graph has high chromatic number, is it always possible to certify this fact from some “local information”?
Surprisingly, the answer is no. The following ingenious construction shows that a graph can be “locally tree-like” while still having high chromatic number.
The girth of a graph is the length of its shortest cycle.
Theorem 3.4.1 (Erdős 1959) For all 𝑘, ℓ, there exists a graph with girth > ℓand chromatic number > 𝑘.
Proof. Let 𝐺∼𝐺(𝑛, 𝑝) with 𝑝= (log 𝑛)2/𝑛(the proof works whenever log 𝑛/𝑛≪ 𝑝≪𝑛−1+1/ℓ). Here 𝐺(𝑛, 𝑝) is Erdős–Rényi random graph (𝑛vertices, every edge appearing with probability 𝑝independently).
Let 𝑋be the number of cycles of length at most ℓin 𝐺. By linearity of expectations, as there are exactly 𝑛 𝑖 (𝑖−1)!/2 cycles of length 𝑖in 𝐾𝑛for each 3 ≤𝑖≤𝑛, we have (recall that ℓis a constant) E𝑋= ℓ ∑︁ 𝑖=3 𝑛 𝑖 (𝑖−1)!
2 𝑝𝑖≤ ℓ ∑︁ 𝑖=3 𝑛𝑖𝑝𝑖= ℓ(log 𝑛)2𝑖= 𝑜(𝑛).
By Markov’s inequality P(𝑋≥𝑛/2) ≤E𝑋 𝑛/2 = 𝑜(1).
(This allows us to get rid of all short cycles.) How can we lower bound the chromatic number 𝜒(·)? Note that 𝜒(𝐺) ≥|𝑉(𝐺)|/𝛼(𝐺), where 𝛼(𝐺) is the independence number (the size of the largest independent set).
With 𝑥= (3/𝑝) log 𝑛= 3𝑛/log 𝑛, P(𝛼(𝐺) ≥𝑥) ≤ 𝑛 𝑥 (1 −𝑝)(𝑥 2) < 𝑛𝑥𝑒−𝑝𝑥(𝑥−1)/2 = (𝑛𝑒−𝑝(𝑥−1)/2)𝑥= 𝑛−Θ(𝑛) = 𝑜(1).
32 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 3.5 Random greedy coloring Let 𝑛be large enough so that P(𝑋≥𝑛/2) < 1/2 and P(𝛼(𝐺) ≥𝑥) < 1/2. Then there is some 𝐺with fewer than 𝑛/2 cycles of length ≤ℓand with 𝛼(𝐺) ≤3𝑛/log 𝑛.
Remove a vertex from each cycle to get 𝐺′. Then |𝑉(𝐺′)| ≥𝑛/2, girth > ℓ, and 𝛼(𝐺′) ≤𝛼(𝐺) ≤3𝑛/log 𝑛, so 𝜒(𝐺′) ≥|𝑉(𝐺′)| 𝛼(𝐺′) ≥ 𝑛/2 3𝑛/log 𝑛= log 𝑛 6 > 𝑘 if 𝑛is sufficiently large.
□ Remark 3.4.2. Erdős (1962) also showed that in fact one needs to see at least a linear number of vertices to deduce high chromatic number: for all 𝑘, there exists 𝜀= 𝜀𝑘 such that for all sufficiently large 𝑛there exists an 𝑛-vertex graph with chromatic number > 𝑘but every subgraph on ⌊𝜀𝑛⌋vertices is 3-colorable. (In fact, one can take 𝐺∼𝐺(𝑛, 𝐶/𝑛); see "Probabilistic Lens: Local coloring" in Alon–Spencer) 3.5 Random greedy coloring In Section 1.3, we saw a simple argument showing that every 𝑘-uniform hypergraph with than 2𝑘−1 edges is 2-colorable (meaning that we can color the vertices red/blue without no monochromatic edge). Take a moment to remember the proof.
In this section, we improve this result. The next result gives the current best known bound.
Theorem 3.5.1 (Radhakrishnan and Srinivasan (2000)) There is some constant 𝑐> 0 so that every 𝑘-uniform hypergraph with at most 𝑐 √︃ 𝑘 log 𝑘2𝑘edges is 2-colorable.
Recall from Section 1.3 that there exists a non-2-colorable 𝑘-uniform hypergraph on 𝑘2 vertices and 𝑂(𝑘22𝑘) edges, via a random construction.
Here we present a simpler proof, based on a random greedy coloring, due to Cherkashin and Kozik (2015), following an approach of Pluhaár (2009).
Proof. Consider a 𝑘-graph with 𝑚edges.
Let us order the vertices using a uniformly random chosen permutation.
Color vertices greedily from left to right: color a vertex blue unless it would create a monochromatic edge, in which case color it red (i.e., every red vertex is the final vertex in an edge with all earlier 𝑘−1 vertices have already been colored blue).
33 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 3 Alterations The resulting coloring has no blue edges. The greedy coloring succeeds if it does not create a red edge.
Analyzing a greedy coloring is tricky, since the color of a single vertex may depend on the entire history. Instead, we identify a specific feature that necessarily results from a unsuccessful coloring.
If there is a red edge, then there must be two edges 𝑒, 𝑓so that the last vertex of 𝑒 is the first vertex of 𝑓. Call such pair (𝑒, 𝑓) conflicting (note that whether (𝑒, 𝑓) is conflicting depends on the random ordering of the vertices, but not on how we assigned colors).
What is the probability of seeing a conflicting pair? Here is the randomness comes from the random ordering of vertices.
Each pair of edges with exactly one vertex in common conflicts with probability (𝑘−1)!2 (2𝑘−1)! = 1 2𝑘−1 2𝑘−2 𝑘−1 −1 ≍𝑘−1/22−2𝑘. Summing over all ≤𝑚2 pairs of edges that share a unique vertex, we find that the expected number of conflicting pairs is at most ≲𝑚2𝑘−1/22−2𝑘, which is < 1 for some 𝑚≍𝑘1/42𝑘.
In this case, there is some ordering of vertices creating no conflicting pairs, in which case the greedy coloring always succeeds.
The above argument, due to Pluhaár (2009), yields 𝑚≲𝑘1/42𝑘. Next we will refine the argument to obtain a better bound of √︃ 𝑘 log 𝑘2𝑘as claimed.
Instead of just considering a random permutation, let us map each vertex to [0, 1] independently and uniformly at random. This map induces an ordering of the vertices, but it comes with further information that we will use.
Write [0, 1] = 𝐿∪𝑀∪𝑅where (𝑝to be decided) 𝐿:= 0, 1 −𝑝 2 , 𝑀:= 1 −𝑝 2 , 1 + 𝑝 2 , 𝑅:= 1 + 𝑝 2 , 1 .
The probability that a given edge lands entirely in 𝐿is ( 1−𝑝 2 )𝑘, and likewise with 𝑅.
Taking a union bound over all edges, P(some edge lies in 𝐿or 𝑅) ≤2𝑚 1 −𝑝 2 𝑘 .
Suppose that no edge of 𝐻lies entirely in 𝐿or entirely in 𝑅. If (𝑒, 𝑓) conflicts, then their unique common vertex 𝑥𝑣∈𝑒∩𝑓must lie in 𝑀. So the probability that (𝑒, 𝑓) 34 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 3.5 Random greedy coloring conflicts is (here we use 𝑥(1 −𝑥) ≤1/4) ∫ (1+𝑝)/2 (1−𝑝)/2 𝑥𝑘−1(1 −𝑥)𝑘−1 𝑑𝑥≤𝑝4−𝑘+1.
Taking a union bound over all ≤𝑚2 pairs of edges, we find that P(some conflicting pair has the common vertex in 𝑀) ≤𝑚2𝑝4−𝑘+1.
Thus P(there is a conflicting pair) ≤P(some edge lies in 𝐿or 𝑅) + P(some conflicting pair has the common vertex in 𝑀) ≤2𝑚 1 −𝑝 2 𝑘 + 𝑚2𝑝4−𝑘+1 < 2−𝑘+1𝑚𝑒−𝑝𝑘+ (2−𝑘+1𝑚)2𝑝 set 𝑝= log(2𝑘−1𝑘/𝑚)/𝑘to minimize the right-hand side to get = 𝑚2 4𝑘−1𝑘+ 𝑚2 4𝑘−1𝑘log 2𝑘−2𝑘 𝑚 which is < 1 for 𝑚= 𝑐2𝑘√︁ 𝑘/log 𝑘with 𝑐> 0 being a sufficiently small constant (we should assume that is 𝑘large enough to ensure 𝑝∈[0, 1]; smaller values of 𝑘can be handled in the theorem exceptionally by later reducing the constant 𝑐).
□ Food for thought: what is the source of the gain in the the 𝐿∪𝑀∪𝑅argument? The expected number of conflicting pairs is unchanged. It must be that we are somehow clustering the bad events by considering the event when some edge lies in 𝐿or 𝑅.
It remains an intriguing open problem to improve this bound further.
Exercises 1. Using the alteration method, prove the Ramsey number bound 𝑅(4, 𝑘) ≥𝑐(𝑘/log 𝑘)2 for some constant 𝑐> 0.
2. Prove that every 3-uniform hypergraph with 𝑛vertices and 𝑚≥𝑛edges contains an independent set (i.e., a set of vertices containing no edges) of size at least 35 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 3 Alterations 𝑐𝑛3/2/√𝑚, where 𝑐> 0 is a constant.
3. Prove that every 𝑘-uniform hypergraph with 𝑛vertices and 𝑚edges has a transver-sal (i.e., a set of vertices intersecting every edge) of size at most 𝑛(log 𝑘)/𝑘+𝑚/𝑘.
4. Zarankiewicz problem. Prove that for every positive integers 𝑛≥𝑘≥2, there exists an 𝑛× 𝑛matrix with {0, 1} entries, with at least 1 2𝑛2−2/(𝑘+1) 1’s, such that there is no 𝑘× 𝑘submatrix consisting of all 1’s.
5. Fix 𝑘. Prove that there exists a constant 𝑐𝑘> 1 so that for every sufficiently large 𝑛> 𝑛0(𝑘), there exists a collection F of at least 𝑐𝑛 𝑘subsets of [𝑛] such that for every 𝑘distinct 𝐹1, . . . , 𝐹𝑘∈F , all 2𝑘intersections Ñ𝑘 𝑖=1 𝐺𝑖are nonempty, where each 𝐺𝑖is either 𝐹𝑖or [𝑛] \ 𝐹𝑖.
6. Acute sets in R𝑛. Prove that, for some constant 𝑐> 0, for every 𝑛, there exists a family of at least 𝑐(2/ √ 3)𝑛subsets of [𝑛] containing no three distinct members 𝐴, 𝐵, 𝐶satisfying 𝐴∩𝐵⊆𝐶⊆𝐴∪𝐵.
Deduce that there exists a set of at least 𝑐(2/ √ 3)𝑛points in R𝑛so that all angles determined by three points from the set are acute.
Remark. The current best lower and upper bounds for the maximum size of an “acute set” in R𝑛(i.e., spanning only acute angles) are 2𝑛−1 + 1 and 2𝑛−1 respectively.
7. ★Covering complements of sparse graphs by cliques a) Prove that every graph with 𝑛vertices and minimum degree 𝑛−𝑑can be written as a union of 𝑂(𝑑2 log 𝑛) cliques.
b) Prove that every bipartite graph with 𝑛vertices on each side of the ver-tex bipartition and minimum degree 𝑛−𝑑can be written as a union of 𝑂(𝑑log 𝑛) complete bipartite graphs (assume 𝑑≥1).
8. ★Let 𝐺= (𝑉, 𝐸) be a graph with 𝑛vertices and minimum degree 𝛿≥2. Prove that there exists 𝐴⊆𝑉with |𝐴| = 𝑂(𝑛(log 𝛿)/𝛿) so that every vertex in 𝑉\ 𝐴 contains at least one neighbor in 𝐴and at least one neighbor not in 𝐴.
9. ★Prove that every graph 𝐺without isolated vertices has an induced subgraph 𝐻on at least 𝛼(𝐺)/2 vertices such that all vertices of 𝐻have odd degree. Here 𝛼(𝐺) is the size of the largest independent set in 𝐺.
36 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment 4.1 Does a typical random graph contain a triangle?
We begin with the following motivating question. Recall that the Erdős–Rényi random graph 𝐺(𝑛, 𝑝) is the 𝑛-vertex graph with edge probability 𝑝.
Question 4.1.1 For which 𝑝= 𝑝𝑛does 𝐺(𝑛, 𝑝) contain a triangle with probability 1 −𝑜(1)?
(We sometimes abbreviate “with probability 1 −𝑜(1) by “with high probability” or simply “whp”. In some literature, this is also called “asymptotically almost surely” or “a.a.s.”) By computing E𝑋(also known as the first moment), we deduce the following.
Proposition 4.1.2 If 𝑛𝑝→0, then 𝐺(𝑛, 𝑝) is triangle-free with probability 1 −𝑜(1).
Proof. Let 𝑋be the number of triangles in 𝐺(𝑛, 𝑝).
We know from linearity of expectations that E𝑋= 𝑛 3 𝑝3 ≍𝑛3𝑝3 = 𝑜(1).
Thus, by Markov’s inequality, P(𝑋≥1) ≤E𝑋= 𝑜(1).
In other words, 𝑋= 0 with probability 1 −𝑜(1).
□ In other words, when 𝑝≪1/𝑛, 𝐺(𝑛, 𝑝) is triangle-free with high probaiblity (recall that 𝑝≪1/𝑛means 𝑝= 𝑜(1/𝑛); see asymptotic notation guide at the beginning of these notes).
What about when 𝑝≫1/𝑛? Can we conclude that 𝐺(𝑛, 𝑝) contains a triangle with high probability? In this case E𝑋→∞, but this is not enough to conclude that 37 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment P(𝑋≥1) = 1 −𝑜(1), since we have not ruled out the probability that 𝑋is almost always zero but extremely large with some tiny probability.
An important technique in probabilistic combinatorics is to show that some random variable is concentrated around its mean. This would then imply that outliers are unlikely.
We will see many methods in this course on proving concentrations of random vari-ables. In this chapter, we begin with the simplest method. It is usually easiest to execute and it requires not much hypotheses. The downside is that it only produces relatively weak (though still useful enough) concentration bounds.
Second moment method: show that a random variable is concentrated near its mean by bounding its variance.
Definition 4.1.3 (Variance) The variance of a random variable 𝑋is Var[𝑋] := E[(𝑋−E𝑋)2] = E[𝑋2] −E[𝑋]2.
The covariance of two random variables 𝑋and 𝑌(jointly distributed) is Cov[𝑋,𝑌] := E[(𝑋−E𝑋)(𝑌−E𝑌)] = E[𝑋𝑌] −E[𝑋]E[𝑌].
(Exercise: check the second equality in the definitions of variance and covariance above).
Remark 4.1.4 (Notation convention). It is standard to use the Greek letter 𝜇for the mean, and 𝜎2 for the variance. Here 𝜎≥0 is the standard deviation.
The following basic result provides a concentration bound based on the variance.
Theorem 4.1.5 (Chebyshev’s inequality) Let 𝑋be a random variable with mean 𝜇and variance 𝜎2. For any 𝜆> 0 P(|𝑋−𝜇| ≥𝜆𝜎) ≤𝜆−2.
Proof. By Markov’s inequality, 𝐿𝐻𝑆= P(|𝑋−𝜇|2 ≥𝜆2𝜎2) ≤E[(𝑋−𝜇)2] 𝜆2𝜎2 = 1 𝜆2 .
□ Remark 4.1.6. Concentration bounds that show small probability of deviating from the mean are called tail bounds (more precisely: upper tail for 𝑋≥𝜇+𝑎and lower tail 38 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.1 Does a typical random graph contain a triangle?
for P(𝑋≤𝜇−𝑎)). Chebyshev’s inequality gives tail bounds that decays quadratically.
Later on we will see tools that give much better decay (usually exponential) provided additional assumptions on the random variable (e.g., independence).
We are often interested in upper bounding the probability of non-existence, i.e., P(𝑋= 0). Chebyshev’s inequality yields the following bound.
Corollary 4.1.7 (Chebyshev bound on the probability of non-existence) For any random variable 𝑋, P(𝑋= 0) ≤Var 𝑋 (E𝑋)2 .
Proof. By Chebyshev inequality, writing 𝜇= E𝑋, P(𝑋= 0) ≤P(|𝑋−𝜇| ≥|𝜇|) ≤Var 𝑋 𝜇2 .
□ Corollary 4.1.8 If E𝑋> 0 and Var 𝑋= 𝑜(E𝑋)2, then 𝑋> 0 and 𝑋∼E𝑋with probability 1 −𝑜(1).
Remark 4.1.9 (Asymptotic statements). The above statement is really referring to not a single random variable, but a sequence of random variables 𝑋𝑛. It is saying that if E𝑋𝑛> 0 and Var 𝑋𝑛= 𝑜(E𝑋𝑛)2, then P(𝑋𝑛> 0) →1 as 𝑛→∞, and for any fixed 𝛿> 0, P(|𝑋𝑛−𝐸𝑋𝑛| > 𝛿E𝑋𝑛) →0 as 𝑛→∞.
In many situations, it is not too hard to compute the second moment.
We have Var[𝑋] = Cov[𝑋, 𝑋]. Also, covariance is bilinear, i.e., for random variables 𝑋1, . . .
and 𝑌1, . . . (no assumptions needed on their independence, etc.) and constants 𝑎1, . . .
and 𝑏1, . . . , one has Cov "∑︁ 𝑖 𝑎𝑖𝑋𝑖, ∑︁ 𝑗 𝑏𝑗𝑌𝑗 # = ∑︁ 𝑖,𝑗 𝑎𝑖𝑏𝑗Cov[𝑋𝑖,𝑌𝑗].
We are often dealing with 𝑋being the cardinality of some random set. We can usually write this as a sum of indicator functions, such as 𝑋= 𝑋1 + · · · + 𝑋𝑛, so that Var 𝑋= Cov[𝑋, 𝑋] = 𝑛 ∑︁ 𝑖,𝑗=1 Cov[𝑋𝑖, 𝑋𝑗] = 𝑛 ∑︁ 𝑖=1 Var 𝑋𝑖+ 2 ∑︁ 𝑖< 𝑗 Cov[𝑋𝑖, 𝑋𝑗].
39 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment We have Cov[𝑋,𝑌] = 0 if 𝑋and 𝑌are independent. Thus in the sum we only need to consider dependent pairs (𝑖, 𝑗).
Example 4.1.10 (Sum of independent Bernoulli). Suppose 𝑋= 𝑋1 + · · · + 𝑋𝑛with each 𝑋𝑖being an independent Bernoulli random variables with P(𝑋𝑖= 1) = 𝑝and P(𝑋𝑖= 0) = 1 −𝑝. Then 𝜇= 𝑛𝑝and 𝜎2 = 𝑛𝑝(1 −𝑝) (note that Var[𝑋𝑖] = 𝑝−𝑝2 and Cov[𝑋𝑖, 𝑋𝑗] = 0 if 𝑖≠𝑗). If 𝑛𝑝→∞, then 𝜎= 𝑜(𝜇), and thus 𝑋= 𝜇+ 𝑜(𝜇) whp.
Note that the above computation remains identical even if we only knew that the 𝑋𝑖’s are pairwise uncorrelated (much weaker than assuming full independence).
Here the “tail probability” (the bound hidden in “whp”) decays polynomially in the deviation. Later on we will derive much sharper rates of decay (exponential) using more powerful tools such as the Chernoff bound when the r.v.’s are independent.
Let us now return to the problem of determining when 𝐺(𝑛, 𝑝) contains a triangle whp.
Theorem 4.1.11 If 𝑛𝑝→∞, then 𝐺(𝑛, 𝑝) contains a triangle with probability 1 −𝑜(1).
Proof. Label the vertices by [𝑛]. Let 𝑋𝑖𝑗be the indicator random variable of the edge 𝑖𝑗, so that 𝑋𝑖𝑗= 1 if the edge is present, and 𝑋𝑖𝑗= 0 if the edge is not present in the random graph. Let us write 𝑋𝑖𝑗𝑘:= 𝑋𝑖𝑗𝑋𝑖𝑘𝑋𝑗𝑘.
Then the number of triangles in 𝐺(𝑛, 𝑝) is given by 𝑋= ∑︁ 𝑖< 𝑗<𝑘 𝑋𝑖𝑗𝑋𝑖𝑘𝑋𝑗𝑘.
Now we compute Var 𝑋. Note that the summands of 𝑋are not all independent.
If 𝑇1 and 𝑇2 are each 3-vertex subsets, then Cov[𝑋𝑇1, 𝑋𝑇2] = E[𝑋𝑇1𝑋𝑇2] −E[𝑋𝑇1]E[𝑋𝑇2] = 𝑝𝑒(𝑇1∪𝑇2) −𝑝𝑒(𝑇1)+𝑒(𝑇2) = 0 if |𝑇1 ∩𝑇2| ≤1, 𝑝5 −𝑝6 if |𝑇1 ∩𝑇2| = 2, 𝑝3 −𝑝6 if 𝑇1 = 𝑇2.
40 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.1 Does a typical random graph contain a triangle?
The number of pairs (𝑇1,𝑇2) of triangles sharing exactly on edge is 𝑂(𝑛4). Thus Var 𝑋= ∑︁ 𝑇1,𝑇2 Cov[𝑋𝑇1, 𝑋𝑇2] = 𝑂(𝑛3)(𝑝3 −𝑝6) + 𝑂(𝑛4)(𝑝5 −𝑝6) ≲𝑛3𝑝3 + 𝑛4𝑝5 = 𝑜(𝑛6𝑝6) as 𝑛𝑝→∞.
Thus Var 𝑋= 𝑜(E𝑋)2, and hence 𝑋> 0 whp by Corollary 4.1.8.
□ Here is what we have learned so far: for 𝑝= 𝑝𝑛and as 𝑛→∞, P(𝐺(𝑛, 𝑝) contains a triangle) → ( 0 if 𝑛𝑝→0, 1 if 𝑛𝑝→∞.
We say that 1/𝑛is a threshold for containing a triangle, in the sense that if 𝑝grows asymptotically faster than this threshold, i.e., 𝑝≫1/𝑛, then the event occurs with probability 1 −𝑜(1), while if 𝑝≪1/𝑛, then the event occurs with probability 𝑜(1).
Note that the definition of a threshold ignores leading constant factors (so that it is also correct to say that 2/𝑛is also a threshold for containing a triangle). Determining the thresholds of various properties in random graphs (as well as other random settings) is a central topic in probabilistic combinatorics. We will discuss thresholds in more depth later in this chapter.
What else might you want to know about the probability that 𝐺(𝑛, 𝑝) contains a triangle?
Remark 4.1.12 (Poisson limit). What if 𝑛𝑝→𝑐> 0 for some constant 𝑐> 0? It turns out in this case that the number of triangles of 𝐺(𝑛, 𝑝) approaches a Poisson distribution with constant mean. You will show this in the homework. It will be done via the method of moments: if 𝑍is some random variable with sufficiently nice properties (known as “determined by moments”, which holds for many common distributions such as the Poisson distribution and the normal distribution), and 𝑋𝑛is a sequence of random variables such that E𝑋𝑘 𝑛→E𝑍𝑘for all nonnegative integers 𝑘, then 𝑋𝑛converges in distribution to 𝑍.
Remark 4.1.13 (Asymptotic normality). Suppose 𝑛𝑝→∞. From the above proof, we also deduce that 𝑋∼E𝑋, i.e., the number of triangles is concentrated around its mean. In fact, we know much more. It turns out that the number 𝑋of triangles in 𝐺(𝑛, 𝑝) is asymptotically normal, meaning that it satisfies a central limit theorem: (𝑋− E𝑋)/ √ Var 𝑋converges in distribution to the standard normal 𝑁(0, 1) in distribution.
This was shown by Rucinski (1988) via the method of moments, by computing the 𝑘-th moment of (𝑋−E𝑋)/ √ Var 𝑋in the limit, and showing that it agrees with the 𝑘-th moment of the standard normal.
41 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment In the homework, you will prove the asymptotic normality of 𝑋using a later-found method of projections. The idea is to show that that 𝑋close to another random variable that is already known to be asymptotically normal by checking that their difference has negligible variance. For triangle counts, when 𝑝≫𝑛−1/2, we can compare the number of triangles to the number of edges after a normalization. The method can be further modified for greater generality. See the §6.4 in the book Random Graphs by Janson, Łuczak, and Rucinski (2000).
Remark 4.1.14 (Better tail bounds). Later on we will use more powerful tools (in-cluding martingale methods and Azuma-Hoeffding inequalities, and also Janson in-equalities) to prove better tail bounds on triangle (and other subgraph) counts.
4.2 Thresholds for fixed subgraphs In the last section, we determined the threshold for 𝐺(𝑛, 𝑝) to contain a triangle. What about other subgraphs instead of a triangle? In this section, we give a complete answer to this question for any fixed subgraph.
Question 4.2.1 What is the threshold for containing a fixed 𝐻as a subgraph?
In other words, we wish to find some sequence 𝑞𝑛so that: • (0-statement) if 𝑝𝑛/𝑞𝑛→0 (i.e., 𝑝𝑛≪𝑞𝑛), then 𝐺(𝑛, 𝑝𝑛) contains 𝐻with probability 𝑜(1); • (1-statement) if 𝑝𝑛/𝑞𝑛→∞(i.e., 𝑝𝑛≫𝑞𝑛), then 𝐺(𝑛, 𝑝𝑛) contains 𝐻with probability 1 −𝑜(1).
(It is not a priori clear why such a threshold exists in the first place. In fact, threshold always exist for monotone properties, as we will see in the next section.) Building on our calculations for triangles from previous section, let us consider a more general setup for estimating the variance so that we can be more organized in our calculations.
Setup 4.2.2 (for variance bound with few dependencies) Suppose 𝑋= 𝑋1 + · · · + 𝑋𝑚where 𝑋𝑖is the indicator random variable for event 𝐴𝑖.
Write 𝑖∼𝑗if 𝑖≠𝑗and the pair of events (𝐴𝑖, 𝐴𝑗) are not independent. Define 𝚫∗:= max 𝑖 ∑︁ 𝑗:𝑗∼𝑖 P 𝐴𝑗 𝐴𝑖 .
42 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.2 Thresholds for fixed subgraphs Remark 4.2.3.
(a) For many applications with an underlying symmetry between the events, the sum in the definition of Δ∗does not actually depend on 𝑖.
(b) In the definition of the dependency graph (𝑖∼𝑗) above, we are only considering pairwise dependence. Later on when we study the Lovász Local Lemma, we will need a strong notion of a dependency graph.
(c) This method is appropriate for a collection of events with few dependencies. It is not appropriate for where there are many weak dependencies (e.g., Section 4.5 on the Hardy–Ramanujan theorem on the number of distinct prime divisors).
We have the bound Cov[𝑋𝑖, 𝑋𝑗] = E[𝑋𝑖𝑋𝑗] −E[𝑋𝑖]E[𝑋𝑗] ≤E[𝑋𝑖𝑋𝑗] = P[𝐴𝑖𝐴𝑗] = P(𝐴𝑖)P(𝐴𝑗|𝐴𝑖).
(Here 𝐴𝑖𝐴𝑗is the shorthand for 𝐴𝑖∧𝐴𝑗, meaning that both events occur.) Also Cov[𝑋𝑖, 𝑋𝑗] = 0 if 𝑖≠𝑗and 𝑖≁𝑗.
Thus Var 𝑋= 𝑚 ∑︁ 𝑖,𝑗=1 Cov[𝑋𝑖, 𝑋𝑗] ≤ 𝑚 ∑︁ 𝑖=1 P(𝐴𝑖) + 𝑚 ∑︁ 𝑖=1 P(𝐴𝑖) ∑︁ 𝑗:𝑗∼𝑖 P(𝐴𝑗|𝐴𝑖) ≤E𝑋+ (E𝑋)Δ∗.
Recall from Corollary 4.1.8 that E𝑋> 0 and Var 𝑋= 𝑜(E𝑋)2 imply 𝑋> 0 and 𝑋∼E𝑋whp. So we have the following.
Lemma 4.2.4 In the above setup, if E𝑋→∞and Δ∗= 𝑜(E𝑋), then 𝑋> 0 and 𝑋∼E𝑋whp.
Let us now determine the threshold for containing 𝐾4.
Theorem 4.2.5 The threshold for containing 𝐾4 is 𝑛−2/3.
Proof. Let 𝑋denote the number of copies of 𝐾4 in 𝐺(𝑛, 𝑝). Then E𝑋= 𝑛 4 𝑝6 ≍𝑛4𝑝6.
If 𝑝≪𝑛−2/3 then E𝑋= 𝑜(1), and thus 𝑋= 0 whp by Markov’s inequality.
43 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment Now suppose 𝑝≫𝑛−2/3, so E𝑋→∞. For each 4-vertex subset 𝑆, let 𝐴𝑆be the event that 𝑆is a clique in 𝐺(𝑛, 𝑝).
For each fixed 𝑆, one has 𝐴𝑆∼𝐴𝑆′ if and only if |𝑆∩𝑆′| ≥2.
• The number of 𝑆′ that share exactly 2 vertices with 𝑆is 6 𝑛−2 2 = 𝑂(𝑛2), and for each such 𝑆′ one has P(𝐴𝑆′|𝐴𝑆) = 𝑝5 (as there are 5 additional edges not in the 𝑆-clique that need to appear clique to form the 𝑆′-clique).
• The number of 𝑆′ that share exactly 3 vertices with 𝑆is 4(𝑛−4) = 𝑂(𝑛), and for each such 𝑆′ one has P(𝐴𝑆′|𝐴𝑆) = 𝑝3.
Summing over all above 𝑆′, we find Δ∗= ∑︁ 𝑆′:|𝑆′∩𝑆|∈{2,3} P(𝐴𝑆′|𝐴𝑆) ≲𝑛2𝑝5 + 𝑛𝑝3 ≪𝑛4𝑝6 ≍E𝑋.
Thus 𝑋> 0 whp by Lemma 4.2.4.
□ For both 𝐾3 and 𝐾4, we saw that any choice of 𝑝= 𝑝𝑛with E𝑋→∞one has 𝑋> 0 whp. Is this generally true?
Example 4.2.6 (First moment is not enough). Let 𝐻= . We have E𝑋𝐻≍𝑛5𝑝7.
If E𝑋= 𝑜(1) then 𝑋= 0 whp. But what if E𝑋→∞, i.e., 𝑝≫𝑛−5/7?
We know that if 𝑛−5/7 ≪𝑝≪𝑛−2/3, then 𝑋𝐾4 = 0 whp, so 𝑋𝐻= 0 whp since 𝐾4 ⊆𝐻.
On the other hand, if 𝑝≫𝑛−2/3, then whp can find 𝐾4, and pick an arbitrary edge to extend to 𝐻(we’ll prove this).
Thus the threshold for 𝐻= is actually 𝑛−2/3, and not 𝑛−5/7 as one might have naively predicted from the first moment alone.
Why didn’t E𝑋𝐻→∞give 𝑋𝐻> 0 whp in our proof strategy? In the calculation of Δ∗, one of the terms is ≍𝑛𝑝(from two copies of 𝐻with a 𝐾4-overlap), and 𝑛𝑝3 𝑛5𝑝7 ≍E𝑋𝐻if 𝑝≪𝑛−2/3.
The above example shows that the threshold is not always necessarily determined by the expectation. For the property of containing 𝐻, the example suggests that we should look at the “densest” subgraph of 𝐻rather than containing 𝐻itself.
44 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.2 Thresholds for fixed subgraphs Definition 4.2.7 Define the edge-vertex ratio of a graph 𝐻by 𝝆(𝑯) := 𝑒𝐻 𝑣𝐻 .
(This is the same as half the average degree.) Define the maximum edge-vertex ratio of a subgraph of 𝐻: 𝒎(𝑯) := max 𝐻′⊆𝐻𝜌(𝐻′).
Example 4.2.8. Let 𝐻= . We have 𝜌(𝐻) = 7/5 whereas 𝜌(𝐾4) = 3/2 > 7/5.
It is not hard to check that 𝑚(𝐻) = 𝜌(𝐾4) = 3/2 as 𝐾4 is the subgraph of 𝐻with the maximum edge-vertex ratio.
Remark 4.2.9 (Algorithm). Goldberg (1984) found a polynomial time algorithm for computing 𝑚(𝐻) via network flow algorithms.
The next theorem completes determines the threshold for containing some fixed graph 𝐻. Basically, it is determined by the expected number of copies of 𝐻′, where 𝐻′ is the “denest” subgraph of 𝐻(i.e., with the maximum edge-vertex ratio).
Theorem 4.2.10 (Threshold for containing a fixed graph: Bollobás 1981) Fix a graph 𝐻. Then 𝑝= 𝑛−1/𝑚(𝐻) is a threshold for containing 𝐻has a subgraph.
Proof. Let 𝐻′ be a subgraph of 𝐻achieving the maximum edge-vertex ratio, i.e., 𝜌(𝐻′) = 𝑚(𝐻). Let 𝑋𝐻denote the number of copies of 𝐻in 𝐺(𝑛, 𝑝).
If 𝑝≪𝑛−1/𝑚(𝐻), then E𝑋𝐻′ ≍𝑛𝑣𝐻′ 𝑝𝑒𝐻′ = 𝑜(1), so 𝑋𝐻′ = 0 whp, hence 𝑋𝐻= 0 whp.
Now suppose 𝑝≫𝑛−1/𝑚(𝐻).
Let us count labeled copies of the subgraph 𝐻in 𝐺(𝑛, 𝑝). Let 𝐽be a labeled copy of 𝐻in 𝐾𝑛, and let 𝐴𝐽denote the event that 𝐽appears in 𝐺(𝑛, 𝑝). We have, for fixed 𝐽, Δ∗= ∑︁ 𝐽′∼𝐽 P (𝐴𝐽′ | 𝐴𝐽) = ∑︁ 𝐽′∼𝐽 𝑝|𝐸(𝐽′)\𝐸(𝐽)| For any 𝐽′ ∼𝐽, we have 𝑛|𝑉(𝐽′)\𝑉(𝐽)|𝑝|𝐸(𝐽′)\𝐸(𝐽)| ≪𝑛|𝑉(𝐽)|𝑝|𝐸(𝐽)| 45 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment since 𝑝≫𝑛−1/𝑚(𝐻) ≥𝑛−1/𝜌(𝐽∩𝐽′) = 𝑛−|𝑉(𝐽)∩𝑉(𝐽′)|/|𝐸(𝐽)∩𝐸(𝐽′)|.
It then follows, after considering all the possible ways that 𝐽′ can overlap with 𝐽, that Δ∗≪𝑛|𝑉(𝐽)|𝑝|𝐸(𝐽)| ≍E𝑋𝐻. So Lemma 4.2.4 yields the result.
□ Remark 4.2.11. The proof also gives that if 𝑝≫𝑛−1/𝑚(𝐻), then the number 𝑋𝐻of copies of 𝐻is concentrated near its mean, i.e., with probability 1 −𝑜(1), 𝑋𝐻∼E𝑋𝐻= 𝑛 𝑣𝐻 𝑣𝐻!
aut(𝐻) 𝑝𝑒𝐻∼𝑛𝑣𝐻𝑝𝑒𝐻 aut(𝐻) .
4.3 Thresholds Previously, we computed the threshold for containing a fixed 𝐻as a subgraph. In this section, we take a detour from the discussion of the second moment method and discuss thresholds in more detail.
We begin by discussing the concept more abstractly by first defining the threshold of any monotone property on subsets. Then we show that thresholds always exist.
Thresholds form a central topic in probabilistic combinatorics. For any given property, it is natural to ask the following questions: 1. Where is the threshold?
2. Is the transition sharp? (And more precisely, what is width of the transition window?) We understand thresholds well for many basic graph properties, but for many others, it can be a difficult problem. Also, one might think that one must first understand the location of the threshold before determining the nature of the phase transition, but surprisingly this is actually not always the case. There are powerful results that can sometimes show a sharp threshold without identifying the location of the threshold.
Here is some general setup, before specializing to graphs.
Let Ω be some finite set (ground set). Let Ω𝑝be a random subset of Ω where each element is included with probability 𝑝independently.
An increasing property, also called monotone property, on subsets of Ω is some binary property so that if 𝐴⊆Ω satisfies the property, any superset of 𝐴automatically satisfies the property.
A property is trivial if all subsets of Ω satisfy the property, or if all subsets of Ω do not satisfy the property. From now on, we only consider non-trivial monotone properties.
46 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.3 Thresholds A graph property is a property that only depends on isomorphism classes of graphs.
Whether the random graph 𝐺(𝑛, 𝑝) satisfies a given property can be cast in our setup by viewing 𝐺(𝑛, 𝑝) as Ω𝑝with Ω = [𝑛] 2 .
Here are some examples of increasing properties for subgraphs of a given set of vertices: • Contains some given subgraph • Connected • Has perfect matching • Hamiltonian • non-3-colorable A family F ⊆P(Ω) of subsets of Ω is called an up-set if whenever 𝐴∈F and 𝐴⊆𝐵, then 𝐵∈F . Increasing property is the same as being an element of an up-set. We will use these two terms interchangeably.
Definition 4.3.1 (Threshold) Let Ω = Ω(𝑛) be a finite set and F = F (𝑛) an monotone property of subsets of Ω. We say that 𝑞𝑛is a threshold for F if, P(Ω𝑝𝑛∈F ) → ( 0 if 𝑝𝑛/𝑞𝑛→0, 1 if 𝑝𝑛/𝑞𝑛→∞.
Remark 4.3.2. The above definition is only for increasing properties. We can similarly define the threshold for decreasing properties by an obvious modification. An example of a non-monotone property is sontaining some 𝐻as an induced subgraph. Some (but not all) non-monotone properties also have thresholds, though we will not discuss it here.
Remark 4.3.3. From the definition, we see that if 𝑟𝑛and 𝑟′ 𝑛are both thresholds of the same property, then they must be within a constant factor of each other (exercise: check this). Thus it makes sense to say “the threshold” rather than “a threshold.” Existence of threshold Question 4.3.4 (Existence of threshold) Does every non-trivial monotone property have a threshold?
47 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment How would a monotone property not have a threshold?
Perhaps one could have P(Ω1/𝑛∈F ) and P(Ω(log 𝑛)/𝑛∈F ) ∈[1/10, 9/10] for all sufficiently large 𝑛?
Before answer this question, let us consider an even more elementary claim.
Theorem 4.3.5 (Monotonicity of satisfying probability) Let Ω be a finite set and F a non-trivial monotone property of subsets of Ω. Then 𝑝↦→P(Ω𝑝∈F ) is a strictly increasing function of 𝑝∈[0, 1].
Let us give two related proofs of this basic fact. Both are quite instructive. Both are based on coupling of random processes.
Proof 1. Let 0 ≤𝑝< 𝑞≤1. Consider the following process to generate two random subsets of Ω. For each 𝑥, generate uniform 𝑡𝑥∈[0, 1] independently at random. Let 𝐴= {𝑥∈Ω : 𝑡𝑥≤𝑝} and 𝐵= {𝑥∈Ω : 𝑡𝑥≤𝑞} .
Then 𝐴has the same distribution as Ω𝑝and 𝐵has the same distribution as Ω𝑞.
Furthermore, since 𝑝< 𝑞, we always have 𝐴⊆𝐵. Since F is monotone, 𝐴∈F implies 𝐵∈F . Thus P(Ω𝑝∈F ) = P(𝐴∈F ) ≤P(𝐵∈F ) = P(Ω𝑞∈F ).
To see that the inequality strict, we simply have to observe that with positive probability, one has 𝐴∉F and 𝐵∈F (e.g., if all 𝑡𝑥∈(𝑝, 𝑞], then 𝐴= ∅and 𝐵= Ω).
□ In the second proof, the idea is to reveal a random subset of Ω in independent random stages.
Proof 2. (By two-round exposure) Let 0 ≤𝑝< 𝑞≤1. Note that 𝐵= Ω𝑞has the same distribution as the union of two independent 𝐴= Ω𝑝and 𝐴′ = Ω𝑝′, where 𝑝′ is chosen to satisfy 1 −𝑞= (1 −𝑝)(1 −𝑝′) (check that the probability that each element occurs is the same in the two processes). Thus P(𝐴∈F ) ≤P(𝐴∪𝐴′ ∈F ) = P(𝐵∈F ).
Like earlier, to observe that the inequality is strict, one observes that with positive probability, one has 𝐴∉F and 𝐴∪𝐴′ ∈F .
□ The above technique (generalized from two round exposure to multiple round ex-posures) gives a nice proof of the following theorem (originally proved using the Kruskal–Katona theorem).1 1(Thresholds for random subspaces of F𝑛 𝑞) The proof of the Bollobás–Thomason paper using the 48 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.3 Thresholds Theorem 4.3.6 (Existence of thresholds: Bollobás and Thomason 1987) Every sequence of nontrivial monotone properties has a threshold.
The theorem follows from the next non-asymptotic claim.
Lemma 4.3.7 (Multiple round exposure) Let Ω be a finite set and F some non-trivial monotone property. If 𝑝∈[0, 1] and 𝑚 is nonnegative integer. Then P(Ω𝑝∉F ) ≤P(Ω𝑝/𝑚∉F )𝑚.
Proof. Consider 𝑚independent copies of Ω𝑝/𝑚, and let 𝑌be their union. Since F is monotone increasing, if 𝑌∉F , then none of the 𝑚copies lie in F . Hence P(𝑌∉F ) ≤P(Ω𝑝/𝑚∉F )𝑚.
Note that 𝑌has the same distribution as Ω𝑞for some 𝑞≤𝑝. So P(Ω𝑝∉F ) ≤ P(Ω𝑞∉F ) = P(𝑌∉F ) by Theorem 4.3.5. Combining the two inequalities gives the result.
□ Proof of Theorem 4.3.6. Since 𝑝↦→P(Ω𝑝∈F ) is a continuous strictly increasing function from 0 to 1 as 𝑝goes from 0 to 1 (in fact it is a polynomial in 𝑝), there is some unique “critical probability” 𝑝𝑐so that P(Ω𝑝𝑐∈F ) = 1/2.
It remains to check for every 𝜀> 0, there is some 𝑚= 𝑚(𝜀) (not depending on the property) so that P(Ω𝑝𝑐/𝑚∉F ) ≥1 −𝜀 and P(Ω𝑚𝑝𝑐∉F ) ≤𝜀.
Kruskal–Katona theorem is still relevant. For example, there is an interesting analog of this concept for properties of subspaces of F𝑛 𝑞, i.e., random linear codes instead of random graphs. The analogue of the Bollobás–Thomason theorem was proved by Rossman (2020) via the the Kruskal–Katona approach. The multiple round exposure proof does not seem to work in the random subspace setting, as one cannot write a subspace as a union of independent copies of smaller subspaces.
As an aside, I disagree with the use of the term “sharp threshold” in Rossman’s paper for describing all thresholds for subspaces—one really should be looking at the cardinality of the subspaces rather than their dimensions. In a related work by Guruswami, Mosheiff, Resch, Silas, and Wootters (2022), they determine thresholds for random linear codes for properties that seem to be analogous to the property that a random graph contains a given fixed subgraph. Here again I disagree with them calling it a “sharp threshold.” It is much more like a coarse threshold once you parameterize by the cardinality of the subspace, which gives you a much better analogy to the random graph setting.
Thresholds for random linear codes seems to an interesting topic that has only recently been studied. I think there is more to be done here.
49 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment (here we write Ω𝑡= Ω if 𝑡> 1.) Indeed, applying Lemma 4.3.7 with 𝑝= 𝑝𝑐, we have P(Ω𝑝𝑐/𝑚∉F ) ≥P(Ω𝑝𝑐∉F )1/𝑚= 2−1/𝑚≥1 −𝜀 if 𝑚≥(log 2)/𝜀.
Applying Lemma 4.3.7 again with 𝑝= 𝑚𝑝𝑐, we have P(Ω𝑚𝑝𝑐∉F ) ≤P(Ω𝑝𝑐∉F )𝑚= 2−𝑚≤𝜀 if 𝑚≥log2(1/𝜀).
Thus 𝑝𝑐is a threshold for F .
□ Examples We will primarily be studying monotone graph properties. In the previous notation, Ω = [𝑛] 2 , and we are only considering properties that depend on the isomorphism class of the graph.
Example 4.3.8 (Containing a triangle). We saw earlier in the chapter that the threshold for containing a triangle is 1/𝑛: P(𝐺(𝑛, 𝑝) contains a triangle) → 0 if 𝑛𝑝→0, 1 −𝑒−𝑐3/6 if 𝑛𝑝→𝑐∈(0, ∞) 1 if 𝑛𝑝→∞.
In this case, the threshold is determined by the expected number of triangles Θ(𝑛3𝑝3), and whether this quantity goes to zero or infinity (in the latter case, we used a second moment method to show that the number of triangles is positive with high probability).
What if 𝑝= Θ(1/𝑛)? If 𝑛𝑝→𝑐for some constant 𝑐> 0, then (you will show in the homework via the method of moments) that the number of triangles is asymptotically Poisson distributed, and in particular the limit probability of containing a triangle increases from 0 to 1 as a continuous function of 𝑐∈(0, ∞). So, in particular, as 𝑝increases, it goes through a “window of transition” of width Θ(1/𝑛) in order for P(𝐺(𝑛, 𝑝) contains a triangle) to increase from 0.01 to 0.99. The width of this window is on the same order as the threshold. In this case, we call it a coarse transition.
Example 4.3.9 (Containing a subgraph). Theorem 4.2.10 tells us that the threshold for containing a fixed subgraph 𝐻is 𝑛−1/𝑚(𝐻).
Here the threshold is not always determined by the expected number of copies of 𝐻.
Instead, we need to look at the “densest subgraph” 𝐻′ ⊆𝐻with the largest edge-vertex ratio (i.e., equivalent to largest average degree). The threshold is determined by whether the expected number of copies of 𝐻′ goes to zero or infinity.
Similar to the triangle case, we have a coarse threshold.
50 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.3 Thresholds The analysis can also be generalized to containing one of several fixed subgraphs 𝐻1, . . . , 𝐻𝑘.
Remark 4.3.10 (Monotone graph properties are characterized by subgraph con-tainment). Every monotone graph property can be characterized as containing some element of H for some H that could depend on the vertex set 𝑛. For example, the property of connectivity corresponds to taking H to be all spanning trees.
More generally, one can take H to be the set of all minimal graphs satisfying the property.
When elements of H are unbounded in size, the problem of thresholds become quite interesting and sometimes difficult.
The original Erdős–Rényi (1959) paper on random graphs already studied several thresholds, such as the next two examples.
Example 4.3.11 (No isolated vertices). With 𝑝= log 𝑛+ 𝑐𝑛 𝑛 , P (𝐺(𝑛, 𝑝) has no isolated vertices) → 0 if 𝑐𝑛→−∞ 1 −𝑒𝑒−𝑐 if 𝑐𝑛→𝑐 1 if 𝑐𝑛→∞ It is a good exercise (and included in the problem set) to check the above claims. The cases 𝑐𝑛→−∞and 𝑐𝑛→∞can be shown using the second moment method. More precisely, when 𝑐𝑛→𝑐, by comparing moments one can show that the number of isolated vertices is asymptotically Poisson.
In this example, the threshold is at (log 𝑛)/𝑛. As we see above, the transition window is Θ(1/𝑛), much narrower the magnitude of the threshold. In particular, the event probability goes from 𝑜(1) to 1 −𝑜(1) when 𝑝increases from (1 −𝛿)(log 𝑛)/𝑛to (1 + 𝛿)(log 𝑛)/𝑛for any fixed 𝛿> 0. In this case, we say that the property has a sharp threshold at (log 𝑛)/𝑛(here the leading constant factor is relevant, unlike the earlier example of a coarse threshold).
Example 4.3.12 (Connectivity). With 𝑝= log 𝑛+ 𝑐𝑛 𝑛 P (𝐺(𝑛, 𝑝) is connected) → 0 if 𝑐𝑛→−∞ 1 −𝑒𝑒−𝑐 if 𝑐𝑛→𝑐 1 if 𝑐𝑛→∞ In fact, a much stronger statement is true, connecting the above two examples: consider a process where one adds a random edges one at a time, then with probability 1−𝑜(1), 51 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment the graph becomes connected as soon as there are no more isolated vertices. Such stronger characterization is called a hitting time result.
A similar statement is true if we replace “is connected” by “has a perfect matching” (assuming 𝑛even).
Example 4.3.13 (Perfect matching in a random hypergraph: Shamir’s problem).
Let 𝐺(3)(𝑛, 𝑝) be a random hypergraph on 𝑛vertices, where each triple of vertices appears as an edge with probability 𝑝. Assume that 𝑛is divisible by 3. What is the threshold for the existence of a perfect matching (i.e., a set of 𝑛/3 edges covering all vertices)?
It is easy to check that the property of having no isolated vertices has a sharp threshold at 𝑝= 2𝑛−2 log 𝑛. Is this also a threshold for having a perfect matching? So for smaller 𝑝, one cannot have a perfect matching due to having an isolated vertex. What about larger 𝑝? This turns out to be a difficult problem known as “Shamir’s problem”.
A difficult result by Johansson, Kahn, and Vu (2008) (this paper won a Fulkerson Prize) showed that there is some constant 𝐶> 0 so that if 𝑝≥𝐶𝑛−2 log 𝑛then 𝐺(3)(𝑛, 𝑝) contains a perfect matching with high probability. They also solved the problem much generally for 𝐻-factors in random 𝑘-uniform hypergraphs.
Recent exciting breakthroughs on the Kahn–Kalai conjecture (2007) by Frankston, Kahn, Narayanan, and Park (2021) and Park and Pham (2024) provide new and much shorter proofs of this threshold for Shamir’s problem.
Recently, Kahn (2022) proved a sharp threshold result, and actually an even stronger hitting time version, of Shamir’s problem, showing that with high probability, one has a perfect matching as soon as there are no isolated vertices.
Sharp transition In some of the examples, the probability that 𝐺(𝑛, 𝑝) satisfies the property changes quickly and dramatically as 𝑝crosses the threshold (physical analogy: similar to how the structure of water changes dramatically as the temperature drops below freezing).
For example, while for connectivity, while 𝑝= log 𝑛/𝑛is a threshold, we see that 𝐺(𝑛, 0.99 log 𝑛/𝑛) is whp not connected and 𝐺(𝑛, 1.01 log 𝑛/𝑛) is whp connected, unlike the situation for containing a triangle earlier. We call this the sharp threshold phenomenon.
52 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.3 Thresholds i S o o Pt P Igoe P Dpet Coarse threshold Sharp threshold Figure 4.1: Examples of coarse and sharp thresholds. The vertical axis is the probability that 𝐺(𝑛, 𝑝) satisfies the property.
Definition 4.3.14 (Sharp thresholds) We say that 𝑟𝑛is a sharp threshold for some property F on subsets of Ω if, for every 𝛿> 0, P(Ω𝑝𝑛∈F ) → ( 0 if 𝑝𝑛/𝑟𝑛≤1 −𝛿, 1 if 𝑝𝑛/𝑟𝑛≥1 + 𝛿.
On the other hand, if there is some fixed 𝜀> 0 and 0 < 𝑐< 𝐶so that P(Ω𝑝𝑛∈F ) ∈ [𝜀, 1 −𝜀] for whenever 𝑐≤𝑝𝑛/𝑟𝑛≤𝐶, then we say that 𝑟𝑛is a coarse threshold.
As in Figure 4.1, the sharp/coarseness of a thresholds is about how quickly P(Ω𝑝∈F ) transitions from 𝜀to 1−𝜀as 𝑝increases. How wide is the transition window for 𝑝? By the Bollobás–Thomason theorem (Theorem 4.3.6) on the existence of thresholds, this transition window always has width 𝑂(𝑟𝑛). If the transition window has width Θ(𝑟𝑛) for some 𝜀> 0, then we have a coarse threshold. On the other hand, if the transition window has width 𝑜(𝑟𝑛) for every 𝜀> 0, then we have a sharp threshold.
From earlier examples, we saw coarse thresholds for the “local” property of containing some given subgraph, as well as sharp thresholds for “global” properties such as connectivity. It turns out that this is a general phenomenon.
Friedgut’s sharp threshold theorem (1999), a deep and important result, completely characterizes when a threshold is coarse versus sharp. We will not state Friedgut’s theorem precisely here since it is rather technical (and actually not always easy to apply). Let us just give a flavor. Roughly speaking, the theorem says that: All monotone graph properties with a coarse threshold may be approxi-mated by a local property.
In other words, informally, if a monotone graph property P has a coarse threshold, then there is finite list of graph 𝐺1, . . . , 𝐺𝑚such that P is “close to” the property of containing one of 𝐺1, . . . , 𝐺𝑚as a subgraph.
53 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment We need “close to” since the property could be “contains a triangle and has at least log 𝑛edges”, which is not exactly local but it is basically the same as “contains a triangle.” There is some subtlety here since we can allow very different properties depending on the value of 𝑛. E.g., P could be the set of all 𝑛-vertex graphs that contain a 𝐾3 if 𝑛is odd and 𝐾4 if 𝑛is even. Friedgut’s theorem tells us that if there is a threshold, then there is a partition N = N1 ∪· · · ∪N𝑘such that on each N𝑖, P is approximately the form described in the previous paragraph.
In the last section, we derived that the property of containing some fixed 𝐻has threshold 𝑛−1/𝑚(𝐻) for some rational number 𝑚(𝐻).
It follows as a corollary of Friedgut’s theorem that every coarse threshold must have this form.
Corollary 4.3.15 (of Friedgut’s sharp threshold theorem) Suppose 𝑟(𝑛) is a coarse threshold of some graph property. Then there is a partition of N = N1 ∪· · · ∪N𝑘and rationals 𝛼1, . . . , 𝛼𝑘> 0 such that 𝑟(𝑛) ≍𝑛−𝛼𝑗for every 𝑛∈N𝑗.
In particular, if (log 𝑛)/𝑛is a threshold of some monotone graph property (e.g., this is the case for connectivity), then we automatically know that it must be a sharp threshold, even without knowing anything else about the property. Likewise if the threshold has the form 𝑛−𝛼for some irrational 𝛼.
The exact statement of Friedgut’s theorem is more cumbersome. We refer those who are interested to Friedgut’s original 1999 paper and his later survey for details and applications. This topic is connected more generally to an area known as the analysis of boolean functions.
Also, it is known that the transition window of every monotone graph property is (log 𝑛)−2+𝑜(1) (Friedgut––Kalai (1996), Bourgain–Kalai (1997)).
Curiously, tools such as Friedgut’s theorem sometimes allow us to prove the existence of a sharp threshold without being able to identify its exact location. For example, it is an important open problem to understand where exactly is the transition for a random graph to be 𝑘-colorable.
Conjecture 4.3.16 (𝑘-colorability threshold) For every 𝑘≥3 there is some real constant 𝑑𝑘> 0 such that for any constant 𝑑> 0, P(𝐺(𝑛, 𝑑/𝑛) is 𝑘-colorable) → ( 1 if 𝑑< 𝑑𝑘, 0 if 𝑑> 𝑑𝑘.
We do know that there exists a sharp threshold for 𝑘-colorability.
54 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.4 Clique number of a random graph Theorem 4.3.17 (Achlioptas and Friedgut 2000) For every 𝑘≥3, there exists a function 𝑑𝑘(𝑛) such that for every 𝜀> 0, and sequence 𝑑(𝑛) > 0, P 𝐺 𝑛, 𝑑(𝑛) 𝑛 is 𝑘-colorable → ( 1 if 𝑑(𝑛) < 𝑑𝑘(𝑛) −𝜀, 0 if 𝑑(𝑛) > 𝑑𝑘(𝑛) + 𝜀.
On the other hand, it is not known whether lim𝑛→∞𝑑𝑘(𝑛) exists, which would imply Conjecture 4.3.16. Further bounds on 𝑑𝑘(𝑛) are known, e.g. the landmark paper of Achlioptas and Naor (2006) showing that for each fixed 𝑑> 0, whp 𝜒(𝐺(𝑛, 𝑑/𝑛) ∈ {𝑘𝑑, 𝑘𝑑+ 1} where 𝑘𝑑= min{𝑘∈N : 2𝑘log 𝑘> 𝑑}. Also see the later work of Coja-Oghlan and Vilenchik (2013).
4.4 Clique number of a random graph The clique number 𝝎(𝑮) of a graph is the maximum number of vertices in a clique of 𝐺.
Question 4.4.1 What is the clique number of 𝐺(𝑛, 1/2)?
Let 𝑋be the number of 𝑘-cliques of 𝐺(𝑛, 1/2). Define 𝑓(𝑛, 𝑘) := E𝑋= 𝑛 𝑘 2−(𝑘 2).
Let us first do a rough estimate to see what is the critical 𝑘to get 𝑓(𝑛, 𝑘) large or small.
Recall that 𝑛 𝑒𝑘 𝑘≤ 𝑛 𝑘 ≤ 𝑒𝑛 𝑘 𝑘. We have log2 𝑓(𝑛, 𝑘) = 𝑘 log2 𝑛−log2 𝑘−𝑘 2 + 𝑂(1) .
And so the transition point is at some 𝑘∼2 log2 𝑛in the sense that if 𝑘≥(2+𝛿) log2 𝑛, then 𝑓(𝑛, 𝑘) →0 while if 𝑘≤(2 −𝛿) log2 𝑛, then 𝑓(𝑛, 𝑘) →∞.
The next result gives us a lower bound on the typical clique number.
55 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment Theorem 4.4.2 (Second moment bound for clique number) Let 𝑘= 𝑘(𝑛) be some sequence of positive integers.
(a) If 𝑓(𝑛, 𝑘) →0, then 𝜔(𝐺(𝑛, 1/2)) < 𝑘whp.
(b) If 𝑓(𝑛, 𝑘) →∞, then 𝜔(𝐺(𝑛, 1/2)) ≥𝑘whp.
Proof sketch. The first claim follows from Markov’s inequality as P(𝑋≥1) ≤E𝑋.
For the second claim, we bound the variance using Setup 4.2.2. For each 𝑘-element subset 𝑆of vertices, let 𝐴𝑆be the event that 𝑆is a clique. Let 𝑋𝑆be the indicator random variable for 𝐴𝑆. Recall Δ∗:= max 𝑖 ∑︁ 𝑗:𝑗∼𝑖 P 𝐴𝑗 𝐴𝑖 .
For fixed 𝑘-set 𝑆, consider all 𝑘-set 𝑇with |𝑆∩𝑇| ≥2: Δ∗= ∑︁ 𝑇∈([𝑛] 𝑘) 2≤|𝑆∩𝑇|≤𝑘−1 P(𝐴𝑇|𝐴𝑆) = 𝑘−1 ∑︁ 𝑖=2 𝑘 𝑖 𝑛−𝑘 𝑘−𝑖 2(𝑖 2)−(𝑘 2) omitted ≪ E𝑋= 𝑛 𝑘 2−(𝑘 2).
It then follows from Lemma 4.2.4 that 𝑋> 0 (i.e., 𝜔(𝐺) ≥𝑘) whp.
□ We can a two-point concentration result for the clique number of 𝐺(𝑛, 1/2). This result is due to Bollobás–Erdős 1976 and Matula 1976.
Theorem 4.4.3 (Two-point concentration for clique number) There exists a 𝑘= 𝑘(𝑛) ∼2 log2 𝑛such that 𝜔(𝐺(𝑛, 1/2)) ∈{𝑘, 𝑘+ 1} whp.
Proof. For 𝑘∼2 log2 𝑛, 𝑓(𝑛, 𝑘+ 1) 𝑓(𝑛, 𝑘) = 𝑛−𝑘 𝑘+ 12−𝑘= 𝑛−1+𝑜(1).
Let 𝑘0 = 𝑘0(𝑛) ∼2 log2 𝑛be the value with 𝑓(𝑛, 𝑘0) ≥𝑛−1/2 > 𝑓(𝑛, 𝑘0 + 1).
Then 𝑓(𝑛, 𝑘0−1) →∞and 𝑓(𝑛, 𝑘0+1) = 𝑜(1). By Theorem 4.4.2, the clique number of 𝐺(𝑛, 1/2) is whp in {𝑘0 −1, 𝑘0}.
□ Remark 4.4.4. By a more careful analysis, one can show that outside a very sparse 56 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.5 Hardy–Ramanujan theorem on the number of prime divisors subset of integers, one has 𝑓(𝑛, 𝑘0) →∞, in which case one has one-point concentra-tion 𝜔(𝐺(𝑛, 1/2)) = 𝑘0 whp.
By taking the complement of the graph, we also get a two-point concentration result about the independence number of 𝐺(𝑛, 1/2). Bohman and Hofstad (2024) extended the two-point concentration result for the independence number of 𝐺(𝑛, 𝑝) to all 𝑝≥𝑛−2/3+𝜀.
Remark 4.4.5 (Chromatic number). Since the chromatic number satisfies 𝜒(𝐺) ≥ 𝑛/𝛼(𝐺), we have 𝜒(𝐺(𝑛, 1/2)) ≥(1 + 𝑜(1)) 𝑛 2 log2 𝑛 whp.
In Theorem 8.3.2, using more advanced methods, we will prove 𝜒(𝐺(𝑛, 1/2)) ∼ 𝑛/(2 log2 𝑛) whp, a theorem due to Bollobás (1987).
In Section 9.3, using martingale concentration, we will show that 𝜒(𝐺(𝑛, 𝑝)) is tightly concentrated around its mean without a priori needing to know where the mean is located.
4.5 Hardy–Ramanujan theorem on the number of prime divisors Let 𝝂(𝒏) denote the number of distinct primes dividing 𝑛(not counting multiplicities).
The next theorem says that “almost all” 𝑛have (1 + 𝑜(1)) log log 𝑛prime factors Theorem 4.5.1 (Hardy and Ramanujan 1917) For every 𝜀> 0, there exists 𝐶such that for all sufficiently large 𝑛, all but 𝜀-fraction of 𝑥∈[𝑛] satisfy |𝜈(𝑥) −log log 𝑛| ≤𝐶 √︁ log log 𝑛 The original proof of Hardy and Ramanujan was quite involved. Here we show a “probabilistic” proof due to Turán (1934), which played a key role in the development of probabilistic methods in number theory.
Proof. Choose 𝑥∈[𝑛] uniformly at random. For prime 𝑝, let 𝑋𝑝= ( 1 if 𝑝|𝑥, 0 otherwise.
57 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment Set 𝑀= 𝑛1/10, and (the sum is taken over primes 𝑝).
𝑋= ∑︁ 𝑝≤𝑀 𝑋𝑝.
We have 𝜈(𝑥) −10 ≤𝑋(𝑥) ≤𝜈(𝑥) since 𝑥cannot have more than 10 prime factors > 𝑛1/10. So it suffices to analyze 𝑋.
Since exactly ⌊𝑛/𝑝⌋positive integers ≤𝑛are divisible by 𝑝, we have E𝑋𝑝= ⌊𝑛/𝑝⌋ 𝑛 = 1 𝑝+ 𝑂 1 𝑛 .
We quote Merten’s theorem from analytic number theory: ∑︁ 𝑝≤𝑛 1/𝑝= log log 𝑛+ 𝑂(1).
(A more precise result says that 𝑂(1) error term converges to the Meissel–Mertens constant.) So E𝑋= ∑︁ 𝑝≤𝑀 1 𝑝+ 𝑂 1 𝑛 = log log 𝑛+ 𝑂(1).
Next we compute the variance. The intuition is that divisibility by distinct primes should behave somewhat independently. Indeed, if 𝑝𝑞divides 𝑛, then 𝑋𝑝and 𝑋𝑞are independent (e.g., by the Chinese Remainder Theorem). If 𝑝𝑞does not divide 𝑛, but 𝑛is large enough, then there is some small covariance contribution. (In contrast to the earlier variance calculations in random graphs, here we have many weak dependices.) If 𝑝≠𝑞, then 𝑋𝑝𝑋𝑞= 1 if and only if 𝑝𝑞|𝑥. Thus Cov[𝑋𝑝, 𝑋𝑞] = E[𝑋𝑝𝑋𝑞] −E[𝑋𝑝]E[𝑋𝑞] = ⌊𝑛/𝑝𝑞⌋ 𝑛 −⌊𝑛/𝑝⌋ 𝑛 ⌊𝑛/𝑞⌋ 𝑛 = 1 𝑝𝑞+ 𝑂 1 𝑛 − 1 𝑝+ 𝑂 1 𝑛 1 𝑞+ 𝑂 1 𝑛 = 𝑂 1 𝑛 .
Thus ∑︁ 𝑝≠𝑞 Cov[𝑋𝑝, 𝑋𝑞] ≲𝑀2 𝑛 ≲𝑛−4/5.
58 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.5 Hardy–Ramanujan theorem on the number of prime divisors Also, Var 𝑋𝑝= E[𝑋𝑝] −(E𝑋𝑝)2 = (1/𝑝)(1 −1/𝑝) + 𝑂(1/𝑛). Combining, we have Var 𝑋= ∑︁ 𝑝≤𝑀 Var 𝑋𝑝+ ∑︁ 𝑝≠𝑞 Cov[𝑋𝑝, 𝑋𝑞] = ∑︁ 𝑝≤𝑀 1 𝑝+ 𝑂(1) = log log 𝑛+ 𝑂(1) ∼E𝑋.
Thus by Chebyshev’s inequality, for every constant 𝜆> 0, P |𝑋−log log 𝑛| ≥𝜆 √︁ log log 𝑛 ≤ Var 𝑋 𝜆2(log log 𝑛) = 1 𝜆2 + 𝑜(1).
Finally, recall that |𝑋−𝜈| ≤10. So the same asymptotic bound holds with 𝑋replaced by 𝜈.
□ The main idea from the above proof is that the number of prime divisors 𝑋= Í 𝑝𝑋𝑝 (from the previous proof) behaves like a sum of independent random variables.
A sum of independent random variables often satisfy a central limit theorem (i.e., asymptotic normality, convergence to Gaussian), assuming some mild regularity hy-potheses.
In particular, we have the following corollary of the Lindenberg–Feller central limit theorem (see Durrett, Theorem 3.4.10): Theorem 4.5.2 (Central limit theorem for sums of independent Bernoullis) If 𝑋𝑛is a sum of independent Bernoulli random variables, and Var 𝑋𝑛→∞as 𝑛→∞, then (𝑋𝑛−E𝑋𝑛)/ √ Var 𝑋converges to the normal distribution.
(Note that the divergent variance hypothesis is necessary and sufficient.) In the setting of prime divisibility, we do not have genuine independence. Nevertheless, it is natural to expect that 𝜈(𝑥) still satisfies a central limit theorem. This is indeed the case, and can be proved by comparing moments against a genuine sum of independent random Bernoulli random variables.
Theorem 4.5.3 (Asymptotic normality: Erdős and Kac 1940) With 𝑥∈[𝑛] uniformly chosen at random, 𝜈(𝑥) is asymptotically normal, i.e., for every 𝜆∈R, lim 𝑛→∞P𝑥∈[𝑛] 𝜈(𝑥) −log log 𝑛 √︁ log log 𝑛 ≥𝜆 !
= 1 √ 2𝜋 ∫∞ 𝜆 𝑒−𝑡2/2 𝑑𝑡 The original proof of Erdős and Kac verifies the above intuition using some more involved results in analytic number theory. Simpler proofs have been subsequently given, and we outline one such proof below, which is based on computing the moments 59 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment of the distribution. The idea of computing moments for this problem was first used by Delange (1953), who was apparently not aware of the Erdős–Kacs paper. Also see a more modern account by Granville and Soundararajan (2007).
The following tool from probability theory allows us to verify asymptotic normality from convergence of moments.
Theorem 4.5.4 (Method of moments) Let 𝑋𝑛be a sequence of real valued random variables such that for every positive integer 𝑘, lim𝑛→∞E[𝑋𝑘 𝑛] equals to the 𝑘-th moment of the standard normal distribution. Then 𝑋𝑛converges in distribution to the standard normal, i.e., lim𝑛→∞P(𝑋𝑛≤𝑎) = P(𝑍≤ 𝑎) for every 𝑎∈R, where 𝑍is a standard normal.
Remark 4.5.5. The same conclusion holds for any probability distribution that is “determined by its moments,” i.e., there are no other distributions sharing the same moments. Many common distributions that arise in practice, e.g., the Poisson distri-bution, satisfy this property. There are various sufficient conditions for guaranteeing this moments property, e.g., Carleman’s condition tells us that any probability distri-bution whose moments do not increase too quickly is determined by its moments. (See Durrett §3.3.5).
Proof of Erdős–Kacs Theorem 4.5.3. We compare higher moments of 𝑋= 𝜈(𝑥) with that of an idealized 𝑌treating the prime divisors as truly random variables.
Set 𝑀= 𝑛1/𝑠(𝑛) where 𝑠(𝑛) →∞sufficiently slowly. As earlier, 𝜈(𝑥) −𝑠(𝑛) ≤𝜈(𝑥) ≤ 𝑣(𝑥).
We construct a “model random variable” mimicking 𝑋. Let 𝑌= Í 𝑝≤𝑀𝑌𝑝, where 𝑌𝑝∼Bernoulli(1/𝑝) independently for all primes 𝑝≤𝑀. We can compute: 𝜇:= E𝑌∼E𝑋∼log log 𝑛 and 𝜎2 := Var𝑌∼Var 𝑋∼log log 𝑛.
Let e 𝑋= (𝑋−𝜇)/𝜎and e 𝑌= (𝑌−𝜇)/𝜎.
A consequence of the Lindeberg–Feller central limit theorem is that a sum of inde-pendent Bernoulli random variables with divergent variance satisfies the central limit theorem. So e 𝑌→𝑁(0, 1) in distribution. In particular, E[e 𝑌𝑘] ∼E[𝑍𝑘] (asymptotics as 𝑛→∞) where 𝑍is a standard normal.
Let us compare e 𝑋and e 𝑌. It suffices to show that for every fixed 𝑘, E[e 𝑋𝑘] ∼E[e 𝑌𝑘].
60 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.6 Distinct sums For every set of distinct primes 𝑝1, . . . 𝑝𝑟≤𝑀, E[𝑋𝑝1 · · · 𝑋𝑝𝑟−𝑌𝑝1 · · ·𝑌𝑝𝑟] = 1 𝑛 𝑛 𝑝1 · · · 𝑝𝑟 − 1 𝑝1 · · · 𝑝𝑟 = 𝑂 1 𝑛 .
Comparing expansions of e 𝑋𝑘in terms of the 𝑋𝑝’s (𝑛𝑜(1) terms), we get E[e 𝑋𝑘−e 𝑌𝑘] = 𝑛−1+𝑜(1) = 𝑜(1).
It follows that e 𝑋is asymptotically normal.
□ 4.6 Distinct sums What is the largest subset of [𝑛] all of whose subsets have distinct sums? Equivalently: Question 4.6.1 For each 𝑘, what is the smallest 𝑛so that there exists 𝑆⊆[𝑛] with |𝑆| = 𝑘and all 2𝑘 subset sums of 𝑆are distinct?
E.g., 𝑆= {1, 2, 22, . . . , 2𝑘−1} (the greedy choice).
We begin with an easy pigeonhole argument. Since all 2𝑘sums are distinct and are at most 𝑘𝑛, we have 2𝑘≤𝑘𝑛. Thus 𝑛≥2𝑘/𝑘.
Erdős offered $300 for a proof or disproof of the following. It remains open.
Conjecture 4.6.2 (Erdős) 𝑛≳2𝑘 Let us use the second moment to give a modest improvement on the earlier pigeonhole argument. The main idea here is that, by second moment, most of the subset sums lie within an 𝑂(𝜎)-interval, so that we can improve on the pigeonhole estimate ignoring outlier subset sums.
Theorem 4.6.3 If there is a 𝑘-element subset of [𝑛] with distinct subset sums. Then 𝑛≳2𝑘/ √ 𝑘.
Proof. Let 𝑆= {𝑥1, . . . , 𝑥𝑘} be a 𝑘-element subset of [𝑛] with distinct subset sums.
Set 𝑋= 𝜀1𝑥1 + · · · + 𝜀𝑘𝑥𝑘 61 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment where 𝜀𝑖∈{0, 1} are chosen uniformly at random independently. We have 𝜇:= E𝑋= 𝑥1 + · · · + 𝑥𝑘 2 and 𝜎2 := Var 𝑋= 𝑥2 1 + · · · + 𝑥2 𝑘 4 ≤𝑛2𝑘 4 .
By Chebyshev’s inequality, P(|𝑋−𝜇| ≥2𝜎) ≤1 4, and thus P(|𝑋−𝜇| < 𝑛 √ 𝑘) = P(|𝑋−𝜇| < 2𝜎) ≥3 4.
Since 𝑋takes distinct values for every (𝜀1, . . . , 𝜀𝑘) ∈{0, 1}𝑘, we have P(𝑋= 𝑥) ≤2−𝑘 for all 𝑥. Since there are ≤2𝑛 √ 𝑘elements in the interval (𝜇−𝑛 √ 𝑘, 𝜇+ 𝑛 √ 𝑘), we have P(|𝑋−𝜇| < 𝑛 √ 𝑘) ≤2𝑛 √ 𝑘2−𝑘.
Putting the upper and lowers bounds on P(|𝑋−𝜇| < 𝑛 √ 𝑘) together, we get 2𝑛 √ 𝑘2−𝑘≤3 4.
So 𝑛≳2𝑘/ √ 𝑘.
□ Dubroff, Fox, and Xu (2021) gave another short proof of this result by applying Harper’s vertex-isoperimetric inequality on the cube (this is an example of “concentration of measure”, which we will explore more later this course).
Consider the graph representing the 𝑛-dimensional boolean cube, with vertex set {0, 1}𝑛 with an edge between every pair of 𝑛-tuples that differ in exactly one coordinate. Given 𝐴⊆{0, 1}𝑛, write 𝜕𝐴for the set of all vertices outside 𝐴that is adjacent to some vertex of 𝐴.
Theorem 4.6.4 (Vertex-isomperimetric inequality on the hypercube: Harper 1966) Every 𝐴⊆{0, 1}𝑘with |𝐴| = 2𝑘−1 has |𝜕𝐴| ≥ 𝑘 ⌊𝑘/2⌋ .
62 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.7 Weierstrass approximation theorem Remark 4.6.5. A stronger form of Harper’s theorem gives the precise value of min 𝐴⊆{0,1}𝑛:|𝐴|=𝑚|𝜕𝐴| for every (𝑛, 𝑚). Basically, the minimum is achieved when 𝐴is a a Hamming ball, or, if 𝑚is not exactly the size of some Hamming ball, then 𝐴consists of the lexicographically first 𝑚elements of {0, 1}𝑛.
Theorem 4.6.6 (Dubroff–Fox–Xu 2021) If there is a 𝑘-element subset of [𝑛] with distinct subset sums, then 𝑛≥ 𝑘 ⌊𝑘/2⌋ = √︂ 2 𝜋+ 𝑜(1) !
2𝑘 √ 𝑘 .
Remark 4.6.7. The above bound has the currently best known leading constant factor, matching an earlier result by Aliev (2009).
Proof. Let 𝑆= {𝑥1, . . . , 𝑥𝑘} be a subset of [𝑛] with distinct sums. Let 𝐴= n (𝜀1, . . . , 𝜀𝑘) ∈{0, 1}𝑘: 𝜀1𝑥1 + · · · + 𝜀1𝑥𝑘< 𝑥1 + · · · + 𝑥𝑘 2 o .
Note that due to the distinct sum hypothesis, one can never have 𝜀1𝑥1 + · · · + 𝜀𝑘𝑥𝑘= (𝑥1 + · · · + 𝑥𝑛)/2. It thus follows by symmetry that |𝐴| = 2𝑘−1.
Note that every element of 𝜕𝐴corresponds to some sum of the form 𝑧+ 𝑥𝑖> (𝑥1 + · · · + 𝑥𝑙)/2 for some 𝑧< (𝑥1 + · · · + 𝑥𝑘)/2, and thus 𝑧+ 𝑥𝑖lies in the open interval 𝑥1 + · · · + 𝑥𝑘 2 , 𝑥1 + · · · + 𝑥𝑘 2 + max 𝑆 .
Since all subset sums are distinct, we must have 𝑛≥|𝜕𝐴| ≥ 𝑘 ⌊𝑘/2⌋ by Harper’s theorem (Theorem 4.6.4).
□ 4.7 Weierstrass approximation theorem We finish off the chapter with an application to analysis.
The Weierstrass approximation theorem says that every continuous real function on an interval can be uniformly approximated by a polynomial.
63 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment Theorem 4.7.1 (Weierstrass approximation theorem 1885) Let 𝑓: [0, 1] →R be a continuous function. Let 𝜀> 0. Then there is a polynomial 𝑝(𝑥) such that |𝑝(𝑥) −𝑓(𝑥)| ≤𝜀for all 𝑥∈[0, 1].
Proof. (Bernstein 1912) The idea is to approximate 𝑓by a sum of polynomials that look like “bumps”: 𝑃𝑛(𝑥) = 𝑛 ∑︁ 𝑖=0 𝐸𝑖(𝑥) 𝑓(𝑖/𝑛) where 𝐸𝑖(𝑥) = P(Binomial(𝑛, 𝑥) = 𝑖) = 𝑛 𝑖 𝑥𝑖(1 −𝑥)𝑛−𝑖 for 0 ≤𝑖≤𝑛 is chosen as some polynomials peaks at 𝑥= 𝑖/𝑛and then decays as 𝑥moves away from 𝑖/𝑛.
For each 𝑥∈[0, 1], the binomial distribution Binomial(𝑛, 𝑥) has mean 𝑛𝑥and variance 𝑛𝑥(1 −𝑥) ≤𝑛. By Chebyshev’s inequality, ∑︁ 𝑖:|𝑖−𝑛𝑥|>𝑛2/3 𝐸𝑖(𝑥) = P(|Binomial(𝑛, 𝑥) −𝑛𝑥| > 𝑛2/3) ≤𝑛−1/3.
(In the next chapter, we will see a much better tail bound.) Since [0, 1] is compact, 𝑓is uniformly continuous and bounded. By multiplying by a constant, we assume that | 𝑓(𝑥)| ≤1 for all 𝑥∈[0, 1]. Also there exists 𝛿> 0 such that | 𝑓(𝑥) −𝑓(𝑦)| ≤𝜀/2 for all 𝑥, 𝑦∈[0, 1] with |𝑥−𝑦| ≤𝛿.
Take 𝑛> max{64𝜀−3, 𝛿−3}. Then for every 𝑥∈[0, 1] (note that Í𝑛 𝑗=0 𝐸𝑗(𝑥) = 1), |𝑃𝑛(𝑥) −𝑓(𝑥)| ≤ 𝑛 ∑︁ 𝑖=0 𝐸𝑖(𝑥)| 𝑓(𝑖/𝑛) −𝑓(𝑥)| ≤ ∑︁ 𝑖:|𝑖/𝑛−𝑥|<𝑛−1/3<𝛿 𝐸𝑖(𝑥)| 𝑓(𝑖/𝑛) −𝑓(𝑥)| + ∑︁ 𝑖:|𝑖−𝑛𝑥|>𝑛2/3 2𝐸𝑖(𝑥) ≤𝜀 2 + 2𝑛−1/3 ≤𝜀.
□ Exercises 1. Let 𝑋be a nonnegative real-valued random variable. Suppose P(𝑋= 0) < 1.
Prove that P(𝑋= 0) ≤Var 𝑋 E[𝑋2] .
64 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.7 Weierstrass approximation theorem 2. Let 𝑋be a random variable with mean 𝜇and variance 𝜎2. Prove that for all 𝜆> 0, P(𝑋≥𝜇+ 𝜆) ≤ 𝜎2 𝜎2 + 𝜆2 .
3. Threshold for 𝑘-APs. Let [𝑛] 𝑝denote the random subset of {1, . . . , 𝑛} where every element is included with probability 𝑝independently.
For each fixed integer 𝑘≥3, determine the threshold for [𝑛] 𝑝to contain a 𝑘-term arithmetic progression.
4. What is the threshold function for 𝐺(𝑛, 𝑝) to contain a cycle?
5. Show that, for each fixed positive integer 𝑘, there is a sequence 𝑝𝑛such that P(𝐺(𝑛, 𝑝𝑛) has a connected component with exactly 𝑘vertices) →1 as 𝑛→∞.
Hint: Make the random graph contain some specific subgraph but not some others.
6. Poisson limit. Let 𝑋be the number of triangles in 𝐺(𝑛, 𝑐/𝑛) for some fixed 𝑐> 0.
a) For every nonnegative integer 𝑘, determine the limit of E 𝑋 𝑘 as 𝑛→∞.
b) Let 𝑌∼Binomial(𝑛, 𝜆/𝑛) for some fixed 𝜆> 0. For every nonnegative integer 𝑘, determine the limit of E 𝑌 𝑘 as 𝑛→∞, and show that it agrees with the limit in (a) for some 𝜆= 𝜆(𝑐).
We know that 𝑌converges to the Poisson distribution with mean 𝜆. Also, the Poisson distribution is determined by its moments.
c) Compute, for fixed every nonnegative integer 𝑡, the limit of P(𝑋= 𝑡) as 𝑛→∞.
(In particular, this gives the limit probability that 𝐺(𝑛, 𝑐/𝑛) contains a triangle, i.e., lim𝑛→∞P(𝑋> 0). This limit increases from 0 to 1 continu-ously when 𝑐ranges from 0 to +∞, thereby showing that the property of containing a triangle has a coarse threshold.) 7. Central limit theorem for triangle counts. Find a real (non-random) sequence 𝑎𝑛 so that, letting 𝑋be the number of triangles and 𝑌be the number of edges in the random graph 𝐺(𝑛, 1/2), one has Var(𝑋−𝑎𝑛𝑌) = 𝑜(Var 𝑋).
Deduce that 𝑋is asymptotically normal, that is, (𝑋−E𝑋)/ √ Var 𝑋converges to the normal distribution.
(You can solve for the minimizing 𝑎𝑛directly similar to ordinary least squares linear regression, 65 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4 Second Moment or first write the edge indicator variables as 𝑋𝑖𝑗= (1 + 𝑌𝑖𝑗)/2 and then expand. The latter approach likely yields a cleaner computation.) 8. Isolated vertices. Let 𝑝𝑛= (log 𝑛+ 𝑐𝑛)/𝑛.
a) Show that, as 𝑛→∞, P(𝐺(𝑛, 𝑝𝑛) has no isolated vertices) → ( 0 if 𝑐𝑛→−∞, 1 if 𝑐𝑛→∞.
b) Suppose 𝑐𝑛→𝑐∈R, compute, with proof, the limit of LHS above as 𝑛→∞, by following the approach in 6.
9. ★Is the threshold for the bipartiteness of a random graph coarse or sharp?
(You are not allowed to use any theorems that we did not prove in class/notes.) 10. Triangle packing.
Prove that, with probability approaching 1 as 𝑛→∞, 𝐺(𝑛, 𝑛−1/2) has at least 𝑐𝑛3/2 edge-disjoint triangles, where 𝑐> 0 is some constant.
Hint: rephrase as finding a large independent set 11. Nearly perfect triangle factor. Prove that, with probability approaching 1 as 𝑛→∞, a) 𝐺(𝑛, 𝑛−2/3) has at least 𝑛/100 vertex-disjoint triangles.
b) Simple nibble. 𝐺(𝑛, 𝐶𝑛−2/3) has at least 0.33𝑛vertex-disjoint triangles, for some constant 𝐶.
Hint: view a random graph as the union of several independent random graphs & iterate (a) 12. Permuted correlation. Recall that the correlation of two non-constant random variables 𝑋and 𝑌is defined to be corr(𝑋,𝑌) := Cov[𝑋,𝑌]/ √︁ (Var 𝑋)(Var𝑌).
Let 𝑓, 𝑔∈[𝑛] →R be two non-constant functions.
Prove that there exist permutations 𝜋and 𝜏of [𝑛] such that, with 𝑍being a uniform random element of [𝑛], corr( 𝑓(𝜋(𝑍)), 𝑔(𝑍)) −corr( 𝑓(𝜏(𝑍)), 𝑔(𝑍)) ≥ 2 √ 𝑛−1 .
Furthermore, show that equality can be achieved for even 𝑛.
Hint: Compute the variance of the correlation for a random permutation.
13. Let 𝑣1 = (𝑥1, 𝑦1), . . . , 𝑣𝑛= (𝑥𝑛, 𝑦𝑛) ∈Z2 with |𝑥𝑖| , |𝑦𝑖| ≤2𝑛/2/(100√𝑛) for all 𝑖∈[𝑛]. Show that there are two disjoint sets 𝐼, 𝐽⊆[𝑛], not both empty, such that Í 𝑖∈𝐼𝑣𝑖= Í 𝑗∈𝐽𝑣𝑗.
66 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 4.7 Weierstrass approximation theorem 14. ★Prove that there is an absolute constant 𝐶> 0 so that the following holds. For every prime 𝑝and every 𝐴⊆Z/𝑝Z with |𝐴| = 𝑘, there exists an integer 𝑥so that {𝑥𝑎: 𝑎∈𝐴} intersects every interval of length at least 𝐶𝑝/ √ 𝑘in Z/𝑝Z.
15. ★Prove that there is a constant 𝑐> 0 so that every hyperplane containing the origin in R𝑛intersects at least 𝑐-fraction of the 2𝑛closed unit balls centered at {−1, 1}𝑛.
67 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5 Chernoff Bound The Chernoff bound is an extremely useful bound on the tails of a sum of independent random variables. It is proved by bounding the moment generating function. This proof technique is interesting and important in its own right. We will see this proof method come up again later on when we prove martingale concentration inequalities.
The method allows us to adapt the proof of the Chernoff bound to other distributions.
Let us give the proof in the most basic case for simplicity and clarity.
Theorem 5.0.1 (Chernoff bound) Let 𝑆𝑛= 𝑋1 + · · · + 𝑋𝑛where 𝑋𝑖∈{−1, 1} uniformly iid. Let 𝜆> 0. Then P(𝑆𝑛≥𝜆√𝑛) ≤𝑒−𝜆2/2.
In contrast, Chebyshev’s inequality gives a weaker bound P(𝑆𝑛≥𝜆√𝑛) ≤1/𝜆2. On the other hand, Chebyshev’s inequality is application in wider settings as it only requires pairwise independence (for the second moment) as opposed to full independence.
Proof. Let 𝑡≥0. Consider the moment generating function E 𝑒𝑡𝑆𝑛 = E 𝑒𝑡Í 𝑖𝑋𝑖 = E "Ö 𝑖 𝑒𝑡𝑋𝑖 # = Ö 𝑖 E 𝑒𝑡𝑋𝑖 = 𝑒−𝑡+ 𝑒𝑡 2 𝑛 .
By comparing Taylor series, we have 𝑒−𝑡+ 𝑒𝑡 2 = ∑︁ 𝑘≥0 𝑥2𝑘 (2𝑘)! ≤ ∑︁ 𝑘≥0 𝑥2𝑘 𝑘!2𝑘= 𝑒𝑡2/2.
By Markov’s inequality, P(𝑆𝑛≥𝜆√𝑛) ≤E 𝑒𝑡𝑆 𝑒𝑡𝜆√𝑛 ≤𝑒−𝑡𝜆√𝑛+𝑡2𝑛/2.
Setting 𝑡= 𝜆/√𝑛gives the bound.
□ Remark 5.0.2. The technique of considering the moment generating function can be thought morally as taking an appropriately high moment.
Indeed, E[𝑒𝑡𝑆] = 69 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5 Chernoff Bound Í 𝑛≥0 E[𝑆𝑛]𝑡𝑛/𝑛! contains all the moments data of the random variable.
The second moment method (Chebyshev + Markov) can be thought of as the first iteration of this idea. By taking fourth moments (now requiring 4-wise independence of the summands), we can obtain tail bounds of the form ≲𝜆−4. And similarly with higher moments.
In some applications, where one cannot assume independence, but can estimate some high moments, the above philosophy can allow us to prove good tail bounds as well.
Also by symmetry, P(𝑆𝑛≤−𝜆√𝑛) ≤𝑒−𝜆2/2. Thus we have the following two-sided tail bound.
Corollary 5.0.3 With 𝑆𝑛as before, for any 𝜆≥0, P(|𝑆𝑛| ≥𝜆√𝑛) ≤2𝑒−𝜆2/2.
Remark 5.0.4. It is easy to adapt the above proof so that each 𝑋𝑖is a mean-zero random variable taking [−1, 1]-values, and independent (but not necessarily identical) across all 𝑖. Indeed, by convexity, we have 𝑒𝑡𝑥≤1−𝑥 2 𝑒−𝑡+ 1+𝑥 2 𝑒𝑡for all 𝑥∈[−1, 1] by convexity, so that E[𝑒𝑡𝑋] ≤𝑒𝑡+𝑒−𝑡 2 . In particular, we obtain the following tail bounds on the binomial distribution.
Theorem 5.0.5 (Chernoff bound with bounded variables) Let each 𝑋𝑖be an independent random variable taking values in [−1, 1] and E𝑋𝑖= 0.
Then 𝑆𝑛= 𝑋1 + · · · + 𝑋𝑛satisfies P(𝑆𝑛≥𝜆√𝑛) ≤𝑒−𝜆2/2.
Corollary 5.0.6 Let 𝑋be a sum of 𝑛independent Bernoulli’s (with not necessarily identitical proba-bility). Let 𝜇= E𝑋and 𝜆> 0. Then P(𝑋≥𝜇+ 𝜆√𝑛) ≤𝑒−𝜆2/2 and P(𝑋≤𝜇−𝜆√𝑛) ≤𝑒−𝜆2/2 The Chernoff bound compares well to that of the normal distribution. For the standard normal 𝑍∼𝑁(0, 1), one has E[𝑒𝑡𝑍] = 𝑒𝑡2/2 and so P(𝑍≥𝜆) = P(𝑒𝑡𝑍≥𝑒𝑡𝜆) ≤𝑒−𝑡𝜆E[𝑒𝑡𝑋] = 𝑒−𝑡𝜆+𝑡2/2.
70 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5.1 Discrepancy Set 𝑡= 𝜆and get P(𝑍≥𝜆) ≤𝑒−𝜆2/2.
And this is actually pretty tight, as, for 𝜆→∞, P(𝑍≥𝜆) = 1 √ 2𝜋 ∫∞ 𝜆 𝑒−𝑡2/2 𝑑𝑡∼𝑒−𝜆2/2 √ 2𝜋𝜆 .
The same proof method allows you to prove bounds for other sums of random variables, which you can adjust based on the distributions. See the Alon–Spencer textbook, Appendix A, for examples of bounds and proofs.
For example, for a sum of independent Bernoulli random variables with small means, we can improve on the above estimates as follows.
Theorem 5.0.7 Let 𝑋be the sum of independent Bernoulli random variables (not necessarily the same probability). Let 𝜇= E𝑋. For all 𝜀> 0, P(𝑋≥(1 + 𝜀)𝜇) ≤𝑒−((1+𝜀) log(1+𝜀)−𝜀)𝜇≤𝑒−𝜀2 1+𝜀𝜇 and P(𝑋≤(1 −𝜀)𝜇) ≤𝑒−𝜀2𝜇/2.
Remark 5.0.8. The bounds for upper and lower tails are necessarily asymmetric when the probabilities are small. Why? Think about what happens when 𝑋∼Bin(𝑛, 𝑐/𝑛), which, for a constant 𝑐> 0, converges as 𝑛→∞to a Poisson distribution with mean 𝑐, whose value at 𝑘is 𝑒−𝑐𝑐𝑘/𝑘! = 𝑒−Θ(𝑘log 𝑘) and not the sub-Gaussian decay 𝑒−Ω(𝑘2) as one might naively predict by an incorrect application of the Chernoff bound formula.
Nonetheless, both formulas tell us that both tails exponentially decay like 𝜀2 for small values of 𝜀∈[0, 1].
5.1 Discrepancy Given a hypergraph (i.e., set family), can we color the vertices red/blue so that every edge has roughly the same number of red versus blue vertices? (Contrast this problem to 2-coloring hypergraphs from Section 1.3.) 71 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5 Chernoff Bound Theorem 5.1.1 Let F be a collection of 𝑚subsets of [𝑛].
Then there exists some assignment [𝑛] →{−1, 1} so that the sum on every set in F is 𝑂( √︁ 𝑛log 𝑚) in absolute value.
Proof. Put ±1 iid uniformly at random on each vertex. On each edge, the probability that the sum exceeds 2 √︁ 𝑛log 𝑚in absolute value is, by Chernoff bound, less than 2𝑒−2 log 𝑚= 2/𝑚2. By union bound over all 𝑚edges, with probability greater than 1 −2/𝑚≥0, no edge has sum exceeding 2 √︁ 𝑛log 𝑚.
□ Remark 5.1.2. In a beautiful landmark paper titled Six standard deviations suffice, Spencer (1985) showed that one can remove the logarithmic term by a more sophisti-cated semirandom assignment algorithm.
Theorem 5.1.3 (Six standard deviations suffice: Spencer 1985) Let F be a collection of 𝑛subsets of [𝑛]. Then there exists some assignment [𝑛] → {−1, 1} so that the sum on every set in F is at most 6√𝑛in absolute value.
More generally, if F be a collection of 𝑚≥𝑛subsets of [𝑛], then we can replace 6√𝑛 by 𝑂( √︁ 𝑛log(2𝑚/𝑛)).
Remark 5.1.4. More generally, Spencer proves that the same holds if vertices have [0, 1]-valued weights.
The idea, very roughly speaking, is to first generalize from {−1, 1}-valued assign-ments to [−1, 1]-valued assignments. Then the all-zero vector is a trivially satisfying assignment. We then randomly, in iterations, alter the values from 0 to other values in [−1, 1], while avoiding potential violations (e.g., edges with sum close to 6√𝑛in absolute value), and finalizing a color of a color when its value moves to either −1 and 1.
Spencer’s original proof was not algorithmic, and he suspected that it could not be made efficiently algorithmic. In a breakthrough result, Bansal (2010) gave an efficient algorithm for producing a coloring with small discrepancy. Lovett and Meka (2015) provided another element algorithm with a beautiful proof.
Here is a famous conjecture on discrepancy.
72 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5.2 Nearly equiangular vectors Conjecture 5.1.5 (Komlós) There exists some absolute constant 𝐾so that for any 𝑣1, . . . , 𝑣𝑚∈R𝑛all lying in the unit ball, there exist 𝜀1, . . . , 𝜀𝑚∈{−1, 1} such that 𝜀1𝑣1 + · · · + 𝜀𝑚𝑣𝑚∈[−𝐾, 𝐾]𝑛.
Banaszczyk (1998) proved the bound 𝐾= 𝑂( √︁ log 𝑛) in a beautiful paper using deep ideas from convex geometry.
Spencer’s theorem’s implies the special case of Komlós conjecture where all vec-tors 𝑣𝑖have the form 𝑛−1/2(±1, . . . , ±1) (or more generally when all coordinates are 𝑂(𝑛−1/2)). The deduction is easy when 𝑚≤𝑛. When 𝑚> 𝑛, we use the following observation.
Lemma 5.1.6 Let 𝑣1, . . . , 𝑣𝑚∈R𝑛. Thenthereexists 𝑎1, . . . , 𝑎𝑚∈[−1, 1]𝑚with |{𝑖: 𝑎𝑖∉{−1, 1}}| ≤ 𝑛such that 𝑎1𝑣1 + · · · + 𝑎𝑚𝑣𝑚= 0 Proof. Find (𝑎1, . . . , 𝑎𝑚) ∈[−1, 1]𝑚satisfying and as many 𝑎𝑖∈{−1, 1} as possi-ble. Let 𝐼= {𝑖: 𝑎𝑖∉{−1, 1}}. If |𝐼| > 𝑛, then we can find some nontrivial linear combination of the vectors 𝑣𝑖, 𝑖∈𝐼, allowing us to to move (𝑎𝑖)𝑖∈𝐼’s to new values, while preserving 𝑎1𝑣1 + · · · + 𝑎𝑚𝑣𝑚= 0, and end up with at one additional 𝑎𝑖taking {−1, 1}-value.
□ Let us explain how to deduce the special caes of Kómlos conjecture as stated earlier.
Let 𝑎1, . . . , 𝑎𝑚and 𝐼= {𝑖: 𝑎𝑖∉{−1, 1}} as in the Lemma. Take 𝜀𝑖= 𝑎𝑖for all 𝑖∉𝐼, and apply a corollary of Spencer’s theorem to find 𝜀𝑖∈{−1, 1}𝑛, 𝑖∈𝐼with ∑︁ 𝑖∈𝐼 (𝜀𝑖−𝑎𝑖)𝑣𝑖∈[−𝐾, 𝐾]𝑛, which would yield the desired result. The above step can be deduced from Spencer’s theorem by first assuming that each 𝑎𝑖∈[−1, 1] has finite binary length (a compactness argument), and then rounding it off one digit at a time during Spencer’s theorem, starting from the least significant bit (see Corollary 8 in Spencer’s paper for details).
5.2 Nearly equiangular vectors How many vectors can one place in R𝑑so that pairwise make equal angles?
73 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5 Chernoff Bound Let 𝑆= {𝑣1, . . . , 𝑣𝑚} be a set of unit vectors in R𝑛whose pairwise inner products all equal to some 𝛼∈[−1, 1). How large can 𝑆be?
The Gram matrix of 𝑆, defined as the matrix of pairwise inner products, has 1’s on the diagonal and 𝛼off diagonal. So © « | · · · | 𝑣1 ...
𝑣𝑚 | · · · | ª ® ® ¬ ⊺ © « | · · · | 𝑣1 ...
𝑣𝑚 | · · · | ª ® ® ¬ = © « 𝑣1 · 𝑣1 · · · 𝑣1 · 𝑣𝑚 .
.
.
...
.
.
.
𝑣𝑚· 𝑣1 · · · 𝑣𝑚· 𝑣𝑚 ª ® ® ¬ = (1 −𝛼)𝐼𝑚+ 𝛼𝐽𝑚 (here 𝐼𝑚and 𝐽𝑚are the 𝑚× 𝑚identity and all-ones matrix respectively). Since the eigenvalues of 𝐽𝑚are 𝑚(once) and 0 (repeated 𝑚−1 times), the eigenvalues of 𝐼𝑚+ (𝛼−1)𝐽𝑚are (𝑚−1)𝛼+1 (once) and 1−𝛼(𝑚−1 times). Since the Gram matrix is positive semidefinite, all its eigenvalues are nonnegative, and so 𝛼≥−1/(𝑚−1).
• If 𝛼≠−1/(𝑚−1), then this 𝑚× 𝑚matrix is non-singular, and since its rank is at most 𝑛(as 𝑣𝑖∈R𝑛), we have 𝑚≤𝑛.
• If 𝛼= −1/(𝑚−1), then this matrix has rank 𝑚−1, and we conclude that 𝑚≤𝑛+ 1.
It is left as an exercise to check all these bounds are tight.
Exercise: given 𝑚unit vectors in R𝑛whose pairwise inner products are all ≤−𝛽, one has 𝑚≤1 + ⌊1/𝛽⌋. (A bit more difficult: show that for 𝛽= 0, one has 𝑚≤2𝑛).
What if instead of asking for exactly equal angles, we ask for approximately the same angle. It turns out that we can get many more vectors.
Theorem 5.2.1 (Exponentially many approximately equiangular vectors) For every 𝛼∈(0, 1) and 𝜀> 0, there exists 𝑐> 0 so that for every 𝑛, one can find at least 2𝑐𝑛unit vectors in R𝑛whose pairwise inner products all lie in [𝛼−𝜀, 𝛼+ 𝜀].
Remark 5.2.2. Such a collection of vectors is a type of “spherical code.” Also, by examining the volume of spherical caps, one can prove an upper bound of the form 2𝐶𝛼,𝜀𝑛.
Proof. Let 𝑝= (1 + √𝛼)/2, and let 𝑣1, . . . , 𝑣𝑚∈{−1, 1}𝑛be independent random vectors which each coordinate independently is +1 with probability 𝑝and −1 with probability 1 −𝑝. Then for 𝑖≠𝑗, the dot product 𝑣𝑖· 𝑣𝑗is a sum of 𝑛independent ±1-valued random variables each with mean 𝑝2 + (1 −𝑝)2 −2𝑝(1 −𝑝) = (𝑝−(1 −𝑝))2 = (2𝑝−1)2 = 𝛼.
74 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5.3 Hajós conjecture counterexample Applying Chernoff bound in the form of Theorem 5.0.5 (after linear transformation on each variable to make each term taking value in [−1, 1] and mean centered at zero), we get P |𝑣𝑖· 𝑣𝑗−𝑛𝛼| ≥𝑛𝜀 ≤2𝑒−Ω(𝑛𝜀2).
By the union bound, the probability that |𝑣𝑖·𝑣𝑗−𝑛𝛼| > 𝑛𝜀for some𝑖≠𝑗is < 𝑚2𝑒Ω(𝑛𝜀2), which is < 1 for some 𝑚at least 2𝑐𝑛. So with positive probability, so such pair occurs, and then 𝑣1/√𝑛, . . . , 𝑣𝑚/√𝑛is a collection of unit vectors in R𝑛whose pairwise inner products all lie in [𝛼−𝜀, 𝛼+ 𝜀].
□ Remark 5.2.3 (Equiangular lines with a fixed angle). Given a fixed angle 𝜃, for large 𝑛, how many lines in R𝑛through the origin can one place whose pairwise angles are all exactly 𝜃? This problem was solved by Jiang, Tidor, Yao, Zhang, Zhao (2021). This is the same as asking for a set of unit vectors in R𝑛whose pairwise inner products are ±𝛼. It turns out that for fixed 𝛼, the maximum number of lines grows linearly with the dimension 𝑛, and the rate of growth depends on properties of 𝛼in relation to spectral graph theory. We refer to the cited paper for details.
5.3 Hajós conjecture counterexample We begin by reviewing some classic result from graph theory. Recall some definitions: • 𝐻is an induced subgraph of 𝐺if 𝐻can be obtained from 𝐺by removing vertices; • 𝐻is a subgraph if 𝐺if 𝐻can be obtained from 𝐺by removing vertices and edges; • 𝐻is a subdivision of 𝐺if 𝐻can be obtained from a subgraph of 𝐺by contracting induced paths to edges; • 𝐻is a minor of 𝐺if 𝐻can be obtained from a subgraph of 𝐺by by contracting edges to vertices.
Kuratowski’s theorem (1930). Every graph without 𝐾3,3 and 𝐾5 as subdivisions as subdivision is planar.
Wagner’s theorem (1937). Every graph free of 𝐾3,3 and 𝐾5 as minors is planar.
(There is a short argument shows that Kuratowski and Wagner’s theorems are equiva-lent.) Four color theorem (Appel and Haken 1977) Every planar graph is 4-colorable.
Corollary: Every graph without 𝐾3,3 and 𝐾5 as minors is 4-colorable.
75 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5 Chernoff Bound The condition on 𝐾5 is clearly necessary, but what about 𝐾3,3? What is the “real” reason for 4-colorability?
Hadwidger’s conjecture, below, remains a major conjectures in graph theory.
Conjecture 5.3.1 (Hadwiger 1936) For every 𝑡≥1, every graph without a 𝐾𝑡+1 minor is 𝑡-colorable.
• 𝑡= 1 trivial • 𝑡= 2 nearly trivial (if 𝐺is 𝐾3-minor-free, then it’s a tree) • 𝑡= 3 elementary graph theoretic arguments • 𝑡= 4 is equivalent to the 4-color theorem (Wagner 1937) • 𝑡= 5 is equivalent to the 4-color theorem (Robertson–Seymour–Thomas 1994; this work won a Fulkerson Prize) • 𝑡≥6 remains open Let us explore a variation of Hadwiger’s conjecture: Hajós conjecture. (1961) Every graph without a 𝐾𝑡+1-subdivision is 𝑡-colorable.
Hajós conjecture is true for 𝑡≤3. However, it turns out to be false in general. Catlin (1979) constructed counterexamples for all 𝑡≥6 (𝑡= 4, 5 are still open).
It turns out that Hajós conjecture is not just false, but very false.
Erdős–Fajtlowicz (1981) showed that almost every graph is a counterexample (it’s a good idea to check for potential counterexamples among random graphs!) Theorem 5.3.2 With probability 1 −𝑜(1), 𝐺(𝑛, 1/2) has no 𝐾𝑡-subdivision with 𝑡= 10√𝑛 .
From Theorem 4.4.3 we know that, with high probability, 𝐺(𝑛, 1/2) has independence number ∼2 log2 𝑛and hence chromatic number ≥(1 + 𝑜(1) 𝑛 2 log2 𝑛. Thus the above result shows that 𝐺(𝑛, 1/2) is whp a counterexample to Hajós conjecture.
Proof. If 𝐺had a 𝐾𝑡-subdivision, say with 𝑆⊆𝑉, |𝑆| = 𝑡. Each pair of vertices of 𝑆 are connected via a path, whose intermediate vertices are outside 𝑆, and distinct for different pairs of vertices.
At most 𝑛of the 𝑡 2 pairs of vertices in 𝑆can be joined this way using a path of at least two edges, since each uses up a vertex outside 𝑆. Thus at ≥ 𝑡 2 −𝑛of the pairs of vertices of 𝑆form edges.
76 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5.3 Hajós conjecture counterexample By Chernoff bound, for fixed 𝑡-vertex subset 𝑆 P 𝑒(𝑆) ≥ 𝑡 2 −𝑛 ≤P 𝑒(𝑆) ≥3 4 𝑡 2 ≤𝑒−𝑡2/10.
Taking a union bound over all 𝑡-vertex subsets 𝑆, and noting that 𝑛 𝑡 𝑒−𝑡2/10 < 𝑛𝑡𝑒−𝑡2/10 ≤𝑒−10𝑛+𝑂(√𝑛log 𝑛) = 𝑜(1) we see that whp no such 𝑆exists, so that this 𝐺(𝑛, 1/2) whp has no 𝐾𝑡-subdivision □ Remark 5.3.3 (Quantitative question). One can ask the following quantitative ques-tion regarding Hadwidger’s conjecture: Can every graph without a 𝐾𝑡+1-minor can be properly colored with a small number of colors?
Wagner (1964) showed that every graph without 𝐾𝑡+1-minor is 2𝑡−1 colorable.
Here is the proof: assume that the graph is connected. Take a vertex 𝑣and let 𝐿𝑖be the set of vertices with distance exactly 𝑖from 𝑣. The subgraph induced on 𝐿𝑖has no 𝐾𝑡-minor, since otherwise such a 𝐾𝑡-minor would extend to a 𝐾𝑡+1-minor with 𝑣. Then by induction 𝐿𝑖is 2𝑡−2-colorable (check base cases), and using alternating colors for even and odd layers 𝐿𝑖yields a proper coloring of 𝐺.
This bound has been improved over time. Delcourt and Postle (2021+) showed that every graph with no 𝐾𝑡-minor is 𝑂(𝑡log log 𝑡)-colorable.
For more on Hadwiger’s conjecture, see Seymour’s survey (2016).
Exercises 1. Prove that with probability 1 −𝑜(1) as 𝑛→∞, every bipartite subgraph of 𝐺(𝑛, 1/2) has at most 𝑛2/8 + 10𝑛3/2 edges.
2. Unbalancing lights.
Prove that there is a constant 𝐶so that for every positive integer 𝑛, one can find an 𝑛× 𝑛matrix 𝐴with {−1, 1} entries, so that for all vectors 𝑥, 𝑦∈{−1, 1}𝑛, |𝑦⊺𝐴𝑥| ≤𝐶𝑛3/2.
3. Prove that there exists a constant 𝑐> 1 such that for every 𝑛, there are at least 𝑐𝑛points in R𝑛so that every triple of points form a triangle whose angles are all less than 61◦.
77 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 5 Chernoff Bound 4. Planted clique. Give a deterministic polynomial-time algorithm for the follow-ing task so that it succeeds over the random input with probability approaching 1 as 𝑛→∞.
Input: some unlabeled 𝑛-vertex 𝐺created as the union of 𝐺(𝑛, 1/2) and a clique on 𝑡= j 100 √︁ 𝑛log 𝑛 k vertices.
Output: a clique in 𝐺of size 𝑡.
5. Weighing coins. You are given 𝑛coins, each with one of two known weights, but otherwise indistinguishable. You can use a scale that outputs the combined weight of any subset of the coins. You must decide in advance which subsets 𝑆1, . . . , 𝑆𝑘⊆[𝑛] of the coins to weigh. We wish to determine the minimum number of weighings needed to identify the weight of every coin. (Below, 𝑋 and 𝑌represent two possibilities for which coins are of the first weight.) a) ★Prove that if 𝑘≤1.99𝑛/log2 𝑛and 𝑛is sufficiently large, then for every 𝑆1, . . . , 𝑆𝑘⊆[𝑛], there are two distinct subsets 𝑋,𝑌⊆[𝑛] such that |𝑋∩𝑆𝑖| = |𝑌∩𝑆𝑖| for all 𝑖∈[𝑘].
(There is a neat solution to part (a) using information theory, though here you are explicitly asked to solve it using the Chernoff bound.) b) ★Show that there is some constant 𝐶such that (a) is false if 1.99 is replaced by 𝐶. (What is the best 𝐶you can get?) 78 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma The Lovász local lemma (LLL) was introduced in the paper of Erdős and Lovász (1975). It is a powerful tool in the probabilistic method.
In many problems, we wish to avoid a certain set of “bad events.” Here are two easy to handle scenarios: • (Complete independence) All the bad events are independent and have probabil-ity less than 1.
• (Union bound) The sum of the bad event probabilities is less than 1.
The local lemma deals with an intermediate situation where there is a small amount of local dependencies.
We saw an application of the Lovász local lemma back in Section 1.1, where we used it to lower bound Ramsey numbers. This chapter explores the local lemma and its applications in depth.
6.1 Statement and proof Definition 6.1.1 (Independence from a set of events) Here we say that an event 𝐴0 is independent from events 𝐴1, . . . , 𝐴𝑚if 𝐴0 is indepen-dent of every event of the form 𝐵1 ∧· · · ∧𝐵𝑚(we sometimes omit the “logical and” symbol ∧) where each 𝐵𝑖is either 𝐴𝑖or 𝐴𝑖, i.e., P(𝐴0𝐵1 · · · 𝐵𝑚) = P(𝐴0)P(𝐵1 · · · 𝐵𝑚), or, equivalently, using Bayes’s rule: P(𝐴0|𝐵1 · · · 𝐵𝑚) = P(𝐴0).
Given a collection of events, we can associate to it a dependency graph. This is a slightly subtle notion, as we will explain. Technically speaking, the graph can be a directed graph (=digraph), but for most applications, it will be sufficient (and easier) to use undirected graphs.
79 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma Definition 6.1.2 (Dependency (di)graph) Let 𝐴1, . . . , 𝐴𝑛be events (the “bad events” we wish to avoid). Let 𝐺be a (directed) graph with vertex set [𝑛]. We say that 𝐺is a dependency (di)graph for the events 𝐴1, . . . , 𝐴𝑛if, for for every 𝑖, 𝐴𝑖is independent from all {𝐴𝑗: 𝑗∉𝑁(𝑖) ∪{𝑖}} (𝑁(𝑖) is the set of (out)neighbors of 𝑖in 𝐺).
Remark 6.1.3 (Non-uniqueness). Given a collection of events, there can be more than one valid dependency graphs. For example, the complete graph is always a valid dependency graph.
Remark 6.1.4 (Important!). Independence ≠pairwise independence The dependency graph is not made by joining 𝑖∼𝑗whenever 𝐴𝑖and 𝐴𝑗are not independent (i.e., P(𝐴𝑖𝐴𝑗) ≠P(𝐴𝑖)P(𝐴𝑗)).
Example: suppose one picks 𝑥1, 𝑥2, 𝑥3 ∈Z/2Z uniformly and independently at random and set, for each 𝑖= 1, 2, 3 (indices taken mod 3), 𝐴𝑖the event that 𝑥𝑖+1 + 𝑥𝑖+2 = 0.
Then these events are pairwise independent but not independent. So the empty graph on three vertices is not a valid dependency graph (on the other hand, having at least two edges makes it a valid dependency graph).
In practice, it is not too hard to construct a valid dependency graph, since most applications of the Lovász local lemma use the following setup (which we saw in Section 1.1).
Setup 6.1.5 (Random variable model / hypergraph coloring) Let {𝑥𝑖: 𝑖∈𝐼} be a collection of independent random variables. Let 𝐸1, . . . , 𝐸𝑛be events where each 𝐸𝑖depends only on the variables indexed by some subset 𝐵𝑖⊆𝐼of variables. A canonical dependency graph for the events 𝐸1, . . . , 𝐸𝑛has vertex set [𝑛] and an edge 𝑖𝑗whenever 𝐵𝑖∩𝐵𝑗≠∅.
It is easy to check that the canonical dependency graph above is indeed a valid depen-dency graph.
Example 6.1.6 (Boolean satisfiability problem (SAT)). Given a CNF formula (con-junctive normal form, i.e., and-of-or’s), e.g., (∧= and; ∨= or) (𝑥1 ∨𝑥2 ∨𝑥3) ∧(𝑥1 ∨𝑥2 ∨𝑥4) ∧(𝑥2 ∨𝑥4 ∨𝑥5) ∧· · · the problem is to find a satisfying assignment with boolean variables 𝑥1, 𝑥2, . . . . Many problems in computer science can be modeled using this way. This problem can be viewed as in Setup 6.1.5, where 𝐴𝑖is the event that the 𝑖-th clause is violated.
80 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.1 Statement and proof The following formulation of the local lemma is easiest to apply and is the most commonly used. It applies to settings where the dependency graph has small maximum degree.
Theorem 6.1.7 (Lovász local lemma; symmetric form) Let 𝐴1, . . . , 𝐴𝑛be events, with P[𝐴𝑖] ≤𝑝for all 𝑖. Suppose that each 𝐴𝑖is independent from a set of all other 𝐴𝑗except for at most 𝑑of them. If 𝑒𝑝(𝑑+ 1) ≤1, then with some positive probability, none of the events 𝐴𝑖occur.
Remark 6.1.8. The constant 𝑒is best possible (Shearer 1985). In most applications, the precise value of the constant is unimportant.
Theorem 6.1.9 (Lovász local lemma; general form) Let 𝐴1, . . . , 𝐴𝑛be events. For each 𝑖∈[𝑛], let 𝑁(𝑖) be such that 𝐴𝑖is independent from {𝐴𝑗: 𝑗∉{𝑖} ∪𝑁(𝑖)}. If 𝑥1, . . . , 𝑥𝑛∈[0, 1) satisfy P(𝐴𝑖) ≤𝑥𝑖 Ö 𝑗∈𝑁(𝑖) (1 −𝑥𝑗) for all 𝑖∈[𝑛], then P(none of the events 𝐴𝑖occur) ≥ 𝑛 Ö 𝑖=1 (1 −𝑥𝑖).
Proof that the general form implies the symmetric form. Set 𝑥𝑖= 1/(𝑑+ 1) < 1 for all 𝑖. Then 𝑥𝑖 Ö 𝑗∈𝑁(𝑖) (1 −𝑥𝑗) ≥ 1 𝑑+ 1 1 − 1 𝑑+ 1 𝑑 > 1 (𝑑+ 1)𝑒≥𝑝 so the hypothesis of general local lemma holds.
□ Here is another corollary of the general form local lemma, which applies if the total probability of any neighborhood in a dependency graph is small.
Corollary 6.1.10 In the setup of Theorem 6.1.9, if P(𝐴𝑖) < 1/2 and Í 𝑗∈𝑁(𝑖) P(𝐴𝑗) ≤1/4 for all 𝑖, then with positive probability none of the events 𝐴𝑖occur.
81 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma Proof. In Theorem 6.1.9, set 𝑥𝑖= 2P(𝐴𝑖) for each 𝑖. Then 𝑥𝑖 Ö 𝑗∈𝑁(𝑖) (1 −𝑥𝑗) ≥𝑥𝑖© « 1 − ∑︁ 𝑗∈𝑁(𝑖) 𝑥𝑗ª ® ¬ = 2P(𝐴𝑖) © « 1 − ∑︁ 𝑗∈𝑁(𝑖) 2P(𝐴𝑖)ª ® ¬ ≥P(𝐴𝑖).
(The first inequality is by “union bound.”) □ In some applications, one may need to apply the general form local lemma with carefully chosen values for 𝑥𝑖.
Proof of Lovász local lemma (general case). We will prove that P © « 𝐴𝑖 Û 𝑗∈𝑆 𝐴𝑗ª ® ¬ ≤𝑥𝑖 whenever 𝑖∉𝑆⊆[𝑛].
(6.1) Once (6.1) has been established, we then deduce that P(𝐴1 · · · 𝐴𝑛) = P(𝐴1)P 𝐴2 𝐴1 P 𝐴3 𝐴1𝐴2 · · · P 𝐴𝑛 𝐴1 · · · 𝐴𝑛−1 ≥(1 −𝑥1)(1 −𝑥2) · · · (1 −𝑥𝑛), which is the conclusion of the local lemma.
Now we prove (6.1) by induction on |𝑆|. The base case |𝑆| = 0 is trivial.
Let 𝑖∉𝑆. Let 𝑆1 = 𝑆∩𝑁(𝑖) and 𝑆2 = 𝑆\ 𝑆1. We have P © « 𝐴𝑖 Û 𝑗∈𝑆 𝐴𝑗ª ® ¬ = P 𝐴𝑖 Ó 𝑗∈𝑆1 𝐴𝑗 Ó 𝑗∈𝑆2 𝐴𝑗 P Ó 𝑗∈𝑆1 𝐴𝑗 Ó 𝑗∈𝑆2 𝐴𝑗 (6.2) For the RHS of (6.2), using that 𝐴𝑖is independent of 𝑗∈𝑆2 : 𝐴𝑗 , numerator ≤P © « 𝐴𝑖 Û 𝑗∈𝑆2 𝐴𝑗ª ® ¬ = P(𝐴𝑖) ≤𝑥𝑖 Ö 𝑗∈𝑁(𝑖) (1 −𝑥𝑖), (6.3) 82 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.2 Coloring hypergraphs and, denoting the elements of 𝑆1 by 𝑆1 = { 𝑗1, . . . , 𝑗𝑟}, denominator = P © « 𝐴𝑗1 Û 𝑗∈𝑆2 𝐴𝑗ª ® ¬ P © « 𝐴𝑗2 𝐴𝑗1 Û 𝑗∈𝑆2 𝐴𝑗ª ® ¬ · · · P © « 𝐴𝑗𝑟 𝐴𝑗1 · · · 𝐴𝑗𝑟−1 Û 𝑗∈𝑆2 𝐴𝑗ª ® ¬ ≥(1 −𝑥𝑗1) · · · (1 −𝑥𝑗𝑟) [by induction hypothesis] ≥ Ö 𝑗∈𝑁(𝑖) (1 −𝑥𝑖) Thus (6.2) ≤𝑥𝑖, thereby finishing the induction proof of (6.1).
□ Remark 6.1.11. We used the independence assumption only at step (6.3) of the proof.
Upon a closer examination, we see that we only need to know correlation inequalities of the form P 𝐴𝑖 Ó 𝑗∈𝑆2 𝐴𝑗 ≤P(𝐴𝑖) for 𝑆2 ⊆𝑁(𝑖), rather than independence. This observation allows a strengthening of the local lemma, known as a lopsided local lemma, that we will explore later in the chapter.
6.2 Coloring hypergraphs Previously, in Theorem 1.3.1, we saw that every 𝑘-uniform hypergraph with fewer than 2𝑘−1 edges is 2-colorable. The next theorem gives a sufficient local condition for 2-colorability.
Theorem 6.2.1 A 𝑘-uniform hypergraph is 2-colorable if every edge intersects at most 𝑒−12𝑘−1 −1 other edges Proof. For each edge 𝑓, let 𝐴𝑓be the event that 𝑓is monochromatic. Then P(𝐴𝑓) = 𝑝:= 2−𝑘+1. Each 𝐴𝑓is independent from all 𝐴𝑓′ where 𝑓′ is disjoint from 𝑓. Since at most 𝑑:= 𝑒−12𝑘−1 −1 edges intersect every edge, and 𝑒(𝑑+ 1)𝑝≤1, so the local lemma implies that with positive probability, none of the events 𝐴𝑓occur.
□ Corollary 6.2.2 For 𝑘≥9, every 𝑘-uniform 𝑘-regular hypergraph is 2-colorable.
(Here 𝑘-regular means that every vertex lies in exactly 𝑘edges.) Proof. Every edge intersects ≤𝑑= 𝑘(𝑘−1) other edges. And 𝑒(𝑘(𝑘−1)+1)2−𝑘+1 < 1 for 𝑘≥9.
□ 83 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma Remark 6.2.3. The statement is false for 𝑘= 2 (triangle) and 𝑘= 3 (Fano plane) but actually true for all 𝑘≥4 (Thomassen 1992).
Here is an example where the symmetric form of the local lemma is insufficient (why?).
Theorem 6.2.4 Let 𝐻be a (non-uniform) hypergraph where every edge has size ≥3. Suppose ∑︁ 𝑓∈𝐸(𝐻){𝑒}:𝑒∩𝑓≠∅ 2−| 𝑓| ≤1 8, for each edge 𝑒, then 𝐻is 2-colorable.
Proof. Consider a uniform random 2-coloring of the vertices. Let 𝐴𝑒be the event that edge 𝑒is monochromatic. Then P(𝐴𝑒) = 2−|𝑒|+1 ≤1/4 since |𝑒| ≥3. Also, ∑︁ 𝑓∈𝐸(𝐻){𝑒}:𝑒∩𝑓≠∅ P(𝐴𝑓) = ∑︁ 𝑓∈𝐸(𝐻){𝑒}:𝑒∩𝑓≠∅ 2−| 𝑓|+1 ≤1/4.
Thus by Corollary 6.1.10 one can avoid all events 𝐴𝑒, and hence 𝐻is 2-colorable.
□ Remark 6.2.5. A sign to look beyond the symmetric local lemma is when there are bad events of very different nature (e.g., having very different probabilities).
Compactness argument Now we highlight an important compactness argument that allows us to deduce the existence of an infinite object, even though the local lemma itself is only applicable to finite systems.
Theorem 6.2.6 Let 𝐻be a (non-uniform) hypergraph on a possibly infinite vertex set, such that each edge is finite, has at least 𝑘vertices, and intersect at most 𝑑other edges. If 𝑒2−𝑘+1(𝑑+ 1) ≤1, then 𝐻has a proper 2-coloring.
Proof. From a vanilla application of the symmetric local lemma, we deduce that for any finite subset 𝑋of vertices, there exists an 2-coloring 𝑋so that no edge contained in 𝑋is monochromatic (color each vertex iid uniformly, and consider the bad event 𝐴𝑒 that the edge 𝑒⊆𝑋is monochromatic).
Next we extend the coloring to the entire vertex set 𝑉by a compactness argument. The set of all colorings is 𝑉. By Tikhonov’s theorem (which says a product of a possibly 84 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.2 Coloring hypergraphs infinite collection of compact topological spaces is compact), 𝑉is compact under the product topology.
For each finite subset 𝑋, let 𝐶𝑋⊆𝑉be the subset of colorings where no edge contained in 𝑋is monochromatic. Earlier from the local lemma we saw that 𝐶𝑋≠∅.
If 𝑌⊆𝑋, then 𝐶𝑌⊇𝐶𝑋. Thus 𝐶𝑋1 ∩· · · ∩𝐶𝑋ℓ⊇𝐶𝑋1∪···∪𝑋ℓ, so {𝐶𝑋: |𝑋| < ∞} is a collection of closed subsets of 𝑉with the finite intersection property (i.e., the intersection of any finite subcollection is nonempty).
Recall from point-set topology the following basic fact (a defining property): a space is compact if and only if every family of closed subsets having the finite intersection property has non-empty intersection.
Hence by compactness of 𝑉, the intersection of 𝐶𝑋taken over all finite 𝑋is non-empty. Any element of this intersection corresponds to a valid coloring of the hypergraph.
□ More generally, the above compactness argument yields the following.
Lemma 6.2.7 (Compactness argument) Consider a variation of the random variable model (Setup 6.1.5) where each variable has only finitely many choices but there can be possibly infinitely many events (each event depends on a finite subset of variables). If it is possible to avoid any finite subset of events, then it is possible to avoid all the events.
□ Remark 6.2.8. Note the conclusion may be false if we do not assume the random variable model (why?).
The next application appears in the paper of Erdős and Lovász (1975) where the local lemma originally appears.
Consider 𝑘-coloring the real numbers, i.e., a function 𝑐: R →[𝑘]. We say that 𝑇⊆R is multicolored with respect to 𝑐if all 𝑘colors appear in 𝑇.
Question 6.2.9 For each 𝑘is there an 𝑚so that for every 𝑆⊆R with |𝑆| = 𝑚, one can 𝑘-color R so that every translate of 𝑆is multicolored?
The following theorem shows that this can be done whenever 𝑚> (3 + 𝜀)𝑘log 𝑘and 𝑘> 𝑘0(𝜀) sufficiently large.
85 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma Theorem 6.2.10 The answer to the above equation is yes if 𝑒(𝑚(𝑚−1) + 1)𝑘 1 −1 𝑘 𝑚 ≤1.
Proof. Each translate of 𝑆is not multicolored with probability 𝑝≤𝑘(1 −1/𝑘)𝑚, and each translate of 𝑆intersects at most 𝑚(𝑚−1) other translates. Consider a bad event for each translate of 𝑆contained in 𝑋. The symmetric local lemma tells us that it is possible to avoid any finite collection of bad events. By the compactness argument, it is possible to avoid all the bad events.
□ Coloring arithmetic progressions Here is an application where we need to apply the asymmetric local lemma.
Theorem 6.2.11 (Beck 1980) For every 𝜀> 0, there exists 𝑘0 and a 2-coloring of Z with no monochromatic 𝑘-term arithmetic progressions with 𝑘≥𝑘0 and common difference less than 2(1−𝜀)𝑘.
Proof. We pick a uniform random color for each element of Z.
For each 𝑘-term arithmetic progression in Z with 𝑘≥𝑘0 and common difference less than 2(1−𝜀)𝑘, consider the event that this 𝑘-AP is monochromatic. By the compactness argument, it suffices to check that we can avoid any finite subset of events.
The event that a particular 𝑘-AP is monochromatic has probability exactly 2−𝑘+1.
(Since this number depends on 𝑘, we should use the asymmetric local lemma.) Recall that in the asymmetric local lemma (Theorem 6.1.9), we need to select 𝑥𝑖∈[0, 1) for each bad event 𝐴𝑖so that P(𝐴𝑖) ≤𝑥𝑖 Ö 𝑗∈𝑁(𝑖) (1 −𝑥𝑗) for all 𝑖∈[𝑛].
It is usually a good idea to select 𝑥𝑖to be somewhat similar to P(𝐴𝑖). In this case, if 𝐴𝑖 is the event corresponding to a 𝑘-AP, then we take 𝑥𝑖= 2−(1−𝜀/2)𝑘= P(𝐴𝑖) 2 1−𝜀/2 (with the same 𝜀as in the statement of the theorem).
86 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.2 Coloring hypergraphs Fix a 𝑘-AP 𝑃in Z with 𝑘≥𝑘0. The number of ℓ-APs with ℓ≥𝑘0 and common difference less than 2(1−𝜀)ℓthat intersects 𝑃is at most 𝑘ℓ2(1−𝜀)ℓ(one choice for the element of 𝑘, a choice of the position of the ℓ-AP, and at most 2(1−𝜀)ℓchoices for the common difference). So to apply the local lemma, it suffices to check that 2−𝜀𝑘/2+1 ≤ Ö ℓ≥𝑘0 1 −2−(1−𝜀/2)ℓ𝑘ℓ2(1−𝜀)ℓ .
Note that 1 −𝑥≥𝑒−2𝑥for 𝑥∈[0, 1/2]. So 𝑅𝐻𝑆≥exp − ∑︁ ℓ≥𝑘0 21−(1−𝜀/2)ℓ· 𝑘ℓ2(1−𝜀)ℓ !
= exp −𝑘 ∑︁ ℓ≥𝑘0 ℓ21−𝜀ℓ/2 !
By making 𝑘0 = 𝑘0(𝜀) large enough, we can ensure that Í ℓ≥𝑘0 ℓ21−𝜀ℓ/2 < 𝜀/4, and so continuing, · · · ≥𝑒−𝜀𝑘/4 ≥2−𝜀𝑘/2+1 provided that 𝑘≥𝑘0(𝜀). So we can apply the local lemma to conclude.
□ Decomposing coverings We say that a collection of disks in R𝑑is a covering if their union is R𝑑. We say that it is a 𝒌-fold covering if every point of R𝑑is contained in at least 𝑘disks (so 1-fold covering is the same as a covering).
We say that a 𝑘-fold covering is decomposable if it can be partitioned into two cover-ings.
In R𝑑, is a every 𝑘-fold covering by unit balls decomposable if 𝑘is sufficiently large?
A fun exercise: in R1, every 𝑘-fold covering by intervals can be partitioned into 𝑘 coverings.
Mani-Levitska and Pach (1986) showed that every 33-fold covering of R2 is decom-posable.
What about higher dimensions?
Surprising, they also showed that for every 𝑘, there exists a 𝑘-fold indecomposable covering of R3 (and similarly for R𝑑for 𝑑≥3).
However, it turns out that indecomposable coverings must cover the space quite un-evenly: 87 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma Theorem 6.2.12 (Mani-Levitska and Pach 1986) Every 𝑘-fold nondecomposable covering of R3 by open unit balls must cover some point ≳2𝑘/3 times.
Remark 6.2.13. In R𝑑, the same proof gives ≥𝑐𝑑2𝑘/𝑑.
We will need the following combinatorial geometric fact: Lemma 6.2.14 A set of 𝑛≥2 spheres in R3 cut R3 into at most 𝑛3 connected components.
Proof. Let us first consider the problem in one dimension lower. Let 𝑓(𝑚) be the maximum number of connected regions that 𝑚circles on a sphere in R3 can cut the sphere into.
We have 𝑓(𝑚+ 1) ≤𝑓(𝑚) + 2𝑚for all 𝑚≥1 since adding a new circle to a set of 𝑚 circles creates at most 2𝑚intersection points, so that the new circle is divided into at most 2𝑚arcs, and hence its addition creates at most 2𝑚new regions.
Combined with 𝑓(1) = 2, we deduce 𝑓(𝑚) ≤𝑚(𝑚−1) + 2 for all 𝑚≥1.
Now let 𝑔(𝑚) be the maximum number of connected regions that 𝑚spheres in R3 can cut R3 into. We have 𝑔(1) = 2, and 𝑔(𝑚+ 1) ≤𝑔(𝑚) + 𝑓(𝑚) ≤𝑔(𝑚) by a similar argument as earlier. So 𝑔(𝑚) ≤𝑓(𝑚−1) + 𝑓(𝑚−2) + · · · + 𝑓(1) + 𝑔(0) ≤𝑚3.
□ Proof. Suppose for contradiction that every point in R3 is covered by at most 𝑡≤𝑐2𝑘/3 unit balls from 𝐹(for some sufficiently small 𝑐that we will pick later).
Construct an infinite hypergraph 𝐻with vertex set being the set of balls and edges having the form 𝐸𝑥= {balls containing 𝑥} for some 𝑥∈R3. Note that |𝐸𝑥| ≥𝑘since we have a 𝑘-fold covering.
Also, note that if 𝑥, 𝑦∈R3 lie in the same connected component in the complement of the union of all the unit spheres, then 𝐸𝑥= 𝐸𝑦(i.e., the same edge).
Claim: every edge of intersects at most 𝑑= 𝑂(𝑡3) other edges Proof of claim: Let 𝑥∈R3. If 𝐸𝑥∩𝐸𝑦≠∅, then |𝑥−𝑦| ≤2, so all the balls in 𝐸𝑦lie in the radius-4 ball centered at 𝑥. The volume of the radius-4 ball is 43 times the unit ball. Since every point lies in at most 𝑡balls, there are at most 43𝑡balls appearing among those 𝐸𝑦intersecting 𝑥, and these balls cut the radius-2 centered at 𝑥into 𝑂(𝑡3) connected regions by the earlier lemma, and two different 𝑦’s in the same region produce the same 𝐸𝑦. So 𝐸𝑥intersects 𝑂(𝑡3) other edges. ■ 88 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.3 Independent transversal With 𝑡≤𝑐2𝑘/3 and 𝑐sufficiently small, and knowing 𝑑= 𝑂(𝑡3) from the claim, we have 𝑒2−𝑘+1(𝑑+1) ≤1. It then follows by Theorem 6.2.6 (local lemma + compactness argument) that this hypergraph is 2-colorable, which corresponds to a decomposition of the covering, a contradiction.
□ 6.3 Independent transversal The application of the local lemma in this section is instructive in that it is not obvious at first what to choose as bad events (even if you are already told to apply the local lemma). It is worth trying different possibilities.
Every graph with maximum degree Δ contains an independent set of size ≥|𝑉|/(Δ+1) (choose the independent set greedily). The following lemma shows that by decreasing the desired size of the independent set by a constant factor, we can guarantee an independent set that is also a transversal to a vertex set partition.
Theorem 6.3.1 Let 𝐺= (𝑉, 𝐸) be a graph with maximum degree Δ and let 𝑉= 𝑉1 ∪· · · ∪𝑉𝑟be a partition, where each |𝑉𝑖| ≥2𝑒Δ. Then there is an independent set in 𝐺containing one vertex from each 𝑉𝑖.
Proof. The first step in the proof is simple yet subtle: we may assume that |𝑉𝑖| = 𝑘:= ⌈2𝑒Δ⌉for each 𝑖, or else we can remove some vertices from 𝑉𝑖(if we do not trim the vertex sets now, we will run into difficulties later).
Pick 𝑣𝑖∈𝑉𝑖uniformly at random, independently for each 𝑖.
This is an instance of the random variable model (Setup 6.1.5), where 𝑣1, . . . , 𝑣𝑟are the random variables.
We would like to design a collection of “bad events” so that if we avoid all of them, then {𝑣1, . . . , 𝑣𝑟} is guaranteed to be independent set.
What do we choose as bad events? It turns out that some choices work better than others.
Attempt 1: For each 1 ≤𝑖< 𝑗≤𝑟where there exists an edge between 𝑉𝑖and 𝑉𝑗, let 𝐴𝑖,𝑗be the event that 𝑣𝑖is adjacent to 𝑣𝑗.
We find that P(𝐴𝑖,𝑗) ≤Δ/𝑘.
The canonical dependency graph has 𝐴𝑖,𝑗∼𝐴𝑖′,𝑗′ if and only if the two sets {𝑖, 𝑗} and {𝑖′, 𝑗′} intersect. This dependency graph has max degree ≤2Δ𝑘(starting from (𝑖, 𝑗), 89 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma look at the neighbors of all vertices in 𝑉𝑖∪𝑉𝑗). The max degree is too large compared to the bad event probabilities.
Attempt 2: For each edge 𝑒∈𝐸, let 𝐴𝑒be the event that both endpoints of 𝑒are picked.
We have P(𝐴𝑒) = 1/𝑘2.
The canonical dependency graph has 𝐴𝑒∼𝐴𝑓if some 𝑉𝑖intersects both 𝑒and 𝑓.
This dependency graph has max degree ≤2𝑘Δ (if 𝑒is between 𝑉𝑖and 𝑉𝑗, then 𝑓must be incident to 𝑉𝑖∪𝑉𝑗).
We have 𝑒(1/𝑘2)(2𝑘Δ + 1) ≤1, so the local lemma implies the with probability no bad event occurs, in which case {𝑣1, . . . , 𝑣𝑟} is an independent set.
□ Remark 6.3.2. Alon (1988) introduced the above result as lemma in his near resolution of the still-open linear arboricity conjecture (see the Alon–Spencer textbook §5.5).
Alon’s approach makes heavy use of the local lemma.
Haxell (1995, 2001) relaxed the hypothesis to |𝑉𝑖| ≥2Δ for each 𝑖. The statement becomes false if 2Δ is replaced by 2Δ −1 (Szabó and Tardos 2006).
6.4 Directed cycles of length divisible by 𝑘 A directed graph is 𝒅-regular if every vertex has in-degree 𝑑and out-degree 𝑑.
Theorem 6.4.1 (Alon and Linial 1989) For every 𝑘there exists 𝑑so that every 𝑑-regular directed graph has a directed cycle of length divisible by 𝑘.
Corollary 6.4.2 For every 𝑘there exists 𝑑so that every 2𝑑-regular graph has a cycle of length divisible by 𝑘.
Proof. Every 2𝑑-regular graph can be made into a 𝑑-regular digraph by orientating its edges according to an Eulerian tour. And then we can apply the previous theorem.
□ We will prove the following more general statement.
90 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.4 Directed cycles of length divisible by 𝑘 Theorem 6.4.3 (Alon and Linial 1989) Every directed graph with min out-degree 𝛿and max in-degree Δ contains a cycle of length divisible by 𝑘∈N as long as 𝑘≤ 𝛿 1 + log(1 + 𝛿Δ) .
Proof. By deleting edges, can assume that every vertex has out-degree exactly 𝛿.
Assign every vertex 𝑣an element 𝑥𝑣∈Z/𝑘Z iid uniformly at random.
We will look for directed cycles where the labels increase by 1 (mod 𝑘) at each step.
These cycles all have length divisible by 𝑘.
For each vertex 𝑣, let 𝐴𝑣be the event that there is nowhere to go from 𝑣(i.e., if no outneighbor is labeled 𝑥𝑣+ 1 (mod 𝑘)). We have P(𝐴𝑣) = (1 −1/𝑘)𝛿≤𝑒−𝛿/𝑘.
Since 𝐴𝑣depends only on {𝑥𝑤: 𝑤∈{𝑣} ∪𝑁+(𝑣)}, where 𝑁+(𝑣) denotes the out-neighbors of 𝑣and 𝑁−(𝑣) the in-neighbors of 𝑣, the canonical dependency graph has 𝐴𝑣∼𝐴𝑤if {𝑣} ∪𝑁+(𝑣) intersects {𝑤} ∪𝑁+(𝑤).
The maximum degree in the dependency graph is at most Δ+𝛿Δ (starting from 𝑣, there are (1) at most Δ choices stepping backward (2) 𝛿choices stepping forward, and (3) at most 𝛿(Δ−1) choices stepping forward and then backward to land somewhere other than 𝑣).
So an application of the local lemma shows that we are done as long as 𝑒1−𝛿/𝑘(1+Δ+𝛿Δ), i.e., 𝑘≤𝛿/(1 + log(1 + Δ + 𝛿Δ)).
This is almost, but not quite the result (though, for most applications, we would be perfectly happy with such a bound).
The final trick is to notice that we actually have an even smaller valid dependency digraph: 𝐴𝑣is independent of all 𝐴𝑤where 𝑁+(𝑣) is disjoint from 𝑁+(𝑤) ∪{𝑤}.
91 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma Indeed, even if we fix the colors of all vertices outside 𝑁+(𝑣), the conditional proba-bility that 𝐴𝑣is still (1 −1/𝑘)𝛿.
The number of 𝑤such that 𝑁+(𝑣) intersects 𝑁+(𝑤) ∪{𝑤} is at most 𝛿Δ (no longer need to consider (1) in the previous count). And we have 𝑒𝑝(𝛿Δ + 1) ≤𝑒1−𝛿/𝑘(𝛿Δ + 1) ≤1.
So we are done by the local lemma.
□ 6.5 Lopsided local lemma Let us move beyond the random variable model, and consider a collection of bad events in the general setup of the local lemma. Instead of requiring that each event is independent of its non-neighbors (in the dependency graph), what if we assume that avoiding some bad events make it easier to avoid some others? Intuitively, it seems that it would only make it easier to avoid bad events.
We can make this notion precise by re-examining the proof of the local lemma. Where did we actually use the independence assumption in the hypothesis of the local lemma?
It was in the following step, Equation (6.3): numerator ≤P © « 𝐴𝑖 Û 𝑗∈𝑆2 𝐴𝑗ª ® ¬ = P(𝐴𝑖) ≤𝑥𝑖 Ö 𝑗∈𝑁(𝑖) (1 −𝑥𝑖).
If we had changed the middle = to ≤, the whole proof would remain valid. This observation allows us to weaken the independence assumption. Therefore we have the following theorem, which was used by Erdős and Spencer (1991) to give an application to Latin transversals that we will see shortly.
92 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.5 Lopsided local lemma Theorem 6.5.1 (Lopsided local lemma) Let 𝐴1, . . . , 𝐴𝑛be events. For each 𝑖, let 𝑁(𝑖) ⊆[𝑛] be such that P © « 𝐴𝑖 Û 𝑗∈𝑆 𝐴𝑗ª ® ¬ ≤P(𝐴𝑖) for all 𝑖∈[𝑛] and 𝑆⊆[𝑛] \ (𝑁(𝑖) ∪{𝑖}) (6.1) Suppose there exist 𝑥1, . . . , 𝑥𝑛∈[0, 1) such that P(𝐴𝑖) ≤𝑥𝑖 Ö 𝑗∈𝑁(𝑖) (1 −𝑥𝑗) for all 𝑖∈[𝑛].
Then P(none of the events 𝐴𝑖occur) ≥ 𝑛 Ö 𝑖=1 (1 −𝑥𝑖).
Like earlier, by setting 𝑥𝑖= 1/(𝑑+ 1), we deduce a symmetric version that is easier to apply.
Corollary 6.5.2 (Lopsided local lemma; symmetric version) In the previous theorem, if |𝑁(𝑖)| ≤𝑑and P(𝐴𝑖) ≤𝑝for every 𝑖∈[𝑛], and 𝑒𝑝(𝑑+1) ≤ 1, then with positive probability none of the events 𝐴𝑖occur.
The (di)graph where 𝑁(𝑖) is the set of (out-)neighbors of 𝑖is called a negative depen-dency (di)graph.
Remark 6.5.3 (Important!). Just as with the usual local lemma, the negative depen-dency graph is not constructed by simply checking pairs of events.
The hypothesis of Theorem 6.5.1 seems annoying to check. Fortunately, many appli-cations of lopsided local lemma fall within a model that we will soon describe, where there is a canonical negative dependency graph that is straightforward to construct.
This is analogous to the random variable model for the usual local lemma, where the canonical dependence graph has two events adjacency if they share variables.
Random injection model We describe a random injection model where there is an easy-to-construct canonical negative dependency graph (Lu and Székely 2007).
Recall that a matching in a graph is a subset of edges with no two sharing a vertex.
In a bipartite graph with vertex parts 𝑋and 𝑌, a complete matching from 𝑋to 𝑌is a matching where every vertex of 𝑋belongs to an edge of the matching.
93 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma Setup 6.5.4 (Random injection model) Let 𝑋and 𝑌be finite sets with |𝑋| ≤|𝑌|.
Let 𝑓: 𝑋→𝑌be an injection chosen uniformly at random. We can also represent 𝑓 by a complete matching 𝑀from 𝑋to 𝑌in 𝐾𝑋,𝑌(the complete bipartite graph between 𝑋and 𝑌). We will speak interchangeably of the injection 𝑓and matching 𝑀.
For a given matching 𝐹(not necessarily complete) in 𝐾𝑋,𝑌, let 𝐴𝐹denote the event that 𝐹⊆𝑀.
Let 𝐹1, . . . , 𝐹𝑛be matchings in 𝐾𝑋,𝑌. The canonical negative dependency graph for the vents 𝐴𝐹1, . . . , 𝐴𝐹𝑛has one vertex for each event, and an edge between the events 𝐴𝐹𝑖and 𝐴𝐹𝑗(𝑖≠𝑗) if 𝐹𝑖and 𝐹𝑗are not vertex disjoint.
The following result shows that the above canonical negative dependency graph is a valid for the lopsided local lemma (Theorem 6.5.1).
Theorem 6.5.5 (Nonnegative dependence for random injections) In Setup 6.5.4, let 𝐹0 be a matching in 𝐾𝑋,𝑌such that 𝐹0 is vertex disjoint from 𝐹1 ∪· · · ∪𝐹𝑘. Then P 𝐴𝐹0 𝐴𝐹1 · · · 𝐴𝐹𝑘 ≤P(𝐴𝐹0).
Proof. Let 𝑋0 ⊆𝑋and 𝑌0 ⊆𝑌be the set of endpoints of 𝐹0.
For each matching 𝑇in 𝐾𝑋,𝑌, let M𝑇= {complete matchings from 𝑋to 𝑌containing 𝑇but not containing any of 𝐹1, . . . , 𝐹𝑘} .
For the desired inequality, note that 𝐿𝐻𝑆= P 𝐴𝐹0 𝐴𝐹1 · · · 𝐴𝐹𝑘 = M𝐹0 |M∅| = M𝐹0 Í 𝑇: 𝑋0↩→𝑌|M𝑇| where the sum is taken over all |𝑌| (|𝑌| −1) · · · (|𝑌| −|𝑋| + 1) complete matchings 𝑇 from 𝑋0 to 𝑌(which we denote by 𝑇: 𝑋0 ↩→𝑌), and 𝑅𝐻𝑆= P(𝐴𝐹0) = 1 |{𝑇: 𝑋0 ↩→𝑌}| .
Thus to show that 𝐿𝐻𝑆≤𝑅𝐻𝑆, it suffices to prove M𝐹0 ≤|M𝑇| for every 𝑇: 𝑋0 ↩→𝑌.
94 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.5 Lopsided local lemma It suffices to construct an injection M𝐹0 ↩→M𝑇. Let 𝑌1 be the set of endpoints of 𝑇 in 𝑌. Fix a permutation 𝜎of 𝑌such that • 𝜎fixes all elements of 𝑌outside 𝑌0 ∪𝑌1; and • 𝜎sends 𝐹0 to 𝑇.
Then 𝜎induces a permutation on the set of complete matchings from 𝑋to 𝑌. It remains to show that if we start with a matching in M𝐹0, so that it avoids 𝐹𝑖for all 𝑖≥1, then it is sent to a matching that also avoids 𝐹𝑖for all 𝑖≥1 (and hence lies in M𝑇). Indeed, this follows from the hypothesis that none of the edges in 𝐹𝑖use any vertex from 𝑋0 or 𝑌0.
□ As an example, here is a quick application.
Corollary 6.5.6 (Derangement lower bound) The probability that a uniform random permutation of [𝑛] has no fixed points is at least (1 −1/𝑛)𝑛.
Proof. In the random injection model, let 𝑋= 𝑌= [𝑛]. Let 𝑓: 𝑋→𝑌be a uniform random permutation. For each 𝑖∈[𝑛], let 𝐹𝑖be the single edge (𝑖, 𝑖), i.e., 𝐴𝐹𝑖is the even that 𝑓(𝑖) = 𝑖. Note that the canonical negative dependency graph is empty since no two 𝐹𝑖’s share a vertex. Since P(𝐴𝑖) = 1 −1/𝑛, we can set 𝑥𝑖= 1 −1/𝑛for each 𝑖 in the lopsided local lemma to obtain the conclusion P( 𝑓has no fixed points) = P(𝐴1 · · · 𝐴𝑛) ≥ 1 −1 𝑛 𝑛 .
□ Remark 6.5.7. A fixed-point free permutation is called a derangement.
Using inclusion-exclusion, one can deduceanexactanswertothe abovequestion: Í𝑛 𝑖=0(−1)𝑖/𝑖!.
This quantity converges to 1/𝑒as 𝑘→∞, and the above lower bound (1 −1/𝑛)𝑛also converges to 1/𝑒and so is asymptotically optimal.
Latin transversal A Latin square of order 𝑛is an 𝑛× 𝑛array filled with 𝑛symbols so that every symbol appears exactly once in every row and column. Example: 1 2 3 2 3 1 3 1 2 These objects are called Latin squares because they were studied by Euler (1707–1783) who used Latin symbols to fill the arrays.
95 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma Given an 𝑛× 𝑛array, a transversal is a set of 𝑛entries with one in every row and column. A Latin transversal is a transversal with distinct entries. Example: 1 2 3 2 3 1 3 1 2 Here are is a famous open conjecture about Latin transversals.1 (Do you see why the hypothesis on parity is necessary?) Conjecture 6.5.8 (Ryser 1967) Every odd order Latin square has a transversal.
The conjecture should be modified for even order Latin squares.
Conjecture 6.5.9 (Ryser-Brualdi-Stein conjecture) Every even order Latin square has a transversal containing all but at most one symbol.
Remark 6.5.10. Keevash, Pokrovskiy, Sudakov and Yepremyan (2022) proved that every order 𝑛Latin square contains a transversal containing all but 𝑂(log 𝑛/log log 𝑛) symbols, improving an earlier bound of 𝑂(log2 𝑛) by Hatami and Shor (2008).
Recently, Montgomery announced a proof of the conjecture for all sufficiently large even 𝑛. The proof uses sophisticated techniques combining the semi-random method and the absorption method.
The next result is the original application of the lopsided local lemma.
Theorem 6.5.11 (Erdős and Spencer 1991) Every 𝑛×𝑛array where every entry appears at most 𝑛/(4𝑒) times has a Latin transversal.
Proof. Pick a transversal uniformly at random. This is the same as picking a permuta-tion 𝑓: [𝑛] →[𝑛] uniformly at random. In Setup 6.5.4, the random injection model, transversals correspond to perfect matchings.
For each pair of equal entries in the array not both lying in the same row or column, consider the bad event that the transversal contains both entries.
The canonical negative dependency graph is obtained by joining an edge between two bad events if the four entries involved share some row or column.
1Not to be confused with another conjecture also known as Ryser’s conjecture concerning an inequality between the covering number and the matching number of multipartite hypergraphs, as a generaliza-tion of König’s theorem. See Best and Wanless (2018) for a historical commentary and a translation of Ryser’s 1967 paper.
96 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.6 Algorithmic local lemma Let us count neighbors in this negative dependency graph. Fix a pair of equal entries in the array. Their rows and columns span fewer than 4𝑛entries, and for each such entry 𝑧, there are at most 𝑛/(4𝑒) −1 choices for another entry equal to 𝑧. Thus the maximum degree in the canonical negative dependence graph is ≤(4𝑛−4) 𝑛 4𝑒−1 ≤𝑛(𝑛−1) 𝑒 −1.
We can now apply the symmetric lopsided local lemma to conclude that with positive probability, none of the bad events occur.
□ 6.6 Algorithmic local lemma Consider an instance of a problem in the random variable setting (e.g., 𝑘-CNF) for which the local lemma guarantees a solution. Can one find a satisfying assignment efficiently?
The local lemma tells you that some good configuration exists, but the proof is non-constructive. The probability that a random sample avoids all the bad events is often very small (usually exponentially small, e.g., in the case of a set of independent bad events). It had been an open problem for a long time whether the local lemma can be made algorithmic.
Moser (2009), during his PhD, achieved a breakthrough by coming up with the first efficient algorithmic version of the local lemma for finding a satisfying assignment for 𝑘-CNF formulas. Moser and Tardos (2010) later extended the algorithm for the general local lemma in the random variable model.
Remark 6.6.1 (Too hard in general). The Moser–Tardos algorithm works in the ran-dom variable model (there are subsequent work that concern other models such as the random injection model). Some assumption on the model is necessary since the problem can be computationally hard in general.
For example, let 𝑞= 2𝑘, and 𝑓: [𝑞] →[𝑞] be some fixed bijection (with an explicit description and easy to compute). Consider the computational task of inverting 𝑓: given 𝑦∈[𝑞], find 𝑥such that 𝑓(𝑥) = 𝑦(we would like an algorithm with running time polynomial in 𝑘).
If 𝑥∈[𝑞] is chosen uniformly, then 𝑓(𝑥) ∈[𝑞] is also uniform. For each 𝑖∈[𝑘], let 𝐴𝑖be the event that 𝑓(𝑥) and 𝑦disagree on 𝑖-th bit. Then 𝐴1, . . . , 𝐴𝑘are independent events. Also, 𝑓(𝑥) = 𝑦if and only if no event 𝐴𝑖occurs. So a trivial version of the local lemma (with empty dependency graph) implies the existence of some 𝑥such that 𝑓(𝑥) = 𝑦.
97 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma On the other hand, it is believed that there exist functions 𝑓that is easy to compute but hard to invert. Such functions are called one-way functions, and they are a fundamental building block in cryptography. For example, let 𝑔be a multiplicative generator of F𝑞, and let 𝑓: F𝑞→F𝑞be given by 𝑓(0) = 0 and 𝑓(𝑥) = 𝑔𝑥and for 𝑥≠0. Then inverting 𝑓is the discrete logarithm problem, which is believed to be computationally difficult.
The computational difficulty of this problem is the basis for the security of important public key cryptography schemes, such as the Diffie–Hellman key exchange.
Moser–Tardos algorithm The algorithm is surprisingly simple.
Algorithm 6.6.2 (Moser–Tardos “fix-it”) input : a set of variables and events in the random variable model output : an assignment of variables avoiding all bad events Initialize by setting all variables to arbitrary values; while there is some violated event do Pick an arbitrary violated event and uniformly resample its variables; (We can make the algorithm more precise by specifying a way to pick an “arbitrary” choice, e.g., the lexicographically first choice.) Theorem 6.6.3 (Moser and Tardos 2010) In Algorithm 6.6.2, letting 𝐴1, . . . , 𝐴𝑛denote the bad events, suppose there are 𝑥1, . . . , 𝑥𝑛∈[0, 1) such that P(𝐴𝑖) ≤𝑥𝑖 Ö 𝑗∈𝑁(𝑖) (1 −𝑥𝑗) for all 𝑖∈[𝑛], then for each 𝑖, E[number of times that 𝐴𝑖is chosen for resampling] ≤ 𝑥𝑖 1 −𝑥𝑖 .
We won’t prove the general theorem here. The proof in Moser and Tardos (2010) is beautifully written and not too long. I highly recommend it reading it. In the next subsection, we will prove the correctness of the algorithm in a special case using a neat idea known as entropy compression.
Remark 6.6.4 (Las Vegas versus Monte Carlo). Here are some important classes of randomized algorithms: 98 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.6 Algorithmic local lemma • Monte Carlo algorithm (MC): a randomized algorithm that terminates with an output, but there is a small probability that the output is incorrect; • Las Vegas algorithm (LV): a randomized algorithm that always returns a correct answer, but may run for a long time (or possibly forever).
The Moser–Tardos algorithm is a LV algorithm whose expected runtime is bounded by Í 𝑖𝑥𝑖/(1 −𝑥𝑖), which is usually at most polynomial in the parameters of the problem.
We are usually interested in randomized algorithms whose running time is small (e.g., at most a polynomial of the input size).
We can convert an efficient LV algorithm into an efficient MC algorithm as follows: suppose the LV algorithm has expected running time 𝑇, and now we run the algorithm but if it takes more than 𝐶𝑇time, then halt and declare a failure. Markov’s inequality then shows that the algorithm fails with probability ≤1/𝐶.
However, it is not always possible to convert an efficient MC algorithm into an efficient LV algorithm. Starting with an MC algorithm, one might hope to repeatedly run it until a correct answer has been found. However, there might not be an efficient way to check the answer.
For example, consider the problem of finding a Ramsey coloring, specifically, 2-edge-coloring of 𝐾𝑛without a monochromatic clique of size ≥100 log2 𝑛. A uniform random coloring works with overwhelming probability, as can be checked by a simple union bound (see Theorem 1.1.2). However, we do not have an efficient way to check whether the random edge-coloring indeed has the desired property. It is a major open problem to find an LV algorithm for finding such an edge-coloring.
Entropy compression argument We now give a simple and elegant proof for a special case of the above algorithm, due to Moser (2009). Actually, the argument in his paper is quite a bit more complicated.
Moser presented a version of the proof below in a conference, and his ideas were popularized by Fortnow and Tao. (Fortnow called Moser’s talk “one of the best STOC talks ever”). Tao introduced the phase entropy compression argument to describe Moser’s influential idea. (We won’t use the language of entropy here, and instead use a more elementary argument involving counting and the pigeonhole principle. We will discuss entropy in Chapter 10.) To keep the argument simple, we work in the setting of 𝑘-CNFs. Recall from Ex-ample 6.1.6 that a 𝒌-CNF formula (conjunctive normal form) consist of a logical conjunction (i.e., and, ∧) of clauses, where each clause is a disjunction (i.e., or, ∨) of exactly 𝑘literals. We shall require that the 𝑘literals of each clause use distinct 99 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma variables (𝑥1, . . . , 𝑥𝑁), and each variable appears either in its positive 𝑥𝑖or negative form 𝑥𝑖. For example, here is a 3-CNF with 4 clauses on 6 variables: (𝑥1 ∨𝑥2 ∨𝑥3) ∧(𝑥1 ∨𝑥2 ∨𝑥4) ∧(𝑥2 ∨𝑥4 ∨𝑥5) ∧(𝑥3 ∨𝑥5 ∨𝑥6).
The problem is to find a satisfying assignment with boolean variables so that the expression output to TRUE.
Algorithm 6.6.5 (Moser “fix-it”) input : a 𝑘-CNF output : a satisfying assignment 1 Initialize by setting all variables to arbitrary values; 2 while there is some violated clause 𝐶do 3 fix (𝐶); 4 Subroutine fix (clause 𝐶) : 5 Resample the variables in 𝐶uniformly at random; 6 while there is some violated clause 𝐷that shares a variable with 𝐶do 7 fix (𝐷); (We can make the algorithm more well defined by specifying a way to pick an “arbitrary” choice, e.g., the lexicographically first choice. Also, in Line 6, we allow taking 𝐷= 𝐶.) Theorem 6.6.6 (Correctness of Moser’s algorithm) Given a 𝑘-CNF where every clause shares variables with at most 2𝑘−3 other clauses, Algorithm 6.6.5 output a satisfying assignment with expected running time at most polynomial in the number of variables and clauses.
Note that the Lovász local lemma guarantees the existence of a solution if each clause shares variables with at most 2𝑘/𝑒−1 clauses (each clause is violated with probability exactly 2−𝑘in a uniform random assignment of variables). So the theorem above is tight up to an unimportant constant factor.
Lemma 6.6.7 (Outer while loop) Each clause of the 𝑘-CNF appears at most once as a violated clause in the outer while loop (Line 2).
Proof. Given an assignment of variables, by calling fix(𝐶) for any clause 𝐶, any clause that was previously satisfied remains satisfied after the completion of the execu-tion of fix(𝐶). Furthermore, 𝐶becomes satisfied after the function call. Thus, once fix(𝐶) is called, 𝐶can never show up again as a violated clause in Line 2.
□ 100 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.6 Algorithmic local lemma Lemma 6.6.8 (The number of recursive calls to fix) Fix a 𝑘-CNF on 𝑛variables where every clause shares variables with at most 2𝑘−3 other clauses. Also fix a clause 𝐶0 and some assignment of variables. Then, in an execution of fix(𝐶0), for any positive integer ℓ, P(there are at least ℓrecursive calls to fix in Line 7) ≤2−ℓ+𝑛+1.
It follows that the expected number of recursive calls to fix is 𝑛+ 𝑂(1). Thus, in the Moser algorithm (Algorithm 6.6.5), the expected total number of calls to fix is 𝑚𝑛+ 𝑂(𝑚), where 𝑛is the number of variables and 𝑚is the number of clauses. This proves the correctness of the algorithm (Theorem 6.6.6).
Proof. Let us formalize the randomness in the algorithm by first initializing a random string of bits. Specifically, let 𝑥∈{0, 1}𝑘ℓbe generated uniformly at random. When-ever the a clause in resampled in Line 5, one replaces the variables in the clause by the next 𝑘bits from 𝑥. Furthermore, if the line Line 7 is called for the ℓ-th time, we halt the algorithm and declare a failure (as we would have run out of random bits to resample had we continued).
At the same time, we keep an execution trace which keeps track of which clauses got called fix, and also when the inner while loop Line 6 ends. Note that the very first call to fix(𝐶0) is not included in the execution trace since it is already given as fixed and so we don’t need to include this information. Here is an example of an execution trace, writing C7 for the 7th clause in the 𝑘-CNF: fix(C7) called fix(C4) called fix(C7) called while loop ended fix(C2) called while loop ended while loop ended ...
For illustration, here is the example of how clause variables could intersect: C2: C4: C7: It is straightforward to deduce which while loop ended corresponds to which fix call by reading the execution trace and keeping track of a first-in-first-out stack.
101 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma Encoding the execution trace as a bit string. We fix at the beginning some canonical order of all clauses (e.g., lexicographic). It would be too expensive to refer to each clause in its absolute position in this order (this is an important point!). Instead, we note that every clause shares variables with at most 2𝑘−3 other clauses, and only these ≤2𝑘−3 could be called in the inner while loop in Line 6. So we can record which one got called using a 𝑘−3 bit string.
• fix(𝐷) called: suppose this was called inside an execution of fix(𝐶), and 𝐷is the 𝑗-th clause among all clauses sharing a variable with 𝐶, then record in the execution trace bit string 0 followed by exactly ℓ−3 bits giving the binary representation of 𝑗(prepended by zeros to get exactly ℓ−3 bits).
• while loop ended: record 1 in the execution trace bit string.
Note that one can recover the execution trace from the above bit string encoding.
Now, suppose the algorithm terminates as a failure due to fix being called the ℓ-th time. Here is the key claim.
Key claim (recovering randomness). At the moment right before the ℓ-th recur-sive call to fix on Line 7, we can completely recover 𝑥from the current variable assignments and the execution trace.
Note that all ℓ𝑘random bits in 𝑥have been used up at this point.
To see the key claim, note that from the execution trace, we can determine which clauses were resampled and in what order. Furthermore, if fix(𝐷) was called on Line 7, then 𝐷must have been violated right before the call, and there is a unique possibility for the violating assignment to 𝐷right before the call (e.g., if 𝐷= 𝑥1 ∨𝑥2 ∨𝑥3, then the only violating assignment is (𝑥1, 𝑥2, 𝑥3) = (0, 0, 1)). We can then rewind history, and put the reassigned values to 𝐷back into the random bit string 𝑥to complete recover 𝑥.
How long can the execution bit string be? It has length ≤ℓ(𝑘−1). Indeed, each of the ≤ℓrecursive calls to fix produces 𝑘−2 bits for the call to fix and 1 bit for ending the while loop. So the total number of possible execution strings is ≤2ℓ(𝑘−1)+1 (the +1 accounts for variable lengths, though it can removed with a more careful analysis).
Thus, the key claim implies that each 𝑥∈{0, 1}ℓ𝑘that leads to a failed execution produces a unique pair (variable assignment, execution bit string). Thus P(≥ℓrecursive calls to fix) 2ℓ𝑘= |{𝑥∈{0, 1}𝑛leading to failure}| ≤2𝑛2ℓ(𝑘−1)+1.
Therefore, the failure probability is ≤2−ℓ+𝑛+1.
□ Remark 6.6.9 (Entropy compression). Tao use the phrase “entropy compression” to describe this argument. The intuition is that the recoverability of the random bit string 102 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.6 Algorithmic local lemma 𝑥means that we are somehow “compressing” a ℓ𝑘-bit random string into a shorter length losslessly, but that would be impossible. Each call to fix uses up 𝑘random bits and converts it to 𝑘−1 bits to the execute trace (plus at most 𝑛bits of extra information, namely the current variables assignment, and this is viewed as a constant amount of information), and this conversion is reversible. So we are “compressing entropy.” The conservation of information tells us that we cannot losslessly compress 𝑘random bits to 𝑘−1 bits for very long.
Remark 6.6.10 (Relationship between the two proofs of the local lemma?). The above proof, along with extensions of these ideas in Moser and Tardos (2010), seems to give a completely different proof of the local lemma than the one we saw at the beginning of the chapter. Is there some way to relate these seemingly completely different proofs? Are they secretly the same proof? We do not know. This is an interesting open-ended research problem.
Exercises 1. Show that it is possible to color the edges of 𝐾𝑛with at most 3√𝑛colors so that there are no monochromatic triangles.
2. Prove that it is possible to color the vertices of every 𝑘-uniform 𝑘-regular hyper-graph using at most 𝑘/log 𝑘colors so that every color appears at most 𝑂(log 𝑘) times on each edge.
3. ★Hitting thin rectangles. Prove that there is a constant 𝐶> 0 so that for every sufficiently small 𝜀> 0, one can choose exactly one point inside each grid square [𝑛, 𝑛+ 1) × [𝑚, 𝑚+ 1) ⊂R2, 𝑚, 𝑛∈Z, so that every rectangle of dimensions 𝜀by (𝐶/𝜀) log(1/𝜀) in the plane (not necessarily axis-aligned) contains at least one chosen point.
4. List coloring. Prove that there is some constant 𝑐> 0 so that given a graph and a set of 𝑘acceptable colors for each vertex such that every color is acceptable for at most 𝑐𝑘neighbors of each vertex, there is always a proper coloring where every vertex is assigned one of its acceptable colors.
5. Prove that, for every 𝜀> 0, there exist ℓ0 and some (𝑎1, 𝑎2, . . . ) ∈{0, 1}N such that for every ℓ> ℓ0 and every 𝑖> 1, the vectors (𝑎𝑖, 𝑎𝑖+1, . . . , 𝑎𝑖+ℓ−1) and (𝑎𝑖+ℓ, 𝑎𝑖+ℓ+1, . . . , 𝑎𝑖+2ℓ−1) differ in at least ( 1 2 −𝜀)ℓcoordinates.
6. Avoiding periodically colored paths.
Prove that for every Δ, there exists 𝑘so that every graph with maximum degree at most Δ has a vertex-coloring using 𝑘 colors so that there is no path of the form 𝑣1𝑣2 . . . 𝑣2ℓ(for any positive integer 103 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6 Lovász Local Lemma ℓ) where 𝑣𝑖has the same color as 𝑣𝑖+ℓfor each 𝑖∈[ℓ]. (Note that vertices on a path must be distinct.) 7. Prove that every graph with maximum degree Δ can be properly edge-colored using 𝑂(Δ) colors so that every cycle contains at least three colors.
(An edge-coloring is proper if it never assigns the same color to two edges sharing a vertex.) 8. ★Prove that for every Δ, there exists 𝑔so that every bipartite graph with maximum degree Δ and girth at least 𝑔can be properly edge-colored using Δ + 1 colors so that every cycle contains at least three colors.
9. ★Prove that for every positive integer 𝑟, there exists 𝐶𝑟so that every graph with maximum degree Δ has a proper vertex coloring using at most 𝐶𝑟Δ1+1/𝑟colors so that every vertex has at most 𝑟neighbors of each color.
10. Vertex-disjoint cycles in digraphs. (Recall that a directed graph is 𝑘-regular if all vertices have in-degree and out-degree both equal to 𝑘. Also, cycles cannot repeat vertices.) a) Prove that every 𝑘-regular directed graph has at least 𝑐𝑘/log 𝑘vertex-disjoint directed cycles, where 𝑐> 0 is some constant.
b) ★Prove that every 𝑘-regular directed graph has at least 𝑐𝑘vertex-disjoint directed cycles, where 𝑐> 0 is some constant.
Hint: split in two and iterate 11.
a) Generalization of Cayley’s formula. Using Prüfer codes, prove the identity 𝑥1𝑥2 · · · 𝑥𝑛(𝑥1 + · · · + 𝑥𝑛)𝑛−2 = ∑︁ 𝑇 𝑥𝑑𝑇(1) 1 𝑥𝑑𝑇(2) 2 · · · 𝑥𝑑𝑇(𝑛) 𝑛 where the sum is over all trees 𝑇on 𝑛vertices labeled by [𝑛] and 𝑑𝑇(𝑖) is the degree of vertex 𝑖in 𝑇.
b) Let 𝐹be a forest with vertex set [𝑛], with components having 𝑓1, . . . , 𝑓𝑠 vertices so that 𝑓1 + · · · + 𝑓𝑠= 𝑛. Prove that the number of trees on the vertex set [𝑛] that contain 𝐹is exactly 𝑛𝑛−2 Î𝑠 𝑖=1( 𝑓𝑖/𝑛𝑓𝑖−1).
c) Independence property for uniform spanning tree of 𝐾𝑛. Show that if 𝐻1 and 𝐻2 are vertex-disjoint subgraphs of 𝐾𝑛, then for a uniformly random spanning tree 𝑇of 𝐾𝑛, the events 𝐻1 ⊆𝑇and 𝐻2 ⊆𝑇are independent.
d) ★Packing rainbow spanning trees. Prove that there is a constant 𝑐> 0 so that for every edge-coloring of 𝐾𝑛where each color appears at most 𝑐𝑛times, there exist at least 𝑐𝑛edge-disjoint spanning trees, where each spanning tree has all its edges colored differently.
104 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 6.6 Algorithmic local lemma (In your submission, you may assume previous parts without proof.) The next two problems use the lopsided local lemma.
12. Packing two copies of a graph. Prove that there is a constant 𝑐> 0 so that if 𝐻 is an 𝑛-vertex 𝑚-edge graph with maximum degree at most 𝑐𝑛2/𝑚, then one can find two edge-disjoint copies of 𝐻in the complete graph 𝐾𝑛.
13. ★Packing Latin transversals. Prove that there is a constant 𝑐> 0 so that every 𝑛× 𝑛matrix where no entry appears more than 𝑐𝑛times contains 𝑐𝑛disjoint Latin transversals.
105 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 7 Correlation Inequalities 7.1 Harris–FKG inequality Recall that 𝐴⊆{0, 1}𝑛is called an increasing event (also: increasing property, up-set) if 𝐴is upwards-closed, meaning that whenever 𝑥is in 𝐴, then everything above 𝑥 in the boolean lattice also lies in 𝐴. In other words, if 𝑥∈𝐴and 𝑥≤𝑦(coordinatewise), then 𝑦∈𝐴.
Similarly, a decreasing event is defined by a downward closed collection of subset of {0, 1}𝑛. A subset 𝐴⊆{0, 1}𝑛is increasing if and only if its complement 𝐴⊆{0, 1}𝑛 is decreasing.
The main theorem of this chapter tells us that increasing events of independent variables are positively correlated .
Theorem 7.1.1 (Harris 1960) If 𝐴and 𝐵are increasing events of independent boolean random variables, then P(𝐴𝐵) ≥P(𝐴)P(𝐵).
Equivalently, we can write P (𝐴| 𝐵) ≥P(𝐴).
Remark 7.1.2 (Independence assumption). It is important that the boolean random variables are independent, also they do not have to be identically distributed.
There are other important settings where the independence assumption can be relaxed.
This is important for certain statistical physics models, where much of this theory originally arose. Indeed, the above inequality is often called the FKG inequality, attributed to Fortuin, Kasteleyn, Ginibre (1971), who proved a more general result in the setting of distributive lattices, which we will not discuss here (see Alon–Spencer).
Remark 7.1.3 (Percolation). Many of such inequalities were initially introduced for the study of percolations. A classic setting of this problem takes place in infinite grid with vertices Z2 with edges connecting adjacent vertices at distance 1. Suppose 107 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 7 Correlation Inequalities we keep each edge of this infinite grid with probability 𝑝independently, what is the probability that the origin is part of an infinite component (in which case we say that there is “percolation”)? This is supposed to an idealized mathematical model of how a fluid permeates through a medium. Harris showed that with probability 1, percolation does not occur for 𝑝≤1/2. A later breakthrough of Kesten (1980) shows that percolation occurs with probability 1 for all 𝑝> 1/2. Thus the “bond percolation threshold” for Z2 is exactly 1/2. Such exact results are extremely rare.
Example 7.1.4. Here is a quick application of Harris’ inequality to a random graph 𝐺(𝑛, 𝑝): P(planar | connected) ≤P(planar).
Indeed, being planar is a decreasing property, whereas being connected is an increasing property.
We state and prove a more general result, which says that independent random variables possess positive association.
Let each Ω𝑖be a linearly ordered set (i.e., {0, 1}, R) and 𝑥𝑖∈Ω𝑖with respect to some probability distribution independent for each 𝑖. We say that a function 𝑓(𝑥1, . . . , 𝑥𝑛) is monotone increasing if 𝑓(𝑥) ≤𝑓(𝑦) whenever 𝑥≤𝑦coordinatewise.
Theorem 7.1.5 (Harris) If 𝑓and 𝑔are monotone increasing functions of independent random variables, then E[ 𝑓𝑔] ≥(E 𝑓)(E𝑔).
This version of Harris inequality implies the earlier version by setting 𝑓= 1𝐴and 𝑔= 1𝐵.
Proof. We use induction on 𝑛.
For 𝑛= 1, for independent 𝑥, 𝑦∈Ω1, we have 0 ≤E[( 𝑓(𝑥) −𝑓(𝑦))(𝑔(𝑥) −𝑔(𝑦))] = 2E[ 𝑓𝑔] −2(E 𝑓)(E𝑔).
So E[ 𝑓𝑔] ≥(E 𝑓)(E𝑔).
(The one-variable case is sometimes called Chebyshev’s inequality. It can also be deduced using the rearrangement inequality).
Nowassume 𝑛≥2. Let ℎ= 𝑓𝑔: Ω1×· · ·×Ω𝑛→R. Define marginals 𝑓1, 𝑔1, ℎ1 : Ω1 → 108 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 7.1 Harris–FKG inequality R by 𝑓1(𝑦1) = E[ 𝑓|𝑥1 = 𝑦1] = E(𝑥2,...,𝑥𝑛)∈Ω2×···×Ω𝑛[ 𝑓(𝑦1, 𝑥2, . . . , 𝑥𝑛)], 𝑔1(𝑦1) = E[𝑔|𝑥1 = 𝑦1] = E(𝑥2,...,𝑥𝑛)∈Ω2×···×Ω𝑛[𝑔(𝑦1, 𝑥2, . . . , 𝑥𝑛)], ℎ1(𝑦1) = E[ℎ|𝑥1 = 𝑦1] = E(𝑥2,...,𝑥𝑛)∈Ω2×···×Ω𝑛[ℎ(𝑦1, 𝑥2, . . . , 𝑥𝑛)].
Note that 𝑓1 and 𝑔1 are 1-variable monotone increasing functions on Ω1.
For every fixed 𝑦1 ∈Ω1, the function (𝑥2, . . . , 𝑥𝑛) ↦→𝑓(𝑦1, 𝑥2, . . . , 𝑥𝑛) is monotone increasing, and likewise with 𝑔. So applying the induction hypothesis for 𝑛−1, we have ℎ1(𝑦1) ≥𝑓1(𝑦1)𝑔1(𝑦1).
(7.1) Thus E[ 𝑓𝑔] = E[ℎ] = E[ℎ1] ≥E[ 𝑓1𝑔1] [by (7.1)] ≥(E 𝑓1)(E𝑔1) [by the 𝑛= 1 case] = (E 𝑓)(E𝑔).
□ Corollary 7.1.6 (Decreasing events and multiple events) Let 𝐴and 𝐵be events on independent random variables.
(a) If 𝐴and 𝐵are decreasing, then P(𝐴∧𝐵) ≥P(𝐴)P(𝐵).
(b) If 𝐴is increasing and 𝐵is decreasing, then P(𝐴∧𝐵) ≤P(𝐴)P(𝐵).
If 𝐴1, . . . , 𝐴𝑘are all increasing (or all decreasing) events on independent random variables, then P(𝐴1 · · · 𝐴𝑘) ≥P(𝐴1) · · · P(𝐴𝑘).
Proof. For the second inequality, note that the complement 𝐵is increasing, so P(𝐴𝐵) = P(𝐴) −P(𝐴𝐵) Harris ≤ P(𝐴) −P(𝐴)P(𝐵) = P(𝐴)P(𝐵).
The proof of the first inequality is similar. For the last inequality we apply the Harris inequality repeatedly.
□ 109 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 7 Correlation Inequalities 7.2 Applications to random graphs Triangle-free probability Question 7.2.1 What’s the probability that 𝐺(𝑛, 𝑝) is triangle-free?
Harris inequality will allow us to prove a lower bound. In the next chapter, we will use Janson inequalities to derive upper bounds.
Theorem 7.2.2 P(𝐺(𝑛, 𝑝) is triangle-free) ≥(1 −𝑝3)(𝑛 3) Proof. For each triple of distinct vertices 𝑖, 𝑗, 𝑘∈[𝑛], the event that 𝑖𝑗𝑘does not form a triangle is a decreasing event (here the ground set is the set of edges of the complete graph on 𝑛). So by Harris’ inequality, P(𝐺(𝑛, 𝑝) is triangle-free) = P © « Û 𝑖< 𝑗<𝑘 {𝑖𝑗𝑘not a triangle}ª ® ¬ ≥ Ö 𝑖< 𝑗<𝑘 P(𝑖𝑗𝑘not a triangle) = (1 −𝑝3)(𝑛 3).
□ Remark 7.2.3. How good is this bound? For 𝑝≤0.99, we have 1 −𝑝3 = 𝑒−Θ(𝑝3), so the above bound gives P(𝐺(𝑛, 𝑝) is triangle-free) ≥𝑒−Θ(𝑛3𝑝3).
Here is another lower bound P(𝐺(𝑛, 𝑝) is triangle-free) ≥P(𝐺(𝑛, 𝑝) is empty) = (1 −𝑝)(𝑛 2) = 𝑒−Θ(𝑛2𝑝).
The bound from Harris is better when 𝑝≪𝑛−1/2. Putting them together, we obtain P(𝐺(𝑛, 𝑝) is triangle-free) ≳ ( 𝑒−Θ(𝑛3𝑝3) if 𝑝≲𝑛−1/2 𝑒−Θ(𝑛2𝑝) if 𝑛−1/2 ≲𝑝≤0.99 (note that the asymptotics agree at the boundary 𝑝≍𝑛−1/2). In the next chapter, we will prove matching upper bounds using Janson inequalities.
110 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 7.2 Applications to random graphs Maximum degree Question 7.2.4 What’s the probability that the maximum degree of 𝐺(𝑛, 1/2) is at most 𝑛/2?
For each vertex 𝑣, deg(𝑣) ≤𝑛/2 is a decreasing event with probability just slightly over 1/2. So by Harris inequality, the probability that every 𝑣has deg(𝑣) ≤𝑛/2 is at least ≥2−𝑛.
It turns out that the appearance of high degree vertices is much more correlated than the independent case. The truth is exponentially more than the above bound.
Theorem 7.2.5 (Riordan and Selby 2000) P(maxdeg 𝐺(𝑛, 1/2) ≤𝑛/2) = (0.6102 · · · + 𝑜(1))𝑛 Instead of giving a proof, we consider an easier continuous model of the problem that motivates the numerical answer. Building on this intuition, Riordan and Selby (2000) proved the result in the random graph setting, although this is beyond the scope of this class.
In a random graphs, we assign independent Bernoulli random variables on edges of a complete graph. Instead, let us assign independent standard normal random variables to each edge of the complete graph.
Proposition 7.2.6 (Max degree with normal random edge labels) Assign an independent standard normal random variable 𝑍𝑢𝑣to each edge of 𝐾𝑛. Let 𝑊𝑣= Í 𝑢≠𝑣𝑍𝑢𝑣be the sum of the labels of the edges incident to a vertex 𝑣. Then P(𝑊𝑣≤0 ∀𝑣) = (0.6102 · · · + 𝑜(1))𝑛 The event 𝑊𝑣≤0 is supposed to model the event that the degree at vertex 𝑣is less than 𝑛/2. Of course, other than intuition, there is no justification here that these two models should behave similarly We have P(𝑊𝑣≤0) = 1/2.
Since each {𝑊𝑣≤0} is a decreasing event of the independent edge labels, Harris’ inequality tells us that P(𝑊𝑣≤0 ∀𝑣) ≥2−𝑛.
The truth turns out to be significantly greater.
Proof sketch of Proposition 7.2.6. The tuple (𝑊𝑣)𝑣∈[𝑛] has a joint normal distribution, with each coordinate variance 𝑛−1 and pairwise covariance 1. So (𝑊𝑣)𝑣∈[𝑛] has the 111 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 7 Correlation Inequalities same distribution as √ 𝑛−2(𝑍′ 1, 𝑍′ 2, . . . , 𝑍′ 𝑛) + 𝑍′ 0(1, 1, . . . , 1) where 𝑍′ 0, . . . , 𝑍′ 𝑛are iid standard normals.
Let Φ be the pdf and cdf of the standard normal 𝑁(0, 1).
Thus P(𝑊𝑣≤0 ∀𝑣) = P 𝑍′ 𝑖≤− 𝑍′ 0 √ 𝑛−2 ∀𝑖∈[𝑛] = 1 √ 2𝜋 ∫∞ −∞ 𝑒−𝑧2/2Φ −𝑧 √ 𝑛−2 𝑛 𝑑𝑧 where the final step is obtained by conditioning on 𝑍′ 0. Substituting 𝑧= 𝑦√𝑛, the above quantity equals to = √︂𝑛 2𝜋 ∫∞ −∞ 𝑒𝑛𝑓(𝑦) 𝑑𝑦 where 𝑓(𝑦) = −𝑦2 2 + log Φ 𝑦 √︂ 𝑛 𝑛−2 .
We can estimate the above integral for large 𝑛using the Laplace method (which can be justified rigorously by considering Taylor expansion around the maximum of 𝑓). We have 𝑓(𝑦) ≈𝑔(𝑦) := −𝑦2 2 + log Φ (𝑦) and we can deduce that lim 𝑛→∞ 1 𝑛log P(max 𝑣∈[𝑛] 𝑊𝑣≤0) = lim 𝑛→∞ 1 𝑛log ∫ 𝑒𝑛𝑓(𝑦) 𝑑𝑦= max 𝑔= log 0.6102 · · · .
□ Exercises 1. Let 𝐺= (𝑉, 𝐸) be a graph. Color every edge with red or blue independently and uniformly at random. Let 𝐸0 be the set of red edges and 𝐸1 the set of blue edges. Let 𝐺𝑖= (𝑉, 𝐸𝑖) for each 𝑖= 0, 1. Prove that P(𝐺0 and 𝐺1 are both connected) ≤P(𝐺0 is connected)2.
2. A set family F is intersecting if 𝐴∩𝐵≠∅for all 𝐴, 𝐵∈F . Let F1, . . . , F𝑘 each be a collection of subsets of [𝑛] and suppose that each F𝑖is intersecting.
Prove that Ð𝑘 𝑖=1 F𝑖 ≤2𝑛−2𝑛−𝑘.
3. Percolation. Let 𝐺𝑚,𝑛be the grid graph on vertex set [𝑚] × [𝑛] (𝑚vertices wide and 𝑛vertices tall). A horizontal crossing is a path that connects some left-most vertex to some right-most vertex. See below for an example of a horizontal crossing in 𝐺7,5.
112 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 7.2 Applications to random graphs Let 𝐻𝑚,𝑛denote the random subgraph of 𝐺𝑚,𝑛obtained by keeping every edge with probability 1/2 independently.
Let RSW(𝑘) denote the following statement: there exists a constant 𝑐𝑘> 0 such that for all positive integers 𝑛, P(𝐻𝑘𝑛,𝑛has a horizontal crossing) ≥𝑐𝑘.
a) Prove RSW(1).
b) Prove that RSW(2) implies RSW(100).
c) ★★(Very challenging) Prove RSW(2).
4. Let 𝐴and 𝐵be two independent increasing events of independent random variables. Prove that there are two disjoint subsets 𝑆and 𝑇of these random variables so that 𝐴depends only on 𝑆and 𝐵depends only on 𝑇.
5. Let 𝑈1 and 𝑈2 be increasing events and 𝐷a decreasing event of independent Boolean random variables. Suppose 𝑈1 and 𝑈2 are independent. Prove that P(𝑈1|𝑈2 ∩𝐷) ≤P(𝑈1|𝑈2).
6. Coupon collector. Let 𝑠1, . . . , 𝑠𝑚be independent random elements in [𝑛] (not necessarily uniform or identically distributed; chosen with replacement) and 𝑆= {𝑠1, . . . , 𝑠𝑚}. Let 𝐼and 𝐽be disjoint subsets of [𝑛]. Prove that P(𝐼∪𝐽⊆ 𝑆) ≤P(𝐼⊆𝑆)P(𝐽⊆𝑆).
7. ★Prove that there exist 𝑐< 1 and 𝜀> 0 such that if 𝐴1, . . . , 𝐴𝑘are increasing events of independent Boolean random variables with P(𝐴𝑖) ≤𝜀for all 𝑖, then, letting 𝑋denote the number of events 𝐴𝑖that occur, one has P(𝑋= 1) ≤𝑐.
(Give your smallest 𝑐. It is conjectured that any 𝑐> 1/𝑒works.) 8. ★Disjoint containment. Let S and T each be a collection of subsets of [𝑛].
Let 𝑅⊆[𝑛] be a random subset where each element is included independently (not necessarily with the same probability). Let 𝐴be the event that 𝑆⊆𝑅for some 𝑆∈S. Let 𝐵be the event that 𝑇⊆𝑅for some 𝑇∈T. Let 𝐶denote the event there exist disjoint 𝑆,𝑇⊆𝑅with 𝑆∈S and 𝑇∈T. Prove that P(𝐶) ≤P(𝐴)P(𝐵).
113 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8 Janson Inequalities We present a collection of inequalities, known collectively as Janson inequalities (Janson 1990). These tools allow us to estimate lower tail large deviation probabilities.
A typical application of Janson’s inequality allows us to upper bound the probability that a random graph 𝐺(𝑛, 𝑝) does not contain any copy of some subgraph. Compared to the second moment method from Chapter 4, Janson inequalities (which is applicable in more limited setups) gives much better bounds, usually with exponential decays.
8.1 Probability of non-existence The following setup should be a reminiscent of both the second moment method as well as Lovász local lemma (the random variable model).
Setup 8.1.1 (for Janson’s inequality: counting containments) Let 𝑅be a random subset of [𝑁] with each element included independently (possibly with different probabilities).
Let 𝑆1, . . . , 𝑆𝑘⊆[𝑁]. Let 𝐴𝑖be the event that 𝑆𝑖⊆𝑅. Let 𝑋= ∑︁ 𝑖 1𝐴𝑖 be the number of sets 𝑆𝑖contained in the same set 𝑅. Let 𝜇= E[𝑋] = ∑︁ 𝑖 P(𝐴𝑖).
Write 𝑖∼𝑗if 𝑖≠𝑗and 𝑆𝑖∩𝑆𝑗≠∅. Let (as in the second moment method) Δ = ∑︁ (𝑖,𝑗):𝑖∼𝑗 P(𝐴𝑖𝐴𝑗) = ∑︁ (𝑖,𝑗):𝑖∼𝑗 P(𝑆𝑖∪𝑆𝑗⊆𝑅) (note that (𝑖, 𝑗) and ( 𝑗, 𝑖) is each counted once).
The following inequality appeared in Janson, Łuczak, and Ruciński (1990).
115 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8 Janson Inequalities Theorem 8.1.2 (Janson inequality I) Assuming Setup 8.1.1, P(𝑋= 0) ≤𝑒−𝜇+Δ/2.
This inequality is most useful when Δ = 𝑜(𝜇).
Remark 8.1.3. When P(𝐴𝑖) = 𝑜(1) (which is the case in a typical application), Harris’ inequality gives us P(𝑋= 0) = P 𝐴1 · · · 𝐴𝑘 ≥ 𝑘 Ö 𝑖=1 P 𝐴𝑖 = 𝑘 Ö 𝑖=1 (1 −P(𝐴𝑖)) = exp −(1 + 𝑜(1)) 𝑘 ∑︁ 𝑖=1 P(𝐴𝑖) !
= 𝑒−(1+𝑜(1))𝜇.
In the setting where Δ = 𝑜(𝜇), two bounds match to give P(𝑋= 0) = 𝑒−(1+𝑜(1)𝜇.
Proof. Let 𝑟𝑖= P(𝐴𝑖|𝐴1 · · · 𝐴𝑖−1).
We have P(𝑋= 0) = P(𝐴1 · · · 𝐴𝑘) = P(𝐴1)P(𝐴2|𝐴1) · · · P(𝐴𝑘|𝐴1 · · · 𝐴𝑘−1) = (1 −𝑟1) · · · (1 −𝑟𝑘) ≤𝑒−𝑟1−···−𝑟𝑘 It suffices now to prove that: Claim. For each 𝑖∈[𝑘] 𝑟𝑖≥P(𝐴𝑖) − ∑︁ 𝑗<𝑖:𝑗∼𝑖 P(𝐴𝑖𝐴𝑗).
Summing the claim over 𝑖∈[𝑘] would then yield 𝑘 ∑︁ 𝑖=1 𝑟𝑖≥ ∑︁ 𝑖 P(𝐴𝑖) −1 2 ∑︁ 𝑖 ∑︁ 𝑗∼𝑖 P(𝐴𝑖𝐴𝑗) = 𝜇−Δ 2 and thus P(𝑋= 0) ≤exp − ∑︁ 𝑖 𝑟𝑖 !
≤exp −𝜇+ Δ 2 116 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8.1 Probability of non-existence Proof of claim. Recall that 𝑖is given and fixed. Let 𝐷0 = Û 𝑗<𝑖:𝑗≁𝑖 𝐴𝑗 and 𝐷1 = Û 𝑗<𝑖:𝑗∼𝑖 𝐴𝑗 Then 𝑟𝑖= P(𝐴𝑖|𝐴1 · · · 𝐴𝑖−1) = P(𝐴𝑖|𝐷0𝐷1) = P(𝐴𝑖𝐷0𝐷1) P(𝐷0𝐷1) ≥P(𝐴𝑖𝐷0𝐷1) P(𝐷0) = P(𝐴𝑖𝐷1|𝐷0) = P(𝐴𝑖|𝐷0) −P(𝐴𝑖𝐷1|𝐷0) = P(𝐴𝑖) −P(𝐴𝑖𝐷1|𝐷0) [by independence] Since 𝐴𝑖and 𝐷1 are both increasing events, and 𝐷0 is a decreasing event, by Harris’ inequality (Corollary 7.1.6), P(𝐴𝑖𝐷1|𝐷0) ≤P(𝐴𝑖𝐷1) = P 𝐴𝑖∧ Ü 𝑗<𝑖:𝑗∼𝑖 𝐴𝑗 !
≤ ∑︁ 𝑗<𝑖:𝑗∼𝑖 P(𝐴𝑖𝐴𝑗) This concludes the proof of the claim, and thus the proof of the theorem.
□ Remark 8.1.4 (History). Janson’s original proof was via analytic interpolation. The above proof is based on Boppana and Spencer (1989) with a modification by Warnke (personal communication). It has some similarities to the proof of Lovász local lemma from Section 6.1.
The above proof incorporates ideas from Riordan and Warnke (2015), who extended Janson’s inequality from principal up-set to general up-sets.
Indeed, the above proof only requires that the events 𝐴𝑖are increasing, whereas earlier proofs of the result (e.g., the proof in Alon–Spencer) requires the full assumption of Setup 8.1.1, namely that each 𝐴𝑖is an event of the form 𝑆𝑖⊆𝑅𝑖(i.e., a principal up-set).
Question 8.1.5 What is the probability that 𝐺(𝑛, 𝑝) is triangle-free?
In Setup 8.1.1, let [𝑁] with 𝑁= 𝑛 2 be the set of edges of 𝐾𝑛, and let 𝑆1, . . . , 𝑆(𝑛 3) be 3-element sets where each 𝑆𝑖is the edge-set of a triangle. As in the second moment calculation in Section 4.2, we have 𝜇= 𝑛 3 𝑝3 ≍𝑛3𝑝3 and Δ ≍𝑛4𝑝5.
(where Δ is obtained by considering all appearances of a pair of triangles glued along an edge).
117 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8 Janson Inequalities If 𝑝≪𝑛−1/2, then Δ = 𝑜(𝜇), in which case Janson inequality I (Theorem 8.1.2 and Remark 8.1.3) gives the following.
Theorem 8.1.6 If 𝑝= 𝑜(𝑛−1/2) , then P(𝐺(𝑛, 𝑝) is triangle-free) = 𝑒−(1+𝑜(1))𝜇= 𝑒−(1+𝑜(1))𝑛3𝑝3/6.
Corollary 8.1.7 For a constant 𝑐> 0, lim 𝑛→∞P(𝐺(𝑛, 𝑐/𝑛) is triangle-free) = 𝑒−𝑐3/6.
In fact, the number of triangles in 𝐺(𝑛, 𝑐/𝑛) converges to a Poisson distribution with mean 𝑐3/6.
On the other hand, when 𝑝≫1/𝑛, the number of triangles is asymptotically normal.
What about if 𝑝≫𝑛−1/2, so that Δ ≫𝜇. Janson inequality I does not tell us anything nontrivial. Do we still expect the triangle-free probability to be 𝑒−(1+𝑜(1))𝜇, or even ≤𝑒−𝑐𝜇?
As noted earlier in Remark 7.2.3, another way to obtain a lower bound on the probability triangle-freeness is to consider the probability the 𝐺(𝑛, 𝑝) is empty (or contained in some fixed complete bipartite graph), in which case we obtain P(𝐺(𝑛, 𝑝) is triangle-free) ≥(1 −𝑝)Θ(𝑛2) = 𝑒−Θ(𝑛2𝑝) (the second step assumes that 𝑝is bounded away from 1. If 𝑝≫𝑛−1/2, so the above lower bound better than the previous one: 𝑒−Θ(𝑛2𝑝) ≫𝑒−(1+𝑜(1))𝜇.
Nevertheless, we’ll still use Janson to bootstrap an upper bound on the triangle-free probability. More generally, the next theorem works in the complement region of the Janson inequality I, where now Δ ≥𝜇.
Theorem 8.1.8 (Janson inequality II) Assuming Setup 8.1.1, if Δ ≥𝜇, then P(𝑋= 0) ≤𝑒−𝜇2/(2Δ).
The proof idea is to applying the first Janson inequality on a randomly sampled subset of events. This sampling technique might remind you of some earlier proofs, e.g., the proof of the crossing number inequality (Theorem 2.6.2), where we first proved a 118 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8.1 Probability of non-existence “cheap bound” that worked in a more limited range, and then used sampling to obtain a better bound.
Proof. For each 𝑇⊆[𝑘], let 𝑋𝑇:= Í 𝑖∈𝑇1𝐴𝑖denote the number of occurring events in 𝑇. We have P(𝑋= 0) ≤P(𝑋𝑇= 0) ≤𝑒−𝜇𝑇+Δ𝑇/2 where 𝜇𝑇= ∑︁ 𝑖∈𝑇 P(𝐴𝑖) and Δ𝑇= ∑︁ (𝑖,𝑗)∈𝑇2:𝑖∼𝑗 P(𝐴𝑖𝐴𝑗) Choose 𝑇⊆[𝑘] randomly by including every element with probability 𝑞∈[0, 1] independently. We have E𝜇𝑇= 𝑞𝜇 and EΔ𝑇= 𝑞2Δ and so E(−𝜇𝑇+ Δ𝑇/2) = −𝑞𝜇+ 𝑞2Δ/2.
By linearity of expectations, thus there is some choice of 𝑇⊆[𝑘] so that −𝜇𝑇+ Δ𝑇/2 ≤−𝑞𝜇+ 𝑞2Δ/2 so that P(𝑋= 0) ≤𝑒−𝑞𝜇+𝑞2Δ/2 for every 𝑞∈[0, 1]. Since Δ ≥𝜇, we can set 𝑞= 𝜇/Δ ∈[0, 1] to get the result.
□ To summarize, the first two Janson inequalities tell us that P(𝑋= 0) ≤ ( 𝑒−𝜇+Δ/2 if Δ < 𝜇 𝑒−𝜇2/(2Δ) if Δ ≥𝜇.
Remark 8.1.9. If 𝜇→∞and Δ ≪𝜇2, then Janson inequality II implies P(𝑋= 0) = 𝑜(1), which we knew from second moment method. However Janson’s inequality gives an exponentially decaying tail bound, compared to only a polynomially decaying tail via the second moment method. The exponential tail will be important in an application below to determining the chromatic number of 𝐺(𝑛, 1/2).
Let us revisit the example of estimating the probability that 𝐺(𝑛, 𝑝) is triangle-free, 119 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8 Janson Inequalities now in the regime 𝑝≫𝑛−1/2. We have 𝑛3𝑝3 ≍𝜇≪Δ ≍𝑛4𝑝5.
So so for large enough 𝑛, Janson inequality II tells us P(𝐺(𝑛, 𝑝) is triangle-free) ≤𝑒−𝜇2/(2Δ) = 𝑒−Θ(𝑛2𝑝) Since P(𝐺(𝑛, 𝑝) is triangle-free) ≥P(𝐺(𝑛, 𝑝) is empty) ≥(1 −𝑝)(𝑛 2) = 𝑒−Θ(𝑛2𝑝) where the final step assumes that 𝑝is bounded away from 1, we conclude that P(𝐺(𝑛, 𝑝) is triangle-free) = 𝑒−Θ(𝑛2𝑝) We summarize the results below (strictly speaking we have not yet checked the case 𝑝≍𝑛−1/2, which we can verify by applying Janson inequalities; note that the two regimes below match at the boundary).
Theorem 8.1.10 Suppose 𝑝= 𝑝𝑛≤0.99. Then P(𝐺(𝑛, 𝑝) is triangle-free) = ( exp −Θ(𝑛2𝑝) if 𝑝≳𝑛−1/2 exp −Θ(𝑛3𝑝3) if 𝑝≲𝑛−1/2 Remark 8.1.11. Sharper results are known. Here are some highlights.
1. The number of triangle-free graphs on 𝑛vertices is 2(1+𝑜(1))𝑛2/4. In fact, an even stronger statement is true: almost all (i.e., 1−𝑜(1) fraction) 𝑛-vertex triangle-free graphs are bipartite (Erdős, Kleitman, and Rothschild 1976).
2. If 𝑚≥𝐶𝑛3/2√︁ log 𝑛for any constant 𝐶> √ 3/4 (and this is best possible), then almost all all 𝑛-vertex 𝑚-edge triangle-free graphs are bipartite (Osthus, Prömel, and Taraz 2003). This result has been extended to 𝐾𝑟-free graphs for every fixed 𝑟(Balogh, Morris, Samotij, and Warnke 2016).
3. For 𝑛−1/2 ≪𝑝≪1, (Łuczak 2000) −log P(𝐺(𝑛, 𝑝) is triangle-free) ∼−log P(𝐺(𝑛, 𝑝) is bipartite) ∼𝑛2𝑝/4.
This result was generalized to general 𝐻-free graphs using the powerful recent method of hypergraph containers (Balogh, Morris, and Samotij 2015).
120 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8.2 Lower tails 8.2 Lower tails Previously we looked at the probability of non-existence.
Now we would like to estimate lower tail probabilities. Here is a model problem.
Question 8.2.1 Fix a constant 0 < 𝛿≤1. Let 𝑋be the number of triangles of 𝐺(𝑛, 𝑝). Estimate P(𝑋≤(1 −𝛿)E𝑋).
We will bootstrap Janson inequality I, P(𝑋= 0) ≤exp(−𝜇+ Δ/2), to an upper bound on lower tail probabilities.
Theorem 8.2.2 (Janson inequality III) Assume Setup 8.1.1. For any 0 ≤𝑡≤𝜇, P(𝑋≤𝜇−𝑡) ≤exp −𝑡2 2(𝜇+ Δ) Note that setting 𝑡= 𝜇we basically recover the first two Janson inequalities (up to an unimportant constant factor in the exponent): P(𝑋= 0) ≤exp −𝜇2 2(𝜇+ Δ) .
(8.1) (Note that this form of the inequality conveniently captures Janson inequalities I & II.) Proof. (by Lutz Warnke1) We start the proof similarly to the proof of the Chernoff bound, by applying Markov’s inequality on the moment generating function. To that end, let 𝜆≥0 to be optimized later. Let 𝑞= 1 −𝑒−𝜆.
By Markov’s inequality, P(𝑋≤𝜇−𝑡) = P 𝑒−𝜆𝑋≥𝑒−𝜆(𝜇−𝑡) ≤𝑒𝜆(𝜇−𝑡) E 𝑒−𝜆𝑋 ≤𝑒𝜆(𝜇−𝑡) E[(1 −𝑞)𝑋].
1Personal communication 121 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8 Janson Inequalities For each 𝑖∈[𝑘], let 𝑊𝑖∼Bernoulli(𝑞) independently. Consider the random variable 𝑌= 𝑘 ∑︁ 𝑖=1 1𝐴𝑖𝑊𝑖.
Conditioned on the value of 𝑋, the probability that𝑌= 0 is (1−𝑞)𝑋(i.e., the probability that 𝑊𝑖= 0 for each of the 𝑋events 𝐴𝑖that occurred). Taking expectation over 𝑋, we have P(𝑌= 0) = E[P(𝑌= 0|𝑋)] = E[(1 −𝑞)𝑋].
Note that 𝑌fits within Setup 8.1.1 by introducing 𝑘new elements to the ground set [𝑁], where each new element is included according to 𝑊𝑖, and enlarging each 𝑆𝑖to include this new element. The relevant parameters of 𝑌are 𝜇𝑌:= E𝑌= 𝑞𝜇 and Δ𝑌:= ∑︁ (𝑖,𝑗):𝑖∼𝑗 E[1𝐴𝑖𝑊𝑖1𝐴𝑗𝑊𝑗] = 𝑞2Δ.
Then Janson inequality I applied to 𝑌gives P(𝑌= 0) ≤𝑒−𝜇𝑌+Δ𝑌/2 = 𝑒−𝑞𝜇+𝑞2Δ/2.
Therefore, E[(1 −𝑞)𝑋] = P(𝑌= 0) ≤𝑒−𝑞𝜇+𝑞2Δ/2.
Continuing the moment calculation at the beginning of the proof, and using that 𝜆−𝜆2 2 ≤𝑞≤𝜆, we have P(𝑋≤−𝜇+ 𝑡) ≤𝑒𝜆(𝜇−𝑡) E[(1 −𝑞)𝑋] ≤exp 𝜆(𝜇−𝑡) −𝑞𝜇+ 𝑞2Δ/2 ≤exp 𝜆(𝜇−𝑡) − 𝜆−𝜆2 2 𝜇+ 𝜆2Δ 2 = exp −𝜆𝑡+ 𝜆2 2 (𝜇+ Δ) We optimize by setting 𝜆= 𝑡/(𝜇+ Δ) to obtain ≤exp −𝑡2 2(𝜇+Δ) .
□ Example 8.2.3 (Lower tails for triangle counts). Let 𝑋be the number of triangles in 122 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8.2 Lower tails 𝐺(𝑛, 𝑝). We have 𝜇≍𝑛3𝑝3 and Δ ≍𝑛4𝑝5. Fix a constant 𝛿∈(0, 1]. Let 𝑡= 𝛿E𝑋.
We have P(𝑋≤(1 −𝛿)E𝑋) ≤exp −Θ −𝛿2𝑛6𝑝6 𝑛3𝑝3 + 𝑛4𝑝5 = ( exp −Θ𝛿(𝑛2𝑝) if 𝑝≳𝑛−1/2, exp −Θ𝛿(𝑛3𝑝3) if 𝑝≲𝑛−1/2.
The bounds are tight up to a constant in the exponent, since P(𝑋≤(1 −𝛿)E𝑋) ≥P(𝑋= 0) = ( exp −Θ(𝑛2𝑝) if 𝑝≳𝑛−1/2, exp −Θ(𝑛3𝑝3) if 𝑝≲𝑛−1/2.
Example 8.2.4 (No corresponding Janson inequality for upper tails). Continuing with 𝑋being the number of triangles of 𝐺(𝑛, 𝑝), from on the above lower tail results, we might expect P(𝑋≥(1 + 𝛿)E𝑋) ≤exp(−Θ𝛿(𝑛2𝑝)), but actually this is false!
By planting a clique of size Θ(𝑛𝑝), we can force 𝑋≥(1 + 𝛿)E𝑋. Thus P(𝑋≥(1 + 𝛿)E𝑋) ≥𝑝Θ𝛿(𝑛2𝑝2) which is much bigger than exp −Θ(𝑛2𝑝). The above is actually the truth (Kahn– DeMarco 2012 and Chatterjee 2012): P(𝑋≥(1 + 𝛿)E𝑋) = 𝑝Θ𝛿(𝑛2𝑝2) if 𝑝≳log 𝑛 𝑛 , but the proof is much more intricate. Recent results allow us to understand the exact constant in the exponent though new developments in large deviation theory. The current state of knowledge is summarized below.
Theorem 8.2.5 (Harel, Mousset, Samotij 2022) Let 𝑋be the number of triangles in 𝐺(𝑛, 𝑝) with 𝑝= 𝑝𝑛satisfying 𝑛−1/2 ≪𝑝≪1, −log P(𝑋≥(1 + 𝛿)E𝑋) ∼min 𝛿 3, 𝛿2/3 2 𝑛2𝑝2 log(1/𝑝), and for 𝑛−1 log 𝑛≪𝑝≪𝑛−1/2, −log P(𝑋≥(1 + 𝛿)E𝑋) ∼𝛿2/3 2 𝑛2𝑝2 log(1/𝑝).
Remark 8.2.6. The leading constants were determined by Lubetzky and Zhao (2017) by solving an associated variational problem. Earlier results, starting with Chatter-jee and Varadhan (2011) and Chatterjee and Dembo (2016) prove large deviation 123 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8 Janson Inequalities frameworks that gave the above theorem for sufficiently slowly decaying 𝑝≥𝑛−𝑐.
For the corresponding problem for lower tails, see Kozma and Samotij (2023) for an approach using relative entropy that reduces the rate problem to a variational problem.
The exact leading constant is known only for sufficiently small 𝛿> 0, where the answer is given by “replica symmetry”, meaning that the exponential rate is given by a uniform decrement in edge densities for the random graph. In contrast, for 𝛿close to 1, we expect (though cannot prove) that the typical structure of a conditioned random graph is close to a two-block model (Zhao 2017).
8.3 Chromatic number of a random graph Question 8.3.1 What is the chromatic number of 𝐺(𝑛, 1/2)?
In Section 4.4, we used the second moment method to find the clique number 𝜔of 𝐺(𝑛, 1/2). We saw that, with probability 1 −𝑜(1), the clique number is concentrated on two values, and in particular, 𝜔(𝐺(𝑛, 1/2)) ∼2 log2 𝑛 whp.
The independence number 𝜶(𝑮) is the size of the largest independent set in 𝐺. The independence number 𝛼(𝐺) is the equal to the clique number the complement of 𝐺. Since 𝐺(𝑛, 1/2) and its graph complement have the same distribution, we have 𝛼(𝐺(𝑛, 1/2)) ∼2 log2 𝑛whp as well.
Using the following lower bound on the chromatic number 𝜒(𝐺): 𝜒(𝐺) ≥|𝑉(𝐺)| 𝛼(𝐺) (since each color class is an independent set), we obtain that 𝜒(𝐺(𝑛, 1/2)) ≥(1 + 𝑜(1))𝑛 log2 𝑛 whp.
The following landmark theorem shows that the above lower bound on 𝜒(𝐺(𝑛, 1/2)) is asymptotically tight.
124 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8.3 Chromatic number of a random graph Theorem 8.3.2 (Chromatic number of a random graph — Bollobás 1988) With probability 1 −𝑜(1), 𝜒(𝐺(𝑛, 1/2)) ∼ 𝑛 2 log2 𝑛.
Recall that 𝜔(𝐺(𝑛, 1/2)) is typically concentrated around the point 𝑘where the ex-pected number of 𝑘-cliques 𝑛 𝑘 2−(𝑘 2) is neither too large nor too close to zero. The next lemma show that this probability drops very quickly when we decrease 𝑘even by a constant.
Lemma 8.3.3 Let 𝑘0 = 𝑘0(𝑛) be the largest possible integer 𝑘so that 𝑛 𝑘 2−(𝑘 2) ≥1. Then P(𝛼(𝐺(𝑛, 1/2)) < 𝑘0 −3) ≤𝑒−𝑛2−𝑜(1) Note that there is a trivial lower bound of 2−(𝑛 2) coming from an empty graph.
Proof. Let us prove the equivalent claim P(𝜔(𝐺(𝑛, 1/2)) < 𝑘0 −3) ≤𝑒−𝑛2−𝑜(1).
Let 𝜇𝑘:= 𝑛 𝑘 2−(𝑘 2). For 𝑘∼𝑘0(𝑛) ∼2 log2 𝑛, we have 𝜇𝑘+1 𝜇𝑘 = 𝑛 𝑘+1 𝑛 𝑘 2−𝑘∼𝑛 𝑘2−(2+𝑜(1)) log2 𝑛= 1 𝑛1−𝑜(1) .
Let 𝑘= 𝑘0 −3 and applying Setup 8.1.1 for Janson inequality with 𝑋being the number of 𝑘-cliques, we have 𝜇= 𝜇𝑘> 𝑛3−𝑜(1) and (details of the computation omitted) Δ ∼𝜇2 𝑘4 𝑛2 = 𝑛4−𝑜(1).
So Δ > 𝜇for sufficiently large 𝑛, and we can apply Janson inequality II: P(𝜔(𝐺(𝑛, 1/2)) < 𝑘) = P(𝑋= 0) ≤𝑒−𝑛2−𝑜(1).
□ Proof of Theorem 8.3.2. The lower bound proof was discussed before the theorem statement. For the upper bound we will give a strategy to properly color the random graph with (2 + 𝑜(1)) log2 𝑛colors. We will proceed by taking out independent sets of 125 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8 Janson Inequalities size ∼2 log2 𝑛iteratively until 𝑜(𝑛/log 𝑛) vertices remain, at which point we can use a different color for each remaining vertex.
Note that after taking out the first independent set of size ∼2 log2 𝑛, we cannot claim that the remaining graph is still distributed as 𝐺(𝑛, 1/2). It is not. Our selection of the vertices was dependent on the random graph. We are not allowed to “resample” the edges after the first selection.
The strategy is to apply the previous lemma to see that every large enough subset of vertices has an independent set of size ∼2 log2 𝑛.
Let 𝐺∼𝐺(𝑛, 1/2). Let 𝑚= 𝑛/(log 𝑛)2 , say. For any set 𝑆of 𝑚vertices, the induced subgraph 𝐺[𝑆] has the distribution of 𝐺(𝑚, 1/2). By Lemma 8.3.3, for 𝑘= 𝑘0(𝑚) −3 ∼2 log2 𝑚∼2 log2 𝑛, we have P(𝛼(𝐺[𝑆]) < 𝑘) = 𝑒−𝑚2−𝑜(1) = 𝑒−𝑛2−𝑜(1).
Taking a union bound over all 𝑛 𝑚 < 2𝑛such sets 𝑆, P(there is an 𝑚-vertex subset 𝑆with 𝛼(𝐺[𝑆]) < 𝑘) < 2𝑛𝑒−𝑛2−𝑜(1) = 𝑜(1).
So the following statement is true in 𝐺(𝑛, 1/2) with probability 1 −𝑜(1): () Every 𝑚-vertex subset contains a 𝑘-vertex independent set.
Assume that 𝐺has property (). Now we execute our strategy at the beginning of the proof: 1. While ≥𝑚vertices remain: i. Find an independent set of size 𝑘, and let it form its own color class ii. Remove these 𝑘vertices 2. Color the remaining < 𝑚vertices each with a new color.
The result is a proper coloring. The number of colors used is 𝑛 𝑘+ 𝑚∼ 𝑛 2 log2 𝑛.
□ Exercises 1. 3-AP-free probability. Determine, for all 0 < 𝑝≤0.99 (𝑝is allowed to depend on 𝑛), the probability that [𝑛] 𝑝does not contain a 3-term arithmetic progression, up to a constant factor in the exponent. (The form of the answer should be similar 126 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 8.3 Chromatic number of a random graph to the conclusion in class about the probability that 𝐺(𝑛, 𝑝) is triangle-free. See 3 for notation.) 2. Prove that with probability 1 −𝑜(1), the size of the largest subset of vertices of 𝐺(𝑛, 1/2) inducing a triangle-free subgraph is Θ(log 𝑛).
3. Nearly perfect triangle factor, again. Using Janson inequalities this time, give another solution to Problem 11 in the following generality.
a) Prove that for every 𝜀> 0, there exists 𝐶𝜀> 0 such that such that with probability 1 −𝑜(1), 𝐺(𝑛, 𝐶𝜀𝑛−2/3) contains at least (1/3 −𝜀)𝑛vertex-disjoint triangles.
b) (Optional) Compare the the dependence of the optimal 𝐶𝜀on 𝜀you obtain using the method in Problem 11 versus this problem (don’t worry about leading constant factors).
4. ★Threshold for extensions. Show that for every constant 𝐶> 16/5, if 𝑛2𝑝5 > 𝐶log 𝑛, then with probability 1 −𝑜(1), every edge of 𝐺(𝑛, 𝑝) is contained in a 𝐾4.
(Be careful, this event is not increasing, and so it is insufficient to just prove the result for one specific 𝑝.) 5. Lower tails of small subgraph counts. Fix graph 𝐻and 𝛿∈(0, 1]. Let 𝑋𝐻denote the number of copies of 𝐻in 𝐺(𝑛, 𝑝). Prove that for all 𝑛and 0 < 𝑝< 0.99, P(𝑋𝐻≤(1 −𝛿)E𝑋𝐻) = 𝑒−Θ𝐻, 𝛿(Φ𝐻) where Φ𝐻:= min 𝐻′⊆𝐻:𝑒(𝐻′)>0 𝑛𝑣(𝐻′) 𝑝𝑒(𝐻′).
Here the hidden constants in Θ𝐻,𝛿may depend on 𝐻and 𝛿(but not on 𝑛and 𝑝).
6. ★List chromatic number of a random graph. Show that the list chromatic number of 𝐺(𝑛, 1/2) is (1 + 𝑜(1)) 𝑛 2 log2 𝑛with probability 1 −𝑜(1).
The list-chromatic number (also called choosability) of a graph 𝐺is defined to the minimum 𝑘 such that if every vertex of 𝐺is assigned a list of 𝑘acceptable colors, then there exists a proper coloring of 𝐺where every vertex is colored by one of its acceptable colors.
127 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure 9.1 Bounded differences inequality Recall that the Chernoff bound allows to prove exponential tail bounds for sums of independent random variables. For example, if 𝑍is a sum of 𝑛independent Bernoulli random variables, then P(|𝑍−E𝑍| ≥𝑡) ≤2𝑒−2𝑡2/𝑛.
In this chapter, we develop tools for proving similar tail bounds for other random variables that do not necessarily arise as a sum of independent random variables.
The next theorem says: A Lipschitz function of many independent random variables is con-centrated.
We will prove the following important and useful result, known by several names: McDiarmid’s inequality, Azuma–Hoeffding inequality, and bounded differences inequality.
Theorem 9.1.1 (Bounded differences inequality) Let 𝑋1 ∈Ω1, . . . , 𝑋𝑛∈Ω𝑛be independent random variables. Suppose 𝑓: Ω1 × · · · × Ω𝑛→R satisfies 𝑓(𝑥1, . . . , 𝑥𝑛) −𝑓(𝑥′ 1, . . . , 𝑥′ 𝑛) ≤1 (9.1) whenever (𝑥1, . . . , 𝑥𝑛) and (𝑥′ 1, . . . , 𝑥′ 𝑛) differ on exactly one coordinate. Then the random variable 𝑍= 𝑓(𝑋1, . . . , 𝑋𝑛) satisfies, for every 𝜆≥0, P(𝑍−E𝑍≥𝜆) ≤𝑒−2𝜆2/𝑛 and P(𝑍−E𝑍≤−𝜆) ≤𝑒−2𝜆2/𝑛.
In particular, we can apply the above inequality to 𝑓(𝑥1, . . . , 𝑥𝑛) = 𝑥1 + · · · + 𝑥𝑛to recover the Chernoff bound. The theorem tells us that the window of fluctuation of 𝑍 has length 𝑂(√𝑛).
Example 9.1.2 (Coupon collector). Let 𝑠1, . . . , 𝑠𝑛∈[𝑛] chosen uniformly and inde-129 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure pendently at random. Denote the number of “missing” elements by 𝑍= |[𝑛] \ {𝑠1, . . . , 𝑠𝑛}|.
Note that changing one of the 𝑠1, . . . , 𝑠𝑛changes 𝑍by at most 1, so we have P (|𝑍−E𝑍| ≥𝜆) ≤2𝑒−2𝜆2/𝑛, with E𝑍= 𝑛 1 −1 𝑛 𝑛 ∈ 𝑛−1 𝑒 , 𝑛 𝑒 .
Theorem 9.1.1 holds more generally allowing the bounded difference to depend on the coordinate.
Theorem 9.1.3 (Bounded differences inequality) Let 𝑋1 ∈Ω1, . . . , 𝑋𝑛∈Ω𝑛be independent random variables. Suppose 𝑓: Ω1 × · · · × Ω𝑛→R satisfies 𝑓(𝑥1, . . . , 𝑥𝑛) −𝑓(𝑥′ 1, . . . , 𝑥′ 𝑛) ≤𝑐𝑖 (9.2) whenever (𝑥1, . . . , 𝑥𝑛) and (𝑥1, . . . , 𝑥𝑛) differ only on the 𝑖-th coordinate.
Here 𝑐1, . . . , 𝑐𝑛are constants.
Then the random variable 𝑍= 𝑓(𝑋1, . . . , 𝑋𝑛) satisfies, for every 𝜆≥0, P(𝑍−E𝑍≥𝜆) ≤exp −2𝜆2 𝑐2 1 + · · · + 𝑐2 𝑛 !
and P(𝑍−E𝑍≤−𝜆) ≤exp −2𝜆2 𝑐2 1 + · · · + 𝑐2 𝑛 !
.
We will prove these inequality using martingales.
9.2 Martingales concentration inequalities Definition 9.2.1 A martingale is a random real sequence 𝑍0, 𝑍1, . . . such that for every 𝑍𝑛, E|𝑍𝑛| < ∞ and E[𝑍𝑛+1|𝑍0, . . . , 𝑍𝑛] = 𝑍𝑛.
(To be more formal, we should talk about filtrations of a probability space . . . ) Example 9.2.2 (Random walks with independent steps). If (𝑋𝑖)𝑖≥0 is a sequence of 130 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.2 Martingales concentration inequalities independent random variables with E𝑋𝑖= 0 for all 𝑖, then the partial sums 𝑍𝑛= Í 𝑖≤𝑛𝑋𝑖 is a Martingale.
Example 9.2.3 (Betting strategy). Betting on a sequence of fair coin tosses. After round, you are allow to change your bet. Let 𝑍𝑛be your balance after the 𝑛-th round.
Then 𝑍𝑛is always a martingale regardless of your strategy.
Originally, the term “martingale” referred to the betting strategy where one doubles the bet each time until the first win and then stop betting. Then, with probability 1, 𝑍𝑛= 1 for all sufficiently large 𝑛. (Why does this “free money” strategy not actually work?) The next example is especially important to us.
Example 9.2.4 (Doob martingale). Let 𝑋1, . . . , 𝑋𝑛be a random sequence (not nec-essarily independent, though they often are independent in practice).
Consider a function 𝑓(𝑋1, . . . , 𝑋𝑛). Let 𝑍𝑖be the expected value of 𝑓after “revealing” (exposing) 𝑋1, . . . , 𝑋𝑖, i.e., 𝑍𝑖= E[ 𝑓(𝑋1, . . . , 𝑋𝑛)|𝑋1, . . . , 𝑋𝑖].
So 𝑍𝑖is the expected value of the random variable 𝑍= 𝑓(𝑋1, . . . , 𝑋𝑛) after seeing the first 𝑖arguments, and letting the remaining arguments be random. Then 𝑍0, . . . , 𝑍𝑛is a martingale (why?). It satisfies 𝑍0 = E𝑍(a non-random quantity) and 𝑍𝑛= 𝑍(the random variable that we care about), and thereby offering a way to interpolate between the two.
Example 9.2.5 (Edge-exposure martingale). We can reveal therandom graph𝐺(𝑛, 𝑝) by first fixing an order on all unordered pairs of [𝑛] and then revealing in order whether each pair is an edge. For any graph parameter 𝑓(𝐺) we can produce a martingale 𝑋0, 𝑋1, . . . , 𝑋(𝑛 2) where 𝑍𝑖is the conditional expectation of 𝑓(𝐺(𝑛, 𝑝)) after revealing whether there are edges for first 𝑖pairs of vertices. See Figure 9.1 for an example.
Example 9.2.6 (Vertex-exposure martingale). Similar to the previous example, ex-cept that we now first fix an order on the vertex set, and, at the 𝑖-th step, with 0 ≤𝑖≤𝑛, we reveal all edges whose endpoints are contained in the first 𝑖vertices. See Figure 9.1 for an example.
Sometimes it is better to use the edge-exposure martingale and sometimes it is better to use the vertex-exposure martingale. It depends on the application. There is a trade-off between the length of the martingale and the control on the bounded differences.
The main result is that a martingale with bounded differences must be concen-trated. The following fundamental result is called Azuma’s inequality or the Azuma– Hoeffding inequality.
131 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure 3 2 2 2 2 2 2 1 2.5 2 2 1.5 2.25 1.75 2 3 2 2 2 2 2 2 1 2.25 1.75 2 Figure 9.1: The edge-exposure martingale (left) and vertex-exposure martingale (right) for the chromatic number of 𝐺(𝑛, 1/2) with 𝑛= 3.
The martingale is obtained by starting at the leftmost point, and splitting at each branch with equal probability.
Theorem 9.2.7 (Azuma’s inequality) Let 𝑍0, 𝑍1, . . . , 𝑍𝑛be a martingale satisfying |𝑍𝑖−𝑍𝑖−1| ≤1 for each 𝑖∈[𝑛].
Then for every 𝜆> 0, P(𝑍𝑛−𝑍0 ≥𝜆√𝑛) ≤𝑒−𝜆2/2.
Note that this is the same bound that we derived in Chapter 5 for 𝑍𝑛= 𝑋1 + · · · 𝑋𝑛 where 𝑋𝑖∈{−1, 1} uniform and iid.
More generally, allowing different bounds on different steps of the martingale, we have the following.
Theorem 9.2.8 (Azuma’s inequality) Let 𝑍0, 𝑍1, . . . , 𝑍𝑛be a martingale satisfying |𝑍𝑖−𝑍𝑖−1| ≤𝑐𝑖 for each 𝑖∈[𝑛].
For any 𝜆> 0, P(𝑍𝑛−𝑍0 ≥𝜆) ≤exp −𝜆2 2(𝑐2 1 + · · · + 𝑐2 𝑛) !
.
The above formulations of Azuma’s inequality can be used to recover the bounded differences inequality (Theorems 9.1.1 and 9.1.3) up to a usually unimportant constant in the exponent. To obtain the exact statement of Theorem 9.1.3, we state the following 132 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.2 Martingales concentration inequalities strengthening of Azuma’s inequality. (You are welcome to ignore the next statement if you do not care about the constant factor in the exponent — and really, you should not care.) Theorem 9.2.9 (Azuma’s inequality for Doob martingales) Consider a Doob martingale 𝑍𝑖= E[ 𝑓(𝑋1, . . . , 𝑋𝑛)|𝑋1, . . . , 𝑋𝑖] as in Example 9.2.4.
Suppose, conditioned on any value of (𝑋1, . . . , 𝑋𝑖−1), the possibilities for 𝑍𝑖lies in an interval of length 𝑐𝑖(here 𝑐𝑖is non-random, but the location of the interval may depend on 𝑋1, . . . , 𝑋𝑖−1). Then for any 𝜆> 0, P(𝑍𝑛−𝑍0 ≥𝜆) ≤exp −2𝜆2 𝑐2 1 + · · · + 𝑐2 𝑛 !
.
Remark 9.2.10. Applying the inequality to the martingale with terms −𝑍𝑛, we obtain the following lower tail bound: P(𝑍𝑛−𝑍0 ≤−𝜆) ≤exp −2𝜆2 𝑐2 1 + · · · + 𝑐2 𝑛 !
.
And we can put them together as P(|𝑍𝑛−𝑍0| ≥𝜆) ≤2 exp −2𝜆2 𝑐2 1 + · · · + 𝑐2 𝑛 !
.
Remark 9.2.11. Theorem 9.2.8 is a special case of Theorem 9.2.9, since we can take (𝑋1, . . . , 𝑋𝑛) = (𝑍1 . . . , 𝑍𝑛) and 𝑓(𝑋1, . . . , 𝑋𝑛) = 𝑋𝑛. Note that the |𝑍𝑖−𝑍𝑖−1| ≤𝑐𝑖 condition in Theorem 9.2.8 implies that 𝑍𝑖lies in an interval of length 2𝑐𝑖if we condition on (𝑋1, . . . , 𝑋𝑖−1).
Lemma 9.2.12 (Hoeffding’s lemma) Let 𝑋be a real random variable contained in an interval of length ℓ. Suppose E𝑋= 0.
Then E[𝑒𝑋] ≤𝑒ℓ2/8.
Proof. Suppose 𝑋∈[𝑎, 𝑏] with 𝑎≤0 ≤𝑏and 𝑏−𝑎= ℓ. Then since 𝑒𝑥is convex, using a linear upper bound on the interval [𝑎, 𝑏], we have (note that RHS below is linear in 𝑥) 𝑒𝑥≤𝑏−𝑥 𝑏−𝑎𝑒𝑎+ 𝑥−𝑎 𝑏−𝑎𝑒𝑏, for all 𝑥∈[𝑎, 𝑏].
133 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Since E𝑋= 0, we obtain E𝑒𝑋≤ 𝑏 𝑏−𝑎𝑒𝑎+ −𝑎 𝑏−𝑎𝑒𝑏.
Let 𝑝= −𝑎/(𝑏−𝑎). Then 𝑎= −𝑝ℓand 𝑏= (1 −𝑝)ℓ. So log E𝑒𝑋≤log (1 −𝑝)𝑒−𝑝ℓ+ 𝑝𝑒(1−𝑝)ℓ = −𝑝ℓ+ log(1 −𝑝+ 𝑝𝑒ℓ).
Fix 𝑝∈[0, 1]. Let 𝜑(ℓ) := −𝑝ℓ+ log(1 −𝑝+ 𝑝𝑒ℓ).
It remains to show that 𝜑(ℓ) ≤ℓ2/8 for all ℓ≥0, which follows from 𝜑(0) = 𝜑′(0) = 0 and 𝜑′′(ℓ) ≤1/4 for all ℓ≥0, as 𝜑′′(ℓ) = 𝑝 (1 −𝑝)𝑒−𝑝ℓ+ 𝑝 1 − 𝑝 (1 −𝑝)𝑒−𝑝ℓ+ 𝑝 ≤1 4, since 𝑡(1 −𝑡) ≤1/4 for all 𝑡∈[0, 1].
□ Proof of Theorem 9.2.9. Let 𝑡≥0 be some constant to be decided later. Conditional on any values of (𝑋1, . . . , 𝑋𝑖−1), the random variable 𝑍𝑖−𝑍𝑖−1 has mean zero and lies in an interval of length 𝑐𝑖. So Lemma 9.2.12 gives E[𝑒𝑡(𝑍𝑖−𝑍𝑖−1)|𝑋1, . . . , 𝑋𝑖−1] ≤𝑒𝑡2𝑐2 𝑖/8.
Then the moment generating function satisfies E[𝑒𝑡(𝑍𝑛−𝑍0)] = E h 𝑒𝑡(𝑍𝑖−𝑍𝑖−1)𝑒𝑡(𝑍𝑖−1−𝑍0)i = E h E h 𝑒𝑡(𝑍𝑖−𝑍𝑖−1) 𝑋1, . . . , 𝑋𝑖−1 i 𝑒𝑡(𝑍𝑖−1−𝑍0)i = 𝑒𝑡2𝑐2 𝑛/8E h 𝑒𝑡(𝑍𝑖−1−𝑍0)i .
Iterating, we obtain E h 𝑒𝑡(𝑍𝑛−𝑍0)i ≤𝑒𝑡2(𝑐2 1+···𝑐2 𝑛)/8.
By Markov, P(𝑍𝑛−𝑍0 ≥𝜆) ≤𝑒−𝑡𝜆E h 𝑒𝑡(𝑍𝑛−𝑍0)i ≤𝑒−𝑡𝜆+ 𝑡2 8 (𝑐2 1+···𝑐2 𝑛).
Setting 𝑡= 4𝜆/(𝑐2 1 + · · · + 𝑐2 𝑛) yields the theorem.
□ Now we apply Azuma’s inequality to deduce the bounded differences inequality.
Proof of the bounded differences inequality (Theorem 9.1.3). ConsidertheDoobmar-134 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.3 Chromatic number of random graphs tingale 𝑍𝑖= E[ 𝑓(𝑋1, . . . , 𝑋𝑛)|𝑋1, . . . , 𝑋𝑖]. The hypothesis of Theorem 9.1.3 implies that the hypothesis of Theorem 9.2.9 is satisfied. The same conclusion then follows.
□ Remark 9.2.13. Azuma’s inequality (Theorem 9.2.9) is more versatile than (Theo-rem 9.1.3). For example, while changing 𝑋𝑖might change 𝑓(𝑋1, . . . , 𝑋𝑛) by a lot in the worst case over all possible (𝑋1, . . . , 𝑋𝑛), it might not change it by much in expec-tation over random choices of (𝑋𝑖+1, . . . , 𝑋𝑛). And so the 𝑐𝑖in Theorem 9.2.9 could potentially be smaller than in Theorem 9.1.3. This will be useful in some applications, including one that we will see later in the chapter.
9.3 Chromatic number of random graphs Concentration of the chromatic number Even before Bollobás (1988) showed that 𝜒(𝐺(𝑛, 1/2)) ∼ 𝑛 2 log2 𝑛whp (Theorem 8.3.2), using the bounded difference inequality, it was already known that the chromatic number of a random graph must be concentrated in a 𝑂(√𝑛) window around its mean.
The following application shows that one can prove concentration around the mean without even knowing where is the mean!
Theorem 9.3.1 (Shamir and Spencer 1987) For every 𝜆≥0, the chromatic number of a random graph 𝑍= 𝜒(𝐺(𝑛, 𝑝)) satisfies P(|𝑍−E𝑍| ≥𝜆 √ 𝑛−1) ≤2𝑒−2𝜆2.
Proof. Let 𝑉= [𝑛], and consider each vertex labeled graph as an element of Ω2 × · · · × Ω𝑛where Ω𝑖= {0, 1}𝑖−1 and its coordinates correspond to edges whose larger coordinate is 𝑖(cf. the vertex-exposure martingale Example 9.2.6). If two graphs 𝐺 and 𝐺′ differ only in edges incident to one vertex 𝑣, then |𝜒(𝐺) −𝜒(𝐺′)| ≤1 since, given a proper coloring of 𝐺using 𝜒(𝐺) colors, one can obtain a proper coloring of 𝐺′ using 𝜒(𝐺) + 1 colors by using a new color for 𝑣. Theorem 9.1.3 implies the result.
□ Remark 9.3.2 (Non-concentration of the chromatic number). Heckel (2021) showed that the 𝜒(𝐺(𝑛, 1/2)) is not concentrated on any interval of length 𝑛𝑐for any constant 𝑐< 1/4.
This was the opposite of what most experts believed in.
It has been conjectured that width of the window of concentrations fluctuates between 𝑛1/4+𝑜(1) to 𝑛1/2+𝑜(1) depending on 𝑛.
135 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Clique number, again Previously in Section 8.3, we used Janson inequalities to prove the following ex-ponentially small bound on the probability that 𝐺(𝑛, 1/2) has small clique num-ber. This was a crucial step in the proof of Bollobás’ theorem (Theorem 8.3.2) that 𝜒(𝐺(𝑛, 1/2)) ∼𝑛/(2 log2 𝑛) whp. Here we give a different proof using the bounded difference inequality instead of Janson inequalities. The proof below in fact was the original approach of Bollobás (1988).
Lemma 9.3.3 (Same as Lemma 8.3.3) Let 𝑘0 = 𝑘0(𝑛) ∼2 log2 𝑛be the largest positive integer so that 𝑛 𝑘0 2−( 𝑘0 2 ) ≥1. Then P(𝜔(𝐺(𝑛, 1/2)) < 𝑘0 −3) = 𝑒−𝑛2−𝑜(1).
A naive approach might be to estimate the number of 𝑘-cliques in 𝐺(this is the approach taken with Janson inequalities. The issue is that this quantity can change too much when we modify one edge of 𝐺. We will use a more subtle function on graphs.
Note that we only care about whether there exists a 𝑘-clique or not.
Proof. Let 𝑘= 𝑘0 −3. Let 𝑌= 𝑌(𝐺) be the maximum number of edge-disjoint set of 𝑘-cliques in 𝐺. Then as a function of 𝐺, 𝑌changes by at most 1 if we change 𝐺by one edge. (Note that the same does not hold if we change 𝐺by one vertex, e.g., when 𝐺consists of many 𝑘-cliques glued along a common vertex.) So by the bounded differences inequality, for 𝐺∼𝐺(𝑛, 1/2), P(𝜔(𝐺) < 𝑘) = P(𝑌= 0) ≤P(𝑌−E𝑌≤−E𝑌) ≤exp −2(E𝑌)2 𝑛 2 !
.
(9.1) It remains to show that E𝑌≥𝑛2−𝑜(1). Create an auxiliary graph H whose vertices are the 𝑘-cliques in 𝐺, with a pair of 𝑘-cliques adjacent if they overlap in at least 2 vertices. Then 𝑌= 𝛼(H). We would like to lower bound the independence number of this graph based on its average degree. Here are two ways to proceed: 1. Recall the Caro–Wei inequality (Corollary 2.3.5): for every graph 𝐻with average degree 𝑑, we have 𝛼(𝐻) ≥ ∑︁ 𝑣∈𝑉(𝐻) 1 1 + 𝑑𝑣 ≥|𝑉(𝐻)| 1 + 𝑑 = |𝑉(𝐻)|2 |𝑉(𝐻)| + 2 |𝐸(𝐻)| .
2. Let 𝐻′ be the induced subgraph obtained from 𝐻by keeping every vertex 136 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.3 Chromatic number of random graphs independently with probability 𝑞. We have 𝛼(𝐻) ≥𝛼(𝐻′) ≥|𝑉(𝐻′)| −|𝐸(𝐻′)| .
Taking expectations of both sides, and noting that E |𝑉(𝐻′)| = 𝑞|𝑉(𝐻)| and E |𝐸(𝐻′)| = 𝑞2 |𝐸(𝐻)| by linearity of expectations, we have 𝛼(𝐻) ≥𝑞E |𝑉(𝐻)| −𝑞2 |𝐸(𝐻)| for every 𝑞∈[0, 1].
Provided that |𝐸(𝐻)| ≥|𝑉(𝐻)| /2, we can take 𝑞= |𝑉(𝐻)| /(2 |𝐸(𝐻)|) ∈[0, 1] and obtain 𝛼(𝐻) ≥|𝑉(𝐻)|2 4 |𝐸(𝐻)| if |𝐸(𝐻)| ≥1 2 |𝑉(𝐻)| .
(This method allows us to recover Turán’s theorem up to a factor of 2, whereas the Caro–Wei inequality recovers Turán’s theorem exactly.
For the present application, we do not care about these constant factors.) By a second moment argument (details again omitted, like in the proofs of Theo-rem 4.4.2 and Lemma 8.3.3), we have, with probability 1 −𝑜(1), that the number of 𝑘-cliques in 𝐺is |𝑉(H)| ∼E |𝑉(H)| = 𝑛 𝑘 2−(𝑘 2) = 𝑛3−𝑜(1) and the number of unordered pairs of edge-overlapping 𝑘-cliques in 𝐺is E |𝐸(H)| = 𝑛4−𝑜(1).
Thus, with probability 1 −𝑜(1), we can apply either of the above lower bounds on independent sets to obtain E𝑌≳E|𝑉(H)|2 |𝐸(H)| ≳E 𝑛6−𝑜(1) |𝐸(H)| ≥ 𝑛6−𝑜(1) E |𝐸(H)| = 𝑛2−𝑜(1).
Together with (9.1), this completes the proof that P(𝜔(𝐺) < 𝑘) = 𝑒−𝑛2−𝑜(1).
□ Chromatic number of sparse random graphs Let us show that 𝐺(𝑛, 𝑝) is concentrated on a constant size window if 𝑝is small enough.
137 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Theorem 9.3.4 (Shamir and Spencer 1987) Let 𝛼> 5/6 be fixed. Then for 𝑝< 𝑛−𝛼, 𝜒(𝐺(𝑛, 𝑝)) is concentrated on four values with probability 1 −𝑜(1). That is, there exists 𝑢= 𝑢(𝑛, 𝑝) such that, as 𝑛→∞, P(𝑢≤𝜒(𝐺(𝑛, 𝑝)) ≤𝑢+ 3) = 1 −𝑜(1).
Proof. It suffices to show that for all 𝜀> 0, there exists 𝑢= 𝑢(𝑛, 𝑝, 𝜀) so that, provided 𝑝< 𝑛−𝛼and 𝑛is sufficiently large, P(𝑢≤𝜒(𝐺(𝑛, 𝑝)) ≤𝑢+ 3) ≥1 −3𝜀.
Let 𝑢be the least integer so that P(𝜒(𝐺(𝑛, 𝑝)) ≤𝑢) > 𝜀.
Now we make a clever choice of a random variable.
Let 𝐺∼𝐺(𝑛, 𝑝). Let 𝑌= 𝑌(𝐺) denote the minimum size of a subset 𝑆⊆𝑉(𝐺) such that 𝐺−𝑆is 𝑢-colorable.
Note that 𝑌changes by at most 1 if we change the edges around one vertex of 𝐺. Thus, by applying Theorem 9.1.1 with respect to vertex-exposure (Example 9.2.6), we have P(𝑌≤E𝑌−𝜆√𝑛) ≤𝑒−2𝜆2 and P(𝑌≥E𝑌+ 𝜆√𝑛) ≤𝑒−2𝜆2.
We choose 𝜆= 𝜆(𝜀) > 0 so that 𝑒−2𝜆2 = 𝜀.
First, we use the lower tail bound to show that E𝑌must be small. We have 𝑒−2𝜆2 = 𝜀< P(𝜒(𝐺) ≤𝑢) = P(𝑌= 0) = P(𝑌≤E𝑌−E𝑌) ≤exp −2(E𝑌)2 𝑛 .
Thus E𝑌≤𝜆√𝑛.
Next, we apply the upper tail bound to show that 𝑌is rarely large. We have P(𝑌≥2𝜆√𝑛) ≤P(𝑌≥E𝑌+ 𝜆√𝑛) ≤𝑒−2𝜆2 = 𝜀.
Each of the following three events occur with probability at least 1−𝜀, for large enough 𝑛, • By the above argument, there is some 𝑆⊆𝑉(𝐺) with |𝑆| ≤2𝜆√𝑛and 𝐺−𝑆 138 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.4 Isoperimetric inequalities: a geometric perspective may be properly 𝑢-colored.
• By the next lemma, one can properly 3-color 𝐺[𝑆].
• 𝜒(𝐺) ≥𝑢(by the minimality of 𝑢at the beginning of the proof).
Thus, with probability at least 1 −3𝜀, all three events occur, and so we have 𝑢≤ 𝜒(𝐺) ≤𝑢+ 3.
□ Lemma 9.3.5 Fix 𝛼> 5/6 and 𝐶. Let 𝑝≤𝑛−𝛼. Then with probability 1 −𝑜(1) every subset of at most 𝐶√𝑛vertices of 𝐺(𝑛, 𝑝) can be properly 3-colored.
Proof. Let 𝐺∼𝐺(𝑛, 𝑝). Assume that 𝐺is not 3-colorable. Choose minimum size 𝑇⊆𝑉(𝐺) so that the induced subgraph 𝐺[𝑇] is not 3-colorable.
We see that 𝐺[𝑇] has minimum degree at least 3, since if deg𝐺𝑇 < 3, then 𝑇−𝑥 cannot be 3-colorable either (if it were, then can extend coloring to 𝑥), contradicting the minimality of 𝑇.
Thus 𝐺[𝑇] has at least 3|𝑇|/2 edges. The probability that 𝐺has some induced subgraph on 𝑡≤𝐶√𝑛vertices and ≥3𝑡/2 edges is, by a union bound, (recall 𝑛 𝑘 ≤(𝑛𝑒/𝑘)𝑘) ≤ 𝐶√𝑛 ∑︁ 𝑡=4 𝑛 𝑡 𝑡 2 3𝑡/2 𝑝3𝑡/2 ≤ 𝐶√𝑛 ∑︁ 𝑡=4 𝑛𝑒 𝑡 𝑡𝑡𝑒 3 3𝑡/2 𝑛−3𝑡𝛼/2 ≤ 𝐶√𝑛 ∑︁ 𝑡=4 𝑂(𝑛1−3𝛼/2√ 𝑡) 𝑡 ≤ 𝐶√𝑛 ∑︁ 𝑡=4 𝑂(𝑛1−3𝛼/2+1/4) 𝑡 .
The sum is 𝑜(1) provided that 𝛼> 5/6.
□ Remark 9.3.6. Theorem 9.3.4 was subsequently improved (by a refinement of the above techniques) by Łuczak (1991) and Alon and Krivelevich (1997). We now know that the chromatic number of 𝐺(𝑛, 𝑛−𝛼) has two-point concentration for all 𝛼> 1/2.
9.4 Isoperimetric inequalities: a geometric perspective We shall explore the following connection, which are two sides of the same coin: Probability Geometry Concentration of Lipschitz functions Isoperimetric inequalities 139 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Milman recognized the importance of the concentration of measure phenomenon, which he heavily promoted in the 1970’s.
The subject was have been since then extensively developed. It plays a central role in probability theory, the analysis of Banach spaces, and it also has been influential in theoretical computer science.
Euclidean space The classic isoperimetric theorem in R𝑛says that that among all subset of R𝑛of given volume, the ball has the smallest surface volume. (The word “isoperimetric” refers to fixing the perimeter; equivalently we fix the surface area and ask to maximize volume.) This result (at least in two-dimensions) was known to the Greeks, but rigorous proofs were only found in towards the end of the nineteenth century.
Let (𝑋, 𝑑𝑋) be a metric space.
Let 𝐴⊆𝑋.
For any 𝑥∈𝑋, write 𝑑𝑋(𝑥, 𝐴) := inf𝑎∈𝐴𝑑𝑋(𝑥, 𝑎) for the distance from 𝑥to 𝐴.
Denote the set of all points within distance 𝑡from 𝐴by 𝐴𝑡:= {𝑥∈𝑋: 𝑑𝑋(𝑥, 𝐴) ≤𝑡} (9.1) This is also known as the radius-𝒕neighborhood of 𝑨. One can visualize 𝐴𝑡by “expanding” 𝐴by distance 𝑡.
Theorem 9.4.1 (Isoperimetric inequality in Euclidean space) Let 𝐴⊆R𝑛be a measurable set, and let 𝐵⊆R𝑛be a ball vol(𝐴) = vol(𝐵). Then, for all 𝑡≥0, vol 𝐴𝑡≥vol 𝐵𝑡.
Remark 9.4.2. A clean way to prove the above inequality is via the Brunn–Minkowski theorem.
Classically, the isoperimetric inequality is stated as (here 𝜕𝐴is the boundary of 𝐴) vol𝑛−1 𝜕𝐴≥vol𝑛−1 𝜕𝐵.
These two formulations are equivalent. Indeed, assuming Theorem 9.4.1, we have vol𝑛−1 𝜕𝐴= 𝑑 𝑑𝑡 𝑡=0 vol𝑛𝐴𝑡= lim 𝑡→0 vol 𝐴𝑡−vol 𝐴 𝑡 ≥lim 𝑡→0 vol 𝐵𝑡−vol 𝐵 𝑡 = vol𝑛−1 𝜕𝐵.
Conversely, we can obtain the neighborhood version from the boundary version by integrating (noting that 𝐵𝑡is always a ball).
140 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.4 Isoperimetric inequalities: a geometric perspective The cube We have an analogous result in the {0, 1}𝑛with respect to Hamming distance.In Hamming cube, Harper’s theorem gives the exact result. Below, for 𝐴⊆{0, 1}𝑛, we write 𝐴𝑡as in (9.1) for 𝑋= {0, 1}𝑛and 𝑑𝑋being the Hamming distance.
Theorem 9.4.3 (Isoperimetic inequality in the Hamming cube; Harper 1966) Let 𝐴⊆{0, 1}𝑛. Let 𝐵⊆{0, 1}𝑛be a Hamming ball with |𝐴| ≥|𝐵|. Then for all 𝑡≥0, |𝐴𝑡| ≥|𝐵𝑡|.
Remark 9.4.4. The above statement is tight when 𝐴has the same size as a Hamming ball, i.e., when |𝐴| = 𝑛 0 + 𝑛 1 + · · · + 𝑛 𝑘 for some integer 𝑘. Actually, more is true.
For any value of |𝐴| and 𝑡, the size of 𝐴𝑡is minimized by taking 𝐴to be an initial segment of {0, 1}𝑛according to the simplicial ordering: first sort by Hamming weight, and for ties, sort by lexicographic order. For more on this topic, particularly extremal set theory, see the book Combinatorics by Bollobás (1986).
Combined with the isoperimetic inequality on the cube, we obtain the following surprising consequence. Suppose we start with just half of the cube, and then expand it by a bit (recall that the diameter of the cube is 𝑛, and we will be expanding it by 𝑜(𝑛)), then resulting expansion occupies nearly all of the cube.
Theorem 9.4.5 (Rapid expansion from half to 1 −𝜀) Let 𝑡> 0. For every 𝐴⊆{0, 1}𝑛with |𝐴| ≥2𝑛−1, we have |𝐴𝑡| > (1 −𝑒−2𝑡2/𝑛)2𝑛.
Proof. Let 𝐵= {𝑥∈{0, 1}𝑛: weight(𝑥) < 𝑛/2}, so that |𝐵| ≤2𝑛−1 ≤|𝐴|. Then by Harper’s theorem (Theorem 9.4.3), |𝐴𝑡| ≥|𝐵𝑡| = |{𝑥∈{0, 1}𝑛: weight(𝑥) < 𝑛/2 + 𝑡}| > (1 −𝑒−2𝑡2/𝑛)2𝑛 by the Chernoff bound.
□ In fact, using the above, we can deduce that even if we start with a small fraction (e.g., 1%) of the cube, and expand it slightly, then we would cover most of the cube.
141 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Theorem 9.4.6 (Rapid expansion from 𝜀to 1 −𝜀) Let 𝜀> 0 and 𝐶= √︁ 2 log(1/(𝜀)). If 𝐴⊆{0, 1}𝑛with |𝐴| ≥𝜀2𝑛, then 𝐴𝐶√𝑛 ≥(1 −𝜀)2𝑛.
First proof via Harper’s isoperimetric inequality. Let𝑡= √︁ log(1/𝜀)𝑛/2so that 𝑒−2𝑡2/𝑛= 𝜀. Applying Theorem 9.4.5 to 𝐴′ = {0, 1}𝑛\ 𝐴𝑡, we see that |𝐴′| < 2𝑛−1 (or else 𝐴′ 𝑡 > (1 −𝜀)2𝑛, so 𝐴′ 𝑡would intersect 𝐴, which is impossible since the distance be-tween 𝐴and 𝐴′ is greater than 𝑡). Thus |𝐴𝑡| ≥2𝑛−1, and then applying Theorem 9.4.5 yields |𝐴2𝑡| ≥(1 −𝜀)2𝑛.
□ Let us give another proof of Theorem 9.4.6 without using Harper’s exact isoperimetric theorem in the Hamming cube, and instead use the bounded differences inequality that we proved earlier.
Second proof via the bounded differences inequality. Pick a uniform random 𝑥∈ {0, 1}𝑛and let 𝑋= dist(𝑥, 𝐴). Note that 𝑋changes by at most 1 if a single coor-dinate of 𝑥is changed. Applying the bounded differences inequality, Theorem 9.1.1, we have the lower tail P(𝑋−E𝑋≤−𝜆) ≤𝑒−2𝜆2/𝑛 for all 𝜆≥0 We have 𝑋= 0 if and only if 𝑥∈𝐴, so 𝜀≤P(𝑥∈𝐴) = P(𝑋= 0) = P(𝑋−E𝑋≤−E𝑋) ≤𝑒−2(E𝑋)2/𝑛.
Thus E𝑋≤ √︂ log(1/𝜀)𝑛 2 = 𝐶√𝑛 2 .
Now we apply the upper tail of the bounded differences inequality P(𝑋−E𝑋≥𝜆) ≤𝑒−2𝜆2/𝑛 for all 𝜆≥0 to yield P(𝑥∉𝐴𝐶√𝑛) = P(𝑋> 𝐶√𝑛) ≤P 𝑋≥E𝑋+ 𝐶√𝑛 2 ≤𝜀.
□ Isoperimetry versus concentration The above two proofs illustrate the link between geometric isoperimetric inequalities and probabilistic concentration inequalities. Let know now state a simple result that 142 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.4 Isoperimetric inequalities: a geometric perspective formalizes this connection.
Definition 9.4.7 (Lipschitz functions) Given two metric spaces (𝑋, 𝑑𝑋) and (𝑌, 𝑑𝑌), we say that a function 𝑓: 𝑋→𝑌is 𝑪-Lipschitz if 𝑑𝑌( 𝑓(𝑥), 𝑓(𝑥′)) ≤𝐶𝑑𝑋(𝑥, 𝑥′) for all 𝑥, 𝑥′ ∈𝑋.
So the bounded differences inequality applies to Lipschitz functions with respect to the Hamming distance. In particular, it tells us that if 𝑓: {0, 1}𝑛→R is 1-Lipschitz (with respect to the Hamming distance on {0, 1}𝑛), it must be concentrated around its mean with respect to the uniform measure on {0, 1}𝑛: P(| 𝑓−E 𝑓| ≥𝑛𝜆) ≤2𝑒−2𝑛𝜆2.
So 𝑓is almost constant almost everywhere. This is a counterintuitive high dimensional phenomenon.
Theorem 9.4.8 (Equivalence between notions of concentration of measure) Let 𝑡, 𝜀≥0. In a probability space (Ω, P) equipped with a metric. The following are equivalent: (a) (Expansion/approximate isoperimetry) If 𝐴⊆Ω with P(𝐴) ≥1/2, then P(𝐴𝑡) ≥1 −𝜀.
(b) (Concentration of Lipschitz functions) If 𝑓: Ω →R is 1-Lipschitz and 𝑚∈R satisfies P( 𝑓≤𝑚) ≥1/2, then P( 𝑓> 𝑚+ 𝑡) ≤𝜀.
Remark 9.4.9 (Median). In (b), we often take 𝑚to be a median of 𝑓, which is defined to be a value such that P( 𝑓≥𝑚) ≥1/2 and P( 𝑓≤𝑚) ≥1/2 (the median always exists but is not necessarily unique). For distributions with good concentration properties, the median and mean are usually close to each other. For example, we leave it as an exercise to check that if there is some 𝑚such that P(| 𝑓−𝑚| ≥𝑡) ≤2𝑒−𝑡2/2 for all 𝑡≥0, then the mean and the medians of 𝑓all lie within 𝑂(1) of 𝑚.
Proof. (a) =⇒(b): Let 𝐴= {𝑥∈Ω : 𝑓(𝑥) ≤𝑚}. So P(𝐴) ≥1/2. Since 𝑓is 1-Lipschitz, we have 𝑓(𝑥) ≤𝑚+ 𝑡for all 𝑥∈𝐴𝑡. Thus by (a) P( 𝑓> 𝑚+ 𝑡) ≤P(𝐴𝑡) ≤𝜀.
143 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure (b) =⇒(a): Let 𝑓(𝑥) = dist(𝑥, 𝐴) and 𝑚= 0. Then P( 𝑓≤0) = P(𝐴) ≥1/2. Also 𝑓 is 1-Lipschitz. So by (b), P(𝐴𝑡) = P( 𝑓> 𝑚+ 𝑡) ≤𝜀.
□ Informally, we say that a space (or rather, a sequence of spaces), has concentration of measure if 𝜀decays rapidly as a function of 𝑡in the above theorem (the notion of “Lévy family” makes this precise). Earlier we saw that the Hamming cube exhibits has concentration of measure. Other notable spaces with concentration of measure include the sphere, Gauss space, orthogonal and unitary groups, postively-curved manifolds, and the symmetric group.
The sphere We discuss analogs of the concentration of measure phenomenon in high dimensional geometry. This is rich and beautiful subject. An excellent introductory to this topic is the survey An Elementary Introduction to Modern Convex Geometry by Ball (1997).
Recall the isoperimetric inequality in R𝑛says: If 𝐴⊆R𝑛has the same measure as ball 𝐵, then vol(𝐴𝑡) ≥vol(𝐵𝑡) for all 𝑡≥0.
Analogous exact isoperimetric inequalities are known in several other spaces. We already saw it for the boolean cube (Theorem 9.4.3). The case of sphere and Gaussian space are particularly noteworthy. The following theorem is due to Lévy (∼1919).
Theorem 9.4.10 (Lévy’s isoperimetric inequality on the sphere) On a sphere in R𝑛, let 𝐴be a measurable subset and 𝐵a spherical cap with vol𝑛−1(𝐴) = vol𝑛−1(𝐵). Then for all 𝑡≥0, vol𝑛−1(𝐴𝑡) ≥vol𝑛−1(𝐵𝑡).
We have the following upper bound estimate on the size of spherical caps.
Theorem 9.4.11 (Upper bound on spherical cap size) Let 𝑥= (𝑥1, . . . , 𝑥𝑛) ∈R𝑛be a uniform random unit vector in R𝑛. Then for any 𝜀≥0, P(𝑥1 ≥𝜀) ≤𝑒−𝑛𝜀2/2.
The following proof (including figures) is taken from Tokz (2012), building on the method by Ball (1997).
144 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.4 Isoperimetric inequalities: a geometric perspective Proof. Let 𝐶denote the spherical cap consisting of unit vectors 𝑥with 𝑥1 ≥𝜀. Write e 𝐶for the convex hull of 𝐶with the origin, i.e., the conical sector spanned by 𝐶. The idea is to contain e 𝐶in a ball of radius 𝑟≤𝑒−𝜀2/2. Writing 𝐵(𝑟) for a ball of radius 𝑟 in R𝑛so that, we have vol𝑛−1 𝐶 vol𝑛−1 𝑆𝑛−1 = vol𝑛e 𝐶 vol𝑛𝐵𝑛(1) = vol𝑛𝐵(𝑟) vol𝑛𝐵(1) = 𝑟𝑛≤𝑒−𝜀2𝑛/2.
Case 1: 𝜀∈[0, 1/ √ 2].
Bn(0,1) Cone P p 1 −"2 As shown above, e 𝐶is contained in a ball of radius 𝑟= √ 1 −𝜀2 ≤𝑒−𝜀2/2.
Case 2: 𝜀∈[1/ √ 2, 1].
Cone Q r 1 Then e 𝐶is contained in a ball of radius 𝑟as shown above. Using similar triangles, we find that 𝑟/(1/2) = 1/𝜀. So 𝑟= 1/(2𝜀) ≤𝑒−𝜀2/2, where final inequality is equivalent to 𝑒𝑥2/2 ≤2𝑥for all [1/ √ 2, 1], which, by convexity, only needs to be checked at the endpoints of the interval.
□ Combining the above two theorems, we deduce the following concentration of measure results.
145 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Corollary 9.4.12 (Concentration of measure on the sphere) Let 𝐴be a measurable subset of the unit sphere in R𝑛, equipped with the metric inherited from R𝑛. If 𝐴⊆𝑆𝑛−1 has vol𝑛−1(𝐴)/vol𝑛−1(𝑆𝑛−1) ≥1/2, then vol𝑛−1(𝐴𝑡) vol𝑛−1(𝑆𝑛−1) ≥1 −𝑒−𝑛𝑡2/4.
Remark 9.4.13. See §14 in Barvinok’s notes for a proof of the sharper estimate with 𝑒−𝑛𝑡2/4 replaced by √︁ 𝜋/8𝑒−𝑛𝑡2/2, where now we are using the geodesic distance on the sphere.
Corollary 9.4.14 (Concentration of measure on the sphere) Let 𝑆𝑛−1 denote the unit sphere in R𝑛. If 𝑓: 𝑆𝑛−1 →R is a 1-Lipschitz measurable function, then there is some real 𝑚so that, for the uniform measure on the sphere, P(| 𝑓−𝑚| > 𝑡) ≤2𝑒−𝑛𝑡2/4.
Informally: every Lipschitz function on a high dimensional sphere is almost constant almost everywhere.
This is a rather counterintuitive high-dimensional phenomenon.
Gauss space Another related setting is the Gauss space, which is R𝑛equipped with the the proba-bility measure 𝛾𝑛induced by the Gaussian random vector whose coordinates are 𝑛iid standard normals, i.e., the normal random vector in R𝑛with covariance matrix 𝐼𝑛. Its probability density function of 𝛾𝑛at 𝑥∈R𝑛is (2𝜋)−𝑛𝑒−|𝑥|2/2. The metric on R𝑛is the usual Euclidean metric.
What would an isoperimetric inequality in Gauss space look like?
Although earlier examples of isoperimetric optimizers were all balls, for the Gauss space, the answer is actually a half-spaces, i.e., points on one side of some hyperplane.
The Gaussian isoperimetric inequality, below, was first shown independently by Borell (1975) and Sudakov and Tsirel’son (1974).
Theorem 9.4.15 (Gaussian isoperimetric inequality) If 𝐴, 𝐻⊆R𝑛, 𝐻a half-space, and 𝛾(𝐴) = 𝛾(𝐻), then 𝛾(𝐴𝑡) ≥𝛾(𝐻𝑡) for all 𝑡≥0, where 𝛾is the Gauss measure.
146 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.4 Isoperimetric inequalities: a geometric perspective If 𝐻= {𝑥1 ≤0}, then 𝐻𝑡= {𝑥1 ≤𝑡}, which has Gaussian measure ≥1 −𝑒−𝑡2/2. Thus: Corollary 9.4.16 (Concentration of measure for Gaussian vectors) If 𝑓: R𝑛→R is 1-Lipschitz, and 𝑍is a vector of i.i.d. standard normals, then 𝑋= 𝑓(𝑍) satisfies, for some 𝑚, P(|𝑋−𝑚| ≥𝑡) ≤2𝑒−𝑡2/2.
Here is a rather handwavy explanation why the half-space is a reasonable answer.
Consider {−1, 1}𝑚𝑛, where both 𝑚and 𝑛are large. Let us group the coordinates of {−1, 1}𝑚𝑛into block of length 𝑚. The sum of entries in each block (after normalizing by √𝑚) approximates normal random variable by the central limit theorem.
In the Hamming cube, Harper’s theorem tells us Hamming balls are isoperimetric optimizers. Since a Hamming ball in {−1, 1}𝑚𝑛is given by all points whose sum of coordinates is below a certain threshold, we should look at the analogous subset in the Gauss space, which would then consist of all points whose sum of coordinates is below a certain threshold. The set of all points whose of coordinate sum is below a certain threshold is half-space. Note also that the Gaussian measure is radially symmetric.
The sphere as approximately a sum of independent Gaussians. The Gauss space is a nice space to work with because a standard normal vector simultaneously possesses two useful properties (and it is essentially the only such random vector to have both properties): (a) Rotational invariance (b) Independence of coordinates The squared-length of a random Gaussian vector is 𝑍2 1 + · · · + 𝑍2 𝑛with iid 𝑍1, . . . , 𝑍𝑛∈ 𝑁(0, 1). It has mean 𝑛and a 𝑂(√𝑛) window of concentration (e.g., by a straightforward adaptation of the Chernoff bound proof). Since √︁ 𝑛+ 𝑂(√𝑛) = √𝑛+ 𝑂(1), the length of Gaussian vector is concentrated in a 𝑂(1) window around √𝑛(the concentration can also be deduced from the above corollary for 𝑓(𝑥) = |𝑥|). So most of the distribution in the Gauss space lies within a constant distance of a sphere of radius √𝑛. Due to rotational invariance, we see that a Gaussian distribution approximates the uniform distribution on sphere of radius √𝑛in high dimensions. In other words: random Gaussian vector ≈√𝑛· random unit vector.
Random Gaussian vectors often yield easier calculations due to coordinate indepen-dence, and so they often give an accessible way to analyze random unit vectors.
Note that how a half-space in the Gauss space intersect the sphere in a spherical cap, with both italicized objects being isoperimetric optimizers in their respective spaces.
147 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Sub-Gaussian distributions We introduce some terminology that captures notions we have seen so far. It will also be convenient for later discussions.
Definition 9.4.17 (Sub-Gaussian distribution) We say that a random variable 𝑋is 𝑲-subGaussian about its mean if P(|𝑋−E𝑋| ≥𝑡) ≤2𝑒−𝑡2/𝐾2 for all 𝑡≥0.
Remark 9.4.18. This definition is not standard. Some places say 𝜎2-subGaussian for what we mean by 𝜎-subGaussian.
Usually we will not worry about constant factors. Thus, saying that a family of random variables 𝑋𝑛is 𝑂(𝐾𝑛)-subGaussian about its mean is the same as saying that there exist constant 𝐶, 𝑐> 0 such that P (|𝑋𝑛−E𝑋𝑛| ≥𝑡) ≤𝐶𝑒−𝑐𝑡2/𝐾2 𝑛 for all 𝑡≥0 and 𝑛.
Also note that, up to changing the constants 𝑐, 𝐶, the definition does not change if we replace E𝑋𝑛by a median of 𝑋𝑛above.
Example 9.4.19. The concentration inequalities so far can be rephrased in terms of subGaussian distributions. Below is summary of results of the form: if 𝑋is a random point drawn from the given space, and 𝑓is a 1-Lipschitz function, then 𝑓(𝑋) is 𝐾-subGaussian.
space distance -subGaussian reference {0, 1}𝑛 Hamming 𝑂(√𝑛) bounded diff. ineq. (Thm. 9.1.1) 𝑆𝑛−1 Euclidean 𝑂(1/√𝑛) Lévy concentration (Cor. 9.4.14) Gauss space R𝑛 Euclidean 𝑂(1) Gaussian isoperimetric ineq. (Cor. 9.4.16) The following lemma shows that for subGaussian random variables, it does not matter much if we define the tails around its median, mean, or root-mean-square.
148 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.4 Isoperimetric inequalities: a geometric perspective Lemma 9.4.20 (Median vs. mean for subGaussian distributions) There exists a constant 𝐶> 0 so that the following holds for any real random variable 𝑋satisfying, for some constants 𝑚and 𝐾, P(|𝑋−𝑚| ≥𝑡) ≤2𝑒−𝑡2/𝐾2 for all 𝑡≥0.
(a) Every median M𝑋of 𝑋satisfies |M𝑋−𝑚| ≤𝐶𝐾.
(b) The mean of 𝑋satisfies |E𝑋−𝑚| ≤𝐶𝐾.
(c) For any 𝑝≥1, writing ∥𝑋∥𝑝:= (E |𝑋|𝑝)1/𝑝for the 𝐿𝑝norm of 𝑋, ∥𝑋∥𝑝−𝑚 ≤𝐶𝐾√𝑝.
(d) For every constant 𝐴there exists a constant 𝑐> 0 so that if |𝑚′ −𝑚| ≤𝐴𝐾, then P(|𝑋−𝑚′| ≥𝑡) ≤2𝑒−𝑐𝑡2/𝐾2 for all 𝑡≥0.
Proof. By considering 𝑋/𝐾instead of 𝑋, we may assume that 𝐾= 1 for convenience.
(a) For any 𝑡> √︁ 2 log 2, we have P(|𝑋−𝑚| ≥𝑡) ≤2𝑒−𝑡2 < 1/2. So every median of 𝑋lies within √︁ 2 log 2 of 𝑚.
(b) We have |E𝑋−𝑚| ≤E |𝑋−𝑚| = ∫∞ 0 P(|𝑋−𝑚| ≥𝑡) 𝑑𝑡 ≤ ∫∞ 0 2𝑒−𝑡2 𝑑𝑡= √𝜋.
(c) Using the triangle inequality on the 𝐿𝑝norm, we have ∥𝑋∥𝑝−𝑚 ≤∥𝑋−𝑚∥𝑝= (E |𝑋−𝑚|𝑝)1/𝑝= ∫∞ 0 P(|𝑋−𝑚|𝑝≥𝑡) 𝑑𝑡 1/𝑝 ≤ ∫∞ 0 2𝑒−𝑡2/𝑝𝑑𝑡 1/𝑝 = 21/𝑝Γ 1 + 𝑝 2 1/𝑝 = 𝑂(√𝑝).
(c) We can make 𝑐small enough so that 𝑅𝐻𝑆= 2𝑒−𝑐𝑡2 ≥1 for 𝑡≤2𝐴. For 𝑡> 2𝐴, 149 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure we note that P(|𝑋−𝑚′| ≥𝑡) ≤P(|𝑋−𝑚| ≥𝑡/2) ≤2𝑒−𝑡2/4.
□ Remark 9.4.21 (Equivalent characterization of subGaussian distributions). Given a real random variable 𝑋, if any of the below is true for some 𝐾𝑖, then the other conditions are true for some 𝐾𝑗≤𝐶𝐾𝑖for some absolute constant 𝐶.
(a) (Tails) P(|𝑋| ≥𝑡) ≤2𝑒−𝑡2/𝐾2 1 for all 𝑡≥0.
(b) (Moments) ∥𝑋∥𝐿𝑝≤𝐾2√𝑝for all 𝑝≥1.
(c) (MGF of 𝑋2) E𝑒𝑋2/𝐾2 3 ≤2.
We leave the proofs as exercises. Also see §2.5.1 in the textbook High-Dimensional Probability by Vershynin (2018), which gives a superb introduction to the subject.
Johnson–Lindenstrauss Lemma Given a set of 𝑁points in high-dimensional Euclidean space, the next result tells us that one can embed them in 𝑂(𝜀−2 log 𝑁) dimensions without sacrificing pairwise distances by more than 1 ± 𝜀factor. This is known as dimension reduction. It is an important tool in many areas, from functional analysis to algorithms.
Theorem 9.4.22 (Johnson and Lindenstrauss 1982) There exists a constant 𝐶> 0 so that the following holds. Let 𝜀> 0. Let 𝑋be a set of 𝑁points in R𝑚. Then for any 𝑑> 𝐶𝜀−2 log 𝑁, there exists 𝑓: 𝑋→R𝑑so that (1 −𝜀) |𝑥−𝑦| ≤| 𝑓(𝑥) −𝑓(𝑦)| ≤(1 + 𝜀) |𝑥−𝑦| for all 𝑥, 𝑦∈𝑋.
Remark 9.4.23. Here the requirement 𝑑> 𝐶𝜀−2 log 𝑁on the dimension is optimal up to a constant factor (Larsen and Nelson 2017).
We will take 𝑓to be √︁ 𝑚/𝑑times an orthogonal projection onto a 𝑑-dimensional subspace chosen uniformly at random. The theorem then follows from the following lemma together with a union bound.
150 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.4 Isoperimetric inequalities: a geometric perspective Lemma 9.4.24 (Random projection) There exists a constant 𝐶> 0 so that the following holds.
Let 𝑚≥𝑑and let 𝑃: R𝑚→R𝑑denote the orthogonal projection onto the subspace spanned by the first 𝑑coordinates. Let 𝑧be a uniform random point on the unit sphere in R𝑚. Let 𝑦= 𝑃𝑧 and 𝑌= |𝑦|. Then, for all 𝑡≥0, P 𝑌− √︂ 𝑑 𝑚 ≥𝑡 !
≤2𝑒−𝑐𝑚𝑡2.
To prove the Theorem 9.4.22, for each pair of distinct points 𝑥, 𝑥′ ∈𝑋, set 𝑧= 𝑥−𝑥′ |𝑥−𝑥′|, so that √︂𝑚 𝑑𝑌= | 𝑓(𝑥) −𝑓(𝑥′)| |𝑥−𝑥′| .
Then the length of the projection of 𝑧onto a uniform random 𝑑-dimensional subspace has the same distribution as 𝑌in the lemma. So setting 𝑡= 𝜀 √︁ 𝑑/𝑚, we find that P √︂𝑚 𝑑𝑌−1 ≥𝜀 ≤2𝑒−𝑐𝜀𝑑< 2𝑁−𝑐𝐶.
Provided that 𝐶> 1/𝑐, we can take a union bound over all 𝑁 2 < 𝑁2/2 pairs of points of 𝑋to show that with some positive probability, the random 𝑓works.
Proof of the lemma. We have 𝑧2 1 + · · · + 𝑧2 𝑛= 1 and each 𝑧𝑖has the same distribution, so E[𝑧2 𝑖] = 1/𝑚for each 𝑖. Thus E[𝑌2] = E 𝑧2 1 + · · · + 𝑧2 𝑑 = 𝑑 𝑚.
Note that 𝑃is 1-Lipschitz on the unit sphere.
By Lévy’s concentration measure theorem on the sphere, letting M𝑌denote the median of 𝑌, P (|𝑌−M𝑌| ≥𝑡) ≤2𝑒−𝑚𝑡2/4.
The result then follows by Lemma 9.4.20, using that ∥𝑌∥2 = √︁ 𝑑/𝑚.
□ Here is a cute application of Johnson–Lindenstrauss (this is related to a homework problem on the Chernoff bound).
Corollary 9.4.25 There is a constant 𝑐> 0 so that for every positive integer 𝑑, there is a set of 𝑒𝑐𝜀2𝑑 points in R𝑑whose pairwise distances are in [1 −𝜀, 1 + 𝜀].
151 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Proof. Applying Theorem 9.4.22 a regular simplex with unit edge lengths with 𝑁 vertices in R𝑁−1 to yield 𝑁points in R𝑑for 𝑑= 𝑂(𝜀−2 log 𝑁) and pairwise distances in [1 −𝜀, 1 + 𝜀].
□ 9.5 Talagrand’s inequality Talagrand (1995) developed a powerful concentration inequality. It is applicable to many combinatorial optimization problems on independent random inputs. The most general form of Talagrand’s inequality can be somewhat difficult to grasp. So we start by discussing a special case with an easier geometric statement. Though, to obtain the full power of Talagrand’s inequality with combinatorial consequences, we will need the full statement to be given later.
We omit the proof of Talagrand’s inequality (see the Alon–Spencer textbook or Tao’s blog post) and instead focus on explaining the theorem and its applications.
Distance to a subspace We start with a geometrically motivated question.
Problem 9.5.1 Let 𝑉be a fixed 𝑑-dimensional subspace. Let 𝑥∼Unif{−1, 1}𝑛. How well is dist(𝑥,𝑉) concentrated?
Let 𝑃= (𝑝𝑖𝑗) ∈R𝑛×𝑛be the matrix giving the orthogonal projection onto 𝑉⊥. We have tr 𝑃= dim𝑉⊥= 𝑛−𝑑. Then dist(𝑥,𝑉)2 = |𝑥· 𝑃𝑥| = ∑︁ 𝑖,𝑗 𝑥𝑖𝑥𝑗𝑝𝑖𝑗.
So E[dist(𝑥,𝑉)2] = ∑︁ 𝑖 𝑝𝑖𝑖= tr 𝑃= 𝑛−𝑑.
How well is dist(𝑥,𝑉) concentrated around √ 𝑛−𝑑?
Some easier special cases (codimension-1): • If 𝑉is a coordinate subspace, then dist(𝑥,𝑉) is a constant not depending on 𝑥.
• If 𝑉= (1, 1, . . . , 1)⊥, then dist(𝑥,𝑉) = |𝑥1 + · · · + 𝑥𝑛|/√𝑛which converge to |𝑍| for 𝑍∼𝑁(0, 1). In particular, it is 𝑂(1)-subGaussian.
152 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.5 Talagrand’s inequality • More generally, if fora hyperplane𝑉= 𝛼⊥for some unitvector 𝛼= (𝛼1, . . . , 𝛼𝑛) ∈ R𝑛, one has dist(𝑥,𝑉) = |𝛼· 𝑥|. Note that flipping 𝑥𝑖changes |𝛼· 𝑥| by at most 2|𝛼𝑖|. So by the bounded differences inequality Theorem 9.1.3, for every 𝑡≥0, P(|dist(𝑥,𝑉) −E dist(𝑥,𝑉)| ≥𝑡) ≤2 exp −2𝑡2 4(𝛼2 1 + · · · + 𝛼2 𝑛) !
≤2𝑒−𝑡2/2.
So again dist(𝑥,𝑉) is 𝑂(1)-subGaussian.
What about higher codimensional subspaces 𝑉? Then dist(𝑥,𝑉) = sup 𝛼∈𝑉⊥ |𝛼|=1 |𝛼· 𝑥| .
It is not clear how to apply the bounded difference inequality to all such 𝛼in the above supremum simultaneously.
The bounded difference inequality applied to the function 𝑥∈{−1, 1}𝑛↦→dist(𝑥,𝑉), which is 2-Lipschitz (with respect to Hamming distance), gives P (|dist(𝑥,𝑉) −E dist(𝑥,𝑉)| ≥𝑡) ≤2𝑒−𝑡2/(2𝑛), showing that dist(𝑥,𝑉) is 𝑂(√𝑛)-subGaussian—but this is a pretty bad result, as |dist(𝑥,𝑉)| ≤√𝑛(half the length of the longest diagonal of the cube).
Perhaps the reason why the above bound is so poor is that the bounded difference inequality is measuring distance in {−1, 1}𝑛using the Hamming distance (ℓ1) whereas we really care about the Euclidean distance (ℓ2).
If, instead of sampling 𝑥∈{−1, 1}𝑛, we took 𝑥to be a uniformly random point on the radius √𝑛sphere in R𝑛(which contains {−1, 1}𝑛), then Lévy concentration on the sphere (Corollary 9.4.14) implies that dist(𝑥,𝑉) is 𝑂(1)-subGaussian. Perhaps a similar bound holds when 𝑥is chosen from {−1, 1}𝑛?
Here is a corollary of Talagrand’s inequality, which we will state in its general form later.
Theorem 9.5.2 Let 𝑉be a fixed 𝑑-dimensional subspace in R𝑛. For uniformly random 𝑥∈{−1, 1}𝑛, one has P | dist(𝑥,𝑉) − √ 𝑛−𝑑| ≥𝑡 ≤𝐶𝑒−𝑐𝑡2, where 𝐶, 𝑐> 0 are some constants.
153 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Convex Lipschitz functions of independent random variables Let us now state Talagrand’s inequality, first in a special case for convex functions, and then more generally. Below dist(·, ·) means Euclidean distance.
Theorem 9.5.3 (Talagrand) Let 𝐴⊆R𝑛be convex. Let 𝑥∼Unif{0, 1}𝑛. Then for any 𝑡≥0, P(𝑥∈𝐴)P(dist(𝑥, 𝐴) ≥𝑡) ≤𝑒−𝑡2/4.
Remark 9.5.4.
(1) Note that 𝐴is a convex body in R𝑛and not simply a set of points in 𝐴.
(2) The bounded differences inequality gives us an upper bound of the form 𝑒−𝑐𝑡2/𝑛, which is much worse than Talagrand’s bound.
Example 9.5.5 (Talagrand’s inequality fails for nonconvex sets). Let 𝐴= n 𝑥∈{0, 1}𝑛: wt(𝑥) ≤𝑛 2 −√𝑛 o (here 𝐴is a discrete set of points and not their convex hull). Then for every 𝑦∈{0, 1}𝑛 with wt(𝑦) ≥𝑛/2, one has dist(𝑦, 𝐴) ≥𝑛1/4 (note that this is Euclidean distance and not Hamming distance).
Using the central limit theorem, we have, for some constant 𝑐> 0 and sufficiently large 𝑛, for 𝑥∼Uniform({−1, 1}𝑛), P(𝑥∈𝐴) ≥𝑐 and P(wt(𝑥) ≥𝑛/2) ≥1/2, so the conclusion of Talagrand’s inequality is false for 𝑡= 𝑛1/4, in the case of this nonconvex 𝐴.
By an argument similar to our proof of Theorem 9.4.8 (the equivalence of notions of concentration of measure), one can deduce the following consequence.
Corollary 9.5.6 (Talagrand) Let 𝑓: R𝑛→R be a convex and 1-Lipschitz function (with respect to Euclidean distance on R𝑛). Let 𝑥∼Unif{0, 1}𝑛. Then for any 𝑟∈R and 𝑡≥0, P( 𝑓(𝑥) ≤𝑟)P( 𝑓(𝑥) ≥𝑟+ 𝑡) ≤𝑒−𝑡2/4.
Remark 9.5.7. The proof below shows that the assumption that 𝑓is convex can be weakened to 𝑓being quasiconvex, i.e., { 𝑓≤𝑎} is convex for every 𝑎∈R.
Proof that Theorem 9.5.3 and Corollary 9.5.6 are equivalent. Theorem 9.5.3 implies Corollary 9.5.6: take 𝐴= {𝑥: 𝑓(𝑥) ≤𝑟}.
We have 𝑓(𝑥) ≤𝑟+ 𝑡whenever 154 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.5 Talagrand’s inequality dist(𝑎, 𝐴) ≤𝑡since 𝑓is 1-Lipschitz. So P( 𝑓(𝑥) ≤𝑟) = P(𝑥∈𝐴) and P( 𝑓(𝑥) ≥ 𝑟+ 𝑡) ≤P(dist(𝑥, 𝐴) ≥𝑡).
Corollary 9.5.6 implies Theorem 9.5.3: 𝑟= 0 and take 𝑓(𝑥) = dist(𝑥, 𝐴), which is a convex function since 𝐴is convex.
□ Let us write M𝑋to be a median for the random variable 𝑋, i.e., a non-random real so that P(𝑋≥M𝑋) ≥1/2 and P(𝑋≤M𝑋) ≥1/2.
Corollary 9.5.8 (Talagrand) Let 𝑓: R𝑛→R be a convex and 1-Lipschitz function (with respect to Euclidean distance on R𝑛). Let 𝑥∼Unif({0, 1}𝑛). Then P(| 𝑓(𝑥) −M 𝑓(𝑥)| ≥𝑡) ≤4𝑒−𝑡2/4.
Proof. Setting 𝑟= M 𝑓(𝑥) in Corollary 9.5.6 yields P( 𝑓(𝑥) ≥M 𝑓(𝑥) + 𝑡) ≤2𝑒−𝑡2/4.
Setting 𝑟= M 𝑓(𝑥) −𝑡in Corollary 9.5.6 yields P( 𝑓(𝑥) ≤M 𝑓(𝑥) −𝑡) ≤2𝑒−𝑡2/4.
□ Combining the two tail bounds yields the corollary.
Theorem 9.5.2 then follows. Indeed, Corollary 9.5.8 shows that dist(𝑥,𝑉) (which is a convex 1-Lipschitz function of 𝑥∈R𝑛) is 𝑂(1)-subGaussian, which immediately implies Theorem 9.5.2.
Example 9.5.9 (Operator norm of a random matrix). Let 𝐴be a random matrix whose entries are uniform iid from {−1, 1}. Viewing 𝐴↦→∥𝐴∥op as a function R𝑛2 →R, we see that it is convex (since the operator norm is a norm) and 1-Lipschitz (using that ∥·∥op ≤∥·∥HS, where the latter is the Hilbert–Schmidt norm, also known as the Frobenius norm, i.e., the ℓ2-norm of the matrix entries). It follows by Talagrand’s inequality (Corollary 9.5.8) that ∥𝐴∥op is 𝑂(1)-subGaussian about its mean.
Convex distance Talagrand’s inequality has a much more general form, which has far-reaching combi-natorial applications. We need a define a more subtle notion of distance.
We consider Ω = Ω1 × · · · × Ω𝑛with product probability measure (i.e., independent random variables).
155 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Weighted hamming distance: given 𝛼= (𝛼1, . . . , 𝛼𝑛) ∈R𝑛 ≥0, 𝑥, 𝑦∈Ω, we set 𝒅𝜶(𝒙, 𝒚) := ∑︁ 𝑖:𝑥𝑖≠𝑦𝑗 𝛼𝑖 For 𝐴⊆Ω, 𝒅𝜶(𝒙, 𝑨) := inf 𝑦∈𝐴𝑑𝛼(𝑥, 𝑦).
Talagrand’s convex distance between 𝑥∈Ω and 𝐴⊆Ω is defined by 𝒅𝑻(𝒙, 𝑨) := sup 𝛼∈R𝑛 ≥0 |𝛼|=1 𝑑𝛼(𝑥, 𝐴).
Here |𝛼| denotes Euclidean length: |𝛼|2 := 𝛼2 1 + · · · + 𝛼2 𝑛.
Example 9.5.10 (Euclidean distance to convex hull). If 𝐴⊆{0, 1}𝑛and 𝑥∈{0, 1}𝑛, then 𝑑𝑇(𝑥, 𝐴) is the Euclidean distance from 𝑥to the convex hull of 𝐴.
Let us give another interpretation of convex distance. For 𝑥, 𝑦∈Ω, let 𝜙𝑥(𝑦) = (1𝑥1≠𝑦1, 1𝑥2≠𝑦2, . . . , 1𝑥𝑛≠𝑦𝑛) ∈{0, 1}𝑛 be the vector of coordinatewise disagreements between 𝑥and 𝑦. Write 𝜙𝑥(𝐴) = {𝜙𝑥(𝑦) : 𝑦∈𝐴} ⊆{0, 1}𝑛.
Then for any 𝛼∈R𝑛 ≥0, 𝑑𝛼(𝑥, 𝐴) = 𝑑𝛼(® 0, 𝜙𝑥(𝐴)), where the LHS is the weighted Hamming distance in Ω whereas the RHS takes place in {0, 1}𝑛. Taking the supremum over 𝛼∈R𝑛 ≥0 with |𝛼| = 1, and using the Example 9.5.10, we deduce 𝑑𝑇(𝑥, 𝐴) = dist(® 0, ConvexHull 𝜙𝑥(𝐴)).
The general form of Talagrand’s inequality says the following. Note that it reduces to the earlier special case Theorem 9.5.3 if Ω = {0, 1}𝑛.
156 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.5 Talagrand’s inequality Theorem 9.5.11 (Talagrand’s inequality: general form) Let 𝐴⊆Ω = Ω1 × · · · × Ω𝑛, with Ω equipped with a product probability measure. Let 𝑥∈Ω be chosen randomly with independent coordinates. Let 𝑡≥0. Then P(𝑥∈𝐴)P(𝑑𝑇(𝑥, 𝐴) ≥𝑡) ≤𝑒−𝑡2/4.
Let us see how Talagrand’s inequality recovers a more general form of our geometric inequalities from earlier, extending from independent boolean random variables to independent bounded random variables.
Lemma 9.5.12 (Convex distance upper bounds Euclidean distance) Let 𝐴⊆[0, 1]𝑛and 𝑥∈[0, 1]𝑛. Then dist(𝑥, ConvexHull 𝐴) ≤𝑑𝑇(𝑥, 𝐴).
Proof. For any 𝛼∈R𝑛, and any 𝑦∈[0, 1]𝑛, we have |(𝑥−𝑦) · 𝛼| ≤ 𝑛 ∑︁ 𝑖=1 |𝛼𝑖| |𝑥𝑖−𝑦𝑖| ≤ 𝑛 ∑︁ 𝑖:𝑥𝑖≠𝑦𝑖 |𝛼𝑖| .
First taking the infimum over all 𝑦∈𝐴, and then taking the supremum over unit vectors 𝛼, the LHS becomes dist(𝑥, ConvexHull 𝐴) and the RHS becomes 𝑑𝑇(𝑥, 𝐴).
□ Corollary 9.5.13 (Talagrand’s inequality: convex sets and convex Lipschitz func-tions) Let 𝑥= (𝑥1, . . . , 𝑥𝑛) ∈[0, 1]𝑛be independent random variables (not necessarily identical). Let 𝑡≥0. Let 𝐴⊆[0, 1]𝑛be a convex set. Then P(𝑥∈𝐴)P(dist(𝑥, 𝐴) ≥𝑡) ≤𝑒−𝑡2/4 where dist is Euclidean distance. Also, if 𝑓: [0, 1]𝑛→R is a convex 1-Lipschitz function, then P(| 𝑓−M 𝑓| ≥𝑡) ≤4𝑒−𝑡2/4.
Here is a form of Talagrand’s inequality that is useful for combinatorial applications.
Below, one should think of 𝑓(𝑥) as the value of some optimization problem on some random input 𝑥. There is a hypothesis on how much 𝑓(𝑥) can change if we alter 𝑥.
An example that we will examine in the next section is the length of the shortest tour through 𝑛random points in the unit square (the Euclidean traveling salesman problem).
157 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Theorem 9.5.14 (Talagrand’s inequality — functions with weighted certificates) Let Ω = Ω1 × · · · × Ω𝑛equipped with the product measure. Let 𝑓: Ω →R be a function. Suppose for every 𝑥∈Ω, there is some 𝛼(𝑥) = (𝛼1(𝑥), . . . , 𝛼𝑛(𝑥)) ∈R𝑛 ≥0 such that 𝑓(𝑦) ≥𝑓(𝑥) − ∑︁ 𝑖:𝑥𝑖≠𝑦𝑖 𝛼𝑖(𝑥) for all 𝑦∈Ω.
Then, for every 𝑡≥0, (recall |𝛼|2 = Í𝑛 𝑖=1 𝛼𝑖(𝑥)2) P (| 𝑓−M 𝑓| ≥𝑡) ≤4𝑒−𝑡2/𝐾2 where 𝐾= 2 sup 𝑥∈Ω |𝛼(𝑥)| .
Remark 9.5.15. By considering −𝑓instead of 𝑓, we can change the hypothesis on 𝑓 to 𝑓(𝑦) ≤𝑓(𝑥) + ∑︁ 𝑖:𝑥𝑖≠𝑦𝑖 𝛼𝑖(𝑥) for all 𝑦∈Ω.
Note that 𝑥and 𝑦play asymmetric roles.
Remark 9.5.16. The vector 𝛼(𝑥) measures the resilience of 𝑓(𝑥) under changing some coordinates of 𝑥. It is important that we can choose a different weight 𝛼(𝑥) for each 𝑥. In fact, if we do not let 𝛼(𝑥) change with 𝑥, then Theorem 9.5.14 recovers the bounded differences inequality Theorem 9.1.3 up to an unimportant constant factor in the exponent of the bound.
Proof. Let 𝑟∈R. Let 𝐴= {𝑦∈Ω : 𝑓(𝑦) ≤𝑟−𝑡}. Consider an 𝑥∈Ω with 𝑓(𝑥) ≥𝑟.
By hypothesis, there is some 𝛼(𝑥) ∈R𝑛 ≥0 such that 𝑑𝛼(𝑥)(𝑥, 𝑦) ≥𝑓(𝑥) −𝑓(𝑦) ≥𝑡 for all 𝑦∈𝐴.
Taking infimum over 𝑦∈𝐴, we find |𝛼(𝑥)| 𝑑𝑇(𝑥, 𝐴) ≥𝑡.
So 𝑑𝑇(𝑥, 𝐴) ≥ 𝑡 |𝛼(𝑥)| ≥2𝑡 𝐾.
And hence by Talagrand’s inequality Theorem 9.5.11, P( 𝑓≤𝑟−𝑡)P( 𝑓≥𝑟) ≤P(𝑥∈𝐴)P 𝑑𝑇(𝑥, 𝐴) ≥2𝑡 𝐾 ≤𝑒−𝑡2/𝐾2.
158 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.5 Talagrand’s inequality Taking 𝑟= M 𝑓+ 𝑡yields P( 𝑓≥M 𝑓+ 𝑡) ≤2𝑒−𝑡2/𝐾2, and taking 𝑟= M 𝑓yields P( 𝑓≤M 𝑓−𝑡) ≤2𝑒−𝑡2/𝐾2.
Putting them together yields the final result.
□ Largest eigenvalue of a random matrix Theorem 9.5.17 Let 𝐴= (𝑎𝑖𝑗) be an 𝑛× 𝑛symmetric random matrix with independent entries in [−1, 1]. Let 𝜆1(𝑋) denote the largest eigenvalue of 𝐴. Then P(|𝜆1(𝐴) −M𝜆1(𝐴)| ≥𝑡) ≤4𝑒−𝑡2/32.
Proof. We shall verify the hypotheses of Theorem 9.5.14. We would like to come up with a good choice of a weight vector 𝛼(𝐴) for each matrix 𝐴so that for any other symmetric matrix 𝐵with [−1, 1] entries, 𝜆1(𝐵) ≥𝜆1(𝐴) − ∑︁ 𝑖≤𝑗:𝑎𝑖𝑗≠𝑏𝑖𝑗 𝛼𝑖,𝑗.
(9.1) Note that in a random symmetric matrix we only have 𝑛(𝑛+ 1)/2 independent random entries: the entries below the diagonal are obtained by reflecting the upper diagonal entries.
Let 𝑣= 𝑣(𝐴) be the unit eigenvector of 𝐴corresponding to the eigenvalue 𝜆1(𝐴).
Then, by the Courant–Fischer characterization of eigenvalues, 𝑣⊺𝐴𝑣= 𝜆1(𝐴) and 𝑣⊺𝐵𝑣≤𝜆1(𝐵).
Thus 𝜆1(𝐴) −𝜆1(𝐵) ≤𝑣⊺(𝐴−𝐵)𝑣≤ ∑︁ 𝑖,𝑗:𝑎𝑖𝑗≠𝑏𝑖𝑗 |𝑣𝑖||𝑣𝑗| 𝑎𝑖𝑗−𝑏𝑖𝑗 ≤ ∑︁ 𝑖,𝑗:𝑎𝑖𝑗≠𝑏𝑖𝑗 2|𝑣𝑖||𝑣𝑗|.
Thus (9.1) holds for the vector 𝛼(𝐴) = (𝛼𝑖𝑗)𝑖≤𝑗defined by 𝛼𝑖𝑗= ( 4|𝑣𝑖||𝑣𝑗| if 𝑖< 𝑗 2|𝑣𝑖|2 if 𝑖= 𝑗.
159 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure We have ∑︁ 𝑖≤𝑗 𝛼2 𝑖𝑗≤8 ∑︁ 𝑖,𝑗 |𝑣𝑖|2|𝑣𝑗|2 = 8 ∑︁ 𝑖 |𝑣𝑖|2 !2 = 8.
So Theorem 9.5.14 yields the result.
□ Remark 9.5.18. If 𝐴has mean zero entries, then a moments computation shows that E𝜆1(𝐴) = 𝑂(√𝑛) (the constant can be computed as well). A much more advanced fact is that, say for uniform {−1, 1} entries, the true scale of fluctuation is 𝑛−1/6, and when normalized, the distribution converges to something known as the Tracy–Widom distribution. This limiting distribution is “universal” in the sense that it occurs in many naturally occurring problems, including the next example.
Certifiable functions and longest increasing subsequence An increasing subsequence of a permutation 𝜎= (𝜎1, . . . , 𝜎𝑛) is defined to be some (𝜎𝑖1, . . . , 𝜎𝑖ℓ) for some 𝑖1 < · · · < 𝑖ℓ.
Question 9.5.19 How well is the length 𝑋of the longest increasing subsequence of uniform random permutation concentrated?
While the entries of 𝜎are not independent, we can generate a uniform random permu-tation by taking iid uniform 𝑥1, . . . , 𝑥𝑛∼Unif[0, 1] and let 𝜎record the ordering of the 𝑥𝑖’s. This trick converts the problem into one about independent random variables.
We leave it as an exercise to deduce that 𝑋is Θ(√𝑛) whp.
Changing one of the 𝑥𝑖’s changes LIS by at most 1, so the bounded differences inequality tells us that 𝑋is 𝑂(√𝑛)-subGaussian. Can we do better?
The assertion that a permutation has an increasing permutation of length 𝑠can be checked by verifying 𝑠coordinates of the permutation. Talagrand’s inequality tells us that in such situations the typical fluctuation should be on the order 𝑂( √ M𝑋), or 𝑂(𝑛1/4) in this case.
Definition 9.5.20 Let Ω = Ω1 × · · · × Ω𝑛. Let 𝐴⊆Ω. We say that 𝐴is 𝒔-certifiable if for every 𝑥∈𝐴, there exists a set 𝐼(𝑥) ⊆[𝑛] with |𝐼| ≤𝑠such that for every 𝑦∈Ω with 𝑥𝑖= 𝑦𝑖for all 𝑖∈𝐼(𝑥), one has 𝑦∈𝐴.
For example, for a random permutation as earlier, having an increasing subsequence of length ≥𝑠is 𝑠-certifiable (namely by the indices of the length 𝑠increasing subse-quence).
160 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.5 Talagrand’s inequality Theorem 9.5.21 (Talagrand’s inequality for certifiable functions) Let Ω = Ω1 × · · · × Ω𝑛be equipped with a product measure. Let 𝑓: Ω →R be 1-Lipschitz with respect to Hamming distance on Ω. Suppose that { 𝑓≥𝑟} is 𝑠-certifiable. Then, for every 𝑡≥0, P( 𝑓≤𝑟−𝑡)P( 𝑓≥𝑟) ≤𝑒−𝑡2/(4𝑠).
Proof. Let 𝐴, 𝐵⊆Ω be given by 𝐴= {𝑥: 𝑓(𝑥) ≤𝑟−𝑡} and 𝐵= {𝑦: 𝑓(𝑦) ≥𝑟}. For every 𝑦∈𝐵, let 𝐼(𝑦) ⊆[𝑛] denote a set of ≤𝑠coordinates that certify 𝑓≥𝑟. Due to 𝑓being 1-Lipschitz, we see that every 𝑥∈𝐴disagrees with 𝑦on ≥𝑡coordinates of 𝐼(𝑦).
For every 𝑦∈𝐵, let 𝛼(𝑦) be the indicator vector for 𝐼(𝑦) normalized in length to a unit vector. Then for any 𝑥∈𝐴, 𝑑𝛼(𝑥, 𝑦) = |{𝑖∈𝐼(𝑦) : 𝑥𝑖≠𝑦𝑖}| √︁ |𝐼| ≥ 𝑡 √𝑠.
Thus 𝑑𝑇(𝑦, 𝐴) ≥𝑡/√𝑠. Thus P( 𝑓≤𝑟−𝑡)P( 𝑓≥𝑟) ≤P(𝐴)P(𝐵) ≤P(𝑥∈𝐴)P(𝑑𝑇(𝑥, 𝐴) ≥𝑡/√𝑠) ≤𝑒−𝑡2/(4𝑠) by Talagrand’s inequality (Theorem 9.5.11).
□ Corollary 9.5.22 (Talagrand’s inequality for certifiable functions) Let Ω = Ω1 × · · · × Ω𝑛be equipped with a product measure. Let 𝑓: Ω →R be 1-Lipschitz with respect to Hamming distance on Ω. Suppose { 𝑓≥𝑟} is 𝑟-certifiable for every 𝑟. Then for every 𝑡≥0, P( 𝑓≤M 𝑓−𝑡) ≤2 exp −𝑡2 4M 𝑓 and P( 𝑓≥M 𝑓+ 𝑡) ≤2 exp −𝑡2 4(M 𝑓+ 𝑡) .
Proof. Applying the previous theorem, we have, for every 𝑟∈R and every 𝑡≥0, P( 𝑓≤𝑟−𝑡)P(𝑋≥𝑟) ≤exp −𝑡2 4𝑟 .
161 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Setting 𝑟= M 𝑓, we obtain the lower tail.
P( 𝑓≤M 𝑓−𝑡) ≤2 exp −𝑡2 4M 𝑓 .
Setting 𝑟= M 𝑓+ 𝑡, we obtain the upper tail P(𝑋≥M 𝑓+ 𝑡) ≤2 exp −𝑡2 4(M 𝑓+ 𝑡) .
□ We can apply the above corollary to [0, 1]𝑛with 𝑓being the length of the longest subsequence. Then 𝑓≥𝑟is 𝑟-certifiable. It is also easy to deduce that M 𝑓= 𝑂(√𝑛).
The above tail bounds give us a concentration window of width 𝑂(𝑛1/4).
Corollary 9.5.23 (Longest increasing subsequence) Let 𝑋be the length of the longest increasing subsequence of a random permutation of [𝑛]. Then for every 𝜀> 0 there exists 𝐶> 0 so that P(|𝑋−M𝑋| ≤𝐶𝑛1/4) ≥1 −𝜀.
Remark 9.5.24. The distribution of the length 𝑋of longest increasing subsequence of a uniform random permutation is now well understood through some deep results.
Vershik and Kerov (1977) showed that E𝑋∼2√𝑛.
Baik, Deift, and Johansson (1999) showed that the correct scaling factor is 𝑛1/6, and furthermore, 𝑛−1/6(𝑋−2√𝑛) converges to the Tracy–Widom distribution, the same distribution for the top eigenvalue of a random matrix.
9.6 Euclidean traveling salesman problem Given points 𝑥1, . . . , 𝑥𝑛∈[0, 1]2, let 𝐿(𝑥1, . . . , 𝑥𝑛) = 𝐿({𝑥1, . . . , 𝑥𝑛}) denote the length of the shortest tour through all given points and returns to its starting point.
Equivalently, 𝐿(𝑥1, . . . , 𝑥𝑛) is the minimum of |𝑥𝜎(1) −𝑥𝜎(2)| + |𝑥𝜎(2) −𝑥𝜎(3)| + · · · + |𝑥𝜎(𝑛) −𝑥𝜎(1)| as 𝜎ranges over all permutations of [𝑛]. This Euclidean traveling salesman problem is NP-hard to solve exactly, although there is a (1 + 𝜀)-factor approximation algorithm with running polynomial time for any constant 𝜀> 0 (Arora 1998).
Let 𝐿𝑛= 𝐿(𝑥1, . . . , 𝑥𝑛) with i.i.d. 𝑥1, . . . , 𝑥𝑛∼Unif([0, 1]2) 162 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.6 Euclidean traveling salesman problem The Mona Lisa TSP challenge.
A tour of 1000 random points.
Exercise: E𝐿𝑛= Θ(√𝑛) Beardwood, Halton, and Hammersley (1959) showed that whp 𝐿𝑛/√𝑛converges to some constant as 𝑛→∞(the exact value of the constant is unknown).
We shall focus on the concentration of 𝐿𝑛.
We will present two methods that illustrate different techniques from this chapter.
Martingale methods The following simple monotonicity property will be important for us: for any 𝑆and 𝑥∈[0, 1]2, 𝐿(𝑆) ≤𝐿(𝑆∪{𝑥}) ≤𝐿(𝑆) + 2 dist(𝑥, 𝑆).
Here is the justification for the second inequality. Let 𝑦be the closest point in 𝑆to 𝑥. Consider a shortest tour through 𝑆of length 𝐿(𝑆). Let us modify this tour by first traversing through it, and when we hit 𝑦, we take a detour excursion from 𝑦to 𝑥and then back to 𝑦. The length of this tour, which contains 𝑆∪{𝑥}, is 𝐿(𝑆) + 2 dist(𝑥, 𝑆), and thus the shortest tour through 𝑆∪{𝑥} has length at most 𝐿(𝑆) + 2 dist(𝑥, 𝑆).
If we simply apply the bounded difference inequality, we find that changing a single 𝑥𝑖might change 𝐿(𝑥1, . . . , 𝑥𝑛) by 𝑂(1) in the worse case, and so 𝐿𝑛is 𝑂(√𝑛)-subGaussian about its mean. This is a trivial result since 𝐿𝑛is typically Θ(√𝑛).
To do better, we apply Azuma’s inequlality to the Doob martingale. The key obser-vation is that the initially revealed points do not affect the conditional expectations by much even in the worst case.
163 © Robert Bosch. All rights reserved. This content is excluded from our Creative Commons license. For more information, see
MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Theorem 9.6.1 (Rhee and Talagrand 1987) 𝐿𝑛is 𝑂( √︁ log 𝑛)-subGaussian about its mean. That is, P(|𝐿𝑛−E𝐿𝑛| ≥𝑡) ≤exp −𝑐𝑡2 log 𝑛 for all 𝑡> 0, where 𝑐> 0 is some constant.
We need the following estimate.
Lemma 9.6.2 Let 𝑆be a set of 𝑘random points chosen independently and uniformly in [0, 1]2. For any (non-random) point 𝑦∈[0, 1]2, one has E dist(𝑦, 𝑆) ≲1 √ 𝑘 .
Proof. We have E dist(𝑦, 𝑆) = ∫√ 2 0 P(dist(𝑦, 𝑆) ≥𝑡) 𝑑𝑡 = ∫√ 2 0 1 −area 𝐵(𝑦, 𝑡) ∩[0, 1]2 𝑘 𝑑𝑡 ≤ ∫√ 2 0 exp −𝑘area 𝐵(𝑦, 𝑡) ∩[0, 1]2 𝑑𝑡 ≤ ∫∞ 0 exp −Ω(𝑘𝑡2) 𝑑𝑡≲1 √ 𝑘 .
□ Proof of Theorem 9.6.1. Let 𝐿𝑛,𝑖(𝑥1, . . . , 𝑥𝑖) = E [𝐿𝑛(𝑥1, . . . , 𝑥𝑛) | 𝑥1, . . . , 𝑥𝑖] be the expectation of 𝐿𝑛conditional on the first 𝑖points (and averaging over the remaining 𝑛−𝑖points).
Claim. 𝐿𝑛,𝑖is 𝑂 1 √ 𝑛−𝑖+1 -Lipschitz with respect to Hamming distance.
We have 𝐿(𝑥1, . . . , 𝑥𝑖, . . . 𝑥𝑛) ≤𝐿(𝑥1, . . . , 𝑥′ 𝑖, . . . 𝑥𝑛) + 2 dist(𝑥𝑖, {𝑥1, . . . , 𝑥𝑖−1, 𝑥𝑖+1, . . . , 𝑥𝑛}) ≤𝐿(𝑥1, . . . , 𝑥𝑖, . . . 𝑥𝑛) + ( 2 dist(𝑥𝑖, {𝑥𝑖+1, . . . , 𝑥𝑛}) if 𝑖< 𝑛 𝑂(1) if 𝑖= 𝑛.
164 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.6 Euclidean traveling salesman problem Taking expectation over 𝑥𝑖+1, . . . , 𝑥𝑛, and applying the previous lemma, we find that 𝐿𝑛,𝑖(𝑥1, . . . , 𝑥𝑖) ≤𝐿𝑛,𝑖(𝑥1, . . . , 𝑥𝑖−1, 𝑥′ 𝑖) + 𝑂 1 √ 𝑛−𝑖+ 1 .
This proves the claim. Thus the Doob martingale 𝑍𝑖= E [𝐿𝑛(𝑥1, . . . , 𝑥𝑛) | 𝑥1, . . . , 𝑥𝑖] = 𝐿𝑛,𝑖(𝑥1, . . . , 𝑥𝑖) satisfies |𝑍𝑖−𝑍𝑖−1| ≲ 1 √ 𝑛−𝑖+ 1 for each 1 ≤𝑖≤𝑛.
Now we apply Azuma’s inequality (Theorem 9.2.8). Since 𝑛 ∑︁ 𝑖=1 1 √ 𝑛−𝑖+ 1 2 = 𝑂(log 𝑛), we deduce that 𝑍𝑁= 𝐿𝑛is 𝑂( √︁ log 𝑛)-subGaussian about its mean.
□ Talagrand’s inequality Using Talagrand’s inequality, we will prove the following stronger estimate.
Theorem 9.6.3 (Rhee and Talagrand 1989) 𝐿𝑛is 𝑂(1)-subGaussian about its mean. That is, P(|𝐿𝑛−E𝐿𝑛| ≥𝑡) ≤𝑒−𝑐𝑡2 for all 𝑡> 0, where 𝑐> 0 is some constant.
Remark 9.6.4. Rhee (1991) showed that this tail bound is sharp.
The proof below, following Steele (1997), applies the “space-filling curve heuristic.” A space-filling curve is a continuous surjection from [0, 1] to [0, 1]2. Peano (1890) constructed the first space-filling curve. Hilbert (1891) constructed another space-filling curve known as the Hilbert curve. We will not give a precise description of the Hilbert curve here. Intuitively, the Hilbert curve is the pointwise limit of a sequence of piecewise linear curves illustrated in Figure 9.2. I recommend this 3Blue1Brown video on YouTube for a beautiful animation of the Hilbert curve along with applications.
We will only need the following property of the Hilbert space filling curve.
165 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure Figure 9.2: The Hilbert space-filling curve is the limit of discrete curves illus-trated.
Definition 9.6.5 (Hölder continuity) Given two metric spaces (𝑋, 𝑑𝑋) and (𝑌, 𝑑𝑌), we say that a map 𝑓: 𝑋→𝑌is Hölder continuous with exponent 𝜶if there is some constant 𝐶(depending on 𝑓) so that 𝑑𝑌( 𝑓(𝑥), 𝑓(𝑥′)) ≤𝐶𝑑𝑋(𝑥, 𝑥′)𝛼 for all 𝑥, 𝑥′ ∈𝑋.
Remark 9.6.6. Hölder continuity with exponent 1 is the same as Lipschitz continuity.
Often 𝑋has bounded diameter, in which case if 𝑓is Hölder continuous with exponent 𝛼, then it is so with any exponent 𝛼′ < 𝛼.
Theorem 9.6.7 The Hilbert curve 𝐻: [0, 1] →[0, 1]2 is Hölder continuous with exponent 1/2.
Proof sketch. The Hilbert space-filling curve 𝐻sends every interval of the form [(𝑖−1)/4𝑛, 𝑖/4𝑛] to a square of the form [( 𝑗−1)/2𝑛, 𝑗/2𝑛] × [(𝑘−1)/2𝑛, 𝑘/2𝑛].
Indeed, for each fixed 𝑛, the discrete curves eventually all have this property.
Let 𝑥, 𝑦∈[0, 1], and let 𝑛be the largest integer so that 𝑥, 𝑦∈[(𝑖−1)/4𝑛, (𝑖+ 1)/4𝑛] for some integer 𝑖. Then |𝑥−𝑦| = Θ(4−𝑛), and |𝐻(𝑥) −𝐻(𝑦)| ≲2−𝑛. Thus |𝐻(𝑥) −𝐻(𝑦)| ≲|𝑥−𝑦|1/2.
□ 166 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.6 Euclidean traveling salesman problem Remark 9.6.8. If a space filling space is Hölder continuous with exponent 𝛼, then 𝛼≤1/2. Indeed, the images of the intervals [(𝑖−1)/𝑘, 𝑖/𝑘], 𝑖= 1, . . . , 𝑘, cover the unit square, and thus one intervals must have image diameter ≳1/ √ 𝑘.
Lemma 9.6.9 (Space-filling curve heuristic) Let 𝑥1, . . . , 𝑥𝑛∈[0, 1]2. There is a permutation of 𝜎of [𝑛] with (indices taken mod 𝑛) 𝑛 ∑︁ 𝑖=1 𝑥𝜎(𝑖) −𝑥𝜎(𝑖+1) 2 = 𝑂(1).
Proof. Order the points as they appear on the Hilbert space filling curve 𝐻: [0, 1] → [0, 1]2 (since 𝐻is not injective, there is more than one possible order). Then, there exist 0 ≤𝑡1 ≤𝑡2 ≤· · · ≤𝑡𝑛≤1 so that 𝐻(𝑡𝑖) = 𝑥𝜎(𝑖) for each 𝑖. Using that 𝐻is Hölder continuous with exponent 1/2, we have 𝑛 ∑︁ 𝑖=1 𝑥𝜎(𝑖) −𝑥𝜎(𝑖+1) 2 = 𝑛 ∑︁ 𝑖=1 |𝐻(𝑡𝑖) −𝐻(𝑡𝑖+1)|2 ≲ 𝑛 ∑︁ 𝑖=1 |𝑡𝑖−𝑡𝑖+1| ≤2.
□ Remark 9.6.10. We leave it as an exercise to find an elementary proof of the lemma without invoking the existence of a space-filling curve. Hint: consider a finite approx-imation of the Hilbert curve.
Using Talagrand’s inequality in the form of Theorem 9.5.14, to prove Theorem 9.6.3 that 𝐿𝑛is 𝑂(1)-subGaussian, it suffices to prove the following lemma.
Lemma 9.6.11 Let Ω = ([0, 1]2)𝑛be the space of 𝑛-tuples of points in [0, 1]2. There exists a map 𝛼: Ω →R𝑛 ≥0 so that for all 𝑥∈Ω, 𝛼(𝑥) = (𝛼1(𝑥), . . . , 𝛼𝑛(𝑥)) ∈R𝑛 ≥0 satisfies 𝐿(𝑥) ≤𝐿(𝑦) + ∑︁ 𝑖:𝑥𝑖≠𝑦𝑖 𝛼𝑖(𝑥) for all 𝑥, 𝑦∈Ω (9.1) and sup 𝑥∈Ω 𝑛 ∑︁ 𝑖=1 𝛼𝑖(𝑥)2 = 𝑂(1).
(9.2) Proof. Let 𝑥= (𝑥1, . . . , 𝑥𝑛) ∈Ω, and let 𝜎be the permutation of [𝑛] given by Lemma 9.6.9, the space-filling curve heuristic. Then 𝜎induces a tour of 𝑥1, . . . , 𝑥𝑛.
Let 𝛼𝑖(𝑥) equal twice the sum of the lengths of the two edges incident to 𝑥𝑖in this tour 167 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure (indices taken mod 𝑛): 𝛼𝑖(𝑥) = 2 𝑥𝑖−𝑥𝜎(𝜎−1(𝑖)+1) + 𝑥𝑖−𝑥𝜎(𝜎−1(𝑖)−1) .
Intuitively, this quantity captures “difficulty to serve” 𝑥𝑖.
Now we prove (9.1). First we take care of the first case when 𝑥𝑖≠𝑦𝑖for all 𝑖: (9.1) follows from 𝐿(𝑥) ≤ 𝑛 ∑︁ 𝑖=1 𝑥𝜎(𝑖) −𝑥𝜎𝑖+1 = 1 2 𝑛 ∑︁ 𝑖=1 𝛼𝑖(𝑥).
Now suppose that 𝑥𝑖= 𝑦𝑖for at least one 𝑖. Suppose we have a tour through 𝑦of length 𝐿(𝑦). Consider, for each 𝑖with 𝑥𝑖≠𝑦𝑖, the point 𝑥𝑖along with the two segments incident to 𝑥𝑖in the 𝜎-induced tour through 𝑥(these are the “new edges”). Starting with an optimal tour through 𝑦, and by making various detours/excursions on the new edges, we can reach all the points of 𝑥, traversing each new edge at most twice. The length of the new tour is at most 𝐿(𝑦) + Í 𝑖:𝑥𝑖≠𝑦𝑖𝛼𝑖(𝑥). This proves (9.1).
Finally, it remains to prove (9.2). By Lemma 9.6.9, 𝑛 ∑︁ 𝑖=1 𝛼𝑖(𝑥)2 ≤4 𝑛 ∑︁ 𝑗=1 𝑥𝜎( 𝑗) −𝑥𝜎( 𝑗+1) + 𝑥𝜎( 𝑗) −𝑥𝜎( 𝑗+1) 2 ≲ 𝑛 ∑︁ 𝑗=1 𝑥𝜎( 𝑗) −𝑥𝜎( 𝑗+1) 2 = 𝑂(1).
□ Exercises 1. Sub-Gaussian tails.
For each part, prove there is some constant 𝑐> 0 so that, for all 𝜆> 0, P(|𝑋−E𝑋| ≥𝜆 √ Var 𝑋) ≤2𝑒−𝑐𝜆2.
a) 𝑋is the number of triangles in 𝐺(𝑛, 1/2).
b) 𝑋is the number of inversions of a uniform random permutation of [𝑛] (an inversion of 𝜎∈𝑆𝑛is a pair (𝑖, 𝑗) with 𝑖< 𝑗and 𝜎(𝑖) > 𝜎( 𝑗)).
2. Prove that for every 𝜀> 0 there exists 𝛿> 0 and 𝑛0 such that for all 𝑛≥𝑛0 and 𝑆1, . . . , 𝑆𝑚⊂[2𝑛] with 𝑚≤2𝛿𝑛and |𝑆𝑖| = 𝑛for all 𝑖∈[𝑚], there exists a function 𝑓: [2𝑛] →[𝑛] so that (1 −𝑒−1 −𝜀)𝑛≤| 𝑓(𝑆𝑖)| ≤(1 −𝑒−1 + 𝜀)𝑛for all 𝑖∈[𝑚].
3. Simultaneous bisections. Fix Δ. Let 𝐺1, . . . , 𝐺𝑚with 𝑚= 2𝑜(𝑛) be connected graphs of maximum degree at most Δ on the same vertex set 𝑉with |𝑉| = 𝑛.
Prove that there exists a partition𝑉= 𝐴∪𝐵so that every 𝐺𝑖has (1+𝑜(1))𝑒(𝐺𝑖)/2 168 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.6 Euclidean traveling salesman problem edges between 𝐴and 𝐵.
4. ★Prove that there is some constant 𝑐> 0 so that for every graph 𝐺with chromatic number 𝑘, letting 𝑆be a uniform random subset of 𝑉and 𝐺[𝑆] the subgraph induced by 𝑆, one has, for every 𝑡≥0, P(𝜒(𝐺[𝑆]) ≤𝑘/2 −𝑡) ≤𝑒−𝑐𝑡2/𝑘.
5. ★Prove that there is some constant 𝑐> 0 so that, with probability 1 −𝑜(1), 𝐺(𝑛, 1/2) has a bipartite subgraph with at least 𝑛2/8 + 𝑐𝑛3/2 edges.
6. Let 𝑘≤𝑛/2 be positive integers and 𝐺an 𝑛-vertex graph with average degree at most 𝑛/𝑘. Prove that a uniform random 𝑘-element subset of the vertices of 𝐺 contains an independent set of size at least 𝑐𝑘with probability at least 1 −𝑒−𝑐𝑘, where 𝑐> 0 is a constant.
7. ★Prove that there exists a constant 𝑐> 0 so that the following holds. Let 𝐺be a 𝑑-regular graph and 𝑣0 ∈𝑉(𝐺). Let 𝑚∈N and consider a simple random walk 𝑣0, 𝑣1, . . . , 𝑣𝑚where each 𝑣𝑖+1 is a uniform random neighbor of 𝑣𝑖. For each 𝑣∈𝑉(𝐺), let 𝑋𝑣be the number times that 𝑣appears among 𝑣0, . . . , 𝑣𝑚. For that for every 𝑣∈𝑉(𝐺) and 𝜆> 0 P © « 𝑋𝑣−1 𝑑 ∑︁ 𝑤∈𝑁(𝑣) 𝑋𝑤 ≥𝜆+ 1ª ® ¬ ≤2𝑒−𝑐𝜆2/𝑚 Here 𝑁(𝑣) is the neighborhood of 𝑣.
8. Prove that for every 𝑘there exists a 2(1+𝑜(1))𝑘/2-vertex graph that contains every 𝑘-vertex graph as an induced subgraph.
9. ★Tighter concentration of chromatic number a) Prove that with probability 1 −𝑜(1), every vertex subset of 𝐺(𝑛, 1/2) with at least 𝑛1/3 vertices contains an independent set of size at least 𝑐log 𝑛, where 𝑐> 0 is some constant.
b) Prove that there exists some function 𝑓(𝑛) and constant 𝐶such that for all 𝑛≥2, P( 𝑓(𝑛) ≤𝜒(𝐺(𝑛, 1/2)) ≤𝑓(𝑛) + 𝐶√𝑛/log 𝑛) ≥0.99.
10. Show that for every 𝜀> 0 there exists 𝐶> 0 so that every 𝑆⊆𝑛with |𝑆| ≥𝜀4𝑛contains four elements with pairwise Hamming distance at least 𝑛−𝐶√𝑛apart.
169 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9 Concentration of Measure 11. Concentration of measure in the symmetric group. Let 𝑈⊆𝑆𝑛be a set of at least 𝑛!/2 permutations of [𝑛]. Let 𝑈𝑡denote the set of permutations that can be obtained starting from some element of 𝑈and then applying at most 𝑡 transpositions. Prove that |𝑈𝑡| ≥(1 −𝑒−𝑐𝑡2/𝑛)𝑛!
for every 𝑡≥0, where 𝑐> 0 is some constant.
Hint: Apply Azuma to a Doob martingale that reveals a random permutation For the remaining exercises in this section, use Talagrand’s inequality 12. Let 𝑄be a subset of the unit sphere in R𝑛. Let x ∈[−1, 1]𝑛be a random vector with independent random coordinates. Let 𝑋= supq∈𝑄⟨x, q⟩. Let 𝑡> 0. Prove that P(|𝑋−M𝑋| ≥𝑡) ≤4𝑒−𝑐𝑡2 where 𝑐> 0 is some constant.
13. First passage percolation. Prove that there are constants 𝑐, 𝐶> 0 so that the following holds. Let 𝐺be a graph, and let 𝑢and 𝑤be two distinct vertices with distance at most ℓbetween them. Every edge of 𝐺is independently assigned some random weight in [0, 1] (not necessarily uniform or identically distributed).
The weight of a path is defined to be the sum of the weights of its edges. Let 𝑋 be the minimum weight of a path from 𝑢to 𝑤using at most ℓedges. Prove that there is some 𝑚∈R so that P(|𝑋−𝑚| ≥𝑡) ≤𝐶𝑒−𝑐𝑡2/ℓ.
14. ★Second largest eigenvalue of a random matrix. Let 𝐴be an 𝑛× 𝑛random symmetric matrix whose entries on and above the diagonal are independent and in [−1, 1]. Show that the second largest eigenvalue 𝜆2(𝐴) satisfies P(|𝜆2(𝐴) −E𝜆2(𝐴)| ≥𝑡) ≤𝐶𝑒−𝑐𝑡2, for every 𝑡≥0, where 𝐶, 𝑐> 0 are constants.
Hint in white: use the Courant–Fischer characterization of the second eigenvalue 15. Longest common subsequence. Let (𝑎1, . . . , 𝑎𝑛) and (𝑏1, . . . , 𝑏𝑚) be two random sequences with independent entries (not necessarily identically distributed). Let 𝑋denote the length of the longest common subsequence, i.e., the largest 𝑘such that there exist 𝑖1 < · · · < 𝑖𝑘and 𝑗1 < · · · < 𝑗𝑘with 𝑥𝑖1 = 𝑦𝑗1, . . . , 𝑥𝑖𝑘= 𝑦𝑗𝑘.
170 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 9.6 Euclidean traveling salesman problem Show that, for all 𝑡≥0, P(𝑋≥M𝑋+ 𝑡) ≤2 exp −𝑐𝑡2 M𝑋+ 𝑡 and P(𝑋≤M𝑋−𝑡) ≤2 exp −𝑐𝑡2 M𝑋 where 𝑐> 0 is some constant.
171 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy My greatest concern was what to call it. I thought of calling it “informa-tion,” but the word was overly used, so I decided to call it “uncertainty.” When I discussed it with John von Neumann, he had a better idea. Von Neumann told me, “You should call it entropy, for two reasons. In the first place your uncertainty function has been used in statistical mechanics under that name, so it already has a name. In the second place, and more important, nobody knows what entropy really is, so in a debate you will always have the advantage.” Claude Shannon, 1971 In this chapter, we look at some neat and powerful applications of entropy to combi-natorics. For a standard introduction to information theory, see the textbook by Cover and Thomas.
10.1 Basic properties We define the (binary) entropy of a discrete random variable as follows.
Definition 10.1.1 Given a discrete random variable 𝑋taking values in 𝑆, with 𝑝𝑠:= P(𝑋= 𝑠), its entropy (or binary entropy to emphasis the base-2 logarithm) is defined to be 𝑯(𝑿) := ∑︁ 𝑠∈𝑆 −𝑝𝑠log2 𝑝𝑠 (by convention if 𝑝𝑠= 0 then the corresponding summand is set to zero).
Remark 10.1.2 (Base of the logarithm). It is also fine to use another base for the logarithm, e.g., the natural log, as long as we are consistent throughout. There is some combinatorial preference for base-2 due to its interpretation as counts bits. For certain results, such as Pinsker’s inequality (which we will unfortunately not cover here), the choice of the base does matter.
173 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy Remark 10.1.3 (Information theoretic interpretation). Intuitively, 𝐻(𝑋) measures the amount of “surprise” in the randomness of 𝑋. It can also be interpreted as the amount of information learned by seeing the random variable 𝑋. A more rigorous interpretation of this intuition is given by the Shannon source coding theorem, which, informally, says that the minimum number of bits needed to encode 𝑛iid copies of 𝑋 is 𝑛𝐻(𝑋) + 𝑜(𝑛).
Here are some basic properties. Throughout we only consider discrete random vari-ables.
The proofs are all routine calculations. It will useful to understand the information theoretic interpretations of these properties.
Lemma 10.1.4 (Uniform bound) 𝐻(𝑋) ≤log2 | support(𝑋)|, with equality if and only if 𝑋is uniformly distributed.
Proof. Let function 𝑓(𝑥) = −𝑥log2 𝑥is concave for 𝑥∈[0, 1]. Let 𝑆= support(𝑋).
Then 𝐻(𝑋) = ∑︁ 𝑠∈𝑆 𝑓(𝑝𝑠) ≤|𝑆| 𝑓 1 |𝑆| ∑︁ 𝑠∈𝑆 𝑝𝑠 !
= |𝑆| 𝑓 1 |𝑆| = log2 |𝑆| .
□ We write 𝐻(𝑋,𝑌) for the entropy of the joint random variables (𝑋,𝑌). In other words, letting 𝑍= (𝑋,𝑌), 𝑯(𝑿,𝒀) := 𝐻(𝑍) = ∑︁ (𝑥,𝑦) −P(𝑋= 𝑥,𝑌= 𝑦) log2 P(𝑋= 𝑥,𝑌= 𝑦).
We can similarly write 𝐻(𝑋1, . . . , 𝑋𝑛) for joint entropy.
Lemma 10.1.5 (Independence) If 𝑋and 𝑌are independent random variables, then 𝐻(𝑋,𝑌) = 𝐻(𝑋) + 𝐻(𝑌).
174 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.1 Basic properties Proof.
𝐻(𝑋,𝑌) = ∑︁ (𝑥,𝑦) −P(𝑋= 𝑥,𝑌= 𝑦) log2 P(𝑋= 𝑥,𝑌= 𝑦) = ∑︁ (𝑥,𝑦) −𝑝𝑥𝑝𝑦log2(𝑝𝑥𝑝𝑦) = ∑︁ (𝑥,𝑦) −𝑝𝑥𝑝𝑦(log2 𝑝𝑥+ log2 𝑝𝑦) = ∑︁ 𝑥 −𝑝𝑥log2 𝑝𝑥+ ∑︁ 𝑦 −𝑝𝑦log2 𝑝𝑦= 𝐻(𝑋) + 𝐻(𝑌).
□ Definition 10.1.6 (Conditional entropy) Given jointly distributed random variables 𝑋and 𝑌, define 𝑯(𝑿|𝒀) := E𝑦[𝐻(𝑋|𝑌= 𝑦)] = ∑︁ 𝑦 P(𝑌= 𝑦)𝐻(𝑋|𝑌= 𝑦) = ∑︁ 𝑦 P(𝑌= 𝑦) ∑︁ 𝑥 −P(𝑋= 𝑥|𝑌= 𝑦) log2 P(𝑋= 𝑥|𝑌= 𝑦) (each line unpacks the previous line.
In the summations, 𝑥and 𝑦range over the supports of 𝑋and 𝑌respectively).
Intuitively, the conditional entropy 𝐻(𝑋|𝑌) measures the amount of additional in-formation in 𝑋not contained in 𝑌.
This is intuition is also captured by the next lemma.
Some important special cases: • If 𝑋= 𝑌, or 𝑋= 𝑓(𝑌), then 𝐻(𝑋|𝑌) = 0.
• If 𝑋and 𝑌are independent, then 𝐻(𝑋|𝑌) = 𝐻(𝑋) • If 𝑋and 𝑌are conditionally independent on 𝑍, then 𝐻(𝑋,𝑌|𝑍) = 𝐻(𝑋|𝑍) + 𝐻(𝑌|𝑍) and 𝐻(𝑋|𝑌, 𝑍) = 𝐻(𝑋|𝑍).
Lemma 10.1.7 (Chain rule) 𝐻(𝑋,𝑌) = 𝐻(𝑋) + 𝐻(𝑌|𝑋) Proof. Writing 𝑝(𝑥, 𝑦) = P(𝑋= 𝑥,𝑌= 𝑦), etc., we have by Bayes’s rule 𝑝(𝑥|𝑦)𝑝(𝑦) = 𝑝(𝑥, 𝑦), 175 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy and so 𝐻(𝑋|𝑌) := E𝑦[𝐻(𝑋|𝑌= 𝑦)] = ∑︁ 𝑦 −𝑝(𝑦) ∑︁ 𝑥 𝑝(𝑥|𝑦) log2 𝑝(𝑥|𝑦) = ∑︁ 𝑥,𝑦 −𝑝(𝑥, 𝑦) log2 𝑝(𝑥, 𝑦) 𝑝(𝑦) = ∑︁ 𝑥,𝑦 −𝑝(𝑥, 𝑦) log2 𝑝(𝑥, 𝑦) + ∑︁ 𝑦 𝑝(𝑦) log2 𝑝(𝑦) = 𝐻(𝑋,𝑌) −𝐻(𝑌).
□ Lemma 10.1.8 (Subadditivity) 𝐻(𝑋,𝑌) ≤𝐻(𝑋) + 𝐻(𝑌), and more generally, 𝐻(𝑋1, . . . , 𝑋𝑛) ≤𝐻(𝑋1) + · · · + 𝐻(𝑋𝑛).
Proof. Let 𝑓(𝑡) = log2(1/𝑡), which is convex. Then 𝐻(𝑋) + 𝐻(𝑌) −𝐻(𝑋,𝑌) = ∑︁ 𝑥,𝑦 −𝑝(𝑥, 𝑦) log2 𝑝(𝑥) −𝑝(𝑥, 𝑦) log2 𝑝(𝑦) + 𝑝(𝑥, 𝑦) log2 𝑝(𝑥, 𝑦) = ∑︁ 𝑥,𝑦 𝑝(𝑥, 𝑦) log2 𝑝(𝑥, 𝑦) 𝑝(𝑥)𝑝(𝑦) = ∑︁ 𝑥,𝑦 𝑝(𝑥, 𝑦) 𝑓 𝑝(𝑥)𝑝(𝑦) 𝑝(𝑥, 𝑦) ≥𝑓 ∑︁ 𝑥,𝑦 𝑝(𝑥, 𝑦) 𝑝(𝑥)𝑝(𝑦) 𝑝(𝑥, 𝑦) !
= 𝑓(1) = 0 More generally, by iterating the above inequality for two random variables, we have 𝐻(𝑋1, . . . , 𝑋𝑛) ≤𝐻(𝑋1, . . . , 𝑋𝑛−1) + 𝐻(𝑋𝑛) ≤𝐻(𝑋1, . . . , 𝑋𝑛−2) + 𝐻(𝑋𝑛−1) + 𝐻(𝑋𝑛) ≤· · · ≤𝐻(𝑋1) + · · · + 𝐻(𝑋𝑛).
□ Remark 10.1.9 (Mutual information). The nonnegative quantity 𝐼(𝑋;𝑌) := 𝐻(𝑋) + 𝐻(𝑌) −𝐻(𝑋,𝑌) is called mutual information. Intuitively, it measures the amount of common infor-mation between 𝑋and 𝑌.
176 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.1 Basic properties Lemma 10.1.10 (Dropping conditioning) 𝐻(𝑋|𝑌) ≤𝐻(𝑋) and more generally, 𝐻(𝑋|𝑌, 𝑍) ≤𝐻(𝑋|𝑍).
Proof. By chain rule and subadditivity, we have 𝐻(𝑋|𝑌) = 𝐻(𝑋,𝑌) −𝐻(𝑌) ≤𝐻(𝑋).
The inequality conditioning on 𝑍follows since the above implies that 𝐻(𝑋|𝑌, 𝑍= 𝑧) ≥𝐻(𝑋|𝑍= 𝑧) holds for every 𝑧, and taking expectation of 𝑧yields 𝐻(𝑋|𝑌, 𝑍) ≤𝐻(𝑋|𝑍).
□ Remark 10.1.11. A related theorem is the data processing inequality: 𝐻(𝑋| 𝑓(𝑌)) ≥ 𝐻(𝑋|𝑌) for any function 𝑓. More generally, 𝑓can be random. In other words, if 𝑋→𝑌→𝑍is a Markov chain, then 𝐻(𝑋|𝑍) ≥𝐻(𝑋|𝑌) (exercise: prove this).
Here are some simple applications of entropy to tail bounds.
Let us denote the entropy of a Bernoulli random variable by 𝐻(𝑝) := 𝐻(Bernoulli(𝑝)) = −𝑝log2 𝑝−(1 −𝑝) log2(1 −𝑝).
0 𝑝 1 0 1 𝐻(𝑝) (This notation 𝐻(·) is standard but unfortunately ambiguous: 𝐻(𝑋) versus 𝐻(𝑝). It is usually clear from context which is meant.) Theorem 10.1.12 If 0 < 𝑘≤𝑛/2, then ∑︁ 0≤𝑖≤𝑘 𝑛 𝑖 ≤2𝐻(𝑘/𝑛)𝑛= 𝑛 𝑘 𝑘 𝑛 𝑛−𝑘 𝑛−𝑘 .
177 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy This bound can be established using our proof technique for Chernoff bound by applying Markov’s inequality to the moment generating function: ∑︁ 0≤𝑖≤𝑘 𝑛 𝑖 ≤(1 + 𝑥)𝑛 𝑥𝑘 for all 𝑥∈[0, 1].
The infimum of the RHS over 𝑥∈[0, 1] is precisely 2𝐻(𝑘/𝑛)𝑛.
Now let us give a purely information theoretic proof to get some practice with entropy.
Proof. Let (𝑋1, . . . , 𝑋𝑛) ∈{0, 1}𝑛be chosen uniformly conditioned on 𝑋1 +· · ·+ 𝑋𝑛≤ 𝑘. Then log2 ∑︁ 0≤𝑖≤𝑘 𝑛 𝑖 = 𝐻(𝑋1, . . . , 𝑋𝑛) ≤𝐻(𝑋1) + · · · + 𝐻(𝑋𝑛).
Each 𝑋𝑖is a Bernoulli with probability P(𝑋𝑖= 1). Note that conditioned on 𝑋1 + · · · + 𝑋𝑛= 𝑚, one has P(𝑋𝑖= 1) = 𝑚/𝑛. Varying over 𝑚≤𝑘≤𝑛/2, we find P(𝑋𝑖= 1) ≤𝑘/𝑛, so 𝐻(𝑋𝑖) ≤𝐻(𝑘/𝑛). Hence log2 ∑︁ 0≤𝑖≤𝑘 𝑛 𝑖 ≤𝐻(𝑘/𝑛)𝑛.
□ Remark 10.1.13. One can extend the above proof to bound the tail of Binomial(𝑛, 𝑝) for any 𝑝. The result can be expressed in terms of the relative entropy (also known as the Kullback–Leibler divergence between two Bernoulli random variables). More concretely, for 𝑋∼Binomial(𝑛, 𝑝), one has log P(𝑋≤𝑛𝑞) 𝑛 ≤−𝑞log 𝑞 𝑝−(1 −𝑞) log 1 −𝑞 1 −𝑝 for all 0 ≤𝑞≤𝑝, and log P(𝑋≥𝑛𝑞) 𝑛 ≤−𝑞log 𝑞 𝑝−(1 −𝑞) log 1 −𝑞 1 −𝑝 for all 𝑝≤𝑞≤1.
10.2 Permanent, perfect matchings, and Steiner triple systems Permanent We define the permanent of an 𝑛× 𝑛matrix 𝐴by per 𝐴:= ∑︁ 𝜎∈𝑆𝑛 𝑛 Ö 𝑖=1 𝑎𝑖,𝜎(𝑖).
178 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.2 Permanent, perfect matchings, and Steiner triple systems The formula for the permanent is simply that of the determinant without the sign factor: det 𝐴:= ∑︁ 𝜎∈𝑆𝑛 sgn(𝜎) 𝑛 Ö 𝑖=1 𝑎𝑖𝜎𝑖.
We’ll consider {0, 1}-valued matrices. If 𝐴is the bipartite adjacency matrix of a bipartite graph, then per 𝐴= the number of perfect matchings.
The following theorem gives an upper bound on the number of perfect matchings of a bipartite graph with a given degree distribution. It was conjectured by Minc (1963) and proved by Brégman (1973).
Theorem 10.2.1 (Brégman–Minc inequality) Let 𝐴= (𝑎𝑖𝑗) ∈{0, 1}𝑛×𝑛, whose 𝑖-th row has sum 𝑑𝑖. Then per 𝐴≤ 𝑛 Ö 𝑖=1 (𝑑𝑖!)1/𝑑𝑖 Note that equality is attained when 𝐴consists diagonal blocks of 1’s (corresponding to perfect matchings in a bipartite graph of the form 𝐾𝑑1,𝑑1 ⊔· · · ⊔𝐾𝑑𝑡,𝑑𝑡).
Let 𝜎be a uniform random permutation of [𝑛] conditioned on 𝑎𝑖𝜎𝑖= 1 for all 𝑖∈[𝑛].
Then log2 per 𝐴= 𝐻(𝜎) = 𝐻(𝜎1, . . . , 𝜎𝑛) = 𝐻(𝜎1) +𝐻(𝜎2|𝜎1) +· · ·+𝐻(𝜎𝑛|𝜎1, . . . , 𝜎𝑛−1).
We have 𝐻(𝜎𝑖|𝜎1, . . . , 𝜎𝑖−1) ≤𝐻(𝜎𝑖) ≤log2 |support 𝜎𝑖| = log2 𝑑𝑖, but this step would be too lossy. In fact, what we just did amounts to a naive worst case counting argument.
The key new idea is to reveal the chosen entries in a uniform random order.
Proof. (Radhakrishnan 1997) Let 𝜎be as earlier. Consider a permutation of 𝜏repre-senting an ordering of the rows of the matrix. Say that 𝑖appears before 𝑗if 𝜏𝑖< 𝜏𝑗.
Let 𝑁𝑖= 𝑁𝑖(𝜎, 𝜏) be the number of ones on row 𝑖that does not lie in the same column as some entry ( 𝑗, 𝜎𝑗) that comes before 𝑖. (Intuitively, 𝑁𝑖is the number of “greedily available” choices for 𝜎𝑖before it is revealed.) 179 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy For any 𝜏, the chain rule gives 𝐻(𝜎) = 𝑛 ∑︁ 𝑖=1 𝐻 𝜎𝑖 𝜎𝑗: 𝑗comes before 𝑖 , and the uniform bound gives 𝐻 𝜎𝑖 𝜎𝑗: 𝑗comes before 𝑖 ≤E𝜎log2 𝑁𝑖.
Let 𝜏vary uniformly over all permutations. Then, 𝐻(𝜎) ≤ 𝑛 ∑︁ 𝑖=1 E𝜎,𝜏log2 𝑁𝑖.
For any fixed 𝜎, as 𝜏varies uniformly over all permutations of [𝑛], 𝑁𝑖varies uniformly over [𝑑𝑖]. (Why?) Thus E𝜏log2 𝑁𝑖= log2 1 + · · · + log2 𝑑𝑖 𝑑𝑖 = log2(𝑑𝑖!) 𝑑𝑖 .
Taking expectation over 𝜎and summing over 𝑖yields log2 per 𝐴= 𝐻(𝜎) ≤ 𝑛 ∑︁ 𝑖=1 E𝜎,𝜏log2 𝑁𝑖≤ 𝑛 ∑︁ 𝑖=1 log2(𝑑𝑖!) 𝑑𝑖 .
□ Corollary 10.2.2 (Kahn and Lovász) Let 𝐺be a graph. Let 𝑑𝑣denote the degree of 𝑣. Then the number pm(𝐺) of perfect matchings of 𝐺satisfies pm(𝐺) ≤ Ö 𝑣∈𝑉(𝐺) (𝑑𝑣!)1/(2𝑑𝑣) = Ö 𝑣∈𝑉(𝐺) pm(𝐾𝑑𝑣,𝑑𝑣)1/(2𝑑𝑣).
Proof. (Alon and Friedland 2008) Brégman’s theorem implies the statement for bipar-tite graphs 𝐺(by considering a bipartition on 𝐺⊔𝐺). For the extension of non-bipartite 𝐺, one can proceed via a combinatorial argument that pm(𝐺⊔𝐺) ≤pm(𝐺× 𝐾2), which is left as an exercise.
□ 180 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.2 Permanent, perfect matchings, and Steiner triple systems The maximum number of Hamilton paths in a tournament Question 10.2.3 What is the maximum possible number of directed Hamilton paths in an 𝑛-vertex tournament?
Earlier we saw that a uniformly random tournament has 𝑛!/2𝑛−1 Hamilton paths in expectation, and hence there is some tournament with at least this many Hamilton paths. This result, due to Szele, is the earliest application of the probabilistic method.
Using Brégman’s theorem, Alon proved a nearly matching upper bound.
Theorem 10.2.4 (Alon 1990) Every 𝑛-vertex tournament has at most 𝑂(𝑛3/2 · 𝑛!/2𝑛) Hamilton paths.
Remark 10.2.5. The upper bound has been improved to 𝑂(𝑛3/2−𝛾𝑛!/2𝑛) for some small constant 𝛾> 0 (Friedgut and Kahn 2005), while the lower bound 𝑛!/2𝑛−1 has been improved by a constant factor (Adler, Alon, and Ross 2001, Wormald 2004). It remains open to close this 𝑛𝑂(1) factor gap.
We first prove an upper bound on the number of Hamilton cycles.
Theorem 10.2.6 (Alon 1990) Every 𝑛-vertex tournament has at most 𝑂(√𝑛· 𝑛!/2𝑛) Hamilton cycles.
Proof. Let 𝐴be an 𝑛× 𝑛matrix whose (𝑖, 𝑗) entry is 1 if 𝑖→𝑗is an edge of the tournament and 0 otherwise. Let 𝑑𝑖be the sum of the 𝑖-th row. Then per 𝐴counts the number of 1-factors (spanning disjoint unions of directed cycles) of the tournament.
So by Brégman’s theorem, we have number of Hamilton cycles ≤per 𝐴≤ 𝑛 Ö 𝑖=1 (𝑑𝑖!)1/𝑑1.
One can check (omitted) that the function 𝑔(𝑥) = (𝑥!)1/𝑥is log-concave, i.e, 𝑔(𝑛)𝑔(𝑛+ 2) ≥𝑔(𝑛+ 1)2 for all 𝑛≥0. Thus, by a smoothing argument, among sequences (𝑑1, . . . , 𝑑𝑛) with sum 𝑛 2 , the RHS above is maximized when all the 𝑑𝑖’s are within 1 of each other, which, by Stirling’s formula, gives 𝑂(√𝑛· 𝑛!/2𝑛).
□ Theorem 10.2.4 then follows by applying the above bound with the following lemma.
181 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy Lemma 10.2.7 Given an 𝑛-vertex tournament with 𝑃Hamilton paths, one can add a new vertex to obtain a (𝑛+ 1)-vertex tournament with at least 𝑃/4 Hamilton cycles.
Proof. Add a new vertex and orient its incident edges uniformly at random. For every Hamilton path in the 𝑛-vertex tournament, there is probability 1/4 that it can be closed up into a Hamilton cycle through the new vertex. The claim then follows by linearity of expectation.
□ Steiner triple systems Definition 10.2.8 (Steiner triple system) A Steiner triple system (STS) of order 𝑛is a 3-uniform hypergraph on 𝑛vertices where every pair of vertices is contained in exactly one triple.
Equivalently: an STS is a decomposition of a complete graph 𝐾𝑛into edge-disjoint triangles.
Example: the Fano plane is an STS of order 7.
It is a classic result that an STS of order 𝑛exists if and only if 𝑛≡1 or 3 mod 6. It is not hard to see that this is necessary, since if an STS of order 𝑛exsits, then 𝑛 2 should be divisible by 3, and 𝑛−1 should be divisible by 2. Keevash (2014+) obtained a significant breakthrough proving the existence of more general designs.
Question 10.2.9 How many STS are there on 𝑛labeled vertices?
We shall prove the following result.
Theorem 10.2.10 (Upper bound on the number of STS — Linial and Luria 2013) The number of Steiner triple systems on 𝑛labeled vertices is at most 𝑛 𝑒2 + 𝑜(1) 𝑛2 .
Remark 10.2.11. Keevash (2018) proved a matching lower bound when 𝑛≡1, 3 (mod 6).
Proof. As in the earlier proof, the idea is to reveal the triples in a random order.
182 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.2 Permanent, perfect matchings, and Steiner triple systems Let 𝑋denote a uniformly chosen STS on 𝑛vertices. We wish to upper bound 𝐻(𝑋).
We encode 𝑋as a tuple (𝑋𝑖𝑗)𝑖< 𝑗∈𝑛 where 𝑋𝑖𝑗is the label of the unique vertex that forms a triple with 𝑖and 𝑗in the STS. Here whenever we write 𝑖𝑗we mean the unordered pair {𝑖, 𝑗}, i.e., an edge of 𝐾𝑛.
Let 𝑦= (𝑦𝑖𝑗)𝑖< 𝑗∈0, 1, and we order the edges of 𝐾𝑛in decreasing 𝑦𝑖𝑗: 𝑘𝑙≺𝑖𝑗 if 𝑦𝑘𝑙> 𝑦𝑖𝑗.
By the chain rule, 𝐻(𝑋) = ∑︁ 𝑖𝑗 𝐻 𝑋𝑖𝑗 𝑋𝑘𝑙: 𝑘𝑙≺𝑖𝑗 .
Let 𝑁𝑖𝑗= 𝑁𝑖𝑗(𝑋, 𝑦) = the number of possibilities for 𝑋𝑖𝑗after revealing 𝑋𝑘𝑙for all 𝑘𝑙≺𝑖𝑗.
By the uniform bound, we have 𝐻(𝑋) ≤ ∑︁ 𝑖𝑗 E𝑋log2 𝑁𝑖𝑗.
Now let 𝑦= (𝑦𝑖𝑗)𝑖< 𝑗∈0, 1 be chosen uniformly at random. We have 𝐻(𝑋) ≤ ∑︁ 𝑖𝑗 E𝑋E𝑦log2 𝑁𝑖𝑗.
