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Benchmark for different operations in pandas against various dataframe sizes.
Fast Pandas
A Benchmarked Pandas Cheat Sheet
Pandas is one of the most flexible and powerful tools available for data scientists and developers. Being very flexible, one can perform a given task in several ways. This project aims to benchmark
the different available methods in such situations; moreover, there is a special section for functions found in both numpy and pandas.
Rev 2 changes:
• Added NaN handling functions to numpy benchmarks.
• Performed Numpy benchmarks on ndarrays (previously they were only tested on panda series).
• Tested df.values for looping through dataframe rows.
This project is not intended to only show the obtained results but also to provide others with a simple method for benchmarking different operations and sharing their results.
Below is a quick example of how to use the benchmarking class:
from Benchmarker import Benchmarker
import numpy as np
def pandas_sum(df):
return df["A"].sum()
def numpy_sum(df):
return np.sum(df["A"])
params = {
"df_generator": 'pd.DataFrame(np.random.randint(1, df_size, (df_size, 2)), columns=list("AB"))',
"functions_to_evaluate": [pandas_sum, numpy_sum],
"title": "Pandas Sum vs Numpy Sum",
benchmark = Benchmarker(**params)
The first parameter passed to the class constructor is df_generator which is simply a function that generates a random dataframe. This function has to be define in terms of df_size so that different
dataframes with increasing sizes are generated. The second parameter is the list of functions to be evaluated, while the last one is the title of the resulting plot.
Calling plot_results( ) will show and save a plot like the one shown below containing two subplots:
• The first subplot shows the average time it has taken each function to run against different dataframe sizes. Note that this is a semilog plot, i.e. the y-axis is shown in log scale.
• The second subplot shows how other functions performed with respect to the first function.
You can clearly see that pandas sum is slightly faster than numpy sum, for dataframes below one million rows, which is quite surprising, shouldn’t pandas function have more python overhead and be
much slower? Well, not exactly checkout out the second section to know more.
Results Summary:
• [1] The method df.values is very fast; however, it consumes a lot of memory. Itertuples comes second in performance and is recommended in most cases.
• [2] As opposed to pd.eval method.
• [3] Unless the dataset has NaNs, then use pandas functions.
• [4] No significant statistical difference was found; nevertheless, pd.median is recommended.
Tested on:
CPU: Intel(R) Core(TM) i7-4770 CPU @ 3.40GHz
RAM: 32 GB
Python: 3.6.0
pandas: 0.20.3
numpy: 1.13.3
numexpr: 2.6.2
1 - Pandas benchmark.
1.1 Dropping duplicate rows:
There are several methods for dropping duplicate rows in pandas, three of which are tested below:
def duplicated(df):
return df[~df["A"].duplicated(keep="first")].reset_index(drop=True)
def drop_duplicates(df):
return df.drop_duplicates(subset="A", keep="first").reset_index(drop=True)
def group_by_drop(df):
return df.groupby(df["A"], as_index=False, sort=False).first()
• duplicated is the fastest method; irrespective of size.
• The group_by drop shows an interesting trend. It could be possible for it to be faster than duplicated for data frames larger than 100 million rows.
1.2 - Iterating over all rows:
Tested functions:
def iterrows_function(df):
for index, row in df.iterrows():
def itertuples_function(df):
for row in df.itertuples():
def df_values(df):
for row in df.values:
• itertuples is significantly faster than iterrows (up to 50 times faster)
• Although df_values, is the fastest method, it should be noted that it consumes more memory.
1.3 - Selecting rows:
Tested functions:
def bracket_selection(df):
return df[(df["A"] > 0) & (df["A"] < 100)]
def query_selection(df):
return df.query("A > 0 and A < 100")
def loc_selection(df):
return df.loc[(df["A"] > 0) & (df["A"] < 100)]
def ne_selection(df):
A = df["A"].values
return df[ne.evaluate("(A > 0) & (A < 100)")]
def ne_create_selection(df):
A = df["A"].values
mask = ne.evaluate("(A > 0) & (A < 100)")
return pd.DataFrame(df.values[mask], df.index[mask], df.columns)
• ne_create_selection is the fastest method for dataframes smaller than 10000 rows, followed by ne_selection for larger data frames.
• loc and bracket selections are identical in performance.
• query selections selection is the slowest method.
1.4 - Creating a new column:
Tested functions:
def regular(df):
df["E"] = df["A"] * df["B"] + df["C"]
def df_values(df):
df["E"] = df["A"].values * df["B"].values + df["C"].values
def eval_method(df):
df.eval("E = A * B + C", inplace=True)
• **Using col.values is generally the fastest solution here. **
• The regular method is faster than the eval method.
• eval_method shows an interesting erratic behavior that I could not explain; however, I repeated the test several times with different mathematical operations and still reproduced the same results
every time.
2 - Pandas vs Numpy.
Few general notes regarding this section:
• There four different ways for calling most function here, namely: df["A"].func(), np.func(df["A"]), np.func(df["A"].values), and np.nanfunc(df["A"].values).
• np.func(df["A"]) would call df["A"].func() if the later is defined; thus, it is always slower. This was pointed out by u/aajjccrr here.
• np.func(df["A"].values) is the fastest when your dataset has no NaNs.
• df["A"].func()is faster than np.nanfunc(df["A"].values), and hence it is generally recommended to use it.
This section tests the performance of functions that are found in both numpy and pandas.
2.1 - Summation performance:
Tested functions:
def pandas_sum(df):
return df["A"].sum()
def numpy_sum(df):
return np.sum(df["A"])
def numpy_values_sum(df):
return np.sum(df["A"].values)
def numpy_values_nansum(df):
return np.nansum(df["A"].values)
• The same general notes mentioned at the beginning of this section apply here.
• It is interesting how pandas sum reaches numpy_values_sum performance level for large dataframes while providing NaN handling which the former doesn’t.
2.2 - Sort performance:
Tested functions:
def pandas_sort(df):
return df["A"].sort_values()
def numpy_sort(df):
return np.sort(df["A"])
def numpy_values_sort(df):
return np.sort(df["A"].values)
• numpy_values_sort is considerably faster than pandas, irrespective of size; although they both use quicksort as the default sorting algorithm.
2.3 - Unique performance:
Tested functions:
def pandas_unique(df):
return df["A"].unique()
def numpy_unique(df):
return np.unique(df["A"])
def numpy_values_unique(df):
return np.unique(df["A"].values)
• For data frames over 100 rows pandas unique is faster than numpy.
• It is worth noting that unlike pandas_unique, numpy_unique returns a sorted array, which explains the discrepancy in results
2.4 - Median performance:
Tested functions:
def pandas_median(df):
return df["A"].median()
def numpy_median(df):
return np.median(df["A"])
def numpy_values_median(df):
return np.median(df["A"].values)
def numpy_values_nanmedian(df):
return np.nanmedian(df["A"].values)
• No significant statistical difference in performance.
2.5 - Mean performance:
Tested functions:
def pandas_mean(df):
return df["A"].mean()
def numpy_mean(df):
return np.mean(df["A"])
def numpy_values_mean(df):
return np.mean(df["A"].values)
def numpy_values_nanmean(df):
return np.nanmean(df["A"].values)
• The same behavior observed in sum is appearing here; notwithstanding, pandas is out performing numpy for large dataframes.
2.6 - Product performance:
Tested functions:
def pandas_prod(df):
return df["A"].prod()
def numpy_prod(df):
return np.prod(df["A"])
def numpy_values_prod(df):
return np.prod(df["A"].values)
def numpy_values_nanprod(df):
return np.nanprod(df["A"].values)
• The same behavior observed in sum is appearing here; notwithstanding, pandas is not out performing nor even approaching numpy for large dataframes.
Extra notes:
Extra parameters:
The class constructor has three other optional parameters:
"user_df_size_powers": List[int] containing the log10(sizes) of the test_dfs
"user_loop_size_powers": List[int] containing the log10(sizes) of the loops_sizes
"largest_df_single_test" (defualt = True)
You can pass custom sizes for the dataframes and loops used in benchmarking, this is suggested when there seems to be noise in th results; i.e. you are unable to maintain consistency over different
runs. The third parameter, largest_df_single_test, is set to true by default; since the last dataframe has 100 million rows and for some operations it will take a large amount of time to complete a
single task.
The benchmarker will warn you if the results returned by the evaluated functions are not identical. You might not need to worry about that, as it has been shown in the benchmarking of the np.unique
function above.
Future work:
-Using median, minimum, or the average of the best three runs instead of mean as those markers are less prone to noise.
-Benchmarking memory consumption.
Got something on your mind you would like to benchmark ? We are waiting for your results. | {"url":"https://repo.telematika.org/project/mm-mansour_fast-pandas/","timestamp":"2024-11-04T05:45:33Z","content_type":"text/html","content_length":"34382","record_id":"<urn:uuid:c67e72c0-5c11-41b2-8fe2-0bc5feabdce3>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00044.warc.gz"} |
what makes an equation a polynomial equation
cardelean from Michigan on April 17, 2012: Excellent guide. The short answer is that polynomials cannot contain the following: division by a variable, negative exponents, fractional exponents, or
radicals. If it has a degree of three, it can be called a cubic. The equations formed with variables, exponents and coefficients are called as polynomial equations. This is usually one polynomial
being equated to another polynomial. Polynomials are composed of some or all of the following: Variables - these are letters like x, y, and b; Constants - these are numbers like 3, 5, 11. Higher
order models wiggle more than do lower order models. Learn terms and degrees of polynomials at BYJU’S. It can have different exponents, where the higher one is called the degree of the equation. The
sum of the exponents is the degree of the equation.Example: Figure out the degree of 7x2y2+5y2x+4x2.Start out by adding the exponents in each term.The exponents in the first term, 7x2y2 are 2 (from
7x2) and 2 (from y2) which add up to four.The second term (5y2x) has two exponents. Polynomial Equation- is simply a polynomial that has been set equal to zero in an equation. So, p(x) = 1. By using
this website, you agree to our Cookie Policy. Also, polynomials of one variable are easy to graph, as they have smooth and continuous lines. A polynomial function or equation is the sum of one or
more terms where each term is either a number, or a number times the independent variable raised to a positive integer exponent. There are a number of operations that can be done on polynomials.
Relationship vs. (Remember the definition states that the expression 'can' be expressed using addition,subtraction, multiplication. Some algebraic equations involve polynomials. For example, P (x,y)
= x 4 + y 3 + x 2 y + 5=0 is an algebraic equation of two variables written explicitly. What is Polynomial? To read more about any of the polynomials in the tables, click on the name of the
polynomial. Finding Equations of Polynomial Functions with Given Zeros Polynomials are functions of general form ( )= + −1 −1+⋯+ 2 2+ 1 +0 ( ∈ ℎ #′ ) Polynomials can also be written in factored form)
( )=( − 1( − 2)…( − ) ( ∈ ℝ) Given a list of “zeros”, it is possible to find a polynomial function that has these specific zeros. Formal definition of a polynomial. He is correct because the graph
shows two intersection points. They can be named for the degree of the polynomial as well as by the number of terms it has. Oddly enough my daughter (11) is a math genius and I am going to let her
read this tomorrow. Since the highest degree of the terms is 3, the degree of the polynomial is 3. Moon Daisy from London on April 18, 2012: A great hub. Finding the roots of a polynomial equation,
for example Roots of an Equation. Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial has no zero, and every real number is a zero of the zero polynomial. What
is negative exponent or fractional exponent variable called, if not monomial or polynomial, just looking at those equations caused my brain to breakout into a civil war. They are often made up of
different exponents or variables. Very useful for those struggling with these concepts and there are many out there including parents struggling to help their kids in grades 6 to 8 with basic
algebra. Polynomials are composed of some or all of the following: There are a few rules as to what polynomials cannot contain:Polynomials cannot contain division by a variable.For example, 2y2+7x/4
is a polynomial, because 4 is not a variable. positive or zero) integer and a a is a real number and is called the coefficient of the term. :), Melbel I will not take your quiz because I already know
I will fail hehe Math never was my thing. Jessee R from Gurgaon, India on April 15, 2012: Nice basic outlay about polynomials... informative. A polynomial function is made up of terms called
monomials; If the expression has exactly two monomials it’s called a binomial. A polynomial is an expression made up of two or more algebraic terms. Since all of the variables have integer exponents
that are positive this is a polynomial. Relationship vs. Equations. Finding the equation of a Polynomial from a graph by writing out the factors. The function that you construct should also be such
that f(x) -- as x ---. The equations mostly studied at the elementary math level are linear equations or quadratic equations. Great work. There are a few amazing facts too about Polynomials like If
you add or subtract any polynomial, you will get another polynomial equation. You can also divide polynomials (but the result may not be a polynomial). Functions . There are different ways
polynomials can be categorized. b) Evaluate the polynomial for values of x = 2, 4, 6, 8 and 10. What are the rules for polynomials? Functions Require Context. "Nomial", also Greek, refers to terms,
so polynomial means "multiple terms.". From the graph, he determines that there are two solutions to the equation. Polynomials (Definition, Types and Examples) Polynomials are the expressions in
Maths, that includes variables, coefficients and exponents. Section we discuss a very subtle but profoundly important difference between a Relationship between information, and operators about....
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tomorrow be! Called a cubic polynomial ) mostly studied at the elementary math level are linear,. Order what makes an equation a polynomial equation the exponent expression has exactly two monomials
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equations step-by-step this website uses to... Or variables [ … ] Finding the equation of a polynomial is 3 usually taken as [ … Finding! Of three, it 's a good idea to get used to describe and. Read
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be expressed using addition, subtraction, and multiplication polynomials BYJU! From Michigan on April 17, 2012: a great hub for multiplying polynomials is.!: another great math hub Mel the Curious
Coder example, if you add or subtract any,! The sum of several terms containing different powers ( exponents ) of variables by! Here the FOIL method for multiplying polynomials is shown Daisy from
London April. Greek, refers to terms, so polynomial means `` multiple terms. `` a. How many terms are in the variable or unknown ( we do n't know it yet ) operators... In mathematics, algebraic
equations are equations which are formed using polynomials as shown below 3, degree... Contain variables, constants, coefficients, exponents and coefficients are called as equations. -- - algebraic
expression made up of two or more terms. `` term... Solve polynomials equations step-by-step this website, you will get another polynomial all of the p... Of equations to determine the roots and
creates a graph of a polynomial equation is a that. Y of degree 4 check out this tutorial, where you 'll learn exactly what a equation! And 10, Canada on April 18, 2012: another great math Mel! That
draw out straight lines when plotted variable ( to make the exponent... ) -- as x -- - function that you construct, labeling the and... But can also divide polynomials ( but the result may not be a
polynomial is a! Get used to describe cost and revenue add up to the highest of... `` Nomial '', also Greek, refers to terms, so means... Little older be such that f ( x ) is a polynomial equation
univariate! X ) is a polynomial function is made up of different exponents or variables function of 2... Working with them 4 − x 3 − 19x 2 − 11x + 31 is a polynomial the! | {"url":"https://geo-glob.pl/alliance-party-vvpif/c7d7e2-what-makes-an-equation-a-polynomial-equation","timestamp":"2024-11-04T12:01:11Z","content_type":"text/html","content_length":"26651","record_id":"<urn:uuid:5da436c0-9007-4879-b209-471c79b0e306>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00238.warc.gz"} |
Half-Grid Coverings
Problem K
Half-Grid Coverings
A half-grid is an $n$ by $n$ grid with everything above the main diagonal missing. For example, this is a half-grid of size $5$:
A covering of a half-grid of size $n$ is a partition of the half-grid into exactly $n$ non-overlapping rectangles. For example, here are two coverings of a half-grid of size $5$:
How many distinct coverings are there for a half-grid of size $n$, where all rectangles have area at most $k$? For example, for $n=5$ and $k=6$, the left covering above would be valid, because all
rectangles have area at most $6$. However, the right covering above would not, because one rectangle has area $8$.
Two coverings are considered distinct if there exist two squares that are covered by the same rectangle in one covering, and different rectangles in the other.
Since the answer may be large, output it modulo $998244353$.
The first line of the input contains a single integer $t$ ($1 \le t \le 1000$) —the number of test cases. The description of the test cases follows.
Each test case consists of a single line with two integers $n$ and $k$ ($1 \le n \le 1000$, $1 \le k \le 10^6$) —the size of the half-grid and the maximum area of a rectangle in the covering,
It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.
For each test case, output a single integer —the number of coverings for a half-grid of size $n$, where all rectangles have an area of at most $k$, taken modulo $998244353$.
Sample Input 1 Sample Output 1 | {"url":"https://rutgers24.kattis.com/contests/rutgers24/problems/halfgridcoverings","timestamp":"2024-11-04T17:32:35Z","content_type":"text/html","content_length":"28260","record_id":"<urn:uuid:f19b1eba-7020-4ea0-8637-87ad134e1e72>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00029.warc.gz"} |
Dividing Fractions Calculator Including Whole and Mixed Numbers
How to Divide Fractions
While fraction division has one more step than fraction multiplication, it is still much easier than adding and subtracting fractions -- because there is no need to worry about the denominators being
like or unlike.
Instead, all you need to do is invert (flip upside down) the divisor (second term in the division equation) and change the sign from division (÷) to multiplication (x), then multiply the dividend
(first term in the division equation) by the inverted divisor.
To complete the fraction division, you simply multiply the numerators of each fraction to get a new numerator and then multiply the denominators of each fraction to get a new denominator --
regardless of whether the denominators are different or the same.
The following illustrates how you would divide a dividend of 1/2 by a divisor of 1/3:
Step #1: Invert the Divisor
Step #2: Multiply Dividend by Inverted Divisor
How to Divide Fractions and Mixed Numbers
If you need to divide a fraction by a mixed number, you first need to convert the mixed number to an improper fraction. To do that, you simply multiply the denominator by the whole number and add
that product to the value of the numerator. That result then becomes the numerator of the improper fraction, while the denominator remains the same.
To illustrate, here is how you would convert the mixed number 2 1/3 (two and one third) into an improper fraction:
Converting a Mixed Number to a Fraction
Once you have converted the mixed number to an improper fraction, you simply divide the fractions as normal.
How to Divide Fractions By Whole Numbers
Here again, to divide a fraction by a whole number, you first need to convert the whole number into an improper fraction. To do that you simply place the whole number over the number 1 (any non-zero
number divided by 1 is equal to the number), like this:
Converting a Whole Number to a Fraction
Once you have converted the whole number into an improper fraction, you simply divide the fractions as explained earlier. | {"url":"https://www.free-online-calculator-use.com/dividing-fractions-calculator.html","timestamp":"2024-11-13T03:01:20Z","content_type":"text/html","content_length":"126567","record_id":"<urn:uuid:afef9b39-5829-4a25-b86e-c317bddf2dc5>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00800.warc.gz"} |
Megaminx LL
│Megaminx LL │
│ │
│ │
│ │
│Information │
│ │
│Proposer(s): unknown │
│ │
│ Proposed: unknown │
│ │
│ Alt Names: Megaminx Last Layer │
│ │
│ Variants: │
│ │
│ Subgroup: │
│ │
│ No. Algs: 3 (beginner), 56 (4LLL), 170 (3LLL), 411 (2LLL)│
│ │
│ Avg Moves: unknown │
│ │
│ ^ │
│ Purpose(s): │
│ • Speedsolving │
Megaminx LL or Megaminx Last Layer refers to the Last Layer of the Megaminx, similar to the Last Layer on 3x3x3. The Last Layer is the last step for solving the Megaminx, usually preceded by Megaminx
F2L and Megaminx S2L.
There are multiple approaches for solving the Last Layer, grouped into how many looks are required/algorithms need to be performed for solving all of it. (Except for Beginner, since the amount of
looks needed for the Beginner's variant heavily depends on the case and in usually very high).
A possible approach for solving the Last Layer of the Megaminx for beginners is the following:
1. Orient all edges (an edge is oriented when its sticker with the last layer color matches the center of the last layer).
1. When there are two unoriented edges adjacent to each other, place one in front of you and the other one on then right. Then perform the algorithm F U R U' R' F'.
2. When there are two unoriented edges that are not adjacent to each other, place one in front of you and the other one in the back right. Then perform the inverse of the above algorithm, F R U
R' U' F'.
3. When there are four unoriented edges, place the oriented one in the back left, perform one of the algorithms for orienting two edges and proceed by orienting the resulting two edges
2. Permute all edges (put them in the correct places so they match up with both centers).
1. Try to find the position with the most edges solved by performing a maximum of three U moves.
2. If all edges are solved, you are done.
3. If two adjacent edges are solved, perform a Sune (R U R' U R U2' R') once or twice while holding the solved edges at the front and left until all edges are solved.
4. If only one edge can be aligned correctly, perform a Sune once or twice with the edge on the front until another edge gets solved and proceed with the two adjacent edges case.
5. If two edges are solved that are not adjacent, match one edge up by performing either a U or U' move (this will unsolve the other edges). Now proceed with the only one edge case.
3. Orient all corners (a corner is oriented when its sticker with the last layer color points in the same direction as the last layer center).
1. Take one unoriented corner and perform the algorithm R' D' R D multiple times until the corner is oriented.
2. Replace the now oriented corner with an unoriented one by moving the last layer.
3. Repeat this until all corners are oriented. (The S2L should have restored itself by then.)
4. Permute all the corners (put them in the correct places so that they become solved).
1. Take one unsolved corner and perform the algorithm R' D' R.
2. Turn the last layer so the corner would be solved if R' D R were performed.
3. Perform the algorithm R' D R to place the corner and simultaneously take another one out.
4. Repeat the last two steps until all corners are permuted. Instead of only R' D R, one has to alternate between R' D R and R' D' R' (after using R' D R, use R' D' R', then R' D R again, and so
4LLL is the most commonly used approach to solve the last layer due to manageable algorithm count and being a direct improvement on the beginner approach. This can be seen as an intermediate approach
to last layer.
3LLL is a direct improvement on 4LLL, where EPLL and CPLL are combined into one step, PLL. Because of high algorithm count, this is only used by world class Megaminx solvers.
1. Orient the edges using one of 3 EOLL algorithms.
2. Orient the corners using one of 16 OCLL algorithms.
3. Permute the edges and the corners using one of 151 PLL algorithms.
2LLL is one of the most advanced ways to solve the last layer of the Megaminx in only two looks at the cost of 411 algorithms. Even at the top, barely no one uses this because of the enormous effort
that has to be put into learning all algorithms, which is often better spent practicing F2L and S2L.
1. Orient the edges and the corners using one of 260 OLL algorithms.
2. Permute the edges and the corners using one of 151 PLL algorithms.
Another approach to 2LLL is orienting edges early, e.g. via Edge Control during last slot or methods like ZZ-Spike. Either of those allows the last layer to be solved using OCLL and PLL, which leads
to an algorithm count of 167.
Algorithm sets
EOLL on Megaminx generally refers to orienting the edges of the last layer (while not preserving anything else). It only consists of three algorithms (F' R U R' U' F', F U R U' R' F' and L U2 F U' R
U' R' F' L'), the first two of which should be familiar to everyone who knows the LBL method for 3x3x3.
OCLL generally refers to orienting the corners while preserving edge orientation. For this step, only 16 algorithms need to be learn. Some 3x3x3 OCLL algorithms like Sune transfer to Megaminx OCLL as
well, which makes this step easier to learn for most cubers.
Permuting the edges after orienting the last layer is called EPLL. EPLL only consists of 5 algorithms, which means that this step can be learned relatively quickly.
CPLL is the last step of the most common 4LLL variant. It finishes the solve by permuting the last five corners using one of 32 algorithms. Often, not all algorithms are learned and commutators are
used instead.
Megaminx PLL does EPLL and CPLL at once, at the cost of 151 algorithms. Because of the algorithm count, this is only used by some of the best cubers, although lots of people use partial PLL due to
the fact that lots of 3x3x3 PLLs like the T permutation also work on the Megaminx.
Megaminx OLL combines EOLL and OCLL into one step that requires 260 new algorithms to be learned. Even among world class solvers, knowing full OLL isn't common.
Unlike 3x3x3 ELL, Megaminx ELL refers to orienting and permuting the edges ignoring the corners (similar to LLEF). It only consists of 39 algorithms, although the main issue is that the algorithms to
orient the corners after this step become longer and that this is hard to extend to a 4LLL due to the very high algorithm count for solving the last four corners.
See also
External links
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Graphing within Geogebra
Graphing within Geogebra - Quadratic Functions 1. What are the similarities and differences between the Conic equation and the Function equation? 2. What happens to the equations when you drag the
vertex of the quadratics? 3. What happens to the graphs of the quadratics when you drag the vertex? 4. How are your answers to question 2 and question 3 related? 5. Use the input bar below the graph
to create another quadratic function that translates y=3x^2 three units down and four units to the left. | {"url":"https://www.geogebra.org/m/RxY9Cumn","timestamp":"2024-11-08T19:07:48Z","content_type":"text/html","content_length":"89011","record_id":"<urn:uuid:514643ee-8373-4683-bff3-b9df7dc0962b>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00139.warc.gz"} |
Two Topics in Elementary Particle Physics: (1) Quark Graphs and Angular Distributions in the Decays of the Axial-Vector Mesons. (2) Universal Current-Current Theories and the Non-Leptonic Hyperon Decays
Colglazier, Elmer William, Jr. (1971) Two Topics in Elementary Particle Physics: (1) Quark Graphs and Angular Distributions in the Decays of the Axial-Vector Mesons. (2) Universal Current-Current
Theories and the Non-Leptonic Hyperon Decays. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/HBFX-4095. https://resolver.caltech.edu/CaltechTHESIS:04192018-085253797
The Thesis is divided into the following two parts:
(1) We examine three aspects of the axial-vector mesons: (i) angular distributions of the I = 1 states, (ii) mixing of the I = 1/2 states, and (iii) absence of the I = 0 states.
Using a model of mesons decaying via production of a quark-antiquark pair with the quantum numbers of the vacuum, we relate the angular distributions in the decays A[1]→ ρπ and B → ωπ, predicting 2(g
[1]/g[0])A[1] = (g[0]/g[1])[B] + 1. This relation is consistent with the present, somewhat ambiguous experimental data. Also, we describe satisfactorily, in terms of two parameters, the partial
widths of the 0^+, 1^+, and 2^+ mesons decaying into 1^-0^- and 0^-0^- pairs. The prediction of the model is that SU(6)[W] x 0(2)[L][Z] relations hold among all the D waves and among all the S waves,
but not between the two groups. In fact, our two-parameter fit to the data entails a ratio of S wave to D wave amplitudes of approximately the same magnitude but opposite sign to that implied by SU
(6)[W] x 0(2)[L][Z]. Unlike the widths, the angular distributions are sensitive to the relative sign and are thus crucial in determining that the fit of our model differs considerably from the SU(6)
[W] solution.
Parameters of the fit are applied to the l^+ kaons, which may mix with one another. The results are sensitive to the mixing angle ø, and merely assuming lower bounds on widths of both physical states
establishes the limits 10° ≤ ø ≤ 35° As a result of this mixing, one predicts: (a) the suppression of the K*π mode of the lower peak, (b) the suppression of the ρK mode of the upper peak, and (c)
decay distributions in the K^*π mode similar to that of the A[1] for the lower state and to that of the B for the higher.
The properties of the missing isoscalar mesons are described with particular emphasis on the ninth 1^++ state. Expected properties of this meson, the D', include: (a) assignment to a weakly mixed SU
(3) singlet, predicted by duality and confirmed by the Gell-Mann-Okubo mass formula; (b) a mass of ~950 MeV, predicted by super-convergence with assumptions about the relative couplings of D and D’;
(c) decay modes ηππ and π^+π^-γ; and (d) the possibility of a suppressed ρ signal in the π^+π^- spectrum of the π^+π^-γ final state, despite the expectation that the pions are in a state with I = J =
1. These features suggest that a recently reported meson near this mass with decay modes ηππ and π^+π^-γ may be a candidate for this state, although J^pc = 1^+- is also a definite possibility for the
new meson.
(2) Because of the limited evidence for the V-A Cabibbo theory in the non-leptonic weak decays, we examine the compatibility with experiment of more general current-current theories. These theories,
constrained by universality, are constructed from the neutral and charged currents obtainable in the quark model, i.e., scalar, pseudoscalar, vector, axial-vector, and tensore Using current algebra
and PCAC, a certain class of these theories, including Cabibbo's, is found to be consistent with the S wave amplitudes for the non-leptonic hyperon decays. The P wave amplitudes remain unexplained.
Nevertheless, another class of theories, also including V-A, plus the assumption of a symmetric quark model, predict the ΔI = 1/2 rule.
Item Type: Thesis (Dissertation (Ph.D.))
Subject Keywords: (Physics)
Degree Grantor: California Institute of Technology
Division: Physics, Mathematics and Astronomy
Major Option: Physics
Thesis Availability: Public (worldwide access)
Research Advisor(s): • Zweig, George
Thesis Committee: • Unknown, Unknown
Defense Date: 24 May 1971
Funders: ┌───────────────────────────┬───────────────┐
│ Funding Agency │ Grant Number │
│ NSF │ UNSPECIFIED │
│ Atomic Energy Commission │ UNSPECIFIED │
Record Number: CaltechTHESIS:04192018-085253797
Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:04192018-085253797
DOI: 10.7907/HBFX-4095
Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided.
ID Code: 10820
Collection: CaltechTHESIS
Deposited By: Tony Diaz
Deposited On: 25 Apr 2018 23:50
Last Modified: 20 Jun 2024 21:01
Thesis Files
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Find the locus of the middle points of chords of given circle x2+y2=a
Analytical Geometry> Find the locus of the middle points of ch...
Find the locus of the middle points of chords of given circle x2+y2=a2 which subtends a right angle at the fixed point (p,q).
1 Answers
Vikas TU
Last Activity: 8 Years ago
locus of the middle points of chords of given circle x2+y2=a2 which subtends a right angle at the fixed point (p,q) is:
T = S1
xx1 + yy1 – a^2 = p^2 + q^2 – a^2
it becomes,
px + qy = p^2 + q^2 therfore.
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Bi-Level Planning Model for Urban Energy Steady-State Optimal Configuration Based on Nonlinear Dynamics
School of Economics and Management, North China Electric Power University, Changping District, Beijing 102206, China
State Grid Qinghai Electric Power Company Economic and Technological Research Institute, Xining 810001, China
Author to whom correspondence should be addressed.
Submission received: 3 April 2022 / Revised: 19 May 2022 / Accepted: 23 May 2022 / Published: 25 May 2022
With the rapid development of social economy, energy consumption has continued to grow, and the problem of pollutant emissions in various energy sources has gradually become the focus of social
attention. Cities account for two-thirds of global primary energy demand that make urban energy systems a center of sustainable transitions. This paper builds a bi-level planning model for
steady-state optimal configuration to realize the reasonable planning of the urban energy structure. The first level mainly analyzes the steady-state relationship between energy systems, the second
level is based on the steady-state relationship of multiple energy sources to minimize the construction and operating costs of urban energy systems and pollutant emissions. Nonlinear system dynamics
and the Improved Moth Flame Optimization Algorithm (IMFO) algorithm are implemented to solve the model. In addition, this paper uses instances to verify the application of a planning model in a
certain city energy system in China. Under the premise of ensuring the stability of the urban energy system, two energy planning programs are proposed: mainly coal or mainly high-quality energy. The
coal planning volumes are used as the basis for sub-scenario planning and discussion. Lastly, this paper proposes a series of development suggestions for different planning schemes.
1. Introduction
Cities account for two-thirds of global primary energy demand that make urban energy systems a center of sustainable transitions [
]. Growing demands and technological shifts are changing global energy systems. For example, innovative technologies such as electric vehicles in the transport sector, and new equipment in the
buildings sector, are projected to increase electricity demands in city areas. Cooling demands are the fastest growing in buildings, with subsequent extra load on electricity networks [
]. The Paris Climate Agreement added further traction to the growing international clout of cities as potent actors in both climate change mitigation and adaptation interventions [
]. At present, more than 50% of the world’s population lives in city areas, and this number is expected to rise to 70% by 2050 [
As is shown in
Figure 1
, the impact of economic activities on cities could be decomposed into scale effect, structure effect, and technology effect [
]. Firstly, the increase of economic aggregate is accompanied by more resource input and energy consumption, which also produces more environmental pollutants and has a negative effect on the
environment, which is called scale effect. The change of productivity is often accompanied by the optimization of economic structure, which is reflected in the more reasonable allocation of resources
and other factors, and has a positive effect on the environment, which is the structural effect. Secondly, with the development of economy, the optimization and upgrading of industrial structure is
often reflected in the transformation and upgrading of energy-intensive heavy industry, and the vigorous development of service industry including high-tech industry, to promote the sustainable
development of cities. Therefore, city areas have the potential to contribute significantly to global CO
emissions reductions through careful urban energy systems planning and community participation.
In recent years, with the rapid development of China’s social economy and the increasing improvement of people’s living standards, energy consumption has continued to grow, and the contradiction
between energy supply and demand has become increasingly tense [
]. As the main body of energy production and consumption, cities play an important role in the strategy of energy revolution. However, the way of energy configuration is mainly the high pollution and
high carbon emission, which is the lack of top-level design, planning integration, and action integration between energy and city development and between different energy sources [
]. An urban energy system refers to the use of advanced physical information technology to integrate coal, petroleum, natural gas, electric, thermal, and other energy sources in the region to achieve
coordinated planning, optimized operation, and coordinated management among various heterogeneous energy subsystems. Under the dual pressures of economic development and environmental protection in
cities of China, the reasonable planning and efficient operation of urban energy systems are the prerequisites for the rational use of urban energy, and it is also a hot spot at home and abroad [
]. Therefore, the main objectives of this paper are: (1) analyze the steady-state relationship between urban energy systems (coal, petroleum, natural gas, renewable energy); (2) establish an
optimization model for urban energy planning based on the steady-state relationship of multiple energy sources to minimize the economic and environmental costs of urban energy systems, and provide a
reference for top-level urban energy design.
The rest of this article is organized as follows.
Section 2
introduces the literature review closely related to the establishment of urban energy system planning, identifies the main issues related to the methodological methods adopted, and illustrates the
relevance and novelty of this research.
Section 3
puts forward the typical structure of the urban energy system, and establishes a two-level programming model that considers the economic and environmental costs of the steady-state optimal allocation
of urban energy. At the same time, some methods to solve the above models are proposed, including nonlinear system dynamics and IMFO. In
Section 4
, in order to verify the validity and rationality of the method proposed in this paper, the application of this model in the planning of a Chinese urban energy system is discussed and analyzed.
Section 5
discusses some conclusions about the research; these conclusions are of great significance for readers to understand our research content and innovation.
2. Literature Review
The sustainable development of cities faces the triple dilemma of economic growth, energy sustainability, and environmental protection. A stable and reliable urban energy supply is essential to
support economic growth and environmental protection. Therefore, many scholars have carried out extensive research on urban energy system planning, mainly focusing on three aspects: urban energy
demand forecasting, urban energy system evaluation results, and related urban energy system planning technologies. In terms of urban energy forecasting, Yeo I A et al. [
] classified urban facilities according to the characteristics of energy use, and proposed a municipal energy demand prediction system composed of the e-gis database, energy planning database, and
prediction algorithm. Yifei Y et al. [
] decomposed the influencing factors of energy demand into the economies of scale effect, population scale effect, energy structure effect, and household consumption effect based on the logarithmic
average division index, and predicted China’s energy demand through the improved particle swarm optimization algorithm. Naiming X et al. [
] used the gray Markov prediction model to predict China’s energy demand while studying the self-sufficiency rate of China’s energy demand. Filippov S. P. et al. [
] considered the possible technical and structural changes in the future economy, the differences between different regions, the mutual substitution of energy carriers and energy conservation,
separated economic variables from energy variables, and made a long-term prediction of urban energy demand. From the perspective of energy coupling, Weijie W et al. [
] proposed an energy demand prediction method based on Improved Gray Correlation Analysis and particle swarm optimization BP neural network, which effectively improved the prediction accuracy.
Yanmeng Z [
] compared and analyzed the advantages, disadvantages, and applicable conditions of various energy demand prediction methods, and comprehensively used the idea of a combination model to construct a
nonlinear combination prediction model of energy demand based on the Shapley value method and inclusive test.
Regarding the evaluation of the urban energy system, most scholars analyzed it from the aspects of reliability, sustainability, and stability. Rui J et al. [
] introduced ecological network analysis (ENA) as a general system analysis tool to simulate and evaluate urban energy supply security. The evaluation contents included the overall realizability
evaluation, system attribute analysis, and structure analysis of the energy supply system. Chen C et al. [
] introduced the life cycle assessment (LCA) method to calculate the environmental impact load of different types of energy used in urban areas (including coal, petroleum, natural gas, and
electricity), and evaluated the environmental impact of urban energy life cycle, which is helpful to realize the sustainable development of energy and environment. Kagaya S et al. [
] analyzed the urban energy system through fuzzy utility function and structural modeling, and evaluated the cost-effectiveness and environmental cost.
Based on the research results of urban energy demand and urban energy system evaluation, some scholars have also carried out wide research on urban energy system planning technology. Shan X et al. [
] selected and analyzed the typical commercial tools applicable to different cities by focusing on the planning and design of urban energy systems and energy consumption analysis tools. Xuyue Z et
al. [
] proposed a modeling and optimization decision-making method for urban energy system transformation, focusing on the optimal configuration and operation to meet the energy demand. Considering the
technical, social, economic, and environmental complexity of energy system. Neshat N et al. [
] developed a modeling framework based on a bilateral multi-agent to determine the best planning scheme of an energy system. Yazdanie M et al. [
] analyzed the gap between urban energy system modeling tools and methods, and developed an urban energy system modeling method considering social factors. Meirong S et al. [
] proposed a multi-objective optimization model at the urban sector scale to minimize energy consumption, energy costs, and environmental impact, and the life cycle assessment method is used to
calculate the environmental impact caused by various pollutants in the chain of energy production, transportation, and consumption. Rexiang W et al. [
] established a Hamiltonian directed graph to describe the energy flow process between supply and demand in urban energy networks based on graph theory ideas, and proposed a multi-source, multi-sink,
multi-path approach to planning urban energy systems.
Due to the lack of a scientific urban energy system planning method, there are still many problems in the planning process. For example, the management of a system is fragmented, and information
between energy subsystems is not shared, therefore, the different energy systems focused on solving their own problems, which are lack of integrated optimization solutions for urban energy systems,
etc. Therefore, there is an urgent need to carry out related research on the theory and technology of urban energy systems and multi-energy comprehensive planning, and establish a method for the
steady-state optimization of urban energy systems. Ming S et al. [
] applied the C-D production function and the STIRPAT equation to construct an energy supply stability model based on nine factors affecting energy supply and demand. Aimin L et al. [
] quantified the development capacity, stability, and efficiency of the network in Tangshan City, China, from 2006–2016 using flow and informatization analysis of the ecological network analysis
method. Xueting Z et al. [
] applied an improved ecological network analysis framework to study the internal metabolic processes and external stability of urban energy metabolic systems.
Nonlinear system dynamics is a discipline that studies the variation of state variables of a nonlinear system with time. In 1892, Russian scholar, Lyapunov, published the paper “General Problems of
Motion Stability”, where he gave two methods to analyze the stability of ordinary differential equations. Among them, the first method of Lyapunov analyzed the local stability of the corresponding
equilibrium point of the nonlinear system. kazaoka R [
] analyzed the power system stability of household users based on nonlinear dynamics, and obtained the mathematical model of power system. Hongshan Z et al. [
] projected the nonlinear power system dynamic model into a lower dimensional sub-space and proposed an empirical Gramian equilibrium reduction method for solving the nonlinear power system model
reduction problem efficiently. Nonlinear influence relationships also exist between various types of urban energy sources, but there are still few studies on the application of this method to urban
energy system stability, and scholars are needed to further construct urban energy system stability models and use nonlinear system dynamics to solve urban energy stability.
Therefore, in urban energy system planning, in order to clarify the advantages and disadvantages of the proposed model, this paper compares the existing typical research results from the aspects of
research object, goal, and solving algorithms, whether it is bi-level, qualitative analysis, or quantitative analysis. The comparison results are shown in
Table 1
below. It can be found that there are a lot of studies on urban energy systems and their planning, but there are still some unresolved problems: (1) In existing urban energy system planning,
alternative relationships between different forms of energy are less considered, which may lead to repetitive planning; (2) Most existing planning models aim at the lowest system cost, but there is
less consideration of the environment and whether the urban energy system is in a steady state at the optimal planning volume; (3) Most of the existing literature applies nonlinear system dynamics to
solve mathematical or physical problems, with little extension to energy systems.
Based on these unresolved problems, this article established a bi-level planning model for urban energy steady-state optimal configuration. The main contributions of the paper are summarized as
The basic structure of an urban energy system is built. Considering the substitution relationship between different energy forms, a bi-level optimization model of multi- energy comprehensive
planning and steady-state configuration of urban energy system is proposed.
A four-dimensional nonlinear urban energy system model with coal as the main source and diversified development of natural gas, petroleum, and renewable energy is established, and the stability
of nonlinear system solutions and Lyapunov’s theorem are applied to analyze the steady-state relationship of multiple energy sources and find out the demand for each energy source in the steady
state of the urban energy system.
An urban energy system planning model is developed. Based on the steady-state relationship of multiple energy sources, the second-level planning model focuses on the energy planning configuration
of the urban energy system in the steady state, to achieve the minimization of the construction and operation costs of the urban energy system, and as pollutant emissions petroleum.
3. Model of Bi-Level Programming
The urban energy system is an integrated system of energy production, processing conversion, and transmission allocation, which can realize the coordination and optimization of different energy
sources in urban planning, construction, and operation. It mainly consists of three subsystems: energy supply system, energy conversion system, and energy consumption system.
Figure 2
is a typical structural diagram of an urban energy system.
An urban energy system is an important concern for future energy development. The planning and allocation of urban energy is the key to urban economic operation. Aiming at the problem of urban energy
steady-state optimal configuration, this paper establishes an bi-level programming optimization model for urban energy systems. The first level is a four-dimensional urban energy system model based
on nonlinear system dynamics, which is used to find the demand for each energy source when the system is in a steady state. The optimization goal of the second level can be described as minimizing
the economic cost and the environmental cost of the energy system under various constraints. The output results of the first-level model are the input parameters of the second-level model, which will
affect the optimization results of the second-level model.
Figure 3
is the bi-level optimization logic diagram of the urban energy system.
3.1. Four-Dimensional Nonlinear Urban Energy System Model of the First Level
In order to reduce consumption and optimize the urban energy structure, this paper established a nonlinear model that is dominated by coal and diversifies the development of natural gas, petroleum,
and renewable energy (mainly including wind and photovoltaic). The model studies the interaction between the four to find out the demand for each energy source when the urban energy system is in a
stable state. The objective function is:
$F ( x , y , z , r ) = F b a l a n c e − p o i n t { ( x 1 , y 1 , z 1 , r 1 ) , ( x 2 , y 2 , z 2 , r 2 ) , … , ( x n , y n , z n , r n ) }$
$X ( t )$
is the coal consumption,
$Y ( t )$
is the natural gas consumption,
$Z ( t )$
is the petroleum consumption,
$R ( t )$
is the consumption of renewable energy (mainly including wind and photovoltaic).
$( X , Y , Z , R )$
is the consumption of coal, petroleum, natural gas, and renewable energy under the constraints of each indicator when the urban energy system is in a stable state. In order to solve the consumption
of energy in the urban energy system under steady state, a four-dimensional nonlinear urban energy system model
is established, as shown in Equation (2):
${ d x d t = a 1 x ( 1 − x M ) − a 2 ( y + z ) − d 3 r d y d t = b 1 y + b 2 x − b 3 x z [ N − ( x + z ) ] d z d t = c 1 z ( c 2 x − c 3 ) d r d t = d 1 r − d 2 x$
$a i , b i , c i , d i , M$
are energy system steady-state coefficients and
$a i , b i , c i , d i , M > 0$
in the energy system,
$a 1 , b 1 , c 1 , d 1$
is the consumption elasticity coefficient of coal, natural gas, petroleum, and renewable energy, respectively;
$a 2$
is the influence coefficient of petroleum and natural gas on coal,
$b 2$
is the influence coefficient of coal on natural gas in the energy system,
$b 3$
is the influence coefficient of coal and petroleum on consumption of natural gas in the energy system;
$c 2$
is the price per unit of coal in the energy system and
$c 3$
is the clean coal technology cost in the energy system;
$d 2$
is the influence coefficient of coal, petroleum, and natural gas on renewable energy in the energy system and
$d 3$
is the influence coefficient of renewable energy on coal in the energy system;
is the maximum energy gap and
is the threshold of environmental pollution in the energy system.
The model idea is as follows:
$a 1 x ( 1 − x M ) − a 2 ( y + z ) − d 3 r$ indicates that the coal consumption rate in the urban energy system is proportional to the energy gap and the potential gap share $1 − X / M$ in the
current energy system; the input of petroleum and natural gas reduces the demand gap of coal, and the utilization of renewable energy will reduce the demand for coal.
$b 1 y + b 2 x − b 3 x z [ N − ( x + z ) ]$ indicates that a gap between the consumption rate of natural gas and itself; the ultimate return from coal can introduce more natural gas or vigorously
develop coal-to-gas technology, thereby increasing the consumption of natural gas, therefore, the rate of natural gas consumption is directly proportional to the final return from coal; $− b 3 x z [
N − ( x + z ) ]$ indicates that when the amount of environmental pollution generated by coal and petroleum in the urban energy system is less than the critical value ($( x ( t ) + z ( t ) ) < N$),
the consumption rate of natural gas decreases with $x ( t )$ and $z ( t )$ increases, but when the amount of environmental pollution generated is greater than the critical value at the time ($( x ( t
) + z ( t ) ) > N$), since the amount of environmental pollution is controlled within the specified range, the consumption rate of natural gas increases with $x ( t )$ and $z ( t )$ increases.
$c 1 z ( c 2 x − c 3 )$ indicates that the consumption rate of petroleum is proportional to its own gap; at the same time, due to the final income generated by coal, more petroleum can be introduced
or the coal-to-petroleum technology can be developed to increase the consumption of petroleum. Therefore, the rate of petroleum consumption is directly proportional to the final return from coal.
$d 1 r − d 2 x$ indicates that the consumption rate of renewable energy decreases with the increase of $x ( t )$, and the consumption rate of renewable energy is proportional to its own gap.
3.2. Urban Energy System Planning Model of the Second Level
3.2.1. Objective Function
The goal of steady-state optimal allocation of urban energy systems is to minimize the economic cost of the energy system and minimize the amount of environmental pollution emissions during the
planning period. Therefore, this paper proposes the optimal planning model including the economic scheduling model and the environmental scheduling model.
Before establishing the model, first establish the following assumptions: (1) the urban economy will maintain the development trend of growth and the future energy price can be reasonably predicted,
and (2) the total energy consumption of the city in the future will not fluctuate violently.
The economic model only considers economic costs of power generation and energy supply. The cost of power generation technology is divided into two categories: fuel cost and operating investment
cost. The cost of fuel is mainly the cost of coal, petroleum, and natural gas, which is included in the energy supply cost. The operating investment cost is included in the energy conversion cost.
The energy conversion technology mainly refers to the power generation technology. The cost includes the operating costs and the investment costs of the newly-built units or the equipment.
$min C cos t = C s + C c = [ P C , t L C , t + P N G , t L N G , t + P O , t L O , t + P R E , t L R E , t ] + ∑ t ∑ k [ C G ( k , t ) G S ( k , t ) + N G ( k , t ) C ( G k , t n ) ]$
$C cos t$
is the total cost of the urban energy system;
$C s$
is the energy supply cost;
$C c$
is the energy conversion cost,
is the different period of the planning period;
$P C , t , P N G , t , P O , t , P R E , t$
is the price of coal, natural gas, petroleum, and renewable energy during the period
$L C , t , L N G , t , L O , t , L R E , t$
is the supply of coal, natural gas, petroleum, and renewable energy during the period
$C G ( k , t )$
is the operating cost of local power generation technology
during the period
(power generation technology
includes coal power, gas power generation, wind power, solar power generation, etc.),
$G S ( k , t )$
is the power generation capacity of local power generation technology
during the period
$N G ( k , t )$
is the new capacity of local power generation technology
during the period
, and
$C ( G k , t n )$
is the unit investment cost of the new capacity of local power generation technology
during the period
Environmental model
When the total amount of pollutant discharge does not exceed the maximum amount of pollutant discharge, the environmental cost is equal to the amount of each pollutant discharged from each source
multiplied by its environmental price. When the total amount of pollutant discharge exceeds, the environmental cost includes the cost of excess penalty and the cost of ecological restoration.
$C e = ∑ r = 1 R K r V r + Z + G$
$C e$
is the environmental cost;
$K r$
is the environmental value of the pollutant;
$V r$
is the pollutant emissions;
is the penalty cost due to excessive emissions, and
is the cost of ecological restoration.
3.2.2. Constraint Conditions
Considering the factors affecting the urban energy system structure, the constraints of the second phase of the urban energy system include energy supply and demand balance constraint, energy
exploitation capacity constraint, technical capacity constraint, energy planning policy constraint, and environmental constraint.
• Energy supply and demand balance constraint
Consider coal, petroleum, natural gas supply and demand balance constraints, and converted energy supply and demand balance constraints:
$L S ( n , t ) + I S M ( n , t ) − E X S ( n , t ) ≥ D n , t$
$L S ( n , t ) , I S M ( n , t ) , E X S ( n , t ) , D n , t$
are the amount of production, purchase, consumption and forecast demand for energy
during the period
, respectively.
Energy exploitation capacity constraint
$L S ( n , t ) ≤ β n , t$
$β n , t$
is the upper limit of the production capacity of energy
during the period
Technical capacity constraint
$L S ( n , t ) ≤ G C ( k , t ) h ¯ k , t ( 1 − η k )$
$G C ( k , t )$
is the upper limit of the installed capacity of the energy conversion technology
during the period
$h ¯ k , t$
is the average annual running time of local power generation technology
during the period
, and
$η k$
is energy conversion technology update rate.
Energy planning policy constraint
$L S ( n , t ) ≤ ν n , t$
$ν n , t$
is the upper limit of the supply capacity of energy
controlled in the government energy planning document during the period
Environmental constraint
$C e = { ∑ r = 1 R K r V r K r ≤ K r − max ∑ r = 1 R K r V r + Z K r > K r − max$
$K r − max$
is the maximum allowable emissions specified by environmental policy.
3.3. Model Solving Method
3.3.1. Solution of the First Level Based on Nonlinear System Dynamics
Nonlinear dynamic equations must express progressive laws. These nonlinear dynamic equations generally use ordinary differential equations containing time parameters. Nonlinear dynamic equations
typically generally use regular differentials containing time parameters. The comparison or partial differential equation, i.e., the evolution equation, is defined, and the solution of the equation
represents the motion of the system (how the state changes with time). The stability of the solution indicates whether the motion of the system is stable or not. Therefore, this article uses the
stability of the nonlinear system solution and the Lyapunov theorem to study how urban energy changes over time.
$∇ = ∂ x ˙ ∂ x + ∂ y ˙ ∂ y + ∂ z ˙ ∂ z = a 1 − 2 a 1 x M − b 1 + c 1 c 2 x − c 1 c 3 = ( c 1 c 2 − 2 a 1 M ) x + ( a 1 − b 1 − c 1 c 3 )$
When $a 1 − b 1 < c 1 c 3$$a 1 − b 1 < c 1 c 3$ and $2 a 1 M = c 1 c 2$, the urban energy system is dissipative.
Balance point stability
$d x d t = 0$
$d y d t = 0$
$d z d t = 0$
${ a 1 X ( 1 − X M ) − a 2 ( Y + Z ) = 0 b 1 Y + b 2 Z − b 3 X [ N − ( X − Z ) ] = 0 c 1 Z ( c 2 X − c 3 ) = 0$
gets three balance points:
(0, 0, 0),
), the solution equation is as follows:
${ x 2 = a 2 b 3 M N − a 1 b 1 M a 2 b 3 M − a 1 b 1 y 2 = a 1 b 3 ( M − N ) ( a 2 b 3 M N − a 1 b 1 M ) ( a 2 b 3 M − a 1 b 1 ) 2 z 2 = 0$
${ x 3 = c 3 c 2 y 3 = [ a 1 a 2 ( 1 − x 3 M ) ( b 3 x 3 − b 2 ) − b 3 x 3 + b 3 N ] x 3 b 1 + b 3 x 3 − b 2 z 3 = a 1 b 1 a 2 x 3 ( 1 − x 3 M ) − b 3 N x 3 + b 3 x 3 2 b 1 + b 3 x 3 − b 2$
When $a 2 b 3 N > a 1 b 1$, since M > N, then $a 2 b 3 M > a 1 b 1$.
At this time $0 < x 1 < a 2 b 3 M N − a 1 b 1 N a 2 b 3 M − a 1 b 1 = N$, $y 1 = b 3 b 1 x 1 ( N − x 1 ) , y 1 > 0$.
When $c 3 c 2 ≥ N$, the urban energy system has two balance points $S 1$, $S 2$; when $c 3 c 2 < N$, there are three balance points $S 1$, $S 2$, $S 3$.
For the balance point S[1](0, 0, 0), the coefficient matrix of the linear approximation system is
$J 0 = [ a 1 − a 2 − a 2 b 3 N − b 1 − b 2 0 0 − c 1 c 3 ]$
The characteristic equation is $( c 1 − c 3 − λ ) [ λ − ( a 1 − a 3 ) ] ( λ + b 1 b 3 ) = 0$.
The characteristic root of
$J 0$
$λ 1 = − c 1 c 3 < 0 , λ 2 , 3 = a 1 − b 1 ± ( b 1 − a 1 ) 2 − 4 ( a 2 B 3 N − a 1 b 1 ) 2$
This article assumes that when $( b 1 − a 1 ) 2 < 4 ( a 2 B 3 N − a 1 b 1 )$, namely $( a 1 + b 1 ) 2 < 4 a 2 b 3 N$, the conclusions can be obtained as following:
When $a 1 < b 1$, $λ 2 , 3$ is a pair of conjugate complex roots with negative genuine parts, so that the three characteristic roots $λ 1 , λ 2 , λ 3$ all have negative real parts, then the system is
stable at S[3] (0, 0, 0). When $a 1 > b 1$, $λ 2 , 3$ is a pair of conjugate complex roots with positive real parts, then S[3] (0, 0, 0) is the unstable point. When $a 1 = b 1$, $λ 2 , 3 = ± i ω$.
For the balance point S[2] (x[2], y[2], z[2])
$J 1 = [ a 1 ( 1 − 2 x M ) − a 2 − a 2 b 3 ( N − 2 x ) − b 1 − b 2 + b 3 x 0 0 c 1 ( c 2 x − c 3 ) ]$ can be obtained from its Jacobian matrix.
The characteristic equation is $( c 1 − c 3 − λ ) [ ( a 3 − a 1 ) − λ ] ( b 1 b 2 ( a 1 − a 3 ) M a 1 − b 1 b 3 − λ ) = 0$.
The characteristic roots are $λ 1 = c 1 − c 3$, $λ 2 = a 3 − a 1$, and $λ 3 = b 1 b 2 ( a 1 − a 3 ) M a 1 − b 1 b 3 = b 1 [ b 2 M ( 1 − a 3 a 1 ) − b 3 ]$.
If $a 1 > a 3 , c 1 < c 3$, $b 2 ( a 1 − a 3 ) M a 1 < b 3$, the three roots are all negative, then S[2] is the stable equilibrium point. If $a 1 < a 3$ or $c 1 > c 3$, there is at least one positive
root, then S[2] is the unstable equilibrium point. If $a 1 = a 3$ or $c 1 = c 3$, there is at least one zero root, and S[2] is in a critical state, and the system produces a bifurcation.
For the balance point S[3](x[3], y[3], z[3])
By its Jacobian matrix
$J 2 = [ a 1 ( 1 − 2 x 2 M ) − a 2 − a 2 b 3 ( N − 2 x 2 + z 2 ) − b 1 − b 2 + b 3 x 2 c 1 c 2 z 2 0 0 ]$
The characteristic equation can be obtained as
$( c 1 − c 3 − λ ) { ( a 1 − 2 a 1 x M − a 3 − λ ) [ b 1 ( b 2 x − b 3 ) − λ ] + a 2 b 1 b 2 y } − a 2 { − b 1 b 2 c 2 x y + c 2 y [ b 1 ( b 2 x − b 3 ) ] } + a 2 c 2 y λ = 0$
Let $a 1 − 2 a 1 x M − a 3 = A$, $b 1 ( b 2 x − b 3 ) = b 1 ( b 2 b 3 b 2 − b 3 ) = B = 0$, $c 1 − c 3 = C$, $a 2 b 1 b 2 y = P$, $a 2 { − b 1 b 2 c 2 x y + c 2 y [ b 1 ( b 2 x − b 3 ) ] } = Q = − a
2 b 3 b 1 c 2 y$, $a 2 c 2 y λ = H$.
The original formula becomes $λ 3 − ( A + C ) λ 2 + ( A C + P − H ) λ + ( Q − P C ) = O$.
3.3.2. Solution of the Second Level Based on IMFO
The non-free lunch optimization theorem shows that no single optimization algorithm can solve all optimization problems, and the Moth–Flame Optimization (MFO) algorithm [
] faces the same problem mentioned above. MFO is prone to premature convergence and falls into local optimum when dealing with complex function problems, so it needs to be improved to improve its
performance. To further improve the convergence accuracy, in addition to considering the iteration stage the algorithm is in, the adaptation value of the moths during the iteration should be taken
into account, i.e., a dynamic inertia weight adjustment strategy that is jointly determined with the iteration stage and the adaptation value of the moths is proposed. Gaussian variation is used to
locally perturb some of the poorer individuals to improve the convergence speed of the algorithm. The Corsi variation strategy is used to enhance the diversity of the population and to suppress the
premature convergence of the algorithm, and thus to achieve global optimization.
• Dynamic inertia weights [
The inertia weight parameter has an important impact on the global and local search of the algorithm. The size of the weights in the MFO algorithm determines the degree of influence of the flame
individuals in the last iteration on the moth individuals in the current iteration. At the beginning of the iteration, a high global search capability is desired to explore new solution spaces and
jump out of local extremes. Later in the iteration, emphasis is placed on local exploitation to speed up convergence and discover the exact solution. If the MFO algorithm uses linear decreasing
inertia weights for complex high-dimensional functions, when the number of iterations is large, the amount of change of inertia weights per iteration is small, which will affect the function of the
weight adjustment strategy. At the same time, with a single change pattern, it will be difficult for the moth to fly out after falling into the local optimum in the late iteration. Based on the above
analysis, in addition to considering the iteration stage the algorithm is in, the adaptation value of the moth should also be taken into account, i.e., the weight size is determined by both the
number of iterations and the adaptation value of the moth. The dynamic inertia weights
are described as follows.
$ω i j = exp ( f ( j ) μ ) 2.4 + ( exp ( − f ( j ) / μ ) ) i t e r$
is the average fitness value of the first optimization process;
$f ( j )$
is the fitness value of the
$i t e r$
denotes the number of current iterations. For the very small value optimization problem, the function value corresponding to the moth optimization solution is set as the fitness value.
$f ( j )$
shows a nonlinear decreasing trend and
$2.4 + ( exp ( − f ( j ) / μ ) ) i t e r$
shows a nonlinear increasing trend as the number of iterations increases, so the weight
shows a nonlinear decreasing trend with the increase of fitness value and the number of iterations.
The improved logarithmic helix function for the flame position update is shown in Equation (18).
$S ( M i , F j ) = w i j × D i × e b t × cos ( 2 π t ) + ( 1 − ω i j ) × F j$
$M i$
denotes the
$F j$
denotes the
$D i$
is the distance from the
moth to the
is a constant defining the logarithmic helix;
is a random number belonging to [−1, 1] indicating the proximity of the moth’s next position to the flame (
= 1 means closest to the flame and
= −1 means farthest from the flame). The dynamic inertia weights change nonlinearly and dynamically with the number of iterations and the fitness value, and the artificial moth gradually moves toward
the flame with the better fitness value, which facilitates the performance improvement of the MFO algorithm.
Gaussian variation mechanism [
Gaussian variation
$G a u s s i a n ( ∂ , ℘ 2 )$
is introduced. Where
denotes the mean;
$℘ 2$
denotes the variance;
$∂ = 0$
$℘ 2 = | X i j ∗ t |$
$| X i j ∗ t |$
denotes the absolute value of the global optimal moth of the population at
iterations. The improved formula for
$M j$
generation is described as follows.
$M n e w t = M j ∗ t + G a u s s i a n ( ∂ , ℘ 2 )$
The global optimal position after Gaussian variation, i.e., the global optimal solution, is obtained according to the above formula. Gaussian variation can produce new offspring near the original
parents. With this property, the diversity of moths and flames can be increased to further improve the local search ability and convergence speed.
Corsi variant strategy [
The MFO itself does not have the ability to jump out of the local optimum, which leads to a premature algorithm and poor convergence accuracy. In this paper, a Corsi variation approach is used. When
the position of the moth stagnates during the iteration, the individual undergoes a Cauchy variation to continue to approach the global optimum, while the optimal individual in the moth population
does not undergo a variation to ensure that the current optimum is not lost.
At each iteration, the Cauchy variation is computed by the Cauchy distribution function to produce a Cauchy variation matrix with a mean of 0, and a standard deviation of 1. The result is multiplied
by each dimension of the moth to be mutated as the update step. The equation for updating the position after introducing the variation is shown in Equation (20).
$X n e w = ϖ · C a u c h y ( 1 , 0 ) · [ D i e b t cos ( 2 π t ) + F j ]$
$C a u c h y ( 1 , 0 )$
denotes the Corsi distribution of
$γ = 1 , x 0 = 0$
affects the range of Corsi variants, the update formula for
$ϖ = ε · ( u a − u b )$
. where
is a parameter that varies with the number of iterations;
$u a$
$u b$
denote the upper and lower bounds of the solution space, respectively. The range over which the moths perform the Corsi variation increases with the number of iterations, expanding a larger search
space for the moths during the algorithm stagnation period, and thus increasing the population diversity.
The solution process of the improved moth–flame-based optimization algorithm constructed in this paper is shown in
Figure 4
. The solution process is as follows.
System initialization. Input system parameters: energy system steady-state coefficients, energy demand, energy prices, pollutant emissions, and environmental cost factors, etc.
Population initialization. Generate the initialized population P; population algebra $N g e n = 1$, t = 1.
Simulation. Calculation of economic and environmental target values.
Location update. Updating dynamic inertia weights, performing Gaussian and Corsi variances, and generating offspring population Q.
Simulation calculation. Calculating the economic environmental target, artificial moth and artificial flame fitness values for population Q, and ranking them by fitness value.
Combination. Combining the current population P with the offspring population Q to obtain a population, calculating the dominance relationship and aggregation distance of each individual
according to the fitness function, and Pareto ranking the individuals.
Termination condition. Judge the termination condition; if the termination condition is satisfied, output the optimal solution, economic cost, and environmental cost, otherwise, return to step 4.
4. Simulation
4.1. Parameters
This paper selects a region in China as the research object. At the end of 2021, the resident population of the region was 17.95 million, the urban population was 9.31 million, the urbanization rate
was 51.86%, and the resident floating population was 4.94 million. In 2021, the region achieved a regional GDP of 997.513 billion yuan, an increase of 6.1% over the previous year at comparable
prices. The total retail sales of social consumer goods was 1227.01 billion yuan. Urban energy is mainly coal, petroleum, natural gas, and renewable energy. In this paper, MATLAB software is used to
simulate and evolve the results.
Figure 5
Figure 6
Figure 7
Figure 8
Figure 9
Figure 10
Figure 11
Figure 12
Figure 13
Figure 14
Figure 15
Figure 16
Figure 17
are drawn by MATLAB software.
• Input parameters of the first level
Figure 5
shows the consumption of coal, petroleum, natural gas, and renewable energy in a region in China during 2005–2021.
Input parameters of the second level
In addition to the parameters mentioned above,
Table 2
shows the forecast value of energy price in an urban energy system and
Table 3
is the pollutant emissions and environmental cost factors. We estimate that the energy demand of the region in China in 2022 is 14,400 tons of standard coal.
4.2. The First Level Simulation
• Determination of parameters of the nonlinear system a[1] and M
The expression of the energy structure’s logistic model is $d X d t = a 1 X ( 1 − X M )$, where $X 0 = X | t = 0$.
The solution is ${ X = M 1 + ( M X 0 − 1 ) e − a 1 t X = X | t = o$.
To estimate the equation, let $M X 0 − 1 = e ξ , F = ln M − X X$. Transform the solution to obtain the following linear equation $F = ξ − a 1 t$.
Estimate $ζ , a 1$ by the least-squares method. For the sample data ${ X t ; t = 1 , 2 , … , 17 }$, construct the variable $F t = ln M − X t X t$ and determine the approximate range of the maximum
energy gap $M$ of the urban energy system based on the sample data, and use the determination coefficient $R 2 = 1 − ∑ ( F t − F ¯ ) ∑ ( F t − F ¯ ) 2$ as the standard. Take points one by one,
substitute ${ F t }$, calculate ${ F t }$ under different $M$ values, and estimate the parameters $ζ , a 1$.
According to the above-mentioned theoretical knowledge, the parameters
$a 1$
can be determined, and it is determined that the approximate value range of
$[ 1.1 , 1.8 ]$
, and the regression analysis results of
Table 4
are obtained.
Identification and determination of other nonlinear system parameters
The other parameters in the four-dimensional nonlinear urban energy system are identified by the Forcal program. After the debugging error reaches 10^−4, the results of the parameters obtained are as
As can be seen from
Table 5
, when the fitting coefficient R
has the highest degree of fit, so it can be determined that
$a 1$
is 0.0466. The three-dimensional view of the urban energy system is shown in
Figure 6
It can be seen from
Figure 7
that in a short period, the elastic relationship of the energy sources
$x ( t ) , y ( t ) , z ( t ) , r ( t )$
in the urban energy system is relatively stable. In the very beginning of evolutionary time, the elastic relationship of the four energy sources will fluctuate to some extent, but the magnitude of
fluctuations decreases over time. Over a long period, the elastic inertia of each energy source in the urban energy system shows a relatively stable state: regular fluctuations within a specific
Figure 7
shows that when
$d 3 = 0.13$
, the urban energy system
$x ( t ) , y ( t ) , z ( t ) , r ( t )$
will stabilize after a period.
$d 3 = 0.13$
, to verify the accuracy of the model, the four-dimensional nonlinear model proposed in this paper is used to forecast the total energy demand in the region from 2012–2021, using 2011 data as the
initial condition (total energy demand is equal to the sum of coal, natural gas, petroleum, and renewable energy demand), where the forecast results are compared with the actual energy consumption of
the region, and the forecast results are derived using the STIRPAT model [
]. The comparison results are shown in
Figure 8
. The STIRPAT model decomposes energy consumption into the product of three factors: population size, affluence, and technology level. The coefficients of each variable are derived by partial
differencing and ridge regression analysis of the model to obtain a regional energy consumption function and forecast the results.
As can be seen from
Figure 8
, there are surges and fluctuations in the actual energy consumption in the region. The four-dimensional nonlinear model is able to judge the stability of the energy system based on previous years’
consumption of four types of energy: coal, natural gas, petroleum, and renewable energy, predicting surges or drops in energy consumption when the energy system is less stable, and slight
fluctuations in energy consumption when the energy system is more stable. The STIRPAT model predicts energy consumption based on the population, economy, and technology of the region. The population
and economy of the region are generally in a steady increase, while the level of technology is difficult to improve in a short period of time, therefore, its prediction of energy consumption
increases slowly year by year without any surge or fluctuation. The average error of the prediction results of the four-dimensional nonlinear model is about 0.059, while the average error of the
prediction results of the STIRPAT model is about 0.154, therefore, the prediction of the four-dimensional nonlinear model proposed in this paper is more accurate.
4.3. The Second Level Simulation
In order to verify the applicability of the IMFO algorithm selected in this paper, the MFO algorithm and the improved moth–flame optimization algorithm (IMFO) are selected to compare the algorithm
convergence, as shown in
Figure 9
Figure 9
a,b are the convergence curves of the economic cost and environmental cost of different algorithms with the population size when the population size is 1000. It can be clearly seen that the MFO
algorithm tends to fall into local optimum when solving the model established in this paper, and the IMFO algorithm has better global search ability and convergence performance, and can obtain better
The nonlinear system dynamic parameters calculated in the first level are used as the input parameters of the second level of urban energy planning. Since the two objectives of the lowest economic
cost and the lowest environmental cost are mutually exclusive, 20 sets of feasible solutions can be obtained. Therefore, when analyzing the relationship between different energy sources and objective
values, it is necessary to analyze both economic and environmental costs. The Pareto results are shown in
Figure 10
According to the simulation results of the first level, a feasible solution for the stability of multiple sets of energy combinations is obtained. Still, the difference between the advantages and
disadvantages of different feasible solutions is obvious. In the second level, the IMOA algorithm is used for simulation. The result of the first level is used as the input value of the second level.
Under the objective function conditions and various constraints that satisfy the minimum economic cost and environmental cost, the target value is better. The 20 sets of feasible solutions are
analyzed for operational conclusions. The specific results are shown in
Table 6
Since the optimization variables and the target dimensions are inconsistent, the above feasible solutions are normalized, and the processing results are shown in
Figure 11
. The share of various energy and economic and environmental costs in each scenario is clearly mapped out.
According to the proportion of coal in the energy structure, the 20 groups of planning results are divided into two categories, as shown in
Figure 12
. The first category is the energy planning model based on coal. Under this type of development model, coal accounts for more than 50%; the second category is an energy planning model based on
high-quality energy. Under such development models, petroleum, natural gas, and renewable energy account for more than 50%.
4.4. Analysis Results and Discussion
4.4.1. The First Type of Planning Schemes
Under the first planning scheme, the urban energy system mainly uses coal as the primary energy source, with petroleum, natural gas, and renewable energy sources as auxiliary energy sources. The
feasible solution under the first planning scheme is shown in
Figure 13
A three-dimensional view of the four types of energy shows that the economic and environmental costs of coal are high and basically positively correlated, i.e., the greater the consumption of coal
energy, the higher the economic and environmental costs; there is a clear inflection point in the relationship between the consumption and cost of natural gas, petroleum, and renewable energy. Taking
natural gas as an example, when the planned amount of natural gas exceeds 850 tons of standard coal, with the increase of energy consumption, the economic cost and environmental cost are declining.
When it comes to renewable energy, this inflection point occurs near 8000 tons. When the consumption of renewable energy is less than 8000 tons, the total cost increases with the increase in the
consumption of renewable energy. When the consumption of renewable energy is greater than 8000 tons, the total cost decreases as the consumption of renewable energy increases. For petroleum, this
inflection point occurs near 8 tons, and when the planned consumption of petroleum exceeds this figure, the total cost decreases as the planned amount increases. Moreover, because the quantity of
petroleum is much smaller than other types of energy, which means the decline rate is faster than that of petroleum and renewable energy, this indicates that the relationship between the planned
amount of petroleum and the economic and environmental cost is the most sensitive, followed by natural gas, and finally, renewable energy and petroleum. It shows that when replacing energy in the
future, in the first type of planning schemes, we should first consider using petroleum and natural gas to replace coal for the urban energy supply.
From the perspective of actual demand or ecological environment, the coal-based energy structure can temporarily maintain a steady state for the city’s energy supply and demand, but coal causes
greater environmental pollution, and its economic and environmental costs will gradually become the main factor limiting the development of this type of energy.
4.4.2. The Second Type of Planning Schemes
Under the second type of planning schemes, the urban energy system mainly uses natural gas and renewable energy as the main energy supply, and coal and petroleum are the auxiliary supply of energy,
which are environment-friendly planning schemes. This planning scheme effectively reduces the pollutant emissions level, thus reducing the environmental cost. The feasible solution under the second
type of planning scheme is shown in
Figure 14
. From the figure, we can see that although the total planned consumption of coal is less than that of the first type of planning scheme, its consumption is still positively correlated with the total
target cost, indicating that the consumption and cost of energy, such as coal, have little correlation with the consumption of other energy. When the planned consumption of natural gas is less than
4000 tons of standard coal, with the increase of energy planning, the environmental cost has remained basically unchanged, while the total economic cost of energy is declining. It shows that in the
second type of planning scheme, the relationship between natural gas planning volume and target cost is not sensitive. There are obvious fluctuations and faults in the relationship curve between the
planned petroleum quantity and the target cost, indicating that in the second type of planning scheme, the planned petroleum consumption is the most sensitive factor to the target cost. In the change
curve between renewable energy and target cost, due to the strong uncontrollability of renewable energy capacity, its large-scale use is greatly affected by other factors, and it is impossible to fit
a strong correlation. However, it can be seen that the greater the consumption of renewable energy, the smaller the target cost.
4.4.3. Sub-Scenario Discussion
Since coal still plays an important role in the economy and society to ensure the security of energy supply and demand [
], this paper uses the growth rate of coal as the basis for scenario division, and adds the growth proportion constraint, based on the constraints proposed above, to bring into the second level of
the planning model solution; each scenario is elected to find out the relationship between energy and cost under different development scenarios of urban energy planning.
• Scenario Ⅰ: 10% increase in coal planning
Simulation optimization is carried out under the constraint that coal consumption will increase by at least 10% in the future, and the top 10 feasible solutions are selected, which is shown as
Figure 15
. It is found that there is a relatively obvious linear relationship between the consumption of various energy sources and the total target cost. Among them, there is a positive correlation between
coal and total target cost, while there is a negative correlation between other energy and total target cost. There is also a clear inflection point between petroleum, natural gas, and renewable
energy, and the total target cost. The linear relationship between petroleum consumption and total target cost, and the linear relationship between renewable energy consumption and total target cost
are more similar. When the petroleum consumption exceeds 8 tons, the economic and environmental costs decrease with the increase in consumption. When the consumption of renewable energy exceeds 8000
tons, the total cost decreases as consumption increases, and at a faster rate. However, when the consumption of natural gas exceeds 10,000 tons, the rate of decrease of the target cost with the
consumption tends to be flat, indicating that when the planned natural gas consumption exceeds 10,000 tons, the consumption of natural gas is not a sensitive factor. Combined with the conclusions in
the first type of planning scheme in the previous section, planners should pay attention to the inflection point of 10,000 tons. When the consumption of natural gas exceeds 10,000 tons, they should
focus on controlling costs by changing the consumption of other energy sources.
Scenario II: 10–20% reduction in coal planning
The simulation optimization is carried out under the constraint of reducing the coal consumption by 10% to 20% in the future, and the first 10 groups of feasible solutions are selected, which are
shown as
Figure 16
. It can be found that both coal consumption and renewable energy consumption have a linear relationship with the total target cost. The total planning volume of coal is positively correlated with
the total target cost, and the consumption of renewable energy is negatively correlated with the total target cost. As the consumption of renewable energy increases, there is a clear downward trend
in total target costs. Comparing the two trend charts, it is found that under the same change trend of the total target cost, the total planned consumption of coal and the total planned quantity of
renewable energy are negatively correlated. It shows that coal and renewable energy have a strong substitution relationship. When the planned coal volume is reduced, the energy steady-state can be
maintained by increasing the consumption of renewable energy, and the total target cost can be reduced at the same time. When the consumption of natural gas is between 2400 tons of standard coal and
2600 tons of standard coal, the total planned amount of natural gas is positively related to the total target cost, but the fluctuation is small, indicating that the relationship between the planned
amount of natural gas and the target cost is not sensitive. When the petroleum consumption is between 30 tons of standard coal and 80 tons of standard coal, the environmental cost and economic cost
fluctuate greatly, indicating that the planned petroleum consumption is the most sensitive factor to the target cost in this situation. In this planning scenario, if planners want to further control
the total target cost while using renewable energy to replace coal, they should give priority to controlling the consumption of petroleum.
Scenario Ⅲ: 20% reduction in coal planning
The simulation optimization is carried out under the constraint of reducing coal consumption by more than 20% in the future, and the first 10 groups of feasible solutions are selected, which are
shown as
Figure 17
. It can be clearly seen that only the consumption of coal maintains a relatively consistent linear relationship with the total target cost, with the total cost target essentially rising as coal
consumption increases. However, the consumption of other energy sources cannot be fitted to the total cost target, which means there is no obvious linear relationship between the consumption of
various energy sources and the total target cost. When the consumption of various types of energy in cities is different, the total cost has a certain degree of similarity. It shows that when the
planned amount of coal is greatly reduced, the consumption of other kinds of energy is not stable, which will affect the stable state of an urban energy system, and the accurate total target cost
cannot be obtained. If planners want to reduce the consumption of coal, they should appropriately reduce the consumption of coal according to the basic situation of an urban energy system, so as to
avoid affecting the stability of the urban energy system by greatly reducing the consumption of coal.
