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11101 | https://brainly.com/question/44397091 | [FREE] What is the formula to find the number of consecutive multiples of a given number within an inclusive - brainly.com
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What is the formula to find the number of consecutive multiples of a given number within an inclusive range?
A) k b−a+1
B) k b−a
C) k b+a
D) (b−a)k
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The number of consecutive multiples of a given number within an inclusive range is calculated using formula A) (b - a)/k + 1. This takes into account the start and end of the range and the specific multiple in question.
Explanation
To find the number of consecutive multiples of a given number within an inclusive range, we use the formula: (b - a)/k + 1, where a is the starting point of the range, b is the ending point of the range, and k is the specific multiple we are counting within the range. This formula counts how many times the number k fits into the range from a to b inclusively.
For example: Suppose we want to count how many multiples of 3 there are between 5 and 20. We subtract 5 from 20 to get 15, divide 15 by 3 to get 5, and then add 1 to get a total of 6 multiples of 3 within the range [5, 20]. The correct choice from the options given is A) (b - a)/k + 1.
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The formula to find the number of consecutive multiples of a given number within an inclusive range is k b−a+1. This formula accounts for the starting point, the ending point, and the specific multiple. Therefore, the correct answer is A) k b−a+1.
Explanation
To determine the number of consecutive multiples of a given number within an inclusive range, we use the formula:
k b−a+1
where:
a is the starting point of the range,
b is the ending point of the range, and
k is the specific multiple we are counting within the range.
This formula effectively calculates how many times the number k divides into the range from a to b inclusively.
Step-by-Step Explanation:
Identify the range: Determine the start (a) and end (b) values of your range.
Determine the multiple: Establish which number (k) you want to find the multiples of within that range.
Compute the difference: Subtract a from b to find how large the range is (this gives you the total distance between the two points).
Divide by the multiple: Divide the result from step 3 by k to find out how many times k fits into that difference.
Add one: Finally, add 1 to account for inclusivity, meaning you should also count the number at position a if it is a multiple of k.
Example:
Suppose you want to count how many multiples of 4 are between 8 and 32 (inclusive).
Let a=8 and b=32.
Your multiple k=4.
Calculate the total range: b−a=32−8=24.
Divide by the multiple: 24/4=6.
Add 1 for inclusivity: 6+1=7.
Thus, there are 7 multiples of 4 between 8 and 32, which are 8, 12, 16, 20, 24, 28, and 32.
Examples & Evidence
An example is counting multiples of 5 between 10 and 50. Here, let a=10, b=50, and k=5. The computation would show that there are 9 multiples of 5 in that range: 10, 15, 20, 25, 30, 35, 40, 45, and 50.
This method follows the principles of arithmetic and number theory, which establish how multiplication and division relate within integer ranges, ensuring accurate counting of intervals.
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11102 | https://math.mit.edu/~gs/books/sec3-2.pdf | 3.2 The Nullspace of A: Solving Ax = 0 11 THE NULLSPACE OF A: SOLVING AX = 0 3.2 This section is about the space of solutions to Ax = 0. The matrix A can be square or rectangular. One immediate solution is x = 0—and for invertible matrices this is the only solution.
For other matrices, not invertible, there are nonzero solutions to Ax = 0. Each solution x belongs to the nullspace of A. We want to find all solutions and identify this very important subspace.
DEFINITION The nullspace of A consists of all solutions to Ax = 0. These vectors x are in Rn. The nullspace containing all solutions x is denoted by N(A).
Check that the solution vectors form a subspace. Suppose x and y are in the nullspace (this means Ax = 0 and Ay = 0). The rules of matrix multiplication give A(x + y) = 0 + 0. The rules also give A(cx) = c0. The right sides are still zero. Therefore x + y and cx are also in the nullspace N(A). Since we can add and multiply without leaving the nullspace, it is a subspace.
To repeat: The solution vectors x have n components. They are vectors in Rn, so the nullspace is a subspace of Rn. The column space C(A) is a subspace of Rm.
If the right side b is not zero, the solutions of Ax = b do not form a subspace.
The vector x = 0 is only a solution if b = 0. When the set of solutions does not include x = 0, it cannot be a subspace. Section 3.4 will show how the solutions to Ax = b (if there are any solutions) are shifted away from the origin by one particular solution.
Example 1 The equation x + 2y + 3z = 0 comes from the 1 by 3 matrix A = [ 1 2 3 ].
This equation produces a plane through the origin. The plane is a subspace of R3. It is the nullspace of A. The solutions to x + 2y + 3z = 6 also form a plane, but not a subspace.
Example 2 Describe the nullspace of A = 1 2 3 6 .
Solution Apply elimination to the linear equations Ax = 0: x1 + 2x2 = 0 3x1 + 6x2 = 0 → x1 + 2x2 = 0 0 = 0 There is really only one equation. The second equation is the first equation multiplied by 3. In the row picture, the line x1 + 2x2 = 0 is the same as the line 3x1 + 6x2 = 0.
That line is the nullspace N(A).
To describe this line of solutions, the efficient way is to give one point on it (one special solution). Then all points on the line are multiples of this one. We choose the second component to be x2 = 1 (a special choice). From the equation x1 + 2x2 = 0, the June 5, 1998 12 Chapter 3 Vector Spaces and Subspaces first component must be x1 = −2. Then the special solution produces the whole nullspace: The nullspace N(A) contains all multiples of s = −2 1 .
This is the best way to describe the nullspace, by computing special solutions to Ax = 0.
The nullspace consists of all combinations of those special solutions. This example has one special solution and the nullspace is a line.
For the plane in Example 1 there are two special solutions: 1 2 3 x y z = 0 has the special solutions s1 = −2 1 0 and s2 = −3 0 1 .
Those vectors s1 and s2 lie on the plane x + 2y + 3z = 0, which is the nullspace of A = 1 2 3 . All vectors on the plane are combinations of s1 and s2.
Notice what is special about s1 and s2 in this example. They have ones and zeros in the last two components. Those components are “free” and we choose them specially.
Then the first components −2 and −3 are determined by the equation Ax = 0.
The first column of A = 1 2 3 contains the pivot, so the first component of x is not free. We only make a special choice (one or zero) of the free components that correspond to columns without pivots. This description of special solutions will be completed after one more example.
Example 3 Describe the nullspaces of these three matrices: A = 1 2 3 8 and B = A 2A = 1 2 3 8 2 4 6 16 and C = A 2A = 1 2 2 4 3 8 6 16 .
Solution The equation Ax = 0 has only the zero solution x = 0. The nullspace is Z, containing only the single point x = 0 in R2. To see this we use elimination: 1 2 3 8 x1 x2 = 0 0 yields 1 2 0 2 x1 x2 = 0 0 and x1 = 0 x2 = 0 .
The square matrix A is invertible. There are no special solutions. The only vector in its nullspace is x = 0.
The rectangular matrix B has the same nullspace Z. The first two equations in Bx = 0 again require x = 0. The last two equations would also force x = 0. When we add more equations, the nullspace certainly cannot become larger. When we add extra rows to the matrix, we are imposing more conditions on the vectors x in the nullspace.
The rectangular matrix C is different. It has extra columns instead of extra rows.
The solution vector x has four components. Elimination will produce pivots in the first June 5, 1998 3.2 The Nullspace of A: Solving Ax = 0 13 two columns, but the last two nonpivot columns are “free”: C = 1 2 2 4 3 8 6 16 becomes U = 1 2 2 4 0 2 0 4 ↑ ↑ ↑ ↑ pivot columns free columns For the free variables x3 and x4, we make the special choices of ones and zeros. Then the pivot variables x1 and x2 are determined by the equation Ux = 0. We get two special solutions in the nullspace of C (and also the nullspace of U).
The special solutions are: s1 = −2 0 1 0 and s2 = 0 −2 0 1 ←pivot ← variables ←free ← variables One more comment to anticipate what is coming soon. Elimination will not stop at the upper triangular U! We continue to make this matrix simpler, in two ways: 1.
Produce zeros above the pivots, by eliminating upward.
2.
Produce ones in the pivots, by dividing the whole row by its pivot.
Those steps don’t change the zero vector on the right side of the equation. The nullspace stays the same. This nullspace becomes easy to see when we reach the re-duced row echelon form R: U = 1 2 2 4 0 2 0 4 becomes R = 1 0 2 0 0 1 0 2 .
↑ ↑ pivot columns contain I I subtracted row 2 of U from row 1, and then I multiplied row 2 by 1 2. The original two equations have simplified to x1 + 2x3 = 0 and x2 + 2x4 = 0. Those are the equations Rx = 0 with the identity matrix in the pivot column.
The special solutions are still the same s1 and s2. They are much easier to find from the reduced system Rx = 0.
Before moving to m by n matrices A and their nullspaces N(A) and the special solutions in the nullspace, allow me to repeat one comment. For many matrices, the only solution to Ax = 0 is x = 0. Their nullspaces contain only that single vector x = 0. The only combination of the columns that produces b = 0 is then the “zero June 5, 1998 14 Chapter 3 Vector Spaces and Subspaces combination” or “trivial combination”. The solution is trivial (just x = 0) but the idea is not trivial.
This case of a zero nullspace Z is of the greatest importance. It says that the columns of A are independent. No combination of columns gives the zero vector except the zero combination. All columns have pivots and no columns are free. You will see this idea of independence again . . . .
Solving Ax = 0 by Elimination This is important. We solve m equations in n unknowns—and the right sides are all zero. The left sides are simplified by row operations, after which we read off the so-lution (or solutions). Remember the two stages in solving Ax = 0: 1.
Forward elimination from A to a triangular U (or its reduced form R).
2.
Back substitution in Ux = 0 or Rx = 0 to find x.
You will notice a difference in back substitution, when A and U have fewer than n pivots. We are allowing all matrices in this chapter, not just the nice ones (which are square matrices with inverses).
Pivots are still nonzero. The columns below the pivots are still zero. But it might happen that a column has no pivot. In that case, don’t stop the calculation. Go on to the next column. The first example is a 3 by 4 matrix: A = 1 1 2 3 2 2 8 10 3 3 10 13 .
Certainly a11 = 1 is the first pivot. Clear out the 2 and 3 below that pivot: A → 1 1 2 3 0 0 4 4 0 0 4 4 (subtract 2 × row 1) (subtract 3 × row 1) The second column has a zero in the pivot position. We look below the zero for a nonzero entry, ready to do a row exchange. The entry below that position is also zero.
Elimination can do nothing with the second column. This signals trouble, which we expect anyway for a rectangular matrix. There is no reason to quit, and we go on to the third column.
The second pivot is 4 (but it is in the third column). Subtracting row 2 from row 3 clears out that column below the pivot. We arrive at U = 1 1 2 3 0 0 4 4 0 0 0 0 (only two pivots) (the last equation became 0 = 0) June 5, 1998 3.2 The Nullspace of A: Solving Ax = 0 15 The fourth column also has a zero in the pivot position—but nothing can be done.
There is no row below it to exchange, and forward elimination is complete. The matrix has three rows, four columns, and only two pivots. The original Ax = 0 seemed to involve three different equations, but the third equation is the sum of the first two. It is automatically satisfied (0 = 0) when the first two equations are satisfied. Elimination reveals the inner truth about a system of equations.
Now comes back substitution, to find all solutions to Ux = 0. With four unknowns and only two pivots, there are many solutions. The question is how to write them all down.
A good method is to separate the pivot variables from the free variables.
P The pivot variables are x1 and x3, since columns 1 and 3 contain pivots.
F The free variables are x2 and x4, because columns 2 and 4 have no pivots.
The free variables x2 and x4 can be given any values whatsoever. Then back substitu-tion finds the pivot variables x1 and x3. (In Chapter 2 no variables were free. When A is invertible, all variables are pivot variables.) The simplest choices for the free vari-ables are ones and zeros. Those choices give the special solutions.
Special Solutions – Set x2 = 1 and x4 = 0.
By back substitution x3 = 0 and x1 = −1.
– Set x2 = 0 and x4 = 1.
By back substitution x3 = −1 and x1 = −1.
These special solutions solve Ux = 0 and therefore Ax = 0. They are in the nullspace.
The good thing is that every solution is a combination of the special solutions.
Complete Solution x = x2 −1 1 0 0 + x4 −1 0 −1 1 = −x2 −x4 x2 −x4 x4 .
special special complete (1) Please look again at that answer. It is the main goal of this section. The vector s1 = (−1, 1, 0, 0) is the special solution when x2 = 1 and x4 = 0. The second special solution has x2 = 0 and x4 = 1. All solutions are linear combinations of s1 and s2.
The special solutions are in the nullspace N(A), and their combinations fill out the whole nullspace.
The MATLAB code nulbasis computes these special solutions. They go into the columns of a nullspace matrix N. The complete solution to Ax = 0 is a combination of those columns. Once we have the special solutions, we have the whole nullspace.
There is a special solution for each free variable. If no variables are free—this means there are n pivots—then the only solution to Ux = 0 and Ax = 0 is the trivial June 5, 1998 16 Chapter 3 Vector Spaces and Subspaces solution x = 0. All variables are pivot variables. In that case the nullspaces of A and U contain only the zero vector. With no free variables, and pivots in every column, the output from nulbasis is an empty matrix.
Example 4 Find the nullspace of U = 1 5 7 0 0 9 .
The second column of U has no pivot. So x2 is free. The special solution has x2 = 1.
Back substitution into 9x3 = 0 gives x3 = 0. Then x1 + 5x2 = 0 or x1 = −5. The solutions to Ux = 0 are multiples of one special solution: x = x2 −5 1 0 The nullspace of U is a line in R3.
It contains multiples of the special solution.
One variable is free, and N = nulbasis (U) has one column.
In a minute we will continue elimination on U, to get zeros above the pivots and ones in the pivots. The 7 is eliminated and the pivot changes from 9 to 1. The final result of this elimination will be R: U = 1 5 7 0 0 9 reduces to R = 1 5 0 0 0 1 .
This makes it even clearer that the special solution is s = (−5, 1, 0).
Echelon Matrices Forward elimination goes from A to U. The process starts with an m by n matrix A. It acts by row operations, including row exchanges. It goes on to the next column when no pivot is available in the current column. The m by n “staircase” U is an echelon matrix.
Here is a 4 by 7 echelon matrix with the three pivots highlighted in boldface: U = x x x x x x x 0 x x x x x x 0 0 0 0 0 x x 0 0 0 0 0 0 0 Three pivot variables x1, x2, x6 Four free variables x3, x4, x5, x7 Four special solutions in N(U) Question What are the column space and the nullspace for this matrix?
Answer The columns have four components so they lie in R4. (Not in R3!) The fourth component of every column is zero. Every combination of the columns—every vector in the column space—has fourth component zero. The column space C(U) consists of all vectors of the form (b1, b2, b3, 0). For those vectors we can solve Ux = b by back substitution. These vectors b are all possible combinations of the seven columns.
The nullspace N(U) is a subspace of R7. The solutions to Ux = 0 are all the combinations of the four special solutions—one for each free variable: June 5, 1998 3.2 The Nullspace of A: Solving Ax = 0 17 1.
Columns 3, 4, 5, 7 have no pivots. So the free variables are x3, x4, x5, x7.
2.
Set one free variable to 1 and set the other free variables to zero.
3.
Solve Ux = 0 for the pivot variables x1, x2, x6.
4.
This gives one of the four special solutions in the nullspace matrix N.
The nonzero rows of an echelon matrix come first. The pivots are the first nonzero entries in those rows, and they go down in a staircase pattern. The usual row operations (in the Teaching Code plu) produce a column of zeros below every pivot.
Counting the pivots leads to an extremely important theorem. Suppose A has more columns than rows. With n > m there is at least one free variable. The system Ax = 0 has at least one special solution. This solution is not zero!
3B If Ax = 0 has more unknowns than equations (A has more columns than rows), then it has nonzero solutions.
In other words, a short wide matrix (n > m) always has nonzero vectors in its nullspace.
There must be at least n −m free variables, since the number of pivots cannot exceed m.
(The matrix only has m rows, and a row never has two pivots.) Of course a row might have no pivot—which means an extra free variable. But here is the point: When there is a free variable, it can be set to 1. Then the equation Ax = 0 has a nonzero solution.
To repeat: There are at most m pivots. With n > m, the system Ax = 0 has a free variable and a nonzero solution. Actually there are infinitely many solutions, since any multiple cx is also a solution. The nullspace contains at least a line of solutions.
With two free variables, there are two special solutions and the nullspace is even larger.
The nullspace is a subspace. Its “dimension” is the number of free variables. This central idea—the dimension of a subspace—is defined and explained in this chapter.
The Reduced Echelon Matrix R From the echelon matrix U we can go one more step. Continue onward from U = 1 1 2 3 0 0 4 4 0 0 0 0 .
We can divide the second row by 4. Then both pivots equal 1. We can subtract 2 times this new row 0 0 1 1 from the row above. That produces a zero above the second pivot as well as below. The reduced row echelon matrix is R = 1 1 0 1 0 0 1 1 0 0 0 0 .
June 5, 1998 18 Chapter 3 Vector Spaces and Subspaces R has 1’s as pivots. It has 0’s everywhere else in the pivot columns. Zeros above pivots come from upward elimination.
If A is invertible, its reduced row echelon form is the identity matrix R = I.
This is the ultimate in row reduction.
The zeros in R make it easy to find the special solutions (the same as before): 1.
Set x2 = 1 and x4 = 0. Solve Rx = 0. Then x1 = −1 and x3 = 0.
2.
Set x2 = 0 and x4 = 1. Solve Rx = 0. Then x1 = −1 and x3 = −1.
The numbers −1 and 0 are sitting in column 2 of R (with plus signs). The numbers −1 and −1 are sitting in column 4 (with plus signs). By reversing signs we can read off the special solutions from the matrix R. The general solution to Ax = 0 or Ux = 0 or Rx = 0 is a combination of those two special solutions: The nullspace N(A) = N(U) = N(R) contains x = x2 −1 1 0 0 + x4 −1 0 −1 1 = (complete solution of Ax = 0).
The next section of the book moves firmly from U to R. The MATLAB command [ R, pivcol ] = rref(A) produces R and also a list of the pivot columns.
REVIEW OF THE KEY IDEAS 1.
The nullspace N(A) contains all solutions to Ax = 0.
2.
Elimination produces an echelon matrix U, or a row reduced R, with pivot columns and free columns.
3.
Every free column leads to a special solution to Ax = 0. The free variable equals 1 and the other free variables equal 0.
4.
The complete solution to Ax = 0 is a combination of the special solutions.
5.
If n > m then A has at least one column without pivots, giving a special solution.
So there are nonzero vectors x in the nullspace of this A. |
11103 | https://av1611.com/kjbp/kjv-dictionary/covet.html | COVET - Definition from the KJV Dictionary
Thou hast magnified thy word above all thy name. —Psalm 138:2,KJV
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KJV Dictionary
C
covet
« covert
cow »
KJV Dictionary Definition: covet
covet
COVET, v.t.
To desire or wish for, with eagerness; to desire earnestly to obtain or possess; in a good sense.
Covet earnestly the best gifts. 1 Corinthians 12.
To desire inordinately; to desire that which it is unlawful to obtain or possess; in a bad sense.
Thou shalt not covet thy neighbors house, wife or servant. Exodus 20.
COVET, v.i. To have an earnest desire. 1 Timothy 6.
covetable
COVETABLE, a. That may be coveted.
coveted
COVETED, pp. Earnestly desired; greatly wished or longed for.
coveting
COVETING, n. Inordinate desire.
covetous
COVETOUS, a.
Very desirous; eager to obtain; in a good sense; as covetous of wisdom, virtue or learning.
Inordinately desirous; excessively eager to obtain and possess; directed to money or goods, avaricious.
A bishop must not be covetous. 1 Timothy 3.
covetously
COVETOUSLY, adv. With a strong or inordinate desire to obtain and possess; eagerly; avariciously.
covetousness
COVETOUSNESS, n.
A strong or inordinate desire of obtaining and possessing some supposed good; usually in a bad sense, and applied to an inordinate desire of wealth or avarice.
Out of the heart proceedeth covetousness. Mark 7.
Mortify your members--and covetousness which is idolatry. Colossians 3.
Strong desire; eagerness.
Definitions from Webster's American Dictionary of the English Language, 1828.
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11104 | https://cse.buffalo.edu/~xinhe/cse191/Classnotes/note04-1x2.pdf | Sets and Set Operations Class Note 04: Sets and Set Operations Computer Sci & Eng Dept SUNY Buffalo c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 1 / 45 Sets Definition: A Set is a collection of objects that do NOT have an order.
Each object is called an element.
We write e ∈S if e is an element of S; and e ̸∈S if e is not an element of S.
Set is a very basic concept used in all branches of mathematics and computer science.
How to describe a set: Either we list all elements in it, e.g., {1, 2, 3}.
Or we specify what kind of elements are in it, e.g., {a | a > 2, a ∈R}.
(Here R denotes the set of all real numbers).
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 3 / 45 Example sets N = {0, 1, 2, . . .}: the set of natural numbers.
(Note: in some books, 0 is not considered a member of N.) Z = {0, −1, 1, −2, 2, . . .}: the set of integers.
Z+ = {1, 2, 3, . . .}: the set of positive integers.
Q = {p/q | p ∈Z, q ∈Z, q ̸= 0}: the set of rational numbers.
Q+ = {x | x ∈Q, x > 0}: the set of positive rational numbers.
R: the set of real numbers.
R+ = {x | x ∈R, x > 0}: the set of positive real numbers.
Definition: The empty set, denoted by ∅, is the set that contains no elements.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 4 / 45 More example sets A={Orange, Apple, Banana} is a set containing the names of three fruits.
B={Red, Blue, Black, White, Grey} is a set containing five colors.
{x | x takes CSE191 at UB in Spring 2014} is a set of 220 students.
{N,Z,Q,R} is a set containing four sets.
{x | x ∈{1, 2, 3} and x > 1 } is a set of two numbers.
Note: When discussing sets, there is a universal set U involved, which contains all objects under consideration. For example: for A, the universal set might be the set of names of all fruits. for B, the universal set might be the set of all colors.
In many cases, the universal set is implicit and omitted from discussion. In some cases, we have to make the universal set explicit.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 5 / 45 Equal sets Definition: Two sets are equal if and only if they have the same elements.
Note that the order of elements is not a concern since sets do not specify orders of elements.
We write A = B, if A and B are equal sets.
Example: {1,2,3} = {2,1,3} {1, 2, 3, 4} = {x ∈Z and 1 ≤x < 5} c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 6 / 45 Subset Definition: A set A is a subset of B if every element of A is also in B.
We write A ⊆B if A is a subset of B.
Clearly, for any set A, the empty set ∅(which does not contain any element) and A itself are both subsets of A.
Definition: If A ⊆B but A ̸= B, then A is a proper subset of B, and we write A ⊂B.
Fact: Suppose A and B are sets. Then A = B if and only if A ⊆B and B ⊆A.
This fact is often used to prove set identities.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 7 / 45 Cardinality Definition: If a set A contains exactly n elements where n is a non-negative integer, then A is a finite set, and n is called the cardinality of A. We write |A| = n.
For a finite set, its cardinality is just the “size” of A.
Note: ∅is the empty set (containing no element); {∅} is the set containing one element (which is the empty set).
Example: |{x | −2 < x < 5, x ∈Z}| =?
|∅| =?
|{x | x ∈∅and x < 3}| =?
|{x | x ∈{∅}}| =?
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 8 / 45 Cardinality of infinite set Definition: If A is not finite, then it is an infinite set.
What is the cardinality (i.e. the size) of an infinite set?
Do all infinite sets have the same size (i.e ∞)?
Apparently, they do not: It appears that there are more rational numbers than integers and there are more real numbers than rational numbers. (I say appears because, with proper definition, only one of these two statements is true.) But how do we define the notion: “an infinite set contains more elements than another infinite set”?
We shall deal with this later.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 9 / 45 Power set Definition The power set of set A is the set of all subsets of A. We denote it by P(A).
Example: P({1, 2, 3}) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}.
P(∅) = {∅}.
P({∅}) = {∅, {∅}}.
Fact: In general, |P(A)| = 2|A|.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 10 / 45 Ordered tuple Recall that a set does not consider its elements order.
But sometimes, we need to consider a sequence of elements, where the order is important.
An ordered n-tuple (a1, a2, . . . , an) has a1 as its first element, a2 as its second element, . . ., an as its nth element.
The order of elements is important in such a tuple.
Note that (a1, a2) ̸= (a2, a1) but {a1, a2} = {a2, a1}.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 11 / 45 Cartesian product Definition: The Cartesian product of A1, A2, . . . , An, denoted by A1 × A2 × · · · × An, is defined as the set of ordered tuples (a1, a2, . . . , an) where a1 ∈A1, a2 ∈A2, . . . , an ∈An. That is: A1 × A2 × · · · × An = {(a1, a2, . . . , an) | a1 ∈A1, a2 ∈A2, . . . , an ∈An} c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 12 / 45 Example Cartesian products Examples: {1, 2} × {3, 4, 5} = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)}.
{Male, Female} × {Married, Single} × {Student, Faculty} = {(Male, Married, Student), (Male, Married, Faculty), (Male, Single, Student), (Male, Single, Faculty), (Female, Married, Student), (Female, Married, Faculty), (Female, Single, Student), (Female, Single, Faculty)}.
R × R = {(x, y) | x ∈R, y ∈R} is the set of point coordinates in the 2D plane.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 13 / 45 Cardinality of Cartesian product Fact: In general, if Ai’s are finite sets, we have: |A1 × A2 × · · · × An| = |A1| × |A2| × · · · × |An| Example: |{(s, g) | s is a CSE191 student and g is a letter grade}| =?
Solution: |{ CSE 191 student }| = 220; |{ Letter grade }| = 10; So, |{(s, g)|s is a CSE 191 student and g is a letter grade }| = 220 x 10 = 2200.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 14 / 45 Using set notation with quantifiers Sometimes, we restrict the domain of a quantified statement explicitly by using set notations.
We use ∀x ∈S (P(x)) to denote that P(x) holds for every x ∈S.
We use ∃x ∈S(P(x)) to denote that P(x) holds for some x ∈S.
Example: ∀x ∈R (x2 ≥0) means that the square of every real number is greater than or equal to 0.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 15 / 45 Truth set Definition: Consider a domain D and a predicate P(x). The truth set of P is the set of elements x in D for which P(x) holds.
Using the notation of sets, we can write {x ∈D | P(x)}.
Clearly, it is a subset of D. It is equal to D if and only if P(x) holds for all x ∈D.
Example: {x ∈{1, 2, 3} | x > 1.5} = {2, 3}.
{x ∈R | x2 = 0} = {0}.
{x ∈R | x2 ≥0} = R.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 16 / 45 Set operations Recall: We have +, −, ×, ÷, . . . operators for numbers.
We have ∨, ∧, ¬, →. . . operators for propositions.
Question: What kind of operations do we have for sets?
Answer: union, intersection, difference, complement, . . .
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 18 / 45 Set Union Definition: The union of two sets A and B is the set that contains exactly all the elements that are in either A or B (or in both).
We write A ∪B.
Formally, A ∪B = {x | x ∈A or x ∈B}.
Example: {1, 2, 3} ∪{3, 4, 5} = {1, 2, 3, 4, 5} {x | x > 0} ∪{x | x > 1} = {x | x > 0} Venn Diagram of Union Operation: U B A c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 19 / 45 Set intersection Definition: The intersection of two sets A and B is the set that contains exactly all the elements that are in both A and B.
We write A ∩B.
Formally, A ∩B = {x | x ∈A and x ∈B}.
Example: {1, 2, 3} ∩{3, 4, 5} = {3} {x | x > 0} ∩{x | x > 1} = {x | x > 1} Venn Diagram of Intersection Operation: U B A c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 20 / 45 Disjoint set Definition: Two sets A and B are disjoint if A ∩B = ∅.
Example: {1, 2, 3} ∩{4, 5} = ∅, so they are disjoint.
{1, 2, 3} ∩{3, 4, 5} ̸= ∅, so they are not disjoint.
Q ∩R+ ̸= ∅, so they are not disjoint.
{x | x < −2} ∩R+ = ∅, so they are disjoint.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 21 / 45 Cardinality of intersection and union Lemma: For any two sets A and B, we have |A ∪B| = |A| + |B| −|A ∩B| Intuitively, when we count the elements in A and the elements in B separately, those elements in A ∩B have been counted twice. So when we subtract |A ∩B| from |A| + |B|, we get the cardinality of the union.
An extension of this result is called the inclusion-exclusion principle. We will discuss this later.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 22 / 45 Set complement Definition: The complement of set A, denoted by A, is the set that contains exactly all the elements that are not in A.
Formally, A = {x | x ̸∈A}.
Suppose U is the universe. Then, A = U −A.
Example: Let the universe be R.
{0} = {x | x ̸= 0 ∧x ∈R} R+ = {x | x ≤0 ∧x ∈R}.
Venn Diagram of Complement Operation: U _ A A c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 23 / 45 Set difference Definition: The difference of set A and set B, denoted by A −B, is the set that contains exactly all elements in A but not in B.
Formally, A −B = {x | x ∈A ∧x ̸∈B} = A ∩B.
Example: {1, 2, 3} −{3, 4, 5} = {1, 2} R −{0} = {x | x ∈R ∧x ̸= 0} Z −{2/3, 1/4, 5/8} = Z Venn Diagram of Difference Operation: U U A B A−B B A B−A c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 24 / 45 Symmetric Difference Definition: The symmetric difference of set A and set B, denoted by A ⊕B, is the set containing those elements in exactly one of A and B.
Formally: A ⊕B = (A −B) ∪(B −A).
Venn Diagram of Symmetric Difference Operation: U A B A−B B−A c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 25 / 45 Example for set operations Suppose A is the set of students who loves CSE 191, and B is the set of students who live in the university dorm.
A ∩B: the set of students who love CSE 191 and live in the university dorm.
A ∪B: the set of students who love CSE 191 or live in the university dorm.
A −B: the set of students who love CSE 191 but do not live in the university dorm.
B −A: the set of students who live in the university dorm but do not love CSE 191.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 26 / 45 Example for calculating set operations Example: Let A = {1, 2, 3, 4, 5} and B = {1, 2, 3, 4, 5, 6, 7, 8}. Then: A ∩B = {1, 2, 3, 4, 5} A ∪B = {1, 2, 3, 4, 5, 6, 7, 8} A −B = ∅ B −A = {6, 7, 8} c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 27 / 45 Representing Sets in Computer Programs Set is an important data structure in CS. How to represent sets in computer programs?
Let U = {s1, s2, . . . , sn} be the universal set.
We can use an array S of n-bits to represent the sets in U and the set operations. Let S[i] be the ith bit in S. Each S[i] is either 0 or 1.
To represent a subset A ⊆S, we use: SA[i] = 0 if si ̸∈A 1 if si ∈A This is called the bit map representation of sets (discussed in CSE250).
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 29 / 45 Example of bit map representations Example: U = {a, b, c, d, e, f, g}.
A = ∅: SA = [0, 0, 0, 0, 0, 0, 0].
A = U: SA = [1, 1, 1, 1, 1, 1, 1].
A = {b, d, f}: SA = [0, 1, 0, 1, 0, 0, 1].
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 30 / 45 Logic Operators in Java and C++ Logic Operators in Java and C++ Bitwise and: & 0 1 0 1 0 1 & 1 1 1 0 0 1 0 1 0 0 0 1 Bitwise or: | 0 1 0 1 0 1 | 1 1 1 0 0 1 1 1 1 1 0 1 Bitwise exclusive or:ˆ 0 1 0 1 0 1 ˆ 1 1 1 0 0 1 1 0 1 1 0 0 c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 31 / 45 Using C++ operators to calculate set operations Using C++ operators to calculate set operations: Let A and B be two subsets of U. Let SA and SB be the bit map representation of A and B, respectively.
SA∩B = SA & SB; SA∪B = SA | SB; SA⊕B = SAˆSB; SA = SAˆ[1, 1, . . . , 1]; c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 32 / 45 Set Identities: distributivity Set operations satisfy several laws. If we consider: ∩similar to ∧; ∪similar to ∨; A similar to ¬A; The universal set U similar to T; The empty set ∅similar to F; then, they are very similar to the logic laws.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 34 / 45 Set Identities: distributivity Distributivity laws: Just like the distributivity in logical equivalence, for sets we have: A ∩(B ∪C) = (A ∩B) ∪(A ∩C) A ∪(B ∩C) = (A ∪B) ∩(A ∪C) Example: {red, blue} ∩({red, black, white} ∪{blue}) = {red, blue} ∩{red, black, white, blue} = {red,blue} = {red} ∪{blue} = ({red,blue} ∩{red,black,white}) ∪({red,blue} ∩{blue}) c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 35 / 45 Set Identities: distributivity Venn Diagram Proof for: A ∩(B ∪C) = (A ∩B) ∪(A ∩C) A ^ (B U C) B U C U U A ^ B A ^ C A B C A B C c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 36 / 45 Set identities: DeMorgan law DeMorgan laws: Just like the DeMorgan laws in logic, we have: A ∩B = A ∪B A ∪B = A ∩B Example: Let the universe be {0,1,2,3}.
{0, 1} ∩{1, 2} = {1} = {0, 2, 3} = {2, 3} ∪{0, 3} = {0, 1} ∪{1, 2}.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 37 / 45 Other identities Table: Set Identifies Identify Name A ∩U = A A ∪∅= A Identity laws A ∪U = U A ∩∅= ∅ Domination laws A ∪A = A A ∩A = A Idempotent laws (A) = A Complementation law A ∪B = B ∪A A ∩B = B ∩A Commutative laws (A ∪B) ∪C = A ∪(B ∪C) (A ∩B) ∩C = A ∩(B ∩C) Associative laws A ∪(B ∩C) = (A ∪B) ∩(A ∪C) A ∩(B ∪C) = (A ∩B) ∪(A ∩C) Distributive laws A ∪B) = A ∩B A ∩B = A ∪B De Morgan’s laws A ∪(A ∩B) = A A ∩(A ∪B) = A Absorption laws A ∪A = U A ∩A = ∅ Complement laws You should get familiar with these laws, so that you can use them to prove set identities.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 38 / 45 How to prove set identities To prove a set identity: X = Y it is necessary and sufficient to show two things: X ⊆Y and Y ⊆X Equivalently, it is necessary and sufficient to show two things: x ∈X →x ∈Y and x ∈Y →x ∈X c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 39 / 45 Example for proving set identities Example: Show that if A, B, C are sets, then: (A −B) −C = (A −C) −(B −C) Proof: First we show x ∈LHS →x ∈RHS: x ∈(A −B) −C (by the definition of “set difference”) ⇒x ∈(A −B) but x ̸∈C. Hence x ∈A, x ̸∈B and x ̸∈C.
So x ∈A −C and x ̸∈B −C. This means x ∈(A −C) −(B −C) = RHS.
Next we show x ∈RHS →x ∈LHS: x ∈(A −C) −(B −C) (by definition of “set difference”) ⇒x ∈(A −C) but x ̸∈(B −C). Hence: x ∈A, x ̸∈C and x ̸∈(B −C).
Here: x ̸∈B −C means either x ̸∈B or x ∈C.
Since the latter contradicts x ̸∈C, we must have x ̸∈B.
This implies x ∈(A −B) −C = LHS.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 40 / 45 Prove set identities by truth table We can also prove the identify by using membership table (which is similar to truth table): A B C A A −B A −C B −C (A −B) −C (A −C) −(B −C) T T T F F F F F F T T F F F F T F F T F T F F F F F F T F F F F F F F F F T T T F F F F F F T F T F T T F F F F T T T F F F F F F F T T T F T T Each row specifies membership conditions. For example, the row 1 is {x |x ∈A, x ∈B, x ∈C}; the row 2 is {x |x ∈A, x ∈B, x ̸∈C}.
The last two columns are identical. So the two sets are the same.
c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 41 / 45 Generalized Union The previously studied union operation applies to only two sets.
We can generalize it to n sets.
Generally speaking, the union of a collection of sets is the set that contains exactly those elements that are in at least one of the sets in the collection.
We write: A1 ∪A2 ∪. . . ∪An = n [ i=1 Ai That is: n [ i=1 Ai = {x | x ∈A1 ∨x ∈A2 ∨. . . ∨x ∈An} c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 42 / 45 Example for generalized union Example 1: Suppose that Ai = {1, 2, . . . , i} for all positive integer i. Then n [ i=1 Ai = {1, 2, . . . , n} = An Example 2: Suppose that Ai = {i + 1, i + 2, . . . , 2i} for all positive integer i. Then: Sn i=1 Ai = {2} ∪{3, 4} ∪{4, 5, 6} ∪. . . ∪{n + 1, n + 2, . . . , 2n} = {2, 3, 4, . . . , 2n} = {x | 2 ≤x ≤2n, x ∈Z} c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 43 / 45 Generalized intersection Similarly, we can generalize intersection to n sets.
Generally speaking, the intersection of a collection of sets is the set that contains exactly those elements that are in all of the sets in the collection.
We write: A1 ∩A2 ∩. . . ∩An = n \ i=1 Ai That is: n \ i=1 Ai = {x | x ∈A1 ∧x ∈A2 ∧. . . ∧x ∈An} c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 44 / 45 Example for generalized intersection Example 1: Suppose that Ai = {1, 2, . . . , i} for all positive integer i. Then: n \ i=1 Ai = {1} = A1 Example 2: Suppose that Ai = {i + 1, i + 2, . . . , 2i} for all positive integer i. Then: Tn i=1 Ai = {2} ∩{3, 4} ∩{4, 5, 6} ∩. . . ∩{n + 1, n + 2, . . . , 2n} = ∅ c ⃝Xin He (University at Buffalo) CSE 191 Discrete Structures 45 / 45 |
11105 | https://www.math.uci.edu/~mathcircle/materials/M6L2.pdf | FUNCTIONAL EQUATIONS ZHIQIN LU 1. What is a functional equation An equation contains an unknown function is called a functional equation.
Example 1.1 The following equations can be regarded as functional equations f(x) = −f(−x), odd function f(x) = f(−x), even function f(x + a) = f(x), periodic function, if a , 0 Example 1.2 The Fibonacci sequence an+1 = an + an−1 defines a functional equation with the domain of which being nonnegative integers.
We can also represent the sequence is f(n + 1) = f(n) + f(n −1).
Example 1.3 (Radioactive decay) Let f(x) represent a measurement of the number of a specific type of radioactive nuclei in a sample of material at a given time x.
We assume that initially, there is 1 gram of the sample, that is, f(0) = 1. By the physical law, we have f(x)f(y) = f(x + y).
Can we determine which function this is?
2. Substitution method Example 2.1 Let a , 1. Solve the equation af(x) + f(1 x) = ax, where the domain of f is the set of all non-zero real numbers.
Date: November 7, 2016.
Lecture notes for the Math Circle, Irvine. Partially supported by the NSF grant DMS-1510232.
1 2 ZHIQIN LU, DEPARTMENT OF MATHEMATICS Solution: Replacing x by x−1, we get af(1 x) + f(x) = a x.
We therefore have (a2 −1)f(x) = a2x −a x, and hence f(x) = a2x −a x a2 −1 .
□ Exercise 2.2 Solving the functional equation (a2 , b2) a f(x −1) + bf(1 −x) = cx.
FUNCTIONAL EQUATIONS 3 Exercise 2.3 Finding a function f : R{0} →R such that (1 + f(x−1))( f(x) −(f(x))−1) = (x −a)(1 −ax) x , where a ∈(0, 1).
3. Recurrence Relations Example 3.1 (Fibonacci Equations) Let f(n + 2) = f(n + 1) + f(n) with f(0) = 0, f(1) = 1. Find a general formula for the sequence.
Solution: We consider the solution of the form f(n) = βn for some real number β. Then we have βn+2 = βn+1 + βn from which we conclude that β2 = β + 1. Therefore β1 = 1 + √ 5 2 , β2 = 1 − √ 5 2 .
4 ZHIQIN LU, DEPARTMENT OF MATHEMATICS A general solution of the sequence can be written as f(n) = c1 1 + √ 5 2 n + c2 1 − √ 5 2 n , where c1, c2 are coefficients determined by the initial values. By the initial condi-tions, we have c1 + c2 = 0 c1 1 + √ 5 2 + c2 1 − √ 5 2 = 1 Thus we have c1 = 1 √ 5 , c2 = −1 √ 5 .
Thus f(n) = 1 √ 5 1 + √ 5 2 n − 1 − √ 5 2 n .
It is interesting to see that the above expression provide all positive integers for any n.
□ Exercise 3.2 Solving the sequence defined by an = 3an−1 −2an−2 for n ≥2 with the initial condition a0 = 0, a1 = 1.
FUNCTIONAL EQUATIONS 5 4. The Cauchy’s Method Example 4.1 Assume that f is a continuous function on R. Assume that for any x, y ∈R f(x + y) = f(x) + f(y).
Find the function f(x).
Solution: First, we have f(0) = 0. Let c = f(1).
Using the math induction, we have f(x1 + · · · + xn) = f(x1) + · · · + f(xn).
Let x1 = · · · = xn = x. Then we get f(nx) = nf(x) for any positive integer n. Let x = 1/m where m is a nonzero integer. Then we have f( n m) = nf( 1 m).
On the other hand, mf( 1 m) = f(1) = c.
Thus we have f( n m) = nf( 1 m) = c n m.
The conclusion here is that for any rational number γ, we have f(γ) = cγ.
If f is continuous, then we conclude that for any real number x, f(x) = cx = x · f(1).
For those of you who are not familiar with the concept of continuity, the as-sumption can be weakened to the boundedness of the function. Assume that f is bounded. Let x be any real number. For any ϵ > 0, we choose a rational number γ such that |x −γ| < ϵ. Let N be the integer part of the 1/ √ϵ. Then |f(N(x −γ))| ≤C because the function is bounded. Thus we have |f(x) −f(1)γ| = |f(x) −f(γ)| ≤C N .
If we choose ϵ to be so small, then we must have f(x) = f(1)x.
□ 6 ZHIQIN LU, DEPARTMENT OF MATHEMATICS Exercise 4.2 (Radioactive Decay) Solve f(x + y) = f(x)f(y), where f is continuous/bounded.
5. Using functional equation to define elementary functions One of the applications of functional equations is that they can be used to char-acterizing the elementary functions. In the following, you are provided exercises for the functional equations for the functions ax, loga x, tan x, sin x, and cos x. Can you setup the functional equations for cot x, sec x, csc x, and hyper-trigonometric functions?
Exercise 5.1 The logarithmic function satisfies the property log(xy) = log x + log y for any positive real numbers. Solve the equation f(xy) = f(x) + f(y), where both x, y are positive real numbers, where f is continuous/bounded.
FUNCTIONAL EQUATIONS 7 Exercise 5.2 Solve the equation f(x + y) = f(x)f(y) where x, y are any real numbers, where f is continuous/bounded.
Exercise 5.3 Solve the equation f(x + y) = f(x) + f(y) 1 −f(x)f(y) for any real numbers x, y, where f is continuous/bounded.
8 ZHIQIN LU, DEPARTMENT OF MATHEMATICS Exercise 5.4 Solve the equations for continuous/bounded functions f(x), g(x) f(x + y) = f(x)g(y) + f(y)g(x) g(x + y) = g(x)g(y) −f(x)f(y) for any real numbers x, y.
FUNCTIONAL EQUATIONS 9 6. Additional Problems Exercise 6.1 Assume that a0 = 1, a1 = 2, and an = 4an−1 −an−2 for n ≥2. Find a prime factor of a2015.
This problem is, obviously, obsolete. By th year 2016 is also special. It is the dimensional of the space of all 26 × 26 skew-symmetric matrices. Just FYI.
Exercise 6.2 Solving f(x) + f( x −1 x ) = 1 + x, x , 0, 1 Exercise 6.3 Exercise 2.3 revisited: is the solution unique?
Exercise 6.4 If we drop the assumption that f is continuous or boundedness in Example 4.1, is the function f still linear?
10 ZHIQIN LU, DEPARTMENT OF MATHEMATICS 7. Solutions of the Exercises Solution of Exercise 2.2: Replacing x −1 by x, we obtain a f(x) + bf(−x) = c(x + 1).
In the above equation, if we replace x by −x, we get a f(−x) + bf(x) = c(−x + 1).
Thus we solve f(x) = ac(x + 1) −bc(−x + 1) a2 −b2 .
After verification, the above is indeed the solution.
□ Solution of Exercise 2.3: By replacing x with x−1, we obtain (1 + f(x))(f(x−1) −(f(x−1))−1) = (x −a)(1 −ax) x = (1 + f(x−1))(f(x) −(f(x))−1).
Thus we have f(x) = f(x−1), and hence f is determined by the equation (1 + f(x))( f(x) −(f(x))−1) = (x −a)(1 −ax) x .
□ Solution of Exercise 3.2: an = 2n −1.
□ Solution of Exercise 4.2: We first conclude that f(x1 + · · · + xn) = f(x1) · · · f(xn).
By taking all xi to be the same, we obtain f(nx) = (f(x))n.
In particular, f( 1 m) = (f(1)) 1 m .
And therefore we have f( n m) = (f(1)) n m .
The continuity/boundedness ensures that the above equation is also valid for irra-tional numbers.
□ FUNCTIONAL EQUATIONS 11 Solution of Exercise 5.1: Let g(x) = f(ex). Then g(x + y) = f(ex · ey) = f(ex) + f(ey) = g(x) + g(y).
The problem is reduced to Example 4.1.
□ Solution of Exercise 5.2: Either f(x) ≡0 or f(x) is positive. So we can take g(x) = log f(x) in case f(x) is positive. Then we reduce to Example 4.1.
□ Solution of Exercise 5.3: Take g(x) = arctan f(x).
□ Solution of Exercise 5.4: Take h(x) = g(x) + √ −1 f(x). Then we obtain h(x + y) = h(x)h(y).
Then using Exercise 4.2.
□ 8. Solutions of the additional problems Solution of Exercise 7.1: an = 1 2((2 + √ 3)n + (2 − √ 3)n).
Since (2 + √ 3)5n + (2 − √ 3)5n = ((2 + √ 3)5 + (2 − √ 3)5) · integer, and since (2 + √ 3)5 + (2 − √ 3)5 = 724 = 181, a2015 is divisible by the prime number 181.
□ Solution of Exercise 7.2: Replacing x by x−1 x , we obtain f( x −1 x ) + f(− 1 x −1) = 1 + x −1 x .
Replacing x by −1 x−1, we obtain f(− 1 x −1) + f(x) = 1 − 1 x −1.
Thus we have f(x) = 1 2(1 + x + 1 − 1 x −1 −1 −x −1 x ) = 1 2(x + 1 x − 1 x −1).
□ Solution of Exercise 7.3: Yes, the solution is unique, even in general a cubic equa-tion has three real roots.
This is because if x is small, the cubic equation only has one real root.
12 ZHIQIN LU, DEPARTMENT OF MATHEMATICS □ Solution of Exercise 7.4: The set of real numbers R can be considered as a vector space over rational numbers Q. A basis for such a vector space is called a Hamel basis. That is, there is a set of real numbers uα such that for any real numbers β, there is a unique representation of the form β = N X i=1 riuαi.
The existence of such a basis is equivalent to the Axiom of Choice. Without assum-ing the continuity or boundedness, we can still prove that f is a linear transforma-tion when regarding R as a vector space over Q. However, one can prescribe any value on the Hamel basis to construct a (unbounded) solution. So in conclusion, the solution is not unique without the additional assumption.
□ Professor, Department of Mathematics, University of California Irvine, E-mail address: zlu@uci.edu |
11106 | https://www.physicsclassroom.com/class/energy/lesson-1/definition-and-mathematics-of-work | Definition and Mathematics of Work
Hold down the T key for 3 seconds to activate the audio accessibility mode, at which point you can click the K key to pause and resume audio. Useful for the Check Your Understanding and See Answers.
In the first three units of The Physics Classroom, we utilized Newton's laws to analyze the motion of objects. Force and mass information were used to determine the acceleration of an object. Acceleration information was subsequently used to determine information about the velocity or displacement of an object after a given period of time. In this manner, Newton's laws serve as a useful model for analyzing motion and making predictions about the final state of an object's motion. In this unit, an entirely different model will be used to analyze the motion of objects. Motion will be approached from the perspective of work and energy. The effect that work has upon the energy of an object (or system of objects) will be investigated; the resulting velocity and/or height of the object can then be predicted from energy information. In order to understand this work-energy approach to the analysis of motion, it is important to first have a solid understanding of a few basic terms. Thus, Lesson 1 of this unit will focus on the definitions and meanings of such terms as work, mechanical energy, potential energy, kinetic energy, and power.
When a force acts upon an object to cause a displacement of the object, it is said that work was done upon the object. There are three key ingredients to work - force, displacement, and cause. In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement. There are several good examples of work that can be observed in everyday life - a horse pulling a plow through the field, a father pushing a grocery cart down the aisle of a grocery store, a freshman lifting a backpack full of books upon her shoulder, a weightlifter lifting a barbell above his head, an Olympian launching the shot-put, etc. In each case described here there is a force exerted upon an object to cause that object to be displaced.
Read the following five statements and determine whether or not they represent examples of work. Then click on the See Answer button to view the answer.
Statement | Answer with Explanation
A teacher applies a force to a wall and becomes exhausted. | See AnswerNo.This is not an example of work. The wall is not displaced. A force must cause a displacement in order for work to be done.
A book falls off a table and free falls to the ground. | See AnswerYes.This is an example of work. There is a force (gravity) which acts on the book which causes it to be displaced in a downward direction (i.e., "fall").
A waiter carries a tray full of meals above his head by one arm straight across the room at constant speed. (Careful! This is a very difficult question that will be discussed in more detaillater.) | See AnswerNo.This is not an example of work. There is a force (the waiter pushes up on the tray) and there is a displacement (the tray is moved horizontally across the room). Yet the force does not cause the displacement. To cause a displacement, there must be a component of force in the direction of the displacement.
A rocket accelerates through space. | See AnswerYes.This is an example of work. There is a force (the expelled gases push on the rocket) which causes the rocket to be displaced through space.
A teacher applies a force to a wall and becomes exhausted.
No.
This is not an example of work. The wall is not displaced. A force must cause a displacement in order for work to be done.
A book falls off a table and free falls to the ground.
Yes.
This is an example of work. There is a force (gravity) which acts on the book which causes it to be displaced in a downward direction (i.e., "fall").
A waiter carries a tray full of meals above his head by one arm straight across the room at constant speed. (Careful! This is a very difficult question that will be discussed in more detail later.)
No.
This is not an example of work. There is a force (the waiter pushes up on the tray) and there is a displacement (the tray is moved horizontally across the room). Yet the force does not cause the displacement. To cause a displacement, there must be a component of force in the direction of the displacement.
A rocket accelerates through space.
Yes.
This is an example of work. There is a force (the expelled gases push on the rocket) which causes the rocket to be displaced through space.
Work Equation
Mathematically, work can be expressed by the following equation.
where F is the force, d is the displacement, and the angle (theta) is defined as the angle between the force and the displacement vector. Perhaps the most difficult aspect of the above equation is the angle "theta." The angle is not just any 'ole angle, but rather a very specific angle. The angle measure is defined as the angle between the force and the displacement. To gather an idea of it's meaning, consider the following three scenarios.
To Do Work, Forces Must Cause Displacements
Let's consider Scenario C above in more detail. Scenario C involves a situation similar to the waiter who carried a tray full of meals above his head by one arm straight across the room at constant speed. It was mentioned earlier that the waiter does not do work upon the tray as he carries it across the room. The force supplied by the waiter on the tray is an upward force and the displacement of the tray is a horizontal displacement. As such, the angle between the force and the displacement is 90 degrees. If the work done by the waiter on the tray were to be calculated, then the results would be 0. Regardless of the magnitude of the force and displacement, Fdcosine 90 degrees is 0 (since the cosine of 90 degrees is 0). A vertical force can never cause a horizontal displacement; thus, a vertical force does not do work on a horizontally displaced object!!
It can be accurately noted that the waiter's hand did push forward on the tray for a brief period of time to accelerate it from rest to a final walking speed. But once up to speed, the tray will stay in its straight-line motion at a constant speed without a forward force. And if the only force exerted upon the tray during the constant speed stage of its motion is upward, then no work is done upon the tray. Again, a vertical force does not do work on a horizontally displaced object.
The equation for work lists three variables - each variable is associated with one of the three key words mentioned in the definition of work (force, displacement, and cause). The angle theta in the equation is associated with the amount of force that causes a displacement. As mentioned in a previous unit, when a force is exerted on an object at an angle to the horizontal, only a part of the force contributes to (or causes) a horizontal displacement. Let's consider the force of a chain pulling upwards and rightwards upon Fido in order to drag Fido to the right. It is only the horizontal component of the tension force in the chain that causes Fido to be displaced to the right. The horizontal component is found by multiplying the force F by the cosine of the angle between F and d. In this sense, the cosine theta in the work equation relates to the cause factor - it selects the portion of the force that actually causes a displacement.
The Meaning of Theta
When determining the measure of the angle in the work equation, it is important to recognize that the angle has a precise definition - it is the angle between the force and the displacement vector. Be sure to avoid mindlessly using any 'ole angle in the equation. A common physics lab involves applying a force to displace a cart up a ramp to the top of a chair or box. A force is applied to a cart to displace it up the incline at constant speed. Several incline angles are typically used; yet, the force is always applied parallel to the incline. The displacement of the cart is also parallel to the incline. Since F and d are in the same direction, the angle theta in the work equation is 0 degrees. Nevertheless, most students experienced the strong temptation to measure the angle of incline and use it in the equation. Don't forget: the angle in the equation is not just any 'ole angle. It is defined as the angle between the force and the displacement vector.
The Meaning of Negative Work
On occasion, a force acts upon a moving object to hinder a displacement. Examples might include a car skidding to a stop on a roadway surface or a baseball runner sliding to a stop on the infield dirt. In such instances, the force acts in the direction opposite the objects motion in order to slow it down. The force doesn't cause the displacement but rather hinders it. These situations involve what is commonly called negative work. The negative of negative work refers to the numerical value that results when values of F, d and theta are substituted into the work equation. Since the force vector is directly opposite the displacement vector, theta is 180 degrees. The cosine(180 degrees) is -1 and so a negative value results for the amount of work done upon the object. Negative work will become important (and more meaningful) in Lesson 2 as we begin to discuss the relationship between work and energy.
Units of Work
Whenever a new quantity is introduced in physics, the standard metric units associated with that quantity are discussed. In the case of work (and also energy), the standard metric unit is the Joule (abbreviated J). One Joule is equivalent to one Newton of force causing a displacement of one meter. In other words,
In fact, any unit of force times any unit of displacement is equivalent to a unit of work. Some nonstandard units for work are shown below. Notice that when analyzed, each set of units is equivalent to a force unit times a displacement unit.
Non-standard Units of Work:
foot•pound | kg•(m/s2)•m | kg•(m2/s2)
In summary, work is done when a force acts upon an object to cause a displacement. Three quantities must be known in order to calculate the amount of work. Those three quantities are force, displacement and the angle between the force and the displacement.
Investigate!
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11107 | https://stackoverflow.com/questions/32277959/sum-consecutive-even-numbers-and-consecutive-odd-numbers-in-an-array | java - Sum consecutive even numbers and consecutive odd numbers in an array - Stack Overflow
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Sum consecutive even numbers and consecutive odd numbers in an array
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Asked 10 years, 1 month ago
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I have a code that sums the consecutive even numbers and consecutive odd numbers, then adds them to an arraylist. This process should be repeated until there are no more consecutive odd or even numbers in the list. Then returns the size of the arraylist.
I used nested for loops and the problem is the loops check the same index which doesn't make sense.
Here's my code:
java
public static int SumGroups(int[] arr) {
ArrayList<Integer> arl = new ArrayList<Integer>();
int even = 0, odd = 0;
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length; j++) {
if (arr[i] % 2 == 0) {
even += arr[i];
if (arr[j] % 2 == 0) {
even += arr[j];
} else {
arl.add(even);
even = 0;
break;
}
} else {
odd += arr[i];
if (arr[j] % 2 != 0) {
odd += arr[j];
} else {
arl.add(odd);
odd = 0;
break;
}
}
}
}
return arl.size();
}
My Question is: How to prevent loops from checking the same index ? in other words, how to make my code sums the consecutive even numbers and consecutive odd numbers ?
Input:
java
int arr[]={2, 1, 2, 2, 6, 5, 0, 2, 0, 5, 5, 7, 7, 4, 3, 3, 9};
Output:
java
6 // [2, 1, 10, 5, 30, 15]
java
arrays
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edited Aug 28, 2015 at 21:03
PKuhn
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asked Aug 28, 2015 at 18:53
David MatriguetDavid Matriguet
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I updated the post to explain my question David Matriguet –David Matriguet 2015-08-28 18:59:06 +00:00 Commented Aug 28, 2015 at 18:59
can you provide sample input and desired output?Iłya Bursov –Iłya Bursov 2015-08-28 18:59:31 +00:00 Commented Aug 28, 2015 at 18:59
2 @DavidMatriguet you're output is incorrect, it should be 8 [2, 1, 10, 5, 2, 24, 4, 15}Iłya Bursov –Iłya Bursov 2015-08-28 19:08:01 +00:00 Commented Aug 28, 2015 at 19:08
@Lashane No, Because 2,24, and 4 are 3 consecutive even numbers David Matriguet –David Matriguet 2015-08-28 19:18:36 +00:00 Commented Aug 28, 2015 at 19:18
1 @DavidMatriguet 2,24(5+5+7+7),4 are not numbers in input array, they are result sums, do you mean that we should run new array through the same function again?Iłya Bursov –Iłya Bursov 2015-08-29 02:39:45 +00:00 Commented Aug 29, 2015 at 2:39
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``java
I think the following code should solve the problem, if you do not want to output the size simply returnsumsinstead ofsums.size()`
public static int sumGroupsRecursively(int[] arr) {
List numbersToSum = IntStream.of(arr).boxed().collect(Collectors.toList());
List currentSumList = sumSublist(numbersToSum);
List nextSumList = sumSublist(currentSumList);
while (currentSumList.size() != nextSumList.size()) {
currentSumList = nextSumList;
nextSumList = sumSublist(currentSumList);
}
return nextSumList.size();
}
public static List sumSublist(List list) {
int current = list.get(0);
int currentSum = 0;
List sums = new ArrayList<>();
for (int i = 0; i < list.size(); i++) {
if (current % 2 == list.get(i) % 2) {
currentSum += list.get(i);
} else {
sums.add(currentSum);
current = list.get(i);
currentSum = current;
}
}
sums.add(currentSum);
return sums;
}
```
If you need to do this in one function what I would discourage because it is harder to read you could use code like this.
java
public static Integer sumSublist(int[] arr) {
List<Integer> sums = new ArrayList<>();
sums.add(0);
int i = 0;
while (i < arr.length - 1) {
int current = arr[i];
int currentSum = 0;
while (current % 2 == arr[i] % 2) {
currentSum += arr[i];
if (i >= arr.length - 1) {
break;
}
i++;
}
if (currentSum % 2 == sums.get(sums.size()-1) % 2) {
sums.set(sums.size() - 1, sums.get(sums.size()-1) + currentSum);
} else {
sums.add(currentSum);
}
}
return sums.size();
}
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edited Aug 28, 2015 at 20:29
answered Aug 28, 2015 at 19:09
PKuhnPKuhn
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PKuhn
PKuhnOver a year ago
if this solved your question, think about marking the answer as acccepted
2015-08-28T19:20:00.243Z+00:00
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You are entering your first for loop passing in arr. Inside the first for loop you enter a second for loop passing in arr a second time. This means that you enter the second for loop as many times as there are elements in arr and transverse arr in the second for loop every single time.
for example, if arr.length() was 2 you would transverse arr 3 times. Once in your outer for loop and twice (once for each element in arr) in your inner loop.
Second, by adding both the odd and even numbers to your arraylist, you are doing nothing but reconstructing arr but in an arraylist rather than array. Therefor, returning arl.size() is the exact same as returning arr.length() which is already known and much easier to do.
Despite that, here is how I would calculate the sum of the odd and evens. I add both to different arraylists. You'll need to figure out exactly what you need to return though because your description is off.
```java
public void test(){
int[] arr = new int;
arr = 1;
arr = 2;
arr = 3;
arr = 4;
arr = 5;
int testOfEven = 6;
int testOfOdd = 9;
int sumOfEven = 0;
int sumOfOdd = 0;
ArrayList evens = new ArrayList();
ArrayList odds = new ArrayList();
for(int i = 0; i < arr.length; i++)
{
if ((arr[i]%2) == 0)
{
evens.add(arr[i]);
sumOfEven += arr[i];
}
else
{
odds.add(arr[i]);
sumOfOdd += arr[i];
}
}
assertEquals(testOfEven, sumOfEven);
assertEquals(testOfOdd, sumOfOdd);
```
}
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answered Aug 28, 2015 at 19:22
GMLewisIIGMLewisII
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after playing some time, here is my version:
java
public static int SumGroups(final int[] arr) {
if (arr.length > 0) {
int n, sum, psum;
psum = sum = n = arr & 1; // parity of first number in sequence
int s = 1; // at least one element in array
int f = 0; // discard first parity change
for (int i = 1; i < arr.length; i++) {
if (n == (arr[i] & 1)) {
sum = (sum + n) & 1; // both even or odd, just increase sum
} else {
s += (psum ^ sum) & f; // compare sums parity
psum = sum; // store current sum's parity
sum = n = arr[i] & 1; // new first number in sequence
f = 1; // do not discard sums parity next time
}
}
s += (psum ^ sum) & f; // array ended, check parity of last sum
return s;
}
return 0;
}
I've put comments, but still some additional notes:
basic idea is the same as @PKuhn, just checked for some edge cases (empty array, integer overflow)
we don't need to have array of sums, we need just previous sum and check parity of it with newly calculated one
sum = (sum + n) & 1 - we don't need to calculate whole sum, we need just parity of the sum
s += (psum ^ sum) & f - we need to increase swap counter only if parity changed, xor helps us to get 1 if changed and 0 if not
Here is the list of tests which I've used:
java
Assert.assertEquals(6, SumGroups(new int[] { 2, 1, 2, 2, 6, 5, 0, 2, 0, 5, 5, 7, 7, 4, 3, 3, 9 }));
Assert.assertEquals(6, SumGroups(new int[] { 0, 0, 0, 0, 2, 1, 2, 2, 6, 5, 0, 2, 0, 5, 5, 7, 7, 4, 3, 3, 9 }));
Assert.assertEquals(1, SumGroups(new int[] { 2, 3, 3 }));
Assert.assertEquals(1, SumGroups(new int[] { 2 }));
Assert.assertEquals(1, SumGroups(new int[] { 2, 2 }));
Assert.assertEquals(1, SumGroups(new int[] { 2, 3, 3, 3, 3, 2 }));
Assert.assertEquals(2, SumGroups(new int[] { 3, 2, 2 }));
Assert.assertEquals(2, SumGroups(new int[] { 1, 3, 3, 2, 2 }));
Assert.assertEquals(2, SumGroups(new int[] { 1, 2, 3, 3, 2, 3, 3, 2 }));
Assert.assertEquals(1, SumGroups(new int[] { 3, 3, 2, 2 }));
Assert.assertEquals(1, SumGroups(new int[] { Integer.MAX_VALUE, Integer.MAX_VALUE }));
Assert.assertEquals(1, SumGroups(new int[] { Integer.MAX_VALUE, Integer.MAX_VALUE, 2 }));
Assert.assertEquals(1, SumGroups(new int[] { Integer.MAX_VALUE, Integer.MAX_VALUE, 3 }));
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answered Aug 29, 2015 at 4:17
Iłya BursovIłya Bursov
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public void findEvenOdd(int a[]){
```java
Boolean flip = false;
int sum = 0, i, m = 0;
for (i = 0; i < a.length; i++) {
if (flip) {
System.out.print(sum + "\t");
sum = a[i];
flip = !flip;
if (i + 1 < a.length && (a[i] % 2 != a[i + 1] % 2))
flip = !flip;
m++;
} else {
sum += a[i];
if (i + 1 < a.length && (a[i] % 2 != a[i + 1] % 2))
flip = !flip;
m++;
}
}
if(m!=a.length-1)
System.out.print(a[a.length-1] + "\t");
}
```
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answered Dec 14, 2017 at 5:04
shivam mehrotrashivam mehrotra
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Stephen Rauch
Stephen RauchOver a year ago
When giving an answer it is preferable to give some explanation as to WHY your answer is the one.
2017-12-14T05:16:37.93Z+00:00
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11108 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10281205/ | A practical approach to the diagnosis and management of chlamydia and gonorrhea - PMC
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A practical approach to the diagnosis and management of chlamydia and gonorrhea
Clara E Van Ommen
Clara E Van Ommen, MD
1 Division of Infectious Diseases (Van Ommen, Grennan), University of British Columbia; British Columbia Centre for Disease Control (Malleson, Grennan), Vancouver, BC
Find articles by Clara E Van Ommen
1, Sarah Malleson
Sarah Malleson, MD
1 Division of Infectious Diseases (Van Ommen, Grennan), University of British Columbia; British Columbia Centre for Disease Control (Malleson, Grennan), Vancouver, BC
Find articles by Sarah Malleson
1, Troy Grennan
Troy Grennan, MD MSc
1 Division of Infectious Diseases (Van Ommen, Grennan), University of British Columbia; British Columbia Centre for Disease Control (Malleson, Grennan), Vancouver, BC
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1,✉
Author information
Article notes
Copyright and License information
1 Division of Infectious Diseases (Van Ommen, Grennan), University of British Columbia; British Columbia Centre for Disease Control (Malleson, Grennan), Vancouver, BC
✉
Correspondence to: Troy Grennan, troy.grennan@bccdc.ca
✉
Corresponding author.
Issue date 2023 Jun 19.
© 2023 CMA Impact Inc. or its licensors
This is an Open Access article distributed in accordance with the terms of the Creative Commons Attribution (CC BY-NC-ND 4.0) licence, which permits use, distribution and reproduction in any medium, provided that the original publication is properly cited, the use is noncommercial (i.e., research or educational use), and no modifications or adaptations are made. See:
PMC Copyright notice
PMCID: PMC10281205 PMID: 37336564
See the reply "The 5 Ps need an update: toward a comprehensive sexual history" on page E1085.
This article is also available in French. See "Approche pratique au diagnostic et à la prise en charge de la chlamydia et de la gonorrhée".
KEY POINTS
The incidence of chlamydia and gonorrhea, 2 common sexually transmitted infections, is increasing.
Annual asymptomatic screening for chlamydia and gonorrhea should be performed in all sexually active patients younger than 30 years, with more frequent screening for higher risk patients.
Nucleic acid amplification testing for chlamydia and gonorrhea should be performed in both asymptomatic and symptomatic patients at sites of sexual exposure, guided by a careful sexual history.
The treatment recommendations for chlamydia and gonorrhea are evolving and clinicians should follow local guidance.
Antimicrobial resistance in gonorrhea is increasing; optimal treatment should be guided by principles of antimicrobial stewardship.
The 2 most frequent reportable bacterial sexually transmitted infections (STIs) worldwide and in Canada are those caused by Chlamydia trachomatis and Neisseria gonorrhoeae.1,2 Rates of both infections have been increasing over the last decade despite public health efforts aimed at prevention, testing and treatment. In 2019, 139 389 cases of chlamydia and 35 443 cases of gonorrhea were reported in Canada, an increase of 33.1% and 181.7%, respectively, since 2010.2 These increases may reflect improved diagnostics, increased screening and contact tracing or a true increase in incidence.2
Sexually transmitted infections have a substantial impact on affected individuals and communities. Chlamydia trachomatis and N. gonorrhoeae are commonly implicated pathogens in pelvic inflammatory disease and, if untreated, can lead to infertility.3 Infection with a bacterial STI is associated with increased risk of HIV acquisition or transmission.4 Perinatal transmission of C. trachomatis and N. gonorrhoeae can lead to ophthalmia neonatorum in infants, among other pathologies.5 Treatment has become more challenging, given the increase in antimicrobial resistance in gonorrhea.6
We summarize the management of chlamydia and gonorrhea in primary care as health care providers work collectively toward the goal of decreasing the frequency of these infections and reducing associated morbidity through appropriate treatment. We draw on evidence from clinical practice guidelines, systematic reviews and meta-analyses (Box 1).
Box 1: Literature review
We conducted a targeted literature search of MEDLINE and Embase from inception to July 2022. Search terms included “Chlamydia trachomatis,” “Neisseria gonorrhoeae,” “sexually transmitted infection,” “STI,” “urethritis,” “cervicitis,” “pelvic inflammatory disease,” “proctitis,” “epididymitis,” “diagnosis,” “screening” and “treatment.” We limited the search to articles in English. Our targeted search focused on identifying clinical practice guidelines, systematic reviews and meta-analyses, although we did not place any formal restriction on article type. We selected relevant articles, and manually reviewed their references for additional articles.
Why is taking a good sexual history important?
Taking a sexual history is essential to comprehensive care in patients presenting with STI symptoms and in asymptomatic people to assess for STI risk, determine the need for screening, address concerns and provide sexual health education.
Patients have reported wanting their health care provider to inquire about sexual health, but many face considerable barriers to self-disclosure of their sexual history.7,8 Stigma is often associated with STIs. Providers conducting a sexual history should do so in a nonjudgmental, patient-centred and trauma-informed manner.9 Syndemics theory describes how disease interacts with social constructs, which can help conceptualize how a person’s unique social, cultural and health context influences how they access STI care.10 Establishing the patient’s pronouns, sexual orientation and gender identity is necessary to create an environment of respect and trust. The components of a sexual history can be remembered by the 5 Ps: partners, practices, protection, past history and pregnancy (Table 1).11
Table 1:
Approach to taking a sexual history11
| Area | Examples of questions |
:---: |
| Partners | Are you currently having sex of any kind? In the last 2 months, how many sexual partners have you had? What is/are the genders of your sexual partners? Do your partners have other sexual partners? What is/are their gender(s)? |
| Practices | To offer the most appropriate testing, can you tell me more about what types of sex you have? What parts of your body are involved when you have sex? Genital sex (penis in the vagina) Anal sex (penis in the anus) Oral sex (mouth on penis, vagina or anus) How do you meet your sexual partners? Have you or any of your partners used drugs? Have you ever exchanged sex to meet your needs (money, housing, food etc.)? |
| Protection | Do you and your partner(s) discuss STI prevention? If you use prevention, what methods do you use? How often do you use these methods (never, sometimes, all of the time)? Have you received the human papillomavirus (HPV), hepatitis B (HBV) or hepatitis A (HAV) vaccine? Have you ever used or considered using HIV pre-exposure prophylaxis (PrEP)? |
| Past history | Have you ever been tested for STIs? Have you ever been diagnosed and/or treated for an STI in the past? Have any of your current or former partners ever been diagnosed or treated for an STI? |
| Pregnancy | How important is it to you to prevent pregnancy? Are you or your partner using contraception or any form of birth control? |
Open in a new tab
Note: STI = sexually transmitted infection.
Based on information contained in Reno H, Park I, Workowski K, et al. A guide to taking a sexual history. Atlanta: Centers for Disease Control and Prevention; reviewed 2022. Available:
What are common clinical presentations?
Most chlamydia and gonorrhea infections cause no symptoms.12 If symptoms develop, the incubation period for gonorrhea is 2–7 days, compared with 2–6 weeks for chlamydia.13 Chlamydia and gonorrhea may have genital or extragenital symptoms, which are generally reflective of the site of infection. The clinical presentations of chlamydia and gonorrhea overlap, and they are usually clinically indistinguishable.
Genital symptoms
Urethritis is the most common syndrome in patients with a penis who are symptomatic. It is characterized by dysuria, urethral pruritis and discharge. Most cases of infectious urethritis are caused by C. trachomatis and N. gonorrhoeae or both. However, in almost half of cases of nongonococcal urethritis, no specific organism is identified despite extensive microbiological investigation (Box 2).14
Box 2: Infectious differential diagnosis of common clinical presentations of sexually transmitted infections
Urethritis
Neisseria gonorrhoeae
Chlamydia trachomatis
Mycoplasma genitalium
Trichomonas vaginalis
Neisseria meningitidis
Hemophilus spp.
Herpes simplex virus
Adenovirus
Cervicitis
Chlamydia trachomatis
Neisseria gonorrhoeae
Trichomoniasis
Herpes simplex virus
Mycoplasma genitalium
Bacterial vaginosis
Proctitis
Neisseria gonorrhoeae
Chlamydia trachomatis (including lymphogranuloma venereum serovars)
Syphilis
Herpes simplex virus
Mpox virus
Epididymitis
Neisseria gonorrhoeae
Chlamydia trachomatis
Enteric organisms (e.g., Escherichia coli)
Patients can develop acute epididymitis from chlamydia or gonorrhea, which is characterized by unilateral, posterior testicular pain and swelling, often accompanied by symptoms of urethritis. Among men younger than 35 years, C. trachomatis and N. gonorrhoeae are the most common causative organisms, but among older men and men who engage in insertive anal intercourse, causative agents can include enteric organisms like Escherichia coli.15
Although cervicitis is often asymptomatic, symptoms may occur and include abnormal vaginal discharge or intermenstrual bleeding.16 Findings on physical examination include purulent endocervical discharge or sustained endocervical bleeding. Most cases of cervicitis have no identified cause. In as many as 25% of cases, C. trachomatis or N. gonorrhoeae is identified.17 In around 15% of female patients, pelvic inflammatory disease can develop, characterized by abdominal or pelvic pain, dyspareunia or abnormal uterine bleeding, with findings of cervical motion or adnexal tenderness on physical examination.18 Patients may have infertility as a consequence of pelvic inflammatory disease. An uncommon complication of pelvic inflammatory disease is Fitz–Hugh–Curtis syndrome, characterized by right upper quadrant pain related to inflammation of the liver capsule.17
Extragenital symptoms
Proctitis caused by chlamydia or gonorrhea may present with tenesmus, anorectal pain, bleeding and mucopurulent discharge. These infections typically occur in patients who engage in receptive anal sex, but can also be transmitted from the vagina to the anal canal.19Chlamydia trachomatis and N. gonorrhoeae are the most commonly identified pathogens in cases of infectious proctitis.20
The lymphogranuloma venereum (LGV) serovars (L1, L2, L3) of C. trachomatis can cause invasive infections that preferentially affect lymphatic tissue. Lymphogranuloma venereum can present as small painless ulcers or painful hemorrhagic proctitis, with complications including anal fistulae and strictures.21 In the last 2 decades, LGV has emerged as an important cause of proctitis among men who have sex with men (MSM) in North America and Europe.22
Oropharyngeal infections with gonorrhea are commonly asymptomatic, although patients can present with sore throat, pharyngeal exudate or cervical lymphadenitis.23 Chlamydia is not an important cause of pharyngitis.24
Although uncommon, gonorrhea infection can cause bacteremia, leading to septic arthritis or disseminated gonococcal infection, with tenosynovitis, dermatitis or polyarthralgias.23 Reactive arthritis — characterized by polyarthritis, conjunctivitis or uveitis, and urethritis or cervicitis — can occur after an infection with chlamydia or gonorrhea, although chlamydia is the more common inciting infection.25
Who should be screened for infection?
Opportunistic screening is critical in identifying asymptomatic chlamydia and gonorrhea infections. The Canadian Task Force on Preventive Health Care recommends annual opportunistic screening for chlamydia and gonorrhea in all sexually active people younger than 30 years.26 Although based on low-quality evidence, an opportunistic approach to screening is likely to increase the number of STIs diagnosed and destigmatize sexual health conversations.
More frequent screening should be offered to people at higher risk of acquiring STIs, although little evidence exists to guide the optimal frequency of screening. Among MSM, current guidance suggests, at minimum, anatomic site-based screening for chlamydia and gonorrhea annually.13,24 More frequent screening (i.e., every 3–6 months) is recommended for at-risk people of any gender within groups who may be disproportionately affected by STIs, including those taking HIV pre-exposure prophylaxis (PrEP), those who have recently had an STI, those living with HIV or those with multiple sexual partners.13,24,27 One cohort study of 557 MSM and transgender women taking HIV PrEP found that semiannual STI screening would have led to delayed diagnosis in more than 30% of patients with chlamydia or gonorrhea, compared with quarterly screening.28 Pregnant patients should be screened at their first prenatal visit, with rescreening in the third trimester if they initially test positive for or are at ongoing risk of STIs.13,24
Clinicians should determine appropriate anatomic sites for screening based on information from the sexual history, although they should consider screening extragenital sites (i.e., rectum and oropharynx), even in the absence of either reported symptoms or sexual exposures. Studies of people attending STI clinics have found that a considerable proportion of STIs are missed when STI testing is conducted only for patients with reported symptoms or on sites with known exposure, or when testing includes only urine.29,30 Testing for gender-diverse patients will depend on their specific anatomy.
How should patients be tested?
In asymptomatic patients, approaches to sample collection for nucleic acid amplication testing (NAAT) for chlamydia and gonorrhea include a first-void urine (first 10–20 mL, any time of day, at least 1 hour since previous void) or vaginal swab; other options include a urethral or cervical swab (Table 2). In patients with a vagina, a vaginal swab is preferred over first-void urine, as urine testing may detect 10% fewer infections.31 Those with a neovagina or gender-affirming penile reconstruction should provide a urine sample for NAAT. Extragenital testing options include a pharyngeal or rectal swab for chlamydia and gonorrhea NAAT. In symptomatic patients, first-void urine and swabs of sites of reported symptoms should be collected for chlamydia and gonorrhea NAAT, and for gonorrhea culture and sensitivity testing. Patient-collected swabs are acceptable, as studies have shown equivalence between self-and clinician-collected oral, vaginal and rectal swabs for chlamydia and gonorrhea testing.32,33 Self-collection may also improve uptake of STI screening.13,24
Table 2:
Testing for chlamydia and gonorrhea
| Site | Approach for asymptomatic patients or screening | Approach for symptomatic patients |
:---:
| Penile urethra | First-void urine for NAAT for chlamydia and gonorrhea | Urethral swab for gonorrhea culture and sensitivity testing, and first-void urine for NAAT for chlamydia and gonorrhea |
| Cervix or vagina | Vaginal swab (preferred), cervical swab or first-void urine for NAAT for chlamydia and gonorrhea | Cervical swab for gonorrhea culture and sensitivity testing, and for NAAT for chlamydia and gonorrhea, or Vaginal swab for gonorrhea culture and sensitivity testing, and for NAAT for chlamydia and gonorrhea, or First-void urine for NAAT for chlamydia and gonorrhea |
| Throat | Throat swab for NAAT for chlamydia and gonorrhea | Throat swab for gonorrhea culture and sensitivity testing, and for NAAT for chlamydia and gonorrhea |
| Rectum | Rectal swab for NAAT for chlamydia and gonorrhea | Rectal swab for gonorrhea culture and sensitivity testing, and for NAAT for chlamydia and gonorrhea |
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Note: NAAT = nucleic acid amplification test.
For patients with a neovagina, a first-void urine is the preferred screening test. In symptomatic patients, efforts should be made to conduct gonorrhea culture and sensitivity testing, as well as NAAT for chlamydia and gonorrhea. When culture and sensitivity testing is not possible, either a cervical or vaginal swab for NAAT or a first-void urine is appropriate.
Clinicians should refer to their local microbiology laboratories for recommendations on collection and transport protocols in their region. First-void urine can be collected in a sterile urine container for chlamydia and gonorrhea NAAT. The swabs contained within chlamydia and gonorrhea NAAT kits can be used on the cervix, urethra, vagina, throat or rectum; swabs from these sites can also be sent for gonorrhea culture. Bacterial culture for chlamydia is not routinely performed in Canada.13
Genotyping of LGV serovars can be requested if a patient presents with a syndrome consistent with LGV.13 Some Canadian jurisdictions will automatically test all positive rectal chlamydia swabs for LGV serovars. However, it is important to indicate suspicion for LGV on laboratory requisitions, as automatic LGV testing is not universal, and nonrectal specimens (e.g., genital ulcers) are not automatically tested.
How should patients be treated?
Gonorrhea
Treatment of gonorrhea is challenging, as it readily develops antimicrobial resistance, and guidelines are not congruent in their recommendations. The Canadian STI guideline recommends dual therapy with ceftriaxone or cefixime, plus azithromycin or doxycycline (Table 3).13 The STI treatment guideline from the United States Centers for Disease Control and Prevention (CDC) increased the previously recommended ceftriaxone dose (Table 3).24 The CDC also recommended against dual therapy based on increasing antimicrobial resistance, and concern for impacts on the microbiome and selective pressure on other pathogens.24 It is likely that this approach will be adopted by guidelines from other jurisdictions in the future. If monotherapy with ceftriaxone is used, an increased dose of ceftriaxone is recommended, compared with that used in dual therapy (Table 3).24 Currently, given varying recommendations, clinicians should follow local guidance, which will be based on resistance patterns in their area.
Table 3:
Treatment of chlamydia and gonorrhea
| Pathogen | Canadian guideline13 | CDC guideline24 | Test of cure | Follow-up |
:---: :---:
| Chlamydia trachomatis | Preferred treatment Doxycycline (100 mg orally, twice daily for 7 d) or azithromycin (1 g orally, once) LGV: doxycycline (100 mg orally, twice daily for 21 d) Alternative treatment Levofloxacin (500 mg orally, daily for 7 d) Treatment for pregnant patients Azithromycin (1 g orally, once) | First-line treatment Doxycycline (100 mg orally, twice daily for 7 d) LGV: doxycycline (100 mg orally, twice daily for 21 d) Second-line treatment Azithromycin (1 g orally, once) or levofloxacin (500 mg orally, daily for 7 d) Treatment for pregnant patients Azithromycin (1 g orally, once) (preferred), or amoxicillin (500 mg orally, 3 times daily for 7 d) | Indications Suspected treatment failure Suspected poor adherence Nonpreferred regimen used Pregnancy Approach Swab for NAAT for chlamydia and gonorrhea 4 wk after therapy completed | Re-screen 3 mo after treatment completed |
| Neisseria gonorrhoeae | Preferred treatment Ceftriaxone (250 mg IM, once) and azithromycin (1 g orally, once), or Cefixime (800 mg orally, once) and azithromycin (1 g orally, once); this is considered an alternative regimen for pharyngeal infections and treatment of MSM Alternative treatment Ceftriaxone (250 mg IM, once) or cefixime (800 mg orally, once), and doxycycline (100 mg orally, twice daily for 7 d), or Azithromycin (2 g orally, once) and gentamicin (240 mg IM, once); this regimen should be considered only if severe allergy or documented resistance to cephalosporins Treatment for pregnant patients Ceftriaxone (250 mg IM, once) or cefixime (800 mg orally, once), and azithromycin (1 g orally, once) | First-line treatment Ceftriaxone (500 mg IM, once, if patient weighs < 150 kg; 1 g IM, once, if patient weighs > 150 kg) Second-line treatment Cefixime (800 mg orally, once) or gentamicin (240 mg IM, once), and azithromycin (2 g orally, once) Treatment for pregnant patients Same as above | Consider for all positive sites Indications Suspected treatment failure Suspected poor adherence Nonpreferred regimen used Pregnancy Pharyngeal infection Documented antimicrobial resistance Approach Swab for gonorrhea culture and sensitivity test 3–7 d after treatment (preferred) or swab for NAAT for chlamydia and gonorrhea 4 wk after treatment | Re-screen 3 mo after treatment completed |
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Note: CDC = Centers for Disease Control and Prevention, IM = intramuscularly, MSM = men who have sex with men, NAAT = nucleic acid amplification test.
Doxycycline is contraindicated in pregnancy.
Chlamydia
The Canadian STI guideline recommends doxycycline or azithromycin as the first-line (preferred) treatment for chlamydia,13 whereas the CDC recommends doxycycline as first-line treatment, with azithromycin as a second-line (alternate) regimen (Table 3).24 The preference for doxycycline is based on a systematic review and meta-analysis comparing treatment with azithromycin and doxycycline for chlamydia, which found that treatment failed more often with azithromycin, particularly among men with rectal chlamydia.34,35 Thus, doxycycline is the preferred agent for treating rectal chlamydia. If adherence to therapy is a concern, single-dose azithromycin may be preferred. For pregnant patients, azithromycin is the first-line treatment.13 For patients with suspected or confirmed LGV, treatment with doxycycline should be continued for 21 days.13
Other treatment considerations
Given the potential complexity of cases and the evolving treatment landscape, providers should consult with an expert in STI management when necessary. All patients being treated for chlamydia or gonorrhea should be strongly advised to abstain from sexual activity for 7 days after treatment and until all partners have been treated.13 Sexual partners from the previous 60 days should be tested and treated, or offered expedited partner treatment (i.e., clinicians can provide empiric treatment for the patient to give to their partner), which has been found to reduce the rates of recurrent or persistent infection.36 Details around indications and timing of tests of cure are discussed in Table 3. Tests of cure and repeat screening recommendations are often not followed, although they remain important for the appropriate care of the patient and to decrease transmission.37
What about antimicrobial resistance?
Globally and in Canada, rates of antimicrobial resistance in N. gonorrhoeae are increasing, with decreasing susceptibility to cephalosporins and azithromycin.6,38 In Canada, between 2012 and 2016, the proportion of multidrug resistant N. gonorrhoeae increased from 6.2% to 8.9%, with most isolates identified in Ontario and Quebec.39 Actions that clinicians can take to combat antimicrobial resistance are to perform gonorrhea culture and sensitivity testing when possible to limit unnecessary antimicrobial use, and to forgo dual therapy for gonorrhea when chlamydia is excluded. Whether the widespread discontinuation of dual therapy for gonorrhea would negatively affect clinical outcomes or prevent antimicrobial resistance has not yet been established, however. Treatment can be delayed until test results are available in situations where reliable patient follow-up is likely. In cases of confirmed or suspected multidrug-resistant N. gonorrhoeae, clinicians should consider consulting an expert in the management of STIs.
Conclusion
Chlamydia and gonorrhea are the most common bacterial STIs in Canada, and their incidence is increasing.2 Most infections are asymptomatic, which highlights the importance of routine screening for people who are sexually active.26 Screening and diagnostic testing in symptomatic patients should be guided by a comprehensive sexual health history, which also provides an opportunity for patient education around sexual health. However, the optimal screening frequency in different populations remains unclear. With increasing rates of antimicrobial resistance, treatment should be guided by adherence to the principles of antimicrobial stewardship.
Acknowledgements
The authors gratefully acknowledge that they live and work on the ancestral, traditional and unceded territory of the Coast Salish peoples, including the Musqueam, Squamish and Tsleil-Waututh Nations.
Footnotes
Competing interests: Troy Grennan is Vice-chair of the Public Health Agency of Canada’s National Advisory Committee on Sexually Transmitted and Blood-Borne Infections and holds a Health Professional Investigator Award from Michael Smith Health Research BC. No other competing interests were declared.
This article has been peer reviewed.
Contributors: All of the authors contributed to the conception and design of the work. Clara Van Ommen drafted the manuscript. All of the authors revised it critically for important intellectual content, gave final approval of the version to be published and agreed to be accountable for all aspects of the work.
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11109 | https://www.gneet.com/biology/b25/cyclic-and-non-cyclic-photophosphorylation.html | Cyclic and Non Cyclic Photophosphorylation|Photosynthesis in Higher Plants
Cyclic and Non Cyclic Photophosphorylation
⬅️ Back to Biology MCQ Index 📘 Need a Clue?
Total Questions: 40 | Time Left: 0h 29m 56s
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Q1) Synthesis of ADP + Pi → ATP in grana/photosynthesis is ... ..... [AIIMS 1993]
Phosphorylation
Photophosphorylation
Oxidative phosphorylation
Photolysis
2
Correct Option (2)
📘 Concept: ATP Formation During Light Reactions – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Photophosphorylation
🧠 Explanation:
During the light reactions of photosynthesis in the grana/thylakoid membranes, light energy is used to generate ATP through a process called:
Photophosphorylation – Synthesis of ATP from ADP and Pi using the energy of sunlight.
It occurs via cyclic or non-cyclic electron transport chains.
This ATP is then used in the Calvin cycle (dark reaction) for carbon fixation.
❌ Why Other Options Are Incorrect:
Option 1:Phosphorylation — General term; not specific to light-driven ATP synthesis.
Option 3:Oxidative phosphorylation — Occurs in mitochondria during cellular respiration, not photosynthesis.
Option 4:Photolysis — Refers to the splitting of water (H₂O) using light energy, releasing O₂, not ATP formation.
💡 NEET Tip: "ATP from sunlight = Photophosphorylation ☀️💥💾"
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q2) The scientist to discover photophosphorylation was ... .. [BHU 1993]
Arnon
Hill
Willstatter and Stoll
Park and Biggins
1
Correct Option (1)
📘 Concept: Discovery of Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Arnon
🧠 Explanation:
Daniel I. Arnon (1954) was the first to demonstrate photophosphorylation, the light-driven synthesis of ATP in isolated chloroplasts.
He showed that chloroplasts can synthesize ATP in the presence of light, ADP, and inorganic phosphate (Pi).
This experiment proved that ATP synthesis could be coupled to the electron transport system driven by light.
❌ Why Other Options Are Incorrect:
Option 2:Hill — Known for the Hill reaction (photolysis of water), not photophosphorylation.
Option 3:Willstatter and Stoll — They worked on the structure and chemistry of chlorophyll, not ATP synthesis.
Option 4:Park and Biggins — Known for studies on light harvesting complexes, not ATP generation.
💡 NEET Tip: “Arnon = ATP from Light 💡 ➝ Photophosphorylation 🔁⚡”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q3) Where are the molecules of the electron transport chain found in plant cells?
thylakoid membranes of chloroplasts
stroma of chloroplasts
inner membrane of mitochondria
matrix of mitochondria
1
Correct Option (1)
📘 Concept: Electron Transport in Photosynthesis – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: thylakoid membranes of chloroplasts
🧠 Explanation:
In plant cells, the electron transport chain (ETC) related to photosynthesis is located in the thylakoid membranes of the chloroplasts.
Light energy excites electrons in the photosystems embedded in the thylakoid membrane (PS II and PS I).
Electrons are transported through molecules like plastoquinone, cytochrome b₆f, plastocyanin, ferredoxin, and finally to NADP⁺ reductase.
This transfer pumps protons into the thylakoid lumen, generating a proton gradient used to synthesize ATP.
❌ Why Other Options Are Incorrect:
Option 2:Stroma — Site of Calvin cycle; no ETC located here.
Option 3:Inner membrane of mitochondria — This is true for cellular respiration, not photosynthesis.
Option 4:Matrix of mitochondria — This is where the Krebs cycle occurs, not the ETC.
💡 NEET Tip: “ETC for photosynthesis = Thylakoid membrane 🌿☀️; for respiration = Mitochondrial inner membrane 🔋”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q4) Cyclic electron flow may be photoprotective (protective to light-induced damage). Which of the following experiments could provide information on this phenomenon?
Using mutated organisms that can grow but that cannot carry out cyclic flow of electrons and compare their abilities to photosynthesize in different light intensities
Using plants that can carry out both linear and cyclic electron flow, or only one or another of these processes, and measuring their light absorbance
Using bacteria that have only cyclic flow and look for their frequency of mutation damage
Using bacteria with only cyclic flow and measuring the number and types of photosynthetic pigments they have in their membranes
1
Correct Option (1)
📘 Concept: Role of Cyclic Electron Flow in Photoprotection – Advanced Understanding in Photosynthesis
✅ Correct Option: Using mutated organisms that can grow but that cannot carry out cyclic flow of electrons and compare their abilities to photosynthesize in different light intensities
🧠 Explanation:
Cyclic electron flow is thought to play a photoprotective role by preventing over-reduction and damage to the photosynthetic apparatus under high light intensity.
An effective experimental approach is to use mutants incapable of cyclic flow and compare them with wild types under various light conditions.
If such mutants perform poorly or show damage under intense light, it would suggest that cyclic flow protects against photodamage.
❌ Why Other Options Are Incorrect:
Option 2: Measuring light absorbance doesn't directly relate to photoprotection or damage response.
Option 3: Mutation frequency doesn’t directly assess photosynthetic damage or performance.
Option 4: Pigment measurement reveals light-capturing ability, not photoprotective function.
💡 NEET Insight: Cyclic flow helps protect PSI and thylakoid membrane from overexcitation – especially under stress or high light 🌞🛡️
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q5) Type of photophosphorylation in which PS I alone participates:
Cyclic photophosphorylation
Non-cyclic photophosphorylation
Oxidative photophosphorylation
Substrate level photophosphorylation
1
Correct Option (1)
📘 Concept: Cyclic vs Non-Cyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Cyclic photophosphorylation
🧠 Explanation:
There are two types of photophosphorylation in plants:
Non-cyclic photophosphorylation: Involves both PS I and PS II. Electrons move linearly and are used to reduce NADP⁺ to NADPH.
Cyclic photophosphorylation: Involves only Photosystem I (PS I). Electrons are cycled back to PS I, producing ATP but not NADPH or O₂.
Cyclic photophosphorylation occurs under conditions where:
There is a higher demand for ATP than NADPH
PS II is inactive or damaged
Low CO₂ levels restrict Calvin cycle activity
❌ Why Other Options Are Incorrect:
Option 2:Non-cyclic photophosphorylation — Requires both PS I and PS II.
Option 3:Oxidative photophosphorylation — Not a real term; likely confusion with oxidative phosphorylation in mitochondria.
Option 4:Substrate level photophosphorylation — Happens in glycolysis/Krebs cycle, not photosynthesis.
💡 NEET Tip: “PS I only? ➝ It’s Cyclic! 🔁⚡”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q6) Donor and acceptor of electrons is the same chlorophyll molecule in:
Cyclic photophosphorylation
Photorespiration
Substrate level phosphorylation
Non-cyclic photophosphorylation
1
Correct Option (1)
📘 Concept: Electron Flow in Photosystems – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Cyclic photophosphorylation
🧠 Explanation:
In cyclic photophosphorylation, only Photosystem I (PS I) is involved.
Light excites electrons in the chlorophyll a molecule (P700), which pass through an electron transport chain and eventually return to the same chlorophyll molecule.
This forms a closed loop where the donor and acceptor are the same chlorophyll, enabling ATP production without NADPH or oxygen evolution.
❌ Why Other Options Are Incorrect:
Option 2:Photorespiration — Involves RuBisCO and occurs in chloroplast, peroxisome, and mitochondria; unrelated to chlorophyll electron transfer.
Option 3:Substrate level phosphorylation — Occurs in glycolysis/Krebs cycle without light or chlorophyll involvement.
Option 4:Non-cyclic photophosphorylation — Electrons move from water to NADP⁺ and do not return to the same chlorophyll; O₂ is released.
💡 NEET Tip: "Cyclic ➝ Same chlorophyll; Non-cyclic ➝ Water donates, NADP⁺ accepts ☀️🔁"
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q7) Cyclic photophosphorylation is confined to:
Photosystem I
Photosystem II
Both a & b
None of the above
1
Correct Option (1)
📘 Concept: Cyclic vs Non-Cyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Photosystem I
🧠 Explanation:
In cyclic photophosphorylation, electrons excited in Photosystem I (P700) return to the same chlorophyll molecule after passing through an electron transport chain.
Only PS I is involved.
No PS II → therefore, no photolysis of water and no O₂ evolution.
ATP is produced, but NADPH is not formed.
❌ Why Other Options Are Incorrect:
Option 2:Photosystem II — Not involved in cyclic flow; PS II participates in non-cyclic flow.
Option 3:Both a & b — Incorrect; only PS I is active in cyclic photophosphorylation.
Option 4:None of the above — Incorrect; PS I is definitely involved.
💡 NEET Tip: “Cyclic = PS I only 🔁 | Non-cyclic = PS I + PS II 🔀”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q8) Which of the following does not occur in the cyclic photophosphorylation?
Excitation of electron by light energy
Photolysis of water
ATP formation
Electron transport chain
2
Correct Option (2)
📘 Concept: Key Features of Cyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Photolysis of water
🧠 Explanation:
In cyclic photophosphorylation, electrons ejected from Photosystem I (PS I) return to the same chlorophyll molecule via the electron transport chain. The cycle has the following features:
✅ Light excites the electron → excitation occurs
✅ Electrons pass through an electron transport chain (ETC)
✅ This flow leads to ATP formation
❌ But no photolysis of water occurs → hence, no oxygen is released, and no NADPH is formed
❌ Why Other Options Are Incorrect:
Option 1:Excitation of electron — Occurs when PS I absorbs light.
Option 3:ATP formation — Yes, ATP is formed via chemiosmosis.
Option 4:Electron transport chain — Electrons pass through ETC before returning to PS I.
💡 NEET Tip: "Cyclic = No photolysis, No NADPH, No O₂ ➝ Only PS I and ATP 🔁⚡"
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q9) In light reaction, plastoquinone facilitates the transfer of electrons from ... [NEET 2020]
PS-I to NADP⁺
PS-I to ATP synthase
PS-II to Cytb₆f complex
Cytb₆f complex to PS-I
3
Correct Option (3)
📘 Concept: Electron Transport Chain in Light Reaction – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: PS-II to Cytb₆f complex
🧠 Explanation:
In the light reaction of photosynthesis, Photosystem II (PS-II) absorbs light and gets excited.
Electrons are transferred from P680 to plastoquinone (PQ).
Plastoquinone (PQ) is a mobile electron carrier that transports electrons from PS-II to the cytochrome b₆f complex.
Cytochrome b₆f then transfers electrons to Plastocyanin (PC), which eventually donates them to PS-I.
❌ Why Other Options Are Incorrect:
Option 1:PS-I to NADP⁺ — This transfer is done by ferredoxin and NADP⁺ reductase, not plastoquinone.
Option 2:PS-I to ATP synthase — ATP synthase is driven by proton gradient, not electron transfer from PS-I.
Option 4:Cytb₆f complex to PS-I — This step is carried out by plastocyanin, not plastoquinone.
💡 NEET Tip: “Plastoquinone = PS-II ➝ Cytb₆f | Plastocyanin = Cytb₆f ➝ PS-I 💡⚡”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q10) Noncyclic photophosphorylation produces ..... ..... [Odisha 2003]
NAD⁺
NADH
NADPH
NADP⁺
3
Correct Option (3)
📘 Concept: Products of Noncyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: NADPH
🧠 Explanation:
In noncyclic photophosphorylation, both Photosystem II (PS-II) and Photosystem I (PS-I) are involved.
Electrons from water are excited in PS-II, passed through an electron transport chain, and ultimately transferred to NADP⁺ by the enzyme NADP⁺ reductase.
This forms NADPH, a reducing molecule required for the Calvin cycle.
Additionally, ATP and O₂ are also produced in this process.
❌ Why Other Options Are Incorrect:
Option 1:NAD⁺ — Involved in mitochondrial respiration, not photosynthesis.
Option 2:NADH — Produced during respiration (glycolysis/Krebs cycle), not in photosynthesis.
Option 4:NADP⁺ — Is the oxidized form that is reduced to NADPH during light reactions.
💡 NEET Tip: “Noncyclic ➝ ATP + NADPH + O₂ ➝ All needed for Calvin Cycle 🔄🍃”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q11) In electron transport system, the last electron acceptor of photophosphorylation is ... ... [CPMT 2003]
NADPH
NADP⁺
Atomic oxygen
Chlorophyll a
2
Correct Option (2)
📘 Concept: Final Electron Acceptor in Noncyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: NADP⁺
🧠 Explanation:
In noncyclic photophosphorylation, electrons move from Photosystem II ➝ ETC ➝ Photosystem I ➝ Ferredoxin ➝ NADP⁺.
At the end of the electron transport chain, NADP⁺ accepts electrons and a proton (H⁺) to become NADPH, which is used in the Calvin cycle.
Thus, NADP⁺ is the final electron acceptor in the light reactions of photosynthesis.
❌ Why Other Options Are Incorrect:
Option 1:NADPH — This is the product, not the electron acceptor.
Option 3:Atomic oxygen — Not involved in the light reaction electron pathway.
Option 4:Chlorophyll a — Donates electrons, especially P680 and P700, but does not act as the terminal acceptor.
💡 NEET Tip: “Light Reaction: NADP⁺ accepts last e⁻ ➝ becomes NADPH ➝ fuels Calvin Cycle 💡🔋”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q12) The sequential order in electron transport from PS II to PS I of photosynthesis is ... ... [Tamil Nadu 2001]
FeS, PQ, PC and Cytochrome
FeS, PQ, Cytochromes and PC
PQ, Cytochromes, PC and FeS
PC, Cytochromes, FeS, PQ
3
Correct Option (3)
📘 Concept: Electron Flow Between Photosystems – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: PQ, Cytochromes, PC and FeS
🧠 Explanation:
In non-cyclic photophosphorylation, the correct sequence of electron carriers from Photosystem II (PS II) to Photosystem I (PS I) is:
Electrons are excited in PS II and passed to plastoquinone (PQ).
PQ transfers them to the cytochrome b₆f complex.
From there, electrons are carried by plastocyanin (PC) to Photosystem I (PS I).
Once excited in PS I, electrons move to ferredoxin (FeS) and eventually to NADP⁺.
❌ Why Other Options Are Incorrect:
Option 1 & 2: Begin with FeS (ferredoxin), which is only involved after PS I, not between PS II and PS I.
Option 4: Randomized and incorrect sequence; PC and FeS are placed before PQ, which is not biologically accurate.
💡 NEET Tip: “PSII ➝ PQ ➝ Cyt b₆f ➝ PC ➝ PSI ➝ FeS ➝ NADP⁺” — memorize this flow like a relay race! 🏃♂️⚡
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q13) Which of the following statement is true with regard to light reaction of photosynthetic mechanism in plants? [Kerala 2006]
Chlorophyll a occurs with peak absorption at 680 nm in PSI and 700 nm in PSII
Mg and Na are associated with photolysis of H₂O
O₂ is evolved during cyclic photophosphorylation
PS I and PS II are both involved in noncyclic photophosphorylation
4
Correct Option (4)
📘 Concept: Photosystems and Light Reactions – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: PS I and PS II are both involved in noncyclic photophosphorylation
🧠 Explanation:
In noncyclic photophosphorylation, electrons flow from Photosystem II (PS II) → Electron Transport Chain → Photosystem I (PS I) → NADP⁺.
This linear flow generates:
ATP via chemiosmosis,
NADPH via ferredoxin-NADP⁺ reductase, and
O₂ via photolysis of water at PS II.
Thus, both PS I and PS II must function together.
❌ Why Other Options Are Incorrect:
Option 1:Chlorophyll a occurs with peak absorption at 680 nm in PSI and 700 nm in PSII — This is reversed. PS II has P680, PS I has P700.
Option 2:Mg and Na — Only Mn (manganese) plays a key role in the oxygen-evolving complex; Na is unrelated.
Option 3:O₂ is evolved during cyclic photophosphorylation — False. O₂ evolution happens only in noncyclic when water is split in PS II.
💡 NEET Tip: “Cyclic = PS I only, No O₂, No NADPH ➝ Noncyclic = PS I + PS II, ATP + NADPH + O₂” 🌿🔁📈
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q14) Which one is required for photosynthetic evolution of oxygen? ... .. [MHTCET 2008]
F-enzyme
Z-enzyme
Co-enzyme
N-enzyme
2
Correct Option (2)
📘 Concept: Water-Splitting Complex and Oxygen Evolution – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Z-enzyme
🧠 Explanation:
The Z-enzyme refers to the oxygen-evolving complex (OEC) located on the inner side of the thylakoid membrane, attached to Photosystem II (PS II).
It contains manganese ions (Mn²⁺) and is responsible for the photolysis of water:
2H₂O → 4H⁺ + 4e⁻ + O₂
This reaction releases electrons for the electron transport chain, protons for chemiosmosis, and molecular oxygen (O₂) as a byproduct.
❌ Why Other Options Are Incorrect:
Option 1:F-enzyme — Not associated with photosynthetic O₂ evolution.
Option 3:Co-enzyme — A general term; not specific to photosystem water-splitting.
Option 4:N-enzyme — No such designation in photosynthesis pathways.
💡 NEET Tip: “Z-complex = Oxygen-evolving complex = Water split → O₂ + electrons” ⚡💧🌿
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q15) Final electron acceptor of PS II is ... ..[MHTCET 2008]
Phaeophytin
Plastoquinone
Cytochrome b
Plastocyanin
4
Correct Option (4)
📘 Concept: Electron Flow from PS II to PS I – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Plastocyanin
🧠 Explanation:
In the light reaction of photosynthesis, electrons excited in Photosystem II (PS II) follow this pathway:
Phaeophytin – The first electron acceptor in PS II (acts immediately after P680⁺).
Plastoquinone (PQ) – Accepts electrons from phaeophytin and transfers them to the cytochrome b₆f complex.
Cytochrome b₆f – Passes electrons to Plastocyanin (PC).
Plastocyanin – A mobile copper-containing protein that carries electrons from PS II to Photosystem I (PS I).
Therefore, the final electron acceptor in the PS II chain is plastocyanin, which delivers electrons to PS I.
❌ Why Other Options Are Incorrect:
Option 1:Phaeophytin — First acceptor within PS II, not the final one in the whole PS II pathway.
Option 2:Plastoquinone — Intermediate carrier; receives electrons early after phaeophytin.
Option 3:Cytochrome b — Part of the complex that transfers electrons to plastocyanin, not the final carrier.
💡 NEET Tip: “PS II ➝ PQ ➝ Cyt b₆f ➝ PC (Plastocyanin) ➝ PS I ➝ Ferredoxin ➝ NADP⁺” 🔁🔋
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q16) Non-cyclic electron flow in chloroplast results in production of ... ..
ATP
ATP and NADPH
ATP, NADPH, O₂
NADPH
3
Correct Option (3)
📘 Concept: Products of Non-Cyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: ATP, NADPH, O₂
🧠 Explanation:
In non-cyclic electron flow (also called the Z-scheme), electrons travel in a linear path from water to NADP⁺. This flow involves both Photosystem II (PS II) and Photosystem I (PS I) and leads to:
💡 ATP production via chemiosmosis
🔋 NADPH formation (final electron acceptor is NADP⁺)
🌬️ O₂ evolution due to photolysis of water at PS II
❌ Why Other Options Are Incorrect:
Option 1:ATP — Incomplete; NADPH and O₂ are also formed.
Option 2:ATP and NADPH — O₂ production is missing; occurs via water splitting in PS II.
Option 4:NADPH — Again incomplete; ATP and O₂ are also essential outputs.
💡 NEET Tip: “Non-Cyclic = ATP + NADPH + O₂ | Cyclic = ATP only (No O₂, No NADPH)” 🔁🌞
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q17) Electrons from excited chlorophyll of PS II are accepted first by ... ... [CBSE 2008]
Quinone
Ferredoxin
Cytochrome b
Cytochrome f
1
Correct Option (1)
📘 Concept: Primary Electron Acceptor of PS II – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Quinone
🧠 Explanation:
In Photosystem II (PS II), the chlorophyll molecule P680 absorbs light energy and becomes excited.
It releases high-energy electrons, which are accepted first by a primary electron acceptor known as pheophytin (a modified chlorophyll), and then transferred to plastoquinone (PQ), a type of quinone.
Plastoquinone then passes the electrons to the cytochrome b₆f complex in the electron transport chain.
❌ Why Other Options Are Incorrect:
Option 2:Ferredoxin — Accepts electrons from PS I, not PS II.
Option 3:Cytochrome b — Receives electrons downstream after quinone.
Option 4:Cytochrome f — Part of the cytochrome b₆f complex; not the first acceptor.
💡 NEET Tip: PS II ➝ P680 ➝ Pheophytin ➝ Quinone (PQ) ➝ Cytochromes ➝ PC ➝ PS I ➝ NADP⁺ 🔁
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q18) NADP⁺ is reduced to NADPH in ...
PS I
PS II
Calvin cycle
Noncyclic photophosphorylation
4
Correct Option (4)
📘 Concept: Formation of NADPH – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Noncyclic photophosphorylation
🧠 Explanation:
In the light reactions, electrons move through both PS II and PS I in a linear path known as noncyclic photophosphorylation.
After excitation in PS I, electrons are transferred to ferredoxin, which then donates electrons to NADP⁺ reductase.
This enzyme reduces NADP⁺ to NADPH, using the electrons provided by the flow from water via PS II and PS I.
❌ Why Other Options Are Incorrect:
Option 1:PS I — While PS I provides the electrons, the actual reduction of NADP⁺ occurs downstream via NADP⁺ reductase.
Option 2:PS II — It initiates the electron flow by photolysis of water but does not participate directly in NADP⁺ reduction.
Option 3:Calvin cycle — Uses NADPH; does not generate it.
💡 NEET Tip: “NADPH is formed only in noncyclic photophosphorylation, not cyclic!” 🔁🌿
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q19) Which of the following is false regarding non-cyclic electron transport?
Electrons excited from PS II go to PS I and then to Ferredoxin, and there is negligible chance to go to NADP but most probably electrons return back to PS I.
Electrons from PS II go to PS I and from PS I to Ferredoxin, and there is a chance to go to PS I but most probably electrons go to NADP.
Electrons from PS II go back to PS I and from PS I to Ferredoxin and strictly go to NADP, and there is no chance to return back to PS I.
Electrons from PS II go to PS I and from PS I to Ferredoxin, and there is equal chance to go to NADP and PS I.
3
Correct Option (3)
📘 Concept: Non-Cyclic vs Cyclic Electron Flow – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Electrons from PS II go back to PS I and from PS I to Ferredoxin and strictly go to NADP, and there is no chance to return back to PS I.
🧠 Explanation:
This statement is false because it incorrectly describes the behavior of electrons in non-cyclic photophosphorylation:
In non-cyclic flow, electrons move in one direction: PS II ➝ PS I ➝ Ferredoxin ➝ NADP⁺.
Once electrons reduce NADP⁺ to NADPH, they do not return to PS I — this is unidirectional flow.
The statement falsely says “electrons from PS II go back to PS I” and implies they strictly go to NADP, leaving no flexibility — this contradicts the core idea of flexibility in electron flow (non-cyclic vs cyclic).
❌ Why Other Options Are Acceptable or Less False:
Option 1: Describes more of a cyclic flow, which may occur when NADP⁺ is low — not applicable to non-cyclic but logically plausible in regulation.
Option 2: Reflects correct path with proper possibility of electron fate depending on cellular conditions.
Option 4: While not strictly true, shows understanding that electrons may either reduce NADP⁺ or cycle back depending on ATP/NADPH demand.
💡 NEET Tip: “Non-cyclic ➝ Linear path ➝ Electrons do not return. Cyclic ➝ Electrons return to PS I 🔁⚡”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q20) NADP is:
An electron acceptor
A proton acceptor
Proton pumping channel
Both electron and proton acceptor
4
Correct Option (4)
📘 Concept: NADP⁺ Reduction in Photosynthesis – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Both electron and proton acceptor
🧠 Explanation:
NADP⁺ (Nicotinamide Adenine Dinucleotide Phosphate) is a coenzyme that plays a central role in light-dependent reactions of photosynthesis.
It acts as the final electron acceptor in the electron transport chain.
NADP⁺ combines with two electrons (e⁻) and one proton (H⁺) to form NADPH:
NADP⁺ + 2e⁻ + H⁺ → NADPH
Thus, NADP⁺ acts as both an electron and a proton acceptor.
It carries reducing power to the Calvin cycle where it helps in CO₂ fixation.
❌ Why Other Options Are Incorrect:
Option 1:Electron acceptor — Partially correct, but incomplete.
Option 2:Proton acceptor — Also true, but incomplete.
Option 3:Proton pumping channel — Describes proteins like cytochrome b₆f, not NADP⁺.
💡 NEET Tip: “NADP⁺ + 2e⁻ + H⁺ → NADPH” → Carries reducing power to stroma 💪
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q21) Which of the following statements is incorrect?
Electrons to PS-I are provided by PS-II
Electrons to PS-II are provided by splitting of water
Electrons to PS-I are provided by splitting of water
PS-II is physically located on the inner side of the membrane of thylakoid
3
Correct Option (3)
📘 Concept: Electron Transport Pathway and Thylakoid Membrane Structure – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Electrons to PS-I are provided by splitting of water
🧠 Explanation:
This statement is incorrect because:
Electrons released by photolysis of water are first accepted by Photosystem II (PS II).
PS II then passes these electrons through the electron transport chain to Photosystem I (PS I).
Therefore, PS I receives electrons from PS II, not directly from the splitting of water.
✔️ Why Other Options Are Correct:
Option 1: True — PS I receives electrons from PS II via plastoquinone, cytochrome b₆f, and plastocyanin.
Option 2: True — PS II receives electrons generated by the photolysis of water (OEC or Z-complex).
Option 4: True — PS II is located on the inner (appressed) side of the grana thylakoid membrane.
💡 NEET Tip: “Water → PS II → ETC → PS I → Ferredoxin → NADP⁺ → Calvin cycle” 🔁🌞
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q22) Where does exactly the NADP reductase enzyme located?
Thylakoid lumen
Thylakoid luminal side of membrane
Stromal side of thylakoid membrane
Cytoplasm
3
Correct Option (3)
📘 Concept: NADP⁺ Reduction and Enzyme Localization – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Stromal side of thylakoid membrane
🧠 Explanation:
The enzyme NADP⁺ reductase is involved in the final step of the light reaction, where it catalyzes the reduction of NADP⁺ to NADPH.
This enzyme is specifically located on the stromal side of the thylakoid membrane, where it has access to:
Electrons from ferredoxin (which operates in the stroma), and
Free protons (H⁺) and NADP⁺ available in the stroma.
❌ Why Other Options Are Incorrect:
Option 1:Thylakoid lumen — This is the internal space where protons accumulate but is not where NADP⁺ reductase is located.
Option 2:Thylakoid luminal side — Incorrect; the enzyme faces the stroma, not the lumen.
Option 4:Cytoplasm — Photosynthesis enzymes operate in the chloroplast, not the cytoplasm.
💡 NEET Tip: “NADP⁺ → NADPH ➝ happens in stroma → enzyme faces stroma” 💡🧪
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q23) What are the products of linear photophosphorylation?
Heat and fluorescence
ATP and P700
ATP and NADPH
ADP and NADP
3
Correct Option (3)
📘 Concept: Products of Linear (Non-Cyclic) Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: ATP and NADPH
🧠 Explanation:
In linear (non-cyclic) photophosphorylation, electrons move from Photosystem II (PS II) to Photosystem I (PS I) and finally reduce NADP⁺ to NADPH. This process produces:
⚡ ATP — via chemiosmotic phosphorylation during electron transport
🔋 NADPH — by reduction of NADP⁺ using electrons from PS I
🌬️ O₂ — as a byproduct from the splitting of water in PS II
❌ Why Other Options Are Incorrect:
Option 1:Heat and fluorescence — These are loss mechanisms in overexcited chlorophyll, not useful products of photophosphorylation.
Option 2:ATP and P700 — P700 is the special chlorophyll molecule in PS I; it's not a product.
Option 4:ADP and NADP — These are substrates used to form ATP and NADPH.
💡 NEET Tip: “Linear = ATP + NADPH + O₂ | Cyclic = ATP only 🔁⚡”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q24) In non-cyclic photophosphorylation:
ATP is generated
Both PSI and PSII are involved
Electron flow is unidirectional
All the above
4
Correct Option (4)
📘 Concept: Non-Cyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: All the above
🧠 Explanation:
Non-cyclic photophosphorylation (also called the Z-scheme) is the primary light-dependent pathway in photosynthesis. It includes:
ATP generation: Electrons pass through the electron transport chain, driving proton pumps and ATP synthase.
Involvement of both PS II and PS I: Electrons originate from water (split at PS II), then move to PS I and finally reduce NADP⁺ to NADPH.
Unidirectional electron flow: Electrons move from water ➝ PS II ➝ PS I ➝ NADP⁺ (they do not cycle back).
❌ Why Other Options Are Less Complete Alone:
Option 1: True but does not mention other key features.
Option 2: Also true but incomplete.
Option 3: Correct fact but lacks full context.
💡 NEET Tip: “Non-cyclic = ATP + NADPH + O₂, via linear e⁻ flow through PSII and PSI” ⚡☀️
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q25) Non-cyclic photophosphorylation is the main photochemical reaction in green plants because
It produces assimilatory power and oxygen
It initiates photolysis of water for production of oxygen
It utilizes more energy for oxygen production
There is no other cycle available
1
Correct Option (1)
📘 Concept: Role of Non-Cyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: It produces assimilatory power and oxygen
🧠 Explanation:
Non-cyclic photophosphorylation is the principal light-dependent process in green plants because it performs two essential roles:
⚡ Produces assimilatory power — that is, ATP and NADPH, which are used in the Calvin cycle for carbon fixation.
🌬️ Generates oxygen — through photolysis of water at Photosystem II (PS II), releasing O₂ as a byproduct.
This dual production supports both the energy and reducing power needs of the plant and maintains atmospheric oxygen levels.
❌ Why Other Options Are Incorrect:
Option 2: Partially true, but does not mention ATP and NADPH formation (assimilatory power), which is the key point.
Option 3: Incorrect — efficiency is not judged by energy use but by product yield.
Option 4: Incorrect — Cyclic photophosphorylation is also available, though less productive.
💡 NEET Tip: “Non-cyclic ➝ Main source of ATP + NADPH + O₂ ➝ Powers the Calvin cycle 🌿”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q26) In non-cyclic photophosphorylation, the electron emitted by P680 is replaced by electron from
NADP
Water
Ferredoxin
Chlorophyll-a
2
Correct Option (2)
📘 Concept: Water Splitting and Electron Flow in PS II – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Water
🧠 Explanation:
In non-cyclic photophosphorylation, light energy excites electrons in P680, the special chlorophyll molecule in Photosystem II (PS II).
These excited electrons are transferred to the electron transport chain and ultimately to NADP⁺ to form NADPH.
To replace the lost electrons from P680, electrons are obtained from water molecules (H₂O) through the process of photolysis catalyzed by the oxygen-evolving complex (Z-complex).
This also leads to the release of oxygen (O₂) and protons (H⁺):
2H₂O → 4H⁺ + 4e⁻ + O₂
❌ Why Other Options Are Incorrect:
Option 1:NADP — It is the final electron acceptor, not a donor to P680.
Option 3:Ferredoxin — It accepts electrons from PS I, not PS II.
Option 4:Chlorophyll-a — P680 itself is a type of chlorophyll-a; it doesn’t replace its own lost electrons.
💡 NEET Tip: “Water splits at PS II → electrons to P680 → O₂ released 🌊⚡🌿”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q27) In non-cyclic photophosphorylation, all the participants act as electron donor and acceptor except
Chl-a of PS I
Chl-a of PS II
NADP
Both (A) and (B)
3
Correct Option (3)
📘 Concept: Electron Transfer Roles in Non-Cyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: NADP
🧠 Explanation:
In non-cyclic photophosphorylation, several components donate and receive electrons in a chain:
Chlorophyll-a of PS II (P680): accepts electrons from photolysis of water and donates them to pheophytin and the ETC.
Chlorophyll-a of PS I (P700): accepts electrons from plastocyanin and donates them to ferredoxin.
NADP⁺: is the final electron acceptor only. It is not an electron donor within the chain. It becomes reduced to NADPH and exits the system.
❌ Why Other Options Are Incorrect:
Option 1:Chl-a of PS I — Acts as both an acceptor (from plastocyanin) and a donor (to ferredoxin).
Option 2:Chl-a of PS II — Accepts electrons from water and donates to the ETC.
Option 4:Both A and B — Incorrect; both participate as donor and acceptor during the electron flow.
💡 NEET Tip: “NADP⁺ = Only Electron Acceptor ➝ Converts to NADPH ➝ Exits the chain 🔋”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q28) Find out the incorrect event in non-cyclic photosynthesis:
(i) Only one Photosystem participates.
(ii) ATP and NADPH₂ are not produced.
(iii) Photolysis of water takes place.
(iv) It is only concerned with the production of ATP.
(v) This could be happening in stroma lamellae.
(i) only
(i) and (ii) only
(i), (ii) and (v) only
(i), (ii), (iv) and (v)
4
Correct Option (4)
📘 Concept: Events in Non-Cyclic Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: (i), (ii), (iv) and (v)
🧠 Explanation:
Let’s evaluate each statement for non-cyclic photophosphorylation:
(i) ❌ Only one Photosystem participates → False; both PS II and PS I are involved.
(ii) ❌ ATP and NADPH₂ are not produced → False; both are key products.
(iii) ✅ Photolysis of water takes place → True; occurs at PS II.
(iv) ❌ Only concerned with ATP production → False; also produces NADPH and O₂.
(v) ❌ This could be happening in stroma lamellae → False; non-cyclic photophosphorylation occurs in grana thylakoids, not stroma lamellae.
Thus, statements (i), (ii), (iv), and (v) are all incorrect in the context of non-cyclic photosynthesis.
💡 NEET Tip: “Non-cyclic ➝ PS II + PS I, ATP + NADPH + O₂, occurs in grana thylakoids 🌿☀️”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q29) Condition favouring cyclic photophosphorylation is ... .. [CPMT 1999]
Anaerobic environment
Aerobic and low light intensity
Aerobic and optimum light
Anaerobic and low light intensity
4
Correct Option (4)
📘 Concept: Regulation of Photophosphorylation – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Anaerobic and low light intensity
🧠 Explanation:
Cyclic photophosphorylation is favored under:
🌫️ Anaerobic conditions, where NADP⁺ is not available or insufficient to act as the terminal electron acceptor.
🌥️ Low light intensity, which limits energy to drive full non-cyclic flow through both photosystems.
🧮 High ATP demand, where the plant shifts to cyclic flow to generate ATP without producing NADPH or O₂.
Only Photosystem I is involved in cyclic flow. The electrons cycle back from ferredoxin to plastoquinone, enhancing the proton gradient for ATP synthesis, but no NADPH or O₂ is produced.
❌ Why Other Options Are Incorrect:
Option 1: Anaerobic alone isn’t sufficient; low light adds to the condition.
Option 2 & 3: Aerobic conditions generally favor non-cyclic photophosphorylation.
💡 NEET Tip: “Low light + No NADP⁺ ➝ Cyclic kicks in ➝ ATP only 🔁⚡”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q30) Quinones are
mobile electron carriers
enzymes of oxidative phosphorylation
enzymes of Krebs cycle
none of the above
1
Correct Option (1)
📘 Concept: Role of Quinones in Electron Transport – NCERT Class 11: Photosynthesis in Higher Plants & Class 12: Respiration in Plants
✅ Correct Option: mobile electron carriers
🧠 Explanation:
Quinones (such as Plastoquinone in chloroplasts and Ubiquinone or Coenzyme Q in mitochondria) are lipid-soluble molecules that move freely within the membrane.
They act as mobile electron carriers in the electron transport chains of both photosynthesis and respiration.
They transfer electrons between protein complexes, e.g.:
In chloroplasts: from Photosystem II to the cytochrome b₆f complex
In mitochondria: from complex I/II to complex III
❌ Why Other Options Are Incorrect:
Option 2:Enzymes of oxidative phosphorylation — Quinones are not enzymes, they are electron carriers.
Option 3:Enzymes of Krebs cycle — Quinones are not involved in the Krebs cycle; they operate in the ETC.
Option 4:None of the above — Incorrect; Option 1 is correct.
💡 NEET Tip: “Plastoquinone and CoQ = Lipid-based taxis for electrons in the membrane 🛻🔋”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q31) Light reaction in stroma lamellae of chloroplast results in the production of ... .. [RPMT 2011]
ATP
NADPH₂
ATP + NADPH₂
O₂
1
Correct Option (1)
📘 Concept: Cyclic Photophosphorylation in Stroma Lamellae – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: ATP
🧠 Explanation:
Stroma lamellae are the unstacked membranes connecting the grana in chloroplasts.
They lack Photosystem II (PS II) and the NADP⁺ reductase enzyme.
Therefore, cyclic photophosphorylation occurs here — involving only Photosystem I.
This leads to the production of ATP only, without generating NADPH₂ or O₂.
❌ Why Other Options Are Incorrect:
Option 2:NADPH₂ — Not formed because NADP⁺ is not reduced in cyclic flow.
Option 3:ATP + NADPH₂ — Only ATP is formed; NADPH₂ requires PS II and NADP⁺ reductase.
Option 4:O₂ — O₂ evolves from photolysis of water at PS II, which is absent in stroma lamellae.
💡 NEET Tip: “Grana = Non-cyclic = ATP + NADPH + O₂, Stroma lamellae = Cyclic = ATP only 🔁🌞”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q32) Following is needed in the ionic form during non-cyclic photophosphorylation but not during cyclic photophosphorylation in higher plants:
Copper
Iron
Manganese
Zinc
3
Correct Option (3)
📘 Concept: Role of Manganese in Photolysis – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Manganese
🧠 Explanation:
During non-cyclic photophosphorylation, the photolysis of water occurs in Photosystem II (PS II).
This process requires the oxygen-evolving complex (OEC), also called the Z-complex, which contains manganese (Mn²⁺) ions.
Manganese plays a critical role in the splitting of water molecules to release electrons, protons, and molecular oxygen:
2H₂O → 4H⁺ + 4e⁻ + O₂
This reaction does not occur in cyclic photophosphorylation, which involves only PS I and does not require water-splitting or oxygen evolution.
❌ Why Other Options Are Incorrect:
Option 1:Copper — Involved in other metabolic pathways but not directly in photolysis.
Option 2:Iron — Present in cytochromes and ferredoxin, used in both cyclic and non-cyclic pathways.
Option 4:Zinc — Not directly involved in photophosphorylation reactions.
💡 NEET Tip: “Manganese = Essential for O₂ evolution in PS II (non-cyclic only) ⚡💧”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q33) Some photosynthetic bacteria (e.g., purple sulfur bacteria) have photosystem I but not II, while others (e.g., cyanobacteria) have both PSI and PSII. Which of the following might this observation imply?
Photosystem II must have been selected against in some species.
Photosystem I must be more ancestral.
Photosystem II may have evolved to be more photoprotective.
Cyclic flow must be more primitive than linear flow of electrons.
2
Correct Option (2)
📘 Concept: Evolutionary Origin of Photosystems – NCERT Chapter 13: Photosynthesis in Higher Plants + Evolutionary Biology Concepts
✅ Correct Option: Photosystem I must be more ancestral.
🧠 Explanation:
Purple sulfur bacteria and green sulfur bacteria possess only Photosystem I (PS I) and perform anoxygenic photosynthesis, meaning they do not produce oxygen.
Cyanobacteria, on the other hand, have both PS I and PS II and are capable of oxygenic photosynthesis (like higher plants).
This suggests that:
Photosystem I evolved first and is more primitive/ancestral.
Photosystem II likely evolved later, allowing organisms to extract electrons from water, thereby releasing oxygen as a by-product.
❌ Why Other Options Are Incorrect:
Option 1:Selected against — There is no evidence for active loss of PS II due to selection pressure in most bacteria lacking it.
Option 3:Photoprotective role of PS II — PS II is not primarily for photoprotection; that’s more relevant to carotenoids and cyclic flow.
Option 4:Cyclic flow is more primitive — While cyclic flow is simpler, this statement does not address the presence/absence of photosystems, which is the core of the question.
💡 NEET Tip: “Bacteria with only PSI → older. Cyanobacteria with PSI + PSII → evolved oxygenic photosynthesis 🌍💧”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q34) How many of the following are true?
(i) Photosynthesis cannot continue for long if during light reaction only cyclic photophosphorylation takes place. This is because only ATP is formed.
(ii) Photosystem II comprises equal amount of Chlorophyll a and chlorophyll b.
(iii) In non–cyclic electron transport, only one ATP is formed during downhill electron transport because of the absence of Ferrodoxin.
(iv) PS II is a strong oxidant.
One
Two
Three
Four
3
Correct Option (3)
📘 Concept: Multiple concepts from NCERT Chapter 11 – Photosynthesis in Higher Plants
✅ Correct Option: Three
🧠 Statement-wise Analysis:
(i) ✅ True — Cyclic photophosphorylation forms only ATP, not NADPH or O₂. Photosynthesis cannot continue long without NADPH (needed for CO₂ reduction in Calvin cycle).
(ii) ❌ False — Photosystem II has more Chlorophyll a than Chlorophyll b; Chlorophyll a is the reaction center, and b is an accessory pigment.
(iii) ✅ True — In non-cyclic flow, ATP is formed mainly during the downhill electron transport from PQ to PS I. Ferredoxin is involved after PS I, in reducing NADP⁺, not in ATP generation.
(iv) ✅ True — P680⁺, the oxidized form of PS II reaction center, is one of the strongest biological oxidants known. It pulls electrons from water molecules during photolysis.
❌ Incorrect Statement: (ii)
💡 NEET Tip: “Cyclic = ATP only; PS II = P680 = super oxidant; Chl-a is dominant in both PS I and PS II 🌿”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q35) As a research scientist, you measure the amount of ATP and NADPH consumed by the Calvin cycle in 1 hour. You find 30,000 molecules of ATP consumed, but only 20,000 molecules of NADPH. Where did the extra ATP molecules come from?
Photosystem II
Photosystem I
Cyclic electron flow
Linear electron flow
3
Correct Option (3)
📘 Concept: Balancing ATP/NADPH Ratio in the Calvin Cycle – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: Cyclic electron flow
🧠 Explanation:
The Calvin cycle uses more ATP than NADPH — specifically, 3 ATP and 2 NADPH per CO₂ fixed.
Linear (non-cyclic) photophosphorylation produces ATP and NADPH in a 1:1 ratio, which is insufficient to meet the Calvin cycle’s higher ATP demand.
To compensate for the ATP shortfall, the plant uses cyclic electron flow, which:
Involves only Photosystem I (PS I)
Generates ATP only — no NADPH or O₂
Helps meet the additional ATP demand for regeneration of RuBP in the Calvin cycle
❌ Why Other Options Are Incorrect:
Option 1: PS II is involved in linear flow; it does not generate ATP alone.
Option 2: PS I participates in both cyclic and linear flow, but ATP generation alone happens during cyclic flow.
Option 4: Linear flow generates equal ATP and NADPH; it cannot explain excess ATP production.
💡 NEET Tip: “Extra ATP ➝ Cyclic Flow ➝ PS I only ➝ No NADPH/O₂ ➝ Only ATP 🌀”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q36) Which one does not take part in electron transfer ... .... [KCET 2001]
CoQ
FeS
ATP
NAD⁺
3
Correct Option (3)
📘 Concept: Role of Electron Carriers in Cellular and Photosynthetic Electron Transport – NCERT Class 11 & 12
✅ Correct Option: ATP
🧠 Explanation:
ATP (Adenosine Triphosphate) is an energy currency, not an electron carrier.
It stores and provides energy for biochemical processes, but does not carry electrons in the electron transport chains.
In contrast, the following are true electron carriers:
CoQ (Coenzyme Q): Also called ubiquinone, part of the mitochondrial electron transport chain.
FeS (Iron-sulfur proteins): Present in both mitochondrial and photosynthetic ETC, involved in electron shuttling.
NAD⁺ (Nicotinamide adenine dinucleotide): A key electron acceptor, reduced to NADH after accepting electrons.
❌ Why Other Options Are Incorrect:
Option 1:CoQ — True electron carrier in mitochondria.
Option 2:FeS — Transfers electrons between ETC components.
Option 4:NAD⁺ — Accepts electrons and protons to become NADH.
💡 NEET Tip: “ATP = energy donor 💥, NAD⁺/FAD/FeS = electron carriers 🔋”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q37) Cytochromes are ... .... [AFMC 1994]
O₂ acceptors
H₂ acceptors
Electron acceptors
H₂O acceptor
3
Correct Option (3)
📘 Concept: Role of Cytochromes in Electron Transport – NCERT Class 11: Chapter 13 (Photosynthesis in Higher Plants) and Class 12: Respiration in Plants
✅ Correct Option: Electron acceptors
🧠 Explanation:
Cytochromes are heme-containing proteins involved in electron transport chains (ETC) in both:
🧬 Cellular respiration (mitochondria)
🌿 Photosynthesis (chloroplast thylakoid membranes)
They alternate between reduced and oxidized states by accepting and donating electrons, helping shuttle electrons through the ETC.
They do not accept molecular hydrogen (H₂), water, or oxygen directly — they are strictly electron carriers.
❌ Why Other Options Are Incorrect:
Option 1:O₂ acceptors — Oxygen is the final electron acceptor in aerobic respiration, but not cytochromes themselves.
Option 2:H₂ acceptors — Hydrogen gas is not involved; protons (H⁺) and electrons are handled separately.
Option 4:H₂O acceptor — Water is a product (in respiration) or a substrate (in photosynthesis), not accepted by cytochromes.
💡 NEET Tip: “Cytochrome = Electron taxi 🚕 on the ETC highway 🔋”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q38) Ferredoxin is a constituent of ... ... [CBSE 1991]
PS I
PS II
Hill reaction
P₆₈₀⁻
1
Correct Option (1)
📘 Concept: Components of Photosystem I – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: PS I
🧠 Explanation:
Ferredoxin is a soluble iron-sulfur (Fe-S) protein located on the stromal side of the thylakoid membrane.
It acts as the terminal electron acceptor from Photosystem I (PS I) in the light reactions of photosynthesis.
After accepting electrons from PS I, ferredoxin transfers them to NADP⁺ reductase, which reduces NADP⁺ to NADPH.
❌ Why Other Options Are Incorrect:
Option 2:PS II — PS II transfers electrons to plastoquinone, not ferredoxin.
Option 3:Hill reaction — Refers to oxygen evolution from water splitting; ferredoxin is not directly involved here.
Option 4:P₆₈₀⁻ — The reaction center of PS II, not related to ferredoxin function.
💡 NEET Tip: “PS I ➝ Ferredoxin ➝ NADP⁺ ➝ NADPH 🔋”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q39) Which is not an electron carrier ... .. [Kerala 2004]
CoQ
Cyt c
Cyt a
H₂O
4
Correct Option (4)
📘 Concept: Electron Transport Chain Components – NCERT Class 12: Respiration in Plants
✅ Correct Option: H₂O
🧠 Explanation:
Electron carriers are molecules that receive and donate electrons during redox reactions in pathways like photosynthesis and respiration.
CoQ (ubiquinone), cytochrome c (Cyt c), and cytochrome a (Cyt a) are well-known electron carriers within the electron transport chain (ETC).
H₂O (water), however, is not an electron carrier. In photosynthesis, it acts as the initial electron donor at PS II and undergoes photolysis, releasing electrons, protons, and oxygen:
2H₂O → 4H⁺ + 4e⁻ + O₂
Thus, H₂O is a source of electrons, not a carrier that transfers them along a chain.
❌ Why Other Options Are Electron Carriers:
Option 1:CoQ — Mobile electron carrier in mitochondrial ETC.
Option 2:Cyt c — Small protein that shuttles electrons between complexes III and IV.
Option 3:Cyt a — Part of cytochrome oxidase (complex IV), key in ETC.
💡 NEET Tip: “H₂O = Electron donor 💧, not a carrier 🚫. Cytochromes & CoQ = true shuttlers 🚕”
📝 Your Selected Option:Not Answered
💬 Remark:Correct
Q40) Which of the following electron carriers are present in the inner surface of thylakoid membrane?
(i) Pheophytin
(ii) Ferredoxin
(iii) Plastocyanin
(iv) Cytochrome b₆
(v) Plastoquinone – B
(vi) Cytochrome f
(i), (ii) and (iii)
(iii), (iv) and (vi)
All except (v)
All except (i)
2
Correct Option (2)
📘 Concept: Orientation of Electron Carriers in Thylakoid Membranes – NCERT Chapter 11: Photosynthesis in Higher Plants
✅ Correct Option: (iii), (iv) and (vi)
🧠 Explanation:
The thylakoid membrane has distinct surfaces: the inner surface (facing the thylakoid lumen) and the outer stromal side.
Electron carriers localized on the inner (lumenal) surface include:
🔹 (iii) Plastocyanin (PC): A mobile copper-containing protein in the lumen.
🔹 (iv) Cytochrome b₆: A transmembrane protein, part of the cytochrome b₆f complex, oriented partly toward the lumen.
🔹 (vi) Cytochrome f: Another component of the b₆f complex exposed to the lumenal side.
Other components:
(i) Pheophytin: Bound to PS II on the stromal side.
(ii) Ferredoxin: Soluble protein on the stromal side of the thylakoid membrane, not inner surface.
(v) Plastoquinone – B (PQ-B): Lipid-soluble carrier within the membrane, not restricted to lumen-facing surface.
💡 NEET Tip: “PC, Cyt b₆, and Cyt f = lumen side; Ferredoxin = stroma side” 🧠
📝 Your Selected Option:Not Answered
💬 Remark:Correct
×
🧠 Need a Clue?
📘 Quick Revision: Light Reaction & Photophosphorylation (Q1–Q43)
🌞 Light Reaction Overview
Location: Occurs in the thylakoid membranes (mostly grana) of chloroplasts.
Products: ATP, NADPH, and O₂ (from water splitting).
Photosystems:
PS II (P680): Located in grana; initiates non-cyclic photophosphorylation; splits water.
PS I (P700): Found in stroma lamellae; active in both cyclic and non-cyclic flow.
🔁 Cyclic vs Non-Cyclic Photophosphorylation
Cyclic:
Involves only PS I
Produces: ATP only
No photolysis, no O₂ evolution, no NADPH
Favored: In low light or anaerobic conditions
Non-Cyclic:
Involves both PS II and PS I
Produces: ATP, NADPH, and O₂
Electrons flow from water → PS II → PS I → NADP⁺
⚡ Key Components & Electron Carriers
P680: Reaction center of PS II; powerful oxidant, pulls electrons from H₂O.
P700: Reaction center of PS I; donates electrons to ferredoxin.
Ferredoxin: Accepts electrons from PS I, reduces NADP⁺ to NADPH.
Plastoquinone & Plastocyanin: Mobile electron carriers within the membrane.
Cytochrome b₆f complex: Pumps protons into lumen; helps ATP synthesis.
ATP Synthase: Enzyme that makes ATP using proton gradient.
🌊 Photolysis & Oxygen Evolution
Occurs at PS II via the oxygen-evolving complex (Z-enzyme).
Requires Mn²⁺ ions for splitting water:
Equation: 2H₂O → 4H⁺ + 4e⁻ + O₂
🔋 NADP⁺ & Energy Balance
NADP⁺: Final electron acceptor in non-cyclic flow; forms NADPH.
NADP⁺ reductase: Located on stromal side of thylakoid membrane.
Calvin cycle consumes more ATP than NADPH, extra ATP made by cyclic photophosphorylation.
🧪 Structures & Membrane Localization
Grana: Site of PS II and non-cyclic flow.
Stroma lamellae: Lacks PS II; site of cyclic photophosphorylation.
Electron carriers on inner thylakoid surface: Plastocyanin, cytochrome b₆, cytochrome f.
❗ Common Misconceptions (Corrected)
ATP is not an electron carrier.
H₂O is an electron donor, not a carrier.
Cytochromes and quinones (like plastoquinone) are true electron carriers.
Photosystem I is more ancestral than Photosystem II.
🔁 Back to full chapter overview: Molecular Basis of Inheritance – Chapter Page |
11110 | https://optimization-online.org/wp-content/uploads/2015/06/4964.pdf | On the steepest descent algorithm for quadratic functions Cl´ ovis C. Gonzaga∗ Ruana M. Schneider† July 9, 2015 Abstract The steepest descent algorithm with exact line searches (Cauchy algorithm) is inefficient, generating oscillating step lengths and a sequence of points converging to the span of the eigen-vectors associated with the extreme eigenvalues. The performance becomes very good if a short step is taken at every (say) 10 iterations. We show a new method for estimating short steps, and propose a method alternating Cauchy and short steps. Finally, we use the roots of a certain Chebyshev polynomial to further accelerate the method.
1 Introduction We study the quadratic minimization problem (Pz) minimize z∈I Rn ¯ f(z) = cTz + 1 2zT Hz, where c ∈I Rn and H ∈I Rn×n is symmetric with eigenvalues 0 < d1 < d2 < . . . < dn, and condition number C = dn/d1. The problem has a unique solution z∗∈I Rn.
The steepest descent algorithm, also called gradient method, is a memoryless method defined by z0 ∈I Rn given, zk+1 = zk −λk∇f(zk), λk > 0.
(1) The only distinction among different steepest descent algorithms is in the choice of the step lengths λk.
For the analysis, the problem may be simplified by assuming that z∗= 0, and so f(z) = zT Hz/2.
The matrix H may be diagonalized by setting z = Mx, where M has orthonormal eigenvectors of H as columns. Then the function becomes f(x) = 1 2 xT Dx, D = diag(d1, d2, . . . , dn), (2) so that for z ∈I Rm, ¯ f(z) = f(MTz). M defines a similarity transformation, and hence for z = Mx, ∥z∥= ∥x∥, ∇¯ f(z) = ∥∇f(x)∥, ∇¯ f(z) = M∇f(x), using throughout the paper the 2-norm ∥z∥2 = zTz.
2Department of Mathematics, Federal University of Santa Catarina, Florian´ opolis, SC, Brazil; e-mail : ccgonzaga1@gmail.com. The author was partially supported by CNPq under grant 308413/2009-1.
2Department of Mathematics, Federal University of Santa Catarina, Florian´ opolis, SC, Brazil; e-mail: ruanamaira@gmail.com.
1 2 We define the diagonalized problem (P) minimize x∈I Rn f(x) = 1 2 xT Dx.
The steepest descent iterations with step lengths λk for minimizing respectively f(·) from the initial point x0 = MTz0, and ¯ f(·) from the initial point z0, are related by zk = Mxk. Thus, we may restrict our study to the diagonalized problem.
The steepest descent method, was devised by Augustine Cauchy [?] in 1847. He studied the quadratic minimization problem, using in each iteration the “Cauchy step” λk = argmin λ≥0 f(xk −λ∇f(xk)).
(3) The steepest descent method with Cauchy steps will be called Cauchy algorithm. Steepest descent is the most basic algorithm for the unconstrained minimization of continuously differentiable func-tions, with step lengths computed by a multitude of line search schemes.
The quadratic problem is the simplest non-trivial non-linear programming problem. Being able to solve it is a pre-requisite for any method for more general problems, and this is the first reason for the great effort dedicated to its solution. A second reason is that the optimal solution of (Pz) is the solution of the linear system Hz = −c.
It was soon noticed that the Cauchy algorithm generates inefficient zig-zagging sequences. This phenomenon was established by Akaike [?] in 1959, and further developed by Forsythe [?]. A clear explanation of its consequences is found in Nocedal, Sartenaer and Zhu [?]. For some time the steepest descent method was displaced by methods using second order information.
In the last years gradient methods returned to the scene due to the need to tackle large scale prob-lems, with millions of variables, and due to novel methods for computing the step lengths. Barzilai and Borwein [?] proposed a new step length computation with surprisingly good properties, which was further extended to non-quadratic problems by Raydan [?], and studied by Dai [?], Raydan and Svaiter [?], Birgin, Mart´ ınez and Raydan [?], among others. In another line of research, several methods were developed to enhance the Cauchy algorithm by breaking its zig-zagging pattern. For the latest developments of this subject, see Asmundis et al. [?, ?].
This paper has two goals. In section 2 we write complete proofs for the main properties of the Cauchy algorithm, including the classical results by Akaike, hopefully simplifying the treatment. In section 3 we develop new ways of breaking the oscillatory behavior of the algorithm and compare several enhancements.
We do not attempt to generalize these procedures to non-quadratic or to constrained problems.
2 The Cauchy algorithm In this section we intend to state and prove the main asymptotic properties of the Cauchy algorithm.
These properties will be summarized in Theorem ?? below, and then the whole section is dedicated to prove it and describe some other properties of the sequences generated by the algorithm. The reader may well accept the theorem without going through the proofs, and proceed to the next section. The reason for proving these results, which are not original, is that the original Akaike and Forsythe papers, not being aimed exclusively at these results, are frequently not considered easy to read. We also prove simplified versions of results in [?, ?, ?], with a unified notation.
Given a point x ∈I Rn, with g = ∇f(x), the Cauchy step from x defined as x+ = (I −λD)x, g+ = (I −λD)g.
where λ = argmin σ≥0 f(x −σg) = gTg gT Dg.
3 If follows from the definition of λ that gTg+ = 0.
We shall frequently use the value µ = 1/λ = (gT Dg)/ ∥g∥2. The following sequences will be associated with an application of the steepest descent algorithm from an initial point x0 with g0 = ∇f(x0): (xk): iterates: xk+1 = (I −λkD)xk.
(gk): gradients: gk+1 = (I −λkD)gk.
(λk): Cauchy step length from xk.
(µk): µk = 1/λk.
(yk): normalized gradients yk = gk/ gk .
Assumption: We assume that g0 has no zero components. This is done because if a component is null, then it will remain null forever.
It is well known that xk →0 and gk →0. The main goal of this section is the study of the asymptotic properties of the sequence of normalized gradients yk, following Akaike [?] and Forsythe [?].
2.1 The main theorem The theorem below summarizes some of the main results of the cited references. It is stated, com-mented, and then proved in the following section.
Theorem 1. Consider the sequences with elements xk, gk, yk ∈I Rn, λk and µk = 1/λk generated by the Cauchy algorithm from an initial point x0 ∈I Rn, assuming that n > 1 and x0 1, x0 n , 0. Then there exist µ, µ′ ∈(d1, dn), r, r′ ∈I Rn and α ∈(0, 1) such that (i) µ2k →µ, µ2k+1 →µ′, with µ + µ′ = d1 + dn.
(ii) yk i →0 for i = 2, 3, . . . , n −1.
(iii) y2k →r, y2k+1 →r′, with r, r′ ∈L(e1, en), where L(e1, en) denotes the subspace generated by e1 and en.
(iv) lim k→∞ gk+2 i gk i ≤1 for all i = 1, . . . , n such that gk i , 0 for k ∈I N.
(v) lim k→∞ gk+2 i gk i = lim k→∞ ∥gk+2∥ ∥gk∥ = α for i = 1 and i = n, with α ≥1 −2 d1dn ˜ d2 −δ2 , where ˜ d = (d1 + dn)/2 and δ = min{|di −˜ d| | i = 1, . . . , n}.
(vi) The limiting values for µ, µ′ are bounded by µmin = ˜ d − q ( ˜ d2 + δ2)/2 ≤µ, µ′ ≤˜ d + q ( ˜ d2 + δ2)/2 = µmax.
(vii) For large k, |gk n| oscillates around the values of |gk 1|, with r1 rn = r′ n r′ 1 = µ −d1 µ′ −d1 .
One of the scopes of this paper is to present a complete proof of this theorem. The proof will be postponed to the next section, after a qualitative analysis of the asymptotic properties of the algorithm.
For this discussion, assume that the condition number C is large, say, C >> 10, and that the space dimension is n > 2, possibly large. The variables will be loosely classified as light, medium and heavy, associated respectively with small, medium and large eigenvalues.
4 In a typical step, g+ i = (1 −λdi)gi, λ = 1/µ. Let us comment on step sizes Short steps: safe and frequent. We call ’short’ a step with µ ≥dn/2. Short steps are: • Harmless: all |gi| decrease.
• Efficient for heavy variables.
• Inefficient for light variables: if λ << 1/di, then |g+ i |/|gi| = |1 −λdi| is near 1.
We conclude that short steps reduce heavy variables, with little effect on light variables. About one half of the steps will be short.
Large steps: dangerous but needed. We call ’large’ a step with µ < dn/2. Large steps are: • Dangerous: the heavy variables (|gn|) increase.
• Reasonably efficient for light variables. Very light variables (di near d1) can only be reduced by very large steps. For example, if d1 = 0.001, dn = 1, a step λ = 500 >> 2/dn produces g+ 1 = g1/2 and g+ n = −499g1.
We conclude that only very large steps are efficient for reducing |g1|, and they increase very much the heavy variables. “Medium” steps are also inefficient for the light variables.
Cauchy steps: safe and essential. Cauchy steps satisfy µ ∈(d1, dn), and hence with g+ i = (1 − di/µ)gi, g1 always decreases in absolute value, keeping the sign unchanged; gn changes sign at all iterations, and increases in absolute value if the step is large.
The Cauchy algorithm generates a sequence of steps λk = 1/µk that oscillate, with (µk, µk+1) converging to the two limit points µ, µ′, with µ + µ′ = d1 + dn. Then (say) µ < (d1 + dn)/2 and µ′ > (d1 + dn)/2 > dn/2. So, for large k, the steps alternate between short and large, Cauchy steps are in general “medium short”, i.e., they are not very large.
The sequence of normalized gradients (yk) zig-zags, so that the pairs of consecutive iterations (y2k, y2k+1) converge to a pair of vectors (r, r′) in L(e1, en). In the original space, this is the subspace generated by the eigenvectors associated with the two extreme eigenvalues d1, dn. The step lengths also zig-zags, so that (λ2k, λ2k+1) →(λ, λ′), or equivalently (µ2k, µ2k+1) →(µ, µ′), with µ + µ′ = d1 + dn.
For large k, the absolute values of all gradient components must decrease in each pair of iter-ations. In fact, gk 1 never changes sign and decreases in absolute value in all iterations; gk n changes size in all iterations, while |gk n| oscillates around |gk 1| for large k.
Let us follow the iterates of the Cauchy algorithm for an example with 50 variables, with eigen-values 0.01 ≤di ≤1, x0 i = 1/ √di.
Here are some numerical values for this example: µmin = 0.148, µ = 0.467, µ′ = 0.543, µmax = 0.862.
α = 0.922, |rn/r1| = 0.926, |r′ n/r′ 1| = 1.080.
The value α in item (v) deserves some attention: if C is large and δ is small (there exists an eigen-value near ˜ d), then α > 1 −8d1 = 1 −8/C, very near the worst case convergence rate known for the Cauchy algorithm, given by ((C −1)/(C + 1))2 ≈1 −4/C, counting pairs of iterations.
The figures show the values of |gi|, with the eigenvalues in the horizontal axis. Each figure shows three iterates gk, gk+1, gk+2, with vertical lines at the points µk, µk+1 for k = 0, k = 5 and k = 12. One sees that the gradient components associated with eigenvalues near µk are substantially reduced in the iteration k. The last figure reproduces the third one using a logarithmic scale in the horizontal axis.
In the beginning iterations the heavy variables are reduced (in absolute value), and soon the oscillatory pattern is achieved. The values of µk, µk+1 converge to the values µ, µ′ as in the theorem, 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 10 −10 10 −8 10 −6 10 −4 10 −2 10 0 eigenvalues |g| Iterations 0,1,2 k−1 k k+1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 10 −10 10 −8 10 −6 10 −4 10 −2 10 0 eigenvalues |g| Iterations 5,6,7 k−1 k k+1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 10 −10 10 −8 10 −6 10 −4 10 −2 10 0 eigenvalues |g| Iterations 12,13,14 k−1 k k+1 10 −2 10 −1 10 0 10 −10 10 −8 10 −6 10 −4 10 −2 10 0 eigenvalues |g| Iterations 12,13,14 k−1 k k+1 Figure 1: Absolute values of gradient components in consecutive iterations, showing the values of µk = 1/λk.
the medium gradient components are reduced and |gn| oscillates around |g1|, being slowly reduced at each pair of iterations. In the last figure we see that the algorithm affects the “medium heavy” variables, with little effect of the light variables. Large steps (small values of µ) are needed to reduce the very light components, and they never happen.
Fig. ?? plots the evolution of µk (left) and of |g1| and |gn|, showing the oscillatory behavior.
The step lengths stabilize very fast. As |g0 n| >> |g0 1|, |gn| decreases (in an oscillatory fashion) in the beginning iterations to match |g1|.
0 5 10 15 20 25 30 35 40 45 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 Step sizes k step length 0 5 10 15 20 25 30 35 40 45 10 −2 10 −1 10 0 First and last components of |g| k |gi| Figure 2: Behavior of λk and of gk 1 , gk n . λk oscillates, gk 1 decreases and gk n oscillates.
6 2.2 The Akaike-Forsythe results: proof of the main theorem The results in this subsection are a simplified version of those in Forsythe [?], reduced to the case studied in this paper. We follow his proofs almost step by step, hopefully simplifying the treatment.
The study is simplified by the introduction of the following sequence: (wk): defined by w0 = g0, wk+1 = (µkI −D)wk.
It is easy to see that wk = gk Qk−1 i=0 µk. The sequence (wk) does not seem to have much interest in itself, but as we shall see it is very handy for the study of each iteration of the algorithm. Note that the normalized gradients satisfy yk = gk/ gk = wk/ wk .
Notation. We are mostly interested in the sequence of gradient vectors. When studying an iteration, we use the following maps, defined for a vector g ∈I Rn, g , 0: µ(g) = gT Dg gTg ∈[d1, dn], g+(g) = (I − 1 µ(g)D)g, y+(g) = g+(g) ∥g+(g)∥.
(4) Whenever no confusion is possible, we omit the argument. If g+ , 0, we also define µ+ = µ(g+) and g++. The fact that µ ∈[d1, dn] is a standard result in linear algebra. If g has at least two non-zero components, then µ(g) ∈(d1, dn).
When dealing with the sequence (wk), we use the following notation: w′ = (µI −D)w, w′′ = (µ+I −D)w′, (5) when w′ , 0. Note that µ(gk) = µ(wk) = µ(yk).
By construction of the Cauchy steps, gTg+ = wTw′ = yTy+ = 0.
Let us begin by isolating a special case (and later prove that it never happens): Lemma 1. Suppose that ¯ w = αep for some p ∈{1, 2, . . . , n}, α , 0. Then µ( ¯ w) = dp, ¯ w′ = 0, lim w→¯ w w′(w) ∥w∥ = 0.
Proof. Clearly, µ( ¯ w) = eT pDep = dp, and then ¯ w′ = α(dpI −D)ep = 0, because Dep = dpep. The last result follows from the continuity of µ(·) (and hence of w′(·)) at ¯ w , 0.
□ Lemma 2. Let p < q be the first and last non-zero components of a vector w ∈I Rn. Then µ(w) ∈ (dp, dq) and w′ p , 0, w′ q , 0. For the Cauchy algorithm, if g0 1 , 0 and g0 n , 0, then the same is true for all wk and gk.
Proof. The result follows from (??) and the lemma above: µ(w) = dp (or µ(w) = dq) can only occur if w1 (or wn) is its only non-zero component. The result on the sequence (gk) follows by induction.
□ From the lemmas above we see that the behavior of the Cauchy algorithm is anomalous at points g on the coordinate axes. Let us call such vectors “scalar”, and eliminate them by defining the sets Ω= {g ∈I Rn | g is not scalar}, Z = {y ∈Ω| ∥y∥= 1}.
Continuous maps: now all the maps defined above (µ, µ+, g+, g++, . . .) are continuous on Ω.
We use the simplified notation µ = µ(w), . . .. We assume that the Cauchy algorithm is applied to the problem (P) from a point g0 ∈Ωwith no null component.
Theorem 2. Let (wk) be generated by the Cauchy algorithm. Then for k = 0, 1, . . ., ∥wk+1∥ ∥wk∥ = cos ψk ∥wk+2∥ ∥wk+1∥≤∥wk+2∥ ∥wk+1∥, where ψk is the angle between wk and wk+2.
7 Proof. Using our simplified notation for w = wk, we first prove that ∥w+∥2 = wTw++ : We have wTw++ = wT(µI −D)(µ+I −D)w ∥w+∥2 = wT(µI −D)(µI −D)w.
Subtracting, we obtain wTw++ −∥w+∥2 = wT(µI −D)(µ+ −µ)w = (µ+ −µ)wTw+ = 0.
Now, using this and Cauchy-Schwartz, wTw++ = ∥w∥∥w++∥cos(ψ) = ∥w+∥2.
which divided by ∥w∥∥w+∥gives the desired result, completing the proof.
□ Theorem 3. Let (wk) be generated by the Cauchy algorithm from g0 ∈Ω, and let ψk be the angle between wk and wk+2. Then: (i) The the sequence (φk), with φk = ∥wk+1∥/∥wk∥increases towards a limit L > 0.
(ii) limk→∞ψk = 0 and limk→∞ yk −yk+2 = 0.
(iii) All limit points of (yk) are in Z.
Proof. (i) The sequence increases as a consequence of Theorem ??. Let us show that it is bounded: given k ∈I N, wk+1 = (µkI −D)wk, and then ∥wk+1∥≤∥µkI −D∥∥wk∥.
We know that µk ∈(d1, dn), and hence ∥µkI −D∥≤max{|µk −di| | i = 1, 2, . . . , n} ≤dn −d1, φk = wk+1 wk ≤dn −d1.
Thus the sequence is increasing and bounded, implying that it converges to some value L.
(ii) From Theorem ??, φk = cos(ψk)φk+1, and taking limits, cos(ψk) = φk φk+1 →1, showing that ψk →0. As ∥yk∥= ∥yk+2∥= 1 and ψk is the angle between these vectors, (ii) holds.
(iii) Assume that yk K →r, with K ⊂I N. If r < Z, then necessarily r = ±ep for some p = 1, . . . , n.
Then, by Lemma ??, lim yk→r w′(yk) ∥yk∥ = 0.
This contradicts (i), because ∥wk+1∥ ∥wk∥ = ∥w′(yk)∥ ∥yk∥ , which converges to L > 0, completing the proof.
□ Summing up, the algorithm generates a sequence (yk) with yk = gk ∥gk∥= wk ∥wk∥, whose limit points are all in the set Z. Then all transformations like µ(y), w+(y), . . . are continuous at any limit point r of (yk).
8 For instance, assuming that yk K →r, and defining r′ = w′(r) = (µ(r)I −D)r, r′′ = w′′(r) = (µ+(r)I −D)r′, r+ = r′/ ∥r′∥, r++ = w′′/ ∥w′′∥, we deduce that yk+1 K →r+ and yk+2 K →r++ = r. This last equality follows from Theorem ??. Also from this theorem, we see that r′′ = L2r.
Remarks. There should be no confusion: given r ∈Z with unit norm, r′ = (µI −D)r has ∥r′∥= L, and r+ has unit norm. Also r++ = r.
The sequence (y2k) satisfies yk+2 −yk →0. It is then easy to prove that either it is convergent or it has no isolated limit point (actually, the set of limit points is a continuum).
Theorem 4. Let (yk) be the sequence of normalized gradients generated by the Cauchy algorithm from y0 = g0/∥g0∥. Then (y2k) converges to a non-scalar point r ∈L(e1, en), the linear space generated by the basis vectors e1 and en.
Proof. We already know that any limit point of (yk) is non-scalar. Let r be a limit point of (y2k).
The proof will be done in three steps.
(i) There exist two indices p < q such that r, r+ ∈L(ep, eq).
Let r be a limit point of (yk), and define the vectors r′, r′′ as above. We have r′′ i = (µ −di)(µ+ −di)ri = L2ri, i = 1, . . . , n, where µ, µ+ are fixed. For each i, either ri = 0 or (µ −di)(µ+ −di) = L2. This is a second degree equation, which will have two real solutions, say, di = dp and di = dq, because Theorem ??(iii) prevents a single solution.
If r ∈L(ep, eq), then trivially r′ = (µI −D)r ∈L(ep, eq).
(ii) r is the unique limit of (y2k).
Assume by contradiction that r is not a unique limit point, and hence a non-isolated limit point of (y2k). Then there exist an infinite number of limit points ˜ r satisfying ∥˜ r −r∥< min{|rp|, |rq|}. All such points must belong to L(ep, eq), for the following reason: any limit point ˜ r must belong to some space L(es, et), where s, t are indices. If {s, t} , {p, q}, then either ˜ rp = 0 or ˜ rq = 0, and then ∥˜ r −r∥≥min |rp|, |rq|.
Let us examine a point r ∈L(ep, eq). We have µ = r2 pdp + r2 qdq, and immediately, µ −dp = r2 q(dp −dq), µ −dq = r2 p(dq −dp).
We also know that ∥r′∥= L, where r′ p = (µ −dp)rp, r′ q = (µ −dq)rq.
So, L = (r′ p)2 + (r′ q)2 = (dq −dp)2(r4 qr2 p + r4 pr2 q) = (dq −dp)2r2 pr2 q(r2 q + r2 p) = (dq −dp)2r2 pr2 q.
Hence, we have the equations r2 pr2 q = L/(dq −dp)2 r2 p + r2 q = 1, a system with a finite number of isolated solutions, establishing a contradiction and proving (ii).
(iii) r ∈L(e1, en).
9 Let us prove that q = n. The proof that p = 1 is similar. Assume by contradiction that q < n. We know that both r and r+ belong to L(ep, eq), and by Lemma ??, µ(r) < dq < dn and µ(r+) < dq.
Since r, r+ are the unique limit points of the sequence (yk), for k sufficiently large, say, k ≥¯ k, µ(yk) < dq. For such k, gk+1 q gk q = |µ(yk) −dq| < |µ(yk) −dq| = gk+1 n gk n .
Hence for all k > ¯ k gk q g¯ k q < gk n g¯ k n , contradicting the fact that yk n →0 and yq →rq , 0.
□ 2.3 Properties of the limiting points Now we know that the study of asymptotic properties of the sequences generated by the Cauchy algorithm may be reduced to the study of the limit points. This leads to simple results that reproduce properties described by Nocedal, Sartenaer and Zhu [?] and by De Asmundis et al [?, ?].
Consider an application of the Cauchy algorithm as above, starting from x0 with no null compo-nent, and define r = limk→∞y2k, with yk = gk/ gk . We know that the only non-zero components of r are r1 and rn. From the analysis above, we know that ∥r∥ = 1 (6) µ = d1r2 1 + dnr2 n, λ = 1/µ (7) r′ = (µI −D)r, r′ = L (8) µ′ = (r′)T Dr/ r′ 2 , λ′ = 1/µ′ (9) r′′ = (µ′I −D)r′ = L2r, r′′ = L2.
(10) Lemma 3. µ + µ′ = d1 + dn.
Proof. As r′′ = (µI −D)(µ′I −D)r = L2r, L2 = (µ −d1)(µ′ −d1) = (µ −dn)(µ′ −dn).
Then −d1(µ + µ′) + d2 1 = −dn(µ + µ′) + d2 n (dn −d1)(µ + µ′) = d2 n −d2 1.
The result follows by dividing by (dn −d1), completing the proof.
□ This completes the proofs of the items (i)-(iii) of Theorem ??. Let us prove the remaining items, studying a double iteration for large k: lim k→∞ gk+2 i gk i = (1 −di µ )(1 −di µ′ ) ≤(1 −dn µ )(1 −dn µ′ ), (11) because di ≤dn. As gk i →0, we must have lim k→∞ gk+2 i gk i ≤1, 10 proving (iv). In view of the inequality above, this is reduced to (1 −di µ )(1 −di µ′ ) > −1, i = 1, . . . , n.
(12) Developing (??), we obtain immediately µµ′ ≥(µ + µ′)di −d2 i 2 = (d1 + dn)di −d2 i 2 , µ + µ′ = d1 + dn.
(13) Let us define ˜ d = (d1 + dn)/2, δi = ˜ d −di, i = 1, . . . , n and δ = argmin{|δi| | i = 1, . . . , n}.
It is straightforward to check that the numerator of (??) satisfies (d1 + dn)di −d2 i = ˜ d2 −δ2 i , and hence we know that µ, µ′ must satisfy µµ′ ≥ ˜ d2 −δ2 i 2 , µ + µ′ = 2 ˜ d, i = 1, . . . , n, which is equivalent to µµ′ ≥ ˜ d2 −δ2 2 , µ + µ′ = 2 ˜ d.
(14) The components |g1|, |gn| are reduced by α = (1 −dn/µ)(1 −dn/µ′). Developing this expression and substituting µ + µ′ = d1 + dn we obtain α = 1 −d1dn µµ′ .
(15) From (??) and this expression we conclude that, α ≥1 −2 d1dn ˜ d2 −δ2 , (16) proving item (v) of Theorem ??.
Bounds for the inverse steps µ, µ′ are obtained by solving the system (??), which reduces to a simple second degree equation whose solution is ¯ µ, ¯ µ′ = ˜ d ± q ( ˜ d2 + δ2)/2, (17) and thus ¯ µ ≤µ, µ′ ≤¯ µ′, proving the item (vi).
The last item of the main theorem is proved in the following lemma: Lemma 4. There exists c > 0 such that c = rn r1 = r′ 1 r′ n = s µ −d1 µ′ −d1 , with rn r1 = − r′ 1 r′ n .
Proof. By construction of the Cauchy step, r ⊥r′, i.e., r1r′ 1 + rnr′ n = 0. The first equalities follow by dividing this by r1rn.
Using this result, consider r+ = (I −D/µ). We have r+ n = (1 −dn/µ)rn, r+ 1 = (1 −d1µ)r1, and so r+ n r+ 1 = (1 −dn/µ) (1 −d1/µ) rn r1 = µ −dn µ −d1 rn r1 = −r1 rn .
11 From this last equality, it follows that c2 = r2 n r2 1 = −µ −d1 µ −dn = µ −d1 µ′ −d1 , using the fact that µ + µ′ = d1 + dn, completing the proof.
□ Some interesting relations proved in [?] are now straightforward: from |rn| = c|r1| and r2 1+r2 n = 1 and then r2 1 + c2r2 1 = 1 we obtain: r2 1 = 1 1 + c2 , r2 n = c2 1 + c2 , (18) µ = d1 + c2dn 1 + c2 = d1 1 + c2C 1 + c2 , (19) µ′ = c2d1 + dn 1 + c2 = d1 c2 + C 1 + c2 .
(20) Let us now relate the constants L and c. We have L2 = (µ −d1)(µ′ −d1), µ −d1 = d1 1 + c2C 1 + c2 −d1 = d1 c2(C −1) 1 + c2 µ′ −d1 = d1 c2 + C 1 + c2 −d1 = d1 C −1 1 + c2 .
Simplifying these expressions and using Lemma ?? we obtain L2 = c2(dn −d1)2 (1 + c2)2 = µµ′ −d1dn.
(21) 3 Breaking the cycle The Cauchy algorithm reduces the medium variables and acts very slowly on the light and heavy variables. The cycle may be broken by enforcing either a very short (µ near dn) or a very large (µ near d1) step. It is difficult to estimate the value d1, and a very large step will cause a great increase in the heavy variables (in absolute values), and will possibly increase the function value. A very large step should only be allowed if it is obtained by a Cauchy iterate, which always reduces the function value.
A very short step is harmless, and reduces the heavy variables. If the medium variables are already small, the function will then be dominated by the light variables, and the next Cauchy step will be large, breaking the cycle.
This is done by Asmundis et al. [?]: if µk, µk+1 are near µ, µ′, then µk + µk+1 ≈d1 + dn, and then the step 1/(µk + µk+1) will be short. They check the evolution of µk and decide to periodically estimate and apply short steps by this method. Their algorithm was named “steepest descent with alignment” (SDA).
We shall follow their approach, but with a different way of estimating short steps: all steps will be Cauchy steps taken at some point, making sure that in all iterations satisfy µk ∈(d1, dn).
In the algorithms below we call λ−gradient step a steepest descent step with step length λ. The algorithm below is the Cauchy algorithm modified by periodically applying short steps.
Stopping rule. Each step must be followed by the statement k = k + 1 and by testing a stopping rule. The usual test is gk ≤ϵ, for a given ϵ > 0. In our examples with diagonal Hessian matrices we test the function values, because the optimal value is zero.
12 Algorithm 1. Cauchy-short Algorithm Data: x0 ∈I Rn, KI, KC, Ks ∈I N (respectively 10, 6, 2 in our choice), k = 0.
Take KI Cauchy steps.
repeat Compute an estimated short step λs.
Take Ks λs−gradient steps.
Take KC Cauchy steps.
Remark. The choice of KI, KC, Ks is somewhat arbitrary, but we noticed in many tests that the oscillatory pattern is achieved in about 10 Cauchy iterations initially, and in about 6 Cauchy itera-tions after each set of short steps. Two consecutive short steps are sufficient to cause a substantial reduction in the large variables. See Fig. ?? for an example. Note that even if the estimated short step is not very near 1/dn, the short steps are harmless.
Estimating a short step: a “mock” large step. We intend to state an algorithm in which all iterations take Cauchy steps, and so all steps will satisfy d1 < µk < dn. As the cycle pattern is established, an artificially very large step will cause a great increase in the heavy variables, and consequently the next Cauchy step will be short. The very large step will be computed but not applied. Let S be a very large step length (S >> 1/d1). At an iteration k we compute the following short step: ˜ g = (I −S D)gk (22) λs = ˜ gT ˜ g ˜ gT D˜ g.
(23) The reason why this will be a short step is the following: for a large value of S , ˜ gi = (1 −S di)gk i ≈−S digi, i = 1, . . . , n.
Hence the following Cauchy step will satisfy λs = ˜ gT ˜ g ˜ gT D˜ g = Pn−1 i=1 g2 1d2 i + g2 nd2 n Pn−1 i=1 g2 1d3 i + g2 nd3 n If the medium components are small, then for i = 1, . . . , n−1, |gidi| << |gndn|, and then λs ≈1/dn. A safeguard may be used to enforce short steps, by setting λs = min{λs, λk−1, λk−2}, as min{λk−1, λk−2} should be a short step.
Fig.?? shows iterates before and after the short steps for our example. Fig. ?? shows the behav-ior of the step lengths for the Cauchy-short algorithm: the cyclic pattern is broken by periodically forcing a short step, which is naturally followed by a large Cauchy step.
Alternating Cauchy and short steps. It is known that about one half of the steps will be short.
The algorithm below uses the computation of short steps as above, but applies them after all Cauchy steps: Algorithm 2. Alternated Cauchy-short steps.
Data: x0 ∈I Rn, KI, KC, Ks (respectively 10, 6, 2 in our choice), k = 0.
Take KI Cauchy steps.
repeat Compute an estimated short step λs.
Take Ks λs−gradient steps.
Take KC iterations composed of a Cauchy step followed by one λs−gradient step.
13 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 10 −10 10 −8 10 −6 10 −4 10 −2 10 0 eigenvalues |g| Short step, followed by a large step k−1 k k+1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 10 −10 10 −8 10 −6 10 −4 10 −2 10 0 eigenvalues |g| Following iteration k−1 k k+1 Figure 3: Before and after 2 short steps in CS algorithm.
0 50 100 150 200 250 10 −12 10 −10 10 −8 10 −6 10 −4 10 −2 10 0 First and last components of |g| k |gi| 0 10 20 30 40 50 60 70 10 0 10 1 10 2 10 3 Step sizes k step length Figure 4: Behavior of λk and of gk 1 , gk n in an application of the CS algorithm.
The algorithm has a stronger effect on the very heavy variables. Consequently the Cauchy and eventually the short steps increase. Fig.?? shows the behavior of the function values on an example with C = 1000 and n = 1000, eigenvalues uniformly distributed between 0.001 and 1, x0 i = 1/ √di.
The figure shows the evolution of the Barzilai-Borwein (BB), the Cauchy-short (CS), the steepest descent with alignment (SDA), and the Alternated Cauchy-short (ACS) algorithms. It also shows the effect of using Chebyshev roots to correct the step lengths, as described below.
Using Chebyshev roots. The paper [?] has the following result: if bounds l ≤d1 and u ≥dn are known, then the stopping condition |xk i | ≤ϵ|x0 i | for i = 1, . . . , n is achieved by the following steepest descent scheme: For k = 0 to K −1, gk+1 = (I −ΛkD)gk, where, using C = u/l, K = K(C) = cosh−1 2 ε !
cosh−1 1 + 2 C −1 !
≈ √ C 2 log 2 ε ! , and the set of step lengths is Λ = {1/µk | k = 0, . . . , K −1}, µk = u −l 2 cos 1 + 2k 2K π !
+ u + l 2 , (24) 14 Iteration 0 50 100 150 200 250 300 350 f(x) 10-8 10-6 10-4 10-2 100 102 104 Convergence history BB CS ACS CS using Chebyshev roots (adaptative) SDA Figure 5: Function values for the BB, CS, ACS, SDA and CS with steps corrected to Chebyshev roots using adaptive bounds for the eigenvalues.
taken in any order. If the bounds l and u are exact, this number of iterations coincides almost exactly with the worst case performance of the Krylov space method, the best possible.
This means that if good bounds for the eigenvalues are known, then the number of iterations of any gradient algorithm may be limited to K(C) if the step lengths are corrected to match elements of Λ. This is the resulting scheme: Let C = u/l, compute K(C) and the set Λ by (??).
Take any steepest descent method, and execute the following command after the computation of each step length λk; Set λk = argmin{|ν −λk | ν ∈Λ}.
Remove the element λk from the set Λ.
Adaptive scheme. This scheme depends on reliable lower and upper bounds l and u for the eigenvalues. When they are not available, we construct them by adding the following procedure to Algorithms ?? and ??: Initialization: choose an integer k0 > Ki + Ks, take the first k0 iterations of the algorithm and set u = 1.2 maxk=0,...,k0−1 1/λk, l = 0.25 mink=0,...,k0−1 1/λk, C = u/l, and compute Λ by (??).
Adaptation: at all steps with k > k0, perform the following adaptation of the set Λ: if u < 1/λk, set u = 1.2 u, C=u/l and reset Λ by (??).
if l > 1/λk, set l = l/4, C=u/l and reset Λ by (??).
The initial value of u will hopefully satisfy u > dn, because the short step computation is efficient, both using our algorithms and ASD (with k0 properly chosen). The lower bound will probably be updated, and each time l changes the size of Λ doubles.
We applied this scheme to our test problems, using the Cauchy-short algorithm. The result is shown in Fig. ??. This scheme is only applicable to quadratic functions, but it reduces all variables to |gK i | ≤ϵ|g0 i |, which may be important in the solution of linear systems of equations and least squares problems.
Performance profile. Finally, Fig. ?? is a performance profile, computed as described in [?], using 120 quadratic problems with 1000 variables, three values of the condition number C = 1000, 10000, 100000, and each problem with a different randomly generated initial point. The eigenvalues are randomly generated with four different distributions, exemplified by Fig. ??.
It shows that all new schemes are efficient, and seem to be superior to the Barzilai-Borwein method, with the advantage of generating monotonically decreasing function values.
15 0 200 400 600 800 1000 0 0.2 0.4 0.6 0.8 1 Uniform dist.
0 200 400 600 800 1000 0 0.2 0.4 0.6 0.8 1 Logarithmic dist.
0 200 400 600 800 1000 0 0.2 0.4 0.6 0.8 1 Sinusoidal dist.
0 200 400 600 800 1000 0 0.2 0.4 0.6 0.8 1 2 Blocks dist.
Figure 6: Eigenvalue distributions: a point (k, s) means that dk = sdn.
1 1.5 2 2.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Barzilai-Borwein CS ACS CS using Chebyshev roots and adaptative C SDA Figure 7: Performance profile for the algorithms Barzilai-Borwein, Algorithm (??) (CS), Algo-rithm (??) (ACS), steepest descent with alignment (SDA) and CS with adaptive computation of Chebyshev steps.
Conclusion. We believe that a good understanding of algorithms for quadratic problems is fundamental for the design of methods for more general problems. The steepest descent method, mainly using approximated Cauchy steps, is the most well known of all methods. Its poor per-formance, which contradicts the simple intuition of greedily computing the maximum function reduction at each iteration, has always been frustrating. It was explained by Akaike, and only re-cently a simple cure has been found: just take a short step once in a while. In this paper we hope to have summarized the classical proofs of this behavior, and proposed new procedures for quadratic problems. We did not, for the time being, try to apply similar schemes to non-quadratic functions, but we hope that these ideas may lead to general methods without the need of strategies for dealing with non-monotonicity.
16 References H. Akaike. On a successive transformation of probability distribution and its application to the analysis of the optimum gradient method. Ann. Inst. Statist. Math. Tokyo, 11:1–17, 1959.
J. Barzilai and J. M. Borwein. Two-point step size gradient methods. IMA J. Numer. Anal., 8:141–148, 1988.
E. G. Birgin, J. M. Mart´ ınez, and M. Raydan.
Spectral Projected Gradient Methods.
In C. A. Floudas and P. M. Pardalos, editors, Encyclopedia of Optimization, pages 3652–3659.
Springer, 2009.
A. Cauchy. M´ ethode g´ en´ erale pour la r´ esolution des syst` emes d’´ equations simultan´ ees. Comp.
Rend. Acad. Sci. Paris, 25:536–538, 1847.
Y. H. Dai. Alternate step gradient method. Optimization, 52(4-5):395–415, 2003.
R. de Asmundis, D. di Serafino, W. Hager, G. Toraldo, and H. Zhang. An efficient gradient method using the Yuan steplength. Technical report, Sapienza University of Rome, Italy, 2014.
R. de Asmundis, D. di Serafino, R. Riccio, and G. Toraldo. On spectral properties of steepest descent mehtods. IMA J. Numer. Anal., 33:1416–1435, 2013.
E. D. Dolan and J. J. Mor´ e. Benchmarking optimization software with performance profiles.
Mathematical Programming, 91(2):201–213, 2002.
G. E. Forsythe. On the asymptotic directions of the s-dimensional optimum gradient method.
Numerische Mathematik, 11:57–76, 1968.
C. C. Gonzaga. Optimal performance of the steepest descent algorithm for quadratic functions.
Technical report, Federal University of Santa Catarina, Florianopolis, Brazil, 2014.
J. Nocedal, A. Sartenaer, and C. Zhu. On the behavior of the gradient norm in the steepest descent method. Computational Optimization and Applications, 22:5–35, 2002.
M. Raydan. The Barzilai and Borwein gradient method for large scale unconstrained mini-mization problem. SIAM Journal on Optimization, 7:26–33, 1997.
M. Raydan and B. Svaiter. Relaxed steepest descent and Cauchy-Barzilai-Borwein method.
Computatinal Optimization and Applications, 21:155–167, 2002. |
11111 | https://wikieducator.org/images/4/41/Logrithm_Table.pdf | LOGARITHM TABLE (for numbers 1 to 5.49) No.
0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 1.0 0.000 0.004 0.009 0.013 0.017 0.021 0.025 0.029 0.033 0.037 1.1 0.041 0.045 0.049 0.053 0.057 0.061 0.064 0.068 0.072 0.076 1.2 0.079 0.083 0.086 0.090 0.093 0.097 0.100 0.104 0.107 0.111 1.3 0.114 0.117 0.121 0.124 0.127 0.130 0.134 0.137 0.140 0.143 1.4 0.146 0.149 0.152 0.155 0.158 0.161 0.164 0.167 0.170 0.173 1.5 0.176 0.179 0.182 0.185 0.188 0.190 0.193 0.196 0.199 0.201 1.6 0.204 0.207 0.210 0.212 0.215 0.217 0.220 0.223 0.225 0.228 1.7 0.230 0.233 0.236 0.238 0.241 0.243 0.246 0.248 0.250 0.253 1.8 0.255 0.258 0.260 0.262 0.265 0.267 0.270 0.272 0.274 0.276 1.9 0.279 0.281 0.283 0.286 0.288 0.290 0.292 0.294 0.297 0.299 2.0 0.301 0.303 0.305 0.307 0.310 0.312 0.314 0.316 0.318 0.320 2.1 0.322 0.324 0.326 0.328 0.330 0.332 0.334 0.336 0.338 0.340 2.2 0.342 0.344 0.346 0.348 0.350 0.352 0.354 0.356 0.358 0.360 2.3 0.362 0.364 0.365 0.367 0.369 0.371 0.373 0.375 0.377 0.378 2.4 0.380 0.382 0.384 0.386 0.387 0.389 0.391 0.393 0.394 0.396 2.5 0.398 0.400 0.401 0.403 0.405 0.407 0.408 0.410 0.412 0.413 2.6 0.415 0.417 0.418 0.420 0.422 0.423 0.425 0.427 0.428 0.430 2.7 0.431 0.433 0.435 0.436 0.438 0.439 0.441 0.442 0.444 0.446 2.8 0.447 0.449 0.450 0.452 0.453 0.455 0.456 0.458 0.459 0.461 2.9 0.462 0.464 0.465 0.467 0.468 0.470 0.471 0.473 0.474 0.476 3.0 0.477 0.479 0.480 0.481 0.483 0.484 0.486 0.487 0.489 0.490 3.1 0.491 0.493 0.494 0.496 0.497 0.498 0.500 0.501 0.502 0.504 3.2 0.505 0.507 0.508 0.509 0.511 0.512 0.513 0.515 0.516 0.517 3.3 0.519 0.520 0.521 0.522 0.524 0.525 0.526 0.528 0.529 0.530 3.4 0.531 0.533 0.534 0.535 0.537 0.538 0.539 0.540 0.542 0.543 3.5 0.544 0.545 0.547 0.548 0.549 0.550 0.551 0.553 0.554 0.555 3.6 0.556 0.558 0.559 0.560 0.561 0.562 0.563 0.565 0.566 0.567 3.7 0.568 0.569 0.571 0.572 0.573 0.574 0.575 0.576 0.577 0.579 3.8 0.580 0.581 0.582 0.583 0.584 0.585 0.587 0.588 0.589 0.590 3.9 0.591 0.592 0.593 0.594 0.595 0.597 0.598 0.599 0.600 0.601 4.0 0.602 0.603 0.604 0.605 0.606 0.607 0.609 0.610 0.611 0.612 4.1 0.613 0.614 0.615 0.616 0.617 0.618 0.619 0.620 0.621 0.622 4.2 0.623 0.624 0.625 0.626 0.627 0.628 0.629 0.630 0.631 0.632 4.3 0.633 0.634 0.635 0.636 0.637 0.638 0.639 0.640 0.641 0.642 4.4 0.643 0.644 0.645 0.646 0.647 0.648 0.649 0.650 0.651 0.652 4.5 0.653 0.654 0.655 0.656 0.657 0.658 0.659 0.660 0.661 0.662 4.6 0.663 0.664 0.665 0.666 0.667 0.667 0.668 0.669 0.670 0.671 4.7 0.672 0.673 0.674 0.675 0.676 0.677 0.678 0.679 0.679 0.680 4.8 0.681 0.682 0.683 0.684 0.685 0.686 0.687 0.688 0.688 0.689 4.9 0.690 0.691 0.692 0.693 0.694 0.695 0.695 0.696 0.697 0.698 5.0 0.699 0.700 0.701 0.702 0.702 0.703 0.704 0.705 0.706 0.707 5.1 0.708 0.708 0.709 0.710 0.711 0.712 0.713 0.713 0.714 0.715 5.2 0.716 0.717 0.718 0.719 0.719 0.720 0.721 0.722 0.723 0.723 5.3 0.724 0.725 0.726 0.727 0.728 0.728 0.729 0.730 0.731 0.732 5.4 0.732 0.733 0.734 0.735 0.736 0.736 0.737 0.738 0.739 0.740 LOGARITHM TABLE (for numbers from 5.5 to 10) No.
0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 5.5 0.740 0.741 0.742 0.743 0.744 0.744 0.745 0.746 0.747 0.747 5.6 0.748 0.749 0.750 0.751 0.751 0.752 0.753 0.754 0.754 0.755 5.7 0.756 0.757 0.757 0.758 0.759 0.760 0.760 0.761 0.762 0.763 5.8 0.763 0.764 0.765 0.766 0.766 0.767 0.768 0.769 0.769 0.770 5.9 0.771 0.772 0.772 0.773 0.774 0.775 0.775 0.776 0.777 0.777 6.0 0.778 0.779 0.780 0.780 0.781 0.782 0.782 0.783 0.784 0.785 6.1 0.785 0.786 0.787 0.787 0.788 0.789 0.790 0.790 0.791 0.792 6.2 0.792 0.793 0.794 0.794 0.795 0.796 0.797 0.797 0.798 0.799 6.3 0.799 0.800 0.801 0.801 0.802 0.803 0.803 0.804 0.805 0.806 6.4 0.806 0.807 0.808 0.808 0.809 0.810 0.810 0.811 0.812 0.812 6.5 0.813 0.814 0.814 0.815 0.816 0.816 0.817 0.818 0.818 0.819 6.6 0.820 0.820 0.821 0.822 0.822 0.823 0.823 0.824 0.825 0.825 6.7 0.826 0.827 0.827 0.828 0.829 0.829 0.830 0.831 0.831 0.832 6.8 0.833 0.833 0.834 0.834 0.835 0.836 0.836 0.837 0.838 0.838 6.9 0.839 0.839 0.840 0.841 0.841 0.842 0.843 0.843 0.844 0.844 7.0 0.845 0.846 0.846 0.847 0.848 0.848 0.849 0.849 0.850 0.851 7.1 0.851 0.852 0.852 0.853 0.854 0.854 0.855 0.856 0.856 0.857 7.2 0.857 0.858 0.859 0.859 0.860 0.860 0.861 0.862 0.862 0.863 7.3 0.863 0.864 0.865 0.865 0.866 0.866 0.867 0.867 0.868 0.869 7.4 0.869 0.870 0.870 0.871 0.872 0.872 0.873 0.873 0.874 0.874 7.5 0.875 0.876 0.876 0.877 0.877 0.878 0.879 0.879 0.880 0.880 7.6 0.881 0.881 0.882 0.883 0.883 0.884 0.884 0.885 0.885 0.886 7.7 0.886 0.887 0.888 0.888 0.889 0.889 0.890 0.890 0.891 0.892 7.8 0.892 0.893 0.893 0.894 0.894 0.895 0.895 0.896 0.897 0.897 7.9 0.898 0.898 0.899 0.899 0.900 0.900 0.901 0.901 0.902 0.903 8.0 0.903 0.904 0.904 0.905 0.905 0.906 0.906 0.907 0.907 0.908 8.1 0.908 0.909 0.910 0.910 0.911 0.911 0.912 0.912 0.913 0.913 8.2 0.914 0.914 0.915 0.915 0.916 0.916 0.917 0.918 0.918 0.919 8.3 0.919 0.920 0.920 0.921 0.921 0.922 0.922 0.923 0.923 0.924 8.4 0.924 0.925 0.925 0.926 0.926 0.927 0.927 0.928 0.928 0.929 8.5 0.929 0.930 0.930 0.931 0.931 0.932 0.932 0.933 0.933 0.934 8.6 0.934 0.935 0.936 0.936 0.937 0.937 0.938 0.938 0.939 0.939 8.7 0.940 0.940 0.941 0.941 0.942 0.942 0.943 0.943 0.943 0.944 8.8 0.944 0.945 0.945 0.946 0.946 0.947 0.947 0.948 0.948 0.949 8.9 0.949 0.950 0.950 0.951 0.951 0.952 0.952 0.953 0.953 0.954 9.0 0.954 0.955 0.955 0.956 0.956 0.957 0.957 0.958 0.958 0.959 9.1 0.959 0.960 0.960 0.960 0.961 0.961 0.962 0.962 0.963 0.963 9.2 0.964 0.964 0.965 0.965 0.966 0.966 0.967 0.967 0.968 0.968 9.3 0.968 0.969 0.969 0.970 0.970 0.971 0.971 0.972 0.972 0.973 9.4 0.973 0.974 0.974 0.975 0.975 0.975 0.976 0.976 0.977 0.977 9.5 0.978 0.978 0.979 0.979 0.980 0.980 0.980 0.981 0.981 0.982 9.6 0.982 0.983 0.983 0.984 0.984 0.985 0.985 0.985 0.986 0.986 9.7 0.987 0.987 0.988 0.988 0.989 0.989 0.989 0.990 0.990 0.991 9.8 0.991 0.992 0.992 0.993 0.993 0.993 0.994 0.994 0.995 1.00 9.9 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 10.0 1.00 |
11112 | https://www.gauthmath.com/solution/1717149889847301/What-is-the-solution-to-the-system-of-equations-p-2n-and-p-n-2-3-n-3-p-6-n-1-p-2 | Solved: What is the solution to the system of equations p =-2n and p=n^2-3 ? n=-3, p=6 n=1, p=-2 n [Math]
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What is the solution to the system of equations p =-2n and p=n^2-3 ? n=-3, p=6 n=1, p=-2 n=2, p=-3 n=1, p=-2 and n=-3, p=6 n=1, p=-2 and n=2, p=-3
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Answer
$$\begin{equation}\begin{cases}n = 1\p = - 2\end{cases}\end{equation}$${n=1 p=−2 or $$\begin{equation}\begin{cases}n = - 3\p = 6\end{cases}\end{equation}$${n=−3 p=6
Explanation
Substitute into one of the equations:$$- 2 n = n^{2} - 3$$−2 n=n 2−3
Rearrange all nonzero terms to the left side of the equation:$$- 2 n - n^{2} + 3 = 0$$−2 n−n 2+3=0
Rearrange the terms in descending order:$$- n^{2} - 2 n + 3 = 0$$−n 2−2 n+3=0
Make the leading coefficient positive:$$n^{2} + 2 n - 3 = 0$$n 2+2 n−3=0
Split the middle term into two terms:$$n^{2} - n + 3 n - 3 = 0$$n 2−n+3 n−3=0
Factor the first two terms and the last two terms separately:$$n (n -1) + 3 (n -1) = 0$$n(n−1)+3(n−1)=0
Factor out:$$(n -1) (n + 3) = 0$$(n−1)(n+3)=0
Apply Zero Product Property:$$n -1 = 0$$n−1=0 or $$n + 3 = 0$$n+3=0
Rearrange unknown terms to the left side of the equation:$$n = 1$$n=1
Rearrange unknown terms to the left side of the equation:$$n = - 3$$n=−3
Combine the results:$$n = 1$$n=1 or $$n = - 3$$n=−3
Substitute into one of the equations:$$p = - 2$$p=−2
Substitute into one of the equations:$$p = - 2 \times (- 3)$$p=−2×(−3)
Determine the sign for multiplication or division:$$p = 2 \times 3$$p=2×3
Calculate the product or quotient:$$p = 6$$p=6
The solution of the system is:$$\begin{equation}\begin{cases}n = 1\p = - 2\end{cases}\end{equation}$${n=1 p=−2 or $$\begin{equation}\begin{cases}n = - 3\p = 6\end{cases}\end{equation}$${n=−3 p=6
Answer: $$\begin{equation}\begin{cases}n = 1\p = - 2\end{cases}\end{equation}$${n=1 p=−2 or $$\begin{equation}\begin{cases}n = - 3\p = 6\end{cases}\end{equation}$${n=−3 p=6
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11113 | https://bmcimmunol.biomedcentral.com/articles/10.1186/1471-2172-2-8 | Role of CD28/B7 costimulation and IL-12/IL-10 interaction in the radiation-induced immune changes | BMC Immunology | Full Text
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Role of CD28/B7 costimulation and IL-12/IL-10 interaction in the radiation-induced immune changes
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Published: 07 August 2001
Role of CD28/B7 costimulation and IL-12/IL-10 interaction in the radiation-induced immune changes
Shu-Zheng Liu1,
Shun-Zi Jin1,
Xiao-Dong Liu1&
…
Yi-Min Sun1
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BMC Immunologyvolume 2, Article number:8 (2001) Cite this article
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Abstract
Background
The present paper aims at studying the role of B7/CD28 interaction and related cytokine production in the immunological changes after exposure to different doses of ionizing radiation.
Results
The stimulatory effect of low dose radiation (LDR) on the proliferative response of lymphocytes to Con A was found to require the presence of APCs. The addition of APCs obtained from both low- and high-dose-irradiated mice to splenic lymphocytes separated from low-dose-irradiated mice caused stimulation of lymphocyte proliferation. B7-1/2 expression on APCs was up-regulated after both low and high doses of radiation. There was up-regulation of CD28 expression on splenic and thymic lymphocytes after LDR and its suppression after high dose radiation (HDR), and cytotoxic T lymphocyte-associated antigen 4 (CTLA-4) expression showed changes in the opposite direction. IL-12 secretion by macrophages was stimulated after both low and high doses of radiation, but IL-10 synthesis by splenocytes was suppressed by low dose radiation and up-regulated by high dose radiation.
Conclusion
The status of CD28/CTLA-4 expression on T lymphocytes in the presence of up-regulated B7 expression on APCs determined the outcome of the immune changes in response to radiation, i.e., up-regulation of CD28 after LDR resulted in immunoenhancement, and up-regulation of CTLA-4 associated with down-regulation of CD28 after HDR led to immunosuppression. Both low and high doses of radiation up-regulated B7-1/2 expression on APCs. After LDR, the stimulated proliferative effect of increased IL-12 secretion by APCs, reinforced by the suppressed secretion of IL-10, further strengthened the intracellular signaling induced by B7-CD28 interaction.
Background
It is well known that the immune system responds to ionizing radiation with distinct characteristics depending on the dose and dose-rate [1, 2]. The prominent suppressive effect of lethal and sublethal doses of ionizing radiation on immunity and hemopoiesis constitutes the basis of the chief clinical manifestations of acute radiation syndrome while the stimulatory effect of low dose radiation (LDR) as disclosed in recent years has even greater significance in understanding the health effects of environmental low level radiation [2–7]. This is the reason why the studies on the mechanisms of the stimulatory effect of LDR have attracted the attention of radiobiologists and health physics workers. It has been demonstrated that LDR enhances T cell activation by facilitating signal transduction and modulating the expression of a number of genes involved in cytokine expression and cell survival in the immune organs . The current model of T cell activation requires two signals [8, 9]. The first signal is specific, requiring T cell receptor recognition and binding to MHC/antigen presented by antigen-presenting cells (APCs). The second signal is nonspecific, resulting from the binding of B7 ligand on the APC with its receptor, CD28, on the T cell. If both signals are provided, the T cell will proliferate and secrete cytokines. Recently, it has been shown that CTLA-4, another receptor for B7 that is up-regulated following T cell activation, can deliver an inhibitory signal, down-regulating T cell proliferation [10, 11]. The role of B7/CD28 interaction and related cytokine production in the immunological changes after exposure to different doses of ionizing radiation remains to be elucidated. In the present paper it is demonstrated for the first time that the cellular interactions between the APCs and lymphocytes play an important role in the distinct changes in the immune response following exposure of mice to different doses of radiation with possible involvement of the costimulatory factors and cytokines in such changes.
Results
Role of APCs in the activation of T cells induced by LDR
The importance of the interaction between the APCs and lymphocytes in normal immune response is well known . But the role of APCs in radiation-induced immunological changes has not been fully elucidated. Figure 1 illustrates the importance of the interactions between the APCs and lymphocytes in the proliferative response induced by Con A in normal as well as irradiated states. In this figure there are 5 groups of experiments (A to E) and in each group there are 3 columns marked as (1) lymphocytes from sham-irradiated mice, (2) lymphocytes from 0.075Gy-irradiated mice and (3) lymphocytes from 2Gy-irradiated mice, respectively. In the analysis to follow, these will be designated by the group (A, B, C, D or E) followed by a column number (1, 2 or 3) for comparison of the changes. As illustrated in Group A of figure 1, the proliferation of splenocytes in response to Con A after whole-body irradiation (WBI) with 0.075 Gy was significantly up-regulated to a value more than doubling that of the sham-irradiated control (P < 0.002, A1 vs A2), while WBI with 2 Gy caused a profound depression of this reaction to about one third of the control (P < 0.01, A1 vs A3). When splenic cells were depleted of their adherent APCs, the proliferative response to Con A was significantly suppressed (P < 0.001, B1 vs A1). In this case no stimulatory effect was observed for the lymphocytes from the 0.075Gy-irradiated mice (P > 0.05, B1 vs B2; P < 0.001, A2 vs B2), while the suppression of proliferation by 2 Gy was still marked (P < 0.05, B1 vs B3), being more marked than that in the control group (P < 0.001, A3 vs B3). When APCs obtained from the control mice were added to the separated lymphocytes, the proliferation compared to group A was largely recovered for the lymphocytes from the control (P > 0.05, A1 vs C1) and 0.075Gy-irradiated mice (P > 0.05, A1 vs C2), and fully recovered for the lymphocytes from the 2Gy-irradiated mice (P > 0.05, A3 vs C3). But in group C, neither the stimulation from LDR nor the suppression from HDR was significant (P > 0.05, C1 vs C2 and C1 vs C3). When APCs from both the 0.075Gy-irradiated and 2Gy-irradiated mice were added to the separated lymphocytes (groups D and E in figure 1), the proliferative response of lymphocytes from the control mice recovered (P > 0.05, A1 vs Dl and A1 vs E1) and that of lymphocytes from the 0.075 Gy-irradiated mice was greater than the control of each group (P < 0.02, D1 vs D2; P < 0.05, E1 vs E2), and this response was also greater than the normal control (P < 0.05, A1 vs D2 and A1 vs E2). The proliferative response in groups D and E after addition of APCs from the 2Gy-irradiated mice was lower than the control of each group (P < 0.001, D1 vs D3 and El vs E3) and the magnitude of response was about the same as that in group A (P > 0.05, A3 vs D3 and A3 vs E3). A similar phenomenon was observed when secretion of IL-2 was measured as an end point (data not shown) and peritoneal macrophages were used instead of splenic adherent cells as the APCs for both the proliferative response (figure 2 ) and IL-2 secretion (data not shown).
Figure 1
Effect of splenic APCs on the proliferative response of splenic lymphocytes to Con A A. Whole cell suspension of spleen; B. Splenocytes depleted of adherent cells (APCs); C. "B" added with APCs obtained from normal mice; D. "B" added with APCs obtained from 0.075Gy-irradiated mice; E. "B" added with APCs obtained from 2Gy-irradiated mice. Mean+/-S.E., n = 5
Full size image
Figure 2
Effect of peritoneal macrophages on the proliferative response of splenic lymphocytes to Con A A. Whole cell suspension of spleen; B. Splenocytes depleted of adherent cells (APCs); C. "B" added with peritoneal macrophages obtained from normal mice; D. "B" added with peritoneal macrophages obtained from 0.075Gy-irradiated mice; E. "B" added with peritoneal macrophages obtained from 2Gy-irradiated mice. Mean+/-S.E., n = 5
Full size image
Expression of B7 molecules on APCs in response to ionizing radiation
The time course of the expression of B7-1 (CD80) and B7-2 (CD86) on peritoneal macrophages was studied after WBI with 0.075 Gy and 2 Gy. It can be seen from the left panel of figure 3 that the expression of B7-1 was markedly up-regulated on peritoneal macrophages after WBI with both low and high doses, with the start of up-regulation being earlier after 2 Gy (8h) than after 0.075 Gy (12h), but peaking at 24 h for both doses. The expression of B7-2 on macrophages was also stimulated (right panel), reaching the peak earlier after LDR (8 h) than after HDR (12 h), and the up-regulation sustained for a longer time after LDR (P < 0.05 for the 24 h point in the 0.075 Gy experiment, but P > 0.05 for the 24 h point in the 2 Gy experiment). The dose-response relationship of the changes in the expression of these molecules showed up-regulation following doses up to 4 Gy with a distinct peak at 0.075 Gy followed by a second peak at 1.0 Gy for B7-1 and at 4 Gy for B7-2. These are illustrated in figure 4 and figure 5 in which both the response of B7-1 and B7-2 on mouse peritoneal macrophages to WBI and that of B7-1 and B7-2 on J774A.1 cells to in vitro irradiation are shown. It should be emphasized that the up-regulation of expression of these two molecules on the peritoneal macrophages after WBI with doses of 0.075 and 2 Gy was statistically significant at multiple time points (P < 0.05 or P < 0.01).
Figure 3
Effect of whole-body X-irradiation on CD80/CD86 expression on peritoneal macrophages Mean+/-S.E., n = 5
Full size image
Figure 4
Dose-effect relationship of CD80 (B7-1) expression on peritoneal macrophages 12 h after WBI and on J774A.1 cells 24 h after in vitro irradiation Mean+/-S.E., n = 4 for peritoneal macrophages and n = 5 for J774A.1 cells
Full size image
Figure 5
Dose-effect relationship of CD86 (B7-2) expression on peritoneal macrophages 12 h after WBI and on J774A.1 cells 24 h after in vitro irradiation Mean+/-S.E., n = 4 for peritoneal macrophages and n = 5 for J774A.1 cells
Full size image
Expression of CD28/CTLA-4 on lymphocytes
The CD28 and CTLA-4 molecules were detected on the surface of splenic and thymic lymphocytes by flow cytometry (FCM) with double immunofluorescence after WBI with different doses. It is seen in figure 6 that WBI with 0.075 Gy caused up-regulation of CD28 expression with opposite changes in CTLA-4 expression on splenocytes (panel A) while WBI with 2 Gy suppressed CD28 expression with opposite changes in the expression of CTLA-4 (panel B). Dose-effect relationship of the expression of these two molecules showed opposite direction of the changes both for the splenocytes 4 h after WBI (panel C) and for the thymocytes 72 h after WBI (panel D).
Figure 6
Effect of whole-body X-irradiation on CD28/CTLA-4 expression on lymphocytes Closed circle: CD28; Open circle: CTLA-4; Mean+/-S.E., n = 5
Full size image
Secretion of IL-12p70 by peritoneal macrophages and synthesis of IL-10 in lymphocytes
As shown in figure 7, the secretion of IL-12p70 as measured by ELISA was stimulated by both low and higher doses (left panel) with an abrupt increase 4 h after WBI with 0.075 Gy followed by its rapid return to the basal level and a sustained increase after WBI with 2 Gy during the observation period of 48 h. The synthesis of IL-10 in the splenocytes detected by FCM, and expressed as a percentage of fluorescence stained positive cells, was significantly suppressed after WBI with 0.075 Gy and markedly up-regulated after WBI with 2 Gy. (right panel).
Figure 7
Effect of whole-body X-irradiation on IL-12p70 secretion by peritoneal macrophages and IL-10 synthesis in splenic lymphocytes Mean+/-S.E., n = 5
Full size image
Discussion
Macrophages, thymocytes and splenocytes were chosen in this study because they are the most important elements in immune responses. The macrophage is a representative of the APC system and the thymus and spleen are most important central and peripheral lymphoid organs, respectively. The realization of an effective immune response depends upon the interaction between the APCs and lymphocytes, via the expression of surface molecules and secretion of cytokines and other growth factors. In the present paper B7-1/2 expression on macrophages and CD28/CTLA-4 expression on lymphocytes as well as the secretion of IL-12 by macrophages and synthesis of IL-10 in splenocytes were selected as parameters to disclose the possible regulatory roles of these molecules from different cell types in the immunological changes after different doses of WBI. It has been established that the B7 molecules presented by APCs as ligands reacting with their receptors CD28/CTLA-4 on T cells are necessary for the realization of competent immune reactions [8, 12]. The B7-1 (CD80) and B7-2 (CD86) molecules function to drive the helper and cytotoxic responses in the presence of ligation of TCR with MHC-Ag complexes, the latter determining the specificity of the reaction . Such reactions are associated with the action of cytokines [11, 13, 14]. It has been well documented that low dose radiation (LDR) stimulates and high dose radiation (HDR) suppresses immunity [1, 2, 6, 7]. But the role of these molecules in the immune reactions induced by different doses of ionizing radiation is unknown.
It is demonstrated in the present paper that depletion of the APCs from the cell suspension of the spleen abrogates the normal proliferative response of splenic lymphocytes to Con A (P < 0.002 for B1 vs A1 in figure 1) as well as the stimulatory effect of LDR on this proliferative reaction (P > 0.05 for B1 vs B2). The suppressive effect of 2 Gy exposure on proliferation of lymphocytes was even more marked after depletion of APCs (P < 0.05 for B1 vs B3 and P < 0.001 for A3 vs B3). This is apparently due to damage of the lymphocytes by this dose since the cells were collected 24 h after WBI. Replacement of the APCs separated from the spleen of normal mice restored the normal proliferative reactions of lymphocytes obtained from sham-irradiated mice (P > 0.05 for C1 vs A1). However, the stimulatory effect of LDR was not restored (P > 0.05 for C1 vs C2) which speaks for the requirement of up-regulation of costimulatory molecules of the APCs for immune stimulation. The suppressive effect of HDR was not significant either (P > 0.05 for C1 vs C3) which may be due to the fact that the recovery of the response of normal lymphocytes had not reached its full strength (C1 being 75.3% of A1) though there was no statistically significant difference between C1 and A1. It is interesting to see that addition of APCs obtained from mice irradiated with either 0.075 Gy or 2 Gy to lymphocytes separated from normal mice also restored the proliferative reaction to the control level (P > 0.05 for D1 vs A1 and E1 vs A1) and stimulated the proliferative response of lymphocytes separated from the 0.075Gy-irradiated mice (P < 0.02 for D1 vs D2 and P < 0.05 for E1 vs E2). This would be readily explained by the observation that the expression of CD80 and CD86 was up-regulated by both low and high doses (figure 3) and that expression of CD28 on splenic lymphocytes was increased after LDR (figure 6, panel A). The observation that the proliferative response of the lymphocytes obtained from 2Gy-irradiated mice remained low no matter whether APCs from either normal or irradiated mice were added could be understood from the up-regulated expression of CTLA-4 and down-regulated expression of CD28 on the lymphocytes after WBI with 2 Gy (figure 6, panels B and C ).
The effective regulation of T cell responses is dependent on opposing signals transmitted through two related cell-surface receptors, CD28 and CTLA-4. Dimerization of CTLA-4 is required for the formation of high-avidity complexes with B7 ligands and for transmission of signals that attenuate T cell activation. CTLA-4 belongs to the immunoglobulin superfamily and displays a strand topology similar to vα domains, with an unusual mode of dimerization that places the B7 binding sites distal to the dimerization interface. This organization allows each CTLA-4 dimer to bind two bivalent B7 molecules and suggests that a periodic arrangement of these components within the immunological synapse may contribute to the regulation of T cell responsiveness . In the experimental design we usually first examine the time course of the changes after WBI with 0.075 Gy as LDR and 2 Gy as HDR since in previous studies it has been repeatedly observed that 0.075 Gy usually elicited a stimulatory effect in the immune system and 2 Gy caused immunlogical suppression. Peaking times are then selected to examine the dose-effect relationship. In the CD28 study relatively higher CD28 expression on lymphocytes was observed at 4 h and 72 h after LDR. Therefore, dose-effect studies were made at an early (4 h, figure 6, panel C) and at a late (72 h, figure 6, panel D) stage after WBI. The CD28 and CTLA-4 molecules on the lymphocytes as disclosed by double immunofluorescence with FCM showed opposite changes both in the spleen (figure 6, panel C) and thymus (figure 6, panel D) in the dose-response study with WBI doses from 0.05 to 6 Gy (altogether 10 dose points). A correlation analysis showed that the changes of CD28 and CTLA-4 expression 4 h after WBI with doses from 0.05 to 6 Gy were inversely correlated both for the splenocytes (r=-0.9356, figure 6,panel C ) and the thymocytes (r=-0.9876, data not shown). Only the changes in the thymus were found to be still inversely correlated when examined 72 h after WBI with the same doses (r=-0.8118, figure 6, panel D). A dose of 2 Gy and above suppresses the expression of CD28 with up-regulation of the expression of CTLA-4 (figure 6, panels C and D) and leads to anergy of the T cells despite the up-regulated expression of B7 molecules on the APCs after this dose. This observation in the analysis of the effect of radiation on the interactions between the lymphocytes and the APCs suggests that the difference in radiosensititivity of these two families of the immune cells may determine the outcome of the immune reaction. With LDR the B7 molecules on the APCs and CD28 on the lymphocyte are up-regulated, resulting in enhanced immune response. With HDR the down-regulation of CD28 and up-regulation of CTLA-4 on lymphocytes suppresses their proliferation even with the up-regulated expression of B7 molecules on the APCs. It is known that the immunosuppressive agent dexamethasone enhances CTLA-4 expression during T cell activation . To the knowledge of the present authors there have not been reports on the effect of different doses of ionizing radiation on CD28/CTLA-4 expression on T cells.
In addition to the interactions between the surface molecules of APCs and lymphocytes, the nature of the production of different cytokines in response to different doses of radiation is also of importance in understanding the outcome of the immune response. As illustrated in figure 7, the production of IL-10 in the spleen is suppressed after WBI with 0.075 Gy and stimulated after WBI with 2 Gy. On the other hand, the secretion of IL-12 by the macrophages is stimulated after both doses, with a more sustained high level after HDR. It is known that IL-12 facilitates the differentiation of the Th0 cells in the direction of development of Th1 subset while IL-10 opposes this function in favor of the differentiation of the Th0 cells to the Th2 type . The changes in IL-10 and IL-12 at protein level was also confirmed by studies at transcription level with Northern blot after WBI with different doses (reported separately). The observations are in accordance with data from in vitro studies . It has been reported that IL-12 synergizes with B7/CD28 interaction in inducing efficient proliferation and cytokine production of human T cells . Willems et al. stated that IL-10 may exert an inhibitory effect on B7 expression on human monocytes, but this was not observed in the present study on mouse peritoneal macrophages after WBI. However, it could not be excluded that in the high-dose-irradiated mice the increased secretion of IL-10 might facilitate the interaction between the B7 molecules with CTLA-4 resulting in immuosuppression since the expression of both of them are up-regulated and this would inhibit the B7/CD28 interaction. Cell elements within the immune system display differences in radiosensitivity with APCs being much more resistant to ionizing radiation than lymphocytes, as demonstrated in surface molecule expression and cytokine production in the present study. These differences are important factors in the analysis of the immunological changes following different doses of radiation.
Conclusion
The different effects of low and high doses of radiation on T cell activation are closely related to the interaction between the surface molecules of APCs and lymphocytes as well as that between cytokines. The following schematic diagram is given to briefly outline the interrelationship of the experimental results obtained in the present study (figure 8). It is understood from the experimental results together with literature analysis that LDR and HDR both stimulate the CD80/86 expression on and IL-12 secretion by the APCs. On the other hand, LDR up-regulates the expression of CD28 and down-regulates that of CTLA-4 on the T cells together with suppressed production of IL-10, thus leading to immunoenhancement, while HDR down-regulates CD28 expression and up-regulates CTLA-4 expression on the T cells together with stimulated IL-10 production resulting in immunosuppression.
Figure 8
A schematic diagram illustrating the interaction of surface molecules on and production of interleukins by the immune cells in response to LDR and HDR LDR=low dose radiation; HDR=high dose radiation; APC=antigen presenting cells; TLC=T lymphocytesUpward and downward red arrows on the right side of the symbols separately stand for up-regulation and down-regulation and the blue arrows between the symbols indicate facilitation or stimulation of the process.
Full size image
Materials and methods
Animals
Male Kunming mice, 6–7 weeks of age, 20–22 g in body weight, obtained from the Department of Experimental Animals of Norman Bethune Medical Center, Jilin University, were used in all experiments. The mice were fed with routine laboratory chow and water ad lib.
Cells and culture
Collection of peritoneal macrophages, separation of splenic lymphocytes and adherent cells and the co-culture of APCs with lymphocytes were performed 24 h after WBI as previously reported . Briefly, cell suspensions were made from spleens in RPMI1640 medium in glass Petri dishes under sterile condition. The separation of splenic lymphocytes from the adherent APCs was performed (after lysing the red blood cells and washing twice with Hank's solution) by incubation of the re-suspended cells in RPMI1640 with 10% NBS at 37°C in a CO 2 incubator in glass Petri dishes for 1 h. This would allow the adherence of the majority of the APCs to the glass wall. This process was repeated 5 times before the non-adherent cells were collected as the separated lymphocytes for experimental use with a purity of >98%. The adherent cells collected after the first incubation were used as the APCs in the co-culture. The co-culture of the lymphocytes with the APCs was performed in 96-well microplates with the addition of 1 × 10 5 lymphocytes in 100 μl, 1 × 10 4 APCs in 50 μl and Con A 1 μg in 50 μl in each well. The microplates containing the cell mixtures were incubated under 37°C and 5% CO 2 for 72 h with addition of 18.5 kBq 3 H-TdR 6 h before termination of incubation. The cells were then collected on glass fiber filters with a cell harvester and the incorporated 3 H-TdR was measured in an LKB1214 scintillation counter with the results expressed in cpm per 1 × 10 5 lymphocytes. The co-cultures from each animal were made in triplicates and 5 mice were used for each data point. Thymocytes and splenocyte for flow cytometric analysis were sampled in the same way but without addition of Con A or 3-HTdR. Treatment of the cells with monoclonal antibodies adhered strictly to the manufacturer's instructions. The peritoneal macrophages for measurement of IL-12p70 secretion were obtained from sham-irradiated and irradiated mice at different time intervals after exposure by injection of 5 ml RPMI1640 into the peritoneal cavity and collection of the cell suspension after gentle massage of the abdomen for 1 min. The cell suspensions were centrifuged and the cells were re-suspended in RPMI1640 with 10% NBS before incubation in 6-well plates. 3 × 10 6 cells in 2 ml (containing LPS 1 μg/ml) were added into each well and incubated at 37°C with 5% CO 2 for 72 h before the collection of the supernatant for ELISA of IL-12p70. Samples were made in triplicates and 5 animals were used for each data point. Peritoneal macrophages for immunohistochemical examination of B7-1/2 expression were collected in the same way but without LPS stimulation.