Write 𝑦−𝑖𝑗∈0, 1−1 to mean 𝑦with the 𝑖𝑗-coordinate removed. Let us bound E𝑦−𝑖𝑗log2 𝑁𝑖𝑗as a function of 𝑦𝑖𝑗.
We define 𝑖𝑗shows up first in its triple to be the event that 𝑖𝑗≺𝑖𝑘, 𝑗𝑘where 𝑘= 𝑋𝑖𝑗.
We have, for any fixed 𝑋, P𝑦−𝑖𝑗(𝑖𝑗shows up first in its triple) = P𝑦−𝑖𝑗(𝑖𝑗≺𝑖𝑘, 𝑗𝑘) = P𝑦−𝑖𝑗(𝑦𝑖𝑗> 𝑦𝑖𝑘, 𝑦𝑗𝑘) = 𝑦2 𝑖𝑗.
If 𝑖𝑗does not show up first in its triple, then 𝑋𝑖𝑗has exactly one possibility (namely 𝑘) by the time it gets revealed, and so 𝑁𝑖𝑗= 1 and log2 𝑁𝑖𝑗= 0. Thus E𝑦−𝑖𝑗log2 𝑁𝑖𝑗= 𝑦2 𝑖𝑗E𝑦−𝑖𝑗 log2 𝑁𝑖𝑗 𝑖𝑗shows up first in its triple ≤𝑦2 𝑖𝑗log2 E𝑦−𝑖𝑗 𝑁𝑖𝑗 𝑖𝑗shows up first in its triple .
Now we use linearity of expectations (over 𝑦−𝑖𝑗with fixed 𝑋). For each 𝑠∈[𝑛] \ {𝑖, 𝑗, 𝑘}, if 𝑠is available as a possibility for 𝑋𝑖𝑗by the time 𝑋𝑖𝑗is revealed, then none of the six edges of 𝐾𝑛consisting of the two triangle 𝑖𝑠𝑋𝑖𝑗and 𝑗𝑠𝑋𝑗𝑠may occur before 183 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy 𝑋𝑖𝑗; the latter event occurs with probability 𝑦6 𝑖𝑗. So E𝑦−𝑖𝑗 𝑁𝑖𝑗 𝑖𝑗shows up first in its triple ≤1 + (𝑛−3)𝑦6 𝑖𝑗.
Thus E𝑦log2 𝑁𝑖𝑗≤ ∫1 0 𝑦2 𝑖𝑗log2(1 + (𝑛−3)𝑦3 𝑖𝑗) 𝑑𝑦𝑖𝑗= 1 3 ∫1 0 log2(1 + (𝑛−3)𝑡2) 𝑑𝑡.
This integralactually hasa closed-formantiderivative (e.g., check Mathematica/Wolfram Alpha), but it suffices for us to obtain the asymptotics. We have ∫1 0 log2 1 𝑛−3 + 𝑡2 𝑑𝑡→ ∫1 0 log2(𝑡2) 𝑑𝑡= −2 log2 𝑒 as 𝑛→∞by the monotone convergence theorem. Thus E𝑦log2 𝑁𝑖𝑗≤log2(𝑛/𝑒2) + 𝑜(1) 3 .
It follows therefore that the log-number of STS on 𝑛vertices is 𝐻(𝑋) ≤ ∑︁ 𝑖𝑗 E𝑋E𝑦log2 𝑁𝑖𝑗≤ 𝑛 2 log2(𝑛/𝑒2) + 𝑜(1) 3 = 𝑛2 6 log2 𝑛 𝑒2 + 𝑜(1) . □ Remark 10.2.12 (Guessing the formula). Here is perhaps how we might have guessed the formula for the number of STSs. Suppose we select 1 3 𝑛 2 triangles in 𝐾𝑛indepen-dently at random. What is the probability that every edge is contained in exactly one triangle? Each edge is contained one triangle on expectation, and so by the Poisson approximation, the probability that a single fixed edge is contained in exactly one tri-angle is 1/𝑒+ 𝑜(1). Now let us pretend as if all the edges behave independently (!) — the probability that every edge is contained in exactly one triangle is (1/𝑒+ 𝑜(1))(𝑛 2).
This would then lead us to guessing that the number of STSs being 1 1 3 𝑛 2 !
𝑛 3 1 3 (𝑛 2) 1 𝑒+ 𝑜(1) (𝑛 2) = 𝑛2 6𝑒 −𝑛2/6 𝑛3 6 𝑛2/6 1 𝑒 𝑛2/2!1+𝑜(1) = 𝑛 𝑒2 + 𝑜(1) 𝑛2/3 .
Here is another heuristic for getting the formula, and this time this method can actually be turned into a proof of matching lower bound on the number of STSs, though with a lot of work (Keevash 2018). Suppose we remove triangles from 𝐾𝑛one at a time.
After 𝑘triangles have been removed, the number of edges remaining is 𝑛 2 −3𝑘. Let us pretend that the remaining edges were randomly distributed. Then the number of 184 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.3 Sidorenko’s inequality triangles should be about 𝑛 3 1 −3𝑘 𝑛 2 !3 ∼36 𝑛3 1 3 𝑛 2 −𝑘 3 If we multiply the above quantity over 0 ≤𝑘< 1 3 𝑛 2 , and then divide by 1 3 𝑛 2 ! to account for the ordering of the triangles, we get 36 𝑛3 𝑛2/6 1 3 𝑛 2 !3 1 3 𝑛 2 !
≈ 𝑛 𝑒2 + 𝑜(1) 𝑛2/3 .
10.3 Sidorenko’s inequality Given graphs 𝐹and 𝐺, a graph homomorphism from 𝐹to 𝐺is a map 𝜙: 𝑉(𝐹) → 𝑉(𝐺) of vertices that sends edges to edges, i.e., 𝜙(𝑢)𝜙(𝑣) ∈𝐸(𝐺) for all 𝑢𝑣∈𝐸(𝐹).
Let hom(𝐹, 𝐺) = the number of graph homomorphisms from 𝐹to 𝐺.
Define the homomorphism density (the 𝑯-density in 𝑮) by 𝑡(𝐹, 𝐻) = hom(𝐹, 𝐺) 𝑣(𝐺)𝑣(𝐹) = P(a uniform random map 𝑉(𝐹) →𝑉(𝐺) is a graph homomorphism 𝐹→𝐺) In this section, we are interested in the regime of fixed 𝐹and large 𝐺, in which case almost all maps 𝑉(𝐹) →𝑉(𝐺) are injective, so that there is not much difference between homomorphisms and subgraphs. More precisely, hom(𝐹, 𝐺) = aut(𝐹)(#copies of 𝐹in 𝐺as a subgraph) + 𝑂𝐹(𝑣(𝐺)𝑣(𝐹)−1).
where aut(𝐹) is the number of automorphisms of 𝐹.
Inequalities between graph homomorphism densities is a central topic in extremal graph theory. For example, see Chapter 5 of my book Graph Theory and Additive Combinatorics. Much of the rest of this chapter is adapted from §5.5 of the book.
Question 10.3.1 Given a fixed graph 𝐹and constant 𝑝∈[0, 1], what is the minimum possible 𝐹-density in a graph with edge density at least 𝑝?
185 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy The 𝐹-density in the random graph 𝐺(𝑛, 𝑝) is 𝑝𝑒(𝐹) +𝑜(1). Here 𝑝is fixed and 𝑛→∞.
Can one do better?
If 𝐹is non-bipartite, then the complete bipartite graph 𝐾𝑛/2,𝑛/2 has 𝐹-density zero.
(The problem of minimizing 𝐹-density is still interesting and not easy; it has been solved for cliques.) Sidorenko’s conjecture (1993) (also proposed by Erdős and Simonovits (1983)) says for any fixed bipartite 𝐹, the random graph asymptotically minimizes 𝐹-density. This is an important and well-known conjecture in extremal graph theory.
Conjecture 10.3.2 (Sidorenko) For every bipartite graph 𝐹, and any graph 𝐺, 𝑡(𝐹, 𝐺) ≥𝑡(𝐾2, 𝐺)𝑒(𝐹).
The conjecture is known to hold for a large family of graphs 𝐹.
The entropy approach to Sidorenko’s conjecture was first introduced by Li and Szegedy (2011) and later further developed in subsequent works. Here we illustrate the entropy approach to Sidorenko’s conjecture with several examples.
We will construct a probability distribution 𝜇on Hom(𝐹, 𝐺), the set of all graph homomorphisms 𝐹→𝐺. Unlike earlier applications of entropy, here we are trying to prove a lower bound on hom(𝐹, 𝐺) instead of an upper bound. So instead of taking 𝜇to be a uniform distribution (which automatically has entropy log2 hom(𝐹, 𝐺)), we actually take 𝜇to be carefully constructed distribution, and apply the upper bound 𝐻(𝜇) ≤log2 |support 𝜇| = log2 hom(𝐹, 𝐺).
We are trying to show that hom(𝐹, 𝐺) 𝑣(𝐺)𝑣(𝐹) ≥ 2𝑒(𝐺) 𝑣(𝐺)2 𝑒(𝐹) .
So we would like to find a probability distribution 𝜇on Hom(𝐹, 𝐺) satisfying 𝐻(𝜇) ≥𝑒(𝐹) log2(2𝑒(𝐺)) −(2𝑒(𝐹) −𝑣(𝐹)) log2 𝑣(𝐺).
(10.1) Theorem 10.3.3 (Blakey and Roy 1965) Sidorenko’s conjecture holds if 𝐹is a three-edge path.
Proof. We choose randomly a walk 𝑋𝑌𝑍𝑊in 𝐺as follows: 186 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.3 Sidorenko’s inequality • 𝑋𝑌is a uniform random edge of 𝐺(by this we mean first choosing an edge of 𝐺uniformly at random, and then let 𝑋be a uniformly chosen endpoint of this edge, and then 𝑌the other endpoint); • 𝑍is a uniform random neighbor of 𝑌; • 𝑊is a uniform random neighbor of 𝑍.
Key observation: 𝑌𝑍is distributed as a uniform random edge of 𝐺, and likewise with 𝑍𝑊 Indeed, conditioned on the choice of 𝑌, the vertices 𝑋and 𝑍are both independent and uniform neighbors of 𝑌, so 𝑋𝑌and 𝑌𝑍are uniformly distributed.
Also, the conditional independence observation implies that 𝐻(𝑍|𝑋,𝑌) = 𝐻(𝑍|𝑌) and 𝐻(𝑊|𝑋,𝑌, 𝑍) = 𝐻(𝑊|𝑍) and futhermore both quantities are equal to 𝐻(𝑌|𝑋) since 𝑋𝑌,𝑌𝑍, 𝑍𝑊are each dis-tributed as a uniform random edge.
Thus 𝐻(𝑋,𝑌, 𝑍, 𝑊) = 𝐻(𝑋) + 𝐻(𝑌|𝑋) + 𝐻(𝑍|𝑋,𝑌) + 𝐻(𝑊|𝑋,𝑌, 𝑍) [chain rule] = 𝐻(𝑋) + 𝐻(𝑌|𝑋) + 𝐻(𝑍|𝑌) + 𝐻(𝑊|𝑍) [cond indep] = 𝐻(𝑋) + 3𝐻(𝑌|𝑋) = 3𝐻(𝑋,𝑌) −2𝐻(𝑋) [chain rule] ≥3 log2(2𝑒(𝐺)) −2 log2 𝑣(𝐺) In the final step we used 𝐻(𝑋,𝑌) = log2(2𝑒(𝐺)) since 𝑋𝑌is uniformly distributed among edges, and 𝐻(𝑋) ≤log2 |support(𝑋)| = log2 𝑣(𝐺). This proves (10.1) and hence the theorem for a path of 4 vertices. (As long as the final expression has the “right form” and none of the steps are lossy, the proof should work out.) □ Remark 10.3.4. See this MathOverflow discussion for the history as well as alternate proofs.
The above proof easily generalizes to all trees. We omit the details.
Theorem 10.3.5 Sidorenko’s conjecture holds if 𝐹is a tree.
187 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy Theorem 10.3.6 Sidorenko’s conjecture holds for all complete bipartite graphs.
Proof. Following the same framework as earlier, let us demonstrate the result for 𝐹= 𝐾2,2. The same proof extends to all 𝐾𝑠,𝑡.
𝑥1 𝑥2 𝑦1 𝑦2 We will pick a random tuple (𝑋1, 𝑋2,𝑌1,𝑌2) ∈𝑉(𝐺)4 with 𝑋𝑖𝑌𝑗∈𝐸(𝐺) for all 𝑖, 𝑗as follows.
• 𝑋1𝑌1 is a uniform random edge; • 𝑌2 is a uniform random neighbor of 𝑋1; • 𝑋2 is a conditionally independent copy of 𝑋1 given (𝑌1,𝑌2).
The last point deserves more attention. Note that we are not simply uniformly randomly choosing a common neighbor of 𝑌1 and 𝑌2 as one might naively attempt. Instead, one can think of the first two steps as generating a distribution for (𝑋1,𝑌1,𝑌2)—according to this distribution, we first generate (𝑌1,𝑌2) according to its marginal, and then produce two conditionally independent copies of 𝑋1 (the second copy is 𝑋2).
As in the previous proof (applied to a 2-edge path), we see that 𝐻(𝑋1,𝑌1,𝑌2) = 2𝐻(𝑋1,𝑌1) −𝐻(𝑋1) ≥2 log2(2𝑒(𝐺)) −log2 𝑣(𝐺).
So we have 𝐻(𝑋1, 𝑋2,𝑌1,𝑌2) = 𝐻(𝑌1,𝑌2) + 𝐻(𝑋1, 𝑋2|𝑌1,𝑌2) [chain rule] = 𝐻(𝑌1,𝑌2) + 2𝐻(𝑋1|𝑌1,𝑌2) [conditional independence] = 2𝐻(𝑋1,𝑌1,𝑌2) −𝐻(𝑌1,𝑌2) [chain rule] ≥2(2 log2(2𝑒(𝐺)) −log2 𝑣(𝐺)) −2 log2 𝑣(𝐺).
[prev. ineq. and uniform bound] = 4 log(2𝑒(𝐺)) −4 log2 𝑣(𝐺).
So we have verified (10.1) for 𝐾2,2.
□ 188 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.3 Sidorenko’s inequality Theorem 10.3.7 (Conlon, Fox, Sudakov 2010) Sidorenko’s conjecture holds for a bipartite graph that has a vertex adjacent to all vertices in the other part.
Proof. Let us illustrate the proof for the following graph. The proof extends to the general case.
𝑥0 𝑦1 𝑦2 𝑦3 𝑥1 𝑥2 Let us choose a random tuple (𝑋0, 𝑋1, 𝑋2,𝑌1,𝑌2,𝑌3) ∈𝑉(𝐺)6 as follows: • 𝑋0𝑌1 is a uniform random edge; • 𝑌2 and 𝑌3 are independent uniform random neighbors of 𝑋0; • 𝑋1 is a conditionally independent copy of 𝑋0 given (𝑌1,𝑌2); • 𝑋2 is a conditionally independent copy of 𝑋0 given (𝑌2,𝑌3).
(as well as other symmetric versions.) Some important properties of this distribution: • 𝑋0, 𝑋1, 𝑋2 are conditionally independent given (𝑌1,𝑌2,𝑌3); • 𝑋1 and (𝑋0,𝑌3, 𝑋2) are conditionally independent given (𝑌1,𝑌2); • The distribution of (𝑋0,𝑌1,𝑌2) is identical to the distribution of (𝑋1,𝑌1,𝑌2).
We have 𝐻(𝑋0, 𝑋1, 𝑋2,𝑌1,𝑌2,𝑌3) = 𝐻(𝑋0, 𝑋1, 𝑋2|𝑌1,𝑌2,𝑌3) + 𝐻(𝑌1,𝑌2,𝑌3) [chain rule] = 𝐻(𝑋0|𝑌1,𝑌2,𝑌3) + 𝐻(𝑋1|𝑌1,𝑌2,𝑌3) + 𝐻(𝑋2|𝑌1,𝑌2,𝑌3) + 𝐻(𝑌1,𝑌2,𝑌3) [cond indep] = 𝐻(𝑋0|𝑌1,𝑌2,𝑌3) + 𝐻(𝑋1|𝑌1,𝑌2) + 𝐻(𝑋2|𝑌2,𝑌3) + 𝐻(𝑌1,𝑌2,𝑌3) [cond indep] = 𝐻(𝑋0,𝑌1,𝑌2,𝑌3) + 𝐻(𝑋1,𝑌1,𝑌2) + 𝐻(𝑋2,𝑌2,𝑌3) −𝐻(𝑌1,𝑌2) −𝐻(𝑌2,𝑌3).
[chain rule] The proof of Theorem 10.3.3 actually lower bounds the first three terms: 𝐻(𝑋0,𝑌1,𝑌2,𝑌3) ≥3 log2(2𝑒(𝐺)) −2 log2 𝑣(𝐺) 𝐻(𝑋1,𝑌1,𝑌2) ≥2 log2(2𝑒(𝐺)) −log2 𝑣(𝐺) 𝐻(𝑋2,𝑌2,𝑌3) ≥2 log2(2𝑒(𝐺)) −log2 𝑣(𝐺).
189 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy We can apply the uniform support bound on the remaining terms.
𝐻(𝑌1,𝑌2) = 𝐻(𝑌2,𝑌3) ≤2 log2 𝑣(𝐺).
Putting everything together, we have 𝐻(𝑋0, 𝑋1, 𝑋2,𝑌1,𝑌2,𝑌3) ≥7 log2(2𝑒(𝐺)) −8 log2 𝑣(𝐺), thereby verifying (10.1).
□ To check that you understand the above proof, where did we use the assumption that 𝐹has a vertex complete to the other part?
Many other graphs can be proved by extending this method.
Remark 10.3.8 (Möbius graph). An important open case (and the smallest in some sense) of Sidorenko conjecture is when 𝐹is the following graph, known as the Möbius graph. It is 𝐾5,5 with a 10-cycle removed. The name comes from it being the face-vertex incidence graph of the simplicial complex structure of the Möbius strip, built by gluing a strip of five triangles.
Möbius graph = 𝐾5,5 \ 𝐶10 = 10.4 Shearer’s lemma Shearer’s entropy lemma extends the subadditivity property of entropy. Before stating it in full generality, let us first see the simplest instance of Shearer’s lemma.
Theorem 10.4.1 (Shearer’s lemma, special case) 2𝐻(𝑋,𝑌, 𝑍) ≤𝐻(𝑋,𝑌) + 𝐻(𝑋, 𝑍) + 𝐻(𝑌, 𝑍) Proof. Using the chain rule and conditioning dropping, we have 𝐻(𝑋,𝑌) = 𝐻(𝑋) + 𝐻(𝑌|𝑋) 𝐻(𝑋, 𝑍) = 𝐻(𝑋) + 𝐻(𝑍|𝑋) ≥𝐻(𝑋) + 𝐻(𝑍|𝑋, 𝑍) 𝐻(𝑌, 𝑍) = 𝐻(𝑌) + 𝐻(𝑍|𝑌) ≥ 𝐻(𝑌|𝑋) + 𝐻(𝑍|𝑋,𝑌) Applying conditioning dropping, we see that their sum is at at least 2𝐻(𝑋) + 2𝐻(𝑌|𝑋) + 2𝐻(𝑍|𝑋,𝑌) = 2𝐻(𝑋,𝑌, 𝑍).
□ 190 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.4 Shearer’s lemma Question 10.4.2 What is the maximum volume of a body in R3 that has area at most 1 when projected to each of the three coordinate planes?
The cube [0, 1]3 satisfies the above property and has area 1. It turns out that this is the maximum.
To prove this claim, first let us use Shearer’s inequality to prove a discrete version.
Theorem 10.4.3 Let 𝑆⊆R3 be a finite set, and 𝜋𝑥𝑦(𝑆) be its projection on the 𝑥𝑦-plane, etc. Then |𝑆|2 ≤ 𝜋𝑥𝑦(𝑆) |𝜋𝑥𝑧(𝑆)| 𝜋𝑦𝑧(𝑆) Proof. Let (𝑋,𝑌, 𝑍) be a uniform random point of 𝑆. Then 2 log2 |𝑆| = 2𝐻(𝑋,𝑌, 𝑍) ≤𝐻(𝑋,𝑌) + 𝐻(𝑋, 𝑍) + 𝐻(𝑌, 𝑍) ≤log2 𝜋𝑥𝑦(𝑆) + log2 𝜋𝑥𝑧(𝑆) + log2 𝜋𝑦𝑧(𝑆).
□ By approximating a body using cubes, we can deduce the following corollary.
Corollary 10.4.4 Let 𝑆be a body in R3. Then vol(𝑆)2 ≤area(𝜋𝑥𝑦(𝑆)) area(𝜋𝑥𝑧(𝑆)) area(𝜋𝑦𝑧(𝑆)).
Let us now state the general form of Shearer’s lemma. (Chung, Graham, Frankl, and Shearer 1986) Theorem 10.4.5 (Shearer’s lemma) Let 𝐴1, . . . , 𝐴𝑠⊆[𝑛] where each 𝑖∈[𝑛] appears in at least 𝑘sets 𝐴𝑗’s. Writing 𝑋𝐴:= (𝑋𝑖)𝑖∈𝐴, 𝑘𝐻(𝑋1, . . . , 𝑋𝑛) ≤ ∑︁ 𝑗∈[𝑠] 𝐻(𝑋𝐴𝑗).
The proof of the general form of Shearer’s lemma is a straightforward adaptation of the proof of the special case earlier.
Like earlier, we can deduce an inequality about sizes of projections. (Loomis and Whitney 1949) 191 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy Corollary 10.4.6 (Loomis–Whitney inequality) Writing 𝜋𝑖for the projection from R𝑛onto the hyperplane 𝑥𝑖= 0, we have for every 𝑆⊆R𝑛, |𝑆|𝑛−1 ≤ 𝑛 Ö 𝑖=1 |𝜋𝑖(𝑆)| Corollary 10.4.7 Let 𝐴1, . . . , 𝐴𝑠⊆Ω where each 𝑖∈Ω appears in at least 𝑘sets 𝐴𝑗. Then for every family F of subsets of Ω, |F |𝑘≤ Ö 𝑗∈[𝑠] F |𝐴𝑗 where F |𝐴:= {𝐹∩𝐴: 𝐹∈F }.
Proof. Each subsetof Ω corresponds toa vector (𝑋1, . . . , 𝑋𝑛) ∈{0, 1}𝑛. Let (𝑋1, . . . , 𝑋𝑛) be a random vector corresponding to a uniform element of F . Then 𝑘log2 |F | = 𝑘𝐻(𝑋1, . . . , 𝑋𝑛) ≤ ∑︁ 𝑗∈[𝑠] 𝐻(𝑋𝐴𝑗) = log2 F |𝐴𝑗 .
□ Triangle-intersecting families We say that a set G of labeled graphs on the same vertex set is triangle-intersecting if 𝐺∩𝐺′ contains a triangle for every 𝐺, 𝐺′ ∈G.
Question 10.4.8 What is the largest triangle-intersecting family of graphs on 𝑛labeled vertices?
The set of all graphs that contain a fixed triangle is triangle-intersecting, and they form a 1/8 fraction of all graphs.
An easy upper bound: the edges form an intersecting family, so a triangle-intersecting family must be at most 1/2 fraction of all graphs.
The next theorem improves this upper bound to < 1/4. It is also in this paper that Shearer’s lemma was introduced.
Theorem 10.4.9 (Chung, Graham, Frankl, and Shearer 1986) Every triangle-intersecting family of graphs on 𝑛labeled vertices has size < 2(𝑛 2)−2.
192 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.4 Shearer’s lemma Proof. Let G be a triangle-intersecting family of graphs on vertex set [𝑛] (viewed as a collection of subsets of edges of 𝐾𝑛) For 𝑆⊆[𝑛] with |𝑆| = ⌊𝑛/2⌋, let 𝐴𝑆= 𝑆 2 ∪ [𝑛]\𝑆 2 (i.e., 𝐴𝑆is the union of the clique on 𝑆and the clique on the complement of 𝑆). Let 𝑟= |𝐴𝑆| = ⌊𝑛/2⌋ 2 + ⌈𝑛/2⌉ 2 ≤1 2 𝑛 2 .
For every 𝑆, every triangle has an edge in 𝐴𝑆, and thus G restricted to 𝐴𝑆must be an intersecting family. Hence G|𝐴𝑆 ≤2|𝐴𝑆|−1 = 2𝑟−1.
Each edge of 𝐾𝑛appears in at least 𝑘= 𝑟 𝑛 2 𝑛 ⌊𝑛/2⌋ different 𝐴𝑆with |𝑆| = ⌊𝑛/2⌋(by symmetry and averaging).
Applying Corol-lary 10.4.7, we find that |G|𝑘≤ 2𝑟−1( 𝑛 ⌊𝑛/2⌋) .
Therefore |G| ≤2(𝑛 2)−(𝑛 2) 𝑟 < 2(𝑛 2)−2.
□ Remark 10.4.10. A tight upper bound of 2(𝑛 2)−3 (matching the construction of taking all graphs containing a fixed triangle) was conjectured by Simonovits and Sós (1976) and proved by Ellis, Filmus, and Friedgut (2012) using Fourier analytic methods.
Berger and Zhao (2023) gave a tight solution for 𝐾4-intersecting families. The general conjecture for 𝐾𝑟-intersecting families is open.
The number of independent sets in a regular bipartite graph Question 10.4.11 Fix 𝑑. Which 𝑑-regular graph on a given number of vertices has the most number of independent sets? Alternatively, which graph 𝐺maximizes 𝑖(𝐺)1/𝑣(𝐺)?
(Note that the number of independent sets is multiplicative: 𝑖(𝐺1⊔𝐺2) = 𝑖(𝐺1)𝑖(𝐺2).) Alon and Kahn conjectured that for graphs on 𝑛vertices, when 𝑛is a multiple of 2𝑑, a disjoint union of 𝐾𝑑,𝑑’s maximizes the number of independent sets.
193 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy Alon (1991) proved an approximate version of this conjecture. Kahn (2001) proved it assuming the graph is bipartite. Zhao (2010) proved it in general.
Theorem 10.4.12 (Kahn, Zhao) Let 𝐺be an 𝑛-vertex 𝑑-regular graph. Then 𝑖(𝐺) ≤𝑖(𝐾𝑑,𝑑)𝑛/(2𝑑) = (2𝑑+1 −1)𝑛/(2𝑑) where 𝑖(𝐺) is the number of independent sets of 𝐺.
Proof assuming 𝐺is bipartite. (Kahn) Let us first illustrate the proof for 𝐺= 𝑥1 𝑥2 𝑥3 𝑦1 𝑦2 𝑦3 Amongallindependentsetsof𝐺, choose one uniformlyatrandom, andlet (𝑋1, 𝑋2, 𝑋3,𝑌1,𝑌2,𝑌3) ∈ {0, 1}6 be its indicator vector. Then 2 log2 𝑖(𝐺) = 2𝐻(𝑋1, 𝑋2, 𝑋3,𝑌1,𝑌2,𝑌3) = 2𝐻(𝑋1, 𝑋2, 𝑋3) + 2𝐻(𝑌1,𝑌2,𝑌3|𝑋1, 𝑋2, 𝑋3) [chain rule] ≤𝐻(𝑋1, 𝑋2) + 𝐻(𝑋1, 𝑋3) + 𝐻(𝑋2, 𝑋3) + 2𝐻(𝑌1|𝑋1, 𝑋2, 𝑋3) + 2𝐻(𝑌2|𝑋1, 𝑋2, 𝑋3) + 2𝐻(𝑌3|𝑋1, 𝑋2, 𝑋3) [Shearer] = 𝐻(𝑋1, 𝑋2) + 𝐻(𝑋1, 𝑋3) + 𝐻(𝑋2, 𝑋3) + 2𝐻(𝑌1|𝑋1, 𝑋2) + 2𝐻(𝑌2|𝑋1, 𝑋3) + 2𝐻(𝑌3|𝑋2, 𝑋3) [cond indep] Here we are using that (a) 𝑌1,𝑌2,𝑌3 are conditionally independent given (𝑋1, 𝑋2, 𝑋3) and (b) 𝑌1 and (𝑋3,𝑌2,𝑌3) are conditionally independent given (𝑋1, 𝑋2). A more general statement is that if 𝑆⊆𝑉(𝐺), then the restrictions to the different connected components of 𝐺−𝑆are conditionally independent given 𝑋𝑆.
It remains to prove that 𝐻(𝑋1, 𝑋2) + 2𝐻(𝑌1|𝑋1, 𝑋2) ≤log2 𝑖(𝐾2,2) and two other analogous inequalities. Let 𝑌′ 1 be conditionally independent copy of 𝑌1 given (𝑋1, 𝑋2). Then (𝑋1, 𝑋2,𝑌1,𝑌′ 1) is the indictor vector of an independent set of 𝐾2,2 (though not necessarily chosen uniformly).
𝑥1 𝑥2 𝑦1 𝑦′ 1 194 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.4 Shearer’s lemma Thus we have 𝐻(𝑋1, 𝑋2) + 2𝐻(𝑌1|𝑋1, 𝑋2) = 𝐻(𝑋1, 𝑋2) + 𝐻(𝑌1|𝑋1, 𝑋2) + 𝐻(𝑌′ 1|𝑋1, 𝑋2) = 𝐻(𝑋1, 𝑋2,𝑌1,𝑌′ 1) [chain rule] ≤log2 𝑖(𝐺) [uniform bound] This concludes the proof for 𝐺= 𝐾2,2, which works for all bipartite 𝐺. Here are the details.
Let 𝑉= 𝐴∪𝐵be the vertex bipartition of 𝐺. Let 𝑋= (𝑋𝑣)𝑣∈𝑉be the indicator function of an independent set chosen uniformly at random. Write 𝑋𝑆:= (𝑋𝑣)𝑣∈𝑆. We have 𝑑log2 𝑖(𝐺) = 𝑑𝐻(𝑋) = 𝑑𝐻(𝑋𝐴) + 𝑑𝐻(𝑋𝐵|𝑋𝐴) [chain rule] ≤ ∑︁ 𝑏∈𝐵 𝐻(𝑋𝑁(𝑏)) + 𝑑 ∑︁ 𝑏∈𝐵 𝐻(𝑋𝑏|𝑋𝐴) [Shearer] ≤ ∑︁ 𝑏∈𝐵 𝐻(𝑋𝑁(𝑏)) + 𝑑 ∑︁ 𝑏∈𝐵 𝐻(𝑋𝑏|𝑋𝑁(𝑏)) [drop conditioning] For each 𝑏∈𝐵, we have 𝐻(𝑋𝑁(𝑏)) + 𝑑𝐻(𝑋𝑏|𝑋𝑁(𝑏)) = 𝐻(𝑋𝑁(𝑏)) + 𝐻(𝑋(1) 𝑏, . . . , 𝑋(𝑑) 𝑏|𝑋𝑁(𝑏)) = 𝐻(𝑋(1) 𝑏, . . . , 𝑋(𝑑) 𝑏, 𝑋𝑁(𝑏)) ≤log2 𝑖(𝐾𝑑,𝑑) where 𝑋(1) 𝑏, . . . , 𝑋(𝑑) 𝑏 are conditionally independent copies of 𝑋𝑏given 𝑋𝑁(𝑏). Sum-ming over all 𝑏yields the result.
□ Now we give the argument from Zhao (2010) that removes the bipartite hypothesis.
The following combinatorial argument reduces the problem for non-bipartite 𝐺to that of bipartite 𝐺.
Starting from a graph 𝐺, we construct its bipartite double cover 𝐺× 𝐾2 (see Fig-ure 10.1), which has vertex set 𝑉(𝐺) × {0, 1}. The vertices of 𝐺× 𝐾2 are labeled 𝑣𝑖 for 𝑣∈𝑉(𝐺) and 𝑖∈{0, 1}. Its edges are 𝑢0𝑣1 for all 𝑢𝑣∈𝐸(𝐺). Note that 𝐺× 𝐾2 is always a bipartite graph.
Lemma 10.4.13 Let 𝐺be any graph (not necessarily regular). Then 𝑖(𝐺)2 ≤𝑖(𝐺× 𝐾2).
Once we have the lemma, Theorem 10.4.12 then reduces to the bipartite case, which we already proved. Indeed, for a 𝑑-regular 𝐺, since 𝐺× 𝐾2 is bipartite, the bipartite 195 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy 2𝐺 𝐺× 𝐾2 𝐺× 𝐾2 Figure 10.1: The bipartite swapping trick in the proof of Lemma 10.4.13: swap-ping the circled pairs of vertices (denoted 𝐴in the proof) fixes the bad edges (red and bolded), transforming an independent set of 2𝐺 into an independent set of 𝐺× 𝐾2.
case of the theorem gives 𝑖(𝐺)2 ≤𝑖(𝐺× 𝐾2) ≤𝑖(𝐾𝑑,𝑑)𝑛/𝑑, Proof of Lemma 10.4.13. Let 2𝐺denote a disjoint union of two copies of 𝐺. Label its vertices by 𝑣𝑖with 𝑣∈𝑉and 𝑖∈{0, 1} so that its edges are 𝑢𝑖𝑣𝑖with 𝑢𝑣∈𝐸(𝐺) and 𝑖∈{0, 1}. We will give an injection 𝜙: 𝐼(2𝐺) →𝐼(𝐺× 𝐾2). Recall that 𝐼(𝐺) is the set of independent sets of 𝐺. The injection would imply 𝑖(𝐺)2 = 𝑖(2𝐺) ≤𝑖(𝐺× 𝐾2) as desired.
Fix an arbitrary order on all subsets of 𝑉(𝐺). Let 𝑆be an independent set of 2𝐺. Let 𝐸bad(𝑆) := {𝑢𝑣∈𝐸(𝐺) : 𝑢0, 𝑣1 ∈𝑆}.
Note that 𝐸bad(𝑆) is a bipartite subgraph of 𝐺, since each edge of 𝐸bad has exactly one endpoint in {𝑣∈𝑉(𝐺) : 𝑣0 ∈𝑆} but not both (or else 𝑆would not be independent).
Let 𝐴denote the first subset (in the previously fixed ordering) of 𝑉(𝐺) such that all edges in 𝐸bad(𝑆) have one vertex in 𝐴and the other outside 𝐴. Define 𝜙(𝑆) to be the subset of 𝑉(𝐺) × {0, 1} obtained by “swapping” the pairs in 𝐴, i.e., for all 𝑣∈𝐴, 𝑣𝑖∈𝜙(𝑆) if and only if 𝑣1−𝑖∈𝑆for each 𝑖∈{0, 1}, and for all 𝑣∉𝐴, 𝑣𝑖∈𝜙(𝑆) if and only if 𝑣𝑖∈𝑆for each 𝑖∈{0, 1}. It is not hard to verify that 𝜙(𝑆) is an independent set in 𝐺× 𝐾2. The swapping procedure fixes the “bad” edges.
It remains to verify that 𝜙is an injection.
For every 𝑆∈𝐼(2𝐺), once we know 𝑇= 𝜙(𝑆), we can recover 𝑆by first setting 𝐸′ bad(𝑇) = {𝑢𝑣∈𝐸(𝐺) : 𝑢𝑖, 𝑣𝑖∈𝑇for some 𝑖∈{0, 1}}, so that 𝐸bad(𝑆) = 𝐸′ bad(𝑇), and then finding 𝐴as earlier and swapping the pairs of 𝐴 back. (Remark: it follows that 𝑇∈𝐼(𝐺× 𝐾2) lies in the image of 𝜙if and only if 𝐸′ bad(𝑇) is bipartite.) □ 196 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.4 Shearer’s lemma The entropy proof of the bipartite case of Theorem 10.4.12 extends to graph homo-morphisms, yielding the following result.
Theorem 10.4.14 (Galvin and Tetali 2004) Let 𝐺be an 𝑛-vertex 𝑑-regular bipartite graph. Let 𝐻be any graph allowing loops.
Then hom(𝐺, 𝐻) ≤hom(𝐾𝑑,𝑑, 𝐻)𝑛/(2𝑑) Some important special cases: • hom(𝐺, ) = 𝑖(𝐺), the number of independent sets of 𝐺; • hom(𝐺, 𝐾𝑞) = the number of proper 𝑞-colorings of 𝐺.
The bipartite hypothesis in Theorem 10.4.14 cannot be always be removed.
For example, if 𝐻= , then log2 hom(𝐺, 𝐻) is the number of connected components of 𝐺, so that the maximizers of log2 hom(𝐺, 𝐻)/𝑣(𝐺) are disjoint unions of 𝐾𝑑+1’s.
For 𝐻= 𝐾𝑞, corresponding to the proper 𝑞-colorings, the bipartite hypothesis was recently removed.
Theorem 10.4.15 (Sah, Sawhney, Stoner, and Zhao 2020) Let 𝐺be an 𝑛-vertex 𝑑-regular graph. Then 𝑐𝑞(𝐺) ≤𝑐𝑞(𝐾𝑑,𝑑)𝑛/(2𝑑) where 𝑐𝑞(𝐺) is the number of 𝑞-colorings of 𝐺.
Furthermore, it was also shown in the same paper that in Theorem 10.4.14, the bipartite hypothesis on 𝐺can be weakened to triangle-free. Furthermore triangle-free is the weakest possible hypothesis on 𝐺so that the claim is true for all 𝐻.
For more discussion and open problems on this topic, see the survey by Zhao (2017).
Exercises The problems in this section should be solved using entropy arguments or results derived from entropy arguments.
1. Submodularity. Prove that 𝐻(𝑋,𝑌, 𝑍) + 𝐻(𝑋) ≤𝐻(𝑋,𝑌) + 𝐻(𝑋, 𝑍).
2. Let F be a collection of subsets of [𝑛]. Let 𝑝𝑖denote the fraction of F that 197 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10 Entropy contains 𝑖. Prove that |F | ≤ 𝑛 Ö 𝑖=1 𝑝−𝑝𝑖 𝑖 (1 −𝑝𝑖)−(1−𝑝𝑖).
3. ★Uniquely decodable codes. Let [𝑟]∗denote the set of all finite strings of elements in [𝑟]. Let 𝐴be a finite subset of [𝑟]∗and suppose no two distinct concatenations of sequences in 𝐴can produce the same string. Let |𝑎| denote the length of 𝑎∈𝐴. Prove that ∑︁ 𝑎∈𝐴 𝑟−|𝑎| ≤1.
4. Sudoku. A 𝑛2 × 𝑛2 Sudoku square (the usual Sudoku corresponds to 𝑛= 3) is an 𝑛2 × 𝑛2 array with entries from [𝑛2] so that each row, each column, and, after partitioning the square into 𝑛× 𝑛blocks, each of these 𝑛2 blocks consist of a permutation of [𝑛2]. Prove that the number of 𝑛2 × 𝑛2 Sudoku squares is at most 𝑛2 𝑒3 + 𝑜(1) 𝑛4 .
5. Prove Sidorenko’s conjecture for the following graph.
6. ★Triangles versus vees in a directed graph. Let 𝑉be a finite set, 𝐸⊆𝑉×𝑉, and △= (𝑥, 𝑦, 𝑧) ∈𝑉3 : (𝑥, 𝑦), (𝑦, 𝑧), (𝑧, 𝑥) ∈𝐸 (i.e., cyclic triangles; note the direction of edges) and ∧= (𝑥, 𝑦, 𝑧) ∈𝑉3 : (𝑥, 𝑦), (𝑥, 𝑧) ∈𝐸 .
Prove that △≤∧.
7. ★Box theorem.
Prove that for every compact set 𝐴⊆R𝑑, there exists an axis-aligned box 𝐵⊆R𝑑with vol 𝐴= vol 𝐵 and vol 𝜋𝐼(𝐴) ≥vol 𝜋𝐼(𝐵) for all 𝐼⊆[𝑛].
198 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 10.4 Shearer’s lemma Here 𝜋𝐼denotes the orthogonal projection onto the 𝐼-coordinate subspace.
(For the purpose of the homework, you only need to establish the case when 𝐴is a union of grid cubes. It is optional to give the limiting argument for compact 𝐴.) 8. Let G be a family of graphs on vertices labeled by [2𝑛] such that the intersection of every pair of graphs in G contains a perfect matching. Prove that |G| ≤2(2𝑛 2 )−𝑛.
9. Loomis–Whitney for sumsets.
Let 𝐴, 𝐵, 𝐶be finite subsets of some abelian group. Writing 𝐴+ 𝐵:= {𝑎+ 𝑏: 𝑎∈𝐴, 𝑏∈𝐵}, etc., prove that |𝐴+ 𝐵+ 𝐶|2 ≤|𝐴+ 𝐵| |𝐴+ 𝐶| |𝐵+ 𝐶| .
10. ★Shearer for sums. Let 𝑋,𝑌, 𝑍be independent random integers. Prove that 2𝐻(𝑋+ 𝑌+ 𝑍) ≤𝐻(𝑋+ 𝑌) + 𝐻(𝑋+ 𝑍) + 𝐻(𝑌+ 𝑍).
199 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 11 Containers Many problems in combinatorics can be phrased in terms of independent sets in hypergraphs.
For example, here is a model question: Question 11.0.1 How many triangle-free graphs are there on 𝑛vertices?
By taking all subgraphs of 𝐾𝑛/2,𝑛/2, we obtain 2𝑛2/4 such graphs. It turns out this gives the correct exponential asymptotic.
Theorem 11.0.2 (Erdős, Kleitman, and Rothschild 1973) The number of triangle-free graphs on 𝑛vertices is 2𝑛2/4+𝑜(𝑛2).
Remark 11.0.3. It does not matter here whether we consider vertices to be labeled, it affects the answer up to a factor of at most 𝑛! = 𝑒𝑂(𝑛log 𝑛).
Remark 11.0.4. Actually the original Erdős–Kleitman–Rothschild paper showed an even stronger result: 1 −𝑜(1) fraction of all 𝑛-vertex triangle-free graphs are bipartite.
The above asymptotic can be then easily deduced by counting subgraphs of complete bipartite graphs. The container methods discussed in this section are not strong enough to prove this finer claim.
We can convert this asymptotic enumeration problem into a problem about independent sets in a 3-uniform hypergraph 𝐻: • 𝑉(𝐻) = [𝑛] 2 • The edges of 𝐻are triples of the form {𝑥𝑦, 𝑥𝑧, 𝑦𝑥}, i.e., triangles We then have the correspondence: • A subset of 𝑉(𝐻) = a graph on vertex set [𝑛] • An independent set of 𝑉(𝐻) = a triangle-free graph 201 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 11 Containers (Here an independent set in a hypergraph is a subset of vertices containing no edges.) Naively applying first moment/union bound does not work—there are too many events to union bound over.
For example, Mantel’s theorem tell us the maximum number of edges in an 𝑛-vertex triangle-free graph is 𝑛2/4 , obtained by 𝐾⌊𝑛/2⌋,⌈𝑛/2⌉. With a fixed triangle-free graph 𝐺, the number of subgraphs of 𝐺is 2𝑒(𝐺), and each of them is triangle-free. Perhaps we could union bound over all maximal triangle-free graphs? It turns out that there are 2𝑛2/8+𝑜(𝑛2) such maximal triangle-free graphs, so a union bound would be too wasteful.
In many applications, independent sets are clustered into relatively few highly corre-lated sets. In the case of triangle-free graphs, each maximal triangle-free graph is very “close” to many other maximal triangle-free graphs.
Is there a more efficient union bound that takes account of the clustering of independent sets?
The container method does exactly that.
Given some hypergraph with controlled degrees, one can find a collection of containers satisfying the following properties: • Each container is a subset of vertices of the hypergraph.
• Every independent set of the hypergraph is a subset of some container.
• The total number of containers in the collection is relatively small.
• Each container is not too large (in fact, not too much larger than the maximum size of an independent set) We can then union bound over all such containers. If the number of containers is not too small, then the union bound is not too lossy.
Here are some of the most typical and important applications of the container method: • Asymptotic enumerations: – Counting 𝐻-free graphs on 𝑛vertices – Counting 𝐻-free graphs on 𝑛vertices and 𝑚edges – Counting 𝑘-AP-free subsets of [𝑛] of size 𝑚 • Extremal and Ramsey results in random structures: – The maximum number of edges in an 𝐻-free subgraph of 𝐺(𝑛, 𝑝) – Szemeŕedi’s theorem in a 𝑝-random subset of [𝑛] • List coloring in graphs/hypergraphs 202 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 11.1 Containers for triangle-free graphs The method of hypergraph containers is one of the most exciting developments in this past decade. Some references and further reading: • The graph container method was developed by Kleitman and Winston (1982) (for counting 𝐶4-free graphs) and Sapozhenko (2001) (for bounding the number of independent sets in a regular graph, giving an earlier version of Theorem 10.4.12) • The hypergraph container theorem was proved independently by Balogh, Morris, and Samotij (2015), and Saxton and Thomason (2015).
• See the 2018 ICM survey of Balogh, Morris, and Samotij for an introduction to the topic along with many applications • See Samotij’s survey article (2015) for an introduction to the graph container method • See Morris’ lecture notes (2016) for a gentle introduction to the proof and applications of hypergraph containers.
11.1 Containers for triangle-free graphs The number of triangle-free graphs We will prove Theorem 11.0.2 that the number of triangle-free graphs on 𝑛vertices is 2𝑛2/4+𝑜(𝑛2).
Theorem 11.1.1 (Containers for triangle-free graphs) For every 𝜀> 0, there exists 𝐶> 0 such that the following holds.
For every 𝑛, there is a collection C of graphs on 𝑛vertices, with |C| ≤𝑛𝐶𝑛3/2 such that (a) every 𝐺∈C has at most ( 1 4 + 𝜀)𝑛2 edges, and (b) every triangle-free graph is contained in some 𝐺∈C.
Proof of upper bound of Theorem 11.0.2. We want to show that the number of 𝑛-vertex triangle-free graphs is at most 2𝑛2/4+𝑜(𝑛2).
Let 𝜀> 0 be any real number (arbitrarily small). Let C be produced by Theorem 11.1.1.
Then every 𝐺∈C has at most ( 1 4 + 𝜀)𝑛2 edges, and every triangle-free graph is 203 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 11 Containers contained in some 𝐺∈C. Hence the number of triangle-free graphs is |C| 2( 1 4+𝛿)𝑛2 ≤2( 1 4+𝜀)𝑛2+𝑂𝜀(𝑛3/2 log 𝑛).
Since 𝜀> 0 can be made arbitrarily small, the number triangle-free graphs on 𝑛 vertices is 2( 1 4+𝑜(1))𝑛2.
□ The same proof technique, with an appropriate container theorem, can be used to count 𝐻-free graphs.
We write ex(𝑛, 𝐻) for the maximum number of edges in an 𝑛-vertex graph without 𝐻 as a subgraph.
Theorem 11.1.2 (Erdős–Stone–Simonovits) Fix a non-bipartite graph 𝐻. Then ex(𝑛, 𝐻) = 1 − 1 𝜒(𝐻) −1 + 𝑜(1) 𝑛 2 .
Note that for bipartite graphs 𝐻, the above theorem just says 𝑜(𝑛2), though more precise estimates are available. Although we do not know the asymptotic for ex(𝑛, 𝐻) for all 𝐻, e.g., it is still open for 𝐻= 𝐾4,4 and 𝐻= 𝐶8.
Theorem 11.1.3 Fix a non-bipartite graph 𝐻.
Then the number of 𝐻-free graphs on 𝑛vertices is 2(1+𝑜(1)) ex(𝑛,𝐻).
The analogous statement for bipartite graphs is false. The following conjecture remains of great interest, and it is known for certain graphs, e.g., 𝐻= 𝐶4.
Conjecture 11.1.4 Fix a bipartite graph 𝐻with a cycle. The number of 𝐻-free graphs on 𝑛vertices is 2𝑂(ex(𝑛,𝐻)).
Mantel’s theorem in random graphs Theorem 11.1.5 If 𝑝≫1/√𝑛, then with probability 1 −𝑜(1), every triangle-free subgraph of 𝐺(𝑛, 𝑝) has at most ( 1 4 + 𝑜(1))𝑝𝑛2 edges.
204 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 11.1 Containers for triangle-free graphs Remark 11.1.6. In fact, a much stronger result is true: the triangle-free subgraph of 𝐺(𝑛, 𝑝) with the maximum number of edges is whp bipartite (DeMarco and Kahn 2015).
Remark 11.1.7. The statement is false for 𝑝≪1/√𝑛. Indeed, in this case, then the expected number of triangles is 𝑂(𝑛3𝑝3), whereas there are whp 𝑛2𝑝/2 edges, and 𝑛3𝑝3 ≪𝑛2𝑝, so we can remove 𝑜(𝑛2𝑝) edges to make the graph triangle-free.
Proof. We prove a slightly weaker result, namely that the result is true if 𝑝≫ 𝑛−1/2 log 𝑛. The version with 𝑝≫𝑛−1/2 can be proved via a stronger formulation of the container lemma (using “fingerprints” as discussed later).
Let 𝜀> 0 be aribtrarily small. Let C be a set of containers for 𝑛-vertex triangle-free graphs in Theorem 11.1.1. For every 𝐺∈C, 𝑒(𝐺) ≤ 1 4 + 𝜀 𝑛2, so by an application of the Chernoff bound, P 𝑒(𝐺∩𝐺(𝑛, 𝑝)) > 1 4 + 2𝜀 𝑛2𝑝 ≤𝑒−Ω𝜀(𝑛2𝑝) Since every triangle-free graph is contained in some 𝐺∈C, by taking a union bound over C, we see that P 𝐺(𝑛, 𝑝) has a triangle-free subgraph with > 1 4 + 2𝜀 𝑛2𝑝edges ≤ ∑︁ 𝐺∈C P 𝑒(𝐺∩𝐺(𝑛, 𝑝)) > 1 4 + 2𝜀 𝑛2𝑝 ≤|C| 𝑒−Ω𝜀(𝑛2𝑝) ≤𝑒𝑂𝜀(𝑛3/2 log 𝑛)−Ω𝜀(𝑛2𝑝) = 𝑜(1) provided that 𝑝≫𝑛−1/2 log 𝑛.
□ 205 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 11 Containers 11.2 Graph containers Theorem 11.2.1 (Container theorem for independent sets in graphs) For every 𝑐> 0, there exists 𝛿> 0 such that the following holds.
Let 𝐺= (𝑉, 𝐸) be a graph with average degree 𝑑and maximum degree at most 𝑐𝑑.
There exists a collection C of subsets of 𝑉, with |C| ≤ |𝑉| ≤2𝛿|𝑉| /𝑑 such that 1. Every independent set 𝐼of 𝐺is contained in some 𝐶∈C.
2. |𝐶| ≤(1 −𝛿) |𝑉| for every 𝐶∈C.
Each 𝐶∈C is called a “container.” Every independent set of 𝐶is contained in some container.
Remark 11.2.2. The requirement |𝐶| ≤(1 −𝛿) |𝑉| looks quite a bit weaker than in Theorem 11.1.1, where each container is only slightly larger than the maximum independent set. In a typical application of the container method, one iteratively applies the (hyper)graph container theorem (e.g., Theorem 11.2.1 and later Theorem 11.3.1) to the subgraphs induced by the slightly smaller containers in the previous iteration.
One iterates until the containers are close to their minimum possible size.
For this iterative application of container theorem to work, one usually needs a super-saturation result, which, roughly speaking, says that every subset of vertices that is slightly larger than the independence number necessarily induces a lot of edges. This property is common to all standard applications of the container method.
The container theorem is proved using The graph container algorithm (for a fixed given graph 𝐺) Input: a maximal independent set 𝐼⊆𝑉.
Output: a “fingerprint” 𝑆⊆𝐼of size ≤2𝛿|𝑉| /𝑑, and a container 𝐶⊇𝐼which depends only on 𝑆.
Throughout the algorithm, we will maintain a partition 𝑉= 𝐴∪𝑆∪𝑋, where • 𝐴, the “available” vertices, initially 𝐴= 𝑉 • 𝑆, the current fingerprint, initially 𝑆= ∅ • 𝑋, the “excluded” vertices, initially 𝑋= ∅.
206 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 11.2 Graph containers The max-degree order of 𝐺[𝐴] is an ordering of 𝐴in by the degree of the vertices in 𝐺[𝐴], with the largest first, and breaking ties according to some arbitrary predeter-mined ordering of 𝑉.
While |𝑋| < 𝛿|𝑉|: 1. Let 𝑣be the first vertex of 𝐼∩𝐴in the max-degree order on 𝐺[𝐴].
2. Add 𝑣to 𝑆.
3. Add the neighbors of 𝑣to 𝑋.
4. Add vertices preceding 𝑣in the max-degree order on 𝐺[𝐴] to 𝑋.
5. Remove from 𝐴all the new vertices added to 𝑆∪𝑋.
Claim: when the algorithm terminates, we obtain a partition 𝑉= 𝐴∪𝑆∪𝑋such that |𝑋| ≥𝛿|𝑉| and |𝑆| ≤2𝛿|𝑉| /𝑑.
Proof idea: due to the degree hypotheses, in every iteration, at least ≥𝑑/2 new vertices are added to 𝑋(provided that 𝑑≤2𝛿|𝑉|). See Morris’ lecture notes for details.
Key facts: • Two different maximal independent sets 𝐼, 𝐼′ ⊆𝑉that produce the same finger-print 𝑆in the algorithm necessarily produces the same partition 𝑉= 𝐴∪𝑆∪𝑋 • The final set 𝑆∪𝐴contains 𝐼(since only vertices not in 𝐼are ever moved to 𝐼) Therefore, the total number possibilities for containers 𝑆∪𝐴is at most the number of sets 𝑆⊆𝑉. Since |𝑆| ≤2𝛿|𝑉| /𝑑and |𝐴∪𝑆| ≤(1 −𝛿) |𝑉|, this concludes the proof of the graph container lemma.
The fingerprint obtained by the proof actually gives us a stronger consequence that will be important for some applications.
207 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 11 Containers Theorem 11.2.3 (Graph container theorem, with fingerprints) For every 𝑐> 0, there exists 𝛿> 0 such that the following holds.
Let 𝐺= (𝑉, 𝐸) a graph with average degree 𝑑and maximum degree at most 𝑐𝑑.
Writing I for the collection of independent sets of 𝐺, there exist functions 𝑆: I →2𝑉 and 𝐴: 2𝑉→2𝑉 (one only needs to define 𝐴(·) on sets in the image of 𝑆) such that, for every 𝐼∈I, • 𝑆(𝐼) ⊆𝐼⊆𝑆(𝐼) ∪𝐴(𝑆(𝐼)) • |𝑆(𝐼)| ≤2𝛿|𝑉| /𝑑 • |𝑆(𝐼) ∪𝐴(𝑆(𝐼))| ≤(1 −𝛿) |𝑉| 11.3 Hypergraph container theorem An independent set in a hypergraph is a subset of vertices containing no edges.
Given an 𝑟-uniform hypergraph 𝐻and 1 ≤ℓ< 𝑟, we write Δℓ(𝐻) = max 𝐴⊆𝑉(𝐻):|𝐴|=ℓthe number of edges containing 𝐴 Theorem 11.3.1 (Container theorem for 3-uniform hypergraph) For every 𝑐> 0 there exists 𝛿> 0 such that the following holds.
Let 𝐻be a 3-uniform hypergraph with average degree 𝑑≥𝛿−1 and Δ1(𝐻) ≤𝑐𝑑 and Δ2(𝐻) ≤𝑐 √ 𝑑.
Then there exists a collection C of subsets of 𝑉(𝐻) with |C| ≤ 𝑣(𝐻) ≤𝑣(𝐻)/ √ 𝑑 such that • Every independent set of 𝐻is contained in some 𝐶∈C, and • |𝐶| ≤(1 −𝛿)𝑣(𝐻) for every 𝐶∈C.
208 MIT OCW: Probabilistic Methods in Combinatorics — Yufei Zhao 11.3 Hypergraph container theorem Like the graph container theorem, the hypergraph container theorem is proved by designing an algorithm to produce, from an independent set 𝐼⊆𝑉(𝐻), a fingerprint 𝑆⊆𝐼and a container 𝐶⊃𝐼.
The hypergraph container algorithm is more involved compared to the graph container algorithm.
In fact, the 3-uniform hypergraph container algorithm calls the graph container algorithm.
Container algorithm for 3-uniform hypergraphs (a very rough sketch): Throughout the algorithm, we will maintain • A fingerprint 𝑆, initially 𝑆= ∅ • A 3-uniform hypergraph 𝐴, initially 𝐴= 𝐻 • A graph 𝐺of “forbidden” pairs on 𝑉(𝐻), initially 𝐺= ∅ While |𝑆| ≤𝑣(𝐻)/ √ 𝑑−1: • Let 𝑢be the first vertex in 𝐼in the max-degree order on 𝐴 • Add 𝑢to 𝑆 • Add 𝑥𝑦to 𝐸(𝐺) whenever 𝑢𝑥𝑦∈𝐸(𝐻) • Remove from 𝑉(𝐴) the vertex 𝑢as well as all vertices proceeding 𝑢in the max-degree order on 𝐴 • Remove from 𝑉(𝐴) every vertex whose degree in 𝐺is larger than 𝑐 √ 𝑑.
• Remove from 𝐸(𝐴) every edge that contains an edge of 𝐺.
Finally, it is will be the case that either • We have removed many vertices from 𝑉(𝐴) • Or the final graph 𝐺has at least Ω( √ 𝑑𝑛) edges and has maximum degree 𝑂( √ 𝑑), so that we can apply the graph container lemma to 𝐺.
In either case, the algorithm produces a container with the desired properties. Again see Morris’ lecture notes for details.
209 MIT OpenCourseWare 18.226 Probabilistic Methods in Combinatorics Fall 2022 For information about citing these materials or our Terms of Use, visit: |
190770 | https://www.hsbc.com/-/files/hsbc/investors/results-and-announcements/stock-exchange-announcements/2015/march/sea-150319-c-hghq-ara-2014-esp.pdf | 聯繫客戶創先機 助握商情百五載 滙豐控股有限公司 2014年報及賬目 策略報告 1 概覽 2 有關前瞻性陳述之提示聲明 3 摘要 4 集團主席報告 7 集團行政總裁之回顧 9 策略目標 12 業務模式 26 優先策略 28 成果 董事會報告 財務回顧 40 財務概要 63 環球業務 78 地區 106 其他資料 111 風險 238 資本 企業管治 263 企業管治報告 264 董事及高級管理層簡歷 270 董事會 276 董事會屬下委員會 288 內部監控 290 持續經營 291 僱員 董事薪酬報告 300 董事薪酬報告 財務報表 328 董事之責任聲明 329 獨立核數師報告 334 財務報表 345 財務報表附註 股東參考資料 458 股東參考資料 466 簡稱 470 詞彙 478 索引 本文件收錄滙豐控股有限公司及其附屬公司的 《2014年報及賬目》 ，並載有英國 《2006年公司法 （策 略報告及董事會報告） 2013年規例》 所規定的策略 報告、董事會報告、董事薪酬報告與財務報表，以 及獨立核數師就財務報表發出之報告。載於第1至 39頁的策略報告、第40至299頁的董事會報告及第 300至327頁的董事薪酬報告是根據英國法律的規 定而編製，而相關的任何責任亦由英國法律所規 管。 有關進一步的資料 （包括2013年與2012年比較之評 述） ，請參閱提交美國證券交易委員會 （ 「美國證交 會」 ） 存檔的20-F表格，亦可登入www.hsbc.com及 www.sec.gov網站瀏覽。 若干界定用語 除文義另有所指外， 「滙豐控股」 乃指滙豐控股有 限公司，而 「滙豐」 、 「集團」 或 「我們」 則指滙豐控 股及其附屬公司。在本文件內，中華人民共和國香 港特別行政區簡稱為 「香港」 。當使用 「股東權益」 及 「股東權益總額」 等用語時， 「股東」 指滙豐控股 的普通股及由滙豐控股發行並分類為股東權益的 優先股及資本證券之持有人。 「百萬美元」 及 「十億 美元」 分別指百萬及十億 （數以千計之百萬） 美元。 財務報表 滙豐的綜合財務報表及滙豐控股的獨立財務報表 均已根據由國際會計準則委員會 （ 「IASB」 ） 頒布並 由歐洲聯盟 （ 「歐盟」 ） 正式通過之 《國際財務報告準 則》 （ 「IFRS」 ） 編製。若於任何時間，歐盟並無正式 通過新訂或經修訂的IFRS，則歐盟正式通過之IFRS 可能與IASB頒布之IFRS 有所不同。於2014年12月 31日，並無任何於截至2014年12月31日止年度內生 效而未經歐盟正式通過之準則對上述綜合及獨立 財務報表有任何影響。就適用於滙豐之IFRS而言， 經歐盟正式通過之IFRS與由IASB頒布之IFRS並無差 異。因此，滙豐是根據由IASB頒布之IFRS，編製截 至2014年12月31日止年度的財務報表。 我們以美元為列賬貨幣，是由於美元及與其掛鈎 之各種貨幣所屬區域，是我們進行交易及為業務 營運提供資金之主要貨幣區。除另有說明外，本文 件呈列的資料已按照IFRS編製。 在列表或評述中提及 「經調整」 時，比較資料乃按 固定匯率列示 （請參閱第40頁） ，亦已撇除滙豐本身 債務的信貸息差變動引致的公允值變動之影響， 並已調整其他重大項目的影響 （見第44頁的對賬） 。 經調整風險加權資產的回報於第62頁界定及對賬。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 1 概覽 我們的背景 滙豐是世界上規模最大的銀行及金融 服務機構之一。 客戶總數： 5,100萬名 員工總數： 266,000名 （257,600名等同全職僱員） 經營四大環球業務： －零售銀行及財富管理 －工商金融 －環球銀行及資本市場 －環球私人銀行 營業地點： 73個 國家及地區 橫跨五個地區： －歐洲 －亞洲 －中東及北非 －北美洲 －拉丁美洲 辦事處總數： 超過6,100個 環球總部： －倫敦 市值： 1,820億美元 於下列證券交易所上市： －倫敦 －香港 －紐約 －巴黎 －百慕達 股東總數： 216,000名，遍及127個 國家及地區 我們的宗旨 我們的宗旨是把握市場的增長機遇、努力建立聯 繫以幫助客戶開拓商機、推動工商企業茁壯成長 及各地經濟蓬勃發展，而最終目標是讓客戶實現 理想。 我們的優先策略 滙豐銳意成為世界領先及最受推崇的國際銀行。 為此，我們將專注滿足客戶及滙豐所在社會的需 要，為所有相關群體帶來長遠可持續的價值。 為落實有關策略，我們於2013年公布了2014至2016 年的三個環環相扣且同等重要的優先範疇： －推動業務及股息增長； －實施環球標準；及 －簡化流程及程序。 這些優先範疇相輔相成，並以我們日常業務所執 行的各項措施為基礎。它們為滙豐的客戶及股東 創造價值，並促進滙豐的長遠持續發展。 我們如何衡量業績表現 我們採用一系列財務及非財務衡量指標或集團表 現關鍵指標，監察落實策略的進展。由2015年起， 我們已重新檢討目標，使之更配合不斷轉變的監 管及經營環境。 2014年摘要載於第3頁。 有關新目標的進一步資料，請參閱第32頁。 獎勵表現 集團全體員工 （包括執行董事） 的薪酬，乃根據財務 及非財務表現目標的達標程度釐定。這些配合集 團策略的目標已詳列於員工的個人年度表現評分 紀錄內。只有全面恪守滙豐價值觀的員工，方獲考 慮授予浮動酬勞獎勵。 有關滙豐價值觀的詳情，請參閱第10頁。 滙豐控股有限公司 2 概覽 （續） 有關前瞻性陳述之提示聲明 《2014年報及賬目》 包含若干對於滙豐財政狀況、 經營業績、資本狀況及業務的前瞻性陳述。 前瞻性陳述並非過往事實的陳述，而是包括描述 滙豐信念及期望的陳述。某些字詞例如 「預料」 、 「期望」 、 「打算」 、 「計劃」 、 「相信」 、 「尋求」 、 「估 計」 、 「潛在」 及 「合理地可能」 ，這些字詞的其他 組合形式及類似措辭，均顯示相關文字為前瞻性 陳述。這些陳述乃基於現行計劃、估計及預測而 作出，故不應對其過份倚賴。前瞻性陳述中所作 表述僅以截至有關陳述作出當日為準。滙豐並無 承諾會修訂或更新任何前瞻性陳述，以反映作出 有關前瞻性陳述當日之後所發生或存在之事件 或情況。 書面及╱或口述形式之前瞻性陳述，亦可能載於 向美國證交會提交之定期匯報、致股東之財務報 表摘要、委託聲明、售股通函及章程、新聞稿及 其他書面資料，以及由滙豐董事、主管人員或僱 員向包括財務分析員在內的第三方以口述形式 作出的陳述。 前瞻性陳述涉及內在風險及不明朗因素。務請注 意，多種因素均可導致實際結果偏離任何前瞻性 陳述所預期或隱含的狀況，在某些情況下甚至會 出現重大偏差。這些因素包括 （但不限於） ： ‧ 滙豐業務所在市場的整體經濟環境產生變化， 例如經濟衰退持續或惡化及就業市場波動超 出一致預測者；匯率及利率變動；股市波動； 批發融資市場流通性不足；國家的房地產市 場流通性不足及出現價格下調壓力；各央行 為金融市場提供流動資金支持的政策出現不 利變動；市場對過度借貸的國家╱地區的主 權信用憂慮加劇；私人及公營機構的界定福 利退休金的資金狀況出現不利變動；及消費 者如何理解持續可用信貸，以及滙豐提供服 務所在市場的價格競爭情況； ‧ 政府政策及規例有變，包括各國央行及其他 監管機構在貨幣、利率及其他方面的政策； 世界各個主要市場的金融機構面對更嚴格的 監管，因而需要採取措施改變金融機構的規 模、業務範疇及其相互聯繫；經修訂的資本 及流動資金基準，使銀行的資產負債表達致 減債，並使現時業務模式及貸款組合可取得 的回報下降；為改變業務組合成分及承受風 險水平而徵收的徵費或稅項；向消費市場提 供服務的金融機構的慣例、定價或責任；有 關資產的剝奪、收歸國有、充公，以及有關外 資擁有權的法例變更；滙豐業務所在主要市 場的破產法例有變及其後果；政府政策出現 整體變化，可能會嚴重影響投資者的決定； 當前市場動盪引致的特殊政府措施；政治或 外交事態出現其他不利發展，造成社會不穩 或法律上的不明朗因素，繼而影響對滙豐旗 下產品及服務的需求；產品監管機構作出檢 討、採取行動或提出訴訟 （包括要求遵守額外 規定） 引致的費用、影響及結果；及滙豐業務 所在市場競爭環境的影響，包括非銀行金融 服務公司 （包括證券行） 造成更激烈的競爭； 及 ‧ 有關滙豐的特定因素，包括審慎管理風險加 權資產增長，以及能夠充分識別所面對的風 險，例如貸款損失或拖欠事件，並有效管理 該等風險 （透過賬項管理、對沖及其他方式） 。 有效的風險管理有賴於 （其中包括） 滙豐能否 透過壓力測試及其他方式，設法防範所用統 計模型無法偵測的事件；以及滙豐能否成功 應付營運、法律及 監管和訴訟 （尤其是遵守美 國延後起訴協議） 方面的挑戰。 提示聲明╱摘要 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 3 除稅前利潤 資本實力 每股普通股股息 （列賬基準） （資本指引4過渡基準普通股權 （就本年度派付） 3 一級比率） 1 187億美元 10.9% 0.50美元 19.0 21.9 20.6 22.6 18.7 2010年 2011年 2012年 2013年 2014年 2010年 2011年 2012年 2013年 2014年 10.8 10.9 2013年 2014年 0.36 0.41 0.45 0.49 0.50 成本效益比率 平均普通股股東權益回報 4 股價 （列賬基準） 2 （於12月31日） 67.3% 7.3% 6.09英鎊 55.2 57.5 62.8 59.6 67.3 9.5 10.9 8.4 9.2 7.3 6.51 4.91 6.47 6.62 6.09 2010年 2011年 2012年 2013年 2014年 2010年 2011年 2012年 2013年 2014年 2010年 2011年 2012年 2013年 2014年 有關列賬基準與經調整業績之間差異的說明，請參閱第40頁。 有關註釋，請參閱第39頁。 摘要 － 列賬基準之除稅前利潤下跌 17%至187億美元。經調整除稅 前利潤 （不包括重大項目及貨 幣換算的影響） 為228億美元， 大致維持不變。 － 派發股息增至96億美元，乃因 運用資本實力促成有機增長， 使我們能夠增加派息。 113億英鎊 1,450億港元 （表內單位：十億美元） 於12月31日 74.00港元 美國預託股份 47.23美元 － 滙豐的資本實力更形鞏固。資 本指引4過渡基準普通股權一 級比率為10.9%，2013年底為 10.8%。 滙豐控股有限公司 4 集團主席報告 概覽 （續） 在大部分國家和地區，經濟活動 仍未達到可以重建消費者信心的 水平，亦不足以帶動投資支出增 長。受此影響最嚴重的國家，因而 推出更多刺激經濟的措施，主要 央行亦將利率維持在前所未見的 低水平。對通縮趨勢的憂慮進一 步升溫，尤以歐元區為甚。雖然 中國內地的增長速度顯著高於其 他主要經濟體，但預期未來增長 將會放緩，嚴重打擊市場情緒， 導致商品價格大幅下跌，進一步 削弱全球投資支出。 處身於這個環境，集團收入的增長 機會主要來自亞洲業務是意料中 事，其中貸款和債務資本融資業 務均錄得增長。集團於全球各地 的支出繼續上升，很大原因在於 實施監管改革和加強風險控制， 特別是金融系統完整性和操守方 面的支出。簡化架構的措施僅能 抵銷部分開支增長。由於過往缺 失所引致的客戶賠償和監管處罰 繼續出現，董事會決心繼續投入 必要資源，優先改善相關系統和 監控措施，以避免重蹈覆轍。 很明顯，現時社會、監管機構和公 共政策對銀行業的期望，令業界 的長期成本結構產生變化。數據 分析技術 （包括 「海量數據」 ） 的發 展，為我們提供更精密的工具， 有助提升集團保護金融系統和防 範不法分子的能力。隨著越來越 多客戶使用互聯網和流動設備進 行交易，我們亦積極投資於網絡 安全，以保障集團和客戶免受威 脅。為建立所需的數據分析能力， 我們須於系統和維護客戶資料方 面作出大規模投資，以確保客戶 資料準確無誤和經常更新。由於 系統過時，加上全球各地歷來均 採用不同的數據標準，要重新整 理客戶和交易數據，使之配合數 碼時代的需要，無疑是一項浩大 工程。然而，加強客戶盡職審查 能力及提升系統安全，是滙豐作 為系統重要性銀行的核心責任， 有助我們更積極地做好守護金融 系統的職責。 在銀行業按公共政策和監管指令 進行重整之際，我們現時需要透 過清晰的業務模式，向社會展示 集團規模和多元化的價值。我們 絕不能忘記，投資者有權選擇投 資對象，員工亦有權選擇其事業 前途。因此，我們要證明滙豐能 為其服務的社會作出積極貢獻， 以建立利於營商的環境，這樣的 環境被公認為對經濟增長和繁榮 至關重要。 過去150年來，滙豐一直緊隨貿易 和投資流向，致力幫助客戶實現 財務目標。時至今日，全球經濟 已由相互關連變成唇齒相依，無 論是大中小型企業還是個人，其 財務前景都與多個國家和地區的 經濟活動息息相關，而我們的業 務模式與這些企業和個人的關係 亦變得更為密切。 工商金融業務最能體現上述關係。 由於供應鏈管理方案的業務擴展 和跨境付款服務的交易量增加， 工商金融業務於去年刷新業績紀 錄。全球的付款活動中，超過85% 來自我們業務網絡覆蓋的國家和 地區，成為支持集團業務模式的 主要元素。投資方面，企業投資繼 續通過集團業務網絡流向增長較 快的新興市場。與此同時，中國內 地對外投資加速增長，反映內地 主要企業有意作多元化發展，同 時爭取外界的技術資源和市場。 滙豐在主要金融中心建立的業務 規模和覆蓋範圍，正好配合上述 發展趨勢，可為客戶提供債務及 股權融資方案、專門設計的流動 資金和交易銀行服務，並主要按 客戶的利率和匯兌風險水平提供 風險管理解決方案。這些業務屢 次贏得業內獎項，反映集團於相 關領域的表現備受肯定 （主要獎項 請參閱 「集團行政總裁之回顧」 ） 。 此外，集團的零售銀行及財富管 理業務繼續構建 「以客為先」 的可 持續發展業務模式，全面取消前 線員工績效薪酬與產品銷售額掛 鈎的制度，並擴展電子銀行和流 動銀行服務。 集團主席報告 2014年的業績表現，反映年內的地緣政治和經濟逆風，很多都出 乎年初時的預期，在此環境下，滙豐繼續重整及強化集團的架構 和業務。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 5 2014年業績 列賬基準除稅前利潤為187億美 元，較2013年減少39億美元或 17%，主要反映出售業務利潤和 重新分類增益減少，以及罰款、 和解開支、英國客戶賠償及相關 準備等重大項目對收入和支出產 生的負面影響。按經調整基準 （用 以評估本年度管理層表現和業務 表現的主要指標之一） 計算，除稅 前利潤為228億美元，與按同類基 準計算的2013年數據相若。 每股盈利為0.69美元，2013年則 為0.84美元。集團資本狀況維持 強勁，年底的過渡基準普通股權 一級比率為10.9%，12個月前則為 10.8%；終點基準比率為11.1%， 12個月前則為10.9%。基於以上資 本實力和集團的資本生成能力， 董事會已批准派發2014年度第四 次股息每股0.2美元，全年每股派 息共計0.5美元 （派息總額達96億美 元，較2013年度增加4億美元） 。 綜合考慮上述財務表現，加上集 團在重整架構、推行監管改革和 實施環球標準等方面取得進一步 進展，董事會認為行政管理層於 加強集團長遠競爭力方面，已於 2014年取得可觀進展。 有關2014年重要業績指標和摘要 的分析，請參閱 「集團行政總裁之 回顧」 。 監管環境較明朗，但仍任重而 道遠 2014年，監管機構在落實監管架 構改革方面取得長足進展，在新 架構全面實施後，滙豐等環球系 統性銀行必須據此營運。我們致 力為環球業務作好部署，以便在 可接受的風險範圍內，達致投資 者所期望的回報。清晰的架構對 實現此一目標非常重要。 尤其在解決 「大到不能倒」 這一難 題方面，監管機構的工作大有進 展，主要是落實以 「自救」 債務的 方式來加強現有的吸納損失能力， 並詳細界定了解決架構的實際運 作方式。兩者均涉及一個關鍵問 題，就是如何處理跨境事項，以 及註冊國和駐在國的監管責任。 然而，尚有大量工作需要完成。 2015年的監管改革議程非常緊湊， 在對銀行業架構影響深遠的多個 領域，仍須等待多項公共政策決 定、監管諮詢及分析研究的結果； 當中包括歐洲商議分隔運作架構 的結論、針對所謂 「影子銀行」 的 其他監管工作 （包括確定系統重 要性非銀行機構） 、探討中央交易 對手的解決架構、最終校準槓桿 比率的計算方法、修訂將予提高 的整體損失吸納能力水平，以及 在全球性金融集團內建立相關能 力。 業內接連有不當行為曝 光，恢復公眾對銀行業的 信任仍是一大挑戰，但我 們必須成功克服這個考驗。 除此之外，業界將開展更多工作， 運用標準風險權數，以恢復監管 機構對內部模型資本計量所失去 的信心。監管機構亦正徹底檢討 交易賬項，重新審視有關活動所 需的資本狀況。該等措施旨在進 一步穩定金融業的架構，約需五 年時間才能落實，足見改革規模 之龐大。 2014年，英國政府亦確認將會長 期徵收英國銀行徵費。此項徵費 於2010年推出，目的之一是補償 納稅人在環球金融危機期間因機 構倒閉而承擔的損失。2014年， 滙豐就此項徵費的支出達11億美 元，較2013年增加2億美元，其中 58%與英國銀行業務無關。 重建信任 業內接連有不當行為曝光，恢復 公眾對銀行業的信任仍是一大挑 戰，但我們必須成功克服這個考 驗。這不僅是為社會，亦是為了員 工。員工把事業前途託付滙豐， 確保他們以此自豪，是滙豐應有 之義。外界憑個別的監控漏洞或 個人的不當行為推斷，質疑滙豐 的企業文化，這對我們的身分認 同無疑是一個沉重打擊。 瑞士私人銀行業務 最近有消息報道集團旗下瑞士私 人銀行業務過往一些不可接受的 經營手法和行為，提醒我們今後 任重而道遠，因為社會對銀行的 責任已有更大期許。事件也提醒 我們需要密切留意監控措施的成 效，以及建立穩健和符合道德的 合規文化。 對於外界所報道的操守及合規缺 失，我們深感遺憾及抱歉。這有 違集團本身的政策，以及公眾對 我們的期望。 因應八年多前資料失竊引發的稅 務調查，集團正徹底改革私人銀 行業務，加強對整個客戶群的盡 職審查及稅務透明度篩查。瑞士 私人銀行業務的客戶及所服務國 家和地區的數目，現時均已縮減 至2007年的大約三分之一。此外， 滙豐已作出行動，落實經合組織 的共同匯報標準及其他措施，務 求提高透明度。我們無法改變過 去，但放眼未來，我們能夠並且 必須持續加強各項監控，提供確 鑿證據證明該等措施行之有效。 這是我們致力實現環球標準的工 作之一，以確保滙豐絕不會在知 情的情況下，與企圖逃稅或利用 金融系統犯罪的交易對手進行業 務往來。 銀行業標準 在更廣泛的層面，繼英國國會銀 行業標準委員會於2013年公布其 建議後，相關的實施工作已取得 重大進展。 《2013年金融服務 （銀 行業改革） 法》 進一步釐清了管理 層和董事會的問責及責任。我們 歡迎Colette Bowe女爵士獲委任為 銀行業標準檢討委員會主席，並 滙豐控股有限公司 6 集團主席報告╱集團行政總裁之回顧 概覽 （續） 承諾會全力支持她的工作。英倫 銀行、英國財政部及金融業操守 監管局正在進行的公平及有效市 場檢討，對於重建批發金融市場 的誠信而言，是非常合時和重要 的措施。 為加強滙豐本身在這些方面的管 治工作，董事會於2014年初設立 行為及價值觀委員會，專注管治 行為的事宜，並已確立為支持董 事會於上述重要領域履行職責的 核心。 董事會人事交替 自中期報告發布以來，我們採取 了進一步措施，加強董事會成員 的專長和經驗，並處理主要職位 的繼任問題。 2015年1月1日，安銘(Phillip Ameen) 加入董事會為獨立非執行董事， 同時出任集團監察委員會成員。安 銘曾擔任通用電氣公司副總裁、 審計長及首席會計師。他在這家 世界領先的國際企業累積了廣泛 的財務及會計經驗，並長期參與 制訂會計準則，具有深厚的技術 知識，必將有助董事會的運作。 憑藉擔任滙豐美國業務董事的經 驗，他亦可就集團董事會的討論 事項提供精闢見解，並加強集團 董事會與主要附屬公司之間既有 的緊密聯繫。 駱耀文爵士此前曾表示，有意於 即將召開的股東周年大會上退任 董事一職。我欣然呈報，駱耀文爵 士已同意留任副主席最少一年。 他除領導各非執行董事履行職責 外，還為我和歐智華提供了有力 的支援。我們很高興能繼續從他 的睿智及經驗中獲益。 150周年 2015年是滙豐在香港及上海創立 的150周年。開業之初，滙豐只是 一家專注於貿易及投資的小型地 區銀行。滙豐仝人對各位先賢心 懷感激及敬意，他們的智慧和遠 見，帶領滙豐成為全球最重要的 金融機構之一，在現今唇齒相依 的世界，為個人及企業提供所需 的金融服務。 展望未來 2015年及往後的種種不明朗因素 和挑戰，多非我們所能控制，尤 其是在經濟復甦乏力和政策支持 措施有限的情況下。因此，我們 必須全盤考慮各種因素和挑戰。 當前地緣政局緊張產生的意外後 果、歐元區成員去留未定、政治變 遷、貨幣及商品價格重新調整、 利率變動及央行非常規政策的成 效，以及眾多其他因素，均可能 嚴重影響經濟狀況、投資信心及 消費決定。作為總部設於英國的 大型金融機構，滙豐更面臨多一 項不明朗經濟因素，就是英國會 否繼續留在歐盟。今天，我們發 表了一份重要研究報告，結論是： 鑑於歐盟市場對英國貿易舉足輕 重，與脫離歐盟相比，合力建成 一體化的服務業市場，並透過改 革來提升歐盟的競爭力，其風險 顯然較低。 目前亦有不少有利的趨勢，支持 我們對未來一年作出正面展望。 中國對外投資呈現增長趨勢、人民 幣有望進一步自由化和國際化， 以及中國經濟由依賴出口轉為倚 靠內需，均令我們深受鼓舞。歐 洲成立資本市場聯盟，而新一屆 歐盟委員會宣布專注發展經濟及 增加就業，亦將帶來業務機會。 美國經濟強勁復甦，加上油價下 跌的效益，可望促進環球經濟增 長。 《跨大西洋貿易與投資夥伴協 定》 及 《泛太平洋夥伴協定》 的談判 進展順利，亦可能令相關群體獲 益良多。現時各方對全球基建投 資資金供應的關注，具有十分重 要的潛在意義。 最後，我謹代表董事會，再次感謝 全球各地的266,000名滙豐同事， 在2014年堅守崗位、勤勉盡力， 協助滙豐繼往開來，與時並進， 再創150年輝煌成就。 集團主席 范智廉 2015年2月23日 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 7 利潤表現未如理想，第四季的艱 難境況，掩蓋了前三季取得的若 干進展。第四季業績面對的挑戰， 有許多亦是行業整體所面對的考 驗。儘管如此，我們仍然看到一 些令人鼓舞的跡象，尤其在工商 金融業務、資金管理服務及人民 幣產品和服務方面。我們亦得以 繼續增加派息。 2014年的列賬基準除稅前利潤為 187億美元，較上年度下降39億美 元，反映出售利潤及重新分類增 益減少，以及其他重大項目的負 面影響，包括總計達37億美元的 罰款、和解開支、英國客戶賠償 及相關準備。 經調整除稅前利潤 （剔除貨幣換 算差額及重大項目按年比較的影 響） 為228億美元，與2013年相若。 亞洲繼續為集團利潤作出重大貢 獻，中東及北非地區2014年的除稅 前利潤更創出新高。亞洲和中東 及北非合共貢獻了集團超過70% 的經調整除稅前利潤。 工商金融業務的列賬基準利潤亦 刷新紀錄，證明我們成功執行策 略。該項業務的收入持續增長， 以香港及英國兩個本位市場的增 幅最為顯著。 環球銀行及資本市場業務在年內 首三季的表現相對較佳，但第四 季則和業內許多機構一樣，表現 未如理想。2014年的收入有所下 降，尤其是資本市場業務，但所 有其他前線業務均錄得按年增長。 零售銀行及財富管理業務的收入 亦告減少，主要原因是我們持續 重整該項業務。然而，在環球投 資管理業務方面，我們繼續推動 集團策略，促進各項環球業務加 強合作，有助吸引290億美元的新 增資金淨額。 環球私人銀行業務的全面改革仍 然持續。是項計劃由2011年開始 加速實施，包括推行多項嚴格措 施，以防範金融犯罪、落實監管 合規和增加稅務透明度。為達到 集團理想的業務模式，我們亦參 考了經六方面考慮得出的結果， 出售若干業務和客戶組合，其中 包括於日本、巴拿馬和盧森堡的 資產。目前，集團旗下的瑞士私 人銀行客戶賬項已較峰值縮減近 七成。2014年內，我們繼續重整 私人銀行的業務模式，包括向LGT Bank出售瑞士的客戶組合，其影 響之一是收入減少。與此同時， 我們亦積極推展符合新模式的業 務，在2014年吸納了新增資金淨 額140億美元，主要來自環球銀行 及資本市場業務和工商金融業務 的客戶。 貸款減值準備下降，反映目前的 經濟環境，以及自2011年以來集 團對業務組合作出的改變。 營業支出上升，乃由於監管及合 規成本上漲、通脹壓力，以及投 資於策略方案以支持業務增長， 其中亞洲及歐洲工商金融業務的 投資尤為重點。重大項目支出 （包 括重組架構成本） 也高於去年的水 平。 有關英國金融業操守監管局和美 國商品期貨交易委員會就外匯市 場進行的查訊，我們已於2014年 達成和解。滙豐的聲譽因少數個 人行為嚴重受損，但這些人並不 代表絕大多數滙豐員工，因為我 們大部分員工均奉行社會對銀行 所期望的價值觀和標準。目前， 事件正由嚴重詐騙調查處妥善處 理。 集團的資產負債狀況保持穩健， 客戶貸款對客戶賬項比率為72%。 若剔除貨幣換算的影響，客戶貸 款於2014年增長280億美元。 於2014年12月31日，過渡基準普通 股權一級比率為10.9%，資本指引 4終點基準普通股權一級比率則 為11.1%。 建立聯繫以助客戶開拓商機 2015年正值滙豐成立150周年。滙 豐於1865年在香港創立，為本地 及國際貿易提供融資，此後迅速 擴展，並把握了亞洲、歐洲及北 美洲之間商貿往來日趨頻繁的商 機。時至今日，為全球各地客戶 建立聯繫的能力，仍然是集團策 略的核心所在。於2014年，我們繼 續開發倚賴國際聯繫的產品，致 力推動相關業務的增長。 我們領先同業的環球貿易及融資 業務仍然表現強勁。在 《全球貿易 評論》 雜誌的 「貿易界領袖」 大獎 中，我們獲評選為 「全球最佳貿易 融資銀行」 和 「中東及北非最佳貿 易融資銀行」 。 集團行政總裁之回顧 2014年是充滿挑戰的一年。年內，我們繼續努力提升業務表現， 同時致力應對營業支出上升所帶來的影響。 滙豐控股有限公司 8 集團行政總裁之回顧╱策略目標 概覽 （續） 資金管理服務方面，客戶委託增 加，客戶服務範圍亦有所改善。 同時，我們在 《歐洲貨幣》 雜誌舉 辦的2014年資金管理調查中，連 續第三年獲評選為 「全球最佳資 金管理銀行」 。 我們於資本融資市場所佔的份額 持續增加，而且在英國和香港兩 個本位市場的債務資本市場，以 及在香港的股票資本市場，均獲 國際市場調查公司Dealogic評選為 世界第一位。在2014年 《國際金融 評論》 雜誌所頒發的大獎中，我們 亦贏得 「年度最佳環球債券行」 、 「年度環球最佳衍生工具機構」 和 「年度亞洲最佳債券行」 多項殊榮。 2014年，我們進一步鞏固在快速 增長的人民幣市場的領先地位。 環球金融訊息系統SWIFT的資料 顯示，在全球最廣泛使用的支付 貨幣中，人民幣已由兩年前的第 十三位，躍升至目前的第五位。 過往12個月內，我們的人民幣產 品收入錄得增長，集團亦繼續在 全球離岸人民幣債券發行機構的 排名中高踞榜首。此外，滙豐在 《亞洲貨幣》 雜誌的 「2014年離岸人 民幣服務調查」 中，獲評選為 「最 佳整體離岸人民幣產品及服務供 應商」 ，並在2014年 《亞洲風險》 雜 誌所頒發的大獎中，贏得 「年度最 佳人民幣服務機構」 的殊榮。 經營環球業務 自2011年集團策略公布以來，經 營環球業務模式的監管相關成本 已顯著上升。 正如集團主席報告所述，監管環 境正持續轉變。 滙豐致力成為世界領先的國際銀 行，意味改善監管合規的實力和 實施環球標準，必然成為我們的 優先要務。自2011年至今，合規部 的職員人數已增逾一倍，但要強 化集團的合規實力，我們仍要再 加努力。 同時，我們所持的資本水平已較 金融危機前增加60%以上。具體而 言，我們已進一步強化資本，以 回應英國審慎監管局的要求。 儘管集團於2011年設定目標時， 已預見日後須按規定持有更多資 本，但此後因種種要求而承擔的 額外支出和資本承諾，卻非當時 所能全面預計。變化的步伐異常 快速，因此集團於2011年設定的 部分目標已不再切合實際。 有鑑於此，我們已重新設定集團 的中期目標，以配合不斷變遷的 經營環境。 我們的股東權益回報率修訂目標 設定為10%以上。新訂目標乃採用 介乎12%至13%的資本指引4終點基 準普通股權一級資本比率以模型 推算。 我們的成本目標是：經調整基準 的收入增長須超過成本增長 （即 收入增長對支出增長的比率為正 數） 。 此外，我們亦重申增加派息的承 諾。具體而言，循序漸進的股息 增長應配合集團整體盈利能力的 提升，並基於我們能夠及時達致 監管規定資本水平的前提來預計。 上述目標能現實地反映滙豐在當 前經營環境下的業務能力。 集團僱員 我十分感謝滙豐全體員工在2014 年的努力、投入和專業表現。 年內，我們為準備接受多個司法 管轄區的壓力測試而進行了大量 工作，結果確定集團具備足夠的 資本實力。滙豐將於2015年面對 更多壓力測試。 我們必須努力確保集團繼續遵守 反洗錢和制裁法例，而在2014年 亦並無鬆懈。 集團管理層和全體員工繼續與監 察員緊密合作，以履行承諾，遵 守我們與美國當局及英國金融業 操守監管局於2012年12月所達成 的和解協議。目前，我們已收到 監察員的第二份年度報告。雖然 報告確認滙豐繼續履行與美國司 法部所訂延後起訴協議下的各種 責任，但一如我們所料，集團往 後仍有大量工作需要完成。 總結及展望 集團維持良好的業務架構，為把 握整體市場趨勢所帶來的機遇而 蓄勢待發。低息環境仍然持續， 宏觀經濟狀況亦維持正面。集團 仍需解決若干過往遺留下來的問 題，但有關事項可望於2015年取 得更大進展。我們亦將繼續進行 2011年開始的簡化架構工作，使 集團更加易於管理和監控。 2014年的業績反映香港業務持續 向好，為集團的整體表現提供有 力支持，亞洲其他地區和中東及 北非的業務亦帶來進一步的重大 貢益。工商金融業務續創佳績， 採用不同方式服務不同類別客戶 的環球銀行及資本市場業務亦展 現出良好韌力，證明集團致力連 繫環球貿易與資金流的策略行之 有效。零售銀行及財富管理業務 的改革仍在進行之中，但減輕業 務風險的工作已於2014年取得較 大進展。此外，環球私人銀行業 務繼續吸引來自其他環球業務的 客戶，並錄得新增資金淨額。我 們亦繼續厲行節約，致力實現淨 額節省，以抵銷因通脹及實施環 球標準而增加的成本。 集團於2015年初的業績令人滿意。 我們將繼續致力推行集團策略， 為廣大股東創造價值。 集團行政總裁 歐智華 2015年2月23日 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 9 策略目標 創建價值及 長期持續發展 我們繼續按照於2011年首次制訂 的滙豐願景，以及有助達成願景 的清晰策略行事。該策略界定我 們競爭的目標範圍與方式。我們 經常根據此策略評估進度，並定 期向相關群體提供最新資料。 我們透過各項主要業務－支付款 項、接受存款、推動貿易、提供 融資和管理風險－在社會和經濟 體系均扮演關鍵角色。我們的目 標是建立和維持可長期持續發展 的業務。 如何創建價值 銀行及銀行內的個別人員以不同 的方式為各方人士創建價值，對 經濟和社會體系均非常重要。 我們讓客戶把儲蓄存入銀行，猶 如放入安全港，既穩妥又方便。 我們直接或透過資本市場，讓存 戶和投資者的資金流向借款人。 借款人運用這些貸款或其他信貸 方式，購入貨品或投資於各種業 務。我們藉此幫助經濟體把個別 而短期的儲蓄匯集成為較長期的 融資。我們把投資者與尋求投資 資金的人集合起來，並發展新的 金融產品。我們亦擔當付款代理 的角色，協助個人和商業客戶在 本國及世界各地進行各項交易。 我們承擔這些活動帶來的風險， 並會加以管理及在價格上反映出 來。 直接貸款包括住宅及商業按揭和 透支，以及有期貸款。我們向經營 國際貿易的進出口商提供融資， 亦向各類公司提供墊款，並以其 客戶的欠款作抵押。 我們亦提供其他金融產品和服 務，包括經紀服務、資產管理、 財務顧問服務、人壽保險、企業 融資、證券服務及另類投資。我 們擔任金融資產的市場莊家，讓 投資者有信心能迅速定價，而且 交投活躍。我們為政府以至大型 及中型企業、中小企、資產豐厚 人士及零售銀行客戶提供以上各 種產品。我們協助客戶在債務及 股票資本市場向外界投資者籌集 資金，又為這些證券造就流通性 和公開價格，讓投資者可在第二 市場進行買賣。我們提供外幣兌 換，以協助國際貿易。 滙豐為廣泛類別客戶提供管理風 險的產品，例如：零售客戶可透 過壽險和退休金產品管理風險， 公司客戶則以應收賬融資或跟單 貿易工具管理風險。企業客戶亦 需要我們的協助，運用滙豐的專 業知識和市場網絡，來管理業 務上的金融風險。 至於資產及負債價值變更及 利率變動引致的風險，則以遠 期、期貨、掉期，以及期權等 衍生工具產品作為主要的管理 工具。因此，我們是活躍的市 場莊家及衍生工具產品交易對 手。客戶利用衍生工具產品來 管理其風險，例如： － 使用遠期外匯合約來對沖 出口銷售的收益或入口原 料的成本； － 使用通脹掉期來對沖日後 的通脹相連負債，例如支 付退休金； － 將不固定的債務利息還款 額轉為定息付款，反之亦 然；或 － 就市場或某些股票價格的 變動為投資者提供對沖。 創建價值 僱員 第三方 政府 股東 僱員 客戶 產品及收益 成本 分派 個人、公司、政府、機構及資產豐厚客戶 及交易對手 分行及辦事處 網絡 市場 貸款/信貸 存款 + 投資 金融服務 利息及費用收益 交易及 其他收益 貸款損失 薪金及其他支出 基礎設施及其他支出 稅項 股息 浮動酬勞 保留利潤 風險及 資本 滙豐控股有限公司 10 策略目標 我們向客戶收取差價，即向客戶 收取的價格與在市場執行對沖的 假設成本之間的差額。若能有效 管理該持倉的風險，在該項交易 期滿時我們便可以保留該差價。 除衍生工具外，我們同時使用其 他金融工具，務使客戶業務產生 的各項風險不會超出風險限額。 在一般情況下，客戶同時買入及 賣出相關工具，在此情況下，我 們會透過與其他交易商或專業交 易對手進行交易，致力管理任何 剩餘風險。若我們未能全面對沖 剩餘風險，則可能獲利或蒙受損 失，因為市況變動會影響組合的 淨值。 我們亦使用壓力測試及其他風險 管理技巧，以確保在不同的潛在 市場境況下，潛在虧損不會超出 我們的承受風險水平。 此外，我們亦管理滙豐集團內部 的風險，包括我們與客戶業務往 來所產生的各項風險。 有關風險的詳情，請參閱第21頁。有關 滙豐如何管理本身風險的詳情，請參閱 第24頁。 長期持續發展 滙豐深明在財務方面的持續成功， 與業務所在地的經濟、環境和社 會狀況息息相關。對我們而言，可 持續發展意味在發展長遠業務的 過程中，所作決策會平衡社會、 環境和經濟方面的考慮因素。這 讓我們推動業務茁壯成長、回饋 股東和僱員、繳納業務所在地的 稅項，以及為社區的健康發展和 成長作出貢獻。我們依賴此基礎 達致股東權益回報持續增長及長 期利潤增長。 我們的營商之道與業務範疇同樣 重要：我們對客戶、員工和股東， 以及廣大社群所肩負的責任，遠 比純粹獲取利潤重要。這包括於 所有業務所在地一致實施最高標 準，以便查察、阻止及防範金融 犯罪。 可持續發展是集團優先策略的基 礎，讓我們可達成目標。我們有 能力識別和處理可能為業務帶來 風險或機遇的環境、社會和道德 發展趨勢，正是我們在財務上取 得成功的要素。我們的可持續發 展決策會構建集團的聲譽、加強 員工的投入感，以及影響業務的 風險狀況，並有助降低成本和開 拓新的收入來源。 滙豐網絡遍布全球，在本位和優 先發展市場的多項業務早已根基 穩固，加上產品和服務組合極其 廣泛，均突顯滙豐比同業競爭對 手優勝，而且業務和營運模式尤 其穩健。這使集團在銀行業最動 盪的時期仍能賺取利潤，我們相 信這些模式對集團的未來續有助 益，並會支持我們實現優先策略。 有關業務及營運模式的詳情，請參閱第 12頁。有關滙豐的可持續發展，詳情請 參閱第36頁。 滙豐價值觀 滙豐的首要目標之一，是在每次 作出決定、與客戶溝通，乃至員 工之間互動交流時，都能貫徹滙 豐價值觀。這也是決定集團營運 方式的要素。 在日常營運中實踐滙豐價值觀， 正是我們文化的底蘊。隨著監管 政策、投資者信心及社會對銀行 的期望不斷演變，秉承滙豐價值 觀尤其重要。我們會依據滙豐價 值觀，挑選、評估、嘉許、酬報和 培訓員工。我們期望集團的行政 人員和僱員在執行職務時，均敢 於以誠信正直行事，做到以下各 點： 滙豐價值觀 穩妥可靠，堅守正道 － 堅持正道、實現承諾、堅毅 穩健及值得信賴； － 勇於負責、具決斷力、善用 判斷力和常識，並鼓勵他人 承擔權責。 坦誠開放，接納不同理念和文 化 － 溝通態度開誠布公、不懼挑 戰，能從錯誤中汲取教訓； － 聽取意見、公平待人、親和 共融，且尊重不同理念。 重視與客戶、社區、監管機構 及員工之間的緊密聯繫 － 建立聯繫網絡、注重對外關 係、跨越領域積極合作； － 關心個人及其發展、給予尊 重、支持和回應。 創建價值及長期持續發展╱滙豐價值觀╱我們的策略 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 11 我們的策略 長遠趨勢 競爭優勢 兩方面的策略 我們訂定的策略配合以下兩項 長遠趨勢： － 全球經濟的聯繫日益緊密， 國際貿易量及跨境資金流的 增長，繼續超越國內生產總 值的平均增幅。在未來10年 內，我們預期貿易及資金流 增長超出國內生產總值的增 長，世界貿易的85％增長將 來自35個市場，而跨境資金 流的集中程度亦不相伯仲。 － 我們預期，受惠於人口結構 與城市化的發展，在全球最 大的30個經濟體中，亞洲、 拉丁美洲、中東及非洲地區 若干經濟體的規模，至2050年 會擴大至大約四倍，尤勝 歐洲及北美洲合計的水平。 到了2050年，最大的30個經 濟體中，預期有18個來自亞 洲、拉丁美洲或中東及非洲。 在目前的環境下，重要的是： － 擁有國際網絡與提供環球產 品的實力，從而把握國際貿 易與資金流所帶來的商機； 以及 － 在增幅最吸引的市場，把握 機遇作有機投資，並維持本 身的投資實力。 滙豐的競爭優勢來自： － 我們在主要的策略市場具備 可觀的業務規模，並且承諾 作長遠的發展； － 我們的業務網絡覆蓋85%以 上的環球貿易與資金流； － 我們擁有均衡的業務組合， 以服務全球客戶為本； － 我們實力雄厚，既可強化資 本基礎，同時向僱員發放具 競爭力的獎勵，並為股東提 供上佳的回報； － 我們的資金基礎穩固，客 戶賬項款額約達14,000億美 元，其中72%已借予客戶；及 － 我們在最有利的金融樞紐， 維持強大的資產負債實力與 交易營運能力。 為應對此等長遠趨勢，我們制 訂了一項兩方面同時並進的策 略，以反映我們的競爭優勢： － 業務網絡聯繫全球。滙豐具 備理想條件把握與日俱增的 國際貿易與資金流。我們的 業務網絡覆蓋全球，而且服 務多元化，是服務客戶的優 越條件，可透過工商金融與 環球銀行及資本市場業務， 協助他們由小企業發展成大 型跨國企業。 － 財富管理及零售銀行業務具 備本土實力。我們的目標為 透過集團的卓越理財方案與 環球私人銀行業務，在橫跨 亞洲、拉丁美洲及中東地區 的優先發展市場，捕捉社會 流動性與創富增值所帶來的 商機。而且，我們預期只會 在業務規模足可為集團帶來 盈利的市場，投資於全方位 的零售銀行業務。 業務模式 滙豐控股有限公司 12 業務模式 市場覆蓋 我們的業務模式是透過國際網 絡，聯繫及服務多個緊密相連的 市場。 我們藉營運附屬及聯營公司，提 供全面的銀行及相關金融服務。 這些服務主要由擁有龐大本土存 款基礎的本土銀行提供。 投資準則 我們循六方面的考慮決定集團作 出投資的時間和範疇。首兩個考 慮因素－國際聯繫和經濟發展－ 確定業務是否符合集團策略。另 外三個考慮因素－盈利能力、成 本效益和流動資金－確定業務的 財務狀況是否具有吸引力。第六 個考慮因素－金融犯罪風險－管 控高風險司法管轄區內的活動， 並在適當情況下藉限制業務範圍 作自我保障。 在決定投入額外資源的範疇時需 考慮以下三個方面： － 策略：我們僅會投資於符合集 團策略的業務，大部分是在我 們的本位和優先發展市場，以 及目標業務與客戶； － 財務：投資必須能為集團增 值，而且回報、收入和成本必 須達到最低限度的要求；以及 － 風險：投資必須符合我們的承 受風險水平。 我們會依據以上六方面的考慮進 行地區及業務組合的年度檢討， 以更新我們的市場及業務的優先 發展範疇。 決策時會作出六方面考慮 與集團策略 有何關係？ 目前回報 是否吸引？ 有何金融犯罪風險？ 1. 聯繫能力 2. 經濟發展 3. 盈利能力 4. 成本效益 5. 流動資金 6. 金融犯罪風險 高 低 投資 轉虧為盈╱改善 維持現狀 高 中╱低 結束╱出售 減低風險 減低風險 減低風險 是 否 是 否 英國和香港是我們的本位市場， 另外19個國家或地區則是我們的 優先發展市場 （見下文） 。這21個 市場佔集團2014年除稅前利潤超 過90%，是我們投放資本的重點所 在。網絡市場富濃厚的國際色彩， 與滙豐的國際網絡相輔相成，主 要透過工商金融業務和環球銀行 及資本市場業務營運。我們的本 位市場、優先發展市場及網絡市 場合共覆蓋全球國際貿易及資金 流約85%。 最後一類是小型市場，包括我們 已在當地建立足以產生盈利的經 營規模或設有代表辦事處的市場。 我們的法律實體受當地監管機構 監管；在英國，整個集團的審慎 監管事宜 （有關安全和穩健） 受審 慎監管局 （ 「PRA」 ） 監管，而操守事 宜 （有關保障消費者及市場） 則受 金融業操守監管局 （ 「FCA」 ） 監管。 滙豐的市場 － － － － － － － － － － － － － － － － － 香港 歐洲 亞洲 中東及北非 北美洲 拉丁美洲 優先發展 市場 網絡市場 小型市場 英國 澳洲 中國內地 印度 印尼 馬來西亞 新加坡 台灣 業務對象主要是工商金融業務和環球銀行及資本市場業務的國際客戶及企業 這些市場連同本位及優先發展市場，合共覆蓋國際貿易及資金流約85% 滙豐已建立具盈利能力的經營規模和重點業務的市場 代表辦事處 法國 埃及 加拿大 阿根廷 巴西 墨西哥 美國 沙地阿拉伯 阿聯酋 德國 瑞士 土耳其 本位市場 － － － － － － － － 市場覆蓋╱組織架構 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 13 組織架構 我們的營運模式以矩陣管理架構 為基礎，並透過各環球業務、地 區及環球部門營運業務。 這個矩陣由一個法律實體架構組 成，並以滙豐控股有限公司為首。 控股公司 滙豐控股為集團的控股公司，是 旗下附屬公司股本的主要來源， 在有需要時也會提供非股權資本。 集團管理委員會 （ 「GMB」 ） 獲滙豐 控股董事會授權，在董事會設定 的承受風險水平內，負責集團的 管理工作及日常運作。集團管理 委員會負責確保集團擁有充足的 現金來源，可向股東支付股息、 向債券持有人支付利息、承擔開 支及繳稅。 滙豐控股並無向任何經營銀行業 務的附屬公司提供核心資金，亦 非該等附屬公司在最後關頭的貸 款人，本身亦無從事任何銀行業 務。滙豐旗下附屬公司均為獲個 別撥資的公司，在營運上實施集 團策略。 環球管理架構 下表詳列我們四大環球業務、五 個地區和11個環球部門，並概述 它們在滙豐管理架構下的職責。 有關主要附屬公司的詳情，請參閱財務 報表附註22。集團的簡明架構圖載於第 462頁。 環球管理架構 滙豐控股有限公司 環球業務 地區 法律實體 環球部門 － 歐洲 － 亞洲 － 中東及北非 － 北美洲 － 拉丁美洲 － 企業傳訊部 － 公司秘書部 － 企業可持續發展部 － 財務部 － 滙豐科技及服務部 － 人力資源部 － 審核部 － 法律事務部 － 市場推廣部 － 風險管理部 （包括合規部） － 策略和規劃部 － 零售銀行及財富管理 － 工商金融 － 環球銀行及資本市場 － 環球私人銀行 負責制訂全球一致的業務策略和營運模式、發出 相關業務的規劃指引，並對其利潤及虧損之表現 負責，以及管理屬下員工數目。 各有其本身的董事會，在運作上為獲個別撥資的實體，負責落實集團策略及提供集團產品和服務。各法律實體須按照集團為其所在國家或地區釐定的承受風險水 平，考慮本身的風險及維持適當的緩衝資本。這些實體會在集團設定的範圍內管理本身的資金和流動資金。 共同承擔落實環球業務所訂策略的責任。各地區 代表集團面向客戶、監管機構、僱員團隊及其他 相關群體；分配資本；按法律實體管理承受風險 水平、流動資金及資金；並對利潤及虧損表現能 否符合環球業務計劃的目標承擔責任。 負責制訂及管理與部門職能有關的一切政策、 程序及服務平台、全面負責部門的全球各地支 出，並管理屬下員工數目。 滙豐控股有限公司 14 業務模式 （續） 組織架構╱管治 結構性改革 銀行業務結構性改革及 復元和解決計劃 全球有多項發展是與銀行業務結 構性改革及推行復元和解決機制 有關。 制訂復元和解決計劃後，一些監 管機構和國家當局亦要求銀行改 變其公司架構，包括要求銀行在 本地註冊成立或將若干業務分隔 運作。英國已頒布分隔運作的法 例，要求將零售及中小企的存款 與交易活動分隔 （見下文） 。在其 他司法管轄區，類似規定已經推 行或正準備推行。 復元和解決計劃的政策背景 經歷金融危機後，20國集團的領 袖要求金融穩定委員會 （ 「FSB」 ） 就 28家 （現時為30家） 指定全球系統 重要性銀行 （ 「G-SIB」 ） 制訂更有效 的復元和解決安排，導致提出一 系列有關復元和解決計劃、跨境 合作協議，以及減低阻礙解決方 案的措施之政策建議。 於2013年12月，審慎監管局為英國 銀行及在英國經營業務的國際銀 行訂立復元和解決規則。此等規 則經修訂後成為由2015年1月起實 施的歐盟 《銀行復元和解決指引》 的一部分。 滙豐的解決策略及公司架構變動 我們一直與英倫銀行、審慎監管 局及集團的其他主要監管機構緊 密合作，為滙豐制訂和協定一項 解決策略。我們認為，解決策略 倘能讓集團於推行解決方案時在 附屬銀行層面分拆 （稱為多點進 入策略） ，而非於推行解決方案時 保留在集團內 （稱為單點進入策 略） ，是最佳的方法，因為這個方 法能配合集團的現有法律和業務 架構。 就如所有全球系統重要性銀行一 樣，我們正與監管機構一起了解 集團內不同業務部門和附屬銀行 實體之間的互相依存性，以提高 可予分拆的程度。 我們已制訂計劃減低或消除不同 業務部門和附屬銀行實體之間的 重大互相依存性，以便進一步協 助集團推行解決方案。具體安排 包括，為消除營運上的互相依存 性 （即由一家附屬銀行向另一家提 供重要服務） ，我們決定將該等重 要服務由附屬銀行轉移至一群獨 立註冊成立的服務公司 （ 「服務公 司集團」 ） 。服務公司集團將獲獨 立撥資，以確保在推行解決方案 過程中持續提供服務。服務公司 集團的主要部分已經存在，此方 案只涉及把仍由附屬銀行負責的 餘下重要服務轉移至服務公司集 團。該等服務此後將由服務公司 集團向附屬銀行提供。 英國的分隔運作安排 2013年12月，英國 《2013年金融服 務 （銀行業改革） 法》 （ 「銀行業改革 法」 ） 獲御准通過。該項法案實施 了銀行業獨立委員會的大部分建 議，其中包括規定大型銀行集團 「分隔」 出英國零售銀行業務，讓 被禁止從事重大交易活動且獨立 註冊成立經營銀行業務的附屬公 司 （ 「分隔運作銀行」 ） 負責。就此 目的而言，英國並不包括皇家屬 地。分隔運作安排將於2019年1月 1日完成。 2014年7月，第二階段立法工作最 終落實，其中包括訂立條文進一 步詳列經參考資產總值而應轉移 至分隔運作銀行的個別客戶，以 及根據營業額、資產及僱員數目 而應轉移的企業。此外，第二階 段立法限制了分隔運作銀行的活 動和地域覆蓋範圍。 2014年10月，審慎監管局公布一份 關於分隔運作規則的諮詢文件， 內容關於法律架構、管治，以及 服務和融資的持續性。審慎監管 局擬進行進一步諮詢，並於適當 時候最終落實分隔運作規則。審 慎監管局亦公布一份有關在解決 方案下持續營運的討論文件。 根據審慎監管局諮詢文件的要 求，我們已於2014年11月向英國的 監管機構提呈一項臨時分隔運作 計劃。該計劃訂明是根據法律規 定於一家獨立附屬公司進行的分 隔運作活動，大致為現時屬英國 滙豐銀行有限公司 （ 「英國滙豐銀 行」 ） 部分業務的零售和中小企服 務。 此外，該計劃又訂明加強服務公 司集團服務的建議，反映審慎監 管局各份諮詢及討論文件對營運 持續性的預期。該計劃仍有待內 部進一步計劃和審批，並最終須 經審慎監管局、英國金融業操守 監管局及其他相關監管機構審批。 歐洲銀行業結構性改革 2014年1月，歐洲委員會公布對歐 洲銀行業結構性改革的立法建議， 包括建議禁止金融工具及商品的 坐盤交易，以及監管機構可酌情 要求若干交易活動在獨立的附屬 公司進行，與接受存款業務分隔。 分隔運作的接受存款公司須與進 行交易活動的公司分別營運，包 括規定分隔資本及管理結構、發 行本身債務及按公平原則進行公 司間交易。 建議的草擬本載有條文，容許已 執行本身結構性改革立法的成員 國減少實施此等規定，但須符合 若干條件。鑑於英國落實銀行業 改革法，此項容許減少實施的安 排或會對英國有利。 該等建議現有待歐洲議會及部長 理事會討論。根據最終規則實施 任何分隔的日期，將須視乎就最 終法例 （如有） 達成協議的日期而 定。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 15 管治 董事會致力在業務所在地確立及 維持最高水平的企業管治。良好 的企業管治對滙豐的長遠成功和 持續發展的能力非常重要。 我們相信，滙豐的持續成功之道， 在於強健而具透明度的企業管治 架構。董事會一直專注加強我們 的企業管治架構，以支持集團成 功實施環球標準。 董事會和委員會的角色 滙豐的策略和承受風險水平由董 事會釐定，而日常業務營運則由 董事會授權集團管理委員會負責。 除了定期會議之外，集團管理委 員會還負責舉行風險管理會議。 董事會設立的非執行董事委員 會之主要任務在下圖說明。各主 要非執行董事委員會的職權範 圍可於滙豐網站www.hsbc.com/ boardcommittees查閱。 有關集團企業管治的詳情，請參閱第 263頁。 滙豐控股董事會的委員會結構及企業管治架構 滙豐控股有限公司 董事會 集團風險管理委員會 集團監察委員會 集團薪酬委員會 提名委員會 集團管理委員會 金融系統風險 防護委員會 行為及價值觀 委員會 主席委員會 慈善及社區投資事務 監察委員會 從高層次角度監督風險相 關事宜及風險管治，並就 此向董事會提供建議的非 執行責任。 監督財務報告相關事宜 及對財務報告進行內部 監控，就此向董事會提 供建議的非執行責任。 為集團的薪酬政策及高級 行政人員的薪酬制訂總體 原則、參數及管治架構的 非執行責任。 掌管董事會的委任程序， 以及物色和提名董事人選 供董事會批核的非執行責 任。 負有監督以下事宜的非 執行責任： (i)監控措施及程序，以 查找可能使滙豐乃至整 個金融系統受到金融犯 罪或濫用系統行為侵襲 的範疇，及 (ii)滙豐的政策及程序是 否足以確保集團持續履 行監管及執法機構所訂 責任。 由董事會直接授權負責滙 豐的管理及日常運作事宜 的執行管理委員會。 監督滙豐政策、程序及 標準，以確保集團以負 責任的態度經營業務， 同時恪守滙豐價值觀， 並就此向董事會提供建 議的非執行責任。 於董事會的預定會議之間 代表董事會行事，以便為 急需董事會通過的臨時不 可預見事務召開緊急董事 會會議。 肩負監督滙豐各項慈善及 社區投資活動的非執行責 任，以支持集團實現企業 可持續發展的目標。 有關註釋，請參閱第39頁。 集團管理委員會執行委員會 集團管理委員會 風險管理會議 透過強大的風險管治架構，就管理整個企業的所有風險提供策略方向和進行監督，特別著重界定承受風險水平和監察風險狀況，包括評估現有和新浮現風險。 建議及╱或批准主要的風險限額、政策和方法，從而管理風險。 制訂及實施各項反映最佳實務的環球標準，以供整個集團採用。 滙豐控股有限公司 16 業務模式 （續） 環球業務 我們的四大環球業務是：零售銀 行及財富管理、工商金融、環球銀 行及資本市場和環球私人銀行。 四大環球業務負責在集團內貫徹 其開拓、實施及管理業務方案，專 注提升盈利能力及效益。該等業 務連同各地區的業務部門，在集 團策略的範圍內制訂本身策略， 負責發出有關該等業務的規劃指 引，對其利潤及虧損負責，並管 理屬下員工數目。 環球業務的主要業務活動，以及該等業 務的產品及服務概述如下。 2014年環球業務的主要業務活動 零售銀行及財富管理 工商金融 環球銀行及資本市場 環球私人銀行 環球業務 負債主導型 資產主導型 費用主導型 及其他 － 存款 － 戶口服務 － 存款 － 資金管理 － 存款 － 資金管理 － 資產負債管理 － 存款 － 戶口服務 － 信貸及貸款 － 信貸及貸款 － 信貸及貸款 － 信貸及貸款 － 資產及貿易融資 － 國際貿易及應收賬融資 － 資產管理 － 匯財策劃及財務策劃 － 經紀業務 － 保險 分銷；制訂壽險產品） 6 － 商業保險及投資 7 8 － 企業融資 － 資本市場 － 證券服務 9 10 7 6 11 － 資產管理 － 財務顧問 － 經紀業務 － 企業融資 （透過環球銀 行及資本市場業務提供） － 另類投資 － 信託及遺產規劃 （ 有關註釋，請參閱第39頁。 零售銀行及財富管理 產品及服務 零售銀行及財富管理業務接受存 款及提供交易銀行服務，讓客戶 管理其日常財務，並為未來儲蓄。 我們提供信貸融資，協助客戶應 付長短期的借貸需求；同時提供 財務顧問、經紀、保險及投資服 務，幫助客戶管理財富，未雨綢 繆。 滙豐開發的產品，可滿足特定客 戶群的需要，包括一系列的不同 服務與服務途徑。 零售銀行及財富管理業務提 供四項主要服務： － 滙豐卓越理財：設專責客戶經理 為中上階層客戶及其直系親屬 提供專業及專門設計的意見。 客戶可於全球各地獲得緊急旅 遊支援、優先電話理財服務， 並可於網上透過 「環球賬戶」 功 能查閱名下所有卓越理財戶口。 － 滙豐運籌理財：協助新冒起的中 上階層客戶管理日常財務，讓 客戶享用一系列優越的產品、 利率和條款。滙豐運籌理財亦是 我們與客戶建立關係的起點， 我們藉此為客戶提供支援及指 引，助其實現理想。 － 匯財策劃及財務策劃：按照個 人客戶特定需要設計的財務策 劃程序，協助客戶保障、增加 和管理財富。我們提供由環球 投資管理、資本市場及滙豐保 險業務與精選第三方供應商制 訂的投資及財富保險產品。 － 個人理財：利用全球產品平台， 按照全球適用的服務標準，提 供既符合環球標準亦切合本土 需求的理財產品與服務。這些 產品與服務不單可靠易明，而 且具有理想的價值。 零售銀行及財富管理業務通過四 個主要途徑提供服務：分行、自助 銀行終端機、電話理財中心及電子 銀行 （互聯網及手機） 。 客戶 零售銀行及財富管理業務為約 5,000萬名客戶服務。我們致力與 客戶建立終身關係，伴隨客戶經 歷人生各個階段，為他們提供專 門設計的產品和服務，以配合其 人生不同目標、願望和鴻圖大計。 當部分客戶面對財務壓力時，我 們的宗旨是以耐心、公平和諒解 的態度給予支持，與客戶共渡難 關。 滙豐時刻以客戶為念，不斷研究和 投入資源，確保客戶可以方便、安 全和可靠地享用我們的服務。我 們一直致力確保出售的產品能切 合客戶所需，而所訂價格對客戶 及股東雙方而言均屬公平合理。 我們亦推行一項新獎勵計劃，這 項計劃與產品銷量並無掛鈎，反 而側重於評估我們能夠滿足客戶 需求的程度。 我們在特定國家或地區對零售銀 行客戶進行獨立市場研究調查， 以衡量客戶的滿意度，並根據 「客 戶推薦度指標」 ，對表現評分。我 們將此項指標的得分，對照每個 市場銀行同業組別的平均得分， 並為各項業務訂立相對於同業競 爭對手的目標。我們預期會在集 團業務所在市場持續改善。我們 的目標是迅速而公平地處理客戶 投訴，監察指標的趨勢，從而進 環球業務 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 17 一步改善服務。 工商金融 產品及服務 工商金融業務提供廣泛系列的銀 行及金融服務，讓客戶可以管理 和擴展其本土和國際業務。我們 透過加強集團內部不同地區和環 球業務之間的合作，協助客戶進 軍各地市場，以達成公認為全球 領先的國際貿易及商業銀行的目 標。全球性的營運模式能夠增加 透明度、令服務貫徹一致、提升 效率，以及提供客戶確實所需的 服務。 工商金融業務為客戶提供的 產品及服務一般包括： － 信貸及貸款：提供範圍廣泛的 本土及跨境融資服務，包括透 支、商務卡、有期貸款，以及銀 團、槓桿、收購及項目融資。在 選定的國家╱地區，我們亦提 供資產融資服務。 － 環球貿易及融資：支援客戶連 繫全球貿易趨勢，並以無與倫 比的經驗，協助客戶應對現今 最複雜的貿易挑戰。我們全面 的產品及服務、信用證、託收、 擔保、應收賬融資、供應鏈方 案、商品及結構融資，以及分 散風險服務，可以結合於我們 的環球方案中，讓企業於整個 貿易程序內易於管理風險、處 理交易及進行融資活動。 － 資金管理： 我們的策略市場覆 蓋全球大部分付款及資金流來 源地。我們透過電子平台提供 本土、地區性和全球性的交易 銀行服務，包括付款、託收、戶 口服務、 電子商貿，以及流動資 金管理，以滿足客戶所需。 － 保險及投資：在選定的國家╱地 區提供業務及財務保障、貿易 保險、僱員福利、企業財富管 理及多種其他商業風險保險產 品。 － 合作：我們的工商金融業務為 環球銀行及資本市場業務、零 售銀行及財富管理業務與環球 私人銀行業務所提供的產品和 服務 （包括外匯、利率、資本市 場和顧問服務、支薪和個人戶 口服務，以及財富管理和財富 過渡服務） ，建立了主要的客戶 基礎。 滙豐正主導以人民幣作為貿易貨 幣的發展，在超過50個市場提供 人民幣服務。 我們的一系列產品、服務和服務 途徑，是專為迎合特定客戶群的 需要而設。 客戶 我們基於客戶的需要和其複雜程 度在工商金融業務的組織架構內， 發展以下三類獨特業務：商務理 財、中型企業及大型企業。 － 商務理財業務現時有兩個按 客戶需要而提供的獨特服務 模式：客戶經理專注為有較複 雜需要的客戶服務；如客戶需 要較簡單及較常見的產品和 服務則進行組合管理。 － 我們已加強專注中型企業客 戶，同時在各個本位和優先發 展市場重組架構及資源，以提 升客戶管理。 － 至於一般有複雜和跨國需要 的大型企業客戶，則由全球管 理的資深客戶服務團隊提供 服務。這些團隊亦能聯繫其他 環球業務部門。 為確保客戶永遠是集團業務的重 心，我們一直十分重視客戶的回 應和投入。今年已是我們第六年 進行客戶關係研究計劃，這項覆 蓋15個市場的環球調查旨在加深 對客戶的了解和鞏固與客戶的關 係。此項計劃連同其他研究計劃， 有助我們識別客戶的重要業務事 宜，讓我們可以為客戶專門設計 更切合所需的解決方案和服務。 與聲譽良好的客戶建立長遠關係， 是集團增長策略和機構價值觀的 核心所在。 環球銀行及資本市場 產品及服務 環球銀行及資本市場業務透過八 個面對客戶的業務部門提供批發 資本市場和交易銀行服務。 環球銀行及資本市場業務的 產品及服務 包括： － 資本市場：在第二市場提供銷 售及交易服務，並根據資產類 別劃分為四項業務。 － 信貸及利率交易：向企業、 金融機構、主權國家、機構 及公共部門發行人等客戶出 售、買賣及分銷定息證券； 並透過利率及信貸衍生工具 協助客戶管理風險，同時透 過回購協議協助客戶融資。 － 外匯：提供現貨及衍生工具 產品，以配合機構投資者的 投資需求；環球銀行及資本 市場業務與工商金融業務的 中小企、中型企業和大型企 業客戶的對沖需要，以及各 分行內的零售銀行及財富管 理業務和環球私人銀行業務 客戶的需要。外匯服務可代 表客戶買賣超過90種貨幣。 － 股票：為客戶提供銷售及買 賣服務，包括讓客戶直接在 市場買賣，以及提供融資和 對沖方案。 － 資本融資：以客戶的資本結構 為重點，提供策略性的融資及 顧問服務。產品包括在第一市 場籌集債務及股本資本、革新 的併購交易顧問與執行，以及 企業借貸和專門的結構融資方 案，例如槓桿和收購融資、資 產及結構融資、房地產、基建 及項目融資，以及出口信貸。 － 資金管理：協助客戶轉移、管 控、獲取及投放所持現金。產 品包括非零售存款，以及國際 性、地區性和本土的資金管理 服務。 － 證券服務：為企業及機構客戶 提供託管和結算服務，同時為 本土和跨境投資者提供資金管 理服務。 － 環球貿易及融資：代表環球銀 行及資本市場業務在整個貿易 程序內為客戶提供貿易服務。 除上文所述者外，資產負債管理 業務負責為集團管理流動資金和 資金，亦負責在資本市場業務的 限額架構下管理結構利率持倉。 滙豐控股有限公司 18 業務模式 （續） 客戶 環球銀行及資本市場業務向全球 各大政府、企業和機構客戶提供 專門設計的理財方案。在滙豐的 管理架構下，這是一項由地區監 督的環球業務，並以建立長期客 戶關係為經營理念，旨在透徹了 解客戶的財務需要和策略目標。 客戶服務以銀行服務為核心，旗 下的客戶經理團隊按行業、地區 和國家劃分，負責了解客戶所需， 同時運用滙豐廣泛系列的產品及 無遠弗屆的環球網絡，為客戶提 供全面的服務方案。 滙豐的目標是要成為優先客戶的 「五大首選」 銀行。我們透過組成 跨地域網絡的客戶服務團隊，與產 品專家緊密合作，致力開發滿足 個別客戶需求的專門設計方案， 務求達成我們的目標。我們的客 戶服務及產品團隊由一個獨有的 客戶關係管理平台及全面的客戶 規劃流程支援。我們的團隊使用 此等平台以更佳服務鞏固全球客 戶關係，讓我們協助客戶掌握國 際增長的機會。 環球私人銀行 產品及服務 我們以滙豐的優勢及市場上最適 合的產品，與客戶合作以提供各 種策劃方案，協助客戶增長、管理 及保障其現有和未來的財富。我 們的產品及服務包括私人銀行、 投資管理和私人匯財策劃。 環球私人銀行業務的產品及 服務包括： － 私人銀行服務包括多種貨幣及 受信存款、戶口服務，以及信 貸與專項借貸。環球私人銀行 業務亦運用滙豐的全面銀行營 運實力，為客戶提供各式產品 及服務，例如信用卡、網上理 財，以至企業銀行及投資銀行 服務方案。 － 投資管理包括顧問及全權代理 投資服務，以及多類資產的經 紀服務。這些服務涵蓋全面的 投資工具、組合管理、證券服 務及另類投資。 － 私人匯財策劃包括信託及遺產 規劃，旨在保障財富及將財富 留給子孫後代。 客戶 環球私人銀行業務在我們的本位 及優先發展市場為資產豐厚和資 產極豐厚的客戶及其家族提供服 務，滿足他們的需要。 在此等廣泛領域內，環球私人銀 行業務設有專責隊伍，服務滙豐 的環球優先客戶 （包括最重要的集 團客戶） ，以及由工商金融業務和 環球銀行及資本市場業務提供私 人銀行方案及服務的其他客戶。 我們的目標是整個集團與此等客 戶建立和加強聯繫，與各環球業 務建立強大的關係，以滿足客戶 需要。我們旨在利用滙豐長久以 來積累的商業銀行經驗和優勢， 成為資產豐厚企業家的領先私人 銀行。 客戶經理是客戶接觸我們的橋樑， 專門設計滿足客戶個人需要的服 務。客戶經理須透徹了解客戶 （包 括客戶的家庭、業務、生活作風 和鴻圖大計） ，然後引薦具有專門 知識的專家，協助客戶制訂最佳 的財務策略。這些專家包括： － 投資顧問，他們根據客戶的投 資和風險概況，與客戶商討相 關投資概念； － 信貸顧問，他們就複雜的流動 資金和借貸需求提供專業意 見；及 － 財富策劃師，他們擁有所需的 知識和專業技能，以管理現有 和傳給子孫後代的財富。 電子平台的使用率與自管客戶的 強勁需求持續同步增長。此等平台 讓客戶可以直接獲取戶口資料、 投資研究和網上交易服務。我們 繼續投資於電子系統，以更佳服 務配合客戶不斷演變的期望和需 要。 僱員 滙豐成功加強價值主導的高績效 文化，對持續實施環球標準極為 重要。無論是作出決定、同事間 的溝通或與客戶互動交流，都時 刻貫徹滙豐價值觀。 － 我們銳意吸納、挽留及激勵最 優秀的人才，並以薪酬政策支 持此目標。 － 我們清晰界定所需的才能，積 極管理繼任計劃，同時發掘別 具才幹的員工，給予適當的事 業及發展機會，讓其於滙豐盡 展所長，以協助達致繼任計劃 的目標。 － 我們提供培訓及發展機會，讓 僱員獲得所需的技術及領導才 能，以發展事業前途。 － 我們致力推行多元及共融文 化，配合多元化的客戶基礎。 － 我們鼓勵僱員投入其工作所在 的社區。 於2014年底，滙豐合共聘用266,000 名全職和兼職僱員，2013年底及 2012年底的數字分別為263,000名 及270,000名。僱員主要集中在以 下國家╱地區 （大約人數） ： 英國 48,000 印度 32,000 香港 30,000 巴西 21,000 中國內地 21,000 墨西哥 17,000 美國 15,000 法國 9,000 其他 73,000 環球業務╱僱員 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 19 管理層結構 於本報告日期，滙豐的高級管理 層包括四名執行董事、11名集團 常務總監及60名集團總經理。在該 等高級管理人員中，有13名(17%) 為女性。此領導團隊的成員分駐 於17個不同國家╱地區，共包含 13個不同國籍，其中71%在滙豐 工作10年以上，而整體平均任期 為20年。 滙豐有13名非執行董事。 員工政策 滙豐價值觀 2014年，我們繼續以入職和其他 培訓課程，教育滙豐各階層的僱 員學習滙豐價值觀。這些課程涵 蓋相關的技術、管理及領導技巧。 我們要求所有僱員奉行崇高的行 為標準。我們致力持續向集團各 階層僱員灌輸有關價值觀及誠信 正直的重要性。例如，我們已更新 僱員入職培訓計劃的內容，以進 一步加強灌輸誠信正直的價值觀 和滿足客戶需要的技巧。此外， 我們對員工遵循滙豐價值觀和 相關的行為表現所作的評估，已 正式成為全體員工的考績評估程 序的一部分。2014年約有145,000名 僱員接受價值觀培訓，而2013年 則有135,000名僱員接受培訓。 預期2015年將有另100,000名僱員 接受培訓。部分僱員因為違反上述 價值觀已離開集團。 僱員發展 僱員發展是集團業務及營運得以 加強及蓬勃發展的重要一環。我 們有系統地物色、培育及調配人 才，確保有大量具適當價值觀、 技能及經驗的優秀人才，可以接 任當前及未來的高級管理層職位。 我們的培訓一貫配合時代步伐， 並定期檢討課程以改善質素，確 保僱員取得營運環球機構所需的 技術和領導才能。我們正規範各 項培訓課程，協助僱員在所有市 場向客戶提供一貫優質的服務， 並支持緩減現有及新浮現風險， 以及推行環球標準計劃。 僱員投入 僱員如能積極投入，會帶來正面 的商業效益及促進業績改善、提 高客戶的滿意度、提升生產力、 加強挽留人才及降低缺勤率。 我們透過 「環球員工意見調查」 ， 評估僱員的投入程度。該調查於 2007至2011年每年進行，此後每兩 年進行一次。最近一次於2013年 進行，集中於評估僱員有否盡力 達成集團的目標，以及僱員認為 自己能否為滙豐達成世界領先國 際銀行的抱負出一分力，以衡量 對價值主導之高績效文化的支持 度。 若與金融服務業及業內最佳標準 比較，我們的僱員投入程度仍然 正面。於2013年，整體投入程度的 得分是68%，較金融服務業的標準 高4個百分點，但較同業最佳標準 低8個百分點。在認知風險 （81%， 且高於同業最佳標準9個百分點） 、 領導能力(67%)，以及貫徹滙豐價 值觀(77%)方面均得分甚高。僱員 發展方面得分顯著改善，從2011年 低於同業最佳標準6個百分點提升 至2013年高於同業最佳標準3個百 分點。需要關注的地方包括自豪 感及擁護，其得分分別低於同業 最佳標準12個百分點及13個百分 點，較2011年下降。下次的 「環球 員工意見調查」 將於2015年進行。 滙豐亦定期進行一項名為Snapshot 的調查，每三個月向集團四分之 一的僱員進行這項調查。透過 Snapshot得出的調查結果適時反 映僱員對集團的想法，包括反映僱 員投入程度的元素。於2014年9月 底，僱員對於特定問題有以下正 面回應：支持滙豐的策略，81%； 在三年內會繼續效力滙豐，74%； 對在滙豐工作感到自豪，79%； 認為滙豐是可以進一步發展事業 的地方而願意向其他資深專業人 才推薦滙豐，68%。需進一步關 注的範疇包括幫助僱員體會滙豐 的優先策略所帶來的正面影響， 62%。 繼任計劃 我們的人才政策旨在確保優秀人 選可以填補主要的職位，並符合 集團全部範疇的業務需要。我們 將繼任計劃與人才管理、個人發 展及事業前途規劃直接配合。繼 任計劃界定滙豐需要的人才，按 角色和專長，來訂定人數、分配和 類別，而具才幹的個人稍後會獲 分配有關角色。此繼任計劃其實 亦界定個人的事業路向及發展。 2014年，我們評估了104名具有潛 質成為領袖的高級僱員，並確定 其事業發展需要。具潛質的繼任 人必須具體展示其對環球標準的 理解，並能體現滙豐價值觀。 我們的人才策略支援我們在新興 市場發展的抱負。於2014年，被界 定為人才的比例為34%。我們會密 切注視被界定為主要領導角色之 短期及中期繼任人的當地人士， 從而提高中期繼任人的本土人士 所佔高級管理層的比例。 多元及共融 滙豐致力推行多元及共融的文化， 讓僱員有信心發表意見；僱員所 關切的問題獲得處理；工作環境 不存在偏袒、歧視和騷擾 （無論性 別、年齡、種族、宗教、性傾向和 殘疾） ，升遷在乎才幹。多元文化 有助支持我們日益多元化的客戶 基礎，同時有助我們吸納、培育 和挽留一群經驗豐富且忠心服務 的僱員。 我們的多元及共融文化事宜由集團 多元化委員會的高級行政人員監 督，並由集團多個附屬的人才╱多 元化委員會輔助有關工作。我們 設有超過55個僱員網絡小組，代表 不同性別、種族、年齡、性傾向、 殘疾、宗教、文化、在職父母、健 康及社區義工服務。上述小組有 助推動共融文化及促使管理層與 僱員之間維持有效溝通。 滙豐控股有限公司 20 業務模式 （續） 男女比例 我們一直關注性別比例，尤其是 集團高層職位的男女比例。我們 致力消除有關聘用、晉升及識別 人才的偏見、擴展導師及贊助計 劃、為重回工作崗位的父母提供 更佳支援，以及增加彈性工作機 會。 於2014年12月31日，滙豐董事和僱 員的男女比例如下： 男女比例 人數 男 女 總計 執行董事 4 － 4 非執行董事 6 6 12 董事 10 6 16 高層僱員 6,719 2,076 8,795 其他僱員 120,496 136,966 257,462 總計 127,225 139,048 266,273 % 男 女 總計 執行董事 100 － 100 非執行董事 50.0 50.0 100 董事 62.5 37.5 100 高層僱員 76.4 23.6 100 其他僱員 46.8 53.2 100 總計 47.8 52.2 100 整體而言，集團於2014年12月31日 的女性僱員佔52.2%，大致維持於 2013年的水平。高級職位的女性 僱員比例從2013年的22.7%上升至 2014年的23.6%，我們的目標是在 2015年前將數字提升至25%。人才 名單中的女性比例已見增加，從 2013年的32.2%上升至2014年12月 的34%；於2014年12月，集團管理 委員會的女性比例為20% （15人中 有三名） 。 集團僱員的平均年齡為36.2歲，平 均已任職8.5年。 不自覺的偏見 人類會在不知不覺間自動為其他 人定型，這種不自覺的行為會削 弱社會的共融程度。我們透過於 工作程序中加入共融行為以應對 此問題，並於2014年持續就 「不自 覺偏見」 向8,700名經理及18,500名 僱員 （2013年：8,300名經理及 50,000名僱員） 提供培訓。 2015年，我們在多元及共融議題 方面的優先工作將繼續包括：透 過目標教育解決不自覺的偏見； 鼓勵有多方面才幹的員工發展事 業 （繼續以男女比例及本土人士發 展機會的範疇為重點） ；以及把共 融文化擴展至涵蓋更廣泛的多元 層面，例如性取向、種族和殘疾 等層面。我們繼續強化不存偏見 的績效管理；優化內部及外界人 選的名單；在全球各地連繫及善 用我們的僱員資源網絡群組，並 維持貫徹一致的全球管治和支持 架構，以便於整個集團推動多元 和共融文化。 健康、福利及安全 滙豐極為重視員工的身心健康、 福利及安全。我們近期推出了環 球職業健康架構，規定必須積極 管理僱員福利，並鼓勵整個集團 共用最佳運作方式。在2012年8月 至2014年底期間，96%的指定滙豐 僱員已進行兩年一次的網上健康 及安全培訓。 我們因應各地具體情況，設有多 個僱員協助計劃，讓僱員使用免 費電話熱線 （每天24小時及一星期 七天） ，聯繫熟練的專業輔導員。 僱員遇到個人或工作相關問題的 壓力，令工作受影響時，可尋求 輔導員協助。我們亦可以安排僱 員、其伴侶及受養人免費接受面 對面輔導。滙豐在英國、香港、 北美洲及印度均提供有關計劃。 舉報 滙豐設有一條全球 「合規披露熱 線」 （電話及電郵） ，讓僱員在無法 或不宜以正常途徑申訴不滿或表 達關注時作披露。 「合規披露熱線」 接受僱員對以下事宜表達關注： 違反法律或法規、涉嫌賄賂及貪 污、未能遵守集團政策、懷疑洗 錢、違反內部監控及詐騙，或在 集團旗下任何公司的財務紀錄上 蓄意犯錯。環球監管合規部負責 「合規披露熱線」 的運作及處理披 露個案。我們會審閱每宗個案及 轉介作適當調查。僱員亦可向高 級行政人員、部門經理、人力資 源部，以及保安及詐騙風險管理 部直接舉報個案。 我們又在多個國家╱地區設立本地 舉報熱線，由保安及詐騙風險管理 部、人力資源部和監管合規部負責 運作。各地舉報熱線接獲的相關披 露資料上報環球監管合規部或防 範金融犯罪部。環球監管合規部 亦負責監察一個對外的電郵地址 （accountingdisclosures@hsbc.com， 在www.hsbc.com的 「投資者關係 及管治」 頁面） ，收取有關會計及內 部財務監控或核數事宜的投訴。 接收到的個案會按情況上報集團 會計總監、集團財務董事或集團 行政總裁。 在獲取及回應會計或審計相關事 宜的舉報披露資料方面，滙豐的 政策和程序由集團監察委員會監 察。與其他舉報披露資料有關的 政策和程序，則由行為及價值觀 委員會監察。 有關金融犯罪合規事宜的披露資 料及所採取的行動，會定期向行 為及價值觀委員會、集團監察委 員會及金融系統風險防護委員會 匯報。 僱員╱風險概覽 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 21 風險概覽 我們所有業務運作均在不同程度 上涉及計量、評估、承擔及管理 風險或多種風險。 我們是提供銀行及金融服務的機 構，積極管理風險是日常活動的 核心工作。我們在整個組織的所 有層面運用一個風險管理架構， 並以堅穩的風險管理文化作為基 礎，通過滙豐價值觀和環球標準 加以鞏固。此一架構訂定我們在 達致策略目標時願意接受的風險 種類和程度，從而確保我們的風 險狀況維持審慎和符合我們的承 受風險水平。 風險及我們的優先策略 集團的三項優先策略反映於我們 的風險管理中。 推動業務及股息增長－我們確保將 風險維持於可接受和適當水平， 但同時又可創造價值和產生利潤。 實施環球標準－我們正透過實施 環球標準，改革我們查察、阻止 及防範金融犯罪的方法，這些標 準管控我們經營業務的方式及選 擇的業務夥伴和客戶。 簡化流程及程序－我們的處置計 劃讓滙豐更易於管理和控制。透 過致力簡化流程及程序，我們得 以降低營運滙豐的複雜程度，同 時創造增長的動力。 我們的業務及營運模式詳見第12頁。 有關環球標準的其他資料請參閱 第26頁。 2014年的風險 2014年內市場仍持續關注已發展及 新興市場的經濟能否持續增長， 與此同時，全球多個地方地緣政 治緊張局勢升級或維持高度緊張。 我們在最有可能承受壓力的範疇 減低風險，以保持穩健的資產負 債結構、充裕的流動資金及雄厚 的資本實力為核心理念，繼續維 持審慎管理的風險狀況： － 我們以審慎抉擇的方式管理主 權債務及銀行交易對手風險， 以確保整體組合的質素維持穩 健； － 我們定期評估較高風險國 家╱地區和行業，並按此調整 風險承受水平、限額和風險承 擔； － 我們在內部及監管計劃中使用 壓力測試來評估風險程度，在 需要時盡早對組合作出調整； － 我們透過六方面考慮的程序 （請參閱第12頁） ，以及集中發 展某些產品或客戶群，繼續重 新部署及退出若干組合； － 我們更嚴謹地挑選客戶，以管 理金融犯罪風險；以及 － 倘我們預測風險 （例如聲譽及 營運風險） 會超越集團的承受 風險水平，便會緩減有關風險。 我們的貸款組合分散至各項環球 業務及各個地區，加上產品類別 眾多，可確保我們不會過份依賴 少數國家╱地區或市場以創造收 益及取得增長。 2014年，我們在處理承受風險水 平的程序中，監察了一系列主要 風險衡量指標，並奉行限額和監 控架構。 承受風險水平的資料詳見第25頁。 我們處理壓力測試的方法於第117頁 說明，監管規定壓力測試計劃載於 第125頁。 業務活動所引致的風險 我們主要的銀行業務風險是：信 貸風險、流動資金及資金風險、 市場風險、營運風險、合規風險、 受信風險、聲譽風險、退休金風 險及可持續發展風險。我們亦會 涉及保險風險。 下頁的圖表展示我們的業務活動 如何反映於集團的風險計量及資 產負債中。第三方資產及負債顯 示各項業務對資產負債的貢獻， 而風險加權資產具體說明了各項 業務涉及的相對風險規模。 有關主要風險的描述，請參閱第114頁。 滙豐控股有限公司 22 業務模式 （續） 首要及新浮現風險 識別和監察首要及新浮現風險是 滙豐管理風險工作的重要任務。 「首要風險」 指當前涉及任何風險 類別、環球業務或區域的已浮現 風險。這些風險可能於一年內形 成並且明確顯露，對我們的財務 業績或聲譽及長期業務模式的可 持續發展可能構成重大影響。 「新 浮現風險」 指其結果可能重大但 尚未明朗的風險，這些風險可能 於一年後才形成並且明確顯露。 若這些風險形成並且明確顯露， 對我們落實長期策略的能力可能 構成重大影響。 我們的首要及新浮現風險架構讓 我們可以識別及管理現有和未來 的風險，以確保我們的承受風險 水平保持恰當。我們透過一套全 面的風險因素，持續評估首要及 新浮現風險，而上述評估可能會 使我們修訂承受風險水平。 於2014年，高級管理層特別關注 多項首要及新浮現風險。我們的 現有風險概述於下頁。 於2014年，我們已對首要及新浮 現風險作出多項變更，以反映此 等風險對滙豐影響之評估。 「來自 新興市場放緩的宏觀經濟風險」 的問題被 「經濟前景和政府干預」 取而代之，因為2014年下半年已 發展經濟體呈現受壓跡象。由於 使用第三方服務供應商 （此等供 應商的透明度可能較低以及更難 於管理或施加影響） 涉及的風險， 「第三方風險管理」 被視為新浮現 的風險。雖然 「人事風險」 已存在 於多項首要或新浮現風險內，但 由於 「人事風險」 持續增加，故此 項風險現被視為一項獨立風險予 以披露。 當下述首要及新浮現風險已超過 或有可能超出我們的承受風險水 平時，我們會採取措施減低有關 風險，包括減少受壓範疇的風險 承擔。鑑於違反美國延後起訴協 議 （ 「US DPA」 ） 對集團的影響，高 級管理層十分關注以下方面：跟 進及監察我們遵守協議規定的情 況，以及改善政策、流程及監控 措施，有助盡量減少違規風險。 有關上述風險的闡述，請參閱第118頁； 有關風險因素的摘要，請參閱第113頁。 各項環球業務涉及的風險 ‧ 滙豐 環球業務 業務活動 資產 負債表 風險 加權資產 風險狀況 － 存款 － 資金管理 － 資產負債管理 － 信貸及貸款 － 資產及貿易融資 － 企業融資 － 資本市場 － 證券服務 － 存款 － 資金管理 － 信貸及貸款 － 國際貿易及應收賬 融資 － 商業保險及投資 － 存款 － 戶口服務 － 信貸及貸款 － 投資管理 － 財務顧問 － 經紀業務 － 企業融資 (透過環球銀行及 資本市場業務提供) － 另類投資 － 信託及遺產規劃 － 滙豐控股及中央 營運部門 十億美元 － 資產 499 － 客戶賬項 581 十億美元 － 資產 1,840 － 客戶賬項 319 十億美元 － 資產 88 － 客戶賬項 85 十億美元 － 資產 165 － 客戶賬項 1 － 資產 373 － 客戶賬項 364 十億美元 － 信貸風險 168 － 營運風險 37 十億美元 － 信貸風險 326 － 交易對手 信貸風險 90 － 營運風險 44 － 市場風險 56 十億美元 － 信貸風險 17 － 營運風險 4 十億美元 － 信貸風險 45 － 營運風險 – 十億美元 十億美元 － 信貸風險 399 － 營運風險 33 流動資金及資金風險 （第163 頁） 、退休金風險 （第200 頁） 、受信風險 （第200 頁） 、聲譽風險 （第199 頁） 、合規風險 （第189 頁） 、 可持續發展風險（第201 頁）及保險風險 （第190 頁） 。後者主要涉及零售銀行及財富管理業務和工商金融業務。 其他 （包括控股公司） 工商金融 環球銀行 及資本市場 環球私人銀行 零售銀行及 財富管理 12 － 存款 － 戶口服務 － 信貸及貸款 － 資產管理 － 匯財策劃及財務策劃 － 經紀業務 － 保險 （分銷；制訂壽險產品） 有關註釋，請參閱第39頁。 有關信貸風險的其他資料，請參閱第127頁；有關資本及風險加權資產的其他資料，請參閱第238頁；有關市場風險 （包 括估計虧損風險） 的其他資料，請參閱第175頁；有關營運風險的其他資料，請參閱第186頁。 風險概覽 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 23 首要及新浮現風險－首╱新 風險 描述 減低方法 宏觀經濟及地緣政治風險 新 經濟前景及政府干預 已發展及新興市場和國家╱地區同時呈現 疲弱的經濟增長，可能對環球貿易及資金 流，以及來自此等國家和地區業務的利潤 造成不利影響。 我們會密切監察主要市場的經濟發展，並按 照情況的變化採取適當行動。 新 地緣政治風險增加 我們的營運須面對若干國家╱地區政局不 穩及社會動亂引發的風險，這些情況可能 對區域穩定及區域與全球經濟造成更廣泛 的影響。 我們會密切監察地緣政治和經濟前景，特別 是我們涉及重大風險及╱或在當地有實體業 務的國家╱地區。 業務模式的宏觀－審慎、監管及法律風險 首 我們的業務模式及集團的盈利 能力受監管環境發展影響 多個政府及監管機構持續制訂各項政策， 因此可能施加新規定，特別是關乎資本及 流動資金管理和業務架構的範疇。 我們會密切聯繫業務所在國家╱地區的政府 及監管機構，以協助確保新規定獲妥善考 慮。 首 監管機構對經營業務方式及金 融犯罪作出的調查、罰款、制 裁，以至相關承諾、同意令及 規定，對集團的業績及品牌構 成負面影響 金融服務供應商須面對因業務經營方式及 金融犯罪問題而遭監管機構制裁或罰款的 風險。 違反美國延後起訴協議，可能令美國有關 當局就有關事宜向滙豐提出檢控。 我們正於所有環球業務及部門推行計劃，致 力加強行為管理。 我們持續採取步驟向相關監管機構徵詢意 見，從而應對美國延後起訴協議及其他同意 令的規定。 首 爭議風險 滙豐在其日常業務營運中會涉及法律訴 訟，可能因而引致財務虧損及重大的聲譽 損害。 我們會識別和監察新浮現的監管和司法趨 勢。 我們正不斷提升在金融犯罪和監管合規方面 的控制措施及資源。 有關業務營運、管治及內部監控制度的風險 首 執行風險提高 項目須符合監管要求的複雜性以及出售業 務和組合引致的風險，均可能影響我們執 行策略的能力。 我們已加強重大項目的優先執行和管治程 序。 首 人事風險 由於監管改革的範圍龐大，現時集團人力 資本方面的需求重大。 我們已檢討薪酬政策，以確保集團具備競爭 力和足以挽留關鍵人才，並會持續增加在主 要領域的專門人才資源水平。 新 第三方風險管理 使用透明度可能較低以及較難於管理或施 加影響的第三方服務供應商會產生風險。 我們現正就使用及監察第三方服務供應商加 強風險管理流程和程序。 首 互聯網罪行及詐騙 由於互聯網和流動服務途徑的使用增加， 滙豐正面對更多詐騙及犯罪活動風險。 我們會持續評估此等威脅的演變，並會調整 控制和防禦措施，以減低此等風險。 首 資訊保安風險 滙豐和其他跨國機構一直是網絡襲擊的目 標。 我們已投入大量資源於員工培訓和提升多個 層面的控制措施，以保護集團的資訊及技術 基礎設施。 首 數據管理 新的監管規定要求我們必須更頻密地提交 精細數據，且必須一致、準確及準時地提 交。 我們的數據策略委員會正於整個集團推動貫 徹一致的數據收集、匯報和管理程序。 首 模型風險 基於錯誤的模型推算結果或基於開發、執 行或使用有欠妥善的模型所作出的決策， 可能會產生不利後果。 對模型的開發、使用和核證加以管治並進行 獨立審查。 滙豐控股有限公司 24 業務模式 （續） 我們如何管理風險 有效管理風險對於執行集團的優 先策略十分重要。 我們的風險管理架構覆蓋整個集 團，有利持續監察風險環境，並 全面評估各類風險及其相互影響 關係。此外，該架構亦確保我們 在整個集團的所有層面和就所有 風險種類制訂一套強而有力且貫 徹一致的風險管理方法。 此架構建基於強而有力的風險管 理文化，有助促使僱員的個人行為 與集團承擔和管理風險的取態相 符，以及確保我們的風險狀況符 合集團的承受風險水平和策略。 我們的風險管理架構通過滙豐價 值觀和環球標準而得以加強。 我們管理風險的方法概述如下。 推動我們的風險管理文化 風險管理架構 批准承受風險水平、計劃及表現目標（請參閱第270頁）。 滙豐董事會的角色 制訂及執行策略和承受風險水平。 負責風險管治和控制措施。 高級管理層的角色 我們會計量、監察及管理來自業務活動的風險 （請參閱第114頁） 。 主要的銀行及保險風險 我們的風險管治架構確保集團對風險管理有適當的監察和問責 （請 參閱第204頁）。 我們的風險管理文化加上薪酬制度，令員工可以為客戶做正確的事 情（請參閱第34頁）。 我們的「三道防線」模式所界定的風險管理角色和責任（請參閱第 186頁）。 獨立的風險管理部確保在風險╱回報的決策上取得必要的平衡（請 參閱第204頁）。 風險管治架構 三道防線 獨立的風險管理部 人事 根據我們的長遠策略、核心風險管理原則、滙豐價值觀，以及風險 管理能力，描述我們準備承受的風險（請參閱第205頁）。 承受風險水平 識別、監察及減低超出承受水平的風險的流程 （請參閱第117頁） 。 首要及新浮現風險 風險圖譜 壓力測試 風險管理流程及程序 集團風險管理委員會就承受風險水平、風險管治和其他高層次風險 相關事宜提供意見（請參閱第280頁）。 金融系統風險防護委員會就金融犯罪和金融系統濫用問題提供意見 （請參閱第282頁）。 行為及價值觀委員會就政策與程序提供意見，確保我們遵循滙豐價 值觀（請參閱第286頁）。 有關董事會屬下其他小組委員會，請參閱第276頁。 集團風險 管理委員會 金融系統風險 防護委員會 行為及價值觀 委員會 董事會屬下委員會的角色 控制措施 流程 以滙豐價值觀作為根基 風險概覽 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 25 承受風險水平 集團的承受風險水平聲明 （ 「RAS」 ） 是風險管理架構的一個主要組成 部分，當中列載我們在達致中長 期策略目標時願意承受的風險類 別及水平。董事會按照集團風險 管理委員會的意見批准承受風險 水平聲明。 承受風險水平是透過集團的承受 風險水平架構建立和監察，該架 構提供一個全球貫徹一致和結構 分明的方法，根據我們的核心風 險管理原則管理、計量和控制風 險。該架構勾劃出應對承受風險 水平的程序、政策、衡量指標及 管治組織和方法，作為日常業務 和風險管理活動的一部分。 承受風險水平聲明透過界定在達 致策略目標時對集團而言屬可取 的前瞻性風險水平，為我們的年 度營運計劃提供指引，並在我們 的六方面考慮流程中扮演重要角 色。我們的承受風險水平可因應 我們對已識別的首要和新浮現風 險之評估而作出修訂。 我們採取定量及定質的衡量指標， 以計量多個主要類別的風險，包 括回報、資本、流動資金及資金、 證券化、風險成本，及集團內部 貸款、風險類別 （例如信貸、市場 及營運風險） 、風險分散與集中程 度，以及防範金融犯罪。這些衡 量指標每年檢討，確保持續有效。 按照以上衡量指標計量，旨在： － 為相關業務活動提供指引，確 保其與承受風險水平聲明所述 者一致； － 監察主要的相關假設，並在有 需要時，在往後的業務策劃周 期作出調整； － 迅速識別減輕風險所需的業務 決策；及 － 通報風險調整報酬。 透過使用重大風險種類的承受風 險能力和限額指標，我們將承受 風險水平包含在日常風險管理決 策內，確保我們的風險狀況與承 受風險水平保持一致，以及平衡 風險與回報。 各環球業務部門及地區的承受風 險水平聲明須與集團的承受風險 水平聲明一致。 我們於2014年計量、監控及每月送 呈集團管理委員會風險管理會議 的部分核心衡量指標表列如下： 主要的承受風險水平衡量指標 2014年 目標 實際 普通股權一級 比率1 ≥ 10% 11.1% 股東權益回報率 趨勢向 上，至 2016年 為12% 至15% 7.3% 風險加權資產 回報率13 2.2%至 2.6% 1.5% 成本效益比率 50至 60% 之中位數 67.3% 貸款對客戶 賬項比率 低於90% 72.2% 風險成本 （貸款減值 準備） 低於營業 收益的 15% 5.4% 有關註釋，請參閱第39頁。 於2014年初，我們對承受風險水 平聲明進行年度檢討，並於2014 年1月及2014年2月分別獲集團管理 委員會的風險管理會議及滙豐控 股董事會批准。承受風險水平聲明 的核心範疇已加入執行董事2014 年的評分紀錄內，有關事項載於 《2013年報及賬目》 第405頁。 我們於2014年在集團的承受風險 水平聲明加入了與防止、查察及 防範金融犯罪核心合規原則有關 的衡量指標，加強了該聲明的效 用。 關於2015年目標的討論，請參閱第32頁。 有關資本指引4規定的詳情，請參閱第 239頁。 風險如何影響我們的表現 風險管理是我們所有活動的主要 組成部分。風險量度了不確定性 及隨後回報的變化。 零售組合中之信貸衡量指標受惠 於持續出售非策略組合、多個市 場的經濟環境改善，以及亞洲和 美國的核心業務增長；而批發組 合則仍然大致維持穩定，主要的 減值指標整體出現有利變動。貸 款減值準備因第29頁所載的原因 而有所下降。 營業虧損受到英國客戶賠償計劃 的準備以及與法律和監管事宜有 關的和解影響而上升。多項因素 會影響法律及監管規定事宜的估 計負債，因此上述事宜的最終罰 款、懲罰及賠償成本仍有很大程 度的不確定性。 在多個司法管轄區，滙豐在其日 常業務營運中會涉及法律訴訟、 調查及監管事宜。於2014年12月31 日，我們關於法律訴訟及監管事 宜的準備及與客戶有關的補救措 施的準備合計40億美元。 滙豐按列賬基準計算的業績反映 編製綜合財務報表所採用的會計 政策、假設及估計，並反映我們 對影響集團風險之財務影響所作 出的評估。 有關重大法律訴訟及監管事宜的詳情， 請參閱第446頁財務報表附註40。 有關法律訴訟及監管事宜以及與客戶有 關的補救措施的準備，於第420頁財務 報表附註29中披露。 有關營業虧損的詳情，請參閱第188頁。 有關關鍵會計估算和判斷的詳情，請參 閱第62頁。 優先策略 滙豐控股有限公司 26 優先策略 我們以往為2014至2016年界定了 三個環環相扣且同等重要的優先 範疇，以協助我們落實集團策略： － 推動業務及股息增長； － 實施環球標準；及 － 簡化流程及程序。 各項優先策略相輔相成，並以日常 業務所執行的各項措施為基礎。 此等優先策略為我們的客戶及股 東創造價值，並促進滙豐的長遠 持續發展。 在這個過程中，我們必須促使業 務維持強健、保持靈活及對環境 而言可持續發展，使客戶抱有信 心、僱員引以為傲，以及我們的 社區可以信賴。 推動業務及股息增長 在推動業務及股息增長方面，我 們的目標是根據本身的有機投資 準則推動風險加權資產增長，循 序漸進地增加股息，同時減低既 有和非策略業務對利潤及風險加 權資產的影響。 我們的策略是從持續增長的國際 貿易和資金流以及創富增值 （特 別是亞洲、中東和拉丁美洲） 中受 惠。我們的目標是運用集團的國 際網絡和客戶基礎，按照策略提 升滙豐在各類產品市場的地位， 從而達致增長。 為促進增長，我們會按照集團的 承受風險水平，將風險加權資產 由集團業務中表現較差的部分， 轉移至表現較佳的部分。 於2014年，我們推行多項優先的 投資策略，以善用我們的全球網 絡和加速有機增長： － 環球貿易及融資：我們投入資 源提升本身的銷售和產品實力 （特別是高增長的產品和貿易 走廊） ，以及擴展貿易中心， 致力加強滙豐於貿易方面的 領導地位。 － 資金管理：我們投資於改進銷 售和客戶服務方案，同時提升 流動銀行技術，從而提供更佳 的客戶服務和產品。 － 外匯：我們致力改良電子交易 平台和實力，藉以改善客戶服 務和營運效率。 － 人民幣：憑著我們的市場領導 地位，我們投入資源推展跨國 人民幣服務，目標是在離岸人 民幣匯兌和資本市場服務方 面爭取更大的市場份額。 行業獎項和市場份額的提升肯定 了我們的策略取得成果。在過去 三年，我們於核心國際連繫產品 方面，例如資金管理、環球貿易 及融資以及外匯等領域，所佔市 場份額均見持續改善。滙豐已連 續三年 （包括於2014年） 在 《歐洲貨 幣》 雜誌舉辦的資金管理調查中獲 選為全球最佳企業及金融機構資 金管理機構。在同一個調查中， 滙豐於2014年連續第二年獲選為 全球最佳非金融機構資金管理機 構。自 《亞洲貨幣》 雜誌於2012年 首辦離岸人民幣服務調查以來， 我們每年均在該項調查中獲得 「最 佳整體產品及服務」 大獎。 我們旨在持續投資於主要的增長 市場，並將全球資源投放於國際 收入迅速增長的城市： － 我們的本位市場英國和香港： 我們的目標是鞏固及發展主 要產品 （例如按揭和個人貸款 等） 在本位市場上的地位。 － 中國：中國內地對滙豐而言繼 續有其策略重要性，而且會帶 來結構性的長遠增長機會。 因此，我們繼續投資於有機增 長，特別在廣東和其他具重要 經濟價值的地區。我們致力投 入資源，率先把握於上海自貿 區等監管變革過程中可能出 現的機會。 － 美國及德國：我們透過擴展企 業業務，持續改善於美國這個 世界最大經濟體及歐洲主要貿 易國的地位。於2014年，我們 提升產品質素、拓闊業務地域 覆蓋和調整承受風險水平，從 而拓展客戶基礎。我們透過加 深與客戶的關係及發展跨業 務的機會，令國際收入增加。 我們的環球銀行業務模式使集團 推動各項環球業務帶來收入。在 2014年，來自各種商機的跨業務 合作所得收入均有增加，惟工商 金融業務的客戶並未為資本市場 業務帶來收入增長，主要是外匯 波幅偏低所致。年內近半的合作 所得收入總額來自向工商金融業 務的客戶提供資本市場及資本融 資業務產品。在環球私人銀行業 務方面，跨業務客戶轉介帶來的 新增資金淨額較2013年高出一倍。 實施環球標準 在滙豐，我們採納最高或最有效 的防範金融犯罪監控措施，並在 每個業務所在地全力推行。 兩項新的環球政策載列反洗錢 （ 「AML」 ） 及制裁方面的監控措施， 是集團的環球標準。 我們矢志成為世界領先的國際銀 行，致力在了解客戶以及查察、 阻止和防範金融犯罪方面訂立行 業標準。為達到此等目的，我們 執行更貫徹一致和全面的方法管 理金融犯罪風險 － 由增加了解客 戶、其活動、其活動所在地和活 動的動機，以至確保其銀行活動 與我們的預期吻合。 在與客戶維持關係的整段期間， 我們將執行防範金融犯罪風險管 理標準：由挑選客戶和展開與客 戶之間的關係，到管理我們與客 戶之間的持續關係，以及監察和 評估銀行不斷變化的風險狀況。 我們新推行的全球反洗錢政策旨 在阻止罪犯透過滙豐洗錢。該項 政策就執行客戶盡職審查、監察 交易和上報對可疑活動的關注列 出全球規定。 推動業務增長╱實施環球標準 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 27 我們新推行的全球制裁政策旨在 確保我們遵守與業務所在地制裁 措施相關的法律與法規，以及聯 合國安全理事會、歐洲聯盟、美 國、英國和香港政府實施的全球 制裁法律。 在許多情況下，我們的政策會高 於法律規定的標準，反映滙豐實 無意與不法份子有任何業務往來。 我們預期環球標準能配合現時和 未來的集團業務經營方式，同時 為我們帶來競爭優勢。預期實施 環球標準能讓我們： － 加強實力以應對持續不斷的 金融犯罪威脅； － 透過劃一以致簡化的程序，監 察及執行集團的嚴格標準； － 加強規範營運方式及客戶對 象的政策和程序；及 － 確保我們一致恪守滙豐價值 觀。 實施環球標準 各環球業務及防範金融犯罪部已 識別需要改進現有程序以符合環 球標準的地方和方法。此等部門 現正配置適當的系統、流程、培 訓和支援，以便於各個業務所在 國家╱地區執行經改進的程序。 這些工作分兩個階段執行： － 按照一個加快的限期執行涉及 有限度基礎設施的政策；及 － 同步執行加強長遠策略性監 控措施和提升相關基礎設施 的方案。 於2014年，我們在多個不同領域 取得重大進展，包括： － 在全球執行客戶篩選政策和 管治方案； － 在收集和核實客戶資料方面 首次執行經改進的客戶盡職 審查程序； － 整合與全球制裁法律有關的 篩選清單，以納入我們的客戶 和交易篩選工具內； － 為最高風險崗位的人員及所 有僱員計劃提供目標明確的 培訓，以提高員工對金融犯罪 風險的關注，同時鼓勵員工向 上級匯報； － 在全球成立金融情報和調查 組，以跟進匯報和預警，以及 識別新浮現的趨勢和事宜；及 － 制訂全球適用的程序和管治 措施，以退出超逾集團金融犯 罪風險承受水平的業務。 管治架構 各環球業務及防範金融犯罪部在 滙豐的科技及服務部支援下，負 責制訂業務程序和監控措施，並 營造相關的經營環境，以便在各 環球業務部門及司法管轄區內執 行新政策。負責監督相關問責情 況的環球標準執行委員會由集團 風險管理總監擔任主席，成員包 括各環球業務部門的行政總裁和 防範金融犯罪部的環球主管。 因應上述情況，以及為促進各業 務部門之間在日常業務過程中更 緊密合作，集團風險管理會議的 經常性議題包括匯報環球標準的 執行情況。金融系統風險防護委 員會及董事會持續省覽環球標準 計劃的定期報告，作為其持續監 督工作的一部分。 承受風險水平 監控金融犯罪風險是集團日常業 務的一部分，我們會按照全球金 融犯罪風險承受水平聲明管治有 關措施。此舉旨在確保集團長期 可持續發展。集團的金融犯罪風 險總體承受水平和應對方法，是 絕不容忍在業務營運中欠缺查察 及防範金融犯罪的系統和監控措 施，而且不會與相信行為不當的 個人或實體有任何業務往來。 企業整體風險評估 我們對遵循制裁和反洗錢法律的 風險與監控措施進行了第二次年 度企業整體評估。評估結果已用 作2015年風險管理規劃、優先處 理工作和資源分配的依據。 監察員 根據與美國司法部、英國金融業 操守監管局 （前身為英國金管局 「FSA」 ） 及美國聯邦儲備局 （ 「聯儲 局」 ） 於2012年訂立的協議 （包括為 期五年的延後起訴協議） ，一名獨 立合規監察員 （ 「監察員」 ） 已獲委 任負責評估集團全面履行責任的 進度，並定期評估集團合規部的 效率。 Michael Cherkasky由2013年7月起展 開監察員的工作，負責評估和匯 報集團的內部監控、政策及程序 對集團持續履行相關反洗錢、制 裁法律、反資助恐怖主義及武器 擴散等責任的成效，為期五年。 滙豐持續就遵循反洗錢和制裁法 律的不足之處，採取協定的補救 措施，並實施環球標準。滙豐亦 正致力執行監察員經2013年檢討 後提出的協定建議。我們明白， 五年的美國延後起訴協議由訂立 至今只有兩年，我們只經歷了其 中部分路程。我們期盼與監察員 及其團隊通力合作，維持密切關 係。 滙豐控股有限公司 28 簡化流程及程序 我們持續改善營運流程、發展環 球部門、執行貫徹的業務模式和 簡化資訊科技。 自2011年起，我們透過推行較精 簡的匯報架構以及建立由環球業 務和部門組成的營運模式，改變 了管理滙豐的方法。此等變化加 上改善開發軟件的生產力、優化 程序和物業組合，使我們實現了 可持續成本節約57億美元，按年 率 （營運率） 計算相當於61億美元。 此成績超出首階段集團策略中優 化架構計劃開展時立下的承諾－ 達到25億至35億美元可持續成本 節約。 持續節省成本來自減少或消除複 雜、欠缺效率或不必要的活動， 並且釋出資本，以供重新投資於 推動業務增長及增加股東回報。 將集團重組為四大環球業務及11個 環球部門，進一步讓我們可以在 全球執行貫徹一致的營運模式。 此舉為我們進行下一階段的簡化 工作奠下基礎。 展望未來，我們會致力投入資源 推動業務增長及合規工作，並且 透過增進效益抵銷通脹的影響。 因此，我們須達致成本淨額減省。 有關工作包括： － 改善端對端優化流程和服務 途徑； － 簡化技術，減少整個集團使用 的應用程式數目；及 － 改進基礎設施，包括優化房地 產的使用率和進行若干活動 的地點。 我們預期將流程、產品、系統及 營運簡易化及全球化，將可收簡 化之效。 「簡易化」 即查找欠缺效 率或過於複雜的流程，然後加以 重訂或優化，使相關流程更易理 解及管理，而且更具效益。 「全球 化」 就是發展標準的環球流程， 並在集團每個環節執行。 成本效益比率 2014年，我們的成本效益比率為 67.3%，高於2013年的59.6%。有 此變化是由於法律、監管規定及 操守方面的和解開支上升、通脹 壓力、持續投資於策略方案，以 及銀行徵費增加所致。部分支出 增幅被13億美元的可持續成本節 約抵銷。 成果 財務表現 業績表現反映出售利潤減少及 其他重大項目的負面影響。 列賬基準之業績 2014年 2013年 2012年 百萬 美元 百萬 美元 百萬 美元 淨利息收益 34,705 35,539 37,672 費用收益淨額 15,957 16,434 16,430 其他收益 10,586 12,672 14,228 營業收益淨額14 61,248 64,645 68,330 貸款減值及其他信貸 風險準備15 (3,851) (5,849) (8,311) 營業收益淨額 57,397 58,796 60,019 營業支出總額 (41,249) (38,556) (42,927) 營業利潤 16,148 20,240 17,092 來自聯營公司 收益16 2,532 2,325 3,557 除稅前利潤 18,680 22,565 20,649 有關註釋，請參閱第39頁。 按列賬基準計算，除稅前利潤為 187億美元，較2013年下跌39億美 元或17%，主要反映出售業務利潤 及重新分類增益減少，以及其他 重大項目 （包括罰款、和解開支、 英國客戶賠償及相關準備） 對收入 及支出的負面影響。 列賬基準之未扣除貸款減值及其 他信貸風險準備之營業收益淨額 （ 「收入」 ） 為610億美元，較2013年 下跌34億美元或5%。於2014年， 出售利潤和重新分類增益 （扣除虧 損淨額） 下降，而2013年的利潤則 包括興業銀行股份有限公司 （ 「興 業銀行」 ） 向第三方發行額外股本 後重新分類為金融投資產生的會 計增益11億美元，以及出售巴拿 馬業務獲得的利潤11億美元。此 外，其他重大項目包括不合資格 對沖的不利公允值變動5億美元， 而2013年則錄得5億美元的有利變 動；因應英國 《消費者信貸法》 持續 合規檢討所產生的準備6億美元； 以及衍生工具合約之借記估值調 整的不利變動淨額4億美元。此等 因素部分因信貸息差變動引致指 定以公允值列賬之本身債務錄得 有利公允值變動4億美元 （2013年則 簡化流程及程序╱財務表現 優先策略╱成果 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 29 為錄得不利變動12億美元） 及2014 年出售所持上海銀行股權錄得利 潤4億元而被抵銷。 貸款減值及其他信貸風險準備為 39億美元，較2013年減少20億美元 或34%，尤以北美洲、歐洲及拉丁 美洲最為明顯。 營業支出為410億美元，較2013年 上升27億美元或7%，主要由於重 大項目較2013年高出9億美元。營 業支出包括與外匯調查有關的和 解開支及準備12億美元，以及與 美國聯邦房屋金融局達成和解協 議相關的準備6億美元。 來自聯營公司的收益為25億美元， 較2013年上升2億美元或9%，主要 反映2013年於越南經營銀行業務 的聯營公司之投資錄得的減值準 備1.06億美元不復再現。 董事會已批准就2014年派發之第四 次股息增加5%至每股0.2美元，較 2013年第四次股息增加0.01美元。 2014年的派息總額為96億美元 （即 每股獲派0.5美元） ，較2013年增加 4億美元。 按過渡基準計算，普通股權一級 比率由2013年底的10.8%上升至 10.9%；而按終點基準計算，則為 11.1%，較2013年底的10.9%有所上 升，此乃由於持續生成資本及採 取管理措施，被風險加權資產增 長、匯兌變動及監管改革的影響 抵銷。 經調整業績 有關非公認會計原則財務衡量指標，經 調整基準的詳情請參閱第40頁，而有形 股本回報的詳情請參閱www.hsbc.com的 資料。 從列賬基準業績至經調整業績 為計算經調整業績，我們已， － 就外幣換算按年比較的影響作出 調整；及 － 就重大項目之影響作出調整。 列賬基準業績與經調整業績的對賬已 載於第44頁。 按經調整基準計算，除稅前利潤 為230億美元，大致維持於2013年 的水平。貸款減值及其他信貸風險 準備下降 （主要在北美洲、歐洲及 拉丁美洲） ，以及收入輕微上升， 大部分被營業支出的增幅抵銷。 下文乃按經調整基準評述。 收入大致維持不變。工商金融業 務錄得增長，尤其於本位市場香 港和英國，惟被零售銀行及財富 管理業務、環球銀行及資本市場 業務和環球私人銀行業務的收入 下降所抵銷 收入上升1億美元至620億美元。 由於香港的貸款平均結欠及存款 平均結餘增加，加上英國的存款 平均結餘不斷上升和借貸息差擴 闊，令工商金融業務的收入有所 增長。收入亦受惠於英國的有期 貸款費用增長。 此等因素大部分被零售銀行及財 富管理業務、環球銀行及資本市 場業務和環球私人銀行業務的收 入減少所抵銷。在零售銀行及財 富管理業務方面，主要是由於縮 減美國的消費及按揭貸款 （ 「CML」 ） 組合所致，而主要零售銀行及財 富管理業務的收入則大致持平。 在環球銀行及資本市場業務方 面，收入減少是由於對若干衍生 工具合約作出資金公允值調整 （ 「FFVA」 ） ，導致產生扣賬額2.63 億美元，加上外匯交易業務收入 減少所致，惟此等減幅部分被資 本融資業務的增幅所抵銷。在環 球私人銀行業務方面，收入下降 反映我們持續為該項業務重新定 位，使客戶資產有序減少，以及 市場波幅收窄。 貸款減值及其他信貸風險準備在 大部分地區錄得跌幅，其中最顯 著的是北美洲、歐洲和拉丁美洲 貸款減值及其他信貸風險準備較 2013年下跌18億美元或31%，跌 幅主要來自北美洲的零售銀行及 財富管理業務，反映消費及按揭 貸款組合的拖欠水平及新增已減 值貸款下降，以及持續縮減組合 和出售貸款令貸款結欠減少。歐 洲的貸款減值及其他信貸風險準 備亦有所下降，主要反映英國工 商金融業務與環球銀行及資本市 場業務的個別評估準備減少，以 及英國環球銀行及資本市場業務 的可供出售資產抵押證券 （ 「ABS」 ） 錄得更高的信貸風險準備撥回淨 額。拉丁美洲的貸款減值及其他 信貸風險準備亦告減少，減幅主 要來自墨西哥及 （其次是） 巴西。 在墨西哥，貸款減值及其他信貸 風險準備下降，主要反映工商金 融業務的個別評估準備減少，而 在巴西，零售銀行及財富管理業 務與工商金融業務均錄得貸款減 值及其他信貸風險準備下降，但 部分被環球銀行及資本市場業務 錄得的增幅抵銷。 經調整除稅前利潤 （十億美元） 22.6 18.7 0.4 23.0 4.1 22.8 經調整利潤 貨幣換算及重大項目 列賬基準 2013年 2014年 普通股股東應佔利潤 （列賬基準） （百萬美元） 2012年 2013年 2014年 13,454 15,631 13,115 每股盈利 （列賬基準） （美元） 2012年 2013年 2014年 0.74 0.84 0.69 有形股本回報 (%) 2012年 2013年 2014年 9.8 11.0 8.5 滙豐控股有限公司 30 成果 （續） 營業支出上升，部分反映監管計 劃及合規成本上升和通脹升温， 惟部分被進一步的可持續成本節 約所抵銷 營業支出為380億美元，較2013年 上升22億美元或6%。由於持續致 力推行環球標準及業界執行更廣 泛的監管改革計劃，以建立必要 的基礎，從而配合現時更高的合 規標準，監管計劃及合規成本有 所增加。 營業支出亦由於通脹壓力增加而 上升。通脹壓力包括工資上漲，主 要來自亞洲和拉丁美洲。此外， 英國的銀行徵費支出較2013年有 所增加，亦是營業支出增加的原 因。我們持續投資於策略方案， 以支持業務的有機增長 （主要在工 商金融業務方面） 。另外，我們亦 增加市場推廣及廣告支出，主要 是支持零售銀行及財富管理業務 賺取收入之活動。 此等因素部分被年內進一步達致 約13億美元的可持續成本節約所 抵銷，主因是重整後勤辦公室的 若干流程。 於2014年底，我們的職員人數 （以 等同全職僱員數目列示） 上升3,500人 或1%。等同全職僱員的平均數目 大致持平，此乃由於可持續成本 節約計劃帶來的減額，被監管計 劃及合規方案以至業務增長帶來 的增額抵銷了。 來自聯營公司收益上升，推動力 主要來自亞洲和中東及北非 來自聯營公司的收益增加，主要 因為來自交通銀行股份有限公司 （ 「交通銀行」 ） 和沙地英國銀行的 貢獻增加，此等增幅主要反映資 產負債增長。 實質稅率為21.3%，而2013年則為 21.1%。 有關集團的財務表現，詳情請參閱第 46頁。 資產負債實力 列賬基準資產總值為26,000億美 元，較2013年12月31日減少1%。 固定匯率基準資產總值增加850億 美元或3%。我們的資產負債仍然 強勁，客戶貸款對客戶賬項比率 為72%。此實力來自我們的業務 模式及保守的承受風險水平，即 以客戶賬項的增長來支持客戶貸 款的增長。 按固定匯率基準計算，貸款增加 280億美元，客戶賬項則增加470億 美元。 有關資產負債的詳情，請參閱第57頁。 有關集團流動資金及資金的詳情，請參 閱第163頁。 資產總值 （十億美元） 2,693 2,671 2,634 2012年 2013年 2014年 平均資產總值之除稅後回報 (%) 0.6 0.7 0.5 2012年 2013年 2014年 客戶貸款 17 （十億美元） 963 992 975 2012年 2013年 2014年 客戶賬項 17 （十億美元） 1,311 1,361 1,351 2012年 2013年 2014年 客戶貸款對客戶存款比率 17 (%) 73.4 72.9 72.2 2012年 2013年 2014年 有關註釋，請參閱第39頁。 財務表現 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 31 資本實力 我們管理集團資本的方針，是確 保資本可以超越當前監管規定的 水平，並作好準備，應對日後預 期的資本需求。 我們採用資本比率和槓桿比率來 監察資本充足程度。資本比率是 用來衡量資本相對於所冒風險 （按 監管規定評估） 之比率；槓桿比率 是用來衡量資本相對於風險承擔 之比率。 2013年6月，歐洲委員會頒布規例 及指引的最終版本 （統稱為資本指 引4） ，促使歐盟的巴塞爾協定3架 構生效。有關規定已於2014年1月 1日落實。 在新機制下，普通股權一級 （ 「CET1」 ） 資本代表最高規格的合 資格監管規定資本，銀行的資本 實力會以此來衡量。2014年，我 們致力管理集團的資本狀況，使 之達到高於10% （按終點基準普通 股權一級比率計算） 的內部目標比 率。自此我們一直修訂該比率， 而於2015年，我們預期採取管理 措施使集團資本達致股東權益回 報率的中期目標10%以上。此目標 是採用介乎12%至13%的普通股權 一級比率 （終點基準） 以模型推算。 槓桿比率 下表為根據審慎監管局指示計算 的估計槓桿比率。其分子採用資 本指引4終點基準一級資本的定義 來計算，而風險承擔的數值則根 據2015年1月公布的歐盟授權法案 （以2014年巴塞爾協定3經修訂定 義為基礎） 來計量。 估計槓桿比率 2014年 十億美元 於12月31日 資本指引4 （終點基準） 下的 一級資本 142 作出監管規定調整後的風險 承擔 2,953 估計槓桿比率 （終點基準） 4.8% 有關槓桿比率的詳情，請參閱第251頁。 有關集團資本及風險加權資產的詳情， 請參閱第239頁。 資本比率和風險加權資產 資本指引4 1 普通股權一級比率 （過渡基準） (%) 普通股權一級比率 （終點基準） (%) 總資本比率 （過渡基準） (%) 風險加權資產 （ 「RWA」 ） （十億美元） 巴塞爾協定2.5 1 核心一級比率 (%) 總資本比率 (%) 風險加權資產 （十億美元） 2013年 2014年 2013年 2014年 2013年 2014年 2013年 2014年 10.8 10.9 14.9 15.6 10.9 11.1 1,215 1,220 12.3 13.6 2012年 2013年 16.1 17.8 2012年 2013年 1,124 1,093 2012年 2013年 有關註釋，請參閱第39頁。 滙豐控股有限公司 32 達致目標 我們設定財務目標，並以此衡量 表現。 於2011年，我們以成為領先的國際 銀行為目標，並設定特定的財務 衡量指標，按此衡量直至2013年 的表現。我們按當時對資本規定 的理解訂立目標，當中包括巴塞 爾協定3下的普通股權一級比率 9.5%至10.5%、股東權益回報率 （ 「ROE」 ） 12%至15%，以及成本效 益比率 （ 「CER」 ） 48%至52%，並以 三年間達致可持續成本節約25億 至35億美元來支持我們的目標。 在直至2013年的一段期間，我們 加強資本狀況，實現可持續成本 節約49億美元，而且按目標向股 東增加派息。 於2013年5月，我們界定了2014 至2016年間的優先策略，並檢討 用以反映業績表現的財務衡量指 標。我們繼續以12%至15%的股東 權益回報率為目標，同時增加另 一個目標，致力使可持續成本節 約達到20億至30億美元。為了在 增長策略上投資，以及反映營運 環球銀行涉及越來越多的規定， 我們將成本效益比率的目標修訂 為50至60的中位數，並加入收入 增長必須高於成本增長的目標 （即 收入增長對支出增長的比率為正 數） 。我們將普通股權一級比率 （終點基準） 的目標設定在10%以 上，並繼續尋求循序漸進地向股 東增派股息。同時，我們亦將貸 存比率上限定為90%。 於2014年，我們的普通股權一級 比率 （終點基準） 為11.1%，並就本 年度宣派總額96億美元的股息。 可持續成本節約遞增13億美元， 貸存比率維持在72%。股東權益 回報率為7.3%，成本效益比率為 67.3%，低於我們的目標。 監管及經營環境不斷變化 於2011年，我們根據高於10%的普 通股權一級比率 （終點基準） 訂立 目標。儘管當中計及可預見的資 本規定，但並未全面預知 （亦無法 全面預知） 未來數年因種種要求而 需承擔的資本承諾及額外支出。 這些目標的考慮因素包括： ‧ 逐步增強資本水平，以回應不 斷提升的資本規定； ‧ 執行監管規定的改革及加強風 險監控措施 （尤其是保障金融 系統健全運作及操守方面） ， 令成本拾級而上； ‧ 銀行徵費增加； ‧ 低息環境持續；及 ‧ 重大項目的影響，尤其是金額 龐大的罰款、和解開支、英國 客戶賠償及相關準備。 因此，我們現時訂立新目標，以 便更配合目前及不斷變遷的經營 環境。 由2015年起，股東權益回報率目 標將更改為10%以上的中期目標。 此目標是採用介乎12%至13%的普 通股權一級比率 （終點基準） 以模 型推算。 同時，我們重申業務收入增長速 度高於營業支出 （按經調整基準計 算） 的目標。 我們亦繼續致力循序漸進地增派 股息。循序漸進的股息增長將配 合集團整體盈利能力的提升，並 基於我們能夠持續達到監管規定 資本水平的前提來預計。 我們的資本仍然穩健，有實力促 成有機增長，並同時為股東帶來 股息回報。 品牌價值 維護滙豐品牌及我們的整體聲譽 一直是集團的優先要務。 我們至今已是第四年採用英國 《銀 行家》 雜誌所公布的Brand Finance 估值法，作為品牌價值的衡量基 準。Brand Finance的計算方法可 以全面衡量品牌實力，以及品牌對 所有業務類別及客戶群的影響。 這是一項完全獨立的計量方法， 其結果並會公開發表。我們的目 標是在銀行同業中位列三甲，且 已成功達標，並以整體價值273億 美元 （較2014年上升2%） 名列第三 位。在本年度的報告中，我們維 持獲評為AAA級 （基於我們的品牌） 公司的地位。 除稅前風險加權資產平均值回報 13 (%) 2012年 2013年 2014年 1.8 2.0 1.5 派息比率 (%) 55.4 57.1 71.0 2012年 2013年 2014年 品牌價值 （十億美元） 22.9 26.9 27.3 第三位 第二位 第三位 2013年2月 2014年2月 2015年2月 有關註釋，請參閱第39頁。 成果 （續） 財務表現 除了採用Brand Finance估值法衡量 以外，我們亦已根據Interbrand於 2014年9月公布的年度最佳全球品 牌報告審視集團的表現。滙豐在 銀行品牌之中稱冠，估值達131億 美元 （高於2013年的120億美元） ； 而在所有金融服務品牌中，滙豐 排名第二位。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 33 經濟效益的分布 2014年2013年2012年 十億 美元 十億 美元 十億 美元 現金稅項流出淨額 7.9 8.6 9.3 向股東及非控股股東 分派 10.6 10.2 8.7 僱員報酬及福利 20.4 19.2 20.5 一般行政開支 （包括物業 及採購） 18.6 17.1 20.0 備考除稅後利潤分配 21 2014年 2013年 % % 保留盈利╱資本 32 53 股息 53 35 浮動酬勞 15 12 截至12月31日 止年度 100 100 有關註釋，請參閱第39頁。 我們相信有此成績，是因為集團近 年建立了強勁的實質品牌資產， 並於2014年全年一直積極推行各 項支持本身品牌的活動。 經濟貢獻 滙豐經營持續發展的業務，即能 以下列方式對經濟作出寶貴的貢 獻：向股東派發股息；向僱員支 付薪金；向供應商付款；及為業 務所在國家和地區的政府帶來稅 收。此外，我們也向許多公司提 供融資，由此而創造就業機會。 滙豐已付稅款淨額18 2014年 2013年 十億美元 十億美元 利得稅 3.6 4.7 僱主稅項 1.6 1.6 英國銀行徵費19 1.0 0.7 不可收回增值稅 0.9 0.8 其他稅項和徵費 0.8 0.8 截至12月31日 止年度 7.9 8.6 有關註釋，請參閱第39頁。 為各地政府徵收的稅項20 2014年 2013年 十億美元 十億美元 地區 英國 1.7 1.5 歐洲其他地區 1.1 1.3 亞洲 2.0 1.5 北美洲 1.0 1.0 拉丁美洲 3.3 3.5 截至12月31日 止年度 9.1 8.8 有關註釋，請參閱第39頁。 市值及股東總回報 收市價 已發行面值0.5美元 之普通股 市值 倫敦 香港 美國預託股份22 192.18億股 1,820億美元 6.09英鎊 74.0港元 47.23美元 2013年：188.30億股 2013年：2,070億美元 2013年：6.62英鎊 2013年：84.15港元 2013年：55.13美元 2012年：184.76億股 2012年：1,940億美元 2012年：6.47英鎊 2012年：81.30港元 2012年：53.07美元 股東總回報23 一年 三年 五年 截至2014年12月31日止 97 144 109 基準： －摩根士丹利資本國際：世界銀行24 100 160 132 有關註釋，請參閱第39頁。 滙豐控股有限公司 34 薪酬 我們的薪酬策略獎勵在商業上 取得成功，並遵守風險管理架 構規定的僱員。 人才的質素及投入程度是集團成 功的基本要素。因此，我們銳意 吸引、挽留和激勵有志在滙豐持 續長遠發展事業的頂尖人才，而 這些人才將履行職責，為股東爭 取長遠利益。 僱員薪酬 我們的薪酬策略力求給予長期持 續表現良好的員工具有競爭力的 回報。滙豐的獎勵由以下四個薪 酬元素組成： － 固定酬勞； － 福利； － 周年獎勵；及 － 集團業績表現股份計劃 （ 「GPSP」 ） 。 集團薪酬委員會監督我們的薪酬 原則及其落實情況，確保我們給予 員工的薪酬能夠配合業務策略， 而且評核員工表現的標準，不僅 限於短期及長期的成果，亦著重 達致成果的方法，因為我們相信 後者對業務的長期可持續發展相 當重要。 集團薪酬政策的詳情，載於我們 的網站 （ 「薪酬政策」 一頁。 行業變更及主要挑戰 新的監管規定，例如花紅上限等， 已對我們向高級行政人員和審慎 監管局識別為對集團風險狀況有 重大影響的員工 （即所謂的 「承受 重大風險人員」 ） 支付薪酬的方法 產生影響。於本年度，有關當局對 企業施行新的規定，確保在2015 年1月1日或以後授出的所有浮動 酬勞獎勵，於授出日期起計最少 七年內可予撤回 （即企業可以追 回已支付及╱或已實際授出的獎 勵） 。在此等規定下，滙豐若要確 保世界各地全部業務所在市場的 員工整體報酬福利維持競爭力 （特 別是相對於不受該等規定限制的 其他銀行而言） ，將要面對挑戰。 展望2015至2016年，預期行政人 員薪酬將要面對更多重大監管改 革，以致我們可能需要變更2016 年的薪酬政策。已經建議及即將 建議的改革數目和規模，使我們 難以向現有及有機會加入集團的 僱員明確解釋適用於他們的薪酬 政策和結構。除此之外，上述情 況亦導致大眾誤解集團政策的運 作方式以及該等政策對僱員表現 的影響。 有關行業變更及主要挑戰的詳情，請參 閱第300頁。 浮動酬勞總額 浮動酬勞總額由2013年的39億美 元減少至2014年的37億美元： 集團 2014年 2013年 百萬美元 百萬美元 浮動酬勞總額25 －總計 3,660 3,920 －佔除稅前利潤 （未計算 浮動酬勞） 的百分比 16% 15% －遞延發放金額 的百分比 14% 18% 有關註釋，請參閱第39頁。 集團薪酬委員會在釐定滙豐的浮 動酬勞總額時考慮很多因素，包 括基於集團的承受風險水平聲明 考慮集團的表現。 此舉確保我們在釐定浮動酬勞總 額時，能顧及各項風險因素，並 從整體角度計及能夠支持我們達 致策略目標的業務、風險及資本 管理方針。 集團薪酬委員會亦會考慮集團的 盈利能力、資本實力、股東回報、 利潤在資本、股息及浮動酬勞之 間的分配、維持市場競爭力的商 業要求，以及集團的整體負擔能 力等因素。 有關釐定浮動酬勞總額的詳情，請參閱 第309頁。 酬勞支出的相對重要性 下圖呈列員工酬勞總額相對於派 息金額的分析。 酬勞支出的相對重要性 （百萬美元） 2014年 2013年 普通股股息26 僱員報酬及福利27 4% 4% 9,600 20,366 9,200 19,626 (430) 19,196 有關註釋，請參閱第39頁。 董事薪酬 執行及非執行董事的薪酬政策已 於2014年5月23日舉行的股東周年 大會上獲股東批准。該等政策的詳 情載於 《2013年報及賬目》 董事薪 酬報告內，並可於以下網站瀏覽： 根據英國 《2008年中大型公司 （賬 目及報告） 規例》 附表8所規定之董 事薪酬的單一數字總計如下： 成果 （續） 薪酬 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 35 執行董事 范智廉 歐智華 麥榮恩 繆思成 2014年 2013年 2014年 2013年 2014年 2013年 2014年 2013年 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 固定酬勞 基本薪金 1,500 1,500 1,250 1,250 700 700 700 － 固定酬勞津貼 － － 1,700 － 950 － 950 － 退休金 750 750 625 625 350 350 350 － 2,250 2,250 3,575 1,875 2,000 1,050 2,000 － 浮動酬勞 周年獎勵 － － 1,290 1,833 867 1,074 1,033 － 集團業績表現股份計劃 － － 2,112 3,667 1,131 2,148 1,131 － － － 3,402 5,500 1,998 3,222 2,164 － 固定及浮動酬勞總額 2,250 2,250 6,977 7,375 3,998 4,272 4,164 － 福利 136 48 589 591 43 33 6 － 非課稅福利 105 102 53 67 28 53 33 － 遞延現金的名義回報 41 27 － － 11 7 36 － 薪酬的單一數字總計 2,532 2,427 7,619 8,033 4,080 4,365 4,239 － 作為集團主席，范智廉不合資格 收取周年獎勵，但符合資格收取 2014年的一次性集團業績表現股 份計劃獎勵。 繆思成為集團風險管理總監，自 2014年1月1日起獲委任為執行董 事，反映風險管理部對滙豐十分 重要、繆思成領導風險管理部的 才能，以及他個人對集團的貢獻。 繆思成2013年的薪酬數字並未披 露。 有關董事薪酬的全部詳情，請參閱第 307頁。 未來的薪酬政策 我們的薪酬政策已於2014年的股 東周年大會上獲股東批准，並將 適用於2015年業績計算年度。下 表概述於2015年如何實施發放酬 勞的各項元素。 對外匯報 有關董事、承受重大風險人員及 集團最高薪僱員的薪酬規定披露 資料，載於第300至323頁董事薪酬 報告內。 目的及與策略的聯繫 薪酬政策在運作及計劃內的修改 固定酬勞 基本薪金 基本薪金水平將維持於2014年的水平，不予改變： ‧ 范智廉：1,500,000英鎊 ‧ 歐智華：1,250,000英鎊 ‧ 麥榮恩：700,000英鎊 ‧ 繆思成：700,000英鎊 固定酬勞津貼28 固定酬勞津貼將維持於2014年的水平，不予改變： ‧ 范智廉：零 ‧ 歐智華：1,700,000英鎊 ‧ 麥榮恩：950,000英鎊 ‧ 繆思成：950,000英鎊 退休金 2015年適用的退休金津貼將為基本薪金的某個百分比，不予改變： ‧ 范智廉：50% ‧ 歐智華：50% ‧ 麥榮恩：50% ‧ 繆思成：50% 福利 福利 並無建議對2015年的福利作出任何修改。 浮動酬勞 周年獎勵28 並無建議對周年獎勵作出任何修改。 集團業績表現股份計劃 並無建議對集團業績表現股份計劃作出任何修改。 有關註釋，請參閱第39頁。 滙豐控股有限公司 36 可持續發展 可持續發展是集團優先策略的 基礎，讓我們可以履行國際銀 行的責任。 對滙豐而言，營商之道與業務範 疇同等重要，可持續發展意味在 發展長遠業務的過程中，所作決 策會平衡社會、環境和經濟等各 方面的考慮因素。此舉有助我們 在推動業務增長的同時，為社區 的健全成長作出貢獻。 達致企業可持續發展的方針 企業可持續發展部由董事會轄下 一個小組委員會 「行為及價值觀 委員會」 管治，該委員會負責監 督一系列事宜，包括緊遵滙豐的 價值觀以及確保我們適當回應社 會和主要相關群體不斷轉變的期 望，並就此等事宜提供意見。 可持續發展的優先策略由環球企 業可持續發展部制訂，有關計劃 亦由該部門帶領執行。滙豐的國 家╱地區業務部門、環球部門及環 球業務會共同合作，確保集團的 業務及營運具備可持續發展的特 性，同時相關措施獲妥善執行。 風險管理部以及滙豐科技及服務 部內的行政人員肩負特定任務， 為集團推展可持續發展計劃的工 作。 我們的可持續發展計劃有三個重 點：可持續發展的融資；可持續 發展的業務營運；以及可持續發 展的社區。 可持續發展的融資 我們預測和管理與氣候、環境和經 濟不斷變化有關的風險和機會。 在一個瞬息萬變的世界，我們必 須確保業務能夠預測環保優先事 務及社會期望的轉變，並為此妥 善準備。 可持續發展風險管理架構 我們致力管理向客戶提供的金融 服務可能對人類或環境帶來不可 接受影響的風險。可持續發展風 險亦可能為客戶帶來商業風險、 為滙豐帶來信貸風險，以及造成 嚴重的聲譽風險。 過去十多年以來，我們一直與企 業客戶合作無間，協助他們了解 及管理與敏感領域和主題有關的 環境及社會影響。我們使用定期 檢討和完善的政策，以評估及支 援客戶。我們制訂涵蓋農業商品、 化工、國防、能源、林業、淡水基 建、礦業和金屬、世界文化遺址 和拉姆薩爾濕地等方面的政策。 除此之外，我們亦採用赤道原則。 我們歡迎非政府組織和活動團體 給予有建設性的意見，並會定期 與他們討論彼此共同關注的事宜。 強而有力的政策、正規的程序，以 及訓練有素和獲得授權的人員， 是我們可持續發展風險管理架構 的基礎。 於2014年，我們為風險及客戶關 係經理提供關於可持續發展風險 的培訓，重點為近年的政策更新 資料和經修訂的流程。我們委派 的可持續發展風險經理向每個地 區的風險管理部、環球銀行及資 本市場業務部門以及工商金融業 務部門的行政人員提供培訓。 我們自2003年起採用赤道原則。 赤道原則的新版本EP3於2013年公 布，滙豐於進行相關培訓和開發 清晰的樣板以確保順利過渡後， 於2014年1月1日採納此等變更。 我們應用赤道原則的資料以及獨立 查證，將於2015年4月上載hsbc.
com的網頁。 2014年的政策檢討及資料更新 2014年，我們公布了由Proforest和 PricewaterhouseCoopers LLP進行獨 立審查的兩份報告，要旨是關於集團 的森林用地及森林產品行業政策的 內容和執行情況。我們亦公布了林 業、農業商品、世界文化遺址和拉姆 薩爾濕地等方面的新政策，以反映我 們採納的建議。此等文件可於hsbc.
com/sus-risk的網頁瀏覽。 林業政策 我們於2014年3月發出的新林業政 策，要求高風險地區的林業客戶於 2014年12月31日前取得由Forest Stewardship Council ( 「FSC」 )或 Programme for the Endorsement of Forest Certification ( 「PEFC」 )發 出的100%認證。該等認證要求客 戶以合法和可持續發展的方式經 營業務。 相關群體對新政策的反應正面。 土耳其和墨西哥等受影響國家的 木材客戶樂意接受新標準，並因 應新規定獲取認證及從建議中受 惠。其他客戶若未能或不願符合 新標準，其客戶關係將於合約條 款容許下盡快結束。 農業商品政策 新的農業商品政策要求棕櫚油 客戶須於2014年6月30日前成為 Roundtable on Sustainable Palm Oil （ 「RSPO」 ） 的成員，並須於2014年 底前最少有一項業務營運獲認證， 並於2018年底前所有業務營運均 獲認證。 未能趕及在期限前符合條件的若 干客戶關係將告結束。其他客戶 已於2014年底前成功加入RSPO及 最少有一項業務營運獲得認證。 其中一個例子為印尼一家加工、 精煉及出口公司。由2014年1月起， 滙豐就相關改革與該公司及其他 公司密切聯繫，並持續提供建議。 該公司的管理層向第三方尋求專 家建議，增加對RSPO認證的了 解，發覺有關過程較預期簡單。 該公司有兩項營運業務於2014年 6月取得RSPO認證，並已計劃再 增加一項認證。 為鼓勵客戶轉向經營可持續發展 的棕櫚油業務，我們向獲認證的 可持續發展棕櫚油貿易提供一項 折扣預付出口融資產品。此項結 構性的特設融資於2014年在新加 坡及印尼推出，並於2015年初在 馬來西亞推出。 使用此項產品的首項融資為一家 大型棕櫚油出口公司，該公司成 為RSPO的成員已達十年，現已全 面獲得認證。該產品提供予現有 和未來的客戶，我們期望可藉此 推動獲認證為可持續發展的棕櫚 油擴展在市場所佔比例。 馬來西亞、印尼、中國內地、台 灣、南韓、泰國、土耳其及墨西 哥的棕櫚油客戶因滙豐的新政策 和期限而決定就本身的業務營運 尋求認證。其他多家公司已獲認 證。有關此等新政策的影響之較 詳盡報告，將於2015年4月上載至 hsbc.com。 成果 （續） 可持續發展 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 37 世界文化遺址及拉姆薩爾濕地政策 世界文化遺址及拉姆薩爾濕地政 策的作用是保護聯合國名單內對 世界十分重要的獨特地點和在國 際上有重要地位的濕地。該政策 牽涉參與大型項目的所有企業客 戶，特別是涉足林業、農業、礦 業、能源、物業及基建發展等領 域的客戶。 此項政策有助滙豐就是否提供融 資予可能影響該等遺址或濕地的 項目，作出平衡和清晰的決定。 滙豐根據該項政策決定不向某些 項目提供融資。 我們管理可持續發展風險的方針載於第 237頁。 氣候業務 我們明白，為應對氣候變化，有 必要轉向發展低碳經濟。我們透 過支持客戶尋找長遠的低碳商業 機會參與 「氣候業務」 ，致力加快 這種轉向的步伐。我們的氣候業 務包括太陽能、風能、生物能源、 能源效益、低碳運輸和水務行業 的客戶。於2014年，我們的氣候 變化研究團隊被譽為業內最佳的 團隊。我們亦是公開市場上為國 際公司安排風能項目股本融資的 領導者，包括在總值14億歐元的 Vestas再融資項目下進行自2010年 以來最大規模的風力渦輪機股本 集資項目。 「綠色債券」 是所得款項專門用於 為氣候或環境項目融資的任何種類 債券工具。於2014年4月，滙豐成 為綠色債券原則國際資本市場協會 執行委員會(International Capital Market Association Executive Committee)的成員。綠色債券原 則是自願性的程序指引，建議透 過釐清發行綠色債券的方法，一 方面提高透明度和加強披露，另 一方面鼓勵發展真正的綠色債券 市場。 於2014年，我們委託氣候債券倡 議組織編製一份題為 《債券及氣候 變化：2014年的的市場狀況》 的報 告，以提高市場對氣候融資的關 注。 滙豐一直站在這個迅速發展領域 的最前線。於2014年，我們擔任 一間亞洲公司發行人Advanced Semiconductor Engineering Inc.
首項綠色債券發行的獨家全球協 調人及聯席牽頭人、經辦人和賬 簿管理人。我們亦擔任歐洲首項 高收益綠色債券 （由Abengoa發行） 的獨家全球協調人，以及加拿大 安大略省市場首項政府債券的聯 席牽頭經辦人兼賬簿管理人。 聯合國環境規劃署金融計劃的保 險業可持續發展原則 作為一項全球可持續發展架構 「保 險業可持續發展原則」 （ 「PSI」 ） 的 簽署機構之一，滙豐的保險業務 致力在其流程中結合各種環境、 社會和管治要素，並每年公開披 露有關工作進展。我們已委任一 名環球計劃經理，負責帶領、協 調和控制全球的保險業可持續發 展策略，並確保相關業務與集團 的方針和PSI原則下的規定保持一 致。有關工作包括在保險業務內 以及與合作夥伴、監管機構和其 他同業推動適當的活動；執行業 內的最佳實務；以及制訂全球性 的可持續發展保險策略。 可持續發展的營運 管理我們在環境方面的政策使集 團業務保持良好效益，同時亦是我 們對社會作出長遠貢獻的一環。 我們同心協力，並與供應商合作， 找出新的方法減低業務營運對環 境造成的影響。我們以更具效益 的方式採購可再生能源、設計和 營運各樓宇和數據中心，並減少 產生廢物。我們致力於2020年前 將碳排放由每名員工每年3.5噸降 低至2.5噸。 可持續發展領導力課程 為達到我們的十項可持續發展目 標，我們自2009年起透過滙豐的可 持續發展領導力課程培訓了847名高 級經理。該課程以滙豐價值觀作 為理念，結合了實況學習和領導 力發展的環節。預期該課程的參 加者在不同業務部門和工作崗位 上作出決策和執行工作計劃時， 會將可持續發展的因素體現出來。 採購可再生能源 於2014年，我們與英國和印度的 可再生能源供應商簽訂了三份能 源購買協議，預期將為滙豐提供 9%的能源。於8月份，位於印度海 得拉巴的10兆瓦太陽能發電廠投 入運作，為集團提供潔淨能源。 該發電廠預期將為印度的三個環 球服務中心及一個科技中心提供 電力。滙豐同意在未來十年按一 個政府保證的固定價格購買該發 電廠的電力，從而在推動該項目 上扮演重要角色。該發電廠將提 供潔淨而可靠的能源。此外，我們 已重訂集團的可再生項目目標， 在新目標下計算的能源，將全部 來自由滙豐委託新建成的可再生 能源來源。 用紙 我們將透過三個方面達到用紙目 標：確保我們是根據集團的採購 用紙政策從可持續發展的來源購 買用紙、減少辦公室和分行的用 紙量，以及向所有零售和商業客 戶提供無紙化銀行服務。我們持 續減少紙張的總採購量，同時促 使更大比重的用紙經核證為來自 FSC及PEFC可持續發展來源。自 2011年起，我們的購紙量減少了 53%，並於2014年底前促使經認 證為源自可持續發展來源的用紙 達到全部用紙量的92%。 滙豐控股有限公司 38 十項可持續營運策略 1. 可持續發展任務：鼓勵僱員 於2020年前提高效率 2. 供應鏈合作：透過提升效率 和創新實現可持續成本節約 3. 滙豐生態效益基金：每年投 資5,000萬美元，根據僱員的 創新概念發展新的工作方式 4. 能源：於2020年前使每名僱員 的每年能源消耗量較2011年 的6.2兆瓦時減少1兆瓦時 5. 廢物：減少消耗及使辦公室 和電子廢物的再用率達到 100% 6. 可再生能源：增加使用可再 生能源，力求於2020年之前 由零提升至25% 7. 綠色建築物：按照最高水平 的國際標準，設計、興建和 營運具能源效益、可持續發 展的樓宇 8. 數據中心：於2020年前達到 1.5級的能源效益 （用電效率） 9. 差旅：減低每名僱員的差旅 碳排放量 10. 用紙：於2020年前向所有零 售及商業客戶提供無紙化銀 行服務及有100%用紙來自 可持續來源 碳排放 滙豐的二氧化碳排放量乃根據我 們於超過28個國家或地區 （涵蓋由 等同全職僱員負責的約93%業務） 的建築物及僱員差旅所用能源計 算。就能源消耗量及旅程距離收 集的數據，會採用下列來源 （如有） 的換算系數換算為二氧化碳排放 量，該等系數按優先次序排列如 下： 1. 由數據╱服務供應商提供的系 數； 2. 由各地公營環保部門提供的系 數。就電力而言，倘未能自上 述兩個來源取得特定系數，我 們將根據 《溫室氣體議定書》 的 建議，採用國際能源機構國家 電網電力的最新可用碳排放 系數；及 3. 至於其他類型能源及差旅，倘 未能自上述兩個來源取得特定 系數，我們會採用英國環境、 食品與農村事務部及╱或英國 能源和氣候變化部提供的最 新可用系數。 為使有關數據涵蓋我們擁有財務 （管理） 控制權的所有業務，計算 出的二氧化碳排放量已根據等同 全職僱員覆蓋率按比例增加，以 計及任何遺漏的數據 （一般少於等 同全職僱員的10%） 。此外，我們 亦已應用排放增加率，以顧及排 放計量和估算值在質量及覆蓋範 圍方面的不明朗因素。電力的比 率為4%、其他能源為10%，而差旅 為6%，此等比率乃根據政府間氣 候變化專門委員會編寫的 《國家溫 室氣體清單優良實務指引和不確 定性管理》 及滙豐對數據覆蓋及質 量的內部分析而計算。 二氧化碳排放量 （噸） 2014年 2013年 總計 752,000 889,000 能源 633,000 755,000 差旅 119,000 134,000 每名等同全職僱員的二氧化碳排放量 （噸） 2014年 2013年 總計 2.92 3.43 能源 2.46 2.91 差旅 0.46 0.52 我們的溫室氣體報告年度由10月至 9月。截至2014年9月30日止年度， 集團業務產生的二氧化碳排放量 為752,000噸。 可持續發展的社區 我們相信，對於一個充滿活力的 社區 （此社區亦為推動經濟和商業 發展的基礎） ，教育及資源 （例如 安全食水和衞生等） 十分重要。 我們在財務上為社區項目作出貢 獻，而全球各地成千上萬的僱員 亦自願貢獻自己的時間和分享他 們的技能，參與義務工作。 員工參與志願服務和捐獻 滙豐遍及世界各地成千上萬的僱 員，每年參與集團各個社區投資項 目的義務工作。有關集團各項義 務計劃的進一步詳情，可到hsbc.
com瀏覽。我們將於2015年4月加 入2014年的更新資料。 2014年，我們合共捐獻了1.14億 美元予社區項目 （2013年：1.17億 美元） 。當中，6,600萬美元在歐 洲捐獻 （2013年：6,400萬美元） ； 2,800萬美元在亞太區捐獻 （2013 年：2,400萬美元） ；300萬美元在 中東捐獻 （2013年：500萬美元） ； 1,000萬美元在北美洲捐獻 （2013 年：1,100萬美元） ；及700萬美元 在拉丁美洲捐獻 （2013年：1,200萬 美元） 。 僱員於工作時間內投入了303,922 小時參與義務工作 （2013年： 255,925小時） 。 人權 我們會直接應用人權考慮因素， 因為這些因素會影響我們的僱員； 我們面對供應商和客戶時 （特別在 項目融資借貸和可持續發展風險 政策方面） ，也會間接考慮這些因 素。與滙豐最直接關連的人權問 題包括：公平而有利的工作條件 及薪酬、同工同酬、成立及加入 工會、休息及享用閑暇的權利， 以及禁止奴役和僱用童工。我們 除了作出多項承諾 （例如自2005年 起已訂定的 《滙豐供應商行為守 則》 、 《滙豐環球標準手冊》 及滙豐 價值觀） 之外，亦已簽署遵守多項 環球承諾及標準，包括 《聯合國全 球契約》 、 《世界人權宣言》 及 《全 球蘇利文原則》 。 有關2014年集團表現的進一步資料，將 於2015年4月底上載至我們的網站，屆 時將一併上載我們應用赤道原則和碳排 放的獨立鑑證。 代表董事會 滙豐控股有限公司 集團主席 范智廉 2015年2月23日 成果 （續） 可持續發展╱註釋 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 39 策略報告註釋 1 資本指引4於2014年1月1日實施，於2014年12月31日的資本及風險加權資產已按此基準計算及呈列。在此之前，資本及風 險加權資產乃按巴塞爾協定2.5的基準計算及呈列。此外，於2013年12月31日的資本及風險加權資產亦按集團對資本指引 4規例最終版本及審慎監管局公布之最終規則的理解而估計。於2012年12月31日，資本指引4估計資本及風險加權資產是 以2011年7月的資本指引4文本為基礎。 2 載於財務報表的股息為一年內就每股普通股宣派的股息，而非與該年度相關的股息。2013年第三次股息為0.1美元，已 於2013年12月11日派發。2013年第四次股息為0.19美元，已於2014年4月30日派發。本公司已分別於2014年7月10日、2014年 10月9日及2014年12月10日派發2014年第一次、第二次及第三次股息，每次派息為每股普通股0.1美元。財務報表附註9提 供2014年宣派股息的更詳細資料。於2015年2月23日，董事會宣布派發2014年第四次股息每股普通股0.2美元，以代替末期 股息。第四次股息將於2015年4月30日以美元、英鎊或港元現金派發予普通股股東 （若為英鎊或港元，將按2015年4月20日 的匯率折算） ，股東亦可選擇以股代息。2014年12月31日的可供分派儲備為488.83億美元。 6.2厘非累積A系列美元優先股的季度股息為每股15.5美元，相等於每股A系列美國預託股份 （每股代表四十分之一股A系列 美元優先股） 派發股息0.3875美元。此股息已於2014年3月17日、2014年6月16日、2014年9月15日及2014年12月15日派發。 8.125厘資本證券的季度票息為每份證券0.508美元，已於2014年1月15日、2014年4月15日、2014年7月15日及2014年10月15日 派發。 8厘資本證券的季度票息為每份證券0.5美元，已於2014年3月17日、2014年6月16日、2014年9月15日及2014年12月15日派發。 3 成本效益比率的定義為營業支出總額除以未扣除貸款減值及其他信貸風險準備之營業收益淨額。 4 平均普通股股東權益回報的定義為母公司普通股股東應佔利潤除以平均普通股股東權益。 5 於2014年12月5日設立。 6 證券、基金及保險產品的中介服務，包括環球銀行及資本市場業務的證券服務。 7 併購、事件及項目融資，以及於環球私人銀行業務的共同投資。 8 包括外匯、利率、信貸及股票交易。 9 包括組合管理。 10 包括私人信託及遺產策劃 （就金融及非金融資產而言） 。 11 包括對沖基金、房地產及私募股本。 12 由於本表並未呈列公司之間的撇銷項目，因此呈列的款額總數與綜合計算的數額並不一致。 13 風險加權資產平均值的除稅前回報以巴塞爾協定2.5為基準，採用截至2013年12月31日 （包括當日） 所有期間的風險加權 資產平均值計算，及按資本指引4終點基準採用由2014年1月1日起所有期間的風險加權資產平均值計算。 14 未扣除貸款減值及其他信貸風險準備之營業收益淨額，亦稱 「收入」 。 15 貸款減值及其他信貸風險準備。 16 應佔聯營及合資公司利潤。 17 自2014年1月1日起，非交易用途反向回購及回購於資產負債表內分行呈列。過往，非交易用途反向回購計入 「同業貸款」 及 「客戶貸款」 ，而非交易用途回購則計入 「同業存放」 及 「客戶賬項」 。比較數字已相應重列。非交易用途反向回購及回 購於資產負債表內分行呈列，使披露方式與市場慣例一致，並提供更具參考價值的貸款相關資料。客戶及同業貸款╱存 放中反向回購及回購的份額載於財務報表附註17。 18 滙豐支付的稅項與滙豐自身的稅務責任有關，並按現金流基準列賬。 19 已付英國銀行徵費反映於歷年向稅務機關支付的款項，並可能與扣取自收益表的確認負債有所不同。 20 徵收的稅項與滙豐在世界各地作為稅務機關代理而有責任支付的稅項有關，其中包括僱員相關稅項，連同由利息付款 預扣的稅項及向客戶提供商品及服務收取的稅項。徵收的稅項按現金流基準列賬。 21 不包括本身債務公允值變動及未計算浮動酬勞資金分派。 22 每股美國預託股份代表5股普通股。 23 股東總回報的定義是有關期間內股份價值及所宣派股息收益的增長。 24 摩根士丹利資本國際世界銀行指數。 25 2014年集團除稅前未計算浮動酬勞利潤的計算方式載於第309頁的董事薪酬報告內。守則職員的遞延浮動酬勞比例為 50%。 26 該年度的每股普通股股息。就2014年而言，包括於2014年派付的第一次、第二次及第三次股息58億美元 （未減代息股份） 及第四次股息38億美元。 27 2013年的僱員報酬及福利總計為191.96億美元，當中包括因改變英國傷病福利計算基準而產生的會計增益4.3億美元。若 不包括此項會計增益，2013年僱員報酬及福利總計為196.26億美元。 28 這方法適用於所有執行董事，但不包括集團主席范智廉，因為他不合資格收取固定酬勞津貼或浮動酬勞獎勵。 董事會報告：財務概要 滙豐控股有限公司 40 採用非公認會計原則之財務衡量指標 財務概要 採用非公認會計原則之財務衡量指標 40 綜合收益表 45 按收支項目列示之集團業績表現 46 淨利息收益 46 費用收益淨額 48 交易收益淨額 49 指定以公允值列賬之金融工具 淨收益╱ （支出） 50 金融投資減除虧損後增益 51 保費收益淨額 51 其他營業收益 52 已支付保險賠償和利益及投保人 負債之變動淨額 53 貸款減值及其他信貸 風險準備 53 營業支出 54 應佔聯營及合資公司利潤 55 稅項支出 56 綜合資產負債表 57 2014年之變動 58 風險加權資產平均值 回報計量之對賬 62 關鍵會計估算及判斷 62 董事會報告： 「財務回顧」 ，連同 「企業管治」 一節 內的 「僱員」 和 「企業可持續發展」 ，以及 「董事薪酬 報告」 中載列的管理層評述，是根據國際會計準則 委員會頒布的國際財務報告準則作業準則 「管理層 評論」 編製。 採用非公認會計原則之 財務衡量指標 集團列賬基準之業績乃根據IFRS編製，詳 情載於自第334頁起呈列的財務報表。我們 在衡量業績時採用的財務衡量指標，包括 源自列賬基準業績的財務衡量指標，以便 撇除會令按年比較資料扭曲的因素。這些 衡量指標被視為非公認會計原則財務衡量 指標。我們採用的主要非公認會計原則財 務衡量指標為 「經調整業績」 。當我們採用 其他非公認會計原則財務衡量指標時，亦 會加以說明，並與最有關連的列賬基準財 務衡量指標進行對賬。 經調整業績 經調整業績的計算方法是，就可能令按年 比較資料扭曲的外幣換算差額及重大項目 之按年計算影響，對列賬基準業績作出調 整。 集團過往採用非公認會計原則之財務衡量 指標 「實際基準之業績表現」 ，計算方法為 就貨幣換算差額、本身信貸息差以及收購、 出售及攤薄投資按年計算的影響，對列賬 基準業績作出調整。於2014年，集團修訂 了相關計算方法，以便更能配合內部檢視 業績的方式及投資者的反饋意見。經調整 業績以實際基準之業績表現為基礎，保持 了貨幣換算差額方面的調整，並於重大項 目的定義內加入本身信貸息差以及收購、 出售及攤薄投資方面的調整。我們採用 「重 大項目」 一詞以綜合反映一組個別調整項 目，這些項目於計算經調整業績時會從列 賬基準之業績撇除。於下文詳述的重大項 目，是管理層和投資者為了加深了解業務 的有關趨勢，一般會於評估業績表現時另 行識別及考慮的項目。 我們相信，經調整業績展示更一致的對內 及對外匯報資料方式，而且識別和量化了 管理層相信屬重大的項目，同時深入反映 管理層如何評估按年計算的業績表現，為 投資者提供了有用的信息。 我們計算經調整業績時，會從列賬基準之 業績撇除以下項目： ‧ 外幣換算差額按年計算的影響。方法是 將2014年列賬基準之業績與按2014年匯 率重新換算的2013年列賬基準業績進 行比較。外幣換算差額反映美元兌大多 數主要貨幣的變動；及 ‧ 扭曲列賬基準業績按年比較資料之重 大項目，因為這些項目令人難以清楚掌 握影響業務營運的相關因素及趨勢。重 大項目包括過往為實際基準衡量指標 一部分的調整，如就本身信貸息差以及 收購、出售及攤薄投資作出的調整。下 頁載列進一步詳情，包括列賬基準業績 與經調整業績之對賬。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 41 在處理附屬公司、聯營公司、合資公司及 各項業務的收購、出售及擁有權變動時， 我們不會計入於有關年度因出售或攤薄投 資而錄得的損益，亦不會計入因重新分類 或確認減值而產生的任何相關損益，同時 亦會在呈報的所有業績年度剔除經收購、 出售或攤薄投資的附屬公司、聯營公司、 影響經調整業績的出售利潤╱ （虧損） 日期 出售 利潤╱ （虧損） 百萬美元 興業銀行股份有限公司向第三方增發股本後，集團將持有的該行 股權重新分類所產生的增益1 2013年1月 1,089 滙豐保險集團 （亞太） 有限公司出售於Bao Viet Holdings的股權1 2013年3月 104 Household Insurance Group Holding Company出售旗下從事制訂保險產品的業務1 2013年3月 (99) HSBC Seguros, S.A. de C.V., Grupo Financiero HSBC出售於墨西哥的財產及 意外傷亡保險業務1 2013年4月 20 英國滙豐銀行有限公司出售於HSBC (Hellas) Mutual Funds Management SA的股權2 2013年4月 (7) 滙豐保險集團 （亞太） 有限公司出售於Hana HSBC Life Insurance Company Limited的股權1 2013年5月 28 英國滙豐銀行有限公司出售HSBC Assurances IARD 2 2013年5月 (4) 香港上海滙豐銀行有限公司出售滙豐人壽保險 （國際） 有限公司的台灣分行業務2 2013年6月 (36) HSBC Markets (USA) Inc.出售附屬公司Rutland Plastic Technologies2 2013年8月 17 HSBC Insurance (Singapore) Pte Ltd出售於新加坡的僱員福利保險業務2 2013年8月 (8) 滙豐投資銀行控股有限公司出售於聯營公司FIP Colorado的投資2 2013年8月 (5) 滙豐投資銀行控股有限公司集團出售於附屬公司Viking Sea Tech的投資1 2013年8月 54 HSBC Latin America Holdings UK Limited出售巴拿馬滙豐銀行2 2013年10月 1,107 HSBC Latin America Holdings UK Limited出售HSBC Bank (Peru) S.A.2 2013年11月 (18) HSBC Latin America Holdings UK Limited出售HSBC Bank (Paraguay) S.A.2 2013年11月 (21) 烟台銀行股份有限公司增加註冊股本後，集團將持有的該行股權 重新分類所產生的虧損1 2013年12月 (38) HSBC Latin America Holdings UK Limited出售HSBC Bank (Colombia) S.A.1 2014年2月 18 於失去對Vietnam Technological & Commercial Joint Stock Bank的重大控制權後， 集團將持有的該行股權重新分類所產生的虧損1 2014年6月 (32) 中東滙豐銀行有限公司出售巴基斯坦的業務1 2014年10月 (27) 有關註釋，請參閱第109頁。 合資公司及各項業務之營業利潤或虧損， 使我們能按對等比較基準檢視業績。除上 述定義包括的投資項目外，出售的策略投 資如屬重大，將載入其他重大項目內。 下列各項收購、出售及業務擁有權變動對 經調整業績產生影響： 外幣換算差額 （ 「固定匯率」 ） 外幣換算差額反映2014年美元兌大多數主 要貨幣的變動。我們在使用固定匯率基準 時未有計入換算差額，乃因固定匯率基準 使我們得以按對等比較基準評估資產負債 表及收益表的表現，從而更深入了解業務 的相關趨勢。 外幣換算差額 2013年外幣換算差額的計算方法，是將經營非美元 業務的分行、附屬公司、合資公司及聯營公司所涉款 額，重新換算為美元： ‧ 2013年的收益表，是按2014年的平均匯率換 算；及 ‧ 於2013年12月31日的資產負債表，則按2014 年12月31日的通行匯率換算。 任何滙豐分行、附屬公司、合資公司或聯營公司的外 幣計值資產及負債於換算為以功能貨幣計值時，並無 對匯率作出調整。凡在列表或評述內提及外幣換算差 額時，按滙豐業務所用功能貨幣呈列的比較數字，均 已按上述基準以本年度內適用的匯率換算。 滙豐控股有限公司 42 董事會報告：財務概要 （續） 採用非公認會計原則之財務衡量指標 其他重大項目 下表詳述其他重大項目於2014年及2013年分別對地區業務及環球業務的影響。 影響經調整業績的其他重大項目－虧損╱ （增益） 2014年 歐洲 亞洲 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 收入 衍生工具合約之借記估值調整 234 69 5 16 8 332 不合資格對沖之公允值變動3 235 4 － 302 － 541 出售數批美國有抵押房地產賬項 所產生的利潤 － － － (168) － (168) 出售所持上海銀行股權所產生的利潤 － (428) － － － (428) 於興業銀行之投資所產生的減值 － 271 － － － 271 因應英國 《消費者信貸法》 持續合規檢討 所產生的準備 632 － － － － 632 1,101 (84) 5 150 8 1,180 營業支出 與美國聯邦房屋金融局達成和解協議 相關的準備 － － － 550 － 550 與外匯調查相關的和解開支及準備 1,187 － － － － 1,187 重組架構及其他相關成本 123 9 2 28 116 278 環球私人銀行業務就監管事宜提撥的準備 16 49 － － － 65 英國的客戶賠償計劃 1,275 － － － － 1,275 2,601 58 2 578 116 3,355 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 收入 衍生工具合約之借記估值調整 － － 332 － － 332 不合資格對沖之公允值變動3 493 (1) 8 1 40 541 出售數批美國有抵押房地產賬項 所產生的利潤 (168) － － － － (168) 出售所持上海銀行股權所產生的利潤 － － － － (428) (428) 於興業銀行之投資所產生的減值 － － － － 271 271 因應英國 《消費者信貸法》 持續合規檢討 所產生的準備 568 24 － 40 － 632 893 23 340 41 (117) 1,180 營業支出 與美國聯邦房屋金融局達成和解協議 相關的準備 17 － 533 － － 550 與外匯調查相關的和解開支及準備 － － 1,187 － － 1,187 重組架構及其他相關成本 88 37 27 6 120 278 環球私人銀行業務就監管事宜提撥的準備 － － － 65 － 65 英國的客戶賠償計劃 992 138 145 － － 1,275 1,097 175 1,892 71 120 3,355 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 43 2013年 歐洲 亞洲 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 收入 完成出售平安保險所產生的利潤淨額 － (553) － － － (553) 衍生工具合約之借記估值調整 (65) (40) (2) 14 (13) (106) 不合資格對沖之公允值變動3 (297) 32 － (246) － (511) 與滙豐控股發行之英鎊債務相關的匯兌增益 (442) － － － － (442) 撇銷與摩納哥環球私人銀行業務相關的 已分配商譽 279 － － － － 279 出售數批美國有抵押房地產賬項所產生的 虧損 － － － 123 － 123 出售美國有抵押非房地產賬項所產生的虧損 － － － 271 － 271 提早終止美國縮減組合內的現金流對沖 所產生的虧損 － － － 199 － 199 出售HFC Bank UK有抵押貸款組合所產生的 虧損 146 － － － － 146 (379) (561) (2) 361 (13) (594) 營業支出 重組架構及其他相關成本 217 86 4 101 75 483 英國的客戶賠償計劃 1,235 － － － － 1,235 與馬多夫事件相關的訴訟費 298 － － － － 298 環球私人銀行業務就監管事宜提撥的準備 317 35 － － － 352 就與卡及零售商戶業務有關的美國客戶 補救措施提撥的準備 － － － 100 － 100 改變於英國提供傷病福利的基準所產生的 會計增益 (430) － － － － (430) 1,637 121 4 201 75 2,038 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 收入 完成出售平安保險所產生的利潤淨額 － － － － (553) (553) 衍生工具合約之借記估值調整 － － (106) － － (106) 不合資格對沖之公允值變動3 (262) － 18 － (267) (511) 與滙豐控股發行之英鎊債務相關的匯兌增益 － － － － (442) (442) 撇銷與摩納哥環球私人銀行業務相關的 已分配商譽 － － － 279 － 279 出售數批美國有抵押房地產賬項所產生的 虧損 123 － － － － 123 出售美國有抵押非房地產賬項所產生的虧損 271 － － － － 271 提早終止美國縮減組合內的現金流對沖 所產生的虧損 199 － － － － 199 出售HFC Bank UK有抵押貸款組合所產生 的虧損 146 － － － － 146 477 － (88) 279 (1,262) (594) 營業支出 重組架構及其他相關成本 167 31 13 73 199 483 英國的客戶賠償計劃 953 148 134 － － 1,235 與馬多夫事件相關的訴訟費 － － 298 － － 298 環球私人銀行業務就監管事宜提撥的準備 － － － 352 － 352 就與卡及零售商戶業務有關的美國客戶 補救措施提撥的準備 100 － － － － 100 改變於英國提供傷病福利的基準所產生的 會計增益 (189) (160) (81) － － (430) 1,031 19 364 425 199 2,038 有關註釋，請參閱第109頁。 滙豐控股有限公司 44 董事會報告：財務概要 （續） 採用非公認會計原則之財務衡量指標╱綜合收益表 下表載列2014年與2013年列賬基準與經調 整之選定項目對賬。有關各項環球業務及 地區業務的同類列表， 可於www.hsbc.com 查閱。 列賬基準項目與經調整項目之對賬 2014年 2013年 變動 5 百萬美元 百萬美元 % 收入4 列賬基準 61,248 64,645 (5) 貨幣換算調整6 (686) 本身信貸息差7 (417) 1,246 收購、出售及攤薄投資 (9) (2,757) 其他重大項目 1,180 (594) 經調整 62,002 61,854 － 貸款減值及其他信貸風險準備 列賬基準 (3,851) (5,849) 34 貨幣換算調整6 168 收購、出售及攤薄投資 － 67 其他重大項目 － － 經調整 (3,851) (5,614) 31 營業支出總額 列賬基準 (41,249) (38,556) (7) 貨幣換算調整6 348 收購、出售及攤薄投資 40 488 其他重大項目 3,355 2,038 經調整 (37,854) (35,682) (6) 經調整成本效益比率 61.1% 57.7% 應佔聯營及合資公司利潤 列賬基準 2,532 2,325 9 貨幣換算調整6 11 收購、出售及攤薄投資 － 87 其他重大項目 － － 經調整 2,532 2,423 4 除稅前利潤 列賬基準 18,680 22,565 (17) 貨幣換算調整6 (159) 本身信貸息差7 (417) 1,246 收購、出售及攤薄投資 31 (2,115) 其他重大項目 4,535 1,444 經調整 22,829 22,981 (1) 有關註釋，請參閱第109頁。 經調整除稅前利潤 2014年 2013年 變動 5 百萬美元 百萬美元 % 按環球業務列示 零售銀行及財富管理 7,648 7,959 (4) 工商金融 8,940 7,910 13 環球銀行及資本市場 8,114 9,208 (12) 環球私人銀行 738 900 (18) 其他 (2,611) (2,996) 13 截至12月31日止年度 22,829 22,981 (1) 按地區列示 歐洲 3,905 4,301 (9) 亞洲8 14,635 14,309 2 中東及北非 1,854 1,673 11 北美洲 2,111 2,048 3 拉丁美洲 324 650 (50) 截至12月31日止年度 22,829 22,981 (1) 有關註釋，請參閱第109頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 45 綜合收益表 綜合收益表五年概要 2014年 2013年 2012年 2011年 2010年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 淨利息收益 34,705 35,539 37,672 40,662 39,441 費用收益淨額 15,957 16,434 16,430 17,160 17,355 交易收益淨額 6,760 8,690 7,091 6,506 7,210 指定以公允值列賬之金融工具淨收益╱ （支出） 2,473 768 (2,226) 3,439 1,220 金融投資減除虧損後增益 1,335 2,012 1,189 907 968 股息收益 311 322 221 149 112 保費收益淨額 11,921 11,940 13,044 12,872 11,146 出售美國分行網絡、美國卡業務及 中國平安保險 （集團） 股份有限公司所產生的利潤 － － 7,024 － － 其他營業收益 1,131 2,632 2,100 1,766 2,562 營業收益總額 74,593 78,337 82,545 83,461 80,014 已支付保險賠償和利益及投保人負債之變動淨額 (13,345) (13,692) (14,215) (11,181) (11,767) 未扣除貸款減值及其他信貸風險準備之 營業收益淨額 61,248 64,645 68,330 72,280 68,247 貸款減值及其他信貸風險準備 (3,851) (5,849) (8,311) (12,127) (14,039) 營業收益淨額 57,397 58,796 60,019 60,153 54,208 營業支出總額 (41,249) (38,556) (42,927) (41,545) (37,688) 營業利潤 16,148 20,240 17,092 18,608 16,520 應佔聯營及合資公司利潤 2,532 2,325 3,557 3,264 2,517 除稅前利潤 18,680 22,565 20,649 21,872 19,037 稅項支出 (3,975) (4,765) (5,315) (3,928) (4,846) 本年度利潤 14,705 17,800 15,334 17,944 14,191 母公司股東應佔利潤 13,688 16,204 14,027 16,797 13,159 非控股股東應佔利潤 1,017 1,596 1,307 1,147 1,032 五年財務資料 2014年 2013年 2012年 2011年 2010年 美元 美元 美元 美元 美元 每股基本盈利 0.69 0.84 0.74 0.92 0.73 每股攤薄後盈利 0.69 0.84 0.74 0.91 0.72 每股普通股股息9 0.49 0.48 0.41 0.39 0.34 % % % % % 派息率10 71.0 57.1 55.4 42.4 46.6 除稅後平均資產總值回報 0.5 0.7 0.6 0.6 0.6 平均普通股股東權益回報 7.3 9.2 8.4 10.9 9.5 外幣兌美元平均換算率： 1美元兌英鎊 0.607 0.639 0.631 0.624 0.648 1美元兌歐元 0.754 0.753 0.778 0.719 0.755 有關註釋，請參閱第109頁。 除另有說明外， 《2014年報及賬目》 內所有列表的數字以列賬基準呈列。 有關滙豐於2014年的財務表現概要，請參閱第28頁。 滙豐控股有限公司 46 董事會報告：財務概要 （續） 按收支項目列示之集團業績表現 按收支項目列示之集團業績表現 淨利息收益 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 利息收益 50,955 51,192 56,702 利息支出 (16,250) (15,653) (19,030) 淨利息收益11 34,705 35,539 37,672 附息資產平均值 1,786,536 1,669,368 1,625,068 總孳息率12 2.85% 3.07% 3.49% 減：資金成本 (1.05%) (1.10%) (1.36%) 淨息差13 1.80% 1.97% 2.13% 淨利息收益率14 1.94% 2.13% 2.32% 有關註釋，請參閱第109頁。 按資產類別劃分的利息收益概要 2014年 2013年 2012年 平均款額 利息收益 收益率 平均款額 利息收益 收益率 平均款額 利息收益 收益率 百萬美元 百萬美元 % 百萬美元 百萬美元 % 百萬美元 百萬美元 % 短期資金及同業貸款27 237,148 3,068 1.29 236,377 2,851 1.21 235,831 3,505 1.49 客戶貸款27 931,311 37,429 4.02 897,322 38,529 4.29 891,699 40,870 4.58 反向回購協議－非交易用途26, 27 198,273 1,800 0.91 114,324 995 0.87 83,105 975 1.17 金融投資 399,816 8,323 2.08 393,309 8,002 2.03 387,329 9,078 2.34 其他附息資產 19,988 335 1.68 28,036 815 2.91 27,104 2,274 8.39 附息資產總值 1,786,536 50,955 2.85 1,669,368 51,192 3.07 1,625,068 56,702 3.49 指定以公允值列賬之交易用途資產及 金融資產15, 16, 26 238,958 5,596 2.34 354,817 5,763 1.62 368,406 6,931 1.88 減值準備 (14,015) (15,954) (17,421) 不附息資產 668,564 683,785 730,901 截至12月31日止年度 2,680,043 56,551 2.11 2,692,016 56,955 2.12 2,706,954 63,633 2.35 有關註釋，請參閱第109頁。 按負債及股東權益類別劃分的利息支出概要 2014年 2013年 2012年 平均款額 利息支出 成本 平均款額 利息支出 成本 平均款額 利息支出 成本 百萬美元 百萬美元 % 百萬美元 百萬美元 % 百萬美元 百萬美元 % 同業存放17, 27 61,217 481 0.79 61,616 555 0.90 78,023 1,001 1.28 指定以公允值列賬之金融負債 －已發行之本身債務18 66,374 837 1.26 72,333 967 1.34 75,016 1,325 1.77 客戶賬項19, 27 1,088,493 9,131 0.84 1,035,500 8,794 0.85 1,012,056 10,650 1.05 回購協議－非交易用途26, 27 190,705 652 0.34 94,410 405 0.43 55,536 387 0.70 已發行債務證券 129,724 4,554 3.51 150,976 4,182 2.77 161,348 4,755 2.95 其他附息負債 10,120 595 5.88 11,345 750 6.61 19,275 912 4.73 11,345 19,275 912 14,024 723 附息負債總額 1,546,633 16,250 1.05 1,426,180 15,653 1.10 1,401,254 19,030 1.36 指定以公允值列賬之交易用途 負債及金融負債 （不包括已發行之 本身債務） 26 178,518 2,856 1.60 301,353 3,027 1.00 318,883 3,445 1.08 不附息往來賬項 185,990 184,370 177,085 各類股東權益及其他不附息負債總額 768,902 780,113 809,732 截至12月31日止年度 2,680,043 19,106 0.71 2,692,016 18,680 0.69 2,706,954 22,475 0.83 有關註釋，請參閱第109頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 47 列賬基準之淨利息收益為350億美元，與 2013年相比下降8.34億美元或2%，包括下表 所概述之重大項目及貨幣換算。 重大項目及貨幣換算 2014年 2013年 百萬美元 百萬美元 重大項目 因應英國 《消費者信貸法》 持續合規檢討所產生的準備 (632) － 收購、出售及攤薄投資 38 386 (594) 386 貨幣換算 － 518 截至12月31日止年度 (594) 904 按列賬基準計算，淨息差及淨利息收益率 同告下降，反映北美洲及歐洲的客戶貸款 收益有所下跌。在北美洲，此乃由於貸款 組合成分有所改變，其中收益率較低的有 抵押資產比重增加，以及縮減消費及按揭 貸款組合所致。在歐洲，此乃主要由於一 個重大項目所致，即因應英國 《消費者信貸 法》 持續合規檢討所產生的準備。此等因素 被資金成本下降所部分抵銷。 若不計及上表詳列的重大項目及貨幣換 算，淨利息收益較2013年增加6.64億美元或 2%，乃因為亞洲錄得增長，部分反映客戶 貸款額增長。 利息收益 列賬基準之利息收益大致持平，乃由於客 戶貸款利息收益減少 （包括就 《消費者信貸 法》 提撥準備的影響） ，被短期資金收益增 加及反向回購交易 （見第48頁） 管理方式改 變帶來的增長所抵銷。 客戶貸款利息收益下降，主要是在北美洲 及拉丁美洲，部分被亞洲收益增長抵銷。 在北美洲，此乃因為出售收益率較高的非 房地產貸款組合，以及由於縮減和出售導 致消費及按揭貸款組合收益減少所致。此 外，零售銀行及財富管理業務和工商金融 業務的新增客戶貸款收益率下降，反映組 合轉向增加收益率較低的第一留置權有抵 押房地產貸款。在拉丁美洲，客戶貸款利 息收益亦下降，反映巴西及墨西哥的收益 率均下跌，惟兩國的有期貸款平均結欠有 所增加。在巴西，收益率下跌反映產品及 客戶組合轉向較穩健、以客戶關係主導的 貸款，而在墨西哥，乃由於中央銀行利率 降低。拉丁美洲地區的利息收益亦受出售 非策略業務所影響。 相對而言，我們在亞洲錄得客戶貸款利息 收益增加，乃因為有期貸款額增加，其次 是年內住宅按揭增加。結欠增加的利好影 響部分被收益率受壓抵銷。在歐洲，若不 計及上述 《消費者信貸法》 相關準備的影響， 客戶貸款利息收益有所上升，原因是按揭 及有期貸款結欠增加。 由於拉丁美洲和亞洲若干國家和地區 （特 別是巴西、阿根廷及中國內地） 的利率上 升，此等地區的短期資金及金融投資的利 息收益有所增加，而平均款額亦告上升。 然而，在歐洲，隨著即將到期的持倉被年 期較長但收益率較低的債券所取代，短期 資金及金融投資的利息收益下跌。 利息支出 本年度列賬基準之利息支出增加。我們於 亞洲及拉丁美洲錄得客戶賬項利息支出 增加，部分被北美洲支出減少抵銷。在亞 洲，支出增加主要來自客戶賬項平均款額 增長。在拉丁美洲，客戶賬項利息支出上 升，乃由於利率上升令資金成本的升幅足 以抵銷平均結欠的減幅有餘，巴西的情況 尤為明顯。然而，因中央銀行利率下跌及 出售非策略業務，墨西哥資金成本下跌部 分抵銷了上述增幅。相反，在北美洲，由 於作出調低存款利率的策略決定，令客戶 存款的利息支出減少。除此之外，由於撥 回一宗稅務狀況不明朗個案的相關應計利 息，其他利息支出因而減少。 已發行債務的利息支出上升。我們的資金 成本上升，部分被整體結欠減少抵銷。拉 丁美洲 （特別是巴西） 的利息支出上升，反 映利率上升及中期貸款票據結欠增加。相 對而言，在北美洲，出售業務導致資金需 求下降。由於票息較高的債務到期及已償 還，資金成本亦告下跌。在歐洲，由於錄 得贖回淨額令未償還的平均結欠下降以及 資金成本下跌，債務的利息支出亦因而減 少。 滙豐控股有限公司 48 董事會報告：財務概要 （續） 按收支項目列示之集團業績表現 回購及反向回購 於2013年最後一季，環球銀行及資本市場 業務改變管理反向回購及回購活動的方 式。由於附息資產及附息負債平均值大幅 上升，導致淨利息收益率下降。此等反向 回購及回購協議分別較我們組合內其餘投 資的收益總額及資金成本為低。 「淨利息收益」 包括內部撥資之交易用途資 產的支出，而相關收入則於 「交易收益淨 額」 項內入賬。該等資產的內部資金成本 減少，原因是交易用途資產平均結餘的跌 幅大於交易用途負債平均結欠。於呈報環 球業務的業績時，此項成本已計入 「交易 收益淨額」 項內。 費用收益淨額 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 戶口服務 3,407 3,581 3,563 管理資金 2,658 2,673 2,561 卡 2,460 2,455 3,030 信貸 1,890 1,907 1,761 經紀業務收益 1,371 1,388 1,350 進出口 1,115 1,157 1,196 單位信託基金 1,005 891 739 包銷 872 866 739 匯款 833 849 819 環球託管 726 698 737 保險 516 551 696 其他 2,692 2,957 2,958 費用收益 19,545 19,973 20,149 減：費用支出 (3,588) (3,539) (3,719) 截至12月31日止年度 15,957 16,434 16,430 列賬基準之費用收益淨額下跌4.77億美元， 主要來自拉丁美洲及北美洲。在拉丁美洲， 跌幅計及貨幣換算的影響，以及繼續重新 定位及出售業務，尤其是於2013年出售巴 拿馬業務。在北美洲，費用收益淨額減少， 乃由於我們與卡及零售商戶 （ 「CRS」 ） 業務買 家訂立的過渡服務協議屆滿，及按揭債務 管理權估值出現不利的調整。 戶口服務費用收益減少，以拉丁美洲及歐 洲尤其明顯。在拉丁美洲，下跌是由於業務 繼續重新定位導致墨西哥客戶數目減少， 以及巴西市場競爭激烈所致。在歐洲，戶 口服務費用減少，主要來自瑞士，是由於 我們的環球私人銀行業務重新定位，而在 英國，則部分反映2013年進行零售分銷檢 討的影響。 相對而言，單位信託基金費用上升，主要 來自亞洲，乃由於香港股票基金的銷售額 增加。 北美洲的其他費用收益下跌，乃由於過渡 服務協議屆滿，以及在拉丁美洲，我們於 2013年出售巴拿馬業務及墨西哥業務持續 重新定位。 此外，費用支出上升，乃由於北美洲的按揭 債務管理權估值出現不利調整，反映2014 年按揭利率下降，而2013年則上升。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 49 交易收益淨額 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 交易活動20 5,419 6,921 5,249 平安保險或有遠期出售合約 － (682) (553) 交易活動之淨利息收益 1,907 2,047 2,683 終止對沖所產生之增益╱ （虧損） 1 (194) － 其他交易收益－低效用對沖： －現金流對沖 34 22 35 －公允值對沖 19 65 (27) 不合資格對沖之公允值變動21 (620) 511 (296) 截至12月31日止年度 6,760 8,690 7,091 有關註釋，請參閱第109頁。 列賬基準之交易收益淨額為68億美元，下 跌19億美元，主要在歐洲。交易收益淨額 減少，部分原因是下表所概述之重大項目 所致。 重大項目及貨幣換算 2014年 2013年 百萬美元 百萬美元 重大項目 計入交易活動內： (332) 548 －衍生工具合約之借記估值調整 (332) 106 －與滙豐控股發行之英鎊債務相關的匯兌增益 － 442 計入其他交易收益淨額： (539) (346) －平安保險或有遠期出售合約22 － (682) －提早終止美國縮減組合內的現金流對沖所產生的虧損 － (199) －不合資格對沖之公允值變動 (541) 511 －收購、出售及攤薄投資 2 24 (871) 202 貨幣換算 － (11) 截至12月31日止年度 (871) 191 有關註釋，請參閱第109頁。 若不計及上表詳列之重大項目及貨幣換算， 來自交易活動的交易收益淨額減少6億美 元，環球銀行及資本市場業務旗下資本市 場業務的減幅尤為明顯，主要由於外匯交 易業務受市場波幅收窄及客戶交易量下降 的影響。此外，股票交易業務收入減少， 原因是2013年受惠於重估增值的增幅，足 以抵銷2014年客戶交易量增加以及衍生工 具收益增長帶來的收入增長有餘。 2014年，我們透過引入資金公允值調整 （ 「FFVA」 ） ，調整了非抵押衍生工具組合的 估算方法，導致交易收益淨額減少2.63億美 元 （主要來自利率交易業務 （1.64億美元） 及 信貸交易業務 （9,700萬美元） 。若不計及資 金公允值調整，信貸交易業務亦受信貸息 差不利變動及既有信貸收入下跌的影響。 相對而言，利率交易業務受有利的市場變 動影響，尤其是在亞洲，同時在結構負債 方面，我們的本身信貸息差引致之公允值 變動極少，而2013年則錄得不利變動。這 些因素部分被歐洲的利率交易業務收益下 跌抵銷。 相比2013年的不利變動，持作經濟對沖項 目用以對沖指定以公允值列賬之外幣債務 的資產錄得有利的匯兌變動，並計入交易 活動所得交易收益淨額項內。此等變動抵 銷了在 「指定以公允值列賬之金融工具淨 收益╱ （支出） 」 內呈列的外幣債務公允值變 動。 此外，交易活動之淨利息收益下跌，乃由於 平均款額下降，主要與反向回購及回購協 議有關，反映環球銀行及資本市場業務改 變管理該等協議的方式。來自這些活動的 淨利息收益現時計入 「淨利息收益」 項內。 滙豐控股有限公司 50 董事會報告：財務概要 （續） 按收支項目列示之集團業績表現 指定以公允值列賬之金融工具淨收益╱ （支出） 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 來自以下各項的淨收益╱ （支出） ： －為應付保單未決賠款及投資合約負債而持有的金融資產 2,300 3,170 2,980 －在投資合約下對客戶之負債 (435) (1,237) (996) －滙豐已發行長期債務及相關衍生工具 508 (1,228) (4,327) －長期債務之本身信貸息差變動 （重大項目） 417 (1,246) (5,215) －公允值之其他變動22 91 18 888 －指定以公允值列賬之其他工具及相關衍生工具 100 63 117 截至12月31日止年度 2,473 768 (2,226) 有關註釋，請參閱第109頁。 產生指定以公允值列賬之金融工具淨收益╱ （支出） 之資產及負債 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 於12月31日指定以公允值列賬之金融資產 29,037 38,430 33,582 於12月31日指定以公允值列賬之金融負債 76,153 89,084 87,720 包括： 為應付下列各項負債╱未決賠款而持有的金融資產： －附有酌情參與條款的保單及投資合約 10,650 10,717 8,376 －單位相連保險及其他保險及投資合約 16,333 25,423 23,655 指定以公允值列賬之已發行長期債務 69,681 75,278 74,768 指定金融工具按公允值列賬之會計政策及 相關收支項目的處理方法，載於財務報表 附註2。 大部分指定以公允值列賬之金融負債屬於 已發行長期定息債務，該等債務已透過掉 期 （屬於明文規定利率管理策略其中一項 措施） 轉為浮息。此等已發行長期債務及 相關對沖項目的公允值變動包括集團本身 信貸息差變動的影響，以及相關掉期與集 團本身債務兩者之間經濟關係可能存在任 何低效情況的影響。我們本身債務之信貸 息差變動及低效情況 （兩者均於收益表中 確認） ，於不同年度會有不同幅度及方向， 但不會改變明文規定利率管理策略中涉及 預期現金流的部分。因此，源自長期債務 之集團本身信貸息差變動的公允值變動， 以及債務和相關衍生工具的其他公允值變 動，在集團內部不會被視為管理業績表現 的一部分，故並未分配至各項環球業務， 而是在 「其他」 項內入賬。集團本身指定以 公允值列賬之債務之信貸息差變動並不計 入經調整業績，而相關公允值變動亦不會 計入監管規定資本內。 2014年，指定以公允值列賬之金融工具錄 得列賬基準淨收益25億美元，而2013年為 7.68億美元。前者包括集團本身長期債務因 信貸息差變動產生的有利公允值變動4.17 億美元，而2013年則錄得不利變動12億美 元。若不計及此重大項目，指定以公允值 列賬之金融工具淨收益增加4,200萬美元。 為應付保單未決賠款及投資合約負債而持 有的金融資產所產生之淨收益為23億美元， 較2013年減少8.7億美元。此乃由於英國及 法國股市市況轉弱，惟部分減幅被香港股 市表現改善及巴西債券組合淨收益上升所 抵銷。 源自股市的投資損益會導致對客戶負債出 現相應變動，而且尤其反映單位相連投保 人於相關資產組合投資表現的參與程度。 若是與為支持投資合約而持有的資產有關 的損益，對客戶負債的相應變動會於 「指定 以公允值列賬之金融工具淨收益╱ （支出） 」 項內列賬。相對而言，若損益與為支持附 有酌情參與條款之保單或投資合約而持有 的資產有關，則對客戶負債的相應變動會 在 「已支付保險賠償和利益及投保人負債 之變動淨額」 項內列賬。 公允值之其他變動反映因利率及匯率低效 用對沖引致的有利變動淨額。但有關數額 被指定以公允值列賬及作為集團整體資金 策略其中一項措施而發行的外幣債務之不 利匯兌變動淨額所部分抵銷 （被 「交易收益 淨額」 項內持作經濟對沖項目的資產抵銷） 。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 51 金融投資減除虧損後增益 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 出售以下項目之利潤╱ （虧損） 淨額： －債務證券 665 491 781 －股權證券 1,037 1,697 823 －其他金融投資 6 (1) 5 1,708 2,187 1,609 可供出售股權證券減值 (373) (175) (420) 截至12月31日止年度 1,335 2,012 1,189 按列賬基準計算，金融投資減除虧損後增 益為13億美元，較2013年下跌6.77億美元。 此項減幅主要反映下文概述之重大項目。 重大項目及貨幣換算 2014年 2013年 百萬美元 百萬美元 重大項目 出售所持上海銀行股權所產生的利潤 428 － 於興業銀行之投資所產生的減值 (271) － 完成出售平安保險所產生的利潤淨額22 － 1,235 收購、出售及攤薄投資 － 5 157 1,240 貨幣換算 － (10) 截至12月31日止年度 157 1,230 有關註釋，請參閱第109頁。 若不計及上表所述之重大項目及貨幣換 算，金融投資減除虧損後增益增加3.96億 美元，主要因為我們積極管理既有信貸業 務組合，令出售債務證券所得利潤淨額增 加。此外，由於市況及相關組合業績改善， 可供出售股權證券錄得更高的出售利潤及 較低的減值額。 保費收益淨額 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 保費收益總額 12,370 12,398 13,602 再保險保費 (449) (458) (558) 截至12月31日止年度 11,921 11,940 13,044 列賬基準之保費收益淨額大致維持不變， 因亞洲保費收益增加大部分抵銷了歐洲及 拉丁美洲的減幅。 在亞洲，保費收益上升，主要來自香港， 乃由於遞延年金、萬用壽險及儲蓄壽險的 新造業務增加。但單位相連合約的新造業 務減少，抵銷了部分增幅。 在歐洲，保費收益減少，主要在英國，反 映2013年下半年外部獨立財務顧問退出分 銷若干相連保單的行列，令銷售額下降。 此減幅部分被法國錄得的升幅抵銷，主要 反映附有酌情參與條款之投資合約銷售額 上升。 拉丁美洲保費收益淨額亦下跌，主要來自 巴西，乃部分由於改變分銷途徑導致銷售 額減少所致。 滙豐控股有限公司 52 董事會報告：財務概要 （續） 按收支項目列示之集團業績表現 其他營業收益 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 已收租金 162 155 210 持作出售用途資產之確認增益╱ （虧損） 220 (729) 485 投資物業之增益 120 113 72 出售物業、機器及設備、無形資產及非金融投資所得利潤 32 178 187 攤薄興業銀行及其他聯營及合資公司權益所得增益╱ （虧損） (32) 1,051 － 出售巴拿馬滙豐銀行所得利潤 － 1,107 － 有效長期保險業務現值變動 261 525 737 其他 368 232 409 截至12月31日止年度 1,131 2,632 2,100 有效長期保險業務現值變動 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 新增業務的價值 百萬美元 111 1,111 新增業務的價值 870 924 1,027 預期回報 (545) (505) (420) 假設變動及經驗差異 (116) 88 69 其他調整 52 18 61 截至12月31日止年度 261 525 737 列賬基準之其他營業收益為11億美元，較 2013年下跌15億美元，主要由於下表概述 的重大項目所致。 重大項目及貨幣換算 2014年 2013年 百萬美元 百萬美元 重大項目 列入持作出售用途資產之確認增益╱ （虧損） ： 168 (772) －撇銷與摩納哥環球私人銀行業務相關的已分配商譽 － (279) －出售美國非房地產組合的利潤╱ （虧損） － (271) －出售美國數批有抵押房地產賬項的利潤╱ （虧損） 168 (123) － Household Insurance Group Holding Company出售旗下 從事制訂保險產品的業務2 － (99) 計入餘下業務項目： (41) 2,193 －興業銀行股份有限公司向第三方增發股本後，集團將於該行持有的 股權重新分類而錄得增益2 － 1,089 － HSBC Latin America Holdings UK Limited出售巴拿馬滙豐銀行3 － 1,107 －滙豐保險集團 （亞太） 有限公司出售於Bao Viet Holdings的股權2 － 104 －出售HFC Bank UK有抵押貸款組合的虧損 － (146) －收購、出售及攤薄投資 (41) 39 貨幣換算 － (18) 截至12月31日止年度 127 1,403 若不計及上表詳列的重大項目及貨幣換算， 其他營業收益較2013年下跌2億美元，主要 由於有效長期保險業務現值 （ 「PVIF」 ） 的有 利變動於2014年減少，以及主要來自香港 的投資物業出售利潤和重估增值下跌。因 我們積極管理組合，減幅部分被英國的環 球銀行及資本市場業務的既有信貸業務錄 得的增益抵銷。 有效長期保險業務資產現值於2014年有利 變動減少，主要由下列因素所致： ‧ 新業務價值下跌，主要來自巴西，是因 為利率上升及交易量減少所致；及 ‧ 假設變動及經驗差異於2014年不利，而 2013年則為有利變動。這主要是因為法 國利率下跌以及香港精算假設更新資 料的不利變動所致，部分被主要來自亞 洲及巴西的利率波動抵銷。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 53 已支付保險賠償和利益及投保人負債之變動淨額 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 已支付保險賠償和利益及投保人負債之變動淨額： －總額 13,723 13,948 14,529 －減再保人應佔份額 (378) (256) (314) 截至12月31日止年度24 13,345 13,692 14,215 有關註釋，請參閱第109頁。 列賬基準之已支付保險賠償和利益及投保 人負債之變動淨額較2013年減少3.47億美 元。 為支持投保人須承擔投資風險之保單而持 有的資產投資回報產生的索償變動金額減 少，反映英國及法國股市表現轉弱，部分 被香港股市表現改善，以及巴西債券組合 淨收益增加所抵銷。為支持該等保單及投 資合約負債而持有的指定以公允值列賬之 金融資產所確認的損益，均列入 「指定以 公允值列賬之金融工具淨收益」 項內。 歐洲及拉丁美洲的新承保業務減少導致賠 償金額下降，大部分被香港錄得的增幅所 抵銷，詳情於 「已賺取保費淨額」 項下闡釋。 貸款減值及其他信貸風險準備 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 貸款減值準備： －已扣除準備撥回額之新撥準備 5,010 7,344 9,306 －收回先前撇賬額 (955) (1,296) (1,146) 4,055 6,048 8,160 個別評估準備 1,780 2,320 2,139 綜合評估準備 2,275 3,728 6,021 可供出售債務證券減值╱ （減值撥回） (319) (211) 99 其他信貸風險準備 115 12 52 截至12月31日止年度 3,851 5,849 8,311 客戶貸款減值準備佔客戶貸款總額平均值之百分比27 0.4% 0.7% 0.9% 有關註釋，請參閱第109頁。 列賬基準之貸款減值及其他信貸風險準備 為39億美元，較2013年減少20億美元，主要 源自北美洲、歐洲及拉丁美洲。減值準備佔 貸款總額平均值之百分比，由2013年12月 31日之0.7%跌至2014年12月31日之0.4%。 個別評估準備減少5.4億美元，主要來自歐 洲，部分被亞洲以及中東及北非的增幅抵 銷。在歐洲，該等減值準備的降幅主要源 自英國工商金融業務 （反映貸款組合質素 以及經濟環境改善） 及環球銀行及資本市 場業務。亞洲方面，香港及中國內地主要 在工商金融業務及環球銀行及資本市場錄 得小幅增長，而在中東及北非，我們錄得 準備淨額，而2013年則為撥回淨額，主要 由於環球銀行及資本市場業務中阿聯酋的 相關風險項目的撥回額下降。 綜合評估準備減少15億美元，主要由於北 美洲及拉丁美洲錄得降幅。在北美洲，減幅 主要出現在零售銀行及財富管理業務，反 映消費及按揭貸款組合中的拖欠率下降和 新增已減值貸款減少。2014年房屋市場復 甦未及2013年明顯，導致相關物業的有利 市場價值調整減少，抵銷了持續縮減組合 及出售貸款令貸款結欠減少的部分影響。 在拉丁美洲，由於2013年巴西零售銀行及 財富管理業務和工商金融業務更改重整貸 款組合的減值模型及修訂有關貸款組合假 設產生的不利影響，綜合評估準備減少。 另外，由於商務理財業務的準備下降，反 映拖欠率改善，加上出售非策略業務的影 響，準備亦因而減少。 信貸風險準備的撥回淨額為2.04億美元， 大致維持不變，原因是歐洲環球銀行及資 本市場業務下可供出售資產抵押證券的撥 回額增加，被拉丁美洲及北美洲的撥備所 抵銷。在拉丁美洲，巴西就環球銀行及資 本市場業務旗下一項擔保提撥準備。在北 美洲，我們在加拿大錄得準備，而2013年 則錄得撥回，有關準備亦主要源自美國， 反映環球銀行及資本市場一項特定貸款的 相關資產價值下跌。 滙豐控股有限公司 54 董事會報告：財務概要 （續） 按收支項目列示之集團業績表現 營業支出 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 按支出類別列示 僱員報酬及福利 20,366 19,196 20,491 物業及設備 （不包括折舊及減值） 4,204 4,183 4,326 一般及行政開支 14,361 12,882 15,657 行政開支 38,931 36,261 40,474 物業、機器及設備折舊與減值 1,382 1,364 1,484 無形資產攤銷及減值 936 931 969 截至12月31日止年度 41,249 38,556 42,927 職員人數 （等同全職僱員） 2014年 2013年 2012年 地區 歐洲 69,363 68,334 70,061 亞洲8 118,322 113,701 112,766 中東及北非 8,305 8,618 8,765 北美洲 20,412 20,871 22,443 拉丁美洲 41,201 42,542 46,556 於12月31日 257,603 254,066 260,591 有關註釋，請參閱第109頁。 重大項目及貨幣換算 2014年 2013年 百萬美元 百萬美元 重大項目 改變於英國提供傷病福利的基準所產生的會計增益 － (430) 與美國聯邦房屋金融局達成和解協議相關的準備 550 － 與馬多夫事件相關的訴訟費 － 298 與外匯調查相關的和解開支及準備 1,187 － 環球私人銀行業務就監管事宜提撥的準備 65 352 英國的客戶賠償計劃 1,275 1,235 就與卡及零售商戶業務有關的美國客戶補救措施提撥的準備 － 100 重組架構及其他相關成本 278 483 收購、出售及攤薄投資 40 488 3,395 2,526 貨幣換算 － 348 截至12月31日止年度 3,395 2,874 列賬基準之營業支出為410億美元，較2013 年增加27億美元或7%。營業支出的增加部 分源自下表所述的重大項目，包括於2014年 第四季錄得有關外匯調查償付及撥備8.09億 美元 （進一步詳情見財務報表附註40） 。 若不計及重大項目及貨幣換算，營業支出 較2013年增加22億美元或6%。 監管計劃及合規成本有所增加，原因是繼 續注重環球標準及業內實施的更廣泛監管 改革計劃，以建設必要的基礎設施，符合 現已加強的合規標準，以及為履行相關義 務而產生實施費用，例如在不同司法管轄 區內進行壓力測試及架構改革。 2014年，我們在集團上下加快推行環球標 準。各環球業務及合規部門已制訂運營程 序，以符合新的環球反洗錢及制裁政策， 該等程序現已於各個國家及地區實施，並 已納入必要的本地規定。2014年，我們亦 投入資源建立我們防範金融犯罪的能力， 並就客戶盡職調查、交易監控及制裁相關 篩查制訂策略性基礎架構解決方案。 我們繼續投資於支持業務有機增長的策略 性計劃，主要是支援工商金融業務在亞洲 的商務理財業務和環球貿易及融資業務， 其次是在歐洲的發展。我們亦增加市場推 廣及廣告開支，為增加收入的活動提供支 援，主要在零售銀行及財富管理業務的卓 越理財及運籌理財兩項核心業務和個人貸 款產品。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 55 有關費用增加亦反映了： ‧ 通脹壓力，包括工資上漲，主要源自亞 洲及拉丁美洲； ‧ 2014年英國銀行徵費支出由2013年的 9.04億美元增至11億美元，主要由於徵 費率上升。兩個年度均已計入與上年度 銀行徵費支出有關的調整(2014年：有 利調整4,500萬美元；2013年：不利調整 1,200萬美元） ；及 ‧ 因為確認的時間性而根據英國的金融 服務賠償計劃繳付徵費。 於2014年，我們進一步達致13億美元的可 持續成本節約，主要由於我們重整後勤部 門的流程所致，並部分抵銷上述投資及通 脹的影響。 全職僱員平均人數大致不變，因透過可持 續節省方案扣減被監管計劃及合規相關的 措施及業務增長抵銷。 列賬基準之成本效益比率 25 2014年 2013年 2012年 % % % 滙豐 67.3 59.6 62.8 地區 歐洲 93.7 84.0 108.4 亞洲8 44.0 40.7 39.4 中東及北非 47.7 51.5 48.0 北美洲 78.9 72.9 60.8 拉丁美洲 71.7 56.1 58.7 環球業務 零售銀行及財富管理 71.2 64.5 58.4 工商金融 45.9 43.1 45.9 環球銀行及資本市場 67.7 51.9 54.2 環球私人銀行 74.8 91.4 67.6 有關註釋，請參閱第109頁。 應佔聯營及合資公司利潤 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 聯營公司 交通銀行股份有限公司 1,974 1,878 1,670 中國平安保險 （集團） 股份有限公司 － － 763 興業銀行股份有限公司 － － 670 沙地英國銀行 455 403 346 其他 64 5 72 應佔聯營公司利潤 2,493 2,286 3,521 應佔合資公司利潤 39 39 36 截至12月31日止年度 2,532 2,325 3,557 按列賬基準計算，滙豐錄得應佔聯營及合 資公司利潤25億美元，上升2.07億美元或 9%，部分乃由於2013年越南一家經營銀行 業務的聯營公司的減值準備1.06億美元不 復再現。若不計此項，我們應佔聯營及合 資公司利潤增加，原因是來自交通銀行及 沙地英國銀行的貢益增加。 我們應佔交通銀行的利潤增加，原因是資 產負債表增長及交易收益增加，惟部分增 幅被營業支出及貸款減值準備上升所抵銷。 於2014年12月31日，我們對交通銀行的投資 進行減值檢討，結論為根據我們的使用價 值計算法 （詳情請參閱財務報表附註20） ， 該等投資並非已減值。 於未來期間，使用價值或會上升或下跌，視 乎模型數據變動的影響而定。預期2015年 的賬面值將會因交通銀行賺取的保留利潤 而有所增加。在賬面值超出使用價值時， 滙豐將繼續確認其應佔交通銀行的損益， 但賬面值將削減至相等於使用價值，並在 收益中作出相應調減，但如市場價值增至 高於賬面值的水平則作別論。 來自沙地英國銀行的利潤增加，反映資產 負債表增長強勁。 滙豐控股有限公司 56 董事會報告：財務概要 （續） 按收支項目列示之集團業績表現╱綜合資產負債表 稅項支出 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 除稅前利潤 18,680 22,565 20,649 稅項支出 (3,975) (4,765) (5,315) 截至12月31日止年度之除稅後利潤 14,705 17,800 15,334 實質稅率 21.3% 21.1% 25.7% 2014年的實質稅率為21.3%，低於本年混合 英國公司稅率21.5%。 本年度之實質稅率反映下列各項產生之經 常性利益：若干集團實體所持政府債券及 股票豁免所得稅、在我們除稅前收益內確 認集團應佔聯營及合資公司除稅後利潤。 此外，實際稅率反映過往期間的即期稅項 減免，而與匯兌調查有關的不可扣稅償付 及撥備抵銷了部分減幅。 2014年的稅項支出減少8億美元至40億美 元，主要由於會計利潤減少及受惠於過往 年度的即期稅項減免。 2014年，集團承擔及已付相關稅務機關的 稅項包括利潤的稅項、銀行徵費及僱主相 關的稅項，為數79億美元 （2013年：86億美 元） 。由於計入除稅前利潤的間接稅項 （如 增值稅及銀行徵費） 及繳納稅項的時間關 係，有關金額有別於收益表內呈列的稅項 支出。 集團在其業務所在司法管轄區亦代表政府 擔當徵稅機構的重要角色。該等稅項包括 僱員相關的稅項，以及從應付存款持有人 的款項中預扣的稅項。於2014年，我們代收 了91億美元 （2013年：88億美元） 的稅款。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 57 綜合資產負債表 綜合資產負債表五年概要 2014年 2013年 2012年 2011年 2010年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 資產 現金及於中央銀行的結餘 129,957 166,599 141,532 129,902 57,383 交易用途資產26 304,193 303,192 408,811 330,451 385,052 指定以公允值列賬之金融資產 29,037 38,430 33,582 30,856 37,011 衍生工具 345,008 282,265 357,450 346,379 260,757 同業貸款27 112,149 120,046 117,085 139,078 141,869 客戶貸款27, 28 974,660 992,089 962,972 899,010 897,847 反向回購協議－非交易用途26, 27 161,713 179,690 70,112 83,328 126,921 金融投資 415,467 425,925 421,101 400,044 400,755 其他資產 161,955 163,082 179,893 196,531 147,094 U N 於12月31日之資產總值 2,634,139 2,671,318 2,692,538 2,555,579 2,454,689 負債及股東權益 負債 同業存放27 77,426 86,507 95,480 95,205 87,221 客戶賬項27 1,350,642 1,361,297 1,311,396 1,223,140 1,190,763 回購協議－非交易用途26, 27 107,432 164,220 40,567 48,402 60,325 交易用途負債26 190,572 207,025 304,563 265,192 300,703 指定以公允值列賬之金融負債 76,153 89,084 87,720 85,724 88,133 衍生工具 340,669 274,284 358,886 345,380 258,665 已發行債務證券 95,947 104,080 119,461 131,013 145,401 保單未決賠款 73,861 74,181 68,195 61,259 58,609 其他負債 121,459 120,181 123,141 134,171 109,954 U N 於12月31日之負債總額 2,434,161 2,480,859 2,509,409 2,389,486 2,299,774 U N 股東權益 股東權益總額 190,447 181,871 175,242 158,725 147,667 非控股股東權益 9,531 8,588 7,887 7,368 7,248 U N 於12月31日之各類股東權益總額 199,978 190,459 183,129 166,093 154,915 U N 於12月31日之各類負債及股東權益總額 2,634,139 2,671,318 2,692,538 2,555,579 2,454,689 五年選錄財務資料 2014年 2013年 2012年 2011年 2010年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 已催繳股本 9,609 9,415 9,238 8,934 8,843 資本來源29, 30 190,730 194,009 180,806 170,334 167,555 無定期後償借貸資本 2,773 2,777 2,778 2,779 2,781 優先證券及定期後償借貸資本31 47,208 48,114 48,260 49,438 54,421 風險加權資產29 1,219,765 1,092,653 1,123,943 1,209,514 1,103,113 財務統計數據 客戶貸款佔客戶賬項百分比27 72.2 72.9 73.4 73.5 75.4 平均股東權益總額對平均資產總值 7.01 6.55 6.16 5.64 5.53 於年底每股普通股資產淨值32 （美元） 9.28 9.27 9.09 8.48 7.94 已發行面值0.50美元普通股 數目 （百萬股） 19,218 18,830 18,476 17,868 17,686 外幣兌美元收市換算率： 1美元兌英鎊 0.642 0.605 0.619 0.646 0.644 1美元兌歐元 0.823 0.726 0.758 0.773 0.748 有關註釋，請參閱第109頁。 綜合資產負債表的其他詳情載於第337頁的財務報表內。 滙豐控股有限公司 58 董事會報告：財務概要 （續） 2014年之變動 按列賬基準計算的資產總值為2.6萬億美元， 較2013年12月31日減少1%。按固定匯率基準 計算，則資產總值增加850億美元或3%。 資產負債狀況保持強勁，客戶貸款對客戶 賬項比率為72%。儘管客戶貸款及客戶賬項 按列賬基準計算均下跌，但按固定匯率基 準計算則有所增加，以亞洲的增幅尤為顯 著。 除另有說明外，下文乃按列賬基準評述。 資產 現金及於中央銀行的結餘減少370億美元， 尤其在歐洲，部分反映回購及反向回購協 議錄得減少淨額。 交易用途資產大致保持不變。若不計及不利 外匯變動180億美元，交易用途資產增加， 主要由於持有亞洲債務證券以支持環球銀 行及資本市場利率交易業務。於歐洲，交 易用途資產大致保持不變，原因為所持股 權證券的大部分增幅被若干其他類別資產 的減幅所抵銷。 指定以公允值列賬之金融資產減少90億美 元，尤其在歐洲，大部分來自將HSBC Life (UK) Limited的英國退休金業務相關結餘轉 撥至 「持作出售用途資產」 。 衍生工具資產增加22%，主要與歐洲的利率 及外匯衍生工具合約有關，反映包括孳息 曲線及外幣匯率變動的市場變化。 客戶貸款微減170億美元或2%，包括不利外 匯變動450億美元。若不計及該等變動，客 戶貸款增加280億美元或3%，主要由於亞 洲增長320億美元，其次是北美洲及拉丁美 洲。與此相反，歐洲結欠減少150億美元， 原因為工商金融和環球銀行及資本市場業 務的有期貸款增加，被少數客戶的企業透 支結欠減少抵銷有餘，於下文進一步詳述。 於亞洲，工商金融和環球銀行及資本市場 業務客戶的有期貸款增加，包括商用物業 及其他物業相關貸款增加。按揭結欠亦增 加，主要於香港。於北美洲，結欠增加乃由 工商金融和環球銀行及資本市場業務貸予 企業及工商客戶的有期貸款增加所帶動， 部分被美國的縮減組合持續減少造成零售 銀行及財富管理業務的客戶貸款減少以及 美國第一留置權按揭結欠轉撥至 「持作出 售用途資產」 所抵銷。拉丁美洲的結欠亦 增加，主要於巴西的工商金融業務及墨西 哥的環球銀行及資本市場業務。 歐洲貸款減少150億美元乃由於企業透支 結欠減少所致。於英國，少數客戶受惠於 可使用其透支與存款作淨利息安排。於本 年度，由於我們就全球資金管理服務採用 更趨一致的處理方式，許多該類客戶增加 結算該等款額之頻率，減少透支及存款數 額，因此結欠額下跌280億美元。其他客戶 貸款增加130億美元，主要在工商金融和環 球銀行及資本市場業務，乃由企業及工商 客戶有期貸款增加所帶動，尤其於2014年 下半年。 反向回購協議減少180億美元，乃由於歐洲 此類協議有序減少所致，原因為我們因應 新監管規定重新評估該等業務活動的整體 回報。該減幅部分被亞洲及北美洲的增幅 所抵銷。 負債 回購協議減少570億美元或35%，乃由歐洲 的減幅所帶動，尤其是英國及法國，反映 上文所述歐洲反向回購協議有序減少。 客戶賬項微降110億美元，包括不利外匯變 動580億美元。若不計及該等變動，相關結 餘增加470億美元或4%，所有地區均錄得增 長，尤其是亞洲達360億美元。亞洲的增長 反映工商金融和環球銀行及資本市場業務 的資金管理業務結餘增長，環球銀行及資 本市場業務證券服務結餘增長，以及零售 銀行及財富管理業務的結餘增長，部分反 映存款推廣活動奏效。於歐洲，相關結餘 略有增加，儘管企業往來賬項減少280億美 元，主要於環球銀行及資本市場業務，與 企業透支結欠降幅一致，以及環球私人銀 行業務客戶存款減少。該等因素被工商金 融業務及其次環球銀行及資本市場業務的 增幅所抵銷有餘，原因為資金管理業務存 款增加及零售銀行及財富管理業務結餘增 加，反映客戶持續偏好將款額存於往來及 儲蓄戶口。 交易用途負債減少160億美元，包括不利外 匯變動120億美元。若不計及此等不利變 動，結欠減少反映客戶需求發生變動。 指定以公允值列賬之金融負債減少130億 美元，主要於歐洲，反映將HSBC Life (UK) Limited的英國退休金業務相關結欠轉撥至 「持作出售用途負債」 。 衍生工具負債的增幅與 「衍生工具資產」 增 幅相一致，原因為相關風險大致匹配。 綜合資產負債表 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 59 股東權益 股東權益總額增加5%，主要來自年內產生 的利潤，部分被已付股息所抵銷。此外， 我們於2014年發行新或有可轉換證券57億 美元，令股東權益增加。有關該等證券的 進一步詳情，請參閱財務報表附註35。該 等變動部分被外匯儲備減少90億美元所抵 銷，反映美元兌英鎊及歐元大幅升值，尤 其是於2014年下半年。 列賬基準與固定匯率基準之綜合資產與負債對賬 2014年12月31日與2013年12月31日比較 2013年 12月31日 2013年 按2014年 2014年 12月31日 貨幣換算 12月31日 12月31日 列賬基準 固定匯率 列賬基準 調整33 匯率計算 列賬基準 變動 基準變動 百萬美元 百萬美元 百萬美元 百萬美元 % % 現金及於中央銀行的結餘 166,599 (9,384) 157,215 129,957 (22) (17) 交易用途資產 303,192 (18,176) 285,016 304,193 7 指定以公允值列賬之金融資產 38,430 (2,467) 35,963 29,037 (24) (19) 衍生工具資產 282,265 (16,582) 265,683 345,008 22 30 同業貸款27 120,046 (4,923) 115,123 112,149 (7) (3) 客戶貸款27 992,089 (45,494) 946,595 974,660 (2) 3 反向回購協議－非交易用途26, 27 179,690 (9,961) 169,729 161,713 (10) (5) 金融投資 425,925 (15,285) 410,640 415,467 (2) 1 其他資產 163,082 (385) 162,697 161,955 (1) － U L 資產總值 2,671,318 (122,657) 2,548,661 2,634,139 (1) 3 同業存放27 86,507 (3,317) 83,190 77,426 (10) (7) 客戶賬項27 1,361,297 (57,673) 1,303,624 1,350,642 (1) 4 回購協議－非交易用途26, 27 164,220 (7,730) 156,490 107,432 (35) (31) 交易用途負債 207,025 (12,208) 194,817 190,572 (8) (2) 指定以公允值列賬之金融負債 89,084 (3,930) 85,154 76,153 (15) (11) 衍生工具負債 274,284 (16,214) 258,070 340,669 24 32 已發行債務證券 104,080 (5,089) 98,991 95,947 (8) (3) 保單未決賠款 74,181 (4,447) 69,734 73,861 6 其他負債 120,181 (4,221) 115,960 121,459 1 5 y N 負債總額 2,480,859 (114,829) 2,366,030 2,434,161 (2) 3 y N 股東權益總額 181,871 (7,720) 174,151 190,447 5 9 非控股股東權益 8,588 (108) 8,480 9,531 11 12 y N 各類股東權益總額 190,459 (7,828) 182,631 199,978 5 9 － 各類負債及股東權益總額 2,671,318 (122,657) 2,548,661 2,634,139 (1) 3 有關註釋，請參閱第109頁。 滙豐控股有限公司 60 董事會報告：財務概要 （續） 貸款及存款綜合分析 26, 27 2014年 2013年 變動 百萬美元 百萬美元 % 客戶－已攤銷成本 客戶貸款 974,660 992,089 (2) 入賬列為 「持作出售用途資產」 之客戶貸款34 577 1,703 (66) 反向回購協議－非交易用途 66,310 88,215 (25) (100) 綜合分析之客戶貸款 1,041,547 1,082,007 (4) 客戶賬項 1,350,642 1,361,297 (1) 入賬列為 「持作出售用途業務組合之負債」 之客戶賬項 145 2,187 (93) 回購協議 －非交易用途 79,556 121,515 (35) 綜合分析之客戶存款 1,430,343 1,484,999 (4) 同業－已攤銷成本 同業貸款 112,149 120,046 (7) 反向回購協議 －非交易用途 95,403 91,475 4 綜合分析之同業貸款 207,552 211,521 (2) 同業存放 77,426 86,507 (10) 回購協議 －非交易用途 27,876 42,705 (35) 綜合分析之同業存放 105,302 129,212 (19) 客戶及同業－公允值 交易用途資產－反向回購 1,297 10,120 (87) －客戶貸款 908 7,180 (87) －同業貸款 389 2,940 (87) 交易用途負債－回購 3,798 17,421 (78) －客戶賬項 898 9,611 (91) －同業存放 2,900 7,810 (63) 有關註釋，請參閱第109頁。 金融投資 2014年 2013年 股權證券 債務證券 總計 股權證券 債務證券 總計 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 資產負債管理 － 306.8 306.8 － 314.4 314.4 保險公司 － 48.5 48.5 － 46.4 46.4 結構公司 0.1 14.9 15.0 0.1 22.6 22.7 資本投資 2.0 － 2.0 2.7 － 2.7 其他 8.6 34.6 43.2 6.3 33.4 39.7 於12月31日 10.7 404.8 415.5 9.1 416.8 425.9 綜合資產負債表 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 61 按國家╱地區分類之客戶賬項 2014年 2013年 百萬美元 百萬美元 歐洲 545,959 581,933 英國 439,313 462,796 法國35 40,750 45,149 德國 15,757 16,615 瑞士 11,058 16,796 土耳其 7,856 7,795 其他 31,225 32,782 亞洲8 577,491 548,483 香港 389,094 365,905 澳洲 19,312 19,812 印度 11,678 11,549 印尼 5,788 5,865 中國內地 46,588 40,579 馬來西亞 16,292 17,093 新加坡 43,731 43,988 台灣 14,901 12,758 其他 30,107 30,934 中東及北非 （不包括沙地阿拉伯） 39,720 38,683 埃及 7,663 7,401 阿聯酋 19,771 18,433 其他 12,286 12,849 北美洲 138,884 140,809 美國 84,894 80,037 加拿大 43,871 47,872 其他 10,119 12,900 拉丁美洲 48,588 51,389 阿根廷 4,384 4,468 巴西 23,204 23,999 墨西哥 18,360 21,529 其他 2,640 1,393 於12月31日 1,350,642 1,361,297 有關註釋，請參閱第109頁。 滙豐控股有限公司 62 董事會報告：財務概要╱環球業務 風險加權資產平均值回報之對賬╱關鍵會計估算及判斷╱概要 風險加權資產平均值回報 計量之對賬 表現管理 於2014年，我們的平均普通股股東權益回報目標 為12%至15%。就內部管理而言，我們根據除稅前 風險加權資產回報率 （此項衡量標準結合股東權益 回報與監管規定資本效益目標） 監控環球業務及各 地區的業務。於2014年，我們的風險加權資產平均 值回報目標為2.2%至2.6%。 除計算風險加權資產平均值回報外，我們 使用經調整風險加權資產平均值回報 （即 經調整除稅前利潤佔就外幣換算差額及重 大項目的影響作出調整的風險加權資產平 均值之百分比） 的非公認會計原則衡量指 標計算內部業績表現。經調整風險加權資 產平均值回報不包括扭曲按年業績表現的 若干項目，詳情請參閱第40頁。 我們亦在進一步調整不視為對集團長期業 績有貢獻的業務的影響之後，呈列按非公 認會計原則衡量指標計算之經調整風險加 權資產平均值回報。當中包括縮減組合以 及於2012年出售的卡及零售商戶業務。 下表的卡及零售商戶業務風險加權資產平 均值為相關營運風險之風險加權資產平均 值，而此等風險加權資產並非於出售時即 時解除，且計算經調整之風險加權資產平 均值回報時未就此作出調整。於2014年底， 有關卡及零售商戶業務組合之剩餘卡及零 售商戶業務營運風險之風險加權資產已悉 數攤銷。 經調整風險加權資產平均值回報之對賬 （不包括縮減組合和卡及零售商戶業務） 2014年 2013年 除稅前回報 風險加權 資產平均值36 風險加權 資產平均值 回報36 除稅前回報 風險加權 資產平均值36 風險加權 資產平均值 回報36 百萬美元 十億美元 % 百萬美元 十億美元 % 列賬基準 18,680 1,209 1.5 22,565 1,104 2.0 經調整37 22,829 1,207 1.9 22,981 1,071 2.1 縮減組合 870 115 0.8 443 121 0.4 環球銀行及資本市場業務之既有信貸業務 172 48 0.4 186 33 0.6 美國消費及按揭貸款及其他38 698 67 1.0 257 88 0.3 卡及零售商戶業務 － － － － 4 － 經調整 （不包括縮減組合 和卡及零售商戶業務） 21,959 1,092 2.0 22,538 946 2.4 + 列賬基準與經調整風險加權資產平均值對賬 截至12月31日止年度 2014年 2013年 變動 十億美元 十億美元 % 列賬基準之風險加權資產平均值36 1,209 1,104 9.5 貨幣換算調整33 － (8) 收購、出售及攤薄投資 (2) (21) 其他重大項目 － (4) 經調整風險加權資產平均值36 1,207 1,071 12.6 有關註釋，請參閱第109頁。 關鍵會計估算及判斷 滙豐的業績會受到編製綜合財務報表所選 擇的會計政策、假設及估算影響。財務報 表附註1及個別附註詳細說明重大會計政策 （包括涵蓋關鍵會計估算及判斷的政策） 。 下文列示涉及重大判斷及估算不確定性並 對財務報表具有重大影響的會計政策： ‧ 貸款減值：第349頁附註1(k)； ‧ 遞延稅項資產：第366頁附註8； ‧ 金融工具之估值：第378頁附註13； ‧ 於聯營公司之權益減值：第403頁附註20； ‧ 商譽減損：第407頁附註21；及 ‧ 準備：第420頁附註29。 鑑於上述項目的確認或計量涉及內在不確 定性及高度主觀成分，下一個財政年度的 結果可能有別於管理層所作估算，因而令 確認及計量與管理層就2014年財務報表所 估計的金額大不相同。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 63 環球業務 概要 63 零售銀行及財富管理 64 工商金融 67 環球銀行及資本市場 70 環球私人銀行 72 其他 75 按環球業務分析 76 概要 滙豐根據多項基準檢討營業活動，包括按 地區以及按環球業務進行分析。 以下評述首先分析環球業務，然後是地區 業務 （請參閱第78頁） 。按此先後次序論述 業績，是因為若干策略主題、業務計劃和 趨勢的影響不限於一個地區。除非另有說 明，否則所有評述均按經調整基準分析 （請 參閱第40頁） ，而列表內的數字除非另有說 明，否則按列賬基準呈列。 編製基準 各項環球業務的業績乃按照編製滙豐綜合財務報 表時所用會計政策呈列。由於滙豐各項業務互相 緊密結合，故呈列環球業務數據時，須就若干收 益及支出項目作出內部分配，其中包括若干後勤 服務及環球部門的成本，但以該等成本確應歸入 相關經營業務範疇為限。儘管該等分配乃以有系 統及貫徹一致的方式進行，但當中難免會涉及部 分主觀判斷。 於適用情況下，呈列的收益及支出金額包括項目 之間調配資金的結果，以及各公司之間和各業務 部門之間相互交易的結果。所有該等交易均按公 平原則進行。 英國銀行徵費支出納入歐洲地區，是因滙豐認為 該徵費是滙豐將總部設在英國的成本。在按環球 業務呈列時，徵費支出會計入 「其他」 項下。 除稅前利潤╱ （虧損） 2014年 2013年 2012年 百萬美元 % 百萬美元 % 百萬美元 % 零售銀行及財富管理 5,651 30.3 6,649 29.5 9,575 46.4 工商金融 8,744 46.8 8,441 37.4 8,535 41.3 環球銀行及資本市場 5,889 31.5 9,441 41.8 8,520 41.3 環球私人銀行 626 3.4 193 0.9 1,009 4.9 其他39 (2,230) (12.0) (2,159) (9.6) (6,990) (33.9) 截至12月31日止年度 18,680 100.0 22,565 100.0 20,649 100.0 資產總值 40 2014年 2013年 百萬美元 % 百萬美元 % 零售銀行及財富管理 499,083 18.9 517,085 19.4 工商金融 372,739 14.2 360,623 13.5 環球銀行及資本市場 1,839,644 69.8 1,975,509 74.0 環球私人銀行 88,342 3.4 97,655 3.7 其他 164,537 6.2 171,812 6.4 滙豐內部項目 (330,206) (12.5) (451,366) (17.0) 於12月31日 2,634,139 100.0 2,671,318 100.0 有關註釋，請參閱第109頁。 風險加權資產 2014年 2013年 十億美元 % 十億美元 % 零售銀行及財富管理 205.1 16.8 233.5 21.4 工商金融 432.4 35.4 391.7 35.8 環球銀行及資本市場 516.1 42.3 422.3 38.6 環球私人銀行 20.8 1.8 21.7 2.0 其他 45.4 3.7 23.5 2.2 於12月31日 1,219.8 100.0 1,092.7 100.0 主要零售銀行及財富管理業務 零售銀行及財富管理業務包括主要零售銀 行及財富管理業務、美國縮減組合和已出 售美國卡及零售商戶業務。我們相信，專 注於主要零售銀行及財富管理業務可讓管 理層更清晰地論述持續經營業務按年產生 重大變動的原因，以及評估業務內預期將 對未來年度有重大影響的因素及趨勢。零 售銀行及財富管理業務與主要零售銀行及 財富管理業務之對賬載於第64頁。按列賬 基準計算與按經調整基準計算之財務衡量 指標的對賬表已上載至www.hsbc.com。 滙豐控股有限公司 64 董事會報告：環球業務 （續） 零售銀行及財富管理 零售銀行及財富管理業務為個人客戶提供 銀行及財富管理服務，協助他們構建美好 的將來和實現夢想。 零售銀行 及財富管理 業務總計 美國卡及 零售商戶 業務 美國 縮減組合 主要零售 銀行及財富 管理業務 百萬美元 百萬美元 百萬美元 百萬美元 2014年 淨利息收益 16,782 － 1,390 15,392 費用收益淨額 6,668 － (4) 6,672 其他收益╱ （支出） 42 1,144 － (49) 1,193 營業收益淨額4 24,594 － 1,337 23,257 貸款減值及其他信貸風險準備43 (1,819) － (30) (1,789) 營業收益淨額 22,775 － 1,307 21,468 營業支出總額 (17,522) － (738) (16,784) 營業利潤 5,253 － 569 4,684 來自聯營公司收益44 398 － － 398 除稅前利潤 5,651 － 569 5,082 風險加權資產平均值回報36 2.6% － 0.8% 3.3% 2013年 淨利息收益 18,339 － 2,061 16,278 費用收益淨額 7,021 － 11 7,010 其他收益╱ （支出） 42 1,380 － (400) 1,780 營業收益淨額4 26,740 － 1,672 25,068 貸款減值及其他信貸風險準備43 (3,227) － (705) (2,522) 營業收益淨額 23,513 － 967 22,546 營業支出總額 (17,248) － (1,166) (16,082) 營業利潤╱ （虧損） 6,265 － (199) 6,464 來自聯營公司收益╱ （支出） 44 384 － (1) 385 除稅前利潤╱ （虧損） 6,649 － (200) 6,849 風險加權資產平均值回報36 2.6% － (0.2%) 4.4% 2012年 淨利息收益 20,298 1,267 2,563 16,468 費用收益淨額 7,205 395 33 6,777 其他收益╱ （支出） 42 6,358 3,155 (200) 3,403 營業收益淨額4 33,861 4,817 2,396 26,648 貸款減值及其他信貸風險準備43 (5,515) (322) (2,569) (2,624) 營業收益 （支出） ╱淨額 28,346 4,495 (173) 24,024 營業支出總額 (19,769) (729) (1,103) (17,937) 營業利潤╱ （虧損） 8,577 3,766 (1,276) 6,087 來自聯營公司收益44 998 － 2 996 除稅前利潤╱ （虧損） 9,575 3,766 (1,274) 7,083 風險加權資產平均值回報36 3.1% 14.7% (1.1%) 4.2% 有關註釋，請參閱第109頁。 主要零售銀行及財富管理業務 的風險加權資產平均值回報為 3.3% 環球流動應用程式下載超過 600萬次 2014年最佳流動理財應用程式 （ 《環球金融》 雜誌） 策略方向 零售銀行及財富管理業務在我們已經或可以採用 具成本效益方式建立具規模目標客戶群的市場， 為個人客戶提供零售銀行及財富管理服務。 我們的三大策略目標重點為： ‧ 透過保險及投資管理業務為零售客戶建立始 終如一、高水平、以客戶需求為主導的財富管 理服務； ‧ 利用集團的環球專長，改善客戶服務及生產 力，迅速向客戶提供高水平的理財方案及服 務；及 ‧ 簡化及重塑零售銀行及財富管理業務組合， 將資本及資源集中於主要市場。 我們的三大優先增長目標為促進目標客戶群增長、 透過財富管理和以關係主導的借貸服務來加強客 戶關係，以及提升服務能力，包括透過電子途徑 提供服務。 實施環球標準、改良風險管理監控及簡化流程， 仍是零售銀行及財富管理業務的當前要務。 列賬基準業績回顧 ‧ 按列賬基準計算，零售銀行及財富管 理業務的除稅前利潤減少10億美元至 57億美元，而主要零售銀行及財富管理 業務的除稅前利潤減少18億美元至51億 美元。零售銀行及財富管理業務利潤 下降，部分反映重大項目 （請參閱第42 頁） 的影響，包括因應英國 《消費者信貸 法》 持續合規檢討而產生的準備5.68億美 元、於2014年不合資格對沖錄得不利變 動4.93億美元 （2013年錄得有利變動2.62 億美元） 、與英國客戶賠償計劃有關的 準備9.92億美元 （2013年為9.53億美元） 及 出售項目。 ‧ 美國縮減組合錄得除稅前利潤，而2013 年則為虧損。收入減少，被貸款減值及 其他信貸風險準備減少抵銷有餘，反映 貸款結欠減少、新增已減值貸款減少及 拖欠水平下降。營業支出亦告下跌，主 要由於集團就過往的卡及零售商戶業 務採取與客戶有關的補救措施而提撥 的準備不復再現，以及退出投資的成本 減少。 零售銀行及財富管理 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 65 經調整業績回顧 45 下文評述反映主要零售銀行及財富管理 46業務的表 現 （請參閱第63頁） 。 除稅前利潤 （百萬美元） 5,082 6,849 1,829 6,911 799 7,648 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整利潤 ‧ 除稅前利潤減少7億美元至69億美元。 收入大致持平，而貸款減值及其他信貸 風險準備的減幅被營業支出的增幅抵 銷有餘。 收入 （百萬美元） 23,257 25,068 728 23,985 (871) 24,197 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整收入 ‧ 儘管推行減輕風險措施，在利率持續低 企及若干主要市場增長停滯不前下，收 入大致持平。往來賬項、儲蓄及存款收 益增加，大致被個人貸款及財富管理產 品的收入減少抵銷。 主要零售銀行及財富管理業務：管理層對 經調整收入的意見 2014年 2013年 百萬美元 百萬美元 往來賬項、 儲蓄及存款 5,839 5,606 財富管理產品 6,201 6,263 －投資產品分銷47 3,456 3,568 －制訂壽險產品 1,603 1,602 －資產管理 1,142 1,093 個人貸款 11,300 11,455 －按揭 3,169 3,182 －信用卡 4,339 4,310 －其他個人貸款48 3,792 3,963 其他49 645 873 營業收益淨額4 23,985 24,197 有關註釋，請參閱第109頁。 ‧ 往來賬項、儲蓄及存款收入增加4%，反 映客戶賬項款額 （主要於香港及英國） 較 2013年增加4%。此外，收入增加反映英 國的儲蓄產品息差擴闊，其次是中國內 地因市場利率上升而令存款息差擴闊。 ‧ 財富管理產品收入減少1%。投資產品分 銷收益下跌，主要由於英國的費用收益 下降，部分原因是2013年進行零售分銷 檢討，而在巴西則反映產品組合改變。 制訂壽險產品的收益大致持平，反映香 港新造業務銷量及投資收益增加，而在 巴西則錄得PVIF資產的有利變動淨額， 但增幅被法國的PVIF資產減幅抵銷，因 為當地的長期收益率下跌使儲蓄業務保 證成本增加。 ‧ 個人貸款收入下跌1%，而按揭及信用卡 收入大致持平。其他個人貸款收益減少 4%，尤以英國的減幅最為明顯，乃由於 不再收取若干透支費用所致。 ‧ 貸款減值及其他信貸風險準備下跌22%， 原因是所有地區均告縮減，在巴西出現 顯著減幅，主要由於2013年就重整貸款 更改減值模型及修訂所用假設，但在 2014年類似情況不再重現。美國及英國 的貸款減值及其他信貸風險準備亦告 下跌，部分反映拖欠水平下跌以及信用 卡及英國貸款未償還結欠減少所致。 營業支出 （百萬美元） 16,784 16,082 (1,099) 15,685 (1,441) 14,641 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整營業支出 ‧ 營業支出上升7%，反映除推行監管計 劃及合規工作的成本增加外，特別是拉 丁美洲的通脹壓力亦造成影響，同時亦 反映確認英國金融服務賠償計劃徵費 的時間，以及各地區推廣支出增加。該 等因素的影響被可持續成本節約逾2億 美元部分抵銷。 滙豐控股有限公司 66 董事會報告：環球業務 （續） 優先增長目標 專注以客戶關係為主導的個人貸款，帶動 資產負債增長 ‧ 於2014年，我們通過持續推行各項減輕 風險方案，繼續致力提高收入質素，收 益因而受壓。這些方案包括為零售銀行 業務前線職員引入新的酌情獎勵架構， 該架構與2013年為財富管理客戶經理推 出的一項獎勵計劃相若。新計劃取消了 前線職員浮動酬勞與產品銷售額掛鈎的 獎勵方式。我們亦繼續簡化產品種類、 提升風險管治水平，以及因應監管環境 變化調整日常工作方式。 ‧ 我們致力深化與現有客戶的關係，並利 用個人貸款業務創造新的商機，從而爭 取各個市場不同類別客戶及產品的份 額。為達此目的，我們繼續使用經改良 的分析方法，以更佳方式支持產品決 策。透過定價及收集客戶意見的措施， 我們已提高來自若干目標客戶群的收 入及業務款額，包括於2014年在17個市 場重新推出運籌理財服務。運籌理財貸 款結欠及存款結餘加上運籌理財客戶 每人帶來的收入較2013年增加。 ‧ 我們繼續嚴守發展貸款業務的紀律，使 之符合集團本位及優先發展市場的承受 風險水平。房屋貸款平均結欠於2014年 增加3%，反映優先發展市場的增長， 尤其顯著的是我們在約半數之該等國 家╱地區中取得雙位數字增長，此乃因 我們重整組合以轉向偏重有抵押貸款， 但組合成分轉變亦使息差收窄。於本位 市場，我們繼續致力增加無抵押貸款， 使平均結欠微升，包括香港信用卡平均 結欠增加，但部分增幅被英國的減幅抵 銷。雖然整體結欠增加，但貸款減值及 其他信貸風險準備仍較2013年為低。 ‧ 2014年，多個市場的客戶推薦度有所改 善，而本年下半年有關產品及服務的 投訴個案總數較2013年同期減少超過 20%。由於客戶的需求不斷變化，我們 仍有需要且會繼續推行進一步工作，以 便提供更佳服務滿足客戶所需。 繼續發展財富管理業務，重點是致力提升 客戶款額 ‧ 我們主要透過卓越理財服務，繼續致力 把握創造財富帶來的商機 （卓越理財客 戶帶來的平均收入，差不多達到非卓越 理財客戶的四倍） 。 ‧ 儘管財富管理產品收入仍低於預期，但 我們會繼續增加財富管理款額 （包括投 資及保險款額） 。與2013年比較，該等款 額 （來自保險、互惠基金及股票交易） 有 所增加。 ‧ 2014年，環球投資管理繼續其加強與其 他環球業務合作的策略，為客戶服務。 這有助吸引290億美元新資金淨額，主 要在定息工具及流動資金產品，尤其來 自環球銀行及資本市場客戶。按價值 計，環球投資管理旗下超過74%合資格 基金的投資表現，均超越市場中位數。 ‧ 2014年，我們透過新的培訓計劃及 工具，改善客戶經理的服務效率。自 2013年起，客戶聯繫及服務覆蓋的比率 均有改善，每名客戶經理約見客戶、進 行財務評估及滿足客戶所需的個案數 目均有所上升。 發展電子銀行服務以支援客戶及節省成本 ‧ 我們繼續開發電子服務途徑及簡化流 程以改進客戶服務，並通過集團的服務 網絡節省成本。 ‧ 2014年，我們的環球流動應用程式 （現 時已加入更多功能） 下載次數超過300萬 次，使下載總數超過600萬次。在2014年 環球最佳網上銀行評選中，滙豐根據應 用程式的全球覆蓋及功能獲 《環球金融》 雜誌評為 「最佳流動理財應用程式」 。 ‧ 此外，我們在英國推出首個網上直通按 揭申請服務，於2014年底，我們全年有 14%的按揭審批經網上處理。我們亦推 出全新的卓越理財平台、電子銀行服務 及平板電腦工具，藉此改進端到端的服 務程序及客戶體驗。我們在各優先發展 市場透過電子服務途徑取得的收入，較 2013年增長18%。 零售銀行及財富管理╱工商金融 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 67 工商金融 工商金融業務向接近60個國家╱地區超過 250萬客戶，包括中小企以至上市公司，提 供全面的工商金融服務及專門設計的解決 方案。 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 淨利息收益 10,506 10,200 10,361 費用收益淨額 4,738 4,717 4,470 其他收益42 1,059 1,448 1,720 營業收益淨額4 16,303 16,365 16,551 貸款減值及其他信貸 風險準備43 (1,675) (2,384) (2,099) 營業收益淨額 14,628 13,981 14,452 營業支出總額 (7,489) (7,049) (7,598) 營業利潤 7,139 6,932 6,854 來自聯營公司收益44 1,605 1,509 1,681 除稅前利潤 8,744 8,441 8,535 風險加權資產平均值回報36 2.1% 2.2% 2.2% 列賬基準除稅前利潤錄得創紀錄的 87億美元 客戶貸款結欠增長 （不包括貨幣換算影響） 10% 連續第三年榮膺 企業及金融機構全球最佳資金管理銀行 （2014年 《歐洲貨幣》 雜誌） 策略方向 工商金融業務的目標，是運用滙豐歴史悠久的豐 富經驗、在國際市場的經營能力及廣泛關係，致 力促進與全球各地的聯繫，成為客戶選擇的理財 夥伴。 我們的四大優先增長目標為： ‧ 以環球客戶群及產品為基石而構建的業務模 式，為客戶提供持續而高效的服務； ‧ 透過獨特的地區網絡，支持及促進環球貿易 及資金流； ‧ 提供優質的核心流向產品－尤其是貿易及資 金管理產品；及 ‧ 加強與其他環球業務的合作。 實施環球標準、改良風險管理監控及簡化流程， 仍是工商金融業務的當前要務。 有關註釋，請參閱第109頁。 列賬基準業績回顧 ‧ 2014年，工商金融業務錄得創紀錄的除 稅前利潤87億美元，較2013年上升4%。 列賬基準除稅前利潤包括若干重大項 目 （請參閱第42頁） 的影響，尤其於2013 年來自出售巴拿馬業務的利潤4.79億美 元。列賬基準除稅前利潤增加亦由貸款 減值及其他信貸風險準備減少所帶動， 但部分被營業支出上升所抵銷。 經調整業績回顧 45 除稅前利潤 （百萬美元） 8,744 8,441 196 8,940 (531) 7,910 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整利潤 ‧ 除稅前利潤增加13%至89億美元，原因 是收入增加和貸款減值及其他信貸風 險準備減少，但部分被營業支出上升所 抵銷。 收入 （百萬美元） 16,303 16,365 9 16,312 (886) 15,479 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整收入 ‧ 收入因信貸及貸款以及資金管理業務 帶動而上升5%，其中本位市場香港及 英國的增幅尤為明顯。此乃由於香港的 貸款平均結欠及存款平均結餘增加，以 及英國的存款平均結餘上升和貸款息 差擴闊，使淨利息收益增加。費用收益 淨額增加乃由於英國有期貸款費用收 益增加所致。 ‧ 儘管與2013年相比貸款息差有所收窄， 但2014年息差維持穩定，並在若干市場 顯示復甦跡象。此外，我們於2014年下 半年錄得英國貸款結欠顯著增長。 滙豐控股有限公司 68 董事會報告：環球業務 （續） 管理層對經調整收入的意見 2014年 2013年 百萬美元 百萬美元 環球貿易及融資業務 2,680 2,625 信貸及貸款業務 6,316 5,938 資金管理服務、往來 賬項及儲蓄存款 5,018 4,709 資本市場業務的產品、 保險及投資和其他51 2,298 2,207 營業收益淨額4 16,312 15,479 有關註釋，請參閱第109頁。 上表已經重列，將外匯交易收入重新分類。於 2014年， 「資本市場業務的產品、保險及投資和其 他」 包括過往計入 「環球貿易及融資業務」 項下的外 匯交易收入2.07億美元 （2013年：2.13億美元） ，及過 往計入 「資金管理服務」 項下的5.16億美元 （2013年： 4.62億美元） 。 ‧ 與2013年比較，環球貿易及融資業務的 收入上升2%。平均數額有所增加，其中 亞洲、歐洲及拉丁美洲同時錄得增長， 但部分增幅因拉丁美洲的息差收窄而被 抵銷，反映巴西的組合變動。2014年， 息差收窄的情況穩定下來，若干市場顯 示復甦跡象。 ‧ 信貸及貸款業務收入較2013年增加6%， 反映香港及美國的平均結欠增加，巴西 的增幅則次之。在英國收入亦增加，乃 由於貸款息差擴闊及費用收益增加，後 者是由於新增業務量上升推動有期貸 款的費用收益增加。但拉丁美洲 （如上 文所述，主要在巴西） 、墨西哥 （因業務 重新定位） 及中國內地的息差收窄，抵 銷了部分增幅。 ‧ 資金管理服務收入較2013年增加7%，反 映存款大幅增長，尤其是在英國及香 港，同時大額資金交易量增加，但部分 增幅被息差收窄抵銷，此情況於歐洲尤 為明顯。 ‧ 資本市場業務的產品、保險及投資和其 他收入增加4%，主要在北美洲。在加拿 大，收入上升反映2013年持作出售用途 投資物業撇減不復再現及2014年出售投 資組合錄得利潤的影響。在美國，收入 增長是由於出售物業組合錄得利潤。 ‧ 貸款減值及其他信貸風險準備減少了 6.63億美元，減幅主要源自歐洲及拉丁 美洲。歐洲的貸款減值及其他信貸風險 準備減少，反映來自英國個別評估的貸 款減值準備減少。拉丁美洲的貸款減值 及其他信貸風險準備減少，乃由於墨西 哥的個別評估準備減少，與房屋建築商 有關的準備減幅尤其顯著，而巴西的綜 合評估減值亦告下降，原因為2013年該 國的商務理財業務組合就重整貸款更改 減值模型並修訂所用假設，在2014年不 再重現。但亞洲的個別評估準備增加， 中國內地及香港尤為明顯，抵銷了上述 部分減幅。 營業支出 （百萬美元） 7,489 7,049 (189) 7,300 (284) 6,765 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整營業支出 ‧ 營業支出增加8%，主要來自歐洲、拉丁 美洲及亞洲。在歐洲及亞洲，成本增加 反映為支持業務發展而在僱員方面的投 資增加及通脹壓力，而拉丁美洲成本上 升，是因為通脹使然，其成因主要源於 巴西及阿根廷的工會協議增薪。此外， 營業支出增加乃由於監管計劃及合規 成本增加所致。 ‧ 來自聯營公司收益增加4%，是由於交 通銀行及沙地英國銀行錄得較佳業績。 優先增長目標 透過面向全球的業務模式提供貫徹一致的 服務 ‧ 我們的業務策略建基於遍及全球的業 務規模及全球貫徹一致的業務模式，專 注服務各個客戶群及應對客戶行為，確 保集團提供專門設計的產品以滿足客 戶所需。我們繼續投入資源，使各類業 務提供覆蓋全球的產品，讓我們可以更 有效率地管理風險。 ‧ 為服務各個客戶群而設立新的高級管 理層職位，並制訂更明確的環球策略， 有助改善我們的客戶服務。2014年，國 際附屬公司銀行業務新聘了一位環球 主管，負責加強投資以支援集團整個業 務網絡的國際客戶。我們亦為主要市場 的國際附屬公司銀行業務建立專責客 戶經理團隊，專注滿足該等附屬公司的 需求及擴大相關收入來源。 ‧ 我們新任的貸款及交易管理部的環球 主管為所有客戶群提供支援。我們設立 這個全球一致的產品組別，是為了在集 團的承受風險水平以內，採用最佳方式 分配資本，同時改善收入的組合成分。 ‧ 2014年，我們重訂大型企業業務的發展 方向，集中服務較少數資產值較高的客 戶。大型企業業務在大多數市場經歷強 工商金融 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 69 勁增長，有利因素包括接受跨越多國的 委託及事件促成的資本市場活動增加。 集團服務更加注重 「環球錢包」 及各地 聯繫，使客戶更能認識我們的業務範疇 和實力，從而與我們建立更穩固的策略 合作夥伴關係。 ‧ 此外，我們在六個主要中型企業業務市 場 （香港、英國、加拿大、美國、墨西哥 及巴西） 增加市場滲透。我們進一步委 任中型企業業務區域及國家╱地區主 管，並提升集團的客戶管理系統。 ‧ 商務理財業務方面，我們於主要市場聘 用更多客戶經理，增加全球培訓，並繼 續為客戶關係管理組合制訂全球一致 的客戶管理系統。我們於2014年推出六 項大型活動，幫助中小企客戶實現增長 目標和擴展海外業務，包括於英國、法 國、美國、加拿大、澳洲及土耳其推出 貸款資金總計達180億美元。 利用我們的地區網絡，支持客戶拓展國際 業務的宏願 ‧ 滙豐的業務網絡覆蓋全球各大貿易走 廊，繼續讓我們為客戶提供增值解決方 案。例如，我們通過提供整體融資及流 動資金解決方案，包括營運資本、貿易 及供應鏈融資，幫助美國其中一家規模 最大的零售商改善其供應鏈管理。 ‧ 資金管理服務方面，工商金融業務繼續 處於有利位置，足可受惠於全球的發展 趨勢，例如跨境付款流量日漸增加，乃 因我們的業務網絡正好覆蓋全球超過 85%付款活動的發源地，擁有強大策略 優勢。例如，客戶委託業務於2013年上 升23%。此外，我們改善了電子銀行服 務，目前為止逾80,000名客戶由使用既 有平台轉用核心電子銀行服務途徑，而 且我們會繼續發展創新的解決方案，包 括提升環球流動資金方案的服務，使中 國內地客戶可以將本身的營運現金與 其環球流動資金架構聯繫起來。 提供優質核心產品 ‧ 滙豐是全世界規模最大的貿易融資銀行 之一，服務範圍覆蓋全球超過85%貿易 與資金流量。我們繼續透過投資於融資 業務及供應鏈，特別是推出新的供應鏈 解決方案平台及將現有應收賬款平台 併入地區業務中心，為客戶提升記賬融 資的服務能力。此舉將讓客戶透過更廣 泛的途徑獲得我們的專業服務及流動 資金，且令我們有能力在新市場迅速推 出各種服務，從而更有效地管理風險， 並降低營業支出。 ‧ 在商品價格不斷下滑的背景下，我們的 商品及結構貿易融資服務資產結餘與 2013年相比錄得雙位數字增長。 加強與其他環球業務的合作 ‧ 我們的焦點繼續集中於擴大集團向客 戶提供的產品種類，加強工商金融業 務與環球銀行及資本市場業務以及環 球私人銀行業務的合作。於2014年，工 商金融業務客戶帶來滙豐合作所得收 入總額的80%以上。由於我們向工商金 融業務客戶銷售的資本市場業務產品 減少 （外匯交易業務方面的減幅尤其明 顯） ，合作所得收入與2013年比較大致 持平，但有關影響因併購及債務資本市 場方面的資本融資業務產品銷量增加 而被抵銷。 滙豐控股有限公司 70 董事會報告：環球業務 （續） 環球銀行及資本市場 環球銀行及資本市場 環球銀行及資本市場業務向全球各主要政 府、大型企業及機構客戶提供專門設計的 理財方案。 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 淨利息收益 7,022 6,766 6,960 費用收益淨額 3,560 3,482 3,329 交易收益淨額50 5,861 6,780 5,690 其他收益42 1,335 2,148 2,294 營業收益淨額4 17,778 19,176 18,273 貸款減值及其他信貸 風險準備43 (365) (207) (670) 營業收益淨額 17,413 18,969 17,603 營業支出總額 (12,028) (9,960) (9,907) 營業利潤 5,385 9,009 7,696 來自聯營公司收益44 504 432 824 除稅前利潤 5,889 9,441 8,520 風險加權資產平均值回報36 1.2% 2.3% 2.1% 儘管外匯交易業務受抑， 股票交易業務的客戶交易量增加 資金管理收入持續增長 年度最佳債券行及衍生工具機構 （2014年 《國際金融評論》 ） 策略方向 環球銀行及資本市場業務已確立良好的業務模式 及策略，目標是使滙豐成為重要客戶以至選定產 品領域及地區的 「前五大」 銀行。 我們的優先增長目標為： ‧ 協助客戶把握環球增長機會； ‧ 繼續妥善部署受益於環球趨勢的產品；及 ‧ 憑藉獨特的國際業務專長及地區網絡，連繫 已發展地區與增長較快的地區。 改良風險管理監控、實施環球標準及與其他環球 業務合作，仍是環球銀行及資本市場業務的當前 要務。 有關註釋，請參閱第109頁。 列賬基準業績回顧 ‧ 環球銀行及資本市場業務的列賬基準除 稅前利潤為59億美元，下跌36億美元， 主要來自歐洲及北美洲，乃由於營業支 出增加及收入減少所致。營業支出增加 及收入減少反映若干重大項目的影響 （請參閱第42頁） 。營業支出包括與外匯 操控調查相關的和解開支及準備12億 美元 （其中8.09億美元於2014年第四季錄 得） ，以及在美國與聯邦房屋金融局達 成和解協議需提撥的準備5.33億美元， 該等支出乃納入重大項目。 經調整業績回顧 45 除稅前利潤 （百萬美元） 5,889 9,441 2,225 8,114 (233) 9,208 3年 1 0 2 4年 1 0 2 列賬基準 貨幣換算及重大項目 經調整利潤 ‧ 除稅前利潤為81億美元，較2013年下降 11億美元，乃因營業支出增加及收入減 少，包括對若干衍生工具合約作出資金 公允值調整，導致產生扣賬額2.63億美 元。 收入 （百萬美元） 17,778 19,176 328 18,106 (644) 18,532 3年 1 0 2 4年 1 0 2 列賬基準 貨幣換算及重大項目 經調整收入 ‧ 收入下跌，主要由於採用資金公允值調 整的影響及外匯交易業務收入減少，減 幅部分被資本融資業務收入增加抵銷。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 71 管理層對經調整收入的意見 2014年 2013年 百萬美元 百萬美元 資本市場52 6,262 6,933 －信貸 567 801 －利率 1,563 1,678 －外匯 2,916 3,140 －股票 1,216 1,314 資本融資 4,066 3,981 資金管理 1,794 1,743 證券服務 1,698 1,653 環球貿易及融資業務 767 723 資產負債管理業務 3,020 3,046 資本投資 531 450 其他53 (32) 3 營業收益總額4 18,106 18,532 有關註釋，請參閱第109頁。 ‧ 下表概述資金公允值調整對業務及經 調整營業收益總額的影響： 資金公允值調整對營業收益總額的影響 2014年 2013年 百萬美元 百萬美元 營業收益總額 18,106 18,532 －利率交易業務的資金公允值調整 (164) － －信貸交易業務的資金公允值調整 (97) － －其他業務的資金公允值調整 (2) － 營業收益總額 （不包括資金公允值調整） 18,369 18,532 －不包括資金公允值調整之 利率交易業務 1,727 1,678 －不包括資金公允值調整之 信貸交易業務 664 801 ‧ 若不包括上述影響，大部分資本市場業 務的收入均告下跌，乃主要由於市場波 幅下降 （2014年上半年尤為明顯） 及客戶 交易量減少，導致外匯交易業務收入下 跌。信貸交易業務收入亦下跌，乃由於 信貸息差出現不利變動及既有信貸組 合減少所致。股票交易業務收入亦告下 跌，乃因2013年錄得較高的重估增值， 抵銷了2014年客戶交易量增加及衍生工 具收益上升所帶來的收入增幅有餘。相 對而言，利率交易業務收入上升，乃由 於市場出現有利變動 （亞洲尤其明顯） ， 以及結構負債的本身信貸息差公允值 變動極小，而2013年則錄得不利變動。 歐洲利率交易業務收入下跌，抵銷了該 等因素的部分影響。 ‧ 資本融資業務方面，收入增加8,500萬美 元，原因是顧問業務、股票資本市場及 貸款產品方面之業務量及市場佔有率 均有所上升，但該等影響部分被息差收 窄及費用減少所抵銷。 ‧ 資金管理業務收入略微增加，乃由於存 款結餘增加 （亞洲尤為明顯） 及高價值的 交易量上升，但息差收窄抵銷了部分增 幅。同時，證券服務 （部分來自歐洲的 新增業務） 與環球貿易及融資業務 （來自 貸款結欠增長） 的收入亦告增加。 ‧ 貸款減值及其他信貸風險準備增加，乃 由於修訂計算綜合評估企業貸款減值 所用的若干估計及個別評估準備增加， 包括在巴西就一項擔保提撥準備 （列作 信貸風險準備） ，但部分增幅因既有組 合中可供出售資產抵押證券的撥回淨 額較2013年增加而被抵銷。 營業支出 （百萬美元） 12,028 9,960 (1,897) 10,131 (398) 9,562 3年 1 0 2 4年 1 0 2 列賬基準 貨幣換算及重大項目 經調整營業支出 ‧ 營業支出增加6%，主要由於監管計劃及 合規相關成本以及職員支出增加，部分 被可持續成本節約逾8,000萬美元抵銷。 優先增長目標 協助客戶把握環球增長機會 ‧ 我們於2013年重整環球銀行及資本市場 業務，作為其中一部分，我們將所有融 資業務併入資本融資業務，包括貸款、 債務資本市場及股票資本市場業務。 隨後，我們繼續致力以更佳方式運用既 有資源滿足客戶所需。我們將客戶群細 分，並設立客戶策略組，確保環球銀行 及資本市場業務的產品、行業及服務專 長可以支援客戶的業務活動增長。我們 近期獲委任就歐洲自2011年以來最大 規模的企業供股計劃擔任聯席全球協 調人及聯席賬簿管理人，足證上述團隊 合作無間。此為我們於過往12個月與該 客戶進行的第五次交易。 ‧ 我們利用環球網絡為已發展市場及增 長較快地區的客戶提供解決方案。我們 在伊斯蘭世界以外的首個伊斯蘭債券 發行項目中，擔任獨家結構設計顧問、 聯席牽頭經辦人及聯席賬簿管理人，突 顯了我們為客戶創造機會的能力。 ‧ 我們繼續提升外匯交易業務的營運實 力，包括致力改善風險管理能力、進一 步發展服務平台及電子定價功能，從而 改善我們的系統及管治，同時讓我們為 客戶提供更加穩健而有效的服務。 滙豐控股有限公司 72 董事會報告：環球業務 （續） 環球銀行及資本市場╱環球私人銀行 繼續妥善部署受益於環球趨勢的產品 ‧ 充分把握人民幣國際化所創造的新商 機，仍然是我們的主要優先增長目標。 近期，我們擔任開創性福爾摩莎債券發 行項目的聯席牽頭經辦人、賬簿管理人 及財務顧問，該債券同時在全球三個證 券交易所上市。2014年，我們連續第三 年在 《亞洲貨幣》 雜誌的離岸人民幣服務 調查中贏得 「最佳整體產品及服務」 大 獎，彰顯我們在海外人民幣服務市場保 持領導地位。 ‧ 2014年11月，我們推出 「滬港通」 服務， 此為上海及香港證券市場交易及結算 的聯繫機制。該項服務讓我們的客戶直 接參與中國A股市場，並可以人民幣購 買股票。 ‧ 大企業擴張業務覆蓋地域及環球貿易 量上升，預期會增加跨境付款及相關服 務的需求。我們在資金管理方面的優勢 備受 《歐洲貨幣》 雜誌肯定，滙豐分別連 續第二年及第三年獲評選為 「全球最佳 非金融機構資金管理機構」 及 「全球最 佳企業和金融機構資金管理機構」 。我 們亦獲得一家尋求在中國內地擴展業 務的環球汽車集團委託，以管理人民幣 資金和進行其他外匯交易及存款業務。 憑藉獨特的國際業務專長及地區網絡，連 繫已發展地區與增長較快的地區 ‧ 我們獨特的地區網絡及環球專業知識， 使我們能為客戶提供真正的環球服務。 我們最近向一個歐洲汽車集團展示了環 球資本市場業務實力的價值，以及在增 長較快地區的領導地位。我們為中國內 地一項本土證券化項目提供顧問服務 及出任聯席分包銷商，該項目的結構設 計旨在吸引國際及本土投資者。我們乃 第一家就中國內地國際評級資產抵押 證券的交易結構提供建議的外資銀行。 ‧ 環球銀行及資本市場業務繼續致力與 其他環球業務合作，並為客戶提供資本 市場及資本融資業務的廣泛產品。2014 年，環球銀行及資本市場業務與工商金 融業務合作所得收入大致持平，乃由於 外匯交易業務的收入減少，但資本融資 業務 （尤其是顧問服務） 收入增長抵銷了 相關減幅。 環球私人銀行 環球私人銀行業務致力在集團優先發展市 場為有複雜和跨地域理財需要的資產豐厚 人士及家族提供服務。 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 淨利息收益 994 1,146 1,294 費用收益淨額 1,056 1,150 1,232 其他收益42 327 143 646 營業收益淨額4 2,377 2,439 3,172 貸款減值及其他信貸 風險準備43 8 (31) (27) 營業收益淨額 2,385 2,408 3,145 營業支出總額 (1,778) (2,229) (2,143) 營業利潤 607 179 1,002 來自聯營公司收益44 19 14 7 除稅前利潤 626 193 1,009 風險加權資產平均值回報36 2.9% 0.9% 4.6% 自2013年12月起 目標增長範疇錄得 140億美元 正數新增資金淨額 表現繼續受客戶基礎 重新定位的安排影響 最佳家族理財服務 （ 《國際私人銀行家》 雜誌全球財富大獎） 策略方向 環球私人銀行業務憑藉滙豐作為商業銀行的悠久 歷史，銳意成為資產豐厚企業擁有人的領先私人 銀行。為此，我們： ‧ 透過集團內部合作，把握本位及優先發展市 場的增長機會，尤其是接觸工商金融業務與 環球銀行及資本市場業務客戶的負責人及主 事人；及 ‧ 將業務重新定位，全力專注在岸市場及少數 目標離岸市場，使之與集團優先增長目標保 持一致。 實施環球標準、改良風險管理監控、提高稅務透 明度及簡化流程，仍是環球私人銀行業務的當前 要務。 有關註釋，請參閱第109頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 73 列賬基準業績回顧 ‧ 列賬基準之除稅前利潤為6.26億美元， 較2013年增長4.33億美元。此乃由於少 數重大項目的影響 （請參閱第42頁） ，其 中2013年撇銷與摩納哥業務有關的已分 配商譽產生虧損2.79億美元及就監管調 查提撥準備3.52億美元的影響尤為明顯。 ‧ 鑑於我們持續重整業務模式，包括減少 非優先發展市場的客戶數目，我們預計 環球私人銀行業務於2015年的業績將受 到客戶資產減少影響。 經調整業績回顧 45 除稅前利潤 （百萬美元） 626 193 112 738 707 900 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整利潤 ‧ 除稅前利潤下跌1.62億美元至7.38億美 元，主要由於業務繼續重新定位致使收 入下降，但部分減幅被營業支出和貸款 減值及其他信貸風險準備減少抵銷。 收入 （百萬美元） 2,377 2,439 41 2,418 286 2,725 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整收入 ‧ 收入較2013年減少11%，乃由於交易收 益及費用收益淨額減少，反映客戶資產 有序縮減及市場波幅收窄。淨利息收益 亦告下跌，主要在歐洲及亞洲，原因分 別為存款結餘及財資業務收益減少，此 兩者均反映我們採取措施將業務重新 定位。此外，貸款息差較2013年收窄。 ‧ 2014年錄得貸款減值撥回淨額，而2013 年則為3,300萬美元提撥額，主要由於在 英國及美國撥回綜合評估減值準備。 營業支出 （百萬美元） 1,778 2,229 (71) 1,707 (423) 1,806 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整營業支出 ‧ 營業支出下降5%，主要由於撥回2012年 確認的一項英國客戶賠償準備、就英國 與瑞士政府簽訂Rubik雙邊稅務協議而 在英國提撥的準備不復再現，以及職員 人數有序減少。 列賬基準之客戶資產 54 2014年 2013年 十億美元 十億美元 於1月1日 382 398 新增資金淨額 (3) (26) 其中：目標增長範疇 14 (7) 價值變動 8 12 出售 (11) (3) 匯兌及其他 (11) 1 於12月31日 365 382 按地區分析列賬基準之客戶資產 2014年 2013年 十億美元 十億美元 歐洲 179 197 亞洲 112 108 北美洲 63 65 拉丁美洲 11 12 於12月31日 365 382 有關註釋，請參閱第109頁。 ‧ 按列賬基準計算，客戶資產 （包括管理 資金及現金存款） 下跌，減幅主要來自 歐洲，是由於出售瑞士客戶組合、出售 盧森堡的HSBC Trinkaus & Burkhardt AG業 務及新增資金淨額錄得負數值構成的 影響。此外，主要在歐洲有不利匯兌變 動，惟部分被有利的市場變動所抵銷。 負數新增資金淨額為30億美元，主要由 於持續將業務重新定位所致，然而，目 標增長範疇 （包括本位及優先發展市場 以及資產豐厚客戶階層） 卻錄得正數新 增資金淨額140億美元。 滙豐控股有限公司 74 董事會報告：環球業務 （續） 環球私人銀行╱其他 ‧ 按列賬基準計算，我們2014年的資產回 報 （定義為收入佔客戶資產平均值的百 分比） 為63個基點，與2013年大致相若。 按經調整基準計算，我們的資產回報於 2014年下跌6個基點，反映業務重新定 位及市場波幅減少的影響。客戶的資產 回報 （不計及財資業務及資本收入） 亦下 跌4個基點。 ‧ 2015年1月，瑞士國家銀行將其貨幣兌 歐元的匯率上限脫鈎，瑞士法郎因而升 值。我們會持續監察匯率波動的影響， 且不預期環球私人銀行業務按列賬基 準計算的業績會受到任何重大影響。 策略方向 把握本位及優先發展市場的增長機會，致 力通過合作提高收入 ‧ 2014年，其他環球業務新轉介的個案帶 來新增資金淨額逾100億美元，較2013年 上升55億美元。總計而言，2014年目標 增長範疇所產生的新增資金淨額中，有 74%來自集團內部轉介客戶，促成因素 包括與其他環球業務合作，採用更協調 及更有系統的方法發掘客戶需求。 ‧ 我們設立企業客戶組，促進環球私人銀 行業務與環球銀行及資本市場業務和 工商金融業務的合作，藉以釐定最佳慣 例的標準及開發專門設計的服務，以更 高效的方式滿足客戶所需，從而增加環 球私人銀行業務與其他業務互相引薦 的客戶。企業客戶組亦負責通過協調方 式加強現有個人及企業客戶的服務。此 外，我們設立環球解決方案組，以便為 資產極豐厚及環球優先客戶提供專門 設計的解決方案。箇中工作涉及環球私 人銀行業務與環球銀行及資本市場業 務和工商金融業務緊密合作，改善我們 為該等資深客戶提供的服務。 ‧ 我們亦成立財富管理客戶組，負責確保 環球私人銀行業務和零售銀行及財富 管理業務採取更一致的業務運作方式 和加強合作，包括利用零售銀行及財富 管理業務的交易銀行服務專長。 ‧ 為支持客戶資產增長，我們擴大產品種 類，提供三種新的另類投資產品，包括 一項私募基金及兩項房地產基金。我們 致力確保現有資產超過500萬美元的多 數客戶可獲專責投資顧問提供服務，以 強化投資業務組別。環球私人銀行業務 與環球銀行及資本市場業務的環球研 究團隊合作，使客戶輕易取得更廣泛類 別的投資研究報告，從而改善我們為客 戶提供的顧問服務。我們計劃於2015年 底前在全球各地提供此項服務。環球私 人銀行業務亦與滙豐證券服務緊密合 作，讓資產極豐厚及家族財富策劃的客 戶使用歐洲、中東及北非的機構環球託 管平台，和享受交易、結算及交收、保 管及投資管理服務。 業務重新定位 ‧ 2014年，環球私人銀行業務透過檢討組 合及努力確保所有客戶均符合環球標 準 （包括遵循防範金融犯罪規例及稅務 透明度的標準） ，繼續重訂業務模式及 客戶基礎的發展方向。 ‧ 我們繼續集中服務本位及優先發展市 場中與集團聯繫較廣的客戶。今年較早 前我們宣布向LGT Bank (Switzerland) Ltd出 售瑞士的客戶組合後，已於2014年下半 年完成轉讓80億美元客戶資產。我們亦 繼續減少非優先發展市場的客戶數目。 ‧ 2014年，我們繼續簡化及優化業務，關 閉若干非策略代表辦事處，並宣布將歐 洲的信託業務併入澤西的地區業務中 心。我們亦開始開發新的環球網上銀行 服務平台。新系統可望將環球私人銀行 業務的多個系統整合成劃一的銀行服 務平台，為客戶提供更高效、更優質和 貫徹一致的服務。我們仍有望於2015年 按計劃落實第一階段的工作。 ‧ 我們在瑞士、摩納哥、盧森堡及根西島 推出新的流動理財應用程式，藉以改善 電子銀行服務，使客戶可以隨時隨地查 看所持投資及各項交易。我們為瑞士的 前線員工引入安全的平板電腦應用程 式，使他們在探訪客戶期間能閱覽電 子文件，同時亦在美國採用視頻會議功 能。該等應用程式和其他程式已計劃於 2015年更廣泛採用。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 75 其他 39 「其他」 項內計入滙豐的控股公司及融資業 務的業績、後勤統籌部門支出及有關收回 額、未分配的投資活動、集中持有之投資 公司、若干物業交易及本身債務的公允值 變動。 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 淨利息支出 (501) (737) (730) 費用收益╱ （支出） 淨額 (65) 64 194 交易收益╱ （支出） 淨額50 (92) 6 (537) 已發行長期債務及相關 衍生工具之公允值變動 508 (1,228) (4,327) 指定以公允值列賬之 其他金融工具之變動 (9) (576) (1,136) 指定以公允值列賬之金融 工具淨收益╱ （支出） 499 (1,804) (5,463) 其他收益 6,524 8,122 8,868 營業收益淨額4 6,365 5,651 2,332 貸款減值及其他信貸風險準備43 － － － 營業收益淨額 6,365 5,651 2,332 營業支出總額 (8,601) (7,796) (9,369) 營業虧損 (2,236) (2,145) (7,037) 來自聯營公司收益╱ （支出） 44 6 (14) 47 除稅前虧損 (2,230) (2,159) (6,990) 有關註釋，請參閱第109頁。 列賬基準業績回顧 列賬基準之除稅前虧損為22億美元，較2013 年上升3%，乃因營業支出增加，部分被收 入增幅抵銷。 除稅前虧損增加7,100萬美元，已計及2014 年本身債務的公允值有利變動4.17億美元， 而2013年則錄得不利變動12億美元。相關 業績亦計及2013年的以下項目： ‧ 撤銷確認興業銀行作為聯營公司所得 增益 （11億美元） ； ‧ 出售中國平安保險 （集團） 股份有限公司 （ 「平安保險」 ） 所產生的利潤淨額 （5.53億 美元） ；及 ‧ 與滙豐控股發行之英鎊債務相關的匯 兌增益 （4.42億美元） ； 及2014年的以下項目： ‧ 於2014年出售所持上海銀行股權所產生 的利潤 （4.28億美元） ；及 ‧ 於興業銀行之投資所產生的減值 （2.71 億美元） 。 有關所有重大項目的其他詳情，請參閱第42頁。 經調整業績回顧 45 除稅前虧損 （百萬美元） (2,230) (2,159) (381) (2,611) (837) (2,996) 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整虧損 ‧ 除稅前虧損下降，反映收入增長，惟部 分增幅被營業支出增加抵銷。 收入 （百萬美元） 6,365 5,651 (501) 5,864 (1,119) 4,532 列賬基準 貨幣換算及重大項目 經調整收入 2014年 2013年 ‧ 收入上升13億美元，主要原因是2014年 對沖主要由滙豐控股及其歐洲附屬公司 所發行指定以公允值列賬之長期債務 時，低效利率及匯率對沖工具錄得有利 變動9,600萬美元，而2013年則錄得不利 變動5.51億美元。此外，環球業務收回 的若干支出增加，反映營業支出增加。 另一方面，我們在歐洲對集團內部融資 交易進行外部對沖錄得利潤。此外，我 們亦撥回美國不確定稅項儲備應計利 息。但部分利好影響因與出售美國卡及 零售商戶業務有關的過渡服務協議屆滿 及亞洲的投資物業收益減少而被抵銷。 營業支出 （百萬美元） 8,601 7,796 (120) 8,481 (261) 7,535 列賬基準 貨幣換算及重大項目 經調整營業支出 2014年 2013年 滙豐控股有限公司 76 董事會報告：環球業務 （續） ‧ 營業支出增加9.46億美元，乃由於監管 計劃及合規成本增加、印度及中國內地 環球資源成本增加，以及2013年亞洲撥 回一項訴訟準備。此外，於2014年英國 的銀行徵費支出11億美元，高於2013年 分析 的9.16億美元徵費，主要由於徵費率增 加。但部分支出增幅因與出售北美洲卡 及零售商戶業務有關的過渡服務協議 屆滿產生減額而被抵銷。 按環球業務分析 滙豐除稅前利潤╱ （虧損） 及資產負債表數據 2014年 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 39 項目之間 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益╱ （支出） 16,782 10,506 7,022 994 (501) (98) 34,705 費用收益╱ （支出） 淨額 6,668 4,738 3,560 1,056 (65) － 15,957 不包括淨利息收益之交易收益╱ （支出） (28) 620 4,063 298 (100) － 4,853 交易活動之淨利息收益╱ （支出） 9 (2) 1,798 (4) 8 98 1,907 交易收益╱ （支出） 淨額50 (19) 618 5,861 294 (92) 98 6,760 已發行長期債務及相關 衍生工具之公允值變動 － － － － 508 － 508 指定以公允值列賬之其他 金融工具淨收益╱ （支出） 1,675 288 12 (1) (9) － 1,965 指定以公允值列賬之 金融工具淨收益╱ （支出） 1,675 288 12 (1) 499 － 2,473 金融投資減除虧損後增益 14 31 1,117 9 164 1,335 － 股息收益 24 18 80 5 184 － 311 保費收益淨額 10,570 1,296 5 50 － － 11,921 其他營業收益 719 248 124 33 6,176 (6,169) 1,131 ) 營業收益總額 36,433 17,743 17,781 2,440 6,365 (6,169) 74,593 保險賠償淨額56 (11,839) (1,440) (3) (63) － － (13,345) 營業收益淨額4 24,594 16,303 17,778 2,377 6,365 (6,169) 61,248 貸款減值準備 （提撥） ╱收回 及其他信貸風險準備 (1,819) (1,675) (365) 8 － － (3,851) 營業收益淨額 22,775 14,628 17,413 2,385 6,365 (6,169) 57,397 職員支出57 (5,038) (2,439) (3,655) (732) (8,502) － (20,366) 其他營業支出 (12,484) (5,050) (8,373) (1,046) (99) 6,169 (20,883) 營業支出總額 (17,522) (7,489) (12,028) (1,778) (8,601) 6,169 (41,249) 營業利潤╱ （虧損） 5,253 7,139 5,385 607 (2,236) － 16,148 應佔聯營及合資公司利潤 398 1,605 504 19 6 － 2,532 除稅前利潤╱ （虧損） 5,651 8,744 5,889 626 (2,230) － 18,680 - % % % % % % 應佔滙豐除稅前利潤 30.3 46.8 31.5 3.4 (12.0) 100.0 成本效益比率 71.2 45.9 67.7 74.8 135.1 67.3 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 359,744 313,999 254,463 44,102 2,352 974,660 資產總值 499,083 372,739 1,839,644 88,342 164,537 (330,206) 2,634,139 客戶賬項27 581,421 363,654 319,121 85,465 981 1,350,642 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 77 2013年 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 39 項目之間 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益╱ （支出） 18,339 10,200 6,766 1,146 (737) (175) 35,539 費用收益淨額 7,021 4,717 3,482 1,150 64 － 16,434 不包括淨利息收益之交易收益╱ （支出） 689 649 4,953 390 (38) － 6,643 交易活動之淨利息收益╱ （支出） (3) － 1,827 4 44 175 2,047 交易收益淨額50 686 649 6,780 394 6 175 8,690 已發行長期債務及相關 衍生工具之公允值變動 － － － － (1,228) － (1,228) 指定以公允值列賬之其他 金融工具淨收益╱ （支出） 1,638 332 599 4 (576) (1) 1,996 指定以公允值列賬之 金融工具淨收益╱ （支出） 1,638 332 599 4 (1,804) (1) 768 金融投資減除虧損後增益 55 1 747 (3) 1,212 － 2,012 股息收益 21 15 129 8 149 － 322 保費收益淨額 10,543 1,375 6 16 － － 11,940 其他營業收益╱ （支出） 544 621 670 (239) 6,761 (5,725) 2,632 營業收益總額 38,847 17,910 19,179 2,476 5,651 (5,726) 78,337 保險賠償淨額56 (12,107) (1,545) (3) (37) － － (13,692) 營業收益淨額4 26,740 16,365 19,176 2,439 5,651 (5,726) 64,645 貸款減值及其他信貸風險準備 (3,227) (2,384) (207) (31) － － (5,849) 營業收益淨額 23,513 13,981 18,969 2,408 5,651 (5,726) 58,796 職員支出57 (5,219) (2,327) (3,549) (776) (7,325) － (19,196) 其他營業支出 (12,029) (4,722) (6,411) (1,453) (471) 5,726 (19,360) 營業支出總額 (17,248) (7,049) (9,960) (2,229) (7,796) 5,726 (38,556) 營業利潤╱ （虧損） 6,265 6,932 9,009 179 (2,145) － 20,240 應佔聯營及合資公司利潤╱ （虧損） 384 1,509 432 14 (14) － 2,325 M 除稅前利潤╱ （虧損） 6,649 8,441 9,441 193 (2,159) － 22,565 % % % % % % 應佔滙豐除稅前利潤 29.5 37.4 41.8 0.9 (9.6) 100.0 成本效益比率 64.5 43.1 51.9 91.4 138.0 59.6 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 375,086 297,852 272,473 44,224 2,454 992,089 資產總值 517,085 360,623 1,975,509 97,655 171,812 (451,366) 2,671,318 客戶賬項27 579,994 354,298 328,800 96,770 1,435 1,361,297 有關註釋，請參閱第109頁。 董事會報告：地區 滙豐控股有限公司 78 地區 概要 78 歐洲 79 亞洲 84 中東及北非 91 北美洲 96 拉丁美洲 101 概要 2014年業績的補充資料，請參閱第40至62頁 「財務概要」 。 在下文按地區分析的利潤及虧損中，營業 收益及營業支出均包括滙豐內部項目，所 涉金額共29.72億美元 （2013年：26.28億美 元；2012年：26.84億美元） 。 自2014年1月1日起， 「亞洲」 地區取代之前按 「香港」 及 「亞太其他地區」 呈列的地區。這 與內部管理業務所用財務資料的變更乃屬 一致。比較數字已相應重列。 除另有說明外，評述均採用經調整基準 （請 參閱第40頁） ，而列表內的數字則按列賬基 準呈列 （除另有說明外） 。 除稅前利潤╱ （虧損） 2014年 2013年 2012年 百萬美元 % 百萬美元 % 百萬美元 % 歐洲 596 3.2 1,825 8.1 (3,414) (16.5) 亞洲8 14,625 78.3 15,853 70.3 18,030 87.3 中東及北非 1,826 9.8 1,694 7.5 1,350 6.5 北美洲 1,417 7.6 1,221 5.4 2,299 11.1 拉丁美洲 216 1.1 1,972 8.7 2,384 11.6 y N 截至12月31日止年度 18,680 100.0 22,565 100.0 20,649 100.0 資產總值 40 2014年 2013年 百萬美元 % 百萬美元 % 歐洲 1,290,926 49.0 1,392,959 52.1 亞洲8 878,723 33.4 831,791 31.1 中東及北非 62,417 2.4 60,810 2.3 北美洲 436,859 16.6 432,035 16.2 拉丁美洲 115,354 4.4 113,999 4.3 滙豐內部項目 (150,140) (5.8) (160,276) (6.0) 於12月31日 2,634,139 100.0 2,671,318 100.0 0 風險加權資產 58 2014年 2013年 十億美元 % 十億美元 % 於12月31日 1,219.8 100.0 1,092.7 100.0 0 歐洲 375.4 30.1 300.1 27.1 亞洲8 499.8 40.0 430.7 38.9 中東及北非 63.0 5.0 62.5 5.7 北美洲 221.4 17.8 223.8 20.2 拉丁美洲 88.8 7.1 89.5 8.1 有關註釋，請參閱第109頁。 概要╱歐洲 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 79 歐洲 集團於歐洲的主要銀行業務包括英國 的英國滙豐銀行有限公司、法國滙豐、 土耳其的HSBC Bank A.S.、滙豐私人銀 行 （瑞士） 有限公司及HSBC Trinkaus & Burkhardt AG。透過該等附屬公司，集 團在整個歐洲大陸為個人、工商及企業 客戶提供一應俱全的銀行、財資和理財 服務。 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 淨利息收益 10,611 10,693 10,394 費用收益淨額 6,042 6,032 6,169 交易收益淨額 2,534 4,423 2,707 其他收益╱ （支出） 2,384 (181) (1,662) 營業收益淨額4 21,571 20,967 17,608 貸款減值及其他 信貸風險準備43 (764) (1,530) (1,921) 營業收益淨額 20,807 19,437 15,687 營業支出總額 (20,217) (17,613) (19,095) 營業利潤╱ （虧損） 590 1,824 (3,408) 來自聯營公司收益╱ （支出） 44 6 1 (6) 除稅前利潤╱ （虧損） 596 1,825 (3,414) < 成本效益比率 93.7% 84.0% 108.4% 風險加權資產 平均值回報36 0.2% 0.6% (1.0%) 年底職員人數 69,363 68,334 70,061 連續第二年 西歐最佳債券行 （ 《歐洲貨幣》 雜誌獎項） 英國最佳貿易銀行 （ 《環球金融》 雜誌） 31億美元 監管機構的罰款、準備、 罰則及英國客戶賠償款項 有關註釋，請參閱第109頁。 經濟背景 英國經濟的復甦步伐持續至2014年下半年， 但至年底時擴張步伐放緩。初步估計顯示， 國內實質生產總值的年增長率為2.6%。失 業率於直至12月份的三個月內下降至5.7%， 工資由極低水平略為加速上升。作為通脹 指標的年度消費物價指數於12月降至14年 來的低位0.5%。繼2013年及2014年最初幾個 月快速增長後，經濟活動及房屋市場的價 格增升於年底均逐漸放緩。英倫銀行維持 銀行利率於0.5厘的穩定水平。 歐元區於2014年的經濟活動復甦緩慢，而 且各個成員國步伐不一。區內實質生產總 值於年內整體增長0.9%。德國及西班牙的 經濟分別增長1.6%及1.5%，而法國的國內 生產總值則較溫和地增長0.4%。歐元區的 通脹率於12月下降至負0.2%，令市場擔心 該區可能走向持續通縮期。為免這種增長 及通脹俱低的情況持續過長時間，歐洲中 央銀行 （ 「歐洲央行」 ） 於9月將主要再融資 利率及存款利率分別調低至0.05厘及負0.2 厘，並推行資產負債擴張政策，開始購買 備兌債券及資產抵押證券。 財務概覽 除稅前利潤 （百萬美元） 2014年 2013年 596 1,825 3,309 3,905 2,476 4,301 列賬基準 貨幣換算及重大項目 經調整利潤 2014年，集團的歐洲業務錄得除稅前利潤 5.96億美元，2013年為18億美元。列賬基準 的除稅前利潤減少，乃受多個重大項目以 及營業支出增加拖累，惟貸款減值及其他 信貸風險準備減少抵銷了部分降幅。前者 包括英國客戶賠償準備13億美元、與監管 機構調查外匯操控問題有關的和解開支及 準備12億美元 （其中8.09億美元於2014年第 四季度錄得） ，及因應英國 《消費者信貸法》 持續合規檢討所產生的準備6.32億美元。有 關所有重大項目的進一步詳情，請參閱第 42頁。 滙豐控股有限公司 80 董事會報告：地區 （續） 按環球業務所在國家╱地區列示之除稅前利潤╱ （虧損） 零售銀行 環球銀行 環球 及財富管理 工商金融 及資本市場 私人銀行 其他 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 英國 589 2,193 (801) 191 (2,228) (56) 法國35 (181) 240 354 － (199) 214 德國 28 71 162 27 (10) 278 瑞士 － 5 2 38 (3) 42 土耳其 (155) 5 92 － (6) (64) 其他 33 34 240 59 (184) 182 截至2014年12月31日止年度 314 2,548 49 315 (2,630) 596 英國 1,471 1,684 1,246 252 (3,493) 1,160 法國35 285 255 351 21 (162) 750 德國 30 70 183 44 (25) 302 瑞士 － 2 2 (291) － (287) 土耳其 (74) 36 108 (1) 1 70 其他 41 41 (89) (190) 27 (170) 截至2013年12月31日止年度 1,753 2,088 1,801 (165) (3,652) 1,825 英國 343 832 (111) 235 (6,355) (5,056) 法國35 135 203 514 (11) (263) 578 德國 29 64 283 40 (72) 344 瑞士 － 2 1 133 － 136 土耳其 (32) 71 104 － 1 144 其他 34 36 195 102 73 440 截至2012年12月31日止年度 509 1,208 986 499 (6,616) (3,414) 有關註釋，請參閱第109頁。 經調整除稅前利潤減少3.96億美元，主要反 映成本增加，惟貸款減值及其他信貸風險 準備減少抵銷了部分降幅；收入與2013年 大致相若。 國家╱地區業務摘要 在英國，整體工商金融業務貸款較2013年 增加7%，未計耗損及攤銷前的新造貸款及 再融資增加38%，而且逾85%的小型企業貸 款申請獲批准。此外，商務理財業務推出 計劃，向中小企客戶提供進一步支援及貸 款。計劃的部分安排為撥出58億英鎊 （99億 美元） 供客戶日後借貸，為英國經濟增長提 供資金支持。環球貿易及融資業務貸款亦 增長3%，因集團進一步鞏固於貿易融資的 市場地位及減少應收賬融資業務現有客戶 流失。 零售銀行及財富管理業務方面，集團向逾 11.8萬名客戶批出新造按揭貸款114億英鎊 （188億美元） ，包括向逾2.75萬名首次置業 人士批出35億英鎊 （58億美元） 貸款。然而， 取用按揭結欠總額略有減少。新造貸款的 貸款估值比率為60%，而按揭組合總額的平 均貸款估值比率為43.7%。於2014年10月， 集團擴大按揭服務途徑以新增一家中介機 構，從而把握英國按揭市場中此類融資需 求所佔比例日益增加所帶來的機遇。 作為2013年重整環球銀行及資本市場業務 的其中一項措施，集團將所有融資業務併 入資本融資業務，包括貸款、債務資本市 場及股票資本市場業務。我們透過委任兩 名英國銀行業務新聯席主管，提升相關業 務的專業實力及地域覆蓋。於2014年，資本 融資業務內顧問及股票資本市場業務的業 務量增長超過市場水平。 在法國的環球銀行及資本市場業務方面， 集團為歐洲其中一宗最大規模併購 （ 「併購」 ） 交易擔任獨家顧問。工商金融業務方面， 資金管理服務運用單一歐元支付區平台為 歐洲各地客戶提供歐元計值信貸轉撥及直 接扣賬付款服務。由此，客戶可採用一套 統一的標準、規則及條件，透過其滙豐戶 口在設有單一歐元支付區平台的34個國家 和地區以歐元支付及收取款項。此外，工 商金融業務於2014年進一步分配15億歐元 （20億美元） 至中小企貸款資金及批核逾20 億歐元 （27億美元） 貸款。零售銀行及財富 管理業務方面，房屋貸款增長強勁。 在德國，為推動業務增長，集團在多特蒙 德、曼海姆及科隆開設三家分行、增加26% 的客戶經理人數，並在法國、中國內地及 英國等國家和地區舉行多次巡迴推介，以 加強德國作為主要國際業務中心的地位。 環球私人銀行業務方面，集團出售盧森堡 的HSBC Trinkaus & Burkhardt AG業務。 在土耳其，監管機構對信用卡及透支利率 設定上限影響收入。儘管如此，於2014年9 月，工商金融業務推出20億土耳其里拉 （9.14 億美元） 的國際基金，為企業發展國際業務 提供持續支持及環球聯繫，其中11億土耳 其里拉 （5.19億美元） 已被取用。 歐洲 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 81 重估增值上升，其影響抵銷了客戶交易量 增加產生的收入增幅及衍生工具收益的升 幅有餘。 零售銀行及財富管理業務收入略有減少， 原因是息差收窄，主要是按揭方面。此外， 費用收益下降，原因是根據合夥協議應付 的費用增加以及投資產品和透支費用收益 減少。該等因素部分被儲蓄產品息差改善 及往來賬項款額增加所抵銷。 在歐洲其他國家╱地區，法國、瑞士及土耳 其的業務收入減少。在法國，收入的減幅主 要源自零售銀行及財富管理業務中的保險 業務，原因是有效長期保險業務現值資產 出現不利變動2.03億美元，反映儲蓄業務長 期收益減少令保證之成本增加，而2013年 則為有利變動4,800萬美元。同時，環球 銀行及資本市場業務的利率交易減少，原 因為市場波幅降低及交易活動減少。在瑞 士，收入下降反映環球私人銀行業務重新 定位及客戶資產減少。土耳其業務的收入 亦下降，主要是零售銀行及財富管理業務 方面，原因為當地監管機構為卡及透支利 率設定上限，惟該降幅被卡費收入增加所 部分抵銷。 貸款減值及其他信貸風險準備減少，主要 是在英國，其次為西班牙。在英國，工商 金融業務的個別評估準備減少，反映業務 組合的質素及經濟狀況改善。環球銀行及 資本市場業務亦錄得貸款減值準備減少， 原因是個別評估準備減少，以及可供出售 資產抵押證券的信貸風險準備撥回淨額增 加。因集團修訂用於計算綜合評估企業貸 款減值的若干估算值而令準備額上升，抵 銷了上述部分減幅。零售銀行及財富管理 業務的貸款減值準備減少，原因是經濟環 境改善令拖欠水平下降及客戶持續減少未 償還信用卡及貸款結欠。西班牙的貸款減 值準備減少，原因是個別評估準備減少。 英國及西班牙的降幅被土耳其及法國的增 長所部分抵銷。土耳其的貸款減值準備增 加，原因是隨著監管規例的變化，卡拖欠 率上升。法國的貸款減值準備增加，主要 是環球銀行及資本市場業務及工商金融業 務方面，原因是個別評估準備增加。 營業支出 （百萬美元） 20,217 17,613 (2,601) 17,616 (1,289) 16,324 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整營業支出 在瑞士，環球私人銀行業務繼續重新定位， 並致力透過服務資產豐厚的客戶推動業務 增長。客戶資產 （包括管理資金及現金存 款） 由於業務重新定位以及出售客戶資產 組合而減少。 於2014年11月，集團出售哈薩克業務，以 配合集團策略。 經調整業績回顧 45 收入增加7,600萬美元，主要在英國，惟法 國、瑞士及土耳其等其他國家╱地區收入 減少抵銷了部分增幅。 收入 （百萬美元） 21,571 20,967 708 22,279 1,236 22,203 列賬基準 貨幣換算及重大項目 經調整收入 2014年 2013年 國家╱地區經調整收入概覽 2014年 2013年 百萬美元 百萬美元 英國 16,080 15,365 法國 2,937 3,097 德國 945 960 瑞士 736 831 土耳其 791 827 其他 790 1,123 截至12月31日止年度 22,279 22,203 在英國，收入增加7.15億美元，原因為於 2014年對沖主要由滙豐控股發行之長期債 務時，相關利率及匯率之低效對沖工具錄 得有利公允值變動2.22億美元，2013年則錄 得不利變動4.8億美元，以及外部對沖集團 內部融資交易產生增益。 工商金融業務收入亦增加，乃由於資金管 理業務的存款量增加，及有期貸款息差擴 闊令淨利息收益增加。此外，費用收益淨 額增加，部分反映大型企業及中型企業的 新造貸款增加。 相對而言，環球銀行及資本市場業務收入 較2013年減少，主要受資本市場業務所影 響，包括對若干衍生工具合約作出資金公 允值調整，產生影響利率及信貸交易業務 的扣賬額。外匯交易業務收入亦減少，反映 市場波幅降低及客戶交易量減少。此外， 股票交易業務收入減少，因2013年受惠於 滙豐控股有限公司 82 董事會報告：地區 （續） 營業支出上升13億美元，主要在英國，反映 所有業務的監管計劃及合規成本增加及職 員支出增加。此外，2014年的11億美元英國 銀行徵費支出較2013年高出2億美元，主要 是由於徵費率上調。支出增加亦因為英國 業務確認金融服務賠償計劃徵費的時間。 該等增幅被可持續成本節約逾3.3億美元所 部分抵銷。 除稅前利潤╱ （虧損） 及資產負債表數據－歐洲 2014年 零售銀行 環球銀行 環球 項目之間 及財富管理 工商金融 及資本市場 私人銀行 其他 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益╱ （支出） 5,196 3,616 1,956 594 (654) (97) 10,611 費用收益╱ （支出） 淨額 2,456 1,900 1,087 626 (27) － 6,042 不包括淨利息收益之 交易收益╱ （支出） (260) 33 1,943 140 (92) － 1,764 交易活動之淨利息收益╱ （支出） 14 2 660 (4) 1 97 770 交易收益╱ （支出） 淨額50 (246) 35 2,603 136 (91) 97 2,534 已發行長期債務及相關衍生工具 之公允值變動 － － － － 614 － 614 指定以公允值列賬之 其他金融工具淨收益╱ （支出） 616 119 14 (1) (11) － 737 指定以公允值列賬之金融工具 淨收益╱ （支出） 616 119 14 (1) 603 － 1,351 金融投資減除虧損後增益 12 10 730 9 11 － 772 股息收益 3 7 50 2 3 － 65 保費收益╱ （支出） 淨額 2,741 217 － 50 － － 3,008 其他營業收益╱ （支出） (127) 45 (3) 29 1,249 (186) 1,007 . 營業收益總額 10,651 5,949 6,437 1,445 1,094 (186) 25,390 保險賠償淨額56 (3,450) (306) － (63) － － (3,819) 營業收益淨額4 7,201 5,643 6,437 1,382 1,094 (186) 21,571 貸款減值準備 （提撥） ╱收回及其他信貸 風險準備 (268) (502) － 4 2 － (764) 營業收益淨額 6,933 5,141 6,437 1,386 1,096 (186) 20,807 營業支出總額 (6,621) (2,594) (6,391) (1,071) (3,726) 186 (20,217) 營業利潤╱ （虧損） 312 2,547 46 315 (2,630) － 590 應佔聯營及合資公司利潤 2 1 3 － － － 6 除稅前利潤╱ （虧損） 314 2,548 49 315 (2,630) － 596 % % % % % % 應佔滙豐除稅前利潤 1.7 13.6 0.3 1.7 (14.1) 3.2 成本效益比率 91.9 46.0 99.3 77.5 340.6 93.7 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 165,112 106,342 113,136 24,766 377 409,733 資產總值 221,679 120,819 948,951 64,676 64,182 (129,381) 1,290,926 客戶賬項27 202,413 135,837 166,075 41,380 254 545,959 歐洲 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 83 2013年 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 項目之間 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益╱ （支出） 5,600 3,353 1,774 722 (694) (62) 10,693 費用收益╱ （支出） 淨額 2,545 1,789 957 744 (3) － 6,032 不包括淨利息收益之 交易收益 206 30 2,181 192 698 － 3,307 交易活動之淨利息收益 2 5 1,013 4 30 62 1,116 交易收益淨額50 208 35 3,194 196 728 62 4,423 已發行長期債務及相關衍生工具之 公允值變動 － － － － (936) － (936) 指定以公允值列賬之 其他金融工具淨收益╱ （支出） 1,059 271 591 4 (570) (1) 1,354 指定以公允值列賬之金融工具 淨收益╱ （支出） 1,059 271 591 4 (1,506) (1) 418 金融投資減除虧損後增益 52 － 344 (17) － － 379 股息收益 4 2 65 4 － － 75 保費收益╱ （支出） 淨額 2,782 361 (1) 16 － － 3,158 其他營業收益╱ （支出） (103) 9 110 (253) 766 － 529 營業收益╱ （支出） 總額 12,147 5,820 7,034 1,416 (709) (1) 25,707 保險賠償淨額56 (4,136) (567) － (37) － － (4,740) 營業收益╱ （支出） 淨額4 8,011 5,253 7,034 1,379 (709) (1) 20,967 貸款減值及 其他信貸風險準備 (329) (935) (242) (24) － － (1,530) 營業收益╱ （支出） 淨額 7,682 4,318 6,792 1,355 (709) (1) 19,437 營業支出總額 (5,934) (2,231) (4,987) (1,519) (2,943) 1 (17,613) 營業利潤╱ （虧損） 1,748 2,087 1,805 (164) (3,652) － 1,824 應佔聯營及合資公司利潤╱ （虧損） 5 1 (4) (1) － － 1 除稅前利潤╱ （虧損） 1,753 2,088 1,801 (165) (3,652) － 1,825 % % % % % % 應佔滙豐除稅前利潤 7.8 9.2 8.0 (0.7) (16.2) 8.1 成本效益比率 74.1 42.5 70.9 110.2 (415.1) 84.0 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 177,357 105,498 145,136 27,289 830 456,110 資產總值 238,499 124,242 1,054,506 75,718 72,174 (172,180) 1,392,959 客戶賬項27 205,288 134,120 191,715 49,789 1,021 581,933 有關註釋，請參閱第109頁。 滙豐控股有限公司 84 董事會報告：地區 （續） 亞洲 7 香港上海滙豐銀行有限公司及恒生銀行有 限公司是滙豐在香港經營銀行業務的主要 附屬公司。前者是在香港註冊成立的最大 銀行，亦是滙豐在亞洲的旗艦銀行。 在中國內地，我們透過於當地註冊的附屬 公司滙豐銀行 （中國） 有限公司及恒生銀行 （中國） 有限公司提供一應俱全的銀行及金 融服務。我們亦透過聯營公司交通銀行， 間接參與中國內地市場。 除香港及中國內地外，我們於亞洲18個國 家及地區經營業務，其中於澳洲、印度、 印尼、馬來西亞及新加坡均有龐大業務網 絡。 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 淨利息收益 12,273 11,432 10,707 費用收益淨額 5,910 5,936 5,418 交易收益淨額 2,622 2,026 2,516 其他收益 2,872 5,038 6,691 ● ● ● 營業收益淨額4 23,677 24,432 25,332 貸款減值及其他 信貸風險準備43 (647) (498) (510) ● ● ● 營業收益淨額 23,030 23,934 24,822 營業支出總額 (10,427) (9,936) (9,980) ● ● ● 營業利潤 12,603 13,998 14,842 來自聯營公司收益44 2,022 1,855 3,188 ● ● ● 除稅前利潤 14,625 15,853 18,030 ● ● ● 成本效益比率 44.0% 40.7% 39.4% 風險加權資產 平均值回報36 3.1% 3.8% 4.4% 年底職員人數 118,322 113,701 112,766 不計及貨幣換算的影響， 客戶貸款結欠增長 10% 在亞洲 （日本除外） 債券 市場佔領導地位 （彭博） 亞洲最佳銀行 （ 《歐洲貨幣》 雜誌 「2014年卓越大獎」 ） 有關註釋，請參閱第109頁。 經濟背景 香港實質本地生產總值增長率於2014年較 2013年放緩，部分歸因於零售銷貨年增長 率放緩導致本地需求疲軟。失業率自歷史 低位回升，勞動市場狀況惡化。訪港旅客 整體上升，中國內地訪客增長令本年度抵 港人次較2013年增長16%。多項政府補貼 到期，抵銷燃油及食品價格通脹下降，令 2014年整體消費物價指數平均上漲率僅高 於4%。 中國內地實質國內生產總值年增長率由 2013年的7.7%放緩至2014年的7.4%，主要由 於建築及製造業投資活動減少，穩健的基 礎設施投資僅抵銷了前述部分減幅。全年 整體消費物價指數上漲率穩步下降至12月 的1.5%，遠低於3.5%的政府目標。中國人民 銀行於11月放鬆貨幣政策，是自2012年7月 以來首次削減政策性利率。一年期存款利 率下調25個基點至2.75厘，而一年期貸款利 率下調40個基點至5.6厘。進一步措施於12 月公布以支持銀行貸款及刺激經濟活動。 日本經濟於2014年經歷大幅波動，消費稅 自4月1日起增加3個百分點。國內生產總值 於2014年第一季以年率計增長5.8%，但於加 稅後大幅放緩，乃由於政府刺激措施及出 口未能抵銷私人消費減少。國內生產總值 於先前季度下降6.7%及1.9%後在第四季度 以2.2%年率上升。日本銀行於2014年10月31 日宣布開始新一輪量化寬鬆政策，促使日 圓進一步貶值。 印度新政府大力改革，為該國經濟長遠前 景提振市場情緒。然而，2014年若干基建項 目因等待政府批准而延遲導致復甦受阻。 國際商品價格於下半年急劇下跌，幫助壓 低商品價格通脹及減少經常賬赤字。於2014 年初調高利率後，央行於全年維持穩定的 貨幣政策。 全球商品價格的下滑趨勢讓印尼及馬來西 亞得以削減昂貴的燃油補貼，預期可降低 對外收支失衡及改善其財政狀況。2014年， 該等國家國內需求仍保持相對強勁，支持 著經濟增長。在新加坡，出口增長放緩及 國內經濟結構調整導致2014年國內生產總 值增長放緩，而金融管理局維持貨幣漸進 升值政策。 亞洲 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 85 按環球業務所在國家╱地區列示之除稅前利潤╱ （虧損） 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 香港 3,727 2,264 1,807 146 198 8,142 澳洲 78 126 232 － (4) 432 印度 4 121 442 11 122 700 印尼 10 53 110 － 25 198 中國內地 292 1,533 954 (3) 175 2,951 馬來西亞 156 122 190 － 28 496 新加坡 129 168 243 57 (8) 589 台灣 19 35 166 － 1 221 其他 57 320 432 － 87 896 截至2014年12月31日止年度 4,472 4,742 4,576 211 624 14,625 香港 3,742 2,110 1,971 208 58 8,089 澳洲 100 131 189 － 26 446 印度 (21) 113 418 7 136 653 印尼 12 106 126 － 36 280 中國內地 223 1,536 842 (4) 1,644 4,241 馬來西亞 148 105 236 － 25 514 新加坡 147 120 262 74 22 625 台灣 7 30 158 － 5 200 其他 61 207 473 (1) 65 805 截至2013年12月31日止年度 4,419 4,458 4,675 284 2,017 15,853 香港 3,694 2,188 1,518 249 (67) 7,582 澳洲 97 38 184 － (44) 275 印度 41 89 497 7 175 809 印尼 29 124 146 － 7 306 中國內地 838 1,724 1,257 (4) 2,525 6,340 馬來西亞 183 131 242 － 8 564 新加坡 201 139 296 97 (65) 668 台灣 62 36 136 － － 234 其他 66 321 567 59 239 1,252 截至2012年12月31日止年度 5,211 4,790 4,843 408 2,778 18,030 在澳洲，實質國內生產總值年增長率於2014年 增加至約2.8%，而失業率基本保持在6.1% 不變。採礦業投資急劇減少，而其他經濟 行業改善僅部分抵銷該減幅。低利率繼續 推動房屋市場活動增加及信貸增長溫和回 升。澳元於年內轉弱，但仍遠高於其長期 平均水平。 在台灣，經濟活動活躍令2014年整個年度國 內生產總值水平上升3.5%。這是自2011年 起最強勁年增長率，較2013年增長率2.1% 有所改善。增長乃由於低失業率及工資上 漲令出口及國內消費強勁。台灣中央銀行 於2014年整個年度保持其主要政策利率於 1.875厘不變，為自2011年以來的水平。 財務概覽 除稅前利潤 （百萬美元） 14,625 15,853 10 14,635 (1,544) 14,309 3年 1 0 2 4年 1 0 2 列賬基準 貨幣換算及重大項目 經調整利潤 於2014年，集團的亞洲業務錄得除稅前利潤 146億美元，較2013年的159億美元減少8%。 減幅反映收入減少，以及貸款減值及其他 信貸風險準備及成本增加，惟部分減幅被 應佔聯營公司利潤的增幅抵銷。收入包括 多個重大項目的影響，尤其於2013年，將興 業銀行重新分類為金融投資產生會計增益 （11億美元） 及完成出售平安保險所得利潤 淨額 （5.53億美元） 。 滙豐控股有限公司 86 董事會報告：地區 （續） 中國內地除稅前利潤╱ （虧損） 的分析 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 交通銀行及其他聯營公司 255 1,421 296 － 1 1,973 中國內地業務 37 112 658 (3) 174 978 興業銀行 － － － － － － 平安保險 － － － － － － 截至2014年12月31日止年度 292 1,533 954 (3) 175 2,951 交通銀行及其他聯營公司 247 1,360 284 － (38) 1,853 中國內地業務 (24) 176 558 (4) 40 746 興業銀行 － － － － 1,089 1,089 平安保險 － － － － 553 553 截至2013年12月31日止年度 223 1,536 842 (4) 1,644 4,241 交通銀行及其他聯營公司 214 1,193 248 － － 1,655 中國內地業務 (52) 176 606 (4) 66 792 興業銀行 54 273 343 － － 670 平安保險 622 82 60 － 2,459 3,223 截至2012年12月31日止年度 838 1,724 1,257 (4) 2,525 6,340 於2014年，重大項目包括出售於上海銀行 的投資所得利潤 （4.28億美元） 及於興業銀 行投資減值 （2.71億美元） 。有關重大項目的 進一步詳情，請參閱第42頁。 按經調整基準計算，除稅前利潤增加3.26億 美元或2%，乃由於收入增加，但部分增幅 被營業支出及貸款減值及其他信貸風險準 備增加所抵銷。 國家╱地區業務摘要 我們繼續在亞洲致力推行各項優先策略， 利用集團的國際網絡推動有機增長及聯繫 各地客戶。我們完成出售於上海銀行的投 資，並落實酌情獎勵架構，取消產品銷售 額與薪酬掛鈎的機制。我們的流動理財應 用程式獲持續採用，無接觸式付款系統擴 展至Android手機，增強了電子銀行營運實 力。 在香港，零售銀行及財富管理業務的按揭平 均結欠增長7%，已取用新造按揭貸款的平 均貸款估值比率為47%，整體組合的平均貸 款估值比率估計為29%。於2014年11月，為 配合啓動滬港通平台，我們推出新服務，協 助零售銀行客戶買賣及投資於上海交易所 上市的合資格股份。我們透過在香港推出 「Visa Signature」 卡產品加強發卡業務，並 繼續於區內建立新商戶夥伴關係。我們亦 向香港及其他九個地區市場新晉中上階層 客戶重新推出運籌理財服務。我們更獲 《亞 洲銀行及金融》 雜誌頒發 「年度國際零售銀 行」 大獎。 工商金融業務方面，我們是率先公布在上 海自由貿易區提供人民幣跨境資金歸集服 務的外資銀行之一。工商金融業務與環球 銀行及資本市場業務於年內繼續合作，已 完成主要市場交易由2013年的122項增加至 2014年的157項，主要為債務資本市場發行 票據及獲委託進行槓桿資產融資交易。此 外，我們於 《金融亞洲》 雜誌2014年成就大獎 中獲評選為 「最佳商業銀行」 。 環球銀行及資本市場業務方面，我們在亞 洲 （日本除外） 的三國集團貨幣及投資級別 債券市場仍然維持領導地位，並且於港元 發債市場領先同業。於2014年，我們參與 了香港股票資本市場五大交易中的其中三 項，以及香港首宗伊斯蘭主權債券發行交 易。此外，我們在香港離岸人民幣發債市場 亦繼續佔領導地位，成為香港金融管理局 離岸人民幣流動資金的主要提供行之一。 我們亦為一家內地中資銀行發行離岸優先 股擔任聯席賬簿管理人，這是中國內地首 項符合巴塞爾協定3規定的額外一級資本 的證券發行。我們繼續透過Custody Plus綜 合平台為利用滬港通的機構投資者提供更 佳服務。 我們繼續於中國內地拓展分行網絡，至年 底，我們在當地共有173個滙豐網點、25個 滙豐村鎮銀行網點及50個恒生銀行網點。 零售銀行及財富管理業務方面，我們是首 批推出與美元兌人民幣匯率掛鈎之人民幣 衍生產品的外資銀行，並連續六年獲 《亞洲 銀行家》 雜誌評選為 「最佳外資零售銀行」 。 於2014年，我們成為首家為新加坡及韓國 的人民幣合格境外機構投資者提供服務的 外資託管銀行。集團亦成為上海黃金交易 所新設立的國際板的成員，國際板是連接 中國內地黃金市場與全球投資者的交易平 台。此外，監管當局已批准我們成為首批 可在中國內地銀行間外匯市場直接買賣人 民幣、歐元及新加坡元的市場莊家。 亞洲 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 87 資金管理業務方面，我們推出環球付款系 統，支援以各種貨幣 （包括人民幣） 進出中 國內地的所有跨境付款活動。環球貿易及 融資業務方面，我們推出貿易連繫措施以 連接中國內地與亞洲其他國家╱地區、德 國及美國，增強國際聯繫及推動重點貿易 通道間的活動。併購業務方面，我們為多 家國有企業的重大海外投資及收購活動擔 任顧問。 於亞洲其他地區，工商金融業務在印度的 資產負債繼續增加，包括有期貸款及資金 管理業務下的存款，尤其是協助英國公司 投資於印度。環球銀行及資本市場業務方 面，我們於2014年在兩項最大規模的併購 交易中擔任顧問。在財富管理業務方面， 我們亦推出多元資產基金系列 「全智選」 產 品。在澳洲，我們在2014年最大規模的採 礦項目融資交易及最大規模的運輸基建項 目中，獲委託出任牽頭安排人。工商金融 業務方面，我們亦宣布設立2.5億澳元 （2.25 億美元） 的國際成長基金，為當地中小企提 供信貸，從而拓展海外商機。 經調整業績回顧 45 收入 （百萬美元） 23,677 24,432 (48) 23,629 (1,978) 22,454 3年 1 0 2 4年 1 0 2 列賬基準 貨幣換算及重大項目 經調整收入 收入增加12億美元或5%，主要來自香港及 中國內地工商金融業務及零售銀行及財富 管理業務資產負債增加，以及環球銀行及 資本市場業務的資產負債管理組合及有期 貸款增長；於印度及澳洲的收入亦有所增 長。 國家╱地區經調整收入概覽 2014年 2013年 百萬美元 百萬美元 香港 13,725 13,211 澳洲 975 898 印度 1,826 1,666 印尼 561 559 中國內地 2,463 1,948 馬來西亞 1,066 1,063 新加坡 1,339 1,319 台灣 491 501 其他 1,183 1,289 截至12月31日止年度 23,629 22,454 香港收入增加4%，主要來自工商金融業務 及零售銀行及財富管理業務，其次為環球 銀行及資本市場業務。工商金融業務收入 增加乃由於若干行業有期貸款增加令淨利 息收益增加，資金管理業務平均存款結餘 增加，匯款量增長令費用增加以及貸款息 差擴闊。 零售銀行及財富管理業務方面，收入增加 乃由於平均貸款結欠 （主要為信用卡及其他 個人貸款） 增加及平均存款結餘增加令淨 利息收益增加，惟息差收窄抵銷了業務量 增加所帶來的部分效益。費用收益淨額亦 增加，主要來自單位信託基金、信用卡交 易及證券經紀業務交易量增加。保險業務 方面，收入增加反映保費收益增長，而保 費收益增長亦對債務證券組合的增長作出 貢獻，惟前述部分增幅被PVIF資產因年度精 算假設更新產生的有利變動減少所抵銷。 環球銀行及資本市場業務收入亦增加，主 要來自資產負債管理業務組合增長及資本 融資業務的有期貸款平均結欠增加。客戶 交易量及資本融資下降令費用收益淨額減 少，抵銷了前述部分增幅，反映費用減少。 中國內地收入較2013年增加26%，環球銀行 及資本市場業務方面，我們錄得資產負債 管理淨利息收益增加，乃由於組合增長及 再投資的息率上升，以及平均有期貸款結 欠增加。此外，債務證券利息收益增加， 加上交易用途債券由於收益率下降推動重 估增益增加，令利率交易收益增加，客戶 交易量增加亦令外匯交易量增加。零售銀 行及財富管理業務收入亦有所增加，主要 由於2014年上半年市場利率上升使存款息 差擴闊，而工商金融業務收入增加乃由於 平均存款結餘及貸款結欠增加。 於亞洲其他地區，印度收入增加10%，主要 來自環球銀行及資本市場業務，乃由於衍 生工具信貸估值調整 （ 「信貸估值調整」 ） 的 有利變動使利率交易業務收益增加，加上 資產負債管理組合增長令淨利息收益增加 所致。在澳洲，我們錄得收入增加9%，主 要來自環球銀行及資本市場業務，乃由於 利率及外匯交易收益增加。2013年縮減零 售銀行及財富管理業務後韓國收入減少， 抵銷了上述部分增幅。 貸款減值及其他信貸風險準備上升1.67億 美元或35%，主要來自環球銀行及資本市場 業務以及工商金融業務，原因是香港及中 國內地少數個別評估減值準備增加，新西 蘭、馬來西亞及越南工商金融業務個別評 估減值準備減少，抵銷了上述部分增幅。 滙豐控股有限公司 88 董事會報告：地區 （續） 營業支出 （百萬美元） 10,427 9,936 (58) 10,369 (320) 9,616 3年 1 0 2 4年 1 0 2 列賬基準 貨幣換算及重大項目 經調整營業支出 營業支出上升7.53億美元，乃由於在區內投 入資源，尤其是監管計劃及合規方面的投 資，以及集團內增加使用環球服務中心。 成本增加亦反映工資上漲及增聘職員，尤 其是香港及中國內地，以支持主要來自工 商金融業務的業務增長，以及市場推廣活 動增加。2014年實現約2.7億美元的可持續 成本節約，抵銷了這些因素的部分影響。 應佔聯營及合資公司利潤上升7,100萬美元， 主要來自交通銀行，反映資產負債增長及 交易收益上升令收入增加，但營業支出及 貸款減值及其他信貸風險準備增加，抵銷 了部分利好影響。 亞洲 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 89 除稅前利潤及資產負債表數據－亞洲 2014年 零售銀行 環球銀行 環球 項目之間 及財富管理 工商金融 及資本市場 私人銀行 其他 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤 淨利息收益╱ （支出） 5,003 3,439 3,579 177 (16) 91 12,273 費用收益淨額 2,792 1,529 1,311 272 6 － 5,910 不包括淨利息收益之 交易收益╱ （支出） 216 382 1,220 142 (5) － 1,955 交易活動之淨利息收益╱ （支出） (13) (9) 771 － 9 (91) 667 交易收益淨額50 203 373 1,991 142 4 (91) 2,622 已發行長期債務及相關衍生工具 之公允值變動 － － － － (4) － (4) 指定以公允值列賬之 其他金融工具淨收益╱ （支出） 543 (6) (2) － 2 － 537 指定以公允值列賬之金融工具 淨收益╱ （支出） 543 (6) (2) － (2) － 533 金融投資減除虧損後增益 1 5 46 － 148 － 200 股息收益 1 － 1 － 177 － 179 保費收益淨額 6,596 794 － － － － 7,390 其他營業收益 516 95 141 3 2,734 (1,158) 2,331 ● ● ● ● ● ● ● 營業收益總額 15,655 6,229 7,067 594 3,051 (1,158) 31,438 保險賠償淨額56 (6,979) (782) － － － － (7,761) ● ● ● ● ● ● ● 營業收益淨額4 8,676 5,447 7,067 594 3,051 (1,158) 23,677 貸款減值準備 （提撥） ╱收回及 其他信貸風險準備 (317) (228) (103) 1 － － (647) ● ● ● ● ● ● ● 營業收益淨額 8,359 5,219 6,964 595 3,051 (1,158) 23,030 營業支出總額 (4,191) (1,897) (2,686) (384) (2,427) 1,158 (10,427) ● ● ● ● ● ● ● 營業利潤 4,168 3,322 4,278 211 624 － 12,603 應佔聯營及合資公司利潤 304 1,420 298 － － － 2,022 ● ● ● ● ● ● ● 除稅前利潤 4,472 4,742 4,576 211 624 － 14,625 ● ● ● ● ● ● ● % % % % % % 應佔滙豐除稅前利潤 23.9 25.4 24.5 1.1 3.4 78.3 成本效益比率 48.3 34.8 38.0 64.6 79.5 44.0 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 115,643 132,509 99,934 12,894 1,975 362,955 資產總值 166,577 158,747 548,865 14,905 79,477 (89,848) 878,723 客戶賬項27 286,670 155,608 104,896 29,847 470 577,491 滙豐控股有限公司 90 董事會報告：地區 （續） 2013年 零售銀行 環球銀行 環球 項目之間 及財富管理 工商金融 及資本市場 私人銀行 其他 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤 淨利息收益╱ （支出） 4,895 3,103 3,245 205 (124) 108 11,432 費用收益╱ （支出） 淨額 2,758 1,518 1,419 249 (8) － 5,936 不包括淨利息收益之 交易收益╱ （支出） 238 377 1,483 175 (739) － 1,534 交易活動之淨利息收益╱ （支出） (16) (6) 608 － 14 (108) 492 交易收益╱ （支出） 淨額50 222 371 2,091 175 (725) (108) 2,026 已發行長期債務及相關衍生工具 之公允值變動 － － － － (1) － (1) 指定以公允值列賬之 其他金融工具淨收益╱ （支出） 315 － 7 － (7) － 315 指定以公允值列賬之金融工具 淨收益╱ （支出） 315 － 7 － (8) － 314 金融投資減除虧損後增益 (1) － 58 14 1,204 － 1,275 股息收益 － 1 6 － 145 － 152 保費收益淨額 6,263 654 1 － － － 6,918 其他營業收益 764 97 163 12 3,871 (1,232) 3,675 ● ● ● ● ● ● ● 營業收益總額 15,216 5,744 6,990 655 4,355 (1,232) 31,728 保險賠償淨額56 (6,609) (687) － － － － (7,296) ● ● ● ● ● ● ● 營業收益淨額4 8,607 5,057 6,990 655 4,355 (1,232) 24,432 貸款減值及 其他信貸風險準備 (347) (144) (3) (4) － － (498) ● ● ● ● ● ● ● 營業收益淨額 8,260 4,913 6,987 651 4,355 (1,232) 23,934 營業支出總額 (4,138) (1,786) (2,560) (367) (2,317) 1,232 (9,936) ● ● ● ● ● ● ● 營業利潤 4,122 3,127 4,427 284 2,038 － 13,998 應佔聯營及合資公司 利潤╱ （虧損） 297 1,331 248 － (21) － 1,855 ● ● ● ● ● ● ● 除稅前利潤 4,419 4,458 4,675 284 2,017 － 15,853 ● ● ● ● ● ● ● % % % % % % 應佔滙豐除稅前利潤 19.6 19.8 20.7 1.3 8.9 70.3 成本效益比率 48.1 35.3 36.6 56.0 53.2 40.7 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 111,769 122,882 89,722 10,904 1,620 336,897 資產總值 158,456 146,898 515,023 12,994 82,453 (84,033) 831,791 客戶賬項27 278,392 141,958 96,546 31,250 337 548,483 有關註釋，請參閱第109頁。 亞洲╱中東及北非 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 91 經濟背景 2014年底中東及北非的地緣政治不穩定因 素加劇及全球油價轉弱，但年內該區經濟 活動仍然強勁。由於區內石油出口國採取 以石油收益刺激經濟及寬鬆貨幣的政策， 促使各國的表現極佳。中東最大的石油出 口國沙地阿拉伯增長強勁，乃由於該國推 進基建及工業擴張計劃。然而，阿拉伯聯 合酋長國 （ 「阿聯酋」 ） 由出口主導的非石油 行業增長及進一步擴張的財政政策，令該 國呈現最強勁的增長動力。儘管經濟增長 速度加快，但海灣地區通脹仍然乏力，維 持於5%以下。 於2014年，埃及出現更多經濟回穩跡象。 儘管仍低於2011年革命前的水平，惟下半 年增長動力再現，乃因進一步獲得經濟援 助，以及於5月總統選舉後政治秩序及決策 均有所改善。通脹上升及預算赤字仍然高 企，且以佔國內生產總值百分比計算，已 連續第三年錄得雙位數赤字。國際儲備於 年底前幾個月內下降，顯示受嚴格管制的 貨幣仍然受壓。 財務概覽 除稅前利潤 （百萬美元） 1,826 1,694 28 1,854 (21) 1,673 2014年 2013年 列賬基準 貨幣換算及重大項目 經調整利潤 儘管受出售業務影響 （包括出售巴基斯坦 業務錄得虧損） ，滙豐的中東及北非業務仍 錄得除稅前利潤18億美元，按列賬基準計 算增加8%。有關重大項目的進一步詳情， 請參閱第42頁。 按經調整基準計算，除稅前利潤增加11%， 乃由於收入增加及來自聯營公司沙地英國 銀行的收益上升。 中東及北非 中東滙豐銀行有限公司的分行網絡，聯 同滙豐的附屬及聯營公司，使滙豐擁有 區內覆蓋範圍廣闊的服務網絡。按資產 總值計，滙豐在沙地阿拉伯的聯營公司 沙地英國銀行 （擁有40%股權） 是該國第 六大銀行。 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 淨利息收益 1,519 1,486 1,470 費用收益淨額 650 622 595 交易收益淨額 314 357 390 其他收益 65 38 (25) 營業收益淨額4 2,548 2,503 2,430 貸款減值及其他 信貸風險準備43 6 42 (286) 營業收益淨額 2,554 2,545 2,144 營業支出總額 (1,216) (1,289) (1,166) 營業利潤 1,338 1,256 978 來自聯營公司收益44 488 438 372 除稅前利潤 1,826 1,694 1,350 成本效益比率 47.7% 51.5% 48.0% 風險加權資產 平均值回報36 2.9% 2.7% 2.2% 年底職員人數 8,305 8,618 8,765 中東最佳投資銀行 （2014年 《歐洲貨幣》 雜誌） 列賬基準之除稅前利潤創新紀錄達 18億美元 配合集團六方面考慮的投資準則， 完成出售約旦及巴基斯坦的業務 有關註釋，請參閱第109頁。 滙豐控股有限公司 92 董事會報告：地區 （續） 國家╱地區業務摘要 在阿聯酋，集團於2013年宣布開展的策略 計劃取得重大進展。零售銀行及財富管理 業務方面，我們增加財富管理產品種類， 包括推出國際債券及組合顧問服務，向卓 越理財客戶提供更多服務。集團引入財務 健全檢查服務以更深入了解客戶需求，並 在阿布達比設立客戶服務組，體現我們以 客為先的服務精神。 工商金融業務方面，我們透過於區內實施 國際附屬公司業務模式，加強為從事國際 貿易的客戶提供的服務，從而更能滿足其 跨境理財需求，鞏固我們的策略合作關係。 集團亦推出第二批國際成長基金10億迪拉 姆 （2.72億美元） 。我們繼續投入資源發展資 金管理業務，包括招聘前線及專責員工，並 獲評選為中東最佳區域資金管理服務供應 商。 環球銀行及資本市場業務方面，我們於一 家主要地區航空公司投資一家歐洲航空公 司時出任顧問，並為杜拜一家大型投資公 司首次發行10億美元債券提供顧問服務。 此外，我們加強與工商金融業務的合作， 特別在資本融資方面，專注服務現有客戶 及利用我們與其他地區的連繫。 油價下跌對我們於阿聯酋的財務業績並無 重大影響。 埃及的零售銀行及財富管理業務擴大了產 品種類、增強產品功能及降低信用卡定價， 並名列客戶推薦度指標排行榜的榜首。環 球銀行及資本市場業務方面，我們擔任一 個政府實體的全球協調人、結構設計銀行、 委託牽頭安排人及信貸代理。這反映我們 致力於支持埃及政府發展該國基礎設施的 計劃。 在沙地阿拉伯，集團透過聯營公司沙地英 國銀行及HSBC Saudi Arabia Limited擔任國 家商業銀行60億美元首次公開招股的聯席 財務顧問、聯席牽頭經辦人及收款銀行。這 是中東地區歷來最大規模及2014年世界第 二大規模的首次公開招股。 經調整業績回顧 45 收入 （百萬美元） 2,548 2,503 (3) 2,545 (101) 2,402 列賬基準 貨幣換算及重大項目 經調整收入 2014年 2013年 集團於大多數市場收入增加，尤以埃及的 所有環球業務及阿聯酋為甚。 國家╱地區經調整收入概覽 2014年 2013年 百萬美元 百萬美元 阿聯酋 1,448 1,401 埃及 531 451 中東及北非其他地區 566 550 截至12月31日止年度 2,545 2,402 在埃及，收入增加8,000萬美元，反映重新 定價導致存款息差擴闊，令零售銀行及財 按環球業務所在國家╱地區列示之除稅前利潤╱ （虧損） 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 埃及 64 94 177 － － 335 阿聯酋 154 190 364 － (46) 662 沙地阿拉伯 91 168 203 19 5 486 其他 14 152 182 － (5) 343 截至2014年12月31日止年度 323 604 926 19 (46) 1,826 埃及 31 37 166 － (29) 205 阿聯酋 142 290 275 1 (72) 636 沙地阿拉伯 82 146 188 15 7 438 其他 3 172 240 － － 415 截至2013年12月31日止年度 258 645 869 16 (94) 1,694 埃及 67 71 157 － (5) 290 阿聯酋 143 235 141 1 (56) 464 沙地阿拉伯 60 120 170 9 18 377 其他 (18) 161 113 － (37) 219 截至2012年12月31日止年度 252 587 581 10 (80) 1,350 中東及北非 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 93 富管理業務淨利息收益增加，及2013年出 售環球銀行及資本市場業務可供出售債務 證券的虧損不復再現。此外，央行於2014年 恢復支付隔夜存款利息，有助於所有環球 業務收入增加。 在阿聯酋，收入增加4,700萬美元，主要來 自環球銀行及資本市場業務，反映項目及 出口融資顧問業務量增加和信貸及貸款業 務重整一項特定貸款錄得增益，令資本融 資業務收入增加。此外，股票及證券業務 因客戶交易量增加令收入上升，部分反映 摩根士丹利資本國際指數將阿聯酋提升至 新興市場的地位。零售銀行及財富管理業 務方面，收入增加但幅度不大，反映按揭 結欠增加令淨利息收益上升及重新定價措 施令存款息差改善，但工商金融業務的收 入因貸款結欠的息差收窄而減少，抵銷了 上述部分增幅，反映市場資金充裕及競爭 激烈，以及資金管理業務外匯交易費減少。 至於區內其他地區，阿曼及卡塔爾收入增 加，而阿爾及利亞收入減少，部分抵銷了上 述增幅。阿曼收入增長部分反映工商金融 業務的客戶貸款增加。卡塔爾收入增加乃 由於環球銀行及資本市場業務費用收入增 加，反映證券服務業務客戶交易量增加， 部分原因是摩根士丹利資本國際指數將卡 塔爾提升至新興市場的地位。阿爾及利亞 收入減少反映監管當局限制對企業客戶交 易收取的外匯差價。 貸款減值撥回淨額減少4,400萬美元，主要 由於環球銀行及資本市場業務中有關阿聯 酋的特定貸款減值撥回額下降所致。 營業支出 （百萬美元） 列賬基準 貨幣換算及重大項目 經調整營業支出 1,216 1,289 (33) 1,183 (75) 1,214 2014年 2013年 營業支出減少3,100萬美元至11.83億美元， 主要由於埃及和阿聯酋減少支出所致。在 埃及，由於我們於2013年就有關稅務法規 詮釋改變錄得若干支出，年內支出相對減 少。在阿聯酋，支出減少乃由於2013年零售 銀行及財富管理業務就海外信用卡交易收 費承擔的客戶賠償計劃準備不復再現。工 資上漲、投資於監管計劃及合規部門、增 加零售銀行及財富管理業務的前線員工， 以及工商金融業務增聘人手提供服務及產 品支援，抵銷了上述部分減幅。 應佔聯營及合資公司利潤上升12%，主要來 自沙地英國銀行，此乃由於資產負債強勁 增長令收入增加。 滙豐控股有限公司 94 董事會報告：地區 （續） 除稅前利潤╱ （虧損） 及資產負債表數據－中東及北非 2014年 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 項目之間 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益 615 467 410 － 3 24 1,519 費用收益╱ （支出） 淨額 152 268 240 － (10) － 650 不包括淨利息收益之 交易收益╱ （支出） 58 68 207 － (5) － 328 交易活動之淨利息 收益 － － 10 － － (24) (14) 交易收益╱ （支出） 淨額50 58 68 217 － (5) (24) 314 指定以公允值列賬之 金融工具淨支出 － － － － (3) － (3) 金融投資減除虧損後增益 1 1 20 － － － 22 股息收益 1 1 12 － － － 14 其他營業收益 8 － 27 － 108 (111) 32 營業收益總額 835 805 926 － 93 (111) 2,548 保險賠償淨額56 － － － － － － － 營業收益淨額4 835 805 926 － 93 (111) 2,548 貸款減值準備 （提撥） ╱ 收回及其他信貸風險準備 (26) (21) 53 － － － 6 營業收益淨額 809 784 979 － 93 (111) 2,554 營業支出總額 (578) (348) (256) － (145) 111 (1,216) 營業利潤╱ （虧損） 231 436 723 － (52) － 1,338 應佔聯營及合資公司利潤 92 168 203 19 6 － 488 除稅前利潤╱ （虧損） 323 604 926 19 (46) － 1,826 % % % % % % 應佔滙豐除稅前利潤 1.7 3.2 5.0 0.1 (0.2) 9.8 成本效益比率 69.2 43.2 27.6 － 155.9 47.7 資產負債表數據40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 6,318 13,104 9,641 － － 29,063 資產總值 7,073 14,911 39,229 77 2,900 (1,773) 62,417 客戶賬項27 18,024 11,809 9,630 － 257 39,720 中東及北非 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 95 2013年 零售銀行 及財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 項目之間 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益 585 486 390 － 4 21 1,486 費用收益╱ （支出） 淨額 161 269 197 － (5) － 622 不包括淨利息收益之 交易收益 59 85 220 － － － 364 交易活動之淨利息 收益 － － 14 － － (21) (7) 交易收益淨額50 59 85 234 － － (21) 357 指定以公允值列賬之 金融工具淨支出 － － － － (2) － (2) 金融投資減除虧損後增益 － － (18) － － － (18) 股息收益 － － 9 － － － 9 其他營業收益 25 30 15 － 99 (120) 49 營業收益總額 830 870 827 － 96 (120) 2,503 保險賠償淨額56 － － － － － － － 營業收益淨額4 830 870 827 － 96 (120) 2,503 貸款減值準備 （提撥） ╱ 收回及其他信貸風險準備 (49) (20) 110 1 － － 42 營業收益淨額 781 850 937 1 96 (120) 2,545 營業支出總額 (606) (350) (256) － (197) 120 (1,289) 營業利潤╱ （虧損） 175 500 681 1 (101) － 1,256 應佔聯營及合資公司利潤 83 145 188 15 7 － 438 除稅前利潤╱ （虧損） 258 645 869 16 (94) － 1,694 % % % % % % 應佔滙豐除稅前利潤 1.1 2.9 3.8 0.1 (0.4) 7.5 成本效益比率 73.0 40.2 31.0 － 205.2 51.5 資產負債表數據40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 6,152 11,814 9,241 － 4 27,211 資產總值 7,016 13,776 39,302 64 3,340 (2,688) 60,810 客戶賬項27 18,771 12,402 7,432 1 77 38,683 有關註釋，請參閱第109頁。 董事會報告：地區 （續） 滙豐控股有限公司 96 北美洲 北美洲 滙豐的北美洲主要業務遍布美國及加拿 大。美國方面，滙豐主要透過美國滙豐 銀行，以及服務網絡跨越全美國的消費 融資公司美國滙豐融資經營業務。HSBC Markets (USA) Inc. 是 （其中包括） HSBC Securities (USA) Inc.的中介控股公司。加 拿大業務乃透過加拿大滙豐銀行經營。 2014年 2013年 2012年 百萬美元百萬美元 百萬美元 淨利息收益 5,015 5,742 8,117 費用收益淨額 1,940 2,143 2,513 交易收益淨額 411 948 507 出售美國分行網絡及卡業務 所得利潤 － － 4,012 其他收益╱ （支出） 786 (30) (456) 營業收益淨額4 8,152 8,803 14,693 貸款減值及其他信貸 風險準備43 (322) (1,197) (3,457) 營業收益淨額 7,830 7,606 11,236 營業支出總額 (6,429) (6,416) (8,940) 營業利潤 1,401 1,190 2,296 來自聯營公司收益44 16 31 3 除稅前利潤 1,417 1,221 2,299 ae 成本效益比率 78.9% 72.9% 60.8% 風險加權資產平均值回報36 0.6% 0.5% 0.8% 年底職員人數 20,412 20,871 22,443 北美洲最佳出口融資安排行 （2014年貿易融資卓越大獎） 工商金融業務客戶貸款結欠增加 11% 按列賬基準計算 貸款減值準備减少 73% 按列賬基準計算 有關註釋，請參閱第109頁。 經濟背景 在美國，2014年實質國內生產總值增長 2.4%，2013年則增長2.2%。消費開支及 商業固定投資於2014年溫和增長，分別上 升2.5%及5.2%。然而，住宅投資增長急速 下滑，由2013年的11.9%放緩至2014年的 1.8%。由於州份及地方政府支出增長被聯 邦政府支出縮減抵銷有餘，因此政府支出 於2014年下降0.2%。失業率由2013年底的 6.7%降至2014年底的5.6%。2014年消費物 價指數上漲率平均為1.6%，2013年則平均 為1.5%。於2014年，聯邦儲備局繼續維持 極度寬鬆的貨幣政策，將聯邦基金利率保 持在零至0.25厘區間內，並於今年首十個 月逐步減少每月購買長期國庫證券及機構 按揭抵押證券，至10月底結束其資產購買 計劃。 加拿大的實質國內生產總值於2014年首三 個季度增長年率為2.4%，較2013年同期的 增幅1.8%有所改善。出口受美國經濟增長 及石油產量增加所支持。商業投資於2014 年大致保持不變。全年消費物價指數上漲 率於2013年底及2014年初升至2.4%高位。 然而，由於油價於年底下跌，燃料價格下 降，全年通脹率因此降至12月的1.5%，低 於加拿大銀行的2%通脹目標。加拿大銀行 自2010年9月起一直維持1厘的政策利率， 於整個2014年貨幣政策仍維持寬鬆。 財務概覽 除稅前利潤 （百萬美元） 1,417 1,221 694 2,111 827 2,048 列賬基準 貨幣換算及重大項目 經調整利潤 2014年 2013年 於2014年，滙豐的北美洲業務錄得除稅前 利潤14億美元，而2013年則為12億美元， 增加了1.96億美元，主要反映了貸款減值 及其他信貸風險準備減少 （減幅主要在於美 國的消費及按揭貸款組合） 。而收入減少， 主要反映了美國持續縮減消費及按揭貸款 以及縮減環球銀行及資本市場業務，抵銷 了上述部分增幅。成本大致保持不變，原 因是組合縮減大致抵銷了有關與聯邦房屋 金融局達成和解的5.5億美元準備。 有關重大項目的詳情，請參閱第42頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 97 經調整除稅前利潤增加6,300萬美元，反映 了貸款減值及其他信貸風險準備及營業支 出減少，惟收入減少而抵銷了部分減幅。 國家╱地區業務摘要 在美國，工商金融業務於2014年增加40億美 元中小企貸款資金，支援從事或有意從事 國際貿易的企業，令有關計劃的可供動用 貸款總額提高至50億美元，並已於2014年 12月31日動用當中的37億美元。有賴於目標 市場擴展持續順利，企業貸款結欠增長， 在中西部及西岸地區的結欠按年增長超過 25%。 在零售銀行及財富管理業務方面，我們繼 續優化按揭辦理流程，以改善客戶體驗及 擴大電子銀行服務。我們重新推出環球卓 越理財客戶計劃，連同其他相關計劃，令 新卓越理財客戶於2014年增加約22,000名， 增長25%。 儘管環球銀行及資本市場業務收入減少， 但我們仍繼續執行增長策略，善用環球銀 行及資本市場業務的獨特客戶網絡、其地 區網絡及產品能力，以聯繫各個市場。除 此之外，與工商金融業務的合作令來自其 客戶的收入增長19%。 在加拿大，工商金融業務繼續專注於吸納新 客戶，對新客戶的貸款達13億美元以上。我 們設立了專責的國際附屬公司銀行團隊， 持續管理及支援國際客戶。環球銀行及資 本市場業務專注於擴大其跨國客戶基礎， 而項目及出口融資業務繼續反映增長。在 零售銀行及財富管理業務方面，我們繼續 投入精力擴大卓越理財客戶基礎，增加按 揭結欠、信用卡結欠及存款結餘，並增加 管理資產。 我們繼續在策略方面取得進展，加快縮減及 出售美國消費及按揭貸款組合。2014年，我 們完成出售數批有抵押房地產賬項，有關 未付本金結欠總額為29億美元，並確認累計 出售利潤1.68億美元。於2014年12月31日， 消費及按揭貸款組合的貸款結欠總額 （包括 持作出售用途貸款） 為250億美元，較2013 年減少58億美元。 經調整業績回顧 45 收入 （百萬美元） 貨幣換算及重大項目 經調整收入 列賬基準 8,152 8,803 116 8,268 536 9,339 2014年 2013年 在美國的零售銀行及財富管理業務收入下 降，部分反映了消費及按揭貸款組合持續 縮減，環球銀行及資本市場業務收入亦下 降。來自加拿大的收入亦減少，主要反映 了消費融資業務持續縮減。 國家╱地區經調整收入概覽 2014年 2013年 百萬美元 百萬美元 美國 6,083 7,071 加拿大 1,921 1,975 其他 264 293 截至12月31日止年度 8,268 9,339 在美國，收入減少9.88億美元，主要見於零 售銀行及財富管理業務。消費及按揭貸款 組合持續縮減及出售貸款令貸款平均結欠 減少，進而令這方面的淨利息收益減少。 此外，貸款收益下降，部分反映了出售高 收益率的消費及按揭貸款非房地產個人貸 款組合，令產品組合出現重大轉向，使低 收益率第一留置權房地產貸款增加。存款 按環球業務所在國家╱地區列示之除稅前利潤╱ （虧損） 零售銀行 環球銀行 環球 及財富管理 工商金融 及資本市場 私人銀行 其他 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 美國 513 400 (403) 82 (60) 532 加拿大 96 514 242 － (23) 829 其他 23 (1) 49 3 (18) 56 截至2014年12月31日止年度 632 913 (112) 85 (101) 1,417 美國 (358) 296 633 53 (350) 274 加拿大 131 506 280 － (3) 914 其他 20 (16) 16 4 9 33 截至2013年12月31日止年度 (207) 786 929 57 (344) 1,221 美國 2,746 637 661 72 (2,901) 1,215 加拿大 207 577 314 (1) (16) 1,081 其他 42 (15) (18) 1 (7) 3 截至2012年12月31日止年度 2,995 1,199 957 72 (2,924) 2,299 董事會報告：地區 （續） 滙豐控股有限公司 98 北美洲 量減少及存款息差收窄亦令收入減少。收 入的部分降幅被解除與之前出售的貸款有 關的按揭貸款回購責任所抵銷，而2013年 則須提撥準備。 環球銀行及資本市場業務的收入減少，原 因在於為風險管理目的而持續將業務組合 重新定位，令可供出售債務證券按列賬基 準計算的出售利潤減少，致使資產負債管 理業務收益減少，以及管理利率風險所用 的經濟對沖表現欠佳。信貸收入亦減少， 主要見於既有信貸組合，部分反映了組合 的不利公允值變動淨額。 與之相反，工商金融業務收入增加，主要 反映了擴展目標市場的貸款結欠增加，及 收購融資活動中的合作增加，令環球銀行 及資本市場業務的收益增加。 在加拿大，收入減少5,400萬美元，主要見 於零售銀行及財富管理業務方面，反映了 淨利息收益下降，原因是持續縮減消費融 資業務令貸款平均結欠減少。若不計及上 述因素，零售銀行及財富管理業務收入增 長，原因是費用增加，部分反映了增加出 售財富管理產品。在工商金融業務方面， 收入亦告增加，大部分是因為持作出售用 途且於2013年確認的投資物業公允值減少 不復再現。與之相反，環球銀行及資本市 場業務收入減少，反映了外匯交易收益減 少及可供出售債務證券按列賬基準計算的 出售利潤減少。 貸款減值及其他信貸風險準備下降，主要見 於消費及按揭貸款組合方面，反映了拖欠 水平下降、新增已減值貸款下跌，以及持續 縮減及出售貸款令貸款結欠減少。2014年 房屋市場復甦幅度未及2013年明顯，導致 對相關物業價格作出不利市場價值調整， 抵銷了上述部分降幅。主要零售銀行及財 富管理業務的貸款減值及其他信貸風險準 備亦下降，主要反映了拖欠水平下降，而 加拿大的工商金融業務亦錄得貸款減值及 其他信貸風險準備下降，原因是個別及綜 合評估減值準備減少。 營業支出 （百萬美元） 列賬基準 貨幣換算及重大項目 經調整營業支出 6,429 6,416 (578) 5,851 (280) 6,136 2014年 2013年 營業支出減少2.85億美元，主要見於美國， 反映了過往卡業務接近完成數據分離流 程，令退出投資支出減少，此外，我們持 續縮減及出售若干批次的消費及按揭貸款 組合，而令平均職員人數及支出下降。此 外，我們亦實現超過1.85億美元的可持續 成本節約，主要得益於優化架構措施。營 業支出的部分降幅被法律成本增加及與監 管計劃及合規有關的成本上升所抵銷，反 映了我們持續投資於環球標準。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 99 除稅前利潤╱ （虧損） 及資產負債表數據－北美洲 2014年 零售銀行 環球銀行 環球 項目之間 及財富管理 工商金融 及資本市場 私人銀行 其他 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益 2,645 1,455 587 204 157 (33) 5,015 費用收益╱ （支出） 淨額 497 572 775 130 (34) － 1,940 不包括淨利息收益之 交易收益╱ （支出） (165) 34 302 13 3 － 187 交易活動之淨利息收益╱ （支出） 7 1 183 － － 33 224 交易收益╱ （支出） 淨額50 (158) 35 485 13 3 33 411 已發行長期債務及相關 衍生工具之公允值變動 － － － － (99) － (99) 指定以公允值列賬之其他 金融工具淨收益 － － － － － － － 指定以公允值列賬之 金融工具淨支出 － － － － (99) － (99) 金融投資減除虧損後增益 － 15 237 － 5 － 257 股息收益 13 8 16 3 4 － 44 保費收益淨額 － － － － － － － 其他營業收益 268 61 101 1 1,872 (1,719) 584 營業收益總額 3,265 2,146 2,201 351 1,908 (1,719) 8,152 保險賠償淨額56 － － － － － － － 營業收益淨額4 3,265 2,146 2,201 351 1,908 (1,719) 8,152 貸款減值準備 （提撥） ╱收回及 其他信貸風險準備 (117) (148) (63) 8 (2) － (322) 營業收益淨額 3,148 1,998 2,138 359 1,906 (1,719) 7,830 營業支出總額 (2,516) (1,101) (2,250) (274) (2,007) 1,719 (6,429) 營業利潤╱ （虧損） 632 897 (112) 85 (101) － 1,401 應佔聯營及合資公司利潤 － 16 － － － － 16 除稅前利潤╱ （虧損） 632 913 (112) 85 (101) － 1,417 % % % % % % 應佔滙豐除稅前利潤 3.4 4.9 (0.6) 0.5 (0.5) 7.6 成本效益比率 77.1 51.3 102.2 78.1 105.2 78.9 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 60,365 41,966 21,110 6,346 － 129,787 資產總值 74,680 48,411 319,819 8,386 16,823 (31,260) 436,859 客戶賬項27 51,258 45,275 30,301 12,050 － 138,884 滙豐控股有限公司 100 董事會報告：地區 （續） 北美洲╱拉丁美洲 2013年 零售銀行 環球銀行 環球 項目之間 及財富管理 工商金融 及資本市場 私人銀行 其他 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益 3,483 1,430 582 195 89 (37) 5,742 費用收益淨額 605 593 741 125 79 － 2,143 不包括淨利息收益之 交易收益 48 40 613 19 7 － 727 交易活動之淨利息收益 11 1 172 － － 37 221 交易收益淨額50 59 41 785 19 7 37 948 已發行長期債務及相關 衍生工具之公允值變動 － － － － (288) － (288) 指定以公允值列賬之其他 金融工具淨收益 － － － － － － － 指定以公允值列賬之 金融工具淨支出 － － － － (288) － (288) 金融投資減除虧損後增益 4 － 282 － 8 － 294 股息收益 12 9 48 4 4 － 77 保費收益淨額 34 － － － － － 34 其他營業收益╱ （支出） (454) － 229 1 1,829 (1,713) (108) A^ A^ A^ A^ A^ A^ A^ 營業收益總額 3,743 2,073 2,667 344 1,728 (1,713) 8,842 保險賠償淨額56 (39) － － － － － (39) 營業收益淨額4 3,704 2,073 2,667 344 1,728 (1,713) 8,803 貸款減值及 其他信貸風險準備 (950) (223) (20) (4) － － (1,197) 營業收益淨額 2,754 1,850 2,647 340 1,728 (1,713) 7,606 營業支出總額 (2,960) (1,096) (1,718) (283) (2,072) 1,713 (6,416) 營業利潤╱ （虧損） (206) 754 929 57 (344) － 1,190 應佔聯營及合資公司利潤╱ （虧損） (1) 32 － － － － 31 除稅前利潤╱ （虧損） (207) 786 929 57 (344) － 1,221 ?