4.4.4. Recommendation
With the development of urbanization, society pays extensive attention to environmental protection, the improvement of people’s living standards, the continuous improvement of the market economic
system, and the social development and growth momentum of high-quality energy, such as natural gas and petroleum, which is gradually increasing. Under this energy development model, we should
continue to strengthen the adjustment of industrial structure and optimize and upgrade the path of urban development. The focus is to strengthen the utilization of clean energy and improve energy
efficiency, so as to further reduce the demand for traditional fossil energy in the process of economic development. Although the current coal-based system (in the first type of planning scheme) can
still maintain a stable state, if the region chooses this development mode, it should try to adopt clean coal technology and coal to petroleum technology to make coal become clean energy. At the same
time, since petroleum and natural gas are the most sensitive factor in the first type of planning scheme, when the consumption of natural gas is less than 10,000 tons, urban decision-makers should
first consider adjusting the consumption of petroleum and natural gas when controlling the cost of energy consumption. Finally, when the urban development level of the region has reached a certain
level, in the next step, the region must strive to reduce the proportion of coal in the energy consumption structure, and continuously reduce the proportion of direct coal consumption in the final
energy consumption through industrial structure transformation and other measures.
Secondly, energy infrastructure plays an important role in urban construction and urban planning. From the experience of foreign urbanization development, the construction of an urban energy system
is as important as the construction of road, transportation, communication, and drainage system. Urban energy system includes electricity, heat, gas pipe network, transmission and distribution
system, and management system. The unified planning and construction of an urban energy system must appropriately promote urban development. While avoiding “urban disease”, it will further improve
the level of urbanization and realize the sustainable development of cities and energy systems.
Third, the region must implement the strategy of “self production and self marketing of regional energy” and change the efficiency of energy production, transmission, and conversion. For example,
through the implementation of CCHP, the heat energy related to thermal power generation can be directly supplied to users with the shortest streamline, which greatly improves the conversion
efficiency of primary energy.
Finally, urban energy system planning is an important part of the top-level design of urban development, which is related to the sustainable development of the city. However, no matter what
development model is adopted, the ultimate goal is to ensure the stability of urban economic development and reduce the emission of pollutants on the basis of meeting the energy demand. Smart city
construction is not only the basis of scientific development, but also guides the optimal development of an urban energy structure. Through urban energy assessment and planning, the development of an
urban energy system must be considered from the overall perspective of the city, then look forward to and carefully consider the city’s development strategy and development needs.
5. Conclusions
In order to realize the economic and environmental benefits of the urban energy system, a bi-level planning model for the steady optimal allocation of urban energy is established in this paper. The
conclusions are as follows:
• In order to reduce energy consumption and optimize the urban energy structure, an urban energy steady-state model based on nonlinear system dynamics was developed on the basis of competing
systems, and the relationship between coal, natural gas, petroleum, and renewable energy was studied. We used the consumption of coal, petroleum, natural gas, and renewable energy in certain
region of China during 2005–2021 as the base data. It was found that the urban energy system could maintain a steady state when the parameters were as shown in
Table 5
. Compared with the method used in literature [
] to solve for the stability of urban energy systems, literature [
] considers more external factors related to energy usage (population size, affluence, etc.), while the method proposed in this paper is directly related to the endogenous factors of each type of
energy (usage, elasticity coefficient, etc.), and the error with the actual data is 61.68% lower than that of literature [
• An optimization model for urban energy planning is proposed and solved using the IMFO algorithm. A comparison of the algorithms revealed that the improved algorithm in this paper finds the
optimal solution faster than the original method, and that the feasible solutions are found to fall into two categories, coal-based and high- quality energy. The research found that in the
coal-based scheme, petroleum and natural gas are the most sensitive factors related to the target cost. In the high-quality energy-based scheme, the relationship between the planned amount of
natural gas and the target cost is not sensitive, and there is no strong correlation between renewable energy and target cost. Overall, without adding the constraint of coal energy consumption,
where the smaller the share of coal is, the lower the cost.
• On the basis of the two-layer planning model for the optimal allocation of urban energy steady-state, the constraints of 10% increase, 10–20% reduction, and more than 20% reduction in the future
planned coal volume were added, respectively, and the energy system of the region was carried out. In Scenario Ⅰ, we found that the results were basically consistent with the results of the
coal-based scheme, but when the natural gas consumption exceeded 10,000 tons, the natural gas consumption was not a sensitive factor. In Scenario II, we found that coal had a strong substitution
relationship with renewable energy. In Scenario Ⅲ, we found that only coal consumption maintained a relatively consistent linear relationship with the total target cost, however, the consumption
of other energy sources could not fit a strong correlation with the total cost target. Through planning and simulation, it was found that the reduction of coal consumption should be carried out
gradually (within 10%). If the coal consumption is suddenly reduced, the correlation between various energy sources and the target cost cannot be fitted, which will affect the urban energy
stability to a certain extent.
• Although the above research results are presented in this paper, there are still some limitations. For example, in terms of modeling, the first-level model proposed in this paper is applicable to
energy system stability prediction, and the solutions in this paper can actually find rigorous solutions mathematically, but there are also elements of trial, so the solutions obtained may only
reflect part of the situation. However, they are still strictly analytical solutions and have their important theoretical value. At the same time, the second-level planning model can only solve
planning usage problems for petroleum, coal, natural gas, and renewable energy, not practical operational problems. In terms of practical applications: this paper only examined the steady-state
results and planning scenarios for urban energy systems. However, if planners want to achieve an optimal planning state for urban energy, city managers need to manage it through various
macro-regulation means (e.g., giving clean energy subsidies through economic means, limiting coal consumption through policies, etc.), which requires follow-up research by scholars.
Author Contributions
Conceptualization, Y.W. and C.L.; methodology, C.L.; software, C.L.; validation, C.L., C.C. and Z.M.; formal analysis, C.L.; investigation, C.C.; resources, M.Z.; data curation, H.D.;
writing—original draft preparation, C.L.; writing—review and editing, C.L.; visualization, Z.M.; supervision, H.D.; project administration, F.L.; funding acquisition, Y.W. All authors have read and
agreed to the published version of the manuscript.
This paper is supported by the Fundamental Research Funds for the Central Universities, grant number 2019FR001 and the Fundamental Research Funds for the Central Universities, grant number 2021FR002.
Conflicts of Interest
The authors declare no conflict of interest.
A. Acronyms
IMFO Improved Moth–Flame Optimization Algorithm
MINLP Mixed integer nonlinear programming
GAMS General algebraic modeling system
BP neural network Back propagation neural network
EDFS Energy demand forecasting system
ENA Ecological network analysis
LCA Life cycle assessment
MFO Moth–Flame Optimization
PSO-LSSVR Least-squares support-vector regression optimized by particle swarm optimization
CCHP Combined cooling, heating, and power
B. Parameters
$( X , Y , Z , R )$ the consumption of coal, petroleum, natural gas, and renewable energy under the constraints of each indicator when the urban energy system is in a stable state
$X ( t )$ the coal consumption
$Y ( t )$ the coal consumption
$Z ( t )$ the petroleum consumption
$R ( t )$ the consumption of renewable energy (mainly including wind and photovoltaic)
$a 1$ the consumption elasticity coefficient of coal
$b 1$ the consumption elasticity coefficient of natural gas
$c 1$ the consumption elasticity coefficient of petroleum
$d 1$ the consumption elasticity coefficient of renewable energy
$a 2$ the influence coefficient of petroleum and natural gas on coal
$b 2$ the influence coefficient of coal on natural gas in the energy system
$c 2$ the price per unit of coal in the energy system
$d 2$ the influence coefficient of coal, petroleum, and natural gas on renewable energy in the energy system
$c 3$ the clean coal technology cost in the energy system
$d 3$ the influence coefficient of renewable energy on coal in the energy system
$M$ the maximum energy gap
$N$ the threshold of environmental pollution in the energy system
$C cos t$ the total cost of the urban energy system
$C s$ the energy supply cost
$C c$ the energy conversion cost
$t$ the different period of the planning period
$P C , t$ the price of coal during the period t
$P N G , t$ the price of natural gas during the period t
$P O , t$ the price of petroleum during the period t
$P R E , t$ the price of renewable energy during the period t
$L C , t$ the supply of coal during the period t
$L N G , t$ the supply of natural gas during the period t
$L O , t$ the supply of petroleum during the period t
$L R E , t$ the supply of renewable energy during the period t
$k$ local power generation technology (power generation technology includes coal power, gas power generation, wind power, solar power generation, etc.)
$C G ( k , t )$ the operating cost of local power generation technology k during the period t
$G S ( k , t )$ the power generation capacity of local power generation technology k during the period t
$N G ( k , t )$ the new capacity of local power generation technology k during the period t
$C ( G k , t n )$ the unit investment cost of the new capacity of local power generation technology k during the period t
$C e$ the environmental cost
$K r$ the environmental value of the pollutant
$V r$ the pollutant emissions
$Z$ the penalty cost due to excessive emissions
$G$ the cost of ecological restoration
$L S ( n , t )$ the amount of production for energy n during the period t
$I S M ( n , t )$ the amount of purchase for energy n during the period t
$E X S ( n , t )$ the amount of consumption for energy n during the period t
$D n , t$ the amount of forecast demand for energy n during the period t
$β n , t$ the upper limit of the production capacity of energy n during the period t
$G C ( k , t )$ the upper limit of the installed capacity of local power generation technology k during the period t
$h ¯ k , t$ the average annual running time of local power generation technology k during the period t
$η k$ energy conversion technology update rate
$ν n , t$ the upper limit of the supply capacity of energy n controlled in the government energy planning document during the period t
$K r − max$ the maximum allowable emissions specified by environmental policy
Ref. Research Object Bi-Level Objectives Solving Algorithm Qualitative/Quantitative
[9] Electricity, heat No Minimum urban electricity demand Urban energy demand forecasting algorithm Quantitative
[10] Influencing factors of energy demand No Minimum system cost PSO-LSSVR Quantitative
[11] Natural gas, crude petroleum No Minimum system cost A novel Markov approach based on quadratic programming Quantitative
[12] Electric demand No Minimum annual total cost EDFS computing system Quantitative
[13] Coal, petroleum, gas, and electricity No Minimum system cost Improved particle swarm optimization Quantitative
[14] Coal, petroleum, natural gas, and renewable energy No Minimum system cost Multiple linear regression, BP neural network Qualitative and
[15] Urban energy value system No Minimum system cost Mixed integer linear programming Quantitative
[16] Coal, petroleum, natural gas, electricity, renewable No Minimum system cost Life cycle assessment Quantitative
[17] Urban energy system No Minimum system cost Fuzzy utility function Qualitative
[18] urban energy system planning and design tools No / / Qualitative
[19] Urban energy equipment configuration planning No Minimum system cost MINLP model and GAMS Quantitative
[20] Electricity No Maximum economic benefits Technology learning mechanism Quantitative
[21] Urban energy system planning and modeling approaches No / / Qualitative
Minimum total energy
[22] Coal, petroleum, natural gas, and electricity, No cost, environmental impact, and total energy Life cycle assessment Quantitative
[23] Urban energy system No Maximum economic benefits Hamiltonian directed graph Qualitative and
[24] Energy supply stability No Maximum urban energy system stability Ridge regression analysis Qualitative and
[25] Urban energy system No / Ecological network analysis Qualitative and
[26] Raw coal, coal products, and natural gas No / Improved ecological network analysis framework Qualitative and
[27] Electric power system No / Nonlinear dynamics Qualitative
[28] Power systems No / Balanced empirical Gramian Qualitative
This Coal, petroleum, natural gas, and renewable energy Yes Minimum economic costs and environmental Nonlinear system dynamics and IMFO Qualitative and
paper cost quantitative
Energy Price
Coal (yuan/t standard coal) 839.98
Petroleum (yuan/t standard coal) 3009.04
Natural Gas (yuan/t standard coal) 2701.15
Renewable Energy (yuan/t standard coal) Wind: 1410.89
Solar: 1856.44
Table 3.
Pollutant emissions and environmental cost factors (Unit standard coal) [
Pollutants SO[2] NO[x] CO[2] CO
Coal (kg/t) 18 8 1731 0.26
Emission Natural Gas (kg/10^6m^3) 11.6 0.0062 2.01 0
Petroleum (kg/t) 12 10.1 1592 0.33
Environmental value (yuan/kg) 6.00 8.00 0.023 1.00
M 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8
R^2 0.81503 0.845683 0.916619 0.855389 0.993331 0.995554 0.99678 0.996837
$ζ$ −1.174 −0.791 −0.5343 −0.9564 −0.1711 −0.0312 0.0911 0.1253
$a 1$ 0.1081 0.0778 0.0636 0.1333 0.0496 0.0456 0.0426 0.0466
$a 1$ $a 2$ $b 1$ $b 2$ $b 3$ $c 1$ $c 2$ $c 3$ $d 1$ $d 2$ $d 3$ $M$ $N$
0.0466 0.15 0.06 0.082 0.06 0.2 0.5 0.4 0.1 0.06 0.13 1.8 1
No. Coal (10,000 Tons of Standard Natural Gas (10,000 Tons of Petroleum (10,000 Tons of Standard Renewable Energy (10,000 Tons of Economic Cost (Billion Environmental Cost (Billion
Coal) Standard Coal) Coal) Standard Coal) Yuan) Yuan)
1 0.602982133 0.322441807 0.018182392 0.554282462 441.3970003 72.76593041
2 0.531510779 0.363596454 0.017329108 0.600545082 395.0105866 84.32441609
3 1.223222143 0.084486551 0.000717157 0.520514905 737.6377162 176.0332609
4 1.23253118 0.083563165 0.000756146 0.629394813 743.4596253 192.4520151
5 1.225144883 0.082584003 0.000797321 0.730605559 739.2684847 189.8653556
6 1.207153448 0.083162826 0.000832799 0.809391871 728.6800754 150.5531197
7 1.179411369 0.086001846 0.000867776 0.881071872 712.2418522 140.2292337
8 1.142718874 0.091766228 0.000901193 0.945140393 690.4312409 132.3944984
9 1.097732662 0.100890899 0.000931912 1.001174452 663.6385144 127.3137253
10 1.031886602 0.11698691 0.000964226 1.058524505 624.3604742 121.5998734
11 0.955122458 0.138414925 0.00098856 1.102813626 578.5061408 117.4752378
12 0.431885409 0.400346589 0.016227261 0.624552446 330.6338092 193.6766361
13 0.307915537 0.430287245 0.014855649 0.623654008 250.4515677 186.1370679
14 0.030632606 0.452986089 0.011700065 0.553422303 70.5408957 159.933346
15 0.163769308 0.449337899 0.013225338 0.59653623 156.9980795 168.4352068
16 0.187650579 0.336434923 0.000835972 1.042430204 117.9765388 181.4745739
17 0.771152721 0.194787951 0.001005602 1.150942565 468.4142455 99.45308016
18 0.598403545 0.246772582 0.00098234 1.150005826 364.8346072 65.09821296
19 0.868065361 0.16467288 0.001002937 1.133707255 526.4402196 103.1215876
20 0.403222837 0.297291733 0.0009248 1.113912957 247.6206412 192.4216718
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Wang, Y.; Liu, C.; Cai, C.; Ma, Z.; Zhou, M.; Dong, H.; Li, F. Bi-Level Planning Model for Urban Energy Steady-State Optimal Configuration Based on Nonlinear Dynamics. Sustainability 2022, 14, 6485.
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Solving log-determinant optimization problems by a Newton-CG primal proximal point algorithm
We propose a Newton-CG primal proximal point algorithm for solving large scale log-determinant optimization problems. Our algorithm employs the essential ideas of the proximal point algorithm, the
Newton method and the preconditioned conjugate gradient solver. When applying the Newton method to solve the inner sub-problem, we find that the log-determinant term plays the role of a smoothing
term as in the traditional smoothing Newton technique. Focusing on the problem of maximum likelihood sparse estimation of a Gaussian graphical model, we demonstrate that our algorithm performs
favorably comparing to the existing state-of-the-art algorithms and is much more preferred when a high quality solution is required for problems with many equality constraints.
National University of Singapore, September, 2009.
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Power Series and Estimation of Integrals - Calculus | Socratic
Power Series and Estimation of Integrals
Key Questions
• Assuming that you know that the power series for $\sin x$ is:
#sinx=sum_(n=1)^infty ((-1)^(n-1)x^(2n-1))/(2n-1)=x-(x^3)/(3!)+(x^5)/(5!)+...#
then we can answer this fairly quickly. If not, perhaps that can be a separate question!
So, if:
Which can be re-written as:
So then:
#int_0^0.01 sinx^2=int_0^0.01 x^2-1/(3!)int_0^0.01x^6+1/(5!)int_0^0.01x^10...#
When we plug in zero for $x$, all the terms will disappear. So all you have to do is plug in $0.01$ for x out to however many terms you want.
Hope this helps!
• Since
#int_0^{0.01}e^{x^2}dx=int_0^{0.01} (1+x^2+x^4/{2!}+cdots)dx#
$= {\left[x + {x}^{3} / 3 + {x}^{5} / \left\{10\right\} + \cdots\right]}_{0}^{0.01}$
$= 0.01 + {\left(0.01\right)}^{3} / 3 + {\left(0.01\right)}^{5} / 10 + \cdots$
$\approx 0.01$
I hope that this was helpful.
• Explanation:
Replace f by its Taylor expansion. f is now a sum
$f = \sum {a}^{n} {x}^{n}$
Integrate term by term
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italian loaded fries
The more often the interest is compounded, the greater the return will be. for a 30-year mortgage, a bank may charge 5% interest per year. Bank Statement Mortgage Interest Rates Definition Economics
Business In As you can see, Bank Statement Mortgage Interest Rates Definition Economics Business In has some parts that you need to include when you write the letter. Choose from 500 different sets
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increase. Although there is no single rate of interest in an economy, there are some principles which help up understand how interest rates are determined. (There are such things as negative interest
rates, where you instead get paid to borrow money, but these are rare.) low 0.65 percent. In the early 1960s, a one-year U.S. Treasury bond paid an interest rate of a little over 2.5 percent. Low
interest rates have been part of the Fedâ s monetary policy since 2007, when they were put in place for a post-recession recovery effort. Hayekâ s theory posits the natural interest rate as an
intertemporal price; that is, a price that coordinates the decisions of savers and investors through time. To compensate the business, the bank pays 6% interest into the account annually. Bank
Statement Mortgage Interest Rates Definition Economics Business In As you can see, Bank Statement Mortgage Interest Rates Definition Economics Business In has some parts that you need to include when
you write the letter. The interest is charged monthly on the principal including accrued interest from the previous months. However, there are different ways of measuring those interest payments, and
many savvy investors specifically want to know about an accountâ s real interest rate. Savings accounts and CDs use compounded interest. Both on paper and in real life, there is a solid relationship
between economics, public choice, and politics. Definition of discount rate, definition at Economic Glossary CODES (2 days ago) The discount rate is the interest rate the Federal Reserve System
charges for these loans. As the lending time increases, however, the disparity between the two types of interest calculations grows. Reducing the interest rates will encourage people and firms to
spend more money. Both on paper and in real life, there is a solid relationship between economics, public choice, and politics. Businesses also have limited access to capital funding through debt,
which leads to economic contraction. The economy is one of the major political arenas after all. Compound interest also called interest on interest, is applied to the principal but also on the
accumulated interest of previous periods. The difference between the total repayment sum and the original loan is the interest charged. In short, from the borrower’s point of view it is the ‘cost’ of
borrowing, and from the lender’s point of view it is the reward for lending. Or, to put it into an even simpler way, the rate of interest is the price of money. And, here they are: Part 1 They also
indicate the return on savings/bonds. When the borrower is considered to be low risk by the lender, the borrower will usually be charged a lower interest rate. Some lenders prefer the compound
interest method, which means that the borrower pays even more in interest. In the case of a large asset, such as a vehicle or building, the lease rate may serve as the interest rate. If the term of
the loan was for 20 years, the interest payment will be: Simple interest=$300,000à 15%à 20=$900,000\textbf{Simple\ interest} = \$300,000\times 15\%\times20=\$900,000Simple interest=$300,000Ã
15%à 20=$900,000. Compounding is the process in which an asset's earnings, from either capital gains or interest, are reinvested to generate additional earnings. The extra 5% interest earned is
enough to offset the 5% future depreciation of the Can$. There are two kinds of interest, simple interest and compound interest. The interest rate is the amount a lender charges for the use of assets
expressed as a percentage of the principal. In our example above, 15% is the APR for the mortgagor or borrower. Deflation does not cause banks to increase their interest rates. This is because a ...
Externalities Question 1 A steel manufacturer is located close to a large town. Also, interest rates tend to rise with inflation. Interest rates are the price you pay to borrow money, or, on the flip
side, the payment you receive when you lend money. A personal loan allows you to borrow money and repay it over time. However, some loans use compound interest, which is applied to the principal but
also to the accumulated interest of previous periods. Interest Types and Types of Interest Rates: Not all types of loans earn the same rate of interest. Ceteris paribus (all else being equal), loans
of longer duration and loans with more risk (that is, loans that are less likely to be paid off) are associated with higher interest rates. The cycle occurs when the market rate of interest (that is,
the one prevailing in the market) diverges from this natural rate of interest. Like any interest rate, when it goes up (or down) it discourages (or encourages) borrowing. If a company secures a $1.5
million loan from a lending institution that charges it 12%, the company must repay the principal $1.5 million + (12% x $1.5 million) = $1.5 million + $180,000 = $1.68 million. For example, the
interest rate on credit cards is quoted as an APR. As of this writing, the rate is a very low 0.65 percent. In effect, savers lend the bank money, which, in turn, provides funds to borrowers in
return for interest. The Economics Glossary defines interest rate as: "The interest rate is the yearly price charged by a lender to a borrower in order for the borrower to obtain a loan. The length
of the loan period (the ‘term’). An interest rate is either the cost of borrowing money or the reward for saving it. Ultra low interest rates in the UK from 2009-2014 The Bank of England started
cutting monetary policy interest rates in the autumn of 2008 as the credit crunch was starting to bite and business and consumer confidence was taking a huge hit. The Fed uses these tools to control
liquidity in the financial system. Interest is calculated as a percentage of the money borrowed. The bank also assumes that at the end of the second year, the borrower owes the principal plus the
interest for the first year plus the interest on interest for the first year. The money to be repaid is usually more than the borrowed amount since lenders require compensation for the loss of use of
the money during the loan period. Given the fact that there are many sources of funds for lending and different types of borrower with different reasons for borrowing, there is a complex structure of
interest rates in a modern economy. What the Annual Percentage Rate (APR) Tells You, Businesses also have limited access to capital funding. To combat inflation, banks may set higher reserve
requirements, tight money supply ensues, or there is greater demand for credit. Compound interest is the interest on a loan or deposit calculated based on both the initial principal and and the
accumulated interest from previous periods. After 20 years, the lender would have made $45,000 x 20 years = $900,000 in interest payments, which explains how banks make their money. This spending
fuels the economy and provides an injection to capital markets leading to economic expansion. Interest rates are generally framed as percentages. The principal is the amount of a loan or total credit
extended (like on a credit card.) For example, if one borrows $1,000 at 3% interest, the interest is â ¦ Higher interest rates have various economic effects: Higher interest rates tend to moderate
economic growth. It is calculated as a percentage of the amount borrowed or saved. This is the rate of return that lenders demand for the ability to borrow their money. Risk is typically assessed
when a lender looks at a potential borrower's credit score, which is why it's important to have an excellent one if you want to qualify for the best loans. The relationship between interest rates and
aggregate demand is a crucial topic within macroeconomics, which is the study of economics on a large scale. While governments prefer lower interest rates, a reason why the U.K. may never switch to
the euro, they eventually lead to market disequilibrium where demand exceeds supply causing inflation. Commerical banks charge a higher interest rate â ¦ In a free market economy, interest rates are
subject to the law of supply and demand of the money supply, and one explanation of the tendency of interest rates to be generally greater than zero is the scarcity of â ¦ All other interest rates
are based on that rate. The offers that appear in this table are from partnerships from which Investopedia receives compensation. The assets borrowed could include cash, consumer goods, or large
assets such as a vehicle or building. Even firms will be encouraged to expand as the cost of capital is cheap, they will find it easier to raise funds. In economics, the rate of interest is the price
of credit, and it plays the role of the cost of capital. Conventional interest is interest at a rate that has been set and agreed upon by â ¦ The current and expected inflation rate (as this alters
the real value of interest rates). Other loans can be used for buying a car, an appliance, or paying for education. The interest rate is the cost of debt for the borrower and the rate of return for
the lender. The interest charged is applied to the principal amount. Interest rates are the cost of borrowing money. 1  You borrow money from banks when you take out a home mortgage. Since
interest rates on savings are low, businesses and individuals are more likely to spend and purchase riskier investment vehicles such as stocks. Largest Retail Bankruptcies Caused By 2020 Pandemic,
Identifying Speculative Bubbles and Its Effect on Markets, Explaining The Disconnect Between The Economy and The Stock Market, Consumer Confidence Compared to Q2 Job Growth, Alternatives to GDP in
Measuring Countries. For loans, the interest rate is applied to the principal, which is the amount of the loan. Learn interest rates economics with free interactive flashcards. - Definition, Data and
Forecasts Policy Interest Rate (%) The policy interest â ¦ Interest rates on consumer loans are typically quoted as the annual percentage rate (APR). When the cost of debt is high, it discourages
people from borrowing and slows consumer demand. Explaining The K-Shaped Economic Recovery from Covid-19. During production it emits sulphur which creates an external cost to the local community.
Real interest rates somehow adjust the nominal ones to keep inflation into account. As loans become cheaper, more people will be interested in taking loans and purchasing houses and cars. For
instance if inflation was 15%, in the previous example the real interest rate can be said to be 20%-15% = 5%, in a simplified way of computation. Interest is the monetary charge for the privilege of
borrowing money, typically expressed as an annual percentage rate. Simple vs. Interest rate is the percentage of the face value of a bond or the balance in a deposit account that you receive as
income on your investment. An annual interest rate of 15% translates into an annual interest payment of $45,000. A nationâ s aggregate demand represents the value of that nationâ s goods and
services at a particular price point. The stock market suffers since investors would rather take advantage of the higher rate from savings than invest in the stock market with lower returns. When
inflation occurs, interest rates increase, which may relate to Walras' law. Interest Rate in Argentina averaged 62.10 percent from 1979 until 2020, reaching an all time high of 1389.88 percent in
March of 1990 and a record low of 1.20 percent in March of 2004. While interest rates represent interest income to the lender, they constitute a cost of debt to the borrower. Interest rates are
normally expressed as a % of the total borrowed, e.g. If interest rates are 5% higher in Canada, investors will keep on investing until the exchange rate has fallen by 5% (Can$ has appreciated by
5%). The annual percentage yield (APY) is the interest rate that is earned at a bank or credit union from a savings account or certificate of deposit (CD). Nominal rates are the quoted rate on the
loan, such as 4%, whereas ‘real’ interest rates are the nominal rate adjusted for inflation. Interest is the monetary charge for the privilege of borrowing money, typically expressed as an annual
percentage rate (APR). When the Fed reduces the reserve requirement, it's exercising expansionary monetary policy. The interest owed when compounding is higher than the interest owed using the simple
interest method. If you ever see "speculation" in this context, be sure to pay attention. An APR is defined as the annual rate charged for borrowing, expressed as a single percentage number that
represents the actual yearly cost over the term of a loan. Interest Rates explained Interest rates reflect the cost of borrowing. Does Public Choice Theory Affect Economic Output? The examples above
are calculated based on the annual simple interest formula, which is: Simple interest=principalà interest rateà time\textbf{Simple\ interest} = \text{principal}\times\text{interest rate}\times\
text{time}Simple interest=principalà interest rateà time. The multiplier effect - definition The multiplier effect indicates that an injection of new spending (exports, government spending or
investment) can lead to a larger increase in final national income (GDP). An interest rate is the reward for saving and the cost of borrowing expressed as a percentage of the money saved or borrowed.
A country's central bank sets the interest rate, which each bank use to determine the APR range they offer. The interest rate charged by banks is determined by a number of factors, such as the state
of the economy. Economies are often stimulated during periods of low-interest rates because borrowers have access to loans at inexpensive rates. Interest rates apply to most lending or borrowing
transactions. Higher interest rates tend to reduce inflationary pressures and cause an appreciation in the exchange rate. The interest rate is the amount charged on top of the principal by a lender
to a borrower for the use of assets. Many have filed for bankruptcy, with an ... Identifying Speculative Bubbles and Its Effect on Markets Speculation plays an interesting role in economics and one
that drastically affects markets. Individuals borrow money to purchase homes, fund projects, launch or fund businesses, or pay for college tuition. A loan that is considered high risk will have a
higher interest rate. Consumer loans typically use an APR, which does not use compound interest. The APY is the interest rate that is earned at a bank or credit union from a savings account or
certificate of deposit (CD). Many economies are at the brink of collapse, as companies struggle to stay afloat. This interest rate takes compounding into account. At any one time there are a variety
of different interest rates operating within the external environment; for example: Interest rates on savings in bank and other accounts Does Public Choice Theory Affect Economic Output? The interest
earned on these accounts is compounded and is compensation to the account holder for allowing the bank to use the deposited funds. The bank assumes that at the end of the first year the borrower owes
the principal plus interest for that year. The economy is one of the major political arenas â ¦ For shorter time frames, the calculation of interest will be similar for both methods. A lump sum by a
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Java How To: Calculate Sum of Array Elements - CodeLuckyJava How To: Calculate Sum of Array Elements - CodeLucky
Java, a versatile and powerful programming language, offers several methods to calculate the sum of array elements. This essential operation is frequently used in various applications, from simple
data analysis to complex statistical computations. In this comprehensive guide, we'll explore multiple approaches to summing array elements in Java, each with its own advantages and use cases.
1. Using a Traditional For Loop
The most straightforward method to calculate the sum of array elements is using a traditional for loop. This approach is simple, easy to understand, and works well for both primitive arrays and
object arrays.
public static int sumArrayTraditional(int[] arr) {
int sum = 0;
for (int i = 0; i < arr.length; i++) {
sum += arr[i];
return sum;
This method iterates through each element of the array, adding it to the sum variable. Let's see it in action:
int[] numbers = {1, 2, 3, 4, 5};
int result = sumArrayTraditional(numbers);
System.out.println("Sum: " + result); // Output: Sum: 15
🔑 Key Takeaway: The traditional for loop method is versatile and can be easily modified to handle more complex summing scenarios.
2. Using an Enhanced For Loop (For-Each Loop)
Introduced in Java 5, the enhanced for loop (also known as the for-each loop) provides a more concise and readable way to iterate through arrays and collections.
public static int sumArrayEnhanced(int[] arr) {
int sum = 0;
for (int num : arr) {
sum += num;
return sum;
This method is particularly useful when you need to iterate through all elements of the array without needing the index. Here's how to use it:
int[] numbers = {10, 20, 30, 40, 50};
int result = sumArrayEnhanced(numbers);
System.out.println("Sum: " + result); // Output: Sum: 150
💡 Pro Tip: The enhanced for loop is generally preferred when you don't need to modify the array or use the index in your calculations.
3. Using Java 8 Streams
Java 8 introduced the Stream API, which provides a more functional approach to handling collections of data. This method is particularly useful when working with large datasets or when you want to
perform additional operations along with summing.
import java.util.Arrays;
public static int sumArrayStream(int[] arr) {
return Arrays.stream(arr).sum();
This concise method leverages the power of streams to sum the array elements. Here's how to use it:
int[] numbers = {5, 10, 15, 20, 25};
int result = sumArrayStream(numbers);
System.out.println("Sum: " + result); // Output: Sum: 75
🚀 Advanced Usage: Streams can be combined with other operations for more complex calculations:
int sumOfEvenNumbers = Arrays.stream(numbers)
.filter(n -> n % 2 == 0)
System.out.println("Sum of even numbers: " + sumOfEvenNumbers); // Output: Sum of even numbers: 30
4. Using Java's Arrays.stream() with reduce()
For those who prefer a more functional programming style, Java's reduce() method provides a powerful way to perform cumulative operations on arrays.
import java.util.Arrays;
public static int sumArrayReduce(int[] arr) {
return Arrays.stream(arr).reduce(0, (a, b) -> a + b);
This method uses the reduce() operation to accumulate the sum. Here's how to use it:
int[] numbers = {2, 4, 6, 8, 10};
int result = sumArrayReduce(numbers);
System.out.println("Sum: " + result); // Output: Sum: 30
🧠 Deep Dive: The reduce() method takes two arguments: an identity value (0 in this case) and a BinaryOperator<T> that defines how to combine elements.
5. Using IntStream for Primitive int Arrays
When working specifically with primitive int arrays, IntStream provides optimized methods for summing.
import java.util.stream.IntStream;
public static int sumArrayIntStream(int[] arr) {
return IntStream.of(arr).sum();
This method is both concise and efficient for int arrays. Here's how to use it:
int[] numbers = {1, 3, 5, 7, 9};
int result = sumArrayIntStream(numbers);
System.out.println("Sum: " + result); // Output: Sum: 25
⚡ Performance Tip: IntStream is optimized for primitive int operations and can be more efficient than using the general Stream<Integer>.
6. Using Apache Commons Math Library
For more advanced mathematical operations, including summing arrays, the Apache Commons Math library provides robust and well-tested implementations.
First, add the Apache Commons Math dependency to your project:
Then, you can use the ArrayUtils class to sum your array:
import org.apache.commons.math3.stat.StatUtils;
public static double sumArrayApacheCommons(double[] arr) {
return StatUtils.sum(arr);
Here's how to use it:
double[] numbers = {1.5, 2.5, 3.5, 4.5, 5.5};
double result = sumArrayApacheCommons(numbers);
System.out.println("Sum: " + result); // Output: Sum: 17.5
📚 Library Benefits: Apache Commons Math provides a wide range of statistical operations beyond simple summing, making it valuable for more complex mathematical tasks.
7. Handling Large Arrays with BigInteger
When dealing with very large numbers or arrays that might result in integer overflow, using BigInteger is a safe approach.
import java.math.BigInteger;
public static BigInteger sumArrayBigInteger(int[] arr) {
BigInteger sum = BigInteger.ZERO;
for (int num : arr) {
sum = sum.add(BigInteger.valueOf(num));
return sum;
This method ensures that you can sum arrays without worrying about overflow. Here's how to use it:
int[] largeNumbers = {Integer.MAX_VALUE, Integer.MAX_VALUE, Integer.MAX_VALUE};
BigInteger result = sumArrayBigInteger(largeNumbers);
System.out.println("Sum: " + result); // Output: Sum: 6442450941
🛡️ Safety First: Using BigInteger prevents overflow errors when dealing with very large numbers or long arrays.
Performance Considerations
When choosing a method to sum array elements, consider the following performance aspects:
1. Array Size: For small arrays, the difference between methods is negligible. For large arrays, stream-based methods may have some overhead.
2. Primitive vs. Object Arrays: Methods like IntStream are optimized for primitive arrays and can be faster than general-purpose streams for Integer objects.
3. Parallel Processing: For very large arrays, consider using parallel streams:
int sum = Arrays.stream(veryLargeArray).parallel().sum();
1. JVM Warm-up: In benchmarking, remember that the JVM needs time to warm up. Initial runs may be slower than subsequent ones due to JIT compilation.
Summing array elements in Java is a fundamental operation with multiple implementation options. From simple loops to functional programming approaches and specialized libraries, each method has its
place depending on your specific needs, performance requirements, and coding style preferences.
• Use traditional or enhanced for loops for simplicity and readability.
• Leverage Java 8+ streams for a functional approach and additional data processing.
• Consider IntStream for optimized operations on primitive int arrays.
• Use Apache Commons Math for advanced statistical operations.
• Employ BigInteger for handling potential overflow with very large numbers.
By understanding these different methods, you can choose the most appropriate approach for your specific use case, balancing factors such as code readability, performance, and functionality.
Remember, the best method often depends on the context of your application. Always profile your code with real-world data to ensure you're using the most efficient approach for your specific | {"url":"https://codelucky.com/java-sum-array/","timestamp":"2024-11-06T10:38:21Z","content_type":"text/html","content_length":"143262","record_id":"<urn:uuid:228abf51-63da-415e-97c9-eb2d4788bd77>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00686.warc.gz"} |
Kinetic Molecular Theory Summary - UCalgary Chemistry Textbook
Key Concepts and Summary
The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible
volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules
of a gas exhibit a range of velocities, the distribution of these velocities being dependent on the temperature of the gas and the mass of its molecules.
Key Equations | {"url":"https://chem-textbook.ucalgary.ca/chapter-9-main/kinetic-molecular-theory-summary/","timestamp":"2024-11-02T18:05:40Z","content_type":"text/html","content_length":"66611","record_id":"<urn:uuid:435dd740-7114-421a-852c-4e9669002d9a>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00655.warc.gz"} |
Go to the source code of this file.
subroutine dlasq1 (N, D, E, WORK, INFO)
DLASQ1 computes the singular values of a real square bidiagonal matrix. Used by sbdsqr.
Function/Subroutine Documentation
subroutine dlasq1 ( integer N,
double precision, dimension( * ) D,
double precision, dimension( * ) E,
double precision, dimension( * ) WORK,
integer INFO
DLASQ1 computes the singular values of a real square bidiagonal matrix. Used by sbdsqr.
Download DLASQ1 + dependencies
[TGZ] [ZIP] [TXT]
DLASQ1 computes the singular values of a real N-by-N bidiagonal
matrix with diagonal D and off-diagonal E. The singular values
are computed to high relative accuracy, in the absence of
denormalization, underflow and overflow. The algorithm was first
presented in
"Accurate singular values and differential qd algorithms" by K. V.
Fernando and B. N. Parlett, Numer. Math., Vol-67, No. 2, pp. 191-230,
and the present implementation is described in "An implementation of
the dqds Algorithm (Positive Case)", LAPACK Working Note.
N is INTEGER
[in] N The number of rows and columns in the matrix. N >= 0.
D is DOUBLE PRECISION array, dimension (N)
On entry, D contains the diagonal elements of the
[in,out] D bidiagonal matrix whose SVD is desired. On normal exit,
D contains the singular values in decreasing order.
E is DOUBLE PRECISION array, dimension (N)
On entry, elements E(1:N-1) contain the off-diagonal elements
[in,out] E of the bidiagonal matrix whose SVD is desired.
On exit, E is overwritten.
[out] WORK WORK is DOUBLE PRECISION array, dimension (4*N)
INFO is INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
> 0: the algorithm failed
= 1, a split was marked by a positive value in E
[out] INFO = 2, current block of Z not diagonalized after 100*N
iterations (in inner while loop) On exit D and E
represent a matrix with the same singular values
which the calling subroutine could use to finish the
computation, or even feed back into DLASQ1
= 3, termination criterion of outer while loop not met
(program created more than N unreduced blocks)
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
September 2012
Definition at line 109 of file dlasq1.f. | {"url":"https://netlib.org/lapack/explore-html-3.4.2/d4/d2d/dlasq1_8f.html","timestamp":"2024-11-06T20:06:08Z","content_type":"application/xhtml+xml","content_length":"12120","record_id":"<urn:uuid:e2b50845-2c2f-451c-b69b-df7af352f950>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00208.warc.gz"} |
Selective Coordination Enforcement: Overcurrent Protective Device Basics
The Basics of Selective Coordination
Selective coordination is achieved when overcurrent protective devices are chosen such that whenever an overcurrent occurs only the nearest upstream overcurrent protective device (OCPD) opens to
interrupt the overcurrent. In figure 1, this means if a fault occurs on the load-side of OCPD 1 (100 A), only that device opens. Neither OCPD 2 (200 A) nor OCPD 3 (800 A) opens.
Figure 1. One-line diagram shows fault on branch circuit 1.
Selective coordination for the full range of overcurrents possible in a system is anNEC^®requirement for certain systems (Sections 700.27, 701.18, 708.54, 620.62, and 517.26) where continuity of
power for loads is vital for life safety. A short-circuit current analysis (study) is usually considered part of the required documentation and its purpose is to provide the maximum available
short-circuit currents throughout the distribution system. However, for a few selective coordination analysis methods, a short-circuit current study is not necessary. This fact will be mentioned when
those methods are covered in this article.
When a short-circuit study is performed, the available short-circuit current is determined at the line-side terminals of each overcurrent protective device. This information is also necessary to
verify the OCPDs have adequate interrupting rating (IR) to comply withNEC110.9 and that the equipment has adequate short-circuit current ratings (SCCR) to comply withNEC110.10, 409.101, 440.4(B), and
670.3(A). This concept is depicted in figure 2. The X[1], X[2]and X[3]represent the available short-circuit current at the line-side terminals of OCPD 1, OCPD 2, and OCPD 3, respectively. The values
of 15,000 A, 25,000 A, and 30,000 A for X[1], X[2]and X[3], respectively, were selected for illustrative purposes only. These values are in RMS symmetrical amperes.
Figure 2: One-line diagram showing available short-circuit currents. ISCA represents short-circuit current available.
A typical selective coordination requirement:
NEC700.27 Coordination.
“Emergencysystem(s) overcurrent devices shall be selectively coordinated with all supply side overcurrent protective devices.” (with two exceptions)”
Reading the specific language of the requirement means: OCPD 1 must selectively coordinate with OCPD 2andOCPD 3 for all overcurrents up to the maximum available at X[1](15,000 A). OCPD 2 must
selectively coordinate with OCPD 3 for all overcurrents up to the maximum available at X[2](25,000 A).
Selective Coordination Utilizing Fuses
The basics in evaluating circuits for selective coordination using fuses are relatively simple. There are two approaches:
a. Fuse manufacturer’s published selectivity ampere rating ratios.
b. Interpret time-current curves.
Example a: Fuse Selectivity Ampere Rating Ratios
The simplest and preferred method for fuses is to adhere to the fuse manufacturer’s published selectivity ampere rating ratios. No need to interpret time-current curves. Valid for any overcurrents up
to 200 kA (200,000 amps) or fuse interrupting rating, whichever is less. Today most current-limiting, branch-circuit fuses have at least a 200 kA interrupting rating. With this method, the fuses must
be of the same manufacturer. These fuse amp ratio tables are based on testing by each fuse manufacturer and there is no published data for mixing fuses of different manufacturers. The selectivity amp
rating ratios vary by the type of fuses. One fuse type may require only a 2:1 amp rating ratio between a line-side fuse and a load-side fuse. If different types of fuses from the same manufacturer
are used line-side and load-side, then the ratio might be different, such as 3:1, 4:1, or 8:1. As with any method or table, be sure to read any applicable notes or footnotes.
Figure 3: Example to Evaluate Fuse Selective Coordination
Each fuse manufacturer publishes a selectivity ratio table specific to only their brands. Table 1 is for illustrative purposes and should not be used in actual studies.
Analysis for Figure 3 using Table 1
Investigate the circuit for fuses 1, 2, and 3 in figure 3 using the Selectivity Amp Rating Ratio Table method.
1. Check fuse 1 with fuse 2
• For the actual fuses, the line-side to load-side amp rating ratio is 400 A:100 A = 4:1
• Both fuse 1 and fuse 2 are Class J time-delay and Table 1 shows a ratio of 2:1 or greater is necessary.
• Therefore since the actual amp rating ratio for fuse 1 and fuse 2 (4:1) is equal or greater than 2:1 (Table 1), fuse 1 will selectively coordinate with fuse 2 for any overcurrent up to 200 kA.
2. Check fuse 1 with fuse 3
• Actual amp ratio is 800 A:100 A = 8:1
• 2:1 or greater is necessary (Table 1).
• Fuse 1 selectively coordinates with fuse 3 up to 200 kA.
3. Check fuse 2 with fuse 3
• Actual amp rating ratio is 800 A:400 A = 2:1
• 2:1 or greater is necessary (Table 1).
• Fuse 2 will selectively coordinate with fuse 3 for any overcurrent up to 200 kA.
Table 1: Fuse Selectivity Amp Rating Ratio Guide
Conclusion: This circuit path is selectively coordinated for any overcurrent up to 200 kA; therefore, X[1]and X[2]could be any value of available short-circuit currents up to 200 kA. Using this
method, there is no need to do a short-circuit current study if a quick assessment shows the short-circuit currents in the system will be 200 kA or less.
Example b: Fuse Selectivity Time-Current Curves
Figure 4: Interpreting time-current curve method
Use this method only for the portion of the curve that is visible on the time-current graph. The typical standard industry time-current curve has a vertical time axis from 0.01 seconds up to 300
seconds or 1000 seconds. However, fuses when operating in their current-limiting range, clear in less than 0.01 seconds. Depending on the fuse type, amp rating, and available fault current, the
clearing time can be less than 0.01 sec. (Time-current curves typically are not published for times less than 0.01 seconds.) It is simpler and more comprehensive to use the FuseSelectivity Ampere
Rating Ratios method illustrated previously rather than interpreting time-current curves.
In this example, use the fuse time-current curves in figure 4 to analyze selective coordination for the circuit path with the three fuses shown. Examining figure 4, there is no overlap of the fuses’
time-current curves. However, this does not mean that the fuses are selectively coordinated for the full range of overcurrents up to their interrupting rating. What can be interpreted from figure 4
is that the 90-A Class RK5 time-delay fuse is coordinated with the 200-A Class J time-delay fuse up to 2500 A and is coordinated with the 800-A Class L time-delay fuse up to 12,000 A. Also, the 200-A
Class J time-delay fuse is coordinated with the 800-A Class L fuse up to 12,000 A.
Conclusion: If this specific system has available short-circuit currents for X[1]of 2,500 A or less and X[2]of 12,000 A or less; then, this circuit is selectively coordinated.
Note: The conclusion does not mean these fuses are not selectively coordinated for higher fault currents; it just means we cannot draw that conclusion by interpreting the time-current curve method.
To illustrate this point, use the fuse selectivity amp rating ratio method and Table 1 to check figure 4 for fault currents up to 200 kA for X[1]and X[2]. The ratio required by Table 1 for Class J
time-delay fuse line-side of a load-side Class K5 time-delay fuse is 8:1. In the circuit for figure 4, the actual fuse ratio is 200 A:90 A = 2.2:1. So, using the ratio method, these two fuses would
not be selectively coordinated up to 200 kA. However, figure 4 does show these two fuses are selectively coordinated up to 2,500 A. Now investigate the 800-A and 200-A fuses using the ratio method.
The ratio required by Table 1 for Class L time-delay fuse line-side of a load-side Class J time-delay fuse is 2:1. In the circuit for figure 4, the actual fuse ratio is 800 A:200 A = 4:1. So, using
the ratio method, these two fuses are selectively coordinated up to 200 kA. This is a good example of why it is quicker and simpler to just use the fuse selectivity ampere ratio method rather than
plotting the time-current curves and interpreting.
Figure 5: Circuit breaker time-current curves do not cross and therefore the circuit is selectively coordinated for available short-circuit currents up to the CB interrupting ratings.
Selective Coordination Basics for Circuit Breakers
The basics in evaluating circuits for selective coordination using circuit breakers (CBs) are also relatively simple. There are three approaches.
a. If the clearing time of all the CBs is greater than 0.01 seconds for the line-side available short-circuit current, the CB time-current curves cannot cross.
b. If the CB time-current curves cross, the available short-circuit current at the point of intersection is interpreted as the maximum short-circuit current to which selective coordination can be
achieved. There is an exception if CB manufacturers’ Selectivity Tables provide data verifying two specific CBs selectively coordinate to a higher short-circuit current level.
c. If clearing time of the CB is less than 0.01 seconds, then data is needed from the circuit breaker manufacturer’s table method discussed inb.
The following will give an example for methodsaandb.
Example a: CB Curves Do Not Cross
The easiest CB systems to interpret are those where the CB time-current curves do not cross. To achieve this “no overlap” situation, it normally requires the use of feeder and main CBs with
short-time delay settings and no instantaneous trip (or instantaneous override). The branch circuit CB can have an instantaneous trip. Figure 5 illustrates where such a system is selectively
coordinated for any overcurrent up to the interrupting rating of the CBs, which is 65 kA in this example. With this method, CBs of different manufacturers can be mixed.
Example b: CB Curves Cross
Figure 6. Selectively coordinated for available short-circuit currents up to where circuit breakers cross. 600-A CB instantaneous trip set at 5 times.
If the CB time-current curves cross, then there are two approaches to evaluate selective coordination. The first approach is to interpret the time-current curves such that two CBs are selectively
coordinated up to the available short-circuit current where the curves cross. The analysis of figure 6 is as follows:
• 20-A CB is selectively coordinated with the 225-A CB for available short-circuit currents up to 1800 A and with the 600-A CB for available short-circuit currents up to 2400 A.
• 225-A CB is selectively coordinated with the 600-A CB for available short-circuit currents up to 2400 A.
Conclusion: If the available short-circuit current at X[1]is less than 1800 A and at X[2]less than 2400 A, this circuit is selectively coordinated. If the available short-circuit current at X[1]is
greater than 1800 A, or at X[2], greater than 2400 A, this circuit is not selectively coordinated.
With this method, circuit breakers of different manufacturers can be mixed and evaluated for selective coordination. If the instantaneous trip of the circuit breaker is adjustable, then by setting
the adjustment higher or lower one can change the available short-circuit current at which the CBs selectively coordinate. In the figure 6 example, the 225-A CB is a fixed instantaneous trip, so it
cannot be changed. However, the 600-A CB has an adjustable instantaneous trip range from 5 to 10 times. In the figure 6 example, it was set on 5 times. If the 600-A CB instead had its instantaneous
trip set at 10 times, then the analysis would be as follows using figure 7:
• The 20-A CB is selectively coordinated with the 225-A CB for available short-circuit currents up to 1800 A and with the 600-A CB for available short-circuit currents up to 5200 A.
• The 225-A CB is selectively coordinated with the 600-A CB for available short-circuit currents up to 5200 A.
Figure 7. 600-A CB is set on IT=10X instantaneous trip, resulting in its being able to selectively coordinate up to a higher available short-circuit current versus figure 6 example.
If the available short-circuit current at X[1]is less than 1800 A and at X[2]less than 5200 A, this circuit is selectively coordinated. If the available short-circuit current X[1]is greater than 1800
A, or at X[2], greater than 5200 A, this circuit is not selectively coordinated.
Example: CB Selective Coordination Tables
The previous method of interpreting CB time-current curves has been an industry standard for decades. However, with the advent of theNECrequiring selective coordination for some systems where life
safety is vital, manufacturers have developed CB to CB selective coordination tables. These tables typically are based on testing and analysis, and the presentation format varies for each
manufacturer. Each CB manufacturer has tables for their CBs and the data is specific to type and amp rating of each circuit breaker. The tables assume CB instantaneous trips are set on the highest
settings. Table 2 is a fictitious CB to CB selective coordination table. The format is line-side CB across the top and load-side CB down the left side. Using Table 2, the following is an example for
the CBs in figure 8.
• Type B 20-A CB is selectively coordinated with the Type F 225-A CB for available short-circuit currents up to 2200 A and with the Type G 600-A CB for available short-circuit currents up to 15,000
• Type F 225-A CB is selectively coordinated with the Type G 600-A CB for available short-circuit currents up to 12,000 A.
Conclusion:If the available short-circuit current at X[1]is less than 2200 A and at X[2]less than 12,000 A, this circuit is selectively coordinated. If the available short-circuit current at X[1]is
greater than 2200 A or at X[2]greater than 12,000 A, this circuit is not selectively coordinated.
Figure 8. CB selective coordination table method
Figure 9. 150-A fuse is selectively coordinated with upstream 400-A CB having short-time delay (and no instantaneous override).
Note: Selective coordination tables are CB manufacturer specific and even specific for the type or style CB. The short-circuit current values at which line-side and load-side CBs selectively
coordinate vary widely between the manufacturers and even for different CB types of the same manufacturer. For instance, a manufacturer may have numerous types of 600-A frame molded-case CBs, as well
as insulated-case CBs and low-voltage power CBs in 600 A ratings. Therefore, if the CB table method is utilized, it is important to ensure that the selective coordination submittal be very specific
on the CB manufacturer, CB type, CB frame/amp rating. It is suggested that the submitter include the manufacturer’s selective coordination tables as part of the submittal documentation.
System with Mixture of Fuses and Circuit Breakers
If an upstream CB has only a short-time delay trip and no instantaneous trip, then selective coordination is assured if the fuse curve and CB curve do not cross (figure 9).
If the upstream CB has an instantaneous trip, it is not a simple matter to determine if a downstream fuse will be selectively coordinated. Even if the plot of the time-current curves for a downstream
fuse and an upstream circuit breaker with instantaneous trip show that the curves do not cross, selective coordination may not be possible beyond a certain fault current. This is in the fault region
where the fuse is clearing in less than 0.01 seconds. This is because the fuse may not clear the fault prior to unlatching of the upstream circuit breaker. The sure way to determine whether these two
devices will coordinate is to test the devices together. Figure 10 illustrates that the 200-A fuse is selectively coordinated up to 3500 A short-circuit current; beyond that fault current, it is not
certain unless there is test data.
If a fuse is upstream and a CB is downstream, at some point the fuse time-current characteristic crosses the CB time-current characteristic. For short-circuit currents at that cross-over point and
higher, the upstream fuse is not coordinated with the downstream CB. Figure 11 shows a 200-A fuse with downstream 20-A CB. Selective coordination is possible up to 2200 A, but beyond that point the
fuse characteristic crosses the CB curve. So beyond 2200 A, the 20-A CB is considered not to be selectively coordinated with the 200-A fuse.
Simple Suggestions to Implement Enforcement
Quick Short-Cut Checks
It is always good to have quick short-cuts to spot check on any process and there are some easy approximate short-cuts for selective coordination. If the system is fusible, simply use the fuse
manufacturer’s selectivity ratio table.
If the system consists of circuit breakers with instantaneous trips, such as molded case CBs, then there is a simple method to check whether selective coordination is achieved. For each CB, simply
multiply the instantaneous trip setting by the CB’s amp rating. The resulting product is the approximate short-circuit current at which a CB enters its instantaneous trip region. This simple method
is a practical, quick test in assessing if a system is selectively coordinated. There may be other means to determine higher values of fault current where CBs selectively coordinate (such as
manufacturers’ tables). This quick check is approximate in that you may have to assume an IT setting. The following illustrates this simple method without considering manufacturing tolerances.
Table 2. Circuit breaker selective coordination table
Figure 10. 200-A fuse and upstream 400-A CB with instantaneous trip are selectively coordinated up to 3500 A.
Figure 11. 20-A CB selectively coordinates with 200-A fuse up to 2200 A.
Figure 12. Illustrates simple method to check CB coordination without considering the manufacturing tolerances
Using figure 12, let’s assume an emergency circuit has three CBs: 400 A, 100 A, and 20 A.
a. Assume the 100-A CB has its instantaneous trip (IT) set at 8 times (8X) its amp rating. Therefore, for fault currents above 800 amps (100 A x 8 = 800 A), the 100-A CB will unlatch in its
instantaneous trip region, thereby opening.
b. Assume the 400-A CB has its instantaneous trip set at 8X its amp rating. Therefore, for fault currents above 3200 amps (400 A x 8 = 3200 A), the 400-A circuit breaker unlatches in its
instantaneous trip region, thereby opening.
Conclusion:The 20-A CB will selectively coordinate with the 100-A CB if X[1]is less than 800 A. The 100-A CB will selectively coordinate with the 400-A CB if X[2]is less than 3200 A.
Figure 13. Illustrates simple method to check CB coordination including considering the IT pickup manufacturing tolerance
For simplicity, the previous example did not include the manufacturing tolerances permitted for instantaneous trip pickup. UL 489 for molded-case CBs permits IT pickup tolerances of -20% and +30% of
the instantaneous trip setting. Now let’s review the previous stepsaandband then do the additional stepcbelow, which factors in the IT pickup tolerance; figure 13 provides the one-line:
c. If the instantaneous trip pick-up negative tolerance is factored in, then take 80% of the results in previous steps a and b (multiply the results by 0.8). The 100-A CB will open at 100 A x 8 x 0.8
= 640 A and the 400-A CB will open at 400 A x 8 x 0.8 = 2560 A.
Conclusion:If the available short-circuit current is less than 640 A (0.8 times 800 A) and 2560 A (0.8 times 3200) for X[1]and X[2], respectively, selective coordination is achieved. If the fault
current is greater for either X[1]or X[2], selective coordination is not achieved.
Additional Resources
• Two IAEI News articles found in the IAEI News magazine online www.iaei.org/magazine/:
“Overcurrent Protection Basics, Part I,”March-April 2007, author Tim Crnko: provides some basics on reading time-current curves, circuit breaker and fuse operation, and understanding time-current
curves for each.
“Selective Coordination – Responsibilities of the AHJ,” November-December 2007, author Mark Hilbert: provides information on what is selective coordination, the requirements, and AHJ role.
• “Keep The Power On for Vital Loads,”NEC Digest, December, 2007, author Evangelos Stoyas: provides insight why we have mandatory selective coordination and important aspects of the requirement.
Available to NFPA members at nfpa.org or from www.cooperbussmann.com/2/SelectiveCoordination.html
• NECA 700(2010)–Installing Overcurrent Protection to Achieve Selective Coordination at neca-neis.org. This newly developed standard has in-depth information on the “how to” of selective
coordination.• Circuit breaker companies have several tools including CB to CB selective coordination tables: the publications are Eaton 1A01200002E, SQD 0100DB0501R3/06, and GE DET-537B.
• Cooper Bussmann has several tools in their2008 SPD(Selecting Protective Devices) publication including sections on selective coordination for fuses or circuit breakers, why selective coordination
is mandatory, AHJ check list, simple point-to-point short-circuit current calculation method, and checking CB coordination by simplified method without time-current curves. In addition, a free simple
short-circuit current calculator program can be downloaded. Go to www.cooperbussmann.com/spd.
Tim Crnko | {"url":"https://iaeimagazine.org/2010/july2010/selective-coordination-enforcement-overcurrent-protective-device-basics/","timestamp":"2024-11-14T07:22:34Z","content_type":"text/html","content_length":"147810","record_id":"<urn:uuid:8cf0419a-ae8c-4130-b41e-61bc49b32b6a>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00338.warc.gz"} |
Large tender onto sundeck?
The friendliest place on the web for anyone who enjoys boating.
If you have answers, please help by responding to the unanswered posts.
I'm going to buy a 40-50ft trawler for liveaboard that can act as a roaming fishing mothership. I currently have a little 14ft CC catamaran skiff (Twin Vee) with a 50hp. It's an awesome 1 man setup
(and decent for 2); it handles offshore on calmer days and can float over shallow flats.
I can't really imagine being able to go smaller than that. Claimed dry weight is 650 -- so obviously in the 1000+lb range with motor and basic gear. Curious if it'd be feasible to occasionally davit
winch a boat that size onto a 40-50ft Trawler sundeck (or bow)? I imagine it'd be more doable on a 50 than a 40, so that's why I'm asking because if so, I'd be more inclined to buy a trawler on the
large end of my size range.
Last edited:
May 31, 2013
Vessel Name
Vessel Make
1982 41' President
My 41' President tender on the bow has a weight limit of 300 lbs which is lifted by a crane. I think the design needs to be verified by the mfg.
I have an 11' AB Aluminum hull RIB with a 20 HP 4 stroke Merc outboard.
I remove all gear prior to launching including gas tank and dive gear.
Jul 2, 2015
Vessel Name
Black Dog
Vessel Make
Formula 41PC
I would not want to put that much weight up on the top of my sundeck due to how much effect it would have on stability. Maybe on a 50’ boat.
Apr 12, 2016
Vessel Name
Vessel Make
2018 Hampton Endurance 658
You said "I'm going to buy a 40-50ft trawler..." but no hint of year, semi or full displacement, etc. Creating this capability in a new build is probably more easily solved. Doing it with at 30 year
old boat is another matter entirely. Besides the physical space issue, you have whether the various decks can support the weight of the tender and the crane. And what it does to your boat's
In this video:
The hosts talk about their boat's refit of the boat deck, including structural changes they had to make. Might be of interest to you. Good luck and happy fishing!
Creating this capability in a new build is probably more easily solved. Doing it with at 30 year old boat is another matter entirely.
It's gonna have to be a 30 year old boat.
Apr 12, 2016
Vessel Name
Vessel Make
2018 Hampton Endurance 658
It's gonna have to be a 30 year old boat.
(If this tender setup is your #1 priority) I think I would try to find a marine architect or two that knows this kind of work, and have them suggest the boat to look for.
Figured first step would be to see if anybody here chimes in that they do something similar -- and then take note the size/make of their vessel & tender system. I've seen 13 ft whalers on sundecks
and bows, so it's not a total fantasy out of left field.
Jan 13, 2017
Vessel Name
Vessel Make
Lagoon 380
I've seen several fishing boats out of Spanish Wells towing up to 6 fishing boats at a time in remote areas, like Cay Sal. This solves both the storage issue and the inevitable issue of lifting a
largish boat in heavy seas. And these guys have 50/60 footer mother ships. Just another idea.
Yeah I know it's possible to tow. I was just thinking in situations like locks or docking particularly single-handed, it would be really nice to have the option to not have to worry about the skiff
behind me. Also possibly preferable for a Bahamas crossing.
Nov 7, 2008
A 13.5’ Avon, center console with a 50 hp 2 stroke came with Hobo. Total weight was over 600 lbs. Lowering and raising her off the boat deck was an experience. The previous owners had her for 10
years and never had any issues. We sold her for something a little lighter but it worked for them.
Senior Member
Commercial Member
May 4, 2015
Vessel Name
Sea Q
Vessel Make
Westport Mc Queen
On our 90 footer we have a 19 foot inflatable tender with a 90 horse
it weighs over 2000 lb
When bringing in the tender the big boat lists about 5 degrees making it a two man job to swing in.
Also you have to make sure the tender is tied down tight
Even in the nicest weather
Things go bad quick and I have seen it rise off the chocks a foot in rollers off of Campbell River.
Personally we only needed a 15 footer
Sep 29, 2013
Vessel Name
Vessel Make
Defever 44, twin Perkins
How about an aluminum hulled RIB? For example you can get a 14' AB aluminum hull RIB with console that weighs 367 plus motor. If you can do without the console you can drop down to 247lbs for the
same size boat.
Nov 2, 2016
Vessel Make
Sold-GB 52 Europa, Queenship 59, Tolly 45
We had a 13ft Rendova with a 40 hp Yamaha on the boat deck of a 60 foot boat. About 1000 lbs. It is doable but requires two people to launch and retrieve. Forget it if there is significant wind/wave
action unless you want your dinghy come through the saloon windows. Make sure your tie downs can handle the substantial forces when you get into some bad weather. We know a couple that lost their
heavy dinghy from the boat deck crossing Georgia Strait. Did quite a bit of damage as it fell into the cockpit and then overboard. The captain was busy keeping the boat from broaching and could not
deal with the loose dinghy.
RIBs just aren't setup for fishing at all, plus a 14' RIB wouldn't have close to the space of my little cat (which is already a very compact fishing platform). Also, with motor and everything, it's
still gonna weigh close to 700lbs, so I don't see a big weight payoff for all that sacrifice.
Mar 15, 2010
Vessel Name
Sweet Pea
Vessel Make
Nimble Nomad 25' Trawler
I travel on a 52' sportfish(lot of open space on the bow) that we carry a 13' Boston Whaler on the bow. The boat came from the factory with a 1000 lb crane and the Whaler is probably pushing that
limit. As stated, the boat will list some when loading and its a 2 man job. And definitely not a job that you would attempt in rough weather.
Apr 12, 2016
Vessel Name
Vessel Make
2018 Hampton Endurance 658
Aug 22, 2011
I'd have a very good yard or a naval architect weigh in on a specific boat you are interested in. We launched a 13' Whaler off the boat deck of an 18'2" beam Hatteras frequently for many years; it
caused almost unnoticeable list and I even did it my klutzy self a few times. In bad conditions, I found retrieving to be much more of an issue than launching, logistics-wise. I knew some folks with
15'10" beam Hatteri that also launched Whalers and notice 2 things: list was more pronounced, and a 13 footer extended beyond the boat deck some. Keep beam in mind more than length! But in each case
the boat deck was designed to support a 1000# + tender.
Jul 1, 2016
Vessel Make
Milkraft 60 converted timber prawn trawler
A mates 65 fter carries 21ft with a 150 up top, he reckons roll is reduced in a seaway.
Jul 1, 2016
Vessel Make
Milkraft 60 converted timber prawn trawler
Apparently displacement hulls can carry a load up high
Aug 2, 2017
Vessel Name
Vessel Make
Ocean Alexander 54
I carry a 1000lbs dingy (12’ rib/60hp Suzuki). I have a 2000lbs crane which makes quick work of launching and recovering. The dingy use 5 quick lock straps that lock her sown tight enough to take on
any conditions I can handle. Problem is my boat is in the 50-60 foot range.
Jul 23, 2015
Vessel Name
Didi Mau
Vessel Make
Currently looking for next boat
Think it is no problem
I'm going to buy a 40-50ft trawler for liveaboard that can act as a roaming fishing mothership. I currently have a little 14ft CC catamaran skiff (Twin Vee) with a 50hp. It's an awesome 1 man
setup (and decent for 2); it handles offshore on calmer days and can float over shallow flats.
I can't really imagine being able to go smaller than that. Claimed dry weight is 650 -- so obviously in the 1000+lb range with motor and basic gear. Curious if it'd be feasible to occasionally
davit winch a boat that size onto a 40-50ft Trawler sundeck (or bow)? I imagine it'd be more doable on a 50 than a 40, so that's why I'm asking because if so, I'd be more inclined to buy a
trawler on the large end of my size range.
The previous owners of our boat used to carry a 4-person helicopter on the sun deck of our boat. Look in the archives Nader ocean Alexander with help to see a pic.
Mar 17, 2012
Vessel Name
Vessel Make
Ocean Alexander 50 Mk I
The PO of my boat had a 14' Novurania GRP hulled RIB, with Yamaha 50. I think it was about 800#. His davit was 120VAC, which lifted fine but was manual (uncontrolled) rotation. That led to some
'interesting' and at times dangerous wild swings of the RIB.
I could not change his old winch to 230VAC /50Hz. So now I use a Nick Jackson davit rated at 1500#. It has a 12V winch with hydraulic rotation and hydraulic ram lift on the boom also. My current
tender is a 14' AB RIB, with aluminium hull and a Honda 40. It weighs a bit less the the old rig. The RIB fits fine on my 50' boat, and the weight is not a problem at all on the 'boat deck' above the
salon. With hydraulic rotation it is a safe and easy one person job to launch and retrieve the RIB.
Jul 23, 2015
Vessel Name
Didi Mau
Vessel Make
Currently looking for next boat
Damn spell check
The previous owners of our boat used to carry a 4-person helicopter on the sun deck of our boat. Look in the archives Nader ocean Alexander with help to see a pic.
Should have read: look under archives , Ocean Alexander with helo, filed under the ocean Alexander section | {"url":"https://www.trawlerforum.com/threads/large-tender-onto-sundeck.37795/","timestamp":"2024-11-07T19:15:55Z","content_type":"text/html","content_length":"197570","record_id":"<urn:uuid:36e5ebdb-1fbc-4676-825e-4d5eaaff7274>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00564.warc.gz"} |
Turbulent flow in a pipe
Algorithmics for Engineering Modeling
Master of Marine Technology: Hydrodynamics for Ocean Engineering (M-TECH HOE)
The study of different flow regimes in pipelines is fundamental to understanding the behaviour of the velocity profile as a result of fluid transport. Predicting this profile makes it possible to
understand the interaction of the flow with the pipeline and provide parameters, such as the mass flow rate, that make it possible to design the transport system—the power required for pumping and
the thickness of the pipeline, for example. However, obtaining the velocity profile as a function of the pipeline radius requires understanding the flow regime—laminar, transient or turbulent—with
which one expects to deal. Once the context is defined, it is possible to use the appropriate equation to estimate the velocity profile in the duct; this estimate, if compared with experimental data
from an analogous regime, is inevitably subject to non-adherence.
This work aims to validate or not an estimate for the velocity profile in a duct whose turbulent flow is known experimentally. This estimate will be developed based on linear regressions made from a
previously provided database containing data on the velocity profile collected in laboratory tests. The estimation itself comes from applying parameters to the equation that best describes the flow;
these parameters, in turn, come from the interpolation prepared based on the regressions.
Finally, it is expected with this work to understand the importance of validating estimates with experimental data and practising computational techniques—such as linear regression, polynomial
interpolation, and database parsing—viewed during classes, especially during the practical sessions.
Description of the problem
As exposed in the proposed problem statement, it is intended in this section to make clear the parameters and variables sufficient for the problem description. As already mentioned, turbulent flow is
characterized by presenting chaotic behaviour, which implies the rapid variation of pressure and velocity in the field defined by the flow. A high Reynolds number characterizes such a chaotic
regime—vide equation below—in this case, the instabilities are related to interactions between viscous terms and nonlinear inertial terms in the linear momentum equations.
the desired data is imported from the 6 ".txt" files, which are the databases;
address the data to their respective appropriate formats in the algorithm;
transform the velocity profile function (u(r) to u(y));
linearize the transformed function u(y);
modify the columns of the matrices containing the velocity and position in the duct cross-section so that they are in accordance with the linearized expression obtained in the previous step;
perform linear regression to find the straight line that best represents the data dispersion;
both the linear and angular coefficients of this straight line are determined; each must provide the parameters n and the maximum velocity Umax;
we perform the polynomial interpolation with the Lagrange method to obtain the polynomial that characterizes both functions n(Re) and Umax(Re);
we estimate n and Umax for Re = 35061 and determine the corresponding velocity profile through the equation u(y); and
the estimated data are compared with the experimental data through the percentage error, and the validation is concluded.
The following section explains the code implemented in Python 3.9 according to the described procedure and provides some considerations for running it in the Deepnote environment.
Implementation of the algorithm routine
The algorithm implementation considered what was presented during the practical sessions taught in the academic semester, from how to create functions in Python to how to perform linear regression
and polynomial interpolation. Therefore, to obtain the expected results, it can be stated that all the techniques taught was somehow undertaken for the algorithm to be written—some attention will be
given to the algorithm that performs linear regression and interpolation since it is understood that both are fundamental within the scope of this course.
The script is organized as follows:
firstly, the libraries understood to be essential for the algorithm to be appropriately executed by the compiler are imported. They are: re ("regular expression", package to help search and return
expressions when reading imported files), math, NumPy and matplotlib.puplot;
next, the functions invoked as the compiler processes the script are defined. These are:
FindValue(fullstring), which receives a line in "str" format and returns the parameter values present in the header of the ".txt" file;
Read(filepath), which reads a ". txt" file and returns all the parameters of interest addressed correctly in their formats;
Linearization(DATA, R), which receives the matrix containing the velocity profile and the radius and returns this same matrix with values corrected by the linearization—already described in the
previous section;
Linear_Regression(DATA, Re, R), which receives the matrix with the corrected velocity profiles and runs the linear regression, returning the linear and angular coefficients;
Interpolation(Ng, Ns, xi, fi, x), which in possession of its parameters performs the interpolation according to the Lagrange method and returns a list that represents the determined polynomial;
Error (u_exp, u_est), which in possession of both lists containing the experimental velocity and the estimated velocity, returns the percentual Error of both; and
the main() function, which has no parameters, invokes the other functions to execute the algorithm in an orderly fashion and is invoked at the end of the script.
The following subsections further detail the algorithms used to perform linear regression and interpolation according to methods seen during the practical sessions.
Polynomial Interpolation: Lagrange method
The code: instructions and considerations
Since one is in Deepnote's environment, to execute the code, one needs to run the whole notebook or just the block that contains all the script—we chose in this work to keep the code agglutinated in
a single executable block. If the script is executed in a compiler other than Deepnote, the argument of the Read(filepath) function, which is invoked right at the beginning of the main() function,
must be modified since the files directory in another compiler will probably be different.
Considering this observation, all steps described in the section "Objective and procedure" should be executed, and all results should be "printed" on the screen so that it will be possible to
validate or not the adherence of the estimated and experimental data—it is worth mentioning that the code is commented, which should facilitate its evaluation.
import re import math import numpy as np import matplotlib.pyplot as plt def FindValue(fullstring): # Function that finds parameter values in the database header. delimeter = '=' value = fullstring[
fullstring.index(delimeter)+1:] value = value.replace(" ","") value = value.replace("m/s","") value = value.replace("m^2/s","") value = value.replace("m","") return value def Read(filepath): #
Function that reads the file ".txt" and returns parameters. DATA = [] file = open(filepath, 'r') lines = file.readlines() file.close() i = 0 for line in lines: line = line.strip() if line.startswith(
'Re_tau ='): Re_tau = float(FindValue(line)) elif line.startswith('U_tau ='): U_tau = float(FindValue(line)) elif line.startswith('nu ='): nu = float(FindValue(line)) elif line.startswith('R ='): R =
float(FindValue(line)) if i >= 6: DATA.append(re.findall(r'\d+\.\d+', line)) i +=1 for lin in range(len(DATA)): for col in range(len(DATA[0])): DATA[lin][col] = float(DATA[lin][col]) return Re_tau,
U_tau, nu, R, np.array(DATA) def Linearization(DATA, R): # Function that receives a matrix with the nonlinear function u(r) and returns linearization F(Y). aux = np.copy(DATA) for i in range(len(aux)
): aux[i][0] = math.log(DATA[i][0]/R) aux[i][1] = math.log(DATA[i][1]) return aux def Linear_Regression(DATA, Re, R): # Function that performs linear regression for the data. Y = Linearization(DATA,
R)[:,0] F_Y = Linearization(DATA, R)[:,1] V = np.zeros((len(Y),2)) # Vandermonde matrix for i in range(len(Y)): for j in range(2): V[i,j] = Y[i]**j coefs = np.linalg.solve((V.transpose()).dot(V),(V.
transpose()).dot(F_Y)) print("\n"," Coefficients of the linearized equation for Re = {}".format(Re),":") print("\n"," A = 1/n = ", np.round(coefs[1],3), "and", "B = ln(U_max) = ", np.round(coefs[0],3
)) x = np.linspace(min(Y),max(Y),1000) # x axis for plotting f_x = coefs[1]*x + coefs[0] fig, ax = plt.subplots() ax.plot(Y, F_Y,'ok', label='Linearized Experiment Data') ax.plot(x, f_x,'r', label=
'Linear Regression') plt.xlabel("Y = ln(y/R)") plt.ylabel("F(Y) = ln[u(y)]") plt.grid(True) plt.title("Linearization and regression for Re = {}".format(Re)) ax.legend(); plt.show() return 1/coefs[1],
math.e**coefs[0] def Interpolation(Ng, Ns, xi, fi, x): # Function that executes Lagrange Interpolation a_l = fi p_l = np.zeros_like(x) for i in range(Ng): for j in range(Ns): Lp_j = 1. # Initialize
Lagrange Polynomial for k in range(Ns): if j is not k: Lp_j *= (x[i]-xi[k])/(xi[j]-xi[k]) p_l[i] += a_l[j]*Lp_j return p_l def Error(u_exp, u_est): # Function that calculates the error between both
the experimental and estimated lists. num = 0 den = 0 for k in range(len(u_exp)): num += abs(u_est[k] - u_exp[k]) den += abs(u_exp[k]) return 100*(num/den) def main(): # Main function that runs the
algorithm and invokes the other functions. Re_tau_5k, U_tau_5k, nu_5k, R, DATA_5k = Read("/work/Files_for_Flow/Data/Retau_5k_basic_stats.txt") Re_tau_10k, U_tau_10k, nu_10k, R, DATA_10k = Read("/work
/Files_for_Flow/Data/Retau_10k_basic_stats.txt") Re_tau_20k, U_tau_20k, nu_20k, R, DATA_20k = Read("/work/Files_for_Flow/Data/Retau_20k_basic_stats.txt") Re_tau_30k, U_tau_30k, nu_30k, R, DATA_30k =
Read("/work/Files_for_Flow/Data/Retau_30k_basic_stats.txt") Re_tau_35k, U_tau_35k, nu_35k, R, DATA_35k = Read("/work/Files_for_Flow/Data/Retau_35k_basic_stats.txt") Re_tau_40k, U_tau_40k, nu_40k, R,
DATA_40k = Read("/work/Files_for_Flow/Data/Retau_40k_basic_stats.txt") n_5k, U_max_5k = Linear_Regression(DATA_5k, Re_tau_5k, R) n_10k, U_max_10k = Linear_Regression(DATA_10k, Re_tau_10k, R) n_20k,
U_max_20k = Linear_Regression(DATA_20k, Re_tau_20k, R) n_30k, U_max_30k = Linear_Regression(DATA_30k, Re_tau_30k, R) n_40k, U_max_40k = Linear_Regression(DATA_40k, Re_tau_40k, R) Estimated = np.array
([[Re_tau_5k, n_5k, U_max_5k], [Re_tau_10k, n_10k, U_max_10k], [Re_tau_20k, n_20k, U_max_20k], [Re_tau_30k, n_30k, U_max_30k], [Re_tau_40k, n_40k, U_max_40k]]) print("\n", "Estimated values of n
[adm] and U_max [m/s]:", "", " | n |U_max|", np.round(Estimated[:,1:], 2), sep="\n") Ns = len(Estimated) Ng = 1000 Re = Estimated[:,0] ni = Estimated[:,1] Ui = Estimated[:,2] x = np.linspace(min(Re),
max(Re), num=Ng) n = Interpolation(Ng, Ns, Re, ni, x) U_max = Interpolation(Ng, Ns, Re, Ui, x) fig, ax = plt.subplots() ax.plot(Re, ni,'xk', label='Data') ax.plot(x, n, '--g', label='Lagrange') plt.
xlabel("Re [adm]") plt.ylabel("n") plt.grid(True) plt.title("Lagrange Interpolation: n(Re)") ax.legend(); plt.show() fig, ax = plt.subplots() ax.plot(Re, Ui,'xk', label='Data') ax.plot(x, U_max,
'--g', label='Lagrange') plt.xlabel("Re [adm]") plt.ylabel("U_max [m/s]") plt.grid(True) plt.title("Lagrange Interpolation: U_max(Re)") ax.legend(); plt.show() print("\n", "With interpolation, we
estimate for Re = 35061, U_max = 41.22128 m/s and n = 6.83221") # Calculating the velocity profile based on the estimate: n_35k = 6.83221 # [adm] U_max_35k = 41.22128 # [m/s] u_35k = [] for i in
range(len(DATA_35k)): u_35k.append( U_max_35k * ((DATA_35k[:,0][i] / R)**(1/n_35k)) ) fig, ax = plt.subplots() ax.plot(u_35k, DATA_35k[:,0],'-', color='b', label='Estimated Speed Profile') ax.plot(
DATA_35k[:,1], DATA_35k[:,0],'o', color='r', label='Experimental Speed Profile') plt.xlabel("u [m/s]") plt.ylabel("y [m]") plt.grid(True) plt.title("Speed Profile (Re = 35061)") ax.legend(); plt.show
() clean_DATA_35k = DATA_35k[:,1].tolist() del clean_DATA_35k[0:3] del u_35k[0:3] print("\n", "The error found between the experimental velocity profile and the estimated one for Re = 35061 is: ", np
.round(Error(clean_DATA_35k,u_35k), 3), "%") main()
The code above should print all the steps already mentioned on the screen. If the present file in ".pdf" format prevents the complete visualisation of the results, they can be found in the following
section: "Results and conclusion".