Irradiation
X-rays of 200 kVp and 10 mA with 0.5 mm Cu and 1 mm A1 as filters were given at a dose-rate of 0.0125 Gy/min for doses below 0.5 Gy and at a dose-rate of 0.287 Gy/min for doses of 0.5 Gy and above as reported . The immunmological parameters were assessed at different time intervals after exposure as indicated in the results.
Parameters
Expression of B7-1/2 on macrophages was assessed with immunohistochemical examination using 5 mice for each time and dose point . Briefly, cell films on glass slides fixed with cold acetone (3 samples from each animal) were treated with 10% normal goat serum to block the endogenous peroxidase, followed successively by addition of first antibody (rat anti-mouse B7-1/2 monoclonal antibody), second antibody (biotin-labeled goat anti-rat-IgG), streptavidin-labeled HRP (horse radish peroxidase) and DAB developing mixture (containing diaminobenzidine and H 2 O 2). The expression of CD28/CTLA-4 on lymphocytes was examined with flow cytometry (FCM) by double immunofluorescence (anti-CD28-PE and anti-CTLA-4-FITC, PharMingen, USA) according to the manufacturer's manual. IL-10 synthesis was measured with FCM by indirect immunofluorescence using monensin to block the release of the cytokine and saponin to perforate the cell membrane for entrance of monoclonal antibodies (PharMingen). The FCM analysis was made with pooled samples of thymocyte or splenocyte from 2 (CD28/CTLA-4 expression) or 3 (IL-10 synthesis) mice and 5 samples (from 10 or 15 mice) were examined for each time or dose point. IL-12p70 was detected by ELISA (PharMingen). These were performed following the manufacturer's instructions in triplicates for each sample and 5 animals were used for each data point. FACScan machine (Becton-Dickinson, USA) was used for the FCM measurements.
Statistics
Student's t and correlation tests were used in the analysis of the results.
References
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Acknowledgment
This work was supported by funds from NSF of China and IRI of Japan. Thanks are due to Jim Muckerheide for his invaluable comments to the manuscript.
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Immunology Laboratory Radiobiology Research Unit, Ministry of Health, PRC Norman Bethune Medical Center, Jilin University, 8 Xinmin Street, Changchun, 130021, China
Shu-Zheng Liu,Shun-Zi Jin,Xiao-Dong Liu&Yi-Min Sun
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Liu, SZ., Jin, SZ., Liu, XD. et al. Role of CD28/B7 costimulation and IL-12/IL-10 interaction in the radiation-induced immune changes. BMC Immunol2, 8 (2001).
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Keywords
Peritoneal Macrophage
Proliferative Response
High Dose Radiation
Splenic Lymphocyte
Immunological Change
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Liu SZ, Bai O: On mechanistic studies of immune responses following low dose ionizing radiation. In: International Meeting on Biological Effects of Low Dose Radiation, Cork, Ireland, 25–26 July 1999 (Edited by Yamada T, Mothersill C, Michael BD and Potten SC). Amsterdam, Elsevier Science,. 2000, 129-135.
Google Scholar
Liu SZ, Bai O, Chen D, Ye F: Genes and protein molecules involved in the cellular activation induced by low dose radiation. Radia. Res Radiat Proc. 2000, 18: 175-186.
CASGoogle Scholar
Liu SZ: Multilevel mechanisms of stimulatory effect of low dose radiation on immunity. In: Low does irradiation and biological defense mechanisms (Edited by Sugahara T, Sagan LA, and Aoyama T). Amsterdam Elsevier Science Publishers,. 1992, 225-232.
Google Scholar
Liu SZ, Xie F: Involvement of the Ca 2+ -protein kinase C and adenylate cyclase signal pathways in the activation of thymocytes in response to whole-body irradiation with low dose X-rays. Chin Med Sci J. 2000, 15: 1-7.
PubMedGoogle Scholar
Liu SZ, Liu WH, Sun JB: Radiation Hormesis: its expression in the immune system. Health Physics. 1987, 52: 579-583.
ArticleCASPubMedGoogle Scholar
Liu SZ, Su X, Zhang YC, Zhao Y: Signal transduction in lymphocytes after low dose radiation. Int J Occup Med Toxicol. 1994, 3: 107-117.
Google Scholar
James SJ, Makinodan T: T cell proliferation in normal and autoimmune-prone mice after extended exposure to low doses of ionizing radiation and/or caloric restriction. Int J Radiat Biol. 1988, 53: 137-152.
ArticleCASGoogle Scholar
Greenfield EA, Nguyen KA, Kuchroo VK: CD28/B7 costimulation: a review. Crit Rev Immunol. 1988, 18: 389-418.
ArticleGoogle Scholar
Agadjanyan MG, Kim JJ, Trivedi N, Wilson DM, Monzavi-Karbassi B, Morrison LD, Nottingham LK, Dentchev T, Tsai A, Dang K, Chalian AA, Maldonado MA, Williams WV, Weiner DB: CD86 (B7-2) can function to drive MHC-restricted antigen-specific CTL responses in vivo. J Immunol. 1999, 162: 3417-3427.
CASPubMedGoogle Scholar
Kuhns MS, Epshteyn V, Sobel RA, Allison JP: Cytotoxic T lymphocyte antigen-4 (CTLA-4) regulates the size, reactivity, and function of a primed pool of CD4+ T cells. Proc NatI Acad Sci USA. 2000, 97: 12711-12716. 10.1073/pnas.220423597.
ArticleCASGoogle Scholar
King C, Hoenger RM, Cleary MM, Murali-Krishna K, Ahmed R, King E, Sarvetnick N: lnterleukin-4 acts at the locus of the antigen-presenting dendritic cell to counter-regulate cytotoxic CD8+ T-cell responses. Nature Medicine. 2001, 7: 206-214. 10.1038/84659.
ArticleCASPubMedGoogle Scholar
Guerder S, Carding SR, Flavell RA: B7 costimulation is necessary for the activation of the lytic function in cytotoxic T lymphocyte precursors. J.Immunol. 1995, 155: 5167-5174.
CASPubMedGoogle Scholar
Creery WD, Diaz-Mitoma F, Filion L, Kumar A: Differential modulation of B7-1 and B7-2 isoform expression on human monocytes by cytokines which influence the development of T helper cell phenotype. Eur. J. Immunol. 1996, 26: 1273-1277.
ArticleCASPubMedGoogle Scholar
Willems F, Marchant A, Delville JP, Gerard C, Delvaux A, Velu T, de Boer M, Goldman M: lnterleukin-10 inhibits B7 and intercellular adhesion molecule-1 expression on human monocytes. Eur. J. Immunol. 1994, 24: 1007-1009.
ArticleCASPubMedGoogle Scholar
Ostrov DA, Shi W, Schwartz J-CD, AImo SC, Nathenson SG: Structure of murine CTLA-4 and its role in modulating T cell responsiveness. Science. 2000, 290: 816-819. 10.1126/science.290.5492.816.
ArticleCASPubMedGoogle Scholar
Xia M, Gasser J, Feige U: Dexamethasone enhances CTLA-4 expression during T cell activation. Cell. Mol. Life Sci. 1999, 55: 1649-1656. 10.1007/s000180050403.
ArticleCASPubMedGoogle Scholar
Du C, Scriram S: Mechanism of inhibition of LPS-induced IL-12p40 production and TGF β in ANA-1 cells. J.Leukoc. Biol. 1998, 64: 92-97.
CASPubMedGoogle Scholar
Kubin M, Kamoun M, Trinchieri G: Interleukin 12 synergizes with B7/CD28 interaction in inducing efficient proliferation and cytokine production of human T cells. J.Exp Med. 1994, 180: 211-222.
ArticleCASPubMedGoogle Scholar
Sun YM, Liu SZ: Interaction of accessory cells with splenic lymphocytes after whole-body X-irradiation of mice. J Radiat Res Radiat Proc. 2000, 18: 108-112.
Google Scholar
Jin SZ, Liu SZ: Effect of X-rays on the expression of B7-1 and B7-2 on peritoneal macrophages. J N Bethune Univ Med Sci. 2001, 27: 111-113.
CASGoogle Scholar
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Find the domain of the polar curve r(θ)=2 cos 2 θ r(θ)=2 cos2 θ
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I am given the following polar function: r(θ)=2 cos 2 θ r(θ)=2 cos2 θ
And I am asked to find the domain of f=r(θ)f=r(θ) if r>0 r>0
I understand that, in this context, finding the domain means to find the allowed values of θ θ when r>0 r>0, am I right?
I plotted the function (which, by the way, is the ultimate objective of this problem):
So what I did was to impose:
r(θ)>0⇔2 cos 2 θ>0⇔2 θ∈(−π 2,π 2)⇔θ∈(−π 4,π 4)r(θ)>0⇔2 cos2 θ>0⇔2 θ∈(−π 2,π 2)⇔θ∈(−π 4,π 4)
But I don't know what this really means (I think it's the rightmost petal in the plot?). The function is well defined everywhere.
My book's solution is:
θ∈[0,π 4]∪[3 π 4,5 π 4]∪[7 π 4,2 π]θ∈[0,π 4]∪[3 π 4,5 π 4]∪[7 π 4,2 π]
Maybe there is something I'm missing. How exactly was this solution obtained? How does it relate to mine? Is it possible to use negative angles in polar coordinates? If the function is well-defined everywhere, what does "domain" mean in this context? I'm still learning, apologies if this question was a bit too elementary. Thank you!
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asked Feb 12, 2018 at 16:08
Jose Lopez GarciaJose Lopez Garcia
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It is usual to take [0,2 π][0,2 π] as the range for θ θ, but you could take [−π,π][−π,π]. Try plotting cos 2 θ cos2 θ. You can easily see that the book's range gives cos 2 θ≥0 cos2 θ≥0. Your plot is wrong if you are trying to plot only the part with r≥0 r≥0 - you only want the horizontal lobes. The vertical ones have r<0 r<0.almagest –almagest 2018-02-12 16:45:00 +00:00 Commented Feb 12, 2018 at 16:45
When you learn about "deceptive Points", you will understand why this question actually is quite important.imranfat –imranfat 2018-02-12 16:46:22 +00:00 Commented Feb 12, 2018 at 16:46
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The book seems to be assuming the domain is a subset of [0,2 π][0,2 π], and is simply removing the values of θ θ for which cos 2 θ cos2 θ is negative. When you require r(θ)≥0 r(θ)≥0, you only get the right- and left-pointing lobes in your graph. The upper and lower lobes come by allowing r(θ)<0 r(θ)<0: the lower lobe is swept out as θ θ runs from π/4 π/4 to 3 π/4 3 π/4, and the upper lobe is swept out as θ θ runs from 5 π/4 5 π/4 to 7 π/4 7 π/4.
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answered Feb 12, 2018 at 16:45
Barry CipraBarry Cipra
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Thanks for the answer. I think I'm having trouble to find when cos(2 θ)>0 cos(2 θ)>0 (how to find the intervals) Would you help me with that?Jose Lopez Garcia –Jose Lopez Garcia 2018-02-12 18:00:28 +00:00 Commented Feb 12, 2018 at 18:00
1 @JoseLopezGarcia, I find it helps to picture which quadrant you land in when you double an angle. For angles up to π/4 π/4 (i.e., 45 45 degrees), you stay in the first quadrant, where the x x coordinate is positive, so that cos 2 θ cos2 θ is positive. For angles from π/4 π/4 to π/2 π/2, you land in the second quadrant, where the x x coordinate is negative, so that cos 2 θ cos2 θ is negative. For angles from π/2 π/2 to 3 π/4 3 π/4, you land in the third quadrant, where the x x coordinate is still negative, etc. I hope all this makes sense.Barry Cipra –Barry Cipra 2018-02-12 18:11:19 +00:00 Commented Feb 12, 2018 at 18:11
It makes a lot of sense, but what if we had a more general angle, like cos(k θ)>0 cos(k θ)>0 for any k>0 k>0. Is there a more systematic way of finding those intervals, as a function of k k? It would be helpful if you could expand your answer or maybe give me a link to another resource, that answers this question. I appreciate all your help @BarryCipra Jose Lopez Garcia –Jose Lopez Garcia 2018-02-12 18:36:53 +00:00 Commented Feb 12, 2018 at 18:36
@JoseLopezGarcia, you might do well to ask that as its own, stand-alone question. (You could link to your question here as a way to provide context.)Barry Cipra –Barry Cipra 2018-02-12 19:09:13 +00:00 Commented Feb 12, 2018 at 19:09
Nevermind, I think I just have to divide the interval [0,π 2]∪[3 π 2,2 π][0,π 2]∪[3 π 2,2 π] by the constant k k right? Thank you for your time Jose Lopez Garcia –Jose Lopez Garcia 2018-02-12 19:53:40 +00:00 Commented Feb 12, 2018 at 19:53
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I may not be qualified to answer your question, and I am not trying to. From what I understand you have already figured it out, but i am posting this to help those like me coming to this topic for the first time. I want to say thanks for all the comments and the answers above they explained it well and helped me understand it but I just want to put my answer in hopes that it will make it easier for beginners like me to understand it.
So the first thing to keep in mind is the domain of a polar equation which is usually [0 0,2 π 2 π] (and this is because of the fact that θ θ would be repeating it self after that point), but it could also some times be [-π π,π π] . it would usually be given or you might have to figure it out depending on the problem. Then what you have to do is find the domain of your equation that is the subset of this domain. Lets take the above example.
r(θ)=2 cos 2 θ r(θ)=2 cos2 θ
First lets find the zeros of this equation. r(θ)=0 r(θ)=0 whenever cos 2 θ=0 cos2 θ=0, and the first zero is at π 4 π 4 and because c o s(2 θ)c o s(2 θ) has a period of 2 π 2 2 π 2 the other zero's are (...,−3 π 4,−π 4,π 4,3 π 4,...)(...,−3 π 4,−π 4,π 4,3 π 4,...) and so on. Because it is easy to graph the c o s c o s function we can identify on which intervals cos(2 θ)>0 cos(2 θ)>0 hence we will know on which intervals r>0 r>0, but for the more harder equation you may need to use first derivative test and the like to find turning points and get a nice sketch of the graph to determine on which intervals it is greater than zero. In our case cos(2 θ)>0 cos(2 θ)>0 for (...,[−5 π 4,−3 π 4],[−π 4,π 4],[3 π 4,5 π 4][7 π 4,9 π 4],...)(...,[−5 π 4,−3 π 4],[−π 4,π 4],[3 π 4,5 π 4][7 π 4,9 π 4],...). Now from this set of intervals, let choose the once that satisfy our first rule. (θ∈(0,2 π)θ∈(0,2 π)
when we do that we can see that θ∈[0,π 4]∪[3 π 4,5 π 4]∪[7 π 4,2 π]θ∈[0,π 4]∪[3 π 4,5 π 4]∪[7 π 4,2 π] so that would be our solution. The graph would then look like ∞∞. The graph if you are interested
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edited May 4, 2020 at 10:07
answered May 4, 2020 at 9:18
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11115 | https://www.ebmedicine.net/topics/toxicology-environmental/toxic-alcohol-poisoning | 0
Toxic Alcohol Ingestion: Prompt Recognition and Management in the Emergency Department (Critical Care Topic and Pharmacology CME)
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Abstract
Identifying patients with potential toxic alcohol exposure and initiating appropriate management is critical to avoid significant patient morbidity. Sources of toxic alcohol exposure include ethylene glycol, methanol, diethylene glycol, propylene glycol, and isopropanol. Treatment considerations include the antidotes fomepizole and ethanol, and hemodialysis for removal of the parent compound and its toxic metabolites. Additional interventions include adjunctive therapies that may improve acidosis and enhance clearance of the toxic alcohol or metabolites. This issue reviews common sources of alcohol exposure, basic mechanisms of toxicity, physical examination and laboratory findings that may guide rapid assessment and management, and indications for treatment.
Case Presentations
A 27-year-old man with a history of depression presents after a reported ingestion of “about 8 ounces” of a workplace industrial solvent that has been identified by a shift supervisor as diethylene glycol. You are familiar with ethylene glycol and wonder if diethylene glycol poisoning results in similar toxicity. The patient has normal vital signs and a normal examination. What laboratory testing is warranted in this patient? Is it possible to obtain a diethylene glycol level? Based on this reported ingestion, what treatment is warranted?
A 54-year-old man you admitted on your last shift for severe alcohol withdrawal remains on a lorazepam infusion that you started 24 hours ago in the ED. When following up his case, you note that the patient has developed a worsening lactic acidosis since his admission. You discuss the case with the intensivist who is caring for the patient in the ICU. His infectious workup has been normal, and the rest of his laboratory testing has been within normal limits. What is the etiology of his lactic acidosis?
A 67-year-old man with a history of alcohol abuse is brought in by EMS after he was found unconscious outside a grocery store. EMS notes that the patient is responding to painful stimuli only and had an episode of blood-tinged emesis on the scene. He arrives to the ED receiving oxygen via nonrebreather mask. An empty bottle of rubbing alcohol is found in his coat pocket. What are your concerns in a patient with likely isopropanol exposure? What laboratory workup is warranted?
Introduction
The term toxic alcohol refers to a group of hydrocarbons that contain a hydroxyl group (-OH group) and are not intended for ingestion. This group includes ethylene glycol, methanol, propylene glycol, and isopropanol.1 Diethylene glycol belongs to a class of glycol ethers that contain hydroxyl groups. It is also discussed here, as it may present in a clinically similar fashion to the toxic alcohols.2
Ethylene glycol is used primarily as an engine coolant, antifreeze, or brake fluid, and it may be unintentionally consumed by children or animals because of its sweet taste.1,3 Methanol, most commonly in the form of windshield-washer fluid, has been implicated in several poisoning epidemics resulting from tainted beverages.1,4-7 Outbreaks have been reported in undeveloped countries as a result of the adulteration of ethanol with methanol. Diethylene glycol, used as an antifreeze and as a solvent in industry and manufacturing, is responsible for multiple mass poisoning events due to its unsafe use as a diluent in medications and as a sweetener in wines, and it has also been used in self-harm attempts.2,8-14 In 1937, 107 deaths followed ingestion of diethylene glycol that was used as a diluent for an elixir of the antibiotic sulfanilamide. This incident led to passage of the 1938 Food, Drug, and Cosmetic Act, which made the declaration that pharmaceutical products must be shown to be safe prior to public marketing.15 Propylene glycol, present in many foods, beverages, and cosmetics, is used as a diluent in many pharmaceuticals and also as an alternative to ethylene glycol in some antifreeze products.16 Isopropanol, or isopropyl alcohol, is typically available as a 70% rubbing alcohol solution and may be abused as an ethanol substitute. It is also a solvent used in household and pharmaceutical products.1
All of the toxic alcohols may cause altered mental status, due to both increased inhibitory gamma-aminobutyric acid (GABA) tone and inhibition of the excitatory N-Methyl-D-aspartic acid (NMDA) glutamate receptors.1 The metabolism of toxic alcohols to their metabolites may result in acidosis or ketosis. Specific end-organ toxicity from toxic alcohol metabolites is possible, including glycolic acid in ethylene glycol exposure, formic acid in methanol exposure, (2-hydroxyethoxy)acetic acid (HEAA) in diethylene glycol exposure, lactate in propylene glycol exposure, and acetone in isopropanol exposure.1,2
The presence of a metabolic acidosis should also prompt consideration of other causes, including poisoning by xenobiotics such as metformin, which can cause a lactic acidosis; cellular poisoning by salicylate, iron, or cyanide; diabetic, alcoholic, or starvation ketosis; renal failure; and multiorgan failure from critical systemic illness such as sepsis.
A large difference in measured osmolarity and calculated osmolality (the osmol gap) should prompt consideration of a toxic alcohol exposure; however, this finding is neither sensitive nor specific and cannot be used to definitively rule in or rule out this exposure.
Disposition of the critically ill patient with toxic alcohol exposure requires consideration of patient needs, including access to antidotal therapies (such as fomepizole or ethanol), the availability of hemodialysis services, if needed, and an inpatient unit capable of providing critical care and resuscitation.
Critical Appraisal Of The Literature
To identify primary relevant literature, standard search strategies were used, including querying MEDLINE® with the search terms toxic alcohol, ethylene glycol, methanol, diethylene glycol, propylene glycol, isopropyl alcohol, isopropanol, and fomepizole. Results were reviewed for clinical and practical relevance. Clinical studies, animal studies, review articles, editorials, commentaries, case reports, and case series were identified for review. The Cochrane Database of Systematic Reviews and the National Guideline Clearinghouse (www.guideline.gov) were also queried. Additional information was obtained from book chapters and Internet material.
Risk Management Pitfalls In Management Of Toxic Alcohol Ingestion
Tables And Figures
References
Evidence-based medicine requires a critical ap¬praisal of the literature based upon study methodology and number of subjects. Not all references are equally robust. The findings of a large, prospective, randomized, and blinded trial should carry more weight than a case report.
To help the reader judge the strength of each reference, pertinent information about the study are included in bold type following the reference, where available. In addition, the most informative references cited in this paper, as determined by the authors, are noted by an asterisk () next to the number of the reference.
Publication Information
Authors
Gillian A. Beauchamp, MD; Matthew Valento, MD
Publication Date
September 1, 2016
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11116 | https://link.springer.com/article/10.2165/00003088-199529030-00002 | Clinical Pharmacokinetics and Pharmacodynamics of Zolpidem | Clinical Pharmacokinetics
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Clinical Pharmacokinetics
Article
Clinical Pharmacokinetics and Pharmacodynamics of Zolpidem
Therapeutic Implications
Review Article
Drug Disposition
Published: 20 October 2012
Volume 29,pages 142–153, (1995)
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Clinical Pharmacokinetics and Pharmacodynamics of Zolpidem
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Joan Costa1
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Summary
Zolpidem is an imidazopyridine which differs in structure from the benzodiazepines and zopiclone. It is a strong sedative with only minor anxiolytic, myorelaxant and anticonvulsant properties, and has been shown to be effective in inducing and maintaining sleep in adults. The available evidence suggests that zolpidem produces no rebound or withdrawal effects, and patients have experienced good daytime alertness. Zolpidem 10mg in non-elderly and a reduced dose of 5mg in elderly individuals are clinically effective.
In humans, the major metabolic routes include oxidation and hydroxylation; none of the metabolites appears to be pharmacologically active. The pharmacological activity of zolpidem results from selective binding to the central benzo-diazepine receptors of the ωl subtype.
Zolpidem is approximately 92% bound to plasma proteins; absolute bioavailability of zolpidem is about 70%. After single 20mg oral doses, typical values of pharmacokinetic variables for zolpidem in humans are: a peak plasma concentration of 192 to 324 µg/L occurring 0.75 to 2.6 hours postdose; a terminal elimination half-life of 1.5 to 3.2 hours; and total clearance of 0.24 to 0.27 ml/min/kg. Zolpidem pharmacokinetics are unchanged during multiple-dose treatment.
Zolpidem pharmacokinetics are not significantly influenced by gender. Clearance of zolpidem in children is 3 times higher than in young adults, and is lower in very elderly people. There are no significant differences in the pharmacokinetic parameters between various racial groups. Dosage reduction appears to be prudent in patients with renal disease, and caution should be exercised when prescribing zolpidem to elderly patients with hepatic impairment.
Coadministration of haloperidol, cimetidine, ranitidine, chlorpromazine, warfarin, digoxin or flumazenil do not alter the pharmacokinetics of zolpidem; flumazenil predictably antagonises the hypnotic effects of zolpidem. Alertness tends to be reduced when cimetidine is combined with zolpidem. Volunteers treated with imipramine plus zolpidem developed anterograde amnesia.
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Pacifici GM, Viani A, Rizzo G, et al. Plasma protein binding of zolpidem in liver and renal insufficiency. Int J Clin Pharmacol Therapeut Toxicol 1988; 26: 439–43
CASGoogle Scholar
Guinebault P, Dubruc C, Hermann Ph, et al. High-performance liquid Chromatographie determination of zolpidem, a new sleep inducer, in biological fluids with fluorimetric detection. J Chromatogr 1986; 383: 206–11
ArticlePubMedCASGoogle Scholar
Ascalone V, Flaminio L, Guinebault P, et al. Determination of zolpidem, a new sleep-inducing agent, and its metabolites in biological fluids: pharmacokinetics, drug metabolism and overdosing investigations in humans. J Chromatogr 1992; 581: 237–50
ArticlePubMedCASGoogle Scholar
Albin H, Vinçon G, Vinçon J, et al. Study of the pharmacokinetics of zolpidem in healthy volunteers after repeated administration: Effect on antipyrine clearance. In: Sauvanet JP, Langer SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press, 1988: 369–70
Google Scholar
Desager JP, Hulhoven R, Harvengt C, et al. Possible interaction between zolpidem, a new sleep inducer, and chlorpromazine, a phenothiazine neuroleptic. Psychopharmacology 1988; 96: 63–6
ArticlePubMedCASGoogle Scholar
Vandel S, Vandel B, Joanne C, et al. Influence of ethnic origin on the clinical pharmacology of zolpidem. In: Sauvanet JP, Langer SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press, 1988: 390–1
Google Scholar
Colle M, Rosenzweig P, Bianchetti G, et al. Nocturnal profile of growth hormone secretion during sleep induced by zolpidem: A double-blind study in young adults and children. Horm Res 1991; 35: 30–4
ArticlePubMedCASGoogle Scholar
Patat A, Trocherie S, Thebault JJ, et al. EEG profile of intravenous zolpidem in healthy volunteers. Psychopharmacology 1994; 114: 138–46
ArticlePubMedCASGoogle Scholar
Priest RG. Ambieu (zolpidem 5 mg & 10 mg). UK product licence application. Paris: Registration Department (L.E.R.S.), 1993
Google Scholar
Harvengt C, Hulhoven R, Desager JP, et al. Drug interactions investigated with zolpidem. In: Sauvanet JP, Langer SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press, 1988: 165–73
Google Scholar
Hulhoven R, Desager JP, Harvengt C, et al. Lack of interaction between zolpidem and H 2 antagonist, cimetidine and ranitidine. Int J Clin Pharmacol Res 1988; 8: 471–6
PubMedCASGoogle Scholar
Danjou P, Forster A, Rosenzweig P, et al. Assessment of pharmacodynamic and pharmacokinetic interactions between intravenous zolpidem and flumazenil in healthy subjects. Biol Psychiatry 1991; 29: 456S–7S
Google Scholar
Kanno O, Watanabe H, Ichikawa I, et al. Effects of zolpidem and triazolam on all night sleep EEG of normal volunteers. Biol Psychiatry 1991; 29: 305S
ArticleGoogle Scholar
Scharf MB, Mayleben DW, McDannold MD, et al. Dose response effects of zolpidem in normal geriatric patients. Sleep Res 1988; 17: 59
Google Scholar
Morselli PL, Larribaud J, Guillet Ph, et al. Daytime residual effects of Zolpidem: a review of available data. In: Sauvauet JP, Lauger SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press, 1988: 183–91
Google Scholar
Gillin JC. Rebound insomnia: a critical review. J Clin Psychopharmacol 1989; 9: 161–72
ArticlePubMedCASGoogle Scholar
Kales A. Rebound insomnia and rebound anxiety: a review. Pharmacology 1983; 26: 121–37
ArticlePubMedCASGoogle Scholar
Jonas JM. Idiosyncratic side effects of short half-life benzodiazepine hypnotics: fact or fancy? Hum Psychopharmacol 1992; 7: 205–16
ArticleGoogle Scholar
Cirignotta F, Mondini S, Zucconi M, et al. Zolpidem-poly-somnographic study of the effect of a new hypnotic drug in sleep apnoea syndrome. Pharmacol Biochem Behav 1988; 29: 807–9
ArticlePubMedCASGoogle Scholar
Kurtz D, Fillius B, Boningen C, et al. Effect of zolpidem on the sleep of healthy elderly subjects: polysomnographic recording assessments and pharmacokinetics. In: Sauvanet JP, Langer SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press 1988: 382
Google Scholar
Scharf MB, Mayleben DW, Kaffeman M, et al. Dose response effects of zolpidem in normal geriatric subjects. J Clin Psychiatry 1991; 52: 77–83
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Salvà, P., Costa, J. Clinical Pharmacokinetics and Pharmacodynamics of Zolpidem. Clin-Pharmacokinet29, 142–153 (1995).
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Google Scholar
Hohens JD, Perry PJ. Zolpidem: A nonbenzodiazepine hypnotic for treatment of insomnia. Clin Pharm 1993; 12: 814–28
Google Scholar
Pacifici GM, Viani A, Rizzo G, et al. Plasma protein binding of zolpidem in liver and renal insufficiency. Int J Clin Pharmacol Therapeut Toxicol 1988; 26: 439–43
CASGoogle Scholar
Guinebault P, Dubruc C, Hermann Ph, et al. High-performance liquid Chromatographie determination of zolpidem, a new sleep inducer, in biological fluids with fluorimetric detection. J Chromatogr 1986; 383: 206–11
ArticlePubMedCASGoogle Scholar
Ascalone V, Flaminio L, Guinebault P, et al. Determination of zolpidem, a new sleep-inducing agent, and its metabolites in biological fluids: pharmacokinetics, drug metabolism and overdosing investigations in humans. J Chromatogr 1992; 581: 237–50
ArticlePubMedCASGoogle Scholar
Albin H, Vinçon G, Vinçon J, et al. Study of the pharmacokinetics of zolpidem in healthy volunteers after repeated administration: Effect on antipyrine clearance. In: Sauvanet JP, Langer SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press, 1988: 369–70
Google Scholar
Desager JP, Hulhoven R, Harvengt C, et al. Possible interaction between zolpidem, a new sleep inducer, and chlorpromazine, a phenothiazine neuroleptic. Psychopharmacology 1988; 96: 63–6
ArticlePubMedCASGoogle Scholar
Vandel S, Vandel B, Joanne C, et al. Influence of ethnic origin on the clinical pharmacology of zolpidem. In: Sauvanet JP, Langer SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press, 1988: 390–1
Google Scholar
Colle M, Rosenzweig P, Bianchetti G, et al. Nocturnal profile of growth hormone secretion during sleep induced by zolpidem: A double-blind study in young adults and children. Horm Res 1991; 35: 30–4
ArticlePubMedCASGoogle Scholar
Patat A, Trocherie S, Thebault JJ, et al. EEG profile of intravenous zolpidem in healthy volunteers. Psychopharmacology 1994; 114: 138–46
ArticlePubMedCASGoogle Scholar
Priest RG. Ambieu (zolpidem 5 mg & 10 mg). UK product licence application. Paris: Registration Department (L.E.R.S.), 1993
Google Scholar
Harvengt C, Hulhoven R, Desager JP, et al. Drug interactions investigated with zolpidem. In: Sauvanet JP, Langer SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press, 1988: 165–73
Google Scholar
Hulhoven R, Desager JP, Harvengt C, et al. Lack of interaction between zolpidem and H 2 antagonist, cimetidine and ranitidine. Int J Clin Pharmacol Res 1988; 8: 471–6
PubMedCASGoogle Scholar
Danjou P, Forster A, Rosenzweig P, et al. Assessment of pharmacodynamic and pharmacokinetic interactions between intravenous zolpidem and flumazenil in healthy subjects. Biol Psychiatry 1991; 29: 456S–7S
Google Scholar
Kanno O, Watanabe H, Ichikawa I, et al. Effects of zolpidem and triazolam on all night sleep EEG of normal volunteers. Biol Psychiatry 1991; 29: 305S
ArticleGoogle Scholar
Scharf MB, Mayleben DW, McDannold MD, et al. Dose response effects of zolpidem in normal geriatric patients. Sleep Res 1988; 17: 59
Google Scholar
Morselli PL, Larribaud J, Guillet Ph, et al. Daytime residual effects of Zolpidem: a review of available data. In: Sauvauet JP, Lauger SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press, 1988: 183–91
Google Scholar
Gillin JC. Rebound insomnia: a critical review. J Clin Psychopharmacol 1989; 9: 161–72
ArticlePubMedCASGoogle Scholar
Kales A. Rebound insomnia and rebound anxiety: a review. Pharmacology 1983; 26: 121–37
ArticlePubMedCASGoogle Scholar
Jonas JM. Idiosyncratic side effects of short half-life benzodiazepine hypnotics: fact or fancy? Hum Psychopharmacol 1992; 7: 205–16
ArticleGoogle Scholar
Cirignotta F, Mondini S, Zucconi M, et al. Zolpidem-poly-somnographic study of the effect of a new hypnotic drug in sleep apnoea syndrome. Pharmacol Biochem Behav 1988; 29: 807–9
ArticlePubMedCASGoogle Scholar
Kurtz D, Fillius B, Boningen C, et al. Effect of zolpidem on the sleep of healthy elderly subjects: polysomnographic recording assessments and pharmacokinetics. In: Sauvanet JP, Langer SZ, Morselli PL, editors. Imidazopyridines in sleep disorders: a novel experimental and therapeutic approach. New York: Raven Press 1988: 382
Google Scholar
Scharf MB, Mayleben DW, Kaffeman M, et al. Dose response effects of zolpidem in normal geriatric subjects. J Clin Psychiatry 1991; 52: 77–83
PubMedCASGoogle Scholar
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11117 | https://learn.hydrologystudio.com/culvert-studio/knowledge-base/mannings-n-values/ | Home
Useful Tables
Manning’s n-values
Manning’s n-values
| Material | Manning's n |
--- |
| PIPES |
| Reinforced concrete | 0.013 |
| Vitrified clay pipe | 0.013 |
| Corrugated metal pipe | 0.023 |
| Polyvinyl chloride (PVC) | 0.010 |
| Smooth welded pipe | 0.011 |
|
| NATURAL CHANNELS |
| Gravel beds, Straight | 0.025 |
| Gravel beds, large boulders | 0.040 |
| Earth, straight, some grass | 0.026 |
| Earth, winding, no vegetation | 0.030 |
| Earth, winding | 0.050 |
|
| EXCAVATED CHANNELS |
| Clean, recently completed | 0.018 |
| With short grass and weeds | 0.027 |
| Earth bottom, rubble banks | 0.030 |
| Stony bottom, weedy banks | 0.035 |
| Not maintained, weeds high as flow depth | 0.080 |
|
| LINED, CONSTRUCTED CHANNELS |
| Concrete, trowel finish | 0.013 |
| Concrete, unfinished | 0.017 |
| Gravel bottom, riprap banks | 0.033 |
|
| OVERLAND "SHEET" FLOW |
| Smooth surfaces, Concrete, asphalt, bare Soil | 0.011 |
|
| Short grass | 0.15 |
| Dense grass | 0.24 |
| Bermuda grass | 0.41 |
|
| Light underbrush woods | 0.40 |
| Dense underbrush woods | 0.80 |
Related Articles
Runoff Coefficients
Standard Elliptical Pipe Dimensions |
11118 | https://www.engineeringtoolbox.com/pipes-series-parallel-d_1787.html | Engineering ToolBox
- Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
Pressure Loss in Pipes connected in Series or Parallel
Calculate pressure loss in pipes connected in series or parallel.
Pipes in Series
For pipes connected in series the pressure loss is the sum of the individual losses:
dp = dp1 + dp2 + .. + dpn (1)
where
dp = total pressure loss (Pa, psi)
dp1..n = individual pressure loss in each pipe (Pa, psi)
The mass flow rate is the same in all pipes:
m = m1 = m2 = .... = mn (1b)
where
m = mass flow (kg/s, lb/s)
Pipes in Parallel
For pipes connected in parallel the pressure loss is the same in all pipes:
dp = dp1 = dp2 = .... = dpn (2)
The total mass flow is the sum of the flow in each pipe:
m = m1 + m2 + .. + mn (2b)
The pressure loss in the equations above can be substituted with a generic expression for pressure loss like the D'Arcy-Weisbach equation.
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11119 | https://www.youtube.com/watch?v=1uvwi6pwUOw | How to Name an Angle | Math with Mr. J
Math with Mr. J
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Posted: 5 May 2020
Welcome to How to Name an Angle with Mr. J! Need help with naming angles? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you need help with how to name angles. Mr. J will go through naming angle examples and explain the steps of naming angles using three points.
About Math with Mr. J: This channel offers instructional videos and mastery checks (practice videos to gauge understanding) that are directly aligned with math standards (5th grade, 6th grade, 7th grade, etc.). Videos can be used to introduce content, reteach content, or as a study tool. Teachers, parents/guardians, and students from around the world have used this channel to help with math content in many different ways. All material is absolutely free.
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Hopefully this video is what you're looking for when it comes to how to name an angle.
Have a great rest of your day and thanks again for watching!
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Transcript:
welcome to math with mr. J in this video I'm going to cover how to name an angle and we're going to go through two examples together in order to get this down so let's jump right into number one here and the first thing we need to do when naming an angle is find the vertex and the vertex is where two lines or line segments meet in order to create the angle now the vertex is named by a letter for number one we have a G here corresponding to our vertex so after we find the vertex we need to find a point on each line segment or line that create the angle so here we have a D this line segment right here creates the angle and this line segment here so we have a D up there and we have a Z down here so once we have the points on our line segments and the vertex identified we are able to name our angle so I'm going to start here with the symbol that means angle and now we start with one of our points we're going to start with D so we start with D here we go down to our vertex G and then down to our other point on our line segment Z so one of our names could be angle D G Z now we can also start at Z here go up the G so start at Z go up to G and then finish by going up to D now notice the order of our letters represents the order of how we're drawing that angle and the vertex is always in the middle always in the middle so we could either do angle D GZ or angle Z G D now I'm going to give you a couple of examples that do not work we cannot do angle G Z D because if we start at G and go down to Z and then up to D that does not give us the angle that we're looking for and we cannot do D start at D no I'm sorry start a G go up to D and then down to Z so we again we cannot start a G go to D and then Z the vertex needs to be in the middle so let's go to number two here now this looks a little different we have two lines that intersect in order to create angles so let's take a look at a down here we need to find angle P F G so this tells us start at P go to our vertex F and end at Point G here so angle p FG would be right here P F G so our angle right here would be p FG now let's go to B see if you can find angle G F a so we started G we go into our vertex F and then up to a so angle G FA would be right here and then lastly see what is another name for angle P F Y so p FY would be right here P F Y another name as long as we have the vertex in the middle we could do Y F P angle YF p start up top here F P so there you have it there's how to name an angle we find two points on the lines or line segments and we also find that vertex and we always put the vertex the letter for the vertex in the middle of our name hopefully that helped thanks so much for watching until next time peace |
11120 | https://rea.ceibal.edu.uy/elp/midiendo-masas/es_lo_mismo_la_masa_y_el_peso.html | Saltar la navegación
Midiendo masas
¿Es lo mismo la masa y el peso?
La masa de un cuerpo es una propiedad característica del mismo, que está relacionada con el número y clase de las partículas que lo forman. Se mide en kilogramos (kg) y también en gramos, toneladas, libras, onzas, etc.
El peso de un cuerpo es la fuerza con que lo atrae la Tierra y depende de la masa del mismo.
Un cuerpo de masa el doble que otro, pesa también el doble. Se mide en Newtons (N) y también en kg-fuerza, dinas, libras-fuerza, onzas-fuerza, etc.
El kg es por tanto una unidad de masa, no de peso.
Sin embargo, muchos aparatos utilizados para medir pesos (básculas, por ejemplo), tienen sus escalas graduadas en kg en lugar de kg-fuerza.
Esto no suele representar, normalmente, ningún problema ya que 1 kg-fuerza es el peso en la superficie de la Tierra de un objeto de 1 kg de masa. Por lo tanto, una persona de 120 kg de masa pesa en la superficie de la Tierra 120 kg-Fuerza. Sin embargo, la misma persona en la Luna pesaría solo 20 kg-fuerza, aunque su masa seguiría siendo de 120 kg. (El peso de un objeto en la Luna, representa la fuerza con que ésta lo atrae)
La costumbre nos ha hecho trabajar desde chicos solo con el concepto de peso, el cual hemos asociado siempre al kilogramo, y nos han habituado a usarlo, sin saberlo nosotros, como sinónimo de masa. Por eso, cuando subimos a una balanza decimos que nos estamos "pesando", cuando en realidad estamos midiendo nuestra cantidad de masa, que se expresa en kilogramos.
Lo que hacemos es usar nuestra medición de MASA como si fuera nuestro "PESO" y al bajar de la balanza decimos "PESÉ 70 KILOS" si la máquina marca esa cantidad, pero el PESO REAL SERÁ 686 Newtons (N) (70 por 9,8 es igual a 686).
Lo concreto es que, en el uso moderno del campo de la mecánica, el peso y la masa son cantidades fundamentalmente diferentes: la masa es una propiedad intrínseca de la materia mientras que el peso es la fuerza que resulta de la acción de la gravedad en la materia.
Sin embargo, el reconocimiento de la diferencia es, históricamente, un descubrimiento relativamente reciente. Es por eso que en muchas situaciones cotidianas la palabra peso continúa siendo usada cuando se piensa en masa.
Por ejemplo, se dice que un objeto pesa un kilogramo cuando el kilogramo es una unidad de masa. |
11121 | https://www.sciencedirect.com/science/article/abs/pii/S0168900207024205 | ACTIVIA: Calculation of isotope production cross-sections and yields - ScienceDirect
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Nuclear Instruments and Methods in Physics Research Section A: Accelerators, Spectrometers, Detectors and Associated Equipment
Volume 586, Issue 2, 21 February 2008, Pages 286-294
ACTIVIA: Calculation of isotope production cross-sections and yields
Author links open overlay panel J.J.Back, Y.A.Ramachers
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Abstract
We present a C++ computer package, ACTIVIA, that can calculate target–product cross-sections and the production and decay yields of isotopes from cosmic ray activation using data tables and semi-empirical formulae. We describe the structure and user interface of the computer code as well as provide comparisons between the calculations and experimental results. We also outline suggestions on how the code can be improved and extended for other applications.
Introduction
There are many observations made in astrophysics, astronomy and particle physics that require knowledge of the cross-sections of nuclear processes at high energies. The physics of cosmic rays is one example. Understanding how they interact with matter allows us to study the composition of the interstellar medium, the Galaxy and the solar system. Estimating the cosmic ray background radiation in spacecraft and isotope production in accelerator facilities are other important applications. For the case of low background experiments, it is necessary to know which isotopes are formed by cosmic ray activation inside target materials before they are shielded underground. Any long-lived isotopes must be accounted for in the signal data analysis, and this requires information about cross-sections and production rates. There are measurements available for use in such studies, but normally only for a restricted energy range which does not cover the total energy spectrum of the input beam. Some important reactions for low background experiments, such as 60 Co produced from natural tellurium by cosmic ray activation, have essentially no experimental data available. Accordingly, a set of semi-empirical formulae have been developed by Silberberg and Tsao , to estimate the cross-sections of various nuclear processes, such as spallation and fission, which can be used to calculate the production yields of nuclear isotopes from cosmic ray activation. The parameters in the formulae, assumed to be the same for proton and neutron beams, have been tuned to best fit the available experimental results.
In this paper we describe a new C++package, ACTIVIA, that uses a combination of semi-empirical formulae and tables of data based on experimental results to calculate the cross-section, production rates and yields of radioactive isotopes from cosmic ray activation. Section 2 describes the Silberberg–Tsao semi-empirical formulae used for the cross-section calculations while Section 3 explains how cross-sections can be evaluated if tables of data (in ASCII text format) are available. Section 4 provides details on how the radioactive decay yields are calculated, Section 5 explains the basic structure of the code, while Section 6 shows comparisons between the calculations and experimental data. Finally, in Section 7 we describe possible modifications and extensions that could be made to the code if other input beam spectra or nuclear physics processes are required.
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Section snippets
Semi-empirical formulae
We use the extensive set of semi-empirical formulae from Silberberg and Tsao , to calculate cross-sections for isotope production when a proton or neutron beam (e.g. cosmic rays) hits a given target. The general form of the cross-section equation for a product (Z,A) from a target (Z t,A t) is σ=σ 0 f(A)f(E)e-P Δ A exp(-R|Z-SA+TA 2+UA 3|ν)Ω η ξ where E is the beam energy (in units of MeV), Δ A is the difference between the target and product mass numbers (A t-A), while f(A) and f(E) are correction
Data tables
There are extensive sets of nuclear data (e.g. the EXFOR database ) that can be used to provide cross-sections as a function of energy for target–product isotope pairs. If present, these (ASCII text) tables are used instead of the semi-empirical formulae. A linear interpolation is performed between data points to provide the cross-section (in mb) at a given beam energy (in MeV). It is important that the data be in the correct units, have enough energy bins to be useful and have a contiguous,
Radioactive decay yields
Consider a target comprised of several isotopes with the same atomic number Z t but having different mass numbers A ti each with a relative abundance fraction f i such that ∑i f i=1. This target is exposed to a beam with an energy spectrum d φ/d E(cm-2 s-1 MeV-1), such as cosmic ray neutrons shown in Fig. 2, which is based on the parameterisation from Armstrong and Gehrels . The production rate (kg-1 day-1) of an isotope j will be Y j=C∑i f i A ti∫d φ d E σ ij(E)d E where C is a normalisation factor and σ ij(E)
Code structure
The code, written in C++ to take advantage of object-oriented design principles, is split into two main sections; the first part performs the calculation of the cross-sections (Eq. (1)) and production rates (Eq. (2)), while the second part calculates the final yields of the radioactive isotope products (Eq. (5)). The basic structure of the code, designed to work in the Unix environment, is described below. Details on how to build and run the code are found in the README file of the computer
Comparisons between measurements and calculations
The calculations of the cross-sections have been benchmarked using experimental data from the EXFOR database for natural tellurium, germanium and copper targets. The results of the calculations agree in general with those shown in various reports from the IDEA collaboration . Fig.4, Fig.5, Fig.6 show the calculated cross-section graphs of several isotopes along with the data points from experimental results . The agreement between calculations and experimental results is not
Code modifications and extensions
The code is structured so that modifications or extensions can be made to several components of the cross-section and yield calculations whilst keeping the overall framework intact.
Different input beam spectra can be used by including classes that inherit from ActBeamSpectrum; the appropriate spectrum must be chosen by the input class. For example, a class describing the flux of an ion beam can be provided, although the formulae for calculating the cross-sections of products from such a beam
Summary
We have presented the ACTIVIA C++ computer package that can be used to calculate target–product cross-sections and the production and decay yields of isotopes from cosmic ray activation. Cross-sections are calculated using a combination of semi-empirical formulae, assumed to be the same for proton and neutron beams, and ASCII tables of (experimental) data.
The basic structure of the code, designed to work in the Unix environment, has been presented. It is structured so that modifications or
Acknowledgments
We thank Tom Latham and Ben Morgan for useful discussions about the code design. We also thank the OECD Nuclear Energy Agency, France, for providing us with the MENDL-2P data tables. This work is supported by the Science and Technology Facilities Council (United Kingdom).
Recommended articles
References (51)
T.W. Armstrong et al.### J. Geophys. Res. (1973)
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E.B. Norman### Nucl. Phys. B Proc. Suppl. (2005)
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S.M. Kormali et al.### J. Radioanal. Nucl. Chem. (1976)
T. Horiguchi### Int. J. Appl. Radiat. Isot. (1983)
S. Takács### Nucl. Instr. and Meth. B (2002)
F. Szelecsényi### Nucl. Instr. and Meth. B (2001)
J. Kuhnhenn, Ph.D. Thesis, University of Köln, Germany,...
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R. Michel### Nucl. Instr. and Meth. B (1995)
M.J. Ozafran### J. Radioanal. Nucl. Chem. (1989)
L. R. Greenwood, R. K. Smither, Prog. Rep. U.S. Department of Energy, Fusion Energy Series, No. 0046, 18 (1984)...
J.M. Sisterson### Nucl. Instr. and Meth. B (2005)
A. Grütter### Nucl. Phys. A (1982)
R. Michel### Nucl. Instr. and Meth. B (1986)
J.E. Cline et al.### Nucl. Phys. A (1971)
S.J. Mills et al.### Int. J. Appl. Radiat. Isot. (1992)
R. Ardito et al., hep-ex/0501010...
C.J. Martoff et al.### Comput. Phys. Commun. (1992)
F.T. Avignone### Nucl. Phys. B Proc. Suppl. (1992)
I. Barabanov### Nucl. Instr. and Meth. B (2006)
J.A. Formaggio et al.### Ann. Rev. Nucl. Part. Sci. (2004)
S. Agostinelli### Nucl. Instr. and Meth. A (2003)
R. Silberberg et al.### Astrophys. J. Suppl. 220 (I) (1973) R. Silberberg et al.### Astrophys. J. Suppl. 220 (II) (1973)
View more references
Cited by (56)
Dual MeV gamma-ray and dark matter observatory - GRAMS Project
2020, Astroparticle Physics Show abstract GRAMS (Gamma-Ray and AntiMatter Survey) is a novel project that can simultaneously target both astrophysical observations with MeV gamma rays and an indirect dark matter search with antimatter. The GRAMS instrument is designed with a cost-effective, large-scale LArTPC (Liquid Argon Time Projection Chamber) detector surrounded by plastic scintillators. The astrophysical observations at MeV energies have not yet been well-explored (the so-called “MeV-gap”) and GRAMS can improve the sensitivity by more than an order of magnitude compared to previous experiments. While primarily focusing on MeV gamma-ray observations, GRAMS is also optimized for cosmic ray antimatter surveys to indirectly search for dark matter. In particular, low-energy antideuterons will provide an essentially background-free dark matter signature. GRAMS will be a next generation experiment beyond the current GAPS (General AntiParticle Spectrometer) project for antimatter survey.
### Direct dark matter detection: The next decade
2012, Physics of the Dark Universe Citation Excerpt :
Available codes such as Activia and Cosmo can also calculate the yields of various isotopes after certain exposure and cooling times. However, only for a few materials such as natural and enriched Ge and Cu the predictions can be compared to actual data [41,43]. More data is need for instance in the case of noble liquids such as Ar and Xe as well as for many other WIMP targets. Show abstract Direct dark matter searches are promising techniques to identify the nature of dark matter particles. I describe the future of this field of research, focussing on the question of what can be achieved in the next decade. I will present the main techniques and R&D projects that will allow to build so-called ultimate WIMP detectors, capable of probing spin-independent interactions down to the unimaginably low cross section of 10−48 cm 2, before the irreducible neutrino background takes over. If a discovery is within the reach of a near-future dark matter experiment, these detectors will be able to constrain WIMP properties such as its mass, scattering cross section and possibly spin. With input from the LHC and from indirect searches, direct detection experiments will hopefully allow to determine the local density and to constrain the local phase-space structure of our dark matter halo.
### Projected WIMP sensitivity of the LUX-ZEPLIN dark matter experiment
2020, Physical Review D
### DarkSide-20k: A 20 tonne two-phase LAr TPC for direct dark matter detection at LNGS
2018, European Physical Journal Plus
### Dark matter direct-detection experiments
2015, Journal of Physics G Nuclear and Particle Physics
### Study of the electromagnetic background in the XENON100 experiment
2011, Physical Review D Particles Fields Gravitation and Cosmology
View all citing articles on Scopus
View full text
Copyright © 2008 Elsevier B.V. All rights reserved.
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11122 | https://proofwiki.org/wiki/Monotone_Convergence_Theorem_(Real_Analysis) | Monotone Convergence Theorem (Real Analysis)
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This article was Featured Proof between 30 June 2009 and 6 July 2009.
This proof is about Monotone Convergence Theorem in the context of Real Analysis. For other uses, see Monotone Convergence Theorem.
Contents
1 Theorem
1.1 Increasing Sequence
1.2 Decreasing Sequence
1.3 Graphical Illustration
2 Examples
2.1 Example: n−1nn−1n
2.2 Example: xnxn for 0<x<10<x<1
3 Sources
Theorem
Let ⟨xn⟩⟨xn⟩ be a bounded monotone sequence sequence in RR.
Then ⟨xn⟩⟨xn⟩ is convergent.
Increasing Sequence
Let ⟨xn⟩⟨xn⟩ be an increasing real sequence which is bounded above.
Then ⟨xn⟩⟨xn⟩ converges to its supremum.
Decreasing Sequence
Let ⟨xn⟩⟨xn⟩ be a decreasing real sequence which is bounded below.
Then ⟨xn⟩⟨xn⟩ converges to its infimum.
Graphical Illustration
The following diagram illustrates the Monotone Convergence Theorem:
Examples
Example: n−1nn−1n
The sequence ⟨an⟩n≥1⟨an⟩n≥1 defined as:
: an=n−1nan=n−1n
is convergent to the limit 11.
Example: xnxn for 0<x<10<x<1
Let x∈Rx∈R such that 0<x<10<x<1.
The sequence ⟨an⟩n≥1⟨an⟩n≥1 defined as:
: an=xnan=xn
is convergent to the limit 00.
Sources
1953: Walter Rudin: Principles of Mathematical Analysis ... (previous): 3.143.14
1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): 11: Review of some real analysis: §1.2§1.2: Real Sequences: Theorem 1.2.61.2.6
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11123 | https://www.geogebra.org/m/JS3VP7DK | Isoceles Triangle Theorems – GeoGebra
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Isoceles Triangle Theorems
Author:Sina Baldwin
Define: Isosceles Triangle
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Drag the vertices of the triangle and observe the angle measurements. What do you notice? Write a conjecture about the base angles of an isosceles triangle.
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Drag a vertex of the triangle until two of the angles are congruent. What do you notice about the sides of the triangle? Write a conjecture about the congruent angles and the sides opposite them.
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Angle Bisector in An Isosceles Triangle
Triangle CAB is an isosceles triangle. The base is segment CB and the legs are segments CA and BA. Segment AD bisects , the vertex angle of the triangle.
Drag a vertex of the traingle. What do you notice about segments CD and DB?
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What can you conclude that segment AD does to segment CB?
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Write a conjecture that summarizes your findings about the angle bisector, segment AD, of the vertex angle in an isosceles triangle and its base, segment CB.
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11124 | https://electronics.stackexchange.com/questions/464628/how-do-you-fix-the-glitches-in-the-following-karnaugh-maps | Stack Exchange Network
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How do you fix the glitches in the following Karnaugh maps?
The following circuit is supposed to output 1 if the input is greater or equal than 1010 (decimal 10). The truth table is given and we are supposed to convert that to POS and SOP using Karnaugh maps.
Here's the truth table with the minterms and the Maxterms:
Here's the Karnaugh map I did for the SOP (Sum of Products)
Same for POS (Product of Sums)
I remember my professor saying that when you make groups on the Karnaugh map if there's adjacent groups there will be a glitch.
I have three questions:
3 Answers 3
The term for this glitch effect is hazard. The Karnaugh map is an easy way to identify and eliminate hazards. A hazard exists when there are adjacent, non-overlapping groups (also called terms) in the K-map. When transitioning from one group to the other, the output shouldn't change according to the logic, but there might be a change or glitch because of uneven delays through the gates. If this is not tolerable, eliminate the hazard by defining an additional, redundant term that overlaps the groups, in effect bridging the gap between them.
For example, in your sum-of-products map, you have two terms, ab and ac. These two overlap; there is no hazard.
An example of a hazard: Consider the latch Q = DE + Q 'E (where 'E = not E). The latch function: when E is high, Q = D. When E goes low, Q maintains its level regardless of subsequent level of D. But there is a hazard: when D and E are high, Q is high and should remain high when E goes low. But both product terms change state, and if the DE term goes low before the Q'E term goes high, Q becomes 0. (The K-map will show that the terms do not overlap.)
To eliminate the hazard, add the additional, overlapping term DQ to the sum: Q = DE + Q'E + DQ
No, you can never make sure there are no glitches in your design.
In real designs you find that glitches are caused by delays in the logic because one signal path is longer then another. You must make sure that glitches have gone by the time the signal get to the input of a register and the clock arrives. (Set-up time).
I had never heard of, lets call them "Karnaugh" glitches, and honestly to me they seem a typical 'academic' problem. To reduce a Karnaugh map you end up with AND, OR, gates and inverters. Each of these will have a different delay and then you have to add to that the delay of the output (The number of inputs to drive, plus the capacitance of the wires leading to them). Pass all that through a cloud of logic and you quickly realize that nobody can avoid glitches.
Karnaugh map are necessary for a fundamental understanding how logic works, However in in HDL design you do not make Karnaugh maps and use them to design an optimise circuits. Just this morning I answered a related question here where somebody did a lot of work to find the logic equations. You can see that the HDL solution ignores all that. What the actual logic is we leave to the synthesis tool.
I admit for the few people who design the synthesis tool, they had better know how to write programs to optimize logic. Also you might have to do some special work e.g. design a 6.4GHz serial I/O circuit. That is where the fundamentals of logic design come back and you have to calculate every gate. But 99.9% of the HDL designers know about Karnaugh maps, they can solve them (or they should be able to solve them...:-) , but they do not use them.
I can't quite reproduce it with this problem (I'm a bit foggy in the head from jet lag though)... But I definitely learned that you can remove glitches by making sure that every selected grouping in your karnaugh map overlaps another group, in such a way as to form a continuous chain. If this can't be done (for example with a simple xor) then you fundamentally have a circuit that will always glitch (like xor).
I have even used this at work, adding a redundant term to a mux that was implemented in and/or gates instead of a mux primitive (I'm an asic designer) to prevent glitches. The problem with the glitchy mux was that if select changes, even if both inputs are 1, the output can glitch 0. An extra term fixed it. Try it with a simple mux to see the difference.
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11125 | https://math.tools/calculator/unit/length/%C2%B5m-to-cm | Micrometer to Centimeter Converter
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11126 | https://www.dictionary.com/e/word-of-the-day/2016/09/19/cosset | Start each day with the Word of the Day in your inbox!
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cosset
verb (used with object)
to treat as a pet; pamper; coddle.
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The verb cosset “to treat as a pet, pamper, coddle” is a derivative verb use of the noun cosset “a lamb raised as a pet.” The noun cosset has no certain etymology, but it has been suggested that it comes from Middle English cot-sēte “cottage dweller, cottager,” from Old English cot-sǣta. Cot-sēte, a rare enough word, is last recorded about 1400. Modern cosset (in the sense “pet lamb”) first appears in English in The Shepheardes Calender (1579) by Edmund Spenser, who uses words and spellings that were already archaic in his time.
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It occurred to me, as I took my bag over, that it might be airline policy to comfort those who were going home for reasons such as mine with an upgrade, to cosset them through the night with quiet sympathy and an extra blanket or something.
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noun
a person noted for courageous acts or nobility of character.
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The English singular noun hero is formed from the plural heroes, which comes from Latin hērōes, the plural of hērōs “(mythical) hero.” Hērōs comes from Greek hḗrōs (plural hḗrōes) “hero,” a very ancient word that meant many things to the Greeks. A compound noun trisērohei, literally “Thrice Hero,” possibly the name of a deity “Clan Ancestor (?),” appears on a Linear B tablet from Pylos, dating to the 13th century b.c. In the Iliad, hḗrōs means “warrior,” and often little more than “man,” and not a semidivine being. In later Greek, hḗrōs was a semidivine being with his own cult, usually local, the only exception being Hercules (Heracles). (Greek Hērākléēs, also spelled Hērāklês, means “Glory of Hera.” Hḗrā is the Greek feminine form of hḗrōs; she is a daughter of Cronus and sister and wife of Zeus. Her name occurs next to the name of Zeus on the same Mycenaean Greek text, which makes likely the assumption that Hera was already honored as the consort of Zeus.) Unfortunately, hḗrōs and its derivative noun Hḗrā, like 60 percent of Greek vocabulary, have no satisfactory etymology. The various etymologies proposed suffer from various degrees of improbability. Hero entered English in the 16th century.
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Amid all the bleak news about the coronavirus pandemic, it’s important to remember that there are so many heroes in America right now.
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force majeure
noun
Law.
an unexpected and disruptive event that may operate to excuse a party from a contract.
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Force majeure, “superior force,” is a legal term in commercial and contract law for an unexpected, disruptive event that may excuse one party or both parties from a contract. The force majeure may be limited to what some jurisdictions term “acts of God,” such as floods, hurricanes, earthquakes, tornadoes, etc. The force majeure may also be broader in scope, including manmade events such as strikes, riots, crime, or other social unrest. Force majeure is unnaturalized in English; even the pronunciation of majeure is at least partly Frenchified. Force comes from Old French force, from Vulgar Latin fortia, a singular feminine noun use of the neuter plural adjective fortia “strong, robust (things),” from the adjective fortis, forte. Many Latin neuter plural nouns and adjectives, which end in –a, become in the Romance languages feminine collective singular nouns, also ending in –a: for instance, the Latin neuter plural gaudia “joys, delights” (singular gaudium) becomes joie in French and gioia in Italian, both feminine singular nouns. Majeure is the normal French development of Latin major– (the inflectional stem of major, majus “greater”). Force majeure first appears in print in A digest of the civil laws now in force in the territory of Orleans…. (1803)–all of the texts, however, are in French. The first appearance of force majeure in English is in Questions and answers on law: Alphabetically arranged, with references to the most approved authorities, Volume 2 (1841).
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What’s more, decisions about whether coronavirus qualifies as a force majeure event will affect entire supply chains, causing a ripple-down effect—one broken obligation, or invocation of the clause, can domino into many others down the line.
All tickets have a force majeure clause, which might get organizers off the hook of paying refunds if the coronavirus is deemed to be “beyond Tokyo 2020’s reasonable control.”
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11127 | https://www.emra.org/emresident/article/pediatric-adrenal-insufficiency | EMRA
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Pediatric Emergency Management of Adrenal Insufficiency and Adrenal Crisis
2/12/2022 Amal Malik , Elizabeth Quaal Hines, MD
Adrenal insufficiency (AI) is a rare diagnosis in childhood and adolescence; however, it can become a major cause of morbidity and mortality in pediatric patients, as 1 in 200 cases of adrenal crisis leads to death.1
AI often presents with nonspecific, variable findings, but early recognition and treatment of these signs and symptoms can prevent detrimental outcomes. If AI goes unrecognized or treatment is delayed, children may progress to significant cardiovascular compromise and collapse.2 In the emergency department, a child may present in adrenal crisis as the initial presentation of disease or a child with a known diagnosis of primary or secondary AI may present requiring resuscitation.
In North America, the most common cause of AI is glucocorticoid withdrawal in patients on prolonged steroid therapy. Therefore, it is extremely important for primary care and emergency providers alike to recognize and take action when the condition arises.
Etiology AI is classified as either primary (adrenal gland dysfunction) or secondary (central, hypothalamus or pituitary gland dysfunction) and includes both congenital and acquired etiologies. Causes of adrenal insufficiency in children are extensive and include enzyme deficiencies in steroidogenesis (Fig 1), metabolic and genetic disorders, autoimmune pathology, hemorrhage or infarction of the adrenal gland secondary to trauma, infection or anticoagulation, or possible drug side effects (ketoconazole, medroxyprogesterone, etomidate, rifampin, phenytoin, barbiturates).3
Congenital adrenal hyperplasia (CAH) due to 21-hydroxylase deficiency is the most common cause of primary adrenal insufficiency with an incidence of 1:10,000 to 1:15,000 people in the United States and Europe.3 All newborns in the United States are screened for 21-hydroxylase deficiency that could lead to classical CAH, a subtype of CAH that can include salt wasting and virilization, especially in female infants. Due to the absence of the 21-hydroxylase enzyme, there is a lack of cortisol, and in the case of salt-wasting CAH, a loss of aldosterone. In severe and untreated forms, this can lead to dehydration, hypotension, and shock.
Additionally, a diversion in the steroidogenesis pathway will result in androgen excess. Clinically, this can present and be diagnosed in female newborns with ambiguous genitalia at birth, whereas males will often present later at 2-3 weeks of age in a salt-wasting crisis.
The most common cause of secondary (central) AI is iatrogenic suppression of the HPA axis following prolonged use of oral glucocorticoid therapy. Long-term glucocorticoid therapy is used to treat numerous pediatric illnesses including autoimmune disease, hematological and oncological disease, inflammatory bowel disease, hematopoietic, and solid organ transplants, to name a few. Chronic glucocorticoid therapy causes suppression of ACTH, which subsequently leads to atrophy of the zona fasciculata of the adrenal cortex, the area of the adrenal gland responsible for the secretion of endogenous glucocorticoids. The duration and dosage of therapy can make a difference as well; however, there are few published studies on the duration of HPA axis suppression after glucocorticoid treatment in children, ranging from 5 days to 9 months.4 This wide variation in the timing of recovery after a course of prolonged glucocorticoid use should make providers aware of the circumstances in which AI can occur.
Diagnosis and Workup As with any patient, paying attention to past medical history, birth history, and medications can be crucial to swift intervention for a deadly, yet oft-overlooked disease process. Adrenal crisis can be tricky to diagnose due to the nonspecific constellation of symptoms that it can present as. Additionally, the onset may be gradual. Triggers may include acute illness, physiological stress, injury, induction of anesthesia, or surgery.5
Clinically, pediatric patients may present with hypotension, tachycardia, fatigue, dizziness, nausea, vomiting, diarrhea, abdominal pain, diaphoresis, or seizures. When considering your initial differential, it is easy to assume these findings could be related to infection, other metabolic syndromes, ingestion, anaphylaxis, or a surgical process.
Initial workup should include obtaining accurate vitals, a fingerstick glucose, and a basic set of electrolytes, as well as a CBC.
In acute AI, hyponatremia is the most common lab finding. Hyperkalemia is seen in primary, but not secondary AI, and can also be accompanied by hypercalcemia and metabolic acidosis.6 Hypoglycemia is more prevalent in neonates and infants in all types of AI. Lab work may also show normocytic anemia, lymphocytosis and eosinophilia. Initial screening for primary AI can be done with a low cortisol level, <140 nmol/L (5 mcg/dL), and a plasma ACTH level greater than two-fold the upper limit of normal for the reference interval. Additionally, according to clinical practice guidelines, the gold standard of diagnostic testing with the cosyntropin test can be done to rule out primary AI in patients with unexplained symptoms as described. However, do not delay the initiation of treatment while awaiting results for cosyntropin testing. This can also be done as a confirmatory test after treatment and patient stabilization.6
Treatment in the EDAs always, first assess your patient’s ABCs, and establish IV access as soon as possible, with lab workup as discussed. However, if the blood draw is proving to be difficult, do not delay treatment.
The initial stress dose of 100 mg/2mL hydrocortisone sodium succinate should 50 mg/m^2 or if BSA is not available, can also be based on the child’s age: children ≤3 years: 25 mg; >3 and <12 years: 50 mg; and older children and adolescents ≥12 years: 100 mg as an initial stress dose. This should be followed by 50–100 mg/m2/day divided into 4 doses given every 6 hours, or as a continuous IV infusion.7,8 If hydrocortisone is not available, you can use prednisolone as an alternative.
Prompt fluid resuscitation for hypovolemic shock should be administered with 20 ml/kg boluses of isotonic fluid such as 0.9% normal saline. Additionally, hypoglycemia should be treated with an initial bolus 2-4 mL/kg D25W infused slowly at 2-3 mL/min or 5-10 ml/kg D10W IV and repeated as necessary. Start 1.5-2x maintenance fluids with D5NS.
For hyperkalemia, always be sure to obtain an EKG to check for peaked T-waves. These can progress to a prolonged PR interval or QRS duration. Usually, hyperkalemia improves with steroid and fluid therapy and rarely requires administration of insulin and glucose. Be sure to monitor and treat any other electrolyte abnormalities.
When treating an adrenal crisis, don’t forget to consider the trigger. If infection or sepsis is suspected, make sure to add on systemic antibiotics accordingly.
Once the patient is stabilized, pediatric endocrinology should then be contacted and the patient should be admitted to ICU level care with continuous hemodynamic monitoring.
Children with known AI should be on maintenance glucocorticoid replacement therapy managed by their pediatrician or endocrinologist in order to help prevent acute AI. Families should also be provided injectable intramuscular hydrocortisone sodium succinate to use at home in the event of stress/illness with vomiting or altered mental status and instructions on when to seek emergent medical care. It is recommended that primary providers also give patients an ED letter, steroid card with appropriate stress dosing, or medical alert bracelet in order to inform EMS or ED providers about their diagnosis and how to quickly initiate treatment for adrenal crisis. These steps can greatly reduce morbidity and mortality.
Conclusion Adrenal insufficiency in the pediatric population may stem from a myriad of causes; however, what is of utmost importance is to consider acute and life-threatening adrenal crisis. This requires timely recognition and initiation of appropriate medical management with steroid therapy in order to avoid further decompensation. ED providers should be aware of this diagnosis and have a high index of suspicion for AI in critically ill patients in the setting of electrolyte abnormalities.
References
Miller BS, Spencer SP, Geffner ME, et al. Emergency management of adrenal insufficiency in children: advocating for treatment options in outpatient and field settings. Journal of Investigative Medicine. 2019;68(1):16-25. doi:10.1136/jim-2019-000999
Shulman DI, Palmert MR, Kemp SF. Adrenal Insufficiency: Still a Cause of Morbidity and Death in Childhood. PEDIATRICS. 2007;119(2):e484-e494.
Speiser PW. Congenital Adrenal Hyperplasia. NORD (National Organization for Rare Disorders). Published June 26, 2018. Accessed April 5, 2021.
Mendoza-Cruz AC, Wargon O, Adams S, Tran H, Verge CF. Hypothalamic-Pituitary-Adrenal Axis Recovery Following Prolonged Prednisolone Therapy in Infants. J Clin Endocrinol Metab. 2013;98(12):E1936-E1940.
Bornstein SR, Allolio B, Arlt W, et al. Diagnosis and Treatment of Primary Adrenal Insufficiency: An Endocrine Society Clinical Practice Guideline. J Clin Endocrinol Metab. 2016;101(2):364-389.
Cortet C, Barat P, Zenaty D, Guignat L, Chanson P. Group 5: Acute adrenal insufficiency in adults and pediatric patients. Annales d'Endocrinologie. 2017;78(6):535-543.
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11128 | https://math.stackexchange.com/questions/3188689/f-is-monotonically-increasing-0-le-f-le-1-and-int-01-fx-x-dx-0 | analysis - $f$ is monotonically increasing, $0 \le f \le 1$ and $\int_0^1 (f(x) - x) dx = 0$ then $\int_0^1|f(x)-x|dx \le \frac{1}{2}$. - Mathematics Stack Exchange
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f f is monotonically increasing, 0≤f≤1 0≤f≤1 and ∫1 0(f(x)−x)d x=0∫0 1(f(x)−x)d x=0 then ∫1 0|f(x)−x|d x≤1 2∫0 1|f(x)−x|d x≤1 2.
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f(x)f(x) is monotonically increasing in [0,1][0,1], 0≤f≤1 0≤f≤1 and ∫1 0(f(x)−x)d x=0∫0 1(f(x)−x)d x=0. Prove that ∫1 0|f(x)−x|d x≤1 2∫0 1|f(x)−x|d x≤1 2.
It's easy if f(x)≥x f(x)≥x in [0,1][0,1]. And even in [a,b][a,b] we have ∫b a|f(x)−x|d x≤(b−a)2 2∫a b|f(x)−x|d x≤(b−a)2 2. But the zero points of f(x)−x f(x)−x may be infinitely many. This is where difficulty exists.
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edited Apr 17, 2019 at 5:40
user173262
asked Apr 15, 2019 at 13:20
XT ChenXT Chen
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2 Look at the region in [0,1]2[0,1]2 between the graphs of f f and y=x y=x. Since f f is increasing then all pieces of that region that lie below y=x y=x can be reflected along y=x y=x and not overlap those pieces that are above y=x y=x. The area of the resulting figure is the integral with the absolute value. At the same time it is contained inside the top triangle in which y=x y=x divides [0,1]2[0,1]2, which has area 1/2 1/2.user647486 –user647486 2019-04-15 13:31:26 +00:00 Commented Apr 15, 2019 at 13:31
@user647486 Can it be turned to an analytical language? Cuz this function need not to be continuous and its area may not be defined.XT Chen –XT Chen 2019-04-15 13:41:10 +00:00 Commented Apr 15, 2019 at 13:41
It is monotonic, it can only have countably many jump discontinuities. Therefore, its integral is defined.user647486 –user647486 2019-04-15 13:45:16 +00:00 Commented Apr 15, 2019 at 13:45
@user647486 I know that integral is defined. But it's not so strict to talk about the area.XT Chen –XT Chen 2019-04-15 13:50:15 +00:00 Commented Apr 15, 2019 at 13:50
Area is, by definition, the integral. Anywhere where you see the word 'area' and don't like it, replace it with the word integral.user647486 –user647486 2019-04-15 13:52:23 +00:00 Commented Apr 15, 2019 at 13:52
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Here we present a bit different, calculus-themed approach. In this answer, we will assume that f:[0,1]→[0,1]f:[0,1]→[0,1] is monotone-increasing. We also write I(f)=∫1 0|f(x)−x|d x I(f)=∫0 1|f(x)−x|d x for brevity.
Step 1 - Proof under extra assumptions. Assume further that f f is piecewise-smooth, f(0)=0 f(0)=0, and f(1)=1 f(1)=1. Then by the formula ∫|x|d x=1 2 x|x|+C∫|x|d x=1 2 x|x|+C, we have
∫1 0|f(x)−x|(f′(x)−1)d x=[1 2|f(x)−x|(f(x)−x)]1 0=0.∫0 1|f(x)−x|(f′(x)−1)d x=[1 2|f(x)−x|(f(x)−x)]0 1=0.
In particular,
I(f)=1 2∫1 0|f(x)−x|(f′(x)+1)d x.I(f)=1 2∫0 1|f(x)−x|(f′(x)+1)d x.
Now pick α∈[0,1]α∈[0,1] so that f(α)+α=1 f(α)+α=1. (This is possible since x↦f(x)+x x↦f(x)+x increases from 0 0 to 2 2. Then by triangle inequality,
∫α 0|f(x)−x|(f′(x)+1)d x≤∫α 0(f(x)+x)(f′(x)+1)d x=1 2.∫0 α|f(x)−x|(f′(x)+1)d x≤∫0 α(f(x)+x)(f′(x)+1)d x=1 2.
Similarly, by writing |f(x)−x|=|(1−f(x))−(1−x)|≤(1−f(x))+(1−x)|f(x)−x|=|(1−f(x))−(1−x)|≤(1−f(x))+(1−x), we get
∫1 α|f(x)−x|(f′(x)+1)d x≤∫1 α(2−f(x)−x)(f′(x)+1)d x=1 2.∫α 1|f(x)−x|(f′(x)+1)d x≤∫α 1(2−f(x)−x)(f′(x)+1)d x=1 2.
Therefore ∫1 0|f(x)−x|(f′(x)+1)d x≤1∫0 1|f(x)−x|(f′(x)+1)d x≤1, which in turn implies I(f)≤1 2 I(f)≤1 2 as required.
Remark. Let γ(t)=(f(t)+t,f(t)−t)γ(t)=(f(t)+t,f(t)−t). Then ∫1 0|f(t)−t|(f′(t)+1)d t=∫γ|y|d x∫0 1|f(t)−t|(f′(t)+1)d t=∫γ|y|d x computes the area between the path γ γ and the horizontal axis. Note that γ γ is essentially the −45∘−45∘-rotation of the graph y=f(x)y=f(x) up to scaling.
Then the above bounds immediately follow from the fact that the graph of γ γ defines a function on [0,2][0,2] which is squeezed between lines y=±x y=±x and y=±(2−x)y=±(2−x).
Step 2 - General case. For the general case, let f n f n be the linear interpolation of the points
(0,0),(1 n,f(1 n)),⋯,(n−1 n,f(n−1 n)),(1,1).(0,0),(1 n,f(1 n)),⋯,(n−1 n,f(n−1 n)),(1,1).
Then by monotonicity,
|I(f n)−I(f)|≤∫1 0|f n(x)−f(x)|d x=∑k=1 n∫k n k−1 n|f n(x)−f(x)|d x≤1 n([f(1 n)−0]+∑k=2 n−1[f(k n)−f(k−1 n)]+[1−f(n−1 n)])=1 n,|I(f n)−I(f)|≤∫0 1|f n(x)−f(x)|d x=∑k=1 n∫k−1 n k n|f n(x)−f(x)|d x≤1 n([f(1 n)−0]+∑k=2 n−1[f(k n)−f(k−1 n)]+[1−f(n−1 n)])=1 n,
hence I(f n)→I(f)I(f n)→I(f) as n→∞n→∞ and the desired inequality I(f)≤1 2 I(f)≤1 2 follows from the previous step.
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edited Apr 17, 2019 at 11:20
answered Apr 15, 2019 at 18:09
Sangchul LeeSangchul Lee
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Do you know if the bound 1/2 1/2 is sharp? I am asking because the same question, only with an upper bound of 1/4 1/4, has been asked here: math.stackexchange.com/q/4436677/42969.Martin R –Martin R 2022-04-26 12:23:04 +00:00 Commented Apr 26, 2022 at 12:23
The upper bound 1/4 1/4 is also claimed in this question math.stackexchange.com/q/3190661/42969, but without proof so far.Martin R –Martin R 2022-04-26 12:37:13 +00:00 Commented Apr 26, 2022 at 12:37
@MartinR, I believe that the bound is by no means sharp, not properly utilizing the condition that ∫1 0(f(x)−x)d x=0∫0 1(f(x)−x)d x=0. Pictures seem to suggest that 1 4 1 4 is indeed the sharp upper bound, attained by f(x)=1 2 f(x)=1 2. I will think about it.Sangchul Lee –Sangchul Lee 2022-04-26 18:01:11 +00:00 Commented Apr 26, 2022 at 18:01
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In case you have some familiarity with measure theory:
Take an increasing sequence of simple functions f k→f f k→f converging pointwise. By the monotone convergence theorem we have lim k→∞∫1 0 f k(x)d x=∫1 0 f(x)d x lim k→∞∫0 1 f k(x)d x=∫0 1 f(x)d x. So it suffices to show the result for simple functions. The functions f k(x)−x f k(x)−x will have only finitely many zeroes, so you know how to do this.
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answered Apr 15, 2019 at 14:22
Lisanne TaamsLisanne Taams
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1 (+1) Just nitpicking, monotone convergence theorem can be replaced by a more elementary argument if needed, thanks to the monotonicity of f f. Indeed, we can realize f k f k's as f k(x)=∑i=1 k f(i−1 k)1[i−1 k<x≤i k]f k(x)=∑i=1 k f(i−1 k)1[i−1 k<x≤i k] so that ∣∣∣∫1 0|f k(x)−x|d x−∫1 0|f(x)−x|d x∣∣∣≤∫1 0|f k(x)−f(x)|d x≤1 k.|∫0 1|f k(x)−x|d x−∫0 1|f(x)−x|d x|≤∫0 1|f k(x)−f(x)|d x≤1 k. Sangchul Lee –Sangchul Lee 2019-04-15 21:21:51 +00:00 Commented Apr 15, 2019 at 21:21
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Here is a sketch of a different argument. Let K K be the set of monotone increasing functions f:[0,1]→[0,1]f:[0,1]→[0,1] that are right continuous and have limits on the left (the càdlàg functions). One may as well restrict the f f of the problem statement to be in K K, as any f f can be modified at its (countably few) discontinuities to make it càdlàg and while preserving all the integrals. Now, K K is convex, and its extreme points are the cumulative distribution functions of the point masses at points a∈[0,1],a∈[0,1], together with the zero function. In particular, the simple functions constant on [0,a)[0,a) and on [a,1][a,1]. (This claim is, in effect, that the point masses and the zero measure are the extreme points of the sub-probability measures on [0,1][0,1].) Now, by a theorem of Dubins (see also), the extreme points of K∩L K∩L, where L L is the set of f f for which ∫1 0 f=1/2∫0 1 f=1/2 are all convex combinations of at most two extreme points of K K. Which is to say, simple functions with at most 2 discontinuities. Since f↦∫1 0|f(x)−x|d x f↦∫0 1|f(x)−x|d x is continuous and convex, the maximum is attained at an extreme point of K∩L K∩L. As the comments indicate, ∫1 0|f(x)−x|d x∫0 1|f(x)−x|d x is equal to the area of the union of finitely many triangles, the sum of whose heights is ≤1≤1 and the sum of whose lengths is ≤1≤1 and hence have total area ≤1/2≤1/2.
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edited Apr 15, 2019 at 16:02
answered Apr 15, 2019 at 15:40
kimchi loverkimchi lover
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Whoops! I'll edit it.kimchi lover –kimchi lover 2019-04-15 16:01:41 +00:00 Commented Apr 15, 2019 at 16:01
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4Prove that ∫1 0|f(x)−x|d x≤1/4∫0 1|f(x)−x|d x≤1/4 if f f is increasing on [0,1][0,1], 0≤f(x)≤1 0≤f(x)≤1 and ∫1 0(f(x)−x)d x=0∫0 1(f(x)−x)d x=0.
1f,g∈[0,1]×[0,1]f,g∈[0,1]×[0,1], ∫f−g d x=0∫f−g d x=0 and are monotonically increasing, then ∫|f−g|d x≤1 2∫|f−g|d x≤1 2
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1f,g∈[0,1]×[0,1]f,g∈[0,1]×[0,1], ∫f−g d x=0∫f−g d x=0 and are monotonically increasing, then ∫|f−g|d x≤1 2∫|f−g|d x≤1 2
0Fixed-point of a monotonically increasing function on an ordered set
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11129 | https://www.geeksforgeeks.org/dsa/program-for-point-of-intersection-of-two-lines/ | Program for Point of Intersection of Two Lines
Last Updated :
16 Jun, 2022
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Given points A and B corresponding to line AB and points P and Q corresponding to line PQ, find the point of intersection of these lines. The points are given in 2D Plane with their X and Y Coordinates. Examples:
```
Input : A = (1, 1), B = (4, 4)
C = (1, 8), D = (2, 4)
Output : The intersection of the given lines
AB and CD is: (2.4, 2.4)
Input : A = (0, 1), B = (0, 4)
C = (1, 8), D = (1, 4)
Output : The given lines AB and CD are parallel.
```
First of all, let us assume that we have two points (x1, y1) and (x2, y2). Now, we find the equation of line formed by these points. Let the given lines be :
a1x + b1y = c1
a2x + b2y = c2
We have to now solve these 2 equations to find the point of intersection. To solve, we multiply 1. by b2 and 2 by b1 This gives us, a1b2x + b1b2y = c1b2 a2b1x + b2b1y = c2b1 Subtracting these we get, (a1b2 - a2b1) x = c1b2 - c2b1 This gives us the value of x. Similarly, we can find the value of y. (x, y) gives us the point of intersection. Note: This gives the point of intersection of two lines, but if we are given line segments instead of lines, we have to also recheck that the point so computed actually lies on both the line segments. If the line segment is specified by points (x1, y1) and (x2, y2), then to check if (x, y) is on the segment we have to just check that
min (x1, x2) <= x <= max (x1, x2)
min (y1, y2) <= y <= max (y1, y2)
The pseudo code for the above implementation:
determinant = a1 b2 - a2 b1
if (determinant == 0)
{
// Lines are parallel
}
else
{
x = (c1b2 - c2b1)/determinant
y = (a1c2 - a2c1)/determinant
}
These can be derived by first getting the slope directly and then finding the intercept of the line.
C++
````
// C++ Implementation. To find the point of
// intersection of two lines
include
using namespace std;
// This pair is used to store the X and Y
// coordinates of a point respectively
define pdd pair
// Function used to display X and Y coordinates
// of a point
void displayPoint(pdd P)
{
cout << "(" << P.first << ", " << P.second
<< ")" << endl;
}
pdd lineLineIntersection(pdd A, pdd B, pdd C, pdd D)
{
// Line AB represented as a1x + b1y = c1
double a1 = B.second - A.second;
double b1 = A.first - B.first;
double c1 = a1(A.first) + b1(A.second);
// Line CD represented as a2x + b2y = c2
double a2 = D.second - C.second;
double b2 = C.first - D.first;
double c2 = a2(C.first)+ b2(C.second);
double determinant = a1b2 - a2b1;
if (determinant == 0)
{
// The lines are parallel. This is simplified
// by returning a pair of FLT_MAX
return make_pair(FLT_MAX, FLT_MAX);
}
else
{
double x = (b2c1 - b1c2)/determinant;
double y = (a1c2 - a2c1)/determinant;
return make_pair(x, y);
}
}
// Driver code
int main()
{
pdd A = make_pair(1, 1);
pdd B = make_pair(4, 4);
pdd C = make_pair(1, 8);
pdd D = make_pair(2, 4);
pdd intersection = lineLineIntersection(A, B, C, D);
if (intersection.first == FLT_MAX &&
intersection.second==FLT_MAX)
{
cout << "The given lines AB and CD are parallel.\n";
}
else
{
// NOTE: Further check can be applied in case
// of line segments. Here, we have considered AB
// and CD as lines
cout << "The intersection of the given lines AB "
"and CD is: ";
displayPoint(intersection);
}
return 0;
}
````
// C++ Implementation. To find the point of
// C++ Implementation. To find the point of
// intersection of two lines
// intersection of two lines
```
include #include
```
using namespace std; using namespace std
// This pair is used to store the X and Y
// This pair is used to store the X and Y
// coordinates of a point respectively
// coordinates of a point respectively
```
define pdd pair#define pdd pair
```
// Function used to display X and Y coordinates
// Function used to display X and Y coordinates
// of a point
// of a point
void displayPoint(pdd P) void displayPoint pdd P
{
cout << "(" << P.first << ", " << P.second cout<<"("<< P first<<", "<< P second
<< ")" << endl;<<")"<< endl
}
pdd lineLineIntersection(pdd A, pdd B, pdd C, pdd D) pdd lineLineIntersection pdd A pdd B pdd C pdd D
{
// Line AB represented as a1x + b1y = c1
// Line AB represented as a1x + b1y = c1
double a1 = B.second - A.second; double a1 = B second - A second
double b1 = A.first - B.first; double b1 = A first - B first
double c1 = a1(A.first) + b1(A.second); double c1 = a1 A first + b1 A second
// Line CD represented as a2x + b2y = c2
// Line CD represented as a2x + b2y = c2
double a2 = D.second - C.second; double a2 = D second - C second
double b2 = C.first - D.first; double b2 = C first - D first
double c2 = a2(C.first)+ b2(C.second); double c2 = a2 C first + b2 C second
double determinant = a1b2 - a2b1; double determinant = a1 b2 - a2 b1
if (determinant == 0) if determinant == 0
{
// The lines are parallel. This is simplified
// The lines are parallel. This is simplified
// by returning a pair of FLT_MAX
// by returning a pair of FLT_MAX
return make_pair(FLT_MAX, FLT_MAX); return make_pair FLT_MAX FLT_MAX
}
else else
{
double x = (b2c1 - b1c2)/determinant; double x = b2 c1 - b1 c2/ determinant
double y = (a1c2 - a2c1)/determinant; double y = a1 c2 - a2 c1/ determinant
return make_pair(x, y); return make_pair x y
}
}
// Driver code
// Driver code
int main() int main
{
pdd A = make_pair(1, 1); pdd A = make_pair 1 1
pdd B = make_pair(4, 4); pdd B = make_pair 4 4
pdd C = make_pair(1, 8); pdd C = make_pair 1 8
pdd D = make_pair(2, 4); pdd D = make_pair 2 4
pdd intersection = lineLineIntersection(A, B, C, D); pdd intersection = lineLineIntersection A B C D
if (intersection.first == FLT_MAX && if intersection first == FLT_MAX&&
intersection.second==FLT_MAX) intersection second == FLT_MAX
{
cout << "The given lines AB and CD are parallel.\n"; cout<<"The given lines AB and CD are parallel.\n"
}
else else
{
// NOTE: Further check can be applied in case
// NOTE: Further check can be applied in case
// of line segments. Here, we have considered AB
// of line segments. Here, we have considered AB
// and CD as lines
// and CD as lines
cout << "The intersection of the given lines AB " cout<< "The intersection of the given lines AB "
"and CD is: ";"and CD is: "
displayPoint(intersection); displayPoint intersection
}
return 0; return 0
}
Java
````
// Java Implementation. To find the point of
// intersection of two lines
// Class used to used to store the X and Y
// coordinates of a point respectively
class Point
{
double x,y;
public Point(double x, double y)
{
this.x = x;
this.y = y;
}
// Method used to display X and Y coordinates
// of a point
static void displayPoint(Point p)
{
System.out.println("(" + p.x + ", " + p.y + ")");
}
}
class Test
{
static Point lineLineIntersection(Point A, Point B, Point C, Point D)
{
// Line AB represented as a1x + b1y = c1
double a1 = B.y - A.y;
double b1 = A.x - B.x;
double c1 = a1(A.x) + b1(A.y);
// Line CD represented as a2x + b2y = c2
double a2 = D.y - C.y;
double b2 = C.x - D.x;
double c2 = a2(C.x)+ b2(C.y);
double determinant = a1b2 - a2b1;
if (determinant == 0)
{
// The lines are parallel. This is simplified
// by returning a pair of FLT_MAX
return new Point(Double.MAX_VALUE, Double.MAX_VALUE);
}
else
{
double x = (b2c1 - b1c2)/determinant;
double y = (a1c2 - a2c1)/determinant;
return new Point(x, y);
}
}
// Driver method
public static void main(String args[])
{
Point A = new Point(1, 1);
Point B = new Point(4, 4);
Point C = new Point(1, 8);
Point D = new Point(2, 4);
Point intersection = lineLineIntersection(A, B, C, D);
if (intersection.x == Double.MAX_VALUE &&
intersection.y == Double.MAX_VALUE)
{
System.out.println("The given lines AB and CD are parallel.");
}
else
{
// NOTE: Further check can be applied in case
// of line segments. Here, we have considered AB
// and CD as lines
System.out.print("The intersection of the given lines AB " +
"and CD is: ");
Point.displayPoint(intersection);
}
}
}
````
Python3
````
Python program to find the point of
intersection of two lines
Class used to used to store the X and Y
coordinates of a point respectively
class Point:
def init(self, x, y):
self.x = x
self.y = y
# Method used to display X and Y coordinates
# of a point
def displayPoint(self, p):
print(f"({p.x}, {p.y})")
def lineLineIntersection(A, B, C, D):
# Line AB represented as a1x + b1y = c1
a1 = B.y - A.y
b1 = A.x - B.x
c1 = a1(A.x) + b1(A.y)
# Line CD represented as a2x + b2y = c2
a2 = D.y - C.y
b2 = C.x - D.x
c2 = a2(C.x) + b2(C.y)
determinant = a1b2 - a2b1
if (determinant == 0):
# The lines are parallel. This is simplified
# by returning a pair of FLT_MAX
return Point(109, 109)
else:
x = (b2c1 - b1c2)/determinant
y = (a1c2 - a2c1)/determinant
return Point(x, y)
Driver code
A = Point(1, 1)
B = Point(4, 4)
C = Point(1, 8)
D = Point(2, 4)
intersection = lineLineIntersection(A, B, C, D)
if (intersection.x == 109 and intersection.y == 109):
print("The given lines AB and CD are parallel.")
else:
# NOTE: Further check can be applied in case
# of line segments. Here, we have considered AB
# and CD as lines
print("The intersection of the given lines AB " + "and CD is: ")
intersection.displayPoint(intersection)
This code is contributed by Saurabh Jaiswal
````
C#
````
using System;
// C# Implementation. To find the point of
// intersection of two lines
// Class used to used to store the X and Y
// coordinates of a point respectively
public class Point
{
public double x, y;
public Point(double x, double y)
{
this.x = x;
this.y = y;
}
// Method used to display X and Y coordinates
// of a point
public static void displayPoint(Point p)
{
Console.WriteLine("(" + p.x + ", " + p.y + ")");
}
}
public class Test
{
public static Point lineLineIntersection(Point A, Point B, Point C, Point D)
{
// Line AB represented as a1x + b1y = c1
double a1 = B.y - A.y;
double b1 = A.x - B.x;
double c1 = a1 (A.x) + b1 (A.y);
// Line CD represented as a2x + b2y = c2
double a2 = D.y - C.y;
double b2 = C.x - D.x;
double c2 = a2 (C.x) + b2 (C.y);
double determinant = a1 b2 - a2 b1;
if (determinant == 0)
{
// The lines are parallel. This is simplified
// by returning a pair of FLT_MAX
return new Point(double.MaxValue, double.MaxValue);
}
else
{
double x = (b2 c1 - b1 c2) / determinant;
double y = (a1 c2 - a2 c1) / determinant;
return new Point(x, y);
}
}
// Driver method
public static void Main(string[] args)
{
Point A = new Point(1, 1);
Point B = new Point(4, 4);
Point C = new Point(1, 8);
Point D = new Point(2, 4);
Point intersection = lineLineIntersection(A, B, C, D);
if (intersection.x == double.MaxValue && intersection.y == double.MaxValue)
{
Console.WriteLine("The given lines AB and CD are parallel.");
}
else
{
// NOTE: Further check can be applied in case
// of line segments. Here, we have considered AB
// and CD as lines
Console.Write("The intersection of the given lines AB " + "and CD is: ");
Point.displayPoint(intersection);
}
}
}
// This code is contributed by Shrikant13
````
JavaScript
````
// Javascript program to find the point of
// intersection of two lines
// Class used to used to store the X and Y
// coordinates of a point respectively
class Point
{
constructor(x, y)
{
this.x = x;
this.y = y;
}
// Method used to display X and Y coordinates
// of a point
displayPoint(p){
document.write("(" + p.x + ", " + p.y + ")");
}
}
function lineLineIntersection(A,B,C,D){
// Line AB represented as a1x + b1y = c1
var a1 = B.y - A.y;
var b1 = A.x - B.x;
var c1 = a1(A.x) + b1(A.y);
// Line CD represented as a2x + b2y = c2
var a2 = D.y - C.y;
var b2 = C.x - D.x;
var c2 = a2(C.x)+ b2(C.y);
var determinant = a1b2 - a2b1;
if (determinant == 0)
{
// The lines are parallel. This is simplified
// by returning a pair of FLT_MAX
return new Point(Number.MAX_VALUE, Number.MAX_VALUE);
}
else
{
var x = (b2c1 - b1c2)/determinant;
var y = (a1c2 - a2c1)/determinant;
return new Point(x, y);
}
}
// Driver code
let A = new Point(1, 1);
let B = new Point(4, 4);
let C = new Point(1, 8);
let D = new Point(2, 4);
var intersection = lineLineIntersection(A, B, C, D);
if (intersection.x == Number.MAX_VALUE && intersection.y == Number.MAX_VALUE){
document.write("The given lines AB and CD are parallel.");
}else{
// NOTE: Further check can be applied in case
// of line segments. Here, we have considered AB
// and CD as lines
document.write("The intersection of the given lines AB " + "and CD is: ");
intersection.displayPoint(intersection);
}
// This code is contributed by shruti456rawal
````
Output:
The intersection of the given lines AB and
CD is: (2.4, 2.4)
Time Complexity: O(1)
Auxiliary Space: O(1)
K
kartik
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Article Tags :
Geometric
DSA
Geometric-Lines
Practice Tags :
Geometric
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11130 | https://ps.uci.edu/~cyu/p214A/LectureNotes/Lecture14/lecture14.pdf | LECTURE 14 Quantized Energy Levels We know from quantum mechanics that the solutions to Schroedinger’s equation Hψ = Eψ (1) have quantized energy levels. For example, a particle of mass m in a box with infinitely high walls, has energy levels given by En = n2π2¯ h2 2ma2 (2) where n = 1, 2, 3, .... A harmonic oscillator is another example. The energy eigenvalues are En = (n + 1 2)¯ hω (3) where n = 0, 1, 2, 3, .... Notice once again that the energy levels are quantized. In this case they are evenly spaced by an amount ∆E = ¯ hω.
Electromagnetic radiation is also quantized.
Light can be described as waves or as particles called photons. A photon has energy hν where ν is the frequency of the electromagnetic wave. Recall that ω = 2πν and that ν = c/λ where c is the speed of light. Often one speaks in terms of the wavenumber k = 2π/λ. If we make it a vector quantity ⃗ k, then we call it a wavevector. This is related to the momentum by ⃗ p = ¯ h⃗ k and to the frequency by ω = ck. So if the electromagnetic wave has a short wavelength, it has a high frequency and the photon carries a lot of energy. Lots of wiggles means lots of energy. Photons are massless and they travel at the speed of light.
Periodic Boundary Conditions:Counting States So energy is quantized into discrete energy levels. Each energy level is associated with a mode or eigenfunction. We have seen that it is often useful to be able to count the number of modes in a box that have energies between E and E + dE.
Suppose we have a 3 dimensional box whose walls are parallel to the x, y, and z axes with lengths Lx, Ly, and Lz. Thus the volume is V = LxLyLz. We can solve this as a particle in a box problem. Inside the box the potential is zero. The eigenmodes are waves. However, let’s choose boundary conditions such that the solution of Schroedinger’s equation are wavefunctions that are plane waves: Ψ = A exp[i(⃗ k · ⃗ r −ωt)] = ψ(⃗ r) exp(−iωt) (4) This is a propagating wave that is never reflected. So our box can’t have hard walls.
Rather let’s imagine that our box is embedded in an infinite set of similar boxes in each of which the physical situation is exactly the same. In other words, each of these boxes is a repeat of the original box.
To describe this situation, we use periodic boundary conditions which we can write as ψ(x + Lx, y, z) = ψ(x, y, z) ψ(x, y + Ly, z) = ψ(x, y, z) ψ(x, y, z + Lz) = ψ(x, y, z) If we require our traveling wave solution ψ(⃗ r) = exp(i⃗ k · ⃗ r) = exp[i(kxx + kyy + kzz)] (5) to satisfy these boundary conditions, then we must require that kx(x + Lx) = kxx + 2πnx (6) where nx is an integer. We can rewrite this as kx = 2π Lx nx (7) Similarly, ky = 2π Ly ny kz = 2π Lz nz Here the numbers nx, ny, and nz are any set of integers: positive, negative, or zero.
We can use p = ¯ hk and E = p2/2m to deduce that E(nx, ny, nz) = ¯ h2 2m(k2 x + k2 y + k2 z) = 2π2¯ h2 m Ãn2 x L2 x + n2 y L2 y + n2 z L2 z !
(8) 2 Once again we see that the energy levels are quantized. Notice that for any kind of macroscopic volume where Lx, Ly, and Lz are large, the energy levels are very closely spaced.
Now we want to count the number of modes or waves that have wavevectors between ⃗ k = (kx, ky, kz) and ⃗ k + d⃗ k = (kx + dkx, ky + dky, kz + dkz). For given values of ky and kz, it follows from (7) that the number ∆nx of possible integers nx for which kx lies in the range between kx and kx + dkx is equal to ∆nx = Lx 2πdkx (9) We see that if Lx is very large, a lot of states can be in the small interval dkx. The same holds true for dky and dkz. So the number of states that lie between ⃗ k and ⃗ k + d⃗ k is ρd3k = ∆nx∆ny∆nz = µLx 2πdkx ¶ µLy 2πdky ¶ µLz 2πdkz ¶ = LxLyLz (2π)3 dkxdkydkz (10) or ρd3k = V (2π)3d3k (11) where d3k ≡dkxdkydkz is an element of volume in “k space.” Notice that the number of states ρ is independent of ⃗ k and proportional to the volume V under consideration. So the “density of states”, i.e., the number of states per unit volume, lying between ⃗ k and ⃗ k + d⃗ k is d3k/(2π)3 which is a constant independent of the magnitude or shape of the volume V . Note that ρ denotes the number of single particle states.
Using the relation ⃗ p = ¯ h⃗ k, we can also deduce that the number of states ρpd3p in the momentum range between ⃗ p and ⃗ p + d⃗ p is ρpd3p = ρd3k = V (2π)3 d3p ¯ h3 = V d3p h3 (12) where h = 2π¯ h is the ordinary Planck’s constant. Notice that V d3p is the volume of the classical 6 dimensional phase space occupied by a particle in a box of volume V and with momentum between ⃗ p and ⃗ p + d⃗ p. Thus (12) shows that subdivision of this phase space into cells of size h3 yields the correct number of quantum states for the particle. If we compare this to our classical expression V d3p/h3 o, we see that our arbitrary length ho is replaced by Planck’s constant h.
If k–space is isotropic, i.e., the same in every direction, then the number of states in a spherical shell lying between radii k and k + dk is ρkdk = ρd3k = V (2π)3(4πk2dk) = V 2π2k2dk (13) If we are considering photons for which ω = ck, then we can plug k = ω/c into (13) to get the number of states lying between ω and ω + dω.
ρωdω = 2 V 2π2c3ω2dω (14) 3 The factor of 2 comes from the fact that there are 2 photon polarizations. The polariza-tion refers to the direction of the electric field vector ⃗ E in the electromagnetic radiation.
Since ⃗ E must be perpendicular to ⃗ k, there are 2 polarization directions. We will use (14) in deriving blackbody radiation. Sometimes the term “density of states” for photons is used to refer to the number of states per unit volume per unit energy: N(ω) = 2 2π2c3ω2 = 1 π2c3ω2 (15) The density of states is very useful for converting sums into integrals as we shall see.
Recap So let’s recap where we are and what we’ve found. If a system with lots of particles has many–particle states R with energy ER, then the average of some quantity A is given by A = 1 Z X R ARe−βER (16) where the partition function Z Z = X R e−βER (17) This is for the canonical ensemble with fixed temperature T and fixed particle number N. We have seen that ln Z is very useful in finding other quantities. For example, F = −kBT ln Z (18) S = kB(ln Z + βE) (19) E = −∂ln Z ∂β (20) CV = ∂E ∂T ¯ ¯ ¯ ¯ ¯ V (21) p = 1 β ∂ln Z ∂V (22) But the problem is that it is very difficult to solve Schroedinger’s equation to get ER: HψR = ERψR (23) It is much easier to solve Schroedinger’s equation to get single particle energies. So we consider systems (gases) of noninteracting particles. If we know how many particles are in each single particle state, then we just sum over all the particles to get the appropriate average, e.g., the mean energy of the whole system. If we have to treat the particles quantum mechanically because their wavefunctions overlap or because the temperature is low, then we need to pay attention to whether the particles are fermions or bosons.
4 Fermions can have at most one particle in a state while bosons can have umpteen particles in a state. So now the mean energy is given by E = X s εsns (24) where the mean number of particles in state s is given by ns = 1 eβ(εs−µ) ± 1 (25) + is for fermions and −is for bosons. It’s usually not easy to do sums, so it would be nice if we could convert the sum into an integral. That’s why we calculated density of single particle states. Then we can do the conversion: X s → Z ρ(ε)dε (26) or X s → Z ρ(ω)dω (27) So the mean energy becomes E = Z dερ(ε)n(ε)ε (28) where n(ε) = 1 eβ(ε−µ) ± 1 (29) Applications Now let’s go do some examples of this strategy. We will cover the following examples: 1. Monatomic Ideal Gas 2. Black Body Radiation 3. Electron Gas (electrons in a metal) 4. Bose–Einstein Condensation (if time permits) Monatomic Ideal Gas Let’s start with our tried and true example of a monatomic ideal gas in the classical limit of low density or high temperature. We want to calculate the partition function.
We found earlier that Z = ζN N!
(30) where ζ is the partition function for one particle.
ζ = X r e−βεr = X kx,ky,kz exp " −β¯ h2 2m ³ k2 x + k2 y + k2 z ´# (31) 5 where kx = 2π Lx nx ky = 2π Ly ny kz = 2π Lz nz (32) Since the number of allowed k states is very large and since these states are very close together with spacing going as 1/L, we can approximate the sum by an integral. This is where the density of states comes in handy.
X ⃗ k → Z ρd3k = V (2π)3 Z d3k = V (2π)3 Z 4πk2dk = V 2π2 Z k2dk (33) So ζ = V 2π2 Z ∞ 0 exp " −β¯ h2k2 2m # k2dk = V 2π2 √π 4 Ã 2m β¯ h2 !3/2 = V h3 (2πmkBT)3/2 (34) and ln Z = ln ÃζN N!
!
= N ln ζ −N ln N + N = N(ln ζ −ln N + 1) = N µ ln V N −3 2 ln β + 3 2 ln 2πm h2 + 1 ¶ (35) Hence E = −∂ln Z ∂β = 3 2NkBT (36) and S = kB(ln Z + βE) = NkB µ ln V N + 3 2 ln T + σo ¶ (37) where σo ≡3 2 ln 2πmkB h2 + 5 2 (38) This is identical to the result obtained for the purely classical ideal gas except that σo now has a well defined value with ho = h = Planck’s constant and the Gibbs paradox has automatically been taken care of.
Blackbody Radiation Blackbody radiation is a nice example of the statistical mechanics that we have been discussing. A black body is a perfect absorber and absorbs all the radiation incident 6 on it. If its temperature is kept constant, then the amount of power it radiates must equal the amount of power it absorbs. Otherwise it would heat up or cool off. We can imagine the black body being kept inside some kind of closed container which is at the same temperature T. The radiation field inside this enclosure is in equilibrium. In other words there is a gas of photons in thermal equilibrium inside the enclosure. By thermal equilibrium, we mean that the average occupation number ns of the single particle states is given by the Planck distribution that we talked about in lecture 13.
ns = 1 eβεs −1 (39) One can imagine making a histogram by counting the photon energy density in each frequency range from ω to ω + dω. It turns out that this distribution of the energy density of blackbody radiation is a universal curve that depends only on the temperature T. In other words if one plots the distribution of the photon energy density (counting both directions of polarization) as a function of photon (angular) frequency ω, the shape of the curve is universal and the position of the peak is a function only of the temperature.
When we say that the curve is universal, we mean that it doesn’t depend on the size or shape of the box, or what the walls are made of. All that matters is the temperature.
Blackbody radiation is historically important in physics for two reasons. The first is that the measurement of the spectral distribution in the late 1800’s led Planck to come up with the idea of energy quantization. He couldn’t explain the distribution unless he postulated that E = hν. This marked the birth of quantum mechanics. The second reason that blackbody radiation is important is that 3 K black body radiation pervades the universe and is the remnant of the Big Bang. This radiation is in the microwave region.
Let’s calculate the distribution of the mean energy density of blackbody radiation.
Since the size and shape of the box don’t matter, let’s imagine a rectangular box of volume V filled with a gas of photons that are in thermal equilibrium. The box has edges with lengths Lx, Ly, and Lz such that each of these lengths is much larger than the longest wavelength of significance. There are 2 factors that determine the energy density at a given frequency. The first is the average energy in each state s which is given by nsεs = εs eβεs −1 (40) If we set εs = ¯ hω and ns = n(¯ hω), we can rewrite this to give: n(¯ hω)¯ hω = ¯ hω eβ¯ hω −1 (41) The second factor is the number of states per unit volume whose frequency lies in the range between ω and ω + dω. This is given by (15) N(ω)dω = 1 π2c3ω2dω (42) 7 So at temperature T the mean energy density u(ω, T)dω contained in the photon gas by photons whose frequencies are between ω and ω + dω is given by the product of the average energy in each single photon state and the density of states which lie in this frequency range: u(ω, T)dω = n(¯ hω)¯ hωN(ω)dω = ω2 π2c3 ¯ hω (eβ¯ hω −1)dω (43) We can rewrite this to give: u(ω, T)dω = ¯ hω3 π2c3 dω (eβ¯ hω −1) (44) This is Planck’s law for the blackbody spectrum.
0e+00 2e+15 4e+15 6e+15 8e+15 1e+16 angular frequency (radians/sec) 0.0e+00 5.0e+06 1.0e+07 1.5e+07 u(ω, T) (K−sec/m 3) Black Body Radiation 3000 K 4000 K 5000 K We can take the high temperature limit to get the classical limit of this spectrum. In the high temperature limit, β is small so we can expand the exponential in the denominator: eβ¯ hω −1 ≈(1 + β¯ hω) −1 = β¯ hω (45) So the high temperature limit of (44) is u(ω, T)dω ≈ ¯ hω3 π2c3(β¯ hω)dω (46) 8 or, using β = 1/kBT, we can write u(ω, T)dω ≈ω2kBT π2c3 dω (47) This is the Rayleigh–Jeans formula for blackbody radiation.
Notice that eqn.
(47) increases as ω2. Therefore the classical spectrum (47) predicts that the energy density goes to infinity as the frequency goes to infinity. By the end of the 1800’s the black body spectrum had been measured and the classical formula had been calculated. There was a clear lack of agreement, so people knew they had a problem. Planck resolved the conflict by proposing that electromagnetic energy was not continuous, but rather was quantized.
He proposed E = ¯ hω (or E = hν) and derived Planck’s law (44). This fit the data very well, and quantum mechanics was born.
We can rewrite (44) in terms of a dimensionless parameter η: η ≡β¯ hω = ¯ hω kBT (48) Planck’s law becomes: u(η, T)dη = ¯ h π2c3 ÃkBT ¯ h !4 η3dη eη −1 (49) If we plot u(η, T) versus η, the maximum occurs around ηmax ≈3.
0 5 10 15 η=hω/kT 0e+00 2e+21 4e+21 6e+21 8e+21 u(η, T) (K−sec/m 3) Scaled Black Body Radiation 3000 K 4000 K 5000 K 9 So if at temperature T1 the maximum occurs at frequency ω1,max, then at some other temperature T2 the maximum occurs at ω2,max. This is because ηmax = ¯ hω1,max kBT1 = ¯ hω2,max kBT2 (50) or ω1,max T1 = ω2,max T2 (51) This is called the Wien displacement law. It says that ωmax ∝T (52) This was initially an empirical relation that was deduced from the experimental data.
We see that it also follows from Planck’s law. It is often useful in physics to express things in terms of dimensionless parameters. The Wien displacement law is an example of useful scaling relations that can result from this.
We can also calculate the total energy density uo(T) contained in the photon gas at temperature T by integrating (44) over frequency: uo(T) = Z ∞ 0 u(ω, T)dω (53) Using (49), we can rewrite this as uo(T) = ¯ h π2c3 ÃkBT ¯ h !4 Z ∞ 0 η3dη eη −1 (54) One can evaluate the integral exactly. The answer is Z ∞ 0 η3dη eη −1 = π4 15 (55) Using this, one finds uo(T) = π2 15 (kBT)4 (¯ hc)3 (56) This is known as the Stefan–Boltzmann law. The important point is that the total energy density goes as the fourth power of the temperature: uo(T) ∝T 4 (57) Finally the mean pressure p exerted on the walls of the enclosure by the radiation is simply related to the total energy density: p = X s ns à −∂εs ∂V !
T = 1 3uo(T) (58) 10 To see where this comes from, recall that εs = ¯ hωs ω = ck (59) Now use kx = 2π Lx nx ky = 2π Ly ny kz = 2π Lz nz (60) to obtain εs = ¯ hck = ¯ hc q k2 x + k2 y + k2 z = ¯ hc µ2π L ¶ q n2 x + n2 y + n2 z = 2π¯ hcV −1/3q n2 x + n2 y + n2 z (61) where V is the volume. So the pressure associated with state s is ps = −∂εs ∂V = 1 32π¯ hcV −4/3q n2 x + n2 y + n2 z = εs 3V (62) So the average pressure for the system is p = X s psns = 1 3V X s nsεs = E 3V (63) or p = 1 3uo (64) (The pressure can also be written as 1 β ³ ∂ln Z ∂V ´ T = −∂F ∂V ¯ ¯ ¯ T.) The “3” in the denominator reflects the fact that the box is 3 dimensional. Radiation pressure is quite small, but it is what gives comets their tails. Solar radiation is what pushes tiny bits of dust and ice that come from the ice ball away from the sun and produces the tail. The comet tail always points away from the sun. The power emitted ∼flux ∼cuo(T).
Principle of Detailed Balance If an object is sitting in a cavity filled with radiation (photons) and is in equilibrium at temperature T, then the • power radiated by body = power absorbed by body If this were not true, the body would be losing or gaining energy and would get cooler or would heat up. As a result, its temperature would no longer be the same as the ambient photons at temperature T; it would no longer be in equilibrium. So it must absorb the same amount of power as it emits in order to stay in equilibrium.
We can make an even stronger statement. Namely, that in equilibrium the power radiated and absorbed by the body must be equal for any particular element of area of the body, for any particular direction of polarization, and for any frequency range.
11 To show that this must be true, one could imagine putting a shield or filter around the object that absorbs all radiation except that, in one small element of area, it is completely transparent to radiation in one direction with one polarization and in one narrow frequency range between ω and ω + dω. In the presence of the shield the body must absorb and emit the same power in order to avoid heating up or cooling off, i.e., in order to stay in equilibrium. So the power radiated and absorbed by the body must be equal for any particular element of area of the body, for any particular direction of polarization, and for any frequency range. This is called the principle of detailed balance.
The principle of detailed balance is a fundamental result that is based on very general arguments. Microscopically it is a result of time reversal invariance and the fundamental assumption of accessible macrostates being equally probable in an isolated system. Con-sider a single isolated system consisting of several weakly interacting parts, e.g., a body and photons. When these parts are not interacting, the system can be in any one of its quantum states labeled by indices r, s, etc. When interactions are present, the interac-tions induce transitions between the states. Let wrs be the transition rate (or transition probability per unit time) from state r to state s. Under time reversal, t →−t, r →r∗ and s →s∗where r∗and s∗are the time reversed states of r and s. For example, if a particle has momentum ⃗ p in state r, then it has momentum −⃗ p in state r∗. If the system is invariant (the same) under time reversal, then ws∗r∗= wrs (65) This expresses the principle of microscopic reversibility. For example, if the body in the cavity emits a photon with wavevector ⃗ k, then the time reversed process is the absorption of a photon with wavevector −⃗ k. Microscopic reversibility asserts that these two processes occur with equal probability.
If we have some initial set A of states labeled by r and some final set B of states labeled by s, the transition probability from A →B is given by WAB = X r X s Prwrs (66) where Pr is the probability of being in state r. The probability of landing in state s is the probability Pr of being in state r multiplied by the transition rate wrs to state s.
Similarly the time reversed process has a transition rate given by WB∗A∗= X s∗ X r∗Ps∗ws∗r∗ (67) But the fundamental postulate of statistical mechanics states that accessible macrostates are equally probable in an isolated system. So Pr = P ∗ s and WB∗A∗= WAB (68) This is the principle of detailed balance.
12 Radiation Emitted by a Body Let us now apply the principle of detailed balance to a body at temperature T in equilibrium with radiation (photon gas) inside an enclosure at this temperature. Let Pi(k, α) be the incident radiation power on a unit area of this body per unit frequency and solid angle range about the vector k with polarization α. Let a(k, α) be the fraction of incident power absorbed, the rest being reflected. By the principle of detailed balance, the power absorbed must equal the power emitted Pe(−k, α) in the opposite direction −k: Pe(−k, α) = a(k, α)Pi(k, α) (69) For a blackbody, a(k, α) = 1; a good absorber is a good emitter and vice-versa.
Let us now calculate explicitly the power Pi(k, α) incident per unit area of a body in an enclosure at temperature T. This is the incident energy flux. Let f(k)d3k be the mean number of photons per unit volume with a given polarization whose wavevector lies between k and k + dk. We found earlier in studying blackbody radiation that f(k)d3k = n(¯ hω) d3k (2π)3 = 1 eβ¯ hω −1 d3k (2π)3 (70) So (c dt cos θ)f(k)d3k photons of a given frequency and polarization strike a unit area of the body in a time dt. Since each photon carries energy ¯ hω, one obtains Pi(k, α)dωdΩ= (¯ hω)(c cos θf(k)d3k) (71) k Area dA θ z c dt Converting d3k to spherical coordinates and using k = ω/c, we obtain d3k = k2dkdΩ= ω2 c3 dωdΩ (72) and Pi(k, α)dωdΩ= ¯ hω3 c2 f(k) cos θ (73) If the body absorbs isotropically, then the fraction of incident radiation absorbed is a(k, α) = a(ω), i.e., a is independent of the direction k. We are also assuming a is 13 independent of the polarization direction. So the power emitted in the direction k′ = −k is Pe(k′, α) = a(ω)Pi(k, α) (74) Now let us find the total power Pe(ω)dω emitted per unit area into the frequency range between ω and dω for both polarization directions by integrating over the solid angle. Using dΩ= sin θdθdφ and multiplying by 2 for both polarizations, we write Pe(ω)dω = 2 Z ΩPe(k′, α)dωdΩ = a(ω)2¯ hω3 c2 f(k)dω à 2π Z π/2 0 cos θ sin θdθ !
= a(ω)2π¯ hω3 c2 f(k)dω (75) The right hand side is proportional to (¯ hω)f(k)d3k, i.e., to the mean radiation density u(ω)dω inside the enclosure. So we can write Pe(ω)dω = a(ω) ·1 4cu(ω)dω ¸ (76) It makes sense to see cu(ω) for the flux. Using Eq. (70), we can write Eq. (75) as Pe(ω)dω = a(ω) ¯ h 4π2c2 ω3dω eβ¯ hω −1 (77) The total power Ptot e emitted per unit area of the body is obtained by integrating Eq. (77) over frequency as we did in Eqs. (54)-(56) to obtain Ptot e = a µ1 4cuo ¶ = a ³ σT 4´ (78) where uo is given by Eq. (56): uo(T) = π2 15 (kBT)4 (¯ hc)3 (79) Eq. (78) is another form of the Stefan-Boltzmann law. The Stefan-Boltzmann constant σ is σ ≡π2 60 k4 B c2¯ h3 = (5.6697 ± 0.0029) × 10−5 erg/(sec cm2 deg4) (80) For a perfect blackbody, a = 1. For something shiny like gold, a ≈0.01.
14 |
11131 | http://princeton.edu/~gasdyn/Papers/Weisberg.pdf | J. Fluid Mech. (1997), vol. 11, pp. 1–000 Copyright c ⃝1997 Cambridge University Press 1 Delaying transition in Taylor-Couette flow with axial motion of the inner cylinder By A R E L Y. W E I S B E R G†, I O A N N I S G. K E V R E K I D I S‡ AND A L E X A N D E R J. S M I T S Department of Mechanical and Aerospace Engineering Princeton University Princeton, New Jersey 08544-0710, U.S.A.
(Received 12 March 1997) Periodic axial motion of the inner cylinder in Taylor-Couette flow is used to delay tran-sition to Taylor vortices. The outer cylinder is fixed. The marginal stability diagram of Taylor-Couette flow with simultaneous periodic axial motion of the inner cylinder is determined using flow visualization. For the range of parameters studied, the degree of enhanced stability is found to be greater than that predicted by Hu & Kelly (1995), and differences in the scaling with axial Reynolds number are found. The discrepancies are attributed to essential differences between the base flow in the open system considered by Hu & Kelly, where mass is conserved over one period of oscillation, and the base flow in the enclosed experimental apparatus, where mass is conserved at all sections at all times.
1. Introduction The central importance of flow between concentric cylinders as a fluid-dynamical paradigm has been well documented in the reviews by Di Prima & Swinney (1985) and Tagg (1994). One reason for the great number of studies of this “Taylor-Couette” flow is the orderly progression of nonlinear flow states observed in the system as it undergoes transition to turbulence. For example, if the outer cylinder is held fixed and the inner one rotates at angular speed Ω, a series of four critical speeds exist between the purely azimuthal Couette flow at low values of Ωand the turbulent Taylor vortices that appear at large values of Ω(Andereck et al., 1986). This orderly progression to turbulence makes Taylor-Couette flow attractive as a model for studying transition to turbulence in more complex systems.
The first transition, from steady azimuthal flow to Taylor vortices, has been the focus of many previous studies. Variations of the basic Taylor-Couette flow experiment which raise the critical angular speed for this transition have also been explored. In particular, Taylor-Couette flow with superposed Poiseuille flow (that is, a non-zero axial through-flow) has been studied experimentally by Cornish (1933), Fage (1938), Kaye & Elgar (1957), Donnelly & Fultz (1960), Snyder (1962), Schwartz et al. (1964), Takeuchi & Jankowski (1981), Buhler & Polifke (1990), Babcock et al. (1991), Lueptow et al. (1992), Tsameret & Steinberg (1994) and Tsameret et al. (1994), and analytically by Goldstein (1937), Chandrasekhar (1960), Di Prima (1960), Krueger & Di Prima (1964), Chung † Present address: SciTec, Inc., 100 Wall St., Princeton NJ 08540 ‡ Department of Chemical Engineering 2 A. Weisberg, A.J. Smits and I. Kevrekidis & Astill (1977), Takeuchi & Jankowski (1981), Buhler & Polifke (1990), Tsameret & Steinberg (1994) and Tsameret et al. (1994). These efforts established that the critical rotation rate increases with increasing axial flow rates.
Taylor-Couette flow with superposed axial Couette flow (that is, where one of the cylinders is moving in the axial direction) has also been studied analytically (Ludweig, 1960, Kiessling, 1963, Wedemeyer, 1967) and experimentally (Ludweig, 1964) but much less extensively than the axial Poiseuille flow case. In all these studies both cylinders were rotating. In addition, a number of researchers examined the general case of both axial Couette and axial Poiseuille flow (Joseph & Munson, 1970, Hung et al., 1972), including a numerical linear stability analysis for a finite flow domain (Ali & Weidman, 1993).
All the previously mentioned investigations were confined to steady axial flows. How-ever, Hu & Kelly (1995) performed a linear stability analysis to include the effects of periodically varying axial pressure gradients and inner cylinder axial speeds, for the case of the inner cylinder rotating, as well as for the cases where the cylinders were co-rotating and counter-rotating. For periodic axial motion of the inner cylinder, with only the inner cylinder rotating, Hu & Kelly found enhanced stability for all axial speeds and oscillation frequencies. In the case of an open flow, corresponding to infinitely long cylinders, the fluid pathlines at subcritical angular speeds consist of constant radius spirals that ascend and descend in response to the inner cylinder’s axial motion. Here, the azimuthal and axial velocity profiles are decoupled, so changes in the axial motion have no effect on the azimuthal velocity field. Once the flow undergoes transition to Taylor vortices the flow field becomes much more complex, but it still instantaneously resembles closely the Taylor-vortex-dominated flow-field which exists when axial motion is not present. This observation, coupled with the fact that the axial and azimuthal subcritical flow fields are independent, yield the supposition that the stability of Taylor-Couette flow with axial motion of the inner cylinder is closely related to the stability of Taylor-Couette flow without axial motion of the inner cylinder.
2. Experiment To study these questions further, the experimental apparatus depicted in Figure 1 was constructed. Angular motion of the inner cylinder is controlled by a stepper motor, through a timing belt drive and a splined shaft. This shaft fits in a splined bushing in the top end cap, which, together with the ball joint in the bottom end cap, enabled the inner cylinder to be simultaneously spun as well as translated in the axial direction. The axial motion was controlled by a separate stepper motor through a variable ratio timing belt drive and a Scotch yoke mechanism, which imparts a sinusoidally varying axial velocity to the inner cylinder. The dimensions of the apparatus are shown in Table 1 and Figure 1.
The top and bottom seals are rigidly attached to the outer cylinder, and are therefore non-rotating. The inner edges of the seals were machined to knife edges but they could not provide perfect dynamic seals, and therefore an overflow reservoir was designed into the top seal, and a second, spring-loaded, rubber lip seal (not shown) was incorporated into the base cylinder.
The working fluid used throughout this work was a 20:1 mixture of distilled water and Kalliroscope AQ-1000 rheoscopic concentrate†. The flakes align themselves with the local shear stress direction thereby making flow patterns visible. When the “face” of a flake is oriented toward the observer incident light is reflected back, “coloring” the † Kalliroscope Corporation, 264 Main St., Box 60, Groton MA 01450, (508) 448-6302 Delaying transition in Taylor-Couette flow 3 local fluid white. When the flake is oriented sideways light is not reflected back to the observer, and the fluid appears darker. These flakes have been used in many previous Taylor-Couette flow experiments (Andereck et al., 1986). When using the water and Kalliroscope mixture, the flow field appears to be uniformly gray at subcritical angular speeds.
Once Taylor vortices ( which are pairs of counter-rotating toroidal vortices) appear, the flow field displays alternating light and dark bands indicating the presence of vortices of opposite sign.
The kinematic viscosity of the mixture as a function of temperature was measured to within 1% by a professional laboratory. The apparatus was placed in a temperature controlled enclosure, resulting in a temperature fluctuation of less than 0.2◦C. The un-certainty in the value of the kinematic viscosity due to temperature and other sources of error is less than 1.4%. Settling of the Kalliroscope flakes can introduce a viscosity gradient, but under our experimental conditions the uncertainty in the viscosity was dominated by temperature variations. When the cylinders are vertical, as in the current experiment, the slowly increasing Kalliroscope flake concentration is known to affect slightly the wavelength of the Taylor vortices (Dominguez-Lermaet al., 1985), but the effect on the critical Taylor number is not known. This issue is addressed further below.
The onset of transition to Taylor vortices was recorded by a black and white CCD video camera connected to a personal computer which acquired and stored the images directly.
The images were processed using a modified version of the public domain image processing program NIH Image† for the Macintosh. Custom stepper motor control circuitry was built which enabled both stepper motors in the experiment to be completely controlled by software (for further details of the image processing procedure and the electronic control circuitry, see Weisberg, 1996).
Three dimensionless parameters determine the state of the system. Following Hu & Kelly, we use a Taylor number, Ta, based on the rotation rate of the inner cylinder, a Reynolds number, Re, based on the maximum axial speed of the inner cylinder, and an oscillation parameter, β, where the axial frequency of oscillation was non-dimensionalized by the viscous time scale: Ta = 4Ω2η2d4 (1 −η2) ν2 , Re = Umaxd ν , β = r ωd2 2ν , where Ωis the angular speed of the inner cylinder, η is the radius ratio ri/ro of the cylinders (= 0.9051), d is the gap size (= ro −ri), ν is the kinematic viscosity of the working fluid, Umax is the amplitude of axial speed of the inner cylinder, and ω is the (angular) frequency of the axial motion of the inner cylinder.
The experimental uncertainties are: δTa = 3.1%, δRe = 2.1%, and δβ = 0.71%.
3. Results To determine the critical value of the Taylor number as a function of Reynolds number and the oscillation parameter, the rotation rate of the inner cylinder was increased very slowly and the flow field was monitored for the presence of Taylor vortices. A number of experiments were conducted to determine quasi-static levels of acceleration, and to determine the threshold at which it was judged that transition had occurred. In most experimental systems, Taylor vortices first appear near the ends of the cylinders at lower † NIH Image developed at the U.S. National Institutes of Health and available from the Inter-net by anonymous FTP from zippy.nimh,nih.gov or on floppy disk from the National Technical Information Service, Springfield, Virginia, part number PB95-500195GEI 4 A. Weisberg, A.J. Smits and I. Kevrekidis than expected values of Ta. The extent of the flow domain dominated by Taylor vortices then spreads from the end walls toward the center of the apparatus, until, for systems with large aspect ratios (cylinder length/gap width), the vortices meet at mid-length at what we expect to correspond to the theoretical critical value of the Taylor number (Lueptow et al., 1992). In contrast to previous experimental work, the vortices observed in this experiment appeared at the lower end first, and propagated up towards the center at progressively higher values of Ta. This sequence is shown in Figure 2. The reason for this discrepancy is not clear. There exists a slight eccentricity of the cylinders at the top of the flow domain but the level of eccentricity (a maximum of 0.005 in, or 0.6% of the gap size) is a factor of 5 less than the level at which the critical Taylor number begins to be reduced (Cole, 1976). For our experiment, transition to Taylor vortices was defined as the value of Ta at which the vortices reached the center of the cylinders. This definition yielded consistent values for the critical Taylor number in a variety of experiments in which the initial speed and the acceleration rate of the inner cylinder was varied. The effects of gravity on the Kalliroscope concentration also seems to be rather small, given the typical duration of an experiment was very much less than the settling time of the flakes.
In fact, using this definition, the critical Taylor number, Tac,0, for ordinary Taylor-Couette flow (Re = 0) was determined to be = 3636 ± 3.6%. This value compares very well with the values found by previous experimental and analytical studies, as indicated in Figure 3. The same method used to obtain Tac,0 was used to determine Tac, the transitional value of Ta at nonzero Re and β. Note that, for our apparatus, Re and β were not independent, and Re ∝β2.
The results for Tac,0 are shown in Figure 4. Two axial drive reduction ratios were used to obtain the two sets of data for Re = 6.35β2. The agreement between these two sets demonstrates that the results are independent of the particular experimental config-uration. Very recently, Marques & Lopez (1996) by a numerical linear stability analysis determined the critical Taylor numbers for the case of an enclosed system similar to that studied in the experiment. The agreement bewteen their results and the experimental data is extremely good, as seen in Figure 4.
Hu & Kelly found that the increase in the critical Taylor number scales with Re2 for Re < 30, and their results for Re = 1 and Re ≪1 very nearly coincide with the results for Re = 30 for the range of β examined here (0 ≤β ≤10). The data in Figure 4, for Re ranging from approximately 10 to 100, was therefore rescaled with Re2 and replotted in Figure 5.
A number of preliminary conclusions can be drawn from Figure 5. The experimental data and analytical results converge at the larger values of the oscillation parameter β.
This is expected, since as the axial frequency is increased, the effects of the oscillat-ing inner cylinder will propagate shorter and shorter distances into the gap due to the damping effects of viscosity. Therefore in the limit of very large values of β, the data and analytical results should approach the critical value of Ta corresponding to Re = 0.
For values of β greater than approximately 1.5, the Re2 scaling is broadly evident in the data although the experimental data and analytical results do not agree. At values of β below 1.5, the Re2 scaling is less evident, and the trends in the experimental data and analytical results differ as well. While Hu & Kelly’s results are nearly constant for β < 1 in the scaling of Figure 5, the experimental data points appear to be increasing with diminishing β. The loss of scaling with Re2 is most likely due to the higher values of Re associated with many of the data points.
While Hu & Kelly computed results for Re = 30, the largest value of Re in the data is close to 100. Further, with the breakdown of the Re2 scaling, the uncertainty Delaying transition in Taylor-Couette flow 5 envelope grows. The growth of the uncertainty in the data, as scaled in Figure 5, is also substantially larger than that seen in Figure 4 due to the presence of the 1/Re2 term.
However, the ratio of parameters on the vertical axis greatly magnifies the experimental uncertainties in the parameters Tac, Tac,0 and Re, and therefore it misrepresents the high degree of accuracy in the measurements. The behavior of the experimental results is more satisfactorily demonstrated in the form given in Figure 4.
Outside the range of Re and β values shown in Figure 4 there are values of Re and β at which a different transition was observed to take place. In these cases vortices did not propagate in an orderly fashion from the bottom end of the apparatus, but appeared in various locations intermittently. At larger values of Ta the vortices became increasingly permanent until they filled the apparatus at all times. These vortices were often tilted with respect to the axis of rotation. Because of the different nature of this transition, those results are not shown here, although they present an interesting direction in which to extend this investigation.
4. Discussion and Conclusions We propose that the discrepancies between the analytical results presented by Hu & Kelly (1995) and the experimental data given here are due to the fundamental differences in the subcritical flow fields (the “base” flows): the experiment was an “enclosed” system, whereas Hu & Kelly considered an “open” system.
In an open flow system, mass is conserved over an entire axial period through any cross section normal to the axis of rotation. In an enclosed system, however, net mass flux through any cross section is zero at all times. At low values of β the base flow in open systems is unidirectional over most of the axial period. In contrast, the base flow in an enclosed system will always consist of equal fluid volumes travelling in opposite directions.
As β is increased from zero, the base flow in the open system develops regions of reversed flow. This can be seen as the growth of the viscous time scale, d2/ν, relative to the axial period. Consequently, in the range of β between approximately 2 and 4, the two types of base flows more closely resemble one another. At larger values of β the axial base flow is almost stationary except close to the inner cylinder wall because viscosity damps out the axial oscillations well before they can propagate into the gap. This is true of the base flow in both the open and enclosed cases.
These differences help to explain why the experimental data and analytical results approach one another at larger values of β and diverge at lower values.
Additional insight can be acquired using just the trend in Hu & Kelly’s results. In particular, the magnitude of azimuthal vorticity in the base flow field, which arises as a consequence of the axial motion, appears to play a key role. At values of β < 1 the axial velocity profile in the open system, which is nearly linear in the case of η ≈1, has very little curvature, and therefore the azimuthal component of the vorticity field is determined by the slope of the axial velocity profile. As β increases, the axial velocity develops regions of reversed flow, and the magnitude of the azimuthal vorticity in the flow, integrated over the gap size, increases. This results in increased stability enhancement when normalized by Re2, as the analytical results in Figure 5 indicate. As β increases further, viscous damping reduces the magnitude of the vorticity in the flow, and the degree of enhanced stability reduces as well.
The role of azimuthal vorticity becomes clearer when the vorticity distribution in supercritical Taylor-Couette flow with Re = 0 is considered.
When the flow field is dominated by Taylor vortices, the azimuthal vorticity changes sign along the axial direc-tion. Consequently, transition from subcritical flow to supercritical flow can be viewed 6 A. Weisberg, A.J. Smits and I. Kevrekidis as the generation of azimuthal vorticity, with equal amounts of fluid acquiring either pos-itive or negative azimuthal vorticity. However, when Re > 0 the fluid already contains azimuthal vorticity that is a function of radial position only. Therefore, transition to Taylor vortices, which have axial variations in azimuthal vorticity, will require that some of the fluid’s azimuthal vorticity undergo a sign reversal. To undergo this change, it is likely that higher than usual values of Ta will be required to reverse the local angular momentum vectors to this extent, which indicates a flow that is centrifugally unstable to a greater degree. For the case of steadily translating cylinders (β = 0) the predicted change in Ta can be readily calculated: ∆Taencl ∝ 1 r0 −ri r0 Z ri ω2 θ,encl rdr ∆Ta∞∝ 1 r0 −ri r0 Z ri ω2 θ,∞rdr where ωθ is the azimuthal vorticity, ∆Ta = Tac −Tac,0, and the subscripts ∞and encl denote the open and enclosed systems, respectively. Inserting the expressions for the azimuthal vorticity for each case we find that: ∆Taencl ∆Ta∞ = r0 R ri ω2 θ,encl rdr r0 R ri ω2 θ,∞rdr = 3.8 After accounting for the lower value of η in the experiments versus the calculations, this value should be 3.5.
This number is in good agreement with the experimental data: extrapolating the data in Figure 5 to β = 0 gives a value of this ratio of about 3 (3.5 falls within the experimental uncertainty). The increased level of the azimuthal vorticity therefore seems to be a good indication of the enhanced stability of the system.
These considerations were first presented by Weisberg (1996). The numerical linear stability analysis by Marques & Lopez (1996) agrees extremely well with the experiment (see Figure 4), providing strong analytical support for the physical arguments presented here and by Weisberg (1996). The divergent trend seen at low β in Figure 5 was also reproduced in their enclosed flow analysis, as was the breakdown of the Re2 scaling for values of Re > 30.
In summary, the marginal stability curve for Taylor-Couette flow has been determined experimentally for the range of parameters: 0.5 < β < 6.0, 10 < Re < 100. The curve traced by the experimental values was found to lie above the curve calculated by Hu & Kelly. It was suggested that this occurs because, in contrast to the system considered by Hu & Kelly, the experiment was an enclosed flow system. Given the essential differences in the subcritical flow fields, as well as the trends in Hu & Kelly’s results, it appears that the integral of the square of azimuthal vorticity over the gap plays an important role in the degree of enhanced stability. The greater the square of the azimuthal vorticity, the greater is the enhanced stability. A physical argument can be made that this enhanced stability results because transition from Taylor-Couette flow with axial motion of the inner cylinder requires a greater change in the azimuthal vorticity of the fluid than in the case without axial motion.
This work was partially supported by: The Guggenheim Fellowship fund, AFOSR Delaying transition in Taylor-Couette flow 7 (Grant numbers F49620-93-1-0427, F49620-93-1-0478, F49620-93-1-0476), DARPA Grants N00014-92-J-1796 and N00014-91-J-1850, NSF Grants CTS-89-57213 and ECS- 90-23362, and ONR/ARPA AASERT proposal N00014-94-1-0149.
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Delaying transition in Taylor-Couette flow 9 Tables and Figures Table 1.
Summary of cylinder dimensions.
Figure 1.
Schematic drawing of the apparatus. All dimensions are in inches.
Figure 2.
From top to bottom, Taylor vortices propagating into the flow domain at subcritical (0.85 Tac,0), critical and supercritical (1.1 Tac,0) values of the Taylor number Ta.
Figure 3.
Critical values of the Taylor number in ordinary Taylor-Couette flow (Re = 0), Tac,0 as a function of the radius ratio, η. 3, present results. The other experimental (×) and theoretical (◦) values are taken from Lueptow et al. (1992) and Cole (1976), and their cited sources. The theoretical values are shown with their best linear curve fit.
Figure 4.
Enhanced stability of Taylor-Couette flow with axial motion of the inner cylinder as a function of Re and β. Experiments: ◦, Re = 1.59β2; 2, Re = 3.18β2; 3, Re = 6.35β2; △, Re = 6.35β2 (in a different experimental configuration); ◦, Re = 12.7β2; 2, Re = 25.6β2; 3, Re = 41.3β2.
The uncertainty in each measurement is between 3% and 4%. The solid lines are the results of the analysis by Marques & Lopez (1996).
Figure 5.
Enhanced stability due to axial motion of the inner cylinder normalized by Re2 as a function of the oscillation parameter β. Symbols are as given in Figure 4.
The range of uncertainty in the measurements is given by the distance between the solid lines. The analytical results by Hu & Kelly (1995) for Re = 30 and β ≤6 are indicated by the × symbol.
10 A. Weisberg, A.J. Smits and I. Kevrekidis Figure 1. Schematic drawing of the apparatus. All dimensions are in inches.
Delaying transition in Taylor-Couette flow 11 Figure 2. From top to bottom, Taylor vortices propagating into the flow domain at subcritical (0.85 T ac,0), critical and supercritical (1.1 T ac,0) values of the Taylor number T a.
Figure 3.
Critical values of the Taylor number in ordinary Taylor-Couette flow (Re = 0), Tac,0 as a function of the radius ratio, η. 3, present results. The other experimental (×) and theoretical (◦) values are taken from Lueptow et al. (1992) and Cole (1976), and their cited sources. The theoretical values are shown with their best linear curve fit.
12 A. Weisberg, A.J. Smits and I. Kevrekidis Figure 4. Enhanced stability of Taylor-Couette flow with axial motion of the inner cylinder as a function of Re and β. Experiments: ◦, Re = 1.59β2; 2, Re = 3.18β2; 3, Re = 6.35β2; △, Re = 6.35β2 (in a different experimental configuration); ◦, Re = 12.7β2; 2, Re = 25.6β2; 3, Re = 41.3β2. The uncertainty in each measurement is between 3% and 4%. The solid lines are the results of the analysis by Marques & Lopez (1996).
Figure 5. Enhanced stability due to axial motion of the inner cylinder normalized by Re2 as a function of the oscillation parameter β. Symbols are as given in Figure 4. The range of uncertainty in the measurements is given by the distance between the solid lines. The analytical results by Hu & Kelly (1995) for Re = 30 and β ≤6 are indicated by the × symbol.. |
11132 | https://physics.stackexchange.com/questions/173976/speed-of-sound-of-a-gas-mixture | thermodynamics - Speed of sound of a gas mixture - Physics Stack Exchange
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Speed of sound of a gas mixture
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Is there a general mixing rule for estimating the speed of sound of a gas mixture, given the speed of sound in the components?
From the ideal gas relation: c = (kRT/M)^{1/2} (where k is the specific heat ratio, R is the gas constant, T is the temperature and M is the molar mass) we can guess that there might be some squares and molar concentration weights involved.
A simple molar weighted average: c_{mix} = x_1c_1+x_2c_2 (where x's are the molar concentrations) works well when the molar masses are similar (for example, oxygen and nitrogen) but very poorly when the molar masses are very different.
For dissimilar molar masses (for example, hydrogen and nitrogen), this relation works better: c_{mix} = ((x_1c_1)^2+(x_2c_2)^2)^{1/2} but it works very poorly when then the molar masses are similar.
Is there a good general way to estimate this?
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asked Apr 3, 2015 at 14:56
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For an ideal gas the sound speed can be written as
c = \sqrt{\frac{k P}{\rho}}
where \rho is the mass density. The constant k is present because the expansion and compression of the gas is supposed to be adiabatic, and k is the adiabatic index. For a gas mixture with n_i moles of the i-th gas we have
k=\frac{\sum_i n_i c_{p,i}}{\sum_i n_i c_{v,i}} = \frac{\sum_i n_i \frac{k_i}{k_i-1}}{\sum_i n_i \frac{1}{k_i-1}}
where k_i is the adiabatic index for the i-th gas. The mass density is
\rho=V^{-1} \sum_i M_i n_i
where M_i is the i-th molar mass. Inserting in the expression for the sound speed we get
c = \sqrt{\left(\frac{1}{\sum_i x_i \frac{1}{k_i-1}}+1\right)\frac{ RT}{\sum_i M_i x_i}}
while
c_i = \sqrt{\frac{k_i RT}{M_i}}
In the general case it is not possible to express c as a function of the ratios k_i/M_i only, so it is not possible to express c as a function of the x_i and c_i only.
A particular case is when all the gases in the mixture have the same adiabatic index (for example, they can be all biatomic). Then
c = \sqrt{\frac{k RT}{\sum_i M_i x_i}}
and
c_i = \sqrt{\frac{k RT}{M_i}}
then
c = \left(\sum_i \frac{x_i}{c_i^2} \right)^{-1/2}
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edited Jun 4, 2020 at 8:53
answered Apr 3, 2015 at 22:22
GCLLGCLL
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1 \begingroup I've used this answer today and investigated in some detail. I think there is a mistake appearing from the second equation. I agree that k is the average cp divided by the average cv, but I do not believe that final step (ratio of sums of fractions involving k) is correct. I was not able to derive it, and in my calculations it gives a different answer. So, I suggest that people calculate k as the average of cp divided by the average cv, as this appears uncontroversial and still easy to calculate.\endgroup Alex Wilber –Alex Wilber 2016-12-14 14:29:20 +00:00 Commented Dec 14, 2016 at 14:29
\begingroup There should be +1 not -1 in fourth equation\endgroup dark32 –dark32 2017-04-04 03:39:04 +00:00 Commented Apr 4, 2017 at 3:39
\begingroup Right, sorry for noting this only now. I corrected the expressions.\endgroup GCLL –GCLL 2020-06-04 08:54:43 +00:00 Commented Jun 4, 2020 at 8:54
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11133 | https://www.ijnrd.org/papers/IJNRD2301108.pdf | © 2023 IJNRD | Volume 8, Issue 1 January 2023 | ISSN: 2456-4184 | IJNRD.ORG IJNRD2301108 International Journal of Novel Research and Development (www.ijnrd.org) b55 ISOLATION OF PURE CULTURE BACTERIA FROM SOIL SAMPLES 1Miss. Vrushali Kailas Fuke, 2Miss. Pratiksha vasant Kulkarni 1Student of Yashodeep Institute of Pharmacy, 2Assistant professor 1B. pharmacy 1Yashodeep Institute of Pharmacy Aurangabad, Maharashtra, India The isolation of soil bacteria from extreme environments represents a major challenge, but also an opportunity to characterize the metabolic potential of soil bacteria that could promote the growth of plants inhabiting these harsh conditions. The aim of this study was to isolate and identify bacteria from soilsoil Samples environments and characterize the beneficial traits for plants through a biochemical approach. Keywords : Culture media ,Soil, Serial dilution technique , Pour Plate technique, Spreads Plate technique 1. INTRODUCTION PURE CULTURE: A culture containing only one species of microbe is called pure culture. Isolation of pure culture - Microorganisms are generally found in nature (air, soil and water) as mixed populations. Even the diseased parts of plants and animals contain a great number of microorganisms, which differ markedly from the microorganisms of other environments. To study the specific role played by a specific microorganism in its environment, one must isolate the same in pure culture. A pure culture theoretically contains a single bacterial species. There are a number of procedures available for the isolation of pure cultures from mixed populations. A pure culture may be isolated by the use of special media with specific chemical or physical agents that allow the enrichment or selection of one organism over another. The differential and selective procedures will be utilized later in this course. Isolation of pure culture method 1. Serial dilution technique 2. Pour Plate technique 3. Spreads Plate technique Culture media - The food material or substances required for growing microorganisms in vitro (outside the body) is called culture medium. © 2023 IJNRD | Volume 8, Issue 1 January 2023 | ISSN: 2456-4184 | IJNRD.ORG IJNRD2301108 International Journal of Novel Research and Development (www.ijnrd.org) b56 Classification of culture media based on physical state: a) solid medium b) semi solid medium c) liquid medium Soil Soil is the thin layer of material covering the earth's surface and is formed from the weathering of rocks. It is made up mainly of mineral particles, organic materials, air, water and living organisms—all of which interact slowly yet constantly. Agriculture soil is a dynamic medium in which a large number of pathogenic and non-patho-genic bacterial and fungal flora live in close association Soil Samples 1. Brown soil 2. Black soil 3. Red soil 2. AIM AND OBJECTIVE Aim- Isolation of pure culture bacteria from soil samples Objective- 1. To demonstrate good aseptic technique in culture transfer or inoculation and in handling sterile materials. 2. To demonstrate skill in isolation of organisms from a mixed culture using selective and differential media. 3. To isolate microorganisms from a wide variety of soil and describe their colonial. 3. SELECTION OF SOILS 1 Brown soil Selected soil from yeshodeep Institute of pharmacy college ground. 2 Black soil Selected soil from farm 3 Red soil Selected soil from play ground © 2023 IJNRD | Volume 8, Issue 1 January 2023 | ISSN: 2456-4184 | IJNRD.ORG IJNRD2301108 International Journal of Novel Research and Development (www.ijnrd.org) b57 Fig.1 Black Soil Fig.2 Red Soil Fig 3.Brown Soil 4. MATERIAL AND METHODS Material 1. Eight 9ml dilution tubes of sterile saline and Distilled water 2. Eight nutrient agar plates 3. 1.0 ml and 0.1 ml pipets 4. Glass spreader aka (hockey stick) 5. 95% ethyl alcohol in glass beaker (WARNING: Keep alcohol away from flame!!) 6. Mixed overnight broth culture of Staphylococcus aureus and Serratia marcescens Methods 1. Serial Dilution technique 2. Pour Plate technique 3. Spreads Plate technique © 2023 IJNRD | Volume 8, Issue 1 January 2023 | ISSN: 2456-4184 | IJNRD.ORG IJNRD2301108 International Journal of Novel Research and Development (www.ijnrd.org) b58 Agar Media Preparation Weigh all the ingredients as per given quantity add all ingredients in proper proportion into water in conical flask apply heat to dissolve and then cover its mouth. Place this conical flask for sterilization at autoclave for sterile of agar solution. After sterilization pour agar solution in petri plate make them solidify. All this process carried out in aseptic area. Tab. 1: Composition of Agar Media Ingredient Amount (gm/ml) Uses Peptone 5gm Growing Bacteria Yeast extract 3gm Additive for culture media. Nacl 8gm Preparing microbiological culture Media Agar 15gm For Growing microorganism Distilled water 1000ml Vehicle Serial dilution Technique 1. Take 9 ml of sterile physiological saline or sterile distilled water in a sterile test tube numbering from 1 to 5 the number of dilution is not fix for any particular bacteria suspension / Specimen. 2. Transfer 1 ml of suspension of a given mixed culture to tast tube to shake the tube well. 3. Transfer 1 ml of suspension for test tube 1 to test tube to shake well. 4. Repeat this same process after the other test tube so has to get the desired diutions 5. Transfer 1 ml of the suspension for each dilution to sterile perti plates, separately to which melted and cooled Agar medium is poured. 6. The plates are rotated gently to solidity and incubated at 35°C for 25 to 48 hours to inverted positions. © 2023 IJNRD | Volume 8, Issue 1 January 2023 | ISSN: 2456-4184 | IJNRD.ORG IJNRD2301108 International Journal of Novel Research and Development (www.ijnrd.org) b59 2. Pour Plate Method 1. Take 9 ml of sterile physiological saline or sterile distilled water in a sterile test tube numbering from 1 to 5 the number of dilution is not fix for any particular bacteria suspension / Specimen. 2. Transfer 1 ml of suspension of a given mixed culture to tast tube to shake the tube well. 3. Transfer 1 ml of 101, 102, 103, 104, 105, 106, 107, 108 suspension for test tube 1 to test tube to shake well. 4. Repeat this same process after the other test tube so has to get the desired dilutions. 5. Transfer 1 ml of the suspension for each dilution to sterile perti plates, separately to which melted and cooled Agar medium is poured. 6. The plates are rotated gently to solidity and incubated at 35°C for 25 to 48 hours to inverted positions. 3. Spreads Plates Techniques 1. Take three nutrient Agar plates and label them with the name of the organism to be inoculated 2. Aseptically inoculate the plates with a loopful of the given organism. © 2023 IJNRD | Volume 8, Issue 1 January 2023 | ISSN: 2456-4184 | IJNRD.ORG IJNRD2301108 International Journal of Novel Research and Development (www.ijnrd.org) b60 3. Place plate 1 on the turn plate table. 4. Sterilize the spreader by putting it first in ethanol (95℅) in a beaker, than on the flame of bunsen burner and cool the rod for the 3 second. 5 . Remove the lid of plate and Spin the turn table. 6 . Touch the Spreader gently on the surface ofi agar and move it for the and back to Spread bacterial cells on the agar surface when the turn table is stanning 7. When turn table stop spinning but the lid over the lower half of petri dish. 8. Sterilize the Spreader again and reapeat the same process for the other two platesplates. 9. Incubate all the plates at 31°c for 24 hours. 4. Colney Count Turn ON the Instrument by pressing the On/Off Switch. Place the Petri Plate on the Glass grid. register a count, there will be a beep and an ink dot will be marked on the Petri Dish. Continue till all the colonies are thus counted. © 2023 IJNRD | Volume 8, Issue 1 January 2023 | ISSN: 2456-4184 | IJNRD.ORG IJNRD2301108 International Journal of Novel Research and Development (www.ijnrd.org) b61 Observation Fig Serial Dilution atechnique Fig Pour Plate technique Fig Pour Plate technique Fig Colney Count © 2023 IJNRD | Volume 8, Issue 1 January 2023 | ISSN: 2456-4184 | IJNRD.ORG IJNRD2301108 International Journal of Novel Research and Development (www.ijnrd.org) b62 Colney Count 10¹ 10² 10³ 10⁴ 105 106 107 108 S1 70 63 50 46 34 32 26 18 S2 61 58 47 40 38 36 29 20 S3 40 31 25 24 22 15 13 11 Identification Test Soil Samples Bacteria Gram Positive Gram Negative Red Soil Steptomycetaceae +ve - Brown Soil Actenomycetes +ve - Black Soil Azatobacter - -ve Result: Isolation of pure culture and identified bacteria from soil samples was performed. Conclusion Isolation of soil bacteria from different places were differnciate. The different soil having different range of bacteria some are useful and some were harmful. This is carried out by the method of Isolation of Pure culture, in this maintain all the sterility for growth of microorganisms in soil. REFERENCE 1) Farquhar J, Bao H, Thiemens M. Atmospheric influence of Earth’s earliest sulfur cycle. sci. 2000; 289(5480): 756-758. 2) Canfield DE. Oxygen: a four billion years history. Princeton University Press. 2014. 3)Falkowski PG. Life’s engines: how microbes made Earth habitable. Princeton University Press 2015; Vol. 24. 4) Hayat R, Ali S, Amara U, Khalid R, Ahmed I. Soil beneficial bacteria and their role in plant growth promotion: a review. Ann Microbiol. 2010; 60(4): 579-598. 5)Sturz AV, Nowak J. Endophytic communities of rhizobacteria and the strategies required to create yield enhancing associations with crops. APPL SOIL ECOL 2000; 15(2): 183-190. 6) Shoebitz M, Ribaudo CM, Pardo MA, Cantore ML, Ciampi L, et al. Plant growth promoting properties of a strain © 2023 IJNRD | Volume 8, Issue 1 January 2023 | ISSN: 2456-4184 | IJNRD.ORG IJNRD2301108 International Journal of Novel Research and Development (www.ijnrd.org) b63 of Enterobacter ludwigii isolated from Lolium perenne rhizosphere. Soil Biol Biochem 2009; 41(9): 1768-1774. 7) Burd GI, Dixon DG, Glick BR. Plant growth-promoting bacteria that decrease heavy metal toxicity in plants. Can J Microbiol. 2000; 46(3): 237-245. 8) Cocking EC. Endophytic colonization of plant roots by nitrogen-fixing bacteria. Plant Soil 2003; 252(1): 169-17 9) Ma Y, Oliveira RS, Wu L, Luo Y, Rajkumar M, Rocha I, Freitas H. Inoculation with metal-mobilizing plant-growth-promoting rhizobacterium Bacillus sp. SC2b and its role in rhizoremediation. J Toxicol Environ. 2015; 78(13-14): 931-944. 10) Awasthi R, Tewari R, Nayyar H. Synergy between plants and P-solubilizing microbes in soils: effects on growth and physiology of crops. Int Res J Microbiol 2011; 2(12): 484-503 11). Chitra Bhattacharyaa, P. Harsha, Sadhana Gupta, Soni Roy. Isolation and characteriazation of bacterial isolates from agricultural soil at Durg District. Indian J Sci. Res. 2014; 4(1):221-226, 12) Collins CH, Lyne PM, Grange GM. Collins and Lyne Microbiological methods, 6th Edition. Butterworth, London, 1989. 13). Diaz E. Bacterial degradation of aromatic pollutants:aparadigm of metabolic versatility. Int. Microbiol, 2004, 173-180. 14). Harold J. Microbiological Applications. Laboratory Manuals in General Microbiology, 8th Edition. McGraw - Hill Higher Education, 2002. 15). Iyengar, S, Bhave PP. In-vessel composting of household wastes. Waste Management, 2005, 1-11. 16). Kummerer K. Resistance in the environment. Journal ofAntimicrobial Chemotherapy. 2004; 45:311-320 |
11134 | https://www.statpearls.com/nursepractitioner/ce/activity/95923 | CE Activity | Mallory-Weiss Syndrome | NPs
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Mallory-Weiss Syndrome
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Mallory-Weiss syndrome is one of the common causes of acute upper gastrointestinal bleeding and is characterized by the presence of longitudinal superficial mucosal lacerations (Mallory-Weiss tears). These tears occur primarily at the gastroesophageal junction and may extend proximally to involve the lower to mid-esophagus or distally to involve the proximal portion of the stomach. This activity reviews the etiology, pathogenesis, evaluation, and management of Mallory-Weiss syndrome and highlights the role of the interprofessional team in evaluating and treating patients with this condition.
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11135 | https://pubmed.ncbi.nlm.nih.gov/23501471/ | Duplications of hox gene clusters and the emergence of vertebrates - PubMed
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. 2013 Jun 15;378(2):194-9.
doi: 10.1016/j.ydbio.2013.03.004. Epub 2013 Mar 15.
Duplications of hox gene clusters and the emergence of vertebrates
Natalia Soshnikova1,Romain Dewaele,Philippe Janvier,Robb Krumlauf,Denis Duboule
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1 Department of Genetics and Evolution, University of Geneva, Sciences III, Quai Ernest-Ansermet 30, 1211 Geneva 4, Switzerland.
PMID: 23501471
DOI: 10.1016/j.ydbio.2013.03.004
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Duplications of hox gene clusters and the emergence of vertebrates
Natalia Soshnikova et al. Dev Biol.2013.
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. 2013 Jun 15;378(2):194-9.
doi: 10.1016/j.ydbio.2013.03.004. Epub 2013 Mar 15.
Authors
Natalia Soshnikova1,Romain Dewaele,Philippe Janvier,Robb Krumlauf,Denis Duboule
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1 Department of Genetics and Evolution, University of Geneva, Sciences III, Quai Ernest-Ansermet 30, 1211 Geneva 4, Switzerland.
PMID: 23501471
DOI: 10.1016/j.ydbio.2013.03.004
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The vertebrate body plan is characterized by an increased complexity relative to that of all other chordates and large-scale gene amplifications have been associated with key morphological innovations leading to their remarkable evolutionary success. Here, we use compound full Hox clusters deletions to investigate how Hox genes duplications may have contributed to the emergence of vertebrate-specific innovations. We show that the combined deletion of HoxA and HoxB leads to an atavistic heart phenotype, suggesting that the ancestral HoxA/B cluster was co-opted to help in diversifying the complex organ in vertebrates. Other phenotypic effects observed seem to illustrate the resurgence of ancestral (plesiomorphic) features. This indicates that the duplications of Hox clusters were associated with the recruitment or formation of novel cis-regulatory controls, which were key to the evolution of many vertebrate features and hence to the evolutionary radiation of this group.
Copyright © 2013 Elsevier Inc. All rights reserved.
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11136 | https://www2.isye.gatech.edu/~nemirovs/MPSerB_Nemirovski.pdf | Mathematical Programming manuscript No.
(will be inserted by the editor) Arkadi Nemirovski Sums of Random Symmetric Matrices and Quadratic Optimization under Orthogonality Constraints Received: date / Accepted: date Abstract Let Bi be deterministic real symmetric m × m matrices, and ξi be independent random scalars with zero mean and “of order of one” (e.g., ξi ∼N(0, 1)). We are interested to know under what conditions “typ-ical norm” of the random matrix SN = N P i=1 ξiBi is of order of 1. An evi-dent necessary condition is E{S2 N} ⪯O(1)I, which, essentially, translates to N P i=1 B2 i ⪯I; a natural conjecture is that the latter condition is sufficient as well. In the paper, we prove a relaxed version of this conjecture, specifi-cally, that under the above condition the typical norm of SN is ≤O(1)m 1 6 : Prob{∥SN∥> Ωm1/6} ≤O(1) exp{−O(1)Ω2} for all Ω> 0 We outline some applications of this result, primarily in investigating the quality of semidef-inite relaxations of a general quadratic optimization problem with orthogo-nality constraints Opt = max Xj∈Rm×m © F(X1, ..., Xk) : XjXT j = I, j = 1, ..., k ª , where F is quadratic in X = (X1, ..., Xk). We show that when F is con-vex in every one of Xj, a natural semidefinite relaxation of the problem is tight within a factor slowly growing with the size m of the matrices Xj: Opt ≤Opt(SDP) ≤O(1)[m1/3 + ln k]Opt.
research was partly supported by the Binational Science Foundation grant #2002038 A. Nemirovski ISYE, Georgia Institute of Technology, 765 Ferst Drive NW Atlanta GA 30332-0205 USA, on leave from the Technion – Israel Institute of Technology.
Tel. +1-404-385-0769 Fax +1-404-894-2301 E-mail: nemirovs@isye.gatech.edu 2 Arkadi Nemirovski Keywords large deviations · random perturbations of linear matrix inequalities · semidefinite relaxations · orthogonality constraints · Procrustes problem Mathematics Subject Classification (2000) 60F10 · 90C22 · 90C25 · 90C59 1 Introduction In this paper, we address the following question: (Q): Let Ξi, 1 ≤i ≤N, be independent random m × m symmetric matrices with zero mean and “light-tail” distributions, and let SN = N P i=1 Ξi. Under what conditions a “typical value” of ∥SN∥is “of order of 1” so that the probability for ∥SN∥to be ≥Ωgoes to 0 exponentially fast as Ω> 1 grows?
Here and in what follows ∥A∥denotes the standard spectral norm (the largest singular value) of a matrix A.
This informal question admits various formal settings; to motivate the one we focus on, we start with describing two applications we intend to consider: tractable approximations of randomly perturbed Linear Matrix Inequalities (LMI) and semidefinite relaxations of nonconvex quadratic minimization un-der orthogonality constraints.
Randomly perturbed LMI’s. Consider a randomly perturbed LMI A0[x] − N X i=1 ξiAi[x] ⪰0, (1) where A0[x], ..., AN[x] are affine functions of the decision vector x taking values in the space Sm of symmetric m×m matrices, and ξi are independent of each other random perturbations (which w.l.o.g. can be assumed to have zero means). Constraints of this type arise in many applications, e.g., in various optimization and control problems with randomly perturbed data. A natural way to treat a randomly perturbed constraint is to pass to its chance form, which in the case of constraint (1) is the deterministic constraint Prob ( ξ = (ξ1, ..., ξN) : A0[x] − N X i=1 ξiAi[x] ⪰0 ) ≥1 −ϵ, (2) where ϵ > 0 is a small tolerance. The resulting chance constraint, however, typically is “heavily computationally intractable” – usually, the probability in the left hand side cannot be computed efficiently, and its reliable estima-tion by Monte-Carlo techniques requires samples of order of ϵ−1, which is prohibitively time-consuming when ϵ is small (like 1.e-6 or 1.e-8). In the rare cases when this difficulty can be circumvented (e.g., when ϵ is not too small), one still has a severe problem: chance constraint (2) defines, in general, a Sums of random symmetric matrices 3 nonconvex set in the space of x-variables, and therefore it is absolutely un-clear how to optimize under this constraint. A natural way to overcome this difficulty is to replace “intractable” chance constraint (2) with its “tractable approximation” – an explicit convex constraint on x such that its validity at a point x implies that x is feasible for (2). Assuming the distribution of ξ symmetric w.r.t. the origin and ϵ < 0.5, an evident necessary condition for x to be feasible for (2) is A0[x] ⪰0; strengthening this necessary condi-tion to A0[x] ≻0, x is feasible for the chance constraint if and only if the SN = N P i=1 ξiA−1/2 0 [x]Ai[x]A−1/2 0 [x] | {z } Ξi ⪯Im with probability ≥1 −ϵ. Assuming, as it is typically the case, that the distributions of ξi are symmetric, this condition is essentially the same as the condition ∥SN∥≤1 with probability ≥1 −ϵ. If we knew how to answer (Q), we could use this answer to build a “tractable” sufficient condition for the relation Prob{∥SN∥≤1} ≥1 −ϵ and thus build a tractable approximation of (2).
Nonconvex quadratic optimization under orthogonality constraints. Here we present a single example – the Procrustes problem, postponing the in-depth considerations till section 4. In the Procrustes problem, one is given matrices a[k], k = 1, ..., K, of the same size m × n and is looking for K orthogonal n × n matrices x[k] minimizing the objective X 1≤k 2, the problem is intractable (for K = 2, there is a closed form solution); it, however, allows for a straightforward semidefinite relaxation.
Let X = X[x, ..., x[K]] be the symmetric matrix defined as follows: the rows and the columns in X are indexed by triples (k, i, j), where k runs from 1 to K and i, j run from 1 to n; the entry Xkij,k′i′j′ in X is xij[k]xi′j′[k′].
Note that X is symmetric positive semidefinite matrix of rank 1. Further, the relation x[k]xT [k] = In is equivalent to a certain system Sk of linear equations on the entries of X, while the relation xT [k]x[k] = In (in fact equivalent to xT [k]x[k] = In) is equivalent to another system Tk of linear equations on the entries of X. Finally, the objective in (P) is a linear function Tr(AX) of X, 4 Arkadi Nemirovski where A is an appropriate symmetric matrix of the same size Kn2 × Kn2 as X. It is immediately seen that (P) is equivalent to the problem max X∈SKn2 {Tr(AX) : X ⪰0, X satisfies Sk, Tk, k = 1, ..., K, Rank(X) = 1} ; removing the only troublemaking constraint Rank(X) = 1, we end up with an explicit semidefinite program max X∈SKn2 {Tr(AX) : X ⪰0, X satisfies Sk, Tk, k = 1, ..., K} (SDP) which is a relaxation of (P), so that Opt(SDP) ≥Opt(P). We shall see in section 4 that an appropriate answer to (Q) allows to prove that Opt(SDP) ≤O(1)(n 1 3 + ln K)Opt(P), (3) and similarly for other problems of quadratic optimization under orthogo-nality constraints. To the best of our knowledge, (3) is the first nontrivial bound on the quality of semidefinite relaxation for problems of this type.
Note that the result we have just mentioned heavily depends on the fact that before passing to the semidefinite relaxation, we represent the orthogo-nality constraint on x[k] “redundantly” – as a pair of (in fact, equivalent to each other) constraints xT [k]x[k] = In, x[k]xT [k] = In. The importance of such a redundant representation of an orthogonality constraint in the relax-ation context was discovered in .
The outlined applications motivate our specific approach to treating (Q).
First, we are interested in the case when the size m of the random matrices in question can be large, and pay primary attention on how this size enters the results (as we shall see, this is the only way to get nontrivial bounds for our second application). In this respect, our goals are similar to those pursued in huge literature on large-scale random matrices inspired by appli-cations in Physics. However, we cannot borrow much from this literature, since the assumptions which are traditional there (appropriate pattern of independence/weak dependence of entries in SN) makes no sense for our ap-plications. What we are interested in when answering (Q), are conditions expressed in terms of distributions of random terms Ξi in SN. Let us try to understand what could be the “weakest possible” condition of this type. In the case when Prob {∥SN∥> Ω} goes rapidly to 0 as Ω> 1 grows, we clearly should have E © S2 N ª ⪯O(1)Im (since S2 N ⪯∥SN∥2Im). Thus, the condition £ E © S2 N ª = ¤ N X i=1 E © Ξ2 i ª ⪯O(1)Im (4) is necessary for ∥SN∥to be “of order of 1”. A natural guess is that this neces-sary condition plus appropriate “light-tail” assumptions on the distributions of Ξi is sufficient for the property in question; we shall see in a while that if this guess were true, it would provide us with all we need in our applications.
Unfortunately, when interpreted literally, the guess fails to be true. First, it is immediately seen that in fact O(1)Im in the right hand side of (4) should Sums of random symmetric matrices 5 be reduced to O(1) 1 ln mIm. Indeed, let Ξi be diagonal matrices with indepen-dent (from position to position and for different i’s) diagonal entries taking values ±αN −1/2 with probabilities 1/2, so that Σ ≡ N X i=1 E © Ξ2 i ª = α2Im.
Here SN is a random diagonal matrix with i.i.d. diagonal entries; by Central Limit Theorem, the distribution of these entries approaches, as N grows, the Gaussian distribution N(0, α2). It follows that when N is large, the typical value of ∥SN∥is the same as the typical value of max i≤m |ζi|, with indepen-dent ζi ∼N(0, α2); in other words, for large N the typical value of ∥SN∥is α √ 2 ln m. In order for this quantity to be of order of 1, α should be of order of (ln m)−1/2, which corresponds to Σ of order of (ln m)−1Im rather than of order of Im. In our context, the consequences of the outlined correction are not that dramatic, since ln m, for all practical purposes, is a moderate constant. A less pleasant observation is that the corrected guess still fails to be true, unless we impose further restrictions on the distributions of Ξi. In-deed, consider the case when m = 2k is even, N = 1, and the random matrix Ξ1 = SN is · ηξT ξηT ¸ , where η is uniformly distributed on the unit sphere in Rk, ξ ∼N(0, Ik) and η, ξ are independent. In this case, direct computa-tion demonstrates that E © Ξ2 1 ª = Im, while ∥Ξ1∥= ∥SN∥= ∥η∥2∥ξ∥2, so that the typical value of ∥SN∥is as large as O(√m). It follows that in order to make our guess valid for the particular case we are considering, the right hand side in (4) should be reduced to O(1)m−1Im. After such a correction, our guess does become valid, but the correction itself turns out to be too bad for our tentative applications. What we intend to do is to try to save the “log-arithmically corrected” guess at the cost of restricting Ξi to be semi-scalar, that is, to be random matrices of the form ξiBi, where Bi are deterministic symmetric matrices and ξi are independent random scalars with zero mean and light-tail distributions. Specifically, we make a conjecture as follows: Conjecture 1 Let Bi, i = 1, ..., N, be deterministic symmetric m×m matrices such that N X i=1 B2 i ⪯Im, (5) and let ξi, i = 1, ..., N, be independent random scalars with zero mean and “of order of 1”, e.g., such that (a) |ξi| ≤1, or (b) ξi ∼N(0, 1), or (c) E © exp{ξ2 i } ª ≤exp{1}. Then Ω≥O(1) √ ln m ⇒Prob ½ ξ = (ξ1, ..., ξN) : ∥ N P i=1 ξiBi∥≥Ω ¾ ≤O(1) exp{−O(1)Ω2} (6) with appropriate positive absolute constants O(1).
6 Arkadi Nemirovski It turns out that (6) would satisfy all the requirements posed by the appli-cations we bear in mind. Unfortunately, for the time being we are unable to prove the conjecture “as it is”. The primary goal of this paper is to prove a weaker statement – the one where √ ln m in the premise of (6) is replaced with m 1 6 , and to use this weaker fact in the applications we have mentioned.
In our opinion, question (Q) in general, and its specialization as presented in Conjecture 1, in particular are quite natural and deserve attention by their own right. Surprisingly, the only, to the best of our knowledge, result in this direction which makes no assumptions on how strong the entries in SN depend on each other, is recent result announced in (for proof, see ) as follows: Proposition 1 Let Ξi be independent symmetric m × m matrices with zero mean such that E © exp{∥Ξi∥2σ−2 i } ª ≤exp{1}, i = 1, ..., N (σi > 0 are deterministic scale factors). Then Prob ∥SN∥≥t v u u t N X i=1 σ2 i ≤O(1) exp{−O(1) t2 ln m} ∀t > 0, (7) with positive absolute constants O(1).
From Proposition 1 it follows that when the premise (5) in Conjecture 1 is strengthened to N P i=1 ∥Bi∥2 ≤1, the conjectured conclusion becomes “nearly true”. Indeed, with P i ∥Bi∥2 ≤1 and independent ξi with zero means and of order of 1, applying Proposition 1 with Ξi = ξiBi, σi = ∥Bi∥and t = Ω √ ln m, we get Prob ( ξ = (ξ1, ..., ξN) : ∥ N X i=1 ξiBi∥≥Ω √ ln m ) ≤O(1) exp{−O(1)Ω2}, which is nearly as good as (6). Unfortunately, in the applications we intend to consider strengthening (5) to the scalar inequality P i ∥Bi∥2 ≤1 is too costly to be of actual use.
The rest of the paper is organized as follows. In section 2, we prove that our conjecture, in its outlined weaker form, indeed is valid. In sections 3 and 4 we apply this result to approximating chance constraints associated with randomly perturbed LMI’s, and to deriving bounds on the quality of semidef-inite relaxations of problems of quadratic approximation under orthogonality constraints.
Sums of random symmetric matrices 7 2 Main result 2.1 Preliminaries: Talagrand’s Inequality We start with the following instrumental fact: Theorem 1 [Talagrand’s Inequality] Let (Ei, ∥· ∥i), i = 1, ..., N, be finite-dimensional normed spaces and µi, i = 1, ..., N, be Borel probability measures on the balls Vi = {xi ∈Ei : ∥xi∥i ≤1/2}. Let us equip the space E = E1 × ... × EN with the norm ∥(x1, ..., xN)∥= s N P i=1 ∥xi∥2 i and with the probability distribution µ which is the product of µ1, ... µN, and let A be a closed convex set in E such that µ(A) > 0. Then Z E exp{ dist2 ∥·∥(x, A) 4 }µ(dx) ≤ 1 µ(A), (8) where dist∥·∥(x, A) = min z∈A ∥x −z∥.
In this form, the Talagrand Inequality is proved in , up to the only dif-ference that in , the supports of µi are assumed to be finite subsets of Vi.
However, finiteness of the supports is of no importance, since a Borel prob-ability measure on Vi can be weakly approximated by probability measures with finite supports contained in Vi.
2.2 Main result Our main result related to question (Q) is as follows: Theorem 2 Let ξ1, ..., ξN be independent random variables with zero mean and zero third moment taking values in [−1, 1], Bi, i = 1, ..., N, be determin-istic symmetric m × m matrices, and Θ > 0 be a real number such that X i B2 i ⪯Θ2I.
(9) Then Ω≥7m1/4 ⇒Prob{∥ N P i=1 ξiBi∥≥ΩΘ} ≤5 4 exp{−Ω2 32 } (a) Ω≥7m1/6 ⇒Prob{∥ N P i=1 ξiBi∥≥ΩΘ} ≤22 exp{−Ω2 32 } (b) (10) Proof. 10. For x ∈RN, let S(x) = N P i=1 xiBi. We start with the following simple observation: Lemma 1 One has ∀(γ > 0, s > 1) : Prob{∥S(ξ)∥> 2sγΘ} ≤ 1 Pr{∥S(ξ)∥≤2γΘ} exp{−(s−1)2γ2 4 }.
(11) 8 Arkadi Nemirovski Proof. Let Q = {x ∈RN : ∥S(x)∥≤Θ}. The set Q is a closed and symmetric w.r.t. the origin convex set. We claim that it contains the unit ∥·∥2-ball centered at the origin. To prove this is exactly the same as to verify that if ∥x∥2 ≤1, then |yT (P i xiBi)y| ≤Θ for all y ∈Rm with yT y ≤1, and here is the required verification: |yT ( X i xiBi)y| ≤ X i |xi|∥Biy∥2 ≤ ÃX i x2 i !1/2 ÃX i yT B2 i y !1/2 ≤Θ.
Now we are ready to prove (11). Let us fix s > 1 and γ > 0, and let A = γQ.
The set A is closed and convex and contains the ∥·∥2-ball of radius γ centered at the origin. It follows that x ̸∈sA ⇒dist∥·∥2(x, A) ≥(s −1)γ.
(12) Indeed, since s > 1, relation x ̸∈sA implies that the set B = x −(s −1)A does not intersect A; since A ⊃{z : ∥z∥2 ≤γ}, we have B ⊃{z : ∥x −z∥2 ≤ (s −1)γ}, and thus dist∥·∥2(x, A) > (s −1)γ due to B ∩A = ∅.
From (12) it follows that Prob{ξ/2 ̸∈sA} exp{(s −1)2γ2 4 } ≤E ( exp{ dist2 ∥·∥2(ξ/2, A) 4 } ) , whence, applying the Talagrand Inequality (Theorem 1) to the distribution of the random vector ζ = ξ/2, Prob{ξ/2 ̸∈sA} ≤ 1 Prob{ξ/2 ∈A} exp{−(s −1)2γ2 4 }.
(13) Recalling the definitions of A and Q, we clearly have Prob{ξ/2 ̸∈sA} = Prob{∥S(ξ)∥> 2sγΘ}, Prob{ξ/2 ∈A} = Prob{∥S(ξ)∥≤2γΘ}, so that (13) implies (11).
⊓ ⊔ 20. Our next observation is as follows: Lemma 2 Let Θ > 0, let Bi ∈Sm be deterministic matrices satisfying (9) and ζi be independent random scalar variables such that E {ζi} = 0, E © ζ2 i ª ≤σ2, E © ζ4 i ª − ¡ E{ζ2 i } ¢2 ≤2σ4.
Let, finally, Sk = k P i=1 ζiBi, 1 ≤k ≤N. Then 1 ≤k ≤N ⇒E © S4 k ª ⪯3σ4Θ4I.
(14) Sums of random symmetric matrices 9 Proof. Setting S0 = 0, Ei = E{S4 i }, σi = ¡ E{ζ2 i } ¢1/2, ωi = ¡ E{ζ4 i } ¢1/4 and taking into account that ζi and Si−1 are independent with zero mean, we have Ei = E © [Si−1 + ζiBi]4ª = E ½ S4 i−1 + σ2 i · Si−1BiSi−1Bi + BiSi−1BiSi−1 | {z } ⪯Si−1B2 i Si−1+BiS2 i−1Bi due to XY T +Y XT ⪯XXT +Y Y T +S2 i−1B2 i + B2 i S2 i−1 +Si−1B2 i Si−1 + BiS2 i−1Bi ¸ + ω4 i B4 i ¾ ⪯E© S4 i−1 + 2σ2 i Si−1B2 i Si−1 + 2σ2 i BiS2 i−1Bi + S2 i−1(σ2 i B2 i ) + (σ2 i B2 i )S2 i−1 +σ4 i B4 i + (ω4 i −σ4 i )B4 i ª = Ei−1 + 2 i−1 P j=1 σ2 j σ2 i BjB2 i Bj + 2 i−1 P j=1 σ2 i σ2 j BiB2 j Bi + i−1 X j=1 σ2 i σ2 j B2 j B2 i + i−1 X j=1 σ2 i σ2 j B2 i B2 j + σ4 i B4 i | {z } = µ i P j=1 σ2 j B2 j ¶2 − µ i−1 P j=1 σ2 j B2 j ¶2 +[ω4 i −σ4 i ]B4 i whence Ek ⪯2 k P i=1 i−1 P j=1 σ2 i σ2 j BjB2 i Bj + 2 k P i=1 i−1 P j=1 σ2 i σ2 j BiB2 j Bi + µ k P j=1 σ2 j B2 j ¶2 + k P i=1 [ω4 i −σ4 i ]B4 i = 2 P 1≤i,j≤k i̸=j σ2 i σ2 j BiB2 j Bi + µ k P j=1 σ2 j B2 j ¶2 + k P i=1 £ ω4 i −σ4 i ¤ | {z } ≤2σ2 B4 i ⪯2 P 1≤i,j≤k i̸=j σ4BiB2 j Bi + µ k P j=1 σ2 j B2 j ¶2 + k P i=1 2σ4B4 i = 2σ4 k P i,j=1 BiB2 j Bi + µ k P j=1 σ2 j B2 j ¶2 = 2σ4 k P i=1 Bi · k P j=1 B2 j ¸ Bi + µ k P j=1 σ2 j B2 j ¶2 ⪯2σ4Θ2 k P i=1 B2 i + µ k X j=1 σ2 j B2 j | {z } =A,0⪯A⪯σ2Θ2I ¶2 ⪯(2σ4Θ4 + σ4Θ4)I, as claimed.
⊓ ⊔ Now we are ready to prove (10.a). Applying Lemma 2 to ζi = ξi (which allows us to take σ = 1), we get E © ∥S(ξ)∥4ª ≤E © Tr(S4(ξ)) ª ≤3mΘ4, whence by Tschebyshev inequality γ > 0 ⇒Prob {∥S(ξ)∥> 2γΘ} < 3m 16γ4 .
(15) 10 Arkadi Nemirovski Setting γ = m1/4 and invoking (11), we get ∀s > 1 : Prob © ∥S(ξ)∥> 2sm1/4Θ ª ≤ 1 Prob{∥S(ξ)∥≤2γΘ} exp{−(s−1)2m1/2 4 } ≤5 4 exp{−(s−1)2m1/2 4 }, (16) where the concluding inequality is given by (15). The resulting inequality (which is valid for all s > 1) immediately implies (10.a). Indeed, given Ω≥ 7m1/4 and setting s = Ω/(2m1/4), we have s ≥7/2, whence also s −1 ≥5 7s.
Therefore Prob{∥S(ξ)∥> ΩΘ} = Prob{∥S(ξ)∥≥2sm1/4Θ} ≤5 4 exp{−(s −1)2m1/2 4 } (see (16)), and at the same time (s−1)2m1/2 4 ≥(5/7)2s2m1/2 4 = 25Ω2 49·16 ≥Ω2 32 , and we arrive at Prob{∥S(ξ)∥> ΩΘ} ≤5 4 exp{−Ω2 32 }, as required in (10.a).
30. Now let us prove (10.b). We start with the following weak analogy to Lemma 2: Lemma 3 Let Bi, i = 1, ..., N, be deterministic symmetric matrices satis-fying (9), and ζi, i = 1, ..., N, be independent scalar random variables with zero mean and zero third moment such that σ2 i ≡E{ζ2 i } ≤σ2, ω4 i ≡E{ζ4 i } ≤ min £ σ4 i + 2σ4, ω4¤ , χ6 i ≡E{ζ6 i } ≤χ6, and let Sk = k P i=1 ζiBi. Then E © Tr(S6 k) ª ≤[45σ6 + 15ω4σ2 + χ6]Θ6m.
(17) Proof. Let φi = E © Tr(S6 i ) ª . Given a multi-index ι = (ι1, ..., ιn) with entries 0, 1 and two symmetric matrices P, Q, let [P, Q]ι stand for the product of n matrices, with ℓ-th factor being P or Q depending on whether ιℓ= 1 or ιℓ= 0 (e.g., P, Q = QP 2). Let I, J be the sets of 6-dimensional multi-indices ι with entries 0,1 such that exactly 4, respectively, 2 of the entries are equal to 1 (so that both I and J contain 15 multi-indices each). Taking into account that Si−1 has zero mean and is independent of ζiBi, and that ζi has zero first and third moments, we have E © S6 i ª = E © S6 i−1 ª + σ2 i X ι∈I E {[Si−1, Bi]ι} + ω4 i X ι∈J E {[Si−1, Bi]ι} + χ6 i B6 i , whence φi = φi−1 + σ2 i P ι∈I E {Tr([Si−1, Bi]ι)} + ω4 i P ι∈J E {Tr([Si−1, Bi]ι)} +χ6 i Tr(B6 i ).
(18) Now let us list all 15 products [Si−1, Bi]ι, ι ∈I; we split these products into groups, all members of the same group being of equal trace in view of the Sums of random symmetric matrices 11 identities Tr(A) = Tr(AT ) and Tr(AB) = Tr(BA). Here are the groups (to simplify notation, we skip indices of Si−1 and Bi) BS4B, S2B2S2, (S4B2, B2S4), (S3B2S, SB2S3) (a) SBS2BS, (S2BS2B, BS2BS2) (b) (S3BSB, BSBS3), (SBS3B, BS3BS), (S2BSBS, SBSBS2) (c) Let the traces of products in the respective groups be Ta = Ta,i(ζ1, ..., ζi−1), Tb = Tb,i(ζ1, ..., ζi−1), Tc = Tc,i(ζ1, ..., ζi−1). We have BS2 |{z} X BS2 |{z} Y T + S2BS2B | {z } Y XT ⪯BS4B | {z } XXT + S2B2S2 | {z } Y Y T , whence Tb ≤Ta, and similarly S2B |{z} X SBS | {z } Y T + SBSBS2 | {z } Y XT ⪯S2B2S2 | {z } XXT + SBS2BS | {z } Y Y T , whence 2Tc ≤Ta +Tb ≤2Ta. The conclusion is that the sum P ι∈I in (18) does not exceed the quantity Ii = 15E © Tr(BiS4 i−1Bi) ª = 15Tr(BiE{S4 i−1}Bi) = 15Tr(BiE{S4 i−1}Bi).
Invoking Lemma 2, we get Ii ≤45σ4Θ4Tr(B2 i ).
Completely similar reasoning as applied to the sum P ι∈J in (18) implies that this sum does not exceed the quantity Ji = 15E © Tr(Si−1B4 i Si−1) ª = 15E © Tr(B2 i S2 i−1B2 i ) ª = 15Tr à B2 i " i−1 P j=1 σ2 j B2 j # B2 i !
≤15Θ2σ2Tr(B4 i ) (see (9)). Thus, (18) implies that φi ≤φi−1 + 45σ6Θ4Tr(B2 i ) + 15ω4 i Θ2σ2Tr(B4 i ) + χ6 i Tr(B6 i ); since P i B2 i ⪯Θ2I, we have Tr(B4 i ) ≤Θ2Tr(B2 i ) and Tr(B6 i ) ≤Θ4Tr(B2 i ).
We arrive at the relation φi ≤φi−1 + Θ4 £ 45σ6 + 15ω4 i σ2 + χ6 i ¤ Tr(B2 i ).
Taking into account that P i B2 i ⪯Θ2I, whence of course P i Tr(B2 i ) ≤Θ2m, we conclude that E © Tr(S6 N) ª ≤[45σ6 + 15ω4σ2 + χ6]Θ6m, as claimed.
⊓ ⊔ 12 Arkadi Nemirovski Now we can derive (10.b) in the same fashion as (10.a). Applying Lemma 3 to ζi = ξi (which allows us to take σ = ω = χ = 1), we get E © ∥S(ξ)∥6ª ≤E © Tr(S6(ξ)) ª ≤61mΘ6, whence by Tschebyshev inequality γ > 0 ⇒Prob {∥S(ξ)∥> 2γΘ} < 61m 64γ6 .
(19) Invoking (11) with γ = m1/6, we get s > 1 ⇒Prob © ∥S(ξ)∥> 2sm1/6Θ ª = Prob {ξ/2 ̸∈sA} ≤ 1 Prob{ξ∈2A} exp{−(s−1)2m1/3 4 } ≤22 exp{−(s−1)2m1/3 4 }, where the concluding inequality is given by (19). The resulting inequality is valid for all s > 1, and (10.b) follows (cf. derivation of (10.a) at the end of item 20).
⊓ ⊔ Corollary 1 Let Ξ1,...,Ξn be independent Gaussian symmetric m × m ran-dom matrices with zero means and Θ > 0 be such that n X i=1 E © Ξ2 i ª ⪯Θ2Im.
(20) Then relations (10) hold true.
Proof. By evident reasons every Gaussian symmetric random matrix Ξi can be represented as M P t=1 ηitBit with independent ηit ∼N(0, 1) and deterministic symmetric matrices Bit; observe that E{Ξ2 i } = P t (Bit)2. Representing in this way every one of the matrices Ξ1, ..., Ξn and taking into account that the resulting Gaussian random variables {ηit} are mutually independent, we conclude that Sn ≡ n X i=1 Ξi = N X i=1 ξiBi with independent ξi ∼N(0, 1) and deterministic symmetric matrices Bi satisfying the relation P i B2 i ⪯Θ2I. Now let {ζij} 1≤i≤N j=1,2,... be a collection of independent random variables taking values ±1 with probabilities 1/2, and let Sn,ν = N X i=1 ν X j=1 ζij 1 √ν Bi.
By Theorem 2, we have Ω≥7m1/4 ⇒Prob {∥Sn,ν∥> ΩΘ} ≤5 4 exp{−Ω2 32 }, (a) Ω≥7m1/6 ⇒Prob {∥Sn,ν∥> ΩΘ} ≤22 exp{−Ω2 32 }. (b) As ν →∞, the distribution of Sn,ν, by Central Limit Theorem, converges weakly to the distribution of Sn, and (10) follows.
⊓ ⊔ Sums of random symmetric matrices 13 2.3 Non-symmetric case Question (Q) makes sense for non-symmetric (and even non-square) random matrices. In this case validity of Conjecture 1 would imply the following statement: (!) Let Ci be deterministic m × n matrices such that N X i=1 CiCT i ⪯Θ2Im, N X i=1 CT i Ci ⪯Θ2In (21) and ξi be independent random scalars with zero mean and of order of 1. Then Ω≥O(1) p ln(m + n) ⇒Prob ½ ξ : ∥ N P i=1 ξiCi∥≥ΩΘ ¾ ≤O(1) exp{−O(1)Ω2}.
(22) Indeed, in order to extract (!) from the assertion proposed by Conjecture 1, it suffices to apply the assertion to our ξi’s and the deterministic symmetric (m + n) × (m + n) matrices Bi = · CT i Ci ¸ .
(23) Utilizing in exactly the same fashion Theorem 2 and Corollary 1, we arrive at the following Proposition 2 Let deterministic m × n matrices Ci satisfy (21), and let ξi be independent random scalars with zero first and third moment and such that either |ξi| ≤1 for all i ≤N, or ξi ∼N(0, 1) for all i ≤N.
Ω≥7(m + n)1/4 ⇒Prob{∥ N P i=1 ξiCi∥≥ΩΘ} ≤5 4 exp{−Ω2 32 }, (a) Ω≥7(m + n)1/6 ⇒Prob{∥ N P i=1 ξiCi∥≥ΩΘ} ≤22 exp{−Ω2 32 }. (b) (24) We are about to add to Proposition 2 a simple additional statement, which allows us to strengthen the result in the case when one of the sizes m, n is much smaller than another: Proposition 3 Let Ci, ξi be as in Proposition 2. Then Ω≥4 p min[m, n] ⇒Prob{∥ N X i=1 ξiCi∥≥ΩΘ} ≤4 3 exp{−Ω2 16 }.
(25) Proof. It suffices to consider the case when |ξi| ≤1; the Gaussian version of the statement can be derived from the one with |ξi| ≤1 in exactly the same fashion as in the proof of Corollary 1.
Let Bi be given by (23). Same as in item 10 of the proof of Theorem 2, setting Q = {x ∈RN : ∥ N P i=1 xiBi∥≤Θ}, we conclude from (21) that the 14 Arkadi Nemirovski closed convex set Q contains the unit Euclidean ball centered at the origin, and that for every γ > 0 one has s > 1 ⇒ Prob ½ ∥ N P i=1 ξiBi∥> 2sγΘ ¾ ≤ exp{−(s −1)2γ2 4 } Prob ½ ξ:∥ N P i=1 ξiBi∥≤2γΘ ¾.
(26) Assume w.l.o.g. that min[m, n] = n. We have N P i=1 CT i Ci ⪯ΘIn, whence, taking traces, N P i=1 ∥Ci∥2 2 ≤nΘ2. It follows that E ( ∥ X i ξiCi∥2 2 ) = X i E © ξ2 i ª ∥Ci∥2 2 ≤ X i ∥Ci∥2 2 ≤nΘ2, whence by Tschebyshev inequality and due to ∥C∥≤∥C∥2 for all t > 0 one has Prob ( ∥ X i ξiCi∥≥tn1/2Θ ) ≤Prob ( ∥ X i ξiCi∥2 ≥tn1/2Θ ) ≤t−2.
Setting γ = n1/2, we conclude from the latter inequality that Prob ( ∥ X i ξiCi∥≥2γΘ ) ≤1/4, whence, in view of ∥P i ξiBi∥= ∥P i ξiCi∥, Prob ( ∥ X i ξiBi∥> 2γΘ ) = Prob {ξ ̸∈2γQ} ≤1/4.
Thus, (26) with γ = n1/2 implies that s > 1 ⇒Prob ( ∥ N X i=1 ξiBi∥> 2sn1/2Θ ) ≤4 3 exp{−(s −1)2n 4 }, and (25) follows (recall that ∥P i ξiCi∥≡∥P i ξiBi∥).
⊓ ⊔ Sums of random symmetric matrices 15 3 Application: Randomly perturbed Linear Matrix Inequality Consider a randomly perturbed Linear Matrix Inequality (LMI) A0[x] − N X i=1 ξiAi[x] ⪰0, (27) where A0[x], ..., AN[x] are symmetric matrices affinely depending on decision vector x, ξi, i = 1, ..., N, are random real perturbations which we assume to be independent with zero means “of order of 1” and with “light tails” (precise formulations of these two assumptions will be given later). We are interested to describe those x for which the randomly perturbed LMI (27) holds true with probability ≥1 −ϵ, where ϵ ≪1. Clearly, for such an x one should have A0[x] ⪰0. We will simplify a little bit our task and focus on points x with A0[x] ≻0. For such an x, setting Bi[x] = A−1/2 0 [x]Ai[x]A−1/2 0 [x], the question becomes to describe those x for which Prob ( N X i=1 ξiBi[x] ⪯I ) ≥1 −ϵ.
(28) Precise description seems to be completely intractable; what we are about to present are verifiable sufficient conditions for (28) to hold true.
3.1 Condition based on Proposition 1 Proposition 4 Let m ≥2, let perturbations ξi be independent with zero means and such that E © exp{ξ2 i } ª ≤exp{1}. Then the condition A0[x] ≻0 & N X i=1 ∥A−1/2 0 [x]Ai[x]A−1/2 0 [x]∥2 ≤ 1 450 exp{1}(ln 3 ϵ )(ln m) (29) is sufficient for (27) to be valid with probability ≥1 −ϵ.
This is a straightforward corollary of Proposition 1 (we use the actual values of absolute constants in (7) presented in ).
A severe shortcoming of (29) is that this condition, although verifiable, in general defines a nonconvex set in the space of decision variables x, which makes it problematic to optimize in x under the conditions. There are, how-ever, two simple cases when the conditions are free of this shortcoming. The first is when Ai[x] are independent of x (“perturbations in the constant term of LMI”); here the “problematic” part of the conditions – the inequality N X i=1 ∥A−1/2 0 [x]Ai[x]A−1/2 0 [x]∥2 ≤τ (∗) on x, τ – can be represented by the system of convex inequalities −A0[x] ⪯µiAi ⪯A0[x], µi > 0, i = 1, ..., N, N X i=1 µ−2 i ≤τ.
16 Arkadi Nemirovski in variables x, µi, τ. The second “good” case is the one when A0[x] ≡A is constant. Here (∗) can be represented by system of convex constraints −λiA ⪯Ai[x] ⪯λiA, i = 1, ..., N, X i λ2 i ≤τ in variables x, λi, τ.
3.2 Conditions based on Theorem 2 and Corollary 1 With these statements in the role of Proposition 1, we arrive at the following statement: Proposition 5 Let perturbations ξi be independent with zero means and zero third moments and either such that |ξi| ≤1, i = 1, ..., N, or such that ξi ∼ N(0, 1), i = 1, ..., N. Let, further, ϵ ∈(0, 1) be such that one of the following two conditions is satisfied: (a) ln ¡ 5 4ϵ ¢ ≥49m1/2 32 (b) ln ¡ 22 ϵ ¢ ≥49m1/3 32 (30) Then the condition A0[x] ≻0 & ∥ N X i=1 (A−1/2 0 [x]Ai[x]A−1/2 0 [x])2∥≤ ( 1 32 ln( 5 4ϵ), case (30.a) 1 32 ln( 22 ϵ ), case (30.b) (31) is sufficient for (27) to be valid with probability ≥1 −ϵ.
Note that condition (31), in contrast to (29), defines a convex domain in the space of design variables. Indeed, this condition is of the form A0[x] ≻0 & N X i=1 Ai[x]A−1 0 [x]Ai[x] ⪯c(ϵ)A0[x], which can be represented by system of LMI’s A0[x] ≻0 & · Yi Ai[x] Ai[x] A0[x] ¸ ⪰0, i = 1, ..., N & N X i=1 Yi ⪯c(ϵ)A0[x] (32) in variables x, Yi. Note also that in Control applications (which are of primary importance for randomly perturbed LMI) m does not exceed few tens, and in this range of values of m the only advantage of (29) as compared with (31), that is, ln(m) in the right hand side of (29) vs. m1/2 and m1/3 in the right hand side of (30), becomes unimportant (in fact, (29), because of large constant factors, in a reasonable range of values of m leads to much more conservative conclusions than (30)).
Sums of random symmetric matrices 17 4 Application: semidefinite relaxation of quadratic minimization under orthogonality constraints 4.1 Problem of interest Consider the following optimization problem: max x∈Mm,n ⟨x, Ax⟩: ⟨x, Bx⟩≤1 (a) ⟨x, Bℓx⟩≤1, ℓ= 1, ..., L (b) Cx = 0 (c) ∥x∥≤1 (d) (P) where – Mm,n is the space of m × n real matrices equipped with the Frobenius inner product ⟨x, y⟩= Tr(xyT ), and ∥x∥= max ξ {∥xξ∥2 : ∥ξ∥2 ≤1} is, as always, the spectral norm of x ∈Mm,n, – the mappings A, B, Bℓare symmetric linear mappings from Mm,n into itself, – B is positive semidefinite of rank 1, – Bℓ, ℓ= 1, ..., L, are positive semidefinite, – C is a linear mapping from Mm,n into RM.
Note that (P) covers a number of problems of quadratic optimization under orthogonality constraints, e.g.
1. Inhomogeneous modification max x∈Mm,n ⟨x, Ax⟩+ 2⟨b, x⟩: ⟨x, Bx⟩≤1 (a) ⟨x, Bℓx⟩≤1, ℓ= 1, ..., L (b) Cx = 0 (c) ∥x∥≤1 (d) (P+) of (P). Indeed, partitioning a matrix y ∈Mn+1,m+1 as ·y00 y01 y10 y11 ¸ with scalar y00, (P+) is equivalent to the problem max y ⟨y, A+y⟩≡⟨y11, Ay11⟩+ 2y00⟨b, y11⟩: ⟨y11, By11⟩≤1; ⟨y11, Bℓy11⟩≤1, ℓ= 1, ..., L; Cy11 = 0, y01 = 0; y10 = 0, ∥y∥≤1.
of the form of (P); 2. Orthogonal relaxation of the quadratic assignment problem (see [12–14] and references therein) max X © Tr(BXAXT ) −2Tr(CX) : X ∈Mm,m, XXT = Im ª (QA) with symmetric m × m matrices A, B. Indeed, the transformation B ← B + bIm converts (QA) into an equivalent problem, thus we can assume 18 Arkadi Nemirovski that B ≻0. Similarly, the transformation A ←A + aIm converts (QA) into equivalent problem, thus we can assume that A ≻0. In the case when B ≻0, A ≻0, representing B = D2 and A = E2 with symmetric D, E, we see that the objective in (QA) is f(X) = Tr([DXE][DXE]T )+2Tr(CX), which is a convex quadratic form of X. Consequently, the maximum of f over the set {X ∈Mm,m : XXT = Im} is exactly the same as the maximum of f over the set {X ∈Mm,m : ∥X∥≤1}, since the former set is exactly the set of extreme points of the latter one1). Thus, (QA) is equivalent to the problem of the form max X © Tr( ¯ BX ¯ AXT ) −2Tr(CX) : ∥X∥≤1 ª , which is of the form of (P+); 3. Procrustes problem which can be posed as (see Introduction) max X[·] X 1≤ℓ<ℓ′≤K Tr(A[ℓ]X[ℓ]XT [ℓ′]AT [ℓ′]) : X[ℓ]XT [ℓ] = In, ℓ= 1, ..., K (Pr) Indeed, the objective in (Pr) is linear in every one of X[ℓ]; thus, we do not affect the problem by relaxing the orthogonality constraints X[ℓ]XT [ℓ] = In to ∥X[ℓ]∥≤1. Indeed, such a relaxation could only increase the optimal value. This, however, does not happen, since given a feasible solution to the problem max X[·] X 1≤ℓ<ℓ′≤K Tr(A[ℓ]X[ℓ]XT [ℓ′]AT [ℓ′]) : ∥X[ℓ]∥≤1 ∀ℓ (Pr+) we can easily convert it into a feasible solution to (Pr) with the same or larger value of the objective.
Indeed, keeping X, ..., X[K] fixed, we can straightforwardly replace X by an orthogonal matrix without spoiling the objective value2).
After X is made orthogonal, we can repeat the outlined procedure 1 We are using the well-known fact: Extreme points P of the ∥· ∥-unit ball B in Mm,n are exactly the orthoprojectors, that is, P T P = In when n ≤m and PP T = Im when n ≥m. Here is the derivation (w.l.o.g., we assume that m ≤n). Let P ∈B.
Applying Singular Value decomposition, we get P = U[Diag{λ}, 0m,n−m]V with orthogonal m×m and n×n matrices U, V and a vector λ ∈Rm satisfying ∥λ∥∞≤1 (the latter - due to P ∈B). If all λj are ±1, then P is an orthoprojector; otherwise ∥λ ± e∥∞≤1 for properly chosen nonzero e ∈Rm, whence P = 1 2[P+ + P−], P± = U[Diag{λ ± e}, 0m,n−m]V , is not an extreme point of B (note that P+ ̸= P−, P+, P−∈B). Thus, every extreme point P of B satisfies PP T = Im. Vice versa, if P ∈Mm,n satisfies PP T = Im, then P ∈B. Assuming that P ± E ∈B for some E, all rows Pi ± Ei in the matrices P ± E must be of ∥· ∥2-norm not exceeding 1, while the rows Pi of P are of ∥· ∥2-norm equal to 1; this is possible only if all rows Ei of E are zero vectors. Thus, and m × n orthoprojector is an extreme point of B.
2 since the objective is linear in X, the remaining variable matrices being fixed, and thus attains its maximum in X varying in the set {X : ∥X∥≤1} at an extreme point of the set, which is an orthogonal matrix; this matrix is easily computable, given X, ..., X[K].
Sums of random symmetric matrices 19 with X in the role of X, and so on. After K steps we end up with a feasible solution to both (Pr+) and (Pr) which is at least as good as the solution we have started with.
It remains to note that problem (Pr+) is of the form of (P) – we can arrange all matrices X[ℓ], ℓ= 1, ..., K, in a large block-diagonal matrix x = X ...
X[K] , thus converting (Pr+) into the equivalent problem max x ( F(x) ≡ X ℓ<ℓ′ Tr(A[ℓ]X[ℓ]XT [ℓ′]AT [ℓ′]) : Cx = 0, ∥x∥≤1 ) where the homogeneous equations Cx = 0 express the fact that x is of the outlined block-diagonal form; the resulting problem is in the form of (P); 4. The problem max X,...,X[K] (X ℓ<ℓ′ ∥A[ℓ]X[ℓ] −A[ℓ′]X[ℓ′]∥2 2 : X[ℓ]XT [ℓ] = In, ℓ= 1, ..., k ) “opposite” to the Procrustes problem. Indeed, since the objective is con-vex in every one of X[ℓ], we, same as above, lose nothing when relaxing the constraints X[ℓ]XT [ℓ] = In to ∥X[ℓ]∥≤1. The resulting problem can be converted to the form of (P) in exactly the same manner as in the previous example. The same argument applies to a general-type prob-lem of quadratic maximization under orthogonality constraints, provided that the objective is convex in every one of the corresponding variable matrices.
Structure of (P). In some of the outlined examples we end up with a par-ticular case of problem (P) where the homogeneous linear constraints (c) in (P) imply that x is a block-diagonal matrix x1 ...
xK with mk × nk diagonal blocks xk, k = 1, .., K. We shall refer to ∆= {(mk, nk)}K k=1 as to the structure of (P). If the homogeneous constraints in (P) do not impose a nontrivial block-diagonal structure on x, then, by definition, B possesses the trivial structure ∆= (m, n) with K = 1.
Semidefinite relaxation of (P). Problem (P), in general, is NP-hard (this is the case already for the generic inhomogeneous (C ̸= 0) orthogonal relaxation of the quadratic assignment problem, see ). At the same time, (P) admits a straightforward semidefinite relaxation as follows. We can identify A in (P) with a symmetric mn × mn matrix A = [Aij,kℓ] with rows and columns indexed by pairs (i, j), 1 ≤i ≤m, 1 ≤j ≤n satisfying the relation [Ax]ij = X k,ℓ Aij,kℓxkℓ 20 Arkadi Nemirovski (from now on, unless stated otherwise, in a sum P p,q , p runs from 1 to m, and q runs from 1 to n). Similarly, B, Bℓcan be identified with symmetric positive semidefinite mn × mn matrices B, Bℓ, with B of rank 1. Finally, C can be identified with a M × mn matrix C = [Cµ,ij]: (Cx)µ = X i,j Cµ,ijxij.
For x ∈Mm,n, let Vec(x) be the mn-dimensional vector obtained from the matrix x by arranging its columns into a single column, and let X(x) ∈Smn + be the matrix Vec(x)VecT (x), that is, the mn × mn matrix [xijxkℓ]. Observe that X(x) ⪰0, and that P i,j cijxij = 0 if and only if 0 = Ã P i,j cijxij !2 ≡ P ij,kℓ cijxijckℓxkℓ= Tr(X(c)X(x)). Further, ⟨x, Ax⟩= X ij,kℓ Aij,kℓxijxkℓ= Tr(AX(x)), and similarly ⟨x, B⟩= Tr(BX(x)), ⟨x, Bℓ⟩= Tr(BℓX(x)).
Finally, ∥x∥≤1 if and only if xxT ⪯Im. The entries in the matrix xxT are linear combinations of the entries in X(x), so that xxT ⪯Im ⇔S(X(x)) ⪯Im, where S is an appropriate linear mapping from Smn to Sm. Similarly, ∥x∥≤1 if and only if xT x ⪯In, which again is a linear restriction on X(x): xT x ⪯In ⇔T (X(x)) ⪯In, where T is an appropriate linear mapping from Smn to Sn. With the above observations, (P) can be rewritten as the problem max x∈Mm,n Tr(AX(x)) : Tr(BX(x)) ≤1 (a) Tr(BℓX(x)) ≤1, ℓ= 1, ..., L (b) Tr(CµX(x)) = 0, µ = 1, ..., M (c) S(X(x)) ⪯Im, T (X(x)) ⪯In (d) , where Cµ ∈Smn + is given by Cµ ij,kℓ= Cµ,ijCµ,kℓ. Since X(x) ⪰0 for all x, the problem max X∈Smn Tr(AX) : Tr(BX) ≤1 (a) Tr(BℓX) ≤1, ℓ= 1, ..., L (b) Tr(CµX) = 0, µ = 1, ..., M (c) S(X) ⪯Im, T (X) ⪯In (d) X ⪰0 (e) (SDP) is a relaxation of (P), so that Opt(P) ≤Opt(SDP). Observe that problem (SDP) is a semidefinite program and as such is computationally tractable.
Sums of random symmetric matrices 21 Remark 1 When (P) possesses a nontrivial structure, the design dimension of relaxation (SDP) can be reduced. Indeed, in this case, as it is immediately seen, (SDP.c) imply that Xij,kℓshould be zero unless both the cells (i, j), (k, ℓ) belong to diagonal blocks in x. Consequently, in fact the decision matrix X in (SDP) can be thought of as a symmetric matrix of the row size K P k=1 mknk rather than of the size mn.
4.2 Quality of the relaxation Our goal is to prove the following Proposition 6 (i) There exists ¯ x ∈Mm,n such that (∗) ⟨¯ x, A¯ x⟩= Opt(SDP) (a) ⟨¯ x, B¯ x⟩≤1 (b) ⟨¯ x, Bℓ¯ x⟩≤Ω2, ℓ= 1, ..., L (c) C¯ x = 0 (d) ∥¯ x∥≤Ω (33) where Ω= max · max 1≤k≤K µk + p 32 ln(132K), p 32 ln(12(L + 1)) ¸ , µk = min h 7(mk + nk) 1 6 , 4 p min[mk, nk] i (34) (ii) In particular, one has Opt(P) ≤Opt(SDP) ≤Ω2Opt(P).
(35) Proof. 00. (ii) is an immediate consequence of (i). Indeed, with ¯ x satisfying (33), the matrix ˜ x = Ω−1x clearly is a feasible solution to (P), and the value of the objective at this solution is Ω−2Opt(SDP) by (33.∗), which gives the right inequality in (35); the left inequality is readily given by the origin of (SDP).
It remains to prove (i). Let Y be an optimal solution to (SDP); then Y ⪰0, so that the matrix S = Y 1/2 is well defined. Let us set SAS = U T ΛU, where Λ is a diagonal mn × mn matrix, and U is an orthogonal mn × mn matrix. Let ξ be a random mn-dimensional vector with independent entries ξij, 1 ≤i ≤m, 1 ≤j ≤n, taking values ±1 with probabilities 1/2, and let random m × n matrix ζ be given by Vec(ζ) = SU T ξ, (36) so that ζ = ζ(ξ) is a deterministic function of ξ.
10. Observe that E {X(ζ)} = Y.
(37) Indeed, E {X(ζ)} = E © Vec(ζ)VecT (ζ) ª = E © SU T ξξT US ª = SU T E © ξξT ª US = SUU T S = Y.
22 Arkadi Nemirovski 20. We have Cζ ≡0.
(38) Indeed, E ½³X Cµ,ijζij ´2¾ = E © Tr(CµVec(ζ)VecT (ζ)) ª = Tr(CµY ) = 0 (we have used (37) and the fact that Y is feasible for (SDP)).
Since the relations Cx = 0 imply that x is block-diagonal with mk × nk diagonal blocks, k = 1, ..., K, we conclude that all realizations of ζ are block-diagonal with mk × nk diagonal blocks ζk, k = 1, ..., K. Recalling (36) and the nature of ξ, we see that all combinations of the columns of the matrix SU T with coefficients ±1 are of the form Vec(z) with block-diagonal, of the block-diagonal structure ∆, m × n matrices z; this is possible if and only if every one of the columns in SU T is of the form Vec(z) with block-diagonal, of the block-diagonal structure ∆, matrices z. Recalling (36), we arrive at ζk = ζk(ξ) = X i,j zk,ijξij, k = 1, ..., K, (39) with deterministic mk × nk matrices zk,ij.
30. We have also (a) ⟨ζ, Aζ⟩≡Opt(SDP) (b) E {⟨ζ, Bζ⟩} ≤1 (b′) E {⟨ζ, Bℓζ⟩} ≤1, ℓ= 1, ..., L (40) Indeed, ⟨ζ, Aζ⟩= Tr(AVec(ζ)VecT (ζ)) = Tr(ASU T ξξT US) = Tr(U(SAS)U T ξξT ) = Tr(U(U T ΛU)U T ξξT ) = Tr(ΛξξT ) = Tr(Λ) = Tr(U T ΛU) = Tr(SAS) = Tr(AY ) = Opt(SDP), as required in (40.a). Further, E {⟨ζ, Bζ⟩} = E © Tr(BVec(ζ)VecT (ζ)) ª = E © Tr(BSU T ξξT US) ª = Tr(BSU T E{ξξT } | {z } =I US) = Tr(BS2) = Tr(BY ) ≤1 where the concluding ≤comes from the fact that Y is feasible for (SDP). We have arrived at (40.b); verification of (40.b′) is completely similar.
40. Finally, we have E © ζζT ª ⪯Im, E © ζT ζ ª ⪯In.
(41) Indeed, by the origin of S and T , we have ζζT = S(X(ζ)), ζT ζ = T (X(ζ)), and (41) follows from (37).
Recalling that ζ = Diag{ζ1, ..., ζK}, we have Sums of random symmetric matrices 23 ζζT = Diag{ζ1ζT 1 , ..., ζKζT K}, ζT ζ = Diag{ζT 1 ζ1, ..., ζT KζK}, and (41) implies that E © ζkζT k ª ⪯Imk, E © ζT k ζk ª ⪯Ink, k = 1, ..., K.
(42) Invoking (39), we have ζk = P i,j zk,ijξij with deterministic mk × nk matrices zk,ij, so that (42) implies X i,j zk,ijzT k,ij ⪯Imk, X i,j zT k,ijzk,ij ⪯Ink, k = 1, ..., K.
(43) Applying Propositions 2, 3 with Θ = 1 and Ω= t and taking into account the definition of µk (see (34)), we deduce from (43) that t > µk ⇒Prob {ξ : ∥ζk(ξ)∥≥t} ≤22 exp ½ −t2 32 ¾ , k = 1, ..., K.
(44) 50. We are basically done; the only additional element we need to complete the proof of (i) is the following simple fact: Lemma 4 One has (a) Prob {ξ : ⟨ζ(ξ), Bζ(ξ)⟩≤1} | {z } Ea ≥1 3 (b) t ≥8 ⇒Prob © ξ : ⟨ζ(ξ), Bℓζ(ξ)⟩> t2ª ≤4 3 exp{−t2 32}, ℓ= 1, ..., L.
(45) Proof. Recall that ⟨ζ(ξ), Bζ(ξ)⟩= Tr(BVec(ζ(ξ))VecT (ζ(ξ))) = Tr(BSU T ξξT US) = Tr((USBSU T )ξξT ).
(46) Since B is positive semidefinite dyadic matrix, so is the matrix USBSU T , that is, USBSU T = ddT for a mn-dimensional deterministic vector d with entries dij, 1 ≤i ≤m, 1 ≤j ≤n. It follows that ⟨ζ(ξ), Bζ(ξ)⟩= (dT ξ)2 = X i,j dijξij 2 .
(47) Applying (40.b), we derive from (47) that P i,j d2 ij = E {⟨ζ(ξ), Bζ(ξ)⟩} ≤1.
Invoking Lemma A.1 in , we conclude that Prob ( | P i,j dijξij| ≤1 ) ≥1 3, and (45.a) follows from (47). Similarly to (46), we have ⟨ζ(ξ), Bℓζ(ξ)⟩= ξT USBℓSU T | {z } Dℓ ξ.