?
?
?
? ?
% % % % % % 應佔滙豐除稅前利潤 (0.9) 3.5 4.1 0.3 (1.6) 5.4 成本效益比率 79.9 52.9 64.4 82.3 119.9 72.9 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 66,192 37,735 18,070 5,956 － 127,953 資產總值 82,530 45,706 313,701 8,542 13,211 (31,655) 432,035 客戶賬項27 53,600 49,225 24,113 13,871 － 140,809 有關註釋，請參閱第109頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 101 拉丁美洲 我們的拉丁美洲業務主要包括巴西滙豐 銀行、墨西哥滙豐及阿根廷滙豐銀行。 除銀行服務外，滙豐在巴西、墨西哥及 阿根廷亦經營保險業務。 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 淨利息收益 5,310 6,186 6,984 費用收益淨額 1,415 1,701 1,735 交易收益淨額 856 936 971 其他收益 691 1,745 1,261 營業收益淨額4 8,272 10,568 10,951 貸款減值及其他信貸 風險準備43 (2,124) (2,666) (2,137) 營業收益淨額 6,148 7,902 8,814 營業支出總額 (5,932) (5,930) (6,430) 營業利潤 216 1,972 2,384 來自聯營公司收益44 － － － 除稅前利潤 216 1,972 2,384 成本效益比率 71.7% 56.1% 58.7% 風險加權資產平均值回報36 0.2% 2.0% 2.4% 年底職員人數 41,201 42,542 46,556 在重新定位巴西及墨西哥 業務方面取得更大進展 年度最佳貸款行及債券行 （ 《拉丁財務》 雜誌，2014年） 阿根廷及墨西哥國內資金管理業務 第一位 （ 《歐洲貨幣》 雜誌資金管理調查，2014年） 有關註釋，請參閱第109頁。 經濟背景 2014年第三季的數據表明，拉丁美洲可能 面臨嚴重經濟下滑，平均實質國內生產總 值增長由2013年的2.6%降至2014年的接近 1.0%。 巴西經濟下滑很好地說明了上述疲弱情 況。繼2013年增長2.5%後，經濟活動水平於 2014年大致不變，但商業信心轉差，以及 商業投資支出相應收縮是經濟下滑的主要 因素。為減低貨幣走弱的通脹壓力，中央 銀行於第四季將其主要政策利率調高75個 基點至11.75厘。 繼2013年取得僅1.1%的實質國內生產總值增 長後，墨西哥經濟增長於2014年加速。2013年 表現疲弱的主要領域消費支出於今年加速 增長，而美國需求上升刺激了出口。通脹 壓力仍弱，墨西哥中央銀行於年初將其主 要政策利率由3.5厘調低至3厘。 阿根廷經濟於2014年收縮，原因是商品價 格下降、巴西經濟沉滯及阿根廷以美元計 值的外債發生技術性違責。2014年初阿根 廷披索大幅貶值使通脹加劇。 財務概覽 除稅前利潤 （百萬美元） 216 1,972 108 324 (1,322) 650 列賬基準 貨幣換算及重大項目 經調整利潤 2014年 2013年 在拉丁美洲，2014年列賬基準除稅前利潤 為2.16億美元，而於2013年則為20億美元。 該減少的原因是收入下降，主要受2013年 巴拿馬業務的出售利潤11億美元不復再現 所致，並被貸款減值及其他信貸風險準備 減少抵銷了部分降幅。 經調整除稅前利潤減少3.26億美元，且包括 在巴西的除稅前虧損。利潤減少主要反映 營業支出增加，主要由於在巴西及阿根廷 的通脹及與工會達成協議調高工資及隨著 我們在業務重新定位方面取得進展令在墨 西哥及巴西的收入減少。在阿根廷的收入 增加以及主要在墨西哥的貸款減值及其他 信貸風險準備減少，部分抵銷了上述因素。 滙豐控股有限公司 102 董事會報告：地區 （續） 拉丁美洲 國家╱地區業務摘要 2014年，在巴西、墨西哥及阿根廷這些優 先發展市場，我們繼續實施策略措施改善 日後回報，並同時面對經濟及通脹壓力。 在巴西，我們在大力推進業務變革方面取 得進展，確保實現長期持續發展。在零售 銀行及財富管理業務方面，我們一直在更 新業務模式，使客戶經理集中服務特定客 戶群，以更好地滿足客戶需要。我們亦更新 貸款產品特點，以提升競爭力，例如延長 部分個人貸款期限，並進一步增強零售信 貸能力，以改善辦理貸款的質素。我們繼 續精簡分行網絡，隨著我們集中精力於收 入快速增長的組合，在增長潛力較低的地 區關閉21家分行，並推出專注銷售及自動 交易的60個客戶服務單位。在工商金融業 務方面，我們增強了中型企業市場業務， 而在零售銀行及財富管理市場方面，在過 去兩年經歷收縮後，貸款錄得4%的增長。 此外，環球銀行及資本市場業務的客戶活 動增加，主要在利率交易業務方面。 在墨西哥，雖然收入溫和，我們仍專注於實 現可持續增長。在零售銀行及財富管理業 務方面，我們引入專責於運籌理財業務的 客戶經理，以提高生產力及改善客戶體驗。 我們推出結欠轉移推廣活動，選擇性地提 高部分低風險客戶的信貸限額，實現按揭 結欠增長5%，反映了具競爭力的定價。在 工商金融業務方面，我們改進商務理財業 務的流程，使客戶經理更好地支援客戶。 在環球銀行及資本市場業務方面，貸款結 欠上升48%，乃由於2014年下半年能源改革 後採取新業務措施。我們在對業務重新定 位方面取得重大進展，減少了客戶數目， 並繼續專注於精簡架構、管理成本及加強 風險管理及監控。 在阿根廷，由於經濟環境仍充滿挑戰，因此 我們繼續審慎管理業務。我們專注於環球 銀行及資本市場業務及工商金融業務企業 客戶的業務增長，於零售銀行及財富管理 業務和商務理財業務，則繼續遵循謹慎貸 款政策。我們在貿易及外匯領域仍保持領 先地位。 經調整業績回顧 45 收入 （百萬美元） 8,272 10,568 (19) 8,253 (2,493) 8,075 列賬基準 貨幣換算及重大項目 經調整收入 2014年 2013年 在阿根廷的收入增加，乃歸因於環球銀行 及資本市場業務的理想業績以及零售銀行 及財富管理業務和工商金融業務增長，而 墨西哥的所有環球業務及主要在巴西的工 商金融業務和環球銀行及資本市場業務的 收入減少抵銷了部分增幅。 國家╱地區經調整收入概覽 2014年 2013年 百萬美元 百萬美元 阿根廷 1,070 718 巴西 4,821 4,932 墨西哥 2,304 2,479 其他 58 (54) 截至12月31日止年度 8,253 8,075 按環球業務所在國家╱地區列示之除稅前利潤╱ （虧損） 零售銀行 環球銀行 環球 及財富管理 工商金融 及資本市場 私人銀行 其他 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 阿根廷 52 135 219 － (22) 384 巴西 (174) (153) 115 (2) (33) (247) 墨西哥 36 (52) 89 (2) (20) 51 其他 (4) 7 27 － (2) 28 截至2014年12月31日止年度 (90) (63) 450 (4) (77) 216 阿根廷 97 142 170 － (1) 408 巴西 (114) (43) 514 5 (11) 351 墨西哥 154 (160) 115 (3) 11 117 其他 289 525 368 (1) (85) 1,096 截至2013年12月31日止年度 426 464 1,167 1 (86) 1,972 阿根廷 209 169 174 － (46) 506 巴西 94 359 696 17 (43) 1,123 墨西哥 338 176 201 2 (18) 699 其他 (33) 47 82 1 (41) 56 截至2012年12月31日止年度 608 751 1,153 20 (148) 2,384 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 103 在阿根廷，收入增長3.52億美元，主要在環 球銀行及資本市場業務，零售銀行及財富 管理業務和工商金融業務亦取得增長。在 環球銀行及資本市場業務方面，收入增加 反映利好的交易業績及資產負債管理業務 的收入增加，因為在波動的市場中與短期 資金有關的業務量及息差上升。 零售銀行及財富管理業務收入增加，主要 原因是投資收益增加令保險收入錄得增 長，反映了債券市場的變動。此外，淨利 息收益增加令收入上升，乃受利率上升令 息差擴闊以及平均存款結餘增加所帶動。 在工商金融業務方面，收入增加，原因是 淨利息收益增長，反映了利率上升令息差 擴闊、貸款平均結欠增加及資金管理服務 下的存款結餘增加。存款額增長亦令資金 管理及貿易產品所得費用增加。 在墨西哥，收入減少1.75億美元，主要在零 售銀行及財富管理業務，其次是在工商金 融業務和環球銀行及資本市場業務。 在零售銀行及財富管理業務方面，收入下 降，主要由於保險業務銷售量減少。收入 亦因我們繼續在對業務重新定位方面取得 進展而受到不利影響。此外，利率下降後， 我們在往來戶口、儲蓄及存款方面面臨負 債息差收窄，雖然此影響被按揭結欠增加 所部分抵銷。 在工商金融業務方面，淨利息收益減少，原 因是資產息差收窄及貸款平均結欠減少， 尤其在商務理財方面，我們繼續為業務重 新定位，少數大型企業提前還款，若干房 屋建築商的部分貸款已被撇銷。存款息差 因利率下降而收窄，亦使淨利息收益受到 不利影響。此外，由於我們繼續對業務重 新定位，戶口服務及資金管理費用減少令 費用收益減少，反映了客戶減少。 在環球銀行及資本市場業務方面，收入下 降，主要是因為市場變動，從而影響了交 易對手信貸息差，導致信貸估值調整準備 增加及可供出售證券的出售利潤減少。 在巴西，工商金融業務和環球銀行及資本 市場業務的收入減少，而零售銀行及財富 管理業務的收入大致保持不變。在工商金 融業務方面，由於組合變動反映低收益率 中型企業業務的增長，儘管整體貸款結欠 增加，收入仍減少。 在環球銀行及資本市場業務方面，資產負 債管理業務的收入減少，雖然在客戶活動 增加的帶動下，部分減幅被利率交易業務 收入增長所抵銷。零售銀行及財富管理業 務的收入大致保持不變。與2013年的不利 變動相比，PVIF資產的有利變動令保險收 入增加，但被若干產品的費用收益減少抵 銷了增幅，部分反映客戶組合轉向更穩健 的低收益率資產及市場競爭激烈。 貸款減值及其他信貸風險準備下降，主要 在墨西哥，其次在巴西。 在墨西哥，貸款減值及其他信貸風險準備 改善，原因是工商金融業務 （尤其是2013年 公共房屋政策改變後對若干房屋建築商評 估的準備下降） 和環球銀行及資本市場業 務 （原因是2013年入賬的大額特別準備不復 再現） 的個別評估準備減少。 在巴西，該下降乃歸因於零售銀行及財富 管理業務和工商金融業務於2013年更改重 整貸款賬項組合的減值模型及修訂所用假 設。此外，工商金融業務的綜合評估減值 減少，尤其在商務理財方面，反映拖欠率 下降，但被個別的評估減值及對擔保所作 準備所帶動的環球銀行及資本市場業務增 幅部分抵銷了上述減幅。 營業支出 （百萬美元） 5,932 5,930 (125) 5,807 (919) 5,011 列賬基準 貨幣換算及重大項目 經調整營業支出 2014年 2013年 營業支出增加7.96億美元，主要在巴西及阿 根廷，很大原因是與工會達成協議調高工 資及通脹壓力。此外，我們在阿根廷的交 易稅項增加，與在該地區的收入增長及基 礎設施成本增加相符。我們亦於2014年在 巴西產生特別成本，涉及零售銀行及財富 管理業務加速折舊準備及無形資產減值。 儘管存在上述因素，我們繼續嚴格控制成 本，並在簡化流程的策略重點上取得進展， 實現可持續成本節約逾1.55億美元。 滙豐控股有限公司 104 董事會報告：地區 （續） 拉丁美洲 除稅前利潤╱ （虧損） 及資產負債表數據－拉丁美洲 2014年 零售銀行 環球銀行 環球 項目之間 及財富管理 工商金融 及資本市場 私人銀行 其他 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益 3,323 1,529 490 19 9 (60) 5,310 費用收益淨額 771 469 147 28 － － 1,415 不包括淨利息收益之 交易收益╱ （支出） 123 103 391 3 (1) － 619 交易活動之淨利息收益╱ （支出） 1 4 174 － (2) 60 237 交易收益╱ （支出） 淨額50 124 107 565 3 (3) 60 856 已發行長期債務及相關衍生工具之 公允值變動 － － － － － － － 指定以公允值列賬之其他金融工具 淨收益 516 175 － － － － 691 指定以公允值列賬之金融工具 淨收益 516 175 － － － － 691 金融投資減除虧損後增益 － － 84 － － － 84 股息收益 6 2 1 － － － 9 保費收益淨額 1,233 285 5 － － － 1,523 其他營業收益 54 47 19 － 213 (184) 149 營業收益總額 6,027 2,614 1,311 50 219 (184) 10,037 保險賠償淨額56 (1,410) (352) (3) － － － (1,765) 營業收益淨額4 4,617 2,262 1,308 50 219 (184) 8,272 貸款減值準備 （提撥） ╱收回及 其他信貸風險準備 (1,091) (776) (252) (5) － － (2,124) 營業收益淨額 3,526 1,486 1,056 45 219 (184) 6,148 營業支出總額 (3,616) (1,549) (606) (49) (296) 184 (5,932) 營業利潤╱ （虧損） (90) (63) 450 (4) (77) － 216 應佔聯營及合資公司利潤 － － － － － － － 除稅前利潤╱ （虧損） (90) (63) 450 (4) (77) － 216 % % % % % % 應佔滙豐除稅前利潤 (0.5) (0.3) 2.4 － (0.5) 1.1 成本效益比率 78.3 68.5 46.3 98.0 135.2 71.7 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 12,306 20,078 10,642 96 － 43,122 資產總值 29,074 29,851 55,827 298 1,155 (851) 115,354 客戶賬項27 23,056 15,125 8,219 2,188 － 48,588 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 105 2013年 零售銀行 環球銀行 環球 項目之間 及財富管理 工商金融 及資本市場 私人銀行 其他 互相撇銷 55 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 除稅前利潤╱ （虧損） 淨利息收益╱ （支出） 3,776 1,828 775 24 (12) (205) 6,186 費用收益淨額 952 548 168 32 1 － 1,701 不包括淨利息收益之 交易收益╱ （支出） 138 117 456 4 (4) － 711 交易活動之淨利息收益 － － 20 － － 205 225 交易收益╱ （支出） 淨額50 138 117 476 4 (4) 205 936 已發行長期債務及相關衍生工具之 公允值變動 － － － － － － － 指定以公允值列賬之其他金融工具 淨收益 264 61 1 － － － 326 指定以公允值列賬之金融工具 淨收益 264 61 1 － － － 326 金融投資減除虧損後增益 － 1 81 － － － 82 股息收益 5 3 1 － － － 9 保費收益淨額 1,464 360 6 － － － 1,830 其他營業收益 312 485 310 1 196 (189) 1,115 營業收益總額 6,911 3,403 1,818 61 181 (189) 12,185 保險賠償淨額56 (1,323) (291) (3) － － － (1,617) 營業收益淨額4 5,588 3,112 1,815 61 181 (189) 10,568 貸款減值及 其他信貸風險準備 (1,552) (1,062) (52) － － － (2,666) 營業收益淨額 4,036 2,050 1,763 61 181 (189) 7,902 營業支出總額 (3,610) (1,586) (596) (60) (267) 189 (5,930) 營業利潤╱ （虧損） 426 464 1,167 1 (86) － 1,972 應佔聯營及合資公司利潤 － － － － － － － 除稅前利潤╱ （虧損） 426 464 1,167 1 (86) － 1,972 % % % % % % 應佔滙豐除稅前利潤 1.9 2.0 5.2 － (0.4) 8.7 成本效益比率 64.6 51.0 32.8 98.4 147.5 56.1 資產負債表數據 40 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 （淨額） 27 13,616 19,923 10,304 75 － 43,918 資產總值 30,584 30,001 52,977 337 634 (534) 113,999 客戶賬項27 23,943 16,593 8,994 1,859 － 51,389 有關註釋，請參閱第109頁。 董事會報告：其他資料 滙豐控股有限公司 106 其他資料 管理資金及託管資產 106 按國家╱地區列示之已付稅款 106 物業 107 資料披露理念 107 按強化信息披露工作組的建議 作出的資料披露 108 管理資金及託管資產 管理資金 59 2014年 2013年 十億美元 十億美元 管理資金 於1月1日 921 910 新增資金淨額 38 (18) 價值變動 40 34 匯兌及其他 (45) (5) 於12月31日 954 921 2014年 2013年 十億美元 十億美元 按業務類別劃分管理資金 環球投資管理 445 420 環球私人銀行 275 282 聯屬機構 5 5 其他 229 214 於12月31日 954 921 有關註釋，請參閱第109頁。 於2014年12月31日，管理資金為9,540億美 元，增加4%，主要由於本年度有利市場的 變動及流入淨額所致。 環球投資管理業務的管理資金增加6%至 4,450億美元，乃由於我們吸納新增資金淨 額290億美元 （來自歐洲及亞洲客戶的定息 產品尤為顯著） ，以及歐洲及北美洲的流 動資金流入淨額。此外，我們轉撥先前於 「其他管理資金」 中呈報的管理資金180億 美元及我們受惠於股票及債券市場的有利 變動。該等增加部分由不利的匯兌變動 （反 映美元兌所有主要貨幣處於強勢） 所抵銷。 環球私人銀行業務的管理資金減少3%至 2,750億美元，原因為我們繼續為客戶基礎 重新定位。按照該等措施，我們向LGT Bank (Switzerland) Ltd出售瑞士資產組合及出售 盧森堡HSBC Trinkaus & Burkhardt AG的業 務 （管理資金合計80億美元） ；該等措施亦 使歐洲新增資金淨額錄得負數值。此外， 管理資金減少亦由於出現不利匯兌變動 ， 尤以歐洲為主。該等因素部分被有利的市 場變動 （歐洲尤為明顯） 及目標增長範疇錄 得正數值的新增資金淨額所抵銷。 其他管理資金增加7%至2,290億美元，主要由 於強勁的淨資金流入及有利的市場變動。 該等因素部分由上文所述管理資金轉撥至 環球投資管理業務所抵銷。 託管 59及管理資產 託管服務是代客戶保管及管理證券及其他金 融資產。於2014年12月31日，我們以託管人 身分持有的資產達6.4萬億美元，較於2013年 12月31日持有的6.2萬億美元增加3%，主要 由於亞洲及歐洲資產流入淨額增加，尤其 受中東及北非所帶動，部分被不利的匯兌 變動所抵銷。 為配合託管業務的營運，我們提供的管理 資產服務還包括提供債券及貸款管理服 務，以及代客戶為證券及其他金融資產組 合估值。於2014年12月31日，集團的管理資 產價值達3.2萬億美元，較2013年12月31日增 加6%，主要由歐洲及亞洲基金業務資產流 入淨額增加所帶動，部分被不利的匯兌變 動所抵銷。 按國家╱地區列示之已付 稅款 下表反映滙豐業務的地理分布，當中的編 製基準與集團履行歐盟資本指引4第89條所 列明按國家╱地區呈報責任之方法一致。 按地區列示之已付稅款分析 60 2014年 2013年 地區 十億美元 十億美元 英國 2.4 2.1 歐洲其他地區 1.3 1.5 亞洲 2.7 2.5 中東及北非 0.2 0.3 北美洲 (0.1) 0.4 拉丁美洲 1.4 1.8 總計 7.9 8.6 有關註釋，請參閱第109頁。 管理資金╱已付稅款╱物業╱資料披露理念 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 107 按國家╱地區列示之已付稅款 60 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 按地區分析已付稅款總額 亞洲 2,687 2,536 2,639 本位及優先發展市場 2,399 2,185 2,225 －香港 1,273 1,248 974 －中國內地 278 207 276 －印度 290 318 349 －澳洲 204 105 209 －馬來西亞 133 106 193 －印尼 76 74 113 －新加坡 101 88 89 －台灣 44 39 22 其他市場 288 351 414 歐洲 3,709 3,570 3,213 本位及優先發展市場 3,466 3,326 3,021 －英國 2,363 2,107 1,906 －法國 790 844 679 －德國 131 151 200 －瑞士 107 142 160 －土耳其 75 82 76 其他市場 243 244 192 中東及北非 210 251 284 優先發展市場 162 213 234 －阿聯酋 102 98 120 －埃及 60 115 114 其他市場 48 38 50 北美洲 (108) 414 1,236 優先發展市場 －美國 (377) 125 798 －加拿大 269 285 434 其他市場 － 4 4 拉丁美洲 1,384 1,836 1,977 優先發展市場 1,338 1,645 1,835 －巴西 804 1,002 1,174 －阿根廷 333 318 391 －墨西哥 201 325 270 其他市場 46 191 142 總計 7,882 8,607 9,349 有關註釋，請參閱第109頁。 物業 於2014年12月31日，我們在全球約有7,885項 營運物業，其中約1,965項位於歐洲、2,500項 位於亞洲、450項位於北美洲、2,700項位 於拉丁美洲及275項位於中東及北非。該等 物業的面積約為5,430萬平方呎 （2013年： 5,660萬平方呎） 。 我們持有永久業權及長期租約的物業，以 及在香港境內的所有租賃土地，已於2014年 進行估值。根據歷史成本法之計量，該等 物業之估值為108億美元 （2012年：103億美 元） ，高於其於綜合資產負債表內之賬面 值。此外，我們亦持有賬面淨值16億美元 （2013年：15億美元） 之物業作投資用途。 我們的營運物業乃按成本 （即歷史成本或 過渡至IFRS當日的公允值 （其設定成本） ） 減 任何減值虧損列賬，並按照物業之估計可 用年期，將有關資產折舊，直至完全撇銷 為止。因收購而擁有的物業於首次列賬時 按公允值確認。 其他詳情載於財務報表附註23。 資料披露理念 滙豐致力於在呈報中維持資料披露的最高 標準。 我們一貫以來都設有政策，提供資料披露 以便投資者和其他相關群體了解集團的表 現、財務狀況及相關變動。根據該政策： ‧ 為使財務報表及其附註更易於理解，我 們已採取措施以提供更為集中的資料 及盡可能刪掉重複內容。因此，我們已 變更附註內用於描述若干會計政策的 內容位置及措辭，刪除若干非重大資料 披露及變更若干章節的次序。於應用財 務報表資料披露之重要性原則時，我們 考慮各項目之金額及性質。2014年財務 報表及隨附的附註呈報的主要變動於 第346及347頁描述。 ‧ 「財務報表附註」 及 「董事會報告」 中所 提供的資料超出會計準則、法定及監管 規定以及上市規則規定的最低水平。尤 其是，我們經考慮強化信息披露工作組 （ 「EDTF」 ） 於2012年10月發表之 《加强銀行 風險披露》 報告的建議後提供額外披露 資料。該報告旨在幫助金融機構識別投 資者強調需要更佳及更透明的銀行風 險資料的領域，及該等風險與表現計量 及呈報有何關係。此外，我們持續根據 相關監管機構及標準制定機關所發布 的良好實務建議，並因應我們的財務報 表使用者的反饋意見，來提升我們的資 料披露工作。 滙豐控股有限公司 108 董事會報告：其他資料 （續） 按強化信息披露工作組的建議作出的資料披露 風險類別 建議 資料披露 頁次 一般 1 業務面對的風險。 112至117 2 我們的承受風險水平及壓力測試。 117至118 3 首要及新浮現風險，以及業績報告期內的變動。 118至124 4 論述影響集團業務模式及盈利能力的未來監管環境 發展，以及相關規例於歐洲的實施情況。 119至120 及252至256 風險管治、風險管理及 業務模式 5 集團風險管理委員會，以及其活動。 280至281 6 風險管理文化及風險管治與責任。 111 7 各項環球業務涉及的風險圖表。 22 8 壓力測試及相關假設。 117至118 資本充足程度及風險加權 資產 9 第一支柱資本規定。 有關第一支柱資本規定之計算方法，請參閱 文件 《2014年第三支柱資料披露》 。 258至259 10 會計基準與監管規定基準資產負債表對賬。 249 11 自前一個業績報告期以來監管規定資本變動流量 表，包括不同級別監管規定資本之變動。 245 12 論述資本水平目標，及達致該目標的計劃。 239 及 252至258 13 按風險類別、環球業務及地區分析風險加權資產， 及市場風險的風險加權資產。 240 14 各巴塞爾資產類別資本規定之分析，請參閱 文件 《2014年第三支柱資料披露》 。 15 各巴塞爾資產類別信貸風險之分析，請參閱 文件 《2014年第三支柱資料披露》 。 16 各風險加權資產類別的風險加權資產變動對賬 流量表。 242至244 17 論述巴塞爾信貸風險模型之表現，請參閱 文件 《2014年第三支柱資料披露》 。 流動資金 18 分析集團的流動資產緩衝。 165至166 資金 19 按資產負債表類別分析的具產權負擔及無產權負擔 資產。 171至173 20 按結算日尚餘合約期限分析的綜合資產、負債及資 產負債表外承諾總額。 426至435 21 分析集團資金來源以及說明我們的資金策略。 168 市場風險 22 按業務範疇分類，交易用途及非交易用途組合與資 產負債表的市場風險計量兩者之間的關係。 179至180 23 論述重大交易及非交易賬項市場風險因素。 176至179 24 估計虧損風險假設、限制及驗證。 223至224 25 論述壓力測試、反向壓力測試及壓力下之估計虧損 風險。 224至225 信貸風險 26 分析信貸風險總額，包括個人貸款及批發貸款詳情。 129至130 27 論述有關識別已減值貸款、界定減值及重議條件貸 款的政策，以及闡釋貸款的暫緩還款政策。 137 及 208至213 28 已減值貸款年初與年底結欠的對賬及年內減值準備。 137 及 142至143 29 分析衍生工具交易產生的交易對手信貸風險。 150至151 30 論述減低信貸風險措施，包括就所有信貸風險來源 持有的抵押品。 146至150 其他風險 31 管理營運風險的量化方法。 187至189 32 論述已公開的風險事件。 118至124 上述32項建議出自金融穩定委員會屬下的強化信息披露工作組於2012年10月所發表的 《加強銀行風險披露》 報 告。 強化信息披露工作組資料披露╱註釋 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 109 第40至108頁註釋 採用非公認會計原則之財務衡量指標 1 除出售損益外，該等出售業務的經營業績已自經調整業績中扣除。 2 該等出售及收購業務的經營業績並未自經調整業績中扣除，因為有關數額並不重大。 3 不包括於同年收益表中列有重大對銷數額的項目。 4 未扣除貸款減值及其他信貸風險準備之營業收益淨額，亦稱為收入。 5 正數為有利，負數為不利。 6 「貨幣換算調整」 為按本年度適用的平均匯率，換算附屬及聯營公司過往年度業績的影響。 7 「本身信貸息差」 包括長期債務因信貸息差產生之公允值變動，而有關變動的淨值於債務到期時將為零。 此數值不包括源自交易用途負債或衍生工具負債之本身信貸風險之公允值變動。 8 自2014年1月1日起， 「亞洲」 地區取代之前按 「香港」 及 「亞太其他地區」 呈列的地區 （有關進一步詳情，請 參閱財務報表附註11） 。比較數字已相應重列。 綜合收益表 9 載於財務報表的股息為一年內就每股普通股宣派的股息，而非與該年度相關的股息。有關更多詳情，請 參閱第39頁註釋3。 10 每股普通股股息乃以佔每股基本盈利之百分比列示。 11 淨利息收益包括交易用途資產之內部資金成本，而相關外來收入列作 「交易收益」 。在滙豐環球業務業績 中，交易用途資產之資金成本乃作為利息支出計入環球銀行及資本市場業務的交易收益淨額中。 12 總孳息率指就附息資產平均值所賺得之平均年息率。 13 淨息差指就附息資產平均值所賺得之平均年息率 （扣除已攤銷溢價及貸款費用） 與就附息資金平均值所 支付之平均年息率兩者間之差額。 14 淨利息收益率指淨利息收益佔附息資產平均值之按年計算百分比。 15 交易用途資產之利息收益於綜合收益表 「交易收益淨額」 項下列賬。 16 指定以公允值列賬之金融資產之利息收益於綜合收益表 「指定以公允值列賬之金融工具淨收益」 項下列 賬。 17 僅計入附息銀行存款。 18 指定以公允值列賬之金融負債之利息支出於綜合收益表 「指定以公允值列賬之金融工具淨收益」 項下列 賬，惟本身債務之利息乃於 「利息支出」 項下列賬。 19 僅計入附息客戶賬項。 20 交易收益淨額包括有利變動1,500萬美元 （2013年：不利變動6,600萬美元；2012年：不利變動6.29億美元） ， 與已發行結構票據及其他混合工具負債之公允值變動有關，此等公允值變動源自滙豐之發債息差變動。 21 交易收益亦包括不合資格對沖工具的變動。該等對沖工具乃根據明文規定利率管理策略而採用的衍生工 具，惟並未或未能採用對沖會計法處理。這些對沖工具主要包括跨貨幣及利率掉期，用以對沖滙豐控股 發行的定息債務及美國滙豐融資發行的浮息債務所涉經濟風險。在收益表中確認的不合資格對沖公允值 變動，於不同年度會有不同幅度及方向，但不會改變明文規定利率管理策略中涉及預計現金流的部分， 此策略適用於該等工具及被對沖經濟風險的相關資產和負債 （如為持至到期日的衍生工具） 。 22 於2013年，我們完成出售平安保險因而錄得利潤淨額5.53億美元，反映撤消確認分類為可供出售及於 「金 融投資減除虧損後增益」 項內入賬的股權證券之增益的影響淨額12.35億美元，部分被 「交易收益淨額」 項 內入賬的或有遠期出售合約之6.82億美元所抵銷。 23 其他公允值變動計入與滙豐已發行長期債務一併管理的衍生工具之公允值變動所產生的損益。 24 已支付保險賠償和利益及投保人負債之變動淨額乃來自壽險及非壽險業務。就非壽險業務而言，呈列的 金額為於年內已付賠償支出及已產生賠償的估計支出。就壽險業務而言，賠償主要包括初期承保時，保 單產生的投保人負債，以及保單生效後該等負債的任何變動，此等變動主要來自有儲蓄成分保單的投資 表現。故此，賠償額與帶有儲蓄成分產品的銷售額及投資市場增長幅度同步上升。 25 成本效益比率的定義為營業支出總額除以未扣除貸款減值及其他信貸風險準備之營業收益淨額。 綜合資產負債表 26 於2013年，環球銀行及資本市場業務改變信貸及利率交易業務旗下反向回購及回購活動的管理方式。此 舉分別導致資產負債表中分類為 「交易用途資產」 及 「交易用途負債」 的金額減少，和按攤銷成本計算分 類為 「非交易用途反向回購」 及按攤銷成本計算分類為 「非交易用途回購」 的金額增加。 27 自2014年1月1日起，非交易用途反向回購及回購於資產負債表內分行呈列。過往，非交易用途反向回購計 入 「同業貸款」 及 「客戶貸款」 ，非交易用途回購計入 「同業存放」 及 「客戶賬項」 。比較數字已相應重列。 非交易用途反向回購及回購於資產負債表內分行呈列，使披露方式與市場慣例一致，並提供更具參考價 值的貸款相關資料。客戶及同業貸款╱存放中反向回購及回購的份額載於財務報表附註17。 28 已扣除減值準備。 29 資本指引4於2014年1月1日實施，於2014年12月31日的資本來源及風險加權資產已按此基準計算及呈列。 2011至2013年的比較數字乃按巴塞爾協定2.5的基準計算及呈列。於2010年比較數字以巴塞爾協定2為基準 計算及呈列。 30 資本來源指監管規定資本總額，其計算方式載於第246頁。 31 包括永久優先證券，詳情載於財務報表附註30。 32 每股普通股資產淨值的定義為股東權益總額減非累積優先股及資本證券，再除以已發行普通股 （不包括 本公司已購回及持作庫存股份之股份） 數目。 董事會報告：其他資料╱風險 滙豐控股有限公司 110 註釋╱風險狀況 33 「貨幣換算調整」 為按本年底適用的匯率，換算附屬及聯營公司上年底資產及負債的影響。 34 計入出售用途業務組合中持作出售用途資產之數額。 35 法國的業務主要包括美國滙豐融資、HSBC Assurances Vie及英國滙豐銀行有限公司的巴黎分行在當地的業 務。 風險加權資產平均值回報計量之對賬 36 風險加權資產及除稅前風險加權資產平均值回報。 37 經調整風險加權資產平均值回報乃按固定匯率基準使用經調整之除稅前回報及列賬基準之風險加權資 產平均值計算，並就重大項目之影響作出調整。 38 「其他」 包括有關美國消費及按揭貸款業務的財資服務及縮減中的商業業務。美國消費及按揭貸款包括 指定列為持作出售用途之縮減業務內的貸款組合。 環球業務及地區 39 在 「其他」 項下呈列之主要項目為滙豐之控股公司及融資業務的業績，包括集中持有之無成本資金所賺 取之利息淨額、總管理處向滙豐提供督導及中央管理服務所涉及之營業支出，以及集團服務中心及內部 服務中心的成本及有關收回額。有關業績亦包括就調查過往未能充分遵守反洗錢及制裁法律而支付的 罰款及罰則 （作為和解的部分安排） 、英國的銀行徵費，以及未分類的投資業務、集團集中持有之投資公 司、攤薄聯營及合資公司權益所得增益及若干物業交易。此外， 「其他」 亦包括指定以公允值列賬之長期 債務之公允值部分變動 （集團本身債務之其餘變動已計入環球銀行及資本市場業務項下） 。 40 按地區及環球業務劃分之資產包括滙豐內部項目。此等項目在適當的情況下會於 「滙豐內部項目」 或 「項 目之間互相撇銷」 （如適用） 下撇銷。 41 就出售的業務而言，包括出售事項之損益及第40頁所述的主要經營業績。 42 就本項目而言，其他收益在適用情況下包括交易收益淨額、指定以公允值列賬之其他金融工具淨收益╱ （支 出） 、金融投資減除虧損後增益、股息收益、保費收益淨額以及其他營業收益減已支付保險賠償和利益 及投保人負債之變動淨額。 43 貸款減值及其他信貸風險準備。 44 應佔聯營及合資公司利潤。 45 請參閱第40頁 「採用非公認會計原則之財務衡量指標」 。經調整業績與列賬基準業績之對賬載於提交美國 證交會的20-F表格，現已上載至www.hsbc.com。 46 主要零售銀行及財富管理業務不計入美國縮減組合及已出售美國卡及零售商戶業務的影響。專注於主要 零售銀行及財富管理業務可讓管理層識別持續經營業務的重大變動，以及評估預期日後將對該等業務有 重大影響的因素及趨勢。主要零售銀行及財富管理業務之列賬基準與經調整財務衡量指標的對賬表已上 載至www.hsbc.com。 47 「投資產品分銷」 涉及的投資項目包括互惠基金 （由滙豐及第三方制訂） 、結構產品及證券交易，而涉及的 財富保險產品分銷則包括由滙豐及第三方制訂的壽險、退休金及投資保險產品。 48 「其他個人貸款」 包括個人非住宅封閉式貸款及個人透支。 49 「其他」 主要包括零售及信貸保障保單的分銷及制訂 （如適用） 。 50 在環球業務的分析中，交易收益╱ （支出） 淨額包括分類為持作交易用途之金融資產及金融負債的公允值 變動所產生之全部損益、相關之外來與內部利息收益及利息支出，以及已收股息。在法定賬項中，內部 利息收益及支出已予撇銷。 51 「資本市場業務的產品、保險及投資和其他」 包括來自外匯交易業務、制訂及分銷保險產品、利率管理措 施及GCF產品之收入。 52 於2014年，資本市場業務包括因結構負債信貸息差擴闊而產生的有利公允值變動1,500萬美元 （2013年：不 利公允值變動6,600萬美元；2012年：不利公允值變動6.29億美元） 。 53 於環球銀行及資本市場一欄中， 「其他」 一項包括由環球業務持有，但未投放於產品的無成本資金所賺 取的利息淨額、所分配資金成本及因出售業務獲得的利潤。在管理層對營業收益總額的意見中，名義稅 項減免額被分配至相關業務，以反映若干活動所產生之經濟利益，而該等經濟利益並未於營業收益中 反映，例如免稅投資所賺取收益之名義稅項減免額 （相關活動的經濟利益於稅項支出中反映） 。為反映以 IFRS為基準計算的營業收益總額，該等稅項減免額的對銷金額計入 「其他」 項內。 54 「客戶資產」 乃按各期末適用的匯率換算，而貨幣換算的影響則另行報告。客戶資產的主要成分為管理資 金 （於2014年12月31日為2,750億美元，並未於集團資產負債表呈報） ，以及客戶存款 （於2014年12月31日為 900億美元，其中850億美元於集團資產負債表呈報及50億美元為資產負債表外存款） 。 55 項目之間互相撇銷包括(i)計入 「其他」 一項並從環球業務收回的集團內部支援服務及集團服務中心成本； 及(ii)環球銀行及資本市場業務進行交易活動的內部資金成本。在環球銀行及資本市場一欄下呈報的滙豐 資產負債管理業務為交易業務提供資金。為了按全面撥資基準呈報環球銀行及資本市場業務的 「交易收 益淨額」 ， 「淨利息收益」 及 「交易活動之淨利息收益╱ （支出） 」 已返計還原，以反映於項目之間互相撇銷 前的內部資金交易。 56 已支付保險賠償和利益及投保人負債之變動淨額。 57 「職員支出」 包括各環球業務直接產生的支出。 「其他」 類別直接產生有關僱員重新分配及分攤的支出及 其他相關支出列於 「其他營業支出」 內。 58 由於集團內之市場風險分散效應，風險加權資產並非各個地區相加的總和。 59 管理資金及託管資產未於集團資產負債表內列報 （除非我們擔任投資經理時被視作以主管而非代理人的 身份行事，且該等資產作為結構公司綜合入賬，請參閱財務報表附註39） 。 60 滙豐已付稅款與滙豐本身稅項負債相關，包括所賺取利潤之稅項、僱主稅項、銀行徵費及其他稅項╱徵 費 （例如印花稅） 。相關數額乃按照現金流基準列賬。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 111 風險 頁次 附錄 1 風險狀況 2 111 管理風險 2 112 風險管理架構 112 風險因素 113 滙豐所管理的風險 114 風險管理流程和程序 117 風險管治 204 承受風險水平 205 首要及新浮現風險 2 118 宏觀經濟及地緣政治風險 118 業務模式的宏觀－審慎、 監管及法律風險 119 監管承諾及同意令 120 有關業務營運、管治及 內部監控制度的風險 122 特別提述部分 2 124 防範金融犯罪及監管合規工作 124 私人銀行業務 124 監管規定壓力測試 125 石油及天然氣價格 126 俄羅斯 126 歐元區 126 信貸風險 4 127 206 流動資金及資金 4 163 215 市場風險 4 175 221 營運風險 2 186 228 合規風險 189 229 法律風險 229 環球保安及詐騙風險 229 系統風險 230 供應商風險管理 231 保險業務風險管理 3 190 231 其他重大風險 2 199 聲譽風險 199 235 受信風險 200 235 退休金風險 200 236 可持續發展風險 201 237 1 風險附錄－風險政策及慣例。 2 未經審核 。 3 經審核。 4 如經審核，則另有說明。 有關滙豐的風險管理及管治政策與慣例，詳見第 204頁風險附錄。 風險狀況 （未經審核） 管理風險狀況 ‧ 保持穩健的資產負債結構仍然是我們的 核心理念。 ‧ 我們的組合繼續配合集團的承受風險 水平和策略。 ‧ 我們的風險管理架構具備預先識別風 險的強勁實力。 維持資本實力及穩健的流動資金狀況 ‧ 我們的過渡基準普通股權一級資本比 率仍然穩健，維持於10.9%。 ‧ 我們在2014年全年一直保持穩健的流 動資金狀況。 ‧ 客戶貸存比率維持在遠低於90%的水 平。 有力管治 ‧ 集團上下設有穩健的風險管治及問責 制度。 ‧ 董事會在集團風險管理委員會的建議 下，審批我們的承受風險水平。 ‧ 環球風險管理營運模式支持我們遵守 全球一致的標準，以及集團內實施的風 險管理政策。 首要及新浮現風險 ‧ 宏觀經濟及地緣政治風險。 ‧ 業務模式的宏觀－審慎、監管及法律風 險。 ‧ 有關業務營運、管治及內部監控制度的 風險。 滙豐控股有限公司 112 董事會報告：風險 （續） 管理風險 （未經審核） 我們是提供銀行及金融服務的機構，積極 管理風險是日常活動的核心工作。 風險管理架構 我們的風險管理架構應用於集團各層面， 載於第24頁。要點於下文論述。 集團的承受風險水平聲明是我們管理風險 的主要組成部分，載於第24頁。 風險管治架構 集團貫徹奉行穩健的風險管治及問責制 度，透過已建立的架構確保在集團各層面 及對所有風險類別，均能就風險管理的成 效作出適當的監督及問責。根據環球標準 及環球風險管理營運模式，滙豐上下必須 遵守一致的標準及風險管理政策。 董事會負有審批滙豐的承受風險水平及風 險管理成效的最終責任。 ‧ 集團風險管理委員會就承受風險水平 及其是否符合策略、風險管治及內部監 控，以及高層次的風險相關事宜，向董 事會提供意見。 ‧ 金融系統風險防護委員會向董事會報 告有關金融犯罪及金融系統濫用的事 宜，並就金融犯罪風險提供前瞻性的意 見。 ‧ 行為及價值觀委員會於2014年1月成立， 負責監督滙豐政策、程序及標準的設計 及應用，以確保集團能以負責任的態度 經營業務，同時恪守滙豐價值觀，並就 此向董事會提供建議。 集團管理委員會的風險管理會議負有持續 監控、評估和管理風險環境及風險管理政 策成效的執行責任。各業務的高級管理層 負責日常風險管理，並由下文 「三道防線」 所述的環球部門協助。 執行及非執行風險管治架構及其相互關係 載於第204頁，各主要營運附屬公司亦有類 似安排。 集團風險管理委員會報告載於第280頁。金融系統 風險防護委員會報告載於第282頁。行為及價值觀 委員會報告載於第286頁。 三道防線 我們在管理風險中採用三道防線模式。 ‧ 第一道防線 － 每名滙豐僱員須負責屬 於其日常工作一部分的風險。第一道防 線確保能透過整體監控環境下適當的 內部監控措施，識別、緩減及監察營運 中的所有主要風險。 ‧ 第二道防線 － 環球風險管理部、環球 財務部及環球人力資源部等環球部門 構成第二道防線。他們負責鑑證、監督 和查核第一道防線進行的活動。 ‧ 第三道防線 － 審核部門構成第三道防 線，就第一及第二道防線向高級管理層 及董事會提供獨立鑑證。 有關營運風險管理架構的詳情，請參閱第186頁。 人才 所有員工均須在其指定職責範圍內識別、 評估及管理風險，因此他們對三道防線的 效用至關重要。滙豐價值觀加強了個人對 環球標準的問責。 清楚及一致地向員工傳遞有關風險的訊 息，可以有效傳達集團的策略，也定下高 級管理層在這個問題上的基調。我們就關 鍵的風險管理及合規課題提供整套強制性 培訓，這不僅能增強員工的相關技能和了 解，更能加強滙豐內部的風險管理文化。這 些培訓能加強員工以集團所期望的行為處 理風險 （如風險管理政策所述） 。有關培訓 會定期更新，闡述集團所承受各類風險的 技術層面，以及如何有效管理上述風險。 員工可循舉報專線，以保密方式提出問題 （請參閱第20頁） 。 集團的薪酬方針加強了我們的風險管理文 化。個人報酬 （包括執行董事的報酬） 是根 據滙豐價值觀的遵守情況，以及與集團承 受風險水平及環球策略一致的財務及非財 務目標達成情況而定。 有關風險及薪酬的其他資料，請參閱第300頁集團 薪酬委員會報告。 獨立的風險管理部 環球風險管理部由集團風險管理總監領導， 負責企業整體風險監察，包括制訂環球政 策、監控風險狀況及進行前瞻性的風險識 別及管理。環球風險管理部亦負責通過風 險管理分支部門的綜合網絡管理風險，支 持滙豐的環球業務及地區。該等分支部門 管理風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 113 壓力情形下進行穩定撥備。 ‧ 多個司法管轄區建議或實施銀行業務 與交易活動分隔運作的架構，可能對業 務營運及經營業績造成重大不利影響。 ‧ 我們於業務所在國家╱地區面對的稅務 相關風險，或會對經營業績造成重大不 利影響。 有關業務、業務營運、管治及內部監控制 度的風險 ‧ 我們實施優先策略面對執行上的風險。 ‧ 我們可能無法獲得策略方案的所有預 期利益。 ‧ 我們在競爭非常激烈的市場經營業務。 ‧ 我們的風險管理措施未必奏效。 ‧ 集團業務存有內在的營運風險。 ‧ 業務營運受欺詐活動威脅。 ‧ 業務營運可能因外界環境影響而中斷。 ‧ 業務營運採用第三方供應商及服務供 應商。 ‧ 業務營運極為依賴資訊科技系統。 ‧ 我們可能無法應監管機構的要求提供 所需資料。 ‧ 業務營運存有內在的聲譽風險。 ‧ 我們可能因僱員行為不當而蒙受損失。 ‧ 我們依賴招聘、挽留及培養合適的高級 管理人員與技術專才。 ‧ 我們的財務報表有部分是依據判斷、估 算及假設編製，而有關依據存在不明朗 因素。 ‧ 滙豐可能因模型的局限性或失效而招 致虧損或需要持有額外資本。 ‧ 第三方可能在我們不知情下，利用我們 （作為中介機構） 進行非法活動，這可能 對集團造成重大不利影響。 ‧ 我們承擔重大的交易對手風險。 ‧ 市場波動可能導致集團的收益或組合 價值下跌。 ‧ 流動資金或隨時可以取得的資金對集 團業務十分重要。 ‧ 滙豐控股、其任何附屬公司或這些附屬 公司的任何債務證券的信貸評級如有下 調，均可能增加資金成本或縮減資金來 源，並對流動資金狀況及利息收益率造 成不利影響。 獨立於集團業務的銷售及交易部門。其獨 立性確保在作出風險╱回報決策時作出必 要的權衡。 我們視環球風險管理為一項業務，確保該 部門具應變能力，以回應相關群體的需求。 滙豐所面對的風險 我們所有業務或多或少均涉及分析、評估、 承擔及管理一種或多種風險。 風險因素 我們已識別出多方面的風險因素，這些因 素涵蓋集團業務面對的多種風險。 多項風險因素有可能影響我們的經營業績 或財務狀況，但不一定會被視作首要或新 浮現風險。然而，該等因素提示我們需要 持續評估首要及新浮現風險，我們的承受 風險水平可能因而需要修訂。該等風險因 素為： 宏觀經濟及地緣政治風險 ‧ 當前經濟及市場狀況或會對業績造成 不利影響。 ‧ 我們面對業務所在地的政治及經濟風 險，包括政府干預的風險。 ‧ 我們可能因歐元區重新燃起的經濟及 主權債務緊張狀況而遭受不利影響。 ‧ 匯率變動可能影響集團業績。 業務模式的宏觀－審慎、監管及法律風險 ‧ 未能履行我們在延後起訴協議下的責 任，可能對業績及業務營運造成重大不 利影響。 ‧ 未能遵守若干監管規定或會對業績及 業務營運造成重大不利影響。 ‧ 未能符合監管規定壓力測試的要求可 能對集團的資本狀況、業務營運、業績 及未來前景造成重大不利影響。 ‧ 我們正面對法律和監管當局採取多項 行動及進行多項調查，相關事件的結果 固然難以預測，但是不利的結果會對經 營業績及品牌造成重大不利影響。 ‧ 法律或監管方面的不利發展，或監管機 構或政府政策的變化，可能對業務營 運、財務狀況及前景造成重大不利影響。 ‧ 滙豐控股及其英國附屬公司可能須根 據 《2009年銀行法》 （經修訂） 在若干重大 滙豐控股有限公司 114 董事會報告：風險 （續） ‧ 集團業務存有涉及借款人信貸質素的 內在風險。 ‧ 保險業務面對與保險索賠率及保險客 戶行為改變有關的風險。 ‧ 滙豐控股是一家控股公司，需依賴附屬 公司償還貸款及派發股息，方能履行責 任。這些責任包括與債務證券有關的責 任，以及提供利潤以便日後向股東支付 股息。 ‧ 我們可能須向退休金計劃作出巨額供 款。 滙豐所管理的風險 與銀行業務及制訂保險產品業務有關之主 要風險於下表闡述。 風險闡述 － 銀行業務 風險 產生自 計量、監控及管理風險 信貸風險 （第127頁） 客戶或交易對手未能 履行合約責任因而產 生之財務虧損風險。 信貸風險主要來自直 接貸款、貿易融資及 租賃業務，但亦會來 自擔保及衍生工具等 若干其他產品。 信貸風險是： ‧ 按客戶或交易對手未能還款因而可能損失的金額計量。倘為衍 生工具，計量風險時會考慮滙豐當前的按市值計價合約價值， 及因市場利率變動導致該價值隨時間推移的預計潛在變動； ‧ 在限額內監控，並由指定授權架構內的人士批准。這些限額是 客戶或交易對手未能履行合約責任時滙豐可能面臨的風險或虧 損最高值；及 ‧ 通過健全的風險監控架構管理風險，而有關架構為風險管理人 員制訂了清晰而一致的政策、原則及指引。 流動資金及資金風險 （第163頁） 我們缺乏足夠財務資 源履行到期時的責任 或只能以過高成本履 行責任之風險。 流動資金風險因現金 流的時間錯配而產生。 資金風險於無法按預 期條款及按需要而取 得流動資金，以便為 流通性不足的資產持 倉提供所需資金時產 生。 流動資金及資金風險是： ‧ 使用內部衡量標準計量，包括受壓的營運現金流預測、覆蓋比 率及貸款對核心資金比率； ‧ 按照集團的流動資金及資金風險管理架構監控，並由地區資產 負債管理委員會 （ 「ALCO」 ） 、集團資產負債管理委員會及風險管 理會議監督；及 ‧ 以獨立形式管理，而不依賴集團旗下任何公司 （除非預先承諾） 或中央銀行，除非已按市場慣例成為既定的常規業務運作。 市場風險 （第175頁） 市場風險因素 （包括 匯率及商品價格、 利率、信貸息差及股 價） 變動導致收益或 組合價值下跌之風 險。 市場風險分為兩個組 合： ‧ 交易用途組合，包 括市場莊家持倉及 源自客戶的持倉。 ‧ 非交易用途組合， 主要包括因對零售 及工商金融業務資 產與負債進行利率 管理而產生的持 倉、指定列為可供 出售及持至到期日 之金融投資，以及 來自保險業務的風 險承擔 （第198頁） 。 市場風險是： ‧ 按估計虧損風險計量，用於估計在指定期間和既定可信程度 內，市場利率和價格變動可能引致風險持倉產生的潛在虧損， 並輔以壓力測試，以評估倘若出現較為極端但有可能發生的事 件或一系列金融變數出現變動時，組合價值所受的潛在影響； ‧ 使用多種衡量標準監控，包括淨利息收益的敏感度及結構匯兌 的敏感度，這些衡量標準適用於各風險類型的市場風險持倉； 及 ‧ 使用集團管理委員會批准的風險限額為滙豐控股及各項環球業 務管理這種風險。這些風險限額單位分配予各業務部門及集團 旗下的法律實體。 滙豐所管理的風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 115 風險 產生自 計量、監控及管理風險 營運風險 （第186頁） 因內部程序、人員及 系統的不足或失效， 或因外圍事件如法律 風險 （連同會計、稅 務、保安及詐騙、人 事、系統、項目、營 運及組織架構變動風 險） 引致虧損之風險。 營運風險產生自日常 營運或外圍事件，且 與集團業務每個環節 均有關連。 下文討論合規風險及 受信風險。其他營運 風險載於 「風險附錄」 （第204頁） 內。 營運風險是： ‧ 使用首要風險分析程序及風險與監控評估程序計量，這些程序 評估風險水平及監控成效； ‧ 使用關鍵指標及其他內部監控活動監控；及 ‧ 主要由環球業務及部門經理管理。管理人員會運用營運風險管 理架構識別及評估風險、執行監控措施以管理此等風險，並監 察該等監控措施的成效。環球營運風險管理部負責此架構並監 督相關業務及部門內的營運風險管理。 合規風險 （第189頁） 我們未能遵守所有相 關法律、守則、規 則、法規及良好市場 慣例準則的條文和精 神，並招致罰款及罰 則且因此蒙受業務虧 損之風險。 合規風險是營運風險的 一部分，源自規則、法 規、其他標準及集團政 策 （包括遵循反洗錢、 反賄賂及貪污、反資助 恐怖主義及武器擴散、 制裁及商業行為的規 定） 。 延後起訴協議及監察員 分別於第120及第27頁 內討論。 合規風險是： ‧ 經參考已識別的衡量標準、對事件的評估 （是否影響滙豐或整體 行業） 、監管當局的回應，以及各環球業務及部門經理的判斷和 評估，從而計量此類風險； ‧ 按照集團合規風險的評估及衡量標準、第二道防線部門 （包括防 範金融犯罪部及監管合規部） 監控活動的成果、內外審核及監管 視察的結果，從而監控此類風險；及 ‧ 通過設立及傳達適當的政策及程序、對僱員進行培訓及監控活 動，以確保僱員遵守政策及程序，從而管理此類風險。如有需 要，我們會積極進行風險監控及╱或補救工作。 其他重大風險 聲譽風險 （第199頁） 集團本身、員工或客 戶或集團代表的非 法、不道德或不當行 為將損害滙豐聲譽， 並有可能引致業務虧 損、罰款或罰則之風 險。 聲譽風險為滙豐自 身、其僱員或關聯人 士的任何事件、行 為、作為或不作為未 能符合相關群體的預 期，致使相關群體對 滙豐有負面看法的風 險。 聲譽風險是： ‧ 經參考滙豐與所有相關群體 （包括媒體、監管機構、客戶及僱 員） 的關係所示的聲譽計量； ‧ 通過聲譽風險管理架構並計及上文概述的合規風險監控活動成 果予以監控；及 ‧ 由各員工管理並納入一系列政策及指引範圍內。集團設立清晰 的委員會架構及指明負責減低聲譽風險的人員，包括集團聲譽 風險政策委員會及地區╱業務的同等組織。 受信風險 （第200頁） 違反受信責任之風 險，該責任界定為滙 豐為第三方持有、管 理、監督資產或就該 等資產承擔責任時， 須在法律及╱或監管 規定下，以最高審慎 標準及真誠的態度行 事的責任。 受信風險是營運風險的 一部分，源自集團以受 信人身分擔任受託人、 投資經理或在法律或法 規授權下從事之業務活 動 （ 「指定業務」 ） 。 受信風險是： ‧ 由各指定業務透過營運風險及監控評估程序 （評估風險水平及主 要監控的成效） 監控他們本身的承受風險水平聲明而計量； ‧ 通過結合測試、關鍵指標以及客戶及監管當局的回應等其他標 準予以監控；及 ‧ 透過已建立的管治架構、全面的政策、程序和培訓計劃在指定 業務範疇內管理。 滙豐控股有限公司 116 董事會報告：風險 （續） 風險 產生自 計量、監控及管理風險 退休金風險 （第236頁） 集團旗下公司及成員 公司的供款未能產生 足夠資金，以應付活 躍成員日後服務的應 計福利支出，以及退 休基金所持資產的表 現，不足以應付現有 退休金負債之風險。 退休金風險源自投資 回報不足、導致公司 倒閉的經濟狀況、利 率或通脹的不利變 動，或成員壽命長於 預期 （長壽風險） 。退 休金風險包括上文所 列的營運風險。 退休金風險是： ‧ 根據有關計劃產生足夠資金以應付應計利益成本的能力而計量； ‧ 透過於集團及地區層面制訂特定的承受風險水平而監控；及 ‧ 通過各地適當的退休金風險管治架構及全球的風險管理會議予 以管理。 可持續發展風險 （第237頁） 提供金融服務所產生 的環境及社會影響超 出經濟利益之風險。 獲提供金融服務的公 司或項目與可持續發 展的需要背道而馳， 即會產生可持續發展 風險。 可持續發展風險是： ‧ 透過評估客戶業務對可持續發展的潛在影響及指定所有高風險 交易的可持續發展風險評級而計量； ‧ 由風險管理會議每季度及集團可持續發展風險管理部的管理層 每月監控；及 ‧ 就項目融資貸款而言，運用可持續發展風險管理政策管理，至 於對環境或社會具有重大影響力的行業，則運用以行業為本的 可持續發展政策管理。 風險闡述－制訂保險產品業務 風險 產生自 計量、監控及管理風險 金融風險 （第194頁） 我們能否將保單未決 賠款與支持該等負 債的資產組合有效配 對，須視乎金融風險 （如市場、信貸及流 動資金風險） 的管理 及投保人承擔該等風 險的程度而定。 單位相連合約下的投 保人負債與相關資產 的價值同步變動，因 此投保人承擔大部分 金融風險。 附有酌情參與條款的 合約讓投保人與股東 按合約類型及特定合 約條款分享相關資產 的表現。 金融風險產生自： ‧ 金融資產公允值變 動或該等資產日後 現金流因變數 （例 如利率、匯率及股 價） 波動而出現變 動的市場風險； ‧ 由於第三方違約未 能履行其責任因而 產生的信貸風險及 潛在財務虧損風 險；及 ‧ 由於並無充足資產 可變現為現金，故 企業未能向投保人 支付到期款項的流 動資金風險。 金融風險是： ‧ 就各類風險獨立計量： － 市場風險根據主要金融變數波動的風險計量； － 信貸風險按客戶或交易對手未能還款因而可能損失的金額 計量；及 － 流動資金風險使用內部衡量標準計量，當中包括受壓的營 運現金流預測。 ‧ 在限額內監控，並由指定授權架構內的人士批准；及 ‧ 通過健全的風險監控架構管理風險，而有關架構為風險管理人 員制訂了清晰及貫徹之政策、原則及指引。倘若附屬公司制訂 附有保證的產品，而承擔的任何市場風險不能以本身所簽發保 單內任何酌情參與 （或紅利） 條款管理，則一般要因此承受市場 利率及股價下跌的風險。 我們的制訂保險產品附屬公司與銀行業務 受到不同的監管機構監管。集團旗下各保 險公司會採用適用於相關保險業務的各種 方法及程序來管理保險業務的風險，但該 等風險仍會在集團層面受到監控。集團的 保險業務亦面對營運風險及上文就銀行業 務呈列的其他重大風險，集團的風險管理 程序已涵蓋該等風險。 風險闡述 － 銀行業務 （續） 滙豐所管理的風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 117 風險 產生自 計量、監控及管理風險 保險風險 （第198頁） 某段時間後獲取保單 成本加上行政開支和 賠償及利益支出的總 額，可能超過所收保 費加投資收益總額之 風險。 賠償及利益支出可能 受多項因素影響，包 括過往的死亡率及發 病率、保單失效率及 退保率，以及 （倘保單 帶有儲蓄成分） 為支持 負債而持有的資產之 表現。 保險風險是： ‧ 按壽險未決賠款計量； ‧ 由零售銀行及財富管理業務風險管理委員會監控，該委員會乃 根據集團管理委員會就保險業務界定的承受風險水平監察保險 業務的風險狀況；及 ‧ 由集團總部及各地部門使用產品設計、承保、再保險及賠償處 理程序管理。 風險管理流程和程序 除承受風險水平外，下列流程亦為滙豐風 險管理的組成部分： ‧ 透過首要及新浮現風險流程識別風險； ‧ 編製風險狀況圖譜；及 ‧ 壓力測試及境況分析。 識別風險 我們持續識別及監察風險。有關流程包括 對風險因素及壓力測試計劃的結果進行分 析，繼而將若干主要風險分類為首要風險 或新浮現風險。如果我們對首要及新浮現 風險作出的評估發生變更，可能需要對業 務策略以至承受風險水平作出調整。 我們當前的首要及新浮現風險於下文論述。 編製風險狀況圖譜 風險乃根據集團環球業務的承受風險水平 由各業務承擔，並在集團、環球業務及地 區層面管理。所有風險透過風險圖譜流程 紀錄及監察，而該流程按類別描述不同地 區及環球業務的風險狀況。 除主要銀行及保險風險外，該風險圖譜流 程亦識別及監察模型、財務管理、資本、伊 斯蘭融資及策略等風險。這些風險透過承 受風險水平架構予以定期評估，並經壓力 測試，以及考慮分類為首要及新浮現風險。 壓力測試 我們按企業基準及主要附屬公司層面進行 不同境況下的集團壓力測試，反映了我們 的業務策略以及所引致的風險。滙豐的壓 力測試及境況分析計劃審視資本計劃的敏 感度及各境況下對監管規定資本的計劃外 需求，並確保適當考慮首要及新浮現風險。 上述境況包括但不限於不利的宏觀經濟事 件、在國家╱地區、行業及交易對手層面 的失敗事件、地緣政治事件及各種預計的 主要營運風險事件。 壓力測試管理委員會由集團財務董事出任 主席，負責壓力測試策略及總管工作。壓 力測試模型通過集團模型監察委員會的架 構批准。集團管理委員會的每次風險管理 會議將獲提供有關壓力測試的最新資料。 集團風險管理委員會獲提供資料，給予意 見及負責審批 （如適用） 。 開發宏觀經濟境況是此流程的關鍵一環。 潛在境況由各環球團隊 （包括環球風險管 理部及環球財務部） 的經濟專家組成的專 家小組界定及制訂。境況的基礎變數及假 設包括多項經濟指標 （例如孳息曲線、匯率 及波幅） ，並由內外團隊加以擴展及完善。 一經管治委員會審批，這些資料將連同測 試指示，提供予地區及環球業務壓力測試 團隊。 我們採用一套壓力測試模型及方法將境況 轉化為財務影響，例如盈利能力預測及風 險加權資產受到的影響。模型須經過獨立 的模型檢討並通過驗證和審批流程。必要 時，可能考慮對模型作全盤管理。 壓力測試結果須經過地區及集團層面的檢 討及反覆驗證流程，然後制訂行動計劃， 力求降低所識別的風險。在發生特定境況 時，該等行動計劃乃根據高級管理層對風 險及其潛在影響的評估而實施，且會顧及 滙豐的承受風險水平。 除集團整體的風險境況外，滙豐各主要附 屬公司亦定期進行針對所屬地區的宏觀經 濟及事件促成之境況分析。這些附屬公司 亦會參與當地監管當局的壓力測試計劃。 壓力測試應用於各種風險，例如市場風險、 流動資金及資金風險以及信貸風險，以評 估壓力境況下組合價值、結構性長期資金 狀況、收益或資本受到的潛在影響。 我們每年按集團及附屬公司實體基準進行 反向壓力測試。此壓力測試假設業務模式 無法運作，反推引起該事件的一系列情況。 多種事件亦可能於銀行資本耗盡前導致發 滙豐控股有限公司 118 董事會報告：風險 （續） 首要及新浮現風險 生無法運作的事件。這些事件包括特有事 件或系統性事件或多項此等事件，及╱或 可能暗示集團的控股公司或集團主要附屬 公司之一無法營運。這些事件不一定表示 所有主要附屬公司同時無法營運。我們使 用反向壓力測試，藉以提早發出預警，通 知管理層採取行動及訂立應變計劃，以減 低集團可能面臨的潛在壓力及風險，從而 提升集團的復元能力。 誠如第125頁概述，滙豐在2014年參與多個 司法管轄區監管規定的壓力測試計劃。 首要及新浮現風險首/ 新 （未經審核） 我們識別和監察首要及新浮現風險的方法 於第22頁載述。 於2014年，高級管理層特別留意多項首要 及新浮現風險。現有的首要及新浮現風險 如下： 宏觀經濟及地緣政治風險 新 經濟前景及政府干預 新 地緣政治風險增加 新 經濟前景及政府干預 在2014年已發展及新興市場國家╱地區的 經濟增長仍然疲弱。 石油及商品價格自2014年中以來大幅下 跌，乃由於全球供求日益失衡所致。能源 價格短時間內急轉直下，改變了風險的性 質及分布情況。該情況令能源出口國面對 更大的財政及融資挑戰，但卻惠及石油進 口國，同時亦更顯出其中部分國家 （尤其是 歐元區國家） 的通縮風險。此外，預期油價 長時間低企亦可能導致勘探投資減少，因 此可能出現未來供應大幅減少的危機。 歐元區的經濟復甦仍面臨風險。儘管貨幣 政策更趨寬鬆，但由於油價低企，通縮壓力 持續。結構改革議程加快短期內亦會加重 通縮壓力。歐元區的情況於第126頁的 「特 別提述部分」 進一步討論。日本在2014年 第三季陷入技術性衰退，而應對政策或不 足以支持經濟活動復甦。美國經濟復甦令 全球經濟有望向上。 新興市場，尤其是國內經濟表現脆弱的市 場，仍然面臨美國貨幣政策正常化且引致更 大的風險規避之影響。儘管中國內地2014年 國內生產總值增長水平按國際標準計算並 不低，但卻為20多年來最低，且最近的預 測顯示未來數年增幅會更低。多年的過度 投資，尤其是物業市場投資，助長金融泡 沫加大，需要實行新的經濟增長模式。 對滙豐帶來的潛在影響 ‧ 滙豐的業績可能會因為低利率或負利 率、通脹水平偏低或通縮及╱或油價低 企的局面曠日持久而受到不利影響。 ‧ 我們來自新興市場業務的利潤所佔比 重較大。我們的業績可能因新興市場經 濟增長長期下滑而受到不利影響。 ‧ 經濟增長疲弱、推行保護主義措施、地 緣政治風險浮現或重新計值風險日增， 可能會令全球貿易及資金流萎縮，從而 亦可能影響我們的盈利能力。 緩減措施 ‧ 我們密切監察主要市場及行業的經濟 發展，旨在確保能洞悉有關趨勢，並評 估有關趨勢對特定客戶、客戶群或組合 的影響，而且會因應事態發展而採取適 當的減低風險措施，包括修訂主要承受 風險水平的指標及限額。 ‧ 我們使用內部及監管規定計劃的壓力 測試，評估經濟狀況改變對我們業務的 影響。監管規定壓力測試於第125頁論 述。 新 地緣政治風險增加 我們的營運面臨多個國家╱地區政治不穩 定及民間動亂的風險，可能對地區穩定及 地區及全球經濟造成更大層面的影響。 2014年地緣政治風險上升。烏克蘭仍有軍 事衝突升級及╱或爆發內戰的可能，而針 對俄羅斯政府、機構及個人的制裁，以及 油價下跌，已對俄羅斯經濟造成不利影響。 在中東，敘利亞的內戰因恐怖組織伊斯蘭國 奪取伊拉克和敘利亞的部分地區而加劇。 至於其他中東國家，利比亞陷入亂局，以 色列和巴勒斯坦的關係持續緊張，伊朗核 計劃談判令人擔憂，在這些因素的綜合影 響下，威脅該地區穩定性的風險增加。在 亞洲，中國內地與鄰國的海洋主權爭議未 見緩和，而印度和巴基斯坦邊境的緊張關 係持續，令人擔憂這兩個擁有核武的相鄰 國家的衝突可能加劇。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 119 2014年，土耳其和香港等多個國家╱地區 出現民眾動亂及示威遊行，亦導致地緣政 治風險隨著政府採取遏制措施而出現。 多個新興及已發展市場將在2015年舉行選 舉，或會導致市場進一步波動。此外，油 價長時間低企可能影響嚴重依賴石油生產 作為主要收入來源的國家╱地區的穩定。 對滙豐帶來的潛在影響 ‧ 我們的業務所在司法管轄區出現不利 的政局發展、幣值波動、社會不穩，以 及政府改變徵用、授權、國際擁有權、 利率上限、外匯轉移及稅務等方面的政 策，或會令我們蒙受虧損，此乃有關業 績的風險。 ‧ 如發生實際衝突，可能會危及員工安全 及╱或損毀實物資產。 緩減措施 ‧ 我們持續監察地緣政治前景，特別是我 們承擔重大風險及╱或設有實體辦事處 的國家╱地區。 ‧ 我們對主權交易對手進行的內部信貸 風險評級會考慮有關因素，並會左右我 們於該等國家╱地區經營業務的意願。 如有需要，我們會調整國家╱地區的風 險限額及風險承擔，以反映我們的承受 風險水平及採取適當的減低風險措施。 業務模式的宏觀－審慎、監管及法律風 險 首 我們的業務模式及集團的盈利能力 受監管環境發展影響 首 監管機構對業務經營方式及金融犯 罪作出的調查、罰款、制裁，以至相 關承諾、同意令及規定，對集團的 業績及品牌構成負面影響 首 爭議風險 在規管及監察方面，金融服務供應商面對 越來越嚴謹且成本高昂的要求，通常涉及 提供大量數據，尤其在資本及流動資金管 理、經營業務方式、營運架構及金融服務 操守等方面。政府對整個行業及個別金融 機構增加干預及加強監控，加上推行減低 系統性風險的措施，對集團部分或所有業 務而言，可能會大幅改變當地、地區及╱或 全球的競爭形勢。有關措施可能以如同上 述各項的方式引入，成為正式的規定，惟 不同監管地區實施的時間表可能有別。 首 我們的業務模式及集團的盈利能力受 監管環境發展影響 監管規例的變更可能影響我們的業務，包 括集團整體及部分或所有主要附屬公司。 這些變更包括： ‧ 英國 《2013年金融服務 （銀行業改革） 法》 規定我們的英國零售銀行業務與批發 銀行業務分隔運作，美國採納的法例及 法規 （包括達德－法蘭克法案下於2013 年12月採納的 《沃爾克規則》 ） 預期其他 業務採用分隔運作架構，法國採納限制 若干交易活動的措施以及歐洲委員會 就有關歐盟較大型銀行提出的架構措 施建議可能實施的其他變革； ‧ 實施跨地域的法例，包括美國 《外國賬 戶稅務合規法案》 （ 「FATCA」 ） 以及其他 共享稅務資料的相關措施，例如經濟合 作開發組織更普遍地推行的措施； ‧ 資本市場運作制度的改變，尤其是根據 達德－法蘭克法案和歐盟的 《歐洲市場 基礎設施規例》 （ 「EMIR」 ） 等強制場外衍 生工具進行中央結算； ‧ 由於監管機構日益關注金融機構如何 經營業務而出現的改變，特別是有關企 業如何公平待客及建立有序╱透明的市 場，以客戶利益為前提下加強有效競爭 （包括英國競爭及市場管理局目前調查 英國個人往來賬戶及中小企銀行市場 的結果及金融業操守監管局最近進一 步專注於英國批發市場的指示） ； ‧ 英倫銀行進行的公平和有效的金融市 場檢討的結果將促使英國批發金融市 場營運方式轉變； ‧ 資本指引4及英國監管規例對薪酬架構 實行限制及加強集團內部對管理層的問 責，以符合英國高級管理人員制度的規 定 （包括英國持續專注於英國國會銀行 業標準委員會就機構 「文化」 、更普遍的 僱員操守和責任 （如舉報等） 相關事宜所 提出更廣泛的推薦建議的落實進展） ； ‧ 實施資本指引4，特別是英國應用緩衝 資本架構及其與第二支柱的關係； ‧ 實施賦予英國金融政策委員會更大權 力、對英國銀行施加槓桿限制的建議； 滙豐控股有限公司 120 董事會報告：風險 （續） 首要及新浮現風險 ‧ 金融穩定委員會的建議，其要求全球系 統重要性銀行根據整體損失吸納能力 持有最低水平的資本和後償債務，但尚 待進行諮詢及納入為國家法例； ‧ 因集團及其個別營運公司解決策略的 安排而須遵守的規定，有關規定在不同 國家╱地區會有不同影響； ‧ 監督機關持續進行壓力測試及這對集 團的資本規定及內部資本轉移的影響； ‧ 多個國家╱地區的監管機構對按揭貸款 及無抵押貸款組合收緊信貸控制；及 ‧ 影響金融服務供應商的稅務規例 （包括 金融交易稅項） 可能進一步修訂的風險 仍然存在。 對滙豐帶來的潛在影響 ‧ 強制場外衍生工具進行中央結算、 《歐 洲市場基礎設施規例》 、分隔運作及類 似規定、 《沃爾克規則》 、復元和解決計 劃、 《外國賬戶稅務合規法案》 ，以及有 關競爭狀況的查詢及調查所得結果等 改革建議及╱或規例的實施，可能影響 我們的業務經營方式及集團架構。 ‧ 對資本或整體損失吸納能力的更高要 求可能增加集團的資金成本及降低我 們的股東權益回報率。 ‧ 強制場外衍生工具進行中央結算亦對 作為結算成員的滙豐帶來新風險，原因 是我們將須為其他結算成員及其客戶 違約致使中央結算交易對手產生的虧 損提供保障。因此，中央結算附加的一 項新元素，是結算成員與客戶之間的相 互聯繫，我們相信可能增加而非降低我 們面臨的系統性風險。 ‧ 當局加強對業務經營方式 （包括獎勵結 構、薪酬、產品管治及銷售過程） 以及 管理層問責的規管，可能對業界在聘用 及挽留僱員、零售及批發市場的產品定 價及盈利能力等方面造成影響。滙豐的 業務可能因有關發展而受到影響。 ‧ 該等措施可能增加經營成本和限制我 們經營的業務類別，導致盈利能力下降 的風險。 緩減措施 ‧ 我們正與業務所在國家╱地區的政府及 監管機構緊密聯繫，以確保新規例經適 當考慮並得以有效實施。 ‧ 我們已制訂並正在實施行為操守管理 的環球方針，並已在董事會轄下成立行 為及價值觀委員會，以監督整個集團的 行為操守管理。 ‧ 我們已加強有關中央結算交易對手的 管治，並已委任專家管理相關的流動資 金及抵押品風險。 ‧ 我們繼續圍繞監管改革管理及規定措 施的落實，提升及加強管治工作及資源 調配，積極應對監管規定持續改革的重 要日程。 首 監管機構對業務經營方式及金融犯罪 作出的調查、罰款、制裁，以至相關承 諾、同意令及規定，對集團的業績及品 牌構成負面影響 金融服務供應商須面對因業務經營方式及 金融犯罪問題而遭監管機構制裁或罰款之 風險。金融服務公司面對監管程序的事件 與日俱增，而因監管調查、制裁或罰款所 涉問題產生或與此有關的民事訴訟亦會增 多。此外，金融機構因 （其中包括） 指稱的 操守、違反反洗錢及制裁條例、反壟斷違 法行為、市場操縱、協助及教唆逃稅及提 供未有執照的跨境銀行業務而遭受刑事檢 控，乃更加司空見慣及可能因媒體更為關 注及檢控方及公眾人士的期望更高致使該 情況的出現更頻繁。再者，金融服務供應 商可能因 （其中包括） 媒體更為關注及監管 機構及公眾人士的期望更高而面臨若干司 法管轄區的類似或更廣的法律訴訟、調查 及監管行動。對滙豐或其一家或以上附屬 公司採取的任何有關起訴或調查或法律訴 訟或監管行動將導致大額罰款、罰則及╱或 沒收，並對集團的業績、業務、財務狀況、 前景及聲譽造成重大不利影響，包括可能 失去關鍵牌照、須撤出若干業務及存戶及 其他相關群體撤回資金。 監管承諾及同意令 2012年12月，滙豐控股、北美滙豐控股有限 公司 （ 「北美滙豐」 ） 及美國滙豐銀行就過往 未能充分遵守反洗錢及制裁法律的調查， 與美國及英國當局達成協議。該等協議包 括滙豐控股及美國滙豐銀行與美國司法部 （ 「司法部」 ） 達成的五年期延後起訴協議 （ 「美 國延後起訴協議」 ） ，以及滙豐控股與紐約 郡地區檢察官訂立的兩年期延後起訴協議 （ 「紐約郡地區檢察官延後起訴協議」 ） 。滙 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 121 豐控股亦與英國金管局訂立承諾 （ 「金融業 操守監管局指令」 ） ，以遵守有關反洗錢及 制裁規定的若干前瞻性責任。另外，滙豐 控股就有關遵守美國反洗錢及制裁規定與 美國聯儲局訂立停止及終止令。 與司法部及聯儲局訂立的協議，以及金融 業操守監管局指令均要求我們委聘一名 獨立監察員，評估我們全面履行責任的進 度，以及定期評估防範金融犯罪部的工作 成效。有關監察員於第27頁論述。 滙豐已達致紐約郡地區檢察官延後起訴協 議施加的所有規定，於2014年12月該協議 兩年期末按其條款屆滿。 我們仍然需要進行大量工作以建立及改善 反洗錢及制裁法律的合規計劃，且我們與 司法部達成的延後起訴協議及其他和解協 議仍然生效，紐約郡地區檢察官延後起訴 協議的屆滿實乃重要里程碑。 美國滙豐銀行亦須遵守於2012年12月與美 國貨幣監理署訂立的協議，即美國 《金融服 務現代化法案》 （ 「GLBA」 ） 協議和其他同意 令。 對滙豐帶來的潛在影響 ‧ 涉及集團業務的監管程序有何結果實 難預計。對我們不利的結果可能會對我 們的聲譽、品牌及業績造成重大不利影 響 （包括損失業務及撤離資金） 。 ‧ 我們在促進國際資金流及貿易上的重 大參與，令集團面臨金融犯罪的風險， 或不經意違反外國資產控制辦公室 （美 國） 及其他監管機構施加的限制及制裁。 ‧ 若我們於協議期內任何時間違反美國 延後起訴協議，司法部可就有關美國延 後起訴協議的事項起訴滙豐控股或美 國滙豐銀行。 ‧ 設計及簽立反洗錢及制裁補救計劃乃 複雜且需要主要投入人才、系統及其他 基礎設施。此複雜性將產生重大的執行 風險，進而影響我們有效管理金融犯罪 風險及準時補救反洗錢及制裁合規的 不足之處的能力。反之，其可影響滙豐 達致監察員規定或遵守美國延後起訴 協議及金融業操守監管局指令或聯儲 局停止和終止令的能力，而可能需要滙 豐日後採取額外的補救措施。 ‧ 未能於同意令或 《金融服務現代化法案》 協議所列期限或另行延長的期限內遵 守有關同意令或 《金融服務現代化法案》 協議的規定，可能會導致監管行動。任 何有關行動可能對滙豐的綜合業績及 營運造成重大不利影響。 緩減措施 ‧ 與相關監管機構商討後，我們已採取或 正在採取多項措施以達致美國延後起 訴協議、金融業操守監管局指令及 《金 融服務現代化法案》 協議的多項規定。 這些措施包括簡化集團的監控架構、委 任新領導層來加強管治架構、修訂主要 政策及實施環球標準以查察、阻止及防 範金融犯罪 （請參閱第26頁） 。此外，我 們於過去數年大幅增加防範金融犯罪 部及監管合規部的開支及人手。 ‧ 在2014年，我們批准一項有關交易監控 的新環球策略。全球適用的標準化反洗 錢調查流程已經制訂，正開始在優先發 展國家╱地區實施。 業務經營方式 英國及其他國家╱地區的監管機構繼續加 強關注有關對客戶公平的結果及有序╱透 明市場的 「行為」 事宜，包括 （例如） 關注銷 售過程及獎勵、產品及投資適切性、產品 管治、僱員活動及問責，以及市場濫用對 基準、指數、其他利率設定流程、擴大交 易活動及更多的業務經營方式的風險。 在英國，金融業操守監管局正加強使用現 有及新增的干預和強制執行權力，包括行 使權力考慮過去從事的業務和執行客戶補 償及賠償計劃，或採取其他可能屬重大的 補救措施。金融業操守監管局如今亦負責 規管之前不受其規管的範疇，例如消費者 信貸，並正考慮所規管市場的競爭問題。 再者，金融業操守監管局及其他監管機構 採取更多措施應對客戶投訴、客戶提出的 負面評價及╱或濫用市場，不論投訴與個 別機構有關或在更多情況下與某項產品有 關。有關可能不當銷售還款保障保險、面 向中小企的利率對沖產品及財富管理產品 方面，已有先例可援。 集團亦仍然牽涉一系列其他監管程序，涉 及多個國家監管機構、保障公平競爭與執 法機關，有關機關正在調查及檢討在釐定 倫敦銀行同業拆息、其他銀行同業拆息及 基準利率的過程中，銀行小組成員過往提 滙豐控股有限公司 122 董事會報告：風險 （續） 交的若干資料及提交資料的程序。集團亦 正接受有關外匯、貴金屬和信貸違責掉期 相關業務的持續調查。該等調查詳情請參 閱財務報表附註40。 對滙豐帶來的潛在影響 ‧ 滙豐可能面臨監管機構的譴責或制裁， 包括罰款及╱或面對法律程序及訴訟。 ‧ 英國及其他國家╱地區的監管機構未來 可能發現可影響集團的業內不當銷售、 市場行為或其他問題。此舉可不時導致 重大直接成本或負債及╱或該等業務的 經營慣例變動。再者，英國金融申訴專員 等機構 （或類似的海外機構） 就客戶投訴 作出的決定，如果應用於更多類別或組 別的客戶，則可能對集團的經營業績、 財務狀況及前景造成重大不利影響。 緩減措施 ‧ 滙豐正在所有環球業務和部門推行加 強業務經營方式管理的計劃。 ‧ 檢討有關管理層和員工的表現管理安 排，注重將獎勵與基於價值觀的行為和 良好操守掛鈎。 ‧ 當局持續加強監督能力及基準利率設 定程序，而滙豐控股及其附屬公司現正 全面配合所有監管調查及檢討工作。 首 爭議風險 滙豐於多個司法管轄區涉及因其日常業務 營運而提出的法律程序及監管事項。有關 進一步詳情載於財務報表附註40。 對滙豐帶來的潛在影響 ‧ 爭議風險可能會造成財務損失及重大 聲譽損害，並可能對客戶及投資者信心 產生不利影響。 緩減措施 ‧ 我們繼續專注於識別新出現的監管及 司法趨勢，以減少日後可能面臨的訴訟 或監管執法行動。 ‧ 我們不斷提升防範金融犯罪及監管合 規的監控及增加有關資源。 有關業務營運、管治及內部監控制度的 風險 首 執行風險提高 首 人事風險 首 互聯網罪行及詐騙 首 資訊保安風險 首 數據管理 首 模型風險 新 第三方風險管理 首 執行風險提高 金融服務業現時處於前所未有的受監視期。 處理監管要求、法律事宜及業務計劃均需 要投入大量的時間和資源。為滿足該等要 求，滙豐須實施龐大而複雜的內部項目， 導致執行風險提高。集團各部門按照策略 執行一些出售交易中的風險亦提高。 對滙豐帶來的潛在影響 ‧ 該等因素可能妨礙集團成功落實優先 策略。 ‧ 出售項目的潛在風險包括違反監管規 定、工業行動、流失主要人員，以及系 統及程序於業務變革過程中受阻。有關 情況均會帶來財務及聲譽方面的影響。 緩減措施 ‧ 我們已加強重大項目的優先次序安排 及管治程序，並已在項目執行及資訊科 技實力方面作出投資。 ‧ 與出售有關的風險會予以仔細評估和 監控，並受管理層緊密監察。 首 人事風險 現時經營環境對集團人力資本方面的需求 是前所未有的。監管改革計劃往往是跨地 域的，且不斷改變，大量消耗人力物力，令 工作量與日俱增。這令員工面對更複雜和 矛盾的需求，而專家一向供不應求，而且在 全球流動性極高。 對滙豐帶來的潛在影響 ‧ 資本指引4中的新規例改變了薪酬政策 及慣例，而有關規定將適用於總部設於 歐盟的銀行的全球僱員。當中主要的變 動是將可支付予任何 「承受重大風險人 員」 （基於歐洲銀行管理局發表的定質和 定量標準） 的浮動酬勞設定上限。有關 首要及新浮現風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 123 變動對滙豐帶來重大挑戰，原因是滙豐 有大量承受重大風險人員均駐於歐盟 以外地區。 ‧ 英國審慎監管局發表的政策聲明將薪 酬守則推展至所有審慎監管局認可機 構，自2015年1月1日起在整個集團對收 取浮動酬勞的承受重大風險人員的已 實際授出╱支付的浮動薪酬實施 「撤回」 機制。 ‧ 審慎監管局和金融業操守監管局在英 國引進高級管理層及認證改革和相關 操守規則 （其詳情現時有待諮詢） ，擬清 楚訂明對高級管理層和較基層僱員問 責和行為的期望。然而，當前該等改革 的確切影響存在若干不確定性，包括對 較高級僱員、非英國僱員和非執行董事 的影響。 ‧ 實施機構變動以支持集團的策略及╱或 實行監管改革方案，可能會令員工加快 流失。 緩減措施 ‧ 新的資本指引4的規定在薪酬方面的變 動，使我們需要檢討薪酬政策，特別是 固定與浮動酬勞的比例，務求令我們在 整體薪酬方面維持競爭力，並且能挽留 具才幹的要員。 ‧ 管理層會密切監察與機構變動以及出 售項目相關的風險。 ‧ 我們繼續增加防範金融犯罪部、監管合 規部以及壓力測試的專家資源，並於監 管機構最終落實新規例時聯絡彼等。 首 互聯網罪行及詐騙 因客戶對互聯網及流動服務的使用日增， 滙豐面對相關詐騙及犯罪活動之風險亦隨 之上升。我們亦會面對流程或程序受阻以 及系統失靈或不可使用的風險，而且業務 可能因完全或部分超越控制範圍的事件 （例 如互聯網罪行及恐怖活動等） 而受阻。 對滙豐帶來的潛在影響 ‧ 互聯網罪行會導致財務損失及╱或客戶 數據與敏感資料被盜用，亦會令滙豐無 法為客戶服務。當我們倚賴外界供應商 向我們及我們的客戶提供服務時，同樣 面對這些風險。 緩減措施 ‧ 隨著互聯網罪行及詐騙風險不斷演變， 我們持續評估這些風險並採取監控措 施以降低風險。 ‧ 我們已加強監察以提高防範能力，並已 實施其他監控措施，例如利用雙重密碼 認證來減低因詐騙而引致損失的可能 性。 首 資訊保安風險 資訊及科技基礎設施保安，對維持銀行應 用程式及運作流程以至保障客戶利益和維 護滙豐的品牌，有舉足輕重的影響。滙豐 及其他跨國機構仍是網絡攻擊的目標，有 關攻擊可能會干擾我們的服務，包括外界 無法使用我們的網站、機構及客戶資料被 盜用或暴露保安漏洞。 對滙豐帶來的潛在影響 ‧ 資訊保安風險可能造成財務損失以及 聲譽受損，對客戶及投資者信心造成不 利影響。遺失客戶資料亦可能違反監管 規例，進而招致罰款及罰則。 緩減措施 ‧ 我們已作出巨額投資，透過加強培訓提 高員工對有關規定的意識，提升多層面 的監控、保障資訊及技術基礎設施，加 強監察及管理可能受到網絡攻擊的潛 在危險以及繼續進行安全評估，管理有 關風險。 首 數據管理 滙豐須制訂清晰的數據策略以應付監管及 其他呈報規定所要求的大量、精細、頻密 及廣泛的工作。作為一家全球系統重要性 銀行，滙豐亦須遵守巴塞爾銀行監管委員 會 （ 「巴塞爾委員會」 ） 頒布文件內訂明的原 則，以實施有效的風險數據匯總及風險報 告。 對滙豐帶來的潛在影響 ‧ 低效的數據管理可能會對我們及時向 監管機構、投資者及高級管理層匯總及 報告完整、準確及統一的數據的能力產 生不利影響。 ‧ 金融機構若未能在規定的最終期限前 履行巴塞爾委員會要求的數據管理責 任，可能受到監管措施制裁。 緩減措施 ‧ 我們自2012年成立數據策略委員會後， 已制訂集團數據策略，並已界定集團層 面的各項原則、標準及政策，以統一匯 總、報告及管理數據。 滙豐控股有限公司 124 董事會報告：風險 （續） 首要及新浮現風險╱特別提述部分 ‧ 我們正開展多個主要計劃和項目，以落 實數據策略，並履行巴塞爾委員會要求 的數據管理責任。 首 模型風險 滙豐在管理業務過程中為配合不同目的使 用各種模型，包括計算監管規定資本及經 濟資本、壓力測試、授出信貸、定價及財 務報告。模型風險指基於錯誤的模型推算 結果及報告來作出決策，或使用該等資料 作其他並非原來指定的用途，而導致不利 後果的潛在風險。模型的開發、實施或運 用欠佳，或管理層因誤解模型的推算結果 而作出不合適的行動，都可能產生模型風 險。監管環境及監管機構對銀行使用內部 模型釐定監管規定資本方面的關注，進一 步增加模型風險。 對滙豐帶來的潛在影響 ‧ 滙豐可能因模型的局限性或出現誤差 而招致虧損或需要持有額外資本。 ‧ 監管機構關注銀行計算監管規定資本 時所使用的各項內部模型及假設，因而 實施風險加權及違責損失率下限。有關 變動有可能令我們的監管規定資本水 平提高及╱或令資本水平更為波動。 緩減措施 ‧ 我們旨在透過對模型的開發、運用及驗 證工作加以適當管治，同時進行獨立檢 討、監察及回應。 新 第三方風險管理 在使用第三方服務供應商方面，我們已加 強風險管理，部分原因是應對全球監管機 構的嚴密審查。這包括如何作出外判決策、 如何管理主要關係，以及對集團使用的所 有第三方採用一致的風險管理。使用第三 方服務供應商的服務所產生的風險，透明 度可能較低，因此，對進行管理或施加影 響帶來更多的挑戰。 對滙豐帶來的潛在影響 ‧ 倘對第三方服務供應商的管理有任何 不足，可能導致多種風險，包括業務中 斷、監管缺失、失去機密資料和引發金 融犯罪。 緩減措施 ‧ 我們正在按照美國貨幣監理署和聯儲 局頒布的指引提升第三方風險管理能 力，加強對使用第三方服務的管控，並 增加這方面的監察和鑑證。 特別提述部分 （未經審核） 2014年，我們已考慮多個特定範疇，原因 是該等範疇可能對集團造成影響。有關範 疇可能已在 「首要及新浮現風險」 一節中確 定，下面詳述我們於年內所採取的相應措 施。 防範金融犯罪及監管合規工作 近年，我們面對的合規風險不斷上升，原 因是監管機構及其他機構就過往的業務活 動展開調查，而我們亦繼續就存在的問題 配合這些機構的工作。有關事項包括就未 能充分遵守反洗錢及制裁法律的調查與美 國當局達成延後起訴協議，以及向英國金 管局作出相關承諾 （ 「金融業操守監管局指 令」 ） 。我們已委任一名監察員評估我們履 行各項責任的進展，該監察員的工作範疇 詳載於第27頁。 我們繼續就金融業操守監管局對英國若干 產品的不當銷售手法，包括銷售還款保障 保險和為中小企設計的利率對沖產品及財 富管理產品，進行的多項調查作出回應。此 外，我們仍然牽涉一系列其他監管程序，包 括多個國家監管機構進行的調查和審查； 保障公平競爭與執法機關就釐定倫敦銀行 同業拆息和其他銀行同業拆息及其他基準 利率的過程，銀行小組成員過往提交的若 干資料及提交資料的程序，進行的調查和 審查。此外，集團亦正接受有關外匯、貴 金屬和信貸違責掉期相關業務的調查。該 等調查及法律程序的詳情，請參閱財務報 表附註40。 憑我們本身及廣泛業界的經驗可以清楚看 到，監管機構及執法機構正就可能違規事 宜展開更多調查活動，且該等違規事宜涉 及的直接及間接成本可能數額不菲。此外， 有關當局不斷推出大量新規例，其中大部 分均帶有某程度的域外效力，而滙豐業務 覆蓋廣泛地域，我們相信，集團在可見將 來面對的固有合規風險的水平會持續高企。 有關我們的合規風險管理工作的詳情，載 於第189頁。 私人銀行業務 近日有媒體報導我們瑞士私人銀行的過往 慣例及該行部分客戶的財務狀況。媒體所 關注的過往事件，反映瑞士業務在八年前 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 125 仍未全面實行我們現今的營運標準。 此後，我們徹底改變滙豐的經營方式並已對 我們客戶設立更嚴謹的中央管控。由7,000 多名合規職員組成的團隊，執行我們訂下 最嚴格的防範金融犯罪、監管合規及稅務 透明度標準。 環球私人銀行業務，尤其是旗下的瑞士私 人銀行業務，已進行徹底改革。我們在過去 數年採取重大步驟，實行改革及不再為不 能符合滙豐所訂嚴謹新標準的客戶服務， 包括令我們關注其稅務合規情況的客戶。 於此重新定位後，滙豐的瑞士私人銀行自 2007年以來客戶數目已減少近70%。 我們全面致力與有關當局交換資訊，並正 積極尋求落實措施，確保客戶稅務具透明 度，甚至比監管或法例規定更早實行。我 們亦正配合有關當局對該等事宜的調查。 監管規定壓力測試 壓力測試是監管機構用以評估銀行業及個 別銀行脆弱程度的重要工具，其結果可能 會對最低資本規定、風險及資本管理慣例 及已計劃的資本措施 （包括往後支付股息） 有重大影響。 我們須在多個司法管轄區進行監管機構規 定的壓力測試。該等測試不僅越來越頻密， 要提交的資料亦更見精細。該等測試包括 英國審慎監管局、聯儲局、歐洲銀行管理 局、歐洲央行、香港金融管理局 （ 「香港金 管局」 ） 和其他監管機構等規定的測試。監 管機構會同時採用定量及定質基準進行評 估，而後者主要聚焦於貸款組合質素、提 供的數據、壓力測試能力、具前瞻性的資 本管理程序以及內部管理程序。 2014年，集團參與了英國審慎監管局的首 次並行壓力測試計劃，涉及英國各大銀行。 該測試在整個企業範圍進行，測試內容包 括歐洲銀行管理局的基本境況，及主要遵 循歐洲銀行管理局的壓力境況，額外疊加 反映英國銀行體系面對的多個不利因素之 變數，包括英鎊幣值、住宅和商用物業價 格以及債券和股票價格大幅下跌，以及經 濟活動下降及失業率上升。滙豐在2014年 6月底向英國審慎監管局提交文件。集 團同時參與一項輔助計劃，根據其公司 層面數據提交架構(Firm Data Submission Framework)向英倫銀行定期提交數據。 英國審慎監管局在2014年12月16日披露 2014年並行壓力測試的結果。滙豐集團的 壓力下普通股權一級資本比率被英國審慎 監管局視為降至最低水平8.7%，並計及經 批准的緩減管理措施。此結果大幅高於目 標最低水平4.5%。 歐洲銀行管理局在2014年上半年進行整個 歐洲範圍的壓力測試，英國銀行的測試通 過英國審慎監管局進行。基本境況涵蓋多 項風險，包括信貸、市場、證券化、主權及 資金風險。不利的宏觀經濟境況包括：特 定國家╱地區對主權債券息差、短期利率 和住宅物業價格的衝擊；以及環球貿易收 縮減、中歐和東歐貨幣貶值及全球生產總 值增長放緩或收縮。 歐洲銀行管理局在2014年10月26日披露壓 力測試的結果，採用該測試的專用模板刊 登滙豐集團的詳細結果。預計我們的壓力 下普通股權一級資本比率於2015年底降至 8.7%的低點，但高於歐洲銀行管理局規定 的最低門檻5.5%。我們的全負荷壓力下普 通股權一級資本比率預計於2016年底為 9.3%，優於歐洲其他主要銀行。 英國審慎監管局和歐洲銀行管理局的測試 結果顯示滙豐的資本實力持續雄厚。 歐洲央行在2014年上半年進行綜合評估， 包括資產質素檢討和歐洲央行的壓力測試 流程，後者採用歐洲銀行管理局制訂的境 況。法國滙豐和馬耳他滙豐都在測試範圍 內，兩者均通過測試，有關結果亦在2014 年10月刊登。法國滙豐的普通股權一級資 本比率預計由2013年的12.9%降至2016年 底的6.6%，仍高於監管規定的最低水平。 降幅反映壓力測試對法國滙豐的業務模式 （包括集團的歐元利率交易業務） 的影響及 歐洲央行的信貸損失基準對貸款組合的影 響。 北美滙豐參與聯儲局的綜合資本分析及 檢討 （ 「CCAR」 ） 和達德－法蘭克壓力測試 （ 「DFAST」 ） 計劃，而美國滙豐銀行則參與 了美國貨幣監理署的DFAST計劃。兩者均 已於2014年1月6日就上述有關計劃提交文 件。於2014年3月26日，聯儲局通知北美滙 豐，基於質量理由，不接納其遞交的資本 計劃，並要求北美滙豐於2015年1月5日前 再次遞交資本計劃及改善壓力測試程序的 方法。然而，聯儲局批准北美滙豐遞交的 CCAR文件中所提出的資本措施，而北美滙 豐獲准可就其本身及附屬公司的已發行優 先股及信託優先證券支付股息。北美滙豐 的壓力下普通股權一級資本比率，由聯儲 局根據經監督的 「嚴重不利」 境況預測，將 降至最低水平9.4%，仍高於監管規定的最 低水平4.5%。北美滙豐提交其2015年CCAR 文件，亦可作為應聯儲局的再提交2014年 CCAR文件要求而作出，而美國滙豐銀行 則於2015年1月5日提交其2015年DFAST計 劃。聯儲局、北美滙豐及美國滙豐銀行將 滙豐控股有限公司 126 董事會報告：風險 （續） 特別提述部分╱信貸風險 於2015年3月披露根據2014年11月刊發的 監管境況進行測試的相關結果。此外，聯 儲局亦將表明是否反對北美滙豐所提交的 2015年CCAR文件中包括的資本計劃及資 本措施。 香港上海滙豐銀行於2014年上半年參與香 港金管局的壓力測試。香港金管局的壓力 境況是預計中國內地的增長大幅放緩及香 港經濟急速收縮。 石油及天然氣價格 石油及商品價格自2014年中以來大幅下 跌，乃由於全球供求日益失衡及市場氣氛 轉變。在2015年及以後，未來價格水平存 在相當大的不確定因素。油價持續受壓將 對國家、行業及個別公司造成不同程度的 影響： ‧ 國家層面：石油淨進口國很可能受惠於 油價下跌。在先進經濟體，這情況很可 能增加消費者的可動用收入，而新興市 場國家的政府財政狀況更有可能受惠。 對石油出口國的影響將取決於來自石 油的收入對財政收入的重要程度、開採 成本及有關國家能夠動用的財政儲備 金額。 ‧ 行業層面：石油及天然氣行業和後勤服 務將受到影響，儘管受影響程度將因相 關細分行業而異。大型綜合生產商很可 能維持穩健。對純生產商而言，例如頁 岩氣及油砂生產商等成本較高的純生產 商很可能承受更大壓力。同樣，基建及 服務供應商很可能因生產商縮減資本 開支而承受壓力。倘價格持續受壓，石 油及天然氣佔成本主要部分的行業 （例 如拖運、運輸及航運） 很可能受惠。 石油及天然氣行業已有一段時間被視為風 險較高的行業，並受到更嚴密的監控，相 關承受風險水平和新造貸款受到更嚴格的 審查。 滙豐對石油及天然氣行業的貸款比較多元 化。環球銀行及資本市場業務的貸款主要 集中於上游業務，並以屬投資級別的大型 環球綜合生產商為主。工商金融業務主要 向服務公司及純生產商提供貸款。其風險 分散於多個國家。 整個組合的風險承擔額約為340億美元，其 中涉及石油服務公司及非綜合生產商的風 險僅逾47%。倘油價受壓持續一至兩年， 被視為可能受此影響的較大型客戶，尤其 是但不限於專注於石油服務的公司，以及 依賴高成本開採方法的生產商，例如頁岩 氣或油砂生產商 （及其供應商） ，滙豐會對 他們作出深入的客戶審查。於進行該等審 查後，約5億美元的貸款被認定為必須給予 充分關注及進行密切管理。這類客戶中雖 然出現信貸疲弱的情況，此階段概無新客 戶被確認為已減值。 俄羅斯 在2014年，俄羅斯聯邦 （ 「俄羅斯」 ） 與西方國 家 （ 「西方」 ） 因烏克蘭問題而令局勢更為緊 張。迄今為止西方作出的反應，是選取一 部分俄羅斯個人、銀行及企業施加制裁， 於2014年實行。有關制裁要求的監控及行 動持續進行，並將對滙豐在俄羅斯的業務 帶來一些限制，儘管對集團的影響預計不 大。我們雖面對在烏克蘭註冊成立及註冊 的交易對手的風險，但認為並不重大。 2014年第四季，俄羅斯盧布及原油價格大 幅下跌，俄羅斯中央銀行多次上調利率。 管理層正在監察這些發展的影響，加上施 加的制裁，意味著俄羅斯的前景仍然極不 明朗，預期俄羅斯經濟在2015年萎縮。 我們在俄羅斯面對的風險主要是貸款。2014 年12月31日，貸款總額為40億美元。 除上文外，若干跨國客戶因擁有俄羅斯實 體 （依賴其供應或出口） 的大部分或少數股 東權益而間接面對俄羅斯風險。倘制裁及 其他行動的範圍及性質擴大或俄羅斯經濟 惡化，該等客戶的營運及業務可能受到不 利影響。再者，我們在俄羅斯的鄰近國家 經營業務，這些國家的金融系統與俄羅斯 經濟緊密連繫。管理層正在監察該等風險 額及潛在嚴重程度。 歐元區 歐盟近年來已修訂一系列的法例，旨在加強 其應對金融危機的能力，以及倘某歐盟成 員國面臨財政困難也可降低連鎖風險。目 前就希臘的紓困援助計劃條款進行磋商， 其結果極不明朗。希臘可能重組債務或可 能就其債務違約，亦有可能最終退出歐元 區。我們在希臘面對的風險主要是貸款及 反向回購。於2014年12月31日該等貸款及 反向回購分別達40億美元及20億美元。貸 款中的20億美元與航運業有關，是以美元 計值及在英國入賬。我們相信，航運業受 希臘經濟的影響較小，因這行業主要依賴 國際貿易。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 127 信貸風險 頁次 附錄 1 列表 頁次 信貸風險 129 206 信貸風險概要 129 五年客戶及同業貸款總額 130 五年貸款減值準備 130 按地區劃分的貸款減值準備 130 按行業劃分的貸款減值準備 130 五年貸款減值準備 130 信貸風險管理 206 信貸風險項目 130 最大信貸風險 130 其他減低信貸風險措施 130 最大信貸風險 131 貸款及其他信貸相關承諾 131 風險集中情況 132 206 金融投資 132 交易用途資產 132 衍生工具 132 貸款 132 按行業及地區劃分的客戶貸款總額 132 金融工具的信貸質素 133 207 信貸質素類別 按信貸質素劃分的金融工具分布情況 134 已逾期但並非已減值之金融 工具總額 136 按地區劃分的已逾期但並非已減值之 金融工具總額 136 已逾期但並非已減值之金融工具總額之 按日賬齡分析 136 已減值貸款 137 按地區劃分的已減值貸款變動 137 重議條件貸款及暫緩還款 138 208 按地區劃分的重議條件客戶貸款 139 按地區劃分的重議條件貸款的變動 140 貸款減值 141 按行業劃分的扣取自收益表之貸款 減值準備 141 按評估類別劃分的扣取自收益表之貸款 減值準備 141 按地區劃分的減值虧損準備佔客戶貸款 總額平均值的百分比 142 按行業及地區劃分的減值準備變動 142 客戶及同業貸款減值準備變動 143 減值評估 212 批發貸款 144 批發貸款總額 144 商用物業 145 商用物業貸款 145 按抵押品水平分析的商用物業貸款 （包括貸款承諾） 147 按抵押品水平 （僅指CRR/EL評級8至 10級） 分析的其他企業、商業及金融 機構 （非銀行） 貸款 （包括貸款承諾） 148 按抵押品水平分析的同業 貸款 （包括貸款承諾） 149 其他信貸風險 149 衍生工具 149 按產品類別劃分的衍生工具名義 合約金額及公允值 150 按類別劃分的場外抵押品協議 150 按地區劃分的反向回購－非交易用途 151 按地區劃分的反向回購－非交易用途 151 貸款管理組 213 滙豐控股有限公司 128 董事會報告：風險 （續） 信貸風險 頁次 附錄 1 列表 頁次 個人貸款 151 個人貸款總額 151 按揭貸款 152 其他個人貸款 153 美國滙豐融資的美國消費及按揭貸款－ 住宅按揭 153 美國滙豐融資：於美國的止贖物業 153 美國違約拖欠兩個月或以上貸款的趨勢 153 美國滙豐融資有抵押房地產貸款結欠的 貸款組合總額 154 美國滙豐融資組合內餘下重議條件 有抵押房地產賬項數目 154 美國滙豐融資修訂貸款條款及 重訂賬齡計劃 154 持有之抵押品及其他強化信貸條件 156 按抵押品水平分析的住宅按揭貸款 （包括貸款承諾） 156 補充資料 157 按行業劃分的五年貸款總額 157 按地區劃分的列賬基準與固定匯率基準 已減值貸款、準備及撥備對賬 158 扣取自收益表之列賬基準與固定匯率 基準之減值準備對賬 158 按行業劃分的五年貸款減值準備 159 減值虧損準備佔客戶貸款總額平均值 的百分比 159 五年減值準備變動 159 按國家╱地區劃分的客戶貸款總額 160 再融資風險 214 滙豐控股 161 滙豐控股－最大信貸風險 161 證券化風險及其他結構產品 161 214 滙豐的整體風險承擔 161 滙豐於綜合計算後所持資產抵押證券 之賬面值 162 資產抵押證券及債務抵押債券的定義 及分類 214 與按揭出售及證券化活動相關的陳述 和保證 162 1 風險附錄－風險政策及慣例。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 129 信貸風險 （未經審核） 信貸風險指一旦客戶或交易對手不能履行 合約責任時產生的財務虧損風險。這種風 險主要來自直接貸款、貿易融資及租賃業 務，也來自其他產品 （例如擔保及信貸衍生 工具） 及持有債務證券形式的資產。 2014年我們的信貸風險管理政策及慣例並 無重大變動。 有關信貸風險的現行政策及慣例，於第204頁風險 附錄內概述。 我們承擔的最大信貸風險載於第131頁，而 信貸質素則載於第133頁。雖然我們大部分 資產負債表項目均會產生信貸風險，但虧 損一般來自貸款及證券化風險項目，以及 其他結構產品。因此，我們的披露集中於 該兩個範疇。 今年，我們重新編製 「信貸風險」 一節，務 求令披露更加清晰，並避免重複。信貸風險 一節首先呈列信貸風險概要，隨後為我們 承擔的風險總額概覽。我們會於分析減值 準備前，說明各項衡量信貸質素的措施， 例如逾期狀況、已減值貸款及重議條件貸 款。年報內有關於批發貸款及個人貸款的 特定章節，提供有關的詳細資料。我們亦 有特別提述部分，提述批發貸款中的商用 物業風險以及個人貸款中的消費及按揭貸 款組合。隨後的章節會概述我們的證券化 風險及其他結構產品。有關我們承擔石油 及天然氣、俄羅斯及希臘的風險詳情，載 於第126頁 「特別提述部分」 。 如第347頁所述於更改資產負債表的呈列 方式後，非交易用途反向回購會在資產負 債表中另行列示，而不再納入為 「客戶貸 款」 及 「同業貸款」 。比較數字已按此重新 呈列。因此，提述客戶或同業貸款的分析 不包括非交易用途的反向回購。向客戶及 同業提供的非交易用途的反向回購金額載 於第151頁。 貸款減值撥備以及貸款減值準備及已減值 貸款均較2013年下跌。 貸款總額於2014年減少280億美元，包括不 利的匯兌變動510億美元，但不包括客戶貸 款增長的變動。 下文乃按固定匯率基準的評述，列表則按 列賬基準呈列。 信貸風險概要 （未經審核） 2014年 2013年 頁次 十億美元 十億美元 年底 最大信貸風險 3,133 3,112 131 貸款總額1 －個人貸款 393 411 132 －批發貸款 706 716 132 總計 1,099 1,127 132 已減值貸款 －個人貸款 15 19 137 －批發貸款 14 18 137 總計 29 37 137 已減值貸款佔貸款總額 百分比 －個人貸款 3.9% 4.6% －批發貸款 2.0% 2.5% －總計 2.7% 3.3% 十億美元 十億美元 減值準備 －個人貸款 4.6 6.6 143 －批發貸款 7.8 8.6 143 總計 12.4 15.2 143 扣除減值準備的貸款1 1,087 1,112 截至12月31日止年度 貸款減值準備 －個人貸款 1.8 3.1 141 －批發貸款 2.3 2.9 141 總計 4.1 6.0 141 有關註釋，請參閱第202頁。 有關固定匯率基準對賬詳情，載於第158頁。有關 按國家╱地區劃分的貸款分析，請參閱第160頁。 批發貸款總額上升210億美元。亞洲上升160億 美元，而北美洲則上升100億美元，增長幅 度較次的是中東及北非和拉丁美洲，但升 幅被歐洲的150億美元跌幅所抵銷。2014年 貸款減值準備較低，原因是不同經濟體系 環境改善，以及低息環境，令我們繼續受 惠。 撇除美國消費及按揭貸款組合按計劃縮減 的影響，個人貸款結欠增長77億美元，主要 受亞洲的按揭及其他貸款增長，以及北美洲 及拉丁美洲的按揭組合增長所帶動，但部分 增長被歐洲的貸款結欠減少所抵銷，原因是 英國的按揭以及信用卡組合還款。消費及按 揭貸款組合年內再縮減57億美元。貸款減值 準備因美國房屋市場好轉，以及我們繼續縮 減消費及按揭貸款組合而下跌。 滙豐控股有限公司 130 董事會報告：風險 （續） 信貸風險 五年客戶及同業貸款總額 1 （十億美元） （未經審核） 批發 個人 393 367 391 392 378 620 647 666 698 692 32 27 24 19 15 15 15 15 18 14 2010 2011 2012 2013 2014 2010 2011 2012 2013 2014 已減值 並非已減值 五年貸款減值準備 （十億美元） （未經審核） 批發 個人 11.2 9.3 5.4 3.1 1.8 2.4 2.2 2.8 2.9 2.3 2010 2011 2012 2013 2014 2010 2011 2012 2013 2014 按地區劃分的貸款減值準備 （十億美元） （未經審核） 歐洲 亞洲 中東及北非 北美洲 拉丁美洲 1.7 0.5 1.2 2.6 1.1 0.6 0.3 2.0 2013年 2014年 – – 按行業劃分的貸款減值準備 （十億美元） （未經審核） 第一 留置權 住宅按揭 其他 個人貸款 商用物業 其他企業 及商業 金融機構 0.6 2.5 0.8 2.2 (0.1) (0.1) 1.9 0.3 2.0 – 2013年 2014年 五年貸款減值準備 （未經審核） 貸款減值準備 （十億美元） 貸款減值準備佔已減值貸款百分比 個人 批發 12.3 9.8 8.2 6.6 4.6 7.9 7.9 8.0 8.6 7.8 38% 37% 35% 35% 30% 53% 51% 53% 48% 55% 2010 2011 2012 2013 2014 2010 2011 2012 2013 2014 有關註釋，請參閱第202頁。 信貸風險項目 最大信貸風險 （經審核） 第131頁列表載列資產負債項目、對銷項目 以及貸款及其他信貸相關承諾的資料。資 產負債表變動的評述載於第58頁。衍生工 具的抵銷額增加，與最大風險額的升幅一 致。 向客戶提供的企業及商業貸款抵銷額下跌 310億美元。跌幅主要來自英國，因為當地 少數客戶能利用其透支額及存款額的淨利 息安排。年內，由於我們調整資金管理服 務方針，更為配合全球一致的策略，故不 少客戶增加還款的頻率，從而減少可供動 用的抵銷額。 「最大信貸風險」 表 （第131頁） 該表呈列資產負債表及資產負債表外金融工具的最 大信貸風險，其中並未計及所持抵押品或其他強化 信貸條件 （除非該等強化信貸條件符合會計對銷規 定） 。在資產負債表確認之金融資產，其最大信貸風 險相等於其賬面值；至於授出的金融擔保及同類合 約，最大信貸風險是對方要求履行擔保時，我們須 支付的最高金額。至於貸款承諾及其他信貸相關承 諾，最大信貸風險一般是信貸承諾所涉的全部金額。 列表中的對銷額，涉及滙豐在交易對手違約下可依 法強制執行之對銷權利，因此存在有關信貸風險的 風險淨額。然而，由於集團在一般情況下無意按淨 額結算該等結餘，因此，該等結餘並未符合按淨額 呈列的會計準則。 至於衍生工具，對銷一欄亦包括以現金及其他金融 資產收取的抵押品。 其他減低信貸風險措施 雖然未於 「最大信貸風險」 列表中披露作對 銷，但我們仍設有其他安排以降低最大信 貸風險，當中包括借款人特定資產 （例如住 宅物業） 的抵押品押記。其他減低信貸風險 措施包括證券短倉及持作相連保險╱投資 合約 （風險主要由投保人承擔） 一部分的金 融資產。此外，我們以未於資產負債表確 認的金融工具形式持有抵押品。 有關若干貸款及衍生工具的抵押品的進一步詳情， 請分別參閱財務報表附註32及第147至156頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 131 最大信貸風險 （經審核） 2014年 2013年 最大風險 對銷 淨額 最大風險 對銷 淨額 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 現金及於中央銀行的結餘 129,957 － 129,957 166,599 － 166,599 向其他銀行託收中之項目 4,927 － 4,927 6,021 － 6,021 香港政府負債證明書 27,674 － 27,674 25,220 － 25,220 交易用途資產 228,944 － 228,944 239,301 (1,777) 237,524 －國庫及其他合資格票據 16,170 － 16,170 21,584 － 21,584 －債務證券 141,532 － 141,532 141,644 － 141,644 －同業貸款 27,581 － 27,581 27,885 － 27,885 －客戶貸款 43,661 － 43,661 48,188 (1,777) 46,411 指定以公允值列賬之金融資產 9,031 － 9,031 12,719 － 12,719 －國庫及其他合資格票據 56 － 56 50 － 50 －債務證券 8,891 － 8,891 12,589 － 12,589 －同業貸款 84 － 84 76 － 76 －客戶貸款 － － － 4 － 4 衍生工具 345,008 (313,300) 31,708 282,265 (252,344) 29,921 按已攤銷成本持有的客戶貸款1 974,660 (67,094) 907,566 992,089 (96,726) 895,363 －個人貸款 388,954 (4,412) 384,542 404,126 (1,348) 402,778 －企業及商業貸款 535,184 (59,197) 475,987 537,922 (90,215) 447,707 －金融機構 （非銀行之 金融機構） 貸款 50,522 (3,485) 47,037 50,041 (5,163) 44,878 按已攤銷成本持有的同業貸款1 112,149 (258) 111,891 120,046 (587) 119,459 反向回購協議－非交易用途 161,713 (5,750) 155,963 179,690 (22,267) 157,423 金融投資 404,773 － 404,773 416,785 － 416,785 －國庫及其他同類票據 81,517 － 81,517 78,111 － 78,111 －債務證券 323,256 － 323,256 338,674 － 338,674 其他資產 35,264 － 35,264 37,324 (22) 37,302 －持作出售用途資產 1,375 － 1,375 3,306 (22) 3,284 －背書及承兌 10,775 － 10,775 11,624 － 11,624 －其他 23,114 － 23,114 22,394 － 22,394 金融擔保及同類合約2 47,078 － 47,078 46,300 － 46,300 貸款及其他信貸相關承諾3 651,380 － 651,380 587,603 － 587,603 於12月31日 3,132,558 (386,402) 2,746,156 3,111,962 (373,723) 2,738,239 有關註釋，請參閱第202頁。 貸款及其他信貸相關承諾 3 （未經審核） 歐洲 亞洲4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 個人貸款 86,247 96,497 2,995 15,636 11,679 213,054 企業及商業貸款 98,045 138,366 20,141 102,911 17,540 377,003 金融機構貸款5 26,605 9,355 711 23,559 1,093 61,323 於2014年12月31日 210,897 244,218 23,847 142,106 30,312 651,380 個人貸款 92,148 74,445 2,940 15,647 9,774 194,954 企業及商業貸款 91,895 120,084 19,045 92,837 21,956 345,817 金融機構貸款5 18,930 8,477 705 17,478 1,242 46,832 於2013年12月31日 202,973 203,006 22,690 125,962 32,972 587,603 有關註釋，請參閱第202頁。 滙豐控股有限公司 132 董事會報告：風險 （續） 信貸風險 風險集中情況 （未經審核） 有關信貸風險集中情況，請參閱第206頁風險附錄。 2014年，我們的貸款組合分散各個地區， 加上環球業務及產品眾多，確保我們不會 過份依賴少數市場帶動增長。我們的業務 遍布不同地區，亦讓我們得以在增長較快 及具有國際聯繫的市場推行增長策略。 金融投資 2014年，我們持有多個不同發行商及地區 之可供出售政府及政府機構債務證券、企 業債務證券、資產抵押證券及其他證券， 其中15%投資於銀行及其他金融機構發行 的證券，72%投資於政府或政府機構債務證 券。我們亦持有有資產支持的保險及投資 合約。 有關金融投資的分析，請參閱財務報表附註18。 交易用途資產 交易用途證券仍佔交易用途資產的最大比 重，為77%，2013年則為75%。政府及政府 機構債務證券佔交易用途證券組合的最大 比重。我們對美國財政部及政府機構債務 證券 （260億美元） 、英國政府債務證券 （93 億美元） 及香港政府債務證券 （69億美元） 均 有重大的風險承擔。 有關持作交易用途的債務及股權證券的分析，請 參閱財務報表附註12。 衍生工具 於2014年12月31日，衍生工具資產為3,450億 美元 （2013年：2,820億美元） 。透過交易所、 中央交易對手以及非中央交易對手結算的 衍生工具金額詳情，載於第150頁。 有關衍生工具的分析，請參閱第150頁及財務報表 附註16。 客戶貸款 以下各表乃根據行業及貸款附屬公司之主 要業務所在地分析各類客戶貸款；如屬香 港上海滙豐銀行、英國滙豐銀行、中東滙 豐銀行有限公司 （ 「中東滙豐銀行」 ） 及美國 滙豐銀行的貸款，則根據貸款分行之所在 地分析。貸款的地區及行業分布與去年相 若。 按國家╱地區劃分的貸款詳情，請參閱第160頁。 按行業及地區劃分的客戶貸款總額 （經審核） 歐洲 亞洲4 中東及 北非 北美洲 拉丁美洲 總計 佔各類 貸款總額 百分比 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 個人貸款 178,531 129,515 6,571 65,400 13,537 393,554 39.9 －第一留置權住宅按揭6 131,000 93,147 2,647 55,577 4,153 286,524 29.0 －其他個人貸款7 47,531 36,368 3,924 9,823 9,384 107,030 10.9 企業及商業貸款 210,585 220,799 20,588 57,862 30,722 540,556 54.8 －製造業貸款 39,456 37,767 2,413 15,299 12,051 106,986 10.9 －國際貿易及服務貸款 76,629 72,814 9,675 13,484 8,189 180,791 18.3 －商用物業貸款 28,187 35,678 579 6,558 2,291 73,293 7.4 －其他與物業有關貸款 7,126 34,379 1,667 8,934 281 52,387 5.3 －政府貸款 2,264 1,195 1,552 164 968 6,143 0.6 －其他商業貸款8 56,923 38,966 4,702 13,423 6,942 120,956 12.3 金融機構貸款 23,103 13,997 3,291 9,034 1,393 50,818 5.1 －非銀行之金融機構貸款 21,867 13,410 3,289 9,034 1,199 48,799 4.9 －結算賬項 1,236 587 2 － 194 2,019 0.2 經重新分類之資產抵押證券 1,938 － － 131 － 2,069 0.2 於2014年12月31日 各類客戶貸款總額(A) 414,157 364,311 30,450 132,427 45,652 986,997 100.0 按地區分類佔A之百分比 42.0% 36.9% 3.1% 13.4% 4.6% 100.0% － 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 133 歐洲 亞洲4 中東及 北非 北美洲 拉丁美洲 總計 佔各類 貸款總額 百分比 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 個人貸款 192,107 124,529 6,484 72,690 14,918 410,728 40.8 －第一留置權住宅按揭6 140,474 92,047 2,451 60,955 3,948 299,875 29.8 －其他個人貸款7 51,633 32,482 4,033 11,735 10,970 110,853 11.0 企業及商業貸款 239,116 203,894 19,760 50,307 30,188 543,265 53.9 －製造業貸款 55,920 30,758 3,180 11,778 12,214 113,850 11.3 －國際貿易及服務貸款 76,700 79,368 8,629 11,676 8,295 184,668 18.3 －商用物業貸款 31,326 34,560 639 5,900 2,421 74,846 7.4 －其他與物業有關貸款 7,308 27,147 1,333 8,716 328 44,832 4.5 －政府貸款 3,340 1,021 1,443 499 974 7,277 0.7 －其他商業貸款8 64,522 31,040 4,536 11,738 5,956 117,792 11.7 金融機構貸款 27,872 9,688 2,532 9,055 1,376 50,523 5.0 －非銀行之金融機構貸款 26,314 9,359 2,532 9,055 1,277 48,537 4.8 －結算賬項 1,558 329 － － 99 1,986 0.2 經重新分類之資產抵押證券 2,578 － － 138 － 2,716 0.3 於2013年12月31日 各類客戶貸款總額(B) 461,673 338,111 28,776 132,190 46,482 1,007,232 100.0 按地區分類佔B之百分比 45.8% 33.6% 2.9% 13.1% 4.6% 100.0% 有關註釋，請參閱第202頁。 金融工具的信貸質素 （經審核） 滙豐金融工具信貸質素的現行政策及慣例於第207 頁的風險附錄內概述。 我們對所有須承擔信貸風險的金融工具評 估信貸質素。 第207頁 （未經審核） 所載定義界定了貸款、 債務證券組合及衍生工具的五類信貸質 素。有關滙豐於綜合計算後所持資產抵押 證券的其他信貸質素資料載於第162頁。 關於下列披露資料，逾期最多90日但根據 披露慣例並無分類為已減值的零售貸款， 不會在相關的預期虧損 （ 「EL」 ） 級別內披露， 而會另行分類為已逾期但並非已減值。 資產整體信貸質素維持於 「穩健」 及 「良好」 類別，佔組合的84%，而屬 「滿意」 、 「低於 標準」 及 「已逾期但未減值」 以及 「已減值」 類別則分別佔13%、2%及1%。 滙豐控股有限公司 134 董事會報告：風險 （續） 信貸風險 按信貸質素劃分的金融工具分布情況 （經審核） 並非逾期或已減值 已逾期 但並非 已減值 已減值 各類 貸款總額 減值 準備 9 總計 穩健 良好 滿意 低於 標準 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 現金及於中央銀行的結餘 127,971 1,438 195 353 129,957 129,957 向其他銀行託收中之項目 4,515 46 365 1 4,927 4,927 香港政府負債證明書 27,674 － － － 27,674 27,674 交易用途資產10 168,521 35,042 24,740 641 228,944 228,944 －國庫及其他合資格票據 13,938 1,641 559 32 16,170 16,170 －債務證券 111,138 17,786 12,305 303 141,532 141,532 －貸款： 同業 17,492 4,961 5,016 112 27,581 27,581 客戶 25,953 10,654 6,860 194 43,661 43,661 指定以公允值列賬之 金融資產10 3,017 4,476 1,207 331 9,031 9,031 －國庫及其他合資格票據 5 － － 51 56 56 －債務證券 3,011 4,476 1,124 280 8,891 8,891 －貸款： 同業 1 － 83 － 84 84 客戶 － － － － － － 衍生工具10 269,490 58,596 15,962 960 345,008 345,008 按已攤銷成本持有的客戶 貸款11 487,734 239,136 196,685 20,802 13,357 29,283 986,997 (12,337) 974,660 －個人貸款 320,678 32,601 15,109 1,130 8,876 15,160 393,554 (4,600) 388,954 －企業及商業貸款 141,375 192,799 171,748 18,986 3,922 13,795 542,625 (7,441) 535,184 －金融機構 （非銀行 之金融機構） 貸款 25,681 13,736 9,828 686 559 328 50,818 (296) 50,522 按已攤銷成本持有的 同業貸款 83,766 19,525 7,945 914 1 47 112,198 (49) 112,149 反向回購協議 －非交易用途 98,470 28,367 33,283 1,593 － － 161,713 － 161,713 金融投資 347,218 27,373 22,600 5,304 － 2,278 404,773 404,773 －國庫及其他同類票據 68,966 6,294 4,431 1,826 － － 81,517 81,517 －債務證券 278,252 21,079 18,169 3,478 － 2,278 323,256 323,256 其他資產 13,015 7,564 12,976 631 210 884 35,280 (16) 35,264 －持作出售用途資產 802 43 79 － 2 465 1,391 (16) 1,375 －背書及承兌 1,507 4,644 4,281 298 34 11 10,775 10,775 －應計收益及其他 10,706 2,877 8,616 333 174 408 23,114 23,114 於2014年12月31日 1,631,391 421,563 315,958 31,530 13,568 32,492 2,446,502 (12,402) 2,434,100 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 135 並非逾期或已減值 已逾期 但並非 已減值 已減值 各類 貸款總額 減值 準備 9 總計 穩健 良好 滿意 低於 標準 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 現金及於中央銀行的結餘 162,017 2,877 265 1,440 166,599 166,599 向其他銀行託收中之項目 5,590 66 286 79 6,021 6,021 香港政府負債證明書 25,220 － － － 25,220 25,220 交易用途資產10 163,444 39,475 34,868 1,514 239,301 239,301 －國庫及其他合資格票據 17,235 3,585 758 6 21,584 21,584 －債務證券 107,831 16,498 16,167 1,148 141,644 141,644 －貸款： 同業 15,804 5,546 6,342 193 27,885 27,885 客戶 22,574 13,846 11,601 167 48,188 48,188 指定以公允值列賬之 金融資產10 6,608 5,183 671 257 12,719 12,719 －國庫及其他合資格票據 50 － － － 50 50 －債務證券 6,490 5,179 664 256 12,589 12,589 －貸款： 同業 68 － 7 1 76 76 客戶 － 4 － － 4 4 衍生工具10 220,711 47,004 13,425 1,125 282,265 282,265 按已攤銷成本持有的客戶 貸款11 488,504 243,077 199,821 23,942 15,460 36,428 1,007,232 (15,143) 992,089 －個人貸款 326,269 39,024 14,882 1,580 10,175 18,798 410,728 (6,602) 404,126 －企業及商業貸款 132,943 194,966 174,905 21,281 5,009 16,877 545,981 (8,059) 537,922 －金融機構 （非銀行 之金融機構） 貸款 29,292 9,087 10,034 1,081 276 753 50,523 (482) 50,041 按已攤銷成本持有的 同業貸款 91,498 21,131 6,266 1,123 11 75 120,104 (58) 120,046 反向回購協議 －非交易用途 111,543 37,878 28,265 2,004 － － 179,690 － 179,690 金融投資 362,799 27,833 17,556 6,089 － 2,508 416,785 416,785 －國庫及其他同類票據 69,364 5,595 1,856 1,296 － － 78,111 78,111 －債務證券 293,435 22,238 15,700 4,793 － 2,508 338,674 338,674 其他資產 12,501 8,028 14,848 1,159 307 592 37,435 (111) 37,324 －持作出售用途資產 1,129 642 1,050 351 89 156 3,417 (111) 3,306 －背書及承兌 1,976 4,824 4,562 225 19 18 11,624 11,624 －應計收益及其他 9,396 2,562 9,236 583 199 418 22,394 22,394 於2013年12月31日 1,650,435 432,552 316,271 38,732 15,778 39,603 2,493,371 (15,312) 2,478,059 有關註釋，請參閱第202頁。 滙豐控股有限公司 136 董事會報告：風險 （續） 信貸風險 已逾期但並非已減值之金融工具總額 （經審核） 已逾期但並非已減值之金融工具總額指雖 然客戶未能根據信貸的合約條款還款，但 未達致第137頁所述已減值貸款標準的貸 款。 整體而言，已逾期但並非已減值貸款結欠 減少22億美元，主要原因是消費及按揭貸 款組合持續縮減及出售貸款。 按地區劃分的已逾期但並非已減值之金融工具總額 （經審核） 歐洲 亞洲4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 按已攤銷成本持有的客戶貸款 2,409 4,260 704 4,634 1,350 13,357 －個人貸款 1,159 2,880 182 3,759 896 8,876 －企業及商業貸款 1,244 1,102 508 623 445 3,922 －金融機構 （非銀行之 金融機構） 貸款 6 278 14 252 9 559 其他金融工具 6 52 31 97 25 211 於2014年12月31日 2,415 4,312 735 4,731 1,375 13,568 按已攤銷成本持有的客戶貸款 2,399 4,211 757 6,453 1,640 15,460 －個人貸款 1,287 2,764 174 4,817 1,133 10,175 －企業及商業貸款 1,092 1,197 580 1,635 505 5,009 －金融機構 （非銀行之 金融機構） 貸款 20 250 3 1 2 276 其他金融工具 45 49 50 101 73 318 於2013年12月31日 2,444 4,260 807 6,554 1,713 15,778 有關註釋，請參閱第202頁。 已逾期但並非已減值之金融工具總額之按日賬齡分析 （經審核） 不多於 29日 30至 59日 60至 89日 90至 179日 180日 及以上 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 按已攤銷成本持有的客戶貸款 10,427 2,057 801 54 18 13,357 －個人貸款 6,477 1,717 676 5 1 8,876 －企業及商業貸款 3,417 328 114 48 15 3,922 －金融機構 （非銀行之 金融機構） 貸款 533 12 11 1 2 559 其他金融工具 130 33 18 12 18 211 於2014年12月31日 10,557 2,090 819 66 36 13,568 按已攤銷成本持有的客戶貸款 11,689 2,587 1,057 76 51 15,460 －個人貸款 7,170 2,124 865 16 － 10,175 －企業及商業貸款 4,290 418 190 60 51 5,009 －金融機構 （非銀行之 金融機構） 貸款 229 45 2 － － 276 其他金融工具 214 55 26 12 11 318 於2013年12月31日 11,903 2,642 1,083 88 62 15,778 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 137 已減值貸款 （經審核） 已減值貸款指符合下列任何標準的貸款： ‧ 分類為客戶風險評級 （ 「CRR」 ） 9級或 CRR10級的批發貸款。當銀行認為客戶 不可能在不追索抵押品下全數償還其 信貸責任，或客戶逾期償還滙豐的任 何重大信貸責任達90日以上，則列入 上述評級。 ‧ 分類為預期虧損 （ 「EL」 ） 9級或EL10級的零 售貸款。零售貸款若已逾期超過90日， 除非已個別評估為並非已減值，否則通 常會列入該等評級。 ‧ 約定現金流會出現變動的重議條件貸 款，而現金流變動是基於還款優惠 （貸 款人原本不會考慮，且倘無還款優惠， 借款人可能無法完全履行合約還款責 任） ，除非還款優惠並不重大，而且貸 款並無其他減值指標則另作別論。重議 條件貸款會繼續分類為已減值，直至有 充分證據顯示日後無法收回現金流的 風險已大幅下降且貸款再無其他減值 指標為止。 至於按綜合基準評估減值的貸款，支持貸 款重新分類不再列為已減值的證據，通常 包括按照原有或經修訂條款履約還款的紀 錄，但有關情況會視乎重議條件的性質和 數量，以及重議條件涉及的信貸風險特性 而定。至於按個別基準評估減值的貸款， 則會就個別情況評估所有取得的證據。 有關CRR及EL等級的其他詳情，請參閱第207頁。 按地區劃分的已減值貸款變動 （未經審核） 歐洲 亞洲4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日的已減值貸款 13,228 1,623 2,285 15,123 4,244 36,503 －個人貸款 2,938 526 317 13,669 1,348 18,798 －企業及商業貸款 9,714 1,082 1,765 1,427 2,889 16,877 －金融機構貸款5 576 15 203 27 7 828 年內分類為已減值 3,367 1,970 346 4,724 3,342 13,749 －個人貸款 1,168 857 193 4,360 1,958 8,536 －企業及商業貸款 2,166 1,113 153 354 1,383 5,169 －金融機構貸款5 33 － － 10 1 44 年內由已減值轉移至未減值 (1,661) (230) (320) (2,609) (730) (5,550) －個人貸款 (282) (184) (178) (2,551) (364) (3,559) －企業及商業貸款 (1,319) (46) (53) (57) (366) (1,841) －金融機構貸款5 (60) － (89) (1) － (150) 撇銷金額 (2,037) (617) (111) (1,369) (2,048) (6,182) －個人貸款 (631) (470) (77) (1,007) (1,371) (3,556) －企業及商業貸款 (1,201) (147) (29) (356) (673) (2,406) －金融機構貸款5 (205) － (5) (6) (4) (220) 還款及其他款項淨額 (2,655) (698) (219) (4,175) (1,443) (9,190) －個人貸款 (649) (238) (13) (3,645) (514) (5,059) －企業及商業貸款 (1,975) (457) (140) (506) (926) (4,004) －金融機構貸款5 (31) (3) (66) (24) (3) (127) 於2014年12月31日的已減值貸款 10,242 2,048 1,981 11,694 3,365 29,330 －個人貸款 2,544 491 242 10,826 1,057 15,160 －企業及商業貸款 7,385 1,545 1,696 862 2,307 13,795 －金融機構貸款5 313 12 43 6 1 375 已減值貸款佔貸款總額百分比 2.3% 0.5% 4.8% 8.4% 6.1% 2.7% －個人貸款 1.4% 0.4% 3.7% 16.6% 7.8% 3.9% －企業及商業貸款 3.5% 0.7% 8.2% 1.5% 7.5% 2.5% －金融機構貸款5 0.7% 0.0% 0.3% 0.0% 0.0% 0.2% 滙豐控股有限公司 138 董事會報告：風險 （續） 信貸風險 歐洲 亞洲4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2013年1月1日的已減值貸款 11,145 1,624 2,474 20,345 3,188 38,776 －個人貸款 2,466 611 368 18,726 1,580 23,751 －企業及商業貸款 8,058 967 1,872 1,592 1,604 14,093 －金融機構貸款5 621 46 234 27 4 932 年內分類為已減值 4,952 1,424 419 6,168 4,333 17,296 －個人貸款 1,176 798 107 5,319 1,872 9,272 －企業及商業貸款 3,726 623 306 837 2,453 7,945 －金融機構貸款5 50 3 6 12 8 79 年內由已減值轉移至未減值 (1,215) (145) (166) (3,198) (642) (5,366) －個人貸款 (265) (137) (68) (3,172) (266) (3,908) －企業及商業貸款 (804) (8) (85) (24) (375) (1,296) －金融機構貸款5 (146) － (13) (2) (1) (162) 撇銷金額 (1,411) (538) (165) (1,706) (1,957) (5,777) －個人貸款 (423) (444) (79) (1,433) (1,456) (3,835) －企業及商業貸款 (927) (91) (75) (270) (499) (1,862) －金融機構貸款5 (61) (3) (11) (3) (2) (80) 還款及其他款項淨額 (243) (742) (277) (6,486) (678) (8,426) －個人貸款 (16) (302) (11) (5,771) (382) (6,482) －企業及商業貸款 (339) (409) (253) (708) (294) (2,003) －金融機構貸款5 112 (31) (13) (7) (2) 59 於2013年12月31日的已減值貸款 13,228 1,623 2,285 15,123 4,244 36,503 －個人貸款 2,938 526 317 13,669 1,348 18,798 －企業及商業貸款 9,714 1,082 1,765 1,427 2,889 16,877 －金融機構貸款5 576 15 203 27 7 828 已減值貸款佔貸款總額百分比 2.7% 0.4% 6.5% 10.9% 7.5% 3.2% －個人貸款 1.5% 0.4% 4.9% 18.8% 9.0% 4.6% －企業及商業貸款 4.0% 0.5% 8.9% 2.8% 9.6% 3.1% －金融機構貸款5 1.1% 0.0% 2.3% 0.2% 0.1% 0.5% 有關註釋，請參閱第202頁。 年內已減值貸款減少72億美元。個人已減 值貸款下降主要由於北美洲的消費及按揭 貸款組合持續縮減及出售。個人貸款中 「還 款及其他款項淨額」 包括29億美元的消費 及按揭貸款組合資產，有關資產已重新分 類為持作出售用途及於年內出售。 已減值批發貸款的跌幅主要來自歐洲，其 次是北美洲及拉丁美洲，乃由於還款及新 的已減值貸款減少，反映於這些市場經濟 狀況改善。有關跌幅被亞洲的增幅所抵銷。 重議條件貸款及暫緩還款 （經審核） 有關重議條件貸款及暫緩還款的現行政策和程序， 載於第208頁風險附錄。 貸款的合約條款可因多種理由修訂，包括 市況的變化、挽留客戶及其他與客戶的現 時或潛在信貸惡化並不相關的因素。 「暫 緩還款」 指因債務人財務困難按貸款合約 條款提供的優惠。若由於我們對借款人履 行合約到期還款責任的能力有重大質疑而 修訂貸款的合約還款條款，我們將在信貸 困境條件下已就其提供優惠的貸款分類為 「重議條件貸款」 並作相應呈報。於重議條 件時，倘取消現有協議並訂立條款大不相 同的新協議，或倘現有協議的條款經過修 訂，以致重議條件貸款已幾乎是完全不相 同之金融工具，則會就會計目的而撤銷確 認有關貸款，並重新確認新貸款。然而， 新確認金融資產將於重議條件貸款分類中 保留。提供予客戶不影響還款結構或還款 基準的貸款優惠 （如豁免財務或抵押契約 條款） 不會直接就客戶償還到期債務的能 力方面為其提供優惠減免，因此並不包括 在此項分類內。 最重大的重議條件貸款組合仍然是北美洲 的組合，其大部分為美國滙豐融資持有的 零售貸款。 下表載列按行業、地區以及信貸質素分類 之集團持有的重議條件客戶貸款的賬面值 總額。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 139 按地區劃分的重議條件客戶貸款 （經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 第一留置權住宅按揭 1,605 94 58 13,540 60 15,357 －並非逾期或已減值 529 63 19 3,695 32 4,338 －已逾期但並非已減值 221 8 1 1,894 5 2,129 －已減值 855 23 38 7,951 23 8,890 其他個人貸款7 324 292 27 1,267 326 2,236 －並非逾期或已減值 184 173 16 453 14 840 －已逾期但並非已減值 40 22 5 214 1 282 －已減值 100 97 6 600 311 1,114 企業及商業貸款 5,469 501 1,439 427 1,324 9,160 －並非逾期或已減值 1,383 102 483 36 303 2,307 －已逾期但並非已減值 68 － 31 1 1 101 －已減值 4,018 399 925 390 1,020 6,752 金融機構貸款5 413 4 323 1 1 742 －並非逾期或已減值 219 － 305 － － 524 －已逾期但並非已減值 － － － － － － －已減值 194 4 18 1 1 218 於2014年12月31日之重議條件貸款 7,811 891 1,847 15,235 1,711 27,495 －並非逾期或已減值 2,315 338 823 4,184 349 8,009 －已逾期但並非已減值 329 30 37 2,109 7 2,512 －已減值 5,167 523 987 8,942 1,355 16,974 重議條件貸款之減值準備 1,458 170 458 1,499 704 4,289 －重議條件貸款佔貸款總額百分比 1.9% 0.2% 6.1% 11.5% 3.7% 2.8% 第一留置權住宅按揭 1,820 117 91 16,853 76 18,957 －並非逾期或已減值 392 78 47 4,332 32 4,881 －已逾期但並非已減值 517 11 3 2,684 4 3,219 －已減值 911 28 41 9,837 40 10,857 其他個人貸款7 431 318 58 1,277 531 2,615 －並非逾期或已減值 253 207 33 503 18 1,014 －已逾期但並非已減值 39 24 17 284 2 366 －已減值 139 87 8 490 511 1,235 企業及商業貸款 7,270 330 1,583 658 2,161 12,002 －並非逾期或已減值 1,796 134 677 47 493 3,147 －已逾期但並非已減值 193 4 126 34 5 362 －已減值 5,281 192 780 577 1,663 8,493 金融機構貸款5 235 2 362 1 1 601 －並非逾期或已減值 93 － 265 － － 358 －已逾期但並非已減值 － － － － － － －已減值 142 2 97 1 1 243 於2013年12月31日之重議條件貸款 9,756 767 2,094 18,789 2,769 34,175 －並非逾期或已減值 2,534 419 1,022 4,882 543 9,400 －已逾期但並非已減值 749 39 146 3,002 11 3,947 －已減值 6,473 309 926 10,905 2,215 20,828 重議條件貸款之減值準備 1,867 101 460 2,285 1,014 5,727 －重議條件貸款佔貸款總額百分比 2.1% 0.2% 7.3% 14.2% 6.0% 3.4% 有關註釋，請參閱第202頁。 下表列示重議條件貸款年內的變動。2014 年，重議條件貸款減少67億美元至270億美 元。個人貸款的重議條件貸款減少40億美 元。計入 「其他」 變動包括19億美元的消費 及按揭貸款組合資產，有關資產已撥入持 作出售用途。新造重議條件貸款及撇銷額 減少，原因是美國房屋市場以及經濟狀況 改善。 批發貸款的重議條件貸款減少27億美元， 跌幅主要集中於歐洲及拉丁美洲，其原因 是撇銷額及還款額增加。 滙豐控股有限公司 140 董事會報告：風險 （續） 信貸風險 按地區劃分的重議條件貸款的變動 （未經審核） 歐洲 亞洲4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日之重議條件貸款 9,756 767 2,094 18,789 2,769 34,175 －個人貸款 2,251 435 149 18,130 607 21,572 －企業及商業貸款 7,270 330 1,583 658 2,161 12,002 －金融機構貸款 235 2 362 1 1 601 年內重議條件且並無撤銷確認之貸款 1,543 371 296 862 725 3,797 －個人貸款 433 83 10 774 310 1,610 －企業及商業貸款 939 288 286 78 415 2,006 －金融機構貸款 171 － － 10 － 181 年內重議條件並確認為新造貸款之貸款 500 5 79 － 92 676 －個人貸款 69 2 － － 28 99 －企業及商業貸款 381 － 61 － 64 506 －金融機構貸款 50 3 18 － － 71 還款 (2,416) (246) (562) (1,518) (1,036) (5,778) －個人貸款 (635) (96) (47) (1,319) (288) (2,385) －企業及商業貸款 (1,757) (149) (445) (189) (747) (3,287) －金融機構貸款 (24) (1) (70) (10) (1) (106) 撇銷金額 (828) (42) (23) (640) (510) (2,043) －個人貸款 (88) (28) (7) (568) (223) (914) －企業及商業貸款 (740) (14) (16) (72) (286) (1,128) －金融機構貸款 － － － － (1) (1) 其他 (744) 36 (37) (2,258) (329) (3,332) －個人貸款 (101) (10) (20) (2,210) (48) (2,389) －企業及商業貸款 (624) 46 (30) (48) (283) (939) －金融機構貸款 (19) － 13 － 2 (4) 於2014年12月31日 7,811 891 1,847 15,235 1,711 27,495 －個人貸款 1,929 386 85 14,807 386 17,593 －企業及商業貸款 5,469 501 1,439 427 1,324 9,160 －金融機構貸款 413 4 323 1 1 742 於2013年1月1日之重議條件貸款 9,974 944 2,389 26,162 2,758 42,227 －個人貸款 2,817 493 190 25,474 781 29,755 －企業及商業貸款 6,829 447 1,859 685 1,975 11,795 －金融機構貸款 328 4 340 3 2 677 年內重議條件且並無撤銷確認之貸款 2,807 49 101 1,727 1,311 5,995 －個人貸款 264 8 16 1,335 507 2,130 －企業及商業貸款 2,541 41 85 391 803 3,861 －金融機構貸款 2 － － 1 1 4 年內重議條件並確認為新造貸款之貸款 105 113 14 － 62 294 －個人貸款 17 76 14 － 25 132 －企業及商業貸款 88 37 － － 37 162 －金融機構貸款 － － － － － － 還款 (2,139) (233) (541) (1,759) (707) (5,379) －個人貸款 (489) (111) (64) (1,387) (353) (2,404) －企業及商業貸款 (1,574) (121) (477) (370) (354) (2,896) －金融機構貸款 (76) (1) － (2) － (79) 撇銷金額 (426) (25) (38) (1,035) (409) (1,933) －個人貸款 (99) (20) (9) (995) (233) (1,356) －企業及商業貸款 (303) (5) (29) (40) (175) (552) －金融機構貸款 (24) － － － (1) (25) 其他 (565) (81) 169 (6,306) (246) (7,029) －個人貸款 (259) (11) 2 (6,297) (120) (6,685) －企業及商業貸款 (311) (69) 145 (8) (125) (368) －金融機構貸款 5 (1) 22 (1) (1) 24 於2013年12月31日 9,756 767 2,094 18,789 2,769 34,175 －個人貸款 2,251 435 149 18,130 607 21,572 －企業及商業貸款 7,270 330 1,583 658 2,161 12,002 －金融機構貸款 235 2 362 1 1 601 有關註釋，請參閱第202頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 141 貸款減值 （經審核） 有關減值評估的現行政策及慣例，於第212頁 風險附錄內概述。按環球業務分類的貸款減值及 其他信貸風險準備分析載於第76頁。 下表分析個別或綜合評估已減值貸款的已 確認減值準備，以及分類為並非已減值貸 款的綜合評估減值準備。 按行業劃分的扣取自收益表之貸款減值準備 （未經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 個人貸款 245 321 25 117 1,095 1,803 －第一留置權住宅按揭 (75) 6 (24) 26 15 (52) －其他個人貸款7 320 315 49 91 1,080 1,855 企業及商業貸款 790 327 6 196 937 2,256 －製造業及國際貿易及服務貸款 520 197 36 116 382 1,251 －商用物業及其他與物業有關貸款 78 29 (28) 27 176 282 －其他商業貸款8 192 101 (2) 53 379 723 金融機構貸款5 44 (4) (32) (13) 1 (4) 截至2014年12月31日止年度 之貸款減值準備總額 1,079 644 (1) 300 2,033 4,055 個人貸款 320 345 46 963 1,522 3,196 －第一留置權住宅按揭 (11) (7) (13) 647 11 627 －其他個人貸款7 331 352 59 316 1,511 2,569 企業及商業貸款 1,467 152 (13) 253 1,115 2,974 －製造業及國際貿易及服務貸款 800 134 37 125 594 1,690 －商用物業及其他與物業有關貸款 432 (2) (5) 79 322 826 －其他商業貸款8 235 20 (45) 49 199 458 金融機構貸款5 (55) (14) (77) 19 5 (122) 截至2013年12月31日止年度 之貸款減值準備總額 1,732 483 (44) 1,235 2,642 6,048 按評估類別劃分的扣取自收益表之貸款減值準備 （未經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 個別評估減值準備 617 351 32 190 590 1,780 －新撥準備 1,112 542 134 298 738 2,824 －撥回不再需要之準備 (486) (171) (95) (88) (90) (930) －收回先前撇賬額 (9) (20) (7) (20) (58) (114) 綜合評估減值準備12 462 293 (33) 110 1,443 2,275 －已扣除準備撥回額之新撥準備 757 426 2 205 1,726 3,116 －收回先前撇賬額 (295) (133) (35) (95) (283) (841) 截至2014年12月31日止年度 之貸款減值準備總額 1,079 644 (1) 300 2,033 4,055 個別評估減值準備 1,376 145 (86) 262 623 2,320 －新撥準備 1,828 316 196 398 702 3,440 －撥回不再需要之準備 (402) (145) (235) (98) (31) (911) －收回先前撇賬額 (50) (26) (47) (38) (48) (209) 綜合評估減值準備12 356 338 42 973 2,019 3,728 －已扣除準備撥回額之新撥準備 943 479 82 1,058 2,253 4,815 －收回先前撇賬額 (587) (141) (40) (85) (234) (1,087) 截至2013年12月31日止年度 之貸款減值準備總額 1,732 483 (44) 1,235 2,642 6,048 有關註釋，請參閱第202頁。 貸款減值準備總額為41億美元，較2013年 減少20億美元，反映主要來自北美洲、歐 洲及拉丁美洲的個人貸款與企業及商業貸 款組合減值準備減少。 滙豐控股有限公司 142 董事會報告：風險 （續） 信貸風險 北美洲的第一留置權按揭及其他個人貸款 的貸款減值準備減少，反映消費及按揭貸 款組合的拖欠水平下降以及新增已減值貸 款減少，以及持續縮減及出售貸款令貸款 結欠減少，但有關減幅因2014年房屋市場 的市況改善情況未如2013年般明顯，令相 關物業的有利市值調整下降而被部分抵銷。 歐洲的貸款減值準備減少，跌幅主要來自企 業及商業貸款的個別評估減值準備減少， 反映組合質素及經濟狀況改善。個人貸款 的貸款減值準備亦有所減少，但跌幅較低， 原因是經濟環境改善令拖欠水平下降，加 上客戶繼續減少信用卡及貸款的未償還結 欠。惟我們修訂計算綜合評估企業貸款減 值的若干估計，令企業及商業貸款的綜合 評估準備上升，而金融業貸款因2013年撥 回準備，本年度則提撥準備，亦令其綜合 評估準備上升，抵銷了上述因素所引致的 部分跌幅。 拉丁美洲的其他個人貸款與企業及商業貸 款組合的貸款減值準備減少，主要反映巴 西去年為重整貸款組合更改減值模型和修 訂所用假設所帶來的不利影響。個別評估 準備大致平穩。在墨西哥，商用物業及其他 與物業有關貸款的貸款減值準備減少，特 別是與若干房屋建築商有關者。在巴西， 個別評估準備因有關一位企業客戶的其他 商業貸款出現減值而上升。 按地區劃分的減值虧損準備佔客戶貸款總額平均值的百分比 （未經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 % % % % % % 已扣除準備撥回額之新撥準備 0.37 0.22 0.14 0.32 5.00 0.53 收回額 (0.08) (0.04) (0.14) (0.09) (0.72) (0.10) 於2014年12月31日之減值虧損準備 總額 0.29 0.18 － 0.23 4.28 0.43 已扣除收回額之撇賬額 0.49 0.13 0.58 0.97 3.59 0.58 已扣除準備撥回額之新撥準備 0.65 0.20 0.15 1.00 5.93 0.81 收回額 (0.17) (0.05) (0.29) (0.09) (0.57) (0.14) 於2013年12月31日之減值虧損準備 總額 0.48 0.15 (0.14) 0.91 5.36 0.67 已扣除收回額之撇賬額 0.42 0.12 0.38 1.10 3.69 0.59 有關註釋，請參閱第202頁。 按行業及地區劃分的減值準備變動 （未經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日之減值準備 5,598 1,214 1,583 4,242 2,564 15,201 撇賬額 個人貸款 (724) (463) (157) (1,030) (1,359) (3,733) －第一留置權住宅按揭 (21) (17) (4) (731) (40) (813) －其他個人貸款7 (703) (446) (153) (299) (1,319) (2,920) 企業及商業貸款 (1,202) (146) (47) (346) (684) (2,425) －製造業及國際貿易及服務貸款 (732) (86) (41) (81) (428) (1,368) －商用物業及其他與物業有關 貸款 (342) (53) (6) (153) (39) (593) －其他商業貸款8 (128) (7) － (112) (217) (464) 金融機構貸款5 (203) － (8) (6) (4) (221) 撇賬總額 (2,129) (609) (212) (1,382) (2,047) (6,379) 收回往年撇賬額 個人貸款 271 143 35 86 283 818 －第一留置權住宅按揭 3 3 － 40 33 79 －其他個人貸款7 268 140 35 46 250 739 企業及商業貸款 29 9 7 25 58 128 －製造業及國際貿易及服務貸款 19 7 7 6 46 85 －商用物業及其他與物業有關 貸款 11 － － 3 1 15 －其他商業貸款8 (1) 2 － 16 11 28 金融機構貸款5 4 1 － 4 － 9 收回往年撇賬總額 304 153 42 115 341 955 扣取自收益表 1,079 644 (1) 300 2,033 4,055 匯兌及其他變動13 (397) (46) (6) (635) (362) (1,446) 於2014年12月31日之減值準備 4,455 1,356 1,406 2,640 2,529 12,386 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 143 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 同業貸款減值準備： －個別評估 31 － 18 － － 49 客戶貸款減值準備： －個別評估 2,981 812 1,110 276 1,016 6,195 －綜合評估12 1,443 544 278 2,364 1,513 6,142 於2014年12月31日之減值準備 4,455 1,356 1,406 2,640 2,529 12,386 於2013年1月1日之減值準備 5,361 1,219 1,811 5,616 2,162 16,169 撇賬額 個人貸款 (876) (461) (107) (1,330) (1,593) (4,367) －第一留置權住宅按揭 (83) (7) (2) (779) (25) (896) －其他個人貸款7 (793) (454) (105) (551) (1,568) (3,471) 企業及商業貸款 (1,264) (96) (78) (277) (514) (2,229) －製造業及國際貿易及服務貸款 (680) (73) (64) (80) (386) (1,283) －商用物業及其他與物業有關 貸款 (289) (7) (2) (141) (23) (462) －其他商業貸款8 (295) (16) (12) (56) (105) (484) 金融機構貸款5 (40) (3) (10) (3) (3) (59) 撇賬總額 (2,180) (560) (195) (1,610) (2,110) (6,655) 收回往年撇賬額 個人貸款 584 153 41 82 237 1,097 －第一留置權住宅按揭 25 4 － 67 23 119 －其他個人貸款7 559 149 41 15 214 978 企業及商業貸款 52 14 46 41 45 198 －製造業及國際貿易及服務貸款 19 7 2 6 27 61 －商用物業及其他與物業有關 貸款 6 4 － 18 1 29 －其他商業貸款8 27 3 44 17 17 108 金融機構貸款5 1 － － － － 1 收回往年撇賬總額 637 167 87 123 282 1,296 扣取自收益表 1,732 483 (44) 1,235 2,642 6,048 匯兌及其他變動13 48 (95) (76) (1,122) (412) (1,657) 於2013年12月31日之減值準備 5,598 1,214 1,583 4,242 2,564 15,201 同業貸款減值準備： －個別評估 35 － 18 5 － 58 客戶貸款減值準備： －個別評估 4,019 634 1,131 410 878 7,072 －綜合評估12 1,544 580 434 3,827 1,686 8,071 於2013年12月31日之減值準備 5,598 1,214 1,583 4,242 2,564 15,201 有關註釋，請參閱第202頁。 客戶及同業貸款減值準備變動 （經審核） 同業 個別評估 客戶 個別評估 綜合評估 總計 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日 58 7,072 8,071 15,201 撇賬額 (6) (2,313) (4,060) (6,379) 收回過往已撇賬之貸款 － 114 841 955 扣取自收益表 4 1,776 2,275 4,055 匯兌及其他變動13 (7) (454) (985) (1,446) 於2014年12月31日 49 6,195 6,142 12,386 減值準備： 客戶貸款 6,195 6,142 12,337 －個人貸款 468 4,132 4,600 －企業及商業貸款 5,532 1,909 7,441 －金融機構貸款 195 101 296 佔貸款的百分比1 0.04% 0.63% 0.62% 1.13% 滙豐控股有限公司 144 董事會報告：風險 （續） 信貸風險 同業 個別評估 客戶 個別評估 綜合評估 總計 百萬美元 百萬美元 百萬美元 百萬美元 於2013年1月1日 57 6,572 9,540 16,169 撇賬額 (4) (1,937) (4,714) (6,655) 收回過往已撇賬之貸款 － 209 1,087 1,296 扣取自收益表 5 2,315 3,728 6,048 匯兌及其他變動13 － (87) (1,570) (1,657) 於2013年12月31日 58 7,072 8,071 15,201 減值準備： 客戶貸款 7,072 8,071 15,143 －個人貸款 589 6,013 6,602 －企業及商業貸款 6,096 1,963 8,059 －金融機構貸款 387 95 482 佔貸款的百分比1 0.05% 0.70% 0.80% 1.35% 有關註釋，請參閱第202頁。 批發貸款 按列賬基準，貸款總額減少110億美元，包 括主要來自歐洲的不利匯兌變動320億美元。 下文乃按固定匯率基準的評述。 批發貸款年內增加210億美元。亞洲方面， 我們繼續憑藉滙豐於新興市場的地位，促 使結欠增長160億美元。北美洲亦錄得100億 美元的強勁增長，原因是我們執行增加核 心產品及服務的範疇，並積極地以在主要 增長市場內需要國際銀行支持的公司為業 務目標。歐洲貸款減少150億美元，主要由 企業透支結欠減少所致。英國有少數客戶 利用透支及存款賬的淨利息安排而受惠。 年內，由於我們調整資金管理服務方針， 配合令其與全球做法更一致，故不少客戶 增加還款的頻率，從而令其透支及存款結 餘減少280億美元。中東及北非和拉丁美洲 的分別增加60億美元及40億美元。 批發貸款總額 （未經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 企業及商業貸款(A) 210,585 220,799 20,588 57,862 30,722 540,556 －製造業貸款 39,456 37,767 2,413 15,299 12,051 106,986 －國際貿易及服務貸款 76,629 72,814 9,675 13,484 8,189 180,791 －商用物業貸款 28,187 35,678 579 6,558 2,291 73,293 －其他與物業有關貸款 7,126 34,379 1,667 8,934 281 52,387 －政府貸款 2,264 1,195 1,552 164 968 6,143 －其他商業貸款8 56,923 38,966 4,702 13,423 6,942 120,956 金融機構 （非銀行之金融機構） 貸款(B) 23,103 13,997 3,291 9,034 1,393 50,818 經重新分類之資產抵押證券 1,938 － － 131 － 2,069 同業貸款(C) 21,978 62,960 10,495 7,405 9,360 112,198 於2014年12月31日之貸款總額(D) 257,604 297,756 34,374 74,432 41,475 705,641 批發貸款減值準備 企業及商業貸款(a) 3,112 1,089 1,171 608 1,461 7,441 －製造業貸款 529 242 141 152 348 1,412 －國際貿易及服務貸款 877 533 536 157 237 2,340 －商用物業貸款 909 44 147 101 476 1,677 －其他與物業有關貸款 203 55 219 57 12 546 －政府貸款 4 － 1 － － 5 －其他商業貸款 590 215 127 141 388 1,461 金融機構 （非銀行之金融機構） 貸款(b) 221 13 21 39 2 296 同業貸款(c) 31 － 18 － － 49 於2014年12月31日之減值準備(d) 3,364 1,102 1,210 647 1,463 7,786 (a)佔(A)的百分比 1.48% 0.49% 5.69% 1.05% 4.76% 1.38% (b)佔(B)的百分比 0.96% 0.09% 0.64% 0.43% 0.14% 0.58% (c)佔(C)的百分比 0.14% － 0.17% － － 0.04% (d)佔(D)的百分比 1.31% 0.37% 3.52% 0.87% 3.53% 1.10% 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 145 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 企業及商業貸款(I) 239,116 203,894 19,760 50,307 30,188 543,265 －製造業貸款 55,920 30,758 3,180 11,778 12,214 113,850 －國際貿易及服務貸款 76,700 79,368 8,629 11,676 8,295 184,668 －商用物業貸款 31,326 34,560 639 5,900 2,421 74,846 －其他與物業有關貸款 7,308 27,147 1,333 8,716 328 44,832 －政府貸款 3,340 1,021 1,443 499 974 7,277 －其他商業貸款8 64,522 31,040 4,536 11,738 5,956 117,792 金融機構 （非銀行之金融機構） 貸款(J) 27,872 9,688 2,532 9,055 1,376 50,523 經重新分類之資產抵押證券 2,578 － － 138 － 2,716 同業貸款(K) 24,273 72,814 6,419 6,420 10,178 120,104 於2013年12月31日之貸款總額(L) 293,839 286,396 28,711 65,920 41,742 716,608 批發貸款減值準備 企業及商業貸款(i) 3,821 918 1,212 769 1,339 8,059 －製造業貸款 618 246 182 89 384 1,519 －國際貿易及服務貸款 1,216 428 502 188 349 2,683 －商用物業貸款 1,116 22 153 202 396 1,889 －其他與物業有關貸款 269 102 236 93 8 708 －政府貸款 3 － 10 1 － 14 －其他商業貸款 599 120 129 196 202 1,246 金融機構 （非銀行之金融機構） 貸款(j) 344 17 60 50 11 482 同業貸款(k) 35 － 18 5 － 58 於2013年12月31日之減值準備(l) 4,200 935 1,290 824 1,350 8,599 (i)佔(I)的百分比 1.60% 0.45% 6.13% 1.53% 4.44% 1.48% (j)佔(J)的百分比 1.23% 0.18% 2.37% 0.55% 0.80% 0.95% (k)佔(K)的百分比 0.14% － 0.28% 0.08% － 0.05% (l)佔(L)的百分比 1.43% 0.33% 4.49% 1.25% 3.23% 1.20% 有關註釋，請參閱第202頁。 商用物業 商用物業貸款 （未經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 並非逾期或已減值 25,860 35,430 333 6,136 1,535 69,294 已逾期但並非已減值 18 170 47 100 28 363 已減值貸款 2,309 78 199 322 728 3,636 於2014年12月31日各類貸款總額 28,187 35,678 579 6,558 2,291 73,293 其中： －重議條件貸款14 1,954 19 183 191 377 2,724 減值準備 909 44 147 101 476 1,677 並非逾期或已減值 28,044 34,433 402 5,400 2,249 70,528 已逾期但並非已減值 95 103 18 29 35 280 已減值貸款 3,187 24 219 471 137 4,038 於2013年12月31日各類貸款總額 31,326 34,560 639 5,900 2,421 74,846 其中： －重議條件貸款14 2,590 20 229 280 461 3,580 減值準備 1,116 22 153 202 396 1,889 有關註釋，請參閱第202頁。 商用物業貸款包括給予企業、機構以及個 別資產豐厚人士 （主要投資於賺取收益資 產，其次是其建設及發展） 的融資。有關業 務主要專注於傳統核心資產類別，例如零 售、辦公室、輕工業以及住宅建築項目。 我們的組合分布全球各地，其中香港、英 國、美國及加拿大佔較大比重。 於發展程度較高的市場，我們的貸款風險 主要包括投資資產的融資、重建現有的庫 存物業以及為支持經濟及人口增長而擴 大商業及住宅市場。在發展程度較低的商 用物業市場，我們的風險包括年期相對較 短的發展資產貸款，特別是用於支持較大 型、資本較雄厚的發展商，供其參與住宅 建設，或用以支持經濟增長的資產。 近年該等市場有不少已不再進行急速的建 滙豐控股有限公司 146 董事會報告：風險 （續） 信貸風險 設，反而專注於投資資產，與發展程度較 高的市場之情況一致。集團承擔的重大貸 款風險主要集中在各個經濟、政治或文化 重點城市。 於2014年12月31日的商用物業貸款總額為 730億美元，減少16億美元，跌幅包括主要 來自歐洲的不利匯兌變動33億美元。 商用物業再融資風險 商用物業貸款往往要求於到期時償還大部 分本金。客戶安排還款時通常會透過獲得 新造貸款以償還現有債務。再融資風險指 無法償還到期債務的客戶無法按商業利率 對債務進行再融資的風險。有關再融資風 險的詳情，請參閱第214頁。我們會密切監 控商用物業組合，評估可能反映潛在再融 資問題的因素。倘若辦理有關貸款反映的 過往市場規律不再適用於現行市場，例如 貸款估值比率或會較高及╱或利息覆蓋率 或會較低，則主要因素為貸款的年限。當 地市場再融資來源的類別亦是重要考慮因 素，倘若貸款人僅限於銀行但銀行流動資 金有限，風險將會上升。此外，相關基本 要素，例如租戶可靠性、出租的能力及物 業本身的狀況亦相當重要，原因是有關因 素會影響物業價值。 就集團的商用物業組合整體而言，市場行 為以及資產質素於2014年並無導致過多疑 慮。於近年引起更大關注的英國，上述因 素現時未足以令管理層加倍注意。 有關英國組合的進一步詳情如下：於2014年 12月31日，商用物業貸款為200億美元 （2013 年：220億美元） ，其中59億美元 （2013年： 68億美元） 於未來12個月須再融資。在該等 結欠中，於集團貸款管理組內受密切監察 的個案之金額達21億美元 （2013年：24億美 元） 。13億美元 （2013年：16億美元） 已披露 為已減值，減值準備6億美元 （2013年：6億 美元） 。貸款不被視為已減值乃因有充分證 據顯示相關約定現金流將能收回，或該等 貸款毋須按我們認為低於一般市場條款再 融資。 貸款抵押品 集團有關使用抵押品的慣例，詳載於第213頁風險 附錄。 就商用物業及其他企業、商業及金融機構 （非銀行） 貸款而持有的抵押品於下文獨立 分析，當中反映與其他貸款的抵押品比較， 商用物業的抵押品表現與償還本金的關係 較為密切。各項分析均包括資產負債表外 貸款承諾，主要為未取用信貸額。 下表計量的抵押品包括房地產固定第一押 記以及現金及有價金融工具的押記。列表 的價值指在公開市場中的預期市值，且並 未對抵押品的預期收回成本作調整。現金 按名義價值估值，而有價證券則按公允值 估值。所呈報的貸款估值比率乃直接將貸 款與個別及專門支持各項信貸的抵押品相 聯繫而計算。倘抵押品資產由多項貸款分 攤，則無論屬於特定或較普遍情況下以全 數抵押，抵押品價值均按比例分攤至受抵 押品保障的各項貸款。 下表並未計量一般就企業及商業貸款接納 的其他類別抵押品，例如無支持的擔保和 客戶業務資產的浮動押記。雖然該等減低 風險措施具有一定價值，且在客戶無力償 債時往往讓貸款人可以行使一定的權利， 但其可獲賦予的價值難以釐定，故就披露 而言並未訂定任何價值。 已減值貸款的抵押品價值不能與已確認的 減值準備直接比較。下表的貸款估值比率 利用未經調整的公開市值為基準，而減值 準備則按不同基準計算，且經考慮其他現 金流量以及就變現抵押品成本而對抵押品 價值作調整，詳情載於第212頁。 商用物業貸款 商用物業抵押品的價值根據專業與內部估 值及實地視察而釐定。由於商用物業抵押 品的估值相當複雜，各地的估值政策會根 據當地市況決定檢討估值的頻密程度。倘 對抵押品或直接債務人的表現提升關注， 則會進行更頻密的重估。倘客戶修訂其銀 行貸款要求，令集團提供更多資金，或令 相關貸款風險或抵押品需作其他重大的重 新安排，導致客戶的風險狀況改變，則亦 會進行重估。因此，CRR1至7評級貸款所使 用的房地產抵押品價值可追溯至最近一次 對抵押品作上述考慮之時。至於CRR 8及9至 10評級的貸款，幾乎所有抵押品曾於最近 三年內進行重估。 香港的市場慣例一般是以擔保作抵押或無 抵押的方式貸款予主要地產公司。在歐洲， 屬營運資金性質的貸款，一般不會以第一 固定押記抵押，因此披露為無抵押。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 147 按抵押品水平分析的商用物業貸款 （包括貸款承諾） （經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 CRR╱EL評級1至7級 無抵押 5,351 16,132 361 87 1,719 23,650 全數抵押 25,873 26,323 23 9,093 556 61,868 部分抵押(A) 1,384 1,599 － 1,819 152 4,954 －A的抵押品價值 1,032 901 － 1,199 47 3,179 32,608 44,054 384 10,999 2,427 90,472 CRR╱EL評級8級 無抵押 34 7 － 9 2 52 全數抵押 568 23 － 30 1 622 貸款估值比率： －低於50% 64 － － 16 1 81 －51%至75% 222 11 － 10 － 243 －76%至90% 132 9 － 4 － 145 －91%至100% 150 3 － － － 153 部分抵押(B) 365 － － 7 － 372 －B的抵押品價值 296 － － 2 － 298 967 30 － 46 3 1,046 CRR╱EL評級9至10級 無抵押 369 48 6 1 499 923 全數抵押 992 15 7 166 178 1,358 貸款估值比率： －低於50% 78 6 7 28 10 129 －51%至75% 593 2 － 91 43 729 －76%至90% 167 2 － 17 53 239 －91%至100% 154 5 － 30 72 261 部分抵押(C) 1,085 15 181 37 50 1,368 －C的抵押品價值 664 5 89 30 13 801 2,446 78 194 204 727 3,649 於2014年12月31日 36,021 44,162 578 11,249 3,157 95,167 CRR╱EL評級1至7級 無抵押 4,865 14,164 192 137 935 20,293 全數抵押 24,154 25,317 21 8,627 1,728 59,847 部分抵押(D) 2,664 2,377 139 704 484 6,368 －D的抵押品價值 1,827 1,688 24 303 292 4,134 31,683 41,858 352 9,468 3,147 86,508 CRR╱EL評級8級 無抵押 109 10 － 1 3 123 全數抵押 793 － 72 68 1 934 貸款估值比率： －低於50% 139 － － 15 － 154 －51%至75% 367 － 72 49 1 489 －76%至90% 173 － － 4 － 177 －91%至100% 114 － － － － 114 部分抵押(E) 360 2 － 13 － 375 －E的抵押品價值 281 1 － 11 － 293 1,262 12 72 82 4 1,432 CRR╱EL評級9至10級 無抵押 564 － 7 4 521 1,096 全數抵押 1,079 12 31 233 286 1,641 貸款估值比率： －低於50% 275 2 7 39 32 355 －51%至75% 436 6 7 110 57 616 －76%至90% 209 3 17 62 62 353 －91%至100% 159 1 － 22 135 317 部分抵押(F) 1,815 5 181 240 56 2,297 －F的抵押品價值 1,284 5 89 115 34 1,527 3,458 17 219 477 863 5,034 於2013年12月31日 36,403 41,887 643 10,027 4,014 92,974 有關註釋，請參閱第202頁。 滙豐控股有限公司 148 董事會報告：風險 （續） 信貸風險 按抵押品水平 （僅指CRR╱EL評級8至10級） 分析的其他企業、商業及金融機構 （非銀行） 貸款 （包 括貸款承諾） （經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 CRR╱EL評級8級 無抵押 2,051 237 15 320 227 2,850 全數抵押 629 56 72 331 11 1,099 貸款估值比率： 13 －低於50% 120 13 － 186 5 324 －51%至75% 293 － － 72 6 371 －76%至90% 51 9 69 46 － 175 －91%至100% 165 34 3 27 － 229 部分抵押(A) 105 44 1 148 6 304 －A的抵押品價值 46 17 1 68 4 136 2,785 337 88 799 244 4,253 CRR╱EL評級9至10級 無抵押 4,185 939 813 62 1,420 7,419 全數抵押 615 143 147 231 124 1,260 貸款估值比率： －低於50% 169 68 25 48 48 358 －51%至75% 136 27 19 39 35 256 －76%至90% 168 16 6 35 26 251 －91%至100% 142 32 97 109 15 395 部分抵押(B) 624 364 547 251 140 1,926 －B的抵押品價值 341 169 92 141 46 789 5,424 1,446 1,507 544 1,684 10,605 於2014年12月31日 8,209 1,783 1,595 1,343 1,928 14,858 CRR╱EL評級8級 無抵押 2,411 185 37 328 456 3,417 全數抵押 259 51 1 227 70 608 貸款估值比率： －低於50% 65 38 1 84 11 199 －51%至75% 103 4 － 47 10 164 －76%至90% 25 8 － 31 5 69 －91%至100% 66 1 － 65 44 176 部分抵押(C) 435 23 528 345 73 1,404 －C的抵押品價值 17 5 398 89 18 527 3,105 259 566 900 599 5,429 CRR╱EL評級9至10級 無抵押 1,467 685 1,089 26 1,615 4,882 全數抵押 1,121 161 49 309 266 1,906 貸款估值比率： －低於50% 124 57 2 24 159 366 －51%至75% 161 21 47 29 49 307 －76%至90% 156 53 － 46 43 298 －91%至100% 680 30 － 210 15 935 部分抵押(D) 1,192 304 770 359 290 2,915 －D的抵押品價值 606 150 102 149 131 1,138 3,780 1,150 1,908 694 2,171 9,703 於2013年12月31日 6,885 1,409 2,474 1,594 2,770 15,132 有關註釋，請參閱第202頁。 其他企業、商業及金融機構 （非銀行） 貸款 於下文獨立分析。其他企業及商業貸款融 資活動方面，抵押品的價值與償還本金的 履約能力並無密切關係。倘若債務人的整 體信貸表現惡化，相關抵押品的價值一般 會重新釐定，如果證明還款需要依賴第二 資金來源，我們須評估有關資金來源可能 具備的履約能力。 因此，下表僅呈列CRR評級8至10級的客戶 所涉貸款價值，因該等貸款一般在較近期 曾進行估值。 同業貸款一般為無抵押。就CRR評級9至10 級客戶 （即分類為已減值） 持有的抵押品價 值獨立披露。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 149 按抵押品水平分析的同業貸款 （包括貸款承諾） （經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 CRR╱EL評級1至8級 無抵押 22,405 64,210 10,472 7,985 9,406 114,478 全數抵押 104 1,587 － － － 1,691 部分抵押(A) 5 － － － － 5 －A的抵押品價值 3 － － － － 3 22,514 65,797 10,472 7,985 9,406 116,174 CRR╱EL評級9至10級 無抵押 102 1 21 － － 124 於2014年12月31日 22,616 65,798 10,493 7,985 9,406 116,298 CRR╱EL評級1至8級 無抵押 21,225 72,986 6,373 7,210 9,837 117,631 全數抵押 3,614 1,376 － － 266 5,256 部分抵押(B) 68 560 － － － 628 －B的抵押品價值 3 389 － － － 392 24,907 74,922 6,373 7,210 10,103 123,515 CRR╱EL評級9至10級 無抵押 153 － 312 14 － 479 於2013年12月31日 25,060 74,922 6,685 7,224 10,103 123,994 有關註釋，請參閱第202頁。 其他信貸風險 除有抵押貸款外，滙豐亦會採用其他強化 信貸條件及方法，降低來自金融資產的信 貸風險，詳情如下： ‧ 政府、銀行及其他金融機構發行的若干 證券，受惠於涵蓋該等資產的政府擔保 提供的其他強化信貸條件。 政府擔保詳載於財務報表附註12、15及18。 ‧ 由銀行及金融機構發行的債務證券包 括資產抵押證券及近似工具，並以相關 金融資產組合支持。與資產抵押證券相 關的信貸風險，透過購入信貸違責掉期 （ 「CDS」 ） 保障而降低。 集團所持資產抵押證券及相關信貸違責掉 期保障的披露資料，載於第162頁。 ‧ 交易用途資產包括因有意用作交易而 持有的貸款，主要包括為符合衍生工具 保證金要求而提供的現金抵押品、結算 賬項、反向回購以及借入股票。由於在 交易對手違約的情況下，所提供的現金 抵押品將用於抵銷有關負債，故該等現 金抵押品的信貸風險有限。反向回購及 借入股票按其性質則屬有抵押。 集團根據該等安排持作擔保的抵押品可予 出售或再質押，詳情載於財務報表附註19。 ‧ 集團承擔的最大信貸風險包括批出的金 融擔保及同類合約，以及貸款及其他信 貸相關承諾。倘擔保被要求履行或貸款 承諾被取用但隨後拖欠還款，則我們對 其他減低信貸風險項目或會有追索權 （視乎安排的條款） 。該等安排的詳情， 請參閱財務報表附註37。 衍生工具 滙豐參與多類令我們承擔交易對手信貸風 險的交易。交易對手信貸風險是指交易對 手未能妥為履行責任而違約所產生的財務 損失風險，主要來自場外衍生工具以及證 券融資交易，在交易及非交易賬項均會計 算。交易的價值會參考利率、匯率或資產 價格等市場因素變動。 呈報衍生工具持倉公允值時，會考慮衍生 工具交易的交易對手風險。公允值的調整 稱為信貸估值調整 （ 「CVA」 ） 。.
有關信貸估值調整的分析，請參閱財務報表附 註13。 下表反映按風險類別分析透過交易所、中 央交易對手及非中央交易對手結算的衍生 工具公允值及名義合約總額。 滙豐控股有限公司 150 董事會報告：風險 （續） 信貸風險 按產品類別劃分的衍生工具名義合約金額及公允值 （未經審核） 2014年 2013年 名義金額 公允值 名義金額 公允值 資產 負債 資產 負債 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 外匯 5,573,415 97,312 95,759 5,291,003 80,914 75,798 －交易所買賣 81,785 229 369 41,384 121 93 －中央交易對手場外結算 18,567 321 349 16,869 415 622 －非中央交易對手場外結算 5,473,063 96,762 95,041 5,232,750 80,378 75,083 利率 22,328,518 473,243 468,152 27,347,918 458,576 452,531 －交易所買賣 1,432,333 112 161 857,562 335 225 －中央交易對手場外結算 15,039,001 261,880 264,509 18,753,836 285,390 285,375 －非中央交易對手場外結算 5,857,184 211,251 203,482 7,736,520 172,851 166,931 股票 568,932 11,694 13,654 589,903 18,389 22,573 －交易所買賣 289,140 2,318 3,201 274,880 8,403 2,949 －非中央交易對手場外結算 279,792 9,376 10,453 315,023 9,986 19,624 信貸 550,197 9,340 10,061 678,256 9,092 8,926 －中央交易對手場外結算 126,115 1,999 2,111 104,532 1,346 1,409 －非中央交易對手場外結算 424,082 7,341 7,950 573,724 7,746 7,517 商品及其他 77,565 3,884 3,508 77,842 2,624 1,786 －交易所買賣 7,015 80 23 6,531 182 6 －非中央交易對手場外結算 70,550 3,804 3,485 71,311 2,442 1,780 場外衍生工具總額 27,288,354 592,735 587,379 32,804,565 560,554 558,341 －中央交易對手結算的場外 衍生工具總額 15,183,683 264,200 266,968 18,875,237 287,151 287,406 －非中央交易對手結算的場外 衍生工具總額 12,104,671 328,535 320,411 13,929,328 273,403 270,935 交易所買賣衍生工具總額 1,810,273 2,739 3,755 1,180,357 9,041 3,273 總額 29,098,627 595,473 591,134 33,984,922 569,595 561,614 對銷 (250,465) (250,465) (287,330) (287,330) 12月31日總額 345,008 340,669 282,265 274,284 滙豐使用衍生工具的目的載於財務報表附註16。 我們傾向以國際掉期業務及衍生投資工具 協會 （ 「ISDA」 ） 總協議作為衍生工具業務的 協議文件。該協議為買賣全線場外交易產 品提供主體合約模式，倘若任何一方違約 或另一宗預先協定可導致合約終止的事件 發生，則合約限定雙方對協議涵蓋的全部 未平倉交易須採用淨額計算。有關雙方於 簽訂ISDA總協議時，亦會簽訂信貸支持附 件 （ 「CSA」 ） ，此乃普遍的做法，亦是我們傾 向選用的做法。根據信貸支持附件，抵押 品會由交易其中一方轉交另一方，以減低 未平倉交易內含的交易對手風險。 我們使用與交易對手的抵押品協議及淨額 計算協議，管理因場外衍生工具合約的市 場風險產生的交易對手風險。雖然我們在 若干情況下可以管理個別風險，但我們目 前並未積極管理於信貸市場的一般場外衍 生工具交易對手風險。 我們一直對抵押品類別設有嚴謹的政策限 制，因此所收取及質押的抵押品類別屬極 為容易變現其價值而且質素較佳，並主要 是現金。倘屬抵押品政策並未涵蓋而需取 得批准的抵押品類別 （包括涉及錯向風險 的抵押品） ，則需要取得三個地區文件審批 委員會的其中一個的批准。該等文件審批 委員會要求地區環球資本市場業務法律事 務及風險管理營運總監的高級代表參與並 簽署。 大部分與我們訂立抵押品協議的交易對手 為歐洲交易對手。而大部分信貸支持附件 的交易對手為金融機構客戶。 基於滙豐的政策，我們訂立的協議主要是 ISDA信貸支持附件，其中大部分以英國法 律編寫。按協議類別劃分的場外抵押品協 議如下： 按類別劃分的場外抵押品協議 （未經審核） 協議數目 ISDA CSA （英國法律） 2,434 ISDA CSA （紐約法律） 1,628 ISDA CSA （日本法律） 18 法國總協議及與CSA對等者15 227 德國總協議及與CSA對等者16 90 其他 205 於2014年12月31日 4,602 有關註釋，請參閱第202頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 151 有關在交易對手違約並已就衍生工具收取抵押品 的情況下可依法強制執行對銷權利的詳情，請參 閱第130頁及財務報表附註32。 按地區劃分的反向回購－非交易用途 第347頁所闡述的資產負債表呈列方式更改 後，非交易用途反向回購會在資產負債表 中另行列示，不再計入 「客戶貸款」 及 「同 業貸款」 。 比較數字已按此重新呈列。因此，於信貸 風險一節中提述客戶或同業貸款的分析不 包括分別對客戶或同業的非交易用途反向 回購。提供予客戶及同業的非交易用途反 向回購金額載列如下以供參考。 按地區劃分的反向回購－非交易用途 （經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 客戶 25,841 5,409 － 35,060 － 66,310 同業 34,748 22,813 19 29,008 8,815 95,403 於2014年12月31日 60,589 28,222 19 64,068 8,815 161,713 客戶 48,091 6,448 － 33,676 － 88,215 同業 49,631 12,973 24 23,744 5,103 91,475 於2013年12月31日 97,722 19,421 24 57,420 5,103 179,690 有關註釋，請參閱第202頁。 個人貸款 我們提供一系列有抵押及無抵押個人貸款 產品，以配合客戶的需要。個人貸款包括向 客戶貸款以供購置資產 （例如住宅物業） ， 而貸款一般以所買資產作抵押。我們亦提 供以現有資產 （例如住宅物業的第一留置 權） 作抵押的貸款及無抵押貸款產品 （例如 透支、信用卡及工資貸款） 。 個人貸款總額 （未經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 第一留置權住宅按揭(A) 131,000 93,147 2,647 55,577 4,153 286,524 其中： －僅須供息 （包括對銷） 按揭 44,163 956 － 276 － 45,395 －負擔能力為本的按揭 （包括可調利率按揭） 337 5,248 － 16,452 － 22,037 其他個人貸款(B) 47,531 36,368 3,924 9,823 9,384 107,030 －其他 34,567 25,695 2,633 4,328 4,846 72,069 －信用卡 12,959 10,289 897 1,050 3,322 28,517 －第二留置權住宅按揭 － 56 2 4,433 － 4,491 －汽車融資 5 328 392 12 1,216 1,953 於2014年12月31日各類貸款總計(C) 178,531 129,515 6,571 65,400 13,537 393,554 個人貸款減值準備 第一留置權住宅按揭(a) 306 46 97 1,644 36 2,129 其他個人貸款(b) 786 208 97 350 1,030 2,471 －其他 438 87 59 43 672 1,299 －信用卡 347 119 33 36 298 833 －第二留置權住宅按揭 － － － 271 － 271 －汽車融資 1 2 5 － 60 68 於2014年12月31日減值準備總計(c) 1,092 254 194 1,994 1,066 4,600 (a)佔A的百分比 0.2% － 3.7% 3.0% 0.9% 0.7% (b)佔B的百分比 1.7% 0.6% 2.5% 3.6% 11.0% 2.3% (c)佔C的百分比 0.6% 0.2% 3.0% 3.0% 7.9% 1.2% 滙豐控股有限公司 152 董事會報告：風險 （續） 信貸風險 2014年12月31日的個人貸款總額由2013年底 的4,110億美元減少至3,940億美元 （按固定 匯率基準為3,920億美元） 。我們繼續縮減北 美洲的消費及按揭貸款組合。年內結欠進 一步減少57億美元。 按固定匯率基準，不包括美國消費及按揭 貸款縮減組合的個人貸款於2014年增加77 億美元，原因是亞洲的按揭及其他貸款增 長，以及美國及巴西的按揭組合增長，但 部分增長被英國的個人貸款減少所抵銷。 按揭貸款 （未經審核） 我們提供種類繁多的按揭產品滿足客戶的 需要，包括償還資本、僅須供息、負擔能 力為本及對銷按揭。 集團的信貸政策規定了可接受的住宅物業 貸款估值比率，新造貸款的最高上限設定 在75%至95%之間。 特定貸款估值比率限額及債務對收入比率 分別於區域及國家層面管理。雖然該等參 數必須符合集團的政策、策略及承受風險 水平，但在不同的業務所在地，這些參數 不盡相同，以反映當地經濟及房屋市場狀 況、規例、組合表現、定價及其他產品特 徵。 下文乃按固定匯率基準評述。 年內個人貸款 （不包括美國消費及按揭貸款 縮減組合） 、按揭貸款結欠增加39億美元。 亞洲的按揭貸款 （不包括於第153頁所討論 重新分類至其他個人貸款的按揭貸款） 增加 48億美元。增幅主要來自香港的結欠持續 增長29億美元，其次是澳洲增加5億美元、 馬來西亞增加4億美元及台灣增加3億美元， 而增長是因為需求強勁以及滙豐在當地提 供具競爭力的產品。亞洲按揭賬項的質素 仍然良好，違約及減值準備極低。香港新 造按揭貸款的平均貸款估值比率為47%， 而整體組合則估計為29%。 北美洲方面，加拿大的按揭結欠於年內上 升5億美元，原因是我們專注於按揭計劃以 及改善程序。美國的卓越理財按揭組合於 2014年上升9億美元，原因是我們繼續專注 於核心組合的增長。由於我們不斷改善按 揭組合的信貸質素，美國業務的綜合評估 減值準備減少。2014年美國的消費及按揭 貸款組合減少57億美元。 巴西的按揭貸款因處理程序及所提供的產 品有所改善以及年內巴西按揭市場的整體 增長而上升5億美元。 在歐洲，貸款減少及還款 （主要是英國按揭 組合） 的影響導致貸款額微跌14億美元或 1%。 英國按揭貸款總額中440億美元為僅須供 息產品，包括First Direct的190億美元對銷按 揭。新造按揭貸款的平均貸款估值比率為 60%，而整體按揭組合則平均為43.7%。英 國按揭組合的信貸質素仍然良好，2014年 的貸款減值準備及拖欠水平同時下降。 我們於法國的按揭貸款賬項因年內需求強 勁而上升6億美元。 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 第一留置權住宅按揭(D) 140,474 92,047 2,451 60,955 3,948 299,875 其中： －僅須供息 （包括對銷） 按揭 49,460 1,115 － 352 － 50,927 －負擔能力為本的按揭 （包括可調利率按揭） 508 5,593 － 16,274 － 22,375 其他個人貸款(E) 51,633 32,482 4,033 11,735 10,970 110,853 －其他 37,126 21,636 2,728 5,309 5,651 72,450 －信用卡 14,496 10,274 915 1,145 3,526 30,356 －第二留置權住宅按揭 － 91 2 5,261 － 5,354 －汽車融資 11 481 388 20 1,793 2,693 於2013年12月31日各類貸款總計(F) 192,107 124,529 6,484 72,690 14,918 410,728 個人貸款減值準備 第一留置權住宅按揭(d) 439 57 124 2,886 32 3,538 其他個人貸款(e) 959 222 169 532 1,182 3,064 －其他 553 93 104 59 881 1,690 －信用卡 403 127 61 47 217 855 －第二留置權住宅按揭 － － － 426 － 426 －汽車融資 3 2 4 － 84 93 於2013年12月31日減值準備總計(f) 1,398 279 293 3,418 1,214 6,602 (d)佔D的百分比 0.3% 0.1% 5.1% 4.7% 0.8% 1.2% (e)佔E的百分比 1.9% 0.7% 4.2% 4.5% 10.8% 2.8% (f)佔F的百分比 0.7% 0.2% 4.5% 4.7% 8.1% 1.6% 有關註釋，請參閱第202頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 153 其他個人貸款 （未經審核） 2014年其他個人貸款上升37億美元，並由亞 洲的個人貸款及循環信貸融資的增長 （主要 來自香港的增長31億美元） 。我們亦將中國 內地的17億美元貸款由 「住宅按揭」 重新分 類為 「其他個人貸款」 ，因為部分正在興建 或已落成物業的抵押品尚未全面登記。增 幅部分因英國的信用卡借貸減少7億美元及 土耳其的信用卡借貸減少3億美元而抵銷， 此乃還款所致。北美洲的有期貸款於年內 下跌7億美元，尤其是加拿大。巴西的汽車 融資交易商縮減組合亦下降了2億美元。 美國滙豐融資的美國消費及按揭貸款－住 宅按揭 17 （未經審核） 2014年 2013年 百萬美元 百萬美元 住宅按揭： －第一留置權 21,915 27,305 其他個人貸款： －第二留置權 2,509 3,014 於12月31日總計(A) 24,424 30,319 減值準備 1,679 3,028 －佔A的百分比 6.9% 10.0% 有關註釋，請參閱第202頁。 美國滙豐融資 2014年，美國滙豐融資的按揭貸款結欠減 少57億美元。我們除繼續出售消費及按揭 貸款組合的貸款外，於年內再轉撥29億美 元的貸款為持作出售用途資產，並預期於 未來12個月在多項交易中出售有關貸款。 減值準備減少，反映新增已減值貸款水平 下降以及未償還貸款結欠減少，原因是持 續變現組合 （包括出售貸款） ，以及由於拖 欠水平及損失的嚴重程度較2013年低，虧 損估算額亦較低。 2014年消費及按揭貸款組合的第一及第二 留置權住宅按揭，違約拖欠兩個月或以上 的貸款結欠減少25億美元至24億美元，反 映持續縮減組合以及出售貸款。 美國滙豐融資：於美國的止贖物業 （未經審核） 2014年 2013年 百萬美元 百萬美元 於年底止贖物業數目 2,139 4,254 於期內加入止贖物業清單的 物業數目 3,716 9,752 出售止贖物業平均 （增 益） ╱虧損額18 (1%) 1% 止贖物業平均虧損總額19 51% 51% 出售止贖物業平均時間 （日 數） 189 154 有關註釋，請參閱第202頁。 於2014年12月31日的止贖物業數目較2013年 底大幅減少，原因是我們於2014年出售的 物業較加入止贖行列的物業多。我們新增 入止贖行列的物業較少，是因為不少未取 得業權的物業已因持續出售消費及按揭貸 款組合的應收賬款而出售。 美國滙豐銀行 美國滙豐銀行方面，2014年我們實施策略， 增加美國卓越理財客源，令按揭貸款結欠增 加9億美元。2014年的信貸質素持續改善， 違約拖欠兩個月或以上的第一留置權住宅 按揭結欠於2014年12月下跌3億美元至11億 美元。我們亦繼續出售所有在第二市場由 代理辦理的新造按揭，以管理我們的利率 風險和改善結構流動性。 美國違約拖欠兩個月或以上貸款的趨勢 （未經審核） 2014年 2013年 百萬美元 百萬美元 美國個人貸款業務 第一留置權住宅按揭 3,271 5,931 －消費及按揭貸款 2,210 4,595 －其他按揭貸款 1,061 1,336 第二留置權住宅按揭 216 406 －消費及按揭貸款 154 276 －其他按揭貸款 62 130 信用卡 17 25 個人非信用卡 7 25 於12月31日總計 3,511 6,387 % % 佔相等貸款及應收賬款結欠的百分比 第一留置權住宅按揭 8.6 14.0 第二留置權住宅按揭 5.0 8.1 信用卡 2.4 3.4 個人非信用卡 1.4 4.9 於12月31日總計 8.1 13.1 滙豐控股有限公司 154 董事會報告：風險 （續） 信貸風險 美國滙豐融資有抵押房地產貸款結欠的貸款組合總額 （未經審核） 重訂賬齡 20 修訂貸款 條款及 重訂賬齡 修訂貸款 條款 重議條件 貸款總額 未重議 條件貸款 總額 各類貸款 總額 減值準備 總額 減值 準備╱ 貸款總額 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 % 於2014年12月31日 6,637 6,581 587 13,805 10,619 24,424 1,679 7 於2013年12月31日 8,167 8,213 768 17,148 13,171 30,319 3,028 10 有關註釋，請參閱第202頁。 美國滙豐融資組合內餘下重議條件有抵押房地產賬項數目 （未經審核） 重議條件貸款數目 （千宗） 貸款 數目總計 （千宗） 重訂賬齡 修訂貸款條款 及重訂賬齡 修訂貸款 條款 總計 於2014年12月31日 85 64 6 155 297 於2013年12月31日 102 78 8 188 352 美國滙豐融資修訂貸款條款及重訂賬齡計 劃 美國滙豐融資設有各類修訂貸款條款及重 訂賬齡 （ 「貸款重議條件」 ） 計劃，以管理客 戶關係、提高收回貸款的機會及避免止贖 （如可能） 。 美國滙豐融資自2006年起一直實行大規模 的貸款重議條件計劃，其大部分貸款組合 已因應美國經濟狀況及美國滙豐融資的客 戶基礎的特點，在客戶關係的某個階段重 議條件。 近年來合資格修訂條款的貸款數量已大幅 縮減。我們預期該趨勢將會持續。由於經 濟狀況改善、我們不再辦理有抵押房地產 及個人非信用卡應收賬款的新造業務，以 及持續縮減消費及按揭貸款組合並出售相 關貸款，故修訂貸款條款的新增個案有所 減少。 合資格標準 賬項必須符合若干標準方合資格重議條 件。然而，美國滙豐融資保留拒絕重議條 件的權利。美國滙豐融資根據現行政策對 合資格賬項重議條件的幅度，會因應其對 當時經濟狀況及其他逐年變化的因素的看 法而有所不同。此外，基於法律或監管協 議或法令，美國滙豐融資可在特定情況下 就政策及慣例作出例外處理。 重議條件的有抵押房地產經過12個月後方 合資格再度進行重議條件，於五年內最多 可重議條件五次。借款人必須獲審批方可 重議條件，且一般於開始重議條件前60日內 須至少有兩次合資格按月還款。於若干情 況下，有關債務於破產程序中已獲重整， 則所規定的還款次數可能較少甚至毋須還 款。涉及破產的有抵押房地產貸款以及借 款人涉及破產法庭第13章還款計劃的賬項， 一般可於收取一次合資格還款後被視為未 過期，而涉及第7章破產保護的借款人，其 賬項可於收取經簽署的再確認協議後重訂 賬齡。此外，對於若干產品，在若干特殊 情況下 （例如，發生自然災害或執行紓困計 劃） ，賬項可在不收取還款下重訂賬齡。 2014年與2013年比較 於2014年12月31日，美國滙豐融資重議條件 之有抵押房地產貸款賬項佔北美洲重議條 件貸款總額91% （2013年：91%） 。80億美元 （2013年：100億美元） 的重議條件之有抵押 房地產貸款分類為已減值。雖然2014年仍然 繼續進行重議貸款條件的工作，但由於消 費及按揭貸款組合縮減及出售貸款，美國 滙豐融資的重議條件貸款的總數有所減少。 美國滙豐融資在集團信貸政策的規限下， 容許貸款在若干情況下多次重議條件。因 此，上表所載的極大部分貸款已多次重訂 賬齡或修訂。就此而言，多次修訂的賬項的 比例維持於相若水平，佔總修訂數目的70% 至75%。 年內進行第二次或後續重議條件的賬項， 未並於統計數據中呈列，而有關統計僅會 將有關貸款作首次重議條件的新增重議條 件貸款處理。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 155 美國滙豐融資貸款重議條件計劃類別 ‧ 暫時修訂條款指變更貸款的合約條款，致使美國 滙豐融資在事先界定的期限內放棄收取約定現金 流的權利。暫時修訂條款，是預期將於修訂條款 期後恢復原有合約條款 （包括計算的利率） ，其中 一個例子是減少支付利息。 美國滙豐融資的大部分修訂條款個案均涉及調低 利率，從而令美國滙豐融資根據合約於日後期間 有權收取的利息收益額減少。過往修訂條款的期 限一般最低為六個月，但最近的修訂條款期限最 低為兩年。 已重訂賬齡的貸款分類為已減值，惟首次貸款於 重訂賬齡時為已逾期少於60日的貸款除外。該等 貸款會繼續分類為已減值，直至有關貸款顯示按 照其原有合約條款有至少12個月的履約還款紀錄 為止。 ‧ 永久修訂條款指變更貸款的合約條款，致使美國 滙豐融資在貸款有效期內放棄收取約定現金流的 權利，其中一個例子是永久調低利率。 由相關財困事件導致永久或長期修訂條款的貸 款，在整段貸款有效期內會一直分類為已減值。 「重訂賬齡」 是一種重議條件安排，在貸款顯示履 約還款能力後，貸款的合約狀況便由拖欠改為未 過期。逾期本金及╱或利息可予遞延，並於較後 日期支付。貸款重訂賬齡使曾有幾次欠交還款的 客戶，得以將貸款狀況由拖欠改為未過期，以使 其信貸評分不受逾期結欠狀況的影響。 已重訂賬齡的貸款會繼續分類為已減值，直至有 關貸款顯示按照其原有合約條款有至少12個月的 履約還款紀錄為止。 儘管貸款可在毋須修訂其原有條款及條件的情況 下重訂賬齡，但暫時或永久修訂條款亦可導致貸 款重訂賬齡。 倘貸款已獲授多次優惠，在上文所述合資格標準 的規限下，由於考慮到借款人的還款能力，優惠 視作已給予，而貸款披露為已減值。貸款自該日 期起仍披露為減值，直至借款人已顯示修訂或重 訂賬齡 （如上所述） 所規定期限的履約還款紀錄為 止。 美國止贖物業的估值 我們獲得房地產的方式，乃通過止贖收取住宅按揭以 擔保形式質押的抵押品。於止贖前，超出公允值減出 售成本的貸款賬面值，會撇減至預期可收回的經折現 現金流 （包括出售物業的現金流） 。 我們會就價格徵詢經紀意見，且每180天更新一次， 同時每季會檢討房地產價格趨勢，以反映任何改善或 進一步惡化的情況。我們會根據當前市況預計可收回 的經折現現金流 （包括出售物業的估計現金流） ，並以 此與經紀提供的最新價格意見作比較，再基於過往內 部與外界評估之間的估計差異進行調整，從而定期核 證我們的方法。止贖物業的公允值於首次列賬時，會 根據經紀就價格提供的意見釐定，然後我們會安排自 止贖起計的90天內，實地檢查物業的室內情況，基於 所得資料釐定更詳盡的物業估值，並視乎情況於賬目 中加入額外準備或撇減額。我們最少每隔45天進行一 次估值更新，直至物業出售為止，估值升跌會透過變 更準備額予以確認。 美國第二留置權按揭 大部分第二留置權住宅按揭由客戶以第三方發出的第 一留置權按揭持有。第二留置權住宅按揭貸款的風險 特徵是貸款估值比率較高，原因是在大多數情況下， 貸款用於完成物業再融資。第二留置權貸款拖欠引致 的虧損嚴重程度，一般接近結欠金額的100%，原因 是物業的權益已用作償還第一留置權貸款。 我們運用滾動率變動分析法，根據過往經驗掌握該等 貸款違約拖欠的傾向，從而釐定該等貸款的減值準備 額。我們一旦相信第二留置權住宅按揭貸款可能逐漸 演變成須予撇銷，便會在確立減值準備時假定消費及 按揭貸款組合的虧損嚴重程度接近100%，而美國滙 豐銀行則超逾80%。 滙豐控股有限公司 156 董事會報告：風險 （續） 信貸風險 持有之抵押品及其他強化信貸條件 （經審核） 按已攤銷成本持有貸款 集團有關抵押品的使用的慣例詳情，載於第213頁 風險附錄。 下表載列我們就借款人的特定資產所持固 定押記之量化價值，而我們就該等抵押品 按抵押品水平分析的住宅按揭貸款 （包括貸款承諾） （經審核） 歐洲 亞洲 4 中東及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 非減值貸款 全數抵押 135,875 99,257 2,431 43,317 3,759 284,639 貸款估值比率： －低於50% 66,075 60,315 1,324 14,003 1,454 143,171 －51%至75% 56,178 31,142 856 20,872 1,777 110,825 －76%至90% 11,856 6,906 212 5,994 480 25,448 －91%至100% 1,766 894 39 2,448 48 5,195 部分抵押： －超過100%貸款估值比率(A) 537 99 60 2,209 167 3,072 －A的抵押品價值 532 81 44 1,999 24 2,680 136,412 99,356 2,491 45,526 3,926 287,711 已減值貸款 全數抵押 906 256 122 8,618 154 10,056 貸款估值比率： －低於50% 232 130 53 1,291 103 1,809 －51%至75% 417 90 29 3,462 35 4,033 －76%至90% 163 32 19 2,471 10 2,695 －91%至100% 94 4 21 1,394 6 1,519 部分抵押： －超過100%貸款估值比率(B) 55 7 31 1,395 2 1,490 －B的抵押品價值 40 5 23 1,181 1 1,250 961 263 153 10,013 156 11,546 於2014年12月31日 137,373 99,619 2,644 55,539 4,082 299,257 非減值貸款 全數抵押 146,326 98,332 2,235 44,125 3,749 294,767 貸款估值比率： －低於50% 55,028 55,479 749 13,172 1,337 125,765 －51%至75% 66,452 34,370 1,095 20,751 1,715 124,383 －76%至90% 21,603 6,836 348 6,933 606 36,326 －91%至100% 3,243 1,647 43 3,269 91 8,293 部分抵押： －超過100%貸款估值比率(C) 1,410 362 42 4,150 59 6,023 －C的抵押品價值 852 307 37 3,681 49 4,926 147,736 98,694 2,277 48,275 3,808 300,790 已減值貸款 全數抵押 1,369 254 90 10,128 160 12,001 貸款估值比率： －低於50% 244 100 15 1,393 97 1,849 －51%至75% 452 96 31 4,250 47 4,876 －76%至90% 320 49 34 2,809 13 3,225 －91%至100% 353 9 10 1,676 3 2,051 部分抵押： －超過100%貸款估值比率(D) 104 17 6 2,548 8 2,683 －D的抵押品價值 91 4 6 2,272 4 2,377 1,473 271 96 12,676 168 14,684 於2013年12月31日 149,209 98,965 2,373 60,951 3,976 315,474 有關註釋，請參閱第202頁。 有強制執行的往績且有能力強制執行，於 借款人未能履行其合約責任時，可以抵押 品清償債務，或該等抵押品為現金或可在 某個既定市場出售從而變現其價值。抵押 品估值並不計及因取得及出售抵押品而作 出的任何調整，特別是以並未抵押或部分 抵押列示的貸款可透過其他形式的減低信 貸風險措施而受惠。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 157 補充資料 按行業劃分的五年貸款總額 （未經審核） 2014年 貨幣換算 調整 變動 2013年 2012年 2011年 2010年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 個人貸款 393,554 (19,092) 1,918 410,728 415,093 393,625 425,320 －第一留置權住宅按揭6 286,524 (12,372) (979) 299,875 301,862 278,963 268,681 －其他個人貸款7 107,030 (6,720) 2,897 110,853 113,231 114,662 156,639 企業及商業貸款 540,556 (24,729) 22,020 543,265 513,229 472,784 445,505 －製造業貸款 106,986 (5,856) (1,008) 113,850 112,149 96,054 91,121 －國際貿易及服務貸款 180,791 (8,232) 4,355 184,668 169,389 152,709 146,567 －商用物業貸款 73,293 (3,270) 1,717 74,846 76,760 73,941 71,880 －其他與物業有關貸款 52,387 (922) 8,477 44,832 40,532 39,539 34,838 －政府貸款 6,143 (395) (739) 7,277 10,785 11,079 8,594 －其他商業貸款8 120,956 (6,054) 9,218 117,792 103,614 99,462 92,505 金融機構貸款 50,818 (2,303) 2,598 50,523 46,871 44,832 41,213 －非銀行之金融機構貸款 48,799 (2,180) 2,442 48,537 45,430 43,888 39,651 －結算賬項 2,019 (123) 156 1,986 1,441 944 1,562 經重新分類之資產抵押證券 2,069 (147) (500) 2,716 3,891 5,280 5,892 各類客戶貸款總額(A) 986,997 (46,271) 26,036 1,007,232 979,084 916,521 917,930 同業貸款總額 112,198 (4,925) (2,981) 120,104 117,142 139,203 142,027 各類貸款總額 1,099,195 (51,196) 23,055 1,127,336 1,096,226 1,055,724 1,059,957 已減值客戶貸款 29,283 (1,538) (5,607) 36,428 38,671 41,584 46,871 －佔A的百分比 3.0% 3.6% 3.9% 4.5% 4.8% 客戶貸款減值準備 12,337 (776) (2,030) 15,143 16,112 17,511 20,083 －佔A的百分比 1.2% 1.5% 1.6% 1.9% 2.2% 貸款減值準備 4,055 (160) (1,833) 6,048 8,160 11,505 13,548 －已 扣除準備撥回額之 新撥準備 5,010 (158) (2,176) 7,344 9,306 12,931 14,568 －收回額 (955) (2) 343 (1,296) (1,146) (1,426) (1,020) 有關註釋，請參閱第202頁。 個人貸款對貸款總額的匯率影響為190億 美元，當中歐洲為130億美元、亞洲26億美 元、拉丁美洲18億美元、北美洲18億美元。 批發貸款對貸款總額的匯率影響為320億 美元，當中歐洲為210億美元、亞洲48億美 元、拉丁美洲47億美元、北美洲15億美元、 中東及北非3億美元。 滙豐控股有限公司 158 董事會報告：風險 （續） 信貸風險 按地區劃分的列賬基準與固定匯率基準已減值貸款、準備及撥備對賬 （未經審核） 2013年 12月31日 列賬基準 貨幣換算 調整 21 2013年 12月31日 按2014年 12月31日 匯率計算 固定匯率 基準變動 2014年 12月31日 列賬基準 列賬基準 變動 22 固定匯率 基準變動 22 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 % % 已減值貸款 歐洲 13,228 (1,011) 12,217 (1,975) 10,242 (23) (16) 亞洲4 1,623 (54) 1,569 479 2,048 26 31 中東及北非 2,285 (8) 2,277 (296) 1,981 (13) (13) 北美洲 15,123 (42) 15,081 (3,387) 11,694 (23) (22) 拉丁美洲 4,244 (425) 3,819 (454) 3,365 (21) (12) 36,503 (1,540) 34,963 (5,633) 29,330 (20) (16) 減值準備 歐洲 5,598 (420) 5,178 (723) 4,455 (20) (14) 亞洲4 1,214 (32) 1,182 174 1,356 12 15 中東及北非 1,583 (4) 1,579 (173) 1,406 (11) (11) 北美洲 4,242 (28) 4,214 (1,574) 2,640 (38) (37) 拉丁美洲 2,564 (294) 2,270 259 2,529 (1) 11 15,201 (778) 14,423 (2,037) 12,386 (19) (14) 貸款減值準備 歐洲 1,732 62 1,794 (715) 1,079 (38) (40) 亞洲4 483 (17) 466 178 644 33 38 中東及北非 (44) － (44) 43 (1) 98 98 北美洲 1,235 (15) 1,220 (920) 300 (76) (75) 拉丁美洲 2,642 (190) 2,452 (419) 2,033 (23) (17) 6,048 (160) 5,888 (1,833) 4,055 (33) (31) 有關註釋，請參閱第202頁。 扣取自收益表之列賬基準與固定匯率基準之減值準備對賬 （未經審核） 2013年 12月31日 列賬基準 貨幣換算 調整 21 2013年 12月31日 按2014年 12月31日 匯率計算 固定匯率 基準變動 2014年 12月31日 列賬基準 列賬基準 變動 22 固定匯率 基準變動 22 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 % % 貸款減值準備 歐洲 1,732 62 1,794 (715) 1,079 (38) (40) －新撥準備 3,082 99 3,181 (736) 2,445 (21) (23) －撥回額 (713) (11) (724) (338) (1,062) (49) (47) －收回額 (637) (26) (663) 359 (304) 52 54 亞洲4 483 (17) 466 178 644 33 38 －新撥準備 953 (31) 922 193 1,115 17 21 －撥回額 (303) 8 (295) (23) (318) (5) (8) －收回額 (167) 6 (161) 8 (153) 8 5 中東及北非 (44) － (44) 43 (1) 98 98 －新撥準備 408 (1) 407 (52) 355 (13) (13) －撥回額 (365) 2 (363) 49 (314) 14 13 －收回額 (87) (1) (88) 46 (42) 52 52 北美洲 1,235 (15) 1,220 (920) 300 (76) (75) －新撥準備 1,640 (17) 1,623 (715) 908 (45) (44) －撥回額 (282) 2 (280) (213) (493) (75) (76) －收回額 (123) － (123) 8 (115) 7 7 拉丁美洲 2,642 (190) 2,452 (419) 2,033 (23) (17) －新撥準備 3,262 (243) 3,019 (312) 2,707 (17) (10) －撥回額 (338) 34 (304) (29) (333) 1 (10) －收回額 (282) 19 (263) (78) (341) (21) (30) 總計 6,048 (160) 5,888 (1,833) 4,055 (33) (31) －新撥準備 9,345 (193) 9,152 (1,622) 7,530 (19) (18) －撥回額 (2,001) 35 (1,966) (554) (2,520) (26) (28) －收回額 (1,296) (2) (1,298) 343 (955) 26 26 有關註釋，請參閱第202頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 159 按行業劃分的五年貸款減值準備 （未經審核） 2014年 2013年 2012年 2011年 2010年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 貸款減值準備提撥╱ （撥回） 個人貸款 1,803 3,196 5,362 9,318 11,187 企業及商業貸款 2,256 2,974 2,802 2,114 2,198 金融機構貸款5 (4) (122) (4) 73 163 截至12月31日止年度 4,055 6,048 8,160 11,505 13,548 有關註釋，請參閱第202頁。 減值虧損準備佔客戶貸款總額平均值的百分比 （未經審核） 2014年 2013年 2012年 2011年 2010年 % % % % % 已扣除準備撥回額之新撥準備 0.53 0.81 1.00 1.34 1.65 收回額 (0.10) (0.14) (0.12) (0.15) (0.12) 減值虧損準備總額 0.43 0.67 0.88 1.19 1.53 已扣除收回額之撇賬額 0.58 0.59 0.93 1.14 2.08 五年減值準備變動 （未經審核） 2014年 2013年 2012年 2011年 2010年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於1月1日之減值準備 15,201 16,169 17,636 20,241 25,649 撇賬額 (6,379) (6,655) (9,812) (12,480) (19,300) －個人貸款 (3,733) (4,367) (6,905) (10,431) (16,458) －企業及商業貸款 (2,425) (2,229) (2,677) (2,009) (2,789) －金融機構貸款5 (221) (59) (230) (40) (53) 收回往年撇賬額 955 1,296 1,146 1,426 1,020 －個人貸款 818 1,097 966 1,175 846 －企業及商業貸款 128 198 172 242 156 －金融機構貸款5 9 1 8 9 18 貸款減值準備 4,055 6,048 8,160 11,505 13,548 匯兌及其他變動13 (1,446) (1,657) (961) (3,056) (676) 於12月31日之減值準備 12,386 15,201 16,169 17,636 20,241 減值準備 －個別評估 6,244 7,130 6,629 6,662 6,615 －綜合評估 6,142 8,071 9,540 10,974 13,626 於12月31日之減值準備 12,386 15,201 16,169 17,636 20,241 已扣除收回額之撇賬額佔客戶 平均貸款總額百分比 0.6% 0.6% 1.0% 1.2% 2.2% 有關註釋，請參閱第202頁。 滙豐控股有限公司 160 董事會報告：風險 （續） 信貸風險 按國家╱地區劃分的客戶貸款總額 （未經審核） 第一 留置權 住宅按揭6 其他 個人貸款7 與物業 有關貸款 商業、 國際貿易及 其他貸款 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 歐洲 131,000 47,531 35,313 200,313 414,157 英國 123,239 21,023 25,927 156,577 326,766 法國 2,914 12,820 7,341 21,834 44,909 德國 6 212 304 7,275 7,797 瑞士 298 8,149 225 614 9,286 土耳其 645 3,389 297 4,244 8,575 其他 3,898 1,938 1,219 9,769 16,824 亞洲 93,147 36,368 70,057 164,739 364,311 香港 56,656 22,891 52,208 82,362 214,117 澳洲 9,154 815 2,130 6,360 18,459 印度 1,235 285 613 5,099 7,232 印尼 64 469 202 5,476 6,211 中國內地 4,238 1,981 6,606 24,875 37,700 馬來西亞 5,201 1,750 1,988 5,217 14,156 新加坡 9,521 5,878 4,210 11,951 31,560 台灣 3,920 626 118 7,057 11,721 其他 3,158 1,673 1,982 16,342 23,155 中東及北非 （不包括沙地阿拉伯） 2,647 3,924 2,246 21,633 30,450 埃及 1 510 98 2,272 2,881 阿聯酋 2,263 1,782 1,545 13,814 19,404 其他 383 1,632 603 5,547 8,165 北美洲 55,577 9,823 15,492 51,535 132,427 美國 37,937 5,482 11,461 38,632 93,512 加拿大 16,236 4,085 3,708 11,825 35,854 其他 1,404 256 323 1,078 3,061 拉丁美洲 4,153 9,384 2,572 29,543 45,652 阿根廷 15 1,169 93 2,119 3,396 巴西 2,067 5,531 1,077 16,814 25,489 墨西哥 1,967 2,642 1,336 9,503 15,448 其他 104 42 66 1,107 1,319 於2014年12月31日 286,524 107,030 125,680 467,763 986,997 歐洲 140,474 51,633 38,634 230,932 461,673 英國 132,174 22,913 28,127 185,534 368,748 法國 2,661 13,840 8,442 23,962 48,905 德國 7 218 127 6,361 6,713 瑞士 364 8,616 269 320 9,569 土耳其 833 4,002 305 4,059 9,199 其他 4,435 2,044 1,364 10,696 18,539 亞洲 92,047 32,482 61,707 151,875 338,111 香港 53,762 19,794 44,904 75,547 194,007 澳洲 9,468 1,236 2,511 7,138 20,353 印度 1,080 297 425 4,231 6,033 印尼 69 447 78 5,361 5,955 中國內地 4,880 300 5,808 22,149 33,137 馬來西亞 5,140 1,994 1,997 5,420 14,551 新加坡 10,283 5,754 3,953 12,188 32,178 台灣 3,797 660 158 5,198 9,813 其他 3,568 2,000 1,873 14,643 22,084 中東及北非 （不包括沙地阿拉伯） 2,451 4,033 1,972 20,320 28,776 埃及 1 477 146 2,232 2,856 阿聯酋 2,082 1,842 1,331 12,344 17,599 其他 368 1,714 495 5,744 8,321 北美洲 60,955 11,735 14,616 44,884 132,190 美國 42,317 6,257 10,174 30,952 89,700 加拿大 17,036 5,116 3,912 13,079 39,143 其他 1,602 362 530 853 3,347 拉丁美洲 3,948 10,970 2,749 28,815 46,482 阿根廷 20 1,425 62 2,103 3,610 巴西 1,811 6,466 1,268 17,132 26,677 墨西哥 2,117 3,079 1,398 8,994 15,588 其他 － － 21 586 607 於2013年12月31日 299,875 110,853 119,678 476,826 1,007,232 有關註釋，請參閱第202頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 161 以上各表乃根據行業及貸款附屬公司之主 要業務所在地分析各類貸款；如屬香港上 海滙豐銀行、英國滙豐銀行、中東滙豐銀 行及美國滙豐銀行的貸款，則根據貸款分 行之所在地分析。 滙豐控股 （經審核） 滙豐控股的風險由滙豐控股資產負債管理 委員會 （ 「HALCO」 ） 監控。滙豐控股所面對的 主要風險為信貸風險、流動資金風險及市 場風險 （反映於利率風險及外匯風險） ，其 中最重大的是信貸風險。 滙豐控股的信貸風險主要來自與集團附屬 公司進行之交易，以及為支持若干集團附 屬公司在日常業務中承擔的責任而提供之 擔保。環球風險管理部負責檢討及管理有 關風險，務使風險不會超出監管機構及滙 豐內部規定的限額。該部門對環球信貸風 險進行高層次的中央監察及管理。 滙豐控股於2014年12月31日之最大信貸風 險表列如下。金融資產主要涉及對集團旗 下歐洲及北美洲附屬公司的債權。 所有衍生工具交易均為與滙豐旗下經營銀 行業務的交易對手進行 （2013年：100%） ， 滙豐控股已就此訂立淨額計算總協議。自 2012年起，信貸風險已按淨額基準管理， 餘下風險淨額則按特定抵押安排以現金方 式處理。 滙豐控股－最大信貸風險 （經審核） 2014年 2013年 最大風險 對銷 信貸風險 （淨額） 最大風險 對銷 信貸風險 （淨額） 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 銀行及庫存現金： －在滙豐旗下業務之結餘 249 － 249 407 － 407 衍生工具 2,771 (2,610) 161 2,789 (2,755) 34 滙豐旗下業務貸款 43,910 － 43,910 53,344 － 53,344 滙豐旗下金融投資 4,073 － 4,073 1,210 － 1,210 金融擔保及同類合約 52,023 － 52,023 52,836 － 52,836 貸款及其他信貸相關承諾 16 － 16 1,245 － 1,245 於12月31日 103,042 (2,610) 100,432 111,831 (2,755) 109,076 貸款及金融投資 （均由集團內部貸款組成） 的信貸質素獲評估為 「穩健」 或 「良好」 風險 級別，100%的貸款均屬並非逾期或已減值 （2013年：100%） 。 證券化風險及其他結構產品 （經審核） 本節載列滙豐就資產抵押證券 （ 「ABS」 ） 所承 擔的風險，而部分資產抵押證券透過綜合 入賬結構公司持有，並概述於下表。 有關滙豐各項風險承擔的性質，於第214頁風險附 錄內概述。 滙豐的整體風險承擔 （經審核） 賬面值23 2014年 2013年 十億美元 十億美元 資產抵押證券 48.9 50.1 －按公允值計入損益賬 3.6 3.1 －可供出售24 29.7 42.7 －持至到期日24 13.4 1.1 －貸款及應收賬款 2.2 3.2 於12月31日 48.9 50.2 有關註釋，請參閱第202頁。 下表按抵押品類別概述滙豐的資產抵押證 券的賬面值，並包括環球銀行及資本市場 業務的既有信貸組合中持有的資產，賬面 值為230億美元 （2013年：280億美元） 。 於2014年12月31日，有關資產抵押證券的可 供出售儲備減值7.77億美元 （2013年：減值 16.43億美元） ，2014年就資產抵押證券的減值 準備撥回為2.76億美元 （2013年：撥回2.89億 美元） 。 滙豐控股有限公司 162 董事會報告：風險 （續） 信貸風險╱流動資金及資金 滙豐於綜合計算後所持資產抵押證券之賬面值 23 （經審核） 交易用途 可供出售 持至 到期日 指定以 公允值 計入 損益賬 貸款及 應收賬款 總計 經綜合 入賬結構 公司持有 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 按揭相關資產： 次優質住宅 122 3,081 － － 308 3,511 2,075 美國Alt-A住宅 96 3,022 11 － 110 3,239 2,411 美國政府機構及資助企業： MBS 82 10,401 13,436 － － 23,919 － 其他住宅 928 1,220 － － 330 2,478 652 商用物業 654 3,627 － － 516 4,797 2,854 槓桿融資相關資產 172 3,660 － － 218 4,050 2,526 學生貸款相關資產 242 3,545 － － 119 3,906 3,284 其他資產 1,264 1,114 － 19 646 3,043 758 於2014年12月31日 3,560 29,670 13,447 19 2,247 48,943 14,560 按揭相關資產： 次優質住宅 178 2,977 － － 403 3,558 2,782 美國Alt-A住宅 101 3,538 18 － 134 3,791 2,926 美國政府機構及資助企業： MBS 178 18,661 1,110 － － 19,949 － 其他住宅 618 1,925 － － 399 2,942 1,513 商用物業 133 5,667 － 104 669 6,573 5,146 槓桿融資相關資產 294 5,011 － － 251 5,556 4,310 學生貸款相關資產 196 3,705 － － 121 4,022 3,495 其他資產 1,271 1,265 － 34 1,186 3,756 989 於2013年12月31日 2,969 42,749 1,128 138 3,163 50,147 21,161 有關註釋，請參閱第202頁。 與按揭出售及證券化活動相關的陳述和 保證 （未經審核） 我們曾參與各種涉及出售住宅按揭及住宅 按揭證券化的活動，這些活動並未於資產 負債表內確認入賬。這些活動包括： ‧ 於2005至2007年間，美國滙豐銀行購入 240億美元由第三方辦理的按揭貸款， 其後由HSBC Securities (USA) Inc.