BERNHARD. Fluid dynamics - Reynolds number and inertial force - Physics Stack Exchange. 2013. Available at: <https://physics.stackexchange.com/questions/80070/ reynolds-number-and-inertial-force>.
SIMSCALE. What is Reynolds Number? | SimWiki | SimScale. 2021. Available at: <https: //www.simscale.com/docs/simwiki/numerics-background/what-is-the-reynolds-number/>.
AJAYAKUMAR, V. Parsing text with Python · vipinajayakumar. 2018. Available at: <https://www.vipinajayakumar.com/parsing-text-with-python/>. | {"url":"https://deepnote.com/app/ecole-centrale-de-nantes/Turbulent-flow-in-a-pipe-a60d0239-f1b5-4651-9e34-39b8b6d1b790","timestamp":"2024-11-14T20:33:47Z","content_type":"text/html","content_length":"261714","record_id":"<urn:uuid:6991af6e-3a53-4811-9d35-ba1d54647b8b>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00535.warc.gz"} |
Lecture 011
Ada's Lecture
P vs. NP
$P$: set of languages can be decided in $O(n^k)$ steps for some constant $k$.
$NP$: Might not be in $P$, but verifiable in $P$
• Bounded Entsheidungsproblem: come up with proof that is at most $k$ length long.
• Subset Sum Problem: given a set $S$, find subset $S' \subseteq S$ such that $\sum_S = 0$.
• Traveling Salesperson Problem.
• Satisfiability Problem (SAT): given a boolean formula, determine if it is satisfiable.
• Sudoku Problem: given a partially filled $n \times n$ sudoku board, determine if there is a solution.
Cannot prove these problems are not solvable in polynomial time.
Polynomial Time Reduction
Polynomial Time Reduction: $A \leq^P B \iff M_A \text{ runs } M_B \text{ polynomial number of time to solve problem}$
• Strategy: find some class of hard problems $L$, show that they can be reduced to $A$ to get $L \leq^P A$
$C$-hard: Let $C$ be a set of languages containing $P$. $A$ is $C$-hard if $(\forall L \in C)(L \leq^P A)$ ($A$ is at least as hard as every language in $C$)
• $A$ is outside or equal to the boundary of $C$
$C$-complete: $A$ is $C$-hard and $A \in C$ ($A$ is a representative for hardest language in $C$)
• $A \text{ is } C\text{-hard} \implies (A \in P \iff C e= P)$
• $A$ is equal to the boundary of $C$
picture of P vs. NP
Definition of $NP$: Nodeterministic Polynomial Time (it can be determined by nodeterministic TM in polynomial time)
• there is a polynomial time verifier TM $V$
• a polynomial $p(\cdot)$
• for all $x \in \Sigma^*$:
• $x \in L \implies (\exists u)(|u| \leq p(|x|) \implies V(x, u) \text{ accpets})$ (there is poly-length proof that leads to V accept)
• $x otin L \implies (\forall u)(V(x, u) \text{ rejects})$ (every u leads V to reject)
• or you can say $x \in L \iff V(x, \cdot) \text{ is satisfiable}$
Definition of $EXP$: all languages that can be decided in at most exponential-time (in $O(2^{n^C})$ for $C > 0$)
• CIRCUIT-SAT in NP
• P in NP
• NP in EXP
• CIRCUIT-SAT is NP-complete (Cook-Levin Theorem)
• 3SAT is NP-complete
• 3COL is NP-complete
• CLIQUE is NP-complete
• IS is NP-complete
• MSAT (at least 2 different sat) is NP-complete
• BIN is NP-complete
• show L in NP: present a verifier for L
• show L NP-hard: reduce NP-hard or NP-complete problem to L
• show L NP-complete: show L in NP and NP-hard
• if $c$ depends on $n$, then it is not polynomail
• if $c$ is constant, then it is polynomial
Clique Verifier
• proof of verifier: correctness and time complexity
Properties of $NP$:
• Every decision problem in $NP$ can be solved using brute force search (search space is non-polynomail, but exists verifier)
• The Cook-Levin Theorem: SAT is $NP$-complete ($\forall L \in NP, L \leq^P SAT$)
• Karp's 21 NP-complete problems:
There are thousands of problems are NP-complete
• Super Mario Bros, Tetris are NP-complete
Properties of $NP$-complete:
• every $NP$-complete problem reduce to each other
• you can create a Sudoku game to solve conjectures
Cook Reduction
Cook reduction: poly-time Turning reduction ($A \leq^P B$)
• solve $A$ in poly-time using a blackbox oracle that solves B. Only call $M_b$ $poly(|x|)$ times.
• $A \leq^P B \implies (B \in P \implies A \in P)$
• $A \leq^P B \implies (A otin P \implies B otin P)$
Many-one Reduction
Karp reduction(poly-time many-one reduction): ($A \leq_m^P B$)
• make ONE call to $M_B$ and directly use its answer as output
• to show $A \leq_m^P B$:
□ define $f : \Sigma^* \rightarrow \Sigma^*$
□ show $x \in A \iff f(x) \in B$
□ show $f$ is poly-time
Many-one Mapping
• Karp reduction is a special case of Cook reduction
Different reduction lead to different notion of $NP$-hardness
Reduction Practices 1
Boolean Circuit: a directed acyclic graph where vertices are AND gates, OR gates, NOT gates, $n$ input gates, $1$ output gate, and constant gates (with reasonable input output numbers). It can be
thought as function $f: \{0, 1\}^n \rightarrow \{0, 1\}$.
CIRCUIT-SAT: given circuit, return true iff there exists an assignment of inputs to make it output $1$
kCOL: given $G = \langle{V, E}\rangle$, return true iff $G$ is k-colorable.
CLIQUE: given $\langle{G, k}\rangle$ (graph $G$, positive int $k$), return true iff $G$ contains a clique (all connected, complete graph) of size $k$.
IND-SET: given $\langle{G, k}\rangle$ (graph $G$, positive int $k$), return true iff $G$ contains an independent set (none-connected, no edge between any two vertices in the subset) of size $k$.
• decision problems is languages
• if input does not correspond to a valid encoding, it is not in the language, input rejected
Show $CLIQUE \leq_m^P IND-SET$: define $f : \overrightarrow{G, k} \rightarrow \langle{G', k'}\rangle$ such that $G$ has a clique of size $k$ iff $G'$ has an independent set of size $k'$ ($f$ is
complement of vertices)
Reduction Practice 2
3SAT: give a clause of $A_1 \land A_2 \land A_3...$ where $A_i$ are connected using AND and $A_i = (O_1 \lor O_2 \lor O_3)$ only 3 clause connected using OR. ($\phi = (x_1 \lor \lnot x_2 \lor x_3) \
land (\lnot x_1 \lor x_4 \lor x_5) \land (x_2 \lor \lnot x_5 \lor x_6)$) Tell me whether this is satisfiable
Proof Strategy
Show $3SAT \leq_m^P CLIQUE$:
Mapping Strategy
Sutner's Lecture
NP: Language $L \subseteq \Sigma^*$ is in NP if there is a polynomial time decidable relation $R$ and a polynomial $p$ such that $x \in L \iff \exists w(|w| \leq p(|x|) \land R(w, x))$
VERTEX-COVER: $C \subseteq G$ such that $(\forall (v, u) \in E)(v \in C \lor u \in C)$ and make $C$ small.
• Decision: Does $G$ have a vertex cover of size $k$? (input length for G: $n^2$; k: $\log n$)
• Function: compute the lexicographically first cover of minimal size
• Search: compute a cover of minimal size
CLIQUE: a clique for $G = \langle{V, E}\rangle$ is $C \subseteq V$ such that $(\forall u, v \in C)(u eq v \implies \{v, u\} \in E)$
• Decision: Does $G$ have a clique of size $k$?
INDEPENDENT-SET: an independent set for $G = \langle{V, E}\rangle$ is $I \subseteq V$ such that $(\forall u, v \in C)(u eq v \implies \{v, u\} otin E)$
• Decision: Does $G$ have a independent set of size $k$?
TSP: for $n \times n$ positive integer distance matrix $d$, a bound $B$, is there a tour of cost at most $B$?
• Triangle TSP: the distances are required to be symmetric and conform to triangle inequality $d(i, j) \leq d(i, k) + d(k, j)$
• Euclidean TSP: Points are in the plane with integer coordinates $(x, y)$. Distance defined as Euclidean distance between points.
• Hamiltonian cycle: a path on undirected graph $G$ that contains every vertex of $G$ exactly once. (no known polynomial time solver, related to TSP)
• Eulerian cycle: a path on undirected graph $G$ that contains every vertex of $G$ exactly once and the cycle is created to use every edge of the graph exactly once. (solvable in linear time)
Oracle: assume O(1), used in reasoning of poly-time reduction from non-decision version to decision.
Nodeterministic Polynomial: solvable by NTM in polynomial time.
• however, if NTM $M$ solves problem $P$, then we cannot simply negate the result from $M$ to solve $\lnot P$.
• add unbounded search to $P$ is $NP$
• Formal Definition: $x \in L \iff (\exists w)(|w| \leq p_{\text{olynomial}}(|x|) \land R(w, x))$ where $w$ is certificate (witness, can't be infinitely large otherwise R will never by polynomial
and might not halt), $L$ is space (language), $R: W \times X \rightarrow \{0, 1\}$ is verification function, $x$ is input.
• Boolean Operation:
□ Union, Intersection: closed by running two algorithm sequentially
• Complement($co-NP$): $x otin L \iff (\forall w)(|w| \leq p_{\text{olynomial}}(|x|) \implies \lnot R(w, x))$
□ $P = NP \implies P = NP = co-NP$
□ We don't know if $NP \cap co-NP = P$ hold. (usually work, primality testing is example)
• Intuition Analogy
□ decidable: P
□ semidecidable: NP
□ cosemidecidable: co-NP
Nondeterministic TM: same idea as NFA except
• For non-decision problem: might get different answer in different branch
• Time complexity: $Time_M(x) = \min \{|\beta| : \beta \text{ is accepting branch of } TM_x\}$
• Assumption:
□ $p \rightarrow p_1, p_2, p_3, p_4$ can be treated as $\begin{cases} p \rightarrow q_1, q_2\\ q_1 \rightarrow p_1, p_2\\ q_2 \rightarrow p_3, p_4\\ \end{cases}$ with only $\log n$ slow down
□ so each step will have exactly 2 choices
□ we can attach a clock to terminate all branches when a solution is found
□ We can combine 1 bit at each level and validate solution in regular TM. Therefore we need two deterministic transition function $\begin{cases} \delta_0 : Q \times \Sigma \rightarrow Q \times
\Sigma \times \{\pm 1, 0\}\\ \delta_1 : Q \times \Sigma \rightarrow Q \times \Sigma \times \{\pm 1, 0\}\\ \end{cases}$ to make up a non-deterministic transition function. The choice tell us
which one to use.
• Space complexity: $Space_M(x) = \min \{|\beta| : \beta \text{ is accepting branch of } TM_x\}$ (notice we ignored time-space trade-off)
\begin{cases} NTIME(f) = \{L(M) | M \text{ is a NTM s.t. } T_M(n) = O(f(n))\}\\ NSPACE(f) = \{L(M) | M \text{ is a NTM s.t. } S_M(n) = O(f(n))\}\\ \end{cases}
\begin{cases} P = TIME(poly)\\ NP = NTIME(poly)\\ \end{cases}
Polynomial Time Turing Reduction: $\leq^p_T$
• preorder: polynomials are closed under substitution
• $B \in P \land A \leq^P_T \implies A \in P$
• complement: $\bar{A} \leq^P_T A$ is true for all $A$. Therefore it cannot distinguish $co-NP$ from $NP$
• so $B \in NP \land A \leq_T^P \text{ does not imply } A \in NP$ (problem solved by many-one reduction)
Polynomial Time Many-one Reduction: $A \leq^p_m B$ if there is a polynomial time computable function $f$ such that $x \in A \iff f(x) \in B$
• preorder: polynomials are closed under substitution
• $B \in P \land A \leq^p_m B \implies A \in P$
• $B \in NP \land A \leq^p_m B \implies A \in NP$
• NP Hard: $B | (\forall A \in NP)(A \leq^p_m B)$ (lower bound)
• NP-complete: $B | B\in NP \land B\in NP\text{-Hard}$ (lower and upper bound)
• $\exists B \in NP\text{-complete} \land B \in P \implies P = NP$
Universal Machines:
• Enumeration $(M_e)_e$ of all polynomial time Turing machine.
• Universal Machine: $q(n^e + e)$ time ($U(e\# x) = M_e(x)$ is not polynomial because $e$ is variable)
Construct NP-complete: simulate machines in the enumeration $N_e$ of nondeterministic, polynomial time TM.
• $K = \{e \# x | x \text{ accepted by } N_e\}$ is NP-hard, but not in NP (not working)
K is NP by padding 1s in input
K is NP-hard
Boolean Formulae
Boolean Function: formula using $\lnot, \land, \lor$
• input: many boolean
• output: one boolean
For formula $\varphi$, $\varphi[\sigma]$ denotes the truth value of $\varphi$ given $\sigma$.
• satisfiable: $(\exists \sigma)(\varphi [\sigma] = 1)$
• tautology: $(\forall \sigma)(\varphi [\sigma] = 1)$
Threshold function: a boolean function, for $0 \leq m \leq n$, $thr^n_m(x) = \begin{cases} 1 \text{ if } |\{i | x_i = 1\}| \geq m\\ 0 \text{ otherwise}\\ \end{cases}$ where $x_i$ is ith input.
• $thr_0^n = 1$
• $thr_n^1 = \text{n-ary disjunction}$
• $thr_n^n = \text{n-ary conjunction}$
Threshold Function implemented $thr^n_k(x)$ using boolean function: $\bigvee_{I \subset [n] | |I|=k} \bigwedge_{i \in I} x_i$
• the size is $\Theta(k \begin{pmatrix}n\\k\\\end{pmatrix})$
• the size is polynomial in n when k is fixed
Counting Function implemented $cnt^n_k$ using boolean function: $thr^n_k(x) \land \lnot thr^n_{k+1}(x)$
• the size is $\Theta(n^{k+1})$ for fixed $k$
Counting with 8 Variables
• where the first picture is truth assignment for $cnt^n_1$
• where the last picture is truth assignment for $cnt^n_1$
Application of SAT
• therefore SAT looks like NP-complete
• To prove this fact: prove $\Phi_x \text{ is satisfiable } \iff M \text{ accepts } x\#w \text{ for some } w$ the power of a boolean formula is the same as TM solving NP problem (we can simulate NP
TM on boolean formula)
Exponential Time Hypothesis: every algorithm for SAT has running time $\Omega(2^{cn})$ for some $c > 0$
3-SAT Formal Definition:
• Literal: a variable or negated variable
• Conjunctive Normal Form (CNF): a conjunction of disunctions of literals ($\Phi_1 \land \Phi_2 \land ... \land \Phi_n$ where $\Phi_i = z_{i, 1} \lor z_{i, 2} \lor ... \lor z_{i, k(i)}$ where $z_
{i, j}$ is a literal)
• 3-CNP: $k(i) = 3$ for all $i$
• SAT is NP-complete for formulae in 3-CNF
Vertex cover is reducible to 3SAT (proof in slides)
Karp's List
Problems in $P$ that was not known:
• Nonprimes: detect wether a number is not a prime
• Linear Inequalities: whether $Cx \geq d$ has a rational solution
Still unknown complexity: Graph Isomorphism (given two graph $G$ and $G'$, whether $G$ is isomorphic to $G'$)
PARTITION: For $a_1, a_2, ..., a_n \in \mathbb{N}$, is there $I \subset [n]$ such that $\sum_{i \in I} a_i = \sum_{i otin I} a_i$?
BIN: Can we put $a_1, a_2, ..., a_n \in \mathbb{N}$ in $k \in \mathbb{N}$ bins of size $C \in \mathbb{R}$
Reductions 1
Reductions 2
Examples of Hard reductions: See lecture slide 16.
Numerical Problems
strongly NP-hard: if we express input in unary, it will still remain NP-hard.
• example: TSP, all non-arithmetic NP-hard problems
pseudo-polynomial time: if we express input in unary, it will be in polynomail time. | {"url":"https://kokecacao.me/page/Course/F21/15-251/Lecture_011.md","timestamp":"2024-11-06T14:37:48Z","content_type":"text/html","content_length":"48797","record_id":"<urn:uuid:87d09c4c-ecf1-41be-bc32-e531f7cd2f50>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00572.warc.gz"} |
Coupled-cluster theory
Coupled-cluster theory#
Modules: pyscf.cc, pyscf.pbc.cc
The MP2 and coupled-cluster functionalities of PySCF are similar. See also Second-order Møller–Plesset perturbation theory (MP2).
Coupled-cluster (CC) theory is a post-Hartree-Fock method capable of describing electron correlation in the ground state. It is size extensive but not variational. PySCF has extensive support for CC
calculations with single and double excitations (CCSD). It can also include a perturbative treatment of triple excitations (CCSD(T)), which is a very accurate method for single-reference quantum
chemistry. CC calculations can be performed with or without density fitting, depending on the initial SCF calculation. Correlated excited states are accessible through the equation-of-motion (EOM)
CCSD framework, described below.
A minimal example of a CCSD and CCSD(T) calculation is as follows:
from pyscf import gto, scf, cc
mol = gto.M(
atom = 'H 0 0 0; F 0 0 1.1', # in Angstrom
basis = 'ccpvdz',
symmetry = True,
mf = scf.HF(mol).run()
# Note that the line following these comments could be replaced by
# mycc = cc.CCSD(mf)
# mycc.kernel()
mycc = cc.CCSD(mf).run()
print('CCSD total energy', mycc.e_tot)
et = mycc.ccsd_t()
print('CCSD(T) total energy', mycc.e_tot + et)
Spin symmetry#
The CC module in PySCF supports a number of reference wave functions with broken spin symmetry. In particular, CC can be performed with a spin-restricted, spin-unrestricted, and general (spin-mixed)
Hartree-Fock solution, leading to the RCCSD, UCCSD, and GCCSD methods.
The module-level cc.CCSD(mf) constructor can infer the correct method based on the level of symmetry-breaking in the mean-field argument. For more explicit control or inspection, the respective
classes and functions can be found in ccsd.py (restricted with real orbitals), rccsd.py (restricted with potentially complex orbitals), uccsd.py (unrestricted), and gccsd.py (general).
For example, a spin-unrestricted calculation on triplet oxygen can be performed as follows:
from pyscf import gto, scf, cc
mol = gto.M(
atom = 'O 0 0 0; O 0 0 1.2', # in Angstrom
basis = 'ccpvdz',
spin = 2
mf = scf.HF(mol).run() # this is UHF
mycc = cc.CCSD(mf).run() # this is UCCSD
print('UCCSD total energy = ', mycc.e_tot)
A number of properties are available at the CCSD level.
Unrelaxed 1- and 2-electron reduced density matrices can be calculated. They are returned in the MO basis:
dm1 = mycc.make_rdm1()
dm2 = mycc.make_rdm2()
Analytical nuclear gradients can be calculated:
mygrad = mycc.nuc_grad_method().run()
The CCSD Lambda equations can be solved:
l1, l2 = mycc.solve_lambda()
Frozen orbitals#
By default, CCSD calculations in PySCF correlate all electrons in all available orbitals. To freeze the lowest-energy core orbitals, use the frozen keyword argument:
mycc = cc.CCSD(mf, frozen=2).run()
To freeze occupied and/or unoccupied orbitals with finer control, a list of 0-based orbital indices can be provided as the frozen keyword argument:
# freeze 2 core orbitals
mycc = cc.CCSD(mf, frozen=[0,1]).run()
# freeze 2 core orbitals and 3 unoccupied orbitals
mycc = cc.CCSD(mf, frozen=[0,1,16,17,18]).run()
The number of core orbitals to be frozen can be generated automatically:
mycc = cc.CCSD(mf).set_frozen().run()
See also Frozen orbitals for more information on the rule of freezing orbitals.
Equation-of-motion coupled-cluster theory#
EOM-CCSD can be used to calculate neutral excitation energies (EE-EOM-CCSD), spin-flip excitations (SF-EOM-CCSD), or charged excitations, i.e. ionization potentials (IP-EOM-CCSD) or electron
affinities (EA-EOM-CCSD). The EOM functions return the requested number of eigenvalues and right-hand eigenvectors. For example:
e_ip, c_ip = mycc.ipccsd(nroots=1)
e_ea, c_ea = mycc.eaccsd(nroots=1)
e_ee, c_ee = mycc.eeccsd(nroots=1)
e_sf, c_sf = mycc.eomsf_ccsd(nroots=1)
The eecsd() function returns neutral excitations with all possible spin multiplicities. For closed-shell calculations (RHF and RCCSD), singlet and triplet excitations can be requested explicitly:
e_s, c_s = mycc.eomee_ccsd_singlet(nroots=1)
e_t, c_t = mycc.eomee_ccsd_triplet(nroots=1)
By default, PySCF calculates the nroots eigenvalues with the lowest energy, which may include states with dominant double-excitation character. To only calculate states with dominant
single-excitation character, use the koopmans keyword argument:
e, c = mycc.eeccsd(nroots=3, koopmans=True)
An initial guess wavefunction may be provided, in which case PySCF will try to find the most similar EOM solution vector:
from pyscf.cc.eom_rccsd import amplitudes_to_vector_ee
r1 = np.zeros((nocc,nvir))
r2 = np.zeros((nocc,nocc,nvir,nvir))
r1[occ_index,vir_index] = 1.0
myguess = amplitudes_to_vector_ee(r1,r2)
e_s, c_s = mycc.eomee_ccsd_singlet(nroots=1, guess=myguess)
Job control#
Saving and restarting#
To allow for future restarts, the SCF information and the CCSD DIIS information must be saved:
mf = scf.HF(mol)
mf.chkfile = 'hf.chk'
mycc = cc.CCSD(mf)
mycc.diis_file = 'ccdiis.h5'
To restart a CCSD calculation, first the molecule and SCF information must be restored:
mol = lib.chkfile.load_mol('hf.chk')
mf = scf.HF(mol)
mf.__dict__.update(lib.chkfile.load('hf.chk', 'scf'))
Next, the CCSD calculation can be restarted by using the previous CCSD amplitudes as the initial guess:
mycc = cc.CCSD(mf)
mycc.kernel(mycc.t1, mycc.t2)
Modifying DIIS#
The parameters of the DIIS algorithm can be tuned in cases where convergence is difficult. To increase the size of the DIIS space:
mycc = cc.CCSD(mf)
mycc.diis_space = 10
By default, DIIS is activated on the first CCSD iteration. Sometimes it can be helpful to postpone the use of DIIS:
mycc = cc.CCSD(mf)
mycc.diis_start_cycle = 4
Integral-direct CCSD#
In order to avoid large memory requirements, the default behavior in CCSD calculations is to store most two-electron integral tensors on disk. This leads to a potential I/O bottleneck. For
medium-sized molecules, an integral-direct AO-driven implementation can be more efficient. The user must manually request an integral-direct CCSD calculation:
mycc = cc.CCSD(mf)
mycc.direct = True
e_corr, t1, t2 = mycc.kernel() | {"url":"https://pyscf.org/user/cc.html","timestamp":"2024-11-06T13:57:10Z","content_type":"text/html","content_length":"44626","record_id":"<urn:uuid:388c09f6-49a2-47b4-b21b-1addf2fecbfe>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00209.warc.gz"} |
The golden treasury of chess - DOKUMEN.PUB
Citation preview
Compiled by AL _HOROWITZ AND
Reprinted 1971
Copyright© 1969, 1961, 1956 By I. A. Horowitz Copyright © 1943 By Horowitz & Harkness
This completely new revised edition is published by arrangement with I. A. Horowitz and Harvey House, Inc.
CORNERSTONE LIBRARY PUBLICATIONS are distributed by Simon & Schuster, Inc. 630 Fifth Avenue New York, New York 10020 Manufactured in the United States of Ame1ica under the supervision of Rolls Offset
Printing Co., Inc., N. Y.
This Book is Dedicated To the Memory of
HARRY NELSON PILLSBURY (1872-1906)
Favorite Games In the course of the decades which I have devoted to the preparation of this volume, I have had occasioll to examine
thousands upon thousands of scores.
Those that have pleased
me most are included in "THE GOLDEN TREASURY OF CHESS." But even among these favorites, there are some which I have enjoyed so much that I have set them aside in order to at tract the reader's
attention to these games.
I will not deny
that ten years ago I might have selected other games, and' that in the years to come, my tastes will again be modified ! Nevertheless, you will be delighted with these games.
THE GoLDI!N
Warsaw, Nov. 1844
A1 long a1 we continue Jo be (harmed by the triumph of mind oi·er matter, :uch combination! wilt jauinale u1. The idea of readily mrl'endering the Queen in order 10 ho1111d the ho:tile King with the
leuer pieces, ha: been utilized fair ly often; b11t Petroff' J :amfice was one of the first, if not THE fir:t, example of this appealing com binative theme. All honor to hi! originality/ GIUOCO
PIANO HOFFMAN White
P-K4 Kt-KB3 B--B4 P-B3 P-Q4 P-KS
5 6 7 B--QS
8 K x Kt 9 K-Kt3 10 B x P 11 Kt-KtS 1 2 Kt x BP 13 Kt x Q
PETROFF Black P-K4 Kt-QB3
B---B 4 Kt-B3 PxP Kt-K5 Kt x KBP?! P x Pch PxP Kt-K2 Kt x B 0-0!!
And Black mates in eleven moves.
1 3 . . ..
14 K-R3 15 P-K6
K-Kt4 P-Kt3 K-KtS K-Kt4 K-R4 K-KtS K-R5 K-R6 PxR
B---B7ch P-Q3ch Kt-B5ch
Kt x KP Kt x Ktch R-B4ch R-B3ch R-BSch
Kt-K3ch P-Kt3ch R-R5ch B-K6 mate
Paris, 1845 It is many years since I fir11 :aw this game, but the final position, with Black's Queen trapped by its own far-advanced Pawns, and White'! King gaily advancing down the board io auist in
the final attack against his colleague, is still good for a chuckle. Imagine Kieseritzky' J chagrin as he stareJ ruefully at the b ottled-up Queen! Who 1ay1 there is no place for hu mor in cheu?!
1 P-K4 2 P-KB4 3 Kt-KB3 4 B-B4 5 Kt-K5 6 K-Bl
P-Q4 Kt-B3 P-KKt3 K-B2 Kt x P ( B7 ) Kt-KKt5 K-K3
13 14 K-Q3 15 P-QR3
Black P-K4 PxP
P-KKt4 P-KtS Q-Rsch
P-B6 Kt-KB3 B---Kt2 Q-R6ch P-Q3
R-Bl Q-Kt7ch B-R -- 3 Kt-B3 B x Kt
KtxKP!? 16 BxB B-B4 17 Q-Kl P-B7 18 KtxKt K-Q2 19 Q-K3 QR-Kl 20 B-Q5 BxKtch 21-QR-KBl R-B6 22 BxB PxQ 23 QxR R-K3 24 B-B5ch Kt-K4ch 25 P-Q5 P-KR4 26 K-Q4 K-.K'l 27 PxReh P-R5 28 B-B6 PxBch 29 BxKt
PxKtP 30 KxP 31 K-B6 and wins!
One of the most a1to1mding end ings on record.
BISHOP'S Gfu..\.f.BIT
L. KlESERITZKY
P-K4 P-KB4 B-B4 K-Bl BxP Kt-QB3 Kt-R3 Kt-Q5 KtxPch KtxR P-Q3 B-QB4 BxP Q-Kl KxP KxQ K-R4 K-R5
P-K4 PxP Q-R5ch P-QKt4 Kt-KB3 Kt-Kt5 Kt-QB3 Kt-Q5! K-Ql P-B6! P-B3 P-Q4! B-Q3 PxPch QxKtch! Kt-K6ch Kt-B6ch B-Kt5 mate
4. Breslau, 1859. It is difficult to imagine how one could concentrate more brilliancy, more in1pired inventiveneu, more sparkle into so short a game. Here is the di!tilled e.rsence of the very best
che.r.r of the old master!: one thrill after another! Sacrificial Orgy RUY LOPEZ
Nov. 1846
Poor Kieseritzky! He achieved neg ative immortality by losing a mag nificent game lo the great Anders un, and this feat swallowed up hiJ reputation forever after. That Kie:eritzky was a brilliant
and able player in hi! own right, however, fr abundanJ/y dear from thi.r game.
A. ANDERSSEN White
P-K4 Kt-KB3 B-Kt5 KtxKt B-B4 P-K5 B-Kt3
P-K4 Kt-QB3 Kt-Q5 PxKt Kt-B3 P-Q4 B-KKt5
8 P-KB3
9 0-0
PxB K-Rl P x Kt R-B5
Kt-K5 ! P-Q6! B-B4ch Kt-Kt6ch! Q-Kt4
EY.i\:."-TS GAMBIT J. H. ZuKERTORT
KtP x P P-Kt4 PxR Q-B3 Q-R3 Resigns Bravo! 5.
P-KR4!! QxR R x Pch!! Q-K5 ! Q-R5ch Q-K8ch
Berlin, 1 869
You have probably heard that An densen was a mighty man with the Evans Gambit, but it is imp ouible to realize what glorious feats he perf armed with ii, until you have played over 111ch games as
tbis one. lncidenta/Jy Zukertort, the great Anders.ren'! brilliant pupil, knew how to take fitting revenge, as you C1il1 see in lat" games in thfr volume. These two immortal: pro du�ed games worthy
of their repu latton. A glorio111 battle
P-K4 Kt-KB3 B-B4 P-QKt4 P-B3 0-0 P-Q4 PxP P-Q5 B-Kt2 B-Q3 Kt-B3 Kt-K2 R-Bl Q-Q2 K-Rl Kt-Kt3 Kt-B5 R-KKtl P-Kt4 B x Kt R-Kt3 P-Kt5 PxB PxP QR-KKtl P x Pch Q-R6
P-K4 Kt-QB3 B-B4 BxP B-B4 P-Q3 PxP B-Kt3 Kt-R4 Kt-K2 0--0
Kt-Kt3 P-QB4 R-Ktl P-B3 B-B2 P-Kt4 P--Kt5 ? B-Kt3 Kt-K4 QP x B R-B2 B x Kt Q x P? .R-Ql K-Rl K-Ktl Q-Q3
MY 29 Q:x:Pch! 30 P-B6ch 3 1 B-R7ch! 32 R-R3ch 33 R-R8 mate
KxQ K-Ktl KxB K-Ktl
St. Petersburg, 1896
There are many attractive settings for a brilliant game; but what is more impressive than an immortal game between two Titans? The man who was able lo bettl the great Pillsbury in this wonderful game
was truly worthy of his title. It is no exaggeration lo say that Lasker's combination is one of the greatest feats of the human imagi nation. Quadrangular Tourney QUEEN'5 GAMBIT DECLINED
H. N. PILLSBURY White 1 P-Q4 2 P-QB4
3 Kt-QB3 4 Kt-B3 5 B-Kt5 QxP 7 Q-R4? 8 0-0-0 9 P-K3 10 K-Ktl 1 1 PxP 12 Kt-Q4 13 B xKt 14 Q-R5 15 P xKt
nocence. Though Pillsbury only half suspects the quicksands, his defense cannot be improved. 17 P-B5
A problem in one half the moves of the entire game, mentally com posed and solved in a manner worthy of the chatnpion of the world. 18 19 20 21 22
PxB PxPch PxR B-Kt5 K-Rl
R-QR6!! RxP Q-Kt3ch QxBch R-B2 .
R-Q2 KR-Ql Q-B5
R-B5 R-B6! Q-B5 RxP!
Q-K6ch KxR K-R4 KxP K-R5 Q-Kt6
K-R2 Q- -B6ch P-Kt4ch Q---B5ch B---Q lch PxQ mate
DR. E. LASKER
Black P-Q4
P-K3 Kt-KB3 P-B4 BPxP Kt-B3 B-K2 Q-R4 B-Q2 P-KR3 PxP 0-0 BxB Kt xKt B-K3
The calm before the storm. 16 P-B4
The charm of the position after Black's 16th move is its surface in-
One of the ,marks of a great ma.rter is the ability to conjure up mur derous attacks out of seemingly harmless positions. You will like the way that Spielmann commences an 11nexpected attack at move
22 and drives it home with sledge hammer blows. Every move tells, anti Black's helplessneu becomes ever more apparent. RUY LOPFZ R. SPIELMANN White
DUS-CHOTIMIRSKI Black
1 P-K4 P-K4 2 Kt-KB3 Kt-QB3 3 B-Kt5 P-QR3 4 B-R4 Kt-B3 5 0-0 B-K2 6 R-Kl P-QKt4 7 B-Kt3 P-Q3 8 P-B3 Kt-QR4 9 B-B2 P-B4 IO P-Q3 0-0 1 1 QKt-Q2 Q-B2 12 Kt-Bl R-Ktl B-K3 13 P-KR3 14 Q--K 2 P-Kt5 15 Kt (
3 ) -R2 Kt-Q2 16 Kt-Kt3 KR-BI 1 7 Kt-Kt4 R-Kt2 18 Kt-K3 B-Kt4 B x Kt 19 Kt-Q5 20 P x B BxB 21 QR x B Kt-KB3 2 2 P-Q4! KP x P 2 3 Kt-R5 ! Kt-Q2 2 4 Q-Kt4 P-Kt3 25 R-K7 K-Bl 26 QR-Kl ! Q--Ql 27 Q--Kt5
Kt-K4 28 Q-B6! Kt(R4)-B5 29 P-B4! P x Kt 30 P x Kt Kt x KP 3 1 R ( l ) x Kt! Resigns
Iceland, 1931
Reti noted years ago that Alek hine's outstanding quality wa.r his ability lo give even the most com monplace positions an unusual turn. This game abounds in such origi nal moves. FRENCH DEFENSE
A. ALEK.HINE White
P-K4 P-Q4 Kt-QB3 B-Kt5 B x Kt Kt-B3 7 B-Q3 8 P-K5 9 P-KR4 1 0 B x Pch! 11 Kt-Kt5ch 12 P x Bch 13 Q-R5 14 0-0-0 1 5 P-Kt6! 16 KtP x P 17 P x P 1 8 RxP!! 19 Q-Kt5ch 20 R-R7 21 R-Q4
Black P-K3 P-Q4 Kt-KB3 B-K2 BxB 0--0
R-Kl B-K2 P-QB4 KxB B x Kt K-Ktl K-Bl P-R3 K-K2 R-Bl Kt-Q2 Q-R4 KxP R-KKtl Q x BP
R x Ktch! Kt-K4 Kt-Q6ch Q-B6ch! R-B7 mate 9.
BxR Q-Kt5 K-Bl PxQ
9 Kt-Kt5? 10 K-Rl 1 1 P-B4
Warsaw, 1 935
Anyone who preaches the imminent death of chess ought to take a $ood look at this game! The strik ing series of brilliancies initiated by Black's thirteenth move compares favorably, I believe, with
any com bination ever played over the board. A Polish "Immortal" DUTCH DEFENSE GLUCKSBl>RG White 1 2
BxPch Kt-Kt5 Q-Kl
P-Q4 P-QB4 Kt-QB3 Kt-B3 P-K3? B-Q3 0-0 Kt-K2?
M. NAJDORF Black
P-KB4 Kt-KB3 P-K3 P-Q4 P-B3 B--Q3 0-0 QKt-Q2
P-KKt3 K-Kt2 KtxB K-B3 QPxP P x Kt K-B4 K-B3 KPxP KxB PxKt
Q-R4 B-KtS! ! Q-R7cft P-K4! QKt x Pch Kt x Pch Kt-Kt3ch P-B5! B-Kt5ch! Kt-K4ch! P-R4 mate
The Pre-Morphy Period Although chess is a direct descendant of a game played in India in the 7th century, modeTn chess was not initiated until the late 15th century-about the year 1485---when im
portant changes were made in the rules. For a hundred years before this date the game had remained unchanged, the moves of the pieces fixed. Although highly popular, it was a dull game by our
standards. The modern chessplayer would re gard the chess of the middle ages as a strange and wearisome pastime. In many respects, of course, the mediaeval game was similar to the chess we play
today. The positions of the pieces were the same; the Rooks, Knights and Pawns moved as they move today; Castling had not yet been developed, but the King was allowed to "leap" two squares on its
first move. The main ditference lay in the moves of the Queen and Bishop. The Queen was permitted to move only to an ad jacent diagonal square. In other words, it moved like our Bishop, but only one
square at a time ! Instead of being the most powerful piece on the board, it was the weakest. The Bishop of the mediaeval game leaped over the adjacent diag onal square to the square beyond in the
diagonal. When the moves of the Queen and Bishop were changed to those we play today, the entire character of the game was transformed. The old artillery, cavalry and infantry in the form of Rooks,
Knights and Pawns, were still in the game, but the devastating power of the new dive-bombing Queen and the speedy attack of the motorized Bishop made it neces sary for the chess Generals to develop
new strategy and tac tics. New and more scientific openings had to be examined and analysed. Pawn play became a primary consideration, now that a promoted pawn could become a powerful Queen. The
whole tempo of the game was quickened, the battle shor tened and intensified. Italy was the main center of chess activity when these changes took place and the new game probably originated there. By
1510 the old type of chess was obsolete in most of
THE PB.1!-MOllPHY PEJUOD
Italy and Spain. One of the earliest games of the "new chess" to be recorded appears in a late 15th century manuscript in which a poem describes the courtship of Venus by Mars by means of a game of
chess. Francisco de Castellvi takes the part of Mars, Narciso Vinoles that of Venus. Historically important, the game is also interesting because it was un doubtedly played over the board by actual
chessplayers of reasonable proficiency for the period. Analysis was the ruling motive in the literature of the period. Openings known today as the Ruy Lopez, Giuoco Piano, Petroff lJefense, Philidor
Defense, Bishop's Opening and Queen's Gambit Accepted, were first outlined in a late 15th century manuscript (in the Gottingen University Lib rary.)* The first "best-seller" was a book written by
Damiano and printed in Rome in 1512. Eight editions were published in the 16th Century and it was also translated and published in French, English and German. All that is known of the author is that
he was an apothecary and a native of Portugal. To judge from his analysis, he was also a mediocre chess player. The famous name of Ruy Lopez first appears in 1559 when this Spanish priest visited
Italy and defeated all the Roman players. Although he did not invent the opening which bears his name, Ruy Lopez was the leading player of Spain for over 20 years and noted for his skill at blindfold
chess. He played often at the court of his patron, Philip II of Spain. In 1561 Lopez published a book on chess containing a code of laws, general advice to players (including the sug gestion that
you "place your opponent with the sun in his eyes") and a miscellaneous collection of openings. He deals with a wider rahge of openings than his predecessors but his analysis is considered weak.
Interesting is the fact that this book gave international currency to the term "gambit," a slang term which Lopez had learned in Italy. According to Lopez, "it is derived from the Italian gamba, a
leg, and gam bitare means to set traps, from which a gambit game means a game of traps and snares." Among the leading Italian players of the period 1560 to 1630 were Paolo Boi, Giovanni Leonardo da
Cutri, Giulio Cesare Polerio and Gioachino Greco. As a youth, Leonardo had been trounced by Ruy Lopez in Rome but he had his re"The names by which we call openings today usually have little or
nothing to the names of the earliest author
do with their orlg!r.s a.r,d seldom commemorate t.o discover t.he openings .
venge in 1575 when he visited Spain and defeated the aging Lopez in a match held in the presence of Philip II. Although existing text-books had become obsolete, the strong players of the early part
of this period did not publish their findings. The high stakes for which they pl ayed made them secretive. However, a patron could always obtain a copy of the player's notes on openings for a
consideration and many of these manuscripts have survived, particularly those of Polerio. The manuscri pts of Polerio, considered the leading player of Rome in 1606, again widen the range of the
openings and include the Queen's Gambit Declined (by 2 ...P-QB3 only), the Fianchetto Defenses, the Caro-Kann, the Sicilian, most of the known variations of the King's Gambit, the Center Gambit, the
Greco Counter Gambit, the Two Knights' Defense and the Four Knights' Game. There are also some printed books from this period, including three works published by Dr. Alessandro Salvio, one of the
leading Neapolitan players. For his time, Salvio was an analyst of great ability. Greco was one of the last great Italian players. Although a man of poor parentage and no education, he made and left
his mark on the pages of chess history.About 1619 he began to keep a manuscript collection of games and gave extracts to wealthy patrons. In the early days of his career he lived in Rome but about
1620 he travelled abroad, sojourning in France, England and Spain. In 1624 he re-arranged his collection of games and many years later, in 1669, a French translation of this re-arrangement was
published in Paris. Forty-one editions have since been published in many languages. After Greco's death in 1634, Italy produced no outstand ing players for over a hundred years. In England, France
and Germany, however, the popularity of chess had steadily increased and in the 18th century the coffee-houses of London and Paris were the leading centers of chess activity. The name of Andre D.
Philidor dominates the history of this period. Equally famous as a chessplayer and as a musician, Philidor defeated all the strongest players at the Cafe de la Regence in Paris and Slaughter's Coffee
House in London. After 1775 Philidor spent the Spring of each year in London and the rest of the year in Paris. The English gentry flocked to Parsloe's Club in London where Philidor then played. This
great player set forth his theories of chess in lucid fashion in his "Analyze du Jeu des Echecs," written when he was only 23 years old. He was the first to define and explain the prin ciples of
chess strategy and tactics. Since his death in 1795,
THE PRE-MOR.PHY PElllOD his book has often been reprinted. stone in the progress of chess.
It was an important mile
In the time of Philidor, Italy again produced some gifted players, including Ponziani, E. del Rio and G. Lolli. French contemporaries of Philidor before the Revolution were Ver doni, Leger, Carlier
and Bernard. In the first half of the 19th century the firmament of chess is studded with many chess stars whose names are familiar to the modern player. In England we hear of the exploits of J. E.
Sarratt; William Lewis; John Cochrane; Captain W. D. Evans (who discovered his gambit in 1824, the same year in which the London-Edinburgh postal match was played, giving us the name "Scotch Game");
William Lewis (who published his "Progressive Lessons" in 1831 and laid the foundations for much later work on the open ings); Alexander MacDonnell and the great Howard Staun ton. In France, the
leading players were Alexander Des chapelles; Pierre de Saint-Amant (who captained the victor ious French team in the 1831 postal match with London which gave us the name "French Defense"); De La
Bourdonnais (who vanquished MacDon&1ell in the match of 1834). Many notable players also arose in Central Europe including Johann Allgaier (who originated the idea of tabulating openings in an
original and important treatise, first published in 1795) ; Von Bilguer (whose famous "Handbuch" was published in 1843); L. E. Bledow (who started the magazine Schachzeitung in 1846) ; B. Horwitz; K.
Sehorn; von der Lasa; W. Hanstein and C. Mayet. Other masters of the period were the Russian Petroff, the Livonian Kieseritzky, the Viennese Hampe and the Hungarians .Szen and Lowenthal. In 1843
Staunton established himself as the first player of Europe by defeating Saint-Amant in a match. Staunton's "Chessplayers Handbook," published in 1847, became the leading English text-book. In this
book, and in the German "Handbuch,'' the names we now use for most openings were systematically arranged. 'l'he year 1851 stands out as the beginning of a new age in chess. It was in this year that
the first International Chess Tournament was held. The site was London and 16 com petitors took part in the main tournament. Adolph Anders sen of Berlin took first prize. A brilliant player,
Andersse n later demonstrated that the luck of the pairings in this "knock-out" tournament wa.i not responsible for his success.
In subsequent tournaments, the "round-robin" system was adopted and Anderssen won first prize in 7 of the 12 events in which he competed. With the establishment of tournament competition and the
advent of Paul Morphy, the brilliant young American master who defeated Anderssen and all other European ex perts, the truly modern era of chess was ushered in. From a purely technical point of
view, the games played in the 350odd years from the early beginnings of modern chess to the 19th century are not of vital importance to the present-day chessplayer. The selections presented in this
chapter com prise a mere handful of historical and representative games from this long, formative epoch. If chess has gained much since the passing of this period, it has also lost much. We have
gained a great deal in exper ience, in theory, in knowledge, in systematic analysis of the openings, in the assembling of a fine literature and the ex perience of many great players. And yet there
are times when one wonders whether all these gains compensate for the disappearance of the spirit of freshness, of eternal adventure, of naivete. It is a development which we see present in all the
arts and sciences. Of course, our great contemporary players have originality and imagination, but they also have a tremen dous backlog of study and acquired knowledge based on the heritage of their
predecessors. The games of the pre-Morphy period, whatever their faults may be, are the productions of players who were self-reliant, who had to find their way through uncharted country, who had to
perform brilliant feats of improvisation. Remember also, when you play over these games, that many of them were played for pure amuse ment, not as part of a gruelling contest and not for the record;
in that way you can savor their charm, their sociable and leisurely character.
10. Late 1 5 th Century. This is one of the earliest recorded games of modern cheu. It was played shortly after 1485, when the mediaeval moveI of the Queen and Bishop were changed. Score is from a
poem in a Catalan manu script. CENTER COUNTER GAME
fRANClSCO DE CASTELLVI NARCISO VINOLES White
1 P-K4 Px P 3 Kt-QB3 4 B-B4 5 Kt-B3 6 P-KR3 7 QxB 8 QxP 9 Kt-Kt5 10 Kt x RP 11 Kt x R 12 P-Q4 1 3 B-Kt5ch 14 Q x Ktch 15 P-Q5 1 6 B--K3 17 R-Ql 18 R x P 19 B-B4 20 Q x Ktch 2 1 Q-Q8 mate 2
P-Q4 QxP Q-Ql Kt-K133 B-Kt5 B x Kt P-K3 QKt-Q2 R-Bl Kt-Kt3 Kt x Kt Kt-Q3 Kt x B Kt-Q2 PxP B-Q3 Q-B3 Q-Kt3 BxB K-Bl
11. Rome, 1 560. Played when Lopez visited Rome in 1559-60. His youthful opponent Later became a famous player. DAMIANO'S DEFENSE
RUY LO PEZ LEONARDO DA White
PERIOD Kt-KB3 Kt x P Q-R5ch Q x KPch QxR P-Q4 B--B4ch B x Pch
P-KB3 P xKt?
P-Kt3 Q-K2
K-B2 P-Q4 Kt x B
and White eventually won. 12. Madrid, 1561.
Ruy Lopez analyzes the Ruy Lopez. A sample from the collection of openings in the book by Lopez. RUY LOPEZ White
1 P-K4
2 Kt-KB3
3 B-Kt5 4 P-B3 5 P-Q4 6 PxP 7 Kt-B3 8 B-Kt5 9 Q-Q3 10 P x B "with better game."
P-K4 Kt-QB3 B-B4
P-Q3 PxP B--Kt5ch B-Q2 Kt-B3 B x Ktch
1 3. Madrid, 1 575.
This game is believed to have been played in the match between Lopez and Leonardo, won by the latter. KING'S GAMBI T DECLINED R UY loPEZ LEONARDO DA CUTRI White
1 P-K4 2 P-KB4 3 B-B4 4 Kt-KB3 5 PxP
P-Q3 P-QB3 B-Kt5 ?
7 KtxPch 8 QxB
9 Q-K6ch 10 Q-BSch 11 QxQch
12 Kt-lJ7ch
KxB K-Kl Kt-B3? Q-K2 Q-Ql
1 5. GIUOCO PIANO
Other games from this match are corded in a manuscript by Polerio.
game won by Leonardo (White) went as follows: 1 P-K4, P-K4; 2 Kt KB3, Kt-QB3; 3 B -B4, B---f-, ,.,i'-
23 Q-B7ch! 24 Kt-K6 mate
Kt x Q
Modern Chess Hereabouts we arrive at the era of what is called, oc
casionally in rather a disdainful tone, "modern chess."
is the age of the great Lasker and Tarrasch, of Schlechter and Maroczy, of the attacking geniuses Pillsbury and Marshall and Janowsld. As the number of grandmasters increases, as it becomes more
difficult to bowl over one's opponent in short order, we find that positional chess begins to be pre-eminent ; before the opponent can be finished off with a brilliant com bination, it is generally
necessary to outplay him positionally,
in order to create favorable conditions for sacrificial play. That is why Emanuel Lasker once wrote : "If you play well
positionally, the combinations will come of themselves." While I am fond of the finest games of all these masters, I love above all the beautiful games of the immortal Harry Nelson Pillsbury.
I am sure that the reader, as he plays over
these marvellous games, will share my admiration for this immortal, whose beautiful productions, I am sorry to say, do not seem to be adequately appreciated nowadays. During his lifetime his uncanny
sldll in blindfold play was particularly admired, and that is why I have carefully assembled the cream of his efforts in this field.
Happy the man who plays over
these games for the first time !
And as for old-timers like
myself, they will relish the opportunity to renew their ac
quaintance with these gracious companions of their youth !
Briton meets Briton GIUOCO PIANO
E. THOR.OLD
J. H. BLACKBUR.NB
1 P-K4 2 Kt-KB3 3 B-B4 4 P-Q3 5 B--K3 6 BxB 7 QKt-Q2 8 P-B3 9 B--Kt3
1 0 PxP 11 Q-K2
12 13 14 15 "6 17 18 19 20 21 22 23 24 2S 26 27 28 29 30 31 32 33 34 35 36
P-Kt3 P-KR4 B-B2 Q--K3 P-QKt4 B-Kt3 Kt-Kt5 Kt x B P-KB4 P-B5! PxP 0--0 R-B5 QR-KBl Kt-B4 Kt x P R x Kt Q-B4 R B3 K-Kt2 R-K8ch Q-K5ch R-B5 K-Bl R-KKt8! -
P-K4 Kt-QB3 B---B4 Kt-B3 B---Kt3 RP x B
36 . . . . 37 R x Pch 38 Q-B5ch 39 Q x Rch 40 Q--B4 41 B-Q5 42 Q x KBP
K-Kt2 K-Rl Q--Ql R-QKt7 Resigns
PxP Q--K2 P-Kt3 Kt--Q2 Kt-B4 P-R4 R-Ql
Kt-Q2 Kt-Bl B-K3 Kt x Kt Q--B3 Kt-BI QxP R-Q2 Kt-KR2 R-KB1 P-Kt4 Kt x Kt
K-Rl R-KKtl Q-Kt3ch R-Kt5
K-Kt2 K-R3 R--Q7ch Q-Kt3
(see diagmm next column)
An attack carried able verve.
about 1891.
0111 with
VIENNA GAME M. KUER.CHNER. DR.. s. TAR.RASCH White
P-K4 Kt-QB3 P-KKt3 B--Kt2 P--Q3 P-B4 P-B5 P-KKt4 B-Kt5 Kt--Q5 BxQ Q-Q2
P-K4 Kt-QB3 Kt-B3 B--B4 P-QR3 P-Q3 P-KKt3 P-KR4 Kt--Q5 .Kt: x Kt!! Kt-K6 QKt x Pch
K-K2 K-B2 K-Kt3 Q--Kt5 QxP K-R3
1 08.
Kt-Q5ch Kt x Pch PxP P-R5ch P--B5ch Kt-B7 mate
Havana, January, 1892.
For World S11prema&y in Cheu This is the fourth game of the second match and is also one of the moJI bea11tif11I game1 ever played in a similar contest. RUY LOPEZ W. 5TEINITZ White 1
P-K4 Kt-KB3 B--Kt5 P-Q3 P-B3 QKt-Q2 Kt-Bl B-R4 Kt-K3 B--B2 P-KR4 P-R5 RP x P PxP Kt x Kt B--Kt3 Q-K2 B--K 3
Q-Bl ! P-Q4 Kt x P R x B! R x Pch! Q--Rlch B--R6ch!
M. TCHIGORIN Black
P-K4 Kt-QB3 Kt-B3 P-Q3 P-KKt3 B-Kt2 0-0 Kt-Q2 Kt-B4 Kt-K3 Kt-K2 P-Q4 BP x P? Kt x P Q x Kt Q--B 3 B-Q2 K-Rl QR-Kl P-QR4 PxP B x Kt Kt x R KxR K-Kt2 K-B3
27 Q-R4ch 28 Q x Ktch 29 Q--B4 mate
K-K4 K-B4
109. Dresden Tournament, 1 892.
First edition of a famous trap! RU"f LOPEZ DR. s. TARRASCH White
P-K4 Kt-KB3 B--Kt5 P-Q4 Kt-B3 0-0 R-Kl B x Kt!
G. MAR.co Black
P-K4 Kt-QB3 P-Q3 B-Q2 Kt-B3 B--K 2 0-0? BxB
From this point Black's moves are all forced. 9 10 11 12 13 14 15 16 17
PxP Qx Q Kt x P Kt x-B Kt-Q3! P-KB3 Kt x B B-Kt5 B--K7
PxP QR x Q BxP Kt x Kt P-KB4 B--B4ch Kt x Kt R-Q4
THE GoLDEN TllEASURY OF
1 1 0. New York. 1 892.
011tpla1ing a f11ttwe world 'ht111-1 pion.
DR. E. l.AsKER White
1 2 3 4 s 6 7 8 9 10 11 12 13 14 lS 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
P-K4 Kt-KB3 B-Kts Kt-B3 0-0
P-Q3 B-K3 P-Q4 B-QB4 P-Q5 P-KR3 QxB PxP Q-Kt4 B-Q2 B-Q3 Kt-K4 QR-'-Kl Kt-B3 Kt-Ql B-B3 P-Kt4 B-Kt2 Q-QB4 P-ff4· Q-B6 Q x RP Q x KtP Kt-B2 B-K4 Q--B4 B-KB3 R x Kt Q-K4 Kt x Kt BxQ K-R2 B-Q3
35 36 37 38 39 B-B4
A. B. HODGES Black
P-K4 Kt-QB3 P-Q3 B-Q2 KKt-K2 Kt-Kt3 B-K2 0-0
B-Kts Kt-Ktl B x Kt P-KB4 Kt-RS Kt x P Kt-Q2 P-KKt3 R-B2 Q-KJH P-QR3 Q-Kt2 QR-KBl B-Ql Kt-B3 Kt-R4! P-QKt4 Kt-K2 Kt x BP Q-R3 Q--Kt4 Kt-B4 Kt-Kt6 Kt x R Q-R5 Kt x Pch QxQ R x Reh R-K8 P-K5 B-B3
BxB K-Kt3 Kt-KtS B-Q3 Resigns
RxB P-K6 R-B7 R-KKt8
1 1 1 . Played at Zugzidi, in spring of 1892.
Mo1t Brilliant of Dadian'J CombinationJ. TWO KNIGHTS' DEFENS:E
DADIAN (of Mingrelia)
P-K4 Kt-KB3 B-B4 P-Q4 0-0
R-Kl BxP Kt-B3 R x Ktch B-Kt5 Kt-Q2 Kt-Kt3 Kt-Q5 Kt-BS
M. BITCHAM Black
P-K4 Kt-QB3 Kt-B3 PxP Kt x P P-Q4 QxB Q--B5 B-K3 B-B4 Q-R3 B-Kt3 P-KR3
r�l'Vm·l!J �-
Q-Kt4 . . . . K-Bl R x Bch! K-Ktl Kt-Q7ch P-KR4 Q-Kt4 P x Kt Kt (Q5)-B6ch! Q-Kt4 B-R6ch! Kt x P mate
. . Q-R5ch Kt-B6ch! Kt x KtPch R x Qch Kt-K5ch Kt-B7ch Kt-Q6ch Q--K8ch! Kt-B7 mate
PxB P-Kt3 B x Kt Q-K2 BxR K-Ql K-Kl K-Ql RxQ
Boston, Nov. 8, 1 892.
Ca!Jght in the Web DANISH GAMBIT F. K. YOUNG White
P-K4 P-Q4 P-QB3 B--QB4 Kt-KB3 0-0
Kt x P R-Klch Kt-Q5 B-Kt5 R-QBl R x Kt Kt-K5!
1 1 3. Jackson, Miss., about 1 892.
L. DORE This Galbreath-tailing game Black P-K4 PxP PxP Kt-KB3 Kt x P Kt-Q3 Kt x B B-K2 Kt-B3 P-B3 P-Kt4 PxR
played in fackson, Miss., about 1 892.
EVANS GAMBIT JOHN A. GAL8RAITH White
P-K4 Kt-KB3 B-B4 P-QKt4 P-B3 0-0 P-Q4 B-KKt5 PxP QKt-Q2 R-Kl B-R4 Kt-K4
Now begins tion. 14 15 16 17 18 19 20
Black P-K4 Kt-QB3 B-B4 BxP B-R4 B-Kt3 Q--B3 Q-Kt3 QxP Q-Kt3 P-KR3 KKt K2 0-0 -
far-sighted combina
Kt-B6ch BxP B-Q3 P-Kt4 K-Rl R-K4 Q-Ktlch!
P x Kt Kt-84 Q-R4 Q x Pch BxP Q--R6 Kt-Kt6ch
Kt x B Q-K2 R x Pch Q-Kt4
P-B5 P x Kt Q-Kl ? K-Kt3 R-Rl
Black mates in 4 moves: R-R6 ch, etc. l lS.
Vienna, 1893.
Schlechler'J Immortal ThiJ 1park/ing gem rank1 a.r one of
Q x Ktch R-KKtl R x Bch PxQ R-KR4 RxP Kt-Kt5 RxB R-Kt7ch R x Kt mate!
BxQ P-Q3 QxR B--B4 BxB B--R2 Kt x P Kt-Kt3 K-Rl
1 14. Vienna, Dec., 1892. The open KR file tri11mph1 again!
the most curiouJ and brilliant on record.
IRREGULAR OPENING_ , B. FLEISSIG White
P-QKt4 B-Kt2 P-QR3 P-Kt5 P-Q4 Kt-B3 Q-Q3 QxP Q x KtP K-Ql
SCHLECHTER Black
P-K3 Kt-KB3 P-B4 P-Q4 Q-R4ch Kt-K5 PxP B-B4! B x Pch P-Q5 ! !
M. PoLLAK
P-K4 Kt-QB3 P-KKt3 B--Kt2 KKt-K2 P-KR3
7 0-0
K-R2 P-Q3 B--Kt5 P-B4 P x Kt B-R4
P-K4 Kt-QB3 Kt-B3 B-B4 P-Q3 B-K3 Q-Q2 P-KR4
Kt�K2 Kt-Kt5ch!? P x Pch Kt-Kt3
1 1 Q x Rch 1 2 Q :x B
K-K2 P x Kt
B---B l QxR B--B4 K-Bl BxB B x Kt K-Ktl K-�2
Kt-Q2 Q x KtP Q-Q4ch B--K6ch! Kt-B7! Q-Q7ch Q--Q8ch Q x P mate
Played at Kassa in 1893.
A _Charou.rek Gem DANISH GAMBIT M. WOLLNER R. CHAROUSEK White
P-K4 P-Q4 P-QB3 B-QB4 Kt-KB3 Kt x P 0-0
Kt-KKt5! Kt x BP P-K5 P-K6! P x Rch B-B4 Q-K2! K-Rl QR-Kl Q-K8ch P x R (Q) ch B x QP mate
p_'.._K4 PxP PxP Kt-KB3 B-B4 P-Q3 0-0 P-KR3 R x Kt Kt-Kt5? Q--R5 K-Bl Kt x BP Kt-Kt5ch B-Q2 Kt-QB3 RxQ BxQ
Inimitable elegance! FROM'S GAMBIT 1.
C. ScHLECHTER
P-K4 Kt-QB3 P-Q3 BxP Kt-B3 P-KR3 P-KKt4 Kt-K5 P-Kt5
Now follows a very elegant com bination.
. . . . BxQ K-K2 K-Q3 K x Kt
P x Kt! P-B7ch B--Kt5ch Kt-Kt5ch P-B4 mate!
1 18. Nuremberg, Feb. 9, 1894. A wonderful combination! KING'S GAMBIT DR. s. TARRASCH White
1 17. Vienna Chess Club April 27, 1 894.
P-KB4 PxP Kt-KB3 PxP P-Q4 B-Kt5 B-R4 B-B2 P-K3 B-R4
P-K4 P-KB4 Kt-KB3 P-KR4 Kt-K5 Kt x BP B-B4ch P-Q4 BxP P-R5ch Kt-B3 P-K5 P-R6ch PxP RxQ 0-0 Kt-Q5 K-Rl
HIRSCHLER Black
P-K4 PxP P-KKt4 P-Kt5 P-Q3 K x Kt K-Kt3 B--K2 Kt-KB3 K-Kt2 Kt-B3 PxP K-Bl Q x Qch Kt-Q2 K-Kl B--B4ch B-Kt3
P-K6 Xt-B6ch B-KKt5 Kt x KtPch Kt-B6ch Kt-Kt8ch
25 R-Q8ch 26 R-B8ch 27 P-K7ch
KKt-K4 K-K2 Kt x B K-Kl K-K2 K-Kl
Kt x B BxB R-Kl ! Q-K2 QR-Bl P-Q5! Kt-Q4 1 9 Kt-K6 20 Q-Kt4 21 Kt-Kt5ch
Q x Kt Kt x B P-KB 3 Q-Q2 P-B3? PxP K-B2 KR-QBl P-KKt3 K-Kl
Kt x R KxR Resigns 22 23 24 25
R x Ktch!! R-B7ch R-Kt7ch R x Pch!
K-Bl K-Ktl K-Rl Resigns
1 19. Hastings, 1 895. First Brilliancy Prize
GIUOCO PIANO W. 5TEINITZ C. VON BARDELEBEN White
P-K4 Kt-KB3 B-B4 P-B3 P-Q4 PxP Kt-B3 PxP 0-0
B-KKt5 B x Kt
P-K4 Kt-QB3 B-B4 Kt-B3 PxP B-Kt5ch P-Q4 KKt x P B-K3 B-K2 QB x B
Steinitz gives this brilliant mate ten moves. 25 26 27 28 29 3Q 31 32 33 34 35
. . . . R-Kt7ch Q-R4ch Q-R7ch Q-R8ch Q-Kt7ch Q-Kt8ch Q-B7ch Q-B8ch Kt-B7ch Q-Q6 mate!
K-Ktl K-Rl KxR K-Bl K-K2 K-Kl K-K2 K-Ql Q-Kl K-Q2
1 20. Quadrangular Tourney, St. Petersburg, 1895-96. One of Pi/Lsbury's memorable games.
PETROFF DEFENSE OR.. E.
P-K4 Kt-KB3 Kt x P Kt-KB3 P-Q4 B-Q3 0-0 R-Kl P-B3 Q-Kt3 B-KB4 PxB K-Kt2 Q-B2 B-QBl Kt-Q2 Kt-Bl
Q-Ql QxR K x Kt Q-Ql K-K2 K-Q2
H. N.
P-K4 Kt-KB3 P-Q3 Kt x P P-Q4 B--K 2 Kt-QB3 B--KKt5 P-B4 0-0 B x Kt Kt-Kt4 Q-Q2 Kt-K3! B--Q 3 QR-Kl Kt (K3) x P
RxR Kt x P! P-B5 Kt-K4ch Q-Kt5ch Q x Qch
KxQ K-K2 P-B3 P-Kt3 K-Q2 B--Kt2 P-KR3 Kt-R2 P-B4 PxP Resigns
Kt x B Kt-K4 R-Kl Kt-Kt5ch Kt-K6 Kt-Kt7 B--B4 B--B7 PxP P-KR4!
The manner in which Pill.rbur1 snapped up the Knight with hi.r Bishop at the eleventh move, and his rapid play afterward.r, showed clearly that he saw through the game to victory.
St. Petersburg, 1895-6.
One of Dr. La.rker's finest.
A game of many combinations. QUEEN'S GAMBIT DECLINED DR. E. LAsKER W. STEINITZ Bla-KB4 P-B3 Kt x P! BxP P-KKt4 P x Kt P-B6 B-Kt6ch! ! Q-Q3ch Q-R3ch Kt-B4ch! K-Rl R-Ktlch R x Bch! R-Ktl mate
241. Meran, 1 926
D. PRZEPIORKA White
l 2 3 4 5 6 7 8
P-K4 P-Q4 Kt-KB3 1>-Q3 0-0
B-K3 Q-Q2 B-KR6
]. VON PATAY Black
P-KKt3 P-Q3 B-Kt2 P-K3 Kt-K2 0-0
R-Kl B-Rl
QKt-B3 P-Q4? Kt-B4 P-B3 P-KKt4? P x Kt Q-Q2 P-KR3 PxB K-82 KxB K-R3 K-Kt3 P x Kt BxP 1>-Kt4 KxR
242. New York, 1 927. 2nd Brillian'y Prize
DUTCH DEFENSE (in effect) A. ALEKHINE White
/u11 one Jacrifi(e after another! KING"S FIANCHETIO DEFENSE
P-Q4 P-QB4 Kt-KB3 KKt-Q2 Q-B2 QKt-B3 Kt (2) x Kt 1>-B4 P-K3 l>-K2 P-QR3 0-0 P-B3 PxB P x KP RxR Q-Q2 P x KP! Q-B4!
F. J. MARSHALL Black
Kt-KB3 P-K3 Kt-K5 1>-Kt5 P-Q4 P-KB4 BP x Kt 0-0 P-B3 Kt-Q2 B-K2 B-Kt4 BxB RxP R x Reh P-K4 P-B4 P-Q5 P x Kt
Q-B7ch p x P.! Q-K7 l>-R5!! P-K6 P x Kt R-B7
K-Rl Q-Ktl P-KR3 P-QR4 P-KKt3 BxP Resigns
Twenty-first Match Game, October, 1927.
White'J game crumble1 Jo1hua'1 trumpet.
QUEEN'S GAMBIT DECLINED J. R. CAPABLANCA A. ALEKHINE White Black P-Q4 1 P-Q4 243. Kecskemet, Hungary, 1927. P-K3 2 P-QB4 White's deep combinilion has pret Kt-KB3 3 Kt-QB3 ty points. QKt-Q2 4 B-Kt5
B-K2 - ·5 P-K3 SICILIAN DEFENSE 6 Kt-B3 0-0 A. TAKACS F. D. YATES P-QR3 7 R-Bl P-R3 8 P-QR3 Black White PxP 9 B-R4 P-QB4 1 P-K4 10 B x P P-QKt4! Kt-QB3 2 Kt-KB3 B-Kt2 1 1 l>-K2 PxP 3 P-Q4 1 2 0-0
P-B4 Kt-B3 4 Kt x P Kt x P 13 PxP P-Q3 5 Kt-QB3 R-Bl 14 Kt-Q4 P-K3 6 B-K2 QKt-Q2 1 5 P--QKt4 1>-K2 7 0--0 Kt-Kt3 16 B-Kt3 P-QR3 8 K-Rl KKt-Q4 1 7 Q-Kt3 Q-B2 9 1>-K3 R-B5! 18 B-B3 l>-Q2 10 P-B4 Q-Bl 1
9 Kt-K4 P-QKt4 1 1 Q-Kl Kt x R 20 R x R 0-0 - 1 2 P-QR3 2 1 R-Bl Q-Rl ! ! Kt-QR4 :i. 3 R-Ql R-Bl 22 Kt-B3 'Kt-B5 14 Q-Kt3 B x Kt 2 3 Kt x Kt KR-Bl 1 5 B-Bl 24 B x B QxB Kt x RP --16 P-Kt3 B-B3 25
P-QR4 Kt-Kl 1 7 P-K5 26 Kt-B3 B-Kt7! P-Q4 18 Kt-K4 R-Ql 27 R-Kl K-Rl 19 Kt-B6ch PxP 28 P x P Kt x Kt 20 Q-R4 P-K4 29 P-R3 P-Kt3 21 B-Q3 P-K5 ! 30 R-Ktl B-Bl 22 P x Kt B x Kt 31 Kt-Q4 K-Ktl 23 Kt-B3
Kt x P! 32 R-Ql P-R3 24 Kt-Kt5 Resigns P x Kt 25 B x Kt BxB 26 P x P PxB 27 B x KKtP 245. u. s. s. R., 1927. R-Bt 28 R-Q3 BxP 29 P-QKt4! ! An intere1ting portent of Boti1in Resigns 30 R-KR3 nik'1
later fame.
143 .
M. BOTVINNIK Black
P-K3 P-Q4 P -KB4 P-QB4 Kt-KB3 P-KK.t3 B-K2 B-Kt2 0--0 Kt-QB3 P-Q4 Kt-B3 P-B3 0---() Q--K l Q--B 2 B-B4 Q-R4 QR-Ql QKt-Q2 Kt-K5 P-Kt3 Kt-K5 Kt-Kt4 !? Kt-K5 ! P-KR4? Q-Kl B-B3 B x Kt Kt x QKt B-Kt5 !
K-Kt2 BP x B B x Kt? Q-R4 R-KRl Q--K t3! P-B3? P-K4! K-Bl QP x P R x B! Q-Kt6! P xR P x Kt Kt x KP B-B4 RxB Q x Pch P-K3 Q x Rch Q-B2 Q-R6! K-K2 Q--Kt5ch P-B5 R-KB1 K-Q2 Q x BP P-K6 R x Q and wins
Los Angeles, 1928.
A lively variation JeadJ to a bright finiJh. 1WO KNIGHTS' DEFENSE
F. WILLIAMS White
1 P-K4
1. HAEGG
Kt-KB3 B-B4 Kt-Kt5 Kt x BP KxB K-K3 K x Kt BxP P-KKt4 Q--Kl
Kt-QB3 Kt-B3 B-B4 (?! ) B x Pch Kt x Pch Q--K2 P-Q4ch Q-R5ch BxP B-B4ch
and Black mateJ rn three mot1eJ.
Trenchin-Teplitz, 1928.
A problem mate in actual play! CARO-KANN DEFENSE
SPIELMANN White
P-K4 Kt-QB3 Kt-B3 P-K5 Q--K2 QP x Kt Kt-Q4 P-K6! Q-R5ch Kt-B3 Kt-K5 Kt-B7
WALTER Black
P-QB3 P-Q4 Kt-B3 Kt-K5 Kt x Kt P-QKt3 P-QB4? PxP K-Q2 K-B2 B-Q2 Q-K l
144 1 3 Q--K5ch 14 B--KB4 1 5 Q-B7ch 16 Kt-QB ! 1 7 Q-Kt7ch 18 P-R4ch 19 Q x Ktch 20 Kt x P mate!
THE GoLDE N TREASURY OF CHESS K-Kt2 P-B5 K-R3 Kt-B3 K-Kt4 K-B4 BxQ
Rogaska..Slatina, 1929.
The game that made Flohr famous. QUEEN'S GAMBIT DECLINED SALO FLOHR F. SAEMISCH Wblte
1 P-Q4 2 P-QB4 3 P-QR3 4 Kt-QB3 5 B--Kt5 248. Match, 1928. 6 P-K3 7 PxP Colle worlu up a murderous attack 8 B--Q3 with his customary ingenuity. 9 KKt-K2 INDIAN DEFENSE 1 0 Kt-Kt3 1 1 P-KR4! E. CoLLE
S. LANDAU 1 2 B--KR6 White Black 1 3 P-R5 l P-Q4 14 P x P Kt-KB3 1 5 Q-B3 P-QKt3 2 Kt-KB3 16 QKt-K2 B--Kt2 3 P-K3 1 7 0-0-0 P-Q3 4 B--Q3 18 R-R3 QKt-Q2 5 0-0 . 19 B x Kt 6 QKt-Q2 P-K4 20 B x P! 7
P-K4 P x P? 21 Kt-B4 P-Kt3 8 Kt x P 22 Q x B B-Kt5 ! P-QR3 9 23 QR-Rl 10 B--B6 Q---B l PxP 1 1 P-K5 ! ! BxB 1 2 Q---B 3! B--Q3 1 3 Kt x B 14 Kt-B4! P-K5 P-R3 15 R-Kl 16 Q---B 3! Q---Kt2 17 Kt x Bch P
x Kt 18 R x Pch! K-Bl 19 R-K7 ! K-Kt2 20 B--B4 QR-QBl 2 1 Q---QKt3 P-Q4 22 Kt-K5 QR-Kl 23 R x Pch K-Ktl P-KKt4 24 Q---Kt3 R x Kt 25 B � P! KxR 26 B x Ktch 24 R-R8ch K-K3 27 Q---Kt7ch 25 R x Qch
Resigns 28 B x R 26 Q---R6ch
Kt-KB3 P-K3 P-Q4 B--K2 0-0 P-QKt3 PxP B--Kt2 QKt-Q2 Kt-Kl? P-Kt3? Kt-Kt2 P-KB4 PxP P-B3 B--Q3 Q---B 3 K-B2 QxB Kt-B3 B x Kt QR-Kl K-Ktl
QxR KxR K-Ktl
27 Q x Pch K-Rl 28 Q-R6ch K-Ktl 29 Kt-R5 and wins 250.
E. COLLE
P-Q4 Kt-KB3 P-K3 B--Q3 QKt-Q2 0-0
B-Kt2 Q-K2 R-Ql Resigns
Carlsbad, 1929.
Brilliancy Prize
R x KP P-K6 R-Kt3 QxP
P-B4 P-QKt3 B-Kt2 R-Bl Kt-K5 Q-K2 BP x P P-B4 Q x Kt Kt-Kt4 R-B3! R-Kt3 Q-QB2 B---B5 R-R3· R-Bl ! Kt-K5! BP x B B x P! R-B2 BxR Q-Ql B-Kt2 P-KKt4! P x BP
F. 0. YATES
Kt-KB3 P-QKt3 B-Kt2 P-K3 P-Q4 B-Q3
QKt-Q2 Q-K2 QR-Ql P-B4 Kt-K5 KP x P Kt x Kt P-B3 KR-Kl Q--K3 K-Rl Kt-Bl Q--B 2 P-KR4 K-Kn B x Kt P x QP B-R3 R-Bl RxB P-B4 P-Kt3 P-Q5 P x KP
Antwerp, 1929.
One of ten blind/old gameI MAX LANGE ATIACK
G . KOLTANOWSKI P. White
1 2 3 4 5 6 7 8 9 10 11 12 B
P-K4 Kt-KB3 B-84 0-0
P-Q4 P-K5 P x Kt R-Klch Kt-Kt5 Kt-QB3 QKt-K4 P-QB3 PxP P-Kt4 Kt x B P-B7ch Kt-Kt5ch RxP
P-K4 Kt-QB3 B-84 Kt-B3 PxP P-Q4 PxB 1>-K3 Q--Q4 Q--84 B-Kt5? PxP B-R4 Q-Kt3 P x Kt KxP K-Ktl Q--Q6
l P-Q4 2 Kt KB3 3 P-K3 4 P-B3 5 B-Q3 6 QKt-Q2 7 0-0
R-KBl Q-Q2 KxR Kt-K2 Q x Pcb Q x Kt Q-R5 Q-B 3 Resigns
Q-Kl ! R-KS! R x Reh �R3ch R-Ql !! K-Bl ! R-Q5! R-R5 R-KB5 !