The matrix Dℓis symmetric positive semidefinite along with Bℓ; setting Fℓ= D1/2 ℓ , we arrive at the identity ⟨ζ(ξ), Bℓζ(ξ)⟩= ∥Fℓξ∥2 2.
(48) 24 Arkadi Nemirovski Invoking (40.b′), we derive from the latter inequality that ∥Fℓ∥2 2 = E {⟨ζ(ξ), Bℓζ(ξ)⟩} ≤1.
(49) We are about to use the following simple fact: Lemma 5 Let bp ∈Rπ, p = 1, ..., P, be deterministic vectors such that P p ∥bp∥2 2 ≤1, and δp, p = 1, ..., P, be independent random scalars taking values ±1 with probabilities 1/2. Then t ≥8 ⇒Prob ( ∥ P X p=1 δpbp∥2 > t ) ≤4 3 exp{−t2 32}.
(50) Proof is similar to the one of Lemma 1. Let Q = {γ ∈RP : ∥P p γpbp∥2 ≤1}, and let µ be the distribution of the random vector γ = (δ1/2, ..., δP /2).
Observe that E ( ∥P p γpbp∥2 2 ) = 1 4 P p ∥bp∥2 2 ≤ 1 4, whence by Tschebyshev inequality µ{γ ̸∈Q} = µ{γ : ∥P p γpbp∥2 > 1} ≤ 1 4, so that µ(Q) ≥ 3 4.
Further, Q clearly is a closed convex set. We claim that this set contains the unit Euclidean ball in RP . Indeed, if u ∈RP and ∥u∥2 ≤1, then ∥ X p upbp∥2 ≤ X p |up|∥bp∥2 ≤ sX p u2 p sX p ∥bp∥2 2 ≤1, so that u ∈Q. Now, if s > 1 and u ∈RP is such that ∥P p upbp∥2 > 2s, then the vector u/2 does not belong to sQ, so that dist∥·∥2(u/2, Q) > s −1 (cf.
the justification of (12) with γ = 1). Applying the Talagrand Inequality to A = Q and the distribution µ, we get s > 1 ⇒ Prob ( δ : ∥P p δpbp∥2 > 2s ) ≤exp{−(s−1)2 4 } R exp{dist 2 ∥·∥2(γ,Q) 4 }µ(dγ) ≤exp{−(s−1)2 4 } µ(Q) ≤4 exp{−(s−1)2 4 } 3 , and (50) follows.
⊓ ⊔ Specifying in Lemma 5 P as mn, bp as the columns of Fℓand δp as the random scalars ξij, we derive from (49) that t ≥8 ⇒Prob {ξ : ∥Fℓξ∥2 > t} ≤4 3 exp{−t2 32}, which combines with (48) to imply (45.b). Lemma 4 is proved.
⊓ ⊔ Sums of random symmetric matrices 25 60. We are ready to complete the proof of (i). Let Ωbe given by (34).
Then clearly Ω≥8, and we can apply (45) with t = Ωto conclude that for every ℓ≤L we have Prob © ξ : ⟨ζ(ξ), Bℓζ(ξ)⟩> Ω2ª ≤ 1 9(L+1), whence Prob © ξ : ⟨ζ(ξ), Bℓζ(ξ)⟩≤Ω2, 1 ≤ℓ≤L ª | {z } Eb ≥8 9.
(51) By (44), with our Ωfor every k ≤K we have also Prob {ξ : ∥ζk(ξ)∥> Ω} ≤ 1 6K , whence Prob ½ ξ : ∥ζ(ξ)∥≡max k≤K ∥ζk(ξ)∥≤Ω ¾ | {z } Ed ≥5 6.
(52) Combining (45.a), (51) and (52), we see that the events Ea, Eb and Ed have a point ξ∗in common. Setting ¯ x = ζ(ξ∗), we see that ¯ x satisfies all the requirements in (33) ((∗) – by (40.a), (a), (b), (d) – due to ξ∗∈Ea ∩Eb ∩Ed, and (c) by (38)).
⊓ ⊔ 4.2.1 Comments A. With norm constraint (d) in (P) eliminated, (P) becomes the purely quadratic program max x∈Mm,n ⟨x, Ax⟩: ⟨x, Bx⟩≤1 (a) ⟨x, Bℓx⟩≤1, ℓ= 1, ..., L (b) Cx = 0 (c) ; (P ′) its semidefinite relaxation (SDP′) is obtained from (SDP) by eliminating constraints (SDP.d). From the proof of Proposition 6 it follows that Opt(P ′) ≤Opt(SDP′) ≤Ω2Opt(P ′), Ω= p 32 ln(12(L + 1)); the resulting statement is a slightly improved version of “Approximate S-Lemma” from and the results of (in the original statements, the role of the number L of quadratic constraints in the formula for Ωwas played by a larger quantity, the total rank of mappings Bℓ, ℓ= 1, ..., L).
B. The proof of Proposition 6 goes along the lines of the proof of Ap-proximate S-Lemma (which, in turn, goes back to ); the crucial new component (bound (44) based on Theorem 2) allows us to treat the norm constraint (P.d).
C. Any further progress towards the proof of Conjecture 1 would result in improving the result of Proposition 6. For example, if Conjecture were true, we would be able to replace the terms 7(mk +nk) 1 6 in (34) with a much nicer, from the theoretical viewpoint, terms O(1) p ln(mk + nk).
D. As it is usually the case with semidefinite relaxations of difficult opti-mization problems, (SDP) not only provides us with an efficiently computable upper bound on the optimal value of (P), but offers as well a randomized 26 Arkadi Nemirovski algorithm for building suboptimal feasible solutions to (P). Such an algo-rithm is suggested by the proof of Proposition 6; specifically, we generate a sample of, say, M = 1000 realizations ζ1, ..., ζM of the random matrix ζ(ξ) (see (36)), choose the largest possible scale factors λp such that the scaled matrices b ζp = λpζp are feasible for (P), thus getting a sample of feasible so-lutions to (P), and then choose among these feasible solutions the one with the best – the largest – value of the objective. Note that the required “feasi-ble scalings” indeed are possible, since every ζ satisfying homogeneous linear equations (P.c) (and thus - every realization of ζ(ξ)) after multiplication by appropriate positive scalar becomes feasible for (P).
E. Under favourable circumstances, the outlined randomized algorithm can be further improved by a kind of “purification”. Specifically, assume that – (P) has no quadratic constraints (P.a-b) (that is, B = 0, Bℓ= 0, ℓ= 1, ..., L); – The linear homogeneous constraints (P.c) say that a feasible solution x to (P) possesses certain block-diagonal structure ∆= {(mk, nk)}K k=1 and impose no further restrictions on x; – The objective f(x) = f(x1, ..., xK) we are maximizing in (P) is convex w.r.t. every one of the diagonal blocks xk in a feasible (and thus block-diagonal) candidate solution x, the remaining components being fixed.
Note that outlined assumptions are satisfied in all problems of quadratic optimization under orthogonality constraints mentioned in the beginning of section 4. Now, given a feasible solution x with components xk, k = 1, ..., K, purification converts x to a feasible solution b x with the same or better value of the objective in such a way that b x is an “extreme point” feasible solution to (P). The latter means that every component b xk of b x satisfies the orthog-onality relation, namely, b xkb xT k = Imk when mk ≤nk and b xT k b xk = Ink when mk ≥nk.
The conversion x 7→b x takes K steps. At the first step, we represent x1 as a convex combination of a moderate number Q of matrices xq 1, q = 1, ..., Q, satisfying the orthogonality constraint (this is possible and can be done efficiently, see below); note that every one of the Q candidate solutions xq = (xq 1, x2, ..., xK) is feasible for (P). We compute the value of f at these solutions and find the one, let it be xq∗, with the largest value of f. Since f(·, x2, ..., xK) is convex and x1 is a convex combination of xq 1, q = 1, ..., Q, we have f(xq∗) ≥f(x). Thus, we have found a feasible solution x(1) = xq∗ to (P) with the same or better value of the objective than the one at x and with the first block satisfying the orthogonality constraint. Now we repeat this procedure with x(1) in the role of x and x2 in the role of x1, thus getting a feasible solution x(2) which is at least as good as x(1) in terms of the objective and has two blocks satisfying the orthogonality constraints. Proceeding in this fashion, we in K steps end up with an extreme point feasible solution b x = x(K) which is at least as good as x.
For the sake of completeness, we present here the standard algorithm for representing a given µ × ν matrix z, ∥z∥≤1, as a convex combination of matrices satisfying the orthogonality constraint. W.l.o.g., let us assume that µ ≥ν, so that the orthogonality constraint is satisfied by a µ×ν matrix w iff wT w = Iν. We first find the singular value decomposition z = UDiag{σ}V T , Sums of random symmetric matrices 27 where V is orthogonal ν × ν matrix, σ ≥0 is the ν-dimensional vector of singular values of z and the µ×ν matrix U satisfies the relation U T U = Iν.
Since ∥z∥≤1, we have 0 ≤σi ≤1. Now observe that whenever γ ∈Rν has entries ±1, the matrix UDiag{γ}V T satisfies the orthogonality constraint.
Thus, all we need is to represent σ as a convex combination P q λqγq of a moderate number of vectors γq with entries ±1, thus inducing the desired representation z = P q λqUDiag{γq}V T . A required representation of σ is immediate. W.l.o.g. we may assume that σ1 ≤σ2 ≤... ≤σν. Let us define ν + 1 ν-dimensional vectors δi as δ1 = (1, ..., 1)T , δ2 = (0, 1, ..., 1)T , δ3 = (0, 0, 1, ..., 1)T , ..., δν+1 = (0, ..., 0)T ; observe that δi is the half-sum of two vectors δi ± with coordinates ±1. The required representation of σ is merely σ = ν+1 X i=1 (σi −σi−1)δi = ν+1 X i=1 σi −σi−1 2 [δi + + δi −] (we have set σ0 = 0, σν+1 = 1). This representation involves Q = 2ν + 1 vectors with coordinates ±1 (note that δ1 + = δ1 −).
F. Finally, when f is affine in every one of xk (as it is the case in the Procrustes problem), the purification can be simplified and improved – here we can at every step easily maximize f in the block to be updated. To simplify notation, consider the first step. Given x2, ..., xK, we can represent the affine function φ(y) = f(y, x2, ..., xK) of m1 × n1 matrix y as φ(y) = Tr(yaT ) + c with a, c readily given by x2, ..., xK. Assuming w.l.o.g. that m1 ≥n1, let us compute the singular value decomposition a = UDiag{σ}V T of a, so that σ ≥0. It is immediately seen that the maximum of φ(y) over y’s satisfying the constraint ∥y∥≤1 is equal to P i σi and is attained at the matrix y∗= UV T satisfying the orthogonality constraint; y∗is clearly the best possible extreme point updating of x1.
4.3 Numerical illustration: Procrustes problem To illustrate the outlined considerations, we are about to present numerical results for the Procrustes problem Opt(Pr) = min x[·] X 1≤k<k′≤K ∥a[k]x[k] −a[k′]x[k′]∥2 2 : x[k]xT [k] = In ∀k , (Pr) a[·] being given N × n matrices. The problem is equivalent to the quadratic problem with orthogonality constraints max x[·] 2 X 1≤k<k′≤K Tr(a[k]x[k]xT [k′]aT [k′]) : x[k]xT [k] = In ∀k ; (53) 28 Arkadi Nemirovski as we have already explained, relaxing the orthogonality constraints in the latter problem to ∥x[k]∥≤1, we preserve the optimal value, so that (Pr) is equivalent to the problem Opt(Pr+) = max x[·] 2 X 1≤k<k′≤K Tr(a[k]x[k]xT [k′]aT [k′]) : ∥x[k]∥≤1 ∀k (Pr+) of the form of (P); the optimal values in (Pr) and (Pr+) are linked by the relation Opt(Pr) = (K −1) K X k=1 Tr(a[k]aT [k]) | {z } C −Opt(Pr+).
(54) In our experiments, we generated random instances of (Pr), solved the semide-finite relaxations (SDP) of the resulting instances of (Pr+), thus obtaining upper bounds on the optimal values of the latter instances (which, in turn, induce via (54) lower bounds on the optimal values in (Pr)), and used the randomized algorithm outlined in item D, section 4.2.1, to get suboptimal solutions to (Pr). The details are as follows.
Generating instances. Given “sizes” K, N, n of (Pr), we generated the data a, ..., a[K] of (Pr) as follows: entries of a were picked at random from the standard Gaussian distribution N(0, 1), while the remaining matrices were generated as a[k] = aUk + ϵQk, 2 ≤k ≤K, with randomly chosen orthogonal matrix Uk and random matrix Qk generated in the same fashion as a. The “closeness parameter” ϵ was chosen at random according to ϵ = exp{ξ} with ξ uniformly distributed in [−3, 3]. The sizes K, n of (Pr) were limited by the necessity to end up with semidefinite relaxation not too difficult for the SDP solver mincx (LMI Toolbox for MATLAB) we used, which means at most 1000 – 1200 free entries in X. This restriction allows us to handle the sizes (K, n) with Kn ≤50 (see below). The column size N of a[·] was always set to 20.
The relaxation of (Pr+) as given by the above construction is the semidefinite problem Opt(SDP) = max X 2Tr(AX) | {z } F (X) : X ≡[Xkij,k′i′j′] k,k′≤K i,i′,j,j′≤n ⪰0 Sk(X) ≡ " n P p=1 Xkpi,kpj # i,j≤n ⪯In, k ≤K Tk(X) ≡ " n P p=1 Xkip,kjp # i,j≤n ⪯In, k ≤K (55) Sums of random symmetric matrices 29 (see (SDP) and Remark 1), where A is the symmetric Kn2 × Kn2 matrix with the entries Akij,k′i′j′ = 1 2 N P ℓ=1 a[k]ℓia[k′]ℓi′, j = j′ and k ̸= k′ 0, j ̸= j′ or k = k′ .
(56) In fact (55) can be significantly simplified. Specifically, let us treat Kn × Kn symmetric matrices Y as K × K block matrices with n × n blocks Y k,k′ = [Y k,k′ ij ]n i,j=1, 1 ≤k, k′ ≤K, and consider the semidefinite program max Y n G(Y ) = Tr(BY ) : Y = {Y k,k′} ∈SKn, Y ⪰0, Y k,k ⪯In ∀k o (57) where B ∈SKn is the block matrix with blocks Bk,k′ = ½ aT [k]a[k′], k ̸= k′ 0, k = k′ of sizes n × n, 1 ≤k, k′ ≤K. Note that the design dimension of (57) is less than the one of (55) by factor ≈n2.
Lemma 6 Problems (55), (57) are equivalent to each other. Specifically, if a matrix X = [Xkij,k′i′j′] is a feasible solution to (55), then the matrix Y = Y[X] ≡{Y k,k′}K k,k′=1 ∈SKn given by Y k,k′ ii′ = n X p=1 Xkip,k′i′p, 1 ≤i, i′ ≤n, 1 ≤k, k′ ≤K (58) is a feasible solution to (57), and F(X) = G(Y ). Moreover, every feasible solution Y to (57) is of the form Y[X] for an appropriate feasible solution X of (55).
Proof. Let X be a feasible solution to (55), Y = Y[X]. Then Y ⪰0. Indeed, since X ⪰0, we have Xkij,k′i′j′ = L P ℓ=1 vℓ kijvℓ k′i′j′ for appropriately chosen L and vℓ kij. It follows that Y ≡Y[X] = L X ℓ=1 { n X j=1 vℓ kijvℓ k′i′j i,i′ }K k,k′=1 | {z } Y ℓ ; it remains to note that the matrices Y ℓare sums of dyadic matrices and thus are symmetric positive semidefinite. Further, we have Tk(X) = Y k,k (see (55)), so that Y is feasible for (57). The relation F(X) = G(Y ) is readily given by (58) and (56).
Now let Y = {Y k,k′}K k,k′=1 be feasible for (57), and let us set Xkij,k′i′j′ = 1 nδj j′Y k,k′ ii′ (59) (δp q is the Kronecker symbol), so that Y = Y[X] by (58). It remains to prove that X is feasible for (55). Indeed, X is the Kronecker product of positive 30 Arkadi Nemirovski semidefinite matrix Y and n−1In and thus is positive semidefinite. Further, by (55) we clearly have Tk(X) = Y k,k ⪯In, and (Sk(X))jj′ = n X p=1 Xkpj,kpj′ = n X p=1 δj j′ 1 nY k,k pp ⇒Sk(X) = Tr(Y k,k) n In ⪯In, where the concluding ⪯is given by Y k,k ⪯In (recall that Y is feasible for (57)).
⊓ ⊔ Remark 2 The origin of (57) is as follows. The objective in (Pr+) is a linear function of the matrix products x[k]xT [k′], k, k′ = 1, ..., K, which are nothing but the blocks Y k,k′ in the positive semidefinite block matrix Y = Y [x] = x .
.
.
x[K] x .
.
.
x[K] T ∈SKn, while the norm bounds in (Pr+) translate into the constraints ∥Y k,k∥≤1. Thus, (57) is obtained from (Pr+) by passing to Y -variable and subsequent eliminating the nonconvex constraint “Y should be Y [x] for some x”.
Note that the outlined “recipe” for simplifying the semidefinite relaxation works in the case of general problem (P), provided that the constraints (P.c) say exactly that the structure of (P) is {(mk = µ, nk = ν)}K k=1 and that the objective and the left hand sides in the constraints (P.a−b) of (P) are linear functions of the matrix products x[k]xT [k′], k, k′ = 1, ..., K. Note also that under the latter assumptions the reasoning completely similar to the one in Lemma 6 demonstrates that the outlined simplification of (55) is in fact equivalent to (55).
Recovering suboptimal solutions to (Pr+), (Pr) was implemented according to the randomized algorithm with purification outlined in section 4.2.1, items D, F. Specifically, after (high-accuracy approximation to) optimal solution Y∗of (57) was found, we “lifted” it, according to (59), to an optimal solution X∗of (55). Then we used X∗to generate a sample of M = 1000 feasible solutions xℓ= {xℓ k}K k=1 to (Pr+) as explained in item 4.2.1.D and purified these solutions as explained in item 4.2.1.F, thus obtaining feasible solutions b xℓto (Pr+) which satisfy the orthogonality constraints and thus are feasible for (53) and (Pr). The resulting suboptimal solution b x to (Pr) was the best (with the smallest value of the objective) of the feasible solutions b xℓ, ℓ= 1, ..., M. The value of the objective of (Pr) at b x is an upper bound Optup(Pr) on Opt(Pr), while the value of the objective of (Pr+) at b x is a lower bound Optlw(Pr+) on the optimal value of (Pr+). Thus, we end up with brackets [L(Pr+), U(Pr+)] ≡[Optlw(Pr+), Opt(SDP)], [L(Pr), U(Pr)] ≡[C −Opt(SDP), Optup(Pr)] on the optimal values of (Pr+), (Pr), respectively, along with a feasible sub-optimal solution b x to (Pr+), (Pr); the values of the objectives of (Pr+), (Pr) at b x are appropriate endpoints of the corresponding brackets.
Sums of random symmetric matrices 31 Sizes of instances. The design dimension of (57) is Kn(Kn+1) 2 ; in order for it to be at most about 1200 (the limitation imposed by the SDP solver we used), Kn should be at most 50. In our experiments, we used Pareto-maximal pairs (K, n) with Kn ≤50, specifically, the 10 pairs (K = ⌊50 n ⌋, n) given by n = 2, 3, 4, 5, 6, 7, 8, 10, 12, 15, and for every pair solved 20 instances of the corresponding sizes (which amounts to the total of 200 instances).
The results of our numerical experiments were surprisingly good, much bet-ter than one could expect looking at the bound (35). Indeed, the latter bound guarantees that Opt(SDP) is at most by factor Ω2 greater than Opt(Pr+), with Ω2 slowly growing with K, n and thus being a “moderate” constant, provided that K, n are not too large3). As about (Pr), Proposition 6 yields no bounds on the ratio of the true optimal value in (Pr) and its efficiently computable lower bound C −Opt(SDP). The outlined theoretical guaran-tees, if any, are not too optimistic, which is in sharp contrast with the actual numerical results we got. In 200 experiments we have run, the largest rela-tive error U(Pr+)−L(Pr+) max[1,U(Pr+)] in solving (Pr+) was as small as 9.0%, while the largest relative error U(Pr)−L(Pr) max[1,U(Pr)] in solving (Pr) was as small as 2.4%. These data correspond to the best of the purified solutions b xℓ. As far as problem (Pr+) is concerned, already the unpurified solutions xℓwere not so bad: the relative error of the best, in terms of the objective, of these solutions e x was at worst 62.2%. Thus, in our experiments we did not observe ratios Opt(SDP)/Opt(Pr+) exceeding 1.09 (cf. with the theoretical upper bound ≈100 on this ratio). The histograms of the relative errors are presented on Fig. 1.
References 1. Anstreicher, K., Wolkowicz, H.: On Lagrangian Relaxation of Quadratic Ma-trix Constraints. SIAM J. Matrix Anal. Appl. 22, 41-55 (2000) 2. Ben-Tal, A., Nemirovski, A., Roos, C.: Robust solutions of uncertain quadratic and conic-quadratic problems. SIAM J. Optimization 13, 535-560 (2002) 3. Browne, M.W.: On oblique Procrustes rotation. Psychometrika 32, 125-132 (1967) 4. Edelman, A., Arias, T. Smith, S.T.: The geometry of algorithms with orthog-onality constraints. SIAM J. Matrix Anal. Appl. 20, 303-353 (1999) 5. Johnson, W.B., Schechtman, G.: Remarks on Talagrand’s deviation inequal-ity for Rademacher functions. In: Odell, E., Rosenthal, H. (eds.) Functional Analysis (Austin, TX 1987/1989), Lecture Notes in Mathematics 1470, 72-22.
Springer (1991) 6. Nemirovski, A.: On tractable approximations of randomly perturbed convex constraints. In: Proceedings of the 42nd IEEE Conference on Decision and Control Maui, Hawaii USA, December 2003, 2419-2422 7. Nemirovski, A., Roos, C., Terlaky, T.: On maximization of quadratic form over intersection of ellipsoids with common center. Math. Prog. 86, 463-473 (1999) 3 For the values of K and m = n we used in our experiments, the bound (35) results in Ω2 varying from ≈243 (K = 3, n = 15) to ≈301 (K = 25, n = 2); more accurate (and more messy) numerics in the proof of Proposition 6 reduces the range of Ω2 for our K, n to [≈84, ≈102].
32 Arkadi Nemirovski 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 10 20 30 40 50 60 70 80 (Pr+), unpurified solutions e x 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0 50 100 150 0 0.005 0.01 0.015 0.02 0.025 0 20 40 60 80 100 120 (Pr+), purified solutions b x (Pr), purified solutions b x Fig. 1 Histograms of relative errors in 200 experiments 8. Nemirovski, A.: Regular Banach spaces and large deviations of random sums.
Working paper 9. Shapiro, A.: Extremal Problems on the Set of Nonnegative Definite Matrices.
Linear Algebra and Appl. 67, 7-18 (1985) 10. Shapiro, A., Botha, J.D.: Dual Algorithms for Orthogonal Procrustes Rota-tions. SIAM J. Matrix Anal. Appl. 9, 378-383 (1988) 11. Ten Berge, J.M.F., Nevels, K.: A general solution to Mosiers oblique Pro-crustes problem. Psychometrika 42, 593-600 (1977) 12. Wolkowicz, H.: Semidefinite Programming Approaches to the Quadratic As-signment Problem. In: Saigal, R., Wolkowitcz, H., Vandenberghe, L. (eds.) Handbook on Semidefinite Programming. Kluwer (2000) 13. Wolkowicz, H., Zhao, Q.: Semidefinite programming relaxations for graph par-titioning problems. Discrete Appl. Math. 96/97, 461-479 (1999) 14. Zhao, Q., Karisch, S. E., Rendl, F., Wolkowicz, H.: Semidefinite programming relaxations for the quadratic assignment problem. J. Comb. Optim. 2, 71-109 (1998) |
11137 | https://www.scribd.com/presentation/26369068/transistor-characteristics | Transistor Characteristics | PDF | Bipolar Junction Transistor | Transistor
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Transistor Characteristics
Pnp npn the terminals are labeled: e - Emitter B - Base C - collector npc. Transistor Operation With external sources, VEE and VCC, connected as shown. Emitter current is the sum of the coll…
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Chapter 3:Bipolar Junction Transistors
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Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458 • All rights reserved.
Electronic Devices and Circuit Theory, 10/e
Robert L. Boylestad and Louis Nashelsky
T ransist or Constr uction
There are two types of transistors:
•
pn p
•
n pn
The terminals are labeled:
•
E - Emitter
•
B - Base
•
C - Collector
pn p n pn
2
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Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458 • All rights reserved.
Electronic Devices and Circuit Theory, 10/e
Robert L. Boylestad and Louis Nashelsky
T ransisto r Ope ration
With the external sources, V
EE
and V
CC
, connected as shown:
•
The emitter-base junction is forward biased
•
The base-collector junction is reverse biased
3
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Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458 • All rights reserved.
Electronic Devices and Circuit Theory, 10/e
Robert L. Boylestad and Louis Nashelsky
Currents in a T ransist or
T
he collector current is comprised of two currents:
B I C I E I
minority CO I majority C I C I
Emitter current is the sum of the collector and base currents:
4
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11138 | https://www.youtube.com/watch?v=R3PNohBOBIw | Integral of 1/cos^2x | Calculus 1 Exercises
Wrath of Math
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We integrate 1/cos^2(x) by recognizing 1/cos^2x = sec^x and then recalling that sec^2x is the derivative of tanx. #calculus #apcalculus
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to integrate 1 / cine 2 x we just have to recall a few simple facts first recall that 1 / cine X is secant x so when we integrate 1 / cine s really we are integrating secant squar and perhaps already you know the answer at this point we recall that the derivative of tangent is secant squared hence the integral of ^ 2 x is tangent of X plus the arbitrary constant that's how to find the integral of 1/ cosine 2 x let me know in the comments if you have any questions and be sure to check out my Calculus 1 course and Calculus 1 exercises playlists in the description for more thanks for watching |
11139 | https://www.janko.at/Raetsel/Rules.htm | Rätsel und Puzzles
The different rule descriptions use common phrases which are
explained in detail on this page.
Neighborship:
A cell (red) has 3, 5 or 8 neighbors (blue and green).
The blue cells are the orthogonal (horizontal and vertical) neighbors of the
red cells.
The green cells are the diagonal neighbors of the red cells
Regions, Areas, Stripes:
A region consists of one or more cells
which are separated by bold lines from other regions.
An area consists of one or more cells which are not separated by bold
lines from other areas, but defined otherwise, e.g. all adjacent cells of the
same color. In the diagram the yellow cells form such an area.
A stripe is a region (green) or an area (red) which is exactly 1 cell
wide and of any length.
Areas of black cells are sometimes called blocks or groups, or
dominoes, triominoes, tetrominoes, pentominoes, or
polyominoes.
Regions as well as areas can be square, rectangular or irregular. (Note: A square
is also a rectangle!)
Snakes
A "snake" is a path through the cells of the diagram which is exactly
one cell wide. In many genres a snake must not touch itself, not even diagonally.
A snake may be open (i.e. the snake has a head and a tail) or closed (i.e. a
loop)
Orthogonally contiguous cells/areas
The white cells form an orthogonally
contiguous area, if it is possible to travel from any white cell to any other
white cell, moving only horizontally and vertically (never diagonally) and never
crossing a black cell.
In the diagram the white cells are colored blue, green and yellow
(actually, they all are white).
• All green cells are orthogonally contiguous: For example, one can travel from
A to B moving only horizontally and vertically, not crossing a black cell.
• Also, all blue cells are orthogonally contiguous.
• The green cells and the blue cells are not orthogonally contiguous. For example,
there is no path from C to D, travelling only horizontally and vertically which
does not cross a black cell.
• Cell E is isolated and is not orthogonally adjacent to any other white cell.
Orthogonally contiguous areas of black cells are defined analogously.
Dominos
A domino is a region of exactly two orthogonally adjacent cells.
Disregarding rotation and reflection there is exactly one domino:
In the diagram you can find 3 Dominos. Yellow and blue have the
same orientation; yellow and green have different orientation. Yellow and green
are orthogonally adjacent; yellow and blue are diagonally adjacent.
Triominoes
A triomino is a region of exactly three orthogonally adjacent
cells. Disregarding rotation and reflection there are exactly two triominoes,
labeled by letters according to their shapes:
In the diagram you can find 3 triominoes. The orange and the blue
triomino together cover a 2x2 area which is not allowed in many puzzle types.
Tetrominoes
A tetromino is a region of exactly four orthogonally adjacent
cells. Disregarding rotation and reflection there are exactly five tetrominoes,
labeled by letters according to their shapes:
In the diagram you can find 3 Tetrominos.
Pentominos
A pentomino is a region of exactly five orthogonally adjacent cells. Disregarding
rotation and reflection there are exactly twelve pentominoes, labeled by letters
according to their shapes:
Polyominoes (N-ominoes)
A polyomino is a region of orthogonally adjacent cells; an N-omino is a region
of exactly N orthogonally adjacent cells. Dominoes, triominoes, tetrominoes and
pentominoes are special cases of N-ominoes (N=2, 3, 4 or 5).
Congruence
Two figures (areas, domains, polyominos) are congruent if they can be made
to coincide by translation, rotation, mirroring, or a combination of these
methods. [Wikipedia]
The following pentominos are all congruent:
Manhattan Distance (Taxi Distance)
The Manhattan Distance is the shortest
link between two cells, travelling only horizontally and vertically between adjacent
cells. The diagram shows three possible paths; all paths are equally
in length.
Formally: The Manhattan Distance between two cells is the sum of the absolute
differences of the coordinates of the cells the path travels through:
In the diagram the Manhattan Distance between the two blue cells
is 7 (= number of cells the path goes through including the target cell). |
11140 | https://www.newscientist.com/article/2495377-gravitational-waves-finally-prove-stephen-hawkings-black-hole-theorem/ | Skip to content
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Gravitational waves finally prove Stephen Hawking's black hole theorem
An exceptionally loud collision between two black holes has been detected by the LIGO gravitational wave observatory, enabling physicists to test a theorem postulated by Stephen Hawking in 1971
By Matthew Sparkes
10 September 2025
Stephen Hawking’s 50-year-old theorem on how black holes merge together has been successfully tested thanks to huge advances in gravitational wave astronomy, which helped astronomers catch the waves caused by an unusually powerful collision as they passed Earth at the speed of light.
Hawking proposed his black hole area theorem in 1971, which states that when two black holes merge, the resulting black hole’s event horizon – the boundary beyond which not even light can escape the clutches of a black hole – cannot have an area smaller than the sum of the two original black holes. The theorem echoes the second law of thermodynamics, which states that the entropy, or disorder within an object, never decreases.
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We are about to hear echoes in the fabric of space for the first time
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Black hole mergers warp the fabric of the universe, producing tiny fluctuations in space-time known as gravitational waves, which cross the universe at the speed of light. Five gravitational wave observatories on Earth hunt for waves 10,000 times smaller than the nucleus of an atom. They include the two US-based detectors of the Laser Interferometer Gravitational-Wave Observatory (LIGO) plus the Virgo detector in Italy, KAGRA in Japan and GEO600 in Germany, operated by an international collaboration known as LIGO-Virgo-KAGRA (LVK).
The recent collision, named GW250114, was almost identical to the one that created the first gravitational waves ever observed in 2015. Both involved black holes with masses between 30 and 40 times the mass of our sun and took place about 1.3 billion light years away.
This time, the upgraded LIGO detectors had three times the sensitivity they had in 2015, so they were able to capture waves emanating from the collision in unprecedented detail. This allowed researchers to verify Hawking’s theorem by calculating that the area of the event horizon was indeed larger after the merger.
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When black holes collide, they produce gravitational waves with overtones like the ringing of a bell, says Laura Nuttall at the University of Portsmouth, UK, a member of the LVK team. Previously these overtones have dissipated too quickly to be observed with enough clarity to calculate the area of the event horizons before and after the collisions, which was necessary to test Hawking’s theory. A 2021 study on the first detected collision backed up the theory with a confidence level of 95 per cent, but the new research raises that confidence to a compelling 99.999 per cent.
In the 10 years that scientists have been observing gravitational waves, they have recorded some 300 black hole collisions. But none have been captured as strongly and clearly as GW250114, which was twice as loud as any other gravitational wave detected to date.
“The ones that are really, really close by – really, really loud in our data – they’re the ones where we can really start probing into the fundamental physics of what’s going on, just because they’re so loud and the uncertainties are so small. So we can start really picking out the nitty gritty details of what’s going on,” says Nuttall. “We’re just waiting for nature to keep giving us those beautiful things.”
The radical idea that space-time remembers could upend cosmology
There are new hints that the fabric of space-time may be made of "memory cells" that record the whole history of the universe. If true, it could explain the nature of dark matter and much more
Only LIGO was operating when the waves from GW250114 reached Earth, not the other detectors monitored by the LVK collaboration. This didn’t affect the test of Hawking’s theory but did mean researchers were unable to pinpoint the origin of the waves in the sky more clearly.
Upgrades to LIGO and other planned observatories due to come online in the future will bring even greater sensitivity and allow us to delve deeper into the physics of black holes, says Ian Harry, also at the University of Portsmouth and part of the LVK team. “We may not get all of them, but we will get an event like this again,” says Harry. “Maybe with the next set of upgrades, maybe in 2028, we see something like this and maybe then it’s at the sensitivity where we can really poke holes.”
The findings pave the way for new research on quantum gravity, through which physicists hope to unite general relativity and quantum physics. Nuttall says the latest results show general relativity and quantum mechanics continue to work well together, but some discrepancy is expected in the future.
“At some point we may start to see that things stop playing nicely, and this will be when we get very close-by signals that just appear extremely loud in our data as the sensitivity of the instruments increases,” says Nuttall.
The latest data from LVK also enabled scientists to confirm mathematician Roy Kerr’s equations from the 1960s which predicted that black holes can be characterised by just two metrics: their mass and their spin. In essence, two black holes with the same mass and spin are mathematically identical. Thanks to the observations of GW250114, we now know that to be true.
Journal reference:
Physical Review Letters
DOI: 10.1103/kw5g-d732
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11141 | https://bpb-us-w2.wpmucdn.com/sites.uwm.edu/dist/8/773/files/2020/07/kk-long-ver2.pdf | Beyond Knights and Knaves Christine T. Cheng1, Andrew McConvey2,⋆, Drew Onderko1,⋆, Nathaniel Shar3,⋆, and Charles Tomlinson4 1 Department of Computer Science, University of Wisconsin-Milwaukee, Milwaukee, WI 53211, USA {ccheng,donderko}@uwm.edu 2 Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA mcconve2@illinois.edu 3 Department of Mathematics, Rutgers University, Piscataway, NJ 08854, USA nshar@math.rutgers.edu 4 Department of Mathematics, University of Nebraska-Lincoln, Lincoln, NE 68588, USA ctomlinson2@math.unl.edu Abstract. In the classic knights and knaves problem, there are n people in a room each of whom is a knight or a knave. Knights always tell the truth while knaves always lie. Everyone in the room knows each other’s identity. You are allowed to ask questions of the form “Person i, is person j a knight?” and you are told that there are more knights than knaves.
What is the fewest number of questions you can ask to determine a knight? How about to determine everyone’s identity?
In this paper, we consider the knights and no-men problem, where a no-man is a person who always answers “no”. Assuming there are at least k knights, we show that n−1 2 −⌊(k−2)(n−1)2 2(k−1) ⌋questions are necessary and sufficient in the worst case to identify a knight. We also show that n −2 questions suffice to identify a no-man, and n−1 2 −⌊(k−2)(n−1)2 2(k−1) ⌋+ n −2 questions suffice to identify everyone in the room.
We then consider a generalization of the knights and knaves problem that captures most of the variants of the knights and knaves problem in the literature. In the agent labeling problem, we wish to identify everyone’s type; in the agent identification problem, we wish to identify an agent having a particular type. We present results with regards to the fewest number of questions needed in the worst case to solve both the agent labeling and agent identification problems.
1 Introduction In the classic knights and knaves problem, there are n people in a room labeled 1 to n. Each person is either a knight or a knave. Knights always tell the truth ⋆Supported in part by the National Science Foundation grant DMS 08-38434 “EMSW21-MCTP: Research Experience for Graduate Students”.
while knaves always lie. Everyone in the room knows each other’s identity. You are allowed to ask questions of the form “Person i, is person j a knight?” and you are told that there are more knights than knaves. What is the fewest number of questions you can ask to determine a knight? How about to determine everyone’s identity? Popular variations to this problem introduce new “types” of people such as spies, who lie or tell the truth arbitrarily, and yes-men, who answer “yes” all the time.
Several equivalent formulations of the knights and knaves problem have been studied by other authors including the problem of classifying computer chips as faulty or reliable , the problem of identifying a coin of the majority weight by performing a series of weight comparisons , the problem of identifying a ball of the most common color by doing a series of color comparisons , , , , , etc.
Saks and Werman studied a version of the knights and knaves problem where knights are in the majority if such a majority exists. Their goal was to identify a knight or declare the nonexistence of such a majority. They showed that n −B(n) questions are necessary and sufficient in the worst case, where B(n) is the number of ones in the binary representation of n. Simpler proofs were given later by Alonso et al. and Wiener . If the number of knights is at least k, and k > n/2, Aigner showed that n−1 −p questions are necessary in the worst case, where p is the highest power of two dividing n−1 k−1 , and that 2(n−k)−B(n−k) questions are necessary and sufficient in the worst case when it is known that a majority exists.
When we want to identify everyone in a room of knights and knaves, Aigner observed that n−1 questions are necessary in the worst case whenever n > 2.
Aigner also showed that when the number of knights is at least k, k > n/2, ⌈(n−k)n+(n−k) n−k+1 ⌉questions are necessary in the worst case, up to a possible error of 1.
The knights and spies problem is just like the knights and knaves problem except that agents which are not knights are spies. Alonso et al. showed that, surprisingly, any strategy that can identify a knight in the knights and knaves setting can also be used to identify a knight in the knights and spies setting.
Therefore, the two problems require the same number of questions to identify a knight in the worst case. Aigner also proved that the same result is true when the number of knights is bounded below by k, k > n/2.
If we wish to identify everyone in a room of knights and spies and there are k > n/2 knights, Blecher and Wildon showed independently that 2n −k −1 questions are necessary and sufficient in the worst case. Thus, it follows that 3n/2 −1 questions are necessary and sufficient in the worst case if it is only known that knights are in the majority.
Most recently, Hanajiri introduced the concept of yes-men and studied the knights and yes-men problem. Suppose there are n people in the room, at most p of whom are knights and at most q of whom are yes-men. Hanajiri showed that n + ⌊log2(p + q −n)⌋questions suffice for identifying everyone in the room.
Additionally, he conjectured that the same number of questions are necessary in the worst case when p, q / ∈{0, n}. He was able to verify his conjecture when p + q ∈{n, n + 1, 2n −1}.
Our Results. Inspired by Hanajiri’s work, we introduce the notion of a no-man, a person who always answers “no”. Assume that there are n people each of whom is a knight or a no-man, and that there are at least k knights. We show that n−1 2 −⌊(k−2)(n−1)2 2(k−1) ⌋questions are necessary and sufficient in the worst case to identify a knight. We also show that n −2 questions suffice to identify a no-man, and n−1 2 −⌊(k−2)(n−1)2 2(k−1) ⌋+ n −2 questions suffice to identify everyone in the room.
Next, we present a generalization of the knights and knaves problem that allows us to consider new types of people and/or put multiple types (not just two types) of people in a room. An instance consists of n people, the m types that each person can have, and a 0-1 matrix Q of order m that describes how the m types of people interact with each other. We use DQ to denote the direct graph on [m] whose adjacency matrix is Q. In the agent labeling problem, we wish to determine the type of every individual in the room; in the agent identification problem, we wish to find an agent having a particular type.
It turns out that even when we have asked the agents all the possible questions we can ask, we may not be able to solve the agent labeling or agent identification problems. When this happens, we call the instance ill-formed; otherwise, it is well-formed. We provide a characterization of the well-formed instances of both the agent labeling and agent identification problems in terms of DQ.
For the agent labeling problem, we prove that when an instance’s DQ has what we call an uninformative partition, identifying everyone’s type requires Ω(n2/m4) questions in the worst case. On the other hand, when the instance’s DQ has no uninformative partitions, everyone’s type can be determined in O(mn) questions. This implies that when m = o(n1/5) there are hard and easy variants of the knights and knaves problem. In particular, when m is a constant, the hard variants need Ω(n2) questions to solve the agent labeling problem in the worst case. The easy variants, however, can solve the agent labeling problem using O(n) questions.
For the agent identification problem, we prove that when an instance’s DQ has uninformative partitions and s∗is an informative type with respect to one of the uninformative partitions, then finding an agent of type s∗requires Ω(n2/m4) questions in the worst case. On the other hand, when the instance’s DQ has no uninformative partition, finding an agent of any type can be solved using O(mn) questions. Thus, setting aside the case when s∗is an uninformative type with respect to every uninformative partition of DQ, our results again imply that when m = o(n1/5), there are easy and hard variants of the knights and knaves problem when it comes to solving the agent identification problem.
2 The Knights and No-Men Problem Let A = {a1, a2, . . . , an} be a set of agents each of whom is a knight or a no-man. We are allowed to ask questions of the form “ai, is aj a knight?” for i ̸= j. A knight always answers truthfully while a no-man always answers “no”.
Additionally, we are told that there are at least k ≥2 knights. Let us consider the problem of finding a knight.
Suppose we asked the agents a set of questions S. Let GS be the undirected graph where V (GS) = A and E(GS) = {{ai, aj} : “ai, is aj a knight?” or “aj, is ai a knight?” is in S}. We shall say that a labeling of the agents f : A → {knight, no-man} is consistent with respect to S if based on the agents’ answers it is plausible that each ai has type f(ai). More specifically, f is a consistent labeling with respect to S if for each question “ai, is aj a knight?” in S, the answer given by ai is the same as the answer an agent of type f(ai) would give if asked about an agent of type f(aj).
Lemma 1. After asking the questions in S, we can conclude that some partic-ular agent is a knight if and only if at least one of these conditions hold: (1) some question’s answer is “yes”, or (2) the answers to all the questions are “no” and there is some ai that is in every independent set of size k in GS.
Proof. First, we argue that the conditions are sufficient. We get a response of “yes” if and only if one knight is asked about another knight. Thus, if condition (1) holds and the answer to the question associated with {ai, aj} ∈E(GS) is a “yes”, we can conclude that both ai and aj are knights. So suppose the answers to the questions in S are all “no”. It must be the case that no knight has been asked about another knight. The set of knights form an independent set of size at least k in GS. If condition (2) holds, some ai is part of every independent set of size k in GS, including the independent sets containing k knights. We can then conclude that ai is a knight.
Next, let us show that the conditions are necessary. Suppose neither condition holds. Since condition (1) does not hold, the responses to the questions in S are all “no”. Once again the set of knights form an independent set I of size at least k in GS. Since condition (2) does not hold, for every ai ∈A, there exists an independent set Iai in GS of size k such that ai ̸∈Iai. Now, for each ai ∈A, let fi be a labeling of the agents that assigns the agents in Iai as knights and all other agents as no-men. Clearly, fi is consistent with respect to S. Since it is possible that each ai is not a knight, we cannot conclude that any particular agent in A is a knight.
⊓ ⊔ The Tur´ an graph T (n, k) is the graph whose n vertices are partitioned into k parts of as equal size as possible, and two vertices are adjacent if and only if they belong to different parts. Tur´ an’s theorem states that among n-vertex graphs without a clique of size k + 1, the graph with the most number of edges is T (n, k). In particular, T (n, k) has ⌊(k−1)n2 2k ⌋edges. The next lemma shows the connection of Lemma 1, condition (2) to the complement of T (n −1, k −1).
Lemma 2. Let G(n, k) contain all graphs G on n vertices such that G has a vertex that is part of every independent set of size k. Let G∗∈G(n, k) so that among all the graphs in G(n, k) it has the fewest number of edges. Then G∗is the complement of T (n −1, k −1) unioned with an isolated vertex.
Proof. Let v be a vertex of G∗that is part of every independent set of size k.
Notice that deg(v) = 0. Otherwise, any edge incident to v can be removed and v will still part of every independent set of size k, contradicting our assumption about the minimality of |E(G∗)|. Next, consider G∗−v. By our assumptions about G∗and v, it must be the case that G∗−v has no independent sets of size k, and among such graphs with n −1 vertices, it has the fewest number of edges. Consequently, its complement has the property that it has no cliques of size k, and among such graphs with n −1 vertices, it has the most number of edges. According to Tur´ an’s theorem , this complement is T (n−1, k −1). Our lemma follows.
⊓ ⊔ Here is our main result.
Theorem 1. Let I(n, k) be the fewest number of questions needed in the worst case to identify a knight in the knights and no-men problem when there are n agents at least k ≥2 of whom are knights. Then I(n, k) = n−1 2 −⌊(k−2)(n−1)2 2(k−1) ⌋, the number of edges in the complement of T (n −1, k −1).
Proof. Let t′(n −1, k −1) denote the number of edges in the complement of T (n −1, k −1). First, we describe a strategy for finding a knight. Start by setting aside an agent. Partition the remaining n −1 agents into k −1 parts of as equal size as possible. For each part, ask every pair {ai, aj} of agents the question “ai, is aj a knight?”. (The order of ai and aj doesn’t matter here.) If the response is “yes”, stop and conclude that that ai and aj are knights. If none of the agents answered “yes”, conclude that (1) there are exactly k knights, (2) each part has exactly one of them, and (3) the agent we had set aside is a knight.
Using the fact that there are at least k knights and at most k −1 parts, it is easy to verify that our strategy’s conclusions are correct. It also asks the most number of questions when the answers to all its questions are “no”. Now, we have chosen the questions so that they form a graph that is isomorphic to G∗so the number of questions our strategy will ask in the worst case is t′(n−1, k −1); i.e., I(n, k) ≤t′(n −1, k −1).
Next, consider an arbitrary strategy. Suppose there are situations where the strategy will conclude that some agent is a knight even though all the answers to its questions are “no”. Let G be the graph formed by these questions. According to Lemma 1, G must have the property described in condition (2). Thus, G ∈ G(n, k). According to Lemma 2, G must have at least t′(n−1, k −1) edges. Now, suppose the strategy will conclude that some agent is a knight only when it receives a “yes” response. Assume that in the worst case it will ask r questions.
This means that in every situation where the strategy has already asked r −1 questions and the answers were all “no”, the rth question must generate a “yes” answer and the two agents that are part of the question are knights. Thus, if the strategy knows about the worst case bound r, it can avoid asking the rth question since after receiving the “no” answer to the (r −1)st question, it can already identify an agent that is a knight.
Applying the reasoning we used in the previous case, we have that r −1 ≥ t′(n −1, k −1) so r > t′(n −1, k −1). We have shown that every strategy for finding a knight where there are n agents and at least k ≥2 knights will need to ask at least t′(n −1, k −1) questions.
⊓ ⊔ Interestingly, it is much easier to find a no-man than a knight as shown in this next theorem.
Theorem 2. Let ˆ I(n, k) be the fewest number of questions needed in the worst case to identify a no-man in the knights and no-men problem when there are n agents, at least k ≥2 knights, and at least 2 no-men. Then ˆ I(n, k) ≤n −2.
Proof. Choose an arbitrary agent a and ask a about n −k agents. If all the answers are “no” then a has to be a no-man because at least one of the n −k agents we asked a about is a knight. But suppose at least one answer is a “yes”, then a is a knight. If there is also a “no” answer, then the agent we asked a about must be a no-man. But suppose all the answers are “yes”. This means that a and the n−k agents are all knights. Then we just ask a about each of the remaining k −1 agents we have not asked a about until we get a “no” response.
This leads us to a no-man using at most n −2 questions.
⊓ ⊔ Next, consider the problem of identifying all n agents as either a knight or a no-man.
Theorem 3. Let L(n, k) be the fewest number of questions needed in the worst case to identify all agents as a knight or no-man when there are n agents at least k ≥2 of whom are knights. Then I(n, k) ≤L(n, k) ≤I(n, k) + n −2.
Proof. Since identifying all agents involves identifying at least one knight, A(n, k) ≥ I(n, k). To prove the upper bound, we describe a strategy for identifying all agents. Start by using the strategy described in the proof of Theorem 1 to iden-tify a knight. Suppose a “yes” answer was obtained. We can immediately deter-mine that two agents are knights. We then ask one of them about the remaining n −2 agents. On the other hand, suppose the answers to the initial strategy for identifying a knight are all “no”. Then we know that the agent that was set aside is a knight, and there are exactly k knights. We then ask the knight about n −2 of the remaining unidentified n −1 agents. If k knights have been identified altogether, the last unidentified agent has to be a no-man; otherwise, he is a knight. In both cases, the number of questions we have asked is at most I(n, k) + n −2.
⊓ ⊔ We suspect that the upper bound given in Theorem 3 is the correct value for A(n, k).
3 Generalizing the Knights and Knaves Problem We now consider a generalization of the knights and knaves problem that encom-passes most of the variants we discussed in the introduction. It has the following set-up: There are n agents in a room each of whom is one of m types. The agents have been together long enough so they know each other’s types. Out-siders are allowed to communicate with the agents by asking only one kind of question. This question is “directed” – it is addressed to some agent ai and is about another agent aj – and has a yes or no answer. Agent ai’s response is a deterministic function of his and aj’s types.5 You happen to come across this room and are told how agents respond to the directed questions as a function of their types. Your goal is to determine the types of all the agents or identify an agent of a particular type using as few questions as possible.
Notice that in our set-up, the actual question itself is not a factor since it is the only question an outsider can ask. What matters instead is the “direction” of the question and the response to it. Assume the types are from the set [m] = {1, 2, . . ., m}. We shall use a (0, 1)-matrix Q of order m to encode the responses of the m types of agents to the directed question. That is, Qst is equal to 0 if the answer to the question addressed to an agent of type s about an agent of type t is “no” and is equal to 1 otherwise. Throughout the paper, we shall represent the structure of agent types in terms of DQ = ([m], Q), the directed graph whose vertex set is [m] and whose adjacency matrix is Q.
Formally, our model has a set of n agents A = {a1, a2, . . . , an}, where t(ai) denotes the type of agent ai. Every t(ai) ∈[m]. As an outsider, we do not know the agents’ types. However, we are allowed to ask one agent about another agent using some standard format. Let q(ai, aj) denote the answer to the question addressed to ai about aj, where ai ̸= aj.6 This answer, given by ai, is Qt(ai),t(aj).
We are interested in the following problems: – The agent labeling problem. Given A, Q, and the ability to query the agents, determine the types of all the agents in A.
– The agent identification problem. Given A, Q, s∗∈[m], and the ability to query the agents, find an agent in A of type s∗.
Our goal is to solve both problems using as few questions as possible.
Some examples. Consider an instance where the agents can have three types: knights, knaves, or yes-men. The table on the left shows how the three types of agents respond to the question “Agent ai, is agent aj a knight?”, the matrix in the middle encodes these responses, while the directed graph on the right has the matrix as its adjacency matrix.
5 Thus, none of the agents are spies since spies lie arbitrarily.
6 We emphasize that ai and aj must be distinct. Otherwise, in the knights and knaves problem, etc., we can just ask each agent if he is a knight – and the agent labeling and agent identification problems become trivial.
Knight Knave Yes-man (v1) Knight Yes No No (v2) Knave No Yes Yes (v3) Yes-man Yes Yes Yes 1 0 0 0 1 1 1 1 1 v1 v3 v2 In the coin partitioning problem, there are n coins which have to be parti-tioned into m distinct weight classes by a series of weight comparisons. That is, there are n agents with m different types. We are allowed to ask the question “Coin ai, is coin aj’s less than or equal to your weight?” The matrix below en-codes the responses, and the associated directed graph is shown on the right.
1 0 0 · · · 0 1 1 0 · · · 0 1 1 1 · · · 0 .
.
. .
.
. .
.
. ... .
.
.
1 1 1 · · · 1 v1 v2 v3 . . .
vm Assumptions. Finally, we make the following assumptions about our model: First, for each s ∈[m], there are at least two agents with type s. This is a technicality we need because we cannot ask an agent about himself. For example, if there is only one agent of a certain type, we will not be able to tell if this agent’s type has a loop or not in DQ.
Second, for any two types s and s′, they either have a different in-neighborhood or a different out-neighborhood in DQ. Otherwise, we have two versions of the same type, and it would be impossible for us to distinguish between these two versions.
Third, we have “infinite computational power”. That is, we can develop strategies and process the agents’ responses using algorithms that may not be efficient. This will allow us to focus solely on analyzing the number of questions we need to ask to solve the agent labeling and agent identification problems.
3.1 Similar Agents and Consistent Labelings In the next section, we will consider the scenario where we asked the agents all the possible questions we can ask. We would like to group them according to how they answered as well as how others answered when asked about them.
Let agent b ̸= ai, aj. We say that ai and aj are similar with respect to b, denoted as ai ∼b aj, if q(ai, b) = q(aj, b) and q(b, ai) = q(b, aj). Additionally, we say that ai and aj are similar, denoted as ai ∼aj, if they are similar with respect to every agent b ̸= ai, aj.
Lemma 3. For any two agents ai and aj, ai ∼aj if and only if t(ai) = t(aj).
Hence, the relation ∼is an equivalence relation over A and has m equivalence classes.
Proof. Let b ̸= ai, aj. When t(ai) = t(aj), q(ai, b) = q(aj, b) and q(b, ai) = q(b, aj) because these answers are simply based on the type of b and the types of ai and aj. Thus, ai ∼aj.
Let us consider the other direction. Suppose ai ∼aj but t(ai) ̸= t(aj). By our second assumption, t(ai) and t(aj) must differ in their in-neighborhoods or their out-neighborhoods in DQ. Assume the former is true. There there is some type s so that Qs,t(ai) ̸= Qs,t(aj). By our first assumption, there is an agent b ̸= ai, aj such that t(b) = s. Then q(b, ai) ̸= q(b, aj), contradicting the assumption that ai ∼aj. The same argument holds when t(ai) and t(aj) have different out-neighborhoods in DQ. It follows that t(ai) = t(aj).
We have shown that for any two agents ai and aj, ai ∼aj if and only if t(ai) = t(aj). This fact immediately implies that the relation ∼is reflexive, symmetric and transitive – i.e., ∼is an equivalence relation. Since there are m types of agents and by our first assumption there is at least two agents of each type, ∼has m equivalence classes.
⊓ ⊔ Another issue we will need to resolve is how we determine the agents’ types.
For this, we extend the notion of consistent labelings described from the previous section to our general setting.
Let S be a set of questions we have asked the agents. Based on their answers, we assign each agent a label from [m]. We say that such a labeling of the agents f : A →[m] is consistent with respect to S if for each question addressed to ai about aj in S, the answer given by ai is the same as the answer an agent of type f(ai) would give if asked about an agent of type f(aj). When S contains all the questions we can ask the agents of A, we simply say that f is a consistent labeling. One such example is the labeling f ∗that assigns each agent their type (i.e., f ∗(a) = t(a) for each a ∈A).
Proposition 1. A labeling f : A →[m] is consistent if for any ordered pair (ai, aj) of agents, Qf(ai),f(aj) = Qt(ai),t(aj). That is, (f(ai), f(aj)) is an edge in DQ if and only if (t(ai), t(aj)) is an edge in DQ.
The next lemma describes the relationships between similar agents and con-sistent labelings.
Lemma 4. For any two agents ai and aj, every consistent labeling f assigns ai and aj the same labels if and only if t(ai) = t(aj).
Proof. Let f be a consistent labeling. Suppose f(ai) = f(aj) but t(ai) ̸= t(aj).
Since t(ai) ̸= t(aj), there must be some type s so that Qs,t(ai) ̸= Qs,t(aj) or Qt(ai),s ̸= Qt(aj),s according to our second assumption. By the first assumption, there is some agent b ̸= ai, aj such that t(b) = s. Then Qf(b),f(ai) = Qs,t(ai) and Qf(b),f(aj) = Qs,t(aj). Since f(ai) = f(aj), Qs,t(ai) = Qs,t(aj). By a similar argument, Qt(ai),s = Qt(aj),s. We have arrived at a contradiction. Hence, t(ai) = t(aj). Thus, if agents have the same labels under f, they also must have the same type.
For each s ∈[m], let bs denote an agent of type s. According to our previous argument, the m agents in {bs : s ∈[m]} are assigned to different labels by f.
Now suppose f assigned an agent of type s, b′ s, to a label different from f(bs).
Then there must be two agents of different types assigned to the same label.
Since this is not possible, f must assign all agents of type s to f(bs).
⊓ ⊔ 4 Ill-Formed Instances and Consistent Labelings When we trade questions and answers with the agents, we are in effect narrowing down the set of consistent labelings of the agents. In the agent labeling problem, our goal is to get to a point where there is only one consistent labeling left so that we are certain about each agent’s type. In the agent identification problem where we want to identify an agent of type s∗, our goal is to reach a point where there is an agent ai, so that in all of the labelings consistent with the set of questions asked so far, ai is assigned the label s∗.
We call (A, Q) an ill-formed instance of the agent labeling problem if there is more than one consistent labeling of the agents. That is, the agent labeling problem cannot be solved even after we have asked all the questions. Similarly, (A, Q, s∗) is an ill-formed instance of the agent identification problem if there is no agent ai so that ai is assigned the label s∗in every consistent labeling of the agents. For both problems, we call an instance well-formed if it is not ill-formed.
As an example, consider an instance (A, Q) of the classic knights and knaves problem. Assume we have asked all the possible questions. Let a be one of the agents. Partition the set of agents into two groups: the first group contains a and all the agents a identified as knights (i.e., a answered “yes” when asked about these agents); the second group contains the remaining agents (i.e., a answered “no” when asked about these agents). Notice that if a is a knight, the first group is composed entirely of knights and the second group entirely of knaves. On the other hand, if a is a knave, the opposite is true. Thus, there are two consistent labelings: f1 assigns the first group as knights and the second group as knaves; f2 does the opposite. In other words, after we have asked all the questions, we can partition the agents into a group of knights and a group of knaves but we cannot tell them apart. This is because knights and knaves behave in the same way – they both say “yes” to agents who have their type and “no” to others. This is captured in DQ where both the knight-type and the knave-type are singleton vertices with a loop. Hence, (A, Q) is an ill-formed instance of the agent labeling problem. Similarly, (A, Q, s∗) where s∗is the type corresponding to a knight (or knave) is an ill-formed instance of the agent identification problem.
The astute reader would note though that we are able to identify a knight in the classic knights and knaves problem. The reason is quite subtle – in the problem’s description, we are told that there are more knights than knaves. Thus, the group with more agents is the knights group. In what follows, we provide a graph-based characterization of the ill-formed instances (and equivalently the well-formed instances) of the agent labeling and agent identification problems.
Lemma 5. Let α be an automorphism of DQ. Then fα such that fα(a) = α(t(a)) for each a ∈A is a consistent labeling of A.
Proof. For any ordered pair of agents (ai, aj) of A, Qf(ai),f(aj) = Qα(t(ai)),α(t(aj)).
But the latter is equal to Qt(ai),t(aj) because α is an automorphism of DQ. Hence, f is a consistent labeling of A.
⊓ ⊔ Lemma 6. Suppose f is a consistent labeling of A. For each s ∈[m], let bs be an agent of type s. Then αf such that αf(s) = f(bs) for each s ∈[m] is an automorphism of DQ.
Proof. Let B = {bs, s ∈[m]}. We note that by Lemma 4, no two agents in B are mapped by f to the same label. Hence, αf is a bijection.
Consider an ordered pair of distinct types (s, s′). By the definition of αf, (αf(s), αf(s′)) is an edge in DQ if and only if (f(bs), f(bs′)) is an edge in DQ.
But the latter is true if and only if (t(bs), t(bs′)) = (s, s′) is an edge in DQ. Thus, αf preserves edges between distinct types in DQ.
Now, consider the ordered pair (s, s). Let b′ s be another agent with label s.
According to Lemma 4, f(bs) = f(b′ s). Hence, (αf(s), αf(s)) is an edge in DQ if and only if (f(bs), f(b′ s)) is an edge in DQ. But the latter is true if and only if (t(bs), t(b′ s)) = (s, s) is an edge in DQ. It follows that αf also preserves edges between the same types in DQ. Hence, αf is an automorphism of DQ.
⊓ ⊔ Theorem 4. An instance (A, Q) of the agent labeling problem is ill-formed if and only if DQ has more than one automorphism.
Proof. Suppose DQ has an automorphism α that is different from the identity map. From Lemma 5, fα is a consistent lebeling of A. Since α is not the identity map, f ̸= f ∗so (A, Q) is ill-formed.
Suppose f ̸= f ∗is another consistent labeling of A because (A, Q) is ill-formed. By Lemma 4, there is some s ∈[m] so that for every agent a of label s, f(a) ̸= s. By Lemma 6, αf is an automorphism of DQ. Since f(bs) ̸= s, αf is not the identity map. Therefore, DQ has more than one automorphism.
⊓ ⊔ Theorem 5. An instance (A, Q, s∗) of the agent identification problem is ill-formed if and only if some automorphism of DQ does not fix s∗(i.e., the auto-morphism does not map s∗to itself).
Proof. Suppose α is an automorphism of DQ that does not fix s∗. By Lemma 5, fα is a consistent labeling of A. Since f assigns all agents of type s∗to a label different from s∗, (A, Q, s∗) is ill-formed.
Suppose (A, Q, s∗) is an ill-formed instance. By Lemma 4, there must be a consistent labeling f that assigns all agents of type s∗to a label s so that s ̸= s∗. According to Lemma 6, αf is an automorphism of DQ. Since f(bs∗) = s, αf(s∗) = s ̸= s∗. That is, αf does not fix s∗.
⊓ ⊔ 4.1 Finding a Consistent Labeling Let us once again assume that we have asked the agents all the possible questions we can ask. How do we go about constructing a consistent labeling of the agents?
Let E1, E2, . . . , Em denote the equivalence classes of ∼. For each Ei, let bi and b′ i be two agents in Ei. Denote by D∼the directed graph whose vertex set is {E1, E2, . . . , Em} and whose edge set is {(Ei, Ej) : q(bi, bj) = 1} ∪{(Ei, Ei) : q(bi, b′ i) = 1}. Since the equivalence classes are in one-to-one correspondence with the m types of agents, D∼is in fact isomorphic to DQ. We now describe a procedure for labeling the agents of A: Label(A, Q, {q(ai, aj) : ai, aj ∈A}) Partition the agents according to ∼and form the equivalence classes of ∼.
Construct the directed graph D∼.
Find an isomorphism α from D∼to DQ.
For each agent a in equivalence class E, set f(a) = α(E). Return f.
Theorem 6. The labeling f returned by Label is consistent. Consequently, (1) when (A, Q) is a well-formed instance of the agent labeling problem, f = f ∗.
(2) when (A, Q, s∗) is a well-formed instance of the agent identification problem, f(a) = s∗if and only if t(a) = s∗for each a ∈A.
Proof. First, consider any ordered pair of agents (a, b) where a ∈Ei, b ∈Ej, a ̸= b but Ei may be the same as Ej. We know that (α(Ei), α(Ej)) is an edge in DQ if and only if (t(a), t(b)) is an edge in DQ. Thus, Qf(a),f(b) = Qα(Ei),α(Ej) = Qt(a),t(b). That is, f is a consistent labeling.
When (A, Q) is a well-formed instance of the agent labeling problem, there is only one consistent labeling of the agents so f must be f ∗. When (A, Q, s∗) is a well-formed instance of the agent identification problem, there is at least one agent a of type s∗so that every consistent labeling assigns a the label s∗. But according to Lemma 4 every consistent labeling must assign all agents of type s∗the label s∗. Furthermore, agents of type s ̸= s∗will not be assigned the label s∗. Since f is a consistent labeling, f(a) = s∗if and only if t(a) = s∗for each a ∈A.
⊓ ⊔ Theorem 6 shows that when we have asked the agents all the questions we can ask and the instances are well-formed, Label will solve the agent labeling and agent identification problems. Starting with the next section onwards, we will assume that the inputs to the agent labeling or agent identification problems are well-formed.
5 Uninformative Partitions Two of our objectives for studying the generalized knights and knaves problem is to understand (1) why solving the agent labeling problem requires significantly more questions in the worst case in the knights and no-men setting than in the knights and yes-men setting, and (2) why finding a knight requires significantly more questions in the worst case than finding a no-man in the knights and no-men setting. In this section, we begin to address these issues by introducing what we call uninformative partitions.
Any two (not necessarily distinct) types s and s′ have four possible relation-ships based on the questions that can be asked between agents of type s and with those of type s′: (1) Qs,s′ = Qs′,s = 1; (2) Qs,s′ = 1, Qs′,s = 0; (3) Qs,s′ = 0, Qs′,s = 1; (4) Qs,s′ = Qs′,s = 0. Keeping this in mind, let P = {P1, P2, . . . , Pr} be a partition of [m] with r < m. Type s is uninformative with respect to P if, for i = 1, . . . , r, the relationship of s with each type in Pi is the same. That is, we cannot use questions addressed to or about agents of type s to distinguish between agents whose types belong to the same part in P. Otherwise, s is infor-mative with respect to P. We call P an uninformative partition of DQ = ([m], Q) if each of its parts contains an uninformative type with respect to P.
For example, in the knights and yes-men problem, the only possible candidate for an uninformative partition is P = {{knight, yes-man}} since we can only consider partitions with just one part. A knight is informative with respect to P because two knights have a relationship that is described by (1) whereas a knight and a yes-man have a relationship that is described by (3). Similarly, a yes-man is also informative with respect to P because the relationship between a yes-man and a knight is described by (2) whereas that between a yes-man and another yes-man is described by (1). Thus, the knights and yes-men problem has no uninformative partitions. On the other hand, in the knights and no-men problem, we leave it up to the reader to verify that the no-man is uninformative with respect to P = {{knight, no-man}} while the knight a informative with respect to the same partition. Hence, the knights and no-man problem has an uninformative partition.
Proposition 2. Let P = {P1, P2, . . . , Pr} be an uninformative partition of DQ.
Each part of P contains exactly one uninformative type with respect to P.
Proof. For each Pi, let si be an uninformative type with respect to P. Without loss of generality, suppose P1 contains two uninformative types: s1 and s′ 1. We will argue that this implies that s1 and s′ 1 have exactly the same in- and out-neighborhoods in DQ, contradicting the second assumption we made about our model. Suppose s ∈Pi is in the in-neighborhood of s1 in DQ. We know that Qsi,s1 = Qsi,s′ 1 since si uninformative. Thus, if s = si, s is also in the in-neighborhood of s′ 1. But suppose s ̸= si. Then Qsi,s1 = Qs,s1 and Qsi,s′ 1 = Qs,s′ 1 since both s1 and s′ 1 are uninformative. Using the first fact we cited, Qs,s1 = Qs,s′ 1 so again s is in the in-neighborhood of s′ 1. It follows that s1 and s′ 1 have the same in-neighborhood. A similar argument shows that they have the same out-neighborhood. Thus, no part of P can have two or more uninformative types with respect to P.
⊓ ⊔ From here on, assume that P = {P1, P2, . . . , Pr} is an uninformative partition of DQ with si as the uninformative type in Pi. An agent a ∈A is uninforma-tive agent with respect to P if t(a) is an uninformative type with respect to P; otherwise, a is informative with respect to P. How difficult is it to find an agent who is informative with respect to P? We note that the agent labeling problem is at least as hard as this problem since knowing each agent’s type allows us to immediately identify an informative agent. The agent identification problem where s∗is an informative type with respect to P is also at least as hard as this problem since every agent of type s∗is informative. Let AP = {A1, A2, . . . , Ar} denote the partition of A such that a ∈Ai if and only if t(a) ∈Pi. We shall now show that even when we know AP, finding an informative agent requires Ω(n2/m4) questions in the worst case. We will prove this result in the same way that we showed that finding a knight in the knights and no-men problem requires Ω(n2) questions in the worst case.
Let a ∈Ai and b ∈Aj. Suppose we asked a about b. We say that the question is unhelpful when q(a, b) = Qsi,sj because we did not gain any definite information about a or b’s type from q(a, b). On the other hand, when q(a, b) ̸= Qsi,sj, we say that the question is helpful because we can conclude that both a and b are informative agents. In the knights and no-man problem, questions with “no” responses were unhelpful while those with “yes” responses were helpful.
Let G = (V, E) be a graph and V = {V1, V2, . . . , Vr} be a partition of V . Let ki be a non-negative integer such that ki ≤|Vi| for i = 1, . . . , r. A (k1, k2, . . . , kr)-independent set I of G is an independent set of G such that |I ∩Vi| = ki for i = 1, . . . , r. We are now ready to present the counterparts to Lemmas 1, 2, and Theorem 1.
Lemma 7. After asking the questions in S, we can conclude that some agent is informative if and only if one of these conditions hold: (1) some question in S is helpful, or (2) every question in S is unhelpful and there is some agent a that is in every (k1, k2, . . . , kr)-independent set in GS where ki = 2(|Pi| −1) for i = 1, . . . , r.
Proof. First, we show that the conditions are sufficient. By definition, a ques-tion is helpful if and only if the two involved agents are informative. Thus, if condition (1) holds, we can immediately identify two informative agents. So sup-pose all questions in S are unhelpful. The set of informative agents must form a (k′ 1, k′ 2, . . . , k′ r)-independent set in GS where k′ i ≥2(|Pi| −1) for i = 1, . . . , r since each informative label in Pi has at least two agents in Ai. If condition (2) holds, a must be part of this independent set so we can conclude that a is an informative agent.
Next, let us show that the conditions are necessary. Suppose neither condition holds. This means that all questions in S are unhelpful so once again the set of informative agents form a (k′ 1, k′ 2, . . . , k′ r)-independent set in S where k′ i ≥ 2(|Pi| −1) for i = 1, . . . , r in GS. Furthermore, for each agent a, there is some (k1, k2, . . . , kr)-independent set Ia in GS that does not contain a. We shall show below that there is a labeling fa that is consistent with respect to S such that f(a) is an uninformative type. This implies that it is plausible that a is an uninformative agent. Since we chose a arbitrarily, we cannot conclude that any particular agent in A is informative.
We now construct fa. For i = 1, . . . , k do the following: (1) for each informa-tive type s′ i in Pi, pick two unlabeled agents from Ia ∩Ai and set their fa-label to s′ i, and (2) assign all agents in Ai −Ia the uninformative type si. The labeling fa is consistent with respect to S because whenever there is an edge between two agents in GS, fa assigned at least one of them an uninformative type.
⊓ ⊔ Lemma 8. Let k = (k1, k2, . . . , kr) such that ki = 2(|Pi| −1) for i = 1, . . . , r.
For each set A of n agents partitioned according to AP, let G(AP, k) contain all graphs G on A that has a vertex that is part of every k-independent set in G.
Then there is some A∗such that every graph in G(A∗ P, k) has Ω(n2/m4) edges.
Proof. Let A∗be a set of n agents such that the size of each part of A∗ P is either ⌊n/r⌋or ⌈n/r⌉. Let G∗∈G(A∗ P, k) so that among all graphs in the set it has the fewest number of edges. Denote the parts in A∗ P as A1, A2, . . . , Ar. Assume a ∈Ai is a vertex of G∗that is part of every k-independent set. Notice that deg(a) = 0; otherwise any edge incident to a can be removed and a will still be part of every k-independent set, contradicting the minimality of |E(G∗)|. Hence, |E(G∗)| = |E(G∗−a)|, and G∗−a has no k-independent sets relative to the partition (A′ 1, A′ 2, . . . , A′ r) of its vertex set, where A′ i = Ai \ {a} and A′ j = Aj for j ̸= i. Let us now bound the size of E(G∗−a).
Suppose that we wish to construct G∗−a on A \ {a} starting with an empty edge set. At this beginning stage, there are X = |A′ 1| k1 |A′ 2| k2 . . .
|A′ r| kr k-independent sets. As we add each edge one by one, some k-independent sets are destroyed until all X of them are completely eliminated. If Y is the maximum number of k-independent sets that the addition of a single edge can destroy, then G∗−a has at least X/Y edges. So let us determine Y . We consider two cases: (1) the added edge e is completely contained within A′ j for some j ∈[r] or (2) e has one endpoint in A′ p and one endpoint in A′ q for some p, q ∈[r], p ̸= q.
For case (1), the maximum number of k-independent sets that can be de-stroyed is Y1 = |A′ 1| k1 |A′ 2| k2 . . .
|A′ j−1| kj−1 |A′ j| −2 kj −2 |A′ j+1| kj+1 . . .
|A′ r| kr .
For case (2), the maximum number of k-independent sets that can be destroyed is Y2 = |A′ 1| k1 . . .
|A′ p−1| kp−1 |A′ p|−1 kp−1 |A′ p+1| kp+1 . . .
. . .
|A′ q−1| kq−1 |A′ q|−1 kq−1 |A′ q+1| kq+1 . . .
|A′ r| kr .
Since Y = max{Y1, Y2}, X/Y = min{X/Y1, X/Y2}. Now, X Y1 = |A′ j| kj |A′ j|−2 kj−2 = |A′ j|2 −|A′ j| k2 j −kj and X Y2 = |A′ p| kp |A′ q| kq |A′ p|−1 kp−1 |A′ q|−1 kq−1 = |A′ p| · |A′ q| kpkq .
Both X/Y1 and X/Y2 are Ω(n2/m4) because A′ j, A′ p, A′ q have Θ(n/r) agents and r ≤m −1 while each ki is at most m. Thus, G∗−a, and consequently G∗, has Ω(n2/m4) edges.
⊓ ⊔ Theorem 7. Given P, {si : i = 1, . . . , r} and AP, finding an informative agent with respect to P requires asking Ω(n2/m4) questions in the worst case.
Proof. Let A∗be the set of n agents described in Lemma 8. Consider an arbitrary strategy for finding an informative agent of A∗. Suppose that there are situations where the strategy will conclude that some agent is informative even though all the questions asked were unhelpful. Let G be the graph formed by these questions. According to Lemma 7, G must have the property described in its second condition. Thus, G ∈G(A∗ P, k). By Lemma 8, G has Ω(n2/m4) edges so this strategy will ask Ω(n2/m4) questions in the worst case.
Now, suppose the strategy will conclude that some agent is informative only when a helpful question has been asked. Assume that in the worst case it uses w questions. It follows that in situations where the strategy has already asked w−1 questions all of which are unhelpful, the wth question must be helpful and the two agents that are part of this question are informative. Thus, if the strategy is aware of the worst case bound w, it can avoid asking the wth question; the first w −1 unhelpful questions are sufficient for identifying an informative agent.
Applying the reasoning we used in the previous case, w −1 ∈Ω(n2/m4). The theorem follows.
⊓ ⊔ Let L(n, Q) denote the fewest number of questions needed in the worst case to solve the agent labeling problem whose input is a well-formed instance (A, Q) with |A| = n. Similarly, let I(n, Q, s∗) be the fewest number of questions needed in the worst case to solve the agent identification problem whose input is a well-formed instance (A, Q, s∗) with |A| = n. In light of our earlier discussion about the difficulty of finding an informative agent in relation to that of the agent labeling and agent identification problems, we have the following result: Corollary 1. When DQ = ([m], Q) has an uninformative partition, L(n, Q) ∈ Ω(n2/m4). Additionally, when s∗is an informative type in some uninformative partition of DQ, I(n, Q, s∗) ∈Ω(n2/m4).
6 Label or Detect Given (A, Q) where |A| = n and Q has order m, we now present an algorithm called LabelOrDetect that either outputs a consistent labeling of the agents in A or detects the presence of an uninformative partition in DQ using O(mn) questions. In the first part of the algorithm, a set of special agents are chosen.
All questions addressed to the special agents or about the special agents are asked during the course of the algorithm. The answers are then used to partition a subset of the agents into parts. The goal is to get to a point where each part has exactly one special agent because the algorithm intends to make a special agent represent the agents in its part. This seems valid because if special agent bi is in part Ai, bi’s answers are just like those of the agents in Ai. In the second part of the algorithm, the answers to all the unanswered questions are then obtained by assuming that, for each part Ai, the agents in it have the same type as bi.
The agents in A are then partitioned according to the ∼relation. If the number of equivalence classes is less than what is expected – which is m – the earlier assumption must be false. The algorithm concludes that it had encountered an uninformative partition. However, if the number of equivalence classes is m, the algorithm passes the answers to all the questions to the procedure Label, which then outputs a labeling f.
Consider any two agents of A. Just like any two types, these two agents have four possible relationships based on how they answer queries about each other.
Let A′ ⊆A, let A′ be a partition of the agents in A′, and let b be an agent not in A′. In our algorithm, we use refine(A′, A′, b) to denote the operation that partitions the agents in each part of A′ according to their relationship with b.
The first part of the algorithm is done through the procedure AskAndPartition, shown in Figure 1. Initially, A′ and A′ are set to A and {A} respectively, and all agents are unmarked. The values q′(a, a′) for each a, a′ ∈A are set to −1 as an indication that the question addressed to a about a′ has not been asked.
The algorithm then enters a while loop. At the beginning of each iteration, the parts in A′ may or may not contain special agents. For those parts A′ i that are missing a special agent, an agent bi ∈A′ i is chosen, marked special and noted as new. All questions between bi and every agent in A′ −bi are asked and then stored in q′. Based on the answers, bi is used to refine A′ i −bi and every other part of A′ later. If bi subdivided A′ i −bi into two or more parts, bi is moved from A′ i to B because it is no longer obvious which part bi should belong to; otherwise, bi stays in A′ i.
Notice that AskAndPartition is type-preserving. That is, if two agents have the same type and neither one was moved to B, they stayed in the same part of A′ throughout the execution of the procedure. Since there are only m types, the number of parts in A′ never exceeds m. Additionally, from one iteration to the next in the while loop of AskAndPartition, the number of parts in A′ either increases or stays the same. Keeping these observations in mind, we now prove some important lemmas about the procedure.
Lemma 9. The algorithm AskAndPartition terminates.
Proof. To prove the lemma, it suffices to show that the while loop in AskAndPar-tition ends. Let A′ (y) denote the set A′ at the beginning of the yth iteration of the while loop. Suppose that the sizes of the A′ (y)’s are always increasing. We note that the while loop will never terminate in this case because at the beginning of each iteration there are always some parts that have no special agents. But this AskAndPartition(A, Q) A′ ←A, A′ ←{A} B ←∅ For every pair of agents a and a′, initialize q′(a, a′) to −1.
Mark all agents in A′ as not special.
while (some parts of A′ has no agent marked special) Denote the parts in A′ as A′ 1, A′ 2, . . . , A′ k.
Bnew ←∅ for each part A′ i in A′ A′ i ←{A′ i} if A′ i does not have an agent marked special choose an arbitrary agent bi ∈A′ i, mark it as special, add it to Bnew ask all questions between bi and every other agent in A′ and record the answers in q′ if A′ i ̸= {bi} A′ i ←refine(A′ i −bi, A′ i −bi, bi) if A′ i contains two or more parts move bi from A′ i to B else A′ i ←{A′ i} endfor for each bi ∈Bnew for each part A′ j that does not contain bi / Questions between bi and A′ j have already been asked earlier./ A′ j ←refine(A′ j, A′ j, bi) endfor endfor A′ ←Sk i=1 A′ i, A′ ←Sk i=1 A′ i endwhile return( A′, A′, B, {q(a, a′) : a, a′ ∈A}) Fig. 1. The procedure AskAndPartition.
will also mean that |A′ (y)| will exceed m at some point, a contradiction. So it must be the case that for some y, |A′ (y)| = |A′ (y+1)|. Choose y so that it has the smallest possible value. Since A′ (y) had more parts than A(y−1), some parts of A′ (y) were assigned new special agents. But since |A′ (y)| = |A′ (y+1)|, none of these special agents refined the parts of A(y). That is, none of them were removed from their part and moved to B. It follows that in the next iteration of the while loop, all parts in A′ (y+1) had special agents so the while loop had to terminate.
⊓ ⊔ Lemma 10. Let A′ f, A′ f, and Bf denote the sets A′, A′ and B at the end of the while loop in AskAndPartition. Either each part in A′ f consists of agents of the same type or DQ has an uninformative partition.
Proof. Let A′ 1, A′ 2, . . . , A′ m′ be the parts in A′ f, and let bi be the special agent in A′ i for i = 1, . . . , m′. Suppose that some part A′ i has two agents whose types are different. To prove the lemma, we will show that DQ has an uninformative par-tition. Let P = {P1, . . . , Pm′, . . . , Pm′′} such that (1) for 1 ≤i ≤m′, Pi contains the types of the agents in A′ i and (2) Pm′+1, . . . , Pm′′ are singleton sets contain-ing the types not found in ∪m′ j=1Pi. (When all the types can be found in ∪m′ j=1Pi, we do not need Pm′+1, . . . , Pm′′.) Since AskAndPartition is type-preserving, no two parts in P contain the same type. Additionally, because ∪m′′ i=1Pi = [m] and some Pi contains at least two types, P is a partition of [m] whose number of parts is strictly less than m.
Next, we claim that if b is a special agent in B, t(b) is an uninformative type with respect to P. Recall that this means that t(b) has the same relationship with each type in Pi for i = 1, . . . , m′′. If Pi contains only one type, this is clearly true. If Pi contains two or more types, this is also true because when b was first marked special and used to refine the parts of A′, it had the opportunity to subdivide Ai – but it never did. Thus, t(b) is an uninformative type with respect to P. Consequently, the single type in Pi is uninformative with respect to P for i = m′ + 1, . . . , m′′.
Let us now consider t(bi) for i = 1, . . . , m′. If there is some special agent b in B with t(b) = t(bi), t(bi) has to be uninformative with respect to P. So suppose no special agent in B has the same type as t(bi). Then all agents of type t(bi) are in Ai. If there is another agent b′ i in Ai such that t(b′ i) ̸= t(bi), then the questions between bi and b′ i were answered in the same way as the questions between bi and every other agent of type t(bi) because bi did not subdivide Ai. This shows that t(bi) has the same relationship with itself and t(b′ i), and in general with every other type in Pi. It also has the same relationship with every type in Pj, j ̸= i, j = 1, . . . , m′ because when bi was first marked special and used to refine the parts of A′, it had the opportunity to subdivide Ai – but again it never did.
So once again t(bi), for i = 1, . . . , m′, is uninformative with respect to P. We have now proven that P is an uninformative partition.
⊓ ⊔ Figure 2 shows the algorithm LabelOrDetect.
Theorem 8. The total number of questions asked in LabelOrDetect is O(mn).
Furthermore, if the algorithm outputs a function f, the function is a consistent LabelOrDetect(A, Q) (A′, A′, B, {q(a, a′) : a, a′ ∈A}) ←AskAndPartition(A, Q) for each pair of distinct agents a, a′ ∈A such that q′(a, a′) = q′(a′, a) = −1 if a and a′ are not marked special let A′ i and A′ j denote the parts of A′ that contain a and a′ respectively if A′ i ̸= A′ j set q′(a, a′) and q′(a′, a) to q′(bi, bj) and q′(bj, bi) respectively else set q′(a, a′) and q′(a′, a) to q′(bi, a′) and q′(a′, bi) respectively endfor Based on {q′(a, a′), a, a′ ∈A}, partition the agents according to the relation ∼.
if there are m equivalence classes f ←Label(A, Q, {q′(a, a′), a, a′ ∈A}) return(f) else return(DQ has an uninformative partition) Fig. 2. The algorithm LabelOrDetect labeling of A. On the other hand, when it concludes that DQ has an uninformative partition, it really has an uninformative partition.
Proof. In AskAndPartition, a special agent is moved to B only when it subdivides the part it came from; that is, it increased the the number of parts in A′ by at least 1. Since the number of parts in A′ never exceeds m, the number of special agents in B is at most m−1. The other special agents can be found in each part of A′ f, one for each part. Hence, there are at most 2m −1 agents marked special in A. Each one asked O(n) questions so the total number of questions asked in AskAndPartition is O(mn).
Next, Lemma 10 states that at the end of AskAndPartition there are two possibilities: (1) each part of A′ f consists of agents of the same type or (2) DQ has an uninformative partition. The first case implies that the way LabelOrDetect filled in the answers to the unanswered questions was correct. That is, q′(a, a′) = q(a, a′) for every pair of distinct agents a, a′. It follows that ∼will have m equivalence classes, and the labeling f produced by Label is consistent.
For the second case, we will argue that LabelOrDetect will discover that ∼ has less than m equivalence classes and conclude that DQ has an uninformative partition. Let P be the uninformative partition described in the proof of Lemma 10. Let Ai be a part of Af that contains agents of different types. Then, there is some b′ i such that t(b′ i) ̸= t(bi) where bi is the special agent in Ai. This means that the algorithm will treat b′ i and all agents of type t(b′ i) in Ai like bi. If after processing {q′(a, a′) : a, a′} the algorithm still noted that ∼has m equivalence classes, then some special agent b ∈B must have t(b) = t(b′ i) – otherwise, there will be no equivalence class representing t(b′ i). But in the proof of Lemma 10, we noted that t(b) is an uninformative type with respect to P. Thus, Ai has two uninformative types with respect to P – t(bi) and t(b′ i) – a contradiction according to Proposition 2. Hence, the algorithm will discover that ∼has less than m equivalence classes and output that DQ has an uninformative partition.
⊓ ⊔ According to Theorem 6, when the instances of the agent labeling and iden-tification problems are well-formed, the consistent labeling produced by Label will solve the problems.
Corollary 2. When DQ = ([m], Q) has no uninformative partitions, L(n, Q) ∈ O(nm) and for each s∗∈[m], I(n, Q, s∗) ∈O(nm).
We emphasize that when DQ has uninformative partitions, it is still possible that LabelOrDetect outputs a consistent labeling. It will do so when it does not come across an uninformative partition. Finally, we also note the following result: Corollary 3. Suppose DQ = ([m], Q) has only one uninformative partition and this partition is {[m]}. When s∗is the uninformative type in {[m]}, I(n, Q, s∗) ∈ O(nm).
Proof. Let (A, Q, s∗) be the instance of the agent identification problem with |A| = n. Run the LabelOrDetect algorithm with (A, Q) as the input. If the algorithm outputs a consistent labeling f of A, use f to find an agent of type s∗.
But suppose the algorithm concludes that DQ has an uninformative partition.
Since DQ has only one uninformative partition and it is {[m]}, it means that the while loop stopped after only one iteration because the chosen special agent b was not able to refine A −b. Thus, b is an uninformative agent with respect to {[m]} and must have type s∗. Hence, using just O(mn) questions, the agent identification problem is solved.
⊓ ⊔ 7 Conclusions Every Q corresponds to a particular variant of the knights and knaves problem because it describes the people that are in the room. According to Corollaries 1 and 2, when m = o(n1/5) there are easy and hard variants of the knights and knaves problem when it comes to solving the agent labeling problem. The easy variants can be solved using O(mn) questions while the hard variants require Ω(n2/m4) questions to solve in the worst case. The “gap” between these two is particularly striking when m is a constant, as stated in the next two corollaries.
Corollary 4. Let (A, Q) be a well-formed instance of the agent labeling problem where |A| = n and m is a constant. When DQ has no uninformative partitions, the agent labeling problem can be solved using O(n) questions. On the other hand, when DQ has an uninformative partition, the agent labeling problem requires Ω(n2) questions to solve.
The above corollary explains why the finding each agent’s type in the knights and yes-men setting was significantly easier than in the knights and no-men setting.
For the agent identification problem, our results are not as complete. Setting aside the case when DQ has uninformative partitions and s∗is an uninformative type for each of these partitions, then Corollaries 1 and 2 show that the agent identification problem also has easy and hard variants of the knights and knaves problem when it comes to solving the agent identification problem. The corollary below summarizes our results when m is a constant.
Corollary 5. Let (A, Q, s∗) be a well-formed instance of the agent identification problem where |A| = n and m is a constant. When (1) DQ has no uninforma-tive partitions or (2) DQ has only one uninformative partition {[m]} and s∗ is uninformative with respect to this partition, the agent identification problem can be solved using O(n) questions. But when DQ has uninformative partition, and s∗is an informative type with respect to one of these partitions, the agent identification problem requires Ω(n2) questions to solve.
The above corollary explains why the finding a no-man was significantly eas-ier than finding a knight in the knights and no-men setting. The knights and no-men problem has only one uninformative partition and it consisted of the agents’ types. The knight was informative with respect to this partition, while the no-man was the uninformative type in this partition.
Acknowledgements This paper is based on results obtained during the Research Experience for Graduate Students (REGS) program in 2012 at the University of Illinois at Urbana-Champaign. The REGS program Combinatorics group was supervised by Doug West. The authors are grateful to Kenta Ozeki for introducing them to the knights and knaves problem and translating relevant work by Hanajiri. The authors would also like to thank Derrek Yager for his valuable comments.
References 1. Martin Aigner. Variants of the majority problem. Discrete Applied Mathematics, 137(1):3 – 25, 2004.
2. Laurent Alonso, Philippe Chassaing, Edward M. Reingold, and Ren´ e Schott. The worst-case chip problem. Information Processing Letters, 89(6):303 – 308, 2004.
3. Laurent Alonso, Edward M. Reingold, and Ren´ e Schott. Determining the majority.
Information Processing Letters, 47(5):253 – 255, 1993.
4. Pavel M. Blecher. On a logical problem. Discrete Mathematics, 43(1):107 – 110, 1983.
5. Zdenˇ ek Dvoˇ r´ ak, V´ ıt Jel´ ınek, Daniel Kr´ al’, Jan Kynˇ cl, and Michael Saks. Three optimal algorithms for balls of three colors. In Volker Diekert and Bruno Durand, editors, STACS 2005, volume 3404 of Lecture Notes in Computer Science, pages 206–217. Springer Berlin / Heidelberg, 2005.
6. A. Hanajiri. On knights, knaves, spies and yes-men problems. Master’s thesis, Keio University, 2012. (In Japenese. Supervisor: Prof. Oda.).
7. Daniel Kr´ al’, Jiˇ r´ ı Sgall, and Tom´ aˇ s Tich´ y. Randomized strategies for the plurality problem. Discrete Applied Mathematics, 156(17):3305 – 3311, 2008.
8. Michael E. Saks and Michael Werman. On computing majority by comparisons.
Combinatorica, 11:383–387, 1991. 10.1007/BF01275672.
9. P´ al Tur´ an. On an extremal problem in graph theory. Mat. ´ es Fiz. Lapok, 48:436– 452, 1941. (In Hungarian.).
10. G´ abor Wiener. Search for a majority element. Journal of Statistical Planning and Inference, 100(2):313 – 318, 2002.
11. Mark Wildon. Knights, spies, games and ballot sequences. Discrete Mathematics, 310(21):2974 – 2983, 2010. |
11142 | https://www.wikiwand.com/en/articles/Lloyd's_mirror | Lloyd's mirror
Optical setup to obtain interference pattern by reflection From Wikipedia, the free encyclopedia
Lloyd's mirror is an optics experiment that was first described in 1834 by Humphrey Lloyd in the Transactions of the Royal Irish Academy., beyond those provided by Thomas Young and Augustin-Jean Fresnel. In the experiment, light from a monochromatic slit source reflects from a glass surface at a small angle and appears to come from a virtual source as a result. The reflected light interferes with the direct light from the source, forming interference fringes..
Setup
Lloyd’s Mirror is used to produce two-source interference patterns that have important differences from the interference patterns seen in Young's experiment.
In a modern implementation of Lloyd's mirror, a diverging laser beam strikes a front-surface mirror at a grazing angle, so that some of the light travels directly to the screen (blue lines in Fig. 1), and some of the light reflects off the mirror to the screen (red lines). The reflected light forms a virtual second source that interferes with the direct light.
In Young's experiment, the individual slits display a diffraction pattern on top of which is overlaid interference fringes from the two slits (Fig. 2). In contrast, the Lloyd's mirror experiment does not use slits and displays two-source interference without the complications of an overlaid single-slit diffraction pattern.
In Young's experiment, the central fringe representing equal path length is bright because of constructive interference. In contrast, in Lloyd's mirror, the fringe nearest the mirror representing equal path length is dark rather than bright. This is because the light reflecting off the mirror undergoes a 180° phase shift, and so causes destructive interference when the path lengths are equal or when they differ by an integer number of wavelengths.
Applications
Interference lithography
The most common application of Lloyd's mirror is in UV photolithography and nanopatterning. Lloyd's mirror has important advantages over double-slit interferometers.
If one wishes to create a series of closely spaced interference fringes using a double-slit interferometer, the spacing d between the slits must be increased. Increasing the slit spacing, however, requires that the input beam be broadened to cover both slits. This results in a large loss of power. In contrast, increasing d in the Lloyd's mirror technique does not result in power loss, since the second "slit" is just the reflected virtual image of the source. Hence, Lloyd's mirror enables the generation of finely detailed interference patterns of sufficient brightness for applications such as photolithography.
Typical uses of Lloyd's mirror photolithography would include fabrication of diffraction gratings for surface encoders and patterning the surfaces of medical implants for improved biofunctionality.
Test pattern generation
High visibility cos2-modulated fringes of constant spatial frequency can be generated in a Lloyd's mirror arrangement using parallel collimated monochromatic light rather than a point or slit source. The uniform fringes generated by this arrangement can be used to measure the modulation transfer functions of optical detectors such as CCD arrays to characterize their performance as a function of spatial frequency, wavelength, intensity, and so forth.
Optical measurement
The output of a Lloyd's mirror was analyzed with a CCD photodiode array to produce a compact, broad range, high accuracy Fourier transform wavemeter that could be used to analyze the spectral output of pulsed lasers.
Radio astronomy
In the late 1940s and early 1950s, CSIRO scientists used a technique based on Lloyd's mirror to make accurate measurements of the position of various galactic radio sources from coastal sites in New Zealand and Australia. As illustrated in Fig. 3, the technique was to observe the sources combining direct and reflected rays from high cliffs overlooking the sea. After correcting for atmospheric refraction, these observations allowed the paths of the sources above the horizon to be plotted and their celestial coordinates to be determined.
Underwater acoustics
An acoustic source just below the water surface generates constructive and destructive interference between the direct path and reflected paths. This can have a major impact on sonar operations.
The Lloyd mirror effect has been implicated as having an important role in explaining why marine animals such as manatees and whales have been repeatedly hit by boats and ships. Interference due to Lloyd's mirror results in low frequency propeller sounds not being discernible near the surface, where most accidents occur. This is because at the surface, sound reflections are nearly 180 degrees out of phase with the incident waves. Combined with spreading and acoustic shadowing effects, the result is that the marine animal is unable to hear an approaching vessel before it has been run over or entrapped by the hydrodynamic forces of the vessel's passage.
See also
References
Further reading
External links
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11143 | https://www.youtube.com/watch?v=gk0lPFtQEh0 | Writing Intervals for Piecewise Functions
WOWmath
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Transcript:
so I want to write a function for this piecewise graph hopefully you notice over here I have four functions that make this full function do you see I have four different functions here and we're gonna describe each one and talk about the interval in which they are they're present so the first one we talked about is read function that read function hopefully you can tell that read function let's call it a polynomial normally you'd have to actually put the equation of the polynomial but right now what I'm just going to do is say the word polynomial as the red function and then we talked about the interval now the interval is from this can you tell this red function goes from all the way over here and just keeps going and stops right here at negative one that's called the interval so the interval for which this red graph is is we call it X is values less than negative one so the polynomial function that red piece is for all values X of less than negative one okay let's go to the nikah graph the next graph is this blue one right here hopefully you can see for that blue graph do you see how it stops right there to there it's just in that little window okay can you tell that blue graph is a line so I'm gonna call it linear all right just to describe it in the future you have to actually give the equation for that line but that's a linear graph and are we okay is that increasing does a decreasing is it flat what can you tell us increasing so I'm just gonna describe it as increasing this is a little bit better description just so I know oh that's a linear graph that's increasing could there's other things going on and the interval for which that is is well isn't that between negative one isn't X between negative 1 and 1 all right okay isn't X between negative 1 and 1 in that the area the space it's in now you might wonder why these lesson signs there that's how you write it I'll describe it some having a few or something but for now they're just less than signs it's saying X is between negative 1 and 1 and I'm gonna add something else don't worry too much about it but I'm gonna put up or equal to here the reason is is since that's negative 1 and that's negative 1 1 of these two graph has to claim negative 1 that's not a big idea right now right now just realize negative 1 this one's less than negative 1 this is from negative 1 to 1 and I did claim this there's a reason not a big part of what we're trying to get at right now so don't let it be something that distracts you next one let's do this green one right here that green part of the graph right there well can you tow it from here to here can you tell it's linear again can you tell it's flat so we're looking at the green one is a flat linear function and can you tell it's from negative 1 the X is between negative 1 oh not negative 1 let's take that back it's not negative 1 it's between what it's between positive 1 & 2 is that correct and we put less than signs and I'm gonna put a line underneath this for a reason because since I did not put a line underneath this one I got to put a line underneath this one there's a reason the key idea is do you understand is from 1 to 2 the blue ones from negative 1 to 1 that's the most important X is between those and the last function well can you see this last function is this purple one right here what kind of function is that well can you tell that purple is another linear and is that linear decreasing or increasing well hopefully you can tell it's decreasing so I'm gonna give it a little bit better description and on what interval is that purple line well isn't it from here on forever so how do we describe that well isn't it X values bigger than 2 and since I didn't claim to here don't have to put a line underneath here because I did not claim it for that one it's going to be values greater than or equal to two so this right here is a piecewise function description of this graph in the future you actually have to find the equations of each equation this line this line this line in that polynomial and find each individual equations a little bit harder but for now we're just kind of hey what are they describing and we're describing the interval now I'm going to give you a new in notation for each of these intervals there's another way to describe them that you also will see this is what's most common for piecewise but this next notation I want you to write down and really understand this next notation is just another way of saying the same thing the firt this one right here is saying negative infinity to negative one are you okay that less than negative one is from negative infinity to negative one hopefully that that's okay the next one can you understand it's from negative one to one the thing is chemic this side a bracket are you okay the next one is from negative one to one now I made this one a bracket because I have a noriko to think of it as grabbing it where the one is just the edge of it let's try it again this next notation it was gonna be from bracket one to two do you see why bracketed one again because it has the ordered equal to and then the last one isn't it going to be from bracket two to infinity again the brackets are just because I have or equal to and all you're doing is writing the same kind of interval here but you're putting in parentheses and brackets and if you're on the edge don't you always go to infinity or negative infinity this is actually the more common notation for the problems we're gonna be doing this right here is more common for piecewise but both are called interval notations they describe the window each graph is inside and the colors help you kind of see okay this graph is in this interval and this is where we're leading with this and piecewise now real quick from this piecewise just a quick little extra can you tell where the functions have maxes and mins and stuff like that there's gonna be a whole bunch of questions we're gonna ask you |
11144 | https://www.youtube.com/watch?v=nXugjkSQtTI | Polyprotic Acid Base Equilibria Problems, pH Calculations Given Ka1, Ka2 & Ka3 - Ice Tables
The Organic Chemistry Tutor
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This acid base equilibrium video tutorial explains how to calculate the pH of a polyprotic acid using ice tables and number lines. It discusses how to calculate the pH of a diprotic acid like H2SO4. It explains what to do when given a concentrated sulfuric acid compared to a dilute solution. In addition, it mentions how to calculate the pH of a triprotic acid like H3PO4 and how to calculate the equilibrium concentrations of H3PO4, H2PO4-, HPO4 2-, and PO4 3- given Ka1, Ka2, and Ka3. In addition, it explains how to calculate the pH of a solution that consist of a mixture of H2PO4- and HPO4 2- which is basically a buffer solution and how to calculate the pH when one of those species predominate using a simple formula. This acids and bases chemistry video contains plenty of examples and practice problems.
Acids-Bases - Free Formula Sheet:
Chapter 15 - Video Lessons:
Chemistry 2 Final Exam Review:
Final Exam and Test Prep Videos:
Acids and Bases - Introduction:
The 7 Strong Acids to Memorize:
Conjugate Acid-Base Pairs:
pH and pOH Calculations:
Estimate The pH Without a Calculator:
Autoionization of Water - Kw:
Which Acid Is Stronger?
Acidic, Basic, & Neutral Salts:
pH of Weak Acids:
Buffer Solutions:
Polyprotic Acid Base Equilibria:
Acid Base Titration Curves:
Acids and Bases - Practice Test:
Ksp - Molar Solubility & Ice Tables:
Complex Ion Equilibria:
Gibbs Free Energy, Entropy & Enthalpy:
Entropy - 2nd Law of Thermodynamics:
Electrochemistry Practice Problems:
Final Exams and Video Playlists:
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Transcript:
in this lesson we're going to focus on solving problems associated with polyprotic acids so sulfuric acid is a polyprotic acid notice that it has two hydrogens per formal unit so it's a diprotic acid now how can we calculate the ph of h2so4 if the concentration of it is relatively large now you need to understand that there's two dissociation reactions that you need to deal with the first hydrogen comes off easily the first ka value which represents the acid dissociation constant for the removal of the first proton it's very large and so it ionizes almost 100 percent so we started with a concentration of three moles per liter so then this is going to become 3m and the same is true for the bisulfate ion now the second proton doesn't come off so easily and so this reaction is an equilibrium and the ka value for the second step is this number it's .012 now here's a question for you the amount of h plus that we get in the solution from the second step is it significant it turns out that when the concentration is high it won't be significant so the ph of the solution is simply going to be negative log of the hydronium ion concentration which is 3 moles per liter and so you can get an estimate which is going to be negative 0.477 so that's an estimation of the ph of the solution now if you want to find the exact answer it's going to be slightly different but we need to take into account the amount of h plus that's released from the bisulfa ion so let's make an ice table so we have the initial concentration of the bisulfate ion that's three moles per liter and the same is true for h plus now sulfate is zero so the reaction has no choice but to shift to the right so this is going to increase by x and the reactant is going to decrease by x so that's going to be 3 minus x 3 plus x and x so now we can write the k a 2 expression so it's going to be the ratio of the products which is going to be these two h plus times the sulfate ion concentration divided by the reactant hso4 minus so ka2 that's 0.012 and that's going to be 3 plus x times x over 3 minus x now if x is small 3 plus x is going to be approximately 3. so we could cancel three plus x and three minus x because three minus x will also be about three so therefore we could see that x is equal to this number point zero one two and keep in mind that is simply an approximation but now to check to see if our approximation makes sense we could plug it back in so let's plug it into this expression and let's see if we get a number that's close to 0.012 so it's going to be 3 plus 0.012 that's 3 plus x times x divided by 3 minus x or 3 minus 0.012 so go ahead and type this in so this is equal to 1 2 .0121 which is very close to this number so the approximation is acceptable now that we have the value of x we can now calculate the atrial plus concentration so based on the ice table it was 3 plus x so that's 3 plus 0.012 so the contribution of h plus from the second step is very small however it will change our answer slightly so now we can calculate the ph of the solution so it's going to be negative log of 3.012 and so that's going to be negative 0.479 so these answers are different if you rounded to the thousandths place but if you round it to the hundreds place it's going to be about the same which is approximately negative 0.48 now let's try this example calculate the ph of a 0.02 molar h2so4 solution so this time the concentration of h2so4 is low so let's see how this problem is going to be different from the last one so the first thing we need to realize just like before is the first proton is going to come off easily so we're going to have 100 ionization so this is going to be 0.02 and this is going to be 0.02 for the first step and now let's move on to the second step so we're going to make a ice table for it now the second step is a reversible reaction so we need to write two arrows as opposed to one and now let's make the ice table so the concentration of bisulfate is .02 initially and the same is true for h plus now because this is zero the reaction has to shift to the right to increase the amount of sulfate that's in the solution so this is going to be plus x and minus x on the left side so now ka2 is going to be the ratio of the products divided by the reactants and ka2 is .012 and h plus i forgot to put .02 plus x very important and sulfate is simply x and then by sulfate .02 minus x now this time we can't just say that x is point zero one two because point zero two and point zero one two are very close so x cannot be ignored it's important this time so what we need to do is we need to use the quadratic equation but first we need to rearrange what we have so let me get rid of this and just keep in mind that h plus is 0.02 plus x we'll need that when we need to calculate the ph later now let's cross multiply so 1 times all of this stuff is just going to be 0.02 plus x times x and if we distribute the x it's going to be 0.02 x plus x squared and then .012 times those two if you multiply 0.012 by .02 that's point zero zero zero two four and then minus point zero one two x now we need to put it in standard form for a quadratic equation so we want to write it like this on the left i already have x squared now i need to take this term move it to this side where it's going to become positive on the left side so it's going to be 0.02 x plus 0.012 x on the left side and so that's going to be 0.032 x and i'm going to move this to the other side where it's going to switch from positive to negative so now let's use the quadratic formula so it's x is equal to negative b plus or minus the square root of b squared minus 4ac divided by 2a so b is the number in front of x which is positive point zero three two and then we're going to have b squared minus four a is the number in front of x squared so that's a one and c is that value negative 0.00024 all divided by 2a now if you type in point zero three two squared minus four times negative point zero zero zero two four you're gonna get point zero zero one nine eight four and then take the square root of that number so that's gonna be point zero four four five four two now if we use the negative sign x is going to become negative which won't make sense so we're going to have to use the positive sign so it's going to be negative 0.032 plus 0.044542 and then take that result divided by two so x is point zero zero six two seven one now let's check it we said that ka2 was going to be .02 plus x that's the h plus concentration times the sulfate concentration divided by the bisulfate concentration which was 0.02 minus x so let's plug in x into this expression and let's see if we get a number that's very close to 0.012 so go ahead and type this in 0.02 plus 0.006271 and then times this number and then divided by point zero two minus point zero zero six two seven one it's always good to make sure if your answer is correct so i got point zero one one nine nine nine eight which we could round it and say that's point zero one and because we have a nine here this can go up to a two so it rounds to point zero one two which means that we have the right x value so now we can calculate the h2o plus concentration so it's point zero two plus 0.006271 so the contribution from the second equation from bisulfate is significant to the h2o plus concentration so it's 0.026271 so the ph which is going to be negative log of this number is 1.58 so that's the ph of the solution now let's say if we made an estimate if we simply use that number if we use negative log point zero two then the ph that we would have gotten would have been one point seven is zero which is significantly different from this answer so this is the more accurate answer so that's the ph of the solution it's 1.58 number three calculate the ph of a two molar h3po4 solution and in addition we need to calculate the equilibrium concentrations of phosphoric acid h2po4 minus hydrogen phosphate and phosphate as well let's start with the first one so phosphoric acid is in equilibrium with h plus and h2po4 minus and so we need an ice table the initial amount of h3po4 that we have is two this is zero and so the reaction has to shift to the right so this is going to be 2 minus x and that's going to be x so ka 1 is going to be the products h plus times h2po4 minus divided by the reactant h3 po4 so ka1 that's 0.0075 and that's going to be equal to h plus which is x h2po4 is x and h3po4 it's 2 minus x now the question is is x small or is it significant well let's assume that it's negligible so we're going to get rid of this value first so what we're going to have to do is cross multiply it's going to be .0075 times 2 which is 0.015 and that's equal to x times x or x squared now if we take the square root of both sides x is going to be .1225 so that might be slightly significant let's plug it back in to see if we'll get a significant answer so 0.1225 squared divided by 2 minus 0.1225 that's going to be 0.008 it changes a little but not by that much so what i'm going to do is use the quadratic formula because i think x is significant in this example so let's get rid of that and let's cross multiply so here we're going to have x squared and that's going to equal .0075 times 2 which is .015 and then times negative x so that's going to be negative .0075 x so let's move everything to the left side so it's going to be x squared plus .0075 x and then minus this is not supposed to have x i don't know why that's there so x is going to be negative b so b is the number in front of x plus or minus the square root of b squared minus four times a which is one and c is negative 0.015 which is that number divided by 2a or 2 times 1. so this is going to be negative 0.0075 and then .0075 squared minus four times negative 0.015 that's about 0.06 and if we take the square root of that number it's going to be plus or minus .245064 divided by 2. now i'm going to use the positive sign because i don't want a negative x value so negative point zero zero seven five plus point two four five zero six four divided by two will give us an x value of point one one eight eight the other x value that we had was 0.1225 so a small difference but still significant so now we can calculate the concentration of h3po4 so that was 2 minus x in the ice table so that's 2 minus 0.1188 so at equilibrium the concentration is going to be 1.881 so that's the concentration of phosphoric acid now the concentration of h2 po4 minus based on the ice table it was equal to x so that's 0.1188 now this is also equal to the h3o plus concentration so we could therefore calculate the ph of the solution so it's going to be negative log of 0.1188 so the ph of the solution is 0.925 now let's calculate the concentration of hydrogen phosphate so let's write this reaction h2po4 minus is an equilibrium with h plus and h po4 2 minus so we can use the expression for ka2 it's going to be h plus times h po4 2 minus divided by h2 po4 minus now we have the value of ka2 it's 6.2 times 10 to the minus 8. and the concentration of h plus is this number so that's 0.1188 and our goal is to calculate the concentration of hpo4 2 minus and h2po4 minus which is here it's also equal to x so that's 0.1188 now because these two numbers are the same it turns out that the concentration of hpo42 minus is equal to ka2 so i'm going to write that here somewhere so hpo4 2 minus is equal to 6.2 times 10 to the minus 8 moles per liter so now we only need to calculate the concentration of phosphate at this point so hpo42 minus is an equilibrium with h plus and a phosphate ion so we can write the expression for ka3 so it's h plus times phosphate divided by hydrogen phosphate so ka3 is 4.8 times 10 to the minus 13. and h plus is the value of x which is 0.1188 now our goal is to calculate the phosphate concentration and we now have hp of 42 minus which is this number so it's going to be 4.8 times 10 to the negative 13 multiplied by 6.2 times 10 to negative 8 and then divided by 0.1188 so the phosphate concentration in this example is 2.51 times 10 to the negative 19. and so that's it for this problem so here's the last problem so we have a solution that has these four species phosphoric acid hydrogen phosphate dihydrogen phosphate and phosphate as well so which is the predominant species at a ph of 1.5 now we're given the ka values so what we need to do is calculate the pka values so let's start with pka1 it's going to be negative log of ka1 so negative log of 7.5 times 10 to the minus 3 that's 2.12 and then pka2 that's going to be negative log of 6.2 times 10 to the minus 8. and so that's 7.21 and finally pka3 that's a negative log of 4.8 times 10 to negative 13. and so that works out to be 12.32 now what we're going to do is make a number line and you want to place the pka values in this number line so this is going to be 2.12 7.21 and 12.32 now what you need to realize is that at a ph below the first pka value the predominant species is going to be phosphoric acid and then take off a hydrogen it's going to be h2 po4 minus and then hpo4 2 minus and then phosphate po4 three minus so considering part a at a ph of 1.5 which is here the predominant species is going to be phosphoric acid in part b at a ph of 5.2 it's going to be dihydrogen phosphate and then at in part c at a ph of 9.4 it's mono hydrogen phosphate and then in part d at a ph of 13.8 it's going to be phosphate which is somewhere in that region and so as the ph increases you need to take off a hydrogen atom the number of hydrogens that's going to be on the phosphate ion will decrease so that's how you can determine the predominant species in a solution if you're given the pka values and if you know the ph of the solution now let's move on to part e what is the ph of a 0.5 molar h2po4 solution now h2po4 is right here so we know that the ph has to be between 7.2 actually that's not h2po4 i take that back this is h2po4 minus so the ph has to be somewhere between 2.12 and 7.21 so if we have a solution that's almost 100 h2po4 minus you can estimate the ph with this equation it's going to be one half the sum of these two pka values so basically it's the average of pka1 and pka2 so it's one half pka1 plus pka2 so that's going to be one half of 2.12 plus 7.21 so 2.12 plus 7.21 that's 9.33 and then if you divide that by 2 we can get an average pk value of 4.67 so the ph of the solution if we only have dihydrogen phosphate is going to be 4.67 now let's move on to part f what is the ph of a solution consistent of 0.5 hydrogen phosphate and 0.5 m phosphate so here we have a buffer we have equal amounts of the weak acid and the weak base and whenever you have equal amounts of acid and base in the buffer solution the ph is equal to the pka so here is the acid hpo42 minus and here's the base and here's the pka that's between them so the ph is going to be equal to the pka so it's 12.32 so let's review what we just learned briefly so let's say if we want to calculate the ph of a solution that's almost 100 percent hpo4 two minus so if you just have one of these to estimate the ph is simply the average of the two pka values so we're focused on this one so we just gotta average 7.21 and 12.32 so it's 7.21 plus 12.32 and then after you add them up divide by two so the ph there is going to be about 9.8 so if you have 100 of one of the intermediates just remember the ph is simply the average of the two pka values that surrounds that intermediate now let's say if we have a buffer solution 50 percent is hvpo4 and the other 50 is uh let's say po4 minus so that's between these two if we have a 50 50 mixture the ph is simply equal to the pka so in this case it's going to be 2.12 now let's say if we have a 50 percent mixture of h2po4 minus and 50 of h po4 2 minus so that's these two the ph is going to equal to this pka which is between those two species so it's going to be 7.21 you |
11145 | https://www.khanacademy.org/test-prep/mcat/chemical-processes/x04f6bc56:rates/v/half-life-of-a-second-order-reaction | Half-life of a second-order reaction (video) | Khan Academy
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Half-life of a second-order reaction
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Deriving half-life equation of a second-order reaction starting from the integrated rate law.
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Vish 9 years ago Posted 9 years ago. Direct link to Vish's post “how did we got the graph ...” more how did we got the graph of [A] VS t?
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studious pigeon 6 years ago Posted 6 years ago. Direct link to studious pigeon's post “why if less concentration...” more why if less concentration makes time go slower in the first order reaction time remains constant?
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Davin V Jones 6 years ago Posted 6 years ago. Direct link to Davin V Jones's post “Time doesn't go slower. I...” more Time doesn't go slower. It has the same half-life regardless of the amount.
For an analogy, imagine having 100 coins and you flip them all once. Approximately 50 will come up heads. If you then take the remaining tails and flip them once more, what do you expect the result will be? Do the coins have any way to know how many other coins there are being flipped?
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mkiwan 5 years ago Posted 5 years ago. Direct link to mkiwan's post “I am struggling in unders...” more I am struggling in understanding something. We usually deal with concentrations that are less than 1 (e.g. 0.1 M, 0.001 M). So I think the whole logic of faster reaction for higher order flips to the opposite so it becomes slower (e.g. when 2nd order R=k[A]^2, hence we square the concentration hence R will be much less than the case if it is a 1st order rxn; of course assuming same k). Am I thinking right?
So in summary, the concentrations being less than or greater than 1 M will have opposite effect with increased order of reaction. Correct?
Also if concentration is 1 M then, it becomes as if 0th order coz the term will always reduce to 1 regardless of the rate of reaction (exponent)?!
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Video transcript
We've already talked about the definition for half-life. Remember half-life is symbolized by t 1/2. The half-life is the time that it takes for the concentration of a reactant to decrease to half of its initial concentration. We've also already talked about the integrated rate law or the integrated rate equation for a second order reaction. So here's one form of it. One over the concentration of A minus one over the initial concentration of A is equal to the rate constant k times the time. So if we're talking about the half-life so that would be when time is equal to t 1/2 so we're going to plug t 1/2 in for the time what's the concentration of A? Well using the definition for half-life, the concentration of A should be half of the initial concentration. So our initial concentration would be this. So it's half of our initial concentration. So we're gonna plug that into here. Alright let's plug those in and see what we have. We have one over the initial concentration of A divided by two minus one over the initial concentration of A is equal to the rate constant k times the half-life. So one the left side this would just be two over the initial concentration of A minus one over the initial concentration of A is equal to k t1/2. So what is two over the initial concentration of A minus one over the initial concentration of A? That would just be one over the initial concentration of A and that's equal to the rate constant k times the half-life. So now we can solve for the half-life. Just divide both sides by k. So we get the half-life is equal to one over k times the initial concentration of A. And so here's our equation for the half-life for a second order reaction. Notice this is very different for the half-life for a first order reaction. For a first order reaction we saw that the half-life was constant but here the half-life isn't constant because the half-life depends on the initial concentration of A. Now let's look at a graph of concentration versus time for a second order reaction so we can understand this concept a little bit better. So when time is equal to zero, alright, this point right here on the graph would be the initial concentration of A. So here is our initial concentration of A. And let's just do a made-up reaction here. Let's say we're starting with eight particles that are eight molecules so here we have one, two, three, four, five, six, seven, eight. Alright, so if we wait for the concentration to decrease to half its initial, what are we left with? We were left with four molecules. Alright so we're left with four molecules so I'll draw those in there and how long did it take to go from eight to four? Alright we can find that on our graph. If this point represents the initial concentration, half that would be right here. That would be the initial concentration of A divided by two. So we find that point on our graph and we drop down to here on the x-axis and it took one second. Alright so our first half-life is one second. So let me write that in here. So this represents our first half-life which is one second. How long does it take for the second half-life? So how long does it take to go from four molecules to two molecules? Let me go ahead and change color and be consistent here. So we're going from four molecules to two molecules. How long does it take to do that? Well if this is our starting concentration now, what's half of that? So on our graph half of that would be right here. And this would represent the initial concentration divided by four now. So here's half of this and if we go over, we can find this point on our graph, right? So that point on our graph would be right there and we drop down and we can see we're at time is equal to three. So how long did it take this time for our half-life? This time our half-life would be two seconds. Right? Our half life is two seconds. It's twice the first half-life and we can understand that by plugging in to our half-life equation over here. Alright so if we say the first half-life is one second, so we say the first half-life is one second here, let's go ahead and plug in for the second half-life. So the second, the second half-life would be equal to this would be one over the rate constant k but now our initial concentration, right, our initial concentration is not this, right? It's this. It's the initial concentration divided by two. Alright so go back over here, this would be the initial concentration of A divided by two. So for the second half-life this would be equal to two over k times the initial concentration of A. And we can see that this, that is twice this up here. If this is our first half-life, which is one second, this is twice that, which of course would be two seconds. So this makes sense, both looking at the graph and also thinking about the equation for our half-life. So our second half-life is twice as long as the first and each half-life is going to be twice as long as the one before it. So if we wait one more half-life, so the third half-life, we're going from two molecules to one molecule so now our starting concentration would be right here so what's half of that? That would be right here so we go over and we find this point on our graph. So we find that point on our graph right about there and we drop down and we can see the time is equal to seven seconds. So what's the third half-life? This would be the third half-life right here on our graph and obviously that would be four seconds, alright? So four seconds which is twice the preceding half-life. Alright so let's see if we can figure this out a little bit more. So what does this mean? This means in the early stages of your reaction you have a higher concentration of your reactant, alright? And the higher concentration means that those molecules can collide better and so therefore the reaction goes faster and so if we have an increased concentration we have an increased reaction rate, right? Therefore the faster the reactant is being consumed and therefore the shorter the half-life, the shorter the time it takes for the concentration of the reactant to decrease to half its initial concentration.
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11146 | https://math.stackexchange.com/questions/2596781/question-about-this-proof-of-the-singleton-bound | coding theory - Question about this proof of the Singleton Bound? - Mathematics Stack Exchange
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Question about this proof of the Singleton Bound?
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I have the following proof of the Singleton Bound (screenshot taken directly from a lecture slide):
This is fairly straightforward to the point where it claims that d(x,y)≤d−1 d(x,y)≤d−1.
However, the proof then goes on to say that since d(x,y)≤d−1 d(x,y)≤d−1, the mapping π:C↦S n−(d−1)π:C↦S n−(d−1) is one-to-one (or injective). Why is this?
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edited Jan 8, 2018 at 13:54
Andrés E. Caicedo
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asked Jan 8, 2018 at 12:47
M SmithM Smith
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x≠y∈C x≠y∈C implies d(x,y)≥d d(x,y)≥d by the definition of an (n,M,d)(n,M,d) code. So d(x,y)≤d−1 d(x,y)≤d−1 is a contradiction.
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answered Jan 8, 2018 at 15:52
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11147 | https://www.jacionline.org/article/S0091-6749(20)31404-4/fulltext | 2020 Focused Updates to the Asthma Management Guidelines: A Report from the National Asthma Education and Prevention Program Coordinating Committee Expert Panel Working Group - Journal of Allergy and Clinical Immunology
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Reviews and feature articleVolume 146, Issue 6p1217-1270 December 2020 Open access
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2020 Focused Updates to the Asthma Management Guidelines: A Report from the National Asthma Education and Prevention Program Coordinating Committee Expert Panel Working Group
Expert Panel Working Group of the National Heart, Lung, and Blood Institute (NHLBI) administered and coordinated National Asthma Education and Prevention Program Coordinating Committee (NAEPPCC)
Expert Panel Working Group of the National Heart, Lung, and Blood Institute (NHLBI) administered and coordinated National Asthma Education and Prevention Program Coordinating Committee (NAEPPCC)
Footnotes
∗
Members of the Expert Panel Working Group: Michelle M. Cloutier, MD (Chair of the Expert Panel Working Group; UCONN Health, Farmington, Conn), Alan P. Baptist, MD, MPH (University of Michigan, Ann Arbor, Mich), Kathryn V. Blake, PharmD (Nemours Children's Health System, Jacksonville, Fla), Edward G. Brooks, MD (University of Texas Health Science Center at San Antonio, San Antonio, Texas), Tyra Bryant-Stephens, MD (Children's Hospital of Philadelphia, Philadelphia, Pa), Emily DiMango, MD (Columbia University Irving Medical Center, New York, NY), Anne E. Dixon, MA, BM, BCh (University of Vermont, Burlington, Vt), Kurtis S. Elward, MD, MPH, FAAFP (Virginia Commonwealth University, Charlottesville, Va), Tina Hartert, MD, MPH (Vanderbilt University School of Medicine, Nashville, Tenn), Jerry A. Krishnan, MD, PhD (University of Illinois Hospital and Health Sciences System, Chicago, Ill), Robert F. Lemanske Jr, MD (University of Wisconsin - Madison, Madison, Wis), Daniel R. Ouellette, MD, FCCP (Henry Ford Health System, Detroit, Mich), Wilson D. Pace, MD, FAAFP (University of Colorado, Aurora, Colo), Michael Schatz, MD, MS (Kaiser Permanente, San Diego, Calif), Neil S. Skolnik, MD (Dr Skolnik withdrew from the Expert Panel on October 1, 2019; Abington - Jefferson Health, Jenkintown, Pa), James W. Stout, MD, MPH (University of Washington, Seattle, Wash), Stephen J. Teach, MD, MPH (George Washington University, Washington, DC), Craig A. Umscheid, MD, MSCE (University of Chicago, Chicago, Ill), and Colin G. Walsh, MD, MA (Dr Walsh withdrew from the Expert Panel on October 29, 2019; Vanderbilt University Medical Center, Nashville, Tenn).
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∗ ∙ Michelle M.Cloutier, MD
Michelle M.Cloutier, MD
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Alan P.Baptist, MD, MPH
Alan P.Baptist, MD, MPH
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ … ∙ Kathryn V.Blake, PharmD
Kathryn V.Blake, PharmD
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Edward G.Brooks, MD
Edward G.Brooks, MD
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Tyra Bryant-Stephens, MD
Tyra Bryant-Stephens, MD
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Emily DiMango, MD
Emily DiMango, MD
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Anne E.Dixon, MA, BM, BCh
Anne E.Dixon, MA, BM, BCh
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Kurtis S.Elward, MD, MPH, FAAFP
Kurtis S.Elward, MD, MPH, FAAFP
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Tina Hartert, MD, MPH
Tina Hartert, MD, MPH
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Jerry A.Krishnan, MD, PhD
Jerry A.Krishnan, MD, PhD
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Robert F.Lemanske, Jr., MD
Robert F.Lemanske, Jr., MD
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Daniel R.Ouellette, MD, FCCP
Daniel R.Ouellette, MD, FCCP
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Wilson D.Pace, MD, FAAFP
Wilson D.Pace, MD, FAAFP
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Michael Schatz, MD, MS
Michael Schatz, MD, MS
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Neil S.Skolnik, MD
Neil S.Skolnik, MD
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ James W.Stout, MD, MPH
James W.Stout, MD, MPH
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Stephen J.Teach, MD, MPH
Stephen J.Teach, MD, MPH
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Craig A.Umscheid, MD, MSCE
Craig A.Umscheid, MD, MSCE
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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∙ Colin G.Walsh, MD, MA
Colin G.Walsh, MD, MA
Affiliations
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
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… Show more Show less
Affiliations & Notes Article Info Linked Articles (5)
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda
∗
Members of the Expert Panel Working Group: Michelle M. Cloutier, MD (Chair of the Expert Panel Working Group; UCONN Health, Farmington, Conn), Alan P. Baptist, MD, MPH (University of Michigan, Ann Arbor, Mich), Kathryn V. Blake, PharmD (Nemours Children's Health System, Jacksonville, Fla), Edward G. Brooks, MD (University of Texas Health Science Center at San Antonio, San Antonio, Texas), Tyra Bryant-Stephens, MD (Children's Hospital of Philadelphia, Philadelphia, Pa), Emily DiMango, MD (Columbia University Irving Medical Center, New York, NY), Anne E. Dixon, MA, BM, BCh (University of Vermont, Burlington, Vt), Kurtis S. Elward, MD, MPH, FAAFP (Virginia Commonwealth University, Charlottesville, Va), Tina Hartert, MD, MPH (Vanderbilt University School of Medicine, Nashville, Tenn), Jerry A. Krishnan, MD, PhD (University of Illinois Hospital and Health Sciences System, Chicago, Ill), Robert F. Lemanske Jr, MD (University of Wisconsin - Madison, Madison, Wis), Daniel R. Ouellette, MD, FCCP (Henry Ford Health System, Detroit, Mich), Wilson D. Pace, MD, FAAFP (University of Colorado, Aurora, Colo), Michael Schatz, MD, MS (Kaiser Permanente, San Diego, Calif), Neil S. Skolnik, MD (Dr Skolnik withdrew from the Expert Panel on October 1, 2019; Abington - Jefferson Health, Jenkintown, Pa), James W. Stout, MD, MPH (University of Washington, Seattle, Wash), Stephen J. Teach, MD, MPH (George Washington University, Washington, DC), Craig A. Umscheid, MD, MSCE (University of Chicago, Chicago, Ill), and Colin G. Walsh, MD, MA (Dr Walsh withdrew from the Expert Panel on October 29, 2019; Vanderbilt University Medical Center, Nashville, Tenn).
Publication History:
Received October 5, 2020; Accepted October 6, 2020
Footnotes:
The 2020 Focused Updates to the Asthma Management Guidelines: A Report from the National Asthma Education and Prevention Program Coordinating Committee Expert Panel Working Group was funded by the National Heart, Lung, and Blood Institute of the National Institutes of Health. No additional funding was received. All authors volunteered their time.
Reported disclosure of potential conflict of interest: Development of this report was funded by the National Heart, Lung, and Blood Institute of the National Institutes of Health. Members of the Expert Panel Working Group (“Expert Panel”) of the National Asthma Education and Prevention Program Coordinating Committee (NAEPPCC) completed financial disclosure forms and disclosed relevant financial interests described as conflicts of interest to each other before their discussions. Members of the Expert Panel were volunteers and received compensation only for travel expenses related to the panel’s in-person meetings. M. M. Cloutier reported that a family member was employed by Regeneron. The Expert Panel did not discuss any Regeneron or related products. M. M. Cloutier was not recused from any Expert Panel discussions. A. Baptist received grant funding from Novartis, AstraZeneca, Teva Pharmaceuticals Industries Ltd (Teva), and Takeda and was recused from drafting, discussing, or voting on sections of the report related to the inhaled corticosteroid recommendations. K. Blake received personal fees from Teva for participation on its Digital Technology Advisory Board and study drugs from Boehringer Ingelheim and GlaxoSmithKline for federally funded clinical trials and was not recused from any Expert Panel discussions. E. G. Brooks served on the Scientific Advisory Board for United Allergy Services until December 2019 and was recused from the initial drafting, discussions, and voting related to the immunotherapy recommendations. However, he participated in the final voting, discussions, and writing of this section of the report. T. Bryant-Stephens received mattress covers and air purifiers from the Asthma and Allergy Foundation of America and was not recused from any Expert Panel discussions. E. DiMango participated in an advisory board meeting for AstraZeneca, reported that a family member was formerly employed by Regeneron, and was not recused from any Expert Panel discussions. A. Dixon received a study medication from MitoQ, also received assistance from a medical writer from Boehringer Ingelheim for a manuscript published in 2019 that was not related to the update topics, and was not recused from any Expert Panel discussions. K. S. Elward had no conflicts of interest and was not recused from any Expert Panel discussions. T. Hartert participated in a Pfizer advisory board for a maternal respiratory syncytial virus vaccine and was not recused from any Expert Panel discussions. J. A. Krishnan participated in a data and safety monitoring committee for Sanofi, received funding from Inogen and ResMed for a study of portable oxygen concentrators, received personal fees from the Critical Path Institute, and was not recused from any Expert Panel discussions. R. F. Lemanske received personal fees from Siolta Therapeutics for a microbiome intervention and from the Food Allergy Research & Education Network and was not recused from any Expert Panel discussions. D. R. Ouellette had no conflicts of interest and was not recused from any Expert Panel discussions. W. Pace participated in a meeting sponsored by Mylan to discuss long-acting muscarinic antagonist (LAMA) therapy, received a study drug from Teva for a Patient-Centered Outcomes Research Institute–funded pragmatic clinical study, received a grant from Boehringer Ingelheim through Optimal Patient Care concerning Chronic Obstructive Pulmonary Disease, and was recused from participation in the final discussion, voting, and section writing related to LAMA therapy. M. Schatz received grant funding from Merck, Teva, and ALK-Abello, Inc, and was recused from the writing, discussion, and voting related to the immunotherapy recommendations. N. S. Skolnik received personal fees and nonfinancial support from AstraZeneca; personal fees from Teva, Eli Lilly and Company, Boehringer Ingelheim, Merck, Sanofi, Janssen Pharmaceuticals, Intarcia Therapeutics, GlaxoSmithKline, and Mylan; was recused from the writing, discussion, and voting related to the inhaled corticosteroids and LAMA recommendations; and withdrew from the Expert Panel on October 1, 2019. J. W. Stout received a training grant from the Patient-Centered Outcomes Research Institute, which distributed pillow and mattress dust covers; nonfinancial support for a new quality review design for the EasyOne Air spirometer (the EasyOne Air build was done with a combination of license revenue from the Spirometry 360 course, and his personal research and training account through his division); nonfinancial support for a smartphone spirometer application from Google, and development of an asynchronous and an abridged asynchronous online spirometer training course (asynchronous training was built with a combination of funding from AHRQ, the National Asthma Control Initiative [NACI], Spirometry 360 license revenue, and unrestricted funding from Astra-Zeneca); and was not recused from any Expert Panel discussions. S. Teach had no conflicts of interest and was not recused from any Expert Panel discussions. C. A. Umscheid received funding from the Agency for Healthcare Research and Quality for systematic reviews on allergen reduction and bronchial thermoplasty and was not recused from any Expert Panel discussions. C. G. Walsh had no conflicts of interest, was not recused from any Expert Panel discussions, and withdrew from the Expert Panel on October 29, 2019.
DOI: 10.1016/j.jaci.2020.10.003 External LinkAlso available on ScienceDirect External Link
Copyright: © 2020 American Academy of Allergy, Asthma & Immunology
User License: Creative Commons Attribution – NonCommercial – NoDerivs (CC BY-NC-ND 4.0) | Elsevier's open access license policy
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Outline
Outline
Abstract
Key words
Abbreviations used
Table of Contents
Preface
Foreword
Section I: Introduction
Section II: Recommendations on the use of F eno testing in the diagnosis and management of asthma
Section III: Recommendations for indoor allergen mitigation in management of asthma
Section IV: Recommendations for the use of intermittent ICS in the treatment of asthma
Section V: Recommendations for the use of LAMAs for asthma
Section VI: The role of subcutaneous and sublingual immunotherapy in the treatment of allergic asthma
Section VII: Recommendations for the use of BT to improve asthma outcomes
Acknowledgments
Supplementary data (1)
References
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Abstract
Key words
Abbreviations used
Table of Contents
Preface
Foreword
Section I: Introduction
Section II: Recommendations on the use of F eno testing in the diagnosis and management of asthma
Section III: Recommendations for indoor allergen mitigation in management of asthma
Section IV: Recommendations for the use of intermittent ICS in the treatment of asthma
Section V: Recommendations for the use of LAMAs for asthma
Section VI: The role of subcutaneous and sublingual immunotherapy in the treatment of allergic asthma
Section VII: Recommendations for the use of BT to improve asthma outcomes
Acknowledgments
Supplementary data (1)
References
Article metrics
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Correction:
Errata
Corrigenda
April 5, 2021
Abstract
The 2020 Focused Updates to the Asthma Management Guidelines: A Report from the National Asthma Education and Prevention Program Coordinating Committee Expert Panel Working Group was coordinated and supported by the National Heart, Lung, and Blood Institute (NHLBI) of the National Institutes of Health. It is designed to improve patient care and support informed decision making about asthma management in the clinical setting. This update addresses six priority topic areas as determined by the state of the science at the time of a needs assessment, and input from multiple stakeholders:
•
Fractional Exhaled Nitric Oxide Testing
•
Indoor Allergen Mitigation
•
Intermittent Inhaled Corticosteroids
•
Long-Acting Muscarinic Antagonists
•
Immunotherapy in the Treatment of Allergic Asthma
•
Bronchial Thermoplasty
A rigorous process was undertaken to develop these evidence-based guidelines. The Agency for Healthcare Research and Quality's (AHRQ) Evidence-Based Practice Centers conducted systematic reviews on these topics, which were used by the Expert Panel Working Group as a basis for developing recommendations and guidance. The Expert Panel used GRADE (Grading of Recommendations, Assessment, Development and Evaluation), an internationally accepted framework, in consultation with an experienced methodology team for determining the certainty of evidence and the direction and strength of recommendations based on the evidence. Practical implementation guidance for each recommendation incorporates findings from NHLBI-led patient, caregiver, and clinician focus groups. To assist clincians in implementing these recommendations into patient care, the new recommendations have been integrated into the existing Expert Panel Report-3 (EPR-3) asthma management step diagram format.
Key words
NHLBI
Asthma Guideline
asthma
fractional exhaled nitric oxide
allergen mitigation
inhaled corticosteroids
long-acting muscarinic antagonist
bronchial thermoplasty
immunotherapy
Abbreviations used
ACP (American College of Physicians)
ACQ (Asthma Control Questionnaire)
ACQ-5 (5-question Asthma Control Questionnaire)
ACQ-6 (6-question Asthma Control Questionnaire)
ACQ-7 (7-question Asthma Control Questionnaire)
ACT (Asthma Control Test)
AHRQ (Agency for Healthcare Research and Quality)
AIR (Asthma Intervention Research)
BELT (Blacks and Exacerbations on LABA vs Tiotropium)
BT (Bronchial thermoplasty)
COI (Conflict of interest)
EPC (Evidence-Based Practice Center)
EPR (Expert Panel Report)
EtD (Evidence to decision)
FDA (US Food and Drug Administration)
F eno (Fractional exhaled nitric oxide)
GRADE (Grading of Recommendations, Assessment, Development, and Evaluation)
HEPA (High-efficiency particulate air (a type of filter))
ICS (Inhaled corticosteroid)
ICS-LABA (Inhaled corticosteroid and long-acting beta 2-agonist combination, typically in a single device)
JACI (Journal of Allergy and Clinical Immunology)
LABA (Long-acting beta 2-agonist)
LAMA (Long-acting muscarinic antagonist)
MID (Minimally important difference)
NAEPP (National Asthma Education and Prevention Program)
NAEPPCC (National Asthma Education and Prevention Program Coordinating Committee)
NHLBAC (National Heart, Lung, and Blood Advisory Council)
NHLBI (National Heart, Lung, and Blood Institute)
OR (Odds ratio)
ppb (Parts per billion)
RCT (Randomized controlled trial)
RISA (Research In Severe Asthma)
RR (Relative risk)
SABA (Short-acting beta 2-agonist)
SCIT (Subcutaneous immunotherapy)
SLIT (Sublingual immunotherapy)
SMART (Single maintenance and reliever therapy)
T2 (Type 2)
Table of Contents
Reported disclosure of potential conflict of interest 1217
ABBREVIATIONS USED1219
PREFACE1219
FOREWORD1220
SECTION I: INTRODUCTION1221
Background and rationale for focused updates1221
Methods1222
Systematic reviews of the literature1222
Updated reviews of the literature1222
Expert panel processes1222
GRADE methodology1224
Focus groups with individuals with asthma and their caregivers1226
2020 focused updates to the 2007 Asthma Guidelines1227
Review and public comment1227
Limitations and research gaps1228
Recommendations1228
Integration of the new recommendations into asthma care1228
Stepwise approach for managing asthma1228
SECTION II: RECOMMENDATIONS ON THE USE OF FenoTESTING IN THE DIAGNOSIS AND MANAGEMENT OF ASTHMA1237
Background1237
Definitions of terms used in this section1237
Question 2.11237
•
Recommendation 1 1237
•
Implementation guidance 1237
•
Summary of the evidence 1238
•
Rationale and discussion 1238
Questions 2.2 and 2.31238
•
Recommendation 2 1238
•
Implementation guidance 1239
•
Summary of the evidence 1239
•
Rationale and discussion 1240
Question 2.41240
•
Recommendation 3 1240
•
Implementation guidance 1240
•
Summary of the evidence 1240
•
Rationale and discussion 1241
Question 2.51241
•
Recommendation 4 1241
•
Implementation guidance 1241
•
Summary of the evidence 1241
•
Rationale and discussion 1241
Future research opportunities1242
SECTION III: RECOMMENDATIONS FOR INDOOR ALLERGEN MITIGATION IN MANAGEMENT OF ASTHMA1242
Background1242
Definitions of terms used in this section1242
Question 3.11243
•
Recommendation 5 1243
•
Recommendation 6 1243
•
Recommendation 7 1243
•
Recommendation 8 1243
•
Implementation guidance 1243
•
Summary of the evidence 1245
•
Single-component allergen mitigation interventions 1245
•
Multicomponent allergen mitigation interventions 1245
•
Rationale and discussion 1246
Future research opportunities1247
SECTION IV: RECOMMENDATIONS FOR THE USE OF INTERMITTENT ICS IN THE TREATMENT OF ASTHMA1247
Background1247
Definitions of terms used in this section1247
Overview of key questions and recommendations for intermittent ICS use1247
Question 4.11248
•
Recommendation 9 1248
•
Implementation guidance 1248
•
Summary of the evidence 1249
•
Rationale and discussion 1249
Question 4.21249
•
Recommendation 10 1249
•
Implementation guidance 1249
•
Summary of the evidence 1249
•
Rationale and discussion 1250
•
Recommendation 11 1250
•
Implementation guidance 1250
•
Summary of the evidence 1250
•
Rationale and discussion 1250
Question 4.31251
•
Recommendation 12 1251
•
Implementation guidance 1251
•
Summary of the evidence 1252
•
Rationale and discussion 1253
•
Recommendation 13 1253
•
Implementation guidance 1253
•
Summary of the evidence 1253
•
Rationale and discussion 1254
Future research opportunities1254
SECTION V: RECOMMENDATIONS FOR THE USE OF LAMAs FOR ASTHMA1254
Background1254
Definitions of terms used in this section1254
Question 5.11255
Question 5.21255
•
Recommendation 14 1255
•
Recommendation 15 1255
•
Implementation guidance 1255
•
Summary of the evidence 1256
•
Rationale and discussion 1256
Question 5.31257
•
Recommendation 16 1257
•
Implementation guidance 1257
•
Summary of the evidence 1257
•
Rationale and discussion 1258
Future research opportunities1258
SECTION VI: THE ROLE OF SUBCUTANEOUS AND SUBLINGUAL IMMUNOTHERAPY IN THE TREATMENT OF ALLERGIC ASTHMA1258
Background1258
Definition of terms used in this section1258
Question 6.11259
•
Recommendation 17 1259
•
Implementation guidance 1259
•
Summary of the evidence 1260
•
Rationale and discussion 1260
Question 6.21260
•
Recommendation 18 1260
•
Implementation guidance 1260
•
Summary of the evidence 1261
•
Rationale and discussion 1261
Overall summary for SCIT and SLIT1261
Future research opportunities1261
SECTION VII: RECOMMENDATIONS FOR THE USE OF BT TO IMPROVE ASTHMA OUTCOMES1262
Background1262
Definitions of terms used in this section1262
Question 7.11262
•
Recommendation 19 1262
•
Implementation guidance 1262
•
Summary of the evidence 1263
•
Rationale and discussion 1263
Future research opportunities1263
REFERENCES1264
APPENDICES
Online SupplementA. Key differences from the Guidelines for the Diagnosis and Management of Asthma (EPR-3)
Online SupplementB. EtD tables
Preface
This report was developed by an Expert Panel Working Group (hereafter referred to as the “Expert Panel”) of the National Asthma Education and Prevention Program (NAEPP) Coordinating Committee (NAEPPCC), presented to the NAEPPCC for the full committee’s consideration, and adopted by the NAEPPCC during a public meeting. The NAEPPCC is coordinated by the National Heart, Lung, and Blood Institute (NHLBI) of the National Institutes of Health.
The NHLBI is pleased to present this update, in which several changes to the approaches used in prior NAEPPCC expert panel reports (EPRs) have been implemented. Specifically:
•
The decision to update Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma (EPR-3) and the selection of topics to update were initiated by engaging the public with a request for information, rather than relying solely on the NAEPP for these decisions.
•
To use the most rigorous methods for gathering information for the focused update, the Agency for Healthcare Research and Quality (AHRQ) conducted systematic reviews.
•
A consultant with expertise in GRADE (Grading of Recommendations, Assessment, Development, and Evaluation) methodology guided the Expert Panel members in their deliberations and development of the recommendations based on the systematic review reports.
In this report, which was adopted by the NAEPPCC, the Expert Panel has included practical implementation guidance for each recommendation that incorporates findings from NHLBI-led focus groups. These focus groups included people with asthma, caregivers, and providers. To assist providers in integrating these recommendations into the care of patients, the new recommendations have been integrated into the EPR-3 step diagram format. Overall, a highly rigorous process was undertaken to facilitate the development of the evidence-based recommendations and supporting information in this report for use by stakeholders to improve asthma management.
This report was developed under the leadership of Dr Michelle Cloutier, Expert Panel chair. The NHLBI is grateful for the tremendous dedication of time and outstanding work of all members of the Expert Panel in developing this report. Appreciation is also extended to the NAEPPCC as well as other stakeholder groups (professional societies, health care organizations, government agencies, consumer and patient advocacy organizations, and companies) for their invaluable comments during the public review period. These comments helped enhance the scientific credibility and practical utility of this document.
Ultimately, broad change in clinical practice depends on the uptake, adoption, and implementation of clinical practice recommendations by primary care providers with input from people who have asthma and their families, as well as support from health care systems. This update can serve as a basis to disseminate and facilitate adoption of the asthma recommendations at all levels and to ensure optimal care and equitable outcomes for all individuals with asthma. We ask for the assistance of every stakeholder in reaching our goal: improving asthma care and the quality of life of every person with asthma.
James P. Kiley, MS, PhD
Director
Division of Lung Diseases
NHLBI George A. Mensah, MD
Director
Center for Translation Research and Implementation Science
NHLBI
Open table in a new tab
Foreword
It has been 13 years since the last revision of the asthma recommendations, and substantial progress has been made since that time in understanding the origins of asthma as well as its pathophysiology and treatment. As members of the pulmonary and allergy provider community and the primary care community that provide more than half of all asthma care in the United States, we now recognize that asthma is not one disease, but it is a syndrome composed of multiple phenotypes. Asthma is much more complex than indicated in the Expert Panel Report: Guidelines for the Diagnosis and Management of Asthma (EPR-1),1
1.
National Asthma Education and Prevention Program
Expert Panel Report: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, 1991
Google Scholar
released in 1991, which characterized asthma as an inflammatory disease that is responsive to corticosteroids.
This document updates selected topics that were identified as high priority by an NHLBI Advisory Council Asthma Expert Working Group based on input from previous guideline developers, NAEPP participant organizations, and the public. The list of these priority topics was published in 2015.2
2.
National Heart, Lung, and Blood Advisory Council Asthma Expert Working Group
Needs Assessment Report for Potential Update of the Expert Panel Report-3 (2007): Guidelines for the Diagnosis and Management of Asthma
National Institutes of Health, National Heart, Lung, and Blood Institute, Bethesda, Md, February 2015
Available at:
Google Scholar
Seventeen topics were suggested initially for updating, and six topics were found to have sufficient new information to warrant an update. Key questions were drafted by the Advisory Council and used by AHRQ Evidence-Based Practice Centers (EPCs) to conduct systematic reviews that were published between October 2017 and March 2018.3-7
3.
D’Anci KE, Lynch MP, Leas BF, Apter AJ, Bryant-Stephens T, Kaczmarek JL, et al. Effectiveness and safety of bronchial thermoplasty in management of asthma. Comparative Effective Review No. 202. (Prepared by the ECRI Institute–Penn Medicine Evidence-based Practice Center under Contract No. 290-2015-00005-I). AHRQ Publication No. 18-EHC0003-EF. Rockville, Md: Agency for Healthcare Research and Quality; December 2017. Available at:
Google Scholar
4.
Leas BF, D’Anci KE, Apter AJ, Bryant-Stephens T, Schoelles K, Umscheid C. Effectiveness of indoor allergen reduction in management of asthma. Comparative Effectiveness Review No. 201. (Prepared by the ECRI Institute–Penn Medicine Evidence-based Practice Center under Contract No. 290-2015-0005-I). AHRQ Publication No. 18-EHC002-EF. Rockville, Md: Agency for Healthcare Research and Quality; February 2018. Posted final reports are located on the Effective Health Care Program search page.
Google Scholar
5.
Lin SY, Azar A, Suarez-Cuervo C, Diette GB, Brigham E, Rice J, et al. The role of immunotherapy in the treatment of asthma. Comparative Effectiveness Review No. 196. (Prepared by the Johns Hopkins University Evidence-based Practice Center under Contract No.290-2015-00006-I). AHRQ Publication No. 17(18)-EHC029-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Posted final reports are located on the Effective Health Care Program search page.
Google Scholar
6.
Sobieraj DM, Baker WL, Weeda ER, Nguyen E, Coleman CI, White CM, et al. Intermittent inhaled corticosteroids and long-acting muscarinic antagonists for asthma. Comparative Effectiveness Review No. 194. (Prepared by the University of Connecticut Evidence-based Practice Center under Contract No. 290-2015-00012-I). AHRQ Publication No. 17(18)-EHC027-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Available at: Posted final reports are located on the Effective Health Care Program search page.
Google Scholar
7.
Wang Z, Pianosi P, Keogh K, Zaiem F, Alsawas M, Alahdab F, et al. The clinical utility of fractional exhaled nitric oxide (FeNO) in asthma management. Comparative Effectiveness Review No. 197. (Prepared by the Mayo Clinic Evidence-based Practice Center under Contract No. 290-2015-00013-I). AHRQ Publication No. 17(18)-EHC030-EF. Rockville, Md: Agency for Healthcare Research and Quality; December 2017. Available at:
Google Scholar
The Expert Panel was then assembled in July 2018 and charged with using these systematic reviews to develop recommendations on these six previously chosen topics.
The Expert Panel updated the literature for the systematic reviews through October 2018 and then developed its recommendations. These recommendations differ from other guidelines in several important ways:
•
The key questions were developed a priori and not after a review of the current literature.
•
The Expert Panel was composed of diverse individuals not only from the asthma specialty community (adult and pediatric pulmonary and allergy specialists) but also from the general medical community (pediatric, internal medicine, family medicine, and emergency medicine providers). Expert Panel members also included health policy and dissemination and implementation experts, and the panel received input from patients and families.
•
The Expert Panel members abided by strict standards for conflicts of interest (COIs) developed by the Institute of Medicine (now the National Academy of Medicine)8
8.
Institute of Medicine Committee on Standards for Developing Trustworthy Clinical Practice Guidelines
Graham, R. ∙ Mancher, M. ∙ Miller Wolman, D. ... (Editors)
Clinical practice guidelines we can trust
National Academies Press, Washington, DC, 2011
Available at:
Google Scholar
and in the spirit of the more recently released recommendations from the American College of Physicians (ACP).9
9.
Qaseem, A. ∙ Wilt, T.J.
Disclosure of interests and management of conflicts of interest in clinical guidelines and guidance statements: methods from the Clinical Guidelines Committee of the American College of Physicians
Ann Intern Med. 2019; 171:354-361
Crossref
Scopus (3)
PubMed
Google Scholar
Individuals with any conflict of interest related to the updated topics recused themselves from discussions of those topics.
•
This was the first time that the NAEPP used the GRADE methodology (discussed later) to provide transparency in the decision-making process.
•
Lastly, but not insignificantly, the Expert Panel sought comments from external groups and individuals, including from the NAEPP Coordinating Committee (whose members represent a diverse group of stakeholders), the public, and federal agencies. Although the panel that developed the Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma (EPR-3) also sought external input, this approach has rarely been used by other asthma guideline committees. The Expert Panel considered this input when it developed the final recommendations and this document.
The methodology framework used for this update, GRADE, is the internationally recommended approach for developing recommendations that clinicians can trust. This framework endorses a systematic and transparent approach to decision making, uses established criteria to rate the certainty of evidence, and determines the strength of the recommendations. Recommendations developed using GRADE combine certainty of evidence with patient values and preferences and weigh the benefits and harms of making treatment recommendations. Importantly, the recommendations are based on the key questions that clinicians, both generalists and specialists, wanted to be answered.
Users of these recommendations may be disappointed by the absence of many strong recommendations—that is, recommendations that clinicians should adhere to for almost all individuals with asthma as the standard of care. This is not, however, surprising given the variations in asthma phenotypes and endotypes and in the outcomes used in the studies reviewed to develop the recommendations. When the GRADE framework is used, randomized controlled trials (RCTs) are initially rated as offering a high certainty of evidence, but issues with study designs (eg, lack of blinding or of a placebo control), heterogeneity of study results, or small numbers of events may result in downgrading the certainty of evidence. For most of the asthma recommendations, the overall certainty of the evidence was downgraded because of inconsistencies in study results, risk of bias, or absence of critical standardized outcome measures. The need to downgrade the evidence should be a clarion call to investigators to use standardized and validated outcome measures that were outlined in the Asthma Outcomes Workshop (2012).10
10.
Busse, W.W. ∙ Morgan, W.J. ∙ Taggart, V. ...
Asthma outcomes workshop: overview
J Allergy Clin Immunol. 2012; 129:S1-S8
Full Text
Full Text (PDF)
PubMed
Google Scholar
This single activity will create more robust evidence to support recommendations in the future.
The working group that identified the six priority topics for this update based its recommendations on information available at that time. This information did not include the subsequent explosion of research and US Food and Drug Administration (FDA) approval of multiple drugs classified as asthma biologics. Any attempt to include biologic agents in this report at the start of this effort would have delayed the release of these recommendations for another 1 to 2 years, and this was felt to be unacceptable. This update also is not a complete revision of EPR-3. Important aspects of care, such as asthma education (including inhaler technique) and assessment tools for asthma control, adherence, and other factors, are not covered. Reasons for these limitations included lack of time, lack of resources, and, for some topics, insufficient new evidence.
Finally, several new features in this update were designed to aid providers and clinicians in addressing these topics with their patients. The biggest of these changes is the addition of an Implementation Guidance section for each recommendation. Each Implementation Guidance section begins with a clinician summary—an expanded statement of the recommendation to quickly assist clinicians in better understanding the recommendation from a user’s perspective. The Implementation Guidance section also provides further clarification of the population to which the recommendation applies, exceptions, and practical aspects of how to use the recommendation in patient care. At the end of each Implementation Guidance section is a list of issues suggested by the Expert Panel to communicate to patients as part of shared decision making about whether to use the therapy or intervention mentioned in the recommendation. Amended step diagrams for asthma management are also provided for the topics being updated. Many of the updated interventions in these diagrams are now preferred first-line treatments.
Moving forward, the process of guideline development needs to be more agile. Creating an ongoing process for developing recommendations that includes individuals with varied expertise and from multiple organizations may facilitate this process. In addition, the structure of the recommendations may need to change. The step diagrams, although useful, are a one-size-fits-all approach. The current recommendations use a patient-centered approach that is critical but not sufficient. In the emerging era of personalized medicine, tailored interventions and treatments customized to particular individuals with specific characteristics will be needed. Discussions about how to address individualized approaches to asthma care and how to incorporate these approaches into the standard of care are needed now so that future recommendations can integrate these new approaches.
Finally, I thank the members of the Expert Panel who voluntarily gave their time and expertise to complete this work. The amount of work that was needed in a compressed period of time from each member was very high. To them, to Drs Kiley and Mensah, whose support was unwavering, and to the NHLBI and Westat staff, thank you.
Michelle M. Cloutier, MD
Chair, Expert Panel
Section I: Introduction
Background and rationale for focused updates
In 1989, the NHLBI created a program, now known as the NAEPP, to address asthma issues in the United States. The NAEPP focuses on raising awareness and ensuring appropriate diagnosis and management of asthma to reduce asthma-related morbidity and mortality and to improve the quality of life of individuals with asthma. To that end, the NAEPP published its first EPR on the diagnosis and management of asthma in 1991.1
1.
National Asthma Education and Prevention Program
Expert Panel Report: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, 1991
Google Scholar
A comprehensive revision, EPR-2, was published in 1997,11
11.
National Asthma Education and Prevention Program
Expert Panel Report 2: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, 1997
Google Scholar
followed by an update of selected topics in 2002 and then a third EPR, EPR-3, in 2007.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
Available at:
Google Scholar
In 2014, the Asthma Expert Working Group of the National Heart, Lung, and Blood Advisory Council (NHLBAC) completed an assessment of the need to revise NAEPP’s Expert Panel Report-3: Guidelines for the Diagnosis and Management of Asthma (EPR-3)12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
Available at:
Google Scholar
and the content of such a revision. After a discussion and review of the responses to a public request for information on the need for and potential content of an update, the NHLBAC Asthma Expert Working Group (which included members of the EPR-3 expert panel) determined that a focused update on six priority topics was warranted. For each of the six priority topics, the NHLBAC Asthma Expert Working Group determined the key questions to address in the systematic reviews. For each key question, the working group of the NHLBAC identified the patient population, intervention, relevant comparators, and outcomes of interest.
The six priority topics identified for systematic review were as follows:
1.
Fractional exhaled nitric oxide (F eno) in diagnosis, medication selection, and monitoring of treatment response in asthma
2.
Remediation of indoor allergens (eg, house-dust mites/pets) in asthma management
3.
Adjustable medication dosing in recurrent wheezing and asthma
4.
Long-acting antimuscarinic agents in asthma management as add-ons to inhaled corticosteroids (ICSs)
5.
Immunotherapy and the management of asthma
6.
Bronchial thermoplasty (BT) in adult severe asthma
The NHLBAC Asthma Expert Working Group recommended that another 11 topics be acknowledged in the update but that no recommendations be developed for these topics because of the lack of sufficient new data for a systematic review of these topics at that time.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
Available at:
Google Scholar
These emerging topics are as follows:
•
Adherence
•
Asthma action plans
•
Asthma heterogeneity
•
Biologic agents
•
Biomarkers (other than F eno)
•
Classification of asthma severity
•
Long-acting beta 2-agonist (LABA) safety
•
Physiological assessments
•
Prevention of asthma onset
•
Role of community health workers in asthma management
•
Step down from maintenance therapy
The AHRQ EPCs conducted systematic reviews of the six priority topics and published the findings from these reviews online between October 2017 and March 2018.3-7
3.
D’Anci KE, Lynch MP, Leas BF, Apter AJ, Bryant-Stephens T, Kaczmarek JL, et al. Effectiveness and safety of bronchial thermoplasty in management of asthma. Comparative Effective Review No. 202. (Prepared by the ECRI Institute–Penn Medicine Evidence-based Practice Center under Contract No. 290-2015-00005-I). AHRQ Publication No. 18-EHC0003-EF. Rockville, Md: Agency for Healthcare Research and Quality; December 2017. Available at:
Google Scholar
4.
Leas BF, D’Anci KE, Apter AJ, Bryant-Stephens T, Schoelles K, Umscheid C. Effectiveness of indoor allergen reduction in management of asthma. Comparative Effectiveness Review No. 201. (Prepared by the ECRI Institute–Penn Medicine Evidence-based Practice Center under Contract No. 290-2015-0005-I). AHRQ Publication No. 18-EHC002-EF. Rockville, Md: Agency for Healthcare Research and Quality; February 2018. Posted final reports are located on the Effective Health Care Program search page.
Google Scholar
5.
Lin SY, Azar A, Suarez-Cuervo C, Diette GB, Brigham E, Rice J, et al. The role of immunotherapy in the treatment of asthma. Comparative Effectiveness Review No. 196. (Prepared by the Johns Hopkins University Evidence-based Practice Center under Contract No.290-2015-00006-I). AHRQ Publication No. 17(18)-EHC029-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Posted final reports are located on the Effective Health Care Program search page.
Google Scholar
6.
Sobieraj DM, Baker WL, Weeda ER, Nguyen E, Coleman CI, White CM, et al. Intermittent inhaled corticosteroids and long-acting muscarinic antagonists for asthma. Comparative Effectiveness Review No. 194. (Prepared by the University of Connecticut Evidence-based Practice Center under Contract No. 290-2015-00012-I). AHRQ Publication No. 17(18)-EHC027-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Available at: Posted final reports are located on the Effective Health Care Program search page.
Google Scholar
7.
Wang Z, Pianosi P, Keogh K, Zaiem F, Alsawas M, Alahdab F, et al. The clinical utility of fractional exhaled nitric oxide (FeNO) in asthma management. Comparative Effectiveness Review No. 197. (Prepared by the Mayo Clinic Evidence-based Practice Center under Contract No. 290-2015-00013-I). AHRQ Publication No. 17(18)-EHC030-EF. Rockville, Md: Agency for Healthcare Research and Quality; December 2017. Available at:
Google Scholar
These systematic reviews provided the evidence used to update the priority topics for this report.
In 2015, the NAEPPCC, which is a federal advisory committee, was created to continue the work of the NAEPP. In 2018, after the systematic reviews on the priority topics were completed, the NAEPPCC established the “Expert Panel,” which was charged with using the published systematic review reports to make recommendations on the key questions that could be implemented by health care providers and people with asthma.
The Expert Panel, composed of 18 members and a chair, included asthma content experts (pediatric and adult pulmonologists and allergists, an emergency room physician, and a pharmacist), primary care clinicians (pediatric, internal medicine, and family medicine providers), health policy experts, and implementation and dissemination experts. The Expert Panel received support from individuals who had experience using the GRADE approach.13
13.
Guyatt, G. ∙ Oxman, A.D. ∙ Akl, E.A. ...
GRADE guidelines, 1: Introduction—GRADE evidence profiles and summary of findings tables
J Clin Epidemiol. 2011; 64:383-394
Full Text
Full Text (PDF)
Scopus (3039)
PubMed
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While the Expert Panel considered its recommendations, the NHLBI convened focus groups made up of diverse asthma management stakeholders, including individuals with asthma, caregivers, and health care providers. These focus groups provided input on participants’ preferences and valuations of various asthma outcomes and interventions. The Expert Panel used summaries of these focus group discussions to inform its recommendations.
The Expert Panel initially presented its draft recommendations for comment and review to the NAEPPCC. The draft recommendations were also issued for public comment as well as for input from federal agencies. The Expert Panel considered all comments received and incorporated many of them into this final report. The NAEPPCC adopted the Expert Panel’s report during a public meeting and recommended the updated guidelines to the US Department of Health and Human Services. Following review and clearance, the US Department of Health and Human Services approved the updated guidelines, which were subsequently published in the Journal of Allergy and Clinical Immunology (JACI). A timeline of the steps completed to produce this report, beginning with the needs assessment, is shown in Fig 1.
Figure viewer
Fig 1 Timeline for 2020 Asthma Guideline Update.
Methods
Four AHRQ EPCs conducted and published systematic review reports on the key questions for the six priority topics. The pharmacologic topics (adjustable medication dosing and long-acting muscarinic antagonists [LAMAs]) were combined into a single systematic review; therefore, five systematic review reports were prepared on the six priority topics:
•
The Clinical Utility of Fractional Exhaled Nitric Oxide (FeNO) in Asthma Management (
•
Effectiveness of Indoor Allergen Reduction in Management of Asthma (
•
Intermittent Inhaled Corticosteroids and Long-Acting Muscarinic Antagonists for Asthma (
•
Role of Immunotherapy in the Treatment of Asthma (
•
Effectiveness and Safety of Bronchial Thermoplasty in Management of Asthma (
Systematic reviews of the literature
The protocols3-7
3.
D’Anci KE, Lynch MP, Leas BF, Apter AJ, Bryant-Stephens T, Kaczmarek JL, et al. Effectiveness and safety of bronchial thermoplasty in management of asthma. Comparative Effective Review No. 202. (Prepared by the ECRI Institute–Penn Medicine Evidence-based Practice Center under Contract No. 290-2015-00005-I). AHRQ Publication No. 18-EHC0003-EF. Rockville, Md: Agency for Healthcare Research and Quality; December 2017. Available at:
Google Scholar
4.
Leas BF, D’Anci KE, Apter AJ, Bryant-Stephens T, Schoelles K, Umscheid C. Effectiveness of indoor allergen reduction in management of asthma. Comparative Effectiveness Review No. 201. (Prepared by the ECRI Institute–Penn Medicine Evidence-based Practice Center under Contract No. 290-2015-0005-I). AHRQ Publication No. 18-EHC002-EF. Rockville, Md: Agency for Healthcare Research and Quality; February 2018. Posted final reports are located on the Effective Health Care Program search page.
Google Scholar
5.
Lin SY, Azar A, Suarez-Cuervo C, Diette GB, Brigham E, Rice J, et al. The role of immunotherapy in the treatment of asthma. Comparative Effectiveness Review No. 196. (Prepared by the Johns Hopkins University Evidence-based Practice Center under Contract No.290-2015-00006-I). AHRQ Publication No. 17(18)-EHC029-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Posted final reports are located on the Effective Health Care Program search page.
Google Scholar
6.
Sobieraj DM, Baker WL, Weeda ER, Nguyen E, Coleman CI, White CM, et al. Intermittent inhaled corticosteroids and long-acting muscarinic antagonists for asthma. Comparative Effectiveness Review No. 194. (Prepared by the University of Connecticut Evidence-based Practice Center under Contract No. 290-2015-00012-I). AHRQ Publication No. 17(18)-EHC027-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Available at: Posted final reports are located on the Effective Health Care Program search page.
Google Scholar
7.
Wang Z, Pianosi P, Keogh K, Zaiem F, Alsawas M, Alahdab F, et al. The clinical utility of fractional exhaled nitric oxide (FeNO) in asthma management. Comparative Effectiveness Review No. 197. (Prepared by the Mayo Clinic Evidence-based Practice Center under Contract No. 290-2015-00013-I). AHRQ Publication No. 17(18)-EHC030-EF. Rockville, Md: Agency for Healthcare Research and Quality; December 2017. Available at:
Google Scholar
that the EPCs used in their systematic reviews describe the prespecified key questions that they addressed (listed in Table IA), the methods they used, and the overall analytic framework.
| Topic | Key question |
--- |
| F eno | What is the diagnostic accuracy of F eno measurement(s) for making the diagnosis of asthma in individuals aged 5 y and older? |
| | What is the clinical utility of F eno measurements in monitoring disease activity and asthma outcomes in individuals with asthma aged 5 y and older? |
| | What is the clinical utility of F eno measurements to select medication options (including steroids) for individuals aged 5 y and older? |
| | What is the clinical utility of F eno measurements to monitor response to treatment in individuals aged 5 y and older? |
| | In children aged 0-4 years with recurrent wheezing, how accurate is F eno testing in predicting the future development of asthma at age 5 y and above? |
| Allergen mitigation | Among individuals with asthma, what is the effectiveness of interventions to reduce or remove exposures to indoor inhalant allergens on asthma control, exacerbations, quality of life, and other relevant outcomes? |
| ICS | What is the comparative effectiveness of intermittent ICS compared to no treatment, pharmacologic therapy, or nonpharmacologic therapy in children aged 0-4 y with recurrent wheezing? |
| | What is the comparative effectiveness of intermittent ICS compared to ICS controller therapy in individuals 5 y and older with persistent asthma? |
| | What is the comparative effectiveness of ICS with LABA used as both controller and quick-relief therapy compared to ICS with or without LABA used as controller therapy in individuals 5 y and older with persistent asthma? |
| LAMA | What is the comparative effectiveness of LAMA compared to other controller therapy as add-on to ICS in individuals aged 12 y and older with uncontrolled, persistent asthma? |
| | What is the comparative effectiveness of LAMA as add-on to ICS controller therapy compared to placebo or increased ICS dose in individuals aged 12 y and older with uncontrolled, persistent asthma? |
| | What is the comparative effectiveness of LAMA as add-on to ICS-LABA compared to ICS-LABA as controller therapy in individuals aged 12 y and older with uncontrolled, persistent asthma? |
| Immunotherapy | What is the evidence for the efficacy of SCIT in the treatment of asthma? |
| | What is the evidence for the safety of SCIT in the treatment of asthma? |
| | What is the evidence for the efficacy of SLIT, in tablet and aqueous form, for the treatment of asthma? |
| | What is the evidence for the safety of SLIT, in tablet and aqueous form, for the treatment of asthma? |
| BT | What are the benefits and harms of using BT in addition to standard treatment for the treatment of individuals aged 18 y and older with asthma? |
Table IA
Systematic review key questions
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When conducting the systematic reviews, the EPCs sought studies that included the prespecified target population(s) and settings and that used the prespecified interventions, comparators, and outcomes. The EPCs excluded articles about studies that did not meet the inclusion criteria listed in the protocols for each systematic review. These inclusion criteria were summarized in the published systematic review reports. (Appendices to the systematic review reports documented the rationales for excluding published articles identified by a broad search of the literature.) The systematic review reports also included the EPCs’ assessments of the risk of bias of each included article and of the strength of evidence for each key question using methods described in the protocols and systematic review reports. The EPCs were not required to use the GRADE methodology to conduct the systematic reviews, but they used a similar framework. After peer review and posting for public comment, the systematic review reports were finalized and published between late 2017 and early 2018.
Updated reviews of the literature
Westat (contract #HHSN268201700020B) conducted a literature search to identify any new articles published between the completion of the EPC’s systematic review literature searches and October 2018, when the Expert Panel began its work. The search strategies and the inclusion and exclusion criteria used in the updated literature searches were as similar as possible to those used in the initial systematic reviews. After reviewing the results of the updated literature searches, the Expert Panel determined that 15 additional articles addressing specific aspects of the key questions should be included in the focused update. The new articles were assessed for risk of bias. The Expert Panel considered the new evidence in conjunction with the evidence from the systematic review reports, but the new evidence was not incorporated into the pooled estimates in the evidence to decision (EtD) tables.
Expert Panel processes
Team structure
The Expert Panel met both in person and via webinar. In addition to their collective efforts, each panel member was assigned to one of six teams to address the topic-specific key questions identified by the NHLBAC Asthma Expert Working Group. Each topic team consisted of at least one content expert, primary care clinician, and individual with implementation expertise; some topic team members had multiple areas of expertise. The Integration and Implementation Team, composed of one representative from each of the topic teams, was tasked with integrating the new recommendations into the step diagrams from EPR-3 to create visual summaries of these steps. The NHLBI assembled and coordinated the Expert Panel. Westat provided technical and support services, including a methodology team with expertise in GRADE.
Disclosure of COIs and conflict management
To identify and manage potential COIs, the Expert Panel complied with the Institute of Medicine (now National Academy of Medicine) recommendations and standards for using systematic, evidence-based reviews to develop trustworthy guidelines.8
8.
Institute of Medicine Committee on Standards for Developing Trustworthy Clinical Practice Guidelines
Graham, R. ∙ Mancher, M. ∙ Miller Wolman, D. ... (Editors)
Clinical practice guidelines we can trust
National Academies Press, Washington, DC, 2011
Available at:
Google Scholar
,14
14.
Institute of Medicine
Conflict of interest in medical research education and practice
National Academies Press, Washington, DC, 2009
Google Scholar
The Expert Panel also followed the spirit of the recommendations for guideline panels that the ACP published in August 2019, midway through the development of these asthma guidelines.9
9.
Qaseem, A. ∙ Wilt, T.J.
Disclosure of interests and management of conflicts of interest in clinical guidelines and guidance statements: methods from the Clinical Guidelines Committee of the American College of Physicians
Ann Intern Med. 2019; 171:354-361
Crossref
Scopus (3)
PubMed
Google Scholar
Where possible, the Expert Panel implemented many of the new ACP guideline panel recommendations.
All Expert Panel members made financial disclosures and reported COIs using the standard author disclosure procedures described by the International Committee of Medical Journal Editors for manuscripts submitted to the JACI; the JACI editors reviewed these COI reports.15
15.
Journal of Allergy and Clinical Immunology. Information for Authors. 2018.
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Expert Panel members disclosed all personal fees, grant support, and nonfinancial support received, including support from entities that could be perceived to have influenced or could potentially have influenced the work of the Expert Panel for the past 36 months. They reported these COIs in writing before the Expert Panel initially convened, before each face-to-face meeting, and at the completion of the guidelines. In keeping with JACI requirements, these disclosure reports did not include sources of research funding, such as government agencies, charitable foundations, or academic institutions.
The Expert Panel chair and JACI editors rated each COI as high, moderate, or low and used a modified version of the ACP recommendations to develop a plan to manage each level of COI. For the Expert Panel, a high COI was defined as multiple interactions with biomedical entities (drug, biotechnology, or medical device companies) and could include interactions that were related or not related to the six priority topics. Participation in any speakers’ bureau of any biomedical entity was also considered a high COI. Individuals with a high COI were excluded from the Expert Panel unless they were able to reduce their level of COI. Expert Panel members who reduced the level of a high COI were then subject to the requirements, including recusals, associated with lower levels of COI.
Interactions related to a specific priority topic with a single biomedical entity were considered moderate COIs. Expert Panel members with a moderate COI related to any of the six priority topics were recused from participating in the writing, discussion, and voting on the recommendations or guideline section for that topic. This recusal process was implemented at the start of the Expert Panel’s work, and the Expert Panel formally recognized these COIs as moderate after the release of the ACP recommendations. Resolution of a moderate COI resulted in reinstatement to full participation in all activities related to that topic. Any report of a previously unreported moderate COI resulted in recusal of the member from activities related to that topic. In addition, members who had no COI discussed the topic again and voted again on the associated recommendations. A low COI was defined as no more than two interactions with a biomedical entity not related to asthma or to the topics under discussion.
As new COIs arose during the guideline-development process, Expert Panel members reported these COIs to the Expert Panel chair, and the chair and the JACI editors reviewed these new COIs and developed a plan to manage them. All Expert Panel members were notified when a member reported a new COI. After the release of the ACP recommendations, Expert Panel members with any new COI were recused from the Expert Panel. All Expert Panel members agreed not to undertake any activities that could result in a new COI for 12 months after the guidelines were released.
GRADE methodology
Overview
GRADE is an internationally accepted framework for determining the quality or certainty of evidence and the direction and strength of recommendations based on this evidence.16
16.
Guyatt, G.H. ∙ Oxman, A.D. ∙ Vist, G.E. ...
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Schunemann, H. ∙ Brozek, J. ∙ Guyatt, G. ... (Editors)
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GRADE Working Group, 2013
Available at: gdt.guidelinedevelopment.org/app/handbook/handbook.html
Accessed September 4, 2019
Google Scholar
A guideline methodologist not involved in the development of the systematic reviews for this update provided training on GRADE methodology to the Expert Panel and ongoing support and consultation throughout the project. The Expert Panel used the GRADE approach to review the evidence, create evidence profiles for critical and important outcomes, develop EtD tables, and write recommendation statements.
Prioritization and rating of asthma outcomes
The Expert Panel discussed asthma outcomes of potential interest and rated the relative importance of each outcome for clinical decision making using the GRADE approach.18
18.
Guyatt, G.H. ∙ Oxman, A.D. ∙ Kunz, R. ...
GRADE guidelines, 2: framing the question and deciding on important outcomes
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During this process, the Expert Panel reviewed the definitions of the outcomes in each of the systematic review reports. The outcomes deemed critical to assess for making recommendations across all topic areas were asthma exacerbations, asthma control, and asthma-related quality of life.
The Expert Panel assessed additional outcomes for specific key questions when these outcomes were relevant to the topic or when data for the three critical outcomes were not available. For example, in some instances, the systematic review reports identified limited or not adequate data on the effect of the interventions listed in the key questions on specific critical outcomes (eg, asthma control). In such cases, the Expert Panel considered available data on a related outcome (eg, asthma symptoms), even though validated outcome instruments were not used in studies or were not available. In this example, the Expert Panel confirmed asthma symptoms as an important outcome based on responses from the focus groups. The Expert Panel then used data on this important outcome to create the evidence profiles and EtD tables for the intervention, based on the available evidence.
After prioritizing the outcomes, the Expert Panel used established thresholds for determining significant improvement, also known as the minimally important difference (MID), for asthma control and asthma-related quality-of-life measures. These MID criteria are listed in Table IB.19-27
19.
Juniper, E.F. ∙ Buist, A.S. ∙ Cox, F.M. ...
Validation of a standardized version of the Asthma Quality of Life Questionnaire
Chest. 1999; 115:1265-1270
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Juniper, E.F. ∙ Gruffydd-Jones, K. ∙ Ward, S. ...
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Juniper, E.F. ∙ Guyatt, G.H. ∙ Cox, F.M. ...
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22.
Juniper, E.F. ∙ Guyatt, G.H. ∙ Feeny, D.H. ...
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Juniper, E.F. ∙ Guyatt, G.H. ∙ Feeny, D.H. ...
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24.
Juniper, E.F. ∙ Guyatt, G.H. ∙ Willan, A. ...
Determining a minimal important change in a disease-specific Quality of Life Questionnaire
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Abstract
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25.
Juniper, E.F. ∙ Svensson, K. ∙ Mork, A.C. ...
Measurement properties and interpretation of three shortened versions of the asthma control questionnaire
Respir Med. 2005; 99:553-558
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26.
Santanello, N.C. ∙ Zhang, J. ∙ Seidenberg, B. ...
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Schatz, M. ∙ Kosinski, M. ∙ Yarlas, A.S. ...
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J Allergy Clin Immunol. 2009; 124:719-723.e1
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For outcomes with no MID established in the literature, such as exacerbations, the Expert Panel reached consensus on clinically important differences that were based in part on a review of effect sizes in RCTs in the literature and on their judgments regarding the clinical relevance of a given change. In keeping with the recommendations from the Asthma Outcomes Workshop (2012),10
10.
Busse, W.W. ∙ Morgan, W.J. ∙ Taggart, V. ...
Asthma outcomes workshop: overview
J Allergy Clin Immunol. 2012; 129:S1-S8
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treatment with systemic (oral and parenteral) corticosteroids, asthma-specific emergency department visits, and hospitalizations were included as core outcome measures for exacerbations. The Expert Panel also included studies that used composite measures of systemic corticosteroids, emergency department visits, and hospitalizations.28
28.
Fuhlbrigge, A. ∙ Peden, D. ∙ Apter, A.J. ...
Asthma outcomes: exacerbations
J Allergy Clin Immunol. 2012; 129:S34-S48
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| Outcome measure | Range (points) | Score interpretation | MID |
--- --- |
| Asthma control |
| ACT | 5-25 | Well controlled: ≥20 Not well controlled: ≤19 | ≥12 y: MID ≥3 points |
| Asthma Control Questionnaire-5 (ACQ-5) Asthma Control Questionnaire-6 (ACQ-6) | 0-6 | Uncontrolled: ≥1.5 Well controlled: <0.75 | ≥18 y: MID ≥0.5 points |
| Asthma Control Questionnaire-7 (ACQ-7) | 0-6 | Uncontrolled: ≥1.5 Well controlled: <0.75 | ≥6 y: MID ≥0.5 points |
| Asthma-related quality of life |
| Asthma Quality of Life Questionnaire Asthma Quality of Life Questionnaire Mini (AQLQ-mini) | 1-7 | Severe impairment= 1 No impairment= 7 | ≥18 y: MID ≥0.5 points |
| Pediatric Asthma Quality of Life Questionnaire | 1-7 | Severe impairment= 1 No impairment= 7 | 7-17 y: MID ≥0.5 points |
| Other |
| Rescue medication use (daytime or nighttime) | Continuous measure of puffs per unit of time | NA | ≥18 y: MID= –0.81 puffs/d |
Table IB
MIDs for asthma-control and asthma-related quality-of-life measures19-27
19.
Juniper, E.F. ∙ Buist, A.S. ∙ Cox, F.M. ...
Validation of a standardized version of the Asthma Quality of Life Questionnaire
Chest. 1999; 115:1265-1270
Full Text
Full Text (PDF)
Scopus (452)
PubMed
Google Scholar
20.
Juniper, E.F. ∙ Gruffydd-Jones, K. ∙ Ward, S. ...
Asthma Control Questionnaire in children: validation, measurement properties, interpretation
Eur Respir J. 2010; 36:1410-1416
Crossref
Scopus (149)
PubMed
Google Scholar
21.
Juniper, E.F. ∙ Guyatt, G.H. ∙ Cox, F.M. ...
Development and validation of the Mini Asthma Quality of Life Questionnaire
Eur Respir J. 1999; 14:32-38
Crossref
Scopus (518)
PubMed
Google Scholar
22.
Juniper, E.F. ∙ Guyatt, G.H. ∙ Feeny, D.H. ...
Measuring quality of life in the parents of children with asthma
Qual Life Res. 1996; 5:27-34
Crossref
Scopus (470)
PubMed
Google Scholar
23.
Juniper, E.F. ∙ Guyatt, G.H. ∙ Feeny, D.H. ...
Measuring quality of life in children with asthma
Qual Life Res. 1996; 5:35-46
Crossref
Scopus (831)
PubMed
Google Scholar
24.
Juniper, E.F. ∙ Guyatt, G.H. ∙ Willan, A. ...
Determining a minimal important change in a disease-specific Quality of Life Questionnaire
J Clin Epidemiol. 1994; 47:81-87
Abstract
Full Text (PDF)
Scopus (1438)
PubMed
Google Scholar
25.
Juniper, E.F. ∙ Svensson, K. ∙ Mork, A.C. ...
Measurement properties and interpretation of three shortened versions of the asthma control questionnaire
Respir Med. 2005; 99:553-558
Full Text
Full Text (PDF)
Scopus (514)
PubMed
Google Scholar
26.
Santanello, N.C. ∙ Zhang, J. ∙ Seidenberg, B. ...
What are minimal important changes for asthma measures in a clinical trial?
Eur Respir J. 1999; 14:23-27
Crossref
Scopus (168)
PubMed
Google Scholar
27.
Schatz, M. ∙ Kosinski, M. ∙ Yarlas, A.S. ...
The minimally important difference of the Asthma Control Test
J Allergy Clin Immunol. 2009; 124:719-723.e1
Full Text
Full Text (PDF)
Scopus (261)
PubMed
Google Scholar
NA, Not applicable/available.
Open table in a new tab
EtD framework
The EtD framework provides a systematic and transparent approach for moving from evidence to recommendations by guideline panels.29
29.
Hultcrantz, M. ∙ Rind, D. ∙ Akl, E.A. ...
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The topic teams developed EtD tables for each key question using the evidence in the systematic review reports and the GRADEpro Guideline Development Tool.30
30.
Brożek, J. ∙ Nowak, A. ∙ Kunstman, P. ...
GRADEpro Guideline Development Tool (G2DT)
Available at: www.guidelinedevelopment.org
Accessed September 4, 2019
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New articles found in the updated literature review were noted in the new evidence sections of the EtD tables, but their data were not incorporated into the pooled estimates. See Table IC for the template used for EtD tables. The EtD tables provided a framework for the Expert Panel to use for assessing the evidence and providing rationales for their judgments on a range of factors that influenced the recommendations, as described in the next section, “Contextualization of judgments.”31
31.
Alonso-Coello, P. ∙ Oxman, A.D. ∙ Moberg, J. ...
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| Content area | Question | Judgment (pick one) | Research evidence | Additional considerations |
--- ---
| Desirable effects | How substantial are the desirable anticipated effects? | Trivial, small, moderate, large, vary, don’t know | | |
| Undesirable effects | How substantial are the undesirable anticipated effects? | Large, moderate, small, trivial, vary, don’t know | | |
| Certainty of evidence | What is the overall certainty of the evidence of the effects? | Very low, low, moderate, high, no included studies | | |
| Values | Is there important uncertainty about or variability in how much people value the main outcomes? | Important uncertainty or variability, possibly important uncertainty or variability, probably no important uncertainty or variability, no important uncertainty or variability | | |
| Balance of effects | Does the balance between desirable and undesirable effects favor the intervention or the comparison? | Favors the comparison, probably favors the comparison, does not favor either the intervention or the comparison, probably favors the intervention, favors the intervention, varies, don’t know | | |
| Acceptability | Is the intervention acceptable to key stakeholders? | No, probably no, probably yes, yes, varies, don’t know | | |
| Feasibility | Is the intervention feasible to implement? | No, probably no, probably yes, yes, varies, don’t know | | |
| Equity | What would be the impact on health equity? | Reduced, probably reduced, probably no impact, probably increased, increased, varies, don’t know | | |
Table IC
EtD table template
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Contextualization of judgments
The Expert Panel members reviewed the summary-of-findings tables in the AHRQ systematic review reports and recorded their judgments about the certainty of the evidence regarding each intervention. See Table ID for explanations of the levels of certainty in the evidence. For each key question, the Expert Panel reviewed the EPCs’ judgments about the risk of bias reported in the systematic review reports. The Expert Panel modified the judgments about the directness or indirectness of, consistency or inconsistency of, precision or imprecision of, and publication bias in the evidence when appropriate to reflect the panel’s contextualized judgments about the certainty of the evidence in the context of clinical practice guidelines.32
32.
Alonso-Coello, P. ∙ Schunemann, H.J. ∙ Moberg, J. ...
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Footnotes in the EtD tables in Appendix B (see this article’s Online Repository at www.jacionline.org) provide detailed explanations of these judgments. When the Expert Panel made a contextualized judgment for a specific outcome (and the opinion of the Expert Panel differed from the judgment of the EPC in the AHRQ systematic review report), the Expert Panel used the following words: “The Expert Panel rated this outcome down for…” Otherwise, the certainty of evidence and risk of bias ratings reflected the EPCs’ judgments from the published systematic review reports, and the Expert Panel identified these ratings by statements that began with “The AHRQ systematic review report rated this outcome down for…”
High We are very confident that the true effect lies close to that of the estimate of the effect.
Moderate We are moderately confident in the effect estimate. The true effect is likely to be close to the estimate of the effect, but there is a possibility that it is substantially different.
Low Our confidence in the effect estimate is limited. The true effect may be substantially different from the estimate of the effect.
Very low We have very little confidence in the effect estimate. The true effect is likely to be substantially different from the estimate of the effect.
Table ID
Certainty of evidence of effects
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Each EtD table includes a summary of the pooled results from the evidence syntheses (in addition to results from any new studies) in relative and absolute terms. The tables also describe any assumptions or evidence on variability in patient values and preferences regarding the intervention; the overall certainty of the evidence; the intervention’s net benefit based on the desirable and undesirable effects; and judgments about the resource requirements, acceptability, feasibility, and equity issues related to that intervention. The Expert Panel members made judgments within these domains and developed clinical recommendations based on the evidence summarized in the EtD tables. Discussions to make these judgments and develop the recommendations took place during online, telephone, and face-to-face meetings. For each recommendation, the Expert Panel indicated its direction (for or against the intervention) and strength, provided accompanying technical remarks and implementation considerations, and identified relevant evidence gaps.
Framing recommendations and coming to consensus
In GRADE, each recommendation has a direction, meaning that the recommendation is either for or against the use of an intervention. Each recommendation is also either strong or conditional, as explained in Table IE. Strong recommendations are those for which, in the judgment of the Expert Panel after it has reviewed all of the evidence and individual judgments, all or almost all people would choose the recommended course of action. Conditional recommendations are those for which, after reviewing all of the evidence and individual judgments, the Expert Panel believes that many informed people are likely to make different decisions about whether to take the recommended course of action. A conditional recommendation implies that engaging in a shared decision-making process is essential for individuals with asthma and their health care providers.31-33
31.
Alonso-Coello, P. ∙ Oxman, A.D. ∙ Moberg, J. ...
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Going from evidence to recommendations
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| Implications | Strong recommendation | Conditional recommendation |
---
| For individuals with asthma | Most individuals in this situation would want the recommended course of action and only a small proportion would not. | Most individuals in this situation would want the suggested course of action, but many would not. |
| For clinicians | Most individuals should receive the intervention. Formal decision aids are not likely to be needed to help individuals make decisions consistent with their values and preferences. | Different choices will be appropriate for individuals consistent with their values and preferences. Use shared decision making. Decision aids may be useful in helping individuals make decisions consistent with their risks, values, and preferences. |
| For policymakers | The recommendation can be adapted as policy or performance measure in most situations. Adherence to this recommendation according to the guideline could be used as a quality criterion or performance indicator. | Policy making will require substantial debate and involvement of various stakeholders. Performance measures should assess whether decision making is documented. |
| For researchers | The recommendation is supported by credible research or other convincing judgments that make additional research unlikely to alter the recommendation. On occasion, a strong recommendation is based on low or very low certainty in the evidence. In such instances, further research may provide important information that alters the recommendations. | The recommendation is likely to be strengthened (for future updates or adaptation) by additional research. An evaluation of the conditions and criteria (and the related judgments, research evidence, and additional considerations) that determined the conditional (rather than strong) recommendation will help identify possible research gaps. |
Table IE
Implications of strong and conditional recommendations∗
∗
Strong recommendations are indicated by statements that lead with “We recommend,” whereas conditional recommendations are indicated by statements that lead with “We conditionally recommend.”
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The Expert Panel drafted, discussed, and revised the recommendations multiple times before all eligible members (those who did not have a COI for the topic) voted on each recommendation. The Expert Panel achieved consensus when more than 90% of the Expert Panel members voted in favor of a recommendation. If less than 90% of members voted in favor of a recommendation, the relevant topic team continued to revise the recommendation until it achieved consensus approval according to these criteria.
Focus groups with individuals with asthma and their caregivers
The NHLBI sponsored focus groups with individuals with asthma and their caregivers to:
•
Identify the types of information and tools that individuals with asthma, their caregivers, and their health care providers would find most helpful in their ongoing efforts to effectively manage asthma and adhere to the new guidelines
•
Ensure that the new asthma guidelines reflect the voices of individuals with asthma and their caregivers
•
Identify potential barriers to uptake by individuals with asthma and their caregivers
Using virtual data-collection methods (ie, telephone and online platforms), the NHLBI conducted 11 in-depth interviews with health care providers who treat individuals with asthma and 10 online focus groups with English- and Spanish-speaking adults with asthma and adult caregivers of children with asthma with household incomes lower than $50,000 per year. In accordance with best practices, both the health care provider in-depth interviews and consumer focus group sessions lasted 75 minutes or less to minimize burden and facilitate engagement. Findings were analyzed using a notes- and transcript-based analysis process similar to that recommended by Krueger34
34.
Krueger, R.A.
Designing and conducting focus group interviews
University of Minnesota, October 2002
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and Patton.35
35.
Patton, M.Q.
Qualitative research and evaluation methods
Sage, Thousand Oaks, Calif, 2002
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The focus groups provided insight into outcomes that individuals with asthma and their caregivers considered most important; factors that affected their treatment choices; preferences for medication type and dosing frequency; and opinions about immunotherapy, allergen reduction, and BT. The Expert Panel considered these insights when developing its recommendations and EtD tables.
Findings of interviews and focus groups
Among both adults with asthma and caregivers of children with asthma, the most desired outcome was relief from symptoms that limit what people with asthma can do. In particular, participants valued symptom relief that would allow individuals with asthma to be more physically active. Caregivers also wanted to reduce the number of hospital visits for individuals with asthma, and Spanish-speaking caregivers sought control of nighttime symptoms. These individuals with asthma and caregiver preferences support the use of asthma symptom relief as an outcome measure when studies did not use validated outcome measurement tools.
Participants stated that cost and insurance coverage, safety, side effects, benefits, success rates, and asthma severity influenced their decisions about asthma treatment. Some participants were concerned that they might become dependent on or addicted to asthma medications (in particular, to pills), and participants with comorbidities expressed concern about drug interactions and contraindications, especially for oral medications.
Individuals with asthma indicated that they preferred inhaled medications over pills or liquids because they perceived inhaled medications to be easier to take or administer, faster acting, and more effective (because the medication is delivered directly to the site where it is needed). Individuals with asthma and caregivers also preferred taking one medication daily at most and viewed a need to take more than two to three medications a day as excessive. Caregivers were concerned about the administration of more medications or more frequent administration of medications to children while they are in school.
Taking medication on a set schedule instead of as needed drew mixed reactions. Perceived benefits of a set schedule included easier adherence, greater effectiveness, and a greater ability to prevent exacerbations (for those with severe asthma). In contrast, taking medication as needed was believed to offer flexibility and potentially reduce side effects. As-needed medications were also described as more appealing to those with mild to moderate asthma and to Spanish-speaking caregivers. Adults with asthma and caregivers were generally receptive to the use of one inhaler to both treat asthma and prevent exacerbations, although they wondered whether medications could do both effectively.
Levels of awareness of immunotherapy were low to moderate in individuals with asthma and caregivers. Some stated that they would consider this type of treatment if it were shown to be effective; others remained skeptical about the value of immunotherapy because of concerns about associated pain, inconvenience, and side effects.
Many participants reported taking action to reduce allergens at home. Most participants said that they used mattress and pillow covers, removed curtains or mold, controlled pests and dust, and vacuumed floors regularly. Some participants who had pets said that the pets were outside most of the time or they vacuumed their floors frequently. Participants also reported keeping windows closed during pollen and wildfire season to reduce the level of allergens and irritants in their home. Very few stated that they would stop their current allergen reduction efforts even if these efforts were proven to be ineffective. Most participants wanted information on cost and level of effort involved to consider making a change.
Spanish-speaking adults with asthma were more receptive to BT than their English-speaking counterparts. However, most participants thought that the procedure was too risky and expressed concerns about the need for anesthesia, multiple hospital visits, and heating of muscle tissue as well as the treatment’s impact on other health conditions. They wanted more information on the therapy’s side effects, risks, complications, and success rates as well as how the procedure is done.
2020 focused updates to the 2007 Asthma Guidelines
After the Expert Panel reached consensus on the recommendations, each topic team drafted a narrative to provide further information on each recommendation. These narratives form the body of this report. Each topic narrative has the following sections:
•
A brief background section that includes definitions of the terms used in the recommendations
•
The key questions addressed
•
The recommendations
•
An Implementation Guidance section that explains the recommendation in greater detail and provides Expert Panel opinion about how to implement the recommendation in clinical practice
•
A summary of the evidence
•
The rationale for the recommendation
•
A discussion of the evidence supporting the recommendation
•
A list of topic-specific research gaps and questions
Differences (if any) between the new recommendations and the recommendations in EPR-3 are discussed in Appendix A (in this article’s Online Repository at www.jacionline.org).
The Implementation Guidance sections are for practicing clinicians, and they contain the following information:
•
Clinician’s summary (more detailed explanation of the recommendation)
•
Population most likely to benefit from the recommendation
•
Any populations to which the recommendation does not apply
•
Topic-specific considerations
•
Issues that clinicians should discuss with their patients as part of the shared decision-making process
Review and public comment
The NAEPPCC reviewed an initial draft report. The NHLBI subsequently made the draft report available for public review and comment from December 2, 2019, to January 17, 2020. Interested stakeholders—including health professionals; representatives of the scientific community, academic institutions, the private sector, professional societies, advocacy groups, and patient communities; and other interested members of the public—were invited to submit comments. The Expert Panel received and reviewed approximately 500 comments from almost 100 individuals and organizations, and the panel used this input to revise the draft report.
One or more individuals and organizational representatives who submitted public comments mentioned almost all of the emerging topics. Of the 11 emerging topics (see list toward the beginning of Section I of this report), biologic agents received the most attention. The first biologic agent for asthma received approval from the US FDA in 2003, but the second biologic agent did not receive approval until November 2015. Between November 2015 and November 2017, four biologic agents received approval, but several others were not shown to be effective in clinical trials. Thus, at the time that the priority topics and key questions were developed, the only biologic agent available for use in the United States was omalizumab, which EPR-3 had addressed. The NHLBAC Asthma Expert Working Group did not believe that this single available biologic agent warranted inclusion in the update and included biologic agents as an emerging topic.
Limitations and research gaps
The Expert Panel identified several limitations in the process it used to identify topics and develop recommendations, including the following:
•
A better mechanism is needed to identify topics that need updating and to decrease the time between updates.
•
The process would benefit from a discussion and development of a plan about how to tailor guideline recommendations in the emerging era of personalized medicine.
•
Expanding engagement with professional societies might benefit both the development and the implementation of new recommendations.
The Expert Panel also identified several overarching research gaps listed below. Research gaps that are specific to individual topics are listed at the end of each topic section.
•
Research studies need to use the core outcome measures identified in the 2012 Asthma Outcomes Workshop.10
10.
Busse, W.W. ∙ Morgan, W.J. ∙ Taggart, V. ...
Asthma outcomes workshop: overview
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Federal agencies that contributed to the 2012 Asthma Outcomes Workshop report should require the studies they fund to measure outcomes as recommended in that report. Because new information on asthma outcomes is now available, the workshop report should be reexamined to determine whether it needs to be revised.
•
The clinical relevance of changes in outcome measures should be formally established to provide MIDs for all asthma outcomes (eg, exacerbations and asthma symptoms) and the cutoffs for tests (eg, F eno). Clinical relevance should be established using a wide range of stakeholder input, especially from individuals with asthma, who should also be included as members of the Expert Panel.
•
Updates are needed to the definitions of asthma severity that incorporate asthma phenotypes and endotypes. The definitions of low-, medium-, and high-dose ICSs also need to be updated.
•
Biologically appropriate subpopulations with asthma should be established and standardized. Although the populations of interest for the focused updates were defined for the systematic reviews, the characterizations of study participants did not reflect current understanding of relevant phenotypes and endotypes (eg, based on asthma severity, allergen-specific sensitization, or airway inflammatory type).
•
Standard reporting of results stratified by race and ethnicity as well as by age groups (0-4 years, 5-11 years, and 12 years and older) is needed to combine results across studies.
•
The vast majority of studies used to inform the guidelines were designed as efficacy studies,36
36.
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which evaluate treatment effects in relatively homogeneous populations and conditions in which fidelity to study protocols is actively promoted. Applicability to real-world clinical and community contexts requires studies with comparative effectiveness designs. Such research would benefit from the use of validated outcome measures and definitions of biologically appropriate subpopulations.
•
Studies need to use measures and outcomes that are important to individuals with asthma. The GRADE methodology gives highest priority to patient-centered outcomes. However, the studies that the Expert Panel used to develop the recommendations often did not measure outcomes that are most relevant or important to individuals with asthma. Research is needed to understand how preferred outcomes vary by race or ethnicity, asthma severity, age (eg, children or older adults), and socioeconomic status.
•
All measures and outcomes relevant to making judgments need to be included in the systematic reviews. For example, although cost-effectiveness data are available for some asthma interventions, the systematic review reports used for the updates did not include these data. Moreover, data regarding the safety of all interventions should be explicitly reported in publications on clinical trials.
Recommendations
In Table IF, all of the Expert Panel’s recommendations are grouped by the six priority topics. Please refer to the topic-specific sections in this report for full discussions of each recommendation, including implementation guidance and a clinician’s summary.
| Topic | Recommendation number∗ | Recommendation | Strength of recommendation† | Certainty of evidence‡ |
--- ---
| F eno | 1 | In individuals aged 5 y and older for whom the diagnosis of asthma is uncertain using history, clinical findings, clinical course, and spirometry, including bronchodilator responsiveness testing, or in whom spirometry cannot be performed, the Expert Panel conditionally recommends the addition of F eno measurement as an adjunct to the evaluation process. | Conditional | Moderate |
| | 2 | In individuals aged 5 y and older with persistent allergic asthma, for whom there is uncertainty in choosing, monitoring, or adjusting anti-inflammatory therapies based on history, clinical findings, and spirometry, the Expert Panel conditionally recommends the addition of F eno measurement as part of an ongoing asthma monitoring and management strategy that includes frequent assessments. | Conditional | Low |
| | 3 | In individuals aged 5 y and older with asthma, the Expert Panel recommends against the use of F eno measurements in isolation to assess asthma control, predict future exacerbations, or assess exacerbation severity. If used, it should be as part of an ongoing monitoring and management strategy. | Strong | Low |
| | 4 | In children aged 0-4 y with recurrent wheezing, the Expert Panel recommends against F eno measurement to predict the future development of asthma. | Strong | Low |
| Allergen mitigation | 5 | In individuals with asthma who do not have sensitization to specific indoor allergens or who do not have symptoms related to exposure to specific indoor allergens, the Expert Panel conditionally recommends against allergen mitigation interventions as part of routine asthma management. | Conditional | Low |
| | 6 | In individuals with asthma who have symptoms related to exposure to identified indoor allergens, confirmed by history taking or allergy testing, the Expert Panel conditionally recommends a multicomponent allergen-specific mitigation intervention. | Conditional | Low |
| | 7 | In individuals with asthma who have sensitization or symptoms related to exposure to pests (cockroaches and rodents), the Expert Panel conditionally recommends the use of integrated pest management alone, or as part of a multicomponent allergen-specific mitigation intervention. | Conditional | Low |
| | 8 | In individuals with asthma who have sensitization or symptoms related to exposure to dust mites, the Expert Panel conditionally recommends impermeable pillow/mattress covers only as part of a multicomponent allergen mitigation intervention, not as a single-component intervention. | Conditional | Moderate |
| ICS | 9 | In children aged 0-4 y with recurrent wheezing triggered by respiratory tract infections and no wheezing between infections, the Expert Panel conditionally recommends starting a short course of daily ICS at the onset of a respiratory tract infection with as-needed SABA for quick-relief therapy compared to as-needed SABA for quick-relief therapy only. | Conditional | High |
| | 10 | In individuals aged 12 y and older with mild persistent asthma, the Expert Panel conditionally recommends either daily low-dose ICS and as-needed SABA for quick-relief therapy or as-needed ICS and SABA used concomitantly. | Conditional | Moderate |
| | 11 | In individuals aged 4 y and older with mild to moderate persistent asthma who are likely to be adherent to daily ICS treatment, the Expert Panel conditionally recommends against a short-term increase in the ICS dose for increased symptoms or decreased peak flow. | Conditional | Low |
| | 12 | In individuals aged 4 y and older with moderate to severe persistent asthma, the Expert Panel recommends ICS-formoterol in a single inhaler used as both daily controller and reliever therapy compared to either: • Higher-dose ICS as daily controller therapy and SABA for quick-relief therapy or • Same-dose ICS-LABA as daily controller therapy and SABA for quick-relief therapy. | Strong | High (ages ≥12 y) Moderate (ages 4-11 y) |
| | 13 | In individuals aged 12 y and older with moderate to severe persistent asthma, the Expert Panel conditionally recommends ICS-formoterol in a single inhaler used as both daily controller and reliever therapy compared to higher-dose ICS-LABA as daily controller therapy and SABA for quick-relief therapy. | Conditional | High |
| LAMA | 14 15 | In individuals aged 12 y and older with uncontrolled persistent asthma, the Expert Panel conditionally recommends against adding LAMA to ICS compared to adding LABA to ICS. If LABA is not used, in individuals aged 12 y and older with uncontrolled persistent asthma, the Expert Panel conditionally recommends adding LAMA to ICS controller therapy compared to continuing the same dose of ICS alone. | Conditional Conditional | Moderate Moderate |
| | 16 | In individuals aged 12 y and older with uncontrolled persistent asthma, the Expert Panel conditionally recommends adding LAMA to ICS-LABA compared to continuing the same dose of ICS-LABA. | Conditional | Moderate |
| Immunotherapy | 17 | In individuals aged 5 y and older with mild to moderate allergic asthma, the Expert Panel conditionally recommends the use of SCIT as an adjunct treatment to standard pharmacotherapy in those individuals whose asthma is controlled at the initiation, build-up, and maintenance phases of immunotherapy. | Conditional | Moderate |
| | 18 | In individuals with persistent allergic asthma, the Expert Panel conditionally recommends against the use of SLIT in asthma treatment. | Conditional | Moderate |
| BT | 19 | In individuals aged 18 y and older with persistent asthma, the Expert Panel conditionally recommends against BT. Individuals aged 18 y and older with persistent asthma who place a low value on harms (short-term worsening symptoms and unknown long-term side effects) and a high value on potential benefits (improvement in quality of life, a small reduction in exacerbations) might consider BT. | Conditional | Low |
Table IF
Expert Panel recommendations
∗
Recommendations are numbered throughout the document for ease of reference.
†
See Table IE on page 1226 for definitions of the strength of recommendations.
‡
See Table ID on page 1225 for definitions of the levels of certainty of evidence of effects.
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Integration of the new recommendations into asthma care
The Expert Panel that produced this 2020 Asthma Guideline Update was asked to address specific questions about six priority topics rather than revise all of EPR-3. The Expert Panel, however, recognized the need to integrate the new evidence-based recommendations into a comprehensive approach to asthma care using the EPR-3 step diagrams.
Stepwise approach for managing asthma
In preparing the step diagrams (Figs 2-4), the Expert Panel used some of the definitions and assumptions from EPR-3. The step diagrams that follow this section retain the EPR-3 recommendations that the Expert Panel did not address in the current report. The Expert Panel encourages readers to review the footnotes in the step diagrams because they offer important information about the use of these diagrams.
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Fig 2 Stepwise approach for management of asthma in individuals aged 0 to 4 years.
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Fig 3 Stepwise approach for management of asthma in individuals aged 5 to 11 years.
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Fig 4 Stepwise approach for management of asthma in individuals aged 12 years and older.
The following conventions apply to Figs 2 to 4:
•
Each figure applies to the care of individuals with asthma in one age group.
–
Fig 2 applies only to ages 0 to 4 years.
–
Fig 3 applies only to ages 5 to 11 years.
–
Fig 4 applies only to ages 12 years and older.
•
Clinicians decide which step of care is appropriate depending on whether the individual is newly diagnosed (ie, is treatment naive) or whether the clinician is adjusting the individual’s therapy to achieve asthma control.
–
For newly diagnosed or treatment-naive individuals, clinicians should first choose the appropriate step diagram for the person’s age and then consider both the individual’s level of asthma impairment and risk when selecting the initial step and treatment.
–
Within a given step, the preferred options are the best management choices supported by the evidence that the Expert Panel reviewed. When the available evidence is insufficient or does not change a previous recommendation, the step diagrams list preferred options from the EPR-3 step diagrams.
–
Within a given step, an alternative option(s) is management strategies that are less effective or have more limited evidence than the preferred options. Clinicians and patients may choose the alternative treatments if individuals with asthma are currently receiving this therapy and their asthma is under control, if the preferred treatments are not available or too costly, or if the individuals with asthma prefer an alternative treatment.
–
Preferred and alternative treatments within a step category are listed alphabetically unless the Expert Panel has established a rank order of preference for the preferred or alternative treatments. A lack of rank order is indicated by “or” between treatment options.
•
In the stepwise approach to therapy for asthma, the clinician escalates treatment as needed (by moving to a higher step) or, if possible, deescalates treatment (by moving to a lower step) once the individual’s asthma is well controlled for at least three consecutive months.
–
For individuals with persistent asthma (ie, who require treatment at Step 2 or above), clinicians should be guided by the current step of treatment and the individual’s response to therapy (in terms of both asthma control and adverse effects) both currently and in the past to decide whether to step up, step down, or continue the current therapy.
–
For individuals with persistent asthma who are using an alternative treatment and have an unsatisfactory or inadequate response to that therapy, the Expert Panel suggests replacing the alternative treatment with the preferred treatment within the same step before stepping up therapy.
•
The Expert Panel did not add management options that the panel recommends against, or for which the evidence is insufficient to determine harms and benefits, to the step diagrams. Instead, these options are listed in Table IF.
•
The guidance provided in the step diagrams is meant to assist and not replace the clinical decision making required for individual patient management12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
Available at:
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and the input from individuals with asthma about their preferences.
Section II: Recommendations on the use of F eno testing in the diagnosis and management of asthma
Background
Nitric oxide can be measured in exhaled breath and can serve as a measure of the level of airway inflammation. In individuals with asthma, F eno may be a useful indicator of type 2 (T2) bronchial or eosinophilic inflammation in the airway. F eno testing requires an expiratory maneuver into a device designed for this purpose.
The Expert Panel addressed key questions on the utility of F eno measurement for asthma diagnosis, management, and prognosis. In this section, the panel discusses factors that confound F eno measurement or the interpretation of F eno test results in the context of the key questions. The evidence in all of these areas reveals important limitations that affect the strength of the recommendations and limit the ability to determine the optimal strategies for F eno measurement. A discussion of the equipment used to measure F eno and how to perform the test is beyond the scope of this update.
Definitions of terms used in this section
Children and adults have allergic asthma if they become symptomatic after acute exposure to something to which they are allergic (eg, a pet) or during a specific season of the year (eg, in the spring, due to tree pollen, or in the fall, due to ragweed pollen).
“Recurrent wheezing” is defined as clinically significant periods of bronchial or respiratory tract wheezing that is reversible or that is consistent with the clinical picture of bronchospasm.
Question 2.1
•
What is the diagnostic accuracy of F eno measurement(s) for making the diagnosis of asthma in individuals aged 5 years and older?
Recommendation 1: In individuals aged 5 years and older for whom the diagnosis of asthma is uncertain using history, clinical findings, clinical course, and spirometry, including bronchodilator responsiveness testing, or in whom spirometry cannot be performed, the Expert Panel conditionally recommends the addition of Fenomeasurement as an adjunct to the evaluation process.
Conditional recommendation, moderate certainty of evidence
Implementation guidance
Clinician’s Summary: The role of an increased level of F eno in the diagnosis of asthma is still evolving, and no definitive test exists for diagnosing asthma. F eno measurement may support a diagnosis of asthma in individuals for whom the diagnosis is uncertain even after a complete history, physical examination, and spirometry testing including bronchodilator responsiveness. Recognition of allergen sensitivity is extremely important for interpreting F eno levels. Allergic rhinitis and atopy, which can be present in individuals with and without asthma, are associated with increased F eno levels, and taking these factors into consideration is critical for accurately interpreting F eno test results.
On the basis of current data on F eno measurement in clinical settings, F eno testing has a supportive role in evaluation when the diagnosis of asthma is uncertain. The Expert Panel makes the following suggestions for use of F eno testing in asthma diagnosis:
•
Individuals in whom a diagnosis of asthma is being considered who may benefit from F eno measurement as part of the evaluation process include the following:
−
Those aged 5 years and older who have an uncertain diagnosis of asthma
−
Those in whom spirometry testing cannot be performed accurately
•
Because the data on the diagnostic accuracy of F eno measurement in children younger than 4 years are not conclusive, F eno measurement in this age group should not be used.
•
F eno test results should not be used alone to diagnose asthma. F eno measurements can serve as an adjunct test that may aid in diagnosing asthma in the appropriate setting. After clinicians consider other conditions that may influence F eno levels, they should perform the test when the results of a thorough clinical assessment, including other appropriate tests, are inconclusive.
•
Clinicians should use the cutoff levels or ranges listed in Table II for F eno measurement when evaluating persons for asthma. The likelihood that individuals aged 5 years and older have asthma increases by 2.8 to 7.0 times when the F eno test result is high. Clinicians who use F eno testing for asthma diagnosis should keep the following considerations in mind:
−
F eno levels of less than 25 parts per billion (ppb) (or <20 ppb in children aged 5-12 years) are inconsistent with T2 inflammation and suggest a diagnosis other than asthma (or that the individual has asthma but their T2 inflammation has been managed with corticosteroids or they have non-T2 inflammation or noneosinophilic asthma).
−
F eno levels greater than 50 ppb (or >35 ppb in children aged 5-12 years) are consistent with elevated T2 inflammation and support a diagnosis of asthma. Individuals who have T2 inflammation are more likely to respond to corticosteroid treatment.
−
F eno levels of 25 ppb to 50 ppb (or 20-35 ppb in children aged 5-12 years) provide little information on the diagnosis of asthma and should be interpreted with caution and attention to the clinical context.
−
The specificity and sensitivity of the F eno testing process depend on the clinical situation. However, in corticosteroid-naive individuals with asthma, F eno measurement is most accurate for ruling out the diagnosis of asthma when the result is less than 20 ppb. In this situation, the test has a sensitivity of 0.79, a specificity of 0.77, and a diagnostic odds ratio (OR) of 12.25.
−
ICS treatment should not be withheld solely based on low F eno levels.
| F eno level |
| <25 ppb (<20 in children aged 5-12 y) | 25-50 ppb (20-35 in children aged 5-12 y) | >50 ppb (>35 in children aged 5-12 y) |
| • Recent or current corticosteroid use • Alternative diagnoses • Phenotype less likely to benefit from ICS • Noneosinophilic asthma • COPD • Bronchiectasis • CF • Vocal cord dysfunction • Rhinosinusitis • Smoking • Obesity | • Evaluate in clinical context • Consider other diagnoses • Consider other factors influencing result • Eosinophilic asthma less likely | • Eosinophilic airways inflammation likely • Phenotype more likely to respond to ICS • Allergic asthma • Eosinophilic bronchitis |
Table II
Interpretations of F eno test results for asthma diagnosis in nonsmoking individuals not taking corticosteroids∗
CF, Cystic fibrosis; COPD, chronic obstructive pulmonary disease.
∗
Reprinted with permission of the American Thoracic Society, ©2019 American Thoracic Society. Dweik RA, Boggs PB, Erzurum SC, Irvin CG, Leigh MW, Lundberg JO, et al. An official ATS clinical practice guideline: interpretation of exhaled nitric oxide levels (FeNO) for clinical applications. Am J Respir Crit Care Med 2011;184:602-15. The American Journal of Respiratory and Critical Care Medicine is an official journal of the American Thoracic Society.
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•
F eno measurements should be performed by appropriately trained personnel who have extensive experience in interpreting the result or who consult experienced clinicians who can interpret the findings accurately. F eno testing can be performed in primary or specialty care settings. However, the costs of testing (ie, for equipment and expendable supplies) may prohibit the test’s adoption in the primary care office setting. Cost and the need for reproducible maneuvers will need to be addressed before home testing can become feasible.
•
What clinicians should discuss with their patients and families: Clinicians should share the following information about F eno testing with individuals suspected of having asthma and caregivers:
−
The F eno measurement process is safe for almost everyone.
−
F eno testing may be helpful in determining whether an individual has asthma, but it cannot be used to diagnose asthma.
−
Clinicians should inform individuals with asthma who have conditions or behaviors (such as smoking) that could affect the interpretation of the F eno test results that these issues could limit the accuracy of diagnostic attempts.
−
F eno test results cannot be used in isolation. Their interpretation must take into account other clinical factors and traditional measures.
−
The evidence favors the use of F eno measurement as an adjunct to other diagnostic methods (including a structured history, clinical findings, and pulmonary function testing) when the results from these other measures are not conclusive.
−
Decisions about treatment with an ICS are not dependent on F eno measurements, but such measurements may help direct stepwise therapeutic choices.
Summary of the evidence
No RCTs could be found to address Question 2.1 (see Appendix B EtD Table I).