（ 「HSI」 ） 予 以證券化； ‧ HSI為第三方發行私營機構按揭抵押證 券 （ 「MBS」 ） （大部分為次優質） 擔任包銷 商，該等證券的原有發行價值為370億 美元；及 ‧ 美國滙豐銀行辦理並出售按揭貸款，出 售對象主要為政府資助企業。 出售按揭貸款及予以證券化時，可能會向 按揭貸款及按揭抵押證券的購買方作出多 項陳述和保證。在購入第三方辦理的按揭 並予以證券化，以及包銷第三方按揭抵押 證券時，一旦出現違反貸款額所涉陳述和 保證的情況，回購貸款的責任便主要由辦 理貸款的機構承擔。 購買後重新包裝整批貸款的美國按揭證券 化市場參與者，如證券化的債務管理人、 辦理機構、包銷商、受託人或保薦人，已 成為訴訟、政府及監管機構調查及研訊的 對象。 於2014年12月31日，集團就美國滙豐銀行 辦理及出售按揭貸款 （對象主要為政府資 助企業） 時作出的多項陳述和保證，確認 一項2,700萬美元的負債 （2013年：9,900萬 美元） 。這些陳述和保證涉及 （其中包括） 該 等貸款的擁有權、留置權的有效性、貸款 的甄選和辦理程序，以及有否遵循有關機 構所訂的辦理準則。若有違反該等陳述和 保證的情況，美國滙豐銀行可能須回購該 等確定有問題的貸款或對買方作出彌償。 該項估計負債是根據下列準則計算：回購 需求水平；要求查看貸款檔案未決個案水 平；以及對迄今已出售按揭的預計日後回 購需求額，該等按揭為已拖欠兩期或以上 還款，或可能按照某個估計轉換率成為拖 欠貸款。於2014年的回購需求額為300萬美 元 （2013年：4,400萬美元） 。 有關法律訴訟及監管事宜的詳情，請參閱財務報 表附註40。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 163 流動資金及資金 頁次 附錄 1 列表 頁次 流動資金及資金 164 215 主要資金來源 215 2014年流動資金及資金 164 客戶存款市場 164 批發優先融資市場 164 流動資金規例 164 流動資金及資金風險管理 165 215 內在流動資金風險分類 215 核心存款 216 貸款對核心資金比率 165 216 貸款對核心資金比率 165 壓力下之償債保障比率 165 216 壓力下之一個月及三個月償債保障比率 165 壓力境況分析 216 滙豐旗下主要營運公司的流動資產 166 217 滙豐旗下主要公司的流動資產 166 約定現金流淨額 166 217 銀行同業及集團內部貸款及存款， 以及反向回購、回購及短倉的 現金流入╱ （流出） 淨額 167 監察批發債務 218 流動資金行為化 218 資金轉移定價 219 貸款承諾產生之或有流動資金風險 167 集團於12月31日之未取用訂約貸款 （根據 或有流動資金風險限額架構進行監察） 167 資金來源 168 回購及借出股票 219 資金來源及用途 168 跨境、集團內部及跨貨幣流動資金及 資金風險 169 按重要貨幣分析的貸款對核心資金比率 169 批發有期債務期限概況 169 按尚餘合約期限列示於金融負債項下 滙豐應付的批發資金現金流 170 具產權負擔及無產權負擔資產 171 220 可動用以支持日後潛在資金及抵押品 需要的資產概要 （資產負債表內及 資產負債表外） 171 抵押品 171 積極管理抵押品的影響 220 就反向回購、借入股票及衍生工具交易 收取並質押的資產負債表外抵押品 171 資產負債表內具產權負擔及 無產權負擔資產分析 171 資產負債表內具產權負擔及 無產權負擔資產分析 172 其他合約責任 173 金融負債合約期限 173 按尚餘合約期限列示於金融負債項下 滙豐應付的現金流 173 跨貨幣流動資金及資金風險管理 221 滙豐控股 174 221 按尚餘合約期限列示於金融負債項下 滙豐控股應付的現金流 174 1 風險附錄 － 風險政策及慣例。 滙豐控股有限公司 164 董事會報告：風險 （續） 流動資金及資金 流動資金及資金 流動資金風險即集團缺乏足夠財務資源履 行到期時的責任，或將要以過高成本履行 責任的風險。此風險因現金流的時間錯配 而產生。 滙豐管理流動資金及資金風險的政策與慣 例，於2014年並無重大變動。 於第347頁所詳述的資產負債表呈列方式變 動後，貸存比率現已不包括與客戶的非交 易用途反向回購以及回購。該變動對所披 露的2013年12月31日貸存比率並無影響。 有關流動資金及資金的現行政策與慣例，於第215 頁風險附錄內概述。 流動資金及資金風險管理架構 我們的流動資金架構旨在讓我們能抵禦極為沉重的流 動資金壓力，並為適應不斷變化的業務模式、市場狀 況及監管規定而設。 我們的流動資金及資金風險管理架構規定： ‧ 旗下營運公司以獨立形式管理流動資金，對集團 或中央銀行並無隱含的倚賴； ‧ 旗下所有營運公司須符合貸款對核心資金比率方 面的限制；及 ‧ 旗下所有營運公司須在集團指定壓力境況下可維 持三個月的正數受壓現金流水平。 2014年流動資金及資金 （未經審核） 2014年集團的流動資金狀況強勁，繼續取 得大量客戶存款，並且能繼續透過批發市 場順利融資。客戶賬項按固定匯率基準計 算上升4% （470億美元） 。按列賬基準計算， 客戶賬項款額微跌1% （110億美元） 。客戶 貸款按固定匯率基準計算上升3% （280億美 元） 。按列賬基準計算，客戶貸款下跌2% （170億美元） 。有關變動令貸存比率微跌至 72% （2013年：73%） 。 於2014年12月31日，英國滙豐的貸款對核 心資金比率 （ 「ACF」 ） 下跌至97% （2013 年： 100%） ，主要由於核心存款的增幅超過貸 款增長，以及出售既有資產。 於2014年12月31日，香港上海滙豐銀行的 貸款對核心資金比率上升至75% （2013年： 72%） ，主要因為貸款的增幅超過核心存款 增長。 於2014年12月31日，美國滙豐的貸款對核心 資金比率上升至100% （2013年：85%） ，主要 由於客戶貸款增長。 英國滙豐、香港上海滙豐銀行及美國滙豐的定義， 請參閱第202頁註釋26至28。貸款對核心資金比率 於第216頁討論。 客戶存款市場 （按固定匯率基準） 零售銀行及財富管理業務 零售銀行及財富管理業務的客戶賬項款額 上升4%，此乃受兩個本位市場 （即英國及香 港） 以及大部分優先發展市場所帶動。 工商金融業務 客戶賬項於2014年上升7%，此乃由於兩個 本位市場的資金管理服務賬項有所增長。 環球銀行及資本市場業務 客戶賬項於2014年上升2%，升幅主要來自 資金管理服務賬項。 環球私人銀行業務 由於滙豐繼續推動環球私人銀行業務重新 定位，並出售客戶業務組合，故環球私人 銀行業務的客戶賬項款額較2013年底下跌 10%。 批發優先融資市場 2014年銀行批發債務市場整體較2013年為 佳，令到銀行以不同資本架構形式在主要 市場的債務發行量增加。然而，市場仍時 有波動，特別是年底前數個月，各界關注 油價下跌以及歐洲經濟增長，再加上多個 令前景不明朗的因素，令市場信心受影響。 2014年，我們透過集團旗下多家公司在公 開資本市場發行不同貨幣、不同期限的有 期優先債務證券，金額相等於200億美元 （2013年：160億美元） 。 流動資金規例 （未經審核） 歐洲於2013年6月頒布文件，採納巴塞爾委 員會的架構 （立法文本稱為 《資本規定規例 及指引》 － 「資本規例╱資本指引4」 ﹚。有關 規定要求自2014年1月起，向歐洲監管機構 呈報流動資金覆蓋比率 （ 「LCR」 ） 及穩定資金 淨額比率 （ 「NSFR」 ） ，但隨後押後至2014年 6月30日實施。資本規例所界定的流動資 金覆蓋比率的呈報及計算，需要作大量詮 釋，原因是某些方面直到2015年1月流動 資金覆蓋比率最終授權法案確定後才能落 實，而有關法案於2015年10月1日始成為監 管標準。巴塞爾委員會於2014年10月提出 最終推薦意見後，歐洲尚未公布穩定資金 淨額比率的微調方案。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 165 流動資金及資金風險管理 （經審核） 滙豐的流動資金及資金風險管理架構 （ 「LFRF」 ） 採用兩項主要衡量標準來界定、 監察及監控旗下各營運公司的流動資金及 資金風險。貸款對核心資金比率用作監察 結構性長期資金狀況，而壓力下之償債保 障比率已納入集團界定壓力境況，用來監 察在嚴重流動資金壓力下的復元力。 下表列示的三家主要公司，佔集團客戶賬 項66% （2013年：66%） 。倘計入其他主要公 司，則百分比為95% （2013年：94%） 。 貸款對核心資金比率 右表顯示滙豐旗下經營銀行業務的主要公 司之客戶貸款，均有可靠及穩定的資金來 源。 於2014年12月31日，滙豐旗下主要營運公司 的貸款對核心資金比率介乎80%至120%之 間。 核心資金指客戶存款的核心部分，以及尚餘 合約期限超過一年的任何有期專業資金。 我們對核心資金的定義不包括資本。 壓力下之償債保障比率 下表所載比率，呈列一個月及三個月內受壓 現金流入量佔受壓現金流出量的百分比。 旗下營運公司須於三個月內維持100%或以 上的比率。 壓力下之償債保障比率的分子中計入的流 入量，乃源自流動資產的流入量 （扣除假設 壓力下之一個月及三個月償債保障比率 25 （經審核） 於以下年度12月31日之 壓力下之一個月償債保障比率 於以下年度12月31日之 壓力下之三個月償債保障比率 2014年 2013年 2014年 2013年 % % % % 英國滙豐26 年底 117 106 109 109 最高 117 114 109 109 最低 102 100 103 101 平均 107 106 104 103 香港上海滙豐銀行27 年底 117 119 112 114 最高 119 131 114 126 最低 114 113 111 109 平均 116 119 112 114 美國滙豐28 年底 111 114 104 110 最高 122 126 111 119 最低 108 110 104 109 平均 115 115 107 112 滙豐旗下其他主要公司總計29 年底 121 121 108 114 最高 121 128 115 119 最低 114 113 108 109 平均 117 120 110 113 有關註釋，請參閱第202頁。 扣減額） ，以及於有關期間按照合約即將到 期資產的相關現金流入量。 客戶貸款一般假設會予以續期，故不會產 生現金流入量。 貸款對核心資金比率 25 （經審核） 於12月31日 2014年 2013年 % % 英國滙豐26 年底 97 100 最高 102 107 最低 97 100 平均 100 104 香港上海滙豐銀行27 年底 75 72 最高 75 77 最低 72 70 平均 74 74 美國滙豐28 年底 100 85 最高 100 85 最低 85 78 平均 95 82 滙豐旗下其他主要公司總計29 年底 92 93 最高 94 93 最低 92 89 平均 93 91 有關註釋，請參閱第202頁。 英國滙豐的壓力下之一個月償債保障比率 上升，原因是之前在一至三個月壓力下被 視為可變現的若干資產，已重估為於一個 月內或三個月後可變現。壓力下之三個月 償債保障比率大致維持不變。 滙豐旗下其他公司的壓力下之償債保障比 率大致維持不變。 滙豐控股有限公司 166 董事會報告：風險 （續） 流動資金及資金 滙豐旗下主要營運公司的流動資產 下表顯示就計算壓力下之三個月償債保障 比率 （按流動資金及資金風險管理架構而 界定） 而言，列作流動類別的資產之估計流 動資金價值 （未計假設扣減） 。 所列報的流動資產水平反映於業績報告日 期的無產權負擔流動資產存量，並經三個 月內到期的反向回購、回購及抵押品掉期 的影響調整，因這些交易的流動資金價值 於約定現金流淨額表內列報為約定現金流。 與剩餘合約期限為三個月以內的反向回購 交易一樣，於三個月內到期的無抵押銀行 同業貸款並不列作流動資產，但被視作約 滙豐旗下主要公司的流動資產 （經審核） 估計流動資金價值30 2014年 12月31日 2013年 12月31日 百萬美元 百萬美元 英國滙豐26 第一級 131,756 168,877 第二級 4,688 1,076 第三級 66,011 63,509 202,455 233,462 香港上海滙豐銀行27 第一級 109,683 108,713 第二級 4,854 5,191 第三級 7,043 7,106 121,580 121,010 美國滙豐28 第一級 51,969 43,446 第二級 15,184 12,709 第三級 197 5,044 其他 9,492 8,000 76,842 69,199 滙豐旗下其他主要公司總計29 第一級 141,659 144,774 第二級 10,419 12,419 第三級 13,038 13,663 165,116 170,856 有關註釋，請參閱第202頁。 流動資產組合中所持的所有資產均屬無產 權負擔資產。 英國滙豐持有的流動資產減少，原因是從 中央銀行儲備調至短期反向回購。約定現 金流淨額表亦顯示英國滙豐的回購現金流 淨額改善。 香港上海滙豐銀行持有的流動資產大致維 持不變。 美國滙豐持有的流動資產主要因短期回購 減少以及按照流動資金及資金風險管理架 構重新分類若干資產為流動資產而上升。 約定現金流淨額 下表就所示之主要公司，量化銀行同業及 定現金流入。 流動資產由獨立營運公司持有及管理。所 示的大部分流動資產均由各營運公司的資 產負債管理部門按流動資金及資金風險管 理架構而直接持有，主要為管理流動資金 風險。 流動資產亦包括就任何其他目的而言，在 資產負債管理業務外持有的任何無產權負 擔的流動資產。流動資金及資金風險管理 架構讓資產負債管理業務對所有無產權負 擔資產及流動資金來源有最終控制權。 流動資產政策概要及下表所示分類的定義，載於 第217頁風險附錄。 集團內部貸款及存款，以及反向回購、回 購 （包括集團內部交易） 及短倉的約定現金 流。該等約定現金流入額及流出額分別在 壓力下之一個月及三個月償債保障比率的 分子和分母中以總額反映，並應與流動資 產的級別一併考慮。 計入壓力下之償債保障比率分母的流出額， 包括第170頁 「按尚餘合約期限列示於金融 負債項下滙豐應付的批發資金現金流」 列 表所呈報批發債務證券的合約期限相關的 主要流出額。 我們政策的概要及下表所示分類的定義，載於第 218頁風險附錄。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 167 銀行同業及集團內部貸款及存款，以及反向回購、回購及短倉的現金流入╱ （流出） 淨額 （經審核） 於2014年12月31日 於2013年12月31日 一個月內 現金流 一至三個月 現金流 一個月內 現金流 一至三個月 現金流 百萬美元 百萬美元 百萬美元 百萬美元 銀行同業及集團內部貸款及存款 英國滙豐26 (14,110) (2,846) (19,033) (5,272) 香港上海滙豐銀行27 (1,277) 6,862 2,314 7,487 美國滙豐28 (18,353) 1,648 (24,268) 729 滙豐旗下其他主要公司總計29 (1,322) 6,158 4,295 10,149 反向回購、回購、借入股票、借出股票及 直接短倉 （包括集團內部） 英國滙豐26 (16,070) 11,551 (39,064) 149 香港上海滙豐銀行27 8,139 8,189 12,662 4,297 美國滙豐28 (4,928) － (11,001) － 滙豐旗下其他主要公司總計29 (22,110) (11,120) (40,223) 9,551 有關註釋，請參閱第202頁。 貸款承諾產生之或有流動資金風險 （經審核） 集團旗下營運公司向不同交易對手提供承 諾。以流動資金風險而言，最重大風險涉 及未取用貸款承諾，因為有關資金可能於 流動資金受壓期間被取用，繼而出現或有 流動資金風險。集團會向客戶提供承諾， 並向以下各方提供貸款承諾：綜合入賬的 多賣方中介機構 （其成立旨在使客戶能運 用靈活的市場資金 （請參閱第443頁） ） 、綜 合入賬證券投資中介機構，以及第三方資 助中介機構。 綜合入賬的證券投資中介機構包括 Solitaire Funding Limited （ 「Solitaire」 ） 及Mazarin Funding Limited （ 「Mazarin」 ） 。該等中介機構 發行由其持有的證券組合作抵押的資產抵 押商業票據。於2014年12月31日，英國滙 豐向該等中介機構提供而未取用的貸款承 諾有110億美元 （2013年：150億美元） ，其中 Solitaire佔95億美元 （2013年：110億美元） ， 而Mazarin則佔餘下16億美元 （2013年：40億 美元） 。雖然英國滙豐提供流動資金信貸 額，但只要滙豐購買已發行的商業票據 （而 其於可預見將來有此意向） ，則Solitaire及 Mazarin毋須動用有關額度。於2014年12月31 日，Solitaire及Mazarin發行的商業票據全部 由英國滙豐持有。由於滙豐控制該等中介 機構持有的證券組合的規模，故並無因該 等未取用的貸款承諾而產生或有流動資金 風險。 下表就五宗最大單一貸款及最大市場類 別，列示向客戶提供而未取用的貸款承諾 水平，以及該等貸款承諾的未取用數額。 集團於12月31日之未取用訂約貸款 （根據或有流動資金風險限額架構進行監察） （經審核） 英國滙豐26 美國滙豐28 加拿大滙豐 香港上海滙豐銀行27 2014年 2013年 2014年 2013年 2014年 2013年 2014年 2013年 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 向中介機構提供的承諾 綜合入賬的多賣方中介 機構 －信貸總額 9.8 10.1 2.3 2.5 0.2 1.0 － － －最大個別信貸額 0.9 0.7 0.5 0.5 0.2 0.7 － － 綜合入賬證券投資中介 機構－信貸總額 11.1 14.8 － － － － － － 第三方資助中介機構－信貸總額 － － 0.1 0.7 － － － － 向客戶提供的承諾 －五大31 2.6 4.4 7.1 6.3 1.7 1.5 1.5 2.4 －最大市場類別32 16.6 9.5 10.0 8.2 3.5 3.4 3.2 2.7 有關註釋，請參閱第202頁。 滙豐控股有限公司 168 董事會報告：風險 （續） 流動資金及資金 資金來源 （經審核） 我們的主要資金來源為客戶往來賬項及客 戶即期或短期通知儲蓄存款。我們發行批 發證券 （有抵押或無抵押） 以補充客戶存款 及改變負債的貨幣組合、期限情況或所在 地。 以下 「資金來源及用途」 列表綜合反映我們 如何為資產負債提供資金，參閱有關資料 時應一併參照流動資金及資金風險管理架 構，該架構要求營運公司以獨立形式管理 流動資金及資金風險。 該表根據主要源自營運業務的資產，以及 主要支持該等業務的資金來源，分析我們 的綜合資產負債表。並非由營運業務產生 的資產及負債，則以結餘淨額來源或資金 投放呈列。 客戶賬項水平繼續超過客戶貸款水平。正 數的資金差額按流動資金及資金風險管理 架構規定，主要投放於現金及於中央銀行 的結餘以及金融投資等流動資產。 應收銀行同業的貸款及其他應收賬款繼續 超過從銀行同業收取的存款。集團繼續是 銀行業內提供無抵押淨貸款的貸款人。 回購及借出股票的資金來源及用途概要，載於第 219頁風險附錄。 資金來源及用途 33 （經審核） 2014年 2013年 2014年 2013年 百萬美元 百萬美元 百萬美元 百萬美元 來源 用途 客戶賬項1 1,350,642 1,361,297 客戶貸款1 974,660 992,089 同業存放1 77,426 86,507 同業貸款1 112,149 120,046 回購協議－非交易用途1 107,432 164,220 回購協議－非交易用途1 161,713 179,690 已發行債務證券 95,947 104,080 交易用途資產 304,193 303,192 後償負債 26,664 28,976 －反向回購 1,297 10,120 指定以公允值列賬之 －借入股票 7,969 10,318 金融負債 76,153 89,084 －結算賬項 21,327 19,435 保單未決賠款 73,861 74,181 －其他交易用途資產 273,600 263,319 交易用途負債 190,572 207,025 金融投資 415,467 425,925 －回購 3,798 17,421 現金及於中央銀行的結餘 129,957 166,599 －借出股票 12,032 12,218 資產負債表其他資產 －結算賬項 17,454 17,428 及負債投放淨額 100,537 118,288 －其他交易用途負債 157,288 159,958 於12月31日 2,198,676 2,305,829 各類股東權益總額 199,979 190,459 於12月31日 2,198,676 2,305,829 有關註釋，請參閱第202頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 169 跨境、集團內部及跨貨幣流動資金及資金 風險 （未經審核） 流動資金及資金風險管理架構規定的營運 公司獨立管理流動資金及資金的方針，限 制營運公司因過於依賴跨境資金而可能產 生的風險。營運公司管理業務所在地的資 金來源，並主要專注於本土客戶存款基礎。 營運公司內的零售銀行及財富管理業務、 工商金融業務及環球私人銀行業務，因客 戶關係產生核心存款，通常反映該營運公 司的本土客戶關係。全球各地參與公開債 市的活動由資產負債管理業務的環球主管 及集團司庫負責統籌，而集團資產負債管 理委員會則每月監察所有已計劃的公開債 務發行。一般原則是營運公司僅可以本土 貨幣發行債務，而且集團會鼓勵這些公司 專注於本土私人配售。以外幣公開發行債 務工具受嚴格控制，並一般僅限於由滙豐 控股及英國滙豐銀行發行。 流動資金及資金風險管理架構獨立方針的 基本原則是，營運公司日後不應依賴集團 旗下其他公司。然而，倘有需要，營運公 司仍可酌情利用其各自由集團旗下其他公 司提供的信貸承諾。此外，國家監管機構 會對各地的個別法律實體施加集團內部大 額風險上限，將法律實體貸予集團旗下其 他公司的無抵押貸款，限於貸款人監管規 定資本的某個百分比以內。 我們的流動資金及資金風險管理架構亦考 慮各公司在以一種貨幣的盈餘應付另一貨 幣的虧絀時，能否於壓力下繼續從外匯市 場取得資金，例如透過外幣掉期市場。在 適用情況下，營運公司須監察非本土貨幣 的壓力下之償債保障比率及貸款對核心資 金比率，並設定限額。透過連續聯繫結算 銀行(Continuous Link Settlement Bank)結算配 對貨幣的外幣掉期市場，被視為極具深度 及流通性極高的市場，並假設參與此等市 場不會存在非系統性風險。下表顯示截至 2014年12月31日止年度按重要貨幣分析的 貸款對核心資金比率。 按重要貨幣分析的貸款對核心資金比率 25 （未經審核） 於2014年 12月31日 % 英國滙豐26 本土貨幣 （英鎊） 98 美元 100 歐元 99 綜合 97 香港上海滙豐銀行27 本土貨幣 （港元） 81 美元 74 綜合 75 美國滙豐28 本土貨幣 （美元） 100 綜合 100 滙豐旗下其他主要公司總計29 本土貨幣 97 美元 101 綜合 92 有關註釋，請參閱第202頁。 對滙豐所有營運公司而言唯一重要外幣 （超 過集團資產負債表負債的5%） 為港元、歐 元、英鎊及美元。 批發有期債務期限情況 （未經審核） 批發有期債務期限情況載於第170頁列表 「按尚餘合約期限列示於金融負債項下滙 豐應付的批發資金現金流」 。 列表所示款額不會與綜合資產負債表的款 額直接對應，因為該表呈列與支付本金有 關的現金流總額，而非資產負債表的賬面 值，當中包括按公允值計量的債務證券及 後償負債。 滙豐控股有限公司 170 董事會報告：風險 （續） 流動資金及資金 按尚餘合約期限列示於金融負債項下滙豐應付的批發資金現金流 （未經審核） 1個月內 到期 1個月 後但 3個月內 到期 3個月 後但 6個月內 到期 6個月 後但 9個月內 到期 9個月 後但 1年內 到期 1年 後但 2年內 到期 2年 後但 5年內 到期 5年 後到期 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 已發行債務證券 17,336 17,161 19,030 9,352 9,055 27,312 40,855 31,928 172,029 －無抵押存款證及商業票據 5,637 9,337 9,237 4,793 3,010 3,506 4,158 185 39,863 －無抵押優先中期票據 1,300 5,679 7,684 2,922 4,794 17,676 23,523 20,715 84,293 －無抵押優先結構票據 1,363 1,082 2,049 1,149 979 4,757 8,444 6,789 26,612 －有抵押備兌債券 － － － 205 － － 2,765 2,942 5,912 －有抵押資產抵押商業票據 8,602 － － － － － － － 8,602 －有抵押資產抵押證券 212 1,063 60 283 272 915 1,562 － 4,367 －其他 222 － － － － 458 403 1,297 2,380 後償負債 － 150 － 3 185 113 5,556 40,487 46,494 －後償債務證券 － 150 － 3 185 113 5,556 34,750 40,757 －優先證券 － － － － － － － 5,737 5,737 於2014年12月31日 17,336 17,311 19,030 9,355 9,240 27,425 46,411 72,415 218,523 已發行債務證券 25,426 9,752 17,942 11,659 10,587 31,839 46,934 31,066 185,205 －無抵押存款證及商業票據 7,589 7,206 9,867 3,239 5,043 4,449 2,749 － 40,142 －無抵押優先中期票據 6,284 71 5,448 4,221 3,062 21,428 33,091 21,433 95,038 －無抵押優先結構票據 987 1,423 1,952 1,689 1,718 3,712 6,036 5,021 22,538 －有抵押備兌債券 － － － 1,250 － 225 2,747 3,317 7,539 －有抵押資產抵押商業票據 10,383 － － － － － － － 10,383 －有抵押資產抵押證券 74 1,052 675 1,260 764 1,861 2,311 － 7,997 －其他 109 － － － － 164 － 1,295 1,568 後償負債 － 28 1,171 144 6 1,460 3,374 41,801 47,984 －後償債務證券 － 28 1,171 144 6 460 3,374 34,899 40,082 －優先證券 － － － － － 1,000 － 6,902 7,902 於2013年12月31日 25,426 9,780 19,113 11,803 10,593 33,299 50,308 72,867 233,189 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 171 具產權負擔及無產權負擔資產 （未經審核） 第172頁 「資產負債表內具產權負擔及無產 權負擔資產分析」 列表概述可用於支持日 後資金及抵押品需要的資產負債表內及資 產負債表外之資產總值，並列示該等資產 現時就此目的而質押的狀況。此項披露的 目的，旨在方便了解為支持日後潛在資金 及抵押品需要的可動用及無限制資產的狀 況。 此項披露並非旨在識別可供應付債權人申 索的資產，或預測在議決或破產的情況下 可分派予債權人的資產。 倘資產已用作某項現有負債的抵押品予以 質押，因而不能提供予集團以擔保融資、 應付抵押品需要或出售以減少資金需求， 則該項資產被界定為具產權負擔。因此， 倘資產未就現有負債予以質押，便可被分 類為無產權負擔。無產權負擔資產再進一 步按四個獨立類別分析： 「可隨時變現資 產」 、 「其他可變現資產」 、 「反向回購╱借入 股票應收賬款及衍生工具資產」 及 「不能作 為抵押品予以質押」 。 於2014年12月31日，集團持有17,700億美元 無產權負擔資產，可用於支持日後潛在資 金及抵押品需要，佔可用於支持資金及抵 押品需要的資產總值 （資產負債表內及資產 負債表外） 85%，其中7,650億美元 （6,840億 美元為資產負債表內） 被評估為可隨時變 現。 可動用以支持日後潛在資金及抵押品需要的資產概要 （資產負債表內及資產負債表外） （未經審核） 2014年 2013年 十億美元 十億美元 資產負債表內資產總值 2,634 2,671 減： 反向回購╱借入股票應收賬款及衍生工具資產 (518) (482) 不能作為抵押品予以質押的其他資產 (281) (255) 可支持資金及抵押品需要的資產負債表內資產總值 1,835 1,934 加資產負債表外資產： 取自反向回購╱借入股票╱衍生工具而可供出售或再質押之抵押品的公允值 257 265 可支持資金及抵押品需要的資產總值 （資產負債表內及資產負債表外） 2,092 2,199 減： 已質押資產負債表內資產 (146) (187) 取自已再質押或出售的反向回購╱借入股票╱衍生工具的資產負債表外抵押品 (176) (187) 於12月31日可動用以支持日後資金及抵押品需要的資產 1,770 1,825 有關抵押品管理政策的概要以及產權負擔的定義， 載於第213頁風險附錄。 抵押品 （未經審核） 就反向回購、借入股票及衍生工具交易收 取並質押的資產負債表外抵押品 於2014年12月31日，獲接納為抵押品並可 以在不違約下出售或再質押之資產的公允 值為2,570億美元 （2013年：2,650億美元） 。 已出售或再質押的該類抵押品的公允值為 1,760億美元 （2013年：1,870億美元） 。我們 須歸還等值證券。該等交易按常規反向回 購、借入股票及衍生工具交易的一般及慣 常條款進行。 就反向回購、借入股票及衍生工具所收取 並再質押的抵押品公允值以總額基準呈 報。如有需要，相關資產負債表的應收及 應付賬款，將根據IFRS對銷準則按淨額基準 呈報。 由於進行反向回購、借入股票及衍生工具 交易，所收取的抵押品可予出售或再質押 但尚未出售或再質押，於2014年12月31日， 我們持有可動用以支持日後潛在資金及抵 押品需要的無產權負擔抵押品為810億美元 （2013年：780億美元） 。 資產負債表內具產權負擔及無產權負擔資 產分析 下表僅呈列資產負債表內項目的分析，並 按流動資金及資金基準呈列具產權負擔的 資產負債表資產的金額。因此，下表並未 呈列就反向回購、借入股票或衍生工具所 收取的任何可動用資產負債表外項目。 滙豐控股有限公司 172 董事會報告：風險 （續） 流動資金及資金 資產負債表內具產權負擔及無產權負擔資產分析 （未經審核） 具產權負擔 無產權負擔 作為 抵押品 予以質押 之資產 可隨時 變現資產 其他可 變現資產 反向 回購╱ 借入股票 應收賬款 及衍生 工具資產 不能 作為抵押品 予以質押 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 現金及於中央銀行的結餘 － 123,990 425 － 5,542 129,957 向其他銀行託收中之項目 － － － － 4,927 4,927 香港政府負債證明書 － － － － 27,674 27,674 交易用途資產 59,162 182,305 17,869 9,266 35,591 304,193 －國庫及其他合資格票據 1,994 14,122 4 － 50 16,170 －債務證券 46,311 94,941 23 － 257 141,532 －股權證券 10,857 62,855 1,497 － 40 75,249 －同業貸款 － 2,530 4,818 2,781 17,452 27,581 －客戶貸款 － 7,857 11,527 6,485 17,792 43,661 指定以公允值列賬之金融資產 － 177 2,330 26,530 29,037 －國庫及其他合資格票據 － － 52 － 4 56 －債務證券 － 177 1,058 － 7,656 8,891 －股權證券 － － 1,139 － 18,867 20,006 －同業及客戶貸款 － － 81 － 3 84 衍生工具 － － － 345,008 － 345,008 同業貸款 178 3,573 74,231 762 33,405 112,149 客戶貸款 24,329 92,238 840,241 1,170 16,682 974,660 反向回購協議－非交易用途 － － － 161,713 － 161,713 金融投資 61,785 275,732 22,780 － 55,170 415,467 －國庫及其他合資格票據 3,176 75,896 2,167 － 278 81,517 －債務證券 58,609 192,411 18,266 － 53,970 323,256 －股權證券 － 7,425 2,347 － 922 10,694 預付款項、應計收益及其他資產 294 6,334 29,780 － 38,768 75,176 本期稅項資產 － － － － 1,309 1,309 於聯營及合資公司之權益 － 22 17,875 － 284 18,181 商譽及無形資產 － － － － 27,577 27,577 遞延稅項 － － － － 7,111 7,111 於2014年12月31日 145,748 684,371 1,005,531 517,919 280,570 2,634,139 現金及於中央銀行的結餘 － 161,240 269 － 5,090 166,599 向其他銀行託收中之項目 － － － － 6,021 6,021 香港政府負債證明書 － － － － 25,220 25,220 交易用途資產 99,326 142,211 14,654 20,438 26,563 303,192 －國庫及其他合資格票據 3,402 17,976 206 － － 21,584 －債務證券 83,563 57,850 － － 231 141,644 －股權證券 8,373 55,156 363 － － 63,892 －同業貸款 1,796 2,813 6,151 5,263 11,861 27,884 －客戶貸款 2,192 8,416 7,934 15,175 14,471 48,188 指定以公允值列賬之金融資產 19 2,706 1,883 － 33,822 38,430 －國庫及其他合資格票據 － － － － 50 50 －債務證券 19 826 776 － 10,968 12,589 －股權證券 － 1,874 1,103 － 22,734 25,711 －同業及客戶貸款 － 6 4 － 70 80 衍生工具 － － － 282,265 － 282,265 同業貸款 162 8,342 80,231 － 31,311 120,046 客戶貸款 32,218 102,203 854,724 65 2,879 992,089 反向回購協議－非交易用途 － － － 179,690 － 179,690 金融投資 54,473 289,093 31,096 － 51,263 425,925 －國庫及其他合資格票據 2,985 72,849 2,052 － 226 78,112 －債務證券 51,488 210,516 25,720 － 50,949 338,673 －股權證券 － 5,728 3,324 － 88 9,140 預付款項、應計收益及其他資產 1,028 16,788 24,619 － 34,407 76,842 本期稅項資產 － － － － 985 985 於聯營及合資公司之權益 － 12 16,356 － 272 16,640 商譽及無形資產 － － － － 29,918 29,918 遞延稅項 － － － － 7,456 7,456 於2013年12月31日 187,226 722,595 1,023,832 482,458 255,207 2,671,318 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 173 按尚餘合約期限列示於金融負債項下滙豐應付的現金流 （經審核） 即期 3個月 內到期 3至 12個月 內到期 1至 5年內 到期 5年後 到期 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 同業存放 52,682 17,337 3,600 3,580 390 客戶賬項 1,088,769 187,207 61,687 15,826 390 回購協議－非交易用途 8,727 91,542 6,180 23 1,057 交易用途負債 190,572 － － － － 指定以公允值列賬之金融負債 365 2,201 9,192 28,260 39,397 衍生工具 335,168 375 1,257 4,231 1,517 已發行債務證券 9 32,513 30,194 37,842 7,710 後償負債 － 737 1,256 10,003 42,328 其他金融負債 41,517 23,228 4,740 1,893 988 1,717,809 355,140 118,106 101,658 93,777 貸款及其他信貸相關承諾 406,561 101,156 64,582 62,312 16,769 金融擔保及同類合約 13,166 6,306 13,753 9,575 4,278 於2014年12月31日 2,137,536 462,602 196,441 173,545 114,824 同業存放 56,198 22,965 3,734 2,819 686 客戶賬項 1,097,159 196,048 57,243 15,520 726 回購協議－非交易用途 37,117 112,621 14,177 － － 交易用途負債 207,025 － － － － 指定以公允值列賬之金融負債 18,689 1,967 3,223 39,554 64,144 衍生工具 269,554 456 1,684 6,099 1,638 已發行債務證券 2,528 35,401 33,695 46,141 6,526 後償負債 55 391 2,687 11,871 44,969 其他金融負債 31,996 30,706 6,564 2,376 1,300 1,720,321 400,555 123,007 124,380 119,989 貸款及其他信貸相關承諾 377,352 79,599 55,124 59,747 16,872 金融擔保及同類合約 18,039 4,796 12,040 7,479 3,988 於2013年12月31日 2,115,712 484,950 190,171 191,606 140,849 金融負債合約期限 （經審核） 下表所示款額不會與綜合資產負債表的款 額直接對應，因為該表按未折現基準計算 所有與本金及未來票息付款有關的現金流 （惟交易用途負債及並非視作對沖用途的 衍生工具除外） 。就對沖用途衍生工具的負 債而應付之未折現現金流，均根據合約期 限而分類。交易用途負債及並非視作對沖 用途的衍生工具，計入 「即期」 一欄而並無 按合約期限列示。 計入交易用途負債的回購及已發行債務證券之期 限分析於財務報表附註31呈列。 此外，貸款及其他信貸相關承諾與金融擔 保及同類合約一般不會在資產負債表中確 認。根據金融擔保及同類合約可能應付之 未折現現金流，則根據其可被提早贖回的 最早日期而分類。 上表呈列為具產權負擔的客戶貸款240億美 元 （2013年：320億美元） 已質押，主要用於 支持發行有抵押債務工具，例如備兌債券 及資產抵押證券，包括由綜合入賬多賣方 中介機構發行的資產抵押商業票據。該項 目亦包括就任何其他形式有抵押借貸所質 押的貸款。 集團主要為客戶之證券融資進行市場莊家活 動而合共質押1,210億美元 （2013年：1,500億 美元） 可轉讓證券。 其他合約責任 根據衍生工具合約 （屬符合國際掉期業務 及衍生投資工具協會的信貸支持附件合約 以及就退休金責任訂立的合約，並不包括 為特設企業訂立的合約及其他終止事件） 下現有抵押品責任的條款，我們估計按照 2014年12月31日的持倉，倘滙豐的信貸評級 被下調一級，我們可能需要額外提供最多 5億美元 （2013年：7億美元） 的抵押品，而倘 信貸評級被下調兩級，有關金額將會上升 至12億美元 （2013年：12億美元） 。 滙豐控股有限公司 174 滙豐控股 （經審核） 滙豐控股資產負債管理委員會負責監督滙 豐控股的流動資金風險。由於滙豐控股有 責任於債務到期時向債務持有人付款，故 產生流動資金風險。滙豐控股透過將債務 責任與內部貸款現金流配對，並維持由滙 豐控股資產負債管理委員會監察的適當流 動資金緩衝，以管理該等現金流的流動資 金風險。 於2014年12月31日，集團有92億美元符合資 本指引4規定的非普通股權資本票據，其中 35億美元分類為二級，57億美元分類為額 外一級﹙年內發行的額外一級票據詳情， 載於財務報表附註35﹚。 下表所示款額不會與滙豐控股資產負債表 的款額直接對應，因為該表按未折現基準 計算所有與本金及未來票息付款有關的現 金流 （惟並非視作對沖用途的衍生工具除 外） 。就對沖用途衍生工具的負債而應付之 未折現現金流，均根據合約期限而分類。 並非視作對沖用途的衍生工具，計入 「即 期」 一欄。 此外，貸款承諾與金融擔保及同類合約一 般不會在資產負債表中確認。根據金融擔 保及同類合約可能應付之未折現現金流， 則根據其可被提早贖回的最早日期而分類。 按尚餘合約期限列示於金融負債項下滙豐控股應付的現金流 （經審核） 即期 3個月 內到期 3至 12個月 內到期 1至5年 內到期 5年 後到期 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 應付滙豐旗下業務款項 1,441 985 42 449 － 指定以公允值列賬之金融負債 － 210 642 6,345 19,005 衍生工具 1,066 － － 103 － 已發行債務證券 － 16 50 263 1,303 後償負債 － 252 770 5,815 28,961 其他金融負債 － 1,132 158 － － 2,507 2,595 1,662 12,975 49,269 貸款承諾 16 － － － － 金融擔保及同類合約 52,023 － － － － 於2014年12月31日 54,546 2,595 1,662 12,975 49,269 應付滙豐旗下業務款項 2,053 1,759 2,315 857 5,654 指定以公允值列賬之金融負債 － 299 671 4,921 26,518 衍生工具 704 － － － － 已發行債務證券 － 37 1,780 279 1,451 後償負債 － 225 676 5,699 24,812 其他金融負債 － 885 284 － － 2,757 3,205 5,726 11,756 58,435 貸款承諾 1,245 － － － － 金融擔保及同類合約 52,836 － － － － 於2013年12月31日 56,838 3,205 5,726 11,756 58,435 董事會報告：風險 （續） 流動資金及資金╱市場風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 175 市場風險 頁次 附錄 1 列表 頁次 2014年市場風險 176 市場風險 176 221 環球業務市場風險概覽 222 按環球業務分類的風險類別 176 市場風險管治 222 市場風險計量 223 監察及限制市場風險 176 223 敏感度分析 176 223 估計虧損風險 176 223 壓力測試 176 224 市場風險壓力測試 176 交易用途組合 176 225 交易用途組合之估計虧損風險 176 單日估計虧損風險 （交易用途組合） 177 交易賬項估計虧損風險 177 回溯測試 177 224 集團交易賬項估計虧損風險與 假設利潤及虧損 比較的回溯測試 177 市場缺口風險 225 脫鈎風險 225 資產抵押證券╱按揭抵押證券風險 225 非交易用途組合 178 225 非交易用途組合之估計虧損風險 178 單日估計虧損風險 （非交易用途組合） 178 非交易賬項估計虧損風險 178 可供出售債務證券之信貸息差風險 （包括證券投資中介機構） 178 226 分類為可供出售的股權證券 179 226 股權證券公允值 179 資產負債表內涉及市場風險的項目 179 計入及不計入交易賬項估計虧損風險 的結餘 179 會計基準資產負債表內涉及市場風險 的項目 180 結構匯兌風險 181 226 非交易賬項利率風險 181 226 利率風險行為化 181 226 資產負債管理業務 227 資產負債管理業務的第三方資產 181 資產負債管理業務的第三方資產 181 淨利息收益的敏感度 181 227 預計淨利息收益的敏感度 182 列賬基準之儲備對利率變動的 敏感度 183 界定福利退休金計劃 183 228 滙豐的界定福利退休金計劃 183 僅適用於母公司的其他市場風險計量 183 228 匯兌風險 183 滙豐控股－匯兌估計虧損風險 183 淨利息收益的敏感度 184 滙豐控股之淨利息收益對 利率變動的敏感度 184 利率重新定價缺口表 184 滙豐控股的重新定價缺口分析 185 1 風險附錄 － 風險政策及慣例。 滙豐控股有限公司 176 董事會報告：風險 （續） 市場風險是指匯率及商品價格、利率、信 貸息差及股價等市場因素出現變動，可能 導致我們的收益或組合價值減少之風險。 滙豐管理市場風險的政策與慣例，於2014年 並無重大改變。 市場風險 市場風險分為兩個組合： ‧ 交易用途組合包括因進行市場莊家活動而持有 及代客戶保管的持倉。滙豐控股發行的定息 證券的利率風險，並未計入集團的估計虧損風 險。管理此項風險的詳情載於第222頁。 ‧ 非交易用途組合包括主要因我們進行零售銀行 及工商金融業務資產與負債利率管理而產生的 持倉、指定列為可供出售及持至到期日之金融 投資，以及來自我們保險業務的風險項目 （請參 閱第225頁） 。 監察及限制市場風險 我們的目標，是管理及監控市場風險，同時保持相 關市場的狀況與我們的承受風險水平相符。 我們利用多種工具監察及限制市場風險，包括： ‧ 敏感度分析包括淨利息收益敏感度分析以及結 構匯兌敏感度分析，用於監察各風險類別的市 場風險概況； ‧ 估計虧損風險 （ 「VaR」 ） 是一種估算方法，用以 估計於指定時限內和既定可信程度下，可能因 市場利率和價格變動而在風險持倉內產生的潛 在虧損；及 ‧ 我們明白估計虧損風險存在局限，因此以壓力 測試加強估計虧損風險的計算，以評估倘若出 現較為極端但有可能發生的事件，或一系列金 融變數的變動時，對組合價值的潛在影響。反 映當前市場關注的境況之例子，為中國內地發 展放緩以及主權債務違約的潛在影響，包括其 更廣泛的連鎖影響。 有關市場風險管理架構 （包括現行政策） 於第221頁 風險附錄內概述。 2014年市場風險 （未經審核） 環球金融市場於年內的特點是低通脹及全 球增長疲弱，主要貨幣管理當局透過低息 率和購買資產等措施以維持寬鬆政策。 由於美國數據顯示國內生產總值增長，美 國聯儲局的購買資產計劃結束。儘管如此， 美元債券收益率進一步下跌。市場目光轉 移至歐洲央行為應對低增長和通縮而採取 的措施。持續通縮會對已陷入衰退且債務 對國內生產總值比率較高的國家╱地區有 極壞影響。2014年的特點是7國集團市場利 率低企，股市表現溫和。 在此情況下，我們繼續在交易業務中整體 維持防禦性風險狀況。防禦性持倉的特點 是未平倉合約淨額低或透過期權交易購買 波幅保障產品。防禦性持倉較低的交易賬 項估計虧損風險因分散程度較低以及監管 機構改變計算估計虧損風險的校準方法導 致估計虧損風險上升而被抵銷。非交易賬 項估計虧損風險於年內下跌，原因是低利 率 （特別是美元） 令非交易用途資產的期限 縮短。 交易用途組合 （經審核） 交易用途組合之估計虧損風險 交易賬項估計虧損風險主要在環球資本市 場業務出現。由於利率交易賬項估計虧損 風險上升、估計虧損以外風險 （ 「RNIV」 ） 的 分散風險作用消退，以及各資產類別組合 分散貢益降低，2014年12月31日的交易賬項 估計虧損風險高於2013年12月31日的水平。 過往一年交易賬項估計虧損風險總額的單 日水平載於下圖。 市場風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 177 單日估計虧損風險 （交易用途組合） ，99% 1日 （百萬美元） （未經審核） 13年12月 14年2月 14年4月 14年5月 14年7月 14年8月 14年10月 14年12月 -50 -30 -10 10 30 50 70 90 交易賬項VaR 利率交易 信貸息差交易 外匯交易 股權交易 分散 （包括RNIV） 集團年內交易賬項估計虧損風險載於下表。 交易賬項估計虧損風險，99% 1日 34 （經審核） 外匯及 商品 利率 股權 信貸息差 組合分散 （包 括估計虧損 以外風險） 35 總計36 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年12月31日 9.8 45.4 7.3 12.5 (14.3) 60.7 平均 16.9 39.5 6.9 13.7 (17.8) 59.2 最高 34.2 50.6 15.6 20.9 77.8 於2013年12月31日 16.0 33.4 9.2 14.2 (20.7) 52.1 平均 15.2 33.4 5.1 16.5 (20.3) 49.9 最高 26.4 71.9 14.1 25.5 81.3 有關註釋，請參閱第202頁。 回溯測試 （未經審核） 於2014年，集團錄得一次特殊虧損情況， 並錄得兩次特殊利潤情況。 特殊虧損情況主要由於部分已發展市場的 匯率及利率波幅增加以及孳息曲線趨向平 坦。 特殊利潤情況來自息差收窄，以及承擔新 興市場的匯率及利率風險。並無證據顯示 模型出現誤差或監控出現問題。 下圖載列於2014年與集團假設利潤及虧損 比較的單日交易賬項估計虧損風險，當中 不包括審慎監管局就監管規定資本目的而 豁免的特殊情況。 集團交易賬項估計虧損風險與假設利潤及虧損比較的回溯測試 （百萬美元） （未經審核） 14年1月 14年2月 14年3月 14年5月 14年6月 14年7月 14年9月 14年10月 14年11月 14年12月 -130 -80 -30 20 70 120 百萬美元 回溯測試特殊情況 假設利潤及虧損 估計虧損風險(99%) 滙豐控股有限公司 178 董事會報告：風險 （續） 非交易用途組合 （經審核） 非交易用途組合之估計虧損風險 集團的非交易賬項估計虧損風險包括來自 所有環球業務的相關風險。非交易用途組 合並無商品風險。於2014年，非交易賬項 估計虧損風險下跌，主要是因為非交易賬 項的期限因利率低 （特別是美元利率） 而縮 減。由於整體持倉減少，加上波幅降低以 及計算估計虧損風險時使用的信貸息差基 線調低，因此信貸息差風險的因素亦令非 交易賬項估計虧損風險降低。此項變動包 括有關集團持有的可供出售債務證券 （不 包括保險業務所持有者） 的信貸息差風險 減少，詳情於第194頁進一步討論。 年內，非交易賬項利率和信貸息差估計虧 損風險降低的影響，因分散的利好影響下 降而被抵銷。 過往一年非交易賬項估計虧損風險總額的 單日水平載於下圖。 單日估計虧損風險 （非交易用途組合） ，99% 1日 （百萬美元） （未經審核） -120 -80 -40 0 40 80 120 160 200 240 280 13年12月 14年2月 14年4月 14年5月 14年7月 14年8月 14年10月 14年12月 非交易賬項VaR 分散 利率非交易 信貸息差非交易 集團年內非交易賬項估計虧損風險載於下表。 非交易賬項估計虧損風險，99% 1日 （經審核） 利率 信貸息差 組合分散 總計 百萬美元 百萬美元 百萬美元 百萬美元 於2014年12月31日 88.2 62.5 (28.5) 122.2 平均 103.3 73.3 (37.4) 139.2 最高 147.7 91.9 189.0 於2013年12月31日 150.6 80.4 (76.4) 154.6 平均 145.7 106.6 (82.1) 170.2 最高 221.7 135.7 252.3 管理銀行賬項利率風險的進一步詳情，包 括資產負債管理業務的角色，載於下文 「非 交易賬項利率風險」 內。 非交易賬項估計虧損風險不包括可供出售 證券的股權風險、結構匯兌風險以及滙豐控 股所發行的定息證券之利率風險，有關風險 的管理詳情載於下文有關章節。該等章節亦 概述滙豐管理非交易賬項市場風險的範圍。 可供出售債務證券之信貸息差風險 （包括 證券投資中介機構） 於2014年12月31日，信貸息差估計虧損風 險變動對集團可供出售債務證券的影響為 8,100萬美元 （2013年：1.13億美元） 。此敏 感度包括根據信貸息差估計虧損風險在資 產負債表內綜合入賬的證券投資中介機構 （ 「SIC」 ） 風險總額。此敏感度並未計及原應 由資本票據持有人承擔的虧損。 2014年12月31日的敏感度低於2013年12月 31日，主要因為年內整體持倉降低，和觀察 所得的波幅減少以及信貸息差基線調低。 市場風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 179 分類為可供出售的股權證券 股權證券公允值 （經審核） 2014年 2013年 十億美元 十億美元 持有的私募股本37 2.0 2.7 配合業務持續發展的投資38 1.2 1.2 其他策略投資 7.5 5.2 於12月31日 10.7 9.1 有關註釋，請參閱第202頁。 分類為可供出售的各類股權證券之公允值 可能大幅波動。上表載列可供出售股權證 券對股東權益可能造成的最大虧損。其他 策略投資增加主要是由於興業銀行的投資 市值上升，抵銷了出售多項直接及私募基 金投資導致持有的私募股本減少的影響。 資產負債表內涉及市場風險的項目 （未經審核） 以下及第180頁載列的資料旨在根據強化信 息披露工作組提出的建議，方便讀者掌握 資產負債表內的項目與市場風險披露資料 所包括的持倉之間的聯繫。 計入及不計入交易賬項估計虧損風險的結餘 （未經審核） 資產負債表 計入交易 賬項估計虧損 風險的結餘 不計入交易 賬項估計虧損 風險的結餘 主要市場 風險敏感度 百萬美元 百萬美元 百萬美元 於2014年12月31日 資產 現金及於中央銀行的結餘 129,957 129,957 B 交易用途資產 304,193 276,419 27,774 A 指定以公允值列賬之金融資產 29,037 29,037 A 衍生工具 345,008 333,880 11,128 A 同業貸款 112,149 112,149 B 客戶貸款 974,660 974,660 B 反向回購協議－非交易用途 161,713 161,713 C 金融投資 415,467 415,467 A 負債 同業存放 77,426 77,426 B 客戶賬項 1,350,642 1,350,642 B 回購協議－非交易用途 107,432 107,432 C 交易用途負債 190,572 170,576 19,996 A 指定以公允值列賬之金融負債 76,153 76,153 A 衍生工具 340,669 334,199 6,470 A 已發行債務證券 95,947 95,947 C 上表列示根據下列資產類別分類涉及若干市場風險的會計項目： A 外匯、利率、股權及信貸息差。 B 外匯及利率。 C 外匯、利率及信貸息差。 上表將資產及負債分為以下兩類： ‧ 計入交易賬項並以估計虧損風險計量； 及 ‧ 不計入交易賬項及╱或並不以估計虧損 風險計量。 計入及不計入交易賬項估計虧損風險的金 融工具分析，展示由風險架構所反映涉及 市場風險的資料。然而，必須強調的是，上 表並未反映我們如何管理市場風險，原因 是我們在估計虧損風險模型中並未區分資 產與負債。 交易收益淨額中大部分的收益，來自計入 交易賬項估計虧損風險的資產與負債。如 第49頁所載，滙豐於2014年的交易收益淨 額為67.6億美元 （2013年：86.9億美元） 。交 易收益的調整 （例如估值調整） 並未納入交 易賬項估計虧損風險模型內。 滙豐控股有限公司 180 董事會報告：風險 （續） 會計基準資產負債表內涉及市場風險的項目 交易用途資產及負債 集團幾乎所有交易用途資產及負債均來自環球銀行及 資本市場業務。倘該等資產及負債主要就短期內出售 或回購目的而購入或產生，或構成一併管理並有證據 顯示近期會採取短期獲利模式的已識別金融工具組合 之一部分，則分類為持作交易用途。該等資產及負債 就市場風險管理而言視作交易賬項風險處理，惟少數 情況除外，這些情況主要在環球銀行業務內出現，其 資產的短期購入及出售與辦理貸款等其他非交易相關 活動有關。 指定以公允值列賬之金融資產 滙豐的指定以公允值列賬之金融資產主要由保險公司 持有。該等資產大部分與單位相連或附有酌情參與條 款的保單及投資合約的投保人負債相連。該等資產的 風險大致抵銷投保人合約負債的市場風險，並以非交 易賬項風險的形式管理。 指定以公允值列賬之金融負債 滙豐的指定以公允值列賬之金融負債主要為滙豐旗下 公司就融資目的而發行的定息證券。倘債務證券按已 攤銷成本入賬，則會出現會計錯配，原因是經濟上對 沖證券市場風險的衍生工具將按公允值入賬，而變動 則於收益表確認。該等負債的市場風險列作非交易賬 項風險處理，而主要風險則為利率及╱或匯兌風險。 我們亦就投資合約承擔對客戶之負債，當中單位相連 合約的負債按單位相連基金資產公允值計算。該等基 金的風險列作非交易賬項風險處理，而主要風險則為 基金相關資產的風險。 衍生工具資產及負債 我們就三項主要目的而進行衍生工具活動：為客戶訂 立風險管理方案；管理來自客戶業務的組合風險；以 及管理及對沖我們本身的風險。我們的衍生工具風險 大部分來自環球銀行及資本市場業務的銷售及交易活 動，並就市場風險管理目的視作交易賬項風險處理。 衍生工具資產及負債中，有若干衍生工具組合並未以 有意作交易用途形式管理風險，該等組合就估計虧損 風險計算目的而言，視作非交易賬項風險處理。出現 此等組合是因為訂立衍生工具是為了管理因非交易賬 項風險項目而產生的風險，當中包括不合資格對沖用 途衍生工具以及合資格採用公允值及現金流對沖會計 法的衍生工具。使用主要風險涉及利率及匯兌風險的 不合資格對沖的詳情，載於第181頁。採用公允值及 現金流對沖會計法之衍生工具的詳情，載於財務報表 附註16。該等工具的主要風險，與利率及匯兌風險有 關。 客戶貸款 客戶貸款內資產的主要風險，為借款人的信貸風險。 該等資產的風險就市場風險管理而言，視作非交易賬 項風險處理。 金融投資 金融投資包括以可供出售及持至到期日形式持有的資 產。集團所持該等證券按會計分類及發行人類別分類 的分析，載於財務報表附註18，而按業務活動分類的 分析則載於第60頁。該等證券主要由環球銀行及資 本市場業務旗下的資產負債管理業務持有。源自管理 結構利率及流動資金風險的持倉，就市場風險管理而 言，視作非交易賬項風險處理。保險公司持有的可供 出售證券視作非交易賬項風險處理，並主要為支持非 相連保險投保人負債而持有。 其他持有的可供出售資產主要為環球銀行及資本市場 業務內既有信貸業務的資產抵押證券，就市場風險管 理而言，視作非交易賬項風險處理，而其主要風險為 債務人信貸風險。 集團之持至到期日證券主要由保險業務持有。持至到 期日資產的風險就風險管理而言，視作非交易賬項風 險處理。 回購及反向回購協議－非交易用途 反向回購協議 （分類為資產） 是一種有抵押貸款。滙豐 在反向回購期內借出現金以換取抵押品 （一般為債券） 。 回購協議 （分類為負債） 與反向回購相反，容許滙豐透 過向貸款人提供抵押品而獲得資金。 兩項交易類別就市場風險管理而言視作非交易賬項風 險處理，主要風險為交易對手信貸風險。 適用於按公允值列賬之金融工具的會計政策詳情，請參閱財務報表附註13。 市場風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 181 結構匯兌風險 （未經審核） 有關我們管理結構匯兌風險的政策及程序，請參 閱風險附錄第226頁。 結構匯兌風險項目詳情請參閱財務報表附註33。 非交易賬項利率風險 （未經審核） 有關非交易賬項利率風險及流動資金與資金風險 的資金轉撥定價程序的政策，請分別參閱風險附 錄第226頁及第219頁。 資產、負債及資本管理部 （ 「ALCM」 ） 在風險 管理會議監督下，負責計量及監控非交易 賬項利率風險，其主要職責為： ‧ 界定非交易賬項利率風險由環球業務 轉移至資產負債管理業務之規管規則； ‧ 界定適用於非交易用途資產╱負債的利 率風險行為化之規管規則 （請參閱下文） ； ‧ 確保所有可抵銷的市場利率風險由環 球業務轉移至資產負債管理業務；及 ‧ 為監管環球業務的剩餘利率風險 （包括 不可抵銷的任何市場風險） 界定規則及 衡量標準。 不同類別的非交易賬項利率風險以及集團 用以量化及限制該等風險的監控方法可分 類如下： ‧ 轉移至資產負債管理業務並在界定的 市場風險授權下由其管理的風險，乃主 要透過使用可供出售組合持有的定息 流動資產 （政府債券） 及╱或構成公允值 對沖或現金流對沖關係一部分的利率 衍生工具管理。此項非交易賬項利率風 險在非交易賬項估計虧損風險及在淨 利息收益 （請參閱下文） 或股東權益經濟 價值 （ 「EVE」 ） 之敏感度反映； ‧ 由於風險不能對沖而保留於資產負債 管理業務外或源自行為轉移定價假設 的風險。此項風險並未在非交易賬項估 計虧損風險中反映，但透過淨利息收益 或股東權益經濟價值之敏感度而掌握， 而相應的上限則構成非交易賬項利率 風險的環球及地區承受風險水平聲明 之一部分。典型例子為主要貨幣利率極 低造成利息收益率受壓； ‧ 於可對沖時轉移至資產負債管理業務 的基差風險。任何於環球業務剩餘的基 差風險均會向資產負債管理委員會匯 報。此項風險並未在非交易賬項估計虧 損風險中反映，但透過淨利息收益或股 東權益經濟價值之敏感度而掌握。典型 例子為按倫敦銀行同業拆息利率曲線 釐定轉換價的管理利率儲蓄產品；及 ‧ 不能透過非交易賬項估計虧損風險、淨 利息收益或股東權益經濟價值之敏感 度而掌握，但透過壓力測試架構控制之 模型風險。典型例子為住宅按揭的提前 還款風險或往後風險。 利率風險行為化 有關我們利率風險行為化的政策，請參閱風險附 錄第226頁。 資產負債管理業務的第三方資產 （未經審核） 有關資產負債管理業務管治架構，請參閱風險附 錄第227頁。 2014年資產負債管理業務第三方資產減少 9%。中央銀行存款減少310億美元，跌幅主 要來自歐洲，原因是回購活動減少，加上 存款利率變為負利率，於歐洲央行的結存 因而減少。同業貸款減少60億美元，跌幅主 要來自香港及亞洲其他地區。金融投資減 少80億美元，原因是外匯變動、香港及南北 美洲的金融投資錄得銷售淨額及到期，但 投放更多資金於亞洲證券市場，抵銷了部 分減幅。 資產負債管理業務的第三方資產 （未經審核） 2014年 2013年 百萬美元 百萬美元 現金及於中央銀行的 結餘 103,008 134,086 交易用途資產 4,610 5,547 指定以公允值列賬之 金融資產 － 72 貸款1： －同業 53,842 59,355 －客戶 1,931 2,146 反向回購協議 59,172 58,968 金融投資 306,763 314,427 其他 2,470 3,700 於12月31日 531,796 578,301 有關註釋，請參閱第202頁。 淨利息收益的敏感度 （未經審核） 下表載述由2015年1月1日起12個月內每季開 始時，假設全球所有市場的孳息曲線平行 上移或下移25個基點，對我們日後會計淨利 息收益 （不包括保險收益） 產生的影響。所 呈列的敏感度顯示假設所有其他非利率風 險變數維持不變，以及管理層不採取任何 行動，在兩種利率境況下可預期的基本情 況預測淨利息收益的變動。於計算基本情 況預測淨利息收益時，所使用的資產及負 債重新定價利率乃來自目前的孳息曲線。 利率敏感度屬指示性，並根據簡化的境況 評估。此項分析的限制於風險附錄第227頁 討論。 假設管理層不作回應，所有孳息曲線連串 上移 （ 「上行震盪」 ） ，會使截至2015年之計劃 滙豐控股有限公司 182 董事會報告：風險 （續） 淨利息收益增加8.85億美元 （2014年：9.38億 美元） ，而所有孳息曲線連串下移 （ 「下行震 盪」 ） ，則會使預計淨利息收益減少20.89億 美元 （2014年：17.34億美元） 。 集團的淨利息收益之敏感度可分為三個主 要部分，即來自四項環球業務 （不包括資產 負債管理業務及資本市場業務） 的結構敏 感度、交易賬項資金 （資本市場業務） 的敏 感度及資產負債管理業務的敏感度。 在利率上升的環境下，結構敏感度為正數， 而在利率下跌的環境中則為負數。交易賬 項資金的敏感度在利率上升的環境下為負 數，而在利率下跌的環境中則為正數。至 於對利潤的影響，淨利息收益變動預期會 因交易收益淨額的相似變動而抵銷。資產 負債管理業務的敏感度將取決於其持倉。 一般而言，假設管理層不作回應，資產負 債管理業務在利率上升的環境下，敏感度 為負數，而在利率下跌的環境中則為正數。 以下的淨利息收益敏感度數字亦加入所應 用任何利率行為化的影響，以及在特定利 率境況下假設產品重新定價的影響，但並 無加入管理層決定改變滙豐資產負債表成 分的影響。 有關利率行為化及資產負債管理業務角色的詳情， 請參閱風險附錄第227頁。 資產負債管理業務的淨利息收益敏感度源 自資產負債管理業務用以緩減被轉移的利 率風險的技術組合，以及用以配合其界定 風險授權的優化收入淨額的方法。下表的 數字並未加入管理層對資產負債管理業務 作出的決定的影響，但現實的情況是管理 層極可能會在有需要時對資產負債管理業 務的持倉作短期調整，以抵銷特定利率境 況下淨利息收益的影響。 源自交易賬項資金的淨利息收益敏感度包 括為交易用途資產提供資金的支出，而來 自該等交易用途資產的收入則在交易收益 淨額中列賬。這情況會導致淨利息收益敏 感度數字出現不一致，並於我們加入淨利 息收益及交易收益淨額的環球業務業績中 沖銷。因此，交易賬項資金對除稅前利潤 的整體影響可能不會如下述數字般顯著。 上行震盪於2014年大致維持不變。下行震 盪上升主要由於資產負債管理業務以美元 計的利率風險變動所致。 預計淨利息收益的敏感度 39 （未經審核） 美元區 美洲其他 貨幣區 港元區 亞洲其他 貨幣區 英鎊區 歐元區 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 2015年因孳息曲線移動產生的 預計淨利息收益變動： 每季初上移25個基點 209 (9) 245 265 321 (146) 885 每季初下移25個基點 (521) (1) (494) (259) (783) (31) (2,089) 2014年因孳息曲線移動產生的 預計淨利息收益變動： 每季初上移25個基點 (107) 12 327 236 598 (128) 938 每季初下移25個基點 (291) (23) (412) (233) (761) (14) (1,734) 有關註釋，請參閱第202頁。 我們會每月評估所有孳息曲線平行上移或 下移100個基點時，可供出售組合及現金流 對沖估值的預期減幅，藉以監察列賬基準 之儲備對利率變動的敏感度。該等特定風 險僅構成集團整體利率風險的一部分。根 據會計處理方法，我們以具經濟效益的方 式對銷下表所示的大部分風險後，餘下利 率風險的重估變動毋須計入儲備項內。 市場風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 183 下表載列我們列賬基準之儲備對所列孳息 曲線變動的敏感度，以及於年內的月底最 高及最低數值。有關敏感度屬指示性質， 並根據簡化的境況評估。列賬基準之儲備 的敏感度變動主要由於可供出售證券組合 減少。 列賬基準之儲備對利率變動的敏感度 （未經審核） 最大影響 最小影響 百萬美元 百萬美元 百萬美元 於2014年12月31日 所有孳息曲線平行上移100個基點 (3,696) (5,212) (3,696) 佔股東權益總額之百分比 (1.9%) (2.7%) (1.9%) 所有孳息曲線平行下移100個基點 3,250 4,915 3,250 佔股東權益總額之百分比 1.7% 2.6% 1.7% 於2013年12月31日 所有孳息曲線平行上移100個基點 (5,762) (5,992) (5,507) 佔股東權益總額之百分比 (3.2%) (3.3%) (3.0%) 所有孳息曲線平行下移100個基點 5,634 5,786 4,910 佔股東權益總額之百分比 3.1% 3.2% 2.7% 界定福利退休金計劃 （經審核） 倘若我們各項界定福利退休金計劃附帶可 確定現金流的資產，不足以悉數配對該等 計劃的責任，便會產生市場風險。 滙豐的界定福利退休金計劃 （經審核） 2014年 2013年 十億美元 十億美元 負債 （現值） 42.1 40.5 % % 資產： 股票 18 18 債務證券 68 70 其他 （包括物業） 14 12 於12月31日 100 100 有關我們各項界定福利計劃之詳情，請參閱財務 報表附註6；而有關退休金風險管理之詳情，請參 閱第200頁。 僅適用於母公司的其他市場風險計量 （經審核） 用於管理市場風險的主要工具為：用於計 算匯兌風險的估計虧損風險；滙豐控股淨 利息收益對未來孳息曲線變動的預計敏感 度；及用於計算利率風險的利率缺口重新 定價表。 匯兌風險 滙豐控股於2014年內部產生的匯兌估計虧 損風險總額如下： 滙豐控股－匯兌估計虧損風險 （經審核） 2014年 2013年 百萬美元 百萬美元 於12月31日 29.3 54.1 平均 42.1 51.1 最低 29.3 46.7 最高 50.0 64.1 匯兌風險主要來自給予附屬公司具資本性 質的貸款，而此等貸款並非以貸方或借方 的功能貨幣計值，並入賬列作金融資產。此 等貸款因匯率差異而產生的賬面值變動， 會直接計入滙豐控股的收益表內。此等貸 款及其大部分相關匯兌風險，均按集團綜 合基準予以撇銷。 滙豐控股有限公司 184 董事會報告：風險 （續） 滙豐控股之淨利息收益對利率變動的敏感度 39 （經審核） 美元區 英鎊區 歐元區 總計 百萬美元 百萬美元 百萬美元 百萬美元 截至12月31日止因孳息曲線移動產生的 預計淨利息收益變動 2014年 每季初上移25個基點 0至1年 78 9 2 89 2至3年 281 17 34 332 4至5年 138 17 24 179 每季初下移25個基點 0至1年 (58) (9) (1) (68) 2至3年 (276) (16) (12) (304) 4至5年 (138) (17) (12) (167) 2013年 每季初上移25個基點 0至1年 104 (14) 2 92 2至3年 382 (93) 38 327 4至5年 245 (101) 38 182 每季初下移25個基點 0至1年 (53) 13 (2) (42) 2至3年 (300) 91 (33) (242) 4至5年 (243) 101 (38) (180) 有關註釋，請參閱第202頁。 上表所列的利率敏感度屬指示性，並根據 簡化的境況評估。上列數字顯示，在我們 預計的孳息曲線境況下、在滙豐控股當前 的利率風險狀況下，及假設在未來五年內 該狀況出現變化下，淨利息收益的假設變 動。於未來五年內涉及風險狀況的假設若 出現變化，可能對該段期間的淨利息收益 敏感度造成重大影響。惟此等數字並未計 及為減輕此項利率風險而可能採取之行動 或會造成的影響。 利率重新定價缺口表 滙豐控股發行的定息證券之利率風險，並 未計入集團的估計虧損風險內，而是按重 新定價缺口基準管理。下列利率重新定價 缺口表，分析滙豐控股資產負債表內於所 有時段利率錯配的結構。 淨利息收益的敏感度 （經審核） 滙豐控股監察淨利息收益在五年期內的敏 感度，此舉可反映適用於金融服務控股公 司的較長期利率風險管理方法。有關敏感 度假設在發行的債項中，倘滙豐控股有權 於未來提早贖回日期償付，則於有關日期 贖回。下表載列由2015年1月1日起12個月內 每季開始時，假設全球所有市場的孳息曲 線平行下移或上移25個基點，對滙豐控股 未來五年期間淨利息收益產生的影響。 假設管理層不採取任何行動，所有孳息曲 線連串上移，會使未來五年之計劃淨利息 收益增加6億美元 （2013年：增加6.02億美 元） ，而所有孳息曲線連串下移，則會使計 劃淨利息收益減少5.39億美元 （2013年：減 少4.64億美元） 。 市場風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 185 滙豐控股的重新定價缺口分析 （經審核） 總計 不超過1年 1至5年 5至10年 10年以上 不附息 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 銀行及庫存現金： －在滙豐旗下業務之結餘 249 － － － － 249 衍生工具 2,771 － － － － 2,771 滙豐旗下業務貸款 43,910 41,603 290 1,093 － 924 滙豐旗下業務金融投資 4,073 3,010 － 731 － 332 於附屬公司之投資 96,264 － － － － 96,264 其他資產 597 － － － － 597 資產總值 147,864 44,613 290 1,824 － 101,137 應付滙豐旗下業務款項 (2,892) (1,877) － － － (1,015) 指定以公允值列賬之金融負債 (18,679) (850) (5,472) (5,400) (4,263) (2,694) 衍生工具 (1,169) － － － － (1,169) 已發行債務證券 (1,009) － － (1,013) － 4 其他負債 (1,415) － － － － (1,415) 後償負債 (17,255) (779) (3,766) (2,000) (10,195) (515) 各類股東權益總額 (105,445) － － － － (105,445) 各類負債及股東權益總額 (147,864) (3,506) (9,238) (8,413) (14,458) (112,249) 對利率敏感的資產負債表外項目 － (21,525) 7,295 7,400 5,763 1,067 於2014年12月31日淨利率風險缺口 － 19,582 (1,653) 811 (8,695) (10,045) 累計利率缺口 － 19,582 17,929 18,740 10,045 － 銀行及庫存現金： －在滙豐旗下業務之結餘 407 357 － － － 50 衍生工具 2,789 － － － － 2,789 滙豐旗下業務貸款 53,344 49,979 290 1,239 645 1,191 滙豐旗下業務金融投資 1,210 300 － 731 － 179 於附屬公司之投資 92,695 － － － － 92,695 其他資產 391 － － － － 391 資產總值 150,836 50,636 290 1,970 645 97,295 應付滙豐旗下業務款項 (11,685) (10,865) － － － (820) 指定以公允值列賬之金融負債 (21,027) (1,928) (4,655) (7,810) (4,325) (2,309) 衍生工具 (704) － － － － (704) 已發行債務證券 (2,791) (1,722) － － (1,069) － 其他負債 (1,375) － － － － (1,375) 後償負債 (14,167) － (3,030) (2,066) (8,912) (159) 各類股東權益總額 (99,087) － － － － (99,087) 各類負債及股東權益總額 (150,836) (14,515) (7,685) (9,876) (14,306) (104,454) 對利率敏感的資產負債表外項目 － (18,620) 4,382 9,876 4,421 (59) 於2013年12月31日淨利率風險缺口 － 17,501 (3,013) 1,970 (9,240) (7,218) 累計利率缺口 － 17,501 14,488 16,458 7,218 － 滙豐控股有限公司 186 董事會報告：風險 （續） 營運風險 （未經審核） 頁次 附錄 1 列表 頁次 營運風險 186 228 營運風險管理架構 186 三道防線 186 營運風險管理架構 187 2014年之營運風險 187 錄得營運風險損失的頻率及金額 188 按風險類別分類的營運風險事件頻率 189 按風險類別分類並按美元計算的營運 風險損失分布 189 合規風險 189 229 法律風險 229 環球保安及詐騙風險 229 系統風險 230 供應商風險管理 231 1風險附錄－風險政策及慣例。 集團各業務環節均涉及營運風險，所涉問 題層面甚廣，尤其是法律、合規、保安及 詐騙。營運風險的定義包括所有因違反法 規及法律、未經授權活動、差錯、遺漏、缺 乏效率、詐騙、系統失靈或因外圍事件而 引致的損失。 將營運風險減至最低是滙豐管理層及職員 的職責。各區域、環球業務、國家╱地區或 業務單位及部門的主管有責任不斷監督其 職責範圍內所有業務及運作的營運風險及 內部監控。 有關營運風險的現行政策及慣例，於第228頁風險 附錄內概述。 營運風險管理架構 集團營運風險管理部及營運風險管理架構 （ 「ORMF」 ） 協助業務管理層履行其職責。 營運風險管理架構界定集團內營運風險與 內部監控的最低標準、程序以及管治架構。 為實施營運風險管理架構，集團採用 「三 道防線」 模式管理風險，詳情如下： 三道防線 第一道 防線 每名滙豐僱員須負責屬於其日常工作 一部分的風險。第一道防線確保能透過 整體監控環境下適當的內部監控措施， 識別、緩減及監察其營運中的所有 主要風險。 第二道 防線 包括環球部門，例如風險管理部、財務 部及人力資源部，負責鑑證、質詢以及 監督第一道防線進行的活動。 第三道 防線 審核部門就第一及第二道防線提供獨立 鑑證。 營運風險管理架構圖載於第187頁。 2014年，我們繼續貫徹應用滙豐的營運風 險管理架構。同時，我們正在簡化營運風 險管理程序，並協調架構組成部分與風險 管理程序。此舉預期將加強營運風險管理 文化，提高洞悉風險的前瞻性，令業務可 釐定重大風險是否控制在集團承受風險水 平範圍內，以及是否需要採取其他行動。 此外，保安及詐騙風險管理部以及防範金 融犯罪部已經設立金融情報組，因應客戶 及業務前景所產生的潛在金融犯罪風險提 供情報，確保集團作出更明智的風險管理 決策。金融情報組提供背景及專業知識， 全面地識別、評估及理解客戶、行業及市 場的金融犯罪風險。 制訂重大營運風險的承受風險水平有助集 團了解滙豐願意承擔的風險水平。集團風 險管理委員會每年批准集團承受營運風險 水平聲明。集團水平計量指標承受風險水 平聲明 （包括承受營運風險水平計量指標） 由滙豐控股董事會批准。我們會定期根據 承受風險水平監察營運風險，並實施風險 承擔程序，務求預先洞悉風險。有關過程 有助管理層決定是否需要採取其他行動。 個別業務單位及部門會進行營運風險及監 控評估 （ 「RCA」 ） 。風險及監控評估程序旨在 讓不同業務及部門能高瞻遠矚地洞悉營運 風險以及評估監控措施的成效，並設有跟 進行動計劃的機制，以便積極管控營運風 險於可接受水平之內。我們至少會每年檢 討並更新風險及監控評估。 營運風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 187 營運風險管理架構 營運風險管理標準 ‧ 監管合規 ‧ 防範金融犯罪 ‧ 受信 ‧ 法律 ‧ 資訊 ‧ 會計 ‧ 稅務 ‧ 外界詐騙 ‧ 內部詐騙 ‧ 人事 ‧ 政治 ‧ 實物 ‧ 業務延續性 ‧ 系統 ‧ 營運 ‧ 項目 風險及監控評估 主要指標 事件╱內部損失 外圍事件 資金模型 呈報及管理措施 風險意識及文化 管治 識別 制訂承受 風險水平 評估 監控 報告 境況分析 ‧ 我們利用風險及監控評估，以協助評估首要風險監控措施的成效。 ‧ 我們利用主要指標協助監察風險及監控措施。 ‧ 境況為管理層提供首要及新浮現營運風險的量化概覽。 ‧ 內部事件用於預測典型損失情況。 ‧ 外部資料來源用於報告極端境況的評估。 我們已考慮適當的緩減及監控措施，包括： ‧ 作出特定改變以加強內部監控環境； ‧ 調查是否有具成本效益的保險保障可 減低風險；及 ‧ 其他使我們免受損失的措施。 此外，我們旗下的重要法律實體已實施經 改良的境況分析程序，藉以改善重大風險 的量化及管理方法。 2014年之營運風險 於2014年，我們的營運風險仍然是以合規 及法律風險為主，詳情載於第118頁 「首要 及新浮現風險」 。我們就過往年度發生的事 件遭受損失。該等事件包括過往可能在英 國不當銷售還款保障保險 （ 「PPI」 ） 產品 （請 參閱財務報表附註29） 。我們繼續採取多項 減低風險的措施，以預防日後出現不當銷 售事件。 金融服務公司面對監管程序及其他法律程 序的事件與日俱增。有關資本及流動資金 的規定、薪酬及╱或稅務的改革建議，或 會增加集團經營業務的成本，並降低未來 的盈利能力。我們仍繼續面對多項監管程 序，包括多個國家監管機構、保障公平競爭 與執法機關調查及檢討在釐定倫敦銀行同 業拆息、其他銀行同業拆息及其他基準利 率的過程中，銀行小組成員過 往提交的若 干資料及提交資料的程序。尚在進行的其 他調查涵蓋匯率、貴金屬以及信貸違責掉 期。我們已就此進行多項部署，包括重組合 規部的分支部門、提升管治及監督水平、 執行措施實施環球標準 （如第26頁所述） 及 制訂其他措施，確保我們擁有適當人員、 流程和程序管理新浮現風險以及新產品及 業務。 詳情請參閱第189頁 「合規風險」 ，而有關調查和法 律程序的詳情，請參閱財務報表附註40。 於2014年11月，英國金融業操守監管局以 及美國商品期貨交易委員會各自宣布，已 就其各自對涉及外匯基準利率交易及其他 行為的調查，與多家銀行 （包括英國滙豐銀 行有限公司） 達成監管方面的和解。根據和 解條款，英國滙豐銀行有限公司同意向英 國金融業操守監管局支付罰款，以及向美 國商品期貨交易委員會支付民事賠償金， 並採取多項補救行動。有關進一步資料， 請參閱財務報表附註40。 滙豐已就英國 《消費者信貸法》 的定額無抵 押貸款協議規定的遵守情況進行審查。我 們已就未有於年度報表中提醒客戶其有權 提早償還部分貸款一事 （即使客戶貸款文 件列載此項權利） 於 「應計項目、遞延收益 及其他負債」 內確認負債，以向有關客戶退 還利息。現時尚未確定我們是否經已符合 滙豐控股有限公司 188 董事會報告：風險 （續） 《消費者信貸法》 的其他技術規定，但我們 已就此評估一項額外的或有負債。有關進 一步詳情，請參閱財務報表附註40。 我們已與美國聯邦房屋金融局就美國聯邦 國民抵押協會 （ 「房利美」 ） 及美國聯邦住宅 貸款抵押公司 （ 「房貸美」 ） 於2005年至2007年 期間購入按揭抵押證券而提出的索償達成 和解。有關進一步資料，請參閱財務報表 附註40。 其他營運風險包括： ‧ 詐騙風險：由或對我們的客戶 （特別是 在零售銀行及工商金融業務） 實施詐騙 的威脅，於經濟環境轉差時可能增加。 我們已加強監察，進行源頭分析及檢討 內部監控措施，以加強抵禦外來攻擊， 以及降低有關範疇的損失。此外，集團 保安及詐騙風險管理部與環球業務緊 密合作，繼續因應環境變化來評估威 脅，並改進我們的監控措施，以降低有 關風險； ‧ 導致營運複雜的變化程度：為確保維持 穩健的內部監控措施，業務管理層與環 球風險管理部推行多項業務改革計劃， 包括參與所有相關的管理委員會。環球 交易團隊已制訂適用於管理出售風險 的加強風險管理架構； ‧ 資訊保安：資訊及科技基礎設施的保安 對維持銀行服務和保障客戶及滙豐的 品牌，有舉足輕重的影響。倘提供有關 保障的監管架構失誤，可能會損及整個 金融業並造成直接財務損失及╱或遺失 客戶資料及其他敏感資料，繼而損害我 們的聲譽以及維繫客戶信心的能力。 我們繼續進行多項工作加強內部保安 監控，防範可未經許可進入我們系統的 行為，以免可能對即時服務造成影響或 導致數據流失或詐騙。一如其他銀行及 跨國企業，我們繼續是 「分布式拒絕服 務」 等日益精密的網絡攻擊的目標，有 關攻擊可對面向客戶提供服務的網站造 成影響。另外，依賴標準互聯網技術、 協議及服務，表示我們可能於有關技術 出現漏洞時，須實施廣泛的糾正措施。 我們從業界過往經歷的襲擊的經驗、其 他金融機構、政府機關以及外界的情報 供應商分享的資料，可更能了解可能面 對的風險，並制訂境況進行測試。上述 各方將繼續專注於進行多項部署，加強 監控環境。我們已作出大量投資，加強 有關資料存取的監控、加強監察可能出 現的網絡襲擊以及繼續培訓員工提升 警覺性，而這些工作是需要我們在營運 程序及應變計劃方面不斷投資的； ‧ 供應商風險管理：我們繼續專注於管理 供應商風險，並在與最主要的供應商執 行供應商表現管理計劃方面取得良好 進展。我們的另一重點是篩選供應商， 使滙豐能夠識別受制裁的供應商，並與 其終止關係。供應商風險管理是第三方 風險管理的核心一環；及 ‧ 遵守監管協議及法令：我們未能履行延 後起訴協議的責任，會對我們的業績及 營運有重大不利影響。法律程序於財務 報表附註40討論，有關合規風險的其他 詳情載於下文。 滙豐亦會透過營運風險管理架構來監察及 管理其他營運風險。 有關該等風險性質的詳情，載於第118頁 「首要及新 浮現風險」 。 錄得營運風險損失的頻率及金額 營運風險事件概況及相關損失於下表概 述，並以發生頻率以及按美元計算的損失 總額列示營運風險事件的分布。 2014年營運虧損上升，乃因有關法律及監 管事項的英國客戶賠償計劃準備及償付所 導致。 如2013年一樣，2014年的營運風險事件包括 高頻率出現但影響較小的事件，亦有影響 較大但發生頻率較低的事件。舉例來說， 外界詐騙事件 （例如信用卡詐騙事件） 較其 他類別事件更頻密發生，但涉及的損失往 往較少。相反，合規類別的營運風險事件 出現頻率相對較低，但卻總代價不菲。 2014年詐騙案件數目大致持平，原因是我 們持續實施嚴謹的監控。 過往重大事件在2014年繼續帶來損失，有 關事件包括在英國可能涉及不當銷售還款 保障保險產品及有關監管事宜，詳情載於 財務報表附註40。 營運風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 189 按風險類別分類的營運風險事件頻率 （個 別虧損>10,000美元） 會計及稅務 防範金融犯罪 監管合規 詐騙 法律 營運及系統 人事 其他 2014年 2013年 1% 0% 6% 39% 4% 23% 24% 2% 1% 0% 5% 39% 7% 22% 25% 1% 按風險類別分類並按美元計算的營運風險 損失分布 會計及稅務 防範金融犯罪 監管合規 詐騙 法律 營運及系統 人事 其他 2% 0% 54% 6% 30% 7% 1% 0% 3% 0% 43% 16% 14% 22% 2% 0% 2014年 2013年 合規風險 （未經審核） 合規風險指集團未能遵守所有相關法律、 守則、規則、法規及良好市場慣例的條文 和精神，並招致罰款及罰則，從而對業務 造成損害的風險。 於2014年，我們完成將環球風險管理部的 合規分支部門重組為兩個新的分支部門： 防範金融犯罪部以及監管合規部，兩者同 時得到合規營運總監、鑑證及聲譽風險管 理團隊的適當支援。我們繼續確保合規分 支部門透過有關部門的運作以及執行集團 策略 （包括實施環球標準的措施） ，作好準 備以面對監管機構及執法機構不斷提升的 監管和審查。另外，我們亦已制訂措施， 確保擁有合適的人才、流程及程序，管理 新浮現風險和新產品及業務。 董事會已於2014年1月批准加入承受風險 水平的強化環球反洗錢及制裁政策。有關 政策採納及尋求施行全球最高或最有效標 準，包括全球貫徹一致的認識客戶方法。 有關政策透過制訂和應用須在全球日常業 務營運加入有關政策的程序而分階段實施。 政策的最重要的目標為使各僱員僅 「以正 確方式進行正確業務」 。 滙豐已履行紐約郡地區檢察官延後起訴協 議所要求的所有規定，故紐約郡地區檢察 官對滙豐的指控已於2014年12月該協議的 兩年期限屆滿後撤銷。若我們於協議期內任 何時間違反美國延後起訴協議，美國司法 部可就美國延後起訴協議的事項起訴滙豐 控股或美國滙豐銀行。詳情請參閱第120頁 「監管承諾及同意令」 。 於2014年5月，董事會通過全球一致的監管 規定行為管理方法，旨在確保我們可為客 戶提供公正的服務，在金融市場上保持行 為有序以及營運透明。有關環球行為操守 將透過環球業務及部門落實，並涵蓋我們 所有業務及經營活動。該等活動的例子於 第121頁 「業務經營方式」 披露。 在可見的未來，我們所面對的固有合規風 險顯然仍會維持於高水平。但在確保我們 準備就緒，足以有效管理此等風險方面， 我們認為已取得良好進展，並會不斷向前。 董事會報告：風險 （續） 滙豐控股有限公司 190 保險業務風險管理 （經審核） 頁次 附錄 1 列表 頁次 滙豐的銀行保險業務模式 190 保險產品概覽 231 風險的性質及程度 232 2014年保險業務風險管理 191 資產與負債配對 191 制訂保險產品附屬公司的資產負債表： －按合約類別列示 191 －按地區列示 193 保險業務股東權益總額變動 193 金融風險 194 232 制訂保險產品附屬公司持有的 金融資產 194 市場風險 194 232 財務回報保證 195 滙豐旗下制訂保險產品附屬公司對 市場風險因素的敏感度 195 信貸風險 196 234 滙豐旗下制訂保險產品附屬公司 持有之國庫票據、其他合資格票據 及債務證券 196 再保人應佔之保單未決賠款 197 流動資金風險 197 234 保單未決賠款的預計期限 197 投資合約負債之尚餘合約期限 197 保險風險 198 235 保險風險分析－保單未決賠款 198 非經濟假設的敏感度 198 敏感度分析 198 1 風險附錄－政策及慣例。 保險業務的大部分風險來自制訂保險產品 活動，並可分為保險風險及金融風險。保 險風險指由保單持有人轉移給發行人 （滙 豐） 的損失風險 （金融風險除外） 。金融風險 包括市場風險、信貸風險及流動資金風險。 2014年，滙豐管理保險業務產生的風險的 政策及慣例沒有重大變動。 有關保險業務的風險管理政策及慣例，以及我們 制訂的主要合約，於第231頁風險附錄內概述。 滙豐的銀行保險業務模式 （未經審核） 我們實行的綜合銀行保險業務模式，是主 要向已與我們建立銀行業務關係的客戶提 供保險產品。我們透過所有環球業務銷售 保險產品，但主要由零售銀行及財富管理 業務及工商金融業務通過全球分行及直接 服務途徑進行銷售。 我們出售的保單，以銀行客戶相關需要為 本，我們從聯絡銷售的途徑及對客戶的認 識，識別有關需要。大部分售出的產品為 儲蓄及投資產品，以及有期及信用壽險產 品。我們透過專注於個人及中小企業務的 需要，獲得最合適數量的保單，並能分散 個別保險風險。 倘我們具備業務規模及承受風險水平，我 們的附屬公司會制訂相關的保險產品，主 要為壽險產品。制訂保險產品讓我們能保 留與簽發保單相關的風險及回報，將部分 承保利潤、投資收益及分銷佣金留在集團 內部。 倘若我們沒有適當的承受風險水平或足以 支持有效制訂保險產品的營運規模，便會 委聘少數具領導地位的外界保險公司制訂 保險產品，然後透過我們的銀行網絡及直 接服務途徑向客戶提供。有關安排一般與 我們的獨家策略合作夥伴共同訂立，而集 團則賺取佣金、費用及利潤分成。 滙豐在各業務所在地分銷保險產品。我們在 七個國家╱地區 （阿根廷、巴西、墨西哥、法 國、英國、香港及新加坡） 擁有制訂保險產 品的核心人壽保險公司，而大部分均為法 定銀行實體的直接附屬公司。我們在中國 內地、馬來西亞及馬耳他亦有制訂人壽保 險產品的附屬公司。 保險業務風險管理 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 191 2014年保險業務風險管理 我們使用經濟資本法計量制訂保險產品業 務的風險狀況，當中資產及負債以市值基準 計算，並且根據業務所承擔的風險持有所 需資本，確保未來一年只有少於二百分之 一的機會出現無力償債的情況。於2014年， 我們調整經濟資本模型中市場、信貸及保 險風險的計量方法，以配合由2016年起適 用的新泛歐保險業資本規則償付能力。 制訂壽險產品業務的風險狀況於2014年內 並無重大變動，壽險保單的投保人負債維 持於740億美元 （2013年：740億美元） 水平。 然而，英國出現一項顯著變動，就是HSBC Life (UK) Ltd已訂立協議出售其退休金業務。 當我們接獲監管機關批准及組合已轉讓予 買方後，始會全面確認相關影響。 資產與負債配對 （經審核） 我們管理金融及保險風險 （尤其是壽險保單 風險） 時，運用的工具主要是資產與負債的 配對。在我們營運的不少市場，我們無法 或不宜實行完美的資產與負債配對策略。 特別是遠期非相連合約，資產與負債的期 限並不配對。因此，我們透過組合結構安 排支持非相連合約的預期債務。 下表按合約類別及地區分類列示於2014年 底的資產及負債組合成分，顯示在各種情 況下資產均足以應付投保人負債。 按合約類別列示制訂保險產品附屬公司的資產負債表 （經審核） 保單 投資合約 附有DPF 單位相連 年金 其他40 附有DPF 41 單位相連 其他 其他資產 42 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 金融資產 29,040 11,278 1,517 6,253 24,238 2,561 4,322 5,732 84,941 －交易用途資產 － － 3 － － － － － 3 －指定以公允值列賬之金融資產 4,304 11,111 533 782 6,346 2,223 1,684 1,713 28,696 －衍生工具 12 1 － 1 101 1 10 73 199 －金融投資 21,152 － 886 5,167 15,677 － 1,807 3,812 48,501 －其他金融資產 3,572 166 95 303 2,114 337 821 134 7,542 再保險資產 190 262 － 617 － － － 2 1,071 PVIF43 － － － － － － － 5,307 5,307 其他資產及投資物業 698 328 23 107 831 7 26 7,383 9,403 資產總值 29,928 11,868 1,540 6,977 25,069 2,568 4,348 18,424 100,722 投資合約負債 － － － － － 2,542 4,155 － 6,697 －指定以公允值列賬 － － － － － 2,542 3,770 － 6,312 －按已攤銷成本列賬 － － － － － － 385 － 385 保單未決賠款 29,479 11,820 1,473 6,021 25,068 － － － 73,861 遞延稅項44 12 － 11 18 － － － 1,180 1,221 其他負債 － － － － － － － 8,577 8,577 U L 負債總額 29,491 11,820 1,484 6,039 25,068 2,542 4,155 9,757 90,356 各類股東權益總額 － － － － － － － 10,366 10,366 A 於2014年12月31日 各類負債及股東權益總額45 29,491 11,820 1,484 6,039 25,068 2,542 4,155 20,123 100,722 董事會報告：風險 （續） 滙豐控股有限公司 192 保單 投資合約 附有DPF 單位相連 年金 其他40 附有DPF 41 單位相連 其他 其他資產42 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 金融資產 26,382 13,348 1,651 4,728 25,676 9,720 4,375 5,846 91,726 －交易用途資產 － － 3 － － － － － 3 －指定以公允值列賬之金融資產 3,850 13,131 532 761 6,867 9,293 1,706 1,757 37,897 －衍生工具 1 3 － － 215 5 － 55 279 －金融投資 19,491 － 959 3,780 16,556 － 1,853 3,745 46,384 －其他金融資產 3,040 214 157 187 2,038 422 816 289 7,163 再保險資產 182 291 522 439 － － － 2 1,436 PVIF43 － － － － － － － 5,335 5,335 其他資產及投資物業 757 284 23 113 791 19 31 546 2,564 資產總值 27,321 13,923 2,196 5,280 26,467 9,739 4,406 11,729 101,061 投資合約負債 － － － － － 9,730 4,209 － 13,939 －指定以公允值列賬 － － － － － 9,730 3,761 － 13,491 －按已攤銷成本列賬 － － － － － － 448 － 448 保單未決賠款 26,920 13,804 2,158 4,872 26,427 － － － 74,181 遞延稅項44 12 － 17 1 － － － 1,163 1,193 其他負債 － － － － － － － 2,048 2,048 負債總額 26,932 13,804 2,175 4,873 26,427 9,730 4,209 3,211 91,361 各類股東權益總額 － － － － － － － 9,700 9,700 B 於2013年12月31日 各類負債及股東權益總額45 26,932 13,804 2,175 4,873 26,427 9,730 4,209 12,911 101,061 有關註釋，請參閱第202頁。 我們最主要的壽險產品為於法國簽發附有 酌情參與條款的投資合約、在香港簽發附 有酌情參與條款的保單，以及在拉丁美洲、 香港及英國簽發的單位相連合約。 我們對來自以上資產負債表的金融風險承 擔，視乎所簽發的合約類別而有所不同。 單位相連合約的投保人須承擔大部分金融 風險，而非相連合約的大部分金融風險則 由股東 （滙豐） 承擔。至於附有酌情參與條 款的合約，倘若有關金融風險不能以所簽 發保單內的任何酌情參與 （或紅利） 條款管 理，則股東要承受該等金融風險。 如上文所述，滙豐於年內訂立協議出售其 英國退休金業務，而相關款額根據IFRS 5呈 報為持作出售用途業務組合 （並因此計入上 表 「其他資產」 一欄內） 。出售用途業務組合 包括負債總額68億美元，即單位相連投資 合約、單位相連保單及年金合約下的負債； 另外亦包括資產總值68億美元，即支持該 等負債的金融及再保險資產，以及與該等 合約相關的有效長期保險業務現值。是次 轉讓須經監管當局批准，且預期於2015年 下半年完成。作為交易一部分，我們亦已 訂立再保險協議，自2014年1月1日起轉讓 該項業務的若干風險及回報予買方，直至 該項交易完成為止。滙豐就訂立此項再保 險協議確認增益4,200萬美元。 保險業務風險管理 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 193 按地區列示制訂保險產品附屬公司的資產負債表 46 （經審核） 歐洲 亞洲 6 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 金融資產 30,178 47,443 7,320 84,941 －交易用途資產 － － 3 3 －指定以公允值列賬之金融資產 10,610 12,497 5,589 28,696 －衍生工具 172 27 － 199 －金融投資 16,947 30,010 1,544 48,501 －其他金融資產 2,449 4,909 184 7,542 再保險資產 308 748 15 1,071 PVIF43 711 4,175 421 5,307 其他資產及投資物業 7,650 1,145 608 9,403 資產總值 38,847 53,511 8,364 100,722 投資合約負債： －指定以公允值列賬 1,585 4,727 － 6,312 －按已攤銷成本列賬 － － 385 385 保單未決賠款 27,312 39,990 6,559 73,861 遞延稅項44 273 806 142 1,221 其他負債 7,932 460 185 8,577 負債總額 37,102 45,983 7,271 90,356 各類股東權益總額 1,745 7,528 1,093 10,366 於2014年12月31日各類負債及股東權益總額45 38,847 53,511 8,364 100,722 金融資產 41,557 42,352 7,817 91,726 －交易用途資產 － － 3 3 －指定以公允值列賬之金融資產 20,742 11,420 5,735 37,897 －衍生工具 272 7 － 279 －金融投資 18,080 26,505 1,799 46,384 －其他金融資產 2,463 4,420 280 7,163 再保險資產 823 596 17 1,436 PVIF43 1,156 3,730 449 5,335 其他資產及投資物業 868 1,101 595 2,564 資產總值 44,404 47,779 8,878 101,061 投資合約負債： －指定以公允值列賬 8,760 4,731 － 13,491 －按已攤銷成本列賬 － － 448 448 保單未決賠款 31,786 35,619 6,776 74,181 遞延稅項44 407 645 141 1,193 其他負債 1,474 371 203 2,048 負債總額 42,427 41,366 7,568 91,361 各類股東權益總額 1,977 6,413 1,310 9,700 於2013年12月31日各類負債及股東權益總額45 44,404 47,779 8,878 101,061 有關註釋，請參閱第202頁。 保險業務各類股東權益總額變動 （經審核） 各類股東權益總額 2014年 2013年 百萬美元 百萬美元 於1月1日 9,700 9,989 長期保險業務PVIF變動43 261 525 資產淨值回報 1,835 848 資本交易 (673) (590) 出售附屬公司╱組合 1 (675) 匯兌差額及其他 (758) (397) 於12月31日 10,366 9,700 有關註釋，請參閱第202頁。 董事會報告：風險 （續） 滙豐控股有限公司 194 金融風險 （經審核） 金融風險的性質以及如何管理該等風險的詳情， 載於第232頁風險附錄。 金融風險可分類為： ‧ 市場風險－金融資產公允值或其日後現 金流因變數 （例如利率、信貸息差、匯 率及股價） 的變動而出現變動的風險； ‧ 信貸風險－由於第三方未能履行其責任 而產生財務損失的風險；及 ‧ 流動資金風險－由於並無充足資產可變 現為現金，故未能向投保人支付到期款 項的風險。 下表按合約類別分析我們旗下各制訂保險 產品附屬公司於2014年12月31日所持的資 產，並可作為金融風險的概覽。單位相連 合約向投保人支付的利益，乃參考用以支 持保單的投資價值而釐定，我們一般會就 指定資產按公允值列賬；非相連合約的資 產，則按相關合約的性質分類。 制訂保險產品附屬公司持有的金融資產 （經審核） 單位相連 合約 47 非相連 合約 48 其他資產 49 總計 百萬美元 百萬美元 百萬美元 百萬美元 交易用途資產 債務證券 － 3 － 3 指定以公允值列賬之金融資產 13,334 13,649 1,713 28,696 國庫票據 － 40 16 56 債務證券 4,589 3,507 618 8,714 股權證券 8,745 10,102 1,079 19,926 金融投資 持至到期日：債務證券 － 21,789 2,494 24,283 可供出售： － 22,899 1,319 24,218 －債務證券 － 22,899 1,290 24,189 －股權證券 － － 29 29 衍生工具 2 124 73 199 其他金融資產49 503 6,905 134 7,542 於2014年12月31日金融資產總值45 13,839 65,369 5,733 84,941 交易用途資產 債務證券 － 3 － 3 指定以公允值列賬之金融資產 22,424 13,716 1,757 37,897 國庫票據 － － 50 50 債務證券 7,809 3,910 546 12,265 股權證券 14,615 9,806 1,161 25,582 金融投資 持至到期日：債務證券 － 21,784 2,142 23,926 可供出售： － 20,855 1,603 22,458 －債務證券 － 20,855 1,594 22,449 －股權證券 － － 9 9 衍生工具 8 216 55 279 其他金融資產49 636 6,238 289 7,163 於2013年12月31日金融資產總值45 23,068 62,812 5,846 91,726 有關註釋，請參閱第202頁。 於2014年12月31日，約有67% （2013年：64%） 的 金融資產投資於債務證券，而24% （2013年： 28%） 則投資於股權證券。 單位相連合約方面，保費收益扣除徵費後， 會投資於一個資產組合。我們透過持有與負 債相連的獨立基金或組合之適當資產，代 投保人管理此類產品的金融風險。於2014年 底，此等資產佔我們旗下制訂保險產品附屬 公司金融資產總值的16% （2013年：25%） 。 支持單位相連合約的資產價值減少93億美 元，主要由於年內將英國退休金業務相關的 63億美元資產分類為持作出售用途﹙請參 閱第192頁﹚以及將支持其他單位相連投資 合約的29億美元資產轉讓予第三方。 餘下金融風險由我們單獨代表股東管理， 倘存在酌情參與條款，則由我們代表股東 及投保人共同管理。 市場風險 （經審核） 倘若產品的負債與支持這些負債的投資資 產出現錯配，即產生市場風險。例如，若 資產與負債的收益率及期限錯配，即會產 生利率風險。 保險業務風險管理 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 195 我們現時的資產組合包括在收益率高於當 前市場水平時發行的債務證券。因此，現 時持有的債務證券的收益率高於當前發行 債務證券的收益率。 長期保險或投資產品可能會包含保證的利 益。固定保證利益 （例如支付年金） 將保留 作為計算保單未決賠款的一部分。 倘為提供保證財務回報的產品，且現行收益 率長期降至低於保證水平，則需要以股東 資本應付投保人負債的風險將會增加。就 保證成本而持有的儲備按模型推測計算。 倘當地規例有所要求，有關儲備會透過投 保人負債持有，而餘額則以有關產品的PVIF 財務回報保證 45, 50 （經審核） 2014年 2013年 保證 之隱含 投資回報 現行 收益率 保證之 成本 保證 之隱含 投資回報 現行 收益率 保證之 成本 % % 百萬美元 % % 百萬美元 資本 0.0 0.0 － 3.5 81 0.0 0.0 － 4.4 57 名義年度回報 0.1 － 2.0 3.6 － 3.6 6 0.1 － 2.0 4.1 － 4.1 9 名義年度回報51 2.1 － 4.0 3.5 － 4.1 646 2.1 － 4.0 4.2 － 4.4 471 名義年度回報 4.1 － 5.0 3.5 － 4.1 30 4.1 － 5.0 4.1 － 4.4 25 實際年度回報52 0.0 － 6.0 4.7 － 7.5 14 0.0 － 6.0 6.4 － 6.4 13 於12月31日 777 575 有關註釋，請參閱第202頁。 減項入賬。下表列出就保證成本而持有的 儲備總額、支持此等產品的資產之投資回 報幅度，以及支持業務履行保證的隱含投 資回報。 部分組合提供的財務保證超過支持有關保 證之資產的現行收益率。保證成本主要因 2014年法國的收益率下跌而增加至7.77億 美元 （2013年：5.75億美元） 。由於收益率下 跌，回報介乎2.1%至4.0%及4.1%至5.0%的 已封閉組合，其保證成本因假設再投資收 益降低而上升。另外，香港亦有已封閉組 合，其保證回報率為5.0%，而現行收益率 為4.1%。我們削減支付予若干附有酌情參 與條款合約的投保人的短期紅利率，以控 制對業務構成的即時負擔。 除上文所述者外，我們就巴西的若干單位 相連退休金產品所附帶的保證年金選擇權 採用以模型計算的成本，從PVIF扣減5,300萬 美元 （2013年：1.34億美元） 。 下表列出選定的利率、股價及匯率境況， 對本年度利潤及制訂保險產品附屬公司各 類股東權益總額的影響。 在適當情況下，我們會於敏感度測試的結 果中，加入壓力對PVIF的影響。利潤及各類 股東權益總額與各項風險因素之間並無直 線關係，因此披露的測試結果不應用以推 算不同壓力水平的敏感度。我們呈列的敏 感度，未有計及為減輕市場利率變動影響 而可能採取的管理措施。上表所列敏感度 已計及投保人行為因市場利率變動而可能 產生的不利變動。 2014年孳息曲線平行上下移100個基點的影 響較2013年上升，主要因為2014年法國的收 益率下降及孳息曲線轉趨平坦。在低孳息 環境下，上述選擇權及保證的預計成本對 孳息曲線變動特別敏感。可供出售債券的 市值亦對孳息曲線變動敏感，因此對股東 權益有較大的反向壓力。 滙豐旗下制訂保險產品附屬公司對市場風險因素的敏感度 （經審核） 2014年 2013年 對各類股東 對各類股東 對除稅後 權益總額 對除稅後 權益總額 利潤的影響 的影響 利潤的影響 的影響 百萬美元 百萬美元 百萬美元 百萬美元 孳息曲線平行上移100個基點 290 (345) 151 (199) 孳息曲線平行下移100個基點53 (549) 214 (230) 139 股價上升10% 180 180 149 149 股價下跌10% (153) (153) (129) (129) 美元兌所有貨幣的匯率上升10% 54 54 21 21 美元兌所有貨幣的匯率下跌10% (54) (54) (21) (21) 董事會報告：風險 （續） 滙豐控股有限公司 196 滙豐旗下制訂保險產品附屬公司持有之國庫票據、其他合資格票據及債務證券 （經審核） 並非逾期或已減值 穩健 良好 滿意 低於標準 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 用作支持非相連保單未決賠款及投資合約負債之工具 交易用途資產－債務證券 3 － － － 3 指定以公允值列賬之金融資產 2,550 530 214 255 3,549 －國庫及其他合資格票據 5 － － 35 40 －債務證券 2,545 530 214 220 3,509 金融投資－債務證券 38,515 4,312 1,662 200 44,689 41,068 4,842 1,876 455 48,241 ' 用作支持股東權益之工具54 指定以公允值列賬之金融資產 214 322 30 69 635 －國庫及其他合資格票據 － － － 16 16 －債務證券 214 322 30 53 619 金融投資－債務證券 3,378 196 154 54 3,782 3,592 518 184 123 4,417 總計45 交易用途資產－債務證券 3 － － － 3 指定以公允值列賬之金融資產 2,764 852 244 324 4,184 －國庫及其他合資格票據 5 － － 51 56 －債務證券 2,759 852 244 273 4,128 金融投資－債務證券 41,893 4,508 1,816 254 48,471 於2014年12月31日 44,660 5,360 2,060 578 52,658 用作支持非相連保單未決賠款及投資合約負債之工具 交易用途資產－債務證券 3 － － － 3 指定以公允值列賬之金融資產 2,780 691 224 215 3,910 －債務證券 2,780 691 224 215 3,910 金融投資－債務證券 36,113 4,596 1,699 231 42,639 38,896 5,287 1,923 446 46,552 D 用作支持股東權益之工具54 指定以公允值列賬之金融資產 191 298 73 34 596 －國庫及其他合資格票據 50 － － － 50 －債務證券 141 298 73 34 546 金融投資－債務證券 3,356 176 139 65 3,736 3,547 474 212 99 4,332 總計45 交易用途資產－債務證券 3 － － － 3 指定以公允值列賬之金融資產 2,971 989 297 249 4,506 －國庫及其他合資格票據 50 － － － 50 －債務證券 2,921 989 297 249 4,456 金融投資－債務證券 39,469 4,772 1,838 296 46,375 於2013年12月31日 42,443 5,761 2,135 545 50,884 有關註釋，請參閱第202頁。 信貸風險 （經審核） 信貸風險可引致來自拖欠的虧損，並可令 收益表及資產負債表的數據隨信貸息差變 動而出現波動，其影響主要涉及支持非相 連合約及股東權益之530億美元債券組合 （2013年：510億美元） 。 我們旗下保險附屬公司的除稅後利潤對信貸 息差上升之資產值影響的敏感度為減少700 萬美元 （2013年：2,100萬美元） 。各類股東權 益總額的敏感度為減少900萬美元 （2013年： 4,600萬美元） 。敏感度相對輕微，因為保 險附屬公司所持有的大部分債務證券均分 類為持至到期日或可供出售，因此，該等 金融投資的公允值若有任何變動 （倘無減 值） ，將不會對除稅後利潤 （或倘為持至到 期日證券，則對各類股東權益總額） 造成影 響。我們根據兩年期內信貸息差的單日變 動計算敏感度，並應用99%的可信程度， 與集團的估計虧損風險所用者一致。 信貸質素 （經審核） 下表乃按信貸質素分析保險業務持有之國 庫票據、其他合資格票據及債務證券。 由於支持單位相連負債的資產所涉金融風 險主要由投保人承擔，因此下表只列出用作 支持非相連保單未決賠款及投資合約負債 及股東權益的資產。下表所列資產的84.8% （2013年：83.4%） 屬 「穩健」 級別的投資。 有關五個信貸質素類別的定義，請參閱第207頁。 保險業務風險管理 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 197 我們將需要承擔的保險風險轉讓予再保人 時，亦會產生信貸風險。根據信貸質素分 析，轉讓予再保人分擔的未決賠款及再保 再保人應佔之保單未決賠款 45 （經審核） 並非逾期或已減值 已逾期 但並非 已減值 總計 穩健 良好 滿意 低於標準 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 單位相連保險 75 185 － － － 260 非相連保險55 751 11 10 － － 772 於2014年12月31日 826 196 10 － － 1,032 再保險債務人 11 6 － － 21 38 單位相連保險 72 218 － － － 290 非相連保險55 1,103 8 7 － － 1,118 於2013年12月31日 1,175 226 7 － － 1,408 再保險債務人 17 1 － － 10 28 有關註釋，請參閱第202頁。 險未決追償額載列如下。第235頁風險附錄 所述我們根據再保險協議承受的第三方風 險亦包括在下表中。 流動資金風險 （經審核） 下表顯示於2014年12月31日，保單未決賠款 的預期未折現現金流及投資合約負債的尚 餘合約期限。就大部分業務而言，流動資 金風險是與投保人共同承擔，而倘為單位 相連業務，則全部由投保人承擔。 於2014年12月31日，保單預計期限的分布情 況與2013年相若。 保單未決賠款的預計期限 45 （經審核） 預期現金流 （未折現） 1年內 1至5年 5至15年 15年以上 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 單位相連保險 709 3,280 9,243 14,544 27,776 非相連保險55 3,504 12,718 29,905 33,108 79,235 於2014年12月31日 4,213 15,998 39,148 47,652 107,011 單位相連保險 1,106 3,609 9,757 13,725 28,197 非相連保險55 3,977 11,731 26,848 31,306 73,862 於2013年12月31日 5,083 15,340 36,605 45,031 102,059 有關註釋，請參閱第202頁。 投資合約負債之尚餘合約期限 （經審核） 制訂保險產品附屬公司所簽發投資合約之負債46 單位相連 投資合約 附有DPF 的投資合約 其他 投資合約 總計 百萬美元 百萬美元 百萬美元 百萬美元 尚餘合約期限： －1年內到期 151 － 389 540 －1至5年內到期 133 － － 133 －5至10年內到期 194 － － 194 －10年後到期 766 － － 766 －無定期56 1,298 25,068 3,765 30,131 於2014年12月31日 2,542 25,068 4,154 31,764 尚餘合約期限： －1年內到期 232 － 454 686 －1至5年內到期 778 － － 778 －5至10年內到期 852 － － 852 －10年後到期 2,254 － － 2,254 －無定期56 5,614 26,427 3,755 35,796 於2013年12月31日 9,730 26,427 4,209 40,366 有關註釋，請參閱第202頁。 董事會報告：風險 （續） 滙豐控股有限公司 198 保險風險 保險風險主要按有效保單未決賠款計量。 我們面對的主要風險是，獲取保單成本及 行政開支加上賠償及利益支出的總額，於 某段時間後會超過所收保費加投資收益的 總額。賠償及利益支出受多項因素影響，包 括過往的死亡率及發病率、保單失效率及 退保率，以及 （倘保單帶有儲蓄成分） 為支 持負債而持有的資產之表現。下表按地區 及業務類別分析我們的壽險風險承擔。保 險風險狀況和相關風險與2013年12月31日 觀察所得大致相若。 保險風險分析－保單未決賠款 46 （經審核） 歐洲 亞洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 非相連保險55 829 34,261 1,883 36,973 附有DPF的保單57 367 29,112 － 29,479 信用人壽 56 87 － 143 年金 71 127 1,275 1,473 其他 335 4,935 608 5,878 單位相連保險 1,415 5,729 4,676 11,820 附有DPF的投資合約41, 57 25,068 － － 25,068 於2014年12月31日保單未決賠款 27,312 39,990 6,559 73,861 非相連保險55 1,383 30,554 2,013 33,950 附有DPF的保單57 380 26,540 － 26,920 信用人壽 130 74 － 204 年金 622 129 1,407 2,158 其他 251 3,811 606 4,668 單位相連保險 3,976 5,065 4,763 13,804 附有DPF的投資合約41, 57 26,427 － － 26,427 於2013年12月31日保單未決賠款 31,786 35,619 6,776 74,181 有關註釋，請參閱第202頁。 我們最主要的壽險產品為在香港簽發附有 酌情參與條款的保單、於法國簽發附有酌 情參與條款的投資合約，以及在拉丁美洲、 香港及英國簽發的單位相連合約。 非經濟假設的敏感度 （經審核） 制訂壽險產品的公司之投保人負債及PVIF， 是經參考包括死亡率及╱或發病率、保單 失效率及支出率等非經濟假設而釐定。下 表載列我們所有制訂保險產品附屬公司的 利潤及各類股東權益總額，對前述非經濟 假設於當日各種合理可能變化的敏感度。 死亡率及發病率風險一般與壽險保單有 關。死亡率或發病率上升對利潤的影響， 視乎承保的業務類別而定。在巴西、法國 及香港，我們面對的死亡率及發病率風險 最大。 保單失效率的敏感度取決於承保的合約類 別。就保單而言，賠償的資金來自收取的保 費及支持未決賠款的投資組合所賺收益。 就定期壽險組合而言，保單失效率上升一 般對利潤造成負面影響，原因是失效保單 日後不再產生保費收益。然而，由於存在 退保費用，故部分保單失效會對利潤有正 面影響。巴西、法國、香港及英國是我們 對保單失效率變動最敏感的地方。 支出率風險是保單管理成本改變所帶來的 風險。若增加的開支未能轉嫁予投保人， 支出率上升會對利潤構成負面影響。 敏感度分析 （經審核） 2014年 2013年 百萬美元 百萬美元 於12月31日對除稅後利潤及 各類股東權益總額的影響 死亡率及╱或發病率上升10% (65) (76) 死亡率及╱或發病率下降10% 72 79 保單失效率上升10%57 (108) (119) 保單失效率下降10%57 122 133 支出率上升10% (106) (101) 支出率下降10% 106 100 有關註釋，請參閱第202頁。 保險業務風險管理╱其他重大風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 199 其他重大風險 頁次 附錄 1 列表 頁次 聲譽風險 199 235 受信風險 200 退休金風險 200 236 主計劃 200 主計劃－資產分配目標 200 支付福利款項 （百萬美元） 201 未來發展 201 界定供款計劃 201 可持續發展風險 201 237 1 風險附錄－風險政策及慣例。 聲譽風險 （未經審核） 聲譽風險是指因滙豐本身、我們的僱員或 與我們有連繫的人士的任何事件、行為、 行動或並未有所行動，未能符合相關群體 的期望，繼而令相關群體對滙豐有負面評 價的風險。 聲譽風險與認知有關，而有關認知未必有 事實理據支持。相關群體的期望會不斷改 變，因此聲譽風險往往經常變動，並會因不 同的地區、群體及個人而異。滙豐作為環 球銀行，會堅定執行集團在業務所在的各 個司法管轄區所設定的崇高標準，並讓市 場知道我們此方面的努力。聲譽風險可能 會造成財務或非財務影響、令相關群體失 去信心、對我們挽留及吸引客戶的能力產 生負面影響等後果。未能符合有關操守、 合規、客戶服務或營運效率的標準，或會 產生聲譽風險。 我們已採取及╱或持續採取多項措施以遵 循美國延後起訴協議或其他規定，加強反 洗錢、制裁及其他監管合規架構。長遠而 言，此等措施應能加強我們的聲譽風險管 理，當中包括以下各項： ‧ 透過逐步執行集團策略以簡化業務，包 括採用環球金融犯罪風險的考慮等，從 而有助規範我們在較高風險國家╱地區 經營業務的方式； ‧ 在我們業務所在的每個地區增加用於 聲譽風險管理工作的資源，並就聲譽風 險管理及客戶關係事宜實施中央個案 管理及跟進程序； ‧ 在各環球業務成立聲譽風險管理及客 戶甄選合併委員會，設立清晰的上報程 序，在合適的層面上解決問題； ‧ 就滙豐價值觀計劃持續開展培訓和溝 通，該計劃界定集團全體成員的行事方 式，並尋求確保將該等價值觀深植於我 們的營運中；及 ‧ 持續發展及實施有關防範金融犯罪的 環球標準，支援我們的業務發展。這包 括確保我們在全球貫徹應用反洗錢和 制裁合規計劃的管治政策。 2014年7月，我們頒布新的聲譽風險管理及 客戶甄選政策，以統一而系統的方式管理 有關風險： ‧ 聲譽風險 （新政策） ：界定聲譽風險並載 列滙豐管理有關風險的方針； ‧ 客戶甄選及吸納業務 （新政策） ：概述於 識別新客戶關係時須考慮的風險因素； ‧ 客戶甄選及終止關係管理：為所有業務 的全部賬戶及業務關係制訂可持續應 用於全球各地的客戶甄選及終止關係 管理方針，並詳述須上報或取得批准的 準則；及 ‧ 第六方面考慮：在高風險司法管轄區營 運的客戶，涉及較高的金融犯罪風險， 可能需要特別審批，或倘此等客戶所涉 風險超逾滙豐的承受風險水平，則集團 可能需要考慮終止相關業務關係。 滙豐絕不容忍在可預見聲譽受損的情況下， 仍無視聲譽損失及不採取措施減少損失， 蓄意進行任何相關業務、活動或聯繫。公 開討論及上報任何可能對集團產生不利影 響的事宜，必須不受阻攔。業務活動的各 個方面均涉及一定程度的風險，但所有業 務決策必須適當考慮其對滙豐良好聲譽的 潛在損害。 要成功查察及防止參與非法行為的人士利 用環球金融系統，必須一直保持警覺，而 我們將繼續與全球各地政府緊密合作，以 實現這個目標。對於我們策略的執行、對 滙豐價值觀，以及對維護與提高我們的聲 譽而言，這是不可或缺的一環。 董事會報告：風險 （續） 滙豐控股有限公司 200 受信風險 （未經審核） 受信風險指集團以受信人身分擔任受託 人、投資經理或在法律或法規授權下行事 時，可能違反受信責任的風險。 受信責任指滙豐為第三方持有、管理、監 督資產或就該等資產負上責任時，須在法 律及╱或監管規定下，以最高審慎標準及 真誠的態度行事的責任。受信人必須以第 三方的最佳利益為依歸作出決定及行事， 並必須以客戶的需要為先，集團的需要為 次。 我們可能因未能按照有關責任行事而承擔 損害賠償或其他罰則。在其他情況下亦會 產生受信責任，例如我們以委託人的代理 身分行事，亦會產生有關責任，除非明確 免除受信責任 （例如根據代理委任合約免 除） ，則作別論。 我們的主要受信業務 （ 「指定業務」 ） 就其多 項受信工作制訂受信風險承受水平聲明， 並已設定監察相關風險的主要指標。 退休金風險 （經審核） 我們在世界各地設有多項界定福利及界定 供款退休金計劃。大部分退休金風險來自 集團的界定福利計劃，其中最大規模的是英 國滙豐銀行 （英國） 退休金計劃 （ 「主計劃」 ） 。 於2014年，滙豐制訂新的環球退休金風險 管理架構，以及管理界定福利及界定供款 計劃相關風險的新環球政策。此外，滙豐 已成立新的環球退休金監察委員會，監督 滙豐在全球營辦的所有退休金計劃的營運。 於2014年12月31日，集團於界定福利退休 金計劃的責任總額為420億美元，資產淨值 為27億美元 （2013年：分別為400億美元及 1億美元） 。資產淨值增加主要由於主計劃 資產的升幅，超過福利責任的升幅。集團 的界定福利責任及計劃的資產淨值中，主 計劃分別佔300億美元及48億美元。主計劃 所佔集團退休金風險最高。 主計劃 （經審核） 主計劃由公司受託人監管，該受託人對退 休金計劃的營運負有受信責任。主計劃包 括界定福利部分及界定供款部分。除另有 說明外，本節述及界定福利部分。 主計劃的投資策略是持有的資產以債券為 主，其餘則為更多元化的投資，包括利率 及通脹掉期組合，以減低利率風險以及通 脹風險 （請參閱財務報表附註41） 。主計劃 於年底的資產分配目標載列如下。滙豐及 受託人已制訂一般架構，使計劃的資產策 略承擔之風險隨著時間推移而降低，進一 步詳情載於下文。 主計劃－資產分配目標 2014年 2013年 % % 股票58 19.4 19.4 債券 64.5 64.5 另類資產59 10.6 10.6 物業 5.5 5.5 現金60 － － 於12月31日 100.0 100.0 有關註釋，請參閱第202頁。 主計劃的最近一次精算估值是在2011年12月 31日由Towers Watson Limited的精算師 C G Singer （英國精算師學會資深會員） 進 行。計劃於當日的資產市值為180億英鎊 （280億美元） （包括有關界定福利計劃、界 定供款計劃及額外自願供款的資產） 。經計 入根據預計單位基數精算成本法得出的預 期日後盈利增長後，計劃資產的市值佔預 期須向成員提供應計福利的所需款額 （按 所採用假設計算） 100%。因此，計劃並未 產生盈餘╱虧損，亦毋須制訂收回計劃。 來自主計劃的預計現金流乃參照2011年12月 31日零售價格指數 （ 「RPI」 ） 的掉期收支平 衡曲線而預測。年度增薪率假設為零售價 格指數加0.5%，而計及通脹的退休金增幅 （最低為0%，最高為5%，就2009年7月1日 起計的服務基數而言，最高為每年3%） 則 假設與零售價格指數一致。預計現金流按 2011年12月31日的倫敦銀行同業拆息掉期 曲線折現，並就投資策略預期回報加上每 年160個基點的邊際差距。是次乃按六年 期間 （2006年至2011年） 主計劃內退休金領 取人的死亡紀錄進行分析，而根據分析結 果，死亡率假設按作出反映退休金領取人 紀錄的調整後的SAPS S1系列死亡率表設 定。有關調查已考慮日後死亡率的改善幅 度，以符合持續死亡率調查核心預測 （即男 性長遠改善率設定為2%，女性為1.5%） 。 預期自2015年起界定福利計劃的應付福利 款項將如下圖所示。 其他重大風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 201 日後支付福利款項 （百萬美元） 0 200 400 600 800 1,000 1,200 1,400 1,600 1,800 2015 2021 2027 2033 2039 2045 2051 2057 2063 2069 2075 2081 2087 2093 2099 2105 2011年12月31日的估值亦計及倘若主計劃 終止及成員福利由保險公司承保 （儘管實際 上對於規模如此龐大的計劃，此情況不大 可能發生） ，或受託人在沒有滙豐支持的情 況下繼續營辦計劃時，償還負債可能需要 的資產金額。根據此方法計算，於2011年 12月31日所需金額估計為260億英鎊 （410億 美元） 。作出此項估計時，所採用的日後死 亡率假設比評估持續經營狀況時所用者更 為審慎，而且假設受託人會將投資策略改 為配對得宜的英國政府債券組合。明確的 支出準備亦已計算在內。 滙豐及受託人已制訂一般架構，使計劃的資 產策略承擔之風險隨著時間推移而降低， 並與預計日後現金流更趨一致，這個組合 稱為目標配對組合 （ 「TMP」 ） 。因此，目標 配對組合擁有充足資產，大部分為類似債 券的資產，與負債狀況更趨一致。我們可 透過取得超過假設的資產回報及╱或提供 額外資金而邁向目標配對組合。於2013年， 滙豐同意作出一般架構注資，分別於2013 年、2014年及2015年每年注資6,400萬英鎊 （1億美元） 及於2016年注資1.28億英鎊 （2億 美元） 。滙豐同意於往後年度在一般架構繼 續實施的情況下繼續注資。 英國滙豐銀行亦正就界定福利部分的成員 應計福利向主計劃供款。自2013年4月起， 滙豐已按可供計算退休金薪金 （扣除成員供 款） 的43%支付供款。滙豐將於進行下一次 精算估值 （其生效日期為2014年12月31日） 時檢討供款水平。有關估值的結果預期將 載於 《2015年報及賬目》 。 未來發展 （未經審核） 滙豐將由2015年6月30日起停止為界定福 利部分的活躍成員累計日後的服務基數。 界定福利部分的所有活躍成員將由2015年 7月1日起成為界定供款部分的成員，而按 照彼等服務年期計算至2015年6月30日之界 定福利退休金，將繼續與退休時的最終薪 金掛鈎 （並按消費物價指數調升） 。因此， 界定福利服務成本將由2015年7月1日起減 至零，而界定供款服務成本將增加。 界定供款計劃 我們的環球策略是視乎地方的法例和新出 現的慣例，由提供界定福利退休金，轉為 提供界定供款退休金。界定供款退休金計 劃的營辦人供款額是已知的，但最終福利 款項一般將視乎僱員作出投資選擇所達致 的投資回報而有差異。雖然界定供款計劃 的市場風險顯著低於界定福利計劃，但我 們仍須承擔營運及聲譽風險。 可持續發展風險 （未經審核） 在我們的整體風險管理程序中，評估向客 戶提供融資所產生的環境及社會影響是重 要一環。 2014年，我們就以往的林業政策進行全面 的內部及外部檢討，隨後頒布有關林業、 農業商品、世界文化遺址及拉姆薩爾濕地 的新政策。過往政策的內容及實施情況的 兩項獨立檢討結果，載於www.hsbc.com。 有關聲譽風險、退休金風險及可持續發展風險的 現行政策及慣例，於第235頁風險附錄內概述。 董事會報告：風險 （續） 滙豐控股有限公司 202 風險註釋 信貸風險 1 自2014年1月1日起，非交易用途反向回購及回購於資產負債表內分行呈列。過往，非交易用途反向回購 計入 「同業貸款」 及 「客戶貸款」 ，而非交易用途回購則計入 「同業存放」 及 「客戶賬項」 。比較數字已相應 重列。 2 於2014年12月31日，金融擔保及同類合約的信貸質素為：170億美元屬 「穩健」 、160億美元屬 「良好」 、120億 美元屬 「滿意」 及20億美元屬 「低於標準」 。 3 貸款承諾金額反映 （如適用） 在致函個人客戶提供預先批核貸款的情況下，預期客戶接受要約所涉金額。 除該等款項外，可能承擔的最大額信貸風險為710億美元 （2013年：340億美元） ，反映全數取用貸款承諾 所涉金額。接受要約的比率一般處於中等水平。於2014年12月31日，貸款及其他信貸相關承諾的信貸質 素為：3,220億美元屬 「穩健」 、1,910億美元屬 「良好」 、1,270億美元屬 「滿意」 、100億美元屬 「低於標準」 及 8億美元屬 「已減值」 。 4 自2014年1月1日起， 「亞洲」 地區取代之前按 「香港」 及 「亞太其他地區」 呈列的地區 （有關進一步詳情，請 參閱財務報表附註23） 。比較數字已重列以反映有關變動。 5 「金融機構貸款」 包括同業貸款。 6 「第一留置權住宅按揭」 包括香港政府 「居者有其屋計劃」 貸款，於2014年12月31日為34億美元 （2013年：32億 美元） 。在相關披露資料中，之前的比較數字為2012年：32億美元；2011年：33億美元；2010年：35億美 元。 7 「其他個人貸款」 包括第二留置權按揭及其他與物業相關貸款。 8 「其他商業貸款」 包括對農業、運輸、能源及公用事業的貸款。 9 並無就金融工具呈報減值準備，金融工具的減值額按直接扣減賬面值的方式呈列，而非透過準備賬呈列。 10 並無就交易用途組合中持有或指定以公允值列賬之資產計量減值，原因為該等組合內的資產乃根據公允 值變動處理，而公允值變動乃直接計入收益表內。因此，我們於 「並非逾期或已減值」 項內呈列所有相關 數額。 11 「客戶貸款」 包括獲外部評級機構評為 「穩健」 （2014年：12億美元；2013年：17億美元） 、 「良好」 （2014年： 2.56億美元；2013年：2.55億美元） 、 「滿意」 （2014年：3.32億美元；2013年：2億美元） 、 「低於標準」 （2014年： 9,400萬美元；2013年：2.83億美元） 及 「已減值」 （2014年：1.28億美元；2013年：2.52億美元） 的資產抵押證 券。 12 「綜合評估減值準備」 乃根據準備之入賬辦事處所在地分配至不同地區的賬項。 13 「匯兌及其他變動」 包括重新分類為持作出售用途的減值準備4億美元 （2013年：2億美元） 。 14 27.24億美元 （2013年：35.8億美元） 的重議條件貸款中，6.08億美元 （2013年：7.16億美元） 為並非逾期或已減 值，100萬美元 （2013年：5,200萬美元） 為已逾期但並非已減值，21.15億美元 （2013年：28.12億美元） 為已減 值。 15 法國銀行業聯合會關於遠期金融工具交易的總協議及等同於信貸支持附件的文件。 16 德國金融衍生工具交易總協議。 17 美國滙豐融資貸款的列示方式是以管理層意見為基準，並包括轉讓予美國滙豐有限公司之貸款，這些貸 款由美國滙豐融資管理。 18 透過止贖取得的物業於首次列賬時按貸款的賬面值或其公允值 （以較低者為準） 減估計出售成本確認 （ 「首 次列賬止贖物業賬面值」 ） 。出售止贖物業平均利潤╱虧損額按下列方式計算：出售所得現金減首次列賬 止贖物業賬面值，除以撇減前未還貸款本金結欠 （不包括任何應計融資收益） ，加若干其他相關償付項目 （根據法例可從所得現金償付的款項，例如預付房地產稅項，且在我們取得物業業權前產生） 。此比率代 表我們取得物業業權後於產生的止贖物業虧損總額中所佔部分。 19 止贖物業平均利潤╱虧損總額包括註釋18所述出售止贖物業的利潤╱虧損，及截至我們取得物業業權時 就貸款確認的累計撇減額。 20 本類別包括已重訂賬齡一次並於重訂賬齡時逾期少於60日的貸款15億美元 （2013年：19億美元） 。就該等貸 款整體而言，因預期有關客戶將於日後履行其原有借貸合約條款，故於重訂賬齡後並無分類為已減值。 21 「貨幣換算」 為按本年度適用的平均匯率，換算附屬及聯營公司上年度業績的影響。 22 負數為有利；正數為不利。 23 本金風險淨額的賬面值。 24 是項總額包括持有由房貸美及房利美發行的資產抵押證券。 流動資金及資金 25 最有利的數字為較低的貸款對核心資金比率及較高的壓力下之一個月及三個月償債保障比率。 26 呈列的英國滙豐旗下公司包括四間法律實體：英國滙豐銀行有限公司 （包括所有海外分行及就財務報表 目的而由英國滙豐銀行有限公司綜合入賬的特設企業） 、Marks and Spencer Financial Services Limited、HSBC Private Bank (UK) Ltd及HSBC Trust Company (UK) Limited，均以單一營運公司模式管理，與應用經英國審慎監 管局同意的英國流動資金規則所規定者一致。 27 香港上海滙豐銀行指集團在香港的業務 （包括其海外分行） 。各分行作為獨立營運公司，接受流動資金及 資金風險方面的監察及監控。 28 呈列的美國滙豐旗下主要公司指美國滙豐有限公司的綜合集團，主要成員為美國滙豐有限公司及美國滙 豐銀行。美國滙豐有限公司的綜合集團是以單一營運公司模式管理。 29 就滙豐旗下其他主要營運公司所示的總計數額，指由集團管理委員會轄下風險管理會議直接監督的所有 其他營運公司的合計水平。 30 估計流動資產價值代表管理層作假設扣減前資產的預期可變現價值。 31 向客戶提供之五大流動資金信貸額承諾的未取用數額，不包括向中介機構提供的信貸額。 註釋 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 203 32 向最大市場類別提供之所有流動資金信貸額承諾總額的未取用數額，不包括向中介機構提供的信貸額。 33 資產負債表的尚餘合約期限狀況，載於財務報表附註31。 市場風險 34 交易用途組合包括因進行市場莊家活動而持有及代客保管的持倉。 35 持有包含不同風險類別的組合分散市場風險的影響，即為組合分散。將多種不同類別風險 （例如利率、 股權及外匯） 納入一個組合內時，可減低非系統性市場風險，此數額即反映減低風險的情況。其計算方 法為個別風險類別的估計虧損風險總和與合計估計虧損風險總額的差額。負數代表組合分散的效益。由 於不同風險類別的最高數額會在不同日期出現，故就有關計量計算組合分散的效益並無意義。就呈列而 言，交易用途組合內的組合分散包括以估計虧損風險為基準的估計虧損以外風險。 36 由於風險分散的影響，估計虧損風險總額並非各類風險的估計虧損風險相加之總和。 37 私募股本投資主要透過管理資金進行，相關投資額會受到限制。我們會對潛在的新承諾進行風險評估， 以確保行業及地域集中程度，在整體組合內維持於可接受水平。我們亦會定期檢討，以核實組合內各項 投資的估值。 38 持作配合業務持續發展的投資包括持有政府資助企業及各地證券交易所的股權。 39 我們並不假設所有利率同步變動，而是將利率風險按貨幣區分類，同一貨幣區的利率被視為可能同步變 動。詳情請參閱 「有關前瞻性陳述之提示聲明」 。 保險業務風險管理 40 其他包括定期壽險、信用壽險、萬用壽險及其餘非壽險。 41 雖然附有酌情參與條款 （ 「DPF」 ） 之投資合約是金融投資，但滙豐在IFRS 4容許的情況下，繼續於賬目中將 該等合約列為保單。 42 其他資產一欄載列股東資產及分類為持作出售用途的資產及負債。保險業務旗下分類為持作出售用途的 大部分資產均列入 「其他資產及投資物業」 ，於2014年12月31日的總額為68億美元 （2013年12月31日：零） 。 該等資產大部分為債務及股權證券。保險業務旗下分類為持作出售用途的所有負債都列入 「其他負債」 ， 於2014年12月31日的總額為68億美元 （2013年12月31日：零） 。該等負債大部分為保單未決賠款及投資合約 負債。 43 附有酌情參與條款之有效長期保單及投資合約的現值。 44 遞延稅項包括因確認PVIF而產生的遞延稅項負債。 45 不包括保險聯營公司SABB Takaful Company，或保險合資公司Canara HSBC Oriental Bank of Commerce Life Insurance Company Limited。 46 滙豐於中東及北非或北美洲並無制訂保險產品的附屬公司。 47 包括單位相連壽險保單及相連長期投資合約。 48 包括非相連保單及非相連長期投資合約。 49 主要包括同業貸款、現金及與其他並非經營保險業務的法律實體之間相互往還的款額。 50 所呈列保證成本的數據包括根據保險附屬公司制訂的產品以模型計算之保證成本，當中包括透過投保人 負債預留的保證成本以及列為PVIF扣減項目的金額。有關數據所提供的資料，較過往期間所披露就保險 附屬公司制訂的保證產品而確立的投保人負債總額更具參考價值。 51 法國一組附帶介乎1.25%至3.72%保證名義年度回報的合約全部以2.1%至4%類別呈報，與該等合約向投保 人提供的平均保證回報2.7%相符。 52 實際年度回報保證為投保人提供超過通脹率的保證回報，並由收益同樣以實際數額列示的通脹相連債務 證券支持。 53 倘孳息曲線平行下移100個基點會產生負利率，對除稅後利潤及各類股東權益總額的影響按最低利率0% 計算。 54 股東權益包括有償債能力及無產權負擔資產。 55 非相連保險包括其餘非壽險業務。 56 在大部分情況下，投保人有權選擇隨時終止保單，並收取其保單的退保金額。此等數額可能大幅低於所 示金額。 57 附有酌情參與條款之保單及投資合約讓投保人在收取保證利益以外，有權按合約收取額外利益。此等額 外利益可能構成合約利益總額之重大部分，但有關金額及支付時間均由滙豐釐定。此等額外利益由合約 條款規定，是按特定的多項合約或資產組合之表現或簽發合約之公司的利潤計算金額。 退休金風險 58 2014年，期權全盤管理策略已予實施，預期可改善股票分配的風險╱回報狀況。 59 另類資產包括資產抵押證券、按揭抵押證券及基礎設施資產。 60 雖然並無現金分配目標，但預期現金比例介乎0-5%，視乎計劃的流動資金規定而定，而現金分配將相應 影響債券的實際分配。 董事會報告：風險 （續） 滙豐控股有限公司 204 風險附錄－政策及慣例 風險附錄 風險政策及慣例 本附錄敍述滙豐在管理信貸風險、流動資金及資金、市場風險、營運風險 （包括合規風險、法律風險及受信 風險） 、保險風險、聲譽風險、退休金風險及可持續發展風險時採用的主要政策及慣例。 風險管治 （未經審核） 我們的風險管治穩健有力，反映董事會及集團風險管理委員會 （ 「GRC」 ） 十分重視制訂集團風 險策略及有效管理風險。我們既設有清晰的風險責任政策架構，亦訂立承受風險水平處理 程序，藉此闡明及監察我們於執行策略時擬接受的風險類別及程度，以及由集團管理委員 會為使業務和風險目標保持一致而制訂的表現評分紀錄，全體員工亦有責任在其指定職責 範圍內識別、評估及管理風險。通過管治架構、強制學習及薪酬制度，落實個人問責，有助 滙豐建立嚴明且有建設性的風險管理及監控文化。 執行及非執行風險管治架構及其相互作用載於下表。各主要營運附屬公司均已設立對監督 風險相關事宜負有非執行責任的董事會屬下委員會，及負責風險相關事宜的執行委員會。 風險管理的管治架構 權力架構 成員 職責包括： 董事會 執行及非執行董事 ‧ 審批集團的承受風險水平、策略及表現目標 ‧ 批准任命附屬公司風險管理總監 ‧ 鼓勵以穩健的風險管治文化，作為集團對風險取態的基礎 集團風險管理 委員會 獨立非執行董事 ‧ 就以下事宜向董事會提供建議： － 承受風險水平及促使此水平與策略保持一致 － 促使薪酬與承受風險水平相配 （透過向集團薪酬委員會提供 建議） － 與建議的策略收購及出售有關的風險 ‧ 從高層次角度監督風險相關事宜 ‧ 檢討集團的風險管理及內部監控 （不包括財務報告） 制度是否 行之有效 ‧ 監督集團維持及發展支援風險管理的文化 金融系統風險防護 委員會 非執行董事 （包括集團薪 酬委員會主席） 及增選 非董事成員 ‧ 監督用以識別金融犯罪或濫用系統之風險範疇的監控措施及程 序 ‧ 監督有關反洗錢、制裁法律、反資助恐怖主義及武器擴散的事 宜 ‧ 檢討各項政策及程序，以確保持續履行監管和執法機構所訂責 任 行為及價值觀委員會獨立非執行董事 ‧ 確保於進行業務時，滙豐以公平的方式對待所有相關群體 ‧ 就滙豐政策、程序及準則向董事會提供建議，以確保集團以負 責任和持續恪守滙豐價值觀的方式進行業務 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 205 權力架構 成員 職責包括： 集團管理委員會 風險管理會議 集團風險管理總監 法律事務總監 集團行政總裁 集團財務董事 所有其他集團常務總監 ‧ 制訂高層次的環球風險管理政策 ‧ 行使獲授的風險管理權限 ‧ 監督集團落實承受風險水平及監控措施 ‧ 監察所有類別的風險及決定適當的減輕風險措施 ‧ 推廣支援風險管理及操守的集團文化 ‧ 在集團上下實施各項環球標準 環球風險管理委員會集團風險管理總監 滙豐環球業務及各地區 的風險管理總監 環球風險管理分支部 門主管 ‧ 支援風險管理會議及集團風險管理總監，為環球風險管理部提 供策略方向，制訂工作的優先次序及監督有關工作 ‧ 監督環球風險管理部採取一貫的方式承擔風險方面的責任，以 及採取減輕風險措施 環球業務風險管理委 員會 環球業務風險管理總監 環球業務行政總裁 環球業務財務總監 環球風險管理分支 部門主管 （如適用） ‧ 評估環球業務活動或經營業務所在市場日後的變動，分析可能 出現的風險的影響並採取適當的行動 ‧ 監督落實環球業務承受風險水平及監控措施 ‧ 監察所有類別的風險及決定適當的減輕風險措施 ‧ 推廣良好的風險管理文化 區域風險管理委員會區域風險管理總監 區域行政總裁 區域財務總監 區域環球業務總監 環球風險管理分支部門 主管 （如適用） ‧ 制訂區域的特定風險政策 ‧ 監督落實區域承受風險水平及監控措施 ‧ 監察所有類別的風險及決定適當的減輕風險措施 ‧ 推廣良好的風險管理文化 負責風險相關事宜的 附屬公司董事會屬 下委員會及環球業 務風險管理委員會 獨立非執行董事及╱或 並無於相關附屬公司 的活動或環球業務具 有職能責任的滙豐僱 員 （如適用） ‧ 就相關附屬公司或業務的風險相關事宜及內部監控措施 （不包 括財務報告） 向集團風險管理委員會或中層風險管理委員會提 供報告 （如有需要） 管治架構亦界定了風險管理分支部門委員會應有的架構、壓力測試及在集團、環球業務、 區域及國家╱地區層面的其他主要範疇。 承受風險水平 （未經審核） 我們的承受風險水平架構具有以下的核心特性。該等核心特性用於在集團整體、環球業務 及區域層面為承受風險水平聲明下定義。 ‧ 良好的資本狀況：在監管及內部層面，普通股權一級比率及整體資本架構穩健，並無過 度的槓桿借貸； ‧ 保守的流動資金管理：多元化的資金結構及保守的紀律；據此，附屬公司根據沒有 「最 後關頭的貸款人」 （不論屬本土監管機構干預或滙豐控股的支持） 的假設規劃其或有流動 資金； ‧ 穩健的資產負債結構：滙豐業務理念的核心，創造穩健的盈利來源； ‧ 卓越的品牌：我們的品牌－ 「世界領先國際銀行」 －至關重要，與集團的聲譽及良好的商 業道德同樣備受重視； ‧ 風險水平必須與回報相配：回報應與所涉風險相配，並應與策略性規劃及風險管理政策 一致； ‧ 集團強健的獨立法律實體架構：法律實體架構提供潛在的防火牆，減輕在危機下流動資 金及資本的連鎖影響； ‧ 環球業務組合應維持長期的盈利增長： 環球業務應該適當地多元化，以提供可預計的穩 定盈利； 董事會報告：風險 （續） 滙豐控股有限公司 206 風險附錄－政策及慣例 ‧ 分散風險：我們的業務分處世界各地，可有效分散風險，我們須持續密切評估風險，並 反映於資本規定內；及 ‧ 金融犯罪風險：我們對金融犯罪風險的整體態度及取向如下：集團不會容讓欠缺能防止 及查察金融犯罪的系統及監控措施的情況出現；如我們認為某些人士或實體從事非法行 為，則不會與有關人士或實體進行業務。 信貸風險 信貸風險管理 （經審核） 環球風險管理部負責進行獨立信貸監控。董事會授予滙豐控股的若干行政人員信貸批核權 限。附屬公司的董事會亦授予有關附屬公司的行政人員類似的信貸批核權限。每家主要附 屬公司的風險管理總監均向當地行政總裁匯報與信貸相關的問題，亦須向環球風險管理部 的集團風險管理總監直接匯報職務。信貸風險管理部的角色及職責，以及管理信貸風險的 政策及程序，詳載於下文。此等方面於2014年並無重大變動。 隸屬環球風險管理部的信貸風險管理部為環球業務信貸風險提供高層次的監督及管理 ‧ 制訂集團的信貸政策。各營運公司除非獲准豁免，否則均須遵循此等政策，並須按照集團的政策制訂當 地的信貸政策； ‧ 向營運公司提供指引，釐清特定市場類別、業務及銀行產品的信貸風險承受水平，並監控在若干較高風 險行業的風險承擔； ‧ 獨立審核及客觀評估風險。所有非銀行商業信貸及風險項目如超出指定限額，須先經環球風險管理部 評估，再決定應否向客戶提供信貸承諾或應否進行交易； ‧ 監察集團內各組合的表現及管理； ‧ 監控源自主權實體、銀行及其他金融機構以及非純粹持作交易用途之債務證券的風險； ‧ 制訂大額信貸風險的集團政策，確保交易對手、行業或地區的風險承擔不會過份集中，以致超出我們資 本基礎的可承擔水平，並將風險維持於內部及監管機構所定限額之內； ‧ 監控我們的跨境風險承擔 （請參閱第207頁） ； ‧ 貫徹執行並完善我們的風險評級架構和制度，其管治整體上由集團模型監察委員會 （ 「MOC」 ） 監察。集團 模型監察委員會每兩個月舉行一次會議，並向風險管理會議匯報。該委員會主席來自風險管理部，其成 員來自環球風險管理及相關環球部門或業務； ‧ 向風險管理會議、集團風險管理委員會及董事會匯報高風險組合、風險集中情況、國家╱地區風險限額 及跨境風險承擔、大額已減值賬項、減值準備、壓力測試結果及提出建議，以及零售組合表現；及 ‧ 擔任主要聯絡人，代表滙豐控股就信貸相關事宜與英倫銀行、審慎監管局、地方監管機構、各評級機 構、分析員及主要銀行和非銀行金融機構的交易對手接觸。 信貸風險管理的主要目標 ‧ 在整個滙豐集團保持嚴謹而負責任的貸款文化，以及穩健的風險管理政策及監控架構； ‧ 與不同業務合作，根據實際及假設境況界定、執行和持續重估承受風險水平，並考究有 關承受風險水平；及 ‧ 確保信貸風險、相關成本及減低風險措施經獨立而專業的審核。 風險集中情況 （經審核） 倘某些交易對手或風險項目具備相若的經濟特點，或該等交易對手從事類似業務，或在同 一地區或行業經營，以致其履行合約責任的整體能力，受同樣的經濟、政治或其他狀況變 動所影響，則出現信貸風險集中的情況。我們採納多項監控和措施，避免組合中行業、國 家╱地區及環球業務的風險過於集中。有關監控和措施包括組合及交易對手限額、審批及 評估監控，以及壓力測試。 錯向風險會在交易對手的風險與其信貸質素成逆向關連時出現。錯向風險共有兩類： ‧ 一般錯向風險會於交易對手的違責或然率與一般風險因素成正相關時產生，如交易對手 居於風險較高的國家╱地區及╱或於風險較高的國家╱地區註冊成立，並尋求出售非當 地貨幣以換取當地貨幣；及 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 207 ‧ 特定錯向風險會於特定交易對手的風險與交易對手的違責或然率成正相關時產生，例如 交易對手反向回購本身債券。滙豐的政策是按個別情況，批准特定錯向交易。 我們使用一系列工具監控錯向風險，包括要求業務在進行預先協定指引以外的錯向風險交 易前，必須事先取得批准。 跨境風險承擔 當考慮跨境風險承擔的減值準備時，我們會評估某些國家╱地區受外幣付款限制 （包括經濟 及政治因素） 的情況。在較易受影響國家╱地區內的所有合資格風險承擔，皆須提撥減值準 備，除非有關風險承擔及內在風險： ‧ 正依期償還，並與貿易有關及期限少於一年； ‧ 除特殊情況外，在相關國家╱地區境外持有可接受的抵押品，令風險減低； ‧ 其形式為持作交易用途的證券，而且相關證券的市場相當流通和活躍，同時證券價值每 天按公允值計量；及 ‧ 屬本金額 （不包括抵押品） 為100萬美元或以下及╱或屬三個月內到期的履約信貸。 金融工具的信貸質素 （經審核） 我們的信貸風險評級制度及程序旨在區別各類風險，藉以突顯風險因素較大和有較大可能 出現嚴重虧損的項目。就批發業務內佔多數的個別大額賬項而言，我們會定期覆核其風險 評級，並會在必要時迅即修訂。我們的零售業務利用多種不同的風險及定價模型取得貸款 組合數據，從而評估及管理風險。 我們的風險評級系統方便集團採納巴塞爾協定架構下的內部評級基準計算法，以支持我們 計算信貸的監管規定最低資本水平。詳情請參閱下文信貸質素類別的定義。 集團會特別關注有問題的風險項目，以便加快採取補救行動。在適當情況下，我們的營運 公司會利用專家小組為客戶提供支援，協助客戶盡可能避免拖欠還款。 集團及地區信貸審核及風險識別小組定期審核風險及程序，以便對滙豐集團整體的信貸風 險提供獨立而嚴謹的評估、加強第二重風險管理監控以及共用最佳運作方式。作為第三重 監控部門，審核部門專注於從環球角度監控風險，以及關注第一及第二重監控的設計和成 效，同時透過抽樣審查環球及地區監控架構進行督導審核工作、對主要或新浮現風險進行 主題審核工作，以及進行項目審核工作，以便評估主要改革計劃的影響。 下文界定的五類信貸質素各自包括一系列涵蓋批發及零售貸款業務的精細內部信貸評級， 以及由外界機構對債務證券所作的外界評級。 信貸質素類別 （未經審核） 債務證券 及其他票據 批發貸款 及衍生工具 零售貸款 外界 信貸評級 內部 信貸評級 12個月違責 或然率(%) 內部 信貸評級1 預期 虧損(%) 質素類別 穩健 A－級及以上 CRR21至CRR2級 0 － 0.169 EL31至EL2級 0 － 0.999 良好 BBB+至BBB－級 CRR3級 0.170 － 0.740 EL3級 1.000 － 4.999 滿意 BB+至B級 及並無評級 CRR4至CRR5級 0.741 － 4.914 EL4至EL5級 5.000 － 19.999 低於標準 B－至C級 CRR6至CRR8級 4.915 － 99.999 EL6至EL8級 20.000 － 99.999 已減值 拖欠 CRR9至CRR10級 100 EL9至EL10級 100+或已拖欠4 1 我們遵守披露慣例，除分類為EL9至EL10級的貸款外，分類為EL1至EL8級而拖欠90日或以上的零售貸款賬 項，亦被視為已減值，但被個別評估為並非已減值者 （請參閱第136頁 「已逾期但並非已減值之金融工具 總額」 ） 除外。 2 客戶風險評級。 3 預期虧損。 4 預期虧損百分比透過結合違責或然率和違責損失率計算得出，如違責損失率高於100%，反映收回款項的 成本，則在此等情況下，預期虧損百分比可能超過100%。 董事會報告：風險 （續） 滙豐控股有限公司 208 風險附錄－政策及慣例 質素類別定義 ‧ 「穩健」 ：大有能力遵守財務承諾、違責或然率極微或甚低及╱或預期虧損水平偏低的風險項目。符合產 品參數及僅於特殊情況下才出現拖欠的零售賬項。 ‧ 「良好」 ：需要較密切監察、有良好能力遵守財務承諾、拖欠機會低的風險項目。一般僅出現短期拖欠情 況，且於採取收回程序後預期虧損極微的零售賬項。 ‧ 「滿意」 ：需要較密切監察，有平均至一般遵守財務承諾的能力、拖欠機會中等的風險項目。一般僅出現 短期拖欠情況，且於採取收回程序後預期虧損輕微的零售賬項。 ‧ 「低於標準」 ：需要不同程度的特別注意及違約風險較令人關注的風險項目。拖欠期較長 （一般為逾期達 90日） 及╱或透過變現抵押品或其他收回程序減低虧損的能力降低，致令預期虧損較高的零售組合賬項。 ‧ 「已減值」 ：被評估為已減值的風險項目。當中包括銀行認為客戶不可能全數償還其信貸責任 （而銀行並 無採取追索行動，例如變現所持抵押品） 或客戶逾期償還任何重大信貸責任達90日以上的批發風險；零 售賬項包括分類為EL9至EL10級的貸款，而分類為EL1至EL8級的貸款逾期超過90日 （除非其被個別評為並非 已減值） ；及已符合規定披露為已減值及並未符合條件重新計入未減值組合中的重議條件貸款 （請參閱 下文） 。 上述客戶風險評級 （ 「CRR」 ） 10級制度概括了一項更精細的23級債務人違責或然率 （ 「PD」 ） 分級 制度。集團會視乎風險項目所用巴塞爾協定2計算法之精密程度，運用10或23分級制度，對 所有滙豐客戶進行評級。 各個客戶風險評級級別都有一個相關的外界評級等級，用作該評級長期違責率的參考 （以發 行人加權過往拖欠率的平均值表示） 。內部及外界評級的配對屬指標性質，經過一段時間後 可能有所不同。 零售業務的預期虧損 （ 「EL」 ） 10級制度概括了一項適用於該等客戶群的更精細預期虧損分級 制度。該制度結合債務人及信貸╱產品風險因素進行綜合計量。 就債務證券及若干其他金融工具而言，根據相關客戶風險評級與外界信貸評級配對的基準， 外界評級已予調整以與該五類信貸質素保持一致。最近期的配對檢討中 「B」 級與CRR5級同 級。因此，現時 「B」 級為 「滿意」 級別，這顯示披露資料的配對級別雖有變，但與交易對手可 信程度的變動無關。 重議條件貸款及暫緩還款 （經審核） 滙豐採取一系列的暫緩還款策略，以便提升客戶關係的管理、力求提高收回貸款的機會， 以及盡可能避免發生違責、止贖或收回抵押品的情況。有關安排包括延長還款期、減低利 息或本金還款額、批准外界債務管理計劃、債務重組、押後止贖，以及其他形式的修訂貸款 條款及重訂賬齡。 我們的政策和慣例所依據的準則，是借款人於業務所在地管理層的判斷下，能否繼續還款。 這些政策和慣例一般會向客戶提供較原先所提供者更優惠的條款和條件。我們只會在客戶 表明還款意願以及預期能履行經修訂還款責任的情況下，為客戶安排暫緩還款。 識別重議條件貸款 貸款的合約條款可因多種理由修訂，包括市況的變化、挽留客戶及其他與客戶的現時或潛 在信貸狀況惡化並不相關的因素。若由於我們對借款人履約到期還款能力有重大質疑而修 訂貸款的合約還款條款，該等貸款會被分類為 「重議條件貸款」 。 就零售貸款而言，我們的信貸風險管理政策已就以下各項制訂多項限制：重議條件的次數 及頻密程度；由有關賬戶開立至考慮重議條件之間的最短期限；須已收取的合資格還款筆 數。是否執行此項政策會視乎市場性質、產品，以及在特殊情況下如何管理客戶關係而定。 考慮客戶履約到期還款能力是否有 「重大質疑」 時，我們會評估客戶的拖欠狀況、賬項行為、 還款紀錄、現時財務狀況及持續還款能力。如客戶不能履約還款，或有證據證明其必須重 議條件方能履約還款，則對其履約還款能力有重大質疑，而貸款亦會披露為已減值，除非 如下文討論授出的優惠並不重大。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 209 就批發貸款的重整貸款而言，有關借款人還款能力的重大質疑指標包括： ‧ 債務人現時已經拖欠任何債務； ‧ 債務人已宣布破產或正進行宣布破產的程序或進入類似程序； ‧ 債務人能否持續經營存在重大疑問； ‧ 債務人現時持有的證券因交易或財政困難被交易所除牌、正進行除牌程序或正面對被除 牌的危機； ‧ 根據估計及預計的資料 （僅按照現時營運業務的能力而作出） ，集團預期債務人的特定企 業現金流，將不足以根據現有協議的合約條款償還債務 （利息及本金） 直至到期為止。在 此情況下，實質拖欠還款的情況可能尚未出現；及 ‧ 如貸款條款未予修訂，除現有債權人外，債務人從其他途徑取得資金的實質利率，不能 等同非財困債務人類似債務的現行市場利率。 如就有關借款人的財政困難而言，修改貸款的合約付款條款在經濟或法律理由上屬一項優 惠，並為我們不另作考慮的一項優惠，則重議條件貸款根據減值貸款披露慣例 （詳情載於第 212頁） 披露為已減值；除非優惠並不重大且並無其他減值指標則作別論。非重大優惠主要 限於美國滙豐融資的消費及按揭貸款組合，當中僅在初步拖欠的階段 （即拖欠少於60日） ，而 一般而言相等於兩次付款的貸款於首次遞延時不會計入減值貸款分類，因合約付款遞延項 目相對貸款到期付款整體而言被視為非重大。有關美國滙豐融資的貸款重議計劃及組合， 請參閱第154至155頁。 重議條件貸款的信貸質素分類 （經審核） 根據IFRS，企業須於每個業績報告期結束時評估是否有任何客觀證據顯示金融資產已減值。 在有客觀證據顯示虧損事件會影響某項貸款的現金流，而所涉數額又能可靠估算時，則會 將該項貸款列作減值，並會確認減值準備。因客戶的財政困難而向客戶授出原本不會考慮 的還款優惠，屬減值的客觀證據，而減值虧損亦按此計量。 在下述情況，重議條件貸款會呈列為已減值： ‧ 因貸款人授出原本不會考慮的還款優惠，導致約定現金流產生變化；及 ‧ 如無還款優惠，借款人很可能會無法全數履行合約付款責任。 除非有關還款優惠涉及數額不大及並無其他減值指標，否則一概如此呈列。 重議條件貸款會繼續披露為已減值，直至有充分證據顯示日後無法收回現金流的風險已大 幅下降且貸款再無其他減值指標為止。就按綜合基準評估減值的貸款而言，有關證據通常 包括按照原有或經修訂條款 （視乎適當情況而定） 顯示一段履約還款紀錄。就按個別基準評 估減值的貸款而言，我們會就不同個案逐一評估所有可得證據。 就零售貸款規定的最短履約還款期視乎組合中的貸款性質而定，但一般不少於六個月。如 屬有較大量暫緩還款活動的組合 （例如美國滙豐融資旗下組合） ，規定的最短履約還款期可 能會大幅度延長 （有關美國滙豐融資的詳情，請參閱第153頁） 。我們會監察履約還款期內的 情況，以確保所定期限仍能配合於組合內觀察所得的累欠情況。除履約期以外，我們還須 在60日期內最少收到客戶兩次還款，客戶才初步合資格重議條件 （若為美國滙豐融資，則在 若干情況下，例如已在破產程序中重整的債務，規定的合資格還款次數可能較少，甚至毋 須計及合資格還款） 。我們規定客戶必須支付合資格還款，以證明重議條件持續適用於借款 人。企業及商業貸款 （其減值須個別評估及較常協議採用非每月還款的方式） 的履約還款紀 錄，會視乎重整貸款時協定的相關還款安排而定。 若對借款人能否履行合約付款條款存在重大質疑因而重議貸款條件，但重議之條件按當前 市價計算，且預期在重議條件後可全數收回約定現金流，則重議條件貸款會歸類為未減值。 未減值的重議條件貸款亦包括先前已減值但已有一段時間履約情況令人滿意或根據所有可 用證據評估再無其他減值指標的重議條件貸款。 董事會報告：風險 （續） 滙豐控股有限公司 210 風險附錄－政策及慣例 若貸款被識別為重議條件貸款，此指定列賬方式將維持至貸款到期或撤銷確認為止。如貸 款作為暫緩還款策略一部分而重整，而重整導致撤銷確認現有貸款，例如在部分債務重組 的情況下，新造貸款將披露為重議條件貸款。 於釐定是否撤銷確認重整的貸款並確認為新貸款時，我們會考慮原合約條款有何種程度的 變動方會產生重議條件貸款，而且是從整體角度作為一項大不相同的金融工具來考量。下 列情況 （個別或共同地） 為可能進行本測試及在會計上應用撤銷確認的例子： ‧ 非抵押貸款變成悉數擔保貸款； ‧ 新增或取消互相擔保條文； ‧ 取消或新增貸款協議附設的轉換條款； ‧ 本金或利息的計值貨幣改變； ‧ 清盤時工具的優先次序或等級改變；或 ‧ 以任何其他方式改動合約，使新合約或經修改合約的條款與原有合約的條款大大不同。 以下例子是我們認為可能指出經修訂貸款屬本質上不同的金融工具的因素 （但非決定性因 素） ： ‧ 規定的擔保或貸款契諾改變； ‧ 抵押品安排的重大變動較輕；或 ‧ 新增還款條文或提前還款保證費條款。 重議條件貸款及確認減值準備 （經審核） 就零售貸款而言，在進行綜合減值評估時，重議條件貸款會與貸款組合內其他部分分開處 理，以反映此類貸款常有的較高虧損率。如有實證數據顯示該等賬項的拖欠率有上升傾向， 且虧損增加 （如美國的重訂賬齡貸款） ，便會使用滾動率方法，就涉及暫緩還款安排的貸款組 合採用特定的滾動率計算，以確保在計算減值準備時已考慮此等因素。若組合規模細小，或 採用滾動率方法所需的資料不足或不可靠，則會採用以過往虧損率為基礎的基本公式法計 算。若有過往證據顯示因結算日前發生的事件，同類貸款組合的貸款 （包括重議條件貸款） 可能會經過各個拖欠階段逐步惡化，並最終確定為無法收回，集團採取滾動率方法便能就 這些組合的貸款確認綜合評估減值準備。不論該等貸款是否按照集團的已減值貸款披露慣 例呈列為已減值，均會採取上述處理方法。當我們認為統計滾動率或過往紀錄未必全面反 映組合內含的其他風險因素時，會對純粹憑統計或過往紀錄計算的減值準備作出調整，以 計入上述風險因素。有關風險因素調整的詳情，請參閱財務報表附註1k。 企業及商業貸款方面，重議條件貸款一般會進行個別評估。減值評估會包含信貸風險評級。 重整財困個案所涉貸款會歸類為已減值貸款。個別減值評估會考慮重議條件貸款內含日後 無法償還現金流的較高風險。 暫緩還款之企業及商業貸款 （未經審核） 企業及商業貸款方面，若按原有條款識別出已有或極可能有還款困難的情況，則會選擇性 地進行暫緩還款活動。下述情況會列作已減值貸款處理： ‧ 客戶因當前的信貸困境在履行對集團的還款責任時存在或極可能存在困難；及 ‧ 集團向客戶提供的經修訂還款安排構成一項還款優惠 （即集團在一般情況下不擬提供的 條件） 。 上述情況稱為重整財困個案。要就符合以上準則的重整個案達成協議，客戶的全部貸款及 交易對手風險均須列作已減值處理。鑑於有此先決條件，當清楚意識到客戶可能需要進行 這種重整的風險正在增加時，有關風險一般會視作低於標準，以反映信貸風險狀況正在惡 化，而當提出重整建議以待審批時，會將有關風險評為已減值，或如有足夠理由懷疑客戶 的還款能力，則應更早將有關風險評為已減值。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 211 為釐定變更客戶協議應否視作重整財困個案處理，以下各種修訂會視作還款優惠： ‧ 客戶轉移應收第三方賬款、房地產或其他資產，以償還全部或部分債務； ‧ 發行或以其他方式授予股本權益，以償還全部或部分債務，除非該項股本權益是根據現 有轉換債務為股本權益的條款而授出；及 ‧ 以下列任何一種或多種方式修訂債務條款： － 調低 （不論肯定或可能調低） 債務原定尚餘期限內的指定利率； － 延長一個或多個到期日，而指定利率低於附帶同類風險的新造債務之當前市場利率； － 調低 （不論肯定或可能調低） 債務的面額或到期應收金額；及 － 調低 （不論肯定或可能調低） 應計利息。 與還款安排無關的修訂 （例如重整抵押品或抵押安排或根據有關文件契諾放棄權利） ，本身 不會被視作出現信貸困境以致影響還款能力的證據。一般而言，訂立契諾是為了賦予集團重 新定價或要求提前還款的權利，但這些權利通常會設定在還款能力不受影響的水平以內，使 集團在信貸惡化初期有權採取行動。這種權利不會直接影響客戶償還原合約債務的能力， 且相關債務不會以重議條件貸款列賬。然而，如客戶要求豁免非還款相關契諾條文，則違 反相關契諾的重要性將與任何其他減值指標一併考慮，同時如顯示信貸困境的嚴重程度會 為還款帶來不確定性，則於釐定虧損事件是否產生時會考慮一切可獲得的證據。然而，豁 免將不會導致將貸款分類為重議條件貸款，因還款的條款並無修改。 如因約定現金流還款方面存有重大質疑而同時作出還款相關及非還款相關的修訂，則貸款 將被視為重整財困個案處理，並披露為重議條件貸款。 在企業及商業貸款而言，就同一客戶的貸款通常會同意於同一時間作出若干類別的修訂。 在英國，最常見的修訂是將貸款改為僅須供息的安排 （不論單獨提供或同時提供其他優惠） ； 而在全球其他地區，延長期限屬較常見的安排，卻較少提供其他優惠 （如調整利率） 。 在評估暫緩還款是否為合理及可以持續採用的策略時，我們會檢討有關客戶的整體風險及 信貸狀況，並會評估其在重議條件下能否履行經修訂責任及其他未經修訂信貸額度的條款。 如此項評估顯示重議條件不能妥善處理客戶還款能力的問題，則可能會選擇採取其他特別 管理措施。有關程序可確定是否有需要特別協助客戶重整其業務營運及活動，使客戶恢復 理想的還款能力。 在考慮可接受的重整條款時，我們必須考慮客戶是否有能力償還經修訂的利息。於考慮修改 償還本金時，我們也會要求客戶能夠遵守經修訂的條款，此乃重整貸款的必要先決條件。如 修改償還本金會導致永久豁免還款，或因其他理由被認為實際上無法再收回未償還本金， 則應撇銷受影響的結欠。如延後償還本金，則預期客戶有能力根據重議條款還款 （包括預期 以再融資來償還延後本金還款的情況） 。在所有情況下，僅於預期客戶有能力遵守經修訂的 條款時，方會授出重議條件貸款。 當客戶需要時間作出還款安排、預計還款能力惡化程度雖嚴峻卻短暫、或客戶需要更多時 間與其他銀行就更長久的安排 （例如銀團貸款通常會涉及多邊磋商） 進行討論時，可暫時修 訂條款。 在進行重整時，如客戶在一段時間內的履約情況令人滿意，只要再無其他減值指標，則有 關貸款可回復至非減值級別 （CRR1-8級） 。若該客戶的任何信貸額度仍有特定減值準備，則有 關貸款不可回復至非減值級別。履約還款期的長度會視乎客戶根據經修訂協議須償還欠款 的頻密度，以及客戶財政狀況的改善程度而定。 董事會報告：風險 （續） 滙豐控股有限公司 212 風險附錄－政策及慣例 減值評估 （經審核） 根據滙豐的政策，當貸款或貸款組合出現客觀的減值證據時，每家營運公司須即時、妥善 地為已減值貸款提撥減值準備。 有關貸款及金融投資減值政策的詳情，請參閱財務報表附註1k。 減值及減低信貸風險 於計算個別評估已減值貸款之減值時，會因有抵押品而受到影響。如我們不再預期可於貸 款到期時或根據原有條款及條件悉數收回本金及利息，則會對貸款作出減值評估。如風險 有抵押品，則於評估是否需要提撥減值準備時，會考慮抵押品的當前可變現淨值。如預期 所有到期金額會於擔保變現時悉數結清，則不會確認減值準備。 由於個人貸款組合一般由大批同類貸款組成，故該等組合一般會按綜合基準評估減值額。 按綜合基準計算準備額之兩種方法為：滾動率方法或根據過往虧損採用較基本的公式法。 2014年，我們檢討了整個集團對零售銀行及小型企業組合所用的減值準備方法，以確保綜 合評估模型採用的假設持續恰當地反映由產生虧損事件至賬項出現拖欠以至最終撇銷所相 距的時間。 ‧ 過往虧損法一般用作計算有抵押或低違責率組合 （如按揭） 的綜合評估減值準備，直至貸 款被個別識別及評為已減值時為止。就使用過往虧損法進行綜合評估的貸款而言，過往 虧損率乃一段特定期間已扣除收回額之平均合約撇賬額。淨合約撇賬率為變現抵押品及 取得收回貸款後的已產生實際虧損金額。 ‧ 當有充足實證數據可供制訂良好的統計模型時，無抵押的組合較常採用滾動率方法。在 某些情況下，如按揭組合在統計數字上有大量違責及虧損，則可產生可靠的滾動率。在 該等情況下，將應用滾動率方法，直至貸款個別而言被識別及評為已減值時為止，同時 各項拖欠的平均虧損率會予以調整，以反映變現抵押品及取得收回貸款後的預期平均虧 損額。預期平均虧損額乃按過往抵押品的平均變現價值計算。 綜合準備評估的性質令個別抵押品價值或貸款估值比率無法納入計算內。然而，綜合評估 採用的虧損率會就抵押品變現的經驗作出調整，並會視乎組合內的貸款估值比率組合成分 而變動。舉例而言，根據過往虧損率方法計算的按揭組合若貸款估值比率較低，過往的虧 損一般會較低，因此淨合約撇賬率亦較低。 就綜合評估的批發貸款而言，則會採用過往虧損法計量已發生但未匯報的虧損事件減值額。 虧損率源自觀察一段特定時間內 （一般不少於60個月） 已扣除收回額之合約撇賬額。淨合約 撇賬率為變現抵押品及取得收回貸款後的已產生實際或預期虧損金額。該等過往虧損率乃 運用經濟因素作出調整，以便過往平均值更準確顯示影響組合的當前經濟情況。為了反映 產生未被識別及評估的虧損事件之可能性，計算過程中將採用生成期的假設，此生成期反 映由出現虧損至識別虧損的相距時間。管理層會就每個已識別的組合估計生成期。可能影 響此項估計之因素包括：經濟及市場狀況、客戶行為、組合管理資料、信貸管理技巧，以及 市場上追收及收回貸款的經驗。生成期按實際經驗定期評估，由於該等因素會改變，所以 生成期會隨著時間而有所不同。 貸款撇銷 （經審核） 有關我們的貸款撇銷政策詳情，請參閱財務報表附註1k。 美國滙豐融資的住宅按揭及第二留置權貸款，會在完成止贖時或之前，或與借款人達成和 解時或之前，將貸款賬面值超出可變現淨值之部分撇銷。若未有合理預期可收回款項並已 採取止贖行動，則貸款一般會於超逾合約期限180日的月份結束前撇銷。我們定期 （每180日） 取得該等有抵押貸款的最新評估，並將賬面值調整至最近期的評估 （不論升跌） ，作為出售 抵押品時將收取的現金流的最佳估計。 無抵押個人信貸，包括信用卡，一般會於逾期150至210日內撇銷，標準撇銷期限為賬項違約 拖欠達180日的月份結束時。撇銷期限可予延長，通常可延至逾期不超過360日。但在極罕見 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 213 情況下，例如少數國家╱地區的規例或法例限制提早撇銷，或有抵押房地產貸款的抵押品 需更多時間方能變現，則撇銷期限可進一步延長。 在零售貸款方面，最終撇銷最遲於出現拖欠情況後60個月內作出。 如牽涉破產或類似法律程序，則可提早於上述期限前撇銷。追收欠款程序可於貸款撇銷後 繼續進行。 減值評估方法 （經審核） 滙豐一般採用業內的標準估值模型來確定可供出售資產抵押證券是否存在減值的客觀證據； 該模型參考相關資產組別後採用有關數據，然後模擬其預計日後現金流。按特定金融資產 水平評估證券的估計日後現金流，是為了釐定現金流任何部分會否因業績報告日期當天或 之前發生的虧損事件而無法收回。 該等模型的主要假設及數據通常包括：相關貸款的拖欠狀況、拖欠貸款發展成違責的或然 率、相關資產的提前還款狀況，以及違責情況下虧損的嚴重程度。然而，該等模型亦會利用 其他與特定抵押品類別相關的變數來預測日後違責情況和收回率。管理層於釐定該等因素 的適當假設時，會根據從外部取得的數據作出判斷。滙豐採用計入過往觀察所得之拖欠累 進率的模型計算法，來確定相關抵押品的預計現金流總額減少會否導致約定現金流短缺。 若屬此等情況，有關抵押品會被視為已減值。 就債務抵押債券而言，我們會評估相關抵押品的預計日後現金流，以確定該債務抵押債券 會否出現約定現金流短缺的情況。 如某種證券享有債券承保公司提供的合約保障，確保支付本金及利息，於釐定該資產抵押 證券預期可得的信貸支持總額時，會評估根據有關合約預期可收回的金額。 貸款管理組 （未經審核） 滙豐貸款管理組 （ 「LMU」 ） 是批發信貸及市場風險管理部的前線客戶聯絡部門，負責管理商業 客戶關係，需要頻繁而緊密地管控銀行的貸款風險。貸款管理組在集團各經營地區運作， 獨立於辦理業務的管理部門。貸款管理組向所屬地區的信貸總監匯報。業務管理層或批發 信貸及市場風險管理部的審批團隊會識別客戶，並將客戶轉介予貸款管理組。 由貸款管理組管理的客戶，通常於集團的承受風險水平之外處理。他們通常面對重大財務 困難，其管理團隊在困境時管理業務的經驗有限，而且向銀行提供的管理及財務資料既不 足夠亦不可靠。 受管理的客戶貸款風險水平以及貸款管理組團隊的規模在各個國家╱地區均有所不同，有 關情況會視乎集團於當地業務的規模而定；然而，貸款管理組通常管理嚴重財困的情況， 即債務超過150萬美元的個別客戶。 貸款管理組的首要任務是透過與客戶共同合作，保障銀行的資本及將虧損減至最低，在可 實現的情況下促進及支持可行的收回欠款策略，最終目標是將客戶交還前線客戶關係管理 團隊跟進。在某些情況下，協助客戶恢復還款能力的措施並不可行，貸款管理組將會考慮 多個方案，保障銀行的風險承擔及客戶的償付能力。貸款管理組有時未必能找到理想的解 決方法，而客戶可能呈請清盤或於當地等同清盤的程序。不論結果如何，貸款管理組亦會 以專業的態度和透明的程序，公平、體諒及正面地對待客戶。 業務部門及貸款管理組可用的補救措施及重組策略，包括向客戶提供不同類型的還款優惠， 同時尋求提升客戶最終向集團還款的能力，包括提高銀行可得的整體抵押。批准還款優惠 的任何決定將視乎區內特定國家╱地區及行業的承受水平、客戶的主要風險衡量指標、市 場環境，以及貸款結構及抵押而定。直接由貸款管理組管理的客戶內部評估會根據相關會 計指引、信貸政策及國家銀行規例，按預定的基準進行。在若干情況下，授予客戶還款優惠 可能使相關貸款分類為重議條件貸款。 持有之抵押品及其他強化信貸條件 （經審核） 按已攤銷成本持有貸款 集團的慣例是參考客戶從其現金流資源償還債務的能力，而非依賴抵押品的價值來決定貸 款的事宜。我們提供的融資額度可能毋須抵押，但須視乎客戶的財政狀況及產品類別而定。 至於其他貸款，則須獲得抵押品作為押記，而有關抵押品亦將用作決定是否提供信貸及相 董事會報告：風險 （續） 滙豐控股有限公司 214 風險附錄－政策及慣例 關定價的考慮因素。倘出現違約，銀行可動用有關抵押品來償還貸款。在降低信貸風險方 面，抵押品可能對財務有重大影響，但仍須視乎抵押品的形式而定。 我們亦會運用其他類別的抵押品及強化信貸條件以管理風險，例如第二押記、其他留置權 及無支持的擔保，但該等減低風險措施的估值較難確定，而其財務影響亦未經量化。 再融資風險 （經審核） 許多類貸款均要求客戶於到期時償還大部分本金。一般而言，客戶的還款機制是通過取得 新貸款以償還現有債務。當客戶於到期日無法償還此項有期債務，或無法以商業利率為債 務再融資，則會產生再融資風險。如有證據證明特定合約可能出現此風險，則滙豐可能需要 以於其他情況下不會考慮的優惠條款為貸款再融資，以便盡最大可能從合約收回現金流， 並可能避免客戶拖欠償還本金。如有足夠證據證明借款人 （基於其現有財政能力） 於到期時 可能無法償還貸款或為貸款再融資，則該等貸款將披露為已減值，並確認相應的減值準備 （如適用） 。 滙豐的證券化及其他結構風險項目之性質 （經審核） 按揭抵押證券 （ 「MBS」 ） 是代表一種集合多組按揭權益的證券，賦予投資者權利收取日後按 揭還款所產生的現金 （利息及╱或本金） 。倘按揭抵押證券需參考不同風險狀況的按揭，其級 別會按其中最高風險級別歸類。 債務抵押債券 （ 「CDO」 ） 是指以一組債券、貸款或其他資產 （例如資產抵押證券 （ 「ABS」 ） ） 作抵 押的證券。倘債務抵押債券的相關資產或參考資產有部分是次優質或Alt-A按揭資產，則可 能要面對此類資產的風險。因為用以支持債務抵押債券的相關抵押品之確實性質通常難以 確定，故所有以住宅按揭相關資產支持的債務抵押債券，一概列為次優質類別。我們持有 的資產抵押證券和債務抵押債券與各類直接貸款，以及各種按揭抵押品和借貸活動概述於 下頁。 我們就非住宅按揭相關資產抵押證券面對的風險，包括採用以下各項相關抵押品作擔保的 證券：商用物業按揭、槓桿融資貸款、學生貸款及其他資產，例如以其他應收賬款相關抵押 品作擔保的證券。 資產抵押證券及債務抵押債券的定義及分類 ABS及CDO類別 定義 分類 次優質 向信貸紀錄有限、收入較低、債務對收入 比率較高，或曾因偶爾拖欠、過往沖銷、 破產或其他信貸相關行動，以致出現信貸 問題的客戶提供的貸款。 就美國按揭而言，主要使用620或以下的 FICO評分以釐定一項貸款是否屬於次優質 類別。就非美國按揭而言，以管理層的判 斷來識別。 美國住宅二按 信貸 （屬於 「次優 質」 類別） 向客戶提供的一種循環信貸融資，絕大多 數以住宅物業的第二留置權或更低級別的 抵押支持。 所持的美國住宅二按信貸均歸類為次優質 項目。 美國Alt-A 這類貸款的風險低於次優質貸款，但與全 面符合標準條件的貸款相比，風險仍屬較 高。 在釐定應否將貸款歸類為Alt-A時，會考慮 美國信貸評分和持有的文件 （例如收入證 明） 是否完備。不屬於美國次優質按揭的 貸款，如不合資格向主要美國政府按揭機 構或資助機構出售，都歸類為Alt-A。 美國政府機構及 政府資助企業按 揭相關資產 由美國政府機構 （例如政府國民抵押協會 （ 「Ginnie Mae」 ） ） 擔保的證券，或由美國政 府資助企業，包括房利美和房貸美擔保的 證券。 所持的美國政府機構及美國政府資助企業 的按揭相關資產，都歸類為優質風險項目。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 215 ABS及CDO類別 定義 分類 英國非標準類型 按揭 （屬於 「次優 質」 類別） 不符合一般借貸條件的英國按揭。例子包 括並無提供預期所需文件 （如自行證明入 息） 的按揭，或借款人信貸紀錄欠佳令風 險增加，進而導致定價較一般借貸利率高 的按揭。 英國非標準類型按揭均列作次優質風險項 目處理。 其他住宅按揭 不符合上述任何分類條件的住宅按揭，包 括優質按揭。 優質住宅按揭相關資產列入此類別。 流動資金及資金 （經審核） 流動資金及資金的管理主要由各地 （按國家╱地區） 的營運公司按照集團流動資金及資金風 險管理架構負責執行，並須遵照集團管理委員會透過風險管理會議設定並由董事會批准的 慣例和限額。該等限額會視乎各公司所屬市場的深度及流通性而改變。我們一貫的政策是， 凡被界定為營運公司者，應當能在本身業務資金需求方面自給自足。倘營運公司之間存在 交易，則兩家公司均會以對等方式反映這些交易。 作為集團資產、負債及資本管理部架構的其中一環，我們在集團、區域及營運公司層面均 設立資產負債管理委員會。各資產負債管理委員會的職權範圍均包括對流動資金及資金的 監察與監控。 根據集團架構及承受風險水平而管理流動資金及資金的主要職責，由各地營運公司的資產 負債管理委員會承擔。我們最重要的營運公司受地區資產負債管理委員會、集團資產負債 管理委員會及風險管理會議監督。其餘較小型的營運公司則受地區資產負債管理委員會監 督，重要問題則會適當上報集團資產負債管理委員會及風險管理會議。 營運公司主要按國家╱地區界定，以反映我們在各地對流動資金及資金進行管理。營運公 司通常界定為單一法律實體。然而，考慮到在某一國家╱地區業務的賬目涵蓋多家附屬公 司或分行的情況： ‧ 營運公司可界定為：在較廣泛層面一群經初步綜合計算的法律實體，而此等法律實體是 在同一國家╱地區註冊成立，流動資金及資金可在實體間自由互換及獲當地監管機構許 可，且該定義能反映流動資金及資金在當地的管理方式；或 ‧ 營運公司可更狹義地界定為在多個國家╱地區營運的較大型法律實體之主要辦事處 （分 行） ，反映流動資金及資金在當地的管理情況。 風險管理會議每年審閱及協定直接監控的公司名單，以及這些公司的組成部分。 主要資金來源 （經審核） 滙豐的資金大部分源自客戶的往來戶口存款及即期或短期通知儲蓄存款，而集團也十分重 視維持其穩定性。存款的穩定性視乎滙豐能否維持存戶對其雄厚資本及流動資金之信心， 以及具競爭力及透明度的定價。 滙豐亦透過發行優先有抵押及無抵押債務證券 （公開及私人） ，以及運用高質素抵押品透過 有抵押回購市場取得借款，從而參與各批發融資市場，以便為經營非銀行業務而不接受存 款的附屬公司取得資金、使資產負債的到期日及貨幣互相配合，並維持在各地批發市場的 影響力。 流動資金及資金風險管理 （經審核） 內在流動資金風險分類 我們將旗下營運公司分為兩個類別 （低及中） ，以反映我們對該等公司的內在流動資金風險 所作評估；有關評估會顧及業務所在國家╱地區的政治、經濟及監管因素，以及有關營運公 司本身的特定因素，例如當地市場、市場份額及資產負債實力等。有關分類工作包含管理 層的判斷，並以營運公司 （相對於集團旗下其他公司而言） 的可見流動資金風險作為判斷的 依據。有關分類方法旨在反映流動資金事件可能產生的影響，而非事件出現的可能性，且 構成集團承受風險水平的一部分，用以確定各營運公司必須有能力承受及設法應對的指定 壓力境況。 董事會報告：風險 （續） 滙豐控股有限公司 216 風險附錄－政策及慣例 核心存款 我們內部架構的主要元素乃客戶存款的分類方式－根據我們預期這些存款在承受流動資金 壓力期間的表現，分為核心和非核心客戶存款。此分類方式顧及辦理存款業務的營運公司 之內在流動資金風險分類、客戶的性質，以及存款的規模和定價。只有根據合約為貸款提 供抵押的存款才會整筆被視為核心存款。各營運公司的核心存款基礎會被視作長期資金來 源，並因此假定在流動資金壓力境況 （我們用以計算主要流動資金風險的衡量指標） 下不會 被提取。 於評估任何營運公司內的存款可否被視為核心存款時，有三方面考慮： ‧ 價格：定價大幅高於市場或基準利率的任何存款，一般會被完全視為非核心存款； ‧ 規模：資金總額超過若干限額的存戶不會計算在內。限額經考慮業務類別及內在流動資 金風險分類方法而設定；及 ‧ 業務類別：經過價格及規模的考慮後，餘下的任何存款成分會按存款涉及的業務類別評 估。根據此方面的考慮可被視為核心存款的任何客戶存款所佔比例為35%至90%。 回購交易及銀行存款不得分類為核心存款。 貸款對核心資金比率 核心客戶存款為向客戶提供貸款的重要資金來源，並可減低對短期批發資金的依賴。營運 公司會受若干限額規限，在核心客戶存款或剩餘期限超過一年的長期債務資金未有相應增 長下，不得向客戶增加貸款。此措施一般稱為 「貸款對核心資金」 比率。 若屬最主要的營運公司，貸款對核心資金比率的限額由風險管理會議設定，若屬較小型的 營運公司，則由地區資產負債管理委員會設定，並由資產、負債及資本管理部團隊監察。該 比率為客戶貸款佔核心客戶存款及剩餘期限超過一年的有期資金兩者總和之比率。一般而 言，客戶貸款乃假設會續期，並計作比率的分子，而不考慮合約到期日。反向回購安排並不 計入貸款對核心資金比率之內。 壓力下之償債保障比率 壓力下之償債保障比率源自受壓現金流境況分析，並以一個月及三個月內受壓現金流入量 佔受壓現金流出量的比率呈列。 受壓現金流入量包括： ‧ 預期從變現流動資產所得 （扣除假設的扣減後） 的流入量；及 ‧ 並未列為動用流動資產的即將到期資產所產生的約定現金流入量。 為配合貸款對核心資金比率的計算方法，一般假設客戶貸款 （不論合約到期日） 在壓力境況 下不會產生任何現金流入量，故不計入壓力下之償債保障比率的分子。 壓力下之償債保障比率如在100%或以上，反映受監察的壓力境況下累計了正數現金流量。 集團旗下營運公司在結合整體市場及滙豐特定情況的壓力境況 （由有關營運公司的內在流 動資金風險分類方法界定） 下，均須能將該比率維持於100%或以上達三個月。 資產、負債及資本管理部團隊會監察營運公司遵守限額的情況，並每月就主要營運公司的 情況向風險管理會議報告，而較小型營運公司的情況則向地區資產負債管理委員會報告。 壓力境況分析 （未經審核） 我們利用數個集團標準壓力境況，以模擬： ‧ 結合整體市場及滙豐特定情況的流動資金危機境況；及 ‧ 整體市場流動資金危機境況。 所有營運公司均會模擬上述境況。作為流動資金及資金風險承受水平審批程序的一部分， 各境況所用假設的適切性會由資產、負債及資本管理部定期檢討，並由風險管理會議及董 事會每年正式批准。 我們將一套標準的指定壓力假設應用於集團的現金流模型，從而釐定受壓現金流出量。我 們的架構規定須運用兩種整體市場境況，以及嚴重程度不斷增加的其他兩種結合整體市場 與滙豐特定情況的壓力境況。除標準的壓力境況外，個別營運公司亦須制訂本身的境況， 以反映所屬市場的具體情況、產品及資金基礎。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 217 較諸整體市場境況，該兩種結合整體市場與滙豐特定情況的境況所模擬的境況更為嚴峻。 由營運公司管理有關結合整體市場與滙豐特定情況的壓力境況，是以其內在流動資金風險 分類方法為基礎。該兩種結合整體市場與滙豐特定情況的壓力境況所包含的主要假設現概 述如下： ‧ 設定非核心存款於三個月內被全數提取 （一個月內被提取80%） ，非核心存款水平會視乎 營運公司的內在流動資金風險分類方法而定； ‧ 在有關境況期間不能取得銀行同業資金，以及無抵押有期債務市場的資金； ‧ 透過流通性不足的資產組合 （證券化及有抵押借貸） 取得資金的能力，只限於過去六個月 發行額或未來六個月預計發行額 （以兩者中的較低者為準） 的25%至75%。有關限制是基於 當前市場情況而制訂，並視乎營運公司的內在流動資金風險分類方法而定； ‧ 在有關境況期間，不能就集團流動資產政策下並未歸類為流動的任何資產，取得回購資 金； ‧ 根據貸款承諾取用貸款須與被模擬的市場壓力嚴重程度相符，並視乎有關營運公司的內 在流動資金風險分類方法而定； ‧ 經界定的長期評級下調觸發流出量。我們就反映此情況的適當等級數目持續進行評估； ‧ 假設客戶貸款於合約期滿時予以續期； ‧ 假設銀行同業貸款及反向回購按合約縮減；及 ‧ 經作出最多20%的界定受壓扣減後，假設界定為流動資產的資產在合約期滿前變現。 滙豐旗下主要營運公司的流動資產 （經審核） 壓力境況分析及償債保障比率的分子包括：經應用適當受壓扣減後變現流動資產所得假定 現金流入量。這些假設乃基於管理層對資產何時被視為可變現的預期而作出。 流動資產指符合集團流動資產定義的無產權負擔資產，且為直接持有，或因剩餘合約期限 超過受監察壓力下之償債保障比率覆蓋時間之反向回購交易而持有。因合約期限不超過受 監察壓力下之償債保障比率覆蓋時間之反向回購交易而持有的任何無產權負擔資產，均不 計入流動資產中，而會列作約定現金流入。 我們的架構將資產類別定義為可在各地評估為高質素及能夠於一個月及一個月至三個月期 內變現。各地的資產負債管理委員會須確認，可根據集團流動資產政策被視為流動的任何 資產，將於受管理的壓力境況下保持為流動資產。 來自動用一個月內流動資產的流入量，計算基礎通常僅限於經確定可提取中央銀行存款或 出售或回購的政府及類似政府貸款 （通常限於以主權實體本土貨幣計值的貸款） 。高質素資 產抵押證券 （主要為美國按揭抵押證券） 及備兌債券亦包括在內，但假設由這些資產所產生 的流入量設有上限。 就大多數流通指數內的高質素非金融及非結構企業債券及股票而言，一個月後的流入量亦 會計及。 內部分類方法 確認現金流入量 資產類別 第一級 一個月內 ‧ 中央政府 ‧ 中央銀行 （包括經確認的可提取儲備） ‧ 超國家金融機構 ‧ 多邊發展銀行 ‧ 硬幣及鈔票 第二級 一個月內但設有上限 ‧ 地方及地區政府 ‧ 公營單位 ‧ 有抵押備兌債券及轉手資產抵押證券 ‧ 黃金 第三級 一個月至三個月 ‧ 無抵押非金融公司證券 ‧ 在認可交易所上市及計入流通指數內的股票 由中央或地方╱區域政府擁有及控制但未獲明確擔保的實體被視為公營單位。 任何獲明確擔保的風險承擔會反映為最終擔保人的風險承擔。 董事會報告：風險 （續） 滙豐控股有限公司 218 風險附錄－政策及慣例 在確定流動資產屬高質素所用準則方面，集團的流動資產政策包括下列額外準則： 1. 以相關主權實體本土貨幣計值並於境內的本土銀行系統持有的中央銀行及中央政府風 險承擔符合條件列為第一級流動資產。 2. 以相關主權實體本土貨幣計值並於境外持有的中央銀行及中央政府風險承擔必須根據 巴塞爾標準風險加權計算法按20%或以下計算風險加權值，方符合條件列為第一級流動 資產。 3. 以相關主權實體以外的貨幣 （即外幣） 計值之中央銀行及中央政府風險承擔必須根據巴 塞爾標準風險加權計算法按20%或以下計算風險加權值，並僅以有限數量的主要貨幣發 行，方符合條件列為第一級流動資產。 至於使用歐元作為本土貨幣的歐元區國家，其處理方式會視乎風險承擔乃於境內的本土 銀行系統持有或於境外持有而定。於境內的本土銀行系統持有的中央銀行及中央政府風 險承擔根據第1項準則符合條件列為第一級流動資產，但於境外持有的中央銀行及中央 政府風險承擔則根據第3項準則被視為以外幣計值。 4. 於境內持有及被地方監管機構認為風險等同中央政府風險承擔的地方╱地區政府風險承 擔，可被視為中央政府風險承擔。 5. 超國家金融機構及多邊發展銀行必須根據巴塞爾標準風險加權計算法按0%計算風險加 權值，方符合條件列為第一級流動資產。 6. 為符合條件列為第二級流動資產，相關風險承擔必須根據巴塞爾標準風險加權計算法按 20%或以下計算風險加權值。 7. 為符合條件列為第三級流動資產，無抵押非金融企業債務的風險承擔必須符合最低的內 部評級規定。 如其他資產在壓力下仍按實況評為流動資產，有關營運公司可按個別情況將該等資產視為 流動資產。該等流動資產會獨立於第一級、第二級及第三級流動資產，而呈報為 「其他」 類 別。 銀行同業及集團內部貸款及存款產生之現金流淨額 根據流動資金及資金風險管理架構，銀行同業及集團內部貸款及存款產生的三個月內之現 金流入淨額，將令流動資產的需求減少。相反，銀行同業及集團內部貸款及存款產生的三 個月內之現金流出淨額，將令流動資產的需求增加。 反向回購、回購、借入股票、借出股票及直接短倉 （包括集團內部） 產生之現金流淨額 現金流入淨額指除流動資產外的其他流動資源，原因是流動資產並不反映由於剩餘合約期 限不超過壓力下之償債保障比率覆蓋時間的反向回購交易而持有之任何無產權負擔資產。 現金流出淨額的影響，取決於在回購到期予以解除時，具產權負擔的相關抵押品是否合資 格列為流動資產。集團大部分回購交易均以流動資產作抵押，因此所列示的現金流出淨額 會因歸還流動資產而被抵銷，故此不包括在以上流動資產列表中。 監察批發債務 （未經審核） 倘為籌集資金而涉足有期債務批發市場，資產負債管理委員會須設定滾動3個月及12個月債 務期限累計限額，以確保到期日不會集中在這些時間範圍內。 流動資金行為化 （未經審核） 流動資金行為化用於反映我們對下列時間的評估：有信心 （即使在嚴峻的流動資金壓力境況 下） 取得負債的預計時間，以及必須假設我們需為集團資產提供資金的預計時間。當合約條 款並不反映預期行為時，便會採用行為化方法。流動資金行為化由各地資產負債管理委員 會遵照風險管理會議制訂的政策進行檢討及審批。我們的流動資金風險管理方法通常指運 用不同的方法管理資產與負債，例如管理層會假定較短的負債年期，以及為資產訂定較長 期的資金需求。集團核心╱非核心及貸款對核心資金架構下的一切核心存款，均假設有多 於一年的流動資金行為化年期，並屬同一核心資金來源。資產行為化的程度更為精細，而 且務求區分我們必須假設將有需要為資產撥資的期間。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 219 資金轉移定價 （未經審核） 我們的資金轉移定價政策設有兩個階段的資金轉移定價方法，反映我們於不同假設下獨立 管理利率風險與流動資金及資金風險。資金轉移定價方法是根據風險管理架構制訂。各營 運公司均須採用集團的轉移定價政策架構，為各種主要貨幣釐定最合適的利率風險轉移定 價曲線、流動資金溢價曲線 （即利率風險轉移定價曲線之上的息差） ，以及流動資金分攤成 本評估 （即利率風險轉移定價曲線之下或之上的息差） 。 利率風險轉移定價政策務求確保因非交易用途 （銀行賬項） 資產及負債結構產生的所有市場 利率風險 （可經外部市場消除或經內部對銷轉移消除） 轉移至資產負債管理業務，以便作為 非交易市場風險集中管理。各營運公司會就各主要貨幣採用單一利率風險轉移定價曲線。 就此目的使用的轉移定價曲線反映各營運公司的資產負債管理業務如何於轉移時最有效地 於市場消除利率風險。如基差風險可於外部資產或外部負債的重新定價基準與利率風險轉 移定價曲線的重新定價基準兩者之間識別，此基差風險可轉移至資產負債管理業務，惟基 差風險必須可於市場消除。 流動資金及資金風險的轉移定價獨立於利率風險，因為營運公司的流動資金及資金風險會 轉移至資產負債管理委員會以便集中管理。資產負債管理委員會負責監察及管理貸款對核 心資金比率，並指派資產負債管理業務管理流動資產組合及執行批發有期債務撥資計劃， 從而幫助資產負債管理委員會確保集團的壓力下之償債保障比率維持於100%以上達三個月。 流動資金及資金風險轉移定價包括兩個部分： ‧ 流動資金分攤成本：持有基準流動資產以應對受壓現金流出的成本 （轉移價格的孳息） 。 基準流動資產由資產負債管理委員會決定，並以投資於第一級流動資產從而可達致目標 的加權平均期限 （剩餘期限最多一年） 為基礎。 ‧ 流動資金溢價：支付有期債務及核心存款的有期資金之經評估成本╱價值 （轉移價格的 孳息） 。 持有流動資產的經評估成本被分配至採用集團內部壓力下之償債保障比率架構模擬得出的 流出金額。 流動資金溢價根據資產的經評估行為化流動資金年期，扣取自影響集團壓力下之三個月償 債保障比率的任何資產；如任何資產會影響集團的貸款對核心資金數值，則須具有至少一 年的最低行為化年期；而現行的流動資金溢價曲線比率由資產負債管理委員會釐定，並根 據集團的修正準則修正。因此，在扣除任何有期資金的成本後，核心存款會平均分攤從存 款所支持的資產扣取的流動資金溢價。 回購及借出股票 環球銀行及資本市場業務向客戶提供有抵押證券融資服務，為客戶提供現金融資或特定證 券。如以證券作抵押品而向客戶提供現金，則所提供的現金在資產負債表中確認為反向回 購。如以現金作抵押品而向客戶提供證券，則所收取的現金在資產負債表中確認為回購， 或倘證券屬股權證券，則確認為借出股票。 各營運公司以中央抵押品組合形式，根據流動資金及資金風險管理架構管理抵押品。倘需 要交付特定證券而有關公司並未能在中央抵押品組合中即時取得有關證券，則會以抵押基 準借入證券。當以現金作抵押品借入證券，所提供的現金在資產負債表中確認為反向回購， 或倘證券屬股權證券，則確認為借入股票。 營運公司亦可利用中央抵押品組合內的證券，以證券作抵押而借入現金。回購及借出股票 可透過此方法為資本市場業務直接擁有證券所產生的現金需求提供資金，以便客戶從事其 業務，亦可滿足為客戶的證券活動提供融資因而產生的現金淨額需求。 倘符合IFRS的對銷準則，則反向回購、借入股票、回購及借出股票以淨額呈報。在若干情況 下，借入或借出證券的交易利用證券作抵押品。這些交易歸類為資產負債表外交易。 任何接納為反向回購或借入股票交易的抵押品的證券，必須具備非常高的質素，而其價值 須作適當扣減。根據反向回購或借入股票交易而借入的證券，於交易期間僅確認為流動資 產緩衝的一部分，並僅於所收取的證券符合流動資金及資金風險管理架構下的流動資產政 策所訂條件始行確認。 我們設有信貸監控措施，以確保所收取的任何抵押品之公允值，相對於因進行抵押而提供 的現金或證券之公允值而言乃屬適當。 董事會報告：風險 （續） 滙豐控股有限公司 220 風險附錄－政策及慣例 積極管理抵押品的影響 抵押品由營運公司管理，與營運公司管理流動資金及資金的方針一致。各營運公司持有的 可供動用抵押品以單一抵押品組合形式管理。各營運公司決定質押哪項抵押品時，會在流 動資金及資金風險管理架構的限制範圍內，盡量善用可供動用的抵押品組合，而不會顧慮 所質押的抵押品是否在資產負債表內確認，或是否就反向回購、借入股票或衍生工具交易 而收取。 以此形式管理抵押品影響我們呈列資產的產權負擔狀況，即使我們並非尋求為已質押的資 產負債表內項目直接融資，但我們仍可以資產負債表內的項目作產權負擔，而同時維持可 供動用無產權負擔的資產負債表外項目。 於量化可轉讓證券的產權負擔水平時，產權負擔乃按個別擔保基準作分析。倘某項擔保具 產權負擔，而我們有權再質押在資產負債表內及資產負債表外所持有的該擔保，則就此項 披露而言，我們假設資產負債表外持有來自第三方的該擔保，在資產負債表內持有的擔保 前已設立產權負擔。 舉例而言，倘我們因反向回購╱借入股票交易而收取特定擔保，但以質押一般抵押品組合 而提供借出的現金，則即使所收取的擔保合資格用於所質押的抵押品組合，亦會產生資產 負債表內具產權負擔及資產負債表外無產權負擔的資產。此外，倘我們因反向回購交易而 收取一般抵押品組合，但以質押特定證券而提供借出的現金，則即使所質押的證券合資格 用於抵押品組合，上述情況同樣會發生。 具產權負擔及無產權負擔資產 「資產負債表內具產權負擔及無產權負擔資產分析」 列表內各類別的定義： ‧ 具產權負擔資產為在資產負債表內已用作某項現有負債的抵押品予以質押的資產，因而有關資產不能 提供予銀行以擔保融資、應付抵押品需要或出售以減少日後潛在資金需求。 ‧ 無產權負擔－可隨時變現資產為銀行視為在其日常業務中可隨時變現的資產，該等資產可用以擔保融 資、應付抵押品需要或出售以減少日後潛在資金需求，且就上述用途不受任何限制。 ‧ 無產權負擔－其他可變現資產為在擔保融資、應付抵押品需要或出售以減少日後潛在資金需求方面並 無限制的資產，但這些資產在現行形式下，不能在日常業務中隨時變現。 ‧ 無產權負擔－反向回購╱借入股票應收賬款及衍生工具資產為特別與反向回購、借入股票及衍生工具交 易相關的資產。該等資產獨立列示，乃因該等資產負債表內資產不能質押，但往往產生不在資產負債表 內確認的非現金資產，可額外用以籌集有抵押融資、應付額外抵押品要求或出售。 ‧ 無產權負擔－不能作為抵押品予以質押為未經質押的資產為並無質押及我們評為不能質押的資產，故 此不能用以擔保融資、應付抵押品需要或出售以減少日後潛在資金需求，例如集團保險附屬公司所持 有，以支持投保人負債並支持該等公司償債能力的資產。 過往，集團概不確認資產的或有流動性價值，惟根據流動資金及資金風險管理架構被界定為流動資產的資 產則除外，而就任何其他受壓的可轉讓工具而言，即使其現為可變現，但我們假設此等工具將於三個月後 方可變現。採納此方針主要因為我們的承受風險水平不會依賴中央銀行。在少數情況下，我們已確認獨立 的資產組合的或有價值，但所涉及金額屬微不足道。因此，我們將大部分客戶貸款及同業貸款呈報為 「其他 可變現資產」 ，原因是管理層須採取其他措施，以令有關資產可轉讓及可隨時變現。 其他資料 財務報表附註19所呈報有關為擔保負債而質押的資產的金額，可能高於第172頁列表中呈列 為具產權負擔的資產的賬面值。產生有關差異的例子包括： ‧ 資產抵押證券及備兌債券，其發行負債加上所需的強制性超額抵押的金額，低於向組合 質押的資產的賬面值。有關差異在上表呈列為 「無產權負擔 － 可隨時變現資產」 ； ‧ 由託管人或結算代理持有的可轉讓證券，以其所持有全部證券作為同日結算負債的浮動 押記，在我們於業績報告日期欠負託管人或結算代理的負債才呈報為具產權負擔，而其 餘則呈報為 「無產權負擔－可隨時變現資產」 ；及 ‧ 預先存於中央銀行或政府機構的資產，在我們以有關抵押品取得資金的情況下才呈報為 具產權負擔。未動用預先存放的抵押品則呈報為 「無產權負擔－可隨時變現資產」 。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 221 資產負債表中所反映就現有負債質押作抵押品或借出的證券，在交易期間反映為具產權負 擔。倘證券以抵押品形式收取或屬借入，以及我們有權出售或再質押此等證券，則於有關 交易期間反映為可供動用及無產權負擔，但已再質押或已出售的證券則除外。有關來自回 購及借出股票的具產權負擔證券，以及因反向回購及借入股票而產生的可動用無產權負擔 資產的進一步分析，載於第171頁題為 「具產權負擔及無產權負擔資產」 一節內。 在日常業務中，我們不會透過不屬於上述環球資本市場業務的有抵押證券融資活動，利用 回購融資作為客戶資產融資的資金來源。 反向回購、借入股票、回購及借出股票的原合約期限屬短期，而且絕大部分交易為期不足 90日。 跨貨幣流動資金及資金風險管理 （未經審核） 我們的流動資金及資金風險管理架構亦會考慮到，當利用某一貨幣的增值來補足另一貨幣 的減值時 （例如利用外幣掉期市場） ，一家公司能否在壓力下繼續涉足外匯市場。在適當情 況下，營運公司亦須監察非本地貨幣在壓力下之償債保障比率及貸款對核心資金比率。 滙豐控股 （經審核） 滙豐控股的主要現金來源是從附屬公司收取的股息、集團內部貸款的利息和還款，及以本 身流動資金賺取的利息。滙豐控股亦透過發行後償及優先債務，從債務資本市場籌集額外 資金。現金主要用作向附屬公司提供資本、向債務持有人支付利息，以及向股東派息。 鑑於滙豐控股已簽發貸款及其他信貸相關承諾與擔保及同類合約，所以亦須承擔或有流動 資金風險。集團僅於仔細考慮滙豐控股為該等承諾及擔保提供融資的能力及出現融資需要 的可能性後，才發出該等承諾及擔保。 滙豐控股積極管理來自其附屬公司之現金流，以善用於控股公司層面持有之現金額。附屬 公司向滙豐控股支付股息或貸出款項的能力，取決於多項因素，包括所屬地區的監管規定 資本水平及銀行業規定、法定儲備，以及財務及營運表現。於2014及2013年，集團的附屬公 司概無於支付股息或償還貸款方面經歷任何重大限制。此外，附屬公司並未預見在支付股 息或償還貸款方面會有任何限制。毋須按監管規定綜合入賬的附屬公司並無低於最低監管 規定水平的資本來源。 市場風險 市場風險是指匯率及商品價格、利率、信貸息差及股價等市場因素出現變動，可能導致收 益或組合價值減少之風險。 市場風險 （包括圖表） 載於由第175頁起的市場風險一節內。 市場風險 （未經審核） 市場風險分為兩個組合： ‧ 交易用途組合包括因進行市場莊家活動而持有及代客戶保管的持倉。 ‧ 非交易用途組合包括主要因進行零售銀行及工商金融業務資產與負債利率管理而產生 的持倉、指定列為可供出售及持至到期日之金融投資，以及來自保險業務的風險項目。 在適用情況下，滙豐將類似風險管理政策及計量技術應用於交易用途及非交易用途組合。 我們的目標在於管理及控制市場風險，以取得理想的風險回報，同時使集團在相關市場中， 維持全球最大規模銀行及金融服務機構之一的地位。 集團推行的各種對沖及減低風險策略，性質等同每個業務所在司法管轄區的可用市場風險 管理工具。該等策略包括使用傳統市場工具 （例如利率掉期） ，以至較複雜的對沖策略，以應 付組合層面產生的多種風險因素。 董事會報告：風險 （續） 滙豐控股有限公司 222 風險附錄－政策及慣例 環球業務市場風險概覽 （未經審核） 下圖說明主要業務範疇的交易及非交易賬項市場風險，以及用以監察及限制風險的市場風 險計量。 風險類別 交易賬項風險 非交易賬項風險 － 外匯及商品 － 結構性匯兌 － 利率 － 利率1 － 信貸息差 － 信貸息差 － 股權 環球業務 環球銀行及資本市場 （包括資產負債管理 （ 「BSM」 ） ） 環球銀行 及資本市 場 （包括 BSM） 環球私人 銀行 工商金融 零售銀行 及財富管 理 風險計量 估計虧損風險 敏感度 壓力測試 估計虧損風險 敏感度 壓力測試 1 滙豐控股發行的定息證券的利率風險並無計入集團估計虧損風險內。此項風險的管理於第181頁載述。 市場風險管治 （經審核） 我們透過集團管理委員會的風險管理會議批予滙豐控股及集團各個環球業務的限額，管理 及監控市場風險。有關限額分配予各業務部門，以及集團的法律實體。 市場風險管理工作主要由環球資本市場業務執行， 當中涉及滙豐 （不包括保險業務） 77%的估計虧損風 險總額及絕大部分交易賬項估計虧損風險，而執 行時採用的風險限額則由集團管理委員會核准。 我們會為組合、產品及風險類別設定估計虧損風 險限額，而且在設定限額水平時，會以市場流動 資金的狀況為主要考慮因素。 集團風險管理部為集團總管理處轄下一個獨立部 門，負責制訂集團的市場風險管理政策及衡量方 法。各主要營運公司均有獨立的市場風險管理及 監控部門，負責根據集團風險管理部制訂的政策， 衡量市場風險，並按規定的限額每日監察及匯報 該等風險。市場風險限額根據左圖說明的框架管 治。 每家營運公司須評估其業務中每項產品產生的市 場風險，並將風險轉移至其業務所在地之資本市 場部門以便管理，或撥入由當地資產負債管理委 員會監督的獨立賬目內加以管理。 此項安排旨在確保所有市場風險統一由具備所需 技巧、工具、管理及管治能力的部門，進行風險管 理。在若干情況下，若市場風險無法全面轉移，我 們會因應任何剩餘風險持倉，識別不同境況對估 值或淨利息收益產生的影響。剩餘風險控制及管 理程序的進一步詳情載於第224頁。 模型風險由模型監察委員會於區域及環球批發信貸及市場風險管理部層面管治，該等委員 會負責直接監察和批准風險計量及管理乃至壓力測試中使用的所有交易風險模型。模型監 察委員會優先制訂集團內交易風險管理所用的模型、方法及實務，並確保此等模型、方法 及實務保持在承受風險水平及業務計劃的範圍內。環球資本市場業務旗下的模型監察委員 會向集團模型監察委員會匯報，而後者則負責監察集團層面的所有模型風險類別。集團模 型監察委員會至少每半年一次將重大事宜通報集團風險管理會議。按照監管規則，風險管 理會議屬集團的 「指定委員會」 ，並已指派環球資本市場業務旗下的模型監察委員會負責所 有交易風險模型的日常管治工作。 主席╱行政總裁 集團管理委員會的風險管理會議 滙豐控股董事會 實體風險管理委員會 主要辦事處經理 業務部門╱業務組╱交易員 集團交易風險管理組 一般措施 特別措施 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 223 我們根據以下政策監控交易用途組合及非交易用途組合的市場風險：由集團風險管理部為 每個業務所在地制訂一份准予交易工具清單，規限每個業務部門的交易僅限於清單上的產 品；執行新產品批核程序；限制較複雜的衍生工具產品交易只可由具備適當產品專業知識 及健全監控系統的辦事處來執行。 市場風險計量 （經審核） 監察及限制市場風險 我們的目標，是管理及監控市場風險，同時保持相關市場的狀況與我們的承受風險水平相 符。 我們利用多種工具監察及限制市場風險，包括敏感度分析、估計虧損風險及壓力測試。 敏感度分析 （未經審核） 敏感度分析計量個別市場因素 （包括利率、匯率及股價） 變動對特定工具或組合的影響，例 如孳息變動1個基點的影響。敏感度衡量方法乃用於監察各個風險類別的市場風險狀況。滙 豐為各組合、產品及風險類別設定敏感度限額，而市場深度是釐定限額水平的重要因素之 一。 估計虧損風險 （經審核） 估計虧損風險 （ 「VaR」 ） 是一種估算方法，用以衡量於指定期間和既定可信程度下，市場利率 和價格的變動引致風險持倉產生的潛在虧損。估計虧損風險的運用融入市場風險管理之中， 我們會為所有交易用途持倉計算估計虧損風險，而不需顧慮我們如何將該等風險資本化。 若沒有認可內部模型，我們會運用各地適當的規則將風險承擔資本化。 此外，我們會為非交易用途組合計算估計虧損風險，以掌握全面的風險狀況。我們的模型 主要以模擬過往經驗為基準。一日持倉期的估計虧損風險乃按99%的可信程度計算。倘我們 並未明確計算估計虧損風險，則會使用下文壓力測試一節所載的市場風險壓力測試表內概 述的其他工具。 我們的估計虧損風險模型利用過往錄得的一系列市場利率及價格，引伸出日後可能出現的 境況，在過程中會考慮不同市場與比率之間的相互關係，如利率與匯率之間的關係。該等 模型亦會計入期權特性對有關風險帶來的影響。 我們採用的歷史模擬模型具有以下特點： ‧ 過往市場利率及價格乃參考匯率及商品價格、利率、股價及相關波幅計算； ‧ 估計虧損風險使用的潛在市場變動乃參考過往兩年的數據計算；及 ‧ 估計虧損風險按99%的可信程度及使用一日持倉期計算。 估計虧損風險模型的性質，意味著相關持倉不變時，觀察所得市場波幅增加將導致估計虧 損風險增加。 滙豐致力持續發展內部風險模型。 估計虧損風險模型的局限 雖然估計虧損風險是衡量風險的一項重要指標，但應留意這些數字有一定的局限。例如： ‧ 採用過往數據作為估計未來事件的準則，未必可以顧及所有可能出現的情況，特別是一 些極端情況； ‧ 持倉期的計算方法，乃假設所有持倉均可以在該段期間套現，或所有風險均可以在該段 期間對沖。這項假設或許未能充分反映當市場流通性極低時，可能因未及在持倉期內全 面套現或對沖所有持倉而產生的市場風險； ‧ 當採用99%的可信程度時，即表示不會考慮在此可信程度以外或會出現的虧損； ‧ 估計虧損風險是以營業時間結束時的未平倉風險作計算基準，因此不一定反映同日內各 種風險；及 ‧ 估計虧損風險不大可能反映只在市場大幅波動時才會出現的潛在虧損。 董事會報告：風險 （續） 滙豐控股有限公司 224 風險附錄－政策及慣例 估計虧損以外風險架構 （未經審核） 集團的估計虧損風險模型旨在反映重要的基差風險，如信貸違責掉期對債券、資產掉期息 差及跨貨幣基準。估計虧損風險未能全面涵蓋的其他基差風險，如倫敦銀行同業拆息期限 基準，由我們的估計虧損以外風險 （ 「RNIV」 ） 之計算補充，並納入資本架構內。 因此，估計虧損以外風險架構旨在反映及運用估計虧損風險模型未能充分涵蓋的主要市場 風險。其中一個例子就是非主要貨幣的倫敦銀行同業拆息隔夜指數掉期基差風險。在此等 情況下，估計虧損以外風險架構利用壓力測試量化資本規定。於2014 年平均計算，來自該 等壓力測試的資本規定佔內部模型為本的市場風險規定資本總額的2.6%。 估計虧損以外風險涵蓋的風險為運用監管機構批准的模型計算的市場風險之風險加權資產 的18%，並包括無法每日觀察的資產類別及產品的相關風險因素所產生的風險，例如股息風 險及相關性風險。 風險因素會定期檢討，並直接計入估計虧損風險模型 （如適用） ，或運用以估計虧損風險為 基準的估計虧損以外風險計算方法或估計虧損以外風險架構內的壓力測試方法量化。各境 況的嚴重程度會按照資本充足程度的規定而修正。以估計虧損風險為基準的估計虧損以外 風險計算結果會納入估計虧損風險的計算以及回溯測試；同時我們亦會就估計虧損風險之 估計虧損以外風險計算方法考慮的風險因素，計算壓力下之估計虧損風險的估計虧損以外 風險。 於2014年，我們按非分散基準修改各風險因素的估計虧損以外風險 （ 「RNIV」 ） 模型，以符合新 的審慎監管局資本指引4實施指引。 第三級資產 交易用途組合內的第三級資產及負債的公允值於第380頁披露，僅佔整體交易用途組合的一 小部分。第三級工具產生的市場風險透過若干市場風險管理方式 （如壓力測試及名義限額） 管理。第384頁的列表顯示第三級金融工具的變動。 回溯測試 我們透過將已結清及假設的利潤及虧損結果，與相應的估計虧損風險數值比較而進行回溯 測試，定期驗證估計虧損風險模型的準確性。假設利潤及虧損不包括非以模型計算之項目， 如同日交易費用、佣金及收入等。 我們預期於一年期內，在99%可信程度下，平均會有兩至三次利潤及兩至三次虧損超出估計 虧損風險。因此，這段期間利潤或虧損超出估計虧損風險的實際次數，可用作衡量該等模 型的效用。為確保計算風險值時以審慎為本，必須留意超出估計虧損風險的利潤僅於回溯 測試模型準確性時才會考慮，而不會用以計算用作風險管理或資本目的之估計虧損風險數 值。 我們會在滙豐作為完整法律實體的不同層面，包括就監管目的而在各地不容許使用估計虧 損風險的集團旗下公司，回溯測試集團的估計虧損風險。 壓力測試 （未經審核） 壓力測試乃融入市場風險管理工具的一項重要工具，用以評定倘若出現較為極端但有可能 發生的事件，或一系列金融變數產生較為極端但有可能出現的變動時，對組合價值的潛在 影響。在此等不尋常情況下，虧損可能較採用估計虧損風險模型所預測者為高。 壓力測試會於法律實體、地區及整體集團層面執行。集團營運的所有地區會持續使用一系 列標準境況。這些境況均經精心設計，以反映各層面的相關事件或市場變動。集團的潛在 壓力虧損風險承受水平按轉介限額釐定及監察。 市場風險壓力測試 敏感度 技術 假設 過往 反向壓力測試 單一風險因素的影 響，例如聯繫匯率 脫鈎 每項風險因素的最大 變動的影響，但不會 考慮任何相關的市場 相關性 潛在的宏觀經濟事件 的影響，例如中國內 地經濟放緩 包含過往對市場變動 的觀察的境況，例如 1987年股市的黑色星 期一事件 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 225 市場風險反向壓力測試乃基於錄得固定虧損的前設進行。壓力測試程序會識別何種境況導 致此項虧損。反向壓力測試的基本原理乃為明暸可能涉及連鎖及系統性影響的正常業務環 境以外的境況。 壓力下之估計虧損風險及壓力測試，連同反向壓力測試及市場缺口風險的管理，使管理層 得悉有關估計虧損風險以外的 「尾端風險」 ，滙豐就此承受的風險有限。 交易用途組合 （經審核） 市場缺口風險 若干產品的結構會增加市場缺口風險，即市場出現缺口事件時產生虧損的風險。市場缺口 事件即市價突然大幅轉變，而隨之並無出現交易機會。舉例說，這種變動可能在發生不利 事件或公布意外消息時出現，市場某部分的變動超出正常的波幅範圍，並一時之間變得不 流通。於2014年，市場缺口風險主要源自不附帶追索權的貸款交易 （主要涉及企業客戶） ，當 中貸款的抵押品限於所提供的股份。於發生市場缺口事件時，股權抵押品的價值可能低於 未償還的貸款金額。 鑑於該等交易別具特性，有關交易難以在估計虧損風險或傳統市場風險敏感度的計量中反 映。我們會利用壓力測試境況掌握該等交易的風險，並持續監察市場缺口風險。於2014年， 我們並未產生任何重大的市場缺口虧損。 脫鈎風險 就若干貨幣而言 （掛鈎貨幣或受限制貨幣） ，現貨匯率會按固定匯率與其他貨幣 （通常為美元 或歐元） 掛鈎，或受限制在接近掛鈎匯率的預定區間內。脫鈎風險是指掛鈎或受限制區間變 動或遭廢除，並轉為浮動制度的風險。 滙豐於處理固定及受限制貨幣制度方面有豐富經驗。透過使用在壓力境況下的現貨匯率， 我們可分析脫鈎事件對滙豐持倉的影響程度。我們會監察掛鈎或受限制貨幣的有關境況， 如港元、人民幣、中東貨幣及瑞士法郎 （於2014年內設下兌歐元的升值上限） 的情況，以及限 制任何可能出現的虧損，補足了例如過往估計虧損風險等傳統市場風險的衡量指標，而傳 統市場風險的衡量指標或不足以全面反映掛鈎或受限制貨幣持倉的風險。過往估計虧損風 險依賴過往事件以釐定潛在損益的可能性。然而，掛鈎或受限制貨幣在過往考慮的時間框 架內，可能未曾經歷過脫鈎事件。 資產抵押證券╱按揭抵押證券風險 交易用途組合內的資產抵押證券╱按揭抵押證券風險，乃透過敏感度及估計虧損風險限額 管理，詳情載於第176頁，並載於上文的壓力測試境況評述內。 非交易用途組合 （經審核） 集團的大部分非交易賬項估計虧損風險與資產負債管理業務 （ 「BSM」 ） 或各地財資管理部門 有關。集團非交易賬項估計虧損風險主要來自所有環球業務的利率及信貸息差風險。非交 易用途組合內並無商品市場風險。 非交易賬項估計虧損風險亦包括環球業務所持有並轉移至資產負債管理業務或各地財資部 門管理的組合之非交易用途金融工具的利率風險。我們在計量、監察及管理非交易用途組 合風險時，估計虧損風險只是所用的其中一項工具。管理銀行賬項利率風險的詳情，包括 資產負債管理業務的角色，載於下文 「非交易賬項之利率風險」 內。 非交易賬項估計虧損風險不包括可供出售證券的股權風險、結構匯兌風險以及滙豐控股所 發行的定息證券之利率風險，有關風險的範圍及管理詳情載於下文有關章節。 非交易用途組合之市場風險監控方法，主要為將資產負債管理業務或環球資本市場業務以 外的非交易用途資產及負債的經評估市場風險轉移至資產負債管理業務管理的賬項，惟市 場風險必須予以消除。風險淨額一般由資產負債管理業務透過採用定息政府債券 （可供出售 賬項內持有的流動資產） 及利率掉期管理。可供出售組合內持有的定息政府債券的利率風險 於集團的非交易賬項估計虧損風險內反映。資產負債管理業務使用的利率掉期一般分類為 公允值對沖或現金流對沖，並計入集團的非交易賬項估計虧損風險。任何未能於市場消除 的市場風險，由各地資產負債管理委員會於獨立的資產負債管理委員會賬項內管理。 董事會報告：風險 （續） 滙豐控股有限公司 226 風險附錄－政策及慣例 可供出售債務工具的信貸息差風險 與信貸息差變動有關的風險，主要透過敏感度限額、壓力測試及估計虧損風險加以管理。 估計虧損風險顯示在兩年期內信貸息差的單日變動對收益的影響，此乃按99%的可信程度計 算。 分類為可供出售的股權證券 滙豐會對潛在的新承諾進行風險評估，以確保行業及地域集中程度，在組合內維持於可接 受水平。滙豐會進行定期檢討，以核實組合內各項投資及持作配合業務持續發展用途之投 資的估值，例如持有政府資助企業及各地證券交易所的股權。 結構匯兌風險 （未經審核） 結構匯兌風險乃指於附屬公司、分行及聯營公司之投資淨額，其功能貨幣為美元以外之其 他貨幣。一家公司的功能貨幣是指其業務所在主要經濟環境的貨幣。 結構風險的匯兌差額於 「其他全面收益」 項內確認。我們以美元為綜合財務報表之列賬貨 幣，因為美元及與美元掛鈎的各種貨幣所屬區域，組成我們進行交易及為業務提供資金的 主要貨幣區。因此，美元與相關附屬公司各種非美元功能貨幣之間的任何匯兌差額，均會 對我們的綜合資產負債表造成影響。 我們只會在有限度的情況下對沖結構匯兌風險。滙豐管理結構匯兌風險的主要目的，是盡 可能保障滙豐之綜合資本比率及經營銀行業務的個別附屬公司之資本比率，基本上免受匯 率變動影響。就各附屬銀行而言，達致上述目標的方法通常是確保特定貨幣的結構風險對 運用該貨幣計值的風險加權資產之比率，大致等於該附屬公司的資本比率。 倘我們認為具結構風險的貨幣重估時可能減值，而且實際上可以進行對沖，便有可能進行 這類對沖。對沖方法是採用遠期外匯合約 （根據IFRS，該等合約在賬目中列為海外業務投資 淨額對沖） ，或以等同所涉功能貨幣的貨幣借款為相關項目提供資金。 非交易賬項利率風險 （未經審核） 非交易用途組合中非交易賬項利率風險，主要來自資產日後的收益與其資金成本因利率變 動出現錯配情況。對某些產品範疇的內含期權性風險 （如按揭提前還款） 必須作出假設，以 及對合約列明須即時償付之負債 （如往來賬項） 的經濟存續期必須作出行為方面的假設，加 上管理利率產品的重新定價行為，均令分析此類風險更為複雜。滙豐的利率風險行為化框 架 （載於下文） 包括了上述行為特徵的假設。 我們旨在透過管理非交易用途組合的市場風險，盡量減低未來利率變動可能導致日後淨利 息收益下降的影響，並同時設法平衡有關對沖活動的成本對當前收入來源淨額產生的影響。 對某些產品範疇的內含期權性風險 （如按揭提前還款） 作出假設，令分析此類風險更為複雜。 我們的資金轉移定價政策產生兩個階段的資金轉移定價方式，詳情請參閱第219頁。 利率風險行為化 與按非常極端壓力境況評估的流動資金風險不同，非交易賬項利率風險以 「照常營業」 基準 評估及管理。在不少情況下，並非由資本市場業務或資產負債管理業務產生的資產╱負債 所帶來的非交易用途資產╱負債合約狀況，並未反映所觀察的行為。 因此，我們會採用行為化模式評估非交易用途資產╱負債的市場利率風險，而此項經評估 的市場風險，乃根據利率風險由環球業務轉移至資產負債管理業務之管治規則，轉移至資 產負債管理業務。 行為化適用於以下三大範疇： ‧ 受管理利率結餘的經評估重新定價頻密程度； ‧ 不附息結餘 （一般是資本賬及往來賬） 的經評估期限；及 ‧ 附帶內含期權性風險的定息結餘按基本情況預期的提前還款行為或往後接納比率。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 227 利率行為化政策須根據集團行為化政策而制訂，並最少每年由各地資產負債管理委員會、 區域資產、負債及資本管理部以及集團資產、負債及資本管理部，聯同各地、區域及集團的 市場風險監察團隊審批一次。 結餘可按行為歸納的程度取決於： ‧ 可根據照常營業狀況評估為 「穩定」 的往來結餘額；及 ‧ 就管理利率結餘而言，乃觀察所得過往市場利率的重新定價行為；或 ‧ 就不附息結餘而言，乃預期結餘停留於照常營業狀況的期限。此項評估往往取決於資產 負債管理業務可運用定息政府債券或利率衍生工具消除風險的再投資期限，而就衍生工 具而言，則取決於可運用現金流對沖的能力。 資產負債管理業務 資產負債管理業務的有效管治，有賴推行雙重匯報機制，即須分別向環球銀行及資本市場 業務的行政總裁及集團司庫匯報。於集團旗下各營運公司，資產負債管理業務部門在當地 資產負債管理委員會 （通常每月舉行一次會議） 的監督下，負責管理流動資金及資金，亦在 環球資本市場業務的限額架構下管理獲轉移之非交易賬項利率持倉。 在代表資產負債管理委員會管理流動資金風險及管理獲轉移之非交易賬項利率持倉時，資 產負債管理業務會按集團的流動資產政策投資於高評級的流動資產。大多數流動資金乃投 放於中央銀行存款，以及投資於政府、超國家金融機構及機構證券，其餘則大多用於短期 銀行同業及中央銀行貸款。 可提取的中央銀行存款入賬列作現金結餘。銀行同業貸款、法定中央銀行儲備及提供予中 央銀行的貸款入賬列作同業貸款。資產負債管理業務持有的證券入賬列作可供出售資產， 其次列作持至到期日之資產。 法定中央銀行儲備不會確認為流動資產。可根據集團的受壓客戶存款流出假設而解除之法 定儲備，列作受壓流入量。 資產負債管理業務獲准運用衍生工具，作為管理利率風險權限的一部分。衍生工具活動主 要透過採用現金流對沖關係或公允值對沖關係其中一部分的常規利率掉期進行。 資產負債管理業務的信貸風險主要限於因銀行同業貸款而產生的短期同業風險，以及中央 銀行及優質主權、超國家金融機構或機構的風險承擔 （此等風險承擔構成資產負債管理業務 流動資金組合的一大部分） 。資產負債管理業務並無管理集團旗下公司任何資產負債的結構 信貸風險。 資產負債管理業務獲准進行單一公司及指數的信貸衍生工具活動，惟有關活動的目的，僅 為了在有限情況下管理其證券組合承擔的特定信貸風險。風險限額所受限制極大，並受到 密切監察。於2014年12月31日及2013年12月31日，資產負債管理業務並無承擔未平倉的信貸 衍生工具指數風險。 我們會計算資產負債管理業務所持有交易及非交易用途持倉的估計虧損風險，並應用資本 市場業務所採用的相同計算方法。有關數據會用作市場風險監控工具。 資產負債管理業務僅在極少數情況下，持有交易用途組合工具。於2014及2013年，持倉及相 關估計虧損風險不大。 淨利息收益的敏感度 （未經審核） 我們管理非交易用途組合之市場風險時，主要集中於監察預計淨利息收益在不同利率境況 （模擬模型） 下的敏感度。此項監察由各地的資產負債管理委員會於公司層面進行。 各地的公司會結合採用與當地業務相關的各種境況及假設，以及滙豐各部門須採用的標準 境況。後者更會予以整合，以顯示對滙豐的綜合淨利息收益造成的合併備考影響。 預計淨利息收益敏感度的數字顯示，在各種預計孳息曲線境況及集團當前的利率風險狀況 下，淨利息收益備考變動帶來的影響，但這些影響未計及資產負債管理業務或有關業務部 門內部為減輕此項利率風險的影響而會採取之行動。實際上，資產負債管理業務會積極改 變利率風險狀況，務求盡量減低虧損及提高收入淨額。淨利息收益敏感度的計算假設所有 期限的利率在 「上行」 境況下均以相同幅度變動。預計數值不假設於 「下行」 境況下利率會降 至負數，就若干貨幣而言，此境況實際上可能造成不平衡變動。此外，淨利息收益敏感度的 計算已計及銀行同業拆息與企業在時間及利率變動方面有酌情權之利率兩者之間的預計變 動差異，對淨利息收益的影響。 董事會報告：風險 （續） 滙豐控股有限公司 228 風險附錄－政策及慣例 界定福利退休金計劃 （經審核） 倘若我們各項界定福利退休金計劃無法以可確定現金流的資產悉數配對該等計劃的責任， 便會產生市場風險。隨著長期利率、通脹、薪酬水平及計劃成員壽命長短的變化，退休金計 劃的責任會出現波動。退休金計劃資產包括股票及債務證券，其現金流會因股票價格及利 率 （及信貸風險） 變動而改變。風險源於市場股價及利率的變動，可能導致資產價值 （連同定 期持續供款） 經過一段時間後不足以應付預期的責任，而該等責任可能會因通脹升溫及成員 壽命延長而增加。管理層及 （在若干情況下） 受託人 （代表退休金計劃受益人行事） 採用由外 界獨立顧問編製的報告評估該等風險，並採取行動，及在適當情況下相應調整投資策略與 供款水平。 滙豐控股 （經審核） 作為一家金融服務控股公司，滙豐控股只參與有限度的市場風險活動。這些活動主要涉及 下列幾方面：保持充裕的資本來源以支持集團的多元化業務；將該等資本來源分配到集團 各項業務；從投放於各項業務的投資中賺取股息及利息收益；向滙豐控股的權益股東支付 股息及向提供債務資本的人士╱機構支付利息；及維持供應短期資本來源以供於特殊情況 下運用。滙豐控股本身並無坐盤交易持倉。 滙豐控股面對的主要市場風險為非交易賬項利率風險及匯兌風險。這些風險來自短期現金 結餘、所持資金水平、提供予附屬公司的貸款、於長期金融資產及金融負債的投資 （包括已 發行債務資本） 。滙豐控股管理市場風險的策略目標是減低上述巿場風險，並將資本來源、 現金流及可供分派儲備的波動盡量減低。滙豐控股的市場風險由滙豐控股資產負債管理委 員會根據其承受風險水平聲明監察。 滙豐控股使用利率掉期及跨貨幣利率掉期管理長期債務發行產生的利率風險及匯兌風險。 營運風險 （未經審核） 我們的營運風險管理目的，是按照集團的承受風險水平 （由集團管理委員會界定） ，以具成 本效益的方式管理及監控營運風險，使之控制在目標範圍內。 營運風險管理部為集團風險管理部內一個特定的風險管理組別，並具備正式的管治架構負 責監督其管理工作。環球營運風險管理部向集團風險管理總監匯報，並支援環球營運風險 管理委員會的工作。該委員會負責建立及維持營運風險管理架構、監察營運損失的程度， 以及監控環境的成效；亦負責集團層面的營運風險匯報，包括編製報告以供風險管理會議 及集團風險管理委員會審閱。環球營運風險管理委員會最少每季舉行一次會議，討論主要 的風險問題及檢討營運風險管理架構是否有效執行。 營運風險管理架構就地區、環球業務及環球部門的營運風險管理及內部監控界定最低標準 及程序，以及相關管治架構。營運風險管理架構已載於嚴格標準手冊，從而成為集團的規 範，並輔以詳盡的政策，當中說明我們識別、評估、監察及控制營運風險的方法，以及發現 缺失時採取緩解措施的指引。 集團的業務經理負責按業務的規模及性質，維持可接受的內部監控水平。他們負責識別及 評估風險、制訂監控措施及監察該等措施的成效。營運風險管理架構界定標準的風險評估 方法，以及提供有系統的營運損失數據匯報工具，有助各經理履行上述職責。 我們使用中央資料庫以紀錄營運風險管理程序的結果。業務部門會紀錄及保留對營運風險 及監控的自我評估結果。各業務及部門管理層和業務風險及控制經理會監察各項已紀錄在 案的行動計劃有何進度，以糾正不足之處。為確保營運風險損失在集團層面作出一致匯報 及監察，集團旗下所有公司均須匯報預期損失淨額超過1萬美元的個別損失，以及總計所有 其他低於1萬美元的營運風險損失。虧損會記入集團營運風險資料庫，並每月向風險管理會 議匯報。 進一步詳情請參閱 《2014年第三支柱資料披露》 報告。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 229 合規風險 （未經審核） 合規風險屬營運風險定義範圍內。集團旗下所有公司均須遵守所有相關法律、守則、規則、 法規及良好市場慣例準則的條文和精神。該等規則、法規、準則及集團政策包括遵循反洗 錢、反賄賂及貪污、反資助恐怖主義及武器擴散、制裁規定、經營業務操守、市場操守及其 他金融法規。 兩個分支部門：防範金融犯罪部 （ 「FCC」 ） 及監管合規部 （ 「RC」 ） 由同一合規營運總監、鑑證及 聲譽風險管理團隊適當支援。防範金融犯罪部及監管合規部的環球主管均向集團風險管理 總監匯報。 我們在經營業務的所有國家╱地區及所有環球業務均設有合規團隊。這些合規團隊主要由 位於歐洲、美國、加拿大、拉丁美洲、亞洲和中東及北非的防範金融犯罪部及監管合規部主 管監察。區域及環球業務合規團隊的效率由鑑證團隊評估。 環球政策及程序規定須及時辨識及向防範金融犯罪部或監管合規部報告所有實際或懷疑違 反任何法律、規則、法規、政策或其他相關規定的行為。上報程序輔以一項規定，要求集團 旗下所有公司及部門於年中及年底須提交合規證書，詳列已知的上述任何違反行為。該等 上報及認證程序的內容，將向風險管理會議、集團風險管理委員會及董事會匯報，並在適 當情況下於 《年報及賬目》 和 《中期業績報告》 內披露。 金融系統風險防護委員會負責向董事會報告有關金融犯罪以及金融系統被濫用的事宜，並 就金融犯罪風險提供高瞻遠矚的意見，使集團在合規及操守方面的工作更為目標明確。另 外，行為及價值觀委員會負責向董事會報告有關負責任的業務操守以及奉行滙豐價值觀的 事宜。 於2014年，董事會通過第189頁 「合規風險」 所述的新訂全球加強反洗錢及制裁政策以及全 球一致的行為管理方針。 法律風險 （未經審核） 各法律部門均須制訂符合集團標準的流程及程序以管理法律風險。 法律風險屬營運風險定義範圍內，包括： ‧ 合約風險，即滙豐的成員公司因其身為訂約方的權利及╱或責任在技術上有缺失而蒙受 財務虧損、面臨法律或監管行動或聲譽受損的風險； ‧ 爭議風險，即滙豐的成員公司因不利的爭議情況或未能就集團成員公司面對或提出的法 律申索採取合適的步驟，以辯護、控告及╱或決定採取實際行動或威脅採取行動，以致 蒙受財務虧損或聲譽受損； ‧ 立法風險，即滙豐旗下公司未有遵從其業務所在司法管轄區法律的風險；及 ‧ 非合約權利風險，即集團成員公司的資產並非正當擁有或受保障或遭他人侵犯，或集團 成員公司侵犯他方權利的風險。 我們的環球法律事務部協助管理層監控法律風險。我們在49個業務所在國家╱地區設有法律 事務部。除環球法律事務部外，我們在歐洲、北美洲、拉丁美洲、中東及北非，以及亞洲，均 設有由區域總法律顧問以及負責各項環球業務的環球總法律顧問主管的區域法律事務部門。 環球保安及詐騙風險 （未經審核） 保安及詐騙風險事宜由環球保安及詐騙風險管理部於集團層面管理。這個部門負責處理資 訊、詐騙、或有風險、金融情報、實質風險，以及地緣政治風險等工作，並已全面納入集團 層面的集團風險管理部。此舉有助管理層識別上述及其他非金融風險的連鎖效應，並減低 各項業務所在司法管轄區內不同業務所受的影響。 資訊保安風險管理部負責制訂策略及政策，並據此保障集團的資訊資產及服務免受內部或 外部人士蓄意或無意泄露、破壞或遺失。該部門會就已制訂或建議制訂的資訊保安監控措 施及慣例是否行之有效，向各業務部門提供獨立意見及指引，並且進行獨立監察。 董事會報告：風險 （續） 滙豐控股有限公司 230 風險附錄－政策及慣例 詐騙風險管理部負責確保就一切形式的詐騙活動 （不論由內部或外部產生） 制訂有效的預防、 偵察及調查措施，並支援各個業務部門。為達成有關目標及集合面對威脅時所需的力量， 一切類別詐騙的風險管理工作 （如有關卡的詐騙、與卡無關的詐騙及內部詐騙，包括調查） 均設於一個管理架構內，並為環球風險管理部的其中一個部門。 我們廣泛使用科技防止及偵測詐騙，例如持續監察客戶的信用卡及扣賬卡消費情況、驗證 可疑交易、檢討網上銀行的時段及以類似方式監察交易，以及篩查所有新開戶口的申請， 以防止詐騙。我們設有詐騙系統策略，旨在提供最低限度的標準，以及更簡易地共用偵測 欺詐及減少誤報的最佳運作方式。 我們已訂立一套完整及有效的反詐騙策略，除使用尖端科技外亦包括防止詐騙的政策及措 施、執行穩健的內部監控、負責回覆調查的團隊，以及於適當情況下與執法部門聯絡。 或有風險管理部負責確保集團的關鍵系統、程序及部門在遇上業務嚴重受阻事件時，有能 力恢復運作，繼續營運。 面對上述影響甚廣的風險，持續營運管理部負責預先制訂復元計劃，盡力降低多項實際或 將會出現的風險對主要業務中斷的不利影響 （不論環球、地區或國家的業務） 。預先制訂的 計劃集中保障客戶服務、員工、收入及數據與文件的完整性，使之符合監管規定。 業務部門各有復元計劃，乃於完成業務影響分析後制訂。此舉釐定了業務在業務中斷的情 況下達致不可接受的虧損前可維持多少時間，即臨界點。有關計劃會每年檢討及測試。計 劃會按集團政策及準則進行，各業務部門會於年度合規證書內確認已達到所有有關標準。 如有例外情況，應予以提出，而集團及地區業務持續營運團隊將監察其短期內的解決方法。 計劃必須可以變通，並涵蓋所有風險，尤其是新浮現風險 （如可能出現的流行病及網絡攻 擊） 。我們使用營運風險管理架構計量對以上風險的復元能力，並已向集團及地區風險管理 委員會確認採用有關架構。 復元能力透過不同的降低風險措施管理，包括與資訊科技部門協調可接受的系統復元時間、 確保我們的主要樓宇均有妥善設施以持續營運、要求重要供應商訂立復元計劃，以及為集 團安排合適的保險，保障範圍須涵蓋業務中斷費用。 金融情報組由保安及詐騙風險部及防範金融犯罪部共同管理。該部門會使用先進分析方法 及專門的知識偵察集團客戶及交易對手在金融犯罪方面的跡象。 實質保安部會制訂可行的實質、電子及營運應對措施，確保集團管理的人員、財產及資產 受到保障並免受罪案、盜竊、攻擊及有損滙豐利益的威脅。 地緣政治風險小組會向業務行政人員及高級保安及詐騙風險管理層定期及特別匯報集團業 務所在國家╱地區的地緣政治風險狀況及所形成的威脅，有助加強策略業務規劃，盡早評 估醞釀中的保安風險。我們亦就差旅制訂保安控制措施及指引。 系統風險 （未經審核） 系統風險為支援集團日常業務運作的自動化平台 （應用系統） ，以及自動化平台所在的系統 基礎設施 （數據中心、網絡及分布式電腦） 出現故障或其他缺陷的風險。 滙豐科技及服務部 （ 「HTS」 ） 監督全球系統風險的管理。我們透過每月的風險管治委員會會議 監察系統風險，會議提供現有及新浮現首要風險的概況。 滙豐科技及服務部管理層使用風險及控制評估及境況分析管理系統風險的控制環境，並使 用主要風險指標以確保地區及業務範圍內業務的風險評估可持續進行。我們透過定期測試 相關的主要控制措施監察重大風險。 業務核心服務透過一個環球監察機構統一識別。我們已就各業務編製定量評分紀錄 （稱為承 受風險水平聲明，用於監察表現） 。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 231 滙豐設有服務復元力及系統持續策劃部，確保系統達致協定的目標服務水平，以及在發生 重大破壞事件時，可在業務部門協定的目標時間內復元。 供應商風險管理 （未經審核） 我們的供應商風險管理 （ 「VRM」 ） 計劃為一項環球架構，以管理第三方供應商風險，尤其是我 們依賴外判協議以向客戶提供關鍵服務涉及的風險。供應商風險管理包括識別重大合約及 其主要風險的嚴格程序，以及確保訂立控制措施管理及降低此等風險。我們已實施環球及 區域監管架構，監察第三方服務供應商。 保險業務風險管理 保險產品概覽 （經審核） 滙豐制訂的合約的主要類別如下： ‧ 附有酌情參與條款 （ 「DPF」 ） 的壽險保單； ‧ 信用壽險業務； ‧ 年金； ‧ 定期壽險及危疾保單； ‧ 相連壽險； ‧ 附有酌情參與條款的投資合約； ‧ 單位相連投資合約；及 ‧ 其他投資合約 （包括在香港承保的退休金合約） 。 此外，我們亦從事小量非壽險業務，主要保障個人及商業財產。 風險的性質及程度 （經審核） 我們的保險業務風險大部分來自制訂保險產品的業務，此等風險可分為金融風險及保險風 險。金融風險包括市場風險、信貸風險及流動資金風險。此外，亦會出現營運及可持續發展 風險，並已納入集團各自的整體營運風險管理程序內。 以下章節描述集團如何管理金融風險及保險風險。制訂保險產品附屬公司的資產已納入第 111至203頁綜合披露之風險資料內，惟第190至198頁的 「保險業務風險管理」 一節亦獨立披露 了制訂保險產品附屬公司的資料。 制訂保險產品的公司會制訂監控程序，而有關程序須符合集團保險業務發出的指引及規定， 以及地方監管規定。國家層面的監管由當地風險管理委員會負責。國家風險管理總監直接 向當地保險業務行政總裁匯報，而職能上則向集團保險業務風險管理總監匯報；集團保險 風險管理總監對環球保險業務的風險管理負全責。集團保險業務執行委員會監察環球保險 業務風險監控架構，並就風險管理事宜向零售銀行及財富管理業務風險管理委員會負責。 此外，各地的資產負債管理委員會監察及審視保險資產與負債在年期和現金流方面的配對 情況。 所有保險產品，不論是由內部或第三方制訂，在推出前都必須依循一套產品審批程序。 金融風險 （經審核） 我們的保險業務承擔多種金融風險，包括市場風險、信貸風險及流動資金風險。市場風險 包括利率風險、股權風險及匯兌風險。下文載述此等風險的性質及管理方法。 制訂保險產品的附屬公司面對各種金融風險，例如金融資產所得款項不足以應付保單及投 資合約產生的責任。在不少司法管轄區，當地的監管規定訂明該等附屬公司必須持有資產 的類別、質素及集中程度，以應付保險未決賠款。此等規定與集團整體政策互補不足。 董事會報告：風險 （續） 滙豐控股有限公司 232 風險附錄－政策及慣例 市場風險 （經審核） 市場風險說明 下文討論由我們旗下制訂保險產品附屬公司制訂而會產生市場風險的產品之主要特點，以 及該等特點令各附屬公司需承受的市場風險。 如資產的收益率低於制訂保險產品附屬公司應付予投保人的保證所隱含的投資回報，即產 生利率風險。當資產的收益率低於保證收益率時，我們或不會向新客戶提供有關產品，又 或會為有關產品重新定價或重組。下表概述滙豐保險合約中各種不同類別的保證。 保證利益的類別 ‧ 年金付款； ‧ 遞延╱即時年金：分為兩個階段，即儲蓄及投資階段和退休收入階段； ‧ 年度回報：年度回報保證不低於某個指定回報率，可能是每年存入投保人戶口的回報，或是於保單有效 期內 （計算截至保單到期日或退保日期） 存入投保人戶口的平均每年回報；及 ‧ 資本：投保人獲保證可收取不少於已付保費加已宣派紅利減支出的款項。 附有酌情參與條款的保險及投資產品的所得款項，主要投資於債券，但其中若干部分會分 配至其他資產類別，以便為客戶提供更高的潛在回報。擁有該等產品組合的附屬公司，將 承受市價下跌的風險，而此風險不能於酌情紅利中充分反映。如市況波動加劇，亦會令向 投保人提供的回報保證價值上升。 長期保險及投資產品一般允許投保人隨時退保或讓保單失效。若退保金額並未與出售相關 支持資產所變現的價值掛鈎，則有關附屬公司便須承擔市場風險，特別在資產價值下跌時 客戶要求退保，公司可能需在虧損情況下出售資產以取得資金贖回保單。 持有長期保險及投資產品組合的附屬公司，特別是持有附有酌情參與條款產品組合之公司， 可能會試圖投資於所在地以外國家╱地區的資產，以期減輕本身承受的本土市場風險。該 等資產可能並非以附屬公司之本土貨幣計值。如與該等資產有關的匯兌風險並未獲對沖， 例如由於為該等資產的匯兌風險進行對冲不符合成本效益，會使附屬公司承受本土貨幣強 於相關資產貨幣的風險。 至於單位相連合約，市場風險主要由投保人承擔，但由於管理該等資產所賺取的費用與相 連資產的市值掛鈎，我們在一般情況下仍要面對市場風險。 資產與負債的配對 資產與負債的年期未必能完全配對，其中一個原因是投保人的行為涉及不確定因素，故未 能確定日後能否收取所有保費以及賠償的時間，另一原因是某些負債的預計還款日期可能 超逾市場上現有最長期投資的年期。 我們使用各種模型評估一系列的未來境況對金融資產和相關負債的價值的影響，然後資產 負債管理委員會應用該等評估結果釐定持有資產以支持負債的最佳結構為何。該等境況包 括應用於影響保險風險的因素 （例如死亡率和保單失效率） 的壓力情景。當中特別重要者是 將現金流入的預期模式與就相關合約的應付利益 （可伸延多年給付） 進行配對。 我們的現有資產組合包括在收益率高於現時市場可見水平時發行的債務證券。因此，現時 持有的債務證券的收益高於現時發行債務證券所得收益。我們降低若干分紅合約向投保人 支付的短期紅利率，從而控制對業務造成的即時壓力。倘利率及孳息曲線維持於低水平， 可能需進一步降低紅利率。 如何管理市場風險 我們所有制訂保險產品的附屬公司設有市場風險授權，指明該等附屬公司獲許可進行投資 的投資工具及其可保留的最高市場風險水平。該等附屬公司會運用下列部分或全部方法管 理市場風險，視乎其所承保合約性質而定。 管理市場風險的方法 ‧ 就附有酌情參與條款的產品而言，透過調整紅利率來管理投保人負債；效果是大部分市場風險由投保 人承擔； ‧ 構建資產組合以支持預計負債現金流； ‧ 有限度使用衍生工具，以免受不利市場變動的影響或更好地配對負債現金流； 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 233 ‧ 設計附有投資保證的新產品時，於釐定保費水平或價格結構時一併考慮成本； ‧ 定期檢討界定為較高風險的產品，這些產品包含與儲蓄及投資產品相連的投資回報保證及內含期權性 風險特性； ‧ 在新產品加入專為減輕市場風險而設的產品特點，例如徵收退保罰款，以彌補因投保人退保所招致的 損失； ‧ 當投資組合的風險達到不能接受的水平時，在可行情況下終止經營該等組合；及 ‧ 重訂向投保人收取的保費價格。 我們透過產品審批程序識別及評估新產品的內含風險。舉例說，當新產品內含期權和回報 保證時，我們會採取盡職審查程序，確保已設定完整和適當的風險管理程序。除最簡單的 保證利益外，所有保證利益由集團保險業務總管理處進行評估。市況特別波動時，管理層 會更加頻密檢討若干風險項目，以確保能及時處理可能出現的任何問題。 如何衡量市場風險 我們制訂保險產品的附屬公司須定期根據規定限額監察風險，並向集團保險業務總管理處 匯報；亦會匯總各地的風險，每季向集團保險業務的高級風險管理會議匯報。 此外，大型的制訂保險產品附屬公司每月均會進行高層次的市場風險評估，評估承受風險 水平；結果呈交予集團保險業務總管理處，而環球評估則會呈報予零售銀行及財富管理業 務風險管理委員會。 用於量化市場風險的標準衡量方法 ‧ 就利率風險而言，資產及預計負債現金流的淨現值 （按總額及按貨幣計算） 相對於計算淨現值的折現曲 線平行移動1個基點時的敏感度； ‧ 就股價風險而言，持有股票的總市值，以及按地區及國家劃分的持股量市值；及 ‧ 就匯兌風險而言，外匯淡倉淨額總計，以及按貨幣劃分的外匯持倉淨額。 雖然使用該等衡量基準計算和匯總風險較為簡單直接，但存在一些限制。最主要的限制是 孳息曲線平行移動1個基點，並不能捕捉若干資產及負債的價值與各項利率之間的非直線關 係，例如因投資回報保證及產品特點 （讓投保人可以退保的條款） 而產生的非直線關係。如 持有以支持保證利益合約的投資收益率低於保證利益隱含的投資回報，則我們須承擔該等 差額。 我們明白該等衡量基準存在此等限制，因此利用壓力測試以補不足；壓力測試旨在考慮到 重大及相關的稅務和會計處理方法後，審視一系列市場利率境況對集團旗下制訂保險產品 附屬公司之年度利潤總額及各類股東權益總額造成的影響。此等測試的結果會每季向集團 保險業務總管理處及各風險管理委員會匯報。 第195頁的 「滙豐旗下制訂保險產品附屬公司對市場風險因素的敏感度」 列表，顯示制訂保險 產品的公司的利潤及各類股東權益總額對市場利率因素的敏感度。 信貸風險 （經審核） 信貸風險說明 滙豐旗下制訂保險產品的公司的信貸風險主要有兩大來源： (i) 為投保人及股東賺取回報而投資保費後，債務證券交易對手違責的風險；及 (ii) 轉移保險風險後，再保險交易對手違責及不就已作出的索償進行賠償的風險。 如何管理信貸風險 我們旗下制訂保險產品的附屬公司負責管理其各自投資組合的信貸風險、質素及表現。我 們對發行人及交易對手信譽可靠度的評估，主要依據國際認可的信貸評級及其他公開資料。 各地制訂保險產品的附屬公司按所定各項上限監察投資信貸風險，並予以匯總，然後向集 團保險信貸風險管理部及集團信貸風險管理部匯報。集團保險業務會利用信貸息差敏感度 以及違責或然率，對投資信貸風險進行壓力測試。 董事會報告：風險 （續） 滙豐控股有限公司 234 風險附錄－政策及慣例 我們使用多種工具管理和監察信貸風險，包括編製信貸預警報告，在報告的預警名單中載 列當前遇上信貸問題的投資，並每隔兩個星期送呈集團保險業務的高級管理層及國家╱地 區風險管理總監，以便識別帶有日後減值風險的投資。 流動資金風險 （經審核） 流動資金風險說明 幾乎所有保單均具有一種內在特質，那就是可能出現的申索賠付金額及賠付申索時間並不 確定，因而產生流動資金風險。 流動資金風險源自三種情況。第一種風險在正常市況下產生，稱為流動資金來源風險；具 體而言亦即是在需要履行付款責任時籌集足夠現金的能力。第二種為市場流動資金風險， 倘若某種工具的持有量十分龐大，導致該等工具無法按接近市價的價格全數出售，即會產 生這種風險。最後是備用流動資金風險，即是在非正常狀況下按條款履行付款責任的能力。 如何管理流動資金風險 制訂保險產品的附屬公司主要以下列現金流入來源應付申索賠付所產生的現金流出： ‧ 新造保單、續保及續繳保費產品的保費； ‧ 投資利息及股息與到期債務投資的本金還款； ‧ 現金來源；及 ‧ 出售投資。 該等附屬公司會運用下列部分或全部方法管理流動資金風險： ‧ 利用特定現金流預測或較常用的資產與負債配對方法，例如年期配對，將現金流入額與 預期現金流出額互相配對； ‧ 保持充足的現金來源； ‧ 投資於信貸質素良好的投資工具及具深度和流通性高的市場，投資額視實際存在的市場 深度和流通性而定； ‧ 監察投資集中情況，在適當情況下施加限制，例如限制發債項目或發行人的集中程度； 及 ‧ 設立或有借貸承諾額度。 上述每種方法均有助減低上文介紹的三類流動資金風險。 制訂保險產品的附屬公司須每季填寫流動資金風險報告，提交予集團保險業務總管理處， 以供核對及審閱。該等報告評估流動資金風險的方法，是計算在一系列壓力境況下，預期 累計現金流淨額的變動，而該等壓力境況旨在確定預期的流動資金減少及現金加快流出的 影響，過程中會假設新造保單數額或續保率偏低，退保或保單失效率亦較預期高。 保險風險 （經審核） 保險風險為損失由保單持有人轉移給發行人 （滙豐） 所產生的風險 （金融風險除外） 。我們制 訂保單時面對的風險，主要為某段時間後獲取保單成本加上行政開支和賠償及利益支出的 總額，可能超過所收保費加投資收益的總額。 賠償和利益支出可能受多項因素影響，包括過往的死亡率及發病率、保單失效率及退保率， 以及 （倘保單帶有儲蓄成分） 為支持負債而持有的資產之表現。 保險風險由集團集中及其地方部門制訂的高層次政策及程序管控，並在適當情況下計及各 地的特定市況及監管規定。相關部門會採用正規的承保、再保險及賠償處理程序，務求確 保符合相關規例，並輔以壓力測試。 除執行承保監控措施外，我們亦運用再保險以減低保險風險。我們透過第三方再保人管理 保險風險時，相關收入和制訂保險產品的利潤會轉讓予再保人。雖然再保險可用於管理保 險風險，該等合約亦使我們承受信貸風險，亦即再保人失責的風險。 導致我們面對保險風險的主要因素於下文論述。長期合約的負債乃參考涉及此等因素的一 系列假設而釐定。此等假設一般反映簽發保單公司本身的經驗。壽險產生的保險風險種類 和數量視乎業務類別而定，不同類別之間有相當的差異。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 235 ‧ 死亡率及發病率：產生此等風險的合約主要為定期壽險、終身壽險、危疾保單、收入保 障合約及年金。風險會受到定期監察，主要的減低風險措施包括承保監控措施和再保 險，以及在若干情況下，保留按過往經驗調整保費的能力； ‧ 保單失效率及退保率：與此等因素有關的風險，一般透過產品設計、設置退保費用及採 取管理措施 （例如管理支付予投保人的紅利水平等） 予以減輕。我們最少每年按產品類別 作出一次詳盡的續保率分析；及 ‧ 支出水平風險透過定價 （例如在若干情況下，保留按經驗調整保費及╱或投保人收費的 能力） 和管理成本的守則予以減輕。 相關假設的變動會對負債造成影響 （請參閱第198頁 「敏感度分析」 ） 。 聲譽風險 （未經審核） 我們定期檢討抵禦聲譽風險的政策及程序。這項循序漸進的過程會顧及各種相關發展趨勢、 行業指引、最佳守則及社會期望。 我們一直致力推行最高的操守標準，並會定期評估我們的業務面對的聲譽風險。聲譽風險 可由多種問題引致。作為一家銀行集團，我們的良好聲譽端賴我們的業務營運方式，但使 用我們的金融服務的客戶以及供應商如何行事，亦會影響我們的聲譽。 防範金融犯罪部及監管合規部的環球主管負責集團的聲譽風險。集團聲譽風險政策委員會 （ 「GRRPC」 ） 主席由集團主席兼任，負責協調制訂政策、維持有效的監控環境以識別、評估、 管理及減輕聲譽風險。聲譽風險政策委員會的主要工作是考慮產生重大聲譽風險的範疇及 活動，並在適當情況下，向集團風險管理會議提供意見，以供考慮需否更改任何政策或程 序，從而減輕有關風險。集團各地區須確保透過其各自的區域風險管理委員會會議的特別 小組或區域聲譽風險政策委員會考慮地區層面的聲譽風險。區域委員會的會議紀錄概要會 呈交集團聲譽風險政策委員會討論。聲譽風險的重大事件會向集團風險管理委員會及滙豐 控股董事會匯報，並會在合適的情況下，向行為及價值觀委員會匯報。 於2014年7月，我們公布了新的聲譽風險及挑選客戶的政策，當中界定了一致而有系統地管 理該等風險的方法。進一步詳情請參閱第199頁的 「聲譽風險」 。各環球業務及部門須設有程 序，以評估和處理提議的業務交易及客戶活動可能引致的聲譽風險。中央團隊會提供支援， 以確保將有關事件於合適的會議上提出、執行有關決定及整理管理資訊並向高級管理層匯 報行動。於2014年，環球業務部設立了一個結合聲譽風險及挑選客戶的委員會，有清晰的程 序提升及處理適當級別的事項。環球部門在現有的營運風險框架內管理聲譽風險及提升其 重要性。 滙豐以及個別附屬公司、業務及部門均已就所有主要業務範疇制訂標準。董事會、集團管 理委員會、風險管理會議、環球標準督導會議、附屬公司董事會、董事會屬下委員會及高級 管理層在制訂政策及訂定我們的標準時，均會考慮及評估聲譽風險 （包括環境、社會及管治 問題） 。這些政策是內部監控制度的重要組成部分 （請參閱第288頁） ，並透過各種工作手冊及 政策聲明，在內部通訊及培訓課程中公布。這些政策詳列我們的承受風險水平及所有涉及 聲譽風險的營運程序，包括反洗錢活動、反資助恐怖主義、環境影響、反賄賂及貪污措施， 以及僱員關係等。政策手冊詳述如何處理風險事宜，集團各部門及各業務部門均要通力合 作，嚴謹奉行我們的風險管理制度及可持續發展慣例。 受信風險 （未經審核） 內含受信風險的業務活動僅獲准於指定範疇的業務內進行。受信風險透過全面政策架構及 監察關鍵指標而於指定業務範疇內管理。集團的主要受信業務及活動 （ 「指定業務及活動」 ） 為： ‧ 滙豐證券服務，透過其資金服務及企業信託業務和貸款代理業務而面對受信風險； ‧ 滙豐環球投資管理業務，因代表客戶進行投資管理活動而面對受信風險； ‧ 滙豐環球私人銀行業務，透過其私人信託部及全權代理投資管理業務而面對受信風險； 董事會報告：風險 （續） 滙豐控股有限公司 236 風險附錄－政策及慣例 ‧ 滙豐保險業務，因提供保險產品及服務時進行投資管理活動而面對受信風險； ‧ 零售銀行及財富管理信託投資產品，按規例規定提供零售銀行及財富管理業務的一般 財富管理產品及服務；及 ‧ 滙豐僱員退休金計劃業務，可能因營運滙豐僱員退休金計劃而進行部分酌情或控制職 能，因而產生受信責任。 集團管理受信風險的規定列於 《環球風險職能指引手冊》 的受信章節內，而有關手冊由環球 營運風險管理部編製。除指定業務外，任何業務均不得在未有通知環球營運風險管理部及 獲具體豁免遵守相關受信政策規定的情況下進行受信活動。 其他有關提供建議的政策，包括提供投資建議及企業顧問服務，以及潛在利益衝突的管理， 亦可降低受信風險。 退休金風險 （經審核） 我們在世界各地設有多項退休金計劃，詳情見第200頁及下文退休金風險一節。 負責資助界定福利計劃的集團旗下公司經諮詢計劃的受託人 （如適用） 後，會按照精算師的 意見作出定期供款，為此等計劃的相關福利提供資金，而在某些情況下僱員亦須供款。界 定福利計劃則將供款投資於各類足可承擔其長期負債的投資項目。 該等供款的水平對滙豐的現金流有直接影響，而且在一般情況下，設定的水平可以確保有 足夠資金，以應付活躍成員日後服務的應計福利支出。但是，倘計劃資產被視為不足以應 付現有退休金負債，則需要作出更多供款。供款率一般每年或每三年修訂一次，情況視乎 不同計劃而定。對主計劃的協定供款則為每三年修訂一次。 界定福利計劃會受多項因素影響而出現虧損，包括： ‧ 投資回報低於計劃所需的預計福利水平。例如股票市值下跌，或長期利率上升，因而導致所持定息證券 跌價； ‧ 當前經濟環境導致公司倒閉，因而觸發資產 （股東權益及債務） 價值撇減； ‧ 利率或通脹變化，導致計劃負債的價值上升；及 ‧ 計劃成員的壽命長於預期 （即長壽風險） 。 一個計劃的投資策略是經計及投資項目的內在市場風險，以及其對日後潛在供款的相應影 響而釐定。滙豐及 （如有關及適用） 受託人兩者對該等計劃的長期投資目標為： ‧ 長遠而言，限制計劃資產未能抵銷計劃負債的風險；及 ‧ 盡量取得與可接受風險水平相符的最大回報，從而控制界定福利計劃的長期支出。 為達致這些長期目標，已設定一項基準，用以將界定福利計劃資產分配到不同資產類別。 此外，每個獲准的資產類別均設有其個別基準，例如股市或物業估值的指數，在相關情況 下，還包括表現超越基準的理想水平等。基準至少每三年在精算估值之日起計18個月內檢 討一次；若當地法律有所規定或情況有此需要，則檢討次數可能更為頻密。檢討過程一般 涉及對資產及負債進行廣泛檢討。 最終投資策略乃受託人的責任，或在若干情況下由管理委員會負責。受託人獨立於滙豐的 程度，因司法管轄區而異。 界定供款計劃使我們面對的市場風險大幅降低，但我們仍須面對營運及聲譽風險，因為有 關計劃需要僱員更直接地負責，讓他們彈性選擇。為管理有關風險，我們會監察界定供款 投資基金的表現，並鼓勵當地僱員積極參與，確保他們就可行的選項，獲提供足夠的資料。 英國的退休金計劃 英國滙豐銀行 （英國） 退休金計劃 （ 「主計劃」 ） 設有界定福利及界定供款的部分。界定福利的 部分佔我們全球界定福利義務總額約72%。於1996年，界定福利部分不再吸納新成員，而由 2015年7月1日起，亦不再為該部分的現有僱員進一步累計；有關僱員將會參加日後退休金的 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 237 界定供款部分。自1996年起，所有新僱員已參加界定供款部分。主計劃由獨立公司受托人監 管，該受托人對退休金計劃的營運負有受信責任。受托人負責監察及管理投資策略及計劃 福利的行政管理。主計劃持有多項投資組合，以滿足累計福利於到期支付時產生的日後現 金流負債。主計劃的受托人須編製投資方針的書面聲明，以監察如何作出投資決策，而在 選擇界定福利部分的資產配置及管理架構時，會考慮到充分分散投資的需要。主計劃的長 壽風險在計量退休金負債的過程中予以評估，並透過計劃的撥資程序管理。 可持續發展風險 （未經審核） 由滙豐提供金融服務的公司或項目若與可持續發展的需要背道而馳，即會產生可持續發展 風險；實際上，有關風險是因環境及社會影響超逾經濟利益而產生。環球企業可持續發展 部為集團總管理處內一個獨立部門，獲授權管理全球各地的可持續發展風險，在適當情況 下會透過各地辦事處開展工作。可持續發展風險經理須負責就地區或國家的環境和社會風 險提供意見，並管理有關風險。環球企業可持續發展部的風險管理責任包括： ‧ 制訂可持續發展風險管理政策，包括監督我們的可持續發展風險標準、應用赤道原則的 情況及可持續發展政策，其中涵蓋農業商品、化工、國防、能源、林業、淡水基建、礦業 及金屬、世界文化遺址及拉姆薩爾濕地；倘交易的可持續發展風險被評估為高風險，則 對交易進行獨立審核，並協助我們的營運公司評估較低程度的同類風險； ‧ 訂立並落實執行以制度為本的程序，以確保政策貫徹一致、減低可持續發展風險的評估 成本，以及紀錄管理資訊以衡量及匯報我們的貸款和投資業務對可持續發展的影響；及 ‧ 在我們的營運公司範疇內，提供培訓並組建有關部門，以確保能按照我們本身的標準、 國際標準或當地法規 （以較嚴格者為準） ，一致地識別及減低可持續發展風險。 董事會報告：資本 （續） 滙豐控股有限公司 238 資本概覽╱風險加權資產 資本 頁次 附錄 1 列表 頁次 資本概覽 239 資本比率 239 資本管理 257 監管規定資本總額及風險加權資產 239 方法及政策 257 壓力測試 257 資本風險 257 風險加權資產目標 257 資本生成 258 資本計量及分配 258 監管規定資本 258 第一支柱資本規定 258 第二支柱資本規定 259 第三支柱資料披露規定 260 風險加權資產 239 按風險類別分析風險加權資產 240 按環球業務分析風險加權資產 240 按地區分析風險加權資產 240 信貸風險的風險加權資產 240 信貸風險承擔－按地區分析風險加權資產 240 信貸風險承擔－按環球業務分析 風險加權資產 240 按主要因素以地區分析風險加權資產 變動－信貸風險－只列示按內部 評級基準計算法計算的數額 242 按主要因素以環球業務分析風險加權 資產變動－信貸風險－只列示按內部 評級基準計算法計算的數額 243 交易對手信貸風險及市場風險的 交易對手信貸風險的風險加權資產 243 風險加權資產 243 按主要因素分析風險加權資產變動 －交易對手信貸風險－高級計算法 243 市場風險的風險加權資產 244 按主要因素分析風險加權資產變動 －市場風險－按內部模式計算法計算 244 按主要因素分析資本及風險加權資產 變動－資本指引4終點基準 245 營運風險的風險加權資產 244 按主要因素分析風險加權資產變動－ 編製基準及補充附註 260 信貸風險因素－定義及量化 260 交易對手風險因素－定義及量化 261 市場風險因素－定義及量化 261 資本結構 245 監管規定資本總額來源及運用 245 監管規定資本組合成分 246 採用過渡基準及轉用估計資本指引4終點基準 計算的監管規定資本之對賬 247 監管規定資產負債表 248 資產負債表對賬－財務會計基準對監管規定 基準綜合計算 249 監管規定與會計綜合基準 248 槓桿比率 251 估計槓桿比率 251 槓桿比率：編製基準 261 監管環境發展 252 監管規定緩衝資本 252 資本規定架構 254 監管規定壓力測試 254 風險加權資產發展 254 槓桿比率建議 255 銀行業結構性改革、復元及解決計劃 255 其他監管規定修訂 256 1 資本附錄。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 239 集團的資本管理目標，是維持適當水平的 資本，一方面配合業務策略，另一方面符 合監管及壓力測試的相關規定。 資本重點 ‧ 我們的過渡基準普通股權一級比率由2013年底 的10.8%上升至10.9%，原因是持續資本生成 及管理措施被風險加權資產增長、外匯變動及 監管規定的改變所抵銷。 ‧ 由於相類的因素，我們的終點基準普通股權一 級比率由2013年底的10.9%上升至11.1%。 資本概覽 （未經審核） 資本比率 （未經審核） 於12月31日 2014年 2013年 % % 資本指引4過渡基準 普通股權一級比率 10.9 10.8 一級比率 12.5 12.0 總資本比率 15.6 14.9 資本指引4終點基準 普通股權一級比率 11.1 10.9 巴塞爾協定2.5基準 核心一級比率 不適用 13.6 一級比率 不適用 14.5 總資本比率 不適用 17.8 監管規定資本總額及風險加權資產 （未經審核） 資本 指引4 過渡基準 於2014年 12月31日 資本指引4 過渡 基準估計 於2013年 12月31日 巴塞爾 協定2.5 基準 於2013年 12月31日 百萬美元 百萬美元 百萬美元 普通股權一級資本 133,200 131,233 核心一級資本 149,051 額外一級資本 19,539 14,408 9,104 二級資本 37,991 35,538 35,854 監管規定資本總額 190,730 181,179 194,009 風險加權資產 1,219,765 1,214,939 1,092,653 資本指引4於2014年1月1日實施，於2014年 12月31日的資本及風險加權資產乃按集團 對審慎監管局所發布資本指引4最終法規 及最終規則的詮釋計算及呈列。於2014年 1月1日前，風險加權資產及資本乃根據過 往資本指引3 （亦稱 「巴塞爾協定2.5」 ） 的制 度計算及呈列。因此，除非另有說明，於 2013年12月31日的資本及風險加權資產的 比較乃按巴塞爾協定2.5為基準。 按資本指引4估算的資本及風險加權資產， 已計入審慎監管局規則手冊所載的審慎監 管局最終規則作為基礎。最終規則將資本 指引4最終法規內多個範疇屬於國家酌情 決定的事項納入英國的法律。在其最終規 則內，審慎監管局並無採用資本指引4的 大部分過渡條文，代之以就普通股權一級 （ 「CET1」 ） 選擇加快採用資本指引4終點基 準有關的定義。然而，資本指引4有關未變 現增益的過渡條文已予應用，故投資物業 及可供出售證券的未變現增益在2015年1月 1日前均未有確認。因此，我們於2014年的 過渡資本比率略低於可比較終點資本比率。 於2014年4月，審慎監管局發表其規則及監 管聲明，實施有關緩衝資本的若干資本指 引4條文，其進一步詳情載於第252頁 「監管 規定緩衝資本」 部分。 2014年6月，審慎監管局亦公布該局修訂對 於主要英國銀行及建屋貸款社的資本比率 的期望，表示自2014年7月1日起，預期我 們達到普通股權一級比率的7%，此數額乃 按照資本指引4終點基準的定義計算。這配 合資本指引4規定。 雖然審慎監管局至今已公布多項規則，但 英國的銀行須持有的資本金額仍繼續存在 不確定性，尤其涉及緩衝資本及第二支柱 的量化和相互關係，審慎監管局現正就第 二支柱的修訂方式、審慎監管局緩衝以及 交流資本指引4緩衝進行諮詢。此外，歐洲 銀行管理局仍有多項草擬和尚未發布的技 術及實施準則將於2015年內公布。 我們管理集團資本的方式，旨在確保我們 超出現有監管規定水平，以及我們尊重資 本提供者的付款優次。我們於2014年全年 符合審慎監管局對於監管規定資本充足的 要求，包括有關壓力測試的要求。我們也 有足夠實力符合預期日後提出的資本要求。 於2014年，我們管理本身的資本狀況，務求 終點基準普通股權一級比率能高於10%的 內部目標。自此我們持續檢討達標情況， 而於2015年，我們預期管理集團資本能達 到10%以上股本回報的中期目標。這是以 12%至13%的終點基準普通股權一級比率 作為參考而制訂。 有關資本管理、計量及分配的政策與慣例概要載 於第257頁資本附錄內。 風險加權資產 （未經審核） 資本指引4提高對資本的要求。所引入主要 變化為： ‧ 證券化持倉過往從核心一級中扣減 50%，以及從資本總額中扣減50%，現 時按1,250%計入風險加權資產值內； ‧ 額外的資本要求，旨在抵補預期由交易 對手風險引起的市值計價虧損風險，稱 為信貸估值調整 （ 「CVA」 ） 風險； ‧ 遞延稅項資產及重大投資須受限額規 限，現採用250%的風險權數； 董事會報告：資本 （續） 滙豐控股有限公司 240 風險加權資產 ‧ 金融機構面臨的風險權數增加指資產值相關 性倍數 （ 「AVC」 ） ；及 ‧ 中央交易對手 （ 「CCP」 ） 風險承擔的新增要求。 通過中央交易對手結算場外衍生工具交易增 加獎勵。 按風險類別分析風險加權資產 （未經審核） 資本指引4過渡基準 及終點基準 巴塞爾 協定 2.5基準 2014年 2013年 2013年 十億美元 十億美元 十億美元 信貸風險 955.3 936.5 864.3 標準計算法 356.9 358.6 329.5 內部評級基準基礎計算法 16.8 13.5 13.6 內部評級基準高級計算法 581.6 564.4 521.2 交易對手信貸風險 90.7 95.8 45.8 標準計算法 25.2 36.6 3.6 高級計算法 65.5 59.2 42.2 市場風險 56.0 63.4 63.4 營運風險 117.8 119.2 119.2 於12月31日 1,219.8 1,214.9 1,092.7 其中： 美國縮減組合 99.2 142.3 104.9 環球銀行及資本市場業務 之既有信貸業務 44.1 63.7 26.4 美國消費及按揭貸款及其他 55.1 78.6 78.5 卡及零售商戶業務1 － 1.1 1.1 有關註釋，請參閱第256頁。 按環球業務分析風險加權資產 （未經審核） 資本指引4 過渡基準及 終點基準 2014年 巴塞爾 協定 2.5基準 2013年 十億美元 十億美元 零售銀行及財富管理 205.1 233.5 工商金融 432.4 391.7 環球銀行及資本市場 516.1 422.3 環球私人銀行 20.8 21.7 其他 45.4 23.5 於12月31日 1,219.8 1,092.7 按地區分析風險加權資產 2 （未經審核） 資本指引4 過渡基準及 終點基準 2014年 巴塞爾 協定 2.5基準 2013年 十億美元 十億美元 歐洲 375.4 300.1 亞洲 499.8 430.7 中東及北非 63.0 62.5 北美洲 221.4 223.8 拉丁美洲 88.8 89.5 於12月31日 1,219.8 1,092.7 有關註釋，請參閱第256頁。 信貸風險的風險加權資產 （未經審核） 信貸風險承擔－按地區分析風險加權資產 歐洲 亞洲3 中東及北非 北美洲 拉丁美洲 總計 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 資本指引4基準 內部評級基準計算法 216.1 213.1 15.6 142.0 11.6 598.4 內部評級基準高級計算法 203.3 213.1 11.6 142.0 11.6 581.6 內部評級基準基礎計算法 12.8 － 4.0 － － 16.8 標準計算法 47.1 186.0 39.0 29.6 55.2 356.9 於2014年12月31日的風險加權資產 263.2 399.1 54.6 171.6 66.8 955.3 巴塞爾協定2.5基準 內部評級基準高級計算法 157.1 182.9 11.2 161.5 8.5 521.2 內部評級基準基礎計算法 9.8 － 3.8 － － 13.6 標準計算法 44.5 165.9 40.0 22.7 56.4 329.5 於2013年12月31日的風險加權資產 211.4 348.8 55.0 184.2 64.9 864.3 有關註釋，請參閱第256頁。 信貸風險承擔－按環球業務分析風險加權資產 主要零售 銀行及 財富管理 零售銀行 及財富管 理 （美國 縮減組合） 整體零售 銀行及 財富管理 工商金融 環球銀行 及資本市場 環球 私人銀行 其他 總計 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 資本指引4基準 內部評級基準計算法 55.9 47.3 103.2 217.4 255.6 10.2 12.0 598.4 內部評級基準高級計算法 55.9 47.3 103.2 209.4 248.1 10.0 10.9 581.6 內部評級基準基礎計算法 － － － 8.0 7.5 0.2 1.1 16.8 標準計算法 60.4 4.8 65.2 181.8 70.1 6.6 33.2 356.9 於2014年12月31日的風險加權資產 116.3 52.1 168.4 399.2 325.7 16.8 45.2 955.3 巴塞爾協定2.5基準 內部評級基準高級計算法 58.4 72.6 131.0 183.2 192.8 10.4 3.8 521.2 內部評級基準基礎計算法 － － 6.3 5.8 0.1 1.4 13.6 標準計算法 60.6 3.1 63.7 169.3 71.6 6.9 18.0 329.5 於2013年12月31日的風險加權資產 119.0 75.7 194.7 358.8 270.2 17.4 23.2 864.3 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 241 信貸風險的風險加權資產乃採用審慎監管 局批准的三種方法計算。為呈報集團的綜 合賬目，我們已為大部分業務採用內部評 級基準 （ 「IRB」 ） 高級計算法；小部分則採用 內部評級基準基礎計算法；而其餘業務則 使用標準計算法。 標準計算法 採用標準計算法處理的組合中，信貸風險 的風險加權資產增加274億美元，反映由於 匯兌變動致令減少136億美元。 在亞洲、歐洲、北美洲及拉丁美洲的企業 增長，包括有期及貿易相關貸款令風險加 權資產增加250億美元，其中聯營公司交通 銀行的增長佔64億美元。 轉用資本指引4基準使於2014年1月1日的風 險加權資產增加71億美元。此變動主要包括 按250%計算風險加權值的資本限額以下的 重大權益及遞延稅項資產總計金額 （283億 美元） ，為匯報目的而重新分類非信貸責任 資產至內部評級基準 （163億美元） 及根據標 準計算法於違責風險承擔按淨額計算綜合 評估減值 （35億美元） ，抵銷了上述部分影 響。 年內，數個個別非重大的組合由內部評級 基準計算法轉為標準計算法，使按標準計 算法計算的風險加權資產增加60億美元， 並使按內部評級基準計算法計算的風險加 權資產減少48億美元。 我們出售約旦、巴基斯坦、哥倫比亞及哈 薩克的業務使風險加權資產減少10億美元。 在亞洲，重大權益的公允值變動 （主要為興 業銀行） 使風險加權資產增加59億美元，但 由於將越南科技及商業股份銀行由聯營公 司重新分類為投資項目，抵銷了上述部分 增幅，風險加權資產因而減少11億美元。 內部評級基準計算法 第242及243頁的列表載列對採用內部評級 基準計算法處理的組合，按主要因素分析 信貸風險的風險加權資產變動。有關信貸 風險、交易對手信貸風險及市場風險的風 險加權資產流量的編製基準，請參閱年報 及賬目第257頁的資本附錄。按內部評級基 準計算法處理的組合中，信貸風險的風險 加權資產增加636億美元，反映由於美元兌 其他貨幣處於強勢的匯兌變動而減少201億 美元。 收購及出售 環球銀行及資本市場業務出售北美洲的資 產抵押證券，使風險加權資產減少42億美 元。此外，環球銀行及資本市場業務繼續 透過出售若干結構性投資中介機構持倉來 降低所持證券化持倉，使歐洲的風險加權 資產減少30億美元。出售哈薩克、哥倫比 亞、巴基斯坦及約旦的業務，令歐洲、拉 丁美洲和中東及北非的風險加權資產減少 12億美元。 賬項規模 賬項規模變動反映企業貸款增加，包括有 期及貿易相關貸款，令工商金融和環球銀 行及資本市場業務在亞洲、歐洲及北美洲 的風險加權資產增加403億美元。環球銀行 及資本市場業務之主權賬項使風險加權資 產增加33億美元，主要於亞洲、拉丁美洲 和中東及北非。 在北美洲，零售銀行及財富管理業務持續 縮減美國消費及按揭貸款業務的零售按揭 組合，令風險加權資產賬項規模減少69億 美元。 賬項質素 美國縮減組合中的風險加權資產減少85億 美元，主要原因是持續縮減，令餘下組合 賬項質素改善。 主要零售銀行及財富管理業務的賬項質素 改善59億美元，乃因重新校準模型，反映 美國物業價格改善，歐洲的組合質素轉佳 令風險加權資產減少。 證券化組合的評級上調，使風險加權資產 減少32億美元。 歐洲、北美洲、中東、北非、亞洲及拉丁美 洲的企業、主權及機構組合的平均客戶信 貸質素不利變動，導致風險加權資產增加 76億美元，抵銷了以上部分減幅。 模型更新 在歐洲，英國企業組合應用違責損失率 （ 「LGD」 ） 下限，導致工商金融和環球銀行 及資本市場業務的風險加權資產增加190億 美元。 北美洲進行模型更新，主要是對美國按揭 縮減組合採用新的風險模型，使風險加權 資產減少62億美元，抵銷了以上部分增幅。 方法及政策改變 方法及政策修訂使風險加權資產增加522億 美元。 資本指引4的影響 該增額涉及於2014年1月1日實施資本指引 4規則，使風險加權資產增加482億美元。 資本指引4主要變動源自證券化持倉，此項 過往從資本中扣減，現計入作為信貸風險 的風險加權資產一部分，並按1250%計算 風險加權值，導致環球銀行及資本市場業 務增加402億美元 （主要位於歐洲） 。資本指 引4亦引入金融交易對手的資產估值相關 性倍數，從而使風險加權資產增加92億美 元，主要為亞洲及歐洲的環球銀行及資本 市場業務。 董事會報告：資本 （續） 滙豐控股有限公司 242 風險加權資產 內部更新 風險加權資產下跌92億美元，是由於可供 出售的負值儲備與環球銀行及資本市場業 務的既有信貸組合的違責風險承擔互相抵 銷所得。 在亞洲，與貿易融資產品相關的內部處理 方法變動，令風險加權資產減少49億美元。 此外，由於個別非重大組合轉用標準計算 法，多個地區的主要零售銀行及財富管理 業務及工商金融業務採用內部評級基準計 算法計算的風險加權資產減少48億美元， 而採用標準計算法計算的風險加權資產則 增加60億美元。 重新分類部分按揭組合使北美洲的風險加 權資產減少45億美元，其中縮減組合佔41億 美元。 外部更新 過往違責率低的所選組合 （主要在歐洲、亞 洲及北美洲） 須遵守引入違責損失率下限 的外部修訂並應用於企業及機構，使風險 加權資產增加98億美元。亞洲零售按揭引 入額外的風險加權資產下限，使風險加權 資產增加17億美元。 非信貸責任資產 非信貸責任資產就列賬而言由標準計算法 重新分類至內部評級基準計算法，使根據 內部評級基準計算法計算的風險加權資產 增加163億美元，同時令根據標準計算法計 算的風險加權資產減少了相同金額。 按主要因素以地區分析風險加權資產變動－信貸風險－只列示按內部評級基準計算法計算 的數額 （未經審核） 歐洲 亞洲 中東及北非 北美洲 拉丁美洲 總計 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 於2014年1月1日按巴塞爾協定 2.5基準計算的風險加權資產 166.9 182.9 15.0 161.5 8.5 534.8 匯兌變動 (11.6) (4.0) (0.2) (2.4) (1.9) (20.1) 收購及出售 (3.5) － (0.7) (4.2) (0.1) (8.5) 賬項規模 11.4 19.5 1.8 2.9 2.0 37.6 賬項質素 (1.5) － (0.8) (10.3) 1.4 (11.2) 模型更新 19.4 0.3 － (6.1) － 13.6 新建╱更新模型 19.4 0.3 － (6.1) － 13.6 方法及政策 35.0 14.4 0.5 0.6 1.7 52.2 內部修訂 (11.7) (5.2) (0.2) (6.4) (0.1) (23.6) 外部修訂 2.2 8.5 (0.2) 0.7 0.1 11.3 資本指引4影響 37.0 5.7 0.4 4.9 0.2 48.2 由標準計算法改為內部評級基準 計算法產生的非信貸責任資產 7.5 5.4 0.5 1.4 1.5 16.3 風險加權資產變動總額 49.2 30.2 0.6 (19.5) 3.1 63.6 於2014年12月31日按資本指引4 基準計算的風險加權資產 216.1 213.1 15.6 142.0 11.6 598.4 於2013年1月1日按巴塞爾協定 2.5基準計算的風險加權資產 150.7 162.3 12.6 187.1 11.2 523.9 匯兌變動 3.3 (4.5) (0.5) (1.9) (1.0) (4.6) 收購及出售 (1.5) － － (8.6) (1.7) (11.8) 賬項規模 2.1 21.2 1.4 (10.6) 0.2 14.3 賬項質素 (1.5) 5.3 1.3 (10.8) (0.3) (6.0) 模型更新 11.6 － 0.1 (0.2) － 11.5 改為採用內部評級基準 計算法的組合 13.4 － － － － 13.4 新建╱更新模型 (1.8) － 0.1 (0.2) － (1.9) 方法及政策 2.2 (1.4) 0.1 6.5 0.1 7.5 內部修訂 (0.2) (7.8) 0.1 (0.6) 0.1 (8.4) 外部修訂 2.4 6.4 － 7.1 － 15.9 風險加權資產變動總額 16.2 20.6 2.4 (25.6) (2.7) 10.9 於2013年12月31日按巴塞爾 協定2.5基準計算的風險加權 資產 166.9 182.9 15.0 161.5 8.5 534.8 有關註釋，請參閱第256頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 243 按主要因素以環球業務分析風險加權資產變動－信貸風險－只列示按內部評級基準計算法 計算的數額 （未經審核） 主要零售 銀行及 財富管理 零售銀行 及財富管 理 （美國 縮減） 整體零售 銀行及 財富管理 工商 金融 環球銀行 及資本 市場 環球 私人銀行 其他 總計 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 於2014年1月1日按巴塞爾協定 2.5基準計算的風險加權資產 58.4 72.6 131.0 189.5 198.5 10.6 5.2 534.8 匯兌變動 (2.6) － (2.6) (8.7) (8.1) (0.2) (0.5) (20.1) 收購及出售 － － － － (8.2) － (0.3) (8.5) 賬項規模 1.8 (6.9) (5.1) 23.2 21.1 (0.5) (1.1) 37.6 賬項質素 (5.7) (8.6) (14.3) 2.8 (0.2) (0.3) 0.8 (11.2) 模型更新 0.6 (6.2) (5.6) 12.2 7.0 － － 13.6 新建╱更新模型 0.6 (6.2) (5.6) 12.2 7.0 － － 13.6 方法及政策 3.4 (3.6) (0.2) (1.6) 45.5 0.6 7.9 52.2 內部修訂 (3.0) (3.9) (6.9) (5.0) (11.2) (0.5) － (23.6) 外部修訂 1.8 － 1.8 2.5 6.3 0.5 0.2 11.3 資本指引4影響 － － － (0.7) 48.6 0.2 0.1 48.2 由標準計算法改為內部評級 基準計算法產生的非信貸 責任資產 4.6 0.3 4.9 1.6 1.8 0.4 7.6 16.3 風險加權資產變動總額 (2.5) (25.3) (27.8) 27.9 57.1 (0.4) 6.8 63.6 於2014年12月31日按資本指引4 基準計算的風險加權資產 55.9 47.3 103.2 217.4 255.6 10.2 12.0 598.4 零售銀行及 財富管理 工商金融 環球銀行及 資本市場 環球 私人銀行 其他 總計 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 於2013年1月1日按巴塞爾 協定2.5基準計算的風險 加權資產 163.1 169.0 177.7 9.6 4.5 523.9 匯兌變動 (0.4) (1.5) (2.7) 0.1 (0.1) (4.6) 收購及出售 (10.1) (0.1) (1.6) － － (11.8) 賬項規模 (12.7) 14.5 13.5 (0.7) (0.3) 14.3 賬項質素 (6.4) 3.5 (3.4) 0.3 － (6.0) 模型更新 (0.2) 10.1 (1.0) 2.6 － 11.5 改為採用內部評級基準 計算法的組合 － 10.0 0.8 2.6 － 13.4 新建╱更新模型 (0.2) 0.1 (1.8) － － (1.9) 方法及政策 (2.3) (6.0) 16.0 (1.3) 1.1 7.5 內部修訂 (2.3) (3.4) (0.6) (2.1) － (8.4) 外部修訂 － (2.6) 16.6 0.8 1.1 15.9 風險加權資產變動總額 (32.1) 20.5 20.8 1.0 0.7 10.9 於2013年12月31日按 巴塞爾協定2.5基準 計算的風險加權資產 131.0 189.5 198.5 10.6 5.2 534.8 交易對手信貸風險及市場風險的風險加權 資產 （未經審核） 交易對手信貸風險的風險加權資產 （未經審核） 按資本 指引4基準 按巴塞爾 協定 2.5基準 2014年 2013年 十億美元 十億美元 高級計算法 65.5 42.2 交易對手信貸風險的內部 評級基準計算法 62.0 42.2 信貸估值調整 3.5 － 標準計算法 25.2 3.5 交易對手信貸風險 標準計算法 4.4 3.5 信貸估值調整 18.0 － 中央交易對手 2.8 － 於12月31日的風險加權資產 90.7 45.7 按主要因素分析風險加權資產變動－交易 對手信貸風險－高級計算法 （未經審核） 按資本 指引4基準 按巴塞爾 協定 2.5基準 2014年 2013年 十億美元 十億美元 於1月1日的風險加權資產 42.2 45.7 賬項規模 1.6 (0.9) 賬項質素 (0.6) (2.7) 模型更新 0.1 － 方法及政策 22.2 0.1 內部修訂 (3.8) 0.1 外部監管規定修訂 9.0 － 資本指引4影響 17.0 － 風險加權資產變動總額 23.3 (3.5) 於12月31日的風險加權資產 65.5 42.2 董事會報告：資本 （續） 滙豐控股有限公司 244 風險加權資產╱資本結構 市場風險的風險加權資產 （未經審核） 按資本 指引4基準 按巴塞爾 協定 2.5基準 2014年 2013年 十億美元 十億美元 按內部模式計算法計算 估計虧損風險 7.3 4.9 壓力下之估計虧損風險 10.4 9.4 遞增風險準備 20.1 23.1 全面風險計量 － 2.6 其他估計虧損風險及 壓力下之估計虧損 風險 6.8 12.2 按內部模式計算法計算 44.6 52.2 標準計算法 11.4 11.2 於12月31日 56.0 63.4 按主要因素分析風險加權資產變動－市場 風險－按內部模式計算法計算 （未經審核） 按資本 指引4基準 按巴塞爾 協定 2.5基準 2014年 2013年 十億美元 十億美元 於1月1日的風險 加權資產 52.2 44.5 收購及出售 (2.2) － 風險水平變動 (4.2) (14.5) 模型更新 － 17.6 方法及政策 (1.2) 4.6 內部修訂 (3.8) 4.6 外部修訂 2.6 － 風險加權資產變動總額 (7.6) 7.7 於12月31日的風險 加權資產 44.6 52.2 交易對手信貸風險的風險加權資產 交易對手信貸風險的風險加權資產於2014年 增加450億美元。按標準計算法計算的風險 加權資產增加217億美元，主要由於2014年 1月1日實施資本指引4，當中引入了信貸估 值調整及中央交易對手的風險加權資產。 高級計算法 賬項規模 賬項規模增加主要由於業務變動，以及美 元兌衍生工具合約的其他市值貨幣處於強 勢所產生的影響。 模型更新 在歐洲，英國企業組合的違責損失率下限 使風險加權資產增加22億美元。由於倫敦 所選組合的內部模式計算法 （ 「IMM」 ） 的模 型更新，使風險加權資產減少20億美元， 抵銷了以上增幅。 方法及政策改變 實施資本指引4時引入的信貸估值調整及 金融交易對手的資產值相關倍數於2014年 1月1日分別使風險加權資產增加68億美元 及102億美元。 在外部監管規定及政策修訂的類別中，所選 組合須遵守審慎監管局違責損失率下限， 使風險加權資產增加75億美元，主要在歐 洲及亞洲。此外，2014年第四季接獲的指 引，就出售期權應用 「潛在日後風險」 ，使 風險加權資產增加15億美元。 內部方法修訂引致風險加權資產減少，主 要由於在進行內部盡職審查及評估後的額 外信貸估值調整豁免的變動，以及在歐洲 更有效分配抵押品，使風險加權資產減少 38億美元。 市場風險的風險加權資產 市場風險的風險加權資產總值於2014年減 少74億美元。 標準計算法 組合市場風險的風險加權資產變動並不在 模型計算法的範圍內，故增加了2億美元。 風險加權資產增加26億美元涉及交易賬項 證券化持倉的資本指引4處理方法，而先前 則從資本扣減。利率持倉風險的風險加權 資產減少，抵銷了25億美元 （主要於拉丁美 洲） ，此乃由於就期權引入情景矩陣方法， 同時拉丁美洲及美國的持倉普遍下跌所致。 按內部模式計算法計算 收購及出售 出售我們相關性交易組合，全面風險計量 風險加權資產減少20億美元。出售哈薩克 的業務使風險加權資產減少2億美元。 風險水平變動 風險水平變動主要反映估計虧損風險及壓 力下之估計虧損風險減少，此乃由於外匯 及股權交易持倉減少所致。 方法及政策改變 外部更新資料的風險加權資產增加主要涉 及就有抵押交易而言引入交易貨幣與抵押 品貨幣之間的基準估計虧損風險計算，而 從估計虧損以外風險 （ 「RNIV」 ） 移除分散效 益，從而增加67億美元。 內部更新資料風險加權資產減少43億美元， 抵銷了部分升幅，主要由於微調股票及利 率業務組的估計虧損以外風險的計算所致。 在獲監管機構批准以模型計算市場風險準 備的綜合計算基準變動後，風險加權資產 減少41億美元，令風險加權資產進一步減 少。 營運風險的風險加權資產 營運風險的風險加權資產減少14億美元， 乃由於2012年5月出售的美國卡及零售商戶 業務組合之營運風險的風險加權資產全數 攤銷，加上三年平均營業收益減少所致。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 245 按主要因素分析資本及風險加權資產變 動－資本指引4終點基準 （未經審核） 普通股權 一級資本 風險加權資產 十億美元 十億美元 於2014年1月1日按資本 指引4終點基準計算4 132.5 1,214.9 期內會計利潤 13.7 － 會計利潤的監管規定調整 (1.0) － 股息 （扣除代息股份） 5 (7.5) － 監管規定變動：違責損 失率下限 38.6 企業貸款增長 － 64.8 管理方案： 2.2 (66.3) －既有業務減少及縮減 2.2 (43.0) －出售組合及實體 － (5.2) －風險加權資產方案 － (18.1) 匯兌差額 (8.4) (33.6) 其他變動 4.5 1.4 於2014年12月31日按資本 指引4終點基準計算 136.0 1,219.8 年內風險加權資產增加，主要來自企業貸 款增長及監管規定變動，惟大部分被管理 方案及外匯變動所抵銷。管理方案包括既 有減省及縮減、出售組合及實體，以及多項 其他方案，包括使估計虧損風險更切合管 理層對風險的意見、改善抵押品的配置、 增加使用內部模式計算法及評估產品與監 管規定類別的配對。 資本結構 監管規定資本總額來源及運用 （經審核） 資本指引4 過渡基準 巴塞爾協定 2.5基準 截至2014年 12月31日 止年度 截至2013年 12月31日 止年度 百萬美元 百萬美元 監管規定資本總額變動 期初普通股權╱核心一級資本4 131,233 138,789 期內利潤對普通股權╱核心一級資本的貢益 12,678 17,124 母公司股東應佔綜合利潤 13,688 16,204 扣除本身信貸息差 （除稅淨額） (328) 920 借記估值調整 254 － 保險公司及特設企業取消綜合入賬 (936) － 股息淨額 （包括可預見股息淨額） 5 (7,541) (6,987) 已扣除根據巴塞爾協定2.5確認之代息股份之股息 － (6,987) 第四次股息收取代息股份超出預計部分的更新 1,108 － 已扣除代息股份之第一次股息 (1,766) － 已扣除代息股份之第二次股息 (1,686) － 已扣除代息股份之第三次股息 (1,835) － 第四次可預見股息 (4,131) － 加回：收取代息股份的預計部分 769 － 已扣減之商譽及無形資產減額 2,424 535 已發行普通股 267 297 外幣換算差額 (8,356) (1,294) 其他，包括監管規定調整 2,495 587 期末普通股權╱核心一級資本 133,200 149,051 期初額外╱其他一級資本4 14,408 12,259 已發行之混合資本證券 （已扣除贖回） 4,961 (1,151) 未綜合入賬之投資 17 (2,004) 其他，包括監管規定調整 153 － 期末一級資本 152,739 158,155 期初其他二級資本4 35,538 29,758 已發行二級資本證券 （已扣除贖回） 2,414 1,609 未綜合入賬之投資 26 6,447 其他，包括監管規定調整 13 (1,960) 期末監管規定資本總額 190,730 194,009 有關註釋，請參閱第256頁。 董事會報告：資本 （續） 滙豐控股有限公司 246 資本結構 監管規定資本組合成分 資本指引4過渡基準 巴塞爾協定 2.5基準 於2014年 12月31日 （經審核） 於2013年 12月31日 估計 （未經審核） 於2013年 12月31日 （經審核） 參考 百萬美元 百萬美元 百萬美元 一級資本 股東權益 166,617 164,057 173,449 按資產負債表之股東權益6 a 190,447 181,871 181,871 可預見股息5 (3,362) (3,005) － 優先股溢價 b (1,405) (1,405) (1,405) 其他股權工具 c (11,532) (5,851) (5,851) 特設企業取消綜合入賬7 a (323) (1,166) (1,166) 保險公司取消綜合入賬 a (7,208) (6,387) － 非控股股東權益 4,640 3,644 4,955 按資產負債表之非控股股東權益 d 9,531 8,588 8,588 優先股非控股股東權益 e (2,127) (2,388) (2,388) 撥入二級資本之非控股股東權益 f (473) (488) (488) 取消綜合入賬附屬公司之非控股股東權益 d (851) (757) (757) 不可計入CET1的非控股股東權益餘額 (1,440) (1,311) － 監管規定會計基準調整 (6,309) (2,230) 480 可供出售債務及股票之未變現 （增益） ╱虧損8 (1,378) － 1,121 本身信貸息差9 767 1,112 1,037 借記估值調整 (197) (451) － 界定福利退休基金調整10 g (4,069) (1,731) (518) 物業重估儲備 (1,375) (1,281) (1,281) 現金流對沖儲備 (57) 121 121 扣減項目 (31,748) (34,238) (29,833) 商譽及無形資產 h (22,475) (24,899) (25,198) 須視乎日後盈利能力的遞延稅項資產 （不包括因暫時差異產生之數額） n (1,036) (680) － 額外估值調整 （稱為PVA） (1,341) (2,006) － 透過持有複合產品 （滙豐為其組成部分） 投資於本身 股份 （交易所買賣基金、衍生工具及指數成分股） (1,083) (677) － 證券化持倉之50% － － (1,684) 預期虧損之稅項減免調整之50% － － 151 計算預期虧損金額產生之負數金額 i (5,813) (5,976) (3,102) 普通股權╱核心一級資本 133,200 131,233 149,051 額外一級資本 扣減前之其他一級資本 19,687 14,573 16,110 優先股溢價 b 1,160 1,160 1,405 優先股非控股股東權益 e 1,955 1,955 2,388 可計入額外一級之非控股股東權益 d 884 731 － 混合資本證券 j 15,688 10,727 12,317 扣減項目 (148) (165) (7,006) 未綜合入賬之投資11 (148) (165) (7,157) 預期虧損之稅項減免調整之50% － － 151 一級資本 152,739 145,641 158,155 普通股權一級資本內有51億美元為內部生 成的資本，即母公司股東應佔利潤 （已就 本身信貸息差、借記估值調整、保險公司 取消綜合入賬作出監管規定調整及扣除股 息） 。2014年第四次股息已扣除收取代息股 份的預計部分。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 247 資本指引4過渡基準 巴塞爾協定 2.5基準 於2014年 12月31日 （經審核） 於2013年 12月31日 估計 （未經審核） 於2013年 12月31日 （經審核） 參考 百萬美元 百萬美元 百萬美元 二級資本 扣減前合資格二級資本總額 38,213 35,786 47,812 物業重估及可供出售股票之未變現增益之儲備 － － 2,755 綜合評估減值準備 k － － 2,616 可計入二級資本的非控股股東權益 d 99 86 － 永久後償債務 l 2,218 2,218 2,777 有期後償債務 m 35,656 33,242 39,364 非控股股東權益佔二級資本 f 240 240 300 源自一級資本以外的扣減項目總額 (222) (248) (11,958) 未綜合入賬之投資11 (222) (248) (7,157) 證券化持倉之50% － － (1,684) 計算預期虧損金額產生之負數金額之50% i － － (3,102) 其他扣減項目 － － (15) 監管規定資本總額 190,730 181,179 194,009 有關註釋，請參閱第256頁。 參考索引(a)至(n)項對照載於第249頁的資產負債表組成項目，該等項目用於計算監管規定資本。 採用過渡基準及轉用估計資本指引4終點基準計算的監管規定資本之對賬 （未經審核） 於2014年 12月31日 於2013年 12月31日 估計 百萬美元 百萬美元 過渡基準普通股權一級資本 133,200 131,233 因物業重估產生的未變現增益 1,375 1,281 可供出售儲備的未變現增益 1,378 － 終點基準普通股權一級資本 135,953 132,514 過渡基準額外一級資本 19,539 14,408 獲豁免工具： 優先股溢價 (1,160) (1,160) 優先股非控股股東權益 (1,955) (1,955) 混合資本證券 (10,007) (10,727) 過渡條文： 可計入額外一級資本之非控股股東權益 (487) (366) 未綜合入賬之投資 148 165 終點基準額外一級資本 6,078 365 終點基準一級資本 142,031 132,879 過渡基準二級資本 37,991 35,538 獲豁免工具： 永久後償債務 (2,218) (2,218) 有期後償債務 (21,513) (21,513) 過渡條文： 非控股股東權益佔二級資本 (240) (240) 可計入二級資本之非控股股東權益 396 345 未綜合入賬之投資 (148) (165) 終點基準二級資本 14,268 11,747 終點基準監管規定資本總額 156,299 144,626 董事會報告：資本 （續） 滙豐控股有限公司 248 監管規定資產負債表 按資本指引4過渡基準呈報的資本狀況，遵 從英國透過審慎監管局於2013年12月在政 策聲明 （ 「PS7/13」 ） 中發布的最終規則所實 施的資本指引4法例，並載入審慎監管局規 則手冊內。 我們編列的數字尚未計入歐洲銀行管理局 技術準則草案的影響。我們的資本狀況及 風險加權資產可能因該等準則而受到進一 步影響。 雖然資本指引4允許由2014年1月1日至2018年 1月1日逐步採納普通股權一級的大部分監 管規定調整及扣減項目，但審慎監管局已大 致決定不採用該等過渡條文。因投資物業 及可供出售證券未變現增益不包括在內， 由2015年1月1日起僅可於普通股權一級資 本項內確認，而審慎監管局加速該等項目 的未變現虧損，我們按過渡基準計算的普 通股權一級資本及比率低於按終點基準計 算者。 至於額外一級及二級資本應用所需監管規 定調整及扣減項目的時間，審慎監管局依循 了資本指引4過渡條文的規定。該等調整項 目的影響由2014年1月1日至2018年1月1日 期間將為每年推進20%。 此外，不符合資本指引4規定的額外一級及 二級工具則受惠於一個豁免期。這項安排 將使合資格金額於2014年1月1日初步減少 20%，然後逐年減少10%，直至2022年1月 1日全數剔除。 根據在英國實施的資本指引4，銀行須符 合以下最低比率：普通股權一級比率為風 險加權資產的4% （自2015年1月1日起升至 4.5%） 、一級比率為風險加權資產的5.5% （自2015年1月1日起升至6%） 及總資本比率 為風險加權資產的8%。除了資本指引4的 規定，自2014年7月1日起，審慎監管局預期 主要英國銀行及建屋貸款社的普通股權一 級比率採用資本指引4終點基準定義計算達 到7%。日後，隨著豁免條文失效，我們擬 按需要發行非普通股權資本，以符合經濟 效益的方式滿足該等監管規定最低比率的 規定。於2014年12月31日，集團有符合資本 指引4規定的非普通股權資本票據198億美 元，其中年內發行的二級為35億美元，額 外一級為57億美元 （年內已發行額外一級工 具的詳情參閱財務報表附註35） 。於2014年 12月31日，集團亦有371億美元符合資本指 引4規定的非普通股權資本票據 （應用上述 的20%減額後，因應用豁免條文而可列作 資本指引4下的合資格資本） 。 監管規定資產負債表 監管規定與會計綜合基準 （未經審核） 符合IFRS的財務會計綜合基準載於財務報 表附註1，該等基準與 《2014年第三支柱資 料披露》 報告第13頁 「監管集團的架構」 所 載監管規定綜合基準並不相同。下表呈列 財務會計基準與監管規定基準綜合計算資 產負債表的對賬。 集團在採用財務會計綜合基準處理於銀行 聯營公司之權益時，會以權益法入賬，但 為符合審慎監管局運用歐盟法例監管規定 而計算時，其風險承擔會按比例綜合計算。 若屬從事保險活動的附屬公司，則不會採 用監管規定綜合計算法處理，投資會按成 本列賬。在過往年度，該等保險附屬公司 的投資按資產淨值列賬。有關處理方法由 2014年1月1日起更改，使之與根據資本指 引4的資本處理方法一致，據此，我們的普 通股權一級資本中不再計入收購後儲備， 使有關投資自按成本估值的普通股權一級 資本中扣除 （受限額所限） 。 集團的監管規定綜合賬目並不包括重大風 險已轉移至第三方的特設企業。此等特設 企業的風險承擔已列為證券化持倉並計算 風險加權值，以符合相關監管規定。 有關所用財務會計綜合基準有別於監管規 定綜合基準的公司，詳情請參閱 《2014年第 三支柱資料披露》 報告之表5。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 249 資產負債表對賬－財務會計基準對監管規定基準綜合計算 （未經審核） 於2014年12月31日 經營銀行 保險╱其他 業務的 監管 會計基準 公司取消 聯營公司 規定基準 資產負債表 綜合入賬 綜合入賬 資產負債表 參考 百萬美元 百萬美元 百萬美元 百萬美元 資產 現金及於中央銀行的結餘 129,957 － 30,731 160,688 向其他銀行託收中之項目 4,927 － 80 5,007 香港政府負債證明書 27,674 － － 27,674 交易用途資產 304,193 (720) 2,357 305,830 指定以公允值列賬之金融資產 29,037 (28,791) 3,312 3,558 衍生工具 345,008 (94) 353 345,267 同業貸款 112,149 (2,727) 7,992 117,414 客戶貸款 974,660 (10,809) 116,484 1,080,335 其中： －採用內部評級基準計算法的組合之減值準備 i (6,942) － － (6,942) －採用標準計算法的組合之減值準備 (5,395) － (2,744) (8,139) 反向回購協議－非交易用途 161,713 (30) 7,510 169,193 金融投資 415,467 (50,420) 33,123 398,170 投入保險及其他公司的資本 － 2,542 － 2,542 本期稅項資產 1,309 (16) － 1,293 預付款項、應計收益及其他資產 75,176 (5,295) 8,501 78,382 其中： －持作出售用途業務組合的商譽及無形資產 h 8 － － 8 －退休福利資產 g (5,028) － － (5,028) －持作出售用途資產的減值準備 (16) － － (16) 其中： －採用內部評級基準計算法的組合 i (16) － － (16) －採用標準計算法的組合 － － － － 於聯營及合資公司之權益 18,181 － (17,479) 702 其中： －收購時產生的商譽正數值 h 621 － (606) 15 商譽及無形資產 h 27,577 (5,593) 571 22,555 遞延稅項資產 n 7,111 163 474 7,748 資產總值 2,634,139 (101,790) 194,009 2,726,358 負債及股東權益 香港紙幣流通額 27,674 － － 27,674 同業存放 77,426 (21) 40,530 117,935 客戶賬項 1,350,642 (535) 141,858 1,491,965 回購協議－非交易用途 107,432 － － 107,432 向其他銀行傳送中之項目 5,990 (3) － 5,987 交易用途負債 190,572 (42) 50 190,580 指定以公允值列賬之金融負債 76,153 (6,317) － 69,836 其中： －有期後償債務佔二級資本 m 21,822 － － 21,822 －混合資本證券佔一級資本 j 1,495 － － 1,495 衍生工具 340,669 37 331 341,037 已發行債務證券 95,947 (7,797) 3,720 91,870 本期稅項負債 1,213 (138) 317 1,392 保單未決賠款 73,861 (73,861) － － 應計項目、遞延收益及其他負債 53,396 (3,659) 5,145 54,882 其中： －退休福利負債 3,208 (2) 56 3,262 －或有負債及合約承諾 234 － － 234 其中： － 採用內部評級基準計算法的組合之信貸相關準備 i 132 － － 132 －採用標準計算法的組合之信貸相關準備 102 － － 102 準備 4,998 (63) － 4,935 遞延稅項負債 1,524 (1,009) 2 517 後償負債 26,664 － 2,056 28,720 其中： －混合資本證券佔一級資本 j 2,761 － － 2,761 －永久後償債務佔二級資本 l 2,773 － － 2,773 －有期後償債務佔二級資本 m 21,130 － － 21,130 董事會報告：資本 （續） 滙豐控股有限公司 250 監管規定資產負債表╱槓桿比率 於2014年12月31日 經營銀行 保險╱其他 業務的 監管 會計基準 公司取消 聯營公司 規定基準 資產負債表 綜合入賬 綜合入賬 資產負債表 參考 百萬美元 百萬美元 百萬美元 百萬美元 股東權益總額 a 190,447 (7,531) － 182,916 其中： －其他股權工具佔一級資本 c, j 11,532 － － 11,532 －優先股溢價佔一級資本 b 1,405 － － 1,405 非控股股東權益 d 9,531 (851) － 8,680 其中： －由附屬公司發行的非累積優先股佔一級資本 e 2,127 － － 2,127 －非控股股東權益佔二級資本，累積優先股 f 300 － － 300 －附 屬公司普通股持有人 應佔非控股股東權益佔二級資本 f, m 173 － － 173 於2014年12月31日的負債及股東權益總額 2,634,139 (101,790) 194,009 2,726,358 於2013年12月31日 經營銀行 保險╱其他 業務的 監管 會計基準 公司取消 聯營公司 規定基準 資產負債表 綜合入賬 綜合入賬 資產負債表 參考 百萬美元 百萬美元 百萬美元 百萬美元 資產 交易用途資產 303,192 32 1,686 304,910 客戶貸款 1,080,304 (13,182) 110,168 1,177,290 其中： －採用內部評級基準計算法的組合之減值準備 i (9,476) － － (9,476) －採用標準計算法的組合之減值準備 k (5,667) － (2,465) (8,132) 金融投資 425,925 (52,680) 31,430 404,675 投入保險及其他公司的資本 － 9,135 － 9,135 於聯營及合資公司之權益 16,640 － (15,982) 658 其中： －收購時產生的商譽正數值 h 608 － (593) 15 商譽及無形資產 h 29,918 (5,369) 631 25,180 其他資產 815,339 (37,634) 57,477 835,182 其中： －持作出售用途業務組合的商譽及無形資產 h 3 － － 3 －退休福利資產 g 2,140 － － 2,140 －持作出售用途資產的減值準備 (111) － － (111) 其中： －採用內部評級基準計算法的組合 i － － － － －採用標準計算法的組合 k (111) － － (111) 於2013年12月31日的資產總值 2,671,318 (99,698) 185,410 2,757,030 負債及股東權益 同業存放 129,212 (193) 33,296 162,315 客戶賬項 1,482,812 (711) 142,924 1,625,025 交易用途負債 207,025 (129) 161 207,057 指定以公允值列賬之金融負債 89,084 (13,471) － 75,613 其中： －有期後償債務佔二級資本 m 18,230 － － 18,230 －混合資本證券佔一級資本 j 3,685 － － 3,685 已發行債務證券 104,080 (9,692) 1,021 95,409 退休福利負債 g 2,931 (11) 56 2,976 後償負債 28,976 2 2,961 31,939 其中： －混合資本證券佔一級資本 j 2,873 － － 2,873 －永久後償債務佔二級資本 l 2,777 － － 2,777 －有期後償債務佔二級資本 m 23,326 － － 23,326 其他負債 436,739 (73,570) 4,991 368,160 其中： －或有負債及合約承諾 177 － － 177 其中： －採用內部評級基準計算法的組合之信貸相關準備 i 155 － － 155 －採用標準計算法的組合之信貸相關準備 k 22 － － 22 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 251 於2013年12月31日 經營銀行 保險╱其他 業務的 監管 會計基準 公司取消 聯營公司 規定基準 資產負債表 綜合入賬 綜合入賬 資產負債表 參考 百萬美元 百萬美元 百萬美元 百萬美元 股東權益總額 a 181,871 (1,166) － 180,705 其中： －其他股權工具佔一級資本 c, j 5,851 － － 5,851 －優先股溢價佔一級資本 b 1,405 － － 1,405 非控股股東權益 d 8,588 (757) － 7,831 其中： －由附屬公司發行的非累積優先股佔一級資本 e 2,388 － － 2,388 －非控股股東權益佔二級資本，累積優先股 f 300 － － 300 － 附屬公司普通股持有人 應佔非控股股東權益佔二級資本 f, m 188 － － 188 於2013年12月31日的負債及股東權益總額 2,671,318 (99,698) 185,410 2,757,030 參考索引(a)至(n)項對照載於第246頁的資產負債表組成項目，該等項目用於計算監管規定資本。 槓桿比率 （未經審核） 有關槓桿比率的編製基準詳情，請參閱第261頁資本附錄。 估計槓桿比率 （未經審核） 於2014年 12月31日 按歐盟授權 法案的基準 於2013年 12月31日 按巴塞爾協定3 2010年的基準 十億美元 十億美元 按會計基準資產負債表列示的資產總值 2,634 2,671 保險╱其他公司取消綜合入賬 (104) － 投資於保險公司之資本 2 － 經營銀行業務的聯營公司綜合入賬 194 － 按監管規定╱會計基準資產負債表列示的資產總值 2,726 2,671 就撥回IFRS准予按淨額計算之貸款及存款而調整 38 93 撥回會計價值： (525) (482) 衍生工具 (345) (282) 回購協議及證券融資 (180) (200) 以應用監管規定規範後的價值代替： 衍生工具： 166 239 按市值計算的價值 81 69 因現金變動保證金而扣減應收資產 (82) － 日後潛在風險額外金額 148 170 因承辦的信貸衍生工具按額外規定處理而產生的風險金額 19 － 回購協議及證券融資： 188 147 證券融資交易資產總值 269 － 證券融資交易資產總值的現金應付及應收款項淨額 (89) － 根據巴塞爾協定3 2010年框架扣除的證券融資交易資產 147 交易對手風險之計量 8 － 加入資產負債表外承諾及擔保： 396 388 擔保及或有負債 67 85 承諾 321 295 其他 8 8 不計入已從資本數值扣減的項目 (36) (28) 作出監管規定調整後的風險數值 2,953 3,028 資本指引4下的一級資本 （終點基準） 142 133 估計槓桿比率 （終點基準） 4.8% 4.4% 於2014年1月，巴塞爾委員會公布最終槓桿 比率架構，連同由2015年1月1日起適用的 公開披露規定，更新其2010年的建議。 於2014年6月，審慎監管局公布修訂當局對 於主要的英國銀行及建屋貸款社的槓桿比 率的期望，即自2014年7月1日起，我們的 董事會報告：資本 （續） 滙豐控股有限公司 252 監管環境發展 終點基準一級槓桿比率應達到3%，當中分 子採用資本指引4資本的定義計算，分母採 用巴塞爾協定2014年的風險數值計算。 於2014年10月，歐洲委員會採納授權法案 為歐洲銀行的槓桿比率設定一般定義 （根 據巴塞爾修訂定義） ，乃於2015年1月在 《歐 盟公報》 公布。 根據資本指引4，槓桿比率的立法建議及最 終校準，預期將取決於巴塞爾委員會修訂 建議的檢討，以及歐洲銀行管理局於2014年 1月1日至2016年6月30日的監察期間對槓桿 比率之影響及有效性所作評估的檢討。 於2015年1月，審慎監管局發出函件，載列 在2014年底披露槓桿比率所採用的計算方 法。比率的分子繼續使用資本指引4終點基 準一級資本的最終定義計算，而風險數值 現時則根據歐盟授權法案 （而非 《2014年中 期業績報告》 所用巴塞爾協定2014年定義） 計算。業績按歐盟授權法案的基準 （而非巴 塞爾協定2014年的定義） 呈報，輕微產生2個 基點的正數差額。 於上文及我們 《2013年報及賬目》 內所披露 的2013年槓桿比率是根據審慎監管局指示 採用的巴塞爾協定2010年文本釐定。呈報 基準由巴塞爾協定2010年文本改為歐盟授 權法案使風險數值增加1,150億美元。主要 的變動包括： ‧ 監管規定基準綜合計算的變動使風險 數值增加1,320億美元。 ‧ 證券融資交易 （SFT） 的淨額計算以會計 準則為基準，而交易對手風險的額外增 加部分使風險數值增加660億美元。 ‧ 按名義金額計值加入承辦的信貸衍生 工具使風險數值增加230億美元。 ‧ 允許以現金變動保證金對銷衍生工具資 產及負債的修訂使風險數值減少650億 美元。 ‧ 應用於資產負債表外風險的信貸換算 因素改變使風險數值減少410億美元。 有關編製基準的進一步詳情，請參閱第261頁。 應當注意的是，金融政策委員會 （ 「FPC」 ） 於2014年10月公布的英國特定槓桿比率建 議，在概念上與巴塞爾協定及資本指引4槓 桿架構不同，且尚未實行。有關英國建議 的進一步資料詳情，請參閱第255頁 「槓桿 比率建議」 一節。 監管環境發展 （未經審核） 監管規定緩衝資本 資本指引4設立數項須以普通股權一級資本 達致的緩衝資本，大致與巴塞爾協定3架構 一致。資本指引4預期上述緩衝將由2016年 1月1日起分階段實施，惟各國家╱地區有 其酌情權。 倘銀行的普通股權一級資本低於資本指 引4綜合緩衝水平 （界定為防護緩衝資本 （ 「CCB」 ） 、反周期緩衝資本 （ 「CCyB」 ） 、全 球系統重要性機構 （ 「G-SII」 ） 緩衝及系統性 風險緩衝 （ 「SRB 」 ） 的總和 （如該等緩衝適 用） ） ，則適用資本分派自動限制。審慎監 管局建議使用審慎監管局緩衝將不會導致 資本分派的自動限制。 於2014年4月，英國財政部公布法定文件 「2014年資本規定 （緩衝資本及宏觀－審慎 措施） 條例」 ，將資本指引4內與緩衝資本相 關的主要條文引入英國法例。惟系統性風 險緩衝屬例外。於2015年1月，英國財政部 已公布該法定文件的修訂本，以納入系統 性風險緩衝。 審慎監管局為監管全球系統重要性機構緩 衝、其他系統重要性機構 （ 「O-SII」 ） 緩衝及 防護緩衝資本的指定機構。於2014年4月， 審慎監管局公布規則及監管聲明，實施資 本指引4有關該等緩衝的主要條文。英倫銀 行為監管反周期緩衝資本及其他宏觀－審 慎措施的指定機構。而審慎監管局為應用 及釐定系統性風險緩衝的指定機構，金融 政策委員會負責設立系統性風險緩衝架構 修正。 全球系統重要性機構緩衝 全球系統重要性機構緩衝 （即歐盟實施的巴 塞爾協定 「全球系統重要性銀行」 緩衝） 須通 過普通股權一級資本達致，並將自2016年 1月1日起分階段實行。2014年10月，最終 技術準則及識別全球系統重要性機構的方 法指引乃於 《歐盟公報》 公布，並將自2015年 1月1日起生效。 於2014年11月，金融穩定委員會及巴塞爾 委員會使用2013年底數據更新全球系統重 要性銀行名單。滙豐先前被設定的額外權 數2.5%維持不變。 遵照審慎監管局於2014年4月在其公布的監 管聲明中對英國銀行發出的指示以及根據 於2014年6月公布的歐洲銀行管理局實施技 術準則終稿及指引，我們於2014年7月公布 歐洲銀行管理局模板，披露用作認定的資 料並重點列明作為指定全球系統重要性銀 行之根本依據的程序。披露規定最終實施 技術準則乃於2014年9月公布，並將構成我 們未來2015年披露全球系統重要性機構指 標的基準。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 253 防護緩衝資本 設立防護緩衝資本旨在確保銀行在受壓 期間以外的時間建立資本，以備一旦出現 虧損時可以取用，並設定為風險加權資產 的2.5%。審慎監管局將於2016年1月1日至 2019年1月1日逐步採用該項緩衝。 反周期及其他宏觀－審慎緩衝 資本指引4如巴塞爾協定3一樣，致力於設 立反周期緩衝，形式為機構特有的反周期 緩衝資本，並將增設規定以應對宏觀－審 慎或系統性風險。 於2014年1月，金融政策委員會發表政策聲 明，說明其有權透過反周期緩衝資本及類 別資本規定 （ 「SCR」 ） 工具，補充資本規定。 儘管當局未設上限，但預期反周期緩衝資 本設定在相關信貸風險承擔的風險加權資 產0-2.5%範圍內。根據英國法例，金融政 策委員會亦須於2016年前決定是否確認其 他歐洲經濟區國家╱地區設定的任何反周 期緩衝資本比率。 於2014年6月，金融政策委員會將英國風 險承擔的反周期緩衝資本比率定為0%。於 2014年9月會議，金融政策委員會將英國風 險承擔的反周期緩衝資本比率維持在0%不 變，並確認挪威及瑞典所引入1%的反周期 緩衝資本比率自2015年10月3日起生效。於 2015年1月，香港金管局公布，香港的風險 承擔反周期緩衝 資本比率採納為0.625%， 自2016年1月1日起適用。按英國法例及審慎 監管局監管聲明PS 3/14，此比率將自2016年 1月1日起直接應用於計算我們的機構特有 的反周期緩衝資本比率。 集團的機構特有的反周期緩衝資本比率將 基於相關信貸風險承擔所在司法管轄區適 用的反周期緩衝資本比率的加權平均值計 算。現時集團的機構特有的反周期緩衝資 本比率為零。類別資本規定工具現不在英 國予以使用。 系統性風險緩衝 除上述措施外，資本指引4為金融業整體， 或一個或多個行業類別設定了系統性風險 緩衝，各歐盟成員國可按需要實施以降低 結構性的宏觀－審慎風險。 於2015年1月，將實施納入系統性風險緩衝 必要的法律規定轉變。系統性風險緩衝將 適用分隔運作銀行及建屋貸款社 （超過若 干限制） ，其共同界定為 （ 「系統性風險緩衝 機構」 ） 。系統性風險緩衝可應用於個別、 初步綜合或綜合基準，並自2019年1月1日 起適用。截至2016年5月31日，金融政策委 員會須設立架構，用以確認系統性風險緩 衝機構未能運作或財困將產生若干長期非 周期系統或宏觀－審慎風險的程度。審慎 監管局將應用此架構以釐定是否特定系統 性風險緩衝機構將按系統性風險緩衝率、 應用緩衝的水平及能否行使監管判斷以釐 定應計利率。倘適用，緩衝率須設定在介 乎1%至3%。緩衝率將適用於所有系統性 風險緩衝機構風險，除非審慎監管局已確 認於另一成員國設定緩衝率則另作別論。 倘按綜合基準應用系統性風險緩衝，則根 據資本指引4，預期將應用全球系統重要性 機構緩衝或系統性風險緩衝，以較高者為 準。 第二支柱及 「審慎監管局緩衝」 根據第二支柱架構，銀行早已規定須因應 內部資本充足程度評估及監管檢討程序持 有資本。此等程序促使審慎監管局最終釐 定第二A支柱及第二B支柱個別資本指引。 第二A支柱過往可以總資本達致要求，但自 2015年1月1日起，按照審慎監管局監管聲 明SS 5/13，須以至少56%普通股權一級資 本達致要求。 第二A支柱指引為對審慎監管局認為銀行 應持有以符合總體財務充足性規則之資本 數額的時間點評估，因此在年度評估及監 管檢討過程中可能出現變動。於2014年， 集團的第二A支柱指引為風險加權資產的 1.5%，其中0.9%符合普通股權一級比率。 於2015年2月，指引修改為風險加權資產的 2.0%，其中1.1%符合普通股權一級比率並 即時生效。 於2015年1月，審慎監管局公布第二支柱架 構諮詢，載列審慎監管局建議使用的方法， 以影響其制訂公司第二支柱資本規定，包 括建議新計算方法以釐定信貸風險、營運 風險、信貸集中風險及退休金責任風險的 第二支柱規定。 作為資本指引4實施的一部分，審慎監管局 建議引入審慎監管局緩衝，取代資本計劃 緩衝 （ 「CPB」 ） （第二B支柱） ，同樣以普通股 權一級資本形式持有。此於近期審慎監管 局的第二支柱架構諮詢中再次得到確認。 其建議審慎監管局緩衝將避免與資本指引 4緩衝重複，並視乎在壓力情況漏洞或審 慎監管局確定風險管理及管治失誤而設定 特定公司緩衝。為解決風險管理及管治弱 點，審慎監管局建議公司普通股權一級資 本第一支柱及第二A支柱資本規定應用等 級。倘審慎監管局認為資本指引4緩衝及 審慎監管局緩衝評估重複，則審慎監管局 建議設定審慎監管局緩衝作為防護緩衝資 本及相關系統性緩衝的超額資本規定。然 而，審慎監管局緩衝將附加於反周期緩衝 資本及類別資本規定。 審慎監管局預期於2015年7月落實第二支柱 架構，並預期自2016年1月1日起實施。在 諮詢落實結果以及修訂規則及指引公布之 前，確切的緩衝率規定及對資本的最終影 響仍存在不明朗因素。 董事會報告：資本 （續） 滙豐控股有限公司 254 監管環境發展 整體資本規定 在出現上文所述的發展後，資本規定架構 的若干成分詳情開始出現。然而，仍未能 精確地確定滙豐的普通股權一級資本的終 點基準資本規定水平。目前已知或量化的 資本規定成分乃載於下圖。時間轉變的元 素 （例如宏觀－審慎工具、第二支柱規定及 系統性緩衝） 可予變動。 資本規定架構 （終點基準） 審慎監管局緩衝 （說明性質） 防護緩衝資本 系統性緩衝 (CET1) (CET1) (CET1) (CET1 ) 審慎監管局 緩衝評估 （取代CPB） 2.5% 2.5% 2.0% 8% （其中CET1為1.1%） (SRB/G-SIIB) （其中CET1為4.5%） 第二A支柱╱個別資本指引 宏觀－審慎監管工具 （反周期緩衝資本╱ 類別資本規定） (CET1、額外 一級及二級） (CET1 、額外 一級及 二級） 第一支柱 除上文表格內的資本規定外，我們須考慮 金融穩定委員會於2014年11月公布有關整 體損失吸納能力 （ 「TLAC」 ） 規定之建議的影 響。有關詳情，請參閱第256頁。 監管規定壓力測試 集團須在多個司法管轄區進行監管機構規 定的壓力測試。監管規定的壓力測試越來 越頻密，並要求更詳盡的結果。因此，壓 力測試為集團的一項工作重點。 於2013年10月，英倫銀行發表首次討論文 件 《英國銀行體系壓力測試架構》 。該架構取 代現行有關資本計劃緩衝的壓力測試 （每年 為不同機構同時進行） ，預期其結果將影響 審慎監管局緩衝、反周期緩衝資本、類別 資本規定的制訂及金融政策委員會向審慎 監管局作出的其他建議。於2014年4月，英 倫銀行公布英國壓力測試活動的詳情，集 團其後參與其中。此活動的結果於2014年 12月公布。 於2014年間，集團參與多個不同司法管轄 區的不同壓力測試活動。有關全部壓力測 試的進一步詳細資料，請參閱第122頁。 風險加權資產發展 於2014年間，監管機構發布一系列建議及 諮詢，旨在修改風險加權資產制度的不同 組成部分，以及增加相關報告及披露。 英國 金融政策委員會於2014年3月公布，其獲提 示於2015年在實際可行情況下盡快建議各 公司採用信貸風險標準計算法申報及披露 資本比率，惟須在對標準計算法進行巴塞 爾審查之後進行。 於2014年6月，審慎監管局公布諮詢文件 CP12/14，建議對信貸風險規則作出兩方面 修訂。首先，建議將有關中央政府、公營單 位、中央銀行及金融業公司使用內部評級基 準高級計算法 （ 「AIRB」 ） 計算的風險承擔，自 2015年6月起根據基礎計算法計算。其次， 建議引入更為嚴格的標準，對位於非歐洲 經濟區國家╱地區內的若干商用物業風險 承擔應用標準化風險權數 （視乎具代表性期 間該等司法管轄區的虧損率而定） 。於10月， 審慎監管局發表政策聲明 （ 「PS10/14」 ） ， 載有第二項建議的最終規則，由2015年4月 起，對非歐洲經濟區的商用物業風險承擔 引入更嚴格的標準。 歐盟 於2014年5月，歐洲銀行管理局就用於計算 信貸及市場風險承擔 （風險加權資產） 之本 身資金需求的內部計算法基準公布諮詢文 件。此乃遵從於2013年進行的一系列基準 行動，以便更好地理解在歐盟各機構的風 險加權資產中觀測到的多種差異的推動因 素。諮詢文件內概述的未來年度基準行動 旨在提高採用內部模式計算法計算的資本 規定之可比較性並由監管機構用於影響其 政策決定。 於2014年6月，歐洲銀行管理局發表有關以 永久部分使用處理的風險應用標準計算法 的限制，以及內部評級基準開展計劃的諮 詢文件。最終監管技術準則 （ 「RTS」 ） 尚未公 布。 於2014年12月， 《歐盟公報》 發表被視為設 有與資本指引4等效的監管制度的非歐洲 經濟區國家╱地區名單。此乃於2015年1月 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 255 1日生效。等效評估影響資本指引4若干不 同方面的風險處理，如第三方國家╱地區 投資公司、信貸機構及交易所風險處理； 向中央政府、中央銀行、地區政府、地方 機構及公營單位應用標準化風險加權計算 法；及按內部評級基準計算法計算企業、 機構、中央政府及中央銀行承擔的風險加 權資產計算。 國際 於2014年間，巴塞爾委員會公布全部第一 支柱風險類型建議，以更新計算資本規定 的標準化、非模型方法，並提供實行資本 下限的基準。 尤其是，於2014年3月，巴塞爾委員會公布 場外交易衍生工具、交易所買賣衍生工具 及長期結算交易中，有關交易對手信貸風 險計算的標準計算法最終建議。此後，於 2014年8月刊發新標準計算法基礎的另一 技術文件。委員會建議以新方法替換現時 風險數值及標準化計算法，預期於2017年 1月1日生效。 於2014年10月，巴塞爾委員會亦公布修訂 計算營運風險的標準計算法的諮詢文件及 定量影響研究 （ 「QIS」 ） 。建議尋求建立新的 統一標準計算法以替換現時非模型基準計 算法，當中包括基本指標計算法及標準計 算法，包括其變量替代標準計算法。實施 日期尚待建議。 於2014年12月，巴塞爾委員會就徹底檢討交 易賬項進行進一步諮詢，當中納入2013年 10月為進行諮詢而公布的市場風險架構修 訂。委員會擬於2015年初進行進一步定量 影響研究，預期於2015年底落實建議。 於2014年12月，巴塞爾委員會公布證券化 風險的修訂架構，將於2018年1月1日生效。 於2014年12月，巴塞爾委員會亦公布信貸 風險標準計算法修訂本的諮詢文件。建議 包括減少依賴外界的信貸評級；增加精細 度及風險敏感度；及更新風險加權修正。 風險加權建議修正乃僅屬指示，並將由負 責定量影響研究諮詢及結果人士進一步公 布。 此外，於2014年12月，巴塞爾委員會公布 諮詢設計資本下限架構，此將替換巴塞爾 協定1下限。然而，下限修正為該諮詢範圍 外。委員會已表明有意於2015年底前公布 最終建議，包括修正及實施時間表。 計算風險規定的標準計算法及引入修訂資 本下限的巴塞爾委員會全部最終建議，於 具法律效力前，須納入歐盟規定。 槓桿比率建議 於2014年10月，金融政策委員會公布英國特 定槓桿比率架構設計及修正的最終建議。 此乃跟隨金融政策委員會早前在2014年7月 的設計架構。金融政策委員會最終建議包 括將於實際可行情況下盡快實施英國全球 系統重要性銀行及主要英國銀行及建屋貸 款社的3%最低槓桿比率，適用於系統重要 性公司的相關風險加權系統性風險緩衝率 35%的補充槓桿比率緩衝，以及相關風險 加權反周期緩衝資本35%的進一步反周期 槓桿比率緩衝 （ 「CCLB」 ） 。最低槓桿比率將 以75%普通股權一級資本達致及25%額外 一級資本達致，而補充槓桿比率緩衝及反 周期槓桿緩衝均以100%普通股權一級資本 達致。金融政策委員會建議英國財政部向 其提供必要權力，指示審慎監管局實施上 述修正及架構以制訂槓桿比率規定。 英國財政部於2014年11月公布諮詢文件， 回應及同意金融政策委員會有關槓桿比率 架構的設計建議。特別是，英國財政部同意 金融政策委員會將獲授權指示審慎監管局 制訂最低規定、額外槓桿比率緩衝 （就環球 系統重要性銀行、主要英國銀行及建屋貸 款社 （包括分隔運作的銀行） ） 及反周期槓桿 比率緩衝。然而，英國財政部並無就修正 提供任何意見。諮詢文件包括立法變動以 向金融政策委員會提供新權力。於2015年 2月，英國財政部公布回覆概要，並於議會 提請草案。 銀行業結構性改革、復元及解決計劃 歐盟落實及於2014年6月公布銀行復元和解 決指引 （ 「BRRD」 ） 。指引於2015年1月1日生 效，並可選擇延遲至2016年1月1日實施自救 機制。儘管設有此選擇，英國自2015年1月 1日引入自救機制權力。英國銀行復元和 解決指引過渡建立在英國已制訂的解決架 構上。於2015年1月，審慎監管局公布載有 復元及解決計劃更新規定的政策聲明，該 聲明修訂自2014年1月1日起生效的審慎監 管局規則。此外，歐洲銀行管理局已制訂 若干監管技術準則，其中部分有待落實， 此將進一步影響銀行復元和解決指引的要 求。 董事會報告：資本 （續） 滙豐控股有限公司 256 監管環境發展╱資本附錄 2013年12月，英國 《2013年金融服務 （銀行 業改革） 法》 獲御准，可以實施銀行業獨立 委員會分隔運作的建議，並以二級立法 （於 2014年7月落實） 加以補充。2014年10月， 審慎監管局公布分隔運作規則的諮詢文件。 審慎監管局擬在適當時進一步諮詢及落實 分隔運作規則，並於2019年1月1日前實施。 於2014年1月，歐洲委員會亦公布交易活動 與存款業務分隔運作以及禁止金融工具及 商品的坐盤交易的立法建議。該等建議現 時有待歐洲議會及部長理事會討論。 政策背景及集團復元和解決計劃方針的進一步詳 情，請參閱第14頁。 整體損失吸納能力建議 於2014年11月，為處理 「大到不能倒」 的問 題，金融穩定委員會公布環球系統重要性 銀行的整體損失吸納能力 （ 「TLAC」 ） 建議。 金融穩定委員會建議包括整體損失吸納能 力最低要求介乎風險加權資產的16-20%， 以及整體損失吸納能力槓桿比率至少為巴 塞爾協定3一級槓桿比率的兩倍。整體損失 吸納能力將根據全球系統重要性銀行的危 機管理小組釐定的個別解決策略應用。定 量影響研究正在進行中，其結果將影響最 終建議。整體損失吸納能力規定的一致期 亦將受定量影響研究影響，但將不會設於 2019年1月1日前。一旦完成，預期任何新 的整體損失吸納能力標準將達致巴塞爾協 定3最低資本規定。 建議草案要求環球系統重要性銀行受最低 整體損失吸納能力規定規限，確切規定將 視乎定量影響研究。根據其解決策略，可 用於達致整體損失吸納能力規定的負債類 別、整體損失吸納能力組成部分及銀行集 團內負債分布方面均設有多項規定。整體 損失吸納能力建議預期於2015年落實，屆 時須於全國立法實施。 其他監管規定修訂 於2015年1月，歐洲銀行管理局發表有關審 慎估值的監管技術準則經修訂最終草稿。 最終規定須由歐洲委員會採納，並於生效 前在 《歐盟公報》 公布。 於2014年6月，歐洲銀行管理局及巴塞爾委 員會各自就第三支柱資料披露刊發諮詢文 件。歐洲銀行管理局最終指引於2014年12月 發布，並涉及有關第三支柱報告額外程序 及管治；以及主要資本、比率、風險加權 資產、槓桿及風險模型資料的半年度或季 度資料披露，超過我們目前中期披露的範 圍。指引由國家監管機構實施，並預期於 2015年內生效。 「經修訂第三支柱披露規定」 的最終巴塞爾 協定標準乃於2015年1月公布。有關標準指 示廣泛使用標準化範本，以提高各銀行披 露間的可比較性，以及規定相當大量的披 露項目須每半年進行一次，而非至今為止 的每年一次。經修訂架構要求最遲於2016年 底與財務報告同步作出披露。 資本註釋 1 根據標準計算法計算的營運風險之風險加權資產乃使用過去三年收入的平均數計算。在出售業務方面， 營運風險的風險加權資產並非於出售時立即解除，而是於一段期間內逐步減少。卡及零售商戶業務之風 險加權資產為業務剩餘之營運風險的風險加權資產。 2 由於集團內之市場風險分散效應，風險加權資產並非各個地區相加的總和。 3 自2014年1月1日起， 「亞洲」 地區取代之前按 「香港」 及 「亞太其他地區」 呈列的地區 （有關進一步詳情，請 參閱財務報表附註23） 。比較數字已重列以反映有關變動。 4 於2013年12月的資本指引4期初結餘是根據集團對資本指引4最終規例及審慎監管局公布的最終規則的詮 釋釐定，有關詳情載於 《2013年報及賬目》 第324頁的編製基準內。 5 包括就普通股宣派的股息、就優先股宣派的季度股息及分類為權益之資本證券的票息。 6 包括截至2014年12月31日止年度由外界核實的利潤。 7 主要包括與特設企業有關的可供出售債務證券之未變現損益。 8 可供出售證券未變現損益已扣除稅項。 9 包括交易用途負債的本身信貸息差。 10 根據巴塞爾協定2.5規則，任何界定福利資產均會撤銷確認，而界定福利負債則可以運用未來五年內支付 予有關計劃的額外資金取代。 11 主要包括於保險公司的投資。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 257 資本附錄 資本管理 （經審核） 方法及政策 我們管理資本的方法，是基於業務所在地的監管、經濟及工商業環境，按集團的策略及組 織架構所需而制訂。除稅前風險加權資產平均值回報是一項衡量營運表現的標準，我們據 此每日管理各項環球業務。該衡量標準結合了股東權益回報率與監管規定下的資本效益目 標。我們的宗旨是維持雄厚的資本，以抵禦旗下各項業務的內在風險，並根據包含六方面 考慮的架構進行投資，使資本於任何時候均高於綜合計算及各地監管規定的資本水平。 資本管理架構是集團資本管理政策的核心所在，讓我們能以貫徹如一的方式管理本身的資 本。此架構經集團管理委員會每年核准，其中包括市值、投入資本、經濟資本及監管規定資 本等多項不同的資本計量標準。鑑於資本指引4已自2014年1月1日起生效，故我們於2014年 按資本指引4普通股權一級終點基準管理內部資本比率目標 （高於10%） 。自此我們持續檢討 達標情況，而於2015年，我們預期管理集團資本能達到10%以上股東權益回報率的中期目 標。這是以12%至13%的普通股權一級比率終點基準作為模型參考而制訂。 資本計量標準 ‧ 市值為滙豐在股市中的價值； ‧ 投入資本為股東投資於滙豐的股本，且已就若干過往攤銷或撇銷的儲備及商譽作出調 整； ‧ 經濟資本為內部計算的資本規定，即我們認為用以支持集團面對風險所需的資本規 定；及 ‧ 監管規定資本是我們遵照審慎監管局就綜合集團所訂規則，以及我們業務所在地監管 機構為集團旗下個別公司所訂規則，因而需要持有的資本。 我們對資本充足程度的評估與對各項風險的評估保持一致，該等風險包括：信貸、市場、營 運、銀行賬項的利率風險、退休金、保險、結構匯兌風險及剩餘風險。 壓力測試 除內部壓力測試外，集團須在多個司法管轄區進行監管機構規定的壓力測試。監管規定的 壓力測試越來越頻密，並要求更精細的結果。該等測試包括審慎監管局、聯邦儲備局、歐洲 銀行管理局、歐洲央行及香港金融管理局規定的測試，以及在多個其他司法管轄區進行的 壓力測試。我們在評估內部資本要求時，會考慮所有該等監管規定的壓力測試結果。 資本風險 除壓力測試架構外，我們會定期評估多項首要及新浮現風險對普通股權一級資本比率的影 響。此外，我們可能發現其他風險，或會影響我們的風險加權資產及╱或資本狀況。該等風 險亦會納入我們對資本風險的評估。我們會根據資本管理目標評估下跌或上升的情況，並 於有需要時採取減低風險的措施。集團管理委員會負責制訂全球資本分配的原則及決策。 我們透過內部管治程序嚴控各項投資及資本分配決策，設法確保投資回報符合集團的管理 目標。我們的策略是根據各項業務及屬下公司達成既定風險加權資產平均值回報目標的能 力、其監管規定資本及經濟資本的規定，將資本分配至不同業務及公司。 風險加權資產目標 我們根據集團的策略方向及承受風險水平，為各項環球業務訂立風險加權資產目標，並透 過集團年度計劃程序批准。由於該等目標是指派予較低層的管理人員，故須制訂行動計劃。 有關計劃可能包括增長策略、積極的組合管理、重組、業務及╱或客戶層面的檢討、風險加 權資產準確性及分配措施，以及減低風險的行動。我們的資本管理程序在經董事會批准的 集團年度資本計劃 （構成年度營運計劃的一部分） 內清楚說明。 透過定期向集團資產負債管理委員會匯報，集團的業績表現會根據該等風險加權資產目標 受到監察。資本扣減項目亦透過風險加權資產的監察架構來管理，方法是計算此等項目的 額外名義扣取額。 董事會報告：資本 （續） 滙豐控股有限公司 258 資本附錄 風險加權資產監察架構採用一系列分析，以識別造成相關持倉水平變動的主要因素，如賬 項規模及賬項質素等，且尤其著重識別及劃分可由日常業務控制的項目，以及受風險模型 或監管規定計算方法改變影響的項目。 資本生成 滙豐控股乃各附屬公司的主要股本提供者，在需要時亦向該等附屬公司提供非股本資本。 該等投資的資金大多數來自滙豐控股本身的資本發行及保留利潤。滙豐控股透過資本管理 程序，力求在本身資本組合成分及對各附屬公司的投資之間，保持審慎平衡。 資本計量及分配 （未經審核） 審慎監管局根據綜合基準監管滙豐，因而可取得滙豐整體資本充足比率之資料，並為集團 釐定整體資本規定。經營銀行業務之個別附屬公司由業務所在地之銀行業監管機構直接監 管，該等機構會釐定有關附屬公司之資本充足比率規定，並監察遵行情況。於2013年，我們 採用巴塞爾協定2的架構 （已按資本指引3修訂，一般稱為巴塞爾協定2.5） 計算集團層面的資 本，並按資本指引4終點基準計算估計資本。自2014年1月1日起，我們在集團層面的資本根 據資本指引4計算並以審慎監管局規則加以補充，以落實指令規定的轉化。 我們在集團層面採用的資本計量及分配政策與慣例，是以資本指引4規則為基礎，但各地監 管機構在實施有關規則方面仍處於不同階段，且若干地區的匯報安排可能仍以巴塞爾協定 1基準為基礎，特別是在美國於2014年內若干機構的風險加權資產之報告。在大部分司法管 轄區，並非經營銀行業務的金融附屬公司亦受當地監管機構監管，並須遵守有關資本規定。 巴塞爾協定3架構類似於巴塞爾協定2，以三個 「支柱」 為基本架構：最低資本規定、監管檢 討程序及市場紀律。歐盟透過資本指引4法例實施巴塞爾協定3，而在英國， 《2013年審慎監 管局規則手冊資本規例公司文書》 將資本指引4法例下由各國家╱地區酌情接納的多項規定 轉化為英國法律。資本指引4及審慎監管局法例於2014年1月1日生效。 監管規定資本 就監管規定而言，我們的資本基礎按特性分為三大類別：普通股權一級、額外一級及二級。 ‧ 普通股權一級資本為最高質素的資本，包括股東股權及相關非控股股東權益 （受到限 制） 。根據資本指引4，該等項目若因資本充足性而以不同方式處理，則須作出若干資本 扣減及監管規定調整－包括就商譽及無形資產、依賴未來盈利能力的遞延稅項資產、根 據內部評級基準計算法計算預期虧損金額所導致的負數、金融業公司的資本證券持倉及 盈餘界定福利退休基金資產所作扣減。 ‧ 額外一級資本包括合資格非普通股權資本證券及任何相關股份溢價；其亦包括附屬公司 發行且受若干限制的合資格證券。金融業公司其他額外一級證券持倉已予扣減。 ‧ 二級資本包括合資格資本證券及任何相關股份溢價以及附屬公司發行且受限制的合資 格二級資本證券。金融業公司二級資本證券持倉已予扣減。 第一支柱資本規定 第一支柱涵蓋信貸風險、市場風險及營運風險的資本來源規定。信貸風險包括交易對手信 貸風險和證券化規定。此等規定均按風險加權資產列示。 信貸風險資本規定 為計算第一支柱信貸風險資本規定，資本指引4採用三個精密程度遞增的計算方法。最基本 的標準計算法規定銀行利用外界的信貸評級，釐定有評級交易對手的風險權數。其他交易 對手會歸入多個廣泛的類別，然後為各個類別釐定標準風險權數。更進一級的內部評級基 準 （ 「IRB」 ） 基礎計算法，則允許銀行根據本身對交易對手違責可能性 （違責或然率 「PD」 ） 所 作的內部評估，計算其信貸風險資本規定水平，但須按照標準的監管規定參數計算違責風 險承擔 （ 「EAD」 ） 及違責損失率 （ 「LGD」 ） 的估計數額。最終一級的內部評級基準高級計算法， 則允許銀行透過內部評估釐定違責或然率，以及量化違責風險承擔和違責損失率。 旨在涵蓋非預期虧損的資本來源規定，乃根據監管規則所定公式計算，當中包括違責或然 率、違責損失率、違責風險承擔，以及相關期限和相關性等其他變數。在採用內部評級基準 計算法計算預期虧損時，會將違責或然率乘以違責風險承擔及違責損失率。倘預期虧損超 出會計基準減值準備總額，則會從資本中扣除超逾的部分。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 259 至於信貸風險，我們已安排大部分組合採用內部評級基準高級計算法，而其他業務則採用 內部評級基準基礎計算法或標準計算法。 根據我們多項推行資本指引4的計劃，集團旗下多家公司及多個組合正過渡至採用內部評級 基準高級計算法。於2014年底，在歐洲、亞洲及北美洲大部分地區，主權、銀行、大型企業 及組合的全球模型均已採用內部評級基準高級計算法。其他地區的業務仍沿用標準或基礎 計算法，因為相關地區的規例仍有待界定或所用模型仍有待批准，或相關公司及業務已獲 豁免採用內部評級基準計算法。在某些情況下，監管機構已允許我們的有限數目組合由高 級計算法過渡至標準計算法。 ‧ 交易對手信貸風險 交易對手信貸風險 （ 「CCR」 ） 來自場外衍生工具及證券融資交易。倘若交易對手在妥善結 算交易前違責，即會產生此類在交易及非交易賬項均會計算的風險。資本指引4訂明三 種計算交易對手信貸風險及釐定所涉風險值的方法：標準計算法、按市值計價計算法及 內部模式計算法。有關風險值會用以釐定根據其中一種信貸風險計算法計算的資本規 定水平，這些計算法即標準計算法、內部評級基準基礎計算法及內部評級基準高級計算 法。 我們採用按市值計價及內部模式計算法計算交易對手信貸風險，而長遠目標是將更多持 倉由按市值計價計算法，改為採用內部模式計算法。 此外，資本指引4對信貸估值調整風險採用一項資本規定。如我們就某項產品同時獲准 使用特定風險的估計虧損風險及內部模式計算法，我們會採用信貸估值調整估計虧損風 險計算法計算信貸估值調整資本要求。如我們未同時取得該兩項批准，則會使用標準計 算法。 ‧ 證券化 交易及非交易賬項均有證券化持倉。就非交易賬項的證券化持倉而言，資本指引4指定 以兩種方法計算信貸風險規定，即標準計算法與內部評級基準計算法。這兩種方法均依 賴評級機構的信貸評級與風險權數 （介乎7%與1250%之間） 的配對。 內部評級基準計算法規定，我們須採用評級基準法計算大部分非交易賬項證券化持倉， 並採用內部評估計算法計算未獲評級的流動資金信貸額，以及資產抵押證券化交易優化 項目下的金額。 為符合相關資本規定，交易賬項內大部分證券化持倉的處理方法，猶如於非交易賬項內 持有並採用標準計算法或內部評級基準計算法計算的持倉。其他買賣的證券化持倉 （即 相關性交易） 則根據審慎監管局批准的內部模式計算法處理。 市場風險資本規定 市場風險資本規定使用審慎監管局批准的內部市場風險模型或歐盟資本規定條例的標準規 則計量。我們的內部市場風險模型包括估計虧損風險、壓力下之估計虧損風險及遞增風險 準備。由於我們於2014年9月出售相關性組合，因此並無與全面風險計量有關的市場風險資 本規定。 營運風險資本規定 資本指引4亦包括營運風險的資本規定，同樣運用三個精密程度不同的計算法計算。根據基 本指標計算法計算的規定資本，是總收入的某個簡單百分比；而根據標準計算法計算的規 定資本，則為分配予八個指定業務範疇的各自營運收益總額減去保險費用所得出的三個不 同百分比中的其中一個。兩個方法均利用過去三個財政年度的收入平均數計算。最終一級 的高級計算法則是利用銀行本身的統計分析，並模擬營運風險數據以計算資本規定水平。 我們已採納標準計算法計算集團營運風險資本規定水平。 第二支柱資本規定 我們進行內部資本充足程度評估程序 （ 「ICAAP」 ） ，按業務策略、風險狀況、承受風險水平及 資本計劃對滙豐的資本規定進行前瞻性評估。此項程序結合了集團的風險管理程序及管治 架構。我們對基礎資本計劃進行一系列壓力測試，並使用經濟資本架構及其他風險管理方 法，以評估滙豐內部的資本充足要求。 內部資本充足程度評估程序經審慎監管局審查，為其監管檢討及評估程序的一部分。審慎 監管局會定期進行監管檢討及評估程序，使之可界定滙豐的個別資本指引或最低資本規定 及資本計劃緩衝 （如有需要） 。 董事會報告：資本 （續） 滙豐控股有限公司 260 資本附錄 第三支柱資料披露規定 巴塞爾協定監管架構的第三支柱涉及市場紀律，目的是要求公司至少每年就其風險、資本 及相關管理公布更多具體詳情，從而增加公司的透明度。滙豐的 《2014年第三支柱資料披露》 已上載至滙豐網站(www.hsbc.com) 「投資者關係」 網頁。 按主要因素分析風險加權資產變動－編製基準及補充附註 （未經審核） 信貸風險因素－定義及量化 為分析風險加權資產變動的因果關係，我們把採用內部評級基準計算法計算的風險加權資 產變動總額分為六個主要因素，下文將詳細論述。首四項屬特定、可識別及可計量的變動。 餘下兩項，即賬項規模及賬項質素，乃經考慮首四項特定因素造成的變動後所衍生。 1. 匯兌變動 滙豐旗下公司所持每個組合的功能貨幣與美元 （即滙豐綜合賬目所用列賬貨幣） 之間的匯率 變動，會導致風險加權資產出現變動。我們管理結構匯兌風險的主要目的，是盡可能保障 滙豐之綜合資本比率及經營銀行業務的各附屬公司之資本比率，使之基本上免受匯率變動 影響。就各附屬銀行而言，達致上述目標的方法通常是確保特定貨幣的結構風險對運用該 貨幣計值的風險加權資產之比率大致相等於該附屬公司的資本比率。我們僅在有限情況下 對冲結構匯兌風險。 2. 收購及出售 風險加權資產會因出售或收購業務而出現變動。所涉業務可能是整項業務或其中一部分。 風險加權資產的變動是根據出售前或收購後之月份的月底信貸風險予以量化。 3. 模型更新 新建╱更新模型 風險加權資產的變動若因採用新建模型及更新現有參數模型而產生，則歸入此因素。有關 數額亦將包括檢討模型假設後產生的變動。如重新校準模型以反映更近期的表現數據，因 此而產生的風險加權資產變動將不歸入此類，但會計入賬項質素項下。 風險加權資產的變動乃根據應用模型前的測試階段所作影響評估而予以估算。有關數值會 用作模擬新建或更新的模型於採用時對組合產生的影響，並假設組合由測試階段至採用階 段並無重大變動。 改為採用內部評級基準計算法的組合 如某個組合由採用標準計算法改為採用內部評級基準計算法，按主要因素分析風險加權資 產變動的報表會顯示，採用內部評級基準計算法計算的風險加權資產有所增加，但不會列 示採用標準計算法計算的風險加權資產的相應減額，因為相關變化僅於採用內部評級基準 計算法時產生。 風險加權資產的變動於採用內部評級基準計算法當日予以量化 （並非如採用新建╱更新模型 般於測試階段量化） 。 4. 方法及政策 內部監管規定修訂 此項目反映因更改風險承擔的內部處理方法而對風險加權資產造成的影響。修訂可能包括 （但不限於） 將組合或其一部分由現有採用內部評級基準模型改為採用標準模型，識別淨額 計算及減低信貸風險的措施。 外部監管規定修訂 外部監管規定修訂具體列明額外施加或更改監管規定的影響。修訂包括 （但不限於） 由監管 機構指定對風險加權資產計算法作出的修訂。此類風險加權資產變動的量化方式，是比較 相關組合根據舊有及新設規定計算的風險加權資產。 5. 賬項規模 由此因素產生的風險加權資產變動，為我們假設在風險狀況穩定下，預期會因某類風險承 擔出現變化而產生的變動 （以違責風險承擔計量） 。該等風險加權資產變動於日常業務過程 中產生，例如信貸風險承擔增加，或因縮減或撇銷而使賬項規模縮減。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 261 風險加權資產的變動按以下方法量化： ‧ 上述四項因素造成的風險加權資產及違責風險承擔變動均不計入期內的變動總額內，以 產生違責風險承擔及風險加權資產經調整的變動。 ‧ 風險加權資產對違責風險承擔的平均百分比於期初的水平，會計入違責風險承擔的經調 整變動內，從而產生估計賬項規模造成的風險加權資產變動，而所用假設是整段期間內 違責風險承擔對風險加權資產的百分比維持不變。 由於上述計算方法依賴平均數值，其結果須視乎組合的總合程度及進行計算的獨立時段數 目而定。就2014年每個季度而言，滙豐旗下公司若持有應採用內部評級基準計算法計算的 組合，均會按環球業務計算每家公司的相關數值，並根據主要的巴塞爾信貸風險類別劃分， 如下表所述： 按照巴塞爾協定所訂類別劃分，滙豐採用內部評級基準計算法計算的信貸風險承擔 中央政府及中央銀行 採用內部評級基準基礎計算法計 算的企業信貸 合資格循環零售信貸風險承擔 機構 採用內部評級基準高級計算法計 算的其他信貸 中小企零售信貸 採用內部評級基準高級計算法計 算的企業信貸 零售按揭 其他零售信貸 按主要因素分析風險加權資產變動的列表內，賬項規模一欄顯示了計算結果的總額。 6. 賬項質素 此項因素造成的變動代表因客戶相關信貸質素變動而產生的風險加權資產變動，其成因為 內部評級基準風險參數因各項活動而產生變動，包括 （但不限於） 重新校準模型、交易對手 外部評級變動、或新造貸款對賬項平均質素的影響。因賬項質素而產生的風險加權資產變 動，是考慮上述所有因素後計算所得的風險加權資產變動數額。 按主要因素分析風險加權資產變動的報表，僅包括根據內部評級基準計算法計算的變動。 信貸風險承擔的若干類別被列為資本扣減項目，因此相關減額不會於報表內呈報。如期內 信貸風險的處理方法由計入風險加權資產值改為列作資本扣減項目，則按主要因素分析風 險加權資產變動的列表內，僅會呈列風險加權資產的減額。在此情況下，風險加權資產減 少不一定顯示資本狀況改善。 交易對手風險因素－定義及量化 按交易對手信貸風險主要因素分析的風險加權資產變動，使用各地區提供的交易層面詳情， 更精細地計算信貸風險因素5及6。 「匯兌變動」 並非交易對手風險因素的列賬層面，原因是 組合內存在跨貨幣淨額計算。 市場風險因素－定義及量化 按市場風險主要因素分析的風險加權資產變動，將信貸風險因素5及6結合成單一因素，稱 為 「風險水平變動」 。 槓桿比率：編製基準 （未經審核） 於計算作為分子的資本數值時，使用自2022年1月1日起適用的一級資本 「終點基準」 定義 （載 於資本指引4最終規則） 。若歐洲銀行管理局的自有資金監管技術準則於業績報告日期已在 歐洲委員會的 《歐盟公報》 中刊發，則以此補充，並在適當時參考 《審慎監管局規則手冊》 。於 計算作為分母的風險數值時，則基於歐洲委員會於2014年10月採納並於2015年1月在 《歐盟 公報》 中刊發的 《槓桿比率授權法案》 基準，其與巴塞爾協定2014年槓桿比率架構完全相符。 該架構沿襲風險資本架構所用監管規定綜合計算的相同範圍，與巴塞爾協定2010年文本不 同；巴塞爾協定2010年文本規定銀行使用其會計基準資產負債表將相關項目入賬。風險數 值一般遵循會計基準數值，並調整如下： ‧ 在資產負債表內，非衍生工具風險乃計入風險數值內，並扣除特定準備或會計估值調整 （例如會計信貸估值調整） ； ‧ 貸款不得於扣減存款後作淨額計算； ‧ 衍生工具淨額計算範圍擴大至我們按監管規定確認淨額計算協議的所有情況； ‧ 以衍生工具資產及負債抵銷現金變動保證金的範圍可予延伸，惟須受限於若干額外條 件，包括保證金須每日兌換，並且保證金貨幣須與衍生工具合約結算的貨幣相同。因 董事會報告：資本╱企業管治 滙豐控股有限公司 262 資本附錄╱集團主席之函件 此，我們已考慮納入可用於支付衍生工具合約、規管合資格淨額計算總協議或其相關信 貸支持附件的任何貨幣。巴塞爾協定2010年文本並不允許此項對銷； ‧ 證券融資交易 （ 「SFT」 ） 的淨額計算方法與IFRS所許可者一致，惟若證券融資交易為有抵 押，則就槓桿比率而言存在額外交易對手信貸風險權數。與巴塞爾協定2010年文本相 比，此規定更為嚴格； ‧ 場外及交易所買賣衍生工具計入日後潛在風險額外權數； ‧ 承辦的信貸衍生工具的名義金額計入風險數值內，並就已購入保障予以對銷。與巴塞爾 協定2010年文本相比，此規定更為嚴格； ‧ 資產負債表外項目透過使用信貸換算因素 （ 「CCF」 ） 換算為信貸風險等值。可根據風險額 的風險類別，應用10%、20%、50%或100%的信貸換算因素。相反，巴塞爾協定2010年文 本規定悉數加入資產負債表外項目，惟滙豐可不作事前通知隨時無條件撤消的承諾除外 （當中只計入10%的風險額） ；及 ‧ 不計及從終點基準一級資本扣減的項目，如商譽及無形資產。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 263 企業管治 頁次 附錄 1 企業管治報告 263 集團主席之函件 263 董事 264 秘書 268 集團常務總監 268 董事會 270 董事 270 企業管治守則 275 董事會屬下委員會 276 集團管理委員會 276 集團監察委員會 277 集團風險管理委員會 280 金融系統風險防護委員會 282 集團薪酬委員會 284 提名委員會 284 行為及價值觀委員會 286 主席委員會 288 慈善及社區投資事務 監察委員會 288 內部監控 288 持續經營 290 僱員 291 獎勵 291 僱員關係 291 多元及共融 291 僱員發展 291 僱用殘疾人士 291 健康與安全 291 薪酬政策 292 僱員股份計劃 292 其他披露 293 股本 294 董事權益 297 股息及股東 298 2014年股東周年大會 293 1 董事會報告附錄。 企業管治報告 第263至333頁所載企業管治常規聲明及提 述的資料構成滙豐控股的企業管治報告。 董事會屬下委員會之報告載於企業管治報 告內。 集團主席之函件 各位股東： 2015年正是滙豐在香港和上海成立第150周年。今 天，很多公司成立不到10年，其市值已上升至市 場的前列，但下跌速度往往亦同樣迅速。由此， 我們值得認真思考一下，滙豐各位創辦人決定以 可持續的方式經營公司，使它由較小的規模，發 展成為全球領先的國際銀行集團之一，這種理念 所蘊含的睿智和遠見。在這段期間，世界不斷轉 變，加上金融市場不時發生嚴重的危機，只有管 治穩固和作風審慎的公司，方能實現如此成就。 董事會的主要職責之一，是確保公司的管治和作 風足以支持公司持續成功。談論何謂良好管治和 審慎作風並非難事；但要懂得執行並且衡量此方 面的表現，則更具挑戰性，而這正是董事會的職 責。 良好管治的重點就是董事會責無旁貸地辦好三大 職務。第一：選擇適當的業務模式，並決定在哪 些國家和地區將模式整個或部分實行；第二：就 每項業務所承擔的各類風險，釐定公司可承受的 適當風險水平；最後亦是最重要的，就是確保管 理團隊的組成，能為所有相關群體帶來最合適的 結果、他們獲發放的獎勵符合股東的利益，以及 努力奠下基礎爭取長期持續的成功，包括為團隊 擬定繼任計劃。 過去四年，銀行業經歷了環球金融危機，並且面 對新制訂的監管規定，董事會與管理層積極互 動，處理這方面的所有問題。歐智華及其管理團 隊建議，逐步重新界定和釐清滙豐的業務模式、 地域分布及承受風險水平，而董事會則透過這個 管治過程支持此項建議。同時，董事會評估最高 管理層的組成和質素；對於管理層的全情投入與 承擔，並且成功實現董事會所設定的目標，董事 會一直深受感動。 董事會亦身負監督管理層表現的重任，包括監督 現正實施的改革方案 （目的是簡化和更有效地管 控集團的管理工作） 。這方案包含滙豐的三大優 先策略︰實施環球標準、推動業務增長，以及簡 化及精簡流程。董事會在每次會議上並透過下設 的委員會檢討落實方案的進展，也按照雙方同意 的各項進度指標，向管理層查究落實的速度，並 會探究各項曾經考慮的其他方案何以被否決。 最後，管治的範疇亦須保證我們能夠從意料之外 的結果、錯誤和管控缺失中汲取教訓，並且適時 有效應對。更重要者，管治必須確保所採取的行 動足以防止我們日後重蹈覆轍，並設置預警機 制，能在相關範疇出現問題時，盡快發出警告。 遺憾的是，2014年再度發生法律及監管訴訟事件， 故需加強對操守和金融犯罪風險管治的監督。 2014年，英國成立銀行業標準檢討委員會、英國 董事會報告：企業管治 （續） 滙豐控股有限公司 264 財政大臣就公平有效市場進行調查，以及英國國 會銀行業標準委員會的前成員近日發表聲明等， 種種事例都是對操守和行為風險的持續關注。 滙豐於2013年成立金融系統風險防護委員會，以 應對金融犯罪事件，又於2014年設立行為及價值 觀委員會，表明高度重視遵守崇高的行為標準和 「堅守正道」 ，並藉此逐步提升集團在這方面的管 治監督能力。上述委員會的報告分別載於第282 至283頁和第286至287頁。 企業管治要取得成功，基本要素在於確保董事會 具備均衡而多元化的才能、知識和經驗。自去年 發出函件以來，董事會於2014年4月14日和9月1日 分別委任施俊仁及苗凱婷，並於2015年1月1日委 任安銘為非執行董事，以加強董事會的實力。各 位新任董事大大加強了滙豐在金融及管治方面的 經驗，而苗凱婷更任職銀行高層超過30年，深入 認識銀行業務。各董事的簡歷載於第264至268頁。 良好管治不只適用於最高管理層，更應貫徹至整 個集團。我們每年舉辦非執行董事論壇，讓各主 要附屬公司的非執行董事聚首一堂，討論管治議 題和分享最佳實務；又為滙豐各主要附屬公司對 監督財務報告和風險相關事宜負有非執行責任的 委員會主席們召開年會，以交流議題及加強實施 一致的標準。 由 《2013年金融服務 （銀行業改革） 法》 引入的新 「高 級人員制度」 ，預料將於今年最終落實，內容將包 括非執行董事的具體職責。同時，英國 《企業管治 守則》 的修訂本亦將於今年生效，其內容涵蓋的 若干改革，將涉及薪酬、股東關係、風險管理和 持續經營的原則和規定。董事會明白良好管治是 持續成功的要素，亦是集團憑藉獨特業務模式捕 捉增長機會之關鍵，因此我們將全力支持最佳實 務的演進。 集團主席 范智廉 2015年2月23日 董事 范智廉 CBE 59歲 集團主席 才能及經驗：擁有廣泛的董事會經驗和管 治知識，其中包括服務滙豐及英國石油有 限公司董事會累積的經驗；在銀行、跨國 財務報告、財資及證券交易運作方面具備 豐富的財務及風險管理知識。1995年加入 滙豐出任集團財務董事。 蘇格蘭特許會計師公會及英國公司司庫公 會成員，亦是英國特許管理會計師公會資 深會員。2006年獲頒授CBE勳銜，表揚他對 金融業的貢獻。 獲委任為董事：1995年。2010年起擔任集團 主席。 現任職位包括：The Hong Kong Association董 事；國際金融學會主席。北京市市長國際 企業家顧問會及上海市市長國際企業家諮 詢會議成員；上海中歐國際工商學院國際 顧問委員會成員；英國政府金融服務貿易 及投資委員會獨立外界成員；英國商務大 使；以及自2014年12月10日起獲委任為彼得 森國際經濟研究所董事。 曾任職位包括：滙豐的集團財務董事、財 務總監兼風險與監管事務執行董事；以及 英國石油有限公司非執行董事及監察委員 會主席。曾於多個就稅務、管治、會計及 風險管理訂定準則、甚具影響力的機構擔 任主席及成員；並曾為KPMG合夥人。 歐智華 55歲 集團行政總裁 集團管理委員會主席 才能及經驗：於1980年加入滙豐，是擁有逾 30年國際業務經驗的銀行家。曾在集團於 全球各地 （包括倫敦、香港、東京、吉隆坡 及阿拉伯聯合酋長國） 的業務機構擔任要 職；在開拓和擴展環球銀行及資本市場業 務方面擔當領導角色。 獲委任為董事：2008年。自2011年起擔任集 團行政總裁。 現任職位包括：香港上海滙豐銀行有限公 司主席及集團管理委員會主席。新加坡金 融管理局國際諮詢委員會成員，以及中國 簡歷 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 265 銀行業監督管理委員會國際諮詢委員會成 員。 曾任職位包括：歐洲、中東及環球業務主 席；英國滙豐銀行有限公司、中東滙豐銀 行有限公司、HSBC Private Banking Holdings (Suisse) SA以及法國滙豐主席；HSBC Trinkaus & Burkhardt AG副主席及其督導委員會成員； 環球銀行及資本市場主管；環球銀行及資 本市場聯席主管；環球資本市場主管；以 及亞太區財資及資本市場主管。 安銘 66歲 獨立非執行董事 自2015年1月1日起出任集團監察委員會成 員。 才能及經驗：註冊會計師，擁有廣泛的財 務及會計經驗。曾任General Electric Capital Co.副總裁、審計長及首席會計師，加入該 公司前為KPMG合夥人。曾參與制訂會計準 則，因而具備淵博的專業知識。 獲委任為董事：2015年1月1日 現任職位包括：北美滙豐控股有限公司、 美國滙豐銀行、美國滙豐融資有限公司 及美國滙豐有限公司非執行董事；Skyonic Corporation及R3 Fusion, Inc.非執行董事；北 卡羅來納州大學商學院諮詢委員會成員。 曾任職位包括：通用電氣公司副總裁、審 計長及首席會計師；Peat Marwick Mitchell & Co （現稱KPMG） 技術審計合夥人。國際會計 準則委員會屬下國際財務報告詮釋委員會 成員、美國註冊會計師協會屬下會計準則 執行委員會成員、財務會計準則委員會新 浮現問題專責小組成員、國際財務主管組 織公司報告委員會主席，以及財務會計基 金會受託人。 祈嘉蓮 48歲 獨立非執行董事 集團監察委員會及金融系統風險防護委員 會成員。 才能及經驗︰具有豐富的金融業監管政策 經驗。曾任美國證券交易委員會委員，在 多項多邊和雙邊監管事務對話中，以及在 20國集團金融穩定委員會和國際證券事務 監察委員會等組織中，擔任該監管機構的 首席代表。 獲委任為董事︰2014年3月1日 現任職位包括︰另類投資管理協會主席、 Patomak Global Partners高級顧問、賓夕法尼 亞州立大學理事會、國會圖書館信託基金 理事會，以及公眾公司會計監督委員會諮 詢委員會成員。 曾任職位包括︰美國參議院銀行、住房和 城市事務委員會行政主管和法律顧問，以 及一名美國參議員的立法事務主任和幕僚 長。 凱芝 53歲 獨立非執行董事 才能及經驗：在國際商務方面深具領導才 能，曾協助Oracle成功轉型為世界最大的商 務管理軟件生產商，以及全球領先的資訊 管理軟件供應商。 獲委任為董事：2008年 現任職位包括：於2014年9月18日獲委任為 Oracle Corporation聯席行政總裁，此前曾任 總裁兼財務總監。於1999年加盟Oracle，至 2001年獲委任為董事。 曾任職位包括：Donaldson, Lufkin & Jenrette 董事總經理。 史美倫 GBS 65歲 獨立非執行董事 自2014年12月5日起出任慈善及社區投資事 務監察委員會主席；亦是行為及價值觀委 員會和提名委員會成員。 才能及經驗：在監管香港與中國內地金融 和證券業及制訂相關政策方面具有豐富經 驗；曾任中國證券監督管理委員會副主席， 是首位獲中華人民共和國國務院委任為副 部級官員的境外人士；獲香港政府頒授金 紫荊及銀紫荊星章，表揚她熱心公職。 獲委任為董事：2011年 現任職位包括：香港上海滙豐銀行有限公 司非執行副主席；香港特區行政會議非官 守議員；第12屆全國人民代表大會的港區 代表；中國電信股份有限公司、Unilever PLC 及Unilever N.V.非執行董事；瑞典資產管理 基金會的高級國際顧問；美國加州律師公 會會員；中國銀行業監督管理委員會國際 諮詢委員會成員；香港特區金融發展局主 董事會報告：企業管治 （續） 滙豐控股有限公司 266 席；以及中國證券監督管理委員會國際顧 問委員會副主席。 曾任職位包括：交通銀行股份有限公司、 寶山鋼鐵股份有限公司、德昌電機控股有 限公司、香港交易及結算所有限公司及印 度塔塔諮詢服務有限公司非執行董事；香 港大學教育資助委員會主席；香港廉政公 署貪污問題諮詢委員會主席；香港證券及 期貨事務監察委員會副主席；以及耶魯大 學管理學院顧問委員會成員。 埃文斯勳爵 57歲 獨立非執行董事 金融系統風險防護委員會主席、行為及價 值觀委員會成員。自2014年12月5日起出任 慈善及社區投資事務監察委員會成員。 才能及經驗︰在國家安全政策和實務方面 具有豐富經驗。為英國國家安全局 （軍情五 處） 前局長，負責領導該局、制訂政策和策 略，包括國際及本土反恐工作、反間諜、 反武器擴散及網絡安全。 獲委任為董事︰2013年 現任職位包括︰英國國家打擊犯罪調查 局非執行總監；Accenture plc高級顧問； Darktrace Limited及Facewatch Limited諮詢委 員會成員。 曾任職位包括︰服務英國國家安全局逾30年， 歷任多項要職，負責監督聯合反恐分析中心 和英國國家基礎設施保護中心，並出席國家 安全委員會的會議。 費卓成 64歲 獨立非執行董事 集團風險管理委員會主席。 才能及經驗：在銀行及資產管理方面具備 豐富國際業務經驗，曾於德國、東京、紐 約及倫敦工作。曾任德盛安聯資產管理行 政總裁及安聯集團管理董事會成員；曾於 花旗集團任職14年，負責交易和項目融資 部工作，並出任歐洲、北美洲及日本資本 市場業務部主管。 獲委任為董事：2012年 現任職位包括：Deutsche Börse AG督導委 員會主席；Joh A. Benckiser SARL股東委員 會主席；Coty Inc.獨立董事；Allianz France S.A.董事；歐洲管理及科技學院諮詢委員 會成員；以及港歐貿易理事會成員。 曾任職位包括：安聯旗下多家附屬公司主 席；Bayerische Börse AG督導委員會成員； OSRAM Licht AG督導委員會成員和監察及 風險管理委員會主席；德國可持續發展理 事會成員；以及西門子集團退休金委員會 諮詢委員會成員。 方安蘭 CBE 53歲 獨立非執行董事 金融系統風險防護委員會及提名委員會成 員。 才能及經驗：在國際工業、出版業、金融業 及經營管理方面具備豐富經驗。曾任金融 時報集團有限公司主席兼行政總裁，負責 其策略、管理及營運；亦曾出任Pearson plc 財務董事，負責監督財務部的日常運作， 並直接負責環球財務報告及管控、稅務與 財資事宜。 獲委任為董事：2004年 現任職位包括：北美滙豐控股有限公司主 席；PepsiCo Inc.非執行董事；英國商務大 使；以及由2014年10月8日起出任英國廣播 公司信託基金主席。 曾任職位包括：Imperial Chemical Industries plc執行副總裁 （策略及集團監控） ；以及 Interactive Data Corporation主席兼董事。曾 任英國政府內閣辦事處委員會成員直至 2014年9月1日；以及The Economist Newspaper Limited非執行董事直至2014年7月1日。 李德麟 59歲 獨立非執行董事 集團薪酬委員會及提名委員會成員。 才能及經驗：具有豐富的國際業務經驗， 尤其熟悉能源業，曾負責管理分布於四大 洲的業務。合資格事務律師，並擁有工商 管理碩士學位。 獲委任為董事：2008年 曾任職位包括：曾任Centrica plc行政總裁及 英國運輸部委員會首席非執行委員，直至 2014年12月31日；Chevron Corporation執行 副總裁；Hanson PLC非執行董事；Enterprise Oil plc行政總裁；Amerada Hess Corporation總 裁兼營運總監；以及英國首相商務顧問小 組成員。 簡歷 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 267 利普斯基 68歲 獨立非執行董事 集團風險管理委員會、提名委員會，以及 集團薪酬委員會成員。 才能及經驗：具有國際業務經驗，曾於智 利、紐約、華盛頓及倫敦工作，並與多國 金融機構、央行及政府往來。曾任國際貨 幣基金組織第一副總裁、署理總裁及特別 顧問。 獲委任為董事：2012年 現任職位包括：約翰霍普金斯大學尼采高 等國際研究院外交政策研究所資深學人； Aspen Institute的世界經濟課程聯席主席；國 家經濟研究局局長及全球發展中心總監； 史丹福經濟政策研究院顧問委員會和外交 關係協會成員；以及世界經濟論壇之全球 議程理事會－國際貨幣體系議題主席。 曾任職位包括：摩根大通投資銀行副主席； 美國對德國事務協會及日本協會理事；紐 約經濟學會受託人；以及Anderson Global Macro, LLC環球政策顧問。 駱美思 69歲 獨立非執行董事 行為及價值觀委員會主席，以及集團監察 委員會及集團風險管理委員會成員。 才能及經驗：在公私營機構均具備豐富經 驗，熟諳英國政府及金融體系的運作情況。 獲委任為董事：2008年 現任職位包括：International Regulatory Strategy Group主席；TheCityUK董事；布 魯塞爾歐洲智庫Bruegel的總監；Arcus European Infrastructure Fund GP LLP及 Heathrow Airport Holdings Limited非執行董 事；倫敦Imperial College校董會成員； Institute of Fiscal Studies院 長；Ditchley Foundation受託人；以及Serco Group plc非執 行董事兼企業責任委員會主席。 曾任職位包括：英倫銀行副行長 （負責貨幣穩 定事務） 及貨幣政策委員會成員；英國政府 運輸部、勞動和退休保障部及威爾斯事務部 常任秘書長；世界銀行副行長兼行長辦公室 主任；以及Reinsurance Group of America Inc.