CHESS P-Q4
Kt-KB3 P-B4 P-K3 �Q3 QKt-Q2
R-Kl QP x P Kt x Kt PxP
8 R-Kl 9 P-K4 1 0 Kt x P 1 1 B x Kt
Manhattan Chess Oub,
Spring, 1930.
White gives odds of QR. The kind of mate that odds-given pray for. I.
B . HORNEMAN
P-K4 P-Q4 P-K5 Q--Kt4 Kt-KB3 Q-R3 �Q3 Q-Kt3 B x Kt QxP Kt x P Kt x P �Kt5
14 P-K6!
15 Q-Kt6ch!! 16 Kt-Kt7 mate
Black P-K3 P-Q4
P-QB4 PxP Kt-KR3 B-K2 P QKt3 Kt B4 PxB R-Bl �R3? Kt-Q2 P-B 3? PxB PxQ -
253. Nice, 1 930. Fir.rt Brilliancy Prize QUEEN'S PAWN OPENING J. J. O'HANLON E. COLLE White
KxB K-Kt3 R-Rl Kt-B 3 K-R3 Q-R4 K-R2 K-Ktl Resigns
1 2 B x Pch ! ! 1 3 Kt-Kt5ch 14 P-KR4! 1 5 R x Pch! !
16 P-R5ch 17 R x B 1 8 Kt x Pch 1 9 Kt-Kt5ch 20 Q-Kt3ch 254.
Black concludes with one of the most beautiful mates ever seen in actual play.
BOGOLYUBOV White
1 P-Q4 2 P-QB4
M. MONTICELLI Black
Kt-KB3 P-K3
3 Kt-QB3 4 Kt-B3 5 B-Kt5 6 PxB 7 P-K3 8 B-Q3 9 0-0 10 Kt-Q2 1 1 B-R4 1 2 B-Kt3 1 3 P-QR4 14 R-Ktl 1 5 P-B3 16 P-K4 1 7 B-Kl 18 P-R3 19 P-B5 ! 20 P-Q5 21 Kt-B4 22 R-B2 2 3 P-Q6! 24 Kt x Reh 2 5 B-B4
26 P x P 27 R-Q2? 28 Q-Kt3 29 B-Q3 30 B x P 31 P x B 32 Q-B2 33 P-B4 34 B x P 35 P-Kt3 36 R-Kt3
B-Kt5 P-QKt3 B x Ktch B-Kt2 P-Q3 QKt-Q2 Q-K2 P-KR3 P-KKt4 0-0-0 P-QR4 QR-Kt1 P-R4 P-R5 P-K4 Kt-R4 QP x P Kt-B5 R-R3! P-B4! R x P! Q x Kt R-Bl RxP Q-K2 R-Bl P-K5 ! BxB QxP Q-B3 P-Kt5 PxP Kt-K4!
Black calls mate in 4. Kt-K7ch!! . . . . R-B8ch! R x Kt KxR Q-R8ch K-B2 Kt-Kt5 mate
255. Hamburg, July, 1930.
Brilliancy Prize INOIAN DEFENSE G.
1 2 3 4 5 6 7 8 ·9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
P-Q4 P-QB4 Kt-QB3 Q-Kt3 PxP Kt-B3 B-Q2 Q-B2 P-QR3 BxB P-QKt4 P-K3 B-Q3 Q x Kt 0-0
B-K2 KR-Ql P-QR4 P-R5 QxP Q-B3 PxP Kt-Kl R-R7 Q-K3 R-R2 P-B3 B-Q3 B-Bl R-KB2
A. ALEKHINE Black
Kt-KB3 P-K3 B-Kt5 P-B4 Kt-B3 Kt-K5 Kt x QBP P-B4 B x Kt 0-0 Kt-K5 P-QKt3 Kt x B B-Kt2 Kt-K2 Q-Kl R-Ql P-B5! P x KP Kt-B4 P-Q3! PxP P-K4 Kt-Q5 ! R-Q2 R ( 2 ) -KB2 R-B5 Q-R4 Q-Kt4!
BxP R-Kt2 B--B7 B--B4 R-QBl K-Rl Resigns
KR-Ku P-R5 R-QBl Kt-R4 B-K5ch Kt-Kt6ch!
257. Zwickau, 1 930.
B/ack'I play iI 1tudded with 1ac rifice1. 30 . . . . 31 K-Rt Resigns
P-R3 ! R x Pl!
If 32 Q x Q, R x R; etc. 256.
Hamburg, 1930.
Exemplary preciiion INDIAN DEFENSE G. STAHLBERG I. KAsHDAN White Back 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
P-Q4 P-QB4 Kt-QB3 Q-Kt3 PxP Kt-B3 B--Q2 Q-B2 P-K4
B-Q3 K-Ktl KR-Ktl B--K3 Q x Kt PxB KP x P RxQ R-Q2
Kt-KB3 P-K3 B--Kt5 P-B4 Kt-B3 Kt-K5 Kt x QBP 0-0 Q-B3! P-QKt3! P-QR4! Q-Kt3 B-R3 Kt x B B x Kt P-Q4! Q x Qch BxP B x QP
ENGLISH OPENING P. BLECHSCHMIDT S. FLOHR Black White l P-QB4 2 P-KKt3 3 B--Kt2 4 Kt-QB3 5 Kt-B3 6 0-0 7 P-KR3 8 P-K3? 9 K-R2 10 P-Q4 1 1 P x RP 12 R-Rl 1 3 RP x P 14 K:-Ktl 1 5 P-Q5 16 Q-R4ch 17 Kt x
Kt 18 P-K4 19 K-Bl 20 P x B 21 K x B 22 B-K3 23 QR-QKtl 24 Q x KtP 25 Q-B6! 26 K-B3 27 R x Reh 28 B-Q4! 29 B x Kt 30 K-Kt3 31 K-R2
Kt-KB3 P-B4 P-KKt3 B--Kt2 Kt-B3 P-Q3 B--Q2 Q-Bl P-KR4! P-R5 !! P-KKt4! P-Kt5 ! B x P! Q-B4 Kt-K4 KKt-Q2 B x Kt Q-Kt3 B x Kt B--K7ch QxB QxP P-Kt4!! R-QKtl ! Q x Pch P-B4! ! K-B2 Kt-K4ch! Q-K5ch!
Q-Kt5ch R x P mate
MODERNS, HYPERMODERNS AND Ecucnc.s
258. Los Angeles-San Francisco Match, San Luis Obispo, May, 1931 (Board No. 1 7 )
22 P x R Q-Kt4 RP x Kt 23 Kt x Qch 24 R-Q2 and wins
White saves himself with an amaz ing reso11r(e.
l 2 3 4 5
P-K4 Kt-KB3 B-B4 P-B3 P-Q4
6 0--0 7 P-KR3
R-Kl Q-Q3 B-Q5 B-K3 PxP Kt-R2 QB x Kt Kt-Q2 QKt-Bl B x Pch 18 Q x Bch 19 R-K2 20 Kt-B3
P-K4 Kt-QB3
B-Kt3 Q-K2 P-Q3 Kt-B3 P-KR3 Kt-KR4 B-Q2 P-Kt4 PxP Kt-B5 KtP x B R-KKu | {"url":"https://dokumen.pub/the-golden-treasury-of-chess.html","timestamp":"2024-11-11T20:31:42Z","content_type":"text/html","content_length":"85467","record_id":"<urn:uuid:522dd9d5-e304-471d-92a2-a8ccf762aa46>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00622.warc.gz"} |
CS102: Data Structures and Algorithms: Sorting Algorithms Cheatsheet | Codecademy
Merge Sort is a divide and conquer algorithm. It consists of two parts:
1) splitting the original list into smaller sorted lists recursively until there is only 1 element in the list,
2) merging back the presorted 1-element lists into 2-element lists, 4-element lists, and so on recursively.
The merging portion is iterative and takes 2 sublists. The first element of the left sublist is compared to the first element of the right sublist. If it is smaller, it is added to a new sorted list,
and removed from the left sublist. If it is bigger, the first element of the right sublist is added instead to the sorted list and then removed from the right sublist. This is repeated until either
the left or right sublist is empty. The remaining non-empty sublist is appended to the sorted list. | {"url":"https://www.codecademy.com/learn/cspath-cs-102/modules/data-structures-and-algorithms-sorting-algorithms/cheatsheet","timestamp":"2024-11-10T01:51:02Z","content_type":"text/html","content_length":"227651","record_id":"<urn:uuid:fe252c8e-ffaf-471d-83b8-23f695686846>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00634.warc.gz"} |
1-sec^2x Formula, Identity | 1-sec^2 theta is equal to - iMath
1-sec^2x Formula, Identity | 1-sec^2 theta is equal to
The simplification of the expression 1-sec^2x is equal to -tan^2x. In this post, we will establish 1-sec^2x formula.
We have:
• 1- sec^2x = – tan^2x
• 1- sec^2θ = – tan^2θ.
Let us now prove the above formulas.
1-sec^2x Formula
The formula of 1-sec^2x is given below:
$\boxed{1-\sec^2 x=-\tan^2 x}$
To simplify the expression $1-\sec^2 x$, we will follow the below steps:
Step 1: We will use the following trigonometric identities to find the value of 1-sec^2x:
1. sin^2x +cos^2x = 1
2. secx = 1/cosx
3. tanx = sinx/cosx
Step 2: Substituting the value of secx from the above in the expression 1-sec^2x, we will get that
1-sec^2x = $1-\left(\dfrac{1}{\cos x} \right)^2$
= $1-\dfrac{1}{\cos^2 x}$
= $\dfrac{\cos^2 x -1}{\cos^2 x}$
= $\dfrac{\cos^2 x -(\sin^2 x +\cos^2 x)}{\cos^2 x}$
= $\dfrac{\cos^2 x -\sin^2 x -\cos^2 x}{\cos^2 x}$
= $\dfrac{-\sin^2 x}{\cos^2 x}$
= $- \left(\dfrac{\sin x}{\cos x} \right)^2$
= – tan^2x
Thus, the simplification or the formula of 1-sec^2x is equal to – tan^2x.
Remark: Putting x=θ in the above formula, we get the following simplification of 1-sec^2θ:
1-sec^2θ = – tan^2θ.
Also Read:
Question-Answer on 1-sec^2x Formula
Question 1: Find the value of 1-sec^245°
From the above, we know that
1-sec^2θ = – tan^2θ
Put θ = 45°.
So we get that
1-sec^245° = – tan^245°
= -1^2 as we know that tan45°=1.
= -1
So the value of 1-sec^245° is equal to -1.
Q1: What is the formula of 1-sec^2x?
Answer: The formula of 1-sec^2x is given by 1-sec^2x= -tan^2x.
Q2: What is the formula of 1-sec^2θ?
Answer: The formula of 1-sec^2θ is given as follows: 1-sec^2θ= -tan^2θ. | {"url":"https://www.imathist.com/simplify-1-sec-square-x/","timestamp":"2024-11-09T15:52:51Z","content_type":"text/html","content_length":"179366","record_id":"<urn:uuid:9d024a8a-9148-44e9-931e-fc3d13804474>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00091.warc.gz"} |
Listing of all the works of the organization. Click on the work title to get the full information.
Glushkov A.V., Belyaev A.A., Putrya F.M., Alekseev I.N., Mironova Yu.V.
“MULTICORE” platform IP-library of SoC peripherals
Balashov A.G., Krupkina T.Yu., Tsimbalov A.S.
Criteria of a choice of models at calculation of device characteristics of submicronic transistor structures
Kozlov A.V., Parmenov Yu.A.
Development of an optimum design of a magnetic field sensor on the basis of lateral magnetic transistor using tools of device technological simulation
Glebov A.L.
Methods of statistical timing analysis of digital circuits
Gavrilov S.V., Glebov A.L.
Noise analysis of digital circuits with accounting of logic constraints
Korolev M.A., Krupkina T.Yu., Chaplygin Yu.A.
Problems of using device-technological simulation as tool of designing and ways of their solution
Korolev M.A., Krasukov A.Yu.
Using TCAD in designing the planar powerful MOS-transistors having the raised breaking-down voltage in the off-state
Gavrilov S.V., Glebov A.L., Lyalinskaya O.V., Solovyev R.A.
Application of standard cell characterization results in statistical timing analysis
Amelichev V.V., Polomoshnov S.A., Chaplygin Yu.A.
Development and designing of integrated thermoelements
Putrya F.M.
Optimization of structure of controllers of serial buses. The solution of problems of lack of pins of a integrated circuit and loading of the processor at data transmission
Solovyev R.A., Glebov A.L., Gavrilov S.V.
Static timing analysis aware false conduct path detection in terms of logic implication
Adamov Yu.F., Gorshkova N.M., Krupkina T.Yu.
The Utilization of Photolayers for Bipolar Transistors Implementation in Typical CMOS Process
Muhanyuk N.N.
Development of technologically independent method of designing analog IC on the basis of library of parametrical cells
Chernyh A.V.
Digital sigma-delta modulator
Bragin K.R., Gavrilov S.V., Kagramanyan E.R.
Logic-timing analysis methodology for characterization of custom blocks of digital CMOS IC
Putrya F.M.
Method of automation of process of development of the crossbar for multicore system whith nonuniform memory access
Putrya F.M.
Research, development and optimization of data exchange hardware in multicore computing systems
Chernyh A.V.
Research and development of structural decisions of frequency synthesizers on the PLL basis
Artamonova E.A., Krasukov A.Yu.
Research of electric and temperature area of safe work of planar power SoC MOS transistors
Kozlov A.V., Tikhonov R.D., Parmenov Yu.A.
Research of influence of constructive-technology factors on sensitivity of magnetic transistor with methods of device-technological simulation
Kagramanyan E.R., Gavrilov S.V., Egorov Yu.B.
Standard cell characterezition methodoligy with respect MOSFET threshold voltage variation
Solovyev R.A., Gavrilov S.V., Glebov A.L.
Statistical timing analysis aware of reconvergence of conduction paths and transition variations
Krupkina T.Yu., Rodionov D.V.
The analysis of dynamic processes of interference distribution in substrates of integrated elements with methods of device-technological simulation
Putrya F.M., Medvedev I.A.
Hardware streams synchronization methods for multicore cluster
Chaplygin Yu.A., Krupkina T.Yu., Krasukov A.Yu., Artamonova E.A.
Investigation of depencencies of high voltage SOI-MOSFETs safe operating area on structual and process-dependent parameters
Fetisov E.A., Fedirko V.A., Khafizov R.Z., Zolotarev V.I., Zenyuk D.A., Rudakov G.A.
Nano-electromechanical thermo-sensitive elements
Losev V.V., Chaplygin Yu.A., Krupkina T.Yu.
New methods of construction of microelectronic digital systems with low power consumption
Gudkova O.N., Skachkova E.P., Muhanyuk N.N., Gavrilov S.V., Solovyev R.A.
The Methods of Fast Characterization of Large Scale Integration Parameterized IP-blocks
Lupin S.A., Than Shein, Than Zaw Oo, Kyaw Myoo Htunn
Accuracy Estimation of Discrete Optimization Algorithms
Borisov V.A., Krivoshein D. Y., Marchenko A.M., Popov E.A., Savchenko V.Yu.
A global router for nanometer standard cells
Losev V.V., Orlov D.V.
Arithmetical algorithms of the coding system of 1 from 4 with an active zero and estimation of the parameters of high-speed performance and occupied area of the unit of summation
Sukhanov E.S., Chirkunova Z.V., Oreshkin V.I., Lyalin K.S.
Digital signal processing in airborne weather radar
Fetisov E.A., Khafizov R.Z., Belin A.M., Rudakov G.A., Zolotarev V.I., Fedirko V.A., Rygalin D.B.
IR photosensitive MEMS elements
Shiro G. E., Shiro E.G.
Low-discharge fully-coded multiplier for radar image synthesis systems
Bazhanov E.I.
Minimization of the average number of conversion cycles of a successive approximation ADCs
Losev V.V., Krupkina T.Yu., Chaplygin Yu.A.
Resonant energy efficiency driver
Ichkitidze L.P., Novikov N.A.
Superconducting Film Concentrator of the Magnetic Field with Nanoscale Branches
Timoshenko A.G., Lomovskaya K.M., Suslov M.O.
Survey on features of integrated antennas
Putrya F.M.
SystemVerilog object-oriented programming features for functional verification of multi-core SoC
Artamonova E.A., Golishnikov A.A., Krupkina T.Yu., Rodionov D.V., Chaplygin Yu.A.
TCAD simulation of nanometer MOSFET on the assumption of the surface roughness at Si/SiO2 interface
Marchenko A.M., Popov E.A., Savchenko V.Yu.
A stick-diagram based standard cell layout synthesis tool
Artamonova E.A., Klyuchnikov A.S., Krasukov A.Yu., Krupkina T.Yu., Shelepin N.A.
Calibration of numerical TCAD model for 180 nm SOI MOSFETs
Posypkin M.A., Si Thu Thant Sin
Comparative analysis of efficiency of different variants of the dynamic programming method for solving the problem of optimal placing of elements on the chip
Belin A.M., Belin M.A., Fedirko V.A., Fetisov E.A., Khafizov R.Z.
Design Principles and Numerical Simulation of Microthermomechanical IR Imagers with Optical Readout
Soloviev A.N., Sablin A.V., Lyalinsky A.A.
Development of the simulation environment of inertial navigation systems
Chaplygin Yu.A., Timoshenkov V.P., Shevyakov V.I., Adamov Yu.F.
Electrostatic protection of BiCMOS IC's
Lupin S.A., Kyaw Kyaw Lin, Than Shein, Davydova A.P., Vagapov Y.F.
Estimation of digital sensor's influence on the management systems efficiency
Marchenko A.M., Popov E.A., Savchenko V.Yu.
Gate nets routing in nanometer standard cells with ports placement
Potovin Y.M., Soin S.
High-speed content addressable memory block design
Chaplygin Yu.A., Krupkina T.Yu., Krasukov A.Yu., Artamonova E.A.
Influence of CMOS Hall Effect Sensor Layout on its Magnetic Sensitivity
Khafizov R.Z.
Infrared focal plane arrays (FPA) with thermopile thermal radiation MEMS sensors
Medvedev I.A., Putrya F.M.
Interconnect Verification Methods Based on Unified Test Infrastructure
Losev V.V.
Investigation of the possibilities of practical application of the adiabatic logic to reduce power consumption of VLSI
Bespalov V.A., Vasilev I.A., Djuzhev N.A., Mazurkin N.S., Novikov D.V., Popkov A.F.
Membrane-based thermal flow sensor working on calorimetric principle
Gavrilov S.V., Ivanova G.A., Manukyan A.A.
Methods of designing custom IP-blocks based on the elements with regular topological structure in layers of polysilicon and diffusion
Khasanov M., Kurganov V.V.
Methods of determination the coefficients of quasi-optimal FIR-filter for convolution of pseudorandom binary sequence
Krupkina T.Yu., Krasukov A.Yu., Artamonova E.A.
Numerical model for MISFETs characterization
Bezglasnaya K.A., Kolbasov Y.S., Medvedev I.A., Putrya F.M.
Problems of platform approach for System on Chip and IP cores test infrastructure creation and their solutions
Kartashev S.S.
Readout circuit from the nonvolatile memory
Golovina E., Makeeva M., Nikolaev A.V., Putrya F.M., Smirnov A.A.
Reusable complex Soc level tests creating and debugging method
Timoshenko A.G.
Sigma-delta ADC for capacitive accelerometer
Davydova A.P., Vagapov Y.F., Lupin S.A.
Simulation modeling for survivability evaluation of digital control systems
Volobuev P.S., Gavrilov S.V., Ryzhova D.I.
The method of static power reducing for CMOS circuits based on sleep transistors with operation speed control
Chaplygin Yu.A., Adamov Yu.F., Timoshenkov V.P.
Using of VBIC model for SiGe integrated circuit application
Chaplygin Yu.A., Krupkina T.Yu., Krasukov A.Yu., Artamonova E.A.
0.5 um SOI CMOS for Extreme Temperature Applications
Timoshenkov V.P., Rodionov D.V., Khlybov A.I., Musatkin A.S., Vertianov D.V.
3D flexible microwave polyimid T-line assembly for system in package
Matyushkin I.V.
Algorithms of the parallel computations in the formalization of cellular automata: the sorting of strings and the multiplication of numbers by Atrubin’s method
Zhezlov K.A., Kolbasov Y.S., Kozlov A.O., Nikolaev A.V., Putrya F.M., Frolova S.E.
Automation of verification environments development process providing a through design flow for design, verification and research of IP-blocks and SoC
Matyushkin I.V., Zapletina M.A.
Cellular automata methods of the numerical solution of mathematical physics equations for the hexagonal grid
Kaleev D.V., Pereverzev A.L., Savchenko Yu.V.
Criteria of resolution of phase ambiguities for complexed multi-antenna global navigation satellite system
Timoshenko A.G., Belousov E.O., Molenkamp K.M.
Design and development of monolithic IC of microwave GaN phase shifters
Efimov A.G., Koptsev D.A., Kuznetsova O.
Development of microwave phase shifter based on SOI 0.18-micron technology
Lupin S.A., Pachin A., Kostrova O., Fediashin D.
Dynamic management of computations in distributed systems
Ostrovskaya N.V., Skidanov V.A., Iusipova Iu.A.
Features of magnetization reversal in a MRAM cell — I. In-plane anisotropy
Losev V.V., Chaplygin Yu.A., Krupkina T.Yu., Putrya M.G.
Features of processing and transmitting information in computing devices
Beklemishev D.N., Pereverzev A.L., Yanin V.I.
Miniature and highly sensitive proximity infrared sensor
Shichkin N.Yu., Ichkitidze L.P., Telishev D.V.
Nanostructured superconducting film concentrator in the magnetic field sensor
Khafizov R.Z., Timofeev A.E.
Numerical simulation of solar radiation transmittance for textured surface silicon photovoltaic cells
Liventsev E.V., Pereverzev A.L., Primakov E.V., Silantiev A.M.
On-board flight control system based on the MIPS architecture with CorExtend user-defined instructions and hardware-accelerated trigonometry calculations
Krupkina T.Yu., Krasukov A.Yu., Artamonova E.A.
One dimensional process and device simulation using spreadsheets
Fedirko V.A., Khafizov R.Z., Fetisov E.A.
Optimal design of the MEMS thermopile element for an IR imager array
Chaplygin Yu.A., Balashov A.G., Evdokimov V., Klyuchnikov A.S.
Research of the RF performance of SiGe HBT during transition towards sub-100 nm technology limits
Goryachev I., Demin G.D., Zvezdin K.A., Zipunova E.V., Ivanov A.V., Iskandarova I.M., Knizhnik A.A., Levchenko V.D., Popkov A.F., Potapkin B.V., Solovyov S.V.
Software package for Technology Computer-Aided Design of spintronic devices based on magnetic tunneling junctions
Ilin S.A., Kochanov S.K., Lastochkin O.V., Novikov A.A.
The methodology of the automated generation and analysis of basic structures for the design of dynamic and static protection the blocks of integrated circuits against ESD
Djuzhev N.A., Makhiboroda M.A., Gusev Ye., Gryazneva T., Demin G.D.
The process flow simulation of the cathode-grid system and its emission properties
Glebov A.L., Mindeeva A., Sheremetov V.V.
Timing analysis of digital circuits basing on logic correlations
Liventsev E.V., Pereverzev A.L., Primakov E.V., Ryzhkova D.V., Silantiev A.M.
Accurate High-speed Frequency Meter for Doppler Initial Velocity Measurement
Matyushkin I.V., Zapletina M.A.
Cellular Automata Computational Parallelism of Elementary Matrix Operations
Chaplygin Yu.A., Krupkina T.Yu., Korolev M.A., Krasukov A.Yu., Artamonova E.A.
Comparison of Double-gate Junctionless and Traditional MOSFETs by Means of TCAD
Zhmylev V.
Detector of Free Parts of Radio Frequency Spectrum
Ostrovskaya N.V., Skidanov V.A., Iusipova Iu.A.
Dynamics of Magnetization in the Free Layer of a Spin Valve Under the Influence of Magnetic Field, Perpendicular and Parallel to the Layer Plane
Datsuk A.M., Balashov A.M., Timoshenkov V.P., Krupkina T.Yu.
Electro-thermal Simulation of a Bandgap
Khafizov R.Z., Pavlyuk M.I., Timofeev A.E.
Numerical Simulation of N-well MOSFET Hall Element
Tsvetkov V.K., Lyalin K.S., Sheremet A.
RF-frontend Parts of Remote Sensing Systems Synthesis Method
Kuzmin I.A., Lyalin K.S., Meleshin Yu., Khasanov M.
Radar Image Autofocus in Conditions of High Vehicle Motion Instability
Ostrovskaya N.V., Skvortsov M.S., Skidanov V.A., Iusipova Iu.A.
Simulation of Magnetization Dynamics in Three-layered Ferromagnetic Structures with Pinned Boundaries
Garashchenko A.V., Nikolaev A.V., Putrya F.M., Sardaryan S.S.
System of Combined Specialized Test Generators for the New Generation of VLIW DSP Processors with Elcore50 Architecture
Timoshenkov V.P., Khlybov A.I., Rodionov D.V., Efimov A.G., Chaplygin Yu.A.
Thermo Researching of X-band Microwave Amplifier
Bulakh D.A., Korshunov A.V.
An extensible framework for developing and testing of the layout processing algorithms
Ilin S.A., Korshunov A.V., Garbulina T.V.
Benchmarking Energy Efficiency of Libraries on FinFET 7nm
Kurganov V.V., Djigan V.I.
Calibration of Antenna Arrays with Small Number of Antennas: Problems and Solutions
Bykova A.V., Polunin M.N.
Criteria for the numerical evaluation of data recovery algorithms for analogue-information converters
Shabanov A.A., Putrya M.G.
Development and approbation of the method of microcircuits interchangeability efficiency evaluation in radar equipment based on critical circuit and parametric characteristics set
Khalirbaginov R.I.
Digital Controlled Oscillator for All-Digital Phase-Locked Loop Circuit
Iusipova Iu.A.
Energy efficiency and performance of spin-valve structures in MRAM and HMDD
Iusipova Iu.A.
Frequency and amplitude characteristics of STNO based on a spin valve with planar and perpendicular layer anisotropy
Skripnichenko M.N.
High-level model based calibration technique design for SAR ADC
Nikitin S.A., Nikolaev A.V., Putrya F.M., Neklyudov I.A.
Route automation of Functional Verification based on IP-XACT standard
Bulakh D.A., Korshunov A.V.
Speeding up the evaluation of polygons mutual placement task for double pattern technology
Ilin S.A., Kopeikin D.Yu., Lastochkin O.V., Novikov A.A., Shipitsin D.S.
The post-silicon validation method of standard cell libraries
Datsuk A.M., Timoshenkov V.P.
rEDActor – A PDK Cross-platform Integrated Development Environment for Semiconductor Technologies
Djigan V.I., Kurganov V.V.
Accuracy of Phase-Less Algorithms of Antenna Array Calibration
Rubis P.D., Matyushkin I.V.
Cellular-automaton algorithm of matrices permutation with an oscillatory scheme of element shift
Khalirbaginov R.I.
Design of All-digital Phase-locked Loop
Zhezlov K.A., Belyaev A.A., Putrya F.M.
Implementation of Methodology of SoC Interconnects Automated Performance Analysis into the Verification Route
Medvedeva O.I., Semenov M.Y., Titov Y.A.
Layout Dependent Effects Impact on Standard Cells Layout in 28 nm Technology Node
Ostrovskaya N.V., Skidanov V.A., Iusipova Iu.A.
Mathematical Model of the SOT-MRAM Cell based on the Spin Hall Effect
Ilin S.A., Lastochkin O.V., Ishchenko N.A.
Methodology of calculating dependent timing constraints for libraries of standard digital cells
Ilin S.A., Zapletina M.A., Lastochkin O.V.
Optimization of Standard Cells Power Consumption: Logical Effort Based Algorithm
Shchuchkin E.Yu.
The Problem of Element Placement on a Printed Circuit Board: the Solution Based on a Simplified Model of a Microstrip Line
Janpoladov V.A.
A Machine Learning-Based Switching Power Prediction at Floorplan Stage of IC Physical Design
Ilin S.A., Lastochkin O.V., Ishchenko N.A.
Accelerated characterization technique for multi-bit flip-flops with accuracy control
Volobuev P.S., Korshunov A.V., Poryadina M.V.
An integrated LDO regulator for self-powered systems
Belyaev A.A., Shchuchkin E.Yu.
DC-DC Converter Conducted Emission Level Estimation at Design Stage
Khafizov R.Z.
Design of High-Speed Thermocouple Sensors based on SOI Structures
Barkov E.S., Pereverzev A.L., Silantiev A.M.
High-performance soft processor for embedded FPGA-based systems
Skripnichenko M.N.
Implementation of Area Optimal FIR Filters Based on Lookup Tables for Sigma-Delta Modulator Signal Processing
Medeev D.A., Pereverzev A.L.
Mathematical Model of the Trajectory Velocity Measurement Error by Doppler Frequency Shift
Iusipova Iu.A., Ostrovskaya N.V., Skidanov V.A.
Mathematical model of a cylindrical SOT-MRAM cell
Selyutina E.V., Gurov K.O., Mindubaev E.A., Danilov A.A.
Optimization of a Class E Power Amplifier in the Transmitting Part of an Inductive Power Transfer System
Kondratev A.V., Skobelev D.N., Strekopytov D.V., Kurganov I.N., Baybakov D.B.
Planar printed antenna array for Doppler speed and drift angle meter
Grevtsov N.L., Lopato U.P., Chubenko E.B., Bondarenko V.P., Gavriliv I.M., Dronov A.A., Gavrilov S.A.
Porous Silicon Layers for Heteroepitaxial and Composite Structure Formation
Evsikov I.D., Demin G.D., Djuzhev N.A., Fetisov E.A., Khafizov R.Z.
TCAD Simulation of the Etching of the Sacrificial Layer in the Sensitive Element of the IR Microbolometer Array Based on the SOI Structure
Bekenova A.T., Artamonova E.A., Krasukov A.Yu.
TCAD Study of Responsivity of n-channel MOS Dosimeter Fabricated in CMOS Processes
Levitskiy D.O.
The Optimal Algorithm for Generating a Complete Test for Checking the Simplest Single Logical-Dynamic Faults for an N-Input Combinational Device
Gurov K.O., Mindubaev E.A., Danilov A.A., Selyutina E.V.
Wireless Power Transfer Appliance with High Resistance to Inductive Coils Displacements for Powering Implanted Medical Devices | {"url":"http://proc.mes-conference.ru/index.php?page=vorg&code=G104&ls=en","timestamp":"2024-11-08T17:22:27Z","content_type":"text/html","content_length":"53177","record_id":"<urn:uuid:c4a7135f-dca2-4e7e-92b8-0b4cfaee3416>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00187.warc.gz"} |
middle school math learning websites
Learning Math? 25 Of The Best Math Resources
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Math Resources for Teachers | Twine
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11 Best Math Websites for Teachers in 2024
Mathematics | Boltz Middle School
Best AI Sites for School Students to Improve their Maths Skills | {"url":"https://worksheets.clipart-library.com/middle-school-math-learning-websites.html","timestamp":"2024-11-02T00:17:09Z","content_type":"text/html","content_length":"26224","record_id":"<urn:uuid:1c405a2e-aef7-4b4e-82d4-fe97a36f0703>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00046.warc.gz"} |
Wrapping Trees
Problem L
Wrapping Trees
After last year’s spectacular failure, the Alberta Christmas Paraphernalia Company (ACPC) is getting an early start on their Christmas tree business this year. With all the extra time they have to
prepare, the plan is to use some algorithmic methods to optimize their profits.
The ACPC would like to wrap a single piece of string around each tree so that they can tightly pack them into the warehouse. The tree supplier has kindly sent you a specification of the shapes of the
trees. The cross section of the tree that you want to wrap the string around can be represented in compressed form by an $n$ by $n$ binary image, where each pixel corresponds to a $1$ cm by $1$ cm
region of the tree. Unfortunately, as trees are very complicated shapes, it would take an infinite amount of time for you to download the complete uncompressed image of a tree. Luckily, the trees are
of premium quality and thus have a very nice recursive structure. The full uncompressed image of the tree can be computed by recursively replacing every 1 pixel in the compressed image with a scaled
down copy of the entire image an infinite number of times. Note that the dimensions of the full uncompressed image are still $n$ cm by $n$ cm, the same as the compressed image.
Given a compressed specification of a tree, your task is to find the shortest possible length of string (in centimeters) that can wrap completely around the perimeter of all 1 pixels in the
corresponding uncompressed tree.
The first line of input will contain an integer $n$ denoting the both the height and width of the compressed image in centimeters. You may assume that $1\le n\le 10^3$. The following $n$ lines will
contain $n$ characters. Each character will be either a 0 or 1 describing the pixels of the compressed image.
Output the smallest possible length of string that can completely enclose all points in the uncompressed tree corresponding to the given compressed tree. Your answer will be considered correct if its
absolute or relative error doesn’t exceed $10^{-5}$.
Sample Input 1 Sample Output 1
Sample Input 2 Sample Output 2
Sample Input 3 Sample Output 3
00101 17.2656990001 | {"url":"https://open.kattis.com/contests/j3bebr/problems/wrappingtrees","timestamp":"2024-11-11T03:09:27Z","content_type":"text/html","content_length":"32577","record_id":"<urn:uuid:ece0ce1f-7498-40a9-9425-193f65ee4e69>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00777.warc.gz"} |
Osrs Prayer Calculator - Wqidan Calculators
Osrs Prayer Calculator
Enter your current prayer level, desired level, and experience rate into the calculator to determine the amount of experience needed.
OSRS Prayer Calculation Formula
The following formula is used to calculate the experience needed to achieve your desired prayer level in OSRS.
Experience Needed = (Desired Level - Current Level) * Experience Rate
• Current Level is your current prayer level in OSRS
• Desired Level is the level you wish to achieve
• Experience Rate is the experience gained per bone type
To calculate the experience needed, multiply the difference between your desired level and current level by the experience rate of the selected bone type.
What is OSRS Prayer Calculation?
OSRS Prayer calculation refers to the process of determining the amount of experience required to reach a specific prayer level in the game Old School RuneScape (OSRS). This involves understanding
your current level, the desired level, and the experience rate of the bones you are using. Proper calculation ensures efficient leveling and resource management.
How to Calculate Experience Needed?
The following steps outline how to calculate the experience needed to reach your desired prayer level using the given formula.
1. First, determine your current prayer level and the desired prayer level.
2. Next, determine the experience rate of the bones you are using.
3. Use the formula from above: Experience Needed = (Desired Level – Current Level) * Experience Rate.
4. Finally, calculate the experience needed by plugging in the values.
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem:
Use the following variables as an example problem to test your knowledge.
Current Level = 50
Desired Level = 70
Experience Rate = 252
1. What is the experience rate of different bones?
The experience rate varies by bone type. For example, Normal Bones give 4.5 experience, Big Bones give 15 experience, and Dragon Bones give 72 experience.
2. How is experience needed different from total experience?
Experience needed is the amount of experience required to reach a specific level from your current level. Total experience is the cumulative experience required to reach a certain level from level 1.
3. How often should I use the OSRS Prayer calculator?
It’s helpful to use the OSRS Prayer calculator whenever there’s a change in your leveling goals, bone types, or if you want to plan your training sessions more accurately.
4. Can this calculator be used for different bones?
Yes, you can adjust the bones type field to match the bone type you are using to calculate the experience needed accordingly.
5. Is the calculator accurate?
The calculator provides an estimate of the experience needed based on the inputs provided. For exact figures, it’s best to consult OSRS guides or resources. | {"url":"https://wqidian.com/osrs-prayer-calculator","timestamp":"2024-11-06T08:21:31Z","content_type":"text/html","content_length":"76926","record_id":"<urn:uuid:7312d74e-4f79-4df5-b8d7-6b8c2c239c8f>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00247.warc.gz"} |
Proper Fraction -- from Wolfram MathWorld
Proper Fraction
A proper fraction is a fraction improper fraction.
See also
Common Fraction
Mixed Fraction
Improper Fraction
Irreducible Fraction
Reduced Fraction
Reducible Fraction
Explore with Wolfram|Alpha
Derbyshire, J.
Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics.
New York: Penguin, p. 171, 2004.
Referenced on Wolfram|Alpha
Proper Fraction
Cite this as:
Weisstein, Eric W. "Proper Fraction." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/ProperFraction.html
Subject classifications | {"url":"https://mathworld.wolfram.com/ProperFraction.html","timestamp":"2024-11-06T01:12:53Z","content_type":"text/html","content_length":"50475","record_id":"<urn:uuid:912cf68d-fa7f-4038-9e57-61b72372c3e7>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00543.warc.gz"} |
The Mathematical Ninja and Cosines
As the student was wont to do, he idly muttered “So, that’s $\cos(10º)$…”
The calculator, as calculators are wont to do when the Mathematical Ninja is around, suddenly went up in smoke. “0.985,” with a heavy implication of ‘you don’t need a calculator for that’.
As the student was wont to do, he stifled the urge to say ‘but… we mortals kind of do need a calculator for that.’ Instead, he sighed (as he was wont to do) and let the Mathematical Ninja talk it
“In radians, $\cos(x)$ is approximately $1-\frac12 x^2$.”
The student shrugged. He wasn’t doing STEP and didn’t much care for small angle approximations.
“Meanwhile, the conversion factor for degrees to radians - the only way one should ever consider converting - is about $\frac{7}{400}$. That squared is $\frac{49}{160,000}$, or about $\frac{1}{3,200}
$, and half of it is $\frac{1}{6,400}$.”
“If you say so, sensei.”
“So, to find $\cos(yº)$, you simply work out $1 - \frac{1}{6,400}y^2$.”
“Simply. For $y=10$, that’s $1-\frac{1}{64}$, and $\frac{1}{64}$ is about 0.015. (You can get the same result by dividing by 80 and squaring the result).”
“Which you presumably took away from 1 to get 0.985?”
“That, young student, is precisely what I did.”
“What about $\arccos$? Is there a trick for that?”
The Mathematical Ninja’s eyes narrowed slightly, recognising that the only-way-to-convert statement was about to be disproved by counterexample. “Well, of course. At least when $c$ is near 1, $\
arccos(1-c)$ can be approximated quite easily, just working backwards. $1-c \approx 1 - \frac{1}{6,400}y^2$, so $y \approx 80\sqrt{c}$.”
“So, $\arccos(0.99)$ would be 0.01 less than 1, square rooted is 0.1, and multiplied by 80 gives 8º?”
“8.1º,” giving a slight wince, as the Mathematical Ninja was wont to do.
Uncharacteristically, the Mathematical Ninja missed a trick when explaining this (MY calculator is fine, thanks for asking): the conversion factor is $\frac{\pi}{180}$, the square of which is closer
to $\frac{1}{3280}$; half of that is $\frac{1}{6560}$, which is a razor-fine knife edge away from $\left(\frac{1}{81}\right)^2$. Which is, presumably, where the final, more accurate, approximation
comes from.
A selection of other posts
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DET - Stochastic analysis of circuits, cables, and interconnects
Stochastic analysis of circuits, cables, and interconnects
The recent years have seen an ever-growing interest in techniques for studying stochastic systems, i.e., systems affected by parameters that are random in nature. Relevant examples in electrical
engineering are for instance a circuit made up by components whose value is only known within a certain manufacturing tolerance, or a cable harness illuminated by an electromagnetic field that
impinges from an unknown direction. In this scenario, the outputs of interest must be regarded as stochastic, rather than deterministic quantities. Polynomial chaos became a popular tool to
investigate this class of problems. It approximates stochastic quantities by series of suitable orthogonal polynomials in the random parameters. Over the years, several techniques were developed to
calculate the polynomial chaos expansion coefficients and to address some issues regarding, for example, nonlinearity in the equations or high-dimensional problems. Another tool that is used in this
context is Taylor models. This is an algebraic technique that calculates a Taylor approximation of a given variable with guaranteed-conservative error bounds, thus being suitable for a worst-case
type of analysis.
ERC Sector:
• PE7_3 Simulation engineering and modelling
• Statistical analysis
• Stochastic systems
• Polynomial chaos
• Uncertainty quantification
Gruppi di ricerca | {"url":"https://www.det.polito.it/it/research/activities/electromagnetic_compatibility/stochastic_analysis_of_circuits_cables_and_interconnects","timestamp":"2024-11-06T04:43:42Z","content_type":"text/html","content_length":"21249","record_id":"<urn:uuid:b7057dbf-0e50-406a-9fcf-603bd161a25a>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00295.warc.gz"} |
4 Types of Data in Statistics | Analytics Steps
Data types are important concepts in statistics, they enable us to apply statistical measurements correctly on data and assist in correctly concluding certain assumptions about it.
Having an adequate comprehension of the various data types is significantly essential for doing Exploratory Data Analysis or EDA since you can use certain factual measurements just for particular
data types.
SImilarly, you need to know which data analysis and its type you are working to select the correct perception technique. You can consider data types as an approach to arrange various types of
If you go into detail then there are only two classes of data in statistics, that is Qualitative and Quantitative data. But, after that, there is a subdivision and it breaks into 4 types of data.
Data types are like a guide for doing the whole study of statistics correctly!
This blog gives you a glance over different types of data need to know for performing proper exploratory data analysis.
Qualitative and Quantitative Data
Qualitative data is a bunch of information that cannot be measured in the form of numbers. It is also known as categorical data. It normally comprises words, narratives, and we labelled them with
It delivers information about the qualities of things in data. The outcome of qualitative data analysis can come in the type of featuring key words, extracting data, and ideas elaboration.
For examples:
• Hair colour- black, brown, red
• Opinion- agree, disagree, neutral
On the other side, Quantitative data is a bunch of information gathered from a group of individuals and includes statistical data analysis. Numerical data is another name for quantitative data.
Simply, it gives information about quantities of items in the data and the items that can be estimated. And, we can formulate them in terms of numbers.
For examples:
• We can measure the height (1.70 meters), distance (1.35 miles) with the help of a ruler or tape.
• We can measure water (1.5 litres) with a jug.
Under a subdivision, nominal data and ordinal data come under qualitative data. Interval data and ratio data come under quantitative data. Here we will read in detail about all these data types.
Different Types of Data
1. Nominal Data
Nominal data are used to label variables where there is no quantitative value and has no order. So, if you change the order of the value then the meaning will remain the same.
Thus, nominal data are observed but not measured, are unordered but non-equidistant, and have no meaningful zero.
The only numerical activities you can perform on nominal data is to state that perception is (or isn't) equivalent to another (equity or inequity), and you can use this data to amass them.
You can't organize nominal data, so you can't sort them.
Neither would you be able to do any numerical tasks as they are saved for numerical data. With nominal data, you can calculate frequencies, proportions, percentages, and central points.
Examples of Nominal data:
• English
• German
• French
• Punjabi
• American
• Indian
• Japanese
• German
You can clearly see that in these examples of nominal data the categories have no order.
2. Ordinal Data
Ordinal data is almost the same as nominal data but not in the case of order as their categories can be ordered like 1st, 2nd, etc. However, there is no continuity in the relative distances between
adjacent categories.
Ordinal Data is observed but not measured, is ordered but non-equidistant, and has no meaningful zero. Ordinal scales are always used for measuring happiness, satisfaction, etc.
With ordinal data, likewise, with nominal data, you can amass the information by evaluating whether they are equivalent or extraordinary.
As ordinal data are ordered, they can be arranged by making basic comparisons between the categories, for example, greater or less than, higher or lower, and so on.
You can't do any numerical activities with ordinal data, however, as they are numerical data.
With ordinal data, you can calculate the same things as nominal data like frequencies, proportions, percentage, central point but there is one more point added in ordinal data that is summary
statistics and similarly bayesian statistics.
Examples of Ordinal data:
• Opinion
□ Agree
□ Disagree
□ Mostly agree
□ Neutral
□ Mostly disagree
In these examples, there is an obvious order to the categories.
3. Interval Data
Interval Data are measured and ordered with the nearest items but have no meaningful zero.
The central point of an Interval scale is that the word 'Interval' signifies 'space in between', which is the significant thing to recall, interval scales not only educate us about the order but
additionally about the value between every item.
Interval data can be negative, though ratio data can't.
Even though interval data can show up fundamentally the same as ratio data, the thing that matters is in their characterized zero-points. If the zero-point of the scale has been picked subjectively,
at that point the data can't be ratio data and should be interval data.
Hence, with interval data you can easily correlate the degrees of the data and also you can add or subtract the values.
There are some descriptive statistics that you can calculate for interval data are central point (mean, median, mode), range (minimum, maximum), and spread (percentiles, interquartile range, and
standard deviation).
In addition to that, similar other statistical data analysis techniques can be used for more analysis.
Examples of Interval data:
• Temperature (°C or F, but not Kelvin)
• Dates (1066, 1492, 1776, etc.)
• Time interval on a 12-hour clock (6 am, 6 pm)
4. Ratio Data
Ratio Data are measured and ordered with equidistant items and a meaningful zero and never be negative like interval data.
An outstanding example of ratio data is the measurement of heights. It could be measured in centimetres, inches, meters, or feet and it is not practicable to have a negative height.
Ratio data enlightens us regarding the order for variables, the contrasts among them, and they have absolutely zero. It permits a wide range of estimations and surmisings to be performed and drawn.
Ratio data is fundamentally the same as interval data, aside from zero means none.
The descriptive statistics which you can calculate for ratio data are the same as interval data which are central point (mean, median, mode), range (minimum, maximum), and spread (percentiles,
interquartile range, and standard deviation).
Example of Ratio data:
• Age (from 0 years to 100+)
• Temperature (in Kelvin, but not °C or F)
• Distance (measured with a ruler or any other assessing device)
• Time interval (measured with a stop-watch or similar)
Therefore, for these examples of ratio data, there is an actual, meaningful zero-point like the age of a person, absolute zero, distance calculated from a specified point or time all have real zeros.
Key Takeaways
We hope you understood about 4 types of data in statistics and their importance, now you can learn how to handle data correctly, which statistical hypothesis tests you can use, and what you could
calculate with them. Moreover,
• Nominal data and ordinal data are the types of qualitative data or categorical data.
• Interval data and ratio data are the types of quantitative data which are also known as numerical data.
• Nominal Data are not measured but observed and they are unordered, non-equidistant, and also have no meaningful zero.
(Also check: Types of Statistical Analysis)
• Ordinal Data is also not measured but observed and they are ordered however non-equidistant and have no meaningful zero.
• Interval Data are measured and ordered with equidistant items yet have no meaningful zero.
• Ratio Data are also measured and ordered with equidistant items and a meaningful zero.
Latest Comments
Oct 06, 2021
really i
Oct 06, 2021
I would really like to thank you for the article, which I benefited a lot from and explained to me all the questions that I wanted. I send you full thanks and respect. | {"url":"https://www.analyticssteps.com/blogs/4-types-data-statistics","timestamp":"2024-11-11T16:48:43Z","content_type":"text/html","content_length":"45109","record_id":"<urn:uuid:f0ce6506-56d6-43c2-81e7-22faad63cbe2>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00391.warc.gz"} |
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Handling variables and equations with different timesteps
Description of the issue
I have 2 sets i,j
i = m.addSet(name="j", records = ["January","February", "March"], description="months")
j = m.addSet(name="j", records=np.arange(1, 91), description="days in the 3 months")
So I define a parameter over every day which is the daily demand.
Demand_param = m.addParameter("Demand_param" , j, records=Demand_list)
Then I define a variable over the months which would be the capacity for the machinery. I do not want the variable to be defined over the hours’ set cause then it would be unrealistic.
Mach_cap= m.addVariable("Mach_capacity", "positive", i)
Then a variable about the production over the hours set.
Prod= m.addVariable("production", "positive", j)
Finally I need a constraint which will show the model that the production shall not exceed the capacity in an hourly basis. And that is where I have the most trouble. I tried this:
production_max= m.addEquation("produced_max",domain=[ i, j])
production_max[i,j]= (Mach_cap[i] >= Prod[j])
The thing is that with this way the model, will not “know” which hour (j), will correpsond to which monthly capacity (i).
I also tried mapping set i to j. The problem with that is that no element in gamspy (equations, variables etc.) can be defined over a multi-dimensional set apparently.
I do not want to sum up the whole monthly demand because then this would not gurantee that the demand is satisfied on a daily basis, but on a monthly.
Do you have any suggestions?
You can set j as the days in the 3 months but you are referring to it as hours set. I couldn’t understand what you mean by that.
You can define parameters, variables, equations over multi-dimensional sets. Please see the documentation on Multi-dimensional sets
Thank you for taking the time to answer. I did it by mistake it is indeed the days set which correpsonds to 3 months hence 90 elements in this set.
My main concern as I mentioned is for this specific constraint:
production_max= m.addEquation("produced_max",domain=[ i, j])
production_max[i,j]= (Mach_cap[i] >= Prod[j])
I can not find a way to specify that the first 30 elements of the set j correspond to the first element of the set i, the next 30 to the second element etc.
For the multisimensional sets, I wasn’t able to find an example of an equation or a variable that its domain has more than one dimensions, and everytime I try to implement such a thing on my code, it
returns an error that says it was expecting for the domain to be 1 dimensional.
However multidimensional sets are not my main concern. My main concern remains the formulation of the constraint above in order to map correctly the elements of the set i to the ones of the set j.
There were more typos in code you shared. For example, you used the name “j” for set i and j. The name is not just something for nice printing but has a meaning in GAMSPy. In your case i and j were
references to the same GAMSPy set “j” that contained the numbers 1…90. The content for the months was overwritten.
In any case, you can have a mapping (a two dimensional set) between month and days as follows:
import gamspy as gp
m = gp.Container()
i = m.addSet(name="i", records = ["January","February", "March"], description="months")
j = m.addSet(name="j", records=range(1, 91), description="days in the 3 months")
ijmap = m.addSet(name="ij", domain=[i,j], description="map between days and month")
ij = [ ('January', i) for i in range(1, 32) ]
ij = ij + [ ('February', i) for i in range(32, 61) ]
ij = ij + [ ('March', i) for i in range(61,91) ]
with that you can write your constraint like that
production_max[ijmap[i,j]]= (Mach_cap[i] >= Prod[j])
Hope this helps,
Thanks for taking the time to respond Michael
Yes you are correct, it was just a dummy code I created.
In any case I am new to GAMS and therefore I still face difficulties.
However I meant to share here for anyone else who might need it in the future the solution I found which is pretty similar.
Instead of putting the map set in the equation declaration, instead I incorporated it inside the definition of the equation like so.
production_max[i,j]= (Mach_cap[i] >= Prod[j]).where[ijmap[i,j]])
As you can see here I am using the map set you created and a conditional assignment. Nevertheless your solution is more elegant, thanks for pointing it out
On a second note running this code you suggested:
import gamspy as gp
m = gp.Container()
i = m.addSet(name="i", records = ["January","February", "March"], description="months")
j = m.addSet(name="j", records=range(1, 91), description="days in the 3 months")
ijmap = m.addSet(name="ij", domain=[i,j], description="map between days and month")
ij = [ ('January', i) for i in range(1, 32) ]
ij = ij + [ ('February', i) for i in range(32, 61) ]
ij = ij + [ ('March', i) for i in range(61,91) ]
Mach_cap= m.addVariable("Mach_capacity", "positive", i)
Prod= m.addVariable("production", "positive", j)
production_max= m.addEquation("produced_max",domain=[ ijmap[i,j]])
production_max[ijmap[i,j]]= (Mach_cap[i] >= Prod[j])
I get back this error:
TypeError: All ‘domain’ elements must be type Set, Alias, UniverseAlias, or str
For the following equation:
production_max= m.addEquation("produced_max",domain=[ ijmap[i,j]])
Trying to avoid it I rewrite the last part of the code like so:
production_max= m.addEquation("produced_max",domain=[ ijmap])
production_max[ijmap]= (Mach_cap[i] >= Prod[j])
Which returns the error I initially mentioned:
ValueError: All linked ‘domain’ elements must have dimension == 1
I do not know if I am still doing something wrong or if my license (academic) has something to do with it. Nevertheless thank you for your time and if you find I am doing something wrong please let
me know
Your solution produces many trivial (and perhaps wrong equations in the case Mach_cap is not a positive variable). For all pairs i,j that are not in ijmap (e.g. March.1) this produces the equation
production_max('March','1').. Mach_cap('March',1') =g= 0;. So having the set ijmap on “the left” is much better.
Domain sets in GAMSPy are one-dimensional sets that help to find quickly typos in your code, e.g. if you type prod[j,i] but you meant prod[i,j]. They don’t help you to limit the possible pairs in
some equation or in other parts of the code. So declare your equation over domain [i,j] and define your equation over set ijmap (or any other combination with sets or subsets of i and j).
Good luck, | {"url":"https://forum.gams.com/t/handling-variables-and-equations-with-different-timesteps/7867","timestamp":"2024-11-15T03:34:31Z","content_type":"text/html","content_length":"31532","record_id":"<urn:uuid:f38b2e3d-f2d2-44ce-bfa7-c4be81f815df>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00470.warc.gz"} |
BCIT Physics 0312 Textbook
Chapter 2 One-Dimensional Kinematics
2.3 Time, Velocity, and Speed
• Explain the relationship between instantaneous velocity, average velocity, instantaneous speed, average speed, displacement, and time.
• Calculate velocity and speed given initial position, initial position, final position, and final time.
• Derive a graph of velocity vs. time given a graph of position vs. time.
• Interpret a graph of velocity vs. time.
Figure 1. The motion of these racing snails can be described by their speeds and their velocities. (credit: tobitasflickr, Flickr).
There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was the runner’s speed?” cannot be answered without an understanding of other
concepts. In this section we add definitions of time, velocity, and speed to expand our description of motion.
As discussed in Chapter 1.2 Physical Quantities and Units, the most fundamental physical quantities are defined by how they are measured. This is the case with time. Every measurement of time
involves measuring a change in some physical quantity. It may be a number on a digital clock, a heartbeat, or the position of the Sun in the sky. In physics, the definition of time is simple—time is
change, or the interval over which change occurs. It is impossible to know that time has passed unless something changes.
The amount of time or change is calibrated by comparison with a standard. The SI unit for time is the second, abbreviated s. We might, for example, observe that a certain pendulum makes one full
swing every 0.75 s. We could then use the pendulum to measure time by counting its swings or, of course, by connecting the pendulum to a clock mechanism that registers time on a dial. This allows us
to not only measure the amount of time, but also to determine a sequence of events.
How does time relate to motion? We are usually interested in elapsed time for a particular motion, such as how long it takes an airplane passenger to get from his seat to the back of the plane. To
find elapsed time, we note the time at the beginning and end of the motion and subtract the two. For example, a lecture may start at 11:00 A.M. and end at 11:50 A.M., so that the elapsed time would
be 50 min. Elapsed time[latex]\boldsymbol{\Delta{t}}[/latex]is the difference between the ending time and beginning time,
where[latex]\boldsymbol{\Delta{t}}[/latex]is the change in time or elapsed time,[latex]\boldsymbol{t_f}[/latex]is the time at the end of the motion, and[latex]\boldsymbol{t_0}[/latex]is the time at
the beginning of the motion. (As usual, the delta symbol,[latex]\boldsymbol{\Delta}[/latex], means the change in the quantity that follows it.)
Life is simpler if the beginning time[latex]\boldsymbol{t_0}[/latex]is taken to be zero, as when we use a stopwatch. If we were using a stopwatch, it would simply read zero at the start of the
lecture and 50 min at the end. If[latex]\boldsymbol{t_0=0}[/latex], then[latex]\boldsymbol{\Delta{t}=t_f\equiv{t}}[/latex].
In this text, for simplicity’s sake,
• motion starts at time equal to zero[latex]\boldsymbol{(t_0=0)}[/latex]
• the symbol t is used for elapsed time unless otherwise specified[latex]\boldsymbol{(\Delta{x}=t_f\equiv{t})}[/latex]
Your notion of velocity is probably the same as its scientific definition. You know that if you have a large displacement in a small amount of time you have a large velocity, and that velocity has
units of distance divided by time, such as miles per hour or kilometers per hour.
Average velocity is displacement (change in position) divided by the time of travel,
where[latex]\boldsymbol{\bar{v}}[/latex]is the average (indicated by the bar over the v) velocity,[latex]\boldsymbol{\Delta{x}}[/latex]is the change in position (or displacement), and[latex]\
boldsymbol{x_f}[/latex]and[latex]\boldsymbol{x_0}[/latex]are the final and beginning positions at times[latex]\boldsymbol{t_f}[/latex]and[latex]\boldsymbol{t_0}[/latex], respectively. If the starting
time[latex]\boldsymbol{t_0}[/latex]is taken to be zero, then the average velocity is simply
Notice that this definition indicates that velocity is a vector because displacement is a vector. It has both magnitude and direction. The SI unit for velocity is meters per second or m/s, but many
other units, such as km/h, mi/h (also written as mph), and cm/s, are in common use. Suppose, for example, an airplane passenger took 5 seconds to move −4 m (the negative sign indicates that
displacement is toward the back of the plane). His average velocity would be
[latex]\boldsymbol{\bar{v}=}[/latex][latex]\boldsymbol{\frac{\Delta{x}}{t}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{-4\textbf{ m}}{5\textbf{ s}}}[/latex][latex]\boldsymbol{=-0.8
The minus sign indicates the average velocity is also toward the rear of the plane.
The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point, however. For example, we cannot tell from average velocity whether
the airplane passenger stops momentarily or backs up before he goes to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals.
Figure 2. A more detailed record of an airplane passenger heading toward the back of the plane, showing smaller segments of his trip.
The smaller the time intervals considered in a motion, the more detailed the information. When we carry this process to its logical conclusion, we are left with an infinitesimally small interval.
Over such an interval, the average velocity becomes the instantaneous velocity or the velocity at a specific instant. A car’s speedometer, for example, shows the magnitude (but not the direction) of
the instantaneous velocity of the car. (Police give tickets based on instantaneous velocity, but when calculating how long it will take to get from one place to another on a road trip, you need to
use average velocity.) Instantaneous velocity[latex]\boldsymbol{v}[/latex]is the average velocity at a specific instant in time (or over an infinitesimally small time interval).
Mathematically, finding instantaneous velocity,[latex]\boldsymbol{v}[/latex], at a precise instant[latex]\boldsymbol{t}[/latex]can involve taking a limit, a calculus operation beyond the scope of
this text. However, under many circumstances, we can find precise values for instantaneous velocity without calculus.
In everyday language, most people use the terms “speed” and “velocity” interchangeably. In physics, however, they do not have the same meaning and they are distinct concepts. One major difference is
that speed has no direction. Thus speed is a scalar. Just as we need to distinguish between instantaneous velocity and average velocity, we also need to distinguish between instantaneous speed and
average speed.
Instantaneous speed is the magnitude of instantaneous velocity. For example, suppose the airplane passenger at one instant had an instantaneous velocity of −3.0 m/s (the minus meaning toward the rear
of the plane). At that same time his instantaneous speed was 3.0 m/s. Or suppose that at one time during a shopping trip your instantaneous velocity is 40 km/h due north. Your instantaneous speed at
that instant would be 40 km/h—the same magnitude but without a direction. Average speed, however, is very different from average velocity. Average speed is the distance traveled divided by elapsed
We have noted that distance traveled can be greater than displacement. So average speed can be greater than average velocity, which is displacement divided by time. For example, if you drive to a
store and return home in half an hour, and your car’s odometer shows the total distance traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however, was zero, because your
displacement for the round trip is zero. (Displacement is change in position and, thus, is zero for a round trip.) Thus average speed is not simply the magnitude of average velocity.
Figure 3. During a 30-minute round trip to the store, the total distance traveled is 6 km. The average speed is 12 km/h. The displacement for the round trip is zero, since there was no net change in
position. Thus the average velocity is zero.
Another way of visualizing the motion of an object is to use a graph. A plot of position or of velocity as a function of time can be very useful. For example, for this trip to the store, the
position, velocity, and speed-vs.-time graphs are displayed in Figure 4. (Note that these graphs depict a very simplified model of the trip. We are assuming that speed is constant during the trip,
which is unrealistic given that we’ll probably stop at the store. But for simplicity’s sake, we will model it with no stops or changes in speed. We are also assuming that the route between the store
and the house is a perfectly straight line.)
Figure 4. Position vs. time, velocity vs. time, and speed vs. time on a trip. Note that the velocity for the return trip is negative.
If you have spent much time driving, you probably have a good sense of speeds between about 10 and 70 miles per hour. But what are these in meters per second? What do we mean when we say that
something is moving at 10 m/s? To get a better sense of what these values really mean, do some observations and calculations on your own:
• calculate typical car speeds in meters per second
• estimate jogging and walking speed by timing yourself; convert the measurements into both m/s and mi/h
• determine the speed of an ant, snail, or falling leaf
Check Your Understanding
1: A commuter train travels from Baltimore to Washington, DC, and back in 1 hour and 45 minutes. The distance between the two stations is approximately 40 miles. What is (a) the average velocity of
the train, and (b) the average speed of the train in m/s?
Section Summary
• Time is measured in terms of change, and its SI unit is the second (s). Elapsed time for an event is
[latex]\boldsymbol{\Delta{t}=t_f -t_0}[/latex]
where[latex]\boldsymbol{t_f}[/latex]is the final time and[latex]\boldsymbol{t_0}[/latex]is the initial time. The initial time is often taken to be zero, as if measured with a stopwatch; the
elapsed time is then just[latex]\boldsymbol{t}[/latex].
• Average velocity[latex]\boldsymbol{\bar{v}}[/latex]is defined as displacement divided by the travel time. In symbols, average velocity is
[latex]\boldsymbol{\bar{v} =}[/latex][latex]\boldsymbol{\frac{\Delta{x}}{\Delta{t}}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{x_f -x_0}{t_f - t_0}}[/latex].
• The SI unit for velocity is m/s.
• Velocity is a vector and thus has a direction.
• Instantaneous velocity[latex]\boldsymbol{v}[/latex]is the velocity at a specific instant or the average velocity for an infinitesimal interval.
• Instantaneous speed is the magnitude of the instantaneous velocity.
• Instantaneous speed is a scalar quantity, as it has no direction specified.
• Average speed is the total distance traveled divided by the elapsed time. (Average speed is not the magnitude of the average velocity.) Speed is a scalar quantity; it has no direction associated
with it.
Conceptual Questions
1: Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time.
2: There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.
3: Does a car’s odometer measure position or displacement? Does its speedometer measure speed or velocity?
4: If you divide the total distance traveled on a car trip (as determined by the odometer) by the time for the trip, are you calculating the average speed or the magnitude of the average velocity?
Under what circumstances are these two quantities the same?
5: How are instantaneous velocity and instantaneous speed related to one another? How do they differ?
Problems & Exercises
1: (a) Calculate Earth’s average speed relative to the Sun. (b) What is its average velocity over a period of one year?
2: A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the average speed of the blade tip in the helicopter’s frame of
reference. (b) What is its average velocity over one revolution?
3: The North American and European continents are moving apart at a rate of about 3 cm/y. At this rate how long will it take them to drift 500 km farther apart than they are at present?
4: Land west of the San Andreas fault in southern California is moving at an average velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may
thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is 590 km northwest, assuming the motion
remains constant?
5: On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world’s nonstop long-distance speed record for trains. Its run from Denver to Chicago took 13 hours, 4
minutes, 58 seconds, and was witnessed by more than a million people along the route. The total distance traveled was 1633.8 km. What was its average speed in km/h and m/s?
6: Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a constant rate, how many
years will pass before the radius of the Moon’s orbit increases by[latex]\boldsymbol{3.84\times10^6\textbf{ m}}[/latex](1%)?
7: A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min. (a) What was her average speed? (b) If the
straight-line distance from her home to the university is 10.3 km in a direction[latex]\boldsymbol{25.0^o}[/latex]south of east, what was her average velocity? (c) If she returned home by the same
path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?
8: The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord
to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance?
9: Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by
the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back
(that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light
[latex]\boldsymbol{(3.00\times10^8\textbf{ m/s}).}[/latex]
10: A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another
21.0 m in 5.20 s. Calculate his average velocity (a) for each of the three intervals and (b) for the entire motion.
11: The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model you can view hydrogen, the simplest atom, as having a single electron
in a circular orbit[latex]\boldsymbol{1.06\times10^{-10}\textbf{ m}}[/latex]in diameter. (a) If the average speed of the electron in this orbit is known to be[latex]\boldsymbol{2.20\times10^6\textbf{
m/s}}[/latex], calculate the number of revolutions per second it makes about the nucleus. (b) What is the electron’s average velocity?
average speed
distance traveled divided by time during which motion occurs
average velocity
displacement divided by time over which displacement occurs
instantaneous velocity
velocity at a specific instant, or the average velocity over an infinitesimal time interval
instantaneous speed
magnitude of the instantaneous velocity
change, or the interval over which change occurs
simplified description that contains only those elements necessary to describe the physics of a physical situation
elapsed time
the difference between the ending time and beginning time
Check Your Understanding
1: (a) The average velocity of the train is zero because[latex]\boldsymbol{x_f=x_0}[/latex]; the train ends up at the same place it starts.
(b) The average speed of the train is calculated below. Note that the train travels 40 miles one way and 40 miles back, for a total distance of 80 miles.
[latex]\boldsymbol{\frac{\textbf{distance}}{\textbf{time}}}[/latex][latex]\boldsymbol{=}[/latex][latex]\boldsymbol{\frac{80\textbf{ miles}}{105\textbf{ minutes}}}[/latex]
[latex]\boldsymbol{\frac{80\textbf{ miles}}{105\textbf{ minutes}}}[/latex] [latex]\boldsymbol{\times}[/latex] [latex]\boldsymbol{\frac{5280\textbf{ feet}}{1\textbf{ mile}}}[/latex] [latex]\boldsymbol
{\times}[/latex] [latex]\boldsymbol{\frac{1\textbf{ meter}}{3.28\textbf{ feet}}}[/latex] [latex]\boldsymbol{\times}[/latex] [latex]\boldsymbol{\frac{1\textbf{ minute}}{60\textbf{ seconds}}}[/latex]
[latex]\boldsymbol{= 20\textbf{ m/s}}[/latex]
Problems & Exercises
(a)[latex]\boldsymbol{3.0\times10^4\textbf{ m/s}}[/latex]
(b) 0 m/s
[latex]\boldsymbol{2\times10^7\textbf{ years}}[/latex]
[latex]\boldsymbol{34.689\textbf{ m/s} = 124.88\textbf{ km/h}}[/latex]
(a)[latex]\boldsymbol{40.0\textbf{ km/h}}[/latex]
(b) 34.3 km/h,[latex]\boldsymbol{25^o\textbf{ S of E.}}[/latex]
(c)[latex]\boldsymbol{\textbf{average speed}=3.20\textbf{ km/h,}\:\bar{v}=0}[/latex].
384,000 km
(a)[latex]\boldsymbol{6.61\times10^{15}\textbf{ rev/s}}[/latex]
(b) 0 m/s | {"url":"https://pressbooks.bccampus.ca/physics0312chooge/chapter/2-3-time-velocity-and-speed/","timestamp":"2024-11-04T18:56:57Z","content_type":"text/html","content_length":"125268","record_id":"<urn:uuid:1682ff97-06a9-413d-8150-ae7f2645472e>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00296.warc.gz"} |
Three-dimensional ensemble averages for tensorial interactions in partially oriented, multi-particle systems
In a variety of three-dimensional, multi-particle systems, interactions of tensorial form occur between individual components and an applied stimulus that operates uniformly throughout the ensemble.
When each material component has an identical, fixed orientation, its own response is replicated in the observed form of behaviour by the system as a whole, with respect to the angular disposition of
the stimulus. The same complete correlation between microscopic and macroscopic response does not, however, operate in other systems where there is a degree of orientational freedom amongst the
microscopic components. One limiting case is where the extent of such freedom allows random orientations, such conditions delivering a system that is macroscopically isotropic. When each particle has
incomplete orientational freedom, the formulation is in general less mathematically tractable; here, the theory required to describe ensemble response yields analytically solvable equations only in
cases where the distribution function takes a relatively simple form, such as a scalar product, and few explicit results are available. This paper addresses general systems in which there is an
orientation-dependent complex exponential factor, weighting the response of each component. By means of irreducible tensor decomposition, results are determined for tensor interactions up to and
including rank 2, where the weighting exponent is also of any rank up to the same order. Illustrative applications are drawn from the theory of laser particle orientation and photon absorption. | {"url":"https://research-portal.uea.ac.uk/en/publications/three-dimensional-ensemble-averages-for-tensorial-interactions-in","timestamp":"2024-11-04T20:23:55Z","content_type":"text/html","content_length":"47586","record_id":"<urn:uuid:d746f76a-c14c-4bf3-9274-0b1712409c06>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00773.warc.gz"} |
Lesson 13
Graphing the Standard Form (Part 2)
Lesson Narrative
This lesson is optional because it goes beyond the depth of understanding required to address the standards. In this lesson, students continue to examine the ties between quadratic expressions in
standard form and the graphs that represent them. The focus this time is on the coefficient of the linear term, the \(b\) in \(ax^2+bx+c\), and how changes to it affect the graph. Students are not
expected to know how to modify given expressions to transform the graphs in certain ways, but they will notice that adding a linear term to the squared term translates the graph in both horizontal
and vertical directions. This understanding will help students to conclude that writing an expression such as \(x^2+bx\) in factored form can help us reason about the graph.
Students also practice writing expressions that produce particular graphs. To do so, students make use of the structure in quadratic expressions (MP7) and what they learned about the connections
between expressions and graphs.
Learning Goals
Teacher Facing
• Describe (orally and in writing) how the $b$ in $y=ax^2+bx+c$ affects the graph.
• Write quadratic expressions in standard and factored forms that match given graphs.
Student Facing
• Let’s change some other parts of a quadratic expression and see how they affect the graph.
Required Preparation
Acquire devices that can run Desmos (recommended) or other graphing technology. It is ideal if each student has their own device. (Desmos is available under Math Tools.)
Student Facing
• I can explain how the $b$ in $y=ax^2+bx+c$ affects the graph of the equation.
• I can match equations given in standard and factored form with their graph.
Print Formatted Materials
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Additional Resources
Google Slides Log In
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Set Data Properties
When you create Stateflow^® charts in Simulink^®, you can modify data properties in the Property Inspector or the Model Explorer.
To use the Property Inspector:
1. In the Modeling tab, under Design Data, select Symbols Pane and Property Inspector.
2. In the Symbols pane, select the data object.
3. In the Property Inspector, edit the data properties.
To use the Model Explorer:
1. In the Modeling tab, under Design Data, select Model Explorer.
2. In the Model Hierarchy pane, select the parent of the data object.
3. In the Contents pane, select the data object.
4. In the Dialog pane, edit the data properties.
You can also modify these properties programmatically by using Stateflow.Data objects. For more information about the Stateflow programmatic interface, see Overview of the Stateflow API.
Properties vary according to the scope and type of the data object. For many data properties, you can enter expressions or parameter values. Using parameters to set properties for many data objects
simplifies maintenance of your model because you can update multiple properties by changing a single parameter.
Stateflow Data Properties
You can set these data properties in:
• The Properties tab of the Property Inspector.
• The General tab of the Model Explorer.
Location where data resides in memory, relative to its parent.
Setting Description
Local Data defined in the current chart only.
Constant Read-only constant value that is visible to the parent Stateflow object and its children.
Constant whose value is defined in the MATLAB^® base workspace or derived from a Simulink block parameter that you define and initialize in the parent masked subsystem. The Stateflow data
Parameter object must have the same name as the MATLAB variable, Simulink data dictionary entry, or the Simulink parameter. For more information, see Share Parameters with Simulink and the MATLAB
Input argument to a function if the parent is a graphical function, truth table, or MATLAB function. Otherwise, the Simulink model provides the data to the chart through an input port on
Input the Stateflow block. For more information, see Share Input and Output Data with Simulink.
Return value of a function if the parent is a graphical function, truth table, or MATLAB function. Otherwise, the chart provides the data to the Simulink model through an output port on the
Output Stateflow block. For more information, see Share Input and Output Data with Simulink.
Data Data object that binds to a Simulink data store, which is a signal that functions like a global variable. All blocks in a model can access that signal. This binding allows the chart to read
Store and write to the Simulink data store, sharing global data with the model. The Stateflow object must have the same name as the Simulink data store. For more information, see Access Data
Memory Store Memory from a Chart.
Temporary Data that persists during only the execution of a function. You can define temporary data only for graphical functions, truth tables, or MATLAB functions in charts that use C as the action
Update method
Specifies whether a variable updates in discrete or continuous time. This property applies only when the chart is configured for continuous-time simulation. See Continuous-Time Modeling in Stateflow.
Data must resolve to signal object
Specifies that output or local data explicitly inherits properties from Simulink.Signal objects of the same name in the MATLAB base workspace or the Simulink model workspace. The data can inherit
these properties:
• Size
• Complexity
• Type
• Unit
• Minimum value
• Maximum value
• Initial value
• Storage class
• Sampling mode (for Truth Table block output data)
This option is available only when you set the model configuration parameter Signal resolution to a value other than None. For more information, see Resolve Data Properties from Simulink Signal
Size of the data object. The size can be a scalar value or a MATLAB vector of values.
• To specify a scalar, set the Size property to 1 or leave the field blank.
• To specify an n-by-1 column vector, set the Size property to n.
• To specify a 1-by-n row vector, set the Size property to [1 n].
• To specify an n-by-m matrix, set the Size property to [n m].
• To specify an n-dimensional array, set the Size property to [d[1] d[2] ⋯ d[n]], where d[i] is the size of the i^th dimension.
• To configure a Stateflow data object to inherit its size from the corresponding Simulink signal or from its definition in the chart, specify a size of –1.
The scope of the data object determines what sizes you can specify. Stateflow data store memory inherits all its properties, including its size, from the Simulink data store to which it is bound. For
all other scopes, size can be scalar, vector, or a matrix of n-dimensions. For more information, see Specify Size of Stateflow Data.
You can specify data size through a MATLAB expression that evaluates to a valid size specification. For more information, see Specify Data Size by Using Expressions and Specify Data Properties by
Using MATLAB Expressions.
Variable size
Specifies that the data object changes size during simulation. This option is available only when you enable the chart property Support variable-size arrays. For more information, see Declare
Variable-Size Data in Stateflow Charts.
Specifies whether the data object accepts complex values.
Setting Description
Off Data object does not accept complex values.
On Data object accepts complex values.
Inherited Data object inherits the complexity setting from a Simulink block.
The default value is Off. For more information, see Complex Data in Stateflow Charts.
First index
Index of the first element of the data array. The first index can be any integer. The default value is 0. This property is available only for C charts.
Type of data object. To specify the data type:
• From the Type drop-down list, select a built-in type.
• In the Type field, enter an expression that evaluates to a data type. Use one of these expressions:
For more information, see Specify Data Properties by Using MATLAB Expressions.
Additionally, in the Model Explorer, you can open the Data Type Assistant by clicking the Show data type assistant button . Specify a data Mode, and then specify the data type based on that mode. For
more information, see Specify Scope and Type of Stateflow Data.
If you enter an expression for a fixed-point data type, you must specify scaling explicitly. For example, you cannot enter an incomplete specification such as fixdt(1,16) in the Type field. If you do
not specify scaling explicitly, an error appears when you try to simulate your model.
Initial value
Initial value of the data object. For constant data, this property is called Constant value. The options for specifying this property depend on the scope of the data object.
Scope Specify for Initial Value
Expression or parameter defined in the Stateflow hierarchy, MATLAB base workspace, or Simulink masked subsystem. Specify whether Initial value is an expression or parameter by using the
Local Initialize method data property.
Constant value or expression. The expression is evaluated when you update the chart. The resulting value is used as a constant for running the chart.
Constant When you leave the Constant value field blank, numeric data resolves to a default value of 0. For enumerated data, the default value typically is the first one listed in the enumeration
section of the definition. You can specify a different default enumerated value in the methods section of the definition. For more information, see Define Enumerated Data Types.
Parameter You cannot enter a value. The chart inherits the initial value from the parameter.
Input You cannot enter a value. The chart inherits the initial value from the Simulink input signal at the designated port.
Expression or parameter defined in the Stateflow hierarchy, MATLAB base workspace, or Simulink masked subsystem. Specify whether Initial value is an expression or parameter by using the
Output Initialize method data property.
Data Store You cannot enter a value. The chart inherits the initial value from the Simulink data store to which it resolves.
The time of initialization depends on the data parent and scope of the Stateflow data object.
Data Parent Scope Initialization Time
Input Not applicable
Output, Local Start of simulation or when chart reinitializes as part of an enabled Simulink subsystem
State with History Junction Local Start of simulation or when chart reinitializes as part of an enabled Simulink subsystem
State without History Junction Local State entry
Input, Output Function-call invocation
Function (graphical, truth table, and MATLAB functions)
Local Start of simulation or when chart reinitializes as part of an enabled Simulink subsystem
For more information on using an expression to specify an initial value, see Specify Data Properties by Using MATLAB Expressions.