More than 50 studies have been conducted, and some of these studies included healthy and symptomatic individuals, smokers and nonsmokers, atopic and nonatopic individuals, and individuals with and without a prior diagnosis of asthma. The protocols for diagnostic F eno assessments varied, and conclusions about the optimal testing protocol remain uncertain.
Based on the Expert Panel’s interpretation of the literature and the systematic review report findings, the overall certainty of evidence for this recommendation is moderate. The Expert Panel considers implementation of the recommendation in a broad population to be appropriate based on the diversity of the populations included in the systematic review report. The imprecision in the studies on the utility of F eno measurement in asthma diagnosis is notable.
Rationale and discussion
In the Expert Panel’s opinion, an additional tool to aid in diagnosing asthma could be beneficial, especially when that tool may help identify specific asthma phenotypes. The Expert Panel considered many facets of harm, risk, opportunity, and benefits in making its recommendation.
The acceptability of F eno measurement to individuals with a potential diagnosis of asthma is likely to be high, given that the test involves minimal effort and does not incur discomfort or side effects. Publications on studies that used F eno testing did not report any overt harms. The Expert Panel noted that most studies conducted F eno measurements only in specialty care research settings, and few data are available on the use of F eno measurement in primary care settings. As with many innovations, the cost of F eno equipment and testing may limit its broader use. These barriers to broader dissemination could have a negative impact on the availability of F eno testing and lead to less equitable care for populations with limited resources.
Questions 2.2 and 2.3
•
What is the clinical utility of F eno measurements to select medication options (including corticosteroids) for individuals aged 5 years and older?
•
What is the clinical utility of F eno measurements to monitor response to treatment in individuals aged 5 years and older?
Recommendation 2: In individuals aged 5 years and older with persistent allergic asthma, for whom there is uncertainty in choosing, monitoring, or adjusting anti-inflammatory therapies based on history, clinical findings, and spirometry, the Expert Panel conditionally recommends the addition of Fenomeasurement as part of an ongoing asthma monitoring and management strategy that includes frequent assessments.
Conditional recommendation, low certainty of evidence
Implementation guidance
Clinician’s Summary: This recommendation is specific to using F eno levels when selecting therapy for individuals with asthma and when monitoring the response to and adjusting the dosage of anti-inflammatory therapies. This recommendation does not apply to individuals taking biologic agents, with the exception of omalizumab, because the systematic review literature searches conducted until October 2018 did not include data on biologic agents other than omalizumab. Clinicians must interpret F eno levels in conjunction with other clinical data because these levels are affected by comorbid conditions, including allergic rhinitis and atopy. The weight of the evidence suggests that when used as part of an asthma management strategy, F eno monitoring is effective in preventing exacerbations only when used frequently (such as every 2-3 months), but even frequent monitoring does not improve asthma control or quality of life in individuals with asthma.
The Expert Panel offers the following suggestions on how to use F eno testing to monitor asthma:
•
Individuals for whom F eno testing may be useful to monitor asthma include the following:
−
Individuals aged 5 years and older with uncontrolled persistent asthma who are currently taking an ICS or an ICS with a LABA, montelukast, or omalizumab
−
Individuals whose symptoms indicate that they might require additional anti-inflammatory therapy
−
Individuals with atopy, especially children
–
Individuals with asthma being treated by providers who agree that frequent (every 2-3 months) assessments of asthma control over the course of a year are warranted
•
F eno levels must be interpreted in conjunction with other clinical data. Current evidence suggests that F eno can prevent exacerbations only when testing is used frequently (eg, every 2-3 months). Cutoff points for adjusting therapy to reduce the risk of exacerbation have not been established.
•
The Expert Panel does not recommend using F eno testing to assess adherence to treatment (mostly for ICSs) because the strength of this evidence is low. Moreover, although F eno levels were associated with adherence to ICSs as measured by electronic or dose counters in two observational studies37
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•
F eno levels are not well correlated with other asthma outcomes (eg, symptoms or control measured by such tools as the Asthma Control Test [ACT] or the Asthma Control Questionnaire [ACQ], prior or subsequent exacerbations, or exacerbation severity; see Recommendation 3). Therefore, clinicians should not use F eno measurement as a substitute for these measures or in isolation. Rather, F eno testing is best used as part of an ongoing asthma monitoring and management strategy that includes frequent assessments.
•
What clinicians should discuss with their patients and families: The Expert Panel suggests that clinicians consider conveying the following information to their patients with asthma as part of shared decision making:
−
F eno measurement is safe for almost everyone.
−
F eno-based asthma monitoring and management strategies are associated with significant reductions in exacerbation frequency, but not with improvements in control (based on ACT or ACQ results) or on quality-of-life measures.
−
To undergo F eno testing, individuals with asthma might need to be referred to a specialty clinic.
−
F eno measurements are used in addition to other evaluations of asthma control, such as lung function testing, symptom assessments, and questions about medication adherence.
−
F eno levels may be affected by multiple conditions in addition to asthma.
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life). The summary of evidence for Recommendation 2 can be found in Appendix B (EtD Table II).
In the Expert Panel’s judgment, the benefit of F eno monitoring is moderate. F eno testing to monitor responses to asthma anti-inflammatory therapies was associated with a meaningful decrease in exacerbations, whereas the average benefit of F eno monitoring for asthma control and quality of life did not achieve the MID (see EtD Table II). The certainty of evidence (for ACT, Pediatric Asthma Quality of Life Questionnaire, or Asthma Quality of Life Questionnaire) is low. The strategies for adjusting anti-inflammatory therapies using F eno test results in conjunction with other assessments varied widely.39-53
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For this reason, no evidence-based F eno cutoff points are available for choosing, monitoring, or adjusting anti-inflammatory therapies, and the Expert Panel has not provided an algorithm to use for this purpose. Most algorithms that have been used in studies involved strict protocols and may not be relevant to typical clinical practices.
The certainty of evidence for the effect of F eno monitoring on exacerbations depends on the definition of an asthma exacerbation. For exacerbations that were defined in terms of a composite end point, the certainty of evidence is high. The composite exacerbation end point used in these studies was defined as any of the following: unscheduled visits to the provider’s office, emergency department visits, hospitalizations, oral corticosteroid use, reductions in FEV 1 or in peak expiratory flow, symptom-associated lung function decline, or Global INitiative for Asthma guideline definitions. The studies that compared an asthma management strategy that includes F eno monitoring to one that does not include 6 RCTs in 1536 adults (OR, 0.62; 95% CI, 0.45-0.86) and 7 RCTs in 733 children (OR, 0.50; 95% CI, 0.31-0.82). Strategies that include F eno monitoring in adults result in an absolute risk reduction of 71 exacerbations per 1000 individuals with asthma (range of 108 to 25 fewer exacerbations). F eno monitoring is also associated with 116 fewer exacerbations per 1000 children with asthma. When only those exacerbations that result in oral corticosteroid use are used (based on 10 RCTs in 1664 adults and children), the certainty of evidence is moderate (OR, 0.67; 95% CI, 0.51-0.90). The absolute risk difference is 67 fewer exacerbations per 1000 individuals with asthma (range of 104 to 19 fewer exacerbations). For exacerbations that result in hospitalization (9 RCTs in 1598 adults and children), the certainty of evidence is low (OR, 0.70; 95% CI, 0.32-1.55). The absolute risk difference is 11 fewer exacerbations per 1000 individuals with asthma (range of 25 fewer to 19 more exacerbations).
The certainty of evidence is low for F eno monitoring to exert a change of at least the established MID using the ACT (MID, 3), Pediatric Asthma Quality of Life Questionnaire (MID, 0.5), or Asthma Quality of Life Questionnaire (MID, 0.5). For each of these outcomes, the mean difference in scores between groups with and without F eno monitoring was less than 0.1.
It is not known whether the recommendation applies to children who do not have allergic asthma because atopy (defined based on a positive skin prick test result or elevated aero-allergen-specific IgE) and allergic asthma were inclusion criteria in most of the pediatric studies, or allergic asthma was highly prevalent in the study populations.39
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Therefore, the evidence supporting this recommendation comes from mixed populations of allergic and nonallergic adults.
Studies evaluating the use of F eno to help select or monitor responses to biologic agents, with the exception of omalizumab, were not available for assessment. Therefore, whether this recommendation applies to other biologic agents is not known.
Rationale and discussion
In making this recommendation, the Expert Panel considered the desirable and undesirable effects of F eno monitoring, including the acceptability of this testing to both individuals with asthma and their providers, the feasibility of testing, and the impact of the use of F eno testing to monitor asthma on health equity. Potential benefits of F eno testing include reducing exacerbations, which is a critical outcome from both the patient and provider perspectives. The undesirable direct effects of F eno testing are expected to be minimal. However, the Expert Panel had concerns about the impact of F eno testing for asthma monitoring on accessibility and equity, as noted below.
F eno levels have been shown to be responsive to changes in anti-inflammatory medications, including ICSs, montelukast, and omalizumab. The Expert Panel did not review the effects on F eno levels of newly available anti-inflammatory biologic therapies for this update.
In the Expert Panel’s judgment, individual preferences and values have an important role in the decision to use F eno monitoring. This monitoring can affect quality of life and exacerbation frequency, and different individuals are likely to place different values on these effects. In addition, the burden (cost, time for appointments, and availability of testing) of frequent monitoring will likely influence an individual’s willingness to undergo regular testing. Therefore, a therapeutic monitoring plan that includes frequent F eno testing requires discussion and agreement between the individual with asthma and the clinician.
The Expert Panel was concerned that if F eno testing is not widely available and its use is restricted by insurance coverage policies, some individuals with asthma might not have the benefit of exacerbation reduction using F eno-based monitoring and management algorithms. As a result, disparities in asthma outcomes would widen. Most of the F eno monitoring studies with cost-effectiveness data were conducted outside the United States44
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and were therefore of limited value for this update. The Expert Panel recommends cost-effectiveness analyses conducted in the United States.
Question 2.4
•
What is the clinical utility of F eno measurements in monitoring disease activity and asthma outcomes in individuals with asthma aged 5 years and older?
Recommendation 3: In individuals aged 5 years and older with asthma, the Expert Panel recommends against the use of Fenomeasurements in isolation to assess asthma control, predict future exacerbations, or assess exacerbation severity. Fenoshould only be used as part of an ongoing monitoring and management strategy.
Strong recommendation, low certainty of evidence
Implementation guidance
Clinician’s Summary: The Expert Panel does not recommend F eno testing on its own to assess asthma control, predict a future asthma exacerbation, or assess the severity of an exacerbation. F eno levels are not well correlated with standard measures of asthma symptoms or control, such as the ACT, ACQ, prior or subsequent exacerbations, or exacerbation severity. Therefore, F eno testing is not a substitute for standard measures and should not be used in isolation to monitor disease activity. F eno measurement, however, may be used in conjunction with an individual’s history, clinical findings, and spirometry as part of an ongoing asthma monitoring and management strategy, which includes frequent assessments as described in Recommendation 2.
•
The Expert Panel recommends against the use of isolated F eno measurement for asthma management and monitoring.
•
F eno measurement should only be used as a part of an ongoing monitoring and management strategy to predict future exacerbations and assess exacerbation severity.
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life).
The Expert Panel considered the use of F eno measurement in adults aged 18 years or older and children aged 5 to 18 years to monitor current asthma control, subsequent and prior exacerbations, and the severity of an ongoing exacerbation. The evidence for these issues comes primarily from correlational studies.
Among adults, F eno levels are weakly associated with asthma control as measured by the ACT and the ACQ.62-65
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This association is even weaker among individuals who smoke, are pregnant, or are taking an ICS. The association between F eno levels and prior or subsequent exacerbations is mixed—depending on the study, this association is strong66
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or weak,67
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or no such association62
62.
Ko, F.W. ∙ Hui, D.S. ∙ Leung, T.F. ...
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exists. Among children and adolescents aged 5 to 18 years, the results are also mixed. For example, two studies showed an association between recent symptoms or uncontrolled asthma and elevated F eno levels.68
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However, another study showed that F eno levels did not correlate with nasal or asthma symptoms.70
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de Bot, C.M. ∙ Moed, H. ∙ Bindels, P.J. ...
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The evidence on the utility of F eno testing to predict exacerbations is inconclusive. These studies assessed different populations and used F eno levels alone as predictors or as part of a strategy that included other tests. For example, two studies showed that F eno levels were moderate predictors of exacerbations.42
42.
Fritsch, M. ∙ Uxa, S. ∙ Horak, Jr., F. ...
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and that F eno levels did not predict future exacerbations among high-risk urban children from minority populations.73
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Among children and adults, F eno levels did not correlate with exacerbation severity.74
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F eno testing was also difficult to perform in children in the acute setting, the results did not correlate with other measures of acute severity,76
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Rationale and discussion
Based on the evidence summarized above, the Expert Panel recommends against the use of F eno measurement to assess asthma control, predict future exacerbations, or assess exacerbation severity unless these measurements are used as part of an ongoing asthma monitoring and management strategy as described in Recommendation 2. Further research is needed to assess the use of F eno as a marker for medication adherence, as well as its impact on asthma outcomes, acceptability, and cost-effectiveness.
Question 2.5
•
In children aged 0 to 4 years with recurrent wheezing, how accurate is F eno testing in predicting the future development of asthma at ages 5 years and above?
Recommendation 4: In children aged 0 to 4 years with recurrent wheezing, the Expert Panel recommends against Fenomeasurement to predict the future development of asthma.
Strong recommendation, low certainty of evidence
Implementation guidance
Clinician’s Summary: In children aged 4 years and younger who have recurrent episodes of wheezing, F eno measurement does not reliably predict the future development of asthma. F eno test results in this population should be interpreted with caution until more data are available. The Expert Panel recommends against using F eno testing to predict future development of asthma in this age group until additional research and clinical practice determinations are available.
Summary of the evidence
The summary of evidence for Recommendation 4 can be found in EtD Table III in Appendix B.
Ten studies addressed the ability of F eno measures in children younger than 5 years to predict the subsequent development of asthma in children aged 5 years and older.78-87
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None of these studies were RCTs; seven studies were nonrandomized longitudinal studies and three were cross-sectional studies. Only four studies investigated the use of F eno measures to predict the diagnosis of asthma (and not wheezing or Asthma Predictive Index score). In one study in children,86
86.
Singer, F. ∙ Luchsinger, I. ∙ Inci, D. ...
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a F eno level indicating an increased risk of asthma had a positive predictive value of 58.0% on a composite measure of wheezing, diagnosis of asthma, or use of an ICS at age 7 years, whereas the negative predictive value was 78.2%. This result was similar to that for the Asthma Predictive Index score without the use of F eno levels. Therefore, although F eno levels appear to reflect eosinophilic bronchial inflammation early in life, the current evidence is insufficient to justify the conclusion that F eno testing in children aged 0 to 4 years reliably predicts a diagnosis of asthma at ages 5 years and above. Future studies may, however, demonstrate otherwise.
Although F eno levels appear to reflect T2 inflammation early in life, T2 inflammation is not specific to asthma. F eno levels in early childhood (ages 0-4 years) strongly correlate with Asthma Predictive Index scores. This correlation is not surprising because of the relationship between atopy and F eno levels and the fact that this index is heavily predicated on an atopic constitution. F eno levels are higher in children with wheezing than in children without a recent history of wheezing and in children with persistent wheezing than in those with transient wheezing. Because most children with transient wheezing stop wheezing by age 3 years,88
88.
Taussig, L.M. ∙ Wright, A.L. ∙ Holberg, C.J. ...
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87.
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is ongoing and may provide new information on this issue.
Rationale and discussion
F eno can be measured in young children who have normal resting breathing, and normal reference values for F eno have been published for children aged 1 to 5 years.90
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Evidence shows that in some preschool children with recurrent coughing and wheezing, an elevated F eno level more than 4 weeks after an upper respiratory tract infection may help predict physician-diagnosed asthma at school age, independently of clinical history or presence of IgE.78-87
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Balinotti, J.E. ∙ Colom, A. ∙ Kofman, C. ...
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However, the studies reviewed for this update had conflicting results, and in the opinion of the Expert Panel, they provided low to moderate certainty for an asthma diagnosis.
A single F eno measurement to predict future asthma is not likely to be physically harmful and is not burdensome. However, unreliable prediction models risk jeopardizing future insurability and could lead to treatment decisions that might rely on inadequate measures. Until better data on the predictive ability of F eno measurement are available for children aged 0 to 4 years, clinicians should inform parents that the data are limited to support the use of F eno measurement for this purpose.
The Expert Panel appreciates the potential value of a noninvasive tool to predict asthma onset, but such testing may cause worry and adversely affect care and treatment if the findings are inaccurate. In the Expert Panel’s judgment, therefore, the acceptability of F eno measurement for predictive purposes is low. Use of this testing is unlikely to change current treatment standards and could actually misdirect care. The feasibility of implementing F eno measurement in this population seems challenging for several reasons, including the likely need for a specialist, not a primary care provider, to do the measuring because of the difficulty of ensuring proper technique and accurate results. In addition, the cost and maintenance requirements of F eno equipment may limit the test’s use.
Given that the Expert Panel recommends against the use of F eno measurement to predict future asthma diagnoses in this population, equity issues are not expected to arise. However, if the test is marketed to patients who have private insurance or who pay for health care out of pocket, it could adversely impact those individuals. Therefore, the Expert Panel believes that the balance of effects does not favor the use of F eno for predicting future asthma diagnoses in young children.
Future research opportunities
The value and potential are clearly high for new methods to evaluate individuals with wheezing, correctly identify those with asthma, select appropriate asthma therapy, and monitor responses to asthma therapy. Research on F eno measurement and its use in asthma has advanced since the Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma was published. To expand this research, further clarify the role of F eno measurement for asthma diagnosis in individuals with wheezing, and use F eno measurement to support the care of individuals with asthma, topics for future research include the following:
•
Use of F eno measurement in the diagnostic process (eg, to determine the point at which F eno testing should be used in relation to other diagnostic tools and which individuals with asthma aged 5 years and older should be tested)
•
Prevalence of asthma in the settings in which the Expert Panel recommends F eno measurement (eg, specialty care settings) to better understand the performance of F eno testing as a diagnostic tool
•
Use of F eno testing to monitor adherence of children and adults to ICSs and other anti-inflammatory treatments
•
Role of F eno measurements in children aged 0 to 5 years who have wheezing or asthma-like symptoms to predict subsequent asthma diagnoses
•
Role of point-of-care F eno measurement to identify children who do not require oral corticosteroid therapy
•
F eno-based asthma management in people with moderate to severe persistent asthma
•
Potential uses of F eno measurement for asthma management in primary care
•
Impact on asthma health disparities of differential access to F eno measurement because of lack of health care coverage
•
Cost-effectiveness of F eno measurement in diverse populations and clinical settings
•
Role of F eno testing in individuals with uncontrolled asthma to predict the benefit of adding T2-directed biologic therapies
•
Refinement and validation of F eno cutoff levels for diagnostic purposes (eg, by determining variations in F eno levels in individuals with different comorbid conditions, physiological determinants of F eno levels, and F eno levels in different ethnic and racial groups)
•
Identification of algorithms for the most useful combination of, and cutoff levels for, objective measures (eg, F eno levels, blood eosinophil levels, spirometry test results, short-acting beta 2-agonist [SABA] use, and symptom scores) for choosing, monitoring, or adjusting anti-inflammatory therapy
•
Refinement of ongoing management strategies that incorporate F eno measurement to better understand the optimal timing and interpretation of F eno levels in a range of asthma phenotypes (eg, eosinophilic vs noneosinophilic asthma)
•
Identification of the populations most likely to benefit from F eno-guided treatment and the optimal frequency of F eno monitoring
Section III: Recommendations for indoor allergen mitigation in management of asthma
Background
Environmental control is one of the four cornerstones of asthma management in Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
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The Expert Panel was tasked with examining the effectiveness of single-component and multicomponent allergen mitigation strategies directed at common, indoor aeroallergens, with the goal of improving asthma outcomes for individuals with asthma. The key questions for this priority topic and the recommendations by the Expert Panel are provided for single-component and multicomponent allergen mitigation strategies.
Not included in the scope of work for this priority topic is an examination of the utility of clinical testing for sensitivity to allergens (eg, using skin prick tests or tests of allergen-specific IgE), mitigation strategies for outdoor allergens, and mitigation of environmental irritants (eg, tobacco smoke). Specific occupational exposures were also outside the scope of work, although the indoor allergens addressed in these recommendations can be encountered in work settings.
Definitions of terms used in this section
An allergen mitigation intervention aims to decrease an individual’s exposure to allergens. The intervention can have a single component or multiple components.
A single-component intervention is an individual mitigation strategy targeted at one or more specific allergens to which an individual is both exposed and sensitized. Single-component allergen mitigation interventions examined in this report include the following:
•
Acaricide: a house-dust mite pesticide that can be applied to carpets, mattresses, and furniture.
•
Air filtration systems and air purifiers, including those with high-efficiency particulate air (HEPA) filters: devices that filter indoor air and remove solid particulates, such as dust, pollen, mold, and bacteria, from the air.
•
Carpet removal: removal of wall-to-wall or area rugs from one or more rooms.
•
Cleaning products: including application of bleach or similar products.
•
HEPA vacuum cleaners: vacuum cleaners that have a HEPA filter.
•
Impermeable pillow and mattress covers: covers placed on mattresses and pillows that are impermeable to dust mites.
•
Integrated pest management: a comprehensive approach to removing and controlling common indoor pests (eg, cockroaches and mice) using, for example, traps, poison, and barriers to influx. The Expert Panel considered integrated pest management to be a single-component intervention even though it may include prevention, mitigation, and removal strategies.
•
Mold mitigation: professional removal, cleaning, sanitization, demolition, or other treatment to remove or prevent mold. The Expert Panel considered mold mitigation to be a single-component intervention even though it may include prevention, mitigation, and removal strategies.
•
Pet removal: complete removal or confinement of furry pets (eg, dogs and cats) to specific rooms in a house.
A “multicomponent intervention” is defined as the use of two or more of the aforementioned single-component interventions at the same time as part of a bundled approach targeted at one or more allergens to which the individual is both sensitized and exposed. An example of a multicomponent intervention is the use of three single-component interventions (eg, air purifiers, impermeable pillow and mattress covers, and HEPA vacuum cleaners) for individuals sensitized and exposed to dust mites and mold.
“Sensitization” is defined in this section as the production of a specific IgE to an aeroallergen whose presence can be confirmed by skin prick testing or assays for a specific IgE.
Question 3.1
•
Among individuals with asthma, what is the effectiveness of interventions (eg, pesticides, air filters/purifiers, mattress covers, and pest control) to reduce or remove indoor inhalant allergens on asthma control, exacerbations, quality of life, and other relevant outcomes?
In some individuals, asthma can have an allergic component. Therefore, clinicians should take a history of the individual’s environmental allergen exposure and pursue testing for specific allergen sensitization, when appropriate. The Expert Panel has several recommendations for this question:
Recommendation 5: In individuals with asthma who do not have sensitization to specific indoor allergens or who do not have symptoms related to exposure to specific indoor allergens, the Expert Panel conditionally recommends against allergen mitigation interventions as part of routine asthma management.
Conditional recommendation, low certainty of evidence
Recommendation 6: In individuals with asthma who have symptoms related to exposure to identified indoor allergens, confirmed by history taking or allergy testing, the Expert Panel conditionally recommends a multicomponent allergen-specific mitigation intervention.
Conditional recommendation, low certainty of evidence
Recommendation 7: In individuals with asthma who have sensitization or symptoms related to exposure to pests (cockroaches and rodents), the Expert Panel conditionally recommends the use of integrated pest management alone, or as part of a multicomponent allergen-specific mitigation intervention.
Conditional recommendation, low certainty of evidence
Recommendation 8: In individuals with asthma who have sensitization or symptoms related to exposure to dust mites, the Expert Panel conditionally recommends impermeable pillow/mattress covers only as part of a multicomponent allergen mitigation intervention, not as a single-component intervention.
Conditional recommendation, moderate certainty of evidence
Implementation guidance
Clinician’s Summary:
For individuals with asthma who do not exhibit any allergy symptoms or for whom testing has not suggested that they have an allergy to certain indoor substances (eg, dust mites or cat dander), the Expert Panel recommends no specific environmental interventions to reduce these allergens within the home.
For individuals with asthma who are exposed to an allergen within the home and who have allergy symptoms or a positive test result suggesting that they have an allergy to certain indoor substances (eg, dust mites or cat dander), the Expert Panel recommends using a multicomponent intervention to try to control the indoor allergen in question. Single-component interventions often do not work.
For individuals with asthma who are exposed to cockroaches or rodents (eg, mice) in the home and who have allergy symptoms or sensitization to these allergens demonstrated by allergy skin testing or a specific IgE, the Expert Panel recommends using integrated pest management to improve asthma outcomes. Integrated pest management can be used alone or with other interventions to reduce exposure to pest-related allergens in the home.
For individuals with asthma who have allergy symptoms or a test result suggesting that they are allergic to dust mites, the Expert Panel recommends using multicomponent interventions to reduce dust mite levels in the home and improve asthma outcomes. Use of pillow and mattress covers alone does not improve asthma outcomes.
Overall, the studies of allergen mitigation strategies provide low certainty of evidence that these strategies are beneficial for key asthma outcomes. Therefore, the Expert Panel recommends tailored allergen intervention strategies only for individuals with asthma who are exposed to these specific allergens and have either symptoms based on clinical history or an allergy to these substances based on allergy testing.
Based on current data on the use of a variety of single-component and multicomponent strategies to reduce exposure to allergens, the Expert Panel makes the following suggestions for implementing allergen exposure reduction strategies:
•
Allergen mitigation strategies can be used in individuals of all ages with asthma of all levels of severity.
•
Clinicians need to tailor mitigation strategies to the individual based on their allergy symptoms, sensitization, and exposures. Clinicians should consider allergen testing when appropriate, before committing individuals to specific allergen mitigation strategies that may be burdensome. See Table IIIA for allergen-specific mitigation interventions addressed in the systematic review report. Table IIIB summarizes the certainty of evidence on various allergen mitigation interventions.
| Intervention assessed in studies in the SR | Allergen |
--- |
| Animal dander | Dust mites | Cockroaches | Mold |
| Acaricide | | ++ | | |
| Air filtration systems and air purifiers | ++ | + | + | ++ |
| Carpet removal | ++ | ++ | | + |
| Cleaning products (eg, bleach) | | | | ++ |
| HEPA vacuum cleaners | ++ | + | + | ++ |
| Impermeable pillow and mattress covers | | ++ | | |
| Integrated pest management | +∗ | | ++ | |
| Mold mitigation | | | | ++ |
| Pet removal | ++ | | | |
Table IIIA
Examples of allergen mitigation interventions and their targeted allergens
SR, Systematic review.
++Primary target allergen(s) for the intervention.
+Secondary target allergen(s) for the intervention.
∗
Dander from rodents.
Open table in a new tab
| Intervention assessed in studies in the SR | EtD table number | Evidence on use as a single-component strategy for allergen mitigation (certainty of evidence) | Evidence on use as part of a multicomponent strategy for allergen mitigation (certainty of evidence)∗ |
--- --- |
| Acaricide | IV | † | Intervention makes no difference (moderate certainty of evidence) |
| Impermeable pillow and mattress covers | V | Intervention makes no difference (moderate certainty of evidence) | Evidence favors intervention (moderate certainty of evidence) |
| Carpet removal | VI | † | Intervention makes no difference (low certainty of evidence) |
| Integrated pest management (for cockroaches and mice) | VII | Evidence favors intervention (low certainty of evidence) | Evidence favors intervention (low certainty of evidence) |
| Air filtration systems and air purifiers | VIII | Intervention makes no difference (low certainty of evidence) | Intervention makes no difference (moderate certainty of evidence) |
| HEPA vacuum cleaners | IX | † | Evidence favors intervention (among children only; moderate certainty of evidence) |
| Cleaning products | X | † | † |
| Mold mitigation | XI | † | Evidence favors intervention (low certainty of evidence) |
| Pet removal | XII | † | † |
Table IIIB
Summary of certainty of evidence on allergen mitigation interventions
SR, Systematic review.
∗
Combination of interventions used in the multicomponent studies varied, and the Expert Panel cannot identify or recommend any particular combination of strategies as optimal at this time.
†
Evidence was insufficient for the Expert Panel to assess the intervention.
Open table in a new tab
•
The Expert Panel recognizes the existing inequities in access to specialists and allergen testing. The panel therefore advises clinicians to, at a minimum, take a clinical history of symptoms and exposures for all individuals with asthma to help determine the need for allergen mitigation.
•
Allergy testing (with a skin prick or allergen-specific IgE test) may have false-positive and false-negative results, and certain allergens (eg, dust) may also act as irritants. For an individual whose symptoms worsen on exposure to specific aeroallergens, the Expert Panel recommends that the clinician consider mitigating that aeroallergen even if the individual’s test result is negative.
•
Some of the interventions examined provide no or low certainty of evidence about their efficacy in improving asthma outcomes (including exacerbations, quality of life, asthma control, and symptoms). The Expert Panel recognizes that some of the interventions, especially integrated pest management and mold mitigation, may have broader public health benefits. However, these interventions do not replace routine good practices, including regular and frequent house cleaning and laundering of bedding materials.
•
Some people are allergic to dander (flakes of skin) or saliva from pets . The few studies on pet removal have had inconclusive results. However, if an individual with asthma experiences symptoms around a pet, the individual should consider removing the pet from the home, keeping the pet outdoors, or, if neither of these options is feasible, keeping the pet out of commonly used rooms. Testing for sensitization to pets may be particularly worthwhile for those with chronic or uncontrolled symptoms and might help support what can be a difficult decision to remove a pet from the home.
•
Some cleaning and integrated pest management interventions may trigger asthma and/or be hazardous. Individuals with asthma need to balance the potential benefits and harms of interventions before implementing them.
•
If an individual with asthma has sensitization to an allergen on skin prick testing and is exposed to that allergen but has no objective evidence of worsened disease control and denies having symptoms, chronic exposure could have led to the development of clinical tolerance to that allergen in that environment. Allergen-specific mitigation strategies could adversely modify this established balanced relationship between the individual and the environment.
•
What clinicians should discuss with their patients and families:
−
Clinicians need to consider the complexity of the patient population and the limitations of the evidence identified. Clinicians may also find it helpful to consider the severity of a patient’s asthma, the small benefit, and the extent of previous symptoms and exacerbations when recommending allergen mitigation interventions.
−
Allergen mitigation interventions may be expensive or difficult for patients to use or maintain. Clinicians should consider the cost implications of certain interventions, especially among those with limited financial resources, and assess the magnitude of the potential value of an intervention in improving an individual’s asthma outcomes.
Summary of the evidence
The Expert Panel specified four outcomes (exacerbations, asthma quality of life, asthma control, and asthma symptoms) as critical outcomes when it reviewed the evidence. The panel considered outcomes related to health care utilization to be important outcomes. The Expert Panel gave higher priority to outcomes measured in studies that used validated outcome instruments than those assessed with nonvalidated outcome measures. When data on validated outcome measures were not available, the Expert Panel used data from nonvalidated outcome measures, such as asthma symptoms. Table IIIB summarizes the Expert Panel’s assessments of the certainty of evidence for each of the allergen mitigation interventions examined, when used as a single-component intervention or as part of a multicomponent intervention. The table also lists the EtD tables for each of the interventions.
Single-component allergen mitigation interventions
For the majority of single-component allergen mitigation interventions, studies to assess the effectiveness of the interventions were limited. For the single-component interventions with enough studies to assess their impact on critical outcomes, the certainty of the evidence was either low or very low, or the results were limited to one or two critical outcomes on which results were inconclusive or that did not improve. The studies included mixed populations, which made it difficult to determine whether better-defined populations might benefit from the intervention. Certainty of evidence was often downgraded because of the limitations of several studies, including those of single-component interventions with acaricides91
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In multicomponent studies that included air filtration systems and air purifiers (three of the four studies used devices with HEPA filters), the results showed no decrease in exacerbations or improvement in quality of life (high certainty of evidence). The results were mixed for asthma control (no benefit, low certainty of evidence) and asthma symptoms (decreased severity or number of reported symptoms in children but not in mixed populations, low certainty of evidence).118
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Most studies did not examine harms, and none reported any important harms from the various allergen mitigation strategies studied. Because of the lack of benefits identified and the potential harms from applications of chemicals, the Expert Panel does not recommend the use of acaricides.
Rationale and discussion
Overall approach for developing allergen mitigation recommendations
When developing each of the four recommendations in this section, the Expert Panel considered the benefits and harms of each of the allergen mitigation interventions and the level of evidence available for assessing the interventions. In addition, the Expert Panel considered the acceptability of the interventions to individuals with asthma and their providers as well as the ease of use, costs, and impact on health equity of each intervention.
Potential harms
Although the identified harms from most of the interventions were minimal, studies rarely examined harms. Therefore, the Expert Panel considered theoretical harms, patient burden, and initial and ongoing costs in its recommendations. For example, the Expert Panel’s judgment was that interventions for mold mitigation and carpet removal could be associated with risks or be costly or difficult to complete. Another Expert Panel determination was that impermeable pillow and mattress covers are low-risk interventions with limited costs but are likely to require frequent cleaning of the bedding above the covers to be effective.
Prioritization of outcomes
Furthermore, the Expert Panel considered the impact of the interventions on asthma symptoms as a critical outcome. The Expert Panel recognized that none of the studies used a validated outcome measure of asthma symptoms, and the definition of asthma symptoms was not standardized across studies. However, asthma symptoms are a relevant patient-centered outcome that was important to individuals with asthma in focus groups and that could be particularly relevant to assess for low-risk interventions.
Heterogeneity of studies
The Expert Panel found the heterogeneity of available studies to be challenging. As outlined in the allergen reduction systematic review report,4
4.
Leas BF, D’Anci KE, Apter AJ, Bryant-Stephens T, Schoelles K, Umscheid C. Effectiveness of indoor allergen reduction in management of asthma. Comparative Effectiveness Review No. 201. (Prepared by the ECRI Institute–Penn Medicine Evidence-based Practice Center under Contract No. 290-2015-0005-I). AHRQ Publication No. 18-EHC002-EF. Rockville, Md: Agency for Healthcare Research and Quality; February 2018. Posted final reports are located on the Effective Health Care Program search page.
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participants’ baseline clinical characteristics were variable, and the findings from these studies suggested that participants were not equally likely to benefit from the interventions reviewed.
In addition, the Expert Panel preserved the systematic review report authors’ distinction between single-component interventions designed to mitigate a single allergen (eg, an acaricide for house-dust mite allergens); single-component interventions that address multiple allergens (eg, air purifiers to control mold and animal dander); and multicomponent interventions, which usually target more than one allergen (see Table IIIA).
Many of the studies available to the Expert Panel examined multicomponent interventions in mixed populations of patients with varying severities of asthma and sensitizations to allergens. Moreover, the combinations of components examined in each study were rarely the same across studies, and most studies did not assess adherence to or use of the interventions. The Expert Panel concurred with the systematic review report authors’ assessment that the interplay between allergen type, intervention type, and individual patient characteristics could have strongly modified the effects of these interventions in these studies.
Targeting recommendations to individuals who are both exposed and allergic to specific allergens
It was the Expert Panel’s judgment that individuals with asthma should not burden themselves with allergen mitigation interventions if they are both not regularly exposed to an allergen and not allergic to a specific allergen. Given that certain populations might not have ready access to allergy specialists and allergen skin prick or IgE testing, the Expert Panel noted that patient histories (eg, symptoms related to exposure to specific indoor allergens) to assess patient sensitivities could suffice. Therefore, the Expert Panel is not recommending allergen mitigation interventions for all individuals with asthma. Instead, the panel is recommending basing decisions about allergen mitigation interventions on a combination of the exposures, symptoms, and sensitization of individuals with asthma.
Single-component interventions are rarely effective
Of the single-component allergen mitigation interventions evaluated in enough studies to assess their impact on critical outcomes, the certainty of the evidence was either low or very low, or the results were limited to one or two critical outcomes, were inconclusive, or demonstrated no improvement. As summarized in Table IIIB, the Expert Panel considered integrated pest management to be a single-component intervention, and it was the only single-component approach with beneficial effects. Single-component dust mite interventions using pillow and mattress covers demonstrated no benefit for any of the critical outcomes, including asthma symptoms. Based on these findings, it was the Expert Panel’s judgment that single-component approaches to mitigating an allergen are rarely effective.
Evidence for multicomponent interventions varies
Across the allergen mitigation interventions examined in this report, it was the Expert Panel’s judgment that mattress and pillow covers, integrated pest management, HEPA vacuum cleaners, and mold mitigation are potentially beneficial when used as part of a multicomponent allergen mitigation strategy, but the benefits are small. Mattress and pillow covers as part of a multicomponent allergen mitigation strategy did not show improvements when validated outcome measures (eg, exacerbations, ACT, or Asthma Quality of Life Questionnaire) were used. The strength of evidence from the studies demonstrating small reductions in symptom days (a nonvalidated outcome measure) and the low risk and relative cost of impermeable pillow and mattress covers resulted in the Expert Panel’s conditional recommendation for use of this intervention only as part of a multicomponent allergen mitigation strategy.
The evidence was stronger on improvements across asthma outcomes for both integrated pest management and HEPA vacuum cleaners used as part of a multicomponent strategy than the evidence on impermeable mattress and pillow covers.
Only three studies examined multicomponent interventions that included mold mitigation.129-131
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The Expert Panel considered the reduction in health care utilization with mold mitigation as well as the broader public health benefit of supporting its use as part of a multicomponent allergen mitigation strategy in making its conditional recommendation.
Additional considerations
For most of these interventions, the certainty of evidence is low, and the benefits are small. It is not the Expert Panel’s intent to suggest that all four of these interventions (mattress and pillow covers, integrated pest management, HEPA vacuum cleaners, and mold mitigation), when used as part of a multicomponent strategy, serve as the optimal allergen mitigation package. Instead, the Expert Panel is indicating that individuals who have symptoms related to exposure to specific allergens should consider using these interventions when appropriate.129
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The Expert Panel recognizes that patients, providers, and other stakeholders generally find mattress and pillow covers to be an acceptable, noninvasive strategy to reduce exposure to dust mites. However, the Expert Panel cautions individuals with asthma not to use these covers as the sole strategy for mitigating dust mites. Studies that applied mattress and pillow covers solely either showed no effect on asthma outcomes or had inconclusive results. It was the Expert Panel’s judgment that mattress and pillow covers should only be applied as part of a multicomponent intervention targeting dust mites.
In summary, the studies of allergen mitigation strategies provided lower certainty of evidence of effectiveness for key asthma outcomes than studies of asthma controller medications. For these reasons, the Expert Panel recommends only tailored allergen intervention strategies for individuals with asthma who have symptoms related to exposure confirmed by allergy testing or clinical history for identified indoor allergens.
Future research opportunities
The Expert Panel has identified the following topics related to allergen mitigation interventions (eg, acaricides, air purifiers, HEPA vacuum cleaners, carpet removal, pet removal, cleaning products, and mold mitigation) that require additional research:
•
Effectiveness of allergen mitigation interventions that use the validated outcome measures recommended by the Asthma Outcomes Workshop10
10.
Busse, W.W. ∙ Morgan, W.J. ∙ Taggart, V. ...
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•
Effectiveness of allergen mitigation interventions in individuals with asthma who have demonstrated exposure and/or sensitization to these allergens at home, school, or work
•
Multicomponent interventions targeted to specific allergens in study populations consisting only of people with demonstrated sensitization and exposure to those allergens
•
Comparisons of different combinations of multicomponent interventions to determine the optimal combination(s) of allergen-specific mitigation strategies that improve outcomes
•
Studies to determine the allergen reduction thresholds for symptoms
•
Interactions and necessity of exposure, sensitization, and symptoms to determine which individuals with asthma will benefit most from allergen mitigation strategies (eg, whether an allergen-specific mitigation strategy is beneficial for an individual with asthma who has sensitization on skin prick testing to an allergen, is exposed to that allergen, and denies having symptoms)
In addition, reports of studies on the effectiveness of allergen mitigation interventions must include details on the intervention studied (eg, the models of air purifiers used) and the protocols for using the intervention (eg, how often the air purifier was turned on, where it was located, and how often the filter was changed). These aspects of the intervention need to be measured, and levels of adherence to the protocol need to be reported.
Section IV: Recommendations for the use of intermittent ICS in the treatment of asthma
Background
Scheduled, daily ICS treatment is the currently preferred pharmacologic controller therapy for persistent asthma in individuals of all ages.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
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Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma (EPR-3), published in 2007, suggested that intermittent ICS dosing schedules may be useful in some settings, but the evidence at that time was insufficient to support a recommendation in favor of this treatment beyond a recommendation based on expert consensus.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
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Definitions of terms used in this section
“Intermittent” ICS dosing in this section includes courses of ICS treatment used for brief periods, usually in response to symptoms or as an add-on with or without a LABA. “Intermittent ICS dosing” does not refer to a single regimen, and its definition is specified in each of the recommendations. Intermittent ICS dosing allows providers to prescribe specific doses, frequencies, and durations of ICS use. When to use intermittent ICS dosing could depend on an individual’s decision (based on need, which is also known as “as-needed” or “pro re nata” dosing), a predefined index showing worsening asthma, or some other predefined criterion.
“Controller therapy” refers to medications that are taken daily on a long-term basis to achieve and maintain control of persistent asthma.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
Available at:
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Both controller therapy and intermittent dosing may involve daily use of a specific dose of an ICS. The terms “ICS-LABA” (inhaled corticosteroid and long-acting beta 2-agonist combination, typically in a single device) and “ICS-formoterol” indicate combination therapy with both an ICS and a LABA, usually and preferably in a single inhaler.
“Quick-relief” therapy refers to medications (eg, an inhaled SABA) used to treat acute symptoms or exacerbations.132
132.
National Asthma Education and Prevention Program. Third Expert Panel on the Diagnosis and Management of Asthma
Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
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In this section, “as-needed” dosing (eg, of a SABA) is intermittent and is based on the patient’s decision (Figs 2-4).
The definitions of “low-,” “medium-,” and “high-dose” ICS are based on the recommendations from EPR-3.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
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The term “puff” refers to a single actuation and inhalation of a medication delivered through any type of inhaler.
“Recurrent wheezing” as used for the studies included in this section is defined as three or more episodes of wheezing triggered by apparent respiratory tract infections in a child’s lifetime or two episodes in the past year.
Overview of key questions and recommendations for intermittent ICS use
Given the range of options for intermittent ICS dosing and the number of comparisons embedded in the three key questions for this priority topic, the Expert Panel made five recommendations for intermittent ICS use to address these key questions. The majority of the studies in the systematic review report6
6.
Sobieraj DM, Baker WL, Weeda ER, Nguyen E, Coleman CI, White CM, et al. Intermittent inhaled corticosteroids and long-acting muscarinic antagonists for asthma. Comparative Effectiveness Review No. 194. (Prepared by the University of Connecticut Evidence-based Practice Center under Contract No. 290-2015-00012-I). AHRQ Publication No. 17(18)-EHC027-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Available at: Posted final reports are located on the Effective Health Care Program search page.
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on this topic used comparative efficacy designs as opposed to comparative effectiveness designs.
Table IV provides an overview of the questions on this topic, interventions and comparators that the Expert Panel considered, and resulting recommendations. As shown, in the opinion of the Expert Panel, the evidence was insufficient to support recommendations for all of the comparators in the questions.
| Question | Intervention | Comparator | Recommendation | Certainty Of evidence |
--- ---
| 4.1 | Short-course daily ICS+ as-needed SABA at start of RTI (Step 1) | As-needed SABA alone | Recommendation 9: Conditional, in favor of the intervention for ages 0-4 y | High |
| | | Daily ICS | No recommendation∗ | |
| | | No therapy | No recommendation∗ | |
| 4.2 | As-needed, concomitantly administered ICS+ SABA | Daily ICS+ as-needed SABA (Step 2) | Recommendation 10: Conditional, in favor of either the intervention or the comparator for ages 12 y and above | Moderate |
| | | | No recommendation∗ for ages 4-11 y | |
| | Intermittent, higher-dose ICS | | Recommendation 11: Conditional, against the intervention for ages 4 y and above | Low |
| 4.3 | Daily and as-needed ICS-formoterol (Steps 3 and 4) | Daily same-dose ICS+ as-needed SABA | No recommendation∗ for ages 4 y and above | |
| | | Daily higher-dose ICS+ as-needed SABA | Recommendation 12: Strong, in favor of the intervention for ages 4 y and above | Moderate for ages 4-11 y |
| | | | | High for ages 12 y and above |
| | | Daily same-dose ICS-LABA+ as-needed SABA | Recommendation 12: Strong, in favor of the intervention for ages 4 y and above | Moderate for ages 4-11 y |
| | | | | High for ages 12 y and above |
| | | Daily higher-dose ICS-LABA+ as-needed SABA | No recommendation∗ for ages 4-11 y | |
| | | | Recommendation 13: Conditional, in favor of the intervention for ages 12 y and above | High for ages 12 y and above |
Table IV
ICS key questions and recommendations
RTI, Respiratory tract infection.
∗
Insufficient evidence.
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In the remainder of this section, each key question is followed by recommendations that are relevant to the question, the evidence that supports the recommendation, and guidance for implementing each recommendation. The Expert Panel did not address the efficacy and safety of the following types of intermittent ICS treatment because they were not mentioned in the key questions:
•
As-needed ICS-formoterol versus as-needed SABA in Step 1 (intermittent asthma) or Steps 5 and 6 (severe asthma) treatment (Figs 2-4)
•
As-needed ICS-formoterol versus low-dose ICS treatment and as-needed SABA in Step 2 (mild persistent asthma) treatment (Figs 2-4)
Question 4.1
•
What is the comparative effectiveness of intermittent ICS compared to no treatment, pharmacologic therapy, or nonpharmacologic therapy in children aged 0 to 4 years with recurrent wheezing?
Recommendation 9: In children aged 0 to 4 years with recurrent wheezing triggered by respiratory tract infections and no wheezing between infections, the Expert Panel conditionally recommends starting a short course of daily ICS at the onset of a respiratory tract infection with as-needed SABA for quick-relief therapy compared to as-needed SABA for quick-relief therapy only.
Conditional recommendation, high certainty of evidence
Implementation guidance
Clinician’s Summary: This recommendation is for children aged 0 to 4 years who have had three or more episodes of wheezing triggered by apparent respiratory tract infections in their lifetime or who have had two such episodes in the past year and are asymptomatic between respiratory tract infections. For this population, the Expert Panel recommends a short (7-10-day) course of ICS daily along with as-needed SABA for quick-relief therapy starting at the onset of signs and symptoms indicating a respiratory tract infection. Respiratory tract infections were not confirmed by culture or PCR in the studies, and no further details on wheezing were provided.
The Expert Panel makes the following suggestions for implementation of intermittent ICS dosing in children aged 0 to 4 years:
•
One regimen used in two studies133
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is budesonide inhalation suspension, 1 mg, twice daily for 7 days at the first sign of respiratory tract infection–associated symptoms.
•
Although the efficacy of intermittent ICS dosing has high certainty of evidence, data regarding effects on growth are conflicting. Clinicians should carefully monitor length or height in children treated with the recommended regimen.
•
Caregivers can initiate intermittent ICS treatment at home without a visit to a health care provider when they have clear instructions. Clinicians should give caregivers written instructions on how to implement the recommended action plan at the onset of a respiratory infection. In addition, clinicians should review the plan with the caregiver at regular intervals.
•
Clinicians should consider this intervention in children who are not taking daily asthma treatment at the first sign of respiratory tract infection–associated symptoms.
•
What clinicians should discuss with caregivers:
–
Caregivers should be confident in the use of the asthma action plan because they will need to decide when to start treatment (ie, at the onset of a respiratory tract infection).
–
The main potential benefit of intermittent ICS use during respiratory tract infections is the reduction in exacerbations requiring systemic corticosteroids. Clinicians should inform caregivers that this treatment could affect growth, and they should carefully monitor growth in children who use this recommended treatment. Clinicians should reconsider implementing this recommended treatment if any evidence shows a reduced growth rate that cannot be attributed to other factors (eg, oral corticosteroid treatment). As part of shared decision making, some parents may weigh the potential benefits and harms differently and may not choose this therapy because of concerns related to their child’s growth.
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life) and one important outcome (rescue medication use) for this question. The summary of evidence for Recommendation 9 is in EtD Table XIII in Appendix B.
Three RCTs with high certainty of evidence133
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Episodic use of an inhaled corticosteroid or leukotriene receptor antagonist in preschool children with moderate-to-severe intermittent wheezing
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did not find an effect on linear growth of budesonide inhalation suspension (1 mg twice daily for 7 days) in comparison with placebo in children with an “identified respiratory tract illness.” Whether these differences in growth effects were due to differences in drugs, doses, duration of treatment, or other factors is not clear.
Rationale and discussion
The main comparator for which data are available is SABA-only therapy. The demonstrated efficacy but conflicting data regarding the effect of a short course of a daily ICS with SABA for quick-relief therapy on growth led the Expert Panel to develop a conditional recommendation for this therapy starting at the onset of an apparent respiratory tract infection for children aged 0 to 4 years with recurrent wheezing. Although one study that compared short ICS courses with regular daily ICS treatment showed no differences in exacerbations requiring systemic corticosteroids with moderate certainty of evidence, the Expert Panel made no recommendation based on this comparison because this study was not adequately powered to demonstrate equivalence.134
134.
Zeiger, R.S. ∙ Mauger, D. ∙ Bacharier, L.B. ...
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No studies produced robust data on comparisons of intermittent ICS use with no treatment or a nonpharmacologic therapy.
Question 4.2
•
What is the comparative effectiveness of intermittent ICS compared to ICS controller therapy in individuals aged 5 years and older with persistent asthma?
Recommendation 10: In individuals aged 12 years and older with mild persistent asthma, the Expert Panel conditionally recommends either daily low-dose ICS and as-needed SABA for quick-relief therapy or as-needed ICS and SABA used concomitantly.
Conditional recommendation, moderate certainty of evidence
Implementation guidance
Clinician’s Summary: For individuals aged 12 years and older with mild persistent asthma, the Expert Panel recommends either of the following two treatments as part of Step 2 therapy: a daily low-dose ICS and as-needed SABA for quick-relief therapy or intermittent as-needed SABA and an ICS used concomitantly (ie, one after the other) for worsening asthma. In this recommendation, “intermittent” ICS dosing is defined as the temporary use of an ICS in response to worsening asthma in an individual with asthma who is not taking ICS controller therapy regularly. This recommendation does not apply to ages 5 to 11 years because this therapy has not been adequately studied in this age group.
The Expert Panel makes the following suggestions for implementation of intermittent ICS dosing in individuals aged 12 years and older:
•
Individuals aged 12 years and older with mild persistent asthma who are not taking asthma treatment may benefit from this therapy. The Expert Panel has made no recommendation for children aged 0 to 4 years or 5 to 11 years with mild persistent asthma because of insufficient evidence.
•
Individuals aged 12 years and older with asthma and a low or high perception of symptoms may not be good candidates for as-needed ICS therapy. Regular low-dose ICS with SABA for quick-relief therapy may be preferred for such patients to avoid ICS undertreatment (low symptom perception) or overtreatment (high symptom perception).
•
Based on the regimen assessed in three of the four studies on intermittent ICS dosing,40
40.
Calhoun, W.J. ∙ Ameredes, B.T. ∙ King, T.S. ...
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,137
137.
Camargos, P. ∙ Affonso, A. ∙ Calazans, G. ...
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,138
138.
Papi, A. ∙ Canonica, G.W. ∙ Maestrelli, P. ...
Rescue use of beclomethasone and albuterol in a single inhaler for mild asthma
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one approach to intermittent therapy is two to four puffs of albuterol followed by 80 to 250 μg of beclomethasone equivalent every 4 hours as needed for asthma symptoms. In these studies, the clinician determined the dosing a priori. Currently, these medications need to be administered sequentially in two separate inhalers, but combination inhalers with albuterol and an ICS may be available in the United States in the future.
•
Individuals who use this type of therapy can initiate intermittent therapy at home. However, they should receive regular follow-up to ensure that the intermittent regimen is still appropriate.
•
What clinicians should discuss with patients and families:
–
Clinicians should inform individuals that the two treatment options do not have different effects on asthma control, asthma quality of life, or the frequency of asthma exacerbations when studied in large groups of people. Similarly, side effects are equally infrequent with daily and intermittent use.
–
Shared decision making will allow the best choice to be made for a particular individual.
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life) and one important outcome (rescue medication use) for this question. The summary of evidence for Recommendation 10 can be found in EtD Table XIV in Appendix B.
The studies showed no differences in asthma control, quality of life, or use of rescue therapy with the two types of intermittent ICS therapy (ICS paired with albuterol in two studies and ICS for worsening asthma symptoms in one study) and daily ICS treatment in three studies with high certainty of evidence in individuals aged 12 years and older.40
40.
Calhoun, W.J. ∙ Ameredes, B.T. ∙ King, T.S. ...
Comparison of physician-, biomarker-, and symptom-based strategies for adjustment of inhaled corticosteroid therapy in adults with asthma: the BASALT randomized controlled trial
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,139
139.
Boushey, H.A. ∙ Sorkness, C.A. ∙ King, T.S. ...
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The three studies also showed no differences in numbers of exacerbations between groups, but the strength of evidence on exacerbations was low. However, none of these studies was powered as an equivalence study, so the Expert Panel issued a conditional recommendation.
The Expert Panel made no recommendation for children aged 4 to 11 years because only low certainty of evidence was available from one small study by Martinez et al140
140.
Martinez, F.D. ∙ Chinchilli, V.M. ∙ Morgan, W.J. ...
Use of beclomethasone dipropionate as rescue treatment for children with mild persistent asthma (TREXA): a randomised, double-blind, placebo-controlled trial
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that addressed this question in this age group (EtD Table XV). Although the systematic review report6
6.
Sobieraj DM, Baker WL, Weeda ER, Nguyen E, Coleman CI, White CM, et al. Intermittent inhaled corticosteroids and long-acting muscarinic antagonists for asthma. Comparative Effectiveness Review No. 194. (Prepared by the University of Connecticut Evidence-based Practice Center under Contract No. 290-2015-00012-I). AHRQ Publication No. 17(18)-EHC027-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Available at: Posted final reports are located on the Effective Health Care Program search page.
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included one study in children aged 5 to 10 years, this study was not included in the EtD tables. In that study, all children received regular ICS treatment for 6 months. For the next 12 months, children were randomized to receive either intermittent ICS treatment or continued daily low-dose ICS treatment. Children in the continuous ICS group experienced significantly fewer exacerbations per individual (0.97) than those in the intermittent group (1.69; P= .008). However, the intermittent group had a greater increase in height after 6 months than the group that maintained regular therapy during months 6 to 18.141
141.
Turpeinen, M. ∙ Nikander, K. ∙ Pelkonen, A.S. ...
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The Expert Panel concluded that the use of regular ICS therapy for 6 months before intermittent therapy made this study’s results difficult to interpret in the context of the key question.
Rationale and discussion
Outcomes did not differ in the groups treated with the two alternate regimens in the three studies40
40.
Calhoun, W.J. ∙ Ameredes, B.T. ∙ King, T.S. ...
Comparison of physician-, biomarker-, and symptom-based strategies for adjustment of inhaled corticosteroid therapy in adults with asthma: the BASALT randomized controlled trial
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,138
138.
Papi, A. ∙ Canonica, G.W. ∙ Maestrelli, P. ...
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,139
139.
Boushey, H.A. ∙ Sorkness, C.A. ∙ King, T.S. ...
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in individuals aged 12 years and older. However, because none of these studies was powered as an equivalence study, the Expert Panel made a conditional recommendation. Although the studies had high certainty of evidence for asthma control and quality of life, they had low certainty of evidence for exacerbations and, taken together, resulted in overall low certainty for the recommendation statement. The Expert Panel made no recommendation based on this comparison for children aged 4 to 11 years because the only small included study in this population had low certainty of evidence, and one additional study had a study design that precluded evaluation for this key question.
Recommendation 11: In individuals aged 4 years and older with mild to moderate persistent asthma who are likely to be adherent to daily ICS treatment, the Expert Panel conditionally recommends against a short-term increase in the ICS dose for increased symptoms or decreased peak flow.
Conditional recommendation, low certainty of evidence
Implementation guidance
Clinician’s Summary: This recommendation addresses temporary increases in the dose of an ICS that is otherwise taken as controller therapy in response to worsening asthma. For this recommendation, a short-term increase in ICS dose refers to a doubling, quadrupling, or quintupling of the regular daily dose. For individuals aged 4 years and older with mild to moderate persistent asthma who are likely to adhere to their daily ICS treatment, the Expert Panel does not recommend doubling, quadrupling, or quintupling the ICS dose for increased symptoms or decreased peak flow. Clinicians can consider quadrupling the regular daily dose for individuals aged 16 years and older whose adherence to daily therapy is not assured (see the Discussion section below).
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life) and one important outcome (rescue medication use) for this question. The summary of evidence for Recommendation 11 can be found in EtD Table XVI in Appendix B.
In children aged 4 to 11 years, increasing the ICS dose temporarily in response to worsening symptoms did not significantly reduce the rate of exacerbations or improve asthma quality of life in one study by Martinez et al.140
140.
Martinez, F.D. ∙ Chinchilli, V.M. ∙ Morgan, W.J. ...
Use of beclomethasone dipropionate as rescue treatment for children with mild persistent asthma (TREXA): a randomised, double-blind, placebo-controlled trial
Lancet. 2011; 377:650-657
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The overall certainty of evidence ranged from low for exacerbations to moderate for quality of life. A more recent study in 254 children by Jackson et al142
142.
Jackson, D.J. ∙ Bacharier, L.B. ∙ Mauger, D.T. ...
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also found no difference in the rate of exacerbations treated with systemic corticosteroids with a quintupling of the ICS dose at early signs of loss of asthma control. In this 48-week study, the growth rate in the intervention group was reduced, although this difference did not reach statistical significance (P= .06). The potential for growth suppression by the intervention and the absence of demonstrated efficacy of the intervention in the articles that the Expert Panel reviewed led to a recommendation against using this intervention in this age group. The Expert Panel rated the recommendation as conditional because of the limited number of studies available in this age group.
In individuals aged 12 years and older (EtD Table XVII), the intervention as implemented did not significantly reduce exacerbations or asthma hospitalizations. The certainty of evidence is low for both outcomes of exacerbations and asthma hospitalizations in the systematic review report. A large, more recent study by McKeever et al143
143.
McKeever, T. ∙ Mortimer, K. ∙ Wilson, A. ...
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showed a modest but significant reduction in time to severe exacerbation and in the rate of use of systemic corticosteroids in individuals with asthma whose action plan included a quadrupling of the ICS dose.143
143.
McKeever, T. ∙ Mortimer, K. ∙ Wilson, A. ...
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However, unlike the studies in the systematic review report, this study did not include a placebo group or use blinding, and the baseline adherence rate was low. Specifically, only 50% of participants in the quadruple-dose group and 42% in the non–quadruple-dose group had good adherence, according to the investigators. Because of the low adherence rate, it was not clear whether the increased ICS dose was effective or whether the initiation of ICS treatment in nonadherent participants influenced the results. Thus, based on the lack of efficacy in the studies in the systematic review report and the possible growth effects, the Expert Panel made a recommendation against a short-term increase in the ICS dose.
In the reviewed studies, the indication for increasing the ICS dose was decreased peak flow and/or increased symptoms. When increased, the ICS dose was doubled, quadrupled, or quintupled.142-146
142.
Jackson, D.J. ∙ Bacharier, L.B. ∙ Mauger, D.T. ...
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143.
McKeever, T. ∙ Mortimer, K. ∙ Wilson, A. ...
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144.
Harrison, T.W. ∙ Oborne, J. ∙ Newton, S. ...
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145.
Lahdensuo, A. ∙ Haahtela, T. ∙ Herrala, J. ...
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146.
Oborne, J. ∙ Mortimer, K. ∙ Hubbard, R.B. ...
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Rationale and discussion
In children aged 4 to 11 years, the intervention did not significantly reduce exacerbations or improve asthma quality of life in one study140
140.
Martinez, F.D. ∙ Chinchilli, V.M. ∙ Morgan, W.J. ...
Use of beclomethasone dipropionate as rescue treatment for children with mild persistent asthma (TREXA): a randomised, double-blind, placebo-controlled trial
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in the systematic review report. The intervention’s potential to suppress growth in a more recent study142
142.
Jackson, D.J. ∙ Bacharier, L.B. ∙ Mauger, D.T. ...
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and the lack of demonstrated efficacy of the intervention in either of the reviewed articles led to the Expert Panel’s recommendation against this intervention in this age group.
In individuals aged 12 years and older, the intervention as implemented also did not significantly reduce exacerbations in three studies144-146
144.
Harrison, T.W. ∙ Oborne, J. ∙ Newton, S. ...
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145.
Lahdensuo, A. ∙ Haahtela, T. ∙ Herrala, J. ...
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146.
Oborne, J. ∙ Mortimer, K. ∙ Hubbard, R.B. ...
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in the evidence summary, but the certainty of evidence is low. The more recent study by McKeever et al143
143.
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showed modest but significant reductions in time to severe exacerbation and rate of ICS use in individuals whose action plan included a quadrupling of the ICS dose. However, unlike the studies in the AHRQ systematic review report, this study did not include a placebo group or use blinding, and the baseline adherence rate was low (42%-50%). The adherence rate in the McKeever et al study might be more similar to the adherence rates in routine clinical practice, whereas adherence rates in the RCTs144-146
144.
Harrison, T.W. ∙ Oborne, J. ∙ Newton, S. ...
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145.
Lahdensuo, A. ∙ Haahtela, T. ∙ Herrala, J. ...
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146.
Oborne, J. ∙ Mortimer, K. ∙ Hubbard, R.B. ...
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were probably higher than in most real-world settings.
Thus, the Expert Panel believes that this recommendation applies most specifically to individuals who are likely to adhere to their daily ICS regimen. An increase in the ICS dose might be a reasonable strategy to include in the action plans of individuals whose adherence rates are less certain. How to assess adherence or the threshold for adequate adherence for this recommendation cannot be determined from the reviewed studies. Based on the study of McKeever et al143
143.
McKeever, T. ∙ Mortimer, K. ∙ Wilson, A. ...
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in individuals aged 12 years and older described in the previous paragraph, the ICS dose could be quadrupled in the short-term in individuals aged 16 years and older in response to an increased need for reliever therapy, greater interference of asthma with sleep, or a peak flow of less than 80% of the individual’s normal level. The potential discrepancy between the efficacy and effectiveness studies described above and the overall low certainty of evidence led to a conditional recommendation for this age group as well.
Question 4.3
•
What is the comparative effectiveness of ICS with LABA used as both controller and quick-relief therapy compared to ICS with or without LABA used as controller therapy in individuals aged 5 years and older with persistent asthma?
Recommendation 12: In individuals aged 4 years and older with moderate to severe persistent asthma, the Expert Panel recommends ICS-formoterol in a single inhaler used as both daily controller and reliever therapy compared to either a higher-dose ICS as daily controller therapy and SABA for quick-relief therapy or the same-dose ICS-LABA as daily controller therapy and SABA for quick-relief therapy.
Strong recommendation, high certainty of evidence for ages 12 years and above, moderate certainty of evidence for ages 4 to 11 years
Implementation guidance
Clinician’s Summary: In individuals aged 4 years and older, the preferred Step 3 (low-dose ICS) and Step 4 (medium-dose ICS) therapy is single-inhaler ICS-formoterol both daily and as needed. In the literature, inhaled ICS-formoterol is referred to as “single maintenance and reliever therapy (SMART).” This form of therapy has only been used with formoterol as the LABA. Formoterol has a rapid onset and a maximum total daily dose that allows it to be used more than twice daily.147
147.
Peters, M.
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The maximum total daily dose of formoterol should not exceed 8 puffs (36 μg) for ages 4 to 11 years and 12 puffs (54 μg) for ages 12 years and above. SMART is administered with a single inhaler containing both formoterol and an ICS (primarily budesonide in the reviewed studies, but one study used beclomethasone). The regimens compared to address this key question required two inhalers: the controller (ICS or ICS-LABA) and the reliever (SABA). The recommended alternate therapy of maintenance ICS-LABA with SABA as quick-relief therapy does not need to be changed if it is providing adequate control. However, patients whose asthma is uncontrolled on such therapy should receive the preferred SMART if possible before moving to a higher step of therapy.
The Expert Panel makes the following suggestions for implementation of daily and intermittent combination ICS-formoterol in individuals aged 4 years and older:
•
No patient characteristics exclude consideration of this option in individuals aged 4 years and older with asthma.
•
The studies demonstrating reduced exacerbations (see below) enrolled individuals with a severe exacerbation in the prior year. The results suggest that such individuals are particularly good candidates for SMART to reduce exacerbations.
•
SMART might not be necessary for individuals whose asthma is well controlled on alternate treatments, such as conventional maintenance ICS-LABA with SABA as quick-relief therapy.
•
SMART is appropriate for Step 3 (low-dose ICS-formoterol) and Step 4 (medium-dose ICS-formoterol) treatment.
•
ICS-formoterol should be administered as maintenance therapy with one to two puffs once to twice daily (depending on age, asthma severity, and ICS dose in the ICS-formoterol preparation) and one to two puffs as needed for asthma symptoms. The maximum number of puffs per day is 12 (54 μg formoterol) for individuals aged 12 years and older and 8 (36 μg formoterol) for children aged 4 to 11 years. Clinicians should advise individuals with asthma or their caregivers to contact their physician if they need to use more than these amounts.
•
The calculation of the dose of formoterol was based on 4.5 μg/inhalation, the most common preparation used in the RCTs reviewed.
•
ICS-formoterol should not be used as quick-relief therapy in individuals taking ICS-salmeterol as maintenance therapy.
•
What clinicians should discuss with their patients and families:
−
Clinicians should inform individuals with asthma and their caregivers that in studies, this intervention consistently reduced asthma exacerbations requiring unscheduled medical visits or systemic corticosteroids. In addition, this intervention improved asthma control and quality of life in some studies.
−
No differences have been documented in harms between this type of therapy and the comparators (ICS or ICS-LABA) in individuals aged 12 years and older. The reductions in exposure to oral corticosteroids and to ICS treatment in most studies suggest that the intervention might reduce future corticosteroid-associated harms.
−
In children aged 4 to 11 years, there may be a lower risk of growth suppression among those taking SMART versus daily higher-dose ICS treatment.
−
This recommendation might not be appropriate for some individuals with asthma for such reasons as cost, formulary considerations, or medication intolerance. However, the additional cost of the medication may be offset by the decrease in exacerbations and the associated improvement in quality of life and reduction in costs to both the patient and the payer.
−
A 1-month supply of ICS-formoterol medication that is sufficient for maintenance therapy may not last a month if the inhaler is used for reliever therapy as well. Providers, individuals with asthma, pharmacists, and payers need to be aware of this possibility and prescribe, plan, dispense, or provide coverage accordingly.
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life) and one important outcome (asthma symptoms) for this question. The summary of evidence for Recommendation 12 can be found in EtD Tables XVIII and XIX in Appendix B.
SMART versus higher-dose ICS treatment in ages 4 years and older (EtD Table XVIII)
Three large RCTs148-150
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Scicchitano, R. ∙ Aalbers, R. ∙ Ukena, D. ...
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(total N= 4662) enrolled individuals aged 12 years and older who were being treated with a low- to medium-dose or medium- to high-dose ICS. Study participants treated with SMART used daily budesonide-formoterol, 160/9 to 320/9 μg, via a dry-powder inhaler. They took up to 10 rescue puffs of budesonide-formoterol (total daily dose of 12 puffs or 54 μg formoterol). The investigators compared this intervention to daily budesonide, 320 to 640 μg, along with SABA for quick-relief therapy. Rabe et al149
149.
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showed a 51% RR reduction in exacerbations, whereas the rates were 35% and 43% RR reduction in Scicchitano et al150
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Scicchitano, R. ∙ Aalbers, R. ∙ Ukena, D. ...
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and O’Byrne et al,148
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respectively. The latter two studies used a composite exacerbation score that included systemic corticosteroid use, hospitalizations, emergency department visits, increase in ICS or other medication doses, and peak expiratory flow less than 70%.148-150
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O’Byrne, P.M. ∙ Bisgaard, H. ∙ Godard, P.P. ...
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Collectively, these RCTs found an RR of 0.6 (range, 0.53-0.68) favoring SMART for asthma exacerbations (high certainty of evidence). The investigators of these studies did not report results from validated outcome measures of quality of life or asthma control. However, results for individual asthma control measures—including total asthma symptom scores, nighttime awakenings, symptom-free days, and asthma control days—significantly favored SMART. The overall doses of inhaled and oral corticosteroids were significantly lower with SMART (two- to fourfold less for inhaled ICS treatments).
Jenkins et al151
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conducted a post hoc analysis of these three studies in 1239 participants aged 12 years and older with milder asthma (daily maintenance ICS dose equal to 400 μg or less budesonide equivalent). The authors confirmed that SMART reduced exacerbations overall. However, in subgroup analyses, participants with the mildest asthma at enrollment (based on rescue SABA use of <1 inhalation/d) showed a marginal and statistically nonsignificant benefit.
Another post hoc analysis of one of the three RCTs (O’Byrne et al148
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) included 224 children aged 4 to 11 years who used medium to high ICS doses (any brand, 200-500 μg daily). The 118 participants in the SMART group were instructed to take budesonide-formoterol, 80/4.5 μg once daily, as their baseline therapy, with up to seven additional rescue puffs (total daily dose of 36 μg formoterol). The other 106 participants took budesonide, 320 μg daily, with rescue SABA. In the SMART group, the RR for a composite exacerbation measure comprised of systemic corticosteroids, hospitalization, emergency department visits, and increase in ICS or other medication dose dropped by 57% (moderate certainty of evidence). The authors did not report on validated outcome measures of quality of life or asthma control, but nighttime awakenings declined significantly with SMART. SMART participants used a lower daily ICS dose (average 127 vs 320 μg/d in the fixed-dose budesonide group) and demonstrated significantly improved growth rates (adjusted mean difference of 1 cm compared with fixed-dose budesonide).152
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SMART versus same-dose ICS-LABA controller therapy for ages 4 years and above (EtD Table XIX)
For ages 12 years and above, the Expert Panel considered four blinded RCTs148
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and two unblinded RCTs156
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for this question. Collectively, these RCTs demonstrated a 32% reduction in exacerbations with SMART148
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(high certainty of evidence). Two of the studies used validated asthma control measures (ACQ-5), and both demonstrated clinically significant improvements with SMART (high certainty of evidence).155
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Three of the blinded studies enrolled a total of 7555 participants with mild to severe persistent asthma. Participants were treated with 160/9 or 320/9 μg budesonide-formoterol daily with up to 10 rescue puffs (total daily dose of 12 puffs or 54 μg formoterol) of budesonide-formoterol (SMART) or rescue SABA.148
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In these three blinded studies, SMART significantly reduced exacerbations.
One of these three studies153
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demonstrated a statistically significant improvement in asthma control (based on ACQ-5). A second blinded study (N= 1748) enrolled participants aged 18 years or older with poorly controlled asthma who took a moderate to high dose of an ICS or ICS-LABA. The SMART group took two puffs daily of beclomethasone-formoterol, 100/6 μg, and up to six puffs of rescue beclomethasone-formoterol per day (total daily dose of 48 μg formoterol). The comparison group used rescue SABA. The investigators actively managed both arms with dose titration. Although severe exacerbations and systemic corticosteroid use were significantly lower with SMART, asthma control scores (ACQ-7) did not differ significantly between groups.154
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An unblinded study, Vogelmeier et al,157
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enrolled 2143 participants from Europe and Asia with poorly controlled asthma taking moderate to high ICS or ICS-LABA doses (500 μg or more of budesonide, fluticasone, or equivalent). They received either daily budesonide-formoterol, 640/18 μg, with budesonide-formoterol rescue (SMART group) or daily fluticasone/salmeterol, 500/100 μg, with SABA for quick-relief therapy. The investigators actively managed both arms with dose titration, and the study was unblinded. With SMART, the RR declined by 20% for exacerbations, defined as emergency department visits, oral corticosteroid days, and hospitalization. SMART also improved asthma control (measured by ACQ-5) and quality of life (measured by Asthma-Related Quality of Life Questionnaire), but these changes were not statistically significant. A reanalysis of these data in 404 participants in China, Korea, Taiwan, and Thailand had similar results; the RR reduction in exacerbation rates was 38%.158
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Vogelmeier, C. ∙ Naya, I. ∙ Ekelund, J.
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enrolled 303 participants in New Zealand who were at risk of severe exacerbations. Participants were treated with budesonide-formoterol, 800/24 μg (by metered-dose inhaler), with one rescue puff of budesonide-formoterol (SMART) or SABA for quick-relief therapy. SMART reduced exacerbations and oral corticosteroid use but increased the use of ICS, and the associated improvement in asthma control (measured by ACQ-7) was not significant.156
156.
Patel, M. ∙ Pilcher, J. ∙ Pritchard, A. ...
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For ages 4 to 11 years, one blinded RCT152
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used budesonide-formoterol, 80/4.5 μg, with up to seven rescue puffs of budesonide-formoterol, 80/4.5 μg (36 μg total daily dose of formoterol; SMART), or SABA as quick-relief therapy. SMART reduced the RR for exacerbations by 72% (moderate certainty of evidence) and showed superiority in one unvalidated outcome measure of asthma control (nighttime awakenings). Growth rates and other safety measures did not differ between treatment groups.
Rationale and discussion
Because the only SMART studied has included formoterol, the Expert Panel’s recommendation favors the use of ICS-LABA combinations containing formoterol rather than those that contain ICS-salmeterol. Daily ICS-salmeterol remains an appropriate therapeutic option for individuals with moderate to severe persistent asthma, but the reviewed data suggest that the use of ICS-formoterol for maintenance and reliever therapy has superior efficacy, ease of use (because it is administered in a single inhaler rather than two separate inhalers), and perhaps safety as a result of reduced corticosteroid exposure. Other LABAs, including newer agents with a rapid onset, may be effective and safe to use for both maintenance and reliever therapy, but their efficacy and safety will need to be demonstrated in clinical studies. The number of studies available and the consistency of the evidence led the Expert Panel to make a strong recommendation to use ICS-formoterol in a single inhaler as both daily controller and reliever therapy.
Data were insufficient to compare ICS-formoterol as SMART with same-dose ICS for daily controller therapy along with SABA for quick-relief therapy in individuals aged 4 years and older. However, multiple studies have demonstrated that adding any LABA to the same ICS dose is more effective than ICS therapy alone.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
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Thus, the lack of comparisons data on ICS-formoterol as SMART versus same-dose ICS and SABA for quick-relief therapy is of minimal clinical importance.
Recommendation 13: In individuals aged 12 years and older with moderate to severe persistent asthma, the Expert Panel conditionally recommends ICS-formoterol in a single inhaler used as both daily controller and reliever therapy compared to higher-dose ICS-LABA as daily controller therapy and SABA for quick-relief therapy.
Conditional recommendation, high certainty of evidence
Implementation guidance
Clinician’s Summary: In individuals aged 12 years and older, the preferred Step 4 therapy is single-inhaler ICS-formoterol used both daily and as needed. The maximum total daily dose of formoterol should not exceed 12 puffs (54 μg) for those aged 12 years and older. The recommended alternate therapy of maintenance ICS-LABA along with SABA as quick-relief therapy does not need to be changed if it is providing adequate control. However, individuals whose asthma is uncontrolled on such therapy should receive the preferred SMART if possible before stepping up their treatment to a higher step of therapy.
In individuals aged 12 years and older with moderate to severe persistent asthma, combination ICS-formoterol used daily and intermittently is more beneficial than an increase in the daily ICS dose if they are already taking combination ICS-LABA (and as-needed SABA). The Expert Panel makes the following suggestions for implementation of daily and intermittent combination ICS-formoterol for individuals aged 12 years and older:
•
This recommendation applies to all individuals with asthma aged 12 years and older.
•
Individuals with asthma should use ICS-formoterol as maintenance therapy with one to two puffs once or twice daily (depending on asthma severity and ICS dose in the ICS-formoterol preparation). The additional rescue dose is 1 to 2 puffs as needed for asthma symptoms, up to a maximum of 12 puffs (54 μg formoterol) per day. Clinicians should advise individuals with asthma to contact their clinician if they need to use more than these amounts.
•
The calculation of the dose of formoterol was based on 4.5 μg/inhalation, the most common preparation used in the RCTs reviewed.
•
Clinicians managing asthma should regularly assess individuals using this therapy.
•
This therapy is appropriate for Step 4.
•
Individuals with asthma should not use ICS-formoterol as reliever therapy if they are taking ICS-salmeterol as maintenance therapy.
•
SMART might not be necessary for individuals whose asthma is well controlled with alternate treatments, such as conventional maintenance ICS-LABA with SABA as quick-relief therapy.
•
For individuals aged 5 to 11 years, the evidence was insufficient to make a recommendation regarding SMART compared to higher-dose ICS-LABA. SMART with low- or medium-dose ICS therapy is preferred for children aged 5 to 11 years as opposed to same-, low-, or medium-dose ICS-LABA plus as-needed SABA as part of Step 3 and Step 4 therapy (Recommendation 12).
•
What clinicians should discuss with their patients and families:
–
Clinicians should inform individuals with asthma and their caregivers that the major demonstrated benefits of combination ICS-formoterol used daily and as-needed are reductions in asthma exacerbations requiring unscheduled medical visits and in use of systemic corticosteroids.
–
Clinicians should also inform individuals with asthma that studies found no difference in documented harms between this type of therapy and daily ICS-LABA.
–
Studies showed that combination ICS-formoterol reduces exposure to corticosteroids, suggesting that the intervention might reduce future corticosteroid-associated harms.
–
This recommendation might not be appropriate for some individuals for such reasons as cost, formulary considerations, or medication intolerance.
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life) for this question. The summary of evidence for Recommendation 13 can be found in EtD Table XIX in Appendix B.
Two blinded RCTs (N= 5481) compared SMART to higher-dose ICS-LABA159
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in individuals with asthma aged 12 years and older. SMART reduced the RR by 25% for exacerbations (high certainty of evidence). SMART also resulted in statistically significant reductions in corticosteroid use but had no significant effect on asthma quality of life or asthma control. As a result, the recommendation was conditional.159
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compared daily budesonide-formoterol (640/18 μg) plus budesonide-formoterol reliever therapy (SMART) in participants aged 12 years and older with daily fluticasone-salmeterol (1000/100 μg) plus SABA for quick-relief therapy, while Kuna et al160
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compared daily budesonide-formoterol (320/9 μg) plus budesonide-formoterol reliever therapy (SMART) with either daily budesonide-formoterol (640/18 μg) or daily fluticasone-salmeterol (500/100 μg) plus SABA for quick-relief therapy. These two studies showed significant reductions in exacerbations in the SMART groups in comparison with maintenance ICS-LABA along with SABA for quick-relief therapy. However, the studies found no differences between groups in asthma control or quality of life, and the lack of differences in these outcomes led to the Expert Panel’s conditional recommendation. Data were insufficient to make a recommendation regarding whether SMART is superior to daily higher-dose ICS-LABA with SABA for quick-relief therapy in children aged 4 to 11 years.
The systematic review report for this topic also included five open-label, real-world clinical trials that compared daily budesonide-formoterol (160-320/4.5-9 μg) plus budesonide-formoterol reliever therapy (SMART) with conventional best practice treatment (total N= 5056).6
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Active management levels varied in these studies. Because of the heterogeneity of the studies and lack of information regarding the type of therapy prescribed and used in the conventional best practice arms, the formal systematic review report did not include these studies. However, the Expert Panel decided to review these studies to compare the potential benefits of SMART with those of diverse approaches in real-world settings. In general, the real-world studies confirmed the results from the RCTs that used SMART.
Future research opportunities
The Expert Panel identified the following topics that would benefit from additional research:
•
Differences by race and ethnicity in benefits and risks of the ICS recommendations
•
Cost-effectiveness of the ICS recommendations
•
Effects on growth of short ICS courses starting at the onset of an apparent respiratory tract infection in children aged 0 to 4 years who have recurrent wheezing triggered only by such infections
•
Optimal short-course ICS regimen to use—on the basis of efficacy, effectiveness, and safety—at the onset of an apparent respiratory tract infection in children aged 0 to 4 years whose recurrent wheezing is triggered by respiratory tract infections
•
Efficacy, effectiveness, and safety of a short ICS course starting at the onset of an apparent respiratory tract infection compared with daily ICS treatment in children aged 0 to 4 years with recurrent wheezing triggered by respiratory tract infections
•
Daily low-dose ICS treatment with SABA for quick relief versus as-needed ICS plus SABA administered concomitantly in children aged 4 to 11 years with mild persistent asthma
•
Optimal dose of albuterol and ICS used for as-needed concomitant therapy in individuals with mild persistent asthma
•
Effectiveness and safety of other rapid-onset LABAs in combination medications used for both daily controller and quick-relief therapy
•
Combination ICS-formoterol as both daily controller and reliever therapy compared with higher-dose ICS-LABA as daily controller therapy and SABA for quick-relief therapy in children aged 4 to 11 years
Other recommended types of research included the following:
•
Confirmation of the efficacy data supporting the ICS recommendations using additional real-world effectiveness studies in clearly defined populations using clearly defined treatment regimens
•
Additional studies powered as equivalence studies to confirm the finding that daily low-dose ICS therapy with SABA for quick relief and concomitant as-needed ICS therapy plus SABA lead to similar outcomes in individuals with mild persistent asthma
•
Real-world studies that monitor growth in children and adherence to evaluate the effectiveness and safety of quadrupling the ICS dose in individuals with mild to moderate persistent asthma taking daily ICS controller therapy who experience early signs of loss of asthma control
Section V: Recommendations for the use of LAMAs for asthma
Background
LAMAs comprise a pharmacologic class of long-acting bronchodilators. The role of LAMAs in the management of asthma was not addressed in Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
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Since that report’s publication in 2007, several trials have investigated LAMAs as controller therapy for individuals with asthma.
The Expert Panel examined the harms and benefits of LAMAs in individuals aged 12 years and older with uncontrolled persistent asthma and addressed three key questions.165
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Mensah, G.A. ∙ Kiley, J.P. ∙ Gibbons, G.H.
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The Expert Panel did not examine the role of LAMA treatment in children aged 6 to 11 years because the key questions and systematic reviews did not address this age group. With the exception of one study that examined the LAMA umeclidinium,166
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the RCTs reviewed by the Expert Panel used tiotropium bromide as the LAMA. At the time this report was written, tiotropium bromide (Respimat) was the only formulation of LAMA with US FDA approval for asthma treatment. The majority of LAMA studies used a comparative efficacy design, and not an effectiveness design, but the key questions were about effectiveness. Therefore, the clinical impact of LAMA treatment in real-world settings is not well understood. Table V provides an overview of the key questions and recommendations on LAMAs.
| Question | Intervention | Comparator | Recommendation | Certainty of evidence |
--- ---
| 5.1 | LAMA as an add-on to ICS controller therapy∗ | LABA as an add-on to same-dose ICS controller therapy∗ | 14: Conditional, against intervention | Moderate |
| | | Montelukast as an add-on to same-dose ICS controller therapy∗ | No recommendation† | |
| 5.2 | LAMA as an add-on to ICS controller therapy∗ | Same-dose ICS controller therapy∗+ placebo | 15: Conditional, in favor of the intervention | Moderate |
| | | Increased ICS dose | No recommendation† | |
| 5.3 | LAMA as an add-on to ICS-LABA | Same-dose ICS-LABA as controller therapy∗ | 16: Conditional, in favor of the intervention | Moderate |
| | | Doubled ICS dose+ LABA | No recommendation† | |
Table V
LAMA key questions and recommendations
∗
ICS controller therapy used daily.
†
Insufficient evidence.
Open table in a new tab
Definitions of terms used in this section
In this section, “controller therapy” refers to medications that are taken daily on a long-term basis to achieve and maintain control of persistent asthma.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
National Heart, Lung, and Blood Institute, National Institutes of Health, Bethesda, Md, August 2007 440
Available at:
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The term “ICS-LABA” indicates therapy with both an ICS and a LABA, usually (and preferably) in a single inhaler.
Question 5.1
•
What is the comparative effectiveness of LAMA compared with other controller therapy as add-on therapy to ICSs in individuals aged 12 years and older with uncontrolled persistent asthma?
Question 5.2
•
What is the comparative effectiveness of LAMA as add-on therapy to ICS controller therapy compared with placebo or increased ICS dose in individuals aged 12 years and older with uncontrolled persistent asthma?
Recommendation 14: In individuals aged 12 years and older with uncontrolled persistent asthma, the Expert Panel conditionally recommends against adding LAMA to ICS compared to adding LABA to ICS.
Conditional recommendation, moderate certainty of evidence
Recommendation 15: If LABA is not used, in individuals aged 12 years and older with uncontrolled persistent asthma, the Expert Panel conditionally recommends adding LAMA to ICS controller therapy compared to continuing the same dose of ICS alone.
Conditional recommendation, moderate certainty of evidence
Implementation guidance
Clinician’s Summary: In individuals with asthma that is not controlled by ICS therapy alone, the Expert Panel recommends adding a LABA rather than a LAMA to an ICS. However, if the individual is not using or cannot use LABA therapy, adding a LAMA to an ICS is an acceptable alternative. Adding a LAMA to ICS controller therapy is more effective than using ICS controller therapy alone in individuals aged 12 years and older with uncontrolled persistent asthma. However, adding a LAMA to ICS controller therapy is not more efficacious than adding a LABA to ICS controller therapy, and adding a LAMA may increase the risk of harm, based on a single real-world study in Blacks.167
167.
Wechsler, M.E. ∙ Yawn, B.P. ∙ Fuhlbrigge, A.L. ...
Anticholinergic vs long-acting beta-agonist in combination with inhaled corticosteroids in black adults with asthma: the BELT randomized clinical trial
JAMA. 2015; 314:1720-1730
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Therefore, the panel recommends preferentially adding LABA over LAMA to ICS. A LABA should not be used when the individual cannot tolerate it, the medication is contraindicated, the device that delivers the LABA is unsuitable for the individual, or the LABA is unavailable for insurance or supply reasons.
The Expert Panel makes the following suggestions on the use of LAMA therapy:
•
A LAMA can be used as an add-on to ICS therapy in individuals aged 12 years and older with uncontrolled asthma therapy as part of Step 4 therapy, but add-on LABA therapy has a more favorable benefit-harm profile.
•
Individuals at risk of urinary retention and those who have glaucoma should not receive LAMA therapy.
•
The small increase in the potential risk of harms from a LAMA may outweigh its benefits in some individuals, particularly in Blacks.
•
LAMA treatment requires appropriate use of specific inhaler devices. Clinicians should teach individuals with asthma how to use these devices appropriately.
•
When clinicians prescribe LAMA therapy, they should prescribe this medication for long-term asthma control in the ambulatory setting. LAMA therapy does not have a role in the management of acute exacerbations of asthma in the ambulatory, emergency department, or inpatient settings.
•
Clinicians should confirm the asthma diagnosis and address factors that often contribute to uncontrolled asthma before they consider intensifying therapy by adding a LAMA. For example, clinicians should identify and suggest ways to mitigate occupational and environmental triggers and ensure that individuals with asthma are using currently prescribed asthma controller therapy appropriately.
•
What clinicians should discuss with their patients about LAMA therapy
–
When discussing the addition of a LAMA versus a LABA for individuals already taking an ICS, clinicians should explain that the LABA is likely to be preferable.
–
Adding a LAMA to ICS controller therapy provides no more benefit than adding a LABA to ICS controller therapy, and may increase the risk of harm, based on a single real-world study in Blacks.
–
Clinicians should tell individuals with asthma that adding a LAMA to an ICS provides a small benefit compared to continuing the same ICS dose if the individual cannot use a LABA for any reason.
–
Individuals with asthma and glaucoma and those at risk of urinary retention should not use LAMA therapy.
Summary of the evidence
The Expert Panel prespecified three critical outcomes (exacerbations, asthma control, and quality of life) and three important outcomes (rescue medication use, adverse events [harms], and mortality). The Expert Panel did not consider lung function (eg, based on spirometry testing) to be a critical or important outcome for the LAMA studies that it reviewed.
The summary of evidence for Recommendation 14 can be found in EtD Table XX in Appendix B. The Expert Panel examined the efficacy of adding a LAMA to ICS therapy in comparison with adding a LABA to ICS therapy in seven RCTs.166-172
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
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J Allergy Clin Immunol. 2011; 128:315-322
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Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
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Rajanandh, M.G. ∙ Nageswari, A.D. ∙ Ilango, K.
Pulmonary function assessment in mild to moderate persistent asthma patients receiving montelukast, doxofylline, and tiotropium with budesonide: a randomized controlled study
Clin Ther. 2014; 36:526-533
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172.
Rajanandh, M.G. ∙ Nageswari, A.D. ∙ Ilango, K.
Assessment of montelukast, doxofylline, and tiotropium with budesonide for the treatment of asthma: which is the best among the second-line treatment? A randomized trial
Clin Ther. 2015; 37:418-426
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Five RCTs166
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
The effect of fluticasone furoate/umeclidinium in adult patients with asthma: a randomized, dose-ranging study
Respir Med. 2015; 109:54-62
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168.
Peters, S.P. ∙ Kunselman, S.J. ∙ Icitovic, N. ...
Tiotropium bromide step-up therapy for adults with uncontrolled asthma
N Engl J Med. 2010; 363:1715-1726
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PubMed
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169.
Bateman, E.D. ∙ Kornmann, O. ∙ Schmidt, P. ...
Tiotropium is noninferior to salmeterol in maintaining improved lung function in B16-Arg/Arg patients with asthma
J Allergy Clin Immunol. 2011; 128:315-322
Full Text
Full Text (PDF)
Scopus (159)
PubMed
Google Scholar
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
Full Text
Full Text (PDF)
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that had a total of 2574 participants found no difference in the exacerbation rate in individuals treated with a LAMA compared with those treated with a LABA (RR= 0.87; 95% CI, 0.53-1.42) as an add-on to an ICS. The exacerbation rate was 4.9% (75 of 1533) in the LAMA group and 5.4% (56 of 1041) in the LABA group (absolute risk difference of 7 fewer per 1000; 95% CI, from 25 fewer to 23 more). The certainty of evidence is moderate for the effect on exacerbations.
Two RCTs170
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
Full Text
Full Text (PDF)
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in 1577 patients detected no differences in asthma control between those treated with a LAMA and those treated with a LABA. The certainty of evidence is high for the lack of improvement in asthma control.
Four RCTs168-170
168.
Peters, S.P. ∙ Kunselman, S.J. ∙ Icitovic, N. ...
Tiotropium bromide step-up therapy for adults with uncontrolled asthma
N Engl J Med. 2010; 363:1715-1726
Crossref
Scopus (418)
PubMed
Google Scholar
169.
Bateman, E.D. ∙ Kornmann, O. ∙ Schmidt, P. ...
Tiotropium is noninferior to salmeterol in maintaining improved lung function in B16-Arg/Arg patients with asthma
J Allergy Clin Immunol. 2011; 128:315-322
Full Text
Full Text (PDF)
Scopus (159)
PubMed
Google Scholar
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
Full Text
Full Text (PDF)
Scopus (125)
PubMed
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in 1982 patients found no differences in asthma-related quality of life between those treated with a LAMA and those treated with a LABA. The certainty of evidence is high for the lack of effect on asthma-related quality of life.
Six RCTs166
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
The effect of fluticasone furoate/umeclidinium in adult patients with asthma: a randomized, dose-ranging study
Respir Med. 2015; 109:54-62
Full Text
Full Text (PDF)
PubMed
Google Scholar
,167
167.
Wechsler, M.E. ∙ Yawn, B.P. ∙ Fuhlbrigge, A.L. ...
Anticholinergic vs long-acting beta-agonist in combination with inhaled corticosteroids in black adults with asthma: the BELT randomized clinical trial
JAMA. 2015; 314:1720-1730
Crossref
Scopus (44)
PubMed
Google Scholar
,169-172
169.
Bateman, E.D. ∙ Kornmann, O. ∙ Schmidt, P. ...
Tiotropium is noninferior to salmeterol in maintaining improved lung function in B16-Arg/Arg patients with asthma
J Allergy Clin Immunol. 2011; 128:315-322
Full Text
Full Text (PDF)
Scopus (159)
PubMed
Google Scholar
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
Full Text
Full Text (PDF)
Scopus (125)
PubMed
Google Scholar
171.
Rajanandh, M.G. ∙ Nageswari, A.D. ∙ Ilango, K.
Pulmonary function assessment in mild to moderate persistent asthma patients receiving montelukast, doxofylline, and tiotropium with budesonide: a randomized controlled study
Clin Ther. 2014; 36:526-533
Full Text
Full Text (PDF)
Scopus (0)
PubMed
Google Scholar
172.
Rajanandh, M.G. ∙ Nageswari, A.D. ∙ Ilango, K.
Assessment of montelukast, doxofylline, and tiotropium with budesonide for the treatment of asthma: which is the best among the second-line treatment? A randomized trial
Clin Ther. 2015; 37:418-426
Full Text
Full Text (PDF)
PubMed
Google Scholar
in 2450 patients found no between-group differences in use of rescue medications. The certainty of evidence is low for the effect on rescue medication use.
Finally, four RCTs166
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
The effect of fluticasone furoate/umeclidinium in adult patients with asthma: a randomized, dose-ranging study
Respir Med. 2015; 109:54-62
Full Text
Full Text (PDF)
PubMed
Google Scholar
,167
167.
Wechsler, M.E. ∙ Yawn, B.P. ∙ Fuhlbrigge, A.L. ...
Anticholinergic vs long-acting beta-agonist in combination with inhaled corticosteroids in black adults with asthma: the BELT randomized clinical trial
JAMA. 2015; 314:1720-1730
Crossref
Scopus (44)
PubMed
Google Scholar
,170
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
Full Text
Full Text (PDF)
Scopus (125)
PubMed
Google Scholar
showed no between-group differences in all-cause mortality rates (OR= 7.50; 95% CI, 0.78-72.27). The mortality rates were 0.2% (3 of 1835) in the LAMA group and 0% (0 of 1135) in the LABA group. The certainty of evidence is low for the effect on mortality.
With respect to harms, data from double-blinded, placebo-controlled RCTs suggest a similar rate of undesirable side effects in individuals treated with ICS-LABA and those treated with an ICS plus a LAMA.166
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
The effect of fluticasone furoate/umeclidinium in adult patients with asthma: a randomized, dose-ranging study
Respir Med. 2015; 109:54-62
Full Text
Full Text (PDF)
PubMed
Google Scholar
,168-170
168.
Peters, S.P. ∙ Kunselman, S.J. ∙ Icitovic, N. ...
Tiotropium bromide step-up therapy for adults with uncontrolled asthma
N Engl J Med. 2010; 363:1715-1726
Crossref
Scopus (418)
PubMed
Google Scholar
169.
Bateman, E.D. ∙ Kornmann, O. ∙ Schmidt, P. ...
Tiotropium is noninferior to salmeterol in maintaining improved lung function in B16-Arg/Arg patients with asthma
J Allergy Clin Immunol. 2011; 128:315-322
Full Text
Full Text (PDF)
Scopus (159)
PubMed
Google Scholar
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
Full Text
Full Text (PDF)
Scopus (125)
PubMed
Google Scholar
However, a real-world comparative effectiveness study167
167.
Wechsler, M.E. ∙ Yawn, B.P. ∙ Fuhlbrigge, A.L. ...
Anticholinergic vs long-acting beta-agonist in combination with inhaled corticosteroids in black adults with asthma: the BELT randomized clinical trial
JAMA. 2015; 314:1720-1730
Crossref
Scopus (44)
PubMed
Google Scholar
that compared the two treatments, the Blacks and Exacerbations on LABA vs Tiotropium (BELT) study, found a 2.6-fold higher rate of asthma-related hospitalizations in the ICS plus LAMA group than in the ICS-LABA group. In addition, the number of hospitalizations in the ICS plus LAMA group in the BELT study (3.6 per 100 hospitalizations/person/y) was higher than in the ICS-LABA groups in the FDA-required safety studies (0.66 per 100 hospitalizations/person/y).173
173.
Busse, W.W. ∙ Bateman, E.D. ∙ Caplan, A.L. ...
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While few asthma-related deaths occurred in the BELT study (2 of 1070 participants), both deaths occurred in the ICS plus LAMA group (2 of 532 [0.38%]). The proportion of asthma-related deaths in the ICS plus LAMA group in the BELT study was 38 times higher than the proportion in an ICS-LABA group in the FDA-required safety studies.173
173.
Busse, W.W. ∙ Bateman, E.D. ∙ Caplan, A.L. ...
Combined analysis of asthma safety trials of long-acting beta2-agonists
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Because of its real-world effectiveness design, the BELT study might better reflect the harms and benefits likely to occur in clinical practice than efficacy studies of the combination of LAMA and ICS therapy. The BELT study included only Blacks, and no similar data are available from real-world trials that assessed harms in other populations. Therefore, the Expert Panel was unable to determine whether these harms are a concern only in Blacks or whether they might occur in other populations.
The summary of evidence for Recommendation 15 can be found in Appendix B (EtD Table XXI). The Expert Panel compared the harms and benefits of adding a LAMA to ICS therapy with adding a placebo to continued ICS therapy in five RCTs (total N= 3036).166
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
The effect of fluticasone furoate/umeclidinium in adult patients with asthma: a randomized, dose-ranging study
Respir Med. 2015; 109:54-62
Full Text
Full Text (PDF)
PubMed
Google Scholar
,169
169.
Bateman, E.D. ∙ Kornmann, O. ∙ Schmidt, P. ...
Tiotropium is noninferior to salmeterol in maintaining improved lung function in B16-Arg/Arg patients with asthma
J Allergy Clin Immunol. 2011; 128:315-322
Full Text
Full Text (PDF)
Scopus (159)
PubMed
Google Scholar
,170
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
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Full Text (PDF)
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PubMed
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174.
Paggiaro, P. ∙ Halpin, D.M. ∙ Buhl, R. ...
The effect of tiotropium in symptomatic asthma despite low- to medium-dose inhaled corticosteroids: a randomized controlled trial
J Allergy Clin Immunol Pract. 2016; 4:104-113.e2
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175.
Hamelmann, E. ∙ Bateman, E.D. ∙ Vogelberg, C. ...
Tiotropium add-on therapy in adolescents with moderate asthma: a 1-year randomized controlled trial
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These trials showed that adding a LAMA to ICS therapy resulted in a slightly smaller rate of exacerbations, 4.2%, than the addition of a placebo to continued ICS therapy, 7.4% (absolute risk difference= 24 fewer per 1000; 95% CI, from 38 fewer to 6 fewer; RR= 0.67; 95% CI, 0.48-0.92). According to these results, 42 patients (95% CI, 26-167) would need treatment to prevent one exacerbation. This effect on exacerbations has moderate certainty of evidence. However, adding a LAMA to ICS therapy did not improve asthma control (measured by the ACQ [ACQ-7, moderate certainty of evidence]).166
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
The effect of fluticasone furoate/umeclidinium in adult patients with asthma: a randomized, dose-ranging study
Respir Med. 2015; 109:54-62
Full Text
Full Text (PDF)
PubMed
Google Scholar
,170
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
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174.
Paggiaro, P. ∙ Halpin, D.M. ∙ Buhl, R. ...
The effect of tiotropium in symptomatic asthma despite low- to medium-dose inhaled corticosteroids: a randomized controlled trial
J Allergy Clin Immunol Pract. 2016; 4:104-113.e2
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Full Text (PDF)
Scopus (75)
PubMed
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175.
Hamelmann, E. ∙ Bateman, E.D. ∙ Vogelberg, C. ...
Tiotropium add-on therapy in adolescents with moderate asthma: a 1-year randomized controlled trial
J Allergy Clin Immunol. 2016; 138:441-450.e8
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Full Text (PDF)
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176.
Ohta, K. ∙ Ichinose, M. ∙ Tohda, Y. ...
Long-term once-daily tiotropium respimat(R) is well tolerated and maintains efficacy over 52 weeks in patients with symptomatic asthma in Japan: a randomised, placebo-controlled study
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The proportion of responders (those with a ≥0.5-point decrease in score) was 67% in the group treated with ICS plus LAMA and was 61% in the group treated with placebo added to continued ICS therapy (RR= 1.08; 95% CI, 0.96-1.21). In addition, adding a LAMA to an ICS did not improve asthma-related quality of life (measured by the Asthma-Related Quality of Life Questionnaire, high certainty of evidence)169
169.
Bateman, E.D. ∙ Kornmann, O. ∙ Schmidt, P. ...
Tiotropium is noninferior to salmeterol in maintaining improved lung function in B16-Arg/Arg patients with asthma
J Allergy Clin Immunol. 2011; 128:315-322
Full Text
Full Text (PDF)
Scopus (159)
PubMed
Google Scholar
,170
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
Full Text
Full Text (PDF)
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PubMed
Google Scholar
and had no effect on rescue medication use (high certainty of evidence).166
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
The effect of fluticasone furoate/umeclidinium in adult patients with asthma: a randomized, dose-ranging study
Respir Med. 2015; 109:54-62
Full Text
Full Text (PDF)
PubMed
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,170
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
Full Text
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Scopus (125)
PubMed
Google Scholar
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174.
Paggiaro, P. ∙ Halpin, D.M. ∙ Buhl, R. ...
The effect of tiotropium in symptomatic asthma despite low- to medium-dose inhaled corticosteroids: a randomized controlled trial
J Allergy Clin Immunol Pract. 2016; 4:104-113.e2
Full Text
Full Text (PDF)
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PubMed
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175.
Hamelmann, E. ∙ Bateman, E.D. ∙ Vogelberg, C. ...
Tiotropium add-on therapy in adolescents with moderate asthma: a 1-year randomized controlled trial
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Full Text (PDF)
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PubMed
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176.
Ohta, K. ∙ Ichinose, M. ∙ Tohda, Y. ...
Long-term once-daily tiotropium respimat(R) is well tolerated and maintains efficacy over 52 weeks in patients with symptomatic asthma in Japan: a randomised, placebo-controlled study
PLoS One. 2015; 10, e0124109
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Harms data are available from six studies that compared the efficacy of adding a LAMA to ICS therapy with adding a placebo to ICS therapy.166
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
The effect of fluticasone furoate/umeclidinium in adult patients with asthma: a randomized, dose-ranging study
Respir Med. 2015; 109:54-62
Full Text
Full Text (PDF)
PubMed
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,170
170.
Kerstjens, H.A. ∙ Casale, T.B. ∙ Bleecker, E.R. ...
Tiotropium or salmeterol as add-on therapy to inhaled corticosteroids for patients with moderate symptomatic asthma: two replicate, double-blind, placebo-controlled, parallel-group, active-comparator, randomised trials
Lancet Respir Med. 2015; 3:367-376
Full Text
Full Text (PDF)
Scopus (125)
PubMed
Google Scholar
,174-176
174.
Paggiaro, P. ∙ Halpin, D.M. ∙ Buhl, R. ...
The effect of tiotropium in symptomatic asthma despite low- to medium-dose inhaled corticosteroids: a randomized controlled trial
J Allergy Clin Immunol Pract. 2016; 4:104-113.e2
Full Text
Full Text (PDF)
Scopus (75)
PubMed
Google Scholar
175.
Hamelmann, E. ∙ Bateman, E.D. ∙ Vogelberg, C. ...
Tiotropium add-on therapy in adolescents with moderate asthma: a 1-year randomized controlled trial
J Allergy Clin Immunol. 2016; 138:441-450.e8
Full Text
Full Text (PDF)
Scopus (94)
PubMed
Google Scholar
176.
Ohta, K. ∙ Ichinose, M. ∙ Tohda, Y. ...
Long-term once-daily tiotropium respimat(R) is well tolerated and maintains efficacy over 52 weeks in patients with symptomatic asthma in Japan: a randomised, placebo-controlled study
PLoS One. 2015; 10, e0124109
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In these studies, the rate of serious adverse events for the addition of a LAMA to ICS therapy was low and was similar to that for the addition of a placebo to ICS therapy. No deaths were reported for any of these studies (see EtD Table XXI). All studies excluded participants with a history of glaucoma or urinary retention. Therefore, whether adding LAMA to ICS therapy is safe in individuals with these conditions is not known.
Rationale and discussion
Outcomes from seven RCTs166-172
166.
Lee, L.A. ∙ Yang, S. ∙ Kerwin, E. ...
The effect of fluticasone furoate/umeclidinium in adult patients with asthma: a randomized, dose-ranging study
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167.
Wechsler, M.E. ∙ Yawn, B.P. ∙ Fuhlbrigge, A.L. ...
Anticholinergic vs long-acting beta-agonist in combination with inhaled corticosteroids in black adults with asthma: the BELT randomized clinical trial
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168.
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showed no significant differences between groups. This evidence therefore provides no basis, based on benefits, for recommending the addition of a LAMA to ICS therapy as opposed to the addition of a LABA to ICS therapy in adults with uncontrolled persistent asthma.
The Expert Panel considered the serious adverse events in African-American adults assigned to the ICS plus LAMA group in the BELT study.167
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The number of asthma-related deaths in this group was higher than expected in African-American adults, and the adjusted rate of asthma-related hospitalizations was statistically higher in the ICS plus LAMA group than in the ICS-LABA group. Although it is difficult for the Expert Panel to draw firm conclusions, in the opinion of the Expert Panel, the balance of the evidence argues against adding a LAMA to an ICS compared with adding a LABA to an ICS because the benefits of added LAMA are trivial, and there is a small concern about the safety of LAMA combined with ICS alone.
In the studies that compared the addition of a LAMA to an ICS with ICS therapy alone, adding a LAMA to an ICS slightly reduced the number of exacerbations166
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or asthma-related quality of life.169
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The Expert Panel’s judgment about the degree of benefit was subjective because no established standards are available for the MID in exacerbations. In addition, individuals with asthma who place a higher value on asthma control and quality of life than on exacerbations may not perceive any benefit from this intervention.
After considerable discussion about the harms found in the BELT study,167
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the Expert Panel concluded that the BELT study did not address the harms of adding a LAMA to an ICS compared with adding placebo to ICS therapy.167
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However, because the BELT study showed a higher adverse event rate in participants assigned to ICS plus LAMA than in those treated with ICS-LABA, the Expert Panel recommends first considering the addition of a LABA to an ICS and considering the addition of a LAMA to an ICS as an alternate approach. This prioritization of therapies may be particularly important in Black adults. The balance of evidence demonstrates that the addition of a LAMA to an ICS offers a small benefit compared with ICS therapy alone, but there is a small concern related to harm.
In addition to the studies described above, the systematic review report compared the efficacy of the addition of a LAMA to ICS controller therapy in individuals aged 12 years and older and adults with uncontrolled, persistent asthma with the efficacy of the addition of montelukast to ICS therapy (EtD Table XXII) and with a doubled ICS dose (EtD Table XXIII).6
6.
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produced findings in participants aged 18 to 60 years after 6 months of treatment in a four-arm, parallel-group, unmasked, active-comparator trial (N= 72 for ICS plus LAMA, N= 68 for ICS plus LABA [formoterol], N= 81 for ICS plus montelukast, and N= 76 for ICS plus doxofylline). A total of 297 of the original 362 participants completed the 6-month study. The study report provided no data on critical outcomes designated by the Expert Panel. The authors reported on only one of the important outcomes (rescue medication use, reported as the difference at day 90 compared with at baseline), and results for this outcome did not differ between groups. In addition, the rate of undesirable effects was similar with both treatments.
After reviewing the available evidence and finding the effect on one noncritical outcome to be inconclusive, the Expert Panel concluded that the data were insufficient to address this question. Therefore, the Expert Panel refrained from making any recommendation regarding the addition of a LAMA to an ICS versus adding montelukast to ICS.
Only one study compared the addition of a LAMA to an ICS with doubling the dose of the ICS. This study found no differences in rates of exacerbations, asthma control, or serious adverse events as well as no differences in asthma-related quality of life between the two groups; no deaths occurred in either group.168
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Although this study showed an improvement in the proportion of control days and in symptom scores of participants assigned to added LAMA treatment, this outcome measure was not validated, and the Expert Panel could not determine the significance of these differences. Therefore, the Expert Panel concluded that the data were insufficient to make a recommendation regarding the addition of a LAMA to an ICS versus doubling the ICS dose.
The Expert Panel also did not make any recommendation regarding the addition of a LAMA to an ICS versus the addition of doxofylline to an ICS because doxofylline is not available in the United States.
Question 5.3
•
What is the comparative effectiveness of LAMA as add-on therapy to ICS plus LABA compared with ICS plus LABA as controller therapy in individuals aged 12 years and older with uncontrolled persistent asthma?
Recommendation 16: In individuals aged 12 years and older with uncontrolled persistent asthma, the Expert Panel conditionally recommends adding LAMA to ICS-LABA compared to continuing the same dose of ICS-LABA.
Conditional recommendation, moderate certainty of evidence
Implementation guidance
Clinician’s Summary: For individuals whose asthma is not controlled with ICS-LABA, the Expert Panel recommends the addition of a LAMA for many individuals.
•
Based on the studies available, the addition of a LAMA to ICS-LABA in individuals aged 12 years and older with uncontrolled persistent asthma offers a small benefit.
•
This therapy is recommended for individuals aged 12 years and older whose asthma is uncontrolled even though they are using ICS-LABA therapy.
•
LAMA therapy should not be used in individuals with glaucoma or urinary retention.
•
Adding a LAMA to ICS-LABA for individuals with uncontrolled asthma who are already taking ICS-LABA improves asthma control and quality of life but has no effect on asthma exacerbations that require systemic corticosteroids or rescue medication.
•
What clinicians should discuss with their patients about adding LAMA therapy to ICS-LABA:
−
Adding LAMA therapy to ICS-LABA requires the use of an additional and different type of inhaler.
−
The addition of a LAMA may improve asthma control and quality of life but may not decrease the frequency of asthma exacerbations, use of oral corticosteroids, or use of rescue medications.
−
Individuals with glaucoma and those at risk of urinary retention should not use LAMA therapy.
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life) and two important outcomes (rescue medication use and mortality). The summary of evidence for Recommendation 16 can be found in EtD Table XXIV in Appendix B.
Two trials (total N= 912) found that the proportion of adults who achieved the MID of 0.5 points on the ACQ-7 for asthma control was higher when tiotropium was added to ICS-LABA than when placebo was added (RR= 1.28; 95% CI, 1.13-1.46); these studies provided moderate certainty of evidence.177
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The single study (N= 388) in youth aged 12 to 17 years found no difference in the proportion whose ACQ-7 scores improved (RR= 1.01; 95% CI, 0.89-1.14).178
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found similar decreases in mean ACQ-7 scores in youths and adults treated with tiotropium and ICS-LABA and in those treated with placebo added to ICS-LABA (mean difference= 0.07 points lower; 95% CI, from 0.31 lower to 0.17 higher); the certainty of evidence is moderate.
Similarly, a higher proportion of adults showed an MID of at least 0.5 points for improved asthma quality of life, as measured by the Asthma-Related Quality of Life Questionnaire, with the addition of a LAMA to ICS-LABA than with the addition of a placebo to continued ICS-LABA (RR= 1.62; 95% CI, 1.34-1.96); the certainty of evidence is high.177
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However, the study did not show a between-group difference in the mean Asthma-Related Quality of Life Questionnaire score (high certainty of evidence). In addition, three trials (total N= 1299)177
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showed no difference in asthma exacerbations requiring treatment with systemic corticosteroids (RR= 0.84; 95% CI, 0.57-1.22; moderate certainty of evidence) or in two trials (N= 907),177
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in exacerbations requiring hospitalization (RR= 0.80; 95% CI, 0.42-1.52; moderate certainty of evidence). The findings showed no between-group difference in the mean number of puffs of rescue medication in 24 hours (95% CI, 0.37/d less to 0.18/d more; moderate certainty of evidence) or mortality rates (no deaths in either group; very low certainty of evidence).
Rationale and discussion
In the studies described above, the desirable effects on asthma control and quality of life of the addition of a LAMA to ICS-LABA compared with the addition of placebo were small, and the risks of asthma exacerbations and of adverse events did not differ between the added LAMA and placebo groups. The Expert Panel believes that the balance of outcomes probably favors adding a LAMA to ICS-LABA instead of continuing the same dose of ICS-LABA alone (moderate certainty of evidence). In addition, the Expert Panel does not believe that the extent to which individuals with asthma value the critical outcomes varies or is uncertain. Thus, the addition of a LAMA to ICS-LABA is probably acceptable. However, individuals with asthma and other stakeholders who place less value on asthma control and quality life than on exacerbations may not find the addition of a LAMA acceptable. Using a LAMA as an add-on therapy is feasible but requires teaching individuals with asthma how to appropriately use devices that deliver the LAMA. The Expert Panel concludes that the use of a LAMA as add-on therapy to ICS-LABA would probably improve health equity because asthma disproportionately affects disadvantaged populations.
The Expert Panel also compared the use of a LAMA as add-on therapy to ICS-LABA with doubling the dose of ICS and continuing the same dose of LABA in individuals aged 12 years and older with uncontrolled persistent asthma (EtD Table XXV). A single, small, open-label RCT randomized 94 individuals who continued to take LABA on a 1:1:1 basis to add-on, once-daily tiotropium bromide 18 μg, montelukast 10 mg, or double-dose ICS.179
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The data were insufficient to support a judgment about the balance of desirable and undesirable effects. The Expert Panel therefore did not find sufficient data to formulate recommendations about the use of a LAMA as add-on therapy to ICS compared with increasing the dose of ICS and continuing the LABA.
Future research opportunities
The Expert Panel offers the following suggestions for future research:
•
Comparative effectiveness studies of LAMA therapy for asthma. Because the majority of LAMA studies were efficacy studies, the clinical impact of LAMA treatment in real-world settings is not well understood.
•
Comparative effectiveness and safety of ICS plus LAMA versus ICS-LABA in ethnically diverse population in studies that are adequately powered to examine the harms and benefits of these two treatment options.
•
Systematic reviews in children with asthma aged 6 to 11 years to inform future guidelines.
•
Comparisons of a LAMA to a leukotriene inhibitor as add-on therapy to ICS-LABA in individuals with uncontrolled persistent asthma.
•
Role of LAMAs other than tiotropium as add-on therapy to ICS therapy in individuals aged 12 years and older with uncontrolled persistent asthma.
Section VI: The role of subcutaneous and sublingual immunotherapy in the treatment of allergic asthma
Background
This section addresses immunotherapy in individuals with allergic asthma. Immunotherapy is the administration of an aeroallergen either subcutaneously (subcutaneous immunotherapy [SCIT]) or sublingually (sublingual immunotherapy [SLIT] in the form of aqueous drops or tablets). The Expert Panel explored the efficacy and safety of the use of both SCIT and SLIT for the treatment of allergic asthma and made two recommendations.
Definition of terms used in this section
“Allergic asthma” refers to asthma that becomes symptomatic after acute exposure to something to which the individual is allergic (eg, a pet) or during a specific season (eg, in the spring, when trees shed pollen, or in the fall, when ragweed pollen disperses through the air). In contrast, the term “allergic asthma” is used in many clinical trials to describe a population of children and adults with asthma who show evidence of allergic sensitization based on immediate hypersensitivity skin testing or in vitro serum IgE testing, regardless of whether they have documented symptoms after relevant exposures. However, more recent trials of immunotherapy have more clearly documented the presence of sensitization and relevant symptoms on exposure to allergens.
“Immunotherapy” (both subcutaneous and sublingual) in this report refers to treatments used to reduce the IgE-mediated allergic clinical response that is associated with asthma. Immunotherapy consists of the therapeutic administration of exogenous aeroallergens to which a person has demonstrable sensitization with the goal of attenuating that individual’s asthmatic response on subsequent exposure to these aeroallergens. Immunotherapy can be administered in two ways: subcutaneously by injection (in individuals aged 5 years or older) or sublingually in either liquid or tablet form. The US FDA has not approved the use of liquid SLIT or tablet forms of immunotherapy for the specific treatment of asthma, but tablet forms do have FDA approval for treatment of allergic rhinitis and conjunctivitis in individuals aged 5 years and older who have sensitization to northern grass and those aged 18 years and older with sensitization to a short ragweed and dust mite mixture.
Before receiving immunotherapy, individuals with asthma must demonstrate allergic sensitization using one of two methods:
1.
Immediate hypersensitivity skin testing followed by an assessment 15 to 20 minutes later for a wheal and flare reaction to the allergens tested
2.
Laboratory testing to measure the level of (aeroallergen) antigen-specific IgE antibody in a blood sample
Question 6.1
•
What is the efficacy and safety of SCIT?
Recommendation 17: In individuals aged 5 years and older with mild to moderate allergic asthma, the Expert Panel conditionally recommends the use of SCIT as an adjunct treatment to standard pharmacotherapy in those individuals whose asthma is controlled at the initiation, build-up, and maintenance phases of immunotherapy.
Conditional recommendation, moderate certainty of evidence
Implementation guidance
Clinician’s Summary: The Expert Panel conditionally recommends SCIT as an adjunctive treatment for individuals who have demonstrated allergic sensitization and evidence of worsening asthma symptoms after exposure to the relevant antigen or antigens either acutely (eg, allergy to pets) or on a seasonal basis (eg, allergy to grass or ragweed) or a chronic basis (eg, allergy to dust mites). Individuals who place a high value on possible small improvements in quality of life, symptom control, and a reduction in long-term and/or quick-relief medication use and a lower value on the risk of systemic reactions of wide-ranging severity might consider SCIT as adjunct therapy.
For individuals with allergic asthma, the Expert Panel makes the following suggestions to implement SCIT:
•
Clinicians can consider SCIT for adults and children (at a developmental stage at which allergic sensitization can be demonstrated) with allergic asthma, a history compatible with a temporal association of worsening symptoms with exposure to aeroallergens, and testing (as described previously) that confirms this sensitization.
•
Clinicians can consider SCIT for individuals whose asthma is not well controlled by their current medical therapy and the treating clinician considers allergen exposure to be a significant contributor to this lack of asthma control. However, clinicians should attempt to optimize asthma control before initiating SCIT to reduce the potential for harm.
•
Clinicians can consider SCIT for individuals whose asthma is well controlled by their current therapy when these individuals and/or their clinicians want to reduce the individuals’ medication burden.
•
In addition to assessing whether an individual with allergic asthma has an appropriate history before considering SCIT, clinicians must formally assess allergic sensitization using either immediate hypersensitivity skin testing or in vitro antigen-specific IgE antibody testing. This evaluation needs to be performed by a trained health care professional skilled in proper testing and result interpretation. The need for these types of specialty evaluations, as with the need for many diagnostic tests and therapeutic interventions, may limit access to care, depending on local availability of these tests and the patient’s health insurance coverage of testing.
•
Clinicians should not administer SCIT in individuals with severe asthma. Furthermore, clinicians should not initiate, increase, or administer maintenance SCIT doses while individuals have asthma symptoms. These individuals should achieve optimal asthma control before beginning SCIT to minimize the harms (systemic reactions) associated with SCIT, which tend to intensify as baseline asthma severity increases.
•
The presence of allergic sensitization is necessary but not sufficient to define the allergic asthma phenotype. A positive test result may not be associated with asthma control over time but might, instead, reflect sensitivity in a different organ (eg, the nose in allergic rhinitis).
•
Allergen exposure could be the only triggering mechanism for allergic asthma symptoms, or it could be just one triggering factor for an individual, and another factor or factors (eg, respiratory tract infections, irritant exposure, or exercise) might also play a role in triggering allergic asthma symptoms. Because of the heterogeneous nature of allergic asthma, determining the precise efficacy of immunotherapy in reducing the allergic component of an individual’s asthma can be difficult.
•
Clinicians should administer SCIT in their offices and provide direct supervision because of the risk of systemic reactions. Such reactions can include a range of anaphylactic symptoms involving the skin (urticaria), respiratory tract (rhinitis and asthma), gastrointestinal tract (nausea, diarrhea, and vomiting), and the cardiovascular system (hypotension and arrhythmias). Although rare, deaths after injections have been reported.
•
Individuals with asthma should not administer SCIT at home.
•
Because clinicians should administer SCIT with direct supervision, personnel with appropriate training should prepare and administer injections for each individual’s dosing schedule, from the build-up to the maintenance phase. Equipment and personnel should be available to treat serious anaphylactic reactions.
•
One of the potential benefits of SCIT is its immunomodulatory effects, which can reduce the allergic inflammatory response in various tissues.180
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•
Before administering each SCIT injection, clinicians should assess individuals with asthma for worsened asthma symptoms that suggest recent loss of asthma control. Physicians should consider withholding SCIT injections temporarily in patients whose asthma symptoms have worsened until their asthma control is restored.
•
What clinicians should discuss with their patients:
–
Clinicians should inform individuals with asthma who are considering SCIT that this treatment has the potential to reduce asthma symptoms and the severity of disease over time.
–
Individuals need to come to their doctor’s office for SCIT because of the associated risk of systemic reactions.
–
Local and systemic reactions of SCIT include a range of anaphylactic symptoms involving the skin (urticaria), respiratory tract (rhinitis and asthma), gastrointestinal tract (nausea, diarrhea, and vomiting), and the cardiovascular system (hypotension and arrhythmias). Although rare, deaths after injections have been reported.
–
Individuals with asthma should not administer SCIT at home.
–
Before initiating immunotherapy, clinicians must review with the individual who has asthma the travel arrangements and time needed to travel to and from the clinic as well as the requirement for at least a 30-minute observational period after each injection. These requirements may complicate compliance. Missed appointments due to scheduling problems are a safety and an efficacy concern because they may increase the likelihood of local and systemic reactions. Missed appointments can also complicate the ability to reach a maintenance dosing regimen that maximizes therapeutic benefit.
–
Delayed systemic reactions (those occurring more than 30 minutes after injection) occur in approximately 15% of individuals after injection.183
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Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life) and three important outcomes (use of quick-relief medication, adverse events [harms], and long-term medication use). Because none of the SCIT studies used validated asthma control outcome measures, the Expert Panel used nonvalidated outcome measures (eg, symptom diaries) as surrogate measures of asthma control when it evaluated 44 studies, but only if the studies used a placebo injection as the comparator.184-226
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The summary of evidence for Recommendation 17 can be found in EtD Table XXVI in Appendix B. Most studies included in the systematic review report evaluated individuals with mild to moderate asthma. The status of asthma control in the studies varied and is classified as controlled, not reported, or uncontrolled. The Expert Panel judged the certainty of evidence for SCIT as low for a small benefit with respect to the critical outcomes of exacerbations, quality of life, and asthma control. Studies on exacerbations were limited. One very small study (N= 29) suggested a decrease in exacerbations (very low certainty of evidence).227
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Both studies used a validated outcome measure but scored the individual domains separately. Two other small studies (N= 57) found no difference in quality of life in individuals treated with SCIT or the comparator.228
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Ohman, Jr., J.L. ∙ Findlay, S.R. ∙ Leitermann, K.M.
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218.
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The Expert Panel noted that when asthma is treated with SCIT, the symptoms of comorbid conditions, such as allergic rhinitis and allergic conjunctivitis, may improve and have a beneficial effect on quality of life.
For the important outcomes, SCIT may reduce use of quick-relief medications214
214.
Olsen, O.T. ∙ Larsen, K.R. ∙ Jacobsan, L. ...
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Lin SY, Azar A, Suarez-Cuervo C, Diette GB, Brigham E, Rice J, et al. The role of immunotherapy in the treatment of asthma. Comparative Effectiveness Review No. 196. (Prepared by the Johns Hopkins University Evidence-based Practice Center under Contract No.290-2015-00006-I). AHRQ Publication No. 17(18)-EHC029-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Posted final reports are located on the Effective Health Care Program search page.
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have found systemic reactions with up to 12% of total injections, during 0.1% of injection visits, and in 80% to 85% of practices. These systemic reactions include pruritus, urticaria, eczema, atopic dermatitis and other forms of eczema, rhinitis, conjunctivitis, nasal congestion, cough, bronchospasm, wheezing, dyspnea, abdominal pain, diarrhea, and hypotension.5
5.
Lin SY, Azar A, Suarez-Cuervo C, Diette GB, Brigham E, Rice J, et al. The role of immunotherapy in the treatment of asthma. Comparative Effectiveness Review No. 196. (Prepared by the Johns Hopkins University Evidence-based Practice Center under Contract No.290-2015-00006-I). AHRQ Publication No. 17(18)-EHC029-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Posted final reports are located on the Effective Health Care Program search page.
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Rates of systemic allergic reactions consistent with anaphylaxis also varied greatly, and RCTs5
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did not have the statistical power to assess such effects. Poorly controlled asthma is a major risk factor for fatal allergic reactions from SCIT. The incidence of fatal and near-fatal anaphylactic reactions ranges from 1 in 20,000 to 1 in 200,000 injections.183
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(low certainty of evidence because of imprecision).
Rationale and discussion
Considering the overall balance between benefits and harms, in the judgment of the Expert Panel, the SCIT recommendation is conditional because individuals may consider SCIT as adjunct therapy if they have the following characteristics:
•
Place a high value on small improvements in quality of life and symptom control
•
Place a high value on reductions in long-term and/or quick-relief medication use
•
Place a lower value on the potential for systemic reactions of wide-ranging severity
The studies available for evaluation tended to have small samples, and study reports did not characterize the races of participants or the social determinants of health that they experienced.5
5.
Lin SY, Azar A, Suarez-Cuervo C, Diette GB, Brigham E, Rice J, et al. The role of immunotherapy in the treatment of asthma. Comparative Effectiveness Review No. 196. (Prepared by the Johns Hopkins University Evidence-based Practice Center under Contract No.290-2015-00006-I). AHRQ Publication No. 17(18)-EHC029-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Posted final reports are located on the Effective Health Care Program search page.
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Studies of SCIT used different protocols and did not use standardized formulations or have a uniform or standardized duration of follow-up. The efficacy of SCIT, which has an acceptable burden of harms, is based on its impact on asthma quality of life and asthma-related symptoms, with low certainty of evidence. Whether to use SCIT should be a shared decision between the individual and the health care provider, and this decision should consider the individual’s asthma severity and willingness to accept the potential harms related to SCIT. Clinicians should administer SCIT in a clinical setting that has the capacity to monitor and treat reactions.
The enthusiasm of the Expert Panel for recommending SCIT for allergic asthma management is reduced by the slight risk of harms and variability in access (because of costs and geographical location); this variability in access can promote health inequities.
Question 6.2
•
What is the efficacy and safety of SLIT?
Recommendation 18: In individuals with persistent allergic asthma, the Expert Panel conditionally recommends against the use of SLIT in asthma treatment.
Conditional recommendation, moderate certainty of evidence
Implementation guidance
Clinician’s Summary: The evidence that the Expert Panel reviewed did not support the use of SLIT specifically for the treatment of allergic asthma. However, the FDA has approved SLIT tablets (but not aqueous preparations) for the treatment of allergic rhinoconjunctivitis. Individuals with this condition who also have asthma might benefit from SLIT and, if so, this benefit is most likely to be in the form of a reduction in the use of quick-relief and/or long-term control medications.
On the basis of the currently available data, the Expert Panel does not recommend SLIT for allergic asthma. SLIT is beneficial for allergic rhinoconjunctivitis.231
231.
Lin SY, Erekosima N, Suarez-Cuervo C, Ramanathan M, Kim JM, Ward D, et al. Allergen-specific immunotherapy for the treatment of allergic rhinoconjunctivitis and/or asthma: comparative effectiveness review. Rockville, Md: Agency for Healthcare Research and Quality (AHRQ). Prepared by the Johns Hopkins University Evidence-based Practice Center under Contract No. 290-2007-10061-I. 2013 Mar (Errata added May and August 2013). Available at: www.effectivehealthcare.ahrq.gov/reports/final.cfm.
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In an individual with comorbid allergic asthma, SLIT for allergic rhinoconjunctivitis might reduce the symptoms of allergic asthma as well (and this potential provides the rationale for making the recommendation conditional). For individuals whose allergic asthma symptoms benefit from SLIT for allergic rhinoconjunctivitis, the Expert Panel offers the following suggestions.
•
The clinician should administer the first dose of SLIT in the office, and the individual with asthma should wait in the office for at least 30 minutes after receiving the dose. If no problems develop, the individual may continue the SLIT dosing at home. Individuals receiving SLIT should ideally have an injectable epinephrine prescription and receive education on how to administer this medication.
•
Currently, only tablet SLIT formulations for short ragweed and dust mite mixture and for northern grass have FDA approval for treatment of allergic rhinitis with and without conjunctivitis. SLIT is not FDA approved specifically for asthma treatment.
•
What clinicians should discuss with their patients:
−
The Expert Panel does not recommend SLIT for the treatment of allergic asthma, but this treatment may benefit individuals with certain comorbid conditions, such as allergic rhinitis with or without conjunctivitis.
−
The FDA has approved the use of SLIT to treat allergic rhinitis and conjunctivitis in response to only a few allergens at this time for individuals aged 5 years and older (for sensitization to northern grass) and in individuals aged 18 years and older (for sensitization to a short ragweed and dust mite mixture).
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life) and three important outcomes (quick-relief medication, adverse events [harms], and long-term medication use). The summary of evidence for Recommendation 18 can be found in EtD Table XXVII in Appendix B.
The evidence shows that SLIT provides a trivial benefit for the critical outcomes of exacerbations,232
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(moderate certainty of evidence). No studies assessed the impact of SLIT on emergency department visits, clinic visits, or hospitalizations. Three studies evaluated exacerbations using different end points. One study did not report the number of exacerbations, but it did report on the time to first exacerbation.233
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SLIT decreased the severity of the first moderate exacerbation, but it did not increase the time to first severe exacerbations requiring systemic corticosteroids. Another study did not provide any raw data or rates of the critical outcomes, and the authors only noted that the results showed no statistically significant improvement in asthma exacerbations.234
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The third study, which enrolled only 60 participants, found a significantly lower number of exacerbations in the treatment group.232
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Four studies (N= 1193) that evaluated asthma control using validated outcome tools (three used the ACQ, and one used the ACT) found no consistent improvement after treatment.233-239
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Finally, multiple studies showed no difference in quality of life in those treated with SLIT or placebo233-235
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(high certainty of evidence).
For important outcomes, SLIT reduced the use of quick-relief medications232
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with moderate certainty of evidence.
The harms were difficult for the Expert Panel to evaluate. Local reactions were frequent and occurred in up to 80% of individuals treated with SLIT, but adverse local reactions were also common in those receiving placebo. The rate of side effects did not differ by the setting of administration (home, clinic, or other), and the relationship between the risk of side effects and the strength of the dose administered was not consistent across studies. None of the RCTs (N= 1772)233
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reported episodes of anaphylaxis. The Expert Panel found no reports of death that was secondary to SLIT.
Rationale and discussion
The 2014-2015 needs assessment report by the NHLBAC Asthma Expert Working Group2
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Needs Assessment Report for Potential Update of the Expert Panel Report-3 (2007): Guidelines for the Diagnosis and Management of Asthma
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included both aqueous and tablet formulations in the research questions on the efficacy and safety of SLIT. For these questions, the systematic review report combined studies of the two types of SLIT, thereby increasing the sample sizes and precision of results for many of the outcomes evaluated.12
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National Asthma Education and Prevention Program
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However, the designs and methodologies of RCTs that used aqueous and drop preparations of SLIT were not as rigorous or standardized as they were for studies that used tablet formulations. In evaluating the data on aqueous or drop and tablet formulations combined, the Expert Panel did not find that SLIT reduced asthma symptoms or improved asthma control or asthma quality of life. Although systemic side effects were common (80% of participants), they were also common in the placebo groups.5
5.
Lin SY, Azar A, Suarez-Cuervo C, Diette GB, Brigham E, Rice J, et al. The role of immunotherapy in the treatment of asthma. Comparative Effectiveness Review No. 196. (Prepared by the Johns Hopkins University Evidence-based Practice Center under Contract No.290-2015-00006-I). AHRQ Publication No. 17(18)-EHC029-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Posted final reports are located on the Effective Health Care Program search page.
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In addition, the limited number of FDA-approved antigens, the costs of SLIT, and the variability in access to this treatment promote health inequities.
Overall summary for SCIT and SLIT
The Expert Panel conditionally recommends SCIT as an adjunct treatment to standard pharmacotherapy for individuals aged 5 years and older with mild to moderate persistent asthma who show clear evidence of a relationship between symptoms and exposure to an allergen to which the individual is sensitive.12
12.
National Asthma Education and Prevention Program
Third Expert Panel on the Diagnosis and Management of Asthma. Expert Panel Report 3: Guidelines for the Diagnosis and Management of Asthma
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The Expert Panel conditionally recommends against the use of SLIT as a treatment specifically for asthma.
The Expert Panel’s immunotherapy recommendations call for shared decision making between the clinician and the individual with asthma. The recommendations also highlight SLIT’s potential to reduce the symptoms of comorbid conditions, such as allergic rhinitis and allergic conjunctivitis, and this potential improvement may be an important consideration for individuals with allergic asthma.5
5.
Lin SY, Azar A, Suarez-Cuervo C, Diette GB, Brigham E, Rice J, et al. The role of immunotherapy in the treatment of asthma. Comparative Effectiveness Review No. 196. (Prepared by the Johns Hopkins University Evidence-based Practice Center under Contract No.290-2015-00006-I). AHRQ Publication No. 17(18)-EHC029-EF. Rockville, Md: Agency for Healthcare Research and Quality; March 2018. Posted final reports are located on the Effective Health Care Program search page.
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Future research opportunities
The Expert Panel identified the following opportunities for additional research:
•
Investigate the safety and efficacy of immunotherapy in individuals with severe asthma, particularly those whose asthma is under control but who want to reduce their medication burden
•
Include only children aged 5 to 11 years in studies of children, or, if a study includes a broader age group, report findings separately for children aged 5 to 11 years and those aged 12 years and older
•
Study more diverse populations to determine whether race or ethnicity influences the efficacy and safety of immunotherapy
•
Study the efficacy and safety of multiple-allergen SCIT or SLIT regimens to assess compliance, adherence, and the effect of these factors on asthma management
•
Standardize methods to report SCIT and SLIT doses used in studies and use validated outcome measurement instruments, such as asthma symptoms and adverse events
Section VII: Recommendations for the use of BT to improve asthma outcomes
Background
The Expert Panel examined studies that compared BT to multicomponent, standard-of-care, medical management, and sham bronchoscopy plus multicomponent medical management. BT is an asthma intervention that was developed over the last decade and was not addressed in previous versions of the asthma guidelines. The Expert Panel made one recommendation on the use of BT for asthma treatment.
Definitions of terms used in this section
Multicomponent medical therapy consists of medium to high doses of ICS treatment, LABAs, omalizumab (in one study), and/or oral corticosteroids. Available studies of BT did not include individuals treated with LAMAs, environmental interventions, and/or newer biologic agents.247-249
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“Life-threatening asthma” is defined as asthma that has resulted in hospitalization in an intensive care unit and/or has been treated with noninvasive ventilation or intubation in the past 5 years.
Question 7.1
•
What are the benefits and harms of using BT in addition to standard treatment for the treatment of individuals aged 18 years and older with asthma?
Recommendation 19: In individuals aged 18 years and older with persistent asthma, the Expert Panel conditionally recommends against BT.
Conditional recommendation, low certainty of evidence
Individuals aged 18 years and older with persistent asthma who place a low value on harms (ie, short-term worsening of symptoms and unknown long-term side effects) and a high value on potential benefits (ie, improvement in quality of life and a small reduction in number of exacerbations) might consider BT.
Implementation guidance
Clinician’s Summary: Most individuals aged 18 years and older with uncontrolled, moderate to severe, persistent asthma should not undergo BT to treat asthma because the benefits are small, the risks are moderate, and the long-term outcomes are uncertain. Some individuals with moderate to severe persistent asthma who have troublesome symptoms may be willing to accept the risks of BT and, therefore, might choose this intervention after shared decision making with their health care provider. Clinicians should offer the procedure in the setting of a clinical trial or a registry study to enable the collection of long-term data on the use of BT for asthma.
The Expert Panel does not recommend BT for individuals aged 18 years and older as part of routine asthma care, even if these individuals have uncontrolled asthma despite using multicomponent medical therapy, because of the small benefit-to-risk ratio. The risks of BT include asthma exacerbations, hemoptysis, and atelectasis during the treatment period. Recognizing, however, that BT is currently being used, the Expert Panel offers the following suggestions for its safe use:
•
BT should not be used in individuals with low lung function (FEV 1 that is <50% or 60% predicted) and life-threatening asthma.
•
BT has not been studied in individuals younger than age 18 years.
•
In the opinion of the Expert Panel, when BT is implemented, it should be used in settings that enroll participants in registries, ongoing clinical trials, or studies that track BT’s long-term safety and effectiveness.
•
For individuals who decide to undergo BT, an experienced specialist (eg, a pulmonologist with training in BT administration) should provide this treatment in a center that has appropriate expertise.
•
Clinicians should optimize asthma treatment and address comorbidities, and they should assess and optimize adherence to existing therapy, before considering BT.
•
In some individuals, BT may provide a small benefit that might last 5 years or longer.250
250.
Thomson, N.C. ∙ Rubin, A.S. ∙ Niven, R.M. ...
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J Allergy Clin Immunol. 2013; 132:1295-1302
Full Text
Full Text (PDF)
PubMed
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•
BT may reduce severe asthma exacerbations in comparison to standard care after treatment.
•
Risks associated with BT include worsening of asthma, respiratory infections, hemoptysis, bronchiectasis, and pulmonary artery complications.252-254
252.
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253.
Facciolongo, N. ∙ Menzella, F. ∙ Lusuardi, M. ...
Recurrent lung atelectasis from fibrin plugs as a very early complication of bronchial thermoplasty: a case report
Multidiscip Respir Med. 2015; 10:9
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PubMed
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254.
Nguyen, D.V. ∙ Murin, S.
Bronchial artery pseudoaneurysm with major hemorrhage after bronchial thermoplasty
Chest. 2016; 149:e95-e97
Full Text
Full Text (PDF)
PubMed
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•
Severe latent or delayed-onset complications have not been reported with BT, but the number of individuals with asthma included in long-term follow-up assessments is very small (fewer than 250 people at the time the systematic review report3
3.
D’Anci KE, Lynch MP, Leas BF, Apter AJ, Bryant-Stephens T, Kaczmarek JL, et al. Effectiveness and safety of bronchial thermoplasty in management of asthma. Comparative Effective Review No. 202. (Prepared by the ECRI Institute–Penn Medicine Evidence-based Practice Center under Contract No. 290-2015-00005-I). AHRQ Publication No. 18-EHC0003-EF. Rockville, Md: Agency for Healthcare Research and Quality; December 2017. Available at:
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on this topic was completed).
•
What clinicians should discuss with their patients about BT:
–
This procedure may reduce severe asthma exacerbations compared with standard care after treatment. Although the benefits could last 5 years or more, only limited data demonstrate that this treatment improves long-term asthma outcomes.
–
The risks associated with BT include worsening of asthma, respiratory infections, hemoptysis, bronchiectasis, and pulmonary artery complications.252-254
252.
Balu, A. ∙ Ryan, D. ∙ Niven, R.
Lung abscess as a complication of bronchial thermoplasty
J Asthma. 2015; 52:740-742
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Scopus (10)
PubMed
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253.
Facciolongo, N. ∙ Menzella, F. ∙ Lusuardi, M. ...
Recurrent lung atelectasis from fibrin plugs as a very early complication of bronchial thermoplasty: a case report
Multidiscip Respir Med. 2015; 10:9
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PubMed
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254.
Nguyen, D.V. ∙ Murin, S.
Bronchial artery pseudoaneurysm with major hemorrhage after bronchial thermoplasty
Chest. 2016; 149:e95-e97
Full Text
Full Text (PDF)
PubMed
Google Scholar
In addition, severe, delayed-onset complications could occur that have not yet been recognized because of the small numbers of individuals who have undergone the procedure.
–
Individuals aged 18 years and older with persistent asthma who place a low value on the harms (short-term worsening symptoms and unknown long-term side effects) and a high value on the potential benefits (improvement in asthma quality of life, small reduction in exacerbations) of BT might consider this treatment.
Summary of the evidence
The Expert Panel specified three critical outcomes (exacerbations, asthma control, and quality of life) and one important outcome (use of rescue medication) for this question. The summary of evidence for Recommendation 19 can be found in Appendix B (EtD Table XXVIII).
The conditional recommendation against the use of BT in individuals aged 18 years and older with poorly controlled asthma after medium- to high-dose ICS treatment paired with a LABA (with or without oral corticosteroids) is based on three RCTs.247-249
247.
Castro, M. ∙ Rubin, A.S. ∙ Laviolette, M. ...
Effectiveness and safety of bronchial thermoplasty in the treatment of severe asthma: a multicenter, randomized, double-blind, sham-controlled clinical trial
Am J Respir Crit Care Med. 2010; 181:116-124
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248.
Cox, G. ∙ Thomson, N.C. ∙ Rubin, A.S. ...
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N Engl J Med. 2007; 356:1327-1337
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Scopus (419)
PubMed
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249.
Pavord, I.D. ∙ Cox, G. ∙ Thomson, N.C. ...
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Am J Respir Crit Care Med. 2007; 176:1185-1191
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All of these trials were funded by the company that markets the BT device.
Two of the studies compared BT with standard care.248
248.
Cox, G. ∙ Thomson, N.C. ∙ Rubin, A.S. ...
Asthma control during the year after bronchial thermoplasty
N Engl J Med. 2007; 356:1327-1337
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PubMed
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,249
249.
Pavord, I.D. ∙ Cox, G. ∙ Thomson, N.C. ...
Safety and efficacy of bronchial thermoplasty in symptomatic, severe asthma
Am J Respir Crit Care Med. 2007; 176:1185-1191
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The Research In Severe Asthma (RISA) study (N= 32)249
249.
Pavord, I.D. ∙ Cox, G. ∙ Thomson, N.C. ...
Safety and efficacy of bronchial thermoplasty in symptomatic, severe asthma
Am J Respir Crit Care Med. 2007; 176:1185-1191
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enrolled individuals treated with a high-dose ICS (>750 μg fluticasone or equivalent) and a LABA (100 μg salmeterol equivalent) with or without daily oral corticosteroids (<30 mg/d prednisone equivalent). The Asthma Intervention Research (AIR)248
248.
Cox, G. ∙ Thomson, N.C. ∙ Rubin, A.S. ...
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N Engl J Med. 2007; 356:1327-1337
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study (N= 112) enrolled individuals taking an ICS (>200 μg/d beclomethasone equivalent) and a LABA (100 μg salmeterol or equivalent). These two studies found improvements in critical outcomes, including decreases in numbers of mild exacerbations not requiring oral or parenteral corticosteroids and in numbers of emergency department visits. The results also showed improved asthma control based on ACQ scores and less rescue medication use (an important outcome).248
248.
Cox, G. ∙ Thomson, N.C. ∙ Rubin, A.S. ...
Asthma control during the year after bronchial thermoplasty
N Engl J Med. 2007; 356:1327-1337
Crossref
Scopus (419)
PubMed
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,249
249.
Pavord, I.D. ∙ Cox, G. ∙ Thomson, N.C. ...
Safety and efficacy of bronchial thermoplasty in symptomatic, severe asthma
Am J Respir Crit Care Med. 2007; 176:1185-1191
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A third study, AIR 2 (N= 288), compared BT with sham bronchoscopy plus standard care.247
247.
Castro, M. ∙ Rubin, A.S. ∙ Laviolette, M. ...
Effectiveness and safety of bronchial thermoplasty in the treatment of severe asthma: a multicenter, randomized, double-blind, sham-controlled clinical trial
Am J Respir Crit Care Med. 2010; 181:116-124
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This study enrolled individuals treated with high-dose ICS (>1000 μg betamethasone or equivalent) plus a LABA. Participants could also continue using leukotriene modifiers and omalizumab if they had used these treatments for at least 1 year. This study found reductions in severe exacerbations requiring oral or parenteral corticosteroid treatment over 12 months in participants treated with BT. Other critical outcomes—such as asthma control, mean asthma quality-of-life scores (measured with the Asthma Quality of Life Questionnaire), and rescue medication use (an important outcome)—did not improve. The percentage of participants with Asthma Quality of Life Questionnaire scores of 0.5 or higher (MID) in the BT group (79%) was significantly different from the corresponding proportion (64%) in the control (sham bronchoscopy) group. The strength of evidence was low for all of these outcomes across the three studies. None of the studies found that BT reduced the number of hospitalizations for asthma over 12 months.247-249
247.
Castro, M. ∙ Rubin, A.S. ∙ Laviolette, M. ...
Effectiveness and safety of bronchial thermoplasty in the treatment of severe asthma: a multicenter, randomized, double-blind, sham-controlled clinical trial
Am J Respir Crit Care Med. 2010; 181:116-124
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248.
Cox, G. ∙ Thomson, N.C. ∙ Rubin, A.S. ...
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N Engl J Med. 2007; 356:1327-1337
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249.
Pavord, I.D. ∙ Cox, G. ∙ Thomson, N.C. ...
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Am J Respir Crit Care Med. 2007; 176:1185-1191
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The AIR extension study followed 69 individuals (45 treated with BT and 24 with control treatment) for 3 years.250
250.
Thomson, N.C. ∙ Rubin, A.S. ∙ Niven, R.M. ...
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The results did not demonstrate any differences in rates of asthma-related events between the two groups over the additional 24 months.
The RISA249
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Pavord, I.D. ∙ Cox, G. ∙ Thomson, N.C. ...
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and AIR248
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studies found increased rates of bronchial irritation, chest discomfort, cough, discolored sputum, dyspnea, night awakenings, and wheezing during the 12-week BT treatment period. The AIR 2 extension study followed 162 of 190 participants treated with BT for up to 5 years after BT treatment.251
251.
Wechsler, M.E. ∙ Laviolette, M. ∙ Rubin, A.S. ...
Bronchial thermoplasty: long-term safety and effectiveness in patients with severe persistent asthma
J Allergy Clin Immunol. 2013; 132:1295-1302
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Long-term results from the RISA extension255
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Safety of bronchial thermoplasty in patients with severe refractory asthma
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and AIR extension250
250.
Thomson, N.C. ∙ Rubin, A.S. ∙ Niven, R.M. ...
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showed ongoing or new dyspnea (9.5% of participants), chest discomfort (4.8%-8.3%), bronchial irritation (2.4%), wheezing (4.8%-8.3%), and cough (4.8%) at the end of the 5-year study period. Hospitalizations during and after the treatment period were more frequent in patients treated with BT in all three studies.247-249
247.
Castro, M. ∙ Rubin, A.S. ∙ Laviolette, M. ...
Effectiveness and safety of bronchial thermoplasty in the treatment of severe asthma: a multicenter, randomized, double-blind, sham-controlled clinical trial
Am J Respir Crit Care Med. 2010; 181:116-124
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N Engl J Med. 2007; 356:1327-1337
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249.
Pavord, I.D. ∙ Cox, G. ∙ Thomson, N.C. ...
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In the AIR 2 study, 16 of 190 patients treated with BT and 2 of 98 patients in the control group were hospitalized during the treatment period. Ten of the 16 patient hospitalizations in patients treated with BT and both of the hospitalizations of patients in the control group were for worsening asthma. In the RISA study, 4 of 15 patients were hospitalized seven times during the 12 months after treatment, whereas none of the 17 patients in the standard care arm was hospitalized.248
248.
Cox, G. ∙ Thomson, N.C. ∙ Rubin, A.S. ...
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In addition to being hospitalized for worsening asthma, participants in the BT arms of the three studies were hospitalized for segmental atelectasis, lower respiratory tract infections, low FEV 1, hemoptysis, and an aspirated prosthetic tooth.247-249
247.
Castro, M. ∙ Rubin, A.S. ∙ Laviolette, M. ...
Effectiveness and safety of bronchial thermoplasty in the treatment of severe asthma: a multicenter, randomized, double-blind, sham-controlled clinical trial
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Scopus (419)
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249.
Pavord, I.D. ∙ Cox, G. ∙ Thomson, N.C. ...
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Twelve case reports and small case series reports252-254
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PubMed
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253.
Facciolongo, N. ∙ Menzella, F. ∙ Lusuardi, M. ...
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254.
Nguyen, D.V. ∙ Murin, S.
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Chest. 2016; 149:e95-e97
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256.
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also described adverse events, including hemoptysis in seven patients, atelectasis in six patients, and lower respiratory tract infections in three patients. One individual in these reports developed a mediastinal hematoma and bloody pleural effusion while on anticoagulation therapy for a pulmonary embolism. The authors of this case report believed that this effect resulted from a pseudoaneurysm of the pulmonary artery caused by the BT. Complications from case reports with one reported occurrence included a lung abscess, an inflammatory bronchial polyp, a pulmonary cyst, and a case of bronchiectasis.252-254
252.
Balu, A. ∙ Ryan, D. ∙ Niven, R.
Lung abscess as a complication of bronchial thermoplasty
J Asthma. 2015; 52:740-742
Crossref
Scopus (10)
PubMed
Google Scholar
253.
Facciolongo, N. ∙ Menzella, F. ∙ Lusuardi, M. ...
Recurrent lung atelectasis from fibrin plugs as a very early complication of bronchial thermoplasty: a case report
Multidiscip Respir Med. 2015; 10:9
Crossref
Scopus (10)
PubMed
Google Scholar
254.
Nguyen, D.V. ∙ Murin, S.
Bronchial artery pseudoaneurysm with major hemorrhage after bronchial thermoplasty
Chest. 2016; 149:e95-e97
Full Text
Full Text (PDF)
PubMed
Google Scholar
,256-264
256.
Cox, G. ∙ Miller, J.D. ∙ McWilliams, A. ...
Bronchial thermoplasty for asthma
Am J Respir Crit Care Med. 2006; 173:965-969
Crossref
Scopus (228)
PubMed
Google Scholar
257.
Doeing, D.C. ∙ Husain, A.N. ∙ Naureckas, E.T. ...
Bronchial thermoplasty failure in severe persistent asthma: a case report
J Asthma. 2013; 50:799-801
Crossref
Scopus (0)
PubMed
Google Scholar
258.
Doeing, D.C. ∙ Mahajan, A.K. ∙ White, S.R. ...
Safety and feasibility of bronchial thermoplasty in asthma patients with very severe fixed airflow obstruction: a case series
J Asthma. 2013; 50:215-218
Crossref
Scopus (0)
PubMed
Google Scholar
259.
Funatsu, A. ∙ Kobayashi, K. ∙ Iikura, M. ...
A case of pulmonary cyst and pneumothorax after bronchial thermoplasty
Respirol Case Rep. 2018; 6, e00286
PubMed
Google Scholar
260.
Han, X. ∙ Zhang, S. ∙ Zhao, W. ...
A successful bronchial thermoplasty procedure in a “very severe” asthma patient with rare complications: a case report
J Asthma. 2019; 56:1004-1007
Crossref
Scopus (2)
PubMed
Google Scholar
261.
Mahajan, A.K. ∙ Hogarth, D.K.
Bronchial thermoplasty: therapeutic success in severe asthma associated with persistent airflow obstruction
J Asthma. 2012; 49:527-529
Crossref
Scopus (3)
PubMed
Google Scholar
262.
Menzella, F. ∙ Lusuardi, M. ∙ Galeone, C. ...
Heat-induced necrosis after bronchial thermoplasty: a new concern?
Allergy Asthma Clin Immunol. 2018; 14:25
Crossref
Scopus (4)
PubMed
Google Scholar
263.
Qiu, M. ∙ Lai, Z. ∙ Wei, S. ...
Bronchiectasis after bronchial thermoplasty
J Thorac Dis. 2018; 10:E721-E726
Crossref
Scopus (4)
PubMed
Google Scholar
264.
Takeuchi, A. ∙ Kanemitsu, Y. ∙ Takakuwa, O. ...
A suspected case of inflammatory bronchial polyp induced by bronchial thermoplasty but resolved spontaneously
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None of the 15 studies reviewed (3 RCTs and 12 case reports and case series) attributed any deaths to BT.
Rationale and discussion
The data on the benefits and harms of BT derive primarily from three RCTs that enrolled a total of 432 patients in both the intervention and treatment arms. Overall, the improvements after BT were small, and the harms of BT were moderate. Long-term follow-up of a sufficient number of patients to fully assess clinical benefits and harms is lacking. The therapy may offer an acceptable benefit-to-harm ratio for some patients after careful shared decision making. Further research that includes randomized trials as well as long-term registry outcomes are desirable.
Future research opportunities
The Expert Panel identified the following research gaps:
•
Identify the population most likely to benefit from BT, such as individuals who have been treated unsuccessfully with different biologic agents.
•
Develop a registry to determine the risk of significant but rare long-term harms, such as bronchiectasis, vascular damage, and other lung complications. Follow both treated and untreated individuals over the long-term to determine whether side effects reported at 5 years in the AIR 2 study247
247.
Castro, M. ∙ Rubin, A.S. ∙ Laviolette, M. ...
Effectiveness and safety of bronchial thermoplasty in the treatment of severe asthma: a multicenter, randomized, double-blind, sham-controlled clinical trial
Am J Respir Crit Care Med. 2010; 181:116-124
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are more common in individuals treated with BT than in a control group.
•
Conduct RCTs and long-term registry studies of BT for asthma treatment, with appropriate controls and a sufficient number of patients, to fully assess the clinical benefits and harms of BT.
Acknowledgments
National Asthma Education and Prevention Program Coordinating Committee
Joseph Kofi Berko Jr, PhD
US Department of Housing and Urban Development
Sheila Brown
US Environmental Protection Agency
Kurtis S. Elward, MD
American Academy of Family Physicians
Anne Mentro Fitzpatrick, PhD, RN
Emory University School of Medicine
Lynn B. Gerald, PhD
University of Arizona
Fernando Holguin, MD, MPH
American Thoracic Society
Joy Hsu, MD
Centers for Disease Control and Prevention
Elliot Israel, MD
Harvard Medical School
Robert F. Lemanske Jr, MD
University of Wisconsin
Kenneth Mendez, MBA
Asthma and Allergy Foundation of America
Giselle Sarah Mosnaim, MD
American Academy of Allergy, Asthma & Immunology
Gary S. Rachelefsky, MD
American Academy of Pediatrics
Lisa M. Wheatley, MD, MPH
National Institute of Allergy and Infectious Disease
Juan P. Wisniesky, DPh, MD
Icahn School of Medicine at Mount Sinai
Darryl C. Zeldin, MD
National Institute of Environmental Health Sciences
National Heart, Lung, and Blood Institute staff
James P. Kiley, PhD
Director
Division of Lung Diseases
Cochair, NAEPPCC
George A. Mensah, MD, FACC
Director
Center for Translation Research and Implementation Science
Cochair, NAEPPCC
Cheryl A. Boyce, PhD
Chief, Implementation Science Branch
Center for Translation Research and Implementation Science
Jennifer Curry, MPH
Program Analyst
Center for Translation Research and Implementation Science
Lenora E. Johnson, DrPH, MPH
Director
Office of Science Policy, Engagement, Education, and Communications
Michelle M. Freemer, MD, MPH
Program Director
Division of Lung Diseases
Susan T. Shero, RN, BSN, MS
Program Officer
Implementation Science Branch
Center for Translation Research and Implementation Science
Executive Secretary, NAEPPCC
Westat staff
Russell E. Mardon, PhD
Project Director
Marguerite Campbell
Meeting and Logistical Support Manager
Sean Chickery, DHSc, MBA, MT(ASCP)
Topic Team Liaison
Susan Hassell, MPH
Project Manager
Nataly Johanson Tello
Topic Team Liaison
Maurice C. Johnson Jr, MPH
Topic Team Liaison
Karen Kaplan
Editorial Manager
Westat subcontractors
Eric Linskens
Health Science Specialist
Roderick MacDonald
Health Science Specialist
Judith Orvos, ELS
Orvos Communications, Senior Editor
Shahnaz Sultan, MD, MHSc
Guideline Methodologist
Supplementary data (1)
Document (1.48 MB)
Online Repository
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11148 | https://www.geogebra.org/m/bWmaQWb4 | The shortest distance between a line and a parallel plane. – GeoGebra
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The shortest distance between a line and a parallel plane.
Author:Mark Willis
1. Move the slider so that the line l is parallel to the plane 3x - y + z = 3; 2. Click to see the normal vector to the plane 3x - y + z = 3; 3. Click to see the parallel plane containing the line l; 4. Click to see the distance; 5. Use "Rotate 3D Graphics View" to view from different angles.
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11149 | https://faculty.cc.gatech.edu/~stasko/papers/pacificvis16-cluster.pdf | To appear in an IEEE VGTC sponsored conference proceedings Interactive Visual Co-Cluster Analysis of Bipartite Graphs Panpan Xu∗ Bosch Research North America Nan Cao† NYU Shanghai Huamin Qu‡ HKUST John Stasko§ Georgia Tech Fig. 1. The interface of our system showing the bipartite relation of U.S. senators’ support of bills and amendments based on roll call vote records. The main view (A) displays clusters of senators in the bottom half and bills in the top half, based on whether the senators support common sets of bills and whether the bills are supported by the same group of senators. The clusters are determined automatically with co-clustering algorithms and displayed via adjacency matrices showing the cohesiveness of the clusters, or a treemap-like space filling layout of the nodes. Using color coded party affiliations, an immediate observation is that the senators mostly vote in accordance with their parties.
Abstract—A bipartite graph models the relation between two different types of entities. It is applicable, for example, to describe persons’ affiliations to different social groups or their association with subjects such as topics of interest. In these applications, it is important to understand the connectivity patterns among the entities in the bipartite graph. For the example of a bipartite relation between persons and their topics of interest, people may form groups based on their common interests, and the topics also can be grouped or categorized based on the interested audiences. Co-clustering methods can identify such connectivity patterns and find clusters within the two types of entities simultaneously. In this paper, we propose an interactive visualization design that incorporates co-clustering methods to facilitate the identification of node clusters formed by their common connections in a bipartite graph. Besides highlighting the automatically detected node clusters and the connections among them, the visual interface also provides visual cues for evaluating the homogeneity of the bipartite connections in a cluster, identifying potential outliers, and analyzing the correlation of node attributes with the cluster structure. The interactive visual interface allows users to flexibly adjust the node grouping to incorporate their prior knowledge of the domain, either by direct manipulation (i.e., splitting and merging the clusters), or by providing explicit feedback on the cluster quality, based on which the system will learn a parametrization of the co-clustering algorithm to better align with the users’ notion of node similarity. To demonstrate the utility of the system, we present two example usage scenarios on real world datasets.
1 INTRODUCTION ∗e-mail: panpan.xu@us.bosch.com †e-mail: nan.cao@nyu.edu ‡e-mail:huamin@cse.ust.hk §e-mail:stasko@cc.gatech.edu Bipartite relations, the connections between two different types of entities, play a key role in gaining insight from data in many application domains.
Examples of bipartite relations include the votes from legislators for the passage of bills and amendments, the involvement of researchers in various topics, and the affiliation of individuals with different social groups .
In bipartite relation data, frequently a person not only wants to know the neighbors and links of individual nodes, but also the commonality in their connections. For example, to what extent do two or more 1 researchers investigate similar topics, or how much do senators agree in their votes? Analyzing the commonality of the connections results in clusters of researchers working on the same topics, or groups of legislators supporting similar sets of bills. Moreover, the analysis of similarity in connections can be applied to both types of nodes in the same manner, which is the notion of duality in bipartite relation analysis . For instance, it is possible to identify not only the senators who have voted for a similar set of bills, but also the group of bills that are supported (or not) by similar groups of senators.
Computational approaches including various clustering and co-clustering algorithms [7, 8, 9, 13, 20] can be applied to identify cohesive node groups with similar bipartite connections. Visually denoting the clusters found by the algorithms in a bipartite graph visualization facili-tates the process of understanding and cluster identification by relieving the analysts of the burden of identifying and comparing the links indi-vidually for each node. By aggregating the nodes, the visual complexity is also reduced, and an overview of the connectivity is possible.
However, we argue that the clusters obtained by running such algorithms should serve as a starting point, rather than the end of analysis, for several reasons. First, the clustering results generated by these algorithms may not be very helpful due to suboptimal parameter settings (e.g., number of clusters [7, 8, 9]). Some clusters might not be very cohesive because nodes that differ significantly in their bipartite connections were grouped together. Second, the ability to explore subspace clusters is desirable in many application scenarios.
For example, in clustering the bills, an analyst may consider grouping them based on the support from a particular political party, instead of all the legislators. In this case, the bill clusters are identified within a subspace, if we consider each bill as a data item and each legislator as a dimension . Third, in bipartite relation data, the nodes are often associated with domain specific features. These features are important for integrating analysts’ prior knowledge, and for generating insight about the correlation of node attributes with node groups.
To address these shortcomings, we advocate for a visual analytics approach. It fits well here because of the need for both automated cluster analysis and the active engagement of analysts to evaluate and refine the clusters, drill down to a subspace, and explore the correlation between node attributes and their bipartite connectivities [29, 44].Often, these visual analysis tasks (e.g., refining the cluster) require taking into account analysts’ notions of node similarities, which might differ from a straightforward comparison of all the bipartite connections for a pair of nodes. Thus, we introduce a prototype visual analytics system that incorporates a semi-supervised clustering approach, which learns similarity metrics for nodes based on user specified constraints. In this way, analysts’ judgments can be included in the automatic clustering procedure. The major contributions of this work are: • The task analysis and design of a visual analytics system for the explorative analysis of connectivity patterns in bipartite graphs through flexible node grouping. Our system employs automatic analysis algorithms, novel visual representations, and advanced interactions to help analysts identify, interpret, compare, and refine co-cluster patterns in bipartite graphs.
• A novel algorithm that applies Laplacian Regularized Metric Learning (LRML) , an efficient metric learning method, for semi-supervised co-clustering analysis of bipartite graphs.
• A flexible visualization designed for representing clusters in bipartite graphs that illustrates both relational and feature patterns of nodes in clusters via an adjacency matrix and treemap respectively.
• The demonstration of the utility of the system through example use scenarios on two datasets: the roll call vote records of US Senators on the passage of bills and amendments, and the topic interest of researchers.
2 BACKGROUND Given a bipartite graph as described in the Introduction, various cluster-ing algorithms can be employed to detect groups of nodes that are in some sense similar in their bipartite connections. For example, given the bipartite graph describing how scientists jointly publish papers, we can derive a co-authorship graph and apply community detection algo-rithms to identify groups of researchers working closely together. For clustering algorithms in general, a good result has high inner-cluster similarity (i.e., the data items within the same cluster are similar) and low inter-cluster similarity (i.e., the data items in different clusters are dissimilar).
Co-clustering is another category of clustering algorithms which is able to detect node groupings on the two modes in a bipartite graph simultaneously. Examples of co-clustering algorithms include [8, 9, 7, 13]. The co-clustering algorithms have been applied in differ-ent contexts such as analyzing gene expression data to find the relations between genes and conditions, grouping documents and words to iden-tify topic groups, and so forth. Most of the co-clustering algorithms can generate row and column clusters (hence clusters in the two modes) given the biadjacency matrix, create a “checkerboard” structure which reflects the common bipartite connections from the nodes within the same row / column clusters. In the prototype, we apply a spectral co-clustering algorithm , although it should be noted that other algorithms can also be integrated in the framework.
3 RELATED WORK 3.1 Graph Visualization Node-link diagrams and adjacency matrices are the two major visualization techniques for graphs [25, 45].
Adjacency matrices generally introduce less visual clutter for denser graphs, but node-link diagrams are more familiar for general users and are arguably more intuitive for understanding a graph’s structure [17, 24]. Hybrid designs combine a node-link diagram and an adjacency matrix by showing the two views simutaneously and synchronizing the interactions on the views , or showing the graph structure at global and community level with varying visual representations [24, 36]. We also adopt a hybrid visual design in the system introduced in this paper, to represent both the bipartite connections and the projected one-mode graph.
3.2 Bipartite and Multimodal Graph Visualization A common approach for visualizing bipartite graphs, or more generally, multimodal graphs, is to allocate separate panes/lists for the different types of nodes. Graphical links connect nodes on different panes, and are displayed selectively based on the user’s current focus. Variations of this approach can be found in semantic substrates , in the List View in Jigsaw , in linked tabular views , and more recently, in PivotPath and MMGraph . Using color, shape or other visual channels to denote the node type instead of spatially separating the different types of nodes, and drawing the graph in a unimodal style is also a possible choice. OntoVis and FacetAtlas are examples.
Ploceus and Orion focus on interactive graph modeling and transformation. One important transformative operation supported by the two systems is projection, which derives, for example, the co-vote relations among legislators given their voting records. For attributed multimodal graphs, GraphTrail and NetLens aggregate along the nodes’ attributes and display views of summary statistics.
3.3 Co-clustering Visualization Co-clustering (or biclustering, subspace clustering) has been widely adopted for analyzing gene expression data in bioinformatics , where the genes and the conditions (which correspond to the two types of entities in bipartite graphs) are clustered simultaneously based on the expression levels. Various visualization systems have been developed to help review and analyze the resulting clusters [1, 18, 22, 37].
Recently, the approach is applied to intelligence analysis, where coordinated relations between sets of entities of different types (people, locations, etc.) are identified and visualized [14, 43]. The entities analyzed can be extracted from textual datasets, connected by their co-occurence in documents. Other work visualizing multimode graphs such as FacetAtlas and SolarMap focuses on one-mode clusters and their interconnections through indirect links.
In this article, we propose an interactive visualization for analyzing bipartite graph data.
Our method combines clustering algorithms and visualization, which facilitates the identification of connectivity 2 To appear in an IEEE VGTC sponsored conference proceedings patterns in bipartite graph data while providing users with enough flexibility to explore based on their domain knowledge and other criteria, thus differentiating it from the previous work.
4 SYSTEM DESIGN AND IMPLEMENTATION In this section, we explain the design and implementation of our system for co-cluster analysis of bipartite graphs. It is based on a set of design requirements that we formulated by reviewing representative analysis tasks on bipartite graphs from a number of different domains. We begin below by listing the requirements. We focus primarily on tasks related to the understanding of connectivity patterns of a bipartite graph.
R1 Determine node connections across the two groups. Connections between nodes across the two groups should be evident. For example, which bills did a senator vote in favor of?
R2 Determine specific attribute values of each type of node. The different attributes of multivariate nodes should be evident. For example, from which political party and state is a particular sena-tor?
R3 Identify similar nodes of each of the two types. The system should determine and present clusters of similar nodes based on the bipartite connection. For example, which senators voted similarly toward all the bills?
R4 Analyze correlation of the connectivity pattern with domain spe-cific node attributes. Mechanisms for selecting and visualizing relevant node attributes should be provided in the system, so that the analysts can inspect their correlation with the bipartite connec-tions to draw insights. For example, which bills did Democratic senators vote for consistently?
R5 Interpret the node clusters through bipartite relations. Given a node cluster, it is important to be able to understand which common neighbors made the nodes similar to each other. Thus, besides the clusters of nodes, the bipartite relation also should be viewable. For example, for a cluster of senators, which votes made them be grouped together?
R6 Explore subspace clusters. The system should provide means for selecting subsets of nodes to manually identify subspace clusters.
The visualization should be updated upon selection to reflect the cohesiveness of node clusters within the subspace. For example, the user should be able to select a subset of senators and observe how similar they are in different ways.
R7 Evaluate and refine node clusters. To assist the search for high quality node groupings, the system needs to provide effective vi-sual cues to communicate the cohesiveness of the clusters, and the degree of separation among them. Moreover, if nodes in a cluster are dissimilar, the visual design should enable the identification of subsets that are more coherent than the others in their bipar-tite connections to suggest potential ways to regroup the nodes.
Analysts should be able to merge and split the existing node clusters by directly interacting with the visual representations.
The system should give immediate feedback on the cohesiveness of updated clusters to effectively guide the exploration process.
Besides providing the visual cues and the interactive tools for the user to regroup the nodes, the system also should ingest the user’s notion of similarity and learn a proper metric to compare the connectivity of two nodes. For example, if the user feels a system-determined cluster of senators is not appropriate, it should be possible to change the clustering and have the system interpret and learn from that.
4.1 System Overview With the above tasks and requirements as a basis, we designed and implemented a system for flexible exploration of bipartitie graphs. Our visual analysis system follows the architectural design illustrated in Fig. 2.
Once a dataset is loaded into the system, the analysis module will detect the co-clusters and send the results to be visualized. Analysts navigate through the visual representations of co-clustering results, and refine the clusters based on their prior knowledge, preferences, and judgments by directly interacting with the visualizations. The analysts’ beliefs will be fed back into a metric learning model, and captured in a distance metric that will be later used for the next round of co-clustering analysis.
Fig. 2. System overview.
The visualization module consists of multiple visualization regions which are coordinated in the user interface shown in Fig. 1. The UI consists of four major components. The first region (A) contains the bipartite graph representation that shows the co-clustering results. It al-lows users to directly interact with and refine the node clusters through splitting and merging, or specify the groups of nodes they consider as similar to each other. This region provides two alternative visual displays of each node cluster: an adjacency matrix of the projected graph and a treemap-like compact packing of the nodes. The second region (B) contains a Table Lens display for visualizing the node attributes. The categorical and numerical attribute values are encoded graphically via the colors and lengths of the rectangles within the table cells. The third region (C) provides a data manipulation panel including tools for users to load different datasets and query the nodes by labels.
The final region (D) contains a control panel for selecting alternative metrics for measuring the similarity of bipartite connections for a pair of nodes, considering either a subspace or the full set of dimensions.
4.2 Analytic Support Before describing details of the system interface and interactions, we explain the metric learning approach applied for co-cluster refinement.
Here, we formally denote a bipartite graph as G = (U,V,E), where U and V are non-overlapping sets corresponding to the two types of entities. E ⊆U ×V is the bipartite relation between the two sets of entities. The edges in E are undirected and can be weighted.
4.2.1 Semi-Supervised Co-Cluster Analysis We introduce a semi-supervised co-clustering method for analyzing bipartite graphs as summarized in Alg. 1, which learns distance metrics between the nodes based on analysts’ input and uses it for co-clustering analysis. It is a combination of two techniques: (1) the co-cluster analysis of bipartite graphs with the learned distance metrics in the form of weightings on the nodes U and V in the bipartite graph, and (2) the semi-supervised metric learning for finding the distance metrics based on the similarity constraints specified by the users.
4.2.2 Metric Learning In cluster analysis, determining a proper distance metric to estimate the similarities among data items is the most important task that directly determines the quality of the cluster results. In most cases, distance metrics are predefined and remain unchanged during the clustering procedure, and the most familiar one is Euclidean distance. However, the metric learning technique suggests an adaptive procedure in which the distance metric is learned based on several examples of the similarities between data items that are manually labeled by the analyzer, thus capturing their prior knowledge and intuition, and making it a good match for interactive visual analysis. Particularly, with information on pairs of nodes that are labeled as ‘similar’ to each other, the metric learning algorithms can find a parametrization of a family of distance functions (e.g., Mahalanobis distances) such that the computed distances between ‘similar’ pairs are small.
3 A bipartite graph G = (U,V,E), can be represented as a biadjacency matrix B∥U∥×∥V∥. The row vectors Bi· and the column vectors B· j correspond to the bipartite connections of node ui ∈U and node vj ∈V.
We employ a semi-supervised metric learning method, Laplacian Regularized Metric Learning (LRML) , to find the distance metrics between two row vectors or two column vectors given a set of similarity constraints on the nodes in either partition specified by the analysts.
In particular, LRML finds a parameterization of the Mahalanobis distance for measuring the distance between a pair of nodes. For example, for two nodes ui0 and ui1 in U, the distance function is: d(ui0,ui1)A = q (Bi0· −Bi1·)T A(Bi0· −Bi1·) (1) where A is a semi-positive definite matrix of size |V|×|V|. When A is the identity matrix, the metric reduces to the Euclidean distance.
When A is a diagonal matrix, the value of the diagonal entry Aj j would reflect the relative importance of the node v j in V for measuring the similarity of the nodes ui0 and ui1 in U. When applying LRML to semi-supervised co-clustering analysis, instead of a full matrix A, we seek for a diagonal matrix as a solution to the optimization problem.
This is due to considerations on both the time complexity for solving the optimization problem and the interpretability of the results. As discussed above, the entries on the diagonal of A correspond to nodes in V and the value of the entries is the importance of the node when comparing the bipartite connection of a pair of nodes in U. In the following discussion, we will simply denote the diagonal of A as Wv.
In the LRML algorithm, given the similarity constraint S for pairs of nodes in U, LRML seeks the matrix A that minimizes the sum of distances for the pairs of nodes in S. A Laplacian regularization term is also included in the optimization goal.
min A⪰0 ∑ (ui0,ui1)∈S d(ui0,ui1)A +λ ∑ ui0∈U ∑ ui1∈U w(ui0,ui1)d(ui0,ui1)A −εlog(det(A)) w(ui0,ui1) = ( 1,if ui0 ∈N(ui1) 0,otherwise N(ui1) is the nearest neighbor list of ui1, which includes the nodes in U with the most similar bipartite connections to ui1 in terms of Euclidean distance. The term containing log(det(A)) is introduced to avoid trivial solutions where all the entries of A are zero.
We respectively apply the above metric learning algorithms for nodes in U and V based on the specified similarity constraints Su for nodes in U and Sv for nodes in V (line 2,3 in Alg. 1), obtaining weights Wu and Wv. These weights indicate how important a node and its cor-responding relationships should be considered during the co-clustering procedure in terms of the given set of similarity constraints.
We “modulate” the biadjacency matrix B by these weight vectors (line 4,5 in Alg. 1), making the entry B′ i,j (i.e., the bipartite link connecting nodes ui and vj) weighted, for example, comparatively less if both ui and vj’s weights found by the metric learning algorithm are small. When compared to links connecting nodes with higher weights, those links will have less effect on the co-clustering results.
The parameter α (line 5 in Alg. 1) controls the extent to which the user specified constraints should affect the clustering results. The co-clustering algorithm used in the prototype is the spectral co-clustering algorithm implemented in python scikit-learn package . It requires the input of a pre-specified number of clusters. After the user input similarity constraints, the co-clustering algorithm is carried out again, with the same specified number of clusters prior to taking any user constraints into account. In this process, the weightings on the nodes learned by the LRML algorithm are used to modulate the weights of the bipartite connections, and this is how the user input is incorporated Input : Bipartite Graph G = (U,V,E), Similarity constraints Su and Sv for nodes in U and V, a k-nearest neighbor graph Gu and Gv for nodes in U and V Output: Grouping of nodes in U and V 1 begin 2 Wu = LRML(Su, G, Gu); 3 Wv = LRML(Sv, G, Gv); 4 B = bijaceny matrix(G); 5 B′ = B+αW T u ×B×Wv; 6 co-clustering(B′); 7 end Algorithm 1: Metric learning for co-cluster refinement in the co-clustering algorithm. Finally, the nodes are clustered by running the co-clustering algorithms on B′ (line 6 in Alg. 1).
There are a number of reasons of choosing LRML instead of other metric learning methods: (1) it is a semi-supervised algorithm which requires only partial labeling of the similar node pairs, thus it imposes lower demand on the users; (2) updating the parametrization of the distance function can be done efficiently without iterative solvers, and the efficiency is desirable for user-facing systems; (3) when the matrix A that parametrizes the Mahalanobis distance function is required to be diagonal, the value of corresponding entries on the diagonal can be interpreted as the weight of the features in the modulated distance metric. By graphically showing the computed weights in the data table (Fig. 1), analysts can reason about which features are more relevant for the specified pairs of nodes to be considered as similar to each other.
4.3 Visual Representation The tasks and requirements described earlier guided our visualization design. We employ two iconic visual representations of clusters, adja-cency matrices and treemaps, inspired by NodeTrix , and Dicon respectively. The representations are switchable for revealing relational (R3, R5, R6, R7) and attribute (R2, R4) patterns of the nodes within the clusters. The main view is divided horizontally with the nodes and clusters of one set on top and the other on the bottom. They are connected by aggregated bipartite links (R5). We also provide a set of interactions for exploring and refining the co-clusters (R6, R7). The details of these designs are explained in this section.
4.3.1 Visualizing Node Clusters as Adjacency Matrices A (a) (b) (c) B C D A B C D 3 3 3 1 1 1 A B C D A B C D Fig. 3. Projection of the bipartite relation onto a group of nodes: (a) the bipartite graph with a cluster consists of node A, B, C and D; (b) a one-mode graph formed by projecting the bipartite relation on the cluster, the weight of the edges is the number of common neighbors of two nodes (i.e., concordance); (c) the adjacency matrix displays the weighted one-mode graph. Nodes A, B, and C have similar bipartite connections.
The process of visualizing node clusters as adjacency matrices con-sists of two major steps: (1) transforming the bipartite relations into weighted one-mode graphs through projection; (2) visualizing the pro-jected graphs in the form of adjacency matrices.
Bipartite graph projection. For the nodes within a cluster inU, key questions are how to visually encode the consistency of their bipartite connections, and how to help analysts identify subsets of nodes with comparatively similar bipartite affiliations (R7). Here, we transform the bipartite relations into a weighted one-mode graph through projection, 4 To appear in an IEEE VGTC sponsored conference proceedings a common operation for bipartite graph analysis. As illustrated in Fig. 3(a) and (b), every two nodes that are affiliated with common neighbors are connected in the projection, and the edge weights reflect the resemblance of the bipartite relations for the pairs.
Either concordance, which is the number of common neighbors of two nodes (Fig. 3(b)), or Jaccard Index, or the Mahalanobis distance found by the aforementioned metric learning algorithm can be used to measure the similarity of the neighborhoods of two nodes. The metric learning procedure obtains weightings on U and V for computing the Mahalanobis distance between any two nodes. The distance obtained reflects the user’s notion of similarities.
Using concordance shows the number of common connections, the viewer can easily read, for example, the number of bills that two senators both voted for, or the number of papers two researchers have co-authored. Using Mahalanobis distance shows the similarity of nodes pairs which are adjusted based on user’s input.
Visualizing the projected one-mode graphs. The projected one-mode graphs are visually represented as adjacency matrices (Fig. 3(c)).
In these matrices, the color intensities of the matrix cells are mapped to the edge weights in the projected graph. With a proper permutation of the rows and columns, blocks in the matrix with higher and more consistent intensity, which signify subsets with more coherent bipartite connections, can be identified. For example, Fig. 3(b) may show a clus-ter of four senators where the first three are more similar to each other than the fourth. In the prototype, a spectral node sequencing algorithm is utilized. It obtains an ordering of the nodes based on the Fielder vector of the Laplacian matrix of the weighted one-mode graph.
Here, we leverage a matrix visualization design for two reasons.
First, with the encoding scheme described above, the matrix view will facilitate the evaluation of cluster qualities (R7). A homogeneous color distribution among matrix cells indicates a cohesive cluster with high intra-group similarity, while a heterogeneous color distribution indicates low cluster quality. Second, the projected one-mode graphs are usually dense, and can be visualized with much less visual clutter with adjacency matrices compared to node-link diagrams .
In the matrix representation, a categorical or numerical attribute of the nodes can be encoded with a colored circle beside each row of the corresponding matrix, as shown in Fig. 1 (A.2). In this example, the party affiliation of the senators is displayed, illustrating its correlation with the votes the senators cast for bills (R4). Other node attributes are displayed in a data table (Fig. 1 (B)), showing additional details (R2).
4.3.2 Visualizing Node Clusters as Treemaps The treemaps consume less space than adjacency matrices (O(N) and O(N2)), and display the distribution of attribute values more effectively (Fig. 1 A.4). In the treemaps, each rectangle corresponds to a node within the cluster. The color intensity or hue of the rectangles encode a selected categorical or numerical attribute of the nodes.
The nodes(/rectangles) are organized with two levels of nesting: the children of the root correspond to all the possible categorical attribute values or intervals of numerical values for the nodes in the cluster, and the nodes are attached to the children by the respective attribute values.
Such nesting structure results in a layout where the distribution of the attribute values can be easily identified (R2). The treemap display also helps analyze the correlation between the bipartite relation to the nodes’ attributes – We can compare if nodes in different clusters have different attribute values more easily with treemaps (R4).
4.3.3 Visualizing Node Attributes A TableLens style visualization displays the details of each data record (node), illustrating their textual, categorical, or numerical attributes (R2). The attribute values are graphically displayed as color coded blocks or bars with varying lengths based on the data attribute types. The Table Lens supports a generalized table row selection scheme: when the user hovers over an entry in the table, the other rows with similar attribute values will be highlighted and they can be selected by simply clicking on the entry. All selections and highlights on the entries in the table lens are linked with the bipartite graph display.
4.3.4 Cluster Layout In the main view of our system, as shown in Fig. 1 (A), we place the two types of node clusters produced by co-clustering algorithms in two horizontally separated regions. The layout of the node clusters works in two steps. First, an initial horizontal ordering of the node clusters on the two display regions is obtained with the barycentric heuristic . Second, we adjust the layout to remove overlap of the clusters by applying a force-directed layout algorithm. In the force directed layout phase, the relative horizontal positions of the nodes are contrained to be the same as from the barycentric algorithm, therefore no more edge crossing should be introduced.
4.3.5 Visualizing Bipartite Connections It is important to show the bipartite connections (R1, R5), but to do so on a one-to-one basis will introduce severe edge clutter. Hence, we de-cided to aggregate the edges for overview and show individual bipartite links through detail-on-demand interaction. The aggregations (bundles) of the bipartite connections are derived based on the node clusters.
The total number of individual edges between two clusters cu and cv is normalized by the maximum number (|cu|×|cv|) of possible edges between the pair of node clusters. The weight of each bundle are then mapped to the width of the links connecting node clusters. Displaying aggregated edges reduces the large amount of crossings and the visual clutter that may result from drawing individual ones.
As the node clusters are vertically spreaded, the node clusters may overlap with the bundled edges and introduce ambiguities. To solve this issue, edge routing techniques can be applied here. Examples of edge routing techniques include and . These techniques can be incor-porated into the current system to improve the readability of the edges.
4.4 User Interaction Fig. 4. The buttons affiliated with each cluster visualization that support cluster representation and refinement.
Based on the design requirements, we implemented the following interactions in the system: Details-on-demand (R1). Besides analyzing the bipartite relations at the level of node clusters, we also design interactions for viewing the details of the interested nodes. For example, as the user hovers on a node in the matrix or treemap, its neighbors will be highlighted and the highlighted state persists after click.
Subspace selection (R6). The analyst can brush on a subset of nodes in one mode, which will be regarded as a subspace in which the similarities among the nodes and the corresponding subspace clusters in the other mode are derived. Here we consider two approaches for specifying a subset of nodes (i.e., the dimensions in the subspace) for analysis. In the first approach, the user directly selects related nodes from the adjacency matrices. In the second approach, the user specifies the value of an attribute and all the nodes with attribute value equal to that will be selected as a subspace. This can be done easily with the generalized selection scheme supported in the Table Lens display.
Cluster refinement (R7).
The user can refine the clusters by directly manipulating the visual data items, and explicitly specify the grouping, or mark their confidence in the cluster quality to specify similarity constraints for the metric learning algorithm. Specifically, users can directly interact with the matrices or treemaps to split selected nodes from an existing cluster, or merge two node clusters into one. Two node clusters are merged when the user drags one matrix or treemap and drops it onto another. Users can split a subset of nodes from node clusters by first selecting the nodes to be separated and 5 then click the scissor button shown in Fig. 4. Analysts can specify similarity constraints by identifying a group of nodes which are similar to each other via clicking on the similarity button shown on top of the corresponding node cluster visualization (Fig. 4).
4.5 Alternative Designs Besides the designs presented above, alternative approaches exist for visually representing a bipartite graph and embedding node group information. One example is a biadjacency matrix with the rows and columns aggregated based on the node cluster they belong to or serialized with matrix reordering algorithms . However, it is relatively difficult with this visualization to compare and access the similarity of the connections for a group of nodes, as the analyst needs to scan over all the rows/column and compare the entries, which can be mentally demanding. It is also possible to use a node-link diagram to represent the bipartite graph and place nodes with similar connections in close proximity, however, it is also hard to compare the connectivities of the nodes directly with it. In our system, we made the decision to visually represent the projected one-mode graph for a node cluster as an adjacency matrix to make these tasks easier. Another possibility is to visualize the entire one-mode projection with node-link diagrams or adjacency matrices, e.g., the co-author graph derived from a bipartite relation between authors and papers. However, the bipartite relation can be important in terms of interpreting how the links and clusters in the projected graphs are formed.
5 SAMPLE USAGE SCENARIOS 5.1 Scenario 1: US Senate Votes We applied the proposed system to analyze the voting behaviours of legislators. We collected the roll-call voting records on the passage of bills and amendments in the U.S. Senate in 2012 from Govtrack.us .
These data include information about whether the senators voted ‘yea’ or ‘nay’ on the bills and amendments. There were 22 bills and 117 amendments voted on by 100 senators, among which 43 bills or amendments were passed or enacted. The bills are further categorized by their subject matter, including economics and public finance, education, and armed forces and national security. A bipartite connection is established if a senator votes for the passage of a bill, in total there are 6962 edges.
Enzi Barrasso Vitter Crapo Risch Shelby Cornyn state hr4310-112 s3254-112 samdt3262-112 samdt3232-112 samdt3158-112 (a) (b) M2’ M3 LA AL TX ID ID WY WY Fig. 5. Investigating the correlation between similarity of bipartite connec-tions and node attributes: (a) a group of bills related to the subject armed forces and national security are supported by a similar set of senators, (b) republicans from the same state vote similarly.
Interpretation of co-clustering results.
After running the co-clustering algorithm on the data and initializing the visualization (Fig. 1), we immediately observed that the clusters of senators (bottom) and bills (top) are reasonably cohesive. In this figure, the color intensity of the matrix cells encodes the measure of concordance, i.e., the number of common voters of two bills/amendments, and the number of bills/amendments supported by both senators. It can be observed that the color is relatively even over most of the matrices, indicating a consensus of opinion for each cluster of senators, and conversely, the bills within each cluster have a common set of supporters. A few senators and bills do not seem to align well with the computed clusters, which can be split out via the supported interactions.
The overall layout of the aggregated bipartite graph shows two distinctive groups of senator clusters (bottom) as well as bill clusters (top). Not surprisingly, the senators are generally grouped by their party affiliations, as indicated by the colored circles on the sides of the matrix showing the party affiliation of each senator, or the colors of the rectangles within the treemaps. In the layout, clusters of bills that lie off the center are those that have received party-line votes, i.e., a majority of a party vote for the bills in the same way in opposition to the other party. We can also identify some clusters of bills that lie in the middle, receiving votes from both parties. Amongst the clusters of senators there were some outliers with senators from both parties (Fig. 1). For example, the cluster M1 in Fig. 1 was composed of one Republican (Scott Brown) and many Democrats. Hovering over the node highlights the bills he supported. We found that his voting preference aligns with the Democrats to some extent.
(a) (b) (c) Fig. 6. A user marked all the Republicans as similar to each other (a), and the cluster of senators (b) found by the semi-supervised co-clustering algorithm, the clustering algorithm found Democrats with relatively neutral political standings (c).
In general, there was no clear correlation between the subject matter of the bills and the degree of overlap of their supporters. Though we can still identify in cluster M2, many of the bills were related to armed forces and national security (Fig. 1). The bills can be separated and form a more coherent cluster (M2’ in Fig. 5(a)). The bipartite links showed that the cluster received support from most senators in both parties. In another cluster (M5) (Fig. 1) supported mostly by Democrats three bills were related to crime and law enforcement. Two of them were about the Foreign Intelligence Surveillance Act.
In cluster M3 (Fig. 5 (b)), we found several pairs of Republicans align better with each other in their voting preferences than others in the cluster. In a further investigation of the reasons by brushing the nodes, and highlight them in the data tables, we found the Republicans were actually paired by their home states.
Semi-supervised co-cluster analysis.
When we applied the semi-supervised co-clustering analysis based on metric learning in the above data, there were more interesting findings. Specifically, we grouped all the Republicans in the dataset into the same cluster (Fig. 6(a)) as a set of similarity constraint fed back into the proposed semi-supervised co-clustering algorithm and ran the algorithm again.
The resulting cluster consisted of all the Republicans and, interestingly, three Democrats, Claire McCaskill, Jay Rockefeller, and Ben Nelson, as shown in Fig. 6(b). After searching on the related news reports and profiles of these senators, we found Jay Rockefeller is the only Democrat in the Rockefeller family dynasty (which was supported by Republicans); Ben Nelson is regarded as ’the least likely Democrat to stand with his party on legislative initiative’ ; Claire McCaskill also had a moderate political standing . These findings demonstrate the power of the proposed algorithm.
6 To appear in an IEEE VGTC sponsored conference proceedings Table 1. Guiding questions for domain expert interviews.
Aim Question Q1 Visual Design Are the matrix/treemap representations of clusters informative to you? and Why?
Q2 Visual Design Does the current layout of bipartite graphs provide a clear view of different types of nodes? and Why?
Q3 Interaction Design Are the current interactions useful and easy to use? and Why?
Q4 Visual Analysis Can you identify different patterns in the matrix/treemap based on different node similarity metrics?
Q5 Visual Analysis Can you interpret the co-clustering results in contexts of different data?
Q6 Visual Analysis Do you think the semi-supervised co-clustering is useful and can help better analysis the bipartite graph?
Q7 General How do you like the system overall in exploring bipartite graphs and identifying insights?
5.2 Scenario 2: Publication Data Our system has also been used for analyzing the publication records in DBLP, a bibliographic database for Computer Science.
We collected the bipartite relation of the authors’ affiliation with different conferences.
In the analysis, we analyze the authors who have published in more than 12 conferences/journals. The filtered bipartite graph extracted from this dataset consists of 146 authors and 77 conferences/journals. In total there are 2010 edges.
Fig. 7. A group of researchers (a) working on algorithms for massive data processing. They publish mostly in the theoretical computer science and data mining conferences (b) as shown by the thick bundles connecting the author cluster and the conference clusters (d). The user marked (a) as a meaningful cluster. The group expanded and new authors who work in similar area are included (c) after running the co-clustering again.
We applied the semi-supervised co-cluster analysis on this dataset.
From the initial clustering result, we identified a group of researchers (Fig. 7(a)) working on algorithms for massive data processing and have publications mostly in the areas of theoretical computer science and data mining (Fig. 7(b)), as indicated by the thick bipartite link bundles connecting the cluster of authors to the two clusters of conferences (Fig. 7(d)). We mark the cluster of researchers as a meaningful one and after using this cluster as the similarity constraint in the semi-supervised co-clustering algorithm, we successfully found more researchers in these research areas who published in similar conferences, and the original group of authors is expanded (Fig. 7(c)).
6 EXPERT REVIEW To evaluate the system’s usability and potential utility, we conducted interviews with expert users (denoted as E1 and E2) who were senior researchers in the area of machine learning and specialized in active learning techniques, and nonexpert users (denoted as E3 and E4) who are graduate students with domain knowledge in the area of social and political science. In each interview, we first demonstrated all the key features of the system to the expert in a tutorial and encouraged the person to use the system on their own based on a preloaded dataset.
They were also encouraged to ask questions whenever they met prob-lems. After they fully explored the tool’s functionality, we conducted a semi-structured interview guided by a set of questions summarized in Table 1. The interviews took about one hour each. We deliberately asked the experts not to be constrained by the questions listed, but to use them as a guide and elaborate their thoughts when using the tool based on their domain knowledge. We recorded the interviews and took notes of the experts’ comments during the interview.
Results. The experts were impressed by the system regardless of their domain expertise area. They believed the system was “efficient for exploring clusters [in bipartite graphs]”. Experts with machine learning background also think the system is a “very useful tool for co-cluster analysis”. We summarize their comments as follows.
Visual and interaction designs.
The experts believed the visu-alization generates “informative” and “aesthetic” results and were impressed by the rich interactions. E2 particularly liked the idea of visualizing clusters in both treemaps and matrices and said: “this [design] is very intuitive ... and now I can find both relational patterns of cluster member and patterns of their feature [inside each cluster]”.
E3 liked the matrix design and said: “with this I can immediately see and compare the senators”, although E3 felt the Treemap is less intuitive. E3 and E4 also commented that the connections between bills and senators are depicted in a clear way with the bundled links.
Visual analysis. Both E1 and E2 confirmed the novelty of the system in terms of its visual analysis functions. In particular, E1 said: “this is a very interesting idea and little work has been done for analyzing bipartite graphs in a semi-supervised approach”. E2 also commented that “this interactive visualization is much better than those active learning systems [developed based on traditional UIs] that I know” and “Interactively editing on those clusters is very interesting [when compared with traditional data labeling procedures.]”. E1 and E2 also believed that the semi-supervised co-clustering algorithm we proposed, although still with room to improve, was a promising idea and generated correct and meaningful results.
Discussion. In addition to the above positive feedback, the experts also identified some limitations of the current system. First, the system may not be scalable enough. However, they mentioned that the query-ing and filtering functions in the system were helpful for navigating through larger datasets. E1 and E2 also commented that all active learn-ing techniques had a common goal of training the underlying analysis module based only on a small set of high quality samples that are pre-cisely labeled by the analyst, instead of based on a large set of imprecise samples. Therefore, the scalability may not be a problem for the system in this case. The second issue they mentioned was that the visualization designs, although very informative, encoded large amounts of infor-mation which took effort to learn. But they both agreed that this issue was primarily due to the complexity of the analysis problem itself and indeed many data properties need to be represented in the visualization.
7 CONCLUSION We introduce an interactive visual analysis system for analyzing clus-ters in a bipartite graph. We employ adjacency matrices and treemaps to visually encode the clusters. We illustrate the proposed techniques 7 via two case studies on real world datasets. Our investigations identify many interesting findings that help to illustrate the usefulness of the system. In the future, we plan to apply the system to analyze bipartite graphs in other application domains, conduct user studies, and extend the functionality of the system for analyzing not only bipartite graphs, but also multimode graphs.
ACKNOWLEDGMENTS This work is partially supported by RGC GRF 618313, HKUST Over-sea Research Grant, and HKUST Postdoctoral Matching Fund.
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11150 | http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magloo.html | | | | |
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| Induced Current from Changing Magnetic Field When a bar magnet is brought toward a loop of wire, it increases the magnetic field in the loop. This change in the magnetic flux in the loop generates a voltage according to Faraday's law. The resulting flow of electric current in the wire loop produces an induced magnetic field that opposes the change of magnetic field (Lenz's Law). | | | Change to the condition of decreasing magnetic field. | | IndexMagnetic field conceptsFaraday's Law concepts |
| | | | --- | | HyperPhysics Electricity and Magnetism | R Nave | | Go Back |
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| Induced Current from Changing Magnetic Field When a bar magnet is moved away from a loop of wire, it decreases the magnetic field in the loop. This change in the magnetic flux in the loop generates a voltage according to Faraday's law. The resulting flow of electric current in the wire loop produces an induced magnetic field that opposes the change of magnetic field (Lenz's Law). | | | Change to the condition of increasing magnetic field. | | IndexMagnetic field conceptsFaraday's Law concepts |
| | | | --- | | HyperPhysics Electricity and Magnetism | R Nave | | Go Back | |
11151 | https://pathwaystochemistry.com/study-guide-general-chemistry-1/electronic-structure-of-atoms/electron-configurations-and-orbital-box-diagrams/ | Pathways to Chemistry
Chemistry for College, Middle and High School, Homeschooled Students, Teachers and Parents
Electron Configurations and Orbital Box Diagrams
To write electron configurations and draw orbital box diagrams, there are three rules that must be applied.
The Aufbau Principle states that electrons are always placed in the lowest energy sublevel that is available.
The Pauli Exclusion Principle states that each orbital may contain a maximum of 2 electrons; electrons must have opposite spins.
Hund’s Rule specifies that when orbitals of equal energy are available, the lowest energy electron configuration has the maximum number of unpaired electrons with parallel spins. In other words, the orbitals in a subshell must be half filled before pairing electrons.
The lowest energy orbitals must be filled first. For example the 1s orbital is filled and then we can fill the 2s orbital and go on to the 2p orbitals. The filling order follows:
1s→2s→2p→3s→3p→4s→3d→4p→5s→4d→5p→6s→4f→5d→6p→7s→5f→6d→7p
Electron configurations and orbital box diagrams can be written right from the periodic table.
The periodic table below, shows the s, p, d, and f-blocks. When reading the periodic table from left to right, one can easily write an electron configuration without memorizing the filling order. You only need to know where the blocks are located on the periodic table.
Hydrogen has 1 electron which goes into the 1s subshell. The electron configuration is 1s1. Helium has 2 electrons and both electrons go into the 1s orbital and the electron configuration is 1s2. Next we go to lithium with 3 electrons; two of the electrons go into the 1s orbital and the third electron into the 2s orbital giving an electron configuration of 1s22s1. The four electrons of Be go into the 1s and the 2s orbitals giving the electron configuration of 1s22s2. Reading the periodic table from left to right, the boron atom has 5 electrons, and the electron configuration is 1s22s22p1. The electron configuration for carbon is 1s22s22p2. An orbital box diagram can be written as well. Boxes, or horizontal lines represent the orbitals, arrows represent the electrons, and if an orbital is full, the electrons must be of opposite spin–one arrow pointing up and the other one pointing down. The orbital box diagrams are listed for the first 20 elements in the figure below.
In the figure below are the electron configurations for the first 20 elements.
Next, we write the electron configuration for Sc. When reading the periodic table from left to right we fill all the way to the 4s orbital and then fill a 3d orbital with the last electron. The electron configuration is 1s22s22p63s23p64s23d1. The electron configuration for Fe is 1s22s22p63s23p64s23d6. The electron configuration for Sn is 1s22s22p63s23p64s23d104p65s24d105p2.
Recall when we fill the 6s orbital, the next orbitals to be filled are the 4f orbitals. The electron configuration for Pb is 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2. Note, when the 4f orbitals were filled, we then fill the 5d orbitals and the remaining electrons go into the 6p orbitals.
The outermost shell is the valence shell. The electrons in the valence shell are the valence electrons. For the main group elements, the number of valence shell electrons are equal to the group number (Groups 1A to 8A). For transition elements, the ns and the (n – 1)d electrons are the valence shell electrons. Sodium, in Group 1A, has 1 valence electron in the 3s oribital and chlorine in Group 7A has 7 valence electrons, with 2 electrons in the 3s and 5 electrons in the 3p orbitals. Iron has 8 valence electrons with 2 electrons in the 4s subshell and 6 electrons in the 3d subshell. The valence electron configurations for all of the elements are in the periodic table below.
There is a shorthand way of writing electron configurations. For example, the electron configuration for sulfur is 1s22s22p63s23p4. Up to the 2sp electrons is the noble gas configuration of neon. We can write the configuration as [Ne]3s23p4. The noble gas that comes before the sulfur atom, Ne, is placed in brackets and the rest of the configuration is written. The shorthand electron configuration for Fe is [Ar]4s23d6. The configuration for Pb can be written as [Xe]6s24f145d106p2.
Anomalous Electron Configurations
There are over 20 elements that do not follow the building up principle. For example, using the building up principle Cr would have an electron configuration of [Ar]4s23d4. Experiments show this is incorrect and the electron configuration for Cr is [Ar]4s13d5. It is more stable to move the electron from the 4s orbital to the 3d orbital and have 2 half filled subshells. The same goes for Cu. The electron configuration is [Ar]4s13d10. It is more stable to move the electron from the 4s orbital into the 3d orbital and have a completely filled d subshell. The same is seen for Mo and Ag. As we move down the periodic table we see there are more anomalous electron configurations as shown in the periodic table above.
Worksheet: Orbital Box Diagrams and Electron Configuration Part 1
Worksheet: Orbital Box Diagrams and Electron Configuration Part 2
Watch the following videos before attempting exercises
Electron Configurations and Valence Electrons
Write Electron Configuration for an Excited State
Exercises
Exercise 1. Draw an orbital box diagram for the electrons in a neutral Br atom. Assign quantum numbers to the electrons in the valence shell.
Exercise 2. Write the full electron configuration for Pd. Write the shorthand notation for Pd.
Exercise 3. Write an electron configuration for Cd.
Exercise 4. Write an electron configuration for Hg.
Exercise 5. Assign quantum numbers to the valence shell electrons in Cr.
Back to Electronic Structure of Atoms
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11152 | https://www.fisicalab.com/apartado/potencial-electrico-punto | Potencial Eléctrico
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A lo largo de este apartado nos centraremos en los siguientes puntos:
¿Qué es el potencial eléctrico en un punto?
¿A que se le llaman superficies equipotenciales?
Potencial eléctrico creado por una carga puntual
Potencial eléctrico creado por varias cargas puntuales
¿Qué es la diferencia de potencial?
¿Cómo se relacionan la intensidad del campo eléctrico y el potencial eléctrico?
¿Cómo se mueven las cargas en el seno de un campo eléctrico?
Potencial Eléctrico
El potencial eléctrico en un punto del espacio es una magnitud escalar que nos permite obtener una medida del campo eléctrico en dicho punto a través de la energía potencial electrostática que adquiriría una carga si la situasemos en ese punto.
El potencial eléctrico en un punto del espacio de un campo eléctrico es la energía potencial eléctrica que adquiere una unidad de carga positiva situada en dicho punto.
donde:
V es el potencial eléctrico en un punto del campo eléctrico. Su unidad en el S.I. es el julio por culombio (J/C) que en honor a Alessandro Volta recibe el nombre de Voltio.
Ep es la energía potencial eléctrica que adquiere una carga testigo positiva q' al situarla en ese punto.
El hecho de que todas las magnitudes sean escalares, permite que el estudio del campo eléctrico sea más sencillo. De esta forma, si conocemos el valor del potencial eléctrico V en un punto, podemos determinar que la energía potencial eléctrica de una carga q situada en él es:
Superficies equipotenciales
Aquellos puntos contiguos donde el valor del potencial eléctrico es el mismo, reciben el nombre de superficie equipotencial. Cada punto de una superficie equipotencial se caracteriza por que:
El campo eléctrico es perpendicular a la superficie en dicho punto y se dirige hacia valores decrecientes de potencial eléctrico
Cada punto solo puede pertenecer a una superficie equipotencial, ya que el potencial eléctrico es un único valor en cada punto.
Potencial eléctrico creado por una carga puntual
Tal y como estudiamos en el apartado de intensidad del campo eléctrico, una única carga q es capaz de crear un campo eléctrico a su alrededor. Si en dicho campo introducimos una carga testigo q' entonces, atendiendo a la definición de energía potencial eléctrica de dos cargas puntuales:
El potencial eléctrico del campo eléctrico creado por una carga puntual q se obtiene por medio de la siguiente expresión:
donde:
V es el potencial eléctrico en un punto. En el S.I. se mide en Voltios (V).
K es la constante de la ley de Coulomb. En el S.I. se mide en N·m2/C2.
q es la carga puntual que crea el campo eléctrico. En el S.I. se mide en culombios (C).
r es la distancia entre la carga y el punto donde medimos el potencial. En el S.I. se mide en metros (m).
Si observas detenidamente la expresión puedes darte cuenta de que:
Si la carga q es positiva, la energía potencial es positiva y el potencial eléctrico V es positivo.
Si la carga q es negativa, la energía el potencial es negativa y el potencial eléctrico V es negativo.
Si no existe carga, la energía potencial y el potencial eléctrico es nulo.
El potencial eléctrico no depende de la carga testigo q' que introducimos para medirlo.
Potencial eléctrico creado por varias cargas puntuales
Si el campo eléctrico es creado por varias cargas puntuales, el potencial eléctrico en un punto sigue el principio de superposición:
El potencial eléctrico originado por n cargas puntuales en un punto de un campo eléctrico es la suma escalar de los potenciales eléctricos en dicho punto creados por cada una de las cargas por separado.
o lo que es lo mismo:
Ejemplo
Dos cargas q1= 3 µC y q2= -6 µC se encuentran en los vértices de un triángulo equilatero de lado 60 cm. Determina el potencial en el vértice libre y la energía potencial que adquiriría una carga q = -5 µC si se situase en dicho punto.
Ver solución
Diferencia de Potencial Eléctrico
Si dos puntos de un campo eléctrico poseen distinto potencial eléctrico, entre ambos puntos existe lo que se denomina una diferencia de potencial o tensión, ΔV. Este valor se encuentra íntimamente relacionado con el trabajo eléctrico. Por definición, el trabajo que debe realizar un campo eléctrico para trasladar una carga q desde un punto A a otro B dentro del campo se obtiene por medio de la siguiente expresión:
Si aplicamos la definición de potencial eléctrico, obtenemos que:
La diferencia de potencial eléctrico entre dos puntos A y B de un campo eléctrico es el opuesto del trabajo realizado por el campo eléctrico para trasladar una unidad de carga positiva desde el punto A al B.
Relación entre Potencial Eléctrico y Campo Eléctrico
Partiendo de la fórmula del trabajo eléctrico es posible obtener una expresión que relacione el campo eléctrico E con el potencial eléctrico V de la siguiente forma:
Teniendo en cuenta la expresión anterior podemos deducir las siguientes cuestiones:
Cuestión 1
Si en una región del espacio no hay campo eléctrico implica que el potencial eléctrico es constante en toda esa región, de forma que tomados dos puntos cualesquiera A y B de dicha región se cumple que VA = VB.
Cuestión 2
El campo eléctrico es perpendicular en cada punto de una superficie equipotencial, ya que VA = VB y esto implica que:
Si el producto escalar es nulo quiere decir que E y dl son perpendiculares:
Cuestión 3
En el seno de un campo eléctrico uniforme, si tomamos su dirección en el eje X (en sentido positivo) se cumple que:
Por tanto, si disponemos de un campo eléctrico uniforme, el potencial eléctrico disminuye uniformemente a medida que nos alejamos en la dirección del propio campo, ya que:
Si consideramos que el desplazamiento es infinitesimal dx, obtenemos que:
Por tanto, su expresión vectorial sería:
De esta forma, hemos obtenido la intensidad del campo eléctrico a partir del potencial eléctrico en el eje x. Sin embargo, lo más común es que el potencial varíe en función de las tres coordenadas x,y,z. De esta forma, podemos expresar la intensidad del campo eléctrico:
Y por tanto:
En un campo eléctrico uniforme, la intensidad del campo électrico en función del potencial eléctrico se puede obtener por medio de la siguiente expresión:
o lo que es lo mismo:
Ejemplo
El potencial eléctrico (en voltios) en una zona del espacio viene dado por la siguiente expresión:
Calcular la expresión general del campo eléctrico y su valor en el punto A(1,0,1)
Ver solución
Cuestión 4
Observando la ecuación anterior, podemos deducir que la intensidad del campo eléctrico puede ser definida con una nueva unidad voltio/metro (V/m).
Potencial Eléctrico y el Movimiento de las Cargas
Si analizamos detenidamente la expresión de la diferencia de potencial:
Tal y como vimos en el tema del trabajo eléctrico, el trabajo realizado por una fuerza eléctrica para desplazar una carga q desde un punto A a otro B, sin presencia de fuerzas externas, es un valor positivo. Si estudiamos que ocurre si la carga q es positiva o negativa, obtenemos que:
| | | |
---
| q<0 | VB-VA>0=>VB>VA | La carga se mueve desde puntos de menor potencial a mayor potencial |
| q>0 | VB-VA<0=>VBA | Se mueve desde puntos de mayor potencial a menor potencial |
Por tanto:
Las cargas positivas se mueven desde zonas de mayor potencial eléctrico a zonas de menor potencial eléctrico.
Las cargas negativas se mueven desde zonas de menor potencial eléctrico a zonas de mayor potencial eléctrico.
Teniendo en cuenta que tal y como estudiamos en el apartado de intensidad del campo eléctrico, las cargas positivas se mueven en el sentido de dicha intensidad entonces, la intensidad de campo eléctrico se dirige siempre desde zonas de mayor potencial a zonas de menor potencial.
La intensidad de campo eléctrico apunta siempre hacia potenciales decrecientes.
Enfoque energético
Desde el punto de vista de la energía mecánica, si sobre las cargas únicamente actúa la fuerza eléctrica, dicha fuerza es conservativa. Esto implica que la energía mecánica de la partícula entre dos posiciones A y B debe ser la misma o dicho de otra forma, la variación de energía mecánica entre A y B es 0 (no cambia).
Esto implica que siempre que una partícula se encuentra en una zona donde existe una diferencia de potencial, esta adquirirá energía cinética.
Ejemplo
Un electrón cuya masa es 9.10·10-31 Kg y su carga q = -1.6·10-19 C que se introduce en un campo eléctrico uniforme adquiere una velocidad de 3·105 m/s. ¿Cuál es la diferencia de potencial que ha provocado dicho incremento de velocidad?
Ver solución
Y ahora... ¡Ponte a prueba!
Sobre el autor
José L. Fernández
José Luis Fernández Yagües es ingeniero de telecomunicaciones, profesor experimentado y curioso por naturaleza. Dedica su tiempo a ayudar a la gente a comprender la física, las matemáticas y el desarrollo web. Ama el queso y el sonido del mar.
Apartados relacionados
Este mismo apartado se encuentra desarrollado en otros niveles educativos. Si sus contenidos no se ajustan al nivel que buscas, prueba a visitar:
Avanzado
Introducción al Concepto de Potencial Eléctrico
Por otro lado, los contenidos de Potencial Eléctrico se encuentran estrechamente relacionados con:
Trabajo de la Fuerza Eléctrica
Concepto de Campo Eléctrico
Energía Potencial Eléctrica
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11153 | https://pubmed.ncbi.nlm.nih.gov/27532368/ | The American Association of Endocrine Surgeons Guidelines for Definitive Management of Primary Hyperparathyroidism - PubMed
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. 2016 Oct 1;151(10):959-968.
doi: 10.1001/jamasurg.2016.2310.
The American Association of Endocrine Surgeons Guidelines for Definitive Management of Primary Hyperparathyroidism
Scott M Wilhelm1,Tracy S Wang2,Daniel T Ruan3,James A Lee4,Sylvia L Asa5,Quan-Yang Duh6,Gerard M Doherty7,Miguel F Herrera8,Janice L Pasieka9,Nancy D Perrier10,Shonni J Silverberg11,Carmen C Solórzano12,Cord Sturgeon13,Mitchell E Tublin14,Robert Udelsman15,Sally E Carty16
Affiliations Expand
Affiliations
1 Department of Surgery, University Hospitals/Case Medical Center, Cleveland, Ohio.
2 Department of Surgery, Medical College of Wisconsin, Milwaukee.
3 Department of Surgery, Brigham and Women's Hospital, Boston, Massachusetts.
4 Department of Surgery, Columbia University College of Physicians and Surgeons, New York, New York.
5 Department of Pathology, University of Toronto, Toronto, Ontario, Canada.
6 Department of Surgery, University of California San Francisco Medical Center, San Francisco.
7 Department of Surgery, Boston University, Boston, Massachusetts.
8 Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico City, Mexico.
9 Department of Surgery, University of Calgary, Calgary, Alberta, Canada.
10 Department of Surgery, MD Anderson Cancer Center, Houston, Texas.
11 Department of Medicine, Columbia University College of Physicians and Surgeons, New York, New York.
12 Department of Surgery, Vanderbilt University Medical Center, Nashville, Tennessee.
13 Department of Surgery, Northwestern University, Chicago, Illinois.
14 Department of Radiology, University of Pittsburgh, Pittsburgh, Pennsylvania.
15 Department of Surgery, Yale University, New Haven, Connecticut.
16 Department of Surgery, University of Pittsburgh, Pittsburgh, Pennsylvania.
PMID: 27532368
DOI: 10.1001/jamasurg.2016.2310
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The American Association of Endocrine Surgeons Guidelines for Definitive Management of Primary Hyperparathyroidism
Scott M Wilhelm et al. JAMA Surg.2016.
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. 2016 Oct 1;151(10):959-968.
doi: 10.1001/jamasurg.2016.2310.