及The Scottish American Investment Company PLC非執行董事。 麥榮恩 53歲 集團財務董事 才能及經驗：2007年加入滙豐，出任北美 滙豐控股有限公司財務總監。具有豐富的 財務及國際業務經驗，曾在倫敦、巴黎、 美國、非洲及亞洲等地工作。蘇格蘭特許 會計師公會成員。 獲委任為董事：2010年 現任職位包括：集團管理委員會成員；於 2014年12月4日獲委任為英國心臟基金會審 核委員會成員。 曾任職位包括：恒生銀行有限公司董事； 通用電氣環球消費融資亞太區財務總監、 副總裁兼財務總監；以及通用電氣醫療保 健－環球診斷影像業務副總裁兼財務總監。 苗凱婷 61歲 獨立非執行董事 自2014年9月1日起出任集團風險管理委員 會和行為及價值觀委員會成員。 才能及經驗：具備豐富的國際銀行及金融 服務經驗。曾任摩根大通集團之國際業務 總裁，負責領導該行的投資銀行、資產管 理和財資及證券服務部門，實施環球擴張 及國際業務策略。 獲委任為董事：2014年9月1日 現任職位包括：First Data Corporation及 General Mills Inc.非執行董事；以及國際財務 報告準則基金會受託人。 曾任職位包括：Merck & Co. Inc.及Progressive Corp非執行董事，直至2014年8月1日；摩根 大通集團財資及證券服務執行副總裁兼行 政總裁；芝加哥第一銀行執行副總裁兼財 務總監；Priceline.com Inc.高級執行副總裁； 以及花旗集團執行副總裁兼財務總監。 繆思成 57歲 集團風險管理總監 才能及經驗：2005年加入滙豐，出任環球 銀行及資本市場財務及風險管理總監。具 有豐富的風險管理及財務經驗。英格蘭及 威爾斯特許會計師公會會員。 獲委任為董事：2014年1月1日 現任職位包括：集團管理委員會成員。滙 豐私人銀行 （瑞士） 有限公司及HSBC Private 董事會報告：企業管治 （續） 滙豐控股有限公司 268 Banking Holdings (Suisse) SA董事。 曾任職位包括：環球銀行及資本市場財務 及風險管理總監；HSBC Insurance (Bermuda) Limited董事；摩根大通歐洲區財務總監；以 及PricewaterhouseCoopers核數業務合夥人。 駱耀文爵士 73歲 副主席兼高級獨立非執行董事 提名委員會及集團薪酬委員會主席，以及 金融系統風險防護委員會成員。 才能及經驗：熟悉國際企業顧問業務，在 收購合併、商人銀行、投資銀行及金融市 場方面的經驗尤其豐富；獲授騎士銜，表 揚他對商界的貢獻；曾在法國、德國、英 國及美國等地工作，擁有豐富的國際業務 經驗。 獲委任為董事：2006年。自2007年起出任高 級獨立非執行董事，自2010年起出任副主 席。 現任職位包括：Simon Robertson Associates LLP創辦股東；Berry Bros. & Rudd Limited、 The Economist Newspaper Limited及Troy Asset Management非執行董事；以及Eden Project Trust 與Royal Opera House Endowment Fund受託人。 曾任職位包括：Rolls-Royce Holdings plc非執行 主席；高盛國際常務董事；Dresdner Kleinwort Benson主席；以及Royal Opera House、Covent Garden Limited及NewShore Partners Limited非 執行董事。 施俊仁 CBE 55歲 獨立非執行董事 自2014年9月1日起出任集團監察委員會主席。 自2014年4月14日起至2014年9月1日出任集團 薪酬委員會成員；亦是行為及價值觀委員會 成員。 才能及經驗︰具有豐富的國際金融服務經 驗，曾在英國、美國及瑞士等地工作。曾任 諾華公司及AstraZeneca plc財務總監。英格蘭 及威爾斯特許會計師公會資深會員。 獲委任為董事︰2014年4月14日 現任職位包括︰英國滙豐銀行有限公司及 Innocoll AG主席；Genomics England Limited及 Proteus Digital Health Inc.非執行董事。 曾任職位包括︰高盛合夥人兼董事總經理； KPMG合夥人；以及Diageo plc非執行董事兼審 核委員會主席。 秘書 馬振聲 48歲 集團公司秘書長 2013年6月加入滙豐，並自2013年7月起擔任 集團公司秘書長。英國特許秘書及行政人 員公會資深會員。曾任職位包括：力拓股 份有限公司及BG Group plc.之集團公司秘書。 集團常務總監 Ann Almeida 58歲 集團人力資源及企業可持續發展主管 （將於2015年5月31日退任） 1992年加入滙豐。自2008年起出任集團常務 總監。曾於滙豐出任：環球銀行及資本市 場、環球私人銀行、環球交易銀行及HSBC Amanah的環球人力資源主管。 安思明 54歲 環球銀行及資本市場行政總裁 1994年加入滙豐。自2011年起出任集團常 務總監。環球金融市場協會和法國滙豐主 席及HSBC Trinkaus & Burkhardt AG董事，自 2014年3月28日起出任英國滙豐銀行有限公 司董事。曾任職位包括：HSBC Global Asset Management Limited及埃及滙豐銀行董事； 環球資本市場主管；以及歐洲、中東及非 洲環球資本市場主管。 貝炳達 59歲 環球私人銀行行政總裁 1975年加入滙豐。自2013年10月起出任集 團常務總監。HSBC Private Bank (Monaco) SA 主席。曾任職位包括：法國滙豐及歐洲大陸業 務行政總裁，以及英國滙豐銀行有限公司、 馬耳他滙豐銀行有限公司及HSBC Trinkaus & Burkhardt AG董事。於2014年9月29日退任 HSBC Global Asset Management Limited董事。 簡歷 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 269 顧世民 47歲 環球工商金融行政總裁 1989年加入滙豐。自2013年10月起出任集團 常務總監，並為環球工商金融行政總裁。 英國滙豐銀行有限公司董事。曾於滙豐出 任：埃及滙豐銀行及HSBC Bank Oman主席； 中東滙豐銀行主席兼行政總裁；韓國滙豐 行政總裁；新加坡滙豐企業及投資銀行主 管；沙地英國銀行及中東滙豐銀行有限公 司董事。 范寧 46歲 零售銀行及財富管理行政總裁 1989年加入滙豐。自2013年1月起出任集 團常務總監。HSBC Private Banking Holdings (Suisse) SA董事。曾任職位包括：加拿大滙 豐銀行董事；集團行政總裁辦公室主任兼 策略和規劃部集團主管；滙豐環球投資管 理行政總裁；集團司庫；以及環球資本市 場副主管。 Pam Kaur 51歲 集團審核部主管 2013年4月加入滙豐，並出任集團常務總監。 英格蘭及威爾斯特許會計師公會理事會增 任會員。曾任職位包括：德意志銀行集團 審核部環球主管；蘇格蘭皇家銀行集團重 組及風險管理部財務總監兼營運總監；駿 𢡟銀行合規及反洗錢部集團主管；以及花 旗集團環球消費金融環球合規總監。 祈偉倫 56歲 英國滙豐銀行有限公司行政總裁 1981年加入滙豐。自2011年起出任集團 常務總監。中東滙豐銀行有限公司、 HSBC Trinkaus & Burkhardt AG及法國滙豐董 事。TheCityUK顧問委員會成員及Bradford University School of Management諮詢委員會 成員。曾任職位包括：環球工商金融環球主 管；HSBC Bank A.S.及HSBC Bank Polska S.A.董 事。 利維 51歲 法律事務總監 2012年加入滙豐，並出任集團常務總監。曾 任職位包括：美國財政部反恐及金融情報 副部長；外交關係協會國家安全及財政健 全高級顧問；美國司法部副部長首席協理； 以及Miller, Cassidy, Larroca & Lewin LLP及 Baker Botts LLP合夥人。 Antonio Losada 60歲 拉丁美洲行政總裁 1973年加入滙豐。自2012年12月起出任集團 常務總監。HSBC Latin America Holdings (UK) Limited、阿根廷滙豐銀行、HSBC Argentina Holdings S.A.、HSBC Mexico, S.A., Institucion de Banca Multiple、Grupo Financiero HSBC及 Grupo Financiero HSBC, S.A. de C.V.及北美滙豐 控股有限公司董事。曾任職位包括：HSBC Argentina行政總裁；巴拿馬滙豐銀行及HSBC Argentina Holdings S.A.主席，以及巴西個人 理財業務副主管。 蘇力宏 59歲 集團營運總監 （將於2015年3月31日退任） 1980年加入滙豐。自2011年起出任集團常 務總監。曾任職位包括：集團科技及服務 總監；英國滙豐銀行有限公司董事兼營運 總監；以及加拿大滙豐銀行營運總監。 王冬勝 63歲 香港上海滙豐銀行有限公司副主席兼行政總裁 2005年加入滙豐。自2010年起出任集團常務 總監。滙豐銀行 （中國） 有限公司董事長及 馬來西亞滙豐銀行有限公司主席；恒生銀 行有限公司、石澳發展有限公司及交通銀 行股份有限公司非執行董事。國泰航空有 限公司獨立非常務董事。曾任職位包括： 越南滙豐銀行有限公司副主席、澳洲滙豐 銀行有限公司董事，以及中國平安保險 （集 團） 股份有限公司董事。 董事會報告：企業管治 （續） 滙豐控股有限公司 270 董事會 董事會 滙豐控股董事會 （ 「董事會」 ） 銳意促進本公 司業務發展的長遠成功，並為股東持續提 供理想之價值。以集團主席為首，董事會 為集團制訂策略及承受風險水平，並審批 管理層為達致策略目標而建議的資本及營 運計劃。落實策略之工作則授權由集團行 政總裁領導之集團管理委員會執行。 董事 滙豐控股設有單一董事會，各董事於董事 會會議行使權力，由董事會集體行事。年內 服務本公司之董事為祈嘉蓮 （於2014年3月 1日獲委任） 、凱芝、史美倫、張建東 （於 2014年8月1日退任） 、顧頌賢 （於2014年 5月23日退任） 、埃文斯勳爵、費卓成、 方安蘭、范樂濤 （於2014年9月1日退任） 、 范智廉、歐智華、何禮泰 （於2014年5月23日 退任） 、李德麟、利普斯基、駱美思、麥榮 恩、苗凱婷 （於2014年9月1日獲委任） 、繆思 成 （於2014年1月1日獲委任） 、駱耀文爵士 及施俊仁 （於2014年4月14日獲委任） 。 安銘於2015年1月1日獲委任。 於通過 《2014年報及賬目》 當日，董事會的 成員包括集團主席、集團行政總裁、集團 財務董事、集團風險管理總監及13位非執 行董事。 董事之姓名及簡歷載於第264至268頁。 執行董事 集團主席、集團行政總裁、集團財務董事 及集團風險管理總監均為滙豐僱員。 非執行董事 非執行董事並非滙豐僱員，不會參與滙豐 之日常業務管理，彼等的立場獨立，以建設 性態度批評和協助制訂策略建議，審視管 理層在實踐既定目標方面的表現，以及監 察集團風險狀況及業績表現的匯報工作。 各非執行董事均具備公營及私營機構的廣 泛經驗，包括曾領導大型而複雜的跨國企 業。 非執行董事委任條款 董事會已確定，非執行董事每年最少投入 的時間預期約為30天。非執行董事投入本 公司的時間可能大幅超出此數，尤其是獲 委任加入董事會屬下委員會的成員。 非執行董事的最初任期為三年，獲股東於 股東周年大會上重選後，一般預期可擔任 兩屆，每屆任期為三年。董事會可邀請董 事延長任期。所有董事須每年接受股東重 選。 載列各非執行董事委任條款的函件可於本 公司的註冊辦事處查閱。 集團主席及集團行政總裁 集團主席和集團行政總裁的職權分立， 分別負責董事會的運作及滙豐業務運作 之行政工作，彼等各司其職，集團主席 和集團行政總裁的角色和職責詳情載於 www.hsbc.com/investor-relations/governance/ board-committees。彼等的主要職責如下： 主要職責 集團主席－范智廉 ‧ 領導董事會，確保其有效履行職責。 ‧ 與政府、監管機構及投資者加強聯繫。 ‧ 領導集團就銀行及金融服務業的公共政策 及監管改革與有關方面溝通。 ‧ 維持企業聲譽和文化。 ‧ 監察集團行政總裁之表現。 集團行政總裁－歐智華 ‧ 制訂業務計劃，並按業務計劃推動集團爭 取表現。 ‧ 制訂集團策略並徵求集團主席同意，然後 向董事會提出建議。 ‧ 以集團管理委員會主席的身分，按照董事 會同意的策略及商業目標推動集團爭取表 現。 副主席及高級獨立非執行董事 由董事會通過，關於副主席及高級獨立非 執行董事的角色和職責詳情載於www.hsbc.
com/investor-relations/governance/board-committees。其主要職責如下： 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 271 主要職責 副主席及高級獨立非執行董事－駱耀文爵士 ‧ 在董事會會議或股東大會上，擔任集團主 席的副手，協助集團主席履行職責。 ‧ 於必要時作為其他非執行董事的中介人。 ‧ 領導非執行董事監察集團主席。 ‧ 確保集團主席和集團行政總裁之間有明確 分工。 董事的委任、退任和重選 董事會可隨時委任願意接受任命者出任董 事，以填補空缺或加入現有董事會出任新 增董事，但董事總數不得少於5位或超過25 位。董事會如上文所述委任之任何董事須 在獲委任後之股東周年大會上告退，惟有 資格參選連任，但不計算在根據公司組織 章程細則必須於該大會上輪流告退之董事 人數內。董事會可任命任何董事擔任任何 受聘職位或行政職位，亦可撤回或終止有 關委任。股東可通過普通決議案委任董事 或在任何董事任期屆滿前將其罷免。董事 會應提名委員會的建議，並為符合英國 《企 業管治守則》 ，決定所有董事須每年接受股 東重選。因此，所有董事將於應屆股東周 年大會上退任並願候選新任董事或參選連 任。 董事會權力 董事會負責監督滙豐全球業務的管理，而 在監督業務管理時，可根據任何相關法例 及規例與公司組織章程細則行使其權力。 董事的職權範圍，可於www.hsbc.com/1/2/ about/board-of-directors瀏覽。董事會將每年 檢討其職權範圍。 尤其值得一提的是，董事會可行使本公司 之所有權力舉債，並將滙豐控股的全部或 任何部分業務、物業或資產 （現時或未來） 按揭或抵押，亦可行使英國 《2006年公司法》 及╱或股東所授予之任何權力。董事會可 於其認為適合之時限內，按其認為適合之 條款將本身任何權力、權限及酌情權 （包括 再轉授的權力） 授予並賦予任何執行董事。 此外，董事會可在任何特定地區設立任何 地區或部門委員會或代理機構，管理滙豐 控股在當地之業務，並於其認為適合之時 限內，按其認為適合之條款將本身任何權 力、權限及酌情權 （包括再轉授的權力） 授予 並賦予任何所委任之地區或部門委員會、 經理或代理。董事會亦可透過授權書或其 他方法委任任何一名或多名人士為滙豐控 股之代理，並於其認為適合之時限內，按 其認為適合之條款將本身任何權力、權限 及酌情權 （包括再轉授的權力） 授予該名或 該等人士。 董事會授權集團管理委員會負責滙豐控股 的日常管理工作，但保留審批若干事項的 權力，包括營運計劃、承受風險水平和表 現目標、營運監控程序、信貸、市場風險限 額、收購、出售、投資、資本支出或變現， 或增設新公司、特定的高層人員委任，以 及資產負債管理政策的任何重大修訂。 滙豐控股於1991年1月17日根據 《公司條例》 第XI部在香港註冊。 董事會會議 董事會於2014年舉行了八次會議及兩次為 期一天的策略會議。每年至少一次董事會 會議在英國以外的主要策略地點舉行。於 2014年在香港及北京舉行了多次董事會會 議。 下表列示各董事於2014年出席董事會會議 的情況。 2014年，非執行董事曾在其他執行董事避 席下與集團主席會面一次。非執行董事亦 曾在集團主席避席下會面四次，包括評核 集團主席的表現。 董事會報告：企業管治 （續） 滙豐控股有限公司 272 董事會 2014年董事會出席紀錄 出席會議 作為董事 有資格出席 之會議 祈嘉蓮1,8 5 5 凱芝 8 8 史美倫 8 8 張建東2 4 5 顧頌賢3 4 4 埃文斯勳爵8 7 7 費卓成 8 8 方安蘭8 7 7 范樂濤4 7 7 范智廉 8 8 歐智華 8 8 何禮泰3 4 4 李德麟 8 8 利普斯基8 7 7 駱美思 8 8 麥榮恩 8 8 苗凱婷5,8 2 2 繆思成6 8 8 駱耀文爵士 8 8 施俊仁7 6 6 於2014年召開的會議8 8 1 於2014年3月1日獲委任為董事。 2 於2014年8月1日退任董事。 3 於2014年5月23日退任董事。 4 於2014年9月1日退任董事。 5 於2014年9月1日獲委任為董事。 6 於2014年1月1日獲委任為董事。 7 於2014年4月14日獲委任為董事。 8 一次會議於短時間通知下召開，未能出席的董 事已於會議前獲得相關簡報。 董事會的架構平衡及董事的獨立性 董事會大多數成員為獨立非執行董事。鑑 於滙豐業務複雜而且地域廣泛，且董事須 投入大量時間，故董事會現有規模乃屬恰 當。 提名委員會定期檢討董事會的架構、規模 及組成 （包括董事的才能、知識、經驗、獨 立性和多元化） ，並向董事會提出任何變革 的建議。 董事會採納了董事會多元化政策，與集團 向來著重僱員種族、年齡及性別多元化的 策略互相呼應。有關董事會多元化政策的 進一步資料載於第285頁。 董事會認為所有非執行董事均具獨立性。 董事會在確定非執行董事的獨立性時，認 為服務年期應由其獲委任為滙豐控股董事 後獲股東推選的日期起計算。方安蘭出任 董事逾九年，若純粹基於這點，其不符合 英國 《企業管治守則》 所載有關獨立性的慣 常準則。儘管方安蘭已服務多年，但經考 慮其對管理事務持續作出具建設性的批 評及參與董事會討論時的重大貢獻，董事 會認為其具有獨立自主的個性及判斷力。 方安蘭將於2015年股東周年大會上參與重 選。我們認為，董事現時及過往在滙豐附 屬公司董事會的服務經驗非常寶貴，且該 經驗不會減低非執行董事的獨立性。董事 會認為並無任何可能影響非執行董事判斷 力的關係或情況，倘可能出現此種關係或 情況，其影響均不屬重大。 按照 《香港聯合交易所有限公司證券上市規 則》 ，每名獲董事會確定為獨立的非執行董 事已提交確認本身獨立性的年度確認書。 資料及支援 董事會定期檢討財務及其他策略目標的進 度表現報告、業務發展報告以及投資者和 對外關係報告。各董事會屬下委員會主席 及集團行政總裁在董事會每次會議上報告 委員會自上一次董事會會議以來的工作情 況。董事會省覽有關環球業務及主要地區 策略及發展的定期報告與匯報，同時亦會 省覽集團承受風險水平、首要及新浮現風 險、風險管理、信貸風險及集團貸款組合、 資產與負債管理、流動資金、訴訟、金融 及監管合規，以及有關聲譽事宜的定期報 告。 各董事可自由及公開與各級管理層聯絡。 出席各地召開的董事會會議及就其他原因 出差時，各非執行董事應順道親自了解當 地業務營運，同時與當地管理層會面。 集團公司秘書長的角色 全體董事均可獲集團公司秘書長提供建議 及服務，集團公司秘書長向董事會負責， 以確保董事會程序及所有適用規則及法規 得以遵守。 根據集團主席的指示，集團公司秘書長的職 責包括須確保董事會及其屬下委員會內， 以及高級管理人員及非執行董事間的資訊 往來保持暢通無阻，以及促進履任啟導並 按需要協助專業發展。 集團公司秘書長負責就企業管治事宜透過 集團主席向董事會作出建議。 會議議程及其他參考文件會在董事會及董 事會屬下委員會會議前預先分派，讓各成 員有充足時間適當審閱，以便在會議上充 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 273 分討論。各董事均可及時取得一切有關資 料，必要時可徵求獨立專業顧問意見，一 切費用由滙豐控股承擔。 履任啟導 新委任董事均獲安排參與正規和特設的履 任啟導計劃。該等計劃乃按個別董事的需 要，並因各董事的技能及經驗而異。一般 的履任啟導計劃包括與其他董事及高級行 政人員進行連串會議，讓新董事加深了解 滙豐業務。董事亦可獲集團公司秘書長提 供有關出任董事之職務及責任的全面指引。 培訓及發展 我們為董事提供培訓及發展，安排課程與 定期董事會會議。透過與集團業務及部門 的高級管理人員日常互動及作出有關簡 報，執行董事可發展和更新他們的才能和 知識。非執行董事可獲得內部培訓與發展 資源及個人化培訓 （如需要） 。所有新委任 董事均參與特設的履任啟導計劃。主席定 期檢討各董事的培訓及發展情況。 年內，董事就以下主題接受培訓︰ ‧ 達德－法蘭克法案； ‧ 英國 《2014年金融服務 （銀行業改革） 法》 ，包括 「高級人員制度」 ；及 ‧ 不斷變化的財務和監管報告環境。 下表顯示各董事於2014年進行的培訓及發 展活動概要︰ 培訓及發展 培訓範圍 最新 監管規定 企業管治 金融 行業發展 董事會屬下 委員會相關 主題簡報 執行董事 范智廉 3 3 3 3 歐智華 3 3 3 3 麥榮恩 3 3 3 3 繆思成 3 3 3 3 非執行董事 祈嘉蓮 3 3 3 凱芝 3 3 史美倫 3 3 3 埃文斯勳爵 3 3 3 費卓成 3 3 3 3 方安蘭 3 3 3 李德麟 3 3 利普斯基 3 3 駱美思 3 3 3 苗凱婷 3 3 駱耀文爵士 3 3 3 施俊仁 3 3 3 董事會表現評估 董事會致力定期評估其本身及其屬下委員 會的績效。於2012及2013年，董事會及其 屬下委員會的績效評估乃由Bvalco Ltd 1 （獨 立第三方公司） 進行，2013年的檢討過程與 2012年的相同，由Bvalco與董事會及數位其 他高級行政人員作深入面談。2013年的檢 討結果已向董事會呈報，於2014年，就此等 事宜制訂行動計劃及向董事會報告計劃進 度。2013年檢討的結論是董事會繼續有效 運作，足可應對集團所面對的挑戰。2013年 檢討所得的主題及所採取行動包括︰ 1 一家獲本公司不時委聘以提供法律服務的律師 事務所持有Bvalco Ltd的20％股權，而Bvalco Ltd 已確認與本公司並無任何其他關連。 董事會報告：企業管治 （續） 滙豐控股有限公司 274 董事會 2013年董事會績效檢討 主題 所採取之行動 確保監管、業務及策略事項於董事會會議中 得到適當平衡。 重要事項得到進一步優先處理。 於董事會會議中撥出更多時間討論有關事項。 為執行及非執行董事提供更多於董事會議室 正式場合以外會面的機會。 在董事會會議前後規劃執行及非執行董事非正式活動，提供更多討 論平台。 非執行董事獲邀出席多個執行董事參加的活動。 安排非執行董事於出差時與當地滙豐辦事處高級管理層成員會面。 確保非執行董事有更多會面時間及機會。 於董事會會議舉行前後安排僅供非執行董事參與的環節。 於年內舉行多個非執行董事非正式活動。 繼續集中制訂繼任計劃。 繼任計劃仍為關注重點，已訂有正式管治程序。 繼任計劃涉及的人選將提呈董事會會議。 Bvalco於年內籌備第三次的最終檢討，以延 續上年評估涉及的主題及跟進有關進展。 董事表現評估 集團主席每年對各非執行董事的個人表現 進行評估。於進行評估時，集團主席會討論 董事的個人貢獻、探討培訓及發展需要、 就董事認為其可以作出更大貢獻的範疇尋 求意見，以及了解董事能否繼續投入所需 的工作時間。根據彼等的個人評估，集團 主席確認全體非執行董事一直有效履行職 責，為滙豐的管治作出正面貢獻及克盡責 任。 各執行董事的個人表現每年均會進行評 估，以作為全體僱員表現管理程序的一部 分，評估結果由集團薪酬委員會考慮，以 釐定浮動酬勞獎勵。 以副主席及高級獨立非執行董事為首的非 執行董事，負責評估集團主席的表現。 董事會將監察各表現評估所提出措施的執 行情況。 董事會擬持續對其本身、其屬下委員會及 個別董事的表現每年進行評估，以及最少 每三年一次由獨立外界機構參與提供意見 （如認為適當） 。 與股東的關係 所有董事應設法了解主要股東的見解。非 執行董事獲邀出席分析員簡報會及與機構 投資者和其代表機構舉行的其他會議。董 事亦每年與機構股東的代表會晤，討論企 業管治事宜。 全體執行董事及若干其他高級行政人員定 期與機構投資者舉行會議。董事會定期省 覽投資者關係活動報告，包括與機構股東 及經紀會面收集的意見、分析員預測、研 究報告資料及股價表現數據。董事會亦定 期省覽我們其中一家公司經紀的報告。 集團的股東通訊政策可於www.hsbc.com/ governance瀏覽。 2014年內，非執行董事 （包括副主席及高級 獨立非執行董事） 曾數次與機構投資者及 其代表會面或溝通，討論企業管治及行政 人員薪酬事宜。 如股東循正常途徑經集團主席、集團行政 總裁、集團財務董事、集團風險管理總監 或其他行政人員未能解決所關注的事宜， 又或不宜與上述人士聯絡，則可聯絡副主 席及高級獨立非執行董事駱耀文爵士。股 東可透過集團公司秘書長 （地址為8 Canada Square, London E14 5HQ） 安排聯絡。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 275 董事的利益衝突、彌償保證及重要合約 董事會已就董事的利益衝突及潛在衝突採 納政策及程序，並可釐定有關情況下的權 限。董事會能夠有效執行批准衝突的權限， 亦有遵守各項程序。董事會每年會檢討所 批准的衝突情況及批准條款。 公司組織章程細則規定，董事倘因若干責 任而遭第三方提出申索，均有權獲得以滙 豐控股資產作出的彌償保證。此等彌償準 備於本財政年度一直存在，但並無董事引 用。全體董事均獲得董事及高級職員責任 保險的保障。 年內或年底時，概無董事直接或間接在滙 豐屬下任何公司所訂立之重要合約中擁有 重大權益。 企業管治守則 滙豐致力恪守高水平的企業管治標準。於 2014年內，滙豐一直遵守(i)英國財務報告評 議會於2012年9月頒布之英國 《企業管治守 則》 的適用守則條文，及(ii) 《香港聯合交易 所有限公司證券上市規則》 附錄14之香港 《企 業管治守則》 的適用守則條文，惟集團風險 管理委員會負責監督內部監控 （而非財務 報告的內部監控） 及風險管理制度 （香港 《企 業管治守則》 第C.3.3條第(f)、(g)及(h)段） 則除 外。倘若並無設立集團風險管理委員會， 該等事宜均會由集團監察委員會負責。英 國 《企業管治守則》 可於www.frc.org.uk瀏覽， 而香港 《企業管治守則》 可於www.hkex.com.
hk瀏覽。 董事會已就董事買賣滙豐集團證券採用一 套交易守則，該守則符合 《英國金融業操守 監管局上市規則》 載述之 《標準守則》 ，以及 《香港聯合交易所有限公司證券上市規則》 所載的 《上市發行人董事進行證券交易的標 準守則》 （ 「香港標準守則」 ） ，但獲香港聯合 交易所有限公司豁免，毋須嚴格遵守香港 標準守則下之若干規定。香港聯合交易所 有限公司授出豁免，主要是考慮到英國公 認的實務守則，特別是有關僱員股份計劃 方面的規定。經具體查詢後，各董事均確 認其於本年度全年內，一直遵守買賣滙豐 集團證券的交易守則。 全體董事定期獲提示彼等於買賣滙豐集團 證券的交易守則下之責任。 董事會報告：企業管治 （續） 滙豐控股有限公司 276 董事會屬下委員會 董事會已設立多個由董事及集團常務總監 組成之委員會，亦已設立包括增選非董事 成員的金融系統風險防護委員會。董事會 屬下委員會的主要職責載於上文。各非執 行董事委員會的主席會在董事會各次會議 上報告委員會自上一次董事會會議以來的 工作情況。 集團管理委員會 角色及成員 集團管理委員會可就本公司及其附屬公司 的管理及日常運作事宜行使董事會的所有 權力、權限及酌情權。 成員 歐智華 （主席） 、麥榮恩及繆思成，均 為執行董事；以及Ann Almeida、安思 明、貝炳達、顧世民、范寧、Pam Kaur （無投票權） 、祈偉倫、利維、Antonio Losada、蘇力宏及王冬勝，均為集團常 務總監。 集團管理委員會主席由集團行政總裁出 任。各環球業務及環球部門主管，以及各 地區行政總裁，以成員身分或獲邀出席集 團管理委員會會議。 集團管理委員會是本公司管理匯報及監控 架構的關鍵部分，所有業務部門均須向集 團管理委員會的成員匯報，又或直接向集 團行政總裁匯報，繼而由集團行政總裁向 集團主席匯報。董事會已為集團管理委員 會制訂目標及衡量準則，使高級行政人員 的工作目標及衡量準則與滙豐整體策略及 營運計劃保持一致。 集團管理委員會主席在每次董事會會議上 報告集團管理委員會的工作。 由集團風險管理總監主持的集團管理委員 會下的風險管理會議定期召開，就企業所 有風險的全面管理提供策略方向及作出監 察，並負責制訂、維持和定期檢討集團內 的風險管理政策及指引。風險管理會議亦 負責檢討環球標準的制訂及實施情況，當 中體現集團上下必須貫徹採納及遵循的最 佳實務。集團防範金融犯罪主管及集團舉 報洗錢活動主管出席這部分的風險管理會 議。 1 於2014年1月17日成立。 2 於2014年12月5日成立。 從高層次角度監督風險相關 事宜及風險管治，並就 此向董事會提供建議的 非執行責任。 監督財務報告相關事宜及 對財務報告進行內部監控， 就此向董事會提供建議 的非執行責任。 滙豐控股有限公司 董事會 集團風險管理 委員會 集團監察 委員會 集團薪酬 委員會 為集團的薪酬政策及高級 行政人員的薪酬制訂總體 原則、參數及管治架構 的非執行責任。 提名委員會 掌管董事會的委任程序， 以及物色和提名董事人選 供董事會批核的 非執行責任。 集團管理 委員會 由董事會直接授權 負責滙豐的管理及 日常運作事宜的 執行管理委員會。 金融系統風險 防護委員會 負有監督以下事宜 的非執行責任： (i)監控措施及 程序，以查找可能使 滙豐乃至整個金融系統 受到金融犯罪或濫用 系統行為侵襲的範疇， 及(ii)滙豐的政策及程序 是否足以確保集團 持續履行監管及 執法機構所訂責任。 行為及價值觀 委員會1 監督滙豐政策、 程序及標準，以確保 集團以負責任的態度 經營業務，同時恪守 滙豐價值觀，並就此 向董事會提供建議的 非執行責任。 慈善及社區投資 事務監察委員會2 肩負監督滙豐各項 慈善及社區投資活動 的非執行責任，以 支持集團實現企業 可持續發展的目標。 主席委員會 於董事會的預定會 議之間代表董事會 行事，以便為急需 董事會通過的臨時 不可預見事務召開 緊急董事會會議。 董事會屬下委員會 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 277 集團監察委員會 主席報告 集團監察委員會的責任是監督財務報 告相關事宜及對財務報告進行內部監 控，就此向董事會提供建議的非執行責 任。本報告載列本年度集團監察委員會 支持此項工作的活動及所面對的事項。 年內，集團監察委員會的主要工作包 括：監督外聘核數師由KPMG Audit Plc （ 「KPMG」 ） 過渡至PricewaterhouseCoopers LLP （ 「羅兵咸永道」 ） 的工作；考慮就操守 相關的法律及監管事宜提撥準備；以及 整合財務報告及內部監控的方法，確保 董事會屬下各委員會採用一致的方法。 2015年，集團監察委員會將專注監察復 元和解決計劃的執行情況，該計劃旨在 確保減低銀行倒閉所帶來的影響，避免 發生嚴重系統性中斷，同時保障相關經 營銀行業務的公司所提供的經濟功能。 集團監察委員會亦會監督財務監控並就 分隔運作英國零售銀行業務和在全球設 立營運公司的影響作出報告。2015年的 另一項重點工作，將是實施經修訂的 《國 際財務報告準則第9號》 「金融工具」 （ 「IFRS 9」 ） ，內容有關金融工具的分類及計量。鑑 於IFRS 9對我們如何分類及計量金融資產 具有潛在影響，因此此項工作特別重要。 祈嘉蓮於2014年3月加入集團監察委員會， 而安銘於2015年1月1日加入集團監察委員 會，他們分別在美國金融業監管政策及會 計準則制訂和報告方面擁有豐富經驗。祈 嘉蓮及安銘之履歷詳載於第265頁。 最後，本人謹此衷心感謝集團監察委員會 的前任主席范樂濤、顧頌賢及已故的張建 東，他們均於本年度退任。他們對委員會 作出的貢獻，本人深表謝意。 集團監察委員會主席 施俊仁 2015年2月23日 角色及成員 集團監察委員會的主要職責包括︰ ‧ 監察財務報表的完整性； ‧ 監督有關財務報告的內部監控制度； ‧ 監察及檢討審核部門的工作成效； ‧ 檢討本公司的財務及會計政策及慣例； 及 ‧ 監督外聘核數師及釐定其費用，並就委 聘外聘核數師向董事會提供意見。 出席會議 次數 有資格出席 之會議 成員1 施俊仁 （主席） 2,7 2 2 顧頌賢3 4 4 范樂濤4 5 5 祈嘉蓮5,7 4 4 張建東6 4 5 駱美思7 7 7 於2014年召開的會議 7 1 所有成員均為獨立非執行董事。 2 於2014年9月1日獲委任為成員及主席。 3 於2014年5月23日退任主席及成員。 4 於2014年5月23日獲委任為主席，並於2014年9月 1日退任成員及主席。 5 於2014年3月1日獲委任為成員。 6 於2014年8月1日退任成員。 7 董事會已確定成員符合美國證交會準則所界 定的獨立性，亦可視為 《Sarbanes-Oxley法案》 第 407條所界定的監察委員會財務專家，並具備 英國 《企業管治守則》 所要求的近期和相關財務 經驗。 管治 下表載列的管治架構，適用於負責監察滙 豐向股東及其他投資者所作報告是否完整 之董事會屬下委員會。各主要營運附屬公 司均已設立對監督財務報告相關事宜負有 非執行責任的董事會屬下委員會。 我們於2014年6月為各主要附屬公司委員會 （負有監督財務報告及風險相關事宜的非 執行責任） 主席舉行研討會，以分享意見及 促使該等附屬公司的委員會以一致的方式 運作。下一個研討會訂於2015年6月舉行。 董事會報告：企業管治 （續） 滙豐控股有限公司 278 監督財務報告的管治架構 權力架構 成員 職責包括： 董事會 執行及非執行董事 ‧ 財務報告 ‧ 委任高級財務主管 披露委員會 環球業務、部門及集團 旗下若干公司的代表 ‧ 檢視集團與投資者的重大溝通事宜 ‧ 協助集團行政總裁及集團財務董事根據 《1934年證券交易所法》 履行與財務報告有關的責任 ‧ 監察及檢討既定監控措施及程序的成效，以確保適當及適時地 披露相關信息 ‧ 向集團行政總裁、集團財務董事及集團監察委員會匯報結果及 提出建議 負責監督財務報告的 附屬公司董事會屬下 委員會及環球業務監 察委員會 獨立非執行董事及╱或 並無於相關附屬公司的 活動或環球業務具有職 能責任的滙豐集團僱員 （如適用） ‧ 就相關附屬公司或業務的財務報表及財務報告之內部監控措施 向集團監察委員會提供報告 （如有需要） 委員會如何履行職責 年內，集團監察委員會定期省覽多個事項 的報告，包括內部審計結果及後續工作、 會計事宜及判斷，以及法律及監管事宜， 並省覽高級管理層多位成員，包括集團財 務董事、集團會計總監及集團審核部主管 的簡報。集團監察委員會主席亦與上述人 士各別舉行會議，為討論特別事項提供另 一個平台。 年內，於管理層避席的情況下，集團監察 委員會與集團審核部主管以及外聘核數師 會面。 集團監察委員會履行以下主要職責： ‧ 監督及查核構成財務報告內部監控架構 的內部監控程序，以及與 《Sarbanes-Oxley 法案》 相關的內部監控程序的成效。於 管理層避席的情況下，集團審核部主管 及外聘核數師亦定期向集團監察委員 會匯報。年內，集團監察委員會確認財 務部門具備足夠資源，以及該部門職員 的資歷及經驗均屬合適。有關內部監控 的進一步詳情，請參閱第288至290頁； ‧ 於2015年全面實施經修訂的反虛假財務 報告委員會發起人委員會 （ 「COSO委員 會」 ） 架構，此乃一個綜合的內部監管架 構，符合我們在 《Sarbanes-Oxley法案》 及 英國和香港企業管治守則下的內部監 控責任； ‧ 檢討滙豐的財務及會計政策，以及其應 用於集團業務及財務表現報告的情況。 有關集團監察委員會此方面工作的進 一步詳情，載於第279頁 「主要活動及審 議的重大事項」 表內； ‧ 監察法律及監管環境；集團監察委員會 定期省覽有關訴訟和應用經修改的法 例、規例、會計政策及慣例的報告，包 括實施IFRS、巴塞爾協定3╱資本指引4 及英國國會銀行業標準委員會建議等 計劃的進度報告，尤其是與會計政策和 財務報告有關者； ‧ 檢討審核部門的工作成效。集團監察委 員會的成效檢討包括審核部門的工作範 圍及內部審計團隊是否具備足夠的技 能。集團監察委員會認為審核部門的工 作仍具成效，此結論獲羅兵咸永道對審 核部門工作進行的質素鑑證審查結果所 支持。年內，集團更新了集團監察委員 會的職權範圍及內部審計章程，以釐清 集團監察委員會監督審核部門管治的責 任，以及集團審核部主管向集團監察委 員會主席匯報的機制。審計章程可於滙 豐網站www.hsbc.com/investor-relations/ governance/internal-control查閱； ‧ 對外聘核數師的工作成效進行年度評 估，包括由集團主要業務地區財務總監 所進行的評估。集團監察委員會考慮審 計時所採用的審查水平及核數師與高 級管理層的相互溝通。於本次檢討後， 集團監察委員會信納KPMG繼續有效履 行外聘核數師的職責；及 ‧ 對外聘核數師的獨立性進行年度檢討。 KPMG於2014年提供的所有服務已經由 集團監察委員會預先審批，並根據集團 監察委員會所制訂的預先審批政策進 董事會屬下委員會 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 279 行。預先審批服務涉及檢討監管政策、 協定程序報告、其他種類的核證報告、 提供意見及美國證交會獨立性規則容 許提供的其他非審計服務。該等服務可 分類為審計服務、審計相關服務、稅務 服務及其他服務。 作出檢討後，集團監察委員會確證其認為 KPMG具獨立性，而KPMG已按照行業標準以 書面向集團監察委員會確認其於截至2014年 12月31日止財政年度的獨立性。集團監察 委員會已批准KPMG的費用及聘用條款， 並建議董事會於截至2014年12月31日止財 政年度重新委聘KPMG。 2013年，集團監察委員會監督審核招標程 序，因而向董事會建議自2015年1月1日開始 的財政年度起，聘用羅兵咸永道為集團的 外聘核數師。集團監察委員會已向董事會 建議於即將舉行的股東周年大會委任羅兵 咸永道。羅兵咸永道已於獲委任前以書面 確認其獨立性。於2014年，委員會與羅兵咸 永道審計團隊定期舉行會議，以協助制訂 2015年的外聘審計計劃。 董事會已根據集團監察委員會的建議，批 准滙豐聘用KPMG及羅兵咸永道前僱員的政 策。集團監察委員會已省覽有關聘用該等 前僱員及擔任要職之前僱員數目的年度報 告。該報告讓集團監察委員會得以考慮外 聘核數師進行審計工作時，其判斷、客觀性 或獨立性有否受損或出現受損跡象。財務 報表附註7披露過去三年來，本公司每年向 KPMG支付的審計及非審計服務費用分析。 除處理上述事項外，集團監察委員會亦審 議下文所述的重大會計事項。集團監察委 員會考慮管理層的判斷及估算是否恰當， 並且在適當情況下與外聘核數師KPMG討論 此等判斷及估算，同時審閱外聘核數師報 告所引述為重大錯誤陳述風險的事宜。 集團監察委員會對其職權範圍及工作成 效進行年度檢討。職權範圍可於滙豐網站 www.hsbc.com/investor-relations/governance/ board-committees查閱。 主要活動及審議的重大事項包括： 主要範疇 所採取的行動 為法律訴訟及監管事 宜提撥準備是否恰當 集團監察委員會省覽管理層就法律訴訟及監管事宜確認及撥備的金額、存在的或有負債及 準備和或有負債相關披露之報告。已處理的特別範疇包括美國聯邦房屋金融局就發售按揭 抵押證券提出的法律訴訟，以及英國金融業操守監管局、美國監管機構及執法機構對外匯 市場買賣活動進行調查而產生的準備。於2015年，集團監察委員會審閱有關多個稅務、監 管及執法機構對滙豐瑞士私人銀行業務展開調查所涉及潛在法律責任的報告及披露文件。 貸款減值準備 集團監察委員會檢討個人及批發貸款的貸款減值準備。重大判斷及估算檢討包括檢討所有 批發貸款組合的虧損生成期、考慮油價下跌對潛在批發貸款減值的影響、批發貸款的個別 大額減值個案及個人貸款組合的綜合評估減值準備是否足夠。 與英國客戶有關的補 救措施 集團監察委員會考慮為不當銷售還款保障保單的賠償、不當銷售利率對沖產品，以及違反 英國 《消費者信貸法》 的責任提撥準備。 金融工具估值 集團監察委員會審議對有關非抵押衍生工具估值計入資金成本的市場慣例發展。為配合不 斷改進的市場慣例，於2014年第四季，我們採納資金公允值調整以計及於計量非抵押衍生 工具估值時加入資金成本的影響。 交通銀行股份有限公 司 （ 「交通銀行」 ） 之減 值測試 年內，集團監察委員會審視滙豐於交通銀行的投資之定期減值檢討，管理層認為該投資並 無減值。當進行於聯營公司之投資的減值測試時，IFRS要求賬面值與公允值及使用價值之較 高者比較。集團監察委員會檢討管理層於此範疇的多項工作，包括減值檢討結果對估計日 後現金流及折現率的估計及假設的敏感度。 商譽減損測試 年度商譽減損測試結果顯示並無減損，而於2014年12月31日進行的減損指標檢討亦無顯示 減損。歐洲環球私人銀行業務的測試結果對主要假設敏感，並須作進一步披露。 確認遞延稅項資產 於考慮集團遞延稅項資產的可收回程度時，集團監察委員會檢討了美國、巴西及墨西哥確 認的遞延稅項資產及日後應課稅收益的相關估計。 非公認會計原則之財 務衡量指標 集團監察委員會考慮在2014年報及賬目中將所呈列的非公認會計原則之財務衡量指標 「實 際基準業績」 改為 「經調整業績」 。 董事會報告：企業管治 （續） 滙豐控股有限公司 280 集團風險管理委員會 主席報告 集團風險管理委員會從高層次角度監督 風險相關事宜及內部監控 （內部財務監控 除外，該等監督工作由集團監察委員會負 責） 並向董事會提供意見。集團風險管理 委員會負責確保集團風險狀況及相關業 務活動與董事會所批准的承受風險水平 一致。 集團高級管理層訂立的基調對有效管理 風險極為重要。年內，集團風險管理委員 會繼續專注於就傳達和加強集團對 「堅守 正道」 的承諾所採取的措施。集團風險管 理委員會向集團薪酬委員會提供有關行 政人員薪酬的意見時，亦反映有關立場。 適用於金融機構的法律及監管架構不斷演 變，其影響已成為一項持續的挑戰。2014年 審慎監管局及歐洲銀行管理局的壓力測 試計劃是集團風險管理委員會年內特別 關注的範疇。2014年法律及監管改革的性 質及速度，亦促使集團風險管理委員會對 集團的承受風險狀況和減輕法律及監管 風險的管理措施作出更嚴格審查。 地緣政治風險仍是集團風險管理委員會 現時的議題。年內，集團風險管理委員會 與集團監察委員會舉行聯合會議，考慮中 國及亞太地區的主要風險。預期於2015年 全年地緣政治風險亦為集團風險管理委 員會的議題。 苗凱婷於2014年9月加入集團風險管理委 員會，她具有豐富的環球金融服務經驗。 苗凱婷曾在多個金融服務機構任職高層， 最近期的是JPMorgan International總裁一 職。苗凱婷之履歷詳載於第267頁。 監管機構於2014年底開始推動整個業界 檢討資訊科技基礎設施，此項工作將延續 至2015年。 集團風險管理委員會主席 費卓成 2015年2月23日 角色及成員 集團風險管理委員會負責： ‧ 從高層次角度就風險相關事宜及風險 管治 （包括現時及前瞻性的風險、日後 風險策略及集團內的風險管理） 向董事 會提供意見； ‧ 就承受風險水平及容忍風險範圍向董 事會提供意見； ‧ 檢討集團風險管理系統架構及內部監 控系統的成效 （但不包括屬集團監察委 員會職責之內部財務監控系統） ； ‧ 監察行政監控及風險管理，包括首要及 新浮現風險；及 ‧ 就促使薪酬與承受風險水平相配而向 集團薪酬委員會提供意見。 集團風險管理委員會由下列獨立非執行董 事組成。 出席會議 次數 有資格出席 之會議 成員 費卓成 （主席） 13 13 顧頌賢1 5 5 利普斯基 13 13 駱美思 13 13 苗凱婷2 4 4 於2014年召開的會議 13 1 於2014年5月23日退任董事及成員。 2 於2014年9月1日獲委任為成員。 於2014年全年，英國滙豐銀行有限公司非 執行董事John Trueman獲邀請出席集團風 險管理委員會的會議。滙豐控股有限公司 非執行董事凱芝出席兩次就資訊科技事宜 向集團風險管理委員會作出的簡報會。 風險管治 滙豐所有業務均涉及計量、評估、承擔及 管理一種或多種風險。董事會按照集團風 險管理委員會的建議，要求及鼓勵以穩健 的風險管治文化，作為集團對風險取態的 基礎。董事會及集團風險管理委員會持續 監察風險環境、集團面對的首要及新浮現 風險，以及計劃和採取的減輕風險措施， 從而監督集團維持及發展強而有力的風險 管理架構。 董事會及其委員會管理風險的管治架構載 於第24頁的表內。集團風險管理委員會負 有監督整個集團的風險之整體非執行責 任。行為及價值觀委員會與金融系統風險 防護委員會負責監督特定範疇的風險，包 括提倡及貫徹滙豐集團價值觀及滙豐集團 原則，並監督有關反洗錢、制裁法律、反 資助恐怖主義及武器擴散的事宜。行為及 價值觀委員會與金融系統風險防護委員會 定期向集團風險管理委員會匯報最新活動 情況。 集團旗下各主要營運附屬公司均已設立對 監督風險相關事宜負有非執行責任的董事 會屬下委員會，及負責風險相關事宜的執 行委員會。集團風險管理委員會已為附屬 公司非執行風險及監察委員會設定主要的 職權範圍。 有關整個集團就管理風險設立的架構之進 一步詳情載於第112至118頁。 集團風險管理委員會如何履行職責 集團風險管理委員會於各次會議中與管理 層討論首要及新浮現風險及集團的風險狀 董事會屬下委員會 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 281 況。於監察首要及新浮現風險時，集團風 險管理委員會省覽集團風險管理總監 （為 執行董事） 、防範金融犯罪環球主管兼集團 舉報洗錢活動主管，以及監管合規環球主 管提交的報告及簡報。年內，高級管理層 的其他成員 （包括集團營運總監、環球風險 管理策略主管及辦公室主任、集團員工表 現及獎勵事務主管及集團資料總監） 出席 了集團風險管理委員會的會議。 集團風險管理總監定期向集團風險管理委 員會提交報告及簡報，包括於各次會議上 提呈的 「風險圖譜」 簡報 （該圖譜按照風險 類別分析環球業務及地區的風險狀況） 、集 團承受風險水平聲明，以及首要及新浮現 風險報告，概述就已識別風險採取的減低 風險措施。集團風險管理委員會要求管理 層就識別作深入考慮的風險相關事宜提交 報告及更新，並定期省覽集團管理委員會 的風險管理會議所討論事宜的報告。 第118頁提供有關集團首要及新浮現風險 的進一步資料。 於本年度全年，集團風險管理委員會主席 根據需要與集團風險管理總監、集團審核 部主管、集團財務董事、法律事務總監及 其他高級行政人員舉行會議。 除處理上述事宜外，集團風險管理委員會 亦重點審議多個主要範疇，包括下表所列 的事項。 主要活動及審議的重大事項包括： 主要範疇 所採取的行動 集團承受風險水平聲 明和根據承受風險水 平聲明監察集團風險 狀況 集團風險管理委員會檢討管理層修訂2014年集團承受風險水平聲明指標的建議。經檢討 後，委員會就集團承受風險水平聲明向董事會建議多項修正，包括成本效益、普通股權 一級資本及主權風險承擔的比率。 集團風險管理委員會根據承受風險水平聲明所載的主要表現標準，定期檢討集團風險狀 況。集團風險管理委員會檢討管理層對風險的評估，並審查管理層建議的減低風險措施。 審慎監管局及歐洲銀 行管理局的並行壓力 測試 集團風險管理委員會監察由審慎監管局及歐洲銀行管理局要求進行的壓力測試，並於提 交壓力測試結果給有關的監管機構前檢討結果。集團風險管理委員會於進行審慎監管局 及歐洲銀行管理局的壓力測試期間省覽相關報告，並於年內舉行三次會議，專門商討壓 力測試相關事宜。集團風險管理委員會於有關會議上檢討審慎監管局及歐洲銀行管理局 所設定的壓力測試境況，以及提升有關境況的方法 （如適用） 。集團風險管理委員會監督 管理層檢討由此壓力測試中汲取的教訓。 審核部評估監管規定壓力測試計劃的進度，並向集團風險管理委員會報告其結論及提出 建議。 執行風險 執行風險是與實施集團策略有關的風險，亦是集團風險管理委員會的常設議程項目。監 察此項風險、查核管理層對執行風險的評估，以及相應的減低風險措施，仍是集團風險 管理委員會的優先處理項目。 除省覽定期報告及對已識別的特別事項進行 「深入檢討」 外，集團風險管理委員會亦要求 審核部就執行工作時識別的議題提交報告。 法律及監管風險 法律及監管環境日趨複雜，對金融服務機構的要求日益提高。 集團風險管理委員會定期省覽法律及監管風險報告，檢討減低該等風險的管理措施，並 考慮此項風險的日後發展對集團的潛在影響。於2015年，這些工作包括審閱有關多個稅 務、監管及執法機構對滙豐瑞士私人銀行業務展開調查所涉風險的報告。監管機構的資 本充足規定尚未明確，仍是集團風險管理委員會特別關注的範疇；集團風險管理委員會 已安排更多時間處理有關事宜。 資訊科技及數據相關 風險 年內，集團風險管理委員會考慮多項資訊科技及數據相關風險，包括互聯網罪行及詐 騙、數據管理及匯總以及資訊保安。集團風險管理委員會檢討管理層對有關風險的評估 及減低風險的管理措施。 預期資訊科技及數據相關風險仍屬集團風險管理委員會於2015年的重點關注範疇。 地緣政治風險 集團風險管理委員會定期省覽地緣政治風險 （包括中東及烏克蘭的危機及南中國海的領海 主權爭端持續緊張） 報告。管理層會因應有關事宜 （包括加強反洗錢、制裁法律及防範金 融犯罪監控） 所採取的減低風險措施定期提供最新報告。集團風險管理委員會亦與集團監 察委員會舉行聯合會議，集中討論於中國內地及亞太區所面對的問題。 有關識別、管理及減低上述風險的進一步資料載於第114至117頁。 董事會報告：企業管治 （續） 滙豐控股有限公司 282 董事會屬下委員會 金融系統風險防護委員會 主席報告 金融系統風險防護委員會監察管理層實 施各項政策，目的是減低滙豐於執行其 策略時面對的金融犯罪及濫用系統的風 險。就此，委員會為集團的監控和流程架 構，提供思想領導、進行管治、監督和制 訂政策指引，以處理滙豐乃至整個金融系 統可能面對的風險。更正式地，委員會負 責監督集團遵循監管法令，包括監督集團 與監察員 1的關係，委員會與監察員定期 會面，並致力確保我們各項優先策略合乎 利益。 我們明白滙豐過去並未貫徹地識別及防 止外界透過我們的網絡誤用及濫用金融 系統。然而，採納最高或最有效的環球合 規標準及最高行為標準是我們策略的一 部分，以找出有關問題再度發生的可能 性，而且有助我們履行監管機構於2012年 訂立的法令下的責任。 如本報告所述，於2014年，委員會在達致 目標方面取得重大進展，包括檢討及採納 有關遵守反洗錢及制裁法律的新環球政 策、協議並制訂關於加強監察交易及客戶 盡職審查系統和程序的里程指標，並由 監察員擔當此項日常工作。於2014年，金 融系統風險防護委員會的另一重要職責 是就金融犯罪風險向集團提供前瞻性觀 點。舉例而言，委員會就減低𣿬豐為客戶 提供產品時處理大量數據涉及的風險， 於2014年曾進行深入檢討，以確保採取有 效措施。金融系統風險防護委員會的五位 主題專家，就識別集團可能面對的風險為 委員會提供寶貴指引及意見，並與我們合 力降低這些風險。 因此，於2015年，金融系統風險防護委員 會將繼續專注實行監控措施及程序，以達 致最高行為和合規標準。穩健的合規文化 對集團策略的成功極為重要，而金融系統 風險防護委員會於年內將繼續關注相關 事宜。 本人藉此機會感謝方安蘭自委員會2013年 初成立以來一直盡心領導。本人於去年5月 欣然接任，帶領委員會確立清晰目標，應 付滙豐面對的挑戰。 金融系統風險防護委員會主席 埃文斯勳爵 2015年2月23日 1 有關監察員的進一步詳情，請參閱第27頁。 角色及成員 金融系統風險防護委員會具有以下非執行 責任： ‧ 就集團的監控和流程架構，進行管治、 監督和制訂政策指引，目的是查找可能 使滙豐乃至整個金融系統受到金融犯 罪或濫用系統行為侵襲的範疇； ‧ 監督涉及反洗錢、制裁法律、資助恐怖 主義及武器擴散的事宜，包括制訂、實 施、維持及加以檢討以確定具備足夠的 政策及程序，足以確保集團持續履行監 管及執法機構所訂責任，以及監督為確 保此等範疇安全而採取的必要行動； ‧ 就落實環球標準計劃提供意見 （如適 用） ；及 ‧ 就金融犯罪風險向董事會提供前瞻性 觀點。 出席 會議次數 有資格 出席 之會議 成員 埃文斯勳爵 （主席） 1 6 7 祈嘉蓮2 5 5 方安蘭3 7 7 費斯域4 7 7 夏力達4 7 7 希爾斯4 7 7 駱耀文爵士 7 7 史蘭克4 7 7 薩維迪4,5 7 7 於2014年召開的會議 7 1 於2014年5月23日獲委任為主席。 2 於2014年3月1日獲委任為非執行董事及成員。 3 於2014年5月23日退任主席。 4 增選非董事成員。 5 亦向北美滙豐控股有限公司董事會提供顧問 服務。 五名增選非董事成員已獲委任為委員會顧 問以協助其工作。彼等之簡歷載列如下： 費斯域，CMG：前英國外交及聯邦事務部 高級官員，專責安全、情報及反恐事務； 2001至2004年間，被借調至英國海關與消 費稅局出任情報 （執法） 主管，致力於國際禁 毒工作和防止稅務及消費稅詐騙；於2009年 獲頒授CMG勳章。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 283 夏力達，CB：前英國稅務海關總署稅務常 務次官；於36年的稅務行政管理工作生涯 中，專門從事稅務政策的制訂、合規和執 法工作，以及處理國際稅務問題；前經合 組織稅務行政管理論壇副主席。 希爾斯，CBE QPM：前英國嚴重及有組織罪 行調查局局長；對有組織罪行的偵破、瓦 解及刑事調查具備豐富國際經驗。 史蘭克：SWIFT前行政總裁。SWIFT是由業界 擁有的環球金融訊息系統。監督SWIFT與美 國財政部以至其他國家╱地區在反恐問題 上的合作。為MIT Corporation受託人理事會 成員。 薩維迪：美國戰略及國際研究中心的高級 顧問；哥倫比亞廣播公司新聞網絡國家安 全高級分析師；哈佛法學院客座法律講師； 國家安全顧問；前美國總統副助理及專責 反恐的副國家安全顧問，負責制訂和實施 美國政府的反恐戰略及政策，以防跨國安 全威脅；負責反資助恐怖主義及金融犯罪 的前美國助理財政部長；檢控恐怖活動團 隊的前聯邦檢察官。 金融系統風險防護委員會如何履行職責 金融系統風險防護委員會專注於滙豐乃至 整個金融系統可能面對的金融犯罪或濫用 系統行為侵襲的風險，而集團風險管理委 員會則仍然負責高層次風險事宜及風險管 治。金融系統風險防護委員會專責範疇包 括：網絡安全；科技及數據系統；監察反 洗錢交易系統；制裁法律及其他金融犯罪 相關風險；以及客戶盡職審查及 「認識你 的客戶」 程序。顧問成員及相關行政人員 就此專責範疇定期向金融系統風險防護委 員會提交報告及更新資料。 金融系統風險防護委員會亦負有監督履行 美國及英國協議的責任，並就滙豐與監察 員的互動情況更新資料。 法律事務總監、集團風險管理總監、防範 金融犯罪環球主管、集團舉報洗錢活動主 管、監管合規環球主管及集團審核部主管 向金融系統風險防護委員會提交報告，內 容包括集團就2012年12月美國和英國監管及 執法機關調查的解決方案及其他一般事項 而作出的合規相關措施，與監管機構進行 會議及向其提交報告的會議之最新資料。 除委員會的例會外，主席會按需要定期與 集團主席、委員會顧問成員及高級行政人 員會面。 年內，金融系統風險防護委員會定期省覽 合規計劃的最新資料，該等資料記錄集團 加強滙豐反洗錢及制裁法律合規工作的策 略，當中包括相關政策、程序和加強培訓。 金融系統風險防護委員會亦會定期收到有 關整個集團的舉報披露及反賄賂及貪污事 宜的報告。 除向董事會作出報告外，金融系統風險防 護委員會亦定期向集團風險管理委員會就 指定事宜提供最新資料，提出供其考慮的 範疇 （如適用） 。 年內，金融系統風險防護委員會專注於下 表所列的多個主要範疇。 主要活動及審議的重大事項包括： 主要範疇 所採取的行動 金融犯罪相關事宜 金融系統風險防護委員會檢討及採納集團的反洗錢政策，該政策現已於滙 豐的所有業務落實。金融系統風險防護委員會定期省覽落實資訊科技策略 的最新資料，作為管理及減低金融犯罪風險工作的一部分，其中重點在於 管理層建議提升集團交易監察系統。 制裁法律 金融系統風險防護委員會檢討及採納集團的遵循制裁法律政策，該政策現 已落實，而集團現時的遵循制裁法律計劃及管理層回應全球制裁擴大的策 略，於年內亦由委員會定期監察。 網絡安全 於2014年，金融系統風險防護委員會檢討網絡安全風險及有關策略，並建 議提升集團的網絡保安能力。這項檢討包括對集團預計、回應及收復網絡 攻擊的能力作出簡報。委員會已經與管理層協定指標及時間表，以監察此 方面的進度。 《外國賬戶稅務合規法案》 及稅務透明 度 金融系統風險防護委員會收取滙豐及整個集團實施 《外國賬戶稅務合規法 案》 項下有關稅務透明度的措施的最新資料。 董事會報告：企業管治 （續） 滙豐控股有限公司 284 董事會屬下委員會 集團薪酬委員會 集團薪酬委員會負責核准薪酬政策。委員 會的職責包括衡量固定酬勞、周年獎勵計 劃、股份計劃、其他長期獎勵計劃、福利 及執行董事與集團其他高層人員個人薪酬 福利之條款。我們在衡量過程中會考慮整 個集團的酬勞及僱用條件。所有董事均沒 有參與釐定本身之薪酬。 出席 會議次數 有資格 出席 之會議 成員1 駱耀文爵士 （主席） 11 11 顧頌賢2 6 6 范樂濤3 7 7 利普斯基4 5 5 李德麟 10 11 施俊仁5 2 3 於2014年召開的會議 11 1 所有成員均為獨立非執行董事。 2 於2014年5月23日退任董事及成員。 3 於2014年9月1日辭任董事及成員。 4 於2014年5月23日獲委任為成員。 5 自2014年4月14日至2014年9月1日為成員。 董事薪酬報告載於第300至327頁。 提名委員會 主席報告 提名委員會的其中一項主要職責，是確保 董事會在技能、知識、經驗、多元化及獨 立性各方面取得適當平衡。經提名委員會 推薦，董事會於2014年委任四名獨立非執 行董事，即安銘、祈嘉蓮、苗凱婷及施俊 仁。他們為董事會帶來不同的專業知識及 經驗。滙豐現已超越董事會本身定下的多 元化政策的目標，當中列明於2020年之前 女性董事會成員的百分比須為30%。 提名委員會的另一項重要職責，是確保執 行董事及高級行政人員的選拔、委任和有 序繼任均可按計劃行事。提名委員會於上 年度召開一次會議，與集團行政總裁就繼 任計劃進行深入檢討，其結論為有關計劃 屬充足及適當，但須每年檢討。 提名委員會繼續注視監管發展，因為有關 發展可能需要對董事會的組成作出更改。 提名委員會已詳細考慮於2014年7月1日生 效的歐盟 《資本規定指引4》 項下的新規定， 該規定限制董事會成員出任多家公司董事 的數目；亦已評估有關新規定對目前董事 會所衍生的後果，並定期檢討有關規定。 提名委員會主席 駱耀文爵士 2015年2月23日 角色及成員 提名委員會掌管董事會的委任程序，並有 物色和提名董事人選供董事會批核的非執 行責任。提名委員會負責董事會成員的繼 任計劃，同時監督高級管理層的繼任計劃。 出席 會議次數 有資格 出席 之會議 成員1 駱耀文爵士 （主席） 4 4 史美倫2 2 2 方安蘭 4 4 何禮泰3 2 2 李德麟2 2 2 利普斯基 4 4 於2014年召開的會議 4 1 所有成員均為獨立非執行董事。 2 於2014年5月23日獲委任為成員。 3 於2014年5月23日退任董事及成員。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 285 提名委員會如何履行職責 提名委員會曾履行以下主要職責： 主要活動及審議的重大事項包括： 主要範疇 所採取的行動 委任新董事 經過嚴格的遴選程序，於2014年提名委員會建議董事會委任四名非執行董 事，即安銘 （自2015年1月1日起生效） 、祈嘉蓮、苗凱婷及施俊仁；及一名 執行董事︰繆思成 （集團風險管理總監） 。 集團就四名非執行董事其中三名 （祈嘉蓮、苗凱婷及施俊仁） 的委任聘用外 界人事顧問MWM Consulting。除作為聘請若干高級行人員的獵頭顧問外， MWM Consulting與滙豐並無其他關連。安銘自2012年起出任北美滙豐控股 有限公司的獨立董事 （擔任審核委員會主席及風險管理委員會成員） 而由委 員會物色。安銘在財務及會計方面擁有廣泛經驗，因他曾在通用電氣任職 多年 （最終擔任職務為通用電氣公司副總裁、審計長及首席會計師） ，亦曾 參與制訂會計準則，掌握湛深的技術知識。 前瞻計劃 提名委員會基於集團業務的需要和發展，以及現任董事的預期退任日期， 採用前瞻性方式，物色有潛質可出任董事會成員的候選人。 董事會及屬下委員會的規模、架構及 組成 提名委員會監督董事會的規模、架構及組成 （包括才能、知識、經驗、多元 化及獨立性） 。 提名委員會已審議在2014年股東周年大會上選舉或重選董事的事宜，並向 董事會建議所有董事在2015年股東周年大會上參與選舉或重選。 監管發展 提名委員會關注可能影響董事會組成的監管發展。於2014年，提名委員會 衡量歐盟 《資本規定指引4》 的企業管治規定及平等及人權委員會對平等法律 架構指引之影響。 多元化 提名委員會相信其中一項重要職責為確保董事會得到適當的平衡，以反映 多元化及其業務的地理性質。董事會將秉持用人唯才，並根據客觀標準考 慮候選人的條件，且會充分顧及董事會多元化的優點。董事會多元化政 策可於www.hsbc.com/investor-relations/governance/corporate-governance-codes查閱。 提名委員會利用以下可量度指標定期監察實施董事會多元化政策的進展： 最少25%的董事會成員應為女性，計劃於2020年前達至30%的目標；提名 委員會僅可委聘已簽署 「獵頭公司自願行為守則」 的外聘人事顧問公司；及 由獵頭公司推薦的非執行董事候選人，最少30%應為女性。我們已遵守此 項規定，於本報告日期，董事會35.3%成員為女性。 董事的培訓與發展 提名委員會已檢討及監察董事與高級管理人員的培訓和持續專業發展。 非執行董事投入的時間及獨立性 提名委員會評估非執行董事的獨立性及所需投入的時間。提名委員會信納 所有非執行董事均有時間履行其監督集團業務的受託責任及為董事會屬下 相關委員會提供服務。所有董事須呈報在外擔任的其他重要職務，並確認 有足夠時間履行作為董事會成員所肩負的責任。 董事會報告：企業管治 （續） 滙豐控股有限公司 286 董事會屬下委員會 行為及價值觀委員會 主席報告 行為及價值觀委員會成立於2014年1月， 目的是替董事會監督集團在提高行為標 準，並深入傳播集團堅信的行為價值方面 所進行的多項工作。帶給客戶公平回報及 維護市場誠信，是任何企業得以持續發展 及賺取盈利的主要推動力。儘管行為風險 並非新概念，但董事會深切明白全球各地 的監管機構以至整個銀行業都越來越關注 這類風險，因此成立委員會以監督如何管 理行為風險乃屬正確之舉。 隨著美國於2011年成立消費者金融保護 局；英國金融業操守監管局開宗明義以確 保客戶受到合適保障及提升金融體系的誠 信操守為其眾多目標之一；以及香港金融 管理局制訂公平待客約章，過去數年這些 跡象清楚顯示有必要加強這方面的工作。 此外，罰款的金額提升，尤其以美國升幅 最顯著，庭外和解的趨勢也逐漸形成。這 些事實一再加強我們的信念：恪守更高的 行為標準對於重建客戶信心及恢復社會對 銀行業的信任實屬必要。 自成立以來，行為及價值觀委員會已採取有 系統的策略，專注審視各項環球業務及環球 部門，並深入分析其本位及優先發展市場 （至目前為止，特別集中檢視英國市場） 。每 次會議議程的討論事項均嚴格與職權範圍聯 繫起來，並確保最少每年一次充分執行各項 主要職責。 本人自行為及價值觀委員會成立以來一直擔 任主席，前企業可持續發展委員會主席史美 倫、埃文斯勳爵、苗凱婷及施俊仁均為成 員。 於2015年，行為及價值觀委員會將繼續專 注落實集團的行為及市場風險管理計劃， 並特別關注僱員培訓及客戶溝通事宜，目 的是以前瞻性角度評估行為風險，並預計 與行為相關的公共政策變動將備受關注。 委員會繼承現已解散的企業可持續發展委員 會的職務，日後將兼負可持續發展責任，以 確保滙豐於其業務所在的社區以負責任的方 式營運。 行為及價值觀委員會主席 駱美思 2015年2月23日 角色及成員 行為及價值觀委員會負責： ‧ 監督滙豐政策、程序及標準，確保集團 以負責任的態度經營業務，同時恪守滙 豐價值觀，目的是配合滙豐的宗旨：努 力建立聯繫以幫助客戶開拓商機、推動 工商企業茁壯成長及各地經濟蓬勃發 展，而最終目標是讓客戶實現理想；及 ‧ 確保於進行業務時，滙豐以公平及公開 的態度對待客戶、以正確的方式與正當 的客戶進行業務往來、是負責任的僱 主、對滙豐營運所在社區以負責任的態 度行事，並且公平對待其他相關群體。 行為及價值觀委員會由獨立非執行董事組 成，載列如下。 出席 會議次數 有資格 出席 之會議 成員 駱美思 （主席） 1 4 4 史美倫2 4 4 埃文斯勳爵2 4 4 苗凱婷3 2 2 施俊仁4 4 4 於2014年召開的會議 4 1 於2014年1月17日獲委任為主席。 2 於2014年1月17日獲委任為成員。 3 於2014年9月1日獲委任為成員。 4 於2014年4月14日獲委任為成員。 管治 行為及價值觀委員會對推廣及貫徹滙豐價 值觀及所須的環球行為成果具有非執行的 監察責任。此外，行為及價值觀委員會將 薪酬與操守掛鈎的理念引入集團薪酬委員 會。委員會定期向董事會就其工作作出報 告。 行為及價值觀委員會如何履行職責 於2014年，行為及價值觀委員會定期省覽 零售銀行及財富管理業務行政總裁、工商 金融業務行政總裁、監管合規環球主管、 集團發展主管、集團企業可持續發展主管 及集團審核部主管的報告及簡報。年內， 曾出席行為及價值觀委員會會議的高級管 理層其他成員包括：環球銀行及資本市場 業務行政總裁、防範金融犯罪環球主管、 企業傳訊環球主管、反賄賂及反貪污環球 主管及市場推廣環球主管。 零售銀行及財富管理業務行政總裁及環球 工商金融業務行政總裁向行為及價值觀委 員會定期提交報告及簡報，包括於各次會 議提交客戶投訴趨勢的分析。行為及價值 觀委員會亦定期省覽舉報個案、內部審核 的結果及集團落實主要價值觀及文化計劃 所採取的措施。 除預定舉行的委員會會議外，主席定期與 集團主席及高級行政人員會面 （如需要） 。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 287 委會員亦有責任向董事會、董事會屬下委 員會及行政管理層就集團的企業可持續發 展政策 （包括環境、社會及道德事宜） 提出 建議。自本年度起，我們的可持續發展政 策及表現的進度於 《策略報告》 及網站www.
hsbc.com/sustainability內報告。 我們不會再刊發獨立的 《可持續發展報告》 ， 以進一步整合可持續發展報告方式。此變 動反映報告的最佳慣例及持續將可持續發 展事宜整合至滙豐的策略及管理中。 年內，行為及價值觀委員會專注於以下多 個主要範疇。 主要活動及審議的重大事項包括： 主要範疇 所採取的行動 行為的環球方針 行為及價值觀委員會核准管理行為的環球方針，並界定及列出所要求的成 果。委員會定期省覽監管合規環球主管提交有關集團如何貫徹管理行為， 以達致所要求的環球行為成果的報告。委員會亦制訂各項旨在改善行為的 計劃及管治工作的情况。於制訂方針的過程中，管理層已考慮策略、業務 模式及決策程序、文化及行為、與客戶互動情況、金融市場活動及管治架 構的影響、監察架構及管理資料。推廣及貫徹滙豐價值觀所進行的工作時 刻保持緊密一致。 價值觀 行為及價值觀委員會對推廣及貫徹滙豐價值觀肩負監察責任。於2013年， 集團推出一項計劃，以便加深理解滙豐價值觀對集團日常行為的影響。這 包括與領導團隊及部門專才面談，各級經理與員工組成的關注小組，以及 檢討管理資料及當地的文件紀錄。 行為及價值觀委員會定期省覽管理層有關此項計劃的報告，並納入其後的 行動計劃中。委員會將於2015年繼續監察文化轉變的落實。 客戶體驗 客戶投訴。行為及價值觀委員會審閱有關客戶體驗、投訴趨勢及處理投訴 的報告，並審視有關改善處理投訴過程的質素，分析根本原因。 銷售過程及獎勵計劃。行為及價值觀委員會審視由零售銀行及財富管理業 務的管理層建立的檢討機制，確保零售銀行及財富管理業務產品的定位滿 足客戶的需要。行為及價值觀委員會亦檢討零售銀行及財富管理與工商金 融業務在銷售程序及銷售獎勵計劃上作出的變動，以及新的品質保證計劃 的成效。此項工作將持續至2015年。 舉報 行為及價值觀委員會肩負集團舉報政策及程序 （包括保障舉報者） 的管治責 任。此責任並無延伸至有關財務報告的事宜及相關的審核事宜 （仍屬於集團 監察委員會的責任） 。行為及價值觀委員會檢討現時舉報過程及披露，並省 覽有關現時進行的提升計劃之報告，該計劃考慮英國國會銀行業標準委員 會的建議、監管指引及業界新興最佳慣例。 僱員參與 行為及價值觀委員會同時監察集團的員工參與度，並省覽於2014年進行 Snapshot員工參與度季度調查結果。委員會將於2015年繼續監察這些調查 結果，以及於年內較後時間進行的集團員工意見調查結果。 董事會報告：企業管治 （續） 滙豐控股有限公司 288 內部監控 主席委員會 角色及成員 主席委員會有權於董事會的預定會議之間 代表董事會行事，以便為急需要董事會通 過的突發事務召開緊急董事會會議。按委 員會的職權範圍所載，委員會決定何時舉 行會議及會議的頻密度，會議法定人數視 乎所討論事項的性質而定。 慈善及社區投資事務監察委員會 角色及成員 慈善及社區投資事務監察委員會於2014年 12月經董事會決議成立，將專注於集團的 慈善活動，包括捐款給慈善機構及員工付 出時間參與義務工作。 委員會具有監察滙豐的慈善及社區投資活 動的非執行責任，以支持集團的企業可持 續發展目標。 委員會將於2015年舉行首次會議，並每年 最少舉行兩次會議。 成員 史美倫1 （主席） 埃文斯勳爵1 Malcolm Grant爵士2,4 Ruth Kelly3,4 Stephen Moss3,4 1 於2014年12月5日獲委任。 2 獨立委員。 3 僱員代表。 4 於2015年2月19日獲委任。 內部監控 程序 董事負責維持及評估風險管理及內部監控 制度的成效，並釐定於實現其策略目標時 願意承擔各項重大風險的性質及程度。為 符合此要求並根據金融業操守監管局手冊 及審慎監管局手冊履行責任，已制訂一系 列程序，以防止資產未經授權而被挪用或 出售、妥善保存會計紀錄，以及確保業務 使用或向外公布之財務資料適用可靠。有 關程序只能提供合理保證，但不能絕對確 保不會出現重大錯誤陳述、錯誤、損失或 詐騙行為。 制訂有關程序是為滙豐提供有效的內部監 控，並遵循英國財務報告評議會於2005年 頒布的董事工作指引的修訂版本。在本年 度及直至2015年2月23日 《2014年報及賬目》 獲通過之日一直實施滙豐程序。英國財務 報告評議會於2013年11月及2014年4月舉行 諮詢後，此項指引已作修訂，風險管理、內 部監控及相關財務及業務報告的指引亦因 而修訂。經修訂指引適用於以2014年10月 1日或以後作為財政年度起始日的公司。 至於在本年度所收購公司的風險管理及內 部監控工作，正根據滙豐所用基準檢討， 並已納入滙豐的程序。 於2014年，集團監察委員會及集團風險管 理委員會授權採納COSO委員會2013年框 架，以監察風險管理及內部監控制度遵守 2002年 《Sarbanes-Oxley法案》 第404條、英國 《企業管治守則》 及香港 《企業管治守則》 的 規定。COSO委員會2013年框架將於2015年 全面實施。截至2014年12月31日止年度， 滙豐繼續根據英國財務報告評議會的 《經修 訂董事內部監控指引》 與1992年原有框架評 估有關財務報告的內部監控情況。 滙豐之主要風險管理及內部監控程序如下： ‧ 集團標準。有關職能、營運、財務報告 及若干管理匯報之標準，由環球部門之 管理委員會訂立，並在滙豐內部全面實 施。除此等標準外，各附屬公司之部門 及當地管理層亦已就業務類別和經營 地域之需要制訂營運標準以作補充。 ‧ 由董事會釐定授予權限。各集團常務總 監獲董事會根據釐定的權限授予管理 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 289 其負責業務或部門日常事務的權力。獲 董事會授權的個別人士，須清晰及恰當 分配重大責任，並監督如何制訂和維持 適用於相關業務或部門的監控制度。滙 豐內部最高層職位之委任須獲董事會 通過。 ‧ 識別及監察風險。集團已制訂各種制度 和程序以識別、監控和匯報滙豐面對的 各項主要風險 （請參閱第21頁） ，包括信 貸、市場、流動資金及資金、資本、財 務管理、模型、聲譽、退休金、策略、 可持續發展、營運 （包括會計、稅務、法 律、監管合規、防範金融犯罪、受信、 保安及詐騙、制度運作、項目及人事風 險） 及保險風險。此等風險由各附屬公 司之風險管理委員會、資產、負債及資 本管理委員會及執行委員會監控，就集 團層面而言，由集團管理委員會之風險 管理會議 （由集團風險管理總監主持） 作 監控。風險管理會議定期召開，以討論 企業上下的風險管理事宜。集團資產負 債管理委員會負責監察資產、負債及資 本管理事宜，並向風險管理會議匯報。 滙豐的營運風險狀況及集團營運風險 管理架構的成效，由環球營運風險管理 委員會監控，該委員會須向風險管理會 議負責。模型風險由模型監察委員會監 控，該委員會亦須向風險管理會議負 責。集團管理委員會會議及風險管理會 議之會議紀錄，須向集團監察委員會、 集團風險管理委員會及董事會成員提 供。 ‧ 市況及市場慣例的改變。目前已有程序 用以識別因市況及市場慣例或客戶行 為改變而引致的新風險，此等風險可能 增加滙豐的虧損風險或聲譽受損風險。 2014年內，集團主要關注： － 經濟前景及政府干預； － 地緣政治風險增加； － 影響我們業務模式及集團的盈利能 力之監管環境發展； － 監管機構對經營業務方式及金融犯 罪作出的調查、罰款、制裁，以至 相關承諾、同意令及規定，對集團 的業績及品牌構成負面影響； － 爭議風險； － 執行風險提高； － 人事風險； － 第三方風險管理； － 互聯網罪行及詐騙； － 資訊保安風險； － 數據管理；及 － 模型風險。 ‧ 策略方案。集團會就各環球業務、環球 部門及納入集團策略架構的若干業務 地區，定期編訂策略方案。滙豐旗下所 有主要營運公司則編訂及採納年度營 運計劃，該計劃詳載有關承受風險水平 的分析，當中載述我們在落實策略時將 面對的風險類別及風險額，並訂定各項 主要業務計劃及列明該等計劃可能產 生的財務影響。 ‧ 披露委員會。披露委員會對滙豐控股 就公開披露的重要資料內任何重大錯 失、失實陳述或遺漏作出檢討。由集團 公司秘書長擔任主席的披露委員會成 員包括環球財務部、法律事務部、風險 管理部 （包括防範金融犯罪組及監管合 規組） 、企業傳訊部、投資者關係及審 核部門的主管，以及來自主要地區及環 球業務的代表。各環球財務及環球風險 管理部門內的架構及程序，有助專門而 慎密地分析及審查財務報告，且經環球 業務、環球部門及若干法律實體的主管 審查核證，確保披露資料正確。 ‧ 財務報告。編製綜合 《2014年報及賬目》 的集團財務報告程序受明文規定的會 計政策及報告格式監控，輔以編製報告 的詳盡指示及列表，並於每個業績報告 期結束前由集團財務部向集團內部所 有匯報公司發出。各匯報公司向集團財 務部提交的財務資料，須由負責的財務 人員證明，並須通過匯報公司及集團的 分析審查程序。 ‧ 風險管理責任。環球業務及環球部門的 管理層主要負責衡量、監察、減低及管 理其業務及部門之風險及監控措施。集 團制訂了流程，輔以三道防線的模型以 確保漏洞已上報高層管理人員並獲得 解決。 ‧ 資訊科技運作。所有資訊科技系統之發 展與運作，均由中央部門監控。在可行 情況下，同類業務運作程序會採用共用 之電腦系統。 ‧ 部門管理。環球部門管理層負責就下 列各類風險制訂政策、程序及準則：信 貸風險、市場風險、流動資金及資金 風險、資本風險、財務管理風險、模型 風險、聲譽風險、退休金風險、策略風 險、可持續發展風險及營運風險 （包括 會計風險、稅務風險、法律風險、防範 董事會報告：企業管治 （續） 滙豐控股有限公司 290 金融犯罪風險、監管合規風險、受信風 險、資訊保安風險、保安及詐騙風險、 制度風險及人事風險） 及保險風險。訂 立信貸風險及市場風險交易的權限授 予集團旗下公司之業務管理層，惟附有 相應限額。然而，具特定較高風險特性 的信貸建議須獲得合適的環球部門同 意。信貸及市場風險會於附屬公司層面 進行計量及報告，並會於集團整體匯總 以分析風險的集中度。 ‧ 內部審核。設立並維持合適的風險管理 和內部監控制度是業務管理層的主要 職責。環球審核部由集團集中管控，該 部門負責就集團的風險管理、監控及管 治流程架構，對整個集團而言其設計的 完密度及運作成效，提供獨立及客觀之 鑑證，並集中處理對滙豐構成最大風險 之範疇 （採用風險基準計算法釐定） 。集 團審核部主管向集團監察委員會主席 匯報，並向集團行政總裁作行政匯報。 ‧ 內部審核建議。行政管理層負責確保由 環球審核部提出之建議，會根據合適及 協定之時間表實行，並向環球審核部作 出有關確認。 ‧ 聲譽風險。董事會及屬下委員會、各附 屬公司董事會及屬下委員會，以及高層 管理人員負責制訂政策，指導各附屬公 司及集團各管理層在經營業務時維護 集團聲譽。聲譽風險可能有不同成因， 包括環境、社會及管治問題，或因營運 風險事件而引致，又或因僱員的行事方 式不符合滙豐的價值觀所致。滙豐的聲 譽乃建基於其經營業務之方式，由滙豐 提供金融服務之客戶，其經營業務之方 式或使用金融產品及服務之方式，亦有 可能影響滙豐的聲譽。 集團監察委員會及集團風險管理委員會的 角色 集團監察委員會代表董事會監督財務報告 的風險管理及內部監控工作，集團風險管 理委員會則負責監督財務報告以外的風險 管理及內部監控工作。 年內，集團風險管理委員會及集團監察委 員會定期檢討此內部監控制度之成效，並 定期向董事會匯報。該等委員會進行檢討 時省覽業務和營運風險的定期評估；定期 省覽集團風險管理總監及集團審核部主管 之匯報；每年省覽檢討滙豐控股的內部監 控架構之報告，當中涵蓋所有財務及非財 務之內部監控工作、由主要附屬公司的審 計及風險委員會每半年就其財務報表是否 已按照集團的政策編製、相關主要附屬公 司是否已按持續經營基準公平呈報事務狀 況向集團監察委員會作出之確認，以及確 認有否因內部監控不足而出現任何重大損 失、或有事故或不明朗狀況、審閱內部審 核報告及外聘核數師報告；審慎檢討經營 運作以及審閱監管機構的報告。集團風險 管理委員會監察首要及新浮現風險的相關 狀況，並考慮已實行的減低風險措施是否 適當。此外，如產生預期以外之虧損或發 生事故，而情況顯示監控架構或遵循集團 政策方面有偏差缺漏，集團風險管理委員 會及集團監察委員會則會審閱管理層所要 求編製的特別報告，分析問題的成因、所 汲取的教訓及管理層就處理有關問題建議 採取的行動。 內部監控成效 董事透過集團風險管理委員會及集團監察 委員會，每年檢討我們的風險管理及內部 監控制度之成效，此制度涵蓋各項重大監控 事宜，包括財務、營運及合規監控工作，以 及各種風險管理制度、會計及財務報告以 及風險管理部門的資源、職員資歷與經驗， 以及職員所接受的培訓課程及有關預算之 充足度。該檢討並未擴展至合資或聯營公 司。我們的風險管理制度及內部監控成效 的年度評估乃參考COSO委員會主要職能， 以指定實體水平控制作為證明。有關各實 體水平控制的成效報告和風險及監控的定 期匯報由若干主要管理委員會層面提升至 集團風險管理委員會及集團監察委員會層 面。 集團風險管理委員會及集團監察委員會已 取得確認，得知行政管理層已經或正就我 們的運作監控架構所識別之任何缺失或漏 洞採取所需補救行動。 持續經營 財務報表乃按持續經營基準編製，因董事 確信集團和母公司擁有足夠資源於可見將 來持續經營業務。 董事於考慮有關目前及未來情況 （包括對 日後盈利能力、現金流及資本來源的預測） 的廣泛資料後作出上述評估。 滙豐的主要業務、業務及經營模式、策略 方向及首要及新浮現風險，載於 「策略報 告」 一節；財務概要，包括綜合收益表及綜 合資產負債表的評述，載於 「財務回顧」 一 內部監控╱持續經營╱僱員 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 291 節；滙豐管理信貸、流動資金及市場風險 的目標、政策及程序，載於 「風險」 一節； 而滙豐管理及分配資本的方式，載於 「資 本」 一節。 僱員 於2014年12月31日，滙豐的全職和兼職僱員 總數為266,000人，2013年底及2012年底的 數字分別為263,000人及270,000人。僱員主 要集中在英國 （約48,000人） 、印度 （32,000 人） 、香港 （30,000人） 、巴西 （21,000人） 、 中國內地 （21,000人） 、墨西哥 （17,000人） 、 美國 （15,000人） 和法國 （9,000人） 。 在目前的環球金融服務營運環境下，表現 卓越及價值主導的員工乃成功關鍵。我們 提倡於制訂決策時坦誠開放地溝通。我們 會就僱傭事宜和影響滙豐業績的財務、經 濟、監管及競爭因素，定期與僱員溝通。 獎勵 我們的獎勵方法是提倡精英管理及市場競 爭力，並通過合乎道德及價值為本的績效 文化，使僱員、股東、監管機構及客戶的 利益趨於一致。 僱員關係 我們會在有需要時與認可的工會組織協商 並徵詢其意見。最多工會成員的五個地區 包括阿根廷、巴西、中國內地、馬耳他及墨 西哥。我們的政策是與所有代表僱員的團 體維持完善的溝通和諮詢。滙豐的業務運 作在過去五年並無因勞資糾紛而嚴重受阻。 多元及共融 滙豐致力建立一個以價值為主導的高績 效文化，而所有僱員和他們的意見均受重 視、尊重。我們一貫致力建立精英管理制 度，提供多元共融的工作環境，讓僱員明 白下情可以上達，所關切的問題可獲得處 理，工作環境不容偏袒、歧視和騷擾 （無 論性別、年齡、種族、宗教、性傾向和殘 疾） ，晉升乃基於客觀標準。我們的兼容並 包文化有助廣納多元化的客源，亦有利吸 納和挽留經驗豐富且負責任的僱員。我們 將透過聘用最優秀的人才及善用他們的意 見、才幹和多元化技能，弘揚我們的文化。 我們的多元共融計劃以及相關活動由集團 多元化委員會的管理人員監督，集團人才 委員會及地方人才╱多元化委員會亦會配 合有關工作。 僱員發展 在成熟市場及新興市場培育僱員發展，是 我們業務長遠實力的核心。我們有完善系 統，用以物色、發展及調派人才，以確保 當前及未來的高級管理職位由具有價值、 技能及經驗的適當人才擔任。 2014年，我們繼續為各個學習課程建立全 球一致的內容，並提高學習課程的實用性 與質素。我們著重領導才能、價值觀及技 術能力方面，並且力求制訂高水準課程。 僱用殘疾人士 我們堅持為所有僱員提供公平就業機會， 包括僱用殘疾人士，而殘疾人士之聘用、 培訓、事業發展及晉升，均根據個人才能而 定。若僱員在受聘於集團期間不幸成為殘 疾人士，我們會盡可能繼續聘用，如有需要 更會提供適當培訓及合理的設備及設施。 健康與安全 滙豐致力為我們的僱員、客戶及訪客提供 安全及健康的環境，並積極管理與業務有 關之健康與安全風險。我們的目標包括遵 守營運所在國家╱地區的健康與安全法律， 識別、消除、減低或控制重大的健康與安 全風險，降低火災、危險事故及僱員、客 戶和訪客發生意外的可能性。 滙豐的企業房地產部對健康與安全事務負 有全部責任，並已制訂環球健康與安全政 策及準則，供滙豐環球業務在其營運所在 地使用。有關國家╱地區的營運總監負責 執行該等政策及準則。 在人身風險及地緣政治風險方面，環球保 安及詐騙風險管理部定期進行保安風險評 估，協助管理層判斷恐怖襲擊及暴力犯罪 的威脅程度。區域保安及詐騙風險管理部 會對集團所有重要樓宇進行兩年一次的保 安檢討，確保有適當措施保障我們的職員、 樓宇、資產及資料免受威脅。 滙豐致力有效管理健康與安全事務，保障 滙豐的僱員、客戶及訪客。 董事會報告：企業管治 （續） 滙豐控股有限公司 292 僱員健康與安全 2014年 2013年 2012年 僱員工作場所意外致命個案宗數 2 1 － － 每100,000名僱員發生的導致缺勤 超過3天的意外宗數 96 101 58 每100,000名僱員的所有意外率 388 2 355 375 1 從事滙豐相關工作的非滙豐僱員。 2 反映呈報率上升。 薪酬政策 僱員的質素和投入程度是集團成功的基本 因素，故董事會銳意吸引、挽留和激勵最 優秀的人才。由於信任和關係是集團業務 的重要元素，故我們的目標是羅致有意在 滙豐長遠發展事業的人才。 為配合這個目標，滙豐的獎勵策略兼顧員 工的短期及持續表現。獎勵策略的目標是 論功行賞，表現不及格者不會得到獎勵， 並確保獎勵與我們的風險管理架構及相關 結果配合。為確保員工薪酬與集團的業務 策略保持一致，集團在釐定員工薪酬時， 將評估其表現能否達致表現評分紀錄中概 述的年度及長期目標，以及能否恪守滙豐 「坦誠開放、重視聯繫及穩妥可靠」 和 「勇 於正直行事」 的價值觀。整體而言，我們不 僅根據員工在短期及長期達致的成果評核 其表現，亦衡量達致成果的方式，因為後 者會影響機構的可持續發展能力。 年度及長期表現評分紀錄內的財務與非財 務表現衡量標準，乃經過深思熟慮後訂立， 以確保符合集團的長遠策略。 有關集團薪酬方案的詳情載於第300頁。 僱員股份計劃 滙豐根據滙豐股份計劃向僱員授出的認股 權及特別股份獎勵，使僱員利益與股東利 僱員╱其他披露╱股東周年大會 益保持一致。以下各頁列表載有尚未行使 認股權之詳情，包括根據 《香港僱傭條例》 界 定為 「持續合約」 之僱傭合約受聘之僱員持 有之認股權。授出認股權均不收取代價。 主要股東、貨品或服務供應商並無獲授認 股權，而授出之認股權亦無超越各項股份 計劃之個別上限。年內滙豐概無註銷認股 權。 2014年內就各計劃授出、行使或失效的認 股權總數概要載於以下列表。根據 《香港聯 合交易所有限公司證券上市規則》 第17章須 予披露的進一步詳情，可透過我們的網站 www.hsbc.com/investor-relations/governance/share-plans 及香港聯合交易所有限公司網站www.
hkex.com.hk查閱或向集團公司秘書長索 取，地址為8 Canada Square, London E14 5HQ。滙豐控股董事所持認股權的詳情載 於第321頁。 財務報表附註6載有以股份為基礎的支出詳情，包 括根據滙豐股份計劃授出特別股份獎勵。 全體僱員股份計劃 集團實行全體僱員優先認股計劃，而合資 格僱員已獲授認股權，以購入滙豐控股普 通股。僱員通常可於三年或五年後行使認 股權。認股權可在若干情況下 （如退休） 提 前行使，或因若干情況延遲行使，例如參與 者身故，則管理參與者遺產的遺囑執行人 可於正常行使期屆滿後六個月內行使有關 認股權。於2014年9月22日，即2014年授出 認股權之前一日，按倫敦證券交易所每日 正式牌價表的每股滙豐控股普通股的收市 中間價為6.58英鎊。集團再無根據滙豐控股 儲蓄優先認股計劃：國際授出認股權。一項 新的國際性全體僱員購股計劃已於2013年 第三季推出。除非董事議決提前終止全體僱 員優先認股計劃，否則有關計劃將於2015 年5月27日終止。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 293 滙豐控股全體僱員優先認股計劃 滙豐控股普通股 授出日期 行使價 行使期 於2014年 1月1日 年內 已授出 年內 已行使1 年內 已失效 於2014年 12月31日 由 至 由 至 由 至 儲蓄優先認股計劃 2008年 2014年 （英鎊） （英鎊） 2013年 2020年 4月30日 9月23日 3.3116 5.9397 8月1日 5月1日 53,950,886 28,688,703 25,097,425 3,798,209 53,743,955 儲蓄優先認股計劃：國際 2008年 2012年 （英鎊） （英鎊） 2013年 2018年 4月30日 4月24日 3.3116 5.9397 8月1日 2月1日 10,022,450 － 5,625,183 683,208 3,714,059 2008年 2012年 （美元） （美元） 2013年 2018年 4月30日 4月24日 4.8876 11.8824 8月1日 2月1日 3,997,069 － 1,528,838 600,903 1,867,328 2008年 2012年 （歐元） （歐元） 2013年 2018年 4月30日 4月24日 3.6361 7.5571 8月1日 2月1日 1,574,652 － 935,177 67,973 571,502 2008年 2012年 （港元） （港元） 2013年 2018年 4月30日 4月24日 37.8797 92.5881 8月1日 2月1日 24,215,341 － 17,206,998 539,561 6,468,782 1 最接近行使認股權日期前，股份之加權平均收市價為6.36英鎊。 特別優先認股計劃 自2005年9月30日以來，滙豐並無根據僱員股份計劃授出特別認股權。 滙豐控股普通股 授出日期 行使價 （英鎊） 行使期 於2014年 1月1日 年內 已行使 年內 已失效 於2014年 12月31日 由 至 由 至 由 至 滙豐控股集團優先認股計劃1, 2 2004年 2005年 2007年 2015年 4月30日 4月20日 7.2181 7.5379 4月30日 4月20日 55,025,868 1,434 48,650,452 6,373,982 滙豐股份計劃1 2005年 2008年 2015年 9月30日 7.9911 9月30日 9月30日 86,046 － － 86,046 1 滙豐控股集團優先認股計劃已於2005年5月26日屆滿，而滙豐股份計劃則於2011年5月27日屆滿，自該等 日期起再無根據該等計劃授出認股權。 2 最接近行使認股權日期前，股份之加權平均收市價為6.09英鎊。 其他披露 有關股本、董事權益、股息及股東，以及 僱員多元化的更多資料載於本節第294頁 附錄。 股東周年大會 第264至268頁載列的所有董事均出席2014年 舉行的股東周年大會，惟分別於2014年9月 1日及2015年1月1日獲委任為董事的苗凱婷 及安銘除外。 滙豐控股的2015年股東周年大會訂於2015年 4月24日 （星期五） 上午11時正假倫敦Queen Elizabeth II Conference Centre舉行，地址 為 Broad Sanctuary, Westminster, London SW1P 3EE。 滙豐控股將於2015年4月20日 （星期一） 下午 4時30分假香港皇后大道中1號舉行股東非 正式會議。 股東周年大會實況將於滙豐網站www.hsbc.
com即場播放。由股東周年大會結束後至 2015年5月22日期間，公眾人士可登入滙豐 網站www.hsbc.com瀏覽大會實況錄影。 代表董事會 滙豐控股有限公司 集團主席 范智廉 註冊編號617987 2015年2月23日 董事會報告：企業管治 （續） 滙豐控股有限公司 294 企業管治附錄－其他披露 股本 已發行股本 滙豐控股於2014年12月31日繳足股款的已發行股本面值為9,608,951,630美元，分為 19,217,874,260股每股面值0.50美元的普通股、1,450,000股每股面值0.01美元的非累積優先 股，以及一股面值0.01英鎊的非累積優先股。 每股面值0.5美元的普通股、每股面值0.01美元的非累積優先股及面值0.01英鎊的非累積優先 股，分別佔滙豐控股於2014年12月31日繳足股款的全部已發行股本面值約99.9998%、0.0002% 及0%。 股份附帶的權利及責任 滙豐控股股本中每類股份所附帶的權利及責任詳載於我們的組織章程細則，惟須受董事會 於配發有關優先股前釐定的每類優先股所附帶的若干權利及責任規限。下文為每類股份所 附帶並與投票、股息、資本及贖回 （如為優先股） 有關的權利及責任的概要。 於登記股份轉讓時，與轉讓文據有關的股份必須已繳足股款，且我們對其並無留置權，同 時轉讓應僅涉及同一面值貨幣的單一類股份。轉讓文據的受益人必須為一名單一承讓人或 不多於四名聯席承讓人，且必須妥為蓋上印花 （如有規定） 。該文據必須連同有關股票或其 他可證明轉讓人擁有權的憑證，送交我們的註冊辦事處或股份登記處。 倘股東或任何擬擁有我們股份權益的人士獲發給根據英國 《2006年公司法》 第793條 （該條款 賦予上市公司權利，向我們知悉或有合理理由相信擁有滙豐股份權益的任何人士索取資料） 規定發出的通告，但未能於通告列明的期限內就任何股份 （ 「違約股份」 ） 提供所需資料，除 非董事會另有決定，否則該股東將無權出席任何股東大會或在會上就違約股份投票或行使 股東獲授予的任何其他權利。倘違約股份佔該類股份已發行股數的面值不少於0.25%，除非 董事會另有決定，否則本公司將扣發任何股息、不計利息、不可作出任何以股代息選擇，以 及除於極少數情況外，該等股東所持有的任何股份將不獲辦理轉讓過戶登記。 普通股 除英國 《2006年公司法》 及組織章程細則另有規定外，滙豐控股可透過普通決議案向普通股 持有人宣派股息，但任何派發的股息不得超過董事會建議的金額。董事會若認為可供分派 的利潤足以派發中期股息，則可作出有關派息決定。所有股息均須按股息相關業績期內任 何一段或多段時間中繳足股款的面值金額比例分配及支付，惟倘於任何發行條款訂明股份 自特定日期起可獲派股息，則該股份可獲派相應股息。除組織章程細則另有規定外，董事 會可在股東通過普通決議案之前事先授權並根據董事會釐定的條款及條件規限下，賦予任 何普通股持有人權利，選擇收取相同或不同貨幣面值並入賬列作繳足股款的普通股，以代 替普通決議案所指定以任何貨幣支付的全部現金股息 （或部分股息，數額由董事會決定） 。 於2012年股東周年大會上，股東授權董事會可提供以股代息選擇，直至2017年股東周年大 會結束或2017年5月24日 （以較早者為準） 為止。 有關董事會就派付普通股各次股息所採納政策的詳情，請參閱第458頁。 優先股 滙豐控股股本中有三類優先股，每股面值0.01美元的非累積優先股 （ 「美元優先股」 ） 、每股面 值0.01英鎊的非累積優先股 （ 「英鎊優先股」 ） 及每股面值0.01歐元的非累積優先股 （ 「歐元優先 股」 ） 。已發行的美元優先股為A系列美元優先股，而已發行的英鎊優先股為A系列英鎊優先 股。並無已發行的歐元優先股。 美元優先股 美元優先股持有人僅於下列情況下方可出席股東大會並於會上投票：董事會在配發有關優先 股前釐定的有關期間仍未全數支付有關優先股的應付股息；或董事會在配發有關優先股前 企業管治附錄 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 295 指定的其他情況，惟須受董事會於配發有關美元優先股前釐定的有關條款規限。對於2015年 2月23日已發行的美元優先股，董事會釐定的有關期間為連續四個派發股息日。每當有關美 元優先股持有人有權於股東大會上就決議案投票時，每名親身或委任代表出席大會的有關 持有人於舉手表決時僅可投一票，而每名親身或委任代表出席大會的有關持有人於投票表 決時則可就所持優先股每股投一票，或根據董事會於配發有關股份前所釐定每股可投的票 數行使投票權。 根據組織章程細則，美元優先股持有人有權按董事會於配發該等優先股前釐定的有關比率 及其他有關條款及條件，於有關日期優先於普通股及滙豐控股已發行的任何其他類別股份 （但不包括(i)其他已發行優先股及就收益而言享有同等權益的任何其他股份；及(ii)依據其發 行條款，就收益而言，所享權益優先於有關優先股的任何股份） 的持有人獲派付任何股息之 前，獲派發非累積優先股股息。於2015年2月23日已發行每股美元優先股的應付年息為每年 62美元。董事會根據全權及絕對酌情權決定每季派付15.50美元股息。 美元優先股並無附帶分享滙豐控股利潤或資產的任何權利 （組織章程細則列明者及英國 《2006年 公司法》 另有規定者除外） ，優先股並未附帶參與發售、供股要約或以其他方式認購額外滙 豐控股股份的任何權利，亦未附帶任何換股權，或參與任何發行紅股或透過儲備資本化發 行股份的權利。 根據相關適用的清盤法例及滙豐控股的組織章程細則，滙豐控股清盤時，有關美元優先股 持有人有權先於普通股持有人及滙豐控股任何已發行的其他類別股份 （但不包括(i)其他相關 優先股及就償還資本而言享有同等權利的任何其他股份；及(ii)依據其發行條款，就償還資 本而言，享有優先於有關優先股的任何股份） 的持有人獲得支付任何款項前，從滙豐控股可 供分派予其股東的資產中，收取一筆款項，其金額相等於根據組織章程細則為應付股息但 未就有關美元優先股支付的任何股息，加上有關美元優先股的繳足股款或入賬列作繳足股 款，以及董事會於配發有關美元優先股前已釐定的有關溢價 （如有） 。就於2015年2月23日已 發行的美元優先股而言，溢價為每股美元優先股999.99美元。 根據組織章程細則及發行與配發有關美元優先股的條款可贖回美元優先股。就於2015年2月 23日已發行的美元優先股而言，滙豐控股可於2010年12月16日當日或之後任何時間全部贖 回有關股份，但須預先取得審慎監管局的同意。 英鎊優先股 英鎊優先股具有與美元優先股於組職章程細則所列者相同的權利及責任，惟有關擬由董事 會於配發相關優先股前釐定的英鎊優先股所附若干權利及責任，以及贖回各類別股份的時 間及所得款項的支付除外。於2015年2月23日已發行的每股英鎊優先股與於2015年2月23日已 發行的每股美元優先股在上節所述範圍內具有相同權利及責任，惟下述者除外： 1 英鎊優先股的持有人無權出席股東大會或於會上投票； 2 英鎊優先股可於董事會可能釐定的任何日期全部贖回；及 3 不宣派或支付股息的例外情況並不適用。 於2015年2月23日已發行每股英鎊優先股的應付股息為每年0.04英鎊。董事會根據全權及絕 對酌情權決定每季派付0.01英鎊股息。 歐元優先股 歐元優先股具有與美元優先股於組織章程細則所列者相同的權利及責任，惟有關擬由董事 會於配發相關優先股前釐定的歐元優先股所附若干權利及責任，以及贖回各類別股份的時 間及所得款項的支付除外。 董事會報告：企業管治 （續） 滙豐控股有限公司 296 2014年股本 年內曾發生下列與滙豐控股普通股股本相關之事項： 代息股份 已發行滙豐控股普通股 面值總額 每股市值 發行代息股份 於 股數 美元 美元 英鎊 2013年第四次股息 2014年4月30日184,047,509 92,023,755 9.9254 5.9788 2014年第一次股息 2014年7月10日 27,302,240 13,651,120 10.3980 6.1996 2014年第二次股息 2014年10月9日 34,787,645 17,393,823 10.6850 6.4478 2014年第三次股息 2014年12月10日 22,338,589 11,169,295 10.1178 6.2750 全體僱員股份計劃 股數 面值總額 行使價 美元 由 至 滙豐控股儲蓄優先認股計劃 以英鎊發行之滙豐普通股 30,722,608 15,361,304 英鎊 3.3116 5.9397 以港元發行之滙豐普通股 17,206,998 8,603,499 港元 37.8797 92.5881 以美元發行之滙豐普通股 1,528,838 764,419 美元 4.8876 11.8824 以歐元發行之滙豐普通股 935,177 467,589 歐元 3.6361 7.5571 已失效可認購滙豐普通股之認股權 5,689,854 2,844,927 回應英國滙豐僱員提出約24,000份申請而於 2014年9月23日授出的可認購滙豐普通股 之認股權 28,688,703 14,344,352 滙豐國際僱員購股計劃 6,470 3,235 英鎊 5.9290 6.5770 儲蓄計劃 為法國滙豐及其附屬公司非英國居民 僱員利益發行的滙豐普通股 1,763,449 881,725 歐元 6.7073 特別股份獎勵計劃 已發行 滙豐控股 普通股 面值總額 美元 行使價 已失效 認股權 由 （英鎊） 至 （英鎊） 根據以下計劃行使的認股權： 滙豐控股集團優先認股計劃 1,434 717 7.2181 7.5379 48,650,452 滙豐股份計劃 已發行滙豐 控股普通股 面值總額 美元 每股市值 由 （英鎊） 至 （英鎊） 根據滙豐股份計劃及2011年滙豐股份 計劃實際授出獎勵 67,226,264 33,613,132 5.9180 6.6040 授權配發及購買股份 於2014年股東周年大會上，股東重新授予董事配發新股的一般授權，配發新股數目不超過 12,576,146,960股普通股、15,000,000股每股面值0.01英鎊之非累積優先股、15,000,000股每 股面值0.01美元之非累積優先股及15,000,000股每股面值0.01歐元之非累積優先股。據此， 董事有權向現有股東以外的人士配發最多943,211,022股普通股以全數換取現金。股東亦重 新授權董事從市場購買不超過1,886,422,044股普通股。董事並無行使此授權。 此外，就滙豐控股或滙豐集團旗下成員公司在指定情況下發行可自動轉換為或兌換為滙豐 控股普通股的或有可轉換證券而言，股東授權董事授予認購或將任何證券轉換為不超過 4,500,000,000股普通股的權利。有關發行或有可轉換證券的詳情載於財務報表附註35。 除上表 「2014年股本」 所述者外，各董事在2014年內並無配發任何股份。 企業管治附錄 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 297 庫存股份 根據香港聯合交易所於2005年12月19日授出之豁免條款，滙豐控股將遵守有關持有任何庫存 股份的相關適用英國法規，並就其可能持有的任何庫存股份遵守豁免條件。根據英國 《2006年 公司法》 第6章，現時並無持有任何庫存股份。 董事權益 根據英國上市規則的規定及滙豐控股遵照香港 《證券及期貨條例》 第352條而保存之董事權益 登記冊所載，滙豐控股董事於2014年12月31日在滙豐控股及其聯營公司之股份及借貸資本 中擁有下列實益 （除非另有說明） 權益： 董事權益－股份及借貸資本 於2014年12月31日 於2014年 配偶或18歲 與另一位 1月1日 實益擁有人 以下之子女 人士共同擁有 受託人 權益總計1 滙豐控股普通股 凱芝3 － 20,045 － － － 20,045 埃文斯勳爵 1,495 5,519 － － － 5,519 費卓成 10,605 24,105 － － － 24,105 方安蘭 21,858 － 76,524 － 76,524 范智廉 392,664 400,748 － － － 400,748 歐智華 2,730,477 2,434,303 176,885 － － 2,611,188 李德麟 35,123 35,352 － － 1,4162 36,768 利普斯基3 15,525 15,820 － － － 15,820 駱美思 － 15,500 － － － 15,500 麥榮恩 65,130 79,933 － － － 79,933 苗凱婷3 － 3,575 － － － 3,575 繆思成 400,753 480,423 － － － 480,423 駱耀文爵士 9,912 22,981 － － － 22,981 施俊仁 － 15,940 4,613 － － 20,553 英國滙豐銀行2015年2.875厘票據 人民幣百萬元 人民幣百萬元 人民幣百萬元 人民幣百萬元 人民幣百萬元 人民幣百萬元 費卓成4 5.1 － － － － 5.1 1 有關執行董事因滙豐控股儲蓄優先認股計劃、滙豐股份計劃及2011年滙豐股份計劃而持有之滙豐控股普 通股的其他權益，載於第320頁的 「董事薪酬報告」 的計劃權益內。於2014年12月31日，根據香港 《證券及 期貨條例》 界定，滙豐控股普通股 （包括來自僱員股份計劃之權益） 的權益總額為：范智廉－405,683股；歐 智華－5,175,003股；繆思成－1,775,461股；及麥榮恩－1,086,284股。各董事的權益總計佔已發行股份不 足0.03%。 2 非實益擁有。 3 凱芝、利普斯基及苗凱婷分別於4,009股、3,164股及715股上市美國預託股份 （ 「ADS」 ） （該等股份根據香港 《證券及期貨條例》 第XV部被歸類為股權衍生工具） 中擁有的權益。每股美國預託股份相當於五股滙豐控 股普通股。 4 於人民幣120萬元的2015年2.875厘票據中擁有的非實益權益。 並無任何董事於滙豐控股及其聯營公司的股份及借貸資本中，持有香港 《證券及期貨條例》 所界定的任何短倉。除上文所述者外，各董事於年初或年底概無擁有滙豐控股或其任何聯 營公司任何股份或債券的權益，而年內各董事或其直系親屬亦無獲授或行使任何可認購滙 豐旗下任何公司股份或債券的權利。 自年底以來，下列董事增持的滙豐控股普通股數目如下： 滙豐控股普通股 范智廉 （實益擁有人） 251 1 透過每月定期供款購入於滙豐控股英國股份獎勵計劃下的股份。 由2014年12月31日至本報告日期止，董事於股份及借貸資本的權益並無其他變動。其後截 至 《股東周年大會通告》 刊發前的最後實際可行日期止的任何變動，將載於該通告的附註。 於2014年12月31日，非執行董事及高層管理人員 （即滙豐控股的執行董事及集團常務總監） ， 合共實益持有17,531,530股滙豐控股普通股之權益 （佔已發行普通股0.09%） 。 董事會報告：企業管治 （續） 滙豐控股有限公司 298 於2014年12月31日，執行董事及高層管理人員持有根據滙豐控股儲蓄優先認股計劃及滙豐 控股集團優先認股計劃授出之認股權，合共可認購28,288股滙豐控股普通股。此等認股權可 於2015至2020年間按每股普通股4.4621英鎊至5.1887英鎊之價格行使。 股息及股東 2014年股息 本公司已分別於2014年7月10日、2014年10月9日及2014年12月10日派發2014年第一、第二及 第三次股息，每次均派發每股普通股0.1美元。財務報表附註9載有關於2014年宣派股息的更 多資料。於2015年2月23日，董事會宣布派發2014年度第四次股息每股普通股0.2美元，以代 替末期股息，並將於2015年4月30日以美元、英鎊或港元現金派發，並按2015年4月20日釐定 的匯率換算，股東亦可選擇以股代息。由於2014年第四次股息於2014年12月31日後宣派，故 並未作為應付賬項計入滙豐資產負債表。2014年12月31日的可供分派儲備為488.83億美元。 6.2厘非累積A系列美元優先股 （ 「A系列美元優先股」 ） 的季度股息為每股15.5美元，相等於每 股A系列美國預託股份 （每股代表四十分之一股A系列美元優先股） 派發股息0.3875美元。此 股息已於2014年3月17日、6月16日、9月15日及12月15日分派。 2015年股息 集團已於2015年2月9日宣派A系列美元優先股季度股息每股15.5美元，相等於每股A系列美國 預託股份 （每股代表四十分之一股A系列美元優先股） 派發股息0.3875美元，並宣派A系列英 鎊優先股季度股息每股0.01英鎊。此等股息訂於2015年3月16日分派。 與股東之溝通 我們十分重視與股東溝通。董事會已採納一項股東溝通政策，詳情可於我們的網站www.hsbc.
com查閱。有關集團業務的廣泛資料於 《年報及賬目》 、 《策略報告》 及 《中期業績報告》 內向股東 提供，並可於我們的網站www.hsbc.com查閱。我們與機構投資者定期會談，亦歡迎個別人士 隨時查詢有關所持股份事宜，以及我們的業務狀況。全部查詢均會盡快處理，同時附以詳盡 資料。集團更歡迎所有股東出席股東周年大會或在香港舉行之股東非正式會議，討論我們業 務之進展。股東可將對董事會的書面查詢發送予集團公司秘書長，地址為HSBC Holdings plc, 8 Canada Square, London E14 5HQ，或將相關電子郵件發送至shareholderquestions@hsbc.
com。 除英國 《2006年公司法》 規定的股東周年大會之外，股東亦可要求董事召開股東大會。召開股 東大會的要求可由佔附有可於滙豐控股股東大會上投票權利之本公司已繳足股本至少5%的 股東提出 （不包括作為庫存股份持有的任何已繳足股本） 。相關要求須說明將於會上處理的 事務的大致性質，並可收錄於會上適當提議及擬將提議的決議案內容。除非決議案 （倘獲通 過） 屬無效 （因不符合任何條例或本公司組織章程文件或其他原因） ；對任何人士構成誹謗； 或屬無關重要或令人困擾，否則可於會上就決議案適當提出動議。相關要求可以印刷本形 式或電子形式提出，並須經提出要求的人士予以核證。相關要求可以書面形式向上一段所 述郵寄地址提出，或將相關電子郵件發送至shareholderquestions@hsbc.com。在應相關要求 召開的任何會議上，概不得處理要求所提述或董事會提議以外的任何事項。 須予公布之股本權益 於2014年12月31日，根據英國金融業操守監管局 《披露規則及透明度規則》 第5條，我們收到 附投票權之主要股權披露資料 （且其後並無更改） 如下： ‧ Legal & General Group Plc於2013年7月10日發出通知，表示該公司於2013年7月9日擁有滙 豐控股普通股之權益低於當日總投票權之3%；及 ‧ BlackRock, Inc.於2009年12月9日發出通知，表示該公司於2009年12月7日擁有以下各項︰ 1,142,439,457股滙豐控股普通股之間接權益，倘獲行使或轉換即可獲得705,100份投票權 的合資格金融工具；及經濟效用類近合資格金融工具，並代表234,880份投票權的金融工 具。該三類權益分別佔當日總投票權之6.56%、0.0041%及0.0013%。 於2014年12月31日，滙豐控股根據香港 《證券及期貨條例》 第336條保存之登記冊顯示： 企業管治附錄 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 299 ‧ JPMorgan Chase & Co.於2014年11月21日發出通知，表示該公司於2014年11月18日擁有 以下滙豐控股普通股之權益：937,591,714股長倉；99,085,113股短倉；及527,117,024股 借貸組合，分別佔當日已發行普通股之4.88%、0.51%及2.74%；及 ‧ BlackRock, Inc.於2014年10月28日發出通知，表示該公司於2014年10月24日擁有以下滙豐控 股普通股之權益：1,238,135,870股長倉及4,572,291股短倉，分別佔當日已發行普通股之 6.45%及0.02%。 遵照 《香港聯合交易所有限公司證券上市規則》 的規定，於2014年內及直至本報告日期的任 何時間內，滙豐控股全部已發行股本中至少有25%由公眾持有。 買賣滙豐控股上市證券 除英國滙豐銀行及香港上海滙豐銀行 （均為歐洲經濟區一間交易所的成員） 以中介機構身分 進行交易外，於截至2014年12月31日止年度內，滙豐控股或其任何附屬公司均無買入、賣出 或贖回其於香港聯合交易所有限公司上市的任何證券。 董事薪酬報告 年度報告 滙豐控股有限公司 300 董事薪酬報告 頁次 附錄 1 集團薪酬委員會主席年度報告 300 2014年薪酬策略及主要決定 300 有關董事薪酬的重大決定 302 未來監管改革 302 董事薪酬政策 303 向下計酬政策 304 與普遍應用於僱員的政策之差異 304 制訂酬勞政策時計及的重大因素 305 調整、扣減及撤回 306 薪酬政策－非執行董事 306 服務合約 306 其他公司董事席位 307 年度薪酬報告 307 薪酬委員會 307 集團浮動酬勞資金 309 薪酬的單一數字 311 薪酬情況及結果 313 集團業績表現股份計劃下之獎勵 314 確定執行董事的年度績效表現 315 退休金權益總額 318 對前董事的付款 318 年內的退任補償 318 2014年授出的計劃權益 318 績效表現概要 319 行政總裁薪酬 319 董事於股份的權益 320 股東背景 322 於2015年執行薪酬政策的情況 322 周年花紅評分紀錄 323 額外資料披露 324 僱員報酬及福利 324 高層管理人員的酬金 324 五位最高薪僱員的酬金 325 八位最高薪高級行政人員的薪酬 325 第三支柱薪酬披露 326 1 董事薪酬報告附錄。 集團薪酬委員會主席年度 報告 各位股東： 本人欣然呈報2014年薪酬報告。於本報告 內，我們會提供滙豐薪酬政策的詳情、我 們於2014年支付予董事的薪酬及理由。 今年乃我們的薪酬政策 （在去年的股東周年 大會上獲批准） 得以落實之首個年度。本人 希望本報告可使閣下理解集團薪酬委員會 （ 「委員會」 ） 於2014年如何落實政策及 （更重 要的是） 行政人員的表現及獲發薪酬與本公 司股東長期利益的相互關聯。 報告分為三節：本人作為委員會主席致閣 下的函件、我們的薪酬政策概要，以及我 們於截至2014年12月31日止年度向董事支 付薪酬的年度報告。其他薪酬相關資料披 露則載於本報告附錄。 2014年薪酬策略及主要決定 我們的薪酬策略力求就長期良好表現提供 具競爭力的報酬，並吸引和激勵有志在集 團持續長遠發展事業並為股東爭取長遠利 益的頂尖人才。 委員會認為支付予人才的薪酬必須與我們 的業務策略相互配合。業績表現不應僅僅基 於所取得的短期及長期成就而進行判斷， 亦須以取得相關成就的方法為依據，因為 我們相信後者會影響業務的長遠可持續發 展能力。 於2014年，當局根據歐盟 《資本規定指引4》 （ 「資本指引4」 ） 引入新的監管規定。英國審 慎監管局薪酬規則的後續變動，已影響我 們向高級行政人員及被審慎監管局識別為 對機構風險狀況產生重大影響的僱員 （即 被界定為承受重大風險人員的人士） 支付 薪酬的方法。 自2014年起，資本指引4引入浮動酬勞上 限，規定歐盟的銀行 （包括滙豐） 將承受重 大風險人員的浮動酬勞獎勵 （如獲股東批 准） 限定最高為固定酬勞的200%。我們已 在去年的股東周年大會上尋求該項權限並 獲股東授予該權限。 要確保我們全球所有業務所在市場的僱員 之整體薪酬福利保持競爭力，資本指引4的 規定為滙豐帶來挑戰，尤其是相對於毋須 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 301 遵守該等規定的其他銀行而言。 因此，為重新平衡執行董事及其他承受重 大風險人員整體薪酬的固定及浮動成分， 我們為彼等推出固定酬勞津貼。委員會認 為，為確保僱員的整體薪酬福利保持競爭 力，推出固定酬勞津貼作為薪酬的一部分 實屬必要。滙豐必須在主要非歐盟市場繼 續挽留及吸引人才，而在該等市場，我們 的國際同業及本土競爭對手毋須遵守資本 指引4酬勞上限。按照資本指引4的規定， 固定酬勞津貼不會與達成任何表現條件掛 鈎，而我們已遵守監管機構頒布的現行指 引。此項給予本公司執行董事及高級行政 人員的津貼為附帶禁售期的股份，目的是 使之緊密配合本公司股東的長遠利益。 於2014年7月，英國審慎監管局引入新的公 司規定，以確保於2015年1月1日或其後授出 的浮動酬勞獎勵，可由授出日期起計至少 七年期間予以撤回 （即公司收回已付及╱或 已實際授出獎勵的能力） 。該項規定增強了 公司在若干情況下實施扣減的能力 （即在 實際授出獎勵前減少或取消未實際授出獎 勵） 。 為遵守英國審慎監管局的新規定，委員會 已制訂一項撤回政策，此項政策將適用於 2015年1月1日或之後授予承受重大風險人 員的所有獎勵。有關可應用 「扣減」 及 「撤 回」 的情況，於本報告下文會提供更詳盡 的資料。 委員會亦已採納一項政策，當其相信集團 在制訂有效的反洗錢及制裁法律合規計劃 方面的年度進展未符理想時，可酌情減少 執行董事及其他高級行政人員的浮動酬勞 獎勵。 於2014年，多宗既有事件產生法律及監管成 本，包括因對外匯市場內若干行為進行調 查而施加的罰則，均已在委員會用以釐定 獎勵金額的利潤水平中全面反映，致使獎 勵金額作出6億美元之調整。此外，我們亦 採取若干行動，包括運用酌情權將2014年 建議給予集團僱員 （包括高級管理層成員） 的浮動酬勞減少2,200萬美元。其他詳情載 於本報告下文。 整體表現概要╱業務背景 滙豐控股有限公司 ‧ 於2014年，集團維持穩健的資產負債狀況及雄厚的資本實力。若不計及貨幣換算的影響，貸款增加280億美 元，而客戶賬項則增加470億美元，客戶貸款對客戶賬項比率為72%。 ‧ 列賬基準之除稅前利潤比2013年有所減少，主要反映於2014年出售業務利潤及重新分類增益下降，以及其他 重大項目的影響，包括就罰款、和解開支及英國客戶賠償提撥的準備37億美元。按經調整基準計算，即撇除 重大項目及貸幣換算的影響，除稅前利潤與2013年大致相若。 ‧ 五個地區當中有三個的經調整除稅前利潤有所增長。 ‧ 工商金融業務於2014年錄得破紀錄的利潤。 ‧ 按經調整基準計算的收入與2013年大致相若，反映工商金融業務收入增長被環球銀行及資本市場業務收入減 少所抵銷，同時零售銀行及財富管理業務以及環球私人銀行業務因業務進行重組而致收入減少。 ‧ 集團於2014年之淨利息收益率保持穩定。 ‧ 貸款減值準備減少，反映自2011年以來集團對業務組合作出的改變。 ‧ 列賬基準成本效益比率由2013年之59.6%升至2014年之67.3%，而按經調整基準計算，則由57.7%升至 2014年之61.1%，主要反映監管計劃及合規成本上漲、通脹壓力、持續投資於策略方案以及銀行徵費上升， 令營業支出增加。該等因素部分被年內可持續成本節約13億美元所抵銷。 ‧ 平均普通股股東權益回報率為7.3%，較2013年之9.2%有所下降，主要反映出售利潤及重新分類增益減少， 以及營業支出增加 （包括就罰款、和解開支及英國客戶賠償提撥的準備） 。 ‧ 2014年度的股息由2013年每股普通股0.49美元增至每股普通股0.5美元。 ‧ 隨著資本指引4過渡基準普通股權一級比率由2013年之10.8%升至10.9%，我們的資本實力於2014年有所增 強。 零售銀行及財富管理業務 ‧ 列賬基準之除稅前利潤有所減少，主要由於美國消 費及按揭貸款組合持續縮減令收入減少，以及主要 零售銀行及財富管理業務的營業支出增加。 工商金融業務 ‧ 工商金融業務的除稅前利潤增長，反映我們於香港及 英國兩個本位市場的收入表現增強，以及貸款減值及 其他信貸風險準備減少 （主要為歐洲及拉丁美洲） 。 環球銀行及資本市場業務 ‧ 環球銀行及資本市場業務之除稅前利潤減少，主要 反映重大項目增加，特別是與外匯調查有關的和解 開支及準備，同時收入減少亦部分反映引入資金公 允值調整之後所作調整及外匯交易業務收入減少。 環球私人銀行業務 ‧ 經調整基準之除稅前利潤減少，主要反映業務繼續重 新定位導致客戶資產有序縮減。儘管客戶資產有所減 少，我們仍在目標增長範疇錄得正數新增資金淨額 140億美元。 董事薪酬報告 （續） 滙豐控股有限公司 302 年度報告╱董事薪酬政策 有關董事薪酬的重大決定 集團風險管理總監繆思成獲委任為執行董 事，自2014年1月1日起生效，反映風險管理 部對滙豐十分重要，以及繆思成具領導風 險管理部的才能及其個人對集團的貢獻。 因此，彼之薪酬符合執行董事的薪酬政策。 於諮詢股東意見之後，集團主席范智廉在 相關政策下符合資格根據集團業績表現股 份計劃收取一次性獎勵。委員會隨後決定， 不會授予集團主席2014年集團業績表現股 份計劃獎勵，而且決定檢討集團主席的基 本薪金，作為向股東提議改變日後政策的 其中一個方案。 委員會已決定於2015年不會增加執行董事 的基本薪金。鑑於本公司部分股東提出的 意見，委員會將檢討執行董事的現金退休 金津貼水平，作為改變日後政策的其中一 個方案。 委員會並非僅憑應用評分紀錄所示比重計 算酬勞，而是運用了酌情權減少執行董事 的整體浮動酬勞。作出是項調整的理據是 考慮整體財務業績以及影響集團的法律、 合規及監管問題 （尤其是與過往事件有關 的問題，包括但不限於外匯問題） 。進一步 詳情載於本報告內。 未來監管改革 展望2015至2016年，鑑於近期英國審慎監 管局及英國金融業操守監管局就 「加強風 險與回報的一致性：新訂薪酬規則」 進行 諮詢，預期行政人員薪酬將面對進一步的 重大監管改革。此外，預期歐洲銀行管理 局將頒布薪酬指引以進行諮詢，當中包括 津貼可視作固定薪酬的準則。 委員會認為，已經建議及即將建議的薪酬 監管改革已太多及規模太大，且令我們難 以向現有及有機會加入集團的僱員明確解 釋適用於他們的薪酬政策及結構。監管方 面的不明朗因素及複雜情況將使大家誤解 我們的薪酬政策運作方式，以及該等政策 對僱員表現的影響。 委員會將考慮這些不同改革的影響及股東 對集團政策的意見。鑑於這些因素，我們 可能需要變更2016年的薪酬政策。 集團薪酬委員會主席 駱耀文爵士 2015年2月23日 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 303 董事薪酬政策 以下章節載列滙豐執行及非執行董事的薪 酬政策概要，有關政策已於2014年5月23 日的股東周年大會上獲股東批准。政策全 文可於 《2013年報及賬目》 內去年的董事薪 酬報告內查閱，亦可登入網站
hsbc.com/investor-relations/financial-and-regulatory-reports下載。 滙豐全體僱員的質素和長期投入程度是成 功的基本要素。因此，我們銳意吸引、挽 留和激勵有志在集團持續長遠發展事業的 頂尖人才，而這些人才將履行職責，為股 東爭取長遠利益。 集團薪酬政策的主要部分，即固定酬勞、 福利及浮動酬勞 （包括周年獎勵及集團業 績表現股份計劃） 列示如下。上述各部分 兼顧獎勵員工短期及長期的良好表現，並 在兩者之間取得平衡，有助實現集團的策 略目標。我們的策略力求論功行賞，要求 僱員薪酬與風險管理架構及風險結果適當 配合。對於我們最高級的僱員，彼等大部 分獎勵均會遞延發放，因而受扣減原則規 限，即是如有事實根據，則未實際授出的 獎勵可以削減或取消。此外，誠如主席報 告所載，自2015年1月1日起授出的浮動酬 勞獎勵可予撤回。 薪酬政策－執行董事 固定酬勞 基本薪金 固定酬勞津貼 退休金 福利 福利 ‧ 每月以現金支付。 ‧ 每年與相關對比組合 對照。 ‧ 於本政策實施期間， 各執行董事的基本薪 金增額，不得高於 2013年董事薪酬報告 所載基本薪金水平之 15%。 ‧ 授予即時實際授出的 股份。 ‧ 發放的股份附帶禁售 期 （20%於緊接財政年 度結束後的3月解除 禁售，80%於首個解 除禁售日期起計的五 年後解除禁售） 。 ‧ 反映董事職務、技能 及經驗，並維持具競 爭力的薪酬福利總 額，以挽留主要人才。 ‧ 不會受扣減或撤回原 則規限。 ‧ 代替退休金的現金津 貼 （最高為基本薪金之 50%） 。 ‧ 透過保持市場競爭 力，吸引及挽留主要 人才。 ‧ 計及所屬市場慣例，並 包括提供醫療保險、收 入保障保險、健康評 估、人壽保險、會所會 籍、稅務協助、使用公 司汽車 （包括有關福利產 生的任何稅項） 及旅遊支 援。 ‧ 歐智華亦獲提供在香港 的住宿及汽車福利。該 項福利應繳納的任何稅 項由滙豐承擔。 ‧ 當行政人員因業務需要 而調派或花費大量時間 在一個以上的司法管轄 區，亦可能獲提供其他 福利。 浮動酬勞 目的及與策略的聯繫 周年獎勵 集團業績表現股份計劃 ‧ 推動及獎賞與策略一致的 表現及符合股東利益。 ‧ 恪守滙豐價值觀乃釐定能 否收取任何浮動酬勞的先 決條件。 ‧ 遞延安排鼓勵長期投入工 作及讓集團可以應用扣減 原則。 ‧ 表現目標乃計及經濟環 境、集團的優先策略及承 受風險水平而設定。 ‧ 最高為固定酬勞的200%。 ‧ 以現金及股份形式發放。最少50% 之獎勵將以股份形式發放。 ‧ 根據周年評分紀錄 （評分基礎包括就 財務和非財務衡量指標設定目標） 而 衡量。評分紀錄因人而異。 ‧ 獎勵總額最少有60%將會遞延，並 於三年期間或委員會釐定的其他期間 內實際授出。 ‧ 最高數額為固定酬勞之67% （即固定 酬勞200%的浮動酬勞上限之三分之 一） 。 ‧ 倘委員會認為不能反映公司的整體狀 況及業績表現，委員會可酌情變更獎 勵。 ‧ 以股份形式發放。 ‧ 獎勵水平乃按照長期表現評分紀錄載 列的持久表現衡量指標，經考慮截至 財政年度結束時的表現而釐定。 ‧ 獎勵於五年期間後實際授出。於實際 授出時，股份於參與者受僱期內必須 一直保留。 ‧ 對於不被視為正常離職的人員，未實 際授出的獎勵將被取消，已實際授出 的股份將於終止受僱日期之每一周年 日或該日前後平均分為三批解除禁 售。 ‧ 最高數額為固定酬勞之133% （即固 定酬勞200%的浮動酬勞上限的三分 之二） 。 ‧ 倘委員會認為不能反映公司的整體狀 況及業績表現，委員會可酌情變更獎 勵。 董事薪酬報告 （續） 滙豐控股有限公司 304 下圖概列根據上述政策集團行政總裁的目 1 源自集團業績表現股份計劃獎勵的股份於參與者受僱期間須予保留不得出售。倘屬被視作正常離職的人 員，其已獲實際授出股份適用的禁售期將於終止受僱時結束。倘屬不被視作正常離職的人員，其已獲實 際授出股份將於終止受僱日期之每一周年日或該日前後平均分為三批解除禁售。 委員會將於2015年對執行董事應用上述政 策。倘監管機構規定於此項政策之外變更 任何固定或浮動薪酬條款，委員會將作出 必要變更以確保遵守監管規定。 向下計酬政策 根據獨立監察員提出的建議，委員會於2014年 引進向下計酬政策，規定在哪些情況下將 對執行董事及其他高級行政人員的任何浮 動酬勞釐定方法作出向下調整。 根據此項政策，用於釐定向下調整的標準 包括： ‧ 在開展有效的反洗錢及制裁法律合規 計劃方面，未取得足夠的年度進展；或 ‧ 未遵守美國延後起訴協議及其他有關 法令。 於決定是否應用任何向下計酬方案及向下 計酬的程度以減少浮動酬勞獎勵時，委員 會將考慮金融系統風險防護委員會的建議。 與普遍應用於僱員的政策之差異 授予僱員的固定及浮動酬勞結合方式與其 職務及經驗以及所屬市場因素相符。 固定酬勞津貼乃授予承受重大風險人員或 按歐盟監管技術準則604/2014所載的定質及 定量標準被識別為對機構風險狀況有重大 影響的個人。倘認為調整其他人士薪酬的 固定及浮動酬勞成分所佔比重乃屬適當， 則亦可授予該等人士固定酬勞津貼。 用於釐定固定酬勞津貼的標準包括所承擔 職務、技能、經驗、技術專長、市場上同類 職務的報酬及僱員於年內可收取的其他薪 酬。 集團常務總監及集團總經理將收取股份形 式之固定酬勞津貼，此等股份的發放條件 與執行董事相同。當津貼低於指定限額時， 所有其他僱員將收取現金固定酬勞津貼。 當固定酬勞津貼高於指定限額時，全部酬 勞將以即時實際授出股份的形式發放。任 何作為固定酬勞津貼一部分而發放的股份 （扣除為繳納任何所得稅及社會保障供款 董事薪酬政策 標表現報酬總額發放情況。 目標報酬總額發放情況 基本薪金 退休金 固定酬勞津貼 周年獎勵 固定酬勞津貼 集團業績表現 股份計劃1 2020年或退休時 （以較後者為準） 2014年 2015年 2016年 2017年 2018年 2019年 2020年 26% 38% 41% 44% 47% 66% 100% 26% 12% 3% 3% 3% 34% 19% 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 305 薪酬元素 執行董事 集團 常務總監 集團總經理 其他承受 重大風險人員 其他僱員 基本薪金 ✓ ✓ ✓ ✓ ✓ 固定酬勞津貼 ✓ ✓ ✓ ✓ ✓ 周年獎勵 ✓ ✓ ✓ ✓ ✓ 集團業績表現股份計劃╱長期獎勵 ✓ ✓ ✓ － － 福利及退休金 ✓ ✓ ✓ ✓ ✓ 制訂酬勞政策時計及的重大因素 委員會於釐定董事的薪酬政策時計及多項因素。 內部因素 資金 ‧ 周年獎勵及集團業績表現股份計劃獎勵的資金來自同一項周年浮動酬勞資金。 ‧ 我們根據集團的盈利能力、資本實力、股東回報、資本、股息與浮動酬勞之間 的利潤分配、承受風險水平聲明、市場競爭力及整體負擔能力，為周年浮動酬 勞資金撥資。 ‧ 本年度浮動酬勞資金的計算詳情請參閱第309頁。 集團內的酬勞和 僱用條件 ‧ 滙豐於釐定其執行董事的薪酬水平時將考慮整個集團的酬勞水平。委員會會比 較各僱員及高級行政人員的酬勞和僱用條件，從而考慮個別人員的獎勵。 ‧ 委員會會邀請集團員工表現及獎勵事務主管就廣大僱員的薪酬提呈建議，並就 薪酬中各種不同元素適用於其他僱員的範圍徵詢意見。 ‧ 於釐定集團薪酬政策時，會顧及僱員投入度調查及滙豐交流會時聽取的意見。 外部因素 監管 ‧ 全球各地規管薪酬架構的本土法規仍有重大分歧，這為業務遍及全球的滙豐帶 來了重大挑戰。 ‧ 為了實現長期的持續良好表現，必須制訂具市場競爭力及於不同地域之間大致 相若的薪酬，以吸引、激勵及挽留世界各地的人才和勤勉盡責的僱員。 ‧ 我們力求薪酬政策與監管措施及股東利益保持一致。 ‧ 滙豐在薪酬方面，已完全遵守於本報告日期適用的金融穩定委員會、英國金融 業操守監管局、審慎監管局及香港金管局指引和規則。 對比組合 ‧ 委員會就執行董事的薪酬福利參考指定薪酬對比組合的市場數據。 ‧ 此組合由十家環球金融服務機構組成，包括：Australia and New Zealand Banking Group Limited （ 「ANZ」 ） 、西班牙國家銀行、美國銀行、巴克萊集團、法國巴黎 銀行集團、花旗集團、德意志銀行、摩根大通公司、渣打集團及瑞士銀行。選 擇此等公司是基於其業務範疇、規模及跨國地域覆蓋與集團大致相若，委員會 並會每年檢討此等公司是否仍可繼續與集團相對比。於組合內加入ANZ乃委員 會於2014年檢討所得結果之一。 ‧ 於釐定薪酬政策時，委員會亦可參考其他公司 （倘相關） 的政策。 股東意見 ‧ 委員會主席、集團員工表現及獎勵事務主管及集團公司秘書長會與主要機構股 東及其他代表機構會面。我們認為此類會議至關重要，因為此類會議可讓我們 就現時及制訂中的薪酬慣例徵集意見，確保集團的獎勵策略繼續與股東的長遠 利益保持一致。 而出售的股份） 將附帶禁售期。該等股份之 40%於授出股份之相關財政年度結束後的 3月解除禁售。其餘60%於首次解除禁售後 每個周年日或該日前後平均分為三批解除 禁售。 集團常務總監同時參與周年獎勵及集團業 績表現股份計劃。集團總經理參與周年獎 勵，並可能收取其他長期獎勵。集團內其 他僱員則合資格參與周年獎勵安排。 董事薪酬報告 （續） 滙豐控股有限公司 306 調整、扣減及撤回 為獎勵表現出色的員工，委員會就風險因 素作出調整後，按財務及非財務表現來釐 定個人獎勵。委員會擁有專屬酌情權可應 用其已採納的扣減及撤回政策，藉此，委 員會經考慮個別員工與有關事件是否有密 切關係及對事件負有的責任後可採取下列 行動。在切實可行的情況下，本年度浮動 酬勞將予調整，其後方會實行扣減，之後 可實行撤回。 此項政策符合審慎監管局的監管規定。 行動類別 受影響的浮動酬勞 獎勵類別 可能適用的情況 （包括但不限於） ： 調整 本年度浮動酬勞 ‧ 不利行為或影響業務名譽的行為，例如有關2014年對外匯市 場內若干行為的調查。 ‧ 涉及造成重大營運虧損的集團層面事件，包括已經或有可能對 滙豐造成嚴重損害的事件。 ‧ 不符合滙豐價值觀及其他強制性規定。 ‧ 就特定個人而言，開展實際的反洗錢及制裁法律合規工作未取 得足夠年度進展或未遵守延後起訴協議及其他有關命令。 扣減 於過往年度提供但未實際 授出的遞延獎勵 ‧ 不利行為或影響業務名譽的行為。 ‧ 過往表現遠遜於最初匯報的水平。 ‧ 任何財務報表重列、更正或修改。 ‧ 風險管理不當或不足。 撤回1 已實際授出或已支付的獎 勵 ‧ 參與對滙豐造成重大損失的行為或須對此類行為負責。 ‧ 未能符合適當的合適及妥當標準。 ‧ 具充分理據將可或應可即時終止僱傭合約的有關行為失當或重 大失誤的合理證據。 ‧ 經參考集團風險管理標準、政策及程序，滙豐或業務單位在風 險管理方面重大失效。 1 撤回僅對於2015年1月1日或之後授出予承受重大風險人員的浮動酬勞獎勵適用，包括但不限於就2014年 業績計算年度發放的獎勵。 薪酬政策－非執行董事 非執行董事並非僱員而就其擔任董事收取 袍金。此外，非執行董事就彼等在履行職 責及繳納任何相關稅項時產生的開支獲得 補償，屬常見慣例。彼等不可收取基本薪 金、固定酬勞津貼、福利、退休金或任何 浮動酬勞。 應付袍金水平反映滙豐控股非執行董事須 投入的時間及承擔的職責。有關袍金乃參 考富時30指數的其他英國公司及銀行以及 其他非英國國際銀行支付的袍金釐定。 董事會定期檢討，以參照非執行董事在角 色、職責及╱或時間投入方面的變化，個 別及綜合評估袍金各部分是否仍具競爭力 及適宜，並確保挽留或可委任擁有適當能 力的人士。董事會 （不包括非執行董事） 可 酌情批准袍金的變動 （在非執行董事薪酬 政策所列明的範圍內） 。董事會亦可能於非 執行董事袍金中引入新的組成部分，惟須 遵守本薪酬政策載列的準則、參數及其他 規定。 本公司於2014年12月5日成立一個新的非執 行董事委員會－慈善及社區投資事務監察 委員會。與其按薪酬政策所規定的權限相 符合，董事會批准該委員會以下袍金水平： 主席－每年25,000英鎊；成員－每年15,000 英鎊。 於此項政策期間概無對非執行董事袍金作 出或建議作出任何其他變動。應付非執行 董事的袍金載於 《2013年報及賬目》 內的去 年董事薪酬報告。 服務合約 執行董事 我們的政策為以通知期為12個月之服務協 議僱用執行董事。為顧及集團的最佳利益， 委員會將盡量減低終止合約的所需支出。 董事可能符合資格收取就法定權利支付的 款項。 董事薪酬政策╱年度薪酬報告 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 307 服務合約 合約生效日期 （自動續約） 通知期 （董事及滙豐） 董事 范智廉 2011年2月14日 12個月 歐智華 2011年2月10日 12個月 麥榮恩 2011年2月4日 12個月 繆思成 2014年11月27日 12個月 除 《2013年報及賬目》 董事薪酬報告內 「董 事薪酬政策」 及 「離職補償政策」 所載內容 外，概無任何其他可能導致須支付薪酬或 離職補償的責任。 非執行董事 非執行董事均按不超過三年的固定任期委 任，經股東周年大會獲股東重選可續約。 非執行董事並無簽訂服務合約，惟受代表 滙豐控股有限公司發出的委任函約束。除 《2013年報及賬目》 的董事薪酬報告內 「薪 酬政策－非執行董事」 所載內容外，非執行 董事委任函內概無任何責任，可能導致須 支付薪酬或離職補償。非執行董事現時任 期將於下列時間屆滿： ‧ 費卓成、方安蘭、利普斯基、駱美思及 駱耀文爵士於2015年屆滿； ‧ 祈嘉蓮、凱芝、史美倫、埃文斯勳爵、 李德麟及施俊仁於2017年屆滿；及 ‧ 苗凱婷及安銘 1於2018年屆滿。 1 自2015年1月1日起獲委任。 其他公司董事席位 執行董事倘獲董事會或提名委員會授權， 即可接受滙豐以外之公司委任為非執行董 事。 董事會或提名委員會在考慮接受非執行董 事的任命要求時，會考慮 （其中包括） 關於 獲提名職務預期需要投入的時間，亦會例 行覆核董事的在外職務所需投入的時間， 以確保該等在外職務不會影響董事對滙豐 的投入。 根據資本指引4，獲審慎監管局批准擔任若 干董事會職務的董事須遵守下列有關其可 擔任的董事職務數目之限制： ‧ 一個執行董事席位連同兩個非執行董 事席位；或 ‧ 四個非執行董事席位。 經審慎監管局同意可另擔任一個非執行董 事席位。 執行董事的任何此類外在職務之應收薪 酬，一般是支付給集團，但獲提名委員會 或董事會另行批准者則除外。 年度薪酬報告 薪酬委員會 角色 委員會在董事會授予之權限範圍內，負責 批准集團的薪酬政策。委員會亦釐定執行 董事、其他高層人員、身居要職的僱員及 其活動已經或可對我們的風險狀況具重大 影響力的僱員之薪酬，並於釐定上述薪酬 時，考慮整個集團的酬勞及條件。執行董 事概不參與釐定本身之薪酬。 成員 於2014年，集團薪酬委員會的成員是駱耀文 爵士 （主席） 、李德麟、利普斯基 （於2014年 5月23日獲委任） 、施俊仁 （於2014年4月14日 獲委任，但於2014年9月1日退出該委員會 以擔任集團監察委員會主席） 、范樂濤 （於 2014年9月1日辭任董事） 及顧頌賢 （於2014 年5月23日退任董事） 。 工作情況 委員會於2014年舉行11次會議。以下為委 員會於2014年的主要工作概要。 董事薪酬報告 （續） 滙豐控股有限公司 308 委員會的主要工作詳情 月份 工作 月份 工作 1月 ‧ 2013年業績計算年度酬勞審查事項 ‧ 制訂新的薪酬政策 ‧ 新股權指引 ‧ 管治事宜 7月 ‧ 2014年股東周年大會的回應 ‧ 2014年業績計算年度酬勞審查事項 ‧ 最新重大事件 ‧ 與整個集團的獎勵有關的事項 ‧ 僱員股份計劃事宜 ‧ 管治事宜 2月 ‧ 2013年業績計算年度酬勞審查事項 ‧ 2014年執行董事集團業績表現股份計 劃及周年評分紀錄 ‧ 制訂新的薪酬政策 ‧ 與全集團獎勵架構有關的事項 ‧ 僱員股份計劃事宜 ‧ 新股權指引 ‧ 呈交監管報告及披露 ‧ 管治事宜 9月 ‧ 2014年業績計算年度酬勞審查事項 ‧ 檢討審慎監管局╱金融業操守監管局 有關風險與回報一致的諮詢 3月 ‧ 就監察員報告作出回應 ‧ 2013年業績計算年度酬勞審查事項 ‧ 檢討審慎監管局有關撤回規則的諮詢 ‧ 最新重大事件 ‧ 與退休福利安排及獎勵計劃有關的事 項 ‧ 呈交監管報告及披露 10月 ‧ 有關薪酬事項的股東回應 ‧ 審慎監管局╱金融業操守監管局有關 風險與回報一致的諮詢之最近更新 11月 ‧ 歐洲銀行管理局報告的最近更新及關 於固定酬勞津貼的意見 ‧ 2014年承受風險水平聲明審查及 《薪 酬守則》 風險評估 ‧ 2014年建議集團浮動酬勞支出及方式 ‧ 批准撤回政策 ‧ 最新重大事件 ‧ 呈交監管報告及披露 ‧ 依據香港金管局的薪酬指引對滙豐獎 勵策略進行獨立審查 4月 ‧ 制訂新的薪酬政策事項 ‧ 籌備2014年股東周年大會 ‧ 與退休福利安排及獎勵計劃有關的事 項 ‧ 呈交監管報告及披露 12月 ‧ 承受風險水平架構及防範金融犯罪最 近更新 ‧ 來自集團風險管理委員會、金融系統 風險防護委員會以及行為及價值觀委 員會的數據 ‧ 2014年業績計算年度酬勞審查事項 ‧ 2015年執行董事集團業績表現股份計 劃及周年評分紀錄 ‧ 最新重大事件 ‧ 呈交監管報告及披露 ‧ 僱員股份計劃事宜 ‧ 管治事宜 5月 ‧ 與實施新薪酬政策有關的事項 ‧ 籌備2014年股東周年大會 ‧ 2014年業績計算年度酬勞審查事項 ‧ 僱員股份計劃事宜 ‧ 管治事宜 年度薪酬報告 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 309 顧問 於2014年，委員會並無委聘任何外界顧問， 且僅在其認為屬必要時方會獨立尋求特定 法律及╱或薪酬意見。 年內，集團行政總裁定期向委員會作出簡 報。此外，委員會亦向集團人力資源及企業 可持續發展主管Ann Almeida、集團員工表 現及獎勵事務主管Alexander Lowen、集團 風險管理總監繆思成及防範金融犯罪環球 主管及集團舉報洗錢活動主管Robert Werner 徵詢意見，以履行滙豐僱員行政的職責。 委員會亦聽取集團風險管理委員會、金融 系統風險防護委員會以及行為及價值觀委 員會就薪酬的風險及合規相關事宜以及實 施向下計酬政策發表的意見及回應。 集團浮動酬勞資金 （未經審核） 釐定浮動酬勞資金 在釐定集團的浮動酬勞資金時，委員會考 慮多項因素。 釐定浮動酬勞資金 表現及承受風險水平聲明 ‧ 委員會在釐定浮動酬勞資金時，在集團承受風險水平聲明的範圍 內，考慮集團的業績表現，有助確保浮動酬勞資金能顧及各項風 險因素及整個集團的重大事件。 ‧ 承受風險水平聲明描述及計量滙豐於落實策略時準備承受的風險 水平及類別，並構成處理業務、風險及資本管理工作的綜合方 針，有助達致集團的目標。集團風險管理總監按照承受風險水平 聲明來考量集團的業績表現，並定期向委員會匯報最新情況。 ‧ 委員會在考慮薪酬時會使用這些最新情況，以確保回報、風險與 薪酬保持一致。 反周期資金分配方法 ‧ 我們使用按下限及上限分類的反周期資金分配方法，當業績表現 有所提升時，派息比率會降低以避免順經濟周期效應之風險。 ‧ 下限確認不論業績表現水平如何，一般均需維持報酬之競爭性保 護。 ‧ 上限則確認於業績表現處於較高水平時可以限制獎勵，原因為毋 須繼續增加浮動酬勞資金，從而限制為提升財務表現而作出不當 行為的風險。 利潤分配 ‧ 此外，我們分配資金的方法顧及資本、股息及浮動酬勞之間的關 係，以確保除稅後利潤在三者之間適當分配 （2014年、2013年及 目標分配詳情請見下頁） 。 商業能力及負擔能力 ‧ 最後，委員會將考慮維持市場競爭力的商業需要和整體負擔能 力。周年獎勵及集團業績表現股份計劃的資金均來自同一項周年 浮動酬勞資金，而個人獎勵亦予以考慮。集團根據集團盈利能 力、資本實力及股東回報的情況為周年浮動酬勞資金撥資。此方 法確保向個別環球業務、環球部門、地區及級別僱員所發放與表 現掛鈎的獎勵乃以整全的方式釐定。 ‧ 市場競爭力乃用於釐定浮動酬勞資金的數據之一。我們可以解決 在與環球同業進行獎勵總額比較時識別的與市場水平的差距。此 乃確認因總部設於英國而產生的挑戰及因此而須應用相比歐盟以 外市場更為嚴格的獎勵方法。影響我們於亞洲、拉丁美洲及美國 在吸引及挽留人才方面具競爭優勢的市場地位的因素為因涉及浮 動酬勞上限、更高及更長期的遞延、扣減以及現時撤回的條例而 產生的對僱員酬勞的折讓。 ‧ 本年度的浮動酬勞資金乃參考集團的列賬基準之除稅前利潤釐 定，當中已作出調整而不計及因信貸息差造成的本身債務公允值 變動、出售事項產生的損益以及借記估值調整。列賬基準之除稅 前利潤計及罰款、罰則以及其他賠償項目支出。 ‧ 經考慮上文所述，委員會決定，鑑於業績表現、競爭性市場環 境、風險因素及其他因素，2014年經調整除稅前未計算浮動酬 勞利潤派付率將為16% （2013年為15%） 。派息率提高反映亞洲及 中東表現強勁，以及更為強調風險及控制職能。 董事薪酬報告 （續） 滙豐控股有限公司 310 集團 環球銀行及 資本市場業務 2014年 2013年 2014年 2013年 浮動報酬獎勵資金佔除稅前 利潤 （未計算浮動酬勞） 的百分比1 16% 15% 15% 13% 遞延浮動酬勞資金的 百分比2 14% 18% 25% 30% 1 2014年集團除稅前未計算浮動酬勞利潤的計 算詳情載於前頁。 2 2014年承受重大風險人員的遞延浮動酬勞百 分比為50%。 備考除稅後利潤分配 （未經審核） 按備考基準，不包括本身債務之公允值變 動及未分派酬勞之前的應佔除稅後利潤， 乃按下圖所示比例分配。集團的目標政策 為將集團的大部分除稅後利潤列為資本及 分配予股東。 酬勞開支的相對比重 （未經審核） 下圖顯示僱員酬勞總額相對於派息款額之 明細。 年度薪酬報告 浮動酬勞資金結果 （百萬美元） （未經審核） 1 該年度的每股普通股股息。就2014年而言，包括 於2014年派付的第一次、第二次及第三次股息 58億美元 （包括代息股份） 以及第四次股息38億 美元。 2 2014年僱員薪酬及福利包括於第303頁概述的 固定酬勞、福利及浮動酬勞。2013年僱員薪酬 及福利總計191.96億美元，包括因英國更改提 供傷病福利的基準而錄得的會計增益4.3億美 元。不計該項會計增益，2013年僱員薪酬及福 利總計196.26億美元。 1 包括向其他股權工具持有人派發之股息，並已扣除發行代息股份 （計算基礎為假設2014年第四季接納20% 代息股份） 。2014年宣派的每股普通股股息為0.50美元，較2013年增加2%。 2 浮動酬勞資金總額，已扣除稅項和將以滙豐股份發放獎勵的部分。 -7% -16% 3,660 1,120 3,920 1,327 2014年 2013年 集團 環球銀行及資本市埸業務 9,600 20,366 9,200 19,626 2014年 2013年 4% 4% (430) 19,196 百萬美元 普通股股息1 僱員薪酬及福利2 32% 股息1 53% 浮動酬勞2 15% 2014年 53% 1 35% 2 12% 2013年 45% 40% 15% 目標 保留盈利╱ 資本 保留盈利╱ 資本 股息 浮動酬勞 保留盈利╱ 資本 2 浮動酬勞 1 股息 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 311 薪酬的單一數字 執行董事 （經審核） 范智廉 歐智華 麥榮恩 繆思成 2014年 2013年 2014年 2013年 2014年 2013年 2014年 2013年 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 固定酬勞 基本薪金 1,500 1,500 1,250 1,250 700 700 700 － 固定酬勞津貼 － － 1,700 － 950 － 950 － 退休金 750 750 625 625 350 350 350 － 2,250 2,250 3,575 1,875 2,000 1,050 2,000 － 浮動酬勞 周年獎勵 － － 1,290 1,833 867 1,074 1,033 － 集團業績表現股份計劃 － － 2,112 3,667 1,131 2,148 1,131 － 3,402 5,500 1,998 3,222 2,164 － 固定及浮動酬勞總額 2,250 2,250 6,977 7,375 3,998 4,272 4,164 － 福利 136 48 589 591 43 33 6 － 非課稅福利 105 102 53 67 28 53 33 － 遞延現金的名義回報 41 27 － － 11 7 36 － 薪酬的單一數字總計 2,532 2,427 7,619 8,033 4,080 4,365 4,239 － 薪酬的單一數字附註 （經審核） 繆思成乃自2014年1月1日起獲委任為執行董事，因此，未披露其2013年數字。 基本薪金 ‧ 年內向執行董事支付的薪金。概無向執行董事支付任何袍金。 固定酬勞津貼 ‧ 年內以即時實際授出的股份授予執行董事的固定酬勞津貼。 ‧ 股份附帶禁售期。20%之股份將於緊隨財政年度結束之後於3月解除禁售。80%於自首次發放日期起計五 年期間之後解除禁售。 ‧ 於禁售期內持有的已實際授出股份將獲派付股息。 退休金 ‧ 金額包括相等於基本年薪50%的津貼，以代替個人退休金安排。 ‧ 執行董事並無收取集團退休金計劃的其他福利。 福利 ‧ 所有應課稅福利 （課稅前總值） 。福利包括提供醫療保險、住宿和汽車、俱樂部會籍、住宿及汽車福利稅 務補償，以及汽車津貼。 ‧ 非課稅福利包括提供壽險及其他保險。 ‧ 上表重大福利的價值如下： 范智廉 歐智華 麥榮恩 繆思成 2014年 2013年 2014年 2013年 2014年 2013年 2014年 2013年 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 汽車福利 （英國及香港） 70 1 － 2 88 1 79 － 2 － 2 － 2 － 於香港的銀行自置居所3 － － 246 229 － － － － 汽車福利及於香港的銀行自置居所稅項支出 58 1 － 2 239 1 266 － 2 － 2 － 2 － 保險福利 （非課稅） 80 78 － 2 54 － 2 － 2 － 2 － 1 2014年提供予范智廉及歐智華的英國汽車福利相比2013年並無變化。汽車福利估值提高乃因其就英國稅 務目的而言不再被視為公司汽車，並計入司機工資、燃油及所有相關費用。 2 范智廉 （2013年度） 、繆思成 （2014年度） 及麥榮恩的汽車福利及汽車福利稅項並不重大，因而並未計入上 表。歐智華及繆思成 （2014年度） 以及麥榮恩的保險福利並不重大，因而亦未計入上表。 3 根據一名外部租賃服務供應商的估計，銀行自置物業的當前市場租金價值，另加水電費、差餉、傢俬的 應課稅價值以及經計及物業的商業用途，住宿的應課稅價值按該等金額總額的70%計算。 周年獎勵 ‧ 因達成相關財政年度的表現衡量基準而獲授的周年獎勵 （包括遞延獎勵） 。60%的獎勵乃遞延。遞延及非 遞延獎勵的50%須以現金支付，餘下50%以股份 （於實際授出時附帶六個月禁售期） 支付。 ‧ 2014年獎勵的遞延部分於三年內發放 （須受服務及扣減條件規限） ：33%於授出日期的第一及第二周年或 董事薪酬報告 （續） 滙豐控股有限公司 312 其前後實際授出及34%於授出日期的第三周年實際授出。就2014年獎勵而言，表現衡量基準及符合表現 條件的結果載於第315至317頁。2013年獎勵的結果載於 《2013年報及賬目》 董事薪酬報告內。 ‧ 遞延股份獎勵亦包括有權收取等同股息額。遞延股份獎勵的等同股息額於相同時間按相同方式以實際授 出的原本遞延獎勵的相關比例以額外股份形式償付。該等等同股息額的預計價值包括在遞延股份獎勵價 值內。 周年獎勵圖解 （未經審核） 2014年 2015年 2016年 2017年 2018年及之後 40% 60% 3月 9月 3月 3月 3月 由薪酬委員會酌情決定發放的獎勵（須受扣減及撤回條文規限） 六個月 禁售期 非遞延部分 遞延部分 表現期 六個月 一年 一年 三年實際授出期 33%於授出後滿 第一周年或前後 實際授出，並附 帶六個月禁售期 34%於授出後滿 第三周年或前後 實際授出，並附 帶六個月禁售期 33%於授出後滿 第二周年或前後 實際授出，並附 帶六個月禁售期 根據2014年表現 釐定的獎勵水平 50%於六 個月之後 以股份 發放 於2015年3月 提供的獎勵， 50%以現金 即時發放 集團業績表現股份計劃 ‧ 因實現持續的長期業績表現而授出集團業績表現股份計劃獎勵。所示數字反映分別於2014及2013年授出 的獎勵面值。 ‧ 獎勵水平乃按照長期表現評分紀錄載列的持久表現衡量基準，考慮表現而釐定。概無訂立授出後表現條 件。 ‧ 獎勵設有五年實際授出期的限制，期間委員會有權取消全部或部分獎勵。於實際授出時，扣除稅項後的 股份必須於參與者受僱期間一直保留。 ‧ 有關2014年獎勵的績效表現條件的結果，載於第314頁 「集團業績表現股份計劃下之獎勵」 一節。2013年 獎勵的結果，載於 《2013年報及賬目》 董事薪酬報告內。 ‧ 集團業績表現股份計劃獎勵亦包括有權於授出日期至實際授出日期期間收取等同股息額。集團業績表現 股份計劃獎勵的等同股息額將於實際授出集團業績表現股份計劃獎勵時支付。2014年並無實際授出集團 業績表現股份計劃獎勵的歷史紀錄。該等等同股息額的預期價值包括在集團業績表現股份計劃獎勵價值 內。 集團業績表現股份計劃圖解 （未經審核） 以往至2014年 2016年 2015年 2017年 2018年 2019年 2020年及之後 表現期 五年實際授出期 附帶禁售期的實際 授出獎勵 由薪酬委員會酌情決定發放的獎勵（須受扣減及撤回條文規限） 根據2014年及長期可持續 表現釐定的獎勵水平 集團業績表現股份計劃獎勵 的價值隨股價表現上下波動 獎勵在五年之後於 2020年3月完全 實際授出 於2015年3月 提供的獎勵 儘管股份已完全實際 授出，執行董事於 受僱期間仍須保留 該等股份 年度薪酬報告 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 313 遞延現金的名義回報 ‧ 周年獎勵的遞延現金獎勵部分亦包括有權於授出日期至實際授出日期期間收取名義回報，乃參考滙豐股 份的股息收益率釐定 （每年釐定） 。 ‧ 名義回報每年根據各實際授出日期實際授出的遞延獎勵的相同比例支付。金額按於支付年度的已付基準 披露。 薪酬情況及結果 下圖列示與實際2014年浮動酬勞結果相比，各執行董事根據現行政策在三種表現情況下的 薪酬之價值及構成。 歐智華 麥榮恩 繆思成 100% 50% 33% 52% 17% 22% 18% 33% 45% 30% 3,575 7,150 10,725 6,977 100% 50% 33% 50% 17% 22% 22% 33% 45% 28% 2,000 4,000 6,000 3,998 100% 50% 33% 48% 17% 22% 25% 33% 45% 27% 2,000 4,000 6,000 4,164 （千英鎊） 固定酬勞1 周年獎勵2 集團業績表現股份計劃3 固定酬勞1 周年獎勵2 集團業績表現股份計劃3 固定酬勞1 周年獎勵2 集團業績表現股份計劃3 目標政策 上限政策 2014年實際值 下限政策 下限政策 目標政策 上限政策 2014年實際值 下限政策 目標政策 上限政策 2014年實際值 （千英鎊） （千英鎊） 浮動酬勞結果 （經審核） 歐智華 麥榮恩 繆思成 固定酬勞 價值 （千英鎊） 3,575 2,000 2,000 周年獎勵 固定酬勞的最高倍數 0.67 0.67 0.67 表現結果 54.1% 65.0% 77.5% 獲授倍數 0.36 0.43 0.52 價值 （千英鎊） 1,290 867 1,033 集團業績表現股份計劃 固定酬勞的最高倍數 1.33 1.33 1.33 表現結果 54.8% 54.8% 54.8% 獲授倍數 0.73 0.73 0.73 酌情授出前價值 （千英鎊） 2,612 1,461 1,461 委員會酌情授出 （千英鎊） (500) (330) (330) 酌情授出後價值 （千英鎊） 2,112 1,131 1,131 浮動酬勞總額 固定酬勞的最高倍數 2.00 2.00 2.00 獲授倍數 0.95 1.00 1.08 價值 （千英鎊） 3,402 1,998 2,164 1 固定酬勞包括年內基本薪金、固定酬勞津貼及 退休金津貼，不包括福利。 2 如薪酬政策內所述的最高獎勵水平。包括獎勵 的遞延部分。目標定義為最高獎勵的50%。最 低獎勵假設並無周年獎勵。 3 如薪酬政策內所述的最高獎勵水平。目標定義 為最高獎勵的50%。集團業績表現股份計劃評 分紀錄並無設定目標或預期的業績表現數值。 最低假設為並無任何集團業績表現股份計劃 獎勵。 董事薪酬報告 （續） 滙豐控股有限公司 314 集團業績表現股份計劃下之獎勵 （經審核） 有關2014年的獎勵乃按 《2013年報及賬目》 刊發而於下文複述之2014年長期評分紀錄 評估。 根據2014年長期評分紀錄的表現評估已計 及財務及非財務目標的達標程度，並於董 事會協定之承受風險水平與策略方針範圍 內設定。 不論績效表現的達標情況或程度如何，僱 員若要符合資格取得集團業績表現股份計 劃的獎勵，須恪守滙豐價值觀，此乃參與 集團業績表現股份計劃的門檻 （已評定所 有執行董事均符合該項標準） 。有關評估過 程及結論的理據概述如下。 年度薪酬報告 年度評估－集團業績表現股份計劃 比重 長期 目標範圍 2014年 實際表現 評估 結果 衡量基準 資本實力 （%） 1 15% >10 11.1 100% 15.0% 累進派息率 （%） 2 15% 40-60 72.5 100% 15.0% 股東權益回報率 （%） 3 15% 12-15 7.2 0% － 成本效益比率 －收入增長對支出增長的比率4 7.5% 收入增長對 支出增長的 比率為正數1 (14.7) 0% － －成本效益比率 （%） 3 7.5% 50%至60% 的中間範圍 67.8 0% － 財務 60% 30.0% 落實策略 20% 判定 不適用 67% 13.3% 風險管理及合規 15% 判定 不適用 50% 7.5% 人才 5% 判定 不適用 80% 4.0% 非財務 40% 24.8% 績效表現總結果 100% 54.8% 1 資本實力界定為普通股權一級資本 （資本指引4終點基準） 。 2 派息率反映年度股息。 3 股東權益回報率及成本效益比率不包括信貸息差變動產生的集團本身指定以公允值列賬之債務公允值 變動影響的回報。 4 收入增長 （不計及信貸息差變動產生的集團本身指定以公允值列賬之債務公允值變動影響） 減營業支出增長。 財務 （60％比重－達到30％） 資本實力 （評估：100%） ：鑑於監管當局就 整體損失吸納能力 （ 「TLAC」 ） 、穩定資金淨 額及槓桿新公布若干重要建議，委員會於 2014年尤其注意集團在準資本及流動資金 標準方面的狀況。對於其業務模式、可取得 的預期回報以至集團的派息能力，集團在 符合該等標準方面的能力將會有重要後果。 委員會亦注意歐洲銀行管理局及審慎監管 局進行的壓力測試結果。集團在作出對資 本實力及股息政策的判斷時，該等測試結 果乃是集團在經濟不景氣及行業不振形勢 下的復元力之獨立證明。委員會得悉該等 壓力測試的結果良好，從而令集團在不利 環境下吸收資本及恢復實力的能力相對於 其同業競爭對手更具優勢。 由於亦注意到年底普通股權一級比率改善， 加上根據資本指引4估計的終點持倉增加， 因此對資本狀況方面的實力給予正面評價。 累進派息率 （評估：100%） ：維持累進派息 政策的預計能力亦獲正面評價，預計累進 政策透過集團強勁的資金狀況、其可供分 派儲備、現金狀況及圍繞未來表現規劃假 設而得以鞏固。儘管部分外部環境下經濟 疲弱，集團的股息仍獲累進增長，顯示集 團多元化業務模式之優勢。 股東權益回報率 （評估：0%） ：集團未實現 之前既定的12%至15%的抱負。儘管在減少 既有持倉方面已取得良好進展，並在必要 情況下減少風險項目，集團的表現仍未達 到其既定目標。於實施新的監管規定後業 務模式已有改變，並加強風險控制以減少 未來發生客戶賠償及操守問題的可能性。 該等措施被視作集團欲實現可持續財務表 現所應確保的必要部分。在此期間，委員會 注意到在部分領域進行的必要結構性變動 很可能制約股東權益的全面回報，並掩蓋 新業務及市場份額提高所得裨益。此外， 集團表現近期繼續面臨不斷演變的監管改 革議程 （包括集團的目標資本比率） 、重大 既有事項之或有法律風險及持續重大客戶 賠償支出產生的不明朗因素。儘管承認於 環球及本地資本規定水平均有增長的背景 下盡力達到12%至15%的股東權益回報率 目標值得嘉許，但委員會仍決定不就此機 會授出任何獎勵。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 315 成本效益比率 （評估：0%） ：根據2014年集 團營業支出的相關情況，已判定不得根據 評分紀錄的成本效益比率部分授出任何獎 勵。委員會注意到，加強監管合規及防範 金融犯罪資源及能力乃費用水平增長的重 要部分。委員會亦確認，該種情況在中期 內不大可能減少。 非財務 （40%比重－達到24.8%） 落實策略 （評估：67%） ：董事會檢討管理層 於2014年在推動定為優先的策略方面所取 得的進展，包括有機增長、落實環球標準 以及簡化流程及程序以進一步提升效率。 在多個重要市場的經濟增長不如預期的背 景下，金融市場活動及流動資金雖因監管 變動引致行業重整而受到進一步制約，但 集團多項全球業務仍錄得相關增長，並維 持重要產品的市場份額。鑑於經濟狀況不 明朗，委員會明白董事會於2014年已就承 受風險水平著重採用審慎方法。 管理層認識到須進一步進行大量工作以全 面展開，並已在執行環球標準方面取得進 一步進展。 就精簡流程而言，集團實現持續節省逾10億 美元，但被用以支持增長計劃以及落實監 管變動、加強風險控制及執行環球標準的 遞增成本所超越。有鑑於此，管理層新推 多項計劃以提高環球業務和部門的效率， 並將持續至2015年。 風險管理及合規 （評估：50%） ：為盡量降低 監管及合規問題對集團聲譽的長期影響， 委員會接收集團風險管理委員會、行為及 價值觀委員會以及金融系統風險防護委員 會有關已取得進展的數據。委員會確信，根 據已接獲的回應，顯然此仍為組織結構內 部的首要任務且於2014年取得進展。委員 會尤其注意在重組集團合規部、投資加強 合規工作、制訂企業整體風險評估方案、開 展加強培訓及持續加強管治方面的工作。 委員會亦注意到於2014年亦有進一步遭受 罰款及罰則事件 （儘管乃與過往期間有關， 畢竟頗令人沮喪） 以及為防止再度發生而 增加的工作。 人才 （評估：80%） ：委員會檢討在人才發 展、繼任計劃、多元共融及部分領域的人 員損耗方面取得的進展。委員會確認持續 取得進展，包括在董事會層面以下，非執 行董事與高級行政人員之間成功開展導師 項目。 此表現評估得出54.8%的整體評分。 儘管如此，委員會隨後酌情將執行董事的 集團業績表現股份計劃獎勵減少下列數額： 集團業績表現股份計劃調整 集團業績表現 股份計劃調整 千英鎊 調整佔浮動 酬勞的百分比 % 董事 歐智華 500 13% 麥榮恩 330 14% 繆思成 330 13% 就歐智華及繆思成而言，根據影響集團的 法律、合規及監管問題 （即使是與過往事件 有關的問題，包括但不限於外匯問題） 之比 重，調整被認為屬適當。就麥榮恩而言， 在按年整體集團盈利能力、獎勵資金及市 場薪酬指標之下，委員會認為調整屬合適。 於2013年，委員會亦酌情將歐智華的整體 浮動酬勞減少18.5%。 確定執行董事的年度績效表現 （經審核） 2014年向執行董事發放的周年獎勵，反映 委員會評估彼等在表現評分紀錄的個人及 公司目標 （由董事會於年初協定） 達標程度。 此衡量方法按照財務及非財務衡量基準衡 量其表現，而有關衡量基準是為反映董事 會認為2014年適當的承受風險水平及優先 策略而設定。此外，根據向下計酬政策， 關於在加強反洗錢及制裁法律合規以及集 團履行根據延後起訴協議及其他有關命令 須承擔的責任方面取得的進展，委員會亦 諮詢金融系統風險防護委員會並考慮該委 員會的回應。 如須根據上述表現評分紀錄授出周年獎 勵，委員會須信納執行董事個人已經恪守 滙豐價值觀和具備推廣滙豐價值觀的領導 能力。此首要測試評估體現滙豐價值觀的 行為，即表現出 「坦誠開放、重視聯繫及穩 妥可靠」 及 「敢於以誠信正直行事」 ，並評 定所有執行董事均達標。 董事薪酬報告 （續） 滙豐控股有限公司 316 儘管在監管方面存在各種困難，整體而言， 執行董事在具挑戰性的市場環境背景下表 現尚佳。 歐智華 年度評估 比重 目標6 表現 評估 結果 衡量基準 除稅前利潤 （十億美元） 1 17.5% 21.6 18.6 86% 15.1% 股東權益回報率(%)2 10% 9.8 7.2 0% － 成本效益比率(%) －收入增長對支出增長的比率3 8.75% 5.6 (14.7) 0% － －成本效益比率(%)2 8.75% 58.5 67.8 0% － 股息(%)4 10% 57.1 72.5 100% 10.0% 資本實力(%)5 5% 10.9 11.1 100% 5.0% 財務 60% 30.1% 落實策略 20% 判定 不適用 70% 14.0% 風險管理及合規 20% 判定 不適用 50% 10.0% 非財務 40% 24.0% 推廣滙豐價值觀 首要測試 達標 總計 100% 54.1% 1 集團列賬基準之除稅前利潤已作調整以剔除因信貸息差產生的本身債務公允值變動、出售損益以及借記 估值調整 （2013年不包括借記估值調整） 。 2 股東權益回報率及成本效益比率不包括信貸息差變動產生的集團本身指定以公允值列賬之債務公允值 變動影響的回報。 3 收入增長 （不計信貸息差變動產生的集團本身指定以公允值列賬之債務公允值變動影響） 減營業支出增 長。 4 派息率反映年度股息。 5 資本實力界定為普通股權一級資本 （資本指引4終點基準） 。 6 按上年度2013年基準。 年度薪酬報告 在具體表現衡量基準下各執行董事的評估 過程於下表概述。 落實策略 （評估：70%） ：董事會檢討於2014年 在推動定為優先的策略方面所取得的進 展，包括有機增長、落實環球標準以及簡 化流程及程序以進一步提升效率。 委員會對實際基準收入反映在落實優先計 劃方面的進展評價正面，包括提高在選定 貿易走廊的市場份額及維持重要產品的市 場地位。 在由設計轉變為落實階段、持續出售及結 束非核心業務及股權方面，委員會注意到 環球標準計劃於整個2014年度取得的進展 （自2011年以來共進行75項交易） 。此外，為 確保年內早前批准的全球反洗錢和制裁政 策得以落實，環球業務將實施營運程序。 委員會注意到，為鞏固防範金融犯罪合規 專業知識並構建客戶盡職審查、交易監察 及制裁篩選的策略性基礎，已繼續作出投 資。故此，委員會獲告知集團已能實現其 2014年主要目標。 委員會對集團透過精簡業務及重整於年內 持續節省逾10億美元評價正面。委員會注 意到，受重大項目以及監管合規及防範金 融犯罪成本增加、通脹壓力、持續投資於 策略方案以及銀行徵費上調影響，2014年 成本有所增加。委員會獲告知，為進一步 提高各環球業務及重要部門的效率，集團 已推出新計劃並將延續至2015年。 風險管理及合規 （評估：50%） ：為強調集 團在該等領域的承擔，該項衡量基準的比 重由2013年之15%升至20%。委員會檢討集 團在增加及提升集團合規部門僱員人數、 推出 「以價值為主導的高績效文化」 計劃、 實施應對行為風險的措施 （如產品範圍審 查及相關產品退出、零售銀行獎勵安排變 動） 以及繼續加強管治方面取得的進展。結 果已受到因持續客戶賠償導致的發生率、 規模及聲譽損害以及於2014年產生的監管 機構罰款及罰則影響。 此表現評估得出54.1%的整體評分。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 317 麥榮恩 年度評估 比重 目標 表現 評估 結果 衡量基準 業務及股息均錄得增長 15% 判定 不適用 85% 13.0% 風險管理及合規，包括環球標準 50% 判定 不適用 75% 37.5% 精簡流程及程序 25% 判定 不適用 45% 11.0% 優先策略 90% 61.5% 人才 10% 判定 不適用 35% 3.5% 推廣滙豐價值觀 首要測試 達標 總計 100% 65.0% 業務及股息均錄得增長 （評估：85%） ：委員 會認可財務部在支持編製、實施及監察業 務用例方面的貢獻，協助取得有機增長， 並使分析員及投資者見面會的清晰程度顯 著改善 （獲分析員及股東正面評價） 。 風險管理及合規，包括環球標準 （評估： 75%） ：委員會注意到若干領域已取得重大 進展。於年內，風險管理及合規衡量基準 以及實施環球標準計劃新的成本及資源監 察程序均獲全面達標。 精簡流程及程序 （評估：45%） ：委員會認 可就對整個集團進行壓力測試的計劃作出 大量投入及其相關成果。委員會亦注意到 費用高於目標水平，而未全面實現持續節 省的目標。同一道理，儘管重整環球財務 部方面正取得進展，但若干計劃仍繼續進 行。 人才 （評估：35%） ：委員會注意到，績效管 理及環球財務部獎勵區分已取得進展，而 提高僱員多元化程度及重整成本則須做更 多工作。 此表現評估得出65%的整體評分。 繆思成 年度評估 比重 目標 表現 評估 結果 衡量基準 業務及股息均錄得增長 20% 判定 不適用 90% 18.0% 風險管理及合規，包括環球標準 50% 判定 不適用 75% 37.5% 精簡流程及程序 20% 判定 不適用 70% 14.0% 優先策略 90% 69.5% 人才 10% 判定 不適用 80% 8.0% 推廣滙豐價值觀 首要測試 達標 總計 100% 77.5% 業務及股息均錄得增長 （評估：90%） ：委 員會認可採用承受風險水平聲明以使業務 可持續發展及提供資源以支持業務發展 （如 風險分析及提升風險程序以使信貸組合的 質素可獲改善） 。 風險管理及合規，包括環球標準 （評估： 75%） ：委員會注意到，實施環球標準、遵 守監管規定及減少組織結構的風險項目取 得進展，而推出反洗錢及制裁合規計劃、 制訂營運風險轉移路線圖以及成功完成審 慎監管局及歐洲銀行管理局的壓力測試即 可資以為證。 精簡流程及程序 （評估：70%） ：委員會以 管理業績、落實主要的精簡措施及重整防 範金融犯罪制度為依據，確認該等目標多 數已告達成。委員會亦注意到，已進行環 球風險數據策略方面的工作以支持審慎監 管局的數據要求 （包括改進風險數據基礎 設施） 。 人才 （評估：80%） ：委員會注意到，已落實 酬勞及表現計劃以及學習及發展計劃 （此 乃環球風險管理部的綜合性人才策略之一 部分） 。 此表現評估得出77.5%的整體評分。 董事薪酬報告 （續） 滙豐控股有限公司 318 非執行董事 袍金及福利 （經審核） 袍金 福利7 總計 2014年 2013年 2014年 2013年 2014年 2013年 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 祈嘉蓮1 129 － 12 － 141 － 凱芝 95 95 4 14 99 109 史美倫2 197 195 22 47 219 242 埃文斯勳爵 167 50 14 － 181 50 費卓成 145 137 10 21 155 158 方安蘭3 494 202 19 6 513 208 李德麟 140 125 － － 140 125 利普斯基 168 150 27 25 195 175 駱美思 205 155 21 8 226 163 苗凱婷4 52 － － － 52 － 駱耀文爵士 260 240 6 1 266 241 施俊仁5 365 － 3 － 368 － 總計6 2,417 1,349 138 122 2,555 1,471 總計 （千美元） 3,979 2,221 229 201 4,208 2,422 1 於2014年3月1日獲委任。 2 包括於2014年擔任香港上海滙豐銀行有限公司董事、副主席及提名委員會成員之袍金57,000英鎊 （2013 年為75,000英鎊） 。 3 包括於2014年擔任北美滙豐控股有限公司非執行主席之袍金334,000英鎊 （於2014年1月1日獲委任） 。 4 於2014年9月1日獲委任。 5 於2014年4月14日獲委任為滙豐控股有限公司非執行董事及英國滙豐銀行有限公司非執行主席，就此收 取袍金247,000英鎊。 6 不包括張建東、顧頌賢、范樂濤及何禮泰 （彼等於2014年12月31日並非董事） 的袍金及福利。張建東於2014 年8月1日辭任。彼於2014年的袍金為113,000英鎊 （2013年為197,000英鎊） （包括擔任恒生銀行有限公司董 事、風險管理委員會主席及審核委員會成員之袍金40,000英鎊） 。彼於2014年的福利為18,000英鎊 （2013 年為45,000英鎊） 。顧頌賢於2014年5月23日退任。彼於2014年的袍金及福利分別為85,000英鎊及5,000英鎊 （2013年分別為205,000英鎊及14,000英鎊） 。范樂濤於2014年9月1日辭任。彼於2014年的袍金及福利分別 為109,000英鎊及10,000英鎊 （2013年分別為145,000英鎊及23,000英鎊） 。何禮泰於2014年5月23日退任。彼 於2014年的袍金及福利分別為50,000英鎊及1,000英鎊 （2013年分別為145,000英鎊及1,000英鎊） 。 7 福利包括與出席董事會及在滙豐控股註冊辦事處舉行的其他會議有關的差旅相關開支。所披露金額為估 計金額，並採用稅率45%返計還原 （倘相關） 。 退休金權益總額 （經審核） 年內概無擔任執行董事的僱員於提早退休 的情況下有權收取滙豐任何最後薪金退休 金計劃款項或有權收取額外福利。集團並 無設定董事退休年齡，惟退休年齡一般為 65歲。 對前董事的付款 （經審核） 本報告並不包括支付予前董事低於本公司 設定的50,000英鎊最低限額的款項資料。 年內的退任補償 （經審核） 於2014年概無向年內或任何過往年度擔任 董事的任何人士支付離職補償。 2014年授出的計劃權益 （未經審核） 下表載列就2013年表現而於2014年授予董 事的計劃權益 （如2013年董事薪酬報告所披 露） 。於本財政年度，概無非執行董事收取 計劃權益。 年度薪酬報告 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 319 2014年計劃獎勵 （未經審核） 獲授的面值 1 就最低表現 獲授的 授出日期 獲授的權益類別 授出獎勵的基準 獎勵日期 千英鎊 可獲百分比 2 股份數目 的股價 1 表現期結束 歐智華 遞延現金 2013年周年獎勵 2014年3月10日 550 0% 不適用 不適用 2013年12月31日 歐智華 有限制股份 2013年周年獎勵 2014年3月10日 550 0% 88,766 6.196英鎊 2013年12月31日 歐智華 有限制股份 2013年集團業績 表現股份計劃 2014年3月10日 3,667 0% 591,779 6.196英鎊 2013年12月31日 麥榮恩 遞延現金 2013年周年獎勵 2014年3月10日 322 0% 不適用 不適用 2013年12月31日 麥榮恩 有限制股份 2013年周年獎勵 2014年3月10日 322 0% 51,997 6.196英鎊 2013年12月31日 麥榮恩 有限制股份 2013年集團業績 表現股份計劃 2014年3月10日 2,147 0% 346,647 6.196英鎊 2013年12月31日 繆思成 遞延現金 2013年周年獎勵 2014年3月10日 322 0% 不適用 不適用 2013年12月31日 繆思成 有限制股份 2013年周年獎勵 2014年3月10日 322 0% 51,992 6.196英鎊 2013年12月31日 繆思成 有限制股份 2013年集團業績 表現股份計劃 2014年3月10日 2,147 0% 346,613 6.196英鎊 2013年12月31日 集團業績表現股份計劃獎勵乃根據截至授出日期前的財政年度結束時的表現而計算，授出後並無附帶其他 表現條件。於授出日期後五年實際授出，且董事於實際授出日期一般須仍為僱員。董事於實際授出日期獲授 之任何股份 （除稅淨額） 須受制於一項禁售規定。 上表不包括作為固定酬勞津貼之一部分而發行的股份之詳情，因該等股份乃即時實際授出且毋須達成任何 服務或表現條件。 1 所用股價為授出日期前最後一個工作日的中間收市價。 2 獎勵根據於截至2013年12月31日止期間達致的表現釐定。倘於截至2013年12月31日止期內僅達成最低表 現，則整體獎勵水平為最高機會的0%。於授出後，獎勵附帶服務及扣減條文。 績效表現概要 （未經審核） 滙豐股東總回報與富時100指數的比較 此圖表顯示於截至2014年12月31日止六年 期內，股東總回報表現與富時100指數的比 較。選用富時100指數，是因為該指數乃市 場公認具廣泛代表性的股市指數，而滙豐 控股亦是富時100指數成分股之一。 2008年12月2009年12月2010年12月2011年12月2012年12月2013年12月2014年12月 滙豐 富時100指數 80% 100% 120% 140% 160% 180% 200% 資料來源：Datastream 行政總裁薪酬 （未經審核） 行政總裁過往薪酬 下表概述過往六年行政總裁薪酬的單一數字，連同各周年獎勵及長期獎勵的結果。 薪酬的單一數字 最高周年獎勵 2 已付周年獎勵 2 最高長期獎勵 4 已付長期獎勵 4 （千英鎊） （佔固定酬勞%） 3 （佔最高金額%） （佔固定酬勞%） 3 （佔最高金額%） 2014年 歐智華 7,619 67% 54.1% 133% 44.3% 2013年 歐智華 8,033 300% 49.0% 600% 49.0% 2012年 歐智華 7,532 300% 52.0% 600% 40.0% 2011年 歐智華 8,047 300% 57.5% 600% 50.0% 2010年1 紀勤 7,932 400% 81.6% 700% 19.1% 2009年1 紀勤 7,580 400% 93.5% 700% 25.4% 1 集團業績表現股份計劃於2011年引入。在此之前，所示價值與根據滙豐股份計劃授出的業績表現股份獎 勵有關。根據此計劃，業績表現股份獎勵於授出後三年實際授出，惟須符合股東總回報、經濟盈利及每 股盈利的表現條件，並最終取決於委員會判定業績是否 「持續改善」 。 2 本表所用歐智華2012年周年獎勵數字包括於2012年董事薪酬報告所披露遞延五年的60%周年獎勵。是否 實際授出此等獎勵，須視乎服務條件及能否在令人滿意的情況下完成延後起訴協議。延後起訴協議條件 於授出日期第五周年當日或其前後終結，除非延後起訴協議延期或以其他方式延續至該日期之後。在此 情況下，獎勵將於延後起訴協議屆滿或以其他方式停止運作當日或其前後實際授出。 3 於2014年，固定酬勞包括年內基本薪金、固定酬勞津貼及退休金津貼，不包括福利。於2013年及之前期 間，固定酬勞僅包括基本薪金。 4 長期獎勵於視作完成大部分表現期間的年度列示。就業績表現股份獎勵而言，為授出日期後第三個財政 年度結束時 （因此，於2010年所示業績表現股份獎勵與於2008年授出的獎勵有關） 。就集團業績表現股份 計劃獎勵而言，為授出日期前一個財政年度結束時 （因此，於2011年至2014年所示集團業績表現股份計 劃獎勵與於2012年至2015年授出的獎勵有關） 。 董事薪酬報告 （續） 滙豐控股有限公司 320 集團行政總裁與全體僱員酬勞比較 下表列示於2013年至2014年期間集團行政總裁酬勞變動與僱員酬勞變動的比較： 薪酬變動百分比 基本薪金 福利 3 周年獎勵 4 集團行政總裁1 0.0% (0.3%) (29.6%) 僱員團隊2 4.4% 5.7% (5.3%) 1 集團行政總裁根據股東批准的新訂薪酬政策自2014年1月1日起收取170萬英鎊之固定酬勞津貼。有關固 定酬勞津貼的進一步詳情載於執行董事薪酬政策一節。 2 僱員團隊包括全球各地的全體僱員，按財務報告所披露的工資及薪金所包括的成本 （不包括固定酬勞津 貼） 及職員人數 （財政年度內等同全職僱員平均數） 計算。 3 僱員團隊僅包括英國僱員 （財政年度內等同全職僱員平均數） ，原因是鑑於不同地方規定，其被視作集團 行政總裁的最佳比較對象。 4 僱員團隊包括全球各地的全體僱員，按財務報告所披露的周年獎勵資金減集團業績表現股份計劃及僱員 人數 （財政年度結束時等同全職僱員） 計算。 董事於股份的權益 （經審核） 指引 為確保與股東利益適當地保持一致，我們 為執行董事、非執行董事及集團常務總監 制訂了股權指引 （以股份數目列示） 。委員 會認為，高級行政人員及非執行董事擁有 股份有助令彼等的利益與股東的利益保持 一致。彼等須持有的股份數目列於下表。 個別人士自2014年或 （倘較後） 彼等獲委任 為執行董事、非執行董事或集團常務總監 起計五年達致推薦的持股量水平。 滙豐就其全體僱員運作一項反對沖政策。 作為此政策的一部分，全體僱員每年須證 實彼等並無訂立任何個人對沖策略 （與彼 等所持的滙豐股份有關） 。 委員會每年查察相關人員是否遵守指引的 持股量規定。如未有遵守規定，委員會可 全權酌情決定施加任何懲罰，當中包括減 少日後授出的集團業績表現股份計劃獎勵 及╱或提高周年浮動酬勞中遞延至股份的 比例。 於2014年身為董事的所有人士於2014年12 月31日或於彼等退任時的持股量 （包括彼等 的關連人士的持股量） 載列於下文。 認股權 （經審核） 行使期 於2014年 於2014年 授出日期 行使價 由1 直至 1月1日 年內已行使 12月31日 范智廉 2012年4月24日 4.4621 2015年8月1日 2016年2月1日 2,016 － 2,016 范智廉 2014年9月23日 5.1887 2019年11月1日 2020年5月1日 － － 2,919 麥榮恩 2014年9月23日 5.1887 2017年11月1日 2018年5月1日 － － 3,469 1 可因若干情況 （如退休） 而提早行使。 年度薪酬報告 滙豐控股儲蓄優先認股計劃為全體僱員的 優先認股計劃。根據此等計劃，合資格之 僱員可獲授認股權，以購入滙豐控股普通 股。僱員可於三年或五年期間每月作出上 限為500英鎊 （或等值金額） 的供款，而這些 供款可由僱員選擇於有關儲蓄合約生效後 三周年或五周年當日或其前後用以行使認 股權。此等計劃之目的為使僱員的利益與 創造股東價值更趨一致。認股權乃以無代 價方式授出，可按最接近要約日期前五個 營業日普通股的平均市值折讓20%行使。 尚未行使之認股權並無附帶任何行使的表 現條件，而自從授出獎勵以來，有關條款 及條件並無任何變更。於2014年12月31日， 每股普通股之市值為6.09英鎊。年內每股 普通股之最高市值為6.81英鎊，最低為5.89 英鎊。市值乃指股份於有關日期根據倫敦 證券交易所每日正式牌價表計算所得的市 場中間價。根據香港 《證券及期貨條例》 ， 該等認股權被歸類為非上市之實物結算股 權衍生工具。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 321 股份 （經審核） 於2014年12月31日或退任日期 計劃權益 持股量 要求 （股數） 1 股份 權益總額 （股數） 認股權 2 須遞延授出的股份 毋須符合 表現條件 3 須符合 表現條件 執行董事 范智廉 400,000 400,748 4,935 － － 歐智華 750,000 2,611,188 － 2,476,808 87,007 麥榮恩 450,000 79,933 3,469 942,732 60,150 繆思成 450,000 480,423 － 1,216,599 58,439 集團常務總監4 250,000 不適用 不適用 不適用 不適用 非執行董事5 凱芝 15,000 20,045 不適用 不適用 不適用 顧頌賢6 15,000 23,845 不適用 不適用 不適用 埃文斯勳爵 15,000 5,519 不適用 不適用 不適用 費卓成 15,000 24,105 不適用 不適用 不適用 方安蘭 15,000 76,524 不適用 不適用 不適用 李德麟 15,000 36,768 不適用 不適用 不適用 利普斯基 15,000 15,820 不適用 不適用 不適用 駱美思 15,000 15,500 不適用 不適用 不適用 苗凱婷7 15,000 3,575 不適用 不適用 不適用 駱耀文爵士 15,000 22,981 不適用 不適用 不適用 施俊仁8 15,000 20,553 不適用 不適用 不適用 1 現行持股量要求不計以未實際授出的股份為基礎的獎勵。 2 所有認股權尚未行使。 3 包括於評估緊接授出日期前截至12月31日止相關期間的表現後授出的集團業績表現股份計劃獎勵，惟附 帶五年的實際授出期間。 4 預期所有集團常務總監於2019年前或自其獲委任日期起計五年內 （以較後者為準） 均將符合其最低持股量 規定。 5 於2014年為非執行董事但未列於上表的人士於2014年12月31日或於彼等退任日期並無直接或透過任何關 連人士持有任何股份。 6 顧頌賢於2014年5月23日退任董事。 7 於2014年9月1日獲委任。 8 於2014年4月14日獲委任。 董事薪酬報告 （續） 滙豐控股有限公司 322 股東背景 （未經審核） 下表顯示於滙豐控股有限公司在2014年5月23日舉行的股東周年大會上與薪酬有關的表決結 果。 投票總數 贊成 反對 棄權 2013年薪酬報告的諮詢票 9,744,121,154 8,180,579,271 (83.95%) 1,563,541,883 (16.05%) 205,528,859 薪酬政策的具約束力股東投票 9,781,954,191 7,762,051,505 (79.35%) 2,019,902,686 (20.65%) 167,509,544 於2015年執行薪酬政策的情況 （未經審核） 下表概述於2015年將如何發放酬勞的各個部分。 目的及與策略的聯繫 薪酬政策在運作及計劃內的修改 固定酬勞 基本薪金 基本薪金水平將維持於2014年的水平，不予改變： 范智廉：1,500,000英鎊 歐智華：1,250,000英鎊 麥榮恩：700,000英鎊 繆思成：700,000英鎊 固定酬勞津貼 1 固定酬勞津貼將維持於2014年的水平，不予改變： 范智廉：零 歐智華：1,700,000英鎊 麥榮恩：950,000英鎊 繆思成：950,000英鎊 退休金 2015年適用的退休金津貼將為基本薪金的某個百分比，不予改變： 范智廉：50% 歐智華：50% 麥榮恩：50% 繆思成：50% 福利 福利 並無建議對2015年的福利作出任何修改。 浮動酬勞1 周年獎勵 並無建議對周年獎勵作出任何修改。 集團業績表現股份計劃 並無建議對集團業績表現股份計劃作出任何修改。 1 本方法適用於全體執行董事，惟集團主席范智廉除外，彼不合資格收取固定酬勞津貼或浮動酬勞獎勵。 年度薪酬報告 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 323 周年花紅評分紀錄 歐智華、麥榮恩及繆思成的2015年周年獎 勵適用的表現衡量基準及比重載於下文。 范智廉並不包括在內，原因是彼不合資格 收取周年獎勵。 委員會認為，周年獎勵的表現目標屬商業 敏感資料，於財政年度開始前披露可能會 損害公司利益。因商業敏感性質，目標將 於相關財政年度結束後於該年度薪酬報告 內予以披露。 2015年周年獎勵評分紀錄 歐智華 麥榮恩 繆思成 衡量基準 比重 與以下各項有關的 功能衡量基準 比重 與以下各項有關的 功能衡量基準 比重 除稅前利潤1 15% 業務及股息均錄得增長 15% 業務及股息均錄得增長 20% 股東權益回報率 15% 環球標準 （包括風險管理 及合規） 50% 環球標準 （包括風險管理 及合規） 50% 收入增長對支出增長的比率2 15% 精簡流程及程序 25% 精簡流程及程序 20% 股息增長3 15% 財務 60% 優先策略 90% 優先策略 90% 落實策略 15% 人才 10% 人才 10% 環球標準 （包括風險管理及合規） 25% 非財務 40% 人才 10% 人才 10% 推廣滙豐價值觀 首要測試 推廣滙豐價值觀 首要測試 推廣滙豐價值觀 首要測試 總計 100% 總計 100% 總計 100% 2015年集團業績表現股份計劃評分紀錄 衡量基準 長期目標範圍 比重 股東權益回報率2 >10% 20% 收入增長對支出增長的比率2 收入增長對支出增長的比率為正數 20% 股息增長3 累進 20% 財務 60% 落實策略 判定 15% 環球標準 （包括風險管理及合規） 判定 25% 非財務 40% 總計 100% 1 除稅前利潤 （就集團浮動酬勞資金所界定者） 。 2 收入增長減營業支出 （按經調整基準） 。 3 年度每股普通股股息 （美元） （按年計量） ；與集團整體盈利能力增長一致，以持續符合監管規定資本水平 的能力為基準。 董事薪酬報告 （續） 滙豐控股有限公司 324 董事薪酬報告附錄 額外資料披露 本附錄提供根據香港法例、香港上市規則、梅林協議、金融業操守監管局 《銀行的審慎措施資料手冊》 ，以及 美國證交會20-F表格披露規定所要求的披露。 僱員報酬及福利 董事酬金 （未經審核） 下表列示於截至2014年12月31日止年度支付予執行董事的酬金詳情。 范智廉 歐智華 麥榮恩 繆思成 千英鎊 千英鎊 千英鎊 千英鎊 基本薪金、津貼及實物福利 2,491 4,217 2,071 2,039 退休金供款 － － － － 獲發或應收與表現掛鈎的報酬 － 3,402 1,998 2,164 獲發或應收加盟獎勵 － － － － 離職補償 － － － － 總計 2,491 7,619 4,069 4,203 總計 （千美元） 4,101 12,545 6,700 6,922 3,752 12,558 6,813 － 2013年總計 （千美元） 3,752 12,558 6,813 － 董事 （包括執行董事及非執行董事） 於截至2014年止年度的酬金總額為34,475,463美元。概無 董事獲支付退休金及離職補償。繆思成自2014年1月1日起獲委任為執行董事，故未披露其 2013年數字。 高層管理人員的酬金 （未經審核） 下表列示支付予高層管理人員 （即滙豐控股執行董事及集團常務總監） 於截至2014年12月31日 止年度或獲委任為董事或集團常務總監期間的酬金詳情。 高層管理人員酬金 高層管理 人員 千英鎊 基本薪金、津貼及實物福利 32,237 退休金供款 433 獲發或應收與表現掛鈎的報酬 23,749 獲發或應收加盟獎勵 － 離職補償 － 總計 56,419 總計 （千美元） 92,893 截至2014年12月31日止年度，高層管理人員之酬金總額為92,892,912美元。高層管理人員之酬 金分為下列各級： 高層管理 人員人數 0 － 1,000,000英鎊 1 1,000,001 － 2,000,000英鎊 1 2,000,001 － 3,000,000英鎊 7 3,000,001 － 4,000,000英鎊 3 4,000,001 － 5,000,000英鎊 3 6,000,001 － 7,000,000英鎊 1 7,000,001 － 8,000,000英鎊 1 截至2014年12月31日止年度，為執行董事及高層管理人員提供退休金、退休或類似福利而 撥出或應計的總額為713,715美元。 附錄 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 325 五位最高薪僱員的酬金 （未經審核） 下表列示截至2014年12月31日止年度內，支付予滙豐五位酬金最高人士 （包括滙豐控股兩位 執行董事及兩位集團常務總監） 的薪酬詳情。 五位最高薪僱員的酬金 五位最高 薪僱員 千英鎊 基本薪金、津貼及實物福利 14,945 退休金供款 128 獲發或應收與表現掛鈎的報酬 11,690 獲發或應收加盟獎勵 － 離職補償 － 總計 26,763 總計 （千美元） 44,066 五位最高薪僱員之酬金分為下列各級： 五位最高薪 僱員人數 4,000,001 － 4,100,000英鎊 1 4,200,001 － 4,300,000英鎊 1 4,700,001 － 4,800,000英鎊 1 6,000,001 － 6,100,000英鎊 1 7,600,001 － 7,700,000英鎊 1 八位最高薪高級行政人員的薪酬 （未經審核） 下表列示八位最高薪之高級行政人員 （包括並非滙豐控股董事的集團管理委員會成員） 薪酬 詳情： 僱員 1 2 3 4 5 6 7 8 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 千英鎊 固定 現金基準 650 606 650 668 741 650 765 561 股份基準 3,016 1,549 904 726 617 710 376 434 固定總額 3,666 2,155 1,554 1,394 1,358 1,360 1,141 995 周年獎勵1 現金 421 345 283 265 224 223 224 199 非遞延股份2 421 345 283 265 224 223 224 199 遞延現金3 631 517 424 398 336 334 336 299 遞延股份3 631 517 424 398 336 334 336 299 周年獎勵總額 2,104 1,724 1,414 1,326 1,120 1,114 1,120 996 集團業績表現 股份計劃 遞延股份 234 191 157 147 125 124 124 111 浮動酬勞總額 2,338 1,915 1,571 1,473 1,245 1,238 1,244 1,107 整體薪酬福利 6,004 4,070 3,125 2,867 2,603 2,598 2,385 2,102 整體薪酬福利 （千美元） 9,887 6,701 5,144 4,720 4,287 4,276 3,928 3,461 1 2014業績計算年度的周年獎勵。 2 實際授出的獎勵附帶六個月的禁售期。 3 於三年期間實際授出的獎勵，授出的第一及第二周年或前後實際授出33%，授出的第三周年或前後實際 授出34%。 董事薪酬報告 （續） 滙豐控股有限公司 326 附錄 第三支柱薪酬披露 下表載列滙豐就2014年向其已識別員工及承受重大風險人員授出的薪酬獎勵。有關人士乃 根據於2014年6月生效的監管技術準則EU604/2014載列的定質及定量準則被識別為承受重大 風險人員。該準則取代之前用於識別守則職員是否遵守審慎監管局及金融業操守監管局 《薪 酬守則》 的準則。 監管技術準則EU604/2014指明用以識別承受重大風險人員之定質及定量準則的範圍遠較過 往年度用於識別守則職員的準則為廣。因此，2014年被識別為承受重大風險人員的人數遠 高於過往年度被識別為守則職員的人數。下表所載2013年的數據乃與2013年被識別為守則職 員的人數有關。 該等披露資料反映金融業操守監管局的 《銀行的審慎措施資料手冊》 的規定。 薪酬支出總額 環球業務合計 零售 銀行及 環球銀行及 環球 非環球 財富管理 工商金融 資本市場 私人銀行 業務合計 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 薪酬支出總額 （2014年承受重大風險人員╱ 2013年守則職員） 1 2014年 94.3 61.7 741.3 70.2 374.4 1,341.9 2013年 39.7 14.6 309.0 44.9 171.2 579.4 1 包括薪金及就2013及2014年表現授出的獎勵 （包括遞延部分） ，以及政策以外的任何退休金或福利。 薪酬－固定及浮動金額－集團整體 2014年 2013年 承受重大 風險人員 守則職員 高級 （非高級 高級 （非高級 管理層1 管理層） 總計 管理層1 管理層） 總計 2014年承受重大風險人員╱ 2013年守則職員數目 98 1,080 1,178 66 264 330 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 固定 現金基準 64.1 517.0 581.1 52.6 101.1 153.7 股份基準 51.8 88.7 140.5 － － － 固定總額 115.9 605.7 721.6 52.6 101.1 153.7 浮動2 現金 18.5 138.9 157.4 19.0 60.1 79.1 非遞延股份3 18.5 132.0 150.5 18.9 56.5 75.4 遞延現金 24.9 119.5 144.4 26.6 79.3 105.9 遞延股份 41.5 126.4 167.9 72.4 92.8 165.2 浮動酬勞總額4 103.4 516.8 620.2 136.9 288.7 425.6 1 2014年高級管理層的定義包括集團管理委員會成員、集團總經理及非執行董事。2013年則僅包括集團管 理委員會成員及集團總經理。 2 浮動酬勞乃視乎2013及2014年的表現。 3 實際授出的股份設有六個月的禁售期。 4 根據於2014年5月23日取得的股東批准，就各承受重大風險人員而言，任何一年內薪酬的浮動金額以承受 重大風險人員薪酬總額的固定金額之200%為限。 薪酬－固定及浮動金額－英國 2014年 2013年 承受重大 風險人員 守則職員 高級 （非高級 （非高級 管理層 管理層） 總計 高級管理層 管理層） 總計 2014年承受重大風險人員╱ 2013年守則職員數目 64 446 510 35 157 192 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 固定總額 73.1 244.5 317.6 30.4 53.7 84.1 浮動酬勞總額1 60.7 205.2 265.9 86.0 120.3 206.3 1 浮動酬勞乃視乎2013及2014年的表現授出。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 327 遞延薪酬 1 2014年 2013年 高級管理層 承受重大 風險人員 （非高級管理層） 總計 高級管理層 守則職員 （非高級管理層） 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於12月31日的遞延薪酬 尚待動用，未實際授出 270.2 691.8 962.0 213.4 331.7 545.1 年內授出 112.6 353.8 466.4 87.0 159.6 246.6 已支付2 33.9 210.3 244.2 110.7 269.9 380.6 扣減額3 － － － 0.4 － 0.4 1 上表提供於2013及2014年業績計算年度所採取的措施詳情。2013及2014年業績計算年度授出的浮動酬勞 獎勵詳情，請參閱上文薪酬表。 2 除於2013年實際授出股份採用實際授出當日的股價計值外，均於有關年度12月31日計值。 3 上表僅披露2014年承受重大風險人員╱2013年守則職員的扣減額情況。年內已對離開集團的其他人士應 用扣減條文。倘適用，將於應用扣減條文前對本年度浮動酬勞作出調整 （有關更多詳情請參閱第306頁） 。 簽約金及遣散費 2014年 2013年 高級管理層 承受重大 風險人員 （非高級管理層） 總計 高級管理層 守則職員 （非高級管理層） 總計 簽約金 年內發放 （百萬美元） 1.9 2.6 4.5 － 3.7 3.7 受益人數目 1 5 6 － 3 3 遣散費 年內發放 （百萬美元） － 4.1 4.1 1.1 1.6 2.7 受益人數目 － 13 13 3 5 8 向個人發放的最高金額 （百萬美元） － 0.5 0.5 0.6 0.6 按薪級分析的承受重大風險人員薪酬 1 2014年承受重大風險人員數目 2013年守則職員數目 高級管理層 承受重大 風險人員 （非高級管理層） 總計 高級管理層 守則職員 （非高級管理層） 總計 0歐元－ 1,000,000歐元 29 829 858 11 139 150 1,000,001歐元－ 1,500,000歐元 20 150 170 19 44 63 1,500,001歐元－ 2,000,000歐元 10 54 64 9 33 42 2,000,001歐元－ 2,500,000歐元 13 23 36 6 19 25 2,500,001歐元－ 3,000,000歐元 10 12 22 7 16 23 3,000,001歐元－ 3,500,000歐元 6 7 13 4 10 14 3,500,001歐元－ 4,000,000歐元 3 3 6 2 1 3 4,000,001歐元－ 4,500,000歐元 2 1 3 3 1 4 4,500,001歐元－ 5,000,000歐元 2 1 3 3 － 3 5,000,001歐元－ 6,000,000歐元 1 － 1 － 1 1 6,000,001歐元－ 7,000,000歐元 － － － － － － 7,000,001歐元－ 8,000,000歐元 1 － 1 1 － 1 8,000,001歐元－ 9,000,000歐元 1 － 1 1 － 1 1 列表根據資本規定規例第450條以歐元編製，乃採用歐洲委員會就其網站上刊發有關呈報年度12月財務 規劃及預算頒布的匯率計算。 董事之責任聲明 滙豐控股有限公司 328 聲明 本聲明應與第329至333頁之核數師報告所載之核數師責任聲明一併閱讀。本聲明旨在向股 東清楚說明董事與核數師對財務報表分別承擔之責任。董事有責任根據適用法律及法規， 編製包括滙豐控股及其附屬公司 （ 「集團」 ） 之綜合財務報表之 《2014年報及賬目》 ，以及滙豐控 股 （ 「母公司」 ） 之控股公司財務報表。於批准本報告當日每位董事均確認，就其本人所悉，並 無任何相關審計資料未為核數師所知，且其本人已採取一切作為董事應採取的步驟，以獲 取任何相關審計資料，並確保核數師知悉該等資料。此項確認乃根據英國 《2006年公司法》 第418條作出，並根據該公司法的條文詮釋和受其規限。 根據公司法，董事會必須於每個財政年度編製集團及母公司財務報表。董事會須根據歐盟 正式通過之IFRS編製集團財務報表，並已選擇按相同基準編製母公司之財務報表。 根據法律及歐盟正式通過之IFRS，集團及母公司之財務報表必須公平呈列集團及母公司之 財政狀況、於相關期間之業績表現及就IFRS而言彼等之現金流。英國 《2006年公司法》 規定， 就該等財務報表而言，公司法相關部分之中述及應提供真實公平意見之財務報表，乃指該 等報表應達致公平之呈列。此外，為符合美國若干規定，滙豐須根據國際會計準則委員會 （ 「IASB」 ） 頒布之IFRS呈列其財務報表。目前，就適用於滙豐的IFRS而言，經歐盟正式通過之IFRS 與由IASB頒布之IFRS並無差異。 在編製集團及母公司各自之財務報表時，董事須： ‧ 選擇合適之會計政策並貫徹應用； ‧ 作出合理及審慎之判斷及估算； ‧ 說明該等財務報表是否根據歐盟正式通過之IFRS編製；及 ‧ 以持續經營基準編製財務報表，除非不適合假設集團及母公司將持續經營業務。由於董 事會信納集團及母公司於可見將來擁有資源持續經營業務，故財務報表將繼續以持續經 營基準編製。 董事有責任保存充分之賬目紀錄，足以反映及說明母公司之交易，以及可隨時相當準確地 披露母公司之財務狀況，並能令董事確保其財務報表符合英國 《2006年公司法》 之規定。董 事須負一般責任採取可供彼等合理選擇之措施，以保障集團之資產，並防止及查察詐騙及 其他不正常情況。 董事有責任根據適用法律及法規編製載於本 《2014年報及賬目》 第1至327頁之策略報告、董事 會報告、董事薪酬報告及企業管治報告，並有責任管理公司網站所載之 《2014年報及賬目》 及 確保其完整性。規管財務報表之編製及發布之英國法律，可能與其他司法管轄區之法律有 差異。 各董事 （其姓名載於 《2014年報及賬目》 第264至268頁 「董事會報告：企業管治」 一節） 均確認： ‧ 盡其所知，已根據IASB頒布及經歐盟正式通過之IFRS編製的綜合財務報表，乃根據適用的 會計準則編製，並真實公平地反映母公司及經綜合計算之旗下各項業務的整體資產、負 債、財務狀況及損益； ‧ 盡其所知，董事會報告所代表的管理層報告，已公平檢視母公司及經綜合計算之旗下各 項業務的整體業務發展、業績表現及狀況，並同時說明其面對的主要風險及不明朗因 素；及 ‧ 彼等認為 《2014年報及賬目》 整體而言屬公平、公正及容易明白，並能為股東提供必要資 料，以評估母公司之業績、業務模式及策略。 代表董事會 集團主席 范智廉 2015年2月23日 獨立核數師僅致滙豐控股有限公司股東之報告 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 329 本核數師的審計意見及結論 我們就財務報表發表的意見為無保留意見 我們已審計第335至457頁所載滙豐控股有限公司截至2014年12月31日止年度的財務報表。 我們認為： ‧ 財務報表真實而公平地反映集團及母公司於2014年12月31日的狀況及集團截至該日止年 度的利潤； ‧ 集團財務報表已根據歐洲聯盟採納的 《國際財務報告準則》 （ 「歐盟採納的IFRS」 ） 適當地編 製； ‧ 母公司財務報表已根據歐盟採納的IFRS及英國 《2006年公司法》 的適用條文適當地編製；及 ‧ 財務報表已根據英國 《2006年公司法》 的規定編製，而集團財務報表則已根據IAS規例 第4條編製。 對重大錯誤陳述風險的評估 我們於達致上述審計意見時，所採取的策略為凡於財務報表中發現有較高風險作出重大錯 誤陳述的地方加強審計程序。 我們於進行風險評估時，會考慮集團和母公司面對的內在風險，包括由各業務模式產生的 風險及集團如何監控該等風險。在此過程中，我們考慮了若干因素，包括：集團的持續經營 能力、詐騙風險、集團為監控風險而制訂和實行的措施及管理層凌駕主要監控措施的風險。 我們測試集團多項主要監控措施的運作成效，包括財務報告的內部監控及具體的反詐騙監 控措施，並對持續經營假設基準進行測試，然後再重新檢視我們的風險評估工作。我們亦 已考慮於編製財務報表時各董事有必要作出判斷及適當披露該等判斷。 經上述評估，對我們的審計工作具有最大影響的重大錯誤陳述最容易出現的地方，是該等 涉及董事須作出重大判斷的範疇，有關風險詳情如下： 風險 我們的回應 貸款減值 請參閱財務報表附註1(k)的關鍵會計估算及判斷、第277至279頁的集團監察委員會報告，以及第111至237頁的 財務回顧中風險一節內經審核部分的信貸風險披露。 貸款減值乃由董事通過判斷及使用非常主 觀的假設後作出估計。 鑑於貸款的重要性 （佔總資產41%） 及相關 估計的不確定性，故貸款減值被視為主要 審計風險。 產生最大不確定性的貸款組合通常是那些 採用綜合模型推算減值的組合，均屬無抵 押或抵押品可能不足。 2014年，我們繼續著重檢視綜合評估減值 的方法，尤其是專注於美國的按揭、工商 金融及環球銀行業務的貸款組合，以及巴 西的個人及企業貸款，主要因其相對規模 或數據和假設出現變動可能產生的影響。 我們亦特別留意那些對演變中或新湧現的 環球經濟趨勢可能較敏感的組合。此外， 我們亦留意那些繼續或已出現，或有可能 個別減值的個別重大風險承擔。 我們的審計程序包括評估對批核、紀錄及 監察貸款所採取的監控措施，以及評估集 團計算綜合評估減值時所用的方法、數據 及假設，以及評估就個別評估貸款所提撥 的減值準備是否足夠。 我們將集團用於綜合及個別評估減值準備 的假設與從外部取得的行業、財務及經濟 數據及我們就相關主要數據的自行評估進 行比較。在比較的過程中，我們會嚴格評 估集團對估計及假設的修訂 （尤其是就工 商金融及環球銀行業務組合的減值模型所 輸入的數據） 以及在使用經濟因素、虧損出 現期間及過往拖欠率觀察期時所作出判斷 是否一致。我們特別挑選了幾項須進行個 別減值評估的貸款，並集中於那些可能對 財務報表產生最重大影響的貸款，我們尤 其質疑集團有關預期日後現金流的假設， 包括根據我們本身的了解及公開市場資料 而得出的可變現抵押品的價值。 我們亦評估財務報表的披露是否適當反映 集團的信貸風險，具體考慮2014年已認定 出現最大風險的該等組合。 獨立核數師僅致滙豐控股有限公司股東之報告 （續） 報告 滙豐控股有限公司 330 風險 我們的回應 訴訟、監管機構採取的行動及與客戶有關的補救措施 請參閱關鍵會計估算及判斷、財務報表附註29、37及40的準備及或有負債披露，以及第277至279頁的集團監 察委員會報告。 就訴訟、監管機構採取的行動及與客戶有 關的補救措施 （統稱 「法律及監管事宜」 ） 確 認和計量準備以及計量和披露相關或有負 債時，須作出重大判斷。鑑於該等事宜的 重要性，以及難以評估及計量任何引致的 責任所涉及的數額，故被視為主要審計風 險。 2014年，我們特別留意那些於期內有顯著進 展或新出現的重大事宜。尤其集中關注英 國客戶賠償計劃及匯兌事宜，以及法國、 美國及瑞士的其他稅務及監管事宜。 我們的審計程序包括評估就識別、評估及 計量因法律及監管事宜引致的潛在責任而 採取的監控措施。 就已識別的事宜，我們根據取得的事實及 情況，考慮是否存在責任、所提撥的準備 及╱或所作披露是否適當。為釐定事實及 情況，我們取得相關監管及訴訟文件，並 進行了評估，同時會見了集團的內部及外 聘法律顧問。我們亦評估了所作的假設及 主要判斷，更根據我們自身經驗及掌握的 市場資訊考慮了可能出現的其他結果。 此外，我們還考慮了集團就估計準備及或 有負債時所作出的判斷的披露是否已充分 反映與法律及監管事宜相關的不確定性。 風險 我們的回應 金融工具的估值 請參閱關鍵會計估算及判斷、財務報表附註12至16、18、24及25的公允值披露、第277至279頁的集團監察委員 會報告，以及第111至237頁的財務回顧中風險一節內經審核部分的市場風險披露。 金融工具的公允值乃透過採用通常涉及董 事所作判斷的估值方法及使用假設及估計 來釐定。 鑑於金融工具的重要性及相關估計的不確 定性，故被視為主要審計風險。於2014年 12月31日，以公允值列賬之金融資產佔資 產總值40%，而以公允值列賬之金融負債 佔負債總額25%。 那些重大估值數據屬不可觀察的金融工具 （即第三級工具） ，其估計的不確定性特別 高。於2014年12月31日，第三級工具佔以 公允值列賬之金融資產1.4%，並佔以公允 值列賬之金融負債1.3%。 2014年，我們繼續集中留意衍生工具公允 值計量方法的發展，尤其是集團就計量無 抵押衍生工具所採用的資金公允值調整。 我們的審計程序包括評估就識別、計量及 管理估值風險所採取的監控措施，以及評 估集團用以釐定公允值的方法、數據及假 設。 就集團的公允值模型而言，我們評估該等 模型及數據是否合適。我們將可觀察數據 與來自獨立源頭及從外部取得的市場數據 進行比較。 我們選取了一些有重大不可觀察估值數據 的工具，在我們內部的估值專家的協助下， 嚴格評估了所用的假設及模型，又或通過 參考我們認為可用的替代方法及主要因素 的敏感度，重新進行獨立估值評估。 我們亦評估了集團在釐定其就無抵押衍生 工具組合所錄得的資金公允值調整時所採 用的方法及數據，並根據我們與類似機構 接觸所認識到的當前市場慣例進行比較。 此外，我們還評估了財務報表就公允值風 險及敏感度所作的披露是否已適當反映集 團所承擔的估值風險。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 331 風險 我們的回應 遞延稅項資產 請參閱關鍵會計估算及判斷、財務報表附註8的遞延稅項披露，以及第277至279頁的集團監察委員會報告。 遞延稅項資產的確認，依賴董事就下列事 項顯著運用判斷：評估日後出現應課稅利 潤的可能性及該等利潤的充足程度、目前 存在的應課稅暫時差異日後的撥回額以及 持續執行的稅務規劃策略。 鑑於集團已確認遞延稅項資產的規模 （74億美元） 及可否收回該等資產涉及的不 確定性，故該等資產被視為主要審計風險。 2014年，我們繼續集中留意來自美國、巴西 及墨西哥業務的最重要遞延稅項資產。我 們尤其關注兩大方面：美國的稅務規劃策 略，該策略繼續依賴母公司的資本支持； 以及管理層對巴西的日後盈利能力所作的 預測，這些預測為遞延稅項資產提供支持。 我們的審計程序包括評估就確認及計量遞 延稅項資產所採取的監控措施，以及評估 用以預測集團在有關司法管轄區的日後應 課稅利潤的假設。我們亦透過評估將採用 的預期稅務規劃策略以及能否取得整體支 持已確認遞延稅項資產進行變現的資金， 質疑集團就繼續在美國投入足夠資本所作 的假設及承諾。 我們將集團用以預測日後利潤所用的主要 數據與從外部取得的數據 （如經濟預測） 及 集團本身的過往數據及業績表現作比較， 並評估結果對各種假設的合理可能變動的 敏感度。我們本身的稅務專家亦嚴格評估 了日後稅務規劃策略的合適程度。 此外，我們還評估了集團就估算已確認及 未確認的遞延稅項資產款額時所作出的判 斷的披露是否適當反映集團的遞延稅項狀 況。 風險 我們的回應 商譽減損 請參閱關鍵會計估算及判斷、財務報表附註21的商譽披露，以及第277至279頁的集團監察委員會報告。 創現單位 （ 「CGU」 ） 的商譽減損測試須依賴 根據估計日後現金流而得出的使用價值估 計數額。 鑑於日後現金流的預測及折現存在不確定 性及集團已確認商譽 （192億美元） 的重要 性，故商譽減損被視為重大風險。 倘創現單位的使用價值與賬面值之間的差 額有限，又或其使用價值對估計日後現金 流非常敏感，則上述不確定性一般處於最 高水平。 2014年，我們留意對日後現金流預測的敏 感度及倚賴程度最高的創現單位，並因近 期過往表現而預期將減少可用額度 （歐洲 環球私人銀行業務尤其如此） 。 我們的審計程序包括評估就集團確認及計 量商譽減損程序所採取的監控措施 （包括 所用假設） 。我們亦測試了構成集團計算使 用價值的基礎的各項主要假設 （包括現金 流預測及折現率） 。 我們評估了現金流預測的合理性，並將主 要數據 （如折現率及增長率） 與從外部取得 的行業、經濟及財務數據，以及集團本身 的過往數據及業績表現作比較。我們在本 身的專家協助下，嚴格評估了用以預測若 干創現單位使用價值的假設及方法，而該 等創現單位的很大部分商譽是對該等假設 的變動敏感。在整體基礎上，我們亦評估 了集團釐定的使用價值總額佔其外部市值 的比率。 此外，我們還審視了集團就估計創現單位 現金流時所作判斷及該等估計結果的敏感 度的披露是否已充分反映相關商譽減損的 風險。 獨立核數師僅致滙豐控股有限公司股東之報告 （續） 報告 滙豐控股有限公司 332 風險 我們的回應 於聯營公司之權益 請參閱關鍵會計估算及判斷、財務報表附註20的於聯營公司之權益披露，以及第277至279頁的集團監察委員 會報告。 滙豐於聯營公司之大部分權益主要有關其 於交通銀行股份有限公司 （ 「交通銀行」 ） 的 19.03%權益。交通銀行於香港及上海證券 交易所上市。 按聯營公司以權益法入賬的會計方式，該 等權益於首次列賬時是按成本入賬，其後 則按收購後滙豐佔該聯營公司資產淨值之 變動減任何減值準備予以調整。 交通銀行的市值長時間低於其賬面值，因 此，其當前賬面值 （146億美元） 繼續取決於 集團在根據交通銀行的使用價值以釐定其 可收回金額時作出的重大判斷。 集團於釐定交通銀行使用價值時採用的預 計日後現金流及折現率會受到估計不明朗 因素及敏感度的影響，我們因此認為這是 一項主要審計風險。 我們的審計程序包括評估集團計算交通銀 行的使用價值時所採用的方法和實際過 程。我們依據交通銀行最近期的財務表現 來評估就現金流所作預測的合理程度，並 考慮所用的主要數據如用以推斷該等現金 流的長期增長率、折現率、風險加權資產 對總資產比率及貸款減值準備對總貸款比 率的合適性，並將該等數據與可取得之行 業、經濟及財務數據以及市場上大多數的 預測作比較。 我們與交通銀行管理層會面以了解該行目 前業務表現及預測，以及集團本身的假設 是否已妥善反映這些表現及預測。我們亦 將計算使用價值的結果，與市場上可取得 的交通銀行及中國內地其他上市銀行的市 盈率作比較，並評估集團對其於交通銀行 的權益的市值與使用價值之間差額所作的 分析。此項評估包括考慮其他市場參與者 採用的估值方法及假設。 此外，我們還考慮了集團就估計可收回金 額時所作的判斷及該等估計結果的敏感度 的披露，能否充分反映聯營公司權益減值 的風險。 我們如何應用重要性原則及審計範圍的概述 集團整體財務報表的重要性水平定為9.3億美元，乃參考集團除稅前利潤的基準釐定，並撇 除信貸息差 （ 「本身信貸息差」 ） 應佔長期債務的公允值變動以取得正常數值，我們認為這是 公司股東評估財務表現時考慮的主要因素之一。重要性水平為集團除稅前利潤的5%，而若 按經調整以撇除本身信貸息差後的除稅前利潤計算，則為5.1%。 我們向集團監察委員會呈報任何數值超過4,500萬美元的已識別錯誤陳述 （已糾正或未糾正） ， 以及基於質量的理由而應當呈報的其他已識別錯誤陳述。 我們劃分組成部分之範圍的方法為：對於部分地區，我們通知地區審計團隊進行全面審計 並向我們匯報，並指定該等地區須由有關地區審計團隊監督進行全面審計的若干組成部分； 在其他地區，我們直接通知組成部分審計團隊進行全面審計並向我們匯報。因此，我們就 下列全部五大地區23個組成部分通知或指定進行全面審計： ‧ 歐洲 （7個組成部分） ‧ 亞洲 （8個組成部分） ‧ 中東及北非 （1個組成部分） ‧ 北美洲 （4個組成部分） ‧ 拉丁美洲 （3個組成部分） 該等審計涵蓋集團營業收益總額的82%；構成集團除稅前利潤的損益總額的78%；以及集團 資產總值的88%。附註11的按類披露載列每個地區的各自重要性。 我們已批准經我們直接通知進行的該等全面審計的重要性水平，由5,000萬美元至7.5億美元 不等，有關水平乃經考慮集團各組成部分的規模和風險狀況兩者組合的影響而釐定。 集團審計團隊曾到訪歐洲、亞洲、北美洲及拉丁美洲的多個經營點，而每個地區的組成部 分都派出團隊參加了一次集團審計規劃會議。集團審計團隊亦定期與上述地區審計團隊及 其他組成部分的核數師舉行電話會議。此外，地區審計團隊也曾到訪其負責地區內各主要 組成部分的經營點。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 333 我們就國際會計準則委員會所頒布IFRS另行發表的意見為無保留意見 誠如集團財務報表附註1(a)已闡釋，集團除為履行其法律責任而應用歐盟採納的IFRS外，亦 應用IASB頒布的IFRS。 我們認為，集團的財務報表符合IASB頒布的IFRS規定。 我們就有關英國 《2006年公司法》 訂明的其他事項的意見為無保留意見 我們認為： ‧ 須予審計的董事薪酬報告部分已根據英國 《2006年公司法》 適當地編製；及 ‧ 就財務報表相關財政年度而編製的策略報告及董事會報告內所提供的資料與財務報表 的資料一致。 現時並無出現狀況，致使須作出例外報告 根據國際審計準則 （英國及愛爾蘭） ，倘若我們根據於審計期間所獲取的資訊，發現年報內 載有其他資料與上述所獲取的資訊或財務報表存在重大分歧、或是對事實有重大錯誤陳述 或屬誤導性質，則我們須向閣下報告。 尤其是下列情況，我們必須向閣下報告： ‧ 如發現在審計過程中獲得的資訊，與董事會發出的聲明，即聲稱彼等認為有關年報及財 務報表整體屬公平、公正及容易明白，並能為股東提供必要資料以評估集團之業績、業 務模式及策略的陳述存在重大分歧；或 ‧ 年報及賬目所載的企業管治部分，所說明的集團監察委員會工作的內容並未有適當回應 我們向集團監察委員會通報的各個事項。 根據英國 《2006年公司法》 ，我們如有下列意見，必須向閣下報告： ‧ 母公司並未備存足夠的會計紀錄，或我們並無造訪的分行未有提供讓我們可以進行充分 審計的報表；或 ‧ 母公司的財務報表及董事薪酬報告中須審計的部分，與會計紀錄及報表不相符；或 ‧ 並無披露若干法律訂明須予披露的董事薪酬資料；或 ‧ 我們並未獲得執行審計工作所需的一切資料及闡釋。 根據上市規則規定，我們必須審閱： ‧ 載於第290及291頁董事會有關持續經營的聲明；及 ‧ 指明須由我們審閱 《企業管治聲明》 中有關公司遵從2012年版英國 《企業管治守則》 十項規 定的情況的部分。 就上述責任而言，我們並無事宜需要報告。 報告的範圍及責任 誠如第328頁董事之責任聲明已詳細闡釋，董事須負責編製財務報表並確信財務報表呈列真 實而公平的意見。 有關財務報表審計範圍的說明，可於英國財務報告評議會網站 auditscopeukprivate查閱。 本報告僅提呈予公司全體股東，並須受我們的網站 所載有關我們責任的重要解釋及免責聲明規限。該等解釋及免責聲明已納入本報告，如同 已全部載列於報告之內。務請詳加閱讀有關內容，以便理解本報告的目的、我們執行的工 作及我們提供意見的依據。 代表法定核數師 KPMG Audit Plc Guy Bainbridge （高級法定核數師） 特許會計師 15 Canada Square London E14 5GL 2015年2月23日 財務報表 綜合收益表 滙豐控股有限公司 334 財務報表 綜合收益表 335 綜合全面收益表 336 綜合資產負債表 337 綜合現金流量表 338 綜合股東權益變動表 339 滙豐控股之資產負債表 341 滙豐控股之現金流量表 342 滙豐控股之股東權益變動表 343 財務報表附註 1 編製基準 345 2 指定以公允值列賬之金融工具 淨收益╱ （支出） 354 3 保費收益淨額 354 4 已支付保險賠償和利益及 投保人負債之變動淨額 355 5 營業利潤 356 6 僱員報酬及福利 356 7 核數師費用 364 8 稅項 365 9 股息 370 10 每股盈利 371 11 按類分析 371 12 交易用途資產 377 13 按公允值列賬之金融工具的 公允值 378 14 非按公允值列賬之金融工具的 公允值 390 15 指定以公允值列賬之金融資產 392 16 衍生工具 394 17 非交易用途反向回購及回購協議 398 18 金融投資 399 19 作為負債擔保而質押之資產、 已轉讓之資產及持作資產擔保 之抵押品 401 20 於聯營及合資公司之權益 403 21 商譽及無形資產 407 22 於附屬公司之投資 413 23 預付款項、應計收益及其他資產 416 24 交易用途負債 417 25 指定以公允值列賬之 金融負債 417 26 已發行債務證券 418 27 應計項目、遞延收益及其他負債 418 28 保單未決賠款 419 29 準備 420 30 後償負債 423 31 資產、負債及資產負債表外承諾 之期限分析 426 32 對銷金融資產及金融負債 434 33 匯兌風險 435 34 非控股股東權益 436 35 已催繳股本及其他股權工具 437 36 現金流量表說明 439 37 或有負債、合約承諾及擔保 441 38 租賃承諾 442 39 結構公司 443 40 法律訴訟及監管事宜 446 41 關連人士交易 455 42 結算日後事項 457 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 335 綜合收益表 截至2014年12月31日止年度 2014年 2013年 2012年 附註 百萬美元 百萬美元 百萬美元 利息收益 50,955 51,192 56,702 利息支出 (16,250) (15,653) (19,030) 淨利息收益 34,705 35,539 37,672 費用收益 19,545 19,973 20,149 費用支出 (3,588) (3,539) (3,719) 費用收益淨額 15,957 16,434 16,430 不包括淨利息收益之交易收益 4,853 6,643 4,408 交易活動之淨利息收益 1,907 2,047 2,683 交易收益淨額 6,760 8,690 7,091 已發行長期債務及相關衍生工具之公允值變動 508 (1,228) (4,327) 指定以公允值列賬之其他金融工具淨收益 1,965 1,996 2,101 指定以公允值列賬之金融工具淨收益╱ （支出） 2 2,473 768 (2,226) 金融投資減除虧損後增益 1,335 2,012 1,189 股息收益 311 322 221 保費收益淨額 3 11,921 11,940 13,044 出售美國分行網絡、美國卡業務及中國平安保險 （集團） 股份有限公司所得利潤 － － 7,024 其他營業收益 1,131 2,632 2,100 營業收益總額 74,593 78,337 82,545 已支付保險賠償和利益及投保人 負債之變動淨額 4 (13,345) (13,692) (14,215) 未扣除貸款減值及其他信貸風險準備之營業收益淨額 61,248 64,645 68,330 貸款減值及其他信貸風險準備 5 (3,851) (5,849) (8,311) 營業收益淨額 57,397 58,796 60,019 僱員報酬及福利 6 (20,366) (19,196) (20,491) 一般及行政開支 (18,565) (17,065) (19,983) 物業、機器及設備折舊與減值 (1,382) (1,364) (1,484) 無形資產攤銷及減值 21 (936) (931) (969) 營業支出總額 (41,249) (38,556) (42,927) 營業利潤 5 16,148 20,240 17,092 應佔聯營及合資公司利潤 20 2,532 2,325 3,557 除稅前利潤 18,680 22,565 20,649 稅項支出 8 (3,975) (4,765) (5,315) 本年度利潤 14,705 17,800 15,334 母公司股東應佔利潤 13,688 16,204 14,027 非控股股東應佔利潤 1,017 1,596 1,307 美元 美元 美元 每股普通股基本盈利 10 0.69 0.84 0.74 每股普通股攤薄後盈利 10 0.69 0.84 0.74 第345至457頁的相關附註構成本期財務報表一部分 1。 有關註釋，請參閱第344頁。 財務報表 （續） 滙豐控股有限公司 336 綜合全面收益表 截至2014年12月31日止年度 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 本年度利潤 14,705 17,800 15,334 其他全面收益╱ （支出） 符合特定條件後，將重新分類至損益賬的項目： 可供出售投資2 2,972 (1,718) 5,070 －公允值增益╱ （虧損） 4,794 (1,787) 6,396 －重新分類至收益表之公允值增益 (1,672) (1,277) (1,872) －重新分類至收益表之減值虧損額 374 286 1,002 －所得稅 (524) 1,060 (456) 現金流對沖 188 (128) 109 －公允值增益 1,512 776 552 －重新分類至收益表之公允值增益 (1,244) (894) (423) －所得稅 (80) (10) (20) 應佔聯營及合資公司其他全面收益╱ （支出） 80 (71) 533 －年內應佔 78 (35) 311 －出售後重新分類納入收益表 2 (36) 222 匯兌差額 (8,903) (1,372) 1,017 －出售一項海外業務後重新分類納入 收益表的匯兌增益 (21) (290) (1,128) －其他匯兌差額 (8,917) (1,154) 2,145 －因匯兌差額產生的所得稅 35 72 － 其後不會重新分類至損益賬的項目： 重新計量界定福利資產╱負債 1,985 (458) (195) －未扣所得稅 2,419 (601) (391) －所得稅 (434) 143 196 本年度其他全面收益 （除稅淨額） (3,678) (3,747) 6,534 本年度全面收益總額 11,027 14,053 21,868 應佔： －母公司股東 9,245 12,644 20,455 －非控股股東 1,782 1,409 1,413 本年度全面收益總額 11,027 14,053 21,868 第345至457頁的相關附註構成本期財務報表一部分 1。 有關註釋，請參閱第344頁。 綜合全面收益表╱綜合資產負債表 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 337 綜合資產負債表 於2014年12月31日 2014年 2013年 附註 百萬美元 百萬美元 資產 現金及於中央銀行的結餘 129,957 166,599 向其他銀行託收中之項目 4,927 6,021 香港政府負債證明書 27,674 25,220 交易用途資產 12 304,193 303,192 指定以公允值列賬之金融資產 15 29,037 38,430 衍生工具 16 345,008 282,265 同業貸款3 112,149 120,046 客戶貸款3 974,660 992,089 反向回購協議－非交易用途 17 161,713 179,690 金融投資 18 415,467 425,925 預付款項、應計收益及其他資產 23 75,176 76,842 當前稅項資產 1,309 985 於聯營及合資公司之權益 20 18,181 16,640 商譽及無形資產 21 27,577 29,918 遞延稅項資產 8 7,111 7,456 於12月31日之資產總值 2,634,139 2,671,318 負債及股東權益 負債 香港紙幣流通額 27,674 25,220 同業存放3 77,426 86,507 客戶賬項3 1,350,642 1,361,297 回購協議－非交易用途 17 107,432 164,220 向其他銀行傳送中之項目 5,990 6,910 交易用途負債 24 190,572 207,025 指定以公允值列賬之金融負債 25 76,153 89,084 衍生工具 16 340,669 274,284 已發行債務證券 26 95,947 104,080 應計項目、遞延收益及其他負債 27 53,396 52,341 當前稅項負債 1,213 607 保單未決賠款 28 73,861 74,181 準備 29 4,998 5,217 遞延稅項負債 8 1,524 910 後償負債 30 26,664 28,976 於12月31日之負債總額 2,434,161 2,480,859 股東權益 已催繳股本 35 9,609 9,415 股份溢價賬 11,918 11,135 其他股權工具 11,532 5,851 其他儲備 20,244 26,742 保留盈利 137,144 128,728 股東權益總額 190,447 181,871 非控股股東權益 34 9,531 8,588 於12月31日之各類股東權益總額 199,978 190,459 於12月31日之各類負債及股東權益總額 2,634,139 2,671,318 第345至457頁的相關附註構成本期財務報表一部分 1。 有關註釋，請參閱第344頁。 集團主席 范智廉 財務報表 （續） 滙豐控股有限公司 338 綜合現金流量表╱綜合股東權益變動表 綜合現金流量表 截至2014年12月31日止年度 2014年 2013年 2012年 附註 百萬美元 百萬美元 百萬美元 營業活動產生之現金流 除稅前利潤 18,680 22,565 20,649 調整下列各項： －投資活動增益淨額 (1,928) (1,458) (2,094) －應佔聯營及合資公司利潤 (2,532) (2,325) (3,557) －出售聯營公司、合資公司、附屬公司及 業務所得 （利潤） ╱虧損 9 (1,173) (7,024) －除稅前利潤包含之其他非現金項目 36 11,262 11,995 19,778 －營業資產之變動 36 25,877 (148,899) (116,521) －營業負債之變動 36 (93,814) 164,757 89,070 －撇銷匯兌差額4 24,571 4,479 (3,626) －已收取聯營公司之股息 757 694 489 －已支付之界定福利計劃供款 (681) (962) (733) －已付稅款 (3,573) (4,696) (5,587) 營業活動產生╱ （所用） 之現金淨額 (21,372) 44,977 (9,156) 投資活動產生之現金流 購入金融投資 (384,199) (363,979) (342,974) 出售金融投資及金融投資到期所得款項 382,837 342,539 329,926 購入物業、機器及設備 (1,477) (1,952) (1,318) 出售物業、機器及設備所得款項 88 441 241 因出售客戶及貸款組合而流入╱ （流出） 之現金淨額 (1,035) 6,518 － 購入無形資產淨值 (903) (834) (1,008) 因出售美國分行網絡及美國卡業務而流入之現金淨額 － － 20,905 出售平安保險所得款項 － 7,413 1,954 因出售其他附屬公司、業務、聯營公司及合資公司 而流入╱ （流出） 之現金淨額 (242) 3,295 (269) 因收購或增持聯營公司股權而流出之現金淨額 (30) (26) (1,804) 投資活動產生╱ （所用） 之現金淨額 (4,961) (6,585) 5,653 融資活動產生之現金流 發行普通股股本 267 297 594 因進行市場莊家活動及投資而出售╱ （購入） 之本身股份淨額 (96) (32) (25) 發行其他股權工具 5,681 － － 贖回優先股份 (234) － － 已發行之後償借貸資本 3,500 1,989 37 已償還之後償借貸資本 (3,163) (1,662) (1,754) 因附屬公司股權變動而流入╱ （流出） 之現金淨額 － － (14) 已付予母公司股東之股息 (6,611) (6,414) (5,925) 已付予非控股股東之股息 (639) (586) (572) 已付予其他股權工具持有人之股息 (573) (573) (573) (573) (573) (573) 融資活動所用之現金淨額 (1,868) (6,981) (8,232) 現金及等同現金項目淨增額╱ （減額） (28,201) 31,411 (11,735) 於1月1日之現金及等同現金項目 346,281 315,308 325,449 現金及等同現金項目之匯兌差額 (16,779) (438) 1,594 (573) (573) (573) 於12月31日之現金及等同現金項目 36 301,301 346,281 315,308 第345至457頁的相關附註構成本期財務報表一部分 1。 有關註釋，請參閱第344頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 339 其他儲備 其他 可供出售 現金流 股東 非控股 各類股東 已催繳 股份 股權 保留 公允值 對沖 匯兌 合併 權益 股東 權益 股本 溢價 工具 11 盈利 5,6 儲備 儲備 儲備 儲備 5,7 總額 權益 總額 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日 9,415 11,135 5,851 128,728 97 (121) (542) 27,308 181,871 8,588 190,459 本年度利潤 － － － 13,688 － － － － 13,688 1,017 14,705 其他全面收益 （除稅淨額） － － － 2,066 2,025 189 (8,723) － (4,443) 765 (3,678) －可供出售投資 － － － － 2,025 － － － 2,025 947 2,972 －現金流對沖 － － － － － 189 － － 189 (1) 188 －重新計量界定福利資產╱負債 － － － 1,986 － － － － 1,986 (1) 1,985 －應佔聯營及合資公司之其他全面收益 － － － 80 － － － － 80 － 80 －匯兌差額 － － － － － － (8,723) － (8,723) (180) (8,903) 本年度全面收益總額 － － － 15,754 2,025 189 (8,723) － 9,245 1,782 11,027 根據僱員薪酬及股份計劃發行之股份 60 917 － (710) － － － － 267 － 267 發行代息股份及因而產生之金額 134 (134) － 2,709 － － － － 2,709 － 2,709 巳發行之資本證券 － － 5,681 － － － － － 5,681 － 5,681 向股東派發之股息8 － － － (9,893) － － － － (9,893) (712) (10,605) 以股份為基礎的支出安排成本 － － － 732 － － － － 732 － 732 其他變動 － － － (176) 21 (10) － － (165) (127) (292) 於2014年12月31日 9,609 11,918 11,532 137,144 2,143 58 (9,265) 27,308 190,447 9,531 199,978 於2013年1月1日 9,238 10,084 5,851 120,347 1,649 13 752 27,308 175,242 7,887 183,129 本年度利潤 － － － 16,204 － － － － 16,204 1,596 17,800 其他全面收益 （除稅淨額） － － － (561) (1,577) (128) (1,294) － (3,560) (187) (3,747) －可供出售投資 － － － － (1,577) － － － (1,577) (141) (1,718) －現金流對沖 － － － － － (128) － － (128) － (128) －重新計量界定福利資產╱負債 － － － (490) － － － － (490) 32 (458) －應佔聯營及合資公司之其他全面收益 － － － (71) － － － － (71) － (71) －匯兌差額 － － － － － － (1,294) － (1,294) (78) (1,372) 本年度全面收益總額 － － － 15,643 (1,577) (128) (1,294) － 12,644 1,409 14,053 根據僱員薪酬及股份計劃發行之股份 60 1,168 － (931) － － － － 297 － 297 發行代息股份及因而產生之金額 117 (117) － 2,523 － － － － 2,523 － 2,523 向股東派發之股息8 － － － (9,510) － － － － (9,510) (718) (10,228) 以股份為基礎的支出安排成本 － － － 630 － － － － 630 － 630 其他變動 － － － 26 25 (6) － － 45 10 55 於2013年12月31日 9,415 11,135 5,851 128,728 97 (121) (542) 27,308 181,871 8,588 190,459 綜合股東權益變動表 截至2014年12月31日止年度 財務報表 （續） 滙豐控股有限公司 340 綜合股東權益變動表╱滙豐控股之資產負債表 其他儲備 其他 可供出售 現金流 股東 非控股 各類股東 已催繳 股份 股權 保留 公允值 對沖 匯兌 合併 權益 股東 權益 股本 溢價 工具 盈利 5,6 儲備 儲備 儲備 儲備 5,7 總額 權益 總額 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2012年1月1日 8,934 8,457 5,851 111,868 (3,361) (95) (237) 27,308 158,725 7,368 166,093 本年度利潤 － － － 14,027 － － － － 14,027 1,307 15,334 其他全面收益 （除稅淨額） － － － 321 5,010 108 989 － 6,428 106 6,534 －可供出售投資 － － － － 5,010 － － － 5,010 60 5,070 －現金流對沖 － － － － － 108 － － 108 1 109 －重新計量界定福利資產╱負債 － － － (212) － － － － (212) 17 (195) －應佔聯營及合資公司之其他全面收益 － － － 533 － － － － 533 － 533 －匯兌差額 － － － － － － 989 － 989 28 1,017 本年度全面收益總額 － － － 14,348 5,010 108 989 － 20,455 1,413 21,868 根據僱員薪酬及股份計劃發行之股份 119 1,812 － (1,337) － － － － 594 － 594 發行代息股份及因而產生之金額 185 (185) － 2,429 － － － － 2,429 － 2,429 向股東派發之股息8 － － － (8,042) － － － － (8,042) (707) (8,749) 以股份為基礎的支出安排成本 － － － 988 － － － － 988 － 988 其他變動 － － － 93 － － － － 93 (187) (94) 於2012年12月31日 9,238 10,084 5,851 120,347 1,649 13 752 27,308 175,242 7,887 183,129 第345至457頁的相關附註構成本期財務報表一部分 1。 有關註釋，請參閱第344頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 341 滙豐控股之資產負債表 於2014年12月31日 2014年 2013年 附註 百萬美元 百萬美元 資產 銀行及庫存現金： －在滙豐旗下業務之結餘 249 407 衍生工具 16 2,771 2,789 滙豐旗下業務貸款 43,910 53,344 滙豐旗下業務金融投資 4,073 1,210 預付款項、應計收益及其他資產 125 133 本期稅項資產 472 245 於附屬公司之投資 22 96,264 92,695 遞延稅項資產 8 － 13 於12月31日之資產總值 147,864 150,836 負債及股東權益 負債 應付滙豐旗下業務款項 2,892 11,685 指定以公允值列賬之金融負債 25 18,679 21,027 衍生工具 16 1,169 704 已發行債務證券 26 1,009 2,791 應計項目、遞延收益及其他負債 1,398 1,327 本期稅項負債 － 48 遞延稅項負債 8 17 － 後償負債 30 17,255 14,167 負債總額 42,419 51,749 股東權益 已催繳股本 35 9,609 9,415 股份溢價賬 11,918 11,135 其他股權工具 11,476 5,828 其他儲備 37,456 37,303 保留盈利 34,986 35,406 各類股東權益總額 105,445 99,087 於12月31日之各類負債及股東權益總額 147,864 150,836 第345至457頁的相關附註構成本期財務報表一部分 1。 有關註釋，請參閱第344頁。 集團主席 范智廉 財務報表 （續） 滙豐控股有限公司 342 滙豐控股之現金流量表╱股東權益變動表 滙豐控股之現金流量表 截至2014年12月31日止年度 2014年 2013年 附註 百萬美元 百萬美元 營業活動產生之現金流 除稅前利潤 6,228 17,725 調整下列各項： －除稅前利潤包含之非現金項目 36 52 74 －營業資產之變動 36 1,854 (10,795) －營業負債之變動 36 (9,914) (1,061) －已收稅款 133 156 營業活動產生╱ （所用） 之現金淨額 (1,647) 6,099 投資活動產生之現金流 因收購或增持附屬公司股權而流出之現金淨額 (1,603) (665) 附屬公司償還資本 3,505 － 投資活動所用之現金淨額 1,902 (665) 融資活動產生之現金流 發行普通股股本 924 1,192 發行其他股權工具 5,635 － 為發給股份獎勵及認股權獎勵而出售之本身股份 － 44 已發行之後償借貸資本 3,500 1,989 已償還之後償借貸資本 (1,654) (1,618) 已償還債務證券 (1,634) － 普通股之已付股息 (6,611) (6,414) 已付予其他股權工具持有人之股息 (573) (573) 融資活動產生╱ （所用） 之現金淨額 (413) (5,380) 現金及等同現金項目淨增額 (158) 54 於1月1日之現金及等同現金項目 407 353 於12月31日之現金及等同現金項目 36 249 407 第345至457頁的相關附註構成本期財務報表一部分 1。 有關註釋，請參閱第344頁。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 343 滙豐控股之股東權益變動表 截至2014年12月31日止年度 其他儲備 可供出售 其他 已催繳 其他 公允值 繳足股款 合併及 股東 股本 股份溢價 股權工具 11 保留盈利 9 儲備 的股本 10 其他儲備 7 權益總額 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日 9,415 11,135 5,828 35,406 124 2,052 35,127 99,087 本年度利潤 － － － 6,527 － － － 6,527 其他全面收益 （除稅淨額） － － － － 116 － － 116 －可供出售投資 － － － － 152 － － 152 －所得稅 － － － － (36) － － (36) 本年度全面收益總額 － － － 6,527 116 － － 6,643 根據僱員股份計劃發行之股份 60 917 － (53) － － － 924 發行代息股份及因而產生之金額 134 (134) － 2,709 － － － 2,709 巳發行之資本證券 － － 5,648 － － － － 5,648 向股東派發之股息8 － － － (9,893) － － － (9,893) 分派之稅項減免 － － － 104 － － － 104 本身股份調整 － － － 103 － － － 103 行使及失效之認股權 － － － (37) － 37 － － 以股份為基礎的支出安排成本 － － － 74 － － － 74 以股份為基礎的支出之所得稅 － － － (2) － － － (2) 根據僱員股份計劃向附屬公司 僱員授出股權投資 － － － 48 － － － 48 於2014年12月31日 9,609 11,918 11,476 34,986 240 2,089 35,127 105,445 於2013年1月1日 9,238 10,084 5,828 24,707 114 1,929 35,127 87,027 本年度利潤 － － － 17,882 － － － 17,882 其他全面收益 （除稅淨額） － － － － 10 － － 10 －可供出售投資 － － － － 2 － － 2 －所得稅 － － － － 8 － － 8 本年度全面收益總額 － － － 17,882 10 － － 17,892 根據僱員股份計劃發行之股份 60 1,168 － (36) － － － 1,192 發行代息股份及因而產生之金額 117 (117) － 2,523 － － － 2,523 向股東派發之股息8 － － － (9,510) － － － (9,510) 分派之稅項減免 － － － 42 － － － 42 本身股份調整 － － － 222 － － － 222 行使及失效之認股權 － － － (123) － 123 － － 以股份為基礎的支出安排成本 － － － 49 － － － 49 以股份為基礎的支出之所得稅 － － － － － － － － 根據僱員股份計劃向附屬公司 僱員授出股權投資 － － － (350) － － － (350) 於2013年12月31日 9,415 11,135 5,828 35,406 124 2,052 35,127 99,087 於2014年12月31日的每股普通股股息為0.49美元 （2013年：0.48美元；2012年：0.41美元） 。 第345至457頁的相關附註構成本期財務報表一部分 1。 有關註釋，請參閱第344頁。 財務報表 （續） 滙豐控股有限公司 344 財務報表註釋 1 「風險」 （第111至237頁） 的經審核部分及 「資本」 （第238至262頁） 的經審核部分，亦屬本期財務報表的一部分。 2 可供出售投資包括分類為 「持作出售用途資產」 之平安保險的投資為零 （2013年：零；2012年：7.37億美元） 。 3 自2014年1月1日起，非交易用途反向回購及回購於資產負債表分行呈列。過往，非交易用途反向回購載入 「同 業貸款」 及 「客戶貸款」 內，而非交易用途回購則載入 「同業存放」 及 「客戶賬項」 內。比較數據已相應重列。非 交易用途反向回購及回購已於資產負債表分行呈列，使披露方式與市場慣例相符，並提供更具參考價值的貸 款相關資料。反向回購及回購所代表的客戶及同業貸款或客戶及同業提供的放款所涉範圍，載於財務報表附 註17。 4 此項調整乃為使期初與期末資產負債表數額之間的變動按平均匯率計算。由於進行逐項調整的細節安排涉及 不合理支出，故並未按此基準作出調整。 5 有關1998年1月1日之前收購的附屬公司，其累計商譽51.38億美元已於儲備內扣除，包括因收購英國滙豐銀行有 限公司而於合併儲備扣除的34.69億美元。餘額16.69億美元已於保留盈利內扣除。 6 保留盈利包括於滙豐保險業務內持有、代投保人於退休基金內持有或代受益人於僱員信託基金內持有 （以備 根據僱員股份計劃或紅利計劃償付股份時提供股份） ，以及環球資本市場業務的市場莊家活動持有之本身股 份85,337,430股 （6.41億美元） （2013年：85,997,271股 （9.15億美元） ；2012年：86,394,826股 （8.74億美元） ） 。 7 根據英國 《1985年公司法》 （ 「公司法」 ） 第131條之規定，已就於1992年收購英國滙豐銀行有限公司、於2000年收購 法國滙豐及於2003年收購美國滙豐融資有限公司取得法定股份溢價寬免，而發行的股份僅以面值記賬。有關 法國滙豐的82.9億美元公允值差額及有關美國滙豐融資有限公司的127.68億美元公允值差額，已於滙豐綜合財 務報表的合併儲備中確認。在進行連串集團內部重組後，為收購美國滙豐融資有限公司而設立之合併儲備其 後已併入HSBC Overseas Holdings (UK) Limited （ 「HOHU」 ） 。於2009年，根據英國 《1985年公司法》 第131條之規定，已就 供股取得法定股份溢價寬免，並於合併儲備中確認157.96億美元。合併儲備包括扣減與供股有關之成本6.14億 美元，其中1.49億美元已於隨後撥入收益表。於該1.49億美元當中，1.21億美元為將與包銷商訂立的協議作為或 有遠期合約入賬而產生的虧損。合併儲備並不包括與對沖供股所得款項相關的遠期外匯合約虧損3.44億美元。 8 包括根據IFRS分類為股東權益之優先股及資本證券支付之分派。 9 保留盈利包括為僱員股份計劃提供資金而持有之本身股份179,419股 （300萬美元） （2013年：330,030股 （500萬 美元） ） 。 10 其他繳足股款的股本因向滙豐控股附屬公司僱員授出的認股權獲行使及失效而產生。 11 於2014年9月，滙豐控股發行22.5億美元、15億美元及15億歐元永久後償或有可轉換資本證券，就此有1,300萬 美元外部發行成本及3,300萬美元集團內部發行成本根據IFRS分類為股東權益。 註釋 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 345 1 編製基準及主要會計政策 (a) 遵守 《國際財務報告準則》 《國際財務報告準則》 （ 「IFRS」 ） 包括國際會計準則委員會 （ 「IASB」 ） 頒布或採納的會計準則及 國際財務報告準則詮釋委員會 （ 「IFRS IC」 ） 頒布或採納的詮釋。 滙豐的綜合財務報表及滙豐控股的獨立財務報表均根據IASB頒布並由歐盟正式通過之IFRS 編製。若於任何時間，新訂或經修訂的IFRS未獲歐盟正式通過，則歐盟正式通過之IFRS可 能與IASB頒布之IFRS有所不同。 於2014年12月31日，並無任何於截至2014年12月31日止年度內生效而未經歐盟正式通過之 準則對上述綜合及獨立財務報表有任何影響。就適用於滙豐之IFRS而言，經歐盟正式通 過之IFRS與由IASB頒布之IFRS並無差異。因此，滙豐是根據由IASB頒布之IFRS，編製截至 2014年12月31日止年度的財務報表。 於截至2014年12月31日止年度採納之準則 於截至2014年12月31日止年度並無應用新準則。 於2014年1月1日，滙豐應用 「對銷金融資產及金融負債 （IAS 32之修訂） 」 。該等修訂澄清了 對銷金融工具之規定，並處理了應用IAS 32 「金融工具：呈列」 的對銷準則時，現行慣例存 在的不一致情況。該等修訂已予追溯應用，但對滙豐的財務報表並未構成重大影響。 於2014年內，滙豐採納多項準則之詮釋及修訂，該等詮釋及修訂對滙豐的綜合財務報表 及滙豐控股的獨立財務報表並無重大影響。 (b) IFRS與 《香港財務報告準則》 之差異 就適用於滙豐之準則而言，IFRS與 《香港財務報告準則》 之間並無重大差異。因此，即使根 據 《香港財務報告準則》 編製財務報表，亦不會有任何重大差異。財務報表附註連同董事 會報告，已涵蓋IFRS及香港的申報規定所要求披露的全部資料。 (c) 會計處理法之未來發展 除下文所述完成金融工具會計處理法之工作計劃外，IASB正推展涉及保險及租賃會計處 理法的工作計劃，這意味會計處理方面的規定日後可能會有重大變更。 由IASB頒布及經歐盟正式通過但於2014年12月31日後生效之準則及修訂 於2014年，歐盟透過2010至2012年周期及2011至2013年周期之IFRS年度改進，正式 通過IASB頒布的修訂以及IAS 19 「僱員福利」 的窄幅修訂。滙豐並未提早應用任何於 2014年12月31日後生效的修訂，並預期該等修訂應用後對滙豐的綜合財務報表及滙豐控 股的獨立財務報表影響甚微。 由IASB頒布但未經歐盟正式通過之準則及修訂 於2014年5月，IASB頒布IFRS 15 「與客戶訂約帶來之收入」 。該準則於2017年1月1日或之後開 始的年度計算期生效，並允許提前應用。IFRS 15提供按原則確認收入的方法，並引入於 履行責任後確認收入的概念。該準則應予追溯應用，並備有若干實用的權宜措施。滙豐 目前正在評估此準則的影響，但於本期財務報表之刊發日期量化有關影響並不切實可 行。 於2014年7月，IASB頒布IFRS 9 「金融工具」 ，此為取代IAS 39 「金融工具：確認及計量」 之全面 準則，並包括金融資產及負債之分類及計量、金融資產減值及對沖會計法的規定。 分類及計量 金融資產之分類及計量將取決於管理實體之業務模式及其約定現金流特性，因此金融資 產會按已攤銷成本入賬、按公允值計入其他全面收益或按公允值計入損益賬。在許多情 況下，分類及計量之結果將與遵照IAS 39相若，但仍有機會出現差異，例如，由於IFRS 9 1－編製基準 財務報表附註 財務報表附註 （續） 滙豐控股有限公司 346 1－編製基準 不會對金融資產應用內含衍生工具會計法，故股權證券將按公允值計入損益賬或 （於有 限情況下） 按公允值計入其他全面收益。應用業務模式及約定現金流特性測試的合併影 響，對於按已攤銷成本或公允值計量的金融資產數目，與遵照IAS 39比較，可能有若干 差異。金融負債的分類大致維持不變，惟按實體本身信貸風險變動之公允值損益計量的 若干負債，將納入其他全面收益內。 減值 減值規定適用於按已攤銷成本計量及按公允值計入其他全面收益的金融資產，以及租賃 應收賬款和若干貸款承諾及金融擔保合約。於初步確認時，須對可能於未來12個月內發 生的違責事件所產生的預期貸款損失 （ 「預期貸款損失」 ） （ 「12個月預期貸款損失」 ） 作出備 抵 （如為承諾及擔保則為準備） 。倘若信貸風險大幅上升，則須對金融工具預期年期內所 有可能發生的違責事件所產生的預期貸款損失 （ 「期限內預期貸款損失」 ） 作出備抵 （或準 備） 。 自首次確認入賬後，每個業績報告期均會評估信貸風險是否大幅上升，方法乃考慮金融 工具餘下期限發生違約風險的變動，而非考慮預期貸款損失的增加。 信貸風險之評估及預期貸款損失之估計，均須不偏不倚，衡量或然率，並應納入與評估 相關之所有可得資料，包括過往事件的資料、當前狀況，以及於業績報告日期對未來事 件及經濟狀況之合理及具支持理據之預測。此外，預期貸款損失之估計應計及資金時間 值。因此，減值之確認及計量擬較根據IAS 39確認及計量者具備較大前瞻性，而所產生 的減值準備有較為波動之傾向。此舉亦有導致減值準備整體水平上升之傾向，因為所有 金融資產將至少評估12個月預期貸款損失，而期限內預期貸款損失所適用的金融資產數 目，很可能多於根據IAS 39被評為具有客觀減值證據的資產數目。 對沖會計法 一般對沖會計法之規定旨在簡化對沖會計法，加強其與風險管理策略之間的聯繫，並允 許前者適用於較多類別的對沖工具及風險。該準則並無明確處理宏觀對沖會計法策略， 因為有關策略正在另一項工作計劃中考慮。為消除現有宏觀對沖會計法慣例與新訂的一 般對沖會計法規定之間存在任何衝突之風險，IFRS 9包括可選擇於會計政策中保留IAS 39 對沖會計法。 過渡 分類、計量及減值之規定透過調整於首次申請日期之期初資產負債表追溯應用，且毋須 重列比較期間的資料。對沖會計法自該日期起普遍適用。 就該準則整體而言強制適用日期為2018年1月1日，但經修訂的呈列方式亦有可能自較早 日期起適用於按公允值計量的若干負債。滙豐擬於歐盟法律批准後，立即修訂與實體本 身信貸風險有關之若干負債公允值損益的呈列方式。倘若此呈列方式於2014年12月31日 適用，基於年內滙豐信貸風險變動導致公允值變動，其將導致除稅前利潤增加，而對其 他全面收益的影響則相反，且對資產淨值並無影響。有關信貸風險 （包括滙豐的信貸風 險） 變動導致公允值變動的進一步詳情，載於附註25。 滙豐正透過自2012年起在集團整體推行的計劃，評估IFRS 9其餘內容對財務報表之影響， 但由於分類及計量、減值以及對沖會計法規定之複雜性和此三者之間的相互關係，現階 段我們無法量化其潛在影響。 (d) 財務報表之呈列及財務報表附註之變動 為使財務報表及其附註更易理解，滙豐已變更附註內描述若干會計政策的位置及用詞、 剔除若干非重大披露及變更若干章節的順序。於衡量財務報表披露之重要性時，我們會 考慮各項目的金額及性質。2014年財務報表之呈列及其附註的主要變動如下： 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 347 ‧ 綜合資產負債表及綜合股東權益變動表：優化若干項目披露資料，使之更加合理， 以側重於重大資料。 ‧ 信貸風險：變更若干披露資料的排序和呈列方式，以剔除重複部分及側重於重大資 料。 ‧ 於2013年，財務報表內加入附註2 「重大會計政策概要」 。於2014年，在可行情況下，於 財務報表相關附註內加入會計政策；變更用詞以便更清楚載列會計政策。用詞的改 變並不表示會計政策有變動。 ‧ 關鍵會計政策：用 「關鍵會計估算及判斷」 替代 「關鍵會計政策」 及將其放入與其相關 的重大會計政策的相關附錄內。新方法符合IAS 1 「財務報表之呈列」 及美國證券交易 委員會的匯報要求。 ‧ 附註6 「僱員報酬及福利」 ：合理調整以剔除重複部分及側重於重大資料。 ‧ 附註38 「租賃承諾」 ：合理調整以側重於重大資料。 ‧ 於2013年，財務報表載列附註13 「以計量基準分析金融資產及負債」 及附註23 「物業、 機器及設備」 。於2014年，該等個別附註已剔除及於其他附註加入相關資料。 ‧ 於2013年，財務報表載列附註20 「轉讓金融資產」 及附註36 「作為負債擔保而質押之 資產及持作資產擔保之抵押品」 。於2014年，該等部分的相關資料已獨立列為附註19 「作為負債擔保而質押之資產、已轉讓之資產及持作資產擔保之抵押品」 。 自2014年1月1日起，滙豐選擇於資產負債表分別呈列非交易用途反向回購及回購。該等 項目就會計目的而言已分類為貸款及應收賬款或按已攤銷成本計量的金融負債。過去， 該等項目按合併計算基準連同按已攤銷成本計量的其他貸款或存款於綜合資產負債表 下列項目內呈列： 「同業貸款」 、 「客戶貸款」 、 「同業存放」 及 「客戶賬項」 。分別呈列使反 向回購及回購之披露方式與市場慣例相符，並提供更具參考價值的貸款相關資料。進一 步說明載於附註17。 自2014年1月1日起， 「亞洲」 地區取代之前按 「香港」 及 「亞太其他地區」 呈列的地區。這能 夠更好地與用於管理業務的集團內部資料保持一致。比較數字已相應重列。更多詳情載 於附註11。 (e) 呈列資料 根據IFRS 4 「保單」 及 IFRS 7 「金融工具：披露」 就保單及金融工具的風險性質及程度披露的 資料，已載於第111至237頁 「董事會報告：風險」 之經審核部分。 根據IAS 1 「財務報表之呈列」 作出的資本披露，已載於第238至262頁 「董事會報告：資本」 之經審核部分。 有關滙豐的證券化活動和結構產品的披露資料，已載於第111至237頁 「董事會報告：風 險」 之經審核部分。 根據滙豐的政策，集團會披露資料讓投資者和其他相關群體了解集團的表現、財務狀況 及相關變動。財務報表附註及董事會報告提供的資料超出會計準則、法定和監管規定， 以及上市規則所規定的最低要求。特別值得一提的是，滙豐經考慮強化信息披露工作組 （ 「EDTF」 ） 於2012年10月發表之 《加強銀行風險披露》 報告所載建議後提供額外披露資料。 報告旨在協助金融機構識別投資者強調需要更佳及更透明的銀行風險資料之範疇，以及 有關風險與衡量表現及報告有何關係。此外，滙豐遵守英國銀行家協會的 《財務報告披 露資料守則》 （ 「英國銀行家協會守則」 ） 。英國銀行家協會守則旨在提升英國銀行披露資 料的質素及可比較程度，當中列明五項披露原則及相關指引。按照英國銀行家協會守則 的原則，滙豐會評估相關監管機構及標準制訂機關不定期頒布的良好實務建議，亦會評 估該等指引對集團是否適用及相關，並會在適當情況下提升披露水平。 在刊發母公司財務報表連同集團的財務報表時，滙豐控股引用英國 《2006年公司法》 第408(3)條的豁免規定，並未呈列其個別收益表及相關附註。 1－編製基準 財務報表附註 （續） 滙豐控股有限公司 348 滙豐之綜合財務報表以美元列賬是由於美元及與其掛鈎之各種貨幣所屬區域，是我們進 行交易及為業務營運提供資金之主要貨幣區。美元亦為滙豐控股之功能貨幣，是因為美 元及與美元掛鈎之各種貨幣與滙豐控股旗下附屬公司之相關交易、事件及狀況關係最密 切，同時其融資活動產生之絕大部分資金亦以該等貨幣計值。 (f) 關鍵會計估算及判斷 編製財務資料須使用有關日後情況之估算及判斷。鑑於確認及計量下文所列事項涉及內 在不確定和高度主觀成分，下個財政年度的結果可能有別於管理層所作估算，導致與管 理層就2014年財務報表得出的結論大不相同。管理層選取的滙豐會計政策 （包括關鍵估 算及判斷） 載列如下，反映政策所應用項目的重要性及所涉及的判斷及估計的高度不確 定性： ‧ 貸款減值：參閱附註1(k)； ‧ 遞延稅項資產：參閱附註8； ‧ 金融工具之估值：參閱附註13； ‧ 於聯營公司之權益減值：參閱附註20； ‧ 商譽減損：參閱附註21； ‧ 撥備：參閱附註29。 (g) 持續經營 財務報表乃按持續經營基準編製，是因為各董事信納集團及母公司擁有足夠資源於可見 將來持續經營業務。於作出此評估時，各董事已考慮有關目前及日後情況的廣泛資料， 包括對日後盈利能力、現金流及資本來源的預測。 (h) 綜合計算及相關披露 當滙豐因參與公司而面對回報有所不同的風險或有權享有回報，並且能夠透過其對公司 的控制權影響相關回報，滙豐實質上控制有關公司，因此會將相關公司綜合入賬。控制 權最初經考慮所有事實及情況而進行評估，其後則於初始情况有任何重大變動時重新評 估。 倘公司受投票權規管，滙豐將於其直接或間接持有必要投票權，足以在公司管治機構通 過決議時就該公司綜合入賬。於其他情況下，控制權的評估更為複雜，並需要判斷其他 因素，包括有權享有可變動回報、對相關活動可行使的權力或作為代理或主事人的權力。 業務合併採用收購法入賬。收購成本按交易日期付出的代價 （包括或有代價） 之公允值計 量。收購相關成本於產生期間在收益表確認為支出。所收購之可識別資產、負債及或有 負債，一般按收購日期之公允值計量。商譽乃按轉讓代價、非控股股東權益金額及滙豐 之前所持有股本權益 （如有） 之公允值的總和，超逾所收購可識別資產與所承擔負債之間 淨額的差額計量。非控股股東權益金額按被收購方可識別資產淨值之公允值或非控股股 東權益所佔比例份額計量。如收購分階段進行，之前所持有股本權益會按收購日期之公 允值重新計量，所產生之損益於收益表內確認。 滙豐內部之一切交易已於綜合入賬時撇銷。 滙豐之綜合財務報表亦包括應佔合資及聯營公司之業績及儲備，計量基準為截至12月 31日的財務報表，或就刊發財務報表日期至12月31日期間發生的任何重大交易或事件按 比例金額作出調整。 (i) 外幣 以外幣進行之交易，均按進行交易當日之通行匯率，以功能貨幣紀錄。以外幣計值之貨 幣資產及負債，均按結算日之匯率換算為功能貨幣。因而產生之任何匯兌差額，均列入 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 349 收益表內。按歷史成本以外幣計量之非貨幣資產及負債，均使用初次進行交易當日之匯 率換算為功能貨幣。以外幣按公允值計量之非貨幣資產及負債，均使用釐定公允值當日 之匯率換算為功能貨幣。非貨幣項目的損益視乎相關非貨幣項目之損益於何處確認而可 於其他全面收益項目或收益表內確認。 於綜合財務報表中，非以美元為功能貨幣之分行、附屬公司、合資公司及聯營公司之資 產及負債，均按結算日之匯率換算為集團之列賬貨幣，而其業績，均按業績報告期之平 均匯率換算為美元。因重新換算期初外幣資產淨值而產生之匯兌差額，以及因重新換算 業績報告期內之業績 （由採用平均匯率改為採用期末通行之匯率） 而產生之匯兌差額， 均在其他全面收益項內確認。屬於海外業務投資淨額一部分之貨幣項目的匯兌差額，於 獨立財務報表內之收益表及於綜合財務報表中其他全面收益中確認。出售海外業務時， 之前於其他全面收益項內確認的匯兌差額，均重新分類納入收益表內，作為重新分類調 整。 (j) 同業及客戶貸款 同業及客戶貸款包括由滙豐辦理，且未分類為持作交易用途或指定以公允值列賬之貸 款。這些貸款於借款人提取現金時確認入賬，並於借款人履行還款責任，或出售貸款， 或已轉讓擁有權附帶的絕大部分風險與回報時撤銷確認。這些貸款首次列賬均按公允值 計量，並計入任何直接相關的交易支出，其後此等貸款將採用實質利率法，按已攤銷成 本扣除減值撥備的方式計量。 貸款將於其符合附註23所呈列的標準時重新分類為 「持作出售用途資產」 ，但將繼續按照 上述政策計量。 滙豐可能會承諾於特定期間內按指定合約條款批核貸款。倘若因借貸承諾而產生之貸款 預期持作交易用途，借貸承諾將入賬列作衍生工具。於取用時，貸款會分類為持作交易 用途。倘若滙豐有意持有貸款，則僅會於滙豐可能蒙受虧損之情況下就貸款承諾提撥準 備。於訂立貸款時，持有之貸款會按公允值列賬，其後則按已攤銷成本計量。就若干交 易而言，如槓桿融資及銀團貸款活動，借出的現金未必是該貸款公允值的最佳證據。就 這些貸款而言，倘若其首次列賬公允值低於借出之現金額，則差額會於收益表自其他營 業收入中扣取。除非貸款已減值，否則該撇減額將透過確認利息收益之方式，於貸款有 效期內收回。 (k) 貸款減值及可供出售金融資產 關鍵會計估算及判斷 貸款減值 貸款減值準備是管理層對結算日的貸款組合虧損所作之最佳估算。在計算個別及綜合評估貸款的貸款減值準備 時，管理層要為所需假設及估算作出判斷。 大部分綜合評估貸款減值準備集中在北美洲，為24億美元，佔集團綜合評估貸款減值準備總額的38% （2013年： 38億美元；47%） 及集團減值準備總額的19% （2013年：25%） 。在北美洲的綜合減值準備中，約71% （2013年： 79%） 與美國消費及按揭貸款組合有關。 綜合減值撥備涉及未能確定的估算，部分原因是組合中有大量個別數額不大的貸款，要確定個別貸款的虧損並 不可行。這些估算方法涉及使用統計分析方法衡量過往資料，亦需要管理層基於當前經濟及信貸狀況，作出重 大判斷，評估已產生虧損的實際水平，會否高於或低於過往經驗之水平。 當經濟、監管或行為等狀況出現變化，令統計模型未能完全反映貸款組合風險因素的最新趨勢，我們會依據各 項風險因素對純粹憑過往虧損經驗計算的減值準備作出調整。 風險因素包括貸款組合增長、產品組合比例、失業率、破產趨勢、地域集中程度、貸款產品特性、經濟狀況 （如 全國及當地的房屋市場走勢） 、利率水平、組合出現周期變化、賬項管理政策及慣例、法律及法規的更改，以 及其他對客戶付款模式的影響。對於不同地區及國家的業務，我們會考慮不同因素，以反映各地不同的經濟狀 況、法律和法規。計算減值虧損所用的方法及假設，均會依據估計虧損與實際虧損兩者之差額予以定期檢討。 舉例而言，滾動率、虧損率及預期日後收回貸款的時間，均會定期與實際結果對照，以確保此等估算仍然適用。 就個別評估貸款而言，則管理層須作出判斷，以確定是否有客觀證據，證明已發生虧損事件，且如發生虧損事 件，則計量減值準備。於確定是否有客觀證據證明已發生虧損事件時，應作出判斷以評估有關減值指標的所有 財務報表附註 （續） 滙豐控股有限公司 350 1－編製基準 相關資料，包括考慮還款是否已超逾合約期限，亦考慮能顯示借款人的財務狀況及前景惡化進而影響彼等還款 能力的其他因素。如借款人有財政困難之跡象，或相關市場類別正承受經濟壓力，尤其是還款的可能性受到再 融資或銷售特定資產前景的影響，則於授出貸款時須作出較高層級的判斷。就有客觀證據顯示存在減值之貸 款，管理層基於一系列的因素 （例如抵押品的可變現價值；清盤或破產時可能收回的清算分配金；客戶的營業模 式是否可行及有關公司能否成功克服財務困難，並創造充足現金流以償還債務） ，以釐定所需的準備金額。 滙豐可能會同意修訂償還貸款的合約條款，讓陷入財政困難的借款人暫緩償還貸款，從而改善客戶關係之管 理、盡量提高收回貸款的機會，或避免發生拖欠或收回抵押品的情況。倘有大量暫緩還款活動，則在釐定其對 貸款減值準備的影響時，將進行較高層級的判斷及不確定性之估算。判斷涉及區別不同暫緩還款情況的信貸風 險特性，包括於重議條件後恢復履約的情況。倘綜合評估貸款組合包括大量貸款暫緩還款，該等組合會分類以 反映具體的信貸風險特性，並估算暫緩還款組合各分類所產生的內在虧損。零售及批發貸款組合內均會發生暫 緩還款活動，但大部分均集中在美國，於美國滙豐融資的消費及按揭貸款組合內。 作出判斷需使用假設，而假設相當主觀且非常容易受風險因素影響，尤其是極易受到多個不同地域的經濟及信 貸狀況變動所影響。許多因素在很大程度上互相影響，而我們的貸款減值準備不會特別容易受某項單一因素影 響。然而，我們的貸款減值準備尤其容易受到北美洲的整體經濟及信貸狀況影響。舉例而言，北美洲的綜合評 估貸款減值準備若上升10%，於2014年12月31日的貸款減值準備便會增加2億美元 （2013年：4億美元） 。 貸款減值 當有客觀證據顯示貸款或貸款組合已出現減值時，便會確認該等已減值貸款之虧損。個 別貸款或綜合評估的貸款組合均會計算減值準備，並於收益表列支及記錄於資產負債表 內減值貸款之賬面值。可能因未來事件而產生之虧損均不會予以確認。 個別評估貸款 在評估減值時，用作決定貸款是否屬於個別大額之考慮因素包括貸款的數額；組合內貸 款的數目；及個別貸款關係的重要性和管理方式。符合該等標準的貸款將會個別評估減 值，惟按拖欠及虧損數額而有充分理由使用綜合評估法處理者除外 （見下文） 。 被視為個別大額之貸款一般為企業及工商客戶之貸款，相關數額一般較大，並按個別情 況管理。對於上述貸款，滙豐會於各結算日逐一考慮是否有客觀證據表明貸款減值。此 評估所用的標準包括： － 借款人遇上已知的現金流困難； － 按合約應付的本金或利息逾期未還超過90日； － 借款人可能破產或進行其他財務變現； － 因涉及借款人財政困難的經濟或法律理由而向借款人授出還款優惠，以致豁免或延 遲收取本金、利息或費用，而且還款優惠並非不重大；及 － 借款人的財務狀況或前景惡化，以致其還款能力受到質疑。 就有客觀證據顯示存在減值之貸款，會在考慮下列因素後釐定減值虧損額： － 滙豐就該客戶承擔的貸款風險總額； － 客戶的營業模式是否可行及有關公司能否成功克服財務困難，並創造充足現金流以 償還債務； － 預期收取款項及收回貸款的數額及時間； － 清盤或破產時可能收回的清算分配金； － 客戶對其他債權人的承諾較滙豐優先或與滙豐等同的程度，以及其他債權人繼續支 持該公司的可能性； － 釐定所有債權人的申索總金額及索償優先次序時涉及的複雜程度，以及法律與保險 方面不明朗因素的明顯程度； － 抵押品 （或其他減低信貸風險措施） 的可變現價值及成功收回的可能性； － 在止贖過程中獲得及出售抵押品可能產生的成本； － 倘若貸款並非以當地貨幣計值，借款人獲得以相關貨幣計值貸款並用以還款的能力； 及 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 351 － 債務的第二市場價格 （如有） 。 進行減值評估時，抵押品的可變現價值乃根據進行評估時的市值釐定。該價值不會因市 價的預計未來變動而作出調整；然而，該價值會為反映各地情況 （如強制出售折扣） 而作 出調整。 減值虧損的計算方法，是按貸款原訂實質利率折現其預計日後現金流 （包括預計日後收 取的合約利息） ，並比較計算所得現值與貸款當前賬面值的差額。個別大額賬項的減值 準備最少每季覆核一次 （有需要時會更頻密） 。 綜合評估貸款 倘為抵補須個別評估的貸款內已產生但尚未識別之虧損，或對於沒有個別大額賬項之同 類貸款組合，則減值按綜合基準評估。由於零售貸款組合一般為大批同類貸款，故該等 組合一般按綜合基準評估減值。 已產生但尚未識別之減值 若貸款經個別評估後並無發現證據顯示已個別識別具體減值，則會按其信貸風險特性合 併處理，以進行綜合減值評估。這些信貸風險特性可能包括產生風險的國家╱地區、涉 及的業務類別、提供的產品種類、獲得的抵押品或其他相關因素。此評估反映滙豐因結 算日前發生之事件而蒙受的減值虧損，滙豐雖未能按個別貸款基準識別這些虧損，但已 能可靠估算其數額。待取得資料可供識別出組合內之個別貸款虧損時，相關貸款會自貸 款組合中剔除，並按個別基準進行評估。 釐定綜合評估減值準備時，會考慮以下因素： － 具有近似信貸風險特性 （例如按行業、貸款級別或產品分類） 的組合過往的虧損經驗； － 由出現減值至識別虧損並確定個別貸款的適當準備額以顯示該項虧損，兩者之間預 計相距的時間；及 － 管理層基於當前經濟及信貸狀況，判斷結算日之內在虧損實際水平，是否高於或低 於過往經驗顯示之水平。 管理層會根據經濟狀況及市場狀況、客戶行為、組合管理資料、信貸管理技巧及市場上 追收及收回貸款的經驗，就每個已識別的貸款組合估計由出現虧損至識別虧損之間的相 距時間。由於有關評估乃憑經驗定期作出，故估計相距時間會隨着該等因素改變而有所 不同。 同類貸款組合 對於沒有個別大額賬項之同類貸款組合，會使用統計法釐定綜合評估減值虧損額。該等 貸款組合之虧損於個別貸款自組合中剔除及撇銷時個別入賬。按綜合基準計算準備額之 方法如下： － 若可取得適當的經驗數據，滙豐會採用滾動率方法。此方法乃運用過往貸款拖欠數 據及經驗的統計分析，以可靠地估算因結算日前發生之事件導致最終撇銷而滙豐未 能按個別基準識別之貸款額。個別貸款按逾期日數範圍劃分為不同系列，並以統計 分析方法估算各系列貸款經過各個拖欠階段逐步惡化並變成無法收回的可能性。此 外，個別貸款按其如上文所述的信貸特徵分類。於應用此項方法時會作出調整，以 估計發生與發現虧損事件 （例如透過未有如期付款） 之間的相距時間 （即生成期） 以及 發現與撇銷之間的相距時間 （即結果期） 。在計算抵補內在虧損所需的合適準備水平 時，亦會評估當前的經濟情況。在若干發展已相當成熟的市場，則會採用更精密的 模型，此等模型會同時考慮行為及賬項管理趨勢 （如破產及債務重整統計數據） 。 － 若組合規模細小，或採用滾動率方法所需的資料不足或不可靠，滙豐會採用根據過往 虧損率經驗或現金流折現模型計算之基本公式法計算。倘若採取基本公式法計算，各 地管理層會明確估計由產生虧損事件至識別虧損之間的相距時間，一般為六至十二 個月。 每個貸款組合的內在虧損乃按照統計模型採用過往數據觀察評估，並會定期更新過往數 據觀察，以反映近期的貸款組合及經濟趨勢。倘統計模型未能完全反映因經濟、監管或 財務報表附註 （續） 滙豐控股有限公司 352 1－編製基準 行為狀況出現變化而形成之最新趨勢，則憑統計模型計算的減值準備會予以調整，以反 映於結算日的該等變動，從而計及該等趨勢。 撇銷貸款 倘收回的機會渺茫，通常會將貸款 （及相關減值準備賬項） 部分或全數撇銷。如貸款有抵 押，則一般會在收取變現抵押品所得任何款項後再撇銷。倘任何抵押品之可變現淨值已 釐定，且並無合理期望可於日後進一步收回款項，則貸款或會提早撇銷。 撥回減值 倘減值虧損額於確認入賬後減少，而減幅客觀而言可能與減值確認後發生之事件相關 連，則相關貸款減值準備賬項的金額會相應撇減，令超額的部分得以撥回。撥回額於收 益表內確認。 交換貸款所得資產 就以貸款交換得來的非金融資產 （作為有秩序變現之部分） 而言，倘該等資產分類為持作 出售用途，則會列為 「持作出售用途資產」 ，並在 「其他資產」 項內列賬。所得資產會於交 換日期按其公允值減出售成本及貸款賬面值 （扣除減值準備額） 兩者中之較低數額入賬。 持作出售用途資產毋須提撥折舊額。減值及過往減值撥回連同出售變現的任何損益，在 收益表內 「其他營業收益」 項下確認。 重議條件貸款 須進行綜合減值評估的貸款，若貸款的條件已經重議，且收到最低規定次數的還款，則 不再視為逾期，而會視為並未逾期之貸款予以計量。為進行綜合減值評估，前述貸款會 與貸款組合之其他部分分開計算，以反映該等貸款的風險狀況。須進行個別減值評估且 其條件已經重議的貸款，須予持續覆核，以決定貸款是否仍屬已減值。被分類為重議條 件之貸款的賬面值將繼續計入此類別，直至到期或撤銷確認為止。 倘取消現有協議並訂立條款大不相同的新協議，或倘現有協議的條款經過修訂，以致重 議條件貸款已是完全不相同之金融工具，則會撤銷確認條件已經重議之貸款。因撤銷確 認事件而訂立的任何新協議將繼續披露為重議條件貸款及按上文評估減值。 可供出售金融資產減值 我們會於每個結算日評估有否任何客觀證據，顯示可供出售金融資產的價值出現減值。 如有客觀證據顯示因金融資產首次確認入賬後發生的一項或以上事件 （ 「虧損事件」 ） 而出 現減值，且虧損事件會影響金融資產的估計日後現金流，而所涉數額又能可靠估算時， 方會確認減值虧損。 如可供出售金融資產出現減值，其收購成本 （扣除任何本金還款及攤銷額） 與當前公允值 （經減除已於收益表確認的任何過往減值虧損後） 之間的差額於收益表內確認。 債務工具的減值虧損於收益表的 「貸款減值及其他信貸風險準備」 項內確認，而股票的 減值虧損則於收益表的 「金融投資減除虧損後增益」 項內確認。可供出售金融資產的減 值方法詳列如下： － 可供出售債務證券。滙豐於評估在業績報告日期是否存在減值的客觀證據時，會考 慮所有可用證據，包括與證券尤其相關之事件的可觀察數據或資料，而該等事件可 能導致日後現金流收回額出現短缺。發行人面臨重大財政困難及其他因素，例如有 關發行人的流動資金、業務及財務風險承擔的資料、同類金融資產之違責程度與趨 勢、國家及本土經濟趨勢與狀況，以及抵押品及擔保之公允值，均可作個別或綜合 考慮，以釐定是否有客觀證據證明債務證券出現減值。 此外，於評估可供出售資產抵押證券是否存在減值的客觀證據時，滙豐會考慮相關 抵押品的表現及市場價格下跌的程度及深度。公允值出現不利變動，以及證券不再 存在交投活躍的市場，均被視為潛在減值的主要指標，而信貸評級的改變為次要。 － 可供出售股權證券。減值的客觀證據，可包括上文所述有關發行人的具體資料，亦 可包括關於技術、市場、經濟或法律領域重大變化的資料，而這些資料可證明股權 證券的成本可能無法收回。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 353 股權的公允值如大幅或長期下跌至低於其成本，亦被視為減值的客觀證據。在評估 公允值是否大幅下跌時，減幅會與首次確認入賬時的資產原有成本作比較。在評估 公允值是否長期下跌時，減幅則按資產公允值持續低於首次確認入賬時原有成本的 時間衡量。 倘已就可供出售金融資產確認減值虧損，其後該項資產公允值變動的會計處理方法會視 乎資產的類型而有所不同： － 就可供出售債務證券而言，倘有進一步客觀證據顯示因該項金融資產的估計日後現 金流進一步減少而導致減值，該工具之公允值其後的減額會於收益表內確認。若無進 一步客觀證據顯示減值，則於其他全面收益項內確認該項金融資產公允值之減額。 倘債務證券之公允值於往後期間增加，而該項增額客觀而言可能與在收益表確認減 值虧損後發生的事件相關連，或工具不再減值，則會從收益表撥回減值虧損； － 就可供出售股權證券而言，該工具之公允值其後一切增額均當作重估處理，並於其 他全面收益項內確認。就股權證券確認的減值虧損不會從收益表撥回。可供出售股 權證券公允值其後之減額於收益表內確認，惟以已產生的進一步累計減值虧損為限。 (l) 資金公允值調整 為符合不斷變化的市場慣例，滙豐通過引入資金公允值調整修訂其評估非抵押衍生工具 組合之方法。資金公允值調整反映日後市場資金成本的估計現值或有關受隔夜指數掉期 外的匯率影響的非抵押衍生工具的溢利。隔夜指數掉期為用於評估抵押衍生工具的基準 利率。採納資金公允值調整於2014年的影響為交易收益淨額減少2.63億美元，反映了計 入與倫敦銀行同業拆息的資金息差。有關更多詳情載於財務報表附註13。 (m) 營業收益 利息收益及支出 除分類為持作交易用途或指定以公允值列賬之金融工具外，所有金融工具 （不包括滙豐 發行之債務證券及與該等債務證券一併管理之衍生工具） 之利息收益及支出，均採用實 質利率法在收益表的 「利息收益」 及 「利息支出」 項內確認。將金融工具預計有效期或在 適當情況下的一段較短期間內估計日後可收取或需支付的現金款額，準確折現至金融資 產或金融負債的賬面淨值時，所用比率即為實質利率。 已減值金融資產的利息，乃採用計量減值虧損時用以折現日後現金流的利率確認。 非利息收益及支出 滙豐向客戶提供廣泛的服務以賺取費用收益。費用收益按以下方式入賬： － 如屬進行一個重要項目所賺取之收益，會於該項目完成時確認為收入，例如為第三 方磋商或參與磋商一項交易 （如安排收購股份或其他證券） 所產生之費用； － 如屬提供服務而賺取之收益，會於提供服務時確認為收入，例如資產管理、資產組 合及其他管理顧問費和服務費；及 － 如屬組成金融工具實質利率不可分割一部分的收益 （例如若干貸款承諾之費用） ，則 確認為對實質利率的調整數額，並在 「利息收益」 項下列賬。 交易收益淨額包括持作交易用途金融資產及金融負債因公允值變動而產生的所有損益， 以及相關的利息收益、支出及股息。 股息收益於確立收取股息付款之權利時確認。有關日期指上市股權證券之除息日，以及 一般為股東已批准非上市股權證券派息的日期。 指定以公允值列賬之金融工具淨收益╱ （支出） 及保費收益淨額的會計政策披露於附註2 及附註3。 財務報表附註 （續） 滙豐控股有限公司 354 2 指定以公允值列賬之金融工具淨收益╱ （支出） 會計政策 指定以公允值列賬之金融工具淨收益╱ （支出） 包括︰ ‧ 指定以公允值計入損益賬之金融資產及負債因公允值變動而產生的所有損益，包括投資合約負債； ‧ 與指定以公允值計入損益賬之金融資產及負債一併管理之衍生工具因公允值變動而產生的所有損益；及 ‧ 有關指定以公允值計入損益賬之金融資產及負債，及與上述各項一併管理的衍生工具的利息收益、利息支出及 股息收益，惟滙豐發行之債務證券及與該等債務證券一併管理的衍生工具產生之利息則不包括在內，該等利息 會於 「利息支出」 項內確認。 指定以公允值列賬之金融工具淨收益╱ （支出） 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 以下因素產生的淨收益╱ （支出） ： －為應付保單未決賠款及投資合約負債而持有的金融資產 2,300 3,170 2,980 －其他指定以公允值列賬之金融資產 131 118 83 －與其他指定以公允值列賬之金融資產 一併管理的衍生工具 (19) (26) 35 2,412 3,262 3,098 －在投資合約下對客戶之負債 (435) (1,237) (996) －滙豐已發行長期債務及相關衍生工具 508 (1,228) (4,327) －長期債務之本身信貸息差變動 417 (1,246) (5,215) －與滙豐的已發行債務證券一併管理的衍生工具 333 (3,743) 431 －公允值之其他變動 (242) 3,761 457 －其他指定以公允值列賬之金融負債 (23) (39) (23) －與其他指定以公允值列賬之金融負債 一併管理的衍生工具 11 10 22 61 (2,494) (5,324) 截至12月31日止年度 2,473 768 (2,226) 滙豐控股 滙豐控股已發行長期債務及相關衍生工具產生之淨收益╱ （支出） 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 以下因素產生的淨收益╱ （支出） ： －長期債務之本身信貸息差變動 339 (695) (2,260) －與滙豐控股的已發行債務證券一併管理的衍生工具 126 (1,558) 456 －公允值之其他變動 (27) 1,213 (474) 截至12月31日止年度 438 (1,040) (2,278) 3 保費收益淨額 會計政策 壽險保單之保費均於應收時入賬，但單位相連保單之保費則於確定未決賠款時入賬。 再保險保費與相關之直接保單保費於同一會計期間入賬。 保費收益淨額 附有 非相連 DPF2之 保險1 相連壽險 投資合約 總計 百萬美元 百萬美元 百萬美元 百萬美元 保費收益總額 7,705 2,195 2,470 12,370 再保人應佔保費收益總額 (441) (8) － (449) 截至2014年12月31日止年度 7,264 2,187 2,470 11,921 2－指定以公允值列賬之金融工具淨收益╱ （支出） ╱3－保費收益淨額╱4－保險賠償淨額 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 355 附有 非相連 DPF2之 保險1 相連壽險 投資合約 總計 百萬美元 百萬美元 百萬美元 百萬美元 保費收益總額 7,002 3,012 2,384 12,398 再保人應佔保費收益總額 (450) (8) － (458) 截至2013年12月31日止年度 6,552 3,004 2,384 11,940 保費收益總額 7,578 3,325 2,699 13,602 再保人應佔保費收益總額 (550) (8) － (558) 截至2012年12月31日止年度 7,028 3,317 2,699 13,044 1 包括非壽險。 2 酌情參與條款。 4 已支付保險賠償和利益及投保人負債之變動淨額 會計政策 壽險保單之保險賠償總額，反映年度內產生之賠償支出總額，包括賠償手續費及任何投保人之紅利 （就預期宣派紅利 而分配） 。 保單期滿申索於到期付款時確認。退保額於付款時確認，或於接獲通知後，保單不再納入相關保險未決賠款的計算 內時提前確認。身故賠償則於接獲通知時確認。 再保險追償金與相關賠償於同一會計期間入賬。 已支付保險賠償和利益及投保人負債之變動淨額 附有 非相連 DPF2之 保險1 相連壽險 投資合約 總計 百萬美元 百萬美元 百萬美元 百萬美元 已支付賠償和利益及負債之變動總額 7,770 2,765 3,188 13,723 － 已支付之賠償、利益及退保額 3,575 1,499 2,215 7,289 － 負債之變動 4,195 1,266 973 6,434 再保人應佔已支付賠償和利益及負債之變動 (411) 33 － (378) － 已支付之賠償、利益及退保額 (176) (88) － (264) － 負債之變動 (235) 121 － (114) 截至2014年12月31日止年度 7,359 2,798 3,188 13,345 已支付賠償和利益及負債之變動總額 6,892 3,379 3,677 13,948 － 已支付之賠償、利益及退保額 3,014 1,976 2,308 7,298 － 負債之變動 3,878 1,403 1,369 6,650 再保人應佔已支付賠償和利益及負債之變動 (367) 111 － (256) － 已支付之賠償、利益及退保額 (164) (426) － (590) － 負債之變動 (203) 537 － 334 截至2013年12月31日止年度 6,525 3,490 3,677 13,692 已支付賠償和利益及負債之變動總額 6,900 3,984 3,645 14,529 － 已支付之賠償、利益及退保額 1,905 1,810 2,525 6,240 － 負債之變動 4,995 2,174 1,120 8,289 再保人應佔已支付賠償和利益及負債之變動 (537) 223 － (314) － 已支付之賠償、利益及退保額 (217) (681) － (898) － 負債之變動 (320) 904 － 584 截至2012年12月31日止年度 6,363 4,207 3,645 14,215 1 包括非壽險。 2 酌情參與條款。 財務報表附註 （續） 滙豐控股有限公司 356 5 營業利潤 營業利潤乃於扣除下列收支、損益以及貸款減值及其他信貸風險準備後列賬： 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 收益 已減值金融資產之確認利息 1,137 1,261 1,261 並非持作交易用途或指定以公允值列賬之 金融資產或負債所賺取的費用 （不包括計算 該等資產及負債之實質利率時包括之費用） 9,438 9,799 10,042 滙豐代客戶持有或投資資產之信託及 其他受信業務所賺取的費用 3,253 3,176 2,897 來自上市投資之收益 6,726 5,432 5,850 來自非上市投資之收益 5,874 6,860 7,677 支出 金融工具之利息 （不包括持作交易用途或 指定以公允值列賬之金融負債之利息） (15,322) (14,610) (17,625) 並非持作交易用途或指定以公允值列賬之 金融資產或負債的應付費用 （不包括計算 該等資產及負債之實質利率時包括之費用） (1,427) (1,396) (1,501) 與滙豐代客戶持有或投資資產之信託及 其他受信業務有關之應付費用 (185) (171) (170) 租賃及分租協議項下之付款 (1,548) (1,425) (1,166) －最低租金款額 (1,199) (1,098) (1,149) －或有租金及分租租金款額 (349) (327) (17) 英國銀行徵費 (1,066) (916) (472) 增益╱ （虧損） 可供出售股權證券減值 (373) (175) (420) 持作出售用途資產之確認增益╱ （虧損） 220 (729) 485 出售巴拿馬滙豐銀行所得利潤 － 1,107 － 攤薄興業銀行以及其他聯營及合資公司 權益所得 （虧損） ╱增益 (32) 1,051 － 貸款減值及其他信貸風險準備 (3,851) (5,849) (8,311) －貸款減值準備淨額 (4,055) (6,048) (8,160) －可供出售債務證券撥回╱ （減值） 319 211 (99) －其他信貸風險準備減值 (115) (12) (52) 6 僱員報酬及福利 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 工資及薪金 17,477 16,879 17,780 社會保障支出 1,666 1,594 1,633 離職後福利 1,223 723 1,078 截至12月31日止年度 20,366 19,196 20,491 滙豐於年內僱用之員工平均數目 2014年 2013年 2012年 歐洲 74,024 75,334 77,204 亞洲 116,492 114,216 116,779 中東及北非 8,616 9,181 8,645 北美洲 21,983 22,568 27,396 拉丁美洲 43,652 47,496 54,162 截至12月31日止年度 264,767 268,795 284,186 5－營業利潤╱6－僱員報酬及福利 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 357 授出獎勵總額與僱員報酬及福利中之獎勵的對賬 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 本年度批准及授出的獎勵總額1 3,660 3,920 3,689 減：本年度已授出的遞延花紅，預期將於未來期間確認 (359) (436) (355) 本年度已授出及確認的獎勵總額 3,301 3,484 3,334 本年度就過往年度遞延花紅的支出 425 427 671 其他 (114) (164) (28) 計入僱員報酬及福利的本年度獎勵總額 3,612 3,747 3,977 1 指已獲批准及授出的集團浮動酬勞資金。已獲集團薪酬委員會批准的集團浮動酬勞資金總額於第310頁 董事薪酬報告內披露。 收益表支出：遞延花紅 本年度 花紅資金 上年度 花紅資金 總計 百萬美元 百萬美元 百萬美元 2014年 於2014年確認之支出 245 425 670 －遞延股份獎勵 147 373 520 －遞延現金獎勵 98 52 150 預期於2015年或之後確認之支出 359 381 740 －遞延股份獎勵 250 334 584 －遞延現金獎勵 109 47 156 2013年 於2013年確認之支出 269 427 696 －遞延股份獎勵 188 354 542 －遞延現金獎勵 81 73 154 預期於2014年或之後確認之支出 436 306 742 －遞延股份獎勵 356 259 615 －遞延現金獎勵 80 47 127 2012年 於2012年確認之支出 277 671 948 －遞延股份獎勵 224 613 837 －遞延現金獎勵 53 58 111 預期於2013年或之後確認之支出 355 376 731 －遞延股份獎勵 315 335 650 －遞延現金獎勵 40 41 81 以股份為基礎的支出 會計政策 滙豐與其僱員訂立股權結算和現金結算以股份為基礎的支出安排，作為僱員提供服務的報酬。與僱員訂立股權結算 以股份為基礎的支出安排之成本，於授出日期參考股權工具之公允值計量，並於實際授出期內以直線基準確認為支 出，同時相應地撥入 「保留盈利」 。 現金結算以股份為基礎的支出安排方面，所獲得之服務及所產生之負債按負債之公允值計量及於僱員提供服務時確 認。負債之公允值於結算時方會重新計量，而公允值變動則在收益表內確認。 公允值乃採用適當的估值模型釐定。實際授出條件包括服務條件及表現條件；安排之任何其他條件均為非實際授出 條件。市場表現條件及非實際授出條件於授出日期估計獎勵之公允值時計入。除市場表現條件外，實際授出條件不 會計入於授出日期首次列賬之公允值估算額內。此等實際授出條件乃透過調整計量交易時所包括之股權工具數目而 計算。 於實際授出期內取消獎勵，會當作提前實際授出處理，並即時確認原應於實際授出期就服務而確認之金額。 若滙豐控股訂立之以股份為基礎的支出安排涉及附屬公司僱員，而相關支出將向該附屬公司分攤，以股份為基礎的支 出安排之成本與預期將予發行以滿足該等安排的股權工具之公允值之間的差額，於實際授出期間確認為 「於附屬公司 之投資」 之調整。 6－僱員報酬及福利 財務報表附註 （續） 滙豐控股有限公司 358 「工資及薪金」 包括以股份為基礎的支出安排之影響，其中7.32億美元以股權結算 （2013年： 6.3億美元；2012年：9.88億美元） ，載列如下： 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 有限制股份獎勵 738 599 912 儲蓄及其他股份獎勵優先認股計劃 36 63 96 截至12月31日止年度 774 662 1,008 滙豐股份獎勵 獎勵 政策 目的 有限制股份獎勵 （包括以股份形式 發放的周年獎勵） 及集團業績表現 股份計劃 ‧ 評估截至12月31日止相關期間的表 現乃用於釐定將要授出的獎勵金額。 ‧ 遞延獎勵通常要求僱員在實際授出 期內仍然受僱用，於授出日期後不 受表現條件所限。 ‧ 遞延周年獎勵一般於三年內實際授 出；集團業績表現股份計劃獎勵於 五年後實際授出。 ‧ 已實際授出股份可能受禁售規定約 束。倘為集團業績表現股份計劃獎 勵，禁售期至僱傭關係終結為止。 ‧ 由2010年起授出之獎勵於實際授出 前須受扣減條文規限。 ‧ 推動及獎勵符合集團策略及股東利 益的表現。 ‧ 遞延授出可激勵僱員長期投入，並 令集團得以應用扣減條文。 滙豐股份獎勵之變動 2014年 2013年 數目 （千） 數目 （千） 於1月1日已授出之有限制股份獎勵 116,932 165,589 本年度增添 82,871 59,261 本年度發放 (78,224) (99,820) 本年度沒收 (5,096) (8,098) 於12月31日已授出之有限制股份獎勵 116,483 116,932 已授出獎勵之加權平均公允值 （美元） 10.18 10.95 滙豐優先認股計劃 主要計劃 政策 目的 儲蓄優先認股計 劃 ‧ 兩項計劃：英國計劃及國際計劃。最 後一次根據國際計劃授出認股權的 時間為2012年。 ‧ 自2014年起合資格僱員每月儲蓄最 多500英鎊 （或就2013年前授出的國 際計劃認股權而言，與250英鎊等值 的美元、港元或歐元金額） ，附有可 使用儲蓄款項購買股份之選擇權。 ‧ 三年期或五年期合約開始後分別於 第三周年或第五周年後六個月內 （或 就2013年前授出的國際計劃認股權 而言，一年期儲蓄合約開始後於第 一周年後三個月內） 可予行使。 ‧ 行使價設定為最接近邀請日期前的 市值折讓20% （2013年：20%） （不包 括根據美國業務轄下計劃於2013年 前授出的一年期認股權，此等認股 權應用15%折讓） 。 ‧ 使全體僱員利益與創造股東價值更 趨一致。 滙豐控股集團優 先認股計劃 ‧ 計劃已於2005年5月終止。 ‧ 於授出日期的第三周年至第十周年 期內可予行使。 ‧ 於2000至2005年期間的長期獎勵計 劃，若干滙豐僱員獲授認股權。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 359 計算公允值 認股權之公允值，乃利用畢蘇數學模型計算。股份獎勵之公允值乃按授出日期之股價為基 準。 滙豐優先認股計劃之變動 儲蓄優先認股計劃 滙豐控股集團優先認股計劃 數目 （千） 加權平均 行使價1 英鎊 數目 （千） 加權平均 行使價1 英鎊 於2014年1月1日尚未行使 93,760 4.04 55,026 7.23 本年度授出2 28,689 5.19 － － 本年度行使3 (50,393) 3.48 (1) 7.22 本年度屆滿 (5,690) 4.81 (48,651) 7.22 於2014年12月31日尚未行使 66,366 4.89 6,374 7.29 加權平均尚餘合約期 （年） 2.66 0.30 於2013年1月1日尚未行使 112,752 4.04 87,173 6.94 本年度授出2 8,679 5.47 － － 本年度行使3 (17,968) 4.56 (17,595) 6.21 本年度屆滿 (9,703) 4.47 (14,552) 4.21 於2013年12月31日尚未行使 93,760 4.04 55,026 7.23 加權平均尚餘合約期 （年） 1.80 0.45 1 加權平均行使價。 2 本年度授出認股權之加權平均公允值為1.9美元 （2013年：2.98美元） 。 3 儲蓄優先認股計劃及滙豐控股集團優先認股計劃於認股權行使日期之加權平均股價分別為9.91美元 （2013年：10.86美元） 及9.49美元 （2013年：10.93美元） 。 離職後福利計劃 會計政策 滙豐在全球各地經營多項退休金及其他離職後福利計劃。該等計劃包括界定福利及界定供款計劃，以及多項其他離 職後福利，例如離職後保健福利。 向界定供款計劃及國家管理退休福利計劃 （滙豐根據該等計劃所承擔之責任與界定供款計劃相等） 支付之款項，均於 僱員提供服務時列作支出扣除。 界定福利退休金支出及界定福利責任之現值，於業績報告日期由計劃之精算師採用預計單位基數精算成本法計算。 扣取自收益表之淨額主要包括服務成本及界定福利資產淨值或負債淨額之利息淨額，並於營業支出項內呈列。 過往服務成本均即時於收益表內扣除，指於過往期間為僱員服務應負的界定福利責任之現值變動，乃由於計劃修訂 （實施或撤銷或更改界定福利計劃） 或削減 （公司大幅削減計劃涵蓋的僱員數目） 。償付指撇銷界定福利計劃下提供的 部分或全部福利應負的所有其他法定和推定責任的交易，向計劃條款所載及納入精算假設的僱員或其代表支付福利 則除外。 重新計量界定福利資產淨值或負債淨額包括精算損益、計劃資產回報 （不包括利息） 及資產上限的影響 （如有，不包 括利息） ，均即時於其他全面收益項內確認。 精算損益包括經驗調整 （先前精算假設與實際情況之間的差異所產生之影響） ，以及精算假設變動之影響。 界定福利資產淨值或負債淨額，指界定福利責任之現值，並減除計劃資產之公允值。任何界定福利盈餘淨額設有上 限，此限額相等於可得退款之現值及日後向計劃供款之扣減數額。 界定福利保健計劃等其他離職後界定福利計劃產生之責任成本，均按界定福利退休金計劃之相同基準入賬。 6－僱員報酬及福利 財務報表附註 （續） 滙豐控股有限公司 360 收益表支出 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 界定福利退休金計劃 469 54 427 界定供款退休金計劃 687 597 599 退休金計劃 1,156 651 1,026 界定福利及供款保健計劃 67 72 52 截至12月31日止年度 1,223 723 1,078 489 1,076 688 在資產負債表內確認之界定福利計劃資產淨值╱ （負債淨額） 計劃資產 之公允值 界定福利 責任現值 計劃盈餘限額 之影響 總計 百萬美元 百萬美元 百萬美元 百萬美元 界定福利退休金計劃 44,824 (42,062) (17) 2,745 界定福利保健計劃 179 (1,104) － (925) 於2014年12月31日 45,003 (43,166) (17) 1,820 僱員福利負債總額 （屬於 「應計項目、遞延收益 及其他負債」 ） (3,208) 僱員福利資產總值 （屬於 「預付款項、應計收益 及其他資產」 ） 5,028 界定福利退休金計劃 40,622 (40,467) (30) 125 界定福利保健計劃 190 (1,106) － (916) ° 於2013年12月31日 40,812 41,573 (30) (791) 僱員福利負債總額 （屬於 「應計項目、遞延收益 及其他負債」 ） (2,931) 僱員福利資產總值 （屬於 「預付款項、應計收益 及其他資產」 ） 2,140 於其他全面收益項內確認的累計精算利潤╱ （虧損） 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 於1月1日 (4,445) (3,844) (3,453) 英國滙豐銀行 （英國） 退休金計劃 2,764 (1,524) 208 其他計劃 (274) 796 (440) 保健計劃 (88) 143 (154) 計劃盈餘限額之影響之變動 17 (16) (5) 於其他全面收益項內確認的精算利潤╱ （虧損） 總計 2,419 (601) (391) 於12月31日 (2,026) (4,445) (3,844) 滙豐退休金計劃 2014年 2013年 2012年 % % % 滙豐僱員之百分比： －於界定供款計劃登記 66 64 62 －於界定福利計劃登記 22 23 23 25 25 27 －受滙豐退休金計劃保障 88 87 85 集團在全球各地經營多項退休金計劃。部分為界定福利計劃，其中最大的為英國滙豐銀行 （英國） 退休金計劃 （ 「主計劃」 ） 。有關主計劃的特點及風險、日後現金流的金額、產生時間及 不確定性，以及相關政策與慣例，載於第200頁退休金風險一節及第236頁風險附錄內。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 361 界定福利退休金計劃 界定福利退休金計劃資產淨值╱ （負債淨額） 計劃資產之公允值 界定福利責任現值 資產上限之影響 界定福利資產 淨值╱ （負債淨額） 英國滙豐銀 行 （英國） 退休金計劃 其他計劃 英國滙豐銀 行 （英國） 退休金計劃 其他計劃 英國滙豐銀 行 （英國） 退休金計劃 其他計劃 英國滙豐銀 行 （英國） 退休金計劃 其他計劃 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日 31,665 8,957 (29,629) (10,838) － (30) 2,036 (1,911) 現時服務成本 － － (228) (257) － － (228) (257) 過往服務成本及因償付而產生的增益╱ （虧損） － (5) (26) 11 － － (26) 6 服務成本 － (5) (254) (246) － － (254) (251) 界定福利資產淨值╱ （負債淨額） 之淨利息收益╱ （支出） 1,386 370 (1,291) (425) － (4) 95 (59) 於其他全面收益項內確認之重新計量影響 4,864 845 (2,100) (1,034) － 17 2,764 (172) －計劃資產回報 （不包括利息收益） 4,864 845 － － － － 4,864 845 －精算虧損 － － (2,317) (987) － － (2,317) (987) －其他變動 － － 217 (47) － 17 217 (30) 匯兌差額 (2,112) (316) 1,838 357 － － (274) 41 滙豐供款 397 278 － － － － 397 278 －正常 265 239 － － － － 265 239 －特別 132 39 － － － － 132 39 僱員供款 38 17 (38) (17) － － － － 已付福利 (954) (543) 954 598 － － － 55 計劃已付的行政開支及稅項出售 (40) (23) 40 23 － － － － 於2014年12月31日 35,244 9,580 (30,480) (11,582) － (17) 4,764 (2,019) 與下列人士有關之界定福利責任現值： － 活躍成員 (9,782) (5,605) － 遞延成員 (8,799) (2,498) － 退休金領取人 (11,899) (3,479) 6－僱員報酬及福利 財務報表附註 （續） 滙豐控股有限公司 362 界定福利退休金計劃資產淨值╱ （負債淨額） （續） 計劃資產之公允值 界定福利責任現值 資產上限之影響 界定福利資產 淨值╱ （負債淨額） 英國滙豐銀 行 （英國） 退休金計劃 其他計劃 英國滙豐銀 行 （英國） 退休金計劃 其他計劃 英國滙豐銀 行 （英國） 退休金計劃 其他計劃 英國滙豐銀 行 （英國） 退休金計劃 其他計劃 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2013年1月1日 29,092 9,015 (26,475) (11,581) － (19) 2,617 (2,585) 現時服務成本 － － (259) (249) － － (259) (249) 過往服務成本及因償付而產生的增益╱ （虧損） 1 － (3) 438 (41) － － 438 (44) 服務成本 － (3) 179 (290) － － 179 (293) 界定福利資產淨值╱ （負債淨額） 之淨利息收益╱ （支出） 1,260 156 (1,127) (229) － － 133 (73) 於其他全面收益項內確認之重新計量影響 817 21 (2,341) 775 － (16) (1,524) 780 －計劃資產回報 （不包括利息收益） 817 21 － － － － 817 21 －精算利潤╱ （虧損） － － (2,453) 829 － － (2,453) 829 －其他變動 － － 112 (54) － (16) 112 (70) 匯兌差額 766 (59) (740) 23 － 5 26 (31) 滙豐供款 605 336 － － － － 605 336 －正常 399 274 － － － － 399 274 －特別 206 62 － － － － 206 62 僱員供款 38 17 (38) (17) － － － － 已付福利 (876) (513) 876 452 － － － (61) 計劃已付的行政開支及稅項 (37) (13) 37 13 － － － － 出售 － － － 16 － － － 16 於2013年12月31日 31,665 8,957 (29,629) (10,838) － (30) 2,036 (1,911) 與下列人士有關之界定福利責任現值： －活躍成員 (8,896) (5,465) －遞延成員 (8,358) (2,144) －退休金領取人 (12,375) (3,229) 1 滙豐向旗下英國僱員宣布，由2015年6月30日起停止為主計劃之界定福利部分的活躍成員累計日後的服 務基數，及界定福利部分的所有活躍成員將由2015年7月1日起成為界定供款部分的成員。有關變動令該 計劃之界定福利責任減少，並因而於2013年錄得相應增益4.3億美元，有關數額已列入上表呈列的 「過往 服務成本及因償付而產生的 （增益） ╱虧損」 內。 滙豐預期於2015年會對界定福利退休金計劃作出5.3億美元的供款。預期在未來五年內每年 從計劃向退休僱員支付的福利，以及其後五年合共支付的福利如下： 預期從計劃支付之福利 2015年 2016年 2017年 2018年 2019年2020-2024年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 英國滙豐銀行 （英國） 退休金計劃1 970 999 1,029 1,060 1,091 5,968 其他計劃1 566 576 595 605 643 3,366 1 根據所採用的披露假設，英國滙豐銀行 （英國） 退休金計劃的界定福利責任之期限為19.8年 （2013年：19.5年） ， 而綜合所有其他計劃則為14.2年 （2013年：13.7年） 。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 363 按資產類別呈列之計劃資產公允值 2014年12月31日 2013年12月31日 價值 於活躍市場 的市場報價 於活躍市場 並無市場報價 滙豐1 價值 於活躍市場 的市場報價 於活躍市場 並無市場報價 滙豐1 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 英國滙豐銀行 （英國） 退休金計劃 計劃資產之公允值 35,244 31,355 3,889 930 31,665 26,520 5,145 2,827 －股票 5,502 4,557 945 － 4,655 3,667 988 － －債券 22,965 22,965 － － 17,708 17,708 － － －衍生工具 1,369 52 1,317 930 2,827 － 2,827 2,827 －其他 5,408 3,781 1,627 － 6,475 5,145 1,330 － 其他計劃 計劃資產之公允值 9,580 6,390 3,190 (13) 8,957 7,731 1,226 574 －股票 2,534 1,778 756 11 2,854 2,789 65 14 －債券 6,376 4,109 2,267 7 4,892 4,409 483 9 －衍生工具 (100) (8) (92) (107) 399 － 399 399 －其他 770 511 259 76 812 533 279 152 1 計劃資產之公允值包括如附註41詳述與英國滙豐銀行有限公司訂立之衍生工具。 離職後界定福利計劃的主要精算財務假設 滙豐經諮詢各地的計劃精算師後，按到期日與界定福利責任相符的優質 （AA級或同級） 債務 工具當前的平均收益率為基準，釐定適用於其責任的折現率。 主計劃的主要精算假設 折現率 通脹率 退休金 增長率 增薪率 利息 信貸利率 % % % % % 英國 於2014年12月31日 3.70 3.20 3.00 3.70 不適用 於2013年12月31日 4.45 3.60 3.30 4.10 不適用 於2012年12月31日 4.50 3.10 2.90 3.60 不適用 主計劃死亡率表及於65歲的平均預期壽命 男性於65歲時 女性於65歲時 死亡率表 的預期壽命： 的預期壽命： 現時65歲 現時45歲 現時65歲 現時45歲 英國 於2014年12月31日 SAPS S11 23.6 25.2 25.0 26.9 於2013年12月31日 SAPS S11 23.6 25.2 24.9 26.8 1 自我管理退休金計劃 （ 「SAPS」 ） 附有死亡率持續研究2014年改善幅度 （2013年：2013年改善幅度） 及1.25%長 期津貼改善幅度。男性退休金領取人適用的低修勻程度列表，附設比率為1.01，女性退休金領取人適用的 低修勻程度列表，附設比率為1.02。 精算假設敏感度 主要假設變動對主計劃的影響 英國滙豐銀行 （英國） 退休金計劃 2014年 2013年 百萬美元 百萬美元 折現率 增加25個基點造成的年底退休金責任變動 (1,420) (1,352) 減少25個基點造成的年底退休金責任變動 1,523 1,450 增加25個基點造成的2015年退休金支出變動 (75) (83) 減少25個基點造成的2015年退休金支出變動 73 79 通脹率 增加25個基點造成的年底退休金責任變動 1,026 994 減少25個基點造成的年底退休金責任變動 (1,184) (1,137) 增加25個基點造成的2015年退休金支出變動 44 53 減少25個基點造成的2015年退休金支出變動 (48) (68) 退休金付款及遞延退休金之增長率 增加25個基點造成的年底退休金責任變動 1,188 1,301 減少25個基點造成的年底退休金責任變動 (1,127) (1,225) 增加25個基點造成的2015年退休金支出變動 50 66 減少25個基點造成的2015年退休金支出變動 (45) (64) 財務報表附註 （續） 滙豐控股有限公司 364 7－核數師費用╱8－稅項 英國滙豐銀行 （英國） 退休金計劃 2014年 2013年 百萬美元 百萬美元 增薪率 增加25個基點造成的年底退休金責任變動 237 212 減少25個基點造成的年底退休金責任變動 (232) (205) 增加25個基點造成的2015年退休金支出變動 12 15 減少25個基點造成的2015年退休金支出變動 (11) (15) 死亡率 假設壽命每增加1年造成的退休金責任變動 768 712 滙豐控股 滙豐控股於2014年的僱員報酬及福利支出為6.81億美元 （2013年：5.42億美元） 。滙豐控股於 2014年聘用的僱員人數平均為2,070人 （2013年：1,525人） 。 屬界定福利退休金計劃成員的滙豐控股僱員主要為英國滙豐銀行 （英國） 退休金計劃或滙豐 國際僱員退休福利計劃的成員。滙豐控股根據計劃受託人釐定之供款時間表為其本身之僱 員向該等計劃支付供款，並於此等供款到期時確認為支出。 董事酬金 根據英國 《2006年公司法》 及 《2008年中大型公司與集團 （賬目及報告） 規例》 計算之滙豐控股 董事酬金總額為： 2014年 2013年 2012年 千美元 千美元 千美元 袍金 4,567 4,027 5,435 薪金及其他酬金 17,812 9,488 10,316 周年獎勵 4,426 7,357 13,983 截至12月31日止年度 26,805 20,872 29,734 實際授出的長期獎勵 － － 5,733 此外，集團根據與前任董事訂立的退休福利協議支付1,269,160美元 （2013年：1,198,744美 元） 。於2014年12月31日，對前任董事未置存基金的退休金責任準備為19,419,524美元 （2013 年：19,729,103美元） 。 年內就董事向退休金計劃供款的總額為零 （2013年：零） 。 薪金及其他酬金金額中包括固定酬勞津貼。授予董事之特別周年獎勵乃經考慮個人表現及 公司業績後發放，並由集團薪酬委員會決定具體金額。各董事之薪酬、認股權及根據滙豐 股份計劃和2011年滙豐股份計劃授出之獎勵，詳情載於第300至327頁 「董事薪酬報告」 內。 7 核數師費用 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 應付KPMG之審計費用1 40.6 43.4 47.2 應付KPMG以外公司之審計費用 1.2 1.1 1.4 截至12月31日止年度 41.8 44.5 48.6 1 就根據法例對滙豐控股賬目進行法定審計及對滙豐附屬公司賬目進行審計應付予KPMG之費用，且包括 就本年度應付之費用。2014年應付KPMG之費用中不包括有關過往年度的應計項目撥回淨額250萬美元及 與轉由羅兵咸永道擔任核數師有關的費用130萬美元。 滙豐應付予集團主要核數師KPMG Audit Plc及其聯營公司 （統稱 「KPMG」 ） 之費用如下： 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 365 滙豐應付予KPMG之費用 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 對滙豐控股賬目進行法定審計之費用1 13.4 12.9 13.2 －與本年度賬目相關之費用 13.4 12.6 12.8 －與上年度賬目相關之費用 － 0.3 0.4 向滙豐提供其他服務之費用 62.5 67.5 67.3 滙豐附屬公司賬目審計2 27.2 30.5 34.0 審計相關之鑑證服務3 22.6 27.4 23.6 稅務相關之服務： －稅務合規服務 1.5 1.3 2.1 －稅務諮詢服務 0.8 1.3 1.3 其他鑑證服務 0.7 0.5 1.1 審計以外之其他服務4 9.7 6.5 5.2 截至12月31日止年度 75.9 80.4 80.5 1 滙豐就KPMG對滙豐之綜合財務報表及滙豐控股之獨立財務報表進行法定審計而應付的費用。有關費用 包括就滙豐控股附屬公司之綜合申報表提供相關服務而應付的款額，該等申報表乃經明確識別對集團的 審計意見可提供支持憑證。2014年費用中不包括與上年度有關的30萬美元及與轉由羅兵咸永道擔任核數 師有關的費用130萬美元。 2 為滙豐旗下附屬公司之財務報表進行法定審計而應付的費用。2014年費用中不包括有關過往年度的應計 項目撥回淨額280萬美元。 3 包括鑑證服務及與法定和監管規定申報有關的其他服務 （如信心保證書及中期業績審閱） 。 4 包括估值及精算服務、翻譯服務、提供非經常性的會計意見、審視財務模型、提供有關資訊科技保安和 持續經營業務及企業融資交易的顧問意見，以及執行經雙方協定的資訊科技測試程序。 滙豐毋須就以下各類服務向KPMG支付費用：內部審計服務，與訴訟、招聘及薪酬相關之服務。 就滙豐之相關退休金計劃應付 KPMG 之費用 2014年 2013年 2012年 千美元 千美元 千美元 對滙豐之相關退休金計劃進行審計 322 379 256 審計相關之鑑證服務 5 5 － 截至12月31日止年度 327 384 256 滙豐之相關退休金計劃毋須就以下各類服務向KPMG支付費用：審計相關之鑑證服務、內部 審計服務、其他鑑證服務、與企業融資交易有關的服務、估值及精算服務、與訴訟、招聘、 薪酬及資訊科技相關的服務。 除上述費用外，據KPMG估算，該公司獲滙豐以外的人士或機構支付費用360萬美元 （2013 年：530萬美元；2012年：330萬美元） ，而滙豐與該等授予合約的人士或機構有關連，因此 可能參與委聘KPMG。該等費用涉及的服務，包括審計滙豐管理的互惠基金及審查向滙豐借 款的企業的財政狀況。 滙豐控股就KPMG提供審計以外的服務而應付的費用並未予以獨立披露，原因是滙豐集團 已按綜合基準披露該等費用。 8 稅項 會計政策 所得稅包括本年度稅項及遞延稅項。所得稅在收益表內確認，但若所得稅與在其他全面收益項內或直接在股東權 益項內確認之項目有關，則所得稅會在相關項目出現的同一份報表內確認。 本年度稅項指預期就本年度應課稅利潤而應繳之稅項，計算基礎為於結算日已頒布或實質頒布之稅率，以及就過 往年度應繳稅項作出之任何調整。滙豐按預計將向稅務機關繳納的稅項金額為可能產生之當前稅項負債提撥準 備。若滙豐有合法之對銷權，且有意按淨額結算，則當前稅項資產及負債會予以對銷。 遞延稅項乃按資產負債表內資產及負債之賬面值與該等資產及負債之課稅值兩者之間的暫時差異予以確認。遞延 稅項負債一般就所有應課稅暫時差異確認入賬；而遞延稅項資產則於日後可能出現應課稅利潤而該等利潤可運用 暫時差異予以扣減時，方會確認入賬。 遞延稅項根據於結算日已頒布或實質頒布之稅率及稅法，採用變現資產或償清負債期間預期適用之稅率計算。若 遞延稅項資產及負債是於同一稅務呈報組別中產生，並與同一稅務機關徵收之所得稅有關，同時滙豐有合法對銷 權，則兩者會互相對銷。 財務報表附註 （續） 滙豐控股有限公司 366 8－稅項 與離職後福利之精算損益有關之遞延稅項，於其他全面收益項內確認。與以股份為基礎的支出交易有關之遞延稅 項直接在股東權益項內確認，但以估計日後稅務扣減金額超出相關累計薪酬支出金額為限。若遞延稅項與重新計 量可供出售投資及現金流對沖工具之公允值有關，該等稅項會扣取自或計入其他全面收益，其後當遞延公允值損 益於收益表內確認時，則遞延稅項亦會於收益表內確認。 關鍵會計估算及判斷 遞延稅項資產 遞延稅項資產的確認視乎對下列各項的評估：日後出現應課稅利潤的可能性及其充足程度；現有應課稅暫時差異 日後的撥回額；以及持續執行的稅務規劃策略。由於缺乏應課稅利潤紀錄，最關鍵的判斷是關於預期日後盈利能 力及稅務規劃策略的適用性，包括企業重組。 美國業務過往出現稅項虧損，但盈利能力預期將會改善，因此，稅務規劃策略支持確認美國遞延的稅項資產，而 美國業務中保留資本為確認遞延稅項資產的關鍵因素。鑑於近期出現稅項虧損，我們於確認巴西遞延的稅項資產 時，已從稅務角度考慮對管理層所作收益預測的倚賴程度，以及運用策略 （如企業重組及其他措施） 改善巴西銀行 業務盈利能力的情況。 稅項支出 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 本期稅項 英國公司稅 69 (8) 250 －本年度稅項 54 103 60 －就過往年度調整 15 (111) 190 海外稅項1 3,881 3,949 5,560 －本年度稅項 4,423 3,947 5,421 －就過往年度調整 (542) 2 139 3,950 3,941 5,810 遞延稅項 25 824 (495) －暫時差異的產生及撥回 (477) 739 (269) －稅率變動之影響 83 93 66 －就過往年度調整 419 (8) (292) 截至12月31日止年度 3,975 4,765 5,315 1 海外稅項包括香港利得稅11.35億美元 （2013年：11.33億美元；2012年：10.49億美元） 。各附屬公司在香 港之應課稅利潤的適用香港稅率為16.5% （2013年：16.5%；2012年：16.5%） 。其他海外附屬公司及海外 分行按各自之業務所在國家╱地區的適用稅率提撥稅項準備。 稅項對賬 倘全部利潤均按英國公司稅率繳稅，所產生的稅項支出與收益表列示之稅項支出並不相同， 其差異如下： 2014年 2013年 2012年 百萬美元 % 百萬美元 % 百萬美元 % 除稅前利潤 18,680 22,565 20,649 稅項支出 按稅率21.5% （2013年：23.25%； 2012年：24.5%） 繳付之稅項 4,016 21.5 5,246 23.25 5,057 24.5 海外利潤按不同稅率繳稅之影響 33 0.2 (177) (0.8) (57) (0.3) 就前期負債所作調整 (108) (0.6) (117) (0.5) 37 0.2 未確認╱ （過往未確認） 之遞延稅項暫時差異 (154) (0.8) 332 1.5 374 1.8 聯營及合資公司利潤之影響 (547) (2.9) (543) (2.4) (872) (4.3) 出售平安保險之稅務影響 － － (111) (0.5) (204) (1.0) 將興業銀行重新分類之稅務影響 － － (317) (1.4) － － 非課稅收益及增益 (668) (3.5) (871) (3.9) (542) (2.6) 永久不可扣稅項目 969 5.1 647 2.9 1,092 5.3 稅率變動 22 0.1 93 0.4 78 0.4 地方稅項及海外預扣稅 434 2.3 551 2.4 581 2.8 其他項目 (22) (0.1) 32 0.1 (229) (1.1) 截至12月31日止年度 3,975 21.3 4,765 21.1 5,315 25.7 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 367 本年度的實質稅率為21.3%，而2013年則為21.1%。本年度的實質稅率反映集團旗下多家公 司持有的政府債券和股票的免稅收益以及在集團除稅前收益項下確認的本集團應佔聯營及 合資公司的除稅後利潤，均帶來經常性利好影響，且過往期間的稅項減免額亦結轉至本期， 惟部分被不可扣稅償付及關於外匯交易業務調查而提撥的準備所抵銷。2013年的實質稅率 較低，是因為撇減遞延稅項資產部分抵銷了非課稅增益的利好影響。 於2014年4月1日，英國公司稅的主要稅率從23%下調至21%，並將於2015年4月1日進一步下 調至20%。調低公司稅率至20%乃透過於2013年7月17日頒布的 《2013年財務法》 立法生效。預 期未來稅率下調將不會對集團構成重大影響。 集團旗下的法律實體須受集團業務所在地的稅務機關定期審查及審計。倘未能確定最終的 稅務處理方法，集團以預期向稅務機關繳付的款項為基準，為可能產生的稅項負債計提準 備。視乎該等稅務事宜的最終處理結果，最終支付的稅款金額與撥備金額可能出現重大差 異。 遞延稅項 下頁附表顯示於資產負債表確認的遞延稅項資產及負債總額，以及於收益表、其他綜合收 益及直接於股東權益內確認入賬的相關款額。 於資產負債表呈列的金額，為在滙豐擁有合法對銷權利及有意按淨額基準結算的情況下， 已對銷資產結餘及負債結欠的款額，故所呈列數字與下頁附表所披露的金額有所差異。於 2014年12月31日，遞延稅項資產淨值合計為56億美元 （2013年：65億美元） 。值得注意之主要 事項如下： 美國 與滙豐美國業務有關的遞延稅項資產淨值為41億美元 （2013年：44億美元） 。此總額所包含 的遞延稅項資產反映：結轉下期之稅項虧損及稅項減免額9億美元 （2013年：7億美元） ；有關 貸款減值準備的可扣減暫時差異8億美元 （2013年：12億美元） ；以及其他暫時差異24億美元 （2013年：25億美元） 。 就計算美國稅項而言，一般會在沖銷或出售 （如較早） 已減值貸款時產生貸款減值扣減額。 稅務扣減額通常會在有關減值已按照會計準則確認入賬後的下一個會計期間出現。因此， 相關遞延稅項資產的金額一般與減值準備額同步變動。 根據既有的證據，包括過往利潤水平、管理層對未來收益的預測，以及滙豐控股為收回遞 延稅項資產而繼續在北美洲保留足夠資本的承諾，我們預期有關業務將會產生足夠的應課 稅收益以供變現該等資產。 管理層目前對美國業務的利潤預測顯示將可於2017年或之前全數收回稅項虧損及稅項減免 額。與貸款減值準備及其他可扣減暫時差異相關的遞延稅項資產現有水平，亦預計於未來 四年期內逐步下降。 鑑於滙豐美國業務近期出現虧損，管理層在分析此等遞延稅項資產的確認情況時，已對來 自美國業務的任何未來預期利潤作大額折扣，並以來自滙豐控股的資本支持作依據，包括 有關該等支持的稅務規劃策略。主要策略是在美國保留高於監管規定正常水平的資金，以 產生未來應課稅利潤，從而減低可扣減資金開支，又或將該等資本另行投放以提高應課稅 收益水平。隨著美國業務的財務表現有所改善，預計在評估未來會計期間的遞延稅項資產 的確認情況時，將會以美國業務的預計日後利潤作依據，理由是相關財務表現已證實可以 持續改善。 巴西 與滙豐巴西業務有關的遞延稅項資產淨值為13億美元 （2013年：10億美元） 。此總計金額所 包含的遞延稅項資產反映：結轉下期之稅項虧損3億美元 （2013年：1億美元） ；與貸款減值準 備有關的可扣減暫時差異7億美元 （2013年：7億美元） ；以及其他暫時差異3億美元 （2013年： 2億美元） 。 就計算巴西稅項而言，一般會在沖銷已減值貸款時產生貸款減值扣減額，並通常會在有關 減值已按照會計準則確認入賬後的下一個會計期間出現。因此，相關遞延稅項資產的金額 一般與減值準備額同步變動。 管理層目前對巴西銀行業務的利潤預測顯示將可於未來五至八年內收回大部份稅項虧損及 其他暫時差異。貸款減值扣減額一般會於按照會計準則確認入賬後兩至三年內在稅務項內 確認。 財務報表附註 （續） 滙豐控股有限公司 368 8－稅項 墨西哥 與滙豐墨西哥業務有關的遞延稅項資產淨值為5億美元 （2013年：5億美元） 。此總額所包含 的遞延稅項資產，主要涉及與計算已減值貸款準備有關的可扣減暫時差異。 管理層在分析此等遞延稅項資產的確認情況時，以出售若干貸款組合的主要策略作依據， 而出售錄得的虧損可於墨西哥用作扣稅。此類稅項扣減會導致撥回就遞延稅項所確認並已 結轉下期的貸款減值準備。遞延稅項餘額將結轉至未來年度且不設限期。 於2013年9月，墨西哥政府提出多項稅務改革，有關改革已於2013年10月獲參議院批准，並 於2013年12月刊登於政府憲報。該等稅務改革就貸款減值準備的稅務扣減額訂立新基準， 准許銀行在某項貸款於資產負債表撇賬時確認稅務扣減額。該等改革亦引入過渡期規則， 准許銀行繼續申報於2013年12月31日的任何未申報扣減額。於2014年7月4日，墨西哥政府頒 布 《雜項稅務解決方案》 的第I.3.22.5條規則，澄清過渡期規則的處理方法，但對墨西哥業務所 持之遞延稅項資產並無影響。 根據既有的證據，包括過往及預測的貸款組合增長水平、貸款減值比率及盈利能力，目前 預期有關業務將於未來五年內變現該等資產。 集團於墨西哥的遞延稅項資產內，並無確認任何結轉下期的重大稅項虧損或稅項減免額。 英國 與滙豐英國業務有關的遞延稅項負債淨額為4億美元 （2013年：資產4億美元） 。此總額所包 含的遞延稅項負債，主要與退休福利有關。 集團於英國的遞延稅項資產內，並無確認任何結轉下期的重大稅項虧損或稅項減免額。 未確認遞延稅項 並無於資產負債表確認遞延稅項資產的暫時差異、未動用稅項虧損及稅項減免額為226億美 元 （2013年：220億美元） 。該等金額包括美國業務在若干州份的未動用稅項虧損額141億美 元 （2013年：173億美元） 。 在未確認的總額中，42億美元 （2013年：50億美元） 並無限期，9億美元 （2013年：10億美元） 將於10年內到期，餘下金額將於10年後到期。 若集團在附屬公司及分行的投資不可能匯出或以其他方法變現，將不會就相關投資確認遞 延稅項，已確定不會出現額外稅項的聯營公司投資和合資公司權益，亦不會確認遞延稅項。 我們並無披露與該等投資相關的未確認遞延稅項款額，亦無披露2014及2013年的相關暫時 差異金額，原因是當任何暫時差異出現撥回時，要釐定應付所得稅款額並不可行。然而，我 們已就聯營公司可供分派儲備提撥1.32億美元 （2013年：2,000萬美元） 遞延稅項準備，因為 於分派該等儲備時將須支付預扣稅。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 369 遞延稅項資產及負債之變動 加速折舊 未動用之 免稅額及 以股份 衍生工具、 貸款減值 稅項虧損及 租予客戶 可供出 現金流 為基礎 FVOD1及 保險 退休福利 準備 稅項減免額 之資產 售投資 對沖 之支出 費用收益 其他投資 技術準備 支出準備 其他 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 資產 274 2,837 978 549 － 211 253 － 1,383 － 1,398 461 8,344 負債 － － － (144) (298) (24) － (59) (213) (840) － (220) (1,798) 於2014年1月1日 274 2,837 978 405 (298) 187 253 (59) 1,170 (840) 1,398 241 6,546 收購及出售 － － － － － － － － － － － － － 收益表 (57) (408) 396 (17) (2) (3) (3) 2 361 (76) (86) (132) (25) 其他全面收益 (438) － － － (203) (87) － － (12) － － 48 (692) 股東權益 － － － － － － (20) － － － － － (20) 匯兌及其他調整 10 (165) (42) (26) 11 1 (21) 1 12 55 (68) 10 (222) 於2014年12月31日 (211) 2,264 1,332 362 (492) 98 209 (56) 1,531 (861) 1,244 167 5,587 資產 － 2,264 1,332 362 － 98 209 － 1,764 － 1,244 167 7,440 負債 (211) － － － (492) － － (56) (233) (861) － － (1,853) 資產 469 3,912 617 473 － 285 305 － 1,530 － 1,457 (22) 9,026 負債 － － － (226) (1,203) (44) － (105) (162) (815) － (10) (2,565) 於2013年1月1日 469 3,912 617 247 (1,203) 241 305 (105) 1,368 (815) 1,457 (32) 6,461 收購及出售 － － (9) － (3) 1 － － － － － (26) (37) 收益表 (419) (985) 399 123 (53) (91) (49) 42 (165) (72) 47 399 (824) 其他全面收益 169 － － － 1,026 38 － － (12) － － － 1,221 股東權益 － － － － － － (2) － － － － － (2) 匯兌及其他調整 55 (90) (29) 35 (65) (2) (1) 4 (21) 47 (106) (100) (273) 於2013年12月31日 274 2,837 978 405 (298) 187 253 (59) 1,170 (840) 1,398 241 6,546 資產 274 2,837 978 549 － 211 253 － 1,383 － 1,398 461 8,344 負債 － － － (144) (298) (24) － (59) (213) (840) － (220) (1,798) 1 本身債務之公允值。 財務報表附註 （續） 滙豐控股有限公司 370 9－股息╱10－每股盈利╱11－按類分析 滙豐控股 遞延稅項資產的變動 加速折舊 免稅額 可供 出售投資 其他投資 以股份為 基礎之支出 其他 短期時差 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日 2 (23) 19 11 4 13 收益表 － － 3 2 1 6 其他全面收益 － (36) － － － (36) 於2014年12月31日 2 (59) 22 13 5 (17) 2013年於1月1日 2 (31) 31 12 － 14 收益表 － － (12) (1) 4 (9) 其他全面收益 － 8 － － － 8 於2013年12月31日 2 (23) 19 11 4 13 並無於資產負債表確認遞延稅項資產的未動用稅項虧損為37.6億美元 （2013年：34.05億美 元） ，其中1,000萬美元 （2013年：900萬美元） 與資本虧損有關。根據既有的證據，包括過往利 潤水平及管理層對未來收益的預測，預期有關業務不能產生足夠的應課稅收益，以供滙豐 控股收回結轉下期的稅項虧損。有關虧損並無限期。 9 股息 派付予母公司股東的股息 2014年 2013年 2012年 每股 總計 以股代息 每股 總計 以股代息 每股 總計 以股代息 美元 百萬美元 百萬美元 美元 百萬美元 百萬美元 美元 百萬美元 百萬美元 就普通股宣派之股息 上年度： －第四次股息 0.19 3,582 1,827 0.18 3,339 540 0.14 2,535 259 本年度： －第一次股息 0.10 1,906 284 0.10 1,861 167 0.09 1,633 748 －第二次股息 0.10 1,914 372 0.10 1,864 952 0.09 1,646 783 －第三次股息 0.10 1,918 226 0.10 1,873 864 0.09 1,655 639 總計 0.49 9,320 2,709 0.48 8,937 2,523 0.41 7,469 2,429 分類為股東權益之優先股的 股息總額 （按季度支付） 62.00 90 62.00 90 62.00 90 分類為股東權益之資本證券的票息總額 2014年 2013年 2012年 首個提早 每份證券 總計 每份證券 總計 每份證券 總計 贖回日 美元 百萬美元 美元 百萬美元 美元 百萬美元 永久後償資本證券1 －22億美元 2013年4月 2.032 179 2.032 179 2.032 179 －38億美元 2015年12月 2.000 304 2.000 304 2.000 304 總計 483 483 483 1 永久後償資本證券的票息按季度支付。 董事會已於本年度終結後就截至2014年12月31日止財政年度宣派第四次股息每股普通股0.2 美元，分派涉及金額約達38.44億美元。第四次股息將於2015年4月30日派付予2015年3月6日 名列英國主要股東名冊、香港或百慕達海外股東分冊之股東。滙豐並無就2014年度第四次 股息而於財務報表內紀錄負債。 於2015年1月15日，滙豐就永久後償資本證券派付另一次票息每份證券0.508美元，分派涉及 金額4,500萬美元。滙豐並無就此次所付票息於2014年12月31日之資產負債表內紀錄負債。 如第438頁所述，滙豐於2014年9月發行三份或有可轉換證券，該等證券根據IFRS分類為股 東權益。或有可轉換證券的票息每半年支付一次及均無於2014年到期。於2015年1月20日， 滙豐就其中之一項或有可轉換證券支付票息每份證券28.125美元，分派涉及金額2,800萬美 元。滙豐並無就此次所付票息於2014年12月31日之資產負債表內紀錄負債。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 371 於2014年12月31日的可供分派儲備為488.83億美元。 10 每股盈利 「每股普通股基本盈利」 的計算方法，為將母公司普通股股東應佔利潤，除以已發行普通股 （不包括集團持有之本身股份） 之加權平均股數。 「每股普通股攤薄後盈利」 的計算方法，為 將基本盈利 （毋須就具攤薄影響之潛在普通股所造成之影響作出調整） 除以下列兩類股份的 加權平均股數總和：(1)已發行普通股 （不包括集團持有之本身股份） ；及(2)因轉換具攤薄影 響之潛在普通股而發行的普通股。 母公司普通股股東應佔利潤 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 母公司股東應佔利潤 13,688 16,204 14,027 分類為股東權益之優先股之應付股息 (90) (90) (90) 分類為股東權益之資本證券之應付票息 (483) (483) (483) 截至12月31日止年度 13,115 15,631 13,454 每股基本及攤薄後盈利 2014年 2013年 2012年 股份 股份 股份 利潤 數目 每股 利潤 數目 每股 利潤 數目 每股 百萬美元 （百萬股） 美元 百萬美元 （百萬股） 美元 百萬美元 （百萬股） 美元 基本1 13,115 18,960 0.69 15,631 18,530 0.84 13,454 18,125 0.74 具攤薄影響之潛在普通股的影響 － 96 － － 124 － － 146 － 攤薄後1 13,115 19,056 0.69 15,631 18,654 0.84 13,454 18,271 0.74 1 已發行 （基本） 或假設已攤薄 （攤薄後） 之普通股加權平均股數。 具攤薄影響之潛在普通股加權平均股數不包括600萬份不具攤薄影響的僱員認股權 （2013年： 6,000萬份；2012年：1.03億份） 。 11 按類分析 會計政策 滙豐採用矩陣式管理架構，滙豐的主要經營決策者為集團管理委員會，為董事會直接授權的全面管理委員會。集 團管理委員會根據若干基準定期檢討營業活動，包括按地區及按環球業務檢討。滙豐認為，地區營業類別為財務 報表閱讀者提供最適當的資料，以就滙豐所從事業務活動的性質及財務影響，以及滙豐經營業務所在的經濟環境 作出最佳評估。此反映地區因素對業務策略及業績表現、分配資本來源和地區管理層在執行策略時所發揮作用的 重要性。因此，滙豐的營業類別被視為按地區劃分。 地區資料乃根據附屬公司之主要業務所在地劃分，但如屬香港上海滙豐銀行、英國滙豐銀行、中東滙豐銀行及美 國滙豐銀行的資料，則根據負責匯報業績或貸出款項之分行所在地劃分。 分類資產、負債、收益及支出均根據集團之會計政策計量。分類收益及支出包括各類別間之轉撥，而該等轉撥乃 按公平原則進行。分佔支出乃按實際分攤數額計入分類賬項內。由於滙豐認為英國銀行徵費為在英國經營業務及 設立總部的成本之一，故有關徵費的支出乃計入歐洲地區類別內。 產品和服務 滙豐在五個地區為其客戶提供全面的銀行及相關金融服務。為客戶提供的產品和服務是按 環球業務分類。 ‧ 零售銀行及財富管理業務提供多種產品和服務，以滿足個別客戶的個人理財及財富管理 需要。滙豐向這類客戶提供的服務一般包括個人理財產品 （往來及儲蓄戶口、按揭及個 人貸款、信用卡、扣賬卡和本地與國際付款服務） 及財富管理服務 （保險與投資產品、環 球資產管理服務及財務策劃服務） 。 ‧ 工商金融業務提供多種產品和服務，以滿足工商客戶 （包括中小企及中型市場企業） 的需 要。有關服務包括信貸及貸款、國際貿易及應收賬融資、財資管理及流動資金方案 （資 財務報表附註 （續） 滙豐控股有限公司 372 11－按類分析 金管理及商業卡） 、商業保險及投資。工商金融業務亦向客戶提供滙豐旗下其他環球業 務 （例如環球銀行及資本市場業務） 的產品和服務，包括外匯產品、債務和股票市場集資 及顧問服務。 ‧ 環球銀行及資本市場業務為全球各主要政府、企業和機構客戶及私人投資者提供專門設 計的財務解決方案。這類以客為本的業務不單提供全面的銀行服務 （包括融資、諮詢顧 問及交易服務） ，並且從事信貸、利率、外匯、股票、貨幣市場和證券服務的資本市場業 務；以及資本投資活動。 ‧ 環球私人銀行業務提供一系列服務，以照顧集團各個優先發展市場內的資產豐厚人士及 家族的複雜及國際理財需要。 營業類別變動 滙豐的營業類別為歐洲、亞洲、中東及北非、北美洲及拉丁美洲。過往，滙豐的營業類別呈 報為歐洲、香港、亞太其他地區、中東及北非、北美洲及拉丁美洲。經考慮向主要經營決策 者呈示的地區財務資料後，香港及亞太其他地區不再被視為獨立呈報營業類別。由2014年 1月1日起，香港及亞太其他地區已由新的營業類別 「亞洲」 取代，該類別在作出業務決策及 分配資源時更能配合用於評估的內部管理資料。主要經營決策者繼續為集團管理委員會， 而計算分類業績的基準並無改變。因此，已重新呈列比較財務資料。 亞洲類別包含的相關業務營運並無改變。截至2014年12月31日止年度，亞洲的列賬基準營 業收益淨額為236.77億美元 （2013年12月31日：244.32億美元；2012年12月31日：253.32億美 元） 。相比香港及亞太其他地區個別計算的列賬基準營業收益淨額之和，上述金額低7.13億 美元 （2013年12月31日：低7.49億美元；2012年12月31日：低6.74億美元） 。營業收益淨額的減 幅被營業支出的相同減幅所抵銷。該差異與兩個地區之間的內部服務費用分攤及所進行的 業務活動有關，該等數額按個別基準呈報時構成兩地的收入或支出，但併入亞洲類別呈報 時會作為 「內部項目」 的活動而撇銷。亞洲的列賬基準除稅前利潤與香港及亞太其他地區個 別計算的除稅前利潤之和並無分別。 本年度利潤╱ （虧損） 滙豐 歐洲 亞洲 中東及北非 北美洲 拉丁美洲 內部項目 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 2014年 淨利息收益 10,611 12,273 1,519 5,015 5,310 (23) 34,705 費用收益淨額 6,042 5,910 650 1,940 1,415 － 15,957 交易收益淨額 2,534 2,622 314 411 856 23 6,760 其他收益 2,384 2,872 65 786 691 (2,972) 3,826 營業收益淨額1 21,571 23,677 2,548 8,152 8,272 (2,972) 61,248 貸款減值準備 （提撥） ╱收回及 其他信貸風險準備 (764) (647) 6 (322) (2,124) － (3,851) 營業收益淨額 20,807 23,030 2,554 7,830 6,148 (2,972) 57,397 僱員報酬及福利 (8,191) (5,862) (676) (3,072) (2,565) － (20,366) 一般及行政開支 (11,076) (3,959) (500) (3,108) (2,894) 2,972 (18,565) 物業、機器及設備折舊與減值 (543) (389) (28) (180) (242) － (1,382) 無形資產之攤銷及減值 (407) (217) (12) (69) (231) － (936) 營業支出總額 (20,217) (10,427) (1,216) (6,429) (5,932) 2,972 (41,249) 營業利潤 590 12,603 1,338 1,401 216 － 16,148 應佔聯營及合資公司利潤 6 2,022 488 16 － － 2,532 除稅前利潤 596 14,625 1,826 1,417 216 － 18,680 稅項支出 (853) (2,542) (339) (195) (46) － (3,975) 本年度利潤╱ （虧損） (257) 12,083 1,487 1,222 170 － 14,705 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 373 本年度利潤╱ （虧損） （續） 滙豐 歐洲 亞洲 中東及北非 北美洲 拉丁美洲 內部項目 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 2013年 淨利息收益 10,693 11,432 1,486 5,742 6,186 － 35,539 費用收益淨額 6,032 5,936 622 2,143 1,701 － 16,434 交易收益淨額 4,423 2,026 357 948 936 － 8,690 其他收益╱ （支出） (181) 5,038 38 (30) 1,745 (2,628) 3,982 營業收益淨額1 20,967 24,432 2,503 8,803 10,568 (2,628) 64,645 貸款減值準備 （提撥） ╱收回 及其他信貸風險準備 (1,530) (498) 42 (1,197) (2,666) － (5,849) 營業收益淨額 19,437 23,934 2,545 7,606 7,902 (2,628) 58,796 僱員報酬及福利 (7,175) (5,666) (634) (3,098) (2,623) － (19,196) 一般及行政開支 (9,479) (3,660) (607) (3,051) (2,896) 2,628 (17,065) 物業、機器及設備折舊與減值 (559) (392) (35) (176) (202) － (1,364) 無形資產之攤銷及減值 (400) (218) (13) (91) (209) － (931) 營業支出總額 (17,613) (9,936) (1,289) (6,416) (5,930) 2,628 (38,556) 營業利潤 1,824 13,998 1,256 1,190 1,972 － 20,240 應佔聯營及合資公司利潤 1 1,855 438 31 － － 2,325 除稅前利潤 1,825 15,853 1,694 1,221 1,972 － 22,565 稅項支出 (1,279) (2,170) (328) (313) (675) － (4,765) 本年度利潤 546 13,683 1,366 908 1,297 － 17,800 2012年 淨利息收益 10,394 10,707 1,470 8,117 6,984 － 37,672 費用收益淨額 6,169 5,418 595 2,513 1,735 － 16,430 交易收益淨額 2,707 2,516 390 507 971 － 7,091 出售美國分行網絡、美國卡業務及 平安保險所得利潤 － 3,012 － 4,012 － － 7,024 其他收益╱ （支出） (1,662) 3,679 (25) (456) 1,261 (2,684) 113 營業收益淨額1 17,608 25,332 2,430 14,693 10,951 (2,684) 68,330 貸款減值及其他信貸風險準備 (1,921) (510) (286) (3,457) (2,137) － (8,311) 營業收益淨額 15,687 24,822 2,144 11,236 8,814 (2,684) 60,019 僱員報酬及福利 (8,070) (5,712) (652) (3,243) (2,814) － (20,491) 一般及行政開支 (10,059) (3,619) (459) (5,413) (3,117) 2,684 (19,983) 物業、機器及設備折舊與減值 (597) (427) (44) (195) (221) － (1,484) 無形資產之攤銷及減值 (369) (222) (11) (89) (278) － (969) 營業支出總額 (19,095) (9,980) (1,166) (8,940) (6,430) 2,684 (42,927) 營業利潤╱ （虧損） (3,408) 14,842 978 2,296 2,384 － 17,092 應佔聯營及合資公司利潤╱ （虧損） (6) 3,188 372 3 － － 3,557 除稅前利潤╱ （虧損） (3,414) 18,030 1,350 2,299 2,384 － 20,649 稅項支出 (173) (2,711) (254) (1,313) (864) － (5,315) 本年度利潤╱ （虧損） (3,587) 15,319 1,096 986 1,520 － 15,334 1 未扣除貸款減值及其他信貸風險準備之營業收益淨額 （亦稱收入） 。 財務報表附註 （續） 滙豐控股有限公司 374 11－按類分析 有關本年度利潤╱ （虧損） 之其他資料 滙豐 歐洲 亞洲 中東及北非 北美洲 拉丁美洲 內部項目 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 2014年 營業收益淨額1 21,571 23,677 2,548 8,152 8,272 (2,972) 61,248 －外來 20,450 22,071 2,524 7,937 8,266 － 61,248 －項目之間 1,121 1,606 24 215 6 (2,972) － 本年度利潤包括 下列重大非現金項目： 折舊、攤銷及減值 950 606 40 182 473 － 2,251 未減收回額及其他信貸風險準備 之貸款減值虧損 1,066 800 37 437 2,466 － 4,806 金融投資減值 (256) 286 － 14 10 － 54 長期債務及相關衍生工具之公允值變動 614 (4) (3) (99) － － 508 重組架構成本 117 7 2 28 57 － 211 2013年 營業收益淨額1 20,967 24,432 2,503 8,803 10,568 (2,628) 64,645 －外來 20,108 22,853 2,497 8,569 10,618 － 64,645 －項目之間 859 1,579 6 234 (50) (2,628) － 本年度利潤包括 下列重大非現金項目： 折舊、攤銷及減值 957 610 48 303 412 － 2,330 未減收回額及其他信貸風險準備 之貸款減值虧損 2,165 665 45 1,321 2,949 － 7,145 金融投資減值 (61) 4 － 15 6 － (36) 長期債務及相關衍生工具之公允值變動 (936) (1) (3) (288) － － (1,228) 重組架構成本 211 79 3 100 42 － 435 2012年 營業收益淨額1 17,608 25,332 2,430 14,693 10,951 (2,684) 68,330 －外來 16,405 23,893 2,455 14,566 11,011 － 68,330 －項目之間 1,203 1,439 (25) 127 (60) (2,684) － 本年度利潤包括 下列重大非現金項目： 折舊、攤銷及減值 966 649 55 363 499 － 2,532 未減收回額及其他信貸風險準備 之貸款減值虧損 2,329 691 361 3,587 2,489 － 9,457 金融投資減值 420 62 1 32 4 － 519 長期債務及相關衍生工具之公允值變動 (3,091) (4) (13) (1,219) － － (4,327) 重組架構成本 292 128 27 219 94 － 760 1 未扣除貸款減值及其他信貸風險準備之營業收益淨額 （亦稱收入） 。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 375 資產負債表資料 滙豐 歐洲 亞洲 中東及北非 北美洲 拉丁美洲 內部項目 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年12月31日 客戶貸款 409,733 362,955 29,063 129,787 43,122 － 974,660 於聯營及合資公司之權益 175 14,958 2,955 83 10 － 18,181 資產總值 1,290,926 878,723 62,417 436,859 115,354 (150,140) 2,634,139 客戶賬項 545,959 577,491 39,720 138,884 48,588 － 1,350,642 負債總額 1,223,371 807,998 52,569 398,356 102,007 (150,140) 2,434,161 已產生之資本支出1 1,168 637 25 208 348 － 2,386 於2013年12月31日 客戶貸款 456,110 336,897 27,211 127,953 43,918 － 992,089 於聯營及合資公司之權益 169 13,822 2,575 74 － － 16,640 資產總值 1,392,959 831,791 60,810 432,035 113,999 (160,276) 2,671,318 客戶賬項 581,933 548,483 38,683 140,809 51,389 － 1,361,297 負債總額 1,326,537 770,938 50,706 393,635 99,319 (160,276) 2,480,859 已產生之資本支出1 907 1,236 32 265 385 － 2,825 於2012年12月31日 客戶貸款 436,141 310,665 28,086 134,475 53,605 － 962,972 於聯營及合資公司之權益 178 15,309 2,262 85 － － 17,834 資產總值 1,389,240 804,709 62,605 490,247 131,277 (185,540) 2,692,538 客戶賬項 535,215 529,754 39,583 141,700 65,144 － 1,311,396 負債總額 1,327,487 749,561 53,498 450,480 113,923 (185,540) 2,509,409 已產生之資本支出1 925 544 102 248 458 － 2,277 1 物業、機器及設備和其他無形資產所產生之支出。不包括透過業務合併而取得之資產及商譽。 其他財務資料 按環球業務列示之營業收益淨額 零售銀行 環球銀行 環球 滙豐 及財富管理 工商金融 及資本市場 私人銀行 其他 1 內部項目 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 2014年 營業收益淨額2 24,594 16,303 17,778 2,377 6,365 (6,169) 61,248 －外來 22,692 16,879 20,055 1,980 (358) － 61,248 －內部 1,902 (576) (2,277) 397 6,723 (6,169) － 2013年 營業收益淨額2 26,740 16,365 19,176 2,439 5,651 (5,726) 64,645 －外來 25,038 17,241 20,767 1,955 (356) － 64,645 －內部 1,702 (876) (1,591) 484 6,007 (5,726) － 2012年 營業收益淨額2 33,861 16,551 18,273 3,172 2,332 (5,859) 68,330 －外來 31,980 17,295 20,410 2,413 (3,768) － 68,330 －內部 1,881 (744) (2,137) 759 6,100 (5,859) － 1 在 「其他」 類別呈列之主要項目，包括若干物業相關業務、未分類的投資業務、集中持有的投資公司、本 身債務的公允值變動及滙豐之控股公司及融資業務。 「其他」 類別亦包括出售若干主要附屬公司或業務 單位所錄得的利潤及虧損。 2 未扣除貸款減值及其他信貸風險準備之營業收益淨額 （亦稱收入） 。 財務報表附註 （續） 滙豐控股有限公司 376 11－按類分析╱12－交易用途資產 按國家╱地區列示之資料 2014年 2013年 2012年 外來營業 非流動 外來營業 非流動 外來營業 非流動 收益淨額 1,2 資產 3 收益淨額 1,2 資產 3 收益淨額 1,2 資產 3 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 英國 14,392 8,671 13,347 17,481 9,149 18,391 香港 12,656 12,376 12,031 12,170 11,307 11,657 美國 5,736 5,685 6,121 4,189 11,779 6,718 法國 2,538 10,301 3,111 11,565 2,881 11,074 巴西 4,817 1,403 5,364 1,715 6,395 2,017 其他國家╱地區 21,109 28,273 24,671 27,879 26,819 30,078 截至12月31日止年度╱於12月31日 61,248 66,709 64,645 74,999 68,330 79,935 1 外來營業收益淨額，乃按負責匯報業績或貸出款項之分行所在地計入相關國家╱地區。 2 未扣除貸款減值及其他信貸風險準備之營業收益淨額 （亦稱收入） 。 3 非流動資產包括物業、機器及設備、商譽、其他無形資產、於聯營及合資公司之權益，以及預期於業績 報告期後12個月以上才可收回之若干其他資產。 按過往地區營業類別呈列的財務資料 亞太 滙豐 歐洲 香港 其他地區 中東及北非 北美洲 拉丁美洲 內部項目 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 營業收益淨額1 2014年 營業收益淨額 21,571 13,844 10,546 2,548 8,152 8,272 (3,685) 61,248 －外來 20,450 12,656 9,415 2,524 7,937 8,266 － 61,248 －項目之間 1,121 1,188 1,131 24 215 6 (3,685) － 2013年 營業收益淨額 20,967 13,203 11,978 2,503 8,803 10,568 (3,377) 64,645 －外來 20,108 12,031 10,822 2,497 8,569 10,618 － 64,645 －項目之間 859 1,172 1,156 6 234 (50) (3,377) － 2012年 營業收益淨額 17,608 12,422 13,584 2,430 14,693 10,951 (3,358) 68,330 －外來 16,405 11,307 12,586 2,455 14,566 11,011 － 68,330 －項目之間 1,203 1,115 998 (25) 127 (60) (3,358) － 除稅前利潤╱ （虧損） 截至下列日期止年度： 2014年12月31日 596 8,142 6,483 1,826 1,417 216 － 18,680 2013年12月31日 1,825 8,089 7,764 1,694 1,221 1,972 － 22,565 2012年12月31日 (3,414) 7,582 10,448 1,350 2,299 2,384 － 20,649 資產負債表資料 於2014年12月31日 資產總值 1,290,926 587,534 359,757 62,417 436,859 115,354 (218,708) 2,634,139 負債總額 1,223,371 556,388 320,178 52,569 398,356 102,007 (218,708) 2,434,161 於2013年12月31日 資產總值 1,392,959 555,413 335,937 60,810 432,035 113,999 (219,835) 2,671,318 負債總額 1,326,537 523,579 306,918 50,706 393,635 99,319 (219,835) 2,480,859 於2012年12月31日 資產總值 1,389,240 518,334 342,269 62,605 490,247 131,277 (241,434) 2,692,538 負債總額 1,327,487 496,640 308,815 53,498 450,480 113,923 (241,343) 2,509,409 1 未扣除貸款減值及其他信貸風險準備之營業收益淨額。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 377 12 交易用途資產 會計政策 倘購入金融資產時，主要目的是在短期內出售，或屬於一併管理的已識別金融工具組合之一部分，並有證據顯示具 有短期獲利的近期模式，則分類為持作交易用途。此等金融資產會於交易日 （即滙豐與交易對手訂立合約安排之日） 確認入賬，並一般會於出售時撤銷確認。其首次列賬均按公允值計量，其交易支出會計入收益表內。其後，其公允 值變動會在收益表的 「交易收益淨額」 項內確認。就交易用途資產而言，利息會列於 「交易收益淨額」 內。 交易用途資產 2014年 2013年 百萬美元 百萬美元 交易用途資產： －交易對手不可能或不會再質押或轉售 247,586 201,492 －交易對手可能再質押或轉售 56,607 101,700 於12月31日 304,193 303,192 國庫及其他合資格票據 16,170 21,584 債務證券 141,532 141,644 股權證券 75,249 63,891 按公允值訂值的交易用途證券 232,951 227,119 同業貸款1 27,581 27,885 客戶貸款1 43,661 48,188 於12月31日 304,193 303,192 1 同業及客戶貸款包括反向回購、結算賬項、借入股票及其他款項。 按公允值訂值的交易用途證券 1 2014年 2013年 百萬美元 百萬美元 美國財政部及美國政府機構2 25,880 23,450 英國政府 9,280 11,591 香港政府 6,946 5,909 其他政府 78,774 86,714 資產抵押證券3 3,494 2,736 企業債務及其他證券 33,328 32,828 股權證券 75,249 63,891 於12月31日 232,951 227,119 1 上述數字包括由銀行及其他金融機構發行之債務證券，其價值為223.99億美元 （2013年：229.89億美元） ， 其中29.49億美元 （2013年：39.73億美元） 由多個政府提供擔保。 2 包括獲美國政府發出明文保證支持之證券。 3 不包括計入美國財政部及美國政府機構內之資產抵押證券。 在認可交易所上市及非上市的交易用途證券 國庫及 其他合資格 票據 債務證券 股權證券 總計 百萬美元 百萬美元 百萬美元 百萬美元 公允值 上市1 1,311 98,028 74,542 173,881 非上市2 14,859 43,504 707 59,070 於2014年12月31日 16,170 141,532 75,249 232,951 公允值 上市1 194 85,821 62,724 148,739 非上市2 21,390 55,823 1,167 78,380 於2013年12月31日 21,584 141,644 63,891 227,119 1 上市投資包括在香港上市之證券，其價值為59.56億美元 （2013年：38.36億美元） 。 2 非上市國庫及其他合資格票據主要包括並非在交易所上市但有流通市場的國庫票據。 財務報表附註 （續） 滙豐控股有限公司 378 13 按公允值列賬之金融工具的公允值 會計政策 所有金融工具首次列賬均按公允值確認。公允值是指市場參與者之間於計量日期在有秩序交易中出售資產將會收取 或轉讓負債將會支付的價格。首次確認入賬之金融工具公允值一般為其交易價格，即已付出或收取之代價的公允 值。然而，有時公允值會按當時同一工具 （未經改動或重新包裝） 的其他可觀察現行市場交易價格來釐定，或根據變 數僅包含可觀察市場數據的估值方法計算，這些市場數據包括利率孳息曲線、期權波幅及匯率。倘存在此等證據， 滙豐於訂約時會確認交易損益 （ 「首日損益」 ） ，即交易價格與公允值之間的差額。當使用重大不可觀察參數時，整筆 首日損益會遞延直至交易到期時、平倉時、估值數據變為可觀察時或於滙豐訂立對銷交易時，於交易有效期內在收 益表內確認。 金融工具之公允值一般按個別基準計量。然而，在滙豐按市場或信貸風險的淨額管理一組金融資產及負債的情況 下，滙豐會按淨額基準計量金融工具組別的公允值，但會於財務報表內個別呈列相關金融資產及負債，惟相關金融 資產及負債符合附註32所述的IFRS對銷準則者則除外。 關鍵會計估算及判斷 金融工具之估值 公允值的最佳證明是於交投活躍主要市場的報價。於交投活躍市場報價的金融工具之公允值，以所持資產的買入價 及所發行負債的賣出價為基準。倘金融工具於交投活躍市場有報價，持有之金融工具總額的公允值會按單位數目乘 以報價計算。判斷市場是否交投活躍時，考慮因素包括 （但不限於） 交易活動的規模及頻密度、價格能否輕易得知及 買賣價差大小等。買賣價差指市場參與者願意購入的價格與願意賣出的價格間的差額。估值方法可能包括假設其他 市場參與者於估值時可能考慮之因素，包括： ‧ 有關工具日後產生現金流之可能性和預期時間。於評估交易對手依照其合約條款履行有關金融工具的責任的能 力時，可能須作出適當判斷。日後現金流亦可能會受到市場利率的變動影響； ‧ 挑選有關工具適用之折現率。於評估市場參與者認為該工具的利率與合適的無風險利率兩者之間的適當息差 時，須作出適當判斷； ‧ 倘選定運用何種模型來計算公允值時涉及特別主觀的考慮 （例如在評估複雜衍生工具產品的價值時） ，決定所選 模型時需要作出的判斷。 我們因應金融工具類別及可得市場數據運用多種估值方法。大部分估值方法乃以現金流折現分析為基礎，據此採 用折現曲線計算及折現預計日後現金流。在考慮信貸風險前，預計日後現金流可能為已知數 （就固定利率的利率掉 期而言） 或可能為不確定及需要預測 （就浮動利率的利率掉期而言） 。 「預測」 利用市場遠期曲線 （如獲提供） 。在期權 模型下，必須考慮不同潛在未來結果的可能性。此外，部分產品的價值視乎一項以上市場因素而定，而在該等情況 下，通常有需要考慮某一市場因素出現變化會如何影響其他市場因素。需要進行有關計算的模型數據，包括利率孳 息曲線、匯率、波幅、相關性、提前還款及拖欠率。倘屬與有抵押交易對手及以大量貨幣進行的利率衍生工具，滙 豐會使用反映隔夜利率 （ 「隔夜指數掉期」 ） 的折現曲線來估值。 大部分估值方法只採用可觀察市場數據。然而，若干金融工具的估值方法，卻包含一項或多項重大不可觀察市場數 據，因此計量這些工具的公允值時牽涉較大程度的判斷。倘管理層認為工具訂約利潤大部分或超過工具估值5%是 依據不可觀察的數據計算，則該工具將全部歸類為按重大不可觀察數據估值。在此情況下， 「不可觀察」 指僅得少量 甚至沒有當前市場數據可用以釐定可能出現的公平交易價格，但一般而言不是指毫無數據可用作釐定公允值的依據 （例如可採用一致定價的數據） 。 監控架構 公允值須符合監控架構的規定，設立該架構是為了確保公允值由承受風險部門以外的部門 釐定或驗證。 至於參考外界報價或輸入模型的可觀察定價數據而釐定公允值的所有金融工具，則使用獨 立定價或驗證。在交投不活躍的市場，滙豐將尋找其他市場資料以驗證金融工具的公允值， 並特別著重被認為較有關連及較為可靠的資料。這方面的考慮因素包括： ‧ 價格預計能反映實際成交價或可交易價格的程度； ‧ 金融工具之間的相似程度； ‧ 不同資料來源的一致程度； ‧ 定價數據提供者採集數據所依循的程序； 13－按公允值列賬之金融工具的公允值 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 379 ‧ 由市場數據相關日期至結算日的時間差距；及 ‧ 獲取數據的方式。 對於以估值模型釐定的公允值，監控架構可能包括 （如適用） 獨立後勤部門對(i)估值模型所 用邏輯；(ii)該等模型所用數據；(iii)估值模型以外任何必要調整；及(iv) （如情況可行） 模型推 算結果的推論或驗證。估值模型須經盡職審查及校準程序方可採用，並會持續根據外界市 場數據作出調整。 公允值變動一般須進行利潤及虧損分析程序。此項程序將公允值變動分為三個高層次類別： (i) 組合變化，例如新交易或到期交易； (ii)市場變動，例如匯率或股價變動；及(iii)其他，例如 公允值調整變動 （詳情請參照下文） 。 大部分按公允值計量的金融工具均來自環球銀行及資本市場業務。下圖以環球銀行及資本 市場業務的公允值管治架構作為例子說明： 集團財務董事 財務部 負責釐定公允值： ‧ 制訂公允值的會計政策 ‧ 制訂規管估值的程序 ‧ 確保遵守所有相關會計 準則 提供 結果 估值委員會 除高級管理人員外， 成員亦包括數個 獨立後勤部門 （產品監控部、市場 風險管理部、量化風險 管理與估值小組及財務部） 的估值專家 由檢討小組 監督並向小組 匯報所有 認為相當 主觀的估值 估值委員會檢討小組 由環球資本市場業務 產品監控部的環球 主管出任主席 成員包括環球 資本市場業務、財務部及 風險管理部的主管 按公允值計量之金融負債 於若干情況下，滙豐會按公允值將已發行本身債務入賬，而公允值則根據有關特定工具於 交投活躍市場的報價 （如有） 釐定。其中一個例子是運用利率衍生工具對沖已發行本身債務 的情況。如缺乏市場報價，已發行本身債務則使用估值方法估值，估值採用的數據會以該 工具於交投不活躍市場的報價為依據，或透過與近似工具於交投活躍市場的報價作比較而 估計。於此兩種情況下，公允值均會計入適用於滙豐負債的信貸息差影響。因集團本身信 貸息差令已發行債務證券公允值產生的變動，乃按下列方式計算：於每個業績報告日期為 每類證券取得外部可核證價格，或以同一發行人的近似證券之信貸息差計算價格。然後， 集團會採用現金流折現法，採用以倫敦銀行同業拆息為基準的折現曲線為每類證券估值。 估值出現差異是由於集團本身的信貸息差。所有證券均一致採用這種計算方法。 已發行結構票據及若干其他混合工具的負債，均列入交易用途負債項內，並按公允值計量。 應用於該等工具的信貸息差源自滙豐發行結構票據時採用的息差。 假若該等債務並非按溢價或折讓償還，因滙豐發行負債之信貸息差變動而產生的損益，將 於債務合約期限內撥回。 財務報表附註 （續） 滙豐控股有限公司 380 公允值等級制 金融資產及金融負債的公允值根據以下等級制釐定： ‧ 第一級－採用市場報價的估值方法：在計量日期有相同工具於滙豐可以進入的交投活躍 市場報價的金融工具。 ‧ 第二級－採用可觀察數據的估值方法：有近似工具於交投活躍市場報價，或有相同或近 似工具於交投不活躍市場報價的金融工具，以及運用所有重要數據均可觀察的模型估值 的金融工具。 ‧ 第三級－採用涉及重大不可觀察數據的估值方法：以使用一項或多項不可觀察重要數據 之估值方法估值的金融工具。 下表按公允值等級制劃分的金融工具。 按公允值列賬之金融工具及估值基準 估值方法 市場報價 第一級 採用 可觀察數據 第二級 涉及重大不 可觀察數據 第三級 總計 百萬美元 百萬美元 百萬美元 百萬美元 於2014年12月31日經常性公允值計量 資產 交易用途資產 180,446 117,279 6,468 304,193 指定以公允值列賬之金融資產 23,697 4,614 726 29,037 衍生工具 4,366 337,718 2,924 345,008 金融投資：可供出售 241,464 131,264 4,988 377,716 負債 交易用途負債 62,385 122,048 6,139 190,572 指定以公允值列賬之金融負債 3,792 72,361 － 76,153 衍生工具 4,649 334,113 1,907 340,669 於2013年12月31日經常性公允值計量 資產 交易用途資產 182,721 115,124 5,347 303,192 指定以公允值列賬之金融資產 30,173 7,649 608 38,430 衍生工具 2,539 277,224 2,502 282,265 金融投資：可供出售 262,836 130,760 7,245 400,841 負債 交易用途負債 88,935 110,576 7,514 207,025 指定以公允值列賬之金融負債 10,482 78,602 － 89,084 衍生工具 4,508 267,441 2,335 274,284 第二級衍生工具款額的增加，反映衍生工具款額整體增加及於附註16詳述。於2014年並無 其他重大變動。 第一級與第二級公允值之間的轉撥 資產 負債 可供出售 持作 交易用途 指定 以公允值 計入損益賬 衍生工具 持作 交易用途 指定 以公允值 計入損益賬 衍生工具 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年12月31日 由第一級轉撥往第二級 2,702 18,149 － － 22,964 － － 由第二級轉撥往第一級 － － － － － － － 公允值等級制中各級之間的轉撥被視作於每半年度業績報告期末出現。由第一級轉撥往第 二級主要反映在重新評估就結算賬項款額應用等級標準後，結算賬項款額及現金抵押品的 重新分類。 13－按公允值列賬之金融工具的公允值 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 381 公允值調整 倘若滙豐認為估值模型並未包括市場參與者將會考慮的其他因素，則會作出公允值調整。 滙豐將公允值調整分類為 「風險相關」 或 「模型相關」 。該等調整大多數與環球銀行及資本市 場業務有關。 公允值調整水平的變動不一定於收益表中確認為利潤或虧損。舉例說，改良模型後，可能 毋須再進行公允值調整。同樣地，相關持倉進行平倉後，公允值調整將減少，但未必會產生 利潤或虧損。 環球銀行及資本市場業務的公允值調整 2014年 2013年 百萬美元 百萬美元 調整類別 風險相關 1,958 1,565 －買賣 539 561 －不確定程度 357 343 －信貸估值調整 871 1,274 －借記估值調整 (270) (616) －資金公允值調整 460 － －其他 1 3 模型相關 57 202 －模型限制 52 199 －其他 5 3 訂約利潤 （首日損益儲備） （附註16） 114 167 於12月31日 2,129 1,934 經常性公允值調整的最大變動是信貸估值調整下降4.03億美元，原因是衍生工具交易對手 風險減低及信貸違責掉期息差整體收窄。滙豐信貸違責掉期息差收窄同樣導致借記估值調 整減少3.46億美元。 資金公允值調整反映提供資金予非按隔夜指數掉期利率計算的非抵押衍生工具組合之潛在 日後成本或貢益。採納資金公允值調整於2014年的影響為交易收益淨額減少2.63億美元，反 映計入倫敦銀行同業拆息的資金息差。資金公允值調整按隔夜指數掉期計量，而資金公允 值調整結餘總額4.6億美元亦反映先前於非抵押衍生工具組合公允值反映之隔夜指數掉期與 倫敦銀行同業拆息的差距。 風險相關調整 買賣 IFRS 13規定使用買賣價差內最能代表公允值的價格。估值模型通常會產生市場中間價值。 買賣調整反映使用現有對沖工具或透過出售持倉或平倉封鎖絕大部分剩餘組合淨額的市場 風險時，將產生的買賣成本。 不確定程度 若干模型數據可能較難基於市場數據釐定，及╱或模型的選擇本身可能較主觀。於該等情 況下，金融工具或市場指標或會假設一系列不同的可能價值，並可能需要作出調整，以反 映於估計金融工具的公允值時，市場參與者可能就不確定的參數及╱或模型假設，採用較 估值模型所用者更為保守的價值。 信貸估值調整 信貸估值調整為對場外衍生工具合約估值作出的調整，藉以於公允值中反映交易對手可能 拖欠還款而滙豐未必可以收取交易的全部市場價值之可能性 （請參閱下文） 。 借記估值調整 借記估值調整為對場外衍生工具合約估值作出的調整，藉以於公允值中反映滙豐可能拖欠 還款而滙豐未必可以支付交易的全部市場價值之可能性 （請參閱下文） 。 財務報表附註 （續） 滙豐控股有限公司 382 資金公允值調整 資金公允值調整採用場外衍生工具組合的任何非抵押部分預期日後資金風險的日後市場資 金息差計算。除全部為非抵押的衍生工具外，其包括抵押衍生工具的非抵押部分。預期日 後資金風險乃按模擬方法 （如適用） 計算。預期日後資金風險會因為可能終止風險項目的事 件 （如滙豐或交易對手違責） 而作出調整。資金公允值調整及借記估值調整各自單獨計算。 模型相關調整 模型限制 用作組合估值之模型或會以一套簡化而並非包含所有重大市場特性的假設為基準。此外， 由於市場演變，於過往足可用作估值的模型可能要加以發展，以包含當前市場狀況的所有 重大市場特性。於該等情況下，會採用模型限制調整。隨著模型進一步發展，估值模型內已 解決模型限制的問題，因此不再需要作出模型限制調整。 訂約利潤 （首日損益儲備） 倘估算公允值時採用的估值模型以一項或以上重大不可觀察數據為基準，將採用訂約利潤 調整。訂約利潤調整的會計處理方法於第378頁討論。遞延首日損益儲備變動之分析，載於 第395頁。 信貸估值調整╱借記估值調整方法 滙豐就每個滙豐旗下法律實體計算獨立的信貸估值調整及借記估值調整，並就每個實體須 因應每個交易對手承受的風險，計算獨立的信貸估值調整及借記估值調整。滙豐將交易對 手的違責或然率 （以滙豐並無違責為條件） 應用於滙豐面對交易對手的預期風險正數值，並 將結果乘以違責時的預期損失，從而計算信貸估值調整。相反，滙豐將其違責或然率 （以交 易對手並無違責為條件） 應用於交易對手面對滙豐的預期風險正數值，並將結果乘以違責時 的預期損失，從而計算借記估值調整。有關計算於潛在風險存續期間進行。 就大部分產品而言，滙豐採用模擬法計算交易對手的預期風險正數值。此方法納入於組合 有效期內與交易對手訂立交易所涉組合的各種潛在風險。模擬方法包括交易對手的淨額計 算協議及與交易對手訂立的抵押品協議等減低信貸風險措施。我們對已發展市場風險普遍 採用60%的標準違責損失率假設，而對新興市場風險則採用75%。在風險性質及可得數據的 支持下，則可能會採用其他違責損失率假設。 至於現時產品之中模擬方法並未支援的若干類非常見衍生工具，或尚未有模擬工具的較小 型交易地點之衍生工具風險，滙豐會採用替代方法。該等方法可能涉及與經模擬工具處理 的近似產品結果進行配對，倘配對方法不適用，則使用通常與模擬方法原則相同的簡化方 法。計算將應用於交易層面，惟淨額計算或抵押品協議等減低信貸風險措施的確認則較模 擬方法所用者更為有限。 該等方法一般並不計入 「錯向風險」 。於進行任何信貸估值調整前，倘衍生工具相關價值與 交易對手的違責或然率成正面的相互關係，便會出現錯向風險。倘出現重大的錯向風險， 將使用特定交易計算法以反映估值內的錯向風險。 除若干中央結算交易對手以外，滙豐將所有第三方交易對手包括於信貸估值調整及借記估 值調整的計算內，而不就滙豐集團旗下公司的有關調整作出淨額計算。滙豐會持續檢討及 改良信貸估值調整及借記估值調整採用之方法。 非抵押衍生工具的估值 過往，滙豐已按基準利率 （一般為倫敦銀行同業拆息或其等值） 折現預計日後現金流從而對 非抵押衍生工具進行估值。為配合行業慣例的發展，滙豐於2014年下半年更改此方法。滙豐 現時將隔夜指數掉期曲線視為所有衍生工具 （抵押及非抵押） 的基準折現曲線，並採納 「資金 公允值調整」 ，從而反映非按隔夜指數掉期利率計算的非抵押衍生工具風險項目的資金。採 納資金公允值調整的影響為交易收入減少2.63億美元。於此方面，行業並未達成一致共識。 滙豐將持續關注行業發展及在必要時調整計算方法。 13－按公允值列賬之金融工具的公允值 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 383 公允值估值基準 採用涉及重大不可觀察數據的估值方法按公允值計量的金融工具－第三級 資產 負債 持作 按公 持作 按公 可供出售 交易用途 允值計量 1 衍生工具 總計 交易用途 允值計量 1 衍生工具 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 私募股本 （包括策略投資） 3,120 164 432 － 3,716 47 － － 47 資產抵押證券 1,462 616 － － 2,078 － － － － 持作證券化用途之貸款 － 39 － － 39 － － － － 結構票據 － 2 － － 2 6,092 － － 6,092 涉及債券承保公司之衍生工具 － － － 239 239 － － 1 1 其他衍生工具 － － － 2,685 2,685 － － 1,906 1,906 其他組合 406 5,647 294 － 6,347 － － － － 於2014年12月31日 4,988 6,468 726 2,924 15,106 6,139 － 1,907 8,046 私募股本 （包括策略投資） 3,729 103 420 － 4,252 － － － － 資產抵押證券 1,677 643 － － 2,320 － － － － 持作證券化用途之貸款 － 83 － － 83 － － － － 結構票據 － 14 － － 14 7,514 － － 7,514 涉及債券承保公司之衍生工具 － － － 320 320 － － － － 其他衍生工具 － － － 2,182 2,182 － － 2,335 2,335 其他組合 1,839 4,504 188 － 6,531 － － － － 於2013年12月31日 7,245 5,347 608 2,502 15,702 7,514 － 2,335 9,849 1 指定以公允值計入損益賬。 第三級金融工具按持續經營業務及既有業務基準呈列。持作證券化用途之貸款、涉及債券 承保公司之衍生工具、若干 「其他衍生工具」 及所有第三級資產抵押證券均為既有業務。滙 豐有能力持有該等倉盤。 私募股本 （包括策略投資） 滙豐的私募股本及策略投資一般歸類為可供出售，且並非於交投活躍市場進行買賣。如某 項投資並無交投活躍的市場，其公允值的估算則基於投資對象的財務狀況及業績、風險狀 況、前景及其他因素的分析，並會參照於交投活躍市場報價的近似企業市價估值，或近似 公司變更擁有權所依據的價格。 資產抵押證券 市場報價一般用以釐定此等證券的公允值，而估值模型則用於核實可得的有限市場數據的 可靠性，以及確認是否需對市場報價作出任何調整。至於包括住宅按揭抵押證券的資產抵 押證券，估值使用行業標準模型，以及根據抵押品類別和履約情況 （如適用） 就提前還款速 度、拖欠率及虧損嚴重程度所作的假設。估值結果會與性質近似的證券之可觀察數據比較， 從而檢測兩者是否一致。 貸款 （包括槓桿融資及持作證券化用途之貸款） 按公允值持有的貸款乃根據經紀報價及╱或市場一致數據提供者 （如有） 提供的數據估值。如 缺乏可觀察市場，則採用其他估值方法釐定公允值，該等估值方法包括現金流折現模型， 此模型包括因應同一或同類機構發行的其他市場工具而假設的貸款的適當信貸息差。 結構票據 結構票據採用涉及重大不可觀察數據的估值方法計算的公允值，源自相關債務證券的公允 值，而內含衍生工具的公允值，則按下文衍生工具一段所述方式釐定。 第三級結構票據主要包括由滙豐發行的股票掛鈎票據 （該等票據向交易對手提供的回報與 若干股權證券表現掛鈎） ，以及其他組合。該等票據因參數不可觀察而歸類為第三級，這些 參數包括遠期股權波幅及股價與股價、股價與利率及利率與匯率之間的相關性等。 財務報表附註 （續） 滙豐控股有限公司 384 衍生工具 場外 （即非交易所買賣） 衍生工具以估值模型估值。估值模型根據 「無套利」 原則計算預計日 後現金流的現值。利率掉期及歐式期權等多種常規衍生工具產品的模型計算法均為業界劃 一採用。對於較複雜的衍生工具產品，市場實際使用的方法可能略有差異。估值模型所用 的數據會盡可能按可觀察市場數據釐定，該等數據包括交易所、交易商、經紀或一致定價 提供者提供的價格。若干數據未必可於市場直接觀察，但可透過模型校準程序按可觀察價 格釐定或按過往數據或其他資料來源估計。不可觀察數據的例子包括不太常見買賣的期權 產品的全部或部分波幅平面，以及匯率、利率及股價等各種市場因素之間的相關性。 以涉及重大不可觀察數據的估值方法估值的衍生工具產品，包括若干類別的相關性產品， 如外匯籃子期權、股票籃子期權、外匯利率混合交易及遠期期權交易。遠期期權交易包括 股票期權、利率及外匯期權，以及若干信貸衍生工具。信貸衍生工具則包括若干分批次的 信貸違責掉期交易。 公允值等級制中第三級公允值計量之對賬 下表詳列採用涉及重大不可觀察數據之估值方法按公允值計量的第三級金融工具於期初與 期末間之結餘變動對賬： 第三級金融工具的變動 資產 負債 指定以公 指定以公 持作 允值計入 持作 允值計入 可供出售 交易用途 損益賬 衍生工具 交易用途 損益賬 衍生工具 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日 7,245 5,347 608 2,502 7,514 － 2,335 於損益賬中確認之 增益╱ （虧損） 總額 174 194 56 959 (25) － (5) －不包括淨利息收益之交易收益 － 194 － 959 (25) － (5) －指定以公允值列賬之其他金融 工具淨收益╱ （支出） － － 56 － － － － －金融投資減除虧損後增益 198 －貸款減值及其他信貸風險準備 (24) 於其他全面收益中確認之 增益╱ （虧損） 總額1 126 (178) (16) (126) (123) － 54 －可供出售投資： 公允值增益╱ （虧損） 208 － － － － － － －現金流對沖： 公允值增益╱ （虧損） － － － (9) － － 34 －匯兌差額 (82) (178) (16) (117) (123) － 20 購入 1,505 705 273 － (31) － － 新發行 － － － － 2,067 － － 出售 (1,237) (481) (149) － － － － 償付 (1,255) (49) (78) 27 (1,655) － (69) 撥出 (3,027) (112) － (544) (1,918) － (527) 撥入 1,457 1,042 32 106 310 － 119 於2014年12月31日 4,988 6,468 726 2,924 6,139 － 1,907 13－按公允值列賬之金融工具的公允值 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 385 資產 負債 指定以公 指定以公 持作 允值計入 持作 允值計入 可供出售 交易用途 損益賬 衍生工具 交易用途 損益賬 衍生工具 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於損益賬中確認與2014年12月 31日所持資產及負債有關之 未變現增益╱ （虧損） (24) 1 46 946 (122) － 134 －不包括淨利息收益之 交易收益 － 1 － 946 (122) － 134 －指定以公允值列賬之 其他金融工具淨收益╱ （支出） － － 46 － － － － －貸款減值及其他信貸風險準備 (24) － － － － － － 於2013年1月1日 8,511 4,378 413 3,059 7,470 － 3,005 於損益賬中確認之 增益╱ （虧損） 總額 (52) 343 36 (205) (747) － 393 －不包括淨利息收益之 交易收益╱ （支出） － 343 － (205) (747) － 393 －指定以公允值列賬之 其他金融工具淨收益 － － 36 － － － － －金融投資減除虧損後增益 (66) － － － － － － －貸款減值及其他信貸風險準備 14 － － － － － － 於其他全面收益中確認之 增益╱ （虧損） 總額1 487 20 － (7) 9 － 57 －可供出售投資：公允值增益 568 － － － － － － －現金流對沖：公允值虧損 － － － (11) － － － －匯兌差額 (81) 20 － 4 9 － 57 購入 1,838 1,293 56 － (482) － － 新發行 － － － － 3,161 － － 出售 (766) (1,821) (4) － (14) － － 償付 (756) (473) (27) (311) (1,150) － (1,004) 撥出 (3,121) (385) (68) (171) (1,051) － (160) 撥入 1,104 1,992 202 137 318 － 44 於2013年12月31日 7,245 5,347 608 2,502 7,514 － 2,335 於損益賬中確認與2013年12月 31日所持資產及負債有關之 未變現增益╱ （虧損） (166) 362 41 (297) (401) － 72 －不包括淨利息收益之交易收益 － 362 － (297) (401) － 72 －指定以公允值列賬之其他 金融工具淨收益 － － 41 － － － － －貸款減值及其他信貸風險準備 (166) － － － － － － 1 計入綜合全面收益表內的 「可供出售投資：公允值增益╱ （虧損） 」 及 「匯兌差額」 。 買賣第三級可供出售資產主要反映資產抵押證券活動，尤其於證券投資中介機構方面的活 動。從第三級可供出售證券撥出反映對若干新興市場企業債務定價的信心加強，而且對若 干資產抵押證券的定價取得更多資料。撥入第三級主要與價格相關資料減少的其他資產抵 押證券有關。新發行交易用途負債反映結構票據發行，主要為股票掛鈎票據。從第三級交 易用途負債中撥出主要與股票掛鈎票據有關，因為若干模型數據變得可觀察。撥入第三級 交易用途資產主要與銀團貸款有關。 財務報表附註 （續） 滙豐控股有限公司 386 重大不可觀察假設改變對合理可行替代假設的影響 下表列示第三級公允值對合理可行替代假設的敏感度： 公允值對合理可行替代假設的敏感度 於損益賬中反映 於其他全面收益中反映 有利變動 不利變動 有利變動 不利變動 百萬美元 百萬美元 百萬美元 百萬美元 衍生工具、交易用途資產及交易用途負債1 296 (276) － － 指定以公允值列賬之金融資產及負債 37 (47) － － 金融投資：可供出售 51 (67) 270 (350) 於2014年12月31日 384 (390) 270 (350) 衍生工具、交易用途資產及交易用途負債1 350 (285) － － 指定以公允值列賬之金融資產及負債 32 (51) － － 金融投資：可供出售 － － 434 (673) 於2013年12月31日 382 (336) 434 (673) 1 衍生工具、交易用途資產及交易用途負債以同一類別呈列，以反映該等金融工具的風險管理方式。 有關衍生工具、交易用途資產及交易用途負債的重大不可觀察數據出現有利及不利變動的 影響減少，主要反映若干新興市場的外匯波幅隨著市場成熟而變得較易掌握。期內有關可 供出售資產的重大不可觀察數據出現有利及不利變動的影響減少，則主要反映第三級工具 的數額下降。 按第三級工具類別劃分的公允值對合理可行替代假設的敏感度 於損益賬中反映 於其他全面收益中反映 有利變動 不利變動 有利變動 不利變動 百萬美元 百萬美元 百萬美元 百萬美元 私募股本 （包括策略投資） 77 (110) 172 (255) 資產抵押證券 49 (22) 60 (55) 持作證券化用途之貸款 1 (1) － － 結構票據 14 (9) － － 涉及債券承保公司之衍生工具 11 (11) － － 其他衍生工具 129 (155) － － 其他組合 103 (82) 38 (40) 於2014年12月31日 384 (390) 270 (350) 私募股本 （包括策略投資） 31 (61) 226 (436) 資產抵押證券 60 (27) 113 (99) 持作證券化用途之貸款 3 (3) － － 結構票據 16 (9) － － 涉及債券承保公司之衍生工具 25 (16) － － 其他衍生工具 212 (164) － － 其他組合 35 (56) 95 (138) 於2013年12月31日 382 (336) 434 (673) 有利及不利變動乃以敏感度分析為基準釐定。敏感度分析旨在計量與應用95%可信程度一 致的公允值範圍。該等方法會考量所採用估值方法的性質，以及可觀察替代法及過往數據 的可獲提供情況及可靠性。當可獲提供數據經不起統計數據分析的驗證，則憑判斷量化不 確定程度，但會維持95%的可信程度。 倘若金融工具的公允值受一個以上不可觀察假設影響，上表反映隨個別假設變化而產生的 最有利或最不利變動。 第三級金融工具之主要不可觀察數據 下表列出第三級金融工具之主要不可觀察數據，並呈列於2014年12月31日該等數據涵蓋之 範圍。主要數據範圍為90%數據所屬的估計範圍。主要不可觀察數據類別之進一步說明載 列如下： 13－按公允值列賬之金融工具的公允值 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 387 有關第三級估值所用重大不可觀察數據的定量資料 公允值 主要不可觀察數據 資產 負債 估值方法 全面數據範圍 核心數據範圍 百萬美元 百萬美元 較低 較高 較低 較高 私募股本 （包括策略投資） 3,716 47 見第388頁所載附註 見第388頁所載附註 不適用 不適用 不適用 不適用 資產抵押證券 2,078 － －CLO╱CDO1 1,122 － 模型－現金流折現 提前還款率 1% 6% 1% 6% 市場替代法 買入報價 0 100 54 85 其他資產抵押證券 956 － 持作證券化用途之貸款 39 － 結構票據 2 6,092 －股票掛鈎票據 － 4,744 模型－期權模型 股權波幅 0.2% 65% 18% 38% 模型－期權模型 股權相關性 27% 92% 44% 79% －基金掛鈎票據 － 562 模型－期權模型 基金波幅 6% 8% 6% 8% －外匯掛鈎票據 2 477 模型－期權模型 外匯波幅 2% 70% 4% 16% －其他 － 309 涉及債券承保公司之衍生工具 239 1 模型－現金流折現 信貸息差 3% 5% 4% 4% 其他衍生工具 2,685 1,906 利率衍生工具： －證券化掉期 449 1,023 模型－現金流折現 提前還款率 0% 50% 6% 18% －遠期利率掉期期權 1,044 152 模型－期權模型 利率波幅 2% 59% 16% 36% －其他 755 151 外匯衍生工具： －外匯期權 89 95 模型－期權模型 外匯波幅 0.1% 70% 4% 14% －其他 7 7 股權衍生工具： －遠期單一認股權 192 256 模型－期權模型 股權波幅 9% 65% 16% 40% －其他 34 162 信貸衍生工具： －其他 115 60 其他組合 6,347 － －結構證 4,420 － 模型－現金流折現 信貸波幅 0.8% 3% 0.8% 3% －新興市場企業債務 372 － 市場替代法 信貸息差 1% 4% 1% 3% 市場替代法 買入報價 58 131 106 130 －其他2 1,555 － 於2014年12月31日 15,106 8,046 1 貸款抵押債券╱債務抵押債券。 2 包括一系列較小規模資產持倉。 財務報表附註 （續） 滙豐控股有限公司 388 私募股本 （包括策略投資） 滙豐的私募股本及策略投資一般歸類為可供出售，且並非於交投活躍市場進行買賣。如某 項投資並無交投活躍的市場，其公允值的估算則依據投資對象的財務狀況及業績、風險狀 況、前景和其他因素之分析，並會參照於交投活躍市場報價的近似企業市價估值，或近似 公司變更擁有權的價格。由於所持各項投資之分析各有不同，列報一系列主要不可觀察數 據並不切實可行。 提前還款率 提前還款率用於計量貸款組合將於到期日前獲提前償還的預計未來速度。提前還款率是資 產抵押證券以模型計算之價值中一項重要數據。如沒有足夠之可觀察市價以供直接釐定市 價，便會使用以模型計算之價格。提前還款率亦是與證券化掛鈎之衍生工具估值中一項重 要數據。例如，所謂證券化掉期具有與證券化中未償還貸款組合規模掛鈎之名義價值，如出 現提前還款則可能下跌。提前還款率會因應貸款組合的性質及對未來市況的預期而有所不 同。例如，普遍預期美國住宅按揭抵押證券的現行提前還款率會隨美國經濟改善而上升。 提前還款率可運用多項證據估算，如替代可觀察證券價格引伸之提前還款率、現時或過往 之提前還款率及宏觀經濟模型等。 市場替代法 倘某項工具沒有特定市場定價，但有具備部分共通特點之工具可提供證據，便可使用市場 替代法定價。在某些情況下，或可識別出特定替代品，但在更多情況下，通常會採用較多種 工具之證據，以了解影響當前市場定價之因素及影響之方式。例如，在貸款抵押債券市場 中，或可確立A級證券存在價格範圍，以及識別出影響範圍內所屬定位之主要因素。應用此 方法估算滙豐組合內之特定A級證券，可以定出一個價格。 因此，用作市場替代定價方法所採納數據之價格範圍可以相當廣泛。該範圍並不反映與個 別證券所得價格相關之不確定程度。 波幅 波幅用於計量市場價格的預計未來變動。在市況受壓的情況下，波幅趨向增加，若市況較 平靜則趨向減少。波幅是為期權定價之重要數據。一般而言，波幅愈大，期權價格愈高。這 反映期權回報率增加之可能性較高，及滙豐對沖與該期權相關之風險可能產生較高的潛在 成本。如期權價格變得更高，滙豐的期權長倉 （即滙豐已購入期權之持倉） 價值將會提高， 而滙豐之期權短倉 （即滙豐沽出期權之持倉） 將蒙受損失。 波幅隨相關參考市價以及期權之行使及到期而變動。波幅亦會隨時間變動。因此，波幅水 平難以一概而論。例如，雖然一般情況下外匯波幅低於股權波幅，但就特定貨幣兌換組合 或特定股權而言，可能出現例外情況。 若干波幅 （通常是期限較長的波幅） 乃不可觀察。不可觀察的波幅因而採用可觀察數據估計。 例如，期限較長的波幅可能由期限較短的波幅推算。第387頁列表中引述的不可觀察波幅範 圍反映經參考市價所得波幅數據變化甚大。例如，掛鈎貨幣的外匯波幅可能甚低，而非受 限制貨幣的外匯波幅可能較高。另一例子是極價內或極價外股票期權的波幅可能會遠高於 平價期權。核心數據範圍遠較全面數據範圍窄，是由於滙豐組合內出現該等波幅極大的例 子較為罕見。就任何單一不可觀察波幅而言，波幅決定因素的不確定程度遠低於上表所列 範圍。 相關性 相關性用於計量兩個市價之間的相互關係，並以介乎-1與1的數字表述。正數相關性暗示兩 個市價趨向往同一方向變動，而相關性為1則暗示兩個市價總是往同一方向變動。負數相 關性暗示兩個市價趨向往相反方向變動，而相關性為-1則暗示兩個市價總是往相反方向變 13－按公允值列賬之金融工具的公允值 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 389 動。相關性用於計算較複雜工具的價值，其派付金額視乎多於一個市價而定。例如，股票籃 子期權的派付金額乃視乎一籃子單一股票的表現，而該等股票價格變動之間的相關性將會 是估值採用的數據。這稱為股票與股票的相關性。有很多種類的工具以相關性作為估值數 據，因此同類資產相關性 （如股票與股票的相關性） 及非同類資產相關性 （如匯率與利率的相 關性） 均會廣泛採用。一般而言，同類資產相關性的變化幅度會較非同類資產相關性為窄。 相關性可能不可觀察。不可觀察相關性可根據多項證據作出估算，包括一致定價服務、滙 豐的交易價格、替代相關性及研究過往價格的關係。 表中所列的不可觀察相關性範圍反映按市價配對組合劃分的相關性數據變化甚大。就任何 單一不可觀察相關性而言，相關性決定因素的不確定程度可能低於上表所列範圍。 信貸息差 信貸息差是市場接納較低信貸質素時要求的基準利率溢價。在現金流折現模型中，信貸息 差增加應用於日後現金流的折現因素，從而降低資產的價值。信貸息差可能從市價引伸。 在流通性較低的市場，未必可以觀察到信貸息差。 主要不可觀察數據之間的相互關係 第三級金融工具之主要不可觀察數據未必相互獨立。如上文所述，市場變數可能具有相關 性。這種相關性通常反映不同市場對宏觀經濟或其他事件傾向採取的對應方式。例如，經 濟狀況改善可能推動市場 「逐險」 ，此時股票及高收益債券等高風險資產的價格將會上升， 而黃金及美國國庫債券等 「避險」 資產的價格將會下跌。此外，不斷轉變的市場變數對滙豐 組合之影響將視乎滙豐涉及各項變數之風險持倉淨額而定。例如，高收益債券價格增加將 使高收益債券長倉得益，但就該等債券持有之任何信貸衍生工具保障的價值將會下跌。 滙豐控股 下表分析於財務報表內按公允值計量之金融資產及金融負債的估值基準： 滙豐控股按公允值計量的金融資產及負債之估值基準 2014年 2013年 百萬美元 百萬美元 採用可觀察數據的估值方法：第二級 於12月31日的資產 衍生工具 2,771 2,789 可供出售 4,073 1,210 於12月31日的負債 指定以公允值列賬 18,679 21,027 衍生工具 1,169 704 財務報表附註 （續） 滙豐控股有限公司 390 14 非按公允值列賬之金融工具的公允值 非按公允值列賬之金融工具的公允值及估值基準 公允值 估值方法 賬面值 市場報價 第一級 採用可 觀察數據 第二級 涉及重大不 可觀察數據 第三級 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年12月31日並非持作出售用途之資產及負債 資產 同業貸款1 112,149 － 109,087 3,046 112,133 客戶貸款1 974,660 － 13,598 959,239 972,837 反向回購協議－非交易用途1 161,713 － 160,600 1,123 161,723 金融投資：債務證券 37,751 1,418 37,671 74 39,163 負債 同業存放1 77,426 － 77,300 98 77,398 客戶賬項1 1,350,642 － 1,336,865 13,730 1,350,595 回購協議－非交易用途1 107,432 － 107,432 － 107,432 已發行債務證券 95,947 146 94,325 1,932 96,403 後償負債 26,664 － 28,806 1,248 30,054 於2013年12月31日並非持作出售用途之資產及負債 資產 同業貸款1 120,046 － 111,297 8,727 120,024 客戶貸款1 992,089 － 10,762 971,520 982,282 反向回購協議－非交易用途1 179,690 － 178,516 1,166 179,682 金融投資：債務證券 25,084 1,432 23,960 25 25,417 負債 同業存放1 86,507 － 86,440 51 86,491 客戶賬項1 1,361,297 － 1,346,343 14,576 1,360,919 回購協議－非交易用途1 164,220 － 164,173 47 164,220 已發行債務證券 104,080 166 101,551 2,941 104,658 後償負債 28,976 － 29,704 1,309 31,013 1 自2014年1月1日起，非交易用途反向回購及回購於資產負債表內分行呈列。過往，非交易用途反向回購 計入 「同業貸款」 及 「客戶貸款」 ，而非交易用途回購則計入 「同業存放」 及 「客戶賬項」 。比較數字已相應 重列。非交易用途反向回購及回購於資產負債表內分行呈列，使披露方式與市場慣例一致，並提供更具 參考價值的貸款相關資料。 公允值乃按照附註13所載的等級制釐定。 其他非按公允值列賬的金融工具主要為短期性質及經常按當前市場利率重新定價。因此， 該等工具的賬面值為其公允值的合理約數。這包括現金及於中央銀行的結餘、向其他銀行 託收╱傳送中之項目、香港政府負債證明書及流通港幣鈔票，全部按攤銷成本計量。 按行業分類之客戶貸款賬面值及公允值 於12月31日的賬面值 並非已減值 已減值 總計 百萬美元 百萬美元 百萬美元 2014年 客戶貸款 954,710 19,950 974,660 －個人貸款 377,154 11,800 388,954 －企業及商業貸款 527,168 8,016 535,184 －金融機構貸款 50,388 134 50,522 2013年 客戶貸款 967,181 24,908 992,089 －個人貸款 390,018 14,108 404,126 －企業及商業貸款 527,483 10,439 537,922 －金融機構貸款 49,680 361 50,041 客戶貸款根據第137頁所述標準分類為並非已減值或已減值。 14－非按公允值列賬之金融工具的公允值 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 391 於12月31日的公允值 並非已減值 已減值 總計 百萬美元 百萬美元 百萬美元 2014年 客戶貸款 954,347 18,490 972,837 －個人貸款 375,615 10,721 386,336 －企業及商業貸款 528,361 7,642 536,003 －金融機構貸款 50,371 127 50,498 2013年 客戶貸款 957,695 24,587 982,282 －個人貸款 379,353 13,774 393,127 －企業及商業貸款 529,029 10,340 539,369 －金融機構貸款 49,313 473 49,786 按地區分類的客戶貸款分析 2014年 2013年 賬面值 公允值 賬面值 公允值 百萬美元 百萬美元 百萬美元 百萬美元 客戶貸款 歐洲 409,733 413,373 456,110 453,331 亞洲 362,955 361,412 336,897 335,132 中東及北非 29,063 28,658 27,211 26,891 北美洲 129,787 126,232 127,953 122,823 拉丁美洲 43,122 43,162 43,918 44,105 於12月31日 974,660 972,837 992,089 982,282 估值 公允值計量為滙豐對市場參與者之間於計量日期在有秩序交易中出售資產將會收取或轉讓 負債將會支付的價格作出的估計，但該金額並不反映滙豐預期於該等工具的預計日後有效 期內，可從該等工具的現金流產生的經濟利益及成本。其他匯報公司可能使用不同的估值 方法及假設，以釐定並無可觀察市場價格之工具的公允值。 下列資產及負債的公允值乃為下文所述披露而估算得出： 同業及客戶貸款 貸款之公允值以可觀察市場交易 （如有） 為依據。如無可觀察市場交易，則以包含一系列假 設數據的估值模型來估計公允值。該等假設可能包括反映場外交易活動由第三方經紀提供 的估計價值；一些前瞻性的現金流折現模型，這些模型運用的假設，與滙豐相信市場參與 者對該等貸款進行估值時所用假設乃屬一致；以及其他市場參與者的交易數據 （包括從觀察 所得的第一及第二市場交易） 。 在可行情況下，貸款會撥歸多個同類組合，並且按類似特性的貸款分層，使預測估值的結 果提高準確度。貸款賬項的分層會考慮所有重大因素，包括年份、辦理時期、估計日後利 率、提前還款速度、拖欠率、貸款估值比率、抵押品質量、違責或然率，以及內部信貸風險 評級。 貸款的公允值反映結算日的貸款減值，以及市場參與者預期貸款有效期內將出現貸款損失 的估算額，以至由辦理貸款至結算日期間重新定價的公允值影響。 北美洲客戶貸款的公允值低於賬面值，主要在美國反映於結算日的市況，原因是過去數年 的嚴峻經濟狀況，包括房價下跌、失業率上升、消費行為改變、折現率變動，以及缺乏支持 購買貸款的可選擇融資。2014年相對公允值上升，主要是由於物業價值上升令樓市好轉， 其次是由於此等類別貸款所需的市場收益率較低而投資者對此等類別貸款的需求增加。 歐洲客戶貸款的公允值相比其賬面值有所提升，主要由於英國按揭市場的競爭加劇及中央 銀行採取刺激貸款的措施，令利率下降及公允值相應上升。 財務報表附註 （續） 滙豐控股有限公司 392 金融投資 上市金融投資之公允值按買入市價釐定。非上市金融投資之公允值則採用估值方法釐定， 當中會考慮同等有報價證券之價格及日後盈利來源。 同業存放及客戶賬項 公允值採用現金流折現法估算，並採用相若尚餘期限之當前存款利率計算。即期存款的公 允值與其賬面值相若。 已發行債務證券及後償負債 公允值乃按結算日之可得市場報價，或參考近似工具之市場報價而釐定。 回購及反向回購協議－非交易用途 公允值按當前利率採用現金流折現法估算。公允值與其賬面值相若，乃由於款額通常為短 期。 滙豐控股 滙豐控股就計量及披露目的而釐定金融工具公允值時採用上述方法。 滙豐控股於資產負債表內非按公允值列賬之金融工具的公允值 2014年 2013年 賬面值 公允值1 賬面值 公允值1 百萬美元 百萬美元 百萬美元 百萬美元 於12月31日的資產 滙豐旗下業務貸款 43,910 45,091 53,344 55,332 於12月31日的負債 應付滙豐旗下業務款項 2,892 2,906 11,685 11,868 已發行債務證券 1,009 1,357 2,791 3,124 後償負債 17,255 20,501 14,167 16,633 1 公允值採用涉及可觀察數據的估值方法 （第二級） 釐定。 15 指定以公允值列賬之金融資產 會計政策 符合下列一項或以上準則的金融工具 （不包括持作交易用途之金融工具） 均會歸入此類別，並會於首次入賬時不可撤 回地指定按此準則列賬。滙豐乃基於下列原因指定金融工具以公允值列賬： ‧ 若按不同基準計量來自相關持倉的金融工具，或確認其損益，會出現前後不一致的計量或確認數值，但採用指 定以公允值列賬的方式則可消除或大幅減少此等不一致情況。根據此標準，滙豐指定的主要類別金融資產為單 位相連保單及單位相連投資合約下的金融資產。在相連合約下對客戶之負債乃按相連基金所持資產的公允值釐 定。倘有關資產並未指定按公允值列賬，則有關資產會分類為可供出售，而其公允值變動將於其他全面收益項 內入賬。相關金融資產及負債乃按公允值基準管理及向管理層呈報。金融資產及相關負債指定以公允值列賬， 可使公允值的變動於收益表內入賬，並於同一項目下呈列； ‧ 金融工具組合若須根據明文規定的風險管理或投資策略按公允值基準管理及評估表現，而且須根據該基準向管 理人員呈報相關金融工具組合的資料，則應指定以公允值列賬，例如若干為應付非相連保單未決賠款而持有之 金融資產。滙豐有明文規定的風險管理及投資策略，目的是於考慮非相連負債後，按淨額基準管理及監察該等 資產的市場風險。公允值的計量亦與保險業務適用規例的監管匯報規定保持一致； ‧ 牽涉的金融工具內含一種或多種並非密切相關的衍生工具。 已指定列賬方式之金融資產會於滙豐與交易對手訂立合約時 （一般為交易日） 按公允值確認，且一般會於出售時撤銷 確認。其後公允值的變動會於收益表之 「指定以公允值列賬之金融工具淨收益」 項內確認。 15－指定以公允值列賬之金融資產 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 393 2014年 2013年 百萬美元 百萬美元 指定以公允值列賬之金融資產： －交易對手不可能或不會再質押或轉售 28,357 38,062 －交易對手可能再質押或轉售 680 368 於12月31日 29,037 38,430 99 91 國庫及其他合資格票據 56 50 債務證券 8,891 12,589 股權證券 20,006 25,711 指定以公允值列賬之證券 28,953 38,350 同業及客戶貸款 84 80 於12月31日 29,037 38,430 指定以公允值列賬之證券 1 2014年 2013年 百萬美元 百萬美元 公允值 美國財政部及美國政府機構2 8 34 英國政府 140 534 香港政府 40 113 其他政府 4,088 4,097 資產抵押證券3 18 140 企業債務及其他證券 4,653 7,721 股權 20,006 25,711 於12月31日 28,953 38,350 1 該等數字包括由銀行及其他金融機構發行之債務證券，其價值為13.88億美元 （2013年：44.19億美元） ，其 中2,400萬美元 （2013年：9,200萬美元） 由多個政府擔保。 2 包括獲美國政府發出明文保證支持之證券。 3 不包括計入美國財政部及美國政府機構類別之資產抵押證券。 在認可交易所上市及非上市的證券 國庫及 其他合資格 票據 債務證券 股權證券 總計 百萬美元 百萬美元 百萬美元 百萬美元 公允值 上市1 5 2,731 13,837 16,573 非上市 51 6,160 6,169 12,380 於2014年12月31日 56 8,891 20,006 28,953 公允值 上市1 － 2,773 18,235 21,008 非上市 50 9,816 7,476 17,342 於2013年12月31日 50 12,589 25,711 38,350 1 上市投資包括在香港認可交易所上市之投資，其價值為13.61億美元 （2013年：11.48億美元） 。 財務報表附註 （續） 滙豐控股有限公司 394 16 衍生工具 會計政策 衍生工具 衍生工具是一種金融工具，其價值來自相關項目 （例如股票、債券、利率、外匯、信貸息差、商品及股票指數或其 他指數） 之價格。 衍生工具首次列賬會按公允值確認，其後亦會按公允值計量。衍生工具之公允值按市場報價計算或採用估值方法釐 定。 在下列情況下，內含衍生工具會與主體合約分別計量：該等工具的經濟特質及風險與主體非衍生工具合約內相關工 具的經濟特質及風險並無明顯及密切的關係；其合約條款在其他方面符合獨立衍生工具的定義；以及合併合約並非 持作交易用途或指定以公允值列賬。分別計量的內含衍生工具會按公允值計量，而公允值的任何變動則於收益表內 確認。 若衍生工具的公允值為正數，會分類為資產，若為負數則分類為負債。 只有符合附註32所列之對銷標準，才可就會計目的對銷不同交易之衍生工具資產及負債。 若為未符合條件採用對沖會計法之衍生工具，其公允值變動所產生之損益 （包括合約利息） ，乃於 「交易收益淨額」 項內列賬。若為與指定以公允值列賬之金融工具一併管理之衍生工具，其損益連同用作對沖相關經濟風險的被對沖 項目之損益，於 「指定以公允值列賬之金融工具淨收益」 項內列賬。如果衍生工具與滙豐發行的指定以公允值列賬 之債務證券一併管理，則合約利息連同已發行債務應付利息一併於 「利息支出」 項內列賬。 對沖會計法 若為指定列作對沖工具之衍生工具，滙豐將其分類為：(i)已確認資產或負債或確實承諾的公允值變動之對沖工具 （ 「公允值對沖」 ） ；(ii)已確認資產或負債或預計交易大有可能產生的日後現金流變動之對沖工具 （ 「現金流對沖」 ） ；或 (iii)海外業務投資淨額的對沖工具 （ 「投資淨額對沖」 ） 。 於訂立對沖關係初期，滙豐會編製文件紀錄對沖工具與被對沖項目之間的關係，以及進行對沖之風險管理目標及策 略。滙豐規定須於開始對沖及持續進行對沖期間，編製文件以評估對沖工具，能否極有效地對銷與對沖風險相關的 被對沖項目公允值或現金流變動。 公允值對沖 指定列作並符合條件列為公允值對沖工具的衍生工具，其公允值如有任何變動，均會連同包括對沖風險的被對沖資 產、負債或組合的公允值變動，在收益表內列賬。如對沖關係不再符合採用對沖會計法的條件，則不再採用對沖會 計法；而被對沖項目的賬面值累計調整，將按重新計算之實質利率於到期前的剩餘期間在收益表內攤銷，但如果被 對沖項目已撤銷確認，上述數額便會即時於收益表內確認。 現金流對沖 指定列作並符合條件列為現金流對沖的衍生工具，其公允值變動的有效部分，會於其他全面收益項內確認；而公允 值變動的低效用部分則會即時在收益表內確認。 在其他全面收益項內確認的累計損益，在被對沖項目影響利潤或虧損的期間，會重新分類納入收益表內。於對沖預 計交易導致確認非金融資產或負債時，先前於其他全面收益項內確認之損益會計入首次列賬時計量之資產或負債 內。 當對沖關係終止時，在其他全面收益項內確認的任何累計損益，仍會繼續保留在股東權益項內，直至預計交易於收 益表內確認為止。如預計交易預期不會落實進行，先前在其他全面收益項內確認的累計損益，會即時重新分類納入 收益表內。 投資淨額對沖 海外業務投資淨額對沖的列賬方式與現金流對沖相若。對沖工具有效部分的損益會於其他全面收益項內確認，而公 允值的剩餘變動則即時於收益表內確認。之前在其他全面收益項內確認的損益，會於出售或部分出售海外業務時重 新分類納入收益表內。 對沖效用測試 要符合條件採用對沖會計法，滙豐規定於開始對沖及在對沖有效期內，每項對沖必須預期能夠持續發揮極大預期及 追溯效用。 就每項對沖關係編製之文件，均會列明如何評估對沖項目之效用，而各企業採納之對沖效用評估方法，將視乎其風 險管理策略而定。就預期效用而言，對沖工具必須在被指定列為對沖工具期間，預期能極有效地對銷與對沖風險有 關之公允值或現金流變動，有效範圍界定為80%至125%。對沖的低效用部分在收益表的 「交易收益淨額」 項內確認。 未符合條件採用對沖會計法的衍生工具 不合資格對沖乃用以對沖資產及負債經濟風險但並無應用對沖會計法之衍生工具。 16－衍生工具 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 395 按產品合約類別劃分滙豐所持衍生工具之公允值 資產 負債 交易用途 對沖用途 總計 交易用途 對沖用途 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 外匯 95,584 1,728 97,312 95,187 572 95,759 利率 471,379 1,864 473,243 463,456 4,696 468,152 股權 11,694 － 11,694 13,654 － 13,654 信貸 9,340 － 9,340 10,061 － 10,061 商品及其他 3,884 － 3,884 3,508 － 3,508 各類公允值總計 591,881 3,592 595,473 585,866 5,268 591,134 對銷 (250,465) (250,465) 於2014年12月31日 345,008 340,669 外匯 78,652 2,262 80,914 75,350 448 75,798 利率 456,282 2,294 458,576 448,434 4,097 452,531 股權 18,389 － 18,389 22,573 － 22,573 信貸 9,092 － 9,092 8,926 － 8,926 商品及其他 2,624 － 2,624 1,786 － 1,786 各類公允值總計 565,039 4,556 569,595 557,069 4,545 561,614 對銷 (287,330) (287,330) 於2013年12月31日 282,265 274,284 2014年內，衍生工具資產增加，乃由於孳息曲線出現變動及外匯市場波幅增加所致。股權 衍生工具資產及負債減少，反映在公允值總值內計入現金結算交易所買賣股權衍生工具的 不同收益率而非予以 「對銷」 。此項改變對衍生工具資產總值並無影響。 按產品合約類別劃分滙豐控股連同附屬公司所持衍生工具之公允值 資產 負債 交易用途 對沖用途 總計 交易用途 對沖用途 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 外匯 680 － 680 1,066 103 1,169 利率 1,607 484 2,091 － － － 於2014年12月31日 2,287 484 2,771 1,066 103 1,169 外匯 1,774 45 1,819 471 － 471 利率 955 15 970 233 － 233 於2013年12月31日 2,729 60 2,789 704 － 704 衍生工具之用途 滙豐基於三個主要目的進行衍生工具交易：為客戶提供風險管理解決方案、管理客戶業務 產生的組合風險，以及管理和對沖滙豐本身之風險。 滙豐從事的衍生工具活動導致衍生工具組合產生大量未平倉合約。此等持倉經專人持續管 理，以確保維持於可接受之風險水平內。訂立衍生工具交易時，滙豐採用的信貸風險管理 架構與傳統貸款所用者相同，以評估及批核潛在信貸風險。 交易用途衍生工具 滙豐大部分衍生工具交易涉及銷售及交易活動。銷售活動包括為客戶設計及向客戶推銷衍 生工具產品，讓客戶可選擇承擔、轉移、修訂或減低當前或預期風險。交易活動包括市場莊 家及風險管理。市場莊家活動涉及向其他市場參與者報價 （提供買入價及賣出價） ，藉差價 和交易量賺取收入。風險管理活動是為了管理客戶交易引致的風險，主要目的是保持客戶 收益率。 其他分類為持作交易用途之衍生工具，包括不合資格對沖用途衍生工具、低效用之對沖用 途衍生工具，以及不評估對沖效用之對沖用途衍生工具組成部分。 滙豐控股與滙豐旗下業務訂立之衍生工具交易，絕大部分均與指定以公允值列賬之金融負 債一併管理。 財務報表附註 （續） 滙豐控股有限公司 396 持作交易用途的衍生工具之名義合約金額，顯示於結算日未平倉交易的面值，並不代表承 擔的風險額。 按產品類別劃分持作交易用途衍生工具的名義合約金額 滙豐 滙豐控股 2014年 2013年 2014年 2013年 百萬美元 百萬美元 百萬美元 百萬美元 外匯 5,548,075 5,264,978 15,595 17,280 利率 22,047,278 27,056,367 8,650 10,304 股權 568,932 589,903 － － 信貸 550,197 678,256 － － 商品及其他 77,565 77,842 － － 於12月31日 28,792,047 33,667,346 24,245 27,584 年內利率衍生工具名義金額減少，反映我們參與整個業界的 「組合收縮」 行動。 信貸衍生工具 滙豐透過其主要交易業務進行信貸衍生工具交易，並作為廣大用戶的主要交易對手、制訂 各種交易為客戶提供各類風險管理產品，或為若干產品進行市場莊家活動。控制風險的基 本方法，是與其他交易對手訂立對銷信貸衍生工具合約。 為了管理來自買賣信貸衍生工具保障之信貸風險，滙豐將相關信貸風險納入有關交易對手 之整體信貸限額架構內。信貸衍生工具交易僅由主要業務中心旗下數個辦事處負責，該等 辦事處具備所需監控基礎設施及市場技巧，可以有效管理產品的內在信貸風險。 信貸衍生工具亦有限度地用以管理集團貸款組合之風險。信貸衍生工具之名義合約金額為 5,500億美元 （2013年：6,780億美元） ，包括買入保障2,720億美元 （2013年：3,390億美元） 及賣出 保障2,780億美元 （2013年：3,390億美元） 。在市場風險管理架構內經營的信貸衍生工具業務， 詳載於第222頁。 採用涉及不可觀察數據的模型估值之衍生工具 首次確認入賬之公允值 （交易價格） ，與假設首次確認入賬時已採用其後計量所用估值方法 計算得出之價值兩者間的差額，減去其後撥回額後所得數額，載列如下： 採用涉及重大不可觀察數據的模型估值之衍生工具的未攤銷數額 2014年 2013年 百萬美元 百萬美元 於1月1日未攤銷數額 167 181 新造交易遞延 177 206 本期於收益表確認： (234) (221) －攤銷 (114) (105) －於數據由不可觀察變為可觀察後 (13) (39) －衍生工具到期、終止或予以對銷 (107) (77) －對沖風險 － － 匯兌差額 4 1 於12月31日未攤銷數額1 114 167 1 此數額尚未於綜合收益表內確認。 對沖會計用途衍生工具 滙豐為管理本身之資產及負債組合與結構持倉而使用衍生工具 （主要為利率掉期） 作對沖用 途。此舉讓滙豐將集團參與債務資本市場之整體成本盡量維持於理想水平，以及減低資產 與負債期限及其他狀況出現結構性失衡因而產生之市場風險。 持作對沖會計用途的衍生工具之名義合約金額，顯示於結算日未平倉交易的面值，並不代 表承擔的風險額。 16－衍生工具 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 397 按產品類別劃分持作對沖會計用途衍生工具的名義合約金額 滙豐 滙豐控股 2014年 2013年 2014年 2013年 現金流對沖 公允值對沖 現金流對沖 公允值對沖 公允值對沖 公允值對沖 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 外匯 25,340 － 25,799 226 1,120 1,120 利率 190,902 90,338 201,197 90,354 5,477 1,977 於12月31日 216,242 90,338 226,996 90,580 6,597 3,097 公允值對沖 滙豐之公允值對沖主要包括利率掉期，其用途乃為保障定息長期金融工具避免因市場利率 變動導致公允值出現變動的風險。 指定列為公允值對沖之衍生工具的公允值 2014年 2013年 資產 負債 資產 負債 百萬美元 百萬美元 百萬美元 百萬美元 滙豐 外匯 － － 5 － 利率 387 4,012 1,163 2,889 於12月31日 387 4,012 1,168 2,889 滙豐控股 外匯 － 103 45 － 利率 484 － 15 － 於12月31日 484 103 60 － 公允值對沖產生之損益 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 滙豐 增益╱ （虧損） ： －對沖工具 (2,542) 1,997 (898) －與對沖風險相關之被對沖項目 2,561 (1,932) 871 截至12月31日止年度 19 65 (27) 滙豐控股 增益╱ （虧損） ： －對沖工具 423 14 － －與對沖風險相關之被對沖項目 (422) (21) － 截至12月31日止年度 1 (7) － 公允值對沖低效用部分之損益，即時於 「交易收益淨額」 項內確認。 現金流對沖 滙豐之現金流對沖主要包括利率掉期、期貨及跨貨幣掉期，這些掉期是用以保障按可變動 利率計息，或預期於日後再撥資或再投資的非交易用途資產及負債，以免因日後利息現金 流出現變動而產生風險。我們會就每個金融資產及負債組合，按其合約條款及其他相關因 素 （包括估計提前還款額及拖欠金額） ，預測日後現金流 （包括本金及利息之流量） 之金額及 產生時間。經過一段時間後，指定列作預計交易現金流對沖之衍生工具有效部分的損益， 可根據所有組合之本金結餘及利息現金流總額予以識別。 指定列為現金流對沖之衍生工具的公允值 2014年 2013年 資產 負債 資產 負債 百萬美元 百萬美元 百萬美元 百萬美元 外匯 1,673 572 2,257 439 利率 1,477 684 1,131 1,208 於12月31日 3,150 1,256 3,388 1,647 財務報表附註 （續） 滙豐控股有限公司 398 預期產生利息現金流之預計本金結餘 3個月以上 1年以上 3個月或以內 至1年 至5年 5年以上 百萬美元 百萬美元 百萬美元 百萬美元 現金流入╱ （流出） 風險淨額 資產 131,694 122,728 79,529 959 負債 (60,814) (46,582) (36,371) (8,169) 於2014年12月31日 70,880 76,146 43,158 (7,210) 現金流入╱ （流出） 風險淨額 資產 135,857 124,670 89,405 2,156 負債 (60,402) (46,990) (38,406) (10,221) 於2013年12月31日 75,455 77,680 50,999 (8,065) 上表反映相關被對沖項目的利率重新定價情況。 指定用作現金流對沖的衍生工具低效用部分之損益即時於 「交易收益淨額」 項內確認。截至 2014年12月31日止年度，因對沖效用低而確認增益3,400萬美元 （2013年：增益2,200萬美元； 2012年：增益3,500萬美元） 。 海外業務投資淨額對沖 集團就若干綜合投資淨額應用對沖會計法。對沖使用遠期外匯合約進行，或利用借入相關 貨幣的貸款提供資金。 於2014年12月31日，指定列為海外業務投資淨額對沖的未平倉金融工具之公允值，為資產 5,500萬美元 （2013年：400萬美元） ，負債100萬美元 （2013年：2,300萬美元） 及名義合約價值 35.25億美元 （2013年：28.4億美元） 。 截至2014年12月31日止年度，於 「交易收益淨額」 項內確認的低效用對沖款額為零 （2013及 2012年：零） 。 17 非交易用途反向回購及回購協議 會計政策 出售之證券如附有按預定價格回購之承諾 （ 「回購」 ） ，會保留於資產負債表內，並會將收取的代價入賬列作負債。根 據轉售承諾而購入之證券 （ 「反向回購」 ） 不會在資產負債表內確認，而初始支付的代價將會列作資產入賬。 非交易用途回購及反向回購按已攤銷成本計量。出售與回購價格兩者之間的差額或購入與轉售價格之間的差額會列 作利息處理，並於協議有效期內在淨利息收益確認。 非交易用途回購及反向回購於資產負債表內作為獨立項目呈列。此獨立呈列方式自2014年1月1日起採納，比較數 字亦相應重列。過往，非交易用途反向回購載於 「同業貸款」 及 「客戶貸款」 項內，而非交易用途回購則載於 「同業 存放」 及 「客戶賬項」 項內。 下表載列與客戶及同業相關之非交易用途反向回購及回購金額。 2014年 2013年 百萬美元 百萬美元 資產 同業 95,403 91,475 客戶 66,310 88,215 於12月31日 161,713 179,690 負債 同業 27,876 42,705 客戶 79,556 121,515 於12月31日 107,432 164,220 17－非交易用途反向回購及回購協議╱18－金融投資 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 399 18 金融投資 會計政策 擬繼續持有的國庫票據、債務證券及股權證券，除非是指定以公允值列賬，否則均分類為可供出售或持至到期日。 此等證券乃於交易日 （即滙豐訂立合約安排以購入該等工具時） 確認，並一般會於出售或贖回證券時撤銷確認。 (i) 可供出售金融資產首次列賬的計量方法，是以其公允值加上直接及遞增交易支出。該等資產其後會按公允值重 新計量，而因此產生的變動則於其他全面收益內確認，直至其已出售或已減值為止。在出售可供出售金融資產 時，過往曾於其他全面收益項內確認的累計損益，均會於收益表內確認為 「金融投資減除虧損後增益」 。 利息收益於債務證券之預計有效期內確認。購入定期債務證券產生的溢價及╱或折讓，均納入已確認之利息 內。股權資產股息則會在確定有權收取相關款項時於收益表內確認。 （ii） 若屬非衍生金融資產，但有固定或可以確定的支付金額及固定的到期日，而滙豐肯定有意及有能力持有直至到 期為止，則列為持至到期日之投資。持至到期日之投資首次列賬會按公允值加上任何直接相關交易的支出計 量，其後則會按已攤銷成本減任何減值虧損予以計量。 有關可供出售證券減值的會計政策於附註1(k)呈列。 倘持有可供出售金融資產至到期的意願或能力因管理該等資產的方式有變而有所改變，則將該等資產重新分類至持 至到期日。重新分類之公允值成為新的攤銷成本，而有關資產隨後將按攤銷成本而非公允值列賬。 金融投資 2014年 2013年 百萬美元 百萬美元 金融投資： －交易對手不可能或不會再質押或轉售 380,419 394,207 －交易對手可能再質押或轉售 35,048 31,718 於12月31日 415,467 425,925 金融投資的賬面值及公允值 2014年 2013年 賬面值 公允值 賬面值 公允值 百萬美元 百萬美元 百萬美元 百萬美元 國庫及其他合資格票據 81,517 81,517 78,111 78,111 －可供出售 81,517 81,517 78,111 78,111 債務證券1 323,256 324,668 338,674 339,007 －可供出售 285,505 285,505 313,590 313,590 －持至到期日 37,751 39,163 25,084 25,417 股權證券 10,694 10,694 9,140 9,140 －可供出售 10,694 10,694 9,140 9,140 於12月31日 415,467 416,879 425,925 426,258 1 年內將110.43億美元可供出售債務證券重新分類至持至到期日之債務證券。 財務報表附註 （續） 滙豐控股有限公司 400 按已攤銷成本及公允值列賬的金融投資 已攤銷成本 1 公允值 2 百萬美元 百萬美元 美國財政部 33,931 34,745 美國政府機構3 18,326 18,516 美國政府資助企業3 9,339 9,761 英國政府 28,680 29,758 香港政府 43,573 43,574 其他政府 159,846 163,401 資產抵押證券4 20,911 19,177 企業債務及其他證券 84,387 87,252 股票 7,421 10,694 於2014年12月31日 406,414 416,878 美國財政部 50,369 50,421 美國政府機構3 19,211 18,771 美國政府資助企業3 5,263 5,445 英國政府 23,565 23,580 香港政府 49,570 49,579 其他政府 153,619 156,208 資產抵押證券4 25,961 24,115 企業債務及其他證券 87,469 88,999 股票 8,081 9,140 於2013年12月31日 423,108 426,258 美國財政部 60,657 61,925 美國政府機構3 22,579 23,500 美國政府資助企業3 5,262 5,907 英國政府 17,018 17,940 香港政府 42,687 42,711 其他政府 146,507 149,179 資產抵押證券4 29,960 26,418 企業債務及其他證券 86,099 89,777 股票 4,284 5,789 於2012年12月31日 415,053 423,146 1 指金融投資的已攤銷成本或成本基準。 2 此等數字包括由銀行及其他金融機構發行之538.77億美元 （2013年：553.03億美元；2012年：599.08億美 元） 債務證券，其中88.12億美元 （2013年：89.46億美元；2012年：69.16億美元） 由不同政府擔保。由銀行 及其他金融機構發行之債務證券的公允值為543.75億美元 （2013年：554.67億美元；2012年：606.16億美 元） 。 3 包括獲美國政府發出明文保證支持之證券。 4 不包括計入美國政府機構及資助企業類別之資產抵押證券。 在認可交易所上市及非上市的金融投資 可供出售之 國庫及 其他合資格 可供出售之 持至到期日 可供出售之 票據 債務證券 之債務證券 股權證券 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 賬面值 上市1 4,101 168,879 6,037 5,928 184,945 非上市2 77,416 116,626 31,714 4,766 230,522 於2014年12月31日 81,517 285,505 37,751 10,694 415,467 賬面值 上市1 1,404 134,473 6,176 3,950 146,003 非上市2 76,707 179,117 18,908 5,190 279,922 於2013年12月31日 78,111 313,590 25,084 9,140 425,925 1 於2014年12月31日，持至到期日之上市債務證券公允值為64.59億美元 （2013年：62.81億美元） 。上市投資包 括在香港認可交易所上市之投資，其價值為37.52億美元 （2013年：28.32億美元） 。 2 可供出售之非上市國庫及其他合資格票據，主要包括並非於交易所上市但有流通市場的國庫票據。 18－金融投資╱19－作為擔保而質押之資產 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 401 按賬面值列賬的債務證券投資期限 1年或以內 1年以上 至5年 5年以上 至10年 10年以上 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 可供出售 68,344 134,815 44,938 37,408 285,505 持至到期日 1,396 9,622 7,087 19,646 37,751 於2014年12月31日 69,740 144,437 52,025 57,054 323,256 可供出售 78,222 146,200 44,556 44,612 313,590 持至到期日 2,993 8,380 6,442 7,269 25,084 於2013年12月31日 81,215 154,580 50,998 51,881 338,674 投資債務證券的合約期限及加權平均收益率 1年內 1年以上至5年 5年以上至10年 10年以上 金額 收益率 金額 收益率 金額 收益率 金額 收益率 百萬美元 % 百萬美元 % 百萬美元 % 百萬美元 % 可供出售 美國財政部 4,136 0.8 20,273 1.0 3,961 2.5 1,490 4.1 美國政府機構 － － 9 4.2 44 3.9 9,704 2.6 美國政府資助機構 － － 1,939 3.2 1,393 3.3 1,138 3.3 英國政府 281 2.2 12,389 1.0 12,541 1.7 － － 香港政府 350 0.4 953 1.0 － － － － 其他政府 46,946 2.2 65,497 2.7 12,806 2.9 2,864 2.4 資產抵押證券 688 1.3 1,172 1.4 4,003 1.4 15,036 1.1 企業債務及其他證券 16,392 2.3 30,687 2.1 7,048 2.7 6,459 3.3 於2014年12月31日之已攤銷成本總額 68,793 132,919 41,796 36,691 賬面總值 68,344 134,815 44,938 37,408 持至到期日 美國財政部 － － 75 4.8 44 4.8 115 4.2 美國政府機構 － － 1 7.6 50 2.6 8,506 2.4 美國政府資助機構 － － 92 1.4 406 2.9 4,370 3.1 香港政府 1 0.5 37 1.3 20 1.8 2 1.2 其他政府 95 4.1 278 4.8 202 5.2 722 4.9 資產抵押證券 － － － － － － 11 6.4 企業債務及其他證券 1,300 3.5 9,139 3.6 6,365 4.0 5,920 4.1 於2014年12月31日之已攤銷成本總額 1,396 9,622 7,087 19,646 賬面總值 1,396 9,622 7,087 19,646 資產抵押證券的期限分布乃按合約到期日於上表呈列。每段期限的加權平均收益率計算方 法，為將截至2014年12月31日止年度的利息收益年率，除以可供出售債務證券於該日的賬 面值。收益率並不包括相關衍生工具的影響。 19 作為負債擔保而質押之資產、已轉讓之資產及持作資產擔保之抵押品 為擔保負債而質押之金融資產 2014年 2013年 百萬美元 百萬美元 國庫票據及其他合資格證券 5,170 6,387 同業貸款 17,294 17,733 客戶貸款 77,960 87,894 債務證券 138,991 190,095 股權 11,373 8,816 其他 6,079 1,035 於12月31日已質押之資產 256,867 311,960 上表列示已按法律及合約基準授出押記以便為負債作擔保之資產。該等資產金額可能大於 為籌集資金或為償付負債而用作抵押品之資產的賬面值。這是證券化及備兌債券之情況， 當中已發行負債金額另加任何強制性超額抵押，乃少於相關資產組合內可用作資金或抵押 品用途之金融資產賬面值。這亦是金融資產存放於託管商或結算代理之情況，當中存放之 所有金融資產均設有浮動押記，以便為結算賬項內任何負債作擔保。 財務報表附註 （續） 滙豐控股有限公司 402 19－作為擔保而質押之資產╱20－於聯營及合資公司之權益 此等交易乃按有抵押交易 （包括 （如適用） 常規借出證券及回購協議） 之一般及慣常條款進行。 已轉讓資產 會計政策 撤銷確認金融資產 當從金融資產收取現金流之合約權利屆滿時，或當滙豐已轉讓其收取金融資產現金流之合約權利時，以及出現下 列其中一種情況時，便會撤銷確認金融資產： ‧ 已轉讓擁有權附帶的絕大部分風險與回報；或 ‧ 滙豐未有保留亦未轉讓絕大部分風險與回報，但亦未保留控制權。 滙豐在日常業務中訂立交易，將其金融資產轉讓予第三方。視乎情況而定，這些轉讓交易 可能導致該等金融資產被撤銷確認或繼續確認。 上文所示的金融資產包括不符合撤銷確認條件已轉讓予第三方的金額，尤其是交易對手根 據回購協議持作抵押品的債務證券及根據借出證券協議借出的股權證券。由於該等交易實 質為有抵押借貸，相關資產抵押品將繼續全數確認，而反映集團於未來日期按固定價格回 購已轉讓資產之責任的相關負債，則於資產負債表確認。由於進行此等交易，集團於交易 有效期內不能使用、出售或質押該等已轉讓資產。集團就此等已質押工具仍然承擔利率風 險及信貸風險。交易對手的追索權並不限於已轉讓資產。 不符合全部撤銷確認條件的已轉讓金融資產及相關金融負債 轉讓前 資產賬面值 已轉讓資產 賬面值 相關負債 賬面值 已轉讓資產 公允值 相關負債 公允值 持倉淨額 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年12月31日 回購協議 78,541 79,141 借出證券協議 13,177 10,643 其他出售安排 （僅可追索 已轉讓資產） 3,775 4,049 4,007 4,018 (11) 按持續參與程度確認的 證券化交易 17,427 11 5 11 5 6 於2013年12月31日 回購協議 125,508 126,175 借出證券協議 9,175 8,884 其他出售安排 （僅可追索 已轉讓資產） 6,707 7,019 6,827 6,707 120 按持續參與程度確認的 證券化交易 17,427 16 8 16 8 8 持作資產擔保之抵押品 滙豐可以在不違約下出售或再質押，與反向回購及借入證券有關之資產抵押品的公允值 為2,690.19億美元 （2013年：2,596.17億美元） 。已出售或再質押之任何有關抵押品的公允值為 1,633.42億美元 （2013年：1,860.13億美元） 。滙豐有責任退回等值證券。 該等交易乃按常規借入證券及反向回購協議之一般及慣常條款進行。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 403 20 於聯營及合資公司之權益 會計政策 滙豐連同一方或多方訂立共同控制之安排以進行經濟活動，此等投資分類為合資公司。滙豐投資於其可施以重大 影響力但並非附屬公司 （附註22） 或合資公司之實體，則分類為聯營公司。 於聯營公司之投資及於合資公司之權益均採用權益法確認。按此方法，該等投資於首次列賬時按成本 （包括應佔商 譽） 計量，其後則會就收購後滙豐所佔資產淨值之變動予以調整。收購合資及聯營公司之權益時，如投資成本高於 滙豐應佔聯營或合資公司可識別資產及負債的公允值淨額，亦會產生商譽。 當有跡象顯示於聯營公司之投資可能出現減值時，集團會測試該投資有否減值。收購合資及聯營公司之權益所產 生的商譽並無就減損分開測試。 滙豐與其聯營及合資公司之間進行交易的利潤，以滙豐於聯營或合資公司各自的權益為限進行撇銷。除非有關交 易可證明所轉讓資產出現減值，否則虧損亦以滙豐於聯營或合資公司的權益為限進行撇銷。 關鍵會計估算及判斷 於聯營公司之權益減值 減值測試涉及於釐定使用價值時，尤其是估計預期自持續持有該投資產生的現金流的現值時作出重大判斷。 最重大的判斷涉及我們於交通銀行的投資的減值測試。下文附註詳述用以估計交通銀行使用價值的主要假設、使 用價值計算對不同假設的敏感度，以及列示主要假設修訂可能將使用價值超過賬面值的差額 （ 「可用額度」 ） 減至零 的敏感度分析。 聯營公司 於2014年12月31日，滙豐於聯營公司之權益的賬面值為179.4億美元 （2013年：164.17億美元） 。 滙豐之主要聯營公司 2014年 2013年 賬面值 公允值1 賬面值 公允值1 百萬美元 百萬美元 百萬美元 百萬美元 上市 交通銀行股份有限公司 14,590 13,140 13,412 9,954 沙地英國銀行 2,811 6,220 2,437 4,693 於12月31日 17,401 19,360 15,849 14,647 1 主要聯營公司於認可證券交易所上市。公允值以所持股份的市場報價為基準 （公允值等級制中的第一級） 。 於2014年12月31日 註冊成立 國家╱地區 及主要 滙豐所佔 營業地點 主要業務 股本權益 已發行股本 交通銀行股份有限公司 中國1 銀行服務 19.03% 人民幣742.63億元 沙地英國銀行 沙地阿拉伯 銀行服務 40.00% 100億沙地阿拉伯里亞爾 1 中華人民共和國。 按英國 《2006年公司法》 第409條規定，滙豐所有聯營及合資公司之詳情，將載於滙豐控股送 交英國公司註冊處的下一份周年報表附錄內。 滙豐於香港上市之聯營公司持有之權益為145.9億美元 （2013年：134.12億美元） 。 財務報表附註 （續） 滙豐控股有限公司 404 交通銀行股份有限公司 （ 「交通銀行」 ） 滙豐於交通銀行的投資，由2004年8月起以權益法入賬。滙豐透過參與交通銀行的董事會確 立對這家銀行施以重大影響力，而根據技術合作及交流計劃，滙豐現正協助交通銀行維持 財務及營運政策，並已調派多名職員到該行協助有關工作。 減值測試 於2014年12月31日，除了2013年一段短暫期間外，在長約32個月的期間內，滙豐於交通銀行 投資的公允值一直低於賬面值。因此，我們對交通銀行的投資賬面值進行減值測試。測試 結果確認，此投資於2014年12月31日並無出現減值。可收回金額為157億美元(2013年：140億 美元） ，較於2014年12月31日的賬面值超出11億美元(2013年：6億美元） （ 「可用額度」 ） 。可用額 度增加乃由於交通銀行的資本狀況改善。 於2014年12月31日 於2013年12月31日 使用價值 賬面值 公允值 使用價值 賬面值 公允值 十億美元 十億美元 十億美元 十億美元 十億美元 十億美元 交通銀行股份有限公司 15.7 14.6 13.1 14.0 13.4 10.0 可收回金額的基準 減值測試的方法，是比較交通銀行的可收回金額 （按使用價值計算方法釐定） 及其賬面值。 使用價值的計算方法，是根據管理層的盈利估計採用現金流折現預測。中短期以後的現金 流，則採用長期增長率推算永久數值。為符合預期監管資本規定，已計入估算維持資本要 求費用且自預測現金流中作出扣減而計算。維持資本要求費用計算方法的主要輸入值包括 估計資產增長、風險加權資產與資產總值比率及預期監管規定資本水平。管理層需要作出 判斷，以估算交通銀行的日後現金流。 使用價值計算方法的主要假設 長期增長率：於2018年後各個期間所採用的增長率為5% （2013年：5%） ，並不超過中國的預 測國內生產總值增長率。 折現率：折現率13% （2013年：13%） 乃透過就交通銀行採用資本資產定價模型計算方法，利 用市場數據得出的價值範圍而計算。管理層對此加以補充，將採用資本資產定價模型計算 方法得出的折現率，與可從外界資料來源獲得的折現率及滙豐評估於中國的投資所採用的 折現率進行比較。所採用折現率處於資本資產定價模型及外界資料來源所示範圍11.4%至 14.2% （2013年：10.5%至15%） 內。 貸款減值準備佔客戶貸款的百分比：所用比率在中短期內介乎0.73%至1% （2013年：0.64% 至1%） 。我們假設長期比率會回復至0.65% （2013年：0.64%） 的過往水平。該等比率處於外 界分析員所披露的中短期預測範圍0.51%至1.08% （2013年：0.55%至1.2%） 內。 風險加權資產佔資產總值的百分比：所用比率在中短期內介乎70%至72%。長期比率回復至 70% （2013年：68.7%） 。 成本收益比率：所用比率在中短期內介乎40%至42.4% （2013年：39.7%至43.2%） 。該等比率 處於外界分析員所披露的中短期預測範圍37.2%至44.5% （2013年：38%至44.2%） 內。 我們已就各主要假設進行敏感度分析，以確定對假設作出合理可能修訂時的影響。計算使 用價值所單獨使用的各項主要假設可能須作出以下修訂，方可將可用額度減至零： 主要假設 為將可用額度減至零而對主要假設作出的修訂 ‧ 長期增長率 ‧ 減少43個基點 ‧ 折現率 ‧ 增加53個基點 ‧ 貸款減值準備佔客戶貸款的百分比 ‧ 增加8個基點 ‧ 風險加權資產佔資產總值的百分比 ‧ 增加3.3% ‧ 成本收益比率 ‧ 增加1.6% 下表說明主要假設出現可能合理變動時使用價值受到的影響。這反映使用價值對各主要假 設本身的敏感度，而多個有利及╱或不利變動有可能同時發生。 20－於聯營及合資公司之權益 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 405 有利變動 當前的模型 不利變動 十億美元 十億美元 十億美元 十億美元 十億美元 於2014年12月31日 賬面值：146億美元 長期增長率 +50個基點 +100個基點 5% -50個基點 -100個基點 使用價值 17.0 18.6 15.7 14.5 13.4 使用價值增加╱ （減少） 1.3 2.9 (1.2) (2.3) 折現率 -50個基點 -100個基點 13% +50個基點 +100個基點 使用價值 16.8 18.1 15.7 14.7 13.9 使用價值增加╱ （減少） 1.1 2.4 (1.0) (1.8) 貸款減值準備佔客戶貸款的百分比 整個期間為0.65% 2014至18年： 0.73%至1% 自2019年起：0.65% 自2014至18年為1% 自2019年起：0.65% 使用價值 16.2 15.7 14.9 使用價值增加╱ （減少） 0.5 (0.8) 風險加權資產佔資產總值的百分比 -100個基點 -200個基點 2014至18年： 70%至72% 自2019年起：70% +100個基點 +200個基點 使用價值 16.0 16.3 15.7 15.4 15.1 使用價值增加╱ （減少） 0.3 0.6 (0.3) (0.6) 成本收益比率 -50個基點 -100個基點 2014至18年： 40.0%至42.4% 自2019年起：42.4% +50個基點 +100個基點 使用價值 16.0 16.3 15.7 15.4 15.1 使用價值增加╱ （減少） 0.3 0.6 (0.3) (0.6) 於2013年12月31日 賬面值：134億美元 長期增長率 +50個基點 +100個基點 5% -50個基點 -100個基點 使用價值 15.4 16.9 14.0 12.9 11.8 使用價值增加╱ （減少） 1.4 2.9 (1.1) (2.2) 折現率 -50個基點 -100個基點 13% +50個基點 +100個基點 使用價值 15.6 17.3 14.0 12.7 11.6 使用價值增加╱ （減少） 1.6 3.3 (1.3) (2.4) 貸款減值準備佔客戶貸款的百分比 整個期間為0.64% 2013至18年： 0.64%至1.00% 自2019年起：0.64% 自2014至18年為1% 使用價值 14.8 14.0 13.5 使用價值增加╱ （減少） 0.8 (0.5) 風險加權資產佔資產總值的百分比 -100個基點 -200個基點 整個期間為68.7% +100個基點 +200個基點 使用價值 14.4 14.7 14.0 13.7 13.4 使用價值增加╱ （減少） 0.4 0.7 (0.3) (0.6) 成本收益比率 -50個基點 -100個基點 2013至18年： 39.7%至43.2% 自2019年起：43.2% +50個基點 +100個基點 使用價值 14.3 14.7 14.0 13.7 13.4 使用價值增加╱ （減少） 0.3 0.7 (0.3) (0.6) 交通銀行的選錄財務資料 交通銀行的法定核算參考日為12月31日。截至2014年12月31日止年度，滙豐以截至2014年9月 30日止12個月的財務報表為入賬基準計入該聯營公司的業績，並已計及其後於2014年10月1日 至2014年12月31日期間所出現且可重大影響相關業績的變動。 於9月30日 2014年 2013年 百萬美元 百萬美元 交通銀行的選錄資產負債表資料 現金及於中央銀行的結餘 150,306 142,209 同業及其他金融機構貸款 79,960 88,049 客戶貸款 547,706 516,161 其他金融資產 178,883 165,521 其他資產 45,140 34,392 資產總值 1,001,995 946,332 同業及其他金融機構存放 209,935 170,916 客戶賬項 663,745 667,588 其他金融負債 28,860 20,564 其他負債 25,361 19,655 負債總額 927,901 878,723 股東權益總額 74,094 67,609 財務報表附註 （續） 滙豐控股有限公司 406 於9月30日 2014年 2013年 百萬美元 百萬美元 於12月31日滙豐的綜合財務報表內交通銀行的 股東權益總額與賬面值之對賬 滙豐應佔股東權益總額 14,040 12,810 加：商譽及其他無形資產 550 602 賬面值 14,590 13,412 截至9月30日止12個月 2014年 2013年 百萬美元 百萬美元 交通銀行的選錄收益表資料 淨利息收益 22,030 20,768 費用及佣金收益淨額 4,792 4,010 貸款減值準備 (3,509) (2,811) 折舊及攤銷 (920) (809) 稅項支出 (3,102) (2,823) 本年度利潤 10,626 10,099 其他全面收益 217 (375) 全面收益總額 10,843 9,724 已收取交通銀行之股息 597 549 所有聯營公司 （不包括交通銀行） 之總體財務資料概要 2014年 2013年 百萬美元 百萬美元 賬面值 3,350 3,005 滙豐分佔╱分攤： －資產總值 20,099 21,007 －負債總額 16,837 18,056 －收入 801 927 －來自持續經營業務的損益 519 408 －其他全面收益 2 9 －全面收益總額 521 417 合資公司 於2014年12月31日，滙豐於合資公司之權益的賬面值為2.41億美元 （2013年：2.23億美元） 。 聯營及合資公司 截至2014年12月31日止年度，滙豐應佔聯營及合資公司之利得稅為6億美元 （2013年：5.56億 美元） ，此數額計入收益表之 「應佔聯營及合資公司利潤」 項內。 於聯營及合資公司之權益的變動 2014年 2013年 百萬美元 百萬美元 於1月1日 16,640 17,834 增添 30 26 出售 (133) (3,148) 應佔業績 2,532 2,325 股息 (757) (694) 匯兌差額 (212) 396 應佔聯營及合資公司其他全面收益╱ （支出） 78 (35) 其他變動 3 (64) 於12月31日1 18,181 16,640 1 包括商譽6.21億美元 （2013年：6.08億美元） 。 20－於聯營及合資公司之權益╱21－商譽及無形資產 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 407 21 商譽及無形資產 2014年 2013年 百萬美元 百萬美元 商譽 19,169 21,179 有效長期保險業務現值 5,307 5,335 其他無形資產 3,101 3,404 於12月31日 27,577 29,918 商譽 會計政策 收購附屬公司時，如所轉讓代價的公允值、任何非控股股東權益的金額，以及先前所持被收購公司任何股本權益 公允值的總和，高於收購所得可識別資產及負債的金額，即會產生商譽。倘收購所得可識別資產及負債的金額較 高，則任何差額會即時於收益表內確認。 商譽會分攤至各個創現單位，以便於最低層面進行減損測試。商譽會在這個層面受到監察，以達致內部管理目 的。滙豐的創現單位乃按地區劃分，再按環球業務細分。減損測試最少每年進行一次，或會在有跡象顯示出現減 值時進行，方法是比較創現單位可收回金額與其賬面值的差距。創現單位賬面值乃按其資產及負債 （包括應佔商 譽） 釐定。創現單位可收回金額為其公允值減出售成本或其使用價值之較高者。使用價值指創現單位預計日後現金 流之現值。倘可收回金額低於賬面值，則於收益表扣取減值虧損。商譽乃按成本減累計減值虧損後於資產負債表 列賬。 於出售業務當日，應佔商譽會於計算出售所得損益時，計入滙豐應佔之資產淨值內。 倘出售用途業務組合為已獲分攤商譽的創現單位或為該創現單位內的業務，商譽會計入出售用途業務組合內。計 入出售用途業務組合之商譽金額乃按所出售業務及所保留創現單位部分之相對價值計量。 關鍵會計估算及判斷 商譽減損 對商譽減損的檢討反映管理層對創現單位的日後現金流及用於折現日後現金流的折現率的最佳估算，而日後現金 流及折現率均受到下列不確定因素的影響： ‧ 創現單位的日後現金流受兩個因素影響：在各段期間可詳細預測的預計現金流；以及就其後可持續產生現金流 的長期規律所作假設。雖然我們會比較預測數字與實際表現和可核證的經濟數據，但這些預測反映管理層於評 估時對未來業務前景的看法；及 ‧ 用以折現預計日後現金流的折現率可能對其估值有重大影響，是根據分攤予個別創現單位的資本成本而釐 定。資本成本比率一般採用資本資產定價模型來推算，該模型包含的數據反映多項財務及經濟變數，包括相關 國家╱地區的無風險利率，以及所評估業務的風險溢價。由於這些變數受管理層無法控制的外界市場利率波動 及經濟情況影響，因此無法完全確定，需要作出重大判斷。 創現單位之預計現金流下降及╱或其資本成本上升，則創現單位之估計可收回金額會隨之減少。若該估計金額低 於創現單位之賬面值，則會於相關年度的收益表中確認商譽減損準備。 遇上市況波動，預計現金流之準確度會非常不確定。於該等市況下，管理層會更頻密地反覆測試商譽的減損數 額，而非只作一年一度的測試，以確保預計現金流所依據的假設仍然反映當時的市況，以及管理層對日後業務前 景所作的最佳估算。 於2014年，我們並無識別任何商譽減損 （2013年：零） 。除於2014年7月1日進行的年度減損測試外，管理層於 2014年12月31日檢討了創現單位的當前及預期表現，並確定分攤予創現單位的商譽並無減損跡象。 財務報表附註 （續） 滙豐控股有限公司 408 商譽對賬 中東 歐洲 亞洲 及北非 北美洲 拉丁美洲 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 總額 於2014年1月1日 14,977 1,016 55 7,861 3,241 27,150 出售 (168) － － － － (168) 匯兌差額 (1,594) (30) (1) 1 (240) (1,864) 重新分類為持作出售用途 (8) － － － 24 16 其他 － 23 － (47) (18) (42) 於2014年12月31日 13,207 1,009 54 7,815 3,007 25,092 累計減值虧損 於2014年1月1日 － － － (5,971) － (5,971) 匯兌差額 － － － 1 － 1 其他 － － － 47 － 47 於2014年12月31日 － － － (5,923) － (5,923) 於2014年12月31日之賬面淨值 13,207 1,009 54 1,892 3,007 19,169 總額 於2013年1月1日 14,660 1,134 60 8,339 3,646 27,839 出售 － － － － (1) (1) 匯兌差額 596 (129) (5) (2) (132) 328 重新分類為持作出售用途1 (611) － － － (272) (883) 因不再列為持作出售用途而重列 332 － － － － 332 其他 － 11 － (476) － (465) 於2013年12月31日 14,977 1,016 55 7,861 3,241 27,150 累計減值虧損 於2013年1月1日 － － － (6,449) － (6,449) 匯兌差額 － － － 2 － 2 其他 － － － 476 － 476 於2013年12月31日 － － － (5,971) － (5,971) 於2013年12月31日之賬面淨值 14,977 1,016 55 1,890 3,241 21,179 1 於2013年，在決定出售摩納哥HSBC Private Bank Holdings (Suisse) S.A.的私人銀行業務後，在歐洲的商譽有6.11 億美元重新分類為持作出售用途資產。在轉撥至持作出售用途項下後，出售用途業務組合錄得2.79億美 元的撇減額並分攤至商譽。其後決定保留該等業務後，資產及負債重新分類不再列作持作出售用途，以 致需要重列餘下的商譽。 減損測試 減損測試之時間 滙豐每年7月1日對分攤至各個創現單位之商譽進行減損測試。 可收回金額之基準 所有已獲分攤商譽之創現單位的可收回金額，相等於2013及2014年各有關測試日期之使用價 值。 就各個重要創現單位而言，使用價值之計算方法是折現管理層對各個創現單位之現金流預 測。計算時採用之折現率，乃根據滙豐分攤至創現單位業務所在國家╱地區之投資的資本 成本而釐定。我們採用長期增長率推算現金流之永久數值，原因是集團內組成創現單位的 業務部門屬於長期性質。對於2014年7月1日進行的商譽減損測試，我們採用管理層對直至 2018年底的現金流預測。 21－商譽及無形資產 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 409 使用價值計算方法之主要假設 超出首次 列賬現金流 於7月1日 預測之名義 之商譽 折現率 增長率 百萬美元 % % 創現單位 零售銀行及財富管理－歐洲 4,298 9.1 4.5 工商金融－歐洲 3,214 10.1 4.2 環球私人銀行－歐洲 3,808 7.1 3.4 環球銀行及資本市場－歐洲 3,296 11.0 4.2 零售銀行及財富管理－拉丁美洲 1,762 12.8 7.9 2014年 16,378 創現單位 零售銀行及財富管理－歐洲 4,135 8.0 3.9 工商金融－歐洲 3,062 10.0 3.8 環球私人銀行－歐洲 3,607 7.3 3.0 環球銀行及資本市場－歐洲 3,101 9.9 3.7 零售銀行及財富管理－拉丁美洲 1,812 11.2 8.6 2013年 15,717 2014年7月1日，分攤至被視為個別而言並非重要創現單位的商譽總額為45.26億美元 （2013年 7月1日：45.5億美元） 。集團創現單位的資產負債表沒有任何無限使用年期之重大無形資產 （商譽除外） 。 名義長期增長率：此增長率反映創現單位業務所在國家╱地區之本地生產總值及通脹率。 增長率以國際貨幣基金組織的預測增長率為依據，是因為該等增長率被視為對未來發展趨 勢的最相關估計。2013及2014年所用的增長率，並未超逾創現單位業務所在國家╱地區之長 期增長率。 折現率：用以折現現金流之折現率，是根據分攤予各個創現單位之資本成本而釐定，此資本 成本乃採用資本資產定價模型計算。資本資產定價模型依賴多項數據，此等數據反映多項 財務及經濟變數，包括無風險利率，以及反映所評估業務內在風險的溢價。這些變數以市 場對經濟變數的評估及管理層的判斷為基準。就2014年7月1日的測試而言，用於釐定各創現 單位折現率之方法已予微調，以更準確地反映創現單位業務所在國家╱地區的通脹率。此 外，測試商譽有否減損時，管理層將採用內部所訂資本資產定價模型計算得出之折現率， 與採用外界資源計算所得之資本成本比率作比較，以補商譽減損測試過程的不足。當管理 層判斷採用外界資源計算所得之資本成本比率能更準確反映當前市場及經濟情況時，滙豐 即會採用這些比率。於2013及2014年，內部計算所得之資本成本比率與採用外界資源計算所 得之比率一致。 管理層估算創現單位現金流時使用之判斷：每個創現單位之現金流預測以集團管理委員會 批准之計劃為基礎。除折現率及名義長期增長率外，就每個重要創現單位所作之主要假設 詳述如下。 環球私人銀行－歐洲：環球私人銀行－歐洲的現金流預測，主要反映該業務重新定位，以集 中服務符合集團優先發展範疇的客戶。環球私人銀行－歐洲的收入主要源自滙豐之客戶關 係，而現金流預測的主要假設，在於策略性重新定位後的管理資產水平及盈利能力。現金 流預測包括環球私人銀行－歐洲的盈利能力提高，在外部環境下，這視乎管理層能否達成 策略性重新定位計劃的目標而定。 於2014年7月1日，環球私人銀行－歐洲的可收回金額超出賬面值18億美元 （ 「可用額度」 ） 。 計算使用價值所用的主要假設可能須作出以下修訂，方可將可用額度減至零： 主要假設 為將可用額度減至零而對主要假設作出的修訂 折現率 增加90個基點 長期增長率 減少102個基點 現金流預測 減少19.7% 下表說明主要假設出現可能合理變動時使用價值受到的影響。這反映使用價值對各主要假 設本身的敏感度，而多個有利及╱或不利變動有可能同時發生。 財務報表附註 （續） 滙豐控股有限公司 410 有利變動 當前的模型 不利變動 十億美元 十億美元 十億美元 於2014年7月1日 創現單位賬面值：73億美元 可收回金額超過賬面值的差額：18億美元 長期增長率 +100個基點 3.4% -100個基點 使用價值 12.2 9.1 7.4 使用價值增加╱ （減少） 3.1 (1.7) 折現率 -100個基點 7.1% +100個基點 使用價值 12.5 9.1 7.2 使用價值增加╱ （減少） 3.4 (1.9) 預計現金流 +20% 378 -20% 使用價值 10.9 9.1 7.3 使用價值增加╱ （減少） 1.8 (1.8) 零售銀行及財富管理－歐洲以及工商金融－歐洲：零售銀行及財富管理－歐洲以及工商金 融－歐洲的現金流預測包括的假設，反映此兩個創現單位業務所在歐洲國家╱地區的經濟 環境及財務前景。主要假設包括利率水平、名義國內生產總值增長率、市場內競爭對手的狀 況和失業率的水平及變動。儘管歐洲目前的經濟狀況仍然嚴峻，但管理層的現金流預測主 要以此等當前情況為基礎。風險包括增長較預期緩慢及監管環境不明朗。零售銀行及財富 管理－歐洲特別易受與客戶有關的進一步補救措施及監管機構採取的進一步行動所影響。 根據結算日的情況，管理層認為上述任何主要假設作出合理可能修訂，均不會導致零售銀 行及財富管理－歐洲或工商金融－歐洲需要確認減值。 環球銀行及資本市場－歐洲：環球銀行及資本市場－歐洲的現金流預測包括的主要假設，為 歐洲市場將會繼續復甦。因此，我們假設歐洲業務收入於至2018年為止的預測期間繼續回 升。利率波動會令歐洲市場收入的回升進一步受壓。環球銀行及資本市場－歐洲實現預計 現金流的能力，或會因預計期間的監管環境變化而受到不利影響，包括 （但不限於） 實施銀 行業獨立委員會最終報告內所載的建議。根據結算日的情況，管理層認為上述任何主要假 設作出合理可能修訂，均不會導致環球銀行及資本市場－歐洲需要確認減值。 零售銀行及財富管理－拉丁美洲：零售銀行及財富管理－拉丁美洲的現金流預測包括的假 設，反映此創現單位所在國家╱地區的經濟環境及財務前景，而巴西和墨西哥為當中最大 的兩個國家。主要假設包括貸款及存款額增長，以及貸款組合的信貸質素。潛在挑戰包括 不利經濟狀況壓抑客戶需求，以及競爭對手爭相降價導致利潤受限。根據結算日的情況， 管理層認為上述任何主要假設作出合理可能修訂，均不會導致零售銀行及財富管理－拉丁 美洲需要確認減值。 無形資產 會計政策 倘無形資產可與商譽分開，或是產生自合約或其他法律權利，且其公允值能夠可靠地計量，則予以確認，而倘若 於業務合併中收購，則與商譽加以區分。 無形資產包括下列各項：有效長期保險業務及附有酌情參與條款之長期投資合約的現值、電腦軟件、商號、按揭 債務管理權、客戶名單、核心存款客戶關係、信用卡客戶關係及商戶或其他貸款客戶關係。電腦軟件包括購買及 內部開發軟件。內部開發軟件的成本，包括創造、生產及預備軟件使其能按管理層擬定的方式運作之所有直接相 關成本。持續保養軟件所產生的成本於產生時即時支銷。 倘若發生任何事件或情況改變，顯示未必可以收回無形資產之賬面值，則須對無形資產進行減值檢討。倘： ‧ 無形資產並無確定可用年期或尚未可供使用，每年均接受減值測試。計算期內確認之無形資產，則於該年度結 束之前接受測試；及倘 ‧ 無形資產有確定可用年期 （PVIF除外） ，均按成本減攤銷及累計減值虧損額列賬，並於其估計可用年期內攤銷。 估計可用年期乃法定期限或預期可用年期兩者之較短者。按揭債務管理權之攤銷列入 「費用收益淨額」 項內。 21－商譽及無形資產 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 411 有確定可用年期的無形資產一般按直線基準於其可用年期內攤銷如下： 商號 10年 按揭債務管理權 一般為5至12年 內部開發軟件 3至5年 購入軟件 3至5年 客戶╱商戶關係 一般為3至10年 其他 一般為10年 有效長期保險業務現值 分類為長期保險業務或附有酌情參與條款 （ 「DPF」 ） 之長期投資合約且於結算日仍然有效之保單，其價值會確認為 一項資產。該項資產代表權益持有人於結算日在相關保險公司因已訂合約而預期產生的利潤中所佔權益之現值。 PVIF是根據權益持有人於當前有效業務預期產生之未來利潤之權益折現值而釐定，方法是對多項評估因素 （例如未 來死亡率、保單失效率及支出水平） 作出適當假設，同時會採用反映有關合約風險溢價的風險折現率來計算。PVIF 包含非市場風險及金融期權及擔保價值之準備。PVIF資產以未計應佔稅項的方式於資產負債表呈列，而PVIF資產 之變動，則以未計稅項的方式列入 「其他營業收益」 項內。 有效長期保險業務現值 我們的壽險業務採用內含價值法入賬，此方法可以提供一個風險及價值評估架構。於2014年 12月31日，我們的PVIF資產為53億美元 （2013年：53億美元） 。 PVIF的變動 2014年 2013年 百萬美元 百萬美元 於1月1日之PVIF 5,335 4,847 本年度新增業務的價值1 870 924 有效業務產生的變動： －預期回報 (545) (505) －經驗差異2 62 (20) －營運假設的變化 (69) 186 投資回報差異 (34) 42 投資假設的變化 (75) (120) 其他調整 52 18 長期保險業務PVIF之變動 261 525 分類為持作出售用途之資產的轉撥3 (122) － 匯兌差額及其他 (167) (37) 於12月31日之PVIF 5,307 5,335 1 本年度新增業務的價值指業務預計所得利潤的現值。 2 經驗差異包括之前計算PVIF所用人口、支出和持續性假設與年內觀察所得實際經驗存在差異所產生的影 響，但以日後業務所得利潤的影響為限。 3 有關於本年度上半年分類為持作出售用途之英國退休金業務。有關詳情，請參閱第191頁。 計算PVIF時，我們會按每項保險業務作出的多種假設而調整，以反映當地市場狀況及管理層 對未來趨勢的判斷，以及應用風險邊際差距以反映相關假設涉及的任何不確定因素後，才 推算預計現金流。所作主要假設乃有關經濟及非經濟假設和投保人的行為。實際經驗的變 化及假設變動，均可導致保險業務的業績出現波動。 PVIF資產的價值變動主要受下列因素影響： ‧ 來自新增業務的預計現金流，此數值已按預計期限及有關投保人行為的假設作出調整 （ 「本年度新增業務的價值」 ） ； ‧ 來自沖抵折現率的預計現金流，並扣減期內預計現金流撥回額 （ 「預期回報」 ） ； ‧ 來自死亡率或失效率等非經濟營運假設變化 （ 「營運假設的變化」 ） 的預計現金流； ‧ 與期初所作假設比較出現營運假設經驗差異，導致預測日後現金流改變之影響 （ 「經驗差 異」 ） 所產生的預計現金流； ‧ 來自未來投資回報變化 （ 「投資假設的變化」 ） 的預計現金流；及 ‧ 與期初之假設比較，實際投資經驗對現有資產之影響 （ 「投資回報差異」 ） 所產生的預計現 金流。 PVIF資產的估值包括使用隨機方法，納入預測假設中非經濟風險的明顯風險邊際差距，以及 金融期權和保證的明顯準備。風險折現率按主動基準參考市場無風險收益率設定。 財務報表附註 （續） 滙豐控股有限公司 412 21－商譽及無形資產╱22－於附屬公司之投資 計算主要壽險業務的PVIF時所用主要假設 釐定經濟假設時，或按照與可觀察市場價值相符的方式進行，或在若干市場 （包括在與保單 未決賠款期限匹配的年期內無風險曲線屬不可觀察的市場） ，我們則使用長期經濟假設。 釐定經濟假設涉及預測長期利率及在發展程度較低的市場的利率將趨近成熟市場慣例的時 限。我們採用集團之經濟研究團隊及外部專家編撰的相關過往數據及研究分析得出假設。 PVIF估值將對該等長期假設的任何變動具敏感度，其方式與對可觀察市場變動的敏感度相 同，而該等變動的影響計入下文所列敏感度內。 2014年 2013年 英國 香港 法國 1 英國 香港 法國 1 % % % % % % 加權平均無風險利率 1.65 1.86 1.21 2.45 2.31 2.38 加權平均風險折現率 2.15 7.42 1.73 2.95 7.41 4.69 支出通脹率 4.67 3.00 2.00 3.39 3.00 2.00 1 就2014年而言，於計算在法國的PVIF時乃假設風險折現率1.73%加風險邊際差距6,300萬美元。就2013年而 言，使用綜合比率4.69%，相等於3.08%加風險邊際差距6,400萬美元。 對經濟假設變動的敏感度 集團制訂PVIF計算方法適用之風險折現率時，先由無風險利率曲線開始，並就已作最佳估算 的現金流模型中未有反映之風險加入明確準備。當股東向投保人提供選擇權及保證時，該 等選擇權及保證之成本即為PVIF的明顯減額，除非已就有關成本提撥準備，作為監管機構所 要求的技術準備明確增額。有關該等保證的進一步詳情，請參閱第195頁。 下表載列倘若所有制訂保險產品附屬公司對主要經濟假設 （即無風險利率） 作出合理可能的 修訂時，PVIF將會受到的影響。由於保單具備若干特點，所以各種因素之間可能並無直線關 係，因此敏感度測試結果不應用作推斷更高程度壓力的境況。下表呈列的敏感度並未考慮 管理層為減輕影響而可能採取的措施，以及投保人的行為因此改變可能帶來的影響。敏感 度於2013至2014年有所提高，主要因為2014年於法國的收益率下降及孳息曲線趨向平坦。 在低收益環境下，PVIF資產對第195頁所述選擇權及擔保之預期成本所帶動之孳息曲線波動 尤其敏感。 2014年 2013年 百萬美元 百萬美元 PVIF於12月31日受到的影響： 無風險利率上移100個基點 320 184 無風險利率下移100個基點1 (589) (289) 1 倘無風險利率下移100個基點將導致負比率，於計算對PVIF的影響時則使用最低比率0%。 對非經濟假設變動的敏感度 制訂壽險產品的公司之投保人負債及PVIF，是經參考包括死亡率及╱或發病率、失效率及支 出率等非經濟假設而釐定。下表載列倘若我們所有制訂保險產品附屬公司對前述非經濟假 設作出合理可能的修訂時，PVIF於當日的敏感度。 2014年 2013年 百萬美元 百萬美元 PVIF於12月31日受到的影響： 死亡率及╱或發病率上升10% (66) (84) 死亡率及╱或發病率下降10% 70 84 失效率上升10% (146) (154) 失效率下降10% 165 173 支出率上升10% (93) (109) 支出率下降10% 94 110 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 413 其他無形資產 無形資產 （不包括商譽及PVIF） 之變動 內部 開發軟件 其他 總計 百萬美元 百萬美元 百萬美元 成本 於2014年1月1日 5,999 2,975 8,974 增添 732 177 909 出售 (35) (80) (115) 撇賬額 (24) (53) (77) 其他變動 (259) (156) (415) 於2014年12月31日 6,413 2,863 9,276 累計攤銷 於2014年1月1日 (3,809) (1,761) (5,570) 本年度費用1 (677) (261) (938) 減值 (11) (54) (65) 出售 32 77 109 撇賬額 24 53 77 其他變動 155 57 212 於2014年12月31日 (4,286) (1,889) (6,175) 於2014年12月31日之賬面淨值 2,127 974 3,101 成本 於2013年1月1日 5,703 3,345 9,048 增添 731 142 873 出售 (117) (196) (313) 撇賬額 (57) (47) (104) 其他變動 (261) (269) (530) 於2013年12月31日 5,999 2,975 8,974 累計攤銷 於2013年1月1日 (3,469) (1,963) (5,432) 本年度費用1 (675) (179) (854) 減值 (39) (4) (43) 出售 111 167 278 撇賬額 57 47 104 其他變動 206 171 377 於2013年12月31日 (3,809) (1,761) (5,570) 於2013年12月31日之賬面淨值 2,190 1,214 3,404 1 本年度攤銷費用在收益表的 「無形資產攤銷及減值」 項內確認，惟按揭債務管理權之攤銷則於 「費用收益淨 額」 項內確認。於2014年，扣除按揭債務管理權之攤銷費用後之重估值為6,700萬美元 （2013年：撥賬3,400萬 美元） 。 22 於附屬公司之投資 會計政策 滙豐將受其控制企業之投資分類為附屬公司。滙豐之綜合入賬政策詳述於附註1(h)。屬於結構公司的附屬公司包含 於附註39。 滙豐控股於附屬公司之投資，乃按成本減減值虧損後列賬。倘若自上次確認減值虧損後，釐定於附屬公司之投資的 可收回金額所用估算數字出現變化，則可從收益表撥回前期確認之減值虧損。 財務報表附註 （續） 滙豐控股有限公司 414 滙豐控股之主要附屬公司 於2014年12月31日 註冊成立 滙豐所佔 或登記 股本權益 已發行 股份 國家╱地區 % 股本 類別 歐洲 HSBC Asset Finance (UK) Limited 英格蘭 100 2.65億英鎊 1英鎊之普通股 HSBC Bank A.S.
土耳其 100 6.52億 土耳其里拉 1土耳其里拉 之A類普通股 1土耳其里拉 之B類普通股 英國滙豐銀行有限公司 英格蘭 100 7.97億英鎊 1英鎊之普通股 1英鎊之優先普通股 0.01美元之系列2 第三貨幣優先股 0.01美元之 第三貨幣優先股 法國滙豐 法國 99.99 3.37億歐元 5.00歐元之股份 HSBC Private Banking Holdings (Suisse) SA 瑞士 100 13.63億 瑞士法郎 1,000瑞士法郎 之普通股 HSBC Trinkaus & Burkhardt AG 德國 80.65 7,540萬歐元 零面值股份 亞洲 恒生銀行有限公司1 香港 62.14 96.58億港元 零面值普通股 澳洲滙豐銀行有限公司 澳洲 100 8.11億澳元 零面值普通股 滙豐銀行 （中國） 有限公司 中國5 100 人民幣154億元 人民幣1.00元之普通股 馬來西亞滙豐銀行有限公司 馬來西亞 100 1.15億馬元 0.50馬元之普通股 滙豐 （台灣） 商業銀行股份有限公司 台灣 100 348億新台幣 10.00新台幣之普通股 滙豐人壽保險 （國際） 有限公司 百慕達 100 41.78億港元 1.00港元之普通股 香港上海滙豐銀行有限公司 香港 100 960.52億港元 零面值普通股 1.00美元之CIP2 1.00美元之CRP3 1.00美元之NIP4 中東及北非 中東滙豐銀行有限公司 澤西 100 9.31億美元 1.00美元之普通股 1.00美元之CRP3 埃及滙豐銀行 埃及 94.53 27.96億埃及鎊 84.00埃及鎊之普通股 北美洲 加拿大滙豐銀行 加拿大 100 12.25億加元 零面值普通股6 美國滙豐銀行 美國 100 200萬美元 100美元之普通股 美國滙豐融資有限公司 美國 100 －6 0.01美元之普通股 HSBC Securities (USA) Inc.
美國 100 －6 0.05美元之普通股 拉丁美洲 阿根廷滙豐銀行 阿根廷 99.99 12.44億 阿根廷披索 1.00阿根廷披索 之普通股-A類 1.00阿根廷披索 之普通股-B類 巴西滙豐銀行 巴西 100 64.02億 巴西雷亞爾 零面值股份 HSBC Mexico, S.A., Institución de Banca Múltiple, Grupo Financiero HSBC 墨西哥 99.99 56.81億 墨西哥披索 2.00墨西哥披索 之普通股 1 在香港上市。 4 非累積不可贖回優先股。 2 累積不可贖回優先股。 5 中華人民共和國。 3 累積可贖回優先股。 6 已發行股本少於100萬美元。 有關主要附屬公司向集團以外人士發行之債務、後償債務及優先股，分別詳載於附註26 「已發行債務證券」 、 附註30 「後償負債」 及附註34 「非控股股東權益」 。 上述各附屬公司均已包括在滙豐綜合財務報表內。 22－於附屬公司之投資 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 415 根據英國 《2006年公司法》 第409條之規定，滙豐旗下各附屬公司之詳情，將載於滙豐控股送 交英國公司註冊處之下一份周年報表附錄內。 除中東滙豐銀行有限公司主要在中東及北非經營及滙豐人壽保險 （國際） 有限公司主要在香 港經營之外，滙豐各公司之主要經營所在國家╱地區與註冊成立所在國家╱地區相同。 組成滙豐架構的網絡，包括地區銀行及在各地註冊成立經營銀行業務的受規管公司。各銀 行根據適用的審慎規定獲個別撥資，並須按照集團就有關國家或地區訂定的承受風險水平 維持適當的緩衝資本。我們的資本管理程序在經董事會批准的集團年度資本計劃內充分說 明。滙豐控股乃各附屬公司的主要股本提供者，有需要時亦會向該等附屬公司提供非股權 資本。該等投資的資金大多數來自滙豐控股的股權及非股權資本發行及保留利潤。滙豐控 股透過資本管理程序，力求在本身資本組合成分及對各附屬公司的投資之間保持平衡。基 於上文所述原則，滙豐控股提供該等投資資金的能力並無即時或可預見的障礙。附屬公司 向滙豐控股派付股息或提供資金的能力，視乎多項因素而定，當中包括所屬地區監管機構 的資本及銀行業規定、法定儲備，以及財務及營運表現。於2014及2013年內，集團旗下附屬 公司在派付股息或償還貸款方面並無受到重大限制。此外，我們的附屬公司預計在派付股 息或償還貸款方面並無可預見限制。 滙豐控股代其他滙豐集團公司提供擔保之金額載於附註37。 滙豐擁有少於50%投票權但仍綜合入賬之結構公司 綜合計算資產 賬面總值 結構公司性質 2014年 2013年 十億美元 十億美元 Solitaire Funding Ltd 9.0 10.2 證券投資中介機構 Mazarin Funding Limited 3.9 7.4 證券投資中介機構 Barion Funding Limited 2.0 3.8 證券投資中介機構 Malachite Funding Limited 1.4 3.0 證券投資中介機構 HSBC Home Equity Loan Corporation I 1.9 2.1 證券化公司 HSBC Home Equity Loan Corporation II 0.9 1.6 證券化公司 Regency Assets Limited 11.0 13.5 中介機構 Bryant Park Funding LLC － 0.4 中介機構 除上述者外，滙豐亦將若干個別並非重要的結構公司綜合入賬，此等公司的資產總值為229億 美元 （2013年：261億美元） 。進一步詳情請參閱附註39。 於上述各情況，當滙豐因參與公司事務而面對可變動回報的風險或有權享有可變動回報， 並且能夠透過對該公司擁有的權力影響相關回報時，滙豐即為控制有關公司，並會將相關 公司綜合入賬。 附有重大非控股股東權益之附屬公司 2014年 2013年 恒生銀行有限公司 非控股股東所持擁有權及投票權比例 37.86% 37.86% 業務所在地 香港 香港 百萬美元 百萬美元 非控股股東應佔利潤 760 1,332 附屬公司非控股股東權益累計 5,765 4,591 已付予非控股股東之股息 513 495 財務資料概要： －資產總值 160,769 145,380 －負債總額 144,642 133,253 －未扣除貸款減值之營業收益淨額 3,687 4,876 －本年度利潤 2,007 3,517 －本年度全面收益總額 4,460 3,145 財務報表附註 （續） 滙豐控股有限公司 416 23 預付款項、應計收益及其他資產 會計政策 持作出售用途資產 倘若出售用途業務組合之資產及負債和非流動資產的賬面值將主要透過出售而非持續使用而收回，便會分類為持 作出售用途。持作出售用途資產一般按其賬面值與公允值減出售成本兩者中的較低數額計量，惟不屬於IFRS 5 「持 作出售用途之非流動資產及終止經營業務」 計量規定範圍內之資產和負債則除外。 最接近首次列賬分類為持作出售用途前，相關資產及負債的賬面值乃根據適用IFRS計量。其後重新計量出售用途業 務組合時，倘若資產和負債不屬於IFRS 5計量規定範圍內但計入分類為持作出售用途之出售用途業務組合內，則其 賬面值乃於釐定出售用途業務組合之公允值減出售成本前，根據適用IFRS重新計量。 物業、機器及設備 土地及樓宇的列賬金額為歷史成本或過渡至IFRS當日之公允值 （ 「設定成本」 ） ，減去減值虧損及按照資產估計可用 年期計算所得折舊額，方法如下： ‧ 永久業權之土地不予折舊； ‧ 永久業權之樓宇按直線基準每年折舊2%，或按尚餘可用年期分攤折舊，並以較高者為準；及 ‧ 租賃土地及樓宇按其尚餘租賃期或尚餘可用年期分攤折舊，並以較短者為準。 設備、裝置及傢具 （包括滙豐為出租人之經營租賃設備） 以成本減去減值虧損及按照資產可用年期計算所得折舊額 列賬，可用年期一般為5至20年。 倘若物業、機器及設備之賬面值未必可以收回，則須對物業、機器及設備進行減值檢討。 滙豐持有若干物業作投資用途，以賺取租金或實現資本增值或以期同時達致此兩項目標，而投資物業乃按公允值 計入資產負債表內。 預付款項、應計收益及其他資產 2014年 2013年 百萬美元 百萬美元 預付款項及應計收益 10,554 11,006 持作出售用途資產 7,647 4,050 黃金 15,726 22,929 背書及承兌 10,775 11,624 再保人應佔之保單未決賠款 （附註28） 1,032 1,408 僱員福利資產 （附註6） 5,028 2,140 其他賬項 13,882 12,838 物業、機器及設備 10,532 10,847 於12月31日 75,176 76,842 預付款項、應計收益及其他資產包括406.22億美元 （2013年：376.35億美元） 金融資產，其中大 部分以已攤銷成本計量。 物業、機器及設備－選錄資料 2014年 2013年 百萬美元 百萬美元 成本或公允值 21,831 21,927 累計折舊及減值 11,299 11,080 於12月31日之賬面淨值 10,532 10,847 按成本增添 1,477 1,980 按賬面淨值出售 69 267 物業、機器及設備包括1： 土地及樓宇 5,234 5,661 －永久業權 1,769 2,062 －長期租賃 1,252 1,266 －中期及短期租賃 2,213 2,333 投資物業2 2,236 1,945 1 在香港的永久業權類別為零 （2013年：零） 、長期租賃類別為13.06億美元 （2013年：13.09億美元） 、中期租賃 類別為26.38億美元 （2013年：24.72億美元） ，而短期租賃類別為零 （2013年：200萬美元） 。 2 投資物業於每年12月31日按市值基準估值，負責估值之獨立專業估價師近期曾為有關地點及類別的物業 估值。滙豐在香港、澳門特別行政區及中國內地的投資物業是由戴德梁行有限公司 （屬下估價師為香港 測量師學會會員） 負責估值，位於上述地點的投資物業佔須予重估之滙豐投資物業總值超過74%。在其他 國家╱地區之物業由不同的獨立具專業資格估價師負責估值，有關物業佔滙豐投資物業總值26%。 23－預付款項、應計收益及其他資產╱24－交易用途負債╱25－指定以公允值列賬之金融負債 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 417 24 交易用途負債 會計政策 若交易用途負債購入或產生的主要目的是在短期內出售或回購，或屬於一併管理的已識別金融工具組合之一部分， 並有證據顯示該等項目近期有短期獲利的規律，則會分類為持作交易用途。此等項目會於交易日確認入賬 （即滙豐 與交易對手訂立合約安排之日） ，並一般於償清時撤銷確認。此等資產首次列賬均按公允值計量，其後公允值變動 及已付利息均會在收益表的 「交易收益淨額」 項內確認。 出售借入證券會分類為交易用途負債。 交易用途負債 2014年 2013年 百萬美元 百萬美元 同業存放1 41,453 43,130 客戶賬項1 50,600 57,688 其他已發行債務證券 （附註26） 33,602 32,155 其他負債－證券短倉淨額 64,917 74,052 於12月31日 190,572 207,025 1 同業存放及客戶賬項包括回購、結算賬項、借出股票及其他款額。 於2014年12月31日，因滙豐的信貸風險變動產生之公允值變動累計金額為虧損7,900萬美元 （2013年：虧損9,500萬美元） 。 25 指定以公允值列賬之金融負債 會計政策 指定工具以公允值列賬的準則及其計量方法詳述於附註15。指定以公允值列賬的方式一經訂明不得撤回。已指定列 賬方式之金融負債會於滙豐與交易對手訂立合約時確認，且一般會於償清時撤銷確認。本節提供該指定方式的例子： ‧ 已發行長期債務。若干已發行定息長期債務證券的應付利息，已與若干利率掉期的利息配對，作為集團明文規 定的利率風險管理策略之一。若該等已發行債務證券按已攤銷成本列賬，則會出現會計上的錯配，而此錯配乃 透過指定以公允值列賬的方式撇銷。 ‧ 在單位相連及非單位相連投資合約下的金融負債 滙豐向客戶簽發合約，當中涉及保險風險、金融風險或同時涉及該兩種風險。滙豐同意向對方承擔非重大保險風險 的合約並不分類為保單，但會入賬列為金融負債。有關滙豐承擔重大保險風險的合約，請參閱附註28。 由保險附屬公司簽發之相連及若干非相連投資合約下對客戶的負債及相應之金融資產，均指定以公允值列賬。負債 至少相等於有關退保額或轉撥價值，該數額乃經參考相關基金或指數之價值而計算。應收保費及提取款項之金額入 賬列作投資合約負債之增額或減額。與獲取新增投資合約或為現有投資合約續期直接有關之遞增成本會遞延入賬， 並於提供投資管理服務期間予以攤銷。 指定以公允值列賬之金融負債－滙豐 2014年 2013年 百萬美元 百萬美元 同業存放及客戶賬項 160 315 在投資合約下對客戶之負債 6,312 13,491 已發行債務證券 （附註26） 46,364 53,363 後償負債 （附註30） 21,822 18,230 優先證券 （附註30） 1,495 3,685 於12月31日 76,153 89,084 於2014年12月31日，指定以公允值列賬之金融負債的賬面值較到期日之合約金額多出58.13億 美元 （2013年：多出43.75億美元） 。因信貸風險變動產生之公允值變動累計金額為虧損8.7億 美元 （2013年：虧損13.34億美元） 。 財務報表附註 （續） 滙豐控股有限公司 418 指定以公允值列賬之金融負債－滙豐控股 2014年 2013年 百萬美元 百萬美元 已發行債務證券 （附註26） ： －應付第三方 8,185 8,106 後償負債 （附註30） ： －應付第三方 9,513 9,760 －應付滙豐旗下業務 981 3,161 於12月31日 18,679 21,027 於2014年12月31日，指定以公允值列賬之金融負債的賬面值較到期日之合約金額多出26.94億 美元 （2013年：多出23.09億美元） 。因信貸風險變動產生之公允值變動累計金額為虧損5.2億 美元 （2013年：虧損8.59億美元） 。 26 已發行債務證券 會計政策 已發行債務證券之金融負債會於滙豐與交易對手訂立合約安排時確認，其首次列賬金額會按公允值計量。在一般情 況下，此金額相等於已收取之代價減除已產生之直接相關交易支出。其後計量金融負債 （不包括按公允值計入損益 賬之金融負債及金融擔保） 的價值時，將按已攤銷成本入賬，計算時會使用實質利率法，按金融工具之預計年期攤 銷下列兩者的差額：扣除已產生之直接相關交易支出後的已收取所得款項與贖回款額之差額。 已發行債務證券－滙豐 2014年 2013年 百萬美元 百萬美元 債券及中期票據 132,539 146,116 其他已發行債務證券 43,374 43,482 175,913 189,598 當中已發行債務證券入賬列作： －交易用途負債 （附註24） (33,602) (32,155) －指定以公允值列賬之金融負債 （附註25） (46,364) (53,363) 於12月31日 95,947 104,080 已發行債務證券－滙豐控股 2014年 2013年 百萬美元 百萬美元 債務證券 9,194 10,897 當中已發行債務證券入賬列作： －指定以公允值列賬之金融負債 （附註25） (8,185) (8,106) 於12月31日 1,009 2,791 27 應計項目、遞延收益及其他負債 2014年 2013年 百萬美元 百萬美元 持作出售用途業務組合之負債 6,934 2,804 應計項目及遞延收益 15,075 16,185 滙豐應付投資者資金綜合計算 782 1,008 融資租賃下之責任 67 252 背書及承兌 10,760 11,614 僱員福利負債 （附註6） 3,208 2,931 其他負債 16,570 17,547 於12月31日 53,396 52,341 應計項目、遞延收益及其他負債包括金融負債438.4億美元 （2013年：462.58億美元） ，其中大 部分按已攤銷成本計量。 26－已發行債務證券╱27－應計項目、遞延收益及其他負債╱28－保單未決賠款 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 419 28 保單未決賠款 會計政策 滙豐向客戶發出合約，當中涉及保險風險、金融風險或同時涉及該兩種風險。保單乃指滙豐同意向對方承擔重大保 險風險的合約，訂明滙豐同意在日後發生某些特定但不確定的事件時，向對方作出補償，並分類為保單。保單亦可 轉移金融風險，惟倘若保險風險重大，則在賬目中仍列作保單。 保單未決賠款 各壽險業務乃根據業務所在地的精算原則，計算非相連壽險保單之未決賠款。單位相連壽險保單之未決賠款至少相 等於有關退保額或轉撥價值，該金額乃經參考相關基金或指數之價值而計算。 保險未決賠款會進行負債充足性測試，以確保根據現時對日後現金流之估算，未決賠款之賬面值為足夠。於進行負 債充足性測試時，會折現所有約定現金流，並與未決賠款之賬面值比較。若識別出短缺額，將會即時自收益表扣 取。 附有DPF之保單的未來利潤 若保單賦予投保人分享酌情參與利潤的利益，保單未決賠款會包括賦予投保人日後酌情利益之準備。該等準備反映 投資組合至今之實際表現、管理層對支持保單的資產在日後表現之預期，以及其他經驗因素，如死亡率、保單失效 率及營運效率 （如適用） 。這項利益可能因保單條款、規例或過往分派政策而產生。 附有DPF之投資合約 儘管附有DPF之投資合約屬金融工具，IFRS 4容許繼續將此等合約當作保單處理。因此，集團將該等合約之保費確 認為收入，並將因而產生之負債賬面值增額確認為支出。 倘該等合約之酌情利益主要反映投資組合之實際表現，就該等合約之未變現投資增益淨額而言，負債之相應增加會 於處理相關資產之未變現增益後，在收益表或其他全面收益項內確認。就未變現虧損淨額而言，遞延參與資產僅會 在該等資產收回機會甚高的情況下方會確認。因相關資產之已變現損益而產生之負債變動，會於收益表內確認。 保單未決賠款 再保人 總額 應佔部分 淨額 百萬美元 百萬美元 百萬美元 非相連保單1 於2014年1月1日 33,950 (1,118) 32,832 已付賠償及利益 (3,575) 175 (3,400) 投保人負債之增額 7,764 (409) 7,355 出售╱撥入持作出售用途 (589) 527 (62) 匯兌差額及其他變動 (577) 53 (524) 於2014年12月31日 36,973 (772) 36,201 附有酌情參與條款之投資合約 於2014年1月1日 26,427 － 26,427 已付賠償及利益 (2,175) － (2,175) 投保人負債之增額 3,188 － 3,188 匯兌差額及其他變動2 (2,372) － (2,372) 於2014年12月31日 25,068 － 25,068 相連壽險保單 於2014年1月1日 13,804 (290) 13,514 已付賠償及利益 (1,499) 88 (1,411) 投保人負債之增額 2,762 33 2,795 出售╱撥入持作出售用途 (2,547) 74 (2,473) 匯兌差額及其他變動3 (700) (165) (865) 於2014年12月31日 11,820 (260) 11,560 於2014年12月31日之投保人負債總額 73,861 (1,032) 72,829 財務報表附註 （續） 滙豐控股有限公司 420 再保人 總額 應佔部分 淨額 百萬美元 百萬美元 百萬美元 非相連保單1 於2013年1月1日 30,765 (952) 29,813 已付賠償及利益 (3,014) 164 (2,850) 投保人負債之增額 6,892 (367) 6,525 出售╱撥入持作出售用途 (52) 13 (39) 匯兌差額及其他變動 (641) 24 (617) 於2013年12月31日 33,950 (1,118) 32,832 附有酌情參與條款之投資合約 於2013年1月1日 24,374 － 24,374 已付賠償及利益 (2,308) － (2,308) 投保人負債之增額 3,677 － 3,677 匯兌差額及其他變動2 684 － 684 於2013年12月31日 26,427 － 26,427 相連壽險保單 於2013年1月1日 13,056 (455) 12,601 已付賠償及利益 (1,976) 426 (1,550) 投保人負債之增額 3,379 111 3,490 匯兌差額及其他變動3 (655) (372) (1,027) 於2013年12月31日 13,804 (290) 13,514 於2013年12月31日之投保人負債總額 74,181 (1,408) 72,773 1 包括非壽險保單未決賠款。 2 包括與應付予投保人酌情參與利潤的利益有關的負債變動，此等利益源自於其他全面收益項內確認的未 變現投資增益淨額。 3 包括於再保險協議下產生之金額。 投保人負債之增額乃指導致年內須額外承擔對投保人負債之所有事項涉及之總額。導致投 保人負債出現變動的主要因素包括身故賠償、退保、失效、於最初訂立保單時即產生的投 保人負債、宣派紅利及投保人應佔其他款額。 29 準備 會計政策 倘若可能需要流出經濟利益，以解決過往事件引致之現有法定或推定責任，而且又能可靠估計相關責任牽涉之數 額，即會確認準備。 關鍵會計估算及判斷 準備 於確定是否存在當前期間需要承擔的責任及估計任何經濟效益外流的可能性、時間及金額時，會涉及判斷。對訴 訟、物業 （包括條件繁苛的合約） 及同類責任作評估時，會採納專家意見。 法律訴訟及監管事宜的準備通常較其他類別的準備需要作出較大程度的判斷。倘若事件處於初步階段，則難以作出 會計判斷，因為確定是否存在當前期間需要承擔的責任、估計任何經濟效益外流的可能性及金額，均有頗多不確定 因素。隨著事件發展，管理層及法律顧問會按持續基準評估應否確認準備，並在適當時修訂先前的判斷及估算。在 更深入的階段，通常可以較容易就更清晰界定且有可能會產生的一系列結果作出判斷及估算。然而，準備額可能極 易受所用假設影響。任何待決法律訴訟、調查或研訊可能出現多種結果。因此，量化個別事件可能產生的多種結 果通常並不可行。為該類準備有意義地總計多種潛在結果所涉數額亦不可行，因為該等事件牽涉多種不同性質及情 況，且包含廣泛類別的不明朗因素。 提撥準備以便採取與客戶有關的補救措施，亦需要作出重大程度的估算及判斷。確認的準備額視乎多項不同假設而 定，例如接獲投訴個案數目、接獲投訴個案數目所涉預計期間、投訴量的下降率、發現屬系統性不當銷售個案所涉 客戶人數，以及每宗客戶投訴個案涉及的保單數目。 29－準備 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 421 準備 重組 架構成本 合約 承諾 法律 訴訟及 監管事宜 與客戶 有關的 補救措施 其他準備 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年1月1日 271 177 1,832 2,382 555 5,217 額外準備╱增撥準備 147 136 1,752 1,440 154 3,629 已動用之準備 (143) (2) (1,109) (1,769) (112) (3,135) 撥回額 (43) (46) (281) (184) (66) (620) 沖抵折現 － 1 43 10 11 65 匯兌差額及其他變動 (35) (32) (53) (48) 10 (158) 於2014年12月31日 197 234 2,184 1,831 552 4,998 於2013年1月1日 251 301 1,667 2,387 646 5,252 額外準備╱增撥準備 179 57 1,209 1,536 230 3,211 已動用之準備 (111) (5) (709) (1,487) (167) (2,479) 撥回額 (65) (66) (340) (94) (126) (691) 沖抵折現 － － 38 7 13 58 匯兌差額及其他變動 17 (110) (33) 33 (41) (134) 於2013年12月31日 271 177 1,832 2,382 555 5,217 有關法律訴訟及監管事宜的進一步詳情載於附註40，當中包括就匯率的調查及訴訟作出的 準備。法律訴訟包括對滙豐旗下公司提出的民事法庭訴訟、仲裁或審裁程序 （不論是以申索 或反申索方式） 或如未能解決即會引起法庭訴訟、仲裁或審裁程序的民事爭議。監管事宜指 監管機構或執法機構就指稱滙豐行為不當而進行或因應有關行動而進行的調查、評估及其 他行動。 與客戶有關的補救措施指滙豐就未能遵照法規或公平對待客戶所引起的相關損失或損害而 補償客戶所作出之行動。與客戶有關的補救措施是滙豐對客戶投訴及╱或業內銷售方式的 發展的回應，不一定由監管機構採取的行動而引致。 還款保障保險 於2014年12月31日，就過往年度可能不當銷售還款保障保單 （ 「PPI」 ） 所涉估計賠償責任而作出 的準備為10.79億美元 （2013年12月31日：9.46億美元） 。於本年度確認增加準備9.6億美元，主 要反映與過往預測相比，申索管理公司接獲投訴的數量增加。現時的接獲投訴數量預測趨 勢顯示賠償計劃將於2018年第一季完成。然而，此時間存在不確定性，因為根據實際經驗， 該趨勢可能隨著時間推移而有所改變。 自2011年上半年的司法覆核判決以來作出的累計準備達42億美元，當中於2014年12月31日已 支付32億美元。 估計須予賠償的責任乃按客戶已付保費總額加單息年利率8厘 （或相關貸款產品本身的利率， 如此為較高者） 計算。計算賠償責任的基準與整付保費及定期保費保單的基準相同。未來估 計賠償水平以每份保單的過往觀察得出賠償為基準。 滙豐自2000年以來已銷售合共約540萬份還款保障保單，產生估計收入為43億美元 （按2014年 的平均匯率計算） 。該等保單的已承保保費總額約為56億美元 （按2014年的平均匯率計算） 。於 2014年12月31日，預期接獲的估計投訴個案總數為190萬宗，相當於已銷售保單總數的36%。 現估計將會聯絡230萬份保單的客戶，相當於已銷售保單總數的42%。這些估計數字包括接 獲投訴以及滙豐就若干保單主動聯絡客戶 （ 「聯絡客戶」 ） 。 下表詳列於2014年12月31日接獲的累計投訴數目及預期日後的申索數目： 財務報表附註 （續） 滙豐控股有限公司 422 累計至 2014年 12月31日 日後預期 接獲投訴1 （按千份保單計） 1,215 344 聯絡客戶 （按千份保單計） 448 291 聯絡客戶所得回應率 51% 51% 每宗申索的平均成立比率2 77% 71% 每宗申索的平均賠償額 （美元） 2,611 3,115 1 不包括投訴人並無持有還款保障保單的無效申索。 2 申索包括接獲投訴及聯絡客戶所得回應。 計算賠償責任涉及的主要假設為接獲投訴個案數目、接獲投訴個案預計期間、投訴量的下 降率、識別為系統性不當銷售個案所涉客戶人數，以及每宗客戶投訴個案涉及的保單數目。 隨著根本原因分析仍在繼續進行，在處理所接獲的客戶投訴方面積累了更多經驗，以及處 理持續聯絡客戶所得回應，主要假設很可能會因時而異。 接獲投訴個案總數增加╱減少100,000宗，將使賠償準備增加╱減少約2.22億美元 （按2014年 的平均匯率計算） 。我們聯絡客戶所得回應率每增加╱減少1%，將使賠償準備增加╱減少約 1,300萬美元。 除了上述因素及假設外，所須作出賠償的範圍亦須視乎個別客戶個案的事實及情況而定。 基於這些原因，截至目前為止，最終賠償支出仍然是未知之數。 利率衍生工具 於2014年12月31日，我們就可能在英國不當銷售利率衍生工具而估計須予賠償的責任持有 準備3.12億美元 （2013年12月31日：7.76億美元） 。有關準備涉及就過往因衍生工具合約支付費 用而應付予客戶的估計賠償、銀行預期撇銷的未平倉衍生工具合約款額，以及估計項目成 本。於本年度，準備增加2.88億美元，反映了最新申索經驗及金融業操守監管局於2015年1月 28日有關延長該計劃至2015年3月31日的公布以及在與受影響客戶溝通後預期將有更多客戶 選擇參與該計劃。 滙豐最終須作出賠償的範圍會視乎檢討期內已聯絡及其他客戶的回應，以及各個別個案的 事實及情況分析而定，包括已接獲的連帶損失申索。基於這些原因，最終就此計劃所涉及 之最終賠償支出目前仍然是未知之數。 英國 《消費者信貸法》 滙豐已就英國 《消費者信貸法》 的定額無抵押貸款協議規定的遵守情況進行審查。於2014年 12月31日，我們已於 「應計項目、遞延收益及其他負債」 內確認3.79億美元，以向客戶退還利 息，此乃主要由於我們未有於年度結單中提醒客戶其有權提早償還部分貸款 （儘管客戶貸款 文件載有此項權利） 。迄今為止之累計負債為5.91億美元，其中2.12億美元已支付予客戶。對 於是否已符合 《消費者信貸法》 的其他技術性要求，亦存在不確定因素，我們已就此作出評 估，或有負債最高可達9億美元。 巴西勞工、民事及財務索償 上文所述的 「法律訴訟及監管事宜」 指勞工、民事及財務訴訟準備5.01億美元 （2013年﹕5億美 元） 。在該等準備中，2.46億美元 （2013年：2.32億美元） 涉及滙豐巴西業務的前度僱員於離職 後提出勞工及超時工作相關訴訟索償。估計負債金額時涉及的主要假設，包括預期離職僱 員人數、個別薪金水平及各個別個案的事實與情況。 29－準備╱30－後償負債 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 423 30 後償負債 滙豐 2014年 2013年 百萬美元 百萬美元 後償負債 按已攤銷成本 26,664 28,976 －後償負債 22,355 24,573 －優先證券 4,309 4,403 指定以公允值列賬 （附註25） 23,317 21,915 －後償負債 21,822 18,230 －優先證券 1,495 3,685 於12月31日 49,981 50,891 後償負債 滙豐控股 25,277 22,308 滙豐內部其他公司 24,704 28,583 於12月31日 49,981 50,891 滙豐之後償負債 後償負債的償債地位低於優先責任，一般計入滙豐的資本基礎。在適用情況下，資本證券可 由滙豐經事先通知審慎監管局及在相關情況下，得到當地銀行監管機構同意後提早贖回。 倘若並無於首個提早贖回日贖回，應付票息可步陞或改為浮息並以銀行同業拆息率為基準。 浮息資本證券利率一般與銀行同業拆息率掛鈎。至於就餘下資本證券支付之利息，以最高 為10.176厘的固定利率計算。 下文所披露的資產負債表金額乃以IFRS為基準計算，並不反映工具對監管規定資本的貢益， 此乃歸因於計入發行成本、監管當局攤銷規定及資本指引4下豁免條文所訂明的監管當局設 定的合資格限額。 滙豐之已發行後償負債 首個提早 贖回日 到期日 2014年 2013年 百萬美元 百萬美元 由滙豐控股有限公司擔保之額外一級資本證券1 14億歐元 5.3687厘非累積步陞永久優先證券2 2014年3月 － 2,022 5億英鎊 8.208厘非累積步陞永久優先證券 2015年6月 779 825 7.5億歐元 5.13厘非累積步陞永久優先證券 2016年3月 979 1,129 9億美元 10.176厘非累積步陞永久優先證券 （系列2） 2030年6月 891 891 2,649 4,867 由英國滙豐銀行有限公司擔保之額外一級資本證券1 3億英鎊 5.862厘非累積步陞永久優先證券 2020年4月 515 534 7億英鎊 5.844厘非累積步陞永久優先證券 2031年11月 1,091 1,157 1,606 1,691 由英國滙豐銀行有限公司發行之二級證券 5億英鎊 4.75厘可提早贖回後償票據3 2015年9月 2020年9月 802 866 3.5億英鎊 5.00厘可提早贖回後償票據4 2018年3月 2023年3月 605 635 3億英鎊 6.50厘後償票據 － 2023年7月 466 494 3.5億英鎊 5.375厘可提早贖回後償步陞票據5 2025年11月2030年11月 620 602 5億英鎊 5.375厘後償票據 － 2033年8月 905 884 2.25億英鎊 6.25厘後償票據 － 2041年1月 349 370 6億英鎊 4.75厘後償票據 － 2046年3月 924 980 5億歐元 可提早贖回後償浮息票據6 2015年9月 2020年9月 588 655 3億美元 7.65厘後償票據 － 2025年5月 400 380 7.5億美元 無定期浮息主資本票據 1990年6月 750 751 5億美元 無定期浮息主資本票據 1990年9月 500 499 3億美元 無定期浮息主資本票據 （系列3） 1992年6月 300 299 7,209 7,415 財務報表附註 （續） 滙豐控股有限公司 424 首個提早 贖回日 到期日 2014年 2013年 百萬美元 百萬美元 由香港上海滙豐銀行有限公司發行之二級證券 4億美元 主資本無定期浮息票據 1990年8月 403 404 4億美元 主資本無定期浮息票據 （第2系列） 1990年12月 401 402 4億美元 主資本無定期浮息票據 （第3系列） 1991年7月 400 400 1,204 1,206 由澳洲滙豐銀行有限公司發行之二級證券 2億澳元 可提早贖回後償浮息票據 2015年11月2020年11月 164 179 164 179 由馬來西亞滙豐銀行有限公司發行之二級證券 5億馬元 4.35厘後償債券 2017年6月 2022年6月 143 152 5億馬元 5.05厘後償債券 2022年11月2027年11月 144 154 287 306 由美國滙豐有限公司發行之二級證券 2億美元 7.808厘資本證券 2006年12月2026年12月 200 200 2億美元 8.38厘資本證券 2007年5月 2027年5月 200 200 1.5億美元 9.50厘後償債務7 － 2014年4月 － 151 1.5億美元 7.75厘資本信託轉手證券 2006年11月2026年11月 150 150 7.5億美元 5.00厘後償票據 － 2020年9月 738 746 2.5億美元 7.20厘後償債券 － 2097年7月 216 215 其他各自少於1.5億美元之後償負債 297 299 1,801 1,961 由美國滙豐銀行發行之二級證券 10億美元 4.625厘後償票據7 － 2014年4月 － 1,000 5億美元 6.00厘後償票據 － 2017年8月 508 513 12.5億美元 4.875厘後償票據 － 2020年8月 1,210 1,262 10億美元 5.875厘後償票據 －2034年11月 1,245 1,081 7.5億美元 5.625厘後償票據 － 2035年8月 934 811 7億美元 7.00厘後償票據 － 2039年1月 676 696 4,573 5,363 由美國滙豐融資有限公司發行之二級證券 10億美元 5.911厘信託優先證券8 2015年11月2035年11月 998 996 29.39億美元 6.676厘優先後償票據9 － 2021年1月 2,185 2,182 3,183 3,178 由巴西滙豐銀行發行之二級證券 3.83億巴西雷亞爾後償存款證 － 2015年2月 144 162 5億巴西雷亞爾 後償浮息存款證 －2016年12月 188 212 其他各自少於1.5億美元之後償負債10 81 224 413 598 由加拿大滙豐銀行發行之二級證券 4億加元 4.80厘後償債券 2017年4月 2022年4月 367 403 2億加元 4.94厘後償債券 2016年3月 2021年3月 172 188 3,900萬加元 浮息債券 1996年10月2083年11月 34 37 573 628 由墨西哥滙豐發行之證券 18.18億墨西哥披索不可轉換後償責任11 2013年9月 2018年9月 124 138 22.73億墨西哥披索不可轉換後償責任11 2013年12月2018年12月 154 173 3億美元 不可轉換後償責任11, 12 2014年6月 2019年6月 240 240 518 551 由滙豐之其他附屬公司發行之證券 其他各自少於2億美元之後償負債11 524 640 滙豐附屬公司發出之後償負債總額 24,704 28,583 1 請參閱下段 「由滙豐控股或英國滙豐銀行擔保」 。 2 於2014年3月，滙豐按面值提早贖回14億歐元5.3687厘非累積步陞永久優先證券。 3 於2015年9月後應付的利率相等於英鎊之三個月倫敦銀行同業拆息加0.82厘。 4 於2018年3月後應付的利率相等於當時之五年期英國金邊證券之總贖回收益率加1.8厘。 5 於2025年11月後應付的利率相等於英鎊之三個月倫敦銀行同業拆息加1.5厘。 6 利息收益率自2015年9月起增加0.5厘。 7 於2014年4月，滙豐按面值贖回10億美元4.625厘後償票據及9.5厘後償債券。 8 分派於2015年11月改為美元之三個月倫敦銀行同業拆息加1.926厘。 9 約25%之優先後償票據由滙豐控股持有。 10 按照於2013年應用的審慎監管局 《一般審慎措施資料手冊》 及於2014年應用的資本指引4規則中之指引， 當中部分證券不符合條件計入滙豐之資本基礎內。 11 按照於2013年應用的審慎監管局 《一般審慎措施資料手冊》 及於2014年應用的資本指引4規則中之指引， 該等證券不符合條件計入滙豐之資本基礎內。 12 約6,000萬美元的後償責任由滙豐控股持有。 30－後償負債 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 425 滙豐控股 2014年 2013年 百萬美元 百萬美元 後償負債： －按已攤銷成本 17,255 14,167 －指定以公允值列賬 （附註25） 10,494 12,921 於12月31日 27,749 27,088 滙豐控股之後償負債 首個提早 贖回日 到期日 2014年 2013年 百萬美元 百萬美元 由滙豐控股有限公司發行之二級證券 應付第三方之金額 4.88億美元 7.625厘後償票據1 － 2032年5月 538 554 2.22億美元 7.35厘後償票據1 －2032年11月 278 278 20億美元 6.5厘後償票據1 － 2036年5月 2,029 2,029 25億美元 6.5厘後償票據1 － 2037年9月 3,278 3,039 15億美元 6.8厘後償票據1 － 2038年6月 1,487 1,487 20億美元 4.25厘後償票據2,5 － 2024年3月 2,069 － 15億美元 5.25厘後償票據2,5 － 2044年3月 1,735 － 9億英鎊 6.375厘可提早贖回後償票據1,3 2017年10月2022年10月 1,558 1,672 6.5億英鎊 5.75厘後償票據2 －2027年12月 1,176 1,158 6.5億英鎊 6.75厘後償票據2 － 2028年9月 1,005 1,066 7.5億英鎊 7.0厘後償票據2 － 2038年4月 1,217 1,288 9億英鎊 6.0厘後償票據2 － 2040年3月 1,379 1,464 16億歐元 6.25厘後償票據2 － 2018年3月 1,950 2,210 17.5億歐元 6.0厘後償票據2 － 2019年6月 2,623 2,884 7億歐元 3.625厘可提早贖回後償票據1,4 2015年6月 2020年6月 878 1,007 15億歐元 3.375厘可提早贖回後償票據1,2,5 2019年1月 2024年1月 1,898 2,075 25,098 22,211 應付滙豐旗下業務之金額 14億歐元 5.3687厘定息╱浮息後償票據6 2014年3月2043年12月 － 2,024 5億英鎊 8.208厘後償步陞累積票據 2015年6月 2040年6月 779 825 7.5億歐元 5.13厘定息╱浮息後償票據 2016年3月2044年12月 981 1,137 9億美元 10.176厘後償步陞累積票據 2030年6月 2040年6月 891 891 2,651 4,877 於12月31日 27,749 27,088 1 應付第三方之金額指按照資本指引4規則下豁免條文作為二級證券計入滙豐資本基礎內之證券。 2 該等證券作為全面遵守資本指引4之二級證券按終點基準計入滙豐資本基礎內。 3 於2017年10月後應付的利率相等於英鎊之三個月倫敦銀行同業拆息加1.3厘。 4 於2015年6月後應付的利率相等於三個月歐洲銀行同業拆息加0.93厘。 5 該等後償票據在滙豐控股按照已攤銷成本計量，利率風險使用公允值對沖進行對沖，而在集團層面則按 公允值計量。 6 於2014年3月，滙豐控股按面值提早贖回14億歐元5.3687厘定息╱浮息後償票據。 額外一級資本證券 額外一級資本證券作為一級資本計入滙豐資本基礎內，此類證券為投資者有權在符合若干 條件的情況下收取非累積分派的永久後償證券。該等證券一般不附有投票權，但在支付票 息及清盤時的地位卻高於普通股。一級證券符合資格的準則於2014年1月1日引入資本指引4 規則時更改。有關額外一級證券的進一步準則指引，請參閱附註35。工具若於資本指引4生 效前發行且完全不符合識別準則，則有足夠條件列為監管規定資本，惟須受限於豁免限額 及逐步退出的規定。於2014年已發行的資本證券按終點基準確認為完全符合資本指引4的額 外一級資本證券，並入賬列為股東權益，詳情見附註35。 由滙豐控股或英國滙豐銀行擔保 由滙豐控股或英國滙豐銀行按後償基準擔保的六批資本證券，均為由Jersey有限責任合夥 公司發行的非累積步陞永久優先證券。發行所得款項透過有限責任合夥公司發行後償票據 轉借各擔保人。此等優先證券因應用豁免條文而根據資本指引4有足夠條件列為滙豐的額外 一級資本，而由英國滙豐銀行擔保的兩批資本證券亦因應用豁免條文而根據資本指引4有足 夠條件列為英國滙豐銀行的額外一級資本 （按個別及綜合基準） 。 財務報表附註 （續） 滙豐控股有限公司 426 此等優先證券連同擔保在內，旨在讓投資者有權享有與假如購入相關發行人的非累積永久 優先股原應享有之權利相同的經濟利益。如支付分派受到英國銀行規例或其他規定所禁制， 或支付分派可引致違反滙豐在資本充足比率方面的規定，或滙豐控股或英國滙豐銀行沒有 足夠的可供分派儲備 （根據有關界定） ，支付分派將受到限制。 滙豐控股及英國滙豐銀行已各自訂下契約，倘若基於若干情況而未能全數支付優先證券之 分派，則其不會就其普通股支付股息或其他分派，亦不會回購或贖回其普通股，直至已全 數支付優先證券之分派為止。 就由滙豐控股擔保之優先證券而言，倘若(i)滙豐的總資本比率低於監管當局所訂之最低比 率，或(ii)鑑於滙豐控股財政狀況惡化，董事預期(i)即將發生，則優先證券可被滙豐控股之優 先股所取代，該等優先股之經濟條款須在所有重大方面與優先證券及其所連帶之擔保相同。 就由英國滙豐銀行擔保之優先證券而言，倘若(i)上述兩次發行之任何優先證券分別於2049年 4月或2048年11月仍未贖回；或(ii)英國滙豐銀行的總資本比率按個別及綜合基準計算，低於 監管當局所訂之最低比率；或(iii)鑑於英國滙豐銀行財政狀況惡化，董事預期(ii)即將發生， 則優先證券可被英國滙豐銀行之優先股所取代，該等優先股之經濟條款須在所有重大方面 與優先證券及其所連帶之擔保相同。 二級資本證券 鑑於應用豁免條文 （符合資本指引4終點規則的已識別滙豐控股證券則除外） ，該等資本證券 根據資本指引4於滙豐資本基礎內列作二級資本。二級資本證券或為永久後償證券或為定期 證券，均有責任支付票息。根據資本指引4，所有二級證券的出資額按監管規定於到期前最 後五年攤銷。 31 資產、負債及資產負債表外承諾之期限分析 第427頁之圖表分析於結算日按剩餘合約期限列示之資產、負債及資產負債表外承諾經綜合 計算之總額。計入期限分析之資產及負債款額如下： ‧ 除反向回購、回購及已發行債務證券外，交易用途資產及負債 （包括交易用途衍生工具） 計入 「1個月內到期」 一欄內，而非按合約期限列示，因為交易用途賬項款額一般只會短 期持有； ‧ 並無合約期限之金融資產及負債 （例如股權證券） 計入 「5年後到期」 一欄內。無定期或永 久工具根據工具交易對手有權給予之合約通知期分類。如無合約通知期，則無定期或永 久合約計入 「5年後到期」 一欄內； ‧ 並無合約期限之非金融資產及負債 （例如物業、機器及設備、商譽及無形資產、流動及 遞延稅項資產及負債，以及退休福利負債） 計入 「5年後到期」 一欄內； ‧ 計入持作出售用途業務組合之資產及負債內之金融工具按相關工具之合約期限分類， 而非按出售交易分類；及 ‧ 保單未決賠款計入 「5年後到期」 一欄內。投資合約負債按照其合約期限分類。無定期投資 合約根據投資者有權給予之合約通知期分類。如無合約通知期，則無定期合約計入 「5年 後到期」 一欄內。 貸款及其他信貸相關承諾按可取用之最早日期分類。 30－後償負債╱31－期限分析 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 427 滙豐 資產及負債之期限分析 於2014年12月31日 1個月內到期 1個月後但 3個月內到期 3個月後但 6個月內到期 6個月後但 9個月內到期 9個月後但 1年內到期 1年後但 2年內到期 2年後但 5年內到期 5年後到期 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 金融資產 現金及於中央銀行的結餘 129,957 － － － － － － － 129,957 向其他銀行託收中之項目 4,927 － － － － － － － 4,927 香港政府負債證明書 27,674 － － － － － － － 27,674 交易用途資產 303,463 － － － － － 730 － 304,193 －反向回購 567 － － － － － 730 － 1,297 －其他交易用途資產 302,896 － － － － － － － 302,896 指定以公允值列賬之金融資產 244 399 417 346 208 1,825 4,634 20,964 29,037 衍生工具 341,558 56 463 220 32 1,003 1,033 643 345,008 －交易用途 341,416 － － － － － － － 341,416 －非交易用途 142 56 463 220 32 1,003 1,033 643 3,592 同業貸款1 73,758 17,649 5,682 1,934 1,850 7,371 1,981 1,924 112,149 客戶貸款1 203,130 76,236 55,018 35,347 37,674 91,300 187,728 288,227 974,660 －個人貸款 42,170 9,673 8,911 7,486 8,672 27,305 54,439 230,298 388,954 －企業及商業貸款 146,250 61,809 41,924 23,720 23,697 56,398 124,796 56,590 535,184 －金融機構貸款 14,710 4,754 4,183 4,141 5,305 7,597 8,493 1,339 50,522 反向回購協議－非交易用途1 116,002 30,490 9,076 2,230 582 868 2,465 － 161,713 金融投資 28,237 50,445 41,503 14,577 17,011 48,392 96,891 118,411 415,467 應計收益及其他金融資產 17,870 7,572 2,415 605 327 748 1,282 9,803 40,622 於2014年12月31日之金融資產 1,246,820 182,847 114,574 55,259 57,684 151,507 296,744 439,972 2,545,407 非金融資產 － － － － － － － 88,732 88,732 於2014年12月31日之資產總值 1,246,820 182,847 114,574 55,259 57,684 151,507 296,744 528,704 2,634,139 財務報表附註 （續） 滙豐控股有限公司 428 31－期限分析 資產及負債之期限分析 （續） 於2014年12月31日 1個月內到期 1個月後但 3個月內到期 3個月後但 6個月內到期 6個月後但 9個月內到期 9個月後但 1年內到期 1年後但 2年內到期 2年後但 5年內到期 5年後到期 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 金融負債 香港紙幣流通額 27,674 － － － － － － － 27,674 同業存放1 66,829 2,890 2,539 511 810 621 2,963 263 77,426 客戶賬項1, 2 1,216,574 57,127 32,925 15,023 13,586 9,278 5,819 310 1,350,642 －個人貸款 572,459 28,580 16,728 10,609 9,625 7,220 3,967 125 649,313 －企業及商業貸款 465,990 21,841 10,688 3,716 2,894 1,615 1,316 150 508,210 －金融機構貸款 178,125 6,706 5,509 698 1,067 443 536 35 193,119 回購協議－非交易用途1 95,243 5,029 4,054 1,392 714 － － 1,000 107,432 向其他銀行傳送中之項目 5,990 － － － － － － － 5,990 交易用途負債 155,604 2,041 2,636 1,439 2,918 5,744 9,603 10,587 190,572 －回購 746 909 224 264 1,249 406 － － 3,798 －已發行債務證券 1,686 1,132 2,412 1,175 1,669 5,338 9,603 10,587 33,602 －其他交易用途負債 153,172 － － － － － － － 153,172 指定以公允值列賬之金融負債 981 912 4,264 972 1,557 8,500 15,037 43,930 76,153 －已發行債務證券：備兌債券 － － － 205 － － 2,705 2,942 5,852 －已發行債務證券：以其他方式抵押 － － － － － － － － － －已發行債務證券：無抵押 942 868 4,242 742 1,409 8,500 9,576 14,233 40,512 －後償負債及優先證券 － 36 － － 18 － 2,623 20,640 23,317 －其他 39 8 22 25 130 － 133 6,115 6,472 衍生工具 335,802 23 86 223 54 621 1,121 2,739 340,669 －交易用途 335,400 － － － － － － － 335,400 －非交易用途 402 23 86 223 54 621 1,121 2,739 5,269 已發行債務證券 14,741 15,424 13,027 7,854 6,050 14,209 19,481 5,161 95,947 －備兌債券 － － － － － － 81 － 81 －以其他方式抵押 8,807 1,063 60 283 272 912 1,562 1,008 13,967 －無抵押 5,934 14,361 12,967 7,571 5,778 13,297 17,838 4,153 81,899 應計項目及其他金融負債 21,084 9,198 3,069 1,221 1,820 1,568 2,225 3,655 43,840 後償負債 － 150 － 3 167 113 3,607 22,624 26,664 於2014年12月31日之金融負債總額 1,940,522 92,794 62,600 28,638 27,676 40,654 59,856 90,269 2,343,009 非金融負債 － － － － － － － 91,152 91,152 於2014年12月31日之負債總額 1,940,522 92,794 62,600 28,638 27,676 40,654 59,856 181,421 2,434,161 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 429 於2013年12月31日 1個月內到期 1個月後但 3個月內到期 3個月後但 6個月內到期 6個月後但 9個月內到期 9個月後但 1年內到期 1年後但 2年內到期 2年後但 5年內到期 5年後到期 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 金融資產 現金及於中央銀行的結餘 166,599 － － － － － － － 166,599 向其他銀行託收中之項目 6,021 － － － － － － － 6,021 香港政府負債證明書 25,220 － － － － － － － 25,220 交易用途資產 296,396 3,098 1,536 2,062 100 － － － 303,192 －反向回購 3,324 3,098 1,536 2,062 100 － － － 10,120 －其他交易用途資產 293,072 － － － － － － － 293,072 指定以公允值列賬之金融資產 1,929 254 494 426 328 2,145 2,819 30,035 38,430 衍生工具 277,747 48 88 389 552 716 1,486 1,239 282,265 －交易用途 277,709 － － － － － － － 277,709 －非交易用途 38 48 88 389 552 716 1,486 1,239 4,556 同業貸款1 76,551 22,107 5,397 1,429 1,290 6,129 2,779 4,364 120,046 客戶貸款1 230,736 73,463 56,053 29,273 32,194 87,942 182,525 299,903 992,089 －個人貸款 43,805 8,929 8,561 6,603 7,578 26,915 58,611 243,124 404,126 －企業及商業貸款 171,713 60,827 43,723 19,505 21,093 53,183 113,897 53,981 537,922 －金融機構貸款 15,218 3,707 3,769 3,165 3,523 7,844 10,017 2,798 50,041 反向回購協議－非交易用途 134,242 35,329 5,287 1,239 2,072 1,136 385 － 179,690 金融投資 34,331 48,053 35,877 22,353 18,816 50,711 105,340 110,444 425,925 應計收益及其他金融資產 18,719 7,684 2,467 1,369 700 1,133 1,027 4,536 37,635 於2013年12月31日之金融資產總值 1,268,491 190,036 107,199 58,540 56,052 149,912 296,361 450,521 2,577,112 非金融資產 － － － － － － － 94,206 94,206 於2013年12月31日之資產總值 1,268,491 190,036 107,199 58,540 56,052 149,912 296,361 544,727 2,671,318 財務報表附註 （續） 滙豐控股有限公司 430 31－期限分析 資產及負債之期限分析 （續） 於2013年12月31日 1個月內到期 1個月後但 3個月內到期 3個月後但 6個月內到期 6個月後但 9個月內到期 9個月後但 1年內到期 1年後但 2年內到期 2年後但 5年內到期 5年後到期 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 金融負債 香港紙幣流通額 25,220 － － － － － － － 25,220 同業存放1 76,298 3,931 1,796 858 318 737 1,922 647 86,507 客戶賬項1, 2 1,229,694 60,683 26,940 13,704 15,384 8,717 5,937 238 1,361,297 －個人貸款 572,514 33,956 16,953 10,544 10,520 6,093 5,138 59 655,777 －企業及商業貸款 494,612 18,084 3,414 1,717 1,786 1,643 674 140 522,070 －金融機構貸款 162,568 8,643 6,573 1,443 3,078 981 125 39 183,450 回購協議－非交易用途 136,137 13,058 6,583 3,711 4,231 － － 500 164,220 向其他銀行傳送中之項目 6,910 － － － － － － － 6,910 交易用途負債 161,231 11,405 4,886 2,844 3,653 6,323 7,979 8,704 207,025 －回購 2,565 9,763 2,715 1,012 1,279 87 － － 17,421 －已發行債務證券 1,217 1,642 2,171 1,832 2,374 6,236 7,979 8,704 32,155 －其他交易用途負債 157,449 － － － － － － － 157,449 指定以公允值列賬之金融負債 4,907 157 92 2,266 68 9,348 21,544 50,702 89,084 －已發行債務證券：備兌債券 － － － 1,268 － 230 2,841 3,257 7,596 －已發行債務證券：以其他方式抵押 － － － － － － － － － －已發行債務證券：無抵押 3,511 5 45 945 11 8,876 18,117 14,256 45,766 －後償負債及優先證券 － 121 － － － 21 － 21,773 21,915 －其他 1,396 31 47 53 57 221 586 11,416 13,807 衍生工具 269,816 33 95 84 61 563 1,978 1,654 274,284 －交易用途 269,739 － － － － － － － 269,739 －非交易用途 77 33 95 84 61 563 1,978 1,654 4,545 已發行債務證券 20,739 8,280 15,734 7,442 8,106 18,552 19,850 5,377 104,080 －備兌債券 － － － － － 6 92 － 98 －以其他方式抵押 10,450 1,051 675 1,260 764 1,857 2,313 1,013 19,383 －無抵押 10,289 7,229 15,059 6,182 7,342 16,689 17,445 4,364 84,599 應計項目及其他金融負債 25,267 10,475 3,278 1,280 1,599 1,831 1,592 936 46,258 後償負債 21 28 1,171 144 6 1,435 3,406 22,765 28,976 於2013年12月31日之金融負債總額 1,956,240 108,050 60,575 32,333 33,426 47,506 64,208 91,523 2,393,861 非金融負債 － － － － － － － 86,998 86,998 於2013年12月31日之負債總額 1,956,240 108,050 60,575 32,333 33,426 47,506 64,208 178,521 2,480,859 1 請參閱第344頁註釋3。 2 包括由保證計劃擔保之3,429.27億美元 （2013年：3,559.80億美元） 。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 431 已取得資產負債表外承諾之期限分析 1個月內到期 1個月後但 3個月內到期 3個月後但 6個月內到期 6個月後但 9個月內到期 9個月後但 1年內到期 1年後但 2年內到期 2年後但 5年內到期 5年後到期 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年12月31日 貸款及其他信貸相關承諾 3,313 － 4,312 607 － － － － 8,232 於2013年12月31日 貸款及其他信貸相關承諾 953 － － － － － － － 953 已作出資產負債表外承諾之期限分析 1個月內到期 1個月後但 3個月內到期 3個月後但 6個月內到期 6個月後但 9個月內到期 9個月後但 1年內到期 1年後但 2年內到期 2年後但 5年內到期 5年後到期 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 於2014年12月31日 貸款及其他信貸相關承諾 455,319 52,398 8,919 14,163 41,500 13,979 48,333 16,769 651,380 其中： －個人貸款 179,088 15,784 452 305 14,036 1,432 1,003 955 213,055 －企業及商業貸款 239,646 34,657 7,595 12,556 23,519 9,926 36,918 12,185 377,002 －金融機構貸款 36,585 1,957 872 1,302 3,945 2,621 10,412 3,629 61,323 於2013年12月31日 貸款及其他信貸相關承諾 404,598 45,255 18,770 16,927 20,242 13,320 46,652 21,839 587,603 其中： －個人貸款 148,541 14,700 454 10,683 12,131 1,273 704 6,469 194,955 －企業及商業貸款 225,333 29,191 17,794 5,662 4,879 9,009 41,851 12,096 345,815 －金融機構貸款 30,724 1,364 522 582 3,232 3,038 4,097 3,274 46,833 財務報表附註 （續） 滙豐控股有限公司 432 31－期限分析 滙豐控股 資產、負債及資產負債表外承諾之期限分析 1個月內到期 1個月後但 3個月內到期 3個月後但 6個月內到期 6個月後但 9個月內到期 9個月後但 1年內到期 1年後但 2年內到期 2年後但 5年內到期 5年後到期 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 金融資產 銀行及庫存現金： －在滙豐旗下業務之結餘 249 － － － － － － － 249 衍生工具 2,287 － － － － － 127 357 2,771 －交易用途 2,287 － － － － － － － 2,287 －非交易用途 － － － － － － 127 357 484 滙豐旗下業務貸款 7,007 858 7,676 － 14 － － 28,355 43,910 滙豐旗下金融投資 26 6 － － － － － 4,041 4,073 應計收益及其他金融資產 8 － － － － － － － 8 於2014年12月31日之金融資產總值 9,577 864 7,676 － 14 － 127 32,753 51,011 非金融資產 － － － － － － － 96,853 96,853 於2014年12月31日之資產總值 9,577 864 7,676 － 14 － 127 129,606 147,864 金融負債 應付滙豐旗下業務款項 2,423 － 32 － 1 436 － － 2,892 指定以公允值列賬之金融負債 － － － － － 1,110 2,623 14,946 18,679 －已發行債務證券 － － － － － 1,110 － 7,075 8,185 －後償負債及優先證券 － － － － － － 2,623 7,871 10,494 衍生工具 1,066 － － － － － 103 － 1,169 －交易用途 1,066 － － － － － － － 1,066 －非交易用途 － － － － － － 103 － 103 已發行債務證券 － － － － － － － 1,009 1,009 應計項目及其他金融負債 924 208 137 21 － － － － 1,290 後償負債 － － － － － － 1,951 15,304 17,255 於2014年12月31日之金融負債總額 4,413 208 169 21 1 1,546 4,677 31,259 42,294 非金融負債 － － － － － － － 125 125 於2014年12月31日之負債總額 4,413 208 169 21 1 1,546 4,677 31,384 42,419 已作出之資產負債表外承諾 未取用之正式備用信貸、 信貸額及其他貸款承諾 16 － － － － － － － 16 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 433 1個月內到期 1個月後但 3個月內到期 3個月後但 6個月內到期 6個月後但 9個月內到期 9個月後但 1年內到期 1年後但 2年內到期 2年後但 5年內到期 5年後到期 總計 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 金融資產 銀行及庫存現金： －在滙豐旗下業務之結餘 407 － － － － － － － 407 衍生工具 2,729 － － － － － － 60 2,789 －交易用途 2,729 － － － － － － － 2,729 －非交易用途 － － － － － － － 60 60 滙豐旗下業務貸款 8,394 4,374 7,595 14 16 － － 32,951 53,344 滙豐旗下金融投資 23 － 8 － － － 330 849 1,210 應計收益及其他金融資產 10 － － － － － － － 10 於2013年12月31日之金融資產總值 11,563 4,374 7,603 14 16 － 330 33,860 57,760 非金融資產 － － － － － － － 93,076 93,076 於2013年12月31日之資產總值 11,563 4,374 7,603 14 16 － 330 126,936 150,836 金融負債 應付滙豐旗下業務款項 3,770 12 2,234 － 1 － 463 5,205 11,685 指定以公允值列賬之金融負債 － － － － － － 1,283 19,744 21,027 －已發行債務證券 － － － － － － 1,283 6,823 8,106 －後償負債及優先證券 － － － － － － － 12,921 12,921 衍生工具 704 － － － － － － － 704 －交易用途 704 － － － － － － － 704 －非交易用途 － － － － － － － － － 已發行債務證券 － － 1,721 － － － － 1,070 2,791 應計項目及其他金融負債 777 109 261 22 － － － － 1,169 後償負債 － － － － － － 2,210 11,957 14,167 於2013年12月31日之金融負債總額 5,251 121 4,216 22 1 － 3,956 37,976 51,543 非金融負債 － － － － － － － 206 206 於2013年12月31日之負債總額 5,251 121 4,216 22 1 － 3,956 38,182 51,749 已作出之資產負債表外承諾 未取用之正式備用信貸、 信貸額及其他貸款承諾 1,245 － － － － － － － 1,245 財務報表附註 （續） 滙豐控股有限公司 434 32 對銷金融資產及金融負債 會計政策 如存在可依法強制執行之權利以對銷已確認金額，並有意按淨額結算或同時變現資產及結算負債 （ 「對銷準則」 ） ，則 可在資產負債表內對銷金融資產及金融負債並呈報淨額。 涉及對銷、可依法強制執行之淨額計算總協議及近似協議之金融資產 已確認 金融資產 之總額 在資產 負債表 內對銷 之總額 在資產 負債表 呈列 之金額 並無在資產負債表內 對銷之金額 淨金額 金融工具 1 已收取現金 抵押品 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 衍生工具2 （附註16） 595,473 (250,465) 345,008 (271,040) (42,260) 31,708 反向回購、借入股票及近似協議3 259,655 (88,676) 170,979 (166,958) (249) 3,772 分類為： －交易用途資產 9,656 (390) 9,266 (9,256) － 10 －非交易用途資產 249,999 (88,286) 161,713 (157,702) (249) 3,762 按已攤銷成本計算之客戶貸款4 101,220 (37,527) 63,693 (55,989) (310) 7,394 於2014年12月31日 956,348 (376,668) 579,680 (493,987) (42,819) 42,874 衍生工具2 （附註16） 569,595 (287,330) 282,265 (215,957) (36,387) 29,921 反向回購、借入股票及近似協議3 288,903 (88,775) 200,128 (197,287) (57) 2,784 分類為： －交易用途資產 39,008 (18,570) 20,438 (20,438) － － －非交易用途資產 249,895 (70,205) 179,690 (176,849) (57) 2,784 按已攤銷成本計算之客戶貸款4 192,437 (92,654) 99,783 (89,419) － 10,364 於2013年12月31日 1,050,935 (468,759) 582,176 (502,663) (36,444) 43,069 1 包括非現金抵押品。 2 包括涉及與不涉及可依法強制執行之淨額計算總協議及近似協議之金額。 3 有關在資產負債表內確認之反向回購、借入股票及近似協議之金額，請參閱第168頁 「資金來源及用途」 之 列表。於以上分析中，在資產負債表呈列之交易用途資產92.66億美元 （2013年12月31日：204.38億美元） 包括 反向回購12.97億美元 （2013年12月31日：101.20億美元） 及借入股票79.69億美元 （2013年12月31日：103.18億 美元） 。 4 於2014年12月31日，按已攤銷成本計算之客戶貸款總額為9,746.60億美元 （2013年12月31日：9,920.89億美 元） ，其中636.93億美元 （2013年12月31日：997.83億美元） 涉及對銷。有關在資產負債表內確認按已攤銷成 本計算之客戶貸款金額，請參閱第168頁 「資金來源及用途」 之列表。 32－對銷金融資產及金融負債╱33－匯兌風險 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 435 涉及對銷、可依法強制執行之淨額計算總協議及近似協議之金融負債 已確認 金融負債 之總額 在資產 負債表內 對銷 之總額 在資產 負債表 呈列 之金額 並無在資產負債表內 對銷之金額 淨金額 金融工具 1 已質押現金 抵押品 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 衍生工具2 （附註16） 591,134 (250,465) 340,669 (272,815) (40,291) 27,563 回購、借出股票及近似協議3 211,938 (88,676) 123,262 (121,722) (105) 1,435 分類為： －交易用途負債 16,220 (390) 15,830 (15,828) － 2 －非交易用途負債 195,718 (88,286) 107,432 (105,894) (105) 1,433 按已攤銷成本計算之客戶賬項4 107,088 (37,527) 69,561 (55,989) － 13,572 於2014年12月31日 910,160 (376,668) 533,492 (450,526) (40,396) 42,570 衍生工具2 （附註16） 561,614 (287,330) 274,284 (216,596) (29,093) 28,595 回購、借出股票及近似協議3 282,634 (88,775) 193,859 (193,354) (81) 424 分類為： －交易用途負債 48,209 (18,570) 29,639 (29,625) － 14 －非交易用途負債 234,425 (70,205) 164,220 (163,729) (81) 410 按已攤銷成本計算之客戶賬項4 195,153 (92,654) 102,499 (89,394) － 13,105 於2013年12月31日 1,039,401 (468,759) 570,642 (499,344) (29,174) 42,124 1 包括非現金抵押品。 2 包括涉及與不涉及可依法強制執行之淨額計算總協議及近似協議之金額。 3 有關在資產負債表內確認之回購、借出股票及近似協議之金額，請參閱第168頁 「資金來源及用途」 之列 表。於以上分析中，在資產負債表呈列之交易用途負債158.3億美元 （2013年12月31日：296.39億美元） 包括回 購37.98億美元 （2013年12月31日：174.21億美元） 及借出股票120.32億美元 （2013年12月31日：122.18億美元） 。 4 於2014年12月31日，按已攤銷成本計算之客戶賬項總額為13,506.42億美元 （2013年12月31日：13,612.97億美 元） ，其中695.61億美元 （2013年12月31日：1,024.99億美元） 涉及對銷。有關在資產負債表內確認按已攤銷 成本計算之客戶賬項金額，請參閱第168頁 「資金來源及用途」 之列表。 計入 「並無在資產負債表內對銷之金額」 一欄之衍生工具及反向回購╱回購、借入╱借出股 票及近似協議包括符合以下條件的交易： ‧ 交易對手與滙豐之間涉及可予對銷的風險，以及存在的淨額計算總協議或近似安排僅有 權在違約、無力償債或破產時對銷，或在其他方面未能符合對銷準則；及 ‧ 已就上述交易收取╱質押現金及非現金抵押品。 至於按已攤銷成本計算之客戶貸款及客戶賬項，上表所載金額通常涉及與企業及商業客戶 就營運資金管理而訂立的交易。於 「並無在資產負債表內對銷之金額」 涉及的交易中，客戶 與滙豐之間涉及可予對銷的風險並已訂立有權對銷的協議，但此等交易在其他方面未能符 合對銷準則。就風險管理而言，該等風險的淨額受制於受若干監控的限額，而相關客戶協 議須在必要時予以檢討及更新，以確保對銷之合法權利仍屬合適。 33 匯兌風險 結構匯兌風險 滙豐之結構匯兌風險乃指集團於採用非美元功能貨幣的附屬公司、分行、合資公司及聯營 公司的外幣股權及後償債務投資的資產淨值。結構匯兌風險的損益均於其他全面收益項內 確認。滙豐之結構匯兌風險管理於第181頁內討論。 財務報表附註 （續） 滙豐控股有限公司 436 34－非控股股東權益╱35－股本及其他股權工具 結構匯兌風險淨額 2014年 2013年 百萬美元 百萬美元 涉及結構風險之貨幣 英鎊1 30,071 28,403 中國人民幣 24,578 20,932 港元 24,028 18,974 歐元 20,378 22,014 墨西哥披索 5,249 5,932 巴西雷亞爾 4,910 5,581 加元 4,187 4,372 印度盧比 3,466 3,222 沙地阿拉伯里亞爾 2,910 2,531 馬元 2,219 2,194 阿聯酋第納爾 2,199 3,069 瑞士法郎 1,864 1,940 台幣 1,721 1,527 澳元 1,516 1,515 土耳其里拉 1,366 1,533 韓圜 1,360 1,373 印尼盾 1,352 1,244 新加坡元 1,185 849 阿根廷披索 1,059 1,067 埃及鎊 868 739 其他，各少於7億美元 5,918 6,157 於12月31日 142,404 135,168 1 於2014年，我們訂立遠期外匯合約16億美元，以管理英鎊結購匯兌風險。 倘歐元及英鎊兌美元匯率下跌5%，股東權益即減少25.22億美元 （2013年：25.21億美元） 。 34 非控股股東權益 2014年 2013年 百萬美元 百萬美元 附屬公司普通股持有人應佔非控股股東權益 7,104 5,900 附屬公司發行之優先證券 2,427 2,688 於12月31日 9,531 8,588 附屬公司發行之優先證券 優先證券為並無責任派付股息及 （如不派付股息） 股息未必可累積之證券。該等證券一般不 附帶投票權，但在支付股息及清盤方面之地位較普通股高。該等證券並無列明到期日，但 可由發行人提早贖回，惟須事先通知審慎監管局及 （在適用情況下） 取得當地銀行監管機構 之同意。浮息優先證券之股息一般與銀行同業拆息有關。 因應用豁免條文，滙豐之資本基礎已計入按照資本指引4規則分類為額外一級資本之非累積 優先證券及分類為二級資本之累積優先證券。 滙豐之附屬公司發行之優先證券 2014年 2013年 首次提早贖回日期 百萬美元 百萬美元 美國滙豐有限公司 1.5億美元 預託股份，每股代表一股可調整股息率之 累積優先股 （D系列） 的25%權益 1999年7月 150 150 1.5億美元 累積優先股 2007年10月 150 150 5.18億美元 浮息非累積優先股 （F系列） 2010年4月 518 518 3.74億美元 浮息非累積優先股 （G系列） 2011年1月 374 374 3.74億美元 6.50厘非累積優先股 （H系列） 2011年7月 374 374 美國滙豐融資有限公司 5.75億美元 6.36厘非累積優先股 （B系列） 2010年6月 559 559 加拿大滙豐銀行 1.75億加元 非累積可贖回類別1優先股 （C系列） 2010年6月 151 164 1.75億加元 非累積類別1優先股 （D系列） 2010年12月 151 164 2.5億加元1 非累積5年重定利率類別1優先股 （E系列） 2014年6月 － 235 於12月31日 2,427 2,688 1 於2014年6月，滙豐以2.34億美元贖回其非累積5年重定利率類別1優先股 （E系列） 。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 437 35 已催繳股本及其他股權工具 會計政策 若無合約責任需要轉撥現金、其他金融資產或發行不定數目之本身股權工具，則已發行金融工具會分類為股東權 益。與發行股權工具直接相關之遞增成本，均於股東權益項內列為所得款項的扣減項目 （除稅淨額） 。 滙豐持有之滙豐控股股份，均於股東權益項內確認為保留盈利的扣減項目，直至股份被註銷為止。若該等股份於其 後出售、重新發行或以其他方式處置，則所收取之任何代價經扣除任何直接相關交易的遞增支出及有關所得稅之影 響後，均會列入股東權益項內。 已發行及繳足股款 滙豐控股每股面值0.50美元之普通股 2014年 2013年 百萬美元 百萬美元 於12月31日之滙豐控股普通股1 9,609 9,415 數目 百萬美元 於2014年1月1日 18,830,007,039 9,415 根據滙豐僱員股份計劃發行之股份 119,391,238 60 發行代息股份 268,475,983 134 於2014年12月31日 19,217,874,260 9,609 於2013年1月1日 18,476,008,664 9,238 根據滙豐僱員股份計劃發行之股份 120,033,493 60 發行代息股份 233,964,882 117 於2013年12月31日 18,830,007,039 9,415 滙豐控股每股面值0.01美元之非累積優先股 數目 百萬美元 於2014年1月1日及2014年12月31日2 1,450,000 － 於2013年1月1日及2013年12月31日 1,450,000 － 1 所有已發行滙豐控股普通股在資本、股息及投票等方面均具有同等權利。 2 因應用豁免條文，已按照資本指引4規則作為額外一級資本計入滙豐之資本基礎內。 滙豐控股之已發行非累積美元優先股 （ 「美元優先股」 ） 之股息由董事會全權酌情決定於每季 派付。如派付股息將導致滙豐控股未能遵守審慎監管局之適用資本充足比率規定，或滙豐 控股可供分派作股息的利潤，不足以使滙豐控股同時就美元優先股及訂於同日派發股息並 享有同等股息權利之任何其他股份悉數支付所有股息，則董事會不會就美元優先股宣派股 息。滙豐控股不可就獲派股息權利遜於美元優先股之任何類別股份宣派或派付股息，亦不 可以任何方式贖回或購回權利相等於或遜於美元優先股之任何其他股份，惟於當時之派息 期已全數支付或已預留一筆款項以供全數支付美元優先股的股息則除外。美元優先股無權 轉換為滙豐控股普通股。如連續四個派息日期並無就美元優先股獲支付全數應付股息，則 美元優先股持有人方有權出席滙豐控股之股東大會，並於會上投票。在該情況下，美元優 先股持有人將有權就股東大會上提呈之所有事項投票，直至滙豐控股全數支付美元優先股 股息為止。滙豐控股可於2010年12月16日或之後任何時間全數贖回美元優先股，但必須事先 通知審慎監管局。 滙豐控股每股面值0.01英鎊之非累積優先股 一股面值0.01英鎊之已發行非累積英鎊優先股 （ 「英鎊優先股」 ） 已自2010年12月29日起發行， 並由滙豐控股之附屬公司持有。英鎊優先股之股息由董事會全權酌情決定於每季派付。英 鎊優先股無權轉換為滙豐控股普通股，亦無出席滙豐控股股東大會並於會上投票之權利。 滙豐控股可選擇於任何時間全數贖回已發行英鎊優先股。 其他股權工具 滙豐之資本基礎已計入的其他股權工具包括額外一級資本證券及屬於額外一級資本的或有 可轉換證券。 財務報表附註 （續） 滙豐控股有限公司 438 35－已催繳股本╱36－現金流量表說明 額外一級資本證券 額外一級資本證券屬永久後償證券，其票息付款可由滙豐控股酌情遞延。倘仍有任何未付 或遞延票息付款，滙豐控股將不會就任何較低或同等級別之證券宣派、支付股息或作出分 派或同類定期付款，或回購、贖回或以其他方式收購任何該等較低或同等級別證券。此類 資本證券一般並無附有投票權，但在支付票息及清盤時的地位高於普通股。因應用豁免條 文，該等證券已根據指本指引4規則計入滙豐之資本基礎內。 滙豐控股可在達成若干條件後酌情決定於任何票息付款日，把資本證券交換為將由滙豐控 股發行之非累積優先股，並在各方面與已發行美元及英鎊優先股享有同等地位。此等優先 股將按面值每股0.01美元及溢價每股24.99美元發行，兩筆款額均獲認購及繳足。滙豐可提早 贖回此等證券，惟須事先通知審慎監管局。 滙豐計入股東權益項內之已發行額外一級資本證券 首次提早 贖回日期 2014年 2013年 百萬美元 百萬美元 22億美元 8.125厘永久後償資本證券 2013年4月 2,133 2,133 38億美元 8.00厘永久後償資本證券 （系列2） 2015年12月 3,718 3,718 於12月31日 5,851 5,851 額外一級資本－或有可轉換證券 於2014年，滙豐發行新的或有可轉換證券，並已按終點基準以此作為完全符合資本指引4的 額外一級資本證券而計入滙豐的資本基礎內。發行所得款項淨額將撥作一般企業用途，以 及根據資本指引4的規定用於進一步增強資本基礎。該等證券按固定利率計息至其首次提早 贖回日期。於首次提早贖回日期後，倘未被贖回，則該等證券的五年期固定息率將按當前 市場利率預先定期釐定。或有可轉換證券的利息僅由滙豐全權酌情決定到期支付，而滙豐 於任何時間均可全權酌情決定以任何理由撤銷支付原應在任何付息日支付的全部或任何部 分利息。如支付分派受到英國銀行規例或其他規定所禁制，或滙豐控股沒有足夠的可供分 派儲備，或滙豐未能滿足證券條款界定的償付能力條件，支付分派將受到限制。 或有可轉換證券屬無定期，且於首次提早贖回日期或首次提早贖回日期後任何五周年之日， 可按滙豐的選擇予以全部償還。此外，該等證券可由滙豐就若干監管或稅務原因而選擇全 部償還。任何償還須獲得審慎監管局事先同意。該等證券與滙豐的美元及英鎊優先股享有 同等地位，並因此優先於普通股。倘滙豐的綜合終點基準普通股權一級比率低於7.0%，或有 可轉換證券將按預定價格轉換為滙豐的普通股。因此，按照證券條款，於自救安排下該等 證券將按轉換價2.70英鎊轉換為普通股，並會作若干反攤薄及匯兌調整，而且將與已發行繳 足股款普通股享有同等地位。 滙豐計入股東權益項內之已發行額外一級資本－或有可轉換證券 首次提早 2014年 2013年 贖回日期 百萬美元 百萬美元 22.5億美元 6.375厘永久後償或有可轉換證券 2024年9月 2,244 － 15億美元 5.625厘永久後償或有可轉換證券 2020年1月 1,494 － 15億歐元 5.25厘永久後償或有可轉換證券 2022年9月 1,943 － 於12月31日 5,681 － 涉及認股權之股份 有關在滙豐控股集團優先認股計劃、滙豐股份計劃及滙豐控股儲蓄優先認股計劃下尚未行 使可用以認購滙豐控股普通股之認股權詳情，請參閱附註6。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 439 在該等計劃下尚未行使之認股權總數 滙豐控股普通股數目 行使期 行使價 2014年12月31日 63,918,042 2014至2020年 3.3116至7.9911英鎊 6,468,782 2014至2018年 37.8797至63.9864港元 571,502 2014至2018年 3.6361至6.0657歐元 1,867,328 2014至2018年 4.8876至8.2094美元 2013年12月31日 119,085,250 2013至2019年 3.3116至7.9911英鎊 24,215,341 2013至2018年 37.8797至92.5881港元 1,574,652 2013至2018年 3.6361至7.5571歐元 3,997,069 2013至2018年 4.8876至11.8824美元 2012年12月31日 159,703,771 2013至2018年 3.3116至7.9911英鎊 31,637,840 2013至2018年 37.8797至94.5057港元 2,180,263 2013至2018年 3.6361至9.5912歐元 6,488,894 2013至2018年 4.8876至12.0958美元 提供滙豐控股普通股的最高責任額 於2014年12月31日，根據上述所有認股權安排，連同根據滙豐股份計劃及╱或2011年滙豐股份 計劃授出之集團業績表現股份計劃獎勵及有限制股份獎勵，須提供滙豐控股普通股的最高 責任額為193,154,512股 （2013年：265,534,885股） 。於2014年12月31日，由職員福利信託基金持有 可供履行此等提供滙豐控股普通股責任的股份，總數為7,943,191股 （2013年：12,068,136股） 。 36 現金流量表說明 計入除稅前利潤之其他非現金項目 滙豐 滙豐控股 2014年 2013年 2012年 2014年 2013年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 折舊、攤銷及減值 2,251 2,330 2,531 39 35 攤薄聯營公司權益所得 （增益） ╱虧損 32 (1,051) － － － 重估投資物業 (120) (113) (72) － － 以股份為基礎的支出 732 630 988 74 49 未減收回額及其他信貸風險準備之 貸款減值虧損 5,125 7,356 9,358 － － 準備 3,074 2,578 5,732 － － 金融投資減值╱ （減值撥回） 54 (36) 519 － － 界定福利計劃開支╱ （撥賬） 535 121 476 － － 折讓增值及溢價攤銷 (421) 180 246 (62) (10) 截至12月31日止年度 11,262 11,995 19,778 51 74 營業資產之變動 滙豐 滙豐控股 2014年 2013年 2012年 2014年 2013年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 滙豐旗下業務貸款之變動 － － － 1,364 (11,669) 交易用途證券及衍生工具淨額之變動 (18,498) (24,870) (36,829) 483 923 同業貸款之變動 5,147 (4,739) 1,174 － － 客戶貸款之變動 12,666 (46,551) (79,388) － － 反向回購協議 （非交易用途） 之變動 18,900 (70,403) 6,678 － － 指定以公允值列賬之金融資產之變動 3,269 (4,922) (2,698) － － 其他資產之變動 4,393 2,586 (5,458) 7 (49) 截至12月31日止年度 25,877 (148,899) (116,521) 1,854 (10,795) 營業負債之變動 滙豐 滙豐控股 2014年 2013年 2012年 2014年 2013年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 同業存放之變動 (9,081) (7,781) 274 － － 客戶賬項之變動 (8,362) 57,365 92,238 － － 回購協議 （非交易用途） 之變動 (56,788) 123,653 (7,834) － － 已發行債務證券之變動 (8,133) (15,381) (11,552) (149) 98 指定以公允值列賬之金融負債之變動 (10,734) 994 2,549 (694) (550) 其他負債之變動 (716) 5,907 13,395 (9,071) (609) 截至12月31日止年度 (93,814) 164,757 89,070 (9,914) (1,061) 財務報表附註 （續） 滙豐控股有限公司 440 現金及等同現金項目 會計政策 現金及等同現金項目包括高度流通的投資，此等投資可隨時轉換為已知數額之現金，且其價值變動之風險極低。此 等投資的到期日一般為收購日期起計三個月內。 現金及等同現金項目 滙豐 滙豐控股 2014年 2013年 2012年 2014年 2013年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 滙豐旗下業務之銀行現金 － － － 249 407 現金及於中央銀行的結餘 129,957 166,599 141,532 － － 向其他銀行託收中之項目 4,927 6,021 7,303 － － 1個月或以下之同業貸款 89,285 96,584 119,400 － － 1個月或以下之同業反向回購協議 68,930 68,007 28,832 － － 3個月以下之國庫票據、 其他票據及存款證 14,192 15,980 25,379 － － 減：向其他銀行傳送中之項目 (5,990) (6,910) (7,138) － － 於12月31日 301,301 346,281 315,308 249 407 1 自2013年起按已攤銷成本計算。 利息及股息 滙豐 滙豐控股 2014年 2013年 2012年 2014年 2013年 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 已付利息 (15,633) (17,262) (18,412) (2,463) (2,705) 已收利息 51,522 50,823 61,112 1,945 1,986 已收股息 1,199 1,133 766 9,077 20,925 於2014年12月31日，滙豐不可動用之現金及等同現金項目金額為437.38億美元 （2013年：380.19億 美元） ，其中298.83億美元 （2013年：218.11億美元） 與中央銀行強制性存款有關。 出售附屬公司及業務 於2014年，我們完成出售中東滙豐有限公司在約旦的銀行業務及在巴基斯坦的業務，產生現金 及等同現金項目流出淨額3.03億美元，該款額已計入第338頁綜合現金流量表的 「投資活動產生 之現金流」 內。 於2013年10月，我們完成出售巴拿馬滙豐銀行，取得現金代價總額22.1億美元，該款額已計入第 338頁綜合現金流量表的 「投資活動產生之現金流」 內。 下表列示於2012年出售附屬公司及業務之影響。 2012年 美國卡業務 美國分行網絡 其他出售項目 總計 百萬美元 百萬美元 百萬美元 百萬美元 資產總值 （不包括現金及等同現金項目） 28,007 2,166 7,302 37,475 負債總額 161 13,206 8,463 21,830 於出售日期之資產淨值總計 （不包括現金及 等同現金項目） 27,846 (11,040) (1,161) 15,645 已出售非控股股東權益 － － (81) (81) 出售利潤 （包括出售成本） 3,148 864 355 4,367 加回：出售成本 72 15 56 143 售價 31,066 (10,161) (831) 20,074 以下列方式支付： 已收╱ （已付） 作為代價之現金及等同現金項目 31,066 (10,091) (542) 20,433 已售現金及等同現金項目 － (70) (321) (391) 直至2012年12月31日已收╱ （已付） 之 現金代價 31,066 (10,161) (863) 20,042 於2012年12月31日尚待收取之現金 － － 32 32 現金代價總額 31,066 (10,161) (831) 20,074 36－現金流量表說明╱37－或有負債、合約承諾及擔保 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 441 37 或有負債、合約承諾及擔保 會計政策 或有負債 或有負債 （包括若干擔保及以附帶擔保形式質押之信用證以及與法律訴訟及監管事宜有關之或有負債 （見附註40） ） 乃因過往事件而可能引致的責任，但是否確實需要承擔這些責任須視乎未來會否發生一宗或多宗無法確定的事件， 而且該等事件乃滙豐無法完全控制；或因過往事件引致之現有責任，但由於不可能為履行該責任而需流出經濟利 益故未予確認，或由於無法可靠計量責任所涉金額故未予確認。或有負債不會在財務報表內確認，但會作出有關披 露，除非需要履行責任的可能性十分低。 金融擔保合約 並未分類為保單之金融擔保合約下的負債 （一般指已收費用或應收費用之現值） ，首次列賬會按公允值計量。於其 後的業績報告期內，金融擔保負債則會按首次列賬公允值減累計攤銷後的數值，或對履行責任所需支出之最佳估計 （以兩者之較高者為準） 予以計量。 滙豐控股有向集團屬下其他公司發出金融擔保及同類合約。滙豐會選擇將若干擔保在滙豐控股的財務報表內入賬列 為保單，在此情況下，該等擔保作為保險未決賠款予以計量及確認。滙豐會按個別合約作出選擇，但不得撤回所作 選擇。 或有負債、合約承諾及擔保 滙豐 滙豐控股 2014年 2013年 2014年 2013年 百萬美元 百萬美元 百萬美元 百萬美元 擔保及或有負債 擔保 86,385 84,554 52,023 52,836 其他或有負債 346 182 － － 於12月31日 86,731 84,736 52,023 52,836 承諾 押匯信用證及短期貿易交易 12,082 12,154 － － 遠期資產購置及遠期有期存款 823 1,005 － － 未取用之正式備用信貸、信貸額及 其他貸款承諾 638,475 574,444 16 1,245 於12月31日 651,380 587,603 16 1,245 上表列示承諾、擔保及其他或有負債之名義本金額。因集團旗下公司成為法律訴訟、監管 及其他事宜的被告而產生的或有負債於附註29及40披露。名義本金額乃指約定金額如被悉 數取用而客戶又拖欠還款時涉及之風險額。由於預期大部分擔保及承諾所涉金額直至期滿 時均不會被取用，故此名義本金總額並不是日後之流動資金需求之參考。 巴西社會保障稅項申索 於2008年4月，最終司法判決宣布滙豐在巴西的保險及租賃公司勝訴，清楚說明屬於社會保 障稅項的利潤參與供款( 「PIS」 )及社會保障資助供款( 「COFINS」 )應僅對銷售貨品及服務所得收 入徵收，而不對保費及金融收入所得收益徵收。根據此判決，滙豐當時相應降低了稅基並 對銷了稅項減免，但後來受到巴西稅務當局的質疑，指稱該等稅基應包括納稅人企業活動 所得的所有收入。隨著新頒布法律自2015年1月1日起生效，PIS及COFINS的稅基已擴大至包括 企業活動所得的所有收入，包括保險及融資收益，因此，滙豐可能承擔的任何支付額外稅 項的責任僅涉及截至2014年底止的稅務年度。上述稅項評估處於不同行政程序階段。根據 目前的已知事實，要求滙豐預測此等事宜的解決時間並不切實可行。 財務報表附註 （續） 滙豐控股有限公司 442 擔保 2014年 2013年 代第三方 提供之擔保 滙豐控股 代滙豐集團 旗下其他公司 提供之擔保 代第三方 提供之擔保 滙豐控股 代滙豐集團 旗下其他公司 提供之擔保 百萬美元 百萬美元 百萬美元 百萬美元 擔保類別 金融擔保1 30,406 36,800 31,224 36,800 信貸相關擔保2 16,672 15,223 15,076 16,036 其他擔保 39,307 － 38,254 － 於12月31日 86,385 52,023 84,554 52,836 1 金融擔保為一種合約，其條款規定滙豐須在指定債務人未能於到期時還款而令持有人蒙受損失時，向持 有人支付指定款項以作補償。 2 信貸相關擔保為一種合約，其特徵與金融擔保合約類似，惟不符合IAS 39對金融擔保合約之定義。 上表披露之金額為名義本金額，反映滙豐根據多項個別擔保承諾所承擔的最大風險。有關 擔保的風險乃根據滙豐的整體信貸風險管理政策及程序予以掌握及管理。上述擔保中約半 數的合約期不足一年。合約期超過一年的擔保須由滙豐每年進行信貸審核。 金融服務賠償計劃 於多家接受存款機構倒閉後，金融服務賠償計劃已向消費者作出賠償。向消費者支付的賠 償目前透過英倫銀行及英國財政部借出的貸款撥付，於2014年12月31日，英倫銀行及英國財 政部借出的貸款約為160億英鎊 （249億美元） 。 為償還預期未能收回的貸款本金額，金融服務賠償計劃向參與計劃之金融機構徵收費用。 於2015年1月，金融服務賠償計劃宣布，2015╱2016計劃年度參與計劃之金融機構之預期徵 費將為3.47億英鎊 （5.41億美元） （2014╱2015計劃年度：3.99億英鎊 （6.6億美元） ） 。 金融服務賠償計劃因金融機構倒閉而最終向業界徵收的徵費，目前無法作出準確估計，因 為徵費視乎多項不確定因素而定，包括金融服務賠償計劃可能收回資產和受保障存款水平 及金融服務賠償計劃當時成員數目的變化。 資本承諾 除第441頁所披露的承諾外，於2014年12月31日，滙豐已訂約但未撥備之資本承諾為6.56億美 元 （2013年：4.01億美元） ，而已核准但未訂約之資本承諾為1.01億美元 （2013年：1.12億美元） 。 聯營公司 於2014年12月31日，滙豐應佔聯營公司之或有負債為475.93億美元 （2013年：465.74億美元） 。 年內並無滙豐須承擔個別責任的事項。 38 租賃承諾 會計政策 協議若轉移資產擁有權附帶的絕大部分風險與回報，則列為融資租賃。 如滙豐為融資租賃之出租人，則租賃下之應收租金於扣減未賺取之費用後，列入 「同業貸款」 或 「客戶貸款」 項內。 如滙豐為融資租賃之承租人，則租賃資產列入 「物業、機器及設備」 項內，而相應負債則列於 「其他負債」 項內。融 資租賃資產及其相應負債首次列賬時按資產之公允值或最低租金款額現值 （以兩者之較低者為準） 確認。 所有其他租賃均分類為經營租賃。若滙豐為出租人，則經營租賃涉及之資產會列於 「物業、機器及設備」 項內。減 值虧損則按賬面值未能全數收回的數額確認。若滙豐為承租人，則不會在資產負債表內確認租賃資產。 融資租賃之融資收益或費用則於租賃期內在 「淨利息收益」 項內確認，以反映固定回報率。經營租賃之應付及應收 租金按租賃期根據直線基準入賬，並於 「一般及行政開支」 及 「其他營業收益」 項內確認。 經營租賃承諾 於2014年12月31日，根據不可撤銷經營租賃就土地、樓宇及設備承擔之日後最低租金款額為 53.72億美元 （2013年：54.96億美元） 。 38－租賃承諾╱39－結構公司 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 443 融資租賃應收賬款 滙豐根據融資租賃向第三方出租各種資產，包括交通資產 （例如飛機） 、物業及一般廠房和 機器。於租賃期結束時，資產可出售予第三方或續租更長年期。租金的計算是用以收回資 產成本減其剩餘價值及賺取融資收益。 2014年 2013年 日後最低 租金總額 未賺取之 融資收益 現值 日後最低 租金總額 未賺取之 融資收益 現值 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 百萬美元 租賃應收賬款： －1年內 3,383 (374) 3,009 3,370 (360) 3,010 －1年後至5年內 8,089 (980) 7,109 7,933 (990) 6,943 －5年後 5,013 (744) 4,269 5,064 (856) 4,208 於12月31日 16,485 (2,098) 14,387 16,367 (2,206) 14,161 39 結構公司 會計政策 結構公司為設計旨在使投票權或類似權利並非決定誰人控制公司的主要因素的公司，例如任何投票權僅與管理工作 有關的情況，且主要活動乃按合約安排指導。結構公司的活動一般受到限制，且具有狹義且明確界定的目標。 結構公司根據附註1(h)所載的會計政策進行綜合入賬評估。 滙豐主要透過金融資產證券化、中介機構及投資基金而涉及結構公司。 滙豐涉及結構公司的安排設立之時均由總部授權，以確保用途適當及監管恰當。由滙豐管 理之結構公司的活動，均由高級管理層密切監察。滙豐涉及已綜合入賬及未綜合入賬的結 構公司，該等結構公司可能由滙豐或第三方成立，詳情見下文。 已綜合入賬結構公司 滙豐按公司類別劃分的已綜合入賬結構公司的資產總值 中介機構 證券化公司 滙豐管理的 基金 其他 總計 十億美元 十億美元 十億美元 十億美元 十億美元 於2014年12月31日 27.2 7.9 11.2 6.7 53.0 於2013年12月31日 38.9 7.1 13.9 8.2 68.1 中介機構 滙豐已成立及管理兩類中介機構：證券投資中介機構及多賣方中介機構。該等公司的設計 旨在使投票權或類似權利並非決定誰人擁有控制權的主要因素；在該等情況下，會以合約 安排的方式指導相關活動。由於滙豐因參與該公司而面對可變動回報的風險或有權享有可 變動回報，並且能夠透過對該公司擁有的權力影響相關回報，因此將中介機構綜合入賬。 證券投資中介機構 滙豐的主要證券投資中介機構Solitaire購入高評級的資產抵押證券以促成切合需要的投資機 會。於2014年12月31日，Solitaire持有80億美元的資產抵押證券 （2013年：90億美元） 。其已載入 第162頁 「經綜合入賬結構公司持有」 的資產抵押證券披露中。滙豐的其他證券投資中介機構 Mazarin、Barion及Malachite，均由2008年重組滙豐的結構投資公司演化而來。 ‧ Solitaire－Solitaire現時由發行予滙豐的商業票據 「CP」 提供全數資金。儘管滙豐繼續提供 流動資金信貸額，只要滙豐繼續購買 （而其亦有意於可見未來繼續購買） 所發行的商業票 據，Solitaire便毋須取用流動資金信貸額。於2014年12月31日，滙豐持有的商業票據達95億 美元 （2013年：110億美元） 。 ‧ Mazarin－滙豐因提供流動資金信貸額而須面對Mazarin資產面值的風險，該等流動資金信 貸額相等於已發行優先債務的已攤銷成本與非拖欠資產的已攤銷成本兩者中之較低者。 於2014年12月31日，滙豐所涉的風險承擔為39億美元 （2013年：74億美元） 。第一損失保障 乃透過Mazarin發行的資本票據提供，該等資本票據絕大部分由第三方持有。 於2014年12月31日，滙豐持有Mazarin發行的資本票據1.2% （2013年：1.3%） ，該等資本票據 的面值為1,000萬美元 （2013年：1,700萬美元） ，賬面值為140萬美元 （2013年：30萬美元） 。 財務報表附註 （續） 滙豐控股有限公司 444 ‧ Barion及Malachite－滙豐就該等證券投資中介機構的主要風險承擔，相當於為支持該等機 構之非現金資產而提供所需債務的已攤銷成本。於2014年12月31日，滙豐就此所涉的風 險承擔為30億美元 （2013年：63億美元） 。第一損失保障乃透過該等機構發行的資本票據 提供，該等資本票據絕大部分由第三方持有。 於2014年12月31日，滙豐持有該等機構發行的資本票據9.9% （2013年：3.8%） ，該等資本票 據的面值為5,480萬美元 （2013年：3,700萬美元） ，賬面值為1,010萬美元 （2013年：330萬美 元） 。 多賣方中介機構 多賣方中介機構成立的目的，是向滙豐客戶提供靈活的市場融資來源。於2014年12月31日， 滙豐承擔的風險相等於向多賣方中介機構提供的特定交易流動資金信貸額，為數154億美元 （2013年：157億美元） 。第一損失保障乃透過特定交易的強化信貸條件而提供。該等保障並 非由滙豐提供，而是由有關資產的辦理機構提供。滙豐以涵蓋整個計劃的信用強化安排形 式提供第二損失保障。 證券化公司 滙豐利用結構公司將本身辦理的客戶貸款證券化，主要為了分散辦理資產的資金來源及提 高資本效益。滙豐將貸款轉讓予結構公司以換取現金或透過信貸違責掉期以人為方式將貸 款轉讓予結構公司，而結構公司則向投資者發行債務證券。 滙豐管理的基金 滙豐已成立多個貨幣市場及非貨幣市場基金。當滙豐在其擔任投資經理時被視為以負責人 而非代理身分行事，滙豐實質上控制該等基金，因此會將該等基金綜合入賬。 其他 滙豐亦已於日常業務中進行多項交易，包括資產和結構融資交易，當中其對結構公司擁有 控制權。此外，滙豐被視為透過其以基金負責人的身分參與而控制多個第三方管理的基金。 未綜合入賬結構公司 「未綜合入賬結構公司」 一詞指所有並非由滙豐控制的結構公司。滙豐會於日常業務中與未 綜合入賬結構公司進行交易，以促成客戶交易及把握特定投資機會。 下表列示滙豐於業績報告日期擁有權益的未綜合入賬結構公司的資產總值及其就該等權益 的最大損失風險承擔。 與滙豐於未綜合入賬結構公司的權益有關的性質及風險 證券化公司 滙豐管理 的基金 非滙豐管理 的基金 其他 總計 十億美元 十億美元 十億美元 十億美元 十億美元 於2014年12月31日 公司的資產總值 11.0 308.5 2,899.9 32.8 3,252.2 滙豐的權益－資產 交易用途資產 － 0.1 0.1 4.6 4.8 指定以公允值列賬之金融資產 － 5.2 2.3 － 7.5 衍生工具 － － － 1.3 1.3 同業貸款 － － － 0.1 0.1 客戶貸款 0.8 － － 1.5 2.3 金融投資 － 2.5 5.9 0.1 8.5 其他資產 － － － 0.1 0.1 與滙豐於未綜合入賬結構公司的 權益有關的資產總值 0.8 7.8 8.3 7.7 24.6 滙豐的權益－負債 其他負債 － － － 0.1 0.1 與滙豐於未綜合入賬結構公司的 權益有關的負債總額 － － － 0.1 0.1 滙豐的最大風險承擔 0.8 7.8 8.3 11.1 28.0 滙豐權益的收益總額1 － 0.1 0.3 0.4 0.8 39－結構公司 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 445 證券化公司 滙豐管理 的基金 非滙豐管理 的基金 其他 總計 十億美元 十億美元 十億美元 十億美元 十億美元 於2013年12月31日 公司的資產總值 9.6 290.3 2,843.3 26.7 3,169.9 滙豐的權益－資產 交易用途資產 － 0.1 0.2 3.8 4.1 指定以公允值列賬之金融資產 － 5.1 1.4 － 6.5 衍生工具 － － － 1.2 1.2 客戶貸款 0.9 － － 1.5 2.4 金融投資 － 2.3 5.4 0.1 7.8 其他資產 － 0.1 － － 0.1 與滙豐於未綜合入賬結構公司的 權益有關的資產總值 0.9 7.6 7.0 6.6 22.1 滙豐的權益－負債 交易用途負債 － － － 0.1 0.1 與滙豐於未綜合入賬結構公司的 權益有關的負債總額 － － － 0.1 0.1 滙豐的最大風險承擔 1.0 7.6 7.0 10.6 26.2 滙豐權益的收益總額1 － 0.1 0.3 0.3 0.7 1 收益包括經常性及非經常性費用、利息、股息、重新計算或撤銷確認結構公司權益的增益或虧損、任何 按淨額結算的市值計價增益╱虧損及向結構公司轉讓資產及負債的增益或虧損。 滙豐就於未綜合入賬結構公司的權益的最大損失風險承擔指因滙豐參與未綜合入賬結構公 司而可能產生的最大損失 （不論產生損失的可能性） 。 ‧ 就承諾及擔保和承辦的信貸違責掉期而言，最大損失風險承擔為潛在未來損失的名義金 額。 ‧ 就於未綜合入賬結構公司的保留及購入投資和未綜合入賬結構公司貸款而言，最大損失 風險承擔為該等權益於資產負債表報告日期的賬面值。 最大損失風險承擔按為減低滙豐的損失風險承擔而訂立的對沖及抵押品安排的總效用列賬。 證券化公司 滙豐透過持有未綜合入賬的證券化公司發行的票據而擁有該等公司的權益。此外，滙豐亦 投資於第三方結構公司發行的資產抵押證券，詳載於第162頁。 滙豐管理的基金 滙豐已成立並管理貨幣市場基金及非貨幣市場投資基金，以為客戶提供投資機會。更多有 關 「管理資金」 的資訊載於第106頁。 滙豐作為基金經理可能有權根據管理資產收取管理及表現費，滙豐可能保留該等基金的單 位。 非滙豐管理的基金 滙豐購入及持有第三方管理的基金的單位，以滿足業務及客戶需要。 此外，滙豐訂立衍生工具合約以促進對非滙豐管理的基金的風險管理方案。於2014年12月31日， 滙豐對非滙豐管理的基金的衍生工具風險之公允值為65億美元。附註16載列有關滙豐所訂 立衍生工具的資料。 其他 滙豐已於日常業務中成立結構公司，如為客戶訂立結構信貸交易以向公營和私營機構基建 項目提供資金，以及為資產和結構融資交易訂立結構信貸交易。 滙豐資助的結構公司 會計政策 如滙豐持續參與另一公司，並在該公司的成立中擔當重要角色或促成相關交易對手進行該公司為之成立的交易，則 滙豐被視為資助該公司。若參與該公司僅屬於行政性質，則滙豐一般不會被視為資助機構。 財務報表附註 （續） 滙豐控股有限公司 446 於2014及2013年，向該等資助公司轉讓的資產及收取的收益金額並不重大。 40 法律訴訟及監管事宜 滙豐在多個司法管轄區內，因日常業務運作而牽涉法律訴訟及監管事宜。除下文所述者外， 滙豐認為此等事宜無一屬重大者。確認準備的方法乃根據附註29所載的會計政策釐定。雖 然法律訴訟及監管事宜的結果存有內在的不明朗因素，但管理層相信，根據所得資料，於 2014年12月31日已就有關法律訴訟及監管事宜提撥適當準備 （請參閱附註29） 。倘個別準備屬 重大，即會註明已提撥準備的事實及其金額。確認任何準備並不代表承認錯誤或承擔法律 責任。若要估計作為或有負債類別之法律訴訟及監管事宜潛在責任所涉總額，並不切實可 行。 證券訴訟 因2002年8月重列前期匯報的綜合財務報表及其他企業事件 （包括於2002年與美國46個州份及 哥倫比亞特區就房地產貸款的手法達成和解） ，Household International及若干前任高級職員於 2002年8月在美國伊利諾伊州北區聯邦地區法院 （ 「伊利諾伊地區法院」 ） 提出的一宗集體訴訟 （Jaffe訴Household International, Inc.及其他被告人） 中被列為被告人。此項申訴乃根據 《美國證 券交易所法》 提出申索，指稱被告人在知情或罔顧後果下，就Household International消費貸款 業務 （包括追收欠款、銷售及貸款手法） 之重要事實，以及上述重列所確證涉及會計實務的 事實，作出虛假及誤導性陳述，有關Household International消費貸款業務的部分陳述最終導 致2002年與各州份達成和解協議。最終一群申訴委託人獲確認為代表於1999年7月至2002年 10月期間購入及出售Household International普通股的所有人士。 陪審團審訊於2009年4月結束，其中一部分判決結果判原告人勝訴。有關裁決的各項法律質 疑已在進行審訊後簡介期間提出。 於2011年12月，於申訴委託人提交申索表格後，法院任命的申索管理人向伊利諾伊地區法 院報告，引起准予申索虧損的申索總數共有45,921宗，而該等申索所涉的總金額約為22億美 元。被告人就集體申訴的依賴推定及遵循申索表格規定方面提出法律質疑，惟伊利諾伊地 區法院於2012年9月否決大部分質疑。伊利諾伊地區法院並指示由法院任命的特委聆案官進 一步審理若干提交申索的問題。 於2013年10月，伊利諾伊地區法院拒絕接受被告人要求依法作出判決或另行重新審訊的額 外審訊後動議，並接受原告人要求作出部分最終判決的動議，以及判給按最優惠利率每年 複合計算的判決前利息。隨後，於2013年10月，伊利諾伊地區法院作出不利於被告人的部分 最終判決，涉及金額約為25億美元 （包括判決前利息） 。除已作出的部分判決外，申索尚餘約 6.25億美元 （未計判處判決前利息） 仍未由伊利諾伊地區法院作出裁決，故仍可遭受反對。 被告人就部分最終判決提交上訴通知書，而美國聯邦上訴法院第七巡迴審判庭 （ 「上訴法院」 ） 已於2014年5月進行口頭辯論聆訊。我們正等待上訴法院的決定。被告人亦已提交相當於部 分最終判決所涉概約金額 （25億美元） 的判決中止執行保證書，以暫緩執行判決等候上訴。 儘管陪審團已作出裁決，伊利諾伊地區法院亦已作出多項裁定，並已作出部分最終判決， 但我們仍然相信提出上訴申請寬免的理據充分。 最終解決此事所需時間及其結果尚未明朗。鑑於實際釐定損害賠償金額涉及複雜和不明朗 的因素 （包括任何上訴結果） ，故此有關事件可能出現多種不同結果。倘上訴法院駁回或僅 部分接納我們提出的論點，根據部分最終判決及其他待決申索和計及該等待決申索的判決 前利息，損害賠償金額可能介乎相對小額至最高36億美元或以上之間。一旦作出判決 （如於 2013年10月作出的部分最終判決約25億美元） ，判決後利息會自判決之日起按相等於聯邦儲 備體系公布的固定一年期國庫票據每周平均收益率的息率累計。滙豐已按管理層對可能流 出資金的最佳估算提撥準備。 40－法律訴訟及監管事宜 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 447 馬多夫證券 於2008年12月，Bernard L. Madoff （ 「馬多夫」 ） 被捕，並最終認罪，承認進行龐氏騙局。馬多夫 承認，他雖然聲稱將客戶的金錢投資於證券，惟實質上從未進行證券投資，而是利用其他 客戶的資金滿足歸還投資的要求。馬多夫的公司Bernard L. Madoff Investment Securities LLC （ 「馬 多夫證券」 ） 正由一名受託人 （ 「受託人」 ） 進行清盤。 滙豐旗下多家非美國公司為若干在美國境外註冊成立的基金提供託管、管理及同類服務， 而該等基金的資產均交由馬多夫證券進行投資。根據馬多夫證券提供的資料，於2008年11月 30日，該等基金聲稱總值84億美元，此數額包括馬多夫申報的虛構利潤。根據滙豐迄今所得 資料，我們估計，在滙豐為該等基金提供服務期間，由該等基金實際轉移至馬多夫證券的 資金減去實際從馬多夫證券提取的資金，合共約為40億美元。滙豐旗下多家公司於馬多夫 詐騙案引致的訴訟中被列為被告人。 於美國╱英國提出的訴訟：受託人於美國破產法院及英國高等法院對滙豐旗下若干公司提 出訴訟。受託人於美國提出的訴訟包括普通法申索，指稱滙豐協助及教唆馬多夫詐騙及違 反受信責任。該等申索已經因缺乏理據而被駁回。受託人於美國的餘下申索根據美國破產 法尋求追回呈請前的轉移款項。該等餘下申索中牽涉滙豐的數額尚未提出或釐定。 作為受託人於美國所提訴訟中的共同被告人，Alpha Prime Fund Ltd （ 「Alpha Prime」 ） 及Senator Fund SPC對滙豐提出交叉申索。上述基金亦在盧森堡起訴滙豐 （討論見下文） 。 受託人於英國提出的訴訟尋求追回由馬多夫證券轉移至滙豐或透過滙豐轉移的非指定數額 款項。滙豐仍未獲送達受託人於英國提出訴訟的傳票。受託人送呈申索傳票的期限已延至 2015年第三季。 將資產交由馬多夫證券進行投資的基金Fairfield Sentry Limited、Fairfield Sigma Limited及Fairfield Lambda Limited （統稱 「Fairfield」 ） ，在美國及英屬處女群島對多名基金股東提出多宗訴訟，該 等基金股東包括為滙豐客戶擔任代名人的滙豐旗下若干公司。Fairfield的訴訟尋求歸還因 贖回股份而支付的款項。Fairfield於美國提出的訴訟目前暫緩審理，等待英屬處女群島有關 Fairfield案件的結果 （討論見下文） 。 於2013年9月，美國聯邦上訴法院第二巡迴審判庭維持駁回馬多夫所投資三個基金的投資者 對滙豐及其他被告人提出的聲稱集體訴訟申索，理由是不方便法院原則。於2014年12月，原 告人向美國最高法院提交移審令狀呈請。最高法院對是否批准移審的覆核預期於2015年上 半年有所決定。 於2014年12月，另有三宗與馬多夫相關的新訴訟提出。第一宗是於2008年12月仍在馬多夫證 券持有投資的直接投資者提出的聲稱集體訴訟，該訴訟提出多項普通法申索，並尋求收回因 指稱滙豐知悉及協助詐騙行為而在馬多夫證券詐騙案中損失的損害賠償。其他兩宗訴訟由 SPV Optimal SUS Ltd （ 「SPV Optimal」 ） 提出，SPV Optimal是馬多夫證券所投資公司Optimal Strategic US Equity Ltd的聲稱受讓人。上述訴訟中，其中一宗在紐約州法院提出，另一宗在美國聯邦 地區法院提出。於2015年1月，SPV Optimal撤銷了控告滙豐的聯邦案件。在州法院對滙豐提出 的訴訟仍待審理。 於英屬處女群島提出的訴訟：自2009年10月起，直接或間接地將資產交由馬多夫證券進行投 資的Fairfield基金在英屬處女群島提出多宗訴訟，控告多名基金股東，包括為滙豐私人銀行業 務客戶及投資於Fairfield基金的其他客戶擔任代名人的滙豐旗下若干公司。Fairfield基金尋求 歸還基金向被告人支付的贖回款項，理據為該等款項依據被誇大的資產淨值而錯誤支付。 於2014年4月，英國樞密院頒布對兩項初步事宜的裁決，裁定英屬處女群島訴訟中的其他被 告人勝訴，並於2014年10月頒布命令。其他被告人亦在英屬處女群島法院提出待審理動議， 質疑Fairfield清盤人在美國尋求申索的權力。英屬處女群島法院已將待審理動議的聆訊押後 至2015年3月。 於百慕達提出的訴訟：於2009年1月，直接或間接地將資產交由馬多夫證券進行投資的Kingate Global Fund Limited及Kingate Euro Fund Limited （統稱 「Kingate」 ） 在百慕達對百慕達滙豐銀行有限 公司提出一宗訴訟，尋求追回在Kingate戶口內所持資金、費用及股息。此訴訟現時仍有待審 理，但於受託人在美國對Kingate及百慕達滙豐銀行有限公司提出的其他訴訟得到解決前， 預期不會有任何進展。 同樣交由馬多夫證券進行投資的基金Thema Fund Limited （ 「Thema」 ） 及Hermes International Fund Limited （ 「Hermes」 ） 亦各自於2009年在百慕達提出三個系列的訴訟。第一個系列的訴訟是控告 財務報表附註 （續） 滙豐控股有限公司 448 HSBC Institutional Trust Services (Bermuda) Limited，尋求追回在滙豐持有之凍結戶口內的資金。 第二個系列的訴訟指稱HSBC Institutional Trust Services (Bermuda) Limited須就過失、追回費用及 違約損害賠償的申索承擔責任。第三個系列的訴訟尋求從百慕達滙豐銀行有限公司及HSBC Securities Services (Bermuda) 退回費用。雖然數年來該等訴訟無甚進展，但Thema及Hermes已於 2015年1月提交意向書表示擬進行上述第二個系列的訴訟。 於開曼群島提出的訴訟：於2013年2月，投資於馬多夫證券的開曼群島基金Primeo Fund對基 金管理人Bank of Bermuda (Cayman)及基金託管人HSBC Securities Services (Luxembourg) （ 「HSSL」 ） 提 出訴訟，指稱違約。Primeo Fund向被告人申索損害賠償，用於補償其聲稱的損失，包括利潤 損失及向受託人承擔的任何負債。審訊已押後至2016年1月。 於盧森堡提出的訴訟：於2009年4月，Herald Fund SPC （ 「Herald」 ） （自2013年7月起正式清盤） 在盧 森堡地方法院對HSSL提出訴訟，尋求歸還 Herald聲稱因馬多夫證券詐騙案而損失的所有現金 及證券或相同數額的損害賠償金。2013年3月，盧森堡地方法院駁回Herald要求退回證券的歸 還申索。Herald要求退回現金及收取損害賠償金的歸還申索則予保留。Herald於2013年5月對 此判決提出上訴。預期提供司法保證金的裁決將於2015年5月下達。至於准予保留的歸還申 索訴訟則暫緩處理，以等待上訴結果。 於2009年10月，Alpha Prime於盧森堡地方法院起訴HSSL，指稱HSSL違約及在委任馬多夫證券擔 任Alpha Prime 資產副託管人時疏忽。Alpha Prime被頒令提供司法保證金。Alpha Prime已要求暫 緩處理有關訴訟，以待與受託人就美國的訴訟進行商議。目前訴訟已暫緩處理。 於2010年3月，Herald (Lux) SICAV （ 「Herald (Lux) 」 ） （由2009年4月起正式清盤） 在盧森堡地方法院 對HSSL提出訴訟，尋求歸還證券或等同現金項目，或以損害賠償金替代。Herald (Lux) 亦要求 歸還HSSL擔任基金託管人及服務代理而獲支付的費用。此訴訟訂於2015年3月進行下一次初 步聆訊。 於2014年12月，Senator Fund SPC於盧森堡地方法院對HSSL提出訴訟，尋求歸還自2008年11 月以來最新資產淨值表所示持有的證券，或以損害賠償金替代。首次初步聆訊的日期訂於 2015年2月。 Primeo Select Fund、Herald、Herald (Lux)以及Hermes基金的股東已對HSSL提出多宗訴訟。這些訴 訟處於不同階段，其中大部分已被駁回、暫緩處理或延期。 於愛爾蘭提出的訴訟：2013年11月，一個投資於馬多夫證券的基金Defender Limited對HSBC Institutional Trust Services (Ireland) Limited （ 「HTIE」 ） 提出訴訟，指稱HTIE違反託管協議，並申索損 害賠償及要求對Defender Limited面對申索招致資金損失作出彌償。該訴訟亦包括在Defender Limited擔任董事及投資經理的四名非滙豐旗下人士。 於2013年7月及12月，Thema International Fund plc （ 「Thema International」 ） 及Alternative Advantage Plc （ 「AA」 ） 分別就HTIE在愛爾蘭高等法院面對的申索達成和解。Thema International個別股東提出 的五宗訴訟仍待審理。 於2014年12月，SPV Optimal對HTIE及HSBC Securities Services (Ireland) Limited 提出新訴訟，指稱該 等公司違反託管協議及申索損害賠償和要求對資金損失作出彌償。 上述與馬多夫相關的多宗法律訴訟可能產生多種不同結果以至最終的財務影響，而這些結 果及影響可能受多項因素左右，包括但不限於訴訟在不同司法管轄區提出及該等訴訟涉及 不同數目的原告人與被告人。基於這些原因 （以及其他原因） ，現階段要求滙豐準確估計在 與馬多夫相關的多宗法律訴訟中所有申索可能產生的負債總額或須承擔的法律責任範圍並 不切實可行，但相關金額可能甚大。 美國按揭相關調查 2011年4月，於完成有關業內的住宅按揭止贖手法的廣泛和全面檢討後，HSBC Bank USA N.A. （ 「美國滙豐銀行」 ） 與美國貨幣監理署 （ 「美國貨監署」 ） 達成同意終止及停止令，而美國滙豐融 資有限公司 （ 「美國滙豐融資」 ） 及北美滙豐控股有限公司 （ 「北美滙豐」 ） 亦與美國聯邦儲備局 （ 「聯儲局」 ） 訂立類似的同意令 （連同與美國貨監署達成的命令，合稱 「債務管理同意令」 ） 。 債務管理同意令要求採取指定行動，以處理經雙方共同檢視後所發現並已在同意令清楚說 明的各種不善之處。美國滙豐銀行、美國滙豐融資及北美滙豐繼續與美國貨監署及聯儲局 合作，務使各自的程序符合同意令的要求，同時正對業務營運模式作出相關的改革。 根據上述債務管理同意令，滙豐已委聘一名獨立顧問對2009年1月至2010年12月期間尚待處 40－法律訴訟及監管事宜 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 449 理或已經完成的止贖個案進行獨立檢討 （ 「止贖程序獨立檢討」 ） ，藉以確定是否有任何借款 人因止贖程序出現錯誤而蒙受財務損失。於2013年2月，美國滙豐銀行與美國貨監署達成協 議，而美國滙豐融資及北美滙豐則與聯儲局達成協議 （統稱 「IFR和解協議」 ） ，止贖程序獨立 檢討由一個覆蓋較廣泛的框架取代。此項協議訂明在該框架下，滙豐和另外12名參與協議 的債務管理人同意合共支付超過93億美元現金並提供其他援助以協助合資格借款人。根據 IFR和解協議，北美滙豐已向一個基金支付9,600萬美元現金，該基金是專為向2009及2010年期 間進入止贖程序的借款人支付款項而設，滙豐亦正提供其他援助 （如修訂貸款條款等） ，以 協助合資格借款人。獲得賠償的借款人將毋須簽立解除責任或放棄權利的文件，亦不會被 阻止提出有關止贖或其他按揭債務管理手法的訴訟。至於參與協議的債務管理人，包括美 國滙豐銀行及美國滙豐融資，若全面履行IFR和解協議的條款即屬已符合債務管理同意令有 關止贖程序獨立檢討的要求，包括逐步結束止贖程序獨立檢討。 債務管理同意令並不會排除銀行監管機構、政府或執法機構 （如美國司法部或州檢察長） 對 美國滙豐銀行、美國滙豐融資或北美滙豐採取其他執法行動。該等行動可能包括就債務管 理同意令內羅列的各種活動而判處民事罰款及實施其他制裁。惟根據與美國貨監署達成的 IFR和解協議，美國貨監署同意，若IFR和解協議條款得以履行，將不會就債務管理同意令內 提及的過往管理按揭債務及有關止贖手法評定任何民事罰款額，或展開任何進一步執法行 動。美國貨監署雖然同意不評定任何民事罰款額但附有額外條件，即北美滙豐須根據可能 與美國司法部達成的有關管理住宅按揭貸款的任何協議支付款項或為借款人提供援助。聯 儲局已同意任何由聯儲局評定的民事罰款額將反映多項調整，包括就消費者濟助撥支的款 項，以及根據可能與美國司法部達成的有關管理住宅按揭貸款的任何協議所支付的款項。 IFR和解協議並不排除仍會有涉及該等處理手法的私人訴訟。 除上文所述債務管理同意令及與止贖程序獨立檢討有關的和解外，於2012年2月，美國五大按 揭債務管理人 （不包括滙豐旗下公司） 就止贖手法及其他按揭債務管理方式與美國司法部、 美國房屋及城市發展部和49個州的州檢察長達成和解。在2012年2月達成和解後，該等相關 政府機構已與其他按揭業債務管理人，包括滙豐、美國滙豐銀行、美國滙豐融資和北美滙 豐與美國的銀行監管機構及其他政府機構商討，探討達成和解的可能性。惟任何有關和解 將不會完全排除各個州或聯邦機構、銀行監管機構或執法機構就止贖手法及其他按揭債務 管理方式 （包括但不限於為投資者提供按揭證券化的有關事宜） 採取其他執法行動。該等手 法過往曾導致私人訴訟，而上文提及可能達成的和解並不能排除仍會有涉及該等手法的其 他私人訴訟。 美國按揭證券化活動及訴訟 美國滙豐銀行曾涉及為用作便利HSBC Securities (USA) Inc.
（ 「HSI」 ） 承銷原始房屋貸款證券化工 具提供的貸款，擔任該等貸款的保薦人╱賣方。於2005至2007年期間，美國滙豐銀行共購買 並向HSI出售此類貸款240億美元，該等貸款隨後被證券化並由HSI出售予第三方。該等貸款於 2014年12月31日的未償還本金結欠約為57億美元。 購買及重新包裝原始房屋貸款的美國按揭證券化市場參與者，包括美國按揭市場的多個群 體，如證券化工具的債務管理人、辦理機構、承銷商、受託人或保薦人，以及這些群體內的 特定參與者，遭到起訴和政府及監管機構的調查和問訊。隨著業內住宅按揭止贖問題持續， 美國滙豐銀行以受託人身分代表多個證券化信託接收的止贖房屋數目不斷增多。作為該等 物業的名義登記擁有人，美國滙豐銀行被各地方政府及租戶控告違反多項法例，包括有關 物業保養維修及租戶權利的法例。儘管滙豐相信且一直堅持相關事宜的責任及法律責任理 當由各信託之債務管理人承擔，惟該等事宜及同類事宜，包括由其他人士以 「滙豐作為受託 人」 的名義辦理的止贖安排，已令滙豐繼續成為媒體關注的焦點，且報導傾向負面。 於2014年6月至12月期間，美國滙豐銀行以250多個按揭證券化信託的受託人身分在紐約州的 州法院及聯邦法院面對多項控訴。該等訴訟乃由一群投資者代表信託提出的衍生訴訟，當 中包括貝萊德及PIMCO基金。其他多個並非滙豐旗下公司的金融機構亦因擔任按揭證券化 組合受託人而於同類訴訟中被起訴。對美國滙豐銀行提出的申訴指稱有關信託已因抵押品 價值下跌而蒙受超過340億美元損失。該等訴訟尋求因涉嫌違反 《美國信託契約法》 、違反受 信責任、疏忽、違約及╱或違反普通法信託責任而產生的未有指明數額損害賠償。滙豐已於 2015年1月向法院提出動議撤銷其中三項訴訟。 滙豐旗下多家公司在若干有關發售住宅按揭抵押證券的訴訟中被列為被告人，該等訴訟大 多指控證券化信託所發行的證券的發售文件載有重大失實陳述並涉及許多遺漏，包括有關 規範相關按揭貸款之承銷準則的陳述。於2011年9月，美國聯邦房屋金融局以美國聯邦國民 財務報表附註 （續） 滙豐控股有限公司 450 抵押協會 （ 「房利美」 ） 及美國聯邦住宅貸款抵押公司 （ 「房貸美」 ） 保護人身分，在美國紐約南區 聯邦地區法院 （ 「紐約地區法院」 ） 提出訴訟，控告美國滙豐銀行、北美滙豐、HSI及HSI Asset Securitization （ 「HASCO」 ） 以至HASCO五名前任及現任高級職員和董事。美國聯邦房屋金融 局尋求損害賠償金或撤銷房利美及房貸美購入由滙豐旗下公司承銷或保薦的按揭抵押證券 的交易。如2014年9月所公布，相關各方已達成最終和解以解決該事宜，滙豐須向美國聯邦 房屋金融局支付合共5.5億美元。 在多項由證券化信託受託人提出的按揭貸款回購訴訟中，美國滙豐銀行、美國滙豐融資及 Decision One Mortgage Company LLC （美國滙豐融資的間接附屬公司） 被列為被告人。概括而言， 該等訴訟尋求被告人 （即滙豐旗下公司） 回購有關按揭貸款，或支付補償性損害賠償以代替 回購，金額合計不少於10億美元。滙豐已就其中兩宗訴訟提呈駁回動議，並已提交全部文 件，目前正在等待進一步處理。 自2010年以來，於滙豐旗下多家公司接獲美國各個機構若干傳票及索取資訊的詢問，要求 提供有關滙豐及其聯屬機構以發行人、保薦人、承銷商、存戶、受託人、託管商或債務管理 人身分參與 （特別是） 私營機構住宅按揭抵押證券交易的文件及資料。滙豐繼續與這些美國 機構合作。於2014 年11月，根據 《金融行業改革、恢復及執行法》 ，北美滙豐代表其本身及旗 下多家附屬公司，包括但不限於美國滙豐銀行、HASCO、HSI、HSI Asset Loan Obligation、HSBC Mortgage Corporation (USA)、美國滙豐融資及Decision One Mortgage Company LLC，接獲科羅拉多 州地區檢察官辦公室的傳票，內容涉及次優質及非次優質住宅按揭的辦理、融資、購買、證 券化及管理。此事宜目前處於初步階段，而滙豐正與相關機構全面合作。 根據目前已知的事實，現階段要求滙豐預測上述事宜的解決方案 （包括解決時間及任何可能 對滙豐造成的影響） ，並不切實可行。 滙豐預期按揭證券化問題將繼續受到關注。因此，滙豐旗下各公司或會因以一組公司其中 一名成員或獨立身分參與美國按揭證券化市場而牽涉其他申索和訴訟，或受到政府或監管 機構的審查。 反洗錢及制裁相關事宜 於2010年10月，美國滙豐銀行與美國貨監署訂立同意停止和終止令，而北美滙豐亦與聯儲局 訂立同意停止和終止令 （ 「同意令」 ） 。該等同意令要求滙豐的所有美國業務採取改善措施， 制訂有效的合規風險管理計劃，涵蓋與美國 《銀行保密法》 及反洗錢合規事宜有關的風險管 理。集團會繼續採取措施應付同意令的要求。 於2012年12月，滙豐控股有限公司 （ 「滙豐控股」 ） 、北美滙豐及美國滙豐銀行就過往未能充分 遵守 《銀行保密法》 、反洗錢和制裁法律，與美國及英國政府機構達成協議。在該等協議中， 滙豐控股及美國滙豐銀行與美國司法部、美國紐約東區檢察官辦公室及美國西維吉尼亞州 北區檢察官辦公室訂立五年期的延後起訴協議 （ 「美國延後起訴協議」 ） ，滙豐控股與紐約郡 地區檢察官訂立兩年期的延後起訴協議 （ 「紐約郡地區檢察官延後起訴協議」 ） ，同時滙豐控 股亦接納聯儲局的停止和終止令，滙豐控股及北美滙豐亦接納聯儲局的民事罰款令。此外， 美國滙豐銀行與金融犯罪執法網絡訂立民事罰款令，並與美國貨監署訂立另一項民事罰款 令。滙豐控股亦與外國資產控制辦公室就過往交易涉及受該辦公室制裁的人士訂立協議， 並與英國金融業操守監管局訂立承諾書，承諾遵守若干前瞻性反洗錢及制裁相關責任。 根據該等協議，滙豐控股及美國滙豐銀行向美國當局支付合共19億美元，並繼續遵守需要 持續承擔的責任。於2013年7月，美國紐約東區聯邦地區法院已批准美國延後起訴協議，並 保留監督該協議實施情況之權力。根據與美國司法部、英國金融業操守監管局及聯儲局訂 立的協議，一名獨立監察員 （就英國金融業操守監管局而言，為 《金融服務及市場法》 第166條 所指的 「技術人員」 ） 現正評價及定期評估滙豐反洗錢及制裁措施合規職能的有效程度及滙 豐根據該等協議履行其補救責任的進展。 於2014年12月，滙豐控股已符合紐約郡地區檢察官延後起訴協議施加的所有規定，該協議按 其條款協議兩年期結束時屆滿。倘若滙豐控股及美國滙豐銀行符合美國延後起訴協議施加 的所有規定，則美國司法部會於該協議五年期結束時撤銷對這些公司的起訴。倘若滙豐控 股或美國滙豐銀行違反美國延後起訴協議的條款，美國司法部可就美國延後起訴協議的主 題事項向滙豐控股或美國滙豐銀行提出檢控。 美國滙豐銀行亦與美國貨監署訂立另一項同意令，要求該行糾正在美國貨監署當時最新的 檢查報告中注意到的情況及狀況，並對美國滙豐銀行直接或間接取得任何新設金融附屬公 40－法律訴訟及監管事宜 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 451 司的控制權或在其中持有權益或該行旗下現有金融附屬公司開展新業務施加若干限制，除 非該行事先獲得美國貨監署批准則另作別論。美國滙豐銀行亦與美國貨監署訂立另一項同 意令，要求該行採納企業整體合規計劃。 與美國及英國機構和解可能引致私人訴訟，也不排除有其他與滙豐遵守適用的 《銀行保密 法》 、反洗錢及制裁法律或其他各項協議未有涵蓋的 《銀行保密法》 、反洗錢、制裁或其他事 宜的監管或執法行動有關的私人訴訟。 於2014年5月，一名滙豐控股股東聲稱代表滙豐控股、美國滙豐銀行、北美滙豐及美國滙豐 有限公司 （ 「名義企業被告人」 ） ，在紐約州最高法院提出控告該等滙豐旗下公司若干現任及 前任董事及高級職員 （ 「個人被告人」 ） 的股東衍生訴訟。申訴指稱個人被告人違反對名義企 業被告人負有的受信責任，及指稱允許及╱或促成涉及美國延後起訴協議的行為而浪費企 業資產。於2014年10月，名義企業被告人動議駁回該訴訟，而已獲送達傳票的個人被告人亦 已回應該申訴。原告人已於2015年2月提出經修訂申訴。 2014年7月，安大略高級法院接獲控告滙豐控股及一位前任僱員的申索，聲稱代表於2006年 7月至2012年7月期間購買滙豐普通股及美國預託股份的一群人士，尋求損害賠償金高達200億 加元。申索人指稱被告人在滙豐控股及其全資附屬公司加拿大滙豐銀行所發出有關滙豐遵 守 《銀行保密法》 、反洗錢、制裁及其他法律的文件上作出法律及普通法下的錯誤陳述。 於2014年11月，替2004年4月至2011年11月在伊拉克死亡或受傷的美國人之代表提出的申訴呈 交美國紐約東區聯邦地區法院。申訴控告滙豐控股、英國滙豐銀行有限公司、美國滙豐銀 行及中東滙豐銀行，以及其他非滙豐旗下銀行及伊朗伊斯蘭共和國 （統稱 「被告人」 ） 。原告 人指稱被告人串謀違反 《美國反恐法》 ，為透過美國處理的交易更改或偽造涉及伊朗、伊朗 相關人士及伊朗銀行的付款信息。被告人的駁回動議將於2015年3月遞交。 該等私人訴訟正處於初步階段。根據目前的已知事實，現階段要求滙豐預測該等私人訴訟 的解決方案 （包括解決時間及任何可能對滙豐造成的影響） ，並不切實可行。 稅務及經紀交易商的調查 滙豐繼續配合美國司法部及美國稅務局持續調查滙豐旗下若干公司及僱員就若干須遵循美 國報稅責任的客戶所採取的行動是否適當。因應該等調查，滙豐私人銀行 （瑞士） 有限公司 （ 「滙豐瑞士私人銀行」 ） 在妥善遵守瑞士法律的情況下，已向美國司法部出示紀錄及其他文 件。於2013年8月，美國司法部通知滙豐瑞士私人銀行，由於原先已批准展開正式調查，故 其並不符合 「瑞士銀行不起訴協議或非目標函件計劃」 的資格。美國司法部已要求滙豐瑞士 私人銀行及其他瑞士銀行就資產撥入及撥出美國人士相關戶口以及管理該等戶口的僱員提 供額外資料。滙豐瑞士私人銀行正以符合瑞士法律的方式擬備此等數據。 滙豐旗下其他公司已收到美國及其他有關當局的傳票及提供資料的要求，包括有關滙豐旗 下一家印度公司的美國客戶事宜。 於2014年11月，滙豐瑞士私人銀行與美國證交會就2003至2011年由滙豐瑞士私人銀行及其前 身公司向美國居民客戶提供的跨境經紀及顧問服務達成最終和解。 此外，比利時、法國、阿根廷、瑞士及印度等全球各地多個稅務管理、監管及執法機關正 就指稱的逃稅或稅務詐騙、洗錢和非法跨境招攬銀行業務，對滙豐瑞士私人銀行進行調查 及檢討。比利時及法國的裁判官已對滙豐瑞士私人銀行進行正式刑事檢查。於2015年2月， 滙豐獲悉法國裁判官認為彼等已完成有關滙豐瑞士私人銀行的調查，並交由檢控官就可能 提出的指控提供意見，同時，裁判官保留繼續對滙豐其他行為展開調查的權利。此外，於 2014年11月，阿根廷稅務當局提出申訴控告滙豐瑞士私人銀行、阿根廷滙豐銀行、美國滙 豐銀行及若干現任及前任滙豐高級職員，指稱彼等之間存在非法合作關係，並指稱相關行 為使滙豐客戶規避阿根廷稅務責任。於2015年2月，瑞士一名檢控官就滙豐瑞士私人銀行展 開調查，印度稅務當局亦向滙豐旗下一家印度公司發出傳票及要求提供資料。 滙豐正配合有關當局，處理仍在進行中的上述各項事宜。根據目前已知事實，該等事宜可 獲解決的條款及解決時間 （包括滙豐可能被判處的罰款、罰則及╱或罰金） 極不明朗，而有關 數額可能甚大。 財務報表附註 （續） 滙豐控股有限公司 452 鑑於近期傳媒對該等事宜的關注，其他稅務管理、監管或執法機關亦可能會開始進行或擴 大類似調查工作或監管訴訟程序。 倫敦銀行同業拆息、歐洲銀行同業拆息及其他基準利率的調查及訴訟 英國、美國、歐盟、瑞士及其他地方等全球各地多個監管機構及保障公平競爭與執法機關， 現正就銀行小組成員所作出的若干過往提呈及有關設定倫敦銀行同業拆息、歐洲銀行同業 拆息及其他基準利率所作出的提呈的過程展開調查及檢討。由於滙豐旗下若干公司為有關 銀行小組成員，滙豐已獲監管機關要求提供資料，而滙豐正配合有關調查及檢討。 於2013年12月，歐洲委員會 （ 「委員會」 ） 宣布已根據卡特爾和解程序，就八家金融機構參與涉 及歐元利率衍生工具及╱或日圓利率衍生工具的非法活動而對該等機構判處罰款。雖然滙 豐並非遭罰款的金融機構之一，但委員會宣布已就純粹涉及歐元利率衍生工具的歐洲銀行 同業拆息相關調查對滙豐展開法律程序。此項調查將根據委員會的標準卡特爾程序繼續進 行。於2014年5月，滙豐收到委員會發出的異議聲明，指稱就歐元利率衍生工具的定價涉及 反競爭手法。該異議聲明列出了委員會的初步意見，而沒有預先判斷其最終調查結果。於 2014年11月，滙豐部分回應了委員會的異議聲明，一旦多項程序事宜得到解決，將有機會於 委員會決定的日期完成回應。 根據目前有關上述各項持續調查的已知事實，持續調查可獲解決的條款及解決時間 （包括罰 款及╱或罰則的數額） 極不明朗，有關金額可能甚大。 此外，在美國提出有關釐定美元倫敦銀行同業拆息的數宗私人訴訟中，滙豐及其他美元倫 敦銀行同業拆息銀行小組成員被列為被告人。該等申訴根據包括美國反壟斷及詐騙法、美 國 《商品交易法》 及州份法例在內的不同美國法律提出申索。該等訴訟包括個人及推定集體 訴訟，當中大部分已移交及╱或合併提交予紐約地區法院進行預審。 於2013年3月，負責監督有關美元倫敦銀行同業拆息之合併法律程序的紐約地區法院，在最 早提出的六宗訴訟中發出決定，駁回原告人提出的全部聯邦和州份反壟斷申索、詐騙申索 及不當得利申索，但容許若干不受適用時效法規所禁制的 《商品交易法》 申索繼續進行。該 等原告人中有部分就紐約地區法院的決定向美國聯邦上訴法院第二巡迴審判庭提出上訴， 審判庭其後駁回上訴。於2015年1月，美國最高法院撤銷上訴法院的決定，將案件退回上訴 法院考慮原告人上訴的理據。 其他原告人尋求在紐約地區法院提出經修訂申訴以堅持額外指稱。於2014年6月，紐約地區 法院頒布決定，其中包括拒絕原告人提出修訂其申訴的要求，有關修訂旨在提出額外理據， 堅稱滙豐及若干非滙豐旗下銀行操控倫敦銀行同業拆息，但允許原告人堅持對另外兩家銀 行提出的有關操控申訴；並批准被告人動議駁回 （基於適用時效法規所禁止） 根據 《商品交易 法》 提出的若干額外申索。有關合併法律程序中所有其他訴訟的法律程序已暫緩處理以等待 該決定。於2014年9月，暫緩審理已被撤銷。經修訂申訴於2014年10月在之前暫緩審理的非集 體訴訟中提交，而經修訂申訴已於2014年11月在多個之前暫緩審理的集體訴訟中提交。駁回 動議已分別於2014年11月及2015年1月提交且待審理。 此外，代表曾買賣與歐洲日圓東京銀行同業拆息相關的歐洲日圓期貨及期權合約之人士在 紐約地區法院提出的推定集體訴訟中，滙豐及其他銀行小組成員亦被列為被告人。該申訴 指稱 （其中包括） 有關歐洲日圓東京銀行同業拆息 （雖然滙豐並非日本銀行家協會歐洲日圓東 京銀行同業拆息銀行小組的成員） 及日圓倫敦銀行同業拆息的不當行為，違反了美國反壟斷 法、美國 《商品交易法》 和州份法例。於2014年3月，紐約地區法院頒布意見，駁回原告人根 據美國反壟斷法及州份法例提出的申索，但維持其根據美國 《商品交易法》 提出的申索。於 2014年6月，原告人動議許可提交第三項經修訂申訴。滙豐已反對該動議。該動議有待審理。 於2013年11月，代表曾買賣與歐洲銀行同業拆息相關的歐元期貨合約及其他金融工具之人士 在紐約地區法院提出的推定集體訴訟中，滙豐及其他銀行小組成員被列為被告人。該申訴 指稱 （其中包括） 有關歐洲銀行同業拆息的不當行為，違反了美國反壟斷法、美國 《商品交易 法》 和州份法例。原告人於2014年5月及2014年10月，分別提出第二項及第三項經修訂申訴。 滙豐擬於法院頒令暫緩審理到期 （現時定於2015年5月） 時，回應第三項經修訂申訴。 40－法律訴訟及監管事宜 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 453 於2014年9月及10月，代表曾執行利率衍生工具交易或買賣與美元ISDAfix利率掛鈎或在每日 ISDAfix設定窗口時間之前、期間或之後執行的金融工具的人士在紐約地區法院提交並合併 審理的多宗推定集體訴訟中，英國滙豐銀行有限公司及其他銀行小組成員被列為被告人。 該申訴指稱 （其中包括） 有關該等活動的不當行為，違反了美國反壟斷法、美國 《商品交易法》 和州份法例。於2014年10月，原告人提交了經修訂合併申訴。駁回該申訴的動議已於2014年 12月提交且待審理。於2015年2月，原告人提交了第二份經修訂合併申訴，並以美國滙豐銀 行取代英國滙豐銀行有限公司。 根據目前的已知事實，現階段要求滙豐預測該等私人訴訟的解決方案 （包括解決時間及任何 可能對滙豐造成的影響） ，並不切實可行。 匯率的調查及訴訟 英國、美國、歐盟及其他地方等全球各地多個監管機構及保障公平競爭與執法機關，現正 就外匯市場交易對多家機構 （包括滙豐） 展開調查及檢討。 於2014年11月，英國滙豐銀行有限公司與英國金融業操守監管局及美國商品期貨交易委員會 （ 「CFTC」 ） 就彼等各自調查滙豐涉及基準匯率的交易及其他操守達成監管和解。根據該等和 解條款，英國滙豐銀行有限公司同意向英國金融業操守監管局支付2.16億英鎊 （3.36億美元） 罰款及向美國商品期貨交易委員會支付2.75億美元民事罰款，並承諾採取若干補救措施。 於2014年12月，香港金融管理局 （ 「香港金管局」 ） 宣布完成對香港上海滙豐銀行有限公司 （ 「香 港上海滙豐銀行」 ） 外匯交易業務的調查。調查並無發現香港上海滙豐銀行操控市場的證據， 且並無判處罰款。香港上海滙豐銀行須採取若干補救措施。 在英國、美國及其他地方其餘的調查及檢討繼續進行。根據目前的已知事實，該等調查及 檢討可獲解決的條款及解決時間 （包括罰款及╱或罰則的數額） 極不明朗，但有關數額可能 甚大。於2014年12月31日，滙豐已就該等事宜確認準備5.5億美元。 此外，於2013年底及2014年初，在紐約地區法院提出的多宗推定集體訴訟中，滙豐控股、英 國滙豐銀行有限公司、北美滙豐、美國滙豐銀行及其他銀行被列為被告人。於2014年3月， 原告人提出經修訂合併申訴，指稱 （其中包括） 被告人透過分享客戶的保密買賣單流量信息 而串謀操控WM/Reuters基準匯率，逼使原告人及其他人士按該等匯率就產品支付虛假及非 競爭性價格，令原告人及其他人士受損 （ 「合併訴訟」 ） 。代表非美國原告人的個別推定集體 訴訟亦已經提出 （ 「外國訴訟」 ） 。被告人動議駁回所有訴訟。於2015年1月，法院否決被告人 駁回合併訴訟的動議，但批准被告人動議駁回外國訴訟。 根據目前的已知事實，現階段要求滙豐預測該等私人訴訟的解決方案 （包括解決時間及任何 可能對滙豐造成的影響） ，並不切實可行。 貴金屬定價相關訴訟及調查 自2014年3月起，有多宗推定集體訴訟在美國紐約南區、新澤西區及加州北區的聯邦地區法 院提出，美國滙豐銀行、英國滙豐銀行有限公司、HSI及倫敦黃金市場定價有限公司多家其 他成員機構被列為被告人。該等申訴指稱由2004年1月至今，被告人串謀於倫敦黃金下午定 價期間操控黃金及黃金衍生工具的價格，以謀取自營交易的利潤。該等訴訟已指定由紐約 地區法院合併處理。經修訂合併集體訴訟申訴已於2014年12月提交，而滙豐已於2015年2月 作出回應。 自2014年7月起，有多宗推定集體訴訟在美國紐約南區及東區聯邦地區法院提出，滙豐控股、 北美滙豐、美國滙豐銀行、美國滙豐有限公司及倫敦銀市場定價有限公司其他成員機構被 列為被告人。該等申訴指稱由2007年1月至今，被告人串謀操控實銀及銀衍生工具的價格， 以謀取共同利益，違反了美國反壟斷法及美國 《商品交易法》 。該等訴訟已指定由紐約地區 法院合併處理。經修訂合併集體訴訟申訴已於2015年1月提交，而滙豐將於2015年3月作出回 應。 於2014年底至2015年初，有多宗推定集體訴訟在紐約地區法院提出，美國滙豐銀行及倫敦鉑 金及鈀金市場定價有限公司其他成員機構被列為被告人。該等申訴指稱，自2007年1月至今， 被告人串謀操控實物鉑族金屬及以鉑族金屬為主的金融產品的價格，以謀取共同利益，違 反了美國反壟斷法及美國 《商品交易法》 。 財務報表附註 （續） 滙豐控股有限公司 454 於2014年11月，美國司法部發出一份文件，要求滙豐控股自願提供有關美國司法部進行的 貴金屬反壟斷刑事調查的若干文件。於2015年1月，美國商品期貨交易委員會向美國滙豐銀 行發出傳票，要求美國滙豐銀行提供有關美國滙豐銀行之貴金屬買賣營運的若干文件。滙 豐正配合美國當局的調查。 該等事宜處於初步階段。根據目前的已知事實，現階段要求滙豐預測該等事宜的解決方案 （包括解決時間或任何可能對滙豐造成的影響） ，並不切實可行。 信貸違責掉期監管調查及訴訟 於2013年7月，滙豐收到委員會發出的異議聲明，內容提及該會正持續調查若干市場參與者 於2006至2009年期間在信貸衍生工具市場的聲稱反競爭活動。該異議聲明列出了委員會的初 步意見，而沒有預先判斷其最終調查結果。滙豐已向委員會提交回應，並於2014年5月出席 口頭聆訊。於口頭聆訊後，委員會決定進入下一步的調查階段，然後才決定是否或如何繼 續處理該個案。滙豐現正配合該個案的進一步調查。根據目前的已知事實，現階段要求滙 豐預測此事宜的解決方案 （包括解決時間或任何可能對滙豐造成的影響） ，並不切實可行。 此外，美國滙豐銀行、滙豐控股及英國滙豐銀行有限公司在 （其中包括） 紐約地區法院及伊 利諾伊地區法院提出的眾多推定集體訴訟中被列為被告人。該等集體訴訟指稱被告人 （包括 ISDA、Markit及若干其他金融機構） 違反美國反壟斷法，串謀透過 （其中包括） 限制進入信貸違 責掉期定價交易所及阻止新參與者進入外匯市場來限制交易，以達到利用人為手段抬高在 美國買賣信貸違責掉期買賣差價的目的及效果。該等訴訟的原告人聲稱代表曾經主要在美 國向被告人買入或出售信貸違責掉期的所有人士提出集體訴訟。 於2013年10月，該等案件在紐約地區法院合併審理。該經修訂合併申訴已於2014年1月提 交，並將美國滙豐銀行、英國滙豐銀行有限公司及非滙豐旗下公司列為被告人。被告人於 2014年3月提出駁回該申訴的初步動議後，原告人提交了第二項經修訂合併申訴，而被告人 亦動議駁回有關申訴。於2014年9月，法院許可部分並否決部分被告人駁回該申訴的動議。 調查仍在進行中。 根據目前的已知事實，現階段要求滙豐預測該等私人訴訟的解決方案 （包括解決時間或任何 可能對滙豐造成的影響） ，並不切實可行。 經濟計劃：巴西滙豐銀行 於1980年代中期及1990年代初期，巴西政府推出若干經濟計劃，以降低不斷攀升的通脹水 平。實施該等計劃對儲蓄戶口持有人造成不利影響，最後數以千計的戶口持有人對巴西的 金融機構 （包括巴西滙豐銀行 （ 「巴西滙豐」 ） ） 展開法律訴訟程序，指稱 （其中包括） 儲蓄戶口 結餘按有別於合約協定的價格指數作出調整，導致該等戶口持有人損失收入。若干該等案 件已送達巴西最高法院 （ 「最高法院」 ） 。最高法院已暫停處理有待下級法院審理的所有相關 案件，直至最高法院就經濟計劃帶來的轉變是否符合憲法作出最終判決為止。預計最高法 院最終判決的結果將為有待下級法院審理的所有相關案件設下判案先例。此外，巴西最高 民事法院 （ 「最高民事法院」 ） 正在考慮 （其中包括） 適用於計算任何收入損失的合約性及懲罰 性利率事宜。 最高法院及最高民事法院的法律程序可獲解決的條款及解決時間 （包括如作出不利判決， 巴西滙豐可能須支付的損失數額） 極不明朗。該等損失可能介乎相對小額至最高8億美元之 間，但我們認為不大可能達此範圍的上限。 消費者 「增值服務產品」 監管審查 美國滙豐融資透過其既有卡及零售商戶業務向客戶提供信貸時，附帶提供有關產品或參與 有關產品的市場推廣、分銷或管理，例如身分盜竊保障及信貸監察產品。美國滙豐融資於 2012年5月前停止提供該等產品。提供及管理上述產品及債務保障產品等其他增值服務產 品使美國消費者金融保護局、美國貨監署及聯邦存款保險公司等監管機關對其他機構採取 執法行動。該等執法行動的結果是頒令向客戶歸還款項及評估大額罰款。我們已就若干增 值服務產品向若干客戶歸還款項，並繼續就監管機關的持續審查與監管機關合作。鑑於監 管機關對其他非滙豐旗下信用卡發行人就其增值服務產品採取的行動，一個或多個監管機 關可能因美國滙豐融資過往提供及管理該等增值服務產品而命令我們向客戶歸還額外款項 40－法律訴訟及監管事宜╱41－關連人士交易 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 455 及╱或判處民事罰款或採取其他濟助措施。此事宜可獲解決的條款及解決時間 （包括可能介 乎零至5億美元的任何額外補救金額） 極不明朗。 41 關連人士交易 集團及滙豐控股的關連人士包括附屬公司、聯營公司、合資公司、為滙豐僱員而設立的離 職後福利計劃、主要管理人員、主要管理人員的近親，及由主要管理人員或彼等近親所控 制或共同控制的企業。 主要管理人員被界定為有權及有責任策劃、督導及管控滙豐控股事務的人士，即滙豐控股 的董事及集團常務總監。 主要管理人員之報酬 滙豐 2014年 2013年 2012年 百萬美元 百萬美元 百萬美元 短期僱員福利 41 38 37 離職後福利 1 2 1 其他長期僱員福利 7 10 10 以股份為基礎的支出 54 35 43 截至12月31日止年度 103 85 91 涉及關連人士之交易、安排及協議 下表乃根據英國 《2006年公司法》 第413條之規定，披露滙豐控股之附屬公司於2014年內與董 事訂立之貸款 （貸款及類似貸款） 、信貸及擔保詳情： 2014年 2013年 百萬美元 百萬美元 於12月31日之貸款及信貸 5 7 下文根據IAS 24之規定，披露與關連人士之交易詳情。下表披露之年底結欠及年度最高結欠 額，均被視為最能反映年度交易金額及未償還結欠額的資料。 2014年 2013年 於12月31日 本年度最高 於12月31日 本年度最高 結欠額 結欠額 結欠額 結欠額 百萬美元 百萬美元 百萬美元 百萬美元 主要管理人員1 貸款及信貸 194 227 146 171 擔保 － － － 8 1 包括主要管理人員、主要管理人員的近親，及由主要管理人員或彼等近親所控制或共同控制的企業。 按 《香港聯合交易所有限公司證券上市規則》 的定義，部分交易為關連交易，惟已獲豁免遵守 該等上市規則之任何披露規定。上述交易乃於日常業務中進行，其條款 （包括利率及抵押） 大致等同與類似背景之人士或 （如適用） 其他僱員進行的可資比較交易。該等交易並不涉及 一般還款風險以外之風險，亦不附帶其他不利條款。 主要管理人員之股權、認股權及其他證券 2014年 2013年 （千） （千） 根據僱員股份計劃持有可認購滙豐控股普通股之認股權數目 28 225 實益及非實益持有之滙豐控股普通股數目 17,533 14,704 實益及非實益持有之滙豐銀行2015年2.875厘票據數目 5 5 於12月31日 17,566 14,934 與滙豐其他關連人士之交易 聯營及合資公司 集團為聯營及合資公司提供若干銀行及金融服務，包括貸款、透支、附息及不附息存款及 往來戶口服務。於聯營及合資公司之權益詳情載於附註20。本年度與聯營及合資公司之交 財務報表附註 （續） 滙豐控股有限公司 456 易及金額載列如下： 2014年 2013年 本年度 於12月31日 本年度 於12月31日 最高金額1 金額1 最高金額1 金額1 百萬美元 百萬美元 百萬美元 百萬美元 應收合資公司款項： －後償 － － 1 － －非後償 205 205 300 300 應收聯營公司款項： －後償 58 － － － －非後償 5,451 4,273 4,884 4,084 於12月31日 5,714 4,478 5,185 4,384 應付合資公司款項 － － 7 7 應付聯營公司款項 650 162 1,178 290 於12月31日 650 162 1,185 297 承諾 17 － 70 17 1 披露之年底金額及年度最高金額，被視為最能反映年內交易情況的資料。 上述未結算金額乃於日常業務中產生，而有關條款 （包括利率及抵押） 大致等同與第三方交 易對手進行的可資比較交易。 離職後福利計劃 於2014年12月31日，有45億美元 （2013年：52億美元） 的滙豐離職後福利計劃資產由滙豐旗下公 司管理。滙豐旗下公司就提供予離職後福利計劃的管理服務賺取之費用為1,200萬美元 （2013年： 2,300萬美元） 。滙豐各項離職後福利計劃有2.23億美元 （2013年：6.2億美元） 存放於滙豐集團 內經營銀行業務之附屬公司，對該等計劃應付之利息為600萬美元 （2013年：100萬美元） 。 上述未結算金額乃於日常業務中產生，而有關條款 （包括利率及抵押） 大致等同與第三方交 易對手進行的可資比較交易。 英國滙豐銀行 （英國） 退休金計劃與滙豐訂立掉期交易，作為管理其負債對通脹及利率敏感 度之部分方法。於2014年12月31日，掉期的名義總值為240億美元 （2013年：380億美元） ，對是 項計劃而言該等掉期的正數公允值為9億美元 （2013年：正數公允值28億美元） ，滙豐已就此 等掉期向該計劃交付價值20億美元 （2013年：38億美元） 的抵押品，滙豐因該等掉期而賺取之 利息為500萬美元 （2013年：3,300萬美元） 。所有掉期均按當前的市場利率並在標準市場買賣 價差範圍內執行交易。年內，該計劃降低了其與滙豐之間掉期交易的水平。 國際僱員退休福利計劃與滙豐訂立掉期交易，以管理其負債及選定資產對通脹及利率的敏 感度。於2014年12月31日，掉期之名義總值為19億美元 （2013年：18億美元） ，對該計劃而言掉 期的負數公允值淨額為1.07億美元 （2013年：正數3.99億美元） 。所有掉期均按當前的市場利 率並在標準市場買賣價差範圍內執行交易。 41－關連人士交易╱42－結算日後事項 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 457 滙豐控股 滙豐控股之主要附屬公司詳情載於附註22。本年度與附屬公司進行之交易及所涉金額如下： 2014年 2013年 本年度 於12月31日 本年度 於12月31日 最高金額1 金額1 最高金額1 金額1 百萬美元 百萬美元 百萬美元 百萬美元 資產 銀行現金 436 249 420 407 衍生工具 3,179 2,771 3,768 2,789 貸款 55,026 43,910 53,344 53,344 金融投資 4,073 4,073 1,220 1,210 於附屬公司之投資 96,264 96,264 92,695 92,695 於12月31日之關連人士資產總值 158,978 147,267 151,447 150,445 負債 應付滙豐旗下業務之款項 12,046 2,892 12,856 11,685 衍生工具 1,169 1,169 1,154 704 後償負債： －按已攤銷成本 1,743 1,670 1,716 1,716 －指定以公允值列賬 3,186 981 4,350 3,161 於12月31日之關連人士負債總額 18,144 6,712 20,076 17,266 擔保 53,180 52,023 52,836 52,836 承諾 1,245 16 1,245 1,245 1 披露之年底金額及年度最高金額，被視為最能反映年內交易情況的資料。上述未結算金額乃於日常業務 中產生，其條款 （包括利率及抵押） 大致等同與第三方交易對手進行的可資比較交易。 滙豐控股部分僱員為英國滙豐銀行 （英國） 退休金計劃成員，此計劃由集團旗下另一公司資 助。滙豐控股為此等僱員承擔的金額，相等於代表彼等向計劃作出之供款。有關計劃之披 露資料載於賬目附註6。 42 結算日後事項 2014年12月31日後，董事會宣布派發2014年度第四次股息每股普通股0.2美元 （派息額約為 38.44億美元） 。 此等賬目已於2015年2月23日經董事會通過並授權公布。 股東參考資料 滙豐控股有限公司 458 2014年第四次股息 董事會已宣布派發2014年第四次股息每股普通股0.20美元。有關以股代息計劃及股東收取現 金股息的貨幣選擇等資料，將於2015年3月20日或該日前後寄予各股東。派息時間表如下： 宣布日期 2015年2月23日 美國預託股份在紐約除息報價 2015年3月4日 股份在倫敦、香港、巴黎及百慕達除息報價 2015年3月5日 倫敦、香港、紐約、巴黎及百慕達紀錄日期1 2015年3月6日 郵寄 《2014年報及賬目》 及╱或 《2014年策略報告》 、股東周年大會通告及股息文件 2015年3月20日 股份登記處接收以股代息選擇表格、投資者中心電子指示 及撤銷以股代息常設指示的最後日期 2015年4月16日 就派發英鎊及港元股息釐定匯率 2015年4月20日 派發日期：寄發股息單、新股票或交易通知書及名義稅單， 並將股份存入CREST股票戶口 2015年4月30日 1 任何人士不得於此日在香港海外股東分冊登記或取消已登記股份。 2015年各次股息 董事會已採納按季派發普通股股息的政策。根據此政策，我們擬派發三次金額相等的股息， 而第四次股息的金額將會不同。預計2015年第一次股息將為每股普通股0.10美元。 股息均以美元為單位宣派，股東可選擇以美元、英鎊或港元或該三種貨幣之組合收取現金 股息，或倘董事會決定就該股息派發代息股份，以發行之新股代替全部或部分現金股息。 股東資料 於2014年12月31日，股東名冊載列以下詳情： 股東數目 所持普通股總數 1股至100股 37,254 1,086,273 101股至400股 28,970 7,109,335 401股至500股 7,269 3,294,206 501股至1,000股 30,675 22,589,447 1,001股至5,000股 70,515 166,806,864 5,001股至10,000股 18,455 130,271,174 10,001股至20,000股 10,964 152,651,126 20,001股至50,000股 6,616 203,096,923 50,001股至200,000股 3,229 296,665,889 200,001股至500,000股 713 223,855,824 500,001股及以上 1,092 18,010,447,199 總計 215,752 19,217,874,260 S 股東參考資料 2014年第四次股息 458 2015年各次股息 458 股東資料 458 2014年股東周年大會 459 經營狀況參考聲明及中期業績 459 股東查詢及通訊 460 股份代號 461 投資者關係 461 有關滙豐的其他資料 461 簡明架構圖 462 股份及股息之稅務事宜 463 簡稱 466 詞彙及索引 470 各次股息╱股東資料╱2014年股東周年大會 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 459 2014年股東周年大會 於2014年5月23日上午11時正在The Barbican Centre, London EC2舉行之2014年股東周年大會上所 有獲考慮之決議案以投票方式通過如下： 總票數 決議案 贊成1 % 反對 % 總數 %2 棄權3 1 接納 《2013年報及賬目》 9,757,667,883 98.77 121,991,215 1.23 9,879,659,098 51.81 67,631,467 2 通過董事薪酬政策 7,762,051,505 79.35 2,019,902,686 20.65 9,781,954,191 51.29 167,509,544 3 通過董事薪酬報告 8,180,579,271 83.95 1,563,541,883 16.05 9,744,121,154 51.10 205,528,859 4 通過浮動酬勞上限 9,722,737,304 98.01 197,867,631 1.99 9,920,604,935 52.06 29,824,365 5 選舉或重選下列人士為董事： (a) 祈嘉蓮 9,919,351,912 99.90 9,703,995 0.10 9,929,055,907 52.07 22,579,886 (b) 凱芝 9,907,586,080 99.79 21,027,161 0.21 9,928,613,241 52.06 22,549,669 (c) 史美倫 9,780,982,617 99.65 34,767,318 0.35 9,815,749,935 51.47 134,402,361 (d) 張建東 9,896,844,476 99.68 32,093,388 0.32 9,928,937,864 52.07 22,444,096 (e) 埃文斯爵士 （現為埃文斯勳爵） 9,919,562,434 99.90 9,445,023 0.10 9,929,007,457 52.07 22,647,584 (f) 費卓成 9,777,856,091 99.57 41,751,739 0.43 9,819,607,830 51.49 130,404,838 (g) 方安蘭 9,850,526,539 99.23 75,986,608 0.77 9,926,513,147 52.05 23,410,104 (h) 范樂濤 8,877,803,961 89.45 1,047,345,520 10.55 9,925,149,481 52.05 26,491,483 (i) 范智廉 9,648,640,713 97.44 253,257,525 2.56 9,901,898,238 51.92 49,764,237 (j) 歐智華 9,901,917,449 99.73 27,013,216 0.27 9,928,930,665 52.07 22,318,409 (k) 李德麟 9,905,841,880 99.76 23,398,827 0.24 9,929,240,707 52.07 22,425,383 (l) 利普斯基 9,916,726,269 99.88 11,952,583 0.12 9,928,678,852 52.06 22,478,974 (m) 駱美思 9,907,427,658 99.78 21,476,877 0.22 9,928,904,535 52.07 22,307,009 (n) 麥榮恩 9,896,296,230 99.67 32,591,433 0.33 9,928,887,663 52.06 22,328,752 (o) 繆思成 9,903,345,079 99.76 24,023,003 0.24 9,927,368,082 52.06 22,605,039 (p) 駱耀文爵士 8,879,523,428 89.43 1,049,218,853 10.57 9,928,742,281 52.06 22,459,665 (q) 施俊仁 9,909,069,564 99.80 19,845,278 0.20 9,928,914,842 52.07 22,747,741 6 重新委聘KPMG Audit Plc為 本公司核數師 9,858,281,428 99.29 70,691,207 0.71 9,928,972,635 52.07 22,181,297 7 授權集團監察委員會釐定 核數師費用 9,899,311,128 99.70 30,047,667 0.30 9,929,358,795 52.07 22,178,082 8 授權董事配發股份 8,960,671,117 90.89 898,368,702 9.11 9,859,039,819 51.70 92,477,179 9 暫不運用優先配股權 9,782,952,816 98.97 101,914,263 1.03 9,884,867,079 51.83 66,129,918 10 授權董事配發回購股份 9,397,626,368 95.18 475,964,720 4.82 9,873,591,088 51.77 77,459,228 11 授權公司購買其本身普通股 9,785,002,326 99.58 41,076,933 0.42 9,826,079,259 51.53 123,932,000 12 授權董事就或有可轉換證券 配發股權證券 9,558,599,010 97.49 246,293,361 2.51 9,804,892,371 51.41 142,376,638 13 暫不運用有關發行或有可轉換 證券的優先配股權 8,915,406,730 89.85 1,007,452,174 10.15 9,922,858,904 52.03 25,542,467 14 批准股東大會 （不包括股東 周年大會） 可在發出至少14 整日之通知後召開 8,798,744,951 88.64 1,127,707,377 11.36 9,926,452,328 52.05 23,437,179 1 包括全權代表之投票。 2 佔已投票之已發行股本百分比。 3 「棄權」 在法律上並不當作 「投票」 ，故在計算投票 「贊成」 及 「反對」 決議案的票數時不予點算。 經營狀況參考聲明及中期業績 經營狀況參考聲明預期將於2015年5月5日及2015年11月2日或前後發表。截至2015年6月30日止 六個月之中期業績預期將於2015年8月3日公布。 股東參考資料 （續） 滙豐控股有限公司 460 股東查詢及通訊 查詢 有關股東名冊所載持股事宜之任何查詢，例如：股份轉讓、轉名、更改地址、報失股票或股 息支票等事項，請致函下列地址之股份登記處。股份登記處提供的 「投資者中心網頁」 為一 項網上服務，讓股東可用電子方式管理其股份。 主要股東名冊： Computershare Investor Services PLC The Pavilions Bridgwater Road Bristol BS99 6ZZ United Kingdom 電話：44 (0) 870 702 0137 透過網站發出電郵： www.investorcentre.co.uk/contactus 投資者中心： www.investorcentre.co.uk 香港海外股東分冊： 香港中央證券登記有限公司 香港 皇后大道東183號 合和中心17樓 1712-1716室 電話：852 2862 8555 電郵： hsbc.ecom@computershare.com.hk 投資者中心： www.investorcentre.com/hk 百慕達海外股東分冊： Investors Relations Team HSBC Bank Bermuda Limited 6 Front Street Hamilton HM 11 Bermuda 電話：1 441 299 6737 電郵： hbbm.shareholder.services@hsbc.bm 投資者中心： www.investorcentre.com/bm 有關美國預託股份之任何查詢，請致函下列存管處： The Bank of New York Mellon Depositary Receipts PO Box 43006 Providence, RI 02940-3006 USA 電話 （美國） ：1 877 283 5786 電話 （國際） ：1 201 680 6825 電郵：shrrelations@bnymellon.com 網站：www.bnymellon.com/shareowner 有關透過法國Euroclear （NYSE Euronext巴黎之結算及中央存託系統） 所持股份之任何查詢，請 致函下列付款代理： HSBC France 103, avenue des Champs Elysées 75419 Paris Cedex 08 France 電話：33 1 40 70 22 56 電郵：ost-agence-des-titres-hsbc-reims.hbfr-do@hsbc.fr 網站：www.hsbc.fr 如閣下獲提名直接從滙豐控股收取一般股東通訊，務請注意閣下就有關處理閣下投資之一 切事宜之主要聯絡人仍為登記股東或代表閣下管理投資之託管商或經紀。因此，有關閣下 個人資料及持股量 （包括任何相關管理事宜） 之任何變更或查詢，必須繼續交由閣下現時之 投資經理或託管商聯絡人處理。滙豐控股不保證會代為處理誤送至本公司之資料。 如需索取 《2014年報及賬目》 印刷本，請致函下列部門： 歐洲、中東及非洲： 亞太區： 美洲： Global Communications 香港皇后大道中1號 Global Publishing Services HSBC Holdings plc 香港上海滙豐銀行有限公司 HSBC - North America 8 Canada Square 企業傳訊部 （亞太區） SC1 Level, 452 Fifth Avenue London E14 5HQ New York, NY 10018 United Kingdom USA 股東查詢╱股份代號╱投資者關係╱有關滙豐的資料 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 461 電子通訊 股東可隨時選擇收取公司通訊的印刷本或收取有關公司通訊已上載至滙豐網站的通知。如 欲以電郵方式收取日後有關公司通訊已上載至滙豐網站的通知，或撤銷或更改以電郵方式 收取該等通知的指示，請登入www.hsbc.com/ecomms。若閣下提供電郵地址以收取滙豐發 出的電子通訊，本公司亦會以電郵通知閣下應得的股息。若閣下收到本文件已上載至滙豐 網站的通知，而欲取閱本文件的印刷本，或欲於日後收取公司通訊的印刷本，請致函或電 郵 （註明股東參考編號） 至適當的股份登記處 （地址見上文） 。印刷本將免費供應。 中譯本 《2014年報及賬目》 備有中譯本，可於2015年3月20日之後向下列股份登記處索閱。 香港中央證券登記有限公司 香港皇后大道東183號 合和中心17樓 1712-1716室 Computershare Investor Services PLC The Pavilions Bridgwater Road Bristol BS99 6ZZ United Kingdom 閣下如欲於日後收取相關文件的中譯本，或已收到本文件的中譯本但不希望繼續收取有關 譯本，均請聯絡股份登記處。 股份代號 滙豐控股普通股以下列股份代號買賣： 倫敦證券交易所 HSBA Euronext巴黎 HSB 香港聯合交易所 5 百慕達證券交易所 HSBC.BH 紐約證券交易所 （美國預託股份） HSBC 投資者關係 如對滙豐的策略或業務有任何查詢，請聯絡： Senior Manager Investor Relations SVP Investor Relations 香港皇后大道中1號 HSBC Holdings plc HSBC North America Holdings Inc. 香港上海滙豐銀行有限公司 8 Canada Square 26525 N Riverwoods Boulevard, Suite 100 亞洲投資者關係主管 London E14 5HQ Mettawa, Illinois 60045 United Kingdom USA 電話：44 (0) 20 7991 3643 1 224 880 8008 852 2822 4908 電郵：investorrelations@hsbc.com investor.relations.usa@us.hsbc.com investorrelations@hsbc.com.hk 有關滙豐的其他資料 如欲瀏覽英文版之 《2014年報及賬目》 ，以及滙豐其他資料，請登入滙豐網站www.hsbc.com。 滙豐控股呈交美國證交會的報告、聲明及資料，均上載至www.sec.gov。投資者亦可致函 美國證交會 （地址：Office of Investor Education and Advocacy, 100 F Street N.E., Washington, DC 20549-0123） 或電郵至PublicInfo@sec.gov索取上述文件的印刷副本，但須支付指定金額 的複印費。投資者如需要進一步協助，可致電美國證交會 （電話：(202) 551 8090） 。投資者 亦可於www.nyse.com （電話：(1) 212 656 3000） 獲取滙豐控股存檔的報告及其他資料。 英國財政部已將資本指引4的規定納入規例，並據此頒布 《2013年資本規定 （按國家） 匯報規 例》 （自2014年1月1日起生效） 。根據該法例，滙豐控股須於2015年7月1日前就截至2014年12月 31日止年度公開額外資料。這些資料屆時將可於滙豐網站www.hsbc.com瀏覽。 股東參考資料 滙豐控股有限公司 462 組織架構圖╱股份及股息之稅務事宜 99.99% 99.99% 99.99% 滙豐控股有限公司簡明架構圖 80.65% 94.53% 62.14% HSBC Asia Holdings (UK) Ltd HSBC Holdings BV HSBC Finance HSBC Private Banking Holdings (Suisse) S.A.
(Netherlands) 巴西滙豐銀行 HSBC Latin America Holdings (UK) Limited HSBC Overseas Holdings (UK) Ltd HSBC Investments (North America) Inc.
HSBC Germany Holdings GmbH HSBC Trinkaus & Burkhardt AG HSBC Asset Finance (UK) Ltd HSBC Bank A.S.
HSBC Securities (USA) Inc.
HSBC México S.A., Institución de Banca Múltiple, Grupo Financiero HSBC 此乃於2014年12月31日集團主要附屬公司的簡明架構圖。 HSBC Asia Holdings BV HSBC Middle East Holdings BV 香港上海滙豐銀行 有限公司 美國滙豐融資 有限公司 美國滙豐銀行 埃及滙豐銀行 北美滙豐控股 有限公司 加拿大滙豐銀行 滙豐控股有限公司 英國滙豐銀行有限公司 法國滙豐 除非另有註明，否則所有附屬公司均屬全資擁有。 本架構圖並未詳列所有中介控股公司。 阿根廷滙豐銀行 中東滙豐銀行 有限公司 滙豐銀行 （中國） 有限公司 恒生銀行有限公司 馬來西亞 滙豐銀行有限公司 澳洲滙豐銀行 有限公司 滙豐人壽保險 （國際） 有限公司 滙豐 （台灣） 商業 銀行股份有限公司 越南滙豐銀行 有限公司 ‧ ‧ ‧ 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 463 股份及股息之稅務事宜 稅務－英國居民 以下摘要概述在現行法例下可能對擁有及 出售滙豐控股普通股有重大影響的英國 稅務之若干考慮因素。本概要無意全面縷 述所有可能涉及股份持有人的稅務考慮因 素。具體而言，本概要主要涉及英國稅務 所指僅於英國居住的股東，並且亦僅限於 持有股份作為投資用途及實益擁有股份的 人士，而並無討論若干類別持有人 （例如證 券交易商） 的稅務處理問題。股份持有人及 可能購買股份的人士應就本身具體情況， 徵詢其顧問有關投資股份的稅務影響 （包 括任何國家、州份或地方法例的影響） 。 股息之稅務事宜 現時滙豐控股派發之股息並無預扣任何稅 項，惟所支付股息附帶之稅項減免額，可用 作抵銷若干股東或有的英國所得稅責任。 現時，該附帶之稅項減免額相等於現金股 息加稅項減免額總和的10% （即現金股息的 九分之一） 。 個人股東如在稅務方面屬於在英國居住， 並須按基本稅率繳納英國所得稅，他在收 取滙豐控股的股息時將毋須另行繳納英國 所得稅。個人股東如須按較高稅率或附加 稅率課稅，則應就其股息加稅項減免額的 總和分別按較高之股息稅率 （目前為32.5%） 及股息附加稅率 （目前為37.5%） 繳納英國所 得稅。該稅項減免額可用於對銷股息較高 稅率及股息附加稅率的責任。屬於在英國 居住的個人股東並不享有任何稅項減免額 退款。 非於英國居住的股東一般不會就所收取的 任何英國股息享有稅項減免額退款，但此 等股東之中，部分人士或可獲退款，條件 為此等人士居住的國家╱地區必須與英國 訂有避免雙重徵稅協議。但事實上，在大 多數情況下，稅項減免額均不會予以退還。 有關取代2013年第四次股息及2014年第一 次、第二次及第三次股息的滙豐控股以股 代息計劃稅務影響的資料，已列載於2014年 3月25日、6月5日、9月3日及11月5日之公司 秘書長致股東函件。在所有情況下所放棄 之現金股息與代息股份市值之間的差額， 概不超過市值的15%，因此，就英國資本增 值稅而言，須繳納稅項的股息收益金額及 滙豐控股普通股的購買價，為已放棄收取 的現金股息。 資本增值稅 須繳納英國資本增值稅的股東於出售滙豐 控股股份時，將會產生資本增值稅責任，而 有關稅項之計算方式頗為複雜，部分視乎 （舉例而言） 股份是否自1991年4月起購入； 或於1991年以香港上海滙豐銀行有限公司 股份換取；又或於1991年後以其他公司的 股份換取。 就資本增值稅而言，普通股的購入成本會 作出調整，以計入其後的供股及資本化發 行。英國公司因出售股份而產生之任何資 本增值亦可予調整，以計及按指數調整的 稅項寬減。股東如有疑問，請諮詢有關的 專業顧問。 印花稅及印花儲備稅 以轉讓書進行的股份轉讓一般須繳交英國 印花稅，稅率為轉讓代價的0.5%，印花稅 一般由承讓人支付。 轉讓股份或其中任何權益的協議，一般須 繳交相等於代價之0.5%的印花儲備稅。但根 據現時英國稅務海關總署的慣例，只要轉 讓股份的文書是根據協議而簽訂，並於須 支付印花儲備稅日期之前正式加蓋印花， 則毋須支付印花儲備稅，亦不必申請取消 此稅項。印花儲備稅一般由承讓人支付。 在英國無紙股份交易系統CREST內進行的 無紙股份轉讓，須繳付代價之0.5%的印花 儲備稅。在CREST交易中，稅款的計算和支 付均自動操作。存入CREST的股份一般毋 須支付印花儲備稅，除非轉入CREST的轉 讓本身是有代價的交易。繼滙豐提交歐洲 法院審理的案件 （案件編號 C-569/07 滙豐 控股有限公司及Vidacos Nominees Ltd訴英 國稅務海關總署署長） 及其後一宗有關預 託證券的案件審理後，英國稅務海關總署 現在接納禁止向預託證券發行人或結算服 務公司發行股份收取1.5%之印花儲備稅。 稅務－美國居民 以下摘要概述在現行法例下英國稅務及美 國聯邦所得稅之主要稅務考慮因素，而就 身為美國居民 （就美國聯邦所得稅而言） 之 持有人 （ 「美國持有人」 ） 而言 （彼等就英國稅 股東參考資料 （續） 滙豐控股有限公司 464 稅務 務而言並非於英國居住） ，該等稅務考慮因 素可能對其擁有及出售股份或美國預託股 份 （ 「美國預託股份」 ） 有重大影響。 本概要無意全面縷述所有可能涉及股份或 美國預託股份持有人的稅務考慮因素。具 體而言，本概要僅涉及持有股份或美國預 託股份作資本資產之美國持有人，而並無 討論須遵守特別稅務規則之持有人之稅務 處理問題，例如銀行、免稅企業、保險公 司、證券或貨幣交易商、持有股份或美國 預託股份作為一項綜合投資 （包括 「馬鞍式 組合」 ，而該項投資是由一股股份或美國預 託股份及一個或多個其他持倉組成） 之一部 分的人士，以及直接或間接擁有滙豐控股 有投票權股份10%或以上之人士。本文的 討論乃以本文件日期生效之法律、條約、 司法判決及監管詮釋為依據，而所有此等 法律、條約、司法判決及監管詮釋或會變 更。 股份或美國預託股份之持有人及可能購買 股份或美國預託股份之人士應就本身具體 情況，徵詢其顧問有關投資股份或美國預 託股份之稅務影響 （包括任何國家、州份或 地方法例之影響） 。 本 《年報及賬目》 所載任何美國聯邦稅務意 見僅供參考，並非旨在或編製以供用作逃 避美國聯邦稅務罰則之用途，亦不得用作 前述用途。 股息之稅務事宜 現時由滙豐控股支付之股息毋須預扣稅 項。就美國稅務而言，美國持有人必須於 其本人或美國預託股份存管機構收取現金 股息當日，將所得股份或美國預託股份之 現金股息，加入一般收益之內，並且按照 收取股息當日的匯率將以英鎊支付的股息 換算為美元。選擇收取股份代替現金股息 的美國持有人，必須於一般收益內計入該 等股份於派發股息日的公允市值，而該等 股份的稅基將相等於該公允市值。 除持倉時間少於61日或持倉已被對沖的若 干例外情況，倘若外國公司被視為 「合資 格外國公司」 （包括就美國聯邦所得稅而言 不被分類為被動海外投資公司的公司） ， 美國個人持有人收取若干股息 （ 「合資格股 息」 ） 通常按優惠稅率繳納美國稅項。根據 公司之經審核財務報表及相關市場及股東 數據，滙豐控股預期不會被歸類為被動海 外投資公司。因此，就股份或美國預託股 份派付之股息，一般應視為合資格股息。 資本增值稅 美國持有人因出售或以其他方式處理股份 或美國預託股份而變現之利潤，通常毋須 繳付英國稅項，除非當時持有人在英國經由 分公司或代理機構或常設機構經營業務、 從事某項專業或工作，以及目前或曾經就 該業務、專業或工作、分公司或代理機構 或常設機構而使用、持有或收購該等股份 或美國預託股份，則另作別論。就美國稅 務而言，該等利潤將計入收益項內，倘若 持有股份或美國預託股份超過一年，則列 為長期資本增值。美國個人持有人所變現 之長期資本增值，一般須按優惠稅率繳付 美國稅項。 遺產稅 就有關遺產及饋贈稅的 《美國英國雙重課 稅條約》 （ 「遺產稅條約」 ） 而言，凡持有股份 或美國預託股份且被認定為美國居民，以 及就此而言並非英國國民的個別人士，倘 若已支付任何應繳納的美國聯邦遺產或饋 贈稅，則毋須因該名人士身故，或於該名 人士在生之時轉讓股份或美國預託股份而 繳納英國遺產稅，但在以下若干情況則另 作別論：倘股份或美國預託股份(i)構成一 項授產安排 （除非於授予該項財產時，有關 財產授予人為美國居民及並非英國國民） ； (ii)是一間企業在英國的常設機構的部分業 務財產；或(iii)與個別人士用作提供獨立個 人服務的英國固定地點有關。在上述情況 下，倘若股份或美國預託股份須同時繳納 英國遺產稅及美國聯邦遺產或饋贈稅，則 遺產稅條約一般會以於英國已付之任何稅 項金額，就美國聯邦稅務責任提供減免。 印花稅及印花儲備稅 － 美國預託股份 倘若股份轉讓至結算服務公司或美國預託 證券 （ 「美國預託證券」 ） 發行人 （包括將股份 轉讓至存管處） ，按照英國稅務海關總署目 前的慣例，將須繳付英國印花稅及╱或印 花儲備稅。印花稅或印花儲備稅一般按轉 讓代價繳付，總計稅率為1.5%。 就有關轉讓應付的印花儲備稅，將會扣減 就同一項轉讓交易已支付的印花稅。 轉讓或協議轉讓美國預託股份均毋須繳付 印花稅，條件是有關美國預託證券及任何 獨立轉讓文件或轉讓協議書均須時刻存放 於英國以外地區，而且任何該等轉讓文件 或轉讓協議書均不得於英國簽立。藉轉讓 美國預託證券方式以達致轉讓或協議轉讓 美國預託股份，一概毋須繳付印花儲備稅。 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 465 美國備用預扣稅及資料申報 於美國境內，又或透過若干金融中介機 構，向美國持有人支付之股份或美國預託 股份分派及出售股份或美國預託股份所得 款項，均須申報資料，並可能須繳付美國 「備用」 預扣稅，但一般而言，倘美國持有 人遵從若干核證程序或屬於獲豁免該項預 扣稅之公司或其他人士，則另作別論。並 非美國人士之持有人一般毋須申報資料或 繳付備用預扣稅，不過可能需要遵從適用 之核證程序，以證明他們並非美國人士， 從而避免因在美國境內或透過若干金融中 介機構收取付款而須作出該等資料申報或 繳付備用預扣稅。 股東參考資料 （續） 滙豐控股有限公司 466 簡稱 所用簡稱 簡要說明 A 澳元 澳元 ABCP 資產抵押商業票據 ABS 1 資產抵押證券 ACF 貸款對核心資金比率 ADR 美國預託證券 ADS 美國預託股份 迪拉姆 阿拉伯聯合酋長國迪拉姆 附息資產平均值 附息資產平均值 ALCM 資產、負債及資本管理部 ALCO 資產負債管理委員會 反洗錢 反洗黑錢 可調利率按揭 1 可調利率按揭 阿根廷披索 阿根廷披索 B 巴塞爾委員會 巴塞爾銀行監管委員會 巴塞爾協定2 1 《2006年巴塞爾資本協定》 巴塞爾協定3 1 巴塞爾委員會為加強環球資本及流動資金規則而推行的改革 英國銀行家協會 英國銀行家協會 交通銀行 全名 「交通銀行股份有限公司」 ，中國規模最大的銀行之一 基點 1 一個基點等於百分之一個百分點 銀行復元和解決指引 1 歐盟 《銀行復元和解決指引》 巴西雷亞爾 巴西雷亞爾 銀行保密法 美國 《銀行保密法》 BSM 資產負債管理業務 C 加元 加拿大元 資本資產定價模型 資本資產定價模型 消費者信貸法 英國 《消費者信貸法》 CCB 1 防護緩衝資本 CCR 1 交易對手信貸風險 CCyB 1 反周期緩衝資本 CD 存款證 CDO 1 債務抵押債券 CDS 1 信貸違責掉期 CET1 1 普通股權一級 創現單位 創現單位 瑞士法郎 瑞士法郎 工商金融 工商金融業務，為一項環球業務 消費及按揭貸款 1 美國消費及按揭貸款 人民幣 人民幣 COSO 美國反虛假財務報告委員會發起人委員會 CP 1 商業票據 CPB 1 資本計劃緩衝 消費物價指數 消費物價指數 資本指引 1 《資本規定指引》 CRR 1 客戶風險評級 CRR╱資本指引4 《資本規定條例及指引》 CRS 卡及零售商戶業務 CVA 1 信貸估值調整 D 紐約郡地區檢察官延後起訴協議 與美國紐約郡地區檢察官訂立的兩年期延後起訴協議 達德－法蘭克法案 美國 《達德－法蘭克華爾街改革及消費者保障法案》 (Dodd-Frank Wall Street Reform and Consumer Protection Act) 司法部 美國司法部 DPA 美國延後起訴協議 DPF 保單及投資合約的酌情參與條款 借記估值調整 1 借記估值調整 簡稱 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 467 所用簡稱 簡要說明 E EAD 1 違責風險承擔 EBA 歐洲銀行管理局 歐洲央行 歐洲中央銀行 EDTF 強化信息披露工作組 埃及鎊 埃及鎊 EL 1 預期虧損 歐洲市場基礎設施規例 歐盟 《歐洲市場基礎設施規例》 歐盟 歐洲聯盟 Euribor 歐元銀行同業拆息 F 房利美 美國聯邦國民抵押協會 FCA 英國金融業操守監管局 金融業操守監管局指令 原先向英國金管局承諾遵守有關反洗錢及制裁規定的若干前 瞻性責任 資金公允值調整 衍生工具合約的資金公允值調整估算方法 First Direct電話理財 英國滙豐銀行有限公司的一個部門 FPC 英國金融政策委員會 聯儲局 美國聯邦儲備局 房貸美 美國聯邦住宅貸款抵押公司 英國金管局 英國金融服務管理局 金融穩定委員會 金融穩定委員會 FSC Forest Stewardship Council 金融系統風險防護委員會 金融系統風險防護委員會 等同全職僱員 等同全職僱員 富時指數 富時股票指數 管理資金 管理資金 G 公認會計原則 公認會計原則 集團監察委員會 集團監察委員會 環球銀行及資本市場 環球銀行及資本市場業務，為一項環球業務 國內╱區內╱本地生產總值 國內╱區內╱本地生產總值 審慎監管局 《一般審慎措施資料手冊》 英國審慎監管局制訂的規則，載於 《一般審慎措施資料手冊》 GLBA 美國 《金融服務現代化法案》 集團管理委員會 集團管理委員會 環球私人銀行 環球私人銀行業務，為一項環球業務 集團業績表現股份計劃 集團業績表現股份計劃 集團風險管理委員會 集團風險管理委員會 集團 滙豐控股連同其附屬業務 全球系統重要性銀行1 全球系統重要性銀行 全球系統重要性機構 全球系統重要性機構 H 恒生銀行 全名恒生銀行有限公司，香港規模最大的銀行之一 港元 港元 北美滙豐 全名北美滙豐控股有限公司 香港 中華人民共和國香港特別行政區 滙豐 滙豐控股連同其附屬業務 英國滙豐銀行 全名HSBC Bank plc （英國滙豐銀行有限公司） 中東滙豐銀行 全名HSBC Bank Middle East Limited （中東滙豐銀行有限公司） 美國滙豐銀行 滙豐在美國的零售銀行HSBC Bank USA, N.A.
加拿大滙豐 由加拿大滙豐銀行、HSBC Trust Company Canada、HSBC Mortgage Corporation Canada及HSBC Securities Canada為流 動資金目的而合併組成的子集團 美國滙豐融資 全名HSBC Finance Corporation （美國滙豐融資有限公司） ，一家 美國消費融資公司 （前稱Household International, Inc.） 法國滙豐 滙豐在法國經營銀行業務的附屬公司，前稱CCF S.A. （法國商業 銀行） 滙豐控股 全名滙豐控股有限公司，滙豐的母公司 股東參考資料 （續） 滙豐控股有限公司 468 所用簡稱 簡要說明 滙豐卓越理財 滙豐的卓越個人環球銀行服務 滙豐私人銀行 （瑞士） 全名HSBC Private Bank (Suisse) SA （滙豐私人銀行 （瑞士） 有限 公司） ，是滙豐在瑞士之私人銀行 美國滙豐 由美國滙豐有限公司 （美國滙豐銀行的控股公司） 及美國滙豐銀行為 流動資金目的而合併組成的子集團 HSI HSBC Securities (USA) Inc.
HSSL HSBC Securities Services (Luxembourg) I IAS 《國際會計準則》 IASB 國際會計準則委員會 ICB 英國銀行業獨立委員會 IFRS 《國際財務報告準則》 興業銀行 全名興業銀行股份有限公司，是中國內地一家全國性股份制 銀行，恒生銀行有限公司持有其股份 IRB 1 內部評級基準 ISDA 國際掉期業務及衍生投資工具協會 K KPMG KPMG Audit Plc及其聯屬機構 L LCR 流動資金覆蓋比率 LFRF 流動資金及資金風險管理架構 LGD 1 違責損失率 Libor 倫敦銀行同業拆息 LIC 貸款減值及其他信貸風險準備 貸款管理組 貸款管理組，屬批發信貸及市場風險管理部的一部分 LTV 1 貸款估值比率 M 馬多夫證券 全名Bernard L Madoff Investment Securities LLC 中國內地 中華人民共和國，不包括香港 資本市場 滙豐的環球銀行及資本市場業務屬下的財資及資本市場業務 Mazarin 全名Mazarin Funding Limited，一家資產抵押商業票據中介機 構 MBS 美國按揭抵押證券 中東及北非 中東及北非 中型市場企業 中型市場企業 債券承保公司 1 債券承保公司 摩根士丹利資本國際 摩根士丹利資本國際指數 中期票據 中期票據 墨西哥披索 墨西哥披索 N 淨利息收益 淨利息收益 NSFR 穩定資金淨額比率 紐約證交所 紐約證券交易所 O OCC 美國貨幣監理署 隔夜指數掉期 隔夜指數掉期 營運風險管理架構 營運風險管理架構 場外 1 場外 P PD 1 違責或然率 PEFC Programme for the Endorsement of Forest Certification 業績表現股份 1 根據企業的業績表現狀況在僱員股份計劃下發放滙豐控股普 通股作為獎勵 平安保險 全名中國平安保險 （集團） 股份有限公司，是中國內地第二大 壽險公司 還款保障保險 還款保障保險產品 審慎監管局 英國審慎監管局 簡稱 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 469 所用簡稱 簡要說明 中國 中華人民共和國 卓越理財 滙豐卓越理財，滙豐的卓越個人環球銀行服務 PVIF 有效長期保險業務及附有酌情參與條款之長期投資合約的現值 羅兵咸永道 PricewaterhouseCoopers LLP及其事務所網絡 Q QIS 定量影響研究 R 零售銀行及財富管理 零售銀行及財富管理業務，為一項環球業務 回購 1 出售及回購交易 有限制股份 有限制股份獎勵界定僱員 （通常於獎勵授出日期起計一至三年 內） 將有權享有的滙豐控股普通股股數 （屆時有關僱員一般須 仍然在職） 反向回購 根據出售承諾而購入的證券 風險管理會議 集團管理委員會屬下會議，監督所有風險的企業整體管理 馬元 馬來西亞令吉 人民幣 人民幣 RMBS 住宅按揭抵押證券 客戶經理 客戶經理 RNIV 估計虧損以外風險 風險加權資產平均值回報 風險加權資產平均值回報 有形股本回報 有形股本回報 RPI 零售價格指數 （英國） RSPO Roundtable on Sustainable Palm Oil RTS 監管技術準則 風險加權資產 1 風險加權資產 S SE 結構公司 美國證交會 美國證券交易委員會 SIC 證券投資中介機構 中小企 中小型企業 Solitaire 全名Solitaire Funding Limited，由滙豐管理的特設企業 SPE 1 特設企業 SRB 1 系統性風險緩衝 T 香港上海滙豐銀行 全名香港上海滙豐銀行有限公司，滙豐的始創成員 TLAC 1 整體損失吸納能力 土耳其里拉 土耳其里拉 TSA 過渡服務協議 （與出售美國的卡及零售商戶業務有關） TSR 股東總回報 U 阿聯酋 阿拉伯聯合酋長國 英國 聯合王國 美元 美元 美國 美利堅合眾國 美國延後起訴協議 與美國司法部及其他機構訂立的五年期延後起訴協議 美國縮減組合 包括我們於美國滙豐融資之消費及按揭貸款、汽車融資及納 稅人理財服務，以及保險、商業、企業及財資業務，按IFRS以 管理層意見為基準計算 V VaR 1 估計虧損風險 Visa 全名Visa Inc.
使用價值 使用價值 1 完整釋義載於第470頁詞彙一節。 股東參考資料 （續） 滙豐控股有限公司 470 詞彙 詞彙 釋義 A 可調利率按揭 （ 「ARM」 ） 利率按某一參考價格定期調整的美國按揭貸款，屬於 「負擔能力為本的 按揭」 。 負擔能力為本的按揭 採用可變動或固定利率，使客戶初期每月還款額按低息率計算，但在優 惠期完結後隨即再重訂為較高息率的按揭貸款產品。 機構風險承擔 接近或類似政府機構承擔的風險，包括由政府全面擁有以進行非商業 活動的公營單位、省政府及地方政府機構、發展銀行及政府成立的基 金。 Alt-A 在美國用以描述風險低於次優質貸款但風險特性高於根據正常標準發 放的貸款。 積欠 倘客戶未能如期履行其責任，以致產生一項未付或逾期的未償還貸款， 即屬積欠 （或處於拖欠狀態） 。倘客戶積欠還款，逾期的未償還貸款總 額則稱為拖欠。 資產抵押證券 （ 「ABS」 ） 代表於相關組別參考資產權益的證券。有關參考組別可包括任何帶來 相關現金流的資產，惟一般屬於住宅或商業按揭組別。 B 回溯測試 一種統計方法，用以監察及評估模型的準確度，以及該模型倘於過往已 採用理應有何表現。 自救債務 自救指企業在無法繼續經營時 （但在破產前） ，對若干銀行負債 （ 「自救債 務」 ） （該等負債在企業仍可持續經營時並不承受虧損風險） 強加損失。 不論以撇減或轉換為股權的方式進行，此舉可使銀行實現資本重組 （儘管不會提供任何新資金） 。 銀行徵費 自2011年1月1日起向英國銀行、建屋貸款社及外資銀行的英國業務收取 之徵費。應付金額乃按集團於12月31日之綜合負債及股東權益扣除若 干項目 （當中最重要的項目為與受保存款結餘、一級資本、保險未決 賠款、優質流動資產相關之項目，以及須受制於可依法強制執行的淨 額結算協議之項目） 後之某個百分比計算。 銀行復元和解決指引 由歐洲委員會頒布並由歐盟成員國採納的歐洲法例組合。該項指引於 2014年7月定稿，而其大部分條文於2015年1月1日生效。該項指引出 台關於政府部門如何介入以處理將會或可能會破產的銀行之歐盟共 同架構。相關架構包括用以防止破產及 （在銀行已破產情況下） 政府部 門用以確保有序解決的早期介入及措施。 巴塞爾協定2 巴塞爾銀行監管委員會於2006年6月以 《統一資本計量和資本標準的國際 協議》 的方式頒布之資本充足架構，其後就更改巿場風險及再證券化 之資本規定而作出修訂 （一般稱為巴塞爾協定2.5） ，自2011年12月31 日起生效。 巴塞爾協定3 2010年12月，巴塞爾委員會發表 《巴塞爾協定3規則：建設更穩健的銀行 及銀行體系之全球監管架構》 及 《流動資金風險計量、標準及監察的國 際架構》 。這兩份文件呈述巴塞爾委員會為加強全球資本及流動資金 規則而提出的改革方案，旨在建設更穩健的銀行業。2011年6月，巴塞 爾委員會發表先前文件的修訂本，列出雙邊交易對手信貸風險之資本 處理方法最終定案。 基點 百分之零點零一 （0.01%） ，因此，100個基點為1%。例如，可用於引述證 券利率或收益的變動。 C 防護緩衝資本 （ 「CCB」 ） 監管機構根據巴塞爾協定3規定的資本緩衝，旨在確保銀行在受壓期間 以外的時間建立資本緩衝，可以在產生虧損時取用。當銀行的資本水 平下降至防護緩衝資本範圍內時，監管機構將限制其資本分派。 資本計劃緩衝 （ 「CPB」 ） 英國審慎監管局根據巴塞爾協定2規定的資本緩衝，旨在確保銀行在受 壓期間以外的時間建立資本緩衝，可以在產生虧損時取用。當銀行的 資本水平下降至資本計劃緩衝範圍內時，將會觸發更嚴厲的監管措 施。 資本規定指引 （ 「CRD」 ） 由歐洲委員會頒布並由歐盟成員國採納的資本充足比率法例組合。首 份資本指引法例組合使巴塞爾協定2的建議於歐盟生效，並於2006年 7月20日實施。資本指引2於2010年12月31日實施，並隨後修訂了資本 票據、大額風險承擔、流動資金風險及證券化的規定。提出其他修訂 的資本指引3，更新了市場風險資本及其他證券化的規定，並於2011 年12月31日實施。 資本指引4組合包括資本規定指引的重訂本及新的 《資本規定規例》 。該 組合落實巴塞爾協定3的資本建議及部分規定的過渡安排。資本指引 4已於2014年1月1日實施。 資本證券 資本證券包括永久後償資本證券及或有可轉換資本證券。 中央交易對手 （ 「CCP」 ） 買方及賣方之間的中介人 （通常為結算所） 。 撤回 已向個人支付的薪酬，在若干情況下須退回予一家機構。 債務抵押債券 （ 「CDO」 ） 由第三方發行的證券，以發行人購買之資產抵押證券及╱或若干其他相 關資產作為參考。債務抵押債券可能因相關資產而須承受次優質按 揭資產風險。 詞彙 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 471 詞彙 釋義 綜合評估減值 按綜合基準對不被視為個別大額賬項之同類貸款組合作出的減值評估， 並用作抵補個別評估貸款所產生但未予識別的虧損。 商業票據 （ 「CP」 ） 由企業發行的無抵押短期債務工具，一般旨在為應收賬款、存貨及應付 短期債務而融資。債務通常按反映當前市場利率的折讓價發行。 商用物業 任何擬用作賺取利潤 （不論來自資本增值或租金收入） 的房地產，包括樓 宇或土地。 普通股權一級資本 （ 「CET1」 ） 巴塞爾協定3下最高質素的監管規定資本，包括已發行普通股及相關股 份溢價、保留盈利及其他儲備 （不包括現金流對沖儲備） ，減特定的監 管規定調整。 普通股權一級比率 巴塞爾協定3的一項衡量指標，以普通股權一級資本佔風險總額的百分 比列示。 共同匯報 （ 「COREP」 ） 在歐洲銀行管理局頒布的資本規定指引下設立的歐洲統一匯報架構。 合規風險 集團未能遵守所有相關法律、守則、規則、法規及良好市場慣例準則的 條文和精神，並招致罰款及罰則，且業務因此受損的風險。 綜合資本分析及檢討 （ 「CCAR」 ） 聯邦儲備局每年進行的綜合資本分析及檢討，以確保機構具備穩健及 前瞻性資本規劃程序，以應付其獨有的風險及有足夠的資金在經濟 及財務受壓的環境中繼續營運。 中介機構 滙豐資助及管理的多賣方中介機構及證券投資中介機構 （ 「SIC」 ） 。多賣 方中介機構於分散的第三方資產組別 （如汽車貸款、貿易應收賬款及 信用卡應收賬款） 內持有權益，並藉發行短期商業票據取得資金及由 流動資金信貸額支持。證券投資中介機構主要持有以商業及住宅按 揭、汽車貸款及信用卡應收賬款為參考的資產抵押證券，並藉發行長 期及短期債務取得資金。 固定匯率 非公認會計原則財務衡量指標，透過比較業績報告期的列賬基準業績 與比較期間按該業績報告期匯率重新換算的列賬基準業績，從而調 整外幣換算差額按年比較的影響。外幣換算差額反映於業績報告期 內美元兌大多數主要貨幣的變動。 固定價格資產淨值基金 （ 「CNAV」 ） 按已攤銷成本將資產定價的基金，惟投資組合的已攤銷賬面值須保持 於其市值的50個基點內。 消費及按揭貸款 （ 「CML」 ） 在美國，消費及按揭貸款組合包括縮減中之消費貸款及按揭業務。 消費貸款業務透過地區分行及直接郵遞提供有擔保及無擔保貸款產品， 如第一及第二留置權按揭貸款、開放式住宅二按貸款及個人非信用卡 貸款。大部分按揭貸款產品用於再融資及債務重組，而非用於置業。 於2009年第一季，我們不再辦理所有消費貸款業務。 於2007年第一季我們不再購入貸款前，按揭業務向第三方非聯屬機構購 買非標準類型第一及第二留置權有抵押房地產貸款。該項業務亦包括 Decision One Mortgage Company （ 「Decision One」 ） 的業務，Decision One過往曾辦理由獨立按揭經紀覓得的按揭貸款，並將此等貸款出售 予第二市場買家。Decision One自2007年9月起不再辦理按揭貸款。 合約期限 任何金融工具的最終款項 （本金或利息） 到期支付之日，屆時須償還所有 仍未償還的本金及利息。 核心一級資本 巴塞爾協定2下最高質素的監管規定資本，包括股東權益總額及相關非 控股股東權益，減商譽及無形資產及若干其他監管規定調整。 核心一級資本比率 巴塞爾協定2的一項衡量指標，以核心一級資本佔風險加權資產總值的 百分比列示。 反周期緩衝資本 （ 「CCyB」 ） 監管機構根據巴塞爾協定3規定的資本緩衝，旨在確保資本規定顧及銀 行業經營所在宏觀金融環境的情況。此緩衝將為銀行業提供額外資 本，於整體金融體系信貸過度增長以致整個體系風險增加時，可保障 銀行業免於蒙受潛在的未來虧損。 交易對手信貸風險 （ 「CCR」 ） 交易對手在妥為結算交易前違責所產生之風險 （於交易及非交易賬項 內） ，即為交易對手信貸風險。 信貸違責掉期 （ 「CDS」 ） 一種衍生工具合約，據此買方向賣方支付費用，以便於相關責任 （不一 定由買方負責） 的界定信貸事件 （例如破產、就一項或多項參考資產拖 欠付款或被評級機構調低評級） 出現時，可收取一筆款項。 強化信貸條件 一種信貸安排，用以強化金融責任的可信程度，並抵補因資產違責而產 生的虧損。 信貸風險 客戶或交易對手未能履行合約責任因而產生的財務虧損風險。這種風 險主要來自直接貸款、貿易融資及租賃業務，但亦會來自擔保、衍生 工具及債務證券等產品。 減低信貸風險 （措施） 運用減低信貸風險措施 （例如抵押品、擔保及信貸保障） ，藉以減低風險 項目所涉信貸風險之方法。 信貸風險息差 市場接納較低信貸質素而要求高於基準或無風險利率的溢價。票息率 及期限結構相同但信貸風險不同的證券之間的收益差距。當信貸評 級下降，收益差距亦會隨之擴大。 信貸息差風險 信貸息差變動將影響金融工具價值之風險。 信貸估值調整 （ 「CVA」 ） 為反映場外衍生工具交易對手的信譽而對場外衍生工具合約的估值作 出的調整。 客戶存款 戶口持有人存放的款項。該等資金入賬列作負債。 股東參考資料 （續） 滙豐控股有限公司 472 詞彙 釋義 與客戶有關的補救措施 滙豐就未能遵照法規所引起的相關虧損或損害為補償客戶而採取之行 動。與客戶有關的補救措施是滙豐為回應客戶投訴而採取的措施，而 非特別由監管機構採取的行動引起。 客戶風險評級 （ 「CRR」 ） 分為23個等級，用以衡量債務人違責或然率。 信貸估值調整風險資本要求 資本指引4下的資本要求，用以抵償衍生工具預期交易對手風險引致的 按市價計值虧損風險。 D 借記估值調整 （ 「DVA」 ） 企業為反映公允值內企業本身的信貸風險而對場外衍生工具負債估值 作出的調整。 債務重整 重整債務協議，據此更改協議的未履行條件及條款。重整通常旨在改善 現金流及借款人償還欠款的能力，並可能涉及更改還款時間表及削 減債務或所收取的利息。 債務證券 集團於資產負債表內的金融資產，包括信貸機構、公共機關或其他公司 的負債證明書，但不包括中央銀行發行的負債證明書。 已發行債務證券 集團向證明書持有人發出的可轉讓負債證明書。該等證明書為集團的 負債，並包括存款證。 沒收契據代替止贖 借款人在不進行止贖程序的情況下向貸款人交出物業契據的安排，而 借款人就有關貸款之任何進一步責任其後得以解除。 界定福利責任 為履行由僱員服務產生的界定福利計劃責任而預計日後所需支付費用 之現值。 同業存放 自本土及外地銀行接收的所有存款，不包括債務證券形式或已就其發 行可轉讓證明書的存款或負債。 下行震盪 該詞用以表述由2015年1月1日起12個月內每季開始時，假設全球所有市 場的孳息曲線平行下移，對我們日後淨利息收益產生的影響 （假設概 無任何管理層回應） 。孳息曲線等量上移則稱為上行震盪。 E 經濟資本 由滙豐內部計算的資本規定，即滙豐認為支持本身面對風險所需的資 本規定。 經濟盈利 股東投入財務資本的回報與該資本成本之間的差額。經濟盈利可以整 數或百分比列示。 股東權益經濟價值 （ 「EVE」 ） 之敏感度 考慮本期資產負債表內所有因重新定價而出現的錯配，且計算一組已 界定之利率波動引致的市值變動。 具產權負擔資產 於資產負債表內已就現有負債質押為抵押品的資產。 強化可變價格資產淨值基金 （ 「ENAV」 ） 按公允值基準為其資產定價的基金。隨後過程可不時變動。 赤道原則 赤道原則乃金融機構用作減少大型項目 （由金融機構提供資金） 對人或 對環境造成的潛在影響。 股權風險 股權或股權相關工具持倉 （不論長倉或短倉） 產生的風險，因為此等持倉 會產生股權或股權工具市價變動的風險。 歐元區 已採用歐元作為其統一貨幣的18個歐洲聯盟國家。該18個國家為奧地 利、比利時、塞浦路斯、愛沙尼亞、芬蘭、法國、德國、希臘、愛爾 蘭、意大利、拉脫維亞、盧森堡、馬耳他、荷蘭、葡萄牙、斯洛伐克、 斯洛文尼亞及西班牙。 預期虧損 （ 「EL」 ） 按照監管規定計算持倉在12個月時限內預期損失的金額，以及衰退損失 估算值。預期虧損的計算方法為將違責或然率 （以百分比列示） 乘以違 責風險承擔 （以金額列示） 及違責損失率 （以百分比列示） 。 風險╱風險項目╱風險承擔 附帶財務損失風險的債權、或有債權或持倉。 違責風險承擔 （ 「EAD」 ） 假若交易對手違責，於採取任何減低信貸風險措施後，預期餘下須承擔 的金額。違責風險承擔反映已取用結欠及未取用承諾金額和或有風 險承擔的準備。 F 公允值調整 使用估值方法 （第2級及第3級） 釐定的金融工具公允值的調整，以涵蓋市 場參與者應會考慮而估值模型並無計及的其他因素。 受信風險 當集團以受信人身分擔任受託人、投資經理或在法律或法規授權下行 事時，便有違反受信責任的風險。 金融業操守監管局 （ 「FCA」 ） 金融業操守監管局於英國規管金融公司的操守及若干公司的審慎標準。 其策略目標為確保相關市場運作良好。 金融政策委員會 （ 「FPC」 ） 英倫銀行的金融政策委員會，主要目標為識別、監督及採取行動，以去 除或減少系統性風險，從而保護及提升英國金融體系的復元力。金融 政策委員會的第二目標為支持英國政府的經濟政策。 財務報告 （ 「FINREP」 ） 由歐盟提議的統一歐洲財務匯報架構，將用於全面了解公司的風險狀 況。 第一留置權 就一項物業授出的擔保權益以保證償還債務，當債務出現違責情況時， 其持有人將可優先收回出售相關抵押品所得還款。 詞彙 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 473 詞彙 釋義 暫緩還款策略 為改善客戶關係管理、盡量提高收回貸款的機會，以及盡可能避免發生 違責、止贖或收回抵押品的情況而採取的策略。有關安排包括延長還 款期、減低利息或本金還款額、批准外界債務管理計劃、債務重組、 押後止贖及修訂其他條件及重訂賬齡。 已撥資風險 交換或已交換合約名義金額而產生的風險。 資金風險 一種流動資金風險，於無法按預期條款及按需要而取得流動資金，以便 為流通性不足的資產持倉提供所需資金時產生。 G 市場缺口風險 當市價大幅變動而無交易機會，即會產生之財務虧損風險。 全球系統重要性銀行 （ 「G-SIB」 ） 金融穩定委員會於2011年11月制定一套方法，根據12項主要指標認定全 球系統重要性銀行。所指定的銀行將應用1%至3.5%的普通股權一級 緩衝，在2019年1月1日前分階段推行。 銀行每年會獲重新評分，而計算方法則每三年檢討，以定期重新評估 全球系統重要性銀行名單。各國監管機構則有酌情權，可以引入高 於最低數額的費用。在資本指引4中，此乃透過全球系統重要性機構 （ 「G-SII」 ） 緩衝實施。 政府資助企業 （ 「GSE」 ） 由美國國會創立的一群金融服務企業，其職能為減低經濟體中若干借 貸機構的資金成本，使之更具效率及更為透明。住宅按揭借貸機構的 例子有房貸美及房利美。政府資助企業獲美國政府實質支持，但並非 美國政府的直接責任。 集團業績表現股份計劃獎勵 用以界定僱員將有權享有 （通常於獎勵授出日期後五年） 的滙豐控股普 通股股數 （個別僱員一般須仍然受僱） 之獎勵。僱員有資格獲得的股份 須受制於一項禁售規定，直至僱傭關係終結為止。 擔保 一方作出之承諾，在債務人未能向債權人付款，該一方會向債權人付 款。 H 扣減 管理層在釐定資產可變現的金額時所採用的折讓；折讓會考慮變現的 方法，包括資產存在交投活躍市場的程度。 過往信用評級變動矩陣 擁有特定評級的交易對手於界定時限內評級出現變動的或然率。 住宅二按信貸 （ 「HELoC」 ） 向美國客戶提供的一種循環信貸融資，絕大多數以住宅物業的第二留 置權或更低級別的抵押支持。所持住宅二按信貸歸類為次優質項目。 I 已減值貸款 集團並不預期收回所有約定現金流的貸款或預期於合約到期後收回的 貸款。 減值準備 管理層對結算日貸款組合產生的虧損所作的最佳估算。 個別評估減值 對不符合綜合評估條件的所有個別大額賬項及所有其他賬項評估的虧 損風險。 制訂保險產品 集團附屬公司獲授權於保險監管範圍內承保保單。滙豐保留承保保險業 務的風險及回報 （或按再保險策略進行再保險） 。保險業務風險管理一 節內呈列的資產負債表分析列示該等實體的合計全面資產負債表。 保險風險 指由合約持有人轉移給保險供應商的風險 （並非金融風險） 。主要的保險 風險為某段時間後賠償支出加上行政開支和獲取保單成本的總額， 可能超過所收保費加投資收益的總額。 內部資本充足程度評估程序 集團從監管規定及經濟資本的角度審查其風險狀況，從而對所需持有 資本水平進行自我評估。 內部模式計算法 巴塞爾協定架構界定的三項計算法之一，用以釐定交易對手信貸風險 的風險承擔價值。 內部評級基準計算法 （ 「IRB」 ） 使用內部而非監管規定的風險參數估算值計算信貸風險資本規定水平 的方法。 投入資本 股東於滙豐投入的股本，就先前已攤銷或撇賬之若干儲備及商譽作出 調整。 投資級別 指類似BBB-級或以上評級的風險狀況 （由外部評級機構界定） 。 內部評級基準高級計算法 （ 「AIRB」 ） 使用內部違責或然率、違責損失率及違責風險承擔模型計算信貸風險 資本規定水平的方法。 內部評級基準基礎計算法 （ 「FIRB」 ） 使用內部違責或然率模型計算信貸風險資本規定水平的方法，但同時 使用監管規定的違責損失率估算值及用以計算違責風險承擔的轉換 因素。 ISDA總協議 由國際掉期業務及衍生投資工具協會制訂用作保護協議的標準合約， 據此訂立雙邊衍生工具合約。 K 主要管理人員 滙豐控股的董事及集團常務總監。 L 環球銀行及資本市場業務之 既有信貸業務 另行管理可以清楚區別的業務，包括Solitaire Funding Limited、證券投資 中介機構、資產抵押證券交易用途組合及相關性信貸組合、與債券承 保公司直接進行的衍生工具交易，以及若干其他結構信貸交易。 股東參考資料 （續） 滙豐控股有限公司 474 詞彙 釋義 法律訴訟 對滙豐旗下公司提出的民事法庭訴訟、仲裁或審裁程序 （不論是以申索 或反申索方式） 或如未能解決即會引起法庭訴訟、仲裁或審裁程序的 民事爭議。 法律風險 合約風險產生的財務虧損、制裁及╱或聲譽受損風險 （即集團旗下成員公 司在合約關係中的權利及╱或責任有欠妥善的風險） ；爭議風險 （即由 於不利爭議局面或管理潛在或實際爭議時須承擔的風險） ；立法風險 （即集團旗下成員公司未有遵從其業務所在司法管轄區法律的風險） ； 及非合約權利風險 （即集團旗下成員公司的資產並非適當擁有或遭他 人侵犯，或集團旗下成員公司侵犯他方權利的風險） 。 第一級－市場報價 於交投活躍市場有相同工具報價的金融工具。 第二級－採用可觀察數據 的估值方法 於交投活躍市場有近似工具報價，或於交投不活躍市場有相同或近似 工具報價的金融工具，以及採用所有重要數據均可觀察的模型估值 的金融工具。 第三級－採用重大不可觀察 數據的估值方法 採用一項或多項重大數據屬不可觀察的估值方法估值的金融工具。 槓桿融資 向負債高於平均水平的企業提供的資金，一般源自次投資級別的收購 項目或事件促成的融資。 槓桿比率 監管機構根據巴塞爾協定3規定的一項衡量指標，即一級資本對風險總 額的比率。風險總額包括資產負債表內項目、資產負債表外項目及衍 生工具，一般須遵循風險的會計方法計量。這項風險資本規定水平的 附加衡量指標旨在約制銀行業過度借貸的情況。 流動資金覆蓋比率 （ 「LCR」 ） 優質流動資產存量對預期往後30日現金流出淨額的比率。優質流動資 產應無產權負擔、於受壓期間在市場上保持流通，及於理想情況下列 為合資格中央銀行資產。巴塞爾協定3的規則規定，該比率自2015年 起將至少為100%。流動資金覆蓋比率仍須經歷一段觀察期及加以檢 討，以應付任何意想不到的後果。 強化流動資金條件 強化流動資金條件使資金可於有需要時基於資產違責以外的原因而取 用，例如為確保準時償還到期的商業票據。 流動資金風險 滙豐沒有足夠財務資源於到期時履行責任，或將要以超支成本履行責 任的風險。此風險因現金流的時間錯配而產生。 修訂貸款條件 暫時或永久改變貸款原有條款及條件的賬項管理措施，這項措施不會 重訂拖欠狀況，除非 「重訂賬齡」 則除外，在有關情況下拖欠狀況亦 會改為未過期。修訂賬項條件可能包括修訂一項或多項貸款條款，包 括 （但不限於） 改動利率、延長攤銷期、降低付款金額及部分豁免償還 或延後償還本金。 重訂貸款賬齡 一種賬項管理措施。於符合若干要求後，若情況顯示貸款預期將按照合 約條款償還，賬項的拖欠狀況將改為未過期。 逾期貸款 逾期償還的貸款。 貸款估值比率 （ 「LTV」 ） 一項將貸款金額列示為抵押價值百分比的數學計算方法。高貸款估值 比率表示，在拖欠還款及未償還貸款結欠被加收利息的情況下，貸款 人所獲得的保障較低，容易受到樓價下跌或貸款增加的影響。 違責損失率 （ 「LGD」 ） 當交易對手違責時，未償還金額的違責風險承擔(EAD)所產生的估計損 失比率 （以百分比列示） 。 虧損嚴重程度 透過與借款人訂立的安排止贖或出售貸款時所產生的已變現虧損金額 （包括結欠的相關款項） 。虧損嚴重程度以佔未償還貸款結欠的百分 比呈列。 M 扣減 （條文） 允許一間機構阻止實際授出因風險結果或表現而延遲授予全部或部分 薪酬獎勵的安排。 市場風險 市場風險因素 （包括匯率及商品價格、利率、信貸息差及股價） 變動導致 收益或組合價值下跌的風險。 中期票據 （ 「MTN」 ） 由企業根據中期票據計劃發行，且定期及持續向投資者發售的各種不 同期限票據。 按揭抵押證券 （ 「MBS」 ） 代表一組按揭權益的證券，可涉及住宅或商用物業。該等證券的投資者 有權就未來按揭付款 （利息及╱或本金） 收取現金。倘按揭抵押證券的 參考按揭有不同的風險狀況，該等按揭抵押證券會按最高風險級別 分類。 按揭相關資產 作為相關按揭的參考。 按揭年份 批出按揭的年份。 N 負資產按揭 資產指物業價值減未償還貸款結欠的數額。當購入物業的價值低於未 償還貸款結欠時，便會產生負資產。 每股資產淨值 股東權益總額減非累積優先股及資本證券，再除以已發行普通股數目。 淨利息收益 已收或應收的資產利息減已付或應付負債利息的金額。 淨利息收益敏感度 考慮本期資產負債表的所有定價錯配，並就資產負債日後的增長作出 適當假設，且計算一組界定利率波動引致的淨利息收益變動。 詞彙 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 475 詞彙 釋義 本金風險淨額 經計及已購買的信貸保障之金融資產本金總額，但不包括對該保障作 出任何交易對手信貸估值調整的影響。當中包括受惠於債券承保公 司保障的資產，但所購保障若附有信貸違責掉期者則除外。 穩定資金淨額比率 （ 「NSFR」 ） 假設於壓力境況下，在一年時間內可獲得的穩定資金對所需穩定資金的 比率。可獲得的穩定資金將包括股本、一年以後到期的優先股及評估 於一年以後到期的負債項目。巴塞爾協定3的規則規定該比率自2018 年起須超過100%。穩定資金淨額比率仍須經歷觀察期及檢討，以應 對任何意想不到的後果。 非標準類型按揭 不符合一般借貸條件的美國按揭。例子包括並無提供預期所需文件 （如 自行證明入息） 的按揭，或借款人信貸紀錄欠佳令風險增加，導致定 價較一般借貸利率高的按揭。 非交易用途組合 包括主要因我們的零售銀行及工商金融業務對資產與負債進行利率管 理而產生的持倉、指定列為可供出售及持至到期日之金融投資，以及 來自我們保險業務的風險項目組合。 非交易賬項風險 非交易用途組合產生的市場風險。 O 對銷按揭 一種靈活的按揭，借款人在相同機構持有之存款結餘，可用作對銷未償 還之按揭結欠。借款人就結欠淨額支付利息，而結欠淨額的計算方 法，是從扣賬額減去撥賬額所得之淨額。對銷按揭之其中一個特點， 是有一個協定融資總額上限，借款人可以在事先協定的上限之下重 新提取貸款。 隔夜指數掉期 （ 「OIS」 ） 折現 有抵押利率衍生工具估值的方法，此方法使用的折現曲線反映就所收 取抵押品一般賺取或支付的隔夜利率。 營運風險 因內部程序、人員及系統的不足或失效，或因外圍事件 （包括法律風險） 引致虧損的風險。 場外 （ 「OTC」 ） 並非在交易所買賣，且採用估值模型估值的雙邊交易 （如衍生工具） 。 P 退休金風險 集團旗下公司及成員公司的供款未能產生足夠資金，以應付活躍成員 日後服務的應計福利支出，以及退休基金所持資產的表現，不足以應 付現有退休金負債之風險。 業績表現股份 根據企業的業績表現狀況按僱員股份計劃獎勵的滙豐控股普通股。 個人貸款 請參閱 「零售貸款」 。 審慎監管局標準規則 於估計虧損風險模型未獲批准時，審慎監管局就計算市場風險資本規 定水平而規定採用的方法。 優質 在美國用以描述向信譽最可靠類別借款人批出的按揭。 私募股本投資 並非於公開交易所報價的營運公司股權證券，通常涉及私人公司的資 本投資或收購上市公司而引致上市股票除牌。 違責或然率 （ 「PD」 ） 債務人於一年內違責的可能性。 利潤參與供款 （ 「PIS」 ） 就巴西法律實體賺取的每月總收入徵收的聯邦稅項，乃僱主對僱員儲 蓄方案作出的強制性供款。 審慎監管局 （ 「PRA」 ） 英國審慎監管局，負責審慎監管及監督銀行、建屋貸款社、信用合作 社、保險公司及主要投資機構。 R 再融資利率 再融資利率由歐洲中央銀行 （ 「歐洲央行」 ） 釐定，為銀行向歐洲央行償還 借款的定價。 監管規定資本 滙豐持有的資本，就綜合集團而言，根據審慎監管局訂立的規則釐定， 以及就集團旗下個別公司而言，根據當地監管機構訂立的規則釐定。 監管事宜 由監管或執法機構就指稱滙豐行為不當而進行的調查、檢討及其他行 動，或因應該等機構的行動而進行的調查、檢討及其他行動。 重議條件貸款 由於對借款人在到期時按合約還款之能力有重大質疑，故貸款的合約 還款條款有所改變。 回購╱反向回購 （或出售及回購協議） 讓借款人向貸款人出售金融資產以取得有抵押貸款的短期融資協議。 作為協議的一部分，借款人承諾於日後回購抵押品，以償還貸款所得 款項。就交易的另一方 （購買抵押品及同意於日後出售） 而言，即是反 向回購協議或反向回購。 聲譽風險 集團本身、員工或客戶或集團代表的非法、不道德或不當行為將損害滙 豐聲譽，並有可能引致業務虧損、罰款或罰則之風險。 有限制股份 用以界定僱員將有權享有 （通常於獎勵授出日期後一至三年） 的滙豐控 股普通股股數 （個別僱員一般須仍然受僱） 之獎勵。僱員有資格獲得之 股份可能須受制於禁售規定。 零售貸款 借予個人而非機構的款項，包括有抵押及無抵押貸款 （例如按揭及信用 卡結欠） 。 股東權益回報 母公司普通股股東應佔利潤除以平均普通股股東權益。 股東參考資料 （續） 滙豐控股有限公司 476 詞彙 釋義 承受風險水平 企業願意於本身承受風險能力以內，為實現策略目標及業務計劃而承 受的風險類別及水平總額。 承受風險能力 企業在違反監管規定資本及流動資金需求與本身責任所決定的限制條 件前，且從行為角度而言，在違反對存戶、投保人、其他客戶及股東 的責任所決定的限制條件前，可承受的最大程度風險。 風險加權資產 （ 「RWA」 ） 計算方法是根據適用的標準或內部評級基準計算法規則，定出風險項 目的風險程度，並以佔風險值的百分比 （風險權數） 列示。 縮減組合 環球銀行及資本市場業務之既有信貸業務、美國消費及按揭貸款組合 及其他美國縮減組合，包括涉及美國消費及按揭貸款業務的財資服 務和縮減中的商業營運部門。縮減組合不再受理新造業務，而結欠則 透過損耗及出售減少。 S 出售及回購協議 請參閱上文回購一節。 第二留置權 就一項物業授出的抵押品權益，以保證償還債務，而與依據相同抵押品 發出的第一留置權債務比較，屬次於第一留置權。倘出現違責情況， 僅會於第一留置權債務獲償還後，此債務才獲還款。 證券化 一項交易或計劃，據此將涉及一項或一組風險承擔的信貸風險分為不 同部分，而於有關交易或計劃中向投資者支付的款項須取決於該項 或該組風險承擔的表現。傳統證券化涉及將證券化風險轉移至一家 發行證券的結構公司。若為組合型資產證券化，則透過使用信貸衍生 工具將風險承擔分為不同部分，而該等風險承擔將不會從辦理機構 的資產負債表剔除。 證券化掉期 證券化過程中與未償還資產組合規模名義上掛鈎的利率或跨貨幣掉期。 證券化掉期一般由證券化公司進行，以對沖資產組合利率風險狀況 與公司所發行證券的利率風險狀況兩者之間出現錯配因而產生的利 率風險。 賠本出售 「賠本出售」 是與信貸風險管理有關的一項安排。據此，銀行准許借款人 以低於貸款協議下未償還金額的價格出售物業。所得款項用作降低 未償還的貸款結欠，而借款人就有關貸款的任何進一步責任其後得 以解除。 單一發行人流動資金信貸額 向企業客戶提供的流動資金或備用金額，且不同於向中介融資公司提 供的同類信貸額。 六方面考慮 為改善集團資本投放而制訂之內部衡量指標。其中五方面的考慮審視 每個地方每項業務與策略之關連 （就聯繫能力及經濟發展而言） 及現 有回報 （就盈利能力、成本效益及流動資金而言） 。第六方面的考慮則 要求遵守環球風險標準。 社會保障資助供款 （ 「COFINS」 ） 就巴西法律實體賺取的每月總收入徵收的聯邦稅項，是資助社會保障 系統的供款。 主權風險承擔 涉及政府、部委、政府部門、大使館、領事館的風險，以及有關存放於 中央銀行的現金結餘及存款之風險。 特設企業 （ 「SPE」 ） 因狹義目的 （包括進行證券化活動） 而成立的企業、信託或其他非銀行公 司。特設企業的結構及業務旨在分隔該公司與辦理機構及證券化活 動實益權益持有人之責任。 結構公司 （ 「SE」 ） 投票或類似權利並非決定公司控制權主要因素的公司，例如投票權僅 與行政工作相關及相關業務由合約安排指定。 標準計算法 （ 「STD」 ） 就信貸風險而言，使用外部信用評估機構 （ 「ECAI」 ） 評級及監管規定風險 權數計算信貸風險資本規定水平的方法。就營運風險而言，則是由八 個指定業務範疇的總收益扣取一個監管規定界定百分比，從而計算 營運風險資本規定水平的方法。 壓力下之估計虧損風險 一項市場風險計量，以交易用途組合持續一年壓力期間的潛在市場變 動為基準。 結構融資╱票據 這類工具的回報乃與特定指數水平或特定資產水平掛鈎。結構票據的 回報可與股票、利率、外匯、商品或信貸掛鈎。結構票據不一定會於 相關指數或資產下跌時提供全部或部分資本保障。 學生貸款相關資產 以學生貸款相關抵押品作擔保的證券。 後償負債 於發行人無力償債或清盤時，權益地位低於其他債權人之負債。 次優質 在美國用以描述高信貸風險的客戶，例如信貸紀錄有限、收入較低、高 債務對收入比率、高貸款估值比率 （有抵押房地產產品） ，或曾因偶爾 拖欠、過往沖銷、破產或其他信貸相關問題，以致出現信貸問題的客 戶。 可持續發展風險 提供金融服務所產生的環境及社會影響超逾經濟利益之風險。 可持續成本節約╱節省 於某個業務活動水平永久縮減成本。可持續成本節約╱節省不包括避免 產生成本以及收入和貸款減值準備的利好因素，因為此等項目並不 表示營運開支減少。因出售業務而節省的成本並不分類為可持續。 詞彙 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 477 詞彙 釋義 系統性風險緩衝 （ 「SRB」 ） 歐盟根據資本指引4規定的資本緩衝，以應對金融業整體或一個或多個 行業類別的風險，各歐盟成員國可按需要實施，以降低結構性的宏 觀－審慎風險，而英國於2015年1月予以採納，並將應用於超過某特 定資本額的分隔運作銀行及建屋貸款社。 系統風險 支援集團日常業務運作的自動化平台及自動化平台所在的系統基礎設 施 （包括數據中心、網絡及分布式電腦） 出現故障或其他缺陷之風險。 T 一級資本 監管規定資本的組成部分，包括普通股權一級資本及額外一級資本。額 外一級資本包括合資格非普通股權資本證券及任何相關股份溢價。 根據巴塞爾協定2，一級資本包括核心一級資本及其他一級資本。其他 一級資本包括合資格的資本票據，例如非累積永久優先股及混合資 本證券。 二級資本 監管規定資本的組成部分，包括合資格資本證券及任何相關股份溢價。 根據巴塞爾協定2，二級資本包括合資格後償借貸資本、相關非控股股 東權益、可列賬綜合評估減值準備及因評估持作可供出售用途股權 工具公允值而產生的未變現增益。二級資本亦包括物業重估儲備。 整體損失吸納能力 （ 「TLAC」 ） 金融穩定委員會為全球系統重要性銀行而提出但尚未落實的建議，內容 為此等銀行須有充足的特定類別負債，以便於銀行推行解決方案時， 吸納損失及重建資本。這些建議旨在促進銀行有序解決損失問題，盡 量減低對金融穩定性的衝擊，確保銀行的主要功能得以持續及避免 納稅人承擔損失。 交易用途組合 因進行市場莊家活動而持有及代客戶保管的持倉。 交易賬項風險 交易用途組合產生的市場風險。 問題債務重整 在美國用以描述債務重整，據此債權人會因涉及債務人財政困難的經濟 或法律理由，向債務人授出原本不會考慮的還款優惠。 U 無產權負擔資產 於資產負債表內並未就現有負債質押為抵押品的資產。 未撥資風險 合約名義金額未交換的風險。 上行震盪 見下行震盪 美國政府機構及美國政府資助企 業的按揭相關資產 由美國政府機構 （例如美國政府國民抵押協會Ginnie Mae） 擔保的證券， 或由美國政府資助企業 （包括房利美和房貸美） 擔保的證券。 V 估計虧損風險 （ 「VaR」 ） 虧損的衡量指標，此種虧損於指定時限內和既定可信程度下，可能因市 場風險因素 （例如利率、價格及波幅） 的不利變動而在風險持倉內產 生。 W 批發貸款 借予主權借款人、銀行、非銀行之金融機構及大型企業的貸款。 撇減╱撇銷 倘金融資產被撇減或撇銷，客戶結餘將分別被部分或全部自資產負債 表剔除。倘現實上收回機會甚微，通常會將部分或全部貸款 （及相關 減值撥備賬項） 撇銷。倘貸款有抵押，則通常是於變現證券收取任何 所得款項後撇銷。於任何抵押品的變現價值淨額已釐定及按合理預 期不能進一步收回的情況下，則可提早作出撇銷。 錯向風險 交易對手違責或然率與相關交易按市值計價的價值之間的不利相關性。 股東參考資料 （續） 滙豐控股有限公司 478 索引 A 簡稱 466 會計 發展 （未來） 345 估算及判斷 349、366、378、403、407、420 政策 （關鍵） 62 政策 345、354、355、357、359、365、 371、377、378、392、394、398、399、 402、403、407、410、413、416、417、 418、419、420、434、437、440、441、 442、443、445 賬目 通過 457 編製基準 63、345 綜合計算及相關披露 348 呈列資料 347 應計項目、遞延收益及其他負債 418 收購及出售 41、241、244、260 精算假設 363 經調整業績 29、40 貸款對核心資金比率 165、169、216 賬齡分析 136 股東周年大會 293、459 周年 （150） 6、8 反洗錢及制裁 450 特別提述部分 124 亞洲 84 經調整業績 87 資產負債表數據 89、375 抵押品 147 固定匯率基準╱列賬基準之對賬 158 國家╱地區業務摘要 86 客戶賬項 82 經濟背景 84 財務概覽 85 已減值貸款 137 貸款 131、132、160 貸款減值準備 141、142 按揭貸款 156 營業支出 88 個人貸款 151 主要業務 84 利潤 84、89、374 按國家╱地區列示之利潤╱ （虧損） 85、86 重議條件貸款 139、140 反向回購 151 風險加權資產 240 職員人數 84 批發貸款 144、145 資產抵押證券 214 資產 資產負債平均值 46 按國家╱地區列示 376 按地區列示 78、82、89、94、99、104、375 按環球業務列示 63、82、89、94、99、104 作為擔保而質押 401 固定匯率基準╱列賬基準之對賬 59 客戶賬項 61 遞延稅項 331、365、366、367、368 具產權負擔╱無產權負擔 171、220、472 財務會計基準╱監管規定基準之對賬 249 五年 57 持作出售用途 349、352、416 託管及管理 106 無形 407、410、413 主要營運公司的流動資產 166 期限分析 426 2014年之變動 58 其他 416 風險加權 31、62、63、78、239至244、476 總計 30、57、59、76、82、89、94、99、 104、337、341、427 交易用途 377 已轉讓 （會計政策） 402 聯營及合資公司 403 會計政策 403 交通銀行 404 或有負債 442 關鍵會計估算及判斷 403 權益 332、406 列賬基準╱經調整基準之對賬 44 應佔利潤 30、55 與其他關連人士之交易 456 核數師 安排 279 費用 364 報告 329 B 回溯測試 177、224、470 資產負債表 平均值 46 綜合 57、337 固定匯率基準╱列賬基準之對賬 59 數據 57、76、82、89、94、99、104 滙豐控股 341 制訂保險產品附屬公司 191 涉及項目 179 2014年之變動 30、58 監管規定 248 資產負債管理業務 181、227 銀行保險業務 190 銀行業標準 5 巴塞爾 31、123、164、218、239、251、470 行為化 181、218、226 董事會 270 架構平衡及獨立性 272 人事交替 6 委員會 15、276 資料及支援 272 會議 271 權力 271 品牌 32 巴西勞工索償 422 巴西社會保障索償 441 緩衝 （資本） 239、252 業務模式 12 C 資本 239 生成 245、258 管理 257 計量及分配 258 2014年監管規定資本變動 245 概覽 239 比率 31、239、471 監管規定 245、247 來源 57 資本風險 257 實力 3、31 結構 245 二氧化碳排放量 38 現金及等同現金項目 440 會計政策 440 現金流 綜合結算表 338 對沖 397 滙豐控股 342 說明 439 於合約期限前應付 173 有關前瞻性陳述之提示聲明 2 主席委員會 15、288 中譯本 461 客戶資產 73 氣候業務 37 抵押品及強化信貸條件 146、150、156、171、213 管理 220 工商金融 16、17、22、67 商用物業 145 委員會 （董事會） 276 與股東之溝通 298、460 合規風險 115、189 風險集中情況 132 行為及價值觀委員會 15、286 中介機構 443 同意令 120 固定匯率 41 《消費者信貸法》 422 目錄封面內頁 索引 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 479 或有負債、合約承諾及擔保 441 會計政策 441 金融負債合約期限 173 企業管治 263 守則報告 275 成本效益比率 3、28、55、79、84、91、96、 101 交易對手信貸風險 243 國家╱地區業務摘要80、86、92 信貸違責掉期監管調查 454 信貸風險項目 130、149 信貸質素 207 類別 207 信貸風險 129、206、258、260 2014年 129 保險 196 管理 206 政策及慣例 206 風險加權資產 240 信貸估值調整 382 關鍵會計估算及判斷 349、366、378、403、 407、420 跨境風險承擔 169 客戶賬項 30、61、82、89、94、99、104 客戶 16、17、18 客戶存款市場 164 客戶貸款及存款 （綜合） 60 客戶服務及滿意度 17 D 買賣滙豐控股有限公司股份 299 借記估值調整 382 已發行債務證券 418 會計政策 418 界定用語封面內頁 存款 綜合分析 60 核心 216 平均款額及平均利率 46 衍生工具 149、382、394、422 會計政策 394 董事 270 委任和重選 271 福利 318 簡歷 264 利益衝突 274 酬金 324、364 執行 270 退任補償 318 袍金 318 履任啟導 273 權益 297、320 離職 318 非執行 270 其他公司董事席位 307 退休金 318 表現評估 274 與股東的關係 274 薪酬 （執行） 34、300、303、311、313 薪酬 （非執行） 306 責任 （聲明） 328 服務合約 306 培訓及發展 273 浮動酬勞 303、309、310 資料披露理念 107 出售利潤╱出售用途業務組合 41 出售440 爭議風險 122 多元及共融 19、291 股息 298、370、458 收益 353 派息比率 32 每股 3、45 捐獻 38 E 每股盈利 29、45、371 經濟背景 亞洲 84 歐洲 79 拉丁美洲 101 中東及北非 91 北美洲 96 經濟貢獻 33 經濟計劃：巴西滙豐銀行 454 僱員 報酬及福利 356 發展 19、291 殘疾 291 多元及共融 19、291 投入 21、288 男女比例 20 健康、福利及安全 20、291 最高薪 325 承受重大風險人員 327 數目 1、18、54、79、84、91、96、101 管理層結構 19 關係 291 薪酬政策 34、292 獎勵 291 風險 112 股份計劃 292 簽約及遣散 327 參與志願服務 37 舉報 20 具產權負擔資產 171、220、472 強化信息披露工作組 108 查詢 （來自股東） 460 股東權益 固定滙率基準╱列賬基準之對賬 59 2014年之變動 59 股權證券 179、226 歐洲 79 經調整業績 81 資產負債表數據 82、375 抵押品 147 固定匯率基準╱列賬基準之對賬 158 國家╱地區業務摘要 80 客戶賬項 61 經濟背景 79 財務概覽 79 已減值貸款 137 貸款 131、132、160 貸款減值準備 141、142 按揭貸款 156 營業支出 81 個人貸款 151 主要業務 79 利潤╱ （虧損） 79、82、374 按國家╱地區列示之利潤╱ （虧損） 80 監管規定修訂 254 重議條件貸款 139、140 反向回購 151 風險加權資產 240 職員人數 79 批發貸款 144、145 歐元區 126 結算日後事項 457 執行風險 122 F 公允值 會計政策 378 調整 381 監控架構 378 對賬 384 估值基準 383 費用收益 （淨額） 48、353 受信風險 115、200 六方面考慮 12 金融資產 392 會計政策 392、402 指定以公允值列賬 181 不符合撤銷確認條件 402 防範金融犯罪及監管合規工作 124 金融擔保合約 442 會計政策 441 金融工具 378、390 會計政策 （公允值） 378 股東參考資料 （續） 滙豐控股有限公司 480 會計政策 （估值） 354 按公允值 330 信貸質素 133、207 關鍵會計估算及判斷 （估值） 378 淨收益 50、354 非按公允值 390 已逾期但並非已減值 136 金融投資 60、399 會計政策 399 減除虧損後增益 51 指定以公允值列賬之金融負債 181、379、417 會計政策 417 合約期限 173 財務概覽 79、85、91、96、101 金融風險 （保險） 194 金融服務賠償計劃 442 金融系統風險防護委員會 15、282 財務報表 334 呈列之變動 346 五年比較 45、57、157、159 固定酬勞 39、40、303、311 註釋 39、109、202、256、344 暫緩還款 139、208、350 止贖 208、449 外幣╱外匯 會計政策 348 風險 435 調查及訴訟 453 換算率 57 換算差額 41 資金公允值調整 353 資金來源 （分散） 168、215 資金轉移定價 219 管理資金 106 G 地區 13、78 環球業務 16、63 環球銀行及資本市場 16、17、22、70、381 環球部門 13 環球員工意見調查 19 環球私人銀行 4、16、18、22、72、124 詞彙 470 持續經營 290、348 商譽 407 會計政策 407 關鍵會計估算及判斷 407 減值 331 管治 15、27 集團監察委員會 15、277、290 集團行政總裁 年度表現 316 簡歷 264 花紅評分紀錄 264 於股份的權益 320 薪酬 313 過往薪酬 319 職責 270 回顧 7 集團主席 簡歷 264 於股份的權益 320 函件 263 職責 270 報告 4 集團風險管理總監 年度表現 317 簡歷 267 花紅評分紀錄 323 於股份的權益 320 薪酬 311、313 集團公司秘書長 簡歷 268 職責 272 集團財務董事 年度表現 317 簡歷 267 花紅評分紀錄 323 於股份的權益 320 薪酬 311、313 集團管理委員會 15、276 集團薪酬委員會 15、284、300、307 集團風險管理委員會 15、280、290 優先增長目標 66、68、71、74 擔保 442 H 健康與安全 291 持作出售用途資產 349、352 會計政策 416 摘要 3 美國滙豐融資 止贖 153 修訂貸款條款 154 滙豐控股有限公司 資產負債表 341 現金流 174、221、342 信貸風險 161 遞延稅項 370 董事酬金 364 股息 370 僱員報酬 364 金融資產及負債 389 非按公允值列賬之金融工具 392 匯兌估計虧損風險 183 流動資金及資金 174 市場風險 183 資產及負債之期限分析 432 金融工具淨收益 354 營運模式 13 關連人士 457 重新定價缺口期限 185 股本 439 股東權益變動表 343 結構匯兌風險 435 後償負債425 人權 38 I 減值 會計政策 350 準備 142、159 評估 212 可供出售金融資產 352 準備 29、53、141、159 固定匯率基準╱列賬基準之對賬 158 關鍵會計估算及判斷 349 商譽 407 已減值貸款 137、329 評估方法 213 按行業及地區劃分的變動 142 列賬基準╱經調整基準之對賬 44 收益表 （綜合） 45、335 有關滙豐的資料 （索取途徑） 461 保險 190 會計政策 354、355 資產與負債配對 191 制訂產品附屬公司的資產負債表 193 銀行保險業務模式 190 已產生賠償 （淨額） 及投保人負債之變動 53、355 2014年 191 已賺取保費淨額 51 保費收益 354 PVIF業務 52 再保人應佔未決賠款 197 風險 116、117、194、198 無形資產 410 會計政策 410 變動 413 （淨） 利息收益╱支出 46、353 會計政策 354 資產負債平均值 46 敏感度 181、184 利率衍生工具 422 經營狀況參考聲明 459 中期業績 459 內部監控 288 互聯網罪行 123 索引 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 481 IFRS與 《香港財務報告準則》 之比較 345 投資準則 12 投資物業 416 投資者關係 461 J 合資公司 403、406、456 K 主要管理人員 455 主要表現指標 29、30、31、32 L 拉丁美洲 101 經調整業績 102 資產負債表數據 104、375 抵押品 147 固定匯率基準╱列賬基準之對賬 158 國家╱地區業務摘要 102 客戶賬項 61 經濟背景 101 財務概覽 101 已減值貸款 137 貸款 131、132、160 貸款減值準備 141、142 按揭貸款 156 營業支出 103 個人貸款 151 主要業務 101 利潤 101、104、374 按國家╱地區列示之利潤╱ （虧損） 102 重議條件貸款 139、140 反向回購 151 風險加權資產 240 職員人數 101 批發貸款 144、145 租賃承諾 442 會計政策 442 法律 訴訟及監管事宜 330、446 風險 229 貸款－綜合分析 60 槓桿融資交易 383、474 槓桿比率 251、255、261 負債 資產負債平均值 46 按地區列示 375 固定匯率基準╱列賬基準之對賬 59 遞延稅項 367 財務會計基準╱監管規定基準之對賬 249 五年 57 期限分析 426 2014年之變動 58 出售用途業務組合 418 其他 418 退休福利 359 後償 423、476 總計57、59、337、341、427 交易用途 417 保單未決賠款 28、419 倫敦銀行同業拆息、歐洲銀行同業拆息及 其他利率的調查 452 流動資金及資金 163、215 資產166 行為化 218 闡述 164 資金轉移定價 219 2014年 164 保險 197 風險管理 114、165 約定現金流淨額 166 政策及程序 215 主要資金來源 168 規例 164 貸款 會計政策 349 按國家╱地區劃分 160 按地區劃分 131 按行業劃分的五年貸款 157 抵押品 146、150、156、171、213 風險集中情況 132 信貸質素 133 美國貸款拖欠率 153 減值 137、141 已逾期但並非已減值 136 重議條件 138 同業貸款 291 客戶貸款 30、132 撇銷 212 貸款管理組 213 M 馬多夫 447 市值 1、33 市場風險 114、175、221、259 資產負債表內涉及的項目 179 闡述 176 管治 222 2014年 176 保險 194 計量指標 223 風險加權資產 244 敏感度分析 181 承受重大風險人員300、327 資產及負債之期限分析 426 最大信貸風險 130 中東及北非 91 經調整業績 92 資產負債表數據 94、375 抵押品 147 固定匯率基準╱列賬基準之對賬 158 國家╱地區業務摘要 92 客戶賬項 61 經濟背景 91 財務概覽 91 已減值貸款 137 貸款 131、132、160 貸款減值準備 141、142 按揭貸款 156 營業支出 93 個人貸款 151 主要業務 91 利潤 91、94、374 按國家╱地區列示之利潤╱ （虧損） 92 重議條件貸款 139、140 風險加權資產 240 職員人數 91 批發貸款 144、145 模型風險 25、122、223 監察員 27 按揭 貸款 152 按揭抵押證券 214、474 美國按揭相關調查 449 N 提名委員會 15、284 非控股股東權益 436 非公認會計原則之衡量指標 40 非利息收益 353 非交易用途組合 178、221、225 北美洲 96 經調整業績 97 資產負債表數據 99、375 抵押品 147 固定匯率基準╱列賬基準之對賬 158 國家╱地區業務摘要 97 客戶賬項 61 美國貸款拖欠趨勢 153 經濟背景 96 財務概覽 96 已減值貸款 137 貸款 131、132、160 貸款減值準備 141、142 按揭貸款 156 營業支出 98 個人貸款 151 主要業務 96 利潤 96、99、374 按國家╱地區列示之利潤╱ （虧損） 97 重議條件貸款 139、140 股東參考資料 （續） 滙豐控股有限公司 482 反向回購 151 風險加權資產 240 職員人數 96 批發貸款 144 O 對銷 130、434 會計政策 434 石油及天然氣價格 125 營業支出 30、42、54 按地區列示 81、88、93、96、103 按環球業務列示 65、68、71、73、75 列賬基準╱經調整基準之對賬 44 營業收益 52、353、375、376 營業利潤 356 營運風險 115、186、259 2014年 187 虧損╱事件 188 普通股 294 組織架構 13 組織架構圖 462 其他 75 展望未來 6 P 用紙 37 還款保障保險 421 退休金計劃 200、237 會計政策 359 界定福利計劃 183、361 董事 318 風險 116、200 人事風險 122 業績 1、5、64 經調整 29、40、65、67、70、73、75、87、92、 97、102 列賬基準 29、64、67、70、73、75 永久後償資本證券 370 個人貸款 151 慈善及社區投資事務監察委員會 15、288 第一、第二及第三支柱 253、258、259、260、326 平安保險 77 離職後福利計劃 359、456 會計政策 359 貴金屬定價相關訴訟及調查 454 優先股 294、437 優先證券 57 預付款項、應計收益及其他資產 416 會計政策 416 產品及服務 16、371 除稅前利潤 3、28、63、76 按國家╱地區列示 80、85、86、92、97、102 按地區列示 44、78、79、82、84、85、89、94、 96、99、101、102、104 按環球業務列示 44、64、65、67、70、72、73、 75、76、82、89、94、99、104 綜合 45 列賬基準╱經調整基準之對賬 44 本年度利潤 28、335、372 物業、機器及設備 107、416 會計政策 416 準備 420 會計政策 420 關鍵會計估算及判斷 420 宗旨 1 PVIF 52、411 R 比率 貸款對核心資金 165、169、216 資本 239 資本實力 3、31 普通股權一級 31 核心一級(CET1) 471 成本效益 3、28、55、79、84、91、96、101、104 客戶貸款對存款 30 派息 32 每股股息 3 每股盈利 29、45、371 槓桿 31、251、255、261 平均普通股股東權益回報 3 平均資產總值回報 30 風險加權資產回報 32、79、84、91、96、101 有形股本回報 29 壓力下之償債保障 165、216 列賬基準項目與經調整項目之對賬 44 風險加權資產平均值回報之對賬 62 復元和解決 14、255 監管規定 資產負債表 248 資本 239、258 緩衝資本 252 資本指引4 470 發展 252 監管環境 5 財務會計對賬 248 審查消費者促進服務產品 455 風險 119、120 壓力測試 125、254、257 銀行業結構性改革 14、255 系統重要性銀行 15、252、473 英國最新的規定 254 關連人士交易 455 薪酬 調整、扣減及撤回 306 福利 311 花紅評分紀錄 323 業務背景 301 委員會 284、300 委員會成員284、307 董事73 離職因素 318 固定酬勞 39、40、303、311 集 團業績表現股份計劃 314 2014年 318 2015年╱2016年 302、322 函件 300 第三支柱薪酬政策 326 政策 35、303、304、305、306 報告 307 獎勵策略 34、300 情況 313 單一數字 311 浮動酬勞 34、309 重議條件貸款 138、154、208 可再生能源 37 陳述和保證 162 重新定價缺口 185 回購及反向回購協議 48、151、219、398 會計政策 398 聲譽風險 115、199 解決策略 14 零售銀行及財富管理 16、22、64 主要零售銀行及財富管理業務 63 收入 29、42 按國家╱地區列示 81、87、92、97、102 按地區列示 81、87、92、97、102 按環球業務列示 65、67、70、73、75 分隔運作 （英國） 14 風險 111 承受水平 25、27、205 銀行風險 114 委員會 16、276、280 合規 115、189 業務經營方式 121 或有流動資金 167 交易對手 150、243、261、471 信貸 114、129、196、206 信貸息差 178、226 跨貨幣 221 數據管理 123 脫鈎 225 爭議 122 經濟前景 118 歐元區 126 執行 122 因素 113 受信 115、200 金融 （保險） 116、194 索引 股東參考資料 財務報表 企業管治 財務回顧 策略報告 滙豐控股有限公司 483 匯兌 183 市場缺口風險 225 地緣政治 118 管治 111、204 2014年 21 資訊保安 123 保險業務 116、117、190、198 利率 181、226 互聯網罪行 123 調查 120 法律 229 流動資金及資金 114、197 管理 24、112、117、215 市場 114 模型 25、122、223 石油及天然氣價格 125 營運 115 概覽 21 人事 122 退休金 116、200 政策及慣例 204 狀況 111、117 再融資 214 監管 119、120 聲譽 115、199 俄羅斯 126 保安及詐騙 187、230 壓力測試 117 可持續發展 116、201 系統 231 第三方 124 首要及新浮現 22、111、118 供應商 190、231 風險加權資產 31、62、260、476 經調整對賬 62 按地區列示 78 按環球業務列示 63 發展 254 五年趨勢 57 2014年之變動 239至244 縮減組合 32 目標257 風險加權資產平均值回報 （計量之對賬） 62、79、84 俄羅斯 126 S 出售及回購協議 398、476 會計政策 398 證券訴訟 446 證券化 風險 161、214、445 訴訟 449 保安及詐騙風險 187、230 按類分析 371 會計政策 371 高層管理人員 簡歷 268 酬金 324 非經濟假設的敏感度 （保險） 198 以股份為基礎的支出 357 會計政策 357 股本 294、437 會計政策 437 五年趨勢 57 2014年 296 須予公布之權益 298 權利及責任 294 庫存股份 297 股份資料 33 認股權 321、359 股份計劃 董事 297、320 僱員 292、296 股東 （通訊） 460 人數 1 資料 458 重大項目 （其他） 42、47、49、51、52、54 資金來源 168 標準 （環球） 26、27 股東權益變動表 339 全面收益表 336 股份代號 461 策略 1、11、26、64、67、70、72 簡化 27 壓力測試 117、125、216、224 壓力下之償債保障比率 165、216 銀行業結構性改革 14、255 結構匯兌風險 181、226、435 結構公司 415、443、476 會計政策 443 後償貸款資本 57 附屬公司 413、440 會計政策 413 持續節省 32 可持續發展 9、36 委員會 286 風險 116、201 系統重要性銀行 15、255、473 系統風險 231、477 T 目標 32 稅項 365 會計政策 365 為各地政府徵收 33 關鍵會計估算及判斷 366 遞延稅項 331、365、366、367、368 支出 56 股份及股息 463 已付 33 按地區及國家╱地區列示之已付 106 對賬 366 稅務及經紀交易商的調查 451 三道防線 112、186 一級資本 258、425、438、477 整體損失吸納能力 256 股東總回報 33、319 交易用途資產 377 會計政策 377 交易收益 （淨額） 49、353 交易用途負債 417 會計政策 417 交易用途組合 176、221、225 U 不可觀察數據 386 V 估計虧損風險 176、178、223 創建價值 9 價值觀 （滙豐） 10、19 供應商風險管理 231 參與志願服務 37 W 舉報 20、287 批發融資 169、218 批發貸款 144 滙豐控股有限公司 484 滙豐控股有限公司 於1959年1月1日在英格蘭註冊成立之 有限公司，受英國 《公司法》 規管 英格蘭註冊編號： 617987 註冊辦事處及集團總管理處 8 Canada Square London E14 5HQ United Kingdom 電話：44 020 7991 8888 傳真：44 020 7992 4880 網站：www.hsbc.com 股份登記處 主要股東名冊 Computershare Investor Services PLC The Pavilions Bridgwater Road Bristol BS99 6ZZ United Kingdom 電話：44 0870 702 0137 電郵：透過網站發出電郵 網站：www.investorcentre.co.uk/contactus 香港海外股東分冊 香港中央證券登記有限公司 香港皇后大道東183號 合和中心17樓 1712-1716室 電話：852 2862 8555 電郵： hsbc.ecom@computershare.com.hk 網站：www.computershare.com/hk/investors 百慕達海外股東分冊 Investor Relations Team HSBC Bank Bermuda Limited 6 Front Street Hamilton HM11 Bermuda 電話：1 441 299 6737 電郵：hbbm.shareholder.services@hsbc.bm 網站：www.computershare.com/investor/bm 美國預託證券存管處 The Bank of New York Mellon Depositary Receipts PO Box 43006 Providence, RI 02940-3006 USA 電話 （美國） ：1 877 283 5786 電話 （國際） ：1 201 680 6825 電郵：shrrelations@bnymellon.com 網站：www.bnymellon.com/shareowner 付款代理人 （法國） HSBC France 103 avenue des Champs Elys é es 75419 Paris Cedex 08 France 電話：33 1 40 70 22 56 電郵：ost-agence-des-titres-hsbc-reims. hbfr-do@hsbc.fr 網站：www.hsbc.fr 股票經紀 Goldman Sachs International Peterborough Court 133 Fleet Street London EC4A 2BB United Kingdom Credit Suisse Securities (Europe) Limited 1 Cabot Square London E14 4QT United Kingdom 英國滙豐銀行有限公司 8 Canada Square London E14 5HQ United Kingdom ©滙豐控股有限公司2015年 版權所有 未經滙豐控股有限公司事先書面許可，不得將本刊 任何部分以任何形式或用任何方法 （無論電子、機 械、複印、錄製或其他形式） 複製、存於檢索系統或 傳送予他人。 出版：滙豐控股有限公司 集團財務部 （倫敦） 封面設計：Black Sun Plc （倫敦） ；內頁設計：Black Sun Plc及滙豐控股有限公司 集團財務部 （倫敦） 中文翻譯：香港上海滙豐銀行有限公司 印務及電 子出版 （香港） 中譯本與英文本如有歧異，概以英文本為準。 承印：DG3 Asia Limited，香港。本刊物以植物油墨 印製，採用Revive 100 White Offset紙。此種紙張在 奧地利製造，成分為100%脫墨用後廢料。紙漿不含 氯。 FSC ®標誌表示產品所含的木料來自管理良好的森林； 該等森林根據Forest Stewardship Council ®的規例獲得 認可。 攝影 封面： （上） 滙豐歷史檔案； （下） 由Matthew Mawson拍攝 集團主席及集團行政總裁：由George Brooks拍攝 GPS 15011902 滙豐控股有限公司 8 Canada Square London E14 5HQ United Kingdom 電話：44 020 7991 8888 www.hsbc.com 滙豐今昔 150年前，滙豐在香港成立，主要為 亞洲與歐洲之間的貿易往來提供融資。 封面圖片展現滙豐隨時代而轉變的歷 程。上圖顯示滙豐於1865年成立後數 年，總行矗立於維多利亞港畔 （圖左） 。 下圖為今日的維多利亞港，左起第六幢 是滙豐總行大廈（部分被遮蓋） 。 香港的地形和經濟均持續轉變，由早期 的貿易中轉站蛻變為國際金融中心， 成為全球矚目的城市。滙豐與香港共同 成長，由一家小型的地區貿易銀行發展 至今，已成為世界最大的銀行及金融服 務機構之一。 與1865年一樣，滙豐的香港總行大 廈依然座落於皇后大道中一號。雖然 現址的總行大廈已是第四代，但滙豐 仍然秉持成立時的價值觀，宗旨是把 握市場的增長機遇、努力建立聯繫以 幫助客戶開拓商機、推動工商企業茁 壯成長及各地經濟蓬勃發展，而最終 目標是讓客戶實現理想。 150年來，我們一直為客戶提供卓越 服務，並為此深感自豪。 HSBC Holdings plc 滙豐控股有限公司 （僅於網站發布） 僱員股份計劃 2014年 12月 31日 僱員股份計劃 滙豐控股有限公司 1 根據香港聯合交易所於 2010年 12月 24日授出豁免的條款，僱員股份計劃下的認股權的全部詳情須根據上市規則第 17.07條及第 17.09條於下文披露。有關披露資料亦可於香港聯合交易所的網站 www.hkex.com.hk查閱，或向集團公司秘書長索取，地址為 8 Canada Square, London E14 5HQ。 僱員股份計劃 滙豐根據滙豐股份計劃授出認股權及特別股份獎勵，以使僱員利益與股東利益保持一致。 第 2至 4頁載有尚未行使之認股權詳情，包括根據《香港僱傭條例》界定為「連續性合約」之僱傭合約受聘之僱員持有之認股權。授出認股權均不收取代價。主要股東、貨品或服務供應商並無獲授認股權，而授出之認股權亦無超越各項股份計劃之個別上限。滙豐年內概無註銷任何認股權。 在僱員股份計劃下可認購的滙豐控股普通股數目受到下列限制。在任何 10年期間，根據 2011年滙豐股份計劃及滙豐控股推行的所有其他僱員股份計劃可予發行或承諾發行的股份，合共不得超過滙豐控股不時已發行普通股的 10%（於 2015年 2月 23日約為 19.22億股滙豐控股普通股）。在任何 10年期間，根據 2011年滙豐股份計劃及滙豐控股採納的所有其他特別股份計劃可予發行或承諾發行的股份，合共不得超過滙豐控股不時已發行普通股的 5%（於 2015年 2月 23日約為 9.61億股滙豐控股普通股）。因行使 2011年滙豐股份計劃項下所有認股權及根據滙豐控股推行的任何其他僱員股份計劃授出的認股權而可予發行的滙豐控股普通股數目，不得超過 1,781,741,789股滙豐控股普通股。根據滙豐控股各項儲蓄優先認股計劃、滙豐股份計劃及滙豐控股集團優先認股計劃，於 2014年 12月 31日尚有可認購 72,825,654股滙豐控股普通股（已發行普通股的 0.38%）之認股權仍未行使。於任何時間，因行使 2011年滙豐股份計劃項下的認股權及根據滙豐控股推行的任何其他僱員股份計劃授出而尚未行使的認股權而可予發行的股份數目，不得超過不時已發行股份的 30%。若根據任何該等計劃授出認股權會導致超過限制，則不得進一步授出認股權。滙豐控股各董事持有可認購滙豐控股股份之認股權詳情，載於《 2014年報及賬目》第 320頁之「董事薪酬報告」內。 授出認股權及股份獎勵（以發行新股的方式支付）對每股盈利之影響，於綜合收益表以每股攤薄後盈利列示，詳情載於《 2014年報及賬目》財務報表附註 10「每股盈利」項內。 全體僱員股份計劃 滙豐控股儲蓄優先認股計劃及滙豐控股儲蓄優先認股計劃：國際均為全體僱員優先認股計劃。根據此等計劃，合資格滙豐僱員獲授認股權，以購入滙豐控股普通股。滙豐不會再根據滙豐控股儲蓄優先認股計劃：國際授出認股權，而最後一批認股權已於 2012年授出。新的國際性全體僱員股份購買計劃則已於 2013年第三季推行。 對於根據滙豐控股儲蓄優先認股計劃授出的認股權，僱員可於三年或五年期間每月供款不超過 500英鎊，並在有關儲蓄合約開始後第三或第五周年期滿的六個月內選擇行使認股權。僱員亦可選擇以現金方式取回儲蓄及（如適用）任何利息或花紅。如屬被裁退、因受傷或健康欠佳而退休、退休、受僱業務轉讓予另一機構，或僱用公司之控制權有變，則認股權可於有關儲蓄合約完成前行使。在若干情況下，根據全體僱員優先認股計劃所授出認股權的行使期可延長，例如倘參與者身故，遺囑執行人可於正常行使期屆滿後六個月內行使有關認股權。 上一段載列的條款亦適用於截至 2012年 4月已根據滙豐控股儲蓄優先認股計劃：國際授出的認股權，惟供款上限為 250英鎊的等值金額。此外，根據滙豐控股儲蓄優先認股計劃：國際獲授認股權的僱員可選擇於一年期間每月供款，並在儲蓄合約開始後第一周年期滿的三個月內行使認股權。 根據滙豐控股儲蓄優先認股計劃及滙豐控股儲蓄優先認股計劃：國際，認股權行使價乃參考最接近要約日期前五個營業日之滙豐控股普通股平均市值折讓 20%釐定（各項計劃之詳情請參閱第 2至 3頁的列表）（不包括根據美國業務轄下計劃授出的一年期認股權，該等認股權折讓 15%﹚。在適用情況下，美元、港元及歐元行使價會按有 僱員股份計劃（續） 滙豐控股有限公司 2 關要約日期前一個工作日的適用匯率從英鎊行使價折算。於 2014年 9月 22日，即 2014年根據滙豐控股儲蓄優先認股計劃授出認股權前一日，每股滙豐控股普通股的收市價為 6.58英鎊。 滙豐不會再根據滙豐控股儲蓄優先認股計劃：國際授出認股權。新的國際性全體僱員股份購買計劃則已於 2013年第三季推行。滙豐控股儲蓄優先認股計劃及滙豐控股儲蓄優先認購計劃：國際將於 2015年 5月 27日終止，除非董事議決提早終止有關計劃則另作別論。 滙豐控股儲蓄優先認股計劃 滙豐控股普通股 行使價 行使期 於 2014年 年內授出 年內行使 年內失效 於 2014年 授出日期 （英鎊） 開始日期 截止日期 1月 1日 之認股權 之認股權 1 之認股權 12月 31日 2008年 4月 30日 5.9397 2013年 8月 1日 2014年 2月 1日 76,623 – 46,621 30,002 – 2009年 4月 29日 3.3116 2014年 8月 1日 2015年 2月 1日 22,639,846 – 21,808,763 204,048 627,035 2010年 4月 21日 5.4573 2013年 8月 1日 2014年 2月 1日 210,336 – 147,385 62,951 – 2010年 4月 21日 5.4573 2015年 8月 1日 2016年 2月 1日 1,268,403 – 44,806 65,067 1,158,530 2011年 4月 20日 5.0971 2014年 8月 1日 2015年 2月 1日 2,870,885 – 2,522,957 123,499 224,429 2011年 4月 20日 5.0971 2016年 8月 1日 2017年 2月 1日 1,438,431 – 39,203 90,802 1,308,426 2012年 4月 24日 4.4621 2015年 8月 1日 2016年 2月 1日 13,491,512 – 426,112 1,035,080 12,030,320 2012年 4月 24日 4.4621 2017年 8月 1日 2018年 2月 1日 3,518,307 – 44,214 246,579 3,227,514 2013年 9月 20日 5.4738 2016年 11月 1日 2017年 5月 1日 6,477,354 – 15,472 1,127,476 5,334,406 2013年 9月 20日 5.4738 2018年 11月 1日 2019年 5月 1日 1,959,189 – 1,892 318,266 1,639,031 2014年 9月 23日 5.1887 2017年 11月 1日 2018年 5月 1日 – 18,553,870 – 367,085 18,186,785 2014年 9月 23日 5.1887 2019年 11月 1日 2020年 5月 1日 – 10,134,833 – 127,354 10,007,479 53,950,886 28,688,703 25,097,425 3,798,209 53,743,955 1 最接近行使認股權日期前，股份之加權平均收市價為 6.36英鎊。 滙豐控股儲蓄優先認股計劃：國際 滙豐控股普通股 行使價 行使期 於 2014年 年內行使 年內失效 於 2014年 授出日期 （英鎊） 開始日期 截止日期 1月 1日 之認股權 1 之認股權 12月 31日 2008年 4月 30日 5.9397 2013年 8月 1日 2014年 2月 1日 53,915 21,930 31,985 – 2009年 4月 29日 3.3116 2014年 8月 1日 2015年 2月 1日 5,521,719 4,985,327 150,469 385,923 2010年 4月 21日 5.4573 2013年 8月 1日 2014年 2月 1日 151,634 93,490 58,144 – 2010年 4月 21日 5.4573 2015年 8月 1日 2016年 2月 1日 218,919 1,094 14,895 202,930 2011年 4月 20日 5.0971 2014年 8月 1日 2015年 2月 1日 685,034 485,431 110,080 89,523 2011年 4月 20日 5.0971 2016年 8月 1日 2017年 2月 1日 186,806 – 34,153 152,653 2012年 4月 24日 4.4621 2015年 8月 1日 2016年 2月 1日 2,669,788 36,791 244,637 2,388,360 2012年 4月 24日 4.4621 2017年 8月 1日 2018年 2月 1日 534,635 1,120 38,845 494,670 10,022,450 5,625,183 683,208 3,714,059 僱員股份計劃（續） 滙豐控股有限公司 3 滙豐控股儲蓄優先認股計劃：國際（續） 滙豐控股普通股 行使價 行使期 於 2014年 年內行使 年內失效 於 2014年 授出日期 （美元） 開始日期 截止日期 1月 1日 之認股權 1 之認股權 12月 31日 2008年 4月 30日 11.8824 2013年 8月 1日 2014年 2月 1日 43,932 – 43,932 – 2009年 4月 29日 4.8876 2014年 8月 1日 2015年 2月 1日 1,497,456 1,222,861 98,833 175,762 2010年 4月 21日 8.1232 2013年 8月 1日 2014年 2月 1日 216,423 59,605 156,818 – 2010年 4月 21日 8.1232 2015年 8月 1日 2016年 2月 1日 129,756 – 14,291 115,465 2011年 4月 20日 8.2094 2014年 8月 1日 2015年 2月 1日 447,961 228,821 73,723 145,417 2011年 4月 20日 8.2094 2016年 8月 1日 2017年 2月 1日 111,396 – 13,979 97,417 2012年 4月 24日 7.1456 2015年 8月 1日 2016年 2月 1日 1,261,942 14,404 142,080 1,105,458 2012年 4月 24日 7.1456 2017年 8月 1日 2018年 2月 1日 288,203 3,147 57,247 227,809 3,997,069 1,528,838 600,903 1,867,328 （歐元） 2008年 4月 30日 7.5571 2013年 8月 1日 2014年 2月 1日 8,329 – 8,329 – 2009年 4月 29日 3.6361 2014年 8月 1日 2015年 2月 1日 843,087 795,030 4,755 43,302 2010年 4月 21日 6.0657 2013年 8月 1日 2014年 2月 1日 15,417 10,363 5,054 – 2010年 4月 21日 6.0657 2015年 8月 1日 2016年 2月 1日 37,852 – 3,287 34,565 2011年 4月 20日 5.7974 2014年 8月 1日 2015年 2月 1日 151,415 125,770 7,364 18,281 2011年 4月 20日 5.7974 2016年 8月 1日 2017年 2月 1日 44,841 – 3,008 41,833 2012年 4月 24日 5.3532 2015年 8月 1日 2016年 2月 1日 384,509 3,828 27,400 353,281 2012年 4月 24日 5.3532 2017年 8月 31日 2018年 2月 1日 89,202 186 8,776 80,240 1,574,652 935,177 67,973 571,502 （港元） 2008年 4月 30日 92.5881 2013年 8月 1日 2014年 2月 1日 13,316 – 13,316 – 2009年 4月 29日 37.8797 2014年 8月 1日 2015年 2月 1日 17,205,856 16,783,784 148,192 273,880 2010年 4月 21日 62.9770 2013年 8月 1日 2014年 2月 1日 82,670 58,016 24,654 – 2010年 4月 21日 62.9770 2015年 8月 1日 2016年 2月 1日 220,985 2,565 18,103 200,317 2011年 4月 20日 63.9864 2014年 8月 1日 2015年 2月 1日 384,366 305,638 8,983 69,745 2011年 4月 20日 63.9864 2016年 8月 1日 2017年 2月 1日 162,426 1,770 10,119 150,537 2012年 4月 24日 55.4701 2015年 8月 1日 2016年 2月 1日 5,466,147 52,198 261,533 5,152,416 2012年 4月 24日 55.4701 2017年 8月 1日 2018年 2月 1日 679,575 3,027 54,661 621,887 24,215,341 17,206,998 539,561 6,468,782 1 最接近行使認股權日期前，股份之加權平均收市價為 6.36英鎊。 特別股份計劃 《 2014年報及賬目》財務報表附註 6載有以股份為基礎的支出詳情，包括根據滙豐股份計劃授出的特別股份獎勵。 有限制股份獎勵可酌情授予高級行政人員，為集團花紅遞延政策的發放機制。有限制股份的有條件獎勵界定僱員有權擁有的股份數目（一般於授出獎勵日期起計不超過三年期間內擁有，且通常有關僱員須仍然受僱）。在若干情況下，僱員的權益取決於其能否達致指定的表現條件。 集團業績表現股份計劃及固定酬勞津貼已納入 2011年滙豐股份計劃，詳情請參閱《 2014年報及賬目》第 303頁。 根據 2011年滙豐股份計劃於任何一年內授予一名僱員之獎勵或認股權之最高價值，為該僱員年薪之 600%。凡於僱員上任時或上任後不久授出，或代替僱員原來有權獲得的全部或任何部分花紅的任何有限制股份獎勵，不會計算在限額之內。 自 2005年 9月以來，集團並無授出任何特別認股權。就行使尚未行使的特別認股權而言，目前並無任何尚未達致的表現條件。認股權一般可於授出日期第三至第十周年期間行使。 根據滙豐股份計劃及滙豐控股集團優先認股計劃所授出認股權之行使價，為認股權授出前五個營業日普通股之平均市值、於認股權授出當日普通股之市值或一股股份之面值，以較高者為準。 僱員股份計劃（續） 滙豐控股有限公司 4 特別優先認股計劃 滙豐控股普通股 行使價 行使期 於 2014年 年內行使 年內失效 於 2014年 授出日期 （英鎊） 開始日期 截止日期 1月 1日 之認股權 之認股權 12月 31日 滙豐控股集團優先認股計劃 1, 2 2004年 4月 30日 7.2181 2007年 4月 30日 2014年 4月 30日 48,256,628 1,434 48,255,194 – 2004年 8月 27日 7.5379 2007年 8月 27日 2014年 8月 27日 281,651 – 281,651 – 2005年 4月 20日 7.2869 2008年 4月 20日 2015年 4月 20日 6,487,589 – 113,607 6,373,982 55,025,868 1,434 48,650,452 6,373,982 滙豐股份計劃 2005年 9月 30日 7.9911 2008年 9月 30日 2015年 9月 30日 86,046 – – 86,046 1 滙豐控股集團優先認股計劃已於 2005年 5月 26日屆滿，而滙豐股份計劃則於 2011年 5月 27日屆滿。概無根據 2011年滙豐股份計劃授出任何認股權。 2 最接近行使認股權日期前，股份之加權平均收市價為 6.09英鎊。 |
190771 | https://math.stackexchange.com/questions/84559/solving-cn-n-2-pn-2-7-n | algebra precalculus - Solving $c(n, n-2) - p(n, 2) = 7 - n$ - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
Solving c(n,n−2)−p(n,2)=7−n c(n,n−2)−p(n,2)=7−n
Ask Question
Asked 13 years, 10 months ago
Modified13 years, 10 months ago
Viewed 130 times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
To help with a friends homework, I was asked how to solve this equation for n n:
c(n,n−2)−p(n,2)=7−n c(n,n−2)−p(n,2)=7−n
Having no further information about what c and p are supposed to be (and they don't seem to know either) I assumed they were probably combination and permutation. But when I expanded that out, I was left with a polynomial n 2−3 n+14 n 2−3 n+14 which has complex solutions. Is there something else obvious I'm missing that c and p could be here?
algebra-precalculus
statistics
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited Nov 22, 2011 at 22:44
Gerry Myerson
187k 13 13 gold badges 235 235 silver badges 408 408 bronze badges
asked Nov 22, 2011 at 13:29
wimwim
806 1 1 gold badge 9 9 silver badges 21 21 bronze badges
3
1 Then your friend should either be asking his/her teacher, or consulting his/her textbook to know the meaning of c and p...J. M. ain't a mathematician –J. M. ain't a mathematician 2011-11-22 13:35:37 +00:00 Commented Nov 22, 2011 at 13:35
I also get that polynomial. Hmm, you might as well ask more details from your friend.Huang –Huang 2011-11-22 13:36:56 +00:00 Commented Nov 22, 2011 at 13:36
1 I get the same; the equation has no solutions if c c and p p are indeed combinations and permutations.David Mitra –David Mitra 2011-11-22 13:47:20 +00:00 Commented Nov 22, 2011 at 13:47
Add a comment|
0
Sorted by: Reset to default
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
algebra-precalculus
statistics
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Report this ad
Related
4Solving x 2−−√3+x−−√3=2 x 2 3+x 3=2
1Interpreting What Data Means
4Can this be solved algebraically? 2 x(6−x)=8 x 2 x(6−x)=8 x
0The black/white hat problem. Is it 50-50?