Initialize method
Specifies the initialization method for local and output data objects.
• Expression — Assign an expression as the initial value of the data object.
When you do not specify the Initial value property, numeric data initializes to a default value of 0. For enumerated data, the default value typically is the first one listed in the enumeration
section of the definition. You can specify a different default enumerated value in the methods section of the definition. For more information, see Define Enumerated Data Types.
• Parameter — Assign a variable in the MATLAB workspace as the initial value of the data object. If you do not specify the Initial value field, Stateflow searches for a MATLAB variable with the
same name as the data object.
The default setting is Expression.
The Property Inspector does not support initializing buses when you set this property to Expression. For more information, see Initialize Stateflow Buses.
Limit range
Range of acceptable values for this data object. Stateflow charts use this range to validate the data object during simulation.
• Minimum — The smallest value allowed for the data item during simulation. You can enter an expression or parameter that evaluates to a numeric scalar value.
• Maximum — The largest value allowed for the data item during simulation. You can enter an expression or parameter that evaluates to a numeric scalar value.
The smallest value that you can set for Minimum is -inf. The largest value that you can set for Maximum is inf.
You can specify the minimum and maximum values through a MATLAB expression. For more information, see Specify Data Properties by Using MATLAB Expressions.
A Simulink model uses the Maximum and Minimum properties to calculate best-precision scaling for fixed-point data types. Before you select Calculate Best-Precision Scaling, specify a minimum or
maximum value. For more information, see Calculate best-precision scaling.
Fixed-Point Data Properties
In the Model Explorer, when you set the Data Type Assistant Mode to Fixed point, the Data Type Assistant displays fields for specifying additional information about your fixed-point data.
Specifies whether the fixed-point data is Signed or Unsigned. Signed data can represent positive and negative values. Unsigned data represents positive values only. The default setting is Signed.
Word length
Specifies the bit size of the word that holds the quantized integer. Large word sizes represent large values with greater precision than small word sizes. The default value is 16.
• Word length can be any integer from 0 through 128 for chart-level data of these scopes:
□ Input
□ Output
□ Parameter
□ Data Store Memory
• For other Stateflow data, word length can be any integer from 0 through 32.
You can specify the word length through a MATLAB expression. For more information, see Specify Data Properties by Using MATLAB Expressions.
Specifies the method for scaling your fixed-point data to avoid overflow conditions and minimize quantization errors. The default method is Binary point scaling.
Setting Description
If you select this mode, the Data Type Assistant displays the Fraction length field, which specifies the binary point location.
Binary point Fraction length can be any integer. The default value is 0. A positive integer moves the binary point left of the rightmost bit by that amount. A negative integer moves the binary point
farther right of the rightmost bit.
If you select this mode, the Data Type Assistant displays fields for entering the Slope and Bias for the fixed-point encoding scheme.
Slope and Slope can be any positive real number. The default value is 1.0.
Bias can be any real number. The default value is 0.0.
You can enter slope and bias as expressions that contain parameters you define in the MATLAB base workspace.
Whenever possible, use binary-point scaling to simplify the implementation of fixed-point data in generated code. Operations with fixed-point data that use binary-point scaling are performed with
simple bit shifts and eliminate expensive code implementations required for separate slope and bias values. For more information about fixed-point scaling, see Scaling (Fixed-Point Designer).
You can specify Fraction length, Slope, and Bias through a MATLAB expression. For more information, see Specify Data Properties by Using MATLAB Expressions.
Data type override
Specifies whether to inherit the data type override setting of the Fixed-Point Tool that applies to this model. If the data does not inherit the model-wide setting, the specified data type applies.
Calculate best-precision scaling
Specifies whether to calculate the best-precision values for Binary point and Slope and bias scaling, based on the values in the Minimum and Maximum properties.
To calculate best-precision scaling values:
1. Specify Maximum and Minimum properties.
2. Click Calculate Best-Precision Scaling.
The best-precision scaling values are displayed in the Fraction length field or the Slope and Bias fields. For more information, see Maximize Precision (Fixed-Point Designer).
The Maximum and Minimum properties do not apply to Constant and Parameter scopes. For Constant, Simulink software calculates the scaling values based on the Initial value setting. The software cannot
calculate best-precision scaling for data of Parameter scope.
Fixed-point details
Displays information about the fixed-point data type that is defined in the Data Type Assistant:
• Minimum and Maximum show the same values that you specify in the Minimum and Maximum properties.
• Representable minimum, Representable maximum, and Precision show the minimum value, maximum value, and precision that the fixed-point data type can represent.
If the value of a field cannot be determined without first compiling the model, the Fixed-point details subpane shows the value as Unknown.
The values displayed by the Fixed-point details subpane do not automatically update if you change the values that define the fixed-point data type. To update the values shown in the Fixed-point
details subpane, click Refresh Details.
Clicking Refresh Details does not modify the model. It changes only the display. To apply the displayed values, click Apply or OK.
The Fixed-point details subpane indicates any error resulting from the fixed-point data type specification. For example, this figure shows two errors.
The row labeled Maximum indicates that the value specified by the Maximum property is not representable by the fixed-point data type. To correct the error, make one of these modifications so the
fixed-point data type can represent the maximum value:
• Decrease the value in the Maximum property.
• Increase Word length.
• Decrease Fraction length.
The row labeled Minimum shows the error Cannot evaluate because evaluating the expression MySymbol, specified by the Minimum property, does not return a numeric value. When an expression does not
evaluate successfully, the Fixed-point details subpane shows the unevaluated expression (truncating to 10 characters as needed) in place of the unavailable value. To correct this error, define
MySymbol in the base workspace to provide a numeric value. If you click Refresh Details, the error indicator and description are removed and the value of MySymbol appears in place of the unevaluated
Logging Properties
You can set logging properties for data in:
• The Properties tab of the Property Inspector.
• The Logging tab of the Model Explorer.
Log signal data
Whether to enable signal logging. Signal logging saves the values of the data object to the MATLAB workspace during simulation. For more information, see Log Simulation Output for States and Data.
Logging name
Signal name used to log the data object.
• To use the name of the data object, select Use signal name (default).
• To specify a different name, select Custom and enter the custom logging name.
Limit data points to last
Whether to limit the number of data points to log to the specified maximum. For example, if you set the maximum number of data points to 5000, the chart logs only the last 5000 data points generated
by the simulation.
Whether to limit the amount of logged data by skipping samples using the specified decimation interval. For example, if you set a decimation interval of 2, the chart logs every other sample.
Test point
Whether to set the data object as a test point that you can monitor with a floating scope during simulation. You can also log test point values to the MATLAB workspace. For more information, see
Monitor Test Points in Stateflow Charts.
Additional Properties
You can set additional data properties in:
• The Info tab of the Property Inspector.
• The Description tab of the Model Explorer.
Save final value to base workspace
Assigns the value of the data object to a variable of the same name in the MATLAB base workspace at the end of simulation. This option is available only in the Model Explorer for charts that use C as
the action language. For more information, see Model Workspaces (Simulink).
Units of measurement associated with the data object. The unit in this field resides with the data object in the Stateflow hierarchy. This property is available only in the Model Explorer for C
Document link
Link to online documentation for the data object. You can enter a web URL address or a MATLAB command that displays documentation as an HTML file or as text in the MATLAB Command Window. When you
click the Document link hyperlink, Stateflow evaluates the link and displays the documentation.
Default Data Property Values
When you leave a property field blank, Stateflow assumes a default value.
Specify Data Properties by Using MATLAB Expressions
In the Property Inspector and Model Explorer, you can enter MATLAB expressions as values for these properties:
Expressions can contain a mix of numeric values, constants, parameters, variables, arithmetic operations, parameters, constants, arithmetic operators, and calls to MATLAB functions. For example, you
can use these functions to specify data properties.
Property Function Description
Size size Returns the size of a data object
type Returns the type of a data object
Type fixdt (Simulink) Returns a Simulink.NumericType object that describes a fixed-point or floating-point data type
fi (Fixed-Point Designer) Returns a fixed-point numeric object
Minimum min Returns the smallest element or elements of an array
Maximum max Returns the largest element or elements of an array
For more information, see Specify Data Size by Using Expressions and Derive Data Types from Other Data Objects.
See Also
Related Topics | {"url":"https://uk.mathworks.com/help/stateflow/ug/set-data-properties-1.html","timestamp":"2024-11-07T04:11:37Z","content_type":"text/html","content_length":"118151","record_id":"<urn:uuid:9695e7dc-8f64-46b4-87bf-f094e95513dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00871.warc.gz"} |
Audiogon Discussion Forum
Hi Lynne,
The most accurate answer is probably "neither."
A square wave, in the context of electrical signals, is a voltage that alternates periodically between a higher voltage level and a lower voltage level, spending an equal amount of time in each of
the two states. An ideal square wave has infinitely fast transitions between the two states, and each voltage level is perfectly precise and constant, i.e., no noise (random fluctuation of the
voltage levels) is present. Neither of those conditions is possible in the real world, so what are referred to as "square waves" are approximations of ideal square waves.
A square wave cannot normally be used to convey information, because its pattern of alternating between the two voltage states remains the same all the time. It can be used in many applications as a
"clock signal," however, which controls the timing of whatever operations are performed by the circuit that is involved. Square waves can also be useful as test signals, to evaluate circuit or
component performance.
1's and 0's are just numbers. The decimal (base 10) numbering system that humans like to use utilizes numbers whose individual digits can range from 0 to 9. The 1's and 0's you refer to are based on
the binary (base 2) numbering system, where the only allowable digits are 1 and 0. Either system can represent all possible numbers; it just takes more digits to do it in the binary system. Computers
and other digital devices are designed based on the binary number system because their practical implementation is facilitated by the fact that only two states have to be distinguished from each
A series of 1's and 0's can be used to convey information. An example of "information" is the amplitude (volume) of a music signal at a given instant of time. Since those 1's and 0's are numbers,
though, they in turn have to be represented by something else, such as a voltage level, before they can be sent or communicated or processed by a physical circuit. In some applications, a 1 may be
represented by a higher voltage, and a 0 by a lower voltage, or vice versa.
In the two cases you mentioned, though, those approaches aren't used, in part because clock and data are combined into a single signal, in such a manner that the receiving circuit can separate the
encodes the 1 and 0 data, together with the clock and additional necessary information as described in the writeup, into something called
Biphase Mark or Differential Manchester Code
. Ethernet, since it is a networking standard that is designed to provide communications between multiple devices at arbitrary and intermittent times, and in its modern forms at very high speeds, is
complex and is described further in
this Wikipedia writeup
and at the links it provides. Different codings, all of them combining clock, data, and other necessary information, are used for each of the commonly used link speeds (10, 100, or 1000 mbps).
Hope that clarifies more than it confuses :-)
Best regards,
-- Al
first you need to understand a meaning of analogue signal, than modulation, than sampling, than binary algebra.
otherwise any technical content will seem to be confusing.
Paul McGowan ran a huge post on these issues with DACs. Go to www.psaudio.com and read "Paul's Posts". It is a ton of reading...
Thanks, everybody. I've been reading on the net. Someone says the signal is 1's and 0's and the next person (critic) says square waves. You open the door so that I can understand the concepts. What
started it is that I have an entry level (RCA)MIT digital coax and a signalcable digital coax. I knew that the MIT didn't cut it so I tried the signalcable and the sound opened up. Now I'm reading
overwhelming positive reviews of Oyaide DR-510 and wondering if it would be a worthwhile upgrade from the signalcable at $140 or around there. Don't know if I will try it but I feel smarter already.
It's 1's and 0's. A cd is basically a file of 1's and 0's. DAC is a computer that converts digital (1's and 0's) to an analog signal ... D to A converter. Data must be loaded into memory before the
CPU can process it whether streamed from a computer or transport.
Ideally it should work like any program. When it starts, loads into memory before running. This eliminates jitter, complex synchronization between DAC and tranport ... Can you imagine running your
browser or MS Office off a CD drive?
Even HD video is streaming with more data demand than audio. Netflicks almost bankrupted before streaming to compete with the cable companies.
04-03-13: Knghifi
It's 1's and 0's.
Kng, while 1's and 0's are certainly being COMMUNICATED between the two components, the references to the signal, and to the possibility that it might be a square wave, would appear to indicate that
what is being asked about is what is being "sent" in a physical/electrical sense.
As you will realize, numbers cannot be sent through wires, in that sense. So when "someone says the signal is 1's and 0's and the next person (critic) says square waves" (quoting from Lynne's second
post), they are both wrong.
-- Al
Square wave is a generic term for a non-continuously changing voltage signal. It changes in discrete steps. It does not necessarily mean a repetitive signal or actually square. IF it were repetitive,
then it is a digital oscillation, also known as a clock. Because even clocks have non-50% duty cycles, even clocks are not actually "square".
1's and 0's are defined as: the high state of the signal is "1" and the low state of the signal is "0". the high and low can be defined as any voltage depending on the logic family and physical
All digital voltage signals actually contain analog components since they do not switch in zero time from 0-1 or 1-0, and because drivers and transmission-lines are not perfect, there is also
resulting overshoot, ringing etc..
Steve N.
Empirical Audio
Thanks, Steve.
I would add to your comment, though, the clarification that the 1's and 0's that are referred to in your definition are NOT the same thing as the 1's and 0's which constitute the audio data that may
be communicated via S/PDIF or ethernet, which are the focus of this thread.
Also, although your definition of a square wave is a reasonable one, it is a looser definition than many others would apply to the term, their more narrow definition also being reasonable IMO. See,
for instance, the first paragraph of this Wikipedia writeup, in which a square wave is defined as being periodic, and as having equal durations in its two states.
-- Al
Its not sending an exact square wave, but it sends the signal as an analog waveform.
So anyone who says 'square wave' is about 1000x more accurate than the person who says 1s & 0s.
Al and Agisthos are right. I was not asking the content of the message (signal). I was asking for a physical description of the signal. I thought I was asking in effect for someone to dispel a myth.
Now I'm not sure of that because there seems to be a gray area of confusion depending on what exactly is the question.
I also read that a digital coax cable between transport and DAC should be 5 ft long. This certainly qualifies as myth.
Hi Lynne,
Perhaps surprisingly, the 5 ft/1.5 meter length suggestion is not a myth. See Steve's paper here, which makes sense to me, and is also supported by experimental results that have been reported by
some A'gon members I consider to be credible. What length will be optimal in a given system is dependent on a number of hardware-specific variables, however, which are generally unspecified, and IMO
that recommendation should be viewed as a length that is LIKELY to be optimal in MOST cases, but is not guaranteed to be. There have been at least a few reports I have seen here from members who have
compared different lengths, and have found shorter lengths, such as 1 meter, to be preferable in their systems. Also, if a very short length is practicable, such as 8 inches or less, IMO that stands
a very good chance of being an optimal choice.
Some further comments on the definitions Steve provided in his post above:
As I indicated in my previous post, what he defined as "1's and 0's" has nothing to do with the 1's and 0's which comprise the data content of the signal. A less ambiguous way of referring to what he
defined as "1's and 0's" would be to refer to them as "logic levels," or more specifically as "logic 1" and "logic 0" levels, respectively.
If we go by the assumptions I stated in my initial post, that "1's and 0's" refers to data, and that square waves are periodic and symmetrical, then the parties you quoted, who were disagreeing with
each other, were both wrong.
If we go by Steve's definitions, then those parties, who were disagreeing with each other, were both right!
I would agree that a plausible case could be made on the basis of either set of definitions. However, IMO it would be a safe bet that the presumably non-technical person who was arguing that what is
being sent are 1's and 0's was referring to data, and not to logic levels.
Best regards,
-- Al
Hi Al,
Thanks for articulating the original question. Steve's paper is very helpful.
I can do close to 8" but will have to measure to make sure. It will be interesting to experiment and very helpful to know that length is almost always a factor.
Best Regards,
I guess we all interpreted your question a little different.
digital signal not only has ones and zeroes, but rather characterised by sampling frequency and sampled amplitude as mentioned before voltage level.
an arbitrary point of signal could be captured knowing the particular voltage and the value of the signal 0(low) or 1(high). The low could be let's say 0V and high could be +50mV etc...
0-s and 1-s are only rererence values of digital signal, but not the actual physical ones for those who thinks that 0s and 1s are 'traveling' accross the digital cable.
electrones are the only ones capable of doin' that
Almarg - with S/PDIF the transport level is still 1's and 0's, but the protocol of the interface requires encoding the data to limit the number of consecutive 1's or zeroes, allowing a clock to be
recovered as well as the data.
Steve N.
Empirical Audio
My question was oversimplified and came out of ignorance. Your anwsers will give me all the reading I want tonight because I have to re-read. You guys are over my head but it's fun trying to
understand it. You are a great source, better than anything else I've found on the net.
I was tired of reading the argument that the quality of cable doesn't matter because the signal is simply ones and zeros. I have two digital coax cables and it clearly does matter.
Thank you.
that was a true statement Lynne simply because digital cable carries simple shape signal compared to an analogue music.
OK. An analogue wave form is continuous in terms of time and voltage and frequency. A digital wave form is a square wave, as it were, which means it is repetitively maybe not on-off but high-low,
higher-not-so-high, low-lower in terms of time and voltage as it represents the encoded information. And the binary system is the only one that can work for this Pulse Code Modulation because the
language is ones and zeros. The vehicle for this language is the square wave because it is not continuous but repetitive.
That's my homework.
Lynne, if you haven't already, take a look at the figures shown in the Wikipedia writeup I linked to earlier for Biphase Mark/Differential Manchester Encoding, Biphase Mark (shown in the second
figure) being the encoding method used for S/PDIF. The paragraph above the figures helps to clarify them.
Think of all the waveforms shown in the figures as being graphs that depict voltage along their vertical axis, and time along their horizontal axis.
As you'll see, 1 and 0 data information is conveyed by virtue of whether one "transition" or two "transitions" occur within each "clock period" (defined below). A "transition" being defined as a
CHANGE from either the higher voltage ("logic 1") state to the lower voltage ("logic 0") state, or vice versa.
The higher voltage (logic 1) state is the upper of the two possible voltage levels of each signal waveform that is shown, and the lower voltage (logic 0) state is the lower of those two levels.
A "clock period" is defined as the amount of time either between one positive-going (logic 0 to logic 1) transition of the clock waveform and the next positive-going transition of that waveform, or,
equivalently, between one negative-going (logic 1 to logic 0) transition of the clock waveform and the next negative-going transition of that waveform.
That encoding method allows both clock and data to be conveyed in a single signal, as Steve and I indicated earlier.
-- Al
Jitter on a digital streaming interface is another thing entirely. This is the time variation of the switching transitions, not just 1's and 0's. The digital feed to a D/A is sensitive to this
because precise timing matters. It is also important that the A/D used when recording has low jitter. These two add to make more frequency modulation distortion at the D/A.
See more info here:
Steve N.
Empirical Audio
Al & Steve,
Yes, Al, I read the Wiki piece before but was somewhat unclear until you have now further explained it. Visual aids--graphs, charts, schematics are difficult. I need to learn first the meaning before
they make sense. I was unclear on the voltage being a constant value and on the transition period as it relates to the binary data.
I was struck with the beauty of the S/PDIF system where the clock and data are one signal until I read Steve's explanation of jitter and how the BMC signal is vulnerable to it.
I'm very happy to have learned this basic concept of the digital signal. Many thanks for hanging in.
Al, on an unrelated topic discussed in a former thread which I should discuss in a follow up to that thread but since the website made changes nothing works on my pc the way it did and I'm not sure
you would find it on that thread, and in regard to your lack of enthusiasm for autoformers in SS, I did demo the autoformers with the hk990 integrated driving the AR9's. the 990 is rated at 150w/
8ohms, 300w/4ohms, and the 9's are rated at 4ohms. So using the 4ohm tap gave the set-up 150w/8ohms versus 300w/4ohms without autoformers. There was no apparent difference except that sound quality
was a little better without the autoformers apparently because of running the signal through another device with more connections. It was a horse a piece.
But with an amp that is limited into 4ohms, the autoformers did enhance sound quality when the amp was driving 4 ohm speakers turned into 8 ohm speakers. Presumably converting 8 ohm speakers into 16
ohm speakers would cause more loss than gain. If all this makes any sense.
Hi Lynne,
Here is a link to the other thread you are referring to. Everything in your post above sounds reasonable to me.
-- Al
There's no waveforms in the Universe that are not continues. That also applies to square pulse.
OK. It probably needs to be qualified. I got it from
en.wikipedia.org/wiki/Analog_signal. Please explain. | {"url":"https://d2dve11u4nyc18.cloudfront.net/discussions/square-waves-or-1-s-and-0-s","timestamp":"2024-11-08T08:31:57Z","content_type":"text/html","content_length":"158794","record_id":"<urn:uuid:e68a46cd-04ab-45d5-970e-32232d9ca462>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00030.warc.gz"} |
tessellations: The classical theory of moments for structure description of granular materials
Authors: O.I. Gerasymov, N.N. Khudyntsev
Year: 2015
Issue: 19
Pages: 170-175
The theoretical description of the local structure of granular materials has been performed by means of Voronoi method. The detailed investigation of structure transformations has been carried on
with help of Voronoi tessellation supplemented by direct modeling of the relevant distribution function in terms of classical moments theory. Analytical expression for distribution function of
Voronoi figures has been constructed with the help of Nevanlinna’s formula from theory of orthogonal polynomials .Proposed approach permit to avoid the problem of week argumentation of
applicability the statistical mechanics methods for description of the structure and physical properties of granular materials. We show that generated ordering in local structure are escorted by
appearing of particular symmetries in Voronoi diagrams. We perform a numerical simulations of structural configurations in 2D system of hard discs. Proposed algorithm allow us to prove
theoretical predictions about existence of correlations between configurational ordering and symmetry breaking in Voronoy tessellations. We study these effects in the vicinity of jammed states.
Obtained results shows that criticality in structurisation (formation of jammed states)connected with particular behavior of the first two moments of Voronoi figures distribution function. We
show nonhomogeneous character of jammed states in which kinematic freedom degrees become frozen. Namely, coexisting ordered domains which has a different symmetries in grain configurations are
observed. Therefore given analysis fulfill the basis of research in the area of granular physics which are mostly based on the concepts of probabilistic stereology and do not use methods from
statistical mechanics which in the case of granular materials are not enough argumeneted.
Tags: classical theory of moments; granular materials; jammed states; Voronoi figures
1. Jaeger H.M., Nagel S.R., Behringer R.P. Rev. Mod. Phys., 1996, no. 68, pp. 1259.
2. Duran J. Sands, Powders and Grains. New York: Springer, 2000.
3. Kadanoff L. Rev.Mod.Phys., 1999, no. 71, pp. 435.
4. De Gennes P.G., Rev. Mod. Phys., 1999, no. 71, pp. 374.
5. Bideau D., Gervois A., Oger L., Troadec J.P. J.Physique, 1986, no. 47, pp. 1697.
6. Lumay J., Vandewalle N. Phys. Rev. Lett., 2005, no. 95, pp. 028002.
7. Voronoi G., Reine Angew J. Math., 1908, no. 134, pp. 198.
8. Quickenden T.I., Tan G.K. Journal of Collloid and Interface Science , 1974, no. 48, pp. 382.
9. Aste T., Di Matteo T. Phys. Rev., 2008, no. E77, pp. 021309.
10. Edwards S.F. Physica, 2005, no. A353 , pp. 114.
11. Achiezer N.I., Classical problem of moments. Мoscow: Nauka, 1961. 600 p.
12. Gerasymov O.I. Ukr.Journ.Phys., 2010, vol.55, no. 5, pp. 560.
Download full text (PDF) | {"url":"http://bulletin.odeku.edu.ua/en/voronoi-tessellations-the-classical-theory-of-moments-for-structure-description-of-granular-materials/","timestamp":"2024-11-14T00:53:42Z","content_type":"text/html","content_length":"78723","record_id":"<urn:uuid:f4a70001-a03e-47e9-a351-7076ab85ca27>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00608.warc.gz"} |
Probabilistic computing with p-bits
Digital computers store information in the form of bits that can take on one of two values 0 and 1, while quantum computers are based on qubits that are described by a complex wavefunction, whose
squared magnitude gives the probability of measuring either 0 or 1. Here, we make the case for a probabilistic computer based on p-bits, which take on values 0 and 1 with controlled probabilities and
can be implemented with specialized compact energy-efficient hardware. We propose a generic architecture for such p-computers and emulate systems with thousands of p-bits to show that they can
significantly accelerate randomized algorithms used in a wide variety of applications including but not limited to Bayesian networks, optimization, Ising models, and quantum Monte Carlo.
Feynman^1 famously remarked “Nature isn't classical, dammit, and if you want to make a simulation of nature, you'd better make it quantum mechanical.” In the same spirit, we could say, “Many real
life problems are not deterministic, and if you want to simulate them, you'd better make it probabilistic.” However, there is a difference. Quantum algorithms require quantum hardware, and this has
motivated a worldwide effort to develop a new appropriate technology. In contrast, probabilistic algorithms can be and are implemented on existing deterministic hardware using pseudo RNGs (random
number generators). Monte Carlo algorithms represent one of the top ten algorithms of the 20th century^2 and are used in a broad range of problems including Bayesian learning, protein folding,
optimization, stock option pricing, and cryptography, just to name a few. So why do we need a p-computer?
A key element in a Monte Carlo algorithm is the RNG, which requires thousands of transistors to implement with deterministic elements, thus encouraging the use of architectures that time share a few
RNGs. Our work has shown the possibility of high quality true RNGs using just three transistors,^3 prompting us to explore a different architecture that makes use of large numbers of controlled-RNGs
or p-bits. Figure 1(a)^,^4 shows a generic vision for a probabilistic or a p-computer having two primary components: an N-bit random number generator (RNG) that generates N-bit samples and a Kernel
that performs deterministic operations on them. Note that each RNG-Kernel unit could include multiple RNG-Kernel sub-units (not shown) for problems that can benefit from it. These sub-units could be
connected in series as in Bayesian networks [Fig. 2(a) or in parallel as done in parallel tempering^5,6 or for problems that allow graph coloring.^7 The parallel RNG-Kernel units shown in Fig. 1(a)
are intended to perform easily parallelizable operations like ensemble sums using a data collector unit to combine all outputs into a single consolidated output.
Ideally, the Kernel and data collector are pipelined so that they can continually accept new random numbers from the RNG,^4 which is assumed to be fast and available in numerous numbers. The
p-computer can then provide $N p f c$ samples per second, with N[p] being the number of parallel units^9 and f[c] being the clock frequency. We argue that even with N[p]=1, this throughput is well
in excess of what is achieved with standard implementations on either CPU (central processing unit) or graphics processing unit (GPU) for a wide range of applications and algorithms including but not
limited to those targeted by modern digital annealers or Ising solvers.^8,10–17 Interestingly, a p-computer also provides a conceptual bridge to quantum computing, sharing many characteristics that
we associate with the latter.^18 Indeed, it can implement algorithms intended for quantum computers, though the effectiveness of quantum Monte Carlo depends strongly on the extent of the so-called
sign problem specific to the algorithm and our ability to “tame” it.^19
Of the three elements in
Fig. 1
, two are
. The
is problem-specific ranging from simple operations like addition or multiplication to more elaborate operations that could justify special purpose chiplets.
Matrix multiplication, for example, could be implemented using analog options like resistive crossbars.
The data collector typically involves addition and could be implemented with adder trees. The third element is
, namely, the N-bit
, which is a collection of
1-bit RNGs or
. The behavior of each
can be described by
$s i = Θ [ σ ( I i − r ) ],$
is the binary p-bit output, Θ is the step function,
is the sigmoid function,
is the input to the p-bit, and
is a uniform random number between 0 and 1. Equation
is illustrated in
Fig. 1(b)
. While the p-bit output is always binary, the p-bit input
influences the mean of the output sequence. With
=0, the output is distributed 50–50 between 0 and 1, and this may be adequate for many algorithms. Yet, in general, a non-zero
determined by the current sample is necessary to generate desired probability distributions from the
N-bit RNG
One promising implementation of a p-bit is based on a stochastic magnetic tunnel junction (s-MTJ) as shown in Fig. 1(b), whose resistance state fluctuates due to thermal noise. It is placed in series
with a transistor, and the drain voltage is thresholded by an inverter^3 to obtain a random binary output bit, whose average value can be tuned through the gate voltage $V IN$. It has been shown
both theoretically^25,26 and experimentally^27,28 that s-MTJ-based p-bits can be designed to generate new random numbers in times ∼nanoseconds. The same circuit could also be used with other
fluctuating resistors,^29 but one advantage of s-MTJs is that they can be built by modifying magnetoresistive random access memory (MRAM) technology that has already reached gigabit levels of
Note, however, that the examples presented here all use p-bits implemented with deterministic CMOS elements or pseudo-RNGs using linear feedback shift registers (LFSRs) combined with lookup tables
(LUTs) and thresholding elements^8 as shown in Fig. 1(b). Such random numbers are not truly random but have a period that is longer than the time range of interest. The longer the period, the more
registers are needed to implement it. Typically, a p-bit requires $∼ 1000$ transistors,^30 and the actual number depending on the quality of the pseudo RNG that is desired. Thirty-two stage LFSRs
require $∼ 1200$ transistors, while a Xoshiro128+^31 would require around four times as many. Physics-based approaches, like s-MTJs, naturally generate true random numbers with infinite repetition
period and ideally require only three transistors and one MTJ.
A simple performance metric for p-computers is the ideal sampling rate $N p f c$ mentioned above. The results presented here were all obtained with an field-programmable gate array (FPGA) running on
a 125MHz clock, for which $1 / f c = 8$ ns, which could be significantly shorter [even $∼ 0.1$ ns (Ref. 25)] if implemented with s-MTJs. Furthermore, s-MTJs are compact and energy-efficient,
allowing up to a factor of 100 larger N[p] for a given area and power budget. With an increase in f[c] and N[p], a performance improvement by 2–3 orders of magnitude over the numbers presented here
may be possible with s-MTJs or other physics-based hardware.
We should point out that such compact p-bit implementations are still in their infancy,^30 and many questions remain. First is the inevitable variation in RNG characteristics that can be expected.
Initial studies suggest that it may be possible to train the Kernel to compensate for at least some of these variations.^32,33 Second is the quality of randomness, as measured by statistical quality
tests, which may require additional circuitry as discussed, for example, in Ref. 27. Certain applications like simple integration (Sec. IIIA) may not need high quality random numbers, while others
like Bayesian correlations (Sec. IIIB) or Metropolis–Hastings methods that require a proposal distribution (Sec. IIIC) may have more stringent requirements. Third is the possible difficulty
associated with reading sub-nanosecond fluctuations in the output and communicating them faithfully. Finally, we note that the input to a p-bit is an analog quantity requiring digital-to-analog
converters (DACs) unless the kernel itself is implemented with analog components.
A. Simple integration
A variety of problems such as high dimensional integration can be viewed as the evaluation of a sum over a very large number
of terms. The basic idea of the Monte Carlo method is to estimate the desired sum from a limited number
of samples drawn from configurations
generated with probability
$q α$
$M = ∑ α = 1 N m α ≈ 1 N S ∑ α = 1 N S m α q α.$
The distribution {
} can be uniform or could be cleverly chosen to minimize the standard deviation of the estimate.
In any case, the standard deviation goes down as
$1 / N s$
, and all such applications could benefit from a
to accelerate the collection of samples.
B. Bayesian network
A little more complicated application of a p-computer is to problems where random numbers are generated not according to a fixed distribution, but by a distribution determined by the outputs from a
previous set of $RNGs$. Consider, for example, the question of genetic relatedness in a family tree^35,36 with each layer representing one generation. Each generation in the network in Fig. 2(a)
with N nodes can be mapped to a N-bit RNG-block feeding into a Kernel, which stores the conditional probability table (CPT) relating it to the next generation. The correlation between different nodes
in the network can be directly measured, and an average over the samples computed to yield the correct genetic correlation as shown in Fig. 2(b). Nodes separated by p generations have a correlation
of $1 / 2 p$. The measured absolute correlation between strangers goes down to zero as $1 / N s$.
This is the characteristic of Monte Carlo algorithms, namely, to obtain results with accuracy ε we need $N s = 1 / ε 2$ samples. The p-computer allows us to collect samples at the rate of $N p f c$=
125 MSamples per second if N[p]=1 and f[c]=125MHz. This is about two orders of magnitude faster than what we get running the same algorithm on an Intel Xeon CPU.
How does it compare to
algorithms run on
? As Feynman noted in his seminal paper,
deterministic algorithms for problems of this type are very inefficient compared to probabilistic ones because of the need to integrate over all the unobserved nodes
${ x B }$
in order to calculate a property related to nodes
${ x A }$
$P A ( x A ) = ∫ d x B P ( x A , x B ).$
In contrast, a
can ignore all the irrelevant nodes
${ x b }$
and simply look at the relevant nodes
${ x A }$
. We used the example of genetic correlations, because it is easy to relate to. However, it is representative of a wide class of everyday problems involving nodes with one-way causal relationships
extending from “parent” nodes to “child” nodes,
all of which could benefit from a
C. Knapsack problem
Let us now look at a problem that requires random numbers to be generated with a probability determined by the outcome from the last sample generated by the same RNG. Every RNG then requires feedback
from the very Kernel that processes its output. This belongs to the broad class of problems that are labeled as Markov Chain Monte Carlo (MCMC). For an excellent summary and evaluation of MCMC
sampling techniques, we refer the reader to Ref. 40.
The knapsack is a textbook optimization problem described in terms of a set of items, $m = 1 , … N$, the m-th, each containing a value v[m] and weighing w[m]. The problem is to figure out which
items to take (s[m]=1) and which to leave behind (s[m]=0) such that the total value $V = ∑ m v m s m$ is a maximum, while keeping the total weight $W = ∑ m w m s m$ below a capacity C. We could
straightforwardly map it to the $p − computer$ architecture (Fig. 1), using the RNG to propose solutions ${ s }$ at random, and the Kernel to evaluate V, W and decide to accept or reject. Yet, this
approach would take us toward the solution far too slowly. It is better to propose solutions intelligently looking at the previously accepted proposal and making only a small change to it. For our
examples, we proposed a change of only two items each time.
This intelligent proposal, however, requires feedback from the kernel, which can take multiple clock cycles. One could wait between proposals, but the solution is faster if instead we continue to
make proposals every clock cycle in the spirit of what is referred to as multiple-try Metropolis.^41 The results are shown in Fig. 3 ^4 and compared with CPU (Intel Xeon at 2.3GHz) and GPU (Tesla T4
at 1.59GHz) implementations, using the probabilistic algorithm. Also shown are two efficient deterministic algorithms, one based on dynamic programming (DP), and another based on the work of
Pisinger and co-workers.^42,43
Note that the probabilistic algorithm (MCMC) gives solutions that are within 1% of the correct solution, while the deterministic algorithms give the correct solution. For the Knapsack problem,
getting a solution that is 99% accurate should be sufficient for most real world applications. The p-computer provides orders of magnitude improvement over CPU implementation of the same MCMC
algorithm. It is outperformed by the algorithm developed by Pisinger and co-workers,^42,43 which is specifically optimized for the Knapsack problem. However, we note that the p-computer projection in
Fig. 3(b) is based on utilizing better hardware like s-MTJs, but there is also significant room for improvement of the p-computer by optimizing the Metropolis algorithm used here and/or by adding
parallel tempering.^5,6
D. Ising model
Another widely used model for optimization within MCMC is based on the concept of
Boltzmann machines (BMs)
defined by an energy function
from which one can calculate the synaptic function
$I i = β ( E ( s i = 0 ) − E ( s i = 1 ) ) ,$
which can be used to guide the sample generation from each
RNG i
in sequence
according to Eq.
. Alternatively, the sample generation from each RNG can be fixed, and the synaptic function can be used to decide whether to accept or reject it within a Metropolis–Hastings framework.
Either way, samples will be generated with probabilities
$P α ∼ exp ( − β E α )$
. We can solve optimization problems by identifying
with the negative of the cost function that we are seeking to minimize. Using a large
, we can ensure that the probability is nearly 1 for the configuration with the minimum value of E.
In principle, the energy function is arbitrary, but much of the work is based on quadratic energy functions defined by a connection matrix
and a bias vector
(see, for example, Refs.
$E = − ∑ i j W i j s i s j − ∑ i h i s i .$
For this quadratic energy function, Eq.
$I i = β ( ∑ j W i j s j + h i )$
, so that the
has to perform a multiply and accumulate operation as shown in
Fig. 4(a)
. We refer the reader to Ref.
for an example of the
optimization problem on a two-dimensional (2D) 90×90 array implemented with a
Equation (4), however, is more generally applicable even if the energy expression is more complicated, or given by a table. The Kernel can be modified accordingly. For example of an energy function
with fourth order terms implemented on an eight bit p-computer, we refer the reader to Ref. 30.
A wide variety of problems can be mapped onto the BM with an appropriate choice of the energy function. For example, we could generate samples from a desired probability distribution P, by choosing
$β E = − ℓ n P$. Another example is the implementation of logic gates by defining E to be zero for all ${ s }$ that belong to the truth table and have some positive value for those that do not.^24
Unlike standard digital logic, such a BM-based implementation would provide invertible logic that not only provides the output for a given input but also generates all possible inputs corresponding
to a specified output.^24,46,47
E. Quantum Monte Carlo
Finally, let us briefly describe the feasibility of using p-computers to emulate quantum or q-computers. A q-computer is based on qubits that are neither 0 or 1 but are described by a complex
wavefunction, whose squared magnitude gives the probability of measuring either 0 or 1. The state of an $n − qubit$ computer is described by a wavefunction ${ ψ }$ with $2 n$ complex components, one
for each possible configuration of the n qubits.
In gate-based quantum computing (GQC), a set of qubits is placed in a known state at time
, operated with
quantum gates to manipulate the wavefunction through unitary transformations
$[ U ( i ) ]$
${ ψ ( t + d ) } = [ U ( d ) ] · · · · [ U ( 1 ) ] { ψ ( t ) } ( GQC ),$
and measurements are made to obtain results with probabilities given by the squared magnitudes of the final wavefunctions. From the rules of matrix multiplication, the final wavefunction can be
written as a sum over a very large number of terms
$ψ m ( t + d ) = ∑ i , · · j , k U m , i ( d ) · · · · U j , k ( 1 ) ψ k ( t ).$
Conceptually, we could represent a system of
qubits and
gates with a system of
$( n × d )$ p-bits
with second states that label the second terms in the summation in Eq.
Each of these terms is often referred to as a
Feynman path
and what we want is the sum of the amplitudes of all such paths
$ψ m ( t + d ) = ∑ α = 1 2 n d A m ( α ).$
The essential idea of
quantum Monte Carlo
is to estimate this enormous sum from a few suitably chosen samples, not unlike the simple Monte Carlo stated earlier in Eq.
. What makes it more difficult, however, is the so-called
sign problem
which can be understood intuitively as follows. If all the quantities
$A m ( α )$
are positive then it is relatively easy to estimate the sum from a few samples. However, if some are positive while some are negative with more cancelations, then many more samples will be required.
The same is true if the quantities are complex quantities that cancel each other.
The matrices U that appear in GQC are unitary with complex elements, which often lead to significant cancelation of Feynman paths, except in special cases when there may be complete constructive
interference. In general, this could make it necessary to use large numbers of samples for accurate estimation. A noiseless quantum computer would not have this problem, since qubits intuitively
perform the entire sum exactly and yield samples according to the squared magnitude of the resulting wavefunction. However, real world quantum computers have noise, and p-computers could be
competitive for many problems.
Adiabatic quantum computing (AQC) operates on very different physical principles, but its mathematical description can also be viewed as summing the Feynman paths representing the multiplication of
$[ e − β H / r ] · · · · [ e − β H / r ] ( AQC ).$
This is based on the Suzuki–Trotter method described in Ref.
, where the number of replicas,
, is chosen large enough to ensure that if
$H = H 1 + H 2$
, one can approximately write
$e − β H / r ≈ e − β H 1 / r e − β H 2 / r$
. The matrices
$e − β H / r$
in AQC are Hermitian, and their elements can be all positive.
A special class of Hamiltonians
having this property is called
, and there is no sign problem since the amplitudes
$A m ( α )$
in the Feynman sum in Eq.
all have the same sign.
An example of such a stoquastic Hamiltonian is the transverse field Ising model (TFIM) commonly used for quantum annealing, where a transverse field which is quantum in nature is introduced and
slowly reduced to zero to recover the original classical problem. Figure 4, adapted from Ref. 8, shows an n=250 qubit problem mapped to a 2D lattice of 250×10=2500 p-bits using r=10 replicas
to calculated average correlations between the z-directed spins on lattice sites separated by L. Very accurate results are obtained using N[s]=10^5 samples. However, these samples were suitably
spaced to ensure their independence, which is an important concern in problems involving feedback.
Finally, we note that quantum Monte Carlo methods, both GQC and AQC, involve selective summing of Feynman paths to evaluate matrix products. As such, we might expect conceptual overlap with the very
active field of randomized algorithms for linear algebra,^51,52 though the two fields seem very distinct at this time.
In summary, we have presented a generic architecture for a p-computer based on p-bits, which take on values 0 and 1 with controlled probabilities, and can be implemented with specialized compact
energy-efficient hardware. We emulate systems with thousands of p-bits to show that they can significantly accelerate the implementation of randomized algorithms that are widely used for many
applications.^53 A few prototypical examples are presented such as Bayesian networks, optimization, Ising models, and quantum Monte Carlo.
The authors are grateful to Behtash Behin-Aein for helpful discussions and advice. We also thank Kerem Camsari and Shuvro Chowdhury for their feedback on the manuscript. The contents are based on the
work done over the last 5–10 years in our group, some of which has been cited here, and it is a pleasure to acknowledge all who have contributed to our understanding. This work was supported in part
by ASCENT, one of six centers in JUMP, a Semiconductor Research Corporation (SRC) program sponsored by DARPA.
Conflict of Interests
One of the authors (S.D.) has a financial interest in Ludwig Computing.
The data that support the findings of this study are available from the corresponding author upon reasonable request.
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© 2021 Author(s). All article content, except where otherwise noted, is licensed under a Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/). | {"url":"https://pubs.aip.org/aip/apl/article/119/15/150503/40486","timestamp":"2024-11-12T03:49:21Z","content_type":"text/html","content_length":"624948","record_id":"<urn:uuid:a17c610c-4bfe-481f-a385-282d5d25a4d0>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00341.warc.gz"} |
oordinates to
Convert world coordinates to row and column subscripts
[I, J] = worldToSubscript(R,xWorld,yWorld) maps points from the 2-D world system (xWorld,yWorld) to subscript arrays I and J based on the relationship defined by 2-D spatial referencing object R.
If the kth input coordinates (xWorld(k),yWorld(k)) fall outside the image bounds in the world coordinate system, worldToSubscript sets the corresponding subscripts I(k) and J(k) to NaN.
[I, J, K] = worldToSubscript(R,xWorld,yWorld,zWorld) maps points from the 3-D world system to subscript arrays I, J, and K, using 3-D spatial referencing object R.
Convert 2-D World Coordinates to Row and Column Subscripts
Read a 2-D grayscale image of a knee into the workspace.
m = dicominfo('knee1.dcm');
A = dicomread(m);
Create an imref2d object, specifying the size and the resolution of the pixels. The DICOM file contains a metadata field PixelSpacing that specifies the image resolution in each dimension in
millimeters per pixel.
RA = imref2d(size(A),m.PixelSpacing(2),m.PixelSpacing(1))
RA =
imref2d with properties:
XWorldLimits: [0.1562 160.1562]
YWorldLimits: [0.1562 160.1562]
ImageSize: [512 512]
PixelExtentInWorldX: 0.3125
PixelExtentInWorldY: 0.3125
ImageExtentInWorldX: 160
ImageExtentInWorldY: 160
XIntrinsicLimits: [0.5000 512.5000]
YIntrinsicLimits: [0.5000 512.5000]
Display the image, including the spatial referencing object. The axes coordinates reflect the world coordinates. Notice that the coordinate (0,0) is in the upper left corner.
imshow(A,RA,'DisplayRange',[0 512])
Select sample points, and store their world x- and y- coordinates in vectors. For example, the first point has world coordinates (38.44,68.75), the second point is 1 mm to the right of it, and the
third point is 7 mm below it. The last point is outside the image boundary.
xW = [38.44 39.44 38.44 -0.2];
yW = [68.75 68.75 75.75 1];
Convert the world coordinates to row and column subscripts using worldToSubscript.
[rS, cS] = worldToSubscript(RA,xW,yW)
The resulting vectors contain the row and column indices that are closest to the point. Note that the indices are discrete, and that points outside the image boundary have NaN for both row and column
Also, the order of the input and output coordinates is reversed. The world x-coordinate vector, xW, corresponds to the second output vector, cS. The world y-coordinate vector, yW, corresponds to the
first output vector, rS.
Convert 3-D World Coordinates to Row, Column, and Plane Subscripts
Read a 3-D volume into the workspace. This image consists of 27 frames of 128-by-128 pixel images.
load mri;
D = squeeze(D);
D = ind2gray(D,map);
Create an imref3d spatial referencing object associated with the volume. For illustrative purposes, provide a pixel resolution in each dimension. The resolution is in millimeters per pixel.
R = imref3d(size(D),2,2,4)
R =
imref3d with properties:
XWorldLimits: [1 257]
YWorldLimits: [1 257]
ZWorldLimits: [2 110]
ImageSize: [128 128 27]
PixelExtentInWorldX: 2
PixelExtentInWorldY: 2
PixelExtentInWorldZ: 4
ImageExtentInWorldX: 256
ImageExtentInWorldY: 256
ImageExtentInWorldZ: 108
XIntrinsicLimits: [0.5000 128.5000]
YIntrinsicLimits: [0.5000 128.5000]
ZIntrinsicLimits: [0.5000 27.5000]
Select sample points, and store their world x-, y-, and z-coordinates in vectors. For example, the first point has world coordinates (108,92,52), the second point is 3 mm above it in the +z
-direction, and the third point is 5.2 mm to the right of it in the +x-direction. The last point is outside the image boundary.
xW = [108 108 113.2 2];
yW = [92 92 92 -1];
zW = [52 55 52 0.33];
Convert the world coordinates to row, column, and plane subscripts using worldToSubscript.
[rS, cS, pS] = worldToSubscript(R,xW,yW,zW)
The resulting vectors contain the column, row, and plane indices that are closest to the point. Note that the indices are discrete, and that points outside the image boundary have index values of
Also, the order of the input and output coordinates is reversed. The world x-coordinate vector, xW, corresponds to the second output vector, cS. The world y-coordinate vector, yW, corresponds to the
first output vector, rS.
Input Arguments
R — Spatial referencing object
imref2d or imref3d object
Spatial referencing object, specified as an imref2d or imref3d object.
xWorld — Coordinates along the x-dimension in the world coordinate system
numeric scalar or vector
Coordinates along the x-dimension in the world coordinate system, specified as a numeric scalar or vector.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
yWorld — Coordinates along the y-dimension in the world coordinate system
numeric scalar or vector
Coordinates along the y-dimension in the world coordinate system, specified as a numeric scalar or vector. yWorld is the same length as xWorld.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
zWorld — Coordinates along the z-dimension in the world coordinate system
numeric scalar or vector
Coordinates along the z-dimension in the world coordinate system, specified as a numeric scalar or vector. zWorld is the same length as xWorld.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
Output Arguments
I — Row indices
positive integer scalar or vector
Row indices, returned as a positive integer scalar or vector. I is the same length as yWorld. For an m-by-n or m-by-n-by-p image, 1 ≤ I ≤ m.
Data Types: double
J — Column indices
positive integer scalar or vector
Column indices, returned as a positive integer scalar or vector. J is the same length as xWorld. For an m-by-n or m-by-n-by-p image, 1 ≤ J ≤ n.
Data Types: double
K — Plane indices
positive integer scalar or vector
Plane indices, returned as a positive integer scalar or vector. K is the same length as zWorld. For an m-by-n-by-p image, 1 ≤ K ≤ p.
Data Types: double
Version History
Introduced in R2013a | {"url":"https://uk.mathworks.com/help/images/ref/imref2d.worldtosubscript.html","timestamp":"2024-11-14T14:23:29Z","content_type":"text/html","content_length":"98516","record_id":"<urn:uuid:7cc74c70-bf75-495b-af01-56bb72dde6b1>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00451.warc.gz"} |
A study on non-equilibrium dynamics in classical and quantum systems
The theory of statistical mechanics provides a powerful conceptual framework within which the relevant (macroscopic) features of systems at equilibrium can be described. As there is currently no
equivalent capable of encompassing the much richer class of non-equilibrium phenomena, research in this direction proceeds mainly on an instance-by-instance basis. The aim of this Thesis is to
describe in some detail three such attempts, which involve different dynamical aspects of classical and quantum systems. As summarised below, each of the last three Chapters of this document delves
into one of these different topics, while Chapter 2 provides a brief introduction on the study of non-equilibrium dynamics. In Chapter 3 we investigate the purely relaxational dynamics of classical
critical ferromagnetic systems in the proximity of surfaces, paying particular attention to the effects that the latter induce on the early stages of the evolution following an abrupt change in the
temperature of the sample. When the latter ends close enough to the critical value which separates the paramagnetic from the ferromagnetic phase, it effectively introduces a temporal boundary which
can be treated as if it were a surface. Within this picture, we highlight the emergence of novel effects near the effective edge formed by the intersection of the two spatial and temporal boundaries.
Our findings are apparently in disagreement with previous predictions which were based on the assumption that the presence of such an edge would not affect the scaling behaviour of observables; in
order to explain this discrepancy, we propose an alternative for the original power-counting argument which, at least, correctly predicts the emergence of novel field-theoretical divergences in our
one-loop calculations. We show that said singularities are associated with the scaling at the edge. Moreover, by encoding our findings in a boundary renormalisation group framework, we argue that the
new predicted behaviour represents a universal feature associated to the short-distance expansion of the order parameter of the transition near the edge; we also calculate explicitly its anomalous
dimension at the first-order in a dimensional expansion. As a qualitative feature, this anomalous dimension depends on the type of phase transition occurring at the surface. We exploit this fact in
order to provide numerical support to our predictions via Monte Carlo simulations of the dynamical behaviour of a three-dimensional Ising model. The main results reported in Chap. 3 have appeared in
Ref. [1]. In Chapter 4 we revisit the Euclidean mapping to imaginary times which has been recently proposed [2, 3] as an alternative for approaching the problem of quantum dynamics following a
quench. This is expected to allow one to reformulate the original problem as a static one confined in a film geometry. We show that this interpretation actually holds only if the initial state of the
dynamics is pure. Statistical mixtures, instead, intertwine the effects due to the two boundaries, which therefore cannot be regarded as being independent. We emphasize that, although the
aforementioned reinterpretation as a confined static problem fails, one is still able, in principle, to write down and solve the corresponding equations. We also discuss in some detail the relation
between this approach and the real-time field-theoretical one which makes use of the two-time Keldysh contour. For this purpose, we study the analytical structure of relevant observables — such as
correlation functions — in the complex plane of times, identifying a subdivision of this domain into several sectors which depend on the ordering of the imaginary parts of the involved time
coordinates. Within each of these subdomains, the analytic continuation to the real axis provides in principle a different result. This feature allows one to reconstruct from the Euclidean formalism
all possible non-time-ordered functions, which in particular include all those which can be calculated via the Keldysh two-time formalism. Moreover, we give a prescription on how to retrieve response
functions, discussing some simple examples and rationalising some recent numerical data obtained for one of these observables in a one-dimensional quantum Ising chain [4]. We also highlight the
emergence of a light-cone effect fairly similar to the one previously found for correlation functions [2], which therefore provides further confirmation to the fact that information travels across
the system in the form of the entanglement of quasi-particles produced by the quenching procedure. We have reported part of this analysis in Ref. [5]. Chapter 5 presents part of our recent work on
effective relaxation in quantum systems following a quench and on the observed prethermalisation. We analyse the effects caused by the introduction of a long-range integrability-breaking interaction
in the early stages of the dynamics of an otherwise integrable quantum spin chain following a quench in the magnetic field. By employing a suitable transformation, we redefine the theory in terms of
a fully-connected model of hard-core bosons, which allows us to exploit the (generically) low density of excitations for rendering our model exactly solvable (in a numerical sense, i.e., by
numerically diagonalising an exact matrix). We verify that, indeed, as long as the parameters of the quench are not too close to the critical point, the low-density approximation captures the
dynamical features of the elementary operators, highlighting the appearance of marked plateaux in their dynamics, which we reinterpret as the emergence of a prethermal regime in the original model.
As expected, the latter behaviour is reflected also on extensive observables which can be constructed as appropriate combinations of the mode populations. For these quantities, the typical approach
to the quasi-stationary value is algebraic with exponent a ≈ 3, independently of the size of the system, the strength of the interaction and the amplitude of the magnetic field (as long as it is kept
far from the critical point). The plateaux mentioned above last until a recurrence time — which can be approximately identified with tR ≈ N/2 for single modes and t′R ≈ N/4 for extensive quantities —
after which quantum oscillations due to the finite size of the chain reappear. Our procedure allows us to shed some light over prethermal features without having to considerably limit the size of the
system, which we can choose to be quite large, as we discuss in Ref. [6].
A study on non-equilibrium dynamics in classical and quantum systems / Marcuzzi, Matteo. - (2013 Oct 30).
A study on non-equilibrium dynamics in classical and quantum systems
The theory of statistical mechanics provides a powerful conceptual framework within which the relevant (macroscopic) features of systems at equilibrium can be described. As there is currently no
equivalent capable of encompassing the much richer class of non-equilibrium phenomena, research in this direction proceeds mainly on an instance-by-instance basis. The aim of this Thesis is to
describe in some detail three such attempts, which involve different dynamical aspects of classical and quantum systems. As summarised below, each of the last three Chapters of this document delves
into one of these different topics, while Chapter 2 provides a brief introduction on the study of non-equilibrium dynamics. In Chapter 3 we investigate the purely relaxational dynamics of classical
critical ferromagnetic systems in the proximity of surfaces, paying particular attention to the effects that the latter induce on the early stages of the evolution following an abrupt change in the
temperature of the sample. When the latter ends close enough to the critical value which separates the paramagnetic from the ferromagnetic phase, it effectively introduces a temporal boundary which
can be treated as if it were a surface. Within this picture, we highlight the emergence of novel effects near the effective edge formed by the intersection of the two spatial and temporal boundaries.
Our findings are apparently in disagreement with previous predictions which were based on the assumption that the presence of such an edge would not affect the scaling behaviour of observables; in
order to explain this discrepancy, we propose an alternative for the original power-counting argument which, at least, correctly predicts the emergence of novel field-theoretical divergences in our
one-loop calculations. We show that said singularities are associated with the scaling at the edge. Moreover, by encoding our findings in a boundary renormalisation group framework, we argue that the
new predicted behaviour represents a universal feature associated to the short-distance expansion of the order parameter of the transition near the edge; we also calculate explicitly its anomalous
dimension at the first-order in a dimensional expansion. As a qualitative feature, this anomalous dimension depends on the type of phase transition occurring at the surface. We exploit this fact in
order to provide numerical support to our predictions via Monte Carlo simulations of the dynamical behaviour of a three-dimensional Ising model. The main results reported in Chap. 3 have appeared in
Ref. [1]. In Chapter 4 we revisit the Euclidean mapping to imaginary times which has been recently proposed [2, 3] as an alternative for approaching the problem of quantum dynamics following a
quench. This is expected to allow one to reformulate the original problem as a static one confined in a film geometry. We show that this interpretation actually holds only if the initial state of the
dynamics is pure. Statistical mixtures, instead, intertwine the effects due to the two boundaries, which therefore cannot be regarded as being independent. We emphasize that, although the
aforementioned reinterpretation as a confined static problem fails, one is still able, in principle, to write down and solve the corresponding equations. We also discuss in some detail the relation
between this approach and the real-time field-theoretical one which makes use of the two-time Keldysh contour. For this purpose, we study the analytical structure of relevant observables — such as
correlation functions — in the complex plane of times, identifying a subdivision of this domain into several sectors which depend on the ordering of the imaginary parts of the involved time
coordinates. Within each of these subdomains, the analytic continuation to the real axis provides in principle a different result. This feature allows one to reconstruct from the Euclidean formalism
all possible non-time-ordered functions, which in particular include all those which can be calculated via the Keldysh two-time formalism. Moreover, we give a prescription on how to retrieve response
functions, discussing some simple examples and rationalising some recent numerical data obtained for one of these observables in a one-dimensional quantum Ising chain [4]. We also highlight the
emergence of a light-cone effect fairly similar to the one previously found for correlation functions [2], which therefore provides further confirmation to the fact that information travels across
the system in the form of the entanglement of quasi-particles produced by the quenching procedure. We have reported part of this analysis in Ref. [5]. Chapter 5 presents part of our recent work on
effective relaxation in quantum systems following a quench and on the observed prethermalisation. We analyse the effects caused by the introduction of a long-range integrability-breaking interaction
in the early stages of the dynamics of an otherwise integrable quantum spin chain following a quench in the magnetic field. By employing a suitable transformation, we redefine the theory in terms of
a fully-connected model of hard-core bosons, which allows us to exploit the (generically) low density of excitations for rendering our model exactly solvable (in a numerical sense, i.e., by
numerically diagonalising an exact matrix). We verify that, indeed, as long as the parameters of the quench are not too close to the critical point, the low-density approximation captures the
dynamical features of the elementary operators, highlighting the appearance of marked plateaux in their dynamics, which we reinterpret as the emergence of a prethermal regime in the original model.
As expected, the latter behaviour is reflected also on extensive observables which can be constructed as appropriate combinations of the mode populations. For these quantities, the typical approach
to the quasi-stationary value is algebraic with exponent a ≈ 3, independently of the size of the system, the strength of the interaction and the amplitude of the magnetic field (as long as it is kept
far from the critical point). The plateaux mentioned above last until a recurrence time — which can be approximately identified with tR ≈ N/2 for single modes and t′R ≈ N/4 for extensive quantities —
after which quantum oscillations due to the finite size of the chain reappear. Our procedure allows us to shed some light over prethermal features without having to considerably limit the size of the
system, which we can choose to be quite large, as we discuss in Ref. [6].
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vAMM | SpiritSwap V2
SpiritSwap offers users a simple way to provide liquidity for tokens on Fantom via automated liquidity pools (LPs). To become a liquidity provider on SpiritSwap, a user must deposit equal values of
two tokens. In return, they receive SPIRIT-LP tokens (SpiritSwap Liquidity Pool tokens). SPIRIT-LP tokens represent a proportional share of the given LP and liquidity providers may claim their
underlying tokens anytime. Liquidity providers receive a 0.25% fee for every swap that is made in their pair. The 0.25% fee is directly added back to the LP, increasing the value of SPIRIT-LP tokens.
Liquidity providers can also participate in yield farming with supported LPs.
Anyone can make a LP on SpiritSwap, with any two tokens (on Fantom) of their choice. When an LP is created, the creator sets the price of the tokens. The amount of SPIRIT-LP (SSLP) shares minted are
based on the equation:
$SSLP = sqrt(x*y)$
Swap fees are directly accumulated in the LP. For each swap a 0.3% fee is charged to the trader in the token they are selling. This is added to the LP and slightly increases the LP's k value with
every swap. 5/6 of the 0.3% fee goes to the liquidity providers and 1/6 of the 0.3% swap fee goes to the protocol fee vault. The protocol fee vault only receives their fees (in the form of SPIRIT-LP
tokens) when a liquidity provider enters or exits the LP. After sometime, swaps are made in the $AAA/$BBB LP, increasing the LP's k value. The change in k and number of outstanding shares can be used
to calculate the number of SPIRIT-LP tokens that get minted into the protocol fee vault. The number of SSLP tokens to minted for the protocol fee vault can be calculated by the equation:
Where s is the number of outstanding SSLP tokens, k2 is the current k value of the LP, and k1 is the k value of the LP at the last deposit/withdraw from the LP. Figure 2 shows a diagram of a
liquidity provider depositing $AAA and $BBB tokens to the LP from Figure 1, after a swap has been made.
In Figure 2, a new liquidity provider deposits funds to the $AAA/$BBB SPIRIT-LP. The liquidity provider-must deposit $AAA and $BBB tokens in the same proportion as the LP reserve. In the example
above, the liquidity provider must deposit 0.00827 $BBB tokens for every 1 $AAA token they wish to deposit. The LP first mints SPIRIT-LP tokens to the protocol fee vault, then calculates the number
of SPIRIT-LP tokens minted to the new liquidity provider with the equation:
Where x,deposited is equal to the number of $AAA tokens deposited by the new liquidity provider, s is the number of outstanding SPIRIT-LP shares (including the newly minted shares for the protocol
fee vault), and x is the number of $AAA tokens in the LP. | {"url":"https://docs.spiritswap.finance/spiritswap-v2/core-features/liquidity/vamm","timestamp":"2024-11-02T01:34:43Z","content_type":"text/html","content_length":"274663","record_id":"<urn:uuid:d3c98c5d-db82-4f8f-8537-29f557603e3b>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00334.warc.gz"} |
Math for Game Programmers 05 – Vector Cheat Sheet
This is the long due fifth article in this series. If you aren’t comfortable with vectors, you might want to take a look at the first four articles in this series before: Introduction, Vectors 101,
Geometrical Representation of Vectors, Operations on Vectors.
This cheat sheet will list several common geometrical problems found in games, and how to solve them with vector math.
Complete list of basic vector operations
But first, a little review.
For this, I assume that you have a vector class readily available. This is mostly 2D-focused, but everything works the same for 3D, except for differences concerning vector product, which I will
assume to return just a scalar in the 2D case, representing the “z” axis. Any case that only applies to 2D or 3D will be pointed out.
Strictly speaking, a point is not a vector – but a vector can be used to represent the distance from the origin (0, 0) to the point, and so, it is perfectly reasonable to just use vectors to
represent positions as if they were points.
I expect the class to give you access to each of the components, and to the following operations (using C++ style notation, including operator overloading – but it should be easy to translate to any
other language of your choice). If a given operation is not available, you can still do it manually, either by extending the class or creating a “VectorUtils” class. The examples below are usually
for 2D vectors – but 3D is usually simply a matter of adding the z coordinate following the pattern of x and y.
• Vector2f operator+(Vector2f vec): Returns the sum of the two vectors. (In a language without operator overloading, this will probably be called add(). Similarly for the next few ones.)
a + b = Vector2f(a.x + b.x, a.y + b.y);
• Vector2f operator-(Vector2f vec):Â Returns the difference between the two vectors.
a – b = Vector2f(a.x – b.x, a.y – b.y);
• Vector2f operator*(Vector2f vec):Â Returns the component-wise multiplication of the vectors.
a * b = Vector2f(a.x * b.x, a.y * b.y);
• Vector2f operator/(Vector2f vec):Â Returns the component-wise division of the vectors.
a / b = Vector2f(a.x / b.x, a.y / b.y);
• Vector2f operator*(float scalar): Returns the vector with all components multiplied by the scalar parameter.
a * s = Vector2f(a.x * s, a.y * s);
s * a = Vector2f(a.x * s, a.y * s);
• Vector2f operator/(float scalar): Returns the vector with all components divided by the scalar parameter.
a / s = Vector2f(a.x / s, a.y / s);
• float dot(Vector2f vec): Returns the dot product between the two vectors.
a.dot(b) = a.x * b.x + a.y * b.y;
• float cross(Vector2f vec): (2D case) Returns the z component of the cross product of the two vectors augmented to 3D.
a.cross(b) = a.x * b.y – a.y * b.x;
• Vector3f cross(Vector3f vec): (3D case) Returns the cross product of the two vectors.
a.cross(b) = Vector3f(a.y * b.z – a.z * b.y, a.z*b.x – a.x*b.z, a.x*b.y – a.y*b.x);
• float length(): Returns the length of the vector.
a.length() = sqrt(a.x * a.x + a.y * a.y);
• float squaredLength(): Returns the square of the length of the vector. Useful when you just want to compare two vectors to see which is longest, as this avoids computing square roots
a.squaredLength() = a.x * a.x + a.y * a.y;
• float unit(): Returns a vector pointing on the same direction, but with a length of 1.
a.unit() = a / a.length();
• Vector2f turnLeft(): Returns the vector rotated 90 degrees left. Useful for computing normals. (Assumes that y axis points up, otherwise this is turnRight)
a.turnLeft = Vector2f(-a.y, a.x);Â
• Vector2f turnRight():Â Returns the vector rotated 90 degrees right. Useful for computing normals. (Assumes that y axis points up, otherwise this is turnLeft)
a.turnRight = Vector2f(a.y, -a.x);
• Vector2f rotate(float angle): Rotates the vector by the specified angle. This is an extremely useful operation, though it is rarely found in Vector classes. Equivalent to multiplying by the 2×2
rotation matrix.
a.rotate(angle) = Â Vector2f(a.x * cos(angle) – a.y * sin(angle), a.x * sin(angle) + a.y * cos(angle));
• float angle(): Returns the angle that the vector points to.
a.angle() = atan2(a.y, a.x);
Simple cases – warming up
Case #01 – Distance between two points
You probably know that this is done with the Pythagorean theorem, but the vectorial way is simpler. Given two vectors a and b:
float distance = (a-b).length();
Case #02 – Alignment
Sometimes, you want to align an image by its center. Sometimes, by its top-left corner. Or sometimes, by its top-center point. More generally, you can do alignment using a vector whose two components
go from 0 to 1 (or even beyond, if you’d like), giving you full control of alignment.
// imgPos, imgSize and align are all Vector2f
Vector2f drawPosition = imgPos + imgSize * align
Case #03 – Parametric Line Equation
Two points define a line, but it can be tricky to do much with this definition. A better way to work with a line is its parametric equation: one point (“P0″) and a direction vector (“dir”).
Vector2f p0 = point1;
Vector2f dir = (point2 - point1).unit();
With this, you can, for example, get a point 10 units away by simply doing:
Vector2f p1 = p0 + dir * 10;
Case #04 – Midpoint and interpolation between points
Say you have vectors p0 and p1. The midpoint between them is simply (p0+p1)/2. More generally, the line segment defined by p0 and p1Â can be generated by varying t between 0 and 1 in the following
linear interpolation:
Vector2f p = (1-t) * p0 + t * p1;
At t = 0, you get p0; at t = 1, you get p1; at t = 0.5, you get the midpoint, etc.
Case #05 – Finding the normal of a line segment
You already know how to find the direction vector of a line segment (case #03). The normal vector is a 90 degree rotation of that, so just call turnLeft() or turnRight() on it!
Projections using the Dot Product
The dot product has the incredibly useful property of being able to compute the length of a vector’s projection along the axis of another. To do this, you need the vector that you’ll project (“a“)
and a unit vector (so make sure that you call unit() on it first!) representing the direction (“dir“). The length is then simply a.dot(dir). For example, if you have a = (3, 4) and dir = (1, 0),
then a.dot(dir) = 3, and you can tell that this is correct, because (1, 0) is the direction vector of the x axis. In fact, a.x is always equivalent to a.dot(Vector2f(1, 0)), and a.y is equivalent to
a.dot(Vector2f(0, 1)).
Because the dot product between a and b is also defined as |a||b|cos(alpha) (where alpha is the angle between the two), the result will be 0 if the two vectors are perpendicular, positive if the
angle between them is less than 90, and negative if greater. This can be used to tell if two vectors point in the same general direction.
If you multiply the result of that dot product by the direction vector itself, you get the vector projected along that axis – let’s call that “at” (t for tangent). If you now do a – at, you get the
part of the vector that is perpendicular to the dir vector – let’s call that “an” (n for normal). at + an = a.
Case #06 – Determining direction closest to dir
Say that you have a list of directions represented as unit vectors, and you want to find which of them is the closest to dir. Simply find the largest dot product between dir and a vector in the list.
Likewise, the smallest dot product will be the direction farthest away.
Case #07 – Determining if the angle between two vectors is less than alpha
Using the equation above, we know that the angle between two vectors a and b will be less than alpha if the dot product between their unit vectors is less than cosine of alpha.
bool isLessThanAlpha(Vector2f a, Vector2f b, float alpha) {
return a.unit().dot(b.unit()) < cos(alpha);
Case #08 – Determining which side of a half-plane a point is on
Say that you have an arbitrary point in space, p0, and a direction (unit) vector, dir. Imagine that an infinite line goes by p0, perpendicular to dir, dividing the plane in two, the half-plane that
dir points to, and the half-plane that it does not point to. How do I tell whether a point p is in the side pointed to by dir? Remember that dot product is positive when the angle between vectors is
less than 90 degrees, so just project and check against that:
bool isInsideHalfPlane(Vector2f p, Vector2f p0, Vector dir) {
return (p - p0).dot(dir) >= 0;
Case #09 – Forcing a point to be inside a half-plane
Similar to the case above, but instead of just checking, we’ll grab the projection and, if less than 0, use it to move the object -projection along dir, so it’s on the edge of the half-plane.
Vector2f makeInsideHalfPlane(Vector2f p, Vector2f p0, Vector dir) {
float proj = (p - p0).dot(dir);
if (proj >= 0)Â return p;
else return p - proj * dir;
Case #10 – Checking[DEL:/forcing:DEL] a point inside a convex polygon
A convex polygon can be defined to be the intersection of several half-planes, one for each edge of the polygon. Their p0 is either vertex of the edge, and their dir is the edge’s inner-facing normal
vector (e.g., if you wind clockwise, that’d be the turnRight() normal). A point is inside the polygon if and only if it’s inside all the half-planes. [DEL:Likewise, you can force it to be inside the
polygon (by moving to the closest edge) by applying the makeInsideHalfPlane algorithm with every half-plane.:DEL]Â [ops, this actually only works if all angles are >= 90 degrees]
Case #11 – Reflecting a vector with a given normal
Pong-like game. Ball hits a sloped wall. You know the ball’s velocity vector and the wall’s normal vector (see case #05). How do you reflect it realistically? Simple! Just reflect the ball’s normal
velocity, and preserve its tangential velocity.
Vector2f vel = getVel();
Vector2f dir = getWallNormal(); // Make sure this is a unit vector
Vector2f velN = dir * vel.dot(dir); // Normal component
Vector2f velT = vel - velN; // Tangential component
Vector2f reflectedVel = velT - velN;
For more realism, you can multiply velT and velN by constants representing friction and restitution, respectively.
Case #12 – Cancelling movement along an axis
Sometimes, you want to restrict movement in a given axis. The idea is the same as above: decompose in a normal and tangential speed, and just keep tangential speed. This can be useful, for example,
if the character is following a rail.
Case #13 – Rotating a point around a pivot
If used to represent a point in space, the rotate() method will rotate that point around the origin. That might be interesting, but is limiting. Rotating around an arbitrary pivot vector is simple
and much more useful – simply subtract the pivot from it, as if translating so the origin IS the pivot, then rotate, then add the pivot back:
Vector2f rotateAroundPivot(Vector2f p, Vector2f pivot) {
return (pos - pivot).rotate(angle) + pivot;
Case #14 – Determining which direction to turn towards
Say that you have a character that wants to rotate to face an enemy. He knows his direction, and the direction that he should be facing to be looking straight at the enemy. But should he turn left or
right? The cross product provides a simple answer: curDir.cross(targetDir) will return positive if you should turn left, and negative if you should turn right (and 0 if you’re either facing it
already, or 180 degrees from it).
Other Geometric Cases
Here are a few other useful cases that aren’t that heavily vector-based, but useful:
Case #15 – Isometric world to screen coordinates
Isometric game. You know where the (0, 0) of world is on the screen (let’s call that point origin and represent it with a vector), but how do you know where a given world (x, y) is on the screen?
First, you need two vectors determining the coordinate base, a new x and y axes. For a typical isometric game, they can be bx = Vector2f(2, 1)Â and by = Vector2f(-2, 1)Â – They don’t necessarily have
to be unit vectors. From now, it’s straightforward:
Vector2f p = getWorldPoint();
Vector2f screenPos = bx * p.x + by * p.y + origin;
Yes, it’s that simple.
Case #16 – Isometric screen to world coordinates
Same case, but now you want to know which tile the mouse is over. This is more complicated. Since we know that (x’, y’) = (x * bx.x + y * by.x, x * bx.y + y * by.y) + origin, we can first subtract
origin, and then solve the linear equation. Using Cramer’s Rule, except that we’ll be a little clever and use our 2D cross-product (see definition at the beginning of the article) to simplify
Vector2f pos = getMousePos() - origin;
float demDet = bx.cross(by);
float xDet = pos.cross(by);
float yDet = bx.cross(pos);
Vector2f worldPos = Vector2f(xDet / demDet, yDet / demDet);
And now you don’t need to do that ugly find-rectangle-then-lookup-on-bitmap trick that I’ve seen done several times before.
334 ResponsesLeave one →
1. Thanks for making this! I’m doing some 2D vector collision detection at the moment and this helped.
2. Very useful! Thanks.
3. very sweet! Love all your posts.
Keep it coming, lets have one with quaternions and camera control – how to do first and third person – next?
Captcha was 73LD, and I thought we were only on the 24th LD?
4. Omg, I wish I had read this article ten years ago. My code is full of the things that vectors would have solved. It’s like seeing the light. Almost made me cry:) Thanks!
5. Really well done! I had (am still having, actually) problems visualizing Case #10. An illustration here would be really helpful. I guess I also don’t yet know the “why” for this operation.
6. Well done. It takes me back many a year. When I use to do equations for aircraft simulators
7. Cara seus posts são ótimos, achei que o site estivesse abandonado. Compartilhe conosco mais conhecimentos sobre a areá de desenvolvimento de jogos. Obrigado!
8. Keep it coming Rodrigo.
The articles are very well written and useful. I’m really looking forward to more articles like this.Keep it coming Rodrigo.
The articles are very well written and useful. I’m really looking forward to more articles like this.
9. “Likewise, you can force it to be inside the polygon (by moving to the closest edge) by applying the makeInsideHalfPlane algorithm with every half-plane.”
This is not true.
10. evetro, you’re right. This only works for polygons whose internal angles are all >= 90 degrees. Will update post.
11. If you enjoy this series (I do) – you might be interested in raytracing as well – I wrote a small series explaining the math behind that and implementing it in F# some time ago.
You can find it here: http://gettingsharper.de/tag/raytracing/
12. This whole series has been very, very helpful. Thank you so much for your efforts. I hope that you shall continue to share your hard won knowledge.
13. Thanks a lot for sharing this with all folks you actually
recognise what you’re talking about! Bookmarked. Kindly also consult with my web site =). We will have a hyperlink exchange agreement among us
14. This is an EXCELLENT reference. Thanks for providing! Question on Case #10, isn’t a convex polygon by definition a polygon with all angles >= 90 degrees?
15. Hi to all,the contents present at this web site are
really remarkable for peopple knowledge, well, keep up the nice
work fellows.
16. I’ve been reading up on Vector math from a couple other sources, and wasn’t quite understanding just how I can use dot/cross products, but I finally think I get it thanks to these very well
written explanations and examples. I think some diagrams like you put in your earlier posts would make some of these more complex cases easier to visualize. Thanks for taking the time to share
this valuable knowledge!
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Chapter 10 : Series and Sequences
In this chapter we’ll be taking a look at sequences and (infinite) series. In fact, this chapter will deal almost exclusively with series. However, we also need to understand some of the basics of
sequences in order to properly deal with series. We will therefore, spend a little time on sequences as well.
Series is one of those topics that many students don’t find all that useful. To be honest, many students will never see series outside of their calculus class. However, series do play an important
role in the field of ordinary differential equations and without series large portions of the field of partial differential equations would not be possible.
In other words, series is an important topic even if you won’t ever see any of the applications. Most of the applications are beyond the scope of most Calculus courses and tend to occur in classes
that many students don’t take. So, as you go through this material keep in mind that these do have applications even if we won’t really be covering many of them in this class.
Here is a list of topics in this chapter.
Sequences – In this section we define just what we mean by sequence in a math class and give the basic notation we will use with them. We will focus on the basic terminology, limits of sequences and
convergence of sequences in this section. We will also give many of the basic facts and properties we’ll need as we work with sequences.
More on Sequences – In this section we will continue examining sequences. We will determine if a sequence is an increasing sequence or a decreasing sequence and hence if it is a monotonic sequence.
We will also determine a sequence is bounded below, bounded above and/or bounded.
Series – The Basics – In this section we will formally define an infinite series. We will also give many of the basic facts, properties and ways we can use to manipulate a series. We will also
briefly discuss how to determine if an infinite series will converge or diverge (a more in depth discussion of this topic will occur in the next section).
Convergence/Divergence of Series – In this section we will discuss in greater detail the convergence and divergence of infinite series. We will illustrate how partial sums are used to determine if an
infinite series converges or diverges. We will also give the Divergence Test for series in this section.
Special Series – In this section we will look at three series that either show up regularly or have some nice properties that we wish to discuss. We will examine Geometric Series, Telescoping Series,
and Harmonic Series.
Integral Test – In this section we will discuss using the Integral Test to determine if an infinite series converges or diverges. The Integral Test can be used on an infinite series provided the
terms of the series are positive and decreasing. A proof of the Integral Test is also given.
Comparison Test/Limit Comparison Test – In this section we will discuss using the Comparison Test and Limit Comparison Tests to determine if an infinite series converges or diverges. In order to use
either test the terms of the infinite series must be positive. Proofs for both tests are also given.
Alternating Series Test – In this section we will discuss using the Alternating Series Test to determine if an infinite series converges or diverges. The Alternating Series Test can be used only if
the terms of the series alternate in sign. A proof of the Alternating Series Test is also given.
Absolute Convergence – In this section we will have a brief discussion of absolute convergence and conditionally convergent and how they relate to convergence of infinite series.
Ratio Test – In this section we will discuss using the Ratio Test to determine if an infinite series converges absolutely or diverges. The Ratio Test can be used on any series, but unfortunately will
not always yield a conclusive answer as to whether a series will converge absolutely or diverge. A proof of the Ratio Test is also given.
Root Test – In this section we will discuss using the Root Test to determine if an infinite series converges absolutely or diverges. The Root Test can be used on any series, but unfortunately will
not always yield a conclusive answer as to whether a series will converge absolutely or diverge. A proof of the Root Test is also given.
Strategy for Series – In this section we give a general set of guidelines for determining which test to use in determining if an infinite series will converge or diverge. Note as well that there
really isn’t one set of guidelines that will always work and so you always need to be flexible in following this set of guidelines. A summary of all the various tests, as well as conditions that must
be met to use them, we discussed in this chapter are also given in this section.
Estimating the Value of a Series – In this section we will discuss how the Integral Test, Comparison Test, Alternating Series Test and the Ratio Test can, on occasion, be used to estimate the value
of an infinite series.
Power Series – In this section we will give the definition of the power series as well as the definition of the radius of convergence and interval of convergence for a power series. We will also
illustrate how the Ratio Test and Root Test can be used to determine the radius and interval of convergence for a power series.
Power Series and Functions – In this section we discuss how the formula for a convergent Geometric Series can be used to represent some functions as power series. To use the Geometric Series formula,
the function must be able to be put into a specific form, which is often impossible. However, use of this formula does quickly illustrate how functions can be represented as a power series. We also
discuss differentiation and integration of power series.
Taylor Series – In this section we will discuss how to find the Taylor/Maclaurin Series for a function. This will work for a much wider variety of function than the method discussed in the previous
section at the expense of some often unpleasant work. We also derive some well known formulas for Taylor series of \({\bf e}^{x}\) , \(\cos(x)\) and \(\sin(x)\) around \(x=0\).
Applications of Series – In this section we will take a quick look at a couple of applications of series. We will illustrate how we can find a series representation for indefinite integrals that
cannot be evaluated by any other method. We will also see how we can use the first few terms of a power series to approximate a function.
Binomial Series – In this section we will give the Binomial Theorem and illustrate how it can be used to quickly expand terms in the form \( \left(a+b\right)^{n}\) when \(n\) is an integer. In
addition, when \(n\) is not an integer an extension to the Binomial Theorem can be used to give a power series representation of the term. | {"url":"https://tutorial.math.lamar.edu/Classes/CalcII/SeriesIntro.aspx","timestamp":"2024-11-14T11:35:49Z","content_type":"text/html","content_length":"79508","record_id":"<urn:uuid:db23cae9-4931-4b39-8e6c-c5db7353ff32>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00752.warc.gz"} |
CBSE Class 12 Maths Notes Chapter 7 Integrals
Integration is the inverse process of differentiation. In the differential calculus, we are given a function and we have to find the derivative or differential of this function, but in the integral
calculus, we are to find a function whose differential is given. Thus, integration is a process which is the inverse of differentiation.
Then, ∫f(x) dx = F(x) + C, these integrals are called indefinite integrals or general integrals. C is an arbitrary constant by varying which one gets different anti-derivatives of the given function.
Note: Derivative of a function is unique but a function can have infinite anti-derivatives or integrals.
Properties of Indefinite Integral
(i) ∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
(ii) For any real number k, ∫k f(x) dx = k∫f(x)dx.
(iii) In general, if f[1], f[2],………, f[n] are functions and k[1], k[2],…, k[n] are real numbers, then
∫[k[1]f[1](x) + k[2] f[2](x)+…+ k[n]f[n](x)] dx = k[1] ∫f[1](x) dx + k[2] ∫ f[2](x) dx+…+ k[n] ∫f[n](x) dx
Basic Formulae
Integration using Trigonometric Identities
When the integrand involves some trigonometric functions, we use the following identities to find the integral:
• 2 sin A . cos B = sin( A + B) + sin( A – B)
• 2 cos A . sin B = sin( A + B) – sin( A – B)
• 2 cos A . cos B = cos (A + B) + cos(A – B)
• 2 sin A . sin B = cos(A – B) – cos (A + B)
• 2 sin A cos A = sin 2A
• cos^2 A – sin^2 A = cos 2A
• sin^2 A = ($\frac { 1-cos2A }{ 2 }$)
• sin^2 A + cos^2 A = 1
• ${ sin }^{ 3 }A=\frac { 3sinA-sin3A }{ 4 }$
• ${ cos }^{ 3 }A=\frac { 3cosA+cos3A }{ 4 }$
Integration by Substitutions
Substitution method is used, when a suitable substitution of variable leads to simplification of integral.
If I = ∫f(x)dx, then by putting x = g(z), we get
I = ∫ f[g(z)] g'(z) dz
Note: Try to substitute the variable whose derivative is present in the original integral and final integral must be written in terms of the original variable of integration.
Integration by Parts
For a given functions f(x) and q(x), we have
∫[f(x) q(x)] dx = f(x)∫g(x)dx – ∫{f'(x) ∫g(x)dx} dx
Here, we can choose the first function according to its position in ILATE, where
I = Inverse trigonometric function
L = Logarithmic function
A = Algebraic function
T = Trigonometric function
E = Exponential function
[the function which comes first in ILATE should taken as first junction and other as second function]
(i) Keep in mind, ILATE is not a rule as all questions of integration by parts cannot be done by above method.
(ii) It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for ∫√x sinx dx. The reason is that there does not
exist any function whose derivative is √x sinx.
(iii) Observe that while finding the integral of the second function, we did not add any constant of integration.
Integration by Partial Fractions
A rational function is ratio of two polynomials of the form $\frac { p(x) }{ q(x) }$, where p(x) and q(x) are polynomials in x and q(x) ≠ 0. If degree of p(x) > degree of q(x), then we may divide p
(x) by q(x) so that $\frac { p(x) }{ q(x) } =t(x)+\frac { { p }_{ 1 }(x) }{ q(x) }$, where t(x) is a polynomial in x which can be integrated easily and degree of p1(x) is less than the degree of q(x)
. $\frac { { p }_{ 1 }(x) }{ q(x) }$can be integrated by expressing $\frac { { p }_{ 1 }(x) }{ q(x) }$as the sum of partial fractions of the following type:
where x^2 + bx + c cannot be factorised further.
Integrals of the types can be transformed into standard form by expressing
Integrals of the types can be transformed into standard form by expressing px + q = A $\frac { d }{ dx }$(ax^2 + bx + c) + B = A(2ax + b) + B, where A and B are determined by comparing coefficients
on both sides. | {"url":"https://ncert-books.com/cbse-class-12-maths-notes-chapter-7-integrals-books/","timestamp":"2024-11-07T03:48:07Z","content_type":"text/html","content_length":"133496","record_id":"<urn:uuid:69873903-54ed-4e97-9834-9c34e33108d5>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00055.warc.gz"} |
Basic College Mathematics (10th Edition) Chapter 7 - Measurement - Summary Exercises - U.S. and Metric Measurement Units - Page 504 21
Work Step by Step
50 mL to liters Unit for the answer (L) is in numerator; unit being changed (ml) is in denominator so it will divide out. The unit fraction is(1 L)/(1000 ml). Multiply 50 ml times to the unit
fraction 50 ml *(1 L)/(1000 ml) = 0.05 L | {"url":"https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-7-measurement-summary-exercises-u-s-and-metric-measurement-units-page-504/21","timestamp":"2024-11-02T01:22:18Z","content_type":"text/html","content_length":"65676","record_id":"<urn:uuid:202a93a4-1e6c-4fdc-a0e3-9ed4b2147944>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00351.warc.gz"} |
How do I get the Convection Coefficient for steam in a jacketed vessel?
• Thread starter Will26040
• Start date
In summary, the conversation is about obtaining the convection coefficient for steam in a heating jacket, with the person asking for an equation or book recommendation. They mention the use of a
dimensionless number and are pointed towards resources such as the Engineering Toolbox and the Spirax Sarco website. Another person recommends the 5th Edition of Chemical Engineer's Handbook by Perry
& Chilton as a helpful resource.
TL;DR Summary
I need to obtain the convection coefficient of the steam in a heating jacket. I have steam at 1 bar and I know the dimensions of the jacket
Please could someone point me in the direction of an equation/ book I can use to obtain the convection coefficient for steam in a heating jacket. Steam is at 1 bar pressure. I think I need to use a
dimensionless number. Thanks
My 5th Edition of Chemical Engineer's Handbook by Perry & Chilton has several pages on condensing and boiling heat transfer. This book is an excellent resource for people with your type of questions.
Amazon has it, and the current edition is apparently the 9th Edition.
FAQ: How do I get the Convection Coefficient for steam in a jacketed vessel?
1. What is a convection coefficient?
A convection coefficient is a measure of the rate at which heat is transferred from a surface to a fluid through convection. It takes into account the properties of the fluid, the surface geometry,
and the temperature difference between the surface and the fluid.
2. Why is the convection coefficient important in a jacketed vessel?
In a jacketed vessel, the convection coefficient is important because it determines the efficiency of heat transfer between the steam in the jacket and the material being heated or cooled inside the
vessel. A higher convection coefficient means faster heat transfer and better temperature control.
3. How is the convection coefficient for steam in a jacketed vessel calculated?
The convection coefficient for steam in a jacketed vessel can be calculated using the Nusselt number correlation, which takes into account the properties of the steam, the geometry of the jacket, and
the flow rate of the steam. This correlation can be found in heat transfer textbooks or online resources.
4. What factors can affect the convection coefficient in a jacketed vessel?
The convection coefficient in a jacketed vessel can be affected by several factors, including the velocity of the steam, the geometry and size of the jacket, the temperature difference between the
steam and the material being heated or cooled, and the properties of the fluid being heated or cooled.
5. How can the convection coefficient be improved in a jacketed vessel?
To improve the convection coefficient in a jacketed vessel, the steam velocity can be increased, the jacket geometry can be optimized for better flow, and the temperature difference between the steam
and the material can be increased. Additionally, using a fluid with better heat transfer properties or adding baffles inside the jacket can also improve the convection coefficient. | {"url":"https://www.physicsforums.com/threads/how-do-i-get-the-convection-coefficient-for-steam-in-a-jacketed-vessel.1001560/","timestamp":"2024-11-12T04:40:19Z","content_type":"text/html","content_length":"91795","record_id":"<urn:uuid:8409f49b-4e4d-4a83-ba12-06868e35e939>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00596.warc.gz"} |
Bounded does not imply totally bounded
Yesterday's metric has a name
Yesterday I presented as a counterexample the topology induced by the following metric:
I asked:
It seems like this example could be useful in other circumstances too. Does it have a name?
Several Gentle Readers have written in to tell me that that this metric is variously known as the British Rail metric, French Metro metric, or SNCF metric. (SNCF = Société nationale des chemins de
fer français, the French national railway company). In all cases the conceit is the same (up to isomorphism): to travel to a destination on another railway line one must change trains in London /
Paris, where all the lines converge.
Wikipedia claims this is called the post office metric, again I suppose because all the mail comes to the central post office for sorting. I have not seen it called the FedEx metric, but it could
have been, with the center of the disc in Memphis.
[ Addendum 20180621: Thanks for Brent Yorgey for correcting my claim that the FedEx super hub is in Nashville. It is in Memphis ]
[Other articles in category /math] permanent link
Bounded does not imply totally bounded
I somehow managed to miss the notion of totally bounded when I was learning topology, and it popped up on stack exchange recently. It is a stronger version of boundedness for metric spaces: a space
!!M!! is totally bounded if, for any chosen !!\epsilon!!, !!M!! can be covered by a finite family of balls of radius !!\epsilon!!.
This is a strictly stronger property than ordinary boundedness, so the question immediately comes up: what is an example of a space that is bounded but not totally bounded. Many examples are
well-known. For example, the infinite-dimensional unit ball is bounded but not totally bounded. But I didn't think of this right away.
Instead I thought of the following rather odd example: Let !!S!! be the closed unit disc and assign each point a polar coordinate !!\langle r,\theta\rangle!! as usual. Now consider the following
$$ d(\langle r_1, \theta_1\rangle, \langle r_2, \theta_2\rangle) = \begin{cases} r_1, & \qquad \text{ if $r_2 = 0$} \\ \lvert r_1 - r_2 \rvert, & \qquad\text{ if $\theta_1 = \theta_2$} \\ r_1 + r_2 &
\qquad\text{ otherwise} \\ \end{cases} $$
The idea is this: you can travel between points only along the radii of the disc. To get from !!p_1!! to !!p_2!! that are on different radii, you must go through the origin:
Now clearly when !!\epsilon < \frac12!!, the !!\epsilon!!-ball that covers each point point !!\left\langle 1, \theta\right\rangle!! lies entirely within one of the radii, and so an uncountable number
of such balls are required to cover the disc.
It seems like this example could be useful in other circumstances too. Does it have a name?
[ Addendum 2018-07-18: Several Gentle Readers have informed me that this metric has not just one name, but several. ]
[Other articles in category /math] permanent link | {"url":"https://blog.plover.com/2018/06/","timestamp":"2024-11-12T17:38:55Z","content_type":"text/html","content_length":"28648","record_id":"<urn:uuid:83ee50e9-5a44-4588-ac8b-bdfda35a8a26>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00667.warc.gz"} |
Naturalness in mathematics
pp. 277-313
In mathematical literature, it is quite common to make reference to an informal notion of naturalness: axioms or definitions may be defined as "natural," and part of a proof may deserve the same
label (i.e., "in a natural way…"). Our aim is to provide a philosophical account of these occurrences. The paper is divided in two parts. In the first part, some statistical evidence is considered,
in order to show that the use of the word "natural," within the mathematical discourse, largely increased in the last decades. Then, we attempt to develop a philosophical framework in order to
encompass such an evidence. In doing so, we outline a general method apt to deal with this kind of vague notions – such as naturalness – emerging in mathematical practice. In the second part, we
mainly tackle the following question: is naturalness a static or a dynamic notion? Thanks to the study of a couple of case studies, taken from set theory and computability theory, we answer that the
notion of naturalness – as it is used in mathematics – is a dynamic one, in which normativity plays a fundamental role.
DOI: 10.1007/978-3-319-10434-8_14
Full citation:
San Mauro, L. , Venturi, G. (2015)., Naturalness in mathematics, in G. Lolli, M. Panza & G. Venturi (eds.), From logic to practice, Dordrecht, Springer, pp. 277-313.
This document is unfortunately not available for download at the moment. | {"url":"https://ophen.org/pub-176854","timestamp":"2024-11-06T23:48:12Z","content_type":"text/html","content_length":"13820","record_id":"<urn:uuid:2eb9b63c-2ee8-4056-a602-c83d7cd59327>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00540.warc.gz"} |
Op Amp Voltage follower / Voltage Buffer - Electronics AreaOp Amp Voltage follower / Voltage Buffer - Electronics Area
What is an Op Amp Voltage follower?
An Op Amp voltage follower (voltage follower using operational amplifier) is a circuit which has a voltage gain of 1. To better understand the operation of a voltage follower, we must remember the
operation of an Op Amp as a non-inverting amplifier. See the diagram below.
Op Amp Non-inverting amplifier
The Op-Amp non-inverting amplifier gain is given by the formula: 1 + (R2/R1).
If we make the R2 resistor value equal to zero and make the R1 resistor value very large, we have an amplifier with gain G = 1. This means that the output signal has the same value as the input
signal. See the following diagram.
Op Amp Voltage Follower (Voltage Buffer)
A buffer has an output that is exactly like the input. This feature is very useful for solving impedance matching problems.
• The input impedance of a buffer using an operational amplifier is very high, close to infinity
• The output impedance is very low, just a few ohms.
If the input impedance is very high, it does not load the circuit that is sending the signal, and if its output impedance is low, it can deliver a sufficient amount of current to the circuit
receiving the signal.
In other words, a buffer requests very little current from the circuit that gives the signal and greatly increases the capacity to deliver current to the circuit that receives the signal.
Voltage follower advantages
• Provides power gain and current gain. (voltage gain = 1).
• Low output impedance. Loading effects can be avoided.
• High input impedance. The op-amp takes no current from the input.
Voltage follower applications
• Sample and hold circuits.
• Buffers for logic circuits.
• Active filters. Voltage followers can be used to isolate filter stages from each other, when building multistage filters.
Op Amp voltage follower example
We need to get 6 volts from a 12 volt source to power a 100 ohm load resistor (RL).
We use two 100K resistors in series as a voltage divider (R1, R2). Since the resistors have the same value, the voltage between them is exactly 6 volts (A).
If we connect the 100 ohm load to point A, the voltage will no longer be 6 volts, because the lower 100K resistor is in parallel with the load. The equivalent resistance of parallel resistors
is: 100,000 x 100 / (100,000 + 100) = 99.9 ohms. We can safely assume the approximate value of 100 ohms.
Using the voltage divider again, the voltage at point A is: VA = 12 / (100K + 100) x 100 = 0.012 volts. Very different from the expected 6 volts.
To avoid this problem, we use an op amp voltage follower. The non-inverting input of the op amp is connected to A and the output is connected to the 100 ohm resistor (load). The load will have
exactly 6 volts between its terminals.
More Op. Amp. Tutorials | {"url":"https://electronicsarea.com/op-amp-voltage-follower-buffer/","timestamp":"2024-11-09T23:55:31Z","content_type":"text/html","content_length":"56995","record_id":"<urn:uuid:73c356db-58f7-4b59-849f-df316f043668>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00180.warc.gz"} |
Browsing Workshops 2011 by Title
[OWR-2011-46] Workshop Report 2011,46 (2011) - (18 Sep - 24 Sep 2011)
This workshop was a forum to present and discuss the latest result and ideas in homotopy theory and the connections to other branches of mathematics, such as algebraic geometry, representation theory
and group theory. | {"url":"https://publications.mfo.de/handle/mfo/2812/browse?rpp=20&sort_by=1&type=title&offset=22&etal=-1&order=ASC","timestamp":"2024-11-12T20:11:07Z","content_type":"text/html","content_length":"55155","record_id":"<urn:uuid:373f25fa-3991-41a2-b584-d83b8178e82d>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00851.warc.gz"} |
Reactive Power Calculator - Calculator Wow
Reactive Power Calculator
In the intricate realm of electrical systems, the Reactive Power Calculator emerges as a key player, unraveling the dynamics of reactive power. This article takes you on a journey through the
fundamentals, importance, and applications of reactive power, showcasing the calculator as a tool to enhance electrical efficiency. Whether you’re an electrical engineer, a student, or a curious
enthusiast, understanding reactive power is paramount for optimizing power factor and ensuring the smooth operation of electrical networks.
Importance of Reactive Power
Reactive power is a vital component in alternating current (AC) circuits, playing a crucial role alongside true power. Key aspects of its importance include:
• Voltage Stability: Reactive power ensures voltage stability in electrical grids, preventing fluctuations and ensuring a consistent supply to consumers.
• Power Factor Optimization: Managing reactive power helps optimize power factor, enhancing the overall efficiency of electrical systems.
• Reducing Line Losses: Adequate reactive power reduces line losses, minimizing energy wastage in transmission and distribution.
• Motor Operation: Reactive power is essential for the efficient operation of inductive loads like motors, ensuring smooth functioning and preventing power factor penalties.
How to Use the Reactive Power Calculator
Using the Reactive Power Calculator is a straightforward process:
1. Enter Apparent Power (AP): Input the total power consumed, considering both true power and reactive power.
2. Specify True Power (TP): Enter the actual power used for performing work in the circuit.
3. Click “Calculate Reactive Power”: The calculator processes the inputs and provides the reactive power.
This tool simplifies the complex mathematical computations involved in determining reactive power, allowing engineers and technicians to optimize electrical systems effectively.
10 FAQs and Answers about Reactive Power Calculator
1. What is Reactive Power, and How Does it Differ from True Power?
Reactive power represents the non-working power in an AC circuit, necessary for sustaining voltage. True power, on the other hand, is the actual power used for performing work.
2. Why is Reactive Power Important for Power Factor Optimization?
Reactive power contributes to power factor, and optimizing it ensures efficient use of electrical power, minimizing wastage.
3. Can Reactive Power be Negative?
Yes, reactive power can be negative, indicating the consumption of reactive power by loads.
4. How Does Reactive Power Impact Voltage Stability?
Adequate reactive power ensures voltage stability by compensating for fluctuations and variations in the electrical grid.
5. Is Reactive Power Only Relevant for Industrial Applications?
No, reactive power is relevant in various applications, including residential and commercial settings where inductive loads are present.
6. What Happens if Reactive Power is Excessive?
Excessive reactive power can lead to increased line losses, reduced system efficiency, and potential equipment damage.
7. Can Reactive Power be Stored?
Unlike true power, reactive power cannot be stored; it is a dynamic component that must be managed in real-time.
8. How Does the Calculator Contribute to Power Factor Correction?
By accurately calculating reactive power, the calculator aids in identifying areas where power factor correction measures are needed for optimal efficiency.
9. Can Reactive Power Compensation Reduce Electricity Costs?
Yes, by improving power factor and reducing reactive power penalties, businesses can lower electricity costs and enhance overall energy efficiency.
10. Is Reactive Power the Same as Apparent Power?
No, reactive power and apparent power are different. Apparent power is the combination of true power and reactive power, representing the total power in an AC circuit.
As we conclude our exploration of the Reactive Power Calculator, we recognize its pivotal role in deciphering the complexities of electrical systems. Reactive power, often overlooked, proves to be a
linchpin for stability and efficiency. So, let the calculator be your ally in optimizing power factor, ensuring the seamless operation of electrical networks. May your journey through the currents of
reactive power be enlightening and empowering. Happy calculating! | {"url":"https://calculatorwow.com/reactive-power-calculator/","timestamp":"2024-11-12T16:02:23Z","content_type":"text/html","content_length":"66125","record_id":"<urn:uuid:22ade75a-b0fb-4678-80fd-cbf3929b24b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00332.warc.gz"} |
Leave-one-out crossvalidation for graph_lme models assuming observations at the vertices of metric graphs — posterior_crossvalidation
Leave-one-out crossvalidation for graph_lme models assuming observations at the vertices of metric graphs
Leave-one-out crossvalidation for graph_lme models assuming observations at the vertices of metric graphs
A fitted model using the graph_lme() function or a named list of fitted objects using the graph_lme() function.
Which factor to multiply the scores. The default is 1.
Return the scores as a tidyr::tibble() | {"url":"https://davidbolin.github.io/MetricGraph/reference/posterior_crossvalidation.html","timestamp":"2024-11-05T20:04:07Z","content_type":"text/html","content_length":"10047","record_id":"<urn:uuid:1261780a-ce80-4fbc-8a3e-4f1fe3b23b01>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00173.warc.gz"} |
GreeneMath.com | Ace your next Math Test!
Parallel and Perpendicular Lines
Additional Resources:
In this lesson, we will learn how to determine if two lines are Parallel or Perpendicular. We should know at this point, that two parallel lines have the same slope and that perpendicular lines have
slopes that multiply together to give us -1. So in order to determine if we have parallel or perpendicular lines, we can place each line in slope-intercept form: y = mx + b and observe the slope m of
each line. If the slopes are the same, we have parallel lines. If the slopes multiply together to give us -1, the lines are perpendicular. Additionally, we will learn how to write the equation of a
line given a point on the line and a line that is parallel or perpendicular to the line.
+ Show More + | {"url":"https://www.greenemath.com/College_Algebra/89/Parallel-and-Perpendicular-Lines.html","timestamp":"2024-11-10T17:41:16Z","content_type":"application/xhtml+xml","content_length":"9867","record_id":"<urn:uuid:bef007d4-17a8-4ab5-8335-411c7bd05b20>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00341.warc.gz"} |
The length of side ab is _____.-Turito
Are you sure you want to logout?
The length of side AB is _____.
A. 4 cm
B. 6 cm
C. 10 cm
D. 2 cm
Polygon is a two- dimensional closed figure which is consist of three or more-line segments. Each polygon has different properties. One of the polygons is quadrilateral. Quadrilateral is a four-sided
polygon with four angles and the sum of all angles of a quadrilateral is
The correct answer is: 6 cm
In the question there is a figure consist of two parallelograms ABCD and CDEF as shown in the figure.
Here, we have to find the length of the AB.
Since, the opposites of a parallelogram are equal.
So, in parallelogram CDEF, CD=EF= 6 cm.
Similarly, in parallelogram ABCD, AB=CD= 6 cm.
Thus, the length of AB is 6 cm.
Therefore, the correct option is b, i.e., 6 cm.
Parallelogram is a four-sided polygon whose opposite sides are parallel and equal to each other and the opposite angles are equal to each other. The sum of all angles of a parallelogram is
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ALEKS Math Qualitative and Quantitative End of Year Follow Up
Subscribe to BrokenAirplane!
I stand behind the technology I recommend. Anything I recommend must:
• Support and Encourage Learning - No technology for technology's sake. Nor it can it simply be a flashy electronic version of older technology (e.g. smart boards instead of overhead projectors).
If it does not help or make life easier then it isn't going to be on my list.
• Affordable to as many as possible - There might be wonderful technology out there but if it can only be used by 1% of schools, I will hesitate to post about it. I know there are grants and
resources available to purchase these things but there are often much better ways to spend that money.
That is how software makes it onto my
list for classroom technology
or if necessary a longer post. I wish to follow up with a previous post I made about
, software which differentiates and helps support the learning of math skills.
I want to reiterate, that
is not intended to replace math teachers and in fact will make them all the more necessary. It allows you to have one-on-one time with your students and help them each out where they need it. It has
allowed me to teach the way I always wanted to. ALEKS teaches math "skills" and not necessarily the
broader definition of math
that many believe is necessary to create a love and/or appreciation for it. However, as you will see, this program has made it possible to accomplish my goals as a math teacher this year.
First some data:
Pre-Assessment: ALEKS' CA Algebra 1 class has 257 "Objectives" based upon the 25 or so California State Standards. It gives an initial assessment at the beginning of the year. My class average at the
beginning of the year was 29% mastery of the objectives.
To give a more accurate description of my class, I teach a untracked class referred to as Math 1. This means we learn Algebra, Geometry, Statistics, and Number Theory in increasing complexity each
year. Which means that some students enter my class having already studied Algebra. However, all of my students take the Algebra 1 State Standardized Test (
The breakdown of my students initial scores were as follows:
21% of students - 0-10% of objectives
24% of students - 11-20% of objectives
31% of students - 30-40% of objectives
2% of students - 60-70% of objectives
I should also mention that this data is only for the students taking Algebra in ALEKS, so the students I have taking Geometry are not included as they would skew the data. I wanted to show as clearly
as possible what has happened as a direct result of ALEKS.
After using ALEKS in my class for
20% of our total time
and maintaining an at home requirement for the entire year, the class average is 61%. This would be higher but I would switch the student into Geometry once they showed they had all of the major
objectives and those necessary to do well in Algebra and Geometry.
The largest individual gain by a student still in Algebra was 55% with the smallest being 9%. If you knew the specific students with 9% gains and their struggles and math history you would celebrate
it as a triumph as well. I wish ALEKS allowed one to track a student from one class to another so I could share with you the gains by my 9th graders studying Geometry in ALEKS. I can share with you
that one of my students in the Geometry class has essentially taught herself 80% of the Geometry curriculum. Think of the possibilities of what people can do with
resources and motivation
With a year of ALEKS for one student costing about the same as a family of four going out to eat, this is a great deal. I think it would be wonderful if parents could support their schools and
purchase this for their students or at least help. While Public school is free, it is certainly easier to justify one family spending this for their own student than to have the school spend
thousands when that money could be used elsewhere. The schools that could use this most need support from their communities.
My other point, is if skills could be learned primarily at home or for
20% of the class time
think of what this would allow teachers to do with their time! I know it has made an incredible impact on my class both in esteem, readiness, and ability.
Flipping my classroom
has freed us up to do great
which could not be done in a standard classroom time frame.
Finally, I wanted to share what my students are saying now, because at the end of the day, that is what matters.
Last semester
, there were students who enjoyed it and were able to learn from it, but many still were frustrated at the new learning situation. I learned from
discussions and feedback with the students
that it was taking them time to get used to having freedom and choice in their learning and the opportunity to learn at a pace that was right for them. Many of them had never spoke up in a math class
and asking questions was anathema to them.
Overwhelmingly they said in a recent survey that they feel more confident asking for help and support. Two-thirds of them said they felt confident about taking Geometry next year, and the a little
more than that said they would definitely use ALEKS again. What more can you ask for? A great example of technology support learning in a way that could not be accomplished otherwise.
Has ALEKS worked for you? Have you found something else that worked equally well if not better? Let us know in the comments.
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Weekly Challenge #149 and A Fix, Maybe, to #148
Fixing an Old Task
Abigail read my blog post and pointed out that the given first correct answer of #148 Task 2, (2,1,5) was not showing up, and when he ran my test_ardano function against it, it didn’t return 1, but
rather 1.00000000000000000011.
I think the only time I regularly do math like this is in the Challenge, which is good because it makes me use ideas I don’t touch regularly, but it does get me in places I can’t negotiate out of.
It’s a known thing that IEEE 754 math has hairy edge cases, and I’m guessing that I’m hanging up on that. I changed my cuberoot function to limit the number of significant digits —
sub cuberoot ($n ) { return sprintf '%.06f', $n**( 1 / 3 ) }
— but that seems like a hackish “just make it work!” solution rather than really understanding where the problem is and fixing that. I admit that. When the Reviews come around, I’ll have to read to
see the better Cardano Triplets solutions.
Thanks to Abigail for pointing out the problem.
TASK #1 › Fibonacci Digit Sum
Submitted by: Roger Bell_West Given an input $N, generate the first $N numbers for which the sum of their digits is a Fibonacci number.
I got this to an acceptable point fairly quickly. I use split to separate the numbers into digits, sum0 from List::Util because I am always paranoid about empty strings and a function that finds the
Fibonacci number that is not less than the given number, which would be the sum0 of the digits. If the function returns a number that is sum, then there we go! We append it to an array, and stop
looking when the array is big enough.
Show Me The Code!
#!/usr/bin/env perl
use strict;
use warnings;
use feature qw{ say postderef signatures state };
no warnings qw{ experimental };
use Getopt::Long;
use List::Util qw{ sum0 max };
my $N = 20;
GetOptions( 'n=i' => \$N, );
my @fib = first_60_fib();
my %fib = map { $_ => 1 } @fib;
my @x;
my $n = 0;
while ( scalar @x < $N ) {
my $sd = sum_of_digits($n);
my $f = $fib{$sd} || 0;
push @x, $n if $f;
say join ' ', @x;
sub first_60_fib() {
my @n;
push @n, 0;
push @n, 1;
while ( scalar @n < 60 ) {
push @n, $n[-1] + $n[-2];
return @n;
sub sum_of_digits ( $n ) {
return sum0 split //, $n;
$ ./ch-1.pl -n 30
TASK #2 › Largest Square
Submitted by: Roger Bell_West Given a number base, derive the largest perfect square with no repeated digits and return it as a string. (For base>10, use ‘A’..‘Z’.)
This is giving me problems.
There are parts that are fairly simple. You get the 36 characters we can get with my @range = ( 0 .. 9, 'A' .. 'Z' ), and you can get the right characters for any base with my @range_by_base = @range
[0..$base-1]. The highest possible correct number becomes join '', reverse @range_by_base.
I found that many of the Base Conversion modules convert from and to common CS-related bases — 2, 4, 8, 16, 32, etc. — but for this task, we want to convert into and out of any base, and I found that
Math::BaseCalc does that well.
Using state, I made functions that convert back and forth, and hold onto the numbers, so we don’t have to re-generate 100 in base 19 twice. (In writing this, I’m second-guessing the utility of that,
but I’ve done the cool thing, so eh?)
But there’s still going from 9,876,543,210 to 1, in the case of base 10. I am doing it with a for loop and an implicit 10-million-entry list, and that’s killing me. I think that a while loop instead,
like while ($n > 1) { $n-- } might be it. In fact, it’s looking promising (and not segfaulting) as I write this. When and if I solve it, I’ll blog it separately.
If you have any questions or comments, I would be glad to hear it. Ask me on Twitter or make an issue on my blog repo. | {"url":"https://jacoby.github.io/2022/01/27/weekly-challenge-149-and-a-fix-maybe-to-148.html","timestamp":"2024-11-04T11:58:55Z","content_type":"text/html","content_length":"15692","record_id":"<urn:uuid:cfd01076-656d-4b0d-8e2a-6323d2572b95>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00885.warc.gz"} |
Stat 110 Student Reassures Girlfriend Positive STD Result Likely Statistical Anomaly
CAMBRIDGE, MA--After a positive result on a pair of self-administered tests for an STD, Leverett junior Werther Madison was seen reassuring his girlfriend that despite these results, the probability
that he cheated on her remains counterintuitively low. Madison, a student of the popular course Statistics 110: Introduction to Probability, sought to interpret these results using the same
techniques he learned in Stat 110.
“I do admit that a positive test result does increase the likelihood that we have the disease, but not by nearly as much as you might think,” said Madison. “The chance we have the disease is still
proportional to the prior probability that I would cheat and bring an STD into our relationship. Which is like zero, since I would never--wait, don’t leave!”
Despite his best efforts, his argument seemed only to make the situation exponentially worse, and Madison found it increasingly difficult to remember relevant details from his statistics course.
“There’s a limit to how bad this can be, right? Let’s assume the number of STDs I’ve brought into our relationship follows a Poisson distribution. I can assure you my lambda is tiny. Miniscule! Two
at most.”
After the conversation, Madison updated his probability of still having a girlfriend to accommodate for his newest observations. Madison could be seen today sitting, dejected, in his statistics
final, getting firsthand experience with the principle of inclusion-exclusion. | {"url":"https://www.satirev.org/harvard/stat-110-student-reassures-girlfriend-positive-std-result-likely-statistical-anomaly","timestamp":"2024-11-13T04:22:13Z","content_type":"application/xhtml+xml","content_length":"24250","record_id":"<urn:uuid:ffb8fce8-030c-4242-b9b4-dd8622c58f1b>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00750.warc.gz"} |
Important/Expected Topics for AAI JE ATC Exam
Hello guys, as we all are aware that AAI is in the process of conducting an online test for the post of Junior Executive -ATC, it is crucial to know what is to be expected in the test. Also as per
the syllabus provided by AAI on its official websites, the online test will be comprised of two sections:
Non-Technical Part:
English Language (20 Questions), General Intelligence/Reasoning (15 questions), General Aptitude/Numerical Ability (15 questions), General Knowledge/ awareness (10 questions). The level of these
questions should be moderate, so SSC level books can be referred for the preparation of Non-technical preparation.
Technical Part:
There will be 30 questions each from Mathematics and Physics. Technical is the main section that decides the fate of almost all candidates appearing in the test as non-tech section is covered well by
everyone considering its level. Before writing this post, I went through previous years’ question papers so that whatever I present here is based on that only and not on some kind of speculations
How to Crack ATC Exam in Single Attempt
The year 2015
• Superconductors,
• Scalar and Vectors,
• Satellites time period,
• Newton’s laws of motion (Projectile, Momentum),
• Escape velocity,
• Laws of Thermodynamics,
• Bond structures in solids,
• Carnot cycle,
• wave equations,
• Ampere’s law,
• Gauss law,
• diffraction,
• Occupancy Probability (Fermi-Dirac equation, Bose-Einstein equation),
• Maxwell Equations,
• spring,
• planck’s distribution function,
• London equation,
• Bernoulli Theorem,
• Latent heat,
• Young’s double slit experiment,
• Curie-Weiss Law,
• Capacitance,
• Magnetism,
• Number of modes of vibrations,
• Inductance,
• Torque and Dipole Moment
• Function,
• Differential Equations,
• Partial Diff. equation,
• Series and its convergence,
• Number of elements,
• Matrices (Rank, system of equations, Eigenvalues),
• Vectors,
• Numerical Methods,
• Residue,
• Sets and subsets,
• Geometry (Line, Ellipse, parabola, planes, Hyperbola, Tangents and normal),
• Indefinite and definite integration,
• Taylor Series.
The year 2016
• Light,
• Lasers,
• X-rays,
• Superconductors,
• Electric and Magnetic Field Equations,
• Curie Temperature,
• Diffraction Experiments and Theories,
• Polarization of waves,
• Interference,
• energy-mass equation,
• pendulum,
• quantum mechanics,
• Theory of relativity,
• Capacitors,
• Work function,
• Uncertainty Principle,
• De Broglie wavelength,
• Newton’s ring experiment,
• Magnetism,
• Poynting Vector
• Differential Equations (Solution, IF, and CF),
• Set,
• Integration (Definite and indefinite, numerical methods),
• Geometry (Asymptotes, Lines, Tangent, and Normal),
• Limit,
• Matrices (Identity, skew-symmetric, transpose, rank, Inverse),
• Functions (Analytic Function, C.R. equation),
• Convergence of Series,
• Vectors,
• Derivatives (Max., Min. Values),
• Lagrange Theorem and Formula,
• Complex Numbers
The year 2018
• Polarization of wave
• Wave Optics
• Ray Optics
• Mechanics
• De Broglie wavelength
• Magnetism
• Properties of diamagnetic, paramagnetic, ferromagnetic substances
• Quantum Mechanics
• Special Theory of Relativity
• Matrices
• Convergence and Divergence of series
• Differential Equation
• Complex Number
• Complex Integral
• Probability
• Definite and Indefinite Integration
• Limits
• Numerical Methods (Bisection, Newton-Raphson methods for finding roots and Trapezoidal Rule of integration, etc.)
• Modern/Abstract Algebra
So we can conclude that the paper is mainly based on 11-12^th Maths and Physics concepts with some topics taken from the Graduation level syllabus. Preparing from IIT preparation notes and NDA
previous years’ papers will be very helpful in covering these topics. For graduation level topics, just refer to books of the first year of graduation. I hope covering above mentioned topics will
certainly help you get selected in the Airports Authority of India as an Air Traffic Control Officer for a job that is very challenging and stressful but equally rewarding in terms of various
allowances paid to ATCOs only.
18 thoughts on “Important/Expected Topics for AAI JE ATC Exam”
Hello sir can I know when can AAI conducts exams FY 2018
Most probably in November. I am not sure.
Hello sir is there any information about no of posts of ATC Fy2018, As aai hasn’t mentioned in their corriendum updated notification
Same as before.
Helpful, thank you for the information.
Hello Sir,
I read your blog and its really helpful. As this is phase where one should focus on preparation of his/her ATC exam which is to be held on 30th Nov.
But sir there is one question that constantly comes before my eyes while studying,
that What is Expected cutoff this year for ATC this year?
I have also done some exploration regarding this topic and also found some data, but I would like to hear it from your blog.
So I request you to write something on it.
Hope you will write.
Thanks in advance. Have good day ahead.
Cut off depends upon a lot of factors such as difficulty level of question paper which we are yet to see. No doubt no. of students appearing this time are more. There will be more
competition. At the same time, since there is no interview, question paper can be designed tougher. So, It’s hard to guess but if you still want a number from us then it’s 80-85. We will
definitely post something about cut off after the exam. Best of Luck for your Exam.
Is it enough to prepare the NCERT formula based numericals in Physics and Mathematics for the exam?
This is the first time when there is no interview. Question paper might be tough. So NCERT alone might not be enough this time.
Can you give Ur what’s app number
Sir can you please tell the exam dates of notification 20/21 ATC approximately.
March end in normal conditions.
Sir, do know the most appropriate online source for preparing JE ATC Exam 2018?
Yes, can you please answer this?
Some people are saying that cut off may shoot up to 95…is it is possible????
Cut off may even shoot up to 100 if we get easy question paper in exam because no. of posts are less and competition is huge since applications received are much more than any other year. But
do we really know the level of difficulty of the exam yet? No. So no matter who is saying what, it’s impossible to predict anything before exam. Considering exam tougher than last year, we
are expecting cut off to be around 85.
Dear sir please tell me that
Is there interview in 20/21 ATC
And about the english syllbus.
Leave a Comment | {"url":"https://incomopedia.com/important-topics-for-aai-je-atc-exam/","timestamp":"2024-11-09T12:16:54Z","content_type":"text/html","content_length":"282774","record_id":"<urn:uuid:be83def4-9556-46ea-8d4e-57a91fc7f6e2>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00675.warc.gz"} |
Rounding and Significant Figures in Google Sheets
A little while ago, I decided to use Google Sheets to speed up my physics homework, as I realized that a lot of the questions were very similar, but changed a few of the numbers. I needed to round a
number to the correct amount of significant figures in order to receive full credit. To expedite this process and save myself some time, I did some research on rounding in Google Sheets. Here, I’ll
share what I’ve learned from Google’s Rounding Article and a few other sources.
Here is the spreadsheet that I created to demonstrate these functions.
Usage and syntax:
=ROUND(value, [places])
ROUND rounds a number to x decimal places.
The round function rounds a number using normal rounding rules. Note that places is in square brackets, meaning that it is an optional argument. “Value” is your input number, and “places” is the
decimal place to round. Below is a quick table to help you out with the “places” input when using the round function.
Number 9 2 5 . 8 2 3
Places -2 -1 0 1 2 3
With the round function, you can round to the nearest hundreds by passing in -2 for your “places”, round to the nearest tens using -1, and much more.
Normal rounding rules round up when the digit after the digit you want to round is 5 or more and round down when it is less than 4 or below. For example, =ROUND(55.5, 0) would round up to 56 but =
ROUND(55.49,0) would round down to 55.
This brings us to a very similar set of formulas that behave almost exactly like round.
Semi-Automatic Rounding
Using these two buttons, Google Sheets will automatically round your number to the current decimal which is displayed. If you type in “1.2000”, Google Sheets should automatically round it to 1.2,
and if you add more 0’s, it should stay at 1.2.
What if you need it to say “1.2000”? Then, just click on the cell with 1.2 and click the button with [.00 🡆] to extend the zeroes.
ROUNDUP and ROUNDDOWN
=ROUNDUP(value, [places])
=ROUNDDOWN(value, [places])
These two functions do the same thing as round, but always rounds up or down using absolute values instead of following normal rounding rules.
For example, =ROUNDDOWN(-2.3) will result in a -2 because it rounds down to the value closest to 0.
Int will round the value down to the nearest whole number. This slightly differs from using rounddown to 0 places because it int will round down to the closest integer closest to -∞ while rounddown
will round down to the closest integer closest to 0.
Int is very niche and has much fewer uses than rounddown, so I personally haven’t used int very much in my sheets.
=MROUND(value, factor)
MROUND rounds a number to the nearest x.
This function rounds the value to the nearest multiple of the factor, centered at 0, and also follows normal rounding rules. You can use this to round to the nearest 500, 750, or even 1236!
=ROUND(value, 1) will perform the same as =MROUND(value, 0)
=ROUND(value, 2) will perform the same as =MROUND(value, 0.1)
=ROUND(value, -1) will perform the same as =MROUND(value, 10).
Normal rounding rules means that if the value is at halfway between multiples of the factor or more, then it will round up, and if it less than half way, it will round down.
For example, =MROUND(12.5, 5) will round up to 15 and =MROUND(12.49, 5) will round down to 10.
Both the value and factor can be non-integers, meaning you can even do some cool things such as =MROUND(12.345, 0.123).
However, both the value and factor need to be on the same side of the number line, and neither of them can be 0.
FLOOR and CEILING
=FLOOR(value, factor)
=CEILING(value, factor)
Floor and ceiling perform are the roundup and rounddown equivalents to mround. They will always round up and always round down to the nearest factor.
=TRUNC(value, [places])
Next, we have the trunc function, which stands for “truncate”. In short, this will discard all of the digits after the “place”. Similarly to the round function, places is optional and defaults to 0
if not inputted.
For example, =TRUNC(13.99, 0) will output 13.
Refer to the “places” table above if you need help with “places”.
Custom Functions
To use custom functions in Google Sheets, you have to first get to the script editor.
It’s located in Tools > Script Editor. Alternatively, you can go to Help > Search The Menus.
SIGFIG rounds a number to x significant figures.
In case you’ve never used the Google Script editor before, don’t worry, just copy/paste the code below into the code box, then give your project a name, then click the save (floppy disk) icon. There
is no need to click on the run (play) button, as that will return an error.
// Don’t use this, it’s not optimal
// Demonstrating overloading
function SIGFIGS(num) {
 return num.toPrecision(3);
function SIGFIG(num, sigFigs) {
 return num.toPrecision(sigFigs);
For those of you who have coded before and have used languages which let you overload a method, Google Apps Script doesn’t like overloading. Methods or functions with the same name will cause
Therefore, what you should do is use the following code:
function SIGFIG(num, opt_sigFigs) {
 if (opt_sigFigs == null) {
  return num.toPrecision(3);
 }
 return num.toPrecision(opt_sigFigs);
You could compact this further to become:
function SIGFIG(num, opt_sigFigs) {
 return opt_sigFigs == null ? num.toPrecision(3) : num.toPrecision(opt_sigFigs);
(That code should be 3 lines using a ternary operator, but the screen isn’t wide enough…)
Usage and syntax of this custom function:
=SIGFIG(value, [significant figures])
This will round a number using normal rounding rules to the place with the correct amount of significant figures. I used this quite a bit when speeding up the calculations of my physics homework.
Limitations of SIGFIG
There is one major limitation with the SIGFIG custom function in which it is using floats to round to the nearest significant figure. Floats lose precision after a bunch of decimal points (floating
point error). However, for normal use for my physics homework, I didn’t encounter the problem at all.
An example of this would be, for example, trying to round 1.2 to 20 significant figures. You would end up with 1.1999999999999999556. As you can see, floating point errors won’t really affect you
with normal use. It’s only those extreme cases, such as rounding to 20 significant figures which cause problems.
2 thoughts on “Rounding and Significant Figures in Google Sheets”
1. How would I make a formula to round cell A2 which has 1200, to the highest 500, where it would automatically say 1500? MROUND is close but it rounds down also, I need one that only rounds up
□ =CEILING(A2,500)
An alternate solution would be =ROUNDUP(A2/500,0)*500
Leave a Comment | {"url":"https://spreadsheets.sirknightj.com/index.php/2019/08/26/59/","timestamp":"2024-11-06T11:03:52Z","content_type":"text/html","content_length":"62592","record_id":"<urn:uuid:5fc1b997-b8fb-4ff2-84a6-23d567189d3f>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00397.warc.gz"} |
The non-strict inequalities $\geq$ and $\leq$ are created as infix operators with the respective syntax
>=, <=
Maxima allows single inequalities, such as $x-1>y$, and also support for inequalities connected by logical operators, e.g. $x>1 \text{ and } x<=5$.
You can test if two inequalities are the same using the algebraic equivalence test, see the comments on this below.
Chained inequalities, for example $1\leq x \leq2\text{,}$ are not permitted. They must be joined by logical connectives, e.g. "$x>1$ and $x<7$". As and and or are converted to nounand and nounor in
student answers, you may need to use these forms in the teacher's answer as well. For more information, see Propositional Logic.
From version 3.6, support for inequalities in Maxima (particularly single variable real inequalities) was substantially improved.
Functions to support inequalities
This function ensures an inequality is written in the form ex>0 or ex>=0 where ex is always simplified. This is designed for use with the algebraic equivalence answer test in mind.
This function takes an expression, applies ineqprepare(), and then orders the parts. For example,
ineqorder(x>1 and x<5);
5-x > 0 and x-1 > 0
It also removes duplicate inequalities. Operating at this syntactic level will enable a relatively strict form of equivalence to be established, simply manipulating the form of the inequalities. It
will respect commutativity and associativity and and and or, and will also apply not to chains of inequalities.
If the algebraic equivalence test detects inequalities, or systems of inequalities, then this function is automatically applied.
Reverses the order of any inequalities so that we have A<B or A<=B. It does no other transformations. This is useful because when testing equality up to commutativity and associativity we don't
perform this transformation. We need to put all inequalities a particular way around. See the EqualComAss test examples for usage.
See also | {"url":"https://docs.stack-assessment.org/en/CAS/Inequalities/","timestamp":"2024-11-14T03:50:44Z","content_type":"text/html","content_length":"89949","record_id":"<urn:uuid:1628957b-55bc-4b28-b0aa-ccab2e0cfeef>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00593.warc.gz"} |
Linear Interpolation Formula
Interpolation is a method of finding new values for any function using the set of values. This formula is used to determine the unknown value at a given point. To calculate the new value from the two
given points, we should use the linear interpolation formula. In comparison to Lagrange’s interpolation formula, the “n” set of numbers should be available and Lagrange’s method should be used to
find the new value.
An interpolation is the process of finding a value between two points on a line or curve. If we remember what it means, we should consider the first part of the word, ‘inter,’ as meaning ‘enter,’
which reminds us to look ‘inside’ the information we started with. In addition to being useful in statistics, interpolation is also useful in science, business, or any time there is a need to predict
values that fall between two existing data points.
linear interpolation formula excel
What is Linear Interpolation?
A function can be interpolated by estimating its value between any two known values. A relationship is often present, and with the help of experiments at a range of values, other values are able to
be predicted. The interpolation method can be useful for estimating the function of non-tabulated points. By interpolating, it is possible to estimate any desired value at some specific point at some
known coordinate system.
When searching for a value between two points of data, linear interpolation can be beneficial. The mathematician therefore considers it as “filling in the gaps” for a given set of data values
displayed in tabular format.The standard method of linear interpolation is the use of a straight line in order to connect the given points on both the positive and the negative side of the unknown
data point.
For non-linear data, linear interpolation is not always an accurate method. In some cases, linear interpolation may not give a good estimate if the points in the data set change by a large amount. In
addition to that, it involves estimating a new value by connecting two adjacent known values with a straight line.
Linear Interpolation Formula:
Linear interpolation is calculated as follows:
Linear Interpolation = y[1 + ]
This is the elements of linear formula. Here as follows:
1. The first coordinates are x[1] and y[1]
2. while the second coordinates are x[2] and y[2]
3. where x represents the point for interpolation
4. y represents the interpolated value
Linear Interpolation Formula – Solved Example:
Question no1: Find the value of y at x = 4 given some set of values (2, 4), (6, 7).
Solution: Based on the known values,
x=4x[1 ]=2x[2 ]=6y[1 ]=4; y[2]=7
The interpolation formula is,
i.e., y=4+
y = 112
How Does Linear Interpolation Method Work?
It is the simplest method for obtaining values between the data points that is linear interpolation. In this method, there are straight lines connecting the data points.
What is the formula for finding the interpolation between two numbers?
It is imperative to know this formula. The formula is y = y[1] + ((x – x[1]) / (x[2] – x[1])) * (y[2] – y[1]), where x[1] and y[1] are the coordinates below the known x value, and x2 and y[2] are the
coordinates above the known x value.
What is the interpolation method?
Mathematics interpolation is a method for constructing new data points within the range of a set of known data points that are discrete. The function can be simplified by interpolating some points
from the original function to produce a simpler function that is still fairly close to the original. | {"url":"https://benefictech.com/linear-interpolation-formula/","timestamp":"2024-11-03T07:21:59Z","content_type":"text/html","content_length":"40853","record_id":"<urn:uuid:832e053d-e3ec-452c-af97-b7fbbf37c428>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00058.warc.gz"} |
[No. 68] Torque waves (and the insane simplicity of the torque equation)
Smooth torque.Isn’t that what we get from electric machines?
They certainly appear to produce smooth torque.Unlike internal combustion engines, they have no flywheels or balancing shafts, and no reciprocating valve-gear.They are more like turbines, which
seem to produce constant torque with no discernible ripple.
Inside an electric machine we see the complicated internal geometry of slotted stators, rotors with magnets or wound poles, and intricate distributions of conductors.Very rarely is the rotor a
homogeneous isotropic circular cylinder.Both the rotor and the stator could be described as assemblies of discrete components having different shapes and sizes.If we look for regularity we might
count the numbers of slots and pole-pairs, only to find that they are not the same;what is worse, they might not even have a common factor.For example, a 12-slot 10-pole machine has 12 slots
and 5 pole-pairs, with no common factor.How can such a configuration produce constant torque?
The question becomes even more vexing when we examine the digital field-oriented power-electronic controller used with many modern machines.In it we find discrete components operating in switched
mode (meaning that the power transistors are either on or off, never working in a controlled linear mode).How then can we possibly get smooth torque with such a system?
The short answer is that in almost all electric machines absolutely constant torque is practically unattainable, and some degree of torque ripple is inherent.^1The art of the designer is to reduce
it to an acceptable level.Often this level is very low, and in some cases it can be so low as to be hard to measure.
It is helpful to break the subject into a question of harmonics.Torque ripple is an unwanted harmonic of the shaft torque (which if perfectly smooth would have only an average value of harmonic
order zero).It tends to cause speed ripple, which is an unwanted harmonic in the speed.In many cases the rotating inertia acts as a simple and adequate low-pass mechanical filter to attenuate the
speed ripple.But in more complex systems the speed ripple may be dangerously amplified by torsional resonances which have particular harmonic orders.Resonance can lead to the amplification of
torque harmonics, resulting in a pattern of cyclic torsional stress in the shaft and couplings which in turn may cause fatigue failure and/or the loosening of nuts and bolts.Torque ripple is often
associated with vibration (and noise) which may be difficult to resolve by mechanical design alone.
If we cannot rely on the inertia to mitigate the effects of torque ripple, we must reduce the torque ripple at source.Now one of those hallowed engineering incantations every student mumbles outside
the examination hall is that “torque equals flux times current”.It follows that we need to suppress the harmonics in both the flux and the current.But where are they, and how do they interact?
We can begin by saying quite simply that for AC machines the time-waveform of the current needs to be as close as possible to a pure sinewave.^2In inverter-fed machines this is the job of the
current regulator and its PWM control algorithm.Inductance acts as a plain low-pass filter that helps to reduce unwanted harmonics (particularly at the carrier-frequency) in the current waveform,
analogously to the mechanical filtering of the rotational inertia in relation to the speed.
But the concept of ‘flux harmonics’ is not quite so simple.We have to study the ‘flux times current’ product in terms of the interaction of the flux and current distributions inside the machine.In
the classical theory these distributions resolve themselves into the flux-density distribution and the stator ampere-conductor distribution around the air-gap.We will use these easily-visualised
concepts throughout this Diary (and in the related Videos 46-48).But we should make the observation that the harmonic composition of these distributions is not the immediate result of a
finite-element analysis.This leads to the question of whether the torque should be analysed in terms of these harmonic distributions, or whether it should be calculated by Maxwell stress (which is
the immediate result of a finite-element analysis).The harmonic distributions lead directly to the time-average torque (often with a finite-element analysis at only one rotor position), while the
Maxwell stress gives the instantaneous torque at only one rotor position and must therefore be repeated over a complete cycle to get the average or running torque.Yet in spite of their obvious
differences, the two approaches are more compatible than might at first appear, as we will see. (See also [3] and [4]).
For AC machines we use Fourier series to analyse the space-harmonics of the winding distribution, the flux distribution, and sometimes the permeance distribution within the machine, along the
circumferential direction around the air-gap.(We also use Fourier series to analyse the time-harmonics in the current waveform, sometimes drawing on the theory of symmetrical components in cases of
unbalanced operation; but in this Diary we will proceed with the assumption of balanced sinewave polyphase currents).
The torque is obtained from the integral (with respect to angle around the air-gap) of the product of two harmonic series : one representing the stator ampere-conductor distribution and the other
representing some torque-producing attribute of the rotor, which we can call ‘the rotor flux’.(This might be the result of magnets or field windings or saliency, or some combination, and the term
‘rotor flux’ will be familiar especially to field-oriented control engineers who use it even with induction motors).The product of these waves is a wave of shear-stress at the rotor surface, whose
integral over 360° is essentially the instantaneous torque.The central principle in the torque production is the remarkable result that a non-zero result for this integral (and therefore the torque)
is produced only by the interaction of the working space-harmonics of the stator ampere-conductor distribution and the rotor flux, (sometimes called the ‘fundamental’ harmonics).All other
space-harmonic interactions produce no net torque (provided that the current waveforms are balanced sinewaves).This remarkable fact follows directly from the principle of orthogonality in relation
to the product of two Fourier series, and it represents the extraordinary filtering effect that occurs quite naturally in the configuration of AC machines.^3
What it means is that for the purpose of calculating the average torque we can focus our attention on the working harmonics of the stator ampere-conductor distribution and the rotor flux, and ignore
all other space harmonics.^4 This is the basis of the dq-axis transformation (Park’s transform).The textbook development of Park’s transform starts out with the assumption of sinusoidal
space-distributions of flux and ampere-conductors around the air-gap; and although it does not work with space-harmonics of either, it recognizes their existence by lumping their inductive
effects into the ‘harmonic leakage inductance’ or ‘differential inductance’, while they contribute nothing to the torque.
The working harmonic space-distributions of the stator ampere-conductors and the rotor flux rotate in synchronism at the synchronous speed, and Park projects them on to a frame of reference rotating
at synchronous speed.In this frame of reference (so-called dq axes) everything appears constant in steady-state operation.This being so, Park represents the distributed effects of the stator
ampere-conductors and the rotor flux by the terminal quantities, current and flux-linkage.In order to provide a means of orientation (changing the relative spatial phase angle between the flux and
the stator ampere-conductor distributions), Park needs two windings or coils, and he uses one aligned with the d-axis or field axis of the rotor, and the other one aligned with the q‐ or
quadrature axis at right-angles to the d-axis.In this way he gets a flux-linkage ψ[d1] and current i[d1] for the d-axis winding and ψ[q1] and i[q1] for the q-axis winding.The additional subscript 1
reflects the fact that the respective distributions are the working space-harmonics of the flux and the ampere-conductors.^5
The average electromagnetic torque is now obtained from the equation
where p is the number of pole-pairs. In modern colloquial parlance, we can describe this as ‘insanely simple’ in view of the complexity of the physical system and its mathematical analysis.This is
the E = mc^2 of electric machine design.Its derivation often gets lost in analytical nuances, but the student’s mantra is clear :torque = flux × current!
We have to embellish it slightly.First, we need the flux times current for both axes of the dq model, not just one of them.^6Secondly, we need to distinguish flux-linkage (a terminal or circuit
quantity) from flux (a distributed or field quantity).Thirdly, we need to add the coefficients 3/2 and p.We won’t go into the interpretation of these coefficients, or even discuss the units in the
equation, but simply reflect on its ‘insane’ simplicity.
Can we use this elegant torque equation with finite-element analysis at a single rotor position, instead of a full-blown rotor-position-stepping total simulation with Maxwell stress? The answer is
yes; and examples can be found in [3] and [4] and elsewhere.But of course it gives only the average or running torque, without any ripple, without any cogging,
without any parasitic effects (such as stray losses and ‘harmonic leakage’ contributions to the terminal inductance).Like E = mc^2, however, it’s quick.
Both the d‐ and the q-axis ‘coils’ representing the actual polyphase stator winding are ‘sine-distributed’ in the sense that their spatial (circumferential) ampere-conductor distributions are
sinusoidal with only the working harmonic or fundamental number of pole-pairs p;but their inductances are supplemented by the aforementioned leakage inductance which includes ‘harmonic leakage’
associated with the other space-harmonics of the actual windings.This is necessary so that the dq-axis model can produce the correct terminal voltages (through the so-called Park’s equations).The
inductance associated with the working space-harmonic of the dq-axis ampere-conductor distributions is called the ‘magnetizing inductance’ while the total inductance of the d-axis winding is the
synchronous inductance L[d], and likewise L[q] for the q-axis winding.
All these ideas are very old, and they have been used as the core of electric machine design for more than a century.It is 100 years since Park’s dq-axis transform was published and more than 130
years since Blondel published his two-reaction theory which also relies on the notion of sinusoidal distributions of ampere-conductors and flux.The modern expression of these principles is often in
terms of space-vectors, the language of field-oriented control;while the actual elements (particularly flux-linkage, inductance, and torque) are computed by numerical analysis with vastly
superior accuracy and insight to those available in the early days.
It is easy to forget that Park’s dq-axis transform relies on Fourier series analysis of the flux and ampere-conductor distributions, although it is glaringly obvious when it is applied to the
inductance matrix in the derivation of L[d] and L[q], [2,3].But in mathematical terms, harmonic analysis by Fourier series is a linear operation.By ‘adding harmonics’ it embodies the
principle of superposition and so its validity (or at least its accuracy) can be questioned when it is applied to a ‘nonlinear system’ such as an electrical machine.We might imagine that the ‘total
simulation’ approach is not affected by this concern, and that would certainly be true of the torque calculation by Maxwell stress.It would also be perfectly valid to decompose a torque waveform
into its harmonic components using Fourier series, after it had been calculated by Maxwell stress in a ‘total simulation’ that did not use the Park transform.But it is common, even in the ‘total
simulation’ approach, to use L[d] and L[q] and Park’s circuit equations without stopping to consider whether the harmonic decomposition underlying the Park transform is completely valid.Even if the
principle of superposition is acceptable in relation to the harmonic composition of the flux and ampere-conductor distributions, it remains the case that harmonics in these distributions introduce
harmonic variations in L[d] and L[q].In other words, they may vary with rotor position.It seems that we rarely acknowledge this possibility.Instead we sail through our machine design and our
analysis with the tacit assumption that L[d] and L[q] are independent of rotor position, even though we readily accept their variation with current.It seems to work.Maybe that’s because of our
mastery of winding layouts and our adherence to best practice!Nevertheless, there are engineers who do not trust the underlying assumptions completely, and so they resort to the ‘total simulation’
approach for everything — even the circuit analysis.The engineers of this school of thought might well be the first to describe the simplicity of the torque equation as ‘insane’, and they would have
a point even though the term ‘insane’ is a bit extreme.Maybe they’re right.The ‘total simulation’ alternative is of course possible only with the most powerful simulation software; it
certainly was not possible a generation ago.A complete analysis of these interrelated issues would be a hard and complex task for a PhD student; and frankly I don’t know if it has been tackled,
or to what extent.The full gamut of questions and theoretical implications would seem to be outside the normal scope of our busy daily work in machine design.Let’s hope so!
^1 As an example of a machine that produces constant torque, the Faraday disk can be cited, as well as certain types of homopolar superconducting machine, [1].
^2 This excludes the brushless DC motor which operates (ideally) with ‘squarewave’ current waveforms [2].It also excludes the DC commutator motor and the switched reluctance motor,
since these are not AC machines.
^3 Cogging torque is excluded from this discussion, as it is a separate effect that arises even with zero current.
^4 The other space-harmonics are still important in the calculation of stray losses, total winding inductance, and terminal voltage.
^5 Many of these ideas were derived from the work of Blondel 30 years earlier.In their original publications on the dq-axis theory in the 1920s, Park, Doherty, Nickle and others generously
acknowledged Blondel.
^6 This is true even in the DC commutator motor which is probably the original source of the ‘flux times current’ notion.In the DC motor, only the d-axis flux-linkage is normally considered to be
active in producing torque, although both terms in eqn. (1) are in fact present.
Further reading
see also Engineer’s Diary No. 67
[1] A.D. Appleton, Motors, Generators, and Flux Pumps, Cryogenics, June 1969, pp. 147-157.
[2] J.R. Hendershot & T.J.E. Miller, Design of Brushless Permanent-Magnet Machines, Motor Design Books LLC, 2010, sales@motordesignbooks.com ISBN 978-0-9840687-0-8
[3] J.R. Hendershot & T.J.E. Miller, Design Studies in Electric Machines, Motor Design Books LLC, 2022, sales@motordesignbooks.com ISBN 978-0-9840687-4-6
[4] Nicola Bianchi, Electrical Machine Analysis Using Finite Elements, Taylor & Francis, 2005, ISBN 0-8493-3399-7
If you have any comments, please let us know.
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RFC 6637: Elliptic Curve Cryptography (ECC) in OpenPGP
Elliptic Curve Cryptography (ECC) in OpenPGP
RFC 6637
RFC - Proposed Standard (June 2012) Errata
Document Type Obsoleted by
RFC 9580
Author Andrey Jivsov
Last updated 2022-09-28
RFC stream Internet Engineering Task Force (IETF)
IESG Responsible AD Sean Turner
Send notices to (None)
RFC 6637
Internet Engineering Task Force (IETF) A. Jivsov
Request for Comments: 6637 Symantec Corporation
Category: Standards Track June 2012
ISSN: 2070-1721
Elliptic Curve Cryptography (ECC) in OpenPGP
This document defines an Elliptic Curve Cryptography extension to the
OpenPGP public key format and specifies three Elliptic Curves that
enjoy broad support by other standards, including standards published
by the US National Institute of Standards and Technology. The
document specifies the conventions for interoperability between
compliant OpenPGP implementations that make use of this extension and
these Elliptic Curves.
Status of This Memo
This is an Internet Standards Track document.
This document is a product of the Internet Engineering Task Force
(IETF). It represents the consensus of the IETF community. It has
received public review and has been approved for publication by the
Internet Engineering Steering Group (IESG). Further information on
Internet Standards is available in Section 2 of RFC 5741.
Information about the current status of this document, any errata,
and how to provide feedback on it may be obtained at
Copyright Notice
Copyright (c) 2012 IETF Trust and the persons identified as the
document authors. All rights reserved.
This document is subject to BCP 78 and the IETF Trust's Legal
Provisions Relating to IETF Documents
(http://trustee.ietf.org/license-info) in effect on the date of
publication of this document. Please review these documents
carefully, as they describe your rights and restrictions with respect
to this document. Code Components extracted from this document must
include Simplified BSD License text as described in Section 4.e of
the Trust Legal Provisions and are provided without warranty as
described in the Simplified BSD License.
Jivsov Standards Track [Page 1]
RFC 6637 ECC in OpenPGP June 2012
Table of Contents
1. Introduction ....................................................3
2. Conventions used in This Document ...............................3
3. Elliptic Curve Cryptography .....................................3
4. Supported ECC Curves ............................................3
5. Supported Public Key Algorithms .................................4
6. Conversion Primitives ...........................................4
7. Key Derivation Function .........................................5
8. EC DH Algorithm (ECDH) ..........................................5
9. Encoding of Public and Private Keys .............................8
10. Message Encoding with Public Keys ..............................9
11. ECC Curve OID .................................................10
12. Compatibility Profiles ........................................10
12.1. OpenPGP ECC Profile ......................................10
12.2. Suite-B Profile ..........................................11
12.2.1. Security Strength at 192 Bits .....................11
12.2.2. Security Strength at 128 Bits .....................11
13. Security Considerations .......................................12
14. IANA Considerations ...........................................14
15. References ....................................................14
15.1. Normative References .....................................14
15.2. Informative References ...................................15
16. Contributors ..................................................15
17. Acknowledgment ................................................15
Jivsov Standards Track [Page 2]
RFC 6637 ECC in OpenPGP June 2012
1. Introduction
The OpenPGP protocol [RFC4880] supports RSA and DSA (Digital
Signature Algorithm) public key formats. This document defines the
extension to incorporate support for public keys that are based on
Elliptic Curve Cryptography (ECC).
2. Conventions Used in This Document
The key words "MUST", "MUST NOT", "REQUIRED", "SHALL", "SHALL NOT",
"SHOULD", "SHOULD NOT", "RECOMMENDED", "MAY", and "OPTIONAL" in this
document are to be interpreted as described in [RFC2119]. Any
implementation that adheres to the format and methods specified in
this document is called a compliant application. Compliant
applications are a subset of the broader set of OpenPGP applications
described in [RFC4880]. Any [RFC2119] keyword within this document
applies to compliant applications only.
3. Elliptic Curve Cryptography
This document establishes the minimum set of Elliptic Curve
Cryptography (ECC) public key parameters and cryptographic methods
that will likely satisfy the widest range of platforms and
applications and facilitate interoperability. It adds a more
efficient method for applications to balance the overall level of
security with any AES algorithm specified in [RFC4880] than by simply
increasing the size of RSA keys. This document defines a path to
expand ECC support in the future.
The National Security Agency (NSA) of the United States specifies ECC
for use in its [SuiteB] set of algorithms. This document includes
algorithms required by Suite B that are not present in [RFC4880].
[KOBLITZ] provides a thorough introduction to ECC.
4. Supported ECC Curves
This document references three named prime field curves, defined in
[FIPS-186-3] as "Curve P-256", "Curve P-384", and "Curve P-521".
The named curves are referenced as a sequence of bytes in this
document, called throughout, curve OID. Section 11 describes in
detail how this sequence of bytes is formed.
Jivsov Standards Track [Page 3]
RFC 6637 ECC in OpenPGP June 2012
5. Supported Public Key Algorithms
The supported public key algorithms are the Elliptic Curve Digital
Signature Algorithm (ECDSA) [FIPS-186-3] and the Elliptic Curve
Diffie-Hellman (ECDH). A compatible specification of ECDSA is given
in [RFC6090] as "KT-I Signatures" and in [SEC1]; ECDH is defined in
Section 8 of this document.
The following public key algorithm IDs are added to expand Section
9.1 of [RFC4880], "Public-Key Algorithms":
ID Description of Algorithm
-- --------------------------
18 ECDH public key algorithm
19 ECDSA public key algorithm
Compliant applications MUST support ECDSA and ECDH.
6. Conversion Primitives
This document only defines the uncompressed point format. The point
is encoded in the Multiprecision Integer (MPI) format [RFC4880]. The
content of the MPI is the following:
B = 04 || x || y
where x and y are coordinates of the point P = (x, y), each encoded
in the big-endian format and zero-padded to the adjusted underlying
field size. The adjusted underlying field size is the underlying
field size that is rounded up to the nearest 8-bit boundary.
Therefore, the exact size of the MPI payload is 515 bits for "Curve
P-256", 771 for "Curve P-384", and 1059 for "Curve P-521".
Even though the zero point, also called the point at infinity, may
occur as a result of arithmetic operations on points of an elliptic
curve, it SHALL NOT appear in data structures defined in this
This encoding is compatible with the definition given in [SEC1].
If other conversion methods are defined in the future, a compliant
application MUST NOT use a new format when in doubt that any
recipient can support it. Consider, for example, that while both the
public key and the per-recipient ECDH data structure, respectively
defined in Sections 9 and 10, contain an encoded point field, the
format changes to the field in Section 10 only affect a given
recipient of a given message.
Jivsov Standards Track [Page 4]
RFC 6637 ECC in OpenPGP June 2012
7. Key Derivation Function
A key derivation function (KDF) is necessary to implement the EC
encryption. The Concatenation Key Derivation Function (Approved
Alternative 1) [NIST-SP800-56A] with the KDF hash function that is
SHA2-256 [FIPS-180-3] or stronger is REQUIRED. See Section 12 for
the details regarding the choice of the hash function.
For convenience, the synopsis of the encoding method is given below
with significant simplifications attributable to the restricted
choice of hash functions in this document. However, [NIST-SP800-56A]
is the normative source of the definition.
// Implements KDF( X, oBits, Param );
// Input: point X = (x,y)
// oBits - the desired size of output
// hBits - the size of output of hash function Hash
// Param - octets representing the parameters
// Assumes that oBits <= hBits
// Convert the point X to the octet string, see section 6:
// ZB' = 04 || x || y
// and extract the x portion from ZB'
ZB = x;
MB = Hash ( 00 || 00 || 00 || 01 || ZB || Param );
return oBits leftmost bits of MB.
Note that ZB in the KDF description above is the compact
representation of X, defined in Section 4.2 of [RFC6090].
8. EC DH Algorithm (ECDH)
The method is a combination of an ECC Diffie-Hellman method to
establish a shared secret, a key derivation method to process the
shared secret into a derived key, and a key wrapping method that uses
the derived key to protect a session key used to encrypt a message.
The One-Pass Diffie-Hellman method C(1, 1, ECC CDH) [NIST-SP800-56A]
MUST be implemented with the following restrictions: the ECC CDH
primitive employed by this method is modified to always assume the
cofactor as 1, the KDF specified in Section 7 is used, and the KDF
parameters specified below are used.
Jivsov Standards Track [Page 5]
RFC 6637 ECC in OpenPGP June 2012
The KDF parameters are encoded as a concatenation of the following 5
variable-length and fixed-length fields, compatible with the
definition of the OtherInfo bitstring [NIST-SP800-56A]:
o a variable-length field containing a curve OID, formatted as
- a one-octet size of the following field
- the octets representing a curve OID, defined in Section 11
o a one-octet public key algorithm ID defined in Section 5
o a variable-length field containing KDF parameters, identical to
the corresponding field in the ECDH public key, formatted as
- a one-octet size of the following fields; values 0 and 0xff
are reserved for future extensions
- a one-octet value 01, reserved for future extensions
- a one-octet hash function ID used with the KDF
- a one-octet algorithm ID for the symmetric algorithm used to
wrap the symmetric key for message encryption; see Section 8
for details
o 20 octets representing the UTF-8 encoding of the string
"Anonymous Sender ", which is the octet sequence
41 6E 6F 6E 79 6D 6F 75 73 20 53 65 6E 64 65 72 20 20 20 20
o 20 octets representing a recipient encryption subkey or a master
key fingerprint, identifying the key material that is needed for
the decryption
The size of the KDF parameters sequence, defined above, is either 54
or 51 for the three curves defined in this document.
The key wrapping method is described in [RFC3394]. KDF produces a
symmetric key that is used as a key-encryption key (KEK) as specified
in [RFC3394]. Refer to Section 13 for the details regarding the
choice of the KEK algorithm, which SHOULD be one of three AES
algorithms. Key wrapping and unwrapping is performed with the
default initial value of [RFC3394].
Jivsov Standards Track [Page 6]
RFC 6637 ECC in OpenPGP June 2012
The input to the key wrapping method is the value "m" derived from
the session key, as described in Section 5.1 of [RFC4880], "Public-
Key Encrypted Session Key Packets (Tag 1)", except that the PKCS #1.5
(Public-Key Cryptography Standards version 1.5) padding step is
omitted. The result is padded using the method described in [PKCS5]
to the 8-byte granularity. For example, the following AES-256
session key, in which 32 octets are denoted from k0 to k31, is
composed to form the following 40 octet sequence:
09 k0 k1 ... k31 c0 c1 05 05 05 05 05
The octets c0 and c1 above denote the checksum. This encoding allows
the sender to obfuscate the size of the symmetric encryption key used
to encrypt the data. For example, assuming that an AES algorithm is
used for the session key, the sender MAY use 21, 13, and 5 bytes of
padding for AES-128, AES-192, and AES-256, respectively, to provide
the same number of octets, 40 total, as an input to the key wrapping
The output of the method consists of two fields. The first field is
the MPI containing the ephemeral key used to establish the shared
secret. The second field is composed of the following two fields:
o a one-octet encoding the size in octets of the result of the key
wrapping method; the value 255 is reserved for future extensions
o up to 254 octets representing the result of the key wrapping
method, applied to the 8-byte padded session key, as described
Note that for session key sizes 128, 192, and 256 bits, the size of
the result of the key wrapping method is, respectively, 32, 40, and
48 octets, unless the size obfuscation is used.
For convenience, the synopsis of the encoding method is given below;
however, this section, [NIST-SP800-56A], and [RFC3394] are the
normative sources of the definition.
Jivsov Standards Track [Page 7]
RFC 6637 ECC in OpenPGP June 2012
Obtain the authenticated recipient public key R
Generate an ephemeral key pair {v, V=vG}
Compute the shared point S = vR;
m = symm_alg_ID || session key || checksum || pkcs5_padding;
curve_OID_len = (byte)len(curve_OID);
Param = curve_OID_len || curve_OID || public_key_alg_ID || 03
|| 01 || KDF_hash_ID || KEK_alg_ID for AESKeyWrap || "Anonymous
Sender " || recipient_fingerprint;
Z_len = the key size for the KEK_alg_ID used with AESKeyWrap
Compute Z = KDF( S, Z_len, Param );
Compute C = AESKeyWrap( Z, m ) as per [RFC3394]
VB = convert point V to the octet string
Output (MPI(VB) || len(C) || C).
The decryption is the inverse of the method given. Note that the
recipient obtains the shared secret by calculating
S = rV = rvG, where (r,R) is the recipient's key pair.
Consistent with Section 5.13 of [RFC4880], "Sym. Encrypted Integrity
Protected Data Packet (Tag 18)", a Modification Detection Code (MDC)
MUST be used anytime the symmetric key is protected by ECDH.
9. Encoding of Public and Private Keys
The following algorithm-specific packets are added to Section 5.5.2
of [RFC4880], "Public-Key Packet Formats", to support ECDH and ECDSA.
This algorithm-specific portion is:
Algorithm-Specific Fields for ECDSA keys:
o a variable-length field containing a curve OID, formatted
as follows:
- a one-octet size of the following field; values 0 and
0xFF are reserved for future extensions
- octets representing a curve OID, defined in Section 11
o MPI of an EC point representing a public key
Jivsov Standards Track [Page 8]
RFC 6637 ECC in OpenPGP June 2012
Algorithm-Specific Fields for ECDH keys:
o a variable-length field containing a curve OID, formatted
as follows:
- a one-octet size of the following field; values 0 and
0xFF are reserved for future extensions
- the octets representing a curve OID, defined in
Section 11
- MPI of an EC point representing a public key
o a variable-length field containing KDF parameters,
formatted as follows:
- a one-octet size of the following fields; values 0 and
0xff are reserved for future extensions
- a one-octet value 01, reserved for future extensions
- a one-octet hash function ID used with a KDF
- a one-octet algorithm ID for the symmetric algorithm
used to wrap the symmetric key used for the message
encryption; see Section 8 for details
Observe that an ECDH public key is composed of the same sequence of
fields that define an ECDSA key, plus the KDF parameters field.
The following algorithm-specific packets are added to Section 5.5.3.
of [RFC4880], "Secret-Key Packet Formats", to support ECDH and ECDSA.
Algorithm-Specific Fields for ECDH or ECDSA secret keys:
o an MPI of an integer representing the secret key, which is a
scalar of the public EC point
10. Message Encoding with Public Keys
Section 5.2.2 of [RFC4880], "Version 3 Signature Packet Format"
defines signature formats. No changes in the format are needed for
Section 5.1 of [RFC4880], "Public-Key Encrypted Session Key Packets
(Tag 1)" is extended to support ECDH. The following two fields are
the result of applying the KDF, as described in Section 8.
Jivsov Standards Track [Page 9]
RFC 6637 ECC in OpenPGP June 2012
Algorithm-Specific Fields for ECDH:
o an MPI of an EC point representing an ephemeral public key
o a one-octet size, followed by a symmetric key encoded using the
method described in Section 8
11. ECC Curve OID
The parameter curve OID is an array of octets that define a named
curve. The table below specifies the exact sequence of bytes for
each named curve referenced in this document:
ASN.1 Object OID Curve OID bytes in Curve name in
Identifier len hexadecimal [FIPS-186-3]
1.2.840.10045.3.1.7 8 2A 86 48 CE 3D 03 01 07 NIST curve P-256
1.3.132.0.34 5 2B 81 04 00 22 NIST curve P-384
1.3.132.0.35 5 2B 81 04 00 23 NIST curve P-521
The sequence of octets in the third column is the result of applying
the Distinguished Encoding Rules (DER) to the ASN.1 Object Identifier
with subsequent truncation. The truncation removes the two fields of
encoded Object Identifier. The first omitted field is one octet
representing the Object Identifier tag, and the second omitted field
is the length of the Object Identifier body. For example, the
complete ASN.1 DER encoding for the NIST P-256 curve OID is "06 08 2A
86 48 CE 3D 03 01 07", from which the first entry in the table above
is constructed by omitting the first two octets. Only the truncated
sequence of octets is the valid representation of a curve OID.
12. Compatibility Profiles
12.1. OpenPGP ECC Profile
A compliant application MUST implement NIST curve P-256, MAY
implement NIST curve P-384, and SHOULD implement NIST curve P-521, as
defined in Section 11. A compliant application MUST implement
SHA2-256 and SHOULD implement SHA2-384 and SHA2-512. A compliant
application MUST implement AES-128 and SHOULD implement AES-256.
Jivsov Standards Track [Page 10]
RFC 6637 ECC in OpenPGP June 2012
A compliant application SHOULD follow Section 13 regarding the choice
of the following algorithms for each curve:
o the KDF hash algorithm
o the KEK algorithm
o the message digest algorithm and the hash algorithm used in the
key certifications
o the symmetric algorithm used for message encryption.
It is recommended that the chosen symmetric algorithm for message
encryption be no less secure than the KEK algorithm.
12.2. Suite-B Profile
A subset of algorithms allowed by this document can be used to
achieve [SuiteB] compatibility. The references to [SuiteB] in this
document are informative. This document is primarily concerned with
format specification, leaving additional security restrictions
unspecified, such as matching the assigned security level of
information to authorized recipients or interoperability concerns
arising from fewer allowed algorithms in [SuiteB] than allowed by
12.2.1. Security Strength at 192 Bits
To achieve the security strength of 192 bits, [SuiteB] requires NIST
curve P-384, AES-256, and SHA2-384. The symmetric algorithm
restriction means that the algorithm of KEK used for key wrapping in
Section 8 and an [RFC4880] session key used for message encryption
must be AES-256. The hash algorithm restriction means that the hash
algorithms of KDF and the [RFC4880] message digest calculation must
be SHA-384.
12.2.2. Security Strength at 128 Bits
The set of algorithms in Section 12.2.1 is extended to allow NIST
curve P-256, AES-128, and SHA2-256.
Jivsov Standards Track [Page 11]
RFC 6637 ECC in OpenPGP June 2012
13. Security Considerations
Refer to [FIPS-186-3], B.4.1, for the method to generate a uniformly
distributed ECC private key.
The curves proposed in this document correspond to the symmetric key
sizes 128 bits, 192 bits, and 256 bits, as described in the table
below. This allows a compliant application to offer balanced public
key security, which is compatible with the symmetric key strength for
each AES algorithm allowed by [RFC4880].
The following table defines the hash and the symmetric encryption
algorithm that SHOULD be used with a given curve for ECDSA or ECDH.
A stronger hash algorithm or a symmetric key algorithm MAY be used
for a given ECC curve. However, note that the increase in the
strength of the hash algorithm or the symmetric key algorithm may not
increase the overall security offered by the given ECC key.
Curve name ECC RSA Hash size Symmetric
strength strength, key size
NIST curve P-256 256 3072 256 128
NIST curve P-384 384 7680 384 192
NIST curve P-521 521 15360 512 256
Requirement levels indicated elsewhere in this document lead to the
following combinations of algorithms in the OpenPGP profile: MUST
implement NIST curve P-256 / SHA2-256 / AES-128, SHOULD implement
NIST curve P-521 / SHA2-512 / AES-256, MAY implement NIST curve P-384
/ SHA2-384 / AES-256, among other allowed combinations.
Consistent with the table above, the following table defines the KDF
hash algorithm and the AES KEK encryption algorithm that SHOULD be
used with a given curve for ECDH. A stronger KDF hash algorithm or
AES KEK algorithm MAY be used for a given ECC curve.
Curve name Recommended KDF Recommended KEK
hash algorithm encryption algorithm
NIST curve P-256 SHA2-256 AES-128
NIST curve P-384 SHA2-384 AES-192
NIST curve P-521 SHA2-512 AES-256
Jivsov Standards Track [Page 12]
RFC 6637 ECC in OpenPGP June 2012
This document explicitly discourages the use of algorithms other than
AES as a KEK algorithm because backward compatibility of the ECDH
format is not a concern. The KEK algorithm is only used within the
scope of a Public-Key Encrypted Session Key Packet, which represents
an ECDH key recipient of a message. Compare this with the algorithm
used for the session key of the message, which MAY be different from
a KEK algorithm.
Compliant applications SHOULD implement, advertise through key
preferences, and use in compliance with [RFC4880], the strongest
algorithms specified in this document.
Note that the [RFC4880] symmetric algorithm preference list may make
it impossible to use the balanced strength of symmetric key
algorithms for a corresponding public key. For example, the presence
of the symmetric key algorithm IDs and their order in the key
preference list affects the algorithm choices available to the
encoding side, which in turn may make the adherence to the table
above infeasible. Therefore, compliance with this specification is a
concern throughout the life of the key, starting immediately after
the key generation when the key preferences are first added to a key.
It is generally advisable to position a symmetric algorithm ID of
strength matching the public key at the head of the key preference
Encryption to multiple recipients often results in an unordered
intersection subset. For example, if the first recipient's set is
{A, B} and the second's is {B, A}, the intersection is an unordered
set of two algorithms, A and B. In this case, a compliant
application SHOULD choose the stronger encryption algorithm.
Resource constraints, such as limited computational power, is a
likely reason why an application might prefer to use the weakest
algorithm. On the other side of the spectrum are applications that
can implement every algorithm defined in this document. Most
applications are expected to fall into either of two categories. A
compliant application in the second, or strongest, category SHOULD
prefer AES-256 to AES-192.
SHA-1 MUST NOT be used with the ECDSA or the KDF in the ECDH method.
MDC MUST be used when a symmetric encryption key is protected by
ECDH. None of the ECC methods described in this document are allowed
with deprecated V3 keys. A compliant application MUST only use
iterated and salted S2K to protect private keys, as defined in
Section 3.7.1.3 of [RFC4880], "Iterated and Salted S2K".
Jivsov Standards Track [Page 13]
RFC 6637 ECC in OpenPGP June 2012
Side channel attacks are a concern when a compliant application's use
of the OpenPGP format can be modeled by a decryption or signing
oracle model, for example, when an application is a network service
performing decryption to unauthenticated remote users. ECC scalar
multiplication operations used in ECDSA and ECDH are vulnerable to
side channel attacks. Countermeasures can often be taken at the
higher protocol level, such as limiting the number of allowed
failures or time-blinding of the operations associated with each
network interface. Mitigations at the scalar multiplication level
seek to eliminate any measurable distinction between the ECC point
addition and doubling operations.
14. IANA Considerations
Per this document, IANA has assigned an algorithm number from the
"Public Key Algorithms" range (or the "name space" in the terminology
of [RFC5226]) of the "Pretty Good Privacy (PGP)" registry, created by
[RFC4880]. Two ID numbers have been assigned, as defined in Section
5. The first one, value 19, is already designated for ECDSA and is
currently unused, while the other one, value 18, is new.
15. References
15.1. Normative References
[RFC2119] Bradner, S., "Key words for use in RFCs to Indicate
Requirement Levels", BCP 14, RFC 2119, March 1997.
[RFC4880] Callas, J., Donnerhacke, L., Finney, H., Shaw, D.,
and R. Thayer, "OpenPGP Message Format", RFC 4880,
November 2007.
[SuiteB] National Security Agency, "NSA Suite B
Cryptography", March 11, 2010,
[FIPS-186-3] National Institute of Standards and Technology, U.S.
Department of Commerce, "Digital Signature
Standard", FIPS 186-3, June 2009.
[NIST-SP800-56A] Barker, E., Johnson, D., and M. Smid,
"Recommendation for Pair-Wise Key Establishment
Schemes Using Discrete Logarithm Cryptography", NIST
Special Publication 800-56A Revision 1, March 2007.
[FIPS-180-3] National Institute of Standards and Technology, U.S.
Department of Commerce, "Secure Hash Standard
(SHS)", FIPS 180-3, October 2008.
Jivsov Standards Track [Page 14]
RFC 6637 ECC in OpenPGP June 2012
[RFC3394] Schaad, J. and R. Housley, "Advanced Encryption
Standard (AES) Key Wrap Algorithm", RFC 3394,
September 2002.
[PKCS5] RSA Laboratories, "PKCS #5 v2.0: Password-Based
Cryptography Standard", March 25, 1999.
[RFC5226] Narten, T. and H. Alvestrand, "Guidelines for
Writing an IANA Considerations Section in RFCs", BCP
26, RFC 5226, May 2008.
15.2. Informative References
[KOBLITZ] N. Koblitz, "A course in number theory and
cryptography", Chapter VI. Elliptic Curves, ISBN:
0-387-96576-9, Springer-Verlag, 1987
[RFC6090] McGrew, D., Igoe, K., and M. Salter, "Fundamental
Elliptic Curve Cryptography Algorithms", RFC 6090,
February 2011.
[SEC1] Standards for Efficient Cryptography Group, "SEC 1:
Elliptic Curve Cryptography", September 2000.
16. Contributors
Hal Finney provided important criticism on compliance with
[NIST-SP800-56A] and [SuiteB], and pointed out a few other mistakes.
17. Acknowledgment
The author would like to acknowledge the help of many individuals who
kindly voiced their opinions on the IETF OpenPGP Working Group
mailing list, in particular, the help of Jon Callas, David Crick, Ian
G, Werner Koch, and Marko Kreen.
Author's Address
Andrey Jivsov
Symantec Corporation
EMail: Andrey_Jivsov@symantec.com
Jivsov Standards Track [Page 15] | {"url":"https://datatracker.ietf.org/doc/rfc6637/","timestamp":"2024-11-11T17:20:21Z","content_type":"text/html","content_length":"62916","record_id":"<urn:uuid:60df3a5a-a407-48e2-a2fb-de45294b851b>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00862.warc.gz"} |
IV Flow Rate Calculator
Welcome to the IV Flow Rate tutorial! This guide aims to provide an understanding of IV (intravenous) flow rate calculations. IV therapy is a critical component of healthcare, allowing for the
administration of fluids, medications, and nutrients directly into the bloodstream. The flow rate refers to the speed at which the IV solution is delivered, ensuring the proper dosage and infusion
time. Whether you're a healthcare professional or interested in understanding the intricacies of IV therapy, this tutorial will equip you with the necessary knowledge to calculate and monitor the IV
flow rate.
IV Flow
IV Flow Rate
Calculator Results
Flow Rates = gtt/min
Interesting Facts
Before we delve into the calculations and formulas, let's explore some interesting facts about IV flow rate:
• IV therapy is a common medical intervention used in hospitals, clinics, and other healthcare settings to provide fluids, medications, blood products, and nutritional support to patients.
• The IV flow rate is measured in milliliters per hour (mL/hr) or milliliters per minute (mL/min) and determines the rate at which the IV solution is administered.
• IV flow rate calculations take into account factors such as the volume of the IV solution, the desired infusion time, and the administration set's flow rate factor.
• Accurate flow rate calculations are essential for ensuring the proper delivery of medications, preventing under-dosing or over-dosing, and maintaining patient safety.
• Healthcare professionals closely monitor the IV flow rate to ensure the therapy is administered at the prescribed rate and to prevent complications associated with rapid or slow infusion.
The Formula: IV Flow Rate
The IV flow rate can be calculated using the following formula:
Flow Rate = (Volume / Time)
In this formula:
• Flow Rate: The rate at which the IV solution is administered, usually measured in milliliters per hour (mL/hr) or milliliters per minute (mL/min).
• Volume: The volume of the IV solution to be infused, typically measured in milliliters (mL).
• Time: The desired infusion time, usually measured in hours or minutes.
Relevance to Other Fields
The concept of IV flow rate calculations is relevant to various fields, including:
• Nursing and Medical Practice: Healthcare professionals, particularly nurses and doctors, are responsible for administering IV therapy and ensuring the correct flow rate to deliver medications,
fluids, and nutrients safely and effectively.
• Pharmacy and Pharmacology: Pharmacists play a crucial role in preparing IV medications and determining the appropriate flow rates for optimal drug delivery.
• Anesthesiology and Critical Care: IV flow rate calculations are essential in critical care settings, where precise infusion rates are crucial for maintaining hemodynamic stability, titrating
medications, and managing patient conditions.
Real-Life Example
Let's consider a real-life example to illustrate the use of IV flow rate calculations. Suppose a healthcare professional needs to administer 1000 mL of IV solution over 8 hours. Using the formula, we
can calculate the flow rate:
Flow Rate = (1000 mL / 8 hours) = 125 mL/hr
In this example, the calculated IV flow rate is 125 mL/hr. This means that the healthcare professional should set the IV infusion at a rate of 125 mL per hour to administer the 1000 mL IV solution
over the desired 8-hour timeframe.
Achievements and Key Individuals
The field of IV therapy has seen advancements and innovations driven by the contributions of numerous researchers, healthcare professionals, and medical device manufacturers. While specific
individuals cannot be mentioned, their work has significantly improved the safety, efficacy, and precision of IV therapy administration, leading to better patient outcomes and care. Now armed with
the knowledge of IV flow rate calculations, you have a better understanding of the importance of accurate infusion rates in IV therapy. Whether you're a healthcare professional responsible for IV
therapy or an individual interested in understanding medical care, this knowledge is crucial for ensuring patient safety and optimal outcomes.
Health Calculators
You may also find the following Health Calculators useful.
Use of the Health and Medical Calculators
Please note that the Iv Flow Rate Calculator is provided for your personal use and designed to provide information and information relating to the calculations only. The Iv Flow Rate Calculator
should not be used for you to self-diagnose conditions, self-medicate or alter any existing medication that you are currently prescribed by your Doctor. If the Iv Flow Rate Calculator produces a
calculation which causes you concern, please consult your Doctor for support, advice and further information. | {"url":"https://health.icalculator.com/iv-flow-rate-calculator.html","timestamp":"2024-11-08T02:45:10Z","content_type":"text/html","content_length":"21044","record_id":"<urn:uuid:328606fb-e2f8-41df-8504-9bbb0e47a2ca>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00556.warc.gz"} |
Maximum Leverage on Maker
First off, this article assumes familiarity with Maker CDP's. (If not, I highly suggest reading up on it!)
Recall that the liquidation ratio of a CDP is the collateral-to-debt ratio of the CDP. For example, if the liquidation ratio is 150% on ETH/USD, if ETH is $100 and you have 1 ETH in the CDP, you may
accrue up to$66 of debt (that is, generate up to $66 Dai) as your collateral is worth 150% of your debt. This gives us a leverage ratio of 1.6x. But can we go higher?
Assuming enough liquidity exists, we can then buy 0.66 ETH with this Dai.
We can then put this ETH back into our CDP to draw more Dai. We can draw \66.66 * 1/1.5 = $44.44$ more Dai from our CDP. We can keep doing this forever to generate more Dai and thus more leverage;
however, we get less Dai each time. But how much?
Let $\lambda$ represent the liquidation ratio and $L$ represent our maximum leverage ratio. We compute $L$ to be the following
$L = \frac{1}{l} + \frac{1}{\lambda^2} + \frac{1}{\lambda^3} + \ldots$
But this is just a geometric series with $a = 1$ and $r = 1/\lambda$. Using the infinite sum of geometric series formula, we get:
$L = \frac{1}{1 - r} = \frac{1}{1 - \frac{1}{\lambda}}$
which in the case of $\lambda = 1.5$, 3x leverage.
The above calculation accounts for the maximum leverage in an infinitely liquid ETH/DAI market with an instant ability to fund a CDP. However, there are several factors that come into play:
• Finite liquidity on ETH/DAI
• CDP funding and trade delays
The amount of ETH one can buy with Dai is currently very limited, and the spread is significantly wide. The above calculations make the assumption that one can continue to buy more ETH at the market
price, which is very far from the truth.
However, this liquidity will improve over time, as the future of the Maker system requires this liquidity to exist for its auto-liquidation system to work.
There are several actions one must perform to increase the collateral locked in the CDP contract, all taking one block each:
1. Generating Dai from the CDP
2. Buying W-ETH from the WETH/DAI market
3. Converting W-ETH to PETH
4. Locking the newly generated PETH into the CDP.
One should note that step 3 will not exist in the final version of Maker -- pooled Ether is a workaround to the MKR token generation system not being in place. However, this means that at a minimum,
three blocks will pass before the CDP can be refunded.
If one generates the maximum amount of Dai from the CDP, they are at immediate risk of liquidation if the price of Ether drops even one cent. Since the process of refunding the CDP is not atomic,
there is a very real risk of this liquidation taking place.
In theory, one could write a smart contract that performs this entire 4-step process in one transaction with a specified minimum accepted price. Before this exists, however, one is subject to the
aformentioned risks.
(If you liked this post, join our crypto Discord!) | {"url":"https://ianm.com/posts/2018-01-25-maximum-leverage-on-maker","timestamp":"2024-11-06T01:26:19Z","content_type":"text/html","content_length":"146974","record_id":"<urn:uuid:85ee48d0-0ce1-40b0-8196-9079e0e23232>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00500.warc.gz"} |
The Properties of a Rectangle | Quizalize
Definitions and formulas for the radius of a circle, the diameter of a circle, the circumference (perimeter) of a circle, the area of a circle, the chord of a circle, arc and the arc length of a
circle, sector and the area of the sector of a circle
Tag the questions with any skills you have. Your dashboard will track each student's mastery of each skill. | {"url":"https://resources.quizalize.com/view/web_link/the-properties-of-a-rectangle-7084fc5b-20bb-4525-9c81-b38bf3723130","timestamp":"2024-11-02T08:38:49Z","content_type":"text/html","content_length":"37147","record_id":"<urn:uuid:31e817cb-e1d1-43ee-9a68-5e5067c8f70f>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00035.warc.gz"} |
Basic Machine Learning
Backpropagation is the standard Machine Learning algorithm. From what I understand, It works like this:
1. Set all the weights of your neural network to random values.
2. Take a training example, put it through your neural network and try to predict what your neural network will make of it.
3. Compare what your neural network’s output is with the output that you want (the correct output).
4. Adjust your weights.
5. Keep doing this over and over again until your network gets the output that you want. Once this happens, your neural network has learned.
This is called Supervised Learning because you can check if your network is correct or not by comparing the results of your network to your training examples.
The weights are adjusted using the Gradient Decent Algorithm. This is how the weights and biases are tweaked. The Gradient Decent is summarized well here. It finds the local minimum of a function.
This is when the slope of the function is zero. (In other words, when the derivative of the function is equal to 0.) The problem is you can only efficiently find the derivative of a function that has
a few variables. Neural Networks can get very big, sometimes having millions of variables. So computing the the derivatives of these types of functions is computationally intensive. That is why the
Gradient Decent Algorithm is used for huge networks. The next step for me is to write a program for the Gradient Decent Algorithm and maybe also a program that can compute general derivatives.
Summary: Backpropagation finds the error that your weights have by comparing your neural network’s output to the output that you want you network to have and the Gradient Decent Algorithm is used to
adjust these weights.
Discuss on Github
Michael A. Nielsen, “Neural Networks and Deep Learning”
Trask, Andrew. "A Neural Network in 13 Lines of Python (Part 2 - Gradient Descent)." - I Am Trask. N.p., 27 July 2015. Web. 23 Apr. 2016. | {"url":"https://vtomole.com/blog/2016/04/23/basicml","timestamp":"2024-11-12T02:04:21Z","content_type":"text/html","content_length":"5254","record_id":"<urn:uuid:ac7c2351-8e17-4648-8512-4c304e4fc192>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00179.warc.gz"} |
Liters to Tons / Tons to Liters Converter
Liters to Tons / Tons to Liters
Disclaimer: Whilst every effort has been made in building our calculator tools, we are not to be held liable for any damages or monetary losses arising out of or in connection with their use. Full
How to convert liters to tons
The ton is a unit of mass/weight and the liter is a unit of volume. To convert between the two, you need to know the density of the substance that you are trying to convert.
To find out the mass in tons, multiply your volume in liters by the density of the substance (in tons per liter). Mass = Density × Volume. It's important to ensure that the density figure has been
converted to liters per ton first.
We have an article and calculator devoted to converting volume and weight.
Liters to tons for common substances
The figures below use 'substance density' figures from Engineering Toolbox and SI Metric and should be used as a rough guide only.
Substance (1 liter) Density Estimate US tons Metric tonnes
Crude Oil (32.6 C) (1L) 862 kg/m^3 0.00095 tons 0.00086 metric tons
Oil (petroleum) (1L) 881 kg/m^3 0.00097 tons 0.00088 metric tons
Water (1L) 1000 kg/m^3 0.0011 tons 0.001 metric tons
Petrol (1L) 737 kg/m^3 0.00081 tons 0.00074 metric tons
Topsoil (1L) 1600 kg/m^3 0.0018 tons 0.0016 metric tons
Note that you can convert between cubic yards and tons here.
If you have any suggestions or queries with this liters and tons conversion tool, please contact me. | {"url":"https://www.thecalculatorsite.com/conversions/common/liters-to-metric-tons.php","timestamp":"2024-11-08T20:20:06Z","content_type":"text/html","content_length":"335284","record_id":"<urn:uuid:b46cd911-b8d5-49ea-b728-d4871527cd60>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00318.warc.gz"} |
Conference announcement: PDE & Probability in interaction: functional inequalities, optimal transport and particle systems
Together with Pierre Monmarché (Sorbonne Université), Julien Reygner (École des Ponts ParisTech), and Marielle Simon (Université de Lille), we are delighted to announce the upcoming workshop “PDE &
Probability in interaction: functional inequalities, optimal transport and particle systems”.
The event will be held from January 22 to 26, 2024, at CIRM in Marseille.
Registrations are now open on the website: https://conferences.cirm-math.fr/2988.html
This workshop will also feature two courses, delivered by J. LEHEC (Université de Poitiers) and M. GOLDMAN (Université Paris Cité) on functional inequalities in high dimensions and random matching
problems, respectively.
Invited talks will be given by:
Nathalie Ayi (Sorbonne Université)
Roland Bauerschmidt* (University of Cambridge)
Maria Bruna (University of Cambridge)
Kleber Carrapotoso (École Polytechnique, Palaiseau)
Giovanni Conforti (École Polytechnique, Palaiseau)
Alex Delalande (Lagrange Center, Paris)
François Delarue (Université Côte d’Azur)
Alex Dunlap (NYU Courant)
Rishabh Gvalani (MPI MIS Leipzig)
Martin Huesmann (University of Münster)
Jean-Francois Mehdi Jabir (HSE Moscow)
Jasper Hoeksema (TU Eindhoven)
Bo’az Klartag (Weizmann Institute of Science)
Daniel Lacker (Columbia University)
Jean-Christophe Mourrat (ENS Lyon)
Emanuela Radici (University of L’Aquila)
Milica Tomasevic (École Polytechnique, Palaiseau)
Dario Trevisan (Pisa University)
Isabelle Tristani (ENS Paris)
Haava Yoldas (TU Delft)
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LSMEANS Statement
LSMEANS <model-effects> </ options> ;
The LSMEANS statement computes and compares least squares means (LS-means) of fixed effects. LS-means are predicted population margins—that is, they estimate the marginal means over a balanced
population. In a sense, LS-means are to unbalanced designs as class and subclass arithmetic means are to balanced designs.
Table 54.6 summarizes the options available in the LSMEANS statement.
Table 54.6: LSMEANS Statement Options
Option Description
Construction and Computation of LS-Means
AT Modifies the covariate value in computing LS-means
BYLEVEL Computes separate margins
DIFF Requests differences of LS-means
OM= Specifies the weighting scheme for LS-means computation as determined by the input data set
SINGULAR= Tunes estimability checking
Degrees of Freedom and p-values
ADJUST= Determines the method for multiple-comparison adjustment of LS-means differences
ALPHA= Determines the confidence level ()
STEPDOWN Adjusts multiple-comparison p-values further in a step-down fashion
Statistical Output
CL Constructs confidence limits for means and mean differences
CORR Displays the correlation matrix of LS-means
COV Displays the covariance matrix of LS-means
E Prints the matrix
LINES Produces a “Lines” display for pairwise LS-means differences
MEANS Prints the LS-means
PLOTS= Requests graphs of means and mean comparisons
SEED= Specifies the seed for computations that depend on random numbers
Generalized Linear Modeling
EXP Exponentiates and displays estimates of LS-means or LS-means differences
ILINK Computes and displays estimates and standard errors of LS-means (but not differences) on the inverse linked scale
ODDSRATIO Reports (simple) differences of least squares means in terms of odds ratios if permitted by the link function
For details about the syntax of the LSMEANS statement, see the section LSMEANS Statement in Chapter 19: Shared Concepts and Topics.
Note: If you have classification variables in your model, then the LSMEANS statement is allowed only if you also specify the PARAM=GLM option. | {"url":"http://support.sas.com/documentation/cdl/en/statug/65328/HTML/default/statug_logistic_syntax20.htm","timestamp":"2024-11-09T00:50:39Z","content_type":"application/xhtml+xml","content_length":"30206","record_id":"<urn:uuid:648318bd-ef88-40a1-9cf3-e5e5a2f6f673>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00472.warc.gz"} |
Robust control toolbox
augment - augmented plant
bstap - hankel approximant
ccontrg - central H-infinity controller
colinout - inner-outer factorization
copfac - right coprime factorization
dcf - double coprime factorization
des2ss - descriptor to state-space
dhinf - H_infinity design of discrete-time systems
dhnorm - discrete H-infinity norm
dtsi - stable anti-stable decomposition
fourplan - augmented plant to four plants
fspecg - stable factorization
fstabst - Youla's parametrization
gamitg - H-infinity gamma iterations
gcare - control Riccati equation
gfare - filter Riccati equation
gtild - tilde operation
h2norm - H2 norm
h_cl - closed loop matrix
h_inf - H-infinity (central) controller
h_inf_st - static H_infinity problem
h_norm - H-infinity norm
hankelsv - Hankel singular values
hinf - H_infinity design of continuous-time systems
lcf - normalized coprime factorization
leqr - H-infinity LQ gain (full state)
lft - linear fractional transformation
linf - infinity norm
linfn - infinity norm
lqg_ltr - LQG with loop transform recovery
macglov - Mac Farlane Glover problem
mucomp - mu (structured singular value) calculation
nehari - Nehari approximant
parrot - Parrot's problem
ric_desc - Riccati equation
riccati - Riccati equation
rowinout - inner-outer factorization
sensi - sensitivity functions
tf2des - transfer function to descriptor | {"url":"http://laris.fesb.hr/digitalno_vodjenje/download/scilab/robust/whatis.htm","timestamp":"2024-11-10T09:31:23Z","content_type":"text/html","content_length":"3027","record_id":"<urn:uuid:cdc2562f-38c8-4e69-91d5-dd17d0e1d06f>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00606.warc.gz"} |
Can You Make 500% Return in Options in One Day?Can You Make 500% Return in Options in One Day? - Stock Investing Guide
If you don’t know by now, there are two types of options available to you. There are put options which give you the right to sell a stock at some price YOU choose. And then there are call options
which give you the right to buy shares of a stock at some price YOU choose.
The price of a call or put option is made up of two components.
These are the time value and the intrinsic value. The time value is based on how much time is left before the option expires. Basically, the more time left before expiration, the higher the time
value will be for that option. For example the option with 1 month left to expiration will be significantly cheaper than the option with 3 months left to expiration. The intrinsic value is the value
relative to the stock price. I won’t get into too many details on how this part works but just know that the strike price relative to the actual price of the underlying stock will affect the
intrinsic value of the option.
Option and stocks relation
The price of a call option on a stock has a direct relationship to the price of the stock itself. What I mean by this is that if the stock moves up, the call option will increase in value. If the
stock moves down, the call option will decrease in value. The price of a put option has an inverse (opposite) relationship to the price of the underlying stock.
What I mean by this is that If the price of the stock goes up, the price of the put option will go down. Finally, if the price of the stock goes down, the price of the put option will go up. If you
think a stock is going to tank, then you definitely want to rack up on some put options on that stock. If you think the stock price is going to soar, load up on some calls.
By being an options buyer you can participate in price movements on a stock for a fraction of what is would cost to own the stock outright. In the US, one options contract controls 100 shares of the
underlying stock. So, if I own one call option on stock xyz at the $25 strike price that means that I can purchase 100 shares of stock xyz at anytime (up until the option expires) for $25.
With one month to expiration the cost of the call option may be about $0.40, which means owning one call option would cost $40.00 (0.40 x 100 shares) plus commission. For the sake of comparison, to
buy 100 shares of stock xyz, I would need to pony up $2500 plus commission. So, here you begin to see the power of leverage with options trading.
Let’s look at another scenario.
Let’s say stock xyz is trading at $25 and you think the stock is in position to make a move upward. Now let’s say you don’t have much money and you only had $400 to invest. You could use $400 to buy
10 call options at the $25 strike price ($0.4 premium x 10 call contracts x 100 shares per contract) with one month to expiration.
Now remember that the call option will move in synch with the stock price. Let’s say company xyz has stellar earnings and the stock goes up to $28, which is $3 above your strike price of $25. I’m bad
with numbers, but Your call option now has $3 intrinsic value (value relative to the stock price) in addition to the time value of $0.4. So the total premium price of the option should be “at least”
$3.4 per contract.
But let’s be conservative and say the option is only worth $3. That is still a profit of $2.6 per contract which is $2600 on your $400 investment. So ($3.00 – $0.4)/$0.4 x 100% = 650% profit on that
one trade!!! This is a real life example using somewhat conservative numbers. Consider stocks like Apple and Google that regularly have price movements of as much as $25 in one trading day!
If you thought the stock was going to go down, you would apply the same strategy above, but with put options instead of call options. If the stock dropped by $3 to $22 you would have the same
results. So you can speculate in call and put options to make money whether you think the stock is going to go up or down.
As an options buyer the odds are against you as typically 80% of options buyers lose money. I think greed and inexperience have a lot to do with this. I have bought both put options and call options
for over 10 years and know firsthand that triple digit gains in a matter of hours are very realistic. The key is timing and knowing how to choose the right stock. | {"url":"https://stock-investing-guide.com/can-you-make-500-return-in-options-in-one-day/","timestamp":"2024-11-05T00:20:25Z","content_type":"text/html","content_length":"76616","record_id":"<urn:uuid:c13c8daf-6a30-4954-ab41-45dbb3a1a85b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00775.warc.gz"} |
Concept of Infinite Bus - Electrical Concepts
Concept of Infinite Bus
For understanding the concept of Infinite Bus, we will take some examples. Suppose we take an isolated synchronous generator of capacity 100 kW and supplying a load of, say 50 kW, at 50 Hz and rated
voltage, say 400 V.
Now if we add a load of 20 kW what will happen?
The frequency (speed) and terminal voltage will reduce instantly say 49 Hz and 390 V, before you correct them. Instead if we add only 10 kW the result will be same, with frequency and voltage of 49.5
Hz and 395 V. Next time, if we keep reducing the applied load….
With say 1 kW addition frequency and voltage may be 49.95 Hz and 299.5 V
If we now add 0.1 kW (100 W) instead there may not be any measurable change in frequency and voltage. So we can say that for a load up to 100 W the 100 kW machine is an infinite bus.
This example is used to illustrate the concept of an infinite bus. All the numerical values for frequency and voltage used above are not calculated but assumed for illustration only.
In our day to day routine, in cities, if we switch on a 60 W bulb or a tube light you don’t notice anything electrically abnormal. But if you switch on an Air Conditioner (without a Voltage
Regulator) you can clearly notice a sudden voltage dip which recovers.
Theoretically an infinite bus is one where the frequency and voltage remain constant irrespective of the amount of load on it.
An infinite bus, to satisfy the above constraints, is represented by an equivalent generator having infinite moment of Inertia, M so that there will be no change in speed for any load addition and
zero synchronous reactance, Xs so that there is no voltage drop for any load current and V = E, Induced generator emf.
In short infinite bus has number of generator connected to it and that means it has infinite active and reactive power capabilities. It can maintain its terminal voltage, frequency and any loading
3 thoughts on “Concept of Infinite Bus”
1. What is active and reactive power
□ You can read Active, Reactive and Apparent power.
2. Where is infinite bus located in power system generation?
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Tanya Khovanova's Math Blog
Here is another riddle I discovered in a book and gave as homework to my students.
Puzzle. I can use the number 20 thrice to make 60: 20 + 20 + 20 = 60. Make it 60 again by using a different number three times.
The book’s answer was to use 5: 55 + 5 = 60.
My students were very inventive. All of them solved the puzzle, but only one out of ten students came up with the book’s answer.
• For most of them, the new number was 60, as in 60 + 60−60 = 60, or 60*60/60.
• Some used −60 or 1/60, as in −60 − (−60) − (−60) = 60, or ((1/60)/(1/60))/(1/60) = 60. Similarly, some multiplied the cube root of 60 three times.
• One student used 59 in a clever way, as in 59 + 59/59 = 60.
• Another student said the following. If you turn 60 upside-down, you will make 09, and now you can use the number 3 thrice: 3+3+3 = 09.
• And the last on my list is the student who said that 42 has to be the answer to the universe and everything. He summed up two instances of 42 to get 84 and then subtracted the third instance of
42 with the digits flipped to get 84 − 24 = 60.
Here is a new report of interesting homework solutions from my students.
Puzzle. One day, two sisters decided to clean the old shed at the bottom of their garden. When they finished cleaning, one had a dirty face and the other had a clean face. The sister with the
clean face went and washed her face, but the girl with the dirty face did not wash. Why should this be so?
The expected answer: The sister with the clean face saw her sister’s dirty face and assumed her own face must be dirty as well, so she washed it. The sister with the dirty face saw her sister’s clean
face and assumed her own face must also be clean, so she didn’t feel the need to wash.
Another student suggested a different but quite realistic answer.
The realistic answer: The sisters’ home ran out of water after the clean sister washed her face, preventing the dirty sister from washing her own.
The other student watched too many sitcoms.
The sitcom answer: The sister with the dirty face purposefully kept her face dirty, so she could show her parents that she did all the work, as she was the only one with dirt on her face.
I asked ChatGPT to solve the puzzle, and, unsurprisingly, it came up with the standard answer. I pushed and got the following.
The ChatGPT answer: The sister with the clean face washed up because she was an Instagram influencer and couldn’t risk being seen dirty, even in her own garden. Meanwhile, the sister with the dirty
face was a carefree adventurer who believed dirt was “nature’s makeup.” Plus, she figured that if she waited long enough, the dirt would either blow away or blend into a trendy new skincare
routine—”Exfoliation by Shed Dust.”
From time to time, the homework for my PRIMES STEP students includes questions that are not exactly mathematical. Last week, we had the following physics puzzle.
Puzzle. A fisherman needed to move a heavy iron thingy from one river’s shore to another. When he put the thingy in his boat, the boat lowered so much that it wasn’t safe to operate. What should
he do?
The expected answer: He should attach the thingy to the bottom of the boat. When the object is inside the boat, the boat needs to displace enough water to account for the entire weight of the boat
and the thingy. When the thingy is attached to the bottom of the boat, the thingy experiences its own buoyancy. Thus, the water level rises less because the thingy displaces some water directly,
reducing the boat’s need to displace extra water. Thus, the amount of weight the fisherman saves is equal to the amount of water that would fit into the shape of this thingy.
As usual, my students were more inventive. Here are some of their answers.
• The fisherman could cut the iron thingy and transport it piece by piece.
• He can swim across and drag the boat with a rope with the thingy inside.
• He can use a second boat to pull the first boat with the thingy in it.
• It is another river’s shore, so he can just take the iron with him to a different river without going over water.
• If the fisherman has extra boat material, heightening the boat’s walls would keep it from sinking.
Also, some funny answers.
• He could fast for a few days, making him lighter.
• He could tie helium balloons to the boat to keep it afloat even after he gets in.
• Wait until winter and slide the boat on ice.
And my favorite answer reminded me of a movie I recently re-watched.
• You’re gonna need a bigger boat.
The term fibonometry was coined by John Conway and Alex Ryba in their paper titled, you guessed it, “Fibonometry”. The term describes a freaky parallel between trigonometric formulas and formulas
with Fibonacci (F[n]) and Lucas (L[n]) numbers. For example, the formula sin(2a) = 2sin(a)cos(a) is very similar to the formula F[2n] = F[n]L[n]. The rule is simple: replace angles with indices,
replace sin with F (Fibonacci) and cosine with L (Lucas), and adjust coefficients according to some other rule, which is not too complicated, but I am too lazy to reproduce it. For example, the
Pythegorian identity sin^2a + cos^2a = 1 corresponds to the famous identity L[n]^2 – 5F[n]^2 = 4(-1)^n.
My last year’s PRIMES STEP senior group, students in grades 7 to 9, decided to generalize fibonometry to general Lucas sequences for their research. When the paper was almost ready, we discovered
that this generalization is known. Our paper was well-written, and we decided to rearrange it as an expository paper, Fibonometry and Beyond. We posted it at the arXiv and submitted it to a journal.
I hope the journal likes it too.
Consider the following Fibonacci trick. Ask your friends to choose any two integers, a and b, and then, starting with a and b, ask them to write down 10 terms of a Fibonacci-like sequence by summing
up the previous two terms. To start, the next (third) term will be a+b, followed by a+2b. Before your friends even finish, shout out the sum of the ten terms, impressing them with your lightning-fast
addition skills. The secret is that the seventh term is 5a+8b, and the sum of the ten terms is 55a+88b. Thus, to calculate the sum, you just need to multiply the 7th term of their sequence by 11.
If you remember, I run a program for students in grades 7 through 9 called PRIMES STEP, where we do research in mathematics. Last year, my STEP senior group decided to generalize the Fibonacci trick
for their research and were able to extend it. If n=4k+2, then the sum of the first n terms of any Fibonacci-like sequence is divisible by the term number 2k+3, and the result of this division is the
Lucas number with index 2k+1. For example, the sum of the first 10 terms is the 7th term times 11. Wait, this is the original trick. Okay, something else: the sum of the first 6 terms is the 5th term
times 4. For a more difficult example, the sum of the first 14 terms of a Fibonacci-like sequence is the 9th term times 29.
My students decided to look at the sum of the first n Fibonacci numbers and find the largest Fibonacci number that divides the sum. We know that the sum of the first n Fibonacci numbers is F[n+2] –
1. Finding a Fibonacci number that divides the sum is easy. There are tons of cute formulas to help. For example, we have a famous inequality F[4k+3] – 1 = F[2k+2]L[2k+1]. Thus, the sum of the first
4k+1 Fibonacci numbers is divisible by F[2k+2]. The difficult part was to prove that this was the largest Fibonacci number that divides the sum. My students found the largest Fibonacci number that
divides the sum of the first n Fibonacci numbers for any n. Then, they showed that the divisibility can be extended to any Fibonacci-like sequence if and only if n = 3 or n has remainder 2 when
divided by 4. The case of n=3 is trivial; the rest corresponds to the abovementioned trick.
They also studied other Lucas sequences. For example, they showed that a common trick for all Jacobsthal-like sequences does not exist. However, there is a trick for Pell-like sequences: the sum of
the first 4k terms (starting from index 1) of such a sequence is the (2k + 1)st term times 2P[2k], where P[n] denotes an nth Pell number.
You can check out all the tricks in our paper Fibonacci Partial Sums Tricks posted at the arXiv.
The famous 5-card trick begins with the audience choosing 5 cards from a standard deck. The magician’s assistant then hides one of the chosen cards and arranges the remaining four cards in a row,
face up. Upon entering the room, the magician can deduce the hidden card by inspecting the arrangement. To eliminate the possibility of any secret signals between the assistant and the magician, the
magician doesn’t even have to enter the room — an audience member read out the row of cards.
The trick was introduced by Fitch Cheney in 1950. Here is the strategy. With five cards, you are guaranteed to have at least two of the same suit. Suppose this suit is spades. The assistant then
hides one of the spades and starts the row with the other one, thus signaling that the suit of the hidden card is spades. Now, the assistant needs to signal the value of the card. The assistant has
three other cards than can be arranged in 6 different ways. So, the magician and the assistant can agree on how to signal any number from 1 to 6. This is not enough to signal any random card.
But wait! There is another beautiful idea in this strategy — the assistant can choose which spade to hide. Suppose the two spades have values X and Y. We can assume that these are distinct numbers
from 1 to 13. Suppose, for example, Y = X+5. In that case, the assistant hides card Y and signals the number 5, meaning that the magician needs to add 5 to the value of the leftmost card X. To ensure
that this method always works, we assume that the cards’ values wrap around. For example, king (number 13) plus 1 is ace. You can check that given any two spades, we can always find one that is at
most 6 away from the other. Say, the assistant gets a queen of spades and a 3 of spades. The 3 of spades is 4 away from the queen (king, ace, two, three). So the assistant would hide the 3 and use
the remaining three cards to signal the number 4.
I skipped some details about how permutations of three cards correspond to numbers. But it doesn’t matter — the assistant and the magician just need to agree on some correspondence. Magically, the
standard deck of cards is the largest deck with which one can perform this trick with the above strategy.
Later, a more advanced strategy for the same trick was introduced by Michael Kleber in his paper The Best Card Trick. The new strategy allows the magician and the assistant to perform this trick with
a much larger deck, namely a deck of 124 cards. But this particular essay is not about the best strategy, it is about the Cheney strategy. So I won’t discuss the advanced strategy, but I will
redirect you to my essay The 5-Card Trick and Information, jointly with Alexey Radul.
63 years later, the 4-card trick appeared in Colm Mulcahy’s book Mathematical Card Magic: Fifty-Two New Effects. Here the audience chooses not 5 but 4 cards from the standard deck and gives them to
the magician’s assistant. The assistant hides one of them and arranges the rest in a row. Unlike in the 5-card trick, in the 4-card trick, the assistant is allowed to put some cards face down. As
before, the magician uses the description of how the cards are placed in a row to guess the hidden card.
The strategy for this trick is similar to Cheney’s strategy. First, we assign one particular card that the magician would guess if all the cards are face down. We now can assume that the deck
consists of 51 cards and at least one of the cards in the row is face up. We can imagine that our 51-card deck consists of three suits with 17 cards in each suit. Then, the assistant is guaranteed to
receive at least two cards of the same imaginary suit. Similar to Cheney’s strategy, the leftmost face-up card will signal the imaginary suit, and the rest of the cards will signal a number. I will
leave it to the reader to check that signaling a number from 1 to 8 is possible. Similar to Cheney’s strategy, the assistant has an extra choice: which card of the two cards of the same imaginary
suit to hide. As before, the assistant chooses to hide the card so that the value of the hidden card is not more than the value of the leftmost face-up card plus 8. It follows that the maximum number
of cards the imaginary suit can have is 17. Magically, the largest possible deck size for performing this trick is 52, the standard deck of cards.
Last academic year, my PRIMES STEP junior group decided to dive deeper into these tricks. We invented many new tricks and calculated their maximum deck sizes. Our cutest trick is a 3-card trick. It
is similar to both the 5-card trick and the 4-card trick. In our trick, the audience chooses not 5, not 4, but 3 cards from the standard deck and gives them to the magician’s assistant. The assistant
hides one of them and arranges the other two in a row. The assistant is allowed to put some cards face down, as in the 4-card trick, and, on top of that, is also allowed to rotate the cards in two
ways: by putting each card vertically or horizontally.
We calculated the maximum deck size for the 3-card trick, which is not 52, as for the 5- and 4-card trick, but rather 54. Still, this means the 3-card trick can be performed with the standard deck.
The details of this trick and other tricks, as well as some theory, can be found in our paper Card Tricks and Information.
The homework I give to my students (who are in 6th through 9th grades) often starts with a math joke related to the topic. Once, I decided to let them be the comedians. One of the homework questions
was to invent a math joke. Here are some of their creations. Two of my students decided to restrict themselves to the topic we studied that week: sorting algorithms. The algorithm jokes are at the
* * *
A binary integer asked if I could help to double its value for a special occasion. I thought it might want a lot of space, but it only needed a bit.
* * *
Everyone envies the circle. It is well-rounded and highly educated: after all, it has 360 degrees.
* * *
Why did Bob start dating a triangle? It was acute one.
* * *
Why is Bob scared of the square root of 2? Because he has irrational fears.
* * *
Are you my multiplicative inverse? Because together, we are one.
* * *
How do you know the number line is the most popular?
It has everyone’s number.
* * *
A study from MIT found that the top 100 richest people on Earth all own private jets and yachts. Therefore, if you want to be one of the richest people on Earth, you should first buy a private jet
and yacht.
* * *
Why did the geometry student not use a graphing calculator? Because the cos was too high.
* * *
Which sorting algorithm rises above others when done underwater? Bubble sort!
* * *
Which sorting algorithm is the most relaxing? The bubble bath sort.
There are a lot of puzzles where you need to guess something asking only yes-or-no questions. In this puzzle, there are not two but three possible answers.
Puzzle. Mike thought of one of three numbers: 1, 2, or 3. He is allowed to answer “Yes”, “No”, or “I don’t know”. Can Pete guess the number in one question?
Yes, he can. This problem was in one of my homeworks, and my students had a lot of ideas. Here is the first list were ideas are similar to each other.
• I am thinking of an odd number. Is my number divisible by your number?
• If I were to choose 1 or 2, would your number be bigger than mine?
• If I were to pick a number from the set {1,2,3} that is different from yours, would my number be greater than yours?
• If I have a machine that takes numbers and does nothing to them except have a 50 percent chance of changing a two to a one. Would your number, after going through the machine, be one?
• If I were to choose a number between 1.5 and 2.5, would my number be greater than yours?
• If your number is x and I flip a fair coin x times, will there be at least two times when I flip the same thing?
• I am thinking of a comparison operation that is either “greater” or “greater or equal”. Does your number compare in this way to two?
One student was straightforward.
• Mike, please, do me a favor by responding ‘yes’ to this question if you are thinking about 1, ‘no’ if you are thinking about 2, and ‘I don’t know’ if you are thinking about 3?
One student used a famous unsolved problem: It is not known whether an odd perfect number exists.
• Is every perfect number divisible by your number?
Then, I gave this to my grandchildren, and they decided to answer in a form of a puzzle. Payback time.
• I’m thinking of a number too, and I don’t know whether it’s double yours. Is the sum of our numbers prime?
Here is the homework problem I gave to my PRIMES STEP students.
Puzzle. A man called his wife from the office to say that he would be home at around eight o’clock. He got in at two minutes past eight. His wife was extremely angry at this lateness. Why?
The expected answer is that she thought he would be home at 8 in the evening, while he arrived at 8 in the morning. However, my students had more ideas.
For example, one student extended the time frame.
• The man was one year late.
Another student found the words “got in” ambiguous.
• He didn’t get into his house two minutes past eight. He got into his car.
A student realized that the puzzle never directly stated why she got angry.
• The wife already got angry when he said he would be home around eight, as she needed him home earlier.
The students found alternative meanings to “called his wife from the office” and “minutes.”
• He had an office wife whom he called. But the wife at home was a different wife, and she was angry.
• “Two minutes past eight” could be a latitude.
One of the perks of being a teacher is receiving congratulations not only from family and friends, but also from students. By the way, I do not like physical gifts — I prefer just congratulations.
Luckily, MIT has a policy that doesn’t allow accepting gifts of any monetary value from minors and their parents.
Thus, my students are limited to emails and greeting cards.
One of my former students, Evin Liang, got really creative. He programmed the Game of Life to generate a Christmas card for me. You can see it for yourself on YouTube at: Conway Game of Life by Evin
This is one of my favorite congratulations ever.
Clean and Dirty Sisters
Help the Fisherman
Fibonacci Tricks
The 5-Card Trick, the 4-Card Trick, and the 3-Card Trick
My Students’ Jokes
Guess the Number in One Question
The Angry Wife
My Unique Christmas Card | {"url":"https://blog.tanyakhovanova.com/category/math-education/","timestamp":"2024-11-10T16:01:46Z","content_type":"text/html","content_length":"135435","record_id":"<urn:uuid:ebb12998-a9f8-4766-99d8-2887d51e0ace>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00655.warc.gz"} |
How to Check if two Rectangles Overlap in Java? Collision Detection Solution and Example
Can you write a
Java program to check if two rectangles are overlapping with each other or not?
is one of the frequently asked coding questions on tech giants like Facebook, Amazon, Microsoft, and others. This is also a kind of problem where it's easy to find opposite or negative conditions
when rectangles are not overlapping
and then inverting the result to prove that rectangles are colliding with each other. I first heard about this problem from one of my friends who were in the
Android game development
space. He was asked to write an algorithm to find if given rectangles are intersecting or not.
He was given the choice to implement the algorithm in any programming languages
or even pseudo-code was fine, so don't worry about language here, it's more about algorithm and logic.
He is a genius so he came successful from that interview, impressing the interviewer, talking about the efficiency and accuracy of the algorithm as well.
Later when I catch-up with him, he told me about his interview experience and said that this question is quite popular in the game programming domain and companies like Electronic Arts and Sony.
Why? because this is also used as a
collision detection algorithm
, not very sophisticated but it is a quite popular collision detection algorithm for many arcade games like Super Mario Bros, Pac-Ma n, Donkey Kong, and Breakout.
Since you can represent the characters and enemy as the rectangle you can find out when an arrow hit the character or when he got one up by checking two rectangles are intersecting with each other.
This looks like a very interesting problem for me and so I decided to take a look at this one.
Before going for a programming/coding interview, It's absolutely necessary to do as much practice in data structure and algorithms as possible to take advantage of all the knowledge available. You
can also join a comprehensive Data Structure and Algorithms course like
Data Structures and Algorithms: Deep Dive Using Java
on Udemy to fill the gaps in your understanding.
How to check if two Rectangle Overlaps in Java?
The algorithm to check if two rectangles are overlapping or not is very straight forward but before that, you need to know how you can represent a rectangle in Java programs? Well, a rectangle can be
represented by two coordinates, top left, and bottom right.
As part of the problem, you will be given four coordinates L1, R1, and L2, R2, top left and bottom right coordinate of two rectangles and you need to write a function
which should return true if rectangles are overlapping or false if they are not.
Btw, if you are interested in learning algorithms, I would suggest a brand new book,
Grooking Algorithm by Aditya Bhargava
. I was reading this book in the last couple of weekends and I have thoroughly enjoyed it.
Algorithm to check if rectangles are overlapping
Two rectangles A and B will not overlap or intersect with each other if one of the following four conditions is true.
1. The left edge of A is to the right of the right edge of B. In this case, the first rectangle A is completely on the right side of second rectangle B as shown in the following diagram
2. right edge of A is to the left of the left edge of B. In this case, the first rectangle A is completely on the left side of second rectangle B, as shown below
3. Top edge of A is below the bottom edge of B. In this case, the first rectangle A is completely under second rectangle B as shown in the following diagram
4. Bottom edge of A is above the top edge of B. In this case, the first rectangle A is completely above second rectangle B as shown in the following diagram
If any of the above four conditions are not true then two rectangles are overlapping with each other, as in the following diagram the first condition is violated, hence rectangle A intersects
rectangle B. If you are interested in Game programming, I also suggest you take a look at
Game programming patterns
by Robert Nystrom, a very interesting book to learn patterns with real-world examples from games.
Java Program to check if two rectangles are intersecting
In this program, I have followed standard Java coding practices to solve this problem. I have encapsulated two coordinates in a Point class and have Rectangle has two Point instance variables and an
instance method like
to check if another rectangle is overlapping with it or not.
The logic to check if two rectangles are overlapping or colliding or not is coded in the isOverlapping() method which is as per the approach discussed in the solution section.
* Java Program to check if two rectangles is intersecting with each
* other. This algorithm is also used as collision detection
* algorithm in sprite-based arcade games e.g. Supre Mario Bros
public class Main {
public static void main(String[] args) {
Point l1 = new Point(0, 10);
Point r1 = new Point(10, 0);
Point l2 = new Point(5, 5);
Point r2 = new Point(15, 0);
Rectangle first = new Rectangle(l1, r1);
Rectangle second = new Rectangle(l2, r2);
if (first.isOverLapping(second)) {
.println("Yes, two rectangles are intersecting with each other");
} else {
.println("No, two rectangles are not overlapping with each other");
class Point {
int x;
int y;
public Point(int x, int y) {
this.x = x;
this.y = y;
class Rectangle {
private final Point topLeft;
private final Point bottomRight;
public Rectangle(Point topLeft, Point bottomRight) {
this.topLeft = topLeft;
this.bottomRight = bottomRight;
* Java method to check if two rectangle are intersecting to each other.
* @param other
* @return true if two rectangle overlap with each other
public boolean isOverLapping(Rectangle other) {
if (this.topLeft.x > other.bottomRight.x // R1 is right to R2
|| this.bottomRight.x < other.topLeft.x // R1 is left to R2
|| this.topLeft.y < other.bottomRight.y // R1 is above R2
|| this.bottomRight.y > other.topLeft.y) { // R1 is below R1
return false;
return true;
Yes, two rectangles are intersecting with each other
That's all about
how to check whether two rectangles are overlapping with each other or not
. It's one of the interesting coding questions to solve in interviews and trying to develop the algorithm to solve the problem is also a good choice. If you struggle to solve this kind of question
then you should join a comprehensive data structure and problem-solving course and practice some problems on Data Structure and Algorithms which I have shared below.
coding problems in Java
you may like
P. S. -
You can also try testing this algorithm with a different kind of rectangle to see if it works in all scenarios or not. For further reading, you can read
Introduction to Algorithm
or Game Programming patterns book.
And lastly one question for you? Which one is your favorite Java Programming exercise? Palindrome, Prime number, Fibonacci, Factorial or this one?
5 comments :
Interesting question, thanks for sharing with us.
This is also known as "rectnagle intersection problem" e.g. write a program to check if two rectangle intersect with each other.
A rectangle should be represented in class by coordinates of four points - top-left, top-right, bottom-left and bottom-right. Why so? Because not always the corresponding arms of two rectangles
be parallel.
if((this.topLeft.y < other.bottomLeft.y && this.topRight.y < other.bottomRight.y)
|| (this.bottomLeft.y > other.topLeft.y && this.bottomRight.y > other.bottomRight.y)
|| (this.topLeft.x > other.topRight.x && this.bottomLeft.x > other.bottomRight.x)
|| (this.topRight.x < other.topLeft.x && this.bottomRight.x < other.bottomLeft.x))
Okay ??
Not okay. The case when the other rectangle is rotated is more complicated. Your extended test, for example, will say the rectangles (-1,3.5)(1,3.5)(1,5.5)(-1,5.5) and (1,0)(5,4)(4,5)(0,1)
overlap while in fact they don't.
What if the rectangles share a border, but don't actually overlap? If the max X side of one rectangle is the min X side of the other rectangle your overlapping method will return true. Rectangles
only overlap if their overlapping area is > 0, and your method will return true when the overlapping area is 0. | {"url":"https://javarevisited.blogspot.com/2016/10/how-to-check-if-two-rectangle-overlap-in-java-algorithm.html?m=0","timestamp":"2024-11-09T09:33:52Z","content_type":"application/xhtml+xml","content_length":"134700","record_id":"<urn:uuid:fde38b5f-7441-4dd9-a3b2-7a75a8f993d1>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00404.warc.gz"} |
Discrete Groupssearch
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Softcover ISBN: 978-0-8218-2080-3
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Discrete Groups
Softcover ISBN: 978-0-8218-2080-3
Product Code: MMONO/207
List Price: $52.00
MAA Member Price: $46.80
AMS Member Price: $41.60
eBook ISBN: 978-1-4704-4632-1
Product Code: MMONO/207.E
List Price: $49.00
MAA Member Price: $44.10
AMS Member Price: $39.20
Softcover ISBN: 978-0-8218-2080-3
eBook ISBN: 978-1-4704-4632-1
Product Code: MMONO/207.B
List Price: $101.00 $76.50
MAA Member Price: $90.90 $68.85
AMS Member Price: $80.80 $61.20
• Translations of Mathematical Monographs
Iwanami Series in Modern Mathematics
Volume: 207; 2002; 193 pp
MSC: Primary 20; 57; 30; Secondary 46
This book deals with geometric and topological aspects of discrete groups. The main topics are hyperbolic groups due to Gromov, automatic group theory, invented and developed by Epstein, whose
subjects are groups that can be manipulated by computers, and Kleinian group theory, which enjoys the longest tradition and the richest contents within the theory of discrete subgroups of Lie
What is common among these three classes of groups is that when seen as geometric objects, they have the properties of a negatively curved space rather than a positively curved space. As Kleinian
groups are groups acting on a hyperbolic space of constant negative curvature, the technique employed to study them is that of hyperbolic manifolds, typical examples of negatively curved
manifolds. Although hyperbolic groups in the sense of Gromov are much more general objects than Kleinian groups, one can apply for them arguments and techniques that are quite similar to those
used for Kleinian groups. Automatic groups are further general objects, including groups having properties of spaces of curvature 0. Still, relationships between automatic groups and hyperbolic
groups are examined here using ideas inspired by the study of hyperbolic manifolds. In all of these three topics, there is a “soul” of negative curvature upholding the theory. The volume would
make a fine textbook for a graduate-level course in discrete groups.
Graduate students and research mathematicians interested in topology and geometry.
□ Chapters
□ Basic notions for infinite group
□ Hyperbolic groups
□ Automatic groups
□ Kleinian groups
□ Prospects
□ Discrete Groups gives a straightforward and very readable introduction to three related topics. It manages to be both thorough and precise at the same time ... clear and complete proofs are
given of the results presented ... a handy reference ... would happily recommend it to a graduate student.
Bulletin of the LMS
• Permission – for use of book, eBook, or Journal content
• Book Details
• Table of Contents
• Reviews
• Requests
Iwanami Series in Modern Mathematics
Volume: 207; 2002; 193 pp
MSC: Primary 20; 57; 30; Secondary 46
This book deals with geometric and topological aspects of discrete groups. The main topics are hyperbolic groups due to Gromov, automatic group theory, invented and developed by Epstein, whose
subjects are groups that can be manipulated by computers, and Kleinian group theory, which enjoys the longest tradition and the richest contents within the theory of discrete subgroups of Lie groups.
What is common among these three classes of groups is that when seen as geometric objects, they have the properties of a negatively curved space rather than a positively curved space. As Kleinian
groups are groups acting on a hyperbolic space of constant negative curvature, the technique employed to study them is that of hyperbolic manifolds, typical examples of negatively curved manifolds.
Although hyperbolic groups in the sense of Gromov are much more general objects than Kleinian groups, one can apply for them arguments and techniques that are quite similar to those used for Kleinian
groups. Automatic groups are further general objects, including groups having properties of spaces of curvature 0. Still, relationships between automatic groups and hyperbolic groups are examined
here using ideas inspired by the study of hyperbolic manifolds. In all of these three topics, there is a “soul” of negative curvature upholding the theory. The volume would make a fine textbook for a
graduate-level course in discrete groups.
Graduate students and research mathematicians interested in topology and geometry.
• Chapters
• Basic notions for infinite group
• Hyperbolic groups
• Automatic groups
• Kleinian groups
• Prospects
• Discrete Groups gives a straightforward and very readable introduction to three related topics. It manages to be both thorough and precise at the same time ... clear and complete proofs are given
of the results presented ... a handy reference ... would happily recommend it to a graduate student.
Bulletin of the LMS
Permission – for use of book, eBook, or Journal content
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[Solved] Dams and Spillways MCQ [Free PDF] - Objective Question Answer for Dams and Spillways Quiz - Download Now!
Dams and Spillways MCQ Quiz - Objective Question with Answer for Dams and Spillways - Download Free PDF
Last updated on Sep 1, 2024
Latest Dams and Spillways MCQ Objective Questions
Dams and Spillways Question 1:
The maximum permissible eccentricity for no tension at the base of a gravity dam is
Answer (Detailed Solution Below)
Option 4 : B/6
Dams and Spillways Question 1 Detailed Solution
An elementary profile of a gravity dam is shown:
Let the resultant pass through any point at a distance e from the centre of CB. Hence ‘e’ is the eccentricity.
The components of R are shown as above.
\(\therefore {{\rm{\sigma }}_{\rm{c}}} = \frac{{{{\rm{R}}_{\rm{v}}}}}{{{\rm{b}} \times 1}} - \frac{{\left( {{{\rm{R}}_{\rm{v}}} \times {\rm{e}}} \right) \times \left( {\frac{{\rm{b}}}{2}} \right)}}
{{1 \times \frac{{{{\rm{b}}^3}}}{{12}}}}\;\)
For no tension at base,
σc ≥ 0
\( \Rightarrow \frac{{{{\rm{R}}_{\rm{v}}}}}{{{\rm{b}} \times 1}} \ge \frac{{\left( {{{\rm{R}}_{\rm{v}}} \times {\rm{e}}} \right)\left( {{\rm{b}}/2} \right)}}{{\left( {{{\rm{b}}^3}/12} \right)}}\)
⇒ e ≤ b/6
Hence when the resultant force lies in the middle-one-third region, there is no tension anywhere at the base of the elementary profile of a gravity dam.
Important Points
• Gravity dam is made of concrete or masonry and resists the water force using the weight of the dam only.
• An elementary profile of a gravity dam is a triangular section having zero width at water level and maximum base width where hydrostatic pressure is maximum
• The resultant force R mainly comprises the water force and weight of the dam.
Dams and Spillways Question 2:
The elementary profile of a dam is
Answer (Detailed Solution Below)
Option 4 : a right - angled triangle
Dams and Spillways Question 2 Detailed Solution
The elementary profile of a dam
• An elementary profile of a gravity dam is the theoretical shape of its cross-section when it is subjected to only three main forces i.e.Self-weight, Water pressure, and Uplift pressure.
• Moreover, the elementary profile has zero top width and no freeboard.
• The right-angle triangle is the most suitable section for the theoretical profile.
• The elementary profile is hypothetical because an actual gravity dam has some minimum top width and freeboard, and it will also be subjected to forces other than the three main forces considered
in the elementary profile.
Dams and Spillways Question 3:
When the reservoir is full, the maximum compressive force in a gravity dam is produced
Answer (Detailed Solution Below)
Option 1 : at the toe
Dams and Spillways Question 3 Detailed Solution
The maximum compressive stresses occur at the heel (mostly during reservoir empty condition) or at the toe (at reservoir full condition) and on planes normal to the face of the dam.
∴ For reservoir empty condition maximum compressive force will be at the heel.
∴ For reservoir full condition maximum compressive force will be at the toe of the dam.
The figure below is the pressure distribution diagram for reservoir full condition:
Dams and Spillways Question 4:
Trap efficiency of reservoir is function of
Answer (Detailed Solution Below)
Option 2 : capacity / inflow ratio
Dams and Spillways Question 4 Detailed Solution
Trap Efficiency:
• Trap efficiency of a reservoir is the ratio of total sediments retained by the reservoir to the total sediments entering into the reservoir (inflow sediments).
• Trap efficiency of the reservoir is the function of the ratio of reservoir capacity and total inflow.
• The reservoir capacity is inversely related to the number of sediments deposited. As time passes, the rate of silting or the rate of deposition of sediments reduces but the amount of sediments
deposited will be increased which in turn reduces the capacity of the reservoir. Hence, trap efficiency is also gets reduced.
Other factors affecting trap efficiency:
• Size of reservoir and stream: A larger reservoir on a small stream has less capacity inflow ratio and hence has higher trap efficiency.
• The velocity of flow of stream: Higher velocity less deposition of sediments, lower is the trap efficiency.
Dams and Spillways Question 5:
Pure clayey soils are generally not preferred for the central impervious cores of zoned type of earthen dams because
Answer (Detailed Solution Below)
Option 3 : clays are susceptible to cracking
Dams and Spillways Question 5 Detailed Solution
Clayey Soils in Central Impervious Cores of Zoned Earthen Dams:
• Pure clayey soils are generally avoided in the central impervious cores of zoned earthen dams.
• One of the main reasons for this is that clays are susceptible to cracking under certain conditions, particularly due to drying or when subjected to differential settlement.
• When cracks develop, they can compromise the impervious nature of the core, leading to potential leakage or failure of the dam.
• Therefore, clays' susceptibility to cracking makes them a less preferred choice for the central impervious core.
Top Dams and Spillways MCQ Objective Questions
In case of non-availability of space due to topography, the most suitable spillway in this condition is ____
Answer (Detailed Solution Below)
Option 3 : Shaft spillway
│Type of Spillway │Suitability │
│Straight drop spillway or free over fall spillway│Suitable for Thin arch dams, Earthen dams │
│Chute spillway/through Spillway │Suitable when the width of the river valley is very narrow. │
│Shaft spillway │Suitable when there is no space to provide for other types of spillways such as ogee spillway, straight drop spillway│
│Ogee spillway │Suitable for Gravity dams, Arch dams, Buttress dams │
│Side channel spillways │Suitable when sufficient width is not available and we need to avoid heavy cutting │
In case of non-availability of space due to topography, the most suitable spillway is
Answer (Detailed Solution Below)
Option 2 : Shaft spillway
Different types of spillways are as follows:
│Type of Spillway │Suitability │
│Straight drop spillway or free over fall spillway│Suitable for Thin arch dams, Earthen dams │
│Chute spillway/through Spillway │Suitable when the width of the river valley is very narrow. │
│Shaft spillway │Suitable when there is no space to provide for other types of spillways such as ogee spillway, straight drop spillway│
│Ogee spillway │Suitable for Gravity dams, Arch dams, Buttress dams │
│Side channel spillways │Suitable when sufficient width is not available and we need to avoid heavy cutting │
When the reservoir is full, the maximum compressive force in a gravity dam is produced
Answer (Detailed Solution Below)
Option 1 : at the toe
The maximum compressive stresses occur at the heel (mostly during reservoir empty condition) or at the toe (at reservoir full condition) and on planes normal to the face of the dam.
∴ For reservoir empty condition maximum compressive force will be at the heel.
∴ For reservoir full condition maximum compressive force will be at the toe of the dam.
The figure below is the pressure distribution diagram for reservoir full condition:
The maximum height of a masonry dam of a triangular section whose base width is b and specific gravity s is
Answer (Detailed Solution Below)
Option 1 : b √s
For no tension Criteria:
\({\rm{B}} = \frac{{\rm{H}}}{{\sqrt {{\rm{S}} - {\rm{C}}} }}\)
Where C = 1, B = Base width of dam, H = Height of dam
No Sliding Criteria:
\({\rm{\;B}} = \frac{{\rm{H}}}{{{\rm{\mu }}\left( {{\rm{S}} - {\rm{C}}} \right)}}\)
Where B’ = Minimum base width for no sliding criteria and S = Specific gravity of material of dam
For C = 0
H = b × √s
The silt load in the stream does not depend upon
Answer (Detailed Solution Below)
Option 4 : alignment of dam
Stream/silt load:
• Stream load is a geologic term referring to the solid matter carried by a stream.
• Erosion and bed shear stress continuously remove the particles which are transported by water either in suspension or as dissolved.
It primarily depends on-
• Nature of the soil in the catchment (as rocks generally don't dissolve, but small earthen particles get dissolved)
• Topography of the catchment as the more the slope more the velocity of the water.
• The intensity of rainfall as more rainfall leads to more runoff, thus increasing the sediment carrying capacity of any stream.
Considering maximum and minimum stress at the base of a dam, it will be correct to assume that:
Answer (Detailed Solution Below)
Option 1 : maximum stress in reservoir empty condition is expected at heel of base
The maximum stress under empty reservoir condition occurs at heel of base because in empty reservoir condition resultant force shift towards heel of base and increases uplift at heel.
Uplift pressure or stress at the base of dam is given by: \(\sigma_{max} =\sum \frac{w}{b}\left ( 1+\frac{6e}{b} \right ) \) , Uplift pressure distribution below gravity dam is shown below:
For safety of a concrete dam against overturning, what must be width of the dam of rectangular cross section of height 10 m, if the height of water storage on one side of it is 9 m? Take unit weight
of water as 10 kN/m^3 and unit weight of concrete as 25 kN/m^3.
(Ignore effect of uplift, friction and any other force)
Answer (Detailed Solution Below)
Option 3 : \(9\dfrac{\sqrt{3}}{5} \ m\)
r[c] = 25 kN/m^3
γ[w] = 10 kW/m^3
let length of Dam is pm
B is width of Dam
Hydrostatic force on dam, \({P_w} = \frac{1}{2}{\gamma_w}H \times H = \frac{1}{2} \times 10 \times 9 \times 9 = 405\;kN\)
Force due to weight of Dam, W = γ[c] × A × L = 25 × B × 10 × 1 = 250 B kN
‘P[w]’ force try to overturn the dam about the and ‘w’ force try to script it.
To safety against overturning,
N[OT] ≤ N[R]
P[w] × ≤ W × B/2
\(405 \times 3 \le 250\;B \times \frac{B}{2}\)
⇒ \(B \ge \sqrt {\frac{{405\; \times\; 2\; \times\; 3}}{{250}}} \)
⇒ \(B \ge \frac{9}{5}\sqrt 3 \;m\)
The flood absorption capacity of a reservoir is:
Answer (Detailed Solution Below)
Option 4 : the storage between FSL and MWL
Reservoirs are meant to absorb a part of flood water and the excess is discharged through a spillway. The difference between Maximum Water Level and Full supply level in a reservoir is called
surcharge storage or flood storage.
A schematic diagram showing different storage zones of a reservoir are shown below:
A) → Free board
B) → Surcharge storage
C) → Active Storage capacity
D) → Inactive storage capacity
E) → Live storage capacity
F) → Buffer storage
NWL → Maximum water level or also called highest flood level (HFL)
FSL → Full supply level
MDDL → Minimum Draw down level
DSL → Dead storage level (No outlet available to drain the water in reservoir by gravity)
The storage of water below the bottom of the lowest sluice way in a reservoir is called:
Answer (Detailed Solution Below)
Option 1 : dead storage
Zones of storage in a Reservoir:
1. Full reservoir Level:- The full reservoir level is the highest water level to which the water surface will rise during normal operating conditions.
2. Maximum water level:- The maximum water level is the maximum level to which the water surface will rise when the design flood passes over the spillway.
3. Minimum pool level:- The minimum pool level is the lowest level up to which the water is withdrawn from the reservoir under ordinary conditions.
4. Dead Storage:- The volume of water held below the minimum pool level is called the dead storage. It is provided to cater for the sediment deposition by the impounding sediment laid in water.
Normally it is equivalent to the volume of sediment expected to be deposited in the reservoir during the design life reservoir.
5. Live/Useful Storage:- The volume of water stored between the full reservoir level and minimum pool level is called useful storage. It assumes the supply of water for a specific period to meet the
6. Bank Storage:- It is developed in voids of soil cover in the reservoir area and becomes available as seepage of water when water levels drops down. It increases the reservoir capacity over and
above that given by elevation storage curves.
7. Valley storage:- The volume of water stored by the natural river channel in its valley up to the top of its banks before constructing of a reservoir is called the valley storage. The valley
storage depends upon the cross-section of the river.
8. Flood/Surcharge storage:- It is storage contained between maximum reservoir level and full reservoir levels. It varies with the spillway capacity of the dam for a given design flood.
To dissipate energy a fall is provided in a canal. A fall which has gradual convex and concave curves for smooth transition of water and to reduce disturbance and impact is a:
Answer (Detailed Solution Below)
Option 1 : ogee fall
Ogee Fall:
• In this type of fall, an ogee curve (a combination of convex curve and concave curve) is provided for carrying the canal water from higher level to lower level.
• This fall is recommended when the natural ground surface suddenly changes to a steeper slope along the alignment of the canal.
• There is a heavy drawdown on the u/s side resulting in lower depth, higher velocities and consequent bed erosion.
• Kinetic energy is not well dissipated due to smooth transition.
Rapid Fall
• The rapid fall is suitable when the slope of the natural ground surface is even and long. It consists of a long sloping glacis with longitudinal slope which varies from 1 in 10 to 1 in 20. It is
nowadays obsolete.
Stepped Fall
• Stepped fall consists of a series of vertical drops in the form of steps. This fall is suitable in places where the sloping ground is very long and requires long glacis to connect the higher bed
level with lower bed level.
• This fall is practically a modification of the rapid fall. The sloping glacis is divided into a number of drops so that the flowing water may not cause any damage to the canal bed. Brick walls
are provided at each of the drops. | {"url":"https://testbook.com/objective-questions/mcq-on-dams-and-spillways--5eea6a0b39140f30f369def5","timestamp":"2024-11-14T04:09:47Z","content_type":"text/html","content_length":"615478","record_id":"<urn:uuid:906147c9-0fb1-423c-b6ca-e354977b7d08>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00186.warc.gz"} |
M14Q3: Combining Reactions and their Equilibrium Constants
• Write an equilibrium constant expression, such as K[c], in terms of reactants and products and the stoichiometry of the reaction.
• Describe how an equilibrium constant for a balanced chemical equation would change if different stoichiometric coefficients were used in the balanced chemical equation, if the reaction were
reversed, or if several equations were added together to give a new net equation, and calculate the new equilibrium constant.
Now that we can write equilibrium reactions and expressions (K[eq]), what happens if we change all the coefficients in the reactions? What about if two reactions are happening to create a third,
unknown reaction, and we want to know if that third reaction is reactant or product favored? In this section, we will be looking at how to manipulate the equilibrium constant in these scenarios.
Equilibrium Constant Expressions and Stoichiometry
Consider the reaction:
Reaction 1: 2 NOCl(g) ⇌ 2 NO(g) + Cl[2](g) K[eq,1] =
This reaction and its resulting equilibrium constant will represent the ratio between the products and reactants at equilibrium. But what if we were more interested in studying one mole of NOCl
reacting to form NO and Cl[2]? If we divide the entire reaction by 2, the equilibrium reaction and equilibrium constant expression would be:
Reaction 2: NOCl(g) ⇌ NO(g) + ½ Cl[2](g) K[eq,2] =
Notice the difference between the two equilibrium constant expressions. The products are still in the numerator and reactants in the denominator, but the exponents that each species is raised to
differs because the coefficients are different. And in fact each coefficient in K[eq,2] is half that of K[eq,1], or each species is raised to the ½ power.
K[eq,2] = (K[eq,1])^½
Equilibrium Constant Expressions and Reversing Reactions
Sometimes we need to flip a reaction in order to properly represent the system we are trying to study. An example would be if we wanted to study:
Reaction 1: 2 NOCl(g) ⇌ 2 NO(g) + Cl[2](g) K[eq,1] =
But we only had information about the opposite reaction:
Reaction 2: 2 NO(g) + Cl[2](g) ⇌ 2 NOCl(g) K[eq,2] =
How do these two equilibrium constant expressions differ?
K[eq,1] = [eq,2] =
When we flip a reaction, the reactants and the products switch sides, so the numerator and the denominator in the equilibrium constant expression also switch and they are inverses of eachother.
K[eq,1] =
Equilibrium Constant Expressions for Multiple Reactions
Suppose we wanted to study the reaction:
Reaction 1: 2 NOCl(g) + O[2](g) ⇌ N[2]O[4](g) + Cl[2](g) K[eq,1] =
Unfortunately for us, this is a completely unstudied reaction and we have no idea if this reaction is reactant- or product-favored, whether it will occur fast or slow, and whether it is safe for us
to even perform the reaction! Luckily, we have two other similar reactions that we can use to give us insight into our desired reaction.
Reaction 2: 2 NOCl(g) ⇌ 2 NO(g) + Cl[2](g) K[eq,2] =
Reaction 3: 2 NO(g) + O[2](g) ⇌ N[2]O[4](g) K[eq,3] =
Notice that if we add Reaction 2 and Reaction 3 together, we are able to get our desired reaction, Reaction 1. But how do we know what the value of K[eq,1] is? Let’s compare K[eq,1], K[eq,2], and K
[eq,3] a little more closely. What do we have to do with K[eq,2] and K[eq,3] in order to get K[eq,1]?
K[eq,1] = [eq,2] = [eq,3] =
Perhaps you noticed that if we multiply K[eq,2] and K[eq,3], it will be equal to K[eq,1].
K[eq,2] × K[eq,3] = [eq,1]
Summary of How to Manipulate K[eq]
1. If you multiply a reaction by a coefficient, you also raise the equilibrium constant expression to that same coefficient. If you divide a reaction by a coefficient, try to reframe the process
into multiplication (i.e. dividing by two is the same as multiplying by ½).
2. If you flip a reaction, you take the inverse of the equilibrium constant expression.
3. If you add reactions together, you multiply the equilibrium constant expressions.
4. If you encounter more than one of these steps in a single problem, steps 1 & 2 can be done in any order, but always perform steps 1 & 2 prior to step 3.
Key Concepts and Summary
If K[eq] is known for a specific reaction, K[eq] can also be calculated if the chemical reaction is modified by changing the coefficients, reversing the reaction, or combining the given reaction with
another reaction. Approaching this as you have approached Hess’ Law in the past is a great strategy, except remember that changing the coefficients requires you to raise K[eq] to that change,
reversing a reaction requires you to take the inverse of K[eq], and combining reactions requires you to multiply the K[eq] for each reaction combined.
Chemistry End of Section Exercises
1. Acetic acid is a weak acid. It reacts with water to a small extent to form acetate and hydronium ions. The equilibrium constant for the reaction is 1.8 x 10^-5.
CH[3]COOH(aq) + H[2]O(ℓ) ⇌ CH[3]COO^–(aq) + H[3]O^+(aq)
What is the K[c] for the following reactions?
a. 2 CH[3]COOH(aq) + 2 H[2]O(ℓ) ⇌ 2 CH[3]COO^–(aq) + 2 H[3]O^+(aq)
b. CH[3]COO^–(aq) + H[3]O^+(aq) ⇌ CH[3]COOH(aq) + H[2]O(ℓ)
c. ½ CH[3]COOH(aq) + ½ H[2]O(ℓ) ⇌ ½ CH[3]COO^–(aq) + ½ H[3]O^+(aq)
d. 4 CH[3]COO^–(aq) + 4 H[3]O^+(aq) ⇌ 4 CH[3]COOH(aq) + 4 H[2]O(ℓ)
2. Given the equations:
2 N[2](g) + O[2](g) ⇌ 2 N[2]O(g) K[c] = 1.2 x 10^-35
N[2]O[4](g) ⇌ 2 NO[2](g) K[c] = 4.6 x 10^-3
½ N[2](g) + O[2](g) ⇌ NO[2](g) K[c] = 4.1 x 10^-9
Calculate the value of K[c] for 2 N[2]O[4](g) ⇌ 2 N[2]O(g) + 3 O[2](g).
3. Determine the equilibrium constant, K[p], for the following reaction,
2 NO(g) + O[2](g) ⇌ 2 NO[2](g) K[p] = ?
by using the two reference equations below:
N[2](g) + O[2](g) ⇌ 2 NO(g) K[p] = 2.3 × 10^–19
½ N[2](g) + O[2](g) ⇌ NO[2](g) K[p] = 8.4 × 10^–7
Answers to Chemistry End of Section Exercises
1. (a) 3.2 × 10^-10; (b) 5.6 × 10^4; (c) 4.2 × 10^-3; (d) 9.5 × 10^18
2. 9.0 × 10^-7
Left-click here to watch Exercise 2 problem solving video.
3. K[p] = 3.1 × 10^6
Please use
this form
to report any inconsistencies, errors, or other things you would like to change about this page. We appreciate your comments. 🙂 | {"url":"https://wisc.pb.unizin.org/chem103and104/chapter/combining-reactions-and-their-equilibrium-constants-m14q3/","timestamp":"2024-11-15T00:47:28Z","content_type":"text/html","content_length":"116913","record_id":"<urn:uuid:66203652-5dff-446b-8af0-6c96d7fb59b5>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00795.warc.gz"} |
Optimal Control of Non-deterministic Systems for a Computationally Efficient Fragment of Temporal Logic
Eric M. Wolff, Ufuk Topcu, and Richard M. Murray
2013 Conference on Decison and Control (CDC)
We develop a framework for optimal control policy synthesis for non-deterministic transition systems subject to temporal logic specifications. We use a fragment of temporal logic to specify tasks
such as safe navigation, response to the environment, persistence, and surveillance. By restricting specifications to this fragment, we avoid a potentially doubly-exponential automaton construction.
We compute feasible con- trol policies for non-deterministic transition systems in time polynomial in the size of the system and specification. We also compute optimal control policies for average,
minimax (bottleneck), and average cost-per-task-cycle cost functions. We highlight several interesting cases when these can be computed in time polynomial in the size of the system and specification.
Additionally, we make connections between computing optimal control policies for an average cost-per-task-cycle cost function and the generalized traveling salesman problem. We give simulation
results for motion planning problems. | {"url":"https://murray.cds.caltech.edu/index.php?title=Optimal_Control_of_Non-deterministic_Systems_for_a_Computationally_Efficient_Fragment_of_Temporal_Logic&oldid=19712","timestamp":"2024-11-11T17:25:38Z","content_type":"text/html","content_length":"19624","record_id":"<urn:uuid:4b704488-6148-4385-8843-fdaeeb9d830d>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00611.warc.gz"} |
Long Short-Term Memory Neural Networks
This topic explains how to work with sequence and time series data for classification and regression tasks using long short-term memory (LSTM) neural networks. For an example showing how to classify
sequence data using an LSTM neural network, see Sequence Classification Using Deep Learning.
An LSTM neural network is a type of recurrent neural network (RNN) that can learn long-term dependencies between time steps of sequence data.
LSTM Neural Network Architecture
The core components of an LSTM neural network are a sequence input layer and an LSTM layer. A sequence input layer inputs sequence or time series data into the neural network. An LSTM layer learns
long-term dependencies between time steps of sequence data.
This diagram illustrates the architecture of a simple LSTM neural network for classification. The neural network starts with a sequence input layer followed by an LSTM layer. To predict class labels,
the neural network ends with a fully connected layer, and a softmax layer.
This diagram illustrates the architecture of a simple LSTM neural network for regression. The neural network starts with a sequence input layer followed by an LSTM layer. The neural network ends with
a fully connected layer.
Classification LSTM Networks
To create an LSTM network for sequence-to-label classification, create a layer array containing a sequence input layer, an LSTM layer, a fully connected layer, and a softmax layer.
Set the size of the sequence input layer to the number of features of the input data. Set the size of the fully connected layer to the number of classes. You do not need to specify the sequence
For the LSTM layer, specify the number of hidden units and the output mode "last".
numFeatures = 12;
numHiddenUnits = 100;
numClasses = 9;
layers = [ ...
For an example showing how to train an LSTM network for sequence-to-label classification and classify new data, see Sequence Classification Using Deep Learning.
To create an LSTM network for sequence-to-sequence classification, use the same architecture as for sequence-to-label classification, but set the output mode of the LSTM layer to "sequence".
numFeatures = 12;
numHiddenUnits = 100;
numClasses = 9;
layers = [ ...
Regression LSTM Networks
To create an LSTM network for sequence-to-one regression, create a layer array containing a sequence input layer, an LSTM layer, and a fully connected layer.
Set the size of the sequence input layer to the number of features of the input data. Set the size of the fully connected layer to the number of responses. You do not need to specify the sequence
For the LSTM layer, specify the number of hidden units and the output mode "last".
numFeatures = 12;
numHiddenUnits = 125;
numResponses = 1;
layers = [ ...
To create an LSTM network for sequence-to-sequence regression, use the same architecture as for sequence-to-one regression, but set the output mode of the LSTM layer to "sequence".
numFeatures = 12;
numHiddenUnits = 125;
numResponses = 1;
layers = [ ...
For an example showing how to train an LSTM network for sequence-to-sequence regression and predict on new data, see Sequence-to-Sequence Regression Using Deep Learning.
Video Classification Network
To create a deep learning network for data containing sequences of images such as video data and medical images, specify image sequence input using the sequence input layer.
Specify the layers and create a dlnetwork object.
inputSize = [64 64 3];
filterSize = 5;
numFilters = 20;
numHiddenUnits = 200;
numClasses = 10;
layers = [
net = dlnetwork(layers);
For an example showing how to train a deep learning network for video classification, see Classify Videos Using Deep Learning.
Deeper LSTM Networks
You can make LSTM networks deeper by inserting extra LSTM layers with the output mode "sequence" before the LSTM layer. To prevent overfitting, you can insert dropout layers after the LSTM layers.
For sequence-to-label classification networks, the output mode of the last LSTM layer must be "last".
numFeatures = 12;
numHiddenUnits1 = 125;
numHiddenUnits2 = 100;
numClasses = 9;
layers = [ ...
For sequence-to-sequence classification networks, the output mode of the last LSTM layer must be "sequence".
numFeatures = 12;
numHiddenUnits1 = 125;
numHiddenUnits2 = 100;
numClasses = 9;
layers = [ ...
Layer Description
sequenceInputLayer A sequence input layer inputs sequence data to a neural network and applies data normalization.
lstmLayer An LSTM layer is an RNN layer that learns long-term dependencies between time steps in time-series and sequence data.
bilstmLayer A bidirectional LSTM (BiLSTM) layer is an RNN layer that learns bidirectional long-term dependencies between time steps of time-series or sequence data. These
dependencies can be useful when you want the RNN to learn from the complete time series at each time step.
gruLayer A GRU layer is an RNN layer that learns dependencies between time steps in time-series and sequence data.
convolution1dLayer A 1-D convolutional layer applies sliding convolutional filters to 1-D input.
maxPooling1dLayer A 1-D max pooling layer performs downsampling by dividing the input into 1-D pooling regions, then computing the maximum of each region.
averagePooling1dLayer A 1-D average pooling layer performs downsampling by dividing the input into 1-D pooling regions, then computing the average of each region.
globalMaxPooling1dLayer A 1-D global max pooling layer performs downsampling by outputting the maximum of the time or spatial dimensions of the input.
flattenLayer A flatten layer collapses the spatial dimensions of the input into the channel dimension.
wordEmbeddingLayer (Text A word embedding layer maps word indices to vectors.
Analytics Toolbox)
Classification, Prediction, and Forecasting
To make predictions on new data, use the minibatchpredict function. To convert predicted classification scores to labels, use the scores2label.
LSTM neural networks can remember the state of the neural network between predictions. The RNN state is useful when you do not have the complete time series in advance, or if you want to make
multiple predictions on a long time series.
To predict and classify on parts of a time series and update the RNN state, use the predict function and also return and update the neural network state. To reset the RNN state between predictions,
use resetState.
For an example showing how to forecast future time steps of a sequence, see Time Series Forecasting Using Deep Learning.
Sequence Padding and Truncation
LSTM neural networks support input data with varying sequence lengths. When passing data through the neural network, the software pads or truncates sequences so that all the sequences in each
mini-batch have the specified length. You can specify the sequence lengths and the value used to pad the sequences using the SequenceLength and SequencePaddingValue training options.
After training the neural network, you can use the same mini-batch size and padding options when you use the minibatchpredict function.
Sort Sequences by Length
To reduce the amount of padding or discarded data when padding or truncating sequences, try sorting your data by sequence length. For sequences where the first dimension corresponds to the time
steps, to sort the data by sequence length, first get the number of columns of each sequence by applying size(X,1) to every sequence using cellfun. Then sort the sequence lengths using sort, and use
the second output to reorder the original sequences.
sequenceLengths = cellfun(@(X) size(X,1), XTrain);
[sequenceLengthsSorted,idx] = sort(sequenceLengths);
XTrain = XTrain(idx);
Pad Sequences
If the SequenceLength training or prediction option is "longest", then the software pads the sequences so that all the sequences in a mini-batch have the same length as the longest sequence in the
mini-batch. This option is the default.
Truncate Sequences
If the SequenceLength training or prediction option is "shortest", then the software truncates the sequences so that all the sequences in a mini-batch have the same length as the shortest sequence in
that mini-batch. The remaining data in the sequences is discarded.
Specify Padding Direction
The location of the padding and truncation can impact training, classification, and prediction accuracy. Try setting the SequencePaddingDirection training options to "left" or "right" and see which
is best for your data.
Because recurrent layers process sequence data one time step at a time, when the recurrent layer OutputMode property is "last", any padding in the final time steps can negatively influence the layer
output. To pad or truncate sequence data on the left, set the SequencePaddingDirection argument to "left".
For sequence-to-sequence neural networks (when the OutputMode property is "sequence" for each recurrent layer), any padding in the first time steps can negatively influence the predictions for the
earlier time steps. To pad or truncate sequence data on the right, set the SequencePaddingDirection option to "right".
Normalize Sequence Data
To recenter training data automatically at training time using zero-center normalization, set the Normalization option of sequenceInputLayer to "zerocenter". Alternatively, you can normalize sequence
data by first calculating the per-feature mean and standard deviation of all the sequences. Then, for each training observation, subtract the mean value and divide by the standard deviation.
mu = mean([XTrain{:}],1);
sigma = std([XTrain{:}],0,1);
XTrain = cellfun(@(X) (X-mu)./sigma,XTrain,UniformOutput=false);
LSTM Layer Architecture
This diagram illustrates the flow of data through an LSTM layer with input $x$ and output $y$ with T time steps. In the diagram, ${h}_{t}$ denotes the output (also known as the hidden state) and ${c}
_{t}$ denotes the cell state at time step t.
If the layer outputs the full sequence, then it outputs ${y}_{1}$, …, ${y}_{T}$, which is equivalent to ${h}_{1}$, …, ${h}_{T}$. If the layer outputs the last time step only, then the layer outputs $
{y}_{T}$, which is equivalent to ${h}_{T}$. The number of channels in the output matches the number of hidden units of the LSTM layer.
The first LSTM operation uses the initial state of the RNN and the first time step of the sequence to compute the first output and the updated cell state. At time step t, the operation uses the
current state of the RNN $\left({c}_{t-1},{h}_{t-1}\right)$ and the next time step of the sequence to compute the output and the updated cell state ${c}_{t}$.
The state of the layer consists of the hidden state (also known as the output state) and the cell state. The hidden state at time step t contains the output of the LSTM layer for this time step. The
cell state contains information learned from the previous time steps. At each time step, the layer adds information to or removes information from the cell state. The layer controls these updates
using gates.
These components control the cell state and hidden state of the layer.
Component Purpose
Input gate (i) Control level of cell state update
Forget gate (f) Control level of cell state reset (forget)
Cell candidate (g) Add information to cell state
Output gate (o) Control level of cell state added to hidden state
This diagram illustrates the flow of data at time step t. This diagram shows how the gates forget, update, and output the cell and hidden states.
The learnable weights of an LSTM layer are the input weights W (InputWeights), the recurrent weights R (RecurrentWeights), and the bias b (Bias). The matrices W, R, and b are concatenations of the
input weights, the recurrent weights, and the bias of each component, respectively. The layer concatenates the matrices according to these equations:
$W=\left[\begin{array}{c}{W}_{i}\\ {W}_{f}\\ {W}_{g}\\ {W}_{o}\end{array}\right],\text{ }R=\left[\begin{array}{c}{R}_{i}\\ {R}_{f}\\ {R}_{g}\\ {R}_{o}\end{array}\right],\text{ }b=\left[\begin{array}
{c}{b}_{i}\\ {b}_{f}\\ {b}_{g}\\ {b}_{o}\end{array}\right],$
where i, f, g, and o denote the input gate, forget gate, cell candidate, and output gate, respectively.
The cell state at time step t is given by
${c}_{t}={f}_{t}\odot {c}_{t-1}+{i}_{t}\odot {g}_{t},$
where $\odot$ denotes the Hadamard product (element-wise multiplication of vectors).
The hidden state at time step t is given by
${h}_{t}={o}_{t}\odot {\sigma }_{c}\left({c}_{t}\right),$
where ${\sigma }_{c}$ denotes the state activation function. By default, the lstmLayer function uses the hyperbolic tangent function (tanh) to compute the state activation function.
These formulas describe the components at time step t.
Component Formula
Input gate ${i}_{t}={\sigma }_{g}\left({W}_{i}{x}_{t}+\text{}{\text{R}}_{i}{h}_{t-1}+{b}_{i}\right)$
Forget gate ${f}_{t}={\sigma }_{g}\left({W}_{f}{x}_{t}+\text{}{\text{R}}_{f}{h}_{t-1}+{b}_{f}\right)$
Cell candidate ${g}_{t}={\sigma }_{c}\left({W}_{g}{x}_{t}+\text{}{\text{R}}_{g}{h}_{t-1}+{b}_{g}\right)$
Output gate ${o}_{t}={\sigma }_{g}\left({W}_{o}{x}_{t}+\text{}{\text{R}}_{o}{h}_{t-1}+{b}_{o}\right)$
In these calculations, ${\sigma }_{g}$ denotes the gate activation function. By default, the lstmLayer function, uses the sigmoid function, given by $\sigma \left(x\right)={\left(1+{e}^{-x}\right)}^
{-1}$, to compute the gate activation function.
[1] Hochreiter, S., and J. Schmidhuber. "Long short-term memory." Neural computation. Vol. 9, Number 8, 1997, pp.1735–1780.
See Also
sequenceInputLayer | lstmLayer | bilstmLayer | gruLayer | dlnetwork | minibatchpredict | predict | scores2label | flattenLayer | wordEmbeddingLayer (Text Analytics Toolbox)
Related Topics | {"url":"https://fr.mathworks.com/help/deeplearning/ug/long-short-term-memory-networks.html","timestamp":"2024-11-06T17:59:02Z","content_type":"text/html","content_length":"122913","record_id":"<urn:uuid:4661d16e-7147-4529-b88e-c4b2df30fa11>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00049.warc.gz"} |
sqrt: square root function - Linux Manuals (3p)
sqrt (3p) - Linux Manuals
sqrt: square root function
This manual page is part of the POSIX Programmer's Manual. The Linux implementation of this interface may differ (consult the corresponding Linux manual page for details of Linux behavior), or the
interface may not be implemented on Linux.
sqrt, sqrtf, sqrtl - square root function
#include <math.h>
double sqrt(double x);
float sqrtf(float x);
long double sqrtl(long double x);
These functions shall compute the square root of their argument x, sqrt(x).
An application wishing to check for error situations should set errno to zero and call feclearexcept(FE_ALL_EXCEPT) before calling these functions. On return, if errno is non-zero or fetestexcept
(FE_INVALID | FE_DIVBYZERO | FE_OVERFLOW | FE_UNDERFLOW) is non-zero, an error has occurred.
Upon successful completion, these functions shall return the square root of x.
For finite values of x < -0, a domain error shall occur, and either a NaN (if supported), or an implementation-defined value shall be returned.
If x is NaN, a NaN shall be returned.
If x is ±0 or +Inf, x shall be returned.
If x is -Inf, a domain error shall occur, and either a NaN (if supported), or an implementation-defined value shall be returned.
These functions shall fail if:
Domain Error
The finite value of x is < -0, or x is -Inf.
If the integer expression (math_errhandling & MATH_ERRNO) is non-zero, then errno shall be set to [EDOM]. If the integer expression (math_errhandling & MATH_ERREXCEPT) is non-zero, then the invalid
floating-point exception shall be raised.
The following sections are informative.
Taking the Square Root of 9.0
#include <math.h>
double x = 9.0;
double result;
result = sqrt(x);
On error, the expressions (math_errhandling & MATH_ERRNO) and (math_errhandling & MATH_ERREXCEPT) are independent of each other, but at least one of them must be non-zero.
Portions of this text are reprinted and reproduced in electronic form from IEEE Std 1003.1, 2003 Edition, Standard for Information Technology -- Portable Operating System Interface (POSIX), The Open
Group Base Specifications Issue 6, Copyright (C) 2001-2003 by the Institute of Electrical and Electronics Engineers, Inc and The Open Group. In the event of any discrepancy between this version and
the original IEEE and The Open Group Standard, the original IEEE and The Open Group Standard is the referee document. The original Standard can be obtained online at http://www.opengroup.org/unix/
online.html .
feclearexcept(), fetestexcept(), isnan(), the Base Definitions volume of IEEE Std 1003.1-2001, Section 4.18, Treatment of Error Conditions for Mathematical Functions, <math.h>, <stdio.h> | {"url":"https://www.systutorials.com/docs/linux/man/3p-sqrt/","timestamp":"2024-11-08T20:27:59Z","content_type":"text/html","content_length":"9619","record_id":"<urn:uuid:3ac1a161-9fc9-4f7a-a02f-273adffe360d>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00238.warc.gz"} |
Bulletin - Courses Home
Course ID: CSCI 4860/6860. 4 hours.
Course Title: Computational Neuroscience
Course Introduction to computational neuroscience. Students will learn basic concepts, algorithms, and software tools for computational neuroscience models. Neural signal processing and neural network
Description: models will be discussed.
Oasis Title: Computational Neuroscience
Prerequisite: CSCI 2720 or CSCI 2725 or permission of department
Semester Course Not offered on a regular basis.
Grading System: A-F (Traditional)
Course This course aims to teach basic concepts and algorithms for computational neuroscience models. The course will cover neural signal processing and neural network modeling. Multidimensional neural signal processing and neural network modeling methods will be discussed in the contexts of neurobiology, brain imaging, cognitive neuroscience, and clinical neuroscience.
Topical 1. Introduction to brain science
Outline: 1.1 Introduction to neurons
1.2 Introduction to axons and their wiring
1.3 Introduction to communication mechanisms among neurons and their networks
1.4 Introduction to basic brain science principles
2. Neural signal processing
2.1 Basic concepts in neural signal formation
2.2 Multidimensional neural signal formation and reconstruction
3. Neural signal representation
3.1 1D signal representation such as EEG and MEG
3.2 2D signal representation such as neurobiological images
3.3 3D signal representation such as neuroimaging data
3.4 4D signal representation such as fMRI data
4. Neural signal transform, modeling, and analysis
4.1 Neural signal transform and filtering
4.2 Model-driven neural signal modeling and analysis
4.3 Data-driven neural signal modeling and analysis
4.4 Hybrid neural signal modeling and analysis
5. Neural network models
5.1 Classical neural network models in neurophysiology
5.2 Structural neural network models
5.3 Functional neural network models
5.4 Multidimensional and multimodal neural network models
5.5 Interface between artificial neural networks and biological neural networks
5.6 Interface between deep learning and neural network models
5.7 Abstraction of common graph models in both artificial and biological neural networks
Syllabus: No Syllabus Available | {"url":"https://bulletin.uga.edu/Link?cid=csci6860","timestamp":"2024-11-07T11:58:42Z","content_type":"application/xhtml+xml","content_length":"16025","record_id":"<urn:uuid:af5c2c16-a3c5-4900-85f7-8745fd55ba26>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00399.warc.gz"} |
trader joe's buttermilk protein pancake mix recipes
We wish now to be able to develop confidence intervals for the population parameter "\(p\)" from the binomial probability density function. Figure \(\PageIndex{9}\) places the mean on the
distribution of population probabilities as \(\mu=np\) but of course we do not actually know the population mean because we do not know the population probability of success, \(p\). The central limit
theorem states that the sampling distribution of the mean approaches a normal distribution as N, the sample size, increases. Which is, a large, properly drawn sample will resemble the population from
which it is drawn. The central limit theorem also states that the sampling distribution will … Graded A. Unlike the case just discussed for a continuous random variable where we did not know the
population distribution of \(X\)'s, here we actually know the underlying probability density function for these data; it is the binomial. 1. That's irrelevant. The mean return for the investment will
be 12% … Every sample would consist of 20 students. Find the population proportion, as well as the mean and standard deviation of the sampling distribution for samples of size n=60. Find study
resources for. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. This theoretical distribution is called the sampling distribution of \(\overline x\)'s. Try dropping a phrase
into casual conversation with your friends and bask in their admiration of you. The Central Limit Theorem tells us that the point estimate for the sample mean, \(\overline x\), comes from a normal
distribution of \(\overline x\)'s. This theoretical distribution is called the sampling distribution of ¯ x 's. This is a parallel question that was just answered by the Central Limit Theorem: from
what distribution was the sample mean, \(\overline x\), drawn? As Central Limit Theorems concern the sample mean, we first define it precisely. In this article, we will be learning about the central
limit theorem standard deviation, the central limit theorem probability, its definition, formula, and examples. Sampling Distribution and CLT of Sample Proportions (This section is not included in
the book, but I suggest that you read it in order to better understand the following chapter. We have assumed that theseheights, taken as a population, are normally distributed with a certain mean
(65inches) and a certain standard deviation (3 inches). We will take that up in the next chapter. Again the Central Limit Theorem provides this information for the sampling distribution for
proportions. The standard deviation of the sampling distribution of sample proportions, \(\sigma_{p^{\prime}}\), is the population standard deviation divided by the square root of the sample size, \
(n\). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . is approximately normal, with mean . Use the Central Limit Theorem for
Proportions to find probabilities for sampling distributions Question In a town, a pediatric nurse is concerned about the number of children who have whooping cough during the winter season. We
concluded that with a given level of probability, the range from which the point estimate comes is smaller as the sample size, \(n\), increases. It is important to remember that the samples that are
taken should be enough by size. The sample size is \(n\) and \(X\) is the number of successes found in that sample. What we have done can be seen in Figure \(\PageIndex{9}\). Watch the recordings
here on Youtube! Before we go in detail on CLT, let’s define some terms that will make it easier to comprehend the idea behind CLT. We saw that once we knew that the distribution was the Normal
distribution then we were able to create confidence intervals for the population parameter, \(\mu\). Graded A (All) Math 225N Week 5 Assignment (2020) - Central Limit Theorem for Proportions. We can
do so by using the Central Limit Theorem for making the calculations easy. Note that the sample mean, being a sum of random variables, is itself a random variable. Central Limit Theory (for
Proportions) Let p be the probability of success, q be the probability of failure. The central limit theorem states that the sampling distribution of the mean of any independent,random variablewill
be normal or nearly normal, if the sample size is large enough. This way, we can get the approximate mean height of all the students who are a part of the sports teams. MATH 225 Statistical Reasoning
for the Health Sciences Week 5 Assignment Central Limit Theorem for Proportions Question Pharmacy technicians are concerned about the rising number of fraudulent prescriptions they are seeing. The
central limit theorem is a result from probability theory.This theorem shows up in a number of places in the field of statistics. This method tends to assume that the given population is distributed
normally. The mean score will be the proportion of successes. Also, all the samples would tend to follow an approximately normal distribution pattern, when all the variances will be approximately
equal to the variance of the entire population when it is divided by the size of the sample. The expected value of the mean of sampling distribution of sample proportions, \(\mu_{p^{\prime}}\), is
the population proportion, \(p\). The central limit theorem is also used in finance to analyze stocks and index which simplifies many procedures of analysis as generally and most of the times you
will have a sample size which is greater than 50. of the 3,492 children living in a town, 623 of them have whooping cough. 00:01. Although the central limit theorem can seem abstract and devoid of
any application, this theorem is actually quite important to the practice of statistics. For instance, what proportion of the population would prefer to bank online rather than go to the bank?
Something called the central limit theorem. Notice the parallel between this Table and Table \(\PageIndex{1}\) for the case where the random variable is continuous and we were developing the sampling
distribution for means. The more closely the original population resembles a normal distrib… 1. But that's what's so super useful about it. Question: A dental student is conducting a study on the
number of people who visit their dentist regularly. If . ), \[\sigma_{\mathrm{p}}^{2}=\operatorname{Var}\left(p^{\prime}\right)=\operatorname{Var}\left(\frac{x}{n}\right)=\frac{1}{n^{2}}(\
operatorname{Var}(x))=\frac{1}{n^{2}}(n p(1-p))=\frac{p(1-p)}{n}\nonumber\]. The Central Limit Theorem for Sample Proportions. To understand the Central Limit Theorem better, let us consider the
following example. The proof of these important conclusions from the Central Limit Theorem is provided below. Example 4 Heavenly Ski resort conducted a study of falls on its advanced run over twelve
consecutive ten minute periods. To explain it in simpler words, the Central Limit Theorem is a statistical theory which states that when a sufficiently larger sample size of a population is given
that has a finite level of variance, the mean value of all the given samples from the same given population is approximately equal to the population mean. The larger the sample, the better the
approximation will be. This is the core principle underlying the central limit theorem. Central limit theorem for proportions We use p as the symbol for a sample proportion. Proportion of population
who would vote for one of the candidates running for the office and so on. Central Limit Theorem for proportions Example: It is believed that college student spends on average 65.5 minutes daily on
texting using their cell phone and the corresponding standard deviation is … Some sample proportions will show high favorability toward the bond issue and others will show low favorability because
random sampling will reflect the variation of views within the population. To do so, we will first need to determine the height of each student and then add them all. Well, this method to determine
the average is too tedious and involves tiresome calculations. MATH 225N Week 5 Assignment: Central Limit Theorem for Proportions. And so I need to explain some concepts in the beginning here to tie
it together with what you already know about the central limit theorem. Below the distribution of the population values is the sampling distribution of \(p\)'s. Formula: Sample mean ( μ x ) = μ
Sample standard deviation ( σ x ) = σ / √ n Where, μ = Population mean σ = Population standard deviation n = Sample size. Central Limit Theorem for proportions & means It’s freaking MAGIC people! The
sampling distribution for samples of size n is approximately normal with mean (1) μ p ¯ = p We take a woman’s height; maybe she’s shorter thanaverage, maybe she’s average, maybe she’s taller. 1. Use
the Central Limit Theorem for Proportions to find probabilities for sampling distributions - Calculator Question According to a study, 60% of people who are murdered knew their murderer. Find the
population proportion, as well as the mean and … In this method of calculating the average, we will first pick the students randomly from different teams and determine a sample. Question: A dental
student is conducting a study on the number of people who visit their dentist regularly.Of the 520 people surveyed, 312 indicated that they had visited their dentist within the past year. =−. 09:07.
Use the Central Limit Theorem for Proportions to find probabilities for sampling distributions Question A kitchen supply store has a total of 642 unique items available for purchase of their
available kitchen items, 260 are kitchen tools. Casual conversation with your friends and bask in their admiration of you dropping a phrase casual... A part of the sample mean, we find the population
values is population! Standard deviation for the sampling distribution of \ ( p\ ) 's to... Normal distribution as the sample size, the more closely the sampling distribution of ‘ s mentioned above
to the... Statistics as well as the mean of these important conclusions from the binomial probability density function of! ●The samples must be met to use the normal the 3,492 children living in a
town, 623 of have. Properties: 1 living in a number of people who visit their dentist within the past year about the proportion... For categorical data, then the parameter we wish to estimate ; p
from binomial... Are useful not only apply to the distribution of ‘ s is \ ( \overline )... Approximate mean height of all these sample mean distribution becomes normal when we calculate it by
repeated sampling you... Too small ( less than 30 ) chapter 8. then add them all at @... With population proportion this further when conducting item inventory is provided below = proportion to
determine the mean and deviation! Too tedious and involves tiresome calculations try dropping a phrase into casual conversation with your and... Running for the random variables, is itself a random
variable probability density.. Repeated sampling the 3,492 children living in a town, 623 of them knew their murderer over twelve ten. Then as you increase the sample mean of a population with
population central limit theorem for proportions the probability of a... Probability theory.This Theorem shows up in a particular state there are currently 50 current cold.. Such as building the
confidence intervals they support the new school bond issue then as you increase sample. Information for the sample proportion can get the approximate mean height of the Limit... Population mean
later, let us consider the following properties: 1 comes from a population population. Apply just to the sample size, the sample proportion for a uniform data set are so many students automatic!
1,500 new prescriptions a month, 28 of which are fraudulent do we it!
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Close to linear space routing schemes
Let (Formula presented.) be an unweighted undirected graph with n vertices and m edges, and let (Formula presented.) be an integer. We present a routing scheme with a poly-logarithmic header size,
that given a source s and a destination t at distance (Formula presented.) from s, routes a message from s to t on a path whose length is (Formula presented.). The total space used by our routing
scheme is (Formula presented.), which is almost linear in the number of edges of the graph. We present also a routing scheme with (Formula presented.) header size, and the same stretch (up to
constant factors). In this routing scheme, the routing table of every(Formula presented.) is at most (Formula presented.), where deg(v) is the degree of v in G. Our results are obtained by combining
a general technique of Bernstein (2009), that was presented in the context of dynamic graph algorithms, with several new ideas and observations.
Bibliographical note
Publisher Copyright:
© 2015, Springer-Verlag Berlin Heidelberg.
Dive into the research topics of 'Close to linear space routing schemes'. Together they form a unique fingerprint. | {"url":"https://cris.biu.ac.il/en/publications/close-to-linear-space-routing-schemes-7","timestamp":"2024-11-11T08:34:40Z","content_type":"text/html","content_length":"52929","record_id":"<urn:uuid:24bc9653-1435-4983-a7de-1dc0aad8ae6f>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00460.warc.gz"} |
Example 24.1 Accumulating Transactional Data into Time Series Data
This example uses the SIMILARITY procedure to illustrate the accumulation of time-stamped transactional data that has been recorded at no particular frequency into time series data at a specific
frequency. After the time series is created, the various SAS/ETS procedures related to time series analysis, similarity analysis, seasonal adjustment and decomposition, modeling, and forecasting can
be used to further analyze the time series data.
Suppose that the input data set WORK.RETAIL contains variables STORE and TIMESTAMP and numerous other numeric transaction variables. The BY variable STORE contains values that break up the
transactions into groups (BY groups). The time ID variable TIMESTAMP contains SAS date values recorded at no particular frequency. The other data set variables contain the numeric transaction values
to be analyzed. It is further assumed that the input data set is sorted by the variables STORE and TIMESTAMP.
The following statements form monthly time series from the transactional data based on the median value (ACCUMULATE=MEDIAN) of the transactions recorded with each time period. The accumulated time
series values for time periods with no transactions are set to zero instead of missing (SETMISS=0). Only transactions recorded between the first day of 1998 (START=’01JAN1998’D ) and last day of 2000
(END=’31JAN2000’D ) are considered and if needed are extended to include this range.
proc similarity data=work.retail out=mseries;
by store;
id timestamp interval=month
end ='31dec2000'd;
target _NUMERIC_;
The monthly time series data are stored in the data WORK.MSERIES. Each BY group associated with the BY variable STORE contains an observation for each of the 36 months associated with the years 1998,
1999, and 2000. Each observation contains the variable STORE, TIMESTAMP, and each of the analysis variables in the input DATA= data set.
After each set of transactions has been accumulated to form the corresponding time series, the accumulated time series can be analyzed by using various time series analysis techniques. For example,
exponentially weighted moving averages can be used to smooth each series. The following statements use the EXPAND procedure to smooth the analysis variable named STOREITEM.
proc expand data=mseries
by store;
id timestamp;
convert storeitem=smooth / transform=(ewma 0.1);
The smoothed series is stored in the data set WORK.SMOOTHED. The variable SMOOTH contains the smoothed series.
If the time ID variable TIMESTAMP contains SAS datetime values instead of SAS date values, the INTERVAL= , START=, and END= options in the SIMILARITY procedure must be changed accordingly, and the
following statements could be used to accumulate the datetime transactions to a monthly interval:
proc similarity data=work.retail
by store;
id timestamp interval=dtmonth
end ='31dec2000:00:00:00'dt;
target _NUMERIC_;
The monthly time series data are stored in the data WORK.TSERIES, and the time ID values use a SAS datetime representation. | {"url":"http://support.sas.com/documentation/cdl/en/etsug/65545/HTML/default/etsug_similarity_examples01.htm","timestamp":"2024-11-02T11:52:12Z","content_type":"application/xhtml+xml","content_length":"16894","record_id":"<urn:uuid:636890f5-c945-4625-8716-58b3219e6b04>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00553.warc.gz"} |
To get more practice in equation, we brought you this problem of the week:
How would you solve the equation \(7+4u+\frac{5}{u}=16\)?
Check out the solution below!
Multiply both sides by \(u\).
Move all terms to one side.
Simplify \(7u+4{u}^{2}+5-16u\) to \(-9u+4{u}^{2}+5\).
Split the second term in \(-9u+4{u}^{2}+5\) into two terms.
Factor out common terms in the first two terms, then in the last two terms.
Factor out the common term \(u-1\).
Solve for \(u\).
Decimal Form: 1, 1.25 | {"url":"https://www.cymath.com/blog/2024-05-13","timestamp":"2024-11-12T22:18:08Z","content_type":"text/html","content_length":"29875","record_id":"<urn:uuid:172e2089-1341-4c1d-b503-de324d36780d>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00715.warc.gz"} |
The application of automated perturbation theory to lattice QCD
Monahan, Christopher John
Predictions of heavy quark parameters are an integral component of precision tests of the Standard Model of particle physics. Experimental measurements of electroweak processes involving heavy
hadrons provide stringent tests of Cabibbo-Kobayashi-Maskawa (CKM) matrix unitarity and serve as a probe of new physics. Hadronic matrix elements parameterise the strong dynamics of these
interactions and these matrix elements must be calculated nonperturbatively. Lattice quantum chromodynamics (QCD) provides the framework for nonperturbative calculations of QCD processes. Current
lattices are too coarse to directly simulate b quarks. Therefore an effective theory, nonrelativistic QCD (NRQCD), is used to discretise the heavy quarks. High precision simulations are required so
systematic uncertainties are removed by improving the NRQCD action. Precise simulations also require improved sea quark actions, such as the highly-improved staggered quark (HISQ) action. The
renormalisation parameters of these actions cannot be feasibly determined by hand and thus automated procedures have been developed. In this dissertation I apply automated lattice pertubartion theory
to a number of heavy quark calculations. I first review the fundamentals of lattice QCD and the construction of lattice NRQCD. I then motivate and discuss lattice perturbation theory in detail,
focussing on the tools and techniques that I use in this dissertation. I calculate the two-loop tadpole improvement factors for improved gluons with improved light quarks. I then compute the
renormalisation parameters of NRQCD. I use a mix of analytic and numerical methods to extract the one-loop radiative corrections to the higher order kinetic operators in the NRQCD action. I then
employ a fully automated procedure to calculate the heavy quark energy shift at two-loops. I use this result to extract a new prediction of the mass of the b quark from lattice NRQCD simulations by
the HPQCD collaboration. I also review the calculation of the radiative corrections to the chromo-magnetic operator in the NRQCD action. This computation is the first outcome of our implementation of
background field gauge for automated lattice perturbation theory. Finally, I calculate the heavy-light currents for highly-improved NRQCD heavy quarks with massless HISQ light quarks and discuss the
application of these results to nonperturbative studies by the HPQCD collaboration.
Lattice QCD, NRQCD, Perturbation theory
Doctor of Philosophy (PhD)
Awarding Institution
University of Cambridge | {"url":"https://www.repository.cam.ac.uk/items/1ced2145-978a-42ee-95ad-adafa55c966b","timestamp":"2024-11-03T01:27:19Z","content_type":"text/html","content_length":"624207","record_id":"<urn:uuid:d5596ff7-1a89-4519-a336-74f357cab208>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00764.warc.gz"} |
What is Legend Receivables Turnover from 2010 to 2024 | Stocks: LEGN - Macroaxis
LEGN Stock USD 40.03 0.92 2.25%
Legend Biotech Receivables Turnover yearly trend continues to be very stable with very little volatility. Receivables Turnover is likely to drop to 1.72. During the period from
2010 to 2024
, Legend Biotech Receivables Turnover quarterly data regression pattern had sample variance of
and median of
View All Fundamentals
First Reported Previous Quarter Current Value Quarterly Volatility
Receivables Turnover
2010-12-31 1.81549207 1.72 335.51264608
Credit Downgrade Yuan Drop Covid
Check Legend Biotech
financial statements
over time to gain insight into future company performance. You can evaluate financial statements to find patterns among Legend Biotech's main balance sheet or income statement drivers, such as
Tax Provision of 268.9 K
, Depreciation And Amortization of 11
Interest Expense of 22.9 M
, as well as many indicators such as
Price To Sales Ratio of 35.3
, Dividend Yield of 0.0 or
PTB Ratio of 8.04
. Legend financial statements analysis is a perfect complement when working with
Legend Biotech Valuation
Legend Receivables Turnover
Check out the analysis of
Legend Biotech Correlation
against competitors.
Latest Legend Biotech's Receivables Turnover Growth Pattern
Below is the plot of the Receivables Turnover of Legend Biotech Corp over the last few years. It is Legend Biotech's Receivables Turnover historical data analysis aims to capture in quantitative
terms the overall pattern of either growth or decline in Legend Biotech's overall financial position and show how it may be relating to other accounts over time.
Receivables Turnover 10 Years Trend
Legend Receivables Turnover Regression Statistics
Arithmetic Mean 87.25
Geometric Mean 0.52
Coefficient Of Variation 384.53
Mean Deviation 161.71
Median 0.10
Standard Deviation 335.51
Sample Variance 112,569
Range 1.3K
R-Value 0.31
Mean Square Error 109,505
R-Squared 0.1
Significance 0.26
Slope 23.33
Total Sum of Squares 1.6M
Legend Receivables Turnover History
About Legend Biotech Financial Statements
Legend Biotech investors utilize fundamental indicators, such as Receivables Turnover, to predict how Legend Stock might perform in the future. Analyzing these trends over time helps investors make
market timing
decisions. For further insights, please visit our
fundamental analysis
Last Reported Projected for Next Year
Receivables Turnover 1.82 1.72
Pair Trading with Legend Biotech
One of the main advantages of trading using pair correlations is that every trade hedges away some risk. Because there are two separate transactions required, even if Legend Biotech position performs
unexpectedly, the other equity can make up some of the losses. Pair trading also minimizes risk from directional movements in the market. For example, if an entire industry or sector drops because of
unexpected headlines, the short position in Legend Biotech will appreciate offsetting losses from the drop in the long position's value.
0.67 ME 23Andme Holding PairCorr
0.86 VALN Valneva SE ADR PairCorr
0.72 DRUG Bright Minds Biosciences PairCorr
0.71 DMAC DiaMedica Therapeutics Earnings Call Tomorrow PairCorr
0.69 VERA Vera Therapeutics PairCorr
0.69 DSGN Design Therapeutics Earnings Call Tomorrow PairCorr
0.66 VCYT Veracyte PairCorr
The ability to find closely correlated positions to Legend Biotech could be a great tool in your tax-loss harvesting strategies, allowing investors a quick way to find a similar-enough asset to
replace Legend Biotech when you sell it. If you don't do this, your portfolio allocation will be skewed against your target asset allocation. So, investors can't just sell and buy back Legend Biotech
- that would be a violation of the tax code under the "wash sale" rule, and this is why you need to find a similar enough asset and use the proceeds from selling Legend Biotech Corp to buy it.
The correlation of Legend Biotech is a statistical measure of how it moves in relation to other instruments. This measure is expressed in what is known as the correlation coefficient, which ranges
between -1 and +1. A perfect positive correlation (i.e., a correlation coefficient of +1) implies that as Legend Biotech moves, either up or down, the other security will move in the same direction.
Alternatively, perfect negative correlation means that if Legend Biotech Corp moves in either direction, the perfectly negatively correlated security will move in the opposite direction. If the
correlation is 0, the equities are not correlated; they are entirely random. A correlation greater than 0.8 is generally described as strong, whereas a correlation less than 0.5 is generally
considered weak.
Correlation analysis
and pair trading evaluation for Legend Biotech can also be used as hedging techniques within a particular sector or industry or even over random equities to generate a better risk-adjusted return on
your portfolios.
Pair CorrelationCorrelation Matching
When determining whether Legend Biotech Corp
offers a strong return on investment
in its stock, a comprehensive analysis is essential. The process typically begins with a thorough review of Legend Biotech's
financial statements
, including income statements, balance sheets, and cash flow statements, to assess its
financial health
. Key financial ratios are used to gauge profitability, efficiency, and growth potential of Legend Biotech Corp Stock.
Outlined below are crucial reports that will aid in making a well-informed decision on Legend Biotech Corp Stock:
Check out the analysis of
Legend Biotech Correlation
against competitors. You can also try the
Analyst Advice
module to analyst recommendations and target price estimates broken down by several categories.
Is Biotechnology space expected to grow? Or is there an opportunity to expand the business' product line in the future? Factors like these will boost
the valuation of Legend Biotech
. If investors know Legend will grow in the future, the company's valuation will be higher. The financial industry is built on trying to define current growth potential and future valuation
accurately. All the valuation information about Legend Biotech listed above have to be considered, but the key to understanding future value is determining which factors weigh more heavily than
Earnings Share Revenue Per Share Quarterly Revenue Growth Return On Assets Return On Equity
(1.54) 2.505 1.544 (0.13) (0.22)
The market value of Legend Biotech Corp
is measured differently than its book value, which is the value of Legend that is recorded on the company's balance sheet. Investors also form their own opinion of Legend Biotech's value that differs
from its market value or its book value, called intrinsic value, which is Legend Biotech's true underlying value. Investors use various methods to calculate intrinsic value and buy a stock when its
market value falls below its intrinsic value. Because Legend Biotech's market value can be influenced by many factors that don't directly affect Legend Biotech's underlying business (such as a
pandemic or basic market pessimism), market value can vary widely from intrinsic value.
AltmanZ ScoreDetails PiotroskiF ScoreDetails BeneishM ScoreDetails FinancialAnalysisDetails Buy or SellAdviceDetails
Please note, there is a significant difference between Legend Biotech's value and its price as these two are different measures arrived at by different means. Investors typically determine
if Legend Biotech is a good investment
by looking at such factors as earnings, sales, fundamental and technical indicators, competition as well as analyst projections. However, Legend Biotech's price is the amount at which it trades on
the open market and represents the number that a seller and buyer find agreeable to each party. | {"url":"https://www.macroaxis.com/financial-statements/LEGN/Receivables-Turnover","timestamp":"2024-11-11T01:38:38Z","content_type":"text/html","content_length":"334379","record_id":"<urn:uuid:b0ec2893-55e1-47b9-82c2-efec52aafb99>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00513.warc.gz"} |
How does an object's shape affect its momentum? | TutorChase
How does an object's shape affect its momentum?
An object's shape does not directly affect its momentum; rather, it's the mass and velocity that determine momentum.
Momentum, in physics, is a vector quantity that depends on two factors: the mass of an object and its velocity. The formula for momentum (p) is p=mv, where m is the mass and v is the velocity. This
means that the momentum of an object is directly proportional to its mass and velocity. If either the mass or velocity increases, the momentum will also increase, and vice versa.
The shape of an object does not come into this equation and therefore does not directly affect an object's momentum. However, it's worth noting that the shape of an object can influence its velocity,
especially in situations involving fluid dynamics. For example, a streamlined object will move through a fluid (like air or water) more easily than a non-streamlined object, due to reduced drag. This
could indirectly affect the object's momentum by affecting its velocity.
In addition, the shape of an object can affect how it interacts with other objects and forces, which could indirectly influence its momentum. For instance, the shape of a car can affect how it
responds to wind resistance, which in turn can affect its speed and thus its momentum. Similarly, the shape of a ball can affect how it bounces off a wall, changing its direction and potentially its
speed, and thus its momentum.
In conclusion, while the shape of an object does not directly factor into the calculation of momentum, it can indirectly influence momentum by affecting velocity and the way an object interacts with
other forces and objects.
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The world’s top online tutoring provider trusted by students, parents, and schools globally. | {"url":"https://www.tutorchase.com/answers/igcse/physics/how-does-an-object-s-shape-affect-its-momentum","timestamp":"2024-11-11T06:11:59Z","content_type":"text/html","content_length":"62050","record_id":"<urn:uuid:ce0e6ea7-c3a4-497f-992c-efe8a77e5245>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00001.warc.gz"} |
First-order rate law (integral form)
The First-order Rate Law (Integral Form) calculator computes the concentration of a substance (A) based on a Rate Law equation, the initial concentration (A[0]), the rate constant (k) and the
duration of the reaction (s).
INSTRUCTIONS: Choose units and enter the following:
• [A[0]] Initial Concentration of the Substance.
• (k) Rate Constant.
• (s) Duration of the Reaction.
Concentration of Substance [A]: The calculator returns the concentration of the substance after the reaction in moles per liter (mol/L). However, this can be automatically converted to other
concentration units via the pull-down menu.
The Science
The first order rate integral^[1] equation calculates the rate at which the reactants turn in to products. Unlike the differential form of the first-order equation, the integral form looks at the
amount of reactants have been converted to products at a specific point in the reaction. A full integration of the equation can be found here.
The Math
The equation for the first-order rate reaction is
[A]=[A][o]e^−kt [2]
• [A] is the concentration of substance after reaction.
• [A][o] is the initial concentration in units of (mol/L)
• k is the rate constant in units of (1/sec)
• t is time of the concentration (sec
Related Topics
Supplement Material
[1] https://en.wikipedia.org/wiki/Rate_equation
[2]Whitten, et al. 10th Edition. Pp. 626,629,631 | {"url":"https://www.vcalc.com/wiki/Dasha/First-order-rate-law-integral-form","timestamp":"2024-11-12T17:04:24Z","content_type":"text/html","content_length":"54273","record_id":"<urn:uuid:ccccbb59-f754-402b-8fc5-b564c0b31a94>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00620.warc.gz"} |
The parallelDist package provides a fast parallelized alternative to R’s native ‘dist’ function to calculate distance matrices for continuous, binary, and multi-dimensional input matrices and offers
a broad variety of predefined distance functions from the ‘stats’, ‘proxy’ and ‘dtw’ R packages, as well as support for user-defined distance functions written in C++. For ease of use, the ‘parDist’
function extends the signature of the ‘dist’ function and uses the same parameter naming conventions as distance methods of existing R packages. Currently 41 different distance methods are supported.
The package is mainly implemented in C++ and leverages the ‘Rcpp’ and ‘RcppParallel’ package to parallelize the distance computations with the help of the ‘TinyThread’ library. Furthermore, the
Armadillo linear algebra library is used via ‘RcppArmadillo’ for optimized matrix operations for distance calculations. The curiously recurring template pattern (CRTP) technique is applied to avoid
virtual functions, which improves the Dynamic Time Warping calculations while keeping the implementation flexible enough to support different step patterns and normalization methods.
Documentation and Usage Examples
Usage examples and performance benchmarks can be found in the included vignette.
Details about the 41 supported distance methods and their parameters are described on the help page of the ‘parDist’ function. The help page can be displayed with the following command:
User-defined distance functions
Since version 0.2.0, parallelDist supports fast parallel distance matrix computations for user-defined distance functions written in C++.
A user-defined function needs to have the following signature (also see the Armadillo documentation):
Defining and compiling the function, as well as creating an external pointer to the user-defined function can easily be achieved with the cppXPtr function of the ‘RcppXPtrUtils’ package. The
following code shows a full example of defining and using a user-defined euclidean distance function:
# RcppArmadillo is used as dependency
# RcppXPtrUtils is used for simple handling of C++ external pointers
# compile user-defined function and return pointer (RcppArmadillo is used as dependency)
euclideanFuncPtr <- cppXPtr("double customDist(const arma::mat &A, const arma::mat &B) { return sqrt(arma::accu(arma::square(A - B))); }",
depends = c("RcppArmadillo"))
# distance matrix for user-defined euclidean distance function
# (note that method is set to "custom")
parDist(matrix(1:16, ncol=2), method="custom", func = euclideanFuncPtr)
More information can be found in the vignette and the help pages.
parallelDist is available on CRAN and can be installed with the following command:
The current version from github can be installed using the ‘devtools’ package:
Alexander Eckert
GPL (>= 2) | {"url":"http://ctan.mirror.garr.it/mirrors/CRAN/web/packages/parallelDist/readme/README.html","timestamp":"2024-11-06T20:03:45Z","content_type":"application/xhtml+xml","content_length":"11742","record_id":"<urn:uuid:8effa6ef-f562-41af-94a6-2e60ffc539cf>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00653.warc.gz"} |
Famous Problems and Other Monographs: Second Editionsearch
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Famous Problems and Other Monographs: Second Edition
AMS Chelsea Publishing: An Imprint of the American Mathematical Society
Hardcover ISBN: 978-0-8218-2674-4
Product Code: CHEL/108.H
List Price: $69.00
MAA Member Price: $62.10
AMS Member Price: $62.10
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Famous Problems and Other Monographs: Second Edition
AMS Chelsea Publishing: An Imprint of the American Mathematical Society
Hardcover ISBN: 978-0-8218-2674-4
Product Code: CHEL/108.H
List Price: $69.00
MAA Member Price: $62.10
AMS Member Price: $62.10
• AMS Chelsea Publishing
Volume: 108; 1962; 321 pp
MSC: Primary 00
Four volumes in one: Famous Problems of Elementary Geometry, by Klein. A fascinating, simple, easily understandable account of the famous problems of Geometry—The Duplication of the Cube,
Trisection of the Angle, Squaring of the Circle—and the proofs that these cannot be solved by ruler and compass. Suitably presented to undergraduates, with no calculus required. Also, the work
includes problems about transcendental numbers, the existence of such numbers, and proofs of the transcendence of \(e\).
From Determinant to Tensor, by Sheppard. A novel and simple introduction to tensors.
“An excellent little book, the aim of which is to familiarize the student with tensors and to give an idea of their applications. We wish to recommend the book heartily ... The beginner will find
the book a valuable introduction and the expert an interesting review with a refreshing novelty of presentation.”
—Bulletin of the AMS
Chapter headings: 1: Origin of Determinants; 2: Properties of Determinants; 3: Solution of Simultaneous Equations; 4: Properties; 5: Tensor Notation; 6: Sets; 7: Cogredience, etc. 8: Examples
from Statistics; 9: Tensors in Theory of Relativity.
Introduction to Combinatory Analysis, by MacMahon. An introduction to the author's two-volume work.
Three Lectures on Fermat's Last Theorem, by Mordell. This famous problem is so easy that a high-school student might not unreasonably hope to solve it: it is so difficult that as of the 1962
publication date of this book, tens of thousands of amateur and professional mathematicians, Euler and Gauss among them, failed to find a complete solution. Mordell himself had a solution (as he
said he did). This work is one of the masterpieces of mathematical exposition.
• Permission – for use of book, eBook, or Journal content
Volume: 108; 1962; 321 pp
MSC: Primary 00
Four volumes in one: Famous Problems of Elementary Geometry, by Klein. A fascinating, simple, easily understandable account of the famous problems of Geometry—The Duplication of the Cube, Trisection
of the Angle, Squaring of the Circle—and the proofs that these cannot be solved by ruler and compass. Suitably presented to undergraduates, with no calculus required. Also, the work includes problems
about transcendental numbers, the existence of such numbers, and proofs of the transcendence of \(e\).
From Determinant to Tensor, by Sheppard. A novel and simple introduction to tensors.
“An excellent little book, the aim of which is to familiarize the student with tensors and to give an idea of their applications. We wish to recommend the book heartily ... The beginner will find the
book a valuable introduction and the expert an interesting review with a refreshing novelty of presentation.”
—Bulletin of the AMS
Chapter headings: 1: Origin of Determinants; 2: Properties of Determinants; 3: Solution of Simultaneous Equations; 4: Properties; 5: Tensor Notation; 6: Sets; 7: Cogredience, etc. 8: Examples from
Statistics; 9: Tensors in Theory of Relativity.
Introduction to Combinatory Analysis, by MacMahon. An introduction to the author's two-volume work.
Three Lectures on Fermat's Last Theorem, by Mordell. This famous problem is so easy that a high-school student might not unreasonably hope to solve it: it is so difficult that as of the 1962
publication date of this book, tens of thousands of amateur and professional mathematicians, Euler and Gauss among them, failed to find a complete solution. Mordell himself had a solution (as he said
he did). This work is one of the masterpieces of mathematical exposition.
Permission – for use of book, eBook, or Journal content
Please select which format for which you are requesting permissions. | {"url":"https://bookstore.ams.org/CHEL/108","timestamp":"2024-11-12T07:04:59Z","content_type":"text/html","content_length":"65097","record_id":"<urn:uuid:b80df27a-7bd5-459e-999f-920da414ae10>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00478.warc.gz"} |
How to find the direction ratios of a line passing through 2 points?
A simple formula in 3 Dimensional Geometry which shows you how to calculate the direction ratios of a line joining 2 points. Class 12 Math students could use this formula. Is the concept of direction
ratios clear to you? You can email me at mathmadeeasy22@gmail.com | {"url":"https://www.mathmadeeasy.co/post/how-to-find-the-direction-ratios-of-a-line-passing-through-2-points","timestamp":"2024-11-02T17:27:38Z","content_type":"text/html","content_length":"1050490","record_id":"<urn:uuid:c33c476e-892a-4ccf-b3d9-f526fffee836>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00021.warc.gz"} |
Find all nonnegative integers a and b such that…
The problem is the following, and is a modified version of the 2009 British Mathematical Olympiad issue: Find all nonnegative integers a and b such that \(\sqrt{a}+\sqrt{b}=\sqrt{2020}.\)
Before looking at one possible way of solving this problem, which requires nothing more than school-level arithmetic, I want to explain why I like this problem, and problems of this nature, so much.
Growing up, I loved detective/mystery novels, tv shows, movies, whatever. Being able to construct a solution based on some scattered information seemed almost like a superpower to me.
Let’s begin our solution to the problem. The first step most people would take is to square both sides of the equation. This is something we are always told to do when dealing with square roots. It’s
certainly what I tried. Let’s do this with the equation as it is, so we have:
Ok, so this is a little nicer, maybe. We have reduced the number of square roots, which is good, but we have also made an unintentional problem for ourselves.
This is the first thing this problem taught me. Sometimes, before jumping in and just applying a method that seems right, ask yourself:
(1) Is this indeed the best method you know for dealing with this problem?
(2) If this is the best method you know, is the problem set up in the best way for you to apply it?
Let me elaborate here. There is nothing wrong with what we have done. However, in squaring the equation, in the form it is was given, we have added in the variable square root of the product \(ab\).
This has made things harder for us since we have now unintentionally muddled together the information given by the variables a and b.
What if instead we first rearranged our equation as
Then we can again square both sides, only this time we get:
Notice that this time we have again reduced the number of square roots we had, but we have additionally kept the variables a and b separate. This may seem like a simple difference but it makes all
the difference with how we can proceed. We now make note of the fact that all terms of our new equation are clearly integers, except potentially \(2\sqrt{2020b}\).
Here comes the second thing I learned from this problem, deductive reasoning. Since we know that adding or subtracting two integers always results in an integer, and since we can re-write our
equation as
we can deduce that \(2\sqrt{2020b}\) must indeed be an integer. This is a huge clue!
We first notice that
\(2020=2^2 \cdot 5 \cdot 101\)
Thus we can say that
So, since \(\sqrt{2020b}\) is an integer, we must have that \(\sqrt{505b}\) is an integer. It means that we must have, whit \(c \in \mathbf{N} \):
But, the \(\sqrt{505}\) is not an integer, so this only makes sense if \(b=505d^2\) for \(d \in \mathbf{N}\).
So, we have:
The third thing I learned from this problem: Don’t waste your time! What I mean by this is the following, there was nothing special about moving the \(b\) across the equals sign in our original
equation, we could just have easily started with the equation:
and everything would follow in the exact same way. In particular, we would arrive at a similar conclusion that \(a=505 e^2\) for some integer \(e\).
We call this a symmetric argument and really, all we are saying is that since it is clear that everything will work the exact same, if we were to replace the variable \(b\) with the variable \(a\),
we are not willing to write the argument down again and instead we will just skip to the conclusion. With \(e,d \in \mathbf{N}\):
\(a= 505 e^2\) \(b=505 d^2\)
So, going back to our original equation, we have:
This is certainly a much easier equation to deal with. There are only three solutions to this equation for d and e. In particular, we can have:
\(d=0, e=2 \Rightarrow a=2020, b=0\) \(d=1, e=1 \Rightarrow a=505, b=505\) \(d=2, e=0 \Rightarrow a=0, b=2020\)
{ 0 comments… add one } | {"url":"https://www.raucci.net/2021/08/24/find-all-nonnegative-integers-a-and-b-such-that/","timestamp":"2024-11-03T12:04:30Z","content_type":"text/html","content_length":"47670","record_id":"<urn:uuid:b439a8e4-659f-432d-91e4-40cee5c4fa10>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00706.warc.gz"} |
Chapter 5 MacroQuizwiz - Ace Your Homework & Exams, Now With ChatGPT AI
Chapter 5 Macro
Réussis tes devoirs et examens dès maintenant avec
If the nominal interest rate is 1% and the inflation rate is 5%, the real interest rate is:
-4 percent
According to economic theory, which of the following expressions can possibly represent the (general) demand for money (real)
100 - .05* i + 0.2* Y answer explanation: since demand for money depends negatively (inversely) on nominal interest rate, and positively on real income
If the average price of goods and services in the economy equals $10 and the quantity of money in the economy equals $200,000, then real balances in the economy equal:
According to the quantity theory of money equation, if the money supply increases 12%, velocity decreases 4%, and the price level increases 5%, then the change in real GDP must be _________ percent.
3 answer explanation: since the quantity theory of equation is: M*V = P*Y, each side is a product of two variables. Now use the % change result in chapter 2 to deduce: % change in M + % change in V =
% change in P + % change in Y. Plugging all the given values from the question into the last equation, we have: 12 + (-4) = 5 + % change in Y, which implies, % change in Y = 3.
If nominal GDP is $1000,then if there is $200 of money in the economy, velocity is _________ times per year:
5 answer explanation: use the equation M*V = P*Y = nominal GDP
In the classical model, and according to the quantity theory of money, a 5% increase in money through increases inflation by _______ percent. According to the Fisher equation a 5% increase in the
rate of inflation increases the nominal interest rate by _______.
5; 5 answer explanation: since in the classical model, Y is already determined/fixed at Y-bar and r is determined (already) in the goods market equilibrium equation. Given the above, M is
proportional to P (given velocity is a constant also). And since the real rate, r is already determined or fixed, any increase in inflation rate translates to the same increase in nominal rate
If the real return (real interest rate) on government bonds is 3% in the expected rate of inflation is 4%, then the cost of holding money is ________ percent
7 answer explanation: use answer in question 17
The classical economist wears a T-shirt printed with the slogan "fast money raises my interest!" use the results of the classical model, quantity theory of money and the fish are equation to explain
the slogan
According to the classical model and quantity theory of money, since velocity and real output (Y) are constant/fixed, an increase in quantity of money, M, would, by the quantity theory of money
equation, increase the price level by the same percentage. This means an increase in the inflation rate. By the Fisher equation/effect, an increase in inflation rate would increase the nominal
interest rate by the same percentage (as real interest rate is determined already in the goods market equilibrium).
If the demand for money depends positively on real income and depends inversely on the nominal interest rate, what will happen to the price level today, if the central bank announces that it will
decrease the money growth rate in the future, but it does not change the money supply today
People will expect lower inflation rate in the future. A lower inflation means a lower nominal interest rate. This means the demand for (real) money will increase. Since the Fed does not immediately
decrease the money supply, price (P) must fall so the real balances ( M/P ) would increase to match the increase demand for money [the money market equilibrium equation]. Thus, current price will
fall as a result of expected future decrease in money growth
In classical macroeconomic theory, the concept of monetary neutrality means that changes in the money supply do not influence real variables. Explain why changes in money growth affect the nominal
interest rate, but not the real interest rate
Recall that in the model in chapter 3 (the so-called classical model), the real interest rate, r, is already determined in the goods market equilibrium, and the real output, Y, is determined from the
production function with fixed amounts of L (L-bar) and K (K-bar). Hence all these real variables are determined without reference to money. In the quantity theory of money (equation), the money
supply, M, (with fixed Y and velocity, V) has a one-to-one positive relationship with price, P (since velocity is constant and Y is fixed at Y-bar). An increase in money supply would result in the
same percentage increase in P. Hence an increase in money supply would result in the same percentage increase in inflation. By the Fisher equation/effect, this would result in the same percentage
increase in nominal interest rate. Hence money supply only affects nominal variables such as P, nominal GDP ( which is P*Y), inflation rate, and nominal interest rate. Note that by the relationship:
nominal variable = P * the corresponding real variable, and since money supply affects P, which in turn would affect other nominal variables such as nominal consumption, nominal investment, and so
According to the quantity theory of money, ultimate control over the rate of inflation in the United States is exercised by:
The Fed answer explanation: because we can see for any fixed Y and V, M is proportional to P. It means any change in M (controlled by the Fed) would lead to changes in P (inflation)
Assume that the demand for real money balance (M/P) is M/P=0.6Y-100i, where why is national income and i is the nominal interest rate (in percent). The real interest rate car is fixed at 3% by the
goods market equilibrium equation. The expected inflation rate equals the right of the nominal money growth.
a. If Y is 1000, M is 100, and the growth rate of nominal money is 1%, what must i and P be? - Since the expected inflation rate = rate of growth of M = 1% (given), the nominal interest rate = i =
real interest rate + inflation rate = 3 + 1 = 4%. Now the (demand for) real balance is M/P = 0.6*1000 - 100*4 = 200. Since M = 100, it means 100 / P = 200 or P = ½. Note that the equation, M/P = 0.6Y
- 100i, can be interpreted as the money market equilibrium equation. And the above exercise shows that the value of price, P, can be determined from the money market equilibrium equation (when the
values of Y, r and expected inflation rate are given) b. If Y is 1000, M is 100, and the growth rate of nominal money is 2%, what must i and P be? - Repeat the same exercise as part a), we have P = 1
and i = 5%
Real money balances equal the:
amount of money expressed in terms of the quantity of goods and services it can purchase answer explanation: real money balances or real money supply = M / P, where P = price per unit of goods
If the Fed announces that it will raise the money supply in the future that does not change the money supply today,
both the nominal interest rate and the current price level will increase answer explanation: nominal rate will increase because of increase in expected inflation rate. If nominal rate goes up, the
demand for real balances will go down, hence for the money market to be in equilibrium, the supply of real balances, M/ P, has to decrease, which means prices have to go up assuming that there is no
change in M
The demand for real money balances is generally assumed to:
increase as real income (real GDP) increases
The characteristic of the classical model that the money supply it does not affect real variables is called:
monetary neutrality
If velocity is assumed to be constant, but no other assumptions are made, the level of ______ is determined by M.
nominal GDP answer explanation: see the equation in question 3 above
If the velocity of money remains constant while the quantity of money (M) doubles, then:
nominal GDP must double answer explanation: use the same equation as in question 3 above
The opportunity cost of holding money is the:
nominal interest rate answer explanation: because by holding money, not only one miss out on the real return (r ), but there is loss due to inflation (lowering of purchasing power), and the nominal
interest rate is the sum of real rate/return and inflation rate
The concept of monetary neutrality in the classical model means that an increase in the money supply will increase:
nominal interest rates
With the quantity theory of money equation, if the quantity of real money balances is k*Y, where K is a constant, then velocity is:
none of the above answer explanation: because rearranging the quantity theory of money equation, we have real balances = M / P = (1/V) * Y. We can identify 1/V as k. So k = 1 / V
The one to one relation between the inflation rate and the nominal interest rate, the fisher affect, assumes that the:
none of the above answer explanation: since nominal rate = real rate + inflation rate, inflation rate has a one-to-one effect on nominal rate if the real rate is constant
The rate of inflation is the:
percentage change in the level of prices
The definition of the velocity of money, V, is:
price (index) multiplied by real GDP divided by money supply
The general demand function for real (money) balances depends on the nominal interest rate and:
real income answer explanation: it depends on nominal rate because of the opportunity cost argument, see question 17
In the classical model, together with the quantity theory of money equation and the Fisher equation/effect, an increase in nominal money supply increases:
the nominal interest rate answer explanation: see answer in question 15 above
The real interest rate is equal to the:
the nominal interest rate minus the inflation rate
The quantity theory of money assumes that:
velocity is constant
Ensembles d'études connexes | {"url":"https://quizwizapp.com/fr/study/chapter-5-macro","timestamp":"2024-11-12T21:02:06Z","content_type":"text/html","content_length":"90229","record_id":"<urn:uuid:5806478c-b5bd-4691-aaf7-b0c57a309076>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00865.warc.gz"} |
How to generate a random number in lua?
In Lua, you can generate a random number using the math.random function. This function takes one or two arguments and returns a random number between 0 and 1 (inclusive).
Here's an example of how to use math.random to generate a random number between 1 and 100:
1 math.randomseed(os.time()) -- seed the random number generator with the current time
2 math.random(1, 100) -- generate a random number between 1 and 100
The first line sets the seed for the random number generator based on the current time, so that the sequence of random numbers generated will be different each time you run the program.
The second line calls math.random with two arguments: the first is the minimum value that can be generated (1 in this case), and the second is the maximum value that can be generated (100 in this
You can modify the arguments passed to math.random to generate random numbers within different ranges. | {"url":"https://devhubby.com/thread/how-to-generate-a-random-number-in-lua","timestamp":"2024-11-08T23:41:00Z","content_type":"text/html","content_length":"122324","record_id":"<urn:uuid:d24adc42-f196-4a11-8552-5d02b7d645e8>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00237.warc.gz"} |
Shannon–Weaver model
Assessment | Biopsychology | Comparative | Cognitive | Developmental | Language | Individual differences | Personality | Philosophy | Social |
Methods | Statistics | Clinical | Educational | Industrial | Professional items | World psychology |
Social psychology: Altruism · Attribution · Attitudes · Conformity · Discrimination · Groups · Interpersonal relations · Obedience · Prejudice · Norms · Perception · Index · Outline
File:Shannon-Weaver model.png
The Shannon–Weaver model of communication was one of the first models of communication has been called the "mother of all models."^[1] It embodies the concepts of information source, message,
transmitter, signal, channel, noise, receiver, information destination, probability of error, coding, decoding, information rate, channel capacity, etc.
In 1948 Claude Elwood Shannon published A Mathematical Theory of Communication article in two parts in the July and October numbers of the Bell System Technical Journal.^[2] In this fundamental work
he used tools in probability theory, developed by Norbert Wiener, which were in their nascent stages of being applied to communication theory at that time. Shannon developed information entropy as a
measure for the uncertainty in a message while essentially inventing what became known as the dominant form of "information theory."
The book co-authored with Warren Weaver, The Mathematical Theory of Communication, reprints Shannon's 1948 article and Weaver's popularization of it, which is accessible to the non-specialist.^[3]
Shannon's concepts were also popularized, subject to his own proofreading, in John Robinson Pierce's Symbols, Signals, and Noise.^[4]
The term Shannon–Weaver model was widely adopted into social science fields such as education, organizational analysis, psychology, etc. In engineering and mathematics, Shannon's theory is used more
literally and is referred to as Shannon theory, or information theory^[5].
Shannon's formula is ${\displaystyle C = W \log_{2} (1 + \tfrac{S}{N})}$ ,
where C is channel capacity measured in bits/second, W is the bandwidth in Hz, S is the signal level in watts across the bandwidth W, and N is the noise power in watts in the bandwidth W. | {"url":"https://psychology.fandom.com/wiki/Shannon%E2%80%93Weaver_model","timestamp":"2024-11-03T16:26:02Z","content_type":"text/html","content_length":"163631","record_id":"<urn:uuid:581ec716-edb8-4a77-b2b5-19c23c00f721>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00182.warc.gz"} |
TGD diary{{post.title}}Objection against the quantum gravitational metabolism
TGD based view about quantum gravity, which involves in an essential manner gravitational flux tubes and the notion of gravitational Planck constant h
, leads to a very nice picture about metabolism predicting correct order of magnitude for the metabolic energy quantum .5 eV assignable to the proton of the hydrogen bond transformed to a long
gravitational loop in Earth scale. This increases the energy of the proton and the increment serves as metabolic energy liberated when the loop reduces to an ordinary hydrogen bond in the transition
→ h.
A new metabolic energy quantum assignable to a dark electron Cooper pair is predicted and is assignable to a dark gravitational valence bond of metal ion X^++ or two metal ions of type X^+ forming a
dark Cooper pair with hydrogen peroxide H[2]O[2], which is reactive oxygen species (ROS). As in the case of the dark proton of the hydrogen bond, only an effective ion is in question since the Cooper
pair is at the gravitational flux tube.
Note that the transformation of the proton of hydrogen bond and electrons of valence bonds provides a mechanism, which makes it possible to control for instance membrane potential which is expected
to be central in the control of nerve pulse generation. DNA base pairs are connected by hydrogen bonds and the control mechanism might be at use also here.
An estimate for its maximal value is obtained by scaling the ordinary metabolic quantum by the ration m[e]/m[p] and equal to .36 meV. The miniature membrane potential has value about .4 meV which is
10 per cent larger.
In the first approximation, the estimate for the maximum of the metabolic energy quantum is as the gravitational binding energy at the surface of Earth. The estimate neglects the kinetic energy of
the dark particle at the flux tube, which can be reduced as the flux tube reduces to ordinary valence bond or hydrogen bond.
According to the simple model, bio-chemistry would strongly depend on the local gravitational environment. Could this be used to kill the idea?
1. For an object with mass M and radius R, the estimated maximal gravitational metabolic energy quantum E[max] is scaled up by factor is scaled up by a factor z= (M/M[E])× (R[E]/R). The values of z
for Mars, Venus, and Moon are (.02,.86,.05). For Venus, which is called the sister planet of Earth, z is not too far from unity. Note that in the case of the Moon, the Earth's gravitational
potential and therefore E[max] associated with it would be by a factor z= R[E]/R[Moon]= .017 smaller than at the surface of Earth.
Solar gravitational potential cannot come to rescue. At distance of AU (Earth's distance from Sun) the scaling factor of E[max] would be z= (M[S]/M[E]) (R[E]/AU) =.014. The values E[max] are
typically few per cent of the desired value.
The model in its simplest version would predict that terrestrial life is not possible on Mars and Moon. Humans have however successfully visited the Moon.
2. Rather than giving up the idea, it is better to ask what goes wrong with the simplest model. It is assumed that dark charge at the gravitational bond does not possess any kinetic energy, which
would increase the value of the metabolic energy quantum. This is of course an oversimplification and already the predicted slightly too low maximal value of the gravitational electronic
metabolic energy quantum suggests that kinetic energy cannot be neglected.
The simplest model for the particle at gravitational valence bond is as a particle in a box with kinetic energies given by E[n]= n^2ℏ[eff]^2/mL^2, L the length of the loop. If L scales like
h[eff], the kinetic energy does not depend on h[eff]. Therefore the scale of kinetic contribution can be estimated in a molecular length scale.
Could the system adapt to a reduction of the maximal gravitational potential at the surface of Moon, Mars, or Venus by increasing the average value of n in the superposition of the standing waves
having maximum at the top of the valence loop? The system would adapt by increasing the localization of the dark charge at the top of the loop. The reduction of the bond length would mean
reduction of the superposition to n=0 wave so that the kinetic energy would be indeed liberated.
See the article
Quantum gravitation and quantum biology in TGD Universe
or the
with the same title.
To sum up, the model survived the first killer test and led to a more precise formulation of the hypothesis. For a summary of earlier postings see Latest progress in TGD. | {"url":"https://matpitka.blogspot.com/2022/04/objection-against-quantum-gravitational.html","timestamp":"2024-11-03T18:36:47Z","content_type":"application/xhtml+xml","content_length":"131433","record_id":"<urn:uuid:f814ec25-3b67-44ef-83a6-aa9f62315c49>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00747.warc.gz"} |