Authors
Scott M Wilhelm1,Tracy S Wang2,Daniel T Ruan3,James A Lee4,Sylvia L Asa5,Quan-Yang Duh6,Gerard M Doherty7,Miguel F Herrera8,Janice L Pasieka9,Nancy D Perrier10,Shonni J Silverberg11,Carmen C Solórzano12,Cord Sturgeon13,Mitchell E Tublin14,Robert Udelsman15,Sally E Carty16
Affiliations
1 Department of Surgery, University Hospitals/Case Medical Center, Cleveland, Ohio.
2 Department of Surgery, Medical College of Wisconsin, Milwaukee.
3 Department of Surgery, Brigham and Women's Hospital, Boston, Massachusetts.
4 Department of Surgery, Columbia University College of Physicians and Surgeons, New York, New York.
5 Department of Pathology, University of Toronto, Toronto, Ontario, Canada.
6 Department of Surgery, University of California San Francisco Medical Center, San Francisco.
7 Department of Surgery, Boston University, Boston, Massachusetts.
8 Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico City, Mexico.
9 Department of Surgery, University of Calgary, Calgary, Alberta, Canada.
10 Department of Surgery, MD Anderson Cancer Center, Houston, Texas.
11 Department of Medicine, Columbia University College of Physicians and Surgeons, New York, New York.
12 Department of Surgery, Vanderbilt University Medical Center, Nashville, Tennessee.
13 Department of Surgery, Northwestern University, Chicago, Illinois.
14 Department of Radiology, University of Pittsburgh, Pittsburgh, Pennsylvania.
15 Department of Surgery, Yale University, New Haven, Connecticut.
16 Department of Surgery, University of Pittsburgh, Pittsburgh, Pennsylvania.
PMID: 27532368
DOI: 10.1001/jamasurg.2016.2310
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Abstract
Importance: Primary hyperparathyroidism (pHPT) is a common clinical problem for which the only definitive management is surgery. Surgical management has evolved considerably during the last several decades.
Objective: To develop evidence-based guidelines to enhance the appropriate, safe, and effective practice of parathyroidectomy.
Evidence review: A multidisciplinary panel used PubMed to review the medical literature from January 1, 1985, to July 1, 2015. Levels of evidence were determined using the American College of Physicians grading system, and recommendations were discussed until consensus.
Findings: Initial evaluation should include 25-hydroxyvitamin D measurement, 24-hour urine calcium measurement, dual-energy x-ray absorptiometry, and supplementation for vitamin D deficiency. Parathyroidectomy is indicated for all symptomatic patients, should be considered for most asymptomatic patients, and is more cost-effective than observation or pharmacologic therapy. Cervical ultrasonography or other high-resolution imaging is recommended for operative planning. Patients with nonlocalizing imaging remain surgical candidates. Preoperative parathyroid biopsy should be avoided. Surgeons who perform a high volume of operations have better outcomes. The possibility of multigland disease should be routinely considered. Both focused, image-guided surgery (minimally invasive parathyroidectomy) and bilateral exploration are appropriate operations that achieve high cure rates. For minimally invasive parathyroidectomy, intraoperative parathyroid hormone monitoring via a reliable protocol is recommended. Minimally invasive parathyroidectomy is not routinely recommended for known or suspected multigland disease. Ex vivo aspiration of resected parathyroid tissue may be used to confirm parathyroid tissue intraoperatively. Clinically relevant thyroid disease should be assessed preoperatively and managed during parathyroidectomy. Devascularized normal parathyroid tissue should be autotransplanted. Patients should be observed postoperatively for hematoma, evaluated for hypocalcemia and symptoms of hypocalcemia, and followed up to assess for cure defined as eucalcemia at more than 6 months. Calcium supplementation may be indicated postoperatively. Familial pHPT, reoperative parathyroidectomy, and parathyroid carcinoma are challenging entities that require special consideration and expertise.
Conclusions and relevance: Evidence-based recommendations were created to assist clinicians in the optimal treatment of patients with pHPT.
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How Best to Approach Surgery for Primary Hyperparathyroidism-Can We All Agree?Sosa JA.Sosa JA.JAMA Surg. 2016 Oct 1;151(10):969. doi: 10.1001/jamasurg.2016.2360.JAMA Surg. 2016.PMID: 27533602 No abstract available.
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11154 | https://www2.math.upenn.edu/~mlazar/math170/notes06-2.pdf | last edited April 30, 2016 6.2 Modular Arithmetic Every reader is familiar with arithmetic from the time they are three or four years old. It is the study of numbers and various ways in which we can combine them, such as through addition and subtraction, multiplication and division.
Since even before they were in grade school, every reader knew that adding 2 and 2 together gives us 4, and can make that calculation now without almost any thinking. And even if the answer is not immediately obvious, every college student (at least in Penn), knows how to add together much larger numbers, such as 4,378,123 and 5,621,877. This is classical arithmetic, and it turns up in countless applications in our everyday lives.
The reader is also likely familiar with another kind of arithmetic, even if we don’t always think of it as such. If it is 4 o’clock now, what will the time be in 25 hours? If we didn’t know from watches and clocks, we would probably have answered 29 o’clock. But we are familiar with watches, clocks, and the standard conventions of time-keeping, and so every reader would probably have answered the answer with 5 o’clock. How can we add 25 to 4 and end up with 5? The reason is that in this system 25 o’clock is the same as 1 o’clock, 26 is the same as 2, and so forth. In many time-keeping systems, we don’t even use numbers larger than 12, and instead use a.m. and p.m. (from the Latin ante meridiem and post meridiem) to denote the earlier and latter halves of a 24-hour period. Such systems, that “wrap around” after hitting some limit, are called modular arithmetic systems, and play an important role both in theoretical and applied mathematics.
Modular arithmetic motivates many questions that don’t arise when study-ing classic arithmetic.
For example, in classic arithmetic, adding a positive number a to another number b always produces a number larger than b. In modular arithmetic this is not always so. For example, if it is now 4 o’clock and we “add” 23 hours, the time will then be 3 o’clock, which doesn’t appear to be larger than 4 o’clock. In fact, it is no longer clear whether it makes sense at all to discuss “larger” and “smaller” in such systems.
Here is another question. Suppose it is now 2 o’clock and we wait for 1 hour and then write down the time. We then wait another hour and mark the time, and repeat this until we eventually mark 2 o’clock again, at which point we stop. It is clear that when we stop, we will have marked down every hour. If we do the same thing but instead wait 2 or 3 hours in between each marking there will be certain hours which we never mark, such as 7 o’clock. But if we wait 5 hours between each marking, then we will eventually mark every hour. This raises the question, for which waiting intervals between marks can we ensure that we will eventually mark every hour?
While this particular example may seem contrived, it should motivate us think, if even momentarily, about modular arithmetic systems and the ways in which they are similar to and di↵erent from the classical arithmetic with which we are familiar. The next several sections will investigate these systems which have a finite number of numbers, and in which numbers “wrap around” after going too high.
72 last edited April 30, 2016 The central definition in studying modular arithmetic systems establishes a relationship between pairs of numbers with respect to a special number m called the modulus: Definition 25. Two integers a and b are congruent modulo m if they di↵er by an integer multiple of m, i.e., b −a = km for some k 2 Z. This equivalence is written a ⌘b (mod m).
Although this definition looks somewhat technical, the idea is very simple.
For some fixed integer m, two numbers are roughly the same if they di↵er by multiples of m. In a sense, this definition generalizes previous discussions of odd and even numbers. In previous sections, we proved theorems such as the square of an even number is even and the square of odd number is odd. As far as even and odds numbers go, and as far as these theorems are concerned, there is no di↵erence between 17 and 2073, as both are odd and behave the same under squaring. In a similar manner, in modular arithmetic, there is no di↵erence between a pair of numbers that di↵er by the modulus m, which could be 2 or could be 15,485,863.
In arithmetic mod 7, for example, there is no di↵erence between 1, 8, and 15, as they all di↵er from one another by multiples of 7. Likewise, 22, 701 and -6 also di↵er from all of these numbers by multiples of 7, and are hence congruent.
Example 1. Every number is congruent to itself for any modulus; that is, a ⌘a (mod m) for any a, m 2 Z. The reason for this is that a −a = 0, which is a multiple of m, since 0 = 0 ⇥m for any m. It might seem a bit silly, but is a consequence of the way in which we defined congruence.
Example 2. Every number is congruent to any other number mod 1; that is, a ⌘b (mod 1) for any a, b 2 Z. The reason for this is that b −a, is a multiple of 1 for any a and b. Again, this might seem a bit silly, but is a consequence of the way in which we defined congruence.
Example 3. Any even numbers are congruent to one another mod 2; likewise, any odd numbers are congruent to one another mod 2. For example, we have 12 ⌘3132 (mod 2) and −7 ⌘19 (mod 3). This is because any pair of even numbers di↵er from one another by a multiple of 2. Likewise, any pair of odd numbers di↵er from one another by a multiple of 2.
Example 4. The numbers 31 and 46 are congruent mod 3 because they di↵er by a multiple of 3. We can write this as 31 ⌘46 (mod 3). Since the di↵erence between 31 and 46 is 15, then these numbers also di↵er by a multiple of 5; i.e., 31 ⌘46 (mod 5).
Example 5. By the definition of congruence, every pair of integers a and b are congruent mod 1, since any pair of integers di↵er by a multiple of 1. In symbols, for all integers a and b, we have a ⌘b (mod 1).
Example 6. In general it is not true that a ⌘−a (mod m), unless m = 2 or else a is a multiple of 2. For example, it is not true that 7 ⌘−7 (mod 3), since the di↵erence between 7 and -7 is 14, which is not a multiple of 3.
73 last edited April 30, 2016 Rules of Modular Arithmetic After considering the basic definition of modular arithmetic, we next consider some of its basic properties. It turns out that modular arithmetic follows many of the same rules of classical arithmetic, thus making it very easy to work with.
In order to highlight what is going on, we try to compare and contrast modular arithmetic to classical arithmetic.
Suppose we have two numbers a and b: a = 5 b = 8.
We all know that in classical arithmetic we can combine these equations to obtain: a + b = 5 + 8 = 13.
More generally, if we have a = c b = d, then we can combine them in many di↵erent ways, to obtain: a + b = c + d, a −b = c −d, a ⇥b = c ⇥d.
Pause to think about this statement, and make sure it aligns with what you know. Of course these are only several ways of combining these equations, and every reader can think of several others. All of the above are “rules” of classical arithmetic. What we would like to do now is consider whether similar rules apply to modular arithmetic as well.
Suppose we have the following two congruence relations: a ⌘ b (mod m) c ⌘ d (mod m).
Are we able to combine these to obtain a + b ⌘ c + d (mod m), a −b ⌘ c −d (mod m), a ⇥b ⌘ c ⇥d (mod m)?
That is, do the rules that govern how we can combine equations in classical arithmetic also govern the ways in which we combine statements in modular arithmetic? In what follows we prove that indeed many of the rules do carry over – the rules of modular arithmetic will be familiar to us.
74 last edited April 30, 2016 Addition The first rule we consider is that associated with addition. Suppose we have two congruence relations: a ⌘b (mod m) and c ⌘d (mod m). In other words, a and b are congruent and c and d are congruent, both mod m. We can add the left sides of these congruent relations, add the right sides, and the results will again be congruent. In symbols, Theorem 15.
If a ⌘ b (mod m) and c ⌘ d (mod m), then a + c ⌘ b + d (mod m).
Proving this result involves nothing more than applying the definition of congruence and some basic algebraic manipulation.
Proof. By the definition of congruence (Definition 25) we know that a and b di↵er by some multiple of m, i.e., b −a = km (64) for some k 2 Z. Likewise we know that c and d also di↵er by some multiple of m, i.e., d −c = jm (65) for some j 2 Z. Note that we use j instead of k since the multiple of m by which c and d di↵er might be di↵erent from the multiple by which a and b di↵er. Next we add these two equations together: (b −a) + (d −c) = km + jm.
(66) We can rewrite this equation as (b + d) −(a + c) = (j + k)m.
(67) By the definition of congruence modulo m, this is the same as saying that a + c is congruent to b+d modulo m, since a+c and b+d di↵er by an integer multiple (j + k) of m. In symbols, we have: a + c ⌘b + d (mod m), (68) as desired.
A similar proof can be used to show that if a ⌘b (mod m) and c ⌘d (mod m), then a −c ⌘b −d (mod m).
These two results allow us to treat all numbers that are congruent modulo m as identical when adding and subtracting numbers. If we know that a ⌘3 (mod 7) and b ⌘4 (mod 7), then we can know that a + b ⌘7 ⌘0 (mod 7).
This is true whether a is 10 or 703, and whether b is 7004, 10000, or 7,000,004.
What a and b actually are does not matter if we only want to determine whether a + b is congruent to 0 or not.
75 last edited April 30, 2016 Multiplication After understanding how addition and subtraction work in modular arithmetic, we turn our attention to understanding multiplication. In classical arithmetic, if a = 2 and b = 5, then of course a⇥b = 2⇥5 = 10. Does a similar relationship also hold in modular arithmetic? In particular, if we know that a ⌘2 (mod m) and b ⌘5 (mod m), do we know that a ⇥b ⌘2 ⇥5 (mod m)?
The following theorem answers this question affirmatively.
Theorem 16.
If a ⌘ b (mod m) and c ⌘ d (mod m), then a ⇥c ⌘ b ⇥d (mod m).
Proof. By the definition of congruence we know that a and b di↵er by a multiple of m, as do c and d: b −a = jm d −c = km for some j, k 2 Z.
Note that we use distinct multiples j and k for the two equations, since a and b might di↵er by one multiple of m, and c and d might di↵er by another multiple of m.
To prove the desired result, we rearrange the equations: b = jm + a d = km + c We multiply both sides by each other to obtain bd = (jm + a)(km + c) = jkm2 + jmc + kma + ac = (jkm + jc + ka)m + ac.
We then subtract ac from both sides to obtain bd −ac = (jkm + jc + ka)m.
Since (jkm + jc + ka)m is an integer multiple of m, then ac and bd di↵er by an integer multiple of m, and so by definition are congruent mod m.
Example 1. If we know that a ⌘3 (mod 7) and we know that b ⌘4 (mod 7), then we can determine that ab ⌘12 ⌘5 (mod 7). This is true whether a is 10, 703, or 7,000,003 and whether b is 7004 or 10000. In any of these cases, the product ab will be congruent to 5 modulo 7.
76 last edited April 30, 2016 Example 2. How can we simplify 20 ⇥21 in arithmetic modulo 19? We first note that 20 ⌘1 (mod 19) and also that 21 ⌘2 (mod 19). Theorem 16 tells us that we can combine these equations to obtain 20 ⇥21 ⌘1 ⇥2 ⌘2 (mod 19).
Example 3. Can we simplify 17753 in arithmetic modulo 9? We first note that 17 ⌘−1 (mod 9), because 17 and -1 di↵er by a multiple of 9. Theorem 16 allows us to then combine this congruence relation as many times as we would like. In particular, by combining 753 copies, we obtain 17753 ⌘(−1)753 (mod 9). Since (−1)n = −1 for any odd integer n, we have 17753 ⌘−1 (mod 9). Finally, if we would like to have a simple, positive answer, then we can add 9 to obtain a final answer of 8.
Theorems 15 and 16 show us that we can treat all numbers that are congruent modulo m as the same, in addition and in multiplication operations. Division is much more complicated, and will not be discussed.
Remainders We take a moment to draw out a connection to division with remainders, an idea we considered briefly in Section 4.1. In particular, back in elementary school we learned about a way of dividing integers by other integers that entirely avoids decimals and fractions. In particular, suppose we divide 7 by 4. In third, fourth, or fifth grade, we learned that we can write this as 1, remainder 3. That is, 4 can 1 time “into” 7, leaving over 3. As we got older, we learned that we could also write the answer as 1.75 or 13⁄4, but we still occasionally deal with situations in which discussing fractions would be silly. If we have 52 playing cards and 5 players, a dealer could give each player 10 cards and then be left with 2 cards.
It makes little sense to say that the dealer should give each player 10.2, or 10 and a fifth, cards.
What is the connection of modular arithmetic to division with remainders?
Suppose that we divide some integer a by another integer m. Notice that the “remainder” is always congruent to a modulo m.
For example, suppose we divide 1031 by 19. We obtain 54, remainder 5. This tells us that 5 is congruent to 1031 modulo 19. Likewise, since the remainder of 7381/57 is 28, we know that 28 ⌘7381 (mod 57).
Why is the remainder after division always congruent to the number we are dividing? One way to think about this is by considering how we can find a remainder without actually doing any division. Suppose we want to know the remainder of 11 after dividing by 3. We can subtract 3 over and over until we obtain a number that is smaller than 3: 11, 8, 5, and eventually 2. Each time we subtract 3, we are realizing that 3 can “go into” 11 one more time; whatever is left at the end is the remainder. At the same time, we got from the original number to the remainder by jumps of 3, so of course the di↵erence between 11 and 2 is divisible by 3, making 11 and 2 congruent. The same idea works for dividing any number a with any other number m.
77 last edited April 30, 2016 Standard Representation We have by now seen that in arithmetic modulo m, there is no di↵erence between writing 1, 1 + m, 1 + 2m, and so forth, at least as far as addition, subtraction, and multiplication are concerned. For this reason, writing 4+11 ⌘15 (mod 13) is “just as correct” as writing 4 + 11 ⌘2 (mod 13), and “just as correct” as writing 4 + 11 ⌘−11 (mod 13). As far as arithmetic modulo 13 is concerned, 2, 15, and -11 are exactly the same number. However, in some applications it is convenient to agree upon a standard way to represent numbers. What is a good way to do this? Which of {. . . , a −2m, a −m, a, a + m, a + 2m, . . .} should we consider the standard representative?
You have likely encountered a similar problem back in your days learning about trigonometric functions. A teacher may have asked you what is the inverse sine of −1, i.e., sin−1(−1). You may have correctly answered 270◦. Or you may have correctly answered −90◦. In fact, any number that can be written 270◦+ n360◦, for any integer n 2 Z, would also be equally correct.
But if each student wrote a di↵erent number on an exam, it could take a long time to determine whether or not every answer is correct. Is 1500◦a correct solution?
Is 1530◦? For this reason, we might specify that we looking for a correct answer between 0◦and 360◦, or else between −180◦and 180◦, since there is exactly one correct answer in each of these ranges.
In the same way, when working in arithmetic modulo 41, the numbers {. . . , −29, 12, 53, 94, 135, . . .} are all the same, yet we might hope to specify one of them to be the standard representation of them. Indeed, in arithmetic modulo m, we refer to the numbers {0, 1, 2, . . . , m−1} as the standard repre-sentations of the integers. If numbers are always represented in this standard form, determining whether or not two numbers are congruent is as easy as look-ing at whether the numbers are equal. Notice also that this set of numbers is also the set of possible remainders after dividing a number by m.
Example 1. Suppose we want to know the remainder of 17 ⇥18 when it is divided by 19. We can do this in two di↵erent ways. First, we can multiply the two numbers directly and obtain 306; some calculation will show that 306 is congruent to 2 modulo 19. Alternatively, we know that 17 ⌘−2 (mod 19) and 18 ⌘−1 (mod 19). Multiplying both sides we see that 17⇥18 ⌘(−2)⇥(−1) ⌘2 (mod 19).
Example 2. Suppose we want to determine the standard form of 172 in mod 19 arithmetic. One way in which we can do this is by considering the square of 17, which is 289, divide that by 19 and then take the remainder. However, since we know that 17 ⌘−2 (mod 19), we can multiply this congruence equation by itself to obtain 172 ⌘−22 ⌘4 (mod 19). We can easily verify that the remainder of 289, when divided by 17, is indeed 4.
Example 3. Suppose we want to determine the standard form of 18489391312 in mod 19 arithmetic. We should first notice that in mod 19 arithmetic, 18 is congruent to −1, and so 18489391312 ⌘(−1)489391312 (mod 19). It is relatively 78 last edited April 30, 2016 easy to see that if n is odd then (−1)n = −1, and if n is even then (−1)n = 1.
Since 489391312 is even, 18489391312 ⌘1 (mod 19).
Dividing by 9 We can use the rules of modular addition and multiplication to prove a theorem you may have once seen. Suppose we have a number, for example 2,383,623, and want to know whether it is divisible by 9. Is there an easy way to figure this out without doing “long division”? You may have learned the following trick: add up the digits of the number (e..g., 2 + 3 + 8 + 3 + 6 + 2 + 3 = 27). If this sum is divisible by 9, then so is the original number; if the sum is not divisible by 9, then neither is the original number. Is this just a miraculous trick, or is it something that we can prove should work?
The rules of modular addition and multiplication (Theorems 15 and 16 above) can help us prove this beautiful result. Let’s begin by proving a sim-pler result about the remainders we get when we divide powers of 10 by 9. In particular, the remainder is always 1.
Lemma 17. For any natural number n, we have 10n ⌘1 (mod 9).
Proof. Recall that if we have two congruences: a ⌘b and c ⌘d (mod m), then we can combine them to form a new congruence relation: ac ⌘bd (mod m).
Since 10 ⌘1 (mod 9), then we can combine the equation with itself to obtain 100 = 10 ⇥10 ⌘1 ⇥1 ⌘1 (mod 9). We can indeed combine this equation with itself as many times as we want (e.g., n times), and therefore have 10n ⌘1n ⌘1 (mod 9) for any natural number n.
Next, let’s consider what happens when we divide numbers such as 300, 5000, and 2,000,000 by 9. What are the remainders? Theorem 16 can help us see that the remainders are 3, 5, and 2 in these examples. To see why this is so, notice that each of these numbers can be written as the product of an integer and a power of 10: 300 = 3 · 102, 5000 = 5 · 103, and 2,000,000= 2 · 106. This leads us to the following theorem.
Lemma 18. For any natural numbers c and n, we have c · 10n ⌘c (mod 9).
Proof. Recall that if we have two congruences: a ⌘b and c ⌘d (mod m), then we can combine them to form a new congruence relation: ac ⌘bd (mod m).
Since c ⌘c and 10n ⌘1 (mod 9) for any n, then we can combine the equations to obtain c · 10n ⌘c · 1 ⌘c (mod 9).
This now leads us to our central theorem: Theorem 19. A number is divisible by 9 if and only if the sum of its digits (written in base 10) is divisible by 9.
Proof. In base 10, every number can be written as a sum of ones, tens, hundreds, thousands, and so forth. For example, 5776 = 5000+700+70+6. More generally, we can write this as n = c0 + c1101 + c2102 + c3103 + . . ., where the ci variables 79 last edited April 30, 2016 are the numbers of ones, tens, hundreds, thousands, and so forth. According to Lemma 18, for each of the ci we have ci · 10n ⌘ci (mod 9). Using Theorem 15, we can combine the congruence relations c0 ⌘ c0 (mod 9), c1 ⌘ c1101 (mod 9), c2 ⌘ c2102 (mod 9), c3 ⌘ c2103 (mod 9), . . .
cn ⌘ c210n (mod 9), to give us c0 + c1101 + c2102 + . . . cn10n ⌘c0 + c1 + c2 + . . . cn (mod 9) (69) In other words, a number n is congruent to the sum of its digits in mod 9. If a number is divisible by 9, i.e., n ⌘0 (mod 9), then so is the sum of its digits.
80 |
11155 | https://www.ck12.org/geometry/sas-triangle-congruence-1501911835.12/lesson/SAS-Triangle-Congruence-GEOM/ | SAS Triangle Congruence 1501911835.12
Two sets of corresponding sides and included angles prove congruent triangles.
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11156 | https://brainly.com/question/29841172 | [FREE] Factor the perfect-square trinomial in: y = (x^2 + 2x + 1) - 1 - 1 y = (x + 1)^2 - 2 - brainly.com
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Factor the perfect-square trinomial in:
y=(x 2+2 x+1)−1−1
y=(x+1)2−2
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The perfect square trinomial in y = ( x 2+2 x+1 ) -1 - 1 will be
⇒ (x+1)2
The perfect square trinomial in y = (x + 2)2 - 1 - 1 will be
⇒ Null
A perfect square trinomial can be expressed as the square of a binomial,
We can write the first expression as,
y = x 2+2 x+1
y = x 2+2 x+1−2
⇒ x 2+x+x+1
⇒ x(x+1)+1(x+1)
⇒ (x + 1) (x + 1)
⇒ (x+1)2
Therefore, according to the first expression, (x 2+2 x+1)−1−1 is a perfect square binomial with the factor = (x+1)2
According to the second expression, y = (x +2)2 − 1 −1
This expression does not show any factors
Therefore, this second expression doesn't have any perfect square trinomial factors.
To learn more about perfect square trinomials,
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Factor the perfect-square trinomial in y = (x2 + 2x + 1) − 1− 1, y = (x +2)2 − 1 −1
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The perfect-square trinomial in the expression can be factored to (x+1)2−2. The trinomial x 2+2 x+1 is recognized as a square of a binomial, which is (x+1)2. Thus, the expression simplifies to this form.
Explanation
To factor the perfect-square trinomial in the given expression
y=(x 2+2 x+1)−1−1 we first simplify the expression inside the parentheses. The term x 2+2 x+1 is a well-known perfect-square trinomial, which can be factored as follows:
Recognize the Form: The expression x 2+2 x+1 fits the pattern (a+b)2 where a=x and b=1. Thus, it can be written as:
(x+1)2
2. Rewrite the Expression: Now substitute this back into the original equation. The expression for y becomes:
y=(x+1)2−2
3. Final Expression: Therefore, we have factored out the trinomial:
y=(x+1)2−2
This result shows that the perfect-square trinomial x 2+2 x+1 can be expressed as (x+1)2, while the −2 is a constant that completes the equation.
Examples & Evidence
For example, if you have z=(x 2+4 x+4), this also can be factored as (x+2)2 since it follows the same pattern of a perfect-square trinomial.
The pattern of perfect-square trinomials is a standard algebraic identity, where a 2+2 ab+b 2=(a+b)2. This is learned in algebra classes and is universally applicable to factoring such expressions.
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### Free Mathematics solutions and answers
Community Answer 8 Form a perfect-square trinomial. y = x2 + 2x + − 1−
Community Answer Find the Perfect Square Trinomial of. y^2 + 1/2y = -1\16
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The probability of event A occurring is (4 m+n)m, and the probability of event B occurring is (4 m+n)n. Find the probability of A or B occurring if the events are mutually exclusive.
6. The operation t is defined by a t b=a+b+2 ab in arithmetic modulo 4. (a) Draw a table for the operation t on the set R={0,1,2,3}.
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\left[\begin{array}{ccc}2 & 3 & 1 \ 4 & 1 & 2 \ 3 & 1 & 2\end{array}\right]\left[\begin{array}{c}2 \ -1 \ 3\end{array}\right]
2 4 53 1 121 22−1 3
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11157 | https://pubmed.ncbi.nlm.nih.gov/17156694/ | Effectiveness of intermittent diazepam prophylaxis in febrile seizures: long-term prospective controlled study - PubMed
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J Child Neurol
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. 2006 Dec;21(12):1036-40.
doi: 10.1177/7010.2006.00221.
Effectiveness of intermittent diazepam prophylaxis in febrile seizures: long-term prospective controlled study
Efterpi Pavlidou1,Maria Tzitiridou,Christos Panteliadis
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1 3rd Department of Pediatrics, Aristotle University of Thessaloniki, Thessaloniki, Greece.
PMID: 17156694
DOI: 10.1177/7010.2006.00221
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Randomized Controlled Trial
Effectiveness of intermittent diazepam prophylaxis in febrile seizures: long-term prospective controlled study
Efterpi Pavlidou et al. J Child Neurol.2006 Dec.
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. 2006 Dec;21(12):1036-40.
doi: 10.1177/7010.2006.00221.
Authors
Efterpi Pavlidou1,Maria Tzitiridou,Christos Panteliadis
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1 3rd Department of Pediatrics, Aristotle University of Thessaloniki, Thessaloniki, Greece.
PMID: 17156694
DOI: 10.1177/7010.2006.00221
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The efficacy of intermittent rectal diazepam prophylaxis is assessed in the prevention of febrile seizures. In a prospective randomized cohort trial, 139 children (77 girls, 62 boys) who experienced a first febrile seizure were allocated to two groups: group A, which received intermittent diazepam (n = 68), and group B, which received no prophylaxis (n = 71). All children had a 3-year follow-up. The inclusion criteria were no personal history of afebrile seizures, normal neurodevelopment, no previous anticonvulsant therapy, and age between 6 months and 3 years. Each group was stratified to low, intermediate, and high risk according to the available clinical data. The 36-month recurrence rates in the no-prophylaxis group were 83% in high-risk patients, 55% in intermediate-risk patients, and 46% in low-risk patients. In the prophylaxis group, the recurrence rates were reduced in all risk groups: 38%, 35%, and 33%, respectively. Intermittent diazepam prophylaxis reduces the recurrence rate mainly in high-risk children provided that sufficient doses are given on time and adequately.
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Recent advances in febrile seizures.Mittal R.Mittal R.Indian J Pediatr. 2014 Sep;81(9):909-16. doi: 10.1007/s12098-014-1532-2. Epub 2014 Aug 8.Indian J Pediatr. 2014.PMID: 25103013 Review.
Acute drug administration in epilepsy: a review.Wolf P.Wolf P.CNS Neurosci Ther. 2011 Oct;17(5):442-8. doi: 10.1111/j.1755-5949.2010.00167.x. Epub 2010 Jul 7.CNS Neurosci Ther. 2011.PMID: 21951369 Free PMC article.Review.
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11158 | https://nrich.maths.org/problems/money-bags | Money Bags | NRICH
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Money bags
Ram divided 15 pennies among four small bags. He could then pay any sum of money from 1p to 15p without opening any bag. How many pennies did Ram put in each bag?
Age
5 to 11
Challenge level
Exploring and noticingWorking systematicallyConjecturing and generalisingVisualising and representingReasoning, convincing and proving
Being curiousBeing resourcefulBeing resilientBeing collaborative
Problem Getting Started Student Solutions Teachers' Resources
Problem
Ram divided 15 pennies among four small bags.
Image
He labelled each bag with the number of pennies inside it.
He could then pay any sum of money from 1p to 15p without opening any bag.
How many pennies did Ram put in each bag?
This problem is based on Money Bags fromMathematical Challenges for Able Pupils in Key Stages 1 and 2, published by DfES.
Getting Started
How would you make 1 p?
What do you need to do now to be able to make 2 p?
Does this help to make 3 p?
Student Solutions
We had quite a few submissions for this challenge and these make interesting reading.
Harrison School sent in a solution from Charlie and Oscar as follows;
To begin, we decided to use post-it-notes as bags and sweets to represent our pennies - this made it easy to visualise the question. Then, we worked out that one could only be added to zero in order to make itself; there was now a bag with one penny in it. After that, we made a bag with two pennies in. We did think that we could make two with two ones yet that would give us a problem later. Next, we worked out that three could already be made using bags one and two. However, four could not be made with bags one and two, therefore we had to make bag three have four pennies in it. We then looked and saw we could make numbers five (bag one and three), six (bags three and two) and seven (bags one, two and three) but eight was unable to be made. Consequently, that means that eight has to be in bag four. This used up all the pennies - there were none left over. Finally, we checked that the numbers nine to fifteen worked, which they did. Nine = bags four and one. Ten = bags four and two. Eleven = bags four, two and one. Twelve = bags four and three. Thirteen = bags four, three and one. Fourteen = bags four, three and two. Fifteen = bags four, three, two and one.
Image
Uma from Westhill Primary School, Aberdeenshire sent in her thoughts and the solution.
The solution for the Money Bags problem is 1,2,4 and 8 Here is the explanation of how I got my answer:
There are four bags so there can only be four different types of coins. The basic idea behind the solution is to find the four different numbers which can add up to give 1p to 15p. The solution for the question is 1,2,4 and 8. The 1st bag holds 1x1p, the 2nd bag contains 2x1p, the 3rd bag holds 4x1p and the 4th bag contains 8x1p. We can get any number between 1 to 15 using any combination of 1,2,4 and 8.
1p = Bag 1; 2p =Bag 2; 3p =Bag 1 +Bag 2; 4p =Bag 3; 5p =Bag 3 +Bag 1
6p =Bag 3 +Bag 2; 7p =Bag 3 +Bag 2 +Bag 1; 8p =Bag 4; 9p =Bag 4 +Bag 1
10p =Bag 4 +Bag 2; 11p =Bag 4 +Bag 2 +Bag 1; 12p =Bag 4 +Bag 3;
13p =Bag 4 +Bag 3 +Bag 1; 14p =Bag 4 +Bag 3 +Bag 2
15p =Bag 4 +Bag 3 +Bag 2 +Bag 1
Mr Howes' Magnificent Maths Group from Pierrepont Gamston Primary School
As part of an intervention group, we were a little confused to start off with - we didn't realise that the question was asking us to split the money into four bags so that the bags' totals could be combined to make any value up to 15p. Once we worked this out we wrote the values 1p to 15p down the side of the page and tried to find a system to solving the problem.
1p = this was easy, it had to be made with a bag with 1p in it!
2p = this could be made by another bag having 1p in it, or a second bag having 2p in it. We went for the latter because it stopped us using up our bags too quickly!
3p = this was also easy: 1p+2p
4p = We could either have another 1p or 2p bag, or a new 4p bag. We used the 4p bag.
At this point we spotted a pattern - 1p, 2p, 4p... would the next bag contain 8p?
5p = 4p+1p; 6p = 4p+2p; 7p = 4p+2p+1p; 8p = We stuck to our prediction and put 8p in the final bag!; 9p = 8p+1p; 10p = 8p+2p; 11p = 8p+2p+1p; 12p = 8p +4p; 13p = 8p+4p+1p
14p = 8p+4p+2p; 15p = 8p+4p+2p+1p
We thought that if the pattern was continued, the next bag should contain 16p ( making the total value 29p).
We wondered if this would be a better system for the values of coins;however, we realised that the values we used help us add up the numbers mentally much faster and make finding change easier!
James, Elliot, Freddie, Anna, Orla and Oliver
Jake from Duke of Norfolk School said
We drew out circles to show the different bags.
We tried to make all the numbers from 1 to 15, and recorded the number of pennies used in the circles.
We started with 1, 2, 3 and 5 coins, but found that we could not make numbers up from 12. We realised that we had to keep 1 and 2, because they were needed to pay 1p and 2p. So we added coins to the other bags,eventually settling on 1, 2, 4 and 8.
Stuart from Dussindale Primary School sent in this work and a good picture illustrating the solution
I started to look at multiples of ten but then realised I needed to use doubles/halves.
My first pair were 2 and 8 and then I saw I needed to have 1 and 2 and 4 and 8.
I used the cubes to make sure that I could make all the numbers from 1 to 15.
Image
Sarah from Cabot Primary School, Bristol sent in the following;
A Y4 maths group from Cabot Primary School, Bristol (Iman, Andrea-Marie,Sakariya and Aisha) worked on this problem together.
First of all, we found 15 play coins and a painting tray to help us think about the investigation and find a solution to the maths problem.
Secondly, we decided that we definitely needed to have one bag with 1p in it (or we would not be able to pay for something that cost 1p!), so we put 1p in one of the pots.
Then we thought about which numbers could make all of the other prices up to 15p. We realised that if we also had a bag with 2p in it, we could make 2p and 3p (1p + 2p), so we put 2p in our second pot. We tried 3p in the third pot and this gave us more prices:
1p (1p); 2p (2p); 3p (3p); 4p (3p + 1p); 5p (3p + 2p); 6p (3p + 2p + 1p)
...but we couldn't make 7p.
So we tried again, this time with 4p in the third pot. We could then make all of the prices up to 7p.
1p (1p); 2p (2p); 3p (2p + 1p); 4p (4p); 5p (4p + 1p); 6p (4p + 2p); 7p (4p + 2p + 1p).
We only had one empty pot left, so we put all of the rest of the coins into it - there were 8 of them. We then found that you could make all of the other numbers:
8p (8p); 9p (8p + 1p); 10p (8p + 2p); 11p (8p + 2p + 1p); 12p (8p + 4p); 13p (8p + 4p + 1p); 14p (8p + 4p + 2p); 15p (8p + 4p + 2p + 1p)
So we knew we had found the solution! We tried all the possibilities and didn't give up because Cabot Primary children are resilient!PS Our paint tray actually has six pots in it. We can see a pattern building up and are now going to think about how to investigate the maximum number of prices we can make if we use all six pots.
Later Sarah wrote and sent in this;
When we looked again at the number of coins in each pot, we noticed that each number was double the previous one. Since double 8 is 16, we put 16 coins into the next empty pot so we had five pots containing 1p, 2p, 4p, 8p and 16p.We knew the maximum price we could now pay would be 31p (our existing 15p plus the extra pot of 16p).We checked that we could now make all of the prices from 15p to 31p (which we could - by adding the new pot of money to each of the combinations that we had already found). We knew that we had found a system and that the sixth pot should therefore contain 32p (which is double 16p) and that we would be able to make all of the prices up to a maximum of 63p.
We then came up with this sequence to show how much money twelve pots (two paint trays) would contain:
1p, 2p, 4p, 8p, 16p, 32p, 64p, £1.28, £2.56, £5.12, £10.24 £20.48
It was good to hear about the different ways that solutions were found. Well done all of you.
Teachers' Resources
Money Bags
Ram divided 15 pennies among four small bags.
Image
He labelled each bag with the number of pennies inside it.
He could then pay any sum of money from 1 p to 15 p without opening any bag.
How many pennies did Ram put in each bag?
This problem is based on Money Bags from 'Mathematical Challenges for Able Pupils Key Stages 1 and 2', published by DfES. You can find out more about this book, including how to order it, on the Standards websitehere .
Why do this problem?
This problem is a good example of a challenge which does not require high-level mathematics, but does need a systematic approach. It also lends itself to a focus on different ways of recording.
Possible approach
In order that children understand the requirements of this problem, it would be a good idea to begin by reversing the situation. Draw two bags on the board and label them 1 p and 3 p. Invite children to suggest the amounts you could pay using these bags if you are not allowed to open either of them [1 p, 3 p and 4 p]. Introduce another bag to go with the first two which contains 5 p. This time, ask the children which amounts of money they wouldn't be able to make using these bags [2 p, 7 p, anything of 10 p and above]. You could give them time in pairs to work on this, perhaps using a mini-whiteboard to make jottings.
Invite pairs to share their solutions and highlight those that have used a systematic approach, for example by starting with the smallest amount of money and working up in steps of a penny at a time. You can then introduce the problem as it stands. Give pairs chance to work together before once again sharing their ideas. Before they give the answer, ask children to explain how they went about finding the solution.
The whole class could then check the solution offered by trying to make all the different amounts from 1 p to 15 p. You may also wish to share different ways of recording used by various pupils. You could discuss the advantages of each method and it would be worth asking whether children would change their chosen method if, for example, they were told someone else was going to read it. This gives you an opportunity to highlight the fact that we might record differently, depending on what the recording is for and who is going to see it.
Key questions
How will you make 1 p?
How will you make 2 p? ...
How will you make sure you can pay for all the amounts from 1 p to 15 p?
Possible extension
You could challenge children to extend this problem by asking what amount they would need in a fifth bag to be able to make as many amounts as possible over 15 p.
Possible support
Having equipment available might help some children grasp this problem. Try to let them choose what they would like to use, but having some small bags handy might be a good idea. Encouraging them to try out ideas is vital.
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11159 | https://oeis.org/A156709 | The OEIS is supported by the many generous donors to the OEIS Foundation.
For all numbers k(n) congruent to -1 or +1 (mod 6) starting with k(n) = {5,7,11,13,...}, a(k(n)) is incremented by the congruence (mod 6) if k(n) is prime and by 0 if k(n) is composite.
4
-1, 0, -1, 0, -1, 0, -1, -1, -2, -1, -1, 0, -1, 0, -1, -1, -2, -2, -3, -2, -2, -1, -2, -1, -1, 0, -1, -1, -2, -2, -2, -1, -2, -1, -2, -1, -2, -2, -2, -2, -2, -1, -2, -2, -3, -2, -2, -2, -3, -2, -2, -1, -1, 0, -1, -1, -2, -2, -3, -2, -2, -2, -3, -2, -3, -2, -2, -2, -2, -1, -1, -1, -1
(list; graph; refs; listen; history; text; internal format)
OFFSET
1,9
COMMENTS
The fact that a(k(n)) is predominantly negative exhibits the Chebyshev Bias (where the congruences that are not quadratic residues generally lead in the prime number races, at least for "small" integers, over the congruences that are quadratic residues).
This bias seems caused (among other causes?) by the presence of all those squares (even powers) coprime to 6 taking away opportunities for primes to appear in the quadratic residue class +1 (mod 6), while the non-quadratic residue class -1 (mod 6) is squarefree.
The density of squares congruent to +1 (mod 6) is 1/(6sqrt(k(n))) since 1/3 of the squares are congruent to +1 (mod 6), while the density of primes in either residue classes -1 or +1 (mod 6) is 1/(phi(6)log(k(n))), with phi(6) = 2.
Here 1 is a quadratic residue mod 6, but 5 (or equivalently -1) is a quadratic non-residue mod 6. All the even powers (included in the squares) map congruences {-1, +1} to {+1, +1} respectively and so contribute to the bias, whereas all the odd powers map {-1, +1} to {-1, +1} respectively and so do not contribute to the bias.
One would then expect the ratio of this bias, if caused exclusively by the even powers, relative to the number of primes in either congruences to asymptotically tend towards to 0 as k(n) increases (since 1/(6sqrt(k(n))) is o(1/(phi(6)log(k(n))))).
The persistence or not of such bias in absolute value then does not contradict The Prime Number Theorem for Arithmetic Progressions (Dirichlet) which states that the asymptotic (relative) ratio of the count of prime numbers in each congruence class coprime to m tends to 1 in the limit towards infinity. (Cf. 'Prime Number Races' link below.)
Also, even if this bias grows in absolute value, it is expected to be drowned out (albeit very slowly) by the increasing fluctuations in the number of primes in each congruence class coprime to 6 since, assuming the truth of the Riemann Hypothesis, their maximum amplitude would be, with x standing for k(n) in our case, h(x) = O(sqrt(x)log(x)) <= Csqrt(x)log(x) in absolute value which gives relative fluctuations of order h(x)/x = O(log(x)/sqrt(x)) <= Clog(x)/sqrt(x) in the densities of primes pi(x, {6, 1})/x and pi(x, {6, 5})/x in either congruence class.
Since 1/(6sqrt(x)) is o(log(x)/sqrt(x)) the bias will eventually be overwhelmed by the "pink noise or nearly 1/f noise" corresponding to the fluctuations in the prime densities in either congruence class. The falsehood of the Riemann Hypothesis would imply even greater fluctuations since the RH corresponds to the minimal h(x).
We get pink noise or nearly 1/f noise if we consider the prime density fluctuations of pi(x, {6, k})/x as an amplitude spectrum over x (with a power density spectrum of (Clog(x)/sqrt(x))^2 = ((Clog(x))^2)/x and see x as the frequency f. This power density spectrum is then nearly 1/x and would have nearly equal energy (although slowly increasing as (Clog(x))^2) for each octave of x. (Cf. 'Prime Numbers: A Computational Perspective' link below.)
Among the positive integers k(n) up to 100000 that are congruent to -1 or +1 (mod 6) [indexed from n = 1 to 33332, with k(n) = 6 ceiling(n/2) + (-1)^n], a tie, where a(k(n)) = 0, is attained or maintained for only 9 integers, and that bias favoring the non-quadratic residue class -1 (mod 6) never gets violated, i.e., a(k(n)) is never +1.
REFERENCES
R. Crandall and C. Pomerance, "Prime Numbers - A Computational Perspective", Second Edition, Springer Verlag 2005, ISBN 0-387-25282-7
LINKS
Daniel Forgues, Table of n, a(n) for n = 1..33332
Andrew Granville and Greg Martin, Prime Number Races, arXiv:math/0408319 [math.NT], 2004.
Eric Weisstein, Chebyshev Bias
Wikipedia, Pink noise
CROSSREFS
Cf. A156706 (whose sum of first n terms gives a(n)).
Cf. A156749 (which exhibits the Chebyshev Bias for congruences -1 or +1 (mod 4)).
Cf. A156707 (whose sum of first n terms gives A156749(n)).
Cf. A075743 (prime characteristic function of numbers congruent to -1 or +1 (mod 6)).
Cf. A101264 (prime characteristic function of numbers congruent to -1 or +1 (mod 4)).
Sequence in context: A124761 A333925 A342595 A081400 A378663 A328194
Adjacent sequences: A156706 A156707 A156708 A156710 A156711 A156712
KEYWORD
sign
AUTHOR
Daniel Forgues, Mar 29 2009
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11160 | https://arxiv.org/pdf/1703.02709 | arXiv:1703.02709v2 [math.NT] 5 Sep 2018
COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE
NASER T. SARDARI
Abstract. By assuming some widely-believed arithmetic conjectures, we show that the task of accepting a number that is representable as a sum of d ≥ 2squares subjected to given congruence conditions is NP-complete. On the other hand, we develop and implement a deterministic polynomial-time algorithm that represents a number as a sum of 4 squares with some restricted congruence conditions, by assuming a polynomial-time algorithm for factoring integers and Conjecture 1.1. As an application, we develop and implement a deterministic polynomial-time algorithm for navigating LPS Ramanujan graphs, under the same assumptions.
Contents
Introduction 12. NP-Completeness 73. Algorithm 10 4. Numerical results 17 References 19 1. Introduction
1.1. Motivtion. We begin by defining Ramanujan graphs. Fix k ≥ 3, and let G
be a k-regular connected graph with the adjacency matrix AG. It follows that k is an eigenvalue of AG. Let λG be the maximum of the absolute value of all the other eigenvalues of AG. By the Alon-Boppana Theorem [LPS88], λG ≥ 2√k − 1 + o(1) ,
where o(1) goes to zero as |G| → ∞ . We say that G is a Ramanujan graph, if
λG ≤ 2√k − 1.
The first construction of Ramanujan Graphs is due to Lubotzky, Phillips and Sarnak [LPS88] and independently by Margulis [Mar88]. We refer the reader to [Sar90, Chapter 3], where a complete history of the construction of Ramanujan graphs and other extremal properties of them are recorded. The LPS construction has the additional property of being strongly explicit. We say that the k-regular graph G is strongly explicit, if there is a polynomial-time algorithm that on inputs
〈v, i 〉 where v ∈ G, 1 ≤ i ≤ k outputs the (index of the) ith neighbor of v. Note that the lengths of the algorithm’s inputs and outputs are O(log |G|), and so it runs in time poly log( |G|). This feature of the LPS Ramanujan graphs is very important in their application to the deterministic error reduction algorithm [AKS87]; see also [HLW06] for other applications of Ramanujan graphs in Computer Science.
Date : September 6, 2018.
12NASER T. SARDARI
The main product of this work is a deterministic polynomial-time algorithm for navigating LPS Ramanujan graphs, by assuming a polynomial-time algorithm for factoring integers and an arithmetic conjecture, which we formulate next. Let Q(t0, t 1) := N
4q2
− (t0 + a0
2q
)2 − (t1 + a1
2q
)2, where q is a prime, N , a0, and a1
are integers, where N ≡ a20 + a21 mod 4 q and gcd( N, 4q) = 1 . Define (1.1) AQ,r := {(t0, t 1) ∈ Z2 : Q(t0, t 1) ∈ Z, |(t0, t 1)| < r, and Q(t0, t 1) ≥ 0},
where r > 0 is some positive real number.
Conjecture 1.1. Let Q and AQ,r be as above. There exists constants γ > 0 and
Cγ > 0, independent of Q and r, such that if |AQ,r | > C γ (log N )γ for some r > 0,then Q expresses a sum of two squares inside AQ,r .
We denote the following assumptions by ( ∗): (1) There exists a polynomial-time algorithm for factoring integers, (2) Conjecture 1.1 holds. The LPS construction is the Cayley graphs of P GL 2(Z/q Z) or P SL 2(Z/q Z) with
p + 1 explicit generators for every prime p and integer q. We denote them by the LPS Ramanujan graph Xp,q , and the p + 1 generators by the LPS generators in this paper. For simplicity for the rest of this paper as in [LPS88], we assume that
q ≡ 1 mod 4 is also a prime, and is a quadratic residue mod p, where p ≡ 1 mod 4 is fixed. By these assumptions, Xp,q is a Cayley graph over PSL 2(Z/q Z); see Section 1.2 for the explicit construction of Xp,q . We say v ∈ Xp,q is a diagonal vertex, if it corresponds to a diagonal matrix in PSL 2(Z/q Z). By a path from u1
to u2, we mean a sequence of vertices 〈v0, . . . , v h〉, where v0 = u1, vh = u2, and vi
is connected to vi+1 for every 0 ≤ i ≤ h − 1.
Theorem 1.2. Assume (∗). We develop and implement a deterministic polynomial-time algorithm in log( q), that on inputs 〈u1, u 2〉, where u1, u 2 ∈ Xp,q are diagonal vertices, outputs a shortest path 〈v0, . . . , v h〉 from u1 to u2. Moreover, for every
α ≥ 0 we have
(1.2) h ≤ max( α, 3 log p(q) + γ log p log( q) + log p(Cγ ) + log p(89)) ,
for all, but at most 89 q4/p (α−1) vertices. In particular, for large enough q the distance of any diagonal vertice from the identity is bounded by
(1.3) (4 /3) log p |Xp,q | + log p(89) .
Remark 1.3. Our algorithm is the q-adic analogue of the Ross and Selinger algo-rithm [RS16], which navigates P SU (2) with a variant of the LPS generators. In their work, the algorithm terminates in polynomial-time under the first assumption in (∗), and some heuristic arithmetic assumptions which are implicit in their work. We formulated Conjecture 1.1, and proved the algorithm terminates in polynomial-time under (∗). Moreover, we give quantitative bounds on the size of the output under (∗). In particular, (1.2) implies that the distance between all but a tiny frac-tion of pairs of diagonal vertices is less than log p(|Xp,q |)+ O(log log |Xp,q |). In order to prove our bounds, we introduce a correspondence between the diagonal vertices of
Xp,q and the index q sublattices of Z2. This is novel in our work; see Section 1.4.
It is known that every pair of vertices of a Ramanujan graph (not necessar-ily an LPS Ramanujan graph) are connected by a logarithmic number of edges. More precisely, for any x, y ∈ G, let d(x, y ) be the length of the shortest path COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE 3
between x and y. Define the diameter of G by diam( G) := sup x,y ∈G d(x, y ). It is easy to check that diam( G) ≥ log k−1 |G|. If G is a Ramanujan graph then diam( G) ≤ 2 log k−1 |G| + O(1); see [LPS88]. Moreover, we [Sar18, Theorem 1.5] showed quantitatively that all but a tiny fraction of the pairs of vertices in G have a distance less than log k−1(|G|) + O(log log |G|). Bounding the diameter of the LPS Ramanujan graph Xp,q is closely related to the diophantine properties of quadratic forms in four variables [Sar15a]. In particular, we showed that for every prime p there exists an infinite sequence of integers {qn},such that diam( Xp,q n ) ≥ (4 /3) log k−1 |Xp,q n |; see [Sar18, Theorem 1.2]. This shows that our upper bound in (1.3) is optimal. In fact, by assuming our conjecture on the optimal strong approximation for quadratic forms in 4 variables [Sar15a, Conjecture 1.3], the diameter of Xp,q is asymptotically (4 /3) log k−1 |Xp,q | as q → ∞ . In our joint work with Rivin [RS17], we gave numerical evidences for this asymptotic. Our navigation algorithm substantially improves the range of our previous numerical results, and gives stronger evidences for [Sar15a, Conjecture 1.3].
Remark 1.4. Sarnak in his letter to Scott Aaronson and Andy Pollington [Sar15b] defined the covering exponent of the LPS generators for navigating P SU (2) . He conjectured that the covering exponent is 4/3; see [Sar15a] and [BKS17]. In par-ticular, this exponent gives the optimal bound on the size of the output of the Ross and Selinger algorithm. lim q→∞ diam (Xp,q )
log p|Xp,q |
is the q-adic analogue of the covering exponent. In fact, [Sar15a, Conjecture 1.3] generalizes Sarnak’s conjecture, and it also implies
lim
q→∞
diam (Xp,q )
log p |Xp,q | = 4 /3.
By assuming ( ∗), we develop a deterministic polynomial-time algorithm that returns a short path between every pair of vertices of Xp,q . This version of the algorithm is not restricted to the diagonal vertices, but it does not necessarily return the shortest possible path; see Remark 1.6.
Theorem 1.5. Assume ( ∗). We develop a deterministic polynomial-time algo-rithm in log( q), that on inputs 〈u1, u 2〉, where u1, u 2 ∈ Xp,q , returns a short path
〈v0, . . . , v h〉 from u1 to u2. Moreover, we have (1.4) h ≤ 16
3 log k−1 |Xp,q | + O(1) .
Furthermore, (1.5) h ≤ 3 log k−1 |Xp,q | + O(log log( |Xp,q |)) for all but O(log( q)−c1 ) fraction of pairs of vertices, where c1 > 0, and the implicit constant in the O notations and c1 are independent of q.We briefly describe our proof in what follows. By [PLQ08, Lemma 1], we express any element of PSL 2(Z/q Z) as a product of a bounded number of LPS generators and four diagonal matrices. This reduces the navigation task to the diagonal case, and so Theorem 1.2 implies (1.4). For proving (1.5), we improve on Lauter, Petit and Quisquater’s diagonal de-composition algorithm. By (1.2), the distance of a typical diagonal element from the identity is less than log p |Xp,q | + O(log p log( |Xp,q |)). So, it suffices to show that all but a tiny fraction of vertices are the product of O(log p log( |Xp,q |)) number of 4 NASER T. SARDARI
LPS generators and three typical diagonal matrices. It is elementary to see that at least 10% of the vertices of Xp,q are the product of a bounded number of LPS generators and three typical diagonal matrices. By the expansion property of the Ramanujan graphs, the distance of all but a tiny fraction of the vertices is less that
O(log p log( |Xp,q |)) from any subset containing more than 10% of vertices. This implies (1.5). We give the dull details of our argument in Section 3.3.
Remark 1.6. By Theorem 1.7 and Corollary 1.9, it follows that finding the shortest path between a generic pair of vertices is essentially NP-complete; see Remark 1.11 for further discussion. The idea of reducing the navigation task to the diagonal case is due Petit, Lauter, and Quisquater [PLQ08], which is crucial in both Ross and Selinger [RS16] and this work. As a result of this diagonal decomposition, the size of the output path is 3 times the shortest possible path for a typical pair of vertices. Improving the constant 3 to 3 − ǫ needs new ideas, and this would have applications in quantum computing.
1.2. Reduction to strong approximation on the sphere. In [LPS88, Section 3], the authors implicitly reduced the task of finding the shortest possible path between a pair of vertices in Xp,q to the task of representing a number as a sum of 4 squares subjected to given congruence conditions, which is the strong approximation on the 3-sphere. We explain this reduction in this section. We begin by explicitly describing Xp,q . Let H(Z) denote the integral Hamiltonian quaternions
H(Z) := {x0 + x1i + x2j + x3k|xt ∈ Z, 0 ≤ t ≤ 3, i 2 = j2 = k2 = −1},
where ij = −ji = k, etc. Let α := x0 + x1i + x2j + x3k ∈ H(Z). Denote ¯ α :=
x0 − x1i − x2j − x3k and Norm( α) := α ¯α = x20 + x21 + x22 + x23. Let (1.6)
Sp := {α ∈ H(Z) : Norm( α) = p, x 0 > 0 is odd and x1, x 2, x 3 are even numbers }.
It follows that Sp = {α1, ¯α1, . . . , α (p+1) /2, ¯α(p+1) /2}. Let Λ′
p
:= {β ∈ H : Norm( β) = ph′
and β ≡ 1 mod 2 }.
Λ′
p
is closed under multiplication. Let Λ p be the set of classes of Λ ′
p
with the relation
β1 ∼ β2 whenever ±pt1 β1 = pt2 β2, where t1, t 2 ∈ Z. Then Λ ′
p
form a group with [β1][ β2] = [ β1β2] and [ β][ ¯β] = .
By [LPS88, Corollary 3.2], Λ p is free on [ α1], . . . , [α(p+1) /2]. Hence, the Cayley graph of Λ p with respect to LPS generator set Sp is an infinite p + 1-regular tree. LPS Ramanujan graphs are associated to the quotient of this infinite p + 1-regular tree by appropriate arithmetic subgroups that we describe in what follows. Let Λp(q) := {[β] ∈ Λp : β = x0 + x1i + x2j + x3k ≡ x0 mod 2 q}.
Λp(q) is a normal subgroup of Λ p. By [LPS88, Proposition 3.3], since q ≡ 1 mod 4 is a prime number and q is a quadratic residue mod p,Λp/Λp(q) = PSL 2(Z/q Z).
The above isomorphism is defined by sending [ α] ∈ Λp, to the following matrix ˜ α
in PSL 2(Z/q Z): (1.7) ˜α := 1
√Norm( α)
[ x0 + ix 1 y + ix 3
−y + ix 3 x0 − ix 1
]
,COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE 5
where i and √p are representatives of square roots of −1 and p mod q. This identifies the finite p + 1-regular graph Λ p/Λp(q) by the Cayley graph of PSL 2(Z/q Z) with respect to ˜Sp (the image of Sp under the above map) that is the LPS Ramanujan graph Xp,q . For v ∈ Xp,q , we denote its associated class in Λ p/Λp(q) by [ v].
Finally, we give a theorem which reduces the navigation task on LPS Ramanujan graphs to an strong approximation problem for the 3-sphere. Since Xp,q is a Cayley graph, it suffices to navigate from the identity vertex to any other vertex of Xp,q .
Theorem 1.7 (Due to Lubotzky, Phillips and Sarnak) . Let v ∈ Xp,q , and a0 +
a1i + a2j + a3k ∈ [v] such that gcd( a0, . . . , a 3, p ) = 1 . There is a bijection between non-backtracking paths (v0, . . . , v h) of length h from v0 = id to vh = v in Xp,q , and the set of integral solutions to the following diophantine equation
x21 + x22 + x23 + x24 = N, xl ≡ λa l mod 2q for 0 ≤ l ≤ 3 and some λ ∈ Z/2qZ,
(1.8)
where N = ph. In particular, the distance between id and v in Xp,q is the smallest exponent h such that (1.8) has an integral solution.
By [Sar15a, Conjecture 1.3], there exists an integral lift if ph ≫ǫ q4+ ǫ and 4 is the optimal exponent. This conjecture implies that diam( Xp,q ) is asymptotically, 4/3 log k−1 |Xp,q |.
1.3. Complexity of strong approximation on the sphere. In this section, we give our main results regarding the complexity of representing a number as a sum of d squares subjected to given congruence conditions. First, we give our result for
d = 2 .
Theorem 1.8. The problem of accepting (N, q, a 0, a 1) such that the diophantine equation
x20 + x21 = N, x0 ≡ a0 and x1 ≡ a1 mod q,
has integral solution (x0, x 1) ∈ Z2 is NP-complete, by assuming GRH and Cramer’s conjecture, or unconditionally by a randomized reduction algorithm.
The above theorem is inspired by a private communication with Sarnak. He showed us that the problem of representing a number as a sum of two squares subjected to inequalities on the coordinates is NP-complete, under a randomized reduction algorithm. The details of this theorem appeared in his joint work with Parzanchevski [PS18, Theorem 2.2]. By induction on d, we generalize our theorem for every d ≥ 2.
Corollary 1.9. Let d ≥ 2. The problem of accepting (N, q, a 0, . . . , a d−1) such that the diophantine equation
x20 + · · · + x2
d−1
= N, X0 ≡ a0 . . . x d−1 ≡ ad−1 mod q,
(1.9)
has integral solution (x0, . . . , x d−1) ∈ Zd is NP-complete, by assuming GRH and Cramer’s conjecture, or unconditionally by a randomized reduction algorithm. 6 NASER T. SARDARI
On the other hand, by assuming ( ∗) and two coordinates of the congruence conditions in (1.9) are zero, we develop and implement a polynomial-time algorithm for this task for d = 4 .
Theorem 1.10. Let q be a prime, and (a0, a 1) ∈ (Z/2qZ)2, where a0 is odd and
a1 is even. Suppose that N = O(qA), gcd( N, 4q) = 1 , and a20 + a21 ≡ N mod 4q. By assuming (∗), we develop and implement a deterministic polynomial-time algorithm in log( q) that finds an integral solution (x0, . . . , x 3) ∈ Z4 to
x20 + · · · + x23 = N, xi ≡ ai mod 2 q,
(1.10)
where a2 = a3 = 0 . If there is no solution to (1.10) , then it returns “No solution”.
By Theorem 1.7, the algorithm in Theorem 1.10 gives the navigation algorithm described in Theorem 1.2.
Remark 1.11. It is possible to generalize our polynomial-time algorithm for any
d ≥ 2, by assuming a variant of (∗) and two coordinates of the congruence con-ditions are zero. On the other hand, by assuming GRH and Cramer conjecture, Corollary 1.9 implies that the complexity of the optimal strong approximation for a generic point on the sphere is NP-complete. Hence, by assuming these widely believed arithmetic assumptions, Corollary 1.9 essentially implies that finding the shortest possible path between a generic pair of vertices in LPS Ramanujan graphs is NP-complete.
1.4. Quantitative bounds on the size of the output. In this section, we give a correspondence between the diagonal vertices of Xp,q and the index q sublattices of Z2. Next, we relate the graph distance between the diagonal vertices (that is a diophantine exponent by Theorem 1.7 ) to the length of the shortest vector of the corresponded sublattice. Let v ∈
[a + ib 00 a − ib
]
∈ Xp,q be a diagonal vertex, and let Lv be the sublattice of Z2 defined by the following congruence equation:
ax + by ≡ 0 mod q.
Let {u1, u 2} be the Gauss reduced basis for Lv, where u1 is a shortest vector in Lv.In the following theorem, we relate the graph distance of v from the identity to the norm of u1.
Theorem 1.12. Assume Conjecture 1.1. Let v, Lv and {u1, u 2} be as above. Suppose that |u2|
|u1|
≥ Cγ log(2 q)γ , then the distance of v from the identity is less than
(1.11) ⌈4 log p(q) − 2 log p |u1| + log p(89) ⌉.
Otherwise, the distance of v from the identity vertex is less than
(1.12) ⌈3 log p(q) + γ log p log( q) + log p(Cγ ) + log p(89) ⌉.
Remark 1.13. In Section 4, we numerically check that the inequality (1.11) is sharp. In particular, the diameter of LPS Ramanujan graphs is asymptotically the longest distance between the diagonal vertices. Moreover, the above theorem implies (1.2) and (1.3) in Theorem 1.2. We also use this theorem in our algorithm COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE 7
in Theorem 1.5, in order to avoid the diagonal vertices with long distance from the identity.
1.5. Further motivations and techniques. Rabin and Shallit [RS85] developed a randomized polynomial-time algorithm that represents any integer as a sum of four squares. The question of representing a prime as a sum of two squares in polynomial-time has been discussed in [Sch85] and [RS85]. Schoof developed a deterministic polynomial-time algorithm that represents a prime p ≡ 1 mod 4 as a sum of two squares by O((log p)6) operations. We use Schoof’s algorithm in our algorithm in Theorem 1.10. Both Ross-Selinger and our algorithm start with searching for integral lattice points inside a convex region that is defined by a simple system of quadratic in-equalities. If the convex region is defined by a system of linear inequalities in a fixed dimension then the general result of Lenstra [Len83] implies this search is polynomially solvable. We use a variant of Lenstra’s argument in the proof of The-orem 1.10. An important feature of our algorithm is that it has been implemented, and it runs and terminates quickly. We give our numerical results in Section 4.
Acknowledgements. I would like to thank my Ph.D. advisor Peter Sarnak for several insightful and inspiring conversations during the course of this work. Fur-thermore, I am very grateful for his letter to me which deals with the Archimedean version of Theorem 1.8. I would like to thank Professor Peter Selinger for providing a public library of his algorithms in [RS16]. This material is partially supported by the National Science Foundation under Grant No. DMS-1440140 while the au-thor was in residence at the Mathematical Sciences Research Institute in Berkeley, California, during the Spring 2017 semester. 2. NP-Completeness
In this section, we prove Theorem 1.8 and Corollary 1.9. We reduce them to the sub-sum problem, which is well-known to be NP-complete. We begin by stating the sub-sum problem, and proving some auxiliary lemmas. The proof of Theorem 1.8 and Corollary 1.9 appear at the end of this section. Let t1, t 2, . . . , t k and t ∈ N with log( t) and log( ti) at most kA.
sub-sum problem 2.1. Are there ǫi ∈ { 0, 1} such that
(2.1)
k
∑
j=1
ǫj tj = t?
Lemma 2.2. By Cramer’s conjecture, there exists a polynomial-time algorithm in
k that returns a prime number q ≡ 3 mod 4 such
(2.2) q > 2k max
1≤i≤k
(ti, t ).
Alternatively, this task can be done unconditionally by a probabilistic polynomial-time algorithm in k.Proof. Let X := 4 k max 1≤i≤k(ti, t ) + 3 . We find q by running the primality test algorithm of Agrawal, Kayal and Saxena [AKS04] on the arithmetic progression
X, X + 4 , . . . . By Cramer’s conjecture this search terminates in O(log( X)2) oper-ations. 8 NASER T. SARDARI
Alternatively, we pick a random number between [ X, 2X] and check by a the primality test algorithm if the number is prime. The expected time of the operations is O(log( X)).
Let (2.3) s := ( q − 1) t +
k
∑
i=1
ti.
By a simple change of variables, solving equation (2.1) is equivalent to solving (2.4)
k
∑
j=1
ξj tj = s,
where ξj ∈ { 1, q }.Let Fq2 be the finite field with q2 elements.
Lemma 2.3. By assuming GRH, there exists a deterministic polynomial-time algo-rithm in log q that returns a finite subset H ⊂ F∗
q2
of size O((log q)8+ ǫ) such that H
contains at least a generator for the cyclic multiplicative group F∗
q2
. Alternatively, this task can be done unconditionally by a probabilistic polynomial algorithm. Proof. Since q ≡ 3 mod 4, Z/q Z[i] is isomorphic to Fq2 , where i2 = −1. By Shoup’s result [Sho92, Theorem 1.2], there is a primitive roots of unity g = a + bi ∈ Z/q Z[i]for the finite field with q2 elements such that a and b has an integral lift of size
O(log( q)4+ ǫ) for any ǫ > 0 result. Hence, the reduction of H : {a + bi : |a|, |b| ≤
log( q)4+ ǫ} mod q has the desired property. Alternatively, this task can be done unconditionally by a probabilistic polynomial algorithm. Because, the density of primitive roots of unity in F∗
q2
is ϕ(q2 − 1) /(q2 −
1), where ϕ is Euler’s totient function. The ratio ϕ(q2 − 1) /(q2 − 1) is well-known to be O(log log q).
Next, we take an element g ∈ H (not necessarily a generator), and let
aj + ib j := gtj for 1 ≤ j ≤ k, a + ib := gs,
(2.5) where tj are given in the subsum problem (2.1), s is defined by equation (2.3) and
aj , b j , a, b ∈ Z/q Z. Next, we find gaussian primes πj ∈ Z[i] such that (2.6) πj ≡ aj + ib j mod q.
Again this is possible deterministically by Cramer’s conjecture. Alternatively, we choose a random integral point ( h1, h 2) ∈ [X, 2X] × [X, 2X], and check by a polynomial-time primality test algorithm if ( h1q + aj )2 + ( h2q + bj )2 is prime in
Z. Set pi := |πj |2 that is a prime in Z and define (2.7) N :=
k
∏
j=1
pj .
Consider the following diophantine equation
X2 + Y 2 = N, X ≡ a and Y ≡ b mod q,
(2.8) COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE 9
where a, b are defined in equation (2.5) and N in (2.7). Our theorem is a conse-quence of the following lemma.
Lemma 2.4. Assume that g ∈ F∗
q2
is a generator. An integral solution (X, Y ) to the diophantine equation (2.8) gives a solution (ξ1, . . . , ξ k) to the equation (2.4) in polynomial-time in log( q).Proof of Lemma 2.4. Assume that the equation (2.8) has an integral solution ( a0, a 1).
A + Bi factors uniquely in Z[i], and we have (2.9) A + iB = ±i
k
∏
j=1
πǫj
j
,
where ǫj ∈ { 0, 1}, and π0
j
= πj and π1
j
= ¯ π (the complex conjugate of πj ). We consider the above equation mod q. Then
A + iB ≡ ± i
k
∏
j=1
πǫj
j
mod q.
By the congruence condition (2.8), A+iB ≡ a+ib mod q, and by (2.6), π0
j
≡ aj +ib j
and π1
j
≡ aj − ib j mod q. By (2.5), we obtain
gs ≡
k
∏
j=1
gξj tj ,
where ξj = 1 if ǫj = 0 and ξj = q if ǫj = 1. Therefore we obtain the following congruence equation k∑
j=1
ξj tj ≡ s mod q(q − 1) .
By the inequality (2.2) and the definition of s in equation (2.3), we deduce that
k
∑
j=1
ξj tj = s.
This completes the proof of our lemma.
Proof of Theorem 1.8. Our theorem is a consequence of Lemma 2.4. For every
g ∈ H we apply Lemma 2.4 and check if ( ξ1, . . . , ξ k ) is a solution to (2.4). Since the size of H is O(log( q)8+ ǫ) and it contains at least a primitive roots of unity, we find a solution to the equation (2.4) in polynomial-time. This concludes our theorem.
Finally, we give a proof for Corollary 1.9.
Proof of Corollary 1.9. We prove this corollary by induction on d. The base case
d = 2 follows from Theorem 1.8. It suffices to reduce the task with d variables to a similar task with d + 1 variables in polynomial-time. The task with d variables is to accept ( N, q, a 1, . . . , a d) such that the following diophantine equation has a solution
X21 + · · · + X2
d
= N, X1 ≡ a1 . . . X d ≡ ad mod q.
(2.10) 10 NASER T. SARDARI
We proceed by taking auxiliary parameters 0 ≤ t, m ∈ Z such that N < q 2t,
m ≤ (1 /3) q2t+1 and gcd( m, q ) = 1. We consider the following diophantine equation
X21 + . . . X 2
d
X2
d+1
= m2 + q2tN, X1 ≡ qta1, . . . , X d ≡ adqt and Xd+1 ≡ m mod qt+1 .
(2.11) Assume that ( X1, . . . , X d+1 ) is a solution to the above equation. Then
Xd+1 ≡ ± m mod q2t+1 .
Since m ≤ (1 /3) q2t+1 , either Xd+1 = ±m or |Xd+1 | ≥ (2 /3) q2t+1 . If |Xd+1 | ≥
(2 /3) q2t+1 , since m ≤ (1 /3) q2t+1 and N < q 2t,
X2
d+1
m 2 + q2tN.
This contradicts with equation (2.11). This shows that Xd+1 = ±m. Hence, the in-tegral solutions to the diophantine equation (2.11) are of the form ( qtX1, . . . , q tXd, ±m)such that ( X1, . . . , X d) is a solution to the equation (2.10). By our induction as-sumption, this problem is NP-complete, and we conclude our corollary.
Algorithm
3.1. Proof of Theorem 1.10. In this section, we prove Theorem 1.10, which is the main ingredient in the navigation algorithms in Theorem 1.2 and Theorem 1.5. Let ( x0, . . . , x 3) ∈ Z4 be a solution to the equation (1.10). We change the variables to ( t0, . . . , t 3) ∈ Z4, where xi = 2 tiq + ai, and |ai| ≤ q. Hence, (3.1) N
4q2 − (t0 + a0/2q)2 − (t1 + a1/2q)2 = t22 + t23.
Let Q(t0, t 1) := N
4q2
− (t0 + a0/2q)2 − (t1 + a1/2q)2. Recall the definition of AQ,r
from (1.1), where r > 0 is some real number. By conjecture 1.1, if |AQ,r | >Cγ (log N )γ then the equation (3.1) has a solution, where ( t0, t 1) ∈ AQ,r .
First, we give a parametrization of ( t0, t 1) ∈ Z2, where Q(t0, t 1) ∈ Z. Let k :=
N−a20−a21
4q
. Since a20 + a21 ≡ N mod 4 q, k ∈ Z. By (3.1), (3.2) a0t0 + a1t1 ≡ k mod q.
Without loss of generality, we assume that a0 6 = 0 mod q. Then a0 has an in-verse mod q, and ( ka −10 , 0) is a solution for the congruence equation (3.2). We lift (ka −10 , 0) ∈ (Z/q Z)2 to the integral vector ( c, 0) ∈ Z2 such that
c ≡ ka −10 mod q and |c| < (q − 1) /2.
The integral solutions of equation (3.2) are the translation of the integral solutions of the following homogenous equation by vector ( c, 0) ∈ Z2
(3.3) a0t1 + a1t1 ≡ 0 mod q.
The integral solutions to equation (3.3) form a lattice of co-volume q that is spanned by the integral basis {v1, v 2} where
v1 := ( q, 0) , and v2 := ( −a1a−10 , 1) .COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE 11
We apply Gauss reduction algorithm on the basis {v1, v 2} in order to find an almost orthogonal basis {u1, u 2} such that span Z〈v1, v 2〉 = span Z〈u1, u 2〉,
|u1| < |u2|,
〈u1, u 2〉 ≤ (1 /2) 〈u1, u 1〉.
(3.4) where span Z〈v1, v 2〉 := {xv 1 + yv 2 : x, y ∈ Z} and 〈u1, u 2〉 ∈ R is the dot product of u1 and u2. Let u0 be a shortest integral vector that satisfies the equation (3.2). We write ( c, 0) as a linear combination of u1 and u2 with coefficients in (1 /q )Z
(0 , c ) = ( h1 + r1/q )u1 + ( h2 + r2/q )u2,
where 0 ≤ r1, r 2 ≤ q − 1. Note that u0 is one of the following 4 vectors
(r1/q − { 0, 1})u1 + (r2/q − { 0, 1})u2.
By triangle inequality,
|u0| < |u2|.
We parametrize the integral solutions ( t0, t 1) of (3.2) by: (3.5) (t0, t 1) = u0 + xu 1 + yu 2,
where x, y ∈ Z. Let u0 = ( u0,0, u 0,1), u1 = ( u1,0, u 1,1) and u2 = ( u2,0, u 2,1). Since
u1 and u2 are solutions to (3.3) and u0 is a solution to (3.2),
u′
0
:= k − a0u0,0 − a1u0,1
q ∈ Z,u′
1
:= a0u1,0 + a1u1,1
q ∈ Z,u′
2
:= a0u2,0 + a1u2,1
q ∈ Z.
Let (3.6) F (x, y ) := u′
0
− xu ′
1
− yu ′
2
− (u0,1 + xu 1,1 + yu 2,1)2 − (u0,2 + xu 1,2 + yu 2,2)2.
By (3.5),
F (x, y ) = Q(t0, t 1).
Hence, Q(t0, t 1) ∈ Z for ( t0, t 1) ∈ Z2, if and only if ( t0, t 1) = u0 + xu 1 + yu 2 for some ( x, y ) ∈ Z2.
Next, we list all the integral points ( x, y ) such that F (x, y ) is positive.
Lemma 3.1. Assume that √N
q|u2|
≥ 14 /3. Let F (x, y ) be as above. Let A :=
√N / (2 q|u1|) − 1 , B := √N / (2 q|u2|) − 1 and
(3.7) C := [ −A, A ] × [−B, B ].
Then F (x, y ) is positive for every (x, y ) ∈ C and negative outside 10 × C.Proof. Recall that ( t0, t 1) = u0 + xu 1 + yu 2 and
F (x, y ) = N/ 4q2 − (t1 + a0/4q)2 − (t2 + a1/2q)2,
where |a0/2q| < 1/2 and |a1/2q| < 1/2. Hence, if |(t0, t 1)| < (√N /q ) − 1, then
F (x, y ) > 0, and if |(t0, t 1)| > (√N/q ) + 1 , then F (x, y ) < 0. By the triangle inequality
|(t0, t 1)| = |u0 + xu 1 + yu 2| ≤ | u0| + |x|| u1| + |y|| u2|.12 NASER T. SARDARI
Since |u0| < |u2| then |(t0, t 1)| ≤ | x|| u1| + (1 + |y|)|u2|. Let A , B and C be as in (3.7). Then for every ( x, y ) ∈ [−A, A ] × [−B, B ], we have
|x|| u1| + (1 + |y|)|u2| ≤ (√N /q ) − | u1| < (√N /q ) − 1.
Hence, F (x, y ) > 0 if ( x, y ) ∈ [−A, A ] × [−B, B ]. Next, we show that F is negative outside 10 × C. By almost orthogonality conditions (3.4), we obtain the following lower bound (3.8) (|x|/2) |u1| + ( |y|/2 − 1) |u2| ≤ | u0 + xu 1 + yu 2|.
The above inequality implies that if x ≥ 10 A, then
|(t0, t 1)| = |u0 + xu 1 + yu 2| > √N /q + (3√N /q |u1| − 10 )|u1|/2.
We assume that √N
q|u2|
≥ 14 /3 and 1 < |u1| < |u2|, then |(t0, t 1)| > √N /q + 1 and hence F (x, y ) is negative. Similarly, if y ≥ 10 B then
|(t0, t 1)| = |u0 + xu 1 + yu 2| > √N /q + (3√N/ (2 q|u2|) − 6)|u2|.
Since √N
q|u2|
≥ 14 /3, it follows that |(t0, t 1)| > √N /q + 1 . Hence, F (x, y ) is negative. Therefore, if ( x, y ) /∈ 10 × C, then F (x, y ) is negative. This concludes our lemma.
In the following lemma, we consider the remaining case, where √N
q|u2|
≤ 14 /3.
Lemma 3.2. Assume that √N
q|u2|
≤ 14 /3 and F (x, y ) > 0 then |y| ≤ 13 .
Proof. Since F (x, y ) > 0 from the first line of the proof of Lemma 3.1, it follows that |(t0, t 1)| < (√N/q ) + 1 . From the the inequality (3.8), we have (|y|/2 − 1) |u2| ≤ | (t0, t 1)| ≤ √N /q + 1 .
Hence,
|y| ≤ 2√N
q|u2| + 4 < 14 .
Since y is an integer, we conclude the lemma.
Proof of Theorem 1.10. Assume that √N
q|u2|
≥ 14 /3. By Lemma 3.1, F (x, y ) is pos-itive inside box C that is defined in (3.7). We list ( x, y ) ∈ C in the order of their distance from the origin. If possible we represents F (x, y ) as a sum of two squares by the following polynomial-time algorithm. We factor F (x, y ) into primes by the poly-nomial algorithm for factoring integers in ( ∗). Next, by Schoof’s algorithm [Sch85]), we write every prime number as a sum of two squares. If we succeed, then we find an integral solution to the equation (1.10), and this concludes the theorem. If the size of box C that is A × B > C γ log( N )γ , then by Conjectrue 1.1 we find a pair ( x, y ) such that F (x, y ) is a sum of two squares in less than O(log( q)O(1) )steps, and the above algorithm terminates. Otherwise, A × B < C γ log( M )γ . By Lemma 3.1, F (x, y ) is negative outside box 10 C and since the size of this box is
O(log( q)γ ) we check all points inside box 10 C in order to represent F (x, y ) as a sum of two squares. If we succeed to represent F (x, y ) as a sum of two squares then we find an integral solution to equation (1.10). Otherwise, the equation (1.10) does not have any integral solution. This concludes our theorem if √N
q|u2|
≥ 14 /3. COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE 13
Finally, assume that √N
q|u2|
≤ 14 /3 then by Lemma 3.2, we have |y| ≤ 13 . We fix
y = l for some |l| < 13 . We note that by equation (3.6),
F (x, l ) = Ax 2 + Bx + C
for some A, B, C ∈ Z. We list x ∈ Z such that F (x, l ) > 0 and then proceed similarly as in the forth line of the first paragraph of the proof. This concludes our theorem.
3.2. Distance of diagonal vertices from the identity. In this section, we give a proof of Theorem 1.12. Then, we give bounds on the size of the outputs in The-orem 1.2 and Theorem 1.5. Recall the notations while formulating Theorem 1.12.
Proof of Theorem 1.12. We proceed by proving (1.11). Assume that
|u2| ≥ Cγ log( q)γ |u1|.
Let (3.9) h := ⌈4 log p(q) − 2 log p |u1| + log p(89) ⌉.
We show that there exists a path from v to the identity of length h. By our assumption p is a quadratic residue mod q. We denote the square root of p mod q
by √p. Set
A := a√ph mod 2 q, B := b√ph mod 2 q.
By Theorem 1.7, there exists a path of length h from v to the identity if and only the following diophantine equation has an integral solution ( t1, t 2, t 3, t 4)(3.10) (2 t1q + A)2 + (2 t2q + Bt )2 + (2 t3q)2 + (2 t4q)2 = ph.
In Theorem 1.10, we developed a polynomial-time algorithm for finding its integral solutions ( t1, t 2, t 3, t 4). We defined the associated binary quadratic form F (x, y )as defined in equation (3.6). By Lemma 3.1, F (x, y ) is positive inside the box [−A, A ] × [−B, B ] where A := √ph/(4 q|u1|) , B := √ph/(4 q|u2|) − 1. By the definition of h in equation (3.9), we have (3.11) ph ≥ 89 q4
|u1|2 .
By the above inequality (3.12) B ≥√89 q2
4q|u1|| u2| − 1.
Since {u1, u 2} is an almost orthogonal basis for a co-volume q lattice then the angle between u1 and u2 is between π/ 3 and 2 π/ 3. Hence, (3.13) |u1|| u2| ≤ 2q/ √3.
We use the above bound on |u1|| u2| in inequality (3.12), and derive
B ≥√3 ∗ 89
8 − 1 > 1.14 NASER T. SARDARI
Next, we give a lower bound on A. Note that
A ≥ |u2|
|u1| B.
By our assumption |u2|
|u1|
≥ Cγ log(2 q)γ , hence
A ≥ Cγ log(2 q)γ B.
Since B > 1,
AB ≥ Cγ log(2 q)γ .
By Conjecture 1.1 and Theorem 1.10, our algorithm returns an integral solution (t1, t 2, t 3, t 4) which gives rise to a path of length h from v to the identity. This concludes the first part of our theorem. Next, we assume that |u1| ≤ | u2| ≤ Cγ log(2 q)γ |u1|. Let (3.14) h′ := ⌈3 log p(q) + γ log p log( q) + log p(Cγ ) + log p(89) ⌉.
We follow the same analysis as in the first part of the theorem. First, we give a lower bound on B := √ph′ /(4 q|u2|) − 1. By the definition of h′ in equation (3.14), we derive (3.15) ph′
≥ 89 Cγ log( q)γ q3.
We multiply both sides of |u2| ≤ Cγ log( q)γ |u1| by |u2| and use the inequality (3.13) to obtain
|u2|2 ≤ Cγ log( q)γ 2q/ √3.
By the above inequality, definition of B and inequality (3.15), we have
B = √ph′ /(4 q|u2|) − 1 ≥√89
4
√
2/√3
− 1 ≥ 1.
Hence,
B ≥ √ph′ /(8 q|u2|).
Next, we use the above inequality and inequality (3.13) and (3.15) to give a lower bound on AB .
AB ≥ ph′
32 q2|u1|| u2|≥ 89 √3Cγ log( q)γ q3
64 q3
C γ log( q)γ .
(3.16) By Conjecture 1.1 and Theorem 1.10, our algorithm returns an integral solution (t1, t 2, t 3, t 4) which gives rise to a path of length h′ from v to the identity. This concludes our theorem.
Finally, we prove (1.2) in Theorem 1.2. We briefly, explain the main idea. We normalize the associated co-volume q lattices Lv, so that they have co-volume 1. These normalized lattices are parametrized by points in SL 2(Z)\H, and it is well-known that they are equidistributed in SL 2(Z)\H with respect to the hy-perbolic measure 1
y2
(dx 2 + dy 2). It follows from this equidistribution and Theo-rem 1.12 that the distance of a typical diagonal matrix from the identity vertex is log( |Xp,q |)+ O(log log( |Xp,q |)). The diagonal points with distance 4 /3 log( |Xp,q |)+ COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE 15
O(log log( |Xp,q |)) from the identity are associated to points x + iy ∈ H with y as big as q.
Proof of (1.2) in Theorem 1.2. Let v be a diagonal vertex with distance h from the identity vertex, where
h ≥ ⌈ 3 log p(q) + γ log p log( q) + log p(Cγ ) + log p(89) ⌉.
Let Lv be the associated lattice of co-volume q and {u1, u 2} be an almost orthogonal basis for Lv. By Theorem 1.12, the distance of v from the identity is less than
⌈4 log p(q) − 2 log p |u1| + log p(89) ⌉.
Therefore,
h ≤ ⌈ 4 log p(q) − 2 log p |u1| + log p(89) ⌉.
Hence, (3.17) |u1|2 ≤ 89 q4/p (h−1) .
Next, we count the number of lattices of co-volume q inside Z2 such that the length of the shortest vector is smaller than r ≤ (1 /2) √q. Let L ⊂ Z2 be a lattice of co-volume q such that L contains a vector of length smaller than (1 /2) √q. It is easy to check that L contain unique vectors ±v := ±(a0, a 1) such they have the shortest length among all vectors inside L. Since q is prime this vector is primitive i.e. gcd( a0, a 1) = 1. On the other hand, the lattice is uniquely determined by
±v := ±(a0, a 1), namely L is the set of all integral points ( x, y ) ∈ Z2 such that
ax + by ≡ 0 mod q.
Therefore, the problem of counting the lattices of co-volume q with shortest vector smaller than r is reduced to counting the projective primitive integral vectors of length smaller than r. The main term of this counting is (3.18) 1/2ζ(2) −1πr 2 = 3
π r2.
By inequality (3.17) and (3.18), we deduce that the number of diagonal vertices with graph distance at least h from the identity in LPS Ramanujan graph Xp,q is less than 89 q4/p (h−1) .
This concludes Theorem 1.2.
3.3. Algorithm for the diagonal decomposition.
Proof of (1.4) in Theorem 1.5. Let M :=
[a bc d
]
∈ PSL 2(Z/q Z) be any element. By [PLQ08, Lemma 1], there exists a polynomial-time algorithm that expresses M
as:
M = D1s1D2s2D3s3D4,
where Di are diagonal matrices for 1 ≤ i ≤ 4 and sj are LPS generators for 1 ≤ j ≤ 3. By Theorem 1.2 and assuming ( ∗), we write each Di as a product of at most 4 /3 log p |Xp,q | + O(1) LPS generators in polynomial time. Therefore, we find a path of size at most 16
3
log k−1 |Xp,q | + O(1) from the identity to M. This concludes (1.4). 16 NASER T. SARDARI
Let
D :=
{ [a 00 a−1
]
∈ PSL 2(Z/q Z)
}
, and R :=
{ [a bb a
]
∈ PSL 2(Z/q Z)
}
.
Define dα :=
[α 00 α−1
]
, and ra,b :=
[a ba b
]
. By the correspondence (1.7) between PSL 2(Z/q Z) and the units of H(Z/q Z), D and R are associated to: ˜D := {a+bi : a, b ∈ Z/q Z, a 2 +b2 = 1 }, and ˜R := {a+bj : a, b ∈ Z/q Z, a 2 +b2 = 1 }.
By Theorem 1.10, there is a polynomial-time algorithm that finds the shortest possible path between the identity and vertices in D or R. Let D1 ⊂ D and R1 ⊂
R be the subset of vertices where their distances from the identity is less than log p |Xp,q | + O(log p log( |Xp,q |)) . By (1.2), in Theorem 1.2,
|R1| ≥ 99% |R| and |D1| ≥ 99% |D|.
Let Y := D1R1D1 ⊂ PSL 2(Z/q Z).
Lemma 3.3. We have
|Y | ≥ 10% |PSL 2(Z/q Z)|.
Proof. Let g ∈ Y. Then,
g = dαra,b dβ
for some a, b, α, β ∈ Z/q Z. We give an upper bound on the number of different ways of expressing g as dαra,b dβ , where ab 6 = 0 . Suppose that dαra,b dβ = dα′ ra′ ,b ′ dβ′ .
Then, it follows that ( α−1α′)2 = ( β−1β′)2 = ±1. This shows that g has only 2 representations as dαra,b dβ . There are only two elements of R with ab = 0, which are R1,0 and R0,i . Since q is a prime, |D| = |R| = q−1
2
, and |PSL 2(Z/q Z)| =
(q−1) q(q+1)
2
. Therefore,
|Y | ≥ 99% 3 (q − 1)( q − 5)( q − 1)
16 .
This concludes our lemma.
Lemma 3.4. Let g ∈ Xp,q . There exists a polynomial-time algorithm in log q that returns a short path of size at most 3 log k−1 |Xp,q | + O(log log( |Xp,q |)) form the identity to g, if g ∈ Y . Otherwise, it returns “Not in Y”. Proof. Let g =
[g1,1 g1,2
g2,1 g2,2
]
. First, we check the solubility of dαra,b dβ = g for some
α, β, a, and b. This is equivalent to the following system of equations:
[ αβa αβ −1bα−1βb α−1β−1a
]
=
[g1,1 g1,2
g2,1 g2,2
]
.
It follows that a2 = g1,1g2,2, b 2 = g1,2g2,1. By the quadratic reciprocity law, we check in polynomial-time algorithm if g1,2g2,1 and g1,1g2,2 are quadratic residue mod q. If either g1,2g2,1 or g1,1g2,2 are quadratic non-residue, the algorithm returns “Not in Y”. Otherwise, by the polynomial-time algorithm for taking square roots in finite fields (e.g [AMM77] or [Sha73] ), we find a and b. Similarly, we find α and
β. By Theorem 1.2, we write dα, d β , and ra,b in terms of the LPS generators and check if they are inside D1 and R1 respectively. This concludes our lemma.
COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE 17
We cite the following proposition form [EMV13, Proposition 2.14].
Proposition 3.5 (Due to Ellenberg, Michel, Venkatesh) . Fix ǫ > 0. For any subset
Y ⊂ Xp,q with |Y | > 10% |Xp,q |, the fraction of non-backtracking paths γ of length
2l satisfying: ∣∣∣ |γ ∩ Y |
2l + 1 − |Y |
|G|
∣∣∣ ≥ ǫ
is bounded by c1exp (−c2l), where c1, c 2 depend only on ǫ.
Proof of (1.5) in Theorem 1.5. It suffices to navigate from the identity to a given vertex v ∈ Xp,q . Recall that Sp is the LPS generator set defined in (1.6). Let W be the set of all words of length at most log log q with letters in Sp. Note that
|W | = O((log q)c), where c = log( p) which only depends on the fixed prime p. By Lemma 3.4, if wv ∈ Y for some w ∈ W , then we find a path that satisfies (1.5). By Proposition 3.5 for ǫ = 9%, it follows that the fraction of the vertices v, such that
wv / ∈ Y for every w ∈ W , is less than c1 exp( −c2 log log q) = O(log( q)−c1 ). This concludes our theorem.
Numerical results
4.1. Diagonal approximation with V-gates. In this section, we give some nu-merical results on the graph distance between diagonal vertices in X5,q (V -gates), which shows that the inequalities (1.2) and (1.3) are sharp. In particular, we nu-merically check that the diameter of X5,q is bigger that (4 /3) log 5 |X5,q | + O(1) .
Let q be a prime number and q ≡ 1, 9 mod 20. The LPS generators associated to p = 5 are called V -gates. V -gates are the following 6 unitary matrices:
V ±
X
:= 1
√5
[ 1 2i
2i 1
]±
, V ±
Y
:= 1
√5
[ 1 2
−2 1
]±
and V ±
Z
:= 1
√5
[1 + 2 i 00 1 − 2i
]±
.
Since q ≡ 1, 9 mod 20, then square root of −1 and 5 exist mod q and we denote them by √5 and i. So, we can realize these matrices inside P SL 2(Z/q Z). The Cayley graph of P SL 2(Z/q Z) with respect to V-gates is a 6-regular LPS Ramanujan graph. We run our algorithm to find the shortest path in V-gates from identity to a given typical diagonal matrix
[a + bi 00 a − bi
]
∈ P SL 2(Z/q Z). By Theorem 1.7, a path of length m from identity to this diagonal element is associated to the integral solution of the following diophantine equation
x2 + y2 + z2 + w2 = 5 m
x ≡ √5m
a mod q, y ≡ √5m
b mod q, z ≡ w ≡ 0 mod q, x ≡ 1 and y ≡ z ≡ w ≡ 0 mod 2 .
(4.1) First, our algorithm in Theorem 1.10 finds an integral solution ( x, y, z, w ) with the least integer m to the equation (4.1). Next from the integral solution ( x, y, z, w )it constructs a path in the Ramanujan graph by factoring x + iy + jz + kw into
V -gates. We give an explicit example. Let q be the following prime number with 100 digits: 18 NASER T. SARDARI
q = 6513516734600035718300327211250928237178281758494 417357560086828416863929270451437126021949850746381 .
For the diagonal matrix let
a = 23147807431234971203978401278304192730471291281 and
b = 1284712970142165365412342134123412341234121234342141234133 The first run of our algorithm returns the following integral lift
x + iy + jz + kw
where
x = −3513686405828860927763754940484616687735954403564689113985383253868329887 073895129393123529043092607930187858085249975614142765081986624258530038940271
y = 3773156548062114482690557548470637380371201820782668326017207890171886678830 601870144317232489264867168831689578223312772963262687237828114002146000356
z = 696150282464006603091186089706225565057448347974579991940267012475009315401865 6570861892918415809962375271929963309479306543335375368842987498287311268
w = 3888519350877870793211628965104035265911619494928178960777970459693109319153422 770196318754816019921662119578623310979387405367017752713898473225295568 and the associated path in the Ramanujan graph by V -gates is
V yV z −1V xV zV xV xV zV zV x −1V x −1V z −1V x −1V zV zV yV z −1V z −1V z −1V yV z −1
V y −1V xV xV zV xV y −1V xV y −1V xV z −1V yV xV zV zV xV z −1V y −1V xV xV zV zV xV x V z −1V xV xV y −1V x −1V zV yV xV zV yV x −1V y −1V y −1V z −1V y −1V zV x −1V z −1V x −1
V x −1V zV y −1V x −1V zV x −1V x −1V z −1V yV zV zV zV yV z −1V xV yV x −1V z −1V x −1V z −1
V x −1V x −1V zV y −1V x −1V x −1V y −1V z −1V xV z −1V x −1V yV yV yV yV yV x −1V zV x −1V z V yV x −1V x −1V yV z −1V xV xV zV y −1V z −1V yV zV x −1V x −1V y −1V z −1V yV x −1V yV z −1V y V zV zV x −1V x −1V y −1V x −1V z −1V x −1V yV zV yV yV x −1V z −1V z −1V yV yV xV yV yV zV z V yV zV xV zV yV zV xV yV z −1V yV x −1V zV xV z −1V y −1V xV xV y −1V xV yV xV y −1V y −1V y −1
V zV xV y −1V zV x −1V z −1V x −1V x −1V z −1V z −1V y −1V xV y −1V x −1V z −1V x −1V zV xV z −1
V y −1V z −1V yV xV zV x −1V y −1V z −1V x −1V z −1V z −1V yV x −1V y −1V z −1V yV z −1V xV zV x V xV yV x −1V x −1V z −1V xV zV y −1V z −1V z −1V y −1V y −1V y −1V x −1V x −1V y −1V z −1
V yV xV xV xV y −1V xV z −1V y −1V zV zV yV zV yV zV zV xV xV y −1V x −1V yV z −1V y −1V x −1
V z −1V x −1V zV xV y −1V x −1V x −1V yV xV yV xV zV y −1V zV zV yV z −1V yV z −1V x −1V x −1
V yV zV x −1V x −1V yV z −1V x −1V y −1V y −1V x −1V yV z −1V y −1V z −1V xV xV yV zV x −1V y −1
V z −1V xV z −1V y −1V y −1V x −1V y −1V y −1V y −1V zV yV x −1V zV x −1V y −1V y −1V x −1V z V x −1V z −1V z −1V yV yV yV x −1V yV yV yV zV yV x −1V y −1V x −1V y −1V zV zV zV y −1V y −1V z V yV zV y −1V xV xV xV y −1V zV zV zV yV z −1V y −1V y −1V y −1V x −1V z −1V x −1V z −1V xV z −1
V y −1V x −1V zV yV x −1V z −1V y −1V x −1V yV xV xV zV xV z −1V xV zV y −1V zV x −1V yV z −1
V z −1V xV z −1V x −1V z −1V x −1V zV xV z −1V x −1V zV yV zV zV yV xV xV y −1V x −1V z −1V xV y COMPLEXITY OF STRONG APPROXIMATION ON THE SPHERE 19
V z −1V z −1V yV z −1V y −1V x −1V zV y −1V z −1V yV x −1V x −1V y −1V y −1V y −1V xV xV z −1
V x −1V y −1V xV yV xV yV x −1V yV x −1V x −1V z −1V xV zV y −1V x −1V y −1V xV yV z −1V z −1V x
That is a path of size 432. The first candidate that our algorithm gives up to factor has 430 letters. It could be a potential path but this means that the distance is optimal up to two letters. We note that the trivial lower bound for a typical element is 3 log 5(q) = 428 .5.
4.2. Lower bound on the diameter of LPS Ramanujan graphs. Let a =0, b = 1 , which is associated to the matrix
[i 00 −i
]
. By our correspondence in Section 1.4, the lattice point associated to this vertex is in the cups neighborhood for every q. In fact, this lattice point has the highest imaginary part among all the other co-volume q lattice point. Let
q = 65135167346000357183003272112509282371782817584944173575600868284168 63929270451437126021949850746381 The length of the shortest path from the identity to
[i 00 −i
]
is 571. Note that 4 log 5(q) = 571 .20 , and recall that [4 log 5(q)] is conjectured to be the asymptotic of the diameter of this Ramanujan graph. We refer the reader to [Sar18, Section 4] for further discussion and more numerical results.
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L-functions , volume 22 of Tata Inst. Fundam. Res. Stud. Math. , pages 119–185. Tata Inst. Fund. Res., Mumbai, 2013. [HLW06] Shlomo Hoory, Nathan Linial, and Avi Wigderson. Expander graphs and their applica-tions. Bull. Amer. Math. Soc. (N.S.) , 43(4):439–561, 2006. [Len83] H. W. Lenstra, Jr. Integer programming with a fixed number of variables. Math. Oper. Res. , 8(4):538–548, 1983. [LPS88] A. Lubotzky, R. Phillips, and P. Sarnak. Ramanujan graphs. Combinatorica , 8(3):261– 277, 1988. [Mar88] G. A. Margulis. Explicit group-theoretic constructions of combinatorial schemes and their applications in the construction of expanders and concentrators. Problemy Peredachi Informatsii , 24(1):51–60, 1988. [PLQ08] Christophe Petit, Kristin Lauter, and Jean-Jacques Quisquater. Full Cryptanalysis of LPS and Morgenstern Hash Functions , pages 263–277. Springer Berlin Heidelberg, Berlin, Heidelberg, 2008. [PS18] Ori Parzanchevski and Peter Sarnak. Super-golden-gates for P U (2). Adv. Math. , 327:869– 901, 2018. 20 NASER T. SARDARI
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11161 | https://chemistry.stackexchange.com/questions/122862/relation-between-the-oxidation-state-of-central-atom-and-its-coordination-number | Relation between the oxidation state of central atom and its coordination number in a complex compound? - Chemistry Stack Exchange
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Relation between the oxidation state of central atom and its coordination number in a complex compound? [closed]
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We know for a complex say [C o(N H X 3)X 3 C l X 3][C o(N H X 3)X 3 C l X 3] we have C o X 3+C o X 3+ and its coordination number here is 6. And for this if we know either oxidation state or coordination number then other is fixed. So is there a way to figure out relation between them and for other compounds like [N i(C O)X 4][N i(C O)X 4], K X 2[P t C l X 4]K X 2[P t C l X 4], and [P t(N H X 3)X 2 C l X 3][P t(N H X 3)X 2 C l X 3].
coordination-compounds
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edited Oct 23, 2019 at 20:56
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asked Oct 23, 2019 at 18:25
Prashant SinghPrashant Singh
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4 Your premise " if we know either oxidation state or coordination number then other is fixed" is completely not true.Mithoron –Mithoron 2019-10-23 19:52:06 +00:00 Commented Oct 23, 2019 at 19:52
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The hexaaquacobalt(II) complex [C o(H X 2 O)X 6]X 2+[C o(H X 2 O)X 6]X 2+ is octahedral.
The tetrachloridocobaltate(II) complex [C o C l X 4]X 2−[C o C l X 4]X 2− is tetrahedral.
knowing the oxidation state does not allow prediction of the coordination number or vice-versa.
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answered Oct 24, 2019 at 3:35
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How [Co(H2O)6]2+ is possible, considering the outer electronic configuration of cobalt (4s2)(3d7)(4p0) if it is in its +2 oxidation state i.e. (4s0)(3d7)(4p0) it's not possible for it to accommodate 6 ligands and thus to have coordination number of 6 Prashant Singh –Prashant Singh 2019-10-24 03:41:40 +00:00 Commented Oct 24, 2019 at 3:41
1 @PrashantSingh Of course it is possible, it is known, isolated and characterised. d 7 d 7 is a perfectly fine electronic configuration of a transition metal. Indeed, you can find all eleven possible configurations from d 0 d 0 to d 10 d 10.Jan –Jan 2019-10-24 03:43:17 +00:00 Commented Oct 24, 2019 at 3:43
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There is no relation between the oxidation state of the central atom and its coordination number. The complex compound you just mentioned is a good example. It was [Co(NH3)6]Cl3 and it contains Cobalt at the oxidation state +3. In water the corresponding ion will be formed : [Co(NH3)6]^3+. But a similar ion may be formed with Cobalt(II) and the same number of 6 NH3 molecules as ligands. It will yield the ion [Co(NH3)6]^2+ and the salt [Co(NH3)6]Cl2.
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answered Oct 23, 2019 at 19:33
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But considering the outer electronic configuration of cobalt (4s2)(3d7)(4p0) if it is in its +2 oxidation state i.e. (4s0)(3d7)(4p0) it's not possible for it to accommodate 6 ligands and thus to have coordination number of 6.Prashant Singh –Prashant Singh 2019-10-24 03:31:23 +00:00 Commented Oct 24, 2019 at 3:31
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11162 | https://testbook.com/question-answer/the-vector-representing-moment-of-a-couple-on-a-ri--617d5054c8b2b4a605343668 | [Solved] The vector representing moment of a couple on a rigid body i
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Work of a Force and Couple
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The vector representing moment of a couple on a rigid body is a ___.
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Option 1 : Free vectors
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Explanation:
Moment of a Couple:
A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance “d".
The moment of a couple is defined as
M o= F.d (using a scalar analysis) or as
M o= r × F (using a vector analysis).
Here r is any position vector from the line of action of F to the line of action of F
The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals F × d.
Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a free vector. It can be moved anywhere on the body and have the same external effect on the body.
Moments due to couples can be added together using the same rules as adding any vectors.
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11163 | https://lrc.ashworthcollege.edu/wp-content/uploads/2018/03/Geometry-Part-II-Lesson-3-Transformations.pdf | Translations, Rotations, Reflections, and Dilations In geometry, a transformation is a way to change the position of a figure.
In some transformations, the figure retains its size and only its position is changed.
Examples of this type of transformation are: translations, rotations, and reflections In other transformations, such as dilations, the size of the figure will change.
TRANSLATION TRANSLATION A translation is a transformation that slides a figure across a plane or through space.
With translation all points of a figure move the same distance and the same direction.
TRANSLATION Basically, translation means that a figure has moved.
An easy way to remember what translation means is to remember… A TRANSLATION IS A CHANGE IN LOCATION.
A translation is usually specified by a direction and a distance.
TRANSLATION What does a translation look like?
A TRANSLATION IS A CHANGE IN LOCATION.
x y Translate from x to y original image TRANSLATION In the example below triangle A is translated to become triangle B.
A B Describe the translation.
Triangle A is slide directly to the right.
TRANSLATION In the example below arrow A is translated to become arrow B.
Describe the translation.
Arrow A is slide down and to the right.
A B ROTATION ROTATION A rotation is a transformation that turns a figure about (around) a point or a line.
The point a figure turns around is called the center of rotation.
Basically, rotation means to spin a shape.
The center of rotation can be on or outside the shape.
ROTATION What does a rotation look like?
A ROTATION MEANS TO TURN A FIGURE center of rotation ROTATION This is another way rotation looks A ROTATION MEANS TO TURN A FIGURE The triangle was rotated around the point.
center of rotation ROTATION If a shape spins 360, how far does it spin?All the way around This is called one full turn .
360 ROTATION If a shape spins 180, how far does it spin?Half of the way around This is called a½ turn .
180 Rotating a shape 180 turns a shape upside down.
ROTATION If a shape spins 90, how far does it spin?One-quarter of the way around This is called a ¼ turn .
90 ROTATION Describe how the triangle A was transformed to make triangle B A B Describe the translation.
Triangle A was rotated right 90 ROTATION Describe how the arrow A was transformed to make arrow B Describe the translation.
Arrow A was rotated right 180 A B ROTATION When some shapes are rotated they create a special situation called rotational symmetry.
to spin a shape the exact same ROTATIONAL SYMMETRY A shape has rotational symmetry if, after you rotate less than one full turn, it is the same as the original shape.
Here is an example… As this shape is rotated 360, is it ever the same before the shape returns to its original direction?
Yes, when it is rotated 90it is the same as it was in the beginning.
So this shape is said to have rotational symmetry.
90 ROTATIONAL SYMMETRY Here is another example… As this shape is rotated 360, is it ever the same before the shape returns to its original direction?
Yes, when it is rotated 180it is the same as it was in the beginning.
So this shape is said to have rotational symmetry.
180 A shape has rotational symmetry if, after you rotate less than one full turn, it is the same as the original shape.
ROTATIONAL SYMMETRY Here is another example… As this shape is rotated 360, is it ever the same before the shape returns to its original direction?
No, when it is rotated 360it is never the same.
So this shape does NOT have rotational symmetry.
A shape has rotational symmetry if, after you rotate less than one full turn, it is the same as the original shape.
ROTATION SYMMETRY Does this shape have rotational symmetry?
120 Yes, when the shape is rotated 120it is the same. Since 120 is less than 360, this shape HAS rotational symmetry REFLECTIONREFLECTION REFLECTION REFLECTIONA REFLECTION IS FLIPPED OVER A LINE.
A REFLECTION IS FLIPPED OVER A LINE.
A reflection is a transformation that flips a figure across a line.
REFLECTIONA REFLECTION IS FLIPPED OVER A LINE.
A REFLECTION IS FLIPPED OVER A LINE.
After a shape is reflected, it looks like a mirror image of itself.
Remember, it is the same, but it is backwards REFLECTION The line that a shape is flipped over is called a line of reflection.A REFLECTION IS FLIPPED OVER A LINE.
A REFLECTION IS FLIPPED OVER A LINE.
Line of reflection Notice, the shapes are exactly the same distance from the line of reflection on both sides.
The line of reflection can be on the shape or it can be outside the shape.
REFLECTION Determine if each set of figures shows a reflection or a translation.A REFLECTION IS FLIPPED OVER A LINE.
A REFLECTION IS FLIPPED OVER A LINE.
A B C A’ B’ C’ REFLECTION Sometimes, a figure has reflectional symmetry.
This means that it can be folded along a line of reflection within itself so that the two halves of the figure match exactly, point by point.
Basically, if you can fold a shape in half and it matches up exactly, it has reflectional symmetry.
REFLECTIONAL SYMMETRY An easy way to understand reflectional symmetry is to think about folding.
Do you remember folding a piece of paper, drawing half of a heart, and then cutting it out?
What happens when you unfold the piece of paper?
REFLECTIONAL SYMMETRY The two halves make a whole heart.
The two halves are exactly the same… They are symmetrical.
Reflectional Symmetry means that a shape can be folded along a line of reflection so the two haves of the figure match exactly, point by point.
The line of reflection in a figure with reflectional symmetry is called a line of symmetry.
Line of Symmetry REFLECTIONAL SYMMETRY The line created by the fold is the line of symmetry.
A shape can have more than one line of symmetry.
Where is the line of symmetry for this shape?
How can I fold this shape so that it matches exactly?NOT THIS WAYI CAN THIS WAY Line of Symmetry REFLECTIONAL SYMMETRY How many lines of symmetry does each shape have?3
4
5 Do you see a pattern?
REFLECTIONAL SYMMETRY Which of these flags have reflectional symmetry?
United States of America Mexico Canada EnglandNoNo CONCLUSION We just discussed three types of transformations.
See if you can match the action with the appropriate transformation.
FLIP SLIDE TURN REFLECTION TRANSLATION ROTATION Translation, Rotation, and Reflection all change the position of a shape, while the size remains the same.
The fourth transformation that we are going to discuss is called dilation.
Dilation changes the size of the shape without changing the shape.
DILATION When you go to the eye doctor, they dilate you eyes. Let’s try it by turning off the lights.
When you enlarge a photograph or use a copy machine to reduce a map, you are making dilations.
Enlarge means to make a shape bigger.
DILATION Reduce means to make a shape smaller.
The scale factor tells you how much something is enlarged or reduced.
DILATION 200% 50% Notice each time the shape transforms the shape stays the same and only the size changes.
ENLARGE REDUCE Look at the pictures below DILATION Dilate the image with a scale factor of 75% Dilate the image with a scale factor of 150% Look at the pictures below DILATION Dilate the image with a scale factor of 100% Why is a dilation of 75% smaller, a dilation of 150% bigger, and a dilation of 100% the same?
Lets try to make sense of all of this TRANSFORMATIONS CHANGE THE POSTION OF A SHAPE CHANGE THE SIZE OF A SHAPE TRANSLATION ROTATION REFLECTION Change in location Turn around a point Flip over a line DILATION Change size of a shape See if you can identify the transformation that created the new shapes TRANSLATION See if you can identify the transformation that created the new shapes REFLECTION Where is the line of reflection?
See if you can identify the transformation that created the new shapes DILATION See if you can identify the transformation that created the new shapes ROTATION See if you can identify the transformation in these pictures?
REFLECTION See if you can identify the transformation in these pictures?
ROTATION See if you can identify the transformation in these pictures?
TRANSLATION See if you can identify the transformation in these pictures?
DILATION See if you can identify the transformation in these pictures?
REFLECTION |
11164 | https://archive.nptel.ac.in/content/syllabus_pdf/112103262.pdf | IC ENGINES AND GAS TURBINES PROF. PRANAB K. MONDAL Department of Mechanical Engineering IIT Guwahati Prof. Vinayak N. Kulkarni Department of Mechanical Engineering IIT Guwahati PRE-REQUISITES : Basic UG-level Fluid Mechanics and Thermodynamics INTENDED AUDIENCE : Undergraduate students of Mechanical/Chemical/Aerospace engg. (5thsemester onwards) and postgraduate students specializing in the thermofluids/Fluid Mechanics/Aerospace/ Automobiles;industry personnel associated with automobile and aerospace engineering;faculty members associated with Mechanical/Chemical/Aero-spacengg. INDUSTRIES APPLICABLE TO : Tata Motors, Ashok Layland, General Electric COURSE OUTLINE : This course deals with the gas power cycles. One part of the course is on IC engines and it focuses on the thermodynamic cycles for different fuels suitable for automobiles. Other part of the course has emphasis on thermodynamic cycle of aircraft engines and the components of the aircraft engine. Thus this course would provide an understanding on electricity generation or transportation application using gas as working medium.
ABOUT INSTRUCTOR : Prof. Pranab K. Mondal is an Associate Professor in the department of Mechanical Engineering at Indian Institute of Technology Guwahati since May 2015. He received his undergraduate and postgraduate degree from Jadavpur University, Kolkata, and completed his Ph.D. from Indian Institute of Technology Kharagpur in 2015. He worked as a Research Associate at IIT Khargpur for nearly one year before joining IIT Guwahati. He has taught several courses, including Fluid Mechanics, Applied Thermodynamics, Thermodynamics, Fundamentals of Microfluidics, Experimental Methods in Fluid Mechanics to both undergraduate and post graduate students at IIT Guwahati. Among His principal research interest, encompassing the broad area of Microfluidics has covered various facets of microscale multiphase transport, electrokinetics, microscale transport of heat and experimental microfluidics. He is currently working on droplet-based microfluidics, magnetofluidics, experimental investigations of capillary filling of bio-fluids. He has co-authored more than 140 referred journal and conference publications. He is a regular reviewer of many reputed international journals and also associated with several sponsored projects pertaining to microscale phenomena.
Dr. Vinayak N. Kulkarni is an Associate Professor in the Department of Mechanical Engineering ofIndian Institute of Technology Guwahati since January 2015. He completed his undergraduate studies inMechanical Engineering in the Shivaji University, Maharashtra, India. His post graduation and PhD isfrom Aerospace Engineering Department of Indian Institute of Science Bangalore. His teachinginterests are basic and applied thermodynamics, gas dynamics, aircraft propulsion and fluid mechanics.His research interests are experimental and computational compressible flows, IC engines and non-conventional energy.
COURSE PLAN : Week 1: Engine Week 2: IC Engines Week 3: Air-standard cycles Week 4: Carburation Week 5: Fuel injection systems Week 6: Combustion in S.I. and C.I.engines Week 7: Introduction to GasTurbines Week 8: Performance analysis of Bryton Cycle Week 9: Aircraft propulsion Week 10: Compressors Week 11: Compressors and Turbines Week 12: Nozzles and Diffusers |
11165 | https://www.harvard.com/book/9781032017204 | Introduction to Number Theory (Textbooks in Mathematics) (Paperback) | Harvard Book Store
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Introduction to Number Theory covers the essential content of an introductory number theory course including divisibility and prime factorization, congruences, and quadratic reciprocity. The instructor may also choose from a collection of additional topics.
Aligning with the trend toward smaller, essential texts in mathematics, the author strives for clarity of exposition. Proof techniques and proofs are presented slowly and clearly.
The book employs a versatile approach to the use of algebraic ideas. Instructors who wish to put this material into a broader context may do so, though the author introduces these concepts in a non-essential way.
A final chapter discusses algebraic systems (like the Gaussian integers) presuming no previous exposure to abstract algebra. Studying general systems helps students to realize unique factorization into primes is a more subtle idea than may at first appear; students will find this chapter interesting, fun and quite accessible.
Applications of number theory include several sections on cryptography and other applications to further interest instructors and students alike.
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11166 | https://blogs.sas.com/content/iml/2016/05/16/babylonian-square-roots.html | # Blogs
The Babylonian method for finding square roots by hand
17
By Rick Wicklin on The DO Loop
When I was in the sixth grade, I learned an iterative procedure for computing square roots by hand. Yes, I said by hand. Scientific calculators with a square root key were not yet widely available, so I and previous generations of children suffered through learning to calculate square roots by hand.
I still remember being amazed when I first saw the iterative square root algorithm. It was the first time that I thought math is magical. The algorithm required that I make an initial guess for the square root. I then applied a "magic formula" a few times. The magic formula improved my guess and estimated the square root that I sought.
The Babylonian square-root algorithm
The iterative method is called the Babylonian method for finding square roots, or sometimes Hero's method. It was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure.
Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following:
Make an initial guess. Guess any positive number x0.
Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).
Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the digits of xn+1 and xn agree to as many decimal places as you desire.
Let's use this algorithm to compute the square root of S = 20 to at least two decimal places.
An initial guess is x0 = 10.
Apply the formula: x1 = (10 + 20/10)/2 = 6. The number 6 is a better approximation to sqrt(20).
Apply the formula again to obtain x2 = (6 + 20/6)/2 = 4.66667. The next iterations are x3 = 4.47619 and x4 = 4.47214.
Because x3 and x4 agree to two decimal places, the algorithm ends after four iterations. An estimate for sqrt(20) is 4.47214.
The Babylonians and Greeks could estimate square roots by hand. Algorithm and #SAS program. Click To Tweet
How do you choose an initial guess?
You can choose any positive value as an initial guess. However, when I was a kid I used to race my friends to see who could find the square root the fastest. I discovered that if you choose a good guess, then you have to compute only a few iterations. I invented a strategy for finding a good guess, which I call "The Rule of Twos and Sevens."
The Rule of Twos and Sevens chooses an initial guess from among the candidates {2, 7, 20, 70, 200, 700, ...}. The Rule use the fact that a square root has about half as many integer digits as the number itself. The Rule follows and is illustrated by using S = 3249.
Count the integer digits in S. For example, S = 3249 has four digits.
Consider the candidates that have half as many digits as S. (If S has an odd number of digits, round up the number of digits.) For example, 20 and 70 are two-digit candidates. Use those candidates when the number S has three or four integer digits.
Use mental arithmetic to decide which candidate is better. For example, 20 is the square root of 400 whereas 70 is the square root of (about) 5000. Because S is closer to 5000, choose 70 as the starting guess. (Challenge: Convince yourself that you can use 20 for all three-digit integers, 200 for all five-digit integers, and so forth.)
In other words, a good guess starts with a 2 or a 7 and has about half as many digits as are in the whole-number part of S. My experience is that The Rule of Twos and Sevens usually converges to a solution (within two decimal places) in four or fewer iterations.
For small numbers, you can choose the initial guess more precisely. If S is less than 225 (=15x15), you can bracket the square root by using the perfect squares {1, 4, 9, ..., 196, 225} and then use the corresponding integer in the range [1, 15] as an initial guess.
Implement the Babylonian square-root algorithm in SAS
In tribute to all students who ever struggled to perform this algorithm by hand, I implemented the Babylonian square-root algorithm in SAS. You can implement the algorithm directly as a DATA step program, but I chose to use PROC FCMP to define new DATA step functions.
| |
| proc fcmp/ Rule of 2s and 7s: Count the number of digits in S. Choose initial guess to have half as many digits (rounded up) and start with a 2 or 7. /()/ provide initial guess / put(floor() 16)/ convert [S] to string / length(())/ count how many digits / ceil(2)/ about half as many digits (round up) / 210(d-1); 710(d-1); if abs(2 - S) < abs(guess72 - S) then return( guess2 ); else return()/ the Babylonian method (aka, Hero's method) for finding square roots /() 100constant("maceps");/ convergence criterion / if 0 then x/ handle negative numbers / else if 0 then x 0/ handle zero / else do x()/ initial guess / 0 do while(abs(x)) x x() 2/ iterate to improve guess / end end return(x) quit/ Functions defined. Compare Babylonian algorithm to modern SQRT function / options/ define location of functions / data input()/ Babylonian algorithm / Sqrt sqrt()/ modern computation / abs(Sqrt)/ compare values / 3 0 1 2 4 10 16 30 100 3249 125348 proc print label label "Babylonian Sqrt" run |
Notice that the Babylonian method produces essentially the same numerical value as the built-in SQRT function in SAS. Of course, the iterative BabylonianSqrt function is not nearly as efficient as the built-in SQRT function.
For more information about the Babylonian algorithm and other algorithms for computing square roots, see Brown (1999) Coll. Math J.
Do you remember learning a square root algorithm in school? Do you still remember it? Are there other "ancient" algorithms that you have learned that are now unnecessary because of modern technology? Leave a comment.
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Rick WicklinDistinguished Researcher in Computational Statistics
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Rick Wicklin, PhD, is a distinguished researcher in computational statistics at SAS and is a principal developer of SAS/IML software. His areas of expertise include computational statistics, simulation, statistical graphics, and modern methods in statistical data analysis. Rick is author of the books Statistical Programming with SAS/IML Software and Simulating Data with SAS.
17 Comments
Anders Sköllermo on
Hi! Third root solution: X(n+1)= (X(n) + S/X(n)2) 1/2 which gives X3= S.
The main question is: Did the Babylonians know this and did they use it.
Reply
2. Ksharp on
Rick,Why not post a IML function ? Actually in IML documentation there is already such a function. I thought it was writen by yourself wasn't it ?
| |
| proc iml begin IML session / start MySqrt(x); begin module / y = 1; initialize y / do until(w<1e-3); begin DO loop / z = y; set z=y / y = 0.5#(z+x/z); estimate square root / w = abs(y-z); compute change in estimate / end; end DO loop / return(y); return approximation / finish; end module / t = MySqrt({3,4,7,9}); call function MySqrt / s = sqrt({3,4,7,9}); compare with true values / diff = t - s; compute differences / print t s diff; print matrices / |
Reply
Rick Wicklin on
Thanks for the SAS/IML function. I didn't use SAS/IML because this operation does not involve vectors or matrices.
Reply
3. Jamie on
Great post Rick, makes me think how lucky we are to have all this computing power at our fingertips; after all, Computer used to be a job title!
Reply
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7. ryan derose on
I think you would get better guesses in your system by using 6 instead of 7.An even-digit number lies in the range [10^(n-1), 10^n), e.g. when a number has 4 digits, it is at least 1,000 and less than 10,000.The geometric mean of these extremes is 10^((2n - 1)/2),The square root of which is 10^((2n-1)/4) = 10^(n/2) 10^(-1/4)When n is odd, you get a power of 10 times 1.778... , which is close to 2but when n is even, you get a power of 10 times 5.623... , which is much closer to 6 than to 7
Reply
8. Dilleesh.k.jose on
8943 this 4 digits number become 0097. 8289. 0618. 5954.. 5619. 7909 how is calculated(8943+483 ) using babilonian methodPlease answer me
Reply
DJ on
Hi, well this is attempting to answer a nearly 3½ year old question. Your question isn't completely clear to me but it looks like you are trying to find the square root of (8943+483) which would be 9426, using the Babylonian Method, also called Hero's method. The closer your first guess, Xo, the less iterations will be necessary.S=9426. Our first guess is 98. Using the formula (Xn+S/Xn) /2, we have (98+9426/98)/2=97.0918367346939=X1Following the formula, we now have (X1+9426/X1)/2 since X1=97.0918367346939, our answer, X2=97.0875894130642Next iteration (X2+9426/X2)/2 =X3X3 = 97.0875893201598X2 matches X3 to 6 decimal places, so we know X2 is accurate to six decimal places.The actual square root of 9426 is the same as X3. You couldn't be certain until you calculated X4 (or checked the true square root with a calculator, Mathematica, etc.) I only went to 13 digits, which is more than sufficient for any real world applications. The square root of 9426 is irrational, so you can perform as many iterations as you like. If you're trying to find the 5,000,000th decimal of the square root of 9426, I would recommend using Mathematica or your preferred computational program. If you don't have Mathematica or Maple, etc., you can enter Mathematica commands into Wolfram Alpha and they will work. For example, if you are using Wolfram Alpha and enterN[Sqrt9426, 5000000]You requested the square root of 9426 to 5,000,000 digits.If you see this so long after asking, I hope it helps and I hope I understood your question correctly.
Reply
9. seth wudel on
My method for finding the initial guess is simply to find the closest perfect square and then estimating how far your number is from that perfect square. I mean, even from 1 to 10 this covers alot..... 1 4, 9, 16, 25, 36, 49, 64, 81, 100,
It's kind of easy for 4 digits because you just count by 10s, so 100s when squared 100, 400, 900 1600, etc
If you wanted to do a number like 550 or something then you are going somewhere between 400 and 900. (answer between 20 and 30 but closer to 20)
I mean I'm really overexplaining here because with your number , 3249, it doesn't take a lot of effort to see that 3600 is the closest perfect square to that number.
Since 3249 is smaller than 3600, you know your square root is going be smaller than 60. You can also somewhat tell that it is not a whole smaller than 60, and in fact if go down to the previous, of 2500, and try to guage where 3249 sits between 2500 and 3600, you would just apply a similar ratio between 50 and 60.
Half way between 2500 and 3600 would be about 3050 so you know 3249 is just over half way between 50 and 60.
I mean you can probably just look at 3249 and not think too hard to see that 56 is probably a fairly reasonable initial guess, but go ahead and try it and tell me how many iterations you need.
Reply
Rick Wicklin on
Thanks for writing. Your method is known as "linear interpolation."
Reply
10. Denis Jaisson on
Do you know a litterature reference for the formula which generates the continues fraction for (1 + x)^0.5, please? There is a lot to be found in the web, regarding that fraction, but I have not been able to find out who came up with it... The formula that I have in mind is(1 + x)^0.5 = 1 + x / (1 + (1 + x)^0.5) (1)On the basis of that formula, I have used(1 + x)^0.5 # (3x + 4) / (x + 4) (2)after approximating(1 + x)^0.5 # 1 + x/2 (3)in the numerator in Eq.1. In other words, I contented myself with a single iteration of Eq.1.Eq. 2 gives (1 + x)^0.5 with a 1% error or less, for 0
Reply
Rick Wicklin on
No, I don't know the name of this equation. It is a tautology, becausesqrt(1+x) = 1 + x/(1 + sqrt(1+x))I think you can derive it by using the continued fraction algorithm. From this equation, you can recursively replace the sqrt(1+x) term on the right side to go to any depth.You might want to study the continued fraction algorithm by starting with the continued fraction for sqrt(2). See See if you can replace '2' with '1+x' and perform the same steps.
Reply
Zane Clark on
I also found a way to do sqrt, group numbers in pairs of 2 and then find closest, but not bigger than, square and double the number you squared and add 1 digit to it and multiply by the same number and do that until you have found the digit you multiply by and add that is closest, but not bigger than the difference between the square and S and then subtract, and add the “digit” to the “number” again and add the “digit” again and keep doing that until you get to the decimal place you desire.(write down “digits” you get for sqrt).
ADDENDUM: Sorry, I missed out a few things, before you subtract square and product from adding “digit” and multiplying “digit”, bring down the next pair to the end of the square and then subtract, if there are no other pairs, then bring down 2 zeros to square, and if you can substitute the digit with 1 and get a bigger number than the square, then add zeros to the square until it is bigger than the “product” and add as many zeros to sqrt and then subtract square and “product”.
Reply
Matthew Lunsford on
When I was a sophomore in high school, a 4-function calculator with memory finally dropped to $20. I bought one and the Babylonian method for square roots is one of the algorithms I learned. In the present-day, I give it to students as a programming exercise after we have covered recursion.
Reply
11. Emil on
I would think that rather than guessing between decimal multiples of 2 and 7, you would extract the even powers of 10 and then look for the square root of numbers between 1 and 100, (or extract all powers of ten, and if its odd (say 2n+1), you calculate the square root of a number between 1 and 10 and then multiply it by tabulated value of sqrt(10) 10^n. Anyways, this was just a thought. One can probably improve the babylonian method in a variety of ways.
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11167 | https://en.wikipedia.org/wiki/Ordered_pair | Jump to content
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Contents
1 Generalities
2 Informal and formal definitions
3 Defining the ordered pair using set theory
3.1 Wiener's definition
3.2 Hausdorff's definition
3.3 Kuratowski's definition
3.3.1 Variants
3.3.2 Proving that definitions satisfy the characteristic property
3.4 Quine–Rosser definition
3.5 Cantor–Frege definition
3.6 Morse definition
4 Category theory
5 See also
6 References
Ordered pair
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From Wikipedia, the free encyclopedia
Pair of mathematical objects
In mathematics, an ordered pair, denoted (a, b), is a pair of objects in which their order is significant. The ordered pair (a, b) is different from the ordered pair (b, a), unless a = b. In contrast, the unordered pair, denoted {a, b}, always equals the unordered pair {b, a}.
Ordered pairs are also called 2-tuples, or sequences (sometimes, lists in a computer science context) of length 2. Ordered pairs of scalars are sometimes called 2-dimensional vectors. (Technically, this is an abuse of terminology since an ordered pair need not be an element of a vector space.) The entries of an ordered pair can be other ordered pairs, enabling the recursive definition of ordered n-tuples (ordered lists of n objects). For example, the ordered triple (a,b,c) can be defined as (a, (b,c)), i.e., as one pair nested in another.
In the ordered pair (a, b), the object a is called the first entry, and the object b the second entry of the pair. Alternatively, the objects are called the first and second components, the first and second coordinates, or the left and right projections of the ordered pair.
Cartesian products and binary relations (and hence functions) are defined in terms of ordered pairs, cf. picture.
Generalities
[edit]
Let and be ordered pairs. Then the characteristic (or defining) property of the ordered pair is:
The set of all ordered pairs whose first entry is in some set A and whose second entry is in some set B is called the Cartesian product of A and B, and written A × B. A binary relation between sets A and B is a subset of A × B.
The (a, b) notation may be used for other purposes, most notably as denoting open intervals on the real number line. In such situations, the context will usually make it clear which meaning is intended. For additional clarification, the ordered pair may be denoted by the variant notation , but this notation also has other uses.
The left and right projection of a pair p is usually denoted by π1(p) and π2(p), or by πℓ(p) and πr(p), respectively. In contexts where arbitrary n-tuples are considered, πni(t) is a common notation for the i-th component of an n-tuple t.
Informal and formal definitions
[edit]
In some introductory mathematics textbooks an informal (or intuitive) definition of ordered pair is given, such as
For any two objects a and b, the ordered pair (a, b) is a notation specifying the two objects a and b, in that order.
This is usually followed by a comparison to a set of two elements; pointing out that in a set a and b must be different, but in an ordered pair they may be equal and that while the order of listing the elements of a set doesn't matter, in an ordered pair changing the order of distinct entries changes the ordered pair.
This "definition" is unsatisfactory because it is only descriptive and is based on an intuitive understanding of order. However, as is sometimes pointed out, no harm will come from relying on this description and almost everyone thinks of ordered pairs in this manner.
A more satisfactory approach is to observe that the characteristic property of ordered pairs given above is all that is required to understand the role of ordered pairs in mathematics. Hence the ordered pair can be taken as a primitive notion, whose associated axiom is the characteristic property. This was the approach taken by the N. Bourbaki group in its Theory of Sets, published in 1954. However, this approach also has its drawbacks as both the existence of ordered pairs and their characteristic property must be axiomatically assumed.
Another way to rigorously deal with ordered pairs is to define them formally in the context of set theory. This can be done in several ways and has the advantage that existence and the characteristic property can be proven from the axioms that define the set theory. One of the most cited versions of this definition is due to Kuratowski (see below) and his definition was used in the second edition of Bourbaki's Theory of Sets, published in 1970. Even those mathematical textbooks that give an informal definition of ordered pairs will often mention the formal definition of Kuratowski in an exercise.
Defining the ordered pair using set theory
[edit]
If one agrees that set theory is an appealing foundation of mathematics, then all mathematical objects must be defined as sets of some sort. Hence if the ordered pair is not taken as primitive, it must be defined as a set. Several set-theoretic definitions of the ordered pair are given below (see also Diepert).
Wiener's definition
[edit]
Norbert Wiener proposed the first set theoretical definition of the ordered pair in 1914: He observed that this definition made it possible to define the types of Principia Mathematica as sets. Principia Mathematica had taken types, and hence relations of all arities, as primitive.
Wiener used {{b}} instead of {b} to make the definition compatible with type theory where all elements in a class must be of the same "type". With b nested within an additional set, its type is equal to 's.
Hausdorff's definition
[edit]
About the same time as Wiener (1914), Felix Hausdorff proposed his definition: "where 1 and 2 are two distinct objects different from a and b."
Kuratowski's definition
[edit]
In 1921 Kazimierz Kuratowski offered the now-accepted definition of the ordered pair (a, b): When the first and the second coordinates are identical, the definition obtains:
Given some ordered pair p, the property "x is the first coordinate of p" can be formulated as: The property "x is the second coordinate of p" can be formulated as: In the case that the left and right coordinates are identical, the right conjunct is trivially true, since is the case.
If then:
This is how we can extract the first coordinate of a pair (using the iterated-operation notation for arbitrary intersection and arbitrary union):
This is how the second coordinate can be extracted:
(if , then the set could be obtained more simply: , but the previous formula also takes into account the case when .)
Note that and are generalized functions, in the sense that their domains and codomains are proper classes.
Variants
[edit]
The above Kuratowski definition of the ordered pair is "adequate" in that it satisfies the characteristic property that an ordered pair must satisfy, namely that . In particular, it adequately expresses 'order', in that is false unless . There are other definitions, of similar or lesser complexity, that are equally adequate:
The reverse definition is merely a trivial variant of the Kuratowski definition, and as such is of no independent interest. The definition short is so-called because it requires two rather than three pairs of braces. Proving that short satisfies the characteristic property requires the Zermelo–Fraenkel set theory axiom of regularity. Moreover, if one uses von Neumann's set-theoretic construction of the natural numbers, then 2 is defined as the set {0, 1} = {0, {0}}, which is indistinguishable from the pair (0, 0)short. Yet another disadvantage of the short pair is the fact that, even if a and b are of the same type, the elements of the short pair are not. (However, if a = b then the short version keeps having cardinality 2, which is something one might expect of any "pair", including any "ordered pair".)
Proving that definitions satisfy the characteristic property
[edit]
Prove: (a, b) = (c, d) if and only if a = c and b = d.
Kuratowski: If. If a = c and b = d, then {{a}, {a, b}} = {{c}, {c, d}}. Thus (a, b)K = (c, d)K.
Only if. Two cases: a = b, and a ≠ b.
If a = b:
: (a, b)K = {{a}, {a, b}} = {{a}, {a, a}} = {{a}}.
: {{c}, {c, d}} = (c, d)K = (a, b)K = {{a}}.
: Thus {c} = {c, d} = {a}, which implies a = c and a = d. By hypothesis, a = b. Hence b = d.
If a ≠ b, then (a, b)K = (c, d)K implies {{a}, {a, b}} = {{c}, {c, d}}.
: Suppose {c, d} = {a}. Then c = d = a, and so {{c}, {c, d}} = {{a}, {a, a}} = {{a}, {a}} = {{a}}. But then {{a}, {a, b}} would also equal {{a}}, so that b = a which contradicts a ≠ b.
: Suppose {c} = {a, b}. Then a = b = c, which also contradicts a ≠ b.
: Therefore {c} = {a}, so that c = a and {c, d} = {a, b}.
: If d = a were true, then {c, d} = {a, a} = {a} ≠ {a, b}, a contradiction. Thus d = b is the case, so that a = c and b = d.
Reverse: (a, b)reverse = {{b}, {a, b}} = {{b}, {b, a}} = (b, a)K.
If. If (a, b)reverse = (c, d)reverse, (b, a)K = (d, c)K. Therefore, b = d and a = c.
Only if. If a = c and b = d, then {{b}, {a, b}} = {{d}, {c, d}}. Thus (a, b)reverse = (c, d)reverse.
Short:
If: If a = c and b = d, then {a, {a, b}} = {c, {c, d}}. Thus (a, b)short = (c, d)short.
Only if: Suppose {a, {a, b}} = {c, {c, d}}. Then a is in the left hand side, and thus in the right hand side. Because equal sets have equal elements, one of a = c or a = {c, d} must be the case.
: If a = {c, d}, then by similar reasoning as above, {a, b} is in the right hand side, so {a, b} = c or {a, b} = {c, d}.
: If {a, b} = c then c is in {c, d} = a and a is in c, and this combination contradicts the axiom of regularity, as {a, c} has no minimal element under the relation "element of."
: If {a, b} = {c, d}, then a is an element of a, from a = {c, d} = {a, b}, again contradicting regularity.
: Hence a = c must hold.
Again, we see that {a, b} = c or {a, b} = {c, d}.
: The option {a, b} = c and a = c implies that c is an element of c, contradicting regularity.
: So we have a = c and {a, b} = {c, d}, and so: {b} = {a, b} \ {a} = {c, d} \ {c} = {d}, so b = d.
Quine–Rosser definition
[edit]
Rosser (1953) employed a definition of the ordered pair due to Quine which requires a prior definition of the natural numbers. Let be the set of natural numbers and define first The function increments its argument if it is a natural number and leaves it as is otherwise; the number 0 does not appear in the range of . As is the set of the elements of not in go on with This is the set image of a set under , sometimes denoted by as well. Applying function to a set x simply increments every natural number in it. In particular, never contains contain the number 0, so that for any sets x and y, Further, define By this, does always contain the number 0.
Finally, define the ordered pair (A, B) as the disjoint union (which is in alternate notation).
Extracting all the elements of the pair that do not contain 0 and undoing yields A. Likewise, B can be recovered from the elements of the pair that do contain 0.
For example, the pair is encoded as provided .
In type theory and in outgrowths thereof such as the axiomatic set theory NF, the Quine–Rosser pair has the same type as its projections and hence is termed a "type-level" ordered pair. Hence this definition has the advantage of enabling a function, defined as a set of ordered pairs, to have a type only 1 higher than the type of its arguments. This definition works only if the set of natural numbers is infinite. This is the case in NF, but not in type theory or in NFU. J. Barkley Rosser showed that the existence of such a type-level ordered pair (or even a "type-raising by 1" ordered pair) implies the axiom of infinity. For an extensive discussion of the ordered pair in the context of Quinian set theories, see Holmes (1998).
Cantor–Frege definition
[edit]
Early in the development of the set theory, before paradoxes were discovered, Cantor followed Frege by defining the ordered pair of two sets as the class of all relations that hold between these sets, assuming that the notion of relation is primitive:
This definition is inadmissible in most modern formalized set theories and is methodologically similar to defining the cardinal of a set as the class of all sets equipotent with the given set.
Morse definition
[edit]
Morse–Kelley set theory makes free use of proper classes. Morse defined the ordered pair so that its projections could be proper classes as well as sets. (The Kuratowski definition does not allow this.) He first defined ordered pairs whose projections are sets in Kuratowski's manner. He then redefined the pair where the component Cartesian products are Kuratowski pairs of sets and where
This renders possible pairs whose projections are proper classes. The Quine–Rosser definition above also admits proper classes as projections. Similarly the triple is defined as a 3-tuple as follows:
The use of the singleton set which has an inserted empty set allows tuples to have the uniqueness property that if a is an n-tuple and b is an m-tuple and a = b then n = m. Ordered triples which are defined as ordered pairs do not have this property with respect to ordered pairs.
Category theory
[edit]
A category-theoretic product A × B in a category of sets represents the set of ordered pairs, with the first element coming from A and the second coming from B. In this context the characteristic property above is a consequence of the universal property of the product and the fact that elements of a set X can be identified with morphisms from 1 (a one element set) to X. While different objects may have the universal property, they are all naturally isomorphic.
See also
[edit]
Cartesian product
Abscissa and ordinate
Tarski–Grothendieck set theory
Trybulec, Andrzej, 1989, "Tarski–Grothendieck Set Theory", Journal of Formalized Mathematics (definition Def5 of "ordered pairs" as { { x,y }, { x } })
References
[edit]
^ Lay, Steven R. (2005), Analysis / With an Introduction to Proof (4th ed.), Pearson / Prentice Hall, p. 50, ISBN 978-0-13-148101-5
^ Devlin, Keith (2004), Sets, Functions and Logic / An Introduction to Abstract Mathematics (3rd ed.), Chapman & Hall / CRC, p. 79, ISBN 978-1-58488-449-1
^ a b Wolf, Robert S. (1998), Proof, Logic, and Conjecture / The Mathematician's Toolbox, W. H. Freeman and Co., p. 164, ISBN 978-0-7167-3050-7
^ Fletcher, Peter; Patty, C. Wayne (1988), Foundations of Higher Mathematics, PWS-Kent, p. 80, ISBN 0-87150-164-3
^ Quine has argued that the set-theoretical implementations of the concept of the ordered pair is a paradigm for the clarification of philosophical ideas (see "Word and Object", section 53). The general notion of such definitions or implementations are discussed in Thomas Forster "Reasoning about theoretical entities".
^ Randall R. Dipert (Jun 1982), "Set-Theoretical Representations of Ordered Pairs and Their Adequacy for the Logic of Relations", Canadian Journal of Philosophy, 12 (2): 353–374, doi:10.1080/00455091.1982.10715803, JSTOR 40231262
^ Wiener's paper "A Simplification of the logic of relations" is reprinted, together with a valuable commentary on pages 224ff in van Heijenoort, Jean (1967), From Frege to Gödel: A Source Book in Mathematical Logic, 1979–1931, Harvard University Press, Cambridge MA, ISBN 0-674-32449-8 (pbk.). van Heijenoort states the simplification this way: "By giving a definition of the ordered pair of two elements in terms of class operations, the note reduced the theory of relations to that of classes".
^ cf introduction to Wiener's paper in van Heijenoort 1967:224
^ cf introduction to Wiener's paper in van Heijenoort 1967:224. van Heijenoort observes that the resulting set that represents the ordered pair "has a type higher by 2 than the elements (when they are of the same type)"; he offers references that show how, under certain circumstances, the type can be reduced to 1 or 0.
^ Kuratowski, Casimir (1921). "Sur la notion de l'ordre dans la Théorie des Ensembles". Fundamenta Mathematicae. 2 (1): 161–171. doi:10.4064/fm-2-1-161-171.
^ This differs from Hausdorff's definition in not requiring the two elements 0 and 1 to be distinct from a and b.
^ Tourlakis, George (2003) Lectures in Logic and Set Theory. Vol. 2: Set Theory. Cambridge Univ. Press. Proposition III.10.1.
^ For a formal Metamath proof of the adequacy of short, see here (opthreg). Also see Tourlakis (2003), Proposition III.10.1.
^ J. Barkley Rosser, 1953. Logic for Mathematicians. McGraw–Hill.
^ Holmes, M. Randall: On Ordered Pairs, on: Boise State, March 29, 2009. The author uses for and for .
^ Holmes, M. Randall (1998) Elementary Set Theory with a Universal Set Archived 2011-04-11 at the Wayback Machine. Academia-Bruylant. The publisher has graciously consented to permit diffusion of this monograph via the web.
^ Frege, Gottlob (1893). "144". Grundgesetze der Arithmetik (PDF). Jena: Verlag Hermann Pohle. Archived from the original (PDF) on 2016-10-21. Retrieved 2017-09-14.
^ Kanamori, Akihiro (2007). Set Theory From Cantor to Cohen (PDF). Elsevier BV. p. 22, footnote 59
^ Morse, Anthony P. (1965). A Theory of Sets. Academic Press.
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11168 | https://www.reddit.com/r/gamedev/comments/cc18wg/rotating_2d_game_pieces_what_kind_of_math/ | Rotating 2d game pieces -- what kind of math? : r/gamedev
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jon_stout
Rotating 2d game pieces -- what kind of math?
Question
Newb game dev here. Let's say I wanted to write a board game where players lay down tiles on a grid. However, the tiles can only (and must) connect to each other at specific points. Meaning that if the players rotate the tiles (which for simplicity's sake let's say they can only do at right angles -- 90 degrees, 180 degrees, 270 degrees, etc.), I need to be able to calculate where exactly the connection points wind up on each piece. So presuming each tile is represented by a 2d array, I want to go from something like this (where c represents a connection point):
[
[ 1, 1, 1, 1 ],
[ c, 1, 1, c ],
[ 1, 1, 1, 1 ]
]
to this:
[
[ 1, c, 1 ],
[ 1, 1, 1 ],
[ 1, 1, 1 ],
[ 1, c, 1 ]
]
Sure, I could just hardcode those states in, the way most Tetris tutorials out there seem to. But what if I wanted to write a generic rotate function that could take the first example and spit out the second? What kind of math would I need to research in order to pull that off? Transformation matrices? Just basic trigonometry? Anyone have suggestions on good libraries (especially in JavaScript) that might do the work for me?
Thanks in advance.
Edit: Thanks again for the help, everyone. If anyone's interested, I put up a free npm library that contains the stuff I learned here. I'm still amazed that one didn't seem to already exist.
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2004 AMC 12A Problems/Problem 22
The following problem is from both the 2004 AMC 12A #22 and 2004 AMC 10A #25, so both problems redirect to this page.
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Video Solution
5 See also
Problem
Three mutually tangent spheres of radius rest on a horizontal plane. A sphere of radius rests on them. What is the distance from the plane to the top of the larger sphere?
Solution 1
The height from the center of the bottom sphere to the plane is , and from the center of the top sphere to the tip is .
We now need the verticalheight of the centers. If we connect centers, we get a rectangularpyramid with an equilateral trianglebase. The distance from the vertex of the equilateral triangle to its centroid can be found by s to be .
By the Pythagorean Theorem, we have . Adding the heights up, we get , or
Solution 2
Connect the centers of the spheres. Note that the resulting prism is a tetrahedron with base lengths of 2 and side lengths of 3. Drop a height from the top of the tetrahedron to the centroid of its equilateral triangle base. Using the Pythagorean Theorem, it is easy to see that the circumradius of the base is . We can use PT again to find the height of the tetrahedron given its base's circumradius and it's leg lengths. Finally, we add the distance from the top of the tetrahedron to the top of the sphere of radius 2 and the distance from the bottom of the ground to get an answer of .
Video Solution
~IceMatrix
See also
2004 AMC 12A (Problems • Answer Key • Resources)
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11170 | https://egrove.olemiss.edu/context/aicpa_exam/article/1141/viewcontent/305183.pdf | University of Mississippi eGrove Examinations and Study American Institute of Certified Public Accountants (AICPA) Historical Collection 1980 Uniform CPA examination questions May 1978 to November 1979 American Institute of Certified Public Accountants. Board of Examiners Follow this and additional works at: Part of the Accounting Commons, and the Taxation Commons This Book is brought to you for free and open access by the American Institute of Certified Public Accountants (AICPA) Historical Collection at eGrove. It has been accepted for inclusion in Examinations and Study by an authorized administrator of eGrove. For more information, please contact egrove@olemiss.edu.
Recommended Citation American Institute of Certified Public Accountants. Board of Examiners, "Uniform CPA examination questions May 1978 to November 1979" (1980). Examinations and Study. 142.
Uniform CPA Examination May 1978 to November 1979 Questions American Institute of Certified Public Accountants Uniform CPA Examination May 1978 to November 1979 Questions Published by the American Institute of Certified Public Accountants 1211 Avenue of the Americas New York, N.Y 10036 Copyright © 1980 American Institute of Certified Public Accountants, Inc. 1211 Avenue of the Americas, New York, N.Y. 10036 1234567890 Ex 89876543210 Foreword This volume, the nineteenth of its kind, includes the texts of the Uniform Certified Public Accountant Examinations from May 1978 to November 1979 prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia. The eighteen prior volumes contain all the examinations since 1917.
While the American Institute of Certified Public Accountants does not publish official answers, it is felt that students and CPA candidates may be benefited by a review of the unofficial answers published twice a year as a supplement to the Journal of Accountancy. Consequently, we are publishing simultaneously with this volume a companion book entitled Unofficial Answers to the Uniform Certified Public. Accountant Examinations, May 1978 to November 1979, which contains unofficial answers to the questions in this volume.
A special note of thanks is extended to John G. Pate, Jr., University of Texas at El Paso, for the comprehensive index included in this volume. A careful reading of this index may benefit candidates in their review when preparing for future examinations.
William C. Bruschi, Vice President-Review and Regulation American Institute of Certified Public Accountants April 1980 Contents Examination Questions Page May 1978 ................................................................... 1 November 1978 ......................................................... 53 Page May 1979 .................................................................. 105 November 1979 .........................................................155 Accounting Practice—Part I May 1978 ................................................................... 1 November 1978 ......................................................... 53 May 1979 ...................................................................105 November 1979 ......................................................... 155 Accounting Practice—Part II May 1978 ................................................................... 12 November 1978 ......................................................... 64 May 1979 ...................................................................115 November 1979 .........................................................166 Auditing May 1978 ................................................................... 22 November 1978 ......................................................... 74 May 1979 .................................................................. 124 November 1979 .........................................................177 Business Law (Commercial Law) May 1978 ................................................................. 31 November 1978 ......................................................... 85 May 1979 .................................................................. 134 November 1979 .........................................................186 Accounting Theory (Theory of Accounts) May 1978 .................................................................. 44 November 1978 ......................................................... 97 May 1979 .................................................................. 146 November 1979 ......................................................... 201 Index ............................................................................................................................................................................ 209 Examination Questions May 1978 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) EXAMINATION IN ACCOUNTING PRACTICE PART I May 3,1978; 1:30 to 6:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: Estimated Minutes All questions are required: No. 1 ..............................
No. 2 ..............................
No. 3 ..............................
No. 4 ..............................
No. 5 ..............................
Total.......................
Minimum Maximum 40 50 40 50 50 60 40 50 50 60 220 270 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 2.
You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. Enclose all scratch sheets. Failure to enclose scratch sheets may result in loss of grading points. Scratch sheets need not have page numbers, but you should show the question number and place them immediately following the question to which they relate.
Fourteen-column sheets, if any, should not be folded until all sheets, both wide and narrow, are placed in the proper sequence and fastened togeth er at the top left comer. All fourteen-column sheets should then be wrapped around the back of the papers.
A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
1 1.
4.
5.
Examination Questions—May 1978 Number 1 (Estimated tim e------40 to 50 minutes) Instructions Select the best answer for each of the following items relating to a variety of financial-accounting problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 97. Gross billings for merchandise sold by Baker Company to its customers last year amounted to $5,260,000; sales returns and allowances reduced the amounts owed by $160,000. How much were net sales last year for Baker Company?
a. $4,800,000.
b. $5,100,000.
c. $5,200,000.
d. $5,260,000.
Answer Sheet 97. Items to be Answered 1. Howard Co. incurred research and development costs in 1977 as follows: Materials used in research and develop ment projects Equipment acquired that will have alternate future uses in future re search and development projects Depreciation for 1977 on above equipment Personnel costs of persons involved in research and development projects Consulting fees paid to outsiders for re search and development projects Indirect costs reasonably allocable to re search and development projects $ 400,000 2 ,000,000 500,000 1,000,000 100,000 200,000 $4,200,000 The amount of research and development costs charged to Howard’s 1977 income statement should be a. $1,500,000.
b. $1,700,000.
c. $2,200,000.
d. $3,500,000.
2. For financial statement reporting, the Lexington Corporation recognizes royalty income in the period earned. For income tax reporting, royalties are taxed when collected. At December 31, 1976, unearned royalties of $400,000 were included in Lexington’s balance sheet. All of these royalties had been collected in 1976. During 1977, royalties of $600,000 were collected. Unearned royalties in Lexington’s December 31, 1977, balance sheet amounted to $350,000. Assuming that the income tax rate was 50%, the amount reported in the provision for deferred income taxes in Lexington’s income statement for the year ended December 31, 1977, should be a a. $25,000 debit.
b. $175,000 credit.
c. $200,000 debit.
d. $300,000 credit.
3. The following information for 1977 has been provided by the Edward Company: Proceeds from short-term borrowings $ 600,000 Proceeds from long-term borrowings 2,000,000 Purchases of fixed assets 1,600,000 Purchases of inventories 4,000,000 Proceeds from sale of Edward’s common stock 1,000,000 Assuming funds are defined as working capital, what is the increase in working capital for the year ended December 31, 1977, as a result of the above informa tion?
a. $400,000.
b. $1,000,000.
c. $1,400,000.
d. $2,000,000.
4. On January 1, 1977, the Robohn Company purchased for cash 40% of the 300,000 shares of voting common stock of the Lowell Company for $1,800,000 when 40% of the underlying equity in the net assets of Lowell was $1,400,000. Robohn amortizes goodwill over a forty-year period with a full year’s amortization taken in the year of the purchase. The amortization is not deductible for income tax reporting. As a result of this transaction, Robohn has the ability to exercise significant influence over the operating and financial policies of Lowell. Lowell’s net income for the year ended December 31, 1977, was $600,000. During 1977, Lowell paid $325,000 in dividends to its stockholders. The income reported by Robohn for its investment in Lowell should be a. $120,000.
b. $130,000.
c. $230,000.
d. $240,000.
2 Accounting Practice— Part I 5. On January 1, 1977, Wilson, Inc., issued 100,000 additional shares of $10 par value voting common stock in exchange for all of Thomson Company’s voting common stock in a business combination appropriately accounted for by the pooling of interests method. Net income for the year ended December 31, 1977, was $400,000 for Thomson and $1,300,000 for Wilson, exclusive of any consideration of Thomson. During 1977, Wilson paid $900,000 in dividends to its stockholders and Thomson paid $250,000 in dividends to Wilson. What should be the consolidated net income for the year ended December 31, 1977?
a. $1,150,000.
b. $1,450,000.
c. $1,550,000.
d. $1,700,000.
6. The Park Company is disposing of a segment of its business. At the measurement date the net loss from the disposal is estimated to be $950,000. Included in the $950,000 are severance pay of $100,000 and employee relocation costs of $50,000, both of which are directly associated with the decision to dispose of the segment; and estimated net losses from operations from the measurement date to the expected disposal date of $200,000. Net losses from operations of $150,000 from the beginning of the year to the measurement date are not included in the estimated net loss from the disposal.Park’s income statement should report a loss on discontinued operations (a separate component of income below the caption “Income From Continuing Operations’’) of a. $750,000.
b. $850,000.
c. $900,000.
d. $1,100,000.
7. At December 31, 1976, the Front Company had 400,000 shares of common stock outstanding. On October 1, 1977, an additional 100,000 shares of common stock were issued. In addition, Front had $10,000,000 of 8% convertible bonds outstanding at December 31, 1976, which are convertible into 225,000 shares of common stock. The bonds were considered common stock equivalents at the time of their issuance and no bonds were converted into common stock in 1977. The net income for the year ended December 31, 1977, was $3,500,000. Assuming the income tax rate was 50%, the primary earnings per share for the year ended December 31, 1977, should be a. $6.00.
b. $6.62.
c. $7.00.
d. $8.24.
8. Information from Lon Company’s balance sheet at December 31, 1977, is as follows: Current assets: Cash $ 3,000,000 Marketable securities, at cost which approximates market 7,000,000 Accounts receivable, net of allowance for doubtful accounts 100,000,000 Inventories, lower of cost or market 130,000,000 Prepaid expenses 2,000,000 Total current assets $242,000,000 $ 4,000,000 40,000,000 30,000,000 1,000,000 6,000,000 $ 81,000,000 $180,000,000 Current liabilities: Notes payable Accounts payable Accrued expenses Income taxes payable Payments due within one year on long-term debt Total current liabilities Long-term debt What is the quick (acid-test) ratio?
a. $110,000,000 ÷ $51,000,000.
b. $110,000,000 ÷ $81,000,000.
c. $112,000,000 $51,000,000.
d. $242,000,000 ÷ $261,000,000.
9. Certain balance sheet accounts in a foreign subsidiary of the Brogan Company at December 31, 1977, have been translated into United States dollars as follows: Translated at Marketable equity securities carried at cost Marketable equity securities carried at current market price Inventories carried at cost Inventories carried at net real izable value Current Rates Historical Rates $100,000 $110,000 120,000 125,000 130,000 132,000 80,000 84,000 $430,000 $451,000 What amount should be shown in Brogan’s balance sheet at December 31, 1977, as a result of the above information?
a. $430,000.
b. $436,000.
c. $442,000.
d. $451,000.
3 Examination Questions— May 1978 10. Information concerning common stock is as follows: the Gas Company’s Per Share Book value at December 31, 1977 Quoted market value on New York Stock Exchange on December 31, 1977 Earnings for 1977 Par value Dividend for 1977 $12.00 9.00 3.00 2.00 1.00 What was the price-earnings ratio on common stock for 1977?
a. 2 to 1 .
b. 2.67 to 1.
c. 3 to 1 .
d. 4 to 1.
11. On January 1, 1977, the Pint Corporation paid $400,000 for 10,000 shares of Quart Company’s com mon stock which represents a 10% investment in Quart. Pint received dividends of $1.00 per share from Quart in 1977. Quart reported net income of $150,000 for the year ended December 31, 1977. The market value of Quart’s common stock on December 31, 1977, was $42 per share. Ignoring income taxes, the amount reported in Pint’s 1977 income statement as a result of Pint’s investment in Quart was a. $10,000.
b. $15,000.
c. $30,000.
d. $35,000.
12. Index Co. was formed on January 1, 1977. Selected balances from the historical-dollar balance sheet at December 31, 1977, were as follows: Cash $60,000 Marketable securities, stocks (purchased January 1, 1977) 70,000 Marketable securities, bonds, (purchased January 1 , 1977 and held for price speculation) 80,000 Long-term receivables 90,000 If the general price-level index was 100 at December 31, 1976, and 110 at December 31, 1977, these selected accounts should be shown in a general price-level balance sheet at December 31, 1977, at Cash a. $60,000 b. $60,000 c. $60,000 d. $60,000 Marketable Securities, Stocks Marketable Securities, Bonds $70,000 $70,000 $77,000 $77,000 $80,000 $80,000 $88,000 $88,000 Long-term Receivables $90,000 $99,000 $90,000 $99,000 13. Fair Value, Inc., paid $1,200,000 in December 1976 for certain of its inventory. In December 1977, one half of the inventory was sold for $1,000,000 when the replacement cost of the original inventory was $1,400,000. Ignoring income taxes, what amount should be shown as the total gain resulting from the above facts in a current fair value accounting income statement for 1977?
a. $200,000.
b. $300,000.
c. $400,000.
d. $500,000.
14. The Carson Company’s marketable equity securi ties portfolio which is appropriately included in current assets is as follows: December 31, 1977 Cost Market Unrealized Gain (Loss) Archer, Inc. Kelly Co.
Pelt Company $100,000 200,000 250,000 $550,000 $100,000 150,000 260,000 $510,000 $ — (50,000) 10,000 $(40,000) December 31, 1976 Cost Market Unrealized Gain (Loss) Archer, Inc. Kelly Co.
Pelt Company $100,000 300,000 200,000 $120,000 260,000 240,000 $20,000 (40,000) 40,000 $600,000 $620,000 $20,000 Ignoring income taxes, what amount should be reported as a charge against income in Carson’s 1977 income statement?
a. $0.
b. $10,000.
c. $40,000.
d. $60,000.
15. In 1977, Sarli, Inc., accrued, for financial state ment reporting, estimated expenses for major repairs of $400,000. The repairs were paid for in 1978. For income tax reporting, these expenses are deducted when paid. Also in 1977, Sarli received $100,000 of interest income on municipal obligations. Assuming that the income tax rate was 50%, the amount reported in the provision for deferred income taxes in Sarli’s income statement for the year ended December 31, 1977, should be a a. $150,000 debit.
b. $150,000 credit.
c. $200,000 credit.
d. $250,000 credit.
4 Accounting Practice—Part I 16. Beth Co. leased equipment to Wolf, Inc., on April 1, 1977. The lease is appropriately recorded as a sale for accounting purposes for Beth. The lease is for an eight-year period expiring March 31, 1985. The first equal annual payment of $500,000 was made on April 1, 1977. Beth had purchased the equipment on January 1, 1977, for $2,800,000. The equipment has an estimated useful life of eight years with no residual value expected. Beth uses straight-line depreciation and takes a full year’s depreciation in the year of purchase. The cash selling price of the equipment is $2,934,000. Assuming an interest rate of 10%, what amount of interest income should Beth record in 1977 as a result of the lease?
a. $0.
b. $182,550.
c. $243,400.
d. $280,000.
17. The Greeley Company was formed on January 1, 1975, and used an accelerated method of depreciation on its machinery until January 1, 1977. At that time, Greeley adopted the straight-line method of deprecia tion for the machinery previously acquired as well as for any new machinery acquired in 1977.
Information concerning depreciation under each method is as follows: amounts Depreciation if accel erated method used $300,000 400,000 450,000 Depreciation if straight- line method used $200,000 250,000 280,000 Assume that the direct effects of this change are limited to the effect on depreciation and the related tax provisions, and that the income tax rate was 50% in each of these years. What should be reported in Greeley’s income statement for the year ended December 31, 1977, as the cumulative effect on prior years of changing to a different depreciation method?
a. $0.
b. $125,000.
c. $210,000.
d. $250,000.
18. On November 1, 1977, Warrior Co. purchased Cosmo, Inc., 10-year, 7%, bonds with a face value of $50,000, for $48,000. An additional $1,167 was paid for the accrued interest. Interest is payable semiannually on January 1 and July 1. The bonds mature on July 1, 1984. Warrior uses the straight-line method of amortization. Ignoring income taxes, the amount reported in Warrior’s 1977 income statement as a result of Warrior’s long-term bond investment in Cosmos was a. $533.
b. $560.
c. $583.
d. $633.
Instructions Select the best answer for each of the following items relating to a variety of managerial-accounting and quantitative-methods problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer ail items. Your grade will be based on your total correct answers.
Items to be Answered Items 19 and 20 are based on the following information: Flemming, Inc., is planning to acquire a new machine at a total cost of $36,000. The estimated life of the machine is six years with no salvage value. The straight-line method of depreciation will be used. Flemming estimates that the annual cash flow from operations, before income taxes, from using this machine will be $9,000. Assume that Flemming’s cost of capital is 8% and the income tax rate is 40%. The present value of $1 at 8% for six years is .630. The present value of an annuity of $1 in arrears at 8% for six years is 4.623.
19. What would the payback period be?
a. 4.0 years.
b. 4.6 years.
c. 5.7 years.
d. 6.7 years.
20. What would the net present value be?
a. $59.
b. $5,607.
c. $10,800.
d. $13,140.
Number 2 (Estimated tim e------40 to 50 minutes) 21. Hayes Company manufactures two models, standard and deluxe. Each product must be processed in each of two departments, grinding and finishing. The standard model requires two hours of grinding and three hours of finishing. The deluxe model requires three hours of grinding and four hours of finishing. The contribution margin is $3.50 for the standard model and $5.00 for the deluxe model. Hayes has four grinding machines and five finishing machines which run sixteen hours a day for six days a week. How would the restriction (constraint) for the finishing department be expressed?
a. 3X + 4Y =5.
b. 3X + 4Y ≤ 480.
c. $3.50X + $5.00Y = 5.
d. $3.50 (3)X + $5.00 (4)Y ≤ 480.
Year 1975 1976 1977 5 Examination Questions— May 1978 Items 22 and 23 are based on the following information: Forward, Inc., manufactures products P, Q, and R from a joint process. Additional information is as follows: Product P Q R Total Units pro duced 4,000 2,000 1,000 7,000 Joint cost $36,000 9 ?
$60,000 Sales value at split-off ?
9 $15,000 $100,000 Additional costs if pro cessed fur ther $ 7,000 $ 5,000 $ 3,000 $ 15,000 Sales value if processed further $70,000 $30,000 $20,000 $120,000 22. Assuming that joint costs are allocated using the relative-sales-value-at-split-off approach, what were the joint costs allocated to products Q and R?
a. $12,000 for Q and $12,000 for R.
b. $14,400 for Q and $9,600 for R.
c. $15,000 for Q and $9,000 for R.
d. $16,000 for Q and $8,000 for R.
23. Assuming that joint costs are allocated using the relative-sales-value-at-split-off approach, what was the sales value at split-off for product P?
a.
$58,333.
b.
$59,500.
c.
$60,000.
d.
$63,000.
Items 25 and 26 are based on the following information: Information concerning Department B of the Toby Company is as follows: Units Costs Beginning work in process 5,000 $ 6,300 Units transferred in 35,000 58,000 40,000 $64,300 Units completed 37,000 Ending work in process 3,000 Costs Trans ferred In Total Materials Conversion Costs Beginning work in pro cess $ 2,900 $ — $ 3,400 $ 6,300 Units trans ferred in 17,500 25,500 15,000 58,000 $20,400 $25,500 $18,400 $64,300 Conversion costs were 20% complete as to the begin ning work in process and 40% complete as to the ending work in process. All materials are added at the end of the process. Toby uses the weighted-average method.
25. The cost per equivalent unit for conversion costs, rounded to the nearest penny, is a. $0.44.
b. $0.46.
c. $0.48.
d. $0.50.
26. The portion of the total cost of ending work in process attributable to transferred-in cost is a. $0.
b. $1,500.
c. $1,530.
d. $1,650.
24. Tice Company is a medium-sized manufacturer of lamps. During the year a new line called “Horolin” was made available to Tice’s customers. The break-even point for sales of Horolin is $200,000 with a contribution margin of 40%. Assuming that the profit for the Horolin line during the year amounted to $100,000, total sales during the year would have amounted to 27. Barter Corporation has been buying Product A in lots of 1,200 units which represents a four months’ supply. The cost per unit is $100; the order cost is $200 per order; and the annual inventory carrying cost for one unit is $25. Assume that the units will be required evenly throughout the year. What is the economic order quantity?
a.
$300,000.
a.
144.
b.
$420,000.
b.
240.
c.
$450,000.
c.
600.
d.
$475,000.
d.
1,200.
6 Accounting Practice— Part I 28. Posa Co. is planning to invest $40,000 in a three-year project. Posa’s expected rate of return is 10%. The present value of $1 at 10% for one year is .909, for two years is .826, and for three years is .751. The cash flow, net of income taxes, will be $15,000 for the first year (present value of $13,635) and $18,000 for the second year (present value of $14,868). Assuming the rate of return is exactly 10%, what would the cash flow, net of income taxes, be for the third year?
a. $8,634.
b. $11,000.
c. $11,497.
d. $15,309.
29. The Lantern Corporation has 1,000 obsolete lan terns that are carried in inventory at a manufacturing cost of $20,000. If the lanterns are remachined for $5,000, they could be sold for $9,000. If the lanterns are scrapped, they could be sold for $1,000. What alternative is more desirable and what are the total relevant costs for that alternative?
Remachine and $5,000.
Remachine and $25,000.
Scrap and $20,000.
Neither, as there is an overall loss under either alternative.
a.
b.
c.
d.
Items 30 and 31 are based on the following information: Beth Company’s budgeted fixed factory overhead costs are $50,000 per month plus a variable factory overhead rate of $4 per direct labor hour. The standard direct labor hours allowed for October production were 18,000. An analysis of the factory overhead indicates that, in October, Beth had an unfavorable budget (controllable) variance of $1,000 and a favorable volume variance of $500. Beth uses a two-way analysis of overhead variances.
30. The actual factory overhead incurred in October is a. $121,000.
b. $122,000.
c. $122,500.
d. $123,000.
32. The following information is available from the Tyro Company: Actual factory overhead $15,000 Fixed overhead expenses, actual $ 7,200 Fixed overhead expenses, budgeted $ 7,000 Actual hours 3,500 Standard hours 3,800 Variable overhead rate per direct labor hour $ 2.50 Assuming that Tyro uses a three-way analysis of over head variances, what is the spending variance?
a. $750 favorable.
b. $750 unfavorable.
c. $950 favorable.
d. $1,500 unfavorable.
Items 33 and 34 are based on the following information: JV Company began its operations on January 1, 1977, and produces a single product that sells for $7.00 per unit. Standard capacity is 100,000 units per year.
100,000 units were produced and 80,000 units were sold in 1977.
Manufacturing costs and selling and administra tive expenses were as follows: Fixed Costs Raw materials Direct labor Factory overhead Selling and ad ministrative $150,000 Variable Costs $1.50 per unit produced 1.00 per unit produced .50 per unit produced 80,000 .50 per unit sold There were no variances from the standard variable costs. Any under- or over-applied overhead is written off directly at year end as an adjustment to cost of goods sold.
33. In presenting inventory on the balance sheet at December 31, 1977, the unit cost under absorption costing is a. $2.50.
b. $3.00.
c. $3.50.
d. $4.50.
34. What is the net income in 1977 under direct costing?
a. $50,000.
b. $80,000.
c. $90,000.
d. $120,000.
31. The applied factory overhead in October is a. $121,000.
b. $122,000.
c. $122,500.
d. $123,000.
7 Examination Questions— May 1978 35. Varsity Co. is preparing its cash budget for the month of May. The following information on accounts receivable collections is available from Varsity’s past collection experience: Current month’s sales 12% Prior month’ s sales 75% Sales two months prior to current month 6% Sales three months prior to current month 4% Cash discounts taken 2% Doubtful accounts 1 % Credit sales are as follows: May — estimated April March February $100,000 90,000 80,000 95,000 What are the estimated accounts receivable collections for May?
a. $85,100.
b. $87,100.
c. $88,100.
d. $90,100.
Number 3 (Estimated tim e------50 to 60 minutes) Number 3 consists of two unrelated parts.
Part a. Howard Corporation is a publicly-owned company whose shares are traded on a national stock exchange. At December 31, 1976, Howard had 25,000,000 shares of $10 par value common stock authorized, of which 15,000,000 shares were issued and 14,000,000 shares were outstanding.
The stockholders’ equity accounts at December 31, 1976, had the following balances: Common stock Additional paid-in capital Retained earnings Treasury stock $150,000,000 80,000,000 50,000,000 18,000,000 During 1977, Howard had the following transactions: • On February 1, 1977, a secondary distribution of 2,000,000 shares of $10 par value common stock was completed. The stock was sold to the public at $18 per share, net of offering costs.
• On February 15, 1977, Howard issued at $110 per share, 100,000 shares of $100 par value, 8% cumulative preferred stock with 100,000 detachable warrants. Each warrant contained one right which with $20 could be exchanged for one share of $10 par value common stock. On February 15, 1977, the market price for one stock right was $1.
• On March 1, 1977, Howard reacquired 20,000 shares of its common stock for $18.50 per share. Howard uses the cost method to account for treasury stock.
• On March 15, 1977, when the common stock was trading for $21 per share, a major stockholder donated 10,000 shares which are appropriately re corded as treasury stock.
• On March 31, 1977, Howard declared a semiannual cash dividend on common stock of $0.10 per share, payable on April 30, 1977, to stockholders of record on April 10, 1977. The appropriate state law prohibits cash dividends on treasury stock.
• On April 15, 1977, when the market price of the stock rights was $2 each and the market price of the common stock was $22 per share, 30,000 stock rights were exercised. Howard issued new shares to settle the transaction.
• On April 30, 1977, employees exercised 100,000 options that were granted in 1975 under a noncompensatory stock option plan. When the options were granted, each option had a preemptive right and entitled the employee to purchase one share of common stock for $20 per share. On April 30, 1977, the market price of the common stock was $23 per share. Howard issued new shares to settle the transaction.
• On May 31, 1977, when the market price of the common stock was $20 per share, Howard declared a 5% stock dividend distributable on July 1, 1977, to stockholders of record on June 1, 1977. The appropriate state law prohibits stock dividends on treasury stock.
• On June 30, 1977, Howard sold the 20,000 treasury shares reacquired on March 1, 1977, and an additional 280,000 treasury shares costing $5,600,000 that were on hand at the beginning of the year. The selling price was $25 per share.
• On September 30, 1977, Howard declared a semiannual cash dividend on common stock of $0.10 per share and the yearly dividend on preferred stock, both payable on October 30, 1977, to stockholders of record on October 10, 1977. The appropriate state law prohibits cash dividends on treasury stock.
• On December 31, 1977, the remaining out standing rights expired.
• Net income for 1977 was $25,000,000. Required: Prepare a work sheet to be used to summarize, for each transaction, the changes in Howard’s stock holders’ equity accounts for 1977. The columns on this work sheet should have the following headings: Date of transaction (or beginning date) Common stock —number of shares Common stock—amount Preferred stock—number of shares Preferred stock—amount Common stock warrants—number of rights Common stock warrants—amount Additional paid-in capital Retained earnings Treasury stock—number of shares Treasury stock—amount Show supporting computations in good form.
8 Accounting Practice— Part I Part b. Tomasco, January 1973 and had Inc., began operations in the following reported net income or loss for each of its five years of operations: 1973 $ 150,000 loss 1974 130,000 loss 1975 120,000 loss 1976 250,000 income 1977 1,000,000 income At December 31, 1977, the Tomasco capital accounts were as follows: Common stock, par value $10 per share; authorized 100,000 shares; issued and outstanding 50,000 shares $ 500,000 4% nonparticipating noncumulative pre ferred stock, par value $100 per share; authorized, issued and outstanding 1,000 shares 100,000 8% fully participating cumulative pre ferred stock, par value $100 per share; authorized, issued and outstanding 10,000 shares 1,000,000 Tomasco has never paid a cash or stock dividend. There has been no change in the capital accounts since Tomasco began operations. The appropriate state law permits dividends only from retained earnings.
Required: Prepare a work sheet showing the maximum amount available for cash dividends on December 31, 1977, and how it would be distributable to the holders of the common shares and each of the preferred shares. Show supporting computations in good form.
Number 4 (Estimated tim e----- 40 to 50 minutes) Number 4 consists of two unrelated parts.
Part a. The Triple C Corporation has engaged you to compute its corporate federal taxable income for the calendar year 1977. You have obtained from them the following information; • Gross margin (sales less cost of sales) $400,000 • Operating expenses not enumerated below $180,000 • Net other income not enumerated below $ 2,000 • Common stock of Wick, Inc., which was pur chased on April 15, 1975, for $70,000 was sold on September 1, 1977, for $94,000.
• Common stock of Erie Co. which was pur chased on May 15, 1975, for $40,000 was sold on September 10, 1977, for $35,000.
• Common stock of Lupo Company which was purchased on June 20, 1977, for $28,000 was sold on December 15, 1977, for $31,000.
• Common stock of Vara Incorporated which was purchased on July 5, 1977, for $16,000 was sold on December 23, 1977, for $14,000.
• Dividends of $12,000 were received from domestic corporations in 1977.
• Dividends of $8,000 were received from a wholly-owned domestic subsidiary in 1977. Triple C does not file a consolidated return with this subsidiary.
• In 1976, Triple C contributed $15,000 more than the maximum deduction allowed for federal in come tax purposes to its qualified pension trust. In 1977, Triple C contributed $130,000 which was $5,000 less than the maximum deduction allowed for federal income tax purposes to its qualified pension trust.
• In 1977, Triple C contributed $85,000 to a profit-sharing trust for its employees which was insti tuted in 1977 and represented 8% of each employee’s salary.
• Triple C made charitable contributions of $10,000 to various qualified organizations in 1977.
• The investment tax credit for Triple C for 1977 was $44,000.
Required: Beginning with gross margin of $400,000, prepare a schedule computing Triple C’s 1977 corporate federal taxable income. Any possible alternative treatments should be resolved in a manner that will minimize taxable income. Show supporting computations in good form.
Part b. Pang Corporation has engaged you to compute its corporate federal income tax due for its calendar year 1977 income tax return. According to Pang’s records, its corporate federal taxable income for the calendar year 1977 amounted to $1,000,000 before considering the following additional information: • Estimated tax payments of $250,000 for the calendar year 1977 were made in 1977.
• An investment credit of $100,000 is allowable for the calendar year 1977 for qualified property bought and placed into service in 1977.
• Pang had a net operating loss of $300,000 for the calendar year 1976. $240,000 of the net operating loss was used to offset profits of prior years, and the appropriate refunds were received in 1977.
Required: Assuming a federal income tax rate of 40%, pre pare a schedule computing Pang’s 1977 corporate federal income tax due. Any possible alternative treat ments should be resolved in a manner that will mini mize taxable income. Show supporting computations in good form.
9 Examination Questions— May 1978 Number 5 (Estimated time----- 50 to 60 minutes) Number 5 consists of two unrelated parts.
Part a. At December 31, 1976, certain accounts included in the property, plant and equipment section of the Townsand Company’s balance sheet had the following balances: Land Buildings Leasehold improvements Machinery and equipment $100,000 800,000 500,000 700,000 During 1977 the following transactions occurred: • Land site number 621 was acquired for $1,000,000. Additionally, to acquire the land Townsand paid a $60,000 commission to a real estate agent. Costs of $15,000 were incurred to clear the land. During the course of clearing the land, timber and gravel were recovered and sold for $5,000.
• A second tract of land (site number 622) with a building was acquired for $300,000. The closing statement indicated that the land value was $200,000 and the building value was $100,000. Shortly after acquisition, the building was demolished at a cost of $30,000. A new building was constructed for $150,000 plus the following costs: Excavation fees Architectural design fees Building permit fee Imputed interest on funds used during construction $11,000 8,000 1,000 6,000 The building was completed and occupied on September 30, 1977.
• A third tract of land (site number 623) was acquired for $600,000 and was put on the market for resale.
• Extensive work was done to a building occupied by Townsand under a lease agreement that expires on December 31, 1986. The total cost of the work was $125,000, which consisted of the following: Painting of ceilings $ 10,000 estimated useful life is one year Electrical work 35,000 estimated useful life is ten years Construction of ex tension to current working area 80,000 estimated useful life is _ thirty years $125,000 The lessor paid one-half of the costs incurred in con nection with the extension to the current working area.
• During December 1977 costs of $65,000 were incurred to improve leased office space. The related lease will terminate on December 31, 1979, and is not expected to be renewed.
• A group of new machines was purchased under a royalty agreement which provides for payment of royalties based on units of production for the machines. The invoice price of the machines was $75,000, freight costs were $2,000, unloading charges were $1,500, and royalty payments for 1977 were $13,000.
Required: 1. Prepare a detailed analysis of the changes in each of the following balance sheet accounts for 1977: Land Buildings Leasehold improvements Machinery and equipment Disregard the related accumulated depreciation accounts.
2. List the items in the fact situation which were not used to determine the answer to 1. above, and indicate where, or if, these items should be included in Townsand’s financial statements.
10 Accounting Practice— Part I Part b. On January 1, 1975, Barth Company, a small machine-tool manufacturer, acquired for $1,000,000 a piece of new industrial equipment. The new equipment was eligible for the investment tax credit and Barth took full advantage of the credit and accounted for the amount using the flow-through method. The new equipment had a useful life of five years and the salvage value was estimated to be $100,000. Barth estimates that the new equipment can produce 10,000 machine tools in its first year. Production is then estimated to decline by 1,000 units per year over the remaining useful life of the equipment.
The following depreciation methods may be used: • Double-declining-balance • Straight-line • Sum-of-the-years-digits • Units-of-output Required: 1. Which depreciation method would result in the maximization of profits for financial statement report ing for the three-year period ending December 31, 1977? Prepare a schedule showing the amount of accumulated depreciation at December 31, 1977, under the method selected. Show supporting computations in good form. Ignore present value, income tax, and deferred income tax considerations in your answer.
2. Which depreciation method would result in the minimization of profits for income tax reporting for the three-year period ending December 31, 1977? Prepare a schedule showing the amount of accumulated depreciation at December 31, 1977, under the method selected. Show supporting computations in good form. Ignore present value considerations in your answer.
11 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXAMINATION IN ACCOUNTING PRACTICE — PART I May 4, 1978; l:30 to 6:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1 No. 2 No. 3 No. 4 No. 5 40 50 40 50 50 60 40 50 50 60 Total 220 270 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. Enclose all scratch sheets. Failure to enclose scratch sheets may result in loss of grading points. Scratch sheets need not have page numbers, but you should show the question number and place them immediately following the question to which they relate.
4. Fourteen-column sheets, if any, should not be folded until all sheets, both wide and narrow, are placed in the proper sequence and fastened togeth er at the top left corner. All fourteen-column sheets should then be wrapped around the back of the papers.
5. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
12 Accounting Practice— Part II Number 1 (Estimated time----- 40 to 50 minutes) Instructions Select the best answer for each of the following items relating to a variety of financial-accounting problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 97. Gross billings for merchandise sold by Baker Company to its customers last year amounted to $5,260,000; sales returns and allowances reduced the amounts owed by $160,000. How much were net sales last year for Baker Company?
a. $4,800,000.
b. $5,100,000.
c. $5,200,000.
d. $5,260,000.
Answer Sheet 97. Items to be Answered 1. In 1976 Dubious Corporation began selling a new line of products that carry a two-year warranty against defects. Based upon past experience with other products, the estimated warranty costs related to dollar sales are as follows: First year of warranty Second year of warranty 2% 5 % 2. The capital accounts of the partnership of Newton, Sharman, and Jackson on June 1, 1977, are presented below with their respective profit and loss ratios: Newton $139,200 1/2 Sharman 208,800 1/3 Jackson 96,000 1/6 $444,000 On June 1, 1977, Sidney was admitted to the partner ship when he purchased, for $132,000, a proportionate interest from Newton and Sharman in the net assets and profits of the partnership. As a result of this trans action, Sidney acquired a one-fifth interest in the net assets and profits of the firm. Assuming that implied goodwill is not to be recorded, what is the combined gain realized by Newton and Sharman upon the sale of a portion of their interests in the partnership to Sidney?
a. $0.
b. $43,200.
c. $62,400.
d. $82,000.
3. During 1977 Suris Corporation acquired a mineral mine for $900,000 of which $150,000 was ascribed to land value after the mineral has been removed. Geo logical surveys have indicated that 15 million units of the mineral could be extracted. During 1977, 1,200,000 units were extracted and 800,000 units were sold. What is the amount of depletion for 1977?
a. $40,000.
b. $48,000.
c. $60,000.
d. $72,000.
4. The stockholders’ equity of Slumber Company at July 31, 1977, is presented below: Sales and actual warranty expenditures for 1976 and 1977 are presented below: Common stock, par value $20, authorized 400,000 shares; issued and outstanding 150,000 shares Capital in excess of par value Retained earnings $3,000,000 140,000 390,000 $3,530,000 1976 1977 On August 1, 1977, the board of directors of Slumber declared a 4% stock dividend on common stock, to be Sales $500,000 $700,000 distributed on September 15th. The market price of Actual warranty expenditures 10,000 30,000 Slumber’s common stock was $35 on August 1, 1977, and $40 on September 15, 1977. What is the amount What is the estimated warranty liability at the end of of the charge to retained earnings as a result of the 1977?
declaration and distribution of this stock dividend?
a.
$39,000.
a. $0.
b.
$44,000.
b. $120,000.
c.
$49,000.
c. $210,000.
d.
$84,000.
d. $240,000.
13 Examination Questions— May 1978 5. On March 1, 1977, Fence Corporation issued $500,000 of 8% nonconvertible bonds at 103, which are due on February 28, 1997. In addition each $1,000 bond was issued with 30 detachable stock warrants, each of which entitled the bondholder to purchase for $50 one share of Fence common stock, par value $25. On March 1, 1977, the fair market value of Fence’s common stock was $40 per share and the fair market value of the warrants was $4. What amount should Fence record on March 1, 1977, as stock warrants out standing?
a. $15,000.
b. $60,000.
c. $375,000.
d. $750,000.
7. The Hastings Company began operations on January 1, 1976, and uses the FIFO method in costing its raw material inventory. Management is contemplat ing a change to the LIFO method and is interested in determining what effect such a change will have on net income. Accordingly, the following information has been developed; Final Inventory FIFO LIFO Net Income (computed under the FIFO method) 1976 $240,000 200,000 1977 $270,000 210,000 120,000 170,000 6. Brower Corporation owns a manufacturing plant in the country of Oust. On December 31, 1977, the plant had a book value of $5,000,000 and an estimated fair market value of $8,000,000. The government of Oust has clearly indicated that it will expropriate the plant during the coming year and will reimburse Brower for 40% of the plant’s estimated fair market value. What journal entry should Brower make on Decem ber 31, 1977, to record the intended expropriation?
a. Estimated loss on expropriation of foreign plant Allowance for estimated loss on foreign plant b. Estimated loss on expropriation of foreign plant Allowance for estimated loss on foreign plant c. Receivable due from foreign government Investment in foreign plant d. Loss on expropriation of foreign plant Receivable due from foreign government Investment in foreign plant Debit $1,800,000 Credit $1,800,000 3,000,000 3,000,000 3,200,000 1,800,000 3,200,000 3,200,000 Based upon the above information, a change to the LIFO method in 1977 would result in net income for 1977 of a. $110,000.
b. $150,000.
c. $170,000.
d. $230,000.
8. The partnership of Jenson, Smith, and Hart share profits and losses in the ratio of 5:3:2, respec tively. The partners voted to dissolve the partnership when its assets, liabilities, and capital were as follows: Assets Cash Other assets Liabilities and Capital Liabilities Jenson, Capital Smith, Capital Hart, Capital $ 40,000 210,000 $250,000 $ 60,000 48,000 72,000 70,000 $250,000 5,000,000 The partnership will be liquidated over a prolonged period of time. As cash is available it will be distributed to the partners. The first sale of noncash assets having a book value of $120,000 realized $90,000. How much cash should be distributed to each partner after this sale?
a. Jenson $0; Smith $28,800; Hart $41,200.
b. Jenson $0; Smith $30,000; Hart $40,000.
c. Jenson $35,(K)0; Smith $21,000; Hart $14,000.
d. Jenson $45,000; Smith $27,000; Hart $18,000.
14 Accounting Practice— Part II 9. The following accounts were abstracted from the trial balance of Marion Company at December 31, 1977; 12. An analysis of stockholders’ equity of Medina Corporation as of January 1, 1977, is as follows: Credit sales Sales discounts Debit $10,000 Credit $500,000 On January 1, 1977, the account Allowance for Uncollectible Accounts Receivable had a credit balance of $12,000. During 1977, $20,000 of accounts receiv able deemed uncollectible were written off.
Historical experience indicates that 3% of gross sales prove uncollectible. What should be the balance in the account Allowance for Uncollectible Accounts Receivable after the current provision is made?
a. $6,700.
b. $7,000.
c. $14,700.
d. $23,000.
10. On March 1, 1977, Smith and Dale formed a part nership with each contributing the following assets: Smith Dale Cash $30,000 $ 70,000 Machinery and equipment 25,000 75,000 Building — 225,000 Furniture and fixtures 10,000 — The building is subject to a mortgage loan of $80,000, which is to be assumed by the partnership. The part nership agreement provides that Smith and Dale share profits and losses 30% and 70%, respectively. On March 1, 1977, the balance in Dale’s capital account should be a. $290,000.
b. $305,000.
c. $314,000.
d. $370,000.
11. On January 1, 1975, Thorp Corporation acquired machinery at a cost of $300,000. Thorp adopted the double-declining-balance method of depreciation for this machinery and had been recording depreciation over an estimated useful life of ten years, with no residual value. At the beginning of 1977 a decision was made to change to the straight-line method of de preciation for the machinery. The cumulative effect of this accounting change, ignoring income tax consider ations, is a. $0.
b. $48,000.
c. $56,400.
d. $108,000.
Common stock, par value $20; authorized 100,000 shares; issued and outstanding 60,000 shares Capital in excess of par value Retained earnings Total $1,200,000 140,000 760,000 $2,100,000 Medina uses the cost method of accounting for treasury stock and during 1977 entered into the following trans actions: • Acquired 1,000 shares of its stock for $35,000.
• Sold 600 treasury shares at $38 per share.
• Retired the remaining treasury shares.
Assuming no other equity transactions occurred during 1977, what should Medina report at December 31, 1977. as capital in excess of par value?
a. $156,800.
b. $150,800.
c. $140,000.
d. $135,800.
13. On January 1, 1973, Ben Corporation issued $600,000 of 5% ten-year bonds at 103. The bonds are callable at the option of Ben at 104. Ben has recorded amortization of the bond premium on the straight-line method (which was not materially different from the interest method).
On December 31, 1977, when the fair market value of the bonds was 97, Ben repurchased $300,000 of the bonds in the open market at 97. Ben has recorded inter est and amortization for 1977. Ignoring income taxes and assuming that the gain is material, Ben should report this reacquisition as a. A gain of $13,500.
b. An extraordinary gain of $13,500.
c. A gain of $21,000.
d. An extraordinary gain of $21,000.
14. On June 30, 1977, Gulch Corporation sold equip ment to an unaffiliated company for $550,000. The equipment had a book value of $500,000 and a remain ing useful life of 10 years. That same day, Gulch leased back the equipment at $1,500 per month for 5 years with no option to renew the lease or repurchase the equipment. Gulch’s equipment rent expense for this equipment for the year ended December 31, 1977, should be a. $4,000.
b. $5,000.
c. $9,000.
d. $11,000.
15 Examination Questions— May 1978 15. On January 2, 1975, Mogul Company acquired equipment to be used in its manufacturing operations. The equipment has an estimated useful life of 10 years and an estimated salvage value of $5,000. The depreci ation applicable to this equipment was $24,000 for 1977, computed under the sum-of-the-years-digits method. What was the acquisition cost of the equip ment?
a. $165,000. l3. $170,000.
c. $240,000.
d. $245,000.
16. In January 1977 Action Corporation entered into a contract to acquire a new machine for its factory. The machine, which had a cash price of $150,000, was paid for as follows: Down payment $ 15,000 Notes payable in 10 equal monthly installments 120,000 500 shares of Action common stock with an agreed value of $50 per share 25,000 Total $160,000 Prior to the machine’s use, installation costs of $4,000 were incurred. The machine has an estimated useful life of 10 years and an estimated salvage value of $5,000. What should Action record as depreciation expense for 1977 under the straight-line method?
a. $15,900.
b. $15,500.
c. $15,000.
d. $14,900.
17. The inventory account of Benson Company at December 31, 1976, included the following items: Inventory Amount Merchandise out on consignment at sales price (including markup of 40% on selling price) Goods purchased, in transit (shipped f.o.b. shipping point) Goods held on consignment by Benson Goods out on approval (sales price $2,500, cost $2,000) $7,000 6,000 4,000 2,500 Based on the above information, the inventory account at December 31, 1976, should be reduced by a.
$7,300.
b.
$12,500.
c.
$13,500.
d.
$19,500.
Number 2 (Estimated time----- 40 to 50 minutes) Instructions Select the best answer for each of the following items relating to the federal income taxation of individuals. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. The answers should be selected in accordance with the current Internal Revenue Code and Tax Regulations. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
Items to be Answered 18. Stanley Brown, a self-employed individual, owned a truck driven exclusively for business use. The truck had an original cost of $4,000 and had an adjusted basis on December 31, 1976, of $1,800. On January 2, 1977, he traded it in for a new truck costing $5,000 and was given a trade-in allowance of $1,000. The new truck will also be used exclusively for business purposes and will be depreciated with no salvage value. The basis of the new truck is a. $4,000.
b. $4,200.
c. $5,000.
d. $5,800.
19. Harold Shore was the owner of a parcel of vacant land that cost him $50,000 when he acquired it in 1970. On January 1, 1977, when the property was valued at $10,000, he made a bona fide gift of the property to his niece, Matilda River. Ms. River held the property until December 15, 1977, and then sold it for $20,000. What is the amount of the gain or loss that Ms. River should include in her adjusted gross income for 1977?
a. $0.
b. $10,000 gain.
c. $30,000 loss.
d. $40,000 loss.
20. On January 1, 1977, Hubert Toast sold stock with a cost of $4,000 to his sister Melba for $3,500, its fair market value. On July 30, 1977, Melba sold the same stock for $4,100 to a friend, in a bona fide transaction. In 1977 as a result of these transactions a. Neither Hubert nor Melba has a recognized gain or loss.
b. Hubert has a recognized loss of $500.
c. Melba has a recognized gain of $100.
d. Melba has a recognized gain of $600.
16 Accounting Practice— Part II 21. In January 1977 Judy Howard was awarded a postgraduate fellowship grant of $4,800 by a tax- exempt educational organization. Ms. Howard is not a candidate for a degree and was awarded the grant to continue her research. The grant is for the period July 1, 1977, through June 30, 1979, and was paid in full on July 1, 1977. What amount should be included in her gross income for 1977?
a. $0.
b. $1,200.
c. $2,400.
d. $4,800.
22. During 1977 Sam Lobell was involved in a collision while driving his automobile. The automobile, which originally cost $4,000 and was used solely for his personal use, had an appraised value of $2,800 for trade-in purposes just before the accident. After the collision, the car was traded in on a new car, but the trade-in value was only $900. Assuming that Lobell carried no collision insurance, what amount can he deduct as a net casualty loss for 1977?
a. $800.
b. $900.
c. $1,800.
d. $1,900.
23. Mr. and Mrs. Brook, both age 62, filed a joint return for 1977. They provided all the support for their son who is 19, legally blind, and who had no income. Their daughter, age 21 and a full-time student at a university, had $4,200 of income and provided 70% of her own support during 1977. How many exemptions should Mr. and Mrs. Brook have claimed on their 1977 joint income tax return?
a. 5.
b. 4.
c. 3.
d. 2.
25. Mort Gage, a cash basis taxpayer, is the owner of an apartment building containing 10 identical apart ments. Gage resides in one apartment and rents out the remaining units. For 1977 the following information was available: Gross rents Fuel Maintenance and repairs (rental apartments) Advertising for vacant apartments Depreciation of building $21,600 2,500 1,200 300 5,000 What amount should Gage report as net rental income for 1977?
a. $12,600.
b. $13,350.
c. $13,500.
d. $17,600.
26. On June 3, 1977, Leon Wren, an electrician, was injured in an accident during the course of his employment. As a result of injuries sustained, he received the following payments during 1977: Damages for personal injuries Workmen’s compensation Reimbursement from his employer’s accident and health plan for medical expenses paid by Wren $8,000 3,000 1,200 included in Wren’s 1977 gross The amount to be income should be a. $0.
b. $1,200.
c. $3,000.
d. $12,200.
27. The following information is available for Rufus Hirsch and his wife Mavis for 1977: 24. In 1977 David Link had adjusted gross income of $30,000. During the year he paid the following medical care expenses: Medicines and drugs Doctors Medical care insurance Crutches for broken leg $ 200 1,100 400 100 What amount can Link deduct as medical expenses in calculating excess itemized deductions for 1977?
a. $1,700.
b. $800.
c. $700.
d. $600.
Rufus Mavis $20,000 $5,000 600 3,000 1,000 100 Adjusted gross income Cash contributions to recog nized charities Cash contributions to political parties What is the maximum amount of the above that they can deduct in calculating excess itemized deductions for 1977?
a. $3,000.
b. $3,300.
c. $3,700.
d. $4,600.
17 Examination Questions— May 1978 28. Ruth Fleming, a widow whose husband died in 1973, maintains her home in which she and her two sons reside. Only one of her sons qualifies as her dependent. What is her filing status for 1977?
a. Single.
b. Surviving spouse.
c. Head of household.
d. Married, filing jointly.
29. On November 1, 1974, Betty Fraser loaned $3,000 to her friend, Irene Cox. The loan was of a non-business nature and was to be repaid on October 30, 1977. During 1977 Ms. Cox filed for and was declared a bankrupt, and the loan was declared totally uncollect ible and worthless. For 1977 Ms. Fraser can report the nonpayment of the loan as a (an) a. Itemized deduction.
b. Nondeductible item.
c. Short-term capital loss.
d. Long-term capital loss.
30. For the year 1977 Morgan Day had $25,000 of ordinary income. In addition he had the following capital transactions: $800 net long-term capital loss $1,500 net short-term capital loss What is the amount of Day’s capital loss deduction for 1977?
a. $1,150.
b. $1,550.
c. $1,900.
d. $2,000.
31. James Robert, a cash basis taxpayer, is a self- employed accountant. During 1977 he established a qualif ied defined-contribution retirement plan of which he will be the only beneficiary. In examining his records for 1977, the following information is available: 32. In examining the records of Tom Ackerson for 1977, a cash-basis sole proprietor, the following in formation was available: Earned income from self-employment Interest income Dividend income Net long-term capital gains Adjusted gross income $40,000 6,000 4,000 10,000 $60,000 What is the maximum amount that Robert can deduct as a contribution to his qualified retirement plan for 1977?
a. $1,500.
b. $6,000.
c. $7,500.
d. $9,000.
Gross receipts Dividend income (on personal investments) Cost of sales Other operating expenses State business taxes paid Federal self-employment tax paid $30,000 200 15,000 3,000 300 800 What amount should Ackerson report as net earnings from self-employment for 1977?
a. $10,900.
b. $11,700.
c. $11,900.
d. $12,000.
33. During 1977 Anthony and Cleo Patra received the following dividends on their jointly held investments: • Dividends of $600 from Hall Corporation, a tax able domestic corporation.
• A liquidating dividend of $500 from Tell Cor poration.
• A dividend of $1,400 from Roe Corporation, a domestic corporation whose earnings con sisted entirely of interest on municipal bonds.
Assuming that the Patras file a joint return for 1977, what amount should they report as dividend income after the allowable exclusion?
a. $900.
b. $1,800.
c. $2,300.
d. $2,500.
34. During 1977 Milton Hanover was granted a divorce from his wife. The divorce decree stipulated that he was to pay both alimony and child support for a specified period of time. The alimony payments were considered to be periodic payments for income tax purposes. In examining his records for 1977, the following information is available: Salary Interest received on bank deposits Interest received on municipal obligations Alimony paid Child support paid $50,000 2,000 1,000 $53,000 $ 3,600 4,800 What is Hanover’s adjusted gross income for 1977?
a.
$44,600.
b.
$48,400.
c.
$49,400.
d.
$52,000.
18 Accounting Practice— Part II Number 3 (Estimated time----- 50 to 60 minutes) The management of Hatfield Corporation, con cerned over a decrease in working capital, has pro vided you with the following comparative analysis of changes in account balances between December 31, 1976, and December 31, 1977; December 31, Increase Debit Balances 1977 1976 (Decrease) Cash Accounts receivable Inventories Securities held for plant expansion purposes Machinery and equipment Leasehold improve ments Patents $ 145,000 $ 186,000 $ (41,000) 253,000 273,000 (20,000) 483,000 538,000 (55,000) 150,000 150,000 927,000 647,000 280,000 87,000 87,000 — 27,800 30,000 (2,200) $2,072,800 $1,761,000 $ 311,800 Credit Balances Allowance for uncollectible accounts receivable Accumulated depreciation of machinery and equipment Allowance for amor tization of lease hold improve ments Accounts payable Cash dividends pay able Current portion of 6% serial bonds payable 6% serial bonds payable Preferred stock Common stock Retained earnings Totals $ 14,000 $ 17,000 $ (3,000) 416,000 372,000 58,000 232,800 49,000 105,000 40,000 — 50,000 50,000 44,000 9,000 127,800 40,000 (50,000) (10,000) ___ 154,000 $2,072,800 $1,761,000 $ 311,800 250,000 90,000 500,000 422,000 300,000 100,000 500,000 268,000 Additional information: During 1977 the following transactions occurred: • New machinery was purchased for $386,000. In addition, certain obsolete machinery, having a book value of $61,000, was sold for $48,000. No other entries were recorded in Machinery and Equipment or related accounts other than provisions for depreciation.
• Hatfield paid $2,000 legal costs in a success ful defense of a new patent. Amortization of patents amounting to $4,200 was recorded.
• Preferred stock, par value $100, was purchas ed at 110 and subsequently cancelled. The premium paid was charged to retained earnings.
• On December 10, 1977, the board of directors declared a cash dividend of $0.20 per share payable to holders of common stock on January 10, 1978.
• A comparative analysis of retained earnings as of December 31, 1977 and 1976, is presented below: December 31, 1977 1976 Balance, January 1 $268,000 $131,000 Net income 195,000 172,000 463,000 303,000 Dividends declared (40,000) (35,000) Premium on preferred stock repurchased (1,000) — $422,000 $268,000 Required: a. Prepare a statement of changes in financial position of Hatfield Corporation for the year ended December 31, 1977, based upon the information pre sented above. The statement should be prepared using a working-capital format.
b. Prepare a schedule of changes in working capital of Hatfield Corporation for the year ended December 31, 1977.
19 Examination Questions— May 1978 You have been engaged to examine the financial statements of the town of Workville for the year ended June 30, 1977. Your examination disclosed that due to the inexperience of the town’s bookkeeper all trans actions were recorded in the General Fund. The follow ing General Fund Trial Balance as of June 30, 1977, was furnished to you.
Town of Workville GENERAL FUND TRIAL BALANCE June 30, 1977 Number 4 (Estimated tim e------40 to 50 minutes) Debit Credit Cash $ 16,800 Short-term investments 40,000 Accounts receivable Taxes receivable-current 11,500 year Tax anticipation notes 30,000 payable $ 50,000 Appropriations 400,000 Expenditures 382,000 Estimated revenue 320,000 Revenues 360,000 General property 85,400 Bonds payable 52,000 Fund balance 127,700 $937,700 $937,700 Your audit disclosed the following additional informa tion; 1. The accounts receivable of $11,500 includes $1,500 due from the town’s water utility for the sale of scrap sold on its behalf. Accounts for the municipal water utility operated by the town are maintained in a separate fund.
2. The balance in Taxes Receivable-Current Year is now considered delinquent, and the town estimates that $24,000 will be uncollectible.
3. On June 30, 1977, the town retired, at face value, 6% General Obligation Serial Bonds totaling $40,000. The bonds were issued on July 1, 1972, at face value of $200,000. Interest paid during the year ended June 30, 1977, was charged to Bonds Payable.
4. In order to service other municipal depart ments, the town at the beginning of the year authorized the establishment of a central supplies warehouse. During the year supplies totaling $128,000 were pur chased and charged to Expenditures. The town chose to conduct a physical inventory of supplies on hand at June 30, 1977, and this physical count disclosed that supplies totaling $84,000 were used.
5. Expenditures for the year ended June 30, 1977, included $11,200 applicable to purchase orders issued in the prior year. Outstanding purchase orders at June 30, 1977, not recorded in the accounts amount ed to $17,500.
6. On June 28, 1977, the State Revenue Depart ment informed the town that its share of a state-collect ed, locally-shared tax would be $34,000.
7. During the year equipment with a book value of $7,900 was removed from service and sold for $4,600. In addition new equipment costing $90,000 was pur chased. The transactions were recorded in General Property.
8. During the year 100 acres of land were do nated to the town for use as an industrial park. The land had a value of $125,000. No recording of this donation has been made.
Required: a. Prepare the formal reclassification, adjust ing, and closing journal entries for the General Fund as of June 30, 1977.
b. Prepare the formal adjusting journal entries for any other funds or groups of accounts as of June 30, 1977.
20 Accounting Practice— Part II Melody Corporation is a manufacturing company that produces a single product known as “Jupiter.” Melody uses the first-in, first-out (FIFO) process cost ing method for both financial statement and internal management reporting.
In analyzing production results, standard costs are used, whereas actual costs are used for financial statement reporting. The standards, which are based upon equivalent units of production, are as follows: Number 5 (Estimated tim e------50 to 60 minutes) Raw material per unit Direct labor per unit Factory overhead per unit 1 pound at $10 per pound 2 hours at $4 per hour 2 hours at $1.25 per hour Budgeted factory overhead for standard hours allowed for April production is $30,000.
Data for the month of April 1977 are presented below; • The beginning inventory consisted of 2,500 units which were 100% complete as to raw material and 40% complete as to direct labor and factory over head.
• An additional 10,000 units were started during the month.
• The ending inventory consisted of 2,000 units which were 100% complete as to raw material and 40% complete as to direct labor and factory overhead.
• Costs applicable to April production are as follows: Actual Cost Standard Cost Raw material used (11,000 pounds) Direct labor (25,000 hours actually worked) Factory overhead $121,000 $100,000 105,575 31,930 82,400 25,750 Required: а. For each element of production for April (raw material, direct labor, and factory overhead) compute the following: 1. Equivalent units of production.
2. Cost per equivalent unit of production at actual and at standard.
Show supporting computations in good form.
b. Prepare a schedule analyzing for April pro duction the following variances as either favorable or unfavorable: 1. Total materials.
2. Materials price.
3. Materials usage.
4. Total labor.
5. Labor rate.
b. Labor efficiency.
7. Total factory overhead.
8. Factory overhead volume.
9. Factory overhead budget.
Show supporting computations in good form.
21 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXAMINATION IN AUDITING May 4, 1978; 8:30 A.M. to 12:00 M NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1 ............................................................................................ 90 No. 2 .............................................................................. 15 No. 3 ............................................................................................ 15 No. 4 .................... 15 No. 5 ............................................................................................
Total........................................................................ ................ 150 105 25 25 25 30 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.
1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
22 Auditing Instructions Select the best answer for each of the following items. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. M ark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 96. One of the generally accepted auditing standards specifies that the auditor a. Inspect all fixed assets acquired during the year.
b. Charge fair fees based on cost.
c. Make a proper study and evaluation of the existing internal control.
d. Count client petty-cash funds.
Answer Sheet 96. Items to be Answered 1. Which of the following material asset accounts would an auditor take exception to in the auditor’s report?
a. Franchise fees paid.
b. Goodwill resulting from revaluation based on an objective appraisal by an expert.
c. Excess cost over the fair value of the assets of a significant subsidiary.
d. Research and development costs that will be billed to a customer at a subsequent date.
2. Auditors often utilize sampling methods when performing tests of compliance. Which of the following sampling methods is most useful when testing for compliance?
a. Attribute sampling.
b. Variable sampling.
c. Unrestricted random sampling with replace ment.
d. Stratified random sampling.
3. A client’s physical count of inventories was lower than the inventory quantities shown in its perpetual records. This situation could be the result of the failure to record.
a. Sales.
b. Sales returns.
c. Purchases.
d. Purchase discounts.
Number 1 (Estimated tim e----- 90 to 105 minutes) 4. An auditor should recognize that the application of auditing procedures may produce evidential matter indicating the possibility of errors or irregularities and therefore should a. Design audit tests to detect unrecorded transactions.
b. Extend the work to audit most recorded trans actions and records of an entity.
c. Plan and perform the engagement with an attitude of professional skepticism.
d. Not depend on internal accounting control features that are designed to prevent or detect errors or irregularities.
5. Which of the following is the most important internal control procedure over acquisitions of prop erty, plant and equipment?
a. Establishing a written company policy distin guishing between capital and revenue expen ditures.
b. Using a budget to forecast and control acqui sitions and retirements.
c. Analyzing monthly variances between au thorized expenditures and actual costs.
d. Requiring acquisitions to be made by user departments.
6. Which of the following is the best audit procedure for the discovery of damaged merchandise in a client’s ending inventory?
a. Compare the physical quantities of slow- moving items with corresponding quantities of the prior year.
b. Observe merchandise and raw materials during the client’s physical inventory taking.
c. Review the management’s inventory repre sentation letter for accuracy.
d. Test overall fairness of inventory values by comparing the company’s turnover ratio with the industry average.
7. The auditor’s program for the examination of long-term debt should include steps that require the a. Verification of the existence of the bond holders.
b. Examination of any bond trust indenture.
c. Inspection of the accounts payable subsidiary ledger.
d. Investigation of credits to the bond interest income account.
8. Preliminary arrangements agreed to by the auditor and the client should be reduced to writing by the auditor. The best place to set forth these arrange ments is in a. A memorandum to be placed in the perma nent section of the auditing working papers.
b. An engagement letter.
c. A client representation letter.
d. A confirmation letter attached to the con structive services letter.
23 Examination Questions— May 1978 9. Which of the following would be the best protec tion for a company that wishes to prevent the “lapping” of trade accounts receivable?
a. Segregate duties so that the bookkeeper in charge of the general ledger has no access to incoming mail.
b. Segregate duties so that no employee has access to both checks from customers and currency from daily cash receipts.
c. Have customers send payments directly to the company’s depository bank.
d. Request that customers’ payment checks be made payable to the company and addressed to the treasurer.
10. The term “special reports” may include all of the following except reports on financial statements a. Of an organization that has limited the scope of the auditor’s examination.
b. Prepared for limited purposes such as a report that relates to only certain aspects of financial statements.
c. Of a not-for-profit organization which follows accounting practices differing in some re spects from those followed by business enter prises organized for profit.
d. Prepared in accordance with a cash basis of accounting.
11. When auditing contingent liabilities, which of the following procedures would be least effective?
a. Abstracting the minutes of the board of directors.
b. Reviewing the bank confirmation letter.
c. Examining customer confirmation replies.
d. Examining invoices for professional services.
12. When verifying debits to the perpetual inventory records of a non-manufacturing company, an auditor would be most interested in examining a sample of purchase a. Approvals.
b. Requisitions.
c. Invoices.
d. Orders.
13. On February 13, 1978, Fox, CPA, met with the audit committee of the Gem Corporation to review the draft of Fox’s report on the company’s financial state ments as of and for the year ended December 31, 1977. On February 16, 1978, Fox completed all remaining field work at the Gem Corporation’s headquarters. On February 17, 1978, Fox typed and signed the final version of the auditor’s report. On February 18, 1978, the final report was mailed to Gem’s audit committee. What date should have been used on Fox’s report?
a. February 13,1978.
b. February 16, 1978.
c. February 17, 1978.
d. February 18, 1978.
14. The AICPA Code of Professional Ethics requires compliance with accounting principles promulgated by the body designated by AICPA Council to establish such principles. The pronouncements comprehended by the code include all of the following except a. Opinions issued by the Accounting Principles Board.
b. AICPA Accounting Research Studies.
c. Interpretations issued by the Financial Ac counting Standards Board.
d. AICPA Accounting Research Bulletins.
15. Which of the following actions should be taken by a CPA who has been asked to examine the financial statements of a company whose fiscal year has ended?
a. Discuss with the client the possibility of an adverse opinion because of the late engage ment date.
b. Ascertain whether circumstances are likely to permit an adequate examination and expres sion of an unqualified opinion.
c. Inform the client of the need to issue a quali fied opinion if the physical inventory has already been taken.
d. Ascertain whether a proper study and evalua tion of internal control can be conducted after completion of the field work.
16. Which of the following is an aspect of scheduling and controlling the audit engagement?
a. Include in the audit program a column for estimated and actual time.
b. Perform audit work only after the client’s books of account have been closed for the period under examination.
c. Write a conclusion on individual working papers indicating how the results thereon will affect the auditor’s report.
d. Include in the engagement letter an esti mate of the minimum and maximum audit fee.
17. One reason why an auditor uses a flowchart is to aid in the a. Evaluation of a series of sequential processes.
b. Study of the system of responsibility account ing.
c. Performance of important, required, dual- purpose tests.
d. Understanding of a client’s organizational structure.
18. The consistency standard does not apply to an accounting change that results from a change in a. An accounting principle that is not generally accepted.
b. An accounting estimate.
c. The reporting entity.
d. An accounting principle inseparable from a change in accounting estimate.
24 Auditing 19. An independent auditor must be without bias with respect to the financial statements of a client in order to a. Comply with the laws established by govern mental agencies.
b. Maintain the appearance of separate interests on the part of the auditor and the client.
c. Protect against criticism and possible litiga tion from stockholders and creditors.
d. Insure the impartiality necessary for an ex pression of the auditor’s opinion.
20. In an electronic data processing system, auto mated equipment controls or hardware controls are designed to a. Arrange data in a logical sequential manner for processing purposes.
b. Correct errors in the computer programs.
c. Monitor and detect errors in source docu ments.
d. Detect and control errors arising from use of equipment.
21. Which of the following is an internal control weakness for a company whose inventory of supplies consists of a large number of individual items?
a. Supplies of relatively little value are ex pensed when purchased.
b. The cycle basis is used for physical counts.
c. The storekeeper is responsible for main tenance of perpetual inventory records.
d. Perpetual inventory records are maintained only for items of significant value.
22. The auditor’s study and evaluation of internal control is done for each of the following reasons except a. To provide a basis for constructive service suggestions.
b. To aid in the determination of the nature, timing and extent of audit tests.
c. To establish a basis for reliance thereon.
d. To provide training and development for staff accountants.
23. Those procedures specifically outlined in an audit program are primarily designed to a. Prevent litigations.
b. Detect errors or irregularities.
c. Test internal systems.
d. Gather evidence.
24. Which of the following is one of the better auditing techniques that might be used by an auditor to detect kiting between inter-company banks?
a. Review composition of authenticated deposit slips.
b. Review subsequent bank statements received directly from the banks.
c. Prepare a schedule of bank transfers.
d. Prepare year-end bank reconciliations.
25. When comparative financial statements are presented, the fourth standard of reporting, which refers to financial statements “taken as a whole,” should be considered to apply to the financial state ments of the a. Periods presented plus the one preceding period.
b. Current period only.
c. Current period and those of the other periods presented.
d. Current and immediately preceding period only.
26. In order to avoid the misappropriation of com pany-owned marketable securities, which of the fol lowing is the best course of action that can be taken by the management of a company with a large portfolio of marketable securities?
a. Require that one trustworthy and bonded employee be responsible for access to the safekeeping area, where securities are kept.
b. Require that employees who enter and leave the safekeeping area sign and record in a log the exact reason for their access.
c. Require that employees involved in the safe keeping function maintain a subsidiary con trol ledger for securities on a current basis.
d. Require that the safekeeping function for securities be assigned to a bank, that will act as a custodial agent.
27. Although the quantity, type, and content of work ing papers will vary with the circumstances, the work ing papers generally would include the a. Copies of those client records examined by the auditor during the course of the engage ment.
b. Evaluation of the efficiency and competence of the audit staff assistants by the partner responsible for the audit.
c. Auditor’s comments concerning the effi ciency and competence of client management personnel.
d. Auditing procedures followed, and the test ing performed in obtaining evidential matter.