4Spivak Question 14, Chapter 1, clarifying a statement.
2Do all real cubic roots have real closed forms?
Hot Network Questions
Storing a session token in localstorage
How different is Roman Latin?
Where is the first repetition in the cumulative hierarchy up to elementary equivalence?
Should I let a player go because of their inability to handle setbacks?
On being a Maître de conférence (France): Importance of Postdoc
How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done?
Another way to draw RegionDifference of a cylinder and Cuboid
Interpret G-code
How to rsync a large file by comparing earlier versions on the sending end?
Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results
Does the curvature engine's wake really last forever?
Riffle a list of binary functions into list of arguments to produce a result
Why are LDS temple garments secret?
How many stars is possible to obtain in your savefile?
Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done?
PSTricks error regarding \pst@makenotverbbox
Who is the target audience of Netanyahu's speech at the United Nations?
Can a GeoTIFF have 2 separate NoData values?
alignment in a table with custom separator
How can the problem of a warlock with two spell slots be solved?
Does a Linux console change color when it crashes?
Gluteus medius inactivity while riding
Does the Mishna or Gemara ever explicitly mention the second day of Shavuot?
Checking model assumptions at cluster level vs global level?
more hot questions
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
190772 | https://arxiv.org/pdf/1202.2986 | arXiv:1202.2986v3 [math.CO] 24 Dec 2014
ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS
NHAN BAO HO
Abstract. We study the periodicity of nim-sequences for subtraction games hav-ing subtraction sets with three elements. In particular, we give solutions in several cases, and we describe how these subtraction sets can be augmented by additional numbers without changing the nim-sequences. The paper concludes with a conjec-ture on ultimately bipartite subtraction games.
Introduction
A subtraction game is a two-player game involving a pile of coins and a finite set
S of positive integers called the subtraction set . The two players move alternately, subtracting some s coins such that s ∈ S. The player who makes the last move wins. Subtraction games provide classical examples of impartial combinatorial games; see [1, 3, 4]. They are completely understood for two-element subtraction sets. For larger subtraction sets, they are known to be all ultimately periodic [6, p.38], but a complete solution of these games is still not known, even for three-element subtraction sets. The purpose of this paper is to report further results in this area. Throughout this paper, the subtraction set S = {s1, s 2, . . . , s k} will be ordered
s1 < s 2 < · · · < s k. The subtraction game with subtraction set S is denoted by S.When we need to specify the subtraction set, we will use S(S) or S(s1, s 2, . . . , s k). For each nonnegative integer n, denote G(n) the Sprague-Grundy value , or nim-value for short, of the single pile of size n of the subtraction game S. The sequence
{G (n)}n≥0 is called the nim-sequence .Suppose for the moment that a subtraction set {s1, s 2, . . . , s k} has d =gcd( s1, s 2, . . . , s k) > 1. Let s′
i
= si/d for 1 ≤ i ≤ k. The nim-sequence for the game S(s1, s 2, . . . , s k) is exactly the d-plicate of that for the game S(s′
1
, s ′
2
, . . . , s ′
k
). That means the former can be obtained from the latter by repeating each value of the latter exactly d times [4, p. 529]. Thus, it suffices to consider subtraction sets whose members are relatively prime.
2000 Mathematics Subject Classification. Primary: 91A46.
Key words and phrases. combinatorial games, subtraction games, bipartite subtraction games, subtraction sets, nim-sequence, periodicity, expansion set.
12NHAN BAO HO
Recall that the sequence {G (n)}n≥0 is said to be ultimately periodic if there exist integers p ≥ 1 and n0 ≥ 1 such that G(n + p) = G(n) for all n ≥ n0. The smallest such numbers n0 and p are called the pre-period length and period length respectively [1, p. 145]. If n0 = 0, the sequence is said to be purely periodic . A purely peri-odic (resp. ultimately periodic) game is called bipartite (resp. ultimately bipartite )if p = 2. (Ultimately bipartite subtraction games ultimately have alternating nim-values 0 , 1, 0, 1, . . . ). The periodicity of subtraction games is discussed in [1, 2, 3, 4]. Bipartite games are first introduced in . Throughout this paper, when saying that a subtraction game has periodic nim-values g1g2 . . . g p, we mean that ultimately, the nim-sequence is the infinite repetition of the subsequence g1, g 2, . . . , g p. The n0 nim-values G(0) , G(1) , . . . , G(n0 −1) is called the pre-periodic nim-values. The omission of commas between the nim-values does not lead to any misunderstanding as all nim-values presented in this paper have exactly one digit. The following lemma gives us a useful idea of how much calculation we need to perform to determine the pre-period and period lengths.
Lemma 1. [1, p. 148] Let sk = max( S). For minimal n0 and p such that G(n + p) =
G(n) for n0 ≤ n < n 0 + sk, the subtraction game S is purely periodic with the pre-period length n0 and the period length p.
Remark 1. [3, p. 84] For a given subtraction set S, if there exists a positive integer
s such that G(n + s) 6 = G(n) for all nonnegative integers n, then s can be adjoined to the subtraction set S without changing the nim-sequence. In this case, for brevity, we will simply say that s can be adjoined to S. The set of all such elements, including elements in S, is called the expansion set of S and denoted by Sex .In reality, to identify such an s, we need to calculate only a small range of nim-values, as much as the calculation needed in Lemma 1. We detail this range as follow.
Theorem 1. Let S be a subtraction game with pre-period length n0 and period length
p.
(i) A number s < n 0 + p can be adjoined to S if and only if G(n + s) 6 = G(n) for all n such that 0 ≤ n < n 0 + p.
(ii ) A number s ≥ n0 + p can be adjoined to S if and only if s − p can be. Proof. The theorem follows immediately from Remark 1 and the definition of ulti-mately periodicity.
It follows from Theorem 1 that if s ≥ n0 + p and s can be adjoined to S then
s = s′ + mp for some m and s′ such that n0 ≤ s′ < n 0 + p and s′ can be adjoined to S.We are going to employ this idea to represent Sex with no more than max( n0 + p, s k)ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 3
elements with sk = max( S). (There is another method to present Sex with no more than n0 + p elements but we avoid its complexity). Let us partition S into two parts: S1 containing elements s of S such that s < n 0
and S2 = S \ S1. Let S′
1
(resp. S′
2
) be the set of all s which can be adjoined to S
such that s / ∈ S and s < n 0 (resp. n0 ≤ s < n 0 + p). Then
Sex = {S1 ∪ S′
1
} ∪ { S2 ∪ S′
2
}∗p
in which T ∗p = {t + mp |t ∈ T, m ≥ 0}. Note that in the formula for Sex , some of the sets S1, S′
1
, S2, and S′
2
may be empty. In particular, S1 ∪ S′
1
= ∅ if S is purely periodic ( n0 = 0).
Example 1. The subtraction game S(1 , 8, 11 , 27) is ultimately periodic with n0 =13, p = 19 and has expansion set Sex = {1, 8, 11 } ∪ { 13 , 20 , 27 }∗19 . Here S1 =
{1, 8, 11 }, S′
1
= ∅, S2 = {27 }, and S′
2
= {13 , 20 }.
Definition 1. If Sex = {S1} ∪ { S2}∗p (equivalently, S′
1
∪ S′
2
= ∅), the set S is said to be non-expandable . Otherwise, it is expandable .Some values of the expansion sets of certain subtraction sets with numbers up to 7 can be found in [3, pp. 84-85]. The following result from shows that for bipartite games, Sex is the set of odd positive integers.
Theorem 2. Let S be a subtraction set with gcd( S) = 1 . The subtraction game S
is bipartite if and only if 1 ∈ S and the elements of S are all odd.
Remark 2. The key results of this paper are the determinations of periodic nim-values and expansion sets in several specific cases. We do not give the proofs for all results. There are considerable similarities in the proofs of various results. We give one proof for the expansion set of the subtraction set {a, b } and one proof for the nim-sequence for the subtraction game S(1 , a, b ). The reason is simply that these two games are addressed early in the paper. The details of proofs for other stated results are similar; some are quite tedious but they are all straightforward. This paper is organized as follows. Section 2 lists our results on 5 classes of subtraction sets: {a, b }, {1, a, b }, {a, b, a + b}, {a, b, a + b + 1 }, and {a, b, a + b + 2 ja }.We consider various cases for each class. Section 3 gives two proofs as mentioned in Remark 2. We provide proofs in Section 3 rather than in Section 2 so that the reader can follow Section 2 easily. In Section 4, we present a family of ultimately bipartite subtraction games and we examine the expansion sets of subtraction sets of ultimately bipartite subtraction games. The section ends with a conjecture that has become apparent during this work. We include a code written in Maple software in the appendix. The code receives input ( S, n ) in which S is the subtraction set and the code will calculate the sequence 4 NHAN BAO HO
{G (i)}0≤i≤n to identify the nim-sequence. If the input n is large enough (at least
n0 + p + sk + 1), the code will return: pre-period and period length, and pre-periodic and periodic nim-values, together with the expansion set Sex . When n is not large enough, the code will return nothing, requiring a larger n.Our results also provide evidential support for the claim that the understanding of the pattern of the periodicity of subtraction games and the expansion sets of their subtraction sets is very far from being complete. For example, slight differences on the subtraction sets in Table 3 would result in considerable modifications of both pre-period and period lengths as well as expansion sets. 2. Results of nim-sequences and expansion sets
We first study the expansion set of the subtraction set {a, b } where a and b are relatively prime. When a = 1, we only consider the case where b is even since the case where a = 1 and b is odd is treated by Theorem 2. Berlekamp et al. [4, p. 530] showed that if b = ta +r such that t ≥ 1 and 0 ≤ r < a ,the subtraction game S(a, b ) is purely periodic with period length a + b and periodic nim-values {
(0 a1a) t
2
0r2a−r1r, if t is even; (0 a1a)t+1
2
2r, otherwise .
Moreover, they also observed that the expansion set Sex is periodic in the sense that if n ∈ Sex , then n + a + b ∈ Sex . We generalize this observation in Theorem 3.
Theorem 3. Let a and b be relatively prime, positive integers such that a < b .Suppose furthermore that if a = 1 , then b is even. Consider the subtraction game
S(a, b ). If a + 1 < b ≤ 2a, then the subtraction set has expansion set
{a, a + 1 , . . . , b }∗(a+b).
If a = 1 , or b = a + 1 , or b > 2a, the subtraction set is non-expandable. Proof. The proof is provided in Section 3.
We now examine the periodicity and the expansion sets of some subtraction games whose subtraction sets are {1, a, b }. For the case where a is odd, Theorem 2 solves the case where b is odd. When b is even, the game is purely periodic. The periodic nim-values and the expansion set are described in Theorem 4.
Theorem 4. Let a be an odd positive integer and let b be an even positive integer. The subtraction game S(1 , a, b ) is purely periodic with the period a+b and the periodic nim-values
(01) b
2
(23) a−1
2
2.ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 5
Moreover, the subtraction set has expansion set
{{ 1, 3, . . . , a } ∪ { b, b + 2 , . . . , b + a − 1}} ∗(a+b).
Proof. The proof is provided in Section 3.
For the case where a is even, we represent b = ka + r where r < a . Our results cover the cases 1 ≤ k ≤ 3. Table 1 (resp. Table 2, Table 3) lists periodic nim-values and expansion sets of those games corresponding to the case k = 1 (resp. k = 2,
k = 3). (The case k = 1 and r = 1 (corresponding to b = a + 1) is analyzed in [4, p. 530].) In our tables, we use “period” rather than “period length”.
Remark 3. To make the tables easily readable, we do not include the pre-period lengths and pre-periodic nim-values of non-purely periodic games. We provide them in B (Tables 10, 11, 12).
Remark 4. We place symbol “ · · · ” in front of nim-sequences of those games that are ultimately periodic to distinguish from purely periodic games. We now examine subtraction game S(a, b, a + b) with 2 ≤ a. Note that b can be represented as ka + r for some integers k and r such that 0 ≤ r ≤ a − 1. Recall from [4, p.531] that when k is odd, the game S(a, b, a + b) is purely periodic with period length ( a + b + ka ) and periodic nim-values (0 a1a)k+1
2
(2 a3a)k−1
2
2a3r.
We give the expansion set of this game in Table 4. When k is even, it is claimed in [4, p. 531] that the game is purely periodic with period length (2 b + r)a. We have not seen a proof of this claim, and at present we do not know the nim-sequence and expansion set for this case. We next give some results for subtraction games S(a, b, a + b + 1) in Table 5. We obtain results for those cases where b = 2 a + r with r ≤ min(2 , a − 1). Pre-period lengths and pre-periodic nim-values for those games that are non-purely periodic are provided in Table 13 of B. Finally, we study subtraction games S(a, b, b + 2 ja ). Note that we can write
b = ka + r for some positive k and r such that r < a . The case where r = 0 is skipped here as it is studied before. We deal with two general cases with j ∈ { 1, 2}
and one special case with j = 3 and r = 1 as shown in Table 6. 6 NHAN BAO HO
Subtraction set
{1, a, b }
a: even
b = a + r
r period p periodic nim-values expansion set
r = 2 even a + 1 (01) a
2 2 non-expandable
r > 2 a + b (01) a
2 2(01) a
2 (23) r
2 −12 {1, a, a + 2 , . . . , b, a + b − 1}∗p
r = 1
odd
a + b − 1 (01) a
2 (23) a
2 {{ 1, 3, . . . , a − 1} ∪ { a}∪{ a + 1 , a + 1 + 2 , . . . , a + b − 2}} ∗p
r < a − 3 a + b · · · 2(32) a−r−3
2 (01) r+1
2 2(01) a−2
2 2(01) r+1
2
{1, a, b } ∪ { b + 2 , a + b + 1 , 2a + b}∗p
r = a − 3 {1, a, b } ∪ { b + 2 }∗p
r = a − 1 b + 1 (01) a
2 2(01) a−2
2 2 non expandable
Table 1. Subtraction games S(1 , a, b ) with b = a + r.
Subtraction set
{1,a,b }
a: even
b = 2 a + r
r period p periodic nim-values expansion set
r = 0, a ≥ 4
even a + b
(01) a
2 2(01) a
2 (23) a−2
2 2 {{ 1} ∪ { a, a + 2 , . . . , b } ∪ { a + b − 1}} ∗p
2 = r < a − 4 · · · 2(32) a
2 −3(01) 22((01) a
2 −1)22(01) 2 non-expandable
4 ≤ r < a − 4 · · · 2(32) a
2 −4(01) 32((01) a
2 2) 2(01) 3
{1, a, b, b + 2 , a + b + 1 , a + 2 b + 2 }
4 < r = a − 4 a − 1 · · · 2(01) a
2 −1
4 ≤ r = a − 2 b + 1 · · · 2(01) a−2
2 2(01) a
2 2(01) a−2
2 {1, a, b } ∪ { b + 2 , a + b + 1 }∗p
r = 1 , 3 odd a + 1 (01) a
2 2 non-expandable (same as S(1 , a ))
5 ≤ r ≤ a − 1 a + b ((01) a
2 2) 2(01) a
2 (23) r−3
2 2 {1, a, a + 2 , b − 2, b, a + b − 1}∗p
Table 2. Subtraction games S(1 , a, b ) with a being even and b = 2 a + r.ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 7
Subtraction set
{1,a,b }
a: even
b = 3 a + r
r period p periodic nim-values expansion set
r = 0, a ≥ 4
even
b + 1 ((01) a
2 2) 2(01) a
2 −12 {1, a, 2a + 1 , b }∗p
r = 2 , 4 a + 1 (01) a
2 2 non-expandable
r = 6
a + b
((01) a
2 2) 4(32) r
2 −2 {1, a, a + 2 , 2a + 3 , b − 2, b, a + b − 1}∗p
r > 6 {1, a, a + 2 , b − 2, b, a + b − 1}∗p
r = 1
odd
((01) a
2 2) 3(32) a
2 −1
3 = r < a − 5
· · · 2(32) a−r−5
2 (01) r+3
2 2((01) a
2 −12) 3(01) r+3
2
non-expandable
3 < r < a − 5
{1, a, b, b + 2 , a + b + 1 , a + 2 b + 2 }5 ≤ r = a − 5 a − 1
5 ≤ r = a − 3 b + 1 · · · 2(01) a
2 −1(2(01) a
2 )22(01) a
2 −1
5 ≤ r = a − 1 {1, a, b } ∪ { b + 2 , a + b + 1 }∗p
Table 3. Subtraction games S(1 , a, b ) with a being even and b = 3 a + r.
Subtraction set k period p periodic nim-values expansion set
{a,b,a+b }
a ≥ 2
b = ka + r
0 < r < a
odd a + 2 b − r (0 a1a)k+1
2 (2 a3a)k−1
2 2a3r
{{sa : s is odd , s ≤ k}∪{ ka + 1 , ka + 2 , . . . , b + a}∪{ b + sa : s is odd , s ≤ k}}∗p
even (2 b + r)a unknown unknown
Table 4. Subtraction games S(a, b, a + b). 8NHAN BAO HO
Subtraction set
{a, b, b + a + 1 }
a ≥ 4
b = 2 a + rr ≤ 2
period p periodic nim-values expansion set
{a, 2a, 3a + 1 }
b + 2 a + 1
· · · 10 a−121 a−102 a−1103 a−22 {a} ∪ { 2a, 3a + 1 , 5a}∗p
{a, 2a + 1 , 3a + 2 } 0a1a02 a−1103 a−221 non-expandable
{a,2a+2,3a+3 } · · · 2a−211003 a−422110 a−121 a−100 {a, 2a + 2 , 3a + 3 } ∪ { 3a + 4 }∗p
and non-expandable if a = 4
Table 5. Subsection games S(a, b, b + a + 1) with a ≥ 4, b = 2 a + r, and 0 ≤ r ≤ 2.
Subtraction set
{a, b, b + 2 ja } period p k periodic nim-values expansion set
{a,b,b+2a } a + b even (0 a1a)k
2 0r2a−r1r
non-expandable
odd (0 a1a)k+1
2 2r
{a,b,b+4a }
k ≥ 3 2b + 4 a even ((0 a1a)k
2 0r2a−r1r)23a−r0r2a−r1r {a, b, b + 2 a, b + 4 a,
2b + 3 a}∗p
odd ((0 a1a)k+1
2 2r)20a−r3r1a−r2r
{a,b,b+6a }
k ≥ 3, r = 1 2b + 6 a odd ((0 a1a)k+1
2 2) 2(0 a−131 a−12) 2 {a, b, b + 2 a, b + 4 a, b + 6 a, 2b + 5 a}∗p
Table 6. Subtraction games S(a, b, b + 2 ja ) with a ≥ 2, b = ka + r.ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 9
k i t n G(n + s) = G(n)
even k < t a + b G(a + b + ka + i) = G(ka + i) = 0 = G(0) = G(a + b)
k = t i < r (t − 1) a G(( t − 1) a + ta + i) = G(( t − 2) a − (r + i) + ta + a + r)= G(( t − 2) a − (r + i) + a + b) = G(( t − 2) a − (r + i)) = 1 = G(( t − 1) a)
odd
k = 1 i > 0 a − 1 G(a − 1 + a + i) = G(2 a + i − 1) = 0 = G(a − 1)
k > 1
i ≤ r
even
b − a − i G(b − a − i + ka + i) = G(a + b + ( k − 2) a) = G(( k − 2) a) = 1 = G(b − a − i)
i > r = 0 b − a G(b − a + ka + i) = G(a + b + ( k − 2) a + i) = G(( k − 2) a + i) = 1 = G(b − a)
i > r > 0 a − i + r − 1 G(a − i + r − 1 + ka + i) = G((k + 1) a + r − 1) = 0 = G(a − i + r − 1)
i < r
odd
ta G(ta + ka + i) = G((k − 1) a − (r − i)) = 1 = G(ta )
i = r = 0 not hold as we require gcd( a, b ) = 1
i = r > 0 a − 1 G(a − 1 + ka + i) = G((k + 1) a + i − 1) = 0 = G(a − 1)
i > r (t + 1) a − 1 G(( t + 1) a − 1 + ka + i) = G(ka + i − r − 1) = 1 = G(( t + 1) a − 1)
Table 7. b = ta + r > 2a; a + 1 ≤ s = ka + i ≤ b − 1. 10 NHAN BAO HO
Two proofs
As outlined in Remark 2, we give in this section two proofs, one for the expansion set of the subtraction set {a, b } claimed in Theorem 3 and another one for the nim-sequence for the subtraction game S(1 , a, b ) claimed in Theorem 4. 3.1. Proof of Theorem 3. Recall that the game S(a, b ) has periodic nim-values
{
(0 a1a) t
2
0r2a−r1r, if t is even; (0 a1a)t+1
2
2r, otherwise .
(1) in which b = ta + r [3, p. 530]. We now prove the expansion set. The proof requires checking several cases, using Theorem 1. The outline of the proof is as follows. (a) For each of the three cases a = 1, b = a + 1, and b > 2a, we show that for every positive integer s, there exists a nonnegative integer n such that G(n + s) = G(n)and so the subtraction set {a, b } is non-expandable. (b) For a + 1 < b ≤ 2a, we show that G(n + s) 6 = G(n) for all nonnegative integers n
if and only if s ∈ { a, a + 1 , . . . , b }∗(a+b) = X and so the subtraction set {a, b } is expandable to X.By Theorem 1, in the two cases (a) and (b) we only need to examine s < a + b with
s / ∈ { a, b }.(a) We first consider the case where a = 1 or b = a + 1 or b > 2a.(i) Consider the case a = 1. We only examine the case b is even. Let s ≤ a+b−1such that s / ∈ { 1, b }. Recall from (1) that the first b nim-values G(n) (for n
from 0 to b − 1) are (01) b
2
. In particular,
G(n) = 1 if n ≤ b − 1 and n is odd .(2)
• If s is even then s ≤ b − 2 as s < b and b is even, implying that s + 1 is odd and s + 1 ≤ b − 1. Let n = 1. By (2), we have G(1 + s) = 1 = G(1).
• If s is odd, let n = b − 1. As s − 2 is odd and 1 ≤ s − 2 ≤ b − 3, G(s − 2) = 1 = G(b − 1) by (2). Also note that G(b + 1 + s − 2) = G(s − 2) since
p = b + 1. Therefore, G(b − 1 + s) = G(b + 1 + s − 2) = G(s − 2) = G(b − 1) .
(ii) Consider the case b = a + 1. We examine two intervals for s ≤ a + b − 1:
{
1 ≤ s ≤ a − 1,a + 2 ≤ s ≤ a + b − 1.ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 11
Recall that the first 4 a + 2 nim-values G(n) (for n from 0 to 4 a + 1) is 0a1a20 a1a2. In other words, for n ≤ 4a + 1, we have
G(n) =
{
0, if either 0 ≤ n ≤ a − 1 or 2 a + 1 ≤ n ≤ 3a
1, if either a ≤ n ≤ 2a − 1 or 3 a + 1 ≤ n ≤ 4a
(3)
• For 1 ≤ s ≤ a − 1, choose n = a. Then a + 1 ≤ a + s ≤ 2a − 1. By (3),
G(a + s) = 1 = G(a).
• For a+2 ≤ s ≤ a+b−1, choose n = a−1. Then 0 ≤ a−1+ s−(a+b) ≤ a−2and so G(a − 1 + s − (a + b)) = 0 = G(a − 1) by (3). By the periodicity,
G(a − 1 + s) = G(a − 1 + s − (a + b)). Therefore, G(a − 1 + s) = G(a − 1). (iii) Consider the case b > 2a. We examine three intervals for s ≤ a + b − 1:
1 ≤ s ≤ a − 1
a + 1 ≤ s ≤ b − 1
b + 1 ≤ s ≤ a + b − 1.
• Consider the case 1 ≤ s ≤ a − 1. Let n = a. Recall that G(m) = 1 for
a ≤ m ≤ 2a−1 by (1) and so G(a+s) = 1 = G(a) as a+1 ≤ a+s ≤ 2a−1.
• Consider the case a + 1 ≤ s ≤ b − 1. We separate this case into many subcases in Table 7. It can be checked that for each s such that a + 1 ≤
s ≤ b − 1, there exists n, as shown in Table 7, such that G(n + a) = G(n). We leave the calculation to the reader.
• Consider the case b + 1 ≤ s ≤ a + b − 1. We have 0 ≤ s − b − 1 ≤ a − 2. Let n = a − 1. Recall that G(m) = 0 for 0 ≤ m ≤ a − 2 by (1), and so
G(a − 1 + s − (a + b)) = G(s − b − 1) = 0 = G(a − 1). By the periodicity,
G(a − 1 + s) = G(a − 1 + s − (a + b)). Therefore, G(a − 1 + s) = G(a − 1). (b) We now consider the case a + 1 < b ≤ 2a. We examine three intervals for s:
1 ≤ s ≤ a − 1,a < s < b, b + 1 ≤ s ≤ a + b − 1.
The case 1 ≤ s ≤ a − 1 and the case b + 1 ≤ s ≤ a + b − 1 can be treated similarly to the case (a).(iii). Consider the case a < s < b . We show that G(n + s) 6 = G(n) for all n. Recall from (1) that the game S(a, b ) with a + 1 < b ≤ 2a has the periodic nim-values
{
0a1a2a, if b = 2 a;0a1a2r, if b = a + r. 12 NHAN BAO HO
It can be checked that if G(n) is in a block 0 ′s, then G(n + s) is either in a block 1′s or in a block 2 ′s; if G(n) is in a block 1 ′s, then G(n + s) is either in a block 2 ′s
or in a block 0 ′s; if G(n) is in a block 2 ′s, then G(n + s) is either in a block 0 ′s
or in a block 1 ′s. Therefore, G(n + s) 6 = G(n) for all s and n such that a < s < b
and n ≥ 0. 3.2. Proof of Theorem 4. We will give the proof for the nim-sequence only. The expansion set can be verified by using a similar technique in Theorem 3. We will show that the first a + 2 b nim-values G(n), for 0 ≤ n ≤ a + 2 b − 1, are (01) b
2
(23) a−1
2
2(01) b
2
and so the condition G(n + a + b) = G(n) holds for 0 ≤ n ≤ b − 1. Lemma 1 then implies that the subtraction game S(1 , a, b ) is purely periodic with the period length
a + b and periodic nim-values (01) b
2
(23) a−1
2
2.
We divide the first a + 2 b nim-values G(n), for n from 0 to a + 2 b − 1, into three blocks: (1) G(0) . . . G(b − 1), (2) G(b) . . . G(a + b − 1), (3) G(a + b) . . . G(a + 2 b − 1). It can be checked by induction on n that the block (1) forms the sequence (01) b
2
,the block (2) forms the sequence (23) a−1
2
2, and the block (3) forms the sequence (01) b
2
. Thus the first a + 2 b nim-values are (01) b
2
(23) a−1
2
2(01) b
2
, as required. 4. More on ultimately bipartite subtraction games
Some ultimately bipartite games were exhibited in . In particular the games with subtraction sets {3, 5, 2k + 1 }, for k ≥ 3, and {a, a + 2 , 2a + 3 }, for odd a ≥ 3, are all ultimately bipartite. In this section, we first introduce another family of ultimately bipartite games with 3-element subtraction sets. We then consider the expansion sets of subtraction sets of ultimately bipartite games, with a conjecture being given at the end.
Theorem 5. For odd integer a ≥ 5, the subtraction game S(a, 2a+1 , 3a) is ultimately bipartite with pre-period length 2a2 − a − 1. The pre-periodic nim-values are shown in Table 8. Proof. Let n0 = 2 a2 − a − 1. We show that for n ≥ n0, G(n) = 0 if n is even and
G(n) = 1 otherwise. ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 13
(1) We place the sequence of first n0 nim-values G(0), G(1), . . . , G(n0 − 1) in Table 8 from left to right and top to bottom with empty cell being empty and the cell with the notation ︸︷︷︸
b
containing b consecutive integers and so on. The nim-values in Table 8 can be checked by induction on n for n < n 0.(Each row from the second to the second last of Table 8 has length 4 a + 2. Observe that on each of these rows, there are exactly a numbers in each two consecutive blocks A and B, B and C, C and D, F and G, G and H.) (2) We next prove that, after the first n0 nim-values in Table 8, the game has al-ternate nim-values 0 , 1, 0, 1, . . . (starting from G(n0).) The proof consists of six steps. (a) Let A1 be the set of all nonnegative integers n < n 0 such that G(n) = 0 in Table 8. Let A2 be the set of all even integers n ≥ n0. Set A = A1 ∪ A2.(b) We then prove the two facts: ((i)) for every position n in A, all the moves from n do not terminate in
A,((ii)) for every position n not in A, there is a move from n that terminates in A.(c) It follows from step (b) and the definition of nim-values that A is the set of positions whose nim-values are zero, by which we have (iii). G(n) = 0 for all even n such that n ≥ n0.(d) For all odd n ≥ n0 + 3 a, we have n − a, n − (2 a + 1), and n − 3a are all in
A. By (iii) and the definition of nim-values, we have (iv). G(n) = 1 for all odd n such that n ≥ n0 + 3 a.(e) We now examine G(n) for odd n such that n0 ≤ n ≤ n0 + 3 a − 1. The 3 a
nim-values G(n) for n0 ≤ n ≤ n0 + 3 a − 1 continuing those in Table 8 is 0101 . . . 01 (the last 3 a values in Table 9, being located in the bottoms of blocks F, G, H, I, and B), by which we have (v). G(n) = 1 for all odd n such that n0 ≤ n ≤ n0 + 3 a − 1. (f) It follows from (iii), (iv), and (v) that the game ultimately has alternate nim-values 0 , 1, 0, 1, . . . with pre-period length n0.Step (a) is the definition of A. Steps (c), (d), and (f) are straightforward. Steps (b) and (e) can be verified by induction on n. Checking several cases is required but is also straightforward and we leave the calculation to the reader.
Remark 5. The relevance of Theorem 5, beyond its general interest, is that it indicates that the ultimately bipartite case, which is the simplest form of ultimate periodicity, can nevertheless entail a high level of complexity at the beginning of the sequence. 14 NHAN BAO HO
Block A Block B Block C Block D Block E Block F Block G Block H Block I
000 . . . 000 ︸ ︷︷ ︸
a
111 . . . 111 ︸ ︷︷ ︸
a
0
22 . . . 22 ︸ ︷︷ ︸
a−1
1 33 . . . 33 ︸ ︷︷ ︸
a−1
0 2 00 . . . 00 ︸ ︷︷ ︸
a−2
10 11 . . . 11 ︸ ︷︷ ︸
a−2
010
2 . . . 2︸ ︷︷ ︸
a−3
101 3 . . . 3︸ ︷︷ ︸
a−3
010 2 0 . . . 0︸ ︷︷ ︸
a−4
1010 1 . . . 1︸ ︷︷ ︸
a−4
01010
... ... ... ... ... ... ... ... ...
2222 101 . . . 01 ︸ ︷︷ ︸
a−4
3333 010 . . . 10 ︸ ︷︷ ︸
a−4
2 000 10 . . . 10 ︸ ︷︷ ︸
a−3
111 010 . . . 10 ︸ ︷︷ ︸
a−2
22 1010 . . . 101 ︸ ︷︷ ︸
a−2
33 0101 . . . 010 ︸ ︷︷ ︸
a−2
2
Table 8. The arrangement of the first (2 a + 1)( a − 1) + 3 a nim-values. G(n) = 0 for n ∈ A .
Block A Block B Block C Block D Block E Block F Block G Block H Block I
... ... ... ... ... ... ... ... ...
2222 101 . . . 01 ︸ ︷︷ ︸
a−4
3333 010 . . . 10 ︸ ︷︷ ︸
a−4
2 000 10 . . . 10 ︸ ︷︷ ︸
a−3
111 010 . . . 10 ︸ ︷︷ ︸
a−2
22 1010 . . . 101 ︸ ︷︷ ︸
a−2
33 0101 . . . 010 ︸ ︷︷ ︸
a−2
2 0 101 . . . 010 ︸ ︷︷ ︸
a−1
1 0101 . . . 010 ︸ ︷︷ ︸
a
10101 . . . 0101 ︸ ︷︷ ︸
a
Table 9. The bottom part of Table 8 with extra 3 a nim-values at the end. ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 15
We now give three partial results that place limitations on the expansion sets of subtraction sets of ultimately bipartite games. Note that S = {s1, s 2, . . . , s k}. First we recall a well known result.
Lemma 2 (Ferguson’s property) . [3, p. 86] G(n) = 0 if and only if G(n + s1) = 1
for all n.
Lemma 3. Let S be an ultimately bipartite game. Then n0 ≥ s1 + sk.Proof. Note that s1 > 1 by Theorem 2. Consider the first s1 + sk nim-values
G(0) , G(1) , . . . , G(s1 + sk − 1). Note that the first s1 nim-values of the subsequence are all zero and so the last s1 nim-values of the subsequence are non-zero as G(m +
sk) 6 = G(m) for all m ≤ s1 − 1. Therefore, the periodicity must start at some
n0 ≥ s1 + sk.
Lemma 4. Let S be an ultimately bipartite game. Then G(n0) = 0 , G(n0 − 1) ≥ 2,
G(n0 − 2) = 0 , and G(n0 − 1 − sk) = 1 .Proof. For the moment let m = n0 if G(n0) = 0 and set m = n0 + 1 if G(n0) = 1, so that G(m) = 0. We have
G(m + i) =
{
0, if i is even; 1, otherwise .
(4) Therefore, G(m − 2 + s1) = 1 and hence by Ferguson’s property, G(m − 2) = 0. Notice that G(m − 1) 6 = 1, as otherwise the periodicity would commence at G(m − 2) or earlier. In particular, G(n0) = 0 and so m = n0. Also by Ferguson’s property,
G(n0 − 1) 6 = 0, as otherwise G(n0 − 1 + s1) = 1 contradicting (4). So G(n0 − 1) ≥ 2. We now show that G(n0 − 1 − sk) = 1. Note that
G(n0 − 1) = mex {G (n0 − 1 − si) : 1 ≤ i ≤ k}.
For each i < k , we have G(n0 − 1 − si) 6 = G(n0 − 1 − si + sk). Moreover, by (4), we have G(n0 − 1 − si + sk) = 1. Therefore, G(n0 − 1 − si) 6 = 1 for all i < k . It follows that G(n0 − 1 − sk) = 1.
Lemma 5. Let S be an ultimately bipartite game. If s can be adjoined to S, then
s < s k.Proof. Assume by contradiction that s > s k and s can be adjoined to S. Note that all si are odd (as the nim-sequence is ultimately 01 . . . 01) and greater than one. We also have s > 1 and s is odd. By Lemma 3, we have n0 ≥ s1 + s. By Lemma 4,
G(n0 − 1 − sk) = 1 and so G(n0 − 1 − sk + s) 6 = 1. This contradicts (4) as s − 1 − sk
is odd.
Conjecture 1. The subtraction set of an ultimately bipartite game is non-expandable. 16 NHAN BAO HO
Acknowledgements
I am grateful to Dr. Grant Cairns, for several suggestions regarding content and exposition.
References
M.H. Albert, R.J. Nowakowski, D. Wolfe, Lessons in play: An introduction to combinatorial game theory, second ed., A K Peters Ltd., Wellesley, MA, 2007. I. Alth¨ ofer, J. B¨ ultermann, Superlinear period lengths in some subtraction games, Theoret. Comput. Sci. 148 (1995), no. 1, 111–119. E.R. Berlecamp, J.H. Conway, R.K. Guy, Winning ways for your mathematical plays, vol. 1, second edition, A.K. Peters, Natick, MA, 2001. E.R. Berlecamp, J.H. Conway, R.K. Guy, Winning ways for your mathematical plays, vol. 3, second edition, A.K. Peters, Natick, MA, 2003. G. Cairns, N.B. Ho, Ultimately bipartite subtraction games, Australas. J. Combin. 48 (2010), 213–220. R.K. Guy, Fair game: How to play impartial combinatorial games, reprinted, Comap, Arling-ton, MA, 1991.
Appendix A. A code, written on Maple, for subtraction game
We include in A.1 the code “sub” whose input is ( S, n ) where S is the subtraction set and n is the largest integer the code needs to calculate G(n) to identify the pre-period length n0 and period length p. If n > n 0 + p + sk the periodicity will be found and the code outputs pre-period length n0, period length p, pre-periodic nim-values
G(0) . . . G(n0 − 1), periodic nim-values G(n0) . . . G(n0 + p − 1), and the expansion set
Sex . If the code returns nothing, it requires a larger n.The code “sub” recalls the subcode “mex” in A.2 and the subcode “F” in A.3. In these codes, for arbitrary sets A, B and element a, ‘union‘(A, B) means A ∪ B,not ‘in‘(a, A) means a / ∈ A, and ‘minus‘(A, B) means A \ B.We provide two examples in A.4. The first example is purely periodic and so it does not have pre-periodic nim-values. A.1. The code “sub”.
sub := proc (S, n) local i, j, k, l, m, s, g, p, T, Y, Z; ############ for i from 0 to min(S) - 1 do g[i] := 0; end do; ############ for i from min(S) to n do T := {}; ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 17
for k in S do if k <= i then T := ‘union‘(T, {g[i-k]}); end if; end do; g[i] := mex(T); end do; ############ for i from 0 to n - max(S) - 1 do ######### for j from i+1 to n - max(S) do for l from 0 to max(S) do if g[i+l] <> g[j+l] then l := 0; break; end if; end do; if l = max(S) + 1 then break; end if; end do; ######### if j < n - max(S) then p := j - i; ###### print(pre_period length) print(i); ###### print(period length); print(p); ###### print(pre_periodic nim_values); if 0 < i then print(seq(g[x], x = 0 .. i - 1)); end if; ###### print(periodic nim_values); print(seq(g[x], x = i .. i + p - 1)); ###### if i = 0 then 18 NHAN BAO HO
Y := S; for s from 1 to p-1 do if not ‘in‘(s, S) then for m from 0 to i - 1 do if g[m+s] = g[m] then break; end if; end do; if m = p then Y: = ‘union‘(Y, {s}); end if; end if; end do; ### print(expansion set); return Y^(p); ### else Y := S intersect F(i-1); Z := ‘minus‘(S, Y); for s from 1 to i + p - 1 do if not ‘in‘(s, S) then for m from 0 to i + p - 1 do if g[m+s] = g[m] then break; end if; end do; if m = i + p then if s < i then Y:= ‘union‘(Y, {s}); else Z:= ‘union‘(Z, {s}); end if; end if; end if; end do; ### print(expansion set); return Y,Z^(p); ### ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 19
end if; ###### break; end if; ######### end do; ############ end proc;
A.2. The subcode “mex”.
mex := proc (S) local i; if S = {} then return 0; else for i from 0 to max(S)+1 do if not ‘in‘(i, S) then return i; end if; end do; end if; end proc;
A.3. The subcode “F”.
F := proc (n) local i, S; S := {}; for i to n do S := ‘union‘(S, {i}); end do; return S; end proc;
A.4. Two examples.
sub({2, 7, 10}, 100); pre_period length 0period length 17 20 NHAN BAO HO
pre_period nim_values periodic nim_values 0, 0, 1, 1, 0, 0, 1, 1, 2, 0, 3, 1, 2, 0, 3, 1, 2 expansion set {2, 7, 10}^17 sub({3, 8, 12}, 100); pre_period length 16 period length 5pre_period nim_values 0, 0, 0, 1, 1, 1, 0, 0, 2, 1, 1, 0, 2, 2, 1, 3 periodic nim_values 0, 0, 2, 1, 1 expansion set {3, 8, 12, 13}, {18}^5
Appendix B. Pre-periodic nim-values of some ultimately periodic games
Subtraction set
{1,a,b }
a: even
b = a + rr: odd
n0 pre-periodic nim-values
3 ≤ r ≤ a − 3 b2 (01) a
2
2(01) r−1
2
23
Table 10. Pre-periodic nim-values of ultimately periodic games
S(1 , a, b ) in Table 1. ON THE EXPANSION OF THREE-ELEMENT SUBTRACTION SETS 21
Subtraction set
{1, a, b }
a: even
b = 2 a + r
r n0 pre-periodic nim-values
2 = r < a − 4
even
a + 2 b + 4 (01) a
2
2(01) a
2
(23) a
2
013 (01) a
2
2(01) a
2−1
20123
4 ≤ r < a − 4 5a + 12 ((01) a
2
2) 2012(32) a
2−2
(01) 22(01) a
2
2(01) a
2−1
2(01) 223
4 < r = a − 4 4a + b ((01) a
2
2) 2(01) a
2−3
2(32) 2(01) a
2−2
2(01) a
2
2(01) a
2−1
2(01) a
2−2
23
4 ≤ r = a − 2 b + 2 (01) a
2
2(01) a
2
2(01) a−4
2
23
Table 11. Pre-periodic nim-values for ultimately periodic games
S(1 , a, b ) in Table 2.
Subtraction set
{1,a,3a+r }
a: even
r n0 pre-periodic nim-values
3 = r < a − 5
odd
2a + 3 b + 6
((01) a
2
2) 3(32) a
2−1
3013 ((01) a
2
2) 2(01) a
2−1
2012(32) a
2−2
(01) 22(01) a
2
2((01) a
2−1
2) 2(01)23
3 < r < a − 5 10 a + b − 8
((01) a
2
2) 3(01) r−3
2
2(32) a
2−2
(01) 22((01) a
2
2) 2(01) a
2−1
2(01) 22(32) a
2−3
(01) a
2−3
2(01) a
2
2((01) a
2−1
2) 2(01) a
2−3
23
5 ≤ r = a − 5 10 a + b − 4
((01) a
2
2) 3(01) r−3
2
2(32) 3(01) a
2−3
2((01) a
2
2) 2(01) a
2−1
2(01) a
2−3
2(32) 2(01) a
2−2
2(01) a
2
2((01) a
2−1
2) 2(01) a
2−2
23
5 ≤ r = a − 3 a + 2 b + 4 ((01) a
2
2) 3(01) r−3
2
2(32) 2(01) r−1
2
2((01) a
2
2) 2(01) a
2−1
2(01) a
2−2
23
5 ≤ r = a − 1 b + 2 ((01) a
2
2) 3(01) a
2−2
23
Table 12. Pre-periodic nim-values for ultimately periodic games
S(1 , a, b ) in Table 3. 22 NHAN BAO HO
Subtraction set
{a, b, b + a + 1 }
a ≥ 4
b = 2 a + rr ≤ 2
n0 pre-periodic nim-values
{a, 2a, 3a + 1 } 4a 0a1a2a03 a−1
{a,2a+2,3a+3 } 6a + 6 0a1a022a−21203 a−322120 a−131 a−1023
Table 13. Pre-periodic nim-values for ultimately periodic games
S(a, b, b + a + 1) in Table 5.
Department of Mathematics, La Trobe University, Melbourne, Australia 3086
E-mail address : nhan.ho@latrobe.edu.au, nhanbaoho@gmail.com |
190773 | https://link.springer.com/article/10.1007/s40735-016-0047-y | Advertisement
Significance of Pourbaix Diagrams to Study the Corrosion Behaviour of Hardfacing Alloys Based on Chromium Carbides at 298 K (25 °C)
14k Accesses
27 Citations
Explore all metrics
Abstract
The Pourbaix diagrams also known as the E-pH diagrams were constructed for hardfacing alloys based on three chromium carbides: Cr7C3, Cr23C6 and Cr3C2 at 298 K (25 °C). Using the thermodynamic data available for the main species at 298 K (25 °C), Pourbaix diagrams for the chromium carbides are constructed at a concentration of 10−6 M of aqueous species. It was found that the diagrams are able to explain the results of experimental work performed on chromium carbides in NaOH. It was found that the stability of carbides is described only by the immune region of the Pourbaix diagram for carbides. Although passive region has been included in the study, it was unclear from the literature if there was a passive film or how protective it was.
Similar content being viewed by others
High-Temperature Corrosion Behavior of T92, TP347HFG and IN625 with Surface Scratching in Carbon Dioxide at 600 °C
Analysis of Film Formation in High Chromium White Iron Hardfacing Alloys in Alkaline Solution using EIS and SIMS
The Effects of Strain on Stability of Passivation in Austenitic Stainless Steels: Comparison with Heat Treatment
Explore related subjects
Avoid common mistakes on your manuscript.
1 Introduction
Weld hardfacing alloys containing chromium carbides are of great interest in mining applications where abrasion resistance is of significance. The selection of hardfacing alloys is described in AS2576 . The standard is probably the most advanced standard in the world as it describes not only the composition and hardness but also the microstructures in order to meet the requirements of service. The hardfacing materials described in the standards are based on carbides of different metals. Amongst these, carbides of chromium which indicate Group 2 alloys in the standards are for abrasion resistance. The matrix in which the carbides are precipitated can be either austenite or martensite. Three chromium carbides which have very significant abrasion resistance are Cr7C3, Cr23C6 and Cr3C2 .
Although carbides are selected on the basis of only their hardness , their corrosion properties are often not considered. In spite of this, hardfacing alloys are used in the alumina industry, the pH of the mining solution is about 14 [3, 4] and in sugarcane crushing the pH is about 3 . Other than this, these alloys are also used in cement, oil and gas drilling, dredging, etc., industries which all operate under corrosive environments. In addition to the above, carbides of chromium have been considered for coating on body implants . Tungsten carbides were considered for use in fuel cell catalysis . Much work has been performed on the kinetics aspects of the various carbides; however, there is very little information on the corrosion thermodynamic behaviour of these materials.
Pourbaix diagrams are useful in predicting the spontaneous direction of electrochemical reactions, identifying the corrosion products and predicting the changes in environment in terms of potential and pH that result in high or low corrosive attack . The present work provides Pourbaix diagrams for chromium carbides that can be used to predict the range of use of these alloys and target electrochemical experiments to areas of interest. It is expected that such analysis will provide guidance for applications and development of hardfacing alloys. The final goal, as future work, is to be able to design hardfacing alloys containing carbides, and other compounds, of different materials by using the Pourbaix diagrams. The authors believe that this would result in more efficient development of hardfacing alloys.
2 Thermodynamic Calculations
The present work provides the details of thermodynamic calculations for the construction of Pourbaix diagrams for chromium carbides, Cr7C3, Cr23C6 and Cr3C2 at 298 K (25 °C). All Pourbaix diagrams were calculated using a concentration of 10−6 M for all the aqueous species. For the dilute solution, the corrosion rate will be higher as the material would be in active state by the absence of passive film . Concentrations of aqueous solutions do affect potentials, and the potentials of various regions are increased with the increase in concentrations of aqueous solutions .
In order to plot the Pourbaix diagrams for chromium carbides at room temperature 298 K (25 °C), thermodynamic calculations were carried out with the help of available thermodynamic data of the chromium carbide species. The standard thermodynamic data of Gibbs free energy change ΔG
0 for the chromium carbide species are obtained from various publications [11–15] and are listed in Table 1 for 298 K (25 °C). The method of constructing Pourbaix diagrams is described in a number of publications [11, 12, 16p
. Accessed 26 Apr 2016")], and these methods are followed in this work. There are several considerations for developing Pourbaix diagram, and these are described below.
2.1 Balancing the Reactions of Two Selected Species
The method of balancing the chemical reactions depending on the chemical species selected is described in this section. The thermodynamic calculations are made for the reduction reactions to construct Pourbaix diagrams using the method described in the literature [11, 12, 16p
. Accessed 26 Apr 2016")]. Balancing a reaction consists of the following four steps which are constructed under chemical equilibrium: (1) the number of all atoms is balanced without considering the oxygen or charge, (2) the oxygen atoms are then added through water (H2O) on the appropriate side of the chemical equation, (3) the hydrogen ions (H+) are added on the appropriate side to balance the number of hydrogen atoms in the chemical equation and finally, (4) the charges are balanced by adding electrons to the appropriate side. The above four steps are illustrated in the following example of balancing the species chromium oxide Cr2O3 and Cr7C3 [16 The nernst equation and Pourbaix diagrams.
. Accessed 26 Apr 2016")] in the chemical reduction reaction of chromium carbide in contact with water.
Step (1): 3.5Cr2O3 + 3C ⇄ Cr7C3 (Balancing the atoms).
Step (2): 3.5Cr2O3 + 3C ⇄ Cr7C3 + 10.5H2O (Balancing the oxygen atoms).
Step (3): 3.5Cr2O3 + 3C + 21H+ ⇄ Cr7C3 + 10.5H2O (Balancing the hydrogen ions).
Step (4): 3.5Cr2O3 + 3C + 21H+ + 21e− ⇄ Cr7C3 + 10.5H2O (Balancing the charge).
All other equilibrium reduction reactions of the species given in Table 1 are also balanced following these steps. The reactions above and E-pH-dependent reactions (7)–(9) in Table 2 show that carbon is formed during the reaction between carbide and water. Although this is a thermodynamic possibility , carbon has been found to be produced at high temperatures or tribological conditions [18, 19]. However, when carbides are in water, carbon dioxide has been found to be generated [7, 20]. It is possible that formation of carbon may be an intermediate step for carbon dioxide generation. However, Andrews et al. had proposed that the formation of carbon can occur when silicon carbide (SiC) reacts with water. The presence of carbon was also detected using Raman spectroscopy . Silicon atoms were found to dissolve in solution leaving carbon on the surface. More work is needed to understand the behaviour of carbides in corrosive conditions for hardfacing applications.
The equilibrium reduction reactions considered here are classified into three categories of reactions: potential (E) dependent, E-pH dependent and pH dependent as described in . The thermodynamic calculations for these three different reactions are calculated at the temperature 298 K (25 °C), and these are described below.
2.2 Thermodynamic Calculations for 298 K (25 °C)
The calculations for the above three types of reactions are briefly described below.
2.2.1 E-pH-Dependent Reactions
The following general E-pH-dependent reaction is used in the thermodynamic calculations for 298 K (25 °C).
where a, m, b and c are the number of moles, respectively, of the species A, H+, B and H2O in the reaction and n is the number of electrons transfer in the electrochemical reaction. In these E-pH-dependent reactions, both electron acceptance and hydrogen ions are involved between the two species A and B .
The electric potential E (V) required for the E-pH-dependent reactions [Eq. (1)] using the Nernst equation used by Beverskog and Puigdomenech :
where E
o is the standard potential in volt, ({\text{pH}} = - \log \left[ {\left( {{\text{H}}^{ + } } \right)} \right]), R is the molar gas constant (8.314 J K−1 mol−1), T is the temperature in Kelvin, m is the number of moles in H+, n is the number of electrons involved in the reaction, F is the Faraday’s constant (96,485.33 C mol−1), and C(A) and C(B) represent the concentration of the species A and B involved in the E-pH-dependent reactions.
For calculating the potential E using Eq. (2), the standard potential E
° is obtained from the Gibbs free energy change ΔG0 of the reaction as :
where ΔG
0 is calculated as in :
In this manner, the E-pH-dependent reactions are calculated from Eq. (2) to construct the Pourbaix diagrams for chromium carbides at 298 K (25 °C) and these are listed in Table 2. The reactions are dependent on both potential (E) and pH and give a sloping straight line in a Pourbaix diagram.
2.2.2 For E-Dependent Reactions
The following general E-dependent reaction is used in the thermodynamic calculations for 298 K (25 °C).
In the above reaction, the electron acceptance takes place between two species A and B without involving hydrogen ions .
The electric potential E (volt) required for the E-dependent reactions [Eq. (4)] using the Nernst equation is given by :
The standard potential E
o is calculated from Eq. (3) and used in the above Nernst equation. This reaction depends on potential (E) which forms a horizontal line in the Pourbaix diagram. Thus, the E-dependent reactions are calculated from Eq. (6) to construct the Pourbaix diagrams for chromium carbides at 298 K (25 °C) and these are listed in Table 2.
2.2.3 For pH-Dependent Reactions
The following general pH-dependent reaction is taken to explain the thermodynamic calculations for 298 K (25 °C).
In the above pH-dependent reaction, the hydrogen ions are involved between two species B and A without transfer of any electrons .
The pH-dependent reaction is obtained from the equilibrium constant K related to the Gibbs free energy as .
By using K = (\frac{{\left[ {C\left( A \right)} \right]^{a} [H^{ + } ]^{m} }}{{\left[ {C\left( B \right)} \right]^{b} }}) and ({\text{pH}} = - \log \left[ {\left( {H^{ + } } \right)} \right]), we get:
The above reaction takes place at any particular pH and hence forms a vertical line in the Pourbaix diagram. Thus, the pH-dependent reactions calculated from Eq. (9) to construct the Pourbaix diagrams for chromium carbides at 298 K (25 °C) are listed in Table 2.
3 Results and Discussion
The procedures described above were used to construct the Pourbaix diagrams for chromium carbides at temperatures 298 K (25 °C) for a concentration of 10−6M for all the aqueous species Figs. 1, 2 and 3. For all E-pH-dependent reactions, the values E calculated from Eq. (2) at temperature 298 K (25 °C) are given in Table 2. For E-dependent reactions, the temperature-dependent Eq. (6) calculated at temperature 298 K (25 °C) is given in Table 2. Likewise, for all pH-dependent reactions, the values of pH calculated from Eq. (9) are given in Table 2 at 298 K (25 °C). Using the E-pH equations given in column 3 of Tables 2, the Pourbaix diagrams are constructed.
Pourbaix diagram for Cr7C3 at 298 K (25 °C) with the concentration of 10−6 M
Pourbaix diagram for Cr23C6 at 298 K (25 °C) with the concentration of 10−6 M
Pourbaix diagram for Cr3C2 at 298 K (25 °C) with the concentration of 10−6 M
Figures 1, 2 and 3 show the immune regions, i.e. where no corrosion can occur, for Cr7C3, Cr23C6 and Cr3C2. The passive region consists of Cr2O3. The corrosion regions are formed from the ionic species Cr2+, Cr3+, Cr(OH)2+, HCrO4
− and CrO4
2−. The Pourbaix diagrams for chromium carbides at room temperature from the Figs. 1, 2 and 3 reveal that the Cr3C2 has the largest immune region followed by Cr7C3 and Cr23C6. There are no data yet in the literature yet to prove this prediction. The stability of immune regions for all the three carbides Cr7C3, Cr23C6 and Cr3C2 is higher in low pH and reduced significantly at higher pH level. The types and morphology of carbides depend on the rate of solidification and chemical composition . The type of carbides also depends on the (\frac{\text{Cr}}{\text{C}}) ratio, and this ratio also determines the wear and corrosion resistance. The Sabet et al. reported that by increasing (\frac{\text{Cr}}{\text{C}} = 6), the carbide volume fraction of Cr7C3 formed and the corrosion resistance for the hardfacing alloys also improved. Thus, the above literature confirmed that Cr7C3 shows better corrosion resistance compare to Cr23C6.
3.1 Significance of Pourbaix Diagrams to Hardfacing Alloys
Pourbaix diagrams for metals consist of three regions. They are regions where corrosion cannot occur, regions where corrosion may occur and regions where corrosion may not occur . Most engineering metals when applied in structures and components are in the passive region. In the case of carbides, immune region is the region where carbides are thermodynamically stable and corrosion cannot occur. It was not clear what a passive region meant for carbides where corrosion may not occur. This is not well covered in the literature and hence this discussion. This is an important consideration as chromium carbides are compounds and it is not clear how passive layers will form in the same way as in stainless steels. In a solid solution, the elements that make up the solid solutions are considered to behave independent of each other . Hence, they can form passive regions and be successful in engineering applications. However, in the case of intermetallic compounds such as carbides if they have to passivate or corrode, the chromium carbide will need to dissociate so an oxide film may form assuming this film provides passivity. Even if passive films could form, the dissociation of chromium carbides may render the alloy unsuitable for wear resistant applications.
Passive films on chromium carbides have been found, but they were found to be very thin in comparison with the matrix. The passive region formed over the carbides was found to be iron dominated . It was found that the silicon carbide exposed in acid solution forms silicon oxide [21, 22]. It was unclear if the film afforded any protection. It is unlikely the passivated surface may actually provide any wear resistance. The investigation by Tylczak et al. observed that the absence of passivity in carbides led to more active potential than matrix. Thus, galvanic cell formed can cause dissolution of the chromium carbide by protecting the matrix. The study conducted by Srisuwan showed that increase in Cr23C6 chromium carbides can reduce the formation of passive film chromium oxide (Cr2O3). In the anodic polarization, studies conducted by Tran and Vargas et al. showed that in alkaline solution, consisting of NaOH, the carbides corroded before the matrix corroded. Tran and Vargas showed that carbides were the first to corrode followed by pitting in matrix at higher anodic potentials and this behaviour followed the trend in the Pourbaix diagrams. Hence, the usefulness of carbides is limited as long as the carbides are held in immune region of potentials.
In general, hardfacing alloys based on chromium carbides have only limited stabilities as the immune region is very small. The only way to improve stability of the chromium carbides is to increase the size of the immune region. This could be attempted by adding elements that form carbides that have better stabilities than chromium carbides. Investigations are under way in the laboratory to study the behaviour of chromium carbide weld deposits through corrosion experiments. Additions of elements such as titanium (Ti), niobium (Nb) and tungsten (W) are being done to improve the corrosion behaviour of hardfacing alloys. The outcome is expected that the results of this paper will provide a better understanding of the application of hardfacing alloys and designing better corrosion resistant hardfacing alloys.
4 Conclusions
The Pourbaix diagrams were constructed for chromium carbides at 298 K (25 °C). It was postulated that the usefulness of hardfacing alloys is mainly in the immune region of the Pourbaix diagrams. The concept of passivity is not well understood for chromium carbides. At room temperature, chromium carbides such as Cr3C2 have the largest immune region followed by Cr7C3 and Cr23C6. The stability of chromium carbides’, Cr7C3, Cr23C6 and Cr3C2, immune regions increases significantly at low pH and reduces at higher pH level.
On behalf of all authors, the Varmaa Marimuthu states that there is no conflict of interest.
References
AS/NZS (2005) AS/NZS 2576:2005, welding consumables for build-up and wear resistance. Australian Standards, Sydney, Australia
Sabet H, Mirdamadi S, Kheirandish S, Masoud G (2013) Effect of volume fraction of (Cr, Fe)7C3 carbides on corrosion resistant of the Fe–Cr–C hardfacing alloys. Assoc Metal Eng Serb 19(2):107–114
Google Scholar
Birss VI, Waudo W (1989) The initial stages of lead oxidation in pH 9–14 aqueous solutions. Can J Chem Eng 67:1098–1104
Article
Google Scholar
Brace X, Matijević E (1977) Coprecipitation of Silica with Aluminum hydroxide. Colloid Polym Sci 255:153–160
Article
Google Scholar
Buchanan V, Shipway P, McCartney D (2007) Microstructure and abrasive wear behaviour of shielded metal arc welding hardfacings used in the sugarcane industry. Wear 263(1):99–110
Article
Google Scholar
Kuhn LT (2005) Introduction to biomedical engineering, 2nd edn. Academic Press, Boston
Google Scholar
Weidman MC, Esposito DV, Hsu IJ, Chen JG (2010) Electrochemical stability of tungsten and tungsten monocarbide (WC) over wide pH and potential ranges. J Electrochem Soc 157(12):F179–F188
Article
Google Scholar
Fontana MG (2005) Corrosion engineering. McGraw-Hill Education, New York
Google Scholar
Salasi M, Stachowiak G, Stachowiak G (2011) Three-body tribocorrosion of high-chromium cast irons in neutral and alkaline environments. Wear 271(9):1385–1396
Article
Google Scholar
Piron DL (2004) Solved problems in electrochemistry for universities and industry. National Library of Canada, Canada
Google Scholar
Hai-Xia Y, Hong-Bin X, Yi Z, Shi-Li Z, Gao YY (2009) Potential-pH diagrams of Cr–H2O system at elevated temperatures. Trans Nonferrous Met Soc China 20:26–31
Google Scholar
Beverskog B, Puigdomenech I (1997) Revised pourbaix diagrams for chromium at 25–300 °C. Corros Sci 39(1):43–57
Article
Google Scholar
Rock PA (1983) Chemical thermodynamics. University Science Books, Mill Valley
Google Scholar
Pankratz L (1995) Thermodynamic properties of carbides, nitrides, and other selected substances. Bulletin 696, US Department of Interior, Bureau of Mines pp 223–225
Chase MW (1998) NIST-JANAF thermochemical tables. J Phys Chem Ref Data 9:1–1951
Google Scholar
DoITPoMS (2014) The nernst equation and Pourbaix diagrams. Accessed 26 Apr 2016
Hogstrom J, Andersson M, Jansson U, Bjorefors F, Nyholm L (2014) On the evaluation of corrosion resistances of amorphous chromium–carbon thin-films. Electrochim Acta 122:224–233
Article
Google Scholar
Jacobson NS, Gogotsi YG, Yoshimura M (1995) Thermodynamic and experimental study of carbon formation on carbides under hydrothermal conditions. J Mater Chem 5(4):595–601
Article
Google Scholar
Kitaoka S, Tsuji T, Katoh T, Yamaguchi Y, Kashiwagi K (1994) Tribological characteristics of SiC ceramics in high-temperature and high-pressure water. J Am Ceram Soc 77(7):1851–1856
Article
Google Scholar
Voorhies JD (1972) Electrochemical and chemical corrosion of tungsten carbide (WC). J Electrochem Soc 119(2):219–222
Article
Google Scholar
Andrews A, Herrmann M, Sephton M, Machio C, Michaelis A (2007) Electrochemical corrosion of solid and liquid phase sintered silicon carbide in acidic and alkaline environments. J Euro Ceram Soc 27(5):2127–2135
Article
Google Scholar
Andrews A (2006) Electrochemical corrosion measurement of solid state sintered silicon carbide (SSiC) and liquid phase sintered silicon carbide (LPSSiC) ceramic materials. MSc Dissertation. University of the Witwatersrand South Africa, pp 49–64
Muñoz-Portero MJ, García-Antón J, Guiñón JL, Pérez-Herranz V (2009) Pourbaix diagrams for chromium in concentrated aqueous lithium bromide solutions at 25 °C. Corros Sci 51(4):807–819
Article
Google Scholar
Chivot J, Mendoza L, Mansour C, Pauporté T, Cassir M (2008) New insight in the behaviour of Co–H2O system at 25–150 °C, based on revised Pourbaix diagrams. Corros Sci 50(1):62–69
Article
Google Scholar
Mahan BH (1963) Elementary chemical thermodynamics. Benjamin, New York
Google Scholar
Wiengmoon A (2011) Carbides in high chromium cast irons. Naresuan Univ Eng J 6(1):64
Google Scholar
McCafferty E (2010) Introduction to corrosion science. Springer, USA
Book
Google Scholar
Salasi M, Stachowiak G, Stachowiak G, Kilburn M (2013) NanoSIMS investigation of passive oxide films on high-Cr cast iron. Corros Sci 67:298–303
Article
Google Scholar
Tylczak J, Friedersdorf F (1997) Selected removal of chromium carbides during wear-corrosion testing. In: Proceedings of materials solution on wear of engineering materials, pp 177–184
Srisuwan N (2010) Investigation of chromium carbide precipitation and corrosion behavior on stainless steel AISI 304 for welding process development. In: RMUTP international conference on Thailand
Tran NV (2015) Corrosion behaviour of hardfacing alloys used in the mining industry. Thesis—the degree of Bachelor of Engineering. Charles Darwin University. Darwin
Vargas M, Kannoorpatti K, Murthy V (2014) Studies on the corrosion behaviour of wear resistant hardfacing alloys. In: ACA annual conference on corrosion and prevention, Darwin
Download references
Author information
Authors and Affiliations
School of Engineering and IT, Charles Darwin University, Darwin, Australia
Varmaa Marimuthu & Krishnan Kannoorpatti
Ecole Nationale Supérieure de Chimie de Lille, 59652, Villeneuve d’Ascq cedex, France
Isabelle Dulac
Search author on:PubMed Google Scholar
Search author on:PubMed Google Scholar
Search author on:PubMed Google Scholar
Corresponding author
Correspondence to
Varmaa Marimuthu.
Rights and permissions
Reprints and permissions
About this article
Cite this article
Marimuthu, V., Dulac, I. & Kannoorpatti, K. Significance of Pourbaix Diagrams to Study the Corrosion Behaviour of Hardfacing Alloys Based on Chromium Carbides at 298 K (25 °C).
J Bio Tribo Corros 2, 17 (2016).
Download citation
Received: 06 March 2016
Revised: 27 April 2016
Accepted: 11 June 2016
Published: 21 June 2016
DOI:
Share this article
Anyone you share the following link with will be able to read this content:
Sorry, a shareable link is not currently available for this article.
Provided by the Springer Nature SharedIt content-sharing initiative
Keywords
Avoid common mistakes on your manuscript.
Advertisement
Search
Navigation
Discover content
Publish with us
Products and services
Our brands
44.213.74.144
Not affiliated
© 2025 Springer Nature |
190774 | https://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html | Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
Radiation Heat Transfer
Heat transfer due to emission of electromagnetic waves is known as thermal radiation.
Heat transfer through radiation takes place in form of electromagnetic waves mainly in the infrared region. Radiation emitted by a body is a consequence of thermal agitation of its composing molecules. Radiation heat transfer can be described by reference to the 'black body'.
The Black Body
The black body is defined as a body that absorbs all radiation that falls on its surface. Actual black bodies don't exist in nature - though its characteristics are approximated by a hole in a box filled with highly absorptive material. The emission spectrum of such a black body was first fully described by Max Planck.
A black body is a hypothetical body that completely absorbs all wavelengths of thermal radiation incident on it. Such bodies do not reflect light, and therefore appear black if their temperatures are low enough so as not to be self-luminous. All black bodies heated to a given temperature emit thermal radiation.
The radiation energy per unit time from a black body is proportional to the fourth power of the absolute temperature and can be expressed with the Stefan-Boltzmann Law as
q = σ T4 A (1)
q = heat transfer per unit time (W)
σ = 5.6703×10-8 (W/m2K4) - The Stefan-Boltzmann Constant
T = absolute temperature in kelvins (K)
A = area of the emitting body (m2)
The Stefan-Boltzmann Constant in Imperial Units
σ = 5.6703×10-8 (W/m2K4)
= 1.714×10-9 ( Btu/(h ft2 oR4))
= 1.19×10-11 ( Btu/(h in2 oR4))
Heat radiation from a black body - surroundings absolute zero (pdf)
Example - Heat Radiation from the surface of the Sun
If the surface temperature of the sun is5800 K and if we assume that the sun can be regarded as a black body the radiation energy per unit area can be expressed by modifying (1) to
q / A = σ T4
= (5.6703×10-8 W/m2K4) (5800 K)4
= 6.42 107 (W/m2)
Gray Bodies and Emissivity Coefficients
For objects other than ideal black bodies ('gray bodies') the Stefan-Boltzmann Law can be expressed as
q = ε σ T4 A (2)
where
ε = emissivity coefficient of the object (one - 1 - for a black body)
For the gray body the incident radiation (also called irradiation) is partly reflected, absorbed or transmitted.
The emissivity coefficient is in the range 0 < ε< 1, depending on the type of material and the temperature of the surface.
oxidized Iron at 390 oF (199 oC) >ε= 0.64
polished Copper at 100 oF (38 oC) >ε= 0.03
emissivity coefficients for some common materials
Net Radiation Loss Rate
If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as
q = ε σ (Th4 - Tc4) Ah (3)
where
Th= hot body absolute temperature (K)
Tc= cold surroundings absolute temperature (K)
Ah = area of the hot object (m2)
Radiation constants for some common building materials
Heat loss from a heated surface to unheated surroundings with mean radiant temperatures are indicated in the chart below.
Download and print Heat Transfer by Radiation chart
Radiation Heat Transfer Calculator
This calculator is based on equation (3) and can be used to calculate the heat radiation from a warm object to colder surroundings.
Note that the input temperatures are in degrees Celsius.
Lambert's cosine law
Heat emission from a surface in an angle β can be expressed with Lambert's cosine law as
qβ = q cos(β) (4)
where
qβ = heat emission in angle β
q = heat emission from the surface
β = angle
Unit Converter
.
Make Shortcut to Home Screen?
Cookie Settings |
190775 | https://www.freshbooks.com/en-au/hub/other/discount-factor?srsltid=AfmBOoqqgUYLK_MyY1DWVDbA5XucGnqAz_y5bbyMH7t1cf24l0RzwEJU | Save up to 500 Hours on Paperwork 🙌 50% Off for 3 Months. BUY NOW & SAVE
50% Off for 3 Months Buy Now & Save
What Is the Discount Factor & How to Calculate It?
When it comes to financial modelling, there is a lot to know and understand. And there can be even more to grasp if you are looking into certain types of investments. How can you determine if an investment is going to be worthwhile for your business or not?
One of the ways that you can do this is by calculating the discount factor. This helps to determine and measure the present value of cash flow and what it might be worth in the future. But what exactly is the discount factor and how can you calculate it?
Here is everything that you need to know about the discount factor.
Here’s What We’ll Cover:
What Is the Discount Factor?
Formula for Calculating the Discount Factor
How to Calculate the Discount Factor
Are There Benefits to Knowing How to Calculate the Discount Factor?
Key Takeaways
What Is the Discount Factor?
To keep things simple, the discount factor is a calculation that you can make to determine the net present value (NPV) of future cash flows. It’s basically a way for you to take cash flow and discount it back to its present value. This can come in handy if your business is looking to invest and wants to determine if the investment is worth it.
One of the best things about calculating the discount factor is that you can see the exact present value of each of your individual cash flows. This is instead of just seeing the total net present value. It’s a simple weighing term that’s commonly used in economics and mathematics.
By multiplying your future income or losses, you can determine today’s net present value. And there isn’t really a limit to where you can apply this. It can get applied to investments, services that you offer or goods that you sell your customers.
It’s frequently used when going through a corporate budget to see if a certain proposal will actually add value. But, any discount factor equation that you use is going to assume that the money you have today will be worth less in the future.
Why? Because you need to factor in things such as inflation. Which will end up giving the discount factor a value that ranges somewhere between zero and one.
Discount Factor and the Net Present Value
An important distinction to remember with the discount factor is that, while related to discount rate, they are different. The discount rate will take a close look at your current value of future cash flow. The discount factor is going to apply to the net present value.
That said, doing these calculations and having these factors handy will allow you to forecast more accurately. You can gain more insights into the expected profits or losses of an investment, for example.
Formula for Calculating the Discount Factor
There can be several different reasons and multiple uses for determining the discount factor. Some of the main reasons include:
And it doesn’t matter which type of business that you operate or the industry that you’re in. The formulas for the discount factor can be used by investors, insurance companies and commercial banks, for example.
There can be two different approaches to take in order to calculate the discount factor. But whichever approach you use, the discount factor is going to be a function of your discount rate and the time period.
The easiest way to think about the discount rate is that it represents the total percentage of return that you can receive from an investment. And this is based on the idea that you would receive that return today, not in the future.
Why does that matter?
Pretend that you had one dollar today. If you do nothing with it, that same dollar that you have today isn’t going to be worth the same three years from now. This is since you would have missed opportunities to invest that one dollar to get a higher return on investment.
The formula to calculate the discount factor would look like this:
Discount Factor = (1 + Discount Rate) – Period Number
You can even rearrange the formula to look like this:
Discount Factor = 1 / (1 x (1 + Discount Rate) Period Number)
The easiest way to calculate the discount factor with these formulas would be by using Excel. But, let’s use the first formula for the examples since it can be a bit easier and a little more convenient to calculate.
How to Calculate the Discount Factor
Here are a few steps that you can take first to help figure out some of your calculations.
For an easy example, let’s say that you invested $1 at a 10% discount rate based over one year. The calculations would look like this:
1 / (1 + 10%) ^ 1 = 0.91
To get the present value (PV), you would multiply the discount factor by your cash flow. But, there’s an important thing to keep in mind here even though the discount rate will stay the same. The discount factor will decrease over time since the period number is going to continue to rise.
But, you can calculate the present value using the same discount factor as above. It would look like this:
$1 x 0.91 = 0.91
You can also have the period number be whatever works best for your calculations. It can be in hours, days, months or even years. The biggest thing to account for is that you have the period number properly aligned with the period of your discount rate.
So what does this all mean?
It means that if you received $1 exactly one year from the current period, the present day value would be worth $0.91.
Are There Benefits to Knowing How to Calculate the Discount Factor?
Once you figure out your discount rate and your discount factor, you can then use those calculations to see the net present value. From here, you can add the present value of all your positive cash flows and then subtract the present value of negative cash flows.
Once you apply the interest rate that you’re using you can finally see the net present value. As mentioned above, you can use Microsoft Excel to make the calculations a bit easier. Plus, there are several discount factor calculators that you can use.
There are always going to be benefits to knowing and understanding certain financial models. And it’s no different when you know how to understand the discount factor. It helps provide a look into what the impacts of compounding would be over time. When the discount rate continues to rise over time, your cash flow is going to decrease.
Ultimately, this is a great way to understand and represent the actual time value of money. And it’s done in a decimal representation. Investors find the discount factor to be extremely helpful. It helps to translate possible future investment returns back to their net present value.
The discount rate is commonly used in discounted cash flow (DCF) analysis. It’s used to help estimate the total value of a future investment. And it’s done based on the expected future cash flows.
If you wanted to look at it in the concept of the time value of money, discount cash flows analysis helps you to determine if an investment is viable or not. And you do this by calculating the present value of any expected future cash flows, all while using a discount rate.
Key Takeaways
It’s always a good thing to gain as much information and insight as you can into an investment. There’s no sense in investing in something if you aren’t going to receive some type of return that makes it worth your while. And future cash flows are going to get reduced by the discount rate.
Basically, having a lower discount rate is going to lead to a higher present value. And it works in the opposite way, as well. Having a higher discount rate is going to lead to a lower present value of future cash flows.
Think about it in a simpler sense. Money in the future is going to be worth less than it is today when the discount rate is higher. Ultimately, this means that it is going to have much less purchasing power.
Before getting started with calculating the discount factor, there are a few things that you should figure out first. The first thing you should do is try and forecast any expected cash flows from the investment you’re going to make. Next, make sure that you choose the right discount rate for the investment.
Then you can put the calculations together to discount the forecasted cash flows all the way back to the present day. There isn’t a perfect recipe for determining the most appropriate discount factor. However, the discount rate that you use is going to depend on the type of analysis you’re going to do.
Knowing and understanding the discount factor can be extremely helpful for potential investments. It allows you to see a visual representation of what will happen over time. This can let you make a much more informed investment decision.
Plus, it helps to calculate your discounted cash flow. Since the discount rate is going to continue to grow over time, your cash flow is going to continue to decrease. This makes it a great way to represent the actual time value of money.
Did you enjoy reading this guide? Head over to our resource hub for more great content!
RELATED ARTICLES |
190776 | https://people.duke.edu/~hpgavin/Risk/mean-example.pdf | Arithmetic Mean, Harmonic Mean and Geometric Mean Henri P. Gavin, Civil & Environmental Engineering, Duke University January 26, 2005 Means The way one determines a mean value of several quantities depends on the application or question one is trying to answer. The appropriate mean answers the question, “If all the quantities had the same value, what would that value have to be in order to achieve the same result?” Arithmetic Mean The arithmetic mean is the sum of the values divided by the total number of values. The arithmetic mean is relevant any time several quantities add together to produce a total.
Geometric Mean The geometric mean of n values is the n-th root of the product of the values, n √Πxi In a formula: the geometric mean of x1, x2, ..., xn is (x1 · x2 · · · xn)1/n, which is n √x1 · x2 · · · xn.
Any time you have a number of factors multiplying together to form a product, and you want to find the “average” factor, the answer is the geometric mean. The geometric mean of a data set is always less than or equal to the set’s arithmetic mean (the two means are equal if and only if all members of the data set are equal). A geometric mean, unlike an arithmetic mean, tends to dampen the effect of extremely high or low values, which might bias the mean if a straight average (arithmetic mean) were calculated. (Show this using numerical examples.) In calculating the geometric mean, the numbers must all be positive.
Determining the average growth rate over a period of many years is a widely-used application of the geometric mean.
Suppose that an investment has these five annual returns: 10%, -20%, 0%, -10%, and 20%. What is the average rate of return? The arithmetic mean of these five returns is exactly 0%. However, the average return is not the arithmetic mean, because what these numbers represent is that on the first year your investment was multiplied (not added to) by 1.10, on the second year it was multiplied by 0.80, the third year it was multiplied by 1.00, and so on. The relevant quantity is the geometric mean of these five numbers. The question about finding the average rate of return can be rephrased as: “By what constant factor would your investment need to be multiplied by each year in order to achieve the same effect as multiplying by 1.10 one year, 0.80 the next, and so on?” The answer is the geometric mean.
The value of the investment is initially V . After the first year it is (1.10)V , after the second year it is (0.80)(1.10)V , after the third year it is (1.00)(0.80)(1.10)V , after the fourth year it is (0.90)(1.00)(0.80)(1.10)V , and after the fifth year it is (1.20)(0.90)(1.00)(0.80)(0.90)V = (0.7776)V If the investment had had the same annual returns, r, for each of the five years, the value after five years would be (r)(r)(r)(r)(r)V = r5V .
Therefore, the average rate of return for this example can be found from r5V = 0.7776V , or r = 5 √ 0.7776 = 0.951, that is, on average over the five year period, the investment lost about 5%. Note that the order does not matter as long as the investment has no contributions or withdrawals during the five years.
If one considers the dollar amount by which the investment grew, then the arithmetic mean would be the correct mean to use. Suppose the investment grew by $100 in the first year, lost $220 in the second year, had no loss or gain in the third year, lost $88 in the fourth year, and gained $158.40 in the fifth year, (corresponding to an initial value of $1000), It would be perfectly correct to say that it has grown by an average of (100-220+0-88+158.4)/5 = -$29.60 yearly, over the five year period.
Another “geometric” example is the “average” dimension of a box that would have the same volume as length × width × height. For example, suppose a box has dimensions 50 × 80 × 100 cm, the “average” dimension of the box is 3 √50 × 80 × 100 = 73.7 cm Any rectangular box with an average dimension of 73.7 cm would enclose the same 400,000 cm3 or 0.4 m3 volume.
1 Harmonic Mean The harmonic mean of n values is the inverse of the sum of the inverses of the n values. In a formula, the harmonic mean of n values, x1, x2, . . . , xn is n/(1/x1 + 1/x2 + · · · + 1/xn), or n/ P 1/xi. Typically, the harmonic mean is appropriate for situations when the average of rates is desired. The harmonic mean answers the question “If the rate was constant over all distance intervals, what would the rate be for the total distance to be covered in the same amount of time.” A common example is averaging a set of speeds tabulated for a set of uniform distances.
For example, suppose that your trip has four 100 km segments.
You travel: 100 km/hr for the first 100 km, 110 km/hr for the second 100 km, 90 km/hr for the third 100 km, and 120 km/hr for the last 100 km.
What is your average speed?
In total, you travel 400 km.
The time it took to travel 400 km is: 100km / 100km/hr + 100km / 110km/hr + 100km / 90km/hr + 100km / 120km/hr, or 1.000 hr + 0.909 hr + 1.111 hr + 0.833 hr = 3.854 hr. Therefore the average speed is 400 km / 3.854 hr = 103.8 km/hr.
For speeds tabulated over uniform distances, (100 km in this example), the average speed may be computed directly from the harmonic mean equation: 4/(1/100+1/110+1/90+1/120)=103.8 km/hr.
As another example, suppose you travel for four hours. In the first hour you drive 100 km/hr, in the second hour you drive 110 km/hr, in the third hour you drive 90 km/hr, and in the fourth hour you drive 120 km/hr. What is your average speed for the trip? In total, you travelled for 4 hours. The total distance you travelled in the four hours is 100+110+90+120 = 420 km. So your average speed for the trip would be 420km/4hr = 105.0 km/hr. Note that in this case, we use the arithmetic mean.
Exercizes 1. Consider two random normally distributed variables, A and B, with means µA = µB = 10 cm and variances σA = 2cm and σB = 5cm. Generate 100 random samples of 30 elements each of the random variables A and B. Compute the arithmetic mean, the harmonic mean, and the geometric mean of each sample. Compute the mean of the sample means and the variance of the sample means. How do the variance of the sample means of A and B compare? You may use the built-in Matlab function randn.m for generating the samples.
2. A bicyle speedometer computes speed by measuring the time required for one wheel revolution and uses the “known” circumference of the wheel. Given the times for n wheel revolutions (n ≫1), t1, . . . , tn, and the circumference of the wheel, c, what is the right way to compute the average speed for the n cycles? How would the values of the arithmetic mean, geometric mean, and hyperbolic mean over-estimate, under-estimate, or correctly-estimate the actual average speed?
References on-line mean mean 2 |
190777 | https://theromefoundation.org/rome-iv/rome-iv-questionnaire/ | Rome IV Diagnostic Questionnaires - Rome Foundation
Skip to content
Register›Online Store›Donate›
Home
News
Contact Us
Online Store
Gut Feelings Book
Gut Feelings: The Patient Story
Communication 101
Communication 101 Program Sample Video
Communication 101.5
Communication 101.5 Program Sample Video
Communication 202
Communication 202 Program Sample Video
Educational Products
Buyer’s Guide
Rome Campus Education
Register Now
Copyright & Licensing
Translations
Search
Search Submit
About
Board
Rome Board Member Publications in 2021
Administration
Rome Foundation Fellows
Rome Foundation Partners Program
Ethics Policy
Sponsors
Standing Committees
Copyright and Licensing Committee
International Liaison Committee
Pediatric Committee
Communication Programs
Award for Communication & Patient Centered Care in DGBI
Communication Videos
Gut Feelings Book Series
Visiting Scholar Program
Rome – AGA Communication Conference Videos
Educational Resources on Communication Skills
Research
Rome Foundation Research Institute
Current and Ongoing Research Institute Studies
Rome Foundation Global Epidemiology Study
Virtual Symposium: Rome Foundation Global Study of DGBI
Rome Foundation Research Program & Awards
Rome – Disorders of Gut-Brain Interaction International Research Awards
The Rome Foundation Douglas Drossman Award for Communication and Patient Centered Care in DGBI
Rome – Aldo Torsoli Research Award
Ray Clouse Award
Ken Heaton Award
Previous Awards
Previous Rome-AGA Pilot Study Awards
Working Teams
Active Working Teams (2020-2023)
Gut-Brain Therapy Behavioral Trial Guidelines Working Team
Overlap in DGBI Working Team
Plausibility Working Team
Completed Working Teams
Food and Diet in DGBI Working Team
Brain-Gut Psychotherapies Working Team
Communications Working Team
Neuromodulators for FGIDs
Post-infection IBS
Brain Imaging in DGBI
Asian Working Team for FGIDs
Food & FGIDs Working Team
Role of Intestinal Flora in FGIDs
Multinational Cross-Cultural Research
Primary Care Committee
Brain Imaging Working Team
Outcomes/Endpoints in Pharmaceutical Clinical Trials
Rome Severity Committee
Rome Working Teams Publications
GastroPsych
Pediatrics
Diet & Nutrition
Rome IV/V & Criteria
Rome V, DGBI 5th Edition
Rome IV Criteria
Rome IV Committees
Chapter Summaries
Gastroenterology Articles (2016)
Rome IV Questionnaire
Rome IV History
Rome IV Launch
Rome IV Project
Rome IV FAQs
What’s New for Rome IV
Rome IV Sponsors
Resources
What is a Disorder of Gut-Brain Interaction (DGBI)
Grand Rounds
News & Updates
Newsletter
Video Library for Providers & Patients
Presentations & Videos
MD Live
Podcasts
Blog
AGA – Rome Lectureship 2021
Pocket Cards
Events
Featured Conference Partners
Upcoming Events
Previous Events
Rome IV Diagnostic Questionnaires
The Rome IV Diagnostic Questionnaire for Functional Gastrointestinal Disorders in Adults (R4DQ) translates the Rome IV diagnostic criteria into questions that can be understood and reported by patients and research subjects. The intended uses of the R4DQ are (1) to provide inclusion criteria for clinical research studies, (2) to serve as case definitions for epidemiological surveys, and (3) to assist clinicians in arriving at a provisional diagnosis. Diagnostic questionnaires were also developed for pediatric FGIDs.
The following Rome IV questionnaires are available upon request. Additional instruments are also available. Please visit ourCopyright and Licensingpage for more information.
For Adults
A self-report integrated questionnaire for diagnosis of all functional gastrointestinal disorders in adults (R4DQ), including alarm symptoms or red flags to alert clinicians to consider testing for alternative medical disorders.
Six modules based on the Rome IV criteria that can be used to screen for the following single diagnoses or groups of diagnoses:
Irritable Bowel Syndrome (IBS)
Bowel Disorders (Includes all bowel disorders and all centrally mediated pain disorders)
Gastroduodenal Disorders
Esophageal Disorders
Gallbladder and Sphincter of Odi Disorders
Anorectal Disorders
Guidelines and Flags for Mental Health Involvement
For Children & Adolescents
Rome IV Diagnostic Questionnaire on Pediatric Functional Gastrointestinal Disorders (R4PDQ) – Child: Self-report form for Children and Adolescents (10 years and older)
R4PDQ – Child: Parent-report form for Children (4 years and older)
R4PDQ – Toddler: Parent-report form for Neonates and Toddlers (0-3 years)
For questions regarding these questionnaires or scoring algorithms, please contact us.
Latest News & Updates
Rachel Sarnoff, MD, MSCR Candidate, Shares Her Visiting Scholar Experience
2024 Annual Digestive Disease Week held in Washington, D.C. advances DGBI research and patient care
Kalin Clark, PsyD, ABPP Shares Her Visiting Scholar Experience
5th Train the Trainer program in Communication Skills to improve the Patient-Provider Relationship
Mailing List
Subscribe to our mailing lists to receive The Rome Foundation's quarterly e-newsletter and free e-mail updates on:
Special Announcements
Educational product special offers
News of upcoming events
Quarterly e-newsletter
Join Mailing List
Rome IV
Rome IV
Buy Online
Rome IV Project
Rome IV FAQs
Rome IV Collection
What’s New for Rome IV
Rome IV Committees
Rome IV Sponsors
Contact Us
14460 Falls of Neuse Rd
Suite 149-116
Raleigh, NC 27614
Email Us
© COPYRIGHT 2021 ROME FOUNDATION. ALL RIGHTS RESERVED. | WEBSITE BY PRECISION MARKETING PARTNERS.
Privacy Policy | Terms & Conditions
Back To Top
Search Submit
Afrikaans Albanian Amharic English Arabic Armenian Azerbaijani Basque Belarusian Bengali Bosnian Bulgarian Catalan Cebuano Chichewa Chinese (Simplified) Chinese (Traditional) Corsican Croatian Czech Danish Dutch Esperanto Estonian Filipino Finnish French Frisian Galician Georgian German Greek Gujarati Haitian Creole Hausa Hawaiian Hebrew Hindi Hmong Hungarian Icelandic Igbo Indonesian Irish Italian Japanese Javanese Kannada Kazakh Khmer Korean Kurdish (Kurmanji) Kyrgyz Lao Latin Latvian Lithuanian Luxembourgish Macedonian Malagasy Malay Malayalam Maltese Maori Marathi Mongolian Myanmar (Burmese) Nepali Norwegian Pashto Persian Polish Portuguese Punjabi Romanian Russian Samoan Scottish Gaelic Serbian Sesotho Shona Sindhi Sinhala Slovak Slovenian Somali Spanish Sundanese Swahili Swedish Tajik Tamil Telugu Thai Turkish Ukrainian Urdu Uzbek Vietnamese Welsh Xhosa Yiddish Yoruba Zulu
English
×Close mobile menu
About
Board
Rome Board Member Publications in 2021
Administration
Rome Foundation Fellows
Rome Foundation Partners Program
Ethics Policy
Sponsors
Standing Committees
Copyright and Licensing Committee
International Liaison Committee
Pediatric Committee
Communication Programs
Award for Communication & Patient Centered Care in DGBI
Communication Videos
Gut Feelings Book Series
Visiting Scholar Program
Rome – AGA Communication Conference Videos
Educational Resources on Communication Skills
Research
Rome Foundation Research Institute
Current and Ongoing Research Institute Studies
Rome Foundation Global Epidemiology Study
Virtual Symposium: Rome Foundation Global Study of DGBI
Rome Foundation Research Program & Awards
Rome – Disorders of Gut-Brain Interaction International Research Awards
The Rome Foundation Douglas Drossman Award for Communication and Patient Centered Care in DGBI
Rome – Aldo Torsoli Research Award
Ray Clouse Award
Ken Heaton Award
Previous Awards
Previous Rome-AGA Pilot Study Awards
Working Teams
Active Working Teams (2020-2023)
Gut-Brain Therapy Behavioral Trial Guidelines Working Team
Overlap in DGBI Working Team
Plausibility Working Team
Completed Working Teams
Food and Diet in DGBI Working Team
Brain-Gut Psychotherapies Working Team
Communications Working Team
Neuromodulators for FGIDs
Post-infection IBS
Brain Imaging in DGBI
Asian Working Team for FGIDs
Food & FGIDs Working Team
Role of Intestinal Flora in FGIDs
Multinational Cross-Cultural Research
Primary Care Committee
Brain Imaging Working Team
Outcomes/Endpoints in Pharmaceutical Clinical Trials
Rome Severity Committee
Rome Working Teams Publications
GastroPsych
Pediatrics
Diet & Nutrition
Rome IV/V & Criteria
Rome V, DGBI 5th Edition
Rome IV Criteria
Rome IV Committees
Chapter Summaries
Gastroenterology Articles (2016)
Rome IV Questionnaire
Rome IV History
Rome IV Launch
Rome IV Project
Rome IV FAQs
What’s New for Rome IV
Rome IV Sponsors
Resources
What is a Disorder of Gut-Brain Interaction (DGBI)
Grand Rounds
News & Updates
Newsletter
Video Library for Providers & Patients
Presentations & Videos
MD Live
Podcasts
Blog
AGA – Rome Lectureship 2021
Pocket Cards
Events
Featured Conference Partners
Upcoming Events
Previous Events
Search Submit |
190778 | https://www.thesaurus.com/browse/prettiness | 19 Synonyms & Antonyms for PRETTINESS | Thesaurus.com
Games
Daily Crossword
Word Puzzle
Word Finder
All games
Featured
Word of the Day
Word of the Year
New words
Language stories
All featured
Culture
Slang
Emoji
Memes
Acronyms
Gender and sexuality
All culture
Writing tips
Writing hub
Grammar essentials
Commonly confused
All writing tips
Games
Featured
Culture
Writing tips
Advertisement
Skip to
Skip to
noun (2)
as in comelinessas in pulchritude
Related WordsQuiz
Advertisement
prettiness
noun as in comeliness
Synonyms Antonyms
Strong matches
beauteousness
beauty
elegance
fairness
handsomeness
loveliness
1
/
Video Player is loading.
Play Video
Unmute
Duration 0:00
/
Current Time 0:00
Advanced Settings
Loaded: 0%
Remaining Time-0:00
Fullscreen Play Rewind 10 Seconds Up Next
This is a modal window.
Beginning of dialog window. Escape will cancel and close the window.
Text Color Transparency Background Color Transparency Window Color Transparency
Font Size Text Edge Style Font Family
Reset restore all settings to the default values Done
Close Modal Dialog
End of dialog window.
Share
Replay the list
TOP ARTICLES
Powered by AnyClip
Privacy Policy
Keyboard Shortcuts
noun as in pulchritude
Synonyms
Strong matches
adorableness
allure
allurement
attraction
elegance
exquisiteness
glamor
handsomeness
loveliness
shapeliness
Weak matches
good looks
physical attractiveness
Advertisement
Discover More
Related Words
Words related to prettiness are not direct synonyms, but are associated with the word prettiness. Browse related words to learn more about word associations.
comeliness
noun as in attractiveness
beauteousness
beauty
elegance
fairness
handsomeness
loveliness
prettiness
pulchritude
noun as in beauty
adorableness
allure
allurement
attraction
elegance
exquisiteness
glamor
good looks
handsomeness
loveliness
physical attractiveness
prettiness
shapeliness
Quiz
Q: Instead of the term COTTON CANDY, British English speakers are more likely to use the word …
candycotton.
candyfloss.
candyfluff.
Take the full quiz.Go to all quizzes
From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group.
Advertisement
Did You Know?
"Muscle" originates from the Latin word for "little mouse," because physicians thought that muscles looked like little mice running under the skin.
Advertisement
Advertisement
Browse
#
aa
bb
cc
dd
ee
ff
gg
hh
ii
jj
kk
ll
mm
nn
oo
pp
qq
rr
ss
tt
uu
vv
ww
xx
yy
zz
About
Careers
Contact us
Cookies, terms, & privacy
Your Privacy Rights
Help
Follow us
Get the Word of the Day every day!
Sign up
By clicking "Sign Up", you are accepting Dictionary.com Terms & Conditions and Privacy Policies.
© 2025 Dictionary.com, LLC |
190779 | https://www.quora.com/What-is-25-as-a-fraction-in-simple-form | What is 25 % as a fraction in simple form? - Quora
Something went wrong. Wait a moment and try again.
Try again
Skip to content
Skip to search
Sign In
Mathematics
Convert
Decimal Fractions
Percentage (%)
Simple Maths
Arithmetic
Basic Mathematics
Fractions (mathematics)
Conversions
5
What is 25 % as a fraction in simple form?
All related (52)
Sort
Recommended
Valentina Veas
Studied at Colegio Saint Gabriels School (Graduated 2023)
·4y
Originally Answered: What would be the correct fraction of 25%? ·
The fraction of 25% is 25/100. Then you simplify it and the final result is 1/4. So the answer is 1/4.
Upvote ·
9 3
Sponsored by Grammarly
Stuck on the blinking cursor?
Move your great ideas to polished drafts without the guesswork. Try Grammarly today!
Download
99 34
Related questions
More answers below
What is .25 written in its simplest fraction form?
What is .25% as fraction?
How do you convert 3.25 into a fraction?
What would be the correct fraction of 25%?
What is 0.25 percent as a fraction in the simplest form?
Lukas Schmidinger
I have graduate CS and my studies included math courses. · Author has 27.7K answers and 14.9M answer views
·6y
Originally Answered: What is 25% as a fraction in the simplest form? ·
What is 25% as a fraction in the simplest form?
1 4 1 4
Remember for a 3 3 digit number
a b c%=a b c 100=a.b c a b c%=a b c 100=a.b c
And 25 100 25 100 is simply reduced to 1 4 1 4 when you factor 25 25 and 100 100.
For 1 or 2 digit numbers simply add leading zeros, for numbers with more than 3 whole digits simple take the 3 next to the decimal point.
Math Antics Percents Video Lessons - YouTube
Upvote ·
9 2
9 2
Assistant
Bot
·
1y
To express 25% as a fraction in simple form, you can follow these steps:
Convert the percentage to a fraction:
25%=25 100 25%=25 100
2. Simplify the fraction:
Divide both the numerator and the denominator by their greatest common divisor, which is 25:
25÷25 100÷25=1 4 25÷25 100÷25=1 4
Thus, 25% as a fraction in simple form is 1 4 1 4.
Upvote ·
Richard Rand
Former Middle School and College Instructor (1965–2007) · Author has 342 answers and 868K answer views
·6y
Originally Answered: What is 25% as a fraction in the simplest form? ·
Per cent means per hundred so 25% means 25 per hundred or 25/100. Now 25/100 is 1/4 as is illustrated below.
Continue Reading
Per cent means per hundred so 25% means 25 per hundred or 25/100. Now 25/100 is 1/4 as is illustrated below.
Upvote ·
9 1
Related questions
More answers below
What is the value of 0.25 as a fraction in simplest form?
What is 2.25 in fraction form?
What is 25 percent as a reduced fraction?
What are the different ways 25% can be a fraction?
What is 2.5% as a fraction in the simplest form?
Stuart Herring
B.S. in Information Technology, American InterContinental University Atlanta (Graduated 2004) · Author has 11.7K answers and 8.2M answer views
·1y
Originally Answered: What is 27% as a fraction? ·
“%” means “per hundred”. The usual way of indicating “per” is with the division symbol: / .
So 27% = 27/100 (or 27 100 27 100).
Upvote ·
9 2
Sponsored by ELEKS
End-to-End e-Excise Setup with System Architects.
Don't risk delays—book a consultation with ELEKS, the team behind Ukraine's e-Excise system.
Learn More
9 4
Charles Holmes
Studied Financial Markets&Mathematics (Graduated 1990) · Author has 16.6K answers and 13.6M answer views
·5y
Originally Answered: What is 24% as a fraction? ·
y=6/25
PREMISES
y=the numeral 24% expressed as a fraction
CALCULATIONS
To convert a percentage to an equivalent fraction begin by removing the percent sign. Divide the remaining number by 100 and reduce the fraction to the lowest terms, i.e., 50% would become 50/100=1/2 in the simplest form.
If the percentage is a decimal percentage, move the express decimal point to the right until the number is an integer. Move the implied decimal point in 100 the same number of places to the right and divide, i.e., 0.5% would become 5/1000=1/200 in the simplest form.
In the instant case,
y=24% as a fraction become
Continue Reading
y=6/25
PREMISES
y=the numeral 24% expressed as a fraction
CALCULATIONS
To convert a percentage to an equivalent fraction begin by removing the percent sign. Divide the remaining number by 100 and reduce the fraction to the lowest terms, i.e., 50% would become 50/100=1/2 in the simplest form.
If the percentage is a decimal percentage, move the express decimal point to the right until the number is an integer. Move the implied decimal point in 100 the same number of places to the right and divide, i.e., 0.5% would become 5/1000=1/200 in the simplest form.
In the instant case,
y=24% as a fraction becomes
y=24/100 (Cancel the common factor 4)
y=
6/25 (Since there are no common factors, no further reduction is possible. The fraction is now in the simplest form.)
C.H.
Upvote ·
9 1
Kaden Slone
A year past senior in high school, now in college · Author has 213 answers and 683.8K answer views
·6y
Originally Answered: What is 25% as a fraction in the simplest form? ·
One over four.
Twenty-five divided by one-hundred is 0.25.
25/100.
One-hundred divided by twenty-five is four.
This would simplify into 1/4.
1
__
4
Upvote ·
9 1
Sponsored by Reliex
Looking to Improve Capacity Planning and Time Tracking in Jira?
ActivityTimeline: your ultimate tool for resource planning and time tracking in Jira. Free 30-day trial!
Learn More
99 24
Lakshmanan Sankaran
expert in solving simple arithmatical problems · Author has 5.3K answers and 8.8M answer views
·5y
Originally Answered: What is the fraction of 24%? ·
24/100=6/25ans
Upvote ·
9 7
Bridgette Kiser
Math Teacher
·5y
Originally Answered: What is the fraction of 24%? ·
24/100
A percent is a part out of 100. Some kids learn to see the percent sign as a one and two zeros, reminding them that a percent is “how many out of 100.”
Your response is private
Was this worth your time?
This helps us sort answers on the page.
Absolutely not
Definitely yes
Upvote ·
9 1
Sponsored by RHUZZALS
You simply can't remain incifferent to the charm of this clothes!
Refresh your look and stay in fashion!
Shop Now
99 21
Amanda Engelmeyer
5y
Originally Answered: What is the fraction of 24%? ·
6/25
Upvote ·
9 2
Charles Holmes
Studied Financial Markets&Mathematics (Graduated 1990) · Author has 16.6K answers and 13.6M answer views
·1y
Originally Answered: What is 27% as a fraction? ·
Y=27/100=0.27 as a decimal fraction
PREMISES
Y=the value of 27% as a proper fraction
CALCULATIONS
Y=27%
Evaluating,
Y=
27/100=0.27 as a decimal fraction
C.H.
Upvote ·
9 2
Sanjay Chakradeo
Engineer, interested in Basic Mathematics. · Author has 3.5K answers and 4.3M answer views
·3y
Originally Answered: Which fraction does 25% represent? ·
25% means 25/100 = 1/4.
So 25% represents the fraction 1/4.
In fact 25% can represent any fraction (1n)/(4n), where n is Natural number like 1,2,3,4, …
Any of these 1/4, 2/8, 3/12, 4/16, …, 25/100, … represent 25%.
Sanjay Chakradeo.
Upvote ·
Related questions
What is .25 written in its simplest fraction form?
What is .25% as fraction?
How do you convert 3.25 into a fraction?
What would be the correct fraction of 25%?
What is 0.25 percent as a fraction in the simplest form?
What is the value of 0.25 as a fraction in simplest form?
What is 2.25 in fraction form?
What is 25 percent as a reduced fraction?
What are the different ways 25% can be a fraction?
What is 2.5% as a fraction in the simplest form?
What is 3.25% as a decimal fraction?
Which fraction does 25% represent?
What is the value of 6/3 + 1/2 - 1.25 in fraction form?
What number is 25% of (2/5) as a fraction?
What is the percentage of a given fraction 14/25?
Related questions
What is .25 written in its simplest fraction form?
What is .25% as fraction?
How do you convert 3.25 into a fraction?
What would be the correct fraction of 25%?
What is 0.25 percent as a fraction in the simplest form?
What is the value of 0.25 as a fraction in simplest form?
Advertisement
About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press ·
© Quora, Inc. 2025 |
190780 | https://zhaozh.xyz/ReviewPartialFractions.pdf | Skills Review: Partial Fractions Motivation: Before we talk about splitting up fractions, let’s review how to combine fractions. You have been doing this since elementary school. To add fractions we need a common denominator.
For example, 2 5 + 7 3 = 6 15 + 35 15 = 41 15. Notice how we multiplied the top and bottom in each fraction in order to get equivalent fractions with the same common denominator. The same procedure works for functions involving fractions.
For example, if you are given f(x) = 1 x−2 + 3 x−5 and you want to combine into one fraction, then you write f(x) = 1 x−2 + 3 x−5 = x−5 (x−2)(x−5) + 3x−6 (x−2)(x−5) = 4x−11 (x−2)(x−5).
As another example, g(x) = x x2+4 + 2 x+3 = x2+3x (x2+4)(x+3) + 2x2+8 (x2+4)(x+3) = 3x2+3x+8 (x2+4)(x+3).
In calculus you learned that integration of rational functions is often easier if you decompose into partial fractions. In other words, you go in the opposite direction of the two examples shown above. That is, if you wanted to integrate 4x−11 (x−2)(x−5), then it is easier to work with the partial fraction form 1 x−2 + 3 x−5.
And if you want to integrate g(x) = 3x2+3x+8 (x2+4)(x+3), then it is easier to work with the partial fraction form x x2+4 + 2 x+3. The method of partial fraction decomposition is used to split up rational functions in this way. In this class, partial fractions will help us solve linear constant coefficient differential equation in a very systematic (algebraic) way.
The Method of Decomposing into Partial Fractions: Given a rational function N(x) D(x) = Numerator Denominator 1. Simplify/Divide, if needed: If anything can be canceled, do it! If the degree (highest power) of the numerator is greater than or equal to the degree of the denominator, divide!
For example: Given x2 x+3, you can divide to rewrite it as x −3 + 9 x+3.
2. Factor Denominator and Complete Squares, if needed: Factor the denominator into linear terms and irreducible quadratics. For the irreducible quadrat-ics, complete the square.
For example: Given 1 (x+1)(x2+10x+28), completing the square gives 1 (x+1)((x+5)2+3).
3. Write the form of the decomposition: Distinct linear terms that look like x −a decompose as A x−a.
Repeated linear terms that look like (x −a)3 decompose as A x−a + B (x−a)2 + C (x−a)3.
Distinct irreducible quadratic terms that look like x2 + a2 decompose as Ax+B x2+a2 .
Repeated irreducible quadratic terms that look like (x2 + a2)2 decompose as Ax+B x2+a2 + Cx+D (x2+a2)2.
Here is an example that has several different types: BLAH (x + 1)(x −3)2((x −2)2 + 7) = A x + 1 + B x −3 + C (x −3)2 + D(x −2) + E (x −2)2 + 7 4. Solve for all the constants: Clear the denominators and equate the two sides. There are sometimes shortcuts (as mentioned in class and in the examples below). But it always works to equate coefficients and solve the corresponding equations until you get all the constants A, B, C, . . ..
5. Write your decomposition: You are done!
Examples: • Give the partial fraction decomposition for x2+2 x3+3x2+2x 1. Simplify/Divide: Done! Degree of numerator smaller than denominator.
2. Factor: x2+2 x(x2+3x+2) = x2+2 x(x+1)(x+2) 3. Form: x2+2 x(x+1)(x+2) = A x + B x+1 + C x+2 4. Solve for constants: Clearing denominators gives x2 + 2 = A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1). Expanding gives: x2 + 2 = (A + B + C)x2 + (3A + 2B + C)x + 2A. So A + B + C = 1, 3A + 2B + C = 0 and 2A = 2. Combining these condition you can solve this to get A, B, and C.
Shortcut: Note that instead of expanding you can plug in strategic values of x to solve for A, B, and C faster (or use the cover up method to be even faster). Here are some strategically chosen values of x: Plugging in x = 0 gives 2 = 2A, so A = 1.
Plugging in x = −1 gives 3 = −B, so B = −3.
Plugging in x = −2 gives 6 = 2C, so C = 3.
5. Done: x2+2 x3+3x2+2x = 1 x + −3 x+1 + 3 x+2.
• Give the partial fraction decomposition for x+1 x3+4x2+8x 1. Simplify/Divide: Done! Degree of numerator smaller than denominator.
2. Factor: x+1 x(x2+4x+8). The expression x2 + 4x + 8 is irreducible (b2 −4ac = −16 < 0).
Completing the square gives x2 + 4x + 4 −4 + 8 = (x + 2)2 + 4 3. Form: x+1 x((x+2)2+4) = A x + B(x+2)+C (x+2)2+4 4. Solve for constants: Expanding gives x + 1 = A((x + 2)2 + 4) + (B(x + 2) + C)x.
Let’s start by plugging strategic values: Plugging in x = 0 gives 1 = 8A, so A = 1/8.
Plugging in x = −2 gives −1 = 4A −2C, so −1 = 1 2 −2C, so C = 3 4.
Thus, we have x + 1 = 1 8((x + 2)2 + 4) + B(x + 2) + 3 4 x. You can expand this all out if you wish. I personally suggest you just look for a coefficient that would involve B. In this case, how are you going to get an x2 term? On the left-hand side the coefficient of x2 is zero. On the right hand side you can see that the only x2 terms you will get are 1 8x2 and Bx2. Thus, 0 = 1 8 + B, so B = −1 8.
5. Done: x+1 x(x2+4x+8) = 1/8 x + −(1/8)(x+2)+3/4 (x+2)2+4 • Give the partial fraction decomposition for 5 (x+1)(x−2)2 1. Simplify/Divide: Done! Degree of numerator smaller than denominator.
2. Factor: Done!
5 (x+1)(x−2)2.
3. Form: 5 (x+1)(x−2)2 = A x+1 + B x−2 + C (x−2)2 4. Solve for constants: Expanding gives 5 = A(x −2)2 + B(x + 1)(x −2)2 + C(x + 1).
Let’s start by plugging strategic values: Plugging in x = −1 gives 5 = 9A, so A = 5/9.
Plugging in x = 2 gives 5 = 3C, so C = 5/3.
Thus, we have 5 = 5 9(x −2)2 + B(x + 1)(x −2)2 + 5 3(x + 1). You can expand this all out if you wish. Again, I personally suggest you look for coefficients involving B. In this situation, looking at coefficients of x2, we have 0 = 5 9x2 + Bx2, so B = −5 9.
5. Done: 5 (x+1)(x−2)2 = 5/9 x+1 + −5/9 x−2 + 5/3 (x−2)2.
• Give the partial fraction decomposition for 1 (x2+1)(x2+4) 1. Simplify/Divide: Done! Degree of numerator smaller than denominator.
2. Factor: Done!
3. Form: 1 (x2+1)(x2+4) = Ax+B x2+1 + Cx+D x2+4 4. Solve for constants: Clearing the denominators gives 1 = (Ax + B)(x2 + 4) + (Cx + D)(x2 + 1).
No good strategic values to plug in (could plug in i or 2i if you want to try), so let’s expand to get 1 = Ax3 + Bx2 + 4Ax + 4B + Cx3 + Dx2 + Cx + D Thus, A + C = 0, B + D = 0, 4A + C = 0, and 4B + D = 1.
Combining A + C = 0 and 4A + C = 0 gives A = 0 and C = 0.
Combining B + D = 0 and 4B + D = 1 gives B = 1/3 and D = −1/3.
5. Done: 1 (x2+1)(x2+4) = 1/3 x2+1 + −1/3 x2+4 |
190781 | https://stackoverflow.com/questions/42568620/find-all-edges-in-a-graph-which-if-removed-would-disconnect-a-pair-of-vertices | Skip to main content
Stack Overflow
About
For Teams
Stack Overflow for Teams
Where developers & technologists share private knowledge with coworkers
Advertising
Reach devs & technologists worldwide about your product, service or employer brand
Knowledge Solutions
Data licensing offering for businesses to build and improve AI tools and models
Labs
The future of collective knowledge sharing
About the company
Visit the blog
Find all edges in a graph which if removed, would disconnect a pair of vertices
Ask Question
Asked
Modified
8 years, 6 months ago
Viewed
4k times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
I have an undirected graph G(V, E). The problem I am trying to solve is broken down into two parts:
Part 1: Given the graph, and a pair of vertices in the graph, find the edges which are contained in all paths between this pair of nodes.
Part 2: Given the graph, find all such edges such that each one is present along all paths between any two vertices in the graph.
An example for Part 2:
Let the graph have nodes {a, b, c, d, e, f, g}.
Let edges be:
```
{a, b},
{b, c},
{c, a},
{c, d},
{d, e},
{e, f}
{f, g},
{g, e}
```
For this graph, {c, d} and {d, e} should be the edges returned.
How can you do it efficiently?
algorithm
graph-algorithm
graph-theory
Share
CC BY-SA 3.0
Improve this question
Follow this question to receive notifications
asked Mar 3, 2017 at 0:41
Diptesh ChatterjeeDiptesh Chatterjee
38511 gold badge55 silver badges1414 bronze badges
3
1
Seems like a homework assignment. What have you tried so far? You've shown no work.
David Makogon
–
David Makogon
03/03/2017 00:44:11
Commented
Mar 3, 2017 at 0:44
It's far from a homework assignment. I am trying to solve this as part of my job. :) The only thing that comes to mind for me is to enumerate all paths and find intersection between them for all pairs of vertices. But unfortunately it is NP complete.
Diptesh Chatterjee
–
Diptesh Chatterjee
03/03/2017 01:49:26
Commented
Mar 3, 2017 at 1:49
Part 2, as formulated, will give no edges for a non-trivial graph. You'd need to consider only pairs of vertices that are in different 2-connected components
voidengine
–
voidengine
03/03/2017 18:05:38
Commented
Mar 3, 2017 at 18:05
Add a comment
|
2 Answers 2
Reset to default
This answer is useful
2
Save this answer.
Show activity on this post.
If an edge E is on a path from vertex A to vertex B, but there is also a path from A to B that does not include E, then then E is in a cycle.
If E is in a cycle, and E is on a path from A to B, then there is a path from A to B through the other edges in the cycle that does not include E.
So, the edges you are looking for in part 2 are the edges that are not in cycles that are also on a path from A to B.
To answer your questions then:
1) Use the Hopcroft/Tarjan algorithm to find all the biconnected components (cycles) in the graph. See
2) Collapse each biconnected component to a single vertex (more precisely, combine any 2 vertexes connected by an edge that is in a cycle until you run out of such edges). This will create a tree corresponding to the block cut tree (also described in the above link). The component that includes A will be A in the new graph, and similarly for B.
3) Since the resulting graph is a tree, it will have only one path from A to B. Find it with DFS. The edges on that path are the answer to part 2.
4) The edges from (3), as well as the edges in every biconnected component that is adjacent to those edges, are the answer to part 1.
Yeah, it kinda sounds like @EmmanuelAC's answer was trying to point in this direction.
Share
CC BY-SA 3.0
Improve this answer
Follow this answer to receive notifications
edited Mar 3, 2017 at 6:07
answered Mar 3, 2017 at 5:37
Matt TimmermansMatt Timmermans
60.6k33 gold badges5555 silver badges105105 bronze badges
1
Thanks Matt. This, combined with MapReduce could be a good enough solution for Part 2 for large graphs. Appreciate the help.
Diptesh Chatterjee
–
Diptesh Chatterjee
03/03/2017 18:25:50
Commented
Mar 3, 2017 at 18:25
Add a comment
|
This answer is useful
1
Save this answer.
Show activity on this post.
there is a algorithm to find "bridges", we call bridge an edge if we remove it the graph is divided in two disjoint graphs. The edges you are looking for are the bridges.
How can I find bridges in an undirected graph?
Part 1. Now that you know what are the bridges, you just need to find a path between the pair of nodes and all the vertices in that path that are also bridges are the solution to part 1.
Part 2. The solution to this are all the bridges in the graph
Share
CC BY-SA 3.0
Improve this answer
Follow this answer to receive notifications
edited May 23, 2017 at 10:29
CommunityBot
111 silver badge
answered Mar 3, 2017 at 1:04
EmmanuelACEmmanuelAC
14133 bronze badges
Add a comment
|
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
algorithm
graph-algorithm
graph-theory
See similar questions with these tags.
The Overflow Blog
Open-source is for the people, by the people
Moving the public Stack Overflow sites to the cloud: Part 1
Featured on Meta
New comment UI experiment graduation
Further Experimentation with Comment Reputation Requirements
Updated design for the new live activity panel experiment
Policy: Generative AI (e.g., ChatGPT) is banned
8 people chatting
Linked
5
How can I find bridges in an undirected graph?
Related
3
Algorithm to find disconnected nodes in a graph
0
Discrete Maths: Check connectivity of a graph after removing a vertex? Efficient Way?
1
Disconnecting a connected graph
2
Finding a vertex whose removal will disconnect two others
2
In undirected connected graph how to find set of vertices removing which graph becomes disconnected?
Disconnect all vertices in a graph - Algorithm
Removing minimum no of edges to disconnect two vertices in a graph
0
Find all pairs of nodes whose deletion disconnects graph
1
Minimum number of vertices whose removal makes the graph no longer connected
2
How to get all the possible set of edges of graph that disconnects the graph ( satisfying minimum cut )
Hot Network Questions
Why does a new Windows 11 PC have repeated paths in %PATH%?
What is "The Self-Proclaimed Wisdom of Elihu"?
What will happen when a pull request includes changes to ignored .gitignore?
Using size of encrypted content to guess corresponding plaintext
If Metamagic is used on a spell, then that spell is stopped by Counterspell, are the Sorcery Points lost as well?
CX order of memory rounds in surface code Y memory experiment
Terminology question: Charge density difference or differential charge density?
Is there a name for the inertial reference frame experienced by something moving at c?
Did LEGO really introduce a new bracket mold that might be out of system?
Which sitcom had the most spin-offs, including spin-offs of the spin-offs?
What does it mean to "rotate the knight"?
How to make use of "1 stopover permitted on the pricing unit"?
Inoptimal Fill Factor on objects in sys schema
How can I force reevaluation of Association values?
What happens if there is long-lasting martial law such that enough MPs die and a quorum is impossible?
Why do I get negative entropy?
Errors in some applications (Excel, Word, Notepad) When Desktop Is Mapped to Network via Registry and Network Is Not Available
How does copyright work when someone owns a language?
Did Napoleon III serve his guests with aluminium plates?
What is difference between nowcasting and forecasting?
How Amstrad CPC games calculated positions for every sprite?
Why are metaphysicians obsessed with language?
What actually carries across runs and decks in Balatro?
Finding the positions of displaced elements in a permutation
Question feed
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. |
190782 | http://www.rohanbyanjankar.com.np/2019/03/cartel-clear-economic-analysis.html | Cartel: A clear economic analysis
About
Macroeconomics
International
Current Account Deficit as a Saving-Investment Gap
TU BBA
8th Semester BBA notes
BBA 8th Semester Notes TU
More than just a blog - this is a journey through questions that challenge and ideas that inspire.
HomeMA Economics
Cartel: A clear economic analysis
by Rohan Byanjankar—March 31, 2019 0
Table of Contents
Cartel
1. Joint Profit Maximization Cartel
Figurative Analysis
2. Market sharing cartel
2.1. Non – price competitive agreements
Figurative Analysis
2.2. Sharing of the market by agreement on quotas
Figurative Analysis
Difficulties in Cartel
Other online materials
Our Posts related to Economics
References
Suggested Readings:
Cartel
Cartel is the formal agreement between two or more oligopolistic firms in terms of price, quantity, and market share in order to reduce uncertainty and risk. In other words, a cartel is a formal agreement between two or more oligopolistic firms to either limit their production and/or to fix prices. Cartel agreements represent the most complete form of collusion among the oligopolistic firms. The cartel having absolute control over its member firms resembles with monopoly.
The cartel operates in both pure and differentiated oligopoly. Organization of Petroleum Exporting Countries (OPEC) is the best example of a cartel that has a significant impact on the world’s fossil fuel market. As of 2018, OPEC with 15 countries accounted for an estimated 44 percent of global oil production and 81.5 percent of the world's "proven" oil reserves, giving OPEC a major influence on global oil prices.
Some of the models of cartels are as follows:
1. Joint Profit Maximization Cartel
The joint profit maximization model is the most stable cartel model in which the firms in cartel aims to maximize the joint profit. This situation is identical to that of a multi-plant monopolist who seeks to maximize his or her profit. This model is based on pure oligopoly. Pure oligopoly is the type of oligopoly where firms produce homogenous products.
Assumptions of this model
There are two firms in the market, Firm 1 and Firm 2 (say)
Firm 1 is assumed to be a low-cost firm and Firm 2 is assumed to be high-cost firm
Pure oligopoly
Appointment of a central agency
The central agency has the authority to decide not only price but also quantity at which the profit is maximized.
Central agency allocates production among the members and distributes the joint profit among the participating firms.
The central agency knows the cost figure of individual firms and can derive individual demand curves.
Central agency acting as the multi-plant monopolist will set the price defined by the intersection of the industry MR and MC curves.
The price and output of the firm under joint profit maximization can be obtained by fulfilling the necessary and sufficient conditions.
Necessary condition: MR = MC [MC = MC 1 + MC 2]
Sufficient condition: Slope of MC > Slope of MR.
Figurative Analysis
Figure 1 depicts the joint profit maximization. Figure 1 (a), figure 1 (b), and figure 1 (c) represents firm 1, firm 2, and the central agency respectively. In all the panels, a, b, and c, x-axis measures quantity and the y-axis measures price.
According to the assumptions of Joint profit maximization model, central agency acting as the multi-plant monopolist set the price defined by the intersection between MR and MC, that is, MR = MC and slope of MC > Slope of MR. Hence, in figure 1 (c), at point e, industry MR intersects with industry MC that determines the common price (P) for all the firms in the cartel. The industry output (X) is also defined by the same intersection. Similarly, the horizontal line is drawn from point 'e' across figure 1 (a) and figure 1 (b) to determine the individual quantity of firm 1 and firm 2.
In figure 1 (a), firm 1 produces X 1 quantity of goods where firm’s MC curve intersects with the horizontal line drawn from the point 'e' of figure 1 (c), which represents industry marginal revenue.
Similarly, in figure 1 (b), firm 2 produces X2 quantity of goods where the firm’s MC curve intersects with the horizontal line. Firm 1 is a low-cost firm and firm 2 is a high-cost firm, so firm 1 earns a higher profit than firm 2; however, both the firms enjoy a supernormal profit.
Hence, the central agency allocates production to the member firms on the basis of cost figures and distributes the joint profit to its member firms.
2. Market sharing cartel
Under this model, the firms are in agreement to share the market. Unlike joint profit maximization, this model does not focus on price or quantity but focuses on market share. Some of the models under market sharing cartel are as follows:
2.1. Non – price competitive agreements
Non – price competitive agreement is the loose form of cartel agreement where member firms agree to offer the product at the common price, at which each of them can sell any quantity demanded. The price is set by bargaining with the low-cost firms pressing for a low price and the high-cost firm for a high price; the agreed price must be such that all the members earn a certain amount of profit.
The firms are not allowed to sell the product below the cartel price but are free to vary the style of their product or their selling activities. In other words, the firms compete with each other on a non-price basis.
Features of Non – price competitive agreement
Firms offer differentiated products
Member firms agree to offer the product at the common price
Firms compete with each other on a non – price basis such as quality, appearance, advertisement, and the like
Member firms are free to sell any quantity demanded
Unlike joint profit maximization cartel, which is a robust form of a cartel agreement, the non – price competitive cartel agreement does not stand long.
Assumptions of this model
There are two firms, Firm A and Firm B (say)
Firm A is assumed to be a high-cost firm and firm B is a low-cost firm
Firms compete on non – price basis
Differentiated oligopoly
Firms are not allowed to sell below the common price set by the central agency or through their negotiation but can sell any quantity demanded
Under the basis of the above assumptions, the price and output can be determined by the marginal approach.
Necessary condition: ∑ MC = ∑ MR
Sufficient condition: Slope of MC > Slope of MR
Figurative Analysis
Figure 2 represents a non – price competitive cartel agreement. Figure 2 (c) represents an industry or central agency. The central agency just sets the common price directed by the intersection of industry MR and industry MC. Industry MR and industry MC are the horizontal summations of firms MR and MC respectively. Hence, P M is the price set forth by the central agency.
Firm A being the high-cost firm readily sells at price set by the cartel agency as shown in figure 2 (a). But the low-cost firm has greater incentive while offering the product at the price below monopoly level, that is at P B, as depicted in figure 2 (b).
The non – competitive agreement is not stable because every firm in a cartel agreement seeks to enlarge its market share. The low-cost firms can split out of the cartel agreement and offer the product at a price lower than monopoly price and can attract considerable numbers of customers from others. Consequently, the demand curve is becoming more elastic and profit increases. Every firm has the incentive to earn higher profit and gradually the member firms leave the cartel agreement. Hence, the cartel is short – lived.
2.2. Sharing of the market by agreement on quotas
Under this model, the firms agree on the quantity that each member may sell at the agreed price. The firms are in agreement to determine the market share in equal proportion because they produce homogenous goods and the price are the same. The quotas are decided on the basis of the past level of sales or production capacity.
Assumptions of this model
There are two firms, Firm 1 and Firm 2.
The firms have identical cost.
Firms produce homogenous products.
The firms have an equal production capacity
Under the basis of the above assumptions, the price and quantity are determined by the marginal approach.
Necessary condition: MR = MC
Such that: Industry MR = Industry MC; Firms’ MR = Firms’ MC
Sufficient condition: Slope of MR = Slope of MC.
Figurative Analysis
Figure 3 represents the sharing of the market by agreement on quotas. According to the assumptions, figure 3 (a), figure 3 (b), and figure 3 (c) have been drawn. Figure 3 (c) represents an industry where industry price (P M) and output (X M) is determined at the point of the intersection, that is, point ‘e’ between industry marginal revenue and industry marginal cost.
The horizontal line is drawn from point ‘e’ to figure 3 (a) and figure 3 (b) represents the marginal revenue. The point of intersection between marginal revenue (horizontal line) and respective marginal cost curves of firms determine the respective output of the firm.
Hence,
X M = X 1 + X 2or X1 = X 2 = (1/2) X M
Difficulties in Cartel
Difficult to estimate market demand
Difficult to estimate marginal cost
A slow process of cartel negotiation
Price stickiness
High-cost firms are in problem
A bluffing attitude of members
Other online materials
Cartel | Business Economics
Our Posts related to Economics
Tax: A deeper analysis with 9 figures
Subsidy: A Deeper Analysis with 6 Figures
Difference between Economic and Econometric Model
References
Kutosoyiannis, A. (1979). Modern Microeconomics. Houndsmill: Macmillan Press Ltd.
Kutosoyiannis, A. (1979). Modern Microeconomics. Houndsmill: Macmillan Press Ltd.
Wikipedia. (2018). OPEC. Retrieved from
Kutosoyiannis, A. (1979). Modern Microeconomics. Houndsmill: Macmillan Press Ltd.
Suggested Readings:
Ahuja. H.L. (1970). Advanced Economic Analysis: Microeconomic Analysis. New Delhi: S. Chand & Company Pvt. Ltd.
Kutosoyiannis, A. (1979). Modern Microeconomics. Houndsmill: Macmillan Press Ltd.
Varian. H.R. (2010). Intermediate Microeconomics - A Modern Approach. W W Norton & Company: New York.
Discuss your questions and confusions with us.
Ask Question
Tags:MA EconomicsMicroeconomicsTU MA Economics First SemesterTu Ma economics notesTU MA microeconomics
Share
Post a Comment
Post a Comment
Previous PostNext Post
Popular
Conversion of Fiscal Year from English to Nepal
Banking
Capital Adequacy: Concept, Tiers, Instruments and requirements in Nepal
Banking
Banking Career: Reference Materials
Econometrics
A Primer on VAR, SVAR and Local Projections
Follow Us
Facebook
Twitter
YouTube
Design by - Black Theme
Home
About
Contact Us
Contact Form
About
Macroeconomics
Econometrics
Data Analysis
TU BBA
Tools
Videos
Home
About
Contact Us |
190783 | https://blog.csdn.net/weixin_45492196/article/details/100887150 | Python解决爱因斯坦的逻辑题_python 解答 1、 英国人住在红房子里 2 、瑞典人养了一条狗 3 、丹麦人喝茶 4 、-CSDN博客
博客
下载
学习
社区
GitCode
InsCodeAI
会议
搜索
AI 搜索
登录
登录后您可以：
复制代码和一键运行
与博主大V深度互动
解锁海量精选资源
获取前沿技术资讯
立即登录
会员·新人礼包
消息
历史
创作中心
创作
Python解决爱因斯坦的逻辑题
最新推荐文章于 2024-07-21 17:30:50 发布
原创 于 2019-09-16 14:41:44 发布·2.5k 阅读
· 8
· 30·
CC 4.0 BY-SA版权
版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。
文章标签：
#Python
Python 专栏收录该内容
1 篇文章
订阅专栏
问题：
有5个具有不同颜色的房间；
每个房间里住着一个不同国籍的人；
每个人都在喝一种特定品牌的饮料；
抽一特定品牌的香烟；
养一特定的宠物；
没有任意的两个人在抽相同品牌的烟、或喝相同品牌饮料、或养相同的宠物。
同时提供下面这些线索：
英国人住在红房子里
瑞典人养狗
丹麦人喝茶
绿房子紧挨着白房子，在白房子的左边
绿房子主人喝咖啡
抽PALL MALL牌香烟的人养鸟
黄房子里的人抽DUNHILL牌的烟
住中间房子的人喝牛奶
挪威人住在第一个房子里（最左边）
抽BLENDS香烟的人和养猫的人相邻
养马的人和抽DUNHILL牌香烟的人相邻
抽BLUEMASTER牌香烟的人喝啤酒
德国人抽PRINCE牌香烟
挪威人和住蓝房子的人相邻
抽BLENDS牌香烟的人和喝矿泉水的人相邻。
问：谁在养鱼？
Solution：
```py
-- coding: GBK --
import time
import itertools
def handle(data):
# 英国人住红色房子
index = data.index('红色')
if data[index] !='英国':
return -1
# 瑞典人养狗
index = data.index('狗')
if data[index] !='瑞典':
return -2
# 丹麦人喝茶
index = data.index('茶')
if data[index] !='丹麦':
return -3
# 抽Pall Mall香烟的人养鸟
index = data.index('Pall Mall')
if data[index] != '鸟':
return -6
# 黄色房子主人抽Dunhill香烟
index = data.index('Dunhill')
if data[index] != '黄色':
return -7
# 抽Blue Master的人喝啤酒
index = data.index('BlueMaster')
if data[index] != '啤酒':
return -12
# 德国人抽Prince香烟
index = data.index('Prince')
if data[index] != '德国':
return -13
# 绿色房子在白色房子左面
index = data.index('绿色')
if index == 4: # 绿色房子在最后
return -4
if data[index + 1] != '白色':
return -4
# 绿色房子主人喝咖啡
if data[index] != '咖啡':
return -5
# 抽Blends香烟的人住在养猫的人隔壁
index = data.index('Blends')
cat_index = data.index('猫')
if cat_index - index != 1 and cat_index - index != -1:
return -10
# 养马的人住抽Dunhill香烟的人隔壁
index = data.index('Dunhill')
horse_index = data.index('马')
if horse_index - index != 1 and horse_index - index != -1:
return -11
# 抽Blends香烟的人有一个喝水的邻居
index = data.index('Blends')
water_index = data.index('水')
if water_index - index != 1 and water_index - index != -1:
return -15
print('找到答案:')
for d_item in data:
print(d_item)
return 0
colour_list_ = list(itertools.permutations(['红色', '黄色', '蓝色', '白色', '绿色'], 5))
country_list_ = list(itertools.permutations(['英国', '丹麦', '挪威', '德国', '瑞典'], 5))
drinks_list_ = list(itertools.permutations(['茶', '水', '咖啡', '啤酒', '牛奶'], 5))
smoke_list_ = list(itertools.permutations(['Pall Mall', 'Dunhill', 'BlueMaster', 'Blends', 'Prince'], 5))
pet_list_ = list(itertools.permutations(['猫', '马', '鱼', '鸟', '狗'], 5))
colour_list = []
country_list = []
drinks_list = []
smoke_list = []
pet_list = []
for colour in colour_list_:
colour = list(colour)
if colour == '蓝色':
colour_list.append(colour)
for country in country_list_:
country = list(country)
if country == '挪威':
country_list.append(country)
for drinks in drinks_list_:
drinks = list(drinks)
if drinks == '牛奶':
drinks_list.append(drinks)
for i in range(120):
smoke_list.append(list(smoke_list_[i]))
for i in range(120):
pet_list.append(list(pet_list_[i]))
~ num = 0
start = time.time()
for country in country_list:
for drinks in drinks_list:
for smoke in smoke_list:
for pet in pet_list:
for colour in colour_list:
data_list = [colour, country, drinks, smoke, pet]
# ~ num += 1
if handle(data_list) == 0:
# ~ print("总运行次数 %d" % num)
exit()
end = time.time()
print("计算用时:%0.4f秒" % (end - start))
```
AI运行代码 py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
Output:
确定要放弃本次机会？
福利倒计时
: :
立减 ¥
普通VIP年卡可用
立即使用
catchy666
关注关注
8点赞
踩
30 收藏 觉得还不错? 一键收藏
4评论
分享复制链接 分享到 QQ 分享到新浪微博 扫一扫
举报 举报
专栏目录
Python 3 趣味系列 题 1 3-----爱因斯坦 问 题
AnFany
07-26 1505
爱因斯坦 问 题：在 一条 街上，有5座房子，喷了不同的5种颜色。每个房子 里 住着不同国籍的 人。每个 人 喝着不同的饮料，抽不同品牌的香烟，养 不同的宠物。 问：谁 养 鱼？ 条件： 1、英国人 住红色房子。 2、瑞典人 养 狗。 3、丹麦 人 喝茶。 4、绿色房子紧挨着白色房子，并在白色房子左边。 5、绿色房子主 人 喝咖啡。 6、抽Pall Mall香烟的 人 养 鸟。 7、黄色房子主 人 抽Dunhill香烟。 8、住在 中间房子的 人...
4 条评论 您还未登录，请先 登录 后发表或查看评论
Einstein 9;s Riddle_einstein riddle
9-23
1、在 一条 街上,有5座房子,喷了5种颜色。 2、每个房 里 住着不同国籍的 人。 3、每个 人 喝不同的饮料,抽不同品牌的香烟,养 不同的宠物。 提示: 1、英国人 住红色房子 2、瑞典人 养 狗 3、丹麦 人 喝茶 4、绿色房子在白色房子左面 5、绿色房子主 人 喝咖啡 6、抽Pall Mall香烟的 人 养 鸟 7、黄色房子主 人 抽Dunhill香...
逻辑 谜 题 解析
9-25
1 有五栋五种颜色的房子 2 每一位房子的主 人 国籍都不同 3 这五个 人 每 人 只喝一种饮料,只抽一种牌子的香烟,只 养 一种宠物 4 没有 人 有相同的宠物,抽相同牌子的香烟,喝相同的饮料 提示: 1 、英国人 住在 红房子 里 2 、瑞典人 养 了 一条 狗 3 、丹麦 人 喝茶 ...
python 逻辑 题
一颗大白菜的博客
01-16 1348
最近开始在找工作了，昨天接了一家公司的笔试。形式是线上笔试。笔试 题 目是 逻辑 题。 题 目如下： 李雷和韩梅梅坐前后排，上课想说话怕被老师发现，所以改为传纸条。为了不被老师发现他们纸条上写的是啥，，他们约定了如下方法传递信息： 将 2 6个大写英文字母，外加空格，一共 2 7个字符分成三组，每组9个。也就是ABCDEFGHI是第一组，JKLMNOPQR是第二组，STUVWXYZ是第三组（此处用代表空格）。 先根...
【爱因斯坦 谜 题】用 python 基础语法 解决 爱因斯坦 谜 题
solemn_silence的博客
09-28 1570
这 里 写自定义目录标 题 欢迎使用Markdown编辑器新的改变功能快捷键合理的创建标 题，有助于目录的生成如何改变文本的样式插入链接与图片如何插入一段漂亮的代码片生成一个适合你的列表创建一个表格设定内容居中、居左、居右SmartyPants创建一个自定义列表如何创建一个注脚注释也是必不可少的KaTeX数学公式新的甘特图功能，丰富你的文章UML 图表FLowchart流程图导出与导入导出导入 欢迎使用Ma...
75道 逻辑 思维 题---会作 1 0道智商就是正常,会作 3 0道就不是凡 人,会作60...
9-22
4 没有 人 有相同的宠物,抽相同牌子的香烟,喝相同的饮料 提示: 1 英国人 住在 红房子 里 2 瑞典人 养 了 一条 狗 3 丹麦 人 喝茶 4 绿房子在白房子左边 5 绿房子主 人 喝咖啡 6抽PALLMALL烟的 人 养 了一只鸟 7 黄房子主 人 抽DUNHILL烟 8 住在 中间那间房子的 人 喝牛奶 ...
学习资料:笔试必看——75道 逻辑 题 答案_笔试 逻辑 测试 题
9-19
1 有五栋五种颜色的房子 2 每一位房子的主 人 国籍都不同 3 这五个 人 每 人 只喝一种饮料,只抽一种牌子的香烟,只 养 一种宠物 4 没有 人 有相同的宠物,抽相同牌子的香烟,喝相同的饮料 提示:1 英国人 住在 红房子 里 2 瑞典人 养 了 一条 狗 3 丹麦 人 喝茶 4 绿房子在白房子左边 ...
python 爱因斯坦 的问 题 _ 爱因斯坦 的思考 题.py
weixin_39715834的博客
12-20 692
"""爱因斯坦 的思考 题 这是一个很有趣的 逻辑 推理 题，传说是 爱因斯坦 提出来的，他宣称世界上只有 2％的 人 能解出这个 题 目，传说不一定属实，但是这个推理 题 还是很有意思的。题 目是这样的，据说有五个不同颜色的房间排成一排，每个房间 里 分别住着一个不同国籍的 人，每个 人 都喝一种特定品牌的饮料，抽一种特定品牌的烟，养 一种宠物，没有任意两个 人 抽相同品牌的香烟，或喝相同品牌的饮料，或 养 相同的宠物，问 题 是谁在 养 鱼作为宠物？为...
巧用 python 求解 逻辑 题，特简单！
Dai_sir_man的博客
09-03 3105
昨晚睡觉前看了看初中买的一本《哈佛学生的600个思维训练》，然后发现有些 题 可以用计算机 解决，便早起随便敲了一下。 第一道 题：1 2 2 猜名字 核心思路：遍历所有 人 的名字，并默认他们就是老师手中写的名字，然后放进同学们说话的情景 里 比对，看看是不是满足只有一个 人 说对的情况。 #1 2 2 代码 list=['a','b','c','d'] for x in list: if (int(x=='c')+int(x!='b')+int(x!='c')+int(x=='a')==1): print(
爱因斯坦 谜 题 解答(三种算法比较)
9-16
爱因斯坦 谜 题 解答(三种算法比较) 爱因斯坦 谜 题: 在 一条 街上有颜色互不相同的五栋房子,不同国籍的 人 分别 住在 这五栋房子力,每 人 抽不同品牌的香烟,喝不同的饮料,养 不同的宠物。已知如下情况: 1. 英国人 住红色房子 里。 2. 瑞典人 养 狗。 3. 丹麦 人 喝茶。
逻辑 训练-爱因斯坦 的推理 题 _ 英国人 住在 红房子 里
9-22
根据条件 1(英国人 住在 红房子 里),2(瑞典人 养 狗),3(丹麦 人 喝茶),1 3(德国 人 抽Prince牌香烟)可以先构造如下表格,然后不断推理完善表格。 看了这张表,感觉空缺太多。那么我们就一步步开始推导吧。 关于房子的顺序,先说结论:黄-蓝-红-绿-白,挪威 人 住黄房子,英国人 喝牛奶。推理过程如下: ...
《算法的乐趣》8.爱因斯坦 的思考 题------python
su.nn.y的博客
07-15 605
题 目如下： 据说有五个不同颜色的房间排成一排，每个房间 里 分别住着一个不同国籍的 人，每个 人 都喝一种特定品牌的饮料，抽一种特定品牌的烟，养 一种宠物，没有任意两个 人 抽相同品牌的香烟，或喝相同品牌的饮料，或 养 相同的宠物。 问 题 是谁在 养 鱼作为宠物？为了寻找答案，爱因斯坦 给出了以下 1 5 条线索。 1.英国人 住在 红色的房子 里； 2.瑞典人 养 狗 作为宠物； 3.丹麦 人 喝茶； 4.绿房子紧挨着白房子，在白房子的左边...
Python 求解 爱因斯坦 的鱼
qq_31232793的博客
08-31 885
问 题： 据说是 爱因斯坦 在 2 0世纪初出的这个谜语 1. 在 一条 街上，有5座房子，喷了5种颜色。 2. 每个房 里 住着不同国籍的 人。 3. 每个 人 喝不同的饮料，抽不同品牌的香烟，养 不同的宠物。 问 题 是：谁 养 鱼？ 提示： 1. 英国人 住红色房子。 2. 瑞典人 养 狗。 3. 丹麦 人 喝茶。 4. 绿色房子在白色房子左面。 5. 绿色房子主 人 喝咖啡。 6. 抽Pall Mall 香烟的 人 养 鸟。...
如何运用 Python 编程和PyCharm工具 解决 爱因斯坦 的五色房子 逻辑 题，并展示三种不同的解 题 方法？
11-08
完成以上步骤后，你将得到一个能够 解决 爱因斯坦 五色房子 逻辑 题 的 Python 程序，并且能够通过不同的算法策略来展示三种不同的解 题 方法。 为了深入学习如何运用 Python 和PyCharm 解决 逻辑 题，你可以参考资源《五色房子...
请描述如何在PyCharm中使用 Python 实现 爱因斯坦 的五色房子 逻辑 题，并提供三种不同的解 题 策略。 最新发布
11-09
在PyCharm中编写程序 解决 爱因斯坦 的五色房子 逻辑 题 是一个很好的练习项目，能够锻炼编程思维和 逻辑 推理能力。首先，我们需要熟悉问 题 的规则：有五座颜色不同的房子，五个 人 分别居 住在 这些房子 里，每个 人 有不同的国籍...
如何使用 Python 在PyCharm中编写程序来 解决 爱因斯坦 的五色房子 逻辑 题，并展示三种不同的解法？
11-09
在 解决 爱因斯坦 的五色房子 逻辑 题 时，首先需要理解问 题 的规则和约束条件，然后将这些 逻辑 转化为程序代码。在PyCharm这样的集成开发环境中，编写和调试代码会更加方便。以下是 解决 这类问 题 的步骤和方法： 参考资源...
Python 爱因斯坦 阶梯问 题
weixin_45688123的博客
03-26 7240
爱因斯坦 阶梯问 题 ·爱因斯坦 阶梯问 题： 设有一阶梯，每步跨 2 阶，最后余 1 阶；每步跨 3 阶，最后余 2 阶；每步跨5阶，最后余 4 阶；每步跨6阶，最后余5阶；只有每步跨7阶时，正好到阶梯顶。 问最少有多少步阶梯？ 要求使用while 循环语句 分析 设 1 000内能输入出最小步阶梯数， 因每步跨 2 阶，最后余 1 阶；每步跨 3 阶，最后余 2 阶；每步跨5阶，最后余 4 阶；每步跨6阶，最后余5阶；只有每步跨7阶时，正好到...
爱因斯坦 难 题 程序
12-03
很早以前就看到 爱因斯坦 的这道推理 题，据说只有 2%的聪明 人 才能做出来，我不属于那 2%，可我想知道答案... 于是就有了这个想法，编程借助计算机计算。 五个 人 分别来自五个国家，那么就有P55=1 2 0种可能，同样五个 人 住五种颜色的房子，也有P55=1 2 0种可能，五个 人 养 五种宠物，P55=1 2 0种可能，五种饮料，P55=1 2 0种可能，五种香烟，P55=1 2 0种可能。那么总共就有 1 2 0×1 2 0×1 2 0×1 2 0×1 2 0=2 4 88 3 2 00000种可能。如何我逐个试过来的话，肯定会得到结果。那么我就逐个试吧...
python 解决 爱因斯坦 阶梯问 题
welcome
04-26 2538
整理以备日后回顾 设有一阶梯，每步跨 2 阶，最后余 1 阶；每步跨 3 阶，最后余 2 阶；每步跨 5 阶，最后余 4 阶；每步跨 5 阶，最后余 4 阶；每步跨 6 阶，最后余 5 阶；每步跨 7 阶，正好到达阶梯顶，问共有多少阶梯 i = 1 while i: if i % 2 == 1 and i % 3 == 2 and i % 5 == 4 and \ ...
python 爱因斯坦 的问 题 _ Python 爬虫笔记二——爬取 爱因斯坦 名言
weixin_39692253的博客
12-20 323
这次的笔记主要和大家分享BeautifulSoup的一些用法。数据定位查找BS一个很大的作用就是可以对HTML中的tag进行定位。其中最常用的函数就是find()和findAll()，这两个函数其实功能相仿，差距在于一个只寻找最近的tag，另一个会查找所有的标签。其主要参数如下：tag : 所要查找的tag，格式为字符串或列表(一系列tag)attributes : 所要查找tag的attribu...
python 的 逻辑 应用
ytym00的博客
04-11 447
python 的 逻辑 语言一般分为条件语句、逻辑 语句，常用的是if判断语句；while、for 逻辑 循环语句；嵌套判断、循环语句。1.使用 python 语句编写99乘法口诀。要实现为下图：查看规律，发现：1.每行都有一个数字相同；2.另外一个数字递增；3.当数字相同是换行。数字递增的话刚好使用for语句加上range函数即可实现，同一个数字下的数字叠加积可以使用for的嵌套实现，加上判断换行时间即可实现乘...
python 编程 爱因斯坦 的问 题 教你如何用编程 _解决 爱因斯坦 的数学问 题
weixin_39717704的博客
12-22 1741
曾经 爱因斯坦 出了一道这样的数学 题：有 一条 长阶梯，若每步跨 2 阶，则最后剩 1 阶，若每步跨 3 阶，则最后剩 2 阶，若每步跨5阶，则最后剩 4 阶，若每步跨6阶则最后剩5阶。只有每次跨7阶，最后才正好一阶不剩。请问在 1~N 内，有多少个数能满足？针对于这个 爱因斯坦 的数学问 题，那么我们用编程知识来如何 解决 它？下面酷叮猫来给您分析一下：首先来问 题 分析，假设用变量x表示阶梯数，则x 应满足：若每步跨 2 阶，则最后剩 1 阶...
python：EinsteinPy 爱因斯坦 的广义相对论和引力物理学
belldeep的专栏
07-21 430
python：EinsteinPy 爱因斯坦 派的广义相对论和引力物理学
Python 每日一 题 - 4 - 爱因斯坦 台阶问 题
qqgg77的博客
04-10 2751
前言： 本 人 因喜欢 python 和c语言等程序设计语言，希望能够在这 里 和大家共同学习共同进步。 因个 人 喜好喜欢做 题，所以想出一个系列专栏关于 python 的习 题 专集，希望大家喜欢。 还是一个小白，接触时间没有很长，如果文章有任何错误，欢迎大家指正。 如果喜欢我的文章还请大家不吝动手给我点赞收藏关注哦，留下你来过的足迹，让我眼熟你。 第四 题 爱因斯坦 台阶问 题 要求： 有 人 走台阶，若以每步走两级则剩下最后一级；若每步走 3 级则剩两级；若每步走 4 级则剩三级；若每步走5级则剩四级；若...
用 Python 爱因斯坦 数学 题
09-18
在 Python 中 解决 爱因斯坦（Einstein）著名的数学谜 题 通常涉及到一些有趣的 逻辑 和条件判断 题 目。比如"不动脑筋的汉斯"（Hans and the Stairs）问 题，这是一种典型的“囚徒困境”（Prisoner 9;s Dilemma）类型的 题 目，也可以通过编程来模拟。 例如，你可以编写一个程序，让两个 人（Hans和朋友）可以选择上楼梯的不同步数，而每一步都有奖励或惩罚。如果他们都选择相同的步数，他们都会得到一定的积分；但如果一个 人 选择更多，他可能会得到更多的积分，但另一个 人 会受到惩罚。这种问 题 是关于合作和自私之间的权衡，类似于 爱因斯坦 对 人 类理性的思考。 以下是简单的 Python 代码示例： ```python def stairs_game(steps, partner_steps): if steps == partner_steps: return "Both got the same reward" elif steps > partner_steps: return f"Hans gets a bigger reward, but his friend pays the price ({steps} vs {partner_steps})" else: return f"His friend gets a bigger reward ({partner_steps} vs {steps})" # 示例 hans_steps = int(input("How many steps does Hans take? ")) friend_steps = int(input("How many steps does his friend take? ")) print(stairs_game(hans_steps, friend_steps))
关于我们
招贤纳士
商务合作
寻求报道
400-660-0108
kefu@csdn.net
在线客服
工作时间 8:30-22:00
公安备案号11010502030143
京ICP备19004658号
京网文〔2020〕1039-165号
经营性网站备案信息
北京互联网违法和不良信息举报中心
家长监护
网络110报警服务
中国互联网举报中心
Chrome商店下载
账号管理规范
版权与免责声明
版权申诉
出版物许可证
营业执照
©1999-2025北京创新乐知网络技术有限公司
catchy666
博客等级
码龄6年
31 原创154 点赞 851 收藏 76 粉丝
关注
私信
亚马逊云科技 性价比之选 - 基于 ARM 架构 Graviton4 处理器实例🚀性能提升30%！广告
热门文章
【深度强化学习】TD3算法：DDPG的进化 41016
【深度强化学习】DDPG算法 14251
DRL实战：DDPG & A3C | Gym环境中经典控制问题Pendulum-v0 12932
Python:下载数据——CSV文件格式&练习题 12735
Python:下载数据——JSON格式 7883
分类专栏
强化学习6篇
Notes11篇
深度学习1篇
目标检测1篇
Python1篇
Python学习18篇
展开全部收起
上一篇： Python:Django入门——建立项目 & 创建应用程序 下一篇： 【Only notes here】20200206
最新评论
DRL实战：DDPG & A3C | Gym环境中经典控制问题Pendulum-v0
CUTESORA:很好奇 理论上说ddpg训练这个 可以收敛到0吗
DRL实战：DDPG & A3C | Gym环境中经典控制问题Pendulum-v0
qq_33925188:我的DDPG平均只要40个episode就收敛了，A3C完全比不了
DRL实战：DDPG & A3C | Gym环境中经典控制问题Pendulum-v0
qq_33925188:图看得有点懵，横坐标到底是step还是episode？感觉总奖励是缩小了10倍吧，但文章没说……
Python:下载数据——CSV文件格式&练习题
小张要当程序员:
Python:使用API——使用Web API
正啊正啊:同，不知道后面有什么影响
最新文章
【目标检测】RCNN系列算法
【转】pytorch与tensorflow未来哪一个会占据更大的用户群体？
【深度强化学习】TD3算法：DDPG的进化
2020年 12篇
2019年 19篇
亚马逊云科技 性价比之选 - 基于 ARM 架构 Graviton4 处理器实例🚀性能提升30%！广告
上一篇： Python:Django入门——建立项目 & 创建应用程序 下一篇： 【Only notes here】20200206
分类专栏
强化学习6篇
Notes11篇
深度学习1篇
目标检测1篇
Python1篇
Python学习18篇
展开全部收起
登录后您可以享受以下权益：
免费复制代码
和博主大V互动
下载海量资源
发动态/写文章/加入社区
×立即登录
评论 4
被折叠的 条评论 为什么被折叠?到【灌水乐园】发言
查看更多评论
添加红包
祝福语
请填写红包祝福语或标题
红包数量
个
红包个数最小为10个
红包总金额
元
红包金额最低5元
余额支付
当前余额 3.43 元 前往充值 >
需支付：10.00 元
取消 确定
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付 元
使用余额支付
点击重新获取
扫码支付
钱包余额 0
抵扣说明：
1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。
余额充值
确定 取消
举报
选择你想要举报的内容（必选）
内容涉黄
政治相关
内容抄袭
涉嫌广告
内容侵权
侮辱谩骂
样式问题
其他
原文链接（必填）
请选择具体原因（必选）
包含不实信息
涉及个人隐私
请选择具体原因（必选）
侮辱谩骂
诽谤
请选择具体原因（必选）
搬家样式
博文样式
补充说明（选填）
取消
确定
点击体验 DeepSeekR1满血版 下载APP 程序员都在用的中文IT技术交流社区 公众号 专业的中文 IT 技术社区，与千万技术人共成长 视频号 关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部 |
190784 | https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/17-7-the-doppler-effect/ | 17.7 The Doppler Effect | University Physics Volume 1
Skip to main content
University Physics Volume 1
17 Sound
Search for:
17.7 The Doppler Effect
Learning Objectives
By the end of this section, you will be able to:
Explain the change in observed frequency as a moving source of sound approaches or departs from a stationary observer
Explain the change in observed frequency as an observer moves toward or away from a stationary source of sound
The characteristic sound of a motorcycle buzzing by is an example of the Doppler effect. Specifically, if you are standing on a street corner and observe an ambulance with a siren sounding passing at a constant speed, you notice two characteristic changes in the sound of the siren. First, the sound increases in loudness as the ambulance approaches and decreases in loudness as it moves away, which is expected. But in addition, the high-pitched siren shifts dramatically to a lower-pitched sound. As the ambulance passes, the frequency of the sound heard by a stationary observer changes from a constant high frequency to a constant lower frequency, even though the siren is producing a constant source frequency. The closer the ambulance brushes by, the more abrupt the shift. Also, the faster the ambulance moves, the greater the shift. We also hear this characteristic shift in frequency for passing cars, airplanes, and trains.
The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer. Although less familiar, this effect is easily noticed for a stationary source and moving observer. For example, if you ride a train past a stationary warning horn, you will hear the horn’s frequency shift from high to low as you pass by. The actual change in frequency due to relative motion of source and observer is called a Doppler shift. The Doppler effect and Doppler shift are named for the Austrian physicist and mathematician Christian Johann Doppler (1803–1853), who did experiments with both moving sources and moving observers. Doppler, for example, had musicians play on a moving open train car and also play standing next to the train tracks as a train passed by. Their music was observed both on and off the train, and changes in frequency were measured.
What causes the Doppler shift? (Figure) illustrates sound waves emitted by stationary and moving sources in a stationary air mass. Each disturbance spreads out spherically from the point at which the sound is emitted. If the source is stationary, then all of the spheres representing the air compressions in the sound wave are centered on the same point, and the stationary observers on either side hear the same wavelength and frequency as emitted by the source (case a). If the source is moving, the situation is different. Each compression of the air moves out in a sphere from the point at which it was emitted, but the point of emission moves. This moving emission point causes the air compressions to be closer together on one side and farther apart on the other. Thus, the wavelength is shorter in the direction the source is moving (on the right in case b), and longer in the opposite direction (on the left in case b). Finally, if the observers move, as in case (c), the frequency at which they receive the compressions changes. The observer moving toward the source receives them at a higher frequency, and the person moving away from the source receives them at a lower frequency.
Figure 17.30 Sounds emitted by a source spread out in spherical waves. (a) When the source, observers, and air are stationary, the wavelength and frequency are the same in all directions and to all observers. (b) Sounds emitted by a source moving to the right spread out from the points at which they were emitted. The wavelength is reduced, and consequently, the frequency is increased in the direction of motion, so that the observer on the right hears a higher-pitched sound. The opposite is true for the observer on the left, where the wavelength is increased and the frequency is reduced. (c) The same effect is produced when the observers move relative to the source. Motion toward the source increases frequency as the observer on the right passes through more wave crests than she would if stationary. Motion away from the source decreases frequency as the observer on the left passes through fewer wave crests than he would if stationary.
We know that wavelength and frequency are related by $$ v=f\lambda , $$ where v is the fixed speed of sound. The sound moves in a medium and has the same speed v in that medium whether the source is moving or not. Thus, f multiplied by $$ \lambda $$ is a constant. Because the observer on the right in case (b) receives a shorter wavelength, the frequency she receives must be higher. Similarly, the observer on the left receives a longer wavelength, and hence he hears a lower frequency. The same thing happens in case (c). A higher frequency is received by the observer moving toward the source, and a lower frequency is received by an observer moving away from the source. In general, then, relative motion of source and observer toward one another increases the received frequency. Relative motion apart decreases frequency. The greater the relative speed, the greater the effect.
The Doppler effect occurs not only for sound, but for any wave when there is relative motion between the observer and the source. Doppler shifts occur in the frequency of sound, light, and water waves, for example. Doppler shifts can be used to determine velocity, such as when ultrasound is reflected from blood in a medical diagnostic. The relative velocities of stars and galaxies is determined by the shift in the frequencies of light received from them and has implied much about the origins of the universe. Modern physics has been profoundly affected by observations of Doppler shifts.
Derivation of the Observed Frequency due to the Doppler Shift
Consider two stationary observers X and Y in (Figure), located on either side of a stationary source. Each observer hears the same frequency, and that frequency is the frequency produced by the stationary source.
Figure 17.31 A stationary source sends out sound waves at a constant frequency $$ {f}{\text{s}}, $$ with a constant wavelength $$ {\lambda }{\text{s}}, $$ at the speed of sound v. Two stationary observers X and Y, on either side of the source, observe a frequency $$ {f}{\text{o}}={f}{\text{s}}$$, with a wavelength $$ {\lambda }{\text{o}}={\lambda }{\text{s}}.$$
Now consider a stationary observer X with a source moving away from the observer with a constant speed $$ {v}{\text{s}}<v $$ ((Figure)). At time $$ t=0$$, the source sends out a sound wave, indicated in black. This wave moves out at the speed of sound _v. The position of the sound wave at each time interval of period $$ {T}{\text{s}} $$ is shown as dotted lines. After one period, the source has moved $$ \text{Δ}x={v}{\text{s}}{T}_{\text{s}} $$ and emits a second sound wave, which moves out at the speed of sound. The source continues to move and produce sound waves, as indicated by the circles numbered 3 and 4. Notice that as the waves move out, they remained centered at their respective point of origin.
Figure 17.32 A source moving at a constant speed $$ {v}{\text{s}} $$ away from an observer X. The moving source sends out sound waves at a constant frequency $$ {f}{\text{s}}, $$ with a constant wavelength $$ {\lambda }{\text{s}}$$, at the speed of sound v. Snapshots of the source at an interval of $$ {T}{\text{s}} $$ are shown as the source moves away from the stationary observer X. The solid lines represent the position of the sound waves after four periods from the initial time. The dotted lines are used to show the positions of the waves at each time period. The observer hears a wavelength of $$ {\lambda }{\text{o}}={\lambda }{\text{s}}+\text{Δ}x={\lambda }{\text{s}}+{v}{\text{s}}{T}_{\text{s}}$$.
Using the fact that the wavelength is equal to the speed times the period, and the period is the inverse of the frequency, we can derive the observed frequency:
$$\begin{array}{ccc}\hfill {\lambda }{\text{o}}& =\hfill & {\lambda }{\text{s}}+\text{Δ}x\hfill \ \hfill v{T}{\text{o}}& =\hfill & v{T}{\text{s}}+{v}{\text{s}}{T}{\text{s}}\hfill \ \hfill \frac{v}{{f}{\text{o}}}& =\hfill & \frac{v}{{f}{\text{s}}}=\frac{{v}{\text{s}}}{{f}{\text{s}}}=\frac{v+{v}{\text{s}}}{{f}{\text{s}}}\hfill \ \hfill {f}{\text{o}}& =\hfill & {f}{\text{s}}(\frac{v}{v+{v}_{\text{s}}}).\hfill \end{array}$$
As the source moves away from the observer, the observed frequency is lower than the source frequency.
Now consider a source moving at a constant velocity $$ {v}{s}, $$ moving toward a stationary observer _Y, also shown in (Figure). The wavelength is observed by Y as $$ {\lambda }{\text{o}}={\lambda }{\text{s}}-\text{Δ}x={\lambda }{\text{s}}-{v}{\text{s}}{T}_{\text{s}}. $$ Once again, using the fact that the wavelength is equal to the speed times the period, and the period is the inverse of the frequency, we can derive the observed frequency:
$$\begin{array}{ccc}\hfill {\lambda }{\text{o}}& =\hfill & {\lambda }{\text{s}}-\text{Δ}x\hfill \ \hfill v{T}{\text{o}}& =\hfill & v{T}{\text{s}}-{v}{\text{s}}{T}{\text{s}}\hfill \ \hfill \frac{v}{{f}{\text{o}}}& =\hfill & \frac{v}{{f}{\text{s}}}-\frac{{v}{\text{s}}}{{f}{\text{s}}}=\frac{v-{v}{\text{s}}}{{f}{\text{s}}}\hfill \ \hfill {f}{\text{o}}& =\hfill & {f}{\text{s}}(\frac{v}{v-{v}_{\text{s}}}).\hfill \end{array}$$
When a source is moving and the observer is stationary, the observed frequency is
$${f}{\text{o}}={f}{\text{s}}(\frac{v}{v\mp {v}_{\text{s}}})\text{‘}$$
where $$ {f}{\text{o}} $$ is the frequency observed by the stationary observer, $$ {f}{\text{s}} $$ is the frequency produced by the moving source, v is the speed of sound, $$ {v}_{\text{s}} $$ is the constant speed of the source, and the top sign is for the source approaching the observer and the bottom sign is for the source departing from the observer.
What happens if the observer is moving and the source is stationary? If the observer moves toward the stationary source, the observed frequency is higher than the source frequency. If the observer is moving away from the stationary source, the observed frequency is lower than the source frequency. Consider observer X in (Figure) as the observer moves toward a stationary source with a speed $$ {v}{\text{o}}$$. The source emits a tone with a constant frequency $$ {f}{\text{s}} $$ and constant period $$ {T}{\text{s}}. $$ The observer hears the first wave emitted by the source. If the observer were stationary, the time for one wavelength of sound to pass should be equal to the period of the source $$ {T}{\text{s}}. $$ Since the observer is moving toward the source, the time for one wavelength to pass is less than $$ {T}{\text{s}} $$ and is equal to the observed period $$ {T}{\text{o}}={T}_{\text{s}}-\text{Δ}t. $$ At time $$ t=0, $$ the observer starts at the beginning of a wavelength and moves toward the second wavelength as the wavelength moves out from the source. The wavelength is equal to the distance the observer traveled plus the distance the sound wave traveled until it is met by the observer:
$$\begin{array}{ccc}\hfill {\lambda }{\text{s}}& =\hfill & v{T}{\text{o}}+{v}{\text{o}}{T}{\text{o}}\hfill \ \hfill v{T}{\text{s}}& =\hfill & (v+{v}{\text{o}}){T}{\text{o}}\hfill \ \hfill v(\frac{1}{{f}{\text{s}}})& =\hfill & (v+{v}{\text{o}})(\frac{1}{{f}{\text{o}}})\hfill \ \hfill {f}{\text{o}}& =\hfill & {f}{\text{s}}(\frac{v+{v}_{\text{o}}}{v}).\hfill \end{array}$$
Figure 17.33 A stationary source emits a sound wave with a constant frequency $$ {f}{\text{s}}$$, with a constant wavelength $$ {\lambda }{\text{s}} $$ moving at the speed of sound v. Observer X moves toward the source with a constant speed $$ {v}{\text{o}}$$, and the figure shows the initial and final position of observer X. Observer X observes a frequency higher than the source frequency. The solid lines show the position of the waves at $$ t=0$$. The dotted lines show the position of the waves at $$ t={T}{\text{o}}$$.
If the observer is moving away from the source ((Figure)), the observed frequency can be found:
$$\begin{array}{ccc}\hfill {\lambda }{\text{s}}& =\hfill & v{T}{\text{o}}-{v}{\text{o}}{T}{\text{o}}\hfill \ \hfill v{T}{\text{s}}& =\hfill & (v-{v}{\text{o}}){T}{\text{o}}\hfill \ \hfill v(\frac{1}{{f}{\text{s}}})& =\hfill & \hfill (v-{v}{\text{o}})(\frac{1}{{f}{\text{o}}})\ \hfill {f}{\text{o}}& =\hfill & {f}{\text{s}}(\frac{v-{v}_{\text{o}}}{v}).\hfill \end{array}$$
Figure 17.34 A stationary source emits a sound wave with a constant frequency $$ {f}{\text{s}}$$, with a constant wavelength $$ {\lambda }{\text{s}} $$ moving at the speed of sound v. Observer Y moves away from the source with a constant speed $$ {v}{\text{o}}$$, and the figure shows initial and final position of the observer Y. Observer Y observes a frequency lower than the source frequency. The solid lines show the position of the waves at $$ t=0$$. The dotted lines show the position of the waves at $$ t={T}{\text{o}}$$.
The equations for an observer moving toward or away from a stationary source can be combined into one equation:
$${f}{\text{o}}={f}{\text{s}}(\frac{v±{v}_{\text{o}}}{v}),$$
where $$ {f}{\text{o}} $$ is the observed frequency, $$ {f}{\text{s}} $$ is the source frequency, $$ {v}{\text{w}} $$ is the speed of sound, $$ {v}{\text{o}} $$ is the speed of the observer, the top sign is for the observer approaching the source and the bottom sign is for the observer departing from the source.
(Figure) and (Figure) can be summarized in one equation (the top sign is for approaching) and is further illustrated in (Figure):
$${f}{\text{o}}={f}{\text{s}}(\frac{v±{v}{\text{o}}}{v\mp {v}{\text{s}}}),$$
| Doppler shift $${f}{\text{o}}={f}{\text{s}}(\frac{v±{v}{\text{o}}}{v\mp {v}{\text{s}}})$$ | Stationary observer | Observer moving towards source | Observer moving away from source |
--- --- |
| Stationary source | $${f}{\text{o}}={f}{\text{s}}$$ | $${f}{\text{o}}={f}{\text{s}}(\frac{v+{v}{\text{o}}}{v})$$ | $${f}{\text{o}}={f}{\text{s}}(\frac{v-{v}{\text{o}}}{v})$$ |
| Source moving towards observer | $${f}{\text{o}}={f}{\text{s}}(\frac{v}{v-{v}{\text{s}}})$$ | $${f}{\text{o}}={f}{\text{s}}(\frac{v+{v}{\text{o}}}{v-{v}{\text{s}}})$$ | $${f}{\text{o}}={f}{\text{s}}(\frac{v-{v}{\text{o}}}{v-{v}{\text{s}}})$$ |
| Source moving away from observer | $${f}{\text{o}}={f}{\text{s}}(\frac{v}{v+{v}{\text{s}}})$$ | $${f}{\text{o}}={f}{\text{s}}(\frac{v+{v}{\text{o}}}{v+{v}{\text{s}}})$$ | $${f}{\text{o}}={f}{\text{s}}(\frac{v-{v}{\text{o}}}{v+{v}{\text{s}}})$$ |
where $$ {f}{o} $$ is the observed frequency, $$ {f}{s} $$ is the source frequency, $$ {v}{w} $$ is the speed of sound, $$ {v}{o} $$ is the speed of the observer, $$ {v}_{s} $$ is the speed of the source, the top sign is for approaching and the bottom sign is for departing.
The Doppler effect involves motion and a video will help visualize the effects of a moving observer or source. This video shows a moving source and a stationary observer, and a moving observer and a stationary source. It also discusses the Doppler effect and its application to light.
Example
Calculating a Doppler Shift
Suppose a train that has a 150-Hz horn is moving at 35.0 m/s in still air on a day when the speed of sound is 340 m/s.
(a) What frequencies are observed by a stationary person at the side of the tracks as the train approaches and after it passes?
(b) What frequency is observed by the train’s engineer traveling on the train?
Strategy
To find the observed frequency in (a), we must use $$ {f}{\text{obs}}={f}{\text{s}}(\frac{v}{v\mp {v}_{\text{s}}}) $$ because the source is moving. The minus sign is used for the approaching train, and the plus sign for the receding train. In (b), there are two Doppler shifts—one for a moving source and the other for a moving observer.
Solution
Enter known values into $$ {f}{\text{o}}={f}{\text{s}}(\frac{v}{v-{v}{\text{s}}}):$$ $${f}{\text{o}}={f}{\text{s}}(\frac{v}{v-{v}{\text{s}}})=(150\,\text{Hz})(\frac{340\,\text{m/s}}{340\,\text{m/s}-35.0\,\text{m/s}}).$$
Calculate the frequency observed by a stationary person as the train approaches:
$${f}_{\text{o}}=(150\,\text{Hz})(1.11)=167\,\text{Hz}\text{.}$$
Use the same equation with the plus sign to find the frequency heard by a stationary person as the train recedes:
$${f}{\text{o}}={f}{\text{s}}(\frac{v}{v+{v}_{\text{s}}})=(150\,\text{Hz})(\frac{340\,\text{m/s}}{340\,\text{m/s}+35.0\,\text{m/s}}).$$
Calculate the second frequency:
$${f}{\text{o}}=(150\,\text{Hz})(0.907)=136\,\text{Hz}\text{.}$$
2. Identify knowns:
It seems reasonable that the engineer would receive the same frequency as emitted by the horn, because the relative velocity between them is zero.
Relative to the medium (air), the speeds are $$ {v}{\text{s}}={v}_{\text{o}}=35.0\,\text{m/s}\text{.}$$
The first Doppler shift is for the moving observer; the second is for the moving source.
Use the following equation:
$${f}_{\text{o}}={f}{\text{s}}(\frac{v±{v}{\text{o}}}{v}).$$
The quantity in the square brackets is the Doppler-shifted frequency due to a moving observer. The factor on the right is the effect of the moving source.
Because the train engineer is moving in the direction toward the horn, we must use the plus sign for $$ {v}{\text{obs}}; $$ however, because the horn is also moving in the direction away from the engineer, we also use the plus sign for $$ {v}{\text{s}}. $$ But the train is carrying both the engineer and the horn at the same velocity, so $$ {v}{\text{s}}={v}{\text{o}}. $$ As a result, everything but $$ {f}_{\text{s}} $$ cancels, yielding
$${f}{\text{o}}={f}{\text{s}}.$$
Significance
For the case where the source and the observer are not moving together, the numbers calculated are valid when the source (in this case, the train) is far enough away that the motion is nearly along the line joining source and observer. In both cases, the shift is significant and easily noticed. Note that the shift is 17.0 Hz for motion toward and 14.0 Hz for motion away. The shifts are not symmetric.
For the engineer riding in the train, we may expect that there is no change in frequency because the source and observer move together. This matches your experience. For example, there is no Doppler shift in the frequency of conversations between driver and passenger on a motorcycle. People talking when a wind moves the air between them also observe no Doppler shift in their conversation. The crucial point is that source and observer are not moving relative to each other.
Check Your Understanding
Describe a situation in your life when you might rely on the Doppler shift to help you either while driving a car or walking near traffic.
Show Solution
If I am driving and I hear Doppler shift in an ambulance siren, I would be able to tell when it was getting closer and also if it has passed by. This would help me to know whether I needed to pull over and let the ambulance through.
The Doppler effect and the Doppler shift have many important applications in science and engineering. For example, the Doppler shift in ultrasound can be used to measure blood velocity, and police use the Doppler shift in radar (a microwave) to measure car velocities. In meteorology, the Doppler shift is used to track the motion of storm clouds; such “Doppler Radar” can give the velocity and direction of rain or snow in weather fronts. In astronomy, we can examine the light emitted from distant galaxies and determine their speed relative to ours. As galaxies move away from us, their light is shifted to a lower frequency, and so to a longer wavelength—the so-called red shift. Such information from galaxies far, far away has allowed us to estimate the age of the universe (from the Big Bang) as about 14 billion years.
Summary
The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer.
The actual change in frequency is called the Doppler shift.
Conceptual Questions
Is the Doppler shift real or just a sensory illusion?
Three stationary observers observe the Doppler shift from a source moving at a constant velocity. The observers are stationed as shown below. Which observer will observe the highest frequency? Which observer will observe the lowest frequency? What can be said about the frequency observed by observer 3?
Show Answer
Observer 1 will observe the highest frequency. Observer 2 will observe the lowest frequency. Observer 3 will hear a higher frequency than the source frequency, but lower than the frequency observed by observer 1, as the source approaches and a lower frequency than the source frequency, but higher than the frequency observed by observer 1, as the source moves away from observer 3.
Shown below is a stationary source and moving observers. Describe the frequencies observed by the observers for this configuration.
Prior to 1980, conventional radar was used by weather forecasters. In the 1960s, weather forecasters began to experiment with Doppler radar. What do you think is the advantage of using Doppler radar?
Show Solution
Doppler radar can not only detect the distance to a storm, but also the speed and direction at which the storm is traveling.
Problems
(a) What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. (b) What frequency does she receive after the ambulance has passed?
Show Solution
a. 878 Hz; b. 735 Hz
(a) At an air show a jet flies directly toward the stands at a speed of 1200 km/h, emitting a frequency of 3500 Hz, on a day when the speed of sound is 342 m/s. What frequency is received by the observers? (b) What frequency do they receive as the plane flies directly away from them?
What frequency is received by a mouse just before being dispatched by a hawk flying at it at 25.0 m/s and emitting a screech of frequency 3500 Hz? Take the speed of sound to be 331 m/s.
Show Solution
$$3.79\,×\,{10}^{3}\,\text{Hz}$$
A spectator at a parade receives an 888-Hz tone from an oncoming trumpeter who is playing an 880-Hz note. At what speed is the musician approaching if the speed of sound is 338 m/s?
A commuter train blows its 200-Hz horn as it approaches a crossing. The speed of sound is 335 m/s. (a) An observer waiting at the crossing receives a frequency of 208 Hz. What is the speed of the train? (b) What frequency does the observer receive as the train moves away?
Show Solution
a. 12.9 m/s; b. 193 Hz
Can you perceive the shift in frequency produced when you pull a tuning fork toward you at 10.0 m/s on a day when the speed of sound is 344 m/s? To answer this question, calculate the factor by which the frequency shifts and see if it is greater than 0.300%.
Two eagles fly directly toward one another, the first at 15.0 m/s and the second at 20.0 m/s. Both screech, the first one emitting a frequency of 3200 Hz and the second one emitting a frequency of 3800 Hz. What frequencies do they receive if the speed of sound is 330 m/s?
Show Solution
The first eagle hears $$ 4.23\,×\,{10}^{3}\,\text{Hz}. $$ The second eagle hears $$ 3.56\,×\,{10}^{3}\,\text{Hz}.$$
Student A runs down the hallway of the school at a speed of $$ {v}{\text{o}}=5.00\,\text{m/s,} $$ carrying a ringing 1024.00-Hz tuning fork toward a concrete wall. The speed of sound is $$ v=343.00\,\text{m/s}\text{.} $$ Student _B stands at rest at the wall. (a) What is the frequency heard by student B? (b) What is the beat frequency heard by student A?
An ambulance with a siren $$ (f=1.00\text{kHz}) $$ blaring is approaching an accident scene. The ambulance is moving at 70.00 mph. A nurse is approaching the scene from the opposite direction, running at $$ {v}_{o}=7.00\,\text{m/s}\text{.} $$ What frequency does the nurse observe? Assume the speed of sound is $$ v=343.00\,\text{m/s}\text{.}$$
Show Solution
$$\begin{array}{cc} {v}{\text{s}}=31.29\,\text{m/s}\hfill \ {f}{\text{o}}=1.12\,\text{kHz}\hfill \end{array}$$
The frequency of the siren of an ambulance is 900 Hz and is approaching you. You are standing on a corner and observe a frequency of 960 Hz. What is the speed of the ambulance (in mph) if the speed of sound is $$ v=340.00\,\text{m/s?}$$
What is the minimum speed at which a source must travel toward you for you to be able to hear that its frequency is Doppler shifted? That is, what speed produces a shift of $$ 0.300\text{%} $$ on a day when the speed of sound is 331 m/s?
Show Solution
An audible shift occurs when $$ \frac{{f}{\text{obs}}}{{f}{\text{s}}}\ge 1.003$$; $$ \begin{array}{cc} {f}{\text{obs}}={f}{\text{s}}\frac{v}{v-{v}{\text{s}}}⇒\frac{{f}{\text{obs}}}{{f}{\text{s}}}=\frac{v}{v-{v}{\text{s}}}⇒\hfill \ {v}_{\text{s}}=0.990\,\text{m/s}\hfill \end{array}$$
Glossary
Doppler effect alteration in the observed frequency of a sound due to motion of either the source or the observer Doppler shift actual change in frequency due to relative motion of source and observer
Candela Citations
CC licensed content, Shared previously
OpenStax University Physics. Authored by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at
Licenses and Attributions
CC licensed content, Shared previously
OpenStax University Physics. Authored by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at
PreviousNext
Privacy Policy |
190785 | https://www.mayoclinic.org/diseases-conditions/proctitis/symptoms-causes/syc-20376933 | Skip to content
Proctitis
Request an appointment
On this page
Overview
Symptoms
When to see a doctor
Causes
Risk factors
Complications
Prevention
Overview
Proctitis is irritation and swelling of the muscular tube connected to the end of the colon, called the rectum. Stool passes through the rectum on its way out of the body.
Proctitis can cause rectal pain, diarrhea, bleeding and rectal discharge, as well as the constant feeling that you need to pass stool. Proctitis symptoms can be temporary or constant.
Proctitis is common in people who have inflammatory bowel disease (IBD), such as ulcerative colitis or Crohn's disease. Sexually transmitted infections are another frequent cause. Proctitis also can be a side effect of radiation treatment.
Products & Services
A Book: Mayo Clinic on Digestive Health
Symptoms
Proctitis symptoms may include:
A frequent or continuous feeling that you need to pass stool.
Rectal bleeding.
Passing of mucus through the rectum.
Rectal pain.
Pain on the left side of the belly.
A feeling of fullness in the rectum.
Diarrhea.
Pain with bowel movements.
When to see a doctor
Make an appointment with a healthcare professional if you have any symptoms of proctitis.
Request an appointment
From Mayo Clinic to your inbox
Sign up for free and stay up to date on research advancements, health tips, current health topics, and expertise on managing health. Click here for an email preview.
We use the data you provide to deliver you the content you requested. To provide you with the most relevant and helpful information, we may combine your email and website data with other information we have about you. If you are a Mayo Clinic patient, we will only use your protected health information as outlined in our Notice of Privacy Practices. You may opt out of email communications at any time by clicking on the unsubscribe link in the email.
Causes
Several conditions, treatments and other factors can cause proctitis, including:
Inflammatory bowel disease (IBD). About 30% of people with IBD have inflammation of the rectum. This is more common in people with ulcerative colitis than in those with Crohn's disease.
Infections. Sexually transmitted infections (STIs) can result in proctitis. These infections happen particularly in people who engage in anal intercourse. Types of STIs that can cause proctitis include gonorrhea, genital herpes, monkeypox and chlamydia. Infections associated with foodborne illness, such as salmonella, shigella and campylobacter infections, also can cause proctitis.
Antibiotics. Sometimes antibiotics used to treat an infection can kill helpful bacteria in the bowels, allowing harmful Clostridioides difficile bacteria to grow in the rectum and the colon.
Surgery. A type of proctitis called diversion proctitis can happen in people after certain small intestine or colon surgeries. In these surgeries, the passage of stool is diverted from the rectum to a surgically created opening, called a stoma.
Food proteins. Food protein-induced proctitis can happen in infants who drink either cow's milk- or soy-based formula. Infants breastfed by mothers who eat dairy products also may develop proctitis.
Buildup of white blood cells. Eosinophilic proctitis happens when a type of white blood cell called an eosinophil builds up in the lining of the rectum. Eosinophilic proctitis more commonly affects children younger than 2 rather than older individuals.
Risk factors
Risk factors for proctitis include:
Sexual activity. Practices that increase the risk of a sexually transmitted infection (STI) can increase the risk of proctitis. Your risk of getting an STI can increase if you have multiple sex partners, have anal sex, don't use condoms and have sex with a partner who has an STI.
Radiation therapy for cancer. Radiation therapy directed at the rectum or nearby areas, such as the prostate, can cause rectal inflammation. Radiation proctitis can begin during radiation treatment and last for a few months after treatment. Or it can occur years after treatment.
Complications
Proctitis that isn't treated or that doesn't respond to treatment may lead to complications, including:
Anemia. Chronic bleeding from the rectum can cause anemia. With anemia, there aren't enough red blood cells to carry oxygen to the tissues. Anemia causes someone to feel tired and possibly have dizziness, shortness of breath, headache, a change in skin color, and irritability.
Ulcers. Long-lasting inflammation in the rectum can lead to open sores, called ulcers, on the inside lining of the rectum.
Fistulas. Sometimes ulcers extend completely through the intestinal wall. This can create a fistula, an irregular connection that forms between different parts of the intestine, between the intestine and skin, or between the intestine and other organs, such as the bladder and vagina.
Prevention
To reduce the risk of proctitis, take steps to protect yourself from sexually transmitted infections (STIs). The surest way to prevent an STI is to abstain from sex, especially anal sex. If you choose to have sex, you can reduce your risk of an STI if you:
Limit your number of sex partners.
Use a latex condom during each sexual contact.
Don't have sex with anyone who has sores or discharge in the genital area.
If you're diagnosed with a sexually transmitted infection, stop having sex until after you've completed treatment. Ask a healthcare professional when it's safe to have sex again.
Request an appointment
By Mayo Clinic Staff
Sep 20, 2025
Print
McNeil CJ, et al. Proctitis: An approach to the symptomatic patient. Medical Clinics of North America. 2024; doi:10.1016/j.mcna.2023.09.002.
Walls RM, et al., eds. Anorectum. In: Rosen's Emergency Medicine: Concepts and Clinical Practice. 10th ed. Elsevier; 2023. Accessed April 19, 2024.
Proctitis. National Institute of Diabetes and Digestive and Kidney Diseases. Accessed April 19, 2024.
Suzuki H, et al. A nationwide survey of non-IgE-mediated gastrointestinal food allergies in neonates and infants. Allergology International. 2024; doi:10.1016/j.alit.2023.10.003.
Proctitis. Merck Manual Professional Version. Accessed April 19, 2024.
Goldman L, et al., eds. Inflammatory bowel disease. In: Goldman-Cecil Medicine. 26th ed. Elsevier; 2020. Accessed April 19, 2024.
How you can prevent sexually transmitted diseases. Centers for Disease Control and Prevention. Accessed April 26, 2024.
Khanna S (expert opinion). Mayo Clinic. April 28, 2024.
Diagnosis & treatment
Diseases & Conditions
Proctitis - Symptoms & causes - Mayo Clinic
Associated Procedures
Colonoscopy
Flexible sigmoidoscopy
Needle biopsy
Products & Services
A Book: Mayo Clinic on Digestive Health
CON-20376915
Final days of 5X Challenge!
5X My Gift!
The challenge ends 10/10.Your gift today can have 5X the impact on AI research and technology.
5X My Gift! |
190786 | https://weather.metoffice.gov.uk/learn-about/weather/atmosphere/global-circulation-patterns | This website uses cookies
The Met Office website uses necessary cookies to make our site work. We also use optional cookies to help us improve your experience, understand how the site is being used for future improvements, and serve personalised advertising.
You can review the third-party vendors and how this information is being used below. You can also customise your consent preferences for cookies and identifying technologies.
218 TCF vendors are seeking consent or wishing to pursue data processing purposes on the basis of their legitimate interests.
Personalised advertising adheres to the IAB Transparency and Consent Framework (TCF).(Opens in a new window)
To agree to the use of these technologies select ‘Accept All’. You can change your preferences at any time by visiting our cookie policy.(Opens in a new window)
Purposes
Measure advertising performance
Information regarding which advertising is presented to you and how you interact with it can be used to determine how well an advert has worked for you or other users and whether the goals of the advertising were reached. For instance, whether you saw an ad, whether you clicked on it, whether it led you to buy a product or visit a website, etc. This is very helpful to understand the relevance of advertising campaigns.
Measure content performance
Information regarding which content is presented to you and how you interact with it can be used to determine whether the (non-advertising) content e.g. reached its intended audience and matched your interests. For instance, whether you read an article, watch a video, listen to a podcast or look at a product description, how long you spent on this service and the web pages you visit etc. This is very helpful to understand the relevance of (non-advertising) content that is shown to you.
Understand audiences through statistics or combinations of data from different sources
Reports can be generated based on the combination of data sets (like user profiles, statistics, market research, analytics data) regarding your interactions and those of other users with advertising or (non-advertising) content to identify common characteristics (for instance, to determine which target audiences are more receptive to an ad campaign or to certain contents).
Develop and improve services
Information about your activity on this service, such as your interaction with ads or content, can be very helpful to improve products and services and to build new products and services based on user interactions, the type of audience, etc. This specific purpose does not include the development or improvement of user profiles and identifiers.
Use limited data to select content
Content presented to you on this service can be based on limited data, such as the website or app you are using, your non-precise location, your device type, or which content you are (or have been) interacting with (for example, to limit the number of times a video or an article is presented to you).
Special Purposes
Ensure security, prevent and detect fraud, and fix errors
Your data can be used to monitor for and prevent unusual and possibly fraudulent activity (for example, regarding advertising, ad clicks by bots), and ensure systems and processes work properly and securely. It can also be used to correct any problems you, the publisher or the advertiser may encounter in the delivery of content and ads and in your interaction with them.
Deliver and present advertising and content
Certain information (like an IP address or device capabilities) is used to ensure the technical compatibility of the content or advertising, and to facilitate the transmission of the content or ad to your device.
Save and communicate privacy choices
The choices you make regarding the purposes and entities listed in this notice are saved and made available to those entities in the form of digital signals (such as a string of characters). This is necessary in order to enable both this service and those entities to respect such choices.
Features
Match and combine data from other data sources
Information about your activity on this service may be matched and combined with other information relating to you and originating from various sources (for instance your activity on a separate online service, your use of a loyalty card in-store, or your answers to a survey), in support of the purposes explained in this notice.
Link different devices
In support of the purposes explained in this notice, your device might be considered as likely linked to other devices that belong to you or your household (for instance because you are logged in to the same service on both your phone and your computer, or because you may use the same Internet connection on both devices).
Identify devices based on information transmitted automatically
Your device might be distinguished from other devices based on information it automatically sends when accessing the Internet (for instance, the IP address of your Internet connection or the type of browser you are using) in support of the purposes exposed in this notice.
Special Features
Use precise geolocation data
With your acceptance, your precise location (within a radius of less than 500 metres) may be used in support of the purposes explained in this notice.
Actively scan device characteristics for identification
With your acceptance, certain characteristics specific to your device might be requested and used to distinguish it from other devices (such as the installed fonts or plugins, the resolution of your screen) in support of the purposes explained in this notice.
Skip to main content
Met Office
Back
Learn
Global circulation patterns
At any time there are many weather systems weaving around the globe, however when averaged over many years a global pattern of air movement emerges.
Differential heating
The reason we have different weather patterns, jet streams, deserts and prevailing winds is all because of the global atmospheric circulation caused by the rotation of the Earth and the amount of heat different parts of the globe receive.
The sun is our main source of heat, and because of the tilt of the Earth, its curvature, our atmosphere, clouds and polar ice and snow, different parts of the world heat up differently. This sets up a big temperature difference between the poles and equator but our global circulation provides a natural air conditioning system to stop the equator becoming hotter and hotter, and poles becoming colder and colder.
The global circulation
Over the major parts of the Earth's surface there are large-scale wind circulations present. The global circulation can be described as the world-wide system of winds by which the necessary transport of heat from tropical to polar latitudes is accomplished.
In each hemisphere there are three cells (Hadley cell, Ferrel cell and Polar cell) in which air circulates through the entire depth of the troposphere. The troposphere is the name given to the vertical extent of the atmosphere from the surface, right up to between 10 and 15 km high. It is the part of the atmosphere where most of the weather takes place.
Hadley cell
The largest cells extend from the equator to between 30 and 40 degrees north and south, and are named Hadley cells, after English meteorologist George Hadley.
Within the Hadley cells, the trade winds blow towards the equator, then ascend near the equator as a broken line of thunderstorms, which forms the Inter-Tropical-Convergence Zone (ITCZ). From the tops of these storms, the air flows towards higher latitudes, where it sinks to produce high-pressure regions over the subtropical oceans and the world's hot deserts, such as the Sahara desert in North Africa.
Ferrel cell
In the middle cells, which are known as the Ferrel cells, air converges at low altitudes to ascend along the boundaries between cool polar air and the warm subtropical air that generally occurs between 60 and 70 degrees north and south. This often occurs around the latitude of the UK which gives us our unsettled weather. The circulation within the Ferrel cell is complicated by a return flow of air at high altitudes towards the tropics, where it joins sinking air from the Hadley cell.
The Ferrel cell moves in the opposite direction to the two other cells (Hadley cell and Polar cell) and acts rather like a gear. In this cell the surface wind would flow from a southerly direction in the northern hemisphere. However, the spin of the Earth induces an apparent motion to the right in the northern hemisphere and left in the southern hemisphere. This deflection is caused by the Coriolis effect and leads to the prevailing westerly and south-westerly winds often experienced over the UK.
Polar cell
The smallest and weakest cells are the Polar cells, which extend from between 60 and 70 degrees north and south, to the poles. Air in these cells sinks over the highest latitudes and flows out towards the lower latitudes at the surface.
The Coriolis effect, winds and UK weather
Now we know about the Hadley, Ferrel and Polar cells, let’s take a look at how all that translates to what we see at the Earth’s surface. As a result of the Earth’s spin, each cell has prevailing winds associated with it, and we also have jet streams, all influenced by something called the Coriolis effect. This explains why air moves in a certain direction around an area of low pressure, and why trade winds exist. It also gives us an idea of why we see certain weather in and around the UK.
Warm moist air from the tropics gets fed north by the surface winds of the Ferrel cell. This then meets cool dry air moving south in the Polar cell. The polar front forms where these two contrasting air mass meet, leading to ascending air and low pressure at the surface, often around the latitude of the UK.
The polar front jet stream drives this area of unstable atmosphere. The UK and many other countries in Europe often experience unsettled weather, which comes from travelling areas of low pressure which form when moist air rises along the polar front.
Weather (or low pressure) systems bearing rain and unsettled conditions move across the Atlantic on a regular basis. The jet stream guides these systems, so its position is important for UK weather.
In summer, the normal position of the jet stream is to be to be north of the UK - dragging those weather systems away from our shores to give us relatively settled weather.
Normally the jet stream runs fairly directly from west to east and pushes weather systems through quite quickly. However, sometimes the steering flow of the jet stream can meander (a bit like a river), curving north and south as it heads east across the Atlantic. This is called a meridional flow, with the more linear west to east flow being called a zonal flow.
During a meridional flow areas of low pressure can become stuck over the UK leading to prolonged periods of rain and strong winds. During the winter the polar front jet stream moves further south leading to a greater risk of unsettled weather, and even snow if cold arctic air masses move south over the UK.
The continued effect of the three circulation cells (Hadley cell, Ferrel cell and Polar cell), combined with the influence of the Coriolis effect results in the global circulation. The net effect is to transfer energy from the tropics towards the poles in a gigantic conveyor belt.
You might also like
Learn
Weather fronts
Weather fronts mark the boundary or transition zone between two air masses and have an important impact upon…
Read more
Learn
Why is the sky blue?
The sky looks blue because the shorter blue light waves are scattered more than other colours in the spectrum…
Read more
Learn
What do satellite pictures show?
They may only show a snapshot of the current weather, but they can tell us a…
Read more
Website feedback
Give feedback on our website |
190787 | https://www.youtube.com/watch?v=xRk9i9_aCfM | Proof of secx^2 = tanx^2 + 1
Alex
9800 subscribers
26 likes
Description
2951 views
Posted: 11 Nov 2017
In this video, I go through a trigonometric proof which is...
1+tan^2=sec^2
The proof is fairly straight forward with some common knowledge of trigonometric functions. This trig identity is used in various academic subjects, so it is good to know where it comes from!
💰 AFFILIATE STUFF (thank you for the support!)
I am a affiliate with Chegg. Although I am an affiliate, I do use Chegg for my studies. It does help for learning problem solving skills in mechanical engineering, and Chegg is also very convenient to have due to the many solutions that are readily available to you from your textbooks! Enjoy!
Get Chegg Study for $14.95/month or $99.95/year!
I am also an Amazon Affiliate and by using these referral links you can support this channel.
▪️HTML & CSS Book (recommended):
▪️Javascript & jQuery Book (recommended):
▪️PHP Book:
▪️DELL XPS 15 (my laptop):
▪️Blue Yetti Microphone (my mic):
▪️Amazon Echo:
▪️Amazon Echo Dot:
Thank you for using these links! ❤️
2 comments
Transcript:
Intro welcome back everybody today we're gonna prove this trigonometric identity which is a pretty simple so the very first step we're going to assume this trig identity already so we're gonna say assume that sine squared theta plus cosine squared theta equals 1 and we're also gonna need to know that sine over theta over cosine of theta equals tangent theta so once we have that we can actually start proving that this right here is true so simply we're gonna Proof take this equation right here and multiply everything by 1 over cosine squared theta so what I'm saying is we're going to take sine squared theta and we're going to multiply that by cosine squared theta plus cosine squared theta over cosine squared theta and that's gonna equal 1 over cosine squared theta so right here this right here is actually just a sine of theta over cosine theta quantity squared and as we've defined earlier by this equation we said that we could say that this is just tangent theta and then we could say that is squared due to this right here Rewriting so once we have that we can rewrite this equation so this is simply gonna go to 1 so anything divided by itself is just 1 and this right here is actually just secant squared theta because we know that 1 over cosine theta is just secant theta and if we just square that this just makes it 1 over secant squared or 1 over cosine squared which equals secant squared so that's also written as cosine squared like that so once we know that we can rewrite this equation so we could say this is equal to tangent squared theta plus 1 equals a secant squared theta and that is going to be our proof so that is the whole proof for this this Conclusion trig identity most of these weird so-called trig functions that are related to each other are defined by this original identity right here so it's just a form of manipulation to derive these kind of formulas are these other trig and trigonometric identities so hopefully this video helped you guys and I'll see you in the next video |
190788 | https://education.ti.com/en/us/~/media/Files/Activities/US/Math/Algebra%20I/Walk%20the%20Line%20Straight%20Line%20Distance%20Graphs/01_WalkTheLine_SE.pdf | Activity 1 Real-World Math Made Easy © 2005 Texas Instruments Incorporated 1 - 1 Walk the Line: Straight Line Distance Graphs When one quantity changes at a constant rate with respect to another, we say they are linearly related. Mathematically, we describe this relationship by defining a linear equation. In real-world applications, some quantities are linearly related and can be represented by using a straight-line graph. In this activity, you will create straight-line, or constant-speed, distance versus time plots using a Motion Detector, and then develop linear equations to describe these plots mathematically. OBJECTIVES • Record distance versus time data for a person walking at a uniform rate. • Analyze the data to extract slope and intercept information. • Interpret the slope and intercept information for physical meaning. MATERIALS TI-83 Plus or TI-84 Plus graphing calculator EasyData application CBR 2 or Go! Motion and direct calculator cable or Motion Detector and data-collection interface Activity 1 1 - 2 © 2005 Texas Instruments Incorporated Real-World Math Made Easy PROCEDURE 1. Set up the Motion Detector and calculator. a. Open the pivoting head of the Motion Detector. If your Motion Detector has a sensitivity switch, set it to Normal as shown. b. Turn on the calculator and make sure it is on the home screen. Connect it to the Motion Detector. (This may require the use of a data-collection interface.) 2. Position the Motion Detector on a table or chair so that the head is pointing horizontally out into an open area where you can walk. There should be no chairs or tables nearby. 3. Set up EasyData for data collection. a. Start the EasyData application, if it is not already running. b. Select from the Main screen, and then select New to reset the application. 4. Stand about a meter from the Motion Detector. When you are ready to collect data, select from the Main screen. Walk away from the Motion Detector at a slow and steady pace. You will have five seconds to collect data. 5. When data collection is complete, a graph of distance versus time will be displayed. Examine the graph. It should show a nearly linearly increasing function with no spikes or flat regions. If you need to repeat data collection, select and repeat Step 4. 6. Once you are satisfied with the graph, select to return to the Main screen. Exit EasyData by selecting from the Main screen and then selecting . ANALYSIS 1. Redisplay the graph outside of EasyData. a. Press [STAT PLOT]. b. Press to select Plot1 and press again to select On. c. Press . d. Press until ZoomStat is highlighted; press to display a graph with the x and y ranges set to fill the screen with data. e. Press to determine the coordinates of a point on the graph using the cursor keys. 2. The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line and b is the y-intercept value. The independent variable is x, which represents time, and y is the dependent variable, which represents distance in this activity. Trace across the graph to the left edge to read the y-intercept. Record this value as b in the Data Table on the Data Collection and Analysis sheet. 3. One way to determine the slope of the distance versus time graph is to guess a value and then check it by viewing a graph of the line with your data. To do this, enter an equation into the calculator, and then enter a value for the y-intercept and store it as variable B. a. Press . b. Press to remove any existing equation. c. Enter the equation MX + B in the Y1 field. Walk the Line Real-World Math Made Easy © 2005 Texas Instruments Incorporated 1 - 3 d. Press until the icon to the left of Y1 is blinking. Press until a bold diagonal line is shown in order to display your model with a thick line. e. Press [QUIT] to return to the home screen. f. Enter your value for the y-intercept and then press B to store the value in the variable B. 4. Now set a value for the slope m, and then look at the resulting graph. To obtain a good fit, you will need to try several values for the slope. Use the steps below to store different values to the variable M. Start with M = 1. Experiment until you find one that provides a good fit. a. Enter a value for the slope m and press M to store the value in the variable M. b. Press to see the data with the model graph superimposed. c. Press [QUIT] to return to the home screen. 5. Record the optimized value for the slope in the Data Table on the Data Collection and Analysis sheet. Use the values of the slope and intercept to write the equation of the line that best fits the distance versus time data. 6. Another way to determine the slope of a line to fit your data is to use two well-separated data points. Use the cursor keys to move along the data points. Choose two points (x1, y1) and (x2, y2) that are not close to each other and record them in the Data Table on the Data Collection and Analysis sheet. 7. Use the points in the table to compute the slope, m, of the distance versus time graph. = − − = 1 2 1 2 x x y y m ⇒ Calculate the slope and answer Question 1 on the Data Collection and Analysis sheet. 8. You can also use the calculator to automatically determine an optimized slope and intercept. a. Press and use the cursor keys to highlight CALC. b. Press the number adjacent to LinReg(ax+b) to copy the command to the home screen. c. Press [L1] [L6] to enter the lists containing your data. d. Press and use the cursor keys to highlight Y-VARS. e. Select Function by pressing . f. Press to copy Y1 to the expression. On the home screen, you will now see the entry LinReg(ax+b) L1, L6, Y1. This command will perform a linear regression using the x-values in L1 as and the y-values in L6. The resulting regression line will be stored in equation variable Y1. g. Press to perform the linear regression. Use the parameters a and b to write the equation of the calculator’s best-fit line, and record it in the Data Table. h. Press to see the graph. ⇒ Answer Questions 2–5 on the Data Collection and Analysis sheet. Activity 1 Real-World Math Made Easy © 2005 Texas Instruments Incorporated 1 - 4 DATA COLLECTION AND ANALYSIS Name ____ Date ____ DATA TABLE y-intercept b optimized slope m optimized line equation x1, y1 x2, y2 regression line equation QUESTIONS 1. How does this value compare with the slope you found by trial and error? 2. How do the values of the slope and intercept as determined by the calculator compare to your earlier values? Would you expect them to be exactly the same? 3. Slope is defined as change in y-values divided by change in x-values. Complete the following statement about slope for the linear data set you collected. In this activity, slope represents a change in ____ divided by a change in ____. 4. Based on this statement, what are the units of measurement for slope in this activity? 5. The y-intercept can be interpreted as the starting position or the starting distance from the Motion Detector. What does the slope represent physically? Hint: Consider the units of measurement for the slope you described in the previous question. |
190789 | https://www.youtube.com/watch?v=cecD1w3-SD0 | R4. Free Body Diagrams
MIT OpenCourseWare
5940000 subscribers
234 likes
Description
30013 views
Posted: 3 Sep 2013
MIT 2.003SC Engineering Dynamics, Fall 2011
View the complete course:
Instructor: J. Kim Vandiver
License: Creative Commons BY-NC-SA
More information at
More courses at
9 comments
Transcript:
the following content is provided under a Creative Commons license your support will help MIT open courseware continue to offer highquality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT open courseware at ocw.mit.edu all right hey you know the drill from the previous couple of weeks get your piece of paper out and write down which you think were important concepts for the last week or so okay let's see what you have on your papers to to okay so s Center of concept of center of mass and things you can use it for so I'm going to generalize that to systems of particles and you're tell what you've just said is Center of mass is one thing that comes from that anybody on this subject of systems of particles anybody else have something angular momentum so what about it yeah so uh this concept of the summation of the hi right equal to kind of the H of the system and that you can use that okay anything else about systems particles that we that you learn Newton's third law between particles yeah Newton's third law kind of kind of helped us supp things out and and a consequence you found Center of mass that's useful but you take the uh the derivative of the center of mass what what do you find out RG here dot times the total mass is what yeah it's momentum but it's momentum of what well it's the momentum of the entire system right and so this is the summation of the pis that's kind of important and finally take another derivative what's that what did you learn about that pardon what kind of forces exter all the external forces on system all of them all of them so powerful things came out of thinking about Center of mass okay all right what something else during the week that wasn't around systems of particles Yes actually shout out give me your say your name when you're when I calling you I'm going to try to learn more of your name Christina Christina right um so when you're looking at equations for example like the for external is equal to um the change in angular momentum um looking at what pieces go where so if you have like an applied moment does it go on this side or that side okay and I put that in the category of free free body diagrams are you thinking of so constructing free body diagrams okay what else Eddie Eddie yeah okay uh the two different methods for finding torque yeah so let's go a little further with torque What do you use the torque for on you what do you want to find torque for to get all right let's generalize this to equation to motion so both in uh for forces and moments two different kinds of problems that we are able to do by being able to find the torqus okay anything else it's important this week Betsy solving motion yeah I mean this isn't a math course yeah you got to be able to solve differential equations but yeah that's maybe um you're you're actually on to something here but let's put it in the context of what kind of equations did we work on a lot on this week what kind of equations of motion I don't mean like second order linear I mean we're doing dynamic we did dynamics of what kind a lot of it acceleration yeah but [Music] also we worked a lot of the examples this week were they rotational problems or translational problems all right kind of heavy emphasis this week on angular momentum right so using okay okay that's a pretty good list my own list had my own list had most of that on them let's see where my list was Center Mass r dot r dble dot I wrote These as equations so this thing in particular the summation of the external torqus with respect to some point for a particle is dhdt plus VA cross p and for we were just beginning to do sums of collections of particles and iate rigid bodies where we have maybe more than one particle the summation of the torqux is DH now Capital H now for the summation of the H's of I DT plus v a n o cross and here's where we learned something up here that you can do down here that was really helpful well it cross with what which P this one up here right and that's actually this one so this is cross P of the center of mass of the system okay good those are uh that's that's pretty good list so now let's do let's do a problem so here's the problem it's a rigid rod with a mass on the end and the Rod's Mass lless just to make it easy for the moment but this is now a rigid body okay there's a torsional spring here here and a torsional spring resists angular displacement with a torque that's equal to the spring constant times the Theta that you push it through okay so it generates a moment a moment restoring moment when you push it away it tries to push this thing back gravity is also acting so it's a pendulum with a rigid as a rigid body and a torsional spring so I want the first step I want you eventually to find the equation of motion and I want you to find do start by doing a free body diagram but before you go to work I'm going to give you a little quick mini lecture on free body diagrams so first thing you do is assign coordinates maybe draw is not really assign the coordinates for the problem determine the number of degrees of freedom we haven't talked about that much yet it will become much more important in the subject the number of degrees of freedom but I'll say it once and we'll do much more with this the number of degrees of freedom is the same thing as the number of independent coordinates you need to completely describe the motion okay and the number of independent coordinates that you need is equal to the number of equations in motion that you'll get so this problem how many coord ordinates does it take and it's so this is planer motion it's confined to the board how many actual coordinates does it take to completely describe the position of this thing I see a one how about anybody else what's the one what would you choose um I think I yeah well what coordinate itself what is the single coordinate you would use to define the motion of the system I here a Theta everybody agree Theta of T if you can figure out Theta of T you know this position of this thing for all time that's all you need it's one degree of Freedom one coordinate and in terms of choosing the coordinate system to use basically you were saying use what Polar Polar coordinates centered here but there's your ntial system but you're going to go with this point this is our a point and is it is it moving so if we use expressions like this what will happen to that term goes away right nope and we'll get to that and if I divert to that I won't finish what I've got planned for the day it's important and we're going to come back to that has to do with constraints um so you assign the coordinates determine the number of degrees of freedom assign positive values to all rotations displacement ments velocities linear velocities and rotational velocities in the problem positive values of those and from that you deduce the direction of the resulting forces and moments and this will help you get the signs correct especially when you have multi-body problems like two masses with a spring in between them which way do the forces go on each Mass right okay so in this problem there's only one coordinate so we're really saying is assign a positive Theta and assume a positive Theta dot what are the resulting forces that end up in your free body diagram so now I want you to right draw the free body diagram for this thing and up here final statement put in all forces and moments don't don't leave out anything because you know they're not going to matter because it'll get you in trouble as the problems get more complicated okay draw your free body okay so you know get in groups of three or four and compare notes Here you guys are kind of a natural natural group and you know can look check each other's stuff and come up with a come up with a final all right um so we were having a debate about whether or not you were asking us for the free body diagram of the system or if you were asking us for the free body diagram of the M two different yeah so are there moments and things involved up here um like that torsional spring when you are including the entire system okay so this is a rigid body I kept saying this is a rigid body it happens to have a single Mass point in it but it's a whole single rigid body and do the free body diagram for the rigid body looking at the can you show me something where you at where are you what's your best shot at this so far diagram we acting on and we say that the spring force is producing a moment so what what's the what's your body that you're working with what's the definition of your rigid body the mass on the end of the okay so you're still thinking about the particle but if you're only thinking about the particle it's kind of hard to get that Toral spring involved so you need to think of this as a sing whole body one whole rigid body and now draw you know just draw a stick in here and put the forces on it it doesn't matter where the mass is for the purpose of the free body diagram right that comes into the couldn't you just use the the the r cross um Force and and get the get the force on the ball from the so you're getting into kind of the equation of motion part of it this is only about assuming this thing has some Dynamic moment in time and draw the external forces and moments that act on the object the object is this full length stick okay so in that case we just have the spring force acting in oppos to or like I guess we can say it's acting positive the weight you have weight for sure you have the spring puts a a moment on the system and there are yet potential other forces I give you a couple minutes because there's a little confusion around whether it's a particle or a rigid body it's a rigid body you got to deal with a whole body all right let's see what you got where's your where's your free body diagram this right here okay so and tell tell me what you have for forces and moments point to each one of them and say what it is um so we have the weight of this end and we assume this was massless yep but it's rigid so it's all one system now yeah one rigid body so um so the center of mass is right here because there's no up that way and then we have reaction forces at the pin and then we have the moment from the spring there okay all right yeah we have so you get convince one another who's right here so it depends on which way the ball so that's why there is a little system there's there's a rubric that I so I recommend you just use the the system this is basically a system for getting equations of motion and this is important that one so then if you know it's positive then you can figure out which direction the reaction is all right I hate stopping all the fun uh you guys are you doing well there's there is Great Value in talking to one another and convincing one another of what the right way to do something is you know you'll have two different points of view and talking it out it's amazing how fast you can make progress so do that do that when you're home doing solving problems so let's uh what was the assignment here find the draw a free body diagram well let's do it so it's a rigid body all one body even though the mass is all concentrated down here the forces uh what are forces that are on this thing give me one wait mg right okay what else noral normal force where top okay so you're talking about a reaction force up here and have you chosen a coordinate system we already kind of decided didn't we on polar coordinates for this right so are the reaction forces known to start with no so what's a what's the cleverest way to to assign them on your free body diagram well and line and and associated with your the coordinates right so I would I which I would have two reaction forces up here one in which one in each of which directions r radial n okay and the positive radial positive tangential like that right so I would say oh well I have an unknown F Theta force and I have an unknown F force and I draw them in their Positive Directions because I have no clue so just make them positive your assume direction for them forces here G there what about the tension on the the tension in the string does it belong in this problem why not the rigid body it's an internal Force it's an little fi J fji thing it's those things that cancel internally right so you don't have to deal with it it's not an external Force to the rigid body so this is great uh we want to find an the next step in this is to find an equation of motion got to be Speedy about this uh so what pardon moment you'd use moment to do it about what point about the point thep all right so if we call that a you'd use to angular momentum and torqux and that sort of thing is that what you're saying moments about it okay no I'm going to have you do it yourself do it so now sit right she's talking about how there's that put the oh we forget that one so which direction is it um clockwise or counterclock all right so there's a moment about this point that is uh we know what it is it's in that direction and it has value KT time Theta so anytime you know what it is right don't just make it an unknown write it out you know it's that and you know it's in this direction okay and we're going to write how many equation of motion what we get and we've chosen polar coordinates you could write you could do this in using forces just Newton's second law but then you have to solve for what if you write a for if you write force balance f equals Ma you have to these appear and you don't know them so the reason to use torqus in this problem is because are these going to appear in your answer no so that's the reason to use torque I saw something yeah is that L this is K sub T to distinguish it as a torsional spring no no it doesn't have an L it's right here it's a spring who gives a moment resistance to being deflected it has KT Theta is units of torque and this isn't units of uh Force per unit displacement this is units of torque per radian okay yeah well you have assigned polar coordinates and my assumption was that I drew it right here there's positive Theta hat right so isn't our torque negative yeah see I've dra but it's negative is accounted for by the sign by the direction of the arrow so let's now write the equation of motion and to do the equation of motion you're going to say that the sum of the what external torqu about what point a is equal to D and I'll use the capital H because we're now working our way on the rigid bodies could have more than one point we could have more than one Mass point dhdt plus v a cross p and what's that in this problem that guy conveniently is zero so uh do it you can now do this problem figure out right down an equation of motion using some of external torques and say you're going to have to figure out an angular momentum and some torqux okay I hate pulling you guys away lots of there's lots of good thinking going on all right drag yourselves away help me out here which groups figured out the external torqus okay what do you got tell me what to do here tell me what it is I want to write the terms down mg sin minus mg L sin Theta in what direction okay others agree with that term wait a second just once a time here the gravity term do we have the sign correct then we have the moment arm correct so it's a this is the moment arm R cross Force looks like minus K to me mg L sin Theta looks pretty good how what other torqus the moment from the moment from the spring okay and tell me how to write that um that's in Direction well then the KT KT Theta also in the K hat Direction and our free body diagram does the forces produce moments okay is everybody happy with that are we done yes any questions about it of why it is the way it is okay we're going to now we need this piece of it so what's an H for rigid body the summation of all the bits how many bits are there just the one and that's some R cross a p can somebody tell me what they got for the p is the momentum of our little Mass here with respect to an inertial frame which is also a because it's not moving and R is l so this is l r Hat Cross M what L Theta dot is its velocity mass times velocity and its direction Theta hat R cross Theta is K so this looks like m l s Theta do K hat we all agreement dhdt this K change direction so the only only time dependent thing I think you're thank you we're going to need that all right the derivative is okay that's equal to that it's all KS get just one equation out of this you don't have to break it down into subc components and you can collect it all on one side m l 2 Theta double dot plus KT theta plus mg L sin Theta equal Z and that's a second order nonlinear ordinary differential equation which you could linearize for small motion it's just a pendulum okay good I got one other one I want to do before the hour is out I got a big piece of cement pipe on the back of a truck it's not tied down the truck's moving at 3 m/ second okay and the uh rotation rate of this piece of pipe is 6 radians per second and the truck's accelerating minus a half a meter perss squared it's actually breaking okay the guys in in trouble right um so the first thing here is Omega is 6 radians per second does it matter with respect to what the truck or the ground I haven't told you talk in your groups and give me a quick answer to that what do you think say again so Kristen thinks it doesn't matter anybody want to counter that why not so you're saying that this could be Omega with respect to the truck or it could be Mega with respect to this fixed frame and the answer will it' be the same in either case so um earlier Christine right Christin okay Christina all right so earlier we had that problem with the with there was an arm that was rotating then there's another one rot on here you you had to know what this Omega was with respect to what because this arm is rotating M also and so with this you have your one XY coordinate system and while the truck is translating the truck's not doing any rotation at so it's staying in the same coordinate plane as it's moving along so your argument is if you were standing on the ground seeing it rotating you'd say it seeing you would see it rotate at the same rate as if you were riding on the truck everybody agree with that does it matter that the truck's accelerating what do you think no it's only whether or not the truck's rotating if the truck's not rotating then rotation rate seen from this Frame or just a translating frame will always seem the same even if it's you're accelerating okay uh so we've done that one now I want you to find the velocity of Point G given this information so whip it down quickly how you would approach this problem see if you can solve it g is the center of mass of that big pipe okay we're going to run it out of time let's uh who's got an answer for me I think they've got it sorted out Kristen velocity of the truck all right so let's be systematic so you do you have to pick some points and do did you use a rotating frame you use equation that requires a rotating reference frame no so I assume though you're talking about this is we're looking at the velocity of G in O is equal to in general the velocity of some point a that you're going to have to pick plus the velocity of G with respect to a and this this thing has two possible components right one which is the derivative of GA DT ignoring the rotation plus okay what's this term in this problem well we haven't picked the point a where you going to pick point a there's dumb points and there's good points uh what put the front of the put front of the truck well it's possible Christina what do you think got your hand up um yeah I put it on the bottom of the disc where it's not the point of contact is that what you're saying anybody else what do you vote for where do you want to put a cont point of contact okay if you're using this equation and you're using these pieces you are using the concept of a rotating frame if there's any Omega in the problem you better have a rotating frame somewhere right unless you're using polar coordinates which is sort of a degenerate rotating frame so if we at this point we're going to make a here and the you have a coordinate system little X1 y1 and attach to a so you have an A X1 y1 here Z1 and is it rotating yeah is it a rotating coordinate system well if it's not then you can't do this you have this assumes this term here is the the velocity at which this point and this point are moving relative to one another as if you were sitting on the pipe ignoring the you you don't you don't see the rotation you're moving with the pipe is the and the pipe it better it's assuming you have then put a reference frame on the pipe that moves with the pipe so it is a little reference frame reference frame axyz move with the pipe it better or you can't use the equation so this this is attached to the pipe and as it rolls that you have to allow that XYZ frame to move with it okay so in that frame what's this term this is zero cuz the two points are fixed on the in the pipe okay and what's this velocity 3 m/ second in the what direction we do it in the Nal frame if we want okay and then what's this term so it's six and we said doesn't matter what frame it is it's all the same rotation rate radians per second in what direction K Hat Cross the radius is capital r what's its direction and now this is at an instant in time at this particular instant in time you've cleverly drawn the XYZ system so it lines up with the master one so it makes the problem easy so at the moment it's either in the big J hat or in the Little J hat it's easiest if you do it in big call it Big J because then there you're done so K cross J I think negative I and so this should work out to be 3 m/s positive I 1 and 1 12 6 is 9 and a minus 9 m/s in the ne in the I direction right so you are at minus 6 m/s isn't Theus to the a wouldn't it be what does it matter if it's negative or oh matters a lot you have to you have to you exercise great care when you pick the the r vector and the r Vector goes from your origin to the point you're talking about so the origin that's your frame and that's the point and when you draw the arrow of the r Vector it goes from a to G that's its direction and it has you know it's length it's length capital r long yep that's that's a really important point you know which way the arrows go okay and so now we're done with that and I see we've got a minute or [Music] two let's talk about the acceler ation let's do the acceleration of this problem acceleration of G in O and this is just an exercise in remembering the terms full 3d vector equation for accelerations what's the what's the first term here how about before we even get into and we're not in polar coordinates in this acceleration all right let's just work it out I start with this one myself cation of a with respect to O what's another term so acceleration of G with respect to a but no rotation right that term plus how many more terms do we have to go three okay what give me one somebody else what's your name St stepen you've got a twoa Omega R and Omega is let's get its unit Vector in here well actually it's not 2 Omega cross R let's just do that okay that's okay what that what's give me another term anybody else Omega cross Omega cross R and got another one on the yeah and in fact we better be a little more careful than that so it's really the velocity of G with respect to a and do it have rotation in it or not this is the no rotation this is the speed at which the points are moving apart from one another right Omega cross Omega cross R we got one two I think we need another term Omega dot cross R and the r is r g a this R is GA okay so what's the velocity what's the acceleration of this nope so this one's minus a half uh what's the what's this one so this is your you're in the frame of the pipe now and this is a speed at which G acceleration of G with respect to the a okay it doesn't move it's fixed length right this one what's the velocity of G with respect to a without rotation in involved okay this one is omega K cross and this is our again Omega K cross and this RG is r in the what direction at this instant J right so we can work that out and what about this one what direction is Omega dot in K cross r at this instant J again okay do we know that back value no we have to that might be something we have to figure out so we could you could now crank this out that'd be the acceleration of that point but you'd find out that hm we don't have that now how would you go about finding out what Omega dot is you get the gold star equations of motion you need to come up with equations of motion and um how how many how many equations of motion would we end up writing this is equivalent to asking how many degrees of freedom are there okay I hear two I hear a a one so how many now now so this there's a lot of nuances to this this discussion anytime something is you're given a fixed value for something that that coordinate is constrained that parameter is constrained the uh we're given the motion of the truck it's completely specified the and you'll just you could just substitute numbers in for those things I mean like this it pops up as a half okay the um I think this thing will boil down to one equation and you need to pick some set of coordinates to describe it so I would probably to if I'm looking for Omega dot I'd probably work with torque about that point some of the torqus on then isolate it to just this object and it's a sum of the torqus around this point and see what happens it's got to be equal to the sum of the external torque got to equal to what the HDT and then you have to sort out this term so in this problem you know what's the direction of this velocity I what is the unit Vector associated with the linear momentum of that pipe what direction is the linear momentum of the pipe what's the velocity what is this answer what's the momentum of that pipe mass times that velocity right M remember the velocity of the system the pipe may have has all its mass around the rim but the velocity of the the momentum of the pipe is the total mass of the pipe times the velocity of its Center of mass and that's the velocity of the center of mass so its linear momentum has what unit Vector assigned with it I and what is the unit Vector associated with VA I I cross I is things this is one of those cases where they're parallel paths and that term drops out so you could come up with very quickly and easily a equation of motion okay very good see you on Tuesday next Tuesday lecture time will be review I think it kind of be more like one of these sessions and then quiz Tuesday night |
190790 | https://www.khanacademy.org/math/in-in-grade-11-ncert/x79978c5cf3a8f108:introduction-to-three-dimensional-geometry/x79978c5cf3a8f108:section-formula/e/section-formula | Section formula (practice) | Khan Academy
Skip to main content
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked.
Explore
Browse By Standards
Explore Khanmigo
Math: Pre-K - 8th grade
Math: High school & college
Math: Multiple grades
Math: Illustrative Math-aligned
Math: Eureka Math-aligned
Math: Get ready courses
Test prep
Science
Economics
Reading & language arts
Computing
Life skills
Social studies
Partner courses
Khan for educators
Select a category to view its courses
Search
AI for Teachers FreeDonateLog inSign up
Search for courses, skills, and videos
Help us do more
We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever.
Select gift frequency
One time
Recurring
Monthly
Yearly
Select amount
$10
$20
$30
$40
Other
Give now
By donating, you agree to our terms of service and privacy policy.
Skip to lesson content
Class 11 math (India)
Course: Class 11 math (India)>Unit 11
Lesson 3: Section formula
Section formula
Section formula: applications
Math>
Class 11 math (India)>
Introduction to three dimensional geometry>
Section formula
© 2025 Khan Academy
Terms of usePrivacy PolicyCookie NoticeAccessibility Statement
Section formula
Google Classroom
Microsoft Teams
Problem
Find the ratio in which Y Z plane divides the line segment joining the points A(−2,7,5) and B(3,11,−4).
Choose 1 answer:
Choose 1 answer:
(Choice A) 2:3 internally A 2:3 internally
(Choice B) 3:2 internally B 3:2 internally
(Choice C) 2:3 externally C 2:3 externally
(Choice D) 3:2 externally D 3:2 externally
Report a problem
Do 4 problems
Skip Check
Use of cookies
Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy
Accept All Cookies Strictly Necessary Only
Cookies Settings
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Allow All
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session.
Functional Cookies
[x] Functional Cookies
These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences.
Targeting Cookies
[x] Targeting Cookies
These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service.
Performance Cookies
[x] Performance Cookies
These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Reject All Confirm My Choices |
190791 | https://www.doe.mass.edu/frameworks/math/2017-06.pdf | i MATHEMATICS Grades Pre-Kindergarten to 12 Massachusetts Curriculum Framework – 2017 This document was prepared by the Massachusetts Department of Elementary and Secondary Education Board of Elementary and Secondary Education Members Mr. Paul Sagan, Chair, Cambridge Mr. James Morton, Vice Chair, Boston Ms. Katherine Craven, Brookline Dr. Edward Doherty, Hyde Park Dr. Roland Fryer, Cambridge Ms. Margaret McKenna, Boston Mr. Michael Moriarty, Holyoke Dr. Pendred Noyce, Boston Mr. James Peyser, Secretary of Education, Milton Ms. Mary Ann Stewart, Lexington Mr. Nathan Moore, Chair, Student Advisory Council, Scituate Mitchell D. Chester, Ed.D., Commissioner and Secretary to the Board The Massachusetts Department of Elementary and Secondary Education, an affirmative action employer, is committed to ensuring that all of its programs and facilities are accessible to all members of the public. We do not discriminate on the basis of age, color, disability, national origin, race, religion, sex, or sexual orientation. Inquiries regarding the Department’s compliance with Title IX and other civil rights laws may be directed to the Human Resources Director, 75 Pleasant St., Malden, MA, 02148, 781-338-6105. © 2017 Massachusetts Department of Elementary and Secondary Education. Permission is hereby granted to copy any or all parts of this document for non-commercial educational purposes. Please credit the “Massachusetts Department of Elementary and Secondary Education.” Massachusetts Department of Elementary and Secondary Education 75 Pleasant Street, Malden, MA 02148-4906 Phone 781-338-3000 TTY: N.E.T. Relay 800-439-2370 www.doe.mass.edu Dear Colleagues, I am pleased to present to you the Massachusetts Curriculum Framework for Mathematics adopted by the Board of Elementary and Secondary Education in March 2017. This Framework builds upon the foundation of the 2010 Massachusetts Curriculum Framework for Mathematics as well as versions of the Massachusetts Mathematics Framework published since 1995. The current Framework incorporates improvements suggested by Massachusetts educators after six years of experience in implementing the 2010 standards in their classrooms. These revised pre-kindergarten to grade 12 standards are based on research and effective practice, and will enable teachers and administrators to strengthen curriculum, instruction, and assessment. The 2017 standards draw from the best of prior Massachusetts standards, and represent the input of hundreds of the Commonwealth’s K–12 and higher education faculty. The 2017 standards present the Commonwealth’s commitment to providing all students with a world-class education. This revision of the Framework retains the strengths of previous frameworks and includes these improved features: Increased coherence across the grades and improved clarity of mathematical terms and language to describe expectations for students. Clear expectations for student mastery of basic addition, subtraction, multiplication, and division facts. An enhanced high school section that includes: 1) clearer model course standards; and 2) guidance on making decisions for course sequences and the model Algebra I course, along with options for various course-taking pathways. Guidance for moving students into an Algebra I course by grade 8 and through Calculus in high school; and More detailed descriptions about rigor and aspirations for students with a stronger emphasis on how the content standards, the standards for mathematical practice, and the guiding principles prepare students for college, careers, and civic participation. In the course of revising these standards, the Department received many valuable comments and suggestions. I want to thank everyone who contributed their suggestions and ideas to make these revised standards useful for educators, students, families, and the community. In particular, I am grateful to the members of the Mathematics Standards Review Panel and our Content Advisors for giving their time generously to the project to improve the learning standards for Massachusetts students. I am proud of the work that has been accomplished. We will continue to collaborate with educators to implement the 2017 Massachusetts Curriculum Framework for Mathematics. Thank you again for your ongoing support and for your commitment to achieving the goal of improved student achievement for all our students. Sincerely, Mitchell D. Chester, Ed. D. Commissioner of Elementary and Secondary Education Massachusetts Department of Elementary and Secondary Education 75 Pleasant Street, Malden, Massachusetts 02148-4906 Massachusetts Curriculum Framework for Mathematics 1 Table of Contents Acknowledgments ......................................................................................................................................................5 Introduction ................................................................................................................................................................8 Guiding Principles for Mathematics Programs in Massachusetts........................................................................ 14 The Standards for Mathematical Practice ........................................................................................................... 16 Standards for Mathematical Content Pre-Kindergarten–Grade 8 ................................ 19 Organization of the Pre-Kindergarten to Grade 8 Content Standards ................................................................ 20 Pre-Kindergarten ......................................................................................................................................... 22 Introduction ......................................................................................................................................................... 22 Pre-Kindergarten Overview ................................................................................................................................. 23 Pre-Kindergarten Content Standards .................................................................................................................. 24 Kindergarten ................................................................................................................................................... 25 Introduction ......................................................................................................................................................... 25 Kindergarten Overview ........................................................................................................................................ 26 Kindergarten Content Standards ......................................................................................................................... 27 Grade 1 ................................................................................................................................................................ 29 Introduction ......................................................................................................................................................... 29 Grade 1 Overview ................................................................................................................................................ 30 Grade 1 Content Standards ................................................................................................................................. 31 Grade 2 ............................................................................................................................................................... 33 Introduction ......................................................................................................................................................... 33 Grade 2 Overview ................................................................................................................................................ 34 Grade 2 Content Standards ................................................................................................................................. 35 Grade 3 ............................................................................................................................................................... 37 Introduction ......................................................................................................................................................... 37 Grade 3 Overview ................................................................................................................................................ 38 Grade 3 Content Standards ................................................................................................................................. 39 Grade 4 ............................................................................................................................................................... 42 Introduction ......................................................................................................................................................... 42 Grade 4 Overview ................................................................................................................................................ 43 Grade 4 Content Standards ................................................................................................................................. 44 Grade 5................................................................................................................................................................ 48 Massachusetts Curriculum Framework for Mathematics 2 Introduction ......................................................................................................................................................... 48 Grade 5 Overview ................................................................................................................................................ 49 Grade 5 Content Standards ................................................................................................................................. 50 Grade 6 ............................................................................................................................................................... 54 Introduction ......................................................................................................................................................... 54 Grade 6 Overview ................................................................................................................................................ 56 Grade 6 Content Standards ................................................................................................................................. 57 Grade 7 ................................................................................................................................................................ 61 Introduction ......................................................................................................................................................... 61 Grade 7 Overview ................................................................................................................................................ 62 Grade 7 Content Standards ................................................................................................................................. 63 Grade 8 ............................................................................................................................................................... 67 Introduction ......................................................................................................................................................... 67 Grade 8 Overview ................................................................................................................................................ 68 Grade 8 Content Standards ................................................................................................................................. 69 The High School Standards for Mathematical Content ......................................................... 72 Introduction ......................................................................................................................................................... 73 High School Content Standards by Conceptual Categories ................................................................................. 76 Conceptual Category: Number and Quantity [N] ....................................................................... 77 Introduction ......................................................................................................................................................... 77 Conceptual Category: Number and Quantity Overview [N] ................................................................................ 78 Conceptual Category: Number and Quantity Content Standards [N] ................................................................. 79 Conceptual Category: Algebra [A] ...................................................................................................... 81 Introduction ......................................................................................................................................................... 81 Conceptual Category: Algebra Overview [A] ....................................................................................................... 83 Conceptual Category: Algebra Content Standards [A] ........................................................................................ 84 Conceptual Category: Functions [F] .................................................................................................. 87 Introduction ......................................................................................................................................................... 87 Conceptual Category: Functions Overview [F] .................................................................................................... 88 Conceptual Category: Functions Content Standards [F] ...................................................................................... 89 Conceptual Category: Modeling [] .................................................................................................. 92 Introduction ......................................................................................................................................................... 92 Conceptual Category: Geometry [G] ................................................................................................. 94 Massachusetts Curriculum Framework for Mathematics 3 Introduction ......................................................................................................................................................... 94 Conceptual Category: Geometry Overview [G] ................................................................................................... 96 Conceptual Category: Geometry Content Standards [G] .................................................................................... 97 Conceptual Category: Statistics and Probability [S] .............................................................. 100 Introduction ....................................................................................................................................................... 100 Conceptual Category: Statistics and Probability Overview [S] .......................................................................... 101 Conceptual Category: Statistics and Probability Content Standards [S]............................................................ 102 The Standards for Mathematical Content High School: Model Pathways and Model Courses ............................................................................................... 105 Introduction ....................................................................................................................................................... 106 Model Traditional Pathway: Model Algebra I [AI] .................................................................. 108 Introduction ....................................................................................................................................................... 108 Model Traditional Pathway: Model Algebra I Overview [AI] ............................................................................. 109 Model Traditional Pathway: Model Algebra I Content Standards [AI] .............................................................. 110 Model Traditional Pathway: Model Geometry [GEO] ........................................................... 115 Introduction ....................................................................................................................................................... 115 Model Traditional Pathway: Model Geometry Overview [GEO] ....................................................................... 117 Model Traditional Pathway: Model Geometry Content Standards [GEO] ........................................................ 118 Model Traditional Pathway: Model Algebra II [AII] .............................................................. 122 Introduction ....................................................................................................................................................... 122 Model Traditional Pathway: Model Algebra II Overview [AII] ........................................................................... 123 Model Traditional Pathway: Model Algebra II Content Standards [AII] ............................................................ 125 Model Integrated Pathway: Model Mathematics I [MI] ....................................................... 129 Introduction ....................................................................................................................................................... 129 Model Integrated Pathway: Model Mathematics I Overview [MI] ................................................................... 131 Model Integrated Pathway: Model Mathematics I Content Standards [MI] ..................................................... 133 Model Integrated Pathway: Model Mathematics II [MII] ................................................... 137 Introduction ....................................................................................................................................................... 137 Model Integrated Pathway: Model Mathematics II Overview [MII] ................................................................. 138 Model Integrated Pathway: Model Mathematics II Content Standards [MII] ................................................... 140 Model Integrated Pathway: Model Mathematics III [MIII] ............................................... 146 Introduction ....................................................................................................................................................... 146 Model Integrated Pathway: Model Mathematics III Overview [MIII] ............................................................... 147 Massachusetts Curriculum Framework for Mathematics 4 Model Integrated Pathway: Model Mathematics III Content Standards [MIII] ................................................. 149 Model Advanced Course: Model Precalculus [PC] .................................................................. 154 Introduction ....................................................................................................................................................... 154 Model Advanced Course: Model Precalculus Overview [PC] ............................................................................ 155 Model Advanced Course: Model Precalculus Content Standards [PC] .............................................................. 156 Model Advanced Course: Model Advanced Quantitative Reasoning [AQR] ............ 160 Introduction ....................................................................................................................................................... 160 Model Advanced Course: Model Advanced Quantitative Reasoning Overview [AQR] ..................................... 161 Model Advanced Course: Model Advanced Quantitative Reasoning Content Standards [AQR] ...................... 162 Making Decisions about High School Course Sequences and Algebra I in Grade 8 .................................................................................................................................. 165 Course Sequences and the Model Algebra I Course .......................................................................................... 165 Appendix I: Application of Standards for English Learners and Students with Disabilities ........................... 168 English Learners ................................................................................................................................................. 168 Students with Disabilities ................................................................................................................................... 169 Appendix II: Standards for Mathematical Practice Grade-Span Descriptions: Pre-K–5, 6–8, 9–12 ............. 171 Standards for Mathematical Practice Grades Pre-K–5 ...................................................................................... 171 Standards for Mathematical Practice Grades 6–8 ............................................................................................. 174 Standards for Mathematical Practice Grades 9–12 ........................................................................................... 177 Appendix III: High School Conceptual Category Tables ............................................................................... 180 Number and Quantity [N] .................................................................................................................................. 181 Algebra [A] ......................................................................................................................................................... 183 Functions [F] ...................................................................................................................................................... 185 Statistics and Probability [S] .............................................................................................................................. 187 Geometry [G] ..................................................................................................................................................... 189 Glossary: Mathematical Terms, Tables, and Illustrations ............................................................................... 191 Tables and Illustrations of Key Mathematical Properties, Rules, and Number Sets ......................................... 203 Bibliography and Resources ................................................................................................................ 207 Massachusetts Curriculum Framework for Mathematics 5 Acknowledgments Massachusetts Curriculum Frameworks for English Language Arts and Literacy and Mathematics Review Panel, 2016-2017 Rachel Barlage, Lead English Teacher, Chelsea High School, Chelsea Public Schools Jennifer Berg, Assistant Professor of Mathematics, Fitchburg State University Tara Brandt, Mathematics Supervisor, K–12, Westfield Public Schools Jennifer Camara-Pomfret, English Teacher, Seekonk High School, Seekonk Public Schools Tricia Clifford, Principal, Mary Lee Burbank School, Belmont Public Schools Linda Crockett, Literacy Coach, Grades 6–8, Westfield South Middle School, Westfield Public Schools Linda Dart-Kathios, Mathematics Department Chairperson, Middlesex Community College Linda Davenport, Director of K–12 Mathematics, Boston Public Schools Beth Delaney, Mathematics Coach, Revere Public Schools Lisa Dion, Manager of Curriculum, Data and Assessment, New Bedford Public Schools Tom Fortmann, Community Representative, Lexington Oneida Fox Roye, Director of English Language Arts and Literacy, K–12, Boston Public Schools Andrea Gobbi, Director of Academic Programs, Shawsheen Valley Technical High School Donna Goldstein, Literacy Coach, Coelho Middle School, Attleboro Public Schools Andrea Griswold, Grade 8 English Teacher, Mohawk Trail Regional Middle and High School, Mohawk Trail/Hawlemont Regional School District Susan Hehir, Grade 3 Teacher, Forest Avenue Elementary School, Hudson Public Schools Anna Hill, Grade 6 English Language Arts Teacher, Charlton Middle School, Charlton Public Schools Sarah Hopson, K–4 Math Coach, Agawam Elementary Schools, Agawam Public Schools Nancy Johnson, 7–12 Mathematics Teacher and Department Head, Hopedale Jr.-Sr. High School, Hopedale Public Schools (retired); President, Association of Teachers of Mathematics in Massachusetts Patty Juranovits, Supervisor of Mathematics, K–12, Haverhill Public Schools Elizabeth Kadra, Grades 7 & 8 Mathematics Teacher, Miscoe Hill Middle School, Mendon-Upton Regional School District Patricia Kavanaugh, Middle School Mathematics Teacher, Manchester-Essex Middle and High School, Manchester-Essex Regional School District John Kucich, Associate Professor of English, Bridgewater State University David Langston, Professor of English/Communications, Massachusetts College of Liberal Arts Stefanie Lowe, Instructional Specialist, Sullivan Middle School, Lowell Public Schools Linda McKenna, Mathematics Curriculum Facilitator, Leominster Public Schools Eileen McQuaid, 6–12 Coordinator of English Language Arts and Social Studies, Brockton Public Schools Matthew Müller, Assistant Professor of English, Berkshire Community College Raigen O'Donohue, Grade 5 Teacher, Columbus Elementary School, Medford Public Schools Eileen Perez, Assistant Professor of Mathematics, Worcester State University Laura Raposa, Grade 5 Teacher, Russell Street Elementary School, Littleton Public Schools Danika Ripley, Literacy Coach, Dolbeare Elementary School, Wakefield Public Schools Heather Ronan, Coordinator of Math and Science, PK–5, Brockton Public Schools Fran Roy, Chief Academic Officer/Assistant Superintendent, Fall River Public Schools Melissa Ryan, Principal, Bourne Middle School, Bourne Public Schools Karyn Saxon, K–5 Curriculum Director, English Language Arts and Social Studies, Wayland Public Schools Jeffrey Strasnick, Principal, Wildwood Early Childhood Center and Woburn Street Elementary School, Wilmington Public Schools Kathleen Tobiasson, Grades 6 & 7 English Teacher, Quinn Middle School, Hudson Public Schools Brian Travers, Associate Professor of Mathematics, Salem State University Massachusetts Curriculum Framework for Mathematics 6 Nancy Verdolino, K–6 Reading Specialist and K–6 English Language Arts Curriculum Chairperson, Hopedale Public Schools; President, Massachusetts Reading Association Meghan Walsh, Grade 3 Teacher, John A. Crisafulli Elementary School, Westford Public Schools Rob Whitman, Professor of English, Bunker Hill Community College Kerry Winer, Literacy Coach, Oak Hill Middle School, Newton Public Schools Joanne Zaharis, Math Lead Teacher/Coach, Sokolowski School, Chelsea Public Schools Content Advisors English Language Arts and Literacy Bill Amorosi, ELA/Literacy Consultant Mary Ann Cappiello, Lesley University Erika Thulin Dawes, Lesley University Lorretta Holloway, Framingham State University Brad Morgan, Essex Technical High School Deborah Reck, ELA/Literacy Consultant Jane Rosenzweig, Harvard University Mathematics Richard Bisk, Worcester State University Andrew Chen, EduTron Corporation Al Cuoco, Center for Mathematics Education, EDC Sunny Kang, Bunker Hill Community College Maura Murray, Salem State University Kimberly Steadman, Brooke Charter Schools External Partner Jill Norton, Abt Associates Massachusetts Executive Office of Education Tom Moreau, Assistant Secretary of Education Massachusetts Department of Higher Education Susan Lane, Senior Advisor to the Commissioner Massachusetts Department of Elementary and Secondary Education Jeffrey Wulfson, Deputy Commissioner Heather Peske, Senior Associate Commissioner Center for Instructional Support Alexia Cribbs Lisa Keenan Ronald Noble Office of Literacy and Humanities Rachel Bradshaw, Lead Writer, ELA/Literacy David Buchanan Mary Ellen Caesar Susan Kazeroid Helene Levine Tracey Martineau Lauren McBride Susan Wheltle, Consultant Office of Science, Technology/Engineering, and Mathematics Anne Marie Condike Anne DeMallie Jacob Foster Melinda Griffin Meto Raha Ian Stith Leah Tuckman Cornelia Varoudakis, Lead Writer, Mathematics Barbara Libby, Consultant Office of Educator Development Matthew Holloway Office of English Language Acquisition and Academic Achievement Fernanda Kray Sara Niño Office of Special Education Planning and Policy Teri Williams Valentine Lauren Viviani Office of Planning, Research, and Delivery Matthew Deninger Commissioner’s Office Jass Stewart Massachusetts Curriculum Framework for Mathematics 7 Office of Student Assessment Services Mary Lou Beasley Catherine Bowler Amy Carithers Haley Freeman Simone Johnson Jennifer Malonson Elizabeth Niedzwiecki Jennifer Butler O’Toole Michol Stapel James Verdolino Daniel Wiener Massachusetts Curriculum Framework for Mathematics 8 Introduction The Origin of these Standards: 1993–2010 The Massachusetts Education Reform Act of 1993 directed the Commissioner and the Department of Elementary and Secondary Education1 to create academic standards in a variety of subject areas. Massachusetts adopted its first set of Mathematics standards in 1995 and revised them in 2000. In 2007, the Massachusetts Department of Elementary and Secondary Education (ESE) convened a team of educators to revise its 2000 Mathematics Curriculum Framework. In 2009 the Council of Chief State School Officers (CCSSO) and the National Governors Association (NGA) began a multi-state standards development project called the Common Core State Standards initiative, whereupon the two efforts merged. The pre-kindergarten to grade 12 Massachusetts Curriculum Framework for Mathematics, a new framework that included both the Common Core State Standards and unique Massachusetts standards and features, was adopted by the Boards of Elementary and Secondary Education and Early Education and Care in 2010. A similar process unfolded for the English Language Arts/Literacy Framework. Review of Mathematics and English Language Arts/Literacy Standards, 2016– 2017 In November 2015, the Massachusetts Board of Elementary and Secondary Education voted to move forward with the development of its own next generation student assessment program in mathematics and English Language Arts/Literacy. At the same time, the Board supported a plan to convene review panels comprised of Massachusetts K-12 educators and higher education faculty to review the 2010 Mathematics and English Language Arts/Literacy Curriculum Frameworks. The review panels were also asked to identify any modifications or additions to ensure that the Commonwealth’s standards match those of the most aspirational education systems in the world, thus representing a course of study that best prepares students for the 21st century. In February 2016, the Department appointed a panel of Massachusetts educators from elementary, secondary, and higher education to review the mathematics and ELA/Literacy standards and suggest improvements based on their experiences using the 2010 Framework. The Department sought comment on the standards through a public survey and from additional content advisors in mathematics and ELA/Literacy. The 2017 Massachusetts Curriculum Framework for Mathematics reflects improvements to the prior Framework that have been informed by the review panel, public comments, and the content advisors. In some cases, the standards have been edited to clarify meaning. Some have been eliminated; others added. The Glossary and Bibliography have been updated and the Department’s 2010 document titled, Making Decisions about Course Sequences and the Model Algebra I Course, has been revised and is now included in the high school section of the Framework. The intent is to present multiple pathways involving the compression or enhancement of mathematics standards to provide alternative course-taking sequences for students enabling them to be successful and prepared for various college and career pursuits, including mathematic-intensive majors and careers. 1 At the time, the agency was called the Department of Education. Massachusetts Curriculum Framework for Mathematics 9 The 2017 standards draw from the best of prior Massachusetts standards, and represent the input of hundreds of the Commonwealth’s K-12 and higher education faculty. The 2017 standards present the Commonwealth’s commitment to providing all students with a world-class education. The Mathematically Proficient Person of the Twenty-First Century The standards describe a vision of what it means to be a mathematically proficient person in this century. Students who are college and career ready in mathematics will at a minimum demonstrate the academic knowledge, skills, and practices necessary to enter into and succeed in entry-level, credit bearing courses in College Algebra, Introductory College Statistics, or technical courses. It also extends to a comparable entry-level course or a certificate or workplace training program requiring an equivalent level of mathematics. At the same time, the standards provide for a course of study that will prepare students for a science, technology, engineering, or mathematics career. For example, the level of mathematics preparation necessary to succeed in an engineering program is more ambitious than the preparation needed to succeed in an entry-level, credit-bearing mathematics course as part of a liberal arts program. The standards provide pathways for students who want to pursue a mathematics-intensive career or academic major after high school. The mathematical skills and understanding that students are expected to demonstrate have wide applicability outside the classroom or workplace. Students who meet the standards are able to identify problems, represent problems, justify conclusions, and apply mathematics to practical situations. They gain understanding of topics and issues by reviewing data and statistical information. They develop reasoning and analytical skills and make conclusions based on evidence that is essential to both private deliberation and responsible citizenship in a democratic society. They understand mathematics as a language for representing the physical world. They are able to use and apply their mathematical thinking in various contexts and across subject areas, for example, in managing personal finances, designing a robot, or presenting a logical argument and supporting it with relevant quantitative data in a debate. Students should be given opportunities to discuss math’s relevance to everyday life and their interests and potential careers with teachers, parents, business owners, and employees in a variety of fields such as computer science, architecture, construction, healthcare, engineering, retail sales, and education. From such discussions, students can learn that a computer animator uses linear algebra to determine how an object will be rotated, shifted, or altered in size. They can discover that an architect uses math to calculate the square footage of rooms and buildings, to lay out floor dimensions and to calculate the required space for areas such as parking or heating and cooling systems (kumon.org}. They can investigate how public policy analysts use statistics to monitor and predict state, national or international healthcare use, benefits, and costs. Students who meet the standards develop persistence, conceptual understanding, and procedural fluency; they develop the ability to reason, prove, justify, and communicate. They build a strong foundation for applying these understandings and skills to solve real world problems. These standards represent an ambitious pre-kindergarten to grade 12 mathematics program that ensure that students are prepared for college, careers, and civic life. A Coherent Progression of Learning: Pre-Kindergarten through Grade 12 The mathematics content standards are presented by individual grade levels in pre-kindergarten through grade 8 to provide useful specificity. The pre-kindergarten standards apply to children who are four-year-olds and younger five-year-olds. A majority of these students attend education programs in a variety of settings: Massachusetts Curriculum Framework for Mathematics 10 community-based early care and education centers; family daycare; Head Start programs; and public preschools. In this age group, the foundations of counting, quantity, comparing shapes, adding, and taking apart – and the ideas that objects can be measured – are formed during conversations, play, and with experiences with real objects and situations. At the high school level, the standards are presented in two different ways: 1. Conceptual Categories: These categories portray a coherent view of high school mathematics standards for the grade-span 9–12. These content standards are organized into six categories: Number and Quantity; Algebra; Functions; Modeling; Geometry; and Statistics and Probability. 2. Model High School Courses: Two high school pathways, the Traditional Pathway Model Courses (Algebra I, Geometry, Algebra II) and the Integrated Pathway Model Courses (Mathematics I, II, III), are presented using the content standards in the Conceptual Categories. These model courses were designed to create a smooth transition from the grade-by-grade pre-k–8 content standards to high school courses. All of the college and career ready content standards are included in appropriate locations within each set of three model high school courses. The high school content standards coded with a (+) symbol are optional and identify higher-level mathematics skills and knowledge that students should learn in order to take more advanced mathematics courses. In addition, two Advanced Model High School Courses, Advanced Quantitative Reasoning and Precalculus, are included. Students may choose to take these courses after completing either of the Model High School Pathways. See the section below titled, “Course-Taking Sequences and Pathways for All Students,” for additional advanced mathematics courses and pathways students might pursue in high school. Focus, Clarity, and Rigor In the past, mathematics standards and curricula were often criticized for being “a mile wide and an inch deep” in almost every topic taught each year. The 2010 Framework presented a new design feature for grades Pre-K– 12: a focus on three to five critical focus areas per grade or course. The 2017 Framework continues to concentrate on fewer topics in each grade to allow students to deepen and consolidate their understanding in these areas. These critical focus areas are useful in communicating with families and the community and for designing curricula, support services, and programs. A Balance of Conceptual Understanding, Procedural Fluency, and Application2 The standards strategically develop students’ mathematical understanding and skills. When students are first introduced to a mathematical concept they explore and investigate the concept by using concrete objects, visual models, drawings, or representations to build their understanding. In the early grades they develop number sense while working with numbers in many ways. They learn a variety of strategies to solve problems and use what they have learned about patterns in numbers and the properties of numbers. This serves to develop a strong understanding of number sense, decomposing and composing numbers, the relationship between addition and subtraction, and multiplication and division. In calculations, students are expected to be able to use the most efficient and accurate way to solve a problem based on their understanding and knowledge of place value and properties of numbers. Students reach fluency by building understanding of mathematical concepts – 2 Sealey, Cathy. Balance is Basic, A 21st Century View of a Balanced Mathematical Program Massachusetts Curriculum Framework for Mathematics 11 this lays a strong foundation that prepares students for more advanced math work – and by building automaticity in the recall of basic computation facts, such as addition, subtraction, multiplication, and division. As students apply their mathematical knowledge and skills to solve real-world problems, they also gain an understanding of the importance of mathematics throughout their lives. To achieve mathematical understanding, students should be actively engaged in meaningful mathematics. The content and practice standards focus on developing students in the following areas: Conceptual understanding – make sense of the math, reason about and understand math concepts and ideas Procedural fluency – know mathematical facts, compute and do the math Capacity – solve a wide range of problems in various contexts by reasoning, thinking, and applying the mathematics they have learned. Middle and High School Course-Taking Sequences and Pathways for All Students The Massachusetts High School Program of Studies (MassCore) is a recommended program of studies that includes four years of mathematics coursework in grades 9–12. MassCore describes other learning opportunities, such as Advanced Placement (AP) classes, dual enrollment, senior projects, online courses for high school or college credit, and service or work-based learning. The Framework provides an opportunity for districts to revisit and plan course sequences in middle and high school mathematics along with educators, middle and high school guidance counselors, parents, college mathematics faculty, and mathematics leaders. This Framework includes a new section titled, Making Decisions about Course Sequences and the Model Algebra I Course. This section presents several options for pathways for students ready to move at an accelerated rate. On Grade Sequence: Students who follow the Framework grade-level sequence pre-K–8 will be prepared for the Traditional or Integrated Model Course high school pathway, beginning with Algebra I or Mathematics I in grade 9. Students following this pathway will be prepared to take a fourth year advanced course in grade 12, such as the Model Precalculus Course, Model Quantitative Reasoning Course, or other advanced courses offered in their district. Model Algebra I in Grade 8 and High School Acceleration: One option for accelerating learning is to take the Model Algebra I course in grade 8. This pathway option compresses the standards for grade 7 with part of the grade 8 standards in grade 7. That allows grade 8 students to complete the grade 8 standards related to algebra and the Algebra I model high school standards in one year (grade 8). Considerations for assigning a student this Massachusetts Curriculum Framework for Mathematics 12 pathway include two factors: grade 8 standards are already rigorous and students are expected to learn the grade 8 standards in order to be prepared for the Algebra I model course. This section also presents pathways for students who are ready to accelerate their learning, beginning in grade 9. Some of these pathways lead to calculus in grade 12 while others offer a sequence to other advanced courses such as Quantitative Reasoning, Statistics, Linear Algebra, AP courses, Discrete Mathematics, or participating in a dual enrollment program. All pathways should aspire to meet the goal of ensuring that no student who graduates from a Massachusetts High School will be placed into a remedial mathematics course in a Massachusetts public college or university. Achieving this goal may require mathematics secondary educators and college faculty to work collaboratively to select or co-develop appropriate 12th grade coursework and assessments. Presenting a variety of course-taking pathways encourages students to persist in their mathematical studies. It also helps them realize that there are multiple opportunities to make course-taking decisions as they continue to advance mathematically and pursue their interests and career and college goals. Mathematics in the Context of a Well-Rounded Curriculum Strong mathematics achievement is a requisite for studying the sciences (including social sciences), technology (including computer science), and engineering. The centrality of mathematics to the pursuit of STEM careers is well documented. In addition, an effective mathematics program builds upon and develops students’ mathematical knowledge and literacy skills. Reading, writing, speaking, and listening skills are necessary elements of learning and engaging in mathematics, as well as in other content areas. The English Language Arts/Literacy standards require that instruction in reading, writing, speaking, listening, and language is a shared responsibility within the school. The pre-K–5 ELA/Literacy standards include expectations for reading, writing, speaking, listening, and language applicable to a range of subjects, including mathematics, social studies, science, the arts, and comprehensive health. Grades 6–12 ELA/Literacy standards are divided into two sections, one for ELA and the other for history/social studies, science, mathematics, and technical subjects. This division reflects the unique, time-honored role of ELA teachers in developing students’ literacy skills, while at the same time recognizing that teachers in other disciplines also contribute in this development. Consistent with emphasizing the importance of an interdisciplinary approach to literacy, the Mathematics Guiding Principles recognize that reading, writing, speaking, and listening skills are necessary elements of learning and engaging in mathematics. Mathematics students learn specialized vocabulary, terms, notations, symbols, representations, and models relevant to the grade level. Being able to read, interpret, and analyze mathematical information from a variety of sources and communicating mathematically in written and oral forms are critical skills to college and career readiness, citizenship, and informed decision-making. In essence, mathematics is a language for describing and understanding the physical world. Notably, the recent revision of the Massachusetts Curriculum Framework for Science and Technology/Engineering (2016) also highlights literacy in its Guiding Principles and Practices. To achieve a well-rounded curriculum at all grade levels, the standards in this Framework are meant to be used with the Massachusetts Curriculum Framework for English Language Arts/Literacy, the Arts, History and Social Science, Science and Technology/Engineering, Comprehensive Health and Physical Education, Foreign Languages. Massachusetts Curriculum Framework for Mathematics 13 In grades 9–12, the standards are also meant to be used with the Framework for Career and Vocational and Technical Education to achieve a truly rich and well-rounded curriculum. What the Mathematics Curriculum Framework Does and Does Not Do The standards define what all students are expected to know and be able to do, not how teachers should teach. While the standards focus on what is most essential, they do not describe all that can or should be taught. A great deal is left to the discretion of teachers and curriculum developers. No set of grade-level standards can reflect the great variety of abilities, needs, learning rates, and achievement levels in any given classroom. The standards define neither the support materials that some students may need, nor the advanced materials that others should have access to. It is also beyond the scope of the standards to define the full range of support appropriate for English learners and for students with disabilities. Still, all students must have the opportunity to learn and meet the same high standards if they are to access the knowledge and skills that will be necessary in their post-high-school lives. The standards should be read as allowing for the widest possible range of students to participate fully from the outset with appropriate accommodations to ensure maximum participation of students with special education needs. For example, for students with disabilities, reading math texts and problems should allow for the use of Braille, screen-reader technology, or other assistive devices. Writing should include the use of a scribe, computer, or speech-to-text technology that includes mathematical terms, notations, and symbols. In a similar manner, speaking and listening should be interpreted broadly to include sign language (see Appendix 1). While the mathematics described herein is critical to college, career, and civic readiness, they do not define the “whole” of readiness. Students require a wide-ranging, rigorous academic preparation and attention to such matters as social, emotional, and physical development and approaches to learning. Document Organization Eight Guiding Principles for Mathematical Programs in Massachusetts follows this introductory section. The Guiding Principles are philosophical statements that underlie the standards and resources in this Framework. Following the Guiding Principles are the eight Standards for Mathematical Practice. These standards describe the varieties of expertise that all mathematics educators at all levels should seek to develop in their students. Following the Standards for Mathematical Practice are the Standards for Mathematical Content, presented in three sections: 1. Pre-kindergarten through grade 8 content standards by grade level 2. High school content standards by conceptual category 3. High school content standards by model high school courses—includes six model courses outlined in two pathways (Traditional and Integrated) and two model advanced courses, Precalculus and Advanced Quantitative Reasoning. As described above, this Framework also includes a section in the high school content standards, entitled “Making Course Decisions about Course Sequences and the Model High School Algebra I Course.” This new section provides options for middle and high school course-taking sequences, including pathways that Massachusetts Curriculum Framework for Mathematics 14 accelerate learning in order to allow students to reach advanced courses, such as Calculus, by the end of grade 12. The supplementary resources that follow the learning standards address both engaging learners in content through the Standards for Mathematical Practice, and guidance in applying the standards for English language learners and students with disabilities. A Glossary of mathematical terms, tables, illustrations, and a list of references is also included. Guiding Principles for Mathematics Programs in Massachusetts The following principles are philosophical statements that underlie the pre-kindergarten through grade 12 mathematics standards and resources presented in this Framework. These principles should guide the design and evaluation of mathematics programs. Programs guided by these principles will prepare students for colleges, careers, and their lives as productive citizens. Guiding Principle 1 Educators must have a deep understanding of the mathematics they teach, not only to help students learn how to efficiently do mathematical calculations, but also to help them understand the fundamental principles of mathematics that are the basis for those operations. Teachers should work with their students to master these underlying concepts and the relationships between them in order to lay a foundation for higher-level mathematics, strengthen their capacity for thinking logically and rigorously, and develop an appreciation for the beauty of math. Guiding Principle 2 To help all students develop a full understanding of mathematical concepts and procedural mastery, educators should provide them with opportunities to apply their learning and solve problems using multiple methods, in collaboration with their peers and independently, and complemented by clear explanations of the underlying mathematics. Guiding Principle 3 Students should have frequent opportunities to discuss and write about various approaches to solving problems, in order to help them develop and demonstrate their mathematical knowledge, while drawing connections between alternative strategies and evaluating their comparative strengths and weaknesses. Guiding Principle 4 Students should be asked to solve a diverse set of real-world and other mathematical problems, including equations that develop and challenge their computational skills, and word problems that require students to design their own equations and mathematical models. Students learn that with persistence, they can solve challenging problems and be successful. Guiding Principle 5 A central part of an effective mathematics curriculum should be the development of a specialized mathematical vocabulary including notations and symbols, and an ability to read and understand mathematical texts and information from a variety of sources. Massachusetts Curriculum Framework for Mathematics 15 Guiding Principle 6 Assessment of student learning should be a daily part of a mathematics curriculum to ensure that students are progressing in their knowledge and skill, and to provide teachers with the information they need to adjust their instruction and differentiate their support of individual students. Guiding Principle 7 Students at all levels should have an opportunity to use appropriate technological tools to communicate ideas, provide a dynamic approach to mathematic concepts, and to search for information. Technological tools can also be used to improve efficiency of calculation and enable more sophisticated analysis, without sacrificing operational fluency and automaticity. Guiding Principle 8 Social and emotional learning can increase academic achievement, improve attitudes and behaviors, and reduce emotional distress. Students should practice self-awareness, self-management, social awareness, responsible decision-making, and relationship skills, by, for example: collaborating and learning from others and showing respect for others’ ideas; applying the mathematics they know to make responsible decisions to solve problems, engaging and persisting in solving challenging problems; and learning that with effort, they can continue to improve and be successful. Massachusetts Curriculum Framework for Mathematics 16 The Standards for Mathematical Practice The Standards for Mathematical Practice describe expertise that mathematics educators at all levels should seek to develop in their students. The Standards for Mathematical Practice describe ways in which developing student practitioners of the discipline of mathematics engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. These practices rest on two sets of important “processes and proficiencies” that have longstanding importance in mathematics education—the National Council of Teachers of Mathematics (NCTM) process standards and the strands of mathematical proficiency specified in the National Research Council’s Report “Adding It Up.” Designers of curricula, assessments, and professional development should endeavor to connect the mathematical practices to mathematical content in instruction. 1. Make sense of problems and persevere in solving them. Mathematically proficient students start by explaining to themselves the meaning of a problem and looking for entry points to its solution. They analyze givens, constraints, relationships, and goals. They make conjectures about the form and meaning of the solution and plan a solution pathway rather than simply jumping into a solution attempt. They consider analogous problems, and try special cases and simpler forms of the original problem in order to gain insight into its solution. They monitor and evaluate their progress and change course if necessary. Older students might, depending on the context of the problem, transform algebraic expressions or change the viewing window on their graphing calculator to get the information they need. Mathematically proficient students can explain correspondences between equations, verbal descriptions, tables, and graphs or draw diagrams of important features and relationships, graph data, and search for regularity or trends. Younger students might rely on using concrete objects or pictures to help conceptualize and solve a problem. Mathematically proficient students check their answers to problems using a different method, and they continually ask themselves, “Does this make sense?” They can understand others’ approaches to solving complex problems and identify correspondences among different approaches. 2. Reason abstractly and quantitatively. Mathematically proficient students make sense of the quantities and their relationships in problem situations. Students bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically, and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meanings of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. 3. Construct viable arguments and critique the reasoning of others. Mathematically proficient students understand and use stated assumptions, definitions, and previously established results in constructing arguments. They make conjectures and build a logical progression of statements to explore the truth of their conjectures. They are able to analyze situations by breaking them into cases, and can recognize and use counterexamples. They justify their conclusions, communicate them to others, and respond to the arguments of others. They reason inductively about data, making plausible arguments that take into account the context from which the data arose. Mathematically proficient students are also able to Massachusetts Curriculum Framework for Mathematics 17 compare the effectiveness of two plausible arguments, distinguish correct logic or reasoning from that which is flawed, and—if there is a flaw in an argument—explain what it is. Elementary students can construct arguments using concrete referents such as objects, drawings, diagrams, and actions. Such arguments can make sense and be correct, even though they are not generalized or made formal until later grades. Later, students learn to determine domains to which an argument applies. Students at all grades can listen or read the arguments of others, decide whether they make sense, and ask useful questions to clarify or improve the arguments. 4. Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation. In middle grades, a student might apply proportional reasoning to plan a school event or analyze a problem in the community. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity of interest depends on another. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose. 5. Use appropriate tools strategically. Mathematically proficient students consider the available tools when solving a mathematical problem. These tools might include pencil and paper, concrete models, a ruler, a protractor, a calculator, a spreadsheet, a computer algebra system, a statistical package, or dynamic geometry software. Proficient students are sufficiently familiar with tools appropriate for their grade or course to make sound decisions about when each of these tools might be helpful, recognizing both the insight to be gained and their limitations. For example, mathematically proficient high school students analyze graphs of functions and solutions generated using a graphing calculator. They detect possible errors by strategically using estimation and other mathematical knowledge. When making mathematical models, they know that technology can enable them to visualize the results of varying assumptions, explore consequences, and compare predictions with data. Mathematically proficient students at various grade levels are able to identify relevant external mathematical resources, such as digital content located on a website, and use them to pose or solve problems. They are able to use technological tools to explore and deepen their understanding of concepts. 6. Attend to precision. Mathematically proficient students try to communicate precisely to others. They try to use clear definitions in discussion with others and in communicating their own reasoning verbally and/or in writing. In problem solving they state the meaning of the symbols they choose, including using the equal sign consistently and appropriately. They are careful about specifying units of measure and labeling axes to clarify the correspondence with quantities in a problem. They calculate accurately and efficiently, expressing numerical answers with a degree of precision appropriate for the problem context. In the elementary grades, students give carefully formulated explanations to each other. By the time they reach high school, they have learned to examine claims and make explicit use of definitions. Massachusetts Curriculum Framework for Mathematics 18 7. Look for and make use of structure. Mathematically proficient students look closely to discern a pattern or structure. Young students, for example, might notice that three and seven more is the same amount as seven and three more, or they may sort a collection of shapes according to how many sides the shapes have. Later, students will see 7 8 equals the well-remembered 7 5 + 7 3, in preparation for learning about the distributive property. In the expression x2 + 9x + 14, older students can see the 14 as 2 7 and the 9 as 2 + 7. They recognize the significance of an existing line in a geometric figure and can use the strategy of drawing an auxiliary line for solving problems. They also can step back for an overview and shift perspective. They can see complicated things, such as some algebraic expressions, as single objects or as being composed of several objects. For example, they can see 5 – 3(x – y)2 as 5 minus a positive number times a square, and use that to realize that its value cannot be more than 5 for any real numbers x and y. 8. Look for and express regularity in repeated reasoning. Mathematically proficient students notice if calculations are repeated, and look both for general methods and for shortcuts. Upper elementary students might notice when dividing 25 by 11 that they are repeating the same calculations over and over again, and conclude they have a repeating decimal. By paying attention to the calculation of slope as they repeatedly check whether points are on the line through (1, 2) with slope 3, middle school students might abstract the equation (y – 2)∕(x – 1) = 3. Noticing the regularity in the way terms cancel when expanding (x – 1)(x + 1), (x – 1)(x2 + x + 1), and (x – 1)(x3 + x2 + x + 1) might lead them to the general formula for the sum of a geometric series. As they work to solve a problem, mathematically proficient students maintain oversight of the process, while attending to the details. They continually evaluate the reasonableness of their intermediate results. Massachusetts Curriculum Framework for Mathematics 19 Standards for Mathematical Content Pre-Kindergarten–Grade 8 ORGANIZATION OF THE K–8 STANDARDS Pre-Kindergarten Kindergarten Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grade 7 Grade 8 Massachusetts Curriculum Framework for Mathematics 20 Organization of the Pre-Kindergarten to Grade 8 Content Standards The pre-kindergarten through grade 8 content standards in this Framework are organized by grade level. Within each grade level, standards are grouped first by domain. Each domain is further subdivided into clusters of related standards. Standards define what students should understand and be able to do. Clusters are groups of related standards. Note that standards from different clusters may sometimes be closely related, because mathematics is a connected subject. Domains are larger groups of related standards. Standards from different domains may sometimes be closely related. The table below shows which domains are addressed at each grade level: Progression of Pre-K–8 Domains Domain Grade Level PK K 1 2 3 4 5 6 7 8 Counting and Cardinality Operations and Algebraic Thinking Number and Operations in Base Ten Number and Operations – Fractions The Number System Ratios and Proportional Relationships Expressions and Equations Functions Measurement and Data Geometry Statistics and Probability Format for Each Grade Level Each grade level is presented in the same format: An introduction and description of the critical areas for learning at that grade. An overview of that grade’s domains and clusters. The content standards for that grade (presented by domain, cluster heading, and individual standard). Massachusetts Curriculum Framework for Mathematics 21 Standards Identifiers/Coding Each standard has a unique identifier that consists of the grade level, (PK, K, 1, 2, 3, 4, 5, 6, 7, or 8), the domain code, and the standard number, as shown in the example below. The first standard above is identified as 1.G.A.1, identifying it as a grade 1 standard in the Geometry Domain, and as the first standard in that domain. Standard 1.G.A.1 is the first standard in this cluster of standards. All of the standards in this Framework use a common coding system. Massachusetts Curriculum Framework for Mathematics 22 Pre-Kindergarten Introduction The pre-kindergarten standards presented by Massachusetts are guideposts to facilitate young children’s underlying mathematical understanding. The Massachusetts pre-kindergarten standards apply to children who are in the age group of older four- and younger five-year olds. The standards—which correspond with the learning activities in the Massachusetts Guidelines for Preschool Learning Experiences (2003)—can be promoted through play and exploration activities, and embedded in almost all daily activities. They should not be limited to “math time.” In this age group, foundations of mathematical understanding are formed out of children’s experiences with real objects and materials. In preschool or pre-kindergarten, activity time should focus on two critical areas: (1) developing an understanding of whole numbers to 10, including concepts of one-to-one correspondence, counting, cardinality (the number of items in a set), and comparison; and (2) recognizing two-dimensional shapes, describing spatial relationships, and sorting and classifying objects by one or more attributes. Relatively more learning time should be devoted to developing children’s sense of number as quantity than to other mathematics topics. 1. Young children begin counting and quantifying numbers up to 10. They begin with oral counting and recognition of numerals and word names for numbers. Experience with counting naturally leads to quantification. Children count objects and learn that the sizes, shapes, positions, or purposes of objects do not affect the total number of objects in the group. One-to-one correspondence matches each element of one set to an element of another set, providing a foundation for the comparison of groups and the development of comparative language such as more than, less than, and equal to. 2. Young children explore shapes and the relationships among them. They identify the attributes of different shapes, including length, area, and weight, by using vocabulary such as long, short, tall, heavy, light, big, small, wide, narrow. They compare objects using comparative language such as longer/shorter, same length, heavier/lighter. They explore and create two- and three-dimensional shapes by using various manipulative and play materials such as Popsicle sticks, blocks, pipe cleaners, and pattern blocks. They sort, categorize, and classify objects and identify basic two-dimensional shapes using the appropriate language. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 23 Pre-Kindergarten Overview Counting and Cardinality A. Know number names and the counting sequence. B. Count to tell the number of objects. C. Compare numbers. Operations and Algebraic Thinking A. Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from. Measurement and Data A. Describe and compare measurable attributes. B. Classify objects and count the number of objects in each category. C. Work with money. Geometry A. Identify and describe shapes (squares, circles, triangles, rectangles). B. Analyze, compare, create, and compose shapes. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 24 Pre-Kindergarten Content Standards Counting and Cardinality PK.CC A. Know number names and the counting sequence. 1. Listen to and say the names of numbers in meaningful contexts. 2. Recognize and name written numerals 0–10. B. Count to tell the number of objects. 3. Understand the relationships between numerals and quantities up to 10. C. Compare numbers. 4. Count many kinds of concrete objects and actions up to ten, using one-to-one correspondence, and accurately count as many as seven things in a scattered configuration. Recognize the “one more,” “one less” patterns. 5. Use comparative language, such as more/less than, equal to, to compare and describe collections of objects. Operations and Algebraic Thinking PK.OA A. Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from. 1. Use concrete objects to model real-world addition (putting together) and subtraction (taking away) problems up through five. Measurement and Data PK.MD A. Describe and compare measurable attributes. 1. Recognize the attributes of length, area, weight, and capacity of everyday objects using appropriate vocabulary (e.g., long, short, tall, heavy, light, big, small, wide, narrow). 2. Compare the attributes of length and weight for two objects, including longer/shorter, same length; heavier/lighter, same weight; holds more/less, holds the same amount. B. Classify objects and count the number of objects in each category. 3. Sort, categorize, and classify objects by more than one attribute. C. Work with money. 4. Recognize that certain objects are coins and that dollars and coins represent money. Geometry PK.G A. Identify and describe shapes (squares, circles, triangles, rectangles). 1. Identify relative positions of objects in space, and use appropriate language (e.g., beside, inside, next to, close to, above, below, apart). 2. Identify various two-dimensional shapes using appropriate language. B. Analyze, compare, create, and compose shapes. 3. Create and represent three-dimensional shapes (ball/sphere, square box/cube, tube/cylinder) using various manipulative materials (such as Popsicle sticks, blocks, pipe cleaners, pattern blocks). Massachusetts Curriculum Framework for Mathematics 25 Kindergarten Introduction In kindergarten, instructional time should focus on two critical areas: (1) representing, relating, and operating on whole numbers, initially with sets of objects; and (2) describing shapes and space. More learning time in kindergarten should be devoted to number than to other topics. 1. Students use numbers, including written numerals, to represent quantities and to solve quantitative problems, such as counting objects in a set; counting out a given number of objects; comparing sets or numerals; and modeling simple joining and separating situations with sets of objects, or eventually with equations such as 5 + 2 = 7 and 7 – 2 = 5. (Kindergarten students should see addition and subtraction equations, and student writing of equations in kindergarten is encouraged, but it is not required.) Students choose, combine, and apply effective strategies for answering quantitative questions, including quickly recognizing the cardinalities of small sets of objects, counting and producing sets of given sizes, counting the number of objects in combined sets, or counting the number of objects that remain in a set after some are taken away. 2. Students describe their physical world using geometric ideas (e.g., shape, orientation, spatial relations) and vocabulary. They identify, name, and describe basic two-dimensional shapes, such as squares, triangles, circles, rectangles, and hexagons, presented in a variety of ways (e.g., with different sizes and orientations), as well as three-dimensional shapes such as cubes, cones, cylinders, and spheres. They use basic shapes and spatial reasoning to model objects in their environment and to construct more complex shapes. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 26 Kindergarten Overview Counting and Cardinality A. Know number names and the counting sequence. B. Count to tell the number of objects. C. Compare numbers. Operations and Algebraic Thinking A. Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from. Number and Operations in Base Ten A. Work with numbers 11–19 to gain foundations for place value. Measurement and Data A. Describe and compare measurable attributes. B. Classify objects and count the number of objects in each category. Geometry A. Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres). B. Analyze, compare, create, and compose shapes. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 27 Kindergarten Content Standards Counting and Cardinality K.CC A. Know number names and the count sequence. 1. Count to 100 by ones and by tens. 2. Count forward beginning from a given number within the known sequence (instead of having to begin at one). 3. Write numbers from 0 to 20. Represent a number of objects with a written numeral 0–20 (with 0 representing a count of no objects). B. Count to tell the number of objects. 4. Understand the relationship between numbers and quantities; connect counting to cardinality. a. When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object. b. Understand that the last number name said tells the number of objects counted. The number of objects is the same regardless of their arrangement or the order in which they were counted. c. Understand that each successive number name refers to a quantity that is one larger. Recognize the one more pattern of counting using objects. 5. Count to answer “how many?” questions about as many as 20 things arranged in a line, a rectangular array, or a circle, or as many as 10 things in a scattered configuration; given a number from 1–20, count out that many objects. C. Compare numbers. 6. Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group for groups with up to 10 objects, e.g., by using matching and counting strategies. 7. Compare two numbers between 1 and 10 presented as written numerals. Operations and Algebraic Thinking K.OA A. Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from. 1. Represent addition and subtraction with objects, fingers, mental images, drawings,3 sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations. 2. Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem. 3. Decompose numbers less than or equal to 10 into pairs in more than one way, e.g., by using objects or drawings, and record each decomposition by a drawing or equation (e.g., 5 = 2 + 3 and 5 = 4 + 1). 4. For any number from 1 to 9, find the number that makes 10 when added to the given number, e.g., by using objects or drawings, and record the answer with a drawing or equation. 5. Fluently add and subtract within 5, including zero. Number and Operations in Base Ten K.NBT A. Work with numbers 11–19 to gain foundations for place value. 1. Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (e.g., 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones. 3 Drawings need not show details, but should show the mathematics in the problem. Massachusetts Curriculum Framework for Mathematics 28 Measurement and Data K.MD A. Describe and compare measurable attributes. 1. Describe measurable attributes of objects, such as length or weight. Describe several measurable attributes of a single object. 2. Directly compare two objects with a measurable attribute in common, to see which object has “more of”/“less of” the attribute, and describe the difference. For example, directly compare the heights of two children and describe one child as taller/shorter. B. Classify objects and count the number of objects in each category. 3. Classify objects into given categories; count the numbers of objects in each category (up to and including 10) and sort the categories by count. Geometry K.G A. Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres). 1. Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. 2. Correctly name shapes regardless of their orientation or overall size. 3. Identify shapes as two-dimensional (lying in a plane, “flat”) or three-dimensional (“solid”). B. Analyze, compare, create, and compose shapes. 4. Analyze and compare two- and three-dimensional shapes, in different sizes and orientations, using informal language to describe their similarities, differences, parts (e.g., number of sides and vertices/“corners”) and other attributes (e.g., having sides of equal length). 5. Model shapes in the world by building shapes from components (e.g., sticks and clay balls) and drawing shapes. 6. Compose simple shapes to form larger shapes. For example, “Can you join these two triangles with full sides touching to make a rectangle?” Massachusetts Curriculum Framework for Mathematics 29 Grade 1 Introduction In grade 1, instructional time should focus on four critical areas: (1) developing understanding of addition, subtraction, and strategies for addition and subtraction within 20; (2) developing understanding of whole number relationships and place value, including grouping in tens and ones; (3) developing understanding of linear measurement and measuring lengths as iterating length units; and (4) reasoning about attributes of, and composing and decomposing geometric shapes. 1. Students develop strategies for adding and subtracting whole numbers based on their prior work with small numbers. They use a variety of models, including discrete objects and length-based models (e.g., cubes connected to form lengths), to model add-to, take-from, put-together, take-apart, and compare situations to develop meaning for the operations of addition and subtraction, and to develop strategies to solve arithmetic problems with these operations. Students understand connections between counting and addition and subtraction (e.g., adding two is the same as counting on two). They use properties of addition to add whole numbers and to create and use increasingly sophisticated strategies based on these properties (e.g., “making tens”) to solve addition and subtraction problems within 20. By comparing a variety of solution strategies, children build their understanding of the relationship between addition and subtraction. 2. Students develop, discuss, and use efficient, accurate, and generalizable methods to add within 100 and subtract multiples of 10. They compare whole numbers (at least to 100) to develop an understanding of and solve problems involving their relative sizes. They think of whole numbers between 10 and 100 in terms of tens and ones (especially recognizing the numbers 11 to 19 as composed of a ten and some ones). Through activities that build number sense, they understand the order of the counting numbers and their relative magnitudes. 3. Students develop an understanding of the meaning and processes of measurement, including underlying concepts such as iterating (the mental activity of building up the length of an object with equal-sized units) and the transitivity principle for indirect measurement.4 4. Students compose and decompose plane or solid figures (e.g., put two triangles together to make a quadrilateral) and build understanding of part-whole relationships as well as the properties of the original and composite shapes. As they combine shapes, they recognize them from different perspectives and orientations, describe their geometric attributes, and determine how they are alike and different, to develop the background for measurement and for initial understandings of properties such as congruence and symmetry. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. 4 Students should apply the principle of transitivity of measurement to make indirect comparisons, but they need not use this technical term. Massachusetts Curriculum Framework for Mathematics 30 Grade 1 Overview Operations and Algebraic Thinking A. Represent and solve problems involving addition and subtraction. B. Understand and apply properties of operations and the relationship between addition and subtraction. C. Add and subtract within 20. D. Work with addition and subtraction equations. Number and Operations in Base Ten A. Extend the counting sequence. B. Understand place value. C. Use place value understanding and properties of operations to add and subtract. Measurement and Data A. Measure lengths indirectly and by iterating length units. B. Tell and write time. C. Represent and interpret data. D. Work with money. Geometry A. Reason with shapes and their attributes. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 31 Grade 1 Content Standards Operations and Algebraic Thinking 1.OA A. Represent and solve problems involving addition and subtraction. 1. Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations (number sentences) with a symbol for the unknown number to represent the problem.5 2. Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem. B. Understand and apply properties of operations and the relationship between addition and subtraction. 3. Apply properties of operations to add. 6 For example, when adding numbers order does not matter. If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known (Commutative property of addition). To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12 (Associative property of addition). When adding zero to a number, the result is the same number (Identity property of zero for addition). 4. Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. C. Add and subtract within 20. 5. Relate counting to addition and subtraction (e.g., by counting on 2 to add 2). 6. Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use mental strategies such as counting on; making 10 (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a 10 (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13). D. Work with addition and subtraction equations. 7. Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2. 8. Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = – 3, 6 + 6 = . Number and Operations in Base Ten 1.NBT A. Extend the counting sequence. 1. Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral. B. Understand place value. 2. Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: a. 10 can be thought of as a bundle of ten ones—called a “ten.” b. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. 5 See Glossary, Table 1. 6 Students need not use formal terms for these properties. Massachusetts Curriculum Framework for Mathematics 32 c. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). 3. Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <. C. Use place value understanding and properties of operations to add and subtract. 4. Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings, and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten. 5. Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used. Identify arithmetic patterns of 10 more and 10 less than using strategies based on place value. 6. Subtract multiples of 10 in the range 10–90 from multiples of 10 in the range 10–90 (positive or zero differences), using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Measurement and Data 1.MD A. Measure lengths indirectly and by iterating length units. 1. Order three objects by length; compare the lengths of two objects indirectly by using a third object. 2. Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps. B. Tell and write time. 3. Tell and write time in hours and half-hours using analog and digital clocks. C. Represent and interpret data. 4. Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another. D. Work with money. 5. Identify the values of all U.S. coins and know their comparative values (e.g., a dime is of greater value than a nickel). Find equivalent values (e.g., a nickel is equivalent to five pennies). Use appropriate notation (e.g., 69¢). Use the values of coins in the solutions of problems (up to 100¢). Geometry 1.G A. Reason with shapes and their attributes. 1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size); build and draw shapes that possess defining attributes. 2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.7 3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. 7 Students do not need to learn formal names such as “right rectangular prism.” Massachusetts Curriculum Framework for Mathematics 33 Grade 2 Introduction In grade 2, instructional time should focus on four critical areas: (1) extending understanding of base-ten notation; (2) building fluency with addition and subtraction; (3) using standard units of measure; and (4) describing and analyzing shapes. 1. Students extend their understanding of the base-ten system. This includes ideas of counting in fives, tens, and multiples of hundreds, tens, and ones, as well as number relationships involving these units, including comparing. Students understand multi-digit numbers (up to 1,000) written in base-ten notation, recognizing that the digits in each place represent amounts of thousands, hundreds, tens, or ones (e.g., 853 is 8 hundreds + 5 tens + 3 ones). 2. Students use their understanding of addition to develop fluency with addition and subtraction within 100. They solve problems within 1,000 by applying their understanding of models for addition and subtraction, and they develop, discuss, and use efficient, accurate, and generalizable methods to compute sums and differences of whole numbers in base-ten notation, using their understanding of place value and the properties of operations. They select and accurately apply methods that are appropriate for the context and the numbers involved to mentally calculate sums and differences for numbers with only tens or only hundreds. 3. Students recognize the need for standard units of measure (centimeter and inch) and they use rulers and other measurement tools with the understanding that linear measure involves an iteration of units. They recognize that the smaller the unit, the more iterations they need to cover a given length. 4. Students describe and analyze shapes by examining their sides and angles. Students investigate, describe, and reason about decomposing and combining shapes to make other shapes. Through building, drawing, and analyzing two- and three-dimensional shapes, students develop a foundation for understanding area, volume, congruence, similarity, and symmetry in later grades. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 34 Grade 2 Overview Operations and Algebraic Thinking A. Represent and solve problems involving addition and subtraction. B. Add and subtract within 20. C. Work with equal groups of objects to gain foundations for multiplication. Number and Operations in Base Ten A. Understand place value. B. Use place value understanding and properties of operations to add and subtract. Measurement and Data A. Measure lengths indirectly and by iterating length units. B. Relate addition and subtraction to length. C. Work with time and money. D. Represent and interpret data. Geometry A. Reason with shapes and their attributes. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 35 Grade 2 Content Standards Operations and Algebraic Thinking 2.OA A. Represent and solve problems involving addition and subtraction. 1. Use addition and subtraction within 100 to solve one- and two-step word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.8 B. Add and subtract within 20. 2. Fluently add and subtract within 20 using mental strategies.9 By end of grade 2, know from memory all sums of two single-digit numbers and related differences. For example, the sum 6 + 5 = 11 has related differences of 11 – 5 = 6 and 11 – 6 = 5. C. Work with equal groups of objects to gain foundations for multiplication. 3. Determine whether a group of objects (up to 20) has an odd or even number of members, e.g., by pairing objects or counting them by 2s; write an equation to express an even number as a sum of two equal addends. 4. Use addition to find the total number of objects arranged in rectangular arrays with up to five rows and up to five columns; write an equation to express the total as a sum of equal addends. Number and Operations in Base Ten 2.NBT A. Understand place value. 1. Understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones; e.g., 706 equals 7 hundreds, 0 tens, and 6 ones. Understand the following as special cases: a. 100 can be thought of as a bundle of ten tens—called a “hundred.” b. The numbers 100, 200, 300, 400, 500, 600, 700, 800, 900 refer to one, two, three, four, five, six, seven, eight, or nine hundreds (and 0 tens and 0 ones). 2. Count within 1,000; skip-count by 5s, 10s, and 100s. Identify patterns in skip counting starting at any number. 3. Read and write numbers to 1,000 using base-ten numerals, number names, and expanded form. 4. Compare two three-digit numbers based on meanings of the hundreds, tens, and ones digits, using >, =, and < symbols to record the results of comparisons. B. Use place value understanding and properties of operations to add and subtract. 5. Fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the relationship between addition and subtraction. 6. Add up to four two-digit numbers using strategies based on place value and properties of operations. 7. Add and subtract within 1,000, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method. Understand that in adding or subtracting three-digit numbers, one adds or subtracts hundreds and hundreds, tens and tens, ones and ones; and sometimes it is necessary to compose or decompose tens or hundreds. 8. Mentally add 10 or 100 to a given number 100–900, and mentally subtract 10 or 100 from a given number 100–900. 9. Explain why addition and subtraction strategies work, using place value and the properties of operations.10 8 See Glossary, Table 1. 9 Strategies such as counting on; making tens; decomposing a number; using the relationship between addition and subtraction; and creating equivalent but easier or known sums. 10 Explanations may be supported by drawings or objects. Massachusetts Curriculum Framework for Mathematics 36 Measurement and Data 2.MD A. Measure and estimate lengths in standard units. 1. Measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes. 2. Measure the length of an object twice, using length units of different lengths for the two measurements; describe how the two measurements relate to the size of the unit chosen. 3. Estimate lengths using units of inches, feet, centimeters, and meters. 4. Measure to determine how much longer one object is than another, expressing the length difference in terms of a standard length unit. B. Relate addition and subtraction to length. 5. Use addition and subtraction within 100 to solve word problems involving lengths that are given in the same units, e.g., by using drawings (such as drawings of rulers) and equations with a symbol for the unknown number to represent the problem. 6. Represent whole numbers as lengths from 0 on a number line diagram with equally spaced points corresponding to the numbers 0, 1, 2, …, and represent whole-number sums and differences within 100 on a number line diagram. C. Work with time and money. 7. Tell and write time from analog and digital clocks to the nearest five minutes, using a.m. and p.m. a. Know the relationships of time, including seconds in a minute, minutes in an hour, hours in a day, days in a week; days in a month and a year and approximate number of weeks in a month and weeks in a year. 8. Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies (up to $10), using $ and ¢ symbols appropriately and whole dollar amounts. For example, if you have 2 dimes and 3 pennies, how many cents do you have? If you have $3 and 4 quarters, how many dollars or cents do you have? (Students are not expected to use decimal notation.) D. Represent and interpret data. 9. Generate measurement data by measuring lengths of several objects to the nearest whole unit, or by making repeated measurements of the same object. Organize and record the data on a line plot (dot plot) where the horizontal scale is marked off in whole-number units. 10. Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems,11 using information presented in a bar graph. Geometry 2.G A. Reason with shapes and their attributes. 1. Recognize and draw shapes having specified attributes, such as a given number of angles or a given number of equal faces.12 Identify triangles, squares, rectangles, rhombuses, trapezoids, pentagons, hexagons, and cubes. 2. Partition a rectangle into rows and columns of same-size squares and count to find the total number of them. 3. Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape. 11 See Glossary, Table 1. 12 Sizes are compared directly or visually, not compared by measuring. Massachusetts Curriculum Framework for Mathematics 37 Grade 3 Introduction In grade 3, instructional time should focus on four critical areas: (1) developing understanding of multiplication and division and strategies for multiplication and division within 100; (2) developing understanding of fractions, especially unit fractions (fractions with numerator 1); (3) developing understanding of the structure of rectangular arrays and of area; and (4) describing and analyzing two-dimensional shapes. 1. Students develop an understanding of the meanings of multiplication and division of whole numbers through activities and problems involving equal-sized groups, arrays, and area models; multiplication is finding an unknown product, and division is finding an unknown factor in these situations. For equal-sized group situations, division can require finding the unknown number of groups or the unknown group size. Students use properties of operations to calculate products of whole numbers, using increasingly sophisticated strategies based on these properties to solve multiplication and division problems involving single-digit factors. By comparing a variety of solution strategies, students learn the relationship between multiplication and division. 2. Students develop an understanding of fractions, beginning with unit fractions. Students view fractions in general as being built out of unit fractions, and they use fractions along with visual fraction models to represent parts of a whole. Students understand that the size of a fractional part is relative to the size of the whole. For example, 1∕2 of the paint in a small bucket could be less paint than 1∕3 of the paint in a larger bucket, but 1∕3 of a ribbon is longer than 1∕5 of the same ribbon because when the ribbon is divided into 3 equal parts, the parts are longer than when the ribbon is divided into 5 equal parts. Students are able to use fractions to represent numbers equal to, less than, and greater than one. They solve problems that involve comparing fractions by using visual fraction models and strategies based on noticing equal numerators or denominators. 3. Students recognize area as an attribute of two-dimensional regions. They measure the area of a shape by finding the total number of same-size units of area required to cover the shape without gaps or overlaps; a square with sides of unit length being the standard unit for measuring area. Students understand that rectangular arrays can be decomposed into identical rows or into identical columns. By decomposing rectangles into rectangular arrays of squares, students connect area to multiplication, and justify using multiplication to determine the area of a rectangle. 4. Students describe, analyze, and compare properties of two-dimensional shapes. They compare and classify shapes by their sides and angles, and connect these with definitions of shapes. Students also relate their fraction work to geometry by expressing the area of part of a shape as a unit fraction of the whole. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 38 Grade 3 Overview Operations and Algebraic Thinking A. Represent and solve problems involving multiplication and division. B. Understand properties of multiplication and the relationship between multiplication and division. C. Multiply and divide within 100. D. Solve problems involving the four operations, and identify and explain patterns in arithmetic. Number and Operations in Base Ten A. Use place value understanding and properties of operations to perform multi-digit arithmetic. Number and Operations—Fractions A. Develop understanding of fractions as numbers. Measurement and Data A. Solve problems involving measurement and estimation of intervals of time, liquid volumes, and masses of objects. B. Represent and interpret data. C. Geometric measurement: understand concepts of area and relate area to multiplication and to addition. D. Geometric measurement: recognize perimeter as an attribute of plane figures and distinguish between linear and area measures. Geometry A. Reason with shapes and their attributes. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 39 Grade 3 Content Standards Operations and Algebraic Thinking 3.OA A. Represent and solve problems involving multiplication and division. 1. Interpret products of whole numbers, e.g., interpret 5 7 as the total number of objects in five groups of seven objects each. For example, describe a context in which a total number of objects can be expressed as 5 7. 2. Interpret whole-number quotients of whole numbers, e.g., interpret 56 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. For example, describe a context in which a number of shares or a number of groups can be expressed as 56 8. 3. Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.13 4. Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 ? = 48, 5 = 3, 6 6 = ?. B. Understand properties of multiplication and the relationship between multiplication and division. 5. Apply properties of operations to multiply.14 For example, when multiplying numbers order does not matter. If 6 4 = 24 is known, then 4 6 = 24 is also known (Commutative property of multiplication); The product 3 5 2 can be found by 3 5 = 15 then 15 2 = 30, or by 5 2 = 10 then 3 10 = 30 (Associative property of multiplication); When multiplying two numbers either number can be decomposed and multiplied; one can find 8 x 7 by knowing that 7 = 5 + 2 and that 8 5 = 40 and 8 2 = 16, resulting in 8 (5 + 2) = (8 5) + (8 2) = 40 + 16 = 56 (Distributive property); When a number is multiplied by 1 the result is the same number (Identity property of 1 for multiplication). 6. Understand division as an unknown-factor problem. For example, find 32 8 by finding the number that makes 32 when multiplied by 8. C. Multiply and divide within 100. 7. Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 5 = 40, one knows 40 5 = 8) or properties of operations. By the end of grade 3, know from memory all products of two single-digit numbers and related division facts. For example, the product 4 x 7 = 28 has related division facts 28 ÷ 7 = 4 and 28 ÷ 4 = 7. D. Solve problems involving the four operations, and identify and explain patterns in arithmetic. 8. Solve two-step word problems using the four operations for problems posed with whole numbers and having whole number answers. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies, including rounding.15 9. Identify arithmetic patterns (including patterns in the addition table or multiplication table) and explain them using properties of operations. For example, observe that 4 times a number is always even, and explain why 4 times a number can be decomposed into two equal addends. 13 See Glossary, Table 2. 14 Students need not use formal terms for these properties. Students are not expected to use distributive notation. 15 Students should know how to perform operations in the conventional order when there are no parentheses to specify a particular order (Order of Operations). Massachusetts Curriculum Framework for Mathematics 40 Number and Operations in Base Ten 3.NBT A. Use place value understanding and properties of operations to perform multi-digit arithmetic. 16 1. Use place value understanding to round whole numbers to the nearest 10 or 100. 2. Fluently add and subtract within 1,000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction. 3. Multiply one-digit whole numbers by multiples of 10 in the range 10–90 (e.g., 9 80, 5 60) using strategies based on place value and properties of operations. Number and Operations—Fractions 3.NF A. Develop understanding of fractions as numbers for fractions with denominators 2, 3, 4, 6, and 8. 1. Understand a fraction 1∕b as the quantity formed by 1 part when a whole (a single unit) is partitioned into b equal parts; understand a fraction a∕b as the quantity formed by a parts of size 1∕b. 2. Understand a fraction as a number on the number line; represent fractions on a number line diagram. a. Represent a unit fraction, 1∕b, on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1∕b and that the fraction 1∕b is located 1∕b of a whole unit from 0 on the number line. b. Represent a fraction a∕b on a number line diagram by marking off a lengths 1∕b from 0. Recognize that the resulting interval has size a∕b and that its endpoint locates the number a∕b on the number line. 3. Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size. a. Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line. b. Recognize and generate simple equivalent fractions, e.g., 1∕2 = 2∕4, 4∕6 = 2∕3. Explain why the fractions are equivalent, e.g., by using a visual fraction model. c. Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. For example, express 3 in the form 3 = 3∕1; recognize that 6∕1 = 6; locate 4∕4 and 1 at the same point of a number line diagram. d. Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. Measurement and Data 3.MD A. Solve problems involving measurement and estimation of intervals of time, liquid volumes, and masses of objects. 1. Tell and write time to the nearest minute and measure time intervals in minutes. Solve word problems involving addition and subtraction of time intervals in minutes, e.g., by representing the problem on a number line diagram. 2. Measure and estimate liquid volumes and masses of objects using standard metric units of grams (g), kilograms (kg), and liters (l).17 Add, subtract, multiply, or divide to solve one-step word problems involving masses or volumes that are given in the same metric units, e.g., by using drawings (such as a beaker with a measurement scale) to represent the problem.18 16 A range of algorithms may be used. 17 Excludes compound units such as cm3 and finding the geometric volume of a container. 18 Excludes multiplicative comparison problems (problems involving notions of “times as much”; see Glossary, Table 2). Massachusetts Curriculum Framework for Mathematics 41 B. Represent and interpret data. 3. Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step “how many more” and “how many less” problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent five pets. 4. Generate measurement data by measuring lengths of objects using rulers marked with halves and fourths of an inch. Record and show the data by making a line plot (dot plot), where the horizontal scale is marked off in appropriate units—whole numbers, halves, or fourths. (See Glossary for example.) C. Geometric measurement: understand concepts of area and relate area to multiplication and to addition. 5. Recognize area as an attribute of plane figures and understand concepts of area measurement. a. A square with side length of one unit, called “a unit square,” is said to have “one square unit” of area, and can be used to measure area. b. A plane figure which can be covered without gaps or overlaps by n unit squares is said to have an area of n square units. 6. Measure areas by counting unit squares (square cm, square m, square in., square ft., and non-standard units). 7. Relate area to the operations of multiplication and addition. a. Find the area of a rectangle with whole-number side lengths by tiling it, and show that the area is the same as would be found by multiplying the side lengths. b. Multiply side lengths to find areas of rectangles with whole-number side lengths in the context of solving real-world and mathematical problems, and represent whole-number products as rectangular areas in mathematical reasoning. c. Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a b and a c. Use area models to represent the distributive property in mathematical reasoning. d. Recognize area as additive. Find areas of rectilinear figures by decomposing them into non-overlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real-world problems. D. Geometric measurement: recognize perimeter as an attribute of plane figures and distinguish between linear and area measures. 8. Solve real-world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters. Geometry 3.G A. Reason with shapes and their attributes. 1. Understand that shapes in different categories (e.g., rhombuses, rectangles, and others) may share attributes (e.g., having four sides), and that the shared attributes can define a larger category (e.g., quadrilaterals). Compare and classify shapes by their sides and angles (right angle/non-right angle). Recognize rhombuses, rectangles, squares, and trapezoids as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories. 2. Partition shapes into parts with equal areas. Express the area of each part as a unit fraction of the whole. For example, partition a shape into four parts with equal areas and describe the area of each part as ¼ of the area of the shape. Massachusetts Curriculum Framework for Mathematics 42 Grade 4 Introduction In grade 4, instructional time should focus on three critical areas: (1) developing understanding and fluency with multi-digit multiplication, and developing understanding of dividing to find quotients involving multi-digit dividends; (2) developing an understanding of fraction equivalence, addition and subtraction of fractions with like denominators, and multiplication of fractions by whole numbers; (3) and understanding that geometric figures can be analyzed and classified based on their properties, such as having parallel sides, perpendicular sides, particular angle measures, and symmetry. 1. Students generalize their understanding of place value to 1,000,000, understanding the relative sizes of numbers in each place. They apply their understanding of models for multiplication (equal-sized groups, arrays, area models), place value, and properties of operations, in particular the distributive property, as they develop, discuss, and use efficient, accurate, and generalizable methods to compute products of multi-digit whole numbers. Depending on the numbers and the context, they select and accurately apply appropriate methods to estimate or mentally calculate products. They develop fluency with efficient procedures for multiplying whole numbers; understand and explain why the procedures work based on place value and properties of operations; and use them to solve problems. Students apply their understanding of models for division, place value, properties of operations, and the relationship of division to multiplication as they develop, discuss, and use efficient, accurate, and generalizable procedures to find quotients involving multi-digit dividends. They select and accurately apply appropriate methods to estimate and mentally calculate quotients, and interpret remainders based upon the context. 2. Students develop understanding of fraction equivalence and operations with fractions. They recognize that two different fractions can be equal (e.g., 15∕9 = 5∕3), and they develop methods for generating and recognizing equivalent fractions. Students extend previous understandings about how fractions are built from unit fractions, composing fractions from unit fractions, decomposing fractions into unit fractions, and using the meaning of fractions and the meaning of multiplication to multiply a fraction by a whole number. 3. Students describe, analyze, compare, and classify two-dimensional shapes. Through building, drawing, and analyzing two-dimensional shapes, students deepen their understanding of properties of two-dimensional objects and the use of them to solve problems involving symmetry. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 43 Grade 4 Overview Operations and Algebraic Thinking A. Use the four operations with whole numbers to solve problems. B. Gain familiarity with factors and multiples. C. Generate and analyze patterns. Number and Operations in Base Ten A. Generalize place value understanding for multi-digit whole numbers less than or equal to 1,000,000. B. Use place value understanding and properties of operations to perform multi-digit arithmetic on whole numbers less than or equal to 1,000,000. Number and Operations—Fractions A. Extend understanding of fraction equivalence and ordering for fractions ordering for fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. B. Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers for fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. C. Understand decimal notation for fractions, and compare decimal fractions. Measurement and Data A. Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit. B. Represent and interpret data. C. Geometric measurement: Understand concepts of angle and measure angles. Geometry A. Draw and identify lines and angles, and classify shapes by properties of their lines and angles. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 44 Grade 4 Content Standards Operations and Algebraic Thinking 4.OA A. Use the four operations with whole numbers to solve problems. 1. Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations. 2. Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison.19 3. Solve multi-step word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. a. Know multiplication facts and related division facts through 12 x 12. B. Gain familiarity with factors and multiples. 4. Find all factor pairs for a whole number in the range 1–100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1–100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1–100 is prime or composite. C. Generate and analyze patterns. 5. Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself. For example, given the rule “Add 3” and the starting number 1, generate terms in the resulting sequence and observe that the terms appear to alternate between odd and even numbers. Explain informally why the numbers will continue to alternate in this way. Number and Operations in Base Ten 4.NBT A. Generalize place value understanding for multi-digit whole numbers less than or equal to 1,000,000. 1. Recognize that in a multi-digit whole number, a digit in any place represents 10 times as much as it represents in the place to its right. For example, recognize that 700 70 = 10 by applying concepts of place value and division. 2. Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. 3. Use place value understanding to round multi-digit whole numbers to any place. B. Use place value understanding and properties of operations to perform multi-digit arithmetic on whole numbers less than or equal to 1,000,000. 4. Fluently add and subtract multi-digit whole numbers using the standard algorithm. 5. Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 6. Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 19 See Glossary, Table 2. Massachusetts Curriculum Framework for Mathematics 45 Number and Operations—Fractions 4.NF A. Extend understanding of fraction equivalence and ordering for fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. 1. Explain why a fraction a∕b is equivalent to a fraction (n a)∕(n b) by using visual fraction models, with attention to how the numbers and sizes of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions, including fractions greater than 1. 2. Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1∕2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. B. Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers for fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. 3. Understand a fraction a∕b with a > 1 as a sum of fractions 1∕b. a. Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. (The whole can be a set of objects.) b. Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using drawings or visual fraction models. Examples: 3∕8 = 1∕8 + 1∕8 + 1∕8 ; 3∕8 = 1∕8 + 2∕8 ; 2 1∕8 = 1 + 1 + 1∕8 = 8∕8 + 8∕8 + 1∕8. c. Add and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction. d. Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using drawings or visual fraction models and equations to represent the problem. 4. Apply and extend previous understandings of multiplication to multiply a fraction by a whole number. a. Understand a fraction a∕b as a multiple of 1∕b. For example, use a visual fraction model to represent 5∕4 as the product 5 (1∕4), recording the conclusion by the equation 5∕4 = 5 (1∕4). b. Understand a multiple of a∕b as a multiple of 1∕b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 (2∕5) as 6 (1∕5), recognizing this product as 6∕5. (In general, n (a∕b) = (n a)∕b.) c. Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat 3∕8 of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie? C. Understand decimal notation for fractions, and compare decimal fractions. 5. Express a fraction with denominator 10 as an equivalent fraction with denominator 100, and use this technique to add two fractions with respective denominators 10 and 100.20 For example, express 3∕10 as 30∕100, and add 3∕10 + 4∕100 = 34∕100. 6. Use decimal notation to represent fractions with denominators 10 or 100. For example, rewrite 0.62 as 62∕100; describe a length as 0.62 meters; locate 0.62 on a number line diagram. 20 Students who can generate equivalent fractions can develop strategies for adding fractions with unlike denominators in general. But addition and subtraction with unlike denominators in general is not a requirement at this grade. Massachusetts Curriculum Framework for Mathematics 46 7. Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual model. Measurement and Data 4.MD A. Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit. 1. Know relative sizes of measurement units within one system of units, including km, m, cm; kg, g; lb, oz.; l, ml; hr, min, sec. Within a single system of measurement, express measurements in a larger unit in terms of a smaller unit. Record measurement equivalents in a two-column table. For example, know that 1 ft. is 12 times as long as 1 in. Express the length of a 4 ft. snake as 48 in. Generate a conversion table for feet and inches listing the number pairs (1, 12), (2, 24), (3, 36), … 2. Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale. 3. Apply the area and perimeter formulas for rectangles in real-world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor. (Note: When finding areas of rectangular regions answers will be in square units. For example, the area of a 1 cm x 1 cm rectangular region will be 1 square centimeter (1 cm2, students are not expected to use this notation.) When finding the perimeter of a rectangular region answers will be in linear units. For example, the perimeter of the region is: 1cm + 1cm + 1cm +1cm = 4 cm or 2(1cm) + 2(1cm) = 4 cm). B. Represent and interpret data. 4. Make a line plot (dot plot) representation to display a data set of measurements in fractions of a unit (1∕2, 1∕4, 1∕8). Solve problems involving addition and subtraction of fractions by using information presented in line plots (dot plots). For example, from a line plot (dot plot) find and interpret the difference in length between the longest and shortest specimens in an insect collection. C. Geometric measurement: Understand concepts of angle and measure angles. 5. Recognize angles as geometric shapes that are formed wherever two rays share a common endpoint, and understand concepts of angle measurement: a. An angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. An angle that turns through 1∕360 of a circle is called a “one-degree angle,” and can be used to measure angles. b. An angle that turns through n one-degree angles is said to have an angle measure of n degrees. 6. Measure angles in whole-number degrees using a protractor. Sketch angles of specified measure. 7. Recognize angle measure as additive. When an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Solve addition and subtraction problems to find unknown angles on a diagram in real-world and mathematical problems, e.g., by using an equation with a symbol for the unknown angle measure. Geometry 4.G A. Draw and identify lines and angles, and classify shapes by properties of their lines and angles. 1. Draw points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines. Identify these in two-dimensional figures. Massachusetts Curriculum Framework for Mathematics 47 2. Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category, and identify right triangles. 3. Recognize a line of symmetry for a two-dimensional figure as a line across the figure such that the figure can be folded along the line into matching parts. Identify line-symmetric figures and draw lines of symmetry. Massachusetts Curriculum Framework for Mathematics 48 Grade 5 Introduction In grade 5, instructional time should focus on four critical areas: (1) developing fluency with addition and subtraction of fractions, and developing understanding of the multiplication of fractions and of division of fractions in limited cases (unit fractions divided by whole numbers and whole numbers divided by unit fractions); (2) extending division to 2-digit divisors, integrating decimal fractions into the place value system and developing understanding of operations with decimals to hundredths, and developing fluency with whole number and decimal operations; and (3) developing understanding of measurement systems and determining volumes to solve problems; and (4) solving problems using the coordinate plane 1. Students apply their understanding of fractions and fraction models to represent the addition and subtraction of fractions with unlike denominators as equivalent calculations with like denominators. They develop fluency in calculating sums and differences of fractions, and make reasonable estimates of them. Students also use the meaning of fractions, of multiplication and division, and the relationship between multiplication and division to understand and explain why the procedures for multiplying and dividing fractions make sense. (Note: this is limited to the case of dividing unit fractions by whole numbers and whole numbers by unit fractions.) 2. Students develop understanding of why division procedures work based on the meaning of base-ten numerals and properties of operations. They finalize fluency with multi-digit multiplication, and division. They apply their understandings of models for decimals, decimal notation, and properties of operations to add and subtract decimals to hundredths. They develop fluency in these computations and make reasonable estimates of their results. Students use the relationship between decimals and fractions, as well as the relationship between finite decimals and whole numbers (i.e., a finite decimal multiplied by an appropriate power of 10 is a whole number), to understand and explain why the procedures for multiplying and dividing finite decimals make sense. They compute products and quotients of decimals to hundredths efficiently and accurately. 3. Students convert among different-sized measurement units within a given measurement system allowing for efficient and accurate problem solving with multi-step real-world problems as they progress in their understanding of scientific concepts and calculations. Students recognize volume as an attribute of three-dimensional space. They understand that volume can be measured by finding the total number of same-size units of volume required to fill the space without gaps or overlaps. They understand that a 1-unit by 1-unit by 1-unit cube is the standard unit for measuring volume. They select appropriate units, strategies, and tools for solving problems that involve estimating and measuring volume. They decompose three-dimensional shapes and find volumes of right rectangular prisms by viewing them as decomposed into layers of arrays of cubes. They measure necessary attributes of shapes in order to determine volumes to solve real-world and mathematical problems. 4. Students learn to interpret the components of a rectangular coordinate system as lines and understand the precision of location that these lines require. Students learn to apply their knowledge of number and length to the order and distance relationships of a coordinate grid and to coordinate this across two dimensions. Students solve mathematical and real world problems using coordinates. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 49 Grade 5 Overview Operations and Algebraic Thinking A. Write and interpret numerical expressions. B. Analyze patterns and relationships. Number and Operations in Base Ten A. Understand the place value system. B. Perform operations with multi-digit whole numbers and with decimals to hundredths. Number and Operations—Fractions A. Use equivalent fractions as a strategy to add and subtract fractions. B. Apply and extend previous understandings of multiplication and division to multiply and divide fractions. Measurement and Data A. Convert like measurement units within a given measurement system. B. Represent and interpret data. C. Geometric measurement: Understand concepts of volume and relate volume to multiplication and to addition. Geometry A. Graph points on the coordinate plane to solve real-world and mathematical problems. B. Classify two-dimensional figures into categories based on their properties. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 50 Grade 5 Content Standards Operations and Algebraic Thinking 5.OA A. Write and interpret numerical expressions. 1. Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols, e.g.,(6 x 30) + (6 x 1∕2). 2. Write simple expressions that record calculations with numbers, and interpret numerical expressions without evaluating them. For example, express the calculation “Add 8 and 7, then multiply by 2” as 2 (8 + 7). Recognize that 3 (18932 + 921) is three times as large as 18932 + 921, without having to calculate the indicated sum or product. B. Analyze patterns and relationships. 3. Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane. For example, given the rule “Add 3” and the starting number 0, and given the rule “Add 6” and the starting number 0, generate terms in the resulting sequences, and observe that the terms in one sequence are twice the corresponding terms in the other sequence. Explain informally why this is so. Number and Operations in Base Ten 5.NBT A. Understand the place value system. 1. Recognize that in a multi-digit number, including decimals, a digit in any place represents 10 times as much as it represents in the place to its right and 1∕10 of what it represents in the place to its left. 2. Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10. 3. Read, write, and compare decimals to thousandths. a. Read and write decimals to thousandths using base-ten numerals, number names, and expanded form, e.g., 347.392 = 3 100 + 4 10 + 7 1 + 3 (1∕10) + 9 (1∕100) + 2 (1∕1000). b. Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. 4. Use place value understanding to round decimals to any place. B. Perform operations with multi-digit whole numbers and with decimals to hundredths. 5. Fluently multiply multi-digit whole numbers. (Include two-digit x four-digit numbers and, three-digit x three-digit numbers) using the standard algorithm. 6. Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 7. Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction and between multiplication and division; relate the strategy to a written method and explain the reasoning used. Number and Operations—Fractions 5.NF A. Use equivalent fractions as a strategy to add and subtract fractions. 1. Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. Massachusetts Curriculum Framework for Mathematics 51 For example, 2∕3 + 5∕4 = 8∕12 + 15∕12 = 23∕12. (In general, a∕b + c∕d = (ad + bc)∕bd.) 2. Solve word problems involving addition and subtraction of fractions referring to the same whole (the whole can be a set of objects), including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2∕5 + 1∕2 = 3∕7, by observing that 3∕7 < 1∕2. B. Apply and extend previous understandings of multiplication and division to multiply and divide fractions. 3. Interpret a fraction as division of the numerator by the denominator (a∕b = a b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret ¾ as the result of dividing 3 by 4, noting that ¾ multiplied by 4 equals 3, and that when three wholes are shared equally among four people each person has a share of size ¾. If nine people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie? 4. Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction. a. Interpret the product (a∕b) q as a parts of a partition of q into b equal parts; equivalently, as the result of a sequence of operations a q b. For example, use a visual fraction model and/or area model to show (2∕3) 4 = 8∕3, and create a story context for this equation. Do the same with (2∕3) (4∕5) = 8∕15 . (In general, (a∕b) (c∕d) = ac∕bd.) b. Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths. Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas. 5. Interpret multiplication as scaling (resizing), by: a. Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication. For example, without multiplying tell which number is greater: 225 or ¾ x 225; 11∕50 or 3∕2 x 11∕50? b. Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b = (n a)/(n b) to the effect of multiplying a∕b by 1. 6. Solve real-world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem. 7. Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.21 a. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1∕3) 4, and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that (1∕3) 4 = 1∕12 because (1∕12) 4 = 1∕3. b. Interpret division of a whole number by a unit fraction, and compute such quotients. For example, create a story context for 4 (1∕5), and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that 4 (1∕5) = 20 because 20 (1∕5) = 4. 21 Students able to multiply fractions in general can develop strategies to divide fractions in general, by reasoning about the relationship between multiplication and division. But division of a fraction by a fraction is not a requirement at this grade. Massachusetts Curriculum Framework for Mathematics 52 c. Solve real-world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions, e.g., by using visual fraction models and equations to represent the problem. For example, how much chocolate will each person get if three people share ½ lb. of chocolate equally? How many 1∕3-cup servings are in two cups of raisins? Measurement and Data 5.MD A. Convert like measurement units within a given measurement system. 1. Convert among different-sized standard measurement units within a given measurement system (e.g., convert 5 cm to 0.05 m), and use these conversions in solving multi-step, real-world problems. B. Represent and interpret data. 2. Make a line plot (dot plot) to display a data set of measurements in fractions of a unit. Use operations on fractions for this grade to solve problems involving information presented in line plot (dot plot). For example, given different measurements of liquid in identical beakers, find the amount of liquid each beaker would contain if the total amount in all the beakers were redistributed equally. C. Geometric measurement: Understand concepts of volume and relate volume to multiplication and to addition. 3. Recognize volume as an attribute of solid figures and understand concepts of volume measurement. a. A cube with side length 1 unit, called a “unit cube,” is said to have “one cubic unit” of volume, and can be used to measure volume. b. A solid figure which can be packed without gaps or overlaps using n unit cubes is said to have a volume of n cubic units. 4. Measure volumes by counting unit cubes, using cubic cm, cubic in., cubic ft., and non-standard units. 5. Relate volume to the operations of multiplication and addition and solve real-world and mathematical problems involving volume. a. Find the volume of a right rectangular prism with whole-number edge lengths by packing it with unit cubes, and show that the volume is the same as would be found by multiplying the edge lengths, equivalently by multiplying the height by the area of the base. Represent threefold whole-number products as volumes, e.g., to represent the associative property of multiplication. b. Apply the formula V = l w h and V = B h (where B stands for the area of the base) for rectangular prisms to find volumes of right rectangular prisms with whole-number edge lengths in the context of solving real-world and mathematical problems. c. Recognize volume as additive. Find volumes of solid figures composed of two non-overlapping right rectangular prisms by adding the volumes of the non-overlapping parts, applying this technique to solve real-world problems. Geometry 5.G A. Graph points on the coordinate plane to solve real-world and mathematical problems. 1. Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the zero on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate). 2. Represent real-world and mathematical problems by graphing points in the first quadrant of the coordinate plane, and interpret coordinate values of points in the context of the situation. Massachusetts Curriculum Framework for Mathematics 53 B. Classify two-dimensional figures into categories based on their properties. 3. Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category. For example, all rectangles have four right angles and squares are rectangles, so all squares have four right angles. 4. Classify two-dimensional figures in a hierarchy based on properties. For example, all rectangles are parallelograms because they are all quadrilaterals with two pairs of opposite sides parallel. Massachusetts Curriculum Framework for Mathematics 54 Grade 6 Introduction In grade 6, instructional time should focus on five critical areas: (1) connecting ratio and rate to whole number multiplication and division, and using concepts of ratio and rate to solve problems; (2) completing understanding of division of fractions and extending the notion of number to the system of rational numbers, which includes negative numbers; (3) writing, interpreting, and using expressions and equations; (4) developing understanding of statistical thinking; and (5) reasoning about geometric shapes and their measurements. 1. Students use reasoning about multiplication and division to solve ratio and rate problems about quantities. By viewing equivalent ratios and rates as deriving from, and extending, pairs of rows (or columns) in the multiplication table, and by analyzing simple drawings that indicate the relative size of quantities, students connect their understanding of multiplication and division with ratios and rates. Thus students expand the scope of problems for which they can use multiplication and division to solve problems, and they connect ratios and fractions. Students solve a wide variety of problems involving ratios and rates. 2. Students use the meaning of fractions, the meanings of multiplication and division, and the relationship between multiplication and division to understand and explain why the procedures for dividing fractions make sense. Students use these operations to solve problems. Students extend their previous understandings of number and the ordering of numbers to the full system of rational numbers, which includes negative rational numbers, and in particular negative integers. They reason about the order and absolute value of rational numbers and about the location of points in all four quadrants of the coordinate plane. 3. Students understand the use of variables in mathematical expressions. They write expressions and equations that correspond to given situations, evaluate expressions, and use expressions and formulas to solve problems. Students understand that expressions in different forms can be equivalent, and they use the properties of operations to rewrite expressions in equivalent forms. Students know that the solutions of an equation are the values of the variables that make the equation true. Students use properties of operations and the idea of maintaining the equality of both sides of an equation to solve simple one-step equations. Students construct and analyze tables, such as tables of quantities that are in equivalent ratios, and they use equations (such as 3x = y) to describe relationships between quantities. 4. Building on and reinforcing their understanding of number, students begin to develop their ability to think statistically. Students recognize that a data distribution may not have a definite center and that different ways to measure center yield different values. The median measures center in the sense that it is roughly the middle value. The mean measures center in the sense that it is the value that each data point would take on if the total of the data values were redistributed equally, and also in the sense that it is a balance point. Students recognize that a measure of variability (interquartile range) can also be useful for summarizing data because two very different sets of data can have the same mean and median yet be distinguished by their variability. Students learn to describe and summarize numerical data sets, identifying clusters, peaks, gaps, and symmetry, considering the context in which the data were collected. 5. Students in grade 6 also build on their work with area in elementary school by reasoning about relationships among shapes to determine area, surface area, and volume. They find areas of right Massachusetts Curriculum Framework for Mathematics 55 triangles, other triangles, and special quadrilaterals by decomposing these shapes, rearranging or removing pieces, and relating the shapes to rectangles. Using these methods, students discuss, develop, and justify formulas for areas of triangles and parallelograms. Students find areas of polygons and surface areas of prisms and pyramids by decomposing them into pieces whose area they can determine. They reason about right rectangular prisms with fractional side lengths to extend formulas for the volume of a right rectangular prism to fractional side lengths. They prepare for work on scale drawings and constructions in grade 7 by drawing polygons in the coordinate plane. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 56 Grade 6 Overview Ratios and Proportional Relationships A. Understand ratio and rate concepts and use ratio reasoning to solve problems. The Number System A. Apply and extend previous understandings of multiplication and division to divide fractions by fractions. B. Compute fluently with multi-digit numbers and find common factors and multiples. C. Apply and extend previous understandings of numbers to the system of rational numbers. Expressions and Equations A. Apply and extend previous understandings of arithmetic to algebraic expressions. B. Reason about and solve one-variable equations and inequalities. C. Represent and analyze quantitative relationships between dependent and independent variables. Geometry A. Solve real-world and mathematical problems involving area, surface area, and volume. Statistics and Probability A. Develop understanding of statistical variability. B. Summarize and describe distributions. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 57 Grade 6 Content Standards Ratios and Proportional Relationships 6.RP A. Understand ratio and rate concepts and use ratio and rate reasoning to solve problems. 1. Understand the concept of a ratio including the distinctions between part:part and part:whole and the value of a ratio; part/part and part/whole. Use ratio language to describe a ratio relationship between two quantities. For example, The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every two wings there was one beak; For every vote candidate A received, candidate C received nearly three votes, meaning that candidate C received three out of every four votes or ¾ of all votes. 2. Understand the concept of a unit rate a/b associated with a ratio a:b with b 0, and use rate language in the context of a ratio relationship, including the use of units. For example, This recipe has a ratio of three cups of flour to four cups of sugar, so there is ¾ cup of flour for each cup of sugar; We paid $75 for 15 hamburgers, which is a rate of five dollars per hamburger.22 3. Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations. a. Make tables of equivalent ratios relating quantities with whole-number measurements. Find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios. b. Solve unit rate problems, including those involving unit pricing, and constant speed. For example, if it took seven hours to mow four lawns, then, at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed? c. Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30∕100 times the quantity); solve problems involving finding the whole, given a part and the percent. d. Use ratio reasoning to convert measurement units within and between measurement systems; manipulate and transform units appropriately when multiplying or dividing quantities. For example, Malik is making a recipe, but he cannot find his measuring cups! He has, however, found a tablespoon. His cookbook says that 1 cup = 16 tablespoons. Explain how he could use the tablespoon to measure out the following ingredients: two cups of flour, ½ cup sunflower seed, and 1¼ cup of oatmeal.23 e. Solve problems that relate the mass of an object to its volume. The Number System 6.NS A. Apply and extend previous understandings of multiplication and division to divide fractions by fractions. 1. Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem. For example, create a story context for (2∕3) (3∕4) and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that (2∕3) (3∕4) = 8∕9 because 3∕4 of 8∕9 is 2∕3. In general, (a∕b) (c∕d) = ad∕bc. How much chocolate will each person get if three people share 1∕2 lb. of chocolate equally? How many 3∕4-cup servings are in 2∕3 of a cup of yogurt? How wide is a rectangular strip of land with length 3∕4 mile and area 1∕2 square mile? B. Compute fluently with multi-digit numbers and find common factors and multiples. 2. Fluently divide multi-digit numbers using the standard algorithm. 3. Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation. 22 Expectations for unit rates in this grade are limited to non-complex fractions. 23 Example is from the Illustrative Mathematics Project: Massachusetts Curriculum Framework for Mathematics 58 4. Use prime factorization to find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two relatively prime numbers. For example, express 36 + 8 as 4(9 + 2). C. Apply and extend previous understandings of numbers to the system of rational numbers. 5. Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, credits/debits, and positive/negative electric charge). Use positive and negative numbers (whole numbers, fractions, and decimals) to represent quantities in real-world contexts, explaining the meaning of zero in each situation. 6. Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates. a. Recognize opposite signs of numbers as indicating locations on opposite sides of 0 on the number line; recognize that the opposite of the opposite of a number is the number itself, e.g., –(–3) = 3, and that zero is its own opposite. b. Understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane; recognize that when two ordered pairs differ only by signs, the locations of the points are related by reflections across one or both axes. c. Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane. 7. Understand ordering and absolute value of rational numbers. a. Interpret statements of inequality as statements about the relative positions of two numbers on a number line diagram. For example, interpret –3 > –7 as a statement that –3 is located to the right of –7 on a number line oriented from left to right. b. Write, interpret, and explain statements of order for rational numbers in real-world contexts. For example, write –3oC > –7oC to express the fact that –3oC is warmer than –7oC. c. Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. For example, for an account balance of –30 dollars, write |–30| = 30 to describe the size of the debt in dollars. d. Distinguish comparisons of absolute value from statements about order. For example, recognize that an account balance less than –30 dollars represents a debt greater than 30 dollars. 8. Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate. Expressions and Equations 6.EE A. Apply and extend previous understandings of arithmetic to algebraic expressions. 1. Write and evaluate numerical expressions involving whole-number exponents. 2. Write, read, and evaluate expressions in which letters stand for numbers. a. Write expressions that record operations with numbers and with letters standing for numbers. For example, express the calculation “Subtract y from 5” as 5 – y. b. Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, and coefficient); view one or more parts of an expression as a single entity. Massachusetts Curriculum Framework for Mathematics 59 For example, describe the expression 2(8 + 7) as a product of two factors; view (8 + 7) as both a single entity and a sum of two terms. c. Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). For example, use the formulas V = s3 and A = 6s2 to find the volume and surface area of a cube with sides of length s = ½ . 3. Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3(2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6(4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y. 4. Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for. B. Reason about and solve one-variable equations and inequalities. 5. Understand solving an equation or inequality as a process of answering a question: Which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true. 6. Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set. 7. Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q, and x are all nonnegative rational numbers. 8. Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams. C. Represent and analyze quantitative relationships between dependent and independent variables. 9. Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. For example, in a problem involving motion at constant speed, list and graph ordered pairs of distances and times, and write the equation d = 65t to represent the relationship between distance and time. Geometry 6.G A. Solve real-world and mathematical problems involving area, surface area, and volume. 1. Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems. 2. Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = lwh and V = Bh to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems. Massachusetts Curriculum Framework for Mathematics 60 3. Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Apply these techniques in the context of solving real-world and mathematical problems. 4. Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface areas of these figures. Apply these techniques in the context of solving real-world and mathematical problems. Statistics and Probability 6.SP A. Develop understanding of statistical variability. 1. Recognize a statistical question as one that anticipates variability in the data related to the question and accounts for it in the answers. For example, “How old am I?” is not a statistical question, but “How old are the students in my school?” is a statistical question because one anticipates variability in students’ ages. 2. Understand that a set of data collected to answer a statistical question has a distribution, which can be described by its center (median, mean, and/or mode), spread (range, interquartile range), and overall shape. 3. Recognize that a measure of center for a numerical data set summarizes all of its values with a single number, while a measure of variation describes how its values vary with a single number. B. Summarize and describe distributions. 4. Display numerical data in plots on a number line, including dot plots, histograms, and box plots. a. Read and interpret circle graphs. 5. Summarize numerical data sets in relation to their context, such as by: a. Reporting the number of observations. b. Describing the nature of the attribute under investigation, including how it was measured and its units of measurement. c. Giving quantitative measures of center (median, and/or mean) and variability (range and/or interquartile range), as well as describing any overall pattern and any striking deviations from the overall pattern with reference to the context in which the data were gathered. d. Relating the choice of measures of center and variability to the shape of the data distribution and the context in which the data were gathered. Massachusetts Curriculum Framework for Mathematics 61 Grade 7 Introduction In grade 7, instructional time should focus on four critical areas: (1) developing understanding of and applying proportional relationships; (2) developing understanding of operations with rational numbers and working with expressions and linear equations; (3) solving problems involving scale drawings and informal geometric constructions, and working with two- and three-dimensional shapes to solve problems involving area, surface area, and volume; and (4) drawing inferences about populations based on samples. 1. Students extend their understanding of ratios and rates and develop understanding of proportionality to solve single- and multi-step problems. Students use their understanding of ratios, rates, and proportionality to solve a wide variety of percent problems, including those involving discounts, interest, taxes, tips, and percent increase or decrease. Students solve problems about scale drawings by relating corresponding lengths between the objects or by using the fact that relationships of lengths within an object are preserved in similar objects. Students graph proportional relationships and understand the unit rate informally as a measure of the steepness of the related line, called the slope. They distinguish proportional relationships from other relationships. 2. Students develop a unified understanding of number, recognizing fractions, decimals (that have a finite or a repeating decimal representation), and percents as different representations of rational numbers. Students extend addition, subtraction, multiplication, and division to all rational numbers, maintaining the properties of operations and the relationships between addition and subtraction, and multiplication and division. By applying these properties, and by viewing negative numbers in terms of everyday contexts (e.g., amounts owed or temperatures below zero), students explain and interpret the rules for adding, subtracting, multiplying, and dividing with negative numbers. They use the arithmetic of rational numbers as they formulate expressions and equations in one variable and use these equations to solve problems. 3. Students continue their work with area from grade 6, solving problems involving the area and circumference of a circle and surface area of three-dimensional objects. In preparation for work on congruence and similarity in grade 8 they reason about relationships among two-dimensional figures using scale drawings and informal geometric constructions, and they gain familiarity with the relationships between angles formed by intersecting lines. Students work with three-dimensional figures, relating them to two-dimensional figures by examining cross-sections. They solve real-world and mathematical problems involving area, surface area, and volume of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms. 4. Students build on their previous work with single data distributions to compare two data distributions and address questions about differences between populations. They begin informal work with random sampling to generate data sets and learn about the importance of representative samples for drawing inferences. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 62 Grade 7 Overview Ratios and Proportional Relationships A. Analyze proportional relationships and use them to solve real-world and mathematical problems. The Number System A. Apply and extend previous understandings of operations with fractions to add, subtract, multiply, and divide rational numbers. Expressions and Equations A. Use properties of operations to generate equivalent expressions. B. Solve real-life and mathematical problems using numerical and algebraic expressions and equations. Geometry A. Draw, construct and describe geometrical figures and describe the relationships between them. B. Solve real-life and mathematical problems involving angle measure, area, surface area, and volume. Statistics and Probability A. Use random sampling to draw inferences about a population. B. Draw informal comparative inferences about two populations. C. Investigate chance processes and develop, use, and evaluate probability models. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 63 Grade 7 Content Standards Ratios and Proportional Relationships 7.RP A. Analyze proportional relationships and use them to solve real-world and mathematical problems. 1. Compute unit rates associated with ratios of fractions, including ratios of lengths, areas, and other quantities measured in like or different units. For example, if a person walks ½ mile in each ¼ hour, compute the unit rate as the complex fraction ½∕¼ miles per hour, equivalently 2 miles per hour. 2. Recognize and represent proportional relationships between quantities. a. Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table, or graphing on a coordinate plane and observing whether the graph is a straight line through the origin. b. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships. c. Represent proportional relationships by equations. For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. d. Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate. 3. Use proportional relationships to solve multi-step ratio, rate, and percent problems. For example: simple interest, tax, price increases and discounts, gratuities and commissions, fees, percent increase and decrease, percent error. The Number System 7.NS A. Apply and extend previous understandings of operations with fractions to add, subtract, multiply, and divide rational numbers. 1. Apply and extend previous understandings of addition and subtraction to add and subtract integers and other rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram. a. Describe situations in which opposite quantities combine to make zero. For example, A hydrogen atom has zero charge because its two constituents are oppositely charged; If you open a new bank account with a deposit of $30 and then withdraw $30, you are left with a $0 balance. b. Understand p + q as the number located a distance |q| from p, in the positive or negative direction depending on whether q is positive or negative. Show that a number and its opposite have a sum of 0 (are additive inverses). Interpret sums of rational numbers by describing real-world contexts. c. Understand subtraction of rational numbers as adding the additive inverse, p – q = p + (–q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts. d. Apply properties of operations as strategies to add and subtract rational numbers. 2. Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide integers and other rational numbers. a. Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (–1)(–1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts. Massachusetts Curriculum Framework for Mathematics 64 b. Understand that integers can be divided, provided that the divisor is not zero, and every quotient of integers (with non-zero divisor) is a rational number. If p and q are integers, then –(p∕q) = (–p)∕q = p∕(–q). Interpret quotients of rational numbers by describing real-world contexts. c. Apply properties of operations as strategies to multiply and divide rational numbers. d. Convert a rational number to a decimal using long division; know that the decimal form of a rational number terminates in 0s or eventually repeats. 3. Solve real-world and mathematical problems involving the four operations with integers and other rational numbers.24 Expressions and Equations 7.EE A. Use properties of operations to generate equivalent expressions. 1. Apply properties of operations to add, subtract, factor, and expand linear expressions with rational coefficients. For example, 4x + 2 = 2(2x +1) and -3(x – 5∕3) = -3x + 5. 2. Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related. For example, a + 0.05a = 1.05a means that “increase by 5%” is the same as “multiply by 1.05.” A shirt at a clothing store is on sale for 20% off the regular price, “p”. The discount can be expressed as 0.2p. The new price for the shirt can be expressed as p – 0.2p or 0.8p. B. Solve real-life and mathematical problems using numerical and algebraic expressions and equations. 3. Solve multi-step real-life and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies. For example, if a woman making $25 an hour gets a 10% raise, she will make an additional 1∕10 of her salary an hour, or $2.50, for a new salary of $27.50. If you want to place a towel bar 9¾ inches long in the center of a door that is 27½ inches wide, you will need to place the bar about 9 inches from each edge; This estimate can be used as a check on the exact computation. 4. Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. a. Solve word problems leading to equations of the form px + q = r and p(x ÷ q) = r, where p, q, and r are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. For example, the perimeter of a rectangle is 54 cm. Its length is 6 cm. What is its width? b. Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem. For example, as a salesperson, you are paid $50 per week plus $3 per sale. This week you want your pay to be at least $100. Write an inequality for the number of sales you need to make, and describe the solutions. c. Extend analysis of patterns to include analyzing, extending, and determining an expression for simple arithmetic and geometric sequences (e.g., compounding, increasing area), using tables, graphs, words, and expressions. 24 Computations with rational numbers extend the rules for manipulating fractions to complex fractions. Massachusetts Curriculum Framework for Mathematics 65 Geometry 7.G A. Draw, construct, and describe geometrical figures and describe the relationships between them. 1. Solve problems involving scale drawings of geometric figures, such as computing actual lengths and areas from a scale drawing and reproducing a scale drawing at a different scale. 2. Draw (freehand, with ruler and protractor, and with technology) two-dimensional geometric shapes with given conditions. Focus on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no triangle. 3. Describe the shape of the two-dimensional face of the figure that results from slicing three-dimensional figures, as in plane sections of right rectangular prisms and right rectangular pyramids. B. Solve real-life and mathematical problems involving angle measure, area, surface area, and volume. 4. Circles and measurement: a. Know that a circle is a two-dimensional shape created by connecting all of the points equidistant from a fixed point called the center of the circle. b. Understand and describe the relationships among the radius, diameter, and circumference of a circle. c. Understand and describe the relationship among the radius, diameter, and area of a circle. d. Know the formulas for the area and circumference of a circle and use them to solve problems. e. Give an informal derivation of the relationship between the circumference and area of a circle. 5. Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write simple equations and use them to solve for an unknown angle in a figure. 6. Solve real-world and mathematical problems involving area, volume, and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms. Statistics and Probability 7.SP A. Use random sampling to draw inferences about a population. 1. Understand that statistics can be used to gain information about a population by examining a sample of the population; Generalizations about a population from a sample are valid only if the sample is representative of that population. Understand that random sampling tends to produce representative samples and support valid inferences. 2. Use data from a random sample to draw inferences about a population with an unknown characteristic of interest. Generate multiple samples (or simulated samples) of the same size to gauge the variation in estimates or predictions. For example, estimate the mean word length in a book by randomly sampling words from the book; predict the winner of a school election based on randomly sampled survey data. Gauge how far off the estimate or prediction might be. B. Draw informal comparative inferences about two populations. 3. Informally assess the degree of visual overlap of two numerical data distributions with similar variabilities, measuring the difference between the centers by expressing it as a multiple of a measure of variability. For example, the mean height of players on the basketball team is 10 cm greater than the mean height of players on the soccer team and both distributions have similar variability (mean absolute deviation) of about 5 cm. The difference between the mean heights of the two teams (10 cm) is about twice the variability (5 cm) on either team. On a dot plot, the separation between the two distributions of heights is noticeable. 4. Use measures of center and measures of variability for numerical data from random samples to draw informal comparative inferences about two populations. For example, decide whether the words in a chapter of a seventh-grade science book are generally longer than the words in a chapter of a fourth-grade science book. C. Investigate chance processes and develop, use, and evaluate probability models. Massachusetts Curriculum Framework for Mathematics 66 5. Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around ½ indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event. 6. Approximate the probability of a chance event by collecting data on the chance process that produces it and observing its long-run relative frequency, and predict the approximate relative frequency given the probability. For example, when rolling a number cube 600 times, predict that a 3 or 6 would be rolled roughly 200 times, but probably not exactly 200 times. 7. Develop a probability model and use it to find probabilities of events. Compare probabilities from a model to observed frequencies; if the agreement is not good, explain possible sources of the discrepancy. a. Develop a uniform probability model by assigning equal probability to all outcomes, and use the model to determine probabilities of events. For example, if a student is selected at random from a class, find the probability that Jane will be selected and the probability that a girl will be selected. b. Develop a probability model (which may not be uniform) by observing frequencies in data generated from a chance process. For example, find the approximate probability that a spinning penny will land heads up or that a tossed paper cup will land open-end down. Do the outcomes for the spinning penny appear to be equally likely based on the observed frequencies? 8. Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation. a. Understand that, just as with simple events, the probability of a compound event is the fraction of outcomes in the sample space for which the compound event occurs. b. Represent sample spaces for compound events using methods such as organized lists, tables, and tree diagrams. For an event described in everyday language (e.g., “rolling double sixes”), identify the outcomes in the sample space which compose the event. c. Design and use a simulation to generate frequencies for compound events. For example, use random digits as a simulation tool to approximate the answer to the question: If 40% of donors have type A blood, what is the probability that it will take at least four donors to find one with type A blood? Massachusetts Curriculum Framework for Mathematics 67 Grade 8 Introduction In grade 8, instructional time should focus on three critical areas: (1) formulating and reasoning about expressions and equations, including modeling an association in bivariate data with a linear equation and solving linear equations and systems of linear equations; (2) grasping the concept of a function and using functions to describe quantitative relationships; and (3) analyzing two- and three-dimensional space and figures using distance, angle, similarity, and congruence, and understanding and applying the Pythagorean Theorem. 1. Students use linear equations and systems of linear equations to represent, analyze, and solve a variety of problems. Students recognize equations for proportions (y/x = m or y = mx) as special linear equations (y = mx + b), understanding that the constant of proportionality (m) is the slope, and the graphs are lines through the origin. They understand that the slope (m) of a line is a constant rate of change, so that if the input or x-coordinate changes by an amount A, the output or y-coordinate changes by the amount m x A. Students also use a linear equation to describe the association between two quantities in bivariate data (such as arm span vs. height for students in a classroom). At this grade, fitting the model and assessing its fit to the data are done informally. Interpreting the model in the context of the data requires students to express a relationship between the two quantities in question and to interpret components of the relationship (such as slope and y-intercept) in terms of the situation. Students strategically choose and efficiently implement procedures to solve linear equations in one variable, understanding that when they use the properties of equality and the concept of logical equivalence, they maintain the solutions of the original equation. Students solve systems of two linear equations in two variables and relate the systems to pairs of lines in the plane; these intersect, are parallel, or are the same line. Students use linear equations, systems of linear equations, linear functions, and their understanding of slope of a line to analyze situations and solve problems. 2. Students grasp the concept of a function as a rule that assigns to each input exactly one output. They understand that functions describe situations where one quantity determines another. They can translate among representations and partial representations of functions (noting that tabular and graphical representations may be partial representations), and they describe how aspects of the function are reflected in the different representations. 3. Students use ideas about distance and angles, how they behave under translations, rotations, reflections, and dilations, and ideas about congruence and similarity to describe and analyze two-dimensional figures and to solve problems. Students show that the sum of the angles in a triangle is the angle formed by a straight line, and that various configurations of lines give rise to similar triangles because of the angles created when a transversal cuts parallel lines. Students understand the statement of the Pythagorean Theorem and its converse, and can explain why the Pythagorean Theorem holds, for example, by decomposing a square in two different ways. They apply the Pythagorean Theorem to find distances between points on the coordinate plane, to find lengths, and to analyze polygons. Students complete their work on volume by solving problems involving cones, cylinders, and spheres. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 68 Grade 8 Overview The Number System A. Know that there are numbers that are not rational, and approximate them by rational numbers. Expressions and Equations A. Work with radicals and integer exponents. B. Understand the connections between proportional relationships, lines, and linear equations. C. Analyze and solve linear equations and pairs of simultaneous linear equations. Functions A. Define, evaluate, and compare functions. B. Use functions to model relationships between quantities. Geometry A. Understand congruence and similarity using physical models, transparencies, or geometry software. B. Understand and apply the Pythagorean Theorem. C. Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres. Statistics and Probability A. Investigate patterns of association in bivariate data. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 69 Grade 8 Content Standards The Number System 8.NS A. Know that there are numbers that are not rational, and approximate them by rational numbers. 1. Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion. For rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number. 2. Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., 2). For example, by truncating the decimal expansion of show that is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations. Expressions and Equations 8.EE A. Work with radicals and integer exponents. 1. Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 32 3−5 = 3−3 = 1∕33 = 1∕27. 2. Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that is irrational. 3. Use numbers expressed in the form of a single digit multiplied by an integer power of 10 to estimate very large or very small quantities, and express how many times as much one is than the other. For example, estimate the population of the United States as 3 108 and the population of the world as 7 109, and determine that the world population is more than 20 times larger. 4. Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology. B. Understand the connections between proportional relationships, lines, and linear equations. 5. Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed. 6. Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane. Derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. C. Analyze and solve linear equations and pairs of simultaneous linear equations. 7. Solve linear equations in one variable. a. Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers). b. Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms. 8. Analyze and solve pairs of simultaneous linear equations. a. Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously. b. Solve systems of two linear equations in two variables algebraically (using substitution and elimination strategies), and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6. Massachusetts Curriculum Framework for Mathematics 70 c. Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair. Functions 8.F A. Define, evaluate, and compare functions. 1. Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output.25 2. Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a linear function represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change. 3. Interpret the equation y = mx + b as defining a linear function whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1, 1), (2, 4) and (3, 9), which are not on a straight line. B. Use functions to model relationships between quantities. 4. Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values. 5. Describe qualitatively the functional relationship between two quantities by analyzing a graph (e.g., where the function is increasing or decreasing, linear or nonlinear). Sketch a graph that exhibits the qualitative features of a function that has been described verbally. Geometry 8.G A. Understand congruence and similarity using physical models, transparencies, or geometry software. 1. Verify experimentally the properties of rotations, reflections, and translations: a. Lines are transformed to lines, and line segments to line segments of the same length. b. Angles are transformed to angles of the same measure. c. Parallel lines are transformed to parallel lines. 2. Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations. Given two congruent figures, describe a sequence that exhibits the congruence between them. 3. Describe the effects of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates. 4. Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations. Given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them. 5. Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the sum of the three angles appears to form a line, and give an argument in terms of transversals why this is so. B. Understand and apply the Pythagorean Theorem. 6. a. Understand the relationship among the sides of a right triangle. b. Analyze and justify the Pythagorean Theorem and its converse using pictures, diagrams, narratives, or models. 7. Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions. 8. Apply the Pythagorean Theorem to find the distance between two points in a coordinate system. 25 Function notation is not required in grade 8. Massachusetts Curriculum Framework for Mathematics 71 C. Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres. 9. Know the formulas for the volumes of cones, cylinders, and spheres, and use them to solve real-world and mathematical problems. Statistics and Probability 8.SP A. Investigate patterns of association in bivariate data. 1. Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities. Describe patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear association. 2. Know that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line and informally assess the model fit by judging the closeness of the data points to the line. 3. Use the equation of a linear model to solve problems in the context of bivariate measurement data, interpreting the slope and intercept. For example, in a linear model for a biology experiment, interpret a slope of 1.5 cm/hr as meaning that an additional hour of sunlight each day is associated with an additional 1.5 cm in mature plant height. 4. Understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. Use relative frequencies calculated for rows or columns to describe possible association between the two variables. For example, collect data from students in your class on whether or not they have a curfew on school nights and whether or not they have assigned chores at home. Is there evidence that those who have a curfew also tend to have chores? Massachusetts Curriculum Framework for Mathematics 72 The High School Standards for Mathematical Content Introduction Conceptual Categories Number and Quantity (N) Algebra (A) Functions (F) Modeling () Geometry (G) Statistics and Probability (S) Massachusetts Curriculum Framework for Mathematics 73 Introduction The organization of the high school sections of this Framework is outlined below, followed by a detailed description of each section. I. High School Content Standards organized by: A. Conceptual Categories B. Model High School Courses 1. Traditional Pathway Algebra I Geometry Algebra II 2. Integrated Pathway Mathematics I Mathematics II Mathematics III 3. Advanced Courses Model Precalculus Model Advanced Quantitative Reasoning II. Guidance for Making Decisions about Course Sequences and the Model Algebra I Course Massachusetts Curriculum Framework for Mathematics 74 I. The High School Content Standards The high school content standards are organized in two ways: (1) by conceptual category and (2) by model courses. The high school content standards specify the mathematics that all students should study in order to be prepared for college, career, and civic engagement. All standards without a (+) symbol should be in the common mathematics curriculum for all college and career ready students. The high school standards indicated by a (+) symbol contain additional mathematical content that high school students should learn in order to be prepared to take advanced courses such as calculus, advanced statistics, or discrete mathematics. Standards with a (+) symbol may also appear in courses intended for all students but, in these cases, are optional and indicate a standard that is beyond college and career readiness. 1. High School Content Standards Organized by Conceptual Category The content of the high school standards presented by conceptual categories portrays a coherent view of learning progressions that develop students’ mathematical knowledge, skills, and understanding through the high school years. For example: a student’s work with the concept of functions crosses a number of traditional course boundaries, potentially up through and including calculus. Similar to the grade level content standards, each conceptual category (except Modeling) is further subdivided into several domains, and each domain is subdivided into clusters. The conceptual categories are: Number and Quantity (N) Algebra (A) Functions (F) Modeling () Geometry (G) Statistics and Probability (S) 2. High School Content Standards Organized by Model Courses The grades 9–12 high school mathematics standards presented by conceptual categories provide guidance on what students are expected to learn in order to be prepared for college and careers. These standards have been configured into eight high school courses. These model high school courses, organized into a Traditional Pathway and Integrated Pathway and advanced course work, represent a smooth transition from the PreK–8 grade standards. All of the content of the college and career ready high school content standards presented by Conceptual Categories is included in the appropriate locations within the three model courses of both Model Pathways. In this Framework, the wording of a Conceptual Category standard may be different in the model courses in which it is covered in order to clarify the content expectations for that particular model course. Note: In the 2010 Framework, footnotes were included in order to clarify content expectations; however, this Framework incorporates the clarifying language directly into the wording of the model course standards for ease of understanding. The grade 8 standards are rigorous; students are expected to learn about linear relationships and equations to begin the study of functions and comparing rational and irrational numbers. In addition, the statistics presented in the grade 8 standards are sophisticated and include connecting linear relations with the representation of bivariate data. The Model Algebra I and Model Mathematics I courses progress from these concepts and skills, and focus on quadratic and exponential functions. Some students may master the grade 8 standards earlier than eighth grade, which would enable these students to take the high school Model Algebra I course or Model Mathematics I course in eighth grade. Students completing either Model Pathway are prepared for additional courses, such as the model-advanced courses that follow the three courses in either model pathway. Model advanced courses are comprised of the higher-level mathematics standards (+) in the conceptual categories. Massachusetts Curriculum Framework for Mathematics 75 Model Pathways and the Model High School Courses presented in this Framework: Model Traditional Pathway o Model Algebra I (AI) o Model Geometry (GEO) o Model Algebra II (AII) Model Integrated Pathway o Model Mathematics I (MI) o Model Mathematics II (MII) o Model Mathematics III (MIII) Advanced Model Courses o Model Precalculus (PC) o Model Advanced Quantitative Reasoning (AQR) The Model Traditional Pathway reflects the approach typically seen in the U.S., consisting of two Model Algebra courses (I and II) with some Statistics and Probability standards included, and a Model Geometry course, with some Number and Quantity standards and some Statistics and Probability standards included. The Model Integrated Pathway reflects the approach typically seen internationally, consisting of a sequence of three model courses, each of which includes Number and Quantity, Algebra, Functions, Geometry, and Statistics and Probability standards. Appendix III provides a table for each conceptual category. Each table shows the distribution of the content standards across the eight Model Courses. Districts and schools maintain the flexibility to distribute standards across courses in other ways; the model courses are not the only possible designs. II. Guidance for Making Decisions about Course Sequences and the Model Algebra I Course Following the Model Course section, this Framework presents information and resources to ground discussions and decision-making about course-taking sequences in middle and high school and presents multiple options for students ready to move at an accelerated rate, including taking the Model High School Algebra I course in grade 8. Connecting Content Standards and the Standards for Mathematical Practice While the Model Pathways and Model Courses organize the Content Standards into possible model pathways to college, career, and civic readiness, the Content Standards must be connected to the Standards for Mathematical Practice to ensure that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise. (Appendix II provides descriptions of the Standards for Mathematical Practice for three grade-spans: Pre-K–5, 6–8, and 9–12.) Massachusetts Curriculum Framework for Mathematics 76 High School Content Standards by Conceptual Categories Content Standards by Conceptual Category Identifiers/Coding The content standards presented by conceptual categories are built on mathematical learning progressions informed by research on cognitive development and by the logical structure of mathematics. These progressions provide the foundation for the grades 9–12 high school content standards. In this section, the standards are organized by Conceptual Categories. The Conceptual Categories are: Number and Quantity (N) Algebra (A) Functions (F) Modeling () Geometry (G) Statistics and Probability (S) The content standards within the Conceptual Categories were then configured into eight high school courses. The code for each high school conceptual category standard begins with the identifier for the conceptual category code (N, A, F, G, S), followed by the domain code, and the standard number, as shown below. The standard highlighted above is identified as N-RN.A.1, identifying it as a standard in the Number and Quantity conceptual category (N-) within that category’s Real Number System domain (RN), and as the first standard in that domain and in that cluster (A.1). All of the standards in this Framework use a common coding system. Massachusetts Curriculum Framework for Mathematics 77 Conceptual Category: Number and Quantity [N] Introduction Numbers and Number Systems During the years from kindergarten to eighth grade, students must repeatedly extend their conception of number. At first, “number” means “counting number”: 1, 2, 3. Soon after that, 0 is used to represent “none” and the whole numbers are formed by the counting numbers together with zero. The next extension is fractions. At first, fractions are barely numbers and tied strongly to pictorial representations. Yet by the time students understand division of fractions, they have a strong concept of fractions as numbers and have connected them, via their decimal representations, with the Base Ten system used to represent the whole numbers. During middle school, fractions are augmented by negative fractions to form the rational numbers. In grade 8, students extend this system once more, augmenting the rational numbers with the irrational numbers to form the real numbers. In high school, students will be exposed to yet another extension of number, when the real numbers are augmented by the imaginary numbers to form the complex numbers. (See Illustration 1 in the Glossary.) With each extension of number, the meanings of addition, subtraction, multiplication, and division are extended. In each new number system—integers, rational numbers, real numbers, and complex numbers—the four operations stay the same in two important ways: They have the Commutative, Associative, and Distributive properties and their new meanings are consistent with their previous meanings. Extending the properties of whole-number exponents leads to new and productive notation. For example, properties of whole-number exponents suggest that (51/3)3 should be 5(1/3)3 = 51 = 5 and that 51/3 should be the cube root of 5. Calculators, spreadsheets, and computer algebra systems can provide ways for students to become better acquainted with these new number systems and their notation. They can be used to generate data for numerical experiments, to help understand the workings of matrix, vector, and complex number algebra, and to experiment with non-integer exponents. Quantities In real-world problems, the answers are usually not numbers but quantities: numbers with units, which involve measurement. In their work in measurement up through grade 8, students primarily measure commonly used attributes such as length, area, and volume. In high school, students encounter a wider variety of units in modeling, e.g., acceleration, currency conversions, derived quantities such as person-hours and heating degree-days, social science rates such as per-capita income, and rates in everyday life such as points scored per game or batting averages. They also encounter novel situations in which they themselves must conceive the attributes of interest. For example, to find a good measure of overall highway safety, they might propose measures such as fatalities per year, fatalities per year per driver, or fatalities per vehicle-mile traveled. Such a conceptual process is sometimes called quantification. Quantification is important for science, as when surface area suddenly “stands out” as an important variable in evaporation. Quantification is also important for companies, which must conceptualize relevant attributes and create or choose suitable measures for them. Massachusetts Curriculum Framework for Mathematics 78 Conceptual Category: Number and Quantity Overview [N] The Real Number System A. Extend the properties of exponents to rational exponents. B. Use properties of rational and irrational numbers. Quantities A. Reason quantitatively and use units to solve problems. The Complex Number System A. Perform arithmetic operations with complex numbers. B. Represent complex numbers and their operations on the complex plane. C. Use complex numbers in polynomial identities and equations. Vector and Matrix Quantities A. Represent and model with vector quantities. B. Perform operations on vectors. C. Perform operations on matrices and use matrices in applications. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 79 Conceptual Category: Number and Quantity Content Standards [N] The Real Number System N-RN A. Extend the properties of exponents to rational exponents. 1. Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 51/3 to be the cube root of 5 because we want (51/3)3 = 5(1/3)3 to hold, so (51/3)3 must equal 5. 2. Rewrite expressions involving radicals and rational exponents using the properties of exponents. B. Use properties of rational and irrational numbers. 3. Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational. Quantities N-Q A. Reason quantitatively and use units to solve problems. 1. Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays. 2. Define appropriate quantities for the purpose of descriptive modeling. 3. Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. a. Describe the effects of approximate error in measurement and rounding on measurements and on computed values from measurements. Identify significant figures in recorded measures and computed values based on the context given and the precision of the tools used to measure. The Complex Number System N-CN A. Perform arithmetic operations with complex numbers. 1. Know there is a complex number i such that i2 = -1, and every complex number has the form a + bi with a and b real. 2. Use the relation i2 = -1 and the Commutative, Associative, and Distributive properties to add, subtract, and multiply complex numbers. 3. (+) Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers. B. Represent complex numbers and their operations on the complex plane. 4. (+) Represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number. 5. (+) Represent addition, subtraction, multiplication, and conjugation of complex numbers geometrically on the complex plane; use properties of this representation for computation. For example, ( )3=8 because ( ) has modulus 2 and argument 120°. 6. (+) Calculate the distance between numbers in the complex plane as the modulus of the difference, and the midpoint of a segment as the average of the numbers at its endpoints. C. Use complex numbers in polynomial identities and equations. 7. Solve quadratic equations with real coefficients that have complex solutions. 8. (+) Extend polynomial identities to the complex numbers. For example, rewrite x 2 + 4 as (x + 2i) (x – 2i). 9. (+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials. Massachusetts Curriculum Framework for Mathematics 80 Vector and Matrix Quantities N-VM A. Represent and model with vector quantities. 1. (+) Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., v, |v|, || v ||, v). 2. (+) Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point. 3. (+) Solve problems involving velocity and other quantities that can be represented by vectors. B. Perform operations on vectors. 4. (+) Add and subtract vectors. a. (+) Add vectors end-to-end, component-wise, and by the parallelogram rule. Understand that (+) the magnitude of a sum of two vectors is typically not the sum of the magnitudes. b. (+) Given two vectors in magnitude and direction form, determine the magnitude and direction of their sum. c. (+) Understand vector subtraction v – w as v + (–w), where –w is the additive inverse of w, with the same magnitude as w and pointing in the opposite direction. Represent vector subtraction graphically by connecting the tips in the appropriate order, and perform vector subtraction component-wise. 5. (+) Multiply a vector by a scalar. a. (+) Represent scalar multiplication graphically by scaling vectors and possibly reversing their direction; perform scalar multiplication component-wise, e.g., as c(vx , vy) = (cvx , cvy). b. (+) Compute the magnitude of a scalar multiple cv using || cv|| = |c|v. Compute the direction of cv knowing that when |c|v ≠ 0, the direction of cv is either along v (for c > 0) or against v (for c < 0). C. Perform operations on matrices and use matrices in applications. 6. (+) Use matrices to represent and manipulate data, e.g., to represent payoffs or incidence relationships in a network. 7. (+) Multiply matrices by scalars to produce new matrices, e.g., as when all of the payoffs in a game are doubled. 8. (+) Add, subtract, and multiply matrices of appropriate dimensions. 9. (+) Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a Commutative operation, but still satisfies the Associative and Distributive properties. 10. (+) Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse. 11. (+) Multiply a vector (regarded as a matrix with one column) by a matrix of suitable dimensions to produce another vector. Work with matrices as transformations of vectors. 12. (+) Work with 2 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area. Massachusetts Curriculum Framework for Mathematics 81 Conceptual Category: Algebra [A] Introduction Expressions An expression is a record of a computation with numbers, symbols that represent numbers, arithmetic operations, exponentiation, and, at more advanced levels, the operation of evaluating a function. Conventions about the use of parentheses and the Order of Operations assure that each expression is unambiguous. Creating an expression that describes a computation involving a general quantity requires the ability to express the computation in general terms, abstracting from specific instances. Reading an expression with comprehension involves analysis of its underlying structure. This may suggest a different but equivalent way of writing the expression that exhibits some different aspect of its meaning. For example, p + 0.05p can be interpreted as the addition of a 5% tax to a price p. Rewriting p + 0.05p as 1.05p shows that adding a tax is the same as multiplying the price by a constant factor. Algebraic manipulations are governed by the properties of operations and exponents and the conventions of algebraic notation. At times, an expression is the result of applying operations to simpler expressions. For example, p + 0.05p is the sum of the simpler expressions p and 0.05p. Viewing an expression as the result of operation on simpler expressions can sometimes clarify its underlying structure. A spreadsheet or a computer algebra system (CAS) can be used to experiment with algebraic expressions, perform complicated algebraic manipulations, and understand how algebraic manipulations behave. Equations and Inequalities An equation is a statement of equality between two expressions, often viewed as a question asking for which values of the variables the expressions on either side are in fact equal. These values are the solutions to the equation. An identity, in contrast, is true for all values of the variables; identities are often developed by rewriting an expression in an equivalent form. The solutions of an equation in one variable form a set of numbers; the solutions of an equation in two variables form a set of ordered pairs of numbers, which can be plotted in the coordinate plane. Two or more equations and/or inequalities form a system. A solution for such a system must satisfy every equation and inequality in the system. An equation can often be solved by successively deducing from it one or more simpler equations. For example, one can add the same constant to both sides without changing the solutions, but squaring both sides might lead to extraneous solutions. Strategic competence in solving includes looking ahead for productive manipulations and anticipating the nature and number of solutions. Some equations have no solutions in a given number system, but have a solution in a larger system. For example, the solution of x + 1 = 0 is an integer, not a whole number; the solution of 2x + 1 = 0 is a rational number, not an integer; the solutions of x2 – 2 = 0 are real numbers, not rational numbers; and the solutions of x2 + 2 = 0 are complex numbers, not real numbers. The same solution techniques used to solve equations can be used to rearrange formulas. For example, the formula for the area of a trapezoid, A = ((b1+b2)∕2)h, can be solved for h using the same deductive process. Inequalities can be solved by reasoning about the properties of inequality. Many, but not all, of the properties of equality continue to hold for inequalities and can be useful in solving them. Massachusetts Curriculum Framework for Mathematics 82 Connections to Functions and Modeling Expressions can define functions, and equivalent expressions define the same function. Asking when two functions have the same value for the same input leads to an equation; graphing the two functions allows for finding approximate solutions of the equation. Converting a verbal description to an equation, inequality, or system of these is an essential skill in modeling. Massachusetts Curriculum Framework for Mathematics 83 Conceptual Category: Algebra Overview [A] Seeing Structure in Expressions A. Interpret the structure of expressions (linear, quadratic, exponential, polynomial, rational). B. Write expressions in equivalent forms to solve problems. Arithmetic with Polynomials and Rational Expressions A. Perform arithmetic operations on polynomials. B. Understand the relationship between zeros and factors of polynomials. C. Use polynomial identities to solve problems. D. Rewrite rational expressions. Creating Equations A. Create equations that describe numbers or relationships. Reasoning with Equations and Inequalities A. Understand solving equations as a process of reasoning and explain the reasoning. B. Solve equations and inequalities in one variable. C. Solve systems of equations. D. Represent and solve equations and inequalities graphically. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 84 Conceptual Category: Algebra Content Standards [A] Seeing Structure in Expressions A-SSE A. Interpret the structure of linear, quadratic, exponential, polynomial, and rational expressions. 1. Interpret expressions that represent a quantity in terms of its context. a. Interpret parts of an expression, such as terms, factors, and coefficients. b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1 + r)n as the product of P and a factor not depending on P. 2. Use the structure of an expression to identify ways to rewrite it. For example, see x4 – y4 as (x2)2 – (y2)2, thus recognizing it as a difference of squares that can be factored as (x2 – y2)(x2 + y2). B. Write expressions in equivalent forms to solve problems. 3. Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. a. Factor a quadratic expression to reveal the zeros of the function it defines. b. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines. c. Use the properties of exponents to transform expressions for exponential functions. For example, the expression 1.15t can be rewritten as (1.151/12)12t ≈ 1.01212t to reveal the approximate equivalent monthly interest rate if the annual rate is 15%. 4. Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems. For example, calculate mortgage payments. Arithmetic with Polynomials and Rational Expressions A-APR A. Perform arithmetic operations on polynomials. 1. Understand that polynomials form a system analogous to the integers, namely, they are closed under certain operations. a. Perform operations on polynomial expressions (addition, subtraction, multiplication, division) and compare the system of polynomials to the system of integers when performing operations. b. Factor and/or expand polynomial expressions, identify and combine like terms, and apply the Distributive property. B. Understand the relationship between zeros and factors of polynomials. 2. Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x). 3. Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. C. Use polynomial identities to solve problems. 4. Prove polynomial identities and use them to describe numerical relationships. For example, the polynomial identity (x2 + y2)2 = (x2 – y2)2 + (2xy)2 can be used to generate Pythagorean triples. 5. (+) Know and apply the Binomial Theorem for the expansion of (x + y)n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal’s Triangle. D. Rewrite rational expressions. 6. Rewrite simple rational expressions in different forms; write a(x)∕b(x) in the form q(x) + r(x)∕b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system. 7. (+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions. Massachusetts Curriculum Framework for Mathematics 85 Creating Equations A-CED A. Create equations that describe numbers or relationships. 1. Create equations and inequalities in one variable and use them to solve problems. (Include equations arising from linear and quadratic functions, and simple root and rational functions and exponential functions.) 2. Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. 3. Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods. 4. Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law V = IR to highlight resistance, R. Reasoning with Equations and Inequalities A-REI A. Understand solving equations as a process of reasoning and explain the reasoning. 1. Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify or refute a solution method. 2. Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise. B. Solve equations and inequalities in one variable. 3. Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. a. Solve linear equations and inequalities in one variable involving absolute value. 4. Solve quadratic equations in one variable. a. Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x – p)2 = q that has the same solutions. Derive the quadratic formula from this form. b. Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b. C. Solve systems of equations. 5. Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions. 6. Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. 7. Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. For example, find the points of intersection between the line y = –3x and the circle x2 + y2 = 3. 8. (+) Represent a system of linear equations as a single matrix equation in a vector variable. 9. (+) Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3 3 or greater). D. Represent and solve equations and inequalities graphically. 10. Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line). Show that any point on the graph of an equation in two variables is a solution to the equation. 11. Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions. Massachusetts Curriculum Framework for Mathematics 86 12. Graph the solutions of a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set of a system of linear inequalities in two variables as the intersection of the corresponding half-planes. Massachusetts Curriculum Framework for Mathematics 87 Conceptual Category: Functions [F] Introduction Functions describe situations where one quantity determines another. For example, the return on $10,000 invested at an annualized percentage rate of 4.25% is a function of the length of time the money is invested. Because we continually make theories about dependencies between quantities in nature and society, functions are important tools in the construction of mathematical models. In school mathematics, functions usually have numerical inputs and outputs and are often defined by an algebraic expression. For example, the time in hours it takes for a car to drive 100 miles is a function of the car’s speed in miles per hour, v; the rule T(v) = 100∕v expresses this relationship algebraically and defines a function whose name is T. The set of inputs to a function is called its domain. We often infer the domain to be all inputs for which the expression defining a function has a value, or for which the function makes sense in a given context. A function can be described in various ways, such as; by a graph (e.g., the trace of a seismograph); by a verbal rule, as in, “I’ll give you a state, you give me the capital city”; by an algebraic expression like f(x) = a + bx; or by a recursive rule. The graph of a function is often a useful way of visualizing the relationship of the function models, and manipulating a mathematical expression for a function can throw light on the function’s properties. Functions presented as expressions can model many important phenomena. Two important families of functions characterized by laws of growth are linear functions, which grow at a constant rate, and exponential functions, which grow at a constant percent rate. Linear functions with a constant term of zero describe proportional relationships. A graphing utility or a computer algebra system can be used to experiment with properties of these functions and their graphs and to build computational models of functions, including recursively defined functions. Connections to Expressions, Equations, Modeling, and Coordinates Determining an output value for a particular input involves evaluating an expression; finding inputs that yield a given output involves solving an equation. Questions about when two functions have the same value for the same input lead to equations, whose solutions can be visualized from the intersection of their graphs. Because functions describe relationships between quantities, they are frequently used in modeling. Sometimes functions are defined by a recursive process, which can be displayed effectively using a spreadsheet or other technology. Massachusetts Curriculum Framework for Mathematics 88 Conceptual Category: Functions Overview [F] Interpreting Functions A. Understand the concept of a function and use function notation. B. Interpret functions that arise in applications in terms of the context (linear, quadratic, exponential, rational, polynomial, square root, cube root, trigonometric, logarithmic). C. Analyze functions using different representations. Building Functions A. Build a function that models a relationship between two quantities. B. Build new functions from existing functions. Linear, Quadratic, and Exponential Models A. Construct and compare linear, quadratic, and exponential models and solve problems. B. Interpret expressions for functions in terms of the situation they model. Trigonometric Functions A. Extend the domain of trigonometric functions using the unit circle. B. Model periodic phenomena with trigonometric functions. C. Prove and apply trigonometric identities. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 89 Conceptual Category: Functions Content Standards [F] Interpreting Functions F-IF A. Understand the concept of a function and use function notation. 1. Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x). 2. Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. For example, given a function representing a car loan, determine the balance of the loan at different points in time. 3. Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers. For example, the Fibonacci sequence is defined recursively by f(0) = f(1) = 1, f(n + 1) = f(n) + f(n 1) for n ≥ 1. B. Interpret functions that arise in applications in terms of the context (linear, quadratic, exponential, rational, polynomial, square root, cube root, trigonometric, logarithmic). 4. For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. 5. Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. 6. Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. C. Analyze functions using different representations. 7. Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. a. Graph linear and quadratic functions and show intercepts, maxima, and minima. b. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. c. Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. d. (+) Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior. e. Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude. 8. Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Use the process of factoring and/or completing the square in quadratic and polynomial functions, where appropriate, to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. b. Use the properties of exponents to interpret expressions for exponential functions. Apply to financial situations such as identifying appreciation and depreciation rate for the value of a house or car some time after its initial purchase. For example, identify percent rate of change in functions such as y = (1.02)t, y = (0.97) t, y = (1.01)12 t, and y = (1.2) t ∕10, and classify them as representing exponential growth or decay. 9. Translate among different representations of functions (algebraically, graphically, numerically in tables, or by verbal descriptions). Compare properties of two functions each represented in a different way. Massachusetts Curriculum Framework for Mathematics 90 For example, given a graph of one polynomial function (including quadratic functions) and an algebraic expression for another, say which has the larger/smaller relative maximum and/or minimum. 10. Given algebraic, numeric and/or graphical representations of functions, recognize the function as polynomial, rational, logarithmic, exponential, or trigonometric. Building Functions F-BF A. Build a function that models a relationship between two quantities. 1. Write a function (linear, quadratic, exponential, simple rational, radical, logarithmic, and trigonometric) that describes a relationship between two quantities. a. Determine an explicit expression, a recursive process, or steps for calculation from a context. b. Combine standard function types using arithmetic operations. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model. c. (+) Compose functions. For example, if T(y) is the temperature in the atmosphere as a function of height, and h(t) is the height of a weather balloon as a function of time, then T(h(t)) is the temperature at the location of the weather balloon as a function of time. 2. Write arithmetic and geometric sequences both recursively and with an explicit formula, use them to model situations, and translate between the two forms. B. Build new functions from existing functions. 3. Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. (Include linear, quadratic, exponential, absolute value, simple rational and radical, logarithmic and trigonometric functions.) Utilize technology to experiment with cases and illustrate an explanation of the effects on the graph. (Include recognizing even and odd functions from their graphs and algebraic expressions for them.) 4. Find inverse functions algebraically and graphically. a. Solve an equation of the form f(x) = c for a simple function f that has an inverse and write an expression for the inverse. (Include linear and simple polynomial, rational, and exponential functions.) For example, f(x) =2x3 or f(x) = (x + 1)∕(x 1) for x ≠ 1. b. (+) Verify by composition that one function is the inverse of another. c. (+) Read values of an inverse function from a graph or a table, given that the function has an inverse. d. (+) Produce an invertible function from a non-invertible function by restricting the domain. 5. (+) Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents. Linear, Quadratic, and Exponential Models F-LE A. Construct and compare linear, quadratic, and exponential models and solve problems. 1. Distinguish between situations that can be modeled with linear functions and with exponential functions. a. Prove that linear functions grow by equal differences over equal intervals, and that exponential functions grow by equal factors over equal intervals. b. Recognize situations in which one quantity changes at a constant rate per unit interval relative to another. c. Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another. 2. Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (including reading these from a table). 3. Observe, using graphs and tables, that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. Massachusetts Curriculum Framework for Mathematics 91 4. For exponential models, express as a logarithm the solution to abct = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology. B. Interpret expressions for functions in terms of the situation they model. 5. Interpret the parameters in a linear or exponential function (of the form f(x) = bx + k) in terms of a context. Trigonometric Functions F-TF A. Extend the domain of trigonometric functions using the unit circle. 1. Understand radian measure of an angle as the length of the arc on the unit circle subtended by the angle. 2. Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle. 3. (+) Use special triangles to determine geometrically the values of sine, cosine, tangent for /3, /4 and /6, and use the unit circle to express the values of sine, cosine, and tangent for x, + x, and 2 x in terms of their values for x, where x is any real number. 4. (+) Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions. B. Model periodic phenomena with trigonometric functions. 5. Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline. 6. (+) Understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed. 7. (+) Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context. C. Prove and apply trigonometric identities. 8. Prove the Pythagorean identity sin2(θ) + cos2(θ) = 1 and use it to find sin(θ), cos(θ), or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant. 9. (+) Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems. Massachusetts Curriculum Framework for Mathematics 92 Conceptual Category: Modeling [] Introduction Modeling links classroom mathematics and statistics to everyday life, work, and decision-making. Modeling is the process of choosing and using appropriate mathematics and statistics to analyze empirical situations, to understand them better, and to improve decisions. Quantities and their relationships in physical, economic, public policy, social, and everyday situations can be modeled using mathematical and statistical methods. When making mathematical models, technology is valuable for varying assumptions, exploring consequences, and comparing predictions with data. A model can be very simple, such as writing total cost as a product of unit price and number bought, or using a geometric shape to describe a physical object like a coin. Even such simple models involve making choices. It is up to us whether to model a coin as a three-dimensional cylinder, or whether a two-dimensional disk works well enough for our purposes. Other situations—modeling a delivery route, a production schedule, or a comparison of loan amortizations—need more elaborate models that use other tools from the mathematical sciences. Real-world situations are not organized and labeled for analysis; formulating tractable models, representing such models, and analyzing them is appropriately a creative process. Like every such process, this depends on acquired expertise as well as creativity. Some examples of such situations might include: Estimating how much water and food is needed for emergency relief in a devastated city of three million people, and how it might be distributed. Planning a table tennis tournament for seven players at a club with four tables, where each player plays against each other player. Designing the layout of the stalls in a school fair so as to raise as much money as possible. Analyzing stopping distance for a car. Modeling savings account balance, bacterial colony growth, or investment growth. Engaging in critical path analysis, e.g., applied to turnaround of an aircraft at an airport. Analyzing risk in situations such as extreme sports, pandemics, and terrorism. Relating population statistics to individual predictions. In situations like these, the models devised depend on a number of factors: How precise an answer do we want or need? What aspects of the situation do we most need to understand, control, or optimize? What resources of time and tools do we have? The range of models that we can create and analyze is also constrained by the limitations of our mathematical, statistical, and technical skills, and our ability to recognize significant variables and relationships among them. Diagrams of various kinds, spreadsheets and other technology, and algebra are powerful tools for understanding and solving problems drawn from different types of real-world situations. One of the insights provided by mathematical modeling is that essentially the same mathematical or statistical structure can sometimes model seemingly different situations. Models can also shed light on the mathematical structures themselves, for example, as when a model of bacterial growth makes more vivid the explosive growth of the exponential function. Massachusetts Curriculum Framework for Mathematics 93 The basic modeling cycle is summarized in the diagram below. It involves: (1) identifying variables in the situation and selecting those that represent essential features; (2) formulating a model by creating and selecting geometric, graphical, tabular, algebraic, or statistical representations that describe relationships between the variables; (3) analyzing and performing operations on these relationships to draw conclusions; (4) interpreting the results of the mathematics in terms of the original situation; (5) validating the conclusions by comparing them with the situation, and then either improving the model or; (6) if it is acceptable, reporting on the conclusions and the reasoning behind them. Choices, assumptions, and approximations are present throughout this cycle. In descriptive modeling, a model simply describes the phenomena or summarizes them in a compact form. Graphs of observations are a familiar descriptive model—for example, graphs of global temperature and atmospheric CO2 over time. Analytic modeling seeks to explain data on the basis of deeper theoretical ideas, albeit with parameters that are empirically based; for example, exponential growth of bacterial colonies (until cut-off mechanisms such as pollution or starvation intervene) follows from a constant reproduction rate. Functions are an important tool for analyzing such problems. Graphing utilities, spreadsheets, computer algebra systems, and dynamic geometry software are powerful tools that can be used to model purely mathematical phenomena (e.g., the behavior of polynomials) as well as physical phenomena. Modeling Standards Modeling is best interpreted not as a collection of isolated topics but rather in relation to other standards. Making mathematical models is a Standard for Mathematical Practice, and specific Modeling standards appear throughout the high school standards indicated by a star symbol (). Massachusetts Curriculum Framework for Mathematics 94 Conceptual Category: Geometry [G] Introduction An understanding of the attributes and relationships of geometric objects can be applied in diverse contexts— interpreting a schematic drawing, estimating the amount of wood needed to frame a sloping roof, rendering computer graphics, or designing a sewing pattern for the most efficient use of material. Although there are many types of geometry, school mathematics is devoted primarily to plane Euclidean geometry, studied both synthetically (without coordinates) and analytically (with coordinates). Euclidean geometry is characterized most importantly by the Parallel Postulate that through a point not on a given line there is exactly one parallel line. (Spherical geometry, in contrast, has no parallel lines.) During high school, students begin to formalize their geometry experiences from elementary and middle school, using more precise definitions and developing careful proofs. Later in college, some students develop Euclidean and other geometries carefully from a small set of axioms. The concepts of congruence, similarity, and symmetry can be understood from the perspective of geometric transformation. Fundamental are the rigid motions: translations, rotations, reflections, and combinations of these, all of which are here assumed to preserve distance and angles (and therefore shapes generally). Reflections and rotations each explain a particular type of symmetry, and the symmetries of an object offer insight into its attributes—as when the reflective symmetry of an isosceles triangle assures that its base angles are congruent. In the approach taken here, two geometric figures are defined to be congruent if there is a sequence of rigid motions that carries one onto the other. This is the principle of superposition. For triangles, congruence means the equality of all corresponding pairs of sides and all corresponding pairs of angles. During the middle grades, through experiences drawing triangles from given conditions, students notice ways to specify enough measures in a triangle to ensure that all triangles drawn with those measures are congruent. Once these triangle congruence criteria (ASA, SAS, and SSS) are established using rigid motions, they can be used to prove theorems about triangles, quadrilaterals, and other geometric figures. Similarity transformations (rigid motions followed by dilations) define similarity in the same way that rigid motions define congruence, thereby formalizing the similarity ideas of “same shape” and “scale factor” developed in the middle grades. These transformations lead to the criterion for triangle similarity that two pairs of corresponding angles are congruent. The definitions of sine, cosine, and tangent for acute angles are founded on right triangles and similarity, and, with the Pythagorean Theorem, are fundamental in many real-world and theoretical situations. The Pythagorean Theorem is generalized to non-right triangles by the Law of Cosines. Together, the Laws of Sines and Cosines embody the triangle congruence criteria for the cases where three pieces of information suffice to completely solve a triangle. Furthermore, these laws yield two possible solutions in the ambiguous case, illustrating that Side-Side-Angle is not a congruence criterion. Analytic geometry connects algebra and geometry, resulting in powerful methods of analysis and problem solving. Just as the number line associates numbers with locations in one dimension, a pair of perpendicular axes associates pairs of numbers with locations in two dimensions. This correspondence between numerical coordinates and geometric points allows methods from algebra to be applied to geometry and vice versa. The solution set of an equation becomes a geometric curve, making visualization a tool for doing and understanding algebra. Geometric shapes can be described by equations, making algebraic manipulation into a tool for geometric understanding, modeling, and proof. Geometric transformations of the graphs of equations correspond to algebraic changes in their equations. Massachusetts Curriculum Framework for Mathematics 95 Dynamic geometry environments provide students with experimental and modeling tools that allow them to investigate geometric phenomena in much the same way as computer algebra systems allow them to experiment with algebraic phenomena. Connections to Equations The correspondence between numerical coordinates and geometric points allows methods from algebra to be applied to geometry and vice versa. The solution set of an equation becomes a geometric curve, making visualization a tool for doing and understanding algebra. Geometric shapes can be described by equations, making algebraic manipulation into a tool for geometric understanding, modeling, and proof. Massachusetts Curriculum Framework for Mathematics 96 Conceptual Category: Geometry Overview [G] Congruence A. Experiment with transformations in the plane. B. Understand congruence in terms of rigid motions. C. Prove geometric theorems and, when appropriate, the converse of theorems. D. Make geometric constructions. Similarity, Right Triangles, and Trigonometry A. Understand similarity in terms of similarity transformations. B. Prove theorems involving similarity. C. Define trigonometric ratios and solve problems involving right triangles. D. Apply trigonometry to general triangles. Circles A. Understand and apply theorems about circles. B. Find arc lengths and areas of sectors of circles. Expressing Geometric Properties with Equations A. Translate between the geometric description and the equation for a conic section. B. Use coordinates to prove simple geometric theorems algebraically. Geometric Measurement and Dimension A. Explain volume formulas and use them to solve problems. B. Visualize relationships between two-dimensional and three-dimensional objects. Modeling with Geometry A. Apply geometric concepts in modeling situations. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 97 Conceptual Category: Geometry Content Standards [G] Congruence G-CO A. Experiment with transformations in the plane. 1. Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, distance along a line, and distance around a circular arc. 2. Represent transformations in the plane using, e.g., transparencies and geometry software; describe transformations as functions that take points in the plane as inputs and give other points as outputs. Compare transformations that preserve distance and angle to those that do not (e.g., translation versus horizontal stretch). 3. Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself. 4. Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. 5. Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another. B. Understand congruence in terms of rigid motions. 6. Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent. 7. Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent. 8. Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions. C. Prove geometric theorems and, when appropriate, the converse of theorems. 9. Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent, and conversely prove lines are parallel; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. 10. Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent, and conversely prove a triangle is isosceles; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; and the medians of a triangle meet at a point. 11. Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. a. Prove theorems about polygons. Theorems include the measures of interior and exterior angles. Apply properties of polygons to the solutions of mathematical and contextual problems. D. Make geometric constructions. 12. Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Constructions include: copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line. 13. Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle. Massachusetts Curriculum Framework for Mathematics 98 Similarity, Right Triangles, and Trigonometry G-SRT A. Understand similarity in terms of similarity transformations. 1. Verify experimentally the properties of dilations given by a center and a scale factor: a. A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged. b. The dilation of a line segment is longer or shorter in the ratio given by the scale factor. 2. Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides. 3. Use the properties of similarity transformations to establish the Angle-Angle (AA) criterion for two triangles to be similar. B. Prove theorems involving similarity. 4. Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity. 5. Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. C. Define trigonometric ratios and solve problems involving right triangles. 6. Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. 7. Explain and use the relationship between the sine and cosine of complementary angles. 8. Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. D. Apply trigonometry to general triangles. 9. (+) Derive the formula A = ½ ab sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side. 10. (+) Prove the Laws of Sines and Cosines and use them to solve problems. 11. (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces). Circles G-C A. Understand and apply theorems about circles. 1. Prove that all circles are similar. 2. Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle. 3. Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral and other polygons inscribed in a circle. 4. (+) Construct a tangent line from a point outside a given circle to the circle. B. Find arc lengths and areas of sectors of circles. 5. Derive using similarity the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector. Expressing Geometric Properties with Equations G-GPE A. Translate between the geometric description and the equation for a conic section. 1. Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation. 2. Derive the equation of a parabola given a focus and directrix. 3. (+) Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant. Massachusetts Curriculum Framework for Mathematics 99 a. (+) Use equations and graphs of conic sections to model real-world problems. B. Use coordinates to prove simple geometric theorems algebraically. 4. Use coordinates to prove simple geometric theorems algebraically including the distance formula and its relationship to the Pythagorean Theorem. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, ) lies on the circle centered at the origin and containing the point (0, 2). 5. Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point). 6. Find the point on a directed line segment between two given points that partitions the segment in a given ratio. 7. Use coordinates to compute perimeters of polygons and areas of triangles and rectangles (e.g., using the distance formula). Geometric Measurement and Dimension G-GMD A. Explain volume formulas and use them to solve problems. 1. Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri’s principle, and informal limit arguments. 2. (+) Give an informal argument using Cavalieri’s principle for the formulas for the volume of a sphere and other solid figures. 3. Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. B. Visualize relationships between two-dimensional and three-dimensional objects. 4. Identify the shapes of two-dimensional cross sections of three-dimensional objects, and identify three-dimensional objects generated by rotations of two-dimensional objects. Modeling with Geometry G-MG A. Apply geometric concepts in modeling situations. 1. Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder). 2. Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot). 3. Apply geometric methods to solve design problems (e.g., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios). 4. Use dimensional analysis for unit conversions to confirm that expressions and equations make sense. Massachusetts Curriculum Framework for Mathematics 100 Conceptual Category: Statistics and Probability [S] Introduction Decisions or predictions are often based on data—numbers in context. These decisions or predictions would be easy if the data always sent a clear message, but the message is often obscured by variability. Statistics provides tools for describing variability in data and for making informed decisions that take it into account. Data are gathered, displayed, summarized, examined, and interpreted to discover patterns and deviations from patterns. Quantitative data can be described in terms of key characteristics: measures of shape, center, and spread. The shape of a data distribution might be described as symmetric, skewed, flat, or bell shaped, and it might be summarized by a statistic measuring center (such as mean or median) and a statistic measuring spread (such as standard deviation or interquartile range). Different distributions can be compared numerically using these statistics or compared visually using plots. Knowledge of center and spread are not enough to describe a distribution. Which statistics to compare, which plots to use, and what the results of a comparison might mean, depend on the question to be investigated and the real-life actions to be taken. Randomization has two important uses in drawing statistical conclusions. First, collecting data from a random sample of a population makes it possible to draw valid conclusions about the whole population, taking variability into account. Second, randomly assigning individuals to different treatments allows a fair comparison of the effectiveness of those treatments. A statistically significant outcome is one that is unlikely to be due to chance alone, and this can be evaluated only under the condition of randomness. The conditions under which data are collected are important in drawing conclusions from the data. In critically reviewing uses of statistics in public media and other reports, it is important to consider the study design, how the data were gathered, and the analyses employed, as well as the data summaries and the conclusions drawn. Random processes can be described mathematically by using a probability model: a list or description of the possible outcomes (the sample space), each of which is assigned a probability. In situations such as flipping a coin, rolling a number cube, or drawing a card, it might be reasonable to assume various outcomes are equally likely. In a probability model, sample points represent outcomes and combine to make up events; probabilities of events can be computed by applying the Addition and Multiplication Rules. Interpreting these probabilities relies on an understanding of independence and conditional probability, which can be approached through the analysis of two-way tables. Technology plays an important role in statistics and probability by making it possible to generate plots, regression functions, and correlation coefficients, and to simulate many possible outcomes in a short amount of time. Connections to Functions and Modeling Functions may be used to describe data; if the data suggest a linear relationship, the relationship can be modeled with a regression line, and its strength and direction can be expressed through a correlation coefficient. Massachusetts Curriculum Framework for Mathematics 101 Conceptual Category: Statistics and Probability Overview [S] Interpreting Categorical and Quantitative Data A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. B. Summarize, represent, and interpret data on two categorical and quantitative variables. C. Interpret linear models. Making Inferences and Justifying Conclusions A. Understand and evaluate random processes underlying statistical experiments. Use calculators, spreadsheets, and other technology as appropriate. B. Make inferences and justify conclusions from sample surveys, experiments, and observational studies. Conditional Probability and the Rules of Probability A. Understand independence and conditional probability and use them to interpret data from simulations or experiments. B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. Using Probability to Make Decisions A. Calculate expected values and use them to solve problems. B. Use probability to evaluate outcomes of decisions. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 102 Conceptual Category: Statistics and Probability Content Standards [S] Interpreting Categorical and Quantitative Data S-ID A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. 1. Represent data with plots on the real number line (dot plots, histograms, and box plots). 2. Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. 3. Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). 4. Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve. B. Summarize, represent, and interpret data on two categorical and quantitative variables. 5. Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and conditional relative frequencies). Recognize possible associations and trends in the data. 6. Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a linear function to the data and use the fitted function to solve problems in the context of the data. Use functions fitted to data or choose a function suggested by the context. Emphasize linear and exponential models. b. Informally assess the fit of a function by plotting and analyzing residuals. c. Fit a linear function for a scatter plot that suggests a linear association. C. Interpret linear models. 7. Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. 8. Compute (using technology) and interpret the correlation coefficient of a linear fit. 9. Distinguish between correlation and causation. Making Inferences and Justifying Conclusions S-IC A. Understand and evaluate random processes underlying statistical experiments. Use calculators, spreadsheets, and other technology as appropriate. 1. Understand statistics as a process for making inferences about population parameters based on a random sample from that population. 2. Decide if a specified model is consistent with results from a given data-generating process (e.g., using simulation). For example, a model says a spinning coin falls heads up with probability 0.5. Would a result of five tails in a row cause you to question the model? B. Make inferences and justify conclusions from sample surveys, experiments, and observational studies. 3. Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization relates to each. 4. Use data from a sample survey to estimate a population mean or proportion; develop a margin of error through the use of simulation models for random sampling. 5. Use data from a randomized experiment to compare two treatments; use simulations to decide if differences between parameters are significant. 6. Evaluate reports based on data. Massachusetts Curriculum Framework for Mathematics 103 Conditional Probability and the Rules of Probability S-CP A. Understand independence and conditional probability and use them to interpret data from simulations or experiments. 1. Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). 2. Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. 3. Understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B. 4. Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. For example, collect data from a random sample of students in your school on their favorite subject among math, science, and English. Estimate the probability that a randomly selected student from your school will favor science given that the student is in tenth grade. Do the same for other subjects and compare the results. 5. Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer. B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. 6. Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A, and interpret the answer in terms of the model. 7. Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model. 8. (+) Apply the general Multiplication Rule in a uniform probability model, P(A and B) = P(A)P(B|A) = P(B)P(A|B), and interpret the answer in terms of the model. 9. (+) Use permutations and combinations to compute probabilities of compound events and solve problems. Using Probability to Make Decisions S-MD A. Calculate expected values and use them to solve problems. 1. (+) Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space; graph the corresponding probability distribution using the same graphical displays as for data distributions. 2. (+) Calculate the expected value of a random variable; interpret it as the mean of the probability distribution. 3. (+) Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value. For example, find the theoretical probability distribution for the number of correct answers obtained by guessing on all five questions of a multiple-choice test where each question has four choices, and find the expected grade under various grading schemes. 4. (+) Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value. For example, find a current data distribution on the number of TV sets per household in the United States, and calculate the expected number of sets per household. How many TV sets would you expect to find in 100 randomly selected households? B. Use probability to evaluate outcomes of decisions. 5. (+) Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values. a. (+) Find the expected payoff for a game of chance. Massachusetts Curriculum Framework for Mathematics 104 For example, find the expected winnings from a state lottery ticket or a game at a fast-food restaurant. b. (+) Evaluate and compare strategies on the basis of expected values. For example, compare a high-deductible versus a low-deductible automobile insurance policy using various, but reasonable, chances of having a minor or a major accident. 6. (+) Use probabilities to make fair decisions (e.g., drawing by lots or using a random number generator). 7. (+) Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, or pulling a hockey goalie at the end of a game and replacing the goalie with an extra skater). Massachusetts Curriculum Framework for Mathematics 105 The Standards for Mathematical Content High School: Model Pathways and Model Courses Introduction Model Traditional Pathway Model Algebra I (AI) Model Geometry (GEO) Model Algebra II (AII) Model Integrated Pathway Model Mathematics I (MI) Model Mathematics II (MII) Model Mathematics III (MIII) Advanced Model Courses Model Precalculus (PC) Model Advanced Quantitative Reasoning (AQR) Massachusetts Curriculum Framework for Mathematics 106 Introduction Organization of the Model High School Courses Each model high school course is presented in three sections: An introduction and description of the critical areas for learning in that course An overview listing the conceptual categories, domains, and clusters included in that course The content standards for that course, presented by conceptual category, domain, and cluster Content Standards by Model Course Identifiers/Coding Standard numbering in the high school model courses is similar to the coding of the high school standards by conceptual category; in addition, a course code has been added at the beginning of each standard to identify the standards in each model course. The illustration below shows a section of content standards from the Traditional Pathway Model Geometry course. The standard highlighted in the illustration is Standard GEO.G-SRT.C.8, identifying it as a standard from the Geometry Model Course; the Geometry conceptual category (G-), in the Similarity, Right Triangles, and Trigonometry domain (SRT), and as the eighth standard in that domain. The star () at the end of the standard indicates that it is a Modeling standard. The star symbol () following the standard in the illustration indicates that it is also a Modeling standard. Modeling is best interpreted not as a collection of isolated topics but in relation to other standards. Making mathematical models is a Standard for Mathematical Practice, and specific modeling standards appear throughout the high school standards indicated by a star symbol (). Massachusetts Curriculum Framework for Mathematics 107 Importance of Modeling in High School Modeling (indicated by a at the end of a standard) is defined as both a conceptual category for high school mathematics and a Standard for Mathematical Practice, and is an important avenue for motivating students to study mathematics, for building their understanding of mathematics, and for preparing them for future success. Development of the Model Pathways into instructional programs will require careful attention to modeling and the mathematical practices. Assessments based on these Model Pathways should reflect both the Standards for Mathematical Content and the Standards for Mathematical Practice. The following Model Pathways and Model Courses are presented in this Section: Model Traditional Pathway o Model Algebra I o Model Geometry o Model Algebra II Model Integrated Pathway o Model Mathematics I o Model Mathematics II o Model Mathematics III Advanced Model Courses o Model Precalculus o Model Advanced Quantitative Reasoning Massachusetts Curriculum Framework for Mathematics 108 Model Traditional Pathway: Model Algebra I [AI] Introduction The fundamental purpose of the Model Algebra I course is to formalize and extend the mathematics that students learned in the middle grades. For the high school Model Algebra I course, instructional time should focus on four critical areas: (1) deepen and extend understanding of linear and exponential relationships; (2) contrast linear and exponential relationships with each other and engage in methods for analyzing, solving, and using quadratic functions; (3) extend the laws of exponents to square and cube roots; and (4) apply linear models to data that exhibit a linear trend. 1. By the end of eighth grade, students have learned to solve linear equations in one variable and have applied graphical and algebraic methods to analyze and solve systems of linear equations in two variables. In Algebra I, students analyze and explain the process of solving an equation and justify the process used in solving a system of equations. Students develop fluency writing, interpreting, and translating among various forms of linear equations and inequalities, and use them to solve problems. They master the solution of linear equations and apply related solution techniques and the laws of exponents to the creation and solution of simple exponential equations. 2. In earlier grades, students define, evaluate, and compare functions, and use them to model relationships between quantities. In Algebra I, students learn function notation and develop the concepts of domain and range. They focus on linear, quadratic, and exponential functions, including sequences, and also explore absolute value, step, and piecewise-defined functions; they interpret functions given graphically, numerically, symbolically, and verbally; translate between representations; and understand the limitations of various representations. Students build on and extend their understanding of integer exponents to consider exponential functions. They compare and contrast linear and exponential functions, distinguishing between additive and multiplicative change. Students explore systems of equations and inequalities, and they find and interpret their solutions. They interpret arithmetic sequences as linear functions and geometric sequences as exponential functions. 3. Students extend the laws of exponents to rational exponents involving square and cube roots and apply this new understanding of number; they strengthen their ability to see structure in and create quadratic and exponential expressions. They create and solve equations, inequalities, and systems of equations involving quadratic expressions. Students become facile with algebraic manipulation, including rearranging and collecting terms and factoring. Students consider quadratic functions, comparing the key characteristics of quadratic functions to those of linear and exponential functions. They select from among these functions to model phenomena. Students learn to anticipate the graph of a quadratic function by interpreting various forms of quadratic expressions. In particular, they identify the real solutions of a quadratic equation as the zeros of a related quadratic function. Students expand their experience with functions to include more specialized functions—absolute value, step, and those that are piecewise-defined. 4. Building upon their prior experiences with data, students explore a more formal means of assessing how a model fits data. Students use regression techniques to describe approximately linear relationships between quantities. They use graphical representations and knowledge of context to make judgments about the appropriateness of linear models. With linear models, they look at residuals to analyze the goodness of fit. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 109 Model Traditional Pathway: Model Algebra I Overview [AI] Number and Quantity The Real Number System A. Extend the properties of exponents to rational exponents. B. Use properties of rational and irrational numbers. Quantities A. Reason quantitatively and use units to solve problems. Algebra Seeing Structure in Expressions A. Interpret the structure of linear, quadratic, and exponential expressions with integer exponents. B. Write expressions in equivalent forms to solve problems. Arithmetic with Polynomials and Rational Expressions A. Perform arithmetic operations on polynomials. Creating Equations A. Create equations that describe numbers or relationships. Reasoning with Equations and Inequalities A. Understand solving equations as a process of reasoning and explain the reasoning. B. Solve equations and inequalities in one variable. C. Solve systems of equations. D. Represent and solve equations and inequalities graphically. Functions Interpreting Functions A. Understand the concept of a function and use function notation. B. Interpret linear, quadratic, and exponential functions with integer exponents that arise in applications in terms of the context. C. Analyze functions using different representations. Building Functions A. Build a function that models a relationship between two quantities. B. Build new functions from existing functions. Linear, Quadratic, and Exponential Models A. Construct and compare linear, quadratic, and exponential models and solve problems. B. Interpret expressions for functions in terms of the situation they model. Statistics and Probability Interpreting Categorical and Quantitative Data A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. B. Summarize, represent, and interpret data on two categorical and quantitative variables. C. Interpret linear models. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 110 Model Traditional Pathway: Model Algebra I Content Standards [AI] Number and Quantity The Real Number System AI.N-RN A. Extend the properties of exponents to rational exponents. 1. Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 51/3 to be the cube root of 5 because we want (51/3)3 = 5(1/3)3 to hold, so (51/3)3 must equal 5. 2. Rewrite expressions involving radicals and rational exponents using the properties of exponents. B. Use properties of rational and irrational numbers. 3. Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational. Quantities AI.N-Q A. Reason quantitatively and use units to solve problems. 1. Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays. 2. Define appropriate quantities for the purpose of descriptive modeling. 3. Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. Algebra Seeing Structure in Expressions AI.A-SSE A. Interpret the structure of linear, quadratic, and exponential expressions with integer exponents. 1. Interpret expressions that represent a quantity in terms of its context. a. Interpret parts of an expression, such as terms, factors, and coefficients. b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1 + r)t as the product of P and a factor not depending on P. 2. Use the structure of an expression to identify ways to rewrite it. For example, see (x + 2)2 – 9 as a difference of squares that can be factored as ((x + 2) + 3)((x + 2 ) – 3). B. Write expressions in equivalent forms to solve problems. 3. Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. a. Factor a quadratic expression to reveal the zeros of the function it defines. b. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines. c. Use the properties of exponents to transform expressions for exponential functions. For example, the expression 1.15t can be rewritten as (1.151/12)12t ≈ 1.01212t to reveal the approximate equivalent monthly interest rate if the annual rate is 15%. Arithmetic with Polynomials and Rational Expressions AI.A-APR A. Perform arithmetic operations on polynomials. 1. Understand that polynomials form a system analogous to the integers, namely, they are closed under certain operations. Massachusetts Curriculum Framework for Mathematics 111 a. Perform operations on polynomial expressions (addition, subtraction, multiplication), and compare the system of polynomials to the system of integers when performing operations. b. Factor and/or expand polynomial expressions, identify and combine like terms, and apply the Distributive property. Creating Equations AI.A-CED A. Create equations that describe numbers or relationships. 1. Create equations and inequalities in one variable and use them to solve problems. (Include equations arising from linear, quadratic, and exponential functions with integer exponents.) 2. Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. 3. Represent constraints by linear equations or inequalities, and by systems of linear equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods. 4. Rearrange formulas to highlight a quantity of interest using the same reasoning as in solving equations (Properties of equality). For example, rearrange Ohm’s law to solve for voltage, V. Manipulate variables in formulas used in financial contexts such as for simple interest, I=Prt . Reasoning with Equations and Inequalities AI.A-REI A. Understand solving equations as a process of reasoning and explain the reasoning. 1. Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify or refute a solution method. B. Solve equations and inequalities in one variable. 3. Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. a. Solve linear equations and inequalities in one variable involving absolute value. 4. Solve quadratic equations in one variable. a. Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x – p)2 = q that has the same solutions. Derive the quadratic formula from this form. b. Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the solutions of a quadratic equation results in non-real solutions and write them as a ± bi for real numbers a and b. C. Solve systems of equations. 5. Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions. 6. Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. 7. Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. For example, find the points of intersection between the line y = –3x and the circle x2 + y2 = 3. D. Represent and solve equations and inequalities graphically. 10. Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line). Show that any point on the graph of an equation in two variables is a solution to the equation. 11. Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using Massachusetts Curriculum Framework for Mathematics 112 technology to graph the functions and make tables of values. Include cases where f(x) and/or g(x) are linear and exponential functions. 12. Graph the solutions of a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set of a system of linear inequalities in two variables as the intersection of the corresponding half-planes. Functions Interpreting Functions AI.F-IF A. Understand the concept of a function and use function notation. 1. Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output (range) of f corresponding to the input x. The graph of f is the graph of the equation y = f(x). 2. Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. For example, given a function representing a car loan, determine the balance of the loan at different points in time. 3. Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers. For example, the Fibonacci sequence is defined recursively by f(0) = f(1) = 1, f(n + 1) = f(n) + f(n 1) for n ≥ 1. B. Interpret linear, quadratic, and exponential functions with integer exponents that arise in applications in terms of the context. 4. For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior. 5. Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. 6. Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. C. Analyze functions using different representations. 7. Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. a. Graph linear and quadratic functions and show intercepts, maxima, and minima. b. Graph piecewise-defined functions, including step functions and absolute value functions. e. Graph exponential functions showing intercepts and end behavior. 8. Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Use the process of factoring and completing the square in a quadratic function to show zeros, maximum/minimum values, and symmetry of the graph, and interpret these in terms of a context. b. Use the properties of exponents to interpret expressions for exponential functions. Apply to financial situations such as identifying appreciation and depreciation rate for the value of a house or car some time after its initial purchase: . For example, identify percent rate of change in functions such as y = (1.02)t, y = (0.97)t, y = (1.01)12t, and y = (1.2) t ∕10, and classify them as representing exponential growth or decay. 9. Translate among different representations of functions (algebraically, graphically, numerically in tables, or by verbal descriptions). Compare properties of two functions each represented in a different way. Massachusetts Curriculum Framework for Mathematics 113 For example, given a graph of one quadratic function and an algebraic expression for another, say which has the larger maximum. Building Functions AI.F-BF A. Build a function that models a relationship between two quantities. 1. Write linear, quadratic, and exponential functions that describe a relationship between two quantities. a. Determine an explicit expression, a recursive process, or steps for calculation from a context. b. Combine standard function types using arithmetic operations. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model. 2. Write arithmetic and geometric sequences both recursively and with an explicit formula them to model situations, and translate between the two forms. B. Build new functions from existing functions. 3. Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f (x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Include linear, quadratic, exponential, and absolute value functions. Utilize technology to experiment with cases and illustrate an explanation of the effects on the graph. 4. Find inverse functions algebraically and graphically. a. Solve an equation of the form f(x) = c for a linear function f that has an inverse and write an expression for the inverse. Linear, Quadratic, and Exponential Models AI.F-LE A. Construct and compare linear, quadratic, and exponential models and solve problems. 1. Distinguish between situations that can be modeled with linear functions and with exponential functions. a. Prove that linear functions grow by equal differences over equal intervals, and that exponential functions grow by equal factors over equal intervals. b. Recognize situations in which one quantity changes at a constant rate per unit interval relative to another. c. Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another. 2. Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (including reading these from a table). 3. Observe, using graphs and tables, that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. B. Interpret expressions for functions in terms of the situation they model. 5. Interpret the parameters in a linear or exponential function (of the form f(x) = bx + k) in terms of a context. Statistics and Probability Interpreting Categorical and Quantitative Data AI.S-ID A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. 1. Represent data with plots on the real number line (dot plots, histograms, and box plots). 2. Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. 3. Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). Massachusetts Curriculum Framework for Mathematics 114 B. Summarize, represent, and interpret data on two categorical and quantitative variables. 5. Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and conditional relative frequencies). Recognize possible associations and trends in the data. 6. Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a linear function to the data and use the fitted function to solve problems in the context of the data. Use functions fitted to data or choose a function suggested by the context (emphasize linear and exponential models). b. Informally assess the fit of a function by plotting and analyzing residuals. c. Fit a linear function for a scatter plot that suggests a linear association. C. Interpret linear models. 7. Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. 8. Compute (using technology) and interpret the correlation coefficient of a linear fit. 9. Distinguish between correlation and causation. Massachusetts Curriculum Framework for Mathematics 115 Model Traditional Pathway: Model Geometry [GEO] Introduction The fundamental purpose of the Model Geometry course is to formalize and extend students’ geometric experiences from the middle grades. In this high school Model Geometry course, students explore more complex geometric situations and deepen their explanations of geometric relationships by presenting and hearing formal mathematical arguments. Important differences exist between this course and the historical approach taken in geometry classes. For example, transformations are emphasized in this course. Close attention should be paid to the introductory content for the Geometry conceptual category. For the high school Model Geometry course, instructional time should focus on six critical areas: (1) establish criteria for congruence of triangles based on rigid motions; (2) establish criteria for similarity of triangles based on dilations and proportional reasoning; (3) informally develop explanations of circumference, area, and volume formulas; (4) apply the Pythagorean Theorem to the coordinate plane; (5) prove basic geometric theorems; and (6) extend work with probability. 1. Students have prior experience with drawing triangles based on given measurements and performing rigid motions including translations, reflections, and rotations. They have used these to develop notions about what it means for two objects to be congruent. In this course, students establish triangle congruence criteria, based on analyses of rigid motions and formal constructions. They use triangle congruence as a familiar foundation for the development of formal proof. Students prove theorems— using a variety of formats including deductive and inductive reasoning and proof by contradiction—and solve problems about triangles, quadrilaterals, and other polygons. They apply reasoning to complete geometric constructions and explain why they work. 2. Students apply their earlier experience with dilations and proportional reasoning to build a formal understanding of similarity. They identify criteria for similarity of triangles, use similarity to solve problems, and apply similarity in right triangles to understand right triangle trigonometry, with particular attention to special right triangles and the Pythagorean Theorem. Students derive the Laws of Sines and Cosines in order to find missing measures of general (not necessarily right) triangles, building on their work with quadratic equations done in Model Algebra I. They are able to distinguish whether three given measures (angles or sides) define 0, 1, 2, or infinitely many triangles. 3. Students’ experience with three-dimensional objects is extended to include informal explanations of circumference, area, and volume formulas. Additionally, students apply their knowledge of two-dimensional shapes to consider the shapes of cross-sections and the result of rotating a two-dimensional object about a line. 4. Building on their work with the Pythagorean Theorem in eighth grade to find distances, students use the rectangular coordinate system to verify geometric relationships, including properties of special triangles and quadrilaterals, and slopes of parallel and perpendicular lines, which relates back to work done in the Model Algebra I course. Students continue their study of quadratics by connecting the geometric and algebraic definitions of the parabola. 5. Students prove basic theorems about circles, with particular attention to perpendicularity and inscribed angles, in order to see symmetry in circles and as an application of triangle congruence criteria. They study relationships among segments on chords, secants, and tangents as an application of similarity. In the Cartesian coordinate system, students use the distance formula to write the equation of a circle when given the radius and the coordinates of its center. Given an equation of a circle, they draw the graph in the coordinate plane, and apply techniques for solving quadratic equations—which relates back to work done in the Model Algebra I course—to determine intersections between lines and circles or parabolas and between two circles. 6. Building on probability concepts that began in the middle grades, students use the language of set theory to expand their ability to compute and interpret theoretical and experimental probabilities for Massachusetts Curriculum Framework for Mathematics 116 compound events, attending to mutually exclusive events, independent events, and conditional probability. Students should make use of geometric probability models wherever possible. They use probability to make informed decisions. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 117 Model Traditional Pathway: Model Geometry Overview [GEO] Number and Quantity Quantities A. Reason quantitatively and use units to solve problems. Geometry Congruence A. Experiment with transformations in the plane. B. Understand congruence in terms of rigid motions. C. Prove geometric theorems and, when appropriate, the converse of theorems. D. Make geometric constructions. Similarity, Right Triangles, and Trigonometry A. Understand similarity in terms of transformations. B. Prove theorems involving similarity. C. Define trigonometric ratios and solve problems involving right triangles. D. Apply trigonometry to general triangles. Circles A. Understand and apply theorems about circles. B. Find arc lengths and area of sectors of circles. Expressing Geometric Properties with Equations A. Translate between the geometric description and the equation for a conic section. B. Use coordinates to prove simple geometric theorems algebraically. Geometric Measurement and Dimension A. Explain volume formulas and use them to solve problems. B. Visualize relationships between two-dimensional and three-dimensional objects. Modeling with Geometry A. Apply geometric concepts in modeling situations. Statistics and Probability Conditional Probability and the Rules of Probability A. Understand independence and conditional probability and use them to interpret data from simulations or experiments. B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 118 Model Traditional Pathway: Model Geometry Content Standards [GEO] Number and Quantity Quantities GEO.N-Q A. Reason quantitatively and use units to solve problems. 3. Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. a. Describe the effects of approximate error in measurement and rounding on measurements and on computed values from measurements. Identify significant figures in recorded measures and computed values based on the context given and the precision of the tools used to measure. Geometry Congruence GEO.G-CO A. Experiment with transformations in the plane. 1. Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, distance along a line, and distance around a circular arc. 2. Represent transformations in the plane using, e.g., transparencies and geometry software; describe transformations as functions that take points in the plane as inputs and give other points as outputs. Compare transformations that preserve distance and angle to those that do not (e.g., translation versus horizontal stretch). 3. Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself. 4. Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. 5. Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another. B. Understand congruence in terms of rigid motions. 6. Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent. 7. Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent. 8. Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions. C. Prove geometric theorems and, when appropriate, the converse of theorems. 9. Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent, and conversely prove lines are parallel; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. 10. Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent, and conversely prove a triangle is isosceles; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. 11. Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. a. Prove theorems about polygons. Theorems include the measures of interior and exterior angles. Apply properties of polygons to the solutions of mathematical and contextual problems. Massachusetts Curriculum Framework for Mathematics 119 D. Make geometric constructions. 12. Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Constructions include: copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line. 13. Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle. Similarity, Right Triangles, and Trigonometry GEO.G-SRT A. Understand similarity in terms of similarity transformations. 1. Verify experimentally the properties of dilations given by a center and a scale factor: a. A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged. b. The dilation of a line segment is longer or shorter in the ratio given by the scale factor. 2. Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides. 3. Use the properties of similarity transformations to establish the Angle-Angle (AA) criterion for two triangles to be similar. B. Prove theorems involving similarity. 4. Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity. 5. Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. C. Define trigonometric ratios and solve problems involving right triangles. 6. Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. 7. Explain and use the relationship between the sine and cosine of complementary angles. 8. Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. D. Apply trigonometry to general triangles. 9. (+) Derive the formula A = ½ ab sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side. 10. (+) Prove the Laws of Sines and Cosines and use them to solve problems. 11. (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces). Circles GEO.G-C A. Understand and apply theorems about circles. 1. Prove that all circles are similar. 2. Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle. 3. Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral and other polygons inscribed in a circle. 4. (+) Construct a tangent line from a point outside a given circle to the circle. B. Find arc lengths and areas of sectors of circles. 5. Derive, using similarity, the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector. Expressing Geometric Properties with Equations GEO.G-GPE Massachusetts Curriculum Framework for Mathematics 120 A. Translate between the geometric description and the equation for a conic section. 1. Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation. 2. Derive the equation of a parabola given a focus and directrix. B. Use coordinates to prove simple geometric theorems algebraically. 4. Use coordinates to prove simple geometric theorems algebraically, including the distance formula and its relationship to the Pythagorean Theorem. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, ) lies on the circle centered at the origin and containing the point (0, 2). 5. Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point). 6. Find the point on a directed line segment between two given points that partitions the segment in a given ratio. 7. Use coordinates to compute perimeters of polygons and areas of triangles and rectangles (e.g., using the distance formula). Geometric Measurement and Dimension GEO.G-GMD A. Explain volume formulas and use them to solve problems. 1. Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri’s principle, and informal limit arguments. 2. (+) Give an informal argument using Cavalieri’s principle for the formulas for the volume of a sphere and other solid figures. 3. Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. B. Visualize relationships between two-dimensional and three-dimensional objects. 4. Identify the shapes of two-dimensional cross-sections of three-dimensional objects, and identify three-dimensional objects generated by rotations of two-dimensional objects. Modeling with Geometry GEO.G-MG A. Apply geometric concepts in modeling situations. 1. Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder). 2. Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot). 3. Apply geometric methods to solve design problems (e.g., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios). 4. Use dimensional analysis for unit conversions to confirm that expressions and equations make sense. Statistics and Probability Conditional Probability and the Rules of Probability GEO.S-CP A. Understand independence and conditional probability and use them to interpret data from simulations or experiments. 1. Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). 2. Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. 3. Understand the conditional probability of A given B as P(A and B)∕P(B), and interpret independence of A and B as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B. Massachusetts Curriculum Framework for Mathematics 121 4. Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. For example, collect data from a random sample of students in your school on their favorite subject among math, science, and English. Estimate the probability that a randomly selected student from your school will favor science given that the student is in tenth grade. Do the same for other subjects and compare the results. 5. Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer. B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. 6. Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A, and interpret the answer in terms of the model. 7. Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model. 8. (+) Apply the general Multiplication Rule in a uniform probability model, P(A and B) = P(A)P(B|A) = P(B)P(A|B), and interpret the answer in terms of the model. 9. (+) Use permutations and combinations to compute probabilities of compound events and solve problems. Massachusetts Curriculum Framework for Mathematics 122 Model Traditional Pathway: Model Algebra II [AII] Introduction Building on their work with linear, quadratic, and exponential functions, students extend their repertoire of functions to include logarithmic, polynomial, rational, and radical functions in the Model Algebra II course. Students work closely with the expressions that define the functions, are facile with algebraic manipulations of expressions, and continue to expand and hone their abilities to model situations and to solve equations, including solving quadratic equations over the set of complex numbers and solving exponential equations using the properties of logarithms. For the high school Model Algebra II course, instructional time should focus on four critical areas: (1) relate arithmetic of rational expressions to arithmetic of rational numbers; (2) expand understandings of functions and graphing to include trigonometric functions; (3) synthesize and generalize functions and extend understanding of exponential functions to logarithmic functions; and (4) relate data display and summary statistics to probability and explore a variety of data collection methods. 1. A central theme of this Model Algebra II course is that the arithmetic of rational expressions is governed by the same rules as the arithmetic of rational numbers. Students explore the structural similarities between the system of polynomials and the system of integers. They draw on analogies between polynomial arithmetic and base-ten computation, focusing on properties of operations, particularly the distributive property. Connections are made between multiplication of polynomials with multiplication of multi-digit integers, and division of polynomials with long division of integers. Students identify zeros of polynomials, including complex zeros of quadratic polynomials, and make connections between zeros of polynomials and solutions of polynomial equations. The Fundamental Theorem of Algebra is examined. 2. Building on their previous work with functions and on their work with trigonometric ratios and circles in the Model Geometry course, students now use the coordinate plane to extend trigonometry to model periodic phenomena. 3. Students synthesize and generalize what they have learned about a variety of function families. They extend their work with exponential functions to include solving exponential equations with logarithms. They explore the effects of transformations on graphs of diverse functions, including functions arising in an application, in order to abstract the general principle that transformations on a graph always have the same effect regardless of the type of the underlying function. They identify appropriate types of functions to model a situation, they adjust parameters to improve the model, and they compare models by analyzing appropriateness of fit and making judgments about the domain over which a model is a good fit. The description of modeling as “the process of choosing and using mathematics and statistics to analyze empirical situations, to understand them better, and to make decisions” is at the heart of this Model Algebra II course. The narrative discussion and diagram of the modeling cycle should be considered when knowledge of functions, statistics, and geometry is applied in a modeling context. 4. Students see how the visual displays and summary statistics they learned in earlier grades relate to different types of data and to probability distributions. They identify different ways of collecting data— including sample surveys, experiments, and simulations—and the role that randomness and careful design play in the conclusions that can be drawn. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 123 Model Traditional Pathway: Model Algebra II Overview [AII] Number and Quantity The Complex Number System A. Perform arithmetic operations with complex numbers. C. Use complex numbers in polynomial identities and equations. Vector and Matrix Quantities A. Represent and model with vector quantities. C. Perform operations on matrices and use matrices in applications. Algebra Seeing Structure in Expressions A. Interpret the structure of exponential, polynomial, and rational expressions. B. Write expressions in equivalent forms to solve problems. Arithmetic with Polynomials and Rational Expressions A. Perform arithmetic operations on polynomials. B. Understand the relationship between zeros and factors of polynomials. C. Use polynomial identities to solve problems. D. Rewrite rational expressions. Creating Equations A. Create equations that describe numbers or relationships. Reasoning with Equations and Inequalities A. Understand solving equations as a process of reasoning and explain the reasoning. D. Represent and solve equations and inequalities graphically. Functions Interpreting Functions B. Interpret functions that arise in applications in terms of the context (polynomial, rational, square root and cube root, trigonometric, and logarithmic functions). C. Analyze functions using different representations. Building Functions A. Build a function that models a relationship between two quantities. B. Build new functions from existing functions. Linear, Quadratic, and Exponential Models A. Construct and compare linear, quadratic, and exponential models and solve problems. Trigonometric Functions A. Extend the domain of trigonometric functions using the unit circle. B. Model periodic phenomena with trigonometric functions. C. Prove and apply trigonometric identities. Statistics and Probability Interpreting Categorical and Quantitative Data Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 124 A. Summarize, represent and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. Making Inferences and Justifying Conclusions A. Understand and evaluate random processes underlying statistical experiments. B. Make inferences and justify conclusions from sample surveys, experiments, and observational studies. Using Probability to Make Decisions B. Use probability to evaluate outcomes of decisions. Massachusetts Curriculum Framework for Mathematics 125 Model Traditional Pathway: Model Algebra II Content Standards [AII] Number and Quantity The Complex Number System AII.N-CN A. Perform arithmetic operations with complex numbers. 1. Know there is a complex number i such that i2 = −1, and every complex number has the form a + bi with x-a and b real. 2. Use the relation i2 = –1 and the Commutative, Associative, and Distributive properties to add, subtract, and multiply complex numbers. C. Use complex numbers in polynomial identities and equations. 7. Solve quadratic equations with real coefficients that have complex solutions. 8. (+) Extend polynomial identities to the complex numbers. For example, rewrite x2 + 4 as (x + 2i)(x – 2i). 9. (+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials. Vector and Matrix Quantities AII.N-VM A. Represent and model with vector quantities. 1. (+) Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., v, |v|, ||v||, v). 3. (+) Solve problems involving velocity and other quantities that can be represented by vectors. C. Perform operations on matrices and use matrices in applications. 6. (+) Use matrices to represent and manipulate data, e.g., to represent payoffs or incidence relationships in a network. 8. (+) Add, subtract, and multiply matrices of appropriate dimensions. 12. (+) Work with 2 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area. Algebra Seeing Structure in Expressions AII.A-SSE A. Interpret the structure of exponential, polynomial, and rational expressions. 1. Interpret expressions that represent a quantity in terms of its context. a. Interpret parts of an expression, such as terms, factors, and coefficients. b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1 + r)n as the product of P and a factor not depending on P. 2. Use the structure of an expression to identify ways to rewrite it. For example, see x4 – y4 as (x2)2 – (y2)2, thus recognizing it as a difference of squares that can be factored as (x2 – y2)(x2 + y2)and further factored (x-y)(x+y)(x-yi)(x+yi). B. Write expressions in equivalent forms to solve problems. 4. Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems. For example, calculate mortgage payments. Arithmetic with Polynomials and Rational Expressions AII.A-APR A. Perform arithmetic operations on polynomials. 1. Understand that polynomials form a system analogous to the integers, namely, they are closed under certain operations. Massachusetts Curriculum Framework for Mathematics 126 a. Perform operations on polynomial expressions (addition, subtraction, multiplication, and division), and compare the system of polynomials to the system of integers when performing operations. B. Understand the relationship between zeros and factors of polynomials. 2. Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x). 3. Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. C. Use polynomial identities to solve problems. 4. Prove polynomial identities and use them to describe numerical relationships. For example, the polynomial identity (x2 + y2)2 = (x2 – y2)2 + (2xy)2 can be used to generate Pythagorean triples. 5. (+) Know and apply the Binomial Theorem for the expansion of (x + y)n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal’s Triangle. D. Rewrite rational expressions. 6. Rewrite simple rational expressions in different forms; write a(x)∕b(x) in the form q(x) + r(x)∕b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system. 7. (+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions. Creating Equations AII.A-CED A. Create equations that describe numbers or relationships. 1. Create equations and inequalities in one variable and use them to solve problems. Include equations arising from simple root and rational functions and exponential functions. 2. Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. 3. Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent equations describing satellites orbiting Earth and constraints on Earth’s size and atmosphere. Reasoning with Equations and Inequalities AII.A-REI A. Understand solving equations as a process of reasoning and explain the reasoning. 2. Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise. D. Represent and solve equations and inequalities graphically. 11. Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are polynomial, rational, and logarithmic functions. Functions Interpreting Functions AII.F-IF B. Interpret functions that arise in applications in terms of the context (polynomial, rational, square root and cube root, trigonometric, and logarithmic functions). 4. For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of Massachusetts Curriculum Framework for Mathematics 127 the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. 5. Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. 6. Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. C. Analyze functions using different representations. 7. Graph functions expressed symbolically and show key features of the graph; by hand in simple cases and using technology for more complicated cases. b. Graph square root and cube root functions. c. Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. e. Graph exponential and logarithmic functions, showing intercepts and end behavior; and trigonometric functions, showing period, midline, and amplitude. 8. Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Use the process of factoring in a polynomial function to show zeros, extreme values, and symmetry of the graph and interpret these in terms of a context. 9. Translate among different representations of functions (algebraically, graphically, numerically in tables, or by verbal descriptions). Compare properties of two functions each represented in a different way. For example, given a graph of one polynomial function and an algebraic expression for another, say which has the larger relative maximum and/or smaller relative minimum. 10. Given algebraic, numeric and/or graphical representations of functions, recognize the function as polynomial, rational, logarithmic, exponential, or trigonometric. Building Functions AII.F-BF A. Build a function that models a relationship between two quantities. 1. Write a function (simple rational, radical, logarithmic, and trigonometric functions) that describes a relationship between two quantities. b. Combine standard function types using arithmetic operations. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model. B. Build new functions from existing functions. 3. Identify the effect on the graph of replacing f (x) by f (x) + k, k f (x), f (kx), and f (x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Include simple rational, radical, logarithmic, and trigonometric functions. Utilize technology to experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them. 4. Find inverse functions algebraically and graphically. a. Solve an equation of the form f(x) = c for a simple function f that has an inverse and write an expression for the inverse. For example, f(x) =2x3 or f(x)= (x + 1)∕(x 1) for x ≠ 1. Linear, Quadratic, and Exponential Models AII.F-LE A. Construct and compare linear, quadratic, and exponential models and solve problems. 4. For exponential models, express as a logarithm the solution to abct = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology. Massachusetts Curriculum Framework for Mathematics 128 Trigonometric Functions AII.F-TF A. Extend the domain of trigonometric functions using the unit circle. 1. Understand radian measure of an angle as the length of the arc on the unit circle subtended by the angle. 2. Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle. B. Model periodic phenomena with trigonometric functions. 5. Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline. C. Prove and apply trigonometric identities. 8. Prove the Pythagorean identity sin2(θ) + cos2(θ) = 1 and use it to find sin(θ), cos(θ), or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant. Statistics and Probability Interpreting Categorical and Quantitative Data AII.S-ID A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. 4. Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve. Making Inferences and Justifying Conclusions AII.S-IC A. Understand and evaluate random processes underlying statistical experiments. 1. Understand statistics as a process for making inferences to be made about population parameters based on a random sample from that population. 2. Decide if a specified model is consistent with results from a given data-generating process, e.g., using simulation. For example, a model says a spinning coin falls heads up with probability 0.5. Would a result of five tails in a row cause you to question the model? B. Make inferences and justify conclusions from sample surveys, experiments, and observational studies. 3. Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization relates to each. 4. Use data from a sample survey to estimate a population mean or proportion; develop a margin of error through the use of simulation models for random sampling. 5. Use data from a randomized experiment to compare two treatments; use simulations to decide if differences between parameters are significant. 6. Evaluate reports based on data. Using Probability to Make Decisions AII.S-MD B. Use probability to evaluate outcomes of decisions. 6. (+) Use probabilities to make fair decisions (e.g., drawing by lots, using a random number generator). 7. (+) Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a game and replacing the goalie with an extra skater). Massachusetts Curriculum Framework for Mathematics 129 Model Integrated Pathway: Model Mathematics I [MI] Introduction The fundamental purpose of the Model Mathematics I course is to formalize and extend the mathematics that students learned in the middle grades. For the high school Model Mathematics I course, instructional time should focus on six critical areas, each of which is described in more detail below: (1) extend understanding of numerical manipulation to algebraic manipulation; (2) synthesize understanding of function; (3) deepen and extend understanding of linear relationships; (4) apply linear models to data that exhibit a linear trend; (5) establish criteria for congruence based on rigid motions; and (6) apply the Pythagorean Theorem to the coordinate plane. 1. By the end of eighth grade students have had a variety of experiences working with expressions and creating equations. Students become facile with algebraic manipulation in much the same way that they are facile with numerical manipulation. Algebraic facility includes rearranging and collecting terms, factoring, identifying and canceling common factors in rational expressions, and applying properties of exponents. Students continue this work by using quantities to model and analyze situations, to interpret expressions, and to create equations to describe situations. 2. In earlier grades, students define, evaluate, and compare functions, and use them to model relationships among quantities. Students will learn function notation and develop the concepts of domain and range. They move beyond viewing functions as processes that take inputs and yield outputs and start viewing functions as objects in their own right. They explore many examples of functions, including sequences; interpret functions given graphically, numerically, symbolically, and verbally; translate between representations; and understand the limitations of various representations. They work with functions given by graphs and tables, keeping in mind that, depending upon the context, these representations are likely to be approximate and incomplete. Their work includes functions that can be described or approximated by formulas as well as those that cannot. When functions describe relationships between quantities arising from a context, students reason with the units in which those quantities are measured. Students build on and informally extend their understanding of integer exponents to consider exponential functions. They compare and contrast linear and exponential functions, distinguishing between additive and multiplicative change. They interpret arithmetic sequences as linear functions and geometric sequences as exponential functions. 3. By the end of eighth grade, students have learned to solve linear equations in one variable and have applied graphical and algebraic methods to analyze and solve systems of linear equations in two variables. Building on these earlier experiences, students analyze and explain the process of solving an equation, and justify the process used in solving a system of equations. Students develop fluency writing, interpreting, and translating among various forms of linear equations and inequalities, and use them to solve problems. They master the solution of linear equations and apply related solution techniques and the laws of exponents to the creation and solution of simple exponential equations. Students explore systems of equations and inequalities, and they find and interpret their solutions. All of this work is grounded on understanding quantities and on relationships among them. 4. Students’ prior experiences with data are the basis for the more formal means of assessing how a model fits data. Students use regression techniques to describe approximately linear relationships among quantities. They use graphical representations and knowledge of the context to make judgments about the appropriateness of linear models. With linear models, they look at residuals to analyze the goodness of fit. 5. In previous grades, students were asked to draw triangles based on given measurements. They also have prior experience with rigid motions: translations, reflections, and rotations, and have used these to develop notions about what it means for two objects to be congruent. Students establish triangle Massachusetts Curriculum Framework for Mathematics 130 congruence criteria, based on analyses of rigid motions and formal constructions. They solve problems about triangles, quadrilaterals, and other polygons. They apply reasoning to complete geometric constructions and explain why they work. 6. Building on their work with the Pythagorean Theorem in eighth grade to find distances, students use a rectangular coordinate system to verify geometric relationships, including properties of special triangles and quadrilaterals and slopes of parallel and perpendicular lines. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 131 Model Integrated Pathway: Model Mathematics I Overview [MI] Number and Quantity Quantities A. Reason quantitatively and use units to solve problems. Algebra Seeing Structure in Expressions A. Interpret the structure of linear and exponential expressions with integer exponents. Creating Equations A. Create equations that describe numbers or relationships. Reasoning with Equations and Inequalities A. Understand solving equations as a process of reasoning and explain the reasoning. B. Solve equations and inequalities in one variable. C. Solve systems of equations. D. Represent and solve equations and inequalities graphically. Functions Interpreting Functions A. Understand the concept of a function and use function notation. B. Interpret linear and exponential functions having integer exponents that arise in applications in terms of the context. C. Analyze functions using different representations. Building Functions A. Build a function that models a relationship between two quantities. B. Build new functions from existing functions. Linear, Quadratic, and Exponential Models A. Construct and compare linear and exponential models and solve problems. B. Interpret expressions for functions in terms of the situation they model. Geometry Congruence A. Experiment with transformations in the plane. B. Understand congruence in terms of rigid motions. D. Make geometric constructions. Expressing Geometric Properties with Equations B. Use coordinates to prove simple geometric theorems algebraically. Statistics and Probability Interpreting Categorical and Quantitative Data Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 132 A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. B. Summarize, represent, and interpret data on two categorical and quantitative variables. C. Interpret linear models. Massachusetts Curriculum Framework for Mathematics 133 Model Integrated Pathway: Model Mathematics I Content Standards [MI] Number and Quantity Quantities MI.N-Q A. Reason quantitatively and use units to solve problems. 1. Use units as a way to understand problems; and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays. 2. Define appropriate quantities for the purpose of descriptive modeling. 3. Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. Algebra Seeing Structure in Expressions MI. A-SSE A. Interpret the structure of linear and exponential expressions with integer exponents. 1. Interpret expressions that represent a quantity in terms of its context. a. Interpret parts of an expression, such as terms, factors, and coefficients. b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1 + r)n as the product of P and a factor not depending on P. Creating Equations MI. A-CED A. Create equations that describe numbers or relationships. 1. Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and exponential functions with integer exponents. 2. Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. 3. Represent constraints by linear equations or inequalities, and by systems of linear equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods. 4. Rearrange formulas to highlight a quantity of interest, using the same reasoning (Properties of equality) as in solving equations. For example, rearrange Ohm’s law, V = IR, to solve for resistance, R. Manipulate variables in formulas used in financial contexts such as for simple interest, I=Prt . Reasoning with Equations and Inequalities MI.A-REI A. Understand solving equations as a process of reasoning and explain the reasoning. 1. Explain each step in solving a simple linear equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify or refute a solution method. B. Solve equations and inequalities in one variable. 3. Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. a. Solve linear equations and inequalities in one variable involving absolute value. C. Solve systems of equations. 5. Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions. 6. Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. Massachusetts Curriculum Framework for Mathematics 134 D. Represent and solve equations and inequalities graphically. 10. Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line). Show that any point on the graph of an equation in two variables is a solution to the equation. 11. Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions and/or make tables of values. Include cases where f(x) and/or g(x) are linear and exponential functions. 12. Graph the solutions of a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set of a system of linear inequalities in two variables as the intersection of the corresponding half-planes. Functions Interpreting Functions MI. F-IF A. Understand the concept of a function and use function notation. 1. Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x). 2. Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. For example, given a function representing a car loan, determine the balance of the loan at different points in time. 3. Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers. For example, the Fibonacci sequence is defined recursively by f(0) = f(1) = 1, f(n + 1) = f(n) + f(n 1) for n ≥ 1. B. Interpret linear and exponential functions having integer exponents that arise in applications in terms of the context. 4. For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior. 5. Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. 6. Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. C. Analyze functions using different representations. 7. Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. a. Graph linear functions and show intercepts. e. Graph exponential functions, showing intercepts and end behavior. 9. Translate among different representations of functions: (algebraically, graphically, numerically in tables, or by verbal descriptions). Compare properties of two functions each represented in a different way. For example, given a graph of one exponential function and an algebraic expression for another, say which has the larger y-intercept. Building Functions MI.F-BF A. Build a function that models a relationship between two quantities. 1. Write linear and exponential functions that describe a relationship between two quantities. Massachusetts Curriculum Framework for Mathematics 135 a. Determine an explicit expression, a recursive process, or steps for calculation from a context. b. Combine standard function types using arithmetic operations. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model. 2. Write arithmetic and geometric sequences both recursively and with an explicit formula, use them to model situations, and translate between the two forms. B. Build new functions from existing functions. 3. Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Include linear and exponential models. (Focus on vertical translations for exponential functions). Utilize technology to experiment with cases and illustrate an explanation of the effects on the graph. Linear, Quadratic, and Exponential Models MI.F-LE A. Construct and compare linear and exponential models and solve problems. 1. Distinguish between situations that can be modeled with linear functions and with exponential functions. a. Prove that linear functions grow by equal differences over equal intervals, and that exponential functions grow by equal factors over equal intervals. b. Recognize situations in which one quantity changes at a constant rate per unit interval relative to another. c. Recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another. 2. Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (including reading these from a table). 3. Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly. B. Interpret expressions for functions in terms of the situation they model. 5. Interpret the parameters in a linear function or exponential function (of the form f(x) = bx + k) in terms of a context. Geometry Congruence MI.G-CO A. Experiment with transformations in the plane. 1. Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, distance along a line, and distance around a circular arc. 2. Represent transformations in the plane using, e.g., transparencies and geometry software; describe transformations as functions that take points in the plane as inputs and give other points as outputs. Compare transformations that preserve distance and angle to those that do not (e.g., translation versus horizontal stretch). 3. Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself. 4. Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. 5. Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another. B. Understand congruence in terms of rigid motions. 6. Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent. 7. Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent. Massachusetts Curriculum Framework for Mathematics 136 8. Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions. D. Make geometric constructions. 12. Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line. 13. Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle. Expressing Geometric Properties with Equations MI.G-GPE B. Use coordinates to prove simple geometric theorems algebraically. 5. Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point). 7. Use coordinates to compute perimeters of polygons and areas of triangles and rectangles (e.g., using the distance formula). Statistics and Probability Interpreting Categorical and Quantitative Data MI.S-ID A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. 1. Represent data with plots on the real number line (dot plots, histograms, and box plots). 2. Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. 3. Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). B. Summarize, represent, and interpret data on two categorical and quantitative variables. 5. Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and conditional relative frequencies). Recognize possible associations and trends in the data. 6. Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a linear function to the data and use the fitted function to solve problems in the context of the data. Use given functions fitted to data or choose a function suggested by the context. Emphasize linear and exponential models. b. Informally assess the fit of a function by plotting and analyzing residuals. c. Fit a linear function for a scatter plot that suggests a linear association. C. Interpret linear models. 7. Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. 8. Compute (using technology) and interpret the correlation coefficient of a linear fit. 9. Distinguish between correlation and causation. Massachusetts Curriculum Framework for Mathematics 137 Model Integrated Pathway: Model Mathematics II [MII] Introduction The focus of the Model Mathematics II course is on quadratic expressions, equations, and functions; comparing their characteristics and behavior to those of linear and exponential relationships from Model Mathematics I. For the high school Model Mathematics II course, instructional time should focus on five critical areas: (1) extend the laws of exponents to rational exponents; (2) compare key characteristics of quadratic functions with those of linear and exponential functions; (3) create and solve equations and inequalities involving linear, exponential, and quadratic expressions; (4) extend work with probability; and (5) establish criteria for similarity of triangles based on dilations and proportional reasoning. 1. Students extend the laws of exponents to rational exponents and explore distinctions between rational and irrational numbers by considering their decimal representations. Students learn that when quadratic equations do not have real solutions, the number system must be extended so that solutions exist; analogous to the way in which extending the whole numbers to the negative numbers allows x + 1 = 0 to have a solution. Students explore relationships between number systems: whole numbers, integers, rational numbers, real numbers, and complex numbers. The guiding principle is that equations with no solutions in one number system may have solutions in a larger number system. 2. Students consider quadratic functions, comparing the key characteristics of quadratic functions to those of linear and exponential functions. They select from among these functions to model phenomena. Students learn to anticipate the graph of a quadratic function by interpreting various forms of quadratic expressions. In particular, they identify the real solutions of a quadratic equation as the zeros of a related quadratic function. When quadratic equations do not have real solutions, students learn that that the graph of the related quadratic function does not cross the horizontal axis. They expand their experience with functions to include more specialized functions—absolute value, step, and those that are piecewise-defined. 3. Students begin by focusing on the structure of expressions, rewriting expressions to clarify and reveal aspects of the relationship they represent. They create and solve equations, inequalities, and systems of equations involving exponential and quadratic expressions. 4. Building on probability concepts that began in the middle grades, students use the language of set theory to expand their ability to compute and interpret theoretical and experimental probabilities for compound events, attending to mutually exclusive events, independent events, and conditional probability. Students should make use of geometric probability models wherever possible. They use probability to make informed decisions. 5. Students apply their earlier experience with dilations and proportional reasoning to build a formal understanding of similarity. They identify criteria for similarity of triangles, use similarity to solve problems, and apply similarity in right triangles to understand right triangle trigonometry, with particular attention to special right triangles and the Pythagorean Theorem. Students develop facility with geometric proof. They use what they know about congruence and similarity to prove theorems involving lines, angles, triangles, and other polygons. They explore a variety of formats for writing proofs. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 138 Model Integrated Pathway: Model Mathematics II Overview [MII] Number and Quantity The Real Number System A. Extend the properties of exponents to rational exponents. B. Use properties of rational and irrational numbers. Quantities A. Reason quantitatively and use units to solve problems. The Complex Number Systems A. Perform arithmetic operations with complex numbers. C. Use complex numbers in polynomial identities and equations. Algebra Seeing Structure in Expressions A. Interpret the structure of quadratic and exponential expressions. B. Write quadratic and exponential expressions in equivalent forms to solve problems. Arithmetic with Polynomials and Rational Expressions A. Perform arithmetic operations on polynomials. Creating Equations A. Create equations that describe numbers or relationships. Reasoning with Equations and Inequalities B. Solve equations and inequalities in one variable. C. Solve systems of equations. Functions Interpreting Functions B. Interpret quadratic and exponential functions with integer exponents that arise in applications in terms of the context. C. Analyze functions using different representations. Building Functions A. Build a function that models a relationship between two quantities. B. Build new functions from existing functions. Linear, Quadratic, and Exponential Models A. Construct and compare linear, quadratic and exponential models and solve problems. Geometry Congruence C. Prove geometric theorems, and when appropriate, the converse of theorems. Similarity, Right Triangles, and Trigonometry A. Understand similarity in terms of similarity transformations. B. Prove theorems involving similarity using a variety of ways of writing proofs, showing validity of underlying reasoning. C. Define trigonometric ratios and solve problems involving right triangles. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 139 Circles A. Understand and apply theorems about circles. B. Find arc lengths and areas of sectors of circles. Expressing Geometric Properties with Equations A. Translate between the geometric description and the equation for a conic section. B. Use coordinates to prove simple geometric theorems algebraically. Geometric Measurement and Dimension A. Explain volume formulas and use them to solve problems. Statistics and Probability Conditional Probability and the Rules of Probability A. Understand independence and conditional probability and use them to interpret data from simulations or experiments. B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. Massachusetts Curriculum Framework for Mathematics 140 Model Integrated Pathway: Model Mathematics II Content Standards [MII] Number and Quantity The Real Number System MII.N-RN A. Extend the properties of exponents to rational exponents. 1. Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 51/3 to be the cube root of 5 because we want (51/3)3 = 5(1/3)3 to hold, so (51/3)3 must equal 5. 2. Rewrite expressions involving radicals and rational exponents using the properties of exponents. B. Use properties of rational and irrational numbers. 3. Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational. Quantities MII.N-Q A. Reason quantitatively and use units to solve problems. 3. Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. a. Describe the effects of approximate error in measurement and rounding on measurements and on computed values from measurements. Identify significant figures in recorded measures and computed values based on the context given and the precision of the tools used to measure The Complex Number System MII.N-CN A. Perform arithmetic operations with complex numbers. 1. Know there is a complex number i such that i2 = −1, and every complex number has the form a + bi with a and b real. 2. Use the relation i2 = –1 and the Commutative, Associative, and Distributive properties to add, subtract, and multiply complex numbers. C. Use complex numbers in polynomial identities and equations. 7. Solve quadratic equations with real coefficients that have complex solutions. Algebra Seeing Structure in Expressions MII.A-SSE A. Interpret the structure of quadratic and exponential expressions. 1. Interpret expressions that represent a quantity in terms of its context. a. Interpret parts of an expression, such as terms, factors, and coefficients. b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1 + r)t as the product of P and a factor not depending on P. 2. Use the structure of an expression to identify ways to rewrite it. For example, see (x + 2)2 – 9 as a difference of squares that can be factored as ((x + 2) + 3)((x + 2) – 3). B. Write quadratic and exponential expressions in equivalent forms to solve problems. 3. Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. a. Factor a quadratic expression to reveal the zeros of the function it defines. Massachusetts Curriculum Framework for Mathematics 141 b. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines. c. Use the properties of exponents to transform expressions for exponential functions. For example, the expression 1.15t can be rewritten as (1.15 1/12)12t ≈ 1.01212t to reveal the approximate equivalent monthly interest rate if the annual rate is 15%. Arithmetic with Polynomials and Rational Expressions MII.A-APR A. Perform arithmetic operations on polynomials. 1. Understand that polynomials form a system analogous to the integers, namely, they are closed under certain operations. a. Perform operations on polynomial expressions (addition, subtraction, multiplication), and compare the system of polynomials to the system of integers when performing operations. b. Factor and/or expand polynomial expressions; identify and combine like terms; and apply the Distributive property. Creating Equations MII.A-CED A. Create equations that describe numbers or relationships. 1. Create equations and inequalities in one variable and use them to solve problems. Include equations arising from quadratic and exponential functions. 2. Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. 4. Rearrange formulas, including formulas with quadratic terms, to highlight a quantity of interest using the same reasoning as in solving equations (Properties of equality). For example, rearrange Ohm’s law to solve for voltage, V. Reasoning with Equations and Inequalities MII.A-REI B. Solve equations and inequalities in one variable. 4. Solve quadratic equations in one variable. a. Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x – p)2 = q that has the same solutions. Derive the quadratic formula from this form. b. Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b. C. Solve systems of equations. 7. Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. For example, find the points of intersection between the line y = –3x and the circle x2 + y2 = 3. Functions Interpreting Functions MII.F-IF B. Interpret quadratic and exponential functions with integer exponents that arise in applications in terms of the context. 4. For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of Massachusetts Curriculum Framework for Mathematics 142 the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior. 5. Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. 6. Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. C. Analyze functions using different representations. 7. Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. a. Graph quadratic functions and show intercepts, maxima, and minima. b. Graph piecewise-defined functions, including step functions and absolute value functions. 8. Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Use the process of factoring and completing the square in a quadratic function to show zeros, minimum/maximum values, and symmetry of the graph and interpret these in terms of a context. b. Use the properties of exponents to interpret expressions for exponential functions. Apply to financial situations such as Identifying appreciation/depreciation rate for the value of a house or car some time after its initial purchase: . For example, identify percent rate of change in functions such as y = (1.02)t, y = (0.97)t, y = (1.01)12t, and y = (1.2) t ∕10, and classify them as representing exponential growth or decay. 9. Translate among different representations of functions (algebraically, graphically, numerically in tables, or by verbal descriptions). Compare properties of two functions each represented in a different way. For example, given a graph of one quadratic function and an algebraic expression for another, say which has the larger maximum. Building Functions MII.F-BF A. Build a function that models a relationship between two quantities. 1. Write linear, quadratic, and exponential functions that describe relationships between two quantities. a. Determine an explicit expression, a recursive process, or steps for calculation from a context. b. Combine standard function types using arithmetic operations. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model. B. Build new functions from existing functions. 3. Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Include exponential, quadratic, and absolute value functions. Utilize technology to experiment with cases and illustrate an explanation of the effects on the graph. 4. Find inverse functions algebraically and graphically. a. Solve an equation of the form f(x) = c for a linear function f that has an inverse and write an expression for the inverse. Linear, Quadratic, and Exponential Models MII.F-LE A. Construct and compare linear, quadratic, and exponential models and solve problems. 3. Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. Massachusetts Curriculum Framework for Mathematics 143 Geometry Congruence MII.G-CO C. Prove geometric theorems and, when appropriate, the converse of theorems. 9. Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent, and conversely prove lines are parallel; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. 10. Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent, and conversely prove a triangle is isosceles; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. 11. Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. a. Prove theorems about polygons. Theorems include the measures of interior and exterior angles. Apply properties of polygons to the solutions of mathematical and contextual problems. Similarity, Right Triangles, and Trigonometry MII.G-SRT A. Understand similarity in terms of similarity transformations. 1. Verify experimentally the properties of dilations given by a center and a scale factor: a. A dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center unchanged. b. The dilation of a line segment is longer or shorter in the ratio given by the scale factor. 2. Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides. 3. Use the properties of similarity transformations to establish the Angle-Angle (AA) criterion for two triangles to be similar. B. Prove theorems involving similarity using a variety of ways of writing proofs, showing validity of underlying reasoning. 4. Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity. 5. Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. C. Define trigonometric ratios and solve problems involving right triangles. 6. Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. 7. Explain and use the relationship between the sine and cosine of complementary angles. 8. Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. Circles MII.G-C A. Understand and apply theorems about circles. 1. Prove that all circles are similar. 2. Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle. 3. Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral and other polygons inscribed in a circle. Massachusetts Curriculum Framework for Mathematics 144 4. (+) Construct a tangent line from a point outside a given circle to the circle. B. Find arc lengths and areas of sectors of circles. 5. Derive, using similarity, the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector. Expressing Geometric Properties with Equations MII.G-GPE A. Translate between the geometric description and the equation for a conic section. 1. Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation. 2. Derive the equation of a parabola given a focus and directrix. B. Use coordinates to prove simple geometric theorems algebraically. 4. Use coordinates to prove simple geometric theorems algebraically including the distance formula and its relationship to the Pythagorean Theorem. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, ) lies on the circle centered at the origin and containing the point (0, 2). 6. Find the point on a directed line segment between two given points that partitions the segment in a given ratio. Geometric Measurement and Dimension MII.G-GMD A. Explain volume formulas and use them to solve problems. 1. Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri’s principle, and informal limit arguments. 2. (+) Give an informal argument using Cavalieri’s principle for the formulas for the volume of a sphere and other solid figures. 3. Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. Statistics and Probability Conditional Probability and the Rules of Probability MII.S-CP A. Understand independence and conditional probability and use them to interpret data from simulations or experiments. 1. Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). 2. Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. 3. Understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B. 4. Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. For example, collect data from a random sample of students in your school on their favorite subject among math, science, and English. Estimate the probability that a randomly selected student from your school will favor science given that the student is in tenth grade. Do the same for other subjects and compare the results. Massachusetts Curriculum Framework for Mathematics 145 5. Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer. B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. 6. Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A, and interpret the answer in terms of the model. 7. Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model. 8. (+) Apply the general Multiplication Rule in a uniform probability model, P(A and B) = P(A)P(B|A) = P(B)P(A|B), and interpret the answer in terms of the model. 9. (+) Use permutations and combinations to compute probabilities of compound events and solve problems. Massachusetts Curriculum Framework for Mathematics 146 Model Integrated Pathway: Model Mathematics III [MIII] Introduction It is in the Model Mathematics III course that students integrate and apply the mathematics they have learned from their earlier courses. For the high school Model Mathematics III course, instructional time should focus on four critical areas: (1) apply methods from probability and statistics to draw inferences and conclusions from data; (2) expand understanding of functions to include polynomial, rational, and radical functions;26 (3) expand right triangle trigonometry to include general triangles; and (4) consolidate functions and geometry to create models and solve contextual problems. 1. Students see how the visual displays and summary statistics they learned in earlier grades relate to different types of data and to probability distributions. They identify different ways of collecting data— including sample surveys, experiments, and simulations—and the roles that randomness and careful design play in the conclusions that can be drawn. 2. The structural similarities between the system of polynomials and the system of integers are developed. Students draw on analogies between polynomial arithmetic and base-ten computation, focusing on properties of operations, particularly the distributive property. Students connect multiplication of polynomials with multiplication of multi-digit integers, and division of polynomials with long division of integers. Students identify zeros of polynomials and make connections between zeros of polynomials and solutions of polynomial equations. Rational numbers extend the arithmetic of integers by allowing division by all numbers except zero. Similarly, rational expressions extend the arithmetic of polynomials by allowing division by all polynomials except the zero polynomial. A central theme of the Model Mathematics III course is that the arithmetic of rational expressions is governed by the same rules as the arithmetic of rational numbers. This critical area also includes exploration of the Fundamental Theorem of Algebra. 3. Students derive the Laws of Sines and Cosines in order to find missing measures of general (not necessarily right) triangles. They are able to distinguish whether three given measures (angles or sides) define 0, 1, 2, or infinitely many triangles. This discussion of general triangles opens up the idea of trigonometry applied beyond the right triangle, at least to obtuse angles. Students build on this idea to develop the notion of radian measure for angles and extend the domain of the trigonometric functions to all real numbers. They apply this knowledge to model simple periodic phenomena. 4. Students synthesize and generalize what they have learned about a variety of function families. They extend their work with exponential functions to include solving exponential equations with logarithms. They explore the effects of transformations on graphs of diverse functions, including functions arising in an application, in order to abstract the general principle that transformations on a graph always have the same effect regardless of the type of underlying function. They identify appropriate types of functions to model a situation; they adjust parameters to improve the model; and they compare models by analyzing appropriateness of fit and making judgments about the domain over which a model is a good fit. The description of modeling as “the process of choosing and using mathematics and statistics to analyze empirical situations, to understand them better, and to make decisions” is at the heart of this Model Mathematics III course. The narrative discussion and diagram of the modeling cycle should be considered when knowledge of functions, statistics, and geometry is applied in a modeling context. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. 26 In this course, rational functions are limited to those whose numerators are of degree at most 1 and denominators are of degree at most 2; radical functions are limited to square roots or cube roots of at most quadratic polynomials. Massachusetts Curriculum Framework for Mathematics 147 Model Integrated Pathway: Model Mathematics III Overview [MIII] Number and Quantity The Complex Number Systems C. Use complex numbers in polynomial identities and equations. Vector and Matrix Quantities A. Represent and model with vector quantities. C. Perform operations on matrices and use matrices in applications. Algebra Seeing Structure in Expressions A. Interpret the structure polynomial and rational expressions. B. Write expressions in equivalent forms to solve problems. Arithmetic with Polynomials and Rational Expressions A. Perform arithmetic operations on polynomials. B. Understand the relationship between zeros and factors of polynomials. C. Use polynomial identities to solve problems. D. Rewrite rational expressions. Creating Equations A. Create equations that describe numbers or relationships. Reasoning with Equations and Inequalities A. Understand solving equations as a process of reasoning and explain the reasoning. D. Represent and solve equations and inequalities graphically. Functions Interpreting Functions B. Interpret functions that arise in applications in terms of the context (rational, polynomial, square root, cube root, trigonometric, logarithmic). C. Analyze functions using different representations. Building Functions A. Build a function that models a relationship between two quantities. B. Build new functions from existing functions. Linear, Quadratic, and Exponential Models A. Construct and compare linear, quadratic and exponential models and solve problems. Trigonometric Functions A. Extend the domain of trigonometric functions using the unit circle. B. Model periodic phenomena with trigonometric functions. C. Prove and apply trigonometric identities. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 148 Geometry Similarity, Right Triangles, and Trigonometry D. Apply trigonometry to general triangles. Geometric Measurement and Dimension B. Visualize relationships between two-dimensional and three-dimensional objects. Modeling with Geometry A. Apply geometric concepts in modeling situations. Statistics and Probability Interpreting Categorical and Quantitative Data A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. Making Inferences and Justifying Conclusions A. Understand and evaluate random processes underlying statistical experiments. B. Make inferences and justify conclusions from sample surveys, experiments, and observational studies. Using Probability to Make Decisions B. Use probability to evaluate outcomes of decisions. Massachusetts Curriculum Framework for Mathematics 149 Model Integrated Pathway: Model Mathematics III Content Standards [MIII] Number and Quantity The Complex Number System MIII.N-CN C. Use complex numbers in polynomial identities and equations. 8. (+) Extend polynomial identities to the complex numbers. For example, rewrite x2 + 4 as (x + 2i)(x – 2i). 9. (+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials. Vector and Matrix Quantities MIII.N-VM A. Represent and model with vector quantities. 1. (+) Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., v, |v|, ||v||, v). 3. (+) Solve problems involving velocity and other quantities that can be represented by vectors. C. Perform operations on matrices and use matrices in applications. 6. (+) Use matrices to represent and manipulate data, e.g., to represent payoffs or incidence relationships in a network. 8. (+) Add, subtract, and multiply matrices of appropriate dimensions. 12. (+) Work with 2 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area. Algebra Seeing Structure in Expressions MIII.A-SSE A. Interpret the structure of polynomial and rational expressions. 1. Interpret expressions that represent a quantity in terms of its context. a. Interpret parts of an expression, such as terms, factors, and coefficients. b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1 + r)n as the product of P and a factor not depending on P. 2. Use the structure of an expression to identify ways to rewrite it. For example, see x4 – y4 as (x2)2 – (y2)2, thus recognizing it as a difference of squares that can be factored as (x2 – y2)(x2 + y2) and as (x-y)(x+y)(x-yi)(x+yi). B. Write expressions in equivalent forms to solve problems. 4. Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems. For example, calculate mortgage payments. Arithmetic with Polynomials and Rational Expressions MIII.A-APR A. Perform arithmetic operations on polynomials. 1. Understand that polynomials form a system analogous to the integers, namely, they are closed under certain operations. a. Perform operations on polynomial expressions (addition, subtraction, multiplication, and division), and compare the system of polynomials to the system of integers when performing operations. B. Understand the relationship between zeros and factors of polynomials. Massachusetts Curriculum Framework for Mathematics 150 2. Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x). 3. Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. C. Use polynomial identities to solve problems. 4. Prove polynomial identities and use them to describe numerical relationships. For example, the polynomial identity (x2 + y2)2 = (x2 – y2)2 + (2xy)2 can be used to generate Pythagorean triples. 5. (+) Know and apply the Binomial Theorem for the expansion of (x + y)n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal’s Triangle. D. Rewrite rational expressions. 6. Rewrite simple rational expressions in different forms; write a(x)∕b(x) in the form q(x) + r(x)∕b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system. 7. (+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions. Creating Equations MIII.A-CED A. Create equations that describe numbers or relationships. 1. Create equations and inequalities in one variable and use them to solve problems. (Include equations arising from simple root and rational functions.) 2. Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. 3. Represent constraints by equations or inequalities and by systems of equations and/or inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent equations describing satellites orbiting earth and constraints on earth’s size and atmosphere. Reasoning with Equations and Inequalities MIII.A-REI A. Understand solving equations as a process of reasoning and explain the reasoning. 2. Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise. D. Represent and solve equations and inequalities graphically. 11. Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are polynomial, rational, and logarithmic functions. Functions Interpreting Functions MIII.F-IF B. Interpret functions that arise in applications in terms of the context (rational, polynomial, square root, cube root, trigonometric, logarithmic). 4. For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. 5. Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. Massachusetts Curriculum Framework for Mathematics 151 For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. 6. Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. C. Analyze functions using different representations. 7. Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. b. Graph square root and cube root functions. c. Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. e. Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and amplitude. 8. Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Use the process of factoring in a polynomial function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. 9. Translate among different representations of functions: (algebraically, graphically, numerically in tables, or by verbal descriptions). Compare properties of two functions each represented in a different way. For example, given a graph of one polynomial function and an algebraic expression for another, say which has the larger relative maximum and/or smaller relative minimum. 10. Given algebraic, numeric and/or graphical representations of functions, recognize the function as polynomial, rational, logarithmic, exponential, or trigonometric. Building Functions MIII.F-BF A. Build a function that models a relationship between two quantities. 1. Write simple rational and radical functions, logarithmic, and trigonometric functions that describes a relationship between two quantities. b. Combine standard function types using arithmetic operations. For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these functions to the model (include simple rational and radical functions, logarithmic, and trigonometric functions). B. Build new functions from existing functions. 3. Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Include simple rational, radical, logarithmic, and trigonometric functions. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them. 4. Find inverse functions algebraically and graphically. a. Solve an equation of the form f(x) = c for a simple function f that has an inverse and write an expression for the inverse. For example, f(x) =2x3 or f(x) = (x + 1)∕(x 1) for x ≠ 1. Linear, Quadratic, and Exponential Models MIII.F-LE A. Construct and compare linear, quadratic, and exponential models and solve problems. 4. For exponential models, express as a logarithm the solution to abct = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology. Trigonometric Functions MIII.F-TF A. Extend the domain of trigonometric functions using the unit circle. Massachusetts Curriculum Framework for Mathematics 152 1. Understand radian measure of an angle as the length of the arc on the unit circle subtended by the angle. 2. Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle. B. Model periodic phenomena with trigonometric functions. 5. Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline. C. Prove and apply trigonometric identities. 8. Prove the Pythagorean identity sin2(θ) + cos2(θ) = 1 and use it to find sin(θ), cos(θ), or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant. Geometry Similarity, Right Triangles, and Trigonometry MIII.G-SRT D. Apply trigonometry to general triangles. 9. (+) Derive the formula A = ½ ab sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side. 10. (+) Prove the Laws of Sines and Cosines and use them to solve problems. 11. (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces). Geometric Measurement and Dimension MIII.G-GMD B. Visualize relationships between two-dimensional and three-dimensional objects. 4. Identify the shapes of two-dimensional cross-sections of three-dimensional objects, and identify three-dimensional objects generated by rotations of two-dimensional objects. Modeling with Geometry MIII.G-MG A. Apply geometric concepts in modeling situations. 1. Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder). 2. Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot). 3. Apply geometric methods to solve design problems (e.g., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios). 4. Use dimensional analysis for unit conversions to confirm that expressions and equations make sense. Statistics and Probability Interpreting Categorical and Quantitative Data MIII.S-ID A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. 4. Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve. Making Inferences and Justifying Conclusions MIII.S-IC A. Understand and evaluate random processes underlying statistical experiments. Massachusetts Curriculum Framework for Mathematics 153 1. Understand statistics as a process for making inferences to be made about population parameters based on a random sample from that population. 2. Decide if a specified model is consistent with results from a given data-generating process, e.g., using simulation. For example, a model says a spinning coin falls heads up with probability 0.5. Would a result of five tails in a row cause you to question the model? B. Make inferences and justify conclusions from sample surveys, experiments, and observational studies. 3. Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization relates to each. 4. Use data from a sample survey to estimate a population mean or proportion; develop a margin of error through the use of simulation models for random sampling. 5. Use data from a randomized experiment to compare two treatments; use simulations to decide if differences between parameters are significant. 6. Evaluate reports based on data. Using Probability to Make Decisions MIII.S-MD B. Use probability to evaluate outcomes of decisions. 6. (+) Use probabilities to make fair decisions (e.g., drawing by lots, using a random number generator). 7. (+) Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a game and replacing the goalie with an extra skater). Massachusetts Curriculum Framework for Mathematics 154 Model Advanced Course: Model Precalculus [PC] Introduction Precalculus combines the trigonometric, geometric, and algebraic techniques needed to prepare students for the study of calculus, and strengthens students’ conceptual understanding of problems and mathematical reasoning in solving problems. Facility with these topics is especially important for students intending to study calculus, physics, and other sciences, and/or engineering in college. Because the standards for this course are (+) standards, students selecting this Model Precalculus course should have met the college and career ready standards. For the high school Model Precalculus course, instructional time should focus on four critical areas: (1) extend work with complex numbers; (2) expand understanding of logarithms and exponential functions; (3) use characteristics of polynomial and rational functions to sketch graphs of those functions; and (4) perform operations with vectors. 1. Students continue their work with complex numbers. They perform arithmetic operations with complex numbers and represent them and the operations on the complex plane. Students investigate and identify the characteristics of the graphs of polar equations, using graphing tools. This includes classification of polar equations; the effects of changes in the parameters in polar equations; conversion of complex numbers from rectangular form to polar form and vice versa; and the intersection of the graphs of polar equations. 2. Students expand their understanding of functions to include logarithmic and trigonometric functions. They investigate and identify the characteristics of exponential and logarithmic functions in order to graph these functions and solve equations and practical problems. This includes the role of e, natural and common logarithms, laws of exponents and logarithms, and the solutions of logarithmic and exponential equations. Students model periodic phenomena with trigonometric functions and prove trigonometric identities. Other trigonometric topics include reviewing unit circle trigonometry, proving trigonometric identities, solving trigonometric equations, and graphing trigonometric functions. 3. Students investigate and identify the characteristics of polynomial and rational functions and use these to sketch the graphs of the functions. They determine zeros, upper and lower bounds, y-intercepts, symmetry, asymptotes, intervals for which the function is increasing or decreasing, and maximum or minimum points. Students translate between the geometric description and equation of conic sections. They deepen their understanding of the Fundamental Theorem of Algebra. 4. Students perform operations with vectors in the coordinate plane and solve practical problems using vectors. This includes the following topics: operations of addition, subtraction, scalar multiplication, and inner (dot) product; norm of a vector; unit vector; graphing; properties; simple proofs; complex numbers (as vectors); and perpendicular components. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 155 Model Advanced Course: Model Precalculus Overview [PC] Number and Quantity The Complex Number System A. Perform arithmetic operations with complex numbers. B. Represent complex numbers and their operations on the complex plane. C. Use complex numbers in polynomial identities and equations. Vector and Matrix Quantities A. Represent and model with vector quantities. B. Perform operations on vectors. C. Perform operations on matrices and use matrices in applications. Algebra Arithmetic with Polynomials and Rational Expressions C. Use polynomial identities to solve problems D. Rewrite rational expressions. Reasoning with Equations and Inequalities C. Solve systems of equations. Functions Interpreting Functions C. Analyze functions using different representations. Building Functions A. Build a function that models a relationship between two quantities. B. Build new functions from existing functions. Trigonometric Functions A. Extend the domain of trigonometric functions using the unit circle. B. Model periodic phenomena with trigonometric functions. C. Prove and apply trigonometric identities. Geometry Similarity, Right Triangles, and Trigonometry D. Apply trigonometry to general triangles. Circles A. Understand and apply theorems about circles. Expressing Geometric Properties with Equations A. Translate between the geometric description and the equation for a conic section. Geometric Measurement and Dimension A. Explain volume formulas and use them to solve problems. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 156 Model Advanced Course: Model Precalculus Content Standards [PC] Number and Quantity The Complex Number System PC.N-CN A. Perform arithmetic operations with complex numbers. 3. (+) Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers. B. Represent complex numbers and their operations on the complex plane. 4. (+) Represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number. 5. (+) Represent addition, subtraction, multiplication, and conjugation of complex numbers geometrically on the complex plane; use properties of this representation for computation. For example, ( )3=8 because ( ) has modulus 2 and argument 120°. 6. (+) Calculate the distance between numbers in the complex plane as the modulus of the difference, and the midpoint of a segment as the average of the numbers at its endpoints. C. Use complex numbers in polynomial identities and equations. 8. (+) Extend polynomial identities to the complex numbers. For example, rewrite x2 + 4 as (x + 2i)(x – 2i). 9. (+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials. Vector and Matrix Quantities PC.N-VM A. Represent and model with vector quantities. 1. (+) Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., v, |v|, ||v||, v). 2. (+) Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point. 3. (+) Solve problems involving velocity and other quantities that can be represented by vectors. B. Perform operations on vectors. 4. (+) Add and subtract vectors. a. (+) Add vectors end-to-end, component-wise, and by the parallelogram rule. Understand that the magnitude of a sum of two vectors is typically not the sum of the magnitudes. b. (+) Given two vectors in magnitude and direction form, determine the magnitude and direction of their sum. c. (+) Understand vector subtraction v – w as v + (–w), where –w is the additive inverse of w, with the same magnitude as w and pointing in the opposite direction. Represent vector subtraction graphically by connecting the tips in the appropriate order, and perform vector subtraction component-wise. 5. (+) Multiply a vector by a scalar. a. (+) Represent scalar multiplication graphically by scaling vectors and possibly reversing their direction; perform scalar multiplication component-wise, e.g., as c(vx, vy) = (cvx, cvy). b. (+) Compute the magnitude of a scalar multiple cv using ||cv|| = |c|v. Compute the direction of cv knowing that when |c|v ≠ 0, the direction of cv is either along v (for c > 0) or against v (for c < 0). C. Perform operations on matrices and use matrices in applications. Massachusetts Curriculum Framework for Mathematics 157 6. (+) Use matrices to represent and manipulate data, e.g., to represent payoffs or incidence relationships in a network. 7. (+) Multiply matrices by scalars to produce new matrices, e.g., as when all of the payoffs in a game are doubled. 8. (+) Add, subtract, and multiply matrices of appropriate dimensions. 9. (+) Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a Commutative operation, but still satisfies the Associative and Distributive properties. 10. (+) Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse. 11. (+) Multiply a vector (regarded as a matrix with one column) by a matrix of suitable dimensions to produce another vector. Work with matrices as transformations of vectors. 12. (+) Work with 2 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area. Algebra Arithmetic with Polynomials and Rational Expressions PC.A-APR C. Use polynomial identities to solve problems. 5. (+) Know and apply the Binomial Theorem for the expansion of (x + y)n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal’s Triangle.27 D. Rewrite rational expressions. 7. (+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions. Reasoning with Equations and Inequalities PC.A-REI C. Solve systems of equations. 8. (+) Represent a system of linear equations as a single matrix equation in a vector variable. 9. (+) Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3 3 or greater). Functions Interpreting Functions PC.F-IF C. Analyze functions using different representations. 7. Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. d. (+) Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior. Building Functions PC.F-BF A. Build a function that models a relationship between two quantities. 1. Write a function that describes a relationship between two quantities. c. (+) Compose functions. 27 The Binomial Theorem can be proved by mathematical induction or by a combinatorial argument. Massachusetts Curriculum Framework for Mathematics 158 For example, if T(y) is the temperature in the atmosphere as a function of height, and h(t) is the height of a weather balloon as a function of time, then T(h(t)) is the temperature at the location of the weather balloon as a function of time. B. Build new functions from existing functions. 4. Find inverse functions. b. (+) Verify by composition that one function is the inverse of another. c. (+) Read values of an inverse function from a graph or a table, given that the function has an inverse. d. (+) Produce an invertible function from a non-invertible function by restricting the domain. 5. (+) Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents. Trigonometric Functions PC.F-TF A. Extend the domain of trigonometric functions using the unit circle. 3. (+) Use special triangles to determine geometrically the values of sine, cosine, tangent for ∕3, ∕4 and ∕6, and use the unit circle to express the values of sine, cosine, and tangent for x, + x, and 2 x in terms of their values for x, where x is any real number. 4. (+) Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions. B. Model periodic phenomena with trigonometric functions. 6. (+) Understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed. 7. (+) Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context. C. Prove and apply trigonometric identities. 9. (+) Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems. Geometry Similarity, Right Triangles, and Trigonometry PC.G-SRT D. Apply trigonometry to general triangles. 9. (+) Derive the formula A = ½ ab sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side. 10. (+) Prove the Laws of Sines and Cosines and use them to solve problems. 11. (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces). Circles PC.G-C A. Understand and apply theorems about circles. 4. (+) Construct a tangent line from a point outside a given circle to the circle. Expressing Geometric Properties with Equations PC.G-GPE A. Translate between the geometric description and the equation for a conic section. 3. (+) Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant. a. (+) Use equations and graphs of conic sections to model real-world problems. Geometric Measurement and Dimension PC.G-GMD Massachusetts Curriculum Framework for Mathematics 159 A. Explain volume formulas and use them to solve problems. 2. (+) Give an informal argument using Cavalieri’s principle for the formulas for the volume of a sphere and other solid figures. Massachusetts Curriculum Framework for Mathematics 160 Model Advanced Course: Model Advanced Quantitative Reasoning [AQR] Introduction Because the standards for this course are (+) standards, students selecting this Model Advanced Quantitative Reasoning course should have met the college and career ready standards. The high school Model Advanced Quantitative Reasoning course is designed as a mathematics course alternative to Precalculus. Through this course, students are encouraged to continue their study of mathematical ideas in the context of real-world problems and decision making through the analysis of information, modeling change, and mathematical relationships. For the high school Model Advanced Quantitative Reasoning course, instructional time should focus on three critical areas: (1) critique quantitative data; (2) investigate and apply various mathematical models; and (3) explore and apply concepts of vectors and matrices to model and solve real-world problems. 1. Students learn to become critical consumers of the quantitative data that surround them every day, knowledgeable decision makers who use logical reasoning, and mathematical thinkers who can use their quantitative skills to solve problems related to a wide range of situations. They link classroom mathematics and statistics to everyday life, work, and decision making, using mathematical modeling. They choose and use appropriate mathematics and statistics to analyze empirical situations, to understand them better, and to improve decisions. 2. Through the investigation of mathematical models from real-world situations, students strengthen conceptual understandings in mathematics and further develop connections between algebra and geometry. Students use geometry to model real-world problems and solutions. They use the language and symbols of mathematics in representations and communication. 3. Students explore linear algebra concepts of matrices and vectors. They use vectors to model physical relationships to define and solve real-world problems. Students draw, name, label, and describe vectors, perform operations with vectors, and relate these components to vector magnitude and direction. They use matrices in relationship to vectors and to solve problems. The Standards for Mathematical Practice complement the content standards so that students increasingly engage with the subject matter as they grow in mathematical maturity and expertise throughout the elementary, middle, and high school years. Massachusetts Curriculum Framework for Mathematics 161 Model Advanced Course: Model Advanced Quantitative Reasoning Overview [AQR] Number and Quantity Vector and Matrix Quantities A. Represent and model with vector quantities. C. Perform operations on matrices and use matrices in applications. Algebra Arithmetic with Polynomials and Rational Expressions C. Use polynomials identities to solve problems. Reasoning with Equations and Inequalities C. Solve systems of equations. Functions Trigonometric Functions A. Extend the domain of trigonometric functions using the unit circle. B. Model periodic phenomena with trigonometric functions. C. Prove and apply trigonometric identities. Geometry Similarity, Right Triangles, and Trigonometry D. Apply trigonometry to general triangles. Circles A. Understand and apply theorems about circles. Expressing Geometric Properties with Equations A. Translate between the geometric description and the equation for a conic section. Geometric Measurement and Dimension A. Explain volume formulas and use them to solve problems. Statistics and Probability Conditional Probability and the Rules of Probability B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. Using Probability to Make Decisions A. Calculate expected values and use them to solve problems. B. Use probability to evaluate outcomes of decisions. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Massachusetts Curriculum Framework for Mathematics 162 Model Advanced Course: Model Advanced Quantitative Reasoning Content Standards [AQR] Number and Quantity Vector and Matrix Quantities AQR.N-VM A. Represent and model with vector quantities. 1. (+) Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., v, |v|, ||v||, v). 2. (+) Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point. 3. (+) Solve problems involving velocity and other quantities that can be represented by vectors. C. Perform operations on matrices and use matrices in applications. 6. (+) Use matrices to represent and manipulate data, e.g., to represent payoffs or incidence relationships in a network. 7. (+) Multiply matrices by scalars to produce new matrices, e.g., as when all of the payoffs in a game are doubled. 8. (+) Add, subtract, and multiply matrices of appropriate dimensions. 9. (+) Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a Commutative operation, but still satisfies the Associative and Distributive properties. 10. (+) Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse. 11. (+) Multiply a vector (regarded as a matrix with one column) by a matrix of suitable dimensions to produce another vector. Work with matrices as transformations of vectors. 12. (+) Work with 2 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area. Algebra Arithmetic with Polynomials and Rational Expressions AQR.A-APR C. Use polynomial identities to solve problems. 5. (+) Know and apply the Binomial Theorem for the expansion of (x + y)n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal’s Triangle.28 Reasoning with Equations and Inequalities AQR.A-REI C. Solve systems of equations. 8. (+) Represent a system of linear equations as a single matrix equation in a vector variable. 9. (+) Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3 3 or greater). 28 The Binomial Theorem can be proved by mathematical induction or by a combinatorial argument. Massachusetts Curriculum Framework for Mathematics 163 Functions Trigonometric Functions AQR.F-TF A. Extend the domain of trigonometric functions using the unit circle. 3. (+) Use special triangles to determine geometrically the values of sine, cosine, tangent for ∕3, ∕4 and ∕6, and use the unit circle to express the values of sine, cosine, and tangent for x, + x, and 2 x in terms of their values for x, where x is any real number. 4. (+) Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions. B. Model periodic phenomena with trigonometric functions. 7. (+) Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context. C. Prove29 and apply trigonometric identities. 9. (+) Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems. Geometry Similarity, Right Triangles, and Trigonometry AQR.G-SRT D. Apply trigonometry to general triangles. 11. (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces). Circles AQR.G-C A. Understand and apply theorems about circles. 4. (+) Construct a tangent line from a point outside a given circle to the circle. Expressing Geometric Properties with Equations AQR.G-GPE A. Translate between the geometric description and the equation for a conic section. 3. (+) Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant. a. (+) Use equations and graphs of conic sections to model real-world problems. Geometric Measurement and Dimension AQR.G-GMD A. Explain volume formulas and use them to solve problems. 2. (+) Give an informal argument using Cavalieri’s principle for the formulas for the volume of a sphere and other solid figures. Statistics and Probability Conditional Probability and the Rules of Probability AQR.S-CP B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. 8. (+) Apply the general Multiplication Rule in a uniform probability model, P(A and B) = P(A)P(B|A) = P(B)P(A|B), and interpret the answer in terms of the model. 29 Advanced Quantitative Reasoning should accept informal proof and focus on the underlying reasoning, and use the theorems to solve problems. Massachusetts Curriculum Framework for Mathematics 164 9. (+) Use permutations and combinations to compute probabilities of compound events and solve problems. Using Probability to Make Decisions AQR.S-MD A. Calculate expected values and use them to solve problems. 1. (+) Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space; graph the corresponding probability distribution using the same graphical displays as for data distributions. 2. (+) Calculate the expected value of a random variable; interpret it as the mean of the probability distribution. 3. (+) Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value. For example, find the theoretical probability distribution for the number of correct answers obtained by guessing on all five questions of a multiple-choice test where each question has four choices, and find the expected grade under various grading schemes. 4. (+) Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value. For example, find a current data distribution on the number of TV sets per household in the United States, and calculate the expected number of sets per household. How many TV sets would you expect to find in 100 randomly selected households? B. Use probability to evaluate outcomes of decisions. 5. (+) Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values. a. (+) Find the expected payoff for a game of chance. For example, find the expected winnings from a state lottery ticket or a game at a fast-food restaurant. b. (+) Evaluate and compare strategies on the basis of expected values. For example, compare a high-deductible versus a low-deductible automobile insurance policy using various, but reasonable, chances of having a minor or a major accident. 6. (+) Use probabilities to make fair decisions (e.g., drawing by lots, using a random number generator). 7. (+) Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a game and replacing the goalie with an extra skater). Massachusetts Curriculum Framework for Mathematics 165 Making Decisions about High School Course Sequences and Algebra I in Grade 8 Course Sequences and the Model Algebra I Course The 2017 Massachusetts Curriculum Framework for Mathematics represents an opportunity to revisit course sequences in middle and high school mathematics. Districts should work with stakeholders, including middle and high school teachers, guidance counselors, external partners and families, to systematically consider the full range of considerations related to course offerings and sequencing in mathematics in light of these revised standards. Students who follow the grade-by-grade pre-kindergarten to grade 8 sequence will be prepared for either the Traditional or Integrated Model Course high school pathways beginning with Algebra I or Mathematics I in grade 9 and will be ready to take a fourth year advanced course in grade 12, such as the Model Precalculus Course, the Model Quantitative Reasoning Course, or other advanced courses offered in the district, such as Statistics. Decisions about secondary students’ course-taking sequences should be made with the goal of identifying each student’s path to success and ensuring that no student who graduates from a Massachusetts High School and enrolls in a Massachusetts public college or university will be placed into a non-credit bearing remedial mathematics course. All students should be encouraged to meet the full expectations of the pre-K to high school standards. There should also be a variety of ways and opportunities for students to take advanced mathematics courses beyond those included in this Framework. Districts are encouraged to work with their mathematics teachers and curriculum coordinators to design pathways that best meet the needs of their students. This section presents information and resources to ground discussions and decision-making about course-taking sequences in three inter-related areas of consideration: The rigor of the grades 6–8 standards and the Model High School Algebra I Course standards. The offering of the Model High School Algebra I Course in grade 8 for students for whom it is appropriate. Options for high school pathways that accelerate starting in grade 9 to allow students to reach advanced mathematics courses, such as Calculus in grade 12. I. Rigor of Grade 8 and the Model High School Algebra I Standards Success in Algebra I is crucial to students’ overall academic success and their continued interest and engagement in mathematics. The pre-kindergarten to grade 8 standards in the 2017 Framework present a tight progression of skills and knowledge that is rigorous and designed to provide a strong foundation for success in Algebra I as defined in the High School Model Algebra I Course. Course Sequences and the Model Algebra I Course The grade 8 standards address foundations of algebra, a more formal treatment of functions, the exploration of irrational numbers and the Pythagorean Theorem; and include geometry standards that relate graphing to algebra and statistics concepts; and skills that are sophisticated and connect linear relations with the representation of bivariate data. Massachusetts Curriculum Framework for Mathematics 166 The Model Algebra I course formalizes and builds on the grade 8 standards. This course begins with more advanced topics and deepens and extends students’ understanding of linear functions, exponential functions and relationships, introduces quadratic relationships, and includes rigorous statistics concepts and skills. II. Offering the Model High School Algebra I course in middle school to grade 8 students for whom it is appropriate (Compacted Pathway) The Mathematics Standards in grades 6–8 are coherent, rigorous, and non-redundant, so the offering of high school coursework in middle school to students for whom it is appropriate requires careful planning to ensure that all content and practice standards are fully addressed. For those students ready to move at a more accelerated pace, one option is to compress the standards for any three consecutive grades and/or courses into an accelerated two-year pathway. Compressing the standards from grade 7, grade 8, and the Model Algebra I (or Model Mathematics I) course into an accelerated pathway for students in grades 7 and 8 could allow students to enter the Model Geometry (or Model Mathematics II) course in grade 9. Selecting and placing students into accelerated opportunities must be done carefully in order to ensure success. Students who follow a compacted pathway will be undertaking advanced work at an accelerated pace. This creates a challenge for these students as well as their teachers, who will be teaching the grade 8 standards and Model Algebra I standards within a compressed timeframe without compromising any of the rigor. Placement decisions should be made based upon a common assessment to be reviewed by a team of stakeholders that includes teachers and administrators. III. Accelerated High School Pathways starting in grade 9 to allow students to reach advanced mathematics courses, such as Calculus, by grade 12 For many students, high school mathematics will culminate during grade 12 with courses such as Model Precalculus and/or Model Advanced Quantitative Reasoning. Although this would represent a robust and rigorous course of study, some students will seek the opportunity to advance to mathematics courses beyond those included in this Framework (for example, Discrete Mathematics, Linear Algebra, AP Statistics and/or AP Calculus). The following models are only some of the pathways by which students’ mathematical needs could be met. Districts are encouraged to work with their mathematics administrators, teachers, and curriculum coordinators to design pathways that best meet the abilities and needs of their students. In high school, compressed and accelerated pathways may follow these models, among others: Students could “double up” by enrolling in the Model Geometry course during the same year as Model Algebra I or Model Algebra II. Standards from the Model Precalculus course could be added to other courses in a high school pathway, allowing students to enter a Calculus course without enrolling in the Model Precalculus course. Standards that focus on a sub-topic such as trigonometry or statistics could be pulled out and taken alongside the Model courses so that students would only need to “double up” for one semester. Standards from the Model Mathematics I, Model Mathematics II, and Mathematics III course could be compressed into an accelerated pathway for students for two years, allowing students to enter the Model Precalculus course in the third year. Massachusetts Curriculum Framework for Mathematics 167 The graphics below depict options for grades 6–12 course sequences: Figure 1 shows a 6–8 grade-by-grade progression followed by the three Model High School Courses culminating in an advanced mathematics course in grade 12. Figure 2–4 depicts three accelerated pathways leading to Calculus. The first accelerated pathway in Figure 2 compresses grades 7, 8, and the High School Model Algebra I course standards in two years. This compacting of standards begins during middle school at the end of grade 6 and ends with Algebra I in grade 8. The last two pathways in Figure 3 and Figure 4 are high school accelerated pathway options, titled “Doubling Up” and “Enhanced Pathway.” Note that the accelerated high school pathways delay decisions about accelerating students until they are in high school while still allowing access to advanced mathematics in grade 12. Massachusetts Curriculum Framework for Mathematics 168 Appendix I: Application of Standards for English Learners and Students with Disabilities English Learners The Massachusetts Department of Elementary and Secondary Education (ESE) strongly believes that all students, including English learners (ELs) should be held to the same high expectations outlined in the Curriculum Framework. English learners may require additional time and support as they work to acquire English language proficiency and content-area knowledge simultaneously. Further, developing proficiency in English takes time, and teachers should recognize that it is possible to meet the standards for mathematical content and practices as students become fluent in English. The structure of programs serving ELs in Massachusetts acknowledges that ELs acquire language while interacting in all classrooms. All educators, including mathematics teachers, are responsible for students’ language development and academic achievement. Collaboration and shared responsibility among administrators and educators are integral to student and program success. ESE uses the term English language development (ELD) to describe all of the language development that takes place throughout a student’s day, both during sheltered content instruction (SCI) in math and in ESL classrooms. Together SCI and ESL comprise a complete program of sheltered English immersion (SEI).30 Districts in Massachusetts must provide EL students with both grade-level academic math content and ESL instruction that is aligned to the World Class Instructional Design and Assessment standards or WIDA and the Curriculum Frameworks as outlined in state guidelines for EL programs. ESE’s Office of English Language Acquisition and Academic Achievement (OELAAA) offers a number of resources to help districts meet these expectations, including a Next-Generation ESL Curriculum Resource Guide, a set of ESL Model Curriculum Units with connections to ESE Model Curriculum Units (MCUs) in various content areas, and a Collaboration tool that supports WIDA standards implementation in conjunction with the Massachusetts Curriculum Frameworks. In partnership with educators, as well as other state and national experts, OELAAA is also developing a suite of updated SEI resources including comprehensive programmatic and curricular guidance for districts and eight new sheltered content immersion MCUs. Regardless of the specific curriculum used, all ELs in formal educational settings must have access to: District and school personnel with the skills and qualifications necessary to support ELs’ growth. Literacy-rich environments where students are immersed in a variety of robust language experiences. Speakers of English who know the language well enough to provide models and support. Yet English learners are a heterogeneous group, with differences in cultural background, home language(s), socioeconomic status, educational experiences, and levels of English language proficiency. Educating ELs effectively requires diagnosing each student instructionally, tailoring instruction to individual needs, and monitoring progress closely and continuously. For example, ELs who are literate in a home language that shares cognates with English can apply home-language vocabulary knowledge when reading in English; likewise, those with extensive schooling can use conceptual knowledge developed in another language when learning academic content in English. Students with limited or interrupted formal schooling (SLIFE) may need to acquire more background knowledge before engaging in the educational task at hand. 30 For more on types of English Learner Education (ELE) programs in Massachusetts, please see Guidance on Identification, Assessment, Placement, and Reclassification of English Language Learners. Massachusetts Curriculum Framework for Mathematics 169 Six key principles should therefore guide instruction for ELs:31 Instruction focuses on providing ELs with opportunities to engage in math-specific practices that build conceptual understanding and language competence in tandem. Instruction leverages ELs’ home language(s), cultural assets, and prior math knowledge. Standards-aligned instruction for ELs is rigorous, grade-level appropriate, and provides deliberate, appropriate, and nuanced scaffolds. Instruction moves ELs forward by taking into account their English proficiency level(s) and prior schooling experiences. Instruction fosters ELs’ autonomy by equipping them with the strategies necessary to comprehend and use language in mathematics classrooms. Responsive diagnostic tools and formative assessment practices measure ELs’ mathematics content knowledge, language competence, and participation in mathematics practices. In sum, the Massachusetts Curriculum Framework for Mathematics articulates rigorous grade-level expectations in the standards for mathematics content and mathematics practice to prepare all students, including ELs, for postsecondary education, careers, and everyday life. This document can be used in conjunction with language development standards designed to guide and monitor ELs’ progress toward English proficiency. Many English learners also benefit from instruction on negotiating situations outside of schooling and career—instruction that enables them to participate on equal footing with English proficient peers in all aspects of social, economic, and civic life. Whether academic, linguistic, or social, support for ELs must be grounded in respect for the great value that multilingualism and multiculturalism add to our society. Students with Disabilities The Massachusetts Curriculum Framework for Mathematics articulates rigorous grade-level expectations. These learning standards identify the mathematical knowledge and skills all students need in order to be successful in college and careers and in everyday life. Students with disabilities—students eligible under the Individuals with Disabilities Education Act (IDEA)—must be challenged to excel within the general mathematics curriculum and be prepared for success in their post-school lives, including college and/or careers. The standards provide an opportunity to improve access to rigorous mathematics content for students with disabilities. The continued development of understanding about research-based instructional practices and a focus on their effective implementation will help improve access to the mathematics content standards and the mathematics practice standards for all students, including those with disabilities. Students with disabilities are a heterogeneous group. Students who are eligible for an Individualized Education Program (IEP) have one or more disabilities and, as a result of the disability/ies, are unable to progress effectively in the general education program without the provision of specially designed instruction, or are unable to access the general mathematics curriculum without the provision of one or more related services (603 CMR 28.05 (2)(a)(1). How these high standards are taught and assessed is of importance in reaching students with diverse needs. In order for students with disabilities to meet high academic standards, their math instruction must incorporate individualized instruction or related services, supports, and accommodations necessary to allow the student to access the general mathematics curriculum. The annual goals included in students’ IEPs must be carefully aligned to and facilitate students’ attainment of grade-level learning standards. Promoting a culture of high expectations for all students is a fundamental goal of the Massachusetts Curriculum Frameworks. In order to participate successfully in the general curriculum, students with disabilities may be provided additional supports and services as identified in their IEPs, including: 31 For more on the Six Key Principles for EL Instruction, please see Principles for ELL Instruction (2013, January). Understanding Language. Massachusetts Curriculum Framework for Mathematics 170 Instructional learning supports based on the principles of Universal Design for Learning (UDL) which foster student engagement by presenting information in multiple ways and allowing for diverse avenues of demonstration, response, action, and expression. UDL is defined by the Higher Education Opportunity Act (PL 110-135) as “a scientifically valid framework for guiding educational practice that (a) provides flexibility in the ways information is presented, in the ways students respond or demonstrate knowledge and skills, and in the ways students are engaged; and (b) reduces barriers in instruction, provides appropriate accommodations, supports, and challenges, and maintains high achievement expectations for all students, including students with disabilities and students who are limited English proficient.” Instructional accommodations (Thompson, Morse, Sharpe & Hall, 2005), such as alternative materials or procedures that do not change the standards or expectations, but allow students to learn within the framework of the general curriculum. Assistive technology devices and services to ensure access to the general education curriculum and the Massachusetts standards for mathematics. Some students with the most significant cognitive disabilities will require substantial supports and accommodations to have meaningful access to certain standards in both instruction and assessment, based on their expressive communication and academic needs. These supports and accommodations must be identified in the students’ IEPs and should ensure that students receive access to multiple means of learning, and opportunities to demonstrate knowledge, but at the same time retain the rigor and high expectations of the Mathematics Curriculum Framework. References: Individuals with Disabilities Education Act (IDEA), 34 CFR §300.34 (a). (2004). Individuals with Disabilities Education Act (IDEA), 34 CFR §300.39 (b)(3). (2004). Thompson, Sandra J., Amanda B. Morse, Michael Sharpe, and Sharon Hall. “Accommodations Manual: How to Select, Administer and Evaluate Use of Accommodations and Assessment for Students with Disabilities,” 2nd Edition. Council for Chief State School Officers, 2005 (Accessed January 29, 2010). Massachusetts Curriculum Framework for Mathematics 171 Appendix II: Standards for Mathematical Practice Grade-Span Descriptions: Pre-K–5, 6–8, 9–12 Standards for Mathematical Practice Grades Pre-K–5 1. Make sense of problems and persevere in solving them. Mathematically proficient elementary students explain to themselves the meaning of a problem, look for entry points to begin work on the problem, and plan and choose a solution pathway. For example, young students might use concrete objects or pictures to show the actions of a problem, such as counting out and joining two sets to solve an addition problem. If students are not at first making sense of a problem or seeing a way to begin, they ask questions that will help them get started. As they work, they continually ask themselves, “Does this make sense?" When they find that their solution pathway does not make sense, they look for another pathway that does. They may consider simpler forms of the original problem; for example, to solve a problem involving multi-digit numbers, they might first consider similar problems that involve multiples of ten or one hundred. Once they have a solution, they look back at the problem to determine if the solution is reasonable and accurate. They often check their answers to problems using a different method or approach. Mathematically proficient students consider different representations of the problem and different solution pathways, both their own and those of other students, in order to identify and analyze correspondences among approaches. They can explain correspondences among physical models, pictures, diagrams, equations, verbal descriptions, tables, and graphs. 2. Reason abstractly and quantitatively. Mathematically proficient elementary students make sense of quantities and their relationships in problem situations. They can contextualize quantities and operations by using images or stories. They interpret symbols as having meaning, not just as directions to carry out a procedure. Even as they manipulate the symbols, they can pause as needed to access the meaning of the numbers, the units, and the operations that the symbols represent. Mathematically proficient students know and flexibly use different properties of operations, numbers, and geometric objects. They can contextualize an abstract problem by placing it in a context they then use to make sense of the mathematical ideas. For example, when a student sees the expression 40-26, the student might visualize this problem by thinking, if I have 26 marbles and Marie has 40, how many more do I need to have as many as Marie? Then, in that context, the student thinks, 4 more will get me to a total of 30, and then 10 more will get me to 40, so the answer is 14. In this example, the student uses a context to think through a strategy for solving the problem, using the relationship between addition and subtraction and decomposing and recomposing the quantities. The student then uses what he/she did in the context to identify the solution of the original abstract problem. Mathematically proficient students can also make sense of a contextual problem and express the actions or events that are described in the problem using numbers and symbols. If they work with the symbols to solve the problem, they can then interpret their solution in terms of the context. 3. Construct viable arguments and critique the reasoning of others. Mathematically proficient elementary students construct verbal and written mathematical arguments—that is, explain the reasoning underlying a strategy, solution, or conjecture—using concrete referents such as objects, drawings, diagrams, and actions. Arguments may also rely on definitions, previously established results, properties, or structures. For example, a student might argue that two different shapes have equal area because it has already been demonstrated that both shapes are half of the same rectangle. Students might also use counterexamples to argue that a conjecture is not true—for example, a rhombus is an example that shows that not all quadrilaterals with 4 equal sides are squares; or, multiplying by 1 shows that a product of two whole Massachusetts Curriculum Framework for Mathematics 172 numbers is not always greater than each factor. Mathematically proficient students present their arguments in the form of representations, actions on those representations, and explanations in words (oral or written). In the elementary grades, arguments are often a combination of all three. Some of their arguments apply to individual problems, but others are about conjectures based on regularities they have noticed across multiple problems (see MP.8). As they articulate and justify generalizations, students consider to which mathematical objects (numbers or shapes, for example) their generalizations apply. For example, young students may believe a generalization about the behavior of addition applies to positive whole numbers less than 100 because those are the numbers with which they are currently familiar. As they expand their understanding of the number system, they may reexamine their conjecture for numbers in the hundreds and thousands. In upper elementary grades, students return to their conjectures and arguments about whole numbers to determine whether they apply to fractions and decimals. Mathematically proficient students can listen to or read the arguments of others, decide whether they make sense, ask useful questions to clarify or improve the arguments, and build on those arguments. They can communicate their arguments both orally and in writing, compare them to others, and reconsider their own arguments in response to the critiques of others. 4. Model with mathematics. When given a problem in a contextual situation, mathematically proficient elementary students can identify the mathematical elements of a situation and create or interpret a mathematical model that shows those elements and relationships among them. The mathematical model might be represented in one or more of the following ways: numbers and symbols; geometric figures, pictures, or physical objects used to abstract the mathematical elements of the situation; a mathematical diagram such as a number line, table, or graph; or students might use more than one of these to help them interpret the situation. For example, when students encounter situations such as sharing a pan of cornbread among six people, they might first show how to divide the cornbread into six equal pieces using a picture of a rectangle. The rectangle divided into 6 equal pieces is a model of the essential mathematical elements of the situation. When the students learn to write the name of each piece in relation to the whole pan as 1 experiments, and observational 6, they are now modeling the situation with mathematical notation. Mathematically proficient students are able to identify important quantities in a contextual situation and use mathematical models to show the relationships of those quantities, particularly in multi-step problems or problems involving more than one variable. For example, if there is a penny jar that starts with three pennies in the jar, and four pennies are added each day, students might use a table to model the relationship between number of days and number of pennies in the jar. They can then use the model to determine how many pennies are in the jar after 10 days, which in turn helps them model the situation with the expression, 4 x 10 + 3. Mathematically proficient students use and interpret models to analyze relationships and draw conclusions. They interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose. As students model situations with mathematics, they are choosing tools appropriately (MP.5). As they decontextualize the situation and represent it mathematically, they are also reasoning abstractly (MP.2). 5. Use appropriate tools strategically. Mathematically proficient elementary students consider the tools that are available when solving a mathematical problem, whether in a real-world or mathematical context. These tools might include physical objects (cubes, geometric shapes, place value manipulatives, fraction bars, etc.); drawings or diagrams (number lines, tally marks, tape diagrams, arrays, tables, graphs, etc.); models of mathematical concepts, paper and pencil, rulers and other measuring tools, scissors, tracing paper, grid paper, virtual manipulatives, appropriate software applications, or other available technologies. Examples: a student may use graph paper to find all the possible rectangles that have a given perimeter or use linking cubes to represent two quantities and then compare the two representations side by side. Proficient students are sufficiently familiar with tools appropriate for their grade and areas of content to make sound decisions about when each of these tools might be helpful, recognizing both the insight to be gained from their use as well as their limitations. Students choose tools that Massachusetts Curriculum Framework for Mathematics 173 are relevant and useful to the problem at hand. These include tools mentioned above, as well as mathematical tools such as estimation or a particular strategy or algorithm. For example, in order to solve 3∕5 – ½, a student might recognize that knowledge of equivalents of ½ is an appropriate tool: since ½ is equivalent to 2½ fifths, the result is ½ of a fifth or 1∕10. This practice is also related to looking for structure (MP.7), which often results in building mathematical tools that can then be used to solve problems. 6. Attend to precision. Mathematically proficient elementary students communicate precisely to others both verbally and in writing. They start by using everyday language to express their mathematical ideas, realizing that they need to select words with clarity and specificity rather than saying, for example, “it works" without explaining what “it" means. As they encounter the ambiguity of everyday terms, they come to appreciate, understand, and use mathematical vocabulary. Once young students become familiar with a mathematical idea or object, they are ready to learn more precise mathematical terms to describe it. In using mathematical representations, students use care in providing appropriate labels to precisely communicate the meaning of their representations. When making mathematical arguments about a solution, strategy, or conjecture (see MP.3), mathematically proficient students learn to craft careful explanations that communicate their reasoning by referring specifically to each important mathematical element, describing the relationships among them, and connecting their words clearly to their representations. Elementary students use mathematical symbols correctly and can describe the meaning of the symbols they use. When measuring, mathematically proficient students use tools and strategies to minimize the introduction of error. Mathematically proficient students specify units of measure; label charts, graphs, and drawings; calculate accurately and efficiently; and use clear and concise notation to record their work. 7. Look for and make use of structure. Mathematically proficient elementary students use structures such as place value, the properties of operations, other generalizations about the behavior of the operations (for example, even numbers can be divided into 2 equal groups and odd numbers, when divided by 2, always have 1 left over), and attributes of shapes to solve problems. In many cases, they have identified and described these structures through repeated reasoning (MP.8). For example, when younger students recognize that adding 1 results in the next counting number, they are identifying the basic structure of whole numbers. When older students calculate 16 x 9, they might apply the structure of place value and the distributive property to find the product: 16 x 9 = (10 + 6) x 9 = (10 x 9) + (6 x 9). To determine the volume of a 3 x 4 x 5 rectangular prism, students might see the structure of the prism as five layers of 3 x 4 arrays of cubes. 8. Look for and express regularity in repeated reasoning. Mathematically proficient elementary students look for regularities as they solve multiple related problems, then identify and describe these regularities. For example, students might notice a pattern in the change to the product when a factor is increased by 1: 5 x 7 = 35 and 5 x 8 = 40—the product changes by 5; 9 x 4 = 36 and 10 x 4 = 40—the product changes by 4. Students might then express this regularity by saying something like, “When you change one factor by 1, the product increases by the other factor." Younger students might notice that when tossing two-color counters to find combinations of a given number, they always get what they call “opposites"—when tossing 6 counters, they get 2 red, 4 yellow and 4 red, 2 yellow and when tossing 4 counters, they get 1 red, 3 yellow and 3 red, 1 yellow. Mathematically proficient students formulate conjectures about what they notice, for example, that when 1 is added to a factor, the product increases by the other factor. As students practice articulating their observations both verbally and in writing, they learn to communicate with greater precision (MP.6). As they explain why these generalizations must be true, they construct, critique, and compare arguments (MP.3). Massachusetts Curriculum Framework for Mathematics 174 Standards for Mathematical Practice Grades 6–8 1. Make sense of problems and persevere in solving them. Mathematically proficient middle school students set out to understand a problem and then look for entry points to its solution. They analyze problem conditions and goals, translating, for example, verbal descriptions into mathematical expressions, equations, or drawings as part of the process. They consider analogous problems, and try special cases and simpler forms of the original in order to gain insight into its solution. For example, to understand why a 20% discount followed by a 20% markup does not return an item to its original price, they might translate the situation into a tape diagram or a general equation; or they might first consider the result for an item priced at $1.00 or $10.00. Mathematically proficient students can explain how alternate representations of problem conditions relate to each other. For example, they can navigate among tables, graphs, and equations representing linear relationships to gain insights into the role played by constant rate of change. Mathematically proficient students check their answers to problems and they continually ask themselves, “Does this make sense?” and “Can I solve the problem in a different way?” While working on a problem, they monitor and evaluate their progress and change course if necessary. They can understand the approaches of others to solving complex problems and compare approaches. 2. Reason abstractly and quantitatively. Mathematically proficient middle school students make sense of quantities and relationships in problem situations. For example, they can apply ratio reasoning to convert measurement units and proportional relationships to solve percent problems. They represent problem situations using symbols and then manipulate those symbols in search of a solution (decontextualize). They can, for example, solve problems involving unit rates by representing the situations in equation form. Mathematically proficient students also pause as needed during problem solving to double-check the meaning of the symbols involved. In the process, they can look back at the applicable units of measure to clarify or inform solution steps (contextualize). Students can integrate quantitative information and concepts expressed in text and visual formats. Quantitative reasoning also entails knowing and flexibly using different properties of operations and objects. For example, in middle school, students use properties of operations to generate equivalent expressions and use the number line to understand multiplication and division of rational numbers. 3. Construct viable arguments and critique the reasoning of others. Mathematically proficient middle school students understand and use assumptions, definitions, and previously established results in constructing verbal and written arguments. They make and explore the validity of conjectures. They can recognize and appreciate the use of counterexamples, for example, using numerical counterexamples to identify common errors in algebraic manipulation, such as thinking that 5 - 2x is equivalent to 3x. Mathematically proficient students can explain and justify their conclusions using numerals, symbols, and visuals, communicate them to others, and respond to the arguments of others. They reason inductively about data, making plausible arguments that take into account the context from which the data arose. For example, they might argue that the great variability of heights in their class is explained by growth spurts, and that the small variability of ages is explained by school admission policies. Mathematically proficient students are also able to compare the effectiveness of two plausible arguments, distinguish correct logic or reasoning from that which is flawed, and if there is a flaw in an argument, explain what it is. Students engage in collaborative discussions, drawing on evidence from texts and arguments of others, follow conventions for collegial discussions, and qualify their own views in light of evidence presented. They consider questions such as “How did you get that?” “Why is that true?” and “Does that always work?” Massachusetts Curriculum Framework for Mathematics 175 4. Model with mathematics. Mathematically proficient middle school students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. This might be as simple as translating a verbal or written description to a drawing or mathematical expression. It might also entail applying proportional reasoning to plan a school event or using a set of linear inequalities to analyze a problem in the community. Mathematically proficient students are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. For example, they can roughly fit a line to a scatter plot to make predictions and gather experimental data to approximate a probability. They are able to identify important quantities in a given relationship such as rates of change and represent situations using such tools as diagrams, tables, graphs, flowcharts and formulas. They can analyze their representations mathematically, use the results in the context of the situation, and then reflect on whether the results make sense while possibly improving the model. 5. Use appropriate tools strategically. Mathematically proficient middle school students strategically consider the available tools when solving a mathematical problem and while exploring a mathematical relationship. These tools might include pencil and paper, concrete models, a ruler, a protractor, a graphing calculator, a spreadsheet, a statistical package, or dynamic geometry software. Proficient students make sound decisions about when each of these tools might be helpful, recognizing both the insights to be gained and their limitations. For example, they use estimation to check reasonableness, graphs to model functions, algebra tiles to see how properties of operations apply to algebraic expressions, graphing calculators to solve systems of equations, and dynamic geometry software to discover properties of parallelograms. When making mathematical models, they know that technology can enable them to visualize the results of their assumptions, to explore consequences, and to compare predictions with data. Mathematically proficient students are able to identify relevant external mathematical resources, such as digital content located on a website, and use them to pose or solve problems. 6. Attend to precision. Mathematically proficient middle school students communicate precisely to others both verbally and in writing. They present claims and findings, emphasizing salient points in a focused, coherent manner with relevant evidence, sound and valid reasoning, well-chosen details, and precise language. They use clear definitions in discussion with others and in their own reasoning and determine the meaning of symbols, terms, and phrases as used in specific mathematical contexts. For example, they can use the definition of rational numbers to explain why a number is irrational and describe congruence and similarity in terms of transformations in the plane. They state the meaning of the symbols they choose, consistently and appropriately, such as inputs and outputs represented by function notation. They are careful about specifying units of measure, and label axes to display the correct correspondence between quantities in a problem. They calculate accurately and efficiently, express numerical answers with a degree of precision appropriate to the context. For example, they accurately apply scientific notation to large numbers and use measures of center to describe data sets. 7. Look for and make use of structure. Mathematically proficient middle school students look closely to discern a pattern or structure. They might use the structure of the number line to demonstrate that the distance between two rational numbers is the absolute value of their difference, ascertain the relationship between slopes and solution sets of systems of linear equations, and see that the equation 3x = 2y represents a proportional relationship with a unit rate of 3/2 = 1.5. They might recognize how the Pythagorean Theorem is used to find distances between points in the coordinate plane and identify right triangles that can be used to find the length of a diagonal in a rectangular prism. They also can step back for an overview and shift perspective, as in finding a representation of consecutive numbers that shows all sums of three consecutive whole numbers are divisible by six. They can see complicated things as Massachusetts Curriculum Framework for Mathematics 176 single objects, such as seeing two successive reflections across parallel lines as a translation along a line perpendicular to the parallel lines or understanding 1.05a as an original value, a, plus 5% of that value, 0.05a. 8. Look for and express regularity in repeated reasoning. Mathematically proficient middle school students notice if calculations are repeated, and look for both general methods and shortcuts. Working with tables of equivalent ratios, they might deduce the corresponding multiplicative relationships and make generalizations about the relationship to rates. By paying attention to the calculation of slope as they repeatedly check whether points are on the line through (1; 2) with slope 3, students might abstract the equation (y – 2)∕(x – 1) = 3. Noticing the regularity with which interior angle sums increase with the number of sides in a polygon might lead them to the general formula for the interior angle sum of an n-gon. As they work to solve a problem, mathematically proficient students maintain oversight of the process while attending to the details. They continually evaluate the reasonableness of their intermediate results. Massachusetts Curriculum Framework for Mathematics 177 Standards for Mathematical Practice Grades 9–12 1. Make sense of problems and persevere in solving them. Mathematically proficient high school students analyze givens, constraints, relationships, and goals. They make conjectures about the form and meaning of the solution and plan a solution pathway rather than simply jumping into a solution attempt. They consider analogous problems, and try special cases and simpler forms of the original problem in order to gain insight into its solution. They monitor and evaluate their progress and change course if necessary. High school students might, depending on the context of the problem, transform algebraic expressions or change the viewing window on their graphing calculator to get the information they need. Mathematically proficient students can explain correspondences between equations, verbal descriptions, tables, and graphs or draw diagrams of important features and relationships, graph and interpret representations of data, and search for regularity or trends. Mathematically proficient students check their answers to problems using different methods of solving, and they continually ask themselves, “Does this make sense?” They can understand the approaches of others to solving complex problems and identify correspondences between different approaches. 2. Reason abstractly and quantitatively. Mathematically proficient high school students make sense of the quantities and their relationships in problem situations. Students bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically, and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Students can write explanatory text that conveys their mathematical analyses and thinking, using relevant and sufficient facts, concrete details, quotations, and coherent development of ideas. Students can evaluate multiple sources of information presented in diverse formats (and media) to address a question or solve a problem. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meanings of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. 3. Construct viable arguments and critique the reasoning of others. Mathematically proficient high school students understand and use stated assumptions, definitions, and previously established results in constructing verbal and written arguments. They make conjectures and build a logical progression of statements to explore the truth of their conjectures. They are able to analyze situations by breaking them into cases, and can recognize and use counterexamples and specific textual evidence to form their arguments. They justify their conclusions, communicate them to others, and respond to the arguments of others. They reason inductively about data, making plausible arguments that take into account the context from which the data arose. Mathematically proficient students are also able to compare the effectiveness of two plausible arguments, distinguish correct logic or reasoning from that which is flawed, and—if there is a flaw in an argument—explain what it is and why. They can construct formal arguments relevant to specific contexts and tasks. High school students learn to determine domains to which an argument applies. Students listen or read the arguments of others, decide whether they make sense, and ask useful questions to clarify or improve the arguments. Students engage in collaborative discussions, respond thoughtfully to diverse perspectives and approaches, and qualify their own views in light of evidence presented. 4. Model with mathematics. Mathematically proficient high school students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity depends on another. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a Massachusetts Curriculum Framework for Mathematics 178 complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose. 5. Use appropriate tools strategically. Mathematically proficient high school students consider the available tools when solving a mathematical problem. These tools might include pencil and paper, concrete models, a ruler, a protractor, a calculator, a spreadsheet, a computer algebra system, a statistical package, or dynamic geometry software. Proficient students are sufficiently familiar with tools appropriate for high school to make sound decisions about when each of these tools might be helpful, recognizing both the insight to be gained and their limitations. For example, mathematically proficient high school students analyze graphs of functions and solutions generated using a graphing calculator. They detect possible errors by strategically using estimation and other mathematical knowledge. They are able to use technological tools to explore and deepen their understanding of concepts. When making mathematical models, they know that technology can enable them to visualize the results of varying assumptions, explore consequences, and compare predictions with data. Mathematically proficient students are able to identify relevant external mathematical resources, such as digital content located on a website, and use them to pose or solve problems. 6. Attend to precision. Mathematically proficient high school students communicate precisely to others both verbally and in writing, adapting their communication to specific contexts, audiences, and purposes. They develop the habit of using precise language, not only as a mechanism for effective communication, but also as a tool for understanding and solving problems. Describing their ideas precisely helps students understand the ideas in new ways. They use clear definitions in discussions with others and in their own reasoning. They state the meaning of the symbols that they choose. They are careful about specifying units of measure, labeling axes, defining terms and variables, and calculating accurately and efficiently with a degree of precision appropriate for the problem context. They develop logical claims and counterclaims fairly and thoroughly in a way that anticipates the audiences’ knowledge, concerns, and possible biases. High school students draw specific evidence from informational sources to support analysis, reflection, and research. They critically evaluate the claims, evidence and reasoning of others and attend to important distinctions with their own claims or inconsistencies in competing claims. Students evaluate the conjectures and claims, data, analysis, and conclusions in texts that include quantitative elements, comparing those with information found in other sources. 7. Look for and make use of structure. Mathematically proficient high school students look closely to discern a pattern or structure. In the expression x2 + 9x + 14, high school students can see the 14 as 2 7 and the 9 as 2 + 7. They recognize the significance of an existing line in a geometric figure and can use the strategy of drawing an auxiliary line for solving problems. They also can step back for an overview and shift perspective. They can see complicated things, such as some algebraic expressions, as single objects or as being composed of several objects. For example, they can see 5 – 3(x – y)2 as 5 minus a positive number times a square, and use that to realize that its value cannot be more than 5 for any real numbers x and y. 8. Look for and express regularity in repeated reasoning. Mathematically proficient high school students notice if calculations are repeated, and look both for general methods and for shortcuts. Noticing the regularity in the way terms sum to zero when expanding (x – 1)(x + 1), (x – 1)(x2 + x + 1), and (x – 1)(x3 + x2 + x + 1) might lead students to the general formula for the sum of a geometric series. As they work to solve a problem, mathematically proficient students maintain oversight of the Massachusetts Curriculum Framework for Mathematics 179 process, while attending to the details and continually evaluating the reasonableness of their intermediate results. Massachusetts Curriculum Framework for Mathematics 180 Appendix III: High School Conceptual Category Tables The mathematical content standards were designed for students to attain mathematical skills and concepts in a progression over time and across grade spans. The progressions were informed by research and by the logic of the mathematics. Conceptual categories are groups of inter-related standards. The tables describe how these conceptual category content standards are distributed across the model courses. Distribution of Content Standards by five Conceptual Categories: Number and Quantity (N) Algebra (A) Functions (F) Statistics and Probability (S) Geometry (G) Each Conceptual Category table shows the distribution of standards across the eight Model Courses: Algebra I (AI) Geometry (GEO) Algebra II (AII) Math I (MI) Math II (MII) Math III (MIII) Precalculus (PC) Advanced Quantitative Reasoning (AQR) Massachusetts Curriculum Framework for Mathematics 181 Number and Quantity [N] A I GEO A II M I M II M III PC AQR The Real Number System (N-RN) A. Extend the properties of exponents to rational exponents. 1 2 B. Use properties of rational and irrational numbers. 3 Quantities (N-Q) A. Reason quantitatively and use units to solve problems. 1 2 3 a The Complex Number System (N-CN) A. Perform arithmetic operations with complex numbers. 1 2 3+ B. Represent complex numbers and their operations on the complex plane. 4+ 5+ 6+ C. Use complex numbers in polynomial identities and equations. 7 8+ 9+ Vector and Matrix Quantities (N-VM) A. Represent and model with vector quantities. 1+ 2+ 3+ B. Perform operations on vectors. 4+ a+ b+ c+ Massachusetts Curriculum Framework for Mathematics 182 A I GEO A II M I M II M III PC AQR 5+ a+ b+ C. Perform operations on matrices and use matrices in applications. 6+ 7+ 8+ 9+ 10+ 11+ 12+ Massachusetts Curriculum Framework for Mathematics 183 Algebra [A] A I GEO A II M I M II M III PC AQR Seeing Structure in Expressions (A-SSE) A. Interpret the structure of expressions. 1 a b 2 B. Write expressions in equivalent forms to solve problems. 3 a b c 4 Arithmetic with Polynomials and Rational Expressions (A-APR) A. Perform arithmetic operations on polynomials. 1 a b B. Understand the relationship between zeros and factors of polynomials. 2 3 C. Use polynomial identities to solve problems. 4 5+ D. Rewrite rational expressions. 6 7+ Creating Equations (A-CED) A. Create equations that describe numbers or relationships. 1 2 3 4 Reasoning with Equations and Inequalities (A-REI) A. Understand solving equations as a process of reasoning and explain the reasoning. 1 2 Massachusetts Curriculum Framework for Mathematics 184 A I GEO A II M I M II M III PC AQR B. Solve equations and inequalities in one variable. 3 a 4 a b C. Solve systems of equations. 5 6 7 8+ 9+ D. Represent and solve equations and inequalities graphically. 10 11 12 Massachusetts Curriculum Framework for Mathematics 185 Functions [F] A I GEO A II M I M II M III PC AQR Interpreting Functions (F-IF) A. Understand the concept of a function and use function notation.
1 2 3 B. Interpret functions that arise in applications in terms of the context (linear, quadratic, exponential, rational, polynomial, square root, cube root, trigonometric, logarithmic).
4 5 6 C. Analyze functions using different representations.
7 a b c d+ e 8 a b 9 10 Building Functions (F-BF) A. Build a function that models a relationship between two quantities.
1 a b c+ 2 B. Build new functions from existing functions.
3 4 a b+ c+ d+ 5+ Linear, Quadratic, and Exponential Models (F-LE) A. Construct and compare linear, quadratic, and exponential models and solve problems.
1 a b c 2 3 4 Massachusetts Curriculum Framework for Mathematics 186 A I GEO A II M I M II M III PC AQR B. Interpret expressions for functions in terms of the situation they model. 5 Trigonometric Functions (F-TF) A. Extend the domain of trigonometric functions using the unit circle. 1 2 3+ 4+ B. Model periodic phenomena with trigonometric functions. 5 6+ 7+ C. Prove and apply trigonometric identities. 8 9+ Massachusetts Curriculum Framework for Mathematics 187 Statistics and Probability [S] A I GEO A II M I M II M III PC AQR Interpreting Categorical and Quantitative Data (S-ID) A. Summarize, represent, and interpret data on a single count or measurement variable. Use calculators, spreadsheets, and other technology as appropriate. 1 2 3 4 B. Summarize, represent, and interpret data on two categorical and quantitative variables. 5 6 a b c C. Interpret linear models. 7 8 9 Making Inferences and Justifying Conclusions (S-IC) A. Understand and evaluate random processes underlying statistical experiments. 1 2 B. Make inferences and justify conclusions from sample surveys, experiments, and observational studies. 3 4 5 6 Conditional Probability and the Rules of Probability (S-CP) A. Understand independence and conditional probability and use them to interpret data. 1 2 3 4 5 B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. 6 7 8+ 9+ Massachusetts Curriculum Framework for Mathematics 188 A I GEO A II M I M II M III PC AQR Using Probability to Make Decisions (S-MD) A. Calculate expected values and use them to solve problems. 1+ 2+ 3+ 4+ B. Use probability to evaluate outcomes of decisions. 5+ a+ b+ 6+ 7+ Massachusetts Curriculum Framework for Mathematics 189 Geometry [G] A I GEO A II M I M II M III PC AQR Congruence (G-CO) A. Experiment with transformations in the plane. 1 2 3 4 5 B. Understand congruence in terms of rigid motions. 6 7 8 C. Prove geometric theorems. 9 10 11 a D. Make geometric constructions. 12 13 Similarity, Right Triangles, and Trigonometry (G-SRT) A. Understand similarity in terms of similarity transformations. 1 a b 2 3 B. Prove theorems involving similarity. 4 5 C. Define trigonometric ratios and solve problems involving right triangles. 6 7 8 D. Apply trigonometry to general triangles. 9+ 10+ 11+ Massachusetts Curriculum Framework for Mathematics 190 A I GEO A II M I M II M III PC AQR Circles (G-C) A. Understand and apply theorems about circles. 1 2 3 4+ B. Find arc lengths and areas of sectors of circles. 5 Expressing Geometric Properties with Equations (G-GPE) A. Translate between the geometric description and the equation for a conic section. 1 2 3+ a+ B. Use coordinates to prove simple geometric theorems algebraically. 4 5 6 7 Geometric Measurement and Dimension (G-GMD) A. Explain volume formulas and use them to solve problems. 1 2+ 3 B. Visualize relationships between two-dimensional and three-dimensional objects. 4 Modeling with Geometry (G-MG) A. Apply geometric concepts in modeling situations. 1 2 3 4 Massachusetts Curriculum Framework for Mathematics 191 Glossary: Mathematical Terms, Tables, and Illustrations This Glossary contains terms found in the 2017 Massachusetts Curriculum Framework for Mathematics, as well as selected additional terms. Glossary Sources (H) Harcourt School Publishers Math Glossary (M) Merriam-Webster Dictionary (MW) MathWords.com (NCTM) National Council of Teachers of Mathematics (OK) Oklahoma State Department of Education A AA similarity. Angle-angle similarity. When two triangles have corresponding angles that are congruent, the triangles are similar. (MW) ASA congruence. Angle-side-angle congruence. When two triangles have corresponding angles and sides that are congruent, the triangles themselves are congruent. (MW) Absolute value. The absolute value of a real number is its (non-negative) distance from 0 on a number line. Addition and subtraction within 5, 10, 20, 100, or 1,000. Addition or subtraction of two whole numbers with whole number answers, and with sum or minuend in the range 0–5, 0–10, 0–20, or 0–100, respectively. Example: 8 + 2 = 10 is an addition within 10, 14 – 5 = 9 is a subtraction within 20, and 55 – 18 = 37 is a subtraction within 100. Additive inverses. Two numbers whose sum is 0 are additive inverses of one another. Example: 3/4 and –3/4 are additive inverses of one another because 3/4 + (–3/4) = (–3/4) + 3/4 = 0. Algorithm/Standard Algorithm: Algorithm. A finite set of steps for completing a procedure, e.g., multi-digit operations (addition, subtraction, multiplication, division). (See standard 3.NBT.A.2.) Standard algorithm. One of the conventional algorithms used in the United States based on place value and properties of operations for addition, subtraction, multiplication, and division. (See standards 4.NBT.B.4, 5.NBT.B.5, and 6.NS.B.2). See Table 5 in the Glossary. Analog. Having to do with data represented by continuous variables, e.g., a clock with hour, minute, and second hands. (M) Analytic geometry. The branch of mathematics that uses functions and relations to study geometric phenomena, e.g., the description of ellipses and other conic sections in the coordinate plane by quadratic equations. Massachusetts Curriculum Framework for Mathematics 192 Argument of a complex number. The angle describing the direction of a complex number on the complex plane. The argument is measured in radians as an angle in standard position. For a complex number in polar form r(cos + i sin ), the argument is . (MW) Associative property of addition. See Table 3 in the Glossary. Associative property of multiplication. See Table 3 in the Glossary. Assumption. A fact or statement (as a proposition, axiom, postulate, or notion) taken for granted. (M) Attribute. A common feature of a set of figures. B Benchmark fraction. A common fraction against which other fractions can be measured, such as ½. Binomial Theorem. A theorem that gives the polynomial expansion for any whole-number power of a binomial. For powers greater than or equal to zero. (OK) Bivariate data. Pairs of linked numerical observations. Example: a list of heights and weights for each player on a football team. Box plot. A graphic method that shows the distribution of data values by using the median, quartiles, and extremes of the data set. A box shows the middle 50% of the data. (DPI) C Calculus. The mathematics of change and motion. The main concepts of calculus are limits, instantaneous rates of change, and areas enclosed by curves. Capacity. The maximum amount or number that can be contained or accommodated, e.g., a jug with a one-gallon capacity; the auditorium was filled to capacity. Cardinal number. A number (such as 1, 5, 15) that is used in simple counting and that indicates how many elements there are in a set. Cartesian plane. A coordinate plane with perpendicular coordinate axes. Cavalieri’s Principle. A method, with formula given below, of finding the volume of any solid for which cross-sections by parallel planes have equal areas. This includes, but is not limited to, cylinders and prisms. Formula: Volume = Bh, where B is the area of a cross-section and h is the height of the solid. (MW) Coefficient. Any of the factors of a product considered in relation to a specific factor. (W) Commutative property. See Table 3 in the Glossary. Compare two treatments. Compare different levels of a variable, imposed as treatments in an experiment, to each other and/or to a control group. Complex fraction. A fraction A/B where A and/or B are fractions (B nonzero). Massachusetts Curriculum Framework for Mathematics 193 Complex number. A number that can be written as the sum or difference of a real number and an imaginary number. See Illustration 1 in the Glossary. (MW) Complex plane. The coordinate plane used to graph complex numbers. (MW) Compose numbers. a) Given pairs, triples, etc. of numbers, identify sums or products that can be computed; b) Each place in the base-ten place value is composed of ten units of the place to the left, i.e., one hundred is composed of ten bundles of ten, one ten is composed of ten ones, etc. Compose shapes. Join geometric shapes without overlaps to form new shapes. Composite number. A whole number that has more than two factors. (H) Computation algorithm. A set of predefined steps applicable to a class of problems that gives the correct result in every case when the steps are carried out correctly. See also: algorithm; computation strategy. Computation strategy. Purposeful manipulations that may be chosen for specific problems, may not have a fixed order, and may be aimed at converting one problem into another. See also: computation algorithm. Congruent. Two plane or solid figures are congruent if one can be obtained from the other by rigid motion (a sequence of rotations, reflections, and translations). Conjugate. The result of writing the sum of two terms as a difference, or vice versa. For example, the conjugate of x – 2 is x + 2. (MW) Coordinate plane. A plane in which a point is represented using two coordinates that determine the precise location of the point. In the Cartesian plane, two perpendicular number lines are used to determine the locations of points. In the polar coordinate plane, points are determined by their distance along a ray through that point and the origin, and the angle that ray makes with a pre- determined horizontal axis. (OK) Cosine. A trigonometric function that for an acute angle is the ratio between a leg adjacent to the angle when the angle is considered part of a right triangle and the hypotenuse. (M) Counting number. A number used in counting objects, i.e., a number from the set 1, 2, 3, 4, 5,. See Illustration 1 in the Glossary. Counting on. A strategy for finding the number of objects in a group without having to count every member of the group. For example, if a stack of books is known to have eight books and three more books are added to the top, it is not necessary to count the stack all over again; one can find the total by counting on—pointing to the top book and saying “eight,” following this with “nine, ten, eleven. There are eleven books now.” D Decimal expansion. Writing a rational number as a decimal. Decimal fraction. A fraction (as 0.25 = 25∕100 or 0.025 = 25∕1000) or mixed number (as 3.025 = 3 25∕1000) in which the denominator is a power of ten, usually expressed by the use of the decimal point. (M) Decimal number. Any real number expressed in base ten notation, such as 2.673. Massachusetts Curriculum Framework for Mathematics 194 Decompose numbers. Given a number, identify pairs, triples, etc. of numbers that combine to form the given number using subtraction and division. Decompose shapes. Given a geometric shape, identify geometric shapes that meet without overlap to form the given shape. Differences between parameters. A difference of numerical characteristics of a population, including measures of center and/or spread. Digit. a) Any of the Arabic numerals 1 to 9 and usually the symbol 0; b) One of the elements that combine to form numbers in a system other than the decimal system. Digital. Having to do with data that is represented in the form of numerical digits; providing a read out in numerical digits, e.g., a digital watch. Dilation. A transformation that moves each point along the ray through the point emanating from a fixed center, and multiplies distances from the center by a common scale factor. Directrix. A parabola is the collection of all points in the plane that are the same distance from a fixed point, called the focus (F), as they are from a fixed line, called the directrix (D). (Lone Star College lonetsar.edu) Discrete mathematics. The branch of mathematics that includes combinatorics, recursion, Boolean algebra, set theory, and graph theory. Dot plot. See: line plot. E Expanded form. A multi-digit number is expressed in expanded form when it is written as a sum of single-digit multiples of powers of ten. For example, 643 = 600 + 40 + 3. Expected value. For a random variable, the weighted average of its possible values, with weights given by their respective probabilities. Exponent. The number that indicates how many times the base is used as a factor, e.g., in 43 = 4 x 4 x 4 = 64, the exponent is 3, indicating that 4 is repeated as a factor three times. Exponential function. A function of the form y = a bx where a > 0 and either 0 < b < 1 or b > 1. The variables do not have to be x and y. For example, A = 3.2 (1.02)t is an exponential function. Expression. A mathematical phrase that combines operations, numbers, and/or variables (e.g., 32 ÷ a). (H) F Fibonacci sequence. The sequence of numbers beginning with 1, 1, in which each number that follows is the sum of the previous two numbers, i.e., 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144. Massachusetts Curriculum Framework for Mathematics 195 First quartile. For a data set with median M, the first quartile is the median of the data values less than M. Example: For the data set {1, 3, 6, 7, 10, 12, 14, 15, 22, 120}, the first quartile is 6.32 See also: median, third quartile, interquartile range. Fluency. Fluency in the grades 1–6 standards is the ability to carry out calculations and apply numerical algorithms quickly and accurately. Fluency in each grade involves a mixture of knowing some answers from memory (instant recall), knowing some answers from patterns (e.g., “adding 0 yields the same number”), and knowing some answers from the use of other strategies. The development of fluency follows a specific progression in these grades that begins with conceptual understanding and eventually requires students to “know from memory their math facts,” use various strategies to arrive at answers, and develop proficiency using the standard algorithm for each operation. (See standards 1.OA.B.3, 2.OA.B.2, 3.OA.B.5, 3.OA.C.7 and 3.NBT.A.2, 4.NBT.B.4, 5.NBT.B.5, 6.NS.B.2 and 6.NS.B.3.) Fraction. A number expressible in the form a/b where a is a whole number and b is a positive whole number. (The word fraction in these standards always refers to a nonnegative number.) See also: rational number. Function. A mathematical relation for which each element of the domain corresponds to exactly one element of the range. (MW) Function notation. A notation that describes a function. For a function ƒ, when x is a member of the domain, the symbol ƒ(x) denotes the corresponding member of the range (e.g., ƒ(x) = x + 3). Fundamental Theorem of Algebra. The theorem that establishes that, using complex numbers, all polynomials can be factored into a product of linear terms. A generalization of the theorem asserts that any polynomial of degree n has exactly n zeros, counting multiplicity. (MW) G Geometric sequence (progression). An ordered list of numbers that has a common ratio between consecutive terms, e.g., 2, 6, 18, 54. (H) H Histogram. A type of bar graph used to display the distribution of measurement data across a continuous range. I Identity property of 0. See Table 3 in the Glossary. Imaginary number. Complex numbers with no real terms, such as 5i. See Illustration 1 in the Glossary. (M) Independently combined probability models. Two probability models are said to be combined independently if the probability of each ordered pair in the combined model equals the product of the original probabilities of the two individual outcomes in the ordered pair. Integer. All positive and negative whole numbers, including zero. (MW) 32 Many different methods for computing quartiles are in use. The method defined here is sometimes called the Moore and McCabe method. See Langford, E., “Quartiles in Elementary Statistics,” Journal of Statistics Education Volume 14, Number 3 (2006). Massachusetts Curriculum Framework for Mathematics 196 Interquartile range. A measure of variation in a set of numerical data, the interquartile range is the distance between the first and third quartiles of the data set. Example: For the data set {1, 3, 6, 7, 10, 12, 14, 15, 22, 120}, the interquartile range is 15 – 6 = 9. See also: first quartile, third quartile. Inverse function. A function obtained by expressing the dependent variable of one function as the independent variable of another; that is the inverse of y = f(x) is x = f –1(y). (NCTM) Irrational number. A number that cannot be expressed as a quotient of two integers, e.g., . It can be shown that a number is irrational if and only if it cannot be written as a repeating or terminating decimal. K Know from Memory. To instantly recall single-digit math facts to use when needed. Note: In the early grades, students develop number sense and fluency in operations. Students are expected to commit single digit math facts to memory by the end of: a) grade 2 for addition and related subtraction facts (see standard 2.OA.B.2); and b) grade 3 for multiplication and related division facts (see standard 3.OA.C.7). L Law of Cosines. An equation relating the cosine of an interior angle and the lengths of the sides of a triangle. (MW) Law of Sines. Equations relating the sines of the interior angles of a triangle and the corresponding opposite sides. (MW) Line plot. A method of visually displaying a distribution of data values where each data value is shown as a dot or mark above a number line. (Also known as a dot plot.) (DPI) Linear association. Two variables have a linear association if a scatter plot of the data can be well approximated by a line. Linear equation. Any equation that can be written in the form Ax + By + C = 0 where A and B cannot both be 0. The graph of such an equation is a line. Linear function. A function with an equation of the form y = mx + b, where m and b are constants Logarithm. The exponent that indicates the power to which a base number is raised to produce a given number. For example, the logarithm of 100 to the base 10 is 2. (M) Logarithmic function. Any function in which an independent variable appears in the form of a logarithm; they are the inverse functions of exponential functions. M Matrix (pl. matrices). A rectangular array of numbers or variables. 2 Massachusetts Curriculum Framework for Mathematics 197 Mean. A measure of center in a set of numerical data, computed by adding the values in a list and then dividing by the number of values in the list.33 Example: For the data set {1, 3, 6, 7, 10, 12, 14, 15, 22, 120}, the mean is 21. Mean absolute deviation. A measure of variation in a set of numerical data, computed by adding the distances between each data value and the mean, then dividing by the number of data values. Example: For the data set {2, 3, 6, 7, 10, 12, 14, 15, 22, 120}, the mean absolute deviation is 20. Measure of variability. A determination of how much the performance of a group deviates from the mean or median, most frequently used measure is standard deviation. Median. A measure of center in a set of numerical data. The median of a list of values is the value appearing at the center of a sorted version of the list; or the mean of the two central values, if the list contains an even number of values. Example: For the data set {2, 3, 6, 7, 10, 12, 14, 15, 22, 90}, the median is 11. Midline. In the graph of a trigonometric function, the horizontal line halfway between its maximum and minimum values. Model. A mathematical representation (e.g., number, graph, matrix, equation(s), geometric figure) for real-world or mathematical objects, properties, actions, or relationships. (DPI) Modulus of a complex number. The distance between a complex number and the origin on the complex plane. The absolute value of a + bi is written |a + bi|, and the formula for |a + bi| is . For a complex number in polar form, r(cos + i sin ), the modulus is r. (MW) Multiplication and division within 100. Multiplication or division of two whole numbers with whole number answers, and with product or dividend in the range 0–100. Example: 72 8 = 9. Multiplicative inverses. Two numbers whose product is 1 are multiplicative inverses of one another. Example: 3∕4 and 4∕3 are multiplicative inverses of one another because 3∕4 4∕3 = 4∕3 3∕4 = 1. N Network. a) A figure consisting of vertices and edges that shows how objects are connected; b) A collection of points (vertices), with certain connections (edges) between them. Non-linear association. The relationship between two variables is nonlinear if the change in the second is not simply proportional to the change in the first, independent of the value of the first variable. Number line diagram. A diagram of the number line used to represent numbers and support reasoning about them. In a number line diagram for measurement quantities, the interval from 0 to 1 on the diagram represents the unit of measure for the quantity. Numeral. A symbol or mark used to represent a number. O Observational study. A type of study in which an action or behavior is observed in such a manner that no interference with, or influence upon, the behavior occurs. 33 To be more precise, this defines the arithmetic mean. Massachusetts Curriculum Framework for Mathematics 198 Order of Operations. Convention adopted to perform mathematical operations in a consistent order. 1. Perform all operations inside parentheses, brackets, and/or above and below a fraction bar in the order specified in steps 3 and 4; 2. Find the value of any powers or roots; 3. Multiply and divide from left to right; 4. Add and subtract from left to right. (NCTM) Ordinal number. A number designating the place (as first, second, or third) occupied by an item in an ordered sequence. (M) P Partition. A process of dividing an object into parts. Pascal’s triangle. A triangular arrangement of numbers in which each row starts and ends with 1, and each other number is the sum of the two numbers above it. (H) Percent rate of change. A rate of change expressed as a percent. Example: if a population grows from 50 to 55 in a year, it grows by 5/50 = 10% per year. Periodic phenomena. Naturally recurring events, for example, ocean tides, machine cycles. Picture graph. A graph that uses pictures to show and compare information. Plane. A flat surface that extends infinitely in a all directions. Polar form. The polar coordinates of a complex number on the complex plane. The polar form of a complex number is written in any of the following forms: rcos + r i sin , r(cos + i sin ), or rcis . In any of these forms, r is called the modulus or absolute value. θ is called the argument. (MW) Polynomial. The sum or difference of terms which have variables raised to positive integer powers and which have coefficients that may be real or complex. The following are all polynomials: 5x3 – 2x2 + x – 13, x2y3 + xy, and (1 + i)a2 + ib2. (MW) Polynomial function. Any function whose value is the solution of a polynomial. Postulate. A statement accepted as true without proof. Prime factorization. A number written as the product of all its prime factors. (H) Prime number. A whole number greater than 1 whose only factors are 1 and itself. Probability distribution. The set of possible values of a random variable with a probability assigned to each. Properties of equality. See Table 4 in the Glossary. Properties of inequality. See Table 6 in the Glossary. Properties of operations. See Table 3 in the Glossary. Massachusetts Curriculum Framework for Mathematics 199 Probability. A number between 0 and 1 used to quantify likelihood for processes that have uncertain outcomes (such as tossing a coin, selecting a person at random from a group of people, tossing a ball at a target, testing for a medical condition). Probability model. A probability model is used to assign probabilities to outcomes of a chance process by examining the nature of the process. The set of all outcomes is called the sample space, and their probabilities sum to 1. See also: uniform probability model. Proof. A proof of a mathematical statement is a detailed explanation of how that statement follows logically from statements already accepted as true. Proportion. An equation that states that two ratios are equivalent, e.g., 4/8 = ½ or 4 : 8 = 1 : 2. Pythagorean Theorem. For any right triangle, the sum of the squares of the measures of the legs equals the square of the measure of the hypotenuse. Q Quadratic equation. An equation that includes only second degree polynomials. Some examples are y = 3x2 – 5x2 + 1, x2 + 5xy + y2 = 1, and 1.6a2 +5.9a – 3.14 = 0. (MW) Quadratic expression. An expression that contains the square of the variable, but no higher power of it. Quadratic function. A function that can be represented by an equation of the form y = ax2 + bx + c, where a, b, and c are arbitrary, but fixed, numbers and a 0. The graph of this function is a parabola. (DPI) Quadratic polynomial. A polynomial where the highest degree of any of its terms is 2. R Radical. The symbol, which is used to indicate square roots or nth roots. (MW) Random sampling. A smaller group of people or objects chosen from a larger group or population by a process giving equal chance of selection to all possible people or objects. (H) Random variable. An assignment of a numerical value to each outcome in a sample space. (M) Ratio. A relationship between quantities such that for every a units of one quantity there are b units of the other. A ratio is often denoted by a:b and read “a to b”. Rational expression. A quotient of two polynomials with a non-zero denominator. Rational number. A number expressible in the form a∕b or – a∕b for some fraction a∕b. The rational numbers include the integers. See Illustration 1 in the Glossary. Real number. A number from the set of numbers consisting of all rational and all irrational numbers. See Illustration 1 in the Glossary. Rectangular array. An arrangement of mathematical elements into rows and columns. Rectilinear figure. A polygon all angles of which are right angles. Massachusetts Curriculum Framework for Mathematics 200 Recursive pattern or sequence. A pattern or sequence wherein each successive term can be computed from some or all of the preceding terms by an algorithmic procedure. Reflection. A type of transformation that flips points about a line, called the line of reflection. Taken together, the image and the pre-image have the line of reflection as a line of symmetry. Relative frequency. The empirical counterpart of probability. If an event occurs N' times in N trials, its relative frequency is N'∕N. (M) Relatively Prime. Two positive integers that share no common divisors greater than 1; that is, the only common positive factor of the two numbers is 1. Remainder Theorem. If f(x) is a polynomial in x then the remainder on dividing f (x) by x − a is f (a). (M) Repeating decimal. A decimal in which, after a certain point, a particular digit or sequence of digits repeats itself indefinitely; the decimal form of a rational number. (M) See also: terminating decimal. Rigid motion. A transformation of points in space consisting of a sequence of one or more translations, reflections, and/or rotations. Rigid motions are here assumed to preserve distances and angle measures. Rotation. A type of transformation that turns a figure about a fixed point, called the center of rotation. S SAS congruence (Side-angle-side congruence). When two triangles have corresponding sides and the angles formed by those sides are congruent, the triangles are congruent. (MW) SSS congruence (Side-side-side congruence). When two triangles have corresponding sides that are congruent, the triangles are congruent. (MW) Sample space. In a probability model for a random process, a list of the individual outcomes that are to be considered. Scatter plot. A graph in the coordinate plane representing a set of bivariate data. For example, the heights and weights of a group of people could be displayed on a scatter plot. (DPI) Scientific notation. A widely used floating-point system in which numbers are expressed as products consisting of a number between 1 and 10 multiplied by an appropriate power of 10, e.g., 562 = 5.62 x 102. (MW) Sequence, progression. A set of elements ordered so that they can be labeled with consecutive positive integers starting with 1, e.g., 1, 3, 9, 27, 81. In this sequence, 1 is the first term, 3 is the second term, 9 is the third term, and so on. Significant figures (digits). A way of describing how precisely a number is written, particularly when the number is a measurement. (MW) Similarity transformation. A rigid motion followed by a dilation. Simultaneous equations. Two or more equations containing common variables. (MW) Massachusetts Curriculum Framework for Mathematics 201 Sine (of an acute angle). The trigonometric function that for an acute angle is the ratio between the leg opposite the angle when the angle is considered part of a right triangle and the hypotenuse. (M) T Tangent. a) Meeting a curve or surface in a single point if a sufficiently small interval is considered. b) The trigonometric function that, for an acute angle, is the ratio between the leg opposite the angle and the leg adjacent to the angle when the angle is considered part of a right triangle. (MW) Tape diagram. A drawing that looks like a segment of tape, used to illustrate number relationships. Also known as a strip diagram, bar model, fraction strip, or length model. Terminating decimal. A decimal is called terminating if its repeating digit is 0. A terminating decimal is the decimal form of a rational number. See also: repeating decimal. Third quartile. For a data set with median M, the third quartile is the median of the data values greater than M. Example: For the data set {2, 3, 6, 7, 10, 12, 14, 15, 22, 120}, the third quartile is 15. See also: median, first quartile, interquartile range. Transformation. A prescription, or rule, that sets up a one-to-one correspondence between the points in a geometric object (the pre-image) and the points in another geometric object (the image). Reflections, rotations, translations, and dilations are particular examples of transformations. Transitivity principle for indirect measurement. If the length of object A is greater than the length of object B, and the length of object B is greater than the length of object C, then the length of object A is greater than the length of object C. This principle applies to measurement of other quantities as well. Translation. A type of transformation that moves every point in a graph or geometric figure by the same distance in the same direction without a change in orientation or size. (MW) Trapezoid. A quadrilateral with at least one pair of parallel sides. (Note: There are two definitions for the term trapezoid. This is the inclusive definition. For more information see commoncoretools.me/wpcontent/uploads/2014/12/ccss_progression_gk6_2014_12_27.pdf). Trigonometric function. A function (as the sine, cosine, tangent, cotangent, secant, or cosecant) of an arc or angle most simply expressed in terms of the ratios of pairs of sides of a right-angled triangle. (M) Trigonometry. The study of triangles, with emphasis on calculations involving the lengths of sides and the measure of angles. (MW) U Uniform probability model. A probability model which assigns equal probability to all outcomes. See also: probability model. Unit fraction. A fraction with a numerator of 1, such as 1/3 or 1/5. V Valid. a) Well-grounded or justifiable; being at once relevant and meaningful, e.g., a valid theory; b) Logically correct. (MW) Massachusetts Curriculum Framework for Mathematics 202 Variable. A quantity that can change or that may take on different values. Refers to the letter or symbol representing such a quantity in an expression, equation, inequality, or matrix. (MW) Vector. A quantity with magnitude and direction in the plane or in space, defined by an ordered pair or triple of real numbers. Visual fraction model. A tape diagram, number line diagram, or area model. W Whole numbers. The numbers 0, 1, 2, 3,. See Illustration 1 in the Glossary. Massachusetts Curriculum Framework for Mathematics 203 Tables and Illustrations of Key Mathematical Properties, Rules, and Number Sets Table 1. Common addition and subtraction situations34 Result Unknown Change Unknown Start Unknown Add to Two bunnies sat on the grass. Three more bunnies hopped there. How many bunnies are on the grass now? 2 + 3 = ? Two bunnies were sitting on the grass. Some more bunnies hopped there. Then there were five bunnies. How many bunnies hopped over to the first two? 2 + ? = 5 Some bunnies were sitting on the grass. Three more bunnies hopped there. Then there were five bunnies. How many bunnies were on the grass before? ? + 3 = 5 Take from Five apples were on the table. I ate two apples. How many apples are on the table now? 5 – 2 = ? Five apples were on the table. I ate some apples. Then there were three apples. How many apples did I eat? 5 – ? = 3 Some apples were on the table. I ate two apples. Then there were three apples. How many apples were on the table before? ? – 2 = 3 Total Unknown Addend Unknown Both Addends Unknown35 Put Together/ Take Apart36 Three red apples and two green apples are on the table. How many apples are on the table? 3 + 2 = ? Five apples are on the table. Three are red and the rest are green. How many apples are green? 3 + ? = 5, 5 – 3 = ? Grandma has five flowers. How many can she put in her red vase and how many in her blue vase? 5 = 0 + 5, 5 = 5 + 0 5 = 1 + 4, 5 = 4 + 1 5 = 2 + 3, 5 = 3 + 2 Difference Unknown Bigger Unknown Smaller Unknown Compare37 (“How many more?” version): Lucy has two apples. Julie has five apples. How many more apples does Julie have than Lucy? (“How many fewer?” version): Lucy has two apples. Julie has five apples. How many fewer apples does Lucy have than Julie? 2 + ? = 5, 5 – 2 = ? (Version with “more”): Julie has three more apples than Lucy. Lucy has two apples. How many apples does Julie have? (Version with “fewer”): Lucy has 3 fewer apples than Julie. Lucy has two apples. How many apples does Julie have? 2 + 3 = ?, 3 + 2 = ? (Version with “more”): Julie has three more apples than Lucy. Julie has five apples. How many apples does Lucy have? (Version with “fewer”): Lucy has three fewer apples than Julie. Julie has five apples. How many apples does Lucy have? 5 – 3 = ?, ? + 3 = 5 34 Adapted from Boxes 2–4 of Mathematics Learning in Early Childhood, National Research Council (2009, pp. 32–33). 35 These take apart situations can be used to show all the decompositions of a given number. The associated equations, which have the total on the left of the equal sign, help children understand that the = sign does not always mean makes or results in but always does mean is the same number as. 36 Either addend can be unknown, so there are three variations of these problem situations. Both Addends Unknown is a productive extension of this basic situation, especially for small numbers less than or equal to 10. 37 For the Bigger Unknown or Smaller Unknown situations, one version directs the correct operation (the version using more for the bigger unknown and using less for the smaller unknown). The other versions are more difficult. Massachusetts Curriculum Framework for Mathematics 204 Table 2. Common multiplication and division situations38 Unknown Product Group Size Unknown (“How many in each group?” Division) Number of Groups Unknown (“How many groups?” Division) 3 6 = ? 3 ? = 18 and 18 ÷ 3 = ? ? 6 = 18 and 18 ÷ 6 = ? Equal Groups There are three bags with six plums in each bag. How many plums are there in all? Measurement example. You need three lengths of string, each six inches long. How much string will you need altogether? If 18 plums are shared equally into three bags, then how many plums will be in each bag? Measurement example. You have 18 inches of string, which you will cut into three equal pieces. How long will each piece of string be? If eighteen plums are to be packed six to a bag, then how many bags are needed? Measurement example. You have 18 inches of string, which you will cut into pieces that are six inches long. How many pieces of string will you have? Arrays,39 Area40 There are three rows of apples with six apples in each row. How many apples are there? Area example. What is the area of a 3 cm by 6 cm rectangle? If 18 apples are arranged into three equal rows, how many apples will be in each row? Area example. A rectangle has area 18 square centimeters. If one side is 3 cm long, how long is a side next to it? If 18 apples are arranged into equal rows of six apples, how many rows will there be? Area example. A rectangle has area 18 square centimeters. If one side is 6 cm long, how long is a side next to it? Compare A blue hat costs $6. A red hat costs three times as much as the blue hat. How much does the red hat cost? Measurement example. A rubber band is 6 cm long. How long will the rubber band be when it is stretched to be three times as long? A red hat costs $18 and that is three times as much as a blue hat costs. How much does a blue hat cost? Measurement example. A rubber band is stretched to be 18 cm long and that is three times as long as it was at first. How long was the rubber band at first? A red hat costs $18 and a blue hat costs $6. How many times as much does the red hat cost as the blue hat? Measurement example. A rubber band was 6 cm long at first. Now it is stretched to be 18 cm long. How many times as long is the rubber band now as it was at first? General a b = ? a ? = p and p a = ? ? b = p and p b = ? 38 The first examples in each cell are examples of discrete things. These are easier for students and should be given before the measurement examples. 39 The language in the array examples shows the easiest form of array problems. A harder form is to use the terms rows and columns: The apples in the grocery window are in three rows and six columns. How many apples are in there? Both forms are valuable. 40 Area involves arrays of squares that have been pushed together so that there are no gaps or overlaps, so array problems include these especially important measurement situations. Massachusetts Curriculum Framework for Mathematics 205 Table 3. The Properties of Operations Here a, b and c stand for arbitrary numbers in a given number system. The properties of operations apply to the rational number system, the real number system, and the complex number system. Associative property of addition (a + b) + c = a + (b + c) Commutative property of addition a + b = b + a Additive identity property of 0 a + 0 = 0 + a = a Existence of additive inverses For every a there exists –a so that a + (–a) = (–a) + a = 0. Associative property of multiplication (a b) c = a (b c) Commutative property of multiplication a b = b a Multiplicative identity property of 1 a 1 = 1 a = a Existence of multiplicative inverses For every a 0 there exists 1∕a so that a 1∕a = 1∕a a = 1. Distributive property of multiplication over addition a (b + c) = a b + a c Table 4. The Properties of Equality Here a, b, and c stand for arbitrary numbers in the rational, real, or complex number systems. Reflexive property of equality a = a Symmetric property of equality If a = b, then b = a. Transitive property of equality If a = b and b = c, then a = c. Addition property of equality If a = b, then a + c = b + c. Subtraction property of equality If a = b, then a – c = b – c. Multiplication property of equality If a = b, then a c = b c. Division property of equality If a = b and c 0, then a c = b c. Substitution property of equality If a = b, then b may be substituted for a in any expression containing a. Table 5. Algorithms and the Standard Algorithms: Addition Example Algorithm Standard Algorithm (for efficiency) 356 +167 400 (Sum of hundreds) 110 (Sum of tens) 13 (Sum of ones) 523 11 356 +167 523 Note: All algorithms have a finite set of steps, are based on place value and properties of operations, and use single-digit computations. Massachusetts Curriculum Framework for Mathematics 206 Table 6. The Properties of Inequality Here a, b, and c stand for arbitrary numbers in the rational or real number systems. Exactly one of the following is true: a < b, a = b, a > b. If a > b and b > c then a > c. If a > b, then b < a. If a > b, then –a < –b. If a > b, then a ± c > b ± c. If a > b and c > 0, then a c > b c. If a > b and c < 0, then a c < b c. If a > b and c > 0, then a c > b c. If a > b and c < 0, then a c < b c. Illustration 1. The Number System The Number System is comprised of number sets beginning with the Counting Numbers and culminating in the more complete Complex Numbers. The name of each set is written on the boundary of the set, indicating that each increasing oval encompasses the sets contained within. Note that the Real Number Set is comprised of two parts: Rational Numbers and Irrational Numbers. Massachusetts Curriculum Framework for Mathematics 207 Bibliography and Resources This bibliography includes samples of works consulted for the development of the previous Massachusetts Mathematics Curriculum Framework and updated resources and references used to create the 2017 Framework. Abt Associates. Final Report. Massachusetts English Language Arts/Literacy and Mathematics Frameworks Review. Boston, 2016. Achieve. Mathematics at Work, 2008. Achieve. Out of Many, One: Towards Rigorous Common Core Standards from the Ground Up, 2008. ACT College and Career Readiness Standards-Mathematics™, 2014. ACT National Curriculum Survey™ 2012. Policy Implications on Preparing for Higher Standards, 2013. ACT Policy Implications of the ACT National Curriculum Survey Results: Aligning Postsecondary Expectations and High School Practice: The Gap Defined, 2005-2006. ACT. The Forgotten Middle: Ensuring that All Students Are on Target for College and Career Readiness before High School. Iowa City, 2008. ACT. Crisis at the Core: Preparing All Students for College and Work, 2005. ACT. On Course for Success: A Close Look at Selected High School Courses That Prepare All Students for College and Work, 2005. ACT. Ready for College and Ready for Work: Same or Different?, 2006. www.act.org. ACT. Rigor at Risk: Reaffirming Quality in the High School Core Curriculum, 2007. ACT. Job Skill Comparison Charts, 2007. Adelman, C. The Toolbox Revisited: Paths to Degree Completion From High School Through College, 2006. American Diploma Project. Ready or Not: Creating a High School Diploma That Counts, 2004. American Federation of Teachers. Sizing Up State Standards, 2008. American Mathematical Association of Two-Year Colleges (AMATYC). Crossroads in Mathematics, 1995 and Beyond Crossroads, 2006. Askey, R., “Knowing and Teaching Elementary Mathematics,” American Educator, Fall 1999. Association for Supervision and Curriculum Development (ASCD), “Math Acceleration for All”. Burris, C., Heubert, J., & Levin, H. Education Leadership. Vol.61(5), 68-71, February 2004. Blum, W., Galbraith, P. L., Henn, H-W. and Niss, M. (Eds) Applications and Modeling in Mathematics Education, ICMI Study 14. Amsterdam: Springer. Boaler, Jo. Fluency Without Fear: Research Evidence on the Best Ways to Learn Math Facts. YouCubed at Stanford University, January 2015. Bransford, J.D., Brown, A.L., and Cocking, R.R., eds. How People Learn: Brain, Mind, Experience, and School. Committee on Developments in the Science of Learning, Commission on Behavioral and Social Sciences and Education, National Research Council, 1999. Brosterman, N. Inventing kindergarten. New York: Harry N. Abrams, 1997. Camara, W.J., Shaw, E., and Patterson, B. First Year English and Math College Coursework. College Board: New York, NY (Available from authors). June 13, 2009. California Department of Education. 2016 California Mathematics Framework for California Public Schools: Kindergarten Through Grade Twelve. Retrieved from www.cde.ca.gov. Carmichael, S.B., W.S. Wilson, Finn, Jr., C.E., Winkler, A.M., and Palmieri, S. Thomas. Stars By Which to Navigate? Scanning National and International Education Standards in 2009. B. Fordham Institute, 2009. Carnevale, Anthony and Desrochers, Donna. Connecting Education Standards and Employment: Course-taking Patterns of Young Workers, 2002. Carpenter, T. P., Fennema, E., Franke, M. L., Levi, L., & Empson, S. B. Children's Mathematics: Cognitively Guided Instruction. Portsmouth, NH: Heinemann, 1999. Clements, D. H., & Sarama, J. Learning and teaching early math: The learning trajectories approach. New York: Routledge, 2009. Massachusetts Curriculum Framework for Mathematics 208 Cobb and Moore, “Mathematics, Statistics, and Teaching,” Amer. Math. Monthly 104(9), pp. 801-823, 1997. College Board. Advanced Placement Calculus, Statistics and Computer Science Course Descriptions, 2016. apcentral.collegeboard.com/apc/public/courses/index.html. Common Core State Standards Initiative. Common Core State Standards for Mathematics: Appendix A: Designing High School Mathematics Courses Based on the Common Core State Standards. Retrieved from College Board. CLEP Precalculus Curriculum Survey: Summary of Results, 2005. College Board. College Board Standards for College Success: Mathematics and Statistics, 2006. Colorado Business Leaders Top Skills, 2006. Confrey, J., “Tracing the Evolution of Mathematics Content Standards in the United States: Looking Back and Projecting Forward.” K12 Mathematics Curriculum Standards conference proceedings, February 5-6, 2007. Conley, D.T. Knowledge and Skills for University Success, 2008. Conley, D.T. Toward a More Comprehensive Conception of College Readiness, 2007. Cronin, J., Dahlin, M., Adkins, D., and Kingsbury, G.G. The Proficiency Illusion; foreword by C.E. Finn, Jr., and M. J. Petrilli. Thomas B. Fordham Institute, 2007. Cross, C.T., Woods, T.A., and Schweingruber, S., eds. Mathematics Learning in Early Childhood: Paths Toward Excellence and Equity. Committee on Early Childhood Mathematics, National Research Council, 2009. Cuoco, A., Goldenberg, E. P., and Mark, J., “Habits of Mind: An Organizing Principle for a Mathematics Curriculum,” Journal of Mathematical Behavior, 15(4), 375-402, 1996. Ginsburg, A., Leinwand, S., and Decker, K., “Informing Grades 1–6 Standards Development: What Can Be Learned from High-Performing Hong Kong, Korea, and Singapore?” American Institutes for Research, 2009. Ginsburg et al., “What the United States Can Learn From Singapore’s World-Class Mathematics System (and what Singapore can learn from the United States),” American Institutes for Research, 2005. Ginsburg et al., “Reassessing U.S. International Mathematics Performance: New Findings from the 2003 TIMMS and PISA,” American Institutes for Research, 2005. Ginsburg, H. P., Lee, J. S., & Stevenson-Boyd, J. (2008). Mathematics education for young children: What it is and how to promote it. Social Policy Report, 22(1), 1-24. Ginsburg, A., & Leinwand, S. Informing Grades 1–6 Mathematics Standards Development: What Can Be Learned From High-Performing Hong Kong, Korea and Singapore? Washington, DC: American Institutes for Research, 2009. Guidelines for Assessment and Instruction in Statistics Education (GAISE) Report: A Pre-K–12 Curriculum Framework. Harel, G., “What is Mathematics? A Pedagogical Answer to a Philosophical Question,” in R. B. Gold and R. Simons (Eds.), Current Issues in the Philosophy of Mathematics from the Perspective of Mathematicians. Mathematical Association of America, 2008. Henry, V. J., & Brown, R. S. First-grade basic facts: An investigation into teaching and learning of an accelerated, high-demand memorization standard. Journal for Research in Mathematics Education, 39, 153-183, 2008. Howe, R., “From Arithmetic to Algebra.” Howe, R., “Starting Off Right in Arithmetic,” International Baccalaureate, Mathematics Standard Level, 2006. Jordan, N. C., Kaplan, D., Ramineni, C., and Locuniak, M. N., “Early math matters: kindergarten number competence and later mathematics outcomes,” Dev. Psychol. 45, 850–867, 2009. Kader, G., “Means and MADS,” Mathematics Teaching in the Middle School, 4(6), 1999, pp. 398-403. Kennelly, L., & Monrad, M. (Eds). Easing the Transition to High School: Research and Best Practices Designed to Support High School Learning. Washington, DC: National High School Center at the American Institutes for Research, 2007. Kilpatrick, J., Mesa, V., and Sloane, F., “U.S. Algebra Performance in an International Context,” in Loveless (ed.), Lessons Learned: What International Assessments Tell Us About Math Achievement. Washington, D.C.: Brookings Institution Press, 2007. Massachusetts Curriculum Framework for Mathematics 209 Kilpatrick, J., Swafford, J., & Findell, B. Adding it up: helping children learn mathematics. Washington, D.C.: National Research Council, 2001. Lamon, S. Teaching fractions and ratios for understanding (3rd ed.). New York, NY: Routledge. 2012. Leinwand, S., and Ginsburg, A., “Measuring Up: How the Highest Performing state (Massachusetts) Compares to the Highest Performing Country (Hong Kong) in Grade 3 Mathematics,” American Institutes for Research, 2009. Ma, Liping, Knowing and Teaching Elementary Mathematics, Mahwah, New Jersey: Lawrence Erlbaum Associates, 1999. Massachusetts Department of Elementary and Secondary Education. Progress Report of the Mathematics Curriculum Framework Revision Panel, 2009. www.doe.mass.edu/boe/docs/0509/item5_report.pdf. Massachusetts Department of Elementary and Secondary Education. Resource Guide to the Massachusetts Curriculum Frameworks for Student with Disabilities: Mathematics, Pre-Kindergarten- Grade 12, Fall 2016. Milken, Lowell, A Matter of Quality: A Strategy for Answering High Caliber of America’s Teachers, Santa Monica, California: Milken Family foundation, 1999. Miller, G.E., Twing, J., and Meyers, J. “Higher Education Readiness Component (HERC) Correlation Study.” Austin, TX: Pearson. Moschkovich, Judit N. “Supporting the Participation of English Language Learners in Mathematical Discussions.” For the Learning of Mathematics 19 (March 1999): 11-19. Moschkovich, J. N. (in press). Language, culture, and equity in secondary mathematics classrooms. To appear in F. Lester & J. Lobato (Ed.), Teaching and Learning Mathematics: Translating Research to the Secondary Classroom, Reston, VA: NCTM. Moschkovich, Judit N. “Examining Mathematical Discourse Practices,” For the Learning of Mathematics 27 (March 2007): 24-30. Moschkovich, Judit N. “Using Two Languages when Learning Mathematics: How Can Research Help Us Understand Mathematics Learners Who Use Two Languages?” Research Brief and Clip, National Council of Teachers of Mathematics, 2009. brief_12_Using_2.pdf (accessed November 25, 2009). Moschkovich, J.N. Bilingual Mathematics Learners: How views of language, bilingual learners, and mathematical communication impact instruction. In N. Nasir and P. Cobb (Eds.), Diversity, Equity, and Access to Mathematical Ideas. New York: Teachers College Press, 89-104, 2007. NAEP Validity Studies Panel, Validity Study of the NAEP Mathematics Assessment: Grades 4 and 8. Daro et al., 2007. National Alliance of Business Study, The American Diploma Project Workplace Study, 2002. National Assessment Governing Board, Mathematics Framework for the 2009 National Assessment of Educational Progress. U.S. Department of Education, 2008. National Center for Education Statistics. Pursuing Excellence: A Study of U.S. Fourth-Grade Mathematics and Science Achievement in the International Context. Accessed June 2000. National Council of Teachers of Mathematics. A Research Companion to Principles and Standards for School Mathematics, 2003. National Council of Teachers of Mathematics. Curriculum Focal Points for Pre-kindergarten through Grade 8 Mathematics: A Quest for Coherence, 2006. National Council of Teachers of Mathematics. Principles and standards for school mathematics. Reston, VA: National Council of Teachers of Mathematics, 2000. National Council of Teachers of Mathematics. Focus in High School Mathematics: Reasoning and Sense Making. Reston, VA: NCTM. National Council on Education and the Disciplines, Steen, L.A. (ed.). Mathematics and Democracy, The Case for Quantitative Literacy, 2001. Massachusetts Curriculum Framework for Mathematics 210 National Governors Association, Council of Chief State School Officers, and Achieve, Inc. Benchmarking for Success: Ensuring U.S. Students Receive a World-Class Education, 2008. National Network of Business and Industry Associations. Common employability skills: A foundation for success in the workplace, 2015. Retrieved from New York State Education Department. Draft Standards for Mathematics, 2016. Retrieved from www.nysed.gov. Niss, M., “Quantitative Literacy and Mathematical Competencies,” in Quantitative Literacy: Why Numeracy Matters for Schools and Colleges, Madison, B. L., and Steen, L.A. (eds.), National Council on Education and the Disciplines. Proceedings of the National Forum on Quantitative Literacy held at the National Academy of Sciences in Washington, D.C., December 1-2, 2001. Northern, Amber, Ph.D. Does Common Core Math Expect Memorization? A candid conversation with Jason Zimba. Fordham Institute, July 2016. Ohio Department of Education. Ohio’s Learning Standards for Mathematics (revised 2017). Retrieved from www.education.ohio.gov. PARCC Model Content Frameworks, Mathematics Grades 3–11 Version 4.0, 2014. Program for International Student Assessment (PISA), 2006. Pratt, C. I learn from children. New York: Simon and Schuster, 1948. Reys, B. (ed.), The Intended Mathematics Curriculum as Represented in State-Level Curriculum Standards: Consensus or Confusion? IAP-Information Age Publishing, 2006. Sarama, J., & Clements, D. H. Early childhood mathematics education research: Learning trajectories for young children. New York: Routledge, 2009. Schleppegrell, M.J. The linguistic challenges of mathematics teaching and learning: A research review. Reading & Writing Quarterly, 23:139–159, 2007. Schmidt, W., Houang, R., and Cogan, L., “A Coherent Curriculum: The Case of Mathematics,” American Educator, Summer 2002, p. 4. Schmidt, W.H. and Houang, R.T., “Lack of Focus in the Intended Mathematics Curriculum: Symptom or Cause?” in Loveless (ed.), Lessons Learned: What International Assessments Tell Us About Math Achievement. Washington, D.C.: Brookings Institution Press, 2007. Schmidt, W.H., McKnight, C.C., Raizen, S.A., et al. U.S. National Research Center for the Third International Mathematics and Science Study. A Splintered Vision: An Investigation of U.S. Science and Mathematics Education. Michigan: State University, 1997. Shih, J., Speer, W. R., & Babbitt, B. C. Instruction: Yesterday, I learned to add; today I forgot. In F. Fennell (Ed.), Achieving fluency: Special education and mathematics (pp. 59-83). Reston, VA: National Council of Teachers of Mathematics, 2011. Siegler, R. S., Carpenter, T., Fennell, F., Geary, D., Lewis, J. Okamoto, Y., Thomas, L., & Wray, J. Developing effective fractions instruction for kindergarten through 8th grade: A practice guide (NCEE, 2010-4039). Retrieved from 2010. Steen, L.A., “From the 2000’s: Facing Facts: Achieving Balance in High School Mathematics.” Mathematics Teacher, Vol. 100 Issue 5, 2005. Tennessee Department of Education. Tennessee’s State Standards for Mathematics, 2014. Retrieved from www.tn.gov. Thompson, S. J., Morse, A.B., Sharpe, M., and Hall, S., “Accommodations Manual: How to Select, Administer and Evaluate Use of Accommodations and Assessment for Students with Disabilities,” 2nd Edition. Council of Chief State School Officers, 2005. The Carnegie Corporation of New York and the Institute for Advanced Study. The Opportunity Equation: Transforming Mathematics and Science Education for Citizenship and the Global Economy, 2009. The Charles A. Dana Center, Treisman, Uri. Dana Center Mathematics Pathways. University of Arizona, Institute for Mathematics and Education. Progressions Documents for the Common Core Math Standards, Massachusetts Curriculum Framework for Mathematics 211 University of Arizona, Institute for Mathematics and Education. The Illustrative Mathematics Project, University of Cambridge International Examinations: General Certificate of Secondary Education in Mathematics, 2009. U.S. Department of Education. Condition of Education:2012. National Center for Education Statistics. U.S. Department of Education, Office of Special Education Programs, “IDEA Regulations: Identification of Students with Specific Learning Disabilities”, 2006. U.S. Department of Education. Foundations for Success: The Final Report of the National Mathematics Advisory Panel. Washington, DC, 2008. Van de Walle, J. A., Karp, K. S, & Bay-Williams, J. M. (2015). Elementary and Middle School Mathematics: Teaching Developmentally (9th edition). Boston, MA: Pearson. Van de Walle, J. A., Karp, K. S, Lovin, L., & Bay-Williams, J. M. Teaching Student-Centered Mathematics Series: Developmentally Appropriate Instruction for Grades Pre-K-2 (vol,1); Grades 3–5 (vol 2); and Grades 6–8 (vol. 3). 2013. Washington Office of the Superintendent of Public Instruction. Guidelines for accelerating students into high school mathematics in grade 8. Retrieved from /Standards/Compression.pdf, 2008. Washington Office of the Superintendent of Public Instruction. Mathematical Practices Progression through Grade Levels. Retrieved from www.k12.wa.us. What Works Clearinghouse. Accelerated middle schools. Washington, DC: Institute of Education Sciences, 2008. Wu, H., “Fractions, decimals, and rational numbers”, 2007, 2008. Wu, H., “Lecture Notes for the 2009 Pre-Algebra Institute,” September 15, 2009. Wu, H., “Preservice professional development of mathematics Teachers,” Wyatt, W.J. & Wiley, A. The development of an index of academic rigor for the SAT. (College Board Research Report). New York: The College Board, 2010. |
190792 | https://www.youtube.com/watch?v=Ynrz3gZWUts | if the perimeter of parallelogram is 21 cm and adjacent side are in the ratio 3:4 find the sides of
Vikas Guru mathematics analysis
7980 subscribers
Description
849 views
Posted: 6 Oct 2023
if the perimeter of parallelogram is 21 cm and adjacent side are in the ratio 3:4 find the sides of the parallelogram
1 comments
Transcript:
तो देखिए क्वेश्चन क्या है इफ द पैरामीटर ऑफ पैरेललोग्राम इज 21 सेंटीमीटर एंड एडजेसेंट साइड्स आर इन द रेशो 34 रेशो का मतलब पहले भी बता चुके इसके मल्टीप्लाई में कुछ छुपा हुआ है ठीक है एडजेसेंट साइड का रेशो दिया है हो सकता है 30 40 हो तो तो देखेंगे आगे एक पहले हम लोग एक डायग्राम ड्र कर लेते हैं इससे थोड़ा सुविधा होता है समझने में चीज ए बी सी एंड डी यह जो है यह 4 एक्स है तो यह है 3 एक् रेश का मतलब है ही छुपा हुआ है एक्स वाई जड आमली कटर बर कुछ भी छुपा हो सकता है कोई भी नंबर छुपे हो सकते हैं इसके आस परस में ना ठीक है यह बात हम लोग मान लिए थे लेट ए बी सीडी वि द पलोग ग्राम ठीक है इन च इन विच इन वि ए बी इ इक्वल टू 4 एकस मान लिए है एंड बी स इक्व टू 3 एक्स तब जाकर के तब हम बोल सकते हैं यह बात ए बी जो होगा इक्वल होगा सीडी के व किसके इक्वल होगा 4 एक्स के एंड बी सी इक्वल होगा एडी के और वह इक्वल होगा 3 एक् के यह बात कैसे जाना हमने कि पैरेललोग्राम के अपोजिट साइड होते हैं इक्वल इस बात से जाना ठीक है और यह बात हम लिखेंगे जो जानते हैं कि पेरीमीटर ऑफ ऑफ पैरेललोग्राम ए बी सीडी पैरेललोग्राम ए बी सीडी इ इक्वल टू क्या बोलेंगे हम लोग एी प्लस बी सी प्लस सीडी प्लस एडी यही तो बोलेंगे जबकि पेरीमीटर ऑफ पैरेललोग्राम क्वेश्चन में देख सकते हो गिवन है 21 सेंटीमीटर इ इक्वल टू ए बोले तो 4 एक् बी स बोले तो 3 एक् सीडी बोले तो 4 एक और एडी बोले तो 3 एक् यानी 21 सेंटीमीटर इ इक्वल टू चती साती 10 च 14 10x दैट मींस x = 21 सेमी बा 14 सेन के टेबल में थ्री टाइम सेन के टेबल में टू टाइम दैट मींस 3/2 सेंटीमीटर x का वैल्यू निकल के आ गया है हमें निकालना क्या है साइड्स निकालने है पैरेललोग्राम के है ना हेंस बोलेंगे हम ए जो है किसके इक्वल था सीडी के व किसके इक्वल था 4x के तो 4 मल्टीप एक्स की जगह पर क्या है 3/2 सेंटीमीटर दैट मींस टू 6 सेंटीमीटर क्लियर है एंड बी स जो था वह इक्वल था किसके एडी के अब वह किसके इक्वल है तो वह इक्वल है 3x के तो 3 न एक पर 3/2 सेंटीमीटर यानी कि 9/2 सेंटीमीटर ना को बोल सकते हैं कितना 4.5 सेंटीमीटर और यही फाइनल आंसर हुआ ठीक है |
190793 | https://www.youtube.com/watch?v=VxJPnqyvrBE | CDF Method: Distribution of Max(U1,U2), U1 and U2 are independent and are both uniform: U(0,1)
Bill Kinney
34600 subscribers
23 likes
Description
3156 views
Posted: 11 Feb 2022
Assuming U1 and U2 are independent uniform random variables on the interval (0,1), find the distribution of Max(U1,U2). Find the CDF first (CDF method to find PDF), then find the PDF, then find the mean, variance, and standard deviation. (Probability and Statistics with Applications: A Problem Solving Text, by Asimow and Maxwell)
probability #transformations #cdfmethod
Links and resources
🔴 Subscribe to Bill Kinney Math:
🔴 Subscribe to my Math Blog, Infinity is Really Big:
🔴 Follow me on Twitter:
🔴 Follow me on Instagram:
🔴 You can support me by buying "Infinite Powers, How Calculus Reveals the Secrets of the Universe", by Steven Strogatz, or anything else you want to buy, starting from this link:
🔴 Check out my artist son Tyler Kinney's website:
AMAZON ASSOCIATE
As an Amazon Associate I earn from qualifying purchases.
Transcript:
let's let a new random variable call it capital v equal what how about the bigger
of u1 and u2 maximum of u1 and u2 i might wonder what's the distribution of v and when i say distribution i'm including both saying find the pdf find the cdf maybe
find the mean find the variance find you know numerical summaries parameters that describe
the distribution of this new random variable what's going on here as as an equation relating
random variables it's a purely abstract thing how would you apply this idea in real life if
you are observing values of u1 and u2 that are uniformly distributed between 0 and 1 maybe
u1 ends up on the first observation u1 is 0.7 and u2 is 0.85 the bigger of those is
0.85 so the observed value of v is 0.85 you do another observation of the two random
variables maybe u1 is 0.94 and u2 is 0.36 the bigger those is 0.94 so the observed
value of v is 0.94 always take the bigger one now uniform distribution on 0 1
obviously has mean at one half center of mass i hope it makes some intuitive
sense that if you're always taking the bigger one probably the mean is going to be bigger
than one half for the maximum right both u1 and u2 have an equal probability of
being less than one-half or greater than one-half but on a random observation most likely or more
likely than not you might say one of them is probably going to be bigger than a half and one's
probably going to be less than half i mean they could could both both be bigger than one half
and both be less than one half that could happen but on average more than 50 percent of the
time probably one of them is going to be bigger than one half and so therefore the
maximum is going to be bigger than one half probably the mean of this
distribution is bigger than one half let's see if we can verify that theoretically
and then we'll try to confirm it with simulation so here's that second application of
simulation confirming theory first application perhaps more important is doing
things when you when the theory is too hard second application still important is
confirming theory when you're unsure about the theory perhaps do a simulation
and see if it confirms the theory let's do the theory how do you find the
distribution of v um same technique as tuesday find the cdf first and then differentiate it to
find the pdf that was the technique of tuesday the cdf of the capital f sub v i guess i should
call its variable little v maybe cursive v so that's a cursive v you could it
sort of looks like the greek letter nu as well and you which we hardly ever use but remember by definition the cdf is
always always always always right a probability that the random variable the capital letter is less than or equal to the
given little quantity which is a real number ordinary variable it's always the case whether
your random is discrete continuous or mixed we haven't done a mixed example yet and i
don't think we're going to get to it today perhaps next tuesday that's always what the cdf is again the
way you compute it depends on whether your variable is discrete continuous or mixed
but in theory this is always what it is now our variables are continuous so in
theory i could compute this perhaps with maybe an integral or something but let's
not worry about integrals for the moment what would be a good goal here perhaps well use that equation somehow and the only obvious way to use it at first
is just to plug it in there for v so do that keep the little v as it is i'm
plugging in for the capital v then you stare at this for a while and you start start on it for longer then you go
get some coffee or today i actually brought water go get some coffee stare at it for a while i don't
know what to do what do you do with this well okay that that's that's an experience many
people experience including me sometimes ask yourself what would it mean if the
bigger of two quantities was less than or equal to some other quantity what
could you say about both quantities if u and u2 are their maximum is less than or
equal to v what could you say about both u1 and u2 if the bigger is less than
b they're both less than v i could rewrite this fact is saying u 1 is
less than or equal to b and i'll go ahead and put the word and in there maybe i should
put intersect u 2 is less than or equal to v it gets a little funny if you put the intersection
symbol because like well what are the sets what are the events well they're they're
continuous events it gets a little confusing with the intersection symbol it's okay if you do this
a little bit more intuitive just put the word and okay i guess that's progress any more any bright ideas for more progress and and where we've seen
and or intersections before chapter one two three four two right chapter two multiplication rule hey and i'm assuming independence assume just what you need to avoid conditional
probabilities i can say this is the product of the probability that u n is
less than or equal to v times the probability that u2 is less than or equal to v this feels like a lot of
progress doesn't it i guess and i know the distributions of
u1 and u2 they're both uniform you know the cdf looks like that these are cdfs and here's where i use the independence
that should be pointed out independence hmm well i mean these things are piecewise
functions is that going to mess things up well you know the the domain of most interest
is when the input is between 0 and 1. if you want and you two are you know they
take on values between the zero and one with nonzero probability so are their
maximum i really need to only assume when i do this calculation that
little v is between zero and one and then i can use the formula for the
cdf what's the formula for the cdf here i haven't used the letter x but
maybe i should call it capital f of x is x when x is between 0 and 1. i mean
maybe i should have called a capital f of u is u but you know the variable name doesn't matter it's
the idea that matters the formula of the recipe that's the function i'm plugging the v into
there and there oh hey it's pretty simple v times v v squared isn't this nice that's that's pretty nice and
that's the cdf what about the pdf pdf of v once again assume v is between zero and one it is sort of
nicest to not worry about being at the endpoints because then you don't have to
worry about the differentiability issues little f sub v of v is the derivative of capital
f sub v of little v which is going to be 2v these valid cdfs and pdfs yeah the pdf has a graph that looks like this and yes it's zero elsewhere that's the number one there it's the area under the pdf is a trying area
of a triangle one half base times height base is one height is two the area is uh one well and we already knew that from the
graph of the cdf the graph of v squared when b is between 0 and 1 looks like that and
goes up to 1. and yes technically it's 0 when v is less than 0 and it's
1 when v is bigger than 1. and now you can see visually this pdf as a density function has a center of
mass that's going to be bigger than 0.5 right the graph is higher when the
input is bigger than a half data simulated data is going to be more
dense close to one then it is close to zero you're going to get a mean bigger
than a half just like x be expected what would the mean be what's the
expected value of v mu sub v if you like got to integrate v times the pdf
over the domain where it's positive integral from 0 to 1 of v
times little f sub v of v dv trying to make my v's cursive
distinguish them from uppercase my little v's so that's integrating 2 times little v squared dv because there's the pdf that's going to give
two-thirds little v cubed evaluated from zero to one and that's going to be two-thirds how about
that yep bigger than one point six repeating how about the variance we'll
find the second moment first we could also do this by finding
moment generating function and differentiating it and plugging in zero and you should be able to do that too by the
way should be able to find the moment generating function here although um to do the integral
for the moment generating function you'd have to use integration by parts i'm not going to
go through the details at least not right now here's the second moment you're ultimately
going to integrate two v cubed from zero to one that'll be one half v to the fourth from zero to
one and that means the second moment is one half and therefore the variance some people were
forgetting the variance is not the second moment so it's easy to make mistake to make
but it's always the second moment minus the square of the first moment get a common denominator of 18. nine
minus eight over eighteen looks like the variance is one eighteenth
but that didn't make a mistake i think that's good you could also label that as sigma
squared sigma sub v squared and then the standard deviation sigma sub v
is the square root of the variance so that'll be square root of 1 18th
which will be 1 over square root of 18 and i don't care if you leave the square
root in the denominator it's okay with me and you also could write this as square root of 9 times square root of 2 which
would be 3 square root of 2. you can leave it like that you can rationalize
the denominator if you want you don't have to multiply the top and bottom by square root of
2 and you write it this way all those are okay exact forms of the answer
for the standard deviation and it's bigger than the variance
right because the variance is less than one take its square root it's about
0.2357 for the standard deviation there you go yeah any of these is okay by the way
if you're going to be planning to be a teacher and you always remember your seventh grade math
teacher saying rationalize the denominator we don't want any radicals in denominators
it's against the law no it's not against the law it's okay dividing by zero is against the
law but you can have roots in denominators it's okay so why do they teach you rationalizing
because it's a useful skill to know sometimes it's helpful to be able to do
to simplify things sometimes so there we go can i clarify anything there you know with normal distributions
we have the 68 95 99.7 rule that for example says about 95 of the data
if you're talking data or about 95 of the probability if you're talking probability is
within two standard deviations of the mean does that rule work approximately even
when your distribution is not normal um very hand wavingly so a very vague rule of thumb is
yeah most of the data or most of the probabilities within two
standard deviations of the mean the mean is 0.6 repeating about 0.67 2 times the
standard deviation is close to what would be 0.47 0.67 plus or minus 0.47 well plus 0.47 brings you
past 1. 0.67 minus 0.47 is 0.2 so yeah most of the data most the probability is between point two and
and one effectively and you can see that it's not a very precise thing to say but certainly can be
said chebyshev's theorem chebyshev's inequality can also be brought to play to get a lower bound
of the probability but let's not worry about that we i said i want to relate this
to simulation so let's do so now and you'll once again be doing something
similar to this in your homework for next time um so again equals rand generates random numbers between zero
and one with a uniform distribution of course every time you copy
and paste changes i do want to do a bigger sample here that's going to
cause me to have to make this small enough that you won't be able to see it real well
i think i'd like to do a sample size 100. oops my my uh mouse pad or trackpad is kinda goofy it's always doing things i don't want
it to do let's go down to line 100 here so there's a hundred observed values
of a uniform zero one distribution i could copy and paste this entire column over which i guess yeah okay and get another 100 observed values of
course again every time you copy and paste everything changes which is why it's nice
to actually also do a copy and paste special so i'm going to copy ctrl c and
over here i'll paste special values only so then we have it now fixed and
so if i copy and paste anything over here again all that stuff changes but
these things are staying fixed just kind of nice to have
something that's staying fixed and yeah i could take the maximum equals max like that it'll take the bigger and those
are fixed numbers so it's going to stay that way so 0.39656 is the first bigger one
for the next one we should get 0.793399 yup um ignore these two columns
now and focus on these columns maybe i should have put a
space in here insert column next one is going to be the 0.84804
etc i'm always taking the bigger number i do equals average here should be close to 0.5 it is next one's
close to 0.5 but the last comment should be close to two-thirds point six seven yeah cool
confirming the theory yay that's what we're after doesn't that feel nice how about the standard deviation
that was about 0.2357 for a uniform distribution based on the exercise you did
it's actually not one of the ones i've graded where was it well okay let's just look at the table uniform distribution constant pdf on a finite
interval zero elsewhere its mean is the midpoint makes intuitive sense its variance is the
length of the interval squared divided by 12. so it's standard deviation is the length of
the interval divided by square root of 12. talking about the uniform one here
not not v 1 over square root of 12 sorry 1 over square root of 12 0.2887 so if i do standard deviation here for these should be close to two 0.29 say one two seven two eight point
two nine somewhere in there yep but with the with v it was closer
it should be smaller closer to 0.24 yep and you could also make histograms of these things which you know histograms are essentially
databased ways of visualizing the pdf so i could insert a chart and uh pick the okay pick the histogram
option for me that's it doesn't always do that so that's a histogram of the first column
which was from uniform distribution of course it's not going to be
perfectly horizontal because it's simulation but you know there's no
clear up or up or down pattern to this it's not an increasing
function clearly or decreasing but for the other one it should be pretty
clearly increasing so the last column insert chart histogram there we
go yeah pretty clearly increasing well that's because the pdf looks like that |
190794 | https://emedicine.medscape.com/article/266799-treatment | Tubal Sterilization Treatment & Management: Surgical Therapy, Preoperative Details, Puerperal Techniques
[x]
X
No Results
No Results
For You
News & Perspective
Tools & Reference
CME/CE
Video
Events
Specialties
Topics
Edition English
Medscape Editions
English
Deutsch
Español
Français
Português
UK
Invitations
About You
Scribe
Professional Information
Newsletters & Alerts
Your Watch List
Formulary Plan Manager
Log Out
Register
Log In
For You
News & Perspective
Tools & Reference
CME/CE
More
Video
Events
Specialties
Topics
EN
Medscape Editions
English
Deutsch
Español
Français
Português
UK
X
Univadis from Medscape
Welcome to Medscape
About YouScribeProfessional InformationNewsletters & AlertsYour Watch ListFormulary Plan ManagerLog Out
RegisterLog In
X
No Results
No Results
close
Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log outCancel
processing....
Tools & Reference>Clinical Procedures
Tubal Sterilization Treatment & Management
Updated: Nov 11, 2024
Author: Robert K Zurawin, MD; Chief Editor: Michel E Rivlin, MDmore...;)
9
Share
Print
Feedback
Close
Facebook
Twitter
LinkedIn
WhatsApp
Email
Sections
Tubal Sterilization
Sections Tubal Sterilization
Overview
Practice Essentials
Background
History of the Procedure
Relevant Anatomy
Contraindications
Prognosis
Show All
Workup
Treatment
Surgical Therapy
Preoperative Details
Puerperal Techniques
Laparoscopic Approaches
Mechanical Techniques
Follow-up
Outcome and Prognosis
Future and Controversies
Show All
References;)
Treatment
Surgical Therapy
Surgical approaches for female sterilization include laparoscopy, hysteroscopy, microlaparoscopy, laparotomy (concurrent with cesarean delivery), minilaparotomy, and vaginal approaches. Although minilaparotomy is the most common approach worldwide, laparoscopy is used most commonly for interval procedures in the United States. Hysteroscopic procedures are rapidly gaining popularity. For postpartum procedures, although some studies describe successful use of laparoscopy, the subumbilical minilaparotomy is used most commonly in the United States and worldwide. Vaginal colpotomy approaches are rarely used because they are associated with a higher incidence of infection and can no longer be recommended.
Local anesthesia is used for more than 75% of sterilizations worldwide. Although some US providers use local anesthesia for laparoscopic sterilization, general anesthesia (for interval procedures) and regional anesthesia (for postpartum procedures) are most common.
For the hysteroscopic approach, local anesthesia is the standard approach, and it may be supplemented by oral or IV sedation as needed.
One meta-analysis identified 9 randomized, controlled trials that compared 5 commonly used methods. Aspects such as training, costs, and maintenance may be important factors in deciding which method to choose.
Note the image below.
Elevation of the fallopian tube through the incision.
Puerperal tubal sterilization
In comparison with interval sterilization, infraumbilical minilaparotomy following delivery in the early puerperium is convenient, simple, and cost effective. However, if maternal or infant complications exist, sterilization should be delayed.
Bilateral tube ligation (BTL) may be performed after closure of the uterine incision during cesarean delivery or following completion of a vaginal delivery within 72 hours. Postpartum BTL is technically simple because the uterine fundus is at the level of the umbilicus, making the fallopian tubes readily accessible through a small periumbilical abdominal incision.
If the procedure is delayed for several days or if the patient has a significantly involuted uterus (as might occur after delivery of a preterm infant), then delaying to an interval procedure is usually prudent, although puerperal laparoscopic BTL has been reported.
Minilaparotomy
Minilaparotomy is defined as a laparotomy with an incision size smaller than 5 cm. The operation can be performed through a suprapubic incision in the interval after pregnancy and through a subumbilical incision within the first 48 hours after delivery.
A 2- to 5-cm periumbilical semilunar incision is made with the skin tented with Allis clamps. Dissection is carried down to the fascia, which is grasped with hemostats or Allis clamps and opened transversely, exposing the peritoneum, which can then be entered sharply. With uterine manipulation and retraction, the tubes can be visualized and grasped with a Babcock clamp. Often, the oviducts can be palpated at their utero-tubal junction and the uterus may then be rotated to position the isthmus of the tube under the incision anteriorly with the aid of Army-Navy retractors. The fallopian tube is "walked" with Babcock clamps until the fimbriated end is identified. A major cause of failure of sterilization is the inadvertent ligation of the round ligament mistakenly identified as the fallopian tube.
After the BTL, the minilaparotomy incision is closed in layers. Closure of the peritoneum is optional. The fascia is closed with running 2-0 or 0 delayed absorbable suture. Subcutaneous closure is optional, and the skin is closed with 3-0 or 4-0 absorbable suture in a subcuticular manner or with acrylic glue.
Laparoscopy
Advantages include small incisions, full access to the oviducts, rapid recovery, and the ability to inspect the pelvis and upper abdomen.
Disadvantages include the need for general anesthesia, the risks of vessel/viscera injury with needle insufflation/trocar entry, and difficulty associated with laparoscopy in patients who are obese or in the presence of abdominal and/or pelvic adhesions. Entry accounts for 30-50% of all laparoscopic sterilization complications.
The failure rate of the laparoscopic approach according to the US Collaborative Review of Sterilization (CREST) ranges from 7.5 per 1000 procedures for unipolar coagulation to a high of 36.5 per 1000 for the spring clip. The Filshie clip was not included in the CREST study, but its failure rate is reported to be between 1 and 2%.
According to one study, the success rate of laparoscopic sterilization on the first attempt is 99%. The same study estimates the success rate of hysteroscopy to be 88% on the first try. These rates account for 6% of women whose first attempt with hysteroscopy is unsuccessful and are ultimately sterilized via laparoscopy.
A French nationwide cohort study by Bouillon et al that included 105,357 women reported that risks of medical outcomes were not significantly increased with hysteroscopic sterilization compared with laparoscopic sterilization. Hysteroscopic sterilization was associated with higher risk of gynecological complications over 1 year and over 3 years compared to laparoscopic sterilization in women undergoing first sterilization.
The patient should always have an examination under anesthesia, and the bladder should be catheterized. A uterine manipulator and the use of the Trendelenburg position enhance exposure.
Microlaparoscopy
Microlaparoscopy involves use of 1.2- to 2-mm microendoscopes with 5- to 7-mm suprapubic ports for bipolar coagulation or mechanical occlusive devices. This surgery is possible because of improved technology in light transmission and fiberoptic bundles.
The theoretical advantages of less pain, less cost, and faster patient recovery have not been assessed through randomized, controlled trials, although several studies have been reported in an office setting. Despite almost 20 years of availability, office microlaparoscopy has not become widely accepted.
Hysteroscopy
Advantages include offering the most cost-effective, minimally invasive approach without the need for abdominal incisions or general anesthesia, thereby avoiding complications associated with trocar injury or thermal burns. Hysteroscopy is regularly performed in an office setting using local anesthesia, thus affording patients a rapid return to normal activity (usually within 24 hours). Compared with laparoscopic techniques, hysteroscopic sterilization studies cite cost savings of more than 50%. Prior abdominal surgery and obesity are not contraindications for hysteroscopy, nor are cardiovascular or anatomic contraindications to general anesthesia.
Essure was shown to be 99.80% effective in preventing pregnancy after 4 years of follow-up, and published papers report placemented rates ranging from 96-99%. [20, 21, 22] The hysteroscopic failure rate is the lowest of any surgical approach with 0.5/1000 at 1 year.
Hysteroscopy requires an FDA-mandated hysterosalpingogram to confirm correct placement and to document tubal occlusion. Patients often find this procedure a reassuring confirmation that the tubes are indeed occluded.
The Essure system was voluntarily removed from the US market on December 31, 2018. As of December 31, 2019, all unused Essure units should have been returned to Bayer so that they are no longer available for implantation.
Next: Preoperative Details
Preoperative Details
Informed consent and preoperative counseling
Inform the patient that a sterilization procedure is intended to be permanent and irreversible and that, as with any form of contraception, a small chance of failure exists. In some languages, sterilization is synonymous with castration. Therefore, the patient must have a very clear grasp of the surgical procedure; diagrams and take-home pamphlets are helpful.
The relative likelihood of an ectopic pregnancy is increased when sterilization failure occurs in all procedures involving occlusion or excision of the isthmic or ampullary segments of the tube, but ectopic pregnancy has not been demonstrated in the hysteroscopic approach. Inform the patient that complications exist, although the incidence of major complications is low. For minilaparotomy and laparoscopic techniques, complications may include injuries to the gastrointestinal and genitourinary tracts, infection, hemorrhage, and complications of anesthesia. For the hysteroscopic approach, the same complications associated with diagnostic hysteroscopy apply — uterine perforation, bleeding, excessive absorption of distention media, and infection.
Alternatives to permanent female sterilization include vasectomy and reversible methods of contraception (eg, injectable and implantable progestins, the levonorgestrel intrauterine system and other commercially available intrauterine devices). Although reversible methods of contraception (eg, injectable and implantable progestins, the levonorgestrel intrauterine system and other commercially available intrauterine devices) are sometimes prescribed in lieu of a permanent solution, these methods are associated with hormonal side effects such as irregular uterine bleeding and are not intended as long-term solutions. For example, Mirena labeling counsels that their product should only be used for up to 5 years for pregnancy prevention.
Screen for risk indicators for regret, including young age, low parity, single parent status, or marital instability. Stress the need to use condoms for protection against sexually transmitted diseases and HIV infection if the patient is at risk of exposure. Document the informed consent process in the patient's medical record. The patient should also receive a copy of the completed consent form to review and retain.
Physicians must be aware of and follow any applicable federal and state requirements regarding informed consent for sterilization procedures, including time intervals between consent and sterilization. If federal funds are used for the procedure, the required US Department of Health and Human Services consent form must be signed 30-180 days prior to surgery. Furthermore, this consent cannot be obtained if the patient is younger than 21 years, in labor, under the influence of drugs or alcohol, mentally incompetent, or having an abortion.
Review the preoperative history and perform a physical examination to determine if any contraindications exist for elective surgery.
Previous
Next: Preoperative Details
Puerperal Techniques
Pomeroy technique
This technique is the simplest and most commonly performed puerperal tubal sterilization.
The mid portion of the oviduct is grasped with a Babcock clamp, creating a loop, which is tied with 2-0 or 0 plain catgut suture, and each limb of the tubal knuckle is cut separately. Specimens are submitted to pathology. The endosalpinx at the cut ends may be cauterized (optional). The ligation sutures are held while the tube is cut to prevent retraction of the tubal stumps into the peritoneal cavity before they can be adequately examined for hemostasis.
The original description consisted of forming a loop of the ampullary segment of the fallopian tube and ligating the base of the loop with a double strand of 1-0 chromic catgut, followed by resection of the top of the ligated loop. The rationale for this technique is based on prompt absorption of the suture ligature with subsequent separation of the cut ends of the tube, which then become sealed by spontaneous reperitonealization and fibrosis. A resultant natural gap of 2-3 cm should occur between the severed proximal and distal segments of the tube.
Many modifications of the Pomeroy technique have been described; the most common involves doubly ligating each loop.
Failure rates are reported to be 1 case in 300-500 patients.
Parkland technique
The Parkland technique is a midsegmental resection similar to the Pomeroy technique, except each leg of the loop is tied separately. The Parkland technique was designed to avoid the intimate approximation of the tubal cut ends, as occurs with the Pomeroy technique, thereby theoretically reducing the risk of subsequent recanalization.
An avascular area in the mesosalpinx directly under the tube is perforated with a hemostat, and the jaws are opened to spread the mesosalpinx, thereby freeing approximately 2.5 cm of tube. The tube is then ligated proximally and distally with a 0 or 00 plain or chromic suture, and a 1- to 2-cm tubal segment is excised and submitted for pathologic confirmation.
Failure rates are reported to be 1 case in 400 patients.
Uchida technique
The mid portion of the oviduct is raised with 2 Babcock clamps. The tubal serosa is hydrodissected from the muscularis by subserosal injection of a dilute (1:100,000) saline solution of epinephrine or isotonic sodium chloride solution. A linear incision is made parallel to the axis of the tube in the ballooning serosa on the antimesosalpingeal aspect with a scalpel, #15 blade. The serosal peritoneum is grasped on either side of the tubal incision with hemostats, and a third hemostat is used to bluntly dissect and reflect the serosa and the surrounding areolar tissue from the tubal muscularis. With the tubal muscularis exposed,
a relatively long (5 cm) segment of tubal muscularis is ligated proximally and distally with a 0 or 0-0 plain catgut suture and resected. The serosal edges are then reapproximated, burying the proximal exposed tubal end within the leaves of the broad ligament, leaving the distal end exposed.
During the puerperium, Uchida modified the sterilization procedure by including fimbriectomy.
Clearly, the excision of such a large segment of tube, combined with a fimbriectomy, accounts for the low rate of failure for this technique. For all practical purposes, it is a salpingectomy.
Irving technique
The Irving technique is designed to be used in conjunction with cesarean delivery.
A mesosalpingeal window is created beneath the tube approximately 4 cm from the uterotubal junction. The tube is doubly ligated with 0 or 00 absorbable suture and severed, with the sutures on the proximal end left long. The proximal tubal stump may require mobilization by dissecting it free from the mesosalpinx. A small nick is made into the serosa on the posterior (or anterior) uterine wall near the uterotubal junction. A hemostat is used to deepen the incision, creating a pocket in the myometrium approximately 1-2 cm deep. The 2 free ends of the proximal stump ligature are then individually threaded onto a curved needle and brought deep into the myometrium tunnel and out through the uterine serosa. Traction on the sutures draws the proximal tubal stump deep into the myometrial tunnel, and the sutures are tied. The serosal opening of the tunnel is then closed around the tube with fine absorbable suture.
An additional option is to bury the distal end of the tube between the leaves of the broad ligament as originally described by Irving.
Failure rates are less than 1 case in 1000 patients.
Previous
Next: Preoperative Details
Laparoscopic Approaches
Electrodesiccation technique
The use of electrosurgery is preferable when the fallopian tube is edematous, thickened, or cannot be mobilized easily for mechanical device placement. This technique should always be readily available during laparoscopic BTL, both as a backup method of sterilization and for control of unexpected bleeding. However, the technique causes greater tubal damage, making
tubal reversal more difficult if the patient regrets her decision.
Bipolar current
Bipolar current is theoretically inherently safer than unipolar current because tissue destruction is essentially confined to the area between and immediately adjacent to the bipolar paddles.
The oviduct is identified and grasped at the mid isthmus region, at least 2.5-3 cm lateral to the uterotubal junction, with the bipolar forceps. The tube is elevated to ensure the forceps are not in contact with any other structure (eg, bowel, sidewall), and current is applied. Older electrosurgical generators do not have active feedback, and excessive tissue destruction is the rule, not the exception. High voltage results in excessive heat and tissue charring, sometimes causing sticking of the electrodes. Extensive damage to the tissue may facilitate future fistula formation and encourage failure. Modern bipolar generators have circuitry that measures tissue impedance, so that when desiccation of the tissue is complete, an audible alert is given. The procedure is repeated 2-3 times for each tube to create a 3-cm contiguous area of desiccation.
Formation of tuboperitoneal fistula, with a subsequent risk of pregnancy (including ectopic pregnancy) or possible pelvic inflammatory disease (PID), is minimized by maintaining the most proximal burn no closer than 2 cm to the uterine cornu. Leaving a 2- to 3-cm pedicle allows enough space for absorption of intrauterine fluid under pressure, such as during menstruation, and minimizes the risk of fistula formation.
However, the downside of this technique becomes apparent if the patient later undergoes endometrial ablation for menorrhagia. Endometrial ablation causes varying degrees of intrauterine synechiae (Asherman syndrome). Since islands of endometrium frequently survive the ablation procedure, small amounts of menstrual fluid are still produced. In most cases, the blood exits the cervix, but occasionally the surviving endometrium is located in the cornual region of the uterus and is surrounded by postablation synechiae. When this happens, the menstrual fluid is forced through the tubal ostium and creates a small hematosalpinx in the surviving segment of fallopian tube. This is accompanied by severe cyclic pain and frequently requires additional surgery, often hysterectomy, to correct.
Monopolar current
In surgery, using monopolar energy, the current passes through the entire patient to complete the circuit between the electrode and the electrosurgical generator. Faulty instrumentation or improper techniques increase the possibility of stray current causing thermal injury to adjacent structures. Details on electrosurgery and its complications may be found in a separate chapter.
The initial popularity of unipolar current occurred during the early years of laparoscopic sterilization when it was often the only instrumentation available. Its use has diminished following many documented bowel injuries. Although all surgeons should have a proper understanding of the principles of electrosurgery that would enable them to avoid complications, unipolar current has largely been replaced with bipolar electrodesiccation.
A return electrode is placed on the patient. The unipolar current applied to the oviduct flows from the electrode through the patient's body and completes the circuit to the generator through the return electrode. The use of a metal trocar sleeve avoids capacitative coupling between the forceps and the sleeve, and any electrical current flowing to the trocar is dispersed through the patient's abdominal wall. An electrocoagulating grasping forceps is placed completely around the isthmic portion of the tube, approximately 4 cm from the uterine cornu. The oviduct is mobilized away from any viscera and the sidewall. A low-voltage generator, with a maximum peak of 600 V and maximum power of 100 W, is used to apply current for approximately 5 seconds, until blanching and swelling of the tube is visible. The highest success rates are achieved when at least 3 cm of tube is destroyed.
Thermal injury to the bowel may occur either from direct current flow via the tube being coagulated or from undetected contact between the forceps or trocar sleeve and bowel. Patients with thermal injuries tend to present with an acute abdomen several days after the procedure.
Previous
Next: Preoperative Details
Mechanical Techniques
Falope (Yoon) ring technique
A nonreactive silicone rubber band measuring 3.6 mm in outer diameter and incorporating 5% barium sulfate for radiographic identification is used. The applicator consists of inner grasping prongs and an outer double-barreled sheath. The Falope ring is stretched around the base of the applicator sheath. Some devices allow for double-loading of the rings so that the applicator
needs to be inserted into the abdominal cavity only once.
The isthmic portion of the fallopian tube is identified. The forceps of the applicator are extended and a segment at least 3 cm from the uterine cornu is grasped, taking care to avoid the proximity of any vessels in the mesosalpinx. Approximately 2.5 cm of tube is gently pulled into the barrel using a slow "milking" technique. This may be difficult with edematous tubes, or in the presence of chronic pelvic adhesions. The larger-diameter outer barrel then pushes the Falope ring over the knuckle of tube, and the ring then returns to its former state, with an inner diameter of 1 mm.
The loop of tube should clearly contain 2 complete lumens of tube. Slowly advancing the entire applicator toward the tube while gradually retracting the tongs and tube into the applicator and avoiding excessive traction on the tube are important.
Failure to do this can result in mesosalpingeal hemorrhage and tubal laceration, which occur in approximately 1-5% of cases. This can be treated with bipolar coagulation, or a Falope ring may be placed on each transected end.
Falope ring application has traditionally been considered more painful postoperatively secondary to ischemia; however, this was not established in a randomized controlled trial. The failure rate is reported to be 3.3 cases per 1000 patients.
Hulka-Clemens clip technique
The clip is designed to be applied at a right angle to the isthmic portion of the tube 2.5-3 cm from the uterotubal junction. When properly applied, only 4 mm of tube and virtually none of the blood supply is destroyed.
The clip consists of 2 toothed jaws of Lexan plastic joined by a stainless steel hinge pin. The lower jaw has a distal hook. A gold-plated spring maintains the clip in an open position. When completely advanced, the spring closes and locks the jaw.
The Hulka applicator is 7 mm in diameter with a 3-ring configuration at the upper end consisting of a central ring (designed to accommodate the thumb for stabilization) and a pair of lower rings (to accommodate the index and middle fingers and control the clip application mechanism). The distal end of the applicator has a fixed lower jaw to accommodate the clip. A mobile upper jaw, when retracted, permits placement of the open clip and, when advanced, closes the clip. When completely advanced, a central piston locks the spring.
Once the oviducts have been identified laparoscopically and deemed suitable for clip sterilization, the Hulka clip applicator is introduced with the clip in the closed position, and the clip is opened after the applicator is intra-abdominal in position. The hook of the lower jaw is placed against the posterior mesosalpinx, the tube is tented slightly upwards, and the clip is applied. The clip may be opened and repositioned repeatedly until the correct position is achieved, at which time the center piston is advanced to permanently lock the clip and unseat it from the applicator. If the clip has not been applied satisfactorily, a second clip is placed immediately alongside the first.
The applicator is withdrawn from the abdomen and reloaded, and the contralateral tube is treated in the same fashion. Failure of the Hulka clip should not exceed 2-3 cases per 1000 patients.
A review by Harrison et al of randomized, double-blind, placebo-controlled studies indicated that postoperative pain following laparoscopic ring or clip tubal ligation can be significantly reduced through the administration of local anesthetic during the procedure. The investigators reported that this strategy substantially decreased pain for up to 8 hours postsurgery.
Filshie clip technique
This technique is widely used in Canada, the United Kingdom, and Australia and was approved for use in the United States in 1997. This technique involves a 12.7-mm long clip of titanium with a silicone rubber lining. The clip is applied laparoscopically with an applicator, much like the Hulka spring clip, at right angles to the isthmus approximately 2-2.5 cm from the uterotubal junction.
Initially, the clip occludes the tubal lumen by pressure. As tubal necrosis occurs, the rubber expands to maintain blockage of the lumen. The tube eventually divides, and the stumps heal closed. The Filshie clip usually remains attached and is eventually covered by peritoneum. Theoretically, because the silicone rubber of the Filshie clip is able to expand and provide continuous pressure even when the tube becomes ischemic, any residual tubal patency, such as may occur with the spring clip, is prevented.
Rare reports of migration of the Filshie clip into the bladder, vagina, peritoneal cavity, and appendix have been published, as have reports of expulsion of Filshie clips from the vagina, urethra, and rectum (occurring at a similar rate as expulsion of the Hulka clip). Migrations and expulsions are usually symptomatic and of little clinical significance. In all cases, the clips were found closed, the tubes were fully occluded, and no long-term adverse sequelae occurred.
Cumulative data at 24 months of follow-up (based on findings of Family Health International, which has conducted 11 studies of the Filshie clip at 43 sites in 10 countries) report a failure rate of 7 cases per 1000 patients.
Hysteroscopic techniques
The Essure microinserts device consists of polyethylene terephthalate (PET) (Dacron) fibers wrapped around a stainless steel core, surrounded by 24 coils of nickel-titanium alloy (Nitinol), a substance widely used for coronary artery stents and cardiovascular devices. The microinserts are dynamic, spring-like devices that are inserted into each fallopian tube. Once deployed, the effectiveness of the Essure microinserts is believed to be due to a combination of the space filling design of the device and a local occlusive, benign tissue response to the PET fibers. This process takes approximately 3 months to form complete occlusion, which is then documented by a low pressure hysterosalpingogram. Hysteroscopy may be performed in the office setting under local anesthesia, which is preferred, although regional or general anesthesia may be selected according to patient or surgeon preference. Successful bilateral placement rates vary from 94.6-99%. [24, 21, 25]
The procedure is basically identical to the Novy cannulation of the fallopian tube for tubal obstruction. A 5-mm operative hysteroscope with a 5-French operating channel is inserted under direct vision through the cervical os, and the uterine cavity is entered. Normal saline is used for the distension medium, which minimizes the risk of fluid overload and virtually eliminates the risk of electrolyte imbalance inherent in the use of isotonic solutions (glycine and sorbitol).
Both tubal ostia are identified. The device is passed through the operating channel and guided into the tubal ostium to the depth of the black indicator on the outer cannula. With the applicator steadied against the hysteroscope, the wheel on the handle is rotated, which causes the outer sheath to retract, exposing the wound-down coils of nickel-titanium. The device is then deployed by pressing a release button and turning the wheel again. The device is then retracted from the operating channel and the procedure is repeated on the contralateral side. Out of the 24 coils, 3-8 coils must be visible trailing in the uterine cavity to confirm proper placement of the device. Since the Essure microinserts are clearly visible via HSG, this provides physicians and patients reassurance regarding proper placement and tubal occlusion.
Long-term 5-year Essure data from the Phase II and Pivotal Trial continue to demonstrate safety and high patient satisfaction with zero reported pregnancies. Since its approval, in the commercial setting, the evaluable performance of Essure is still consistent with the age adjusted effectiveness of 99.85% (internal Conceptus data, with permission). Perforation of the fallopian tube due to forceful insertion has been reported but did not cause intraoperative or postoperative symptoms and was detected only on hysterosalpingogram. The same risks and complications associated with diagnostic hysteroscopy apply to this procedure.
Ultimately, the Essure system was removed from the US market on December 31, 2018. As of December 31, 2019, all unused Essure units should have been returned to Bayer so that they are no longer available for implantation.
Another hysteroscopic device, Adiana, uses radiofrequency energy from a separate electrosurgical generator to desiccate a small segment of the interstitial portion of the fallopian tube, after which a small silicone plug is inserted. This method, using monopolar electrosurgery, requires glycine as the distension medium and introduces the risk of electrolyte imbalance inherent in the use of nonionic solutions as well as the potential risk of thermal injury should the electrode perforate into the abdominal cavity. Since the silicone plug is not radiopaque, its final location cannot be confirmed on HSG to aide in confirmation of occlusion. Twelve-month data for Adiana reported a 1.07% failure rate that included 6 reported pregnancies, 1 of which was ectopic. Two-year data had 3 pregnancies out of 524 pivotal trial patients for a cumulative failure rate of 1.67%. At 4 years there was one pregnancy after 42 months of undetermined origin.
The Adiana system for female sterilization is no longer available in the United States. The manufacturer stopped producing and marketing the system in 2012.
Another device very similar to the Essure microinsert is in trials. The use of quinacrine to cause a chemical occlusion of the fallopian tube is under investigation.
Previous
Next: Preoperative Details
Follow-up
The follow-up visit for open or laparoscopic approaches is 1-2 weeks postoperatively. Instruct the patient to notify her health care provider if she develops fever (38°C or 100.4°F), increasing or persistent abdominal pain, or bleeding or purulent discharge from the incision.
Patients who have undergone hysteroscopic sterilization must be counseled to use an alternate form of contraception for 3 months at which time a low-pressure hysterosalpingogram must be obtained to confirm placement and bilateral tubal occlusion. The importance of the 3-month hysterosalpingogram needs to be communicated to patients at the time of microinsert placement; subsequent office follow-up may be required to ensure patients comply with confirmation test.
Inform all women who have undergone sterilization about the signs and symptoms of pregnancy (eg, amenorrhea, vaginal bleeding/spotting, abdominal pain) and ectopic pregnancy, and advise these women to seek immediate medical attention if such signs occur.
Previous
Next: Preoperative Details
Outcome and Prognosis
Noncontraceptive benefits are as follows:
Ovarian cancer
Several studies report a protective effect of sterilization against ovarian cancer, with a relative risk ranging from 0.2-0.8. Protection is hypothesized to result from reduced exposure of the ovaries to potential environmental carcinogens and infectious sources of malignant transformation (eg, oncogenic viruses).
Pelvic inflammatory disease
Although BTL does not protect against the acquisition of sexually transmitted disease, sterilization has been demonstrated to reduce the spread of organisms from the lower genital tract to the peritoneal cavity and thus protect against PID. Studies have reported that PID is less common in women who are sterilized compared with women who are not sterilized; however, protection is not absolute because uncommon reports exist of PID in sterilized women within 4-6 weeks or several years following surgery.
In cases of infection occurring within weeks of surgery, manipulation of the cervix, uterus, or oviducts is postulated to exacerbate a chronic infection or facilitate bacterial ascent from the lower genital tract at the time of surgery (eg, chlamydial or gonococcal cervicitis). PID occurring years after BTL results from bacteria ascending through a uteroperitoneal fistula or spontaneously recanalized tube. In these situations, if the patient requires surgery for diagnosis or treatment, relegating or excising the oviducts to help prevent future episodes of infection and ectopic pregnancy is prudent. Case reports have also demonstrated abscesses in the stump of the proximal fallopian tube.
Previous
Next: Preoperative Details
Future and Controversies
Future methods
Researchers continue to explore the possibility of using various substances that can be introduced through the cervix to occlude the tubal lumen through sclerosis or mechanical occlusion.
Quinacrine (historically used as an antimalarial drug) used for sterilization is instilled into the oviducts via transcervical application through a modified copper T intrauterine device. Although not approved for sterilization purposes in any country, an estimated 100,000 women have undergone this method of sterilization. The mechanism of action is tubal occlusion caused by inflammation and fibrosis of the intramural portion of the tube. Long-term failure rates, complications, optimal doses, and the need for adjuvants (eg, nonsteroidal anti-inflammatory drugs) are not clear because of nonsystematic investigation of the method and poor follow-up of women who have received it. Toxicologic studies and follow-up data are needed before the initiation of any human trials in the United States.
Post–tubal ligation syndrome
Proposed in 1951, this syndrome is a controversial constellation of symptoms, including pelvic discomfort, ovarian cystic changes, and menorrhagia, which are suggested to occur as a result of disruption of the uteroovarian blood supply, with resultant disturbances of ovulatory function after BTL. Often, these patients have a history of these problems before BTL or have been taking birth control pills, which masked their symptoms.
After extensive study, BTL apparently causes few, if any, menstrual abnormalities within several years after sterilization, regardless of the method of tubal occlusion used. Data from the CREST study indicate that women who are tubally sterilized are no more likely than women who have not undergone the procedure to report intermenstrual bleeding or changes in menstrual cycle length. However, women who are sterilized are more likely to report a reduced number of days of bleeding, less overall bleeding and menstrual pain, and increased cycle irregularity. BTL also seems to help reduce blood flow in women who reported very heavy menstrual bleeding at baseline.
Previous
;)
References
Powell CB, Alabaster A, Simmons S, et al. Salpingectomy for Sterilization: Change in Practice in a Large Integrated Health Care System, 2011-2016. Obstet Gynecol. 2017 Nov. 130 (5):961-7. [QxMD MEDLINE Link].
Mandelbaum RS, Matsuzaki S, Sangara RN, et al. Paradigm shift from tubal ligation to opportunistic salpingectomy at cesarean delivery in the United States. Am J Obstet Gynecol. 2021 Oct. 225 (4):399.e1-399.e32. [QxMD MEDLINE Link].
Wagar MK, Forlines GL, Moellman N, Carlson A, Matthews M, Williams M. Postpartum Opportunistic Salpingectomy Compared With Bilateral Tubal Ligation After Vaginal Delivery for Ovarian Cancer Risk Reduction: A Cost-Effectiveness Analysis. Obstet Gynecol. 2023 Apr 1. 141 (4):819-27. [QxMD MEDLINE Link].
[Guideline] ACOG Committee Opinion No. 774: Opportunistic Salpingectomy as a Strategy for Epithelial Ovarian Cancer Prevention. Obstet Gynecol. 2019 Apr. 133 (4):e279-e284. [QxMD MEDLINE Link].
Uchida H. Uchida tubal sterilization. Am J Obstet Gynecol. 1975 Jan 15. 121(2):153-8. [QxMD MEDLINE Link].
Hulka JF, Fishburne JI, Mercer JP, Omran KF. Laparoscopic sterilization with a spring clip: a report of the first fifty cases. Am J Obstet Gynecol. 1973 Jul 1. 116(5):715-8. [QxMD MEDLINE Link].
Filshie GM, Casey D, Pogmore JR, Dutton AG, Symonds EM, Peake AB. The titanium/silicone rubber clip for female sterilization. Br J Obstet Gynaecol. 1981 Jun. 88(6):655-62. [QxMD MEDLINE Link].
FDA restricts sale and distribution of Essure to protect women and to require that patients receive risk information. U.S. Food and Drug Administration. Available at April 9, 2018; Accessed: August 20, 2018.
US Food & Drug Administration. Essure Permanent Birth Control. FDA. Available at 2021 Jun 14; Accessed: July 25, 2021.
[Guideline] American College of Obstetricians and Gynecologists' Committee on Practice Bulletins—Gynecology. ACOG Practice Bulletin No. 208: Benefits and Risks of Sterilization. Obstet Gynecol. 2019 Mar. 133 (3):e194-e207. [QxMD MEDLINE Link].
US Food and Drug Administration. Laparoscopic Trocar Injuries:. A report from a U.S. Food and Drug Administration (FDA) Center for Devices and Radiological Health (CDRH) Systematic Technology Assessment of Medical Products (STAMP) Committee. 2003;. [Full Text].
Rodriguez MI, Edelman AB, Kapp N. Postpartum sterilization with the titanium clip: a systematic review. Obstet Gynecol. 2011 Jul. 118(1):143-7. [QxMD MEDLINE Link].
Syed R, Levy J, Childers ME. Pain associated with hysteroscopic sterilization. JSLS. 2007 Jan-Mar. 11(1):63-5. [QxMD MEDLINE Link].
Levie M, Chudnoff S, Kaiser B, Levy B, Snyder D. Multicenter Trial of Hysteroscopic Sterilization in the Office Setting Under Local Anesthesia: Patient Assessment of Procedural Pain and Satisfaction. Washington DC. AAGL. 2007.
Schepens JJ, Mol BW, Wiegerinck MA, Houterman S, Koks CA. Pregnancy outcomes and prognostic factors from tubal sterilization reversal by sutureless laparoscopical re-anastomosis: a retrospective cohort study. Hum Reprod. 2011 Feb. 26 (2):354-9. [QxMD MEDLINE Link].
Lawrie TA, Nardin JM, Kulier R, Boulvain M. Techniques for the interruption of tubal patency for female sterilisation. Cochrane Database Syst Rev. 2011 Feb 16. 2:CD003034. [QxMD MEDLINE Link].
Gariepy AM, Creinin MD, Schwarz EB, Smith KJ. Reliability of laparoscopic compared with hysteroscopic sterilization at 1 year: a decision analysis. Obstet Gynecol. 2011 Aug. 118(2 Pt 1):273-9. [QxMD MEDLINE Link].
Bouillon K, Bertrand M, Bader G, Lucot JP, Dray-Spira R, Zureik M. Association of Hysteroscopic vs Laparoscopic Sterilization With Procedural, Gynecological, and Medical Outcomes. JAMA. 2018 Jan 23. 319 (4):375-387. [QxMD MEDLINE Link].
Hopkins MR, Creedon DJ, Wagie AE, Williams AR, Famuyide AO. Retrospective cost analysis comparing Essure hysteroscopic sterilization and laparoscopic bilateral tubal coagulation. J Minim Invasive Gynecol. 2007 Jan-Feb. 14(1):97-102. [QxMD MEDLINE Link].
Kerin JF, Carignan CS, Cher D. The safety and effectiveness of a new hysteroscopic method for permanent birth control: results of the first Essure pbc clinical study. Aust N Z J Obstet Gynaecol. 2001 Nov. 41(4):364-70. [QxMD MEDLINE Link].
Arjona JE, Miño M, Cordón J, Povedano B, Pelegrin B, Castelo-Branco C. Satisfaction and tolerance with office hysteroscopic tubal sterilization. Fertil Steril. 2008 Oct. 90(4):1182-6. [QxMD MEDLINE Link].
Bradley L. Long-Term Follow-Up of Hysteroscopic Sterilization with the Essure Microinsert. Supplement to The Journal of Minimally Invasive Gyenocology. Fertil Steril. 2008. 15(6):S14-S15.
Harrison MS, DiNapoli MN, Westhoff CL. Reducing postoperative pain after tubal ligation with rings or clips: a systematic review and meta-analysis. Obstet Gynecol. 2014 Jul. 124(1):68-75. [QxMD MEDLINE Link].
Essuremd. ESS305 Information for Use (IFU). [Full Text].
Ubeda A, Labastida R, Dexeus S. Essure: a new device for hysteroscopic tubal sterilization in an outpatient setting. Fertil Steril. 2004 Jul. 82(1):196-9. [QxMD MEDLINE Link].
Vancaillie TG, Anderson TL, Johns DA. A 12-month prospective evaluation of transcervical sterilization using implantable polymer matrices. Obstet Gynecol. 2008 Dec. 112(6):1270-7. [QxMD MEDLINE Link].
Carney PI, Lin J, Xia F, Law A. Temporal trend in the use of hysteroscopic vs laparoscopic sterilization and the characteristics of commercially insured and Medicaid-insured females in the US who have had the procedures. Int J Womens Health. 2016. 8:137-44. [QxMD MEDLINE Link].
Media Gallery
Elevation of the fallopian tube through the incision.
of 1
Tables
Back to List
;)
Contributor Information and Disclosures
Author
Robert K Zurawin, MD Associate Professor, Chief, Section of Minimally Invasive Gynecologic Surgery, Department of Obstetrics and Gynecology, Baylor College of Medicine
Robert K Zurawin, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Society for Reproductive Medicine, Association of Professors of Gynecology and Obstetrics, Central Association of Obstetricians and Gynecologists, Society of Laparoscopic and Robotic Surgeons, Texas Medical Association, AAGL, Harris County Medical Society, North American Society for Pediatric and Adolescent Gynecology
Disclosure: Received consulting fee from Ethicon for consulting; Received consulting fee from Bayer for consulting; Received consulting fee from Hologic for consulting.
Coauthor(s)
Avi J Sklar, MD, FACOG, FACS, FRCSC Co-Chief, Division of Gynecology, Department of Obstetrics and Gynecology, Santa Clara Valley Medical Center
Avi J Sklar, MD, FACOG, FACS, FRCSC is a member of the following medical societies: American Society for Colposcopy and Cervical Pathology, American College of Obstetricians and Gynecologists, American College of Surgeons, Phi Beta Kappa, Royal College of Physicians and Surgeons of Canada
Disclosure: Nothing to disclose.
Specialty Editor Board
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Received salary from Medscape for employment.
Richard S Legro, MD Professor, Department of Obstetrics and Gynecology, Division of Reproductive Endocrinology, Pennsylvania State University College of Medicine; Consulting Staff, Milton S Hershey Medical Center
Richard S Legro, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, Society of Reproductive Surgeons, American Society for Reproductive Medicine, Endocrine Society, Phi Beta Kappa
Disclosure: Received honoraria from Korea National Institute of Health and National Institute of Health (Bethesda, MD) for speaking and teaching; Received honoraria from Greater Toronto Area Reproductive Medicine Society (Toronto, ON, CA) for speaking and teaching; Received honoraria from American College of Obstetrics and Gynecologists (Washington, DC) for speaking and teaching; Received honoraria from National Institute of Child Health and Human Development Pediatric and Adolescent Gynecology Research Thi.
Chief Editor
Michel E Rivlin, MD Former Professor, Department of Obstetrics and Gynecology, University of Mississippi School of Medicine
Michel E Rivlin, MD is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Medical Association, Mississippi State Medical Association, Royal College of Surgeons of Edinburgh, Royal College of Obstetricians and Gynaecologists
Disclosure: Nothing to disclose.
Additional Contributors
Anthony Charles Sciscione, DO Professor, Department of Obstetrics and Gynecology, Drexel University College of Medicine; Director, Maternal and Fetal Medicine, Christiana Care Health System; Director, Delaware Center for Maternal and Fetal Medicine
Anthony Charles Sciscione, DO is a member of the following medical societies: American College of Obstetricians and Gynecologists, American Medical Association
Disclosure: Nothing to disclose.
Close;)
What would you like to print?
What would you like to print?
Print this section
Print the entire contents of
Print the entire contents of article
Top Picks For You
encoded search term (Tubal Sterilization) and Tubal Sterilization
What to Read Next on Medscape
Related Conditions and Diseases
Oral Contraceptive Pill (Birth Control Pill) Hormonal Activity
Contraception
Contraceptive Implant Placement
Women's Health and Epilepsy
Patient Simulation: A 36-Year-Old Woman Seeking Long-term Contraceptive Options
Rapid Review Quiz: Hormonal Contraception
News & Perspective
AAP Updates Adolescent Contraception Guidance
Study Questions Safety of Progestins in Contraception
Scottish Community Pharmacies to Provide 'Bridging Contraception'
Advanced Antibiotic Strategies in Hidradenitis Suppurativa
SLE and Reproductive Health: Navigating Risks and Options
Society for Pediatric Dermatology (SPD) 2025 Annual Meeting
Tools
Drug Interaction Checker
Pill Identifier
Calculators
Formulary
Slideshow
HPV Diseases: From Cutaneous to Anogenital to Epidermodysplasia Verruciformis
Sections Tubal Sterilization
Overview
Practice Essentials
Background
History of the Procedure
Relevant Anatomy
Contraindications
Prognosis
Show All
Workup
Treatment
Surgical Therapy
Preoperative Details
Puerperal Techniques
Laparoscopic Approaches
Mechanical Techniques
Follow-up
Outcome and Prognosis
Future and Controversies
Show All
References;)
Recommended
2002 2172310-overview Tables & Protocols Tables & ProtocolsOral Contraceptive Pill (Birth Control Pill) Hormonal Activity
2003 /viewarticle/1002478 Journal Article U.S. Selected Practice Recommendations for Contraceptive Use, 2024 6.0 CME / CE / ABIM MOC Credits Journal Article You are being redirected to Medscape Education Yes, take me there6.0 CME / CE / ABIM MOC U.S. Selected Practice Recommendations for Contraceptive Use, 2024
2002 258507-overview Procedures ProceduresContraception
Medscape
Log in or register for free to unlock more Medscape content
Unlimited access to our entire network of sites and services
Log in or Register
Log in or register for free to unlock more Medscape content
Unlimited access to industry-leading news, research, resources, and more
Email
Continue
or
Log in with Google
Log in with Facebook
Log in with Apple
Policies
Medscape
About
For Advertisers
Privacy PolicyEditorial PolicyAdvertising PolicyYour Privacy ChoicesTerms of UseCookies
News & PerspectivesTools & ReferenceCME/CEVideoEventsSpecialtiesTopicsAccount InformationScribeNewsletters & Alerts
About MedscapeMedscape StaffMarket ResearchHelp CenterContact Us
Advertise with UsAdvertising Policy
Get the Medscape App
Download on the App Store
Get it on Google Play
All material on this website is protected by copyright, Copyright © 1994-2025 by WebMD LLC. This website also contains material copyrighted by 3rd parties.
Close |
190795 | https://www.khanacademy.org/math/ap-calculus-ab/ab-differential-equations-new/ab-7-6/v/identifying-separable-equations | Use of cookies
Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account.
For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session.
Functional Cookies
These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences.
Targeting Cookies
These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission.
We do not use cookies to serve third party ads on our Service.
Performance Cookies
These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates.
For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users.
We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
Cookie List
Consent Leg.Interest
label
label
label |
190796 | https://fa.ewi.tudelft.nl/~hart/37/publications/the_papers/published/what_is_finite.pdf | What is ‘Finite’?
K. P. Hart When you search the website of the Dutch Science Agenda ( for the keywords ‘oneindig’ and ‘oneindigheid’ (infinite and infinity) you will get more hits than for ‘eindig’ and ‘eindigheid’( finite and finiteness): 25 against 6.
Apparently people find the notion of the infinite more difficult, interesting, fascinating, …, than that of the finite.
Mathematically the difference need not be so large: ‘infinite’ is ‘not finite’ and ‘finite’ is ‘not infinite’, so if you know one you know the other. The question then is: “What is ‘finite’?” Once we have an answer to that question we also know what ‘infinite’ is: just put ‘not’ in front of that answer. Unfortunately (or maybe fortunately) it is not quite that simple; it will turn out that what we want from both notions makes it difficult, at first, to make them simply each others negations.
The dictionary It is always interesting to look in a dictionary for the ‘normal’ meaning of a mathematical term; often that will also give an idea why students find it hard to get to grips with a concept: that ‘normal’ meaning quite often has no bearing on what mathematics wants to say.
I use The Chambers Dictionary () when doing crosswords, so I looked there for definitions of ‘finite’ and ‘infinite’.
finite adj having an end or limit; subject to limitations or conditions, opp to infinite. [l. finitus, pap of finire to limit] infinite adj without end or limit; greater than any quantity that can be assigned [maths]; extending to infinity; vast; in vast numbers; inex-haustible; infinitated (logic) infinitate vt to make infinite; to turn into a negative term (logic).
So, the first definitions of ‘finite’ and ‘infinite’ are each other’s negations but then there is a slight divergence. ‘Finite’ is the opposite of ‘infinite’, but not vice versa; also, ‘infinite’ has more variety than finite. I included the verb ‘infinitate’ because I had not seen it before and because it has a nice logical constant as well.
There is one noteworthy bit among the definitions of ‘infinite’: the phrase “greater than any quantity that can be assigned” comes more or less straight out of Euclid’s Elements: Book IX, Proposition 20. The primes are more than any assigned multitude of prime numbers.
Nowadays we formulate this as: “There are infinitely many prime numbers.” Mathematics The problem with dictionary definitions is that they use words that have their own definition and quite often looking up those definitions will lead you to others and, …, after a while you find yourself going round in circles. The definition of ‘infinite’ that goes back to Euclid contains ‘greater’, ‘quantity’, and ‘assigned’. We need to give meaning to these when we want to make the definition unambiguous.
What we will do is define ‘finite’ in such a way that ‘infinite’, as its negation, will let Euclid’s formulation make sense.
Before we do that first an aside: in Analysis you sometimes read about ‘finite intervals’; those are simply intervals with real numbers as limits. For example [0, 1] and (2, 10220] are finite intervals, but not (0, →). This corresponds to the dictionary definition of ‘finite’ but we will not discuss this type of finiteness because it does not occur that often and because these intervals do have an infinite number of elements.
Finite sets So what is the definition of ‘finite set’?
A bit anticlimactic maybe, but a set X is finite if there are a natural number n and a bijection f : n →X.
Before anyone panics, “a bijection between a set and natural number?”: in Set Theory we define N in such a way that n = {i ∈N : i < n}. This is because we want to keep things a simple as possible and this is indeed the simplest possible way to define natural numbers set-theoretically, see . So, for example: 0 = ∅, 1 = {0}, 2 = {0, 1}, …, 7 = {0, 1, 2, 3, 4, 5, 6}, …; in general: n + 1 = n ∪{n}.
It is a good exercise to show that there is at most one such n.
Exercise. Prove, for every n, that there is no bijection between n and n + 1.
The consequence is now that we can define for finite sets what the number of elements is: the unique n for which the desired bijection exists.
Infinite sets Now we also know what an infinite set is: one for which no n as in the definition of ‘finite set’ can be found.
That is quite negative; in fact you are basically empty-handed when your set is infinite: you have no n and no bijection.
What you do have is lots of injections: if a set X is infinite then you can prove, by induction, that for every n ∈N there is an injection from n into X.
This may seem completely self-evident — “just pick some points” — but it takes a bit of work to turn those words into a proper proof, and we will see later why this might need some work.
For now we note that Euclid’s proof of his theorem, that you can look up on-line, see , shows indeed that the set P of prime numbers is infinite according to our definition: it shows that for every n no injection f : n →P is surjective.
Alfred Tarski In the Polish mathematician Alfred Tarski made a thorough investigation of the notion of a finite set. One thing he asked himself was whether you could definite finiteness without referring directly to natural numbers.
Figure 1: Alfred Tarski The answer was “yes, that is possible”.
Nowadays we call a set X Tarski-finite is the following holds: every nonempty family A of subsets of X has a maximal element. Note: ‘maximal’ means that there is no larger element (and that is not the same as being the largest element). For example in the family { {x} : x ∈X } every member is maximal, but there is no maximum (if X has more than one element).
Exercise. Prove, by induction, that every natural number is Tarski-finite (and hence that ever finite set is Tarski-finite).
Once you have done this exercise you should, of course, do the following one too.
Exercise. Prove that every Tarski-finite set is finite.
Here is a hint: let A be the family of all finite subsets of X. It is non-empty because ∅∈A. Next prove that a maximal element of A must be a maximum and indeed that it must be X itself.
This characterization of finiteness, which we might also call The Maximality Principle, is very useful in mathematics, especially in Combinatorics.
Richard Dedekind One of the first mathematicians who sought to define finite and infinite sets was Richard Dedekind. His most famous definition is from . It even made it into the dictionary: infinite set n (maths) a set that can be put into one-one correspondence with part of itself In terms of maps: a set X is Dedekind-infinite is there is an injective map f : X →X that is not surjective. This property has two nice equivalences, especially number 2 is something that we would like to be true about infinite sets.
Exercise. Prove that the following three statements are equivalent.
1. X is Dedekind-infinite 2. there is an injective map f : N →X 3. there is a bijection f : X →X ∪{p} for (some) p not in X Now, by definition, a set X is Dedekind-finite if every injective map f : X → X is surjective.
Exercise. Prove, by induction, that every natural number is Dedekind-finite (and hence that every finite set is Dedekind-finite).
Figure 2: Richard Dedekind We now have a problem; our notion of ‘finite’ is very practical: a finite set comes with an enumeration f : n →X and we can use that in our proofs.
Similarly a Dedekind-infinite set comes with an injective map f : N →X, also quite handy to have.
Unfortunately, ‘(in)finite’ is not the same as ‘Dedekind-(in)finite’.
It seems easy to prove that every infinite set is Dedekind-infinite: pick a point x0, the set is infinite so there is another point x1, there is another point x2, …, and so on. The map that send n to xn is injective.
Well, …, no; there is no way to turn the “…, and so on” into a formal set-theoretic proof that involves only finitely many steps. By ‘formal proof’ I mean a proof as described in the course Logic (TW3520) that you can take every year (recommended).
To give a proof in finitely many steps you need an extra assumption, the Axiom of Choice ().
Cyclic permutations In the foreword to the second edition of Dedekind mentioned another definition of ‘finite’: a set X is Dedekind-finite if there is a map f : X →X such that if A ⊆X is such that f[A] ⊆A then A = ∅or A = X.
It is not hard to show that natural numbers are Dedekind-finite: think of cyclic permutations.
On the other hand: if X is Dedekind-finite then X is in fact finite, and the map f is indeed a cyclic permutation.
Exercise. Prove this. Hint: take some x ∈X and prove that x = fn(x) for some n ≥1.
So, Dedekind had two definitions of ‘finite’ (and of ‘infinite’) and he tried quite hard to prove that they were equivalent but did not succeed. If you want to know why then you should study .
References The Chambers Dictionary, Revised 13th Edition, (2014). Chambers Harrap Publishers Ltd, Edin-burgh.
Richard Dedekind, Was sind und was sollen die Zahlen?, (1888). Vieweg & Sohn, Braunschweig.
D. E. Joyce, Euclid’s Elements.
Thomas J. Jech, The axiom of choice, Studies in Logic and the Foundations of Mathematics, Vol. 75. (1973) North-Holland Publishing Co., Amsterdam.
K. Kunen, Set theory. An introduction to independence proofs, Studies in Logic and the Foundations of Mathematics 102. (1980) North-Holland Publishing Co., Amsterdam.
Alfred Tarski, Sur les ensembles finis. Fundamenta Mathematicae 6.1 (1924): 45-95.
Aad Vijn, Het Keuzeaxiom: Een studie naar de geschiedenis van het keuzeaxioma, de gevolgen, equivalente uitspraken en de onbewijsbaarheid.
Bachelorscriptie, 2012. |
190797 | https://www.youtube.com/watch?v=fAM598n8lvI | The solution of differential equation dx/dt=x^2 with initial condition x(0)=1 will blow up as t is
Physics Vision - by Anuj Sir
1560 subscribers
24 likes
Description
3311 views
Posted: 23 May 2021
Myself Dr. Anuj Gupta (Multiple times Qualified NET/JRF, JEST, GATE, TIFR, CET PG, IIT-JAM etc.). I have teaching experience of several years in various reputed institutes. You will get the content for various competitive exams like UGC-NET, GATE, JEST, College Lecturer, CET (PG), IIT-JAM etc. by subscribing to this channel.
You can always join me and my team for the full course at our app
Telegram Link -
Also, visit our YouTube channel for Mathematics –
Kindly like the videos and subscribe our YouTube channels.
3 comments
Transcript:
ए वेलकम एडिशन चुनें इससे अधिक मीनिंग आफ ए क्वेश्चन इज मिंटू सॉन्ग डिफरेंशियल इक्वेशन इन कंडीशन इन विच विल बे ओक सो लेट्स ऑल द वे एक विचित्र जीवन है डिटेल्स बेदी डिफिकल्ट 2X परहेज़ इक्वेशन एंड विक इन थ्य सौल विद यूजिंग मेज़ इज इक्वल टू वन माइनस प्लस सीट इक्वल टू माइनस वन अपॉन पति प्लस सीट ओके बट लेटेस्ट कैलकुलेटर वैल्यूज पेस्ट ड्रेसेस फॉर नियति को 202 सचिव व सीईओ - - वर्ब - 151 - विल बी ऑप्शन सही आंसर है कि Bigg Boss ट्यून विद द एंड थैंक्स फॉर वाचिंग है |
190798 | https://flexbooks.ck12.org/cbook/ck-12-algebra-i-concepts-honors/section/5.5/primary/lesson/graphs-of-linear-inequalities-alg-i-hnrs/ | Skip to content
Math
Elementary Math
Grade 1
Grade 2
Grade 3
Grade 4
Grade 5
Interactive
Math 6
Math 7
Math 8
Algebra I
Geometry
Algebra II
Conventional
Math 6
Math 7
Math 8
Algebra I
Geometry
Algebra II
Probability & Statistics
Trigonometry
Math Analysis
Precalculus
Calculus
What's the difference?
Science
Grade K to 5
Earth Science
Life Science
Physical Science
Biology
Chemistry
Physics
Advanced Biology
FlexLets
Math FlexLets
Science FlexLets
English
Writing
Spelling
Social Studies
Economics
Geography
Government
History
World History
Philosophy
Sociology
More
Astronomy
Engineering
Health
Photography
Technology
College
College Algebra
College Precalculus
Linear Algebra
College Human Biology
The Universe
Adult Education
Basic Education
High School Diploma
High School Equivalency
Career Technical Ed
English as 2nd Language
Country
Bhutan
Brasil
Chile
Georgia
India
Translations
Spanish
Korean
Deutsch
Chinese
Greek
Polski
EXPLORE
Flexi
A FREE Digital Tutor for Every Student
FlexBooks 2.0
Customizable, digital textbooks in a new, interactive platform
FlexBooks
Customizable, digital textbooks
Schools
FlexBooks from schools and districts near you
Study Guides
Quick review with key information for each concept
Adaptive Practice
Building knowledge at each student’s skill level
Simulations
Interactive Physics & Chemistry Simulations
PLIX
Play. Learn. Interact. eXplore.
CCSS Math
Concepts and FlexBooks aligned to Common Core
NGSS
Concepts aligned to Next Generation Science Standards
Certified Educator
Stand out as an educator. Become CK-12 Certified.
Webinars
Live and archived sessions to learn about CK-12
Other Resources
CK-12 Resources
Concept Map
Testimonials
CK-12 Mission
Meet the Team
CK-12 Helpdesk
FlexLets
Know the essentials.
Pick a Subject
Donate
Sign Up
5.5
Graphs of Linear Inequalities
Written by:Brenda Meery | Kaitlyn Spong
Fact-checked by:The CK-12 Editorial Team
Last Modified: Aug 02, 2025
A linear equation is a line when graphed. What about a linear inequality? Can you represent the following linear inequality in the Cartesian plane?
Graphing Linear Inequalities
When a linear inequality is graphed on a grid, the solution will appear as a shaded area, as opposed to simply a straight line. The general form of a linear inequality with two variables is or , where ‘’ and ‘’ are the coefficients of the variables and are not both equal to zero. The constant is ‘’.
The inequality symbol is read “greater than” and the inequality symbol is read “less than”. The inequality symbol is read “greater than or equal to” and the inequality symbol is read “less than or equal to”. The inequality symbol determines whether the line is dashed or solid.
All inequalities that have the symbol or are graphed with a dashed line.
All inequalities that have the symbol or are graphed with a solid line.
Don't forget that when an inequality is divided or multiplied by a negative number, the direction of the inequality sign is reversed.
Let's practice by graphing the following inequalities:
The above graph represents the graph of the inequality before the solution set region is shaded. The line is a solid line because the inequality symbol is . To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded. The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality.
Yes, seven is less than 12. The point (1, 1) satisfies the inequality. Therefore, the solution set is all of the area below the line that contains the point (1, 1). The ordered pair that satisfies the inequality will lie within the shaded region.
The solution set for the inequality is the entire shaded region shown in the graph. The solid line means that all of the points on the line will satisfy the inequality. In general, you will shade below the line if the inequality is of the form or .
Note that the inequality was divided by negative 4 which caused the inequality sign to reverse its direction.
The above graph represents the graph of the inequality before the solution set region is shaded. The line is a dashed line because the inequality symbol is . To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded. The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality.
No, negative two is greater than negative twelve. The point (1, 1) does not satisfy the inequality. Therefore, the solution set is all of the area above the line that does not contain the point (1, 1). The ordered pair does not satisfy the inequality and will not lie within the shaded region.
The solution set for the inequality is the entire shaded region shown in the graph. The dashed or dotted line means that none of the points on the line will satisfy the inequality. In general, you will shade above the line if the inequality is of the form or .
Now, let's write the appropriate inequality for a given graph:
Begin by determining the slope of the line. The slope of the line is determined by counting 2 units to the right and 3 units upward. The slope of the line for this graph is
The -intercept for the line is (0, –2). The equation of the line in slope-intercept form is
The solution set is found in the shaded region that is above the line. The line is a solid line. Therefore the inequality symbol that must be inserted is greater than or equal to. The inequality that is modeled by the above graph is:
Examples
Example 1
Earlier, you were asked to graph the following linear inequality:
The first step is to rearrange the inequality in slope-intercept form. This process is the same as it is for linear equations.
Note that the inequality was divided by negative 3 which caused the inequality sign to reverse its direction.
The graph of the inequality is done the same as it is for a linear equation. In this case the graph will be a dashed or dotted line because the sign is greater than .
The above graph represents the graph of the inequality before the solution set region is shaded. To determine whether to shade above the line or below the line, choose a point that is not on the line and test its coordinates in the original inequality. If the coordinates of the point satisfy the inequality, then the area above or below the line, containing the point, will be shaded. If the coordinates of the point do not satisfy the inequality, then the area above or below the line, that does not contain the test point, will be shaded.
The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality.
Yes, negative one is less than six. The point (1, 1) satisfies the inequality. Therefore, the solution set is all of the area above the line that contains the point (1, 1). The ordered pair that satisfies the inequality will lie within the shaded region.
The solution set for the inequality is the entire shaded region shown in the graph. The dashed or dotted line means that none of the points on the line will satisfy the inequality.
Example 2
Without graphing, determine if each point is in the shaded region for each inequality.
(2, –3) and
(–3, 5) and
If the point satisfies the inequality, then the point will lie within the shaded region. Substitute the coordinates of the point into the inequality and evaluate the inequality. If the solution is true, then the point is in the shaded area.
Yes, negative six is less than negative five. The point (2, –3) satisfies the inequality. The ordered pair will lie within the shaded region.
No, nine is not greater than sixteen. The point (–3, 5) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.
Example 3
Graph the following inequality:
: Note that the inequality was divided by negative 3 which caused the inequality sign to reverse its direction.
: Is the graph of the inequality shaded correctly?
: The point (1, 1) is not on the graphed line. The point will be tested to determine if the coordinates satisfy the inequality.
: No, two is not greater than or equal to fifteen. The point (1, 1) does not satisfy the inequality. The ordered pair does not satisfy the inequality and will not lie within the shaded region.
Example 4
Determine the inequality that models the following graph:
Begin by determining the slope of the line. The slope of the line is determined by counting 2 units to the left and 4 units downward. The slope of the line for this graph is
: The -intercept for the line is (0, –3). The equation of the line in slope-intercept form is
: The solution set is found in the shaded region that is below the line. The line is a solid line. Therefore the inequality symbol that must be inserted is less than or equal to. The inequality that is modeled by the above graph is:
Review
Without graphing, determine if each point is in the shaded region for each inequality.
(2, 1) and
(–1, 3) and
(–5, –1) and
(6, 2) and
(5, –6) and
Determine the inequality that is modeled by each of the following graphs.
.
.
.
.
.
Graph the following linear inequalities on a Cartesian plane.
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
| Image | Reference | Attributions |
---
Student Sign Up
Are you a teacher?
Having issues? Click here
By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy
Already have an account?
Save this section to your Library in order to add a Practice or Quiz to it.
(Edit Title)29/ 100
This lesson has been added to your library.
|Searching in:
| |
|
Looks like this FlexBook 2.0 has changed since you visited it last time. We found the following sections in the book that match the one you are looking for:
Go to the Table of Contents
No Results Found
Your search did not match anything in . |
190799 | https://abhyasonline.in/contents/CLAT/Quantitative/Percentages/Successive%20Percentage%20Change/ | Self Learning
Guided Learning
Test Series
Job Alerts
Current Affair
M&VP
Help Video?
Help?
Active Courses
Courses
FAQs
FAQs
Trial
Trial
Login
ABOUT ABHYASABOUT ABHYAS About UsMission & VisionQ&AHow to Proceed? Have a LookBLOGFAQStatisticsPrivacy PolicyCancellation & Refund Policy Mrs. Anu Sethi, Post Graduate in Physics, conceptualised ABHYAS Academy with the aim to give conceptual knowledge to the students. We had the vision to give analytical testing platform to the students to understand concepts instead of Rote Learning. We provide bridge course for students of Class KG to 10th to crack future competitions and preparation for Banks / Govt. Exams. We believe in Concept based Learning to enable students to understand concepts instead of Rote Learning. Mental Ability Testing (MAT), Vedic Maths are also part of our curriculum to provide maximum support to student for NTSE. Learn more
WHAT WE PROVIDE?WHAT WE PROVIDE? STUDENT Self LearningGuided LearningTest SeriesJob AlertsCurrent Affair We provide Self Learning, Guided Learning and Test Series to our students enrolled online. Job Alerts and TEst Series are also part of our curriculum to help students to improve their conceptual knowledge for Govt. Exams. PARENTS Features Parents Features Modules Modules For Parents ABHYAS ONLINE keeps you updated about your child’s performance. You can check your ward performance from time to time and keep a track of his/her performance in academics. This increases transparency between parents and their wards. Learn More SCHOOL / INSTITUTE / TUTOR School School FeaturesSchool Modules we Provide Tutor Tutor FeaturesTutor Modules we Provide Institute Institute FeaturesInstitute Modules we Provide ABHYAS ONLINE helps in improving the day - to - day functionalities in the school. The portal generates real time student and teacher performance reports on a single click. Also, ABHYAS ONLINE helps emerging academies and institutes overcoming the problmes of manual student records like fees details, exam marks etc. Moreover, ABHYAS ONLINE clearly understands that role of a tutor involves teaching and organizing academic and financial records of each student. Thus we offer a complete solution to increase your efficiency in teaching by giving an interface to solve other managerial difficulties. Learn More MISSION & VISION PARTNER Mission & Vision Partner Enquiry Be a Mission & Visison Partner with us. Our partners will get all the back-end to Staff, Students, Parents & Management regarding Course scheduling, Time table, Class Notes, Class Response, Analytical Testing and many more. M&VP FeaturesM&VP ModulesM&VP EnquiryPending Enquiry
PROGRAMSPROGRAMS BOARDS CBSE Kinder Garden1st CBSE2nd CBSE3rd CBSE4th CBSE5th CBSE6th CBSE7th CBSE8th CBSE9th CBSE10th CBSE ICSE 6th ICSE7th ICSE8th ICSE9th ICSE10th ICSE DAV 6th DAV7th DAV8th DAV Australia Education Foundation program for Class 9th ABHYAS Classroom Programs are unique to ensure best possible result depending upon the basic potential of the Student and his / her ability to work hard. All Programs OLYMPIADS NTSE 5th Foundation NTSE6th Foundation NTSE7th Foundation NTSE8th Foundation NTSE9th Foundation NTSE10th NTSE IMO-Mathematics Olympiad 6th IMO7th IMO8th IMO9th IMO10th IMO ISO - Science Olympiads 6th ISO7th ISO8th ISO9th ISO10th ISO ABHYAS Programs for Olympiads are challenging as to improves aptitude and competitive spirit among students. These exams help students to cultivate analytical thinking which is useful in any examination. Olympiads not only improve the logical thinking and help in brainstorming but also enhance the analytical and reasoning ability, problem-solving skills, and confidence. Overall, it helps in the development of a child at a young stage.. All Programs ENTRANCE EXAMSNavodaya Entrance 6th & 9th Navodaya Vidyalaya Entrance 6thNavodaya Vidyalaya Entrance 9th Sainik School Entrance Sainik School Entrance 6thSainik School Entrance 9th OTHER ENTRANCE MBA Entrance Common Admission Test Law Entrance Common Law Admission Test ABHYAS Programs for Entrance Exams gives an edge to the Students to sharpen their skill for beter results. All Programs EMPLOYABILITY Central Government Service Program for SSC CGLGeneral Awareness For SSCGeneral Knowledge For SSCSSC CHSLSSC MTSSSC CHSL Test Series State Government Services HCS PrelimsHaryana GKSETCHCS Prelims Test Series Banking And Insurance Bank AO Prelims Test SeriesIBPS Clerk Prelims Test SeriesIBPS RRB Prelims Test SeriesIBPS SO Prelims Test SeriesSBI PO Prelims Test SeriesSBI Clerical Prelims Test SeriesSBI SO General Aptitude Test SeriesBank PO Prelims Test SeriesBank Clerical Prelims Test SeriesIBPS PO Prelims Test SeriesBank SO Prelims Test SeriesQuantitative Aptitude Test SeriesEXIM AO Prelims Test SeriesVerbal Ability Test SeriesBank Of Baroda AO Prelims Test SeriesLIC ADO Prelims Test SeriesLIC AAO Prelims Test SeriesSBI PO PrelimsBank PO PrelimsIBPS PO PrelimsVerbal Ability ProficiencyQuantitative Aptitude ProficiencyIBPS Clerk PrelimsIBPS RRB PrelimsIBPS SO PrelimsSBI SO General AptitudeBank Clerical PrelimsSBI Clerical PrelimsEXIM AO PrelimsBank AO PrelimsBank SO PrelimsBank Of Baroda AO PrelimsLIC ADO PrelimsLIC AAO Prelims Railways & Metro Services Foundation program for RailwaysRRB Technician Grade I SignalRRB Technician Grade IIIRPF ConstableRPF Sub InspectorRPF Constable Test Series Andhra Bank Clerk-cum-CashierAndhra Bank Clerk-cum-Cashier Test Series SKILL English Proficiency 6th English Proficiency7th English Proficiency8th English Proficiency9th English Proficiency10th English ProficiencyEnglish Proficiency For Employment Reasoning Proficiency 6th Reasoning Proficiency7th Reasoning Proficiency8th Reasoning Proficiency9th Reasoning Proficiency10th Reasoning ProficiencyReasoning Proficiency For EmploymentReasoning Proficiency Test Series Computer Proficiency Computer ProficiencyProficiency In C ProgramingProficiency in Computerised Accounting ABHYAS Programs for Govt. Exams bring renaissance for the candidates aspiring for Govt. jobs. We provide quality course materials and analytical testing to know Strong and Weak areas at an early stage. All Programs
ABHYAS SUCCESSABHYAS SUCCESS WISE PerformersSelf Learning PerformersOffline Exam PerformersQuestion of DAY PerformersSelf Study AchiversAlumniPast ResultStudent Monthly Performance Our understanding of the student needs at different stages of their preparation has been the guiding force behind the structure and methodology of each course at ABHYAS and also the secret of success of ABHYAS Students. Learn more
WHY ABHYAS ?WHY ABHYAS ? Our Programs / CoursesPersonalized CoachingStudy ResourcesOur MethodologyTestimonialsOur Testing Pattern ABHYAS ONLINE Is A Concept Based Learning And Testing Platform That Builds Strong Foundation For Future Excellence. We provide an innovative modularized curriculum developed after years of experience in teaching. This pedagogy inculcates the habit of understanding the concept rather than rote learning in the Child. Learn more
REPOSITORYREPOSITORY Whats NewOur ContentsStudy ResourcesSelf Learning VideosObjective/Subjective PapersQuestion BankSearch QuestionsSocial MediaReel / ShortsGallery Reository ABHYAS repository offers a range of program brochures, contents, objective & subjective papers, achievements & student performances as well as a range of study resources for School Programs, Olympiads, Entrance Exams, Govt. Exams, etc. All Downloads
CONTACT USCONTACT US Self Learning abhyasonline.info@gmail.com Guided Learning abhyasonline.info@gmail.com Test Series abhyasonline.info@gmail.com Job Alerts abhyasonline.info@gmail.com General Queries abhyasonline.info@gmail.com
Admission Helpline No.
+91-9416541198
Helpline No.
+91-9461225678 ABHYAS Academy
Near Govt. PG College,
(Behind Pragat Shiv Mandir),
Ambala Cantt.-133001
M.: +91-8168925411
e-mail : abhyasonline.info@gmail.com SEND A MESSAGE
LoginLOGIN M&VPAuthorStudentParentConsultant
BLOG
Loading…
Loading…
Close
Close
Close
Close
×
Close
×
Attention !!!!!
Close
×
Close
Loading…
Loading…
Like us!
Follow us!
Follow us!
Have a Look!
WhatsApp Message
Register/Login
Share Link
Home
Contents
CLAT
Quantitative
Percentages
Successive Percentage Change
Successive Percentage Change
Percentages - Concepts
Express as PercentagePercentage ChangeSuccessive Percentage ChangePercentage Increase or Decrease Using MultiplierPercetage Increase or DecreaseUnitary Method In Percentages
Class - CLAT Subjects
Concept Explanation
Successive Percentage Change
The concept of successive percentage change deals with two or more percentage changes applied to quantity consecutively. In this case, the final change is not the simple addition of the two percentage changes (as the base changes after the first change).
Suppose a number N undergoes a percentage change of x % and then y%, the net change is:
Here,z is the effective percentage change when a number is changed successively by two percentage changes.
Various cases for Percentage Change:
Both percentage changes are positive:
One percentage change is positive and the other is negative:
Both percentage changes are negative:
Percentage Change involving three changes:
If value of an object/number P is successively changed by x%, y% and then z%, then final value.
Illustration 1:A’s salary is increased by 10% and then decreased by 10%. The change in salary is
Solution:As firstly salary increases, thus use positive sign and then salary decreases, so use negative sign
Percentage change formula when x is positive and y is negative is given by
Here, x = 10, y = 10
As negative sign shows a decrease, hence the final salary is decreased by 1%.
Illustration 2:A number is first increased by 10% and then it is further increased by 20%. The original number is increased altogether by?
Solution :Percentage change formula when both x and y are positive =
Here, x = 10 and y = 20 Hence net percentage change
As positive sign shows increase, Hence, The original number is increased altogether by 32 %.
Sample Questions
(More Questions for each concept available in Login)
Question : 1
If value of an object P is successively decreased by 10%, 20% and then 30%, then find final value if P = 10?.
A.
5.04 B.
50.40 C.
0.504 D.
5.400
Right Option : A
View Explanation
Explanation
Here we have
x= -10%, y= -20%, z= -30%
Since P = 10, we have
= 5.04
Hence, the correct option is A.
Question : 2
Ifarticle's price was reduced by 30% and its sale increased by 70%. The net effect on the revenue is ________.
A.
12%B.
19%C.
15%D.
20%
Right Option : B
View Explanation
Explanation
We know that ,when X is positive and Y is negative i.e. (x = 70, y = -30) the formula is :
x = 70, y = -30
40 - 21= 19%
Hence the correct option is B.
Question : 3
Two successive price increase of 20% and 20% of an article are equivalent to a single price increase of __ .
A.
44%B.
40%C.
25%D.
50%
Right Option : A
View Explanation
Explanation
We know that,when X and Y are positive then we use
Given : x =20 ,y =20
40 + 4 % = 44%
so, the correct option is A.
Video Link - Have a look !!!
Language - English
Chapters
Problems On Boats And StreamProblems On TrainsPipe And CisternIntroduction to TrigonometryNumber TheoryNumber SystemRules For DivisibilitySquare and Square RootCube and Cube RootsSimplificationAlgebraAlgebraic IdentitiesSurds And IndicesHCF And LCMAverages FoundationPercentagesProfit Loss and DiscountSimple and Compound InterestRatio and ProportionPartnership and MixtureTime, Speed and DistanceQuadratic EquationsTime and WorkMensurationSurface Area and VolumePolynomialCoordinate GeometryProbabilityPermutation and CombinationSequences and SeriesSet TheoryData InterpretationData Sufficiency
Content / Category
CBSEICSENTSEDAVAustralia EducationIMO-Mathematics OlympiadCentral Government ServiceISO - Science OlympiadsState Government ServicesBanking And InsuranceRailways & Metro ServicesMBA EntranceLaw EntranceEnglish ProficiencyReasoning ProficiencyComputer ProficiencyNavodaya Entrance 6th & 9thSainik School EntranceAndhra Bank
Class / Course
CLAT
Related Videos
Express as Percentage
Language - English
Percentage Change
Language - English
Why ABHYAS?
Self Learning
Sign Up
Course Benefits
Our Areas of Expertise
CBSEICSENTSEDAVAustralia EducationIMO-Mathematics OlympiadCentral Government ServiceISO - Science OlympiadsState Government ServicesBanking And InsuranceRailways & Metro ServicesMBA EntranceLaw EntranceEnglish ProficiencyReasoning ProficiencyComputer ProficiencyNavodaya Entrance 6th & 9thSainik School EntranceAndhra Bank
Students / Parents Reviews
It was a good experience with Abhyas Academy. I even faced problems in starting but slowly and steadily overcomed. Especially reasoning classes helped me a lot.
It was a good experience with Abhyas Academy. I even faced problems in starting but slowly and steadily overcomed. Especially reasoning classes helped me a lot.
Cheshta
10th
Third consective year,my ward is in Abhyas with nice experience of admin and transport support.Educational standard of the institute recumbent at satisfactory level. One thing would live to bring i...
Third consective year,my ward is in Abhyas with nice experience of admin and transport support.Educational standard of the institute recumbent at satisfactory level. One thing would live to bring in notice that last year study books was distributed after half of the session was over,though study ...
Ayan Ghosh
8th
My experience with Abhyas is very good. I have learnt many things here like vedic maths and reasoning also. Teachers here first take our doubts and then there are assignments to verify our weak poi...
My experience with Abhyas is very good. I have learnt many things here like vedic maths and reasoning also. Teachers here first take our doubts and then there are assignments to verify our weak points.
Shivam Rana
7th
Usually we see institutes offering objective based learning which usually causes a lag behind in subjective examinations which is the pattern followed by schools. I think it is really a work of pla...
Usually we see institutes offering objective based learning which usually causes a lag behind in subjective examinations which is the pattern followed by schools. I think it is really a work of planning to make us students grab the advantages of modes of examination, Objective Subjective and Onli...
Anika Saxena
8th
It was good as the experience because as we had come here we had been improved in a such envirnment created here.Extra is taught which is beneficial for future.
It was good as the experience because as we had come here we had been improved in a such envirnment created here.Extra is taught which is beneficial for future.
Eshan Arora
8th
Abhyas is an institute of high repute. Yogansh has taken admission last year. It creates abilities in child to prepare for competitive exams. Students are motivated by living prizes on basis of per...
Abhyas is an institute of high repute. Yogansh has taken admission last year. It creates abilities in child to prepare for competitive exams. Students are motivated by living prizes on basis of performance in Abhyas exams. He is satisfied with institute.
Yogansh Nyasi
7th
Abhyas institute is one of the best coaching institute in the vicinity of Ambala cantt.The institute provides good and quality education to the students.The teachers are well experienced and are v...
Abhyas institute is one of the best coaching institute in the vicinity of Ambala cantt.The institute provides good and quality education to the students.The teachers are well experienced and are very helpful in solving the problems. The major advantages of the institute is extra classes for weak...
Shreya Shrivastava
8th
A marvelous experience with Abhyas. I am glad to share that my ward has achieved more than enough at the Ambala ABHYAS centre. Years have passed on and more and more he has gained. May the centre f...
A marvelous experience with Abhyas. I am glad to share that my ward has achieved more than enough at the Ambala ABHYAS centre. Years have passed on and more and more he has gained. May the centre flourish and develop day by day by the grace of God.
Archit Segal
7th
One of the best institutes to develope a child interest in studies.Provides SST and English knowledge also unlike other institutes. Teachers are co operative and friendly online tests andPPT develo...
One of the best institutes to develope a child interest in studies.Provides SST and English knowledge also unlike other institutes. Teachers are co operative and friendly online tests andPPT develope practical knowledge also.
Aman Kumar Shrivastava
10th
Abhyas Methodology is very good. It is based on according to student and each child manages accordingly to its properly. Methodology has improved the abilities of students to shine them in future.
Abhyas Methodology is very good. It is based on according to student and each child manages accordingly to its properly. Methodology has improved the abilities of students to shine them in future.
Manish Kumar
10th
It has a great methodology. Students here can get analysis to their test quickly.We can learn easily through PPTs and the testing methods are good. We know that where we have to practice
It has a great methodology. Students here can get analysis to their test quickly.We can learn easily through PPTs and the testing methods are good. We know that where we have to practice
Barkha Arora
10th
The experience was nice. I studied here for three years and saw a tremendous change in myself. I started liking subjects like English and SST which earlier I ran from. Extra knowledge gave me confi...
The experience was nice. I studied here for three years and saw a tremendous change in myself. I started liking subjects like English and SST which earlier I ran from. Extra knowledge gave me confidence to overcome competitive exams. One of the best institutes for secondary education.
Aman Kumar Shrivastava
10th
We started with lot of hope that Abhyas will help in better understnding of complex topics of highers classes. we are not disappointed with the progress our child has made after attending Abhyas. T...
We started with lot of hope that Abhyas will help in better understnding of complex topics of highers classes. we are not disappointed with the progress our child has made after attending Abhyas. Though need to mention that we expected a lot more. On a scale of 1-10, we would give may be 7.
Manya
8th
Abhyas institute is one of the best coaching institute in the vicinity of Ambala Cantt area. The teachers of the institute are well experienced and very helpful in solving the problems of the stude...
Abhyas institute is one of the best coaching institute in the vicinity of Ambala Cantt area. The teachers of the institute are well experienced and very helpful in solving the problems of the students.The good thing of the institute is that it is providing extra classes for the students who are w...
Aman Kumar Shrivastava
10th
My experience was very good with Abhyas academy. I am studying here from 6th class and I am satisfied by its results in my life. I improved a lot here ahead of school syllabus.
My experience was very good with Abhyas academy. I am studying here from 6th class and I am satisfied by its results in my life. I improved a lot here ahead of school syllabus.
Ayan Ghosh
8th
My experience with Abhyas academy is very nice or it can be said wonderful. I have been studying here from seven class. I have been completing my journey of three years. I am tinking that I should ...
My experience with Abhyas academy is very nice or it can be said wonderful. I have been studying here from seven class. I have been completing my journey of three years. I am tinking that I should join Abhyas Academy in tenth class as I am seeing much improvement in Maths and English
Hridey Preet
9th
Being a parent, I saw my daughter improvement in her studies by seeing a good result in all day to day compititive exam TMO, NSO, IEO etc and as well as studies. I have got a fruitful result from m...
Being a parent, I saw my daughter improvement in her studies by seeing a good result in all day to day compititive exam TMO, NSO, IEO etc and as well as studies. I have got a fruitful result from my daughter.
Prisha Gupta
8th
About Abhyas metholodology the teachers are very nice and hardworking toward students.The Centre Head Mrs Anu Sethi is also a brilliant teacher.Abhyas has taught me how to overcome problems and has...
About Abhyas metholodology the teachers are very nice and hardworking toward students.The Centre Head Mrs Anu Sethi is also a brilliant teacher.Abhyas has taught me how to overcome problems and has always taken my doubts and suppoeted me.
Shreya Shrivastava
8th
When I have not joined Abhyas Academy, my skills of solving maths problems were not clear. But, after joining it, my skills have been developed and my concepts of science and SST are very well. I a...
When I have not joined Abhyas Academy, my skills of solving maths problems were not clear. But, after joining it, my skills have been developed and my concepts of science and SST are very well. I also came to know about other subjects such as vedic maths and reasoning.
Sharandeep Singh
7th
Abhyas academy is great place to learn. I have learnt a lot here they have finished my fear of not answering.It has created a habit of self studying in me.The teachers here are very supportive and ...
Abhyas academy is great place to learn. I have learnt a lot here they have finished my fear of not answering.It has created a habit of self studying in me.The teachers here are very supportive and helpful. Earlier my maths and science was good but now it has been much better than before.
Barkha Arora
10th
View All
Our Contact
ABHYAS Academy,
Mrs. Anu Sethi,
Near Govt. PG College,
Ambala Cantt.-133 001 (Haryana) India.
VIEW LOCATION MAP
M - +91-9416541198, +91-9416225678
abhyasonline.info@gmail.com
Quick Links
Login
Contact us
Whats NEW
Blog
Why ABHYAS?
What ABHYAS provide?
Parents / Students - Q&A
Testimonials
Social Media Posts
Sample Papers
FAQs
Question Bank
Quick Links
Past Results
ABHYAS Alumni
Gallery Repository
Current Performers - Online
Current Performers - Offline
Current Performers - Question of Day
Self Study Achievers
Students Monthly Performance
ABHYAS Statistics
Our Contents
Mock Tests (Free)
Self Learning Videos
Free TRIAL
You Can avail Free Trial/Subscription to have glimpse of our Methodology. To avail, click here.
Be ABHYAS M&VP
(Mission & Vision Partner)
Any School / Institution / Academy / Tution Centre can be a ABHYAS M&VP.
M&VP Enquiry
Disclaimer | Privacy Policy / Terms of Service | Cancellation & Refund Policy
© Copyright 2020 ABHYAS Academy | All Rights Reserved. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.