28. An independent auditor has the responsibility to plan the audit examination to search for errors and irregularities that might have a material effect on the financial statements. Which of the following, if material, would be an irregularity as defined in Statements on Auditing Standards?
a. Misappropriation of an asset or groups of assets.
b. Clerical mistakes in the accounting data underlying the financial statements.
c. Mistakes in the application of accounting principles.
d. Misinterpretation of facts that existed when the financial statements were prepared.
25 Examination Questions— May 1978 29. A normal audit procedure is to analyze the current year s repairs and maintenance accounts to provide evidence in support of the audit proposition that a. Expenditures for fixed assets have been recorded in the proper period.
b. Capital expenditures have been properly authorized.
c. Noncapitalizable expenditures have been properly expensed.
d. Expenditures for fixed assets have been capitalized.
30. Which of the following best describes the distin guishing feature of statistical sampling?
a. It requires the examination of a smaller number of supporting documents.
b. It provides a means for measuring mathe matically the degree of uncertainty that results from examining only part of a popula tion.
c. It reduces the problems associated with the auditor’s judgment concerning materiality.
d. It is evaluated in terms of two parameters: statistical mean and random selection.
31. With respect to consistency, which of the following should be done by an independent auditor, who has not examined a company’s financial state ments for the preceding year but is doing so in the current year?
a. Report on the financial statements of the current year without referring to consistency.
b. Consider the consistent application of prin ciples within the year under examination but not between the current and preceding year.
c. Adopt procedures, that are practicable and reasonable in the circumstances, to obtain assurance that the principles employed are consistent between the current and pre ceding year.
d. Rely on the report of the prior year’s auditors if such a report does not take exception as to consistency.
32. Which of the following internal control pro cedures will most likely prevent the concealment of a cash shortage resulting from the improper write-off of a trade account receivable?
a. Write-offs must be approved by a responsible officer after review of credit department recommendations and supporting evidence.
b. Write-offs must be supported by an aging schedule showing that only receivables over due several months have been written-off.
c. Write-offs must be approved by the cashier who is in a position to know if the receivables have, in fact, been collected.
d. Write-offs must be authorized by company field sales employees who are in a position to determine the financial standing of the customers.
33. A CPA may reduce the audit work on a first-time audit by reviewing the working papers of the predecessor auditor. The predecessor should permit the successor to review working papers relating to matters of continuing accounting significance such as those that relate to a. Extent of reliance on the work of specialists.
b. Fee arrangements and summaries of pay ments.
c. Analysis of contingencies.
d. Staff hours required to complete the engage ment.
34. Which of the following is not a factor that affects the independent auditor’s judgment as to the quan tity, type, and content of working papers?
a. The timing and the number of personnel to be assigned to the engagement.
b. The nature of the financial statements, sched ules, or other information upon which the auditor is reporting.
c. The need for supervision of the engagement.
d. The nature of the auditor’s report.
35. Addison Corporation is required to but does not wish to prepare and issue a statement of changes in financial position along with its other basic financial statements. In these circumstances the independent auditor’s report on the Addison financial statements should include a. A qualified opinion with a middle paragraph explaining that the company declined to pre sent the required statement.
b. An unqualified opinion with an accurate and complete statement of changes in financial position prepared by the auditor and in cluded in the auditor’s report.
c. An adverse opinion stating that the financial statements, taken as a whole, are not fairly presented because of the omission of the required statement.
d. A disclaimer of opinion with a separate ex planatory paragraph stating why the company declined to present the required statement.
36. When making a limited review of interim financial information the auditor’s work consists primarily of a. Studying and evaluating limited amounts of documentation supporting the interim finan cial information.
b. Scanning and reviewing client-prepared, in ternal financial statements.
c. Making inquiries and performing analytical procedures concerning significant accounting matters.
d. Confirming and verifying significant account balances at the interim date.
26 Auditing 37. If, as a result of a limited review of interim financial information, a CPA concludes that such in formation does not conform with generally accepted accounting principles, the CPA should a. Insist that the management conform the in formation with generally accepted accounting principles and if this is not done, resign from the engagement.
b. Adjust the financial information so that it conforms with generally accepted account ing principles.
c. Prepare a qualified report that makes refer ence to the lack of conformity with generally accepted accounting principles.
d. Advise the board of directors of the respects in which the information does not conform with generally accepted accounting prin ciples.
38. A CPA’s retention of client records as a means of enforcing payment of an overdue audit fee is an action that is a. Considered acceptable by the AICPA Code of Professional Ethics.
b. Ill advised since it would impair the CPA’s independence with respect to the client.
c. Considered discreditable to the profession.
d. A violation of generally accepted auditing standards.
39. Kiting is a technique that might be used to conceal a cash shortage. The auditor can best detect kiting by performing which of the following pro cedures?
a. Examining the details of deposits made to all bank accounts several days subsequent to the balance sheet date.
b. Comparing cash receipts records with the de tails on authenticated bank deposit slips for dates subsequent to the balance sheet date.
c. Examining paid checks returned with bank statements subsequent to the balance sheet date.
d. Comparing year-end balances per the stan dard bank confirmation forms with the like balances on the client’s bank reconciliations.
40. A secondary objective of the auditor’s study and evaluation of internal control is that the study and evaluation provide a. A basis for constructive suggestions concern ing improvements in internal control.
b. A basis for reliance on the system of internal accounting control.
c. An assurance that the records and docu ments have been maintained in accordance with existing company policies and pro cedures.
d. A basis for the determination of the resultant extent of the tests to which auditing proce dures are to be restricted.
41. Effective internal accounting control over the pay roll function should include procedures that segregate the duties of making salary payments to employees and a. Controlling unemployment insurance claims.
b. Maintaining employee personnel records.
c. Approving employee fringe benefits.
d. Hiring new employees.
42. If the auditor believes that required disclosures of a significant nature are omitted from the financial statements under examination, the auditor should decide between issuing a. A qualified opinion or an adverse opinion.
b. A disclaimer of opinion or a qualified opinion.
c. An adverse opinion or a disclaimer of opinion.
d. An unqualified opinion or a qualified opinion.
43. Whenever special reports, filed on a printed form designed by authorities, call upon the indepen dent auditor to make an assertion that the auditor believes is not justified, the auditor should a. Submit a short-form report with explanations.
b. Reword the form or attach a separate report.
c. Submit the form with questionable items clearly omitted.
d. Withdraw from the engagement.
44. If, during an audit examination, the successor auditor becomes aware of information that may indicate that financial statements reported on by the prede cessor auditor may require revision, the successor auditor should a. Ask the client to arrange a meeting among the three parties to discuss the information and attempt to resolve the matter.
b. Notify the client and the predecessor auditor of the matter and ask them to attempt to resolve it.
c. Notify the predecessor auditor who may be required to revise the previously issued financial statements and auditor’s report.
d. Ask the predecessor auditor to arrange a meeting with the client to discuss and resolve the matter.
45. An auditor is planning an audit engagement for a new client in a business that is unfamiliar to the auditor. Which of the following would be the most useful source of information for the auditor during the preliminary planning stage, when the auditor is trying to obtain a general understanding of audit problems that might be encountered?
a. Client manuals of accounts and charts of accounts.
b. AICPA Industry Audit Guides.
c. Prior-year working papers of the predecessor auditor.
d. Latest annual and interim financial state ments issued by the client.
27 Examination Questions— May 1978 46. The primary purpose of tests of compliance is to provide reasonable assurance that a. The accounting and administrative control procedures are adequately designed to assure employee compliance therewith.
b. The accounting and administrative control procedures are being applied as prescribed.
c. The administrative control procedures are being applied as prescribed.
d. The accounting control procedures are being applied as prescribed.
47. Which of the following is not an element of quality control that should be considered by a firm of independent auditors?
a. Assigning personnel to engagements.
b. Consultation with appropriate persons.
c. Keeping records of quality control policies and procedures.
d. Supervision.
48. An investor is reading the financial statements of the Stankey Corporation and observes that the state ments are accompanied by an auditor’s unqualified report. From this the investor may conclude that a. Any disputes over significant accounting issues have been settled to the auditor’s satisfaction.
b. The auditor is satisfied that Stankey is finan cially sound.
c. The auditor has ascertained that Stankey’s financial statements have been prepared accurately.
d. Informative disclosures in the financial state ments but not necessarily in Stankey’s foot notes are to be regarded as reasonably adequate.
49. Audit programs are modified to suit the circum stances on particular engagements. A complete audit program for an engagement generally should be devel oped a. Prior to beginning the actual audit work.
b. After the auditor has completed an evalua tion of the existing internal accounting control.
c. After reviewing the client’s accounting re cords and procedures.
d. When the audit engagement letter is pre pared.
50. To best ascertain that a company has properly included merchandise that it owns in its ending inven tory, the auditor should review and test the a. Terms of the open purchase orders.
b. Purchase cut-off procedures.
c. Contractual commitments made by the purchasing department.
d. Purchase invoices received on or around year end.
51. That segment of an auditor’s internal control work which focuses directly on the purpose of preventing or detecting material errors or irregularities in financial statements is known as a. Compliance with the existing system of in ternal control.
b. Review of the existing system of internal control.
c. Evaluation of the existing system of internal control.
d. Study of the existing system of internal control.
52. Which of the following is an effective audit planning and control procedure that helps prevent misunderstandings and inefficient use of audit person nel?
a. Arrange to make copies, for inclusion in the working papers, of those client supporting documents examined by the auditor.
b. Arrange to provide the client with copies of the audit programs to be used during the audit.
c. Arrange a preliminary conference with the client to discuss audit objectives, fees, timing, and other information.
d. Arrange to have the auditor prepare and post any necessary adjusting or reclassification entries prior to final closing.
53. Under the AICPA Code of Professional Ethics, a CPA is prohibited from performing which of the follow ing actions?
a. Expressing an opinion on interim financial statements.
b. Permitting the CPA’s college alumni maga zine to report that the CPA has opened offices as “ an Accountant.’’ c. Making a direct uninvited solicitation of a a specific potential client.
d. Assisting a client in preparing forecasts of the results of future transactions and events.
54. A CPA is associated with client-prepared financial statements, but is not independent. With respect to the CPA’s lack of independence, which of the following actions by the CPA might confuse a reader of such financial statements?
a. Stamping the word “unaudited” on each page of the financial statements.
b. Disclaiming an opinion and stating that in dependence is lacking.
c. Issuing a qualified auditor’s report explaining the reason for the auditor’s lack of indepen dence.
d. Preparing an auditor’s report that included essential data that was not disclosed in the financial statements.
28 Auditing 55. Which of the following is the most likely first step an auditor would perform at the beginning of an initial audit engagement?
a. Prepare a rough draft of the financial state ments and of the auditor’s report.
b. Study and evaluate the system of internal administrative control.
c. Tour the client’s facilities and review the gen eral records.
d. Consult with and review the work of the pre decessor auditor prior to discussing the en gagement with the client management.
56. When using statistical sampling for tests of compliance an auditor’s evaluation of compliance would include a statistical conclusion concerning whether a. Procedural deviations in the population were within an acceptable range.
b. Monetary precision is in excess of a certain predetermined amount.
c. The population total is not in error by more than a fixed amount.
d. Population characteristics occur at least once in the population.
57. During the course of an audit of a medium-sized manufacturing concern, which of the following areas would be expected to require substantially less audit time than the others?
a. Assets.
b. Liabilities.
c. Revenues.
d. Owners’ equity.
58. The confirmation of the client’s trade accounts receivable is a means of obtaining evidential matter and is specifically considered to be a generally accepted auditing a. Principle.
b. Standard.
c. Procedure.
d. Practice.
59. If an independent auditor’s examination leading to an opinion on financial statements causes the auditor to believe that material errors or irregularities exist the auditor should a. Consider the implications and discuss the matter with appropriate levels of manage ment.
b. Make the investigation necessary to deter mine whether the errors or irregularities have in fact occurred.
c. Request that the management investigate to determine whether the errors or irregu larities have in fact occurred.
d. Consider whether the errors or irregularities were the result of a failure by employees to comply with existing internal control pro cedures.
60. An effective internal accounting control measure that protects against the preparation of improper or inaccurate disbursements would be to require that all checks be a. Signed by an officer after necessary support ing evidence has been examined.
b. Reviewed by the treasurer before mailing.
c. Sequentially numbered and accounted for by internal auditors.
d. Perforated or otherwise effectively cancelled when they are returned with the bank state ment.
Number 2 (Estimated time----- 15 to 25 minutes) Brown, CPA, received a telephone call from Calhoun, the sole owner and manager of a small cor poration. Calhoun asked Brown to prepare the financial statements for the corporation and told Brown that the statements were needed in two weeks for external financing purposes. Calhoun was vague when Brown inquired about the intended use of the statements. Brown was convinced that Calhoun thought Brown’s work would constitute an audit. To avoid confusion Brown decided not to explain to Calhoun that the en gagement would only be to prepare the financial state ments. Brown, with the understanding that a substan tial fee would be paid if the work were completed in two weeks, accepted the engagement and started the work at once.
During the course of the work, Brown discovered an accrued expense account labeled “ professional fees’’ and learned that the balance in the account represented an accrual for the cost of Brown’s services. Brown suggested to Calhoun’s bookkeeper that the account name be changed to “fees for limited audit en gagement.’’ Brown also reviewed several invoices to determine whether accounts were being properly clas sified. Some of the invoices were missing. Brown listed the missing invoice numbers in the working papers with a note indicating that there should be a follow-up on the next engagement. Brown also discovered that the avail able records included the fixed asset values at esti mated current replacement costs. Based on the records available. Brown prepared a balance sheet, income statement and statement of stockholder’s equity. In addition, Brown drafted the footnotes but decided that any mention of the replacement costs would only mislead the readers. Brown suggested to Calhoun that readers of the financial statements would be better informed if they received a separate letter from Calhoun explaining the meaning and effect of the esti mated replacement costs of the fixed assets. Brown mailed the financial statements and footnotes to Calhoun with the following note included on each page: “The accompanying financial statements are submitted to you without complete audit verification.’’ 29 Examination Questions— May 1978 Required: Identify the inappropriate actions of Brown and indicate what Brown should have done to avoid each inappropriate action.
Organize your answer sheet as follows: Inappropriate Action What Brown Should Have Done To Avoid Inappro priate Action Number 3 (Estimated time----- 15 to 25 minutes) Feiler, the sole owner of a small hardware busi ness, has been told that the business should have finan cial statements reported on by an independent CPA. Feiler, having some bookkeeping experience, has per sonally prepared the company’s financial statements and does not understand why such statements should be examined by a CPA. Feiler discussed the matter with Farber, a CPA, and asked Farber to explain why an audit is considered important.
Required: a. Describe the objectives of an independent audit.
b. Identify ten ways in which an independent audit may be beneficial to Feiler.
Number 4 (Estimated time----- 15 to 25 minutes) The following five topics are part of the relevant body of knowledge for CPAs having field work or im mediate supervisory responsibility in audits involving a computer: 1. Electronic data processing (EDP) equipment and its capabilities.
2. Organization and management of the data processing function.
3. Characteristics of computer based systems.
4. Fundamentals of computer programming.
5. Computer center operations.
CPAs who are responsible for computer audits should possess certain general knowledge with respect to each of these five topics. For example, on the subject of EDP equipment and its capabilities, the auditor should have a general understanding of computer equipment and should be familiar with the uses and capabilities of the central processor and the peripheral equipment.
Required: For each of the topics numbered 2 through 5 above, describe the general knowledge that should be possessed by those CPAs who are responsible for computer audits.
Number 5 (Estimated time-----15 to 30 minutes) Antonio, CPA, is conducting the 1977 audit of the financial statements of Johnson Company and has been given the following statement of changes in financial position: Johnson Company STATEMENT OF CHANGES IN FINANCIAL POSITION For the Year Ended December 31, 1977 (Amounts in thousands) Funds Provided: Operations: Net income from operations Add or (deduct) items not requiring outlay of working capital in the current period: Depreciation Deferred income taxes Investment in unconsolidated subsidiary (net of cash dividends paid) From operations prior to extra ordinary loss Extraordinary loss From operations after extra ordinary loss Proceeds from exercise of stock options Total Funds Used: Cash dividends paid on common and preferred stock Increase in long-term investments (net of allowance) Allowance for unrealized losses on non- current marketable equity securities Increase in working capital Total Changes in Working Capital- Increase (Decrease): Cash and short-term investments Accounts receivable Inventories Accounts payable and accrued liabilities Other current assets and liabilities Increase in working capital $46,000 5,500 1,500 (1,700) 51,300 (13,000) 38,300 2,700 $41,000 5,000 18,000 $41,000 $(9,000) 17,000 18,000 (7,000) (1,000) $18,000 Antonio has completed necessary auditing pro cedures on the balance sheet, income statement, and statement of stockholders’ equity and is satisfied that these statements are fairly presented with adequate disclosures in the statements and in the footnotes. Antonio has not yet examined the statement of changes in financial position.
Required: a. What general steps should be followed by Antonio to examine the statement of changes in financial position of the Johnson Company? Do not discuss the presentation or verification of any specific item on the statement.
b. What additional specific steps should be followed by Antonio to verify each of the four items marked with an asterisk () on the statement of changes in financial position of the Johnson Company?
30 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) EXAMINATION IN BUSINESS LAW (Commercial Law) May 5 , 1978; 8:30 A.M. to 12:00 M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: No. 1 ...............................
No. 2 ...............................
No. 3 ...............................
No. 4 ...............................
No. 5 ...............................
Total.......................
Estimated Minutes Minimum Maximum 90 105 25 30 20 25 25 30 15 20 175 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
31 Examination Questions— May 1978 Number 1 (Estimated time----- 90 to 105 minutes) Instructions Select the best answer for each of the following items. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 99. The text of a letter from Bridge Builders, Inc., to Allied Steel Co. follows: We offer to purchase 10,000 tons of No. 4 steel pipe at today’s quoted price for delivery two months from today.
Your acceptance must be received in five days.
What type of contract did Bridge Builders intend to create?
a. Express.
b. Unilateral.
c. Bilateral.
d. Joint.
Answer Sheet 99. Items to be Answered 1. Wilcox works as a welder for Miracle Muffler, Inc. He was specially trained by Miracle in the procedures and safety precautions applicable to installing replace ment mufflers on automobiles. One rule of which he was aware involved a prohibition against installing a muffler on any auto which had heavily congealed oil or grease or which had any leaks. Wilcox disregarded this rule, and as a result an auto caught fire causing extensive property damage and injury to Wilcox. Which of the following statements is correct?
a. Miracle is not liable because its rule pro hibited Wilcox from installing the muffler in question.
b. Miracle is not liable to Wilcox under the workmen’s compensation laws.
c. Miracle is liable irrespective of its efforts to prevent such an occurrence and the fact that it exercised reasonable care.
d. Wilcox does not have any personal liability for the loss because he was acting for and on behalf of his employer.
2. Halliday engaged Fox as her agent. It was mutual ly agreed that Fox would not disclose that he was acting as Halliday’s agent. Instead he was to deal with pro spective customers as if he were a principal acting on his own behalf. This he did and made several contracts for Halliday. Assuming Halliday, Fox, or the customer seeks to avoid liability on one of the contracts involved, which of the following statements is correct?
a. The third party may choose to hold either Fox or Halliday liable.
The third party can avoid liability because he believed he was dealing with Fox as a principal.
Halliday must ratify the Fox contracts in order to be held liable.
Fox has no liability once he discloses that Halliday was the real principal.
b.
c.
d.
3. Smith has been engaged as a general sales agent for the Victory Medical Supply Company. Victory, as Smith’s principal, owes Smith several duties which are implied as a matter of law. Which of the following duties is owed by Victory to Smith?
a. Not to compete.
b. To reimburse Smith for all expenditures as long as they are remotely related to Smith’s employment and not specifically prohibited.
c. Not to dismiss Smith without cause for one year from the making of the contract if the duration of the contract is indefinite. d. To indemnify Smith for liability for acts done in good faith upon Victory’s orders.
4. Winter is a sales agent for Magnum Enterprises. Winter has assumed an obligation to indemnify Magnum if any of Winter’s customers fail to pay. Under these circumstances, which of the following is correct?
a. Winter’s engagement must be in writing regardless of its duration.
b. Upon default, Magnum must first proceed against the delinquent purchaser-debtor.
c. The above facts describe a del credere agency relationship and Winter will be liable in the event his customers fail to pay Magnum.
d. There is no fiduciary relationship on either Winter’s or Magnum’s part.
5. Gladstone has been engaged as sales agent for the Doremus Corporation. Under which of the following circumstances may Gladstone delegate his duties to another?
a. Where an emergency arises and the dele gation is necessary to meet the emergency.
b. Where it is convenient for Gladstone to do so.
c. Only with the express consent of Doremus.
d. If Doremus sells its business to another.
32 Business Law 6. Mr. Jackson owns approximately 40% of the shares of common stock of Triad Corporation. The rest of the shares are widely distributed among 2,000 share holders. Jackson needs funds for other business ventures and would like to raise about $2,000,000 through the sale of some of his Triad shares. He accord ingly approached Underwood & Sons, an investment banking house in which he knew one of the principals, to purchase his Triad shares and distribute the shares to the public at a reasonable price through its offices in the United States. Any profit on the sales could be retained by Underwood pursuant to an agreement reached between Jackson and Underwood. In this situation a. The securities to be sold probably do not need to be registered with the Securities and Exchange Commission.
b. Underwood & Sons probably is not an under writer as defined in the federal securities law.
c. Jackson probably is considered an issuer under federal securities law.
d. Under federal securities law, no prospectus is required to be filed in connection with this contemplated transaction.
7. Over a six-year period, Yeats Corporation ac quired 46% of the outstanding stock of Glick, Inc. More than 40% of the shares so acquired were purchased from Click’s past and present directors. Both Click and Yeats have capital, surplus, and undivided profits ag gregating more than $1,000,000. Yeats’ current directors owned stock in both corporations and were on the board of directors of each. Yeats utilized its ownership control to elect the remaining members of the board of directors and its own slate of officers for Click. Click and Yeats are manufacturers of goods which are in competition with each other throughout the United States. Since Yeats acquired control of Click, Yeats’ percentage share of the nationwide market has remained relatively stable. However, Yeats and Click each by agreement have ceased marketing in certain geographical territories where it is more ad vantageous for the other to sell. Which of the following statements applies to the above situation?
a. There is nothing in the facts as revealed above which would constitute a violation of the federal antitrust laws. b. The interlocking directorate is not illegal because less than 50% of the Click stock is owned by Yeats.
c. The interlocking directorate is a clear viola tion of the federal antitrust laws.
d. There is no current violation of the federal antitrust law because there has been no marked improvement in the competitive position of Yeats or Click.
8. The four largest manufacturers in their industry have had a combined share of the market in excess of 80% each year for several years. As members of a trade association, certain officers of these corporations meet periodically to discuss various topics of mutual interest. Matters discussed include engineering de sign, production methods, product costs, market shares, merchandising policy, and inventory levels. Open discussion of pricing is scrupulously avoided. However, the representatives usually see each other after the association meetings and pricing is frequently discussed. These representatives have maintained prices in accordance with an informal oral agreement terminable at will by any company wishing to with draw. They have never reduced their agreement to a written document or memorandum. The four corpo rations compete with each other in interstate com merce. Which of the following applies?
a. The members of the trade association may validly appoint the trade association as their representative to set minimum prices.
b. If the trade association suggested it, the distributors of the four corporations may legally enter an agreement among them selves to follow the industry leader’s pricing policy.
c. The trade association could legally allocate marketing areas among its members.
d. The four corporations have illegally entered into a price maintenance agreement among themselves.
9. Super Sports, Inc., sells branded sporting goods and equipment throughout the United States. It sells to wholesalers, jobbers, and retailers who in turn re sell the goods to their respective customers. The whole salers and jobbers, who do not sell at retail, are charged lower prices than retailers, but are required to purchase in larger quantities than retailers with the cost savings inherent in such purchases accounting for the lower prices. The retailers are all charged the same prices but receive discounts for quantity purchases based exclusively upon the cost savings resulting from such quantity purchases. Girard, one of Super’s retail cus tomers, has demanded discounts comparable to those available to the wholesalers and jobbers in its vicinity. Super has refused to acquiesce in this demand. There fore, Girard sues Super alleging an illegal price dis crimination in violation of the Robinson-Patman Act. Which defense by Super listed below will be most likely to prevail?
a. Girard does not have the right to sue under the Robinson-Patman Act.
b. The discounts are functional, that is, Super’s wholesalers and jobbers do not compete with retailers such as Girard.
c. Super does not have the requisite intent to discriminate among its purchasers.
d. The prices Super charges are reasonable, and its profit margins are low.
33 Examination Questions— May 1978 10. Futterman operated a cotton factory and employ ed Marra as a general purchasing agent to travel through the southern states to purchase cotton. Futterman telegraphed Marra instructions from day to day as to the price to be paid for cotton. Marra entered a cotton district in which she had not previously done business and represented that she was purchasing cotton for Futterman. Although directed by Futterman to pay no more than 25 cents a pound, Marra bought cotton from Anderson at 30 cents a pound, which was the prevailing offering price at that time. Futterman refused to take the cotton. Under these circumstances, which of the following is correct?
a. The negation of actual authority to make the purchase effectively eliminates any liability for Futterman.
b. Futterman is not liable on the contract.
c. Marra has no potential liability.
d. Futterman is liable on the contract.
11. Under the Securities Act of 1933, which of the following is the most important criterion to determine whether a private placement to a limited number of persons or a public offering has been made?
a. The size of the issuing corporation.
b. The type of security being offered.
c. The prompt resale of the securities by the purchasers.
d. The participating purchasers have signed an investment letter.
12. The Diablo Oven Company entered into agree ments with retail merchants whereby they agreed not to sell beneath Diablo’s minimum “ suggested” retail price of $85 in exchange for Diablo’s agreeing not to sell its ovens at retail in their respective territories. The agreement does not preclude the retail merchants from selling competing ovens. What is the legal status of the agreement?
a. It is legal if the product is a trade name or trademarked item.
b. It is legal if the power to fix maximum prices is not relinquished.
c. It is illegal unless it can be shown that the parties to the agreement were preventing cutthroat competition.
d. It is illegal even though the price fixed is reasonable.
13. Potter orally engaged Arthur as a salesman on A p ril 5, 1978, for exactly one year commencing on May 1, 1978. Which of the following is correct insofar as the parties are concerned?
a. If Arthur refuses to perform and takes an other job on April 14, 1978, he will not be liable if he pleads the statute of frauds.
b. The contract need not be in writing since its duration is exactly one year.
c. Potter may obtain the remedy of specific performance if Arthur refuses to perform.
d. The parol evidence rule applies.
14. Wexford Furniture, Inc., is in the retail furniture business and has stores located in principal cities in the United States. Its designers created a unique cock tail table. After obtaining prices and schedules, Wexford ordered 2,000 tables to be made to its design and specifications for sale as a part of its annual spring sales promotion campaign. Which of the following represents the earliest time Wexford will have an in surable interest in the tables?
a. At the time the goods are in Wexford’s possession.
b. Upon shipment of conforming goods by the seller.
c. When the goods are marked or otherwise designated by the seller as the goods to which the contract refers.
d. At the time the contract is made.
15. Milbank undertook to stage a production of a well-known play. He wired Lucia, a famous actress, offering her the lead in the play at $2,000 per week for six weeks from the specified opening night plus $1,000 for a week of rehearsal prior to opening. The telegram also said, “Offer ends in three days.” Lucia wired an acceptance the same day she received it. The telegram acceptance was temporarily misplaced by the telegraph company and did not arrive until five days after its dispatch. Milbank, not hearing from Lucia, assumed she had declined and abandoned the production. Which of the following is correct if Lucia sues Milbank?
a. The contract was automatically terminated when Milbank decided not to proceed.
b. Lucia has entered into a valid contract and is entitled to recover damages if Milbank fails to honor it.
c. Lucia may not take any other engagement for the period involved if she wishes to recover.
d. Milbank is excused from any liability since his action was reasonable under the circum stances.
16. Higgins orally contracted to pay $3,500 to Clark for $4,000 of thirty-day accounts receivable that arose in the course of Clark’s office equipment leasing busi ness. Higgins subsequently paid the $3,500. What is the legal status of this contract?
a. The contract is unenforceable by Higgins since the statute of frauds requirement has not been satisfied.
b. If Higgins failed to notify the debtors whose accounts were purchased, they will, upon payment in good faith to Clark, have no liability to Higgins.
c. The contract in question is illegal because it violates the usury laws.
d. Higgins will be able to collect against the debtors free of the usual defenses which would be assertable against Clark.
34 Business Law 17. Adams Company purchased a factory and ware house from Martinson for $150,000. Adams obtained a $100,000 real estate mortgage loan from a local bank and was required by the lender to pay for the cost of title insurance covering the bank’s interest in the prop erty. In addition, Adams was required to obtain fire insurance sufficient to protect the bank against loss due to fire. The co-insurance factor has been satisfied. Under these circumstances, which of the following is correct?
a. Adams can purchase only $50,000 of title insurance since it already obtained a $100,000 title policy for the bank equal to the bank loan.
b. The bank could not have independently ob tained a fire insurance policy on the property because Adams has legal title.
c. If Adams obtained a $150,000 fire insurance policy which covered its interest and the bank’s interest in the property and there is an estimated $50,000 of fire loss, the insurer will typically be obligated to pay the owner and the bank the amounts equal to their respective interests as they may appear.
d. If Adams obtained a $100,000 fire insurance policy covering the bank’s interest and $150,000 covering his own interest, each would obtain these amounts upon total de struction of the property.
18. Mario is seeking to collect on a property insurance policy covering certain described property which was destroyed. The insurer has denied recovery based upon Mario’s alleged lack of an insurable interest in the property. In which of the situations described below will the insurance company prevail?
a. The insured property does not belong to Mario, but instead to a corporation which he controls.
b. Mario is not the owner of the insured property but a mere long-term lessee.
c. The insured property belongs to a general trade debtor of Mario and the debt is un secured.
d. The property had been willed to Mario’s father for life and, upon his father’s death, to Mario as the remainderman.
19. Normally, the offer initiates the process by which a contract is created. Therefore, the offer is critical insofar as satisfying basic contract law requirements. Which of the following statements is incorrect?
a. The offer may only be expressed in words.
b. The offer must be communicated to the other party.
c. The offer must be certain enough to deter mine the liability of the parties.
d. The offer must be accepted by the other party.
20. Peters leased a restaurant from Brady with all furnishings and fixtures for a period of five years with an option to renew for two additional years. Peters made several structural improvements and modifica tions to the interior of the building. He obtained a fire insurance policy for his own benefit insuring his in terest in the property for $25,000. The restaurant was totally destroyed by an accidental fire. Peters seeks recovery from his insurer. Subject to policy limits, which of the following is correct?
a. Peters is entitled to recover damages to the extent of the value of his leasehold interest.
b. Peters is entitled to recover for lost profits due to the fire even though the policy is silent on the point.
c. Peters must first seek redress from the owner before he is entitled to recover.
d. Peters will not recover because he lacks the requisite insurable interest in the property.
21. Ames and Bates have agreed that Bates will sell a parcel of land to Ames for $10,000 if the land is re zoned from residential to industrial use within six months of the agreement. Bates agreed to use his best efforts to obtain the rezoning, and Ames agreed to make a $2,000 good-faith deposit with Bates two weeks after the date of the agreement. What is the status of this agreement?
a. No contract results because the event is contingent.
b. The agreement is probably unenforceable because Bates would be required to attempt to influence governmental action.
c. The parties have entered into a bilateral contract subject to a condition.
d. Ames is not obligated to make the deposit at the agreed time even though Bates has by then made an effort to procure a rezoning.
22. The Johnson Corporation sent its only pump to the manufacturer to be repaired. It engaged Travis, a local trucking company, both to deliver the equip ment to the manufacturer and to redeliver it to Johnson promptly upon completion of the repair. Johnson’s entire plant was inoperative without this pump, but the trucking company did not know this. The trucking company delayed several days in its delivery of the repaired pump to Johnson. During the time it expected to be without the pump, Johnson incurred $5,000 in lost profits. At the end of that time Johnson rented a replacement pump at a cost of $200 per day. As a result of these facts, what is Johnson entitled to recover from Travis?
a. The $200 a day cost incurred in renting the pump.
b. The $200 a day cost incurred in renting the pump plus the lost profits.
c. Actual damages plus punitive damages.
d. Nothing because Travis is not liable for damages.
35 Examination Questions— May 1978 23. Robert Cunningham owns a shop in which he repairs electrical appliances. Three months ago Electrical Supply Company sold Cunningham, on credit, a machine for testing electrical appliances and obtained a perfected security interest at that time to secure payment of the balance due. Cunningham’s creditors have now filed an involuntary petition in bankruptcy against him. What is the status of Electrical Supply?
a. Electrical Supply is a secured creditor that has the right, if not paid, to assert its rights against the machine sold to Cunningham to enforce its claim.
b. Electrical Supply must surrender its per fected security interest to the trustee in bank ruptcy and share as a general creditor of the bankrupt’s estate.
c. Electrical Supply’s perfected security interest constitutes a preference and is voidable.
d. Electrical Supply must elect to resort ex clusively to its secured interest or to relin quish it and obtain the same share as a general creditor.
24. A client has joined other creditors of the Ajax Demolition Company in a composition agreement seek ing to avoid the necessity of a bankruptcy proceeding against Ajax. Which statement describes the compo sition agreement?
a. It provides for the appointment of a receiver to take over and operate the debtor’s busi ness.
b. It must be approved by all creditors.
c. It does not discharge any of the debts in cluded until performance by the debtor has taken place.
d. It provides a temporary delay, not to ex ceed six months, insofar as the debtor’s obligation to repay the debts included in the composition.
25. On January 10, 1978, Edwards gave Cantrell a mortgage on his office building to secure a past-due $40,000 obligation which he owed Cantrell. Cantrell promptly recorded the mortgage. On March 15, 1978, a petition in bankruptcy was filed against Edwards. Simpson, the trustee in bankruptcy, desires to prevent Cantrell from qualifying as a secured creditor. In seek ing to set aside the mortgage, which of the following statements is correct?
a. The mortgage cannot be set aside since it is a real property mortgage and recorded.
b. Even if the mortgage is set aside, Cantrell has a priority in respect to the office building.
c. The mortgage can only be set aside if the mortgage conveyance was fraudulent.
d. The mortgage can be set aside if it was taken with knowledge of the fact that Edwards was insolvent in the bankruptcy sense.
26. Franklin engaged in extensive negotiations with Harlow in connection with the proposed purchase of Harlow’s factory building. Which of the following must Franklin satisfy to establish a binding contract for the purchase of the property in question?
a. Franklin must obtain an agreement signed by both parties.
b. Franklin must obtain a formal, detailed, all- inclusive document.
c. Franklin must pay some earnest money at the time of final agreement.
d. Franklin must have a writing signed by Harlow which states the essential terms of the understanding.
27. Olson conveyed real property to his sons, Sampson and David, but the deed was ambiguous as to the type of estate created and the interest each son had in relation to the other. David died intestate shortly after Olson. David’s widow and children are contend ing that they have rights in the property. Which of the following would be the widow’s and children’s best argument to claim valid rights in the real property?
a. The conveyance by Olson created a life estate in Sampson with a contingent remainder in terest in David.
b. The conveyance by Olson created a joint tenancy with a right of survivorship.
c. The conveyance by Olson created a tenancy in common.
d. The widow is entitled to her statutory share.
28. Lutz sold his moving and warehouse business, including all the personal and real property used there in, to Arlen Van Lines, Inc. The real property was en cumbered by a duly-recorded $300,000 first mortgage upon which Lutz was personally liable. Arlen acquired the property subject to the mortgage but did not as sume the mortgage. Two years later, when the out standing mortgage was $260,000, Arlen decided to aban don the business location because it had become unprofitable and the value of the real property was less than the outstanding mortgage. Arlen moved to another location and refused to pay the installments due on the mortgage. What is the legal status of the parties in regard to the mortgage?
a. Lutz must satisfy the mortgage debt in the event that foreclosure yields an amount less than the unpaid balance.
b. If Lutz pays off the mortgage, he will be able to successfully sue Arlen because Lutz is subrogated to the mortgagee’s rights against Arlen.
c. Arlen took the real property free of the mortgage.
d. Arlen breached its contract with Lutz when it abandoned the location and defaulted on the mortgage.
36 Business Law 29. Donaldson, Inc., loaned Watson Enterprises $50,000 secured by a real estate mortgage which in cluded the land, buildings, and “ all other property which is added to the real property or which is con sidered as real property as a matter of law.” Star Company also loaned Watson $25,000 and obtained a security interest in all of Watson’s “inventory, ac counts receivable, fixtures, and other tangible per sonal property.’’ There is insufficient property to satisfy the two creditors. Consequently, Donaldson is attempting to include all property possible under the terms and scope of its real property mortgage. If Donaldson is successful in this regard, then Star will receive a lesser amount in satisfaction of its claim. What is the probable outcome of Donaldson’s action?
a. Donaldson will not prevail if the property in question is detachable trade fixtures.
b. Donaldson will prevail if Star failed to file a financing statement.
c. Donaldson will prevail if it was the first lender and duly filed its real property mortgage.
d. The problem will be decided by taking all of Watson’s property (real and personal) subject to the two secured creditors’ claims and dividing it in proportion to the respec tive debts.
30. Harris is the trustee named in Filmore’s trust. The trust named Filmore as the life beneficiary, re mainder to his children at age 21. The trust consists of stocks, bonds, and three pieces of rental income property. Which of the following statements best de scribes the trustee’s legal relationships or duties?
a. The trustee has legal and equitable title to the rental property.
b. The trustee must automatically reinvest the proceeds from the sale of one of the rental properties in like property.
c. The trustee is a fiduciary with respect to the trust and the beneficiaries.
d. The trustee must divide among all the bene ficiaries any insurance proceeds received in the event the real property is destroyed.
31. In connection with the audit of Fiske & Company, you found it necessary to examine a deed to certain property owned by the client. In this connection, which of the following statements is correct?
a. A deed purporting to convey real property, but which omits the day of the month, is in valid.
b. A deed which lacks the signature of the grantor is valid.
c. A quitclaim deed which purports to transfer to the grantee “whatever title the grantor has’’ is invalid.
d. A deed which purports to convey real prop erty and recites a consideration of $1 and other valuable consideration is valid.
32. You are setting up the accounts for Barkum Enterprises, which operates a retail novelties business. Barkum wants all true liabilities shown. He recently received a discharge in bankruptcy, but the following proved claims were unpaid because of lack of funds. Which would you consider to be a claim against him and his new venture?
a. The unpaid amount owed to a secured credi tor who received less than the full amount after resorting to his security interest.
b. The unpaid amounts owed to his trade sup pliers for goods sold by him in the ordinary course of his prior business.
c. A personal loan by his father made in an attempt to stave off bankruptcy. d. The unpaid amount on the claim of a creditor who extended credit on the strength of a fraudulent financial statement.
33. Kilgore created an irrevocable fifteen-year trust for the benefit of his minor children. At the end of the fifteen years, the principal (corpus) reverts to Kilgore. Kilgore named the Reliable Trust Company as trustee and provided that Reliable would serve without the necessity of posting a bond. In understanding the trust and rules applicable to it, which of the following is correct?
a. The trust is not a separate legal entity for federal tax purposes.
b. The facts indicate that the trust is a separate legal entity for both tax and non-tax pur poses.
c. Kilgore may revoke the trust after eleven years, since he created it, and the principal reverts to him at the expiration of the fifteen years.
d. If Kilgore dies ten years after creation of the trust, it is automatically revoked and the property is distributed to the benefici aries of his trust upon their attaining age 21.
34. Fifteen years ago Madison executed a valid will. He named his son, Walker, as the executor of his will and left two thirds of his estate to his wife and the balance equally to his children. Madison is now dead and the approximate size of his estate is one million dollars. Which of the following statements is correct?
a. The will is invalid because it was executed at a time which is beyond the general statute of limitations.
b. The estate is not recognized as a taxable entity for tax purposes.
c. All the property bequeathed to his wife will be excluded from the decedent’s estate for federal estate tax purposes.
d. Walker must, in addition to being named in the will, be appointed or approved by the appropriate state court to serve as the executor.
37 Examination Questions— May 1978 35. Hacker is considering the creation of either a lifetime (intervivos) or testamentary (by his will) trust. In deciding what to do, which of the following state ments is correct?
a. An intervivos trust must meet the same legal requirements as one created by a will.
b. Property transferred to a testamentary trust upon the grantor’s (creator’s) death is not included in the decedent’s gross estate for federal tax purposes.
c. Hacker can retain the power to revoke an intervivos trust.
d. If the trust is an intervivos trust, the trustee must file papers in the appropriate state office roughly similar to those required to be filed by a corporation.
36. Cutler sent Foster the following offer by mail: “I offer you 150 Rex portable electric type writers, model J-1, at $65 per typewriter, F.O.B. your truck at my warehouse, terms 2/10, net/30. I am closing out this model, hence the substantial discount. Accept all or none. (signed) Cutler” Foster immediately wired back: ‘‘I accept your offer re the Rex electric type writers, but will use Blue Express Company for the pickup, at my expense of course. In addition, if possible, could you have the ship ment ready by Tuesday at 10:00 AM because of the holidays? (signed) Foster” Based on the above correspondence, what is the status of Foster’s acceptance?
a. It is not valid because it states both additional and different terms than those contained in the offer.
b. It represents a counteroffer which will be come a valid acceptance if not negated by Cutler within 10 days.
c. It is valid but will not be effective until re ceived by Cutler.
d. It is valid upon dispatch despite the fact it states both additional and different terms than those contained in the offer.
37. Wilcox Manufacturing repudiated its contract to sell 100 radios to Ready Stores, Inc. What recourse does Ready Stores have?
a. It can obtain specific performance by the seller.
b. It can recover punitive damages.
c. It can “cover” , that is, procure the goods elsewhere and recover any damages.
d. It must await the seller’s performance for a commercially reasonable time after repudi ation if it wishes to recover anything.
38. Haworth Discount Stores mailed its order to Eagle Recordings, Inc., for 100 eight-track cassette record ings of “ Swan Songs” by the Paginations at $5.50 per cassette. Eagle promptly wired its acceptance, delivery to take place within two weeks from date of Haworth’s order and terms of net 30 days. Before de livery was made by Eagle, the retail price of this record ing by the Paginations fell to $4.95. Haworth informed Eagle of this and pleaded with Eagle, “because we have been good customers, give us a break by either reducing the price of $4.95 so we can break even or by allowing us to cancel the order.” Eagle’s sales manager called Haworth the next day and informed them that the price would be $4.95 per cassette, not the price that appeared on the original invoice. Which of the following is correct insofar as the modification of the initial Haworth-Eagle contract?
a. The modification is invalid due to lack of consideration.
b. The modification is voidable by Eagle at any time prior to shipment of the 100 cassettes.
c. The modification must be written and signed by the parties to be valid if there is no con sideration given for the reduced price.
d. The modification need not satisfy the statute of frauds.
39. Badger Corporation sold goods to Watson. Watson has arbitrarily refused to pay the purchase price. Under what circumstances will Badger not be able to recover the price if it seeks this remedy instead of other possible remedies?
a. If Watson refused to accept delivery and the goods were resold in the ordinary course of business.
b. If Watson accepted the goods but seeks to return them.
c. If the goods sold were destroyed shortly after the risk of loss passed to the buyer.
d. If the goods were identified to the contract and Badger made a reasonable effort to re sell them at a reasonable price but was un able to do so.
40. Viscount Appliances sold Conway a refrigerator. Viscount wishes to disclaim the implied warranty of fitness for a particular purpose. Which of the following will effectively disclaim this warranty?
a. The fact that the refrigerator is widely ad vertised and was sold under its brand name.
b. A conspicuous written statement which states that “ any and all warranty protection is hereby disclaimed.” c. A conspicuous written statement indicating that “there are no warranties which extend beyond the description contained in the con tract of sale.” d. An inconspicuous written statement which specifically negates the warranty.
38 Business Law 41. Devoid Manufacturing, Inc., entered into a con tract for the sale to Hillary Company of 2,000 solid-state CB radios at $27.50 each, terms 2/10, N/30, F.O.B. Hillary’s loading platform. After delivery of the first 500 radios, a minor defect was discovered. Although the defect was minor, Hillary incurred costs to correct the defect. Hillary sent Devoid a signed memorandum indicating that it would relinquish its right to recover the costs to correct the defect, provided that the balance of the radios were in conformity with the terms of the contract and the delivery dates were strictly adhered to. Devoid met these conditions. Shortly before the last shipment of radios arrived, a dispute between the parties arose over an unrelated matter. Hillary notified Devoid that it was not bound by the prior generous agreement and would sue Devoid for dam ages unless Devoid promptly reimbursed Hillary. In the event of litigation, what would be the result and the basis upon which the litigation would be decided?
a. Devoid will lose in that Hillary’s relinquish ment of its rights was not supported by a consideration.
b. Devoid will win in that the defect was minor and the substantial performance doctrine applies.
c. Hillary will lose in that the communication constituted a waiver of Hillary’s rights.
d. Hillary will win in that there was a failure to perform the contract, and Hillary suffered damages as a result.
42. Hendrickson is a well-known author of popular detective stories, plays, and film scripts for which he obtained copyrights. As a result of some major declines in the values of certain risk ventures to which he was financially committed, he found himself hard-pressed for cash. Consequently, he decided to assign one of his copyrights on an earlier play to Eureka Enterprises. After protracted and often heated negotiations, Hendrickson and Eureka arrived at an oral agreement whereby Hendrickson would transfer to Eureka the copyright for $4,800. Prior to the actual transfer, one of Hendrickson’s investments paid off spectacularly and he refused to transfer the copyright to Eureka, alleging the contract to be null and void or at least voidable for several reasons. In light of the circum stances, which of the following is correct?
a. If Hendrickson had received a memorandum of the agreement which would bind Eureka and has not objected to it, he is bound.
b. Unless Eureka gave Hendrickson some earnest money at the time of the oral under standing, the contract is invalid.
c. The contract is voidable by Hendrickson if he pleads the statute of frauds.
d. The contract is null and void in that Hendrickson was under extreme mental stress and financial hardship at the time he made the agreement.
43. Wilson Corporation entered into a contract to sell goods to Marvin who has a place of business in the same town as Wilson. The contract was clear with respect to price and quantity, but failed to designate the place of delivery. Which of the following state ments is correct?
a. The contract is unenforceable because of indefiniteness.
b. The place for delivery must be designated by the parties within five days or the con tract is voidable.
c. The seller’s place of business is the proper place for delivery.
d. The buyer’s place of business is the proper place for delivery.
44. Bonnie Brook Wholesalers ordered 10,000 five- pound bags of standard granulated household sugar from Crane Sugar Plantations, Inc., for delivery within two months. Crane underestimated its existing back log of orders and overestimated its inventory. As a result Crane found that it would be either unable or extremely hard-pressed to fill the order on time. Con sequently, Crane assigned the contract to Devon Sugars, Inc., a smaller local producer and jobber, and paid Devon $200. Midway through the performance Devon defaulted because one of its suppliers’ ware houses was destroyed by fire. Bonnie seeks to recover damages for breach of contract from both Crane and Devon. In the event of litigation, which of the following statements is correct?
a. The unforeseen fire which destroyed the sup plier’s warehouse negates any liability on Devon’s part.
b. Devon will prevail because Devon has no contractual duty to Bonnie.
c. Bonnie will not prevail against either Crane or Devon unless Bonnie first exhausts its rights against the supplier whose warehouse was destroyed.
d. Bonnie will prevail against either Crane or Devon but will be entitled to only one re covery.
45. Filbert is the surety on a loan made by Holmes to Watson. Which statement describes Filbert’s legal relationship or status among the respective parties?
a. Filbert is a fiduciary insofar as Holmes is concerned.
b. As between Watson and Filbert, Filbert has the ultimate liability.
c. Filbert is not liable immediately upon default by Watson, unless the agreement so pro vides. d. Upon default by Watson and payment by Filbert, Filbert is entitled to subrogation to the rights of Holmes or to obtain reimburse ment from Watson.
39 Examination Questions— May 1978 46. Safety Surety Company, Inc., issued a surety bond for value received which guaranteed: (1) com pletion of a construction contract Poe had made with Cairns and (2) payment by Poe of his workmen. Poe defaulted and did not complete the contract. The workers were not paid for their last week's work. Poe had in fact become insolvent, and a petition in bank ruptcy was filed two months after the issuance of the bond. What is the effect upon Safety as a result of the above events? a. If Safety pays the workers in full, it is en titled to the same priority in the bankruptcy proceedings that the workers would have had.
b. If Safety pays damages to Cairns as a result of default on the contract, Safety is entitled to recover in the bankruptcy proceedings the entire amount it paid prior to the payment of the general creditors of Poe.
c. If Safety has another separate claim against Cairns, Safety may not set it off against any rights Cairns may have under this contract.
d. As a compensated surety, Safety would be discharged by Poe’s bankruptcy.
47. Barnaby is an employee of the Excelsior Manu facturing Company, a multi-state manufacturer of toys. The plant in which he works is unionized and Barnaby is a dues paying member. Which statement is correct insofar as the Federal Fair Labor Standards Act is concerned?
a. Excelsior is permitted to pay less than the minimum wage to employees since they are represented by a bona fide union.
b. The act allows a piece-rate method to be em ployed in lieu of the hourly rate method where appropriate.
c. The act excludes from its coverage the em ployees of a labor union.
d. The act sets the maximum number of hours that an employee can work in a given day or week.
48. Jones has filed a claim with the appropriate Workmen’s Compensation Board against the Atlas Metal & Magnet Company. Atlas denies liability under the State Workmen’s Compensation Act. In which of the following situations will Jones recover from Atlas or its insurer?
a. Jones intentionally caused an injury to himself.
b. Jones is an independent contractor.
c. Jones is basing the claim upon a disease unrelated to the employment.
d. Jones and another employee of Atlas were grossly negligent in connection with their employment, resulting in injury to Jones.
49. Gladstone Warehousing, Inc., is an independ ent bonded warehouse company. It issued a warehouse receipt for 10,000 bales of cotton belonging to Travis. The word “NEGOTIABLE” was conspicuously printed on the warehouse receipt it issued to Travis. The ware house receipt also contained a statement in large, clear print that the cotton would only be surrendered upon return of the receipt and payment of all storage fees. Travis was a prominent plantation owner engaged in the cotton growing business. Travis pledged the ware house receipt with Southern National Bank in exchange for a $50,000 personal loan. A financing statement was not filed. Under the circumstances, which of the follow ing is correct?
a. Travis’ business creditors cannot obtain the warehouse receipt from Southern National unless they repay Travis’ outstanding loan.
b. The bank does not have a perfected security interest in the cotton since it did not file a financing statement.
c. Travis’ personal creditors have first claim, superior to all other parties, to the cotton in question because the loan was a personal loan and constituted a fraud upon the per sonal creditors.
d. The fact that the word “NEGOTIABLE” and the statement regarding the return of the receipt were conspicuously printed upon the receipt is not binding upon anyone except Travis.
50. Vega Manufacturing, Inc., manufactures and sells hi-fi systems and components to the trade and at re tail. Repossession is frequently made from customers who are in default. Which of the following statements is correct concerning the rights of the defaulting debtors who have had property repossessed by Vega?
a. Vega has the right to retain all the goods repossessed as long as it gives notice and cancels the debt.
b. It is unimportant whether the goods re possessed are defined as consumer goods, inventory, or something else in respect to the debtor’s rights upon repossession.
c. If the defaulting debtor voluntarily signs a statement renouncing his rights in the col lateral, the creditor must nevertheless resell them for the debtor’s benefit.
d. If a debtor has paid sixty percent or more of the purchase price of consumer goods in satisfaction of a purchase money security interest, the debtor has the right to have the creditor dispose of the goods.
40 Business Law Part a. Your CPA firm was engaged to audit the Meglo Corporation. During the audit you examined the following instrument: Number 2 (Estimated time------25 to 30 minutes) April 2 , 1977 Charles Noreen 21 West 21st Street St. Louis, Missouri I, Charles Noreen, do hereby promise to pay to Roger Smith, Two Thousand Dollars ($2,000) one year from date, with 8% in terest upon due presentment.
FOR: Payment for used IBM typewriters. Meglo purchased the instrument from Smith on April 10, 1977, for $1,700. Meglo received the instru ment with Smith’s signature and the words “ Pay to the order of Meglo Corporation’’ on the back. Upon maturity, Meglo presented the instrument to Noreen, who refused to pay. Noreen alleged that the typewriters were defective and did not satisfy certain warranties given in connection with the purchase of the used IBM typewriters which were guaranteed for one year. Noreen had promptly notified Smith of this fact and had told him he would not pay the full amount due.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Is the instrument in question negotiable com mercial paper?
2. Assuming that the instrument is negotiable, does Meglo qualify as a holder in due course entitled to collect the full $2,000?
3. Assuming that the instrument is negotiable, is Noreen’s defense valid against a holder in due course?
4. Assuming that the instrument is nonnego tiable, what is the legal effect of the transfer by Smith to Meglo?
Part b. Marvin Farber cashed a check for Harold Kern which was made to the order of Charles Walker by Marglow Investments & Securities. The check had the following indorsements on the back: 1. 2. without recourse 3. Pay to the order of Harold Kern 4. Pay to the order of Marvin Farber Kern neglected to sign his indorsement when he gave the check to Farber, and Farber did not notice this until the following day. Before Farber could locate Kern and obtain his signature, Farber learned that Walker had fraudulently obtained the check from Marglow (the drawer). Farber finally located Kem and obtained his signature. Farber promptly indorsed the check in blank and cashed it at National Bank. National Bank presented the check for payment through normal banking channels, but it was dishonor ed by Marglow’s bank pursuant to a valid stop order. National Bank contacted Farber and informed him of the situation. Farber repaid the amount and the check was returned to him with National Bank’s blank indorsement on the back.
Required: Answer the following, setting forth for any conclusions stated.
reasons 1. Identify the type of indorsement and indicate the liability for each indorsement numbered 1, 2, and 3 above.
2. Will Farber prevail in a legal action seeking payment of the check by Marglow?
41 Examination Questions— May 1978 Part a. A CPA firm was engaged to examine the financial statements of Martin Manufacturing Corpo ration for the year ending December 31, 1977. The facts revealed that Martin was in need of cash to continue its operations and agreed to sell its common stock invest ment in a subsidiary through a private placement. The buyers insisted that the proceeds be placed in escrow because of the possibility of a major contingent tax liability that might result from a pending government claim. The payment in escrow was completed in late November 1977. The president of Martin told the audit partner that the proceeds from the sale of the subsidi ary’s common stock, held in escrow, should be shown on the balance sheet as an unrestricted current account receivable. The president was of the opinion that the government’s claim was groundless and that Martin needed an “ uncluttered” balance sheet and a “clean” auditor’s opinion to obtain additional working capital from lenders. The audit partner agreed with the presi dent and issued an unqualified opinion on the Martin financial statements which did not refer to the contin gent liability and did not properly describe the escrow arrangement.
The government’s claim proved to be valid, and pursuant to the agreement with the buyers, the pur chase price of the subsidiary was reduced by $450,000. This adverse development forced Martin into bank ruptcy. The CPA firm is being sued for deceit (fraud) by several of Martin’s unpaid creditors who extended credit in reliance upon the CPA firm’s unqualified opinion on Martin’s financial statements.
Required: Answer the following, setting forth reasons for any conclusions stated.
Based on these facts, can Martin’s unpaid credi tors recover from the CPA firm?
Part b. A CPA firm has been named as a defend ant in a class action by purchasers of the shares of stock of the Newly Corporation. The offering was a public offering of securities within the meaning of the Securities Act of 1933. The plaintiffs alleged that the firm was either negligent or fraudulent in connec tion with the preparation of the audited financial state ments which accompanied the registration statement filed with the SEC. Specifically, they allege that the CPA firm either intentionally disregarded, or failed to exercise reasonable care to discover, material facts which occurred subsequent to January 31, 1978, the date of the auditor’s report. The securities were sold to the public on March 16, 1978. The plaintiffs have subpoenaed copies of the CPA firm’s working papers. The CPA firm is considering refusing to relinquish Number 3 (Estimated time------20 to 25 minutes) the papers, asserting that they contain privileged com munication between the CPA firm and its client. The CPA firm will, of course, defend on the merits irre spective of the questions regarding the working papers.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Can the CPA firm rightfully refuse to sur render its working papers?
2. Discuss the liability of the CPA firm in re spect to events which occur in the period between the date of the auditor’s report and the effective date of the public offering of the securities.
Number 4 (Estimated time----- 25 to 30 minutes) Part a. Grace Dawson was actively engaged in the promotion of a new corporation to be known as Multifashion Frocks, Inc. On January 3, 1978, she ob tained written commitments for the purchase of shares totaling $600,000 from a group of 15 potential investors. She was also assured orally that she would be engaged as the president of the corporation upon the commence ment of business. Helen Banks was the principal inves tor, having subscribed to $300,000 of the shares of Multifashion. Dawson immediately began work on the incorporation of Multifashion, made several contracts for and on its behalf, and made cash expenditures of $1,000 in accomplishing these goals. On February 15, 1978, Banks died and her estate has declined to honor the commitment to purchase the Multifashion shares. At the first shareholders’ meeting on April 5, 1978, the day the corporation came into existence, the share holders elected a board of directors. With shareholder approval, the board took the following actions: 1. Adopted some but not all of the contracts made by Dawson.
2. Authorized legal action, if necessary, against the Estate of Banks to enforce Banks’ $300,000 commitment.
3. Declined to engage Dawson in any capacity (Banks had been her main sup porter).
4. Agreed to pay Dawson $750 for those cash outlays which were deemed to be directly beneficial to the corporation and rejected the balance.
Required: Answer the following, setting forth reasons for any conclusions stated.
Discuss the legal implications of each of the above actions taken by the board of directors of Multifashion.
42 Business Law Part b. Duval is the chairman of the board and president of Monolith Industries, Inc. He is also the largest individual shareholder, owning 40 percent of the shares outstanding. The corporation is publicly held, and there is a dissenting minority. In addition to his position with Monolith, Duval owns 85 percent of Variance Enterprises, a corporation created under the laws of the Bahamas. During 1977 Carlton, the presi dent of Apex Industries, Inc., approached Duval and suggested that a tax-free merger of Monolith and Apex made good sense to him and that he was prepared to recommend such a course of action to the Apex board and to the shareholders. Duval studied the proposal and decided that Apex was a most desirable candidate for acquisition. Duval informed the president of Variance about the overture, told him it was a real bar gain, and suggested that Variance pick it up for cash and notes. Not hearing from Duval or Monolith, Carlton accepted an offer from Variance and the business was sold to Variance. Several dissenting shareholders of Monolith learned the facts surrounding the Variance acquisition and have engaged counsel to represent them. The Variance acquisition of Apex proved to be highly profitable.
Required: Answer the following, setting forth reasons for any conclusions stated.
Discuss the rights of the dissenting Monolith shareholders and the probable outcome of a legal action by them.
Number 5 (Estimated time----- 15 to 20 minutes) Part a. Millard rented office space in a building owned by Burbank. Millard was in the import-export business and was desperately in need of additional cash. Therefore, he decided to use Burbank’s name in conjunction with his own as if they were partners in order to obtain credit from several lenders. He placed a nameplate on his door with the legend “ Millard & Burbank’’ and had the same name listed on the directory in the lobby. In addition, he had business cards and stationery printed with the same title. Final ly, he placed an announcement in the local paper that Burbank had joined him in the newly-created partner ship of Millard & Burbank. Burbank’s rental agent saw the partnership name on the door, saw the listing in the lobby directory, and informed Burbank of the situation. Burbank read the notice in the local paper. In response to this misrepresentation, Burbank told his rental agent to remove the listing in the lobby and to tell Millard to stop the nonsense. These instruc tions were not followed. In the interim, Millard was negotiating a $10,000 loan with Easy Credit Corpo ration, one of the other tenants in the office building. Dunlop, one of Easy’s officers, had seen the “partner ship” nameplate on Millard’s door, the listing in the lobby, and the notice in the paper. Therefore, based exclusively upon the credit standing of Burbank, the loan was made. Millard defaulted on the loan and is hopelessly insolvent. Easy demanded payment from Burbank who refused to pay and denied any liability as Millard’s partner.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Are Millard and Burbank partners in fact?
2. Under what legal theory could Easy prevail?
Part b. Idaho Mining & Minerals was formed 10 years ago by five individuals as a general partnership. After several years of losses the partnership finally began to show a modest profit from its operations which has increased steadily. The current picture is one of justified optimism. The most recent year showed a profit in excess of $250,000, and realistic projections indicate profits in excess of $1 million within five years. The partners are considering incorporating the busi ness and going public sometime in the future. The part ners are concerned about the implications of incorpo rating from the standpoint of (1) their liability after incorporation, with particular emphasis on possible liability for claims against the corporation and pre existing claims against the partnership, and (2) the tax implications to them of a transfer of the business to a newly-created corporation.
Their concern is predicated upon the following facts. There is a significant lawsuit against the part nership which is believed to be baseless but which could nevertheless result in serious liability. Further more, the tax basis of their respective partnership interests is low in relation to the current fair market value of the partnership’s assets.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Explain the legal liability and implications of the proposed incorporation to the partners and the corporation.
2. What are the federal income tax implications to the partners as to recognition of gain or loss as a result of a transfer of the partnership assets to a newly- created corporation?
43 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) EXAMINATION IN ACCOUNTING THEORY (Theory of Accounts) May 5, 1978; 1:30 to 5:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: No. 1 .............................
No. 2 .............................
No. 3 .............................
No. 4 .............................
No. 5 .............................
No. 6 .............................
No. 7 .............................
Total.....................
Estimated Minutes Minimum Maximum 175 35 40 35 40 20 25 20 25 20 25 25 30 20 25 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you as indicating inefficiency in accounting work.) are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
44 Accounting Theory Instructions Select the best answer for each of the following items relating to a variety of issues in financial account ing. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked; Item 99. The recommended title for the financial statement which summarizes the results of operations is the a. Earnings statement.
b. Statement of changes in financial condition.
c. Statement of sources and applications of funds.
d. Retained earnings statement.
Answer Sheet 99. Items to be Answered 1. The concept of objectivity is complied with when an accounting transaction occurs that a. Involves an arm’s-length transaction between two independent interests.
b. Furthers the objectives of the company.
c. Is promptly recorded in a fixed amount of dollars.
d. Allocates revenues or expense items in a rational and systematic manner.
2. When translating an amount for fixed assets shown on the statement of financial position of a foreign subsidiary, the appropriate rate of translation is the a. Current exchange rate.
b. Average exchange rate for the current year.
c. Historical exchange rate.
d. Average exchange rate over the life of each fixed asset.
3. Generally, revenues should be recognized at a point when a. Management decides it is appropriate to do so.
b. The product is available for sale to the ulti mate consumer.
c. An exchange has taken place and the earn ings process is virtually complete.
d. An order for a definite amount of merchan dise has been received for shipment FOB destination.
Number 1 (Estimated time----- 35 to 40 minutes) 4. In theory (disregarding any other marketplace variables) the proceeds from the sale of a bond will be equal to a. The face amount of the bond.
b. The present value of the principal amount due at the end of the life of the bond plus the present value of the interest payments made during the life of the bond, each discounted at the prevailing market rate of interest.
c. The face amount of the bond plus the present value of the interest payments made during the life of the bond discounted at the pre vailing market rate of interest.
d. The sum of the face amount of the bond and the periodic interest payments.
5. Why are certain costs of doing business capital ized when incurred and then depreciated or amortized over subsequent accounting cycles?
a. To reduce the federal income tax liability.
b. To aid management in the decision-making process.
c. To match the costs of production with reve nues as earned.
d. To adhere to the accounting concept of con servatism.
6. Which of the following describes a change in reporting entity?
a. A company acquires a subsidiary that is to be accounted for as a purchase.
b. A manufacturing company expands its market from regional to nationwide.
c. A company acquires additional shares of an investee and changes from the equity method of accounting to consolidation of the sub sidiary.
d. A business combination is made using the pooling of interests method.
7. What accounting concept justifies the usage of accruals and deferrals?
a. Going concern.
b. Materiality.
c. Consistency.
d. Stable monetary unit.
8. Under what condition is it proper to recognize revenues prior to the sale of the merchandise?
a. When the concept of internal consistency (of amounts of revenue) must be complied with.
b. When the revenue is to be reported as an installment sale.
c. When the ultimate sale of the goods is at an assured sales price.
d. When management has a long-established policy to do so.
45 Examination Questions— May 1978 9. What is the underlying concept that supports the immediate recognition of a loss?
a. Matching.
b. Consistency.
c. Judgment.
d. Conservatism.
10. A net unrealized loss on a company’s long-term portfolio of marketable equity securities should be re flected in the current financial statements as a. An extraordinary item shown as a direct reduction from retained earnings.
b. A current loss resulting from holding market able equity securities.
c. A footnote or parenthetical disclosure only.
d. A valuation allowance and included in the equity section of the statement of financial position.
11. A net unrealized gain on a company’s long-term portfolio of marketable equity securities should be re flected in the current financial statements as a. An extraordinary item shown as a direct increase to retained earnings.
b. A current gain resulting from holding market able equity securities.
c. A footnote or parenthetical disclosure only.
d. A valuation allowance and included in the equity section of the statement of financial position. 12. In accordance with generally accepted accounting principles, which of the following methods of amorti zation is normally recommended for intangible assets?
a. Sum-of-the-years-digits.
b. Straight-line.
c. Units of production.
d. Double-declining balance.
13. The appropriate valuation of an operating lease on the statement of financial position of a lessee is a. Zero.
b. The absolute sum of the lease payments.
c. The present value of the sum of the lease payments discounted at an appropriate rate.
d. The market value of the asset at the date of the inception of the lease.
14. With respect to the computation of earnings per share, which of the following would be most indicative of a simple capital structure?
a. Common stock, preferred stock, and con vertible securities outstanding in lots of even thousands.
b. Earnings derived from one primary line of business.
c. Ownership interests consisting solely of common stock.
d. Equity represented materially by liquid assets.
15. What are the three types of period costs that a lessee experiences with capital leases?
a. Lease expense, interest expense, amortiza tion expense.
b. Interest expense, amortization expense, executory costs.
Amortization expense, executory costs, lease expense.
expense.
d. Executory costs, interest expense, expense.
lease 16. Which of the following types of transactions or situations would preclude a company from accounting for a business combination as a pooling of interests?
a. Immediately after the combination, the acquiring corporation reacquires the stock issued to effect the combination.
b. The combined company sells assets that were acquired in the combination which represent duplicate facilities.
c. The acquiring corporation acquires only ninety percent of the voting common stock of the other corporation in exchange for its voting common stock.
d. The combination is effected within nine months of the initiation of the plan of com bination.
17. An accrued expense can best be described as an amount a. Paid and currently matched with earnings.
b. Paid and not currently matched with earn ings.
c. Not paid and not currently matched with earnings.
d. Not paid and currently matched with earn ings.
18. Which of the following principles best describes the conceptual rationale for the methods of matching depreciation expense with revenues?
a. Associating cause and effect.
b. Systematic and rational allocation.
c. Immediate recognition.
d. Partial recognition.
19. Which of the following conditions generally exists before market value can be used as the basis for valu ation of a company’s marketable equity securities?
a. Market value must approximate historical cost.
b. Management’s intention must be to dispose of the security within one year.
c. Market value must be less than cost for each security held in the company’s marketable equity security portfolio.
d. The aggregate valuation of a company’s marketable equity security portfolio must be less than the aggregate cost of the portfolio.
46 Accounting Theory 20. Which of the following cost items would be matched with current revenues on a basis other than association of cause and effect?
a. Goodwill.
b. Sales commissions.
c. Cost of goods sold.
d. Purchases on account.
21. Which of the following items is included on a statement of changes in financial position only because of the all-financial-resources concept?
a. Depreciation.
b. Issuance (sale) of common stock.
c. Purchase of treasury stock.
d. Retirement of long-term debt by issuance of preferred stock.
22. Which of the following is the proper accounting treatment of a gain contingency?
a. An accrued amount.
b. Deferred earnings.
c. An account receivable with an additional disclosure explaining the nature of the transaction.
d. A disclosure only.
23. If a contingent loss is probable and can be reason ably estimated to be within a given range, but no amount within the range is a better estimate than any other amount within the range, the amount to be accrued should be a. Zero.
b. The upper limit of the range.
c. The lower limit of the range.
d. The mean of the upper and lower limits of the range.
24. Which of the following items, if material in amount, would normally be considered an extraordi nary item for reporting results of operations?
a. Utilization of a net operating loss carry forward.
b. Gains or losses on disposal of a segment of a business.
c. Adjustments of accruals on long-term con tracts.
d. Gains or losses from a fire.
25. Which of the following is not a basis for the im mediate recognition of a cost during a period?
a. The cost provides no discernible future benefit.
b. The cost recorded in a prior period no longer produces discernible benefits.
c. The federal income tax savings using the immediate write-off method exceed the savings obtained by allocating the cost to several periods.
d. Allocation of the cost on the basis of associ ation with revenue or among several account ing periods is considered to serve no useful purpose.
Instructions Select the best answer for each of the following items relating to a variety of issues in managerial ac counting and municipal accounting. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
Items to be Answered 26. If a credit was made to the fund balance in the process of recording a budget for a governmental unit, it can be assumed that a. Estimated expenses exceed actual revenues.
b. Actual expenses exceed estimated expenses.
c. Estimated revenues exceed appropriations.
d. Appropriations exceed estimated revenues.
27. Which of the following types of governmental revenue would be susceptible to accrual using the modified accrual method of accounting?
a. State sales tax.
b. Property tax.
c. Income tax.
d. Business licenses.
28. Which governmental fund would account for fixed assets in a manner similar to a “for-profit” organization?
a. Enterprise.
b. Capital projects.
c. General fixed asset group of accounts.
d. General.
29. If a governmental unit established a data process ing center to service all agencies within the unit, the data processing center should be accounted for as a (an) a. Capital projects fund.
b. Intragovernmental service fund.
c. Agency fund.
d. Trust fund.
30. The “reserve for encumbrances — past year” account represents amounts recorded by a govern mental unit for a. Anticipated expenditures in the next year.
b. Expenditures for which purchase orders were made in the prior year but disbursement will be in the current year.
c. Excess expenditures in the prior year that will be offset against the current-year budget ed amounts.
d. Unanticipated expenditures of the prior year that become evident in the current year.
Number 2 (Estimated time------35 to 40 minutes) 47 Examination Questions— May 1978 31. Which of the following types of revenue would generally be recorded directly in the general fund of a governmental unit?
a. Receipts from a city-owned parking structure.
b. Property taxes.
c. Interest earned on investments held for re tirement of employees.
d. Revenues from intragovernmental service funds.
32. Which of the following is a relevant factor in the determination of an economic order quantity?
a. Physical plant insurance costs.
b. Warehouse supervisory salaries.
c. Variable costs of processing a purchase order.
d. Physical plant depreciation charges.
33. Which of the following is a useful technique in determining the fixed and variable elements of a semi-variable expense?
a. Linear programming.
b. Queuing theory.
c. Program Evaluation and Review Technique.
d. Simple regression analysis.
34. To obtain the break-even point stated in terms of dollars of sales, total fixed costs are divided by which of the following?
a. Variable cost per unit.
b. (Sales price per unit — variable cost per unit) sales price per unit.
c. Fixed cost per unit.
d. Variable cost per unit sales price per unit.
35. The weighted average method of process costing differs from the first-in, first-out method of process costing in that the weighted average method a. Requires that ending work-in-process inven tory be stated in terms of equivalent units of production.
b. Can be used under any cost-flow assumption.
c. Considers the ending work-in-process inven tory only partially complete.
d. Does not consider the degree of completion of beginning work-in-process inventory when computing equivalent units of production.
36. What is the basic difference between direct cost ing and absorption costing?
a. Direct costing always produces less taxable earnings than absorption costing.
b. Direct costing recognizes fixed costs as a period cost and absorption costing recog nizes fixed costs as a product cost.
c. Direct costing cannot use standards, whereas standards may be used with absorption costing.
d. Direct costing may be used only in situations where production is essentially homogeneous but absorption costing may be used under any manufacturing condition.
37. An equivalent unit of material or conversion cost is equal to a. The amount of material or conversion cost necessary to complete one unit of production.
b. A unit of work-in-process inventory.
c. The amount of material or conversion cost necessary to start a unit of production into work in process.
d. Fifty percent of the material or conversion cost of a unit of finished goods inventory (assuming a linear production pattern).
38. If two or more products share a common process before they are separated, the joint costs should be allocated in a manner that a. Assigns a proportionate amount of the total cost to each product by means of a quanti tative basis.
b. Maximizes total earnings.
c. Minimizes variations in a unit of production cost.
d. Does not introduce an element of estimation into the process of accumulating costs for each product.
39. What factor, related to manufacturing costs, causes the difference in net earnings computed using absorption costing and net earnings computed using direct costing?
a. Absorption costing considers all costs in the determination of net earnings, whereas direct costing considers only direct costs.
b. Absorption costing “inventories” all direct costs, but direct costing considers direct costs to be period costs, c. Absorption costing “ inventories” all fixed costs for the period in ending finished goods inventory, but direct costing expenses all fixed costs.
d. Absorption costing allocates fixed costs be tween cost of goods sold and inventories, and direct costing considers all fixed costs to be period costs.
40. A useful tool in financial statement analysis is termed “ common size financial statements.” What does this tool enable the financial analyst to do?
a. Evaluate financial statements of companies within a given industry of the approximate same value.
b. Determine which companies in a similar industry are at approximately the same stage of development.
c. Compare the mix of assets, liabilities, capital, revenue, and expenses within a company over a period of time or between companies within a given industry without respect to relative size.
d. Ascertain the relative potential of companies of similar size in different industries.
48 Accounting Theory 41. Which of the following would not be used in job- order costing?
a. Standards.
b. Averaging of direct labor and material rates.
c. Direct costing.
d. Factory overhead allocation based on direct labor hours applied to the job.
42. In comparing the current ratios of two companies, why is it invalid to assume that the company with the higher current ratio is the better company?
a. The current ratio includes assets other than cash.
b. A high current ratio may indicate inadequate inventory on hand.
c. A high current ratio may indicate inefficient use of various assets and liabilities.
d. The two companies may define working capital in different terms.
43. Excess direct labor wages resulting from overtime premium will be disclosed in which type of variance?
a. Yield.
b. Quantity.
c. Labor efficiency.
d. Labor rate.
44. If the total materials variance (actual cost of materials used compared with the standard cost of the standard amount of materials required) for a given operation is favorable, why must this variance be further evaluated as to price and usage?
a. There is no need to further evaluate the total materials variance if it is favorable.
b. Generally accepted accounting principles require that all variances be analyzed in three stages.
c. All variances must appear in the annual report to equity owners for proper disclosure.
d. To allow management to evaluate the effi ciency of the purchasing and production functions.
45. How is a labor rate variance computed?
a. The difference between standard and actual rate multiplied by standard hours.
b. The difference between standard and actual hours multiplied by actual rate.
c. The difference between standard and actual rate multiplied by actual hours.
d. The difference between standard and actual hours multiplied by the difference between standard and actual rate.
46. If the sum of the present values of all future cash flows related to a proposed capital expenditure dis counted at the company’s cost of capital is positive, it indicates that the a. Resultant amount is the maximum that should be paid for the asset.
b. Return on the investment exceeds the com pany’s cost of capital.
c. Investment is the best alternative.
d. Discount rate used is not the proper cost of capital for this company.
47. When using a flexible budget, what will occur to fixed costs ( on a per unit basis) as production increases within the relevant range?
a. Fixed costs per unit will decrease.
b. Fixed costs per unit will remain unchanged.
c. Fixed costs per unit will increase.
d. Fixed costs are not considered in flexible budgeting.
48. In setting earnings objectives, management must consider all of these items except a. Sales volume attainable in the present plant.
b. The break-even point.
c. Indices of industrial production.
d. Earnings or losses for given sales-volume levels.
49. In analyzing factory overhead variances, the volume variance is the difference between the a. Amount shown in the flexible budget and the amount shown in the master budget.
b. Master budget application rate and the flex ible budget application rate multiplied by actual hours worked.
c. Budget allowance based on standard hours allowed for actual production for the period and the amount applied during the period.
d. Actual amount spent for overhead items during the period and the amount applied during the period.
50. Why is direct costing not in accordance with generally accepted accounting principles?
a. Fixed manufacturing costs are assumed to be period costs.
b. Direct costing procedures are not well known in industry.
c. Net earnings are always overstated when using direct costing procedures.
d. Direct costing ignores the concept of tower of cost or market when valuing inventory.
49 Examination Questions— May 1978 Part a. The Financial Accounting Standards Board discusses certain terminology essential to both the translation of foreign currency transactions and foreign currency financial statements in its Statement No 8. Included in the discussion is a definition of and distinction between the terms “ measure” and “denominate.” Required: Define the terms “measure” and “denominate” as discussed by the Financial Accounting Standards Board and give a brief example that demonstrates the distinction between accounts measured in a particular currency and accounts denominated in a particular currency.
Part b. There are several methods of translat ing foreign currency transactions or accounts reflected in foreign currency financial statements. Among these methods are: current/noncurrent, monetary/non-monetary. current rate, and the temporal method (the method adopted by the Financial Accounting Stand ards Board).
Required: Define the temporal method of translating foreign currency financial statements. Specifically include in your answer the treatment of the following four ac counts: 1. Long-term accounts receivable.
2. Deferred income.
3. Inventory valued at cost.
4. Long-term debt.
Number 3 (Estimated time------20 to 25 minutes) Number 4 (Estimated time----- 20 to 25 minutes) Part a. Generally accepted accounting principles require that pension costs be accounted for on the accrual basis. The various components of pension ex pense include (but are not limited to): 1. normal cost, 2. past service cost, 3. prior service cost, and 4. interest.
Required: Define each of the four terms designated above, and discuss how each of the costs is accounted for under generally accepted accounting principles.
Part b. The accounting for past service cost has been a controversial issue. Some members of the pro fession advocate the accrual of past service cost only to the extent funded, and others advocate the accrual of past service cost regardless of the amount funded.
Required: 1. What are the arguments in favor of accruing past service cost only to the extent funded?
2. What are the arguments in favor of accruing past service cost regardless of the amount funded?
Number 5 (Estimated time----- 20 to 25 minutes) The Thomas Company is in the process of develop ing a revolutionary new product. A new division of the company was formed to develop, manufacture, and market this new product. As of year end (December 31, 1977) the new product has not been manufactured for resale; however, a prototype unit was built and is in operation.
Throughout 1977 the new division incurred certain costs. These costs include design and engineering studies, prototype manufacturing costs, administrative expenses (including salaries of administrative person nel), and market research costs. In addition, approxi mately $500,000 in equipment (estimated useful life — 10 years) was purchased for use in developing and manufacturing the new product. Approximately $200,000 of this equipment was built specifically for the design development of the new product; the re maining $300,000 of equipment was used to manu facture the pre-production prototype and will be used to manufacture the new product once it is in commercial production.
Required: a. What is the definition of “research” and of “development” as defined in Statement of Financial Accounting Standards No. 2?
b. Briefly indicate the practical and conceptual reasons for the conclusion reached by the Financial Accounting Standards Board on accounting and report ing practices for research and development costs.
c. In accordance with Statement of Financial Accounting Standards No. 2, how should the various costs of Thomas described above be recorded on the financial statements for the year ended December 31, 1977?
50 Accounting Theory Part a. Advocates of current value accounting propose several methods for determining the valuation of assets to approximate current values. Two of the methods proposed are replacement cost and present value of future cash flows.
Required: Describe each of the two methods cited above and discuss the pros and cons of the various procedures used to arrive at the valuation for each method.
Part b. The financial statements of a business entity could be prepared by using historical cost or current value as a basis. In addition, the basis could be stated in terms of unadjusted dollars or dollars restated for changes in purchasing power. The various permu tations of these two separate and distinct areas are shown in the following matrix: Dollars Restated for Changes in Purchasing Power Number 6 (Estimated time------25 to 30 minutes) Unadjusted Dollars Historical cost Current value 1 2 3 4 Block number 1 of the matrix represents the traditional method of accounting for transactions in accounting today, wherein the absolute (unadjusted) amount of dollars given up or received is recorded for the asset or liability obtained (relationship between resources). Amounts recorded in the method described in block number 1 reflect the original cost of the asset or liability and do not give effect to any change in value of the unit of measure (standard of comparison). This method assumes the validity of the accounting concepts of going concern and stable monetary unit. Any gain or loss (including holding and purchasing power gains or losses) resulting from the sale or satisfaction of amounts recorded under this method is deferred in its entirety until sale or satisfaction.
Required: For each of the remaining matrix blocks (2, 3 and 4) respond to the following questions. Limit your dis cussion to nonmonetary assets only.
• How will this method of recording assets affect the relationship between resources and the standard of comparison?
• What is the theoretic justification for using each method?
• How will each method of asset valuation affect the recognition of gain or loss during the life of the asset and ultimately from the sale or abandonment of the asset? Your response should include a discussion of the timing and magnitude of the gain or loss and conceptual reasons for any difference from the gain or loss computed using the traditional method.
Complete your discussion for each matrix block before proceeding to the discussion of the next matrix block.
Sometimes a business entity may change its method of accounting for certain items. The change may be classified as a change in accounting principle, a change in accounting estimate, or a change in report ing entity.
Listed below are three independent, unrelated sets of facts relating to accounting changes.
Situation I A company determined that the depreciable lives of its fixed assets are presently too long to fairly match the cost of the fixed assets with the revenue produced. The company decided at the beginning of the current year to reduce the depreciable lives of all of its existing fixed assets by five years.
Situation II On December 31, 1976, Gary Company owned 51 percent of Allen Company, at which time Gary reported its investment using the cost method due to political uncertainties in the country in which Allen was located. On January 2, 1977, the management of Gary Company was satisfied that the political uncertainties were resolved and the assets of the company were in no danger of nationalization. Accordingly, Gary will pre pare consolidated financial statements for Gary and Allen for the year ended December 31, 1977.
Situation III A company decides in January 1977 to adopt the straight-line method of depreciation for plant equip ment. The straight-line method will be used for new acquisitions as well as for previously-acquired plant equipment for which depreciation had been provided on an accelerated basis.
Number 7 (Estimated time------20 to 25 minutes) Required: For each of the situations described above, provide the information indicated below. Complete your dis cussion of each situation before going on to the next situation.
a. Type of accounting change.
b. Manner of reporting the change under current generally accepted accounting principles in cluding a discussion, where applicable, of how amounts are computed.
c. Effect of the change on the statement of financial position and earnings statement.
d. Footnote disclosures which would be neces sary.
51 Examination Questions November 1978 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all the states, territories, and the District of Columbia.) EXAMINATION IN ACCOUNTING PRACTICE — PART I November 1, 1978; 1:30 to 6:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1............................................................................................ 40 No. 2 ............................................................................................ 40 No. 3 ............................................................................................ 50 No. 4 ............................................................................................ 40 No. 5 ............................................................................................ S O Total.................................................................................... 220 50 50 60 50 60 270 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For in stance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. Enclose all scratch sheets. Failure to enclose scratch sheets may result in loss of grading points. Scratch sheets need not have page numbers, but you should show the question number and place them immediately following the question to which they relate.
4. Fourteen-column sheets, if any, should not be folded until all sheets, both wide and narrow, are placed in the proper sequence and fastened together at the top left corner. All fourteen- column sheets should then be wrapped around the back of the papers.
5. A CPA is continually confronted with the necessi ty of expressing opinions and conclusions in writ ten reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and applica tion of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
53 Examination Questions—November 1978 Number 1 (Estimated time 40 to 50 minutes) Instructions Select the best answer for each of the following items relating to a variety of financial-accounting problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 97. Gross billings for merchandise sold by Baker Company to its customers last year amounted to $5,260,000; sales returns and allowances reduced the amounts owed by $160,000. How much were net sales last year for Baker Company?
a. $4,800,000.
b. $5,100,000.
c. $5,200,000.
d. $5,260,000.
Answer Sheet 97. © © © Items to be Answered 1. Dark Company had interest income on municipal obligations of $150,000 in 1977. For financial statement reporting, Dark included the $150,000 in the other in come section of its income statement. For income tax reporting, the $150,000 was exempt income. Assuming an income tax rate of 50%, what should be reported in the provision for deferred income taxes in Dark’s in come statement for the year ended December 31, 1977?
a. $0.
b. $75,000 credit.
c. $75,000 debit.
d. $150,000 debit.
2. On November 30, 1977, File, Incorporated, pur chased for cash of $25 per share all 3()0,000 of the outstanding common stock of Mooney Company. Mooney’s balance sheet at November 30, 1977, showed a book value of $6,000,000. Additionally, the fair value of Mooney’s property, plant, and equipment on November 30, 1977, was $800,000 in excess of its book value. What amount, if any, will be shown in the balance sheet caption “Goodwill” in the November 30, 1977, consolidated balance sheet of File, Incorporated, and its wholly-owned subsidiary, Mooney Company?
a. $0.
b. $700,000.
c. $800,000.
d. $1,500,000.
3. During 1977, Hollin Company determined, as a result of additional information, that machinery that was previously depreciated over a seven-year life had a total estimated useful life of only five years. An accounting change was made in 1977 to reflect this ad ditional information. If the change had been made in 1976, the allowance for accumulated depreciation would have been $2,600,000 at December 31, 1976, in stead of $2,100,000. As a result of this change, 1977 depreciation expense was $200,000 greater than it would have been if the change had not been made. Assume that the direct effects of this change are limited to the effect on depreciation and the related tax provision, and that the income tax rate was 50% in both years. What should be reported in Hollin’s income statement for the year ended December 31, 1977, as the cumulative effect on prior years of changing the estimated useful life of the machinery?
a. $0.
b. $250,000.
c. $350,000.
d. $500,000.
4. On November 1, 1977, Robby, Inc., issued shares of its voting common stock in exchange for all of the voting common stock of Rudy Company in a business combination appropriately accounted for by the pooling of interests method. Both companies have a December 31 year end. Net income for each company is as follows: Robby Rudy 12 months ended December 31, 1977 2 months ended December 31, 1977 $1,300,000 $800,000 240,000 170,000 Assuming that the net income of Robby given above does not include the equity in the net income of Rudy, the consolidated net income for the year ended December 31, 1977, should be a. $410,000.
b. $1,300,000.
c. $1,470,000.
d. $2,100,000.
5. Tackle Company sells football helmets. In 1977 Tackle discovered a defect in the helmets which has pro duced lawsuits that are reasonably estimated to result in losses of $900,000. Based on its own experience and the experience of other enterprises in the business, Tackle considers it probable that additional lawsuits that are reasonably estimated to result in losses of $1,600,000 will occur even though the particular parties that will bring suit are not identifiable at this time. What amount of loss, if any, should be accrued by a charge to income in 1977?
a. $0.
b. $900,000.
c. $1,600,000.
d. $2,500,000.
54 Accounting Practice — Part I 6. Fountain, Incorporated, has 5,000,000 shares of common stock outstanding on December 31, 1976. An additional 1,000,000 shares of common stock were issued on April 1, 1977, and 500,000 more on July 1, 1977. On October 1, 1977, Fountain issued 10,000, $1,000 face value, 7% convertible bonds. Each bond is convertible into 40 shares of common stock. The bonds were not considered common stock equivalents at the time of their issuance and no bonds were converted into common stock in 1977. What is the number of shares to be used in computing primary earnings per share and fully diluted earnings per share, respectively?
a. 5,750,000 and 5,950,000.
b. 5,750,000 and 6,150,000.
c. 6,000,000 and 6,100,000.
d. 6,000,000 and 6,900,000.
7. Growing, Inc., had net income for 1977 of $10,600,000 and earnings per share on common stock of $5.00. Included in the net income was $1,000,000 of bond interest expense related to its long-term debt. The income tax rate for 1977 was 50%. Dividends on pre ferred stock were $600,000. The dividend-payout ratio on common stock was 40%. What were the dividends on common stock in 1977?
a. $3,600,000.
b. $3,800,000.
c. $4,000,000.
d. $4,240,000.
8. Seed Company has a receivable from a foreign customer which is payable in the local currency of the foreign customer. On December 31, 1976, this receivable was appropriately included in the accounts receivable section of Seed’s balance sheet at $450,000. When the receivable was collected on January 4, 1977, Seed converted the local currency of the foreign customer into $440,000. Seed also owns a foreign sub sidiary in which exchange gains of $45,000 resulted as a consequence of translation in 1977. What amount, if any, should be included as an exchange gain or loss in Seed’s 1977 consolidated income statement?
a. $0.
b. $10,000 exchange loss.
c. $35,000 exchange gain.
d. $45,000 exchange gain.
9. The Lake Company sold some machinery to the View Company on January 1, 1976, for which the cash selling price was $758,200. View entered into an install ment sales contract with Lake at an interest rate of 10%. The contract required payments of $200,000 a year over five years with the first payment due on December 31, 1976. What amount of interest income, if any, should be included in Lake’s 1977 income statement (the second year of the contract), using the “interest method’’?
a. $0.
b. $63,402.
c. $75,820.
d. $100,000.
10. Information from Guard Company’s financial statements is as follows: 1976 1977 Current assets at December 31 $2,000,000 $2,100,000 Current liabilities at December 31 1,000,000 900,000 Stockholders’ equity at December 31 2,500,000 2,700,000 Net sales for year 8,300,000 8,800,000 Cost of goods sold for year 6,200,000 6,400,000 Operating income for year 500,000 550,000 What is the current ratio at December 31, 1977?
a. 1.20 to 1.
b. 2.25 to 1.
c. 2.33 to 1.
d. 7.33 to 1.
11. In May 1978, Roy Company spent $200,000 on an advertising campaign for subscriptions to the school magazine it sells. The subscriptions do not start until September 1978 and the magazine is only sold on a year ly subscription basis. What amount of expense should be included in Roy’s quarterly income statement for the three months ended June 30, 1978, as a result of this ex penditure?
a. $0.
b. $50,000.
c. $66,667.
d. $200,000.
12. On January 1, 1977, Barley Corp. paid $600,000 for 60,000 shares of Oat Company’s common stock which represents a 25% investment in Oat. Barley has the ability to exercise significant influence over Oat. Barley received a dividend of $1 per share from Oat in 1977. Oat reported net income of $320,000 for the year ended December 31, 1977. The balance in Barley’s balance sheet account “ Investment in Oat Company’’ at December 31, 1977, should be a. $600,000.
b. $620,000.
c. $680,000.
d. $740,000.
13. Level, Inc., was formed on January 1, 1977, when common stock of $200,000 was issued for cash of $50,000 and land valued at $150,000. Level did not begin operations until 1978, and no transactions occurred in 1977 except the recording of the issuance of the common stock. If the general price-level index was 100 at December 31, 1976, and 110 at December 31, 1977, what would the general price-level gain or loss be in Level’s 1977 general price-level income statement?
a. $0.
b. $5,000 loss.
c. $5,000 gain.
d. $15,000 gain.
55 Examination Questions—November 1978 14. Mill Company operates in three different in dustries, each of which is appropriately regarded as a reportable segment. Segment No. 1 contributed 60% of Mill Company’s total sales. Sales for Segment No. 1 were $900,000 and traceable costs were $400,000. Total common costs for Mill were $600,000. Mill allocates common costs based on the ratio of a segment’s sales to total sales, an appropriate method of allocation. What should be the operating profit presented for Segment No. 1 for 1977?
a. $140,000.
b. $360,000.
c. $500,000.
d. $540,000.
15. Coleman, Incorporated, purchased a machine on January 1, 1970, for $100,000. Coleman is depreciating the machine on a straight-line basis with no salvage and a ten-year life. At December 31, 1976, the replacement cost (current fair value) of the machine was $32,000. On January 1, 1977, the machine was sold for $35,000. Ignoring income taxes, what amount should be shown as the gain or loss on the sale of the machine in a current fair value accounting income statement for 1977?
a. $2,000 gain.
b. $3,000 loss.
c. $3,000 gain.
d. $5,000 gain.
16. Token Company sold some of its fixed assets during 1977. The original cost of the fixed assets was $750,000 and the allowance for accumulated deprecia tion at the date of sale was $600,000. The proceeds from the sale of the fixed assets were $210,000. Assuming funds are defined as working capital, the information concerning the sale of the fixed assets should be shown on Token’s statement of changes in financial position for the year ended December 31, 1977, as a. A subtraction from net income of $60,000 and a source of $150,000.
b. An addition to net income of $60,000 and a source of $150,000.
c. A subtraction from net income of $60,000 and a source of $210,000.
d. A source of $150,000.
17. Minor Baseball Company had a player contract with Doe that was recorded in its accounting records at $145,000. Better Baseball Company had a player con tract with Smith that was recorded in its accounting records at $140,000. Minor traded Doe to Better for Smith by exchanging each player’s contract. The fair value of each contract was $150,000. What amount should be shown in the accounting records after the ex change of player contracts?
18. Kirt, Incorporated, had net income for 1977 of $3,000,000. Additional information is as follows: Minor Better a.
$140,000 $140,000 b.
$140,000 $145,000 c.
$145,000 $140,000 d.
$150,000 $150,000 Amortization of goodwill Depreciation on fixed assets Long-term debt: Bond discount amortization Interest expense Provision for doubtful accounts: Current receivables Long-term receivables $ 80,000 3,200,000 130,000 2,600,000 700,000 210,000 Assuming funds are defined as working capital, what should be the working capital provided from operations in the statement of changes in financial position for the year ended December 31, 1977?
a. $6,200,000.
b. $6,410,000.
c. $6,620,000.
d. $9,220,000.
19. On July 1, 1977, Round Company issued for $525,000 a total of 5,000 shares of $100 par value, 7% noncumulative preferred stock along with one detachable warrant for each share issued. Each warrant contains a right to purchase one share of Round’s $10 par value common stock for $15 a share. The market price of the rights on July 1, 1977, was $2.25 per right. On October 31, 1977, when the market price of the com mon stock was $19 per share and the market value of the rights was $3.00 per right, 4,000 rights were exercised. As a result of the exercise of the 4,000 rights and the is suance of the related common stock, what journal entry would Round make?
a. Cash Common stock Additional paid-in capital b. Cash Common stock rights outstanding Common stock Additional paid-in capital c. Cash Common stock rights outstanding Common stock Additional paid-in capital d. Cash Common stock rights outstanding Common stock Additional paid-in capital Debit Credit $60,000 $40,000 60,000 20,000 9,000 40,000 60,000 29,000 12,000 40,000 60,000 32,000 16,000 40,000 36,000 56 Accounting Practice — Part I 20. In 1976, Orlando, Inc., issued for $105 per share, 8,000 shares of $100 par value convertible preferred stock. One share of preferred stock can be converted into three shares of Orlando’s $25 par value common stock at the option of the preferred shareholder. In August 1977, all of the preferred stock was converted into common stock. The market value of the common stock at the date of the conversion was $30 per share. What total amount should be credited to additional paid-in capital as a result of the issuance of the pre ferred stock and its subsequent conversion into common stock?
a. $80,000.
b. $120,000.
c. $200,000.
d. $240,000.
Number 2 (Estimated time-----40 to 50 minutes) Instructions Select the best answer for each of the following items relating to a variety of managerial-accounting and quantitative-methods problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
Items to be Answered 21. Helen Corp. manufactures products W, X, Y, and Z from a joint process. Additional information is as follows: Sales If Processed Further Units Product Produced Value at Split-off Additional Sales Costs Value W 6,000 $ 80,000 $ 7,500 $ 90,000 X 5,000 60,000 6,000 70,000 Y 4,000 40,000 4,000 50,000 Z 3,000 20,000 2,500 30,000 18,000 $200,000 $20,000 $240,000 Assuming that total joint costs of $160,000 were allocated using the relative-sales-value at split-off ap proach, what were the joint costs allocated to each pro duct?
W X Y Z a. $40,000 $40,000 $40,000 $40,000 b. $53,333 $44,444 $35,556 $26,667 c. $60,000 $46,667 $33,333 $20,000 d. $64,000 $48,000 $32,000 $16,000 22. Walsh, Inc., is preparing its cash budget for the month of November. The following information is available concerning its inventories: Inventories at beginning of November $180,000 Estimated cost of goods sold for November 900,000 Estimated inventories at end of November 160,000 Estimated payments in November for purchases prior to November 210,000 Estimated payments in November for purchases in November 80% What are the estimated cash disbursements for inven tories in November?
a. $720,000.
b. $914,000.
c. $930,000.
d. $1,042,000.
23. Kyle Company manufactures products S and T from a joint process. The sales value at split-off was $50,000 for 6,000 units of Product S and $25,000 for 2,000 units of Product T. Assuming that the portion of the total joint costs properly allocated to Product S using the relative-sales-value at split-off approach was $30,000, what were the total joint costs?
a. $40,000.
b. $42,500.
c. $45,000.
d. $60,000.
Items 24 and 25 are based on the following information: Taylor, Inc., produces only two products, Acdom and Belnom. These account for 60% and 40% of the total sales dollars of Taylor, respectively. Variable costs (as a percentage of sales dollars) are 60% for Acdom and 85% for Belnom. Total fixed costs are $150,000. There are no other costs.
24. What is Taylor’s breakeven point in sales dollars?
a. $150,000.
b. $214,286.
c. $300,000.
d. $500,000.
25. Assuming that the total fixed costs of Taylor in crease by 30%, what amount of sales dollars would be necessary to generate a net income of $9,000?
a. $204,000.
b. $464,000.
c. $659,000.
d. $680,000.
57 Examination Questions—November 1978 26. Sant Company is planning to invest $40,000 in a machine with a useful life of five years and no salvage value. The straight-line method of depreciation will be used. Sant estimates that the annual cash inflow from operations, net of income taxes, from using this machine will be $10,000. Sant’s desired rate of return on investments of this type is 10%. The present value of an ordinary annuity of $1 for five periods at 10% is 3.791. The present value of $1 for five periods at 10% is 0.621. Using the net present-value method, Sant’s true rate of return on this investment is a. 0%.
b. Less than 10%, but more than 0%.
c. 10%.
d. More than 10%.
27. Horn Corp. invested in a four-year project. Horn’s expected rate of return is 8%. Additional infor mation on the project is as follows; Year 1 2 3 4 Cash inflow from operations, net of income taxes $2,000 2,200 2,400 2,600 Present Value of $1 at 8% .926 .857 .794 .735 Assuming a positive net present value of $500, what was the amount of the original investment?
a. $1,411.
b. $2,411.
c. $7,054.
d. $8,700.
28. Carter Company buys a certain part for its manufacturing process for $20 a part. Ten thousand parts a year are needed. It costs $3 a year to carry one of these parts in inventory. The cost of placing a purchase order for these parts is $15. Assuming that the parts will be required evenly throughout the year, the formula for the economic order quantity is a.
2 X 10,000 X 3 15 2 X 10,000 X 15 3 X 3 c. (l0,000 d. |l0,000 x 15 3 29. Brike Company, which manufactures robes, has enough idle capacity available to accept a special order of 10,000 robes at $8 a robe. A predicted income state ment for the year without this special order is as follows: Per Unit Total Sales $12.50 $1,250,000 Manufacturing costs- variable 6.25 625,000 Manufacturing costs-fixed 1.75 175,000 Manufacturing costs-total 8.00 800,000 Gross profit 4.50 450,000 Selling expenses-variable 1.80 180,000 Selling expenses-fixed 1.45 145,000 Selling expenses-total 3.25 325,000 Operating income $ 1.25 $ 125,000 Assuming no additional selling expenses, what would be the effect on operating income if the special order was accepted?
a. $8,000 increase.
b. $17,500 increase.
c. $32,500 decrease.
d. $40,000 increase.
Items 30 and 31 are based on the following infor mation: Milligan Company manufactures two models, small and large. Each model is processed as follows; Small (X) Large (Y) Machining Department 2 hours 4 hours Polishing Department 1 hour 3 hours The available time for processing the two models is 100 hours a week in the Machining Department and 90 hours a week in the Polishing Department. The con tribution margin expected is $5 for the small model and $7 for the large model.
30. How would the objective function (maximization of total contribution margin) be expressed?
a. 5X + 7Y.
b. 5X + 7Y≤190.
c. 5X(3) + 7Y(7) ≤190.
d. 12X + 1 0 Y.
31. How would the restriction (constraint) for the Machining Department be expressed?
a. 2(5X)+ 4(7Y)≤100.
b. 2X + 4Y.
c. 2X + 4Y≤100.
d. 5X +7Y ≤100.
58 Accounting Practice — Part I 32. Walton, Incorporated, had 8,000 units of work in process in Department A on October 1, 1978. These units were 60% complete as to conversion costs. Materials are added in the beginning of the process. During the month of October, 34,000 units were started and 36,000 units completed. Walton had 6,000 units of work in process on October 31, 1978. These units were 80% complete as to conversion costs. By how much did the equivalent units for the month of October using the weighted-average method exceed the equivalent units for the month of October using the first-in, first-out method?
Materials Conversion Costs a.
0 3,200 b.
0 4,800 c.
8,000 3,200 d.
8,000 4,800 33. Lucas Company adds materials in the beginning of the process in the Forming Department, which is the first of two stages of its production cycle. Information concerning the materials used in the Forming Depart ment in October 1978 is as follows: Materials Units Costs Work in process at October 1, 1978 6,000 $ 3,000 Units started during October 50,000 25,560 Units completed and trans ferred to next Department during October 44,000 Using the weighted-average method, what was the materials cost of work in process at October 31, 1978?
a. $3,060.
b. $5,520.
c. $6,000.
d. $6,120.
34. Matt Company uses a standard cost system. Infor mation for raw materials for Product RBI for the month of October is as follows: Standard unit price Actual purchase price per unit Actual quantity purchased Actual quantity used Standard quantity allowed for actual production $1.60 $1.55 2,000 units 1,900 units 1,800 units What is the materials purchase price variance?
a. $90 favorable.
b. $90 unfavorable.
c. $100 favorable.
d. $100 unfavorable.
35. Golden, Inc., has been manufacturing 5,000 units of Part 10541 which is used in the manufacture of one of its products. At this level of production, the cost per unit of manufacturing Part 10541 is as follows: Direct materials $ 2 Direct labor 8 Variable overhead 4 Fixed overhead applied 6 Total $20 Brown Company has offered to sell Golden 5,000 units of Part 10541 for $19 a unit. Golden has determined that it could use the facilities presently used to manufacture Part 10541 to manufacture Product RAC and generate an operating profit of $4,000. Golden has also determined that two thirds of the fixed overhead applied will continue even if Part 10541 is pur chased from Brown. To determine whether to accept Brown’s offer, the net relevant costs to Golden are a. $70,000.
b. $80,000.
c. $90,000.
d. $95,000.
36. Lab Corp. uses a standard cost system. Direct labor information for Product CER for the month of October is as follows; Standard rate $6.00 per hour Actual rate paid $6.10 per hour Standard hours allowed for actual production 1,500 hours Labor efficiency variance $600 unfavorable What are the actual hours worked?
a. 1,400.
b. 1,402.
c. 1,598.
d. 1,600.
37. Air, Inc., uses a standard cost system. Overhead cost information for Product CO for the month of October is as follows: Total actual overhead incurred $12,600 Fixed overhead budgeted $ 3,300 Total standard overhead rate per direct labor hour $4.00 Variable overhead rate per direct labor hour $3.00 Standard hours allowed for actual production 3,500 What is the overall (or net) overhead variance?
a. $1,200 favorable.
b. $1,200 unfavorable.
c. $1,400 favorable.
d. $1,400 unfavorable.
59 Examination Questions—November 1978 38. Metro, Inc., manufactures dolls. The cost of carrying one doll in inventory for one year is $0.60. Metro manufactures 6,000 dolls evenly throughout the year. Using the economic-order-quantity approach, the optimal production run would be 200 when setup costs are a. $2.00.
b. $3.00.
c. $3.50.
d. $4.00.
39. An office equipment representative has a machine for sale or lease. If you buy the machine, the cost is $7,596. If you lease the machine, you will have to sign a noncancellable lease and make 5 payments of $2,000 each. The first payment will be paid on the first day of the lease. At the time of the last payment you will receive title to the machine. The present value of an or dinary annuity of $1 is as follows: Number of Periods Present Value 10% 12% 16% 1 0.909 0.893 0.862 2 1.736 1.690 1.605 3 2.487 2.402 2.246 4 3.170 3.037 2.798 5 3.791 3.605 3.274 The interest rate implicit in this lease is approximately a. 10%.
b. 12%.
c. Between 10% and 12%.
d. 16%.
Number 3 (Estimated time-----50 to 60 minutes) Number 3 consists of three unrelated parts.
Part a. On January 1, 1978, MyKoo Corporation issued $1,000,000 in five-year, 5% serial bonds to be repaid in the amount of $200,000 on January 1, of 1979, 1980, 1981, 1982, and 1983. Interest is payable at the end of each year. The bonds were sold to yield a rate of 6%. Information on present value and future amount factors is as follows: Present value of an ordinary annuity of $1 for 5 years Future amount of an ordinary annuity of $1 for 5 years 5% 6% 5% 6% 4.3295 4.2124 5.5256 5.6371 Number Present Value of $1 Future amount of $1 of years 5% 6% 5% 6% 1 .9524 .9434 1.0500 1.0600 2 .9070 .8900 1.1025 1.1236 3 .8638 .8396 1.1576 1.1910 4 .8227 .7921 1.2155 1.2625 5 .7835 .7473 1.2763 1.3382 Required: 1. Prepare a schedule showing the computation of the total amount received from the issuance of the serial bonds. Show supporting computations in good form.
2. Assume the bonds were originally sold at a dis count of $26,247. Prepare a schedule of amortization of the bond discount for the first two years after issuance, using the interest (effective rate) method. Show supporting computations in good form.
Part b. On January 1, 1976, when its $30 par value common stock was selling for $80 per share, a cor poration issued $10,000,000 of 4% convertible deben tures due in ten years. The conversion option allowed the holder of each $1,000 bond to convert the bond into five shares of the corporation’s $30 par value common stock. The debentures were issued for $ 1 1 ,000,000. The present value of the bond payments at the time of is suance was $8,500,000 and the corporation believes the difference between the present value and the amount paid is attributable to the conversion feature. On January 1, 1977, the corporation’s $30 par value com mon stock was split 3 for 1. On January 1, 1978, when the corporation’s $10 par value common stock was sell ing for $90 per share, holders of 40% of the convertible debentures exercised their conversion options. The cor poration uses the straight-line method for amortizing any bond discounts or premiums.
Required: 1. Prepare in general journal format the entry to record the original issuance of the convertible deben tures.
2. Prepare in general journal format the entry to record the exercise of the conversion option, using the book value method. Show supporting computations in good form.
Part c. On July 1, 1978, Salem Corporation issued $2,000,000 of 7% bonds payable in ten years. The bonds pay interest semiannually. The bonds include detachable warrants giving the bondholder the right to purchase for $30, one share of $1 par value common stock at any time during the next ten years. The bonds were sold for $2,000,000. The value of the warrants at the time of issuance was $100,000.
Required: Prepare in general journal format the entry to record the issuance of the bonds.
60 Accounting Practice — Part I Number 4 consists of two unrelated parts.
Part a. The Allen Company is a wholesale distributor of automotive replacement parts. Initial amounts taken from Allen’s accounting records are as follows: Number 4 (Estimated tim e------40 to 50 minutes) Inventory at December 31, 1977 (based on physical count of goods in Allen’s ware house on December 31, 1977) Accounts payable at December 31, 1977: Vendor Baker Company Charlie Company Dolly Company Eager Company Full Company Greg Company Sales in 1977 Terms 2% 10 days, net 30 Net 30 Net 30 Net 30 Net 30 Net 30 $1,250,000 Amount $ 265,000 210,000 300,000 225,000 $1,000,000 $9,000,000 Additional information is as follows: 1. Parts held on consignment from Charlie to Allen, the consignee, amounting to $155,000, were in cluded in the physical count of goods in Allen’s warehouse on December 31, 1977, and in accounts payable at December 31, 1977.
2. $22,000 of parts which were purchased from Full and paid for in December 1977 were sold in the last week of 1977 and appropriately recorded as sales of $28,000. The parts were included in the physical count of goods in Allen’s warehouse on December 31, 1977, because the parts were on the loading dock waiting to be picked up by customers.
3. Parts in transit on December 31, 1977, to customers, shipped F.O.B. shipping point, on December 28, 1977, amounted to $34,000. The customers received the parts on January 6, 1978. Sales of $40,000 to the customers for the parts were recorded by Allen on January 2, 1978.
4. Retailers were holding $210,000 at cost ($250,000 at retail), of goods on consignment from Allen, the consignor, at their stores on December 31, 1977.
5. Goods were in transit from Greg to Allen on December 31, 1977. The cost of the goods was $25,000, and they were shipped F.O.B. shipping point on December 29, 1977.
6. A quarterly freight bill in the amount of $2,000 specifically relating to merchandise purchases in December 1977, all of which was still in the inventory at December 31, 1977, was received on January 3, 1978. The freight bill was not included in either the inventory or in accounts payable at December 31, 1977; 7. All of the purchases from Baker occurred during the last seven days of the year. These items have been recorded in accounts payable and accounted for in the physical inventory at cost before discount. Allen’s policy is to pay invoices in time to take advantage of all cash discounts, adjust inventory accordingly, and record accounts payable, net of cash discounts.
Required: Prepare a schedule of adjustments to the initial amounts using the format shown below. Show the effect, if any, of each of the transactions separately and if the transactions would have no effect on the amount shown, state NONE.
Accounts Inventory Payable Sales Initial amounts $1,250,000 $1,000,000 $9,000,000 Adjustments-increase (decrease) 1 2 3 4 5 6 7 Total adjustments Adjusted amounts $ _ $ $ Part b. The Deytyme Construction Company began operations January 1, 1977. During the year, Deytyme entered into a contract with Redbeard Razor Corporation to construct a manufacturing facility. At that time, Deytyme estimated that it would take five years to complete the facility at a total cost of $4,800,000. The total contract price for construction of the facility is $6,000,000. During the year, Deytyme in curred $1,250,000 in construction costs related to the construction project. The estimated cost to complete the contract is $3,750,000. Redbeard was billed and paid 30% of the contract price.
Required: Prepare schedules to compute the amount of gross profit to be recognized for the year ended December 31, 1977, and the amount to be shown as “cost of un completed contract in excess of related billings’’ or “billings on uncompleted contract in excess of related costs’’ at December 31, 1977, under each of the follow ing methods: 1. Completed-contract method.
2. Percentage-of-completion method.
Show supporting computations in good form.
61 Examination Questions—November 1978 Number 5 consists of two unrelated parts.
Part a. Harold Barton is a self-employed manu facturers’ representative. The following schedule of receipts and disbursements for the calendar year 1977 was prepared by his secretary from the cash book that she maintains for his business: Cash receipts: Commissions received $65,100 Rent for office sublease 5,000 Insurance company payment 3,600 Total cash receipts $73,700 Cash disbursements: • Harold Barton’s withdrawals 22,000 Secretary’s salary (net of taxes withheld) 8,000 Commissions paid to outsiders 12,000 Office rent 14,000 Payment to Keogh retire ment plan 3,000 Automobile expenses 1,800 Office expenses 1,160 Country club dues and expenses 3,900 Payroll taxes remitted: Taxes withheld from secretary’s salary 1,262 Employer’s payroll taxes 825 Total cash disburse ments 67,947 Excess of cash receipts over cash disbursements $ 5,753 Additional information is as follows: • Depreciation for 1977 on fixed assets used in Barton’s business was as follows: Automobile $1,000 Office equipment 1,600 $2,600 • The insurance company payment was in settle ment of a court suit that arose from a December 1976 automobile accident involving Mr. Barton. The pay ment was as follows: Number 5 (Estimated tim e------50 to 60 minutes) Mr. Barton had paid $700 for the car repairs in 1976 and had claimed this payment as a business expense in 1976. The car involved in the accident is used 100% for business purposes.
• Of the country club dues and expenses totaling $3,900, Mr. Barton has kept records which show that 60% of the total amount was incurred in connection with his business.
• During 1977 Mr. Barton signed up to par ticipate in a Keogh (HR—10) retirement plan. In December 1977 he made an initial contribution to the plan of $3,000 and in January 1978 made a second con tribution of $1,000. His secretary was not eligible and thus not covered by the plan in 1977.
• Mr. and Mrs. Barton jointly owned 100 shares of common stock of Bramma Corporation which they acquired in 1975 for $6,900. On November 1, 1977, they sold these shares for $3,400.
• During 1977 Mr. and Mrs. Barton gave $600 to recognized charitable organizations and donated old clothes valued at $275 to a needy family in their neighborhood.
• The following taxes were paid in 1977: Real estate (residence) taxes $2,330 Sales taxes 770 State gasoline taxes on automobile used for personal use 102 State income tax (balance 1976) 625 State income tax (estimate 1977) 750 Federal income tax (balance 1976) 460 Federal income tax (estimate 1977) 3,750 Required: 1. Prepare a schedule to compute the adjusted gross income of Harold and Bella Barton for 1977. Prepare a supporting schedule for Harold Barton’s in come from self-employment. Any possible alternative treatments should be resolved in a manner that will minimize income taxes for 1977. Show supporting com putations in good form.
2. Prepare a schedule to compute the total itemized deductions of Harold and Bella Barton for 1977. Any possible alternative treatments should be resolved in a manner that will minimize income taxes for 1977. Show supporting computations in good form.
Compensation for personal injuries sustained Repairs to automobile $2,900 700 $3,600 62 Accounting Practice — Part I Part b. For the calendar year 1977, Agnes Jones, a widow and the mother of a four-year-old son, reported the following types of income: Salary Interest income Dividend income (all qualifying dividends) 1976 state income tax refund $18,000 4,000 600 700 $23,300 An analysis of her cash expenses indicated the following: Doctor bills Drugs prescribed by doctors Medical insurance premiums Interest paid on installment purchases Real estate taxes paid (residence) Child care expenses (paid to the Happy Day Nursery in order for Agnes to be gainfully employed) $2,400 300 800 180 2,000 3,600 Additional information is as follows: • Interest income was comprised of the following amounts: Interest on savings deposits Interest on state obligations $2,200 1,800 $4,000 • Her W-2 form for 1977 stated that $790 was withheld for state income taxes.
• Mrs. Jones included a deduction of $900 for state income taxes in her 1976 itemized deductions.
Required: 1. Prepare a schedule to compute Agnes Jones’s adjusted gross income for 1977.
2. Prepare a schedule to compute Agnes Jones’s total itemized deductions for 1977. Prepare a supporting schedule for Agnes Jones’s deductible medical expenses.
3. Determine the amount of the child care credit that Agnes Jones can apply against her 1977 federal in come tax.
63 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all the states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXAMINATION IN ACCOUNTING PRACTICE — PART II November 2, 1978; 1:30 to 6:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1 No. 2 No. 3 No. 4 No 5 .... Total 40 50 40 50 50 60 40 50 50 60 220 270 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For in stance, if 12 pages are used for your answers they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3 . Enclose all scratch sheets. Failure to enclose scratch sheets may result in loss of grading points. Scratch sheets need not have page numbers, but you should show the question number and place them immediately following the question to which they relate.
4 . Fourteen-column sheets, if any, should not be folded until all sheets, both wide and narrow, are placed in the proper sequence and fastened together at the top left corner. All fourteen- column sheets should then be wrapped around the back of the papers.
5 . A CPA is continually confronted with the necessi ty of expressing opinions and conclusions in writ ten reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and applica tion of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
64 Accounting Practice—Part II Instructions Select the best answer for each of the following items relating to a variety of financial-account ing prob lems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 97. Gross billings for merchandise sold by Baker Company to its customers last year amounted to $5,260,000; sales returns and allowances reduced the amounts owed by $160,000. How much were net sales last year for Baker Company?
a. $4,800,000.
b. $5,100,000.
c. $5,200,000.
d. $5,260,000.
Answer Sheet 97. Items to be Answered 1. The stockholders’ equity section of Sola Corpo ration as of December 31, 1976, was as follows: Common stock, $20 par value, authorized 150,000 shares, issued and outstanding 100,000 shares $2,000,000 Capital in excess of par value 400,000 Retained earnings 200,000 $2,600,000 Number 1 (Estimated tim e------40 to 50 minutes) On March 1, 1977, Sola reacquired 10,000 shares for $240,000. The following transactions occurred in 1977 with respect to treasury stock acquired: June 1 — Sold 3,000 shares for $84,000. August 1 — Sold 2,000 shares for $42,000. September 1 —Retired remaining 5,000 shares.
Sola accounts for treasury stock on the cost method. As a result of these transactions a. Stockholders’ equity remained unchanged.
b. Common stock decreased $100,000 and re tained earnings decreased $14,000.
c. Common stock decreased $100,000 and capital in excess of par decreased $14,000.
d. Common stock decreased $126,000.
2. The Harris Corporation has an incentive com mission plan for its salesmen, entitling them to an ad ditional sales commission when actual quarterly sales exceed budgeted estimates. An analysis of the account “incentive commission expense’’ for the year ended December 31, 1977, follows: For Quarter Amount Ended Date Paid $40,000 December 31, 1976 January 23, 1977 28,000 March 31, 1977 April 24, 1977 15,000 June 30, 1977 July 19, 1977 23,000 September 30, 1977 October 22, 1977 The incentive commission for the quarter ended December 31, 1977, was $27,000. This amount was recorded and paid in January 1978. What amount should Harris record as incentive commission expense for 1977?
a. $133,000.
b. $106,000.
c. $93,000.
d. $66,000.
3. On January 1, 1977, Flip Corporation signed a ten-year noncancellable lease for certain machinery. The terms of the lease called for Flip to make annual payments of $30,000 for ten years with title to pass to Flip at the end of this period. The machinery has an estimated useful life of 15 years and no salvage value. Flip uses the straight-line method of depreciation for all of its fixed assets. Flip accordingly accounted for this lease transaction as an installment purchase of the machinery. The lease payments were determined to have a present value of $201,302 with an effective interest rate of 10%. With respect to this capitalized lease, Flip should record for 1977 a. Lease expense of $30,000.
b. Interest expense of $16,580 and depreciation expense of $13,420.
c. Interest expense of $20,130 and depreciation expense of $13,420.
d. Interest expense of $13,420 and depreciation expense of $16,580.
4. On January 10, 1977, Conroy Corporation ac quired 1,000 shares of Alva Corporation common stock at $70 per share as a short-term investment. On that date Alva had 100,000 shares issued and outstanding. On November 1, 1977, Alva declared and paid cash dividends of $2 per share on its outstanding common stock. On December 31, 1977, the market value of Alva’s common stock was $62 per share. At what value should Conroy report the investment in common stock of Alva on its December 31, 1977, balance sheet?
a. $60,000.
b. $62,000.
c. $68,000.
d. $70,000.
65 Examination Questions—November 1978 5. The Amlin Corporation was incorporated on January 1, 1977, with the following authorized capital ization: 20,000 shares of common stock, no par value, stated value $40 per share.
5,000 shares of 5% cumulative preferred stock, par value $10 per share.
During 1977 Amlin issued 12,000 shares of common stock for a total of $600,000 and 3,000 shares of preferred stock at $16 per share. In addition, on December 20, 1977, subscriptions for 1,000 shares of preferred stock were taken at a purchase price of $17. These subscribed shares were paid for on January 2, 1978. What should Amlin report as total contributed capital on its December 31, 1977, balance sheet?
a. $520,000.
b. $648,000.
c. $665,000.
d. $850,000.
6. On August 1 , 1978, Bamco Corporation purchased a new machine on a deferred payment basis. A down payment of $1,000 was made and 4 monthly install ments of $2,500 each are to be made beginning on September 1, 1978. The cash equivalent price of the machine was $9,500. Bamco incurred and paid instal lation costs amounting to $300. The amount to be capitalized as the cost of the machine is a. $9,500.
b. $9,800.
c. $11,000.
d. $11,300.
7. The owners of the Zoot Suit Clothing Store are contemplating selling the business to new interests. The cumulative earnings for the past 5 years amounted to $450,000 including extraordinary gains of $10,000. The annual earnings based on an average rate of return on investment for this industry would have been $76,000. If excess earnings are to be capitalized at 10%, then implied goodwill should be a. $120,000.
b. $140,000.
c. $440,000.
d. $450,000.
8. On July 1, 1976, Carol Corporation purchased factory equipment for $25,000. Salvage value was estimated to be $1,000. The equipment will be depreciated over ten years using the double-declining- balance method. Counting the year of acquisition as one-half year, Carol should record depreciation expense for 1977 on this equipment of a. $3,840.
b. $4,500.
c. $4,800.
d. $5,000.
9. The Gunther Company acquired a tract of land containing an extractable natural resource. Gunther is required by its purchase contract to restore the land to a condition suitable for recreational use after it has extracted the natural resource. Geological surveys estimate that the recoverable reserves will be 4,000,000 tons, and that the land will have a value of $1,000,000 after restoration. Relevant cost information follows: Land $9,000,000 Estimated restoration costs 1,200,000 If Gunther maintains no inventories of extracted ma terial, what should be the charge to depletion expense per ton of extracted material?
a. $2.00.
b. $2.25.
c. $2.30.
d. $2.55.
10. On February 1, 1977, Reflection Corporation pur chased a parcel of land as a factory site for $50,000. An old building on the property was demolished, and construction began on a new building which was com pleted on November 1, 1977. Costs incurred during this period are listed below: Demolition of old building Architect's fees Legal fees for title investigation and purchase contract Construction costs $ 4,000 10,000 2,000 500,000 (Salvaged materials resulting from demolition were sold for $1,000.) Reflection should record the cost of the land and new building respectively as a. $52,000 and $513,000.
b. $53,000 and $512,000.
c. $53,000 and $510,000.
d. $55,000 and $510,000.
11. On January 1, 1976, Kent Corporation purchased a machine for $50,000. Kent paid shipping expenses of $500 as well as installation costs of $1,200. The machine was estimated to have a useful life of ten years and an estimated salvage value of $3,000. In January 1977 additions costing $3,600 were made to the machine in order to comply with pollution control ordinances. These additions neither prolonged the life of the machine nor did they have any salvage value. If Kent records depreciation under the straight-line method, depreciation expense for 1977 is a. $4,870.
b. $5,170.
c. $5,270.
d. $5,570.
66 Accounting Practice—Part II 12. The following data were available from the records of Johlson’s Department Store for the year ended December 31, 1977: At Cost A t Retail Merchandise inventory, January 1 , 1977 Purchases Markups Markdowns Sales $ 90,000 330,000 $130,000 460,000 10,000 40,000 480,000 Using the retail method, an estimate of the merchan dise inventory at December 31, 1977, valued at the lower of cost or market, would be a. $56,000.
b. $60,000.
c. $80,000.
d. $110,000.
13. Buildit Construction Corporation contracted to construct a building for $400,000. Construction began in 1976 and was completed in 1978. Data relating to the contract are summarized below: Year ended December 31, Costs incurred Estimated costs to complete 1976 $200,000 100,000 1977 $110,000 Buildit uses the percentage-of-completion method as the basis for income recognition. For the years ended December 31, 1976, and 1977, respectively, Buildit should report income of a. $0 and $90,000.
b. $45,000 and $45,000.
c. $66,667 and $23,333.
d. $90,000 and $0.
Items 14 and 15 are based on the following in formation: The Johnson Corporation adopted a pension plan in 1976 on a funded, noncontributory basis. Johnson elected to amortize past service costs over twelve years and to fund past service costs over ten years. Normal costs are to be funded as incurred each year. The following schedule reflects both amortization of the past service cost and funding for the years 1976 and 1977.
1976 1977 12-year accrual $100,000 $100,000 Reduction for interest — 835 Past service pension cost 100,000 99,165 10-year funding 113,909 113,909 Balance sheet—deferred charge: Balance 13,909 28,653 Increase 13,909 14,744 14. If normal cost for 1976 was $70,000, Johnson should record pension expense in 1976 of a. $70,000.
b. $113,909.
c. $170,000.
d. $183,909.
15. If normal cost in 1977 was $75,000, the entry that Johnson should make in 1977 to record pension expense and funding is Debit Credit a.
b.
c.
d.
Pension expense $100,000 Deferred charge — funding in excess of costs 13,909 Cash $113,909 Pension expense (normal cost) $ 75,000 Pension expense (past service) 99,165 Deferred charge — funding in excess of costs 14,744 Cash $188,909 Pension expense $175,000 Deferred charge — funding in excess of costs 13,909 Cash $188,909 Pension expense (normal cost) $ 75,000 Pension expense (past service) 100,000 Deferred charge — funding in excess of costs 14,744 Cash $189,744 67 Examination Questions— November 1978 16. The Avec Company’s account balances at December 31, 1977, for accounts receivable and the related allowance for uncollectible accounts were $800,000 and $40,000 respectively. An aging of ac counts receivable indicated that $71,100 of the December 31 receivables may be uncollectible. The net realizable value of accounts receivable was a. $688,900.
b. $728,900.
c. $760,000.
d. $768,900.
18. In the preparation of a revised comparative state ment of income, Worth should report under the caption “Discontinued Operations’’ for 1977 and 1976, respec tively a. A loss of $320,000 and a loss of $250,000.
b. A loss of $160,000 and a loss of $125,000.
c. Income of $130,000 and a loss of $250,000.
d. Income of $65,000 and a loss of $125,000.
Items 17 and 18 are based on the following in formation: The following condensed statement of income of Worth Corporation, a diversified company, is presented for the two years ended December 31, 1977, and 1976: Net sales Cost of sales Gross profit Operating expenses Operating income Gain on sale of division Income before income taxes Provision for income taxes Net income 1977 1976 $5,000,000 $4,800,000 3,100,000 3,000,000 1,900,000 1,800,000 1,100,000 1,200,000 800,000 600,000 450,000 — 1,250,000 600,000 625,000 300,000 $ 625,000 $ 300,000 On January 1, 1977, Worth entered into an agree ment to sell for $1,600,000 the assets and product line of one of its separate operating divisions. The sale was consummated on December 31, 1977, and resulted in a pre tax gain on disposition of $450,000. This divi sion’s contribution to Worth’s reported operating in come before income taxes for each year was as follows: 1977 1976 $(320,000) loss $(250,000) loss Number 2 (Estimated time----- 40 to 50 minutes) Instructions Select the best answer for each of the following items relating to the federal income taxation of corpora tions and partnerships. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. The answers should be selected in accordance with the cur rent Internal Revenue Code and Tax Regulations. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
Items to be Answered 19. The Tempest Corporation, not a dealer in se curities, had accumulated earnings and profits of $75,000 at the beginning of 1977. The earnings and profits for 1977 were $25,000. On October 15, 1977, Tempest distributed to its shareholders as a dividend, marketable securities having a fair market value of $12,000. The securities had cost $7,000. As a result of the distribution, accumulated earnings and profits were a. Increased by $5,000.
b. Decreased by $5,000.
c. Decreased by $7,000.
d. Decreased by $12,000.
20. In 1977, its first year of operation, the Champion Corporation, not a dealer in securities, realized taxable income of $64,000 from the operation of its business. In addition to its regular business operations, it real ized the following gains and losses from the sale of marketable securities: Assume an income tax rale of 50%.
17. In the preparation of a revised comparative state ment of income, Worth should report income from continuing operations (after income taxes) for 1977 and 1976, respectively, amounting to a. $560,000 and $300,000.
b. $560,000 and $425,000.
c. $625,000 and $300,000.
d. $625,000 and $425,000.
Short-term capital gain Short-term capital loss Long-term capital gain Long-term capital loss $ 5,000 ( 2,000) 6,000 (16,000) What is Champion’s total taxable income for 1977?
a. $57,000.
b. $62,000.
c. $64,000.
d. $75,000.
68 Accounting Practice—Part II 21. The Guardian Corporation, not a dealer in secu rities, owned marketable securities that had an ad justed basis of $150,000. On June 30, 1977, when the market value of these securities rose to $180,000, Guardian distributed these securities to its share holders. As a result of this distribution, Guardian should report a. No gain on the distribution.
b. Ordinary income of $30,000.
c. A capital gain of $30,000.
d. A capital gain of $180,000.
22. In 1977 Jonna Corporation sold some excess property for $2,500 to John Moll, an individual and one of its minority shareholders. On the date of sale the property had a fair market value of $6,000 and an ad justed basis of $4,000. Jonna must report a property distribution of a. $1,500.
b. $2,000.
c. $2,500.
d. $3,500.
23. In January 1977 Marlin and Louis formed a part nership with each contributing $75,000 cash. The partnership agreement provided that Martin would receive a guaranteed salary of $20,000 and that part nership profits and losses (computed after deducting Martin’s salary) would be shared equally. For the year ended December 31, 1977, the partnership’s operations resulted in a loss of $18,000 after payment of Martin’s salary. The partnership had no outstand ing liabilities as of December 31, 1977. What is the amount of Martin’s partnership basis as of December 31, 1977?
a. $46,000.
b. $66,000.
c. $76,000.
d. $86,000.
24. Logo Corporation’s net income per books was $150,000 for the year 1977 after recording amortiza tion of organization costs. Organization costs of $70,000 incurred at the organization date two years earlier are being written off over a ten-year period on the financial statements, and over the minimum period for income tax purposes. Assuming that there were no other recon ciling items, what is Logo’s taxable income for 1977?
a. $136,000.
b. $143,000.
c. $150,000.
d. $157,000.
25. For its tax year ended December 31, 1977, Kind Corporation had taxable income of $90,000 before deducting contributions to charitable organizations. Kind had a contribution carryover of $3,000 from 1976 and actually made contributions of $2,000 during 1977. Assuming that there were no special deductions, what can Kind deduct as contributions for 1977?
a. $4,250.
b. $4,400.
c. $4,500.
d. $5,000.
26. The partnership of Luchen and Kup reported the following elements of net income for the year ended December 31, 1977: Operating income Long-term capital gains Dividends received on investments in domestic corporations Net income $150,000 4,000 2,000 $156,000 Included in the determination of operating in come were charitable contributions of $1,000 and de ductible guaranteed payments to partners of $60,000. What is the ordinary income of the partnership for 1977?
a. $151,000.
b. $155,000.
c. $211,000.
d. $212,000.
27. In order for a corporation to elect subchapter S status, the corporation should a. Have only individuals or corporations as its shareholders.
b. Not have more than one class of stock.
c. Have no more than twenty shareholders.
d. Require that its shareholders adopt the same taxable year as itself.
28. In 1977, its first year of operation, Commerce Corporation had a gross profit from operations of $360,000 and deductions of $500,000 excluding any special deductions. Commerce also received dividends of $100,000 from unaffiliated domestic corporations. What is the net operating loss, for 1977?
a. $34,000.
b. $40,000.
c. $55,000.
d. $125,000.
69 Examination Questions— November 1978 29. The Taggart Corporation, a calendar-year com pany, has operated as a subchapter S corporation for the past two years. At December 31, 1976, Taggart had undistributed taxable income of $25,000. For the year ended December 31, 1977, it had taxable income and current earnings and profits of $130,000. During 1977 Taggart made four distributions in cash to its ten equal stockholders who are also on a calendar-year basis. The cash distributions to stockholders were as follows: January 8 , 1977 March 10, 1977 June 15, 1977 November 1 , 1977 $ 12,000 18,000 35,000 40,000 $105,000 For the calendar year 1977, what amount should be included in each stockholder’s gross income from Taggart?
a. $7,500.
b. $10,500.
c. $13,000.
d. None of the above.
30. Edward owns a 70% interest in the capital and profits of the partnership of Edward and Moore. Dur ing 1977 Edward purchased a piece of surplus machinery from the partnership for $5,000. On the date of sale the machinery had an adjusted basis to the part nership of $8,000. For the year ended December 31, 1977, the partnership’s net income was $50,000 after recording the loss on sale of machinery. Assuming that there were no other partnership items to be specially reported, what is Edward’s distributive share of the partnership’s taxable income for 1977?
a. $35,000.
b. $35,630.
c. $36,470.
d. $37,100.
31. During 1977 Simon Manufacturing Corporation, in need of additional factory space, exchanged 5,000 shares of its common stock with a par value of $50,000 for a building with a fair market value of $60,000. On the date of the exchange the stock had a market value of $65,000. For 1977, how much and what type of gain or loss should Simon report on this transaction?
a. No gain or loss.
b. $5,000 capital loss.
c. $10,000 capital gain.
d. $10,000 section 1231 gain.
32. For the year ended December 31, 1977, the Lewis Corporation reported net income per books of $250,000. The following items were included in the determination of net income: State corporate income tax refunds $ 4,000 Interest income on tax-exempt municipal securities 15,000 Loss on sale of land acquired in 1975 for investment 20,000 Interest expense on loan to purchase tax-exempt municipal securities 8,000 Provision for federal income taxes 200,000 What is the taxable income of Lewis for 1977?
a. $270,000.
b. $450,000.
c. $459,000.
d. $463,000.
33. The accumulated earnings tax imposed on corpo rations for improper accumulation of earnings in ex cess of reasonable needs does not apply to a. Closely-held corporations.
b. Widely-held corporations.
c. Personal holding companies.
d. None of the above.
34. On December 31, 1977, Topic Corporation sold machinery for $24,000. The machinery which had been purchased on January 1, 1973, for $20,000 had an ad justed basis of $14,000 on the date of sale. For 1977 Topic should report a. Ordinary income of $10,000.
b. A section 1231 gain of $10,000.
c. A section 1231 gain of $6,000 and ordinary income of $4,000.
d. A section 1231 gain of $4,000 and ordinary income of $6,000.
35. On January 1, 1977, Dodge Corporation acquired machinery for $30,000. The machinery had an esti mated useful life of ten years and salvage value was estimated at $3,000. If Dodge uses the straight-line method of depreciation, the maximum amount of de preciation that Dodge can deduct for 1977 is a. $4,500.
b. $4,800.
c. $8,100.
d. $8,400.
70 Accounting Practice—Part II Number 3 (Estimated time----- 50 to 60 minutes) The Noble Corporation is in the process of negoti ating a loan for expansion purposes. The books and records have never been audited and the bank has requested that an audit be performed. Noble has pre pared the following comparative financial statements for the years ended December 31, 1977, and 1976: BALANCE SHEET As of December 31, 1977 and 1976 Assets Current Assets Cash Accounts receivable Allowance for uncollectible accounts Marketable securities, at cost Merchandise inventory Total current assets Fixed Assets Property, plant and equipment Accumulated depreciation Total fixed assets Total assets Liabilities and Stockholders’ Equity Liabilities Accounts payable Stockholders’ equity Common stock, par value $10, authorized 50,000 shares, issued and out standing 20,000 shares Retained earnings Total stockholders’ equity Total liabilities and stock holders’ equity 1977 1976 $ 163,000 $ 82,000 392,000 296,000 (37,000) (18,000) 78,000 78,000 207,000 202,000 803,000 640,000 167,000 169,500 (121,600) (106,400) 45,400 63,100 $ 848,400 $ 703,100 $ 121,400 $ 196,100 260,000 260,000 467,000 247,000 727,000 507,000 $ 848,400 $ 703,100 STATEMENT OF INCOME For the Years Ended December 31, 1977 and 1976 1977 1976 Sales $1,000,000 $900,000 Cost of sales 430,000 395,000 Gross profit 570,000 505,000 Operating expenses 210,000 205,000 Administrative expenses 140,000 105,000 350,000 310,000 Net income $ 220,000 $195,000 During the course of the audit, the following ad ditional facts were determined: 1. An analysis of collections and losses on ac counts receivable during the past two years indicates a drop in anticipated losses due to bad debts. After consultation with management it was agreed that the loss experience rate on sales should be reduced from the recorded 2% to 1%, beginning with the year ended December 31, 1977.
2. An analysis of marketable securities revealed that this investment portfolio consisted entirely of short-term investments in marketable equity securities that were acquired in 1976. The total market valuation for these investments as of the end of each year was as follows: December 31, 1976 $81,000 December 31, 1977 $62,000 3. The merchandise inventory at December 31, 1976, was overstated by $4,000 and the merchandise in ventory at December 31, 1977, was overstated by $6, 100.
4. On January 2, 1976, equipment costing $12,000 (estimated useful life of ten years and residual value of $1,000) was incorrectly charged to operating expenses. Noble records depreciation on the straight-line method. In 1977 fully depreciated equipment (with no residual value) that ori ginally cost $17,500 was sold as scrap for $2,500. Noble credited the proceeds of $2,500 to property and equipment.
5. An analysis of 1976 operating expenses revealed that Noble charged to expense a three-year insurance premium of $2,700 on January 15, 1976.
Required: a. Prepare the journal entries to correct the books at December 31, 1977. The books for 1977 have not been closed. Ignore income taxes.
b. Prepare a schedule showing the computation of corrected net income for the years ended December 31, 1977 and 1976, assuming that any adjustments are to be reported on comparative statements for the two years. The first items on your schedule should be the net income for each year. Ignore income taxes. (Do not prepare financial statements.) 71 Examination Questions— November 1978 Number 4 consists of two unrelated parts.
Part a. The Scarborough Corporation manufac tures and sells two products, Thingone and Thingtwo. In July 1977, Scarborough’s budget department gathered the following data in order to project sales and budget requirements for 1978.
Number 4 (Estimated tim e------40 to 50 minutes) 1978 Projected Sales: Product Thingone Thingtwo Units 60,000 40,000 1978 Inventories - in units: Expected Price 70 $100 Desired Product January I, 1978 December 31, 1978 Thingone Thingtwo 20,000 8,000 25,000 9,000 In order to produce one unit of Thingone and Thingtwo, the following raw materials are used: Raw Material Unit Amount used per unit Thingone Thingtwo A lbs.
4 5 B lbs.
2 3 C each 1 Projected data for 1978 with respect to raw materials is as follows: Expected Desired Anticipated Inventories Inventories Raw Purchase January 1, December 31, Material Price 1978 1978 A $8 32,000 lbs.
36,000 lbs.
B $5 29,000 lbs.
32,000 lbs.
C $3 6,000 each 7,000 each Projected direct labor requirements for 1978 and rates are as follows: Product Hours per unit Rate per hour Required: Based upon the above projections and budget re quirements for 1978 for Thingone and Thingtwo, prepare the following budgets for 1978: 1. Sales budget (in dollars) 2. Production budget (in units) 3. Raw materials purchase budget (in quantities) 4. Raw materials purchase budget (in dollars) 5. Direct labor budget (in dollars) 6. Budgeted finished goods inventory at December 31, 1978 (in dollars) Part b. The Felix Manufacturing Company uses a process-cost system to account for the costs of its only product known as “ Nino.” Production begins in the fabrication department where units of raw material are molded into various connecting parts. After fabrication is complete, the units are transferred to the assembly department. There is no material added in the assembly department. After assembly is complete, the units are transferred to the packaging department where the units are packaged for shipment. At the completion of this process the units are complete and they are transferred to the shipping department.
At year end, December 31, 1977, the following inven tory of “ Nino’s” is on hand: • No unused raw material or packing material.
• Fabrication department: 6,000 units, 25% complete as to raw material and 40% complete as to direct labor.
• Assembly department: 10,000 units, 75% com plete as to direct labor.
• Packaging department: 3,000 units, 60% com plete as to packing material and 75% complete as to direct labor.
• Shipping department: 8,000 units.
Required: Prepare in proper form schedules showing the follow ing at December 31, 1977: 1. The number of equivalent units of raw material in all inventories.
2. The number of equivalent units of fabrication department direct labor in all inventories.
3. Number of equivalent units of packaging department material and direct labor in the packaging department inventory.
Thingone Thingtwo $3 $4 Overhead is applied at the rate of $2 per direct labor hour.
72 Accounting Practice—Part II Presented below is the current funds balance sheet of Burnsville University as of the end of its fiscal year ended June 30, 1977 Burnsville University CURRENT FUNDS BALANCE SHEET June 30, 1977 Number 5 (Estimated tim e------50 to 60 minutes) Assets Liabilities and Fund Balances Current Funds: Unrestricted: Cash Accounts receivable- student tuition and fees, less allowance for doubtful ac counts of $9,000 State appropriations receivable $210,000 341,000 75,000 $626,000 Current Funds: Unrestricted: Accounts payable Deferred revenues Fund balances $ 45,000 66,000 515,000 $626,000 Restricted: Cash Investments 7,000 60,000 67,000 Restricted: Fund Balances 67,000 Total current funds $693,000 Total current funds $693,000 The following transactions presented below oc curred during the fiscal year ended June 30, 1978: 1. On July 7, 1977, a gift of $100,000 was re ceived from an alumnus. The alumnus requested that one half of the gift be used for the purchase of books for the university library and the remainder be used for the establishment of a scholarship fund. The alumnus further requested that the income generated by the scholarship fund be used annually to award a scholar ship to a qualified disadvantaged student. On July 20, 1977, the board of trustees resolved that the funds of the newly established scholarship fund would be invested in savings certificates. On July 21, 1977, the savings cer tificates were purchased.
2. Revenue from student tuition and fees ap plicable to the year ended June 30, 1978, amounted to $1,900,000. Of this amount, $66,000 was collected in the prior year and $1,686,000 was collected during the year ended June 30, 1978. In addition, at June 30, 1978, the university had received cash of $158,000 represent ing fees for the session beginning July 1, 1978.
3. During the year ended June 30, 1978, the university had collected $349,000 of the outstanding ac counts receivable at the beginning of the year. The balance was determined to be uncollectible and was written off against the allowance account. At June 30, 1978, the allowance account was increased by $3,000.
4. During the year interest charges of $6,000 were earned and collected on late student fee payments.
5. During the year the state appropriation was received. An additional unrestricted appropriation of $50,000 was made by the state, but had not been paid to the university as of June 30, 1978.
6. An unrestricted gift of $25,000 cash was re ceived from alumni of the university.
7. During the year investments of $21,000 were sold for $26,000. Investment income amounting to $1,900 was received.
8. During the year unrestricted operating ex penses of $1,777,000 were recorded. At June 30, 1978, $59,000 of these expenses remained unpaid.
9. Restricted current funds of $13,000 were spent for authorized purposes during the year.
10. The accounts payable at June 30, 1977, were paid during the year.
11. During the year, $7,000 interest was earned and received on the savings certificates purchased in ac cordance with the board of trustees resolution, as discussed in item 1.
Required: a. Prepare journal entries to record in summary the above transactions for the year ended June 30, 1978. Each journal entry should be numbered to correspond with the transaction described above.
Your answer sheet should be organized as follows: Current Funds Unrestricted Restricted Endowment Fund Accounts Dr.
Cr.
Dr. Cr. Dr.
b. Prepare a statement of changes balances for the year ended June 30, 1978.
Cr.
in fund 73 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining bonds of all the states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXAMINATION IN AUDITING November 2, 1978; 8:30 A.M. to 12:00 M NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: No. 1 .....................
No. 2 .....................
No. 3 .....................
No. 4 .....................
No. 5 .....................
Total........... .
Estimated Minutes Minimum Maximum 90 110 15 25 15 25 15 25 15 25 150 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For in stance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessi ty of expressing opinions and conclusions in writ ten reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and applica tion of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
74 Auditing Instructions Select the best answer for each of the following items. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 96. One of the generally accepted auditing standards specifies that the auditor a. Inspect all fixed assets acquired during the year.
b. Charge fair fees based on cost.
c. Make a proper study and evaluation of the existing internal control.
d. Count client petty-cash funds.
Answer Sheet 96. Items to be Answered 1. Morgan, CPA, is the principal auditor for a multinational corporation. Another CPA has examined and reported on the financial statements of a signifi cant subsidiary of the corporation. Morgan is satisfied with the independence and professional reputation of the other auditor, as well as the quality of the other auditor’s examination. With respect to Morgan’s report on the financial statements, taken as a whole, Morgan a. Must not refer to the examination of the other auditor.
b. Must refer to the examination of the other auditor.
c. May refer to the examination of the other auditor.
d. May refer to the examination of the other auditor, in which case Morgan must include in the auditor’s report on the consolidated financial statements, a qualified opinion with respect to the examination of the other auditor.
2. Which of the following is an effective internal accounting control over cash payments?
a. Signed checks should be mailed under the supervision of the check signer.
b. Spoiled checks which have been voided should be disposed of immediately.
c. Checks should be prepared only by persons responsible for cash receipts and cash dis bursements.
d. A check-signing machine with two signatures should be utilized.
Number 1 (Estimated tim e------90 to 110 minutes) 3. When reporting on comparative financial state ments the auditor’s report should include a consistency exception when the financial statements of the current year include a. A change in the format and presentation of the statement of changes in financial position from a cash concept to a working capital concept.
b. A change in the estimated salvage value of several assets.
c. A change in the income statement classifica tion of an expense account from “Other Expenses’’ to “ Selling Expenses.’’ d. A change in the inventory pricing to correct a mathematical error in the total value of the opening inventory.
4. When considering internal control, an auditor must be aware of the concept of reasonable assurance which recognizes that a. The employment of competent personnel pro vides assurance that the objectives of internal control will be achieved.
b. The establishment and maintenance of a system of internal control is an important responsibility of the management and not of the auditor.
c. The cost of internal control should not exceed the benefits expected to be derived from internal control.
d. The segregation of incompatible functions is necessary to obtain assurance that the in ternal control is effective.
5. Nonaccounting data included in a long-form report have been subjected to auditing procedures. The auditor’s report should state this fact and should explain that the nonaccounting data are presented for analysis purposes. In addition, the auditor’s report should state whether the nonaccounting data are a. Beyond the scope of the normal engagement and therefore, not covered by the opinion on the financial statements.
b. Within the framework of generally accepted auditing standards, which apply to the finan cial statements, taken as a whole.
c. Audited, unaudited, or reviewed on a limited basis.
d. Fairly stated in all material respects in rela tion to the basic financial statements, taken as a whole.
6. Which of the following employees normally would be assigned the operating responsibility for designing an electronic data processing installation, including flowcharts of data processing routines?
a. Computer programmer.
b. Data processing manager.
c. Systems analyst.
d. Internal auditor.
75 Examination Questions— November 1978 7. In pursuing its quality control objectives with respect to assigning personnel to engagements, a firm of independent auditors may use policies and pro cedures such as a. Designating senior qualified personnel to provide advice on accounting or auditing questions throughout the engagement.
b. Requiring timely identification of the staffing requirements of specific engagements so that enough qualified personnel can be made available.
c. Establishing at entry levels a policy for re cruiting that includes minimum standards of academic preparation and accomplishment.
d. Requiring auditing personnel to have current accounting and auditing literature available for research and reference purposes through out the engagement.
8. An auditor should be familiar with a client’s electronic data processing hardware and software. An important element of the client’s software is the program. Another element of software is the a. Cathode ray tube (CRT).
b. Central processing unit (CPU).
c. Magnetic tape drive.
d. Compiler.
9. When a client declines to make essential dis closures in the financial statements or in the footnotes, the independent auditor should a. Provide the necessary disclosures in the auditor’s report and appropriately modify the opinion.
b. Explain to the client that an adverse opinion must be issued.
c. Issue an unqualified report and inform the stockholders of the improper disclosure in an “ unaudited” footnote.
d. Issue an opinion “subject to” the client’s lack of disclosure of supplementary informa tion as explained in a middle paragraph of the report.
10. Of the following procedures which is the most important that an auditor should use when making an overall, review of the income statement?
a. Select sales and expense items and trace amounts to related supporting documents.
b. Compare actual revenues and expenses with the corresponding figures of the previous year and investigate significant differences.
c. Obtain from the proper client representa tives, inventory certificates for the beginning and ending inventory amounts that were used to determine cost of sales.
d. Ascertain that the net income amount in the statement of changes in financial position agrees with the net income amount in the income statement.
11. When an on-line, real-time (OLRT) electronic data processing system is in use, internal control can be strengthened by a. Providing for the separation of duties be tween keypunching and error listing opera tions.
b. Attaching plastic file protection rings to reels of magnetic tape before new data can be entered on the file.
c. Preparing batch totals to provide assurance that file updates are made for the entire input.
d. Making a validity check of an identification number before a user can obtain access to the computer files.
12. A client’s procurement system ends with the assumption of a liability and the eventual payment of the liability. Which of the following best describes the auditor’s primary concern with respect to liabilities resulting from the procurement system?
a. Accounts payable are not materially under stated.
b. Authority to incur liabilities is restricted to one designated person.
c. Acquisition of materials is not made from one vendor or one group of vendors.
d. Commitments for all purchases are made only after established competitive bidding pro cedures are followed.
13. A CPA has a financial interest in a corporation and is associated with that corporation’s unaudited financial statements. Under such circumstances the CPA’s report should state that the CPA is not indepen dent with respect to the corporation and should include a. A statement that the financial statements were unaudited and accordingly the CPA does not express an opinion on the financial statements.
b. A description of the reasons for the CPA’s lack of independence and a disclaimer of opinion on the financial statements.
c. A statement that each page of the financial statements is “ unaudited” and a qualified opinion on the financial statements.
d. A description of the reasons for the CPA’s lack of independence and a qualified opinion on the financial statements.
14. Use of the ratio estimation sampling technique to estimated dollar amounts is Inappropriate when a. The total book value is known and corre sponds to the sum of all the individual book values.
b. A book value for each sample item is un known.
c. There are some observed differences between audited values and book values.
d. The audited values are nearly proportional to the book value.
76 Auditing 15. When a principal auditor decides to make refer ence to the examination of another auditor, the prin cipal auditor’s report should clearly indicate the a. Principal auditor’s qualification on the over all fairness of the financial statements, taken as a whole, “subject to’’ the work and report of the other auditor.
b. Procedures that were performed by the other auditor in connection with the other auditor’s examination.
c. Division of responsibility between that por tion of the financial statements covered by the examination of the principal auditor and that covered by the examination of the other auditor.
d. Procedures that were performed by the prin cipal auditor to obtain satisfaction as to the reasonableness of the examination of the other auditor.
16. When the report of a principal auditor makes reference to the examination made by another auditor, the other auditor may be named if express permission to do so is given and a. The report of the principal auditor names the other auditor in both the scope and opinion paragraphs.
b. The principal auditor accepts responsibility for the work of the other auditor.
c. The report of the other auditor is presented together with the report of the principal auditor.
d. The other auditor is not an associate or cor respondent firm whose work is done at the request of the principal auditor.
17. Which of the following is the best way for an auditor to determine that every name on a company’s payroll is that of a bona fide employee presently on the job?
a. Examine personnel records for accuracy and completeness.
b. Examine employees’ names listed on pay roll tax returns for agreement with payroll accounting records.
c. Make a surprise observation of the company’s regular distribution of paychecks.
d. Visit the working areas and confirm with employees their badge or identification numbers.
18. Governmental auditing often extends beyond examinations leading to the expression of opinion on the fairness of financial presentation and includes audits of efficiency, economy, effectiveness, and also a. Accuracy.
b. Evaluation.
c. Compliance.
d. Internal control.
19. Confirmation of individual accounts receivable balances directly with debtors will, of itself, normally provide evidence concerning the a. Collectibility of the balances confirmed.
b. Ownership of the balances confirmed.
c. Validity of the balances confirmed.
d. Internal control over balances confirmed.
20. Which of the following best describes the objec tive of the fourth standard of reporting, which requires that the auditor’s report shall either contain an ex pression of opinion regarding the financial statements, taken as a whole, or an assertion to the effect that an opinion cannot be expressed?
a. To protect the auditor against allegations that some portion of the financial statements includes a material misstatement.
b. To prevent misinterpretation of the degree of responsibility the auditor is assuming when the auditor’s name is associated with the financial statements.
c. To prevent the reader from assuming that an auditor will detect errors or irregularities that have a material effect on the financial statements.
d. To protect auditors who are not associated with financial statements and who do not lend their names to the financial statements.
21. One example of a “ special report,’’ as defined by Statements on Auditing Standards, is a report issued in connection with a. A feasibility study.
b. A limited review of interim financial informa tion.
c. Price-level basis financial statements.
d. Compliance with a contractual agreement not related to the financial statements.
22. A CPA reviews a client’s payroll procedures. The CPA would consider internal control to be less than effective if a payroll department supervisor was as signed the responsibility for a. Distributing payroll checks to employees.
b. Reviewing and approving time reports for subordinate employees.
c. Hiring subordinate employees.
d. Initiating requests for salary adjustments for subordinate employees.
23. Which of the following best describes the refer ence to the expression “taken as a whole’’ in the fourth generally accepted auditing standard of reporting?
a. It applies equally to a complete set of finan cial statements and to each individual finan cial statement.
b. It applies only to a complete set of financial statements.
c. It applies equally to each item in each finan cial statement.
d. It applies equally to each material item in each financial statement.
77 Examination Questions— November 1978 24. In connection with the audit of a current issue of long-term bonds payable, the auditor should a. Determine whether bondholders are persons other than owners, directors, or officers of the company issuing the bond.
b. Calculate the effective interest rate to see if it is substantially the same as the rates for similar issues.
c. Decide whether the bond issue was made without violating state or local law.
d. Ascertain that the client has obtained the opinion of counsel on the legality of the issue.
25. It would be appropriate for the payroll accounting department to be responsible for which of the following functions?
a. Approval of employee time records.
b. Maintenance of records of employment, dis charges, and pay increases.
c. Preparation of periodic governmental reports as to employees’ earnings and withholding taxes.
d. Temporary retention of unclaimed employee paychecks.
26. In connection with the third generally accepted auditing standard of field work, an auditor examines corroborating evidential matter which includes all of the following except a. Client accounting manuals.
b. Written client representations.
c. Vendor invoices.
d. Minutes of board meetings.
27. Below are the names of four CPA firms and pertinent facts relating to each firm. Unless otherwise indicated, the individuals named are CPAs and part ners, and there are no other partners. Which firm name and related facts indicates a violation of the AICPA Code of Professional Ethics?
a. Arthur, Barry, and Clark, CPAs (Clark died about five years ago; Arthur and Barry are continuing the firm).
b. Dave and Edwards, CPAs (The name of Fredricks, CPA, a third active partner, is omitted from the firm name).
c. Jones & Co., CPAs, P.C. (The firm is a pro fessional corporation and has ten other stock holders who are all CPAs).
d. George and Howard. CPAs (Howard died three years ago; George is continuing the firm as a sole proprietorship).
28. The primary responsibility for the adequacy of disclosure in the financial statements of a publicly held company rests with the a. Partner assigned to the audit engagement.
b. Management of the company.
c. Auditor in charge of the field work.
d. Securities and Exchange Commission.
29. Subsequent events affecting the realization of assets ordinarily will require adjustment of the financial statements under examination because such events typically represent a. The culmination of conditions that existed at the balance sheet date.
b. The final estimates of losses relating to casu alties occurring in the subsequent events period.
c. The discovery of new conditions occurring in the subsequent events period.
d. The preliminary estimate of losses relating to new events that occurred subsequent to the balance sheet date.
30. In which of the following circumstances would a CPA be bound by ethics to refrain from disclosing any confidential information obtained during the course of a professional engagement?
a. The CPA is issued a summons enforceable by a court order which orders the CPA to present confidential information.
b. A major stockholder of a client company seeks accounting information from the CPA after management declined to disclose the requested information.
c. Confidential client information is made avail able as part of a quality review of the CPA’s practice by a review team authorized by the AICPA.
d. An inquiry by a disciplinary body of a state CPA society requests confidential client in formation.
31. The auditor can best verify a client’s bond sinking fund transactions and year-end balance by a. Recomputation of interest expense, interest payable, and amortization of bond discount or premium.
b. Confirmation with individual holders of retired bonds.
c. Confirmation with the bond trustee.
d. Examination and count of the bonds retired during the year.
32. Limitation on the scope of the auditor’s examina tion may require the auditor to issue a qualified opinion or to disclaim an opinion. Which of the following would generally be a limitation on the scope of the auditor’s examination?
a. The unavailability of sufficient competent evidential matter.
b. The engagement of the auditor to report on only one basic financial statement.
c. The examination of a subsidiary’s financial statements by an auditor other than the one who examines and reports on the consoli dated financial statements.
d. The engagement of the auditor after year end.
78 Auditing 33. Which of the following publications does not qualify as a statement of generally accepted account ing principles under the AICPA Code of Professional Ethics?
a. Accounting interpretations issued by the FASB.
b. Accounting interpretations issued by the AICPA.
c. AICPA Accounting Research Bulletins.
d. Statements of Financial Standards issued by the FASB.
34. Which of the following is an internal accounting control weakness related to factory equipment?
a. A policy exists requiring all purchases of factory equipment to be made by the depart ment in need of the equipment.
b. Checks issued in payment of purchases of equipment are not signed by the controller.
c. Factory equipment replacements are gener ally made when estimated useful lives, as in dicated in depreciation schedules, have ex pired.
d. Proceeds from sales of fully depreciated equipment are credited to other income.
35. Which of the following is a basic tool used by the auditor to control the audit work and review the progress of the audit?
a. Time and expense summary.
b. Engagement letter.
c. Progress flowchart.
d. Audit program.
36. With respect to a small company’s system of purchasing supplies, an auditor’s primary concern should be to obtain satisfaction that supplies ordered and paid for have been a. Requested by and approved by authorized individuals who have no incompatible duties.
b. Received, counted, and checked to quan tities and amounts on purchase orders and invoices.
c. Properly recorded as assets and systemat ically amortized over the estimated useful life of the supplies.
d. Used in the course of business and solely for business purposes during the year under audit.
37. When erroneous data are detected by computer program controls, such data may be excluded from processing and printed on an error report. The error report should most probably be reviewed and followed up by the a. Supervisor of computer operations.
b. Systems analyst.
c. EDP control group.
d. Computer programmer.
38. An auditor must obtain written client representa tions that normally should be signed by a. The president and the chairperson of the board.
b. The treasurer and the internal auditor.
c. The chief executive officer and the chief financial officer.
d. The corporate counsel and the audit com mittee chairperson.
39. If, during the course of an annual audit of a publicly held manufacturing company, an independent auditor becomes aware of a material weakness in the company’s internal accounting control, the auditor is required to communicate the material weakness to a. The senior management and the board of directors of the company.
b. The senior management of the company.
c. The board of directors of the company.
d. The audit committee of the board of directors.
40. The AICPA Code of Professional Ethics would be violated if a CPA accepted a fee for services and the fee was a. Fixed by a public authority.
b. Based on a price quotation submitted in competitive bidding.
c. Determined, based on the results of judicial proceedings.
d. Payable after a specified finding was at tained.
41. Which of the following is the best audit procedure for determining the existence of unrecorded liabilities?
a. Examine confirmation requests returned by creditors whose accounts appear on a sub sidiary trial balance of accounts payable.
b. Examine a sample of cash disbursements in the period subsequent to year end.
c. Examine a sample of invoices a few days prior to and subsequent to year end to as certain whether they have been properly re corded.
d. Examine unusual relationships between monthly accounts payable balances and re corded purchases.
42. Which of the following best describes the primary reason for the auditor’s use of flowcharts during an audit engagement?
a. To comply with the requirements of general ly accepted auditing standards.
b. To classify the client’s documents and trans actions by major operating functions, e.g., cash receipts, cash disbursements, etc.
c. To record the auditor’s understanding of the client’s system of internal accounting control.
d. To interpret the operational effectiveness of the client’s existing organizational struc ture.
79 Examination Questions— November 1978 43. Which of the following is the most important consideration of an auditor when examining the stock holders’ equity section of a client’s balance sheet?
a. Changes in the capital stock account are verified by an independent stock transfer agent.
b. Stock dividends and/or stock splits during the year under audit were approved by the stockholders.
c. Stock dividends are capitalized at par or stated value on the dividend declaration date.
d. Entries in the capital stock account can be traced to a resolution in the minutes of the board of directors’ meetings.
44. Which of the following best explains why ac counts payable confirmation procedures are not al ways used?
a. Inclusion of accounts payable balances on the liability certificate completed by the client allows the auditor to refrain from using con firmation procedures.
b. Accounts payable generally are insignificant and can be audited by utilizing analytic re view procedures.
c. The auditor may feel certain that the creditors will press for payment.
d. Reliable externally generated evidence sup porting accounts payable balances is gener ally available for audit inspection on the client’s premises.
45. Which of the following is generally included or shown in the auditor’s working papers?
a. The procedures used by the auditor to verify the personal financial status of members of the client’s management team.
b. Analyses that are designed to be a part of, or a substitute for, the client’s accounting records.
c. Excerpts from authoritative pronouncements that support the underlying generally accept ed accounting principles used in preparing the financial statements.
d. The manner in which exceptions and un usual matters disclosed by the auditor’s pro cedures were resolved or treated.
46. Which of the following would lessen internal control in an electronic data processing system?
a. The computer librarian maintains custody of computer program instructions and de tailed listings.
b. Computer operators have access to operator instructions and detailed program listings.
c. The control group is solely responsible for the distribution of all computer output.
d. Computer programmers write and debug programs which perform routines designed by the systems analyst.
47. A company includes selected interim financial information in a note to its annual financial statements. The independent auditor has made a limited review of the information and is satisfied with its presentation. Under these circumstances, the auditor’s report on the annual financial statements a. Should be modified to make reference to the limited review and the selected interim financial information.
b. Need not be modified to make reference to the limited review but should be modified to make reference to the selected financial information.
c. Should be modified to make reference to the limited review but not the selected interim financial information.
d. Need not be modified to make reference to the limited review or the selected interim financial information.
48. If an auditor wishes to issue a qualified opinion because the financial statements include a departure from generally accepted accounting principles, the auditor’s report should have an explanatory para graph referring to a footnote that discloses the prin cipal effects of the subject matter of the qualification. The qualification should be referred to in the opinion paragraph by using language such as a. “ With the exception of.’’ b. “ When read in conjunction with the foot notes.’’ c. “ With the foregoing explanation.’’ d. “ Subject to the departure explained in the footnotes.’’ 49. When considering the effectiveness of a system of internal accounting control, the auditor should rec ognize that inherent limitations do exist. Which of the following is an example of an inherent limitation in a system of internal accounting control?
a. The effectiveness of procedures depends on the segregation of employee duties.
b. Procedures are designed to assure the exe cution and recording of transactions in ac cordance with management’s authorization.
c. In the performance of most control pro cedures, there are possibilities of errors arising from mistakes in judgment.
d. Procedures for handling large numbers of transactions are processed by electronic data processing equipment.
80 Auditing 50. If the independent auditor decides that the work performed by internal auditors may have a bearing on the independent auditor’s own procedures, the indepen dent auditor should consider the objectivity of the inter nal auditors. One method of judging objectivity is to a. Review the recommendations made in the reports of internal auditors.
b. Examine, on a test basis, documentary evi dence of the work performed by internal auditors.
Inquire of management about the qualifica tions of the internal audit staff.
Consider the client’s practices for hiring, training, and supervising the internal audit staff.
c.
d.
53. If the auditor discovers that the carrying amount of a client’s investments is overstated because of a loss in value which is other than a temporary decline in market value, the auditor should insist that a. The approximate market value of the invest ments be shown on the face of the balance sheet.
b. The investments be classified as long term for balance sheet purposes with full dis closure in the footnotes.
c. The loss in value be recognized in the finan cial statements of the client.
d. The equity section of the balance sheet separately show a charge equal to the amount of the loss.
51. The independent auditor should acquire an un derstanding of the internal audit function as it relates to the independent auditor’s study and evaluation of internal accounting control because a. The audit programs, working papers, and reports of internal auditors can often be used as a substitute for the work of the independ ent auditor’s staff.
b. The procedures performed by the internal audit staff may eliminate the independent auditor’s need for an extensive study and evaluation of internal control.
c. The work performed by internal auditors may be a factor in determining the nature, timing, and extent of the independent audi tor’s procedures.
d. The understanding of the internal audit func tion is an important substantive test to be performed by the independent auditor.
52. Which of the following best describes the distin guishing feature of statistical sampling?
a. It provides for measuring mathematically the degree of uncertainty that results from examining only a part of the data.
b. It allows the auditor to have the same degree of confidence as with judgment sampling but with substantially less work.
c. It allows the auditor to substitute sampling techniques for audit judgment.
d. It provides for measuring the actual misstate ments in financial statements in terms of reliability and precision.
54. Rogers, a CPA, not in public practice, works as an internal auditor for a large conglomerate. The management of the conglomerate asked Rogers to perform an examination and report on a potential ac quisition. Rogers’ report will be used by the manage ment for internal purposes. Under these circumstances, how should Rogers sign the report?
a. Rogers, CPA.
b. Rogers, CPA (Internal Auditor).
c. Rogers, Internal Auditor.
d. Rogers, Internal Auditor (CPA).
55. If an auditor, planning to use statistical sampling, is concerned with the number of a client’s sales in voices that contain mathematical errors, the auditor would most likely utilize a. Random sampling with replacement.
b. Sampling for attributes.
c. Sampling for variables.
d. Stratified random sampling.
56. During the course of an audit engagement an auditor prepares and accumulates audit working papers. The primary purpose of the audit working papers is to a. Aid the auditor in adequately planning his work.
b. Provide a point of reference for future audit engagements.
c. Support the underlying concepts included in the preparation of the basic financial statements.
d. Support the auditor’s opinion.
81 Examination Questions— November 1978 57. The sufficiency and competency of evidential matter ultimately is based on the a. Availability of corroborating data.
b. Generally accepted auditing standards.
c. Pertinence of the evidence.
d. Judgment of the auditor.
60. Which of the following ratios would be the least useful in reviewing the overall profitability of a manu facturing company?
a. Net income to net worth.
b. Net income to total assets.
c. Net income to sales.
d. Net income to working capital.
58. A system of internal accounting control normal ly would include procedures that are designed to pro vide reasonable assurance that a. Employees act with integrity when perform ing their assigned tasks.
b. Transactions are executed in accordance with management’s general or specific au thorization.
c. Decision processes leading to management’s authorization of transactions are sound.
d. Collusive activities would be detected by segregation of employee duties.
59. Management furnishes the independent auditor with information concerning litigation, claims, and assessments. Which of the following is the auditor’s primary means of initiating action to corroborate such information?
a. Request that client lawyers undertake a re consideration of matters of litigation, claims, and assessments with which they were con sulted during the period under examination.
b. Request that client management send a letter of audit inquiry to those lawyers with whom management consulted concerning litigation, claims, and assessments.
c. Request that client lawyers provide a legal opinion concerning the policies and pro cedures adopted by management to identi fy, evaluate, and account for litigation, claims, and assessments.
d. Request that client management engage outside attorneys to suggest wording for the text of a footnote explaining the nature and probable outcome of existing litigation, claims, and assessments.
Number 2 (Estimated time-----15 to 25 minutes) For many years the financial and accounting com munity has recognized the importance of the use of audit committees and has endorsed their formation.
At this time the use of audit committees has become widespread. Independent auditors have become increasingly involved with audit committees and conse quently have become familiar with their nature and function.
Required: a. Describe what an audit committee is.
b. Identify the reasons why audit committees have been formed and are currently in operation.
c. What are the functions of an audit committee?
Number 3 (Estimated time 15 to 25 minutes) During the course of an audit engagement an in dependent auditor gives serious consideration to the concepts of materiality. This concept of materiality is inherent in the work of the independent auditor and is important for planning, preparing, and modifying audit programs. The concept of materiality underlies the ap plication of all the generally accepted auditing stand ards, particularly the standards of field work and re porting.
Required: a. Briefly describe what is meant by the in dependent auditor’s concept of materiality.
b. What are some common relationships and other considerations used by the auditor in judging materiality?
c. Identify how the planning and execution of an audit program might be affected by the independent auditor’s concept of materiality.
82 Auditing Number 4 (Estimated time-----15 to 25 minutes) Long, CPA, has been engaged to examine and report on the financial statements of Maylou Corpora tion. During the review phase of the study of Maylou’s system of internal accounting control over purchases, Long was given the following document flowchart for purchases.
Maylou Corporation DOCUMENT FLO W C HART FOR PURCHASES CONTROLLER VICE PRESIDENT—MANUFACTURING Accounts Payable Purchasing Agent Receiving Stores Required: a. Identify the procedures, relating to purchase requisitions and purchase orders, that Long would ex pect to find if Maylou’s system of internal accounting control over purchases is effective. For example, pur chase orders are prepared only after giving proper con sideration to the time to order, and quantity to order. Do not comment on the effectiveness of the flow of documents as presented in the flowchart or on separa tion of duties.
b. What are the factors to consider in determin ing— 1. The time to order?
2, The quantity to order?
8 3 Examination Questions— November 1978 Number 5 (Estimated time 15 to 25 minutes) An auditor is conducting an examination of the financial statements of a wholesale cosmetics distributor with an inventory consisting of thousands of individual items. The distributor keeps its inventory in its own distribution center and in two public warehouses. An in ventory computer file is maintained on a computer disc and at the end of each business day the file is updated. Each record of the inventory file contains the following data: Item number Location of item Description of item Quantity on hand Cost per item Date of last purchase Date of last sale Quantity sold during year The auditor is planning to observe the distributor’s physical count of inventories as of a given date. The auditor will have available a computer tape of the data on the inventory file on the date of the physical count and a general purpose computer software package.
Required: The auditor is planning to perform basic inventory auditing procedures. Identify the basic inventory auditing procedures and describe how the use of the general purpose software package and the tape of the in ventory file data might be helpful to the auditor in per forming such auditing procedures.
Organize your answer as follows: Basic inventory auditing procedure How general purpose com puter software package and tape of the inventory file data might be helpful I. O bserve the ph ysical count, m aking a n d recording test cou n ts where applicable.
D eterm ining which item s are to be test co u n ted b y selecting a r a n d o m s a m p le o f a represen tative num ber o f item s fro m the in ven tory file as o f the date o f the ph ysical count.
84 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all the states, territories, and the District of Columbia.) EXAMINATION IN BUSINESS LAW (Commercial Law) November 3, 1978; 8:30 AM to 12:00 M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1 No. 2 No. 3 No. 4 No. 5 Total 90 105 25 30 20 25 20 25 20 25 175 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
85 Examination Questions— November 1978 Number 1 (Estimated tim e----- 90 to 105 minutes) Instructions Select the best answer for each of the following items. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 99. The text of a letter from Bridge Builders, Inc., to Allied Steel Co. follows: We offer to purchase 10,000 tons of No. 4 steel pipes at today’s quoted price for delivery two months from today. Your acceptance must be received in five days.
Bridge Builders intended to create a (an) a. Option contract.
b. Unilateral contract.
c. Bilateral contract.
d. Joint contract.
Answer Sheet 99. Items to be Answered 1. Pickwick is a troublesome chain store appliance dealer. He constantly engaged in price cutting on widely advertised name products in order to lure customers to his store so that he could sell them other products. The “big three” manufacturers agreed that Pickwick could no longer sell their products unless he ceased and desisted from such practices. Pickwick refused and the three manufacturers promptly cut off his supply of their branded products. Which of the following is a correct statement?
a. Since the conduct described was unilateral, and Pickwick did not agree to stop his price cutting, the manufacturers’ conduct is legal.
b. The conduct described is a joint boycott, and as such is illegal per se.
c. If the harm to the public was minor, and the products were readily available from other ap pliance dealers in a market marked by free and open competition, there would be no violation of the law.
d. Since a businessman has the freedom to choose with whom he will deal, the conduct in question is not illegal under antitrust law.
2. Magnus Enterprises engaged a CPA firm to per form the annual examination of its financial statements. Which of the following is a correct statement with respect to the CPA firm’s liability to Magnus for negligence?
a. Such liability can not be varied by agreement of the parties.
b. The CPA firm will be liable for any fraudulent scheme it does not detect.
c. The CPA firm will not be liable if it can show that it exercised the ordinary care and skill of a reasonable man in the conduct of his own affairs.
d. The CPA firm must not only exercise reasonable care in what it does, but also must possess at least that degree of accounting knowledge and skill expected of a CPA.
3. The Apex Surety Company wrote a general fidelity bond covering defalcations by the employees of Watson, Inc. Thereafter, Grand, an employee of Watson, embezzled $18,900 of company funds. When his activities were discovered, Apex paid Watson the full amount in accordance with the terms of the fidelity bond, and then sought recovery against Watson’s auditors, Kane & Dobbs, CPAs. Which of the following would be Kane & Dobbs’ best defense?
Apex is not in privity of contract.
The shortages were the result of clever forgeries and collusive fraud which would not be detected by an examination made in accor dance with generally accepted auditing stan dards.
Kane & Dobbs were not guilty either of gross negligence or fraud.
Kane & Dobbs were not aware of the Apex- Watson surety relationship.
a.
b.
c.
d.
4. The CPA firm of Knox and Knox has been sub poenaed to testify and produce its correspondence and working papers in connection with a lawsuit brought against Johnson, one of its clients. Regarding the at tempted resort to the privileged communication rule in seeking to avoid admission of such evidence in the lawsuit, which of the following is correct?
Federal law recognizes such a privilege if the accountant is a Certified Public Accountant. The privilege is available regarding the work ing papers since the accountant is deemed to own them.
The privilege is as widely available as the attorney-client privilege.
In the absence of a specific statutory provi sion, the law does not recognize the existence of the privileged communication rule between an accountant and his client.
a.
b.
c.
d.
86 Business Law 5. The following facts arose during the examination of the financial statements of Western Manufacturing, Inc. Western, a large manufacturer of industrial pro ducts, is attempting to acquire a controlling block of shares of Davis, Inc., a competing manufacturer. Davis’ market share is about 40% in the section of the United States in which Western and Davis compete. Western’s share of this market is about 35%. Four other firms compete for the remaining 25%. Western’s share of the national market is 20% and Davis’ share is 8%. In the event that the Justice Department seeks to prevent the above acquisition, which of the following is the most likely result?
a. The court would grant an injunction pro hibiting the acquisition in question.
b. The court would dismiss the Justice Depart ment’s action in that the relative shares of the two corporations are less than 30% if the en tire United States is taken as the geographic market.
c. The court would permit the acquisition and agree to review the case after a year’s ex perience under the merger to determine its ac tual effect upon competition.
d. The court would dismiss the Justice Depart ment’s action if Western could show that the acquisition would result in significant business efficiencies.
6. Wonder Electronix, Inc., entered into “ reasonable” minimum price-resale-maintenance agreements with the distributors of its trademarked pro ducts. The distributors willingly accepted the minimum price and it was in fact a “ reasonable” price for the pro ducts. What is the legal status of this arrangement?
a. It is subject to the rule of reason as contrasted with the per se illegality rule applicable to such price fixing among competitors.
b. It is subject to the per se illegality rule.
c. It will not be illegal if the product bears the label, trademark, or brand name of Wonder and it competes with like products of Wonder’s competitors.
d. It is exempt from antitrust law if the product is patented.
7. Stein bought an office building valued at $200,000. The fire insurance policy contained a 100% co-insurance clause. Stein insured the building for $120,000. Subsequently, a fire caused damage of $40,000 to the building. Which of the following is the correct amount Stein will recover?
a. $40,000.
b. $24,000.
c. $13,333.
d. Nothing because the building was not insured for 100% of its value.
8. Which of the following defenses by a life insurance company would be unsuccessful by reason of the in contestable clause?
a. A material misstatement of fact by the insured relating to his health.
b. Lack of an insurable interest on the part of the party who insured the deceased.
c. A limitation upon the scope of the causes of death which the policy covers.
d. Cancellation of the policy due to nonpayment of premiums.
Items 9 and 10 are based on the following informa tion: Dexter, Inc., was incorporated in its home state. It expanded substantially and now does 20% of its business in a neighboring state in which it maintains a permanent facility. It has not filed any papers in the neighboring state.
9. Which of the following statements is correct?
a. Since Dexter is a duly-incorporated domestic corporation in its own state, it can transact business anywhere in the United States without further authority as long as its cor porate charter so provides.
b. As long as Dexter’s business activities in the neighboring state do not exceed 25%, it need not obtain permission to do business in the neighboring state.
c. Dexter must create a subsidiary corporation in the neighboring state to continue to do business in that state.
d. Dexter is a foreign corporation in the neighboring state and as such must obtain a certificate of authority or it will not be permit ted to maintain any action or suit in the state with respect to its intrastate business.
10. Which of the following statements is incorrect?
a. Dexter has automatically appointed the secretary of state of the neighboring state as its agent for the purpose of service of legal process if it failed to appoint or maintain a registered agent in that state.
b. Dexter will be able to maintain an action or suit in the neighboring state if it subsequently obtains a certificate of authority.
c. Dexter can not defend against a suit brought against it in the neighboring state’s courts.
d. The attorney general of the neighboring state can recover all back fees and franchise taxes which would have been imposed plus all penalties for failure to pay same.
87 Examination Questions— November 1978 11. A CPA firm is being sued by a third party pur chaser of securities sold in interstate commerce to the public. The third party is relying upon the Securities Act of 1933. The CPA firm had issued an unqualified opin ion on incorrect financial statements. Which of the following represents the best defense available to the CPA firm?
a. The securities sold had not been registered with the SEC.
b. The CPA firm had returned the entire fee it charged for the engagement to the corpora tion.
c. The third party was not in privity of contract with the CPA firm.
d. The action had not been commenced within one year after the discovery of the material misrepresentation.
12. The Ripper Corporation has been found guilty of widespread price fixing along with several of its com petitors in the sale of a certain product. As a result of these facts, Ripper and its officers may be subject to various sanctions, penalties, or liabilities. Which of the following would not be applicable?
a. Fines and imprisonment imposed against the principal officers responsible.
b. Fines imposed against the corporation.
c. Treble damages awarded to third parties who purchased the product from the corporation and suffered damages.
d. A prohibition against the corporation from engaging in the sale of the product in question in interstate commerce for a period of not more than six months.
13. Nabor, Inc., purchased a three-year fire insurance policy from the Fidelity Insurance Company covering its factory and warehouse. Which of the following statements is correct as a general rule of insurance law?
a. The policy will not cover the intentional destruction of the property by a third party.
b. The policy will not cover the destruction of the property if it is caused by the gross negligence of an employee of Nabor.
c. If Nabor sells the insured property to a third party and assigns the insurance policy to the buyer, it continues in effect.
d. If Nabor sells the insured property, but retains the fire insurance policy, it will not be able to collect on the policy in the event of its destruc tion by fire.
14. Which of the following can not properly be received as the consideration for the issuance of shares?
a. Promissory notes.
b. Services actually performed for the corpora tion.
c. Shares of stock of another corporation.
d. Intangible property rights.
15. Watson entered into an agreement to purchase 1,000 shares of the Marvel Corporation, a corporation to be organized in the near future. Watson has since had second thoughts about investing in Marvel. Under the circumstances, which of the following is correct?
a. A written notice of withdrawal of his agree ment to purchase the shares will be valid as long as it is received prior to incorporation.
b. A simple transfer of the agreement to another party will entirely eliminate his liability to pur chase the shares of stock.
c. Watson may not revoke the agreement for a period of six months in the absence of special circumstances.
d. Watson may avoid liability on his agreement if he can obtain the consent of the majority of other individuals committed to purchase shares to release him.
16. For which of the following purposes is a general partnership recognized as an entity by the Uniform Partnership Act?
a. Insulation of the partners from personal lia bility.
b. Taking of title and ownership of property.
c. Continuity of existence.
d. Recognition of the partnership as the employer of its members.
17. In the course of your audit of Harvey Fox, doing business as Harvey’s Apparels, a sole proprietorship, you discovered that in the past year Fox had regularly joined with Leopold Morrison in the marketing of bathing suits and beach accessories. You are concerned whether Fox and Morrison have created a partnership relationship. Which of the following factors is the most important in ascertaining this status?
a. The fact that a partnership agreement is not in existence.
b. The fact that each has a separate business of his own which he operates independently.
c. The fact that Fox and Morrison divide the net profits equally on a quarterly basis.
d. The fact that Fox and Morrison did not intend to be partners.
18. Which of the following is a correct statement con cerning the similarities of a limited partnership and a corporation?
a. Both provide insulation from personal liabili ty for all of the owners of the business.
b. Both can only be created pursuant to a statute and each must file a copy of their respective certificates with the proper state authorities.
c. Both are recognized for federal income tax purposes as taxable entities.
d. Shareholders and limited partners may both participate in the management of the business and retain limited liability.
88 Business Law 19. Derek Corporation decided to acquire certain assets belonging to the Mongol Corporation. As con sideration for the assets acquired, Derek issued 20,000 shares of its no-par common stock with a stated value of $10 per share. The value of the assets acquired subse quently turned out to be much less than the $200,000 in stock issued. Under the circumstances, which of the following is correct?
a. It is improper for the board of directors to ac quire assets other than cash with no-par stock.
b. Only the shareholders can have the right to fix the value of the shares of no-par stock ex changed for assets.
c. In the absence of fraud in the transaction, the judgment of the board of directors as to the value of the consideration received for the shares shall be conclusive.
d. Unless the board obtained an independent ap praisal of the acquired assets’ value, it is liable to the extent of the overvaluation.
20. Maxwell was the head cashier of the Amalgamated Merchants Bank. The Excelsior Surety Company bond ed Maxwell for $200,000. An internal audit revealed a $1,000 embezzlement by Maxwell. Maxwell persuaded the bank not to report him, and he promised to pay the money back within ten days. The bank acquiesced and neither the police nor Excelsior was informed of the theft. Maxwell shortly thereafter embezzled $75,000 and fled. Excelsior refuses to pay. Is Excelsior liable? Why?
a. Excelsior is liable since the combined total of the embezzlements is less than the face amount of the surety bond.
b. Excelsior is liable for $75,000, but not the $1,000 since a separate arrangement was agreed to by Amalgamated with Maxwell.
c. Excelsior is liable since it is a compensated surety and as such assumed the risk.
d. Excelsior is not liable since the failure to give notice of the first embezzlement is a valid defense.
21. Adams, Baker, and Carter are co-sureties on a $250,000 loan by the Wilson National Bank to Marathon Motors, Inc. Adams is a surety for the full amount of the debt; Baker’s obligation is limited to $100,000; and Carter has agreed to pay $50,000 upon default. In the event of default by Marathon on the en tire $250,000 loan, what is the liability of Adams, Baker, and Carter?
a. Baker is liable for the first $100,000; Carter, the next $50,000; and Adams, the balance.
b. Baker and Adams are each liable for $100,000 and Carter for $50,000.
c. If both Baker and Carter know of Adams’ obligation for the full amount, then they are not liable unless Adams can not satisfy the debt.
d. Adams is liable for $156,250; Baker, $62,500; and Carter, $31,250.
22. Hobson, Jones, Carter, and Wolff are all medical doctors who have worked together for several years. They decided to form a corporation and their attorney created a typical professional corporation for them. Which of the following is correct?
a. Such a corporation will not be recognized for federal tax purposes if one of its goals is to save taxes.
b. The state in which they incorporated must have enacted professional corporation statutes permitting them to do so.
c. Upon incorporation, the doctor-shareholder is insulated from personal liability beyond his capital contribution.
d. The majority of states prohibit the creation of professional corporations by doctors.
23. In determining the liability of a partnership for the acts of a partner purporting to act for the partnership without the authorization of his fellow partners, which of the following actions will bind the partnership?
a. A written admission of liability in a lawsuit brought against the partnership.
b. Signing the partnership name as a surety on a note for the purchase of that partner’s sum mer home.
c. An assignment of the partnership assets in trust for the benefit of creditors.
d. The renewal of an existing supply contract which the other partners had decided to ter minate and which they had specifically voted against.
24. Ace Corporation loaned $10,000 to King Enter prises, Inc., one of its best customers. The loan was for three years and was evidenced by a note duly executed by King’s president on behalf of the corporation. In ad dition, Walsh and Paxton, King’s principal shareholders, had orally guaranteed the repayment of the loan. With respect to Walsh and Paxton, which of the following is a correct statement?
a. Unless otherwise indicated, each guaranteed $5,000 of the loan.
b. They will be denied the usual surety defenses.
c. They are co-sureties and, as such, their surety undertaking must be in a signed writing.
d. Some additional consideration, independent of the making of the loan by Ace, must pass directly to Walsh and Paxton.
25. Which of the following contractual prerequisites is not generally necessary to establish a legally enforceable surety relationship?
a. A signed writing.
b. The solvency of the principal debtor.
c. Separate consideration for the surety’s promise.
d. The legal capacity of the surety.
89 Examination Questions— November 1978 26. Which of the following is a correct statement with respect to the rights of a limited partner?
a. The limited partner will only have taxable in come if the limited partnership makes a distribution in the tax year.
b. The partnership is required to purchase the limited partnership interest at the current book value if the limited partner demands this.
c. The limited partner may assign his partnership interest to whomsoever he wishes at any time.
d. The limited partner must first offer his in terest to the partnership before he may sell to another party.
27. Martin Stores, Inc., decided to sell a portion of its eight-acre property. Consequently, the president of Martin wrote several prospective buyers the following letter: Dear Sir: We are sending this notice to several prospective buyers because we are interested in selling four acres of our property located in downtown Metropolis. If you are interested, please communicate with me at the above address. Don’t bother to reply unless you are thinking in terms of at least $100,000.
James Martin, President Under the circumstances, which of the following is cor rect?
a. The statute of frauds does not apply because the real property being sold is the division of an existing tract which had been properly recorded.
b. Markus, a prospective buyer, who telegraphed Martin that he would buy at $100,000 and for warded a $100,000 surety bond to guarantee his performance, has validly accepted.
c. Martin must sell to the highest bidder.
d. Martin’s communication did not constitute an offer to sell.
28. Fashion Swimming Pools, Inc., mailed a letter to Direct Distributors offering a three-year franchise dealership. The offer stated the terms in detail and at the bottom stated that “the offer would not be withdrawn prior to October 1, 1978.’’ Under the cir cumstances, which of the following is correct?
a. The offer is an irrevocable option which can not be withdrawn prior to October 1, 1978.
b. A letter of acceptance from Direct to Fashion sent on October 1, 1978, but not received until October 2, 1978, would not create a valid con tract.
c. The statute of frauds would not apply to the proposed contract.
d. The offer can not be assigned to another party if Direct chooses not to accept.
29. Mayer wrote Jackson and offered to sell Jackson a building for $50,000. The offer stated it would expire 30 days from July 1, 1978. Mayer changed his mind and does not wish to be bound by his offer. If a legal dispute arises between the parties regarding whether there has been a valid acceptance of the offer, which of the following is correct?
a. The offer can not be legally withdrawn for the stated period of time.
b. The offer will not expire prior to the 30 days even if Mayer sells the property to a third per son and notifies Jackson.
c. If Jackson phoned Mayer on August 1 and unequivocally accepted the offer, it would create a contract, provided he had no notice of withdrawal of the offer.
d. If Jackson categorically rejects the offer on July 10th, Jackson can not validly accept within the remaining stated period of time.
30. Exeter Industries, Inc., orally engaged Werglow as one of its district sales managers for an 18-month period commencing April 1, 1978. Werglow commenced work on that date and performed his duties in a highly compe tent manner for several months. On October 1, 1978, the company gave Werglow a notice of termination as of November 1, 1978, citing a downturn in the market for its products. Werglow sues seeking either specific performance or damages for breach of contract. Exeter pleads the statute of frauds and/or a justified dismissal due to the economic situation. What is the probable out come of the lawsuit?
a. Werglow will prevail because the statute of frauds does not apply to contracts such as his.
b. Werglow will prevail because he has partially performed under the terms of the contract.
c. Werglow will lose because the reason for his termination was caused by economic factors beyond Exeter’s control.
d. Werglow will lose because such a contract must be in writing and signed by a proper agent of Exeter.
31. Franklin purchased 100 sets of bed frames from Tully Manufacturing, Inc. Franklin made substantial prepayments on the purchase price. Tully is insolvent and the goods have not been delivered as promised. Franklin wants the bed frames. Under the cir cumstances, which of the following will prevent Franklin from obtaining the bed frames?
a. The fact that he can obtain a judgment for damages.
b. The fact that he was not aware of Tully’s in solvency at the time he purchased the bed frames.
c. The fact that the goods have not been iden tified to his contract.
d. The fact that he did not pay the full price at the time of the purchase even though he has made a tender of the balance and holds it available to Tully upon delivery.
90 Business Law 32. On July 25, 1978, Archer, the president of Post Corporation, with the approval of the board of direc tors, engaged Biggs, a CPA, to examine Post’s July 31, 1978, financial statements and to issue a report in time for the annual stockholders’ meeting to be held on September 5, 1978. Notwithstanding Biggs’ reasonable efforts, the report was not ready until September 7 because of delays by Post’s staff. Archer, acting on behalf of Post, refused to accept or to pay for the report since it no longer served its intended purpose. In the event Biggs brings a legal action against Post, what is the probable outcome?
a. The case would be dismissed because it is unethical for a CPA to sue for his fee.
b. Biggs will be entitled to recover only in quasi contract for the value of the services to the client.
c. Biggs will not recover since the completion by September 5th was a condition precedent to his recovery.
d. Biggs will recover because the delay by Post’s staff prevented Biggs from performing on time and thereby eliminated the timely perfor mance condition.
33. Young owns 200 shares of stock of Victory Manufacturing Company. Victory is listed on a national stock exchange and has in excess of one million shares outstanding. Young claims that Truegood, a Victory director, has purchased and sold shares in violation of the insider trading provisions of the Securities Exchange Act of 1934. Young has threatened legal action. Which of the following statements is correct?
a. Truegood will have a valid defense if he can show he did not have any insider information which influenced his purchases or sales.
b. Young can sue Truegood personally, but his recovery will be limited to his proportionate share of Truegood’s profits plus legal ex penses.
c. In order to prevail, Young must sue for and on behalf of the corporation and establish that the transactions in question occurred within less than six months of each other and at a profit to Truegood.
d. Since Young’s stock ownership is less than 1%, his only recourse is to file a complaint with the SEC or obtain a sufficient number of other shareholders to join him so that the 1% requirement is met.
34. The intestate succession distribution rules a. Do not apply to property held in joint tenan cy.
b. Do not apply to real property.
c. Effectively prevent a decedent from totally disinheriting his wife and children.
d. Apply to situations where the decedent failed to name an executor.
35. Gordon purchased 100 automatic sprinklers from Thompson, a jobber. Conrad was the rightful owner of the sprinklers which had been stolen from his warehouse. He had the sheriff repossess them and has asserted his ownership of them. Gordon’s bill of sale specifically indicated that it made no implied warran ties. The bill of sale did not contain any warranties of title. Which of the following is correct based on the above facts?
a. The title warranties have been effectively negated.
b. It is not possible to disclaim the title warran ties.
c. Gordon’s best course of action is to assert his superior title to the sprinklers since he is a good faith purchaser for value.
d. Thompson is liable in that he warranted that the title conveyed was good and his transfer rightful.
36. Parks furnished specifications and ordered 1,000 specially-constructed folding tables from Metal Manufacturing Company, Inc. The tables were unique in design and had not appeared in the local market. Metal completed the job and delivered the order to Parks. Parks sold about 600 of the tables when Unusual Tables, Inc., sued both Parks and Metal for patent in fringement. If Unusual wins, what is the status of Parks and Metal?
a. Metal is liable to Parks for breach of the war ranty against infringement.
b. Parks is liable to Metal for breach of the war ranty against infringement.
c. The warranty against infringement is not available to either Parks or Metal.
d. Parks and Metal are jointly and severally liable and, as such, must pay the judgment in equal amounts.
37. Which of the following receipts should be allocated by a trustee exclusively to income?
a. A stock dividend.
b. An extraordinary year-end cash dividend.
c. A liquidating dividend whether in complete or partial liquidation.
d. A stock split.
38. The Unity Trust Company is the trustee of a trust which has large real estate investments. Which of the following receipts or charges should be allocated by the trustee to income?
a. Paving assessment for a new street.
b. Prepaid rent received from tenants.
c. A loss on the sale of one of the rental proper ties.
d. The proceeds from an eminent domain pro ceeding.
91 Examination Questions— November 1978 39. An executor named in a decedent’s will a. Can not be the principal beneficiary of the will.
b. Must serve without compensation unless the will provides otherwise.
c. Need not serve if he does not wish to do so.
d. Must consent to serve, have read the will, and be present at the execution of the will.
40. Which of the following receipts or disbursements by a trustee should be credited to or charged against in come?
a. Amortization payment on real property sub ject to a mortgage.
b. Capital gain distributions received from a mutual fund.
c. Stock rights received from the distributing corporation.
d. The discount portion received on redemption of treasury bills.
41. Winslow conveyed a 20-acre tract of land to his two children, George and Martha, “equally as tenants in common.’’ What is the legal effect of this form of conveyance?
a. George and Martha are joint owners with a right of survivorship.
b. Each must first offer the other the right to purchase the property before he or she can sell to a third party.
c. Neither may convey his or her interest in the property unless both join in the conveyance.
d. Each owns an undivided interest in the whole, which he or she may dispose of by deed or by will.
42. Marlin ordered merchandise from Plant to be delivered the following day and gave Plant a check payable to its order drawn on Marlin’s account in First Bank. It was agreed that the check would not be transferred unless delivery was received and accepted. The goods were not delivered and Marlin notified Plant that he exercised his right to rescind. Plant, never theless, negotiated the check for full value to Rose who took it in good faith and without notice of any defense. Rose then negotiated it for full value to Quirk who knew of Plant’s breach of the agreement. Marlin promptly stopped payment on the check and refuses to pay it. Under these circumstances, which of the following statements is correct?
a. Marlin would have a valid defense in a suit by Rose for the amount of the check.
b. Marlin would have a valid defense in a suit by Quirk for the amount of the check.
c. Despite the fact that Quirk can not personally qualify as a holder in due course, he can assert Rose’s standing as such.
d. A stop payment order will not prevent a holder in due course from collecting from the bank.
43. Paxton owned Blackacre, and he obtained a $10,000 loan from a bank secured by a real property mortgage on Blackacre. The mortgage was properly recorded. Paxton subsequently sold Blackacre to Rogers, expressly warranting that there were no mort gages on the property. Rogers was unaware of the bank’s interest in the property. Paxton has disappeared, and the bank has demanded payment from Rogers. Rogers has refused, and the bank is seeking to foreclose its mortgage. Which of the following statements is cor rect?
a. Rogers is personally liable on the mortgage loan.
b. As a bona fide purchaser for value, Rogers will prevail and retain the property free of the mortgage.
c. The bank will prevail in its foreclosure action.
d. The bank must obtain a judgment against Paxton before it can foreclose the mortgage.
44. Your client, Commercial Factors, Inc., discounted a $2,000 promissory note, payable in two years, for $1,500. It paid $500 initially and promised to pay the balance ($1,000) within 30 days. Commercial paid the balance within the 30 days, but before doing so learned that the note had been obtained originally by fraudulent misrepresentations in connection with the sale of land which induced the maker to issue the note. For what amount will Commercial qualify as a holder in due course?
a. None because the 25% discount is presump tive or prima facie evidence that Commercial is not a holder in due course.
b. $500.
c. $1,500.
d. $2,000.
45. An instrument which was indorsed by several subsequent holders reads as follows: PAY One month after date to Richard Williams or bearer One Hundred and no/100 No. 452 April 1, 1978 Dollars Fireman’s Bank Chicago, Illinois Which of the following statements about this instru ment is correct?
a. The instrument is nonnegotiable.
b. Richard Williams’ signature is necessary to negotiate the instrument.
c. The instrument in question is a check.
d. Presentment for payment must be made on the due date or next business day in order to hold the subsequent indorsers liable.
92 Business Law 46. Maxwell is a holder in due course of a check which was originally payable to the order of Clark and has the following indorsements on its back; Pay to the order of White Without Recourse Which of the following statements about this check is correct?
a. If the bank refuses to pay, Maxwell’s only recourse is to sue Dobbins.
b. The instrument was bearer paper in Dobbins’ hands.
c. Clark’s signature was not necessary to negotiate the instrument.
d. White has no warranty liability to Maxwell on the instrument.
47. Davidson bore a remarkable physical resemblance to Ford, one of the town’s most prominent citizens. He presented himself one day at the Friendly Finance Com pany, represented himself as Ford, and requested a loan of $500. The manager mistakenly, but honestly, believed that Davidson was Ford. Accordingly, being anxious to please so prominent a citizen, the manager required no collateral and promptly delivered to Davidson a $500 check payable to the order of Ford. Davidson took the check and signed Ford’s name to it on the back and negotiated it to Robbins, who took in the ordinary course of business (in good faith and for value). Upon learning the real facts, Friendly stopped payment on the check. Robbins now seeks recovery against Friendly. Under these circumstances, which of the following statements is correct?
a. Friendly could not validly stop payment on the check.
b. Davidson’s signature of Ford’s name on the check constitutes a forgery and is a real defense which is valid against Robbins.
Since both Friendly and Robbins were mistaken as to Davidson’s real identity, they will share the loss equally.
Davidson’s signature of Ford’s name on the check is effective and Robbins will prevail against Friendly.
48. Charles is a commercial tenant of Luxor Buildings, Inc. The term of the lease is five years and two years have elapsed. The lease prohibits subletting, but does not contain any provision relating to assign ment. Charles approached Luxor and asked whether Luxor could release him from the balance of the term of the lease for $500. Luxor refused unless Charles would agree to pay $2,000. Charles located Whitney who was interested in renting in Luxor’s building and transferred the entire balance of the lease to Whitney in considera tion of his promise to pay Luxor the monthly rental and otherwise perform Charles’ obligations under the lease. Luxor objects. Which of the following statements is cor rect?
a. The assignment is invalid without Luxor’s consent.
b. The assignment does not extinguish Charles’ obligation to pay the rent if Whitney defaults.
c. The assignment need not be in writing.
d. A prohibition of the right to sublet contained in the lease completely prohibits an assign ment.
49. There are several legally significant differences between a negotiable instrument and a contract right and the transfer of each. Which of the following statements is correct?
a. A negotiable instrument is deemed prima facie to have been issued for consideration whereas a contract is not deemed prima facie to be sup ported by consideration.
b. Generally, the transferee of a negotiable in strument and the assignee of a contract right take free of most defenses.
c. Neither can be transferred without a signed writing or by a delivery.
d. The statute of frauds rules apply to both.
50. An instrument reads as follows: $5,000.00 Boise, Idaho October 1, 1978 Thirty days after date I promise to pay to the order of Cash__ at 120 BROADWAY, New York City c.
d.
Value received with interest at the rate of eight percent per annum.
This instrument arises out of a separate agreement.
No. 20 Due October 31, 1978 Which of the following statements about this instru ment is correct?
a. The instrument is negotiable.
b. The instrument is order paper.
c. The instrument is a time draft.
d. Failure to make a timely presentment will ex cuse Loeb from liability.
93 Examination Questions— November 1978 Part a. National Finance Company engages in a wide variety of secured transactions which may be broken down into three categories.
I. Consumer loans in connection with the pur chase of automobiles, appliances, and furniture. National makes these loans in two ways. First, it makes direct loans to the consumer-borrower who then makes the purchase with the proceeds. Second, it is contacted by the seller and provides the financing for the purchase by the customer. In either case National takes a security interest in the property purchased.
Number 2 (Estimated tim e------25 to 30 minutes) II. Collateralized loans to borrowers who deliver possession of property, such as diamonds, to National to secure repayment of their loans.
III. Loans to merchants to finance their inventory purchases. National takes a security interest in the in ventory and proceeds.
Except for category III, National does not file a financing statement.
work out its financial difficulties.
Disco first approached several banks and attemp ted to mortgage the property. However, the banks would only loan an amount equal to of the value of the property. The maximum funds that Disco could ob tain from a bank was $30,000 which was not enough to forestall bankruptcy. Disco abandoned this approach.
Disco then placed the property on the market. The property was listed with several brokers on a non exclusive basis and was advertised in the local papers. Jolly learned of the proposed sale and, because of unique advantages of the property to Jolly, offered $62,500. This was $2,500 more than any other prospec tive purchaser had offered. The purchase price consisted of cancellation of the $30,000 overdue debt and $32,500 in cash.
The sale was consummated on July 1, 1978. On October 14, 1978, a petition in bankruptcy against Disco was duly filed by a group of its aggrieved creditors. The trustee in bankruptcy has attacked the sale of the land as a voidable preference.
Required: Answer the following, setting forth reasons for any conclusions stated.
Will the trustee prevail?
Required: Answer the following, setting forth reasons for any conclusions stated.
1. When does National’s security interest in the various types of property attach?
2. As a secured creditor, against what parties must National protect itself?
3. Does National have a perfected security in terest in any of the above property? If so, against whom?
4. If the facts indicate that National does not have a perfected security interest against all parties, what should it do?
5. Can National fully protect itself against all subsequent parties who might claim superior rights to the property involved?
Part b. Disco Records, Inc., was in dire financial condition which was widely known in the relevant business community. Disco’s liabilities exceeded its assets by approximately $200,000. Jolly Plastics, Inc., one of Disco’s creditors, was pressing Disco to pay $30,000 of overdue accounts. Disco decided to mort gage or sell its warehouse to improve its cash position, provide funds to forestall creditors, and permit it to Number 3 (Estimated tim e-----20 to 25 minutes) Part a. Rapid Delivery Service, Inc., hired Dolson as one of its truck drivers. Dolson was carefully selected and trained by Rapid. He was specifically instructed to obey all traffic and parking rules and regulations. One day while making a local delivery, Dolson double parked and went into a nearby customer’s store. In do ing so, he prevented a car legally parked at the curb from leaving. The owner of the parked car, Charles, proceeded to blow the horn of the truck repeatedly. Charles was doing this when Dolson returned from his delivery. As a result of a combination of several factors, particularly Charles’ telling him to “ move it’’ and that he was “acting very selfishly and in an unreasonable manner,’’ Dolson punched Charles in the nose, severely fracturing it. When Charles sought to restrain him, Dolson punched Charles again, this time fracturing his jaw. Charles has commenced legal action against Rapid.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Will Charles prevail?
2. What liability, if any, would Dolson have?
94 Business Law Part b. Harold Watts was employed by Superior Sporting Goods as a route salesman. His territory, route, and customers were determined by Superior. He was expected to work from 9:00 AM to 5:00 PM, Mon day through Friday. He received a weekly salary plus time and one-half for anything over 40 hours. He also received a small commission on sales which exceeded a stated volume. The customers consisted of sporting goods stores, department stores, athletic clubs, and large companies which had athletic programs or spon sored athletic teams. Watts used his personal car in making calls or, upon occasion, making a delivery where the customer was in a rush and the order was not large. Watts was reimbursed for the use of the car for company purposes. His instructions were to assume the customer is always right and to accommodate the customer where to do so would cost little and would build goodwill for the company and himself.
One afternoon while making a sales call and dropping off a case of softballs at the Valid Clock Com pany, the personnel director told Watts he was planning to watch the company’s team play a game at a softball field located on the other side of town, but that his car would not start. Watts said, “ Don’t worry, it will be my pleasure to give you a lift and I would like to take in a few innings myself.’’ Time was short and while on the way to the ballpark. Watts ran a light and collided with another car. The other car required $800 of repairs and the owner suffered serious bodily injury.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. What is Superior’s potential liability, if any, to the owner of the other car?
2. What is Valid’s potential liability, if any, to the owner of the other car?
Number 4 (Estimated time 20 to 25 minutes) Part c. Eureka Enterprises, Inc., started doing business in July 1977. It manufactures electronic com ponents and currrently employs 35 individuals. In an ticipation of future financing needs. Eureka has engaged a CPA firm to audit its financial statements. During the course of the examination, the CPA firm discovers that Eureka has no workmen’s compensation insurance, which is in violation of state law, and so in forms the president of Eureka.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. What is the purpose of a state workmen’s com pensation law?
2. What are the legal implications of not having workmen’s compensation insurance?
Part a. Glover Corporation is a small rapidly- expanding manufacturing company. In 1977 Glover made a public offering of its shares for $400,000 in ac cordance with Regulation A, issued by the Securities and Exchange Commission pursuant to the Securities Act of 1933. The shares are not listed on any exchange, but are sometimes bought and sold in interstate com merce. At the end of 1977 Glover had total assets of $900,000, 429 shareholders, and sales of $650,000 for the year.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. What is a Regulation A offering and what are the general requirements which must be met in order to qualify for making such an offering?
2. What difference is there in the potential liabili ty of the parties making an offering under Regulation A as contrasted with a full registration?
3. What are the time limitations in which an ag grieved party may commence an action for failure to comply with the Securities Act of 1933?
4. What are the major provisions of the Securities Exchange Act of 1934 which do not apply to Glover and its officers, directors, and principal shareholders after the public offering and which major provisions do apply?
Part b. Gordon & Groton, CPAs, were the auditors of Bank & Company, a brokerage firm and member of a national stock exchange. Gordon & Groton examined and reported on the financial statements of Bank which were filed with the Securities and Exchange Commis sion.
Several of Bank’s customers were swindled by a fraudulent scheme perpetrated by Bank’s president who owned 90% of the voting stock of the company. The facts establish that Gordon & Groton were negligent but not reckless or grossly negligent in the conduct of the audit and neither participated in the fraudulent scheme nor knew of its existence.
The customers are suing Gordon & Groton under the antifraud provisions of Section 10(b) and Rule 10b-5 of the Securities Exchange Act of 1934 for aiding and abetting the fraudulent scheme of the president. The customers’ suit for fraud is predicated exclusively on the nonfeasance of the auditors in failing to conduct a proper audit, thereby failing to discover the fraudulent scheme.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. What is the probable outcome of the lawsuit?
2. What other theory of liability might the customers have asserted?
95 Examination Questions— November 1978 Part a. Clauson Enterprises, Inc., was considering adding a new product line to its existing lines. The deci sion was contingent upon its being assured of a supply of an electronic component for the product at a certain price and a positive market study which clearly justified the investment in the venture.
Clauson’s president approached Migrane Elec tronics and explained the situation to Migrane’s presi dent. After much negotiation, Migrane agreed to grant Clauson an option to purchase 12,000 of the necessary electronic components at $1.75 each or at the prevailing market price, whichever was tower. Clauson prepared the option below incorporating their understanding.
Number 5 (Estimated tim e------20 to 25 minutes) Option Agreement Clauson Enterprises/Migrane Electronics Migrane Electronics hereby offers to sell Clauson Enterprises 12,000 miniature solid state electronic breakers at $1.75 each or at the existing market price at the time of delivery, whichever is lower, delivery to be made in 12 equal monthly in stallments beginning one month after the exercise of this option. This option is irrevocable for six months from January 1, 1978.
Clauson Enterprises agrees to deliver to Migrane its market survey for the product line in which the component would be used if it elects not to exercise the option.
Both parties signed the option agreement and Migrane’s president signed Migrane’s corporate name alongside the last sentence of the first paragraph. On May 1, 1978, Migrane notified Clauson that it was revoking its offer. The market price for the component had increased to $1.85. On May 15, 1978, Clauson notified Migrane that it accepted the offer and that if Migrane did not perform, it would be sued and held liable for damages. Migrane replied that the offer was not binding and was revoked before Clauson accepted. Furthermore, even if it were binding, it was good for only three months as a matter of law.
Upon receipt of Migrane’s reply, Clauson in stituted suit for damages.
Required: Answer the following, setting forth reasons for any conclusions stated.
Who will prevail? Discuss all the issues and arguments raised by the fact situation.
Part b. On May 30, 1978, Hargrove ordered 1,000 spools of nylon yarn from Flowers, Inc., of Norfolk, Virginia. The shipping terms were “ F.O.B., Nor folk & Western RR at Norfolk.” The transaction was to be a cash sale with payment to be simultaneously ex changed for the negotiable bill of lading covering the goods. Title to the goods was expressly reserved in Flowers. The yarn ordered by Hargrove was delivered to the railroad and loaded in a boxcar on June 1, 1978. Flowers obtained a negotiable bill of lading made out to its own order. The boxcar was destroyed the next day while the goods were in transit. Hargrove refused to pay for the yarn and Flowers sued Hargrove for the pur chase price.
Required: Answer the following, setting forth reasons for any conclusions stated.
Who will prevail?
Part c. Novack, an industrial designer, accepted an offer from Superior Design Corporation to become one of its designers. The contract was for three years and ex pressly provided that it was irrevocable by either party except for cause during that period of time. The con tract was in writing and signed by both parties. After a year, Novack became dissatisfied with the agreed com pensation which he was receiving. He had done a brilliant job and several larger corporations were at tempting to lure him away, Novack, therefore, demanded a substantial raise, and Superior agreed in writing to pay him an additional amount as a bonus at the end of the third year. Novack remained with Superior and performed the same duties he had agreed to perform at the time he initially ac cepted the position. At the end of the three years, Novack sought to collect the additional amount of money promised. Superior denied liability beyond the amount agreed to in the original contract.
Required: Answer the following, setting forth reasons for any conclusions stated.
Can Novack recover the additional compensation from Superior?
Part d. The basic facts are the same as stated in Part c except that one of Superior’s competitors, Dixon Corporation, successfully lured Novack away from Superior by offering a substantially higher salary. Dixon did this with full knowledge of the terms of the original three-year contract between Novack and Superior.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Does Superior have any legal redress against Dixon?
2. Would Superior be successful if it seeks the equitable relief of specific performance (an order by the court compelling Novack to perform his contractual undertaking) for the remaining two years of the con tract?
96 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all the states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXAMINATION IN ACCOUNTING THEORY (Theory of Accounts) November 3, 1978; 1:30 to 5:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1................................................... ...................................... 95 105 No. 2 ................................................... ...................................... 15 20 No. 3 ................................................... ...................................... 15 20 No. 4 ................................................... ...................................... 20 25 No. 5 .................................................... ...................................... 15 20 No. 6 .................................................... ...................................... 15 20 Total............................................ ...................................... 175 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For in stance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessi ty of expressing opinions and conclusions in writ ten reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and applica tion of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
9 7 Examination Questions— November 1978 Instructions Number 1 (Estimated time------95 to 105 minutes) Select the best answer for each of the following items relating to a variety of issues in financial account ing, municipal accounting, and managerial accounting. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 99. The recommended title for the financial statement which summarizes the results of operations is the a. Earnings statement.
b. Statement of changes in financial condition.
c. Statement of sources and applications of funds.
d. Retained earnings statement.
Answer Sheet 99. Items to be Answered 1. The financial statement which has as its primary function the summarization of the financing and investing aspects of all significant transactions affect ing financial position is the a. Retained earnings statement.
b. Earnings statement.
c. Statement of changes in financial position.
d. Statement of financial position.
2. The computation of earnings per share in accor dance with generally accepted accounting principles may involve the consideration of securities deemed common stock equivalents. Common stock equivalents are an example of a. Form over substance.
b. Substance over form.
c. Form over accounting principle.
d. Substance over accounting principle.
3. Which of the following methods of reporting attempts to eliminate the effect of the changing value of the dollar?
a. Discounted net present value of future cash flows.
b. Historical cost restated for changes in the general price level.
c. Replacement value.
d. Retirement value.
4. Accounting changes are often made and the monetary impact is reflected in the financial statements of a company even though, in theory, this may be a violation of the accounting concept of a. Materiality.
b. Consistency.
c. Conservatism.
d. Objectivity.
5. When a company makes a change in accounting principle, prior year financial statements are not gener ally restated to reflect the change. The Accounting Principles Board decided that this procedure would prevent a dilution of public confidence in financial statements but recognized that this procedure conflicts with the accounting concept of a. Materiality.
b. Conservatism.
c. Objectivity.
d. Comparability.
6. During the lifetime of an entity accountants pro duce financial statements at arbitrary points in time in accordance with which basic accounting concept?
a. Objectivity.
b. Periodicity.
c. Conservatism.
d. Matching.
7. Under which of the following conditions would flood damage be considered an extraordinary item for financial reporting purposes?
a. Only if floods in the geographical area are unusual in nature and occur infrequently.
b. Only if floods are normal in the geographical area but do not occur frequently.
c. Only if floods occur frequently in the geo graphical area but have been insured against.
d. Under any circumstance flood damage should be classified as an extraordinary item.
8. Which of the following events would be accounted for as a prior period adjustment?
a. Change in the depreciable lives of fixed assets.
b. Change in amount of opening retained earn ings balance due to an error in a prior period.
c. Change in the method of computing deprecia tion of fixed assets.
d. Change in inventory cost flow assumption from first-in, first-out to weighted average.
9. Which of the following principles best describes the current method of accounting for research and development costs?
a. Immediate recognition as an expense.
b. Associating cause and effect.
c. Systematic and rational allocation.
d. Income tax minimization.
98 Accounting Theory 10. Which of the following is the appropriate basis for valuing fixed assets acquired in a business combination accounted for as a purchase carried out by exchanging cash for common stock?
a. Historic cost.
b. Book value.
c. Cost plus any excess of purchase price over book value of asset acquired.
d. Fair value.
11. If a marketable equity security which was classified as noncurrent in a prior period were to be reclassified as current in the current period, what would be the effect upon the valuation allowance attendant to that security assuming no change in its market value?
a. The valuation allowance should be reclassi fied to current also.
b. The valuation allowance should be recognized as a loss in the current period.
c. The valuation allowance should be adjusted to zero and the security reclassified at cost.
d. The valuation allowance should be recog nized as a gain in the subsequent period.
12. When inventory declines in value below original (historical) cost, and this decline is considered other than temporary, what is the maximum amount that the inventory can be valued at?
a. Sales price net of conversion costs.
b. Net realizable value.
c. Historic cost.
d. Net realizable value reduced by a normal profit margin.
13. A contingency can be accrued when a. It is certain that funds are available to settle the disputed amount.
b. An asset may have been impaired.
c. The amount of the loss can be reasonably estimated and it is evident prior to issuance of financial statements that an asset has been impaired or a liability incurred.
d. It is evident that an asset has been impaired or a liability incurred even though the amount of the loss cannot be reasonably estimated.
15. How should a gain from the sale of treasury stock be reflected when using the cost method of recording treasury stock transactions?
a. As ordinary earnings shown on the earnings statement.
b. As paid-in capital from treasury stock trans actions.
c. As an increase in the amount shown for common stock.
d. As an extraordinary item shown on the earn ings statement.
16. Which of the following costs is not a part of the defined maximum for pension cost determination?
a. Normal cost.
b. Provision for vested benefits.
c. Interest on overfunding.
d. 10% of prior service costs.
17. How should an unusual event not meeting the current criteria for an extraordinary item be disclosed in the financial statements?
a. Shown as a separate item in operating reve nues or expenses a nd supplemented by a footnote if deemed appropriate.
b. Shown in operating revenues or expenses but not shown as a separate item.
c. Shown after ordinary net earnings but before extraordinary items.
d. Shown after extraordinary items net of in come tax but before net earnings.
18. Which of the following is a potential abuse that can arise when a business combination is accounted for as a pooling of interests?
a. Assets of the investee may be overvalued when the price paid by the investor is allo cated among specific assets.
b. Earnings of the pooled entity may be in creased because of the combination only and not as a result of efficient operations.
c. Liabilities may be undervalued when the price paid by the investor is allocated to the specific liabilities.
d. An undue amount of cost may be assigned to goodwill, thus potentially allowing for an overstatement of pooled earnings.
14. Which of the following requires intraperiod tax allocation?
a. That portion of dividends reduced by the divi dends received deduction by corporations under existing federal income tax law.
b. The excess of accelerated depreciation used for tax purposes over straight-line deprecia tion used for financial reporting purposes.
c. Extraordinary gains or losses as defined by the Accounting Principles Board.
d. All differences between taxable income and financial statement earnings.
19. Which of the following research and development related costs should be capitalized and amortized over current and future periods?
a. Research and development general labora tory building.
b. Inventory used for a specific research project.
c. Administrative salaries allocated to research and development.
d. Research findings purchased from another company to aid a particular research project currently in process.
99 Examination Questions— November 1978 20. Which of the following is (are) the proper time period(s) to record a change in accounting estimate?
a. Current period and prospectively.
b. Current period and retroactively.
c. Retroactively only.
d. Current period only.
21. When preparing a statement of changes in finan cial position using the cash basis for defining funds, an increase in ending inventory over beginning inventory will result in an adjustment to reported net earnings because a. Funds were increased since inventory is a current asset.
b. The net increase in inventory reduced cost of goods sold but represents an assumed use of cash.
c. Inventory is an expense deducted in com puting net earnings, but is not a use of funds.
d. All changes in non-cash accounts must be disclosed under the all financial resources concept.
22. Which of the following is not a method of determining the current value of an asset?
a. Replacement cost.
b. Market value.
c. Restatement of cost for changes in general price level.
d. Net present value of expected future cash flows.
23. In addition to a statement of financial position, earnings statement, retained earnings statement, and statement of changes in financial position, a complete set of financial statements must include a. Adequate disclosure of accounting policies.
b. An auditor’s opinion.
c. A ten-year summary of operations.
d. Historical “common size’’ financial sum maries.
24. The par-value method of accounting for treasury stock differs from the cost method in that a. No gains or losses are recognized on the sale of treasury stock using the par-value method.
b. Any gain is recognized upon repurchase of stock but a loss is treated as an adjustment to retained earnings.
c. It reverses the original entry to issue the common stock with any difference between carrying value and proceeds adjusted through paid-in capital and treats a sub sequent resale like a new issuance of common stock.
d. It reverses the original entry to issue the common stock with any difference being shown as an ordinary gain or loss and does not recognize any gain or loss on a subse quent resale of the stock.
25. A change in the foreign currency exchange rate between the date a transaction occurred and the date of the current financial statements would give rise to an exchange gain or loss if a. The asset or liability being translated is car ried at a price in a current purchase or sale exchange.
b. The asset or liability being translated is car ried at a price in a past purchase or sale ex change.
c. The revenue or expense item relates to an asset or liability that is translated at histori cal rates.
d. The revenue or expense item relates to a de ferred asset or liability shown on a previous statement of financial position.
26. Which of the following should be accrued as revenues by the general fund of a local government?
a. Sales taxes held by the state which will be remitted to the local government.
b. Parking meter revenues.
c. Sales taxes collected by merchants.
d. Income taxes currently due.
27. Which account represents the equity of a non enterprise fund?
a. Net assets.
b. Fund balance.
c. Reserves.
d. Unencumbered balance.
28. An “actuarial deficiency’’ would appear in which fund?
a. Agency.
b. Trust.
c. General.
d. Debt service.
29. “ Excess of net billings to departments over costs’’ would appear in the financial statement of which fund?
a. Intragovernmental service.
b. Enterprise.
c. Capital projects.
d. Special revenue.
30. The accounting for special revenue funds is most similar to which type of fund?
a. Capital projects.
b. General.
c. Enterprise.
d. Special assessment.
31. Which type of fund can be either expendable or non-expendable?
a. Debt service.
b. Enterprise.
c. Trust.
d. Special revenues.
100 Accounting Theory 32. “Investment in general fixed assets” accounts would appear in which fund or group of accounts?
a. General fixed assets.
b. Enterprise.
c. Capital projects.
d. General.
33. Which of the following expenditures is normally recorded on the accrual basis in the general fund?
a. Interest.
b. Personal services.
c. Inventory items.
d. Prepaid expenses.
34. When should revenues from interest on assess ments receivable be recorded in a special assessment fund?
a. When legally due.
b. When assessed.
c. When collected in cash.
d. When the amount is known.
35. If the fixed costs attendant to a product increase while variable costs and sales price remain constant what will happen to (1) contribution margin and (2) break-even point?
Contribution Breakeven point margin a. Unchanged b. Unchanged c. Increase d. Decrease increase unchanged decrease increase 36. In a process costing system that assumes that normal spoilage occurs at the end of a process, the cost attributable to normal spoilage should be assigned to a. Ending work-in-process inventory.
b. Cost of goods manufactured and ending work-in-process inventory in the ratio of units worked on during the period to units remaining in work-in-process inventory.
c. Cost of goods manufactured (transferred out).
d. A separate loss account in order to highlight production inefficiencies.
37. Which of the following must be known about a production process in order to institute a direct costing system?
a. The variable and fixed components of all costs related to production.
b. The controllable and non-controllable com ponents of all costs related to production.
c. Standard production rates and times for all elements of production.
d. Contribution margin and break-even point for all goods in production.
38. Under which of the following conditions will the first-in, first-out method of process costing produce the same cost of goods manufactured amount as the weighted-average method?
a. When goods produced are homogeneous in nature.
b. When there is no beginning inventory.
c. When there is no ending inventory.
d. When beginning and ending inventories are each fifty percent complete.
39. A basic tenet of direct costing is that period costs should be currently expensed. What is the basic ratio nale behind this procedure?
a. Period costs are uncontrollable and should not be charged to a specific product.
b. Period costs are generally immaterial in amount and the cost of assigning the amounts to specific products would outweigh the benefits.
c. Allocation of period costs is arbitrary at best and could lead to erroneous decisions by management.
d. Period costs will occur whether or not pro duction occurs and so it is improper to allo cate these costs to production and defer a current cost of doing business.
40. How should the balances of progress billings and construction in progress be shown at reporting dates prior to the completion of a long-term contract?
a. Progress billings as deferred income, con struction in progress as a deferred ex pense.
b. Progress billings as income, construction in progress as inventory.
c. Net, as a current asset if debit balance and current liability if credit balance.
d. Net, as income from construction if credit balance, and loss from construction if debit balance.
41. The purpose of a flexible budget is to a. Allow management some latitude in meeting goals.
b. Eliminate cyclical fluctuations in production reports by ignoring variable costs.
c. Compare actual and budgeted results at virtually any level of production.
d. Reduce the total time in preparing the annual budget.
42. When using full absorption costing, what costs attendant to an element of production (material, labor or overhead) are used in order to compute variances from standard amounts?
a. Total costs.
b. Variable costs.
c. Fixed costs.
d. Controllable costs.
101 Examination Questions— November 1978 43. Which of the following is the most probable reason a company would experience an unfavorable labor rate variance and a favorable labor efficiency variance?
The mix of workers assigned to the particular job was heavily weighted towards the use of higher paid experienced individuals.
The mix of workers assigned to the particular job was heavily weighted towards the use of new relatively low paid unskilled workers. Because of the production schedule workers from other production areas were assigned to assist this particular process.
Defective materials caused more labor to be used in order to produce a standard unit.
a.
b.
c.
d.
44. Each of the following areas must always have a footnote included in the financial statements comment ing on normal transactions within that area except a. Assets acquired by lease.
b. Trade accounts receivable.
c. Pension plans.
d. Employee stock options.
45. Which of the following quantitative techniques is used to determine the fixed and variable elements of a semivariable cost?
a. Queuing theory.
b. Linear programming.
c. Simplex method.
d. Least squares.
46. When using the simplex method to solve a linear programming problem for the maximization of con tribution margin, the optimum solution has been reached when the values in the index row of the matrix are a. All zero.
b. All positive or zero.
c. Equal to zero when added across.
d. Equal to a positive value when added across.
47. Internal reports prepared under the responsi bility-accounting approach should be limited to which of the following costs?
a. Only variable costs of production.
b. Only conversion costs.
c. Only controllable costs.
d. Only costs properly allocable to the cost center under generally accepted accounting principles.
48. Each of the following is a method by which to allocate joint costs except a. Relative sales value.
b. Relative profitability.
c. Relative weight, volume, or linear measure.
d. Average unit cost.
49. Normal spoilage is properly classified as a (an) a. Extraordinary item.
b. Period cost.
c. Product cost.
d. Deferred charge.
50. The minimum return that a project must earn for a company in order to leave the value of the company unchanged is the a. Current borrowing rate.
b. Discount rate.
c. Cost of capital.
d. Capitalization rate.
Number 2 (Estimated time----- 15 to 20 minutes) The earnings per share data required of a company depend on the nature of its capital structure. A corpo ration may have a simple capital structure and only compute “earnings per common share” or may have a complex capital structure and have to compute “ primary earnings per share” and “fully diluted earn ings per share.” Required: a. Define the term “common stock equivalent” and describe what securities would be considered common stock equivalents in the computation of earnings per share.
b. Define the term “complex capital structure” and discuss the disclosures (both financial and explanatory) necessary for earnings per share when a corporation has a complex capital structure.
Number 3 (Estimated time----- 15 to 20 minutes) An important part of managerial accounting is the analysis of the types of costs that a business entity can incur. These types of costs are generally classified as variable, fixed, and semivariable.
Required: a. Define and discuss the identifying character istics of a 1. Variable cost.
2. Fixed cost.
b. With respect to a semivariable cost 1. Define and discuss the identifying charac teristics of a semivariable cost.
2. Discuss the three basic methods employed to “break down” a semivariable cost into its component parts.
102 Accounting Theory Milton Corporation entered into a lease arrange ment with James Leasing Corporation for a certain machine. James's primary business is leasing and it is not a manufacturer or dealer. Milton will lease the machine for a period of three years which is 50% of the machine’s economic life. James will take possession of the machine at the end of the initial three-year lease and lease it to another smaller company that does not need the most current version of the machine. Milton does not guarantee any residual value for the machine and will not purchase the machine at the end of the lease term.
Milton’s incremental borrowing rate is 10% and the implicit rate in the lease is 8½ %. Milton has no way of knowing the implicit rate used by James. Using either rate, the present value of the minimum lease payments is between 90% and 100% of the fair value of the machine at the date of the lease agreement.
Milton has agreed to pay all executory costs directly and no allowance for these costs is included in the lease payments.
James is reasonably certain that Milton will pay all lease payments, and, because Milton has agreed to pay all executory costs, there are no important uncertainties regarding costs to be incurred by James.
Required: a. With respect to Milton (the lessee) answer the following: 1. What type of lease has been entered into? Explain the reason for your answer.
2. How should Milton compute the appro priate amount to be recorded for the lease or asset acquired?
3. What accounts will be created or affected by this transaction and how will the lease or asset and other costs related to the transaction be matched with earnings?
4. What disclosures must Milton make regarding this lease or asset?
b. With respect to James (the lessor) answer the following: 1. What type of leasing arrangement has been entered into? Explain the reason for your answer.
2. How should this lease be recorded by James and how are the appropriate amounts determined?
3. How should James determine the appro priate amount of earnings to be recog nized from each lease payment?
4. What disclosures must James make re garding this lease?
Number 4 (Estimated time------20 to 25 minutes) Part a. The appropriate method of amortizing a premium or discount on issuance of bonds is the effective interest method.
Required: 1. What is the effective interest method of amortization and how is it different from and similar to the straight-line method of amortization?
2. How is amortization computed using the effective interest method, and why and how do amounts obtained using the effective interest method differ from amounts computed under the straight-line method?
Part b. Gains or losses from the early extinguish ment of debt that is refunded can theoretically be accounted for in three ways: • Amortized over remaining life of old debt.
• Amortized over the life of the new debt issue.
• Recognized in the period of extinguishment.
Required: 1. Discuss the supporting arguments for each of the three theoretic methods of accounting for gains and losses from the early extinguishment of debt.
2. Which of the above methods is generally accepted and how should the appropriate amount of gain or loss be shown in a company’s financial state ments?
Number 5 (Estimated time------15 to 20 minutes) Number 6 (Estimated time----- 15 to 20 minutes) Interim financial reporting has become an im portant topic in accounting. There has been consider able discussion as to the proper method of reflecting results of operations at interim dates. Accordingly, the Accounting Principles Board issued an opinion clarifying some aspects of interim financial reporting.
Required: a. Discuss generally how revenue should be recognized at interim dates and specifically how reve nue should be recognized for industries subject to large seasonal fluctuations in revenue and for long term contracts using the percentage-of-completion method at annual reporting dates.
b. Discuss generally how product and period costs should be recognized at interim dates. Also discuss how inventory and cost of goods sold may be afforded special accounting treatment at interim dates.
c. Discuss how the provision for income taxes is computed and reflected in interim financial statements.
103 Examination Questions May 1979 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) EXAMINATION IN ACCOUNTING PRACTICE — PART I May 2, 1979; 1:30 to 6:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: No. 1.....................
No. 2 .....................
No. 3 .....................
No. 4 .....................
No. 5 .....................
Total...........
Estimated Minutes Minimum Maximum 40 50 40 50 50 60 50 60 40 50 220 270 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For in stance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. Enclose all scratch sheets. Failure to enclose scratch sheets may result in loss of grading points. Scratch sheets need not have page numbers, but you should show the question number and place them immediately following the question to which they relate.
4. Fourteen-column sheets, if any, should not be folded until all sheets, both wide and narrow, are placed in the proper sequence and fastened together at the top left corner. All fourteen- column sheets should then be wrapped around the back of the papers.
5. A CPA is continually confronted with the necessi ty of expressing opinions and conclusions in writ ten reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and applica tion of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
105 Examination Questions— May 1979 Instructions Select the best answer for each of the following items relating to a variety of financial-accounting problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 97. Gross billings for merchandise sold by Baker Company to its customers last year amounted to $5,260,000; sales returns and allowances reduced the amounts owed by $160,000. How much were net sales last year for Baker Company?
a. $4,800,000.
b. $5,100,000.
c. $5,200,000.
d. $5,260,000.
Answer Sheet 97. Items to be Answered Number 1 (Estimated time----- 40 to 50 minutes) 1. Tob Corporation purchased certain machinery on January 1, 1973. At the date of acquisition, the machinery had an estimated useful life of ten years with no salvage. The machinery was being depreciated using the double-declining-balance method for both financial statement reporting and income tax reporting. On January 1, 1978, Tob changed to the straight-line method for depreciation of the machinery for financial statement reporting but not for income tax reporting. Assume that Tob can justify the change.
The accumulated depreciation from January 1, 1973, through December 31, 1977, under the double- declining-balance method was $200,000. If the straight- line method had been used, the accumulated deprecia tion from January 1, 1973, through December 31, 1977, would have been $140,000. Assuming that the income tax rate for the years 1973 through 1978 is 50%, the amount shown in the 1978 income statement for the cumulative effect of changing from the double- declining-balance method to the straight-line method would be a. $0.
b. $30,000 credit.
c. $60,000 credit.
d. $60,000 debit.
2. The Raff Company purchased a machine on January 1, 1978, for $5,500,000. The machine has an estimated useful life of ten years with no salvage. The machine is being depreciated using the sum-of-the- years-digits method for income tax reporting and the straight-line method for financial statement reporting. Assuming that the income tax rate is 50%, the amount of deferred taxes charged to Raff’s 1978 income state ment would be a. $225,000.
b. $275,000.
c. $450,000.
d. $550,000.
3. On December 31, 1977, Kim, Inc., had 2,000,000 shares of authorized $10 par value voting common stock of which 1,600,000 were issued and outstanding. On December 1, 1978, Kim issued 250,000 additional shares of its $10 par value voting common stock in exchange for all 100,000 shares of Terry Company’s outstanding $20 par value voting common stock in a business com bination appropriately accounted for by the pooling of interests method. The market value of Kim’s voting common stock was $30 per share on the date of the business combination. Wh at is the total consolidated common stock issued and outstanding for Kim and its subsidiary, Terry, at December 31, 1978?
a. $17,000,000.
b. $18,500,000.
c. $22,500,000.
d. $55,500,000.
4. At December 31, 1978, the Suppa Company had 500,000 shares of common stock issued and out standing, 400,000 of which had been issued and out standing throughout the year and 100,000 of which were issued on October 1, 1978. Net income for the year ended December 31, 1978, was $2,144,000. What should be Suppa’s 1978 earnings per common share, rounded to the nearest penny?
a. $4.29.
b. $4.76.
c. $5.04.
d. $5.36.
5. The Madden Company had 600,000 shares of common stock issued and outstanding at December 31, 1977. During 1978, no additional common stock was issued. On January 1, 1978, Madden issued 400,000 shares of nonconvertible preferred stock. During 1978, Madden declared and paid $200,000 cash dividends on the common stock and $110,000 on the nonconvertible preferred stock. Net income for the year ended December 31, 1978, was $750,000. What should be Madden’s 1978 earnings per common share, rounded to the nearest penny?
a. $0.73.
b. $0.92.
c. $1.07.
d. $1.25.
106 Accounting Practice — Part 1 6. During 1978, Red, Incorporated, purchased $2,000,000 of inventory. The cost of goods sold for 1978 was $2,200,000, and the ending inventory at December 31, 1978, was $400,000. What was the inven tory turnover for 1978?
a. 4.0.
b. 4.4.
c. 5.5.
d. 11.0.
7. An inventory loss from market decline of $600,000 occurred in May 1978. The Kup Company recorded this loss in May 1978 after its March 31, 1978, quarterly report was issued. None of this loss was recovered by the end of the year. How should this loss be reflected in Kup’s quarterly income statements?
Three Months Ended March June September December 31,1978 30,1978 30,1978 31,1978 a. $ 0 b. $ 0 c. $ 0 d. $150,000 $ 0 $200,000 $600,000 $150,000 $ 0 $200,000 $ 0 $150,000 $600,000 $200,000 $ 0 $150,000 8. On October 1, 1978, Mann Company purchased 500 of the $1,000 face value, 8% bonds of Womann, In corporated, for $540,000, which includes accrued interest of $10,000. The bonds, which mature on January 1, 1985, pay interest semiannually on January 1 and July 1. Assuming that Mann uses the straight-line method of amortization and that the bonds are appropriately recorded as a long-term investment, the bonds should be shown on Mann’s December 31, 1978, balance sheet at a. $528,400.
b. $528,800.
c. $530,000.
d. $540,000.
9. On December 1, 1978, Chest Corporation pur chased 200,000 shares representing 45% of the out standing stock of Park Company for cash of $2,500,000. As a result of this purchase, Chest has the ability to exercise significant influence over the operating and financial policies of Park. 45% of the net income of Park amounted to $20,000 for the month of December and $350,000 for the year ended December 31, 1978. The appropriate amount of goodwill amor tization to be recorded by Chest in 1978 as a result of its purchase of Park stock would be $10,000. On January 15, 1979, cash dividends of $0.30 per share were paid to stockholders of record on December 31, 1978. Chest’s long-term investment in Park should be shown in Chest’s December 31, 1978, balance sheet at a. $2,450,000.
b. $2,460,000.
c. $2,500,000.
d. $2,510,000.
10. The France Company owns a foreign subsidiary with 2,400,000 local currency units (LCU) of property, plant, and equipment before accumulated depreciation at December 31, 1978. Of this amount, 1,500,000 LCU were acquired in 1976 when the rate of exchange was 1.5 LCU to $1, and 900,000 LCU were acquired in 1977 when the rate of exchange was 1.6 LCU to $1. The rate of exchange in effect at December 31, 1978, was 1.9 LCU to $1. The weighted average of exchange rates which were in effect during 1978 was 1.8 LCU to $1. Assuming that the property, plant, and equipment are depreciated using the straight-line method over a ten-year period with no salvage value, how much depreciation expense relating to the foreign subsidiary’s property, plant, and equipment should be charged in France’s income statement for 1978?
a. $126,316.
b. $133,333.
c. $150,000.
d. $156,250.
11. The Standard Company leased a piece of equip ment to the Piping Company on July 1, 1977, for a one- year period expiring June 30, 1978, for $90,000 a month. On July 1, 1978, Standard leased this piece of equipment to the Tacking Company for a three-year period expiring June 30, 1981, for $100,000 a month. The original cost of the piece of equipment was $6,000,000. The piece of equipment which has been con tinually on lease since July 1, 1973, is being depreciated on a straight-line basis over an eight-year period with no salvage value. Assuming that both the lease to Piping and the lease to Tacking are appropriately recorded as operating leases for accounting purposes, what is the amount of income (expense) before income taxes that each would record as a result of the above facts for the year ended December 31, 1978?
Standard Piping a. $ 390,000 b. $ 390,000 c. $1,140,000 d. $1,140,000 ($540,000) ($540,000) ($165,000) ($915,000) Tacking ($600,000) ($975,000) ($225,000) ($600,000) 12. On December 1, 1978, the Simpson Company issued at 103, one hundred of its 5%, $1,000 bonds. Attached to each bond was one detachable stock pur chase warrant entitling the holder to purchase 10 shares of Simpson’s common stock. On December 1, 1978, the market value of the bonds, without the stock purchase warrants, was 94, and the market value of each stock purchase warrant was $60. The amount of the proceeds from the issuance that should be accounted for as the in itial carrying value of the bonds payable would be a. $ 94,000.
b. $ 96,820.
c. $ 97,000.
d. $103,000.
107 Examination Questions— May 1979 13. Cash dividends on the $10 par value common stock of Ray Company were as follows: 1st quarter of 1978 $ 800,000 2nd quarter of 1978 900,000 3rd quarter of 1978 1,000,000 4th quarter of 1978 1,100,000 The 4th quarter cash dividend was declared on December 20, 1978, to stockholders of record on December 31, 1978. Payment of the 4th quarter cash dividend was made on January 9, 1979.
In addition, Ray declared a stock dividend on its $10 par value common stock on December 1, 1978, when there were 300,000 shares issued and outstanding and the market value of the common stock was $20 per share. The shares were issued on December 21, 1978.
What was the effect on Ray’s stockholders’ equity accounts as a result of the above transactions?
Common Stock Additional Paid-In Capital Retained Earnings a. $ 0 $ 0 $3,800,000 debit b. $150,000 credit $ 0 $3,950,000 debit c. $150,000 credit $150,000 credit $4,100,000 debit d. $300,000 credit $300,000 debit $3,800,000 debit 14. The Tall Company leased equipment from the Small Company on July 1, 1978, for an eight-year period expiring June 30, 1986. Equal annual payments under the lease are $500,000 and are due on July 1 of each year. The first payment was made on July 1, 1978. The rate of interest contemplated by Tall and Small is 10%. The cash selling price of the equipment is $2,934,000 and the cost of the equipment on Small’s ac counting records was $2,500,000. Assuming that the lease is appropriately recorded as a sale for accounting purposes by Small, what is the amount of profit on the sale and the interest income that Small would record for the year ended December 31, 1978?
a. $0 and $0.
b. $0 and $121,700.
c. $434,000 and $121,700.
d. $434,000 and $133,250.
15. Donahue Corporation purchased a machine in 1976 when the general price-level index was 180. The price-level index was 190 in 1977 and 200 in 1978. The price-level indexes above are stated in terms of the 1958 base year. Donahue prepares supplemental general price-level financial statements (financial statements restated for changes in the general purchasing power of the dollar), as recommended by APB Statement No. 3. Depreciation is $100,000 a year. In Donahue’s general price-level income statement for 1978, the amount of depreciation would be stated as a. $ 90,000.
b. $ 95,000.
c. $105,263.
d. $111,111.
16. The working capital provided from operations in Seat’s statement of changes in financial position for 1978 was $8,000,000. For 1978, depreciation on fixed assets was $3,800,000, amortization of goodwill was $100,000, and dividends on common stock were $2,000,000. Based on the information given above, Seat’s net income for 1978 was a. $2,100,000.
b. $4,100,000.
c. $8,000,000.
d. $11,900,000.
17. Selected information from Basket Company’s 1978 accounting records is as follows: Proceeds from issuance of common stock $8,000,000 Proceeds from issuance of preferred stock 2,000,000 Dividends on common stock 1,000,000 Dividends on preferred stock 400,000 Purchases of treasury stock 300,000 Sales of stock to officers and employees not included above 200,000 Assuming funds are defined as working capital, Basket’s statement of changes in financial position for the year ended December 31, 1978, would show the following sources and uses of funds, based on the infor mation given above Sources Uses a.
$ 9,900,000 $1,400,000 b.
$10,000,000 $1,400,000 c.
$10,000,000 $1,900,000 d.
$10,200,000 $1,700,000 18. Royal Company’s net accounts receivable were $500,000 at December 31, 1977, and $600,000 at December 31, 1978. Net cash sales for 1978 were $200,000. The accounts receivable turnover for 1978 was 5.0. What were Royal’s total net sales for 1978?
a. $2,950,000.
b. $3,000,000.
c. $3,200,000.
d. $5,500,000.
19. On January 1, 1978, the Ben Company formed a foreign subsidiary. On February 15, 1978, Ben’s sub sidiary purchased 100,000 local currency units (LCU) of inventory. 25,000 LCU of the original inventory pur chased on February 15, 1978, made up the entire inven tory on December 31, 1978. The exchange rates were 2.2 LCU to $1 from January 1, 1978, to June 30, 1978, and 2 LCU to $1 from July 1, 1978, to December 31, 1978. The December 31, 1978, inventory balance for Ben’s foreign subsidiary should be translated into United States dollars of a. $10,500.
b. $11,364.
c. $11,905.
d. $12,500.
108 Accounting Practice — Part 1 20. On December 30, 1976, Future, Incorporated, paid $2,000,000 for land. At December 31, 1977, the fair value of the land was $2,200,000. In January 1978, the land was sold for $2,250,000. Ignoring income taxes, by what amount should stockholders’ equity be increased for 1977 and 1978 as a result of the above 22. Bernie Company purchased a new machine with an estimated useful life of five years with no salvage value for $45,000. The machine is expected to produce cash flow from operations, net of income taxes, as follows: facts in current fair value financial statements?
1st year $ 9,000 2nd year 12,000 1977 1978 3rd year 15,000 4th year 9,000 a. $ 0 $ 50,000 5th year 8,000 b.
c.
d.
$ 0 $200,000 $200,000 $250,000 $ 0 $ 50,000 Bernie will use the sum-of-the-years-digits method to depreciate the new machine in its accounting records as follows: Number 2 (Estimated time---- - 40 to 50 minutes) Instructions Select the best answer for each of the following items relating to a variety of managerial-accounting and quantitative-methods problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
Items to be Answered 21. Serven Corporation has estimated its activity for June 1979. Selected data from these estimated amounts are as follows: • Sales $700,000 Gross profit (based on sales) 30% Increase in trade accounts receivable during month $ 20,000 Change in accounts payable during month $ 0 Increase in inventory during month $ 10,000 • Variable selling, general and administrative ex penses (S, G & A) includes a charge for uncollectible ac counts of 1% of sales.
• Total S, G & A is $71,000 per month plus 15% of sales.
• Depreciation expense of $40,000 per month is included in fixed S, G & A.
On the basis of the above data, what are the estimated cash disbursements from operations for June?
a. $619,000.
b. $626,000.
c. $629,000.
d. $636,000.
1st year $15,000 2nd year 12,000 3rd year 9,000 4th year 6,000 5th year 3,000 What is the pay-back period?
a. 2 years.
b. 3 years.
c. 4 years.
d. 5 years.
23. The Bread Company is planning to purchase a new machine which it will depreciate on a straight-line basis over a ten-year period. A full year’s depreciation will be taken in the year of acquisition. The machine is ex pected to produce cash flow from operations, net of in come taxes, of $3,000 in each of the ten years. The ac counting (book value) rate of return is expected to be 10% on the initial increase in required investment. The cost of the new machine will be a. $12,000.
b. $13,500.
c. $15,000.
d. $30,000.
24. O’Connor Company manufactures Product J and Product K from a joint process. For Product J, 4,000 units were produced having a sales value at split-off of $15,000. If Product J were processed further, the addi tional costs would be $3,000 and the sales value would be $20,000. For Product K, 2,000 units were produced having a sales value at split-off of $10,000. If Product K were processed further, the additional costs would be $1,000 and the sales value would be $12,000. Using the relative-sales-value at split-off approach, the portion of the total joint product costs allocated to Product J was $9,000. What were the total joint product costs?
a. $14,400.
b. $15,000.
c. $18,400.
d. $19,000.
109 Examination Questions— May 1979 25. Oxford Company had sales of $3,000,000, variable costs of $1,800,000 and fixed costs of $800,000 for Product Brum. What would be the amount of sales dollars at the break-even point?
a. $2,000,000.
b. $2,400,000.
c. $2,600,000.
d. $2,760,000.
26. The Seahawk Company is planning to sell 200,000 units of Product B. The fixed costs are $400,000 and the variable costs are 60% of the selling price. In order to realize a profit of $100,000, the selling price per unit would have to be a. $3.75.
b. $4.17.
c. $5.00.
d. $6.25.
27. Gene, Inc., invested in a machine with a useful life of six years and no salvage value. The machine was depreciated using the straight-line method and it was ex pected to produce annual cash inflow from operations, net of income taxes, of $2,000. The present value of an ordinary annuity of $1 for six periods at 10% is 4.355. The present value of $1 for six periods at 10% is 0.564. Assuming that Gene used a time-adjusted rate of return of 10%, what was the amount of the original invest ment?
a. $5,640.
b. $8,710.
c. $9,000.
d. $11,280.
28. The Polly Company wishes to determine the amount of safety stock that it should maintain for Pro duct D that will result in the lowest cost.
The following information is available: Stockout cost Carrying cost of safety stock Number of purchase orders $80 per occurrence $2 per unit 5 per year 29. Sybil Company purchased a new machine with a useful life of seven years and no salvage on January 1, 1978, for $400,000. The machine was depreciated using the straight-line method and it was expected to produce annual cash flow from operations, net of income taxes, of $90,000. The present value of an ordinary annuity of $1 for seven periods at 12% is 4.564. The present value of $1 for seven periods at 12% is 0.452. Assuming that Sybil used a time-adjusted rate of return of 12%, what was the net present value?
a. $6,800.
b. $10,760.
c. $182,000.
d. $230,000.
30. The Aron Company requires 40,000 units of Pro duct Q for the year. The units will be required evenly throughout the year. It costs $60 to place an order. It costs $10 to carry a unit in inventory for the year. What is the economic order quantity?
a. 400.
b. 490.
c. 600.
d. 693.
31. Information for the month of May concerning Department A, the first stage of Wit Corporation’s pro duction cycle, is as follows: Conversion Materials costs Work in process, beginning $ 4,000 $ 3,000 Current costs 20,000 16,000 Total costs $24,000 $19,000 Equivalent units based on weighted- average method 100,000 95,000 The available options open to Polly are as follows: Average unit costs $ 0.24 $ 0.20 Units of Safety Stock Probability of Running Out of Safety Stock 10 50% 20 40% 30 30% 40 20% 50 10% 55 5% The number of units of safety stock that will result in the lowest cost are a. 20.
b. 40.
c. 50.
d. 55.
Goods completed Work in process, end 90,000 units 10,000 units Material costs are added at the beginning of the process. The ending work in process is 50% complete as to con version costs. How would the total costs accounted for be distributed, using the weighted-average method?
Goods Work in Completed Process, End a.
$39,600 $3,400 b.
$39,600 $4,400 c.
$43,000 $ 0 d.
$44,000 $3,400 110 Accounting Practice — Part 1 32. Andy Company manufactures products N, P, and R from a joint process. The following information is available: Product N P R Total Units pro duced Sales value 12,000 ?
7 24,000 at split-off ?
7 $50,000 $200,000 Joint costs Sales value if pro cessed fur ther Additional costs if processed further $ 48,000 7 7 $120,000 $110,000 $90,000 $60,000 $260,000 $ 18,000 $14,000 $10,000 $ 42,000 Assuming that joint product costs are allocated using the relative-sales-value at split-off approach, what was the sales value at split-off for products N and P?
Product N Product P a.
$ 66,000 $84,000 b.
$ 80,000 $70,000 c.
$ 98,000 $84,000 d.
$100,000 $50,000 33. The Pauley Company plans to expand its sales force by opening several new branch offices. Pauley has $10,400,000 in capital available for new branch offices. Pauley will consider opening only two types of branches; 20-person branches (Type A) and 10-person branches (Type B). Expected initial cash outlays are $1,300,000 for a Type A branch and $670,000 for a Type B branch. Expected annual cash inflow, net of in come taxes, is $92,000 for a Type A branch and $36,000 for a Type B branch. Pauley will hire no more than 200 employees for the new branch offices and will not open more than 20 branch offices. Linear programming will be used to help decide how many branch offices should be opened.
In a system of equations for a linear programming model, which of the following equations would not represent a constraint (restriction)?
a. A + B ≤ 20.
b. 20 A + 10 B ≤ 200.
c. $92,000 A + $36,000 B ≤ $128,000.
d. $1,300,000 A + $670,000 B ≤ $10,400,000.
34. Bert Company has projected cost of goods sold of $2,0 0 ,000 including fixed costs of $400,000 and variable costs are expected to be 75% of net sales. What will be the projected net sales?
a. $2,133,333.
b. $2,400,000.
c. $2,666,667.
d. $3,200,000.
35. The Wiring Department is the second stage of Flem Company’s production cycle. On May 1, the beginning work in process contained 25,000 units which were 60% complete as to conversion costs. During May, 100,000 units were transferred in from the first stage of Flem’s production cycle. On May 31, the ending work in process contained 20,000 units which were 80% com plete as to conversion costs. Material costs are added at the end of the process. Using the weighted-average method, the equivalent units were Transferred-in costs Materials Conversion costs 125,000 105,000 105,000 125,000 36. Cardinal Company needs 20,000 units of a certain part to use in its production cycle. The following in formation is available: a.
b.
c.
d.
100,000 125,000 125,000 125,000 100,000 105,000 121,000 121,000 Cost to Cardinal to make the part: Direct materials Direct labor Variable overhead Fixed overhead applied Cost to buy the part from the Oriole Company $ 4 16 8 10 $38 $36 If Cardinal buys the part from Oriole instead of making it, Cardinal could not use the released facilities in another manufacturing activity. 60% of the fixed overhead applied will continue regardless of what deci sion is made.
In deciding whether to make or buy the part, the total relevant costs to make the part are a. $560,000.
b. $640,000.
c. $720,000.
d. $760,000.
37. Information on Material Company’s direct- material costs is as follows: Actual units of direct materials used Actual direct-material costs Standard price per unit of direct materials Direct-material efficiency variance-favorable 20,000 $40,000 $ 2.10 $ 3,000 What was Material’s direct-material price variance?
a. $1,000 favorable.
b. $1,000 unfavorable.
c. $2,000 favorable.
d. $2,000 unfavorable.
.1 1 1 Examination Questions— May 1979 38. Boyer Company manufactures basketballs. The forecasted income statement for the year before any special orders is as follows: Sales Manufacturing cost of goods sold Gross profit Selling expenses Operating income Amount $4,000,000 3,200,000 800,000 300,000 Per Unit $10.00 8.00 2.00 .75 $ 500,000 $ 1.25 Fixed costs included in the above forecasted income statement are $1,200,000 in manufacturing cost of goods sold and $100,000 in selling expenses.
A special order offering to buy 50,000 basketballs for $7.50 each was made to Boyer. There will be no ad ditional selling expenses if the special order is accepted. Assuming Boyer has sufficient capacity to manufacture 50,000 more basketballs, by what amount would operating income be increased or decreased as a result of accepting the special order?
a. $25,000 decrease.
b. $62,500 decrease.
c. $100,000 increase.
d. $125,000 increase.
39. Information on Overhead Company’s overhead costs is as follows: Standard applied overhead $80,000 Budgeted overhead based on standard direct-labor hours allowed $84,000 Budgeted overhead based on actual direct-labor hours allowed $83,000 Actual overhead $86,000 What is the total overhead variance?
a. $2,000 unfavorable.
b. $3,000 favorable.
c. $4,000 favorable.
d. $6,000 unfavorable.
40. Cooper plans to invest $2,000 at the end of each of the next ten years. Assume that Cooper will earn interest at an annual rate of 6% compounded annually. The future amount of an ordinary annuity of $1 for ten periods at 6% is 13.181. The present value of $1 for ten periods at 6% is 0.558. The present value of an ordinary annuity of $1 for ten periods at 6% is 7.360. The invest ment after the end of ten years would be a. $14,720.
b. $21,200.
c. $26,362.
d. $27,478.
Number 3 (Estimated tim e------50 to 60 minutes) Number 3 consists of three unrelated parts.
Part a. On September 1, 1977, the Horn Company purchased 200,000 shares representing 45% of the outstanding stock of Mat Company for cash. As a result of the purchase, Horn has the ability to exercise significant influence over the operating and financial policies of Mat. Goodwill of $500,000 was appropriate ly recognized by Horn at the date of the purchase.
On December 1, 1978, Horn purchased 300,000 shares representing 30% of the outstanding stock of Simon Company for cash of $2,500,000. The stockholders’ equity section of Simon’s balance sheet at the date of the acquisition was as follows: Common stock, par value $2.00 a share Additional paid-in capital Retained earnings $2,000,000 1,000,000 4,000,000 $7,000,000 Furthermore, at the date of acquisition, the fair value of Simon’s property, plant, and equipment, net, was $3,800,000 whereas the book value was $3,500,000. For all of the other assets and liabilities of Simon the fair value and book value was equal. As a result of the transaction, Horn has the ability to exercise significant influence over the operating and financial policies of Simon.
Assume that Horn amortizes goodwill over the maximum period allowed and takes a full year’s amortization in the year of purchase.
Required: Prepare a schedule computing the amount of good will and accumulated amortization at December 31, 1978, and the goodwill amortization for the year ended December 31, 1978. Show supporting computations in good form.
Part b. The Barb Company has provided in formation on intangible assets as follows: • A patent was purchased from the Lou Company for $1,500,000 on January 1, 1977. Barb estimated the remaining useful life of the patent to be ten years. The patent was carried in Lou’s accounting records at a net book value of $ 1,250,000 when Lou sold it to Barb.
• During 1978, a franchise was purchased from the Rink Company for $500,000. In addition, 5% of revenue from the franchise must be paid to Rink. Revenue from the franchise for 1978 was $2,000,000. Barb estimates the useful life of the franchise to be ten years and takes a full year’s amortization in the year of purchase.
112 Accounting Practice — Part I • Barb incurred research and development costs in 1978 as follows: Materials and equipment Personnel Indirect costs $120,000 140,000 60,000 $320,000 Barb estimates that these costs will be recouped by December 31, 1981.
• On January 1, 1978, Barb, based on new events that have occurred in the field, estimates that the re maining life of the patent purchased on January 1, 1977, is only five years from January 1, 1978.
Required: 1. Prepare a schedule showing the intangibles section of Barb’s balance sheet at December 31, 1978. Show supporting computations in good form.
2. Prepare a schedule showing the income state ment effect for the year ended December 31, 1978, as a result of the above facts. Show supporting computa tions in good form.
Part c. The Wing Company purchased a machine on January 1, 1975, for $240,000. At the date of acquisition, the machine had an estimated useful life of ten years with an estimated salvage value of $20,000. The machine is being depreciated on a straight-line basis. On January 1, 1978, Wing appropriately adopted the sum-of-the-years-digits method of depreciation for this machine.
Required: 1. Prepare a schedule computing the book value of this machine, net of accumulated depreciation, that would be included in Wing’s balance sheet at December 31, 1978. Show supporting computations in good form.
2. Prepare a schedule computing the cumulative effect on prior years of changing to a different deprecia tion method for the year ended December 31, 1978. Assume that the direct effects of this change are limited to the effect on depreciation and the related tax provi sion, and that the income tax rate was 50% in all years. Show supporting computations in good form.
available facts, the 1978 provision for doubtful accounts is estimated to be 0.7% of net sales. During 1978, uncollectible receivables amounting to $410,000 were written off against the allowance for doubtful ac counts.
Required: Prepare a schedule computing the balance in Master’s allowance for doubtful accounts at December 31, 1978. Show supporting computations in good form.
Part b. The Guide Company requires additional cash for its business. Guide has decided to use its ac counts receivable to raise the additional cash as follows: • On July 1, 1978, Guide assigned $200,000 of ac counts receivable to the Cell Finance Company. Guide received an advance from Cell of 85% of the assigned accounts receivable less a commission on the advance of 3%. Prior to December 31, 1978, Guide collected $150,000 on the assigned accounts receivable, and remitted $160,000 to Cell, $10,000 of which represented interest on the advance from Cell.
• On December 1, 1978, Guide sold $300,000 of net accounts receivable to the Factoring Company for $260,000. The receivables were sold outright on a nonrecourse basis.
• On December 31, 1978, an advance of $100,000 was received from the Domestic Bank by pledging $120,000 of Guide’s accounts receivable. Guide’s first payment to Domestic is due on January 30, 1979.
Required: Prepare a schedule showing the income statement effect for the year ended December 31, 1978, as a result of the above facts. Show supporting computations in good form.
Part c. On January 1, 1977, the Lock Company sold property to the Key Company which originally cost Lock $600,000. Key gave Lock a $900,000 noninterest bearing note payable in six equal annual installments of $150,000, with the first payment due and paid on January 1, 1977. There was no established exchange price for the property and the note has no ready market. The prevailing rate of interest for a note of this type is 12%. The present value of an annuity of $1 in advance for six periods at 12% is 4.605.
Number 4 (Estimated time-----50 to 60 minutes) Number 4 consists of three unrelated parts.
Part a. At January 1, 1978, the credit balance in the allowance for doubtful accounts of the Master Company was $400,000. For 1978, the provision for doubtful accounts is based on a percentage of net sales. Net sales for 1978 were $50,000,000. Based on the latest Required: 1. Prepare a schedule computing the balance in Lock’s net receivables from Key at December 31, 1978, based on the above facts. Show supporting computa tions in good form.
2. Prepare a schedule showing the income or loss before income taxes for the years ended December 31, 1977, and 1978, that Lock should record as a result of the above facts.
113 Examination Questions— May 1979 Number 5 (Estimated tim e------40 to 50 minutes) Number 5 consists of two unrelated parts.
Part a. Left Corporation has engaged you to compute its federal taxable income for the year ended December 31, 1978. The following information was sup plied by Left’s bookkeeper: Gross profit from sales Operating expenses Operating income Other expense Income before income taxes $1,500,000 1,000,000 500,000 50,000 $ 450,000 Additional information, none of which was included above, is as follows: • On January 15, 1978, Left sold for $10,000 marketable securities that were purchased for $8,500 on January 15, 1976.
• On February 15, 1978, Left sold for $8,000 marketable securities that were purchased on October 15, 1977, for $9,000.
• On March 30, 1978, Left purchased 10,000 shares of its own $10 par value common stock for $210,000.
• During 1978, Left received dividends from domestic corporations of $12,000 and dividends from a wholly-owned domestic subsidiary with which Left does not file a consolidated return, of $10,000.
• Left has a pension plan for its employees. Left has set up an employees’ trust which meets the re quirements of Code Section 401. Information concern ing the pension plan activities for 1978 is as follows: Contributions to the trust by Left $25,000 Income from the investments of the trust 7,000 Pension benefits paid to retirees 21,000 • On December 20, 1978, Left distributed hams costing $400 to its employees as a Christmas bonus.
• During 1978, Left made contributions to various recognized charitable organizations of $40,000 and to various indigent persons of $100.
Required: 1. Prepare a schedule, beginning with income before income taxes of $450,000, computing the federal taxable income of Left for the 1978 calendar year. Any possible alternative treatments should be resolved in a manner that will minimize taxable income. Show sup porting computations in good form.
2. List the items in the fact situation which were not used to determine the answer to 1. above. Explana tions are not required.
Part b. Right Corporation appropriately applied the normal and surtax rates to its federal taxable income for the calendar year 1978 and arrived at $500,000. Assume that the effective combined normal and surtax rate is 50%. Additional information is as follows: • The balance due of $20,000 for the 1977 federal income tax return was paid in 1978.
• Based on estimates of its 1978 federal income tax, Right made estimated tax payments of $430,000 in 1978.
• Right’s investment tax credit for 1978 purchases of qualified investments was $30,000. An unused invest ment tax credit from 1977 purchases of qualified in vestments was $15,000.
• Right properly computed its foreign tax credit at $10,000.
• Right has withheld $40,000 of Social Security taxes from employees’ wages in 1978. The employer’s share has already been deducted as an expense in the calculation of the total federal income tax of $500,000 above.
Required: 1. Prepare a schedule computing Right’s federal income tax due for the 1978 calendar year. Any possible alternative treatments should be resolved in a manner that will minimize income taxes.
2. List the items in the fact situation which were not used to determine the answer to 1. above. Explana tions are not required.
114 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXAMINATION IN ACCOUNTING PRACTICE — PART II May 3, 1979; 1:30 to 6:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1.... No. 2 . .. . No. 3 ... . No. 4 .... No. 5 ....
Total 40 50 40 50 50 60 50 60 40 50 220 270 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “ continued” at the bottom of the page. Number pages consecutively. For in stance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2 . Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3 . Enclose all scratch sheets. Failure to enclose scratch sheets may result in loss of grading points. Scratch sheets need not have page numbers, but you should show the question number and place them immediately following the question to which they relate.
4 . Fourteen-column sheets, if any, should not be folded until all sheets, both wide and narrow, are placed in the proper sequence and fastened together at the top left corner. All fourteen- column sheets should then be wrapped around the back of the papers.
5 . A CPA is continually confronted with the necessi ty of expressing opinions and conclusions in writ ten reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and applica tion of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
115 Examination Questions— May 1979 Number 1 (Estimated time----- 40 to 50 minutes) Instructions Select the best answer for each of the following items relating to a variety of financial-accounting problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 97. Gross billings for merchandise sold by Baker Company to its customers last year amounted to $5,260,000; sales returns and allowances reduced the amounts owed by $160,000. How much were net sales last year for Baker Company?
a. $4,800,000.
b. $5,100,000.
c. $5,200,000.
d. $5,260,000.
Answer Sheet 97. Items to be Answered 1. Presented below is the stockholders’ equity section of Caper Corporation at December 31, 1977: Common stock, par value $20; authorized 50,000 shares; issued and outstanding 30,000 shares $600,000 Capital in excess of par value 150,0(X) Retained earnings 230,000 $980,000 During 1978 the following transactions occurred relating to stockholders’ equity: 1,000 shares were reacquired at $28 per share.
900 shares were reacquired at $30 per share.
1,500 shares of treasury stock were sold at $32 per share.
For the year ended December 31, 1978, Caper reported net income of $110,000. The state in which Caper is in corporated places a restriction on the distribution of re tained earnings equal to the cost of treasury stock. Assuming Caper accounts for treasury stock under the cost method, what should it report as total stockholders’ equity on its December 31, 1978, balance sheet?
a. $1,071,000.
b. $1,078,000.
c. $1,083,000.
d. $1,090,000.
2. In an effort to increase sales, Nick Razor Blade Company inaugurated a sales promotional campaign on June 30, 1978, whereby Nick placed a coupon in each package of razor blades sold, the coupons being redeemable for a premium. Each premium costs Nick $.50 and 5 coupons must be presented by a customer to receive a premium. Nick estimated that only 60% of the coupons issued will be redeemed. For the six months ended December 31, 1978, the following information is available: Packages o f Razor Blades Sold 800,000 Premiums Purchased 60,000 Coupons Redeemed 200,000 What is the estimated liability for premium claims outstanding at December 31, 1978?
a. $20,000.
b. $28,000.
c. $36,000.
d. $48,000.
3. On its December 31, 1977, balance sheet, the Noble Corporation reported the following as in vestments in long-term marketable equity securities: Investment in long-term marketable equity securities at cost $300,000 Less allowance to reduce long-term equity securities to market 28,000 $272,000 At December 31, 1978, the market valuation of the port folio was $298,000. What should Noble report on its 1978 Statement of Income as a result of the increase in the market value of the investments in 1978?
a. $0.
b. Unrealized loss of $2,000.
c. Realized gain of $26,000.
d. Unrealized gain of $26,000.
4. Spare Corporation had two issues of securities outstanding: common stock and a 5% convertible bond issue in the face amount of $10,000,000. Interest pay ment dates of the bond issue are June 30th and December 31st. The conversion clause in the bond in denture entitles the bondholders to receive forty shares of $20 par value common stock in exchange for each $1,000 bond. On June 30, 1978, the holders of $900,000 face value bonds exercised the conversion privilege. The market price of the bonds on that date was $1,100 per bond and the market price of the common stock was $35. The total unamortized bond discount at the date of conversion was $500,000. In applying the book value method, what amount should Spare credit to the ac count “capital in excess of par,’’ as a result of this con version?
a. $135,000.
b. $180,000.
c. $460,000.
d. $540,000.
116 Accounting Practice — Part II 5. Dobbin Corporation, a manufacturer of household paints, is preparing annual financial statements at December 31, 1978. Because of a recently proven health hazard in one of its paints, the govern ment has clearly indicated its intention of having Dobbin recall all cans of this paint sold in the last six months. The management of Dobbin estimates that this recall would cost $1,000,000. What accounting recogni tion, if any, should be accorded this situation?
a. No recognition.
b. Footnote disclosure.
c. Operating expense of $1,000,000.
d. Extraordinary loss of $1,000,000.
6. On October 1, 1978, The Ajax Company consigned one hundred television sets to M & R Retailers, Inc. Each television set had a cost of $150. Freight on the shipment was paid by Ajax in the amount of $200.
On December 1, 1978, M & R submitted an account sales stating that it had sold sixty sets and it remitted the $12,840 balance due. The remittance was net of the following deductions from the sales price of the televi sions sold: Commission (20% of sales price) Advertising Delivery and installation charges $500 100 What was the total sales price of the television sets sold by M & R?
a. $13,440.
b. $15,000.
c. $16,800.
d. $17,000.
7. Jackson Corporation provides an incentive com pensation plan under which its president is to receive a bonus equal to 10% of Jackson’s income in excess of $100,000 before deducting income tax but after deduct ing the bonus. If income before income tax and the bonus is $320,000, the amount of the bonus should be a. $44,000.
b. $32,000.
c. $22,000.
d. $20,000.
8. On January 1, 1978, Harry Corporation sold equipment costing $2,000,000 with accumulated depreciation of $500,000 to Anna Corporation, its wholly-owned subsidiary, for $1,800,000. Harry was depreciating the equipment on the straight-line method over twenty years with no salvage value, which Anna continued. In consolidation at December 31, 1978, the cost and accumulated depreciation, respectively, should be a. $1,500,000 and $100,000.
b. $1,800,000 and $100,000.
c. $2,000,000 and $100,000.
d. $2,000,000 and $600,000.
Items 9 and 10 are based on the following informa tion: The following balance sheet is presented for the partner ship of Davis, Wright, and Dover who share profits and losses in the ratio of 5:3:2 respectively: $ 60,000 Cash Other assets Liabilities Davis, Capital Wright, Capital Dover, Capital 540,000 $600,000 $140,000 280,000 160,000 20,000 $600,000 9. Assume that the assets and liabilities are fairly valued on the balance sheet and the partnership decided to admit Hank as a new partner with a one-fifth in terest. No goodwill or bonus is to be recorded. How much should Hank contribute in cash or other assets?
a. $120,000.
b. $115,000.
c. $92,000.
d. $73,600.
10. Assume that instead of admitting a new partner, the partners decided to liquidate the partnership. If the other assets are sold for $400,000, how should the available cash be distributed to each partner?
a. Davis, $280,000; Wright, $160,000; Dover, $20,000.
b. Davis, $210,000; Wright, $118,000; Dover, $8,000.
c. Davis, $206,000; Wright, $114,000; Dover, $0.
d. Davis, $205,000; Wright, $115,000; Dover, $0.
11. On December 1, 1978, Branch Corporation leased office space for 10 years at a monthly rental of $15,000. On that date Branch paid the landlord the following amounts: Rent deposit First month’s rent Last month’s rent Installation of new walls and offices $ 15,000 15,000 15,000 96,000 $141,000 The entire amount of $141,000 was charged to rent ex pense in 1978. What amount should Branch have charged to expense for the year ended December 31, 1978?
a. $15,000.
b. $15,800.
c. $30,800.
d. $96,000.
117 Examination Questions— May 1979 12. Brad Corporation provides an allowance for its doubtful accounts receivable. At December 31, 1977, the allowance account had a credit balance of $4,000. Each month, Brad accrues bad debt expense in an amount equal to 1% of sales on account. Total sales on account during 1978 amounted to $1,000,000. During 1978 uncollectible accounts receivable totaling $16,000 were written off against the allowance account. An aging of accounts receivable at December 31, 1978, in dicates that an allowance of $20,000 should be provided for doubtful accounts as of that date. Accordingly, bad debt expense previously accrued during 1978 should be increased by a. $26,000.
b. $22,000.
c. $20,000.
d. $2,000.
13. A review of the December 31, 1978, financial statements of Rhur Corporation revealed that under the caption “extraordinary losses,” Rhur reported a total of $260,000. Further analysis revealed that the $260,000 in losses was comprised of the following items: 1. Rhur recorded a loss of $50,000 incurred in the abandonment of equipment formerly used in the business.
2. In an unusual and infrequent occurrence, a loss of $75,000 was sustained as a result of hurricane damage to a warehouse.
3. During 1978, several factories were shut down during a major strike by employees. Shut down expenses totaled $120,000.
4. Uncollectible accounts receivable of $15,000 were written off as uncollectible.
Ignoring income taxes, what amount of loss should Rhur report as extraordinary on its 1978 Statement of Income?
a. $50,000.
b. $75,000.
c. $135,000.
d. $260,000.
14. In January 1975 Tracy Corporation purchased a patent for a new consumer product for $180,000. At the time of purchase, the patent was valid for fifteen years. Due to the competitive nature of the product however, the patent was estimated to have a useful life of only ten years. During 1978 the product was permanently removed from the market under governmental order because of a potential health hazard present in the product. What amount should Tracy charge to expense during 1978, assuming amortization is recorded at the end of each year?
a. $12,000.
b. $18,000.
c. $126,000.
d. $144,000.
15. The Fisher Company purchased a machine on November 1, 1969, for $74,000. At the time of acquisi tion, the machine was estimated to have a useful life of ten years and an estimated salvage value of $2,000. Fisher has recorded monthly depreciation using the straight-line method. On July 1, 1978, the machine was sold for $6,500. What should be the loss recognized from the sale of the machine?
a. $0.
b. $2,500.
c. $5,100.
d. $6,500.
16. On January 1, 1974, Hal Company purchased equipment at a cost of $31,000. The equipment was estimated to have a salvage value of $1,000 and it is be ing depreciated over five years under the sum-of-the- years-digits method. What should be the charge for depreciation of this equipment for the year ended December 31, 1978?
a. $1,000.
b. $2,000.
c. $3,000.
d. $6,000.
17. In 1978 Murray Corporation developed a new pro duct that will be marketed in 1979. In connection with the development of this product, the following costs were incurred in 1978.
Research and development departmental costs Materials and supplies consumed Compensation paid to research consultants $200,000 50,000 60,000 $310,000 It is anticipated that these costs will be recovered in 1981. What is the amount of research and development costs that Murray should record in 1978 as a charge to income?
a. $0.
b. $60,000.
c. $250,000.
d. $310,000.
18. The Miller Corporation was established in 1970. In 1978 it adopted a pension plan for its employees. On December 31, 1978, the past service cost was determined to be $500,000. Miller had elected to amortize past service cost over ten years and to fund past service cost over fifteen years. The past service cost of $500,000 as of December 31, 1978, should be accounted for as a charge to a. Prior periods as a prior-period adjustment.
b. Operations in 1978.
c. Operations ratably from 1978 through 1987.
d. Operations ratably from 1978 through 1992.
118 Accounting Practice — Part II 19. On April 1, 1978, Milo Corporation purchased for $270,000 a tract of land on which was located a warehouse and office building. The following data were collected concerning the property: Current Vendor’ s Assessed Original Valuation Cost Land Warehouse Office building $ 87,500 37,500 100,000 $225,000 $ 70,000 40,000 90,000 $200,000 What are the appropriate amounts that Milo should record for the land, warehouse, and office building, respectively?
a. Land, $70,000; warehouse, $40,000; office building, $90,000.
b. Land, $87,500; warehouse, $37,500; office building, $100,000.
c. Land, $94,500; warehouse, $54,000; office building, $121,500.
d. Land, $105,000; warehouse, $45,000; office building, $120,000.
20. During 1978 Boyd Corporation, which uses the allowance method of accounting for uncollectible ac counts, recorded charges to bad debt expense of $50,000 and in addition it wrote off, as uncollectible, accounts receivable of $42,000. As a result of these transactions, working capital was decreased by a. $50,000.
b. $42,000.
c. $8,000.
d. $0.
Number 2 (Estimated time-----40 to 50 minutes) Instructions Select the best answer for each of the following items relating to the federal income taxation of indivi duals. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. The answers should be selected in accordance with the current Internal Revenue Code and Tax Regulations. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
Items to be Answered 21. Mrs. Irma Felton, by herself, maintains her home in which she and her unmarried son reside. Her son, however, does not qualify as her dependent. Mrs. Felton’s husband died in 1977. What is Mrs. Felton’s filing status for 1978?
a. Single.
b. Surviving spouse.
c. Head of household.
d. Married filing jointly.
22. Art Hollender was divorced from his wife Diane in 1977. Under the terms of the divorce decree, he was re quired to make the following periodic payments each month to his former wife who retained custody of their children: Alimony Child support $600 400 For 1978 his only income was his salary of $40,000, and he paid $12,000 to his former wife under the terms of the divorce decree. What is his 1978 adjusted gross in come?
a. $28,000.
b. $32,800.
c. $35,200.
d. $40,000.
23. Jack and Joan Mitchell, married taxpayers and residents of a separate property state, elect to file a joint return for 1978 during which they received the following dividends: Received by Alert Corporation (a qualified, domestic corporation) Canadian Mines, Inc. (a Canadian Company) Eternal Life Mutual In surance Company (divi dends on life insurance policy) Jack $400 Joan $ 50 300 200 For 1978 what amount should the Mitchells report on their joint return as dividend income net of the allowable dividend exclusion?
a. $550.
b. $600.
c. $750.
d. $800.
24. Arthur Mends, age 19, is a full-time student at Gordon College and a candidate for a bachelor’s degree. During 1978 he received the following payments: State scholarship for ten months Loan from college financial aid office Cash support from parents Cash dividends on qualified investments Cash prize awarded in contest What is his adjusted gross income for 1978?
a. $700.
b. $800.
c. $3,200.
d. $6,200.
$2,400 1,000 2,000 500 300 $6,200 119 Examination Questions— May 1979 25. For the year 1978 Fred and Wilma Todd reported the following items of income: Salary Interest income Cash prize won on T.V. game show Fred $40,000 1,000 $41,000 Wilma $ 200 8,800 $9,000 Fred is not covered by any qualified retirement plan and he and Wilma established individual retirement ac counts during the year. Assuming a joint return was filed for 1978, what is the maximum amount that they can be allowed for contributions to their individual retirement accounts?
a. $1,500.
b. $1,750.
c. $6,000.
d. $7,500.
26. James Martin received the following compensa tion and fringe benefits from his employer during 1978: Salary Year-end bonus Medical insurance premiums paid by employer Allowance paid for moving expenses $50,000 10,000 1,000 5,000 What amount of the preceding payments should be in cluded in Martin’s 1978 gross income?
a. $60,000.
b. $61,000.
c. $65,000.
d. $66,000.
27. Which of the following is deductible to arrive at adjusted gross income?
a. Unreimbursed dues to AICPA by an employee of an accounting firm.
b. Unreimbursed union dues by an employee of a company.
c. Medical expenses of a self-employed in dividual.
d. None of the above.
28. On July 1, 1976, the original date of issue David Karp purchased for $9,520, a $10,000 ten-year bond of the Expoxy Corporation. The bond was issued for long term financing. On January 31, 1978, he sold the bond to an unrelated party for $9,800. What amount should Karp report as a long-term capital gain from this trans action?
a. $76.
b. $204.
c. $200.
d. $280.
29. Mr. and Mrs. Morris Benson were 68 years old in June of 1978 when Mr. Benson died. In filing their tax return for 1978, the maximum number of exemptions that can be taken is a. 1.
b. 2.
c. 3.
d. 4.
30. Julian Davis, a cash-basis taxpayer, had adjusted gross income of $30,000 in 1978. During the year he in curred and paid the following medical-related expenses: Drugs and medicines prescribed by doctors $ 200 Health insurance premiums 500 Doctors’ fees 1,500 Eyeglasses and repairs 50 $2,250 Davis received $600 in 1978 as reimbursement for a por tion of the doctors’ fees. If Davis were to itemize his deductions from adjusted gross income, what would his allowable net medical expense deduction be?
a. $550.
b. $750.
c. $1,100.
d. $1,650.
31. In gathering information for her income tax return for 1978, Mabel Herzog listed the following miscel laneous expenses incurred and paid for in 1978: Hobby expenses (not engaged in for profit) $500 Union dues 400 Employment agency fees paid in securing a new job in same profession 200 Legal fees paid in connection with a libel suit 700 What can Herzog report as allowable deductions from adjusted gross income?
a. $600.
b. $1,100.
c. $1,300.
d. $1,800.
32. Gordon Thorn had an adjusted gross income in 1978 of $45,000. During the year his personal summer home was completely destroyed by a cyclone. Pertinent data with respect to the home follows: Cost basis Value before casualty Value after casualty $26,000 30,000 2,000 Thorn was partially insured for his loss and in 1978 he received a $10,000 insurance settlement. What is Thorn’s allowable casualty loss deduction for 1978?
a. $15,900.
b. $16,000.
c. $17,000.
d. $18,000.
120 Accounting Practice — Part II 33. During 1978 George Burke, a salaried taxpayer, paid the following taxes which were not incurred in con nection with a trade or business: Federal income tax (withheld by employer) $1,500 State income tax (withheld by employer) 1 ,000 F.I.C.A. tax (withheld by employer) 700 State sales taxes 900 Federal auto gasoline taxes 200 Federal excise tax on telephone bills 50 What taxes are allowable deductions from Burke’s ad justed gross income for 1978?
a. $4,350.
b. $2,850.
c. $2,550.
d. $1,900.
34. During 1978 Richard Jason was assessed a defi ciency on his 1976 federal income tax return. As a result of this assessment he had to pay $635 determined as follows: Additional tax Penalty Interest $500 50 85 $635 If Jason itemizes his deductions on his 1978 return, this payment of $635 will allow him to claim an allowable deduction of a. $635.
b. $135.
c. $85.
d. $50.
35. During 1978, Albert Mason purchased the follow ing long-term investments at par: $10,000 general obligation bonds of Tulip County (wholly tax exempt) $10,000 debentures of Laxity Corporation He financed these purchases by obtaining a loan from the Community Bank for $20,000. For the year 1978, he paid the following amounts as interest expense: Community Bank Interest on mortgage Interest on installment purchases $1,600 3,000 300 $4,900 What amount can Mason deduct as interest expense in 1978?
36. Mark Berkson loaned a friend $1,000 in 1976 and it had not been repaid in 1978 when the friend died in solvent. For 1978 Berkson should account for the non payment of the loan as a a. Deduction from adjusted gross income.
b. Long-term capital loss.
c. Ordinary loss.
d. Short-term capital loss.
37. For the year 1978 Michael King reported salary and taxable interest income of $40,000. His capital asset transactions during the year were as follows: Long-term capital gains Long-term capital losses Short-term capital gains $2,000 (8,000) 1,000 For 1978 King should report adjusted gross income of a. $35,000.
b. $37,500.
c. $38,000.
d. $39,000.
38. Gilda Bach is a cash basis self-employed consul tant. For the year 1978 she determined that her net in come from self-employment was $80,000. In reviewing her books you determine that the following items were included as business expenses in arriving at the net in come of $80,000: Salary drawn by Gilda Bach $20,000 Estimated federal self-employment and income taxes paid 6,0(X) Malpractice insurance premiums 4,000 Cost of attending professional seminar 1,000 Based upon the above information, what should Gilda Bach report as her net self-employment income for 1978?
a. $91,000.
b. $105,000.
c. $106,000.
d. $110,000.
39. Phil Collins owns numerous oil leases in the Southwest. During 1978 he made several trips to inspect oil wells on the leases and to consult about future oil wells to be drilled on these sites. As a result of these overnight trips, he paid the following expenses: Plane fares Hotels Meals Entertaining lessees $4,000 1,000 800 500 Of the $6,300 in expenses incurred, he can claim as deductible expenses a. $6,300.
b. $5,800.
c. $5,500.
d. $5,000.
a.
$4,900.
b.
$4,100.
c.
$3,600.
d.
$3,300.
121 Examination Questions— May 1979 40. Sidney Fields owns a two-family house which has two identical apartments. Fields lives in one apartment and rents out the other. In 1978, the rental apartment was fully occupied and Fields received $2,400 in rent. For the year ended December 31, 1978, he paid the following amounts; Real estate taxes Painting of rental apartment Annual fire insurance premium $3,200 400 300 In 1978 depreciation for the entire house was deter mined to be $2,500. Fields should include in his adjusted gross income for 1978 a. Income of $2,400.
b. Income of $250.
c. Loss of $800.
d. Loss of $1,000.
Number 3 (Estimated time-----50 to 60 minutes) Number 3 consists of three unrelated parts.
Part a. On June 30, 1978, a flash flood damaged the warehouse and factory of Padway Corporation, completely destroying the work-in-process inventory. There was no damage to either the raw materials or finished goods inventories. A physical inventory taken after the flood revealed the following valuations: Raw materials Work in process Finished goods $ 62,000 - 0-119,000 The inventory on January 1, 1978, consisted of the following: Raw materials Work in process Finished goods $ 30,000 100,000 140.000 $270,000 A review of the books and records disclosed that the gross profit margin historically approximated 25% of sales. The sales for the first six months of 1978 were $340,000. Raw material purchases were $115,000. Direct labor costs for this period were $80,000, and manufacturing overhead has historically been applied at 50% of direct labor.
Required: Compute the value of the work-in-process inven tory lost at June 30, 1978. Show supporting computa tions in good form.
Part b. The Supreme Clothing Store values its in ventory under the retail inventory method at the lower of cost or market. The following data are available for the month of November 1978: Cost Selling Price Inventory, November 1 $ 53,800 $ 80,000 Markdowns 21,000 Markups 29,000 Markdown cancellations 13,000 Markup cancellations 9,000 Purchases 173,200 223,600 Sales 244,000 Purchase returns and allowances 3,000 3,600 Sales returns and allowances 12,000 Required: Based upon the data presented above, prepare a schedule in good form to compute the estimated inven tory at November 30, 1978, at the lower of cost or market under the retail inventory method.
Part c. The Acute Company manufactures a single product. On December 31, 1975, Acute adopted the dollar-value LIFO inventory method. The inventory on that date using the dollar-value LIFO inventory method was determined to be $300,000.
Inventory data for succeeding years are as follows; Year Ended December 31, 1976 1977 1978 Inventory at Respective Year-End Prices $363,000 420,000 430,000 Relevant Price Index (base year 1975) 1.10 1.20 1.25 Required: Compute the inventory amounts at December 31, 1976, 1977, and 1978, using the dollar-value LIFO in ventory method for each year.
122 Accounting Practice — Part II Number 4 (Estimated tim e-----50 to 60 minutes) You are engaged in the audit of the December 31, 1978, financial statements of Spirit Corporation, a manufacturer of a digital watch. You are attempting to verify the costing of the ending inventory of work in process and finished goods which were recorded on Spirit’s books as follows: Work in process (50% complete as to labor and overhead) Finished goods Units 300,000 200,000 Cost $ 660,960 $1,009,800 Materials are added to production at the beginning of the manufacturing process and overhead is applied to each product at the rate of 60% of direct labor costs. There was no finished goods inventory on January 1, 1978. A review of Spirit’s inventory cost records dis closed the following information: Costs Units Materials Labor Work in process January 1, 1978 (80% complete as to labor and over head) Units started in production Material costs Labor costs Units completed 200,000 $ 200,000 $ 315,000 1,000,000 $1,300,000 900,000 $1,995,000 Required: a. Prepare schedules as of December 31, 1978, to compute the following: • Equivalent units of production using the weighted-average method.
• Unit costs of production of materials, labor and overhead.
• Costing of the finished goods inventory and work-in-process inventory.
b. Prepare the necessary journal entry to correct ly state the inventory of finished goods and work in pro cess, assuming the books have not been closed. (Ignore income tax considerations.) Number 5 (Estimated time-----40 to 50 minutes) In a special election held on May 1, 1977, the voters of the city of Nicknar approved a $10,000,000 issue of 6% general obligation bonds maturing in 1997. The pro ceeds of this sale will be used to help finance the con struction of a new civic center. The total cost of the pro ject was estimated at $15,000,000. The remaining $5,000,000 will be financed by an irrevocable state grant which has been awarded. A capital projects fund was established to account for this project and was designated the Civic Center Construction Fund. The formal project authorization was appropriately record ed in a memorandum entry.
The following transactions occurred during the fiscal year beginning July 1, 1977, and ending June 30, 1978: • On July 1, the General Fund loaned $500,000 to the Civic Center Construction Fund for defraying engineering and other expenses.
• Preliminary engineering and planning costs of $320,000 were paid to Akron Engineering Company. There had been no encumbrance for this cost.
• On December 1st, the bonds were sold at 101. The premium on bonds was transferred to the Debt Service Fund.
• On March 15th, a contract for $12,000,000 was entered into with Candu Construction Company for the major part of the project.
• Orders were placed for materials estimated to cost $55,000.
• On April 1st, a partial payment of $2,500,000 was received from the State.
• The materials that were previously ordered were received at a cost of $51,000 and paid.
• On June 15, a progress billing of $2,000,000 was received from Candu Construction for work done on the project. As per the terms of the contract, the city will withhold 6% of any billing until the project is com pleted.
• The General Fund was repaid the $500,000 previously loaned.
Required: Based upon the transactions presented above: a. Prepare journal entries to record the transac tions in the Civic Center Construction Fund for the period July 1, 1977, through June 30, 1978, and the ap propriate closing entries at June 30, 1978.
b. Prepare a balance sheet of the Civic Center Construction Fund as of June 30, 1978.
123 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of a ll states, territories, and the District of Columbia.) EXAMINATION IN AUDITING May 3, 1979; 8:30 AM to 12:00 M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1 No. 2 No.
No.
No.
90 n o 15 25 15 25 15 25 15 25 Total 150 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
124 Auditing Number 1 (Estimated tim e------90 to 110 minutes) Instructions Select the best answer for each of the following items. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each Item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 96. One of the generally accepted auditing standards specifies that the auditor should a. Inspect all fixed assets acquired during the year.
b. Charge fair fees based on cost.
c. Make a proper study and evaluation of the ex isting internal control.
d. Count client petty-cash funds.
Answer Sheet 96. Items to be Answered 1. Which event that occurred after the end of the fiscal year under audit but prior to issuance of the auditor’s report would not require disclosure in the financial statements?
a. Sale of a bond or capital stock issue.
b. Loss of plant or inventories as a result of fire or flood.
c. A major drop in the quoted market price of the stock of the corporation.
d. Settlement of litigation when the event giving rise to the claim took place after the balance- sheet date.
2. Which of the following best describes the primary purpose of audit program procedures?
a. To detect error or irregularities.
b. To comply with generally accepted accounting principles.
c. To gather corroborative evidence.
d. To verify the accuracy of account balances.
3. Which of the following is mandatory if the auditor is to comply with generally accepted auditing standards?
a. Possession by the auditor of adequate technical training.
b. Use of analytical review on audit engagements.
c. Use of statistical sampling whenever feasible on an audit engagement.
d. Confirmation by the auditor of material ac counts receivable balances.
4. A major difference of opinion concerning an ac counting issue has arisen between an assistant on an audit engagement and the auditor with final respon sibility for the engagement. If after consultation the assistant believes it necessary to disassociate himself from the resolution of the matter, both the auditor and his assistant must a. Document details of the disagreement and the basis of resolution.
b. Inform management of the nature of the disagreement.
c. Inform the company’s audit committee of the nature of the disagreement.
d. Refer the matter to the firm’s peer review committee.
5. With respect to errors and irregularities, which of the following should be part of an auditor’s planning of the audit engagement?
a. Plan to search for errors or irregularities that would have a material or immaterial effect on the financial statements.
b. Plan to discover errors or irregularities that are either material or immaterial.
c. Plan to discover errors or irregularities that are material.
d. Plan to search for errors or irregularities that would have a material effect on the financial statements.
6. When an auditor tests a client’s cost accounting system, the auditor’s tests are primarily designed to determine that a. Quantities on hand have been computed based on acceptable cost accounting techniques that reasonably approximate actual quantities on hand.
b. Physical inventories are in substantial agree ment with book inventories.
c. The system is in accordance with generally ac cepted accounting principles and is function ing as planned.
d. Costs have been properly assigned to finished goods, work-in-process and cost of goods sold.
7. Which of the following best describes the opera tional audit?
a. It requires the constant review by internal auditors of the administrative controls as they relate to the operations of the company.
b. It concentrates on implementing financial and accounting control in a newly organized com pany.
c. It attempts and is designed to verify the fair presentation of a company’s results of opera tions.
d. It concentrates on seeking out aspects of operations in which waste would be reduced by the introduction of controls.
125 Examination Questions— May 1979 8. Auditors sometimes use comparison of ratios as audit evidence. For example, an unexplained decrease in the ratio of gross profit to sales may suggest which of the following possibilities?
a. Unrecorded purchases.
b. Unrecorded sales.
c. Merchandise purchases being charged to sell ing and general expense.
d. Fictitious sales.
9. The securities of Ralph Corporation are listed on a regional stock exchange and registered with the Securities and Exchange Commission. The management of Ralph engages a CPA to perform an independent audit of Ralph’s financial statements. The primary ob jective of this audit is to provide assurance to the a. Regional Stock Exchange.
b. Board of Directors of Ralph Corporation.
c. Securities and Exchange Commission.
d. Investors in Ralph securities.
10. A company has additional temporary funds to in vest. The board of directors decided to purchase marketable securities and assigned the future purchase and sale decisions to a responsible financial executive. The best person(s) to make periodic reviews of the in vestment activity should be a. The investment committee of the board of directors.
b. The treasurer.
c. The corporate controller.
d. The chief operating officer.
11. The basic concept of internal accounting control which recognizes that the cost of internal control should not exceed the benefits expected to be derived is known as a. Management by exception.
b. Management responsibility.
c. Limited liability.
d. Reasonable assurance.
12. In estimation sampling for variables, which of the following must be known in order to estimate the ap propriate sample size required to meet the auditor’s needs in a given situation?
a. The total amount of the population.
b. The desired standard deviation.
c. The desired confidence level.
d. The estimated rate of error in the population.
13. Which of the following might be detected by an auditor’s cut-off review and examination of sales jour nal entries for several days prior to and subsequent to the balance sheet date?
a. Lapping year-end accounts receivable.
b. Inflating sales for the year.
c. Kiting bank balances.
d. Misappropriating merchandise.
14. The machine-language program that results when a symbolic-language program is translated is called a (an) a. Processor program.
b. Object program.
c. Source program.
d. Wired program.
15. An auditor is testing sales transactions. One step is to trace a sample of debit entries from the accounts receivable subsidiary ledger back to the supporting sales invoices. What would the auditor intend to establish by this step?
a. Sales invoices represent bona fide sales.
b. All sales have been recorded.
c. All sales invoices have been properly posted to customer accounts.
d. Debit entries in the accounts receivable sub sidiary ledger are properly supported by sales invoices.
16. An independent auditor has concluded that the client’s records, procedures, and representations can be relied upon based on tests made during the year when internal control was found to be effective. The auditor should test the records, procedures, and representations again at year end if a. Inquiries and observations lead the auditor to believe that conditions have changed significantly.
b. Comparisons of year-end balances with like balances at prior dates revealed significant fluctutations.
c. Unusual transactions occurred subsequent to the completion of the interim audit work.
d. Client records are in a condition that facilitate effective and efficient testing.
17. A footnote to a company’s financial statements in cludes an indication that the company’s auditor per formed certain procedures regarding the company’s unaudited replacement cost information. The footnote does not indicate whether the auditor expresses an opin ion on the replacement cost information. Which of the following is appropriate in these circumstances?
a. The auditor’s report oh the audited financial statements should be expanded to include a disclaimer of opinion on the replacement cost information.
b. A separate report on the unaudited replace ment cost information should be rendered and should include a disclaimer of opinion on the replacement cost information.
c. The auditor’s report on the audited financial statements should be qualified because of the replacement cost information.
d. A separate report on the replacement cost in formation should be rendered and should in dicate whether the information is fairly presented in relation to the basic financial statements.
126 Auditing 18. Which of the following sampling plans would be designed to estimate a numerical measurement of a population, such as a dollar value?
a. Numerical sampling.
b. Discovery sampling.
c. Sampling for attributes.
d. Sampling for variables.
19. One of the auditor’s objectives in observing the ac tual distribution of payroll checks is to determine that every name on the payroll is that of a bona fide employee. The payroll observation is an auditing pro cedure that is generally performed for which of the following reasons?
a. The professional standards that are generally accepted require the auditor to perform the payroll observation.
b. The various phases of payroll work are not sufficiently segregated to afford effective in ternal accounting control.
c. The independent auditor uses personal judg ment and decides to observe the payroll distribution on a particular audit.
d. The standards that are generally accepted by the profession are interpreted to mean that payroll observation is expected on an audit unless circumstances dictate otherwise.
20. Internal control is a function of management, and effective control is based upon the concept of charge and discharge of responsibility and duty. Which of the following is one of the overriding principles of internal control?
a. Responsibility for accounting and financial duties should be assigned to one responsible officer.
b. Responsibility for the performance of each duty must be fixed.
c. Responsibility for the accounting duties must be borne by the auditing committee of the company.
d. Responsibility for accounting activities and duties must be assigned only to employees who are bonded.
21. Which of the following is an internal control pro cedure that would prevent a paid disbursement voucher from being presented for payment a second time?
a. Vouchers should be prepared by individuals who are responsible for signing disbursement checks.
b. Disbursement vouchers should be approved by at least two responsible management of ficials.
c. The date on a disbursement voucher should be within a few days of the date the voucher is presented for payment.
d. The official signing the check should compare the check with the voucher and should deface the voucher documents.
22. Which of the following reports is an indication of the changing role of the CPA that calls for an extension of the auditor’s attest function?
a. Report on annual comparative financial statements.
b. Report on internal control based on an audit.
c. Report on separate balance sheet of a holding company.
d. Report on balance sheet and statements of in come, retained earnings, and changes in financial position prepared from incomplete financial records.
23. Which of the following would detect an understatement of a purchase discount?
a. Verify footings and crossfootings of pur chases and disbursement records.
b. Compare purchase invoice terms with disbursement records and checks.
c. Compare approved purchase orders to receiv ing reports.
d. Verify the receipt of items ordered and in voiced.
24. Which of the following best describes an auditor’s report on supplementary financial statements prepared in accordance with a price-level basis of accounting that has substantial support?
a. A supplementary report.
b. An unaudited report.
c. A report on limited review of financial infor mation.
d. A special purpose report.
25. Which of the following best describes what is meant by generally accepted auditing standards?
a. Acts to be performed by the auditor.
b. Measures of the quality of the auditor’s per formance.
c. Procedures to be used to gather evidence to support financial statements.
d. Audit objectives generally determined on audit engagements.
26. Which of the following is not a medium that can normally be used by an auditor to record information concerning a client’s system of internal accounting con trol?
a. Narrative memorandum.
b. Procedures manual.
c. Flowchart.
d. Decision table.
27. Which of the following underlies the application of generally accepted auditing standards, particularly the standards of field work and reporting?
a. The elements of materiality and relative risk.
b. The element of internal control.
c. The element of corroborating evidence.
d. The element of reasonable assurance.
127 Examination Questions— May 1979 28. Which of the following is a conclusion reached by the Commission on Auditors’ Responsibilities, the in dependent commission established by the American In stitute of Certified Public Accountants to study the role and responsibilities of independent auditors?
a. Different auditing standards should apply to audits of publicly owned and private entities.
b. The AICPA Auditing Standards Executive Committee should be replaced by a larger, part-time group.
c. The oversight of professional practice should remain with the accounting profession.
d. “ Safe Harbors’’ should be made available for all work done by an auditor.
29. Which of the following is not a procedure per formed primarily for the purpose of expressing an opin ion on the financial statements, but may bring possible illegal acts to the auditor’s attention?
a. Study and evaluation of internal accounting control.
b. Review of internal administrative control.
c. Tests of transactions.
d. Tests of balances.
30. Transaction authorization within an organization may be either specific or general. An example of specific transaction authorization is the a. Setting of automatic reorder points for material or merchandise.
b. Approval of a detailed construction budget for a warehouse.
c. Establishment of requirements to be met in determining a customer’s credit limits.
d. Establishment of sales prices for products to be sold to any customer.
31. Audit evidence can come in different forms with different degrees of persuasiveness. Which of the following is the least persuasive type of evidence?
a. Documents mailed by outsiders to the auditor.
b. Correspondence between auditor and ven dors.
c. Sales invoices inspected by the auditor.
d. Computations made by the auditor.
32. Which of the following best describes the indepen dent auditor’s approach to obtaining satisfaction con cerning depreciation expense in the income statement?
a. Verify the mathematical accuracy of the amounts charged to income as a result of depreciation expense.
b. Determine the method for computing depreciation expense and ascertain that it is in accordance with generally accepted account ing principles.
c. Reconcile the amount of depreciation expense to those amounts credited to accumulated depreciation accounts.
d. Establish the basis for depreciable assets and verify the depreciation expense.
33. A lawyer’s response to a letter of audit inquiry may be limited to matters that are considered in dividually or collectively material to the financial statements if a. The auditor has instructed the lawyer regard ing the limits of materiality in financial statements.
b. The client and the auditor have agreed on the limits of materiality and the lawyer has been notified.
c. The lawyer and auditor have reached an understanding on the limits of materiality for this purpose.
d. The lawyer’s response to the inquiry explains the legal meaning of materiality limits and establishes quantitative parameters.
34. In connection with the study of internal control, an auditor encounters the following flowcharting symbols: The auditor would conclude that a. A master file has been created by a manual operation.
b. A master file has been created by a computer operation.
c. A document has been generated by a com puter operation.
d. A document has been generated by a manual operation.
35. The AICPA Committee on Management Services has stated its belief that a CPA should not undertake a management advisory service engagement for im plementation of the CPA’s recommendations unless a. The client has made a firm decision to proceed with implementation based on a complete understanding and consideration of alter natives.
b. The client does not understand the nature and implications of the recommended course of action.
c. The client does not have sufficient expertise within its organization to comprehend the significance of the changes being made.
d. The CPA withdraws as independent auditor for the client.
36. An auditor is unable to determine the amounts associated with certain illegal acts committed by a client. In these circumstances the auditor would most likely a. Issue either a qualified opinion or a disclaimer of opinion.
b. Issue only an adverse opinion.
c. Issue either a qualified opinion or an adverse opinion.
d. Issue only a disclaimer of opinion.
128 Auditing 37. Effective internal control over the payroll function would include which of the following?
a. Total time recorded on time-clock punch cards should be reconciled to job reports by employees responsible for those specific jobs.
b. Payroll department employees should be supervised by the management of the person nel department.
c. Payroll department employees should be responsible for maintaining employee person nel records.
d. Total time spent on jobs should be compared with total time indicated on time-clock punch cards.
38. If as a result of auditing procedures an auditor believes that the client may have committed illegal acts, which of the following actions should be taken im mediately by the auditor?
a. Consult with the client’s counsel and the auditor’s counsel to determine how the suspected illegal acts will be communicated to the stockholders.
b. Extend normal auditing procedures to ascer tain whether the suspected illegal acts may have a material effect on the financial statements.
c. Inquire of the client’s management and con sult with the client’s legal counsel or other specialists, as necessary, to obtain an understanding of the nature of the acts and their possible effects on the financial statements.
d. Notify each member of the audit committee of the board of directors of the nature of the acts and request that they give guidance with respect to the approach to be taken by the auditor.
39. Which of the following must accompany unaudited financial statements which are prepared by a CPA?
a. Only a disclaimer of opinion.
b. Either a disclaimer of opinion or adverse opinion.
c. Either a disclaimer of opinion or a qualified opinion.
d. Either a disclaimer of opinion, adverse opin ion, or qualified opinion.
40. As generally conceived, the audit committee of a publicly held company should be made up of a. Representatives from the clients management, investors, suppliers, and customers.
b. The audit partner, the chief financial officer, the legal counsel, and at least one outsider.
c. Representatives of the major equity interests (bonds, preferred stock, common stock).
d. Members of the board of directors who are not officers or employees.
41. Which of the following is not a factor affecting the independent auditor’s judgment as to the quantity, type, and content of audit working papers?
a. The needs in the particular circumstances for supervision and review of the work performed by any assistants.
b. The nature and condition of the client’s records and internal controls.
c. The expertise of client personnel and their ex pected audit participation.
d. The type of the financial statements, schedules, or other information upon which the auditor is reporting.
42. Effective internal control requires organizational independence of departments. Organizational in dependence would be impaired in which of the follow ing situations?
a. The internal auditors report to the audit com mittee of the board of directors.
b. The controller reports to the vice president of production.
c. The payroll accounting department reports to the chief accountant.
d. The cashier reports to the treasurer.
43. Generally accepted auditing standards are ap plicable when an auditor examines and reports on any financial statement. For reporting purposes, the in dependent auditor should consider each of the following types of financial presentations to be a financial state ment except a a. Statement of assets and liabilities arising from cash transactions.
b. Statement of changes in owners’ equity.
c. Statement of forecasted results of operations.
d. Statement of operations by product line.
44. When a CPA is approached to perform an audit for the first time, the CPA should make inquiries of the predecessor auditor. This is a necessary procedure because the predecessor may be able to provide the suc cessor with information that will assist the successor in determining a. Whether the predecessor’s work should be utilized.
b. Whether the company follows the policy of rotating its auditors.
c. Whether in the predecessor’s opinion internal control of the company has been satisfactory.
d. Whether the engagement should be accepted.
45. Totals of amounts in computer-record data fields which are not usually added but are used only for data processing control purposes are called a. Record totals.
b. Hash totals.
c. Processing data totals.
d. Field totals.
129 Examination Questions—May 1979 46. Which of the following client electronic data pro cessing (EDP) systems generally can be audited without examining or directly testing the EDP computer pro grams of the system?
a. A system that performs relatively un complicated processes and produces detailed output.
b. A system that affects a number of essential master files and produces a limited output.
c. A system that updates a few essential master files and produces no printed output other than final balances.
d. A system that performs relatively complicated processing and produces very little detailed output.
47. Which of the following procedures would best detect the theft of valuable items from an inventory that consists of hundreds of different items selling for $1 to $10 and a few items selling for hundreds of dollars?
a. Maintain a perpetual inventory of only the more valuable items with frequent periodic verification of the validity of the perpetual in ventory record.
b. Have an independent CPA firm prepare an in ternal control report on the effectiveness of the administrative and accounting controls over inventory.
c. Have separate warehouse space for the more valuable items with sequentially numbered tags.
d. Require an authorized officer’s signature on all requisitions for the more valuable items.
48. Effective internal control over the purchasing of raw materials should usually include all of the following procedures except a. Systematic reporting of product changes which will affect raw materials.
b. Determining the need for the raw materials prior to preparing the purchase order.
c. Obtaining third-party written quality and quantity reports prior to payment for the raw materials.
d. Obtaining financial approval prior to making a commitment.
49. Which of the following best describes why publicly-traded corporations follow the practice of hav ing the outside auditor appointed by the board of direc tors or elected by the stockholders?
a. To comply with the regulations of the Finan cial Accounting Standards Board.
b. To emphasize auditor independence from the management of the corporation.
c. To encourage a policy of rotation of the in dependent auditors.
d. To provide the corporate owners with an op portunity to voice their opinion concerning the quality of the auditing firm selected by the directors.
50. A firm of CPAs may use policies and procedures such as notifying professional personnel as to the names of audit clients having publicly held securities and con firming periodically with such personnel that prohibited relations do not exist. This is done to achieve effective quality control in which of the following areas?
a. Acceptance and continuance of clients.
b. Assigning personnel to engagements.
c. Independence.
d. Inspection.
51. What is the computer process called when data processing is performed concurrently with a particular activity and the results are available soon enough to in fluence the particular course of action being taken or the decision being made?
a. Batch processing.
b. Realtime processing.
c. Integrated data processing.
d. Random access processing.
52. Which of the following circumstances would not be considered a departure from the auditor’s standard report?
a. The auditor wishes to emphasize a particular matter regarding the financial statements.
b. The auditor’s opinion is based in part on the report of another auditor.
c. The financial statements are affected by a departure from a generally accepted account ing principle.
d. The auditor is asked to report only on the balance sheet but has unlimited access to in formation underlying all the basic financial statements.
53. Before expressing an opinion concerning the results of operations, the auditor would best proceed with the examination of the income statement by a. Applying a rigid measurement standard de signed to test for understatement of net in come.
b. Analyzing the beginning and ending balance sheet inventory amounts.
c. Making net income comparisons to published industry trends and ratios.
d. Examining income statement accounts con currently with the related balance sheet ac counts.
54. The audit working papers often include a client- prepared, aged trial balance of accounts receivable as of the balance sheet date. This aging is best used by the auditor to a. Evaluate internal control over credit sales.
b. Test the accuracy of recorded charge sales.
c. Estimate credit losses.
d. Verify the validity of the recorded receivables.
130 Auditing 55. Which of the following is not a principal objective of the auditor in the examination of revenues?
a. To verify cash deposited during the year.
b. To study and evaluate internal control, with particular emphasis on the use of accrual ac counting to record revenue.
c. To verify that earned revenue has been record ed, and recorded revenue has been earned.
d. To identify and interpret significant trends and variations in the amounts of various categories of revenue.
56. An auditor’s report would be designated as a special report when it is issued in connection with which of the following?
a. Financial statements for an interim period which are subjected to a limited review.
b. Financial statements which are prepared in accordance with a comprehensive basis of accounting other than generally accepted ac counting principles.
c. Financial statements which purport to be in accordance with generally accepted accoun ting principles but do not include a presenta tion of the statement of changes in financial position.
d. Financial statements which are unaudited and are prepared from a client’s accounting records.
57. An auditor decides that it is important and necessary to observe a client’s distribution of payroll checks on a particular audit. The client organization is so large that the auditor cannot conveniently observe the distribution of the entire payroll. In these cir cumstances, which of the following is most acceptable to the auditor?
a. Observation should be limited to one or more selected departments.
b. Observation should be made for all depart ments regardless of the inconvenience.
c. Observation should be eliminated and other alternative auditing procedures should be utilized to obtain satisfaction.
d. Observation should be limited to those depart ments where employees are readily available.
58. During the year under audit, a company has com pleted a private placement of a substantial amount of bonds. Which of the following is the most important step in the auditor’s program for the examination of bonds payable?
a. Confirming the amount issued with the bond trustee.
b. Tracing the cash received from the issue to the accounting records.
c. Examining the bond records maintained by the transfer agent.
d. Recomputing the annual interest cost and the effective yield.
59. Which of the following individuals is the most ap propriate person to be assigned the reponsibility of distributing envelopes that include employee payroll checks?
a. The company paymaster.
b. A member of the accounting department.
c. The internal auditor.
d. A representative of the bank where the com pany payroll account is maintained.
60. The auditor’s judgment concerning the overall fairness of the presentation of financial position, results of operations, and changes in financial position is ap plied within the framework of a. Quality control.
b. Generally accepted auditing standards which include the concept of materiality.
c. The auditor’s evaluation of the audited com pany’s internal control.
d. Generally accepted accounting principles.
Number 2 (Estimated time-----15 to 25 minutes) A partially-completed charge sales systems flowchart follows. The flowchart depicts the charge sales activities of the Bottom Manufacturing Corpora tion.
A customer’s purchase order is received and a six- part sales order is prepared, therefrom. The six copies are initially distributed as follows: Copy No. 1 — Billing copy - to billing depart ment.
Copy No. 2 — Shipping copy - to shipping department.
Copy No. 3 — Credit copy - to credit depart ment.
Copy No. 4 — Stock request copy - to credit department.
Copy No. 5 — Customer copy - to customer. Copy No. 6 — Sales order copy - file in sales order department.
When each copy of the sales order reaches the ap plicable department or destination it calls for specific in ternal control procedures and related documents. Some of the procedures and related documents are indicated on the flowchart. Other procedures and documents are labeled letters a to r.
Required: List the procedures or the internal documents that are labeled letters c to r in the flowchart of Bottom Manufacturing Corporation’s charge sales system.
Organize your answer as follows (note that an ex planation of the letters a and b which appear in the flowchart are entered as examples): Flowchart Symbol Letter Procedures or Internal Document a.
b.
Prepare six-part sales order.
File by order number.
131 BOTTOM MANUFACTURING CORPORATION Flowchart of Credit Sales Activities Examination Questions— May 1979 Sales Order Credit Finished Goods Shipping Billing Accounts Receivable 132 CODE: A = CUSTOMER Auditing Number 3 (Estimated time-----15 to 25 minutes) Number 5 (Estimated time-----15 to 25 minutes) When auditing an electronic data processing (EDP) accounting system, the independent auditor should have a general familiarity with the effects of the use of EDP on the various characteristics of accounting control and on the auditor’s study and evaluation of such control. The independent auditor must be aware of those control procedures that are commonly referred to as “general” controls and those that are commonly referred to as “application” controls. General controls relate to all EDP activities and application controls relate to specific accounting tasks.
Required: a. What are the general controls that should exist in EDP-based accounting systems?
b. What are the purposes of each of the follow ing categories of application controls?
1. Input controls 2. Processing controls 3. Output controls Number 4 (Estimated time-----15 to 25 minutes) During an audit engagement Harper, CPA, has satisfactorily completed an examination of accounts payable and other liabilities and now plans to determine whether there are any loss contingencies arising from litigation, claims, or assessments.
Required: What are the audit procedures that Harper should follow with respect to the existence of loss contingencies arising from litigation, claims, and assessments? Do not discuss reporting requirements.
Lando Corporation is a domestic company with two wholly-owned domestic subsidiaries. Michaels, CPA, has been engaged to examine the financial statements of the parent company and one of the sub sidiaries and to act as the principal auditor. Thomas, CPA, has examined the financial statements of the other subsidiary whose operations are material in rela tion to the consolidated financial statement.
The work performed by Michaels is sufficient for Michaels to serve as the principal auditor and to report as such on the financial statements. Michaels has not yet decided whether to make reference to the examination made by Thomas.
Required: a. There are certain required audit procedures that Michaels should perform with respect to the ex amination made by Thomas, whether or not Michaels decides to make reference to Thomas in Michaels’ auditor’s report. What are these audit procedures?
b. What are the reporting requirements with which Michaels must comply if Michaels decides to name Thomas and make reference to the examination of Thomas?
133 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) EXAMINATION IN BUSINESS LAW (Commercial Law) May 4, 1979; 8:30 AM to 12:00 M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: No. 1 .....................
No. 2 . . . . ..............
No. 3 .....................
No. 4 .....................
No. 5 .....................
Total.............
Estimated Minutes Minimum Maximum 85 105 25 30 25 30 20 25 15 20 170 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
134 Business Law Number 1 (Estimated time-----85 to 105 minutes) Instructions Select the best answer for each of the following items. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 99. The text of a letter from Bridge Builders, Inc., to Allied Steel Co. follows: We offer to purchase 10,000 tons of No. 4 steel pipes at today’s quoted price for delivery two months from today. Your acceptance must be received in five days.
Bridge Builders intended to create a (an) a. Option contract.
b. Unilateral contract.
c. Bilateral contract.
d. Joint contract.
Answer Sheet 99. Items to be Answered 1. Martinson borrowed $50,000 from Wisdom Finance Company. The loan was evidenced by a non- negotiable promissory note secured by a first mortgage on Martinson’s ranch. One of the terms of the note re quired acceleration of repayment in the event that Wisdom “deemed itself insecure.’’ When the value of the property declined, Wisdom notified Martinson that pursuant to the terms of the note, it “deemed itself in secure’’ and demanded that either additional collateral or an acceptable surety be provided. Martinson ar ranged for Clark, a personal friend, to act as surety on the loan. Clark signed the note as an indorser and Wisdom agreed in writing not to accelerate repayment of the loan during the life of the debt. Martinson has defaulted. Which of the following is a correct state ment?
a. Clark’s promise is not supported by con sideration, hence is unenforceable.
b. Clark is a guarantor of collection and his obligation is conditioned upon Wisdom’s first proceeding against Martinson.
c. Release of the mortgage by Wisdom would release Clark to the extent of the value of the property.
d. Wisdom must first foreclose the mortgage before it can proceed against Clark.
2. Wilson obtained a fire insurance policy on his dairy farm from the Columbus Insurance Company. The policy was for $80,000 which was the value of the property. The policy was the standard fire insurance policy sold throughout the United States. A fire oc curred late one night and caused a $10,000 loss. Which of the following will prevent Wilson from recovering the full amount of his loss from Columbus Insurance?
a. The coinsurance clause.
b. Wilson had a similar policy with another in surance company for $40,000.
c. The fact that 50% of the loss was caused by smoke and water damage.
d. The fact that his negligence was the primary cause of the fire.
3. Dobbins insured his life for $100,000, naming his wife as beneficiary. After the policy had been in effect for ten years and had a cash surrender value in excess of $15,000, Dobbins assigned the policy to Suburban National Bank to secure a $20,000 loan. A copy of the assignment was filed with Suburban at its home office. Dobbins has died and his widow and Suburban are seek ing to recover on the $100,000 life insurance policy. Which of the following is correct?
a. Suburban’s recovery is limited to the amount of the loan outstanding plus interest.
b. The assignment to Suburban was void without the beneficiary’s consent.
c. Suburban will be denied recovery due to a lack of an insurable interest.
d. The widow had a vested interest in the in surance policy in question.
4. DuBary and Young were business partners. They agreed that each would insure the life of the other for his own benefit (cross insurance). On the application for insurance, DuBary stated that he had never had any heart trouble when, in fact, he had suffered a mild heart attack some years before. Young’s policy on DuBary’s life contained a two-year incontestability clause. Three years later, after the partnership had been dissolved but while the policy on his life was still in force, DuBary was killed when his car was struck by a car negligently driven by Peters. The insurer has refused to pay the policy pro ceeds to Young, asserting that Young at the time DuBary died had no insurable interest in DuBary’s life and that DuBary’s misrepresentation voided the policy. Which of the following statements is correct?
a. The misrepresentation in the application is a bar to recovery.
b. If DuBary and Young were never partners, recovery will be denied.
c. Dissolution of the partnership eliminated any possible insurable interest and is a bar to recovery.
d. If the insurance Company has to pay, it will be subrogated to DuBary’s rights against Peters.
135 Examination Questions— May 1979 5. Dunlop loaned Barkum $20,000 which was secured by a security agreement covering Barkum’s machinery and equipment. A financing statement was properly filed covering the machinery and equipment. In addition, Delson was a surety on the Barkum loan. Barkum is now insolvent and a petition in bankruptcy has been filed against him. Delson paid the amount owed ($17,000) to Dunlop. The property was sold for $12,000. Which of the following is correct? a. Delson has the right of a secured creditor to the $12,000 via subrogation to Dunlop’s rights and the standing of general creditor for the balance.
b. To the extent Delson is not fully satisfied for the $17,000 he paid Dunlop, his claim against Barkum will not be discharged in bankruptcy.
c. Delson’s best strategy would have been to pro ceed against Barkum in his own right for reim bursement.
d. Delson should have asserted his right of exoneration.
6. Quinn was the sole owner of Sunnydale Farms, Inc. The business was in dire need of additional working capital in order to survive. Click Company was willing to loan Sunnydale $12,000, but only if Click obtained a security interest in Sunnydale’s machinery and equip ment and a promise from Quinn to guarantee repay ment of the loan. Click obtained both. Sunnydale was subsequently adjudged bankrupt. Click filed a reclama tion claim for the machinery and equipment which was denied by the trustee in bankruptcy. The property was sold at public auction for $10,500. Click negotiated a settlement with the trustee whereby it received the $10,500 proceeds on the sale in full settlement of its claim against the bankrupt. Which of the following is a correct statement?
a. Where a surety is the sole owner of the stock of the corporation whose debt he guarantees, he is a compensated surety.
b. Click first had to exhaust his remedies against the property before he could sue Quinn.
c. Quinn must pay Click the $1,500 difference, plus interest. d. The settlement released Quinn from his surety obligation.
7. Crawford and Blackwell separately agreed to act as sureties on a loan of $25,000 by Lux to Factor. Each promised to pay the full $25,000 upon default of Factor. Lux subsequently released Blackwell from his surety undertaking. Which of the following is a correct state ment?
a. The release has no effect upon Crawford’s right to contribution if he is obligated to pay.
b. The release of Blackwell had no effect upon Crawford’s liability.
c. The release of Blackwell also totally released Crawford. d. The release of Blackwell also released Crawford to the extent of $12,500.
8. Arthur sold his house to Michael. Michael agreed to pay the existing mortgage on the house. The Safety Bank, which held the mortgage, released Arthur from liability on the debt. The above declared transaction (relating to the mortgage debt) is a. A delegation.
b. A novation.
c. Invalid in that bank did not receive any addi tional consideration from Arthur.
d. Not a release of Arthur if Michael defaults, and the proceeds from the sale of the mortgaged house are insufficient to satisfy the debt.
9. Williams purchased a heating system from Radiant Heating, Inc., for his factory. Williams insisted that a clause be included in the contract calling for ser vice on the heating system to begin not later than the next business day after Williams informed Radiant of a problem. This service was to be rendered free of charge during the first year of the contract and for a flat fee of $200 per year for the next two years thereafter. During the winter of the second year, the heating system broke down and Williams promptly notified Radiant of the situation. Due to other commitments, Radiant did not send a man over the next day. Williams phoned Radiant and was told that the $200 per year service charge was uneconomical and they could not get a man over there for several days. Williams in desperation promised to pay an additional $100 if Radiant would send a man over that day. Radiant did so and sent a bill for $100 to Williams. Is Williams legally required to pay this bill and why?
a. No, because the pre-existing legal duty rule applies to this situation.
b. No, because the statute of frauds will defeat Radiant’s claim.
c. Yes, because Williams made the offer to pay the additional amount.
d. Yes, because the fact that it was uneconomical for Radiant to perform constitutes economic duress which freed Radiant from its obliga tion to provide the agreed-upon service.
10. Austin is attempting to introduce oral evidence in court to explain or modify a written contract he made with Wade. Wade has pleaded the parol evidence rule. In which of the following circumstances will Austin not be able to introduce the oral evidence?
a. The contract contains an obvious ambiguity on the point at issue.
b. There was a mutual mistake of fact by the par ties regarding the subject matter of the con tract.
c. The modification asserted was made several days after the written contract had been ex ecuted.
d. The contract indicates that it was intended as the “entire contract’’ between the parties and the point is covered in detail.
136 Business Law 11. Montbanks’ son, Charles, was seeking an account executive position with Dobbs, Smith, and Fogarty, Inc., the largest brokerage firm in the United States. Charles was very independent and wished no in terference by his father. The firm, after several weeks deliberation, decided to hire Charles. They made him an offer on April 12, 1979, and Charles readily accepted. Montbanks feared that his son would not be hired. Be ing unaware of the fact that his son had been hired, Montbanks mailed a letter to Dobbs on April 13 in which he promised to give the brokerage firm $50,000 in commission business if the firm would hire his son. The letter was duly received by Dobbs and they wish to en force it against Montbanks. Which of the following statements is correct?
a. Past consideration is no consideration, hence there is no contract.
b. The pre-existing legal duty rule applies and makes the promise unenforceable.
c. Dobbs will prevail since the promise is con tained in a signed writing.
d. Dobbs will prevail based upon promissory estoppel.
12. Philpot purchased the King Pharmacy from Golden. The contract contained a promise by Golden that he would not engage in the practice of pharmacy for one year from the date of the sale within one mile of the location of King Pharmacy. Six months later Golden opened the Queen Pharmacy within less than a mile of King Pharmacy. Which of the following is a correct statement?
a. Golden has not breached the above covenant since he did not use his own name or the name King in connection with the new pharmacy.
b. The covenant is reasonable and enforceable.
c. The contract is an illegal restraint of trade and illegal under federal antitrust laws.
d. The covenant is contrary to public policy and is illegal and void.
13. Keats Publishing Company shipped textbooks and other books for sale at retail to Campus Bookstore. An honest dispute arose over Campus’s right to return cer tain books. Keats maintained that the books in question could not be returned and demanded payment of the full amount. Campus relied upon trade custom which indicated that many publishers accepted the return of such books. Campus returned the books in question and paid for the balance with a check marked “Account Paid in Full to Date.’’ Keats cashed the check. Which of the following is a correct statement?
a. Keats is entitled to recover damages.
b. Keats’ cashing of the check constituted an ac cord and satisfaction.
c. The pre-existing legal duty rule applies and Keats is entitled to full payment for all the books.
d. The custom of the industry argument would have no merit in a court of law.
14. Wilcox Laboratories, Ltd., manufactures medical equipment for sale to medical institutions and retailers. Wilcox also sells directly to consumers in its wholly- owned retail outlets. Wilcox has created a subsidiary, Wilcox Finance Corporation, for the purpose of financ ing the purchase of its products by the various customers. In which of the following situations does Wilcox Finance not have to file a financing statement to perfect its security interest against competing creditors in the equipment sold by Wilcox?
a. Sales made to medical institutions.
b. Sales made to consumers who purchase for their own personal use.
c. Sales made to retailers who in turn sell to buyers in the ordinary course of business.
d. Sales made to any buyer when the equipment becomes a fixture.
15. Kelmore Appliances, Inc., sells various brand name appliances at discount prices. Kelmore maintains a large inventory which it obtains from various manufacturers on credit. These manufacturer-creditors have all filed and taken secured interests in the ap pliances and proceeds therefrom which they have sold to Kelmore on credit. Kelmore in turn sells to hundreds of ultimate consumers; some pay cash but most buy on credit. Kelmore takes a security interest but does not file a financing statement for credit sales. Which of the following is correct?
a. The appliances in Kelmore’s hands are con sumer goods.
b. Since Kelmore takes a purchase money securi ty interest in the consumer goods sold, its security interest is perfected upon attachment.
c. The appliance manufacturers can enforce their secured interests against the appliances in the hands of the purchasers who paid cash for them.
d. A subsequent sale by one of Kelmore’s customers to a bona fide purchaser will be subject to Kelmore’s secured interest.
16. Barkam is starting a new business, Barkam Enter prises, which will be a sole proprietorship selling retail novelties. Barkam recently received a discharge in bankruptcy, but certain proved claims were unpaid because of lack of funds. Which of the following would be a claim against Barkam?
a. The unpaid amounts owed to secured creditors who received less than the full amount after resorting to their security in terest and receiving their bankruptcy divi dend.
b. The unpaid amounts owed to trade suppliers for goods purchased and sold by Barkam in the ordinary course of his prior business.
c. A personal loan by his father made in an at tempt to stave off bankruptcy.
d. The unpaid amount of taxes due to the United States which became due and owing within three years preceding bankruptcy.
137 Examination Questions— May 1979 17. Abacus Corporation sent Frame Company an of fer by a telegram to buy its patent on a calculator. The Abacus telegram indicated that the offer would expire in ten days. The telegram was sent on February 1, 1979, and received on February 2, 1979, by Frame. On February 8 , 1979, Abacus telephoned Frame and in dicated they were withdrawing the offer. Frame telegraphed an acceptance on the 11th of February. Which of the following is correct?
a. The offer was an irrevocable offer, but Frame’s acceptance was too late.
b. Abacus’ withdrawal of the offer was ineffec tive because it was not in writing.
c. Since Frame used the same means of com munication, acceptance was both timely and effective.
d. No contract arose since Abacus effectively revoked the offer on February 8 , 1979.
18. Migrane Financial does a wide variety of lending. It provides funds to manufacturers, middlemen, retailers, consumers, and home owners. In all instances it intends to create a security interest in the loan transac tions it enters into. To which of the following will Article 9 (Secured Transactions) of the the Uniform Commercial Code not apply?
a. A second mortgage on the borrower’s home.
b. An equipment lease.
c. The sale of accounts.
d. Field warehousing.
19. Marigold, Inc., was in extreme financial difficulty. Hargrove, one of its persistent creditors, insisted upon payment of the entire amount due on the shipments of goods to Marigold over the past four months or it would sue Marigold and obtain a judgment against it. In order to dissuade Hargrove from taking such action, Marigold persuaded Hargrove to accept its note which was secured by a second mortgage on Marigold’s ware house. Hargrove filed the mortgage on November 1, 1978, the same day that the note and mortgage were ex ecuted. On February 1, 1979, Marigold concluded that things were hopeless and filed a voluntary petition in bankruptcy. The trustee in bankruptcy is attacking the validity of the mortgage as a voidable preference. Which of the following is correct?
a. The mortgage is not a voidable preference since it was filed the same day it was obtained.
b. The fact that Marigold was delinquent on its payment to Hargrove establishes that Hargrove knew that Marigold was insolvent in the bankruptcy sense.
c. The antecedent indebtedness requirement necessary to establish a voidable preference has not been satisfied under the facts given.
d. Whether Hargrove knew of the fact that Marigold was insolvent in the bankruptcy sense is irrelevant insofar as deciding whether the mortgage constitutes a preference as con trasted with a voidable preference.
20. Bigelow manufactures mopeds and sells them through franchised dealers who are authorized to resell them to the ultimate consumer or return them. Bigelow delivers the mopeds on consignment to these retailers. The consignment agreement clearly states that the agree ment is intended to create a security interest for Bigelow in the mopeds delivered on consignment. Bigelow wishes to protect itself against the other creditors of and purchasers from the retailers who might assert rights against the mopeds. Under the circumstances, Bigelow a. Must file a financing statement and give notice to certain creditors in order to perfect his security interest.
b. Will have rights against purchasers in the or dinary course of business who were aware of the fact that Bigelow had filed.
c. Need take no further action to protect himself, since the consignment is a sale or return and title is reserved in Bigelow.
d. Will have a perfected security interest in the mopeds upon attachment.
21. Johnson loaned money to Visual, Inc., a strug gling growth company, and sought to obtain a security interest in negotiable stock certificates which are traded on a local exchange. To perfect his interest against Visual’s other creditors, Johnson a. Need do nothing further in that his security in terest was perfected upon attachment.
b. May file or take possession of the stock cer tificates.
c. Must take possession of the stock certificates.
d. Must file and give the other creditors notice of his contemplated security interest.
22. Robert Cunningham owns a shop in which he repairs electrical appliances. Three months ago Elec trical Supply Company sold Cunningham, on credit, a machine for testing electrical appliances and obtained a perfected security interest at the time as security for pay ment of the unpaid balance. Cunningham’s creditors have now filed an involuntary petition in bankruptcy against him. What is the status of Electrical in the bankruptcy proceeding?
a. Electrical is a secured creditor and has the right against the trustee if not paid to assert a claim to the electrical testing machine it sold to Cunningham.
b. Electrical must surrender its perfected security interest to the trustee in bankruptcy and share as a general creditor of the bankrupt’s estate.
c. Electrical’s perfected security interest con stitutes a preference and is voidable.
d. Electrical must elect to resort exclusively to its secured interest or to relinquish it and obtain the same share as a general creditor.
138 Business Law 23. Hance, doing business as Hance Fashions, is hopelessly insolvent. As a means of staving off his ag gressive creditors and avoiding bankruptcy, Hance has decided to make a general assignment for the benefit of his creditors. Consequently, he transferred all his nonexempt property to a trustee for equitable distribu tion to his creditors. What are the legal consequences of Hance’s actions?
a. A debtor may not make an assignment for the benefit of creditors if he has been adjudicated a bankrupt and discharged within the preceding six years.
b. All his creditors must participate in the assign ment and distribution of property if a majori ty in number and amount participate.
c. Upon distribution of all his assigned property to the participating creditors, he is discharged from all liability.
d. He has committed an act of bankruptcy and may be petitioned into bankruptcy by his creditors.
24. Markson is a general creditor of Black. Black filed a voluntary petition in bankruptcy. Markson is irate and wishes to have the bankruptcy court either deny Black a general discharge or at least not have his debt dis charged. The discharge will be granted and it will in clude Markson’s debt even if a. It is unscheduled.
b. Markson extended the credit based upon a fraudulent financial statement.
c. Markson was a secured creditor who was not fully satisfied from the proceeds obtained upon disposition of the collateral.
d. Black had received a previous discharge in bankruptcy within six years.
25. Martindale Retail Fish Stores, Inc., purchased a large quantity of fish from the Seashore Fish Wholesalers. The exact amount was not ascertainable at the moment,and Martindale, rather than waiting for the exact amount, gave Seashore a check which was blank as to the amount. Seashore promised not to fill in any amount until it had talked to Martindale’s purchasing agent and had the amount approved. Seashore disregarded this agreement and filled in an amount that was $300 in excess of the correct price. The instrument was promptly negotiated to Clambake & Company, one of Seashore’s persistent creditors, in payment of an ac count due. Martindale promptly stopped payment. For what amount will Martindale be liable to Clambake? Why?
a. Nothing because Martindale can assert the real defense of material alteration.
b. Nothing because Clambake did not give value and the stop order is effective against it.
c. Only the correct amount because the wrongful filling in of the check for the $300 excess amount was illegal.
d. The full amount because the check is in the hands of a holder in due course.
26. Filbert Corporation has in its possession an instru ment which Groves, the maker, assured Filbert was negotiable. The instrument contains several clauses which are not typically contained in such an instrument and Filbert is not familiar with their legal effect. Which of the following will adversely affect the negotiability of the instrument?
a. A promise to maintain collateral and to pro vide additional collateral if the value of ex isting collateral decreases.
b. A term authorizing the confession of judg ment on the instrument if not paid when due.
c. A statement to the effect that the instrument arises out of the November 1, 1978, sale of goods by Filbert to Groves.
d. A statement that it is payable only out of the proceeds from the resale of the goods sold by Filbert to Groves on November 1, 1978.
27. Mitchell sold his ranch to Campbell. Campbell tendered his uncertified check for $35,000 at the closing. Mitchell objected and asserted that the check had to be certified. An examination of the contract revealed that the usual certification requirement had been omitted. Mitchell begrudgingly accepted the check after a phone call to the bank confirmed that “the check is good.’’ He promptly proceeded to Campbell’s bank and requested that it be certified in that he needed a certified check for a closing the following day in connection with the pur chase of another property. Regarding Mitchell’s request for certification, which of the following is correct?
a. The bank is legally obligated to certify the check.
b. If the bank did certify the check, it constituted an acceptance and released Campbell.
c. The bank’s oral statement that “the check is good’’ constituted an implied certification.
d. Once a check has been properly certified, fur ther negoti ation is prohibited.
28. Johnson lost a check that he had received for pro fessional services rendered. The instrument on its face was payable to Johnson’s order. He had indorsed it on the back by signing his name and printing “ for deposit only’’ above his name. Assuming the check is found by Alcatraz, a dishonest person who attempts to cash it, which of the following is correct?
a. Any transferee of the instrument must pay or apply any value given by him for the instru ment consistent with the indorsement.
b. The indorsement is a blank indorsement and a holder in due course who cashed it for Alcatraz would prevail.
c. The indorsement prevents further transfer or negotiation by anyone.
d. If Alcatraz simply signs his name beneath Johnson’s indorsement, he can convert it into bearer paper and a holder in due course would take free of the restriction.
139 Examination Questions— May 1979 29. Archer has in his possession a bearer negotiable in strument. He took it by negotiation from Perth who had stolen it from Cox’s office along with cash and other property. The robbery of Cox’s office had received ap propriate coverage in the local papers in the area in which both Archer and Cox reside. Archer did not know that Perth had stolen the instrument when he purchased it at a 20% discount. Cox refuses to pay and Archer has commenced legal action asserting that he is a holder in due course. Which of the following statements is cor rect?
a. Even if all other requisites are satisfied, Archer’s title is defective in that there was no delivery by Cox of the instrument.
b. Archer is a holder in due course and will prevail.
c. Archer is prevented from qualifying as a holder in due course because there had been general notice published in the community about the robbery.
d. The discount in and of itself prevents Archer from qualifying as a holder in due course or at least prevents him from so qualifying as to the 20%.
30. Path stole a check made out to the order of Marks. Path forged the name of Marks on the back and made the instrument payable to himself. He then negotiated the check to Harrison for cash by signing his own name on the back of the instrument in Harrison’s presence. Harrison was unaware of any of the facts surrounding the theft or forged indorsement and presented the check for payment. Central County Bank, the drawee bank, paid it. Disregarding Path, which of the following will bear the loss?
a. The drawer of the check payable to Marks.
b. Central County Bank.
c. Marks.
d. Harrison.
31. Troy fraudulently induced Casper to make a negotiable instrument payable to the order of Troy in exchange for goods he never intended to deliver. Troy negotiated it to Gorden, who took with notice of the fraud. Gorden in turn negotiated it to Wagner, a holder in due course. Wagner presented it for payment to Casper, who refused to honor it. Wagner contacted Gorden who agreed to reacquire the instrument by negotiation from Wagner. Which of the following statements is correct?
a. Casper would have been liable if Wagner had pursued his rights on the negotiable instru ment.
b. Gorden was initially a holder in due course as a result of the negotiation to him from Troy.
c. Casper is liable to all parties except Troy in that it was his fault that the instrument was issued to Troy.
d. Gorden can assert the rights of his prior holder in due course, Wagner, as a result of the repurchase.
32. Franco & Sons, Inc., was engaged in the furniture manufacturing business. One of its bi-weekly paychecks was payable to Stein, who negotiated it to White in pay ment of a gambling debt. White proceeded to raise the amount of the check from $300 to $800 and negotiated it to Carson, a holder in due course, for cash. Upon presentment by Carson at the drawee bank, the teller detected the raising of the amount and contacted Franco who stopped payment on the check. Franco refuses to pay Carson. Carson is seeking to recover the $800. Under the circumstances, which of the following is a correct statement?
a. Franco is liable, but only for $300.
b. Franco is liable for the $800.
c. Stein is liable for the $800.
d. Franco has no liability to Carson.
33. Strattford Theaters made a contract with Avon, Inc., for the purchase of $450 worth of theater supplies. Delivery was to take place in one month. One week after accepting the order, the price of materials and labor in creased sharply. In fact, to break even on the contract, Avon would have to charge an additional $600. Avon phoned Strattford and informed them of the situation. Strattford was sympathetic and said they were sorry to hear about the situation but that the best they would be willing to do was split the rise in price with Avon. Avon accepted the modification on Strattford’s terms. As a result of the above modification, which of the following is correct?
a. Avon’s continuing to perform the contract after informing Strattford of the price dif ficulty constitutes consideration for the modification of the price.
b. The oral modification is not effective since there was no consideration.
c. The statute of frauds applies to the contract as modified.
d. The contract contained an implied promise that it was subject to price rises.
34. Mara Oil, Inc., had a contract with Gotham Apartments to supply it with its fuel oil needs for the year, approximately 10,000 gallons. The price was fixed at ten cents above the price per gallon that Mara paid for its oil. Due to an exceptionally cold winter, Mara found that its capacity to fulfill this contract was doubt ful. Therefore, it contacted Sands Oil Company and of fered to assign the contract to it for $100. Sands agreed. Which of the following is correct as a result of the above assignment?
a. The contract with Gotham was neither assignable nor delegable.
b. Mara is now released from any further obliga tion to perform the Gotham contract.
c. Mara has effectively assigned to Sands its rights and delegated its duties under the terms of the contract with Gotham.
d. In the event Sands breaches the contract with Gotham, Mara has no liability.
140 Business Law 35. Almovar Electronics was closing out several lines of electronic parts which were becoming outdated. It sent a letter on March 8 to Conduit Sales & Service Company, one of its largest retail customers, offering the entire lot at a substantial reduction in price. The of fer indicated that it was for “ immediate acceptance.’’ The terms were “cash, pick up by your carrier at our loading dock and not later than March 15.’’ It also in dicated that the terms of the offer were not subject to variance. The letter did not arrive until March 10 and Conduit’s letter accepting the offer was not mailed until March 12. The letter of acceptance indicated that Conduit would take the entire lot, would pay in accord ance with the usual terms (2/10, net/30), and would pick up the goods on March 16. Which of the following best describes the legal relationship of the parties?
a. The acceptance was not timely, hence no con tract.
b. The different terms of the acceptance are to be construed as proposals for changes in the con tract.
c. The different terms of the acceptance con stituted a rejection of the offer.
d. Since both parties were merchants and the changes in the acceptance were not material, there is a valid contract.
36. Donaldson sold Randal six bundles of mink skins. The contract contained no specific provision regarding title warranties. It did, however, contain a provision which indicated that the skins were sold “with all faults and defects.’’ Two of the bundles of skins sold to Randal had been stolen and were reclaimed by the rightful owner. Which of the following is a correct state ment?
a. Since there was no express title warranty, Randal assumed the risk.
b. The disclaimer “with all faults and defects’’ effectively negates any and all warranties.
c. The contract automatically contained a war ranty that the title conveyed is good and can only be excluded by specific language.
d. The implied warranty of title is eliminated by the parol evidence rule.
37. Marco Auto Inc., made many untrue statements in the course of inducing Rockford to purchase a used auto for $3,500. The car in question turned out to have some serious faults. Which of the following untrue statements made by Marco should Rockford use in seek ing recovery from Marco for breach of warranty?
a. “ I refused a $3,800 offer for this very same auto from another buyer last week.’’ b. “This auto is one of the best autos we have for sale.’’ c. “At this price the auto is a real steal.’’ d. “ I can guarantee that you will never regret this purchase.’’ 38. If a seller repudiates his contract with a buyer for the sale of 100 radios, what recourse does the buyer have?
a. He can “cover,’’ i.e., procure the goods elsewhere and recover the difference.
b. He must await the seller’s performance for a commercially reasonable time after repudia tion.
c. He can obtain specific performance by the seller.
d. He can recover punitive damages.
39. Duval Liquor Wholesales, Inc., stored its inven tory of goods in the Reliable Warehouse Company. Duval’s shipments would arrive by truck and be deposited with Reliable who would in turn issue negotiable warehouse receipts to Duval. Duval would resell the liquor by transferring the negotiable warehouse receipts to the buyer who was responsible for transporting it to his place of business. In one of the sales of liquor to a retailer, the liquor was badly dam aged and a question has arisen as to who has the risk of loss, Duval or the retailer. If the contract is silent on this point, when did the risk of loss pass to the retailer?
a. When the goods have been placed on the warehouseman’s delivery dock awaiting pick up by the retailer.
b. When the goods have been identified to the contract.
c. On his receipt of the negotiable warehouse receipts covering the goods.
d. When the goods have been properly loaded upon the retailer’s carrier.
40. Dolby was employed as an agent for the Ace Used Car Company to purchase newer model used cars. His authority was limited by a $3,000 maximum price for any car. A wholesaler showed him a 1938 classic car which was selling for $5,000. The wholesaler knew that Ace only dealt in newer model used cars and that Dolby had never paid more than $3,000 for any car. Dolby bought the car for Ace, convinced that it was worth at least $7,000. When he reported this to Williams, Ace’s owner, Williams was furious but he nevertheless authorized processing of the automobile for resale. Williams also began pricing the car with antique car dealers who indicated that the current value of the car was $4,800. Williams called the wholesaler, told him that Dolby had exceeded his authority, that he was returning the car, and that he was demanding repay ment of the purchase price. What is the wholesaler’s best defense in the event of a lawsuit?
a. Dolby had apparent authority to purchase the car. b. Dolby’s purchase was effectively ratified by Ace.
c. Dolby had express authority to purchase the car.
d. Dolby had implied authority to purchase the car.
141 Examination Questions— May 1979 41. Zebra Acquisitions, Inc., has been steadily acquir ing the assets and stock of various corporations manufacturing brass. It also has been purchasing the stock of its customers and others who purchase substan tial quantities of brass. It now has 8% of the brass manufacturing facilities in the United States and 22% in the tri-state area in which it is located. Which of the following claims is the United States Department of Justice likely to assert?
a. The relevant market in question is the entire United States.
b. It is illegal per se to purchase the stock of competitors.
c. It is illegal per se to purchase the stock of customers and other potential buyers.
d. The most recent acquisition substantially lessens competition in the tri-state area.
42. Drummond Company manufactures and sells distinctive clocks. Its best selling item is a reproduction of a rare antique grandfather clock. Adams purchased 100 of the clocks from Drummond at $79.50. Much to Adam’s chagrin it discovered that Young, one of its competitors, had purchased the same clock at $74.50 per clock. Adams has complained and threatened legal action. In the event the issue is litigated a. Drummond will prevail if it can show it did not intend to harm Adams.
b. Adams has a presumption in its favor that it has been harmed by price discrimination.
c. Drummond will prevail if it can show that it sold the clocks at the lower price to all customers such as Young who had been doing business with it continuously for ten years or more.
d. Drummond will prevail if it can establish that there were several other clock companies with which Adams could deal if Adam’s were dissatisfied.
43. Tweed Manufacturing, Inc., plans to issue $5 million of common stock to the public in interstate com merce after its registration statement with the SEC becomes effective. What, if anything, must Tweed do in respect to those states in which the securities are to be sold?
a. Nothing, since approval by the SEC automatically constitutes satisfaction of any state requirements.
b. Make a filing in those states which have laws governing such offerings and obtain their ap proval.
c. Simultaneously apply to the SEC for permis sion to market the securities in the various states without further clearance.
d. File in the appropriate state office of the state in which it maintains its principal office of business, obtain clearance, and forward a certified copy of that state’s clearance to all other states.
44. Major Steel Manufacturing, Inc., signed a con tract on October 2, 1978, with the Hard Coal & Coke Company for its annual supply of coal for three years commencing on June 1, 1979, at a price to be determin ed by taking the average monthly retail price per ton, less a ten cent per ton quantity discount. On March 15, 1979, Major discovered that it had made a bad bargain and that it could readily fulfill its requirements elsewhere at a much greater discount. Major is seeking to avoid its obligation. Which of the following is cor rect?
a. The pricing term is too indefinite and uncer tain hence there is no contract.
b. Since the amount of coal required is unknown at the time of the making of the contract, the contract is too indefinite and uncertain to be valid.
c. Major is obligated to take its normal annual coal requirements from Hard or respond in damages.
d. There is no contract since Major could con ceivably require no coal during the years in question.
45. What fiduciary duty, if any, exists in an agency relationship?
a. The agent owes a fiduciary duty to third parties he deals with for and on behalf of his principal b. The principal owes a fiduciary duty to his agent.
c. The agent owes a fiduciary duty to his prin cipal.
d. There is no fiduciary duty in an agency rela tionship.
46. Research Development Corporation has made a major breakthrough in the development of a super TV antenna. It has patented the product and is seeking to maximize the profit potential. In this effort, Research can legally a. Require its retailer to take stipulated quan tities of its other products in addition to the antennas.
b. Require its retailers to sell only Research’s products, including the antennas, and not sell similar competing products.
c. Sell the product to its retailers upon condition that they do not sell the antennas to the public for less than a stated price.
d. Sell the product at whatever price the traffic will bear even though it has a monopoly.
47. Wilton was grossly negligent in the operation of a drill press. As a result he suffered permanent disability. His claim for workmen’s compensation will be a. Reduced by the percentage share attributable to his own fault.
b. Limited to medical benefits.
c. Denied.
d. Paid in full.
142 Business Law 48. Harvey Wilson is a senior vice president, 15% shareholder and a member of the Board of Directors of Winslow, Inc. Wilson has decided to sell 10% of his stock in the company. Which of the following methods of disposition would subject him to SEC registration re quirements?
a. A redemption of the stock by the corporation.
b. The sale by several brokerage houses of the stock in the ordinary course of business.
c. The sale of the stock to an insurance company which will hold the stock for long-term invest ment purposes.
d. The sale to a corporate officer who currently owns 5% of the stock of Winslow and who will hold the purchased stock for long-term in vestment.
49. The Securities Exchange Act of 1934 holds certain insiders liable for short-swing profits under section 16(b) of the act. Which of the following classes of peo ple would not be insiders in relation to the corporation in which they own securities?
a. An executive vice president.
b. A major debenture holder.
c. An 11 % owner, 8% of which he owns in his or her own name and 3% in an irrevocable trust for his or her benefit for life.
d. A director who owns less than 10% of the shares of stock of the corporation.
50. Jane Sabine was doing business as Sabine Fashions, a sole proprietorship. Sabine suffered finan cial reverses and began to use social security and income taxes withheld from her employees to finance the business. Sabine finally filed a voluntary petition in bankruptcy. Which of the following would not apply to her as a result of her actions?
a. She would remain liable for the taxes due.
b. She is personally liable for fines and imprison ment.
c. She could justify her actions by showing that the use of the tax money was vital to continua tion of the business.
d. She may be assessed penalties up to the amount of taxes due.
Number 2 (Estimated time----- 25 to 30 minutes) Part a. Factory Discount Prices, Inc., is a chain store discount outlet which sells women’s clothes. It has an excessively large inventory on hand and is in urgent need of additional cash. It is bordering on bankruptcy, especially if the inventory has to be liquidated by sale to other stores instead of the public. Furthermore, about 15% of the inventory is not resalable except at a drastic discount below cost. Faced with this financial crisis, Factory approached several of the manufacturers from whom it purchases. Dexter Apparel, Inc., one of the parties approached, indicated a willingness to loan Factory $300,000 under certain conditions. First, Factory was to submit audited financial statements for the express purpose of providing the correct financial condition of the company. The loan was to be predicated upon these financial statements and Factory’s engagement letter with Dunn & Clark, its CPAs, expressly indicated this.
The second condition insisted upon by Dexter was that it obtain a secured position in all unsecured inven tory, accounts, and other related personal property. In due course a security agreement was executed and a financing statement properly filed and recorded.
In preparing the financial statements, Factory valued the inventory at cost which was approximately $100,000 over the current fair market value. Also, Factory failed to disclose two secured creditors to whom substantial amounts are owed and who take priority over Dexter’s security interests.
Dunn & Clark issued an unqualified opinion on the financial statements of Factory which they believed were fairly presented.
Six months later Factory filed a voluntary bankruptcy petition. Dexter received $125,000 as its share of the bankrupt’s estate. It is suing Dunn & Clark for the loss of $175,000. Dunn & Clark deny liability based upon lack of privity and lack of negligence.
Required: Answer the following, setting forth reasons for any conclusions stated.
Is Dexter entitled to recover its loss from Dunn & Clark?
Part b. The CPA firm of Blank, Miller & Tage prepares a significant number of individual and cor porate income tax returns. Jones is a newly hired junior accountant. This is Jones’ first job since graduation from school. Jones’ initial assignment is to work with the tax department in the preparation of clients’ 1978 in come tax returns.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. What is the principal legal basis for potential liability of the CPA firm and Jones to clients in connec tion with the preparation of income tax returns?
2. Give some examples of performance which would result in such liability.
3. What is the basis for determining the amount of damages to be awarded to the client?
Part c. Smith, CPA, is the auditor for Juniper Manufacturing Corporation, a privately-owned com pany which has a June 30 fiscal year. Juniper arranged for a substantial bank loan which was dependent upon the bank receiving, by September 30, audited financial statements which showed a current ratio of at least 2 to 1. On September 25, just before the audit report was to be issued, Smith received an anonymous letter on Juniper’s stationery indicating that a five-year lease by 143 Examination Questions—May 1979 Juniper, as lessee, of a factory building which was ac counted for in the financial statements as an operating lease was in fact a capital lease. The letter stated that there was a secret written agreement with the lessor modifying the lease and creating a capital lease.
Smith confronted the president of Juniper who ad mitted that a secret agreement existed but said it was necessary to treat the lease as an operating lease to meet the current ratio requirement of the pending loan and that nobody would ever discover the secret agreement with the lessor. The president said that if Smith did not issue his report by September 30, Juniper would sue Smith for substantial damages which would result from not getting the loan. Under this pressure and because the working papers contained a copy of the five-year lease agreement which supported the operating lease treatment, Smith issued his report with an unqualified opinion on September 29.
In spite of the fact that the loan was received, Juniper went bankrupt within two years. The bank is su ing Smith to recover its losses on the loan and the lessor is suing Smith to recover uncollected rents.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Is Smith liable to the bank?
2. Is Smith liable to the lessor?
Number 3 (Estimated tim e-----25 to 30 minutes) Part a. The Decimile Corporation is a well- established, conservatively-managed, major company. It has consistently maintained a $3 or more per share dividend since 1940 on its only class of stock, which has a $1 par value. Decimile’s board of directors is deter mined to maintain a $3 per share annual dividend distribution to maintain the corporation’s image in the financial community, to reassure its shareholders, and to prevent a decline in the price of the corporation’s shares which would occur if there were a reduction in the dividend rate. Decimile’s current financial position is not encouraging although the corporation is legally solvent. Its cash flow position is not good and the cur rent year’s earnings are only $0.87 per share. Retained earnings amount to $17 per share. Decimile owns a substantial block of Integrated Electronic Services stock which it purchased at $1 per share in 1950 and which has a current value of $6.50 per share. Decimile has paid dividends of $1 per share so far this year and con templates distributing a sufficient number of shares of Integrated to provide an additional $2 per share.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. May Decimile legally pay the $2 per share divi dend in the stock of Integrated?
2. As an alternative, could Decimile pay the $2 dividend in its own authorized but unissued shares of stock? What would be the legal effect of this action upon the corporation?
3. What are the federal income tax consequences to the noncorporate shareholders— (a) If Decimile distributes the shares of Integrated?
(b) If Decimile distributes its own authorized but unissued stock?
Part b. Clayborn is the president and a director of Marigold Corporation. He currently owns 1,000 shares of Marigold which he purchased several years ago upon joining the company and assuming the presidency. At that time, he received a stock option for 10,000 shares of Marigold at $10 per share. The option is about to ex pire but Clayborn does not have the money to exercise his option. Credit is very tight at present and most of his assets have already been used to obtain loans. Clayborn spoke to the chairman of Marigold’s board about his plight and told the chairman that he is going to borrow $100,000 from Marigold in order to exercise his option. The chairman was responsible for Clayborn’s being hired as the president of Marigold and is a close per sonal friend of Clayborn. Fearing that Clayborn will leave unless he is able to obtain a greater financial in terest in Marigold, the chairman told Clayborn: “ It is okay with me and you have a green light.’’ Clayborn authorized the issuance of a $100,000 check payable to his order. He then negotiated the check to Marigold in payment for the shares of stock.
Required: Answer the following, setting forth reasons for any conclusions stated.
What are the legal implications, problems, and issues raised by the above circumstances?
Part c. Towne is a prominent financier, the owner of 1% of the shares of Toy, Inc., and one of its direc tors. He is also the chairman of the board of Unlimited Holdings, Inc., an investment company in which he owns 80% of the stock. Toy needs land upon which to build additional warehouse facilities. Toy’s president, Arthur, surveyed the land sites feasible for such a pur pose. The best location in Arthur’s opinion from all standpoints, including location, availability, access to transportation, and price, is an eight-acre tract of land owned by Unlimited. Neither Arthur nor Towne wish to create any legal problems in connection with the possi ble purchase of the land.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. What are the legal parameters within which this transaction may be safely consummated?
2. What are the legal ramifications if there were to be a $50,000 payment “ on the side’’ to Towne in order that he use his efforts to “ smooth the way’’ for the proposed acquisition?
144 Business Law Number 4 (Estimated time 20 to 25 minutes) Part a. Strom, Lane, and Grundig formed a part nership on July 1, 1974, and selected “ Big M Associates” as their partnership name. The partnership agreement specified a fixed duration of ten years for the partnership. Business went well for the partnership for several years and it established an excellent reputation in the business community. In 1978, Strom, much to his amazement, learned that Grundig was padding his ex pense accounts by substantial amounts each month and taking secret kick-backs from certain customers for price concessions and favored service. Strom informed Lane of these facts and they decided to seek an account ing of Grundig, a dissolution of the firm by ousting Grundig, and the subsequent continuation of the firm by themselves under the name, “ Big M Associates.” Required: Answer the following, setting forth reasons for any conclusions stated.
1. Were there any filing requirements to be satisfied upon the initial creation of the partnership?
2. What will be the basis for the accounting and dissolution and should such actions be successful?
3. Can Strom and Lane obtain the right to con tinue to use the firm name if they prevail?
Part b. Palmer is a member of a partnership. His per sonal finances are in a state of disarray, although he is not bankrupt. He recently defaulted on a personal loan from the Aggressive Finance Company. Aggressive in dicated that if he did not pay within one month, it would obtain a judgment against him and levy against all his property including his share of partnership pro perty and any interest he had in the partnership. Both Palmer and the partnership are concerned about the effects of this unfortunate situation upon Palmer and the partnership.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Has a dissolution of the partnership occurred?
2. What rights will Aggressive have against the partnership or Palmer concerning Palmer’s share of partnership property or his interest in the partnership?
3. Could Palmer legally assign his interest in the partnership as security for a loan with which to pay off Aggressive?
Number 5 (Estimated time-----15 to 20 minutes) Part a. Hammar Hardware Company, Inc., pur chased all the assets and assumed all the liabilities of JoMar Hardware for $60,000. Among the assets and liabilities included in the sale was a lease of the building in which the business was located. The lessor-owner was Marathon Realty, Inc., and the remaining unexpired term of the lease was nine years. The lease did not con tain a provision dealing with the assignment of the leasehold. Incidental to the purchase, Hammar express ly promised JoMar that it would pay the rental due Marathon over the life of the lease and would hold JoMar harmless from any future liability thereon.
When Marathon learned of the proposed transac tion, it strenuously objected to the assignment of the lease and to the occupancy by Hammar. Later, after this dispute was resolved and prior to expiration of the lease, Hammar abandoned the building and ceased doing business in that area. Marathon has demanded payment by JoMar of the rent as it matures over the balance of the term of the lease.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Was the consent of Marathon necessary in order to assign the lease?
2. Is JoMar liable on the lease?
3. If Marathon were to proceed against Hammar, would Hammar be liable under the lease?
Part b. The Merchants and Mechanics County Bank expanded its services and facilities as a result of the economic growth of the community it serves. In this connection, it provided safe deposit facilities for the first time. A large vault was constructed as a part of the renovation and expansion of the bank building. Mer chants purchased a bank vault door from Foolproof Vault Doors, Inc., for $65,000 and installed it at the vault entrance. The state in which Merchants was located had a real property tax but did not have a per sonal property tax. When the tax assessor appraised the bank building after completion of the renovation and expansion, he included the bank vault door as a part of the real property. Merchants has filed an objection claiming the vault door was initially personal property and remains so after installation in the bank.
There are no specific statutes or regulations deter minative of the issue. Therefore, the question will be decided according to common law principles of proper ty law.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. What is the likely outcome as to the classifica tion of the bank vault door?
2. The above situation involves a dispute between a tax authority and the owner of property. In what other circumstances might a dispute arise with respect to the classification of property as either real or personal property?
145 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXA M IN A TIO N IN ACCOUNTING TH E O R Y (Theory of Accounts) May 4, 1979; 1:30 to 5:00 P.M .
N O TE TO C A N D ID A TES: Suggested time allotments are as follows: All questions are required: No. 1 ......................
No. 2 ......................
No. 3 ......................
No. 4 ......................
No. 5 ..................
No. 6 ......................
To ta l..............
Estimated Minutes Minimum Maximum 90 105 20 25 15 20 15 20 15 20 15 20 170 210 IN STR U C TIO N S TO C A N D ID A TES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “ continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
146 Accounting Theory Number 1 (Estimated time-----90 to 105 minutes) Instructions Select the best answer for each of the following items relating to a variety of issues in accounting. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 99. The financial statement which summarizes the results of operations is the a. Income statement.
b. Balance sheet.
c. Statement of sources and applications of funds.
d. Retained earnings statement.
Answer Sheet 99. Items to be Answered 1. In a business combination what is the appropriate method of accounting for an excess of fair value as signed to net assets over the cost paid for them?
a. Record as negative goodwill.
b. Record as additional paid-in capital from combination on the books of the combined company.
c. Proportionately reduce values assigned to nonmonetary assets and record any remaining excess as a deferred credit.
d. Proportionately reduce values assigned to noncurrent assets and record any remaining excess as a deferred credit.
2. The year-end balance of accounts receivable on the books of a foreign subsidiary should be translated by the parent company for consolidation purposes at the a. Historical rate.
b. Current rate.
c. Negotiated rate.
d. Spot rate.
3. Which of the following methods of inventory valuation is allowable at interim dates but not at year- end?
a. Weighted average.
b. Estimated gross profit rates.
c. Retail method.
d. Specific identification.
4. Which of the following transactions related to a business combination would require that the combina tion be accounted for as a purchase?
a. The combination is to be completed within twelve months from the date the plan was in itiated.
b. Ninety-two percent of one company’s com mon stock is exchanged for only common stock in the other company.
c. The combined company is to retire a portion of the common stock exchanged to effect the combination within twelve months of the combination, d. The combined company will dispose of numerous fixed assets representing duplicate facilities subsequent to the combination.
5. Which of the following sets of conditions would give rise to the accrual of a contingency under current generally accepted accounting principles?
a. Amount of loss is reasonably estimable and event occurs infrequently.
b. Amount of loss is reasonably estimable and occurrence of event is probable.
c. Event is unusual in nature and occurrence of event is probable.
d. Event is unusual in nature and event occurs in frequently.
6. Based solely upon the following sets of cir cumstances indicated below, which set gives rise to a sales type or direct financing lease of a lessor?
Transfers ownership by end of lease?
a. No b. Yes c. Yes d. No Contains bargain purchase provision?
Yes No No Yes Collectibil ity o f lease payments assured?
Yes No No Yes Any important uncertain-ties?
No No Yes Yes 7. A change in accounting entity is actually a change in accounting a. Principle.
b. Estimate.
c. Method.
d. Concept, 8. In which of the following pension instances would the accrual of past service costs have to be reduced for interest presumed earned?
a. When past service costs have been fully ac crued prior to funding.
b. When pension expense exceeds the maximum allowable accrual.
c. When past service costs have been fully fund ed prior to accrual.
d. When interest presumed earned on previously accrued past service cost exceeds interest presumed earned on unaccrued past service cost.
147 Examination Questions— May 1979 9. The amount of income tax applicable to transac tions that must be reported using intraperiod income tax allocation is computed a. By multiplying the item by the effective in come tax rate.
b. As the difference between the tax computed based on taxable income without including the item and the tax computed based on taxable income including the item.
c. As the difference between the tax computed on the item based on the amount used for financial reporting and the amount used in computing taxable income.
d. By multiplying the item by the difference be tween the effective income tax rate and the statutory income tax rate.
10. Which of the following transactions would require the use of the present value of an annuity due concept in order to calculate the present value of the asset obtained or liability owed at the date of incurrence?
a. A capital lease is entered into with the initial lease payment due upon the signing of the lease agreement.
b. A capital lease is entered into with the initial lease payment due one month subsequent to the signing of the lease agreement.
c. A ten-year 8% bond is issued on January 2 with interest payable semi-annually on July 1 and January 1 yielding 7%.
d. A ten-year 8% bond is issued on January 2 with interest payable semi-annually on July 1 and January 1 yielding 9%.
11. Assuming no beginning inventory, what can be said about the trend of inventory prices if cost of goods sold computed when inventory is valued using the FIFO method exceeds cost of goods sold when inventory is valued using the LIFO method?
a. Prices decreased.
b. Prices remained unchanged.
c. Prices increased.
d. Price trend cannot be determined from infor mation given.
12. What is the proper time or time period over which to match the cost of an intangible asset with revenues if it is likely that the benefit of the asset will last for an in determinate but very long period of time?
a. Forty years.
b. Fifty years.
c. Immediately.
d. At such time as diminution in value can be quantitatively determined.
13. The sale of a depreciable asset resulting in a loss, indicates that the proceeds from the sale were a. Less than current market value.
b. Greater than cost.
c. Greater than book value.
d. Less than book value.
14. Interperiod income tax allocation is justified by the basic theory that income taxes should be treated as which of the following?
a. An expense.
b. A distribution of earnings.
c. A distribution of earnings for the current por tion and an expense for the deferred portion.
d. An expense for the current portion and a distribution of earnings for the deferred por tion.
15. A marketable equity security must have a ready market in order to be classified as current, and a. Be available to management for use in short run operations.
b. Be traded on a recognized national exchange.
c. Have a current market value in excess of original cost.
d. Have been owned less than one year.
16. An increase in inventory balance would be reported in a statement of changes in financial position as a a. Use of working capital.
b. Source of working capital.
c. Source of cash.
d. Use of cash.
17. Consolidated financial statements are prepared when a parent-subsidiary relationship exists in recogni tion of the accounting concept of a. Materiality.
b. Entity.
c. Objectivity.
d. Going concern.
18. Dilutive common stock equivalents must be used in the computation of a. Fully diluted earnings per share only.
b. Primary earnings per share only.
c. Fully diluted and primary earnings per share.
d. Other potentially dilutive securities only.
19. Which of the following asset valuation methods is not a violation of the accounting concept of historical cost?
a. Net present value of future cash flows.
b. Replacement cost.
c. Market value.
d. General price-level restatement.
20. What type of direct material variances for price and usage will arise if the actual number of pounds of materials used exceeds standard pounds allowed but ac tual cost was less than standard cost?
Usage Price a.
Unfavorable Favorable b.
Favorable Favorable c.
Favorable Unfavorable d.
Unfavorable Unfavorable 148 Accounting Theory 21. In considering a special order situation that will enable a company to make use of presently idle capaci ty, which of the following costs would be irrelevant?
a. Materials.
b. Depreciation.
c. Direct labor.
d. Variable overhead.
22. In job order costing, payroll taxes paid by the employer for factory employees are preferably ac counted for as a. Direct labor.
b. Factory overhead.
c. Indirect labor.
d. Administrative costs.
23. If a company follows a practice of isolating variances at the earliest point in time, what would be the appropriate time to isolate and recognize a direct material price variance?
a. When material is issued.
b. When material is purchased.
c. When material is used in production.
d. When purchase order is originated.
24. Which of the following cost allocation methods would be used to determine the lowest price that could be quoted for a special order that would utilize idle capacity within a production area?
a. Job order.
b. Process.
c. Variable.
d. Standard.
25. How should a usage variance that is significant in amount be treated at the end of an accounting period?
a. Reported as a deferred charge or credit.
b. Allocated among work-in-process inventory, finished goods inventory, and cost of goods sold.
c. Charged or credited to cost of goods manufactured.
d. Allocated among cost of goods manufac tured, finished goods inventory, and cost of goods sold.
26. Which of the following quantitative methods will separate a semi-variable cost into its fixed and variable components with the highest degree of precision under all circumstances?
a. High-low method.
b. Simplex method.
c. Least squares method.
d. Scattergraph method.
27. A measure of the extent to which two variables are related linearly is referred to as a. Cause-effect ratio.
b. Coefficient of correlation.
c. Sensitivity analysis.
d. Input-output analysis.
28. Which of the following would cause the break even point to change?
a. Sales increased.
b. Total production decreased.
c. Total variable costs increased as a function of higher production.
d. Fixed costs increased due to addition to physical plant.
29. A true process costing system could make use of each of the following except a. Standards.
b. Individual lots.
c. Variable costing.
d. Responsibility accounting.
30. Joint costs are most frequently allocated based upon relative a. Profitability.
b. Conversion costs.
c. Sales value.
d. Prime costs.
31. Which of the following components of production are allocable as joint costs when a single manufacturing process produces several salable products?
a. Materials, labor, overhead.
b. Materials and labor only.
c. Labor and overhead only.
d. Overhead and materials only.
32. Each of the following would affect the breakeven point except a change in the a. Number of units sold.
b. Variable cost per unit.
c. Total fixed costs.
d. Sales price per unit.
33. What is the normal year-end treatment of im material variances recognized in a cost accounting system utilizing standards?
a. Reclassified to deferred charges until all related production is sold.
b. Allocated among cost of goods manufactured and ending work-in-process inventory.
c. Closed to cost of goods sold in the period in which they arose.
d. Capitalized as a cost of ending finished goods inventory.
34. Which of the following items of cost would be least likely to appear in a performance report based on responsibility accounting techniques for the supervisor of an assembly line in a large manufacturing situation?
a. Supervisor’s salary, b. Materials.
c. Repairs and maintenance.
d. Direct labor.
149 Examination Questions— May 1979 35. When using the graphic method of solving a linear programming problem, the optimal solution will always be at a. Minimum value of X.
b. X and Y intercept.
c. A corner point described by the feasible area.
d. Point of inception.
36. The quantitative analysis tool whose primary ob jective is to define a critical path is a. Linear programming.
b. Queuing theory.
c. Program evaluation and review technique.
d. Multiple regression analysis.
37. The method of cost accounting that lends itself to breakeven analysis is a. Variable.
b. Standard.
c. Absolute.
d. Absorption.
38. The basic difference between a master budget and a flexible budget is that a a. Flexible budget considers only variable costs but a master budget considers all costs.
b. Flexible budget allows management latitude in meeting goals whereas a master budget is based on a fixed standard.
c. Master budget is for an entire production facility but a flexible budget is applicable to single departments only.
d. Master budget is based on one specific level of production and a flexible budget can be prepared for any production level within a relevant range.
39. In order to compute equivalent units of produc tion using the FIFO method of process costing, work for the period must be broken down to units a. Completed during the period and units in end ing inventory.
b. Completed from beginning inventory, started and completed during the month, and units in ending inventory.
c. Started during the period and units trans ferred out during the period.
d. Processed during the period and units com pleted during the period.
40. Spoilage occurring during a manufacturing pro cess can be considered normal or abnormal. The proper accounting for each of these costs is Normal Abnormal a.
Product Period b.
Product Product c.
Period Product d.
Period Period 41. Within a governmental unit, three funds that are accounted for in a manner similar to a for-profit entity are a. General, Debt Service, Special Assessment.
b. Special Assessment, Enterprise, Intragovern mental Service.
c. Intragovernmental Service, Enterprise, General.
d. Enterprise, General, Debt Service.
42. In the loan fund of a college or university, each of the following types of loans would be found except a. Student.
b. Staff.
c. Building.
d. Faculty.
43. What is the recommended method of accounting to be used by colleges and universities?
a. Cash.
b. Modified cash.
c. Restricted accrual.
d. Accrual.
44. Why do voluntary health and welfare organiza tions, unlike some not-for-profit organizations, record and recognize depreciation of fixed assets?
a. Fixed assets are more likely to be material in amount in a voluntary health and welfare organization than in other not-for-profit organizations.
b. Voluntary health and welfare organizations purchase their fixed assets, and therefore have a historical cost basis from which to determine amounts to be depreciated.
c. A fixed asset used by a voluntary health and welfare organization has alternative uses in private industry and this opportunity cost should be reflected in the organization’s financial statements.
d. Contributors look for the most efficient use of funds, and since depreciation represents a cost of employing fixed assets, it is appropriate that a voluntary health and welfare organiza tion reflect it as a cost of providing services.
45. Which of the following receipts is properly record ed as restricted current funds on the books of a universi ty?
a. Tuition.
b. Student laboratory fees.
c. Housing fees.
d. Research grants.
46. Which of the following funds of a voluntary health and welfare organization does not have a counterpart fund in governmental accounting?
a. Current unrestricted.
b. Land, building, and equipment.
c. Custodian.
d. Endowment.
150 Accounting Theory 47. A city’s general fund budget for the forthcoming fiscal year shows estimated revenues in excess of ap propriations. The initial effect of recording this will result in an increase in a. Taxes receivable.
b. Fund balance.
c. Reserve for encumbrances.
d. Encumbrances.
48. Under the modified accrual method of accounting used by a local governmental unit, which of the follow ing would be a revenue susceptible to accrual?
a. Income taxes.
b. Business licenses.
Shown below are the financial statements issued by Allen Corporation for its fiscal year ended October 31, 1978: Number 2 (Estimated time----- 20 to 25 minutes) Allen Corporation Statement of Financial Position October 31, 1978 Assets 50. What is the underlying reason a governmental unit uses separate funds to account for its transactions?
a. Governmental units are so targe that it would be unduly cumbersome to account for all transactions as a single unit.
b. Because of the diverse nature of the services offered and legal provisions regarding ac tivities of a governmental unit, it is necessary to segregate activities by functional nature.
c. Generally accepted accounting principles re quire that not-for-profit entities report on a funds basis.
d. Many activities carried on by governmental units are short-lived and their inclusion in a general set of accounts could cause undue probability of error and omission.
Liabilities Accounts payable Accrued expenses Total current liabilities Deferred income tax payable (Note 4) Total liabilities c. Property taxes.
Cash $ 15,000 d. Sales taxes.
Accounts receivable, net 150,000 49. Which of the following accounts is a budgetary ac-Inventory 120,000 count in governmental accounting?
Total current assets 285,000 a. Reserve for inventory of supplies.
b. Fund balance.
Trademark (Note 3) 250,000 c. Appropriations.
d. Estimated uncollectible property taxes.
Land 125,000 Total assets $660,000 $ 80,000 20,000 100,000 80,000 180,000 Stockholders' Equity Common stock, par $1 (Note 5) Additional paid-in capital Retained earnings Total liabilities and stockholders’ equity $100,000 180,000 200,000 480,000 $660,000 151 Examination Questions— May 1979 Allen Corporation Earnings Statement For the fiscal year ended October 31, 1978 Sales $1,000,000 Cost of goods sold 750,000 Gross margin 250,000 Expenses: Bad debt expense $ 7,000 Insurance Lease expenses 13,000 (Note 1) Repairs and 40,000 maintenance 30,000 Pensions (Note 2) 12,000 Salaries Earnings before 60,000 162,000 provision for income tax Provision for income 88,000 tax 28,740 Net earnings $ 59,260 Earnings per common share outstanding $ 0.5926 Allen Corporation Statement of Retained Earnings For the fiscal year ended October 31, 1978 Retained earnings, November 1, 1977 $150,000 Extraordinary gain, net of income tax Net earnings for the fiscal year ended 25,000 October 31, 1978 59,260 234,260 Dividends ($0.3426 per share) 34,260 Retained earnings, October 31, 1978 $200,000 FOOTNOTES Note 1 — Long-Term Lease Under the terms of a 5-year noncancelable lease for buildings and equipment, the Company is obligated to make annual rental payments of $40,000 in each of the next four fiscal years. At the conclusion of the lease period, the Company has the option of purchasing the leased assets for $20,000 (a bargain purchase option) or entering into another 5-year lease of the same property at an annual rental of $5,000.
Note 2 — Pension Plan Substantially all employees are covered by the Com pany’s pension plan. Pension expense is equal to the total of pension benefits paid to retired employees during the year.
Note 3 — Trademark The Company’s trademark was purchased from Apex Corporation on January 1, 1976, for $250,000.
Note 4 — Deferred Income Tax Payable The entire balance in the deferred income tax payable account arose from tax-exempt municipal bonds that were held during the previous fiscal year giving rise to a difference between taxable income and reported net earnings for the fiscal year ended October 31, 1977. The deferred liability amount was calculated on the basis of expected tax rates in future years.
Note 5 — Warrants On January 1, 1977, one common stock warrant was issued to stockholders of record for each common share owned. An additional share of common stock is to be issued upon exercise of ten stock warrants and receipt of an amount equal to par value. For the six months ended October 31, 1978, the average market value for the Company’s common stock was $5 per share and no warrants had yet been exercised.
Note 6 — Contingent Liability On October 31, 1978, the Company was contingently liable for product warranties in an amount estimated to aggregate $75,000.
Required: Review the preceding financial statements and related footnotes. Identify any inclusions or exclusions from them that would be in violation of generally accepted accounting principles, and indicate corrective action to be taken. Do not comment as to format or style. Re spond in the following order: • Statement of Financial Position.
• Footnotes.
• Earnings Statement.
• Statement of Retained Earnings.
• General.
152 Accounting Theory Part a. The most common method of accounting for unconsolidated subsidiaries is the equity method.
Required: Answer the questions shown below with respect to the equity method.
1. Under what circumstances should the equity method be applied?
2. At what amount should the initial investment be recorded and what events subsequent to the initial in vestment (if any) would change this amount?
3. How are investment earnings recognized under the equity method, and how is the amount determined?
Part b. For the past five years Herbert has maintained an investment (properly accounted for and reported upon) in Broome amounting to a 10% interest in the voting common stock of Broome. The purchase price was $700,000 and the underlying net equity in Broome at the date of purchase was $620,000. On January 2 of the current year, Herbert purchased an additional 15% of the voting common stock of Broome for $1,200,000; the underlying net equity of the additional investment at January 2 was $1,000,000. Broome has been profitable and has paid dividends annually since Herbert’s initial acquisition.
Required: Discuss how this increase in ownership affects the ac counting for and reporting upon the investment in Broome. Include in your discussion adjustments, if any, to the amount shown prior to the increase in investment to bring the amount into conformity with generally ac cepted accounting principles. Also include how current and subsequent periods would be reported upon.
Number 4 (Estimated time-----15 to 20 minutes) In accounting for long-term contracts (those taking longer than one year to complete), the two methods commonly followed are the percentage-of-completion method and the completed-contract method.
Required: a. Discuss how earnings on long-term contracts are recognized and computed under these two methods.
b. Under what circumstances is it preferable to use one method over the other?
c. Why is earnings recognition as measured by in terim billings not generally accepted for long-term con tracts?
d. How are job costs and interim billings reflected on the balance sheet under the percentage-of- completion method and the completed-contract method?
Number 3 (Estimated time----- 15 to 20 minutes) Part a. In order to properly understand current generally accepted accounting principles with respect to accounting for and reporting upon segments of a business enterprise, as stated by the Financial Account ing Standards Board in its Statement 14, it is necessary to be familiar with certain unique terminology.
Required: With respect to segments of a business enterprise, ex plain the following terms: 1. Industry segment.
2. Revenue.
3. Operating profit and loss.
4. Identifiable assets.
Part b. A central issue in reporting on industry segments of a business enterprise is the determination of which segments are reportable.
Required: 1. What are the tests to determine whether or not an industry segment is reportable?
2. What is the test to determine if enough in dustry segments have been separately reported upon and what is the guideline on the maximum number of in dustry segments to be shown?
Number 5 (Estimated time----- 15 to 20 minutes) Number 6 (Estimated time-----15 to 20 minutes) Part a. Capital stock is an important area of a cor poration’s equity section. Generally the term “capital stock’’ embraces common and preferred stock issued by a corporation.
Required: 1. What are the basic rights inherent in ownership of common stock, arid how are they exercised?
2. What is preferred stock? Discuss the various preferences afforded preferred stock.
Part b. In dealing with the various equity securities of a corporate entity it is important to understand cer tain terminology related thereto.
Required: Define the following terms.
1. Treasury stock.
2. Legal capital.
3. Stock right.
4. Stock warrant.
153 Examination Questions November 1979 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) EXAMINATION IN ACCOUNTING PRACTICE — PART I November 7, 1979; 1:30 to 6:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1............................................................................................ 50 No. 2 ............................................................................................ 50 No. 3 ........................................................................................... 40 No. 4 ............................................................................................ 40 No. 5 ............................................................................................ 40 Total.................................................................................... 220 INSTRUCTIONS TO CANDIDATES 60 60 50 50 50 270 (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For in stance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. Enclose all scratch sheets. Failure to enclose scratch sheets may result in loss of grading points. Scratch sheets need not have page numbers, but you should show the question number and place them immediately following the question to which they relate.
4. Fourteen-column sheets, if any, should not be folded until all sheets, both wide and narrow, are placed in the proper sequence and fastened together at the top left corner. All fourteen- column sheets should then be wrapped around the back of the papers.
5. A CPA is continually confronted with the necessi ty of expressing opinions and conclusions in writ ten reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and applica tion of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
155 Examination Questions—November 1979 Instructions Select the best answer for each of the following items relating to a variety of financial-accounting pro blems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item Number 1 (Estimated tim e------50 to 60 minutes) 97. Gross billings for merchandise sold by Baker Company to its customers last year amounted to $5,260,000; sales returns and allowances reduced the amounts owed by $160,000. How much were net sales last year for Baker Company?
a. $4,800,000 b. $5,100,000 c. $5,200,000 d. $5,260,000 97.
Answer Sheet Items to be Answered Items 1 and 2 are based on the following informa tion: Bond Company purchased a machine on January 1, 1975, for $3,000,000. At the date of acquisition, the machine had an estimated useful life of six years with no salvage. The machine is being depreciated on a straight- line basis. On January 1, 1978, Bond determined, as a result of additional information, that the machine had an estimated useful life of eight years from the date of acquisition with no salvage. An accounting change was made in 1978 to reflect this additional information.
1. Assuming that the direct effects of this change are limited to the effect on depreciation and the related tax provision, and that the income tax rate was 50% in 1975, 1976, 1977 and 1978, what should be reported in Bond’s income statement for the year ended December 31, 1978, as the cumulative effect on prior years of changing the estimated useful life of the machine?
a. $0 b. $187,500 c. $250,000 d. $375,000 2. What is the amount of depreciation expense on this machine that should be charged in Bond’s income statement for the year ended December 31, 1978?
a. $100,000 b. $300,000 c. $375,000 d. $500,000 3. On December 1, 1978, Drew Company issued shares of its voting common stock in exchange for all of the voting common stock of Art Company in a business combination appropriately accounted for by the pooling of interests method. Net income for each company is as follows: Drew Art 12 months ended December 31, 1978 1 month ended December 31, 1978 $2,000,000 $1,200,000 220,000 115,000 During 1978 Drew paid $900,000 in dividends to its stockholders. Art had paid $500,000 in dividends to its stockholders in September 1978. Assuming that the net income of Drew given above does not include the equity in net income of Art, the consolidated net income for the year ended December 31, 1978, should be a. $ 335,000 b. $2,115,000 c. $2,700,000 d. $3,200,000 4. On July 31, 1979, Light Company purchased for cash of $8,000,000, all of the outstanding common stock of Shirk Company when Shirk’s balance sheet showed net assets of $6,400,000. Shirk’s assets and liabilities had fair values different from the book values as follows: Property, plant, and equipment, net Other assets Long-term debt Book Value Fair Value $10,000,000 $11,500,000 1,000,000 700,000 6,000,000 5,600,000 As a result of the transaction above, what amount, if any, will be shown as goodwill in the July 31, 1979, con solidated balance sheet of Light Company and its wholly-owned subsidiary, Shirk Company?
a. $0 b. $ 100,000 c. $1,200,000 d. $1,600,000 156 Accounting Practice—Part I 5. Faucet Company has 2,500,000 shares of common stock outstanding on December 31, 1977. An additional 500,000 shares of common stock were issued on April 1, 1978, and 250,000 more on July 1 , 1978. On October I, 1978, Faucet issued 5,000, $1,000 face value, 7% con vertible bonds. Each bond is convertible into 40 shares of common stock. The bonds were not considered com mon stock equivalents at the time of their issuance, and no bonds were converted into common stock in 1978. What is the number of shares to be used in computing primary earnings per share and fully diluted earnings per share, respectively, for the year ended December 31, 1978?
a. 2,875,000 and 2,975,000.
b. 2,875,000 and 3,075,000.
c. 3,000,000 and 3,050,000.
d. 3,000,000 and 3,200,000.
6. At December 31, 1977, the Merlin Company had 50,000 shares of common stock issued and outstanding. On April 1, 1978, an additional 10,000 shares of com mon stock were issued. Merlin’s net income for the year ended December 31, 1978, was $172,500. During 1978 Merlin declared and paid $100,000 cash dividends on its nonconvertible preferred stock. The earnings per com mon share, rounded to the nearest penny, for the year ended December 31, 1978, should be a. $1.26 b. $1.32 c. $3.00 d. $3.14 8. Information from Life Company’s balance sheet is as follows: Current assets: Cash $ 4,000,000 Marketable securities 12,500,000 Accounts receivable 96,000,000 Inventories 110,500,000 Prepaid expenses 2,000,000 Total current assets $225,000,000 Current liabilities: Notes payable $ 3,000,000 Accounts payable 39,000,000 Accrued expenses 25,000,000 Income taxes payable 1,000,000 Payments due within one year on long-term debt 7,000,000 Total current liabilities $75,000,000 What is the quick (acid-test) ratio?
a.
1.33 to 1.
b.
1.50 to 1.
c.
1.65 to 1.
d.
3.00 to 1.
7. Selected information from the accounting records of the Code Company is as follows: Cost of goods sold for 1978 $1,200,000 Inventories at December 31, 1977 350,000 Inventories at December 31, 1978 310,000 Assuming a business year consisting of 300 days, what was the number of days’ sales in average inventories for 1978?
a. 36.5 b. 77.5 c. 82.5 d. 87.5 9. Fore Company had a $30,000 exchange loss resulting from the translation of the accounts of its wholly owned foreign subsidiary for the year ended December 31, 1978. Fore also had a receivable from a foreign customer which was payable in the local cur rency of the foreign customer. On December 31, 1977, this receivable for 500,000 local currency units (LCU) was appropriately included in the accounts receivable section of Fore’s balance sheet at $245,000. When the receivable was collected on February 5, 1978, the ex change rate was 2 LCU to $1. What amount should be included as an exchange gain or loss in the 1978 con solidated income statement of Fore Company and its wholly owned foreign subsidiary as a result of the above?
a. $ 5,000 exchange gain.
b. $20,000 exchange loss.
c. $25,000 exchange loss.
d. $30,000 exchange loss.
157 Examination Questions— November 1979 10. The Bay Company sold some machinery to the Inlet Company on January 1, 1976. The cash selling price would have been $379,100. Inlet entered into an installment sales contract which required annual payments of $100,000, including interest at 10%, over five years. The first payment was due on December 31, 1976. What amount of interest income should be includ ed in Bay’s 1978 income statement (the third year of the contract)?
a. $24,180 b. $24,871 c. $37,910 d. $50,000 13. On January 1, 1978, Grade Company paid $300,000 for 20,000 shares of Medium Company’s com mon stock which represents a 15% investment in Medium. Grade does not have the ability to exercise significant influence over Medium. Medium declared and paid a dividend of $1 a share to its stockholders during 1978. Medium reported net income of $260,000 for the year ended December 31, 1978. The balance in Grade’s balance sheet account “ Investment in Medium Company’’ at December 31, 1978, should be a. $280,000 b. $300,000 c. $319,000 d. $339,000 11. In August 1978 Ella Company spent $150,000 on an advertising campaign for subscriptions to the magazine it sells on getting ready for the skiing season. There are only two issues; one in October and one in November. The magazine is only sold on a subscription basis and the subscriptions started in October 1978. Assuming Ella’s fiscal year ends on March 31, 1979, what amount of expense should be included in Ella’s quarterly income statement for the three months ended December 31, 1978, as a result of this expenditure?
a. $ 37,500 b. $ 50,000 c. $ 75,000 d. $150,000 12. In 1978 West Company accrued, for financial statement reporting, estimated losses on disposal of unused plant facilities of $800,000. The facilities were sold in March 1979. Also, in 1978 West paid $100,000 of premiums on officers’ life insurance. Assuming that the effective income tax rate was 40%, the amount reported in the provision for deferred income taxes in West’s in come statement for the year ended December 31, 1978, should be a a. $320,000 credit.
b. $320,000 debit.
c. $360,000 credit.
d. $360,000 debit.
Items 14 and 15 are based on the following infor mation: The Morn Company leased equipment to the Lizard Company on May 1, 1978. At that time the col lectibility of the minimum lease payments was not reasonably predictable. The lease expires on May 1, 1980. Lizard could have bought the equipment from Morn for $900,000 instead of leasing it. Morn’s ac counting records showed a book value for the equip ment on May 1, 1978, of $800,000. Morn’s depreciation on the equipment in 1978 was $200,000. During 1978 Lizard paid $240,000 in rentals to Morn. Morn incurred maintenance and other related costs under the terms of the lease of $18,000 in 1978. After the lease with Lizard expires, Morn will lease the equipment to the Cold Company for another two years.
14. The income before income taxes derived by Morn from this lease for the year ended December 31, 1978, should be a. $ 22,000 b. $100,000 c. $122,000 d. $240,000 15. Ignoring income taxes, the amount of expense in curred by Lizard from this lease for the year ended December 31, 1978, should be a.
$ 22,000 b.
$200,000 c.
$218,000 d.
$240,000 158 Accounting Practice—Part I 16. Chip Company operates in four different in dustries, each of which is appropriately regarded as a reportable segment. Total sales for 1978 for all the segments combined were $1,000,000. Sales for Segment No. 2 were $400,000 and traceable costs were $150,000. Total common costs for all the segments combined were $500,000. Chip allocates common costs based on the ratio of a segment’s sales to total sales, an appropriate method of allocation. The operating profit presented for Segment No. 2 for 1978 should be a. $ 50,000 b. $125,000 c. $200,000 d. $250,000 17. Job Company purchased a machine on January 1, 1977, for $500,000. Job is depreciating the machine on a straight-line basis with no salvage and a five-year life. At December 31, 1977, the replacement cost (current value) of the machine was $410,000. On January 1, 1978, the machine was sold for $412,000. Ignoring in come taxes, what amount should be shown as a result of the sale of the machine in 1978 in Job’s current value statements?
a. $ 2,000 gain.
b. $ 8,000 gain.
c. $10,000 gain.
d. $12,000 gain.
18. The following information on selected cash transactions for 1978 has been provided by the Smith Company: Proceeds from short-term borrowings $1,200,000 Proceeds from long-term borrowings 4,000,000 19. An analysis of the machinery accounts of the Pending Company for 1978 is as follows: Accumu lated deprecia tion Machinery Balance at January 1, 1978 $1,000,000 $400,000 $600,000 Purchases of new machinery in 1978 for cash Depreciation in 500,000 — 500,000 1978 — 250,000 (250,000) Balance at December 31, 1978 $1,500,000 $650,000 $850,000 Assuming funds are defined as working capital, the in formation concerning Pending’s machinery accounts should be shown in Pending’s statement of changes in financial position for the year ended December 31, 1978, as a.
b.
c.
d.
A subtraction from net income of $250,000 and a source of funds of $500,000.
An addition to net income of $250,000 and a use of funds of $500,000.
A source of funds of $250,000.
A use of funds of $500,000.
20. On August 1, 1979, Winston Company reacquired 4,000 shares of its $15 par value common stock for $18 per share. Winston uses the cost method to account for treasury stock. What journal entry should Winston make to record the acquisition of treasury stock?
Purchases of fixed assets 3,200,000 Purchases of inventories 8,000,000 Debit Credit Proceeds from sale of Smith’s a.
Treasury stock $60,000 common stock 2,000,000 Additional paid-in capital 12,000 Cash $72,000 What is the increase in working capital for the year ended December 31, 1978, as a result of the above infor mation?
b.
Treasury stock 60,000 Retained earnings 12,000 Cash 72,000 a. $ 800,000 c.
Retained earnings 72,000 b. $2,000,000 Cash 72,000 c. $2,800,000 d.
Treasury stock 72,000 d. $4,000,000 Cash 72,000 159 Machinery, net of accumul ated depre ciation Examination Questions— November 1979 Instructions Select the best answer for each of the following items relating to a variety of managerial-accounting and quantitative-methods problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
Items to be Answered 21. Terry Company is preparing its cash budget for the month of April. The following information is available concerning its inventories: Number 2 (Estimated time-----50 to 60 minutes) Inventories at beginning of April Estimated purchases for April Estimated cost of goods sold for April Estimated payments in April for purchases in March Estimated payments in April for purchases prior to March Estimated payments in April for purchases in April $ 90,000 440,000 450,000 75,000 20,000 75% 23. The Polar Company is planning to purchase a new machine for $30,000. The pay-back period is expected to be five years. The new machine is expected to pro duce cash flow from operations, net of income taxes, of $7,000 a year in each of the next three years and $5,500 in the fourth year. Depreciation of $5,000 a year will be charged to income for each of the five years of the pay back period. What is the amount of cash flow from operations, net of taxes, that the new machine is ex pected to produce in the last (fifth) year of the pay-back period?
a. $1,000 b. $3,500 c. $5,000 d. $8,500 24. The Fudge Company is planning to purchase a new machine which it will depreciate on a straight-line basis over a ten-year period with no salvage value and a full year’s depreciation taken in the year of acquisition. The new machine is expected to produce cash flow from operations, net of income taxes, of $66,000 a year in each of the next ten years. The accounting (book value) rate of return on the initial investment is expected to be 12%. How much will the new machine cost?
a. $300,000 b. $550,000 c. $660,000 d. $792,000 What are the estimated cash disbursements for inven tories in April?
a. $401,250 b. $405,000 c. $425,000 d. $432,500 22. The Fresh Company is preparing its cash budget for the month of May. The following information is available concerning its accounts receivable: Estimated credit sales for May Actual credit sales for April Estimated collections in May for credit sales in May Estimated collections in May for credit sales in April Estimated collections in May for credit sales prior to April Estimated write-offs in May for uncollectible credit sales Estimated provision for bad debts in May for credit sales in May $200,000 $150,000 20% 70% $ 12,000 $ 8,000 $ 7,000 What are the estimated cash receipts from accounts receivable collections in May?
a. $142,000 b. $149,000 c. $150,000 d. $157,000 25. Sideways Company manufactures products A, B, and C from a joint process. Additional information is as follows: Product Units pro duced Joint costs Sales value at split-off Additional costs if processed further Sales value if pro cessed fur ther A B C Total 8,000 4,000 2,000 14,000 $ 72,000 ?
7 $120,000 7 7 $30,000 $200,000 $ 14,000 $10,000 $ 6,000 $ 30,000 $140,000 $60,000 $40,000 $240,000 Assuming that joint costs are allocated using the relative-sales-value at split-off approach, what was the sales value at split-off for product A?
a. $116,667 b. $119,000 c. $120,000 d. $126,000 160 Accounting Practice—Part I 26. The Rote Company manufactures products C and R from a joint process. The total joint costs are $60,000. The sales value at split-off was $75,000 for 8,000 units of Product C and $25,000 for 2,000 units of Product R. Assuming that total joint costs are allocated using the relative-sales-value at split-off approach, what were the joint costs allocated to Product C?
a. $15,000 b. $30,000 c. $45,000 d. $48,000 27. The Ship Company is planning to produce two products, Alt and Tude. Ship is planning to sell 100,000 units of Alt at $4 a unit and 200,000 units of Tude at $3 a unit. Variable costs are 70% of sales for Alt and 80% of sales for Tude. In order to realize a total profit of $160,000, what must the total fixed costs be?
a. $ 80,000 b. $ 90,000 c. $240,000 d. $600,000 28. The Meredith Company is planning to sell product Z for $5 a unit. Variable costs are $3 a unit and fixed costs are $100,000. What must total sales be to break even?
a. $160,000 b. $166,667 c. $250,000 d. $266,667 29. Cause Company is planning to invest in a machine with a useful life of five years and no salvage value. The machine is expected to produce cash flow from opera tions, net of income taxes, of $20,000 in each of the five years. Cause’s expected rate of return is 10%. Informa tion on present value and future amount factors is as follows: Period Present value of $1 at 10% .909 .826 .751 .683 .621 Present value of annuity of $1 at 10% .909 1.736 2.487 3.170 3.791 Future amount of $1 at 10% 1.100 1.210 1.331 1.464 1.611 Future amount of annuity of $1 at 10% 1.000 2.100 3.310 4.641 6.105 How much will the machine cost?
a. $ 32,220 b. $ 62,100 c. $ 75,820 d. $122,100 30. Heap Company invested in a two-year project. Heap’s expected rate of return is 10%. The present value of $1 for one period at 10% is .909 and for two periods at 10% is .826. The machine is expected to pro duce cash flow from operations, net of income taxes, of $40,000 in the first year and $50,000 in the second year. How much will the project cost?
a. $74,340 b. $77,660 c. $81,810 d. $90,000 31. Politan Company manufactures bookcases. Set up costs are $2.00. Politan manufactures 4,000 bookcases evenly throughout the year. Using the economic-order- quantity approach, the optimal production run would be 200 when the cost of carrying one bookcase in inven tory for one year is a. $0.05 b. $0.10 c. $0.20 d. $0.40 32. Milford Company manufactures two models, medium and large. The contribution margin expected is $12 for the medium model and $20 for the large model. The medium model is processed two hours in the machining department and four hours in the polishing department. The large model is processed three hours in the machining department and six hours in the polishing department. How would the formula for determining the maximization of total contribution margin be ex pressed?
a. 5X + 10Y b. 6X + 9Y c. 12X + 20Y d. 12X(2 + 4) + 20Y(3+ 6) 33. Maurice Company adds materials at the beginning of the process in the Forming Department, which is the first of two stages of its production cycle. Information concerning the materials used in the Forming Depart ment in April 1979 is as follows: Units Materials Costs 12,000 $ 6,000 100,000 $51,120 88,000 Work in process at April 1, 1979 Units started during April Units completed and trans ferred to next depart ment during April Using the weighted-average method, what was the materials cost of the work in process at April 30, 1979?
a. $ 6,120 b. $11,040 c. $12,000 d. $12,240 161 Examination Questions— November 1979 34. The Ace Company had computed the physical flow (of physical units) for Department A, for the month of April 1979 as follows: Units completed: From work in process on April 1, 1979 From April production 10,000 30,000 40,000 Materials are added at the beginning of the process. Units of work in process at April 30, 1979, were 8,000. The work in process at April 1 , 1979, was 80% complete as to conversion costs and the work in process at April 30, 1979, was 60% complete as to conversion costs. What are the equivalent units of production for the month of April 1979 using the FIFO method?
The Tray Company has offered to sell 20,000 units of Part No. 498 to Reno for $60 per unit. Reno will make the decision to buy the part from Tray if there is a savings of $25,000 for Reno. If Reno accepts Tray’s offer, $9 per unit of the fixed overhead applied would be totally eliminated. Furthermore, Reno has deter mined that the released facilities could be used to save relevant costs in the manufacture of Part No. 575. In order to have a savings of $25,000, the amount of rele vant costs that would be saved by using the released facilities in the manufacture of Part No. 575 would have to be a. $ 80,000 b. $ 85,000 c. $125,000 d. $140,000 37. Information on Westcott Company’s direct-labor costs is as follows: Materials Conversion Costs Standard direct-labor rate $3.75 a.
38,000 36,800 Actual direct-labor rate $3.50 b.
38,000 38,000 Standard direct-labor hours 10,000 c.
48,000 44,800 Direct-labor usage (efficiency) d.
48,000 48,000 variance - unfavorable $4,200 35. Woody Company, which manufactures sneakers, has enough idle capacity available to accept a special order of 20,000 pairs of sneakers at $6.00 a pair. The normal selling price is $10.00 a pair. Variable manufac turing costs are $4.50 a pair, and fixed manufacturing costs are $1.50 a pair. Woody will not incur any selling expenses as a result of the special order. What would the effect on operating income be if the special order could be accepted without affecting normal sales?
a. $0.
b. $ 30,000 increase.
c. $ 90,000 increase.
d. $120,000 increase.
36. The Reno Company manufactures Part No. 498 for use in its production cycle. The cost per unit for 20,000 units of Part No. 498 are as follows: Direct materials Direct labor Variable overhead Fixed overhead applied $ 6 30 12 16 $64 What were the actual hours worked, rounded to the nearest hour?
a.
10,714 b.
11,120 c.
11,200 d.
11,914 38. Information on Kennedy Company’s direct-material costs is as follows: Standard unit price $3.60 Actual quantity purchased 1,600 Standard quantity allowed for actual production 1,450 Materials purchase price variance-favorable $240 What was the actual purchase price per unit, rounded to the nearest penny?
a. $3.06 b. $3.11 c. $3.45 d. $3.75 162 Accounting Practice—Part I 39. Information on Fire Company’s overhead costs is as follows: Actual variable overhead $73,000 Actual fixed overhead $17,000 Standard hours allowed for actual production 32,000 Standard variable overhead rate per direct-labor hour $2.50 Standard fixed overhead rate per direct-labor hour $0.50 What is the total overhead variance?
a. $1,000 unfavorable.
b. $6,000 favorable.
c. $6,000 unfavorable.
d. $7,000 favorable.
40. The Ernie Company has provided information concerning its 1979 projections as follows: Net sales $10,000,000 Fixed manufacturing costs 1,000,000 Ernie projects variable manufacturing costs of 60% of net sales. Assuming no change in inventory, what will the projected cost of goods sold be?
a. $5,000,000 b. $6,000,000 c. $7,000,000 d. $8,000,000 Number 3 (Estimated time-----40 to 50 minutes) The Point Corporation has engaged you to com pute its federal taxable income for the year ended December 31, 1978. The income before income taxes furnished from Point’s accounting records was $600,000. Additional information, none of which is in cluded in the $600,000 of income before income taxes, is as follows: • • On January 20, 1978, Point sold for $30,000, common stock of Cullen Company that was purchased for $32,000 on April 1, 1976.
• On February 21, 1978, Point sold for $40,000 common stock of Mary Company that was purchased for $36,000 on December 3, 1976.
• On March 16, 1978, Point sold for $53,000 preferred stock of Mealy Company that was purchased for $52,000 on December 14, 1977.
• On May 29, 1978, Point sold for $60,500, com mon stock of Bin Company that was purchased for $61,000 on April 17, 1978.
• On June 3, 1978, Point purchased for $70,000, common stock of Dorn Company. On December 31, 1978, Point still owned this common stock but the market value of the stock had increased to $73,000.
• During 1978 Point received dividends from domestic corporations of $15,000 and dividends from current earnings of a wholly owned domestic subsidiary, with which Point does not file a consolidated return, of $14,000.
• During 1978 Point received interest income from corporate bonds of $7,000 and from municipal obligations of $6,500.
• Point rents a small portion of the space in its plant. During 1978 Point received $18,000 in rental income from this space. Depreciation and other occupancy costs of this space are included in the income before income taxes of $600,000 above.
• During 1978 Point received $23,000 in royalties from a patent on an invention.
• On December 2 0 , 1978, Point paid a 10% bonus to all of its officers because Point had record sales and earnings in 1978. The bonus amounted to a total of $20,000.
• On December 31, 1978, Point contributed $90,000 to a qualified profit-sharing trust. The con tribution represented 5% of each employee’s salary.
• In 1978 Point had a fire in a shed outside its plant. At the time of the fire, the shed had an adjusted basis of $16,000. The value of the shed before the fire was $18,000; but the shed was only worth $11,000 after the fire. Point’s insurance policy covered $6,000 of the fire loss.
• During 1978 Point paid finance charges of $650 on gasoline credit cards used by Point’s employees for business purposes.
• During 1978 Point made contributions to various recognized charitable organizations of $11,000 and to the local tax-exempt private college of $2,200.
Required: Prepare a schedule, beginning with income before income taxes of $600,000, computing the federal taxable income of Point for the 1978 calendar year. Any possi ble alternative treatments should be resolved in a man ner that will minimize taxable income. Show supporting computations in good form.
163 Examination Questions—November 1979 Number 4 consists of three unrelated parts.
Part a. The F rate Company was formed on December 1, 1978. The following information is available from F rate’s inventory records for Product Ply: Number 4 (Estimated time----- 40 to 50 minutes) Units Unit Cost January 1, 1979 (beginning inventory) 800 $ 9.00 Purchases: January 5, 1979 1,500 $10.00 January 25, 1979 1,200 $10.50 F ebruary 16, 1979 600 $11.00 March 26, 1979 900 $11.50 A physical inventory on March 31, 1979, shows 1,600 units on hand.
Required: Prepare schedules to compute the ending inventory at March 31, 1979, under each of the following inven tory methods: 1. F IF O.
2. L I F O.
3. Weighted average.
Show supporting computations in good form.
Part b. The Red Department Store uses the retail inventory method. Information relating to the computa tion of the inventory at December 31, 1978, is as follows: Part c. On November 21, 1978, a fire at Hodge Company’s warehouse caused severe damage to its en tire inventory of Product Tex. Hodge estimates that all usable damaged goods can be sold for $10,000. The following information was available from Hodge’s ac counting records for Product Tex: Inventory at November 1 , 1978 Purchases from November 1, 1978, to date of fire Net sales from November 1, 1978, to date of fire $100,000 140,000 220,000 Based on recent history, Hodge had a gross margin (profit) on Product Tex of 30% of net sales.
Required: Prepare a schedule to calculate the estimated loss on the inventory in the fire, using the gross margin (profit) method. Show supporting computations in good form.
Inventory at January 1, 1978 Sales Purchases F reight in Markups Markup cancellations Markdowns Markdown cancellations Estimated normal shrink age is 2% of sales.
Cost $ 32,000 270,000 7,m Retail $ 80,000 600,000 590,000 60,000 10,000 25,000 5,000 Required: Prepare a schedule to calculate the estimated ending inventory at the lower of average cost or market at December 31, 1978, using the retail inventory method. Show supporting computations in good form.
164 Accounting Practice—Part I Number 5 consists of three unrelated parts.
Part a. On January 1, 1979, the Hopewell Company sold its bonds that had a face value of $1,000,000. Interest is payable at December 31, each year. The bonds mature on January 1, 1989. The bonds were sold to yield a rate of 10%. The present value of an ordinary annuity of $1 for 10 periods at 10% is 6.1446. The present value of $1 for 10 periods at 10% is 0.3855.
Required: Prepare a schedule to compute the total amount received from the sale of the bonds. Show supporting computations in good form.
Part b. On September 1, 1978, the Junction Company sold at 104, (plus accrued interest) four thou sand of its 9%, ten-year, $1,000 face value, nonconver tible bonds with detachable stock warrants. Each bond carried two detachable warrants; each warrant was for one share of common stock, at a specified option price of $15 per share. Shortly after issuance, the warrants were quoted on the market for $3 each. No market value can be determined for the bonds above. Interest is payable on December 1, and June 1. Bond issue costs of $40,000 were incurred.
Required: Prepare in general journal format the entry to record the issuance of the bonds. Show supporting com putations in good form.
Part c. On December 1, 1976, The Cone Company issued its 7%, $2,000,000 face value bonds for $2,200,000, plus accrued interest. Interest is payable on November 1 and May 1. On December 31, 1978, the book value of the bonds, inclusive of the unamortized premium, was $2,100,000. On July 1, 1979, Cone reac quired the bonds at 98, plus accrued interest. Cone ap propriately uses the straight-line method for the amor tization of bond premium because the results do not materially differ from using the interest method.
Required: Prepare a schedule to compute the gain or loss on this early extinguishment of debt. Show supporting computations in good form.
Number 5 (Estimated time----- 40 to 50 minutes) 165 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXAMINATION IN ACCOUNTING PRACTICE — PART II November 8, 1979; 1:30 to 6:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1........................................................................................... 50 No. 2 ........................................................................................... 50 No. 3 ........................................................................................... 40 No. 4 ........................................................................................... 40 No. 5 ........................................................................................... 40 Total................................................................................... 220 60 60 50 50 50 270 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 3.
1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “Continued” at the bottom of the page. Number pages consecutively. For in stance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
Enclose all scratch sheets. Failure to enclose scratch sheets may result in loss of grading points. Scratch sheets need not have page numbers, but you should show the question number and place them immediately following the question to which they relate.
4. Fourteen-column sheets, if any, should not be folded until all sheets, both wide and narrow, are placed in the proper sequence and fastened together at the top left corner. All fourteen- column sheets should then be wrapped around the back of the papers.
5. A CPA is continually confronted with the necessi ty of expressing opinions and conclusions in writ ten reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and applica tion of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
166 Accounting Practice—Part II Instructions Select the best answer for each of the following items relating to a variety of financial-accounting problems. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 97. Gross billings for merchandise sold by Baker Company to its customers last year amounted to $5,260,000; sales returns and allowances reduced the amounts owed by $160,000. How much were net sales last year for Baker Company?
a. $4,800,000 b. $5,100,000 c. $5,200,000 d. $5,260,000 Answer Sheet 97. Items to be Answered Number 1 (Estimated time----- 50 to 60 minutes) 1. On January 1 , 1977, the Dan Company purchased a new machine for $4,000,000. The new machine has an estimated useful life of eight years and the salvage value was estimated to be $400,000. Depreciation was com puted on the sum-of-the-years-digits method. What amount should be shown in Dan’s balance sheet at December 31, 1978, net of accumulated depreciation, for this machine?
2. The Supple Food Company distributes to con sumers coupons which may be presented (on or before a stated expiration date) to grocers for discounts on cer tain products of Supple. The grocers are reimbursed when they send the coupons to Supple. In Supple’s ex perience, 40% of such coupons are redeemed, and generally one month elapses between the date a grocer receives a coupon from a consumer and the date Supple receives it. During 1978 Supple issued two separate series of coupons as follows: Issued On Total Value 1/1/78 7/1/78 $100,000 120,000 Consumer Expiration Date 6/30/78 12/31/78 Amount Disbursed as o f 12/31/78 $34,000 40,000 The December 31, 1978, balance sheet should include a liability for unredeemed coupons of a. $0 b. $ 8,000 c. $14,000 d. $32,000 3. On November 30, 1978, Eagle, Incorporated, pur chased for cash at $25 per share all 300,000 shares of the outstanding common stock of Perch Company. Perch’s balance sheet at November 30, 1978, showed a book value of $6,000,000. Additionally, the fair value of Perch’s property, plant, and equipment on November 3 0 , 1978, was $800,000 in excess of its book value. What amount, if any, will be shown in the balance sheet cap tion “ Goodwill” in the November 30, 1978, con solidated balance sheet of Eagle, Incorporated, and its wholly-owned subsidiary, Perch Company?
a. $0 b. $ 700,000 c. $ 800,000 d. $1,500,000 4. On January 1 , 1970, Burry Corporation purchased for $76,000, equipment having a useful life of ten years and an estimated salvage value of $4,000. Burry has recorded monthly depreciation of the equipment on the straight-line method. On December 3 1 , 1978, the equip ment was sold for $15,000. As a result of this sale, Burry should recognize a gain of a.
$2,100,000 a.
$0 b.
$2,500,000 b.
$ 3,800 c.
$3,150,000 c.
$ 7,400 d.
$3,300,000 d.
$11,400 167 Examination Questions— November 1979 5. Elton and Don are partners who share profits and losses in the ratio of 7:3, respectively. On November 5, 1978, their respective capital accounts were as follows: Elton $ 70,000 Don 60,000 $130,000 On that date they agreed to admit Kravitz as a partner with a one-third interest in the capital and profits and losses upon his investment of $50,000. The new partner ship will begin with a total capital of $180,000. Im mediately after Kravitz’s admission what are the capital balances of Elton, Don, and Kravitz, respectively?
a. $60,000; $60,000; $60,000.
b. $63,000; $57,000; $60,000.
c. $63,333; $56,667; $60,000.
d. $70,000; $60,000; $50,000.
7. The following condensed balance sheet is presented for the partnership of Fisher, Taylor and Simon who share profits and losses in the ratio of 6:2:2, respectively: Cash Other assets Liabilities Fisher, capital Taylor, capital Simon, capital $ 40,000 140,000 $180,000 $ 70,000 50,000 50,000 10,000 $180,000 The assets and liabilities are fairly valued on the above balance sheet, and it was agreed to by all the partners that the partnership would be liquidated after selling the other assets. What would each of the partners receive at this time if the other assets are sold for $80,000?
6. On June 30, 1978, the balance sheet for the part nership of Williams, Brown and Lowe together with their respective profit and loss ratios was as follows: Fisher Taylor Simon a.
$12,500 $37,500 $0 b.
$13,000 $37,000 $0 c.
$14,000 $38,000 $ 2,000 d.
$50,000 $50,000 $10,000 Assets, at cost $300,000 Williams, loan $ 15,000 Williams, capital (20%) 70,000 Brown, capital (20%) 65,000 Lowe, capital (60%) 150,000 Total $300,000 Williams has decided to retire from the partnership and by mutual agreement the assets are to be adjusted to their fair value of $360,000 at June 30, 1978. It was agreed that the partnership would pay Williams $102,000 cash for his partnership interest exclusive of his loan which is to be repaid in full. No goodwill is to be recorded in this transaction. After Williams’ retire ment what are the capital account balances of Brown and Lowe, respectively?
a. $65,000 and $150,000.
b. $72,000 and $171,000.
c. $73,000 and $174,000.
d. $77,000 and $186,000.
8. Jenny Corporation was organized on January 1, 1978, with an authorization of 500,000 shares of com mon stock with a par value of $5 per share.
During 1978 the corporation had the following capital transactions: January 5 — issued 100,000 shares @ $5 per share April 6 — issued 50,000 shares @ $7 per share June 8 — issued 15,000 shares @ $10 per share July 28 — purchased 25,000 shares @ $4 per share December 31 — sold the 25,000 shares held in treasury @ $8 per share Jenny used the par value method to record the pur chase and reissuance of the treasury shares.
What is the amount of paid-in capital in excess of par value as of December 31, 1978?
a. $175,000 b. $200,000 c. $250,000 d. $275,000 168 Accounting Practice—Part II 9. The Good Trader Company values its inventory by using the retail method (FIFO basis, lower of cost or market). The following information is available for the year 1978.
Beginning inventory Purchases Freight-in Shortages Markups (net) Markdowns (net) Sales Cost $ 80,000 297,000 4,000 Retail $140,000 420,000 8,000 10,000 2,000 400,000 At what amount would The Good Trader Company report its ending inventory?
a. $112,000 b. $113,400 c. $117,600 d. $119,000 10. On December 1, 1978, Dartmouth Corporation ex changed 1,000 shares of its $25 par value common stock held in treasury for a parcel of land to be held for a future plant site. The treasury shares were acquired by Dartmouth at a cost of $40, and on the exchange date the common shares of Dartmouth had a fair market value of $55 per share. Dartmouth received $5,000 for selling scrap when an existing building on the property was removed from the site. Based upon these facts, the land should be capitalized at a. $35,000 b. $40,000 c. $50,000 d. $55,000 11. Delta, Inc., is a retail store operating in a state with a 5% retail sales tax. The state law provides that the retail sales tax collected during the month must be remitted to the state during the following month. If the amount collected is remitted to the state on or before the twentieth of the following month, the retailer may keep 2% of the sales tax collected. On April 10, 1979, Delta remitted $16,905 sales tax to the state tax division for March 1979 retail sales. What was Delta’s March 1979 retail sales subject to sales tax?
a. $331,340 b. $331,480 c. $338,100 d. $345,000 12. The Swenson Company reported the following results for the two years ended December 31, 1978, and 1977, respectively: December 31 Income (per books before income taxes) Taxable income 1978 $1,200,000 1,600,000 1977 $800,000 120,000 The disparity between book income and taxable income is attributable to timing differences. What should Swenson record as income tax expense for the year ended December 31, 1978, assuming an income tax rate of 40% ?
a. $640,000 b. $480,000 c. $368,000 d. $208,000 13. On the December 31, 1978, balance sheet of the Stat Company, the current assets were comprised of the following items: Cash $ 70,000 Accounts receivable 120,000 Inventories 60,000 $250,000 An examination of the accounts revealed that the ac counts receivable were composed of the following items: Accounts receivable Trade accounts $ 93,000 Allowance for uncollectible accounts (2,000) Claim against shipper for goods lost in transit (November 1978) 3,000 Selling price of unsold goods sent by Stat on consignment at 130% of cost (and not included in Stat’s ending inventory) 26,000 $120,000 What is the correct amount of current assets as of December 31, 1978?
a.
$221,000 b.
$224,000 c.
$244,000 d.
$250,000 169 Examination Questions— November 1979 14. The December 31, 1978, general ledger of The North Company contained an account “6% Bonds Payable.” This account had a balance of $95,000 as of that date. Further examination revealed that the bonds had a face value of $100,000, with a yield of 8% and were issued at a discount. The amortization of the bond discount was recorded under the effective interest method. Interest was paid on January 1 and July 1 of each year. On July 1, 1979, several years before their maturity, North retired the bonds at 102, excluding ac crued interest. What is the extraordinary loss that North should record on the early retirement of the bonds on July 1, 1979?
a. $4,200 b. $6,200 c. $7,000 d. $7,800 15. The Park Corporation purchased factory equip ment that was installed and put into service January 2, 1977, at a total cost of $32,000. Salvage value was estimated at $2,000. The equipment is being depreciated over eight years using the double-declining-balance method. For the year 1978, Park should record depreciation expense on this equipment of a. $5,625 b. $6,000 c. $7,500 d. $8,000 16. In January 1978, the Under Mine Corporation purchased a mineral mine for $3,400,000 with removable ore estimated by geological surveys at 4,000,000 tons. The property has an estimated value of $200,000 after the ore has been extracted. The company incurred $800,000 of development costs preparing the mine for production. During 1978 400,000 tons were removed and 375,000 tons were sold. What is the amount of depletion that Under Mine should record for 1978?
a. $375,000 b. $393,750 c. $400,000 d. $420,000 17. During 1978 the Henderson Company purchased the net assets of John Corporation for $800,000. On the date of the transaction, John had no long-term in vestments in marketable securities and had $100,000 of liabilities. The fair value of John’s assets when acquired were as follows: Current assets $ 400,000 Noncurrent assets 600,000 $1,000,000 How should the $100,000 difference between the fair value of the net assets acquired ($900,000) and the cost ($800,000) be accounted for by Henderson?
a. The $100,000 difference should be credited to retained earnings.
The noncurrent assets should be recorded at $500,000.
The current assets should be recorded at $360,000, and the noncurrent assets should be recorded at $540,000.
A deferred credit of $100,000 should be set up and then amortized to income over a period not to exceed forty years.
b.
c.
d.
18. Alan Company purchased the net assets of Barry Company in a business combination accounted for as a purchase. As a result, goodwill was recorded. For tax purposes this combination was considered to be a tax- free merger.
One of Barry’s assets that Alan purchased was a building with an appraised value of $150,000 at the date of the business combination. This asset had a cost of $90,000 which was net of accumulated depreciation and which was based on accelerated depreciation for ac counting purposes. The building had an adjusted tax basis to Barry (and to Alan as a result of the merger) of $100,000. Assuming a 48% income tax rate, at what amount should Alan record this building on its books after the purchase?
a. $100,000 b. $121,200 c. $126,000 d. $150,000 170 Accounting Practice—Part II Items 19 and 20 are based on the following infor mation: 19. On January 1, 1978, Avow, Inc., purchased 30% of the outstanding common stock of Depot Corporation for $129,000 cash. Avow is accounting for this invest ment on the equity method. On the date of acquisition, the fair value of Depot’s net assets was $310,000. Avow has determined that the excess of the cost of the invest ment over its share of Depot’s net assets has an indeter minate life. Depot’s net income for the year ended December 31, 1978, was $90,000. During 1978 Depot declared and paid cash dividends of $10,000. There were no other transactions between the two companies. On January 1, 1978, the investment in Depot should have been recorded as a. $ 93,000 b. $120,000 c. $129,000 d. $165,000 20. Ignoring income taxes, Avow’s statement of in come for the year ended December 31, 1978, should in clude “equity in net income of Depot Corporation’’ in the amount of a. $17,000 b. $26,100 c. $27,000 d. $27,900 Number 2 (Estimated time-----50 to 60 minutes) Instructions Select the best answer for each of the following items relating to the federal income taxation of in dividuals. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. The answers should be selected in accordance with the current Inter nal Revenue Code and Tax Regulations. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
Items to be Answered 21. During 1978 James Lee paid the following interest charges: Interest on home mortgage $2,400 Installment charge accounts 150 Interest on personal loan (proceeds used to purchase tax-exempt bonds) 400 In addition, $9,000 was borrowed from a bank on October 1, 1978, for financing a new business venture. Interest of $900 was deducted by the bank in advance and the loan is being repaid in twelve equal monthly in stallments. Beginning on November 1, 1978, Lee made timely payments on this loan. In itemizing his deduc tions for 1978, Lee can claim an interest expense deduc tion of a. $2,550 b. $2,700 c. $2,900 d. $2,950 22. Mr. and Mrs. Vonce, both age 62, filed a joint return for 1978. They provided all the support for their daughter who is 19, legally blind, and who has no in come. Their son, age 21 and a full-time student at a university, had $4,200 of income and provided 70% of his own support during 1978. How many exemptions should Mr. and Mrs. Vonce have claimed on their 1978 joint income tax return?
a. 2 b. 3 c. 4 d. 5 23. Ben Reed purchased a personal residence for $31,000 in 1971. It had a fair market value of $56,000 on June 2,1978, when it was damaged by a fire. The fair market value of the house after the fire was $45,000 and insurance proceeds totalled $7,000. The net amount of the casualty loss that he can claim is a. $ 3,900 b. $ 4,000 c. $ 6,900 d. $18,000 24. Bud Ace, a self-employed carpenter, reports his in come on the cash basis. During 1978 he completed a job for a customer and sent him a bill for $3,000. The customer was not satisfied with the work and indicated that he would only pay $1,500. Ace agreed to reduce the bill to $2,000 but before payment was made the customer died. Ace could not collect from the customer’s estate and should treat this loss as a. An ordinary business deduction of $3,000.
b. An ordinary business deduction of $2,000.
c. A short-term capital loss of $1,500.
d. A nondeductible loss as no income was reported.
171 Examination Questions— November 1979 25. Victor and Claire Anet, residents of a separate property state, were divorced in February 1978. Specific requirements of the divorce decree and Mr. Anet’s per formance of those requirements follow: • Transfer title in their personal residence to Claire as part of a lump-sum property settle ment. On the day of the transfer, Victor’s basis in the house was $38,000, the fair market value was $42,000, and the property was sub ject to a mortgage of $20,000.
• Make the mortgage payments on the twenty- year mortgage. He paid $2,500 from March 1, 1978, through December 31, 1978.
• Repay to Claire a $3,000 loan which he did on April 1, 1978.
• Pay Claire $700 per month of which $200 is designated as child support. He made ten such payments in 1978.
Assuming that Claire has no other income, her 1978 gross income should be a. $ 7,500 b. $ 9,500 c. $12,500 d. $16,000 26. Mrs. Grant, a widow, elected to receive the pro ceeds of a $50,000 face value insurance policy on the life of her deceased husband in ten annual installments of $6,800 each beginning in 1978. Of the $6,800 received in 1978, the amount subject to income tax is a. $ 800 b. $1,800 c. $5,000 d. $6,800 29. John Abel, whose wife died in December 1977, filed a joint tax return for 1977. He did not remarry but continued to maintain his home in which his two depen dent children lived. In the preparation of his tax return for 1978, Abel should file as a a. Single individual.
b. Surviving spouse.
c. Head of household.
d. Married individual filing separately.
30. Sergio Morris, age 35, single with no dependents, is a self-employed individual. For the year 1978, his business sustained a net loss from operations of $18,000. There was no net operating loss in any prior year. The following additional information was ob tained from his personal records for 1978: Interest income Dividend income (after exclusion) Itemized deductions, including a net casualty loss of $700 Personal exemption $2,000 500 4,000 750 Based upon the above information, what is his net operating loss for 1978?
a. $18,000 b. $18,700 c. $20,500 d. $22,000 27. Joseph Kurtz exchanged land that he held for four years as an investment, with a tax basis of $36,000, for similar land valued at $40,000 which was owned by Adrian Flemming. In connection with this transaction, Kurtz assumed Flemming’s $10,000 mortgage and Flem ming assumed Kurtz’s $12,000 mortgage. As a result of this transaction Kurtz should report a long-term capital gain of a. $0 b. $2,000 c. $4,000 d. $6,000 28. Gilbert Quinn loaned a friend $2,000 in 1976 and it had not been repaid in 1978 when the friend died insol vent. For 1978 Quinn should account for the nonpay ment of the loan as a (an) a. Ordinary loss.
b. Long-term capital loss.
c. Short-term capital loss.
d. Deduction from adjusted gross income.
31. In July 1963 Dan Farley leased a building to Robert Shelter for a period of fifteen years at a monthly rental of $1,000 with no option to renew. At that time the building had a remaining estimated useful life of twenty years.
Prior to taking possession of the building. Shelter made improvements at a cost of $18,000. These im provements had an estimated useful life of twenty years at the commencement of the lease period. The lease ex pired on June 30, 1978, at which point the im provements had a fair market value of $2,000. The amount that Farley, the landlord, should include in his gross income for 1978 is a. $ 6,000 b. $ 8,000 c. $10,500 d. $18,500 172 Accounting Practice—Part II 32. For the year 1978 Roberta Collins, who is di vorced, reported the following items of income: Interest income Wages Earnings from self-employment $ 100 4,000 3,000 She maintains a household for herself and her 5-year- old son who qualifies as her dependent. What is the maximum earned income credit available to her for 1978?
a. $90 b. $310 c. $400 d. $710 35. During 1978 Seth Parker, a self-employed in dividual, paid the following taxes: Federal income tax $5,000 State income tax 2,000 Real estate taxes on land in South America (held as an investment) 900 State sales taxes 500 Federal self-employment tax 800 State unincorporated business tax 200 What amount can Parker claim for 1978 as an itemized deduction for taxes paid?
a. $7,500 b. $4,400 c. $3,600 d. $3,400 33. For the year 1978 George and Mary Kay, residents of a separate property state, reported the following dividends received on their respective investments: George Able Corporation (a domestic publicly listed corporation) $ 70 Garvey Corporation (a Subchapter S corporation paid out of current earnings and profits) 200 Mary Regan Corp. (a foreign corporation) 100 If the Kays file a joint tax return for 1978 what amount should they report as dividend income after the allowable exclusion?
a. $170 b. $200 c. $300 d. $370 36. Hyman and Carmela Foley elected to file a joint tax return for 1978. Their adjusted gross income was $20,000. During the year they incurred the following medical and related expenses: Medical care insurance premiums ($300 each) Medicines prescribed by doctors Doctors Transportation to and from doctors’ offices $ 600 400 1,200 100 In addition they received $500 as reimbursement from their insurance company for doctor bills paid during the year.
In itemizing their deductions for 1978, what amount can they report as a medical expense deduction?
a. $ 900 b. $1,000 c. $1,200 d. $1,500 34. Under the provisions of ERISA, deductible con tributions to a qualified retirement plan on behalf of a self-employed individual whose earned income is $20,000, are limited to 37. During 1978 Vincent Tally gave to the municipal art museum title to his private collection of rare books that was assessed and valued at $60,000. However, he reserved the right to the collection’s use and possession during his lifetime. For 1978 he reported an adjusted gross income of $100,000. Assuming that this was his only contribution during the year, and that there were no carryovers from prior years, what amount can he deduct as contributions for 1978?
a.
$1,500 a.
$0 b.
$2,000 b.
$30,000 c.
$3,000 c.
$50,000 d.
$7,500 d.
$60,000 173 Examination Questions— November 1979 38. Sam Peterson is a plumber employed by a major contracting firm. During 1978 he paid the following miscellaneous personal expenses: Specialized work clothes (required by employer) Union dues Preparation of will Cost of income tax preparation Safe deposit box rental (used only for personal effects) $410 600 150 100 20 If Peterson were to itemize his personal deductions, what amount could he claim as miscellaneous deductible expenses?
a. $ 680 b. $ 770 c. $1,110 d. $1,130 39. On January 1, 1978, Sandy Beech, single and a sole proprietor, purchased a new machine for $40,000 to be used in her business. The machine was estimated to have a useful life of 10 years with a salvage value of $3,000. Beech, who uses the straight-line method of depreciation, elected to record additional first-year depreciation in 1978. What is the total maximum depreciation expense that she can deduct for this machine in 1978?
a. $ 5,800 b. $ 7,400 c. $ 8,000 d. $10,360 40. On September 18, 1978, Dennis Hanes was killed in an automobile accident. In October 1978 his widow received a lump sum death benefit from his employer in the amount of $15,000. For 1978 what amount should his widow include in adjusted gross income assuming a joint tax return were filed?
a. $0 b. $10,000 c. $14,000 d. $15,000 Number 3 (Estimated time 40 to 50 minutes) Selected accounts included in the property, plant and equipment section of the Kingston Corporation’s balance sheet at December 31, 1977, had the following balances: Land Land improvements Buildings Machinery and equipment $175,000 90,000 900,000 850,000 During 1978 the following transactions occurred: • A tract of land was acquired for $125,000 as a potential future building site.
• A plant facility consisting of land and building was acquired from the Nostrand Company in exchange for 10,000 shares of Kingston’s common stock. On the acquisition date, Kingston’s stock had a closing market price of $45 per share on a national stock exchange. The plant facility was carried on Nostrand’s books at $89,000 for land and $130,000 for the building at the exchange date. Current appraised values for the land and building, respectively, are $120,000 and $240,000.
• Items of machinery and equipment were pur chased at a total cost of $300,000. Additional costs were incurred as follows: Freight and unloading Sales and use taxes Installation $ 5,000 12,000 25,000 • Expenditures totaling $75,000 were made for new parking lots, streets and sidewalks at the corpora tion’s various plant locations. These expenditures had an estimated useful life of fifteen years.
• A machine costing $50,000 on January 1, 1970, was scrapped on June 30, 1978. Double-declining- balance depreciation has been recorded on the basis of a ten-year life.
• A machine was sold for $20,000 on July 1, 1978. Original cost of the machine was $36,000 on January 1, 1975, and it was depreciated on the straight- line basis over an estimated useful life of seven years and a salvage value of $1,000.
Required: 1. Prepare a detailed analysis of the changes in each of the following balance sheet accounts for 1978: Land Land improvements Buildings Machinery and equipment Disregard the related accumulated depreciation ac counts.
2. List the items in the fact situation which were not used to determine the answer to 1. above, showing the pertinent amounts and supporting computations in good form for each item. In addition, indicate where, or if, these items should be included in Kingston’s financial statements.
174 Accounting Practice—Part II Number 4 (Estimated time----- 40 to 50 minutes) Part a. The Wing Manufacturing Corporation produces a chemical compound, product X which deteriorates and must be discarded if it is not sold by the end of the month during which it is produced. The total variable cost of the manufactured compound, product X, is $50 per unit and its selling price is $80 per unit. Wing can purchase the same compound from a com peting company at $80 per unit plus $10 freight per unit. Management has estimated that failure to fill orders would result in the loss of 80 percent of customers plac ing orders for the compound. Wing has manufactured and sold product X for the past 20 months. Demand for product X has been irregular and at present there is no consistent sales trend. During this period monthly sales have been as follows: Units Sold per Month 8,000 9,000 10,000 Number o f Months 5 12 3 Required: 1. Compute the probability of sales of product X of 8,000, 9,000, or 10,000 units in any month.
2. Compute what the contribution margin would be if 9,000 units of product X were ordered and either 8,000, 9,000, or 10,000 units were manufactured in that same month, (with additional units, if necessary, being purchased).
3. Compute the average monthly contribution margin that Wing can expect if 9,000 units of product X are manufactured every month and all sales orders are filled.
Part b. In the production of product X, Wing uses a primary ingredient, K-1. This ingredient is pur chased from an outside supplier at a cost of $24 per unit of compound. It is estimated that there is a 70 percent chance that the supplier of K-1 may be shut down by a strike for an indefinite period. A substitute ingredient, K-2 is available at $36 per unit of compound but Wing must contact this alternative source immediately to secure sufficient quantities. A firm purchase contract for either material must now be made for production of the primary ingredient next month. If an order were placed for K-1 and a strike occurred, Wing would be released from the contract and management would pur chase the chemical compound from its competitor. Assume that 9,000 units are to be manufactured and all sales orders are to be filled.
Required: 1. Compute the monthly contribution margin from sales of 8,000, 9,000, and 10,000 units if the substitute ingredient, K-2 is ordered.
2. Prepare a schedule computing the average monthly contribution margin that Wing should expect if the primary ingredient, K-1 is ordered with the existing probability of a strike at the supplier. Assume that the expected average monthly contribution margin from manufacturing will be $130,000 using the primary ingre dient, and the expected average monthly loss from pur chasing product X from the competitor (in case of a strike) will be $45,000.
175 Examination Questions— November 1979 Number 5 (Estimated time----- 40 to 50 minutes) You have been engaged by the Town of Rego to ex amine its June 30, 1978, balance sheet. You are the first CPA to be engaged by the Town and find that accep table methods of municipal accounting have not been employed. The Town clerk stated that the books had not been closed and presented the following pre-closing trial balance of the General Fund as at June 30, 1978: Credit Cash Taxes receivable — current year Estimated losses — current year taxes receivable Taxes receivable — prior year Estimated losses— prior year taxes receivable Estimated revenues Appropriations Donated land Expenditures—building addition constructed Expenditures—Serial bonds paid Other expenditures Special assessment bonds payable Revenues Accounts payable Fund balance Debit $150,000 59,200 8,000 310,000 27,000 50,000 16,000 280,000 100,000 354,000 26,000 ____ 44,000 $900,200 $900,200 $ 18,000 10,200 348,000 Additional information: • • The estimated losses of $18,000 for current year taxes receivable were determined to be a reasonable estimate.
• Included in the Revenues account is a credit of $27,000 representing the value of land donated by the state as a grant-in-aid for construction of a municipal park.
• The Building Addition Constructed account balance is the cost of an addition to the Town Hall building. This addition was constructed and completed in June 1978. The General Fund recorded the payment as authorized.
• The Serial Bonds Paid account reflects the an nual retirement of general obligation bonds issued to finance the construction of the Town Hall. Interest payments of $7,000 for this bond issue are included in expenditures.
• Operating supplies ordered in the prior fiscal year and chargeable to that year were received, recorded and consumed in July 1977. The outstanding purchase orders for these supplies, which were not recorded in the accounts at June 30, 1977, amounted to $8,800. The vendors’ invoices for these supplies totaled $9,400. Ap propriations lapse one year after the end of the fiscal year for which they are made.
• Outstanding purchase orders at June 30, 1978, for operating supplies totaled $2,100. These purchase orders were not recorded on the books.
• The special assessment bonds were sold in June 1978 to finance a street paving project. No contracts have been signed for this project and no expenditures have been made.
• The balance in the Revenues account includes credits for $20,000 for a note issued to a bank to obtain cash in anticipation of tax collections and for $1,000 for the sale of scrap iron from the Town’s water plant. The note was still outstanding at June 30, 1978. The opera tions of the water plant are accounted for in the Water Fund.
Required: a. Prepare the formal adjusting and closing jour nal entries for the General Fund for the fiscal year ended June 30, 1978.
b. The foregoing information disclosed by your examination was recorded only in the General Fund even though other funds or groups of accounts were in volved. Prepare the formal adjusting journal entries for any other funds or groups of accounts involved.
176 Uniform Certified Public Accountant Examination (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) EXAMINATION IN AUDITING November 8, 1979; 8:30 A.M. to 12:00 M NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. 1 ........................................................................................... 90 No. 2 ........................................................................................... 15 No. 3 ........................................................................................... 15 No. 4 ........................................................................................... 15 No. 5 ............................................................................................. 15_ Total.................................................................................... 150 n o 25 25 25 25 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For in stance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessi ty of expressing opinions and conclusions in writ ten reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and applica tion of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
177 Examination Questions— November 1979 Number 1 (Estimated time-----90 to 110 minutes) Instructions Select the best answer for each of the following items. Use a soft pencil, preferably No 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 96. One of the generally accepted auditing standards specifies that the auditor a. Inspect all fixed assets acquired during the year.
b. Charge fair fees based on cost.
c. Make a proper study and evaluation of the ex isting internal control.
d. Count client petty-cash funds.
Answer Sheet 96.
Items to be Answered 3. In connection with a review of the prepaid in surance account, which of the following procedures would generally not be performed by the auditor?
a. Recompute the portion of the premium that expired during the year.
b. Prepare excerpts of insurance policies for audit working papers.
c. Confirm premium rates with an independent insurance broker.
d. Examine support for premium payments.
4. The auditor tests the quantity of materials charged to work-in-process by tracing these quantities to a. Cost ledgers.
b. Perpetual inventory records.
c. Receiving reports.
d. Material requisitions.
5. For internal control purposes, which of the follow ing individuals should preferably be responsible for the distribution of payroll checks?
a. Bookkeeper.
b. Payroll clerk.
c. Cashier.
d. Receptionist.
1. A change from cash to working capital in a state ment of changes in financial position constitutes a change which requires a. Only a disclosure in the auditor’s report.
b. That the auditor’s opinion contain a “subject to’’ qualification as to consistency.
c. That the auditor’s opinion contain an excep tion as to conformity with generally accepted accounting principles.
d. That the auditor’s opinion contain an excep tion as to consistency in the opinion paragraph.
2. Which of the following is a computer test made to ascertain whether a given characteristic belongs to the group?
a. Parity check.
b. Validity check.
c. Echo check.
d. Limit check.
6. Auditors who prefer statistical to judgmental sampling believe that the principal advantage of statistical sampling flows from its unique ability to a. Define the precision required to provide audit satisfaction.
b. Provide a mathematical measurement of uncertainty.
c. Establish conclusive audit evidence with decreased audit effort.
d. Promote a more legally defensible procedural approach.
7. Which of the following, if material, would be an irregularity as defined in Statements on Auditing Stan dards?
a. Errors in the application of accounting prin ciples.
b. Clerical errors in the accounting data underly ing the financial statements.
c. Misinterpretation of facts that existed when the financial statements were prepared.
d. Misappropriation of an asset or groups of assets.
178 Auditing 8. A CPA’s report on a client’s balance sheet, income statement, and statement of changes in financial posi tion was sent to the stockholders. The client now wishes to present only the balance sheet along with an ap propriately modified auditor’s report in a newspaper advertisement. The auditor may a. Permit the publication as requested.
Permit only the publication of the originally issued auditor’s report and accompanying financial statements.
Not permit publication of a modified auditor’s report.
Not permit publication of any auditor’s report in connection with a newspaper advertise ment.
b.
c.
d.
9. After performing all necessary procedures a predecessor auditor reissues a prior-period report on financial statements at the request of the client without revising the original wording. The predecessor auditor should a. Delete the date of the report.
b. Dual-date the report.
c. Use the reissue date.
d. Use the date of the previous report.
10. For the purpose of proper accounting control, postdated checks remitted by customers should be a. Restrictively endorsed.
b. Returned to customer.
c. Recorded as a cash sale.
d. Placed in the joint custody of two officers.
13. If the auditor believes that financial statements which are prepared on a comprehensive basis of ac counting other than generally accepted accounting prin ciples are not suitably titled, the auditor should a. Modify the auditor’s report to disclose any reservations.
b. Consider the effects of the titles on the finan cial statements taken as a whole.
c. Issue a disclaimer of opinion.
d. Add a footnote to the financial statements which explains alternative terminology.
14. Which of the following is the best evidence of real estate ownership at the balance sheet date?
a. Title insurance policy.
b. Original deed held in the client’s safe.
c. Paid real estate tax bills.
d. Closing statement.
15. With respect to contingent liabilities, the Standard Bank Confirmation Inquiry form approved jointly by the AICPA and the Bank Administration Institute re quests information regarding notes receivable a. Held by the bank in a custodial account.
b. Held by the bank for collection.
c. Collected by the bank.
d. Discounted by the bank.
11. An auditor who accepts an audit engagement and does not possess the industry expertise of the business entity, should a. Engage financial experts familiar with the nature of the business entity.
Obtain a knowledge of matters that relate to the nature of the entity’s business.
Refer a substantial portion of the audit to another CPA who will act as the principal auditor.
First inform management that an unqualified opinion can not be issued.
b.
c.
16. The primary reason for preparing a reconciliation between interest-bearing obligations outstanding during the year and interest expense presented in the financial statements, is to a. Evaluate internal control over securities.
b. Determine the validity of prepaid interest ex pense.
c. Ascertain the reasonableness of imputed in terest.
d. Detect unrecorded liabilities.
12. A limitation on the scope of the auditor’s ex amination sufficient to preclude an unqualified opinion will always result when management a. Engages an auditor after the year-end physical inventory count.
b. Refuses to furnish a representation letter.
c. Knows that direct confirmation of accounts receivable with debtors is not feasible.
d. Engages an auditor to examine only the balance sheet.
17. The auditor should insist that a representative of the client be present during the physical examination of securities in order to a. Lend authority to the auditor’s directives.
b. Detect forged securities.
c. Coordinate the return of all securities to pro per locations.
d. Acknowledge the receipt of securities re turned.
179 Examination Questions— November 1979 18. When asked to perform an examination in order to express an opinion on one or more specified elements, accounts or items of a financial statement, the auditor a. May not describe auditing procedures applied.
b. Should advise the client that the opinion will result in a piecemeal opinion.
c. May assume that the first standard of report ing with respect to generally accepted account ing principles does not apply.
d. Should comply with the request only if they constitute a major portion of the financial statements on which an auditor has disclaimed an opinion based on an audit.
19. Which of the following is necessary to audit balances in an on-line EDP system in an environment of destructive updating?
a. Periodic dumping of transaction files.
b. Year-end utilization of audit hooks.
c. An integrated test facility.
d. A well documented audit trail.
20. A CPA while performing tax services for a client may learn of a material error in a previously filed tax return. In such an instance the CPA should a. Prepare an affidavit with respect to the error.
b. Recommend compensating for the prior year’s error in the current year’s tax return where such action will mitigate the client’s cost and inconvenience.
c. Advise the client to file a corrected return regardless of whether or not the error resulted in an overstatement or understatement of tax.
d. Inform the IRS of the error.
21. The primary purpose of performing compliance tests is to provide reasonable assurance that a. Accounting control procedures are being ap plied as prescribed.
b. The flow of transactions through the ac counting system is understood.
c. Transactions are recorded at the amounts ex ecuted.
d. All accounting control procedures leave visi ble evidence.
22. An inventory turnover analysis is useful to the auditor because it may detect a. Inadequacies in inventory pricing.
b. Methods of avoiding cyclical holding costs.
c. The optimum automatic reorder points.
d. The existence of obsolete merchandise.
23. Accounting control procedures within the EDP ac tivity may leave no visible evidence indicating that the procedures were performed. In such instances, the auditor should test these accounting controls by a. Making corroborative inquiries.
Observing the separation of duties of per sonnel.
Reviewing transactions submitted for pro cessing and comparing them to related output. Reviewing the run manual.
b.
c.
24. Which of the following ultimately determines the specific audit procedures necessary to provide an in dependent auditor with a reasonable basis for the ex pression of an opinion?
a. The audit program.
b. The auditor’s judgment.
c. Generally accepted auditing standards.
d. The auditor’s working papers.
25. An auditor’s examination performed in accor dance with generally accepted auditing standards generally should a. Be expected to provide assurance that illegal acts will be detected where internal control is effective.
b. Be relied upon to disclose violations of truth in lending laws.
c. Encompass a plan to actively search for illegalities which relate to operating aspects.
d. Not be relied upon to provide assurance that illegal acts will be detected.
26. With respect to issuance of an audit report which is dual dated for a subsequent event occurring after the completion of fieldwork but before issuance of the auditor’s report, the auditor’s responsibility for events occurring subsequent to the completion of fieldwork is a. Extended to include all events occurring until the date of the last subsequent event referred to.
b. Limited to the specific event referred to.
c. Limited to all events occurring through the date of issuance of the report.
d. Extended to include all events occurring through the date of submission of the report to the client.
27. One reason why the independent auditor makes an analytical review of the client’s operations is to identify probable a. Weaknesses of a material nature in the system of internal control.
b. Unusual transactions.
c. Non-compliance with prescribed control pro cedures.
d. Improper separation of accounting and other financial duties.
180 Auditing 28. In order to safeguard the assets through proper in ternal control, accounts receivable that are written off are transferred to a(an) a. Separate ledger.
b. Attorney for evidence in collection pro ceedings.
c. Tax deductions file.
d. Credit manager since customers may seek to reestablish credit by paying.
29. An auditor evaluates the existing system of inter nal control in order to a. Determine the extent of substantive tests which must be performed.
b. Determine the extent of compliance tests which must be performed.
c. Ascertain whether irregularities are probable.
d. Ascertain whether any employees have incom patible functions.
30. Significant unexpected fluctuations identified by analytical review procedures will usually necessitate a (an) a. Consistency qualification.
b. Review of internal control.
c. Explanation in the representation letter.
d. Auditor investigation.
31. The most efficient and least costly method of dumping information for purposes of maintaining a backup file is from disk to a. Dump.
b. Printout.
c. Cards.
d. Tape.
32. A primary advantage of using generalized audit packages in the audit of an advanced EDP system is that it enables the auditor to a. Substantiate the accuracy of data through self-checking digits and hash totals.
b. Utilize the speed and accuracy of the com puter.
c. Verify the performance of machine operations which leave visible evidence of occurrence.
d. Gather and store large quantities of suppor tive evidential matter in machine readable form.
33. Which of the following statements best describes the auditor’s responsibility regarding the detection of fraud?
a. The auditor is responsible for the failure to detect fraud only when such failure clearly results from nonperformance of audit pro cedures specifically described in the engage ment letter.
b. The auditor must extend auditing procedures to actively search for evidence of fraud in all situations.
The auditor must extend auditing procedures to actively search for evidence of fraud where the examination indicates that fraud may exist.
The auditor is responsible for the failure to detect fraud only when an unqualified opinion is issued.
c.
d.
34. The date of the management representation letter should coincide with the a. Date of the auditor’s report.
b. Balance sheet date.
c. Date of the latest subsequent event referred to in the notes to the financial statements.
d. Date of the engagement agreement.
35. Compliance testing of an advanced EDP system a. Can be performed using only actual transac tions since testing of simulated transactions is of no consequence.
b. Can be performed using actual transactions or simulated transactions.
c. Is impractical since many procedures within the EDP activity leave no visible evidence of having been performed. d. Is inadvisable because it may distort the evidence in master files.
36. Jerome has completed an examination of the financial statements of Bold, Inc. Last year’s financial statements were examined by Smith, CPA. Since last year’s financial statements will be presented for com parative purposes without Smith’s report, Jerome’s report should a. State that the prior year’s financial statements were examined by another auditor.
b. State that the prior year’s financial statements were examined by Smith.
c. Not refer to the prior year’s examination.
d. Refer to Smith’s report only if the opinion was other than unqualified.
37. Failure to detect material dollar errors in the financial statements is a risk which the auditor primarily mitigates by a. Performing substantive tests.
b. Performing compliance tests.
c. Evaluating internal control.
d. Obtaining a client representation letter.
181 Examination Questions— November 1979 38. An auditor selects a preliminary sample of 100 items out of a population of 1,000 items. The sample statistics generate an arithmetic mean of $60, a standard deviation of $6 and a standard error of the mean of $.60. If the sample was adequate for the auditor’s pur poses and the auditor’s desired precision was plus or minus $1,000, the minimum acceptable dollar value of the population would be a. $61,000 b. $60,000 c. $59,000 d. $58,800 39. Which of the following would not be included in a CPA’s report based upon a review of the financial statements of a non-public entity?
a. A statement that the review was in accordance with generally accepted auditing standards.
b. A statement that all information included in the financial statements are the representa tions of management.
c. A statement describing the principal pro cedures performed.
d. A statement describing the auditor’s conclu sions based upon the results of the review.
40. With respect to records in a CPA’s possession, rules of conduct provide that a. Copies of client records incorporated into audit workpapers must be returned to the client upon request.
b. Worksheets in lieu of a general ledger belong to the auditor and need not be furnished to the client upon request.
c. An extensive analysis of inventory prepared by the client at the auditor’s request are workpapers which belong to the auditor and need not be furnished to the client upon re quest.
d. The auditor who returns copies of client records, must return the original records upon request.
41. The auditor who intends to express a qualified opinion should disclose all the substantive reasons in a separate explanatory paragraph of the report, except when the opinion paragraph a. Makes reference to a note in the financial statements which discloses the pertinent facts.
b. Describes a limitation on the scope of the examination.
c. Describes an insufficiency in evidential matter.
d. Has been modified because of a change in ac counting principle.
42. Where no independent stock transfer agents are employed and the corporation issues its own stocks and maintains stock records, cancelled stock certificates should a. Be defaced to prevent reissuance and attached to their corresponding stubs.
b. Not be defaced, but segregated from other stock certificates and retained in a cancelled certificates file.
c. Be destroyed to prevent fraudulent reissuance.
d. Be defaced and sent to the secretary of state.
43. During an examination Wicks learns that the audit client was granted a three month waiver of the repay ment of principal on the installment loan with Blank Bank without an extension of the maturity date. With respect to this loan, the audit program used by Wicks would be least likely to include a verification of the a. Interest expense for the year.
b. Balloon payment.
c. Total liability at year end.
d. Installment loan payments.
44. During 1979, a bookkeeper perpetrated a theft by preparing erroneous W-2 forms. The bookkeeper’s FICA withheld was overstated by $500.00 and the FICA withheld from all other employees was understated. Which of the following is an audit procedure which would detect such a fraud?
a. Multiplication of the applicable rate by the in dividual gross taxable earnings.
b. Utilizing form W-4 and withholding charts to determine whether deductions authorized per pay period agree with amounts deducted per pay period.
c. Footing and crossfooting of the payroll register followed by tracing postings to the general ledger.
d. Vouching cancelled checks to federal tax forms 941.
45. Which of the following is an internal accounting control weakness related to factory equipment?
a. Checks issued in payment of purchases of equipment are not signed by the controller.
b. All purchases of factory equipment are re quired to be made by the department in need of the equipment.
c. Factory equipment replacements are generally made when estimated useful lives, as indicated in depreciation schedules, have expired.
d. Proceeds from sales of fully depreciated equipment are credited to other income.
182 Auditing 46. When financial statements are prepared on the basis of a going concern and the auditor believes that the client may not continue as a going concern, the auditor should issue a. A “ subject to” opinion.
b. An unqualified opinion with an explanatory middle paragraph.
c. An “except for” opinion.
d. An adverse opinion.
51. When auditing a public warehouse, which of the following is the most important audit procedure with respect to disclosing unrecorded liabilities?
a. Confirmation of negotiable receipts with holders.
b. Review of outstanding receipts.
c. Inspection of receiving and issuing pro cedures.
d. Observation of inventory.
47. Any assessment of the operational capabilities of a computer system must consider downtime. Even in a fully-protected system, downtime will exist because of a. Electrical power losses.
b. Unscheduled maintenance.
c. Unauthorized entry.
d. Keypunching errors.
48. The first general standard requires that a person or persons have adequate technical training and proficien cy as an auditor. This standard is met by a. An understanding of the field of business and finance.
b. Education and experience in the field of auditing.
c. Continuing Professional Education.
d. A thorough knowledge of the Statements on Auditing Standards.
49. The principal auditor is satisfied with the in dependence and professional reputation of the other auditor who has audited a subsidiary but wants to in dicate the division of responsibility. The principal auditor should a. Modify the scope paragraph of the report.
b. Modify the scope and opinion paragraphs of the report.
c. Not modify the report except for inclusion of an explanatory middle paragraph.
d. Modify the opinion paragraph of the report.
50. Analytical review procedures are a. Statistical tests of financial information designed to identify areas requiring intensive investigation.
b. Analytical tests of financial information made by a computer.
c. Substantive tests of financial information made by a study and comparison of relation ships among data.
d. Diagnostic tests of financial information which may not be classified as evidential matter.
52. A management information system is designed to ensure that management possesses the information it needs to carry out its functions through the integrated actions of a. Data-gathering, analysis and reporting func tions.
b. A computerized information retrieval and decision-making system.
c. Statistical and analytical review functions.
d. Production-budgeting and sales-forecasting activities.
53. Audit working papers are used to record the results of the auditor’s evidence-gathering procedures. When preparing working papers the auditor should remember that working papers should be a. Kept on the client’s premises so that the client can have access to them for reference pur poses.
b. The primary support for the financial statements being examined.
c. Considered as a part of the client’s accounting records which is retained by the auditor.
d. Designed to meet the circumstances and the auditor’s needs on each engagement.
54. Proper internal control over the cash payroll func tion would mandate which of the following?
a. The payroll clerk should fill the envelopes with cash and a computation of the net wages.
b. Unclaimed pay envelopes should be retained by the paymaster.
c. Each employee should be asked to sign a receipt.
d. A separate checking account for payroll be maintained.
55. A company policy should clearly indicate that defective merchandise returned by customers is to be delivered to the a. Sales clerk.
b. Receiving clerk.
c. Inventory control clerk.
d. Accounts receivable clerk.
183 Examination Questions— November 1979 56. When a contingency is resolved immediately subsequent to the issuance of a report which was qualified with respect to the contingency, the auditor should a. Insist that the client issue revised financial statements.
b. Inform the audit committee that the report can not be relied upon.
c. Take no action regarding the event.
d. Inform the appropriate authorities that the report can not be relied upon.
57. Some firms which dispose of only a small part of their total output by consignment shipments fail to make any distinction between consignment shipments and regular sales. Which of the following would suggest that goods have been shipped on consignment?
a. Numerous shipments of small quantities.
b. Numerous shipments of large quantiti es and few returns.
c. Large debits to accounts receivable and small periodic credits.
d. Large debits to accounts receivable and large periodic credits.
58. An audit program provides proof that a. Sufficient competent evidential matter was obtained.
b. The work was adequately planned.
c. There was compliance with generally accepted standards of reporting.
d. There was a proper study and evaluation of internal control.
59. A CPA reviews a client’s payroll procedures. The CPA would consider internal control to be less than effective if a payroll department supervisor was as signed the responsibility for a. Reviewing and approving time reports for subordinate employees.
b. Distributing payroll checks to employees.
c. Hiring subordinate employees.
d. Initiating requests for salary adjustments for subordinate employees.
60. The objective of quality control mandates that a public accounting firm should establish policies and procedures for professional development which provide reasonable assurance that all entry-level personnel a. Prepare working papers which are standard ized in form and content.
b. Have the knowledge required to enable them to fulfill responsibilites assigned.
c. Will advance within the organization.
d. Develop specialties in specific areas of public accounting.
Number 2 (Estimated time----- 15 to 25 minutes) Jones, CPA, is approached by a prospective client who desires to engage Jones to perform an audit which in prior years was performed by another CPA.
Required: Identify the procedures which Jones should follow in accepting the engagement.
Number 3 (Estimated time-----15 to 25 minutes) Loman, CPA, who has examined the financial statements of the Broadwall Corporation, a publicly held company, for the year ended December 31, 1979, was asked to perform a limited review of the financial statements of Broadwall Corporation for the period ending March 31, 1980. The engagement letter stated that a limited review does not provide a basis for the expression of an opinion.
Required: a. Explain why Loman’s limited review will not provide a basis for the expression of an opinion.
b. What are the review procedures which Loman should perform, and what is the purpose of each pro cedure? Structure your response as follows: Procedure Purpose of Procedure 184 Auditing Number 4 (Estimated time----- 15 to 25 minutes) In the past, the records to be evaluated in an audit have been printed reports, listings, documents and written papers, all of which are visible output. However, in fully computerized systems which employ daily updating of transaction files, output and files are frequently in machine-readable forms such as cards, tapes, or disks. Thus, they often present the auditor with an opportunity to use the computer in performing an audit.
Required: Discuss how the computer can be used to aid the auditor in examining accounts receivable in such a fully computerized system.
Number 5 (Estimated time-----15 to 25 minutes) Internal control comprises the plan of organization and all of the coordinate methods and measures adopted within a business to safeguard its assets, check the accuracy and reliability of its accounting data, pro mote operational efficiency, and encourage adherence to prescribed managerial policies.
Required: a. What is the purpose of the auditor’s study and evaluation of internal control?
b. What are the objectives of a preliminary evaluation of internal control?
c. How is the auditor’s understanding of the system of internal control documented?
d. What is the purpose of tests of compliance?
185 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXAMINATION IN BUSINESS LAW (Commercial Law) November 9, 1979; 8:30 AM to 12:00 M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: No. 1.....................
No. 2 .....................
No. 3 .....................
No. 4 .....................
No. 5 ....................
Estimated Minutes Minimum Maximum 90 105 25 30 20 25 20 25 20 25 Total 175 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
186 Business Law Instructions Select the best answer for each of the following items. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 99. The text of the letter from Bridge Builders, Inc., to Allied Steel Co. follows: We offer to purchase 10,000 tons of No.
4 steel pipes at today’s quoted price for delivery two months from today. Your ac ceptance must be received in five days.
Bridge Builders intended to create a (an) a. Option contract.
b. Unilateral contract.
c. Bilateral contract.
d. Joint contract.
Answer Sheet Number 1 (Estimated time----- 90 to 105 minutes) 2. In the process of negotiating the sale of his manufacturing business to Grand, Sterling made certain untrue statements which Grand relied upon. Grand was induced to purchase the business for $10,000 more than its true value. Grand is not sure whether he should seek relief based upon misrepresentation or fraud. Which of the following is a correct statement?
a. If Grand merely wishes to rescind the contract and get his money back, misrepresentation is his best recourse.
b. In order to prevail under the fraud theory, Grand must show that Sterling intended for him to rely on the untrue statements; whereas he need not do so if he bases his action on misrepresentation.
c. Both fraud and misrepresentation require Grand to prove that Sterling knew the statements were false.
d. If Grand chooses fraud as his basis for relief, the statute of fraud applies.
3. Which of the following will not be sufficient to satisfy the consideration requirement for a contract?
a. The offeree expends both time and money in studying and analyzing the offer.
b. The offeree makes a promise which is a legal detriment to him.
c. The offeree performs the act requested by the offeror.
d. The offeree makes a promise which benefits the offeror.
Items to be Answered 99. 1. Cox Manufacturing repudiated its contract to sell 300 televisions to Ruddy Stores, Inc. What recourse does Ruddy Stores have?
a. It can obtain specific performance by the seller.
b. It can recover punitive damages.
c. It must await the seller’s performance for a commercially reasonable time after repudia tion if it wishes to recover anything.
d. It can “ cover” , that is, procure the goods elsewhere and recover any damages.
4. Marsh and Lennon entered into an all inclusive written contract involving the purchase of a tract of land. Lennon claims that there was a contemporaneous oral agreement between the parties which called for the removal by Marsh of several large rocks on the land. Marsh relies upon the parol evidence rule to avoid having to remove the rocks. Which of the following is correct?
a. The parol evidence rule does not apply to con temporaneous oral agreements.
b. Since the statute of frauds was satisfied in respect to the contract for the purchase of the land, the parol evidence rule does not apply.
c. Since the oral agreement does not contradict the terms of the written contract, the oral agreement is valid despite the parol evidence rule.
d. The parol evidence rule applies and Lennon will be precluded from proving the oral pro mise in the absence of fraud.
187 Examination Questions— November 1979 5. Which of the following represents the basic distinction between a bilateral contract and a unilateral contract?
a. Specific performance is available if the con tract is unilateral whereas it is not if the con tract is bilateral.
b. There is only one promise involved if the con tract is unilateral whereas there are two pro mises if the contract is bilateral.
c. The statute of frauds applied to a bilateral contract but not to a unilateral contract.
d. The rights under a bilateral contract are assignable whereas rights under a unilateral contract are not assignable.
6. The Flick Corporation sold various interrelated products that it manufactured. One of the items was manufactured almost exclusively by Flick and sold throughout the United States. Flick realized the impor tance of this product to its purchasers and decided to capitalize on the situation by requiring all purchasers to take at least two other products in order to obtain the item over which it has almost complete market control. At Flick’s spring sales meeting, its president informed the entire sales force that they were to henceforth sell only to those customers who agreed to take the addi tional products. As a result of this plan, gross sales of the additional items increased by more than $1 million. Which of the following best describes the legality of the above situation?
a. It is illegal only if the products are patented products.
b. It is an illegal tying arrangement.
c. It is legal as long as the price charged to retailers for the other products is competitive.
d. It is legal if the retailers do not complain about purchasing the other products.
7. Master Corporation, a radio and television manufacturer, invited Darling Discount Chain to ex amine several odd lots of discontinued models and make an offer for the entire lot. The odd lots were segregated from the regular inventory but inadvertently included 15 current models. Darling was unaware that Master did not intend to include the 15 current models in the group. Darling made Master an offer of $9,000 for the entire lot, which represented a large discount from the normal sales price. Unaware of the error, Master accepted the offer. Master would not have accepted had it known of the inclusion of the 15 current models. Upon learning of the error, Master alleged mistake as a defense and re fused to perform. Darling sued for breach of contract. Under the circumstances, what is the status of the con tract?
a. There is no contract since Master did not in tend to include the 15 current models in the group of radios to be sold.
b. The contract is voidable because of a unilateral mistake.
c. The contract is voidable because of a mutual mistake.
d. There is a valid and binding contract which in cludes the 15 current-model radios.
8. A salesman for A & C Company called upon the purchasing agent for Major Enterprises, Inc., and of fered to sell Major 1,500 screwdriver sets at $1.60 each. Major’s purchasing agent accepted and the following day sent A & C a purchase order which bore Major’s name and address at the top and also had the purchasing agent’s name and title stamped at the bottom with his initials. The purchase order recited the agreement reached orally the prior day. Subsequently, Major decided it did not want the screwdriver sets since it was overstocked in that item. Major thereupon repudiated the contract and asserted the statute of frauds as a defense. Under the circumstances, which of the follow ing is correct?
a. The statute of frauds does not apply to this transaction since performance is to be com pleted within one year from the date of the making of the contract.
b. Major will lose but only if its purchasing agent’s authority to make the contract was in writing.
c. The fact that an authorized agent of A & C did not sign the purchase order prevents its use by A & C against Major to satisfy the statute of frauds.
d. The purchase order is sufficient to satisfy the statute of frauds even though the purchasing agent never signed it in full.
188 Business Law 9. Lally sent Queen Supply Company, Inc., a telegram ordering $700 of general merchandise. Lally’s telegram indicated that immediate shipment was necessary. That same day Queen delivered the goods to the Red Freight Company. The shipment was delayed due to a breakdown of the truck which was transporting the goods. When the merchandise did not arrive as promptly as expected, Lally notified Queen that it revoked the offer and was purchasing the goods elsewhere. Queen indicated to Lally that the merchan dise had been shipped the same day Lally had ordered it and Lally’s revocation was not good. Which of the following statements best describes the transaction?
a. The statute of frauds will be a defense on any action by Queen to enforce the contract.
b. Prompt shipment of the merchandise by Queen constituted an acceptance.
c. Lally’s revocation of the offer was effective since Lally had not received a notice of accep tance.
d. Lally’s order was an offer to Queen to enter into a bilateral contract which could be ac cepted only by a promise.
12. Charleston, Inc., had its warehouse destroyed by fire. Charleston’s property was insured against fire loss by the Conglomerate Insurance Company. An in vestigation by Conglomerate revealed that the fire had been caused by a disgruntled employee whom Charleston had suspended for one month due to in subordination. Charleston seeks to hold its insurer liable for the $200,000 loss of its warehouse. Which of the following is correct insofar as the dispute between Charleston and the Conglomerate Insurance Company?
a. Since the loss was due to the deliberate destruction by one of Charleston’s employees, recovery will be denied.
b. Conglomerate must pay Charleston, but it will be subrogated to Conglomerate’s rights against the wrongdoing employee.
c. The fact that the employee has been suspend ed for one month precludes recovery against Conglomerate.
d. Arson is excluded from the coverage of most fire insurance policies, and therefore Con glomerate is not liable.
10. Marcross and two business associates own real property as tenants in common that they have invested in as a speculation. The speculation proved to be highly successful, and the land is now worth substantially more than their investment. Which of the following is a cor rect legal incident of ownership of the property?
a. Upon the death of any of the other tenants, the deceased’s interest passes to the survivor(s) unless there is a will.
b. Each of the co-tenants owns an undivided in terest in the whole.
c. A co-tenant can not sell his interest in the property without the consent of the other tenants.
d. Upon the death of a co-tenant, his estate is en titled to the amount of the original invest ment, but not the appreciation.
11. Which of the following omissions will prevent a writing from satisfying the statute of frauds with respect to the sale of goods?
a. It does not indicate that a sale has occurred.
b. It is not signed by both the buyer and seller.
c. The time and place of delivery are not in dicated.
d. The payment terms are not contained in the writing.
13. On March 11, Vizar Sales Corporation tele graphed Watson Company: “Will sell 1,000 cases of coffee for $28 a case for delivery at our place of business on April 15. You may pick them up at our loading platform.’’ Watson telegraphed its acceptance on March 12. On March 20, coffee prices rose to $30 a case. Vizar telegraphed Watson on March 21 that it repudiated the sale and would not make delivery. The telegram was received by Watson on March 22 when the price was $32; Watson could have covered at that price but chose not to do so. On April 15 the coffee was selling at $35 a case. Watson tendered $28,000 to Vizar and indicated it was ready to take delivery. Vizar refused to deliver. What relief, if any, is Watson entitled to?
a. Specific performance, because it made a valid tender of performance.
Nothing, because it failed to cover.
Damages of $4,000 (the difference between the contract price and the fair market value at the time Watson learned of the breach). Damages of $7,000 (the difference between the contract price and the fair market value at the time delivery should have been made).
b.
c.
d.
189 Examination Questions— November 1979 14. Dowling is a promoter and has decided to use a limited partnership for conducting a securities invest ment venture. Which of the following is unnecessary in order to validly create such a limited partnership?
a. All limited partners’ capital contributions must be paid in cash.
b. There must be a state statute which permits the creation of such a limited partnership.
c. A limited partnership certificate must be signed and sworn to by the participants and filed in the proper office in the state.
d. There must be one or more general partners and one or more limited partners.
15. Jon and Frank Clarke are equal partners in the partnership of Clarke & Clarke. Both Jon Clarke and the partnership are bankrupt. Jon Clarke personally has $150,000 of liabilities and $100,000 of assets. The part nership’s liabilities are $450,000 and its assets total $250,000. Frank Clarke, the other partner, is solvent with $800,000 of assets and $150,000 of liabilities. What are the rights of the various creditors of Jon Clarke, Frank Clarke, and the partnership?
a. Jon Clarke must divide his assets equally among his personal creditors and firm creditors.
b. Frank Clarke will be liable in full for the $200,000 partnership deficit.
c. Jon Clarke’s personal creditors can recover the $50,000 deficit owed to them from Frank Clarke.
d. Frank Clarke is liable only for $100,000, his equal share of the partnership deficit.
16. King, Kline and Fox were partners in a wholesale business. Kline died and left to his wife his share of the business. Kline’s wife is entitled to a. The value of Kline’s interest in the partner ship.
b. Kline’s share of specific property of the part nership.
c. Continue the partnership as a partner with King and Fox.
d. Kline’s share of the partnership profits until her death.
17. Which of the following will not result in a dissolu tion of a partnership?
a. The bankruptcy of a partner as long as the partnership itself remains solvent.
b. The death of a partner as long as his will pro vides that his executor shall become a partner in his place.
c. The wrongful withdrawal of a partner in con travention of the agreement between the part ners.
d. The assignment by a partner of his entire part nership interest.
Items 18 and 19 are based on the following infor mation: Teal and Olvera were partners of the T & O Real Estate Investment Company. They decided to seek more capital in order to expand their participation in the booming real estate business in the area. They obtained five individuals to invest $100,000 each in their venture as limited partners.
18. Assuming the limited partnership agreement is silent on the point, which of the following acts may Teal and Olvera engage in without the written consent of all limited partners?
a. Admit an additional person as a general part ner.
b. Continue the partnership business upon the death or retirement of a general partner.
c. Invest the entire amount ($500,000) of con tributions by the limited partners in a single venture.
d. Admit additional limited partners from time to time in order to obtain additional working capital.
19. Which of the following rights would the limited partners not have?
a. The right to have a dissolution and winding up by court decree where such is appropriate.
b. The right to remove a general partner by a majority vote if the limited partners determine that a general partner is not managing the partnership affairs properly.
c. The right upon dissolution to receive their share of profits and capital contributions before any payment is made to the general partners.
d. The right to have the partnership books kept at the principal place of business and to have access to them.
190 Business Law 20. Woodrow died and left a will that named as co- executors the Fundamental Trust Company and Harlow, who is one of the residuary legatees. The will was silent on various points indicated below. Which of the following is correct?
a. If Woodrow’s will was not properly executed, it will not be admitted to probate and his property will be distributed according to the intestate succession laws even though this is contrary to Woodrow’s wishes as stated in the will.
b. Since Harlow is one of the residuary legatees, Harlow can not serve as executor since this would represent a conflict of interest and also would violate Harlow’s fiduciary duty.
c. All taxes paid will be allocated to the residuary estate and not apportioned.
d. The executors have complete discretion in sofar as investing the estate’s assets during the term of their administration.
22. Williamson purchased from Dilworth Hardware a new lathe for his home workshop for cash. Two weeks later, Williamson was called by the Easy Loan Company. Easy explained to Williamson that it had been financing Dilworth’s purchases from the manufac turers and that to protect its interest it had obtained a perfected security interest in Dilworth’s entire inventory of hardware and power tools, including the lathe which Williamson bought. Easy further explained that Dilworth had defaulted on a payment due to Easy, and Easy intended to assert its security interest in the lathe and repossess it unless Williamson was willing to make payment of $200 for a release of Easy’s security interest. If Williamson refuses to make the payment, which of the following statements is correct?
a. Williamson will not take free of Easy’s secu rity interest if he was aware of said interest at the time he purchased the lathe.
b. Even if Williamson had both actual notice and constructive notice via recordation of Easy’s interest, he will prevail if Easy seeks to repossess the lathe.
c. Easy’s security interest in the lathe in question is invalid against all parties unless its filing specifically described and designated the par ticular lathe Williamson purchased.
d. Williamson must pay the $200 or the lathe can be validly repossessed and sold to satisfy the amount Dilworth owes Easy and any excess paid to Williamson.
21. Wanton Corporation, its president, and several other officers of the corporation are found guilty of conspiring with its major competitor to fix prices. Which of the following sanctions would not be ap plicable under federal antitrust laws?
a. Suspension of corporate right to engage in in terstate commerce for not more than one year.
b. Treble damages.
c. Seizure of Wanton’s property illegally shipped in interstate commerce.
d. Fines against Wanton and fines and imprison ment of its president and officers.
23. Two Uniform Commercial Code concepts relating to secured transactions are “attachment” and “perfec tion.” Which of the following is correct in connection with the similarities and differences between these two concepts?
a. They are mutually exclusive and wholly in dependent of each other.
b. Satisfaction of one automatically satisfies the other.
c. Attachment relates primarily to the rights against the debtor and perfection relates primarily to the rights against third parties.
d. It is not possible to have a simultaneous at tachment and perfection.
191 Examination Questions— November 1979 24. Which of the following requirements is not necessary in order to have a security interest attach?
a. There must be a proper filing.
b. The debtor must have rights in the collateral.
c. Value must be given by the creditor.
d. Either the creditor must take possession or the debtor must sign a security agreement which describes the collateral.
25. In respect to obtaining a purchase money security interest, which of the following requirements must be met?
a. The property sold may only be consumer goods.
b. Only a seller may obtain a purchase money security interest.
c. Such a security interest must be filed in all cases to be perfected.
d. Credit advanced to the buyer must be used to obtain the property which serves as the col lateral.
26. Vista Motor Sales, a corporation engaged in sell ing motor vehicles at retail, borrowed money from Sun shine Finance Company and gave Sunshine a properly executed security agreement in its present and future in ventory and in the proceeds therefrom to secure the loan. Sunshine’s security interest was duly perfected under the laws of the state where Vista does business and maintains its entire inventory. Thereafter, Vista sold a new pickup truck from its inventory to Archer and received Archer’s certified check in payment of the full price. Under the circumstances, which of the following is correct?
a. Sunshine must file an amendment to the financing statement every time Vista receives a substantial number of additional vehicles from the manufacturer if Sunshine is to ob tain a valid security interest in subsequently delivered inventory.
b. Sunshine’s security interest in the certified check Vista received is perfected against Vista’s other creditors.
c. Unless Sunshine specifically included pro ceeds in the financing statement it filed, it has no rights to them.
d. The term “proceeds” does not include used cars received by Vista since they will be resold.
27. Taylor is the executive Vice President for Marketing of Reflex Corporation and a member of the Board of Directors. Based on information obtained dur ing the course of his duties, Taylor concluded that Reflex’s profits would fall by 50% for the quarter and 30% for the year. He quietly contacted his broker and disposed of 10,000 shares of his Reflex stock at a profit, some of which he had acquired within 6 months of the sale. In fact, Reflex’s profits did not fall, but its stock price declined for unrelated reasons. Taylor had also ad vised a friend to sell her shares and repurchase the stock later. She followed Taylor’s advice, sold for $21, and subsequently repurchased an equal number of shares at $11. A shareholder has commenced a shareholder derivative action against Taylor and the friend for viola tion of the Securities Exchange Act of 1934. Under these circumstances, which of the following is correct?
a. Taylor is not an insider in relation to Reflex.
b. Taylor must account to the corporation for his short-swing profit.
Taylor and the friend must both account to the corporation for their short-swing profits. Neither Taylor nor the friend has incurred any liability under the 1934 act.
c.
d.
28. Which of the following is exempt from registration under the Securities Act of 1933?
a. First mortgage bonds.
b. The usual annuity contract issued by an in surer.
c. Convertible preferred stock.
d. Limited partnership interests.
29. Under the Securities Act of 1933, an accountant may be held liable for any materially false or misleading financial statements, including an omission of a material fact therefrom, provided the purchaser a. Proves reliance on the registration statement or prospectus.
b. Proves negligence or fraud on the part of the accountant.
c. Brings suit within four years after the security is offered to the public.
d. Proves a false statement or omission existed and the specific securities were the ones of fered through the registration statement.
192 Business Law 30. Which of the following is a correct statement con cerning the similarities of a limited partnership and a corporation?
a. Shareholders and limited partners may both participate in the management of the business and retain limited liability.
b. Both are recognized for federal income tax purposes as taxable entities.
Both can only be created pursuant to a statute and each must file a copy of their respective certificates with the proper state authorities. Both provide insulation from personal liabi lity for all of the owners of the business.
c.
d.
32. Whitworth has been charged by Bonanza Cor poration with violating the Securities Exchange Act of 1934. Whitworth was formerly the president of Bonanza, but he was ousted as a result of a proxy battle. Bonanza seeks to recover from Whitworth any and all of his short-swing profits. Which of the following would be a valid defense to the charges?
a. Whitworth is a New York resident, Bonanza was incorporated in New York, and the trans actions were all made through the New York Stock Exchange; therefore, interstate com merce was not involved.
b. Whitworth did not actually make use of any insider information in connection with the various stock transactions in question.
c. All the transactions alleged to be in violation of the 1934 act were purchases made during February 1979 with the corresponding sales made in September 1979.
d. Whitworth’s motivation in selling the stock was solely a result of the likelihood that he would be ousted as president of Bonanza.
31. Magnus Real Estate Developers, Inc., wanted to acquire certain tracts of land in Marshall Township in order to build a shopping center complex. To ac complish this goal, Magnus engaged Dexter, a sophisticated real estate dealer, to represent them in the purchase of the necessary land without revealing the ex istence of the agency . Dexter began to slowly but steadi ly acquire the requisite land. However, Dexter made the mistake of purchasing one tract outside the description of the land needed. Which of the following is correct under these circumstances?
a. The use of an agent by Magnus, an undis closed principal, is manifestly illegal.
b. Either Magnus or Dexter may be held liable on the contracts for the land, including the land that was not within the scope of the pro posed shopping center.
c. An undisclosed principal such as Magnus can have no liability under the contract since the third party believed he was dealing with Dexter as a principal.
d. An agent for an undisclosed principal assumes no liability as long as he registers his relation ship to the principal with the clerk of the proper county having jurisdiction.
33. Wanamaker, Inc., engaged Anderson as its agent to purchase original oil paintings for resale by Wanamaker. Anderson’s express authority was specifically limited to a maximum purchase price of $25,000 for any collection provided it contained a minimum of five oil paintings. Anderson purchased a seven-picture collection on Wanamaker’s behalf for $30,000. Based upon these facts, which of the following is a correct legal conclusion?
a. The express limitation on Anderson’s autho rity negates any apparent authority.
b. Wanamaker can not ratify the contract since Anderson’s actions were clearly in violation of his contract.
c. If Wanamaker rightfully disaffirms the unauthorized contract, Anderson is personal ly liable to the seller.
d. Neither Wanamaker nor Anderson is liable on the contract since the seller was obligated to ascertain Anderson’s authority.
193 Examination Questions—November 1979 34. Which of the following activities engaged in by a corporation will not be deemed illegal under the an titrust law?
a. A price-fixing agreement with competitors aimed at lowering prices to a reasonable level.
b. The charging of a price aimed at maximizing its profits based upon economic analysis of supply and demand for its products.
c. Participating in a plan suggested by the trade association aimed at territorial allocations of markets to cut costs.
d. The payment of brokerage commissions to the purchasers of goods.
35. The Donner Corporation has obtained a patent on a revolutionary coin-operated washing machine. It is far superior to the existing machines currently in use. Which of the following actions taken by Donner will not result in a violation of federal antitrust law?
a. Maintaining the resale price for machines it sells to distributors.
b. Obtaining a near total monopolization of the market as a result of the patent.
c. Requiring the purchasers of the machines to buy from Donner all their other commonplace supplies connected with the use of the machine.
d. Joining in a boycott with other appliance manufacturers to eliminate a troublesome dis count distributor.
36. Expansion Corporation is an aggressive, large sized conglomerate. It is seeking to obtain control of several additional corporations including Resistance Corporation. Expansion does not currently buy from, sell to, or compete with Resistance. Which of the following statements applies to this proposed takeover?
a. Since Expansion does not buy from, sell to, or compete with Resistance, antitrust laws do not apply.
If Expansion can consummate the acquisition before there is an objection to it, the acquisi tion can not subsequently be set aside.
The acquisition is likely to be declared illegal if there will be reciprocal buying and there is a likelihood that other entrants into the market would be precluded.
The acquisition is legal on its face if there will be cost efficiency resulting from combined marketing and advertising.
b.
c.
37. The Justice Department is contemplating com mencing an action against Lion Corporation for monopolizing the off-shore oil drilling business in viola tion of Section 2 of the Sherman Act. Which of the following would be Lion’s best defense against such an action?
a. Since the drilling is off-shore, interstate com merce is not involved.
b. The monopoly was originally the result of a long since expired patent.
c. Lion had no specific wrongful intent to monopolize.
d. Lion’s market share is such that it does not have the power to fix prices or to exclude com petitors.
38. Glick was the owner of a factory valued at $100,000. He procured a fire insurance policy on the building for $40,000 from Safety Insurance Company, Inc. The policy contained an 80% coinsurance clause. The property was totally destroyed by fire. How much will Click recover from the insurance company?
a. $20,000.
b. $32,000.
c. $40,000.
d. Click will recover nothing because he did not meet the coinsurance requirements.
39. Lantz sold his moving and warehouse business, in cluding all the personal and real property used therein, to Mallen Van Lines, Inc. The real property was en cumbered by a duly-recorded $300,000 first mortgage upon which Lantz was personally liable. Mallen ac quired the property subject to the mortgage but did not assume the mortgage. Two years later, when the outstanding mortgage was $260,000, Mallen decided to abandon the business location because it had become unprofitable and the value of the real property was less than the outstanding mortgage. Mallen moved to another location and refused to pay the installments due on the mortgage. What is the legal status of the parties in regard to the mortgage?
a. Mallen breached its contract with Lantz when it abandoned the location and defaulted on the mortgage.
b. Mallen took the real property free of the mortgage.
c. If Lantz pays off the mortgage, he will be able to successfully sue Mallen because Lantz is subrogated to the mortgagee’s rights against Mallen.
d. Lantz must satisfy the mortgage debt in the event that foreclosure yields an amount less than the unpaid balance.
194 d.
Business Law 40. Carter wished to obtain additional working capital for his construction company. His bankers indicated that they would be willing to lend the company $50,000 if the bank could obtain a first mortgage on the real property belonging to the business. Carter reluctantly acquiesced and mortgaged all his real property to secure repayment of the loan. Unknown to the bank one por tion of the real property was already mortgaged to Johnson for $30,000, but Johnson had neglected to record the mortgage. The bank promptly recorded its mortgage. Which of the following is correct regarding the rights of the parties?
a. Johnson’s failure to record makes the mort gage invalid against Carter.
b. The bank’s mortgage will have a priority over Johnson’s mortgage.
c. Both mortgagees would share the proceeds from any foreclosure on a pro rata basis.
d. The bank will be deemed to have notice of Johnson’s mortgage and will take subject to the mortgage.
42. Nicks is a troublesome chain store furniture dealer. He constantly engaged in price cutting on widely advertised name products in order to lure customers to his store so that he could sell them other products. The “big three’’ manufacturers agreed that Nicks could no longer sell their products unless he ceased and desisted from such practices. Nicks refused and the three manufacturers promptly cut off his supply of their branded products. Which of the following is a correct statement?
Since a businessman has the freedom to choose with whom he will deal, the conduct in question is not illegal under the antitrust laws. If the harm to the public was minor, and the products were readily available from other ap pliance dealers in a market marked by free and open competition, there would be no violation of the law.
The conduct described is a joint boycott, and as such is illegal per se.
Since the conduct described was unilateral, and Nicks did not agree to stop his price cut ting, the manufacturers’ conduct is legal.
a.
b.
d.
41. Dunbar Dairy Farms, Inc., pursuant to an expan sion of its operations in Tuberville, purchased from Moncrief a 140-acre farm strategically located in the general area in which Dunbar wishes to expand. Unknown to Dunbar, Cranston, an adjoining land- owner, had fenced off approximately five acres of the land in question. Cranston installed a well, constructed a storage shed and garage on the fenced-off land, and continuously farmed and occupied the five acres for ap proximately 22 years prior to Dunbar’s purchase. Cranston did this under the mistaken belief that the five acres of land belonged to him. Which of the following is a correct answer in regard to the five acres occupied by Cranston?
a. Under the circumstances Cranston has title to the five acres.
b. As long as Moncrief had properly recorded a deed which includes the five acres in dispute, Moncrief had good title to the five acres.
At best, the only right that Cranston could ob tain is an easement.
If Dunbar is unaware of Cranston’s presence and Cranston has failed to record, Dunbar can oust him as a trespasser.
c.
d.
43. Gail Monet has decided to make certain gifts to her family. Her goal is to reduce her estate and income taxes. Which of the following need not be present in order for Monet to make valid gifts?
a. Monet must be competent to make the gifts in question.
b. Monet must have some purpose or motive other than, or in addition to, the mere saving of taxes.
c. There must be delivery of the gifts to the donees.
d. The gifts must be made voluntarily and with the requisite donative intent.
195 Examination Questions— November 1979 44. James Gordon decided to create an inter vivos trust for the benefit of his grandchildren. He wished to bypass his own children, and to provide an independent income for his grandchildren. He did not, however, wish to completely part with the assets he would transfer to the trust. Therefore, he transferred the assets to the York Trust Company, in trust for the benefit of his grandchildren irrevocably for a period of 12 years. Which of the following is correct regarding the trust?
a. The trust will fail for want of a proper pur pose.
b. The trust income will not be taxable to Gordon during its existence.
c. Gordon retains beneficial title to the property transferred to the trust.
d. If Gordon demands the return of the trust assets prior to the 12 years, York must return them to him since he created the trust and the assets will eventually be his again.
45.
The usual fire insurance policy does not a. Have to meet the insurable interest test if this requirement is waived by the parties.
b. Provide for subrogation of the insurer to the insured’s rights upon payment of the amount of the loss covered by the policy.
c. Cover losses caused by the negligence of the insured’s agent.
Permit assignment of the policy prior to loss without the consent of the insurer.
d.
Items 46 and 47 are based on the following infor mation: Martin is the trustee of the Baker Trust which has assets in excess of $1 million. Martin has engaged the CPA firm of Hardy & Fox to prepare the annual ac counting statement for the allocation of receipts and ex penditures between income and principal. The trust in denture provides that “receipts and expenses are to be allocated to income or principal according to law.’’ 47. Which of the following receipts from real property should be allocated to principal?
a. An unexpected payment of nine months’ arrears in rental payments.
b. A six-month prepayment of rent.
c. Insurance proceeds for the destruction of a garage on one of the properties.
d. Interest on a purchase money mortgage aris ing from the sale of a parcel of the trust’s real property.
48. Yeats Manufacturing is engaged in the manufac ture and sale of convertible furniture in interstate com merce. Yeats’ manufacturing facilities are located in a jurisdiction which has a compulsory workmen’s com pensation act. Hardwood, Yeats’ president, decided that the company should, in light of its safety record, choose to ignore the requirement of providing workmen’s compensation insurance. Instead, Hardwood indicated that a special account should be created to provide for such contingencies. Basset was severely injured as a result of his negligent operation of a lathe which accelerated and cut off his right arm. In assessing the potential liability of Yeats, which of the following is a correct answer?
a. Federal law applies since Yeats is engaged in interstate commerce.
Yeats has no liability, since Basset negli gently operated the lathe.
Since Yeats did not provide workmen’s com pensation insurance, it can be sued by Basset and cannot resort to the usual common law defenses.
Yeats is a self-insurer, hence it has no liability beyond the amount of the money in the in surance fund.
b.
c.
196 46. Which of the following receipts should be allocated to income?
a. Rights to subscribe to shares of the distributing corporation.
b. Sale of rights to subscribe to shares of the distributing corporation.
c. A 2% stock dividend.
d. Rights to subscribe to shares of another cor poration.
Business Law 49. The federal Social Security Act applies in general to both employers and employees. Hexter Manufactur ing is a small business as defined by the Small Business Administration. Regarding Hexter’s relationship to the requirements of the Social Security Act, which of the following is correct?
a. Since Hexter is a small business, it is exempt from the Social Security Act.
b. Social Security payments made by Hexter’s employees are tax deductible for federal in come tax purposes.
c. Hexter has the option to be covered or ex cluded from the provisions of the Social Security Act.
d. The Social Security Act applies to both Hexter and its employees.
50. Ozgood is a principal and Flood is his agent. Ozgood is totally dissatisfied with the agency relation ship and wishes to terminate it. In which of the follow ing situations does Ozgood not have the power to ter minate the relationship?
a. Ozgood and Flood have agreed that their agency is irrevocable.
b. Flood has been appointed as Ozgood’s agent pursuant to a power of attorney.
c. Flood is an agent coupled with an interest.
d. The agency agreement is in writing and pro vides for a specific duration which has not elapsed.
Number 2 (Estimated time-----25 to 30 minutes) Part a. Fairfax Corporation was created on April 2, 1979. Its initial capitalization consisted of (1) 5,000 shares of no-par voting common stock which it sold to subscribers at $20 a share; (2) 1,0 0 shares of cumulative, non-voting, 8%, $100 par value, preferred stock which it sold at $100 per share; and (3) 2,000 20-year debentures with a face value of $1,000 each, in terest at 10%, which it sold at a 5% discount. All securities were sold for cash during April 1979. At a meeting on May 15, 1979, the board of directors voted to increase the capital surplus of the corporation by transferring to capital surplus $19 per share of the $20 per share originally credited to stated capital upon the sale of the no-par common stock.
Required: Answer the following, setting forth reasons for any conclusions stated.
What is the stated capital of the Fairfax Corpora tion as of May 16, 1979?
Part b. The directors of Despard & Company, Inc., are considering several alternatives to their usual declaration of a cash dividend. The cost of borrowing money has become prohibitive and the directors would prefer to retain the cash to further the corporation’s ex pansion plans. The following possibilities have been suggested: • A dividend to each shareholder consisting of 60% treasury stock and 40% cash.
• A stock dividend declared and paid in its own authorized and unissued $1.00 par value common shares.
• A 2-for-1 split-up of the issued shares of the $1.00 par value common shares. Par value would be changed from $1.00 to $0.50.
Required: Answer the following, setting forth reasons for any conclusions stated.
1. Separately analyze and discuss the legal impact of each of these possibilities from the standpoint of the corporate requirements (ignore accounting entries) that must be met and the effect that each would have upon the stated capital of the corporation.
2. What is the federal income tax effects or im plications to the shareholder as to each of the above possibilities?
3. What is the federal income tax consequence if the corporation continuously elects not to pay any cash dividends?
Part c. The United States Justice Department commenced a criminal action against Sky Manufactur ing Corporation and its president, Masterson, for con spiring to fix prices on the sale of certain heavy in dustrial machinery. Both the corporation and Masterson denied the allegations. After a lengthy trial, the jury found that although a conspiracy did exist among certain manufacturers, neither Sky nor Masterson were parties to the illegal conspiracy. The cost to the corporation to defend the action against it was $500,000. Masterson’s individual legal fees and ex penses amounted to $250,000 of which Sky has paid $50,000 directly. Masterson seeks indemnification for the remaining $200,000.
Heinz, a dissenting shareholder of Sky, advised the board of directors that payment by the corporation of any of Masterson’s expenses was improper. In the event no action is taken to recover the $50,000 already ad vanced, Heinz will commence a shareholder derivative action against Masterson. Furthermore, unless the board unequivocally promises not to indemnify Master- son for the unpaid balance of his legal expenses, Heinz will seek injunctive relief.
Required: Answer the following, setting forth reasons for any conclusions stated.
What rights and limitations apply to Sky’s payment of Masterson’s legal fees and expenses in defending the criminal action brought against him?
197 Examination Questions— November 1979 Part a. Glasco Machinery and Manufacturing, Inc., sells industrial machinery to various customers on credit terms of 20% down and three-month promissory notes for the balance.
Glasco was experiencing severe financial difficulty and desperately needed a loan for working capital and to stave off persistent creditors. Its bank insisted upon security for any loan it might make. Glasco agreed to pledge $25,000 of its customer’s promissory notes as collateral for a $20,000 demand loan. The notes pledged included some which Glasco knew had been received on sales of defective machinery and several notes which Glasco’s president forged in anticipation of future shipments to customers.
After a short time Glasco’s president saw that detection was inevitable, withdrew all funds in the bank, and absconded with the cash. The bank is seeking to en force payment of the notes against the various parties.
Required: Answer the following, setting forth reasons for any conclusions stated.
Discuss the bank’s rights, if any, to collection on the various promissory notes.
Part b. Grover had an $80 check payable to the order of Parker that Parker had indorsed in blank. The check was drawn by Madison on State Bank. Grover deftly raised the amount to $800 and cashed it at Friend ly Check Cashing Company. Friendly promptly presented it at State Bank where it was dishonored as an overdraft. Grover has been apprehended by the police and is awaiting trial. He has no known assets. Friendly is seeking collection on the instrument against any or all of the other parties involved.
Required: Answer the following, setting forth reasons for any conclusions stated.
Will Friendly recover against Madison, State Bank, or Parker?
Number 3 (Estimated tim e----- 20 to 25 minutes) Part c. Horn Audio purchased some audio com ponents from Samuels Sounds. The high quality audio components were to be used by Horn in its expensive customized sound systems to be sold to its customers. Samuels fraudulently substituted a large number of reconditioned audio components for the new ones that Horn was shown and believed he had purchased. In pay ment of the purchase, Horn executed and delivered the following instrument to Samuels: January 8 , 1979 For value received, Horn Audio promises to pay Three Thousand Dollars ($3,000.00) to the order of Samuels Sounds, two weeks after their receipt and out of the proceeds from the resale of the audio components this day purchased from Samuels Sounds and used as major components in the customized sound systems sold to our customers. Horn Audio Samuels transferred the instrument to Wilmont for value by signing it on the back and delivering it to him. Wilmont had no knowledge of the fraudulent substitu tion of the audio components by Samuels. Several months later, Wilmont presented the instrument to the maker for payment. Horn refused to pay the instrument alleging fraud and breach of warranty. Furthermore, Horn stated that all the audio components were re turned to Samuels immediately upon discovery of the facts. Wilmont has commenced legal action against Horn on the instrument.
Required: Answer the following, setting forth reasons for any conclusions stated.
Will Wilmont prevail in his legal action against Horn on the instrument?
198 Business Law Part a. The King Surety Company, Inc., wrote a performance bond for Allie Stores, Inc., covering the construction of a department store. Rapid Construction Company, the department store contractor, is a general contractor and is simultaneously working on several buildings. Until the entire building is completed, the bond contained a provision that obligated Allie to withhold 20% of the progress payments to be made to Rapid at various stages of completion. After approx imately two-thirds of the project had been satisfactorily completed, Rapid pleaded with Allie to release the 20% withheld to date. Rapid indicated that he was having a cash flow problem and unless funds were released to satisfy the demands of suppliers, workmen, and other creditors, there would be a significant delay in the com pletion date of the department store. Rapid claimed that if the 20% withheld were released, the project could be completed on schedule. Allie released the amounts withheld. Two weeks later Rapid abandoned the pro ject, citing as its reason rising cost which made the con tract unprofitable. Allie has notified King of the facts and demands that either King complete the project or respond in damages. King denies liability on the surety bond.
Required: Answer the following, setting forth reasons for any conclusions stated.
Who will prevail?
Part b. Barclay Surety, Inc., is the surety on a construction contract that the Gilmore Construction Company made with Shadow Realty, Inc. By the terms of the surety obligation, Barclay is not only bound to Shadow, but also is bound to satisfy materialmen and laborers in connection with the contract. Gilmore defaulted, and Barclay elected to complete the project and pay all claims and obligations in connection with the contract, including all unpaid materialmen and laborers’ claims against Gilmore. The total cost to com plete exceeded the construction contract payments Barclay received from Shadow. Some of the materialmen who were satisfied had either liens or security interests against Gilmore. Gilmore has filed a voluntary bankruptcy petition.
Required: Answer the following, setting forth reasons for any conclusions stated.
What rights does Barclay have as a result of the above facts?
Number 4 (Estimated time----- 20 to 25 minutes) Part a. Marcall is a limited partner of Guarcross, a limited partnership, and is suing a CPA firm which was retained by the limited partnership to perform auditing and tax return preparation services. Guarcross was formed for the purpose of investing in a diversified portfolio of risk capital securities. The partnership agreement included the following provisions: Number 5 (Estimated time----- 20 to 25 minutes) The initial capital contribution of each limited partner shall not be less than $250,000; no partner may withdraw any part of his interest in the partnership, except at the end of any fiscal year upon giving written notice of such intention not less than 30 days prior to the end of such year; the books and records of the partnership shall be audited as of the end of the fiscal year by a certified public accountant designated by the general partners; and proper and complete books of account shall be kept and shall be open to in spection by any of the partners or his or her accredited representative.
Marcall’s claim of malpractice against the CPA firm centers on the firm’s alleged failure to comment, in its audit report, on the withdrawal by the general part ners of $2,000,000 of their $2,600,000 capital invest ment based on back-dated notices, and the lumping together of the $2,000,000 withdrawals with $49,0()0 in withdrawals by limited partners so that a reader of the financial statement would not be likely to realize that the two general partners had withdrawn a major portion of their investments.
The CPA firm’s contention is that its contract was made with the limited partnership, not its partners. It further contends that since the CPA firm had no privity of contract with the third party limited partners, the limited partners have no right of action for, negligence.
Required: Answer the following, setting forth reasons for any conclusions stated.
Discuss the various theories Marcall would rely upon in order to prevail in a lawsuit against the CPA firm.
199 Examination Questions— November 1979 Part b. Farr & Madison, CPAs, audited Glamour, Inc. Their audit was deficient in several respects: • Farr and Madison failed to verify properly cer tain receivables which later proved to be fictitious.
• With respect to other receivables, although they made a cursory check, they did not detect many ac counts which were long overdue and obviously uncol lectible.
• No physical inventory was taken of the securities claimed to be in Glamour’s possession, which in fact had been sold. Both the securities and cash received from the sales were listed on the balance sheet as assets.
There is no indication that Farr & Madison actually believed that the financial statements were false. Subse quent creditors, not known to Farr & Madison, are now suing based upon the deficiencies in the audit described above. Farr and Madison moved to dismiss the lawsuit against it on the basis that the firm did not have actual knowledge of falsity and therefore did not commit fraud.
Required: Answer the following, setting forth reasons for any conclusions stated.
May the creditors recover without demonstrating Farr & Madison had actual knowledge of falsity?
Part c. The Bigelow Corporation decided to liquidate. A board member suggested the possibility of electing a one calendar month liquidation pursuant to section 333 of the Internal Revenue Code. In order to determine whether this type of liquidation was desirable, Bigelow engaged Fanslow & Angelo, CPAs, to perform a tax analysis of the corporation’s data and figures to ascertain the amount of dividend per share that would be taxable as dividend income to the shareholders if this method of liquidation were elected. Such a determination is largely dependent on the amount of earnings and profits present, both current and historical.
In making the computation, Fanslow and Angelo treated retained earnings as stated in the financial statements as earnings and profits for tax purposes. However, on two prior occasions transfers were made from retained earnings to stated capital upon the is suance of stock dividends. The result of failure to adjust earnings and profits to reflect these transfers was to understate the amount of taxable dividend income per share by some $20 per share.
Required: Answer the following, setting forth reasons for any conclusions stated.
Do Fanslow & Angelo have any liability under the above-stated facts?
200 (Prepared by the Board of Examiners of the American Institute of Certified Public Accountants and adopted by the examining boards of all states, territories, and the District of Columbia.) Uniform Certified Public Accountant Examination EXAMINATION IN ACCOUNTING THEORY (Theory of Accounts) November 9, 1979; 1:30 to 5:00 P.M.
NOTE TO CANDIDATES: Suggested time allotments are as follows: All questions are required: Estimated Minutes Minimum Maximum No. I ........................................................................ 90 No. 2 ........................................................................................... 20 No. 3 ........................................................................................... 15 No. 4 ........................................................................................... 15 No. 5 ........................................................................................... 15 No. 6 ...........................................................................................
Total.................................................................................... 170 105 25 20 20 20 20 210 INSTRUCTIONS TO CANDIDATES (Disregard of these instructions may be considered as indicating inefficiency in accounting work.) 1. You must arrange the papers in numerical order of the questions. If more than one page is required for an answer, write “continued” at the bottom of the page. Number pages consecutively. For instance, if 12 pages are used for your answers, they should be numbered 1 through 12. The printed answer sheet provided for the objective-type items should be considered to be Page 1.
2. Answer all objective-type items on the printed answer sheet provided for that purpose. It is to your advantage to attempt all questions even if you are uncertain of the answer. You are likely to get the highest score if you omit no answers.
3. A CPA is continually confronted with the necessity of expressing opinions and conclusions in written reports in clear, unequivocal language. Although the primary purpose of the examination is to test the candidate’s knowledge and application of the subject matter, the ability to organize and present such knowledge in acceptable written language will be considered by the examiners.
201 Examination Questions— November 1979 Instructions Select the best answer for each of the following items relating to a variety of issues in accounting. Use a soft pencil, preferably No. 2, to blacken the appropriate circle on the separate printed answer sheet to indicate your answer. Mark only one answer for each item. Answer all items. Your grade will be based on your total correct answers.
The following is an example of the manner in which the answer sheet should be marked: Item 99. The financial statement which summarizes the results of operations is the a. Income statement.
b. Balance sheet.
c. Statement of changes in financial position.
d. Retained earnings statement.
Number 1 (Estimated time----- 90 to 105 minutes) 3. Information prior to the issuance of the financial statements indicates that it is probable that, at the date of the financial statements, a liability has been incurred for obligations related to product warranties. The amount of the loss involved can be reasonably estimated. Based on the above facts, an estimated loss contingency should be a. Accrued.
b. Disclosed but not accrued.
c. Neither accrued nor disclosed.
d. Classified as an appropriation of retained earnings.
An accrued expense is an expense a. Incurred but not paid.
b. Incurred and paid.
c. Paid but not incurred.
d. Not reasonably estimable.
Answer Sheet 99. Items to be Answered 1. When the interest payment dates of a bond are May 1 and November 1, and a bond issue is sold on June 1, the amount of cash received by the issuer will be a. Decreased by accrued interest from June I to November 1.
b. Decreased by accrued interest from May 1 to June 1.
c. Increased by accrued interest from June 1 to November 1.
d. Increased by accrued interest from May 1 to June 1.
2. Abbot Co. is being sued for illness caused to local residents as a result of negligence on the company’s part in permitting the local residents to be exposed to highly toxic chemicals from its plant. Abbot’s lawyer states that it is probable that Abbot will lose the suit and be found liable for a judgment costing Abbot anywhere from $500,000 to $2,500,000. However, the lawyer states that the most probable cost is $1,000,000. As a result of the above facts, Abbot should accrue a. A loss contingency of $500,000 and disclose an additional contingency of up to $2,000,000.
b. A loss contingency of $1,000,000 and disclose an additional contingency of up to $1,500,000.
c. A loss contingency of $1,000,000 but not disclose any additional contingency.
d. No loss contingency but disclose a contin gency of $500,000 to $2,500,000.
5. Under what condition is it proper to recognize revenues prior to the sale of the merchandise?
a. When the ultimate sale of the goods is at an assured sales price.
b. When the revenue is to be reported as an in stallment sale.
c. When the concept of internal consistency (of amounts of revenue) must be complied with.
d. When management has a long-established policy to do so.
6. When treasury stock is purchased for more than the par value of the stock and the cost method is used to account for treasury stock, what account(s) should be debited?
a. Treasury stock for the par value and addi tional paid-in capital for the excess of the pur chase price over the par value.
b. Additional paid-in capital for the purchase price.
c. Treasury stock for the purchase price.
d. Treasury stock for the par value and retained earnings for the excess of the purchase price over the par value.
7. At the date of the financial statements, common stock shares issued would exceed common stock shares outstanding as a result of the a. Declaration of a stock split.
b. Declaration of a stock dividend.
c. Purchase of treasury stock.
d. Payment in full of subscribed stock.
202 Accounting Theory 8. The vested benefits of an employee in a pension plan represent a. Benefits to be paid to the retired employee in the current year.
b. Benefits to be paid to the retired employee in the subsequent year.
c. Benefits accumulated in the hands of an in dependent trustee.
d. Benefits that are not contingent on the employee’s continuing in the service of the employer.
9. Which of the following principles best describes the current method of accounting for research and development costs?
a. Associating cause and effect.
b. Systematic and rational allocation.
c. Income tax minimization.
d. Immediate recognition as an expense.
10. At what translation rates should the following balance sheet accounts in foreign statements be translated into United States dollars?
Equipment Accumulated Depreciation of Equipment a.
Current Current b.
Current Average for year c.
Historical Current d.
Historical Historical 14. Rent revenue collected one month in advance should be accounted for as a. Revenue in the month collected.
b. A current liability.
c. A separate item in stockholders’ equity.
d. An accrued liability.
15. The percentage-of-completion method of ac counting for long-term construction-type contracts is preferable when a. Estimates of costs to complete and extent of progress toward completion are reasonably dependable.
b. The collectibility of progress billings from the customer is reasonably assured.
c. A contractor is involved in numerous pro jects.
d. The contracts are of a relatively short dura tion.
16. In order to be classified as an extraordinary item in the income statement, an event or transaction should be a. Infrequent and material; but it need not be unusual in nature.
b. Unusual in nature and material; but it need not be infrequent.
c. Unusual in nature, infrequent, and material.
d. Unusual in nature and infrequent; but it need not be material.
11. In accounting for a long-term construction-type contract using the percentage-of-completion method, the gross profit recognized during the first year would be the estimated total gross profit from the contract multiplied by the percentage of the costs incurred during the year to the a. Total costs incurred to date.
b. Total estimated cost.
c. Unbilled portion of the contract price.
d. Total contract price.
12. In a lease that is appropriately recorded as a direct financing lease by the lessor, unearned income a. Should be amortized over the period of the lease using the interest method.
b. Should be amortized over the period of the lease using the straight-line method.
c. Does not arise.
d. Should be recognized at the lease’s expiration.
13. When bad debt expense is estimated on the basis of the percentage of past actual losses from bad debts to past net credit sales, and this percentage is adjusted for anticipated conditions, the accounting concept of a. Matching is being followed.
b. Matching is not being followed.
c. Substance over form is being followed.
d. Going concern is not being followed.
17. Which of the following requires intraperiod tax allocation?
a. Extraordinary gains or losses as defined by the Accounting Principles Board.
That portion of dividends reduced by the dividends received deduction by corporations under existing federal income tax law.
The excess of accelerated depreciation used for tax purposes over straight-line deprecia tion used for financial reporting purposes. All differences between taxable income and financial statement earnings.
b.
c.
18.
the One of the basic features of financial accounting is a. Direct measurement of economic resources and obligations and changes in them in terms of money and sociological and psychological impact.
b. Direct measurement of economic resources and obligations and changes in them in terms of money.
c. Direct measurement of economic resources and obligations and changes in them in terms of money and sociological impact.
d. Direct measurement of economic resources and obligations and changes in them in terms of money and psychological impact.
203 Examination Questions— November 1979 19. Which of the following is an example of the con cept of conservatism?
a. Stating inventories at the lower of cost or market.
b. Stating inventories using the FIFO method in periods of rising prices.
c. Using the percentage-of-completion method in the first year of a long-term construction contract.
d. Using the interest method instead of the straight-line method to record interest in the first year of a long-term receivable.
20. The past service costs in a pension plan a. Should be charged to income in the year of the inception of the pension plan.
b. Should be funded in the year of the inception of the pension plan.
c. Represent pension cost assigned to years prior to the current balance sheet date.
d. Represent pension cost assigned to years prior to the inception of the pension plan.
21. APB Opinion No. 8 sets minimum and maximum limits on the annual provision for pension cost. An amount that is always included in the calculation of both the minimum and maximum limit is a. Amortization of past service cost.
b. Normal cost.
c. Interest on unfunded past and prior service costs.
d. Retirement benefits paid.
22. Which of the following is the appropriate basis for valuing fixed assets acquired in a business combination accounted for as a purchase carried out by exchanging cash for common stock?
a. Fair value.
b. Book value.
c. Cost plus any excess of purchase price over book value of asset acquired.
d. Historic cost.
23. Conditions warranted that a company have a quasi-reorganization. Immediately after the quasi reorganization, the retained earnings account a. Has a zero balance.
b. Remains the same as it was before the quasi reorganization.
c. Is frozen and dated, and subsequent transactions will be shown separately.
d. Has a debit balance equal to the write-down of the assets which were overstated.
24. In current value (fair value) financial statements a. General price-level gains or losses are recognized on net monetary items.
b. Amounts are always stated in common pur chasing power units of measurements.
c. All balance sheet items are different in amount than they would be in a historical-cost balance sheet.
d. Holding gains are recognized.
25. The profitability information that should be reported for each reportable segment of a business enterprise consists of a. An operating profit or loss figure consisting of segment revenues less traceable costs and allocated common costs.
An operating profit or loss figure consisting of segment revenues less traceable costs but not allocated common costs.
An operating profit or loss figure consisting of segment revenues less allocated common costs but not traceable costs.
Segment revenues only.
26. If all other conditions for consolidation are met, how should subsidiaries acquired in a business combina tion be shown under each of the following methods?
b.
c.
d.
Purchase Pooling o f Interests a.
Consolidated Not Consolidated b.
Consolidated Consolidated c.
Not Consolidated Consolidated d.
Not Consolidated Not Consolidated 27. Continuation of an accounting entity in the absence of evidence to the contrary is an example of the basic concept of a. Accounting entity.
b. Consistency.
c. Going concern.
d. Substance over form.
28. Presenting consolidated financial statements this year when statements of individual companies were presented last year is a. A correction of an error.
b. An accounting change that should be reported prospectively.
c. An accounting change that should be reported by restating the financial statements of all prior periods presented.
d. Not an accounting change.
29. When computing primary earnings per share, common stock equivalents are a. Ignored.
b. Recognized whether they are dilutive or anti dilutive.
c. Recognized only if they are anti-dilutive.
d. Recognized only if they are dilutive.
204 Accounting Theory 30. When the market value of a company’s current marketable equity securities portfolio is lower than its cost, the difference should be a. Accounted for as a liability.
b. Disclosed and described in a footnote to the financial statements but not accounted for.
c. Accounted for as a valuation allowance deducted from the asset to which it relates.
d. Accounted for separately in the shareholders’ equity section of the balance sheet.
31. If the fixed costs for a product decrease and the variable costs (as a percentage of sales dollars) decrease, what will be the effect on the contribution margin ratio and the breakeven point, respectively?
Contribution Margin Ratio Breakeven Point a.
Decreased Increased b.
Increased Decreased c.
Decreased Decreased d.
Increased Increased 32. In deciding whether to manufacture a part or buy it from an outside vendor, a cost that is irrelevant to the short-run decision is a. Direct labor.
b. Variable overhead.
c. Fixed overhead that will be avoided if the part is bought from an outside vendor.
d. Fixed overhead that will continue even if the part is bought from an outside vendor.
33. One of the accepted methods of accounting for a by-product is to recognize the value of the by-product as it is produced. Under this method, inventory costs for the by-product would be based on a. An allocation of some portion of joint costs but not any subsequent processing costs.
b. Neither an allocation of some portion of joint costs nor any subsequent processing costs. Subsequent processing costs less an allocation of some portion of joint costs.
An allocation of some portion of joint costs plus any subsequent processing costs.
c.
d.
34. If the amount of spoilage in a manufacturing pro cess is abnormal, it should be classified as a a. Deferred charge.
b. Joint cost.
c. Period cost.
d. Product cost.
35. Factory overhead includes a. All manufacturing costs.
b. All manufacturing costs, except direct materials and direct labor.
c. Indirect materials but not indirect labor.
d. Indirect labor but not indirect materials.
36. A company buys a certain part for its manufactur ing process. In order to determine the optimum size of a normal purchase order, the formula for the economic order quantity (EOQ) is used. In addition to the annual demand, what other information is necessary to com plete the formula?
a. Cost of placing an order, and annual cost of carrying a unit in stock.
b. Cost of the part, and annual cost of carrying a unit in stock.
c. Cost of placing an order.
d. Cost of the part.
37. Simple regression analysis involves the use of a. One variable.
b. Two variables.
c. Three variables.
d. More than three variables.
38. Program evaluation review technique (PERT) is a system which uses a. Least squares method.
b. Linear programming.
c. Economic order quantity formula.
d. Network analysis and critical path methods.
39. On May 1, 1979, a company sold some machinery to another company. The two companies entered into an installment sales contract at a predetermined interest rate. The contract required five equal annual payments with the first payment due on May 1, 1979. What pre sent value concept is appropriate for this situation?
a. Present value of an annuity due of $1 for five periods.
b. Present value of an ordinary annuity of $1 for five periods.
c. Future amount of an annuity of $1 for five periods.
d. Future amount of $1 for five periods.
40. A company manufactures two models, X and Y. Model X is processed 4 hours in the machining depart ment and 2 hours in the polishing department. Model Y is processed 9 hours in the machining department and 6 hours in the polishing department. The available time for processing the two models is 200 hours a week in the machining department and 180 hours a week in the polishing department. The contribution margins ex pected are $10 for Model X and $14 for Model Y. How would the restriction (constraint) for the polishing department be expressed?
a. 2X + 6Y ≤ 180.
b. 6X + 15Y≤180.
c. 2(10X) + 6(14Y) ≤ 180.
d. 10X 14Y≤ 180.
205 Examination Questions— November 1979 41. Repairs that have been made for a governmental unit, and for which a bill has been received, should be recorded in the General Fund as an a.
Appropriation.
b.
Encumbrance.
c.
Expenditure.
d.
Expense.
42. Under the modified accrual basis of accounting, which of the following taxes is usually recorded before it is received in cash?
a. Property.
b. Income.
c. Gross receipts.
d. Gift.
49. When a truck is received by a governmental unit, it should be recorded in the General Fund as a (an) a. Appropriation.
b. Encumbrance.
c. Expenditure.
d. Fixed asset.
50. How should wages that have been earned by the employees of a governmental unit, but not paid, be recorded in the General Fund?
a. Appropriation, b. Encumbrance.
c. Expenditure.
d. Expense.
43. Which fund is not an expendable fund?
a. Capital projects.
b. General.
c. Special revenue.
d. Intragovernmental service.
44. Encumbrances would not appear in which fund?
a. Capital projects.
b. Special revenue.
c. General.
d. Enterprise.
45. Interest expense on bonds payable should be recorded in a Debt Service Fund a. At the end of the fiscal period if the interest due date does not coincide with the end of the fiscal period.
b. When bonds are issued.
c. When legally payable.
d. When paid.
46. An expenditures account appears in a. The general fixed-assets group of accounts.
b. The general long-term debt group of ac counts.
c. A special revenue fund.
d. An intragovernmental service fund.
47. Which of the following steps in the acquisition of goods and services occurs first?
a. Appropriation.
b. Encumbrance.
c. Budget.
d. Expenditure.
48. Which of the following terms refers to an actual cost rather than an estimate?
a. Expenditure.
b. Appropriation.
c. Budget.
d. Encumbrance.
Number 2 (Estimated time-----20 to 25 minutes) Part a. When a company has a policy of making sales for which credit is extended, it is reasonable to expect a portion of those sales to be uncollectible. As a result of this, a company must recognize bad debt expense. There are basically two methods of recognizing bad debt expense: (1) direct write-off method, and (2) allowance method.
Required: 1. Describe fully both the direct write-off method and the allowance method of recognizing bad debt ex pense.
2. Discuss the reasons why one of the above methods is preferable to the other and the reasons why the other method is not usually in accordance with generally accepted accounting principles.
Part b. For numerous reasons a corporation may reacquire shares of its own capital stock. When a com pany purchases treasury stock, it has two options as to how to account for the shares: (1) cost method, and (2) par value method.
Required: Compare and contrast the cost method with the par value method for each of the following: Purchase of shares at a price less than par Purchase of shares at a price greater than par 1.
value.
2.
value.
3. Subsequent resale of treasury shares at a price less than purchase price, but more than par value.
4. Subsequent resale of treasury shares at a price greater than both purchase price and par value.
5. Effect on net income.
206 Accounting Theory Number 3 (Estimated time 15 to 20 minutes) Generally accepted accounting principles require the use of accruals and deferrals in the determination of income.
Required: a. How does accrual accounting affect the deter mination of income? Include in your discussion what constitutes an accrual and a deferral, and give ap propriate examples of each.
b. Contrast accrual accounting with cash accoun ting.
Number 4 (Estimated time-----15 to 20 minutes) Part a. Deferred income taxes are required under generally accepted accounting principles. Accounting Principles Board Opinion No. 11 requires the use of the deferred method of comprehensive interperiod tax allocation. Two ways to account for timing differences under the deferred method are: (1) gross change method, and (2) net change method.
Required: 1. Describe the gross change method.
2. Describe the net change method.
Part b. Timely financial information is important to users of financial statements. As a result, many com panies produce financial information more frequently than annually.
Required: 1. How are revenues, costs and expenses recognized for interim reporting related to those recognized for year-end reporting?
2. How are income taxes recognized at interim dates?
When a business combination is effected by an ex change of common stock, the transaction is accounted for as a purchase or as a pooling of interests, depending on the circumstances. The methods are not optional and each yields significantly different results as to financial position and results of operations.
Required: Discuss the supportive arguments for each of the following: a. Purchase method.
b. Pooling of interests method.
Do not discuss in your answer the rules for distinguishing between a purchase and a pooling of in terests.
Number 6 (Estimated time-----15 to 20 minutes) Part a. Although direct costing is not a current generally accepted method of costing inventory for ex ternal reporting, it is useful for internal purposes.
Required: 1. Describe the difference between direct costing and the current generally accepted method of costing in ventory for external reporting.
2. Describe how a direct costing structure facilitates calculation of the contribution margin and the breakeven point.
Part b. The success of a standard cost system depends on the reliability, accuracy, and acceptance of the standards. Variances from the standards must be analyzed to ascertain why they occurred.
Required: 1. Describe the two standards that must be developed for materials cost and how the variances (both favorable and unfavorable) from these standards are calculated.
2. Describe the two standards that must be developed for labor cost and how the variances (both favorable and unfavorable) from these standards are calculated.
Number 5 (Estimated time----- 15 to 20 minutes) 207 Index HOW TO USE THIS INDEX: This index includes an examination and question number reference in addition to the normal page reference. The examination and question number reference is in parentheses immediately preceding the page reference (e.g., (M78L-3)). The initial letter of the reference identifies the May or November examination, the next two numbers identify the year and are followed by the section of the examination identified as follows: A—Auditing, L—Business Law, T—Theory, PI—Accounting Practice-Part I, and PII— Accounting Practice-Part II. The first number after the dash identifies the question number. In the case of multiple choice questions this is followed by the item and an “m” which means multiple choice.
Accountant’s legal liability Audit Auditor of statements accompanying registrations statement is required to review events between date of report and date of public sale or will be liable (M78L-3) 42 Brokerage firm—customers might have a common law action for negligence of the CPA or might prove liability under sec.
17 of Securities Exchange Act of 1934 (N78L-4) 95 Creditors of a bankrupt can recover from a CPA firm that issued an unqualified opinion on statements that had material inadequate disclosures relative to matters that eventually caused the bankruptcy (M78L-3) 42 Embezzlement—best defense is that the shortages were result of clever forgeries and collusive fraud which would not be detected by examination in accord with GAAS (N78L-l-3m) 86 Fraud Auditor liable to creditor for issuance of false statements (knowingly presenting a capital lease as an operating lease) (M79L-2) 143 Auditor not liable to lessor whose own fraudulent actions led to his loss—secret agreement with lessee so that capital lease might appear to be operating lease (M79L-2) 143 Gross negligence—creditors may recover from CPA firm without showing actual knowledge of falsity; scienter requirement for action by third parties based on fraud satisfied by either reckless disregard for truth or gross negligence, both of which can be shown by failure to check existence of certain receivables, collectibility of others, existence of security investments, and lack of due care (N79L-5) 200 Limited partnership audit—privity in state of change and gradual erosion—in order to prevail against a CPA firm, a limited partner might assert third-party beneficiary doctrine that the duty of care extends to fixed, definable, and contemplated group whose conduct was to be governed by the audit, gross negligence, and general partners signing contract for each member of limited partnership (N79L-5) 199 Negligence See Gross negligence above Auditor liable to creditor for material misstatement of statements (capital lease presented as operating lease) (M79L-2) 143 CPA firm engaged to perform annual audit must not only exercise reasonable care, but also must possess at least that degree of accounting knowledge and skill expected of a CPA (N78L-l-2m) 86 Creditor can recover loss from auditor when engagement letter expressly indicated audit was for loan purposes, and an unqualified opinion was given in spite of facts that inventory was stated at cost (substantially in excess of fair market value), and that two secured creditors to whom material amounts were owed were not disclosed though their claims would take priority over the loan (M79L-2) 143 Accountant’s legal liability (cont.) Privileged communication Law does not recognize its existence between accountant and client in absence of specific statutory provision (N78L-l-4m) 86 Securities Act of 1933 Any materially false or misleading financial statements, including omission of material fact, provided purchaser proves such existed and the specific securities were the ones offered through the registration statement (N79L-l-29m) 192 Unqualified opinion on incorrect statements—best defense of four listed against third party purchaser of securities sold in interstate commerce, action had not been commenced within one year after the discovery of the material misrepresentation (N78L-1-11m) 87 Securities Exchange Act of 1934— antifraud provisions of Sec. KKB) and Rule 10b-5 do not apply in customer suit against negligent CPA audit, because fraud is intentional and requires more than showing negligence (N78L-4) 95 Term “audit” should not be used on nonaudit engagements (M78A-2) 29 Working papers subpoenaed must be surrendered by CPA firm— even where privilege rule is applicable, it can only be claimed by client (M78L-3) 42 Tax preparation and analysis Tax analysis to determine taxable dividend per share if liquidate—treated retained earnings as in statements as earnings and profits for tax purposes without consideration of transfers on two occasions for stock dividends and negligent CPA firm would be liable for any interest and penalties imposed on shareholders and may be held liable for additional taxes incurred by shareholders as result of erroneous advice (N79L-5) 200 Tax return preparation Common examples of negligence (M79L-2) 74 Damages awarded client may include penalties assessed, interest assessed, no-longer-recoverable taxes erroneously paid, other costs directly resulting, and punitive damages (M79L-2) 143 Principal legal basis for liability of the firm and its employee (agent) (M79L-2) 143 Accountant’s report Review of statements of a nonpublic entity Includes statement that all information in statements are representations of management, a statement describing principal procedures performed, and a statement describing conclusions based on the review, but not a statement that the review was in accord with GAAS (N79A-l-39m) 182 Accounting changes Change in accounting principle Book value of machine computation one year after changing from straight-line to sum-of-the-years-digits 209 A Accounting changes Change in accounting principle (cont.) depreciation (M79PI-3) 113 Conflicts with comparability (N78T-l-5m) 98 Cumulative effect of change from double-declining-balance depreciation to straight-line, given the total amount of depreciation for prior years under each method and the income tax rate (M79PI-1-lm) 106 Cumulative effect of change from straight-line to sum-of-the- years-digits depreciation (M79PI-3) 113 Cumulative effect of change in depreciation method for all fixed assets reflected in current statements, comparative statements presented as before, recompute beginning retained earnings, effect on depreciation and accumulated depreciation, disclose effect for current period and on pro forma basis for all comparative statements presented, effect in each instance disclosed for income before extraordinary, net income, and all related per-share amounts, and footnote disclosure of nature and justification (M78T-7) 51 Depreciation Computation of cumulative effect of change from accelerated to straight-line (M78PI-1-17m) 5 Double-declining-balance to straight-line— cumulative effect computation (M78PII-1-11m) 15 Inventory— computation of net income resulting from change from FIFO to LIFO (M78PII-1-7m) 14 Change in estimate Consistency standard does not apply (M78A-1-18m) 24 Depreciable lives are reflected in current and future periods, no cumulative effect shown or pro forma earnings data, affects depreciation and accumulated depreciation, footnote disclosure of effect on income before extraordinary, net income, and related per-share amounts (M78T-7) 51 No cumulative effect on prior years reported (N79PI-1-lm) 156 No cumulative effect on prior years shown on income statement (N78PI-1-3m) 54 Record in current period and prospectively (N78T-1-20m) 100 Remaining book value depreciated over remaining life, computation of first-year depreciation (N79PI-1-2m) 156 Change in reporting entity Actually a change in accounting principle (M79T-7m) 147 Business combination is made using pooling of interests (M78T-1-6m) 45 Presenting consolidated financial statements this year when statements of individual companies were presented last year (N79T-1-28m) 204 Restate all prior-period statements, in initial set of statements after change disclose by footnote nature and reason, effect on consolidated statements, effect on earnings before extraordinary, net earnings, and related per-share amounts disclosed for all periods presented, disclosures not repeated subsequently (M78T-7) 51 Consistency Change in estimate— standard does not apply (M78A-1-18m) 24 Consistency may be violated (N78T-1-4m) 98 Consistency standard applies to change in principle not generally accepted, change in reporting entity, and change in principle inseparable from a change in estimate (M78A-1-18m) 24 Accounting Principles Board (N78T-6) 103 Accounting Principles Board Statements See APB Statements Accounts payable See also Audit procedures Auditor’s primary concern is that payables are not materially understated (N78A-1-12m) 76 Accounts receivable See also Allowance for doubtful accounts See also Allowance for uncollectible accounts See also Internal control Aging Used to estimate credit losses (M79A-1-54m) 130 Allowance adjusting entry as result in loss experience rate (N78PII-3) 71 Allowance for uncollectible accounts Computation of balance given beginning balance, write-offs, sales and percent of sales uncollectible (M78PII-1-9m) 15 Assigned Computation of expenses resulting (M79PI-4) 113 Cash collections estimate during month computation, given estimated credit sales for each of current and previous month, estimated collection percentage for each of the two months, and estimated collections in current month for credit sales prior to previous month (N79PI-2-22m) 160 Confirmation Generally accepted auditing procedure (rather than principle, standard, or practice) (M78A-1-58m) 29 Lapping Payments sent directly to depository bank is best of given procedures to prevent lapping (M78A-1-9m) 24 Large debits and small periodic credits might suggest consignments (N79A-1-57m) 184 Net realizable value would be balance of account less aged estimate of uncollectibles (N78PI-1-16m) 68 Sold Computation of expense (M79PI-4) 113 207 Accrual accounting Contrasted with cash accounting (N79T-3) Going concern concept justifies accruals and deferrals (M78T-1-7m) 45 How it affects the determination of income, including what constitutes an accrual and a deferral with examples of each (N79T-3) 207 Incentive commission expense computation involving addition of amounts applicable to particular year ignoring item accrued in previous year (N78PII-1-2m) 65 Accrued expense Incurred but not paid (N79T-1-4m) Not paid but are currently matched with earnings (M78T-1-17m) 46 202 Accrued interest Collected on issuance of bonds between interest dates (N79T-1-lm) 202 Accumulated depreciation Computation of amount taken under double-declining-balance method over eight-and-one-half years of machine with expected useful life of ten years, amount which should be removed from accounts as machine was scrapped and utilized in determining book value (N79PII-3) 174 Computation of amount taken under straight-line method over three- and-one-half years of machine with expected useful life of seven years, amount which should be removed from accounts as machine was sold and utilized in determining book value (N79PII-3) 174 Accounting Research Bulletins See ARBs Accounting Research Studies See AICPA Accounting Research Studies Advertising expense Expenditures in August promoting subscriptions to skiing magazine with only two issues (October and November) should all be expensed on the quarterly income statement for the three months ended with December (N79PI-1-llm) 158 210 Agency Agent violated express specific limit on authority If principal rightfully disaffirms the contract, agent is personally liable to seller (N79L-1-33m) 193 Authority exceeded Defense of seller who knew is principal effectively ratified purchase by authorizing processing (M79L-1-40m) 141 Customer No liability for agent of seller having auto accident taking personnel director of customer to softball game (N78L-3) 95 D el credere Agent liable to principal in event agent’s customers fail to pay principal (M78L-1-4m) 32 Delegation by agent of agent’s duties Emergency and delegation necessary to meet the emergency (M78L-1-5m) 32 Duties of principal to agent Indemnify agent for liability for acts done in good faith on principal’s orders (M78L-1-3m) 32 Employee liable for his torts Truck driver liable for fist fight when truck was double-parked in course of his job, despite fact employer may also be liable (N78L-3) 94 Employer liable for employee’s accident Employee, apparently following instructions of employer to accommodate customer where it would cost little and would build goodwill, has auto accident taking customer to softball game per customer’s request, and employer is liable (N78L-3) 95 Employer liable for employee’s torts Employer of truck driver probably liable for fist fight of driver when truck was double-parked in course of his job, despite fact that employer is in no way personally at fault or has instructed driver to obey all traffic and parking rules and regulations (N78L-3) 94 Liable for damages from employee’s disregard of instructions irrespective of efforts to prevent or employer’s reasonable care (M78L-1-lm) 32 Fiduciary duty owed by agent to his principal (M79L-1-45m) 142 Liability of principal Purchasing agent went out of usual district and bought at prevailing price, which was above price directed by principal (M78L-1-10m) 34 Undisclosed principal Customer may hold either principal or agent liable (M78L-1-2m) 32 Either liable on contracts for purchase of land within scope of the agency, including land purchased by mistake of agent (N79L-1-31m) 193 Nonexistent principal Promoter of corporation to be organized liable as a party to the contract (M78L-4) 42 Termination Principal has power to terminate even if he and agent had agreed agency was irrevocable, even if agent was appointed pursuant to a power of attorney, and even if agency agreement is in writing and provides for specific duration not elapsed, but does not have such power if agent is agent coupled with an interest (N79L-1-50m) 197 AICPA Accounting Research Studies Ethics does not require compliance (M78A-1-14m) 24 AICPA Code of Professional Ethics Accountant’s working papers not property of client Extensive analysis of inventory prepared by client at auditor’s request are workpapers which belong to auditor and need not be furnished to client upon request (N79A-1-40m) 182 Attestation in public advertising Auditor may permit client to present only the balance sheet along with an appropriately modified auditor’s report in a AICPA Code of Professional Ethics Attestation in public advertising (cont.) newspaper advertisement— CPA has already reported on the basic statements sent to stockholders (N79A-1-8m) 179 Confidential client information Must refrain from disclosing where a major stockholder of client company seeks accounting information from the CPA after management declined to disclose (N78A-1-30m) 78 Not bound by ethics to refrain from disclosing where court orders CPA to present, where made available as part of a peer review authorized by the AICPA, or where a disciplinary body of a state CPA society requests confidential client information (N78A-1-30m) 78 Contingent fees Professional services engagement should not have been accepted on contingent fee arrangement (M78A-2) 29 Discreditable acts Retention of client records as a means of enforcing payment of overdue fee (M78A-1-38m) 27 Employee (internal auditor) not in public practice For internal purposes examines and reports on potential acquisition and should sign report, Rogers, Internal Auditor, rather than Rogers, CPA, or Rogers CPA (Internal Auditor), or Rogers, Internal Auditor (CPA) (N78A-1-54m) 81 Fees Violated if payable after a specified finding was attained, but not violated if a fee is fixed by a public authority, based on a price quotation submitted in competitive bidding, or if based on the results of judicial proceedings (N78A-1-40m) 79 Firm name No violation where named Arthur, Barry, and Clark, CPAs (Clark died about five years ago; Arthur and Barry are continuing the firm), or where named Dave and Edwards, CPAs (the name of Fredricks, CPA, a third active partner, is omitted from the firm name), or where named Jones & Co., CPAs, P.C. (the firm is a professional corporation and has ten other stockholders who are all CPAs) (N78A-1-27m) 78 Violation where named George and Howard, CPAs (Howard died three years ago; George is continuing the firm as a sole proprietorship) (N78A-1-27m) 78 Generally accepted accounting principles Required compliance with APB Opinions, FASB Interpretations, and ARBs, but not with AICPA Accounting Research Studies (M78A-1-14m) 24 Publications that qualify as statements of generally accepted accounting principles include accounting interpretations issued by FASB, AICPA Accounting Research Bulletins, and Statements of Financial Standards issued by the FASB, but not accounting interpretations issued by the AICPA (N78A-1-33m) 79 Solicitation Direct, uninvited solicitation of specific potential client prohibited (M78A-1-53m) 28 Allowance for doubtful accounts See also Allowance for uncollectible accounts Balance of account computation (M79PI-4) 113 Expense previously accrued should be increased by a computed amount as a result of the allowance shown as needed by aging accounts, given beginning balance of allowance, usual percentage of sales added at end of year, charged during year, and aging result (M79PII-1-12m) 118 Allowance for uncollectible accounts See also Allowance for doubtful accounts Adjusting entry as result of change in loss experience rate (N78PII-3) 71 Computation of balance given beginning balance, write-offs, sales and percent of sales uncollectible (M78PII-1-9m) 15 American Institute of Certified Public Accountants (M79A-1-28m) 128, 211 American Institute of Certified Public Accountants (cont.) (M79A-1-35m) 128, (N79A-1-15m) 179 Amortization Goodwill matched on basis other than cause and effect (M78T-1-20m) 47 Schedule for first two years of bond discount on serial bonds using interest (effective rate) method (N78PI-3) 60 Straight-line normally recommended for intangibles (M78T-1-12m) 46 Analytical review See also Audit procedures Substantive tests of finsmcial information by study and comparison of relationships among data (N79A-1-50m) 183 Antitrust Acquisition Acquisition of another corporation that acquiring corporation does not currently buy from, sell to, or compete with, likely to be declared illegal if there will be reciprocal buying and there is likelihood that other entrants into the market would be precluded (N79L-1-36m) 194 Acquisition substantially lessens competition in tri-state area (M79L-1-41m) 142 Court would grant injunction, if Justice Department sought one, where manufacturer with 35% of market attempts to acquire controlling interest in competitor with 40% of market (N78L-1-5m) 87 Boycott Joint—illegal per se where manufacturers agreed not to sell their products to price-cutting dealer (N78L-1-lm) 81 Conspiring with major competitor to fix prices Sanctions include treble damages, seizure of property illegally shipped in interstate commerce, and fines against corporation and fines and imprisonment of its officers, but not suspension of corporate right to engage in interstate commerce (N79L-1-21m) 191 Corporations ceased marketing in geographical territories where more advantageous for the other (M78L-1-7m) 33 Illegalities Include maintaining the resale price for far superior machines sold to distributors, requiring purchasers of the far superior machine to buy front seller all their other commonplace supplies connected with use of that machine, and boycotting with other appliance manufacturers to eliminate a troublesome discounter, but not obtaining a near total monopolization of the market as a result of the patent on the far superior machine (N79L-1-35m) 194 Include price-fixing with competitors, brokerage commissions to purchasers of goods, and participation in a plan suggested by the trade association aimed at territorial allocations of markets to cut costs, but not charging price aimed at maximizing profits based upon economic analysis of supply and demand for its profits (N79L-1-34m) 194 Interlocking directorate violation (M78L-1-7m) 33 Joint boycott of troublesome discounter is illegal per se (N79L-1-42m) 195 Minimum price-resale-maintenance agreements, though reasonable and willingly accepted by distributors, subject to the per se illegality rule (N78L-1-6m) 87 Monopolizing off-shore drilling business Best defense is that defending company’s market share is such that it does not have the power to fix prices or to exclude competitors (N79L-1-37m) 194 Monopoly Company with patented product that is a major breakthrough may sell the product at whatever the traffic will bear even though it has a monopoly (M79L-1-46m) 142 Price discrimination Defense that lower prices to wholesalers and jobbers than to retailers because the discounts are functional is likely to prevail (M78L-1-9m) 33 Antitrust Price discrimination (cont.) Harmful to company when its competitor purchases from same manufacturer at lower price (M79L-1-42m) 142 Price fixing by retailers Illegal even though price is reasonable (M78L-1-12m) 34 Sanctions, penalties, or liabilities to which a corporation and its officer may be subject upon being found guilty of widespread price fixing with competitors include fines and imprisonment imposed against principal officers responsible, fines imposed on the corporation, and treble damages to third parties, but not prohibition from engaging in sale of product in interstate commerce for not more than six months (N78L-1-12m) 88 Tying arrangement illegal Requiring purchasers to buy other products to obtain the product over which seller has almost complete market control (N79L-1-6m) 188 APB Opinions Ethics requires compliance (M78A-1-14m) 24 No. 8 (M78T-4) 50, (N78Pn-1-14m) 67, (N78PII-1-15m) 67, (N78T-1-16m) 99, (M79Pn-1-l8m) 118, (M79T-1-8m) 147, (M79T-2) 151, (N79T-1-8m) 203, (N79T-1-20m) 204, (N79T-1-21m) 204 No. 9 (M79T-2) 152 No. 1 1 (M78T-1-24m) 47, (N78T-1-14m) 99, (M79PI-1-2m) 106, (M79T-1-9m) 148, (M79T-1-14m) 148, (M79T-2) 151, (N79PI-1-12m) 158, (N79PII-1-12m) 169, (N79T-1-17m) 203, (N79T-4) 207 No. 14 (M78PII-1-5m) 14, (N78PI-3) 60, (M79PI-1-12m) 107, (N79PI-5) 165 No. 15 (M78PI-1-7m) 3, (N78T-1-2m) 98, (N78T-2) 102, (M79PM-4m) 106, (M79PI-1-5m) 106, (M79T-1-18m) 148, (M79T-2) 152, (N79PI-1-5m) 157, (N79PI-1-6m) 157, (N79T-1-29m) 204 No. l6(M78PI-1-5m) 3. (N78PI-1-4m) 54, (N78T-1-10m) 99, (N78T-1-18m) 99, (M79PI-1-3m) 106, (M79T-1-lm) 147, (M79T-1-4m) 147, (N79PI-1-3m) 156, (N79PI-1-4m) 156, (N79PII-1-3m) 167, (N79PII-1-17m) 170, (N79PII-1-18m) 170, (N79T-1-22m) 204, (N79T-1-26m) 204, (N79T-5) 207 No. 17 (M78T-1-12m) 46, (M79PI-3) 112, (M79T-1-12m) 148, (M79T-2) 151 No. 18 (M78PI-1-4m) 2, (N78PI-1-12m) 55, (M79?I-1-9m) 107, (M79PI-3) 112, (M79T-3) 153, (N79PI-1-13m) 158, (N79PII-1-19m) 171, (N79PII-1-20m) 171 No. 19 (M78PI-1-3m) 2, (M78A-2) 29, (M78T-1-21m) 47. (N78PI-1-16m) 56, (N78PI-1-18m) 56, (N78T-1-lm) 98, (N78T-1-21m) 100, (M79PI-1-16m) 108, (M79PI-1-17m) 108, (M79T-1-16m) 148, (M79T-2) 151, (N79PI-1-19m) 159 No. 20 (M78PI-1-17m) 5, (M78T-7) 51, (N78PI-1-3m) 54, (N78PII-3) 71, (N78T-1-4m) 98, (N78T-1-5m) 98.
(N78T-1-20m) 100, (M79PI-1-lm) 106, (M79PI-3) 113, (M79T-1-7m) 147, (N79PI-1-lm) 156, (N79PI-1-2m) 156, (N79T-1-28m) 204 No. 21 (N78PI-3) 60, (N78T-5) 103, (N79P1-5) 165 No. 22 (N78T-1-23m) 100, (M79T-2) 152 No. 25 (M78PI-3) 8 No. 26 (M78PII-1-13m) 15, (N78T-5) 103, (N79PI-5) 165, (N79PII-1-14m) 170 212 APB Opinions (cont.) No. 28 (N78T-6) 103, (M79PI-1-7m) 107, (M79T-I-3m) 147, (N79PM-llm) 158, (N79T-4) 207 No. 29 (N78PI-1-17m) 56, (N79PII-1-10m) 169, (N79PII-3) 174 No. 30 (M78PI-1-6m) 3, (M78T-1-24m) 47, (N78PII-1-17m) 68, (N78PIM-18m) 68, (N78T-1-7m) 98, (N78T-1-17m) 99, (N79T-1-16m) 203 APB Statements No. 3 (M78PM-12m) 4, (M78PI-1-13m) 4, (M78T-6) 51, (N78PM-13m) 55, (N78T-1-3m) 98, (N78T-1-22m) 100, (M79PI-1-15m) 108, (M79T-1-19m) 148 No. 4 (M79T-1-15m) 148 ARBs Ethics requires compliance (M78A-1-14m) 24 No. 43, Ch. 4 (M79PI-1-7m) 107 No. 43, Ch. 7 (M79PI-1-13m) 108 No. 43, Ch. 7b, 13b (M78PI-3) 8 No. 45 (N78PI-4) 61, (M79T-4) 153 No. 51 (M79Pn-1-8m) 117 Attest function Extension for report on internal control based on an audit is indication of changing role of the CPA (M79A-1-22m) 127 Audit committees Describe, identify reasons why formed, and tell functions (N78A-2) 82 Members of board of directors who are not officers or employees (M79A-1-40m) 129 Audit engagement Acceptance After fiscal year end, first ascertain whether circumstances are likely to permit an adequate examination and expression of unqualified opinion (M78A-1-15m) 24 Audited by another CPA in prior years—procedures in accepting (N79A-2) 184 Inappropriate actions and what should have been done (M78A-2) 29 Benefits of independent audit (M78A-3) 30 First examination See also Generally accepted auditing standards—Standards of reporting—Consistency Make inquiries of predecessor auditor because may receive information assisting determination of whether engagement should be accepted (M79A-1-44m) 129 Tour of client’s facilities and review of the general records is most likely first step of those given (M78A-I-55m) 29 Major difference of opinion between assistant and auditor in charge concerning accounting issue, assistant believes necessary to disassociate himself from the resolution, both auditor and assistant must document details of disagreement and basis of resolution (M79A-1-4m) 125 Objectives of independent audit (M78A-3) 30 Primary objective to provide assurance to investors of listed company (M79A-1-9m) 126 Quality control Includes assigning personnel, consultation with appropriate persons, and supervision, but not keeping records of quality control policies and procedures (M78A-1-47m) 28 Unfamiliar business—AICPA Industry Audit Guides most useful source of sources given for information during the preliminary planning stage (M78A-1-45m) 27 Audit procedures See also Electronic data processing Audit procedures (cont.) See also Electronic data processing—Programs generally examined or directly tested in audit See also Statement of changes in financial position See also Subsequent events Accounts payable Confirmation—not always used because (best answer of four) reliable externally generated evidence supporting generally available on client’s premises (N78A-1-44m) 80 Accounts receivable Confirmation Evidence of validity of balances confirmed (N78A-1-19m) 77 Large debits and small periodic credits might suggest consignments (N79A-1-57m) 184 Analytical review Analytical review of client’s operations to identify probable unusual transactions (N79A-1-27m) 180 Identification of significant unexpected fluctuations necessitates (usually) auditor investigation (N79A-1-30m) 181 Substantive tests of financial information by study and comparison of relationships among data (N79A-1-50m) 183 Auditor’s judgment Ultimately determines specific procedures necessary to provide reasonable basis for opinion (N79A-1-24m) 180 Bond sinking fund Transactions and year-end balance best verified (of four procedures listed) by confirmation with the bond trustee (N78A-1-31m) 78 Bonds payable Ascertain that client has obtained opinion of counsel on legality of current issue (N78A-1-24m) 78 Private placement—tracing cash received from issue to accounting records (M79A-1-58m) 131 Cash Kiting Best detected of procedures offered by examining paid checks returned with bank statements subsequent to balance sheet date (M78A-1-39m) 27 Schedule of bank transfers is one of better techniques to detect (M78A-1-24m) 24 Confirmation Generally accepted auditing procedure (rather than principle, standard, or practice) (M78A-1-58m) 29 Contingencies Loss from litigation, claims, and assessments (M79A-4) 133 Contingent liabilities Notes receivable discounted—Standard Bank Confirmation Inquiry form approved jointly by AICPA and Bank Administration Institute requests information on these (N79A-1-15m) 179 Cost accounting tests Primarily to determine that costs are properly assigned to finished goods, work-in-process and cost of goods sold (M79A-1-6m) 125 Cut-off review and examination of sales journal entries for several days prior to and subsequent to balance sheet date might detect inflation of sales for the year (M79A-1-13m) 126 Depreciation expense Establish basis for depreciable assets and verify the expense (M79A-1-32m) 128 [ First examination Tour client’s facilities and review the general records is most likely first step of those given (M78A-1-55m) 29 Fixed assets Repairs and maintenance analyzed to provide evidence that expenditures for fixed assets have been capitalized (M78A-1-29m) 26 Flowcharts Record auditor’s understanding of client’s internal accounting control system (N78A-1-42m) 79 213 Fraud may exist indicated by examination Extend procedures to actively search for evidence of fraud (N79A-1-33m) 181 Governmental Extends beyond opinion and includes audits of efficiency, economy, effectiveness, and compliance (N78A-1-18m) 77 Illegal acts Review of internal administrative control not primarily for expressing an opinion, but may bring possible illegal acts to auditor’s attention (M79A-1-29m) 128 Income statement Examine accounts concurrently with related balance sheet accounts (M79A-1-53m) 130 Installment loan to client Three month waiver of repayment of principal without extension of maturity— more likely to verify interest expense, liability at year end, and installment loan payments, than to verify the balloon payment (N79A-1-43m) 182 Internal auditors Objectivity may be judged by a review of the recommendations made in their reports (N78A-1-50m) 81 Internal control Based on tests during year auditor concluded records, procedures, and representations could be relied on, but auditor should test these again at year end if enquiries and observations lead auditor to believe that conditions have changed significantly (M79A-1-16m) 126 Review not primarily for expressing an opinion, but may bring possible illegal acts to auditor’s attention (M79A-1-29m) 128 Study and evaluation of internal accounting control— work performed by internal auditors may be factor in determining nature, timing, and extent of independent auditor’s procedures (N78A-1-52m) 81 Inventory Basic procedures and how general-purpose computer software package and tape of inventory file data might be helpful (N78A-5) 84 Damaged merchandise discovery— observation is best of given procedures (M78A-1-6m) 23 Perpetual (M78A-1-12m) 24 Purchase cut-off—best of given items for auditor to review and test to ascertain merchandise properly included (M78A-1-50m) 28 Turnover analysis may detect obsolete merchandise (N79A-1-22m) 180 Liabilities Unrecorded existence best determined (of four choices) by examining a sample of cash disbursements in period subsequent to year end (N78A-1-41m) 79 Litigation, claims, and assessments (N78A-1-59m) 23 Long-term debt Examination of bond trust indenture (M78A-1-7m) 23 Observation Best of given procedures for discovery of damaged merchandise (M78A-1-6m) 23 Owners’ equity Expected to require substantially less audit time than other areas listed (M78A-1-57m) 29 Payroll Observation of actual distribution of payroll checks is because various phases of payroll work are not sufficiently segregated to afford effective internal accounting control (M79A-1-19m) 127 Observation of distribution of checks of so large a client organization that auditor cannot conveniently observe entire payroll may result in limiting observation to one or more selected departments (M79A-1-57m) 131 Surprise observation of distribution of paychecks is best way of four mentioned to determine that every name is bona fide employee presently on the job (N78A-1-17m) 77 Audit procedures (cont.) Preliminary conference with client Helps prevent misunderstandings and inefficient use of audit personnel (M78A-1-52m) 28 Prepaid insurance Examine support for payments, recompute portion expired, and prepare excerpts of insurance policies for audit working papers, but do not generally confirm premium rates with an independent insurance broker (N79A-1-3m) 178 Primary purpose is to gather corroborative evidence (M79A-1-2m) 125 Principal auditor with respect to other auditor of material subsidiary (M79A-5) 133 Public warehouse Unrecorded liabilities search— inspection of receiving and issuing procedures (N79A-1-51m) 183 Purchase discount Understatement detected by comparing purchase invoice terms with disbursement records and checks (M79A-1-23m) 127 Real estate ownership Paid real estate tax bills is better evidence of ownership at balance sheet date than title insurance policy, original deed in client’s safe, or closing statement (N79A-1-14m) 179 Representation letters Signed by chief executive officer and the chief financial officer (normally) of the client (N78A-1-38m) 79 Revenues Objectives of examination include to study and evaluate internal control, with particular emphasis on accrual, to verify earned revenue recorded, and recorded revenue earned, and to identify and interpret trends and variations in amounts, but not to verify cash deposited (M79A-1-55m) 131 Sales transaction testing Trace debits in accounts receivable subsidiary to supporting sales invoices to establish that such entries are properly supported by sales invoices (M78A-1-15m) 126 Securities physical examination Insist that representative of client be present in order to acknowledge receipt of securities returned (N79A-1-17m) 179 Stockholders’ equity Most important consideration of auditor is that entries in the capital stock account can be traced to a resolution in minutes of the board (N78A-1-43m) 80 Supplies With respect to a small company’s system of purchasing, an auditor’s primary concern should be to obtain satisfaction that supplies ordered and paid for have been received, counted, and checked to quantities and amounts on purchase orders and invoices (N78A-1-36m) 79 Unrecorded liabilities detection Primary reason for reconciling interest-bearing obligations outstanding during year and interest expense (N79A-1-16m) 179 W -2 forms Multiplication of the applicable FICA rate by individual gross taxable earnings would have detected fraud of bookkeeper’s overstated FICA withheld while all other employees’ FICA withheld was understated (N79A-1-44m) 182 Work in process Trace quantity of materials charged to material requisitions (N79A-1-4m) 178 Working papers Include the manner in which exceptions and unusual matters were resolved or treated (N78A-1-45m) 80 Primary purpose is to support the auditor’s opinion (N78A-1-56m) 81 Audit program See also Audit procedures Basic tool to control the audit work and review the Audit procedures (cont.) 214 Audit program See also Audit procedures (cont.) progress (N78A-1-35m) 79 Developed after auditor has completed evaluation of internal accounting control (M78A-1-49m) 28 Installment loan to client Three month waiver of repayment of principal without extension of maturity— more likely to verify interest expense, liability at year end, and installment loan payments, than to verify the balloon payment (N79A-1-43m) 182 Primary purpose of procedures is to gather corroborative evidence (M79A-1-2m) 125 Procedures outlined are primarily to gather evidence (M78A-1-23m) 25 Provides proof that work was adequately planned (N79A-1-58m) 184 Scheduling and controlling the engagement Include column for estimated and actual time (M78A-1-16m) 24 Audit working papers See Working papers Auditor Outside auditor appointed by board of directors or elected by stockholders of publicly-traded corporations to emphasize auditor independence from management (M79A-1-49m) 130 Auditor’s legal liability See Accountant’s legal liability Auditor’s report See also Accountant’s report Adverse Client declines to make essential disclosures and auditor must provide them in his report and appropriately modify the opinion (N78A-1-9m) 76 Balance sheet only Attestation in public advertising— auditor may permit client to present only the balance sheet along with an appropriately modified auditor’s report in a newspaper advertisement— CPA has already reported on the basic statements sent to stockholders (N79A-1-8m) 179 Comparative financial statements If the predecessor auditor’s report is not presented, state that prior year’s statements were examined by another auditor (N79A-1-36m) 181 Comprehensive basis of accounting other than GAAP If statements are not suitably titled, modify report to disclose any reservations (N79A-1-13m) 179 Consistency Exception when there is a change in format and presentation of the statement of changes in financial position from cash concept to working capital (N78A-1-3m) 75 No exception for change in estimated salvage, change in classification of an expense from “other” to “selling expense,” or for change in inventory pricing to correct math error in total of opening inventory (N78A-1-3m) 75 Date Completion of field work (M78A-1-13m) 24 Dual dated f or subsequent event after fieldwork before issuance of report (N79A-1-26m) 180 Departure from standard report Includes emphasis of a particular matter regarding statements, opinion based in part on report of another auditor, and statements affected by a departure from GAAP (M79A-1-52m) 130 Disclaimer Improper—should have referred to incomplete presentation, set forth and discussed unacceptable accounting, and used proper form regarding not audited and no opinion (M78A-2) 29 Fairly presented Applied within framework of GAAP (M79A-1-60m) 131 Illegal acts by clients If unable to determine amounts associated, qualify or disclaim opinion (M79A-1-36m) 128 Independence lacking State “not independent” and include statement that statements were unaudited and accordingly the CPA does not express an opinion (N78A-1-13m) 76 Interim financial information included in annual statements Need not modify report to refer to the limited review of such information or to the information (N78A-1-47m) 80 Long-form If nonaccounting data is included, state that it is presented for analysis purposes, subjected to auditing procedures, it is fairly stated in all material respects in relation to the basic statements taken as a whole (N78A-1-5m) 75 Nature Affects quantity, type and content of working papers (M78A-1-34m) 26 Opinion Consistency— exception if statement of changes in financial position is changed from accounting for cash to accounting for working capital (N79A-1-lm) 178 Exception to GAAP—recording goodwill from revaluation is not acceptable (M78A-1-lm) 23 Inadequate disclosure— decide between qualified and adverse opinion (M78A-1-42m) 27 Qualified Departure from GAAP— explanatory paragraph referring to footnote that discloses principal effects and opinion uses such language as “with the exception of” (N78A-1-48m) 80 “Except for” when opinion paragraph has been modified because of a change in accounting principle, all substantive reasons should be disclosed in a separate explanatory paragraph (N79A-1-41m) 182 Inadequate disclosure Client declines to make essential disclosures and auditor must provide them in his report and appropriately modify the opinion (N78A-1-9m) 76 Omission of statement of changes in financial position (M78A-1-35m) 26 “Subject to”— statements prepared as going concern that auditor believes may not continue (N79A-1-46m) 183 Unqualified Investor may conclude that any disputes over accounting were settled to the auditor’s satisfaction (M78A-1-48m) 28 Precluded when management refuses to furnish a representation letter (N79A-1-12m) 179 Other auditor of subsidiary Modify scope and opinion paragraphs to reflect division of responsibility (N79A-1-49m) 183 Price-level basis supplementary financial statements Special purpose report (M79A-1-24m) 127 Principal auditor May refer to the examination (of a significant subsidiary) of another auditor (N78A-1-lm) 75 Reference to another auditor Clearly indicate division of responsibility between portion covered by principal auditor and that of other auditor (N78A-1-15m) 77 Other auditor may be named if express permission to do so is given and the report of the other auditor is presented together with the report of the principal auditor (N78A-1-16m) 77 Requirements if report names other auditor of material subsidiary and makes reference to his examination (M79A-5) 133 Reissuance without revision at request of client Use date of previous report after performing all necessary Auditor’s report (cont.) 215 Auditor’s report Reissuance without revision (cont.) procedures (N79A-1-9m) 179 Replacement cost information Unaudited— footnote indicates auditor performed certain procedures but does not indicate whether auditor expresses an opinion on this information, with the result that the auditor’s report on audited statements should be expanded to disclaim this information (M79A-1-17m) 126 Report only on balance sheet Not considered a departure from standard report when auditor has unlimited access to information underlying all the basic statements (M79A-1-52m) 130 Scope limitation Sufficient to preclude unqualified opinion if management refuses to furnish a representation letter (N79A-1-12m) 179 Special purpose Price-level basis supplementary financial statements (M79A-1-24m) 127 Special reports Issued in connection with statements prepared in accord with a comprehensive basis of accounting other than GAAP (M79A-1-56m) 131 Subsequent events After fieldwork but before issuance of report, report is dual-dated and auditor’s responsibility for events subsequent to field work limited to specific event referred to (N79A-1-26m) 180 Auditor’s responsibilities Detection of errors or irregularities Plan and perform the engagement with an attitude of professional skepticism (M78A-1-4m) 23 B Bad debt expense Description of direct write-off and allowance methods of recognizing (N79T-2) 206 Matching estimates it on basis of percentage of past actual losses from bad debts to past net credit sales, adjusted for anticipated conditions (N79T-1-13m) 203 Reasons why direct write-off is not usually GAAP and the allowance method is preferable (N79T-2) 206 Bank Administration Institute (N79A-1-15m) 179 Bankruptcy Acts Transfer of all nonexempt property to trustee for equitable distribution to creditors may result in being petitioned into bankruptcy by creditors (M79L-1-23m) 139 Claims not discharged Unpaid taxes due U.S. which became due and owing within three years preceding (M79L-1-16m) 137 Composition agreement Does not discharge any of the debts included until performance by debtor has taken place (M78L-1-24m) 36 Discharges unpaid amount owed secured creditor who received less than full amount after resorting to his security, unpaid amounts owed trade suppliers for goods sold in ordinary course, personal loan by father made in attempt to stave off bankruptcy, but not unpaid amount of creditor who extended credit on strength of fraudulent financial statement (M78L-1-32m) 37 Mortgage can be set aside if taken with knowledge debtor was insolvent in bankruptcy sense (M78L-1-25m) 36 Perfected security interest Creditor is a secured creditor and has the right against the trustee if not paid to assert a claim to the machine sold (M79L-1-22m) 138 Preference Whether creditor knew of fact that debtor was insolvent in bankruptcy sense is irrelevant insofar as deciding whether second mortgage received as security on note of debtor constitutes a preference as contrasted with a voidable preference (M79L-1-19m) 138 Secured creditor Debt discharged even if not fully satisfied from proceeds obtained upon disposition of the collateral (M79L-1-24m) 139 Social security and income taxes withheld from employees Used to finance the business, employer will remain liable for the taxes due, personally liable for fines and imprisonment, and may be assessed penalties up to the amount of taxes due (M79L-1-50m) 143 Surety company pays workers of obligor in full and is entitled to the same priority in bankruptcy of the obligor that the workers would have had (M78L-1-46m) 40 Voidable preference Sale of property of debtor, while insolvent, in part for cancellation of antecedent debt, constitutes a preference, but whether voidable depends on whether purchaser had requisite knowledge of the insolvency at the time (N78L-2) 94 Bankruptcy (cont.) Bond investments See Investments- -Long-term— Bonds Bond sinking fund See Audit procedures Bonds payable See also Audit procedures Amortization of premium or discount Effective interest method computation, and why and how amounts differ from straight-line (N78T-4) 103 Effective interest method explanation, and how it differs from and is similar to straight-line (N78T-4) 103 Collect accrued interest when sold between interest dates (N79T-1-lm) 202 Convertible Computation of amount credited to capital in excess of par given conversion of 9% of convertible bonds outstanding, number and par value of common shares exchanged for each bond, total bonds outstanding, and unamortized bond discount at date of conversion (M79PII-1-4m) 116 Entry upon conversion using book value method after giving effect to stock split (N78PI-3) 60 Entry upon issuance (N78PI-3) 60 Detachable stock purchase warrants Entry for issuance, given sale at face and total value of the warrants at issuance (N78PI-3) 60 Initial carrying value of bonds is total issuance price of bonds multiplied by relative market value of bonds to warrants (M79PI-1-12m) 107 Issuance journal entry involving deferred bond issue costs, accrued interest at date of sale, premium on bonds, and value assigned to the warrants (N79PI-5) 165 Early extinguishment Extraordinary gain computation involving computation of premium on original sale, amortization of premium to date, determining one-half of unamortized premium applies to bonds reacquired and addition of computed discount on reacquisition (M78PII-1-13m) 15 Proceeds Present value of principal and of interest, each discounted at prevailing market rate of interest (M78T-1-4m) 45 Sale receipts computation, given face, contract percentage interest, yield rate, interest paid annually, life of bonds, and present value of $1 and of annuity of $1 for life of bonds at yield rate (N79PI-5) 165 216 Bonds payable (cont.) Serial Amortization schedule for first two years of bond discount using interest (effective rate) method (N78PI-3) 60 Computation of amount received, given face of five-year 5% bonds sold to yield 6% , amount and date of principal payments, interest payment dates, present value of an annuity value at 6%, and present value of $1 at 6% table (N78PI-3) 60 Bonus Computation, given income before tax and bonus, and that bonus is 10% of income in excess of another given amount (M79Pn-1-7m) 117 Breakeven point Affected by variable cost per unit, total fixed costs, and sales price per unit, but not by number of units sold (M79T-1-32m) 149 Computation given that a specified net income is desired, percentage increase in fixed costs, and (in previous problem) total fixed costs having determined variable cost percentage of sales there (N78PI-2-25m) 57 Computation of sales given breakeven sales, contribution margin percentage and profit (M78PI-2-24m) 6 Fixed costs increased due to addition to physical plant would also have caused an increase in the breakeven point (M79T-1-28m) 149 Increases and contribution margin ratio increases if fixed costs decrease and variable costs as a percentage of sales decrease (N79T-1-31m) 205 Increases if fixed costs attendant to a product increase while variable costs and sales price remain constant— and contribution margin would be unchanged (N78T-1-35m) 101 Its total sales computation, given unit sale price, unit variable cost, and total fixed costs (N79PI-2-28m) 161 Relevant in setting earnings objectives (M78T-2-48m) 49 Sales-at-breakeven-point computation, given total costs and variable costs at one sales level (M79PI-2-25m) 110 Sales dollars Total fixed costs divided by (Sales price per unit-variable cost per unit) ÷ sales price per unit (M78T-2-34m) 48 Two products Computation given percentage of total sales dollars for each product, variable cost percentage of sales dollars for each product, and total fixed costs (N78PI-2-24m) 57 Fixed cost amount computation, given number of units and sale price of expectations for each product, variable cost percentage for each product, and desired total profit (N79PI-2-27m) 161 Unit selling price computation, given number of units, fixed costs, variable cost % relationship to sales, and profit amount desired (M79PI-2-26m) 110 Variable cost accounting lends itself to this analysis (M79T-1-37m) 150 Budgets See also Capital budgeting Cash Accounts receivable collections estimate computation (M78PI-2-35m) 8 Computation of disbursements estimate, given estimates of sales, gross profit %, no change expected in accounts payable, increase in inventory, fixed S, G & A expense amount (including depreciation amount), and variable S, G & A % of sales (including % of sales charge for uncollectible accounts) (M79PI-2-21m) 109 Disbursements for inventories in one month, given inventories at beginning and estimated at end, estimated cost of goods sold, estimated percent of purchases paid in month of purchase, and estimated payments during month for Budgets Cash (cont.) purchases in prior months (N78PI-2-22m) 57 Cost-of-goods-sold projected computation, given net sales, fixed manufacturing costs, and variable manufacturing cost percentage of sales (N79PI-2-40m) 162 Flexible budget purpose is to compare actual and budgeted results at virtually any level of production (N78T-1-41m) 101 Master budget and flexible budget basic difference Master based on one specific level of production and flexible prepared for any production level within a relevant range (M79T-1-38m) 150 Preparation of sales, production, raw materials purchases, direct labor, and finished goods inventory budgets in dollars or units or quantities (N78PII-4) 72 Buildings Analysis of account computation, given beginning balance, issuance of stock (quoted market price given) for land and building, and appraised value for the land and the building (N79PII-3) 174 Cost computation (M78PI-5) 10 Cost computation includes architect’s fees and construction costs, and problem includes cost computation of land (N78PII-1-10m) 66 Business combinations See also Consolidated financial statements Excess of fair value assigned to net assets over cost Proportionately reduce values assigned to noncurrent assets and record any remaining excess as a deferred credit (M79T-1-lm) 147 Pooling of interests Consolidated net income computation, given the net income for the full year for each company (N79PI-1-3m) 156 Consolidated net income is total net income for both companies for the full year though combination was on November 1 (N78PI-1-4m) 54 Par value of parent company common stock issued and outstanding before combination added to par value of common stock issued for other company’s shares to obtain total consolidated common stock issued and outstanding (M79PI-1-3m) 106 Potential abuse is that earnings of pooled entity may be increased because of combination only and not as result of efficient operations (N78T-1-18m) 99 Precluded when immediately afterwards the acquiring corporation reacquires the stock issued to effect combination (M78T-1-16m) 46 Supportive arguments (N79T-5) 207 Purchase Building valuation recorded in tax-free merger computation, given appraised value, adjusted tax basis, tax rate, and that goodwill was recorded (N79PII-1-18m) 170 Excess of fair value of net assets acquired over cost reduces book (and fair) valuation of noncurrent assets (N79PII-1-17m) 170 Exchange of cash for common stock— fixed asset valuation is fair value (N79T-1-22m) 204 Fixed asset valuation is fair value (N78T-1-10m) 99 No goodwill results because all excess of cost over net assets can be assigned to assets, considering that the fair value of long-term debt is less than its book value (N79PI-1-4m) 156 Such accounting required by fact that combined company is to retire a portion of the common stock exchanged to effect the combination within twelve months of the combination (M79T-1-4m) 147 Supportive arguments (N79T-5) 207 Subsidiaries consolidated under both purchase and pooling of interests (N79T-1-26m) 204 217 Capital budgeting See also Quantitative methods— Net present value See also Quantitative methods— Payback See also Quantitative methods— Present value Minimum return project must earn to leave value of company unchanged is cost of capital (N78T-1-50m) 102 Pay-back period Computation, given cost of machine and cash flow from operations (net of income taxes) by year (M79PI-2-22m) 109 Positive net present value indicates that return on investment exceeds cost of capital (M78T-2-46m) 49 Rate of return Accounting (book value)— machine cost computation, given percentage rate of return, useful life, straight-line depreciation, no salvage, and cash flow from operations, net of tax, for each of the years of life (M79PI-2-23m) 109, (N79PI-2-24m) 160 Time-adjusted Investment computation, given percentage rate of return, annual cash inflow (net of tax) from operations, and present value of ordinary annuity for useful life at return percentage (M79PI-2-27m) 110 Machine cost computation, given useful life, no salvage, cash flow amount, net of tax, from operations uniform for all years, expected rate of return, and present value of an annuity for useful life at expected return rate (N79PI-2-29m) 161 Project (two-year) investment cost computation, given percentage expected rate of return, cash flow amounts (net of tax) from operations for each of the years, and present values of $1 for one and for two periods at the expected rate of return (N79PI-2-30m) 161 Capital in excess of par Computation of amount credited, given conversion of 9% of convertible bonds outstanding, number and par value of common shares exchanged for each bond, total bonds outstanding, and unamortized bond discount at date of conversion (M79PII-1-4m) 116 Computation of balance, given beginning balance, treasury stock purchase above par, cost method, sale of some treasury shares and retirement of the rest (M78PII-1-12m) 15 Capital stock See Stock Capitalize or expense Match costs with revenues (M78T-1-5m) 45 Cash See also Auditing procedures See also Internal control— Accounting Disbursements estimate for month for inventories Computation, given estimated purchases, estimated percentage payments for purchases of month during month, and estimated payments during month for each of previous two month’s purchases (N79PI-2-21m) 160 Received on issuance of bonds between interest dates includes accrued interest (N79T-1-lm) 202 Cash accounting Contrasted with accrual accounting (N79T-3) 207 Cash budget See also Budgets See also Cash See also Quantitative methods— Payback Accounts-receivable-estimated-cash-collections-during-month computation, given estimated credit sales for current and previous month, estimated collection percentage for each of Cash budget (cont.) the two months, and estimated collections in current month for credit sales prior to previous month (N79PI-2-22m) 160 Cash flow See also Quantitative methods— Present value Operations flow for last payback year of one machine computation, given cost of machine, payback period, and cash flow from operations, net of tax, in each year prior to the last (N79PI-2-23m) 160 Code of Professional Ethics See AICPA Code of Professional Ethics College accounting See Fund accounting See University accounting Commercial paper Acceptance Certification of check by drawee bank on request of holder releases drawer (M79L-1-27m) 139 Bank’s rights to collection of various promissory notes used as collateral, where bank is holder in due course of some and others are forged, and total amount of all notes exceeds amount of loans secured (N79L-3) 198 Bearer paper Check with several types of indorsements was bearer paper in hands of last indorser (N78L-1-46m) 93 Difference between negotiable instrument and contract Negotiable instrument deemed prima facie to have been issued for consideration, whereas contract not so deemed (N78L-1-49m) 93 Example instrument is negotiable (N78L-1-50m) 93 Forged indorsement on stolen check Holder, unaware of the facts, accepted the check, but, disregarding thief, must bear the loss (M79L-1-30m) 140 Holder had notice of defense of maker prior to completion of negotiation but can assert the standing of his transferor and will prevail against maker (M78L-2) 41 Holder in due course Blank (as to amount) check filled in by creditor in excessive amount will have to be paid in full even though payment was stopped promptly (M79L-1-25m) 139 Discounter of note making an initial payment, and paying the balance within the agreed time (but after learning that the note was obtained originally by fraudulent misrepresenta tions), will qualify as a holder in due course for the initial payment only (N78L-1-44m) 92 Maker liable on instrument to such holder even though payee did not perform his part of contract, for which it was issued (M79L-1-31m) 140 Maker of raised check liable for original amount of check (M79L-1-32m) 140 Negotiated check to another who knew of a breach of agreement by payee not to transfer check, with result that receiver of check does not personally qualify as a holder in due course but can assert such holder’s (from whom he received it) standing as such (N78L-1-42m) 92 Not if instrument negotiated is not negotiable (M78L-2) 41 Personal defense that merchandise for which negotiable instrument was issued did not satisfy warranties would not prevail (M78L-2) 41 Purchased instrument at discount without knowledge it was stolen and can collect (M79L-1-29m) 140 Takes note not overdue for value and without notice or knowledge of any defect (M78L-2) 41 Will prevail against finance company that stops payment of check issued to imposter of prominent citizen who then negotiated it to a holder in due course while indorsing citizen’s name (N78L-1-47m) 93 218 Commercial paper (cont.) Indorsement “for deposit only” above name Any transferee must pay or apply any value given by him for the instrument consistent with the indorsement (M79L-1-28m) 139 Indorsement types and liability of each (M78L-2) 41 Negotiability Adversely affected by a statement that it is payable only out of the proceeds from the resale of the goods sold by first party to second party on given date (M79L-1-26m) 139 Nonnegotiable because example instrument neither payable to “order” or to “bearer,” even though indorsement does use such words (M78L-2) 41 Not adversely affected by clauses within the instrument, such as a promise to maintain collateral and to provide additional collateral if value of existing collateral decreases, a term authorizing confession of judgment on the instrument if not paid when due, or a statement that the instrument arises out of a particular sale (M79L-I-26m) 139 Nonnegotiable instrument “Negotiated” merely assigns all of transferor’s rights and leaves personal defenses of maker valid (M78L-2) 41 Not payable at definite time and payable only out of proceeds of particular resale, with result that holder is not a holder in due course and cannot collect from maker who returned the fraudulent goods to original receiver of instrument and owes nothing (N79L-3) 198 Raised check Holder in due course is entitled to collect original amount of check from drawer (unless he can assert and prove drawer was negligent in drafting the instrument and contributed to alteration), or from indorser in blank, but can collect nothing from bank, which rightfully dishonored it (N79L-3) 198 Subsequent indorsers Presentation for payment of example instrument indorsed by several subsequent holders must be made on due date or next business day in order to hold the subsequent indorsers liable (N78L-1-45m) 92 Commission on Auditors’ Responsibilities Concluded that oversight of professional practice should remain with the accounting profession (M79A-1-28m) 128 Committee on Management Services (M79A-1-35m) 128 Comparability Change in accounting principle conflicts with (N78T-1-5m) 98 Comparative statements See Generally accepted auditing standards— Standards of reporting— Fourth standard Confirmation See Audit procedures Conservatism Example is stating inventories at lower of cost or market (N79T-1-19m) 204 Supports immediate recognition of loss (M78T-1-9m) 45 Consignments Total sales price of consigned inventory by consignee computation, given account sales stating balance due net of 20% commission (on sales price) to consignee, advertising, and delivery and installation charges (M79PII-1-6m) 117 Consistency See also Auditor’s report See also Generally accepted auditing standards— Standards of reporting Accounting changes may violate (N78T-1-4m) 98 Consolidated financial statements See also Business combinations See also Goodwill— Consolidated Equipment cost and accumulated depreciation should be same at end of year after sale to subsidiary at a profit as it would have been in hands of parent, computation (M79PII-1-8m) 117 Goodwill Computation upon acquisition of company for 1.5 million more than net assets with .8 million allocable to assets (N78PI-1-2m) 54 Income statement Foreign exchange gain included, computation (N78PI-1-8m) 55 Pooling Income statement Net income computation (M78PI-1-5m) 3 Prepared in recognition of the accounting concept of entity (M79T-1-17m) 148 Contingencies See also Audit procedures See also Expropriation Accrual Amount of the loss can reasonably be estimated and it is evident prior to issuance of statements that an asset has been impaired or a liability incurred (N78T-1-13m) 99 Amount of loss can reasonably be estimated and occurrence of event is probable (M79T-1-5m) 147 Probable loss on product warranties (N79T-1-3m) 202 Gain Disclosure only is proper accounting (M78T-1-22m) 47 Loss from lawsuit On basis of client’s lawyer’s estimate of probable loss of lawsuit and amount of judgment for $500,0()0 to $2,500,000 with most probable cost of $1,000,000 set up loss contingency of $1,000,000 and disclose additional contingency of up to $1,500,000 (N79T-1-2m) 202 Loss is probable Accrue lower limit of range when the loss can reasonably be estimated within range and any amount within range is equally probable (M78T-1-23m) 47 Reasonably estimated losses on lawsuits in progress relative to product defects and on additional lawsuits should be accrued (N78PI-1-5m) 54 Recall expense estimate charged to operating expense currently where government has clearly indicated its intention of requiring such (M79PII-1-5m) 117 Warranty contingency both probable and reasonably estimable and should be accrued as liability and an expense (M79T-2) 151 Contingent liabilities See Audit procedures Contracts Acceptance Constructive receipt— damages available to famous actress who accepted by telegram within time limit specified in offer sent by telegram, even though her telegram was delivered by mistake beyond the time limit (M78L-1-15m) 34 Example is valid upon dispatch despite the fact it states both additional and different terms than those contained in the offer because they relate to routine or mechanical details of execution or request a favor (M78L-1-36m) 38 Accord and satisfaction Cashing of check marked “Account Paid in Full to Date” (M79L-1-13m) 137 Assignment Assigns rights and delegates duties from assignor to assignee (M79L-1-34m) 140 Nonnotification of debtor will allow debtor to pay assignor and have no liability to assignee (M78L-1-16m) 34 219 Bilateral and unilateral difference One promise involved if unilateral whereas two in bilateral (N79L-1-5m) 188 Bilateral subject to a condition Agreement for sale of property, if rezoned, with deposit (M78L-1-21m) 35 Consideration Offeree makes promise that is legal detriment to him, offeree performs act requested by offeror, or offeree makes promise that benefits offeror would all be sufficient, but not if offeree merely expended time and money in studying and analyzing the offer (N79L-1-3m) 187 Past consideration is no consideration, hence there is no contract where a man promised a firm a sizable amount of business if that firm (which already had an employment contract with the man’s son) would hire his son (M79L-1-11m) 137 Covenant not to compete Covenant not to engage in practice of pharmacy for one year from sale of pharmacy within one mile is reasonable and enforceable (M79L-1-12m) 137 Difference between negotiable instrument and contract Negotiable instrument deemed prima facie to have been issued for consideration, whereas contract is not so deemed (N78L-1-49m) 93 Employment Employee with three-year contract demanded raise and it was agreed in writing to give him bonus at end of contract, but the additional compensation is unenforceable because of no consideration (N78L-5) 96 Employee with three-year contract was lured away by another employer with full knowledge of the three-year contract, with result that such employer’s conduct is tortious and allows first employer to recover damages (N78L-5) 96 Employer won’t be successful in seeking equitable relief of specific performance by employee with three-year contract who left after two years, because court of equity will not force a person to fulfill a contract for personal services (N78L-5) 96 Goods not identified to his contract will prevent purchaser from obtaining items for which substantial prepayments were made from an insolvent (N78L-1-31m) 90 Infringement suit Buyer liable to manufacturer-seller for breach of warranty against infringement where buyer furnished specifications for uniquely designed and specially-constructed items and both were sued by another for patent infringement (N78L-1-36m) 91 Modification of written contract by reduction of price to meet current conditions need not satisfy the statute of frauds (M78L-1-38m) 38 Novation Bank released seller of house from mortgage and accepted buyer (M79L-1-8m) 136 Offer Communication example of a willingness to sell and a possible minimum price range did not constitute an offer to sell (N78L-1-27m) 90 Different terms of acceptance constituted a rejection (M79L-1-35m) 141 If offeree categorically rejects offer before expiration date, he cannot validly accept within the remaining time before that expiration date (N78L-1-29m) 90 Letter to specific company offers three-year franchise and shows an expiration date, but such cannot be assigned to another if the offeree chooses not to accept (N78L-1-28m) 90 Must be communicated to the other party, be certain enough to determine the liability of the parties, and be accepted, but need not only be expressed in words (M78L-1-19m) 35 Offer by telegram that was effectively revoked by telephone cannot be accepted thereafter, even before originally specified termination date (M79L-1-17m) 138 Contracts (cont.) Option to purchase Supported by consideration and consequently is a binding contract at a given price or prevailing market, whichever is lower, which is definite and certain (N78L-5) 96 Oral Statute of frauds— eighteen-month contract of employment cannot be enforced because such must be in writing and signed by an agent of the employer (N78L-1-30m) 90 Parol evidence rule Oral evidence to explain or modify a written contract admissable if the contract contains an obvious ambiguity, if there was a mutual mistake of fact regarding the subject matter, or if the modification asserted was made several days after the written contract, but not if the contract indicates that it was intended as the “entire contract” between the parties and the point is covered in detail (M79L-1-10m) 136 Pre-existing legal duty Prepaid service contractor did not meet time commitment for service and was promised more for immediate service, but additional amount was unenforceable (M79L-1-9m) 136 Purchase commitment for annual supply for three years leaves company that thinks it made a bad bargain obligated to take its normal annual requirements from that supplier or respond in damages (M79L-1-44m) 142 Sale of business Misrepresentation is best for seeking relief for buyer if he merely wishes to rescind the contract (N79L-1-2m) 187 Sale of land Parol evidence rule prevents buyer from proving contemporaneous oral agreement that seller would remove large rocks (N79L-1-4m) 187 Sales Automatically contains a warranty that title conveyed is good and can only be excluded by specific language (M79L-1-36m) 141 Breach of warranty recovery should use (of many untrue statements) “I refused a $3,800 offer for this very same auto from another buyer last week” (M79L-1-37m) 141 Negotiable warehouse receipts transferred to buyer who was responsible for transporting goods to his place of business, and risk of loss passes to buyer on his receipt of the negotiable warehouse receipts (M79L-1-39m) 141 Repudiated by seller Buyer can “cover,” that is, procure the goods elsewhere and recover any damages (N79L-1-lm) 187, (M79L-1-38m) 141 Buyer did not “cover,” prices continued to rise, and buyer is entitled to damages equal to difference between contract price and fair market value at time he learned of the breach (N79L-1-13m) 189 Customer can procure the goods elsewhere and recover damages (M78L-1-37m) 38 Risk of loss— shipping terms, F.O.B. shipping point, determine risk even though title was reserved by seller and seller retained negotiable bill of lading for simultaneous exchange for cash— title is irrelevant (N78L-5) 96 Seller may recover the price if customer accepted the goods but seeks to return them, if the goods sold were destroyed shortly after risk of loss passed to buyer, or if the goods were identified to the contract and seller made reasonable effort to resell but was unable, but may not recover this remedy if customer refused to accept delivery and the goods were resold in ordinary course of business (M78L-1-39m) 38 Seller of stolen goods is liable, even though bill of sale specifically indicated that it made no implied warranties, to buyer after repossession by rightful owner in that seller warranted that title conveyed was good and his transfer rightful (N78L-1-35m) 91 Seller’s place of business is proper place of delivery where Contracts (cont.) 220 Contracts Sales (cont.) contract failed to designate place of delivery and seller and buyer are in the same town (M78L-1-43m) 39 Seller wishes to disclaim an implied warranty of fitness for a particular purpose and may do so by a conspicuous written statement, which states that “any and all warranty protection is hereby disclaimed” (M78L-1-40m) 38 Statute of frauds Applies to contract modified from price of $450 to price of $750 (M79L-1-33m) MO Oral contract of employment that could not be performed within one year of date of the contract would not be enforceable if employee took another job (M78L-1-13m) 34 Purchase order with name and address of company at top, reciting oral agreement, and with purchasing agent’s name and title stamped at bottom with his initials is sufficient (N79L-1-8m) 188 Writing satisfies, even if not signed by buyer and seller, even though time and place of delivery not indicated, and even though payment terms not included, but does not satisfy if it does not indicate a sale has occurred (N79L-1-11m) 189 Wholesaler assigned contract to another and it was partially defaulted with result that customer may recover damages once for breach of contract from either (M78L-1-44m) 39 Telegraphed order requiring immediate shipment Shipment same day constituted an acceptance and contract is valid despite fact that breakdown in transportation system significantly delayed delivery (N79L-1-9m) 189 Timely performance CPA can recover fees for examining statements of a company when the delay by the company’s staff prevented CPA from performing on time and thereby eliminated the timely performance condition (N78L-1-32m) 91 Unilateral mistake leaves valid and binding contract (N79L-1-7m) 188 Violation of terms Lack of prompt delivery made trucking company liable for amount spent in minimizing loss (M78L-1-22m) 35 Waiver of rights Customer’s signed memo indicating he would relinquish right to recover costs of correcting minor defects, provided balance of merchandise was good and delivery strictly adhered to (M78L-1-41m) 39 Contributed capital See Stockholders’ equity Contribution margin See Cost accounting— Direct costing Corporations Accumulation of earnings and profits where corp. continuously elects not to pay dividends may be taxed as unreasonable or ambitious expansion plans would normally constitute a bona fide business purpose and prevent such tax (N79L-2) 197 Board of directors valuation of consideration received for shares conclusive in absence of fraud (N78L-1-19m) 89 Directors and officers Cannot benefit in dealings with or for corp.— dissenting minority shareholders may institute legal action in name of corp. to have acquisition (which benefited chairman) of one corp. by another chairman-controlled corp. and have the acquisition transferred to their own corp. with profits of interim, or minority could seek damages from chairman (M78L-4) 43 Indemnification of officer’s legal fees and expenses— rights and limitations applicable where president was defending himself successfully against criminal action brought against him in connection with an antitrust suit against the corporation, and indemnification is proper (N79L-2) 197 Corporations Directors and officers (cont.) Interlocking directorate Land purchase fair and reasonable to the corporation may be made from such company if the facts are known by board or committee that authorizes (not counting votes of interested directors), or stockholders entitled to vote have such knowledge and approve (M79L-3) 144 Payment “on the side” to interlocking director to assist transaction with other company would violate the director’s fiduciary duty, may be criminally illegal, and would constitute grounds to make the transaction voidable (M79L-3) 144 Loan to president (employee) for benefit of company authorized by chairman (should have been board of directors) may leave chairman subject to personal liability (M79L-3) 144 Dividends From authorized but unissued shares legal out of unreserved and unrestricted retained earnings, but where there is a par value, they must be issued at not less than par (M79L-3) 144 From shares of another company legal where enough unreserved and unrestricted retained earnings available (M79L-3) 144 Tax consequences of dividend from authorized but unissued shares is that basis for recipient’s shares must be allocated over a new number of total shares, but the distribution is not taxable (M79L-3) 144 Tax consequences of dividend from shares of another company are that the shareholders must include FMV of shares received as dividend income subject to the dividend exclusion and their tax basis of the shares is FMV (M79L-3) 144 Foreign Incorporated in one state, doing business in a neighboring state, must obtain a certificate of authority or it will not be permitted to maintain any action or suit in the neighboring state with respect to its intrastate business (N78L-1-9m) 87 Not having filed any papers in the neighboring state, cannot defend against a suit brought against it in the neighboring state’s courts (N78L-1-10m) 87 Issuance of shares Proper consideration includes services actually performed, shares of stock of another corporation, and intangible property rights, but not promissory notes (N78L-1-14m) 88 Limited partnership similar in that both can only be created pursuant to statute and each must file a copy of their respective certificates with proper state authorities (N78L-1-18m) 88, (N79L-1-30m) 193 Pre-incorporation contracts Board of directors at inception of corporation is free to either adopt or reject pre-incorp. agreements made on behalf of the corp., including employment contract of promoter (M78L-4) 42 Parties who agree to provide capital vital to corp. creation not permitted to withdraw commitments for six months, and this is enforceable against estate of such party (M78L-4) 42 Promoter entitled to cash expenditures because of implied adoption by board of directors accepting all the benefits or because implied in law based upon unjust enrichment (M78L-4) 42 Subscription to buy shares in a corporation to be organized may not be revoked for six months in absence of special circumstances (N78L-1-15m) 88 Professional State in which doctors incorporated must have enacted professional corporation statutes permitting them to do so (N78L-1-22m) 89 Promoter of corporation to be organized liable as a party to the contract (M78L-4) 42 Seeks recovery of short-swing profits of ousted president 221 Corporations Pre-incorporation contracts (cont.) Valid defense is that all transactions alleged in violation of Securities Exchange Act of 1934 were purchases in February 1979 with corresponding sales in September 1979 (over six months) (N79L-1-32m) 193 Stock dividends Analyze and discuss legal impact from standpoint of corporate requirements that must be met, effect upon stated capital, and federal income tax effects or implications to shareholder (N79L-2) 197 Stock split Analyze and discuss legal impact from standpoint of corporate requirements that must be met, effect upon stated capital, and federal income tax effects or implications to shareholder (N79L-2) 197 Cost See Expense recognition Cost accounting See also Special order Absorption costing Differences from direct costing (M78T-2-36m) 4 8 , (N79T-6) 207 Net earnings difference from direct costing is caused by absorption allocating fixed costs between cost of goods sold and inventories, and direct costing expensing such costs as period costs (M78T-2-39m) 48 Product cost— computation, given standard capacity, fixed factory overhead, and variable unit costs (M78PI-2-33m) 7 Total costs of an element of production (material, labor, or overhead) are used in order to compute variances from standard amounts (N78T-1-42m) 101 Audit tests Primarily to determine that costs are properly assigned to finished goods, work in process, and cost of goods sold (M79A-1-6m) 125 Breakeven See Breakeven point Budgets Preparation of sales, production, raw materials purchases, direct labor, and finished goods inventory budgets in dollars or units or quantities (N78PII-4) 72 By-product Recognize value as produced— inventory based on allocation of joint costs plus subsequent processing costs (N79T-1-33m) 205 Common costs Operating profit computation for a segment, given segment’s percent of total sales, common costs allocated on that basis, sales for the segment, traceable costs for the segment, and total common costs (N78PI-1-14m) 56 Direct costing Contribution margin for various combinations of unit sales and units manufactured (or purchased) of single product (units manufactured and not sold by end of month must be discarded) (N79PII-4) 175 Contribution margin is unchanged and breakeven point increases if fixed costs attendant to a product increase while variable costs and sales price remain constant (N78T-1-35m) 101 Contribution margin monthly computation if given number of units manufactured and all sales orders filled either from those or from other purchased (units manufactured and not sold by end of month must be discarded), and substitute more expensive ingredient must be used because of potential strike (N79PII-4) 175 Contribution margin ratio increases and breakeven point decreases if fixed costs decrease and variable costs as a percentage of sales decrease (N79T-1-31m) 205 Differences from absorption costing (M78T-2-36m) 4 8 , (N79T-6) 207 How it facilitates calculation of contribution margin and breakeven point (N79T-6) 207 Net earnings difference from absorption costing is caused by absorption allocating fixed costs between cost of goods sold and inventories, and direct costing expensing such costs as period costs (M78T-2-39m) 48 Net income computation, given selling price, number of units sold, variable unit costs, and fixed costs (M78PI-2-34m) 7 Not GAAP because fixed manufacturing costs are assumed to be period costs (M78T-2-50m) 49 Operating-income-increase-as-result-of-accepting-special-order computation, given brief absorption costing income statement, fixed costs in manufacturing expenses, number of units and price in special order, no additional selling expenses, and per unit sales price and manufacturing cost (M79PI-2-38m) 112 Operating-income-increase-as-result-of-accepting-special-order computation, given variable manufacturing cost, selling price, and number of units (N79PI-2-35m) 162 Operating income increase from accepting special order for specified number of units, given price, variable manufacturing cost per unit, and no additional selling expenses (N78PI-2-29m) 58 Period costs should be currently expensed because they will occur whether or not production occurs, so it is improper to allocate these costs to production and defer a current cost of doing business— a basic tenet (N78T-1-39m) 101 To institute (install), variable and fixed components of all costs related to production must be known (N78T-1-37m) 101 Economic order quantity See Quantitative methods Factory overhead Includes all manufacturing costs, except direct materials and direct labor (N79T-1-35m) 205 Fixed cost Definition and identifying characteristics (N78T-3) 102 Per unit decrease as production increases (M78T-2-47m) 47 Flexible budgets Fixed costs per unit decrease as production increases (M78T-2-47m) 49 Job order Factory overhead includes payroll taxes paid by employer for factory employees (M79T-1-22m) 149 Used with standards, with direct costing, and with overhead allocation based on direct labor hours, but not with an averaging of direct labor and material rates (M78T-2-41m) 49 Joint costs Allocated on relative sales value, relative weight, or linear measure, and average unit cost, but not relative profitability (N78T-1-48m) 102 Allocated to each of joint products a proportionate amount of total cost by means of a quantitative basis (M78T-2-38m) 48 Allocation-to-one-of-two-products computation, given total joint costs, sales value as split-off for each product, and allocated based on relative-sales-value-at-split- off (N79PI-2-26m) 161 Allocation using relative-sales-value-at-split-off approach, given such value and total joint costs (N78PI-2-21m) 57 Computation of allocation based on relative-sales-value-at-split- off (M78PI-2-22m) 6 Computation of sales value at split-off, given cost allocation, using relative-sales-value-at-split-off (M78PI-2-23m) 6 Computation of total joint cost, given relative-sales-value-at-split- off approach, such sales values for each of two products, and the allocated cost to one (M79PI-2-24m) 109 Cost accounting Direct costing (cont.) 222 Materials, labor, and overhead are allocable (M79T-1-31m) 149 Relative sales value is most frequent allocation base (M79T-1-30m) 149 Sales-value-at-split-off computation for company with three products, given relative-sales-value-at-split-off approach used for allocation of joint costs, sales value at split-off for one product and in total, joint costs total, and joint cost allocation to another product (M79PI-2-32m) 111 Sales value at split-off for one product computation, given joint costs in total and amount allocated to that product, relative- sales-value-at-split-off approach used for allocation of joint costs, and total sales value at split-off for all products (N79PI-2-25m) 160 Total computation, given relative-sales-value allocation method, amount allocated to one product, and sales value at split- off (N78PI-2-23m) 57 Make or buy Relevant costs computation, given per unit materials, labor, variable and fixed overhead costs, reduction in fixed overhead, and number of units (N78PI-2-35m) 59 Relevant costs include direct labor, variable overhead, and fixed overhead avoided if the part is bought, but not fixed overhead that will continue even if the part is bought (N79T-1-32m) 205 Relevant costs including computation of amount that would have to be saved by using released facilities manufacture another part in order to decide to buy, given number of units, variable costs per unit, fixed overhead applied per unit eliminated, cost per unit to buy, and total amount of savings desired to make buy decision (N79PI-2-36m) 162 Relevant costs total computation, given per unit variable costs and fixed costs, number of units, and percentage change of per unit fixed costs that would continue (M79PI-2-36m) 111 Process costing Computation of distribution of total cost to goods completed and work in process for one department, given average equivalent unit cost for materials and for conversion costs, number of units in each inventory, materials added at beginning, and ending work in process 50% complete as to conversion costs (M79PI-2-31m) 110 Equivalent units Computation (M78PII-5) 21 Compute difference between weighted-average and FIFO for materials and for conversion costs (N78PI-2-32m) 59 Conversion costs (M78PI-2-25m) 6 Cost computation (M78PII-5) 21 Direct labor in all inventories (N78PII-4) 72 FIFO— work for the period must be broken down to units completed from beginning inventory, started and completed during the month, and units in ending inventory (M79T-1-39m) 150 Materials and conversion costs— FIFO computation, given units completed from beginning work in process and from the month’s production, units in process at end, percentage complete as to conversion costs of both beginning and ending work in process inventories, and materials added at beginning (N79PI-2-34m) 162 Materials and direct labor in packaging inventory (N78PII-4) 72 Materials or conversion cost— amount necessary to complete one unit of production (M78T-2-37m) 48 Raw materials in all inventories (N78PII-4) 72 Weighted-average method Costing of finished goods and work in process (M79PII-4) 123 Cost accounting Joint costs (cont.) Cost accounting Process costing (cont.) Does not consider degree of completion of beginning work-in-process inventory when computing equivalent units of production, which is different from FIFO method (M78T-2-35m) 48 Entry to correct finished goods and work-in-process inventories (M79PII-4) 123 Materials, labor, and overhead computation (M79PII-4) 123 Transferred-in costs, materials, arid conversion costs computation (M79PI-2-35m) 111 Unit costs of production for materials, labor, and overhead (M79PII-4) 123 FIFO will produce same cost of goods manufactured as weighted-average when there is no beginning inventory (N78T-1-38m) 101 Normal spoilage cost should be assigned to cost of goods manufactured (transferred out) if normal spoilage occurs at end of process (N78T-1-36m) 101 Transferred-in cost— computation of amount in ending work in process (M78PI-2-26m) 6 Weighted-average method Materials-cost-of-work-in-process computation (N78PI-2-33m) 59 Work-in-process-materials-inventory (ending)-cost computation, given materials added at beginning, units and costs in process at beginning, units and costs started, and units completed and transferred (N79PI-2-33m) 161 Would make use of standards, variable costing, responsibility accounting, but not individual lots (M79T-1-29m) 149 Responsibility-accounting Internal reports limited to controllable costs (N78T-1-47m) 102 Performance report— for supervisor of assembly line in large manufacturing situation likely to include materials, repairs and maintenance, direct labor, but not the supervisor’s salary (M79T-1-34m) 149 Sell or process further (M78PI-2-29m) 7 Semivariable cost Definition and identifying characteristics (N78T-3) 102 Three basic methods (scattergraph, high-low, least-squares) “break down’’ into component parts discussion (N78T-3) 102 Spoilage Abnormal (N79T-1-34m) 205 Normal (N78T-1-49m) 102 Normal is a product cost, while abnormal is a period cost (M79T-1-40m) 150 Standards See Cost accounting— Variance analysis Variable cost Definition and identifying characteristics (N78T-3) 102 Variance analysis Absorption costing—total costs of an element of production (material, labor, or overhead) are used in order to compute variances from standard amounts (N78T-1-42m) 101 Computation of all variances, including two-variance overhead (M78PII-5) 21 Direct labor Description of standards that must be developed and how variances from those standards are calculated (N79T-6) 207 Efficiency Actual-hours computation, given standard direct-labor rate, standard direct-labor hours, and usage unfavorable variance amount (N79PI-2-37m) 162 Actual-hours computation, given standard rate, standard hours allowed for actual production, and the variance (N78PI-2-36m) 59 Most probable reason for unfavorable labor rate variance and favorable labor efficiency variance is that the mix of 223 workers assigned was heavily weighted towards use of higher paid, experienced individuals (N78T-1-43m) 102 Rate Computed as difference between standard and actual rate multiplied by actual hours (M78T-2-45m) 49 Would disclose excess direct labor wages from overtime premium (M78T-2-43m) 49 Direct material Description of standards that must be developed and how variances from those standards are calculated (N79T-6) 207 Price Actual-purchase-price-per-unit computation, given standard unit price, actual quantity purchased, and price favorable variance amount (N79PI-2-38m) 162 Computation, given actual units used, actual material costs, and standard price per unit of materials (M79PI-2-37m) 111 Computation, given standard and actual prices, and actual quantity purchased (N78PI-2-34m) 59 Recognize when material is purchased if wish to isolate variance at earliest point in time (M79T-1-23m) 149 Total— evaluated in regard to price and usage to allow management to evaluate the efficiency of purchasing and production (M78T-2-44m) 49 Usage unfavorable and price favorable if actual number of pounds used exceeds standard allowed but actual cost was less than standard cost (M79T-1-20m) 148 Factory overhead (M78T-2-49m) 49 Immaterial variances— normal year-end treatment is closing to cost of goods sold in the period in which they arose (M79T-1-33m) 149 Overhead Overall (or net) computation, given total actual overh ead, total standard overhead rate, and standard hours allowed for actual production (N79PI-2-37m) 59 Three-way spending variance computation, given actual overhead, budgeted fixed overhead, actual hours, and variable overhead rate (M78PI-2-32m) 7 Total computation, given standard applied overhead and actual overhead (M79PI-2-39m) 112 Total overhead variance computation, given actual variable and actual fixed overhead, standard hours allowed for actual production, and standard variable and standard fixed overhead rates per direct-labor hour (N79PI-2-39m) 162 Two-way Computation of actual overhead, given fixed amount, variable rate, and the two variances (M78PI-2-30m) 7 Computation of applied overhead, given fixed amount, variable rate, and the two variances (M78PI-2-31m) 7 Usage— allocated among work-in-process, finished goods, and cost-of-goods-sold inventories at end of accounting period (M79T-1-25m) 149 Current assets Computation of amount, given consignment-out carried in accounts receivable at a stated percentage (higher than cost), and the total current assets as on the books (N79PII-1-13m) 169 Current liability Rent revenue collected one month in advance (N79T-1-14m) 203 Current value accounting See Price-level accounting Cost accounting Variance analysis (cont.) D Data processing See Electronic data processing Debentures See Bonds payable Decision making See also Cost accounting— Direct costing See also Quantitative methods— Probability See also Special order Make or buy Relevant costs include direct labor, variable overhead, and fixed overhead avoided if the part is bought, but not fixed overhead that will continue even if the part is bought (N79T-1-32m) 205 Relevant costs including computation of amount that would have to be saved by using released facilities manufacture another part in order to decide to buy, given number of units, variable costs per unit, fixed overhead applied per unit eliminated, cost per unit to buy, and total amount of savings desired to make buy decision (N79PI-2-36m) 162 Relevant costs total computation, given per unit variable costs and fixed costs, number of units, and percentage change of per unit fixed costs that would continue (M79PI-2-36m) 111 Sell or process further Relevant costs (M78PI-2-29m) 7 Special order Operating income increase as result of accepting computation, given brief absorption costing income statement, fixed costs in manufacturing expenses, number of units and price in special order, no additional selling expenses, and per unit sales price and manufacturing cost (M79PI-2-38m) 112 Operating income increase as result of accepting special order computation, given variable manufacturing cost, selling price, and number of units (N79PI-2-35m) 162 Depletion Computation of amount, given cost of mine, development costs, value after ore extracted, ore amount, and amount of ore removed during year (N79PII-1-16m) 170 Computation of amount, given cost of mine, residual value of land, number of tons that might be extracted in total and number of tons extracted that year (M78PII-1-3m) 13 Computation should include charges for restoring the land after extraction according to contract (N78PII-1-9m) 66 Depreciation Change in accounting principle Computation of cumulative effect of change from accelerated to straight-line (M78PI-1-17m) 5 Computation of cumulative effect of change from double-declining-balance to straight-line (M78PII-1-llm ) 15 Double-declining-balance Computation and comparison with other specified methods for maximization of profits for accounting and for minimization of profits for income tax (M78PI-5) 11 Computation for year after equipment was purchased in midyear, ignores salvage, doubles straight-line rate, and takes and declines balance ½ year in first year (N78PII-1-8m) 66 Computation of amount taken over eight-and-one-half years of machine with expected useful life of ten years (N79PII-3) 174 Second-year-amount computation, given cost and useful life (N79PIM-15m) 170 Methods Conceptual rationale— systematic and rational (M78T-1-18m) 46 224 Depreciation (cont.) Straight-line Computation and comparison with other specified methods for maximization of profits for accounting and for minimization of profits for income tax (M78PI-5) 11 Computation, given cash price, installation costs, useful life and salvage value (M78PII-1-16m) 16 Computation includes price, shipping, installation, salvage value, and useful life for first year, the addition of pollution control items (that neither prolonged life nor had salvage) to cost in second year with undepreciated cost now depreciated over remaining life (N78Pn-1-llm) 66 Computation of amount taken over three-and-one-half years of machine with expected life of 7 years (N79PII-3) 174 Sum-of-the-years-digits Book-value-of-machine computation, given cost, useful life, salvage, and date at end of second year (N79PII-1-1m) 167 Comparison with other specified methods for maximization of profits for accounting and for minimization of profits for income tax (M78PI-5) 11 Computation of cost of equipment, given useful life, salvage value, and the amount of depreciation in the third year (M78Pn-1-15m) 16 Computation of last year’s charge, given cost, useful life, salvage, and this method (M79PII-1-16m) 118 Disagreement between assistant and auditor in charge Major difference of opinion concerning accounting issue, assistant believes necessary to disassociate himself from the resolution, both auditor and assistant must document details of disagreement and basis of resolution (M79A-1-4m) 125 Disclaimer See Auditor’s report Disclosure See also Accounting changes Accounting changes Change in estimate—footnote disclosure of effect on income before extraordinary, net income, and related per-share amounts (M78T-7) 51 Change in reporting entity—in initial set of statements after change disclose by footnote nature and reason, and disclose effect on earnings before extraordinary, net earnings and related per-share amounts for all periods presented, disclosures not repeated subsequently (M78T-7) 51 Accounting policies Basic statements and this included in complete set of statements (N78T-1-23m) 100 Footnote Always on normal transactions of assets acquired by lease, pension plans, and employee stock options, but not trade accounts receivable (N78T-1-44m) 102 Inadequate See also Auditor’s report—Opinion Client declines to make essential disclosures and auditor must provide them in his report and appropriately modify the opinion (N78A-1-9m) 76 Decide between qualified and adverse opinion (M78A-1-42m) 27 Primary responsibility for adequacy rests with management (N78A-1-28m) 78 Discontinued operations (disposal of segment) Computation of income from continuing operations (after income taxes) on revised comparative statements, given operating income in each of two years, loss of the segment in each of the two years, and income tax rate of 50% (N78PI-1-17m) 68 Computation of income or loss from discontinued operations after income taxes on revised comparative statements, given that Discontinued operations Computation of income or loss (cont.) company had income in each of the two years and an income tax rate of 50%, gain on sale of the segment (division), and loss of the segment in each of the two years (N78Pn-1-18m) 68 Net loss includes estimated loss of disposal and loss of segment during year of statement (M78PI-1-6m) 3 Discount on. purchases See Audit procedures—Purchase discount Dividends See also Federal income tax—Corporations See also Stock dividends Computation of total common dividends, given net income, dividends on preferred, and dividend-payout ratio on common (N78PI-1-7m) 55 Computation on shares outstanding (M78PI-3) 8 Worksheet showing maximum amount available for cash dividends and how distributable to common, noncumulative and nonparticipating preferred, and cumulative and fully participating preferred with dividends in arrears (M78PI-3) 9 Early extinguishment of debt See also Extraordinary items—Early extinguishment of debt Extraordinary gain computation (M78PII-1-13m) 15 Extraordinary loss computation (N79PII-1-14m) 170 Gain computation (N79PI-5) 165 Gains or losses Immediate recognition is the only acceptable method and, if material, must be reflected as extraordinary items (N78T-5) 103 Supporting arguments for each of the three theoretic methods (amortized over remaining life of old debt, over life of new debt, recognized in period) of accounting for gains or losses (N78T-5) 103 Earnings per share Accounting changes See also Accounting changes Change in estimate—footnote disclosure of effect (M78T-7) 51 Change in reporting entity—effect disclosed for all periods presented (M78T-7) 51 Capital structure (M78T-1-14m) 46 Common stock equivalent Defi nition and description of securities that would be considered such (N78T-2) 102 Dilutive used in computation of both primary and fully diluted (M79T-1-18m) 148 Example of substance over form (N78T-1-2m) 98 Complex capital structure Definition and disclosures (both financial and explanatory) necessary (N78T-2) 102 Computation, given number of common shares outstanding throughout year, preferred dividend requirement paid, and net income for the year (M79PI-1-5m) 106 Computation, given number of shares outstanding throughout year, additional shares issued at beginning of second quarter, net income, and dividends declared and paid to nonconvertible preferred (N79PI-1-6m) 157 Computation, given number of shares outstanding throughout year, number of shares issued at beginning of last quarter, and the net income (M79PI-1-4m) 106 Fully diluted Number of shares used in computing this and primary, given number of shares outstanding at beginning of year, number of shares and issuance dates of two issues during year, and 225 E Earnings per share Fully diluted (cont.) number of convertible bonds (not common stock equivalents) with conversion ratio issued at beginning of last quarter (N78PI-1-6m) 55 Not properly entitled, computed, or disclosed (M79T-2) 152 Number of shares used in computing primary and fully diluted, given number of shares outstanding throughout year, additional shares issued at beginning of second and of third quarters, and number of convertible bonds (not common equivalents and conversion rate given) issued at beginning of last quarter (N79PI-1-5m) 157 Primary Common stock equivalents recognized only if dilutive (N79T-1-29m) 204 Computation including convertible bonds (M78PI-1-7m) 3 Number of shares used in computing, given number of shares outstanding at beginning of year and number of shares and issuance dates of two issues during year, with additional information for fully diluted on issuance of convertible bonds (not common stock equivalents) at beginning of last quarter (N78PI-1-6m) 55 Economic order quantity See Quantitative methods EDP See Electronic data processing Electronic data processing Accounting control May leave no visible evidence indicating performance: test by reviewing transactions submitted for processing and comparing them to related output (N79A-1-23m) 180 Application controls Purposes of input controls, processing controls, and output controls (M79A-3) 133 Compliance testing Can be performed using actual transactions or simulated transactions (N79A-1-35m) 181 Controls Hardware— detect and control errors arising from use of equipment (M78A-1-20m) 25 Downtime will exist even in fully-protected system because of unscheduled maintenance (N79A-1-47m) 183 Dumping information for maintaining backup file Most efficient and least costly method is from disk to tape (N79A-1-31m) 181 Error reports are most probably reviewed and followed up by the EDP control group (N78A-1-37m) 79 Flowcharts See Internal control— Flowcharts General controls (M79A-3) 133 General knowledge of four given EDP topics that should be possessed by CPAs responsible for computer audits (M78A-4) 30 General purpose computer software package How helpful in inventory audit with tape of inventory file data (N78A-5) 84 Generalized audit packages Primary advantage enables auditor to utilize computer’s speed and accuracy (N79A-1-32m) 181 Hash totals Totals of amounts in computer-record data fields not usually added but used only for data processing control purposes (M79A-1-45m) 129 Internal control is lessened by computer operators having access to operator instructions and detailed program listings, but not by computer librarian maintaining custody of such instructions and listings, or by control group being solely responsible for distribution of output, or by programmers writing and debugging programs which perform routines designed by systems analyst (N78A-1-46m) 80 Electronic data processing Inventory Basic audit procedures and how general purpose computer software package and tape of inventory file data might be helpful (N78A-5) 84 Object program Machine-language program that results when a symbolic-language program is translated (M79A-1-14m) 126 On-line Destructive updating—well documented audit trail necessary to audit balances (N79A-1-19m) 180 On-line, real-time (OLRT) Internal control is strengthened by making a validity check of an identification number before a user can obtain access to the computer files (N78A-1-llm ) 76 Programs generally examined or directly tested in audit Include systems that affect a number of essential master files and produce a limited output, systems that update a few essential master files and produce no printed output other than final balances, and systems that perform relatively complicated processing and produce ver y little detailed output (M79A-1-46m) 130 Programs not generally or directly tested in audit System that performs relatively uncomplicated processes and produces detailed output (M79A-1-46m) 130 Real time processing (M79A-1-51m) 130 Software Compiler (N78A-1-8m) 76 Systems analyst normally assigned operating respo n sibility for designing installation, including flowcharts (N78A-1-6m) 75 Uses of computer to aid auditor’s examination of accounts receivable in fully computerized system (N79A-4) 185 Validity check Ascertains whether given characteristic belongs to the group (N79A-1-2m) 178 Employer and employee relationships See also Agency See also Contracts— Employment Employer liable for damages from employee’s disregard of instructions, irrespective of efforts to prevent or employer’s reasonable care (M78L-1-lm) 32 Federal Fair Labor Standards Act Piece rates may be employed in lieu of hourly rates where appropriate (M78L-1-47m) 40 Social Security Act applies to both employer and employees (N79L-1-49m) 197 Workmen’s compensation Compulsory within jurisdiction, but company did not provide— workman severely injured as result of his own negligence can sue company and company cannot resort to the usual common law defenses (N79L-1-48m) 196 Employee will recover from employer or its insurer if his injury was result of gross negligence on his and another employee’s part, but not if he intentionally injured himself, if he is an independent contractor, or if his claim is based on a disease unrelated to the employment (M78L-1-48m) 40 Grossly negligent workman suffering permanent disability as a result will find his claim paid in full (M79L-1-47m) 142 Legal implications of not having workmen’s comp, include being subject to common law actions in elective jurisdictions, being precluded from certain defenses, additional compensation payments over and above basic amounts, immediate lump-sum payment, fines, and imprisonment (N78L-3) 95 Purpose to provide system of compensation for employees (survivors or dependents) injured, disabled, or killed in course of their employment (N78L-3) 95 226 Engagement letter Best place to reduce preliminary arrangements to writing (M78A-1-8m) 23 Verbal commitment should have been reduced to writing (M78A-2) 29 EPS See Earnings per share Equipment See also Machinery and equipment Adjusting entry for incorrectly expensing in earlier year (N78PII-3) 77 Adjusting entry for incorrectly recording sale (N78PII-3) Book value of machine computation, given cost, useful life, salvage, sum-of-the-years-digits depreciation, and date at end of second year (N79PII-1-1m) 167 Cost computation (M78PI-5) 10 Cost purchased on installment plan is cash price plus installation costs (N78PII-1-6m) 66 Gain-on-sale computation, given purchase date and price, useful life, salvage, straight-line depreciation recorded monthly, and sale date and price (N79PII-1-4m) 167 Loss on sale of machine Computation, given purchase date and cost, salvage, monthly straight-line depreciation, and sale price (M79PII-1-15m) 118 71 Equity method Long-term investments Computation of income reported (M78PI-1-4m) Errors or irregularities See also Auditor’s responsibilities— Errors or irregularities Believed by auditor to exist Consider implications and discuss with appropriate levels of management (M78A-1-59m) 29 Internal control Evaluation of (not compliance with, review of, or study of) existing system focuses directly on purpose of preventing or detecting errors or irregularities (M78A-1-51m) 28 Irregularity Includes misappropriation of an asset or group of assets (M78A-1-28m) 25 Misappropriation of an asset or groups of assets (N79A-I-7m) 178 Missing invoices ignored (M78A-2) 29 Plan audit engagement to search for errors or irregularities that would have a material effect on the statements (M79A-1-5m) 125 Substantive tests Performed to mitigate risk of failure to detect material dollar errors in the statements (N79A-1-37m) 181 Ethics See AICPA Code of Professional Ethics Evidence See Generally accepted auditing standards— Standards of field work Expense recognition Allocation on basis of association with revenue or among periods considered to serve no useful purpose (M78T-1-25m) 47 No discernible future benefit (M78T-1-25m) 47 Not justified by income tax savings of immediate write-off versus allocating over periods (M78T-1-25m) 47 Expropriation Entry and computation of amount given book value, fair market and reimbursement percentage of fair market (M78PII-1-6m) 14 Extension of attest function Report on internal control based on an audit is indication of the changing role of the CPA (M79A-1-22m) 127 Extraordinary items Do not include gains or losses on disposal of a segment, adjustments of accruals on long-term contracts, or gains or losses from a fire (M78T-1-24m) 47 Early extinguishment of debt Extraordinary gain computation (M78PII-1-13m) 15 Ext raordinary loss computation (N79PII-1-14m) 170 Extraordinary gain improperly shown on retained earnings statement rather than income statement (M79T-2) 152 Flood damage considered such only if floods in the geographical area are unusual and occur infrequently (N78T-1-7m) 98 Governmental order to remove patented product from market because of health hazard results in expense (rather than extraordinary item) (M79PII-1-14m) Unusual and infrequent loss from hurricane damage (M79PII-1-13m) 118 Unusual in nature, infrequent, and material (N79T-1-16m) 203 Utilization of a net operating loss carryforward (M78T-1-24m) 47 118 FASB Interpretations Ethics require compliance FASB Statements No. 2 (M78PI-1-lm) (N78T-1-9m) 98.
(M79PI-3) 113, (N79T-1-9m) 203 No. 4 (M78PII-1-13m) No. 5 (M78PII-1-6m) (M78T-1-23m) 47, (N78T-1-13m) 99, (M79A-4) 133, (M79T-2) 151, (M78A-1-14m) 24 2 , (M78T-5) (N78T-1-19m) (M79PII-1-17m) 50, 99, 118, 15 14, (M78T-1-22m) (N78PI-1-5m) 54, (M79PIM-5m) 117, (M79T-1-5m) 147, (N79T-1-2m) 202, 47, (N79T-1-3m) 202 No. 7 (M78T-5) 50 No. 8 (M78PI-1-9m) 3, (M78T-3) 5 0 , (N78PI-1-8m) 55, (N78T-1-25m) 100, (M79PI-1-10m) 107, (M79PI-1-19m) 108, (M79T-1-2m) 147, (N79PI-1-9m) 157, (N79T-1-10m) 203 No. 12 (M78PI-1-■14m) 4, (M78T-1-10m) 4 (M78T-1-19m) 46, (N78Pn-1-4m) 65, (N78PII-3) 71, (N78A-1-53m) 81, (N78T-1-11m) 99, (M79PII-1-3m) 116, (N79T-1-30m) 205 No. 13 (M78PI-1-16m) 5, (M78T-1-13m) 4 (N78PII-1-3m) 65, (N78T-4) 103, (M79PI-1-llm ) 107, (M79PI-1-14m) 108, (M79T-1-6m) 147, (M79T-2) 151, (N79PI-1-14m) 158, (N79PI-1-15m) 158, (N79T-1-12m) 203 No. 14 (M79T-5) 153, (N79PI-1-16m) 159, (N79T-1-25m) 204 No. 16 (N78T-1-8m) 98 46, Federal Fair Labor Standards Act Piece rates may be employed in lieu of hourly rates where appropriate (M78L-1-47m) 40 Federal income tax Allocation Computation (M78PI-1-2m) 227 Federal income tax Allocation (cont.) Computation involving repair expense and interest on municipals (M78PI-1-15m) 4 Deferred income tax liability would not arise from a permanent difference and improper method of computing deferral used in any case (M79T-2) I5I Deferred-taxes-charged-to-income-statement computation, given cost, no salvage, and useful life of machine, sum-of-the- years-digits for income tax and straight-line for accounting, and tax rate (M79PI-1-2m) 106 Income-tax-expense computation, given income per books before income taxes, tax rate, and any differences in book and tax are timing (N79PII-1-12m) 169 Interperiod Deferred method Gross-change-method description (N79T-4) 207 Net-change-method description (N79T-4) 207 Justified by theory that income taxes should be treated as an expense (M79T-1-14m) 148 Intraperiod Computed as difference between tax computed based on taxable income without including the item and tax computed based on taxable income including the item (M79T-1-9m) 148 Extraordinary gains or losses (N79T-1-17m) 203 Extraordinary gains or losses requires intraperiod allocation (N78T-1-14m) 99 Permanent differences No provision for deferred income tax on municipal interest (N78PI-1-lm) 54 Premiums on officers’ life insurance have no effect on provision for deferred income taxes (N79PI-1-12m) 158 Provision-for-deferred-income-tax-in-income-statement computation, given estimated loss on disposal of unused plant facilities and effective income tax rate (N79PI-1-12m) 158 Basis Trade-in (M78PII-2-18m) 16 Corporations Accumulation of earnings and profits where corp. continuously elects not to pay dividends may be taxed as unreasonable or ambitious expansion plans would normally constitute a bona fide business purpose and prevent such tax (N79L-2) 197 Balance due from previous year’s return not deductible (M79PI-5) 114 Bonus to officers (N79PI-3) 163 Charitable contributions (N79PI-3) 163 Carryover (M79PI-5) 114 Deduction limitation (M78PI-4) 9, (M79PI-5) 114 Limit of 5% of taxable income computed before charitable deductions or special deductions made up partly of carryover and t h is year’s contributions (N78PII-2-25m) 69 Christmas bonus to employees deductible (M79PI-5) 114 Contributions to local tax-exempt private college (N79PI-3) 163 Contributions to pension plan trust (M79PI-5) 114 Contributions to profit-sharing trust (N79PI-3) 163 Contributions to various indigents not deductible (M79PI-5) 114 Depreciation (N78PII-2-35m) 70 Dividends From affiliated corporation (M78PI-4) 9 From current earnings of wholly owned domestic subsidiary (N79PI-3) 163 From domestic corporations (M78PI-4) 9, (N79PI-3) 163 From wholly owned domestic subsidiary (M79PI-5) 114 In cash and in treasury stock, effects or implications to shareholder (N79L-2) 197 9.
Federal income tax Corporations (cont.) Marketable securities, not an inventory asset, reduce accumulated earnings and profits by their cost, not their higher market (N78PII-2-19m) 68 Marketable securities, not an inventory asset, with a market value above cost do not result in a gain on distribution (N78PII-2-21m) 69 Dividends received deduction (M78PI-4) 9 Computation (M79PI-5) 114 Involving dividends received from domestic corporations and dividends received from current earnings of wholly owned domestic subsidiary (N79PI-3) 163 Employer’s share of social security taxes is deductible, but was used in a previous calculation in example (M79PI-5) 114 Equipment sale (N78PII-2-34m) 70 Estimated tax payments (M78PI-4) 9, (M79PI-5) 114 Federal-income-tax-due computation (M79PI-5) 114 Finance charges on gasoline credit cards (N79PI-3) 163 Fire loss involving value before and after fire and insurance (N79PI-3) 163 Foreign tax credit (M79PI-5) 114 Income from investments of pension trust not taxable (M79PI-5) 114 Interest income from corporate bonds (N79PI-3) 163 Interest on loan to purchase municipal securities not deductible (N78Pn-2-32m) 70 Investment credit (M78PI-4) 9 Investment tax credit carried over Long-term capital gain (M79PI-5) (N79PI-3) 163 Long-term capital loss (N79PI-3) 163, (M78PI-4) 9 Municipal securities— interest not included in taxable income (N78PII-2-32m) 70 Net capital gain (M78PI-4) 9, (M79PI-5) 114, (N79PI-3) 163 Net long-term capital gain (N79PI-3) 163 Net long-term capital loss Not deductible against ordinary income (N78PII-2-32m) 70 Offset against net short-term capital gain, remainder cannot be deducted, but is available for carryback and carryover, so that total taxable income is the given taxable income from operations (N78PII-2-20m) 68 Net operating loss (N78PII-2-28m) 69 Net short-term capital gain (N79PI-3) 163 Operating loss deduction (M78PI-4) 9 Pension costs (M78PI-4) 9 Pension costs carryover (M78PI-4) 9 Pensions paid to retirees by pension trust not deductible (M79PI-5) 114 Personal holding company— accumulated earnings tax does not apply (N78PII-2-33m) 70 Profit-sharing costs (M78PI-4) 9 Property distribution—difference between fair market value and lower sale price of property to minority shareholder must be reported (N78PII-2-22m) 69 Provision for federal income taxes not deductible in determining taxable income (N78PU-2-32m) 70 Rental income (N79PI-3) 163 163 (M79PI-5) 114 114, (M78PI-4) Royalties from patent on invention (N79PI-3) Short-term capital gain (N79PI-3) 163, (M78PI-4) 9 Short-term capital loss (M79PI-5) 114, (N79PI-3) 163, (M78PI-4) 9 Social security taxes withheld from employees not deductible (M79PI-5) 114 Stock dividends effects or implications to shareholder (N79L-2) 197 Stock dividends f r o m treas ury stock and cash effects or 228 Federal income tax Corporations (cont.) implications to shareholder (N79L-2) 197 Stock split effects or implications to shareholder (N79L-2) 197 Subchapter S Should not have more than one class of stock (N78PII-2-27m) 69 Stockholder’s gross income is his share of taxable income and current earnings and profits reported (N78PII-2-29m) 70 Taxable income computation, given that any alternative treatments should be resolved to minimize such figure (N79PI-3) 163 Taxable income lower than book income because organization costs written off over 5 years for tax and 10 years for books (N78PII-2-24m) 69 Tax-free exchange— no gain or loss on exchange of stock for property (N78PII-2-31m) 70 Treasury stock purchase is not deductible (M79PI-5) 114 Depreciation Computation and comparison of straight-line, double-declining- balance, and sum-of-the-years-digits for minimization of profits for income tax and maximization of profits for accounting (M78PI-5) 11 Dividends from shares of another company included in dividend income at FMV received subject to the dividend exclusion, and their tax basis is FMV (M79L-3) 144 Equipment Trade-in (M78PII-2-18m) 16 Error in previous return discovered Advise client to file a corrected return regardless of whether or not error resulted in overstatement or understatement of tax (N79A-1-20m) 180 Individual Adjusted gross income Computation (N78PI-5) 62 Computation involving 50% deductible net long-term capital loss and short-term capital gain (M79PII-2-37m) 121 Computation of college student (M79PII-2-24m) 119 Computed after deducting alimony (but not child support) (M79PII-2-22m) 119 Includes deduction for loss on rental property resulting from rental income received less ½ of real estate taxes (½ of house was used as personal apartment), of annual fire insurance premium, ½ of depreciation for entire house, and all painting expense of rental area (M79PII-2-40m) 121 Includes salary, interest (except on municipal obligations), and periodic payments of alimony, but not child support payments resulting from divorce (M78PII-2-34m) 18 Not affected by gift or by sale of gift between donor’s basis and fair market value at date of gift (M78PII-2-19m) 16 Allowance for moving expenses included in gross income (M79PII-2-26m) 120 Apartment building owner— lives in one of 10 apartments and must treat as a personal expense one-tenth of fuel and depreciation costs in computing (M78PII-2-25m) 17 Bad debt from noncollection of bill for personal services of carpenter not deductible because no income on this bill was ever reported (on cash basis) (N79PII-2-24m) 171 Bad nonbusiness debt deductible as short-term capital loss (M79PII-2-36m) 121 Bad personal debt from loan of two years results in short-term capital loss (N79PII-2-28m) 172 Bond purchased at a discount must have the basis adjusted each year by the amortized portion of the discount (included income) to arrive at the basis for long-term capital gain on sale computation (M79PII-2-28m) 120 Cash prize awarded in contest included in adjusted gross income (M79PII-2-24m) 119 Federal income tax Individual (cont.) Casualty Loss claim computed by deducting fair market value of personal residence after the fire from fair market value before the fire, less insurance proceeds, and less the $100 limitation (N79PII-2-23m) 171 Loss deduction equals decrease in trade-in value of auto (M78PII-2-22m) 17 Loss-to-home-from-damage-by-cyclone computation, difference between value before and value after casualty exceeds adjusted basis, and insurance proceeds and $100 limitation used to reduce adjusted basis to amount deductible (M79PII-2-32m) 120 Charitable contribution deduction for rare books to museum not allowed where right to use and possession are reserved to the contributor during his lifetime (N79PII-2-37m) 173 Child care credit allowable (N78PI-5) 63 Contribution to Keogh retirement plan, including January contribution (N78PI-5) 62 Country club dues deductible to extent incurred in connection with business (N78PI-4) 62 Damages for personal injuries not included in gross income (M78PII-2-26m) 17 Death benefits paid to widow by employer of husband excludable from adjusted gross income up to $5,000 (N79PII-2-40m) 174 Deductible expenses on overnight business trips include plane fares, hotels, meals, and business entertainment (M79PII-2-39m) 121 Deductions for adjusted gross income do not include unreimbursed dues to AICPA by an employee of an accounting firm, unreimbursed union dues by an employee of a company, or medical expenses of a self-employed individual (M79PII-2-27m) 120 Deductions-from-adjusted-gross-income computation, involving union dues and employment agency fees paid in securing a new job in same profession, but not hobby expenses (not engaged in for profit) or legal fees paid in connection with a libel suit (M79PII-2-31m) 120 Defined-contribution retirement plan—maximum contribution deduction is 15% of earned income from self- employment (M78PII-2-31m) 18 Depreciation deduction, considering 20% of first $10,000 additional first-year deduction, and salvage of less than 10% may be ignored (to get maximum depreciation) under straight-line (N79PII-2-39m) 174 Dividend exclusion does not apply to dividends from foreign corporation (M79PII-2-23m) 119 Dividend exclusion not fully used by one spouse on separately owned stock not available to other spouse on his separately owned stock on joint return (M79PII-2-23m) 119 Dividend income Computation, involving exclusion not applying to Subchapter S corporation dividend paid out of current earnings and profits or to foreign corporation dividends (N79PII-2-33m) 173 Does not include dividends on life insurance policy (M79PII-2-23m) 119 Joint return— liquidating dividend is not dividend income, exclusion of $200 allowed on joint return and dividend from domestic corporation with earnings entirely from municipal bond interest is included (M78PII-2-33m) 18 Less exclusion (N78PI-5) 63 Net of allowable dividend exclusion computation on joint return with each spouse separately owning stock, and foreign company dividends (M79PII-2-23m) 119 Dividends received less dividend exclusion included in adjusted gross income (M79PII-2-24m) 119 Divorcee gross income includes alimony payments (excluding child support) received during year and mortgage payments made by husband on home transferred to her, but not 229 Federal income tax Individual (cont.) repayment of a loan she made to husband (N79PII-2-25m) 172 Earned-income-credit-available computation, involving divorced head of household, interest income, wages, and earnings from self-employment (N79PII-2-32m) 775 Employment agency fees paid in securing a new job in same profession are deductible from adjusted gross income (M79PII-2-31m) 120 Exchange of real estate, both subject to mortgages, results in gain recognized as long-term capital gain to extent of difference in mortgage assumed (N79PII-2-27m) 772 Exemptions— couple may claim exemptions for themselves on joint return and one exemption for blind son they fully support, but no exemption for student daughter who provides 70% of her own support (M78PII-2-23m) 77 Exemptions on joint return for the two taxpayers and the daughter they fully support, but no extra exemption for the daughter’s blindness or for a son who provides most of his own support (N79PII-2-22m) 777 Exemptions on joint return that may be used in year of death of a spouse for each spouse and extra exemptions for each spouse 65 or older (as of date of death, if applicable) (M79PII-2-29m) 120 Fellowship grants are tax exempt and not included in gross income (M78PII-2-21m) 77 Gift— sale in 1977 of property received by gift in 1977 at a price below the donor’s basis (basis for gain) and above the fair market value at the date of the gift (basis for loss) results in zero adjusted gross income (M78PII-2-19m) 76 Gross income computation involving salary, year-end bonus, and allowance for moving expenses, but not medical insurance premiums paid by employer (M79PII-2-26m) 120 Gross income does not include workmen’s compensation, damages for personal injuries, or reimbursement of medical expenses (M78PII-2-26m) 77 Gross-income-on-lease computation includes rental payments received during last year of lease but not fair market value of improvements left by lessor (N79PII-2-31m) 772 Head of household Unmarried woman maintains home in which she and her unmarried son, not a dependent, reside (M79PII-2-21m) 779 Widow for 3-4 years maintains her home in which she and her dependent son reside (M78PII-2-28m) 18 Hobby expenses (not engaged in for profit) not deductible from adjusted gross income (M79PII-2-31m) 120 Individual-retirement-accounts-maximum-contribution computation involving one spouse without compensation (types counted for IRAs), joint return, husband not covered by any qualified retirement plan, and the maximum contribution under such circumstances of $1,750 (M79PII-2-25m) 120 Interest expense (N78PI-5) 63 Deductions include interest on home mortgage, installment charge accounts, and (2 payments of 12 payments) times interest deducted in advance by bank to be repaid in equal installments, but not interest on loan to purchase tax-exempt bonds (N79PII-2-21m) 777 On borrowings to buy corporate bonds, on mortgage, and on installment purchases if deductible, but not interest on borrowings to purchase tax exempt bonds (M79Pn-2-35m) 727 Interest income (N78PI-5) 63 Interest on assessment of deficiency on federal income tax return is deductible, but not the penalty or additional tax (M79Pn-2-34m) 727 Joint return (N78PI-5) 62 Keogh plan deductible contributions limited to 15% of earned income from self-employment (N79PII-2-34m) 173 Legal fees paid in connection with a libel suit not deductible from adjusted gross income (M79PII-2-31m) 120 Federal income tax Individual (cont.) Life insurance dividends not dividend income (M79PII-2-23m) 779 Life insurance proceeds elected to be paid in annual installments, amount subject to tax is annual installment amount less face of policy divided by number of installments, and less $1,000 for spouse of deceased (widow) (N79PII-2-26m) 772 Long-term capital gain in exchange of real estate (N79PII-2-27m) 772 Long-term capital gain on sale of bond (M79Pn-2-28m) 120 Long-term capital loss (N78PI-5) 62 Loss— related interests nondeductible (M78PII-2-20m) 16 Medical deduction Computation (N78PI-5) 63 Computation involves knowledge of deductibility of one-half of medical insurance premium (not exceeding $150), medicines and drugs must exceed 1% of adjusted gross income before the excess may be grouped with other medical expenses for the 3% test, medical expenses include crutches, and excess medical insurance premium, excess medicines and drugs, and other medical expenses are grouped and deductible to extent they exceed 3% of adjusted gross income (M78PII-2-24m) 77 Given adjusted gross income, medical insurance premiums, prescribed medicines, doctors, transportation to and from doctors’ offices, and reimbursement from insurance (N79PII-2-36m) 173 Involving drugs and medicines prescribed, health insurance premiums, doctors’ fees (some reimbursed), and eyeglasses and repairs (M79PII-2-30m) 120 Medical insurance premiums (N78PI-5) 63 Medical insurance premiums paid by employer not included in gross income (M79PII-2-26m) 120 Miscellaneous itemized deductions of plumber include specialized work clothes (required by employer), union dues, and cost of income tax preparation, but not will preparation or safe deposit box for personal items (N79PII-2-38m) 174 Net long-term capital loss (M78PII-2-30m) 18, (M79PII-2-37m) 727 Net operating loss computation, given self-employed net loss, interest and dividend income, itemized deductions including net casualty loss, and personal exemption (N79PII-2-30m) 772 Net short-term capital gain (M79PlI-2-37m) 727 Net short-term capital loss (M78PII-2-30m) 18 Nonbusiness debt becomes worthless and is treated as a short term capital loss (M78PII-2-29m) 18 Real estate taxes (N78PI-5) 62 Reimbursed medical expenses not included in gross income (M78PII-2-26m) 77 Related interests— losses nondeductible (M78PII-2-20m) 76 Rental income (loss) used in computing adjusted gross income computation involving ½ of dwelling used for personal apartment, leaves ½ of real estate taxes, Vi of annual fire insurance premium, ½ of depreciation for entire house, and all of painting expense on rental area deductible from the rent received, resulting in loss (M79Pn-2-40m) 722 Rental income (net)— owner of apartment building lives in one of 10 apartments and must treat as a personal expense one- tenth of fuel and depreciation costs in computing (M78PII-2-25m) 77 Sales tax (N78PI-5) 62 Scholarship excluded from income and adjusted gross income (M79Pn-2-24m) 779 Self-employed Deductible contributions to qualified retirement plan on behalf of a self-employed individual under ERISA limited to 15% of earned income from self- employment (N79PII-2-34m) 173 230 Federal income tax Individual (cont.) Income computation (N78PI-5) 62 Income computation involving nondeductible salary drawn by the self-employed person, nondeductible federal self- employment and income taxes paid, deductible malpractice insurance premiums, and deductible cost of attending professional seminar (M79PII-2-38m) 121 Maximum contribution-to-defined-contribution-retirement-plan deduction is 15% of earned income from self- employment (M78PII-2-31m) 18 Net earnings involves gross receipts, cost of sales, other operating expenses and state business taxes paid, but not dividend income on personal investments or federal self- employment tax paid (M78PII-2-32m) 18 Short-term capital loss results from personal bad debt from loan of two years (N79PII-2-28m) 172 State gasoline taxes (N78PI-5) 62 State income tax, including estimated (N78PI-5) 62 State income tax refund (N78PI-5) 63 Surviving spouse filing of widower who did not remarry but continued to maintain his home in which his two dependent children lived, and had filed a joint return with his wife in the previous year, the year of her death (N79PII-2-29m) 172 Taxes deductible from adjusted gross income include state income tax withheld and state sales taxes paid, but not federal income tax or FICA withheld, federal auto gasoline taxes, or federal excise tax on telephone bills (M79PII-2-33m) 121 Taxes paid deductible as an itemized deduction include state income tax, real estate taxes on land in South America (held as an investment), and state sales taxes, but not federal income tax, federal self-employment tax, or state unincorporated business tax (N79PII-2-35m) 173 Total itemized deductions computation (N78PI-5) 62 Union dues deductible from adjusted gross income (M79PII-2-31m) 120 Workmen’s compensation not included in gross income (M78PII-2-26m) 17 Partnership Basis to partner (N78PII-2-23m) 69 Ordinary income includes operating income but not deduction of charitable contributions, long-term capital gains, or dividends received from domestic corporations (N78PII-2-26m) 69 Sale of machinery to partner below adjusted basis (N78Pn-2-30m) 70 Stock dividends not taxable, but new basis for shares computed by allocation (M79L-3) 144 Trade-in Basis of new truck includes adjusted basis of old plus additional payment required (M78PII-2-18m) 16 Federal securities regulation Accountant’s legal liability Unqualified opinion on incorrect statements— best defense of four listed against third party purchaser of securities sold in interstate commerce, action had not been commenced within one year after the discovery of the material misrepresentation (N78L-1-llm ) 87 Auditor of statements accompanying registrations statement required to review events between date o f report and date of public sale or will be liable (M78L-3) 42 Brokerage firm Customers might have a common law action for negligence, or might prove liability under Sec. 17 of Securities Exchange Act of 1934 (N78L-4) 95 Corporation planning to issue common stock to the public in interstate commerce after its registration statement with the SEC becomes e ff ective must also make a filing in those states (in which the securities are to be sold) that have laws governing such o ff erings and obtain their Federal securities regulation Corporation planning to issue common stock (cont.) approval (M79L-1-43m) 142 Insider includes executive vice president, an 11% owner (8% of which he owns in his own name and 3% in an irrevocable trust for his or her benefit for life), or director who owns less than 10% of the shares of stock of the corporation, but not a major debenture holder (M79L-1-49m) 143 Insider short-swing transaction Vice president must account to corporation for his profit (N79L-1-27m) 192 Insider trading Shareholder must sue for and on behalf of the corporation and establish that the purchases and sales of a director occurred within less than six months of each other and at a profit to the director (N78L-1-33m) 91 Issuer Stockholder is probably an issuer if he sells 40% of his shares through an investment bank under its normal procedures (M78L-1-6m) 33 Private placement Prompt resale by purchasers would probably negate (M78L-1-llm ) 34 Sale by several brokerage houses in ordinary course of business will subject senior vice president, 15% shareholder, and member of Board of Directors selling 10% of his stock in the company to SEC registration requirements (M79L-1-48m) 143 Securities Act of 1933 Accountant’s legal liability— any materially false or misleading financial statements, including omission of material fact, provided purchaser proves such existed and the specific securities were the ones offered through the registration statement (N79L-1-29m) 192 Exempt from registration— usual annuity contract issued by insurer, but not first mortgage bonds, convertible preferred, or limited partnership interests (N79L-1-28m) 192 Statute of limitations— action must be brought within one year of discovery of untrue statement or omission, or after such should have been made by reasonable diligence, but in no event more than three years after security was bought in good faith (N78L-4) 95 Securities and Exchange Commission— Regulation A Offering types and general requirements which must be met to qualify (N78L-4) 95 Same liability for false statement or material omission that applies to a full registration (N78L-4) 95 Securities Exchange Act of 1934 Antifraud provisions of Sec. 10(B) and Rule 10b-5 do not apply in customer suit against negligent CPA audit, because fraud is intentional and requires more than showing negligence (N78L-4) 95 Corporation seeks recovery of short-swing profits of ousted president (N79L-1-32m) 193 Major provisions that do apply and those that do not apply to given corporation and its officers, directors, and principal shareholders after a public offering (N78L-4) 95 Field work See Generally accepted auditing standards— Standards of field work Financial Accounting Standards Board (M78T-3) 50, (M79T-5) 153 Financial Accounting Standards Board Interpretations See FASB Interpretations Financial statement analysis Common size statements— enables financial analyst to compare mix of assets, liabilities, capital, revenue, and expenses within a company or over a period or between companies without respect to relative size (M78T-2-40m) 48 231 Financial statements Identify inclusions or exclusions from illustrated statements that would violate GAAP and indicate corrective action (M79T-2) 151 Statement of forecasted results of operations not considered a financial statement covered by the standards of reporting (M79A-1-43m) 129 Fixed assets See a lso Audit procedures See a lso Equipment See a lso Machinery and equipment Loss on sale would indicate that proceeds were less than book value (M79T-1-13m) 148 Lump-sum purchase Land, warehouse, and office building purchased for given total price allocated on relative current assessed values, rather than vendor’s original costs (M79PII-1-19m) 119 Flowcharts Aids in evaluation of a series of sequential processes (M78A-1-17m) 24 Example symbols mean a document has been generated by a manual operation (M79A-1-34m) 128 Interpret symbols and partially complete a flowchart on charge sales activities telling the internal control procedure or internal document (M79A-2) 131 Record auditor’s understanding of client’s internal accounting control system (N78A-1-42m) 79 Footnotes See a lso Disclosure Initial footnote should describe all significant accounting policies (M79T-2) 152 Forecasts See a lso Budgets Statement of forecasted results of operations not considered a financial statement covered by the standards of reporting (M79A-1-43m) 129 Foreign exchange Foreign currency translation Change in rate between date of transaction and date of statements would give rise to exchange gain or loss if asset or liability being translated is carried at a price in a current purchase or sale exchange (N78T-1-25m) 100 Current rate is used to translate accounts receivable of foreign subsidiary (M79T-1-2m) 147 Denominate Define and distinguish between denominate and measure and give brief example demonstrating the distinction (M78T-3) 5 0 Exchange gain or loss in consolidated income statement computa tion, given amount of exchange loss from translation of wholly owned foreign subsidiary for year, converted amount of receivable from foreign customer payable in his currency at end of previous year, and exchange rate when he paid it early in the current year (N79PI-1-9m) 157 Fixed assets Historical rate used (M78T-1-2m) 4 5 Gain computation, given receivable (in foreign currency) valuation at year end, amount receivable converted into when collected shortly thereafter, and exchange gains from translation of another foreign subsidiary (N78PI-1-8m) 55 Historical rate used for equipment and its accumulated depreciation (N79T-1-10m ) 203 Inventory computation (inventory in local currency units divided by exchange rate at time of purchase of the inventory) (M79PI-1-19m) 108 Marketable equity securities and inventories carried at cost translated at historical rates while such items carried at current market or net realizable value translated at current Foreign exchange Foreign currency translation (con t.) rates (M78PI-1-9m) 3 Measure Define and distinguish between measure and denominate and give brief example demonstrating the distinction (M78T-3) 5 0 Temporal method of translating statements Define and include in answer treatment of long-term accounts receivable, deferred income, inventory valued at cost and long-term debt (M78T-3) 5 0 Use of historical rates to convert depreciation on fixed assets of foreign subsidiary to amount on income statement of parent computation, given acquisition price, date, and rate at time of two acquisitions, straight-line, useful life, and no salvage (M79PI-1-10m) 107 Fraud See Accountant’s legal liability See Audit procedures Fund accounting See also Governmental accounting S ee a lso Municipalities See also State and local government See a lso University accounting S ee also Voluntary health and welfare organizations University accounting Entries to record summary of transactions (N78PII-5) 73 Statement of changes in fund balances preparation (N78PII-5) 73 GAAP See Generally accepted accounting principles GAAS See Generally accepted auditing standards General standards S ee Generally accepted auditing standards Generally accepted accounting principles See also AICPA Code of Professional Ethics— Generally accepted accounting principles Auditor’s judgment concerning overall fairness applied within this framework (M79A-1-60m) 131 Exception Recording goodwill from revaluation is not acceptable (M78A-1-lm) 23 Publications that qualify as statements of such under the AICPA Code of Professional Ethics include accounting interpretations issued by the FASB, AICPA Accounting Research Bulletins, and Statements of Financial Standards issued by the FASB, but not accounting interpretations issued by the AICPA (N78A-1-33m) 79 Generally accepted auditing standards Acceptance of an engagement after fiscal year end First ascertain whether circumstances are likely to permit an adequate examination and expression of unqualified opinion (M78A-1-15m) 24 Detection of illegal acts not assured by examination in accordance with GAAS (N79A-1-25m) 180 General standards Adequate technical training (M79A-1-3m) 125 First— Adequate technical training and proficiency as an auditor is met by education and experience in auditing (N79A-1-48m) 183 Independence must be without bias (M78A-1-19m) 25 Lack of independence (M78A-1-54m) 2 8 232 G Generally accepted auditing standards (cont.) Measures of quality of auditor's performance (M79A-1-25m) 127 Standards of field work Evidence— audit program procedures outlined are primarily to gather evidence (M78A-1-23m) 25 Materiality and relative risk underlie application (M79A-1-27m) 127 Planning Acceptance of engagement after fiscal year end (M78A-1-15m) 24 Audit Program (M78A-1-16m) 24 Preliminary conference with client helps prevent misunderstandings and inefficient use of audit personnel (M78A-1-52m) 28 Tour of client’s facilities and review of the general records is most likely first step of those given (M78A-1-55m) 29 Planning and supervision Audit program is basic tool to control the audit work and review the progress (N78A-1-35m) 79 Audit program provides proof that the work was adequately planned (N79A-1-58m) 184 Need for supervision of engagement affects quantity, type, and content of working papers (M78A-1-34m) 26 New client in unfamiliar business— AICPA Industry Audit Guides most useful source of those given for information during the preliminary planning stage (M78A-1-45m) 27 Obtain a knowledge of matters that relate to nature of entity’s business if accept audit engagement without industry expertise (N79A-1-llm ) 179 Sufficient competent evidential matter Examine evidence that includes written client representations, vendor invoices, and minutes, but not client accounting manuals (N78A-1-26m) 78 Persuasiveness— documents mailed by outsiders to auditor, correspondence between auditor and vendors, and computations made by auditor are all more persuasive than sales invoices inspected by the auditor (M79A-1-31m) 128 Primary purpose of audit program procedures is to gather corroborative evidence (M79A-1-2m) 125 Ultimately based on judgment of auditor (N78A-1-57m) 82 Standards of reporting Consistency Accounting changes Changes in estimate— standard does not apply (M78A-1-18m) 24 Standard applies to change in principle not generally accepted, change in repotting entity, and change in principle inseparable from a change in estimate (M78A-1-18m) 24 Exception when there is a change in format and presentation of the statement of changes in financial position from cash concept to working capital (N78A-1-3m) 75 First examination (M78A-1-31m) 31 No exception for change in estimated salvage, change in classification of an expense from “other” to “selling expense,” or for change in inventory pricing to correct math error in total of opening inventory (N78A-1-3m) 75 Financial statements— include, for reporting purposes, statements of assets and liabilities from cash transactions, statements of changes in owners’ equity, statements of operations by product line, but not statements of forecasted results of operations (M79A-1-43m) 129 First standard does not apply when asked to perform examination to express opinion on one or more specified elements, accounts, or items of a statement (N79A-1-18m) 180 Fourth standard Comparative statements (M78A-1-25m) 24 Prevents misinterpretation of degree of responsibility the Generally accepted auditing standards Standards of reporting (cont.) auditor is assuming when the auditor’s name is associated with the statements (N78A-1-20m) 77 “Taken as a whole” applies equally to a complete set of statements and to each statement (N78A-1-23m) 77 Materiality and relative risk underlie application (M79A-1-27m) 127 Going concern Continuation of accounting entity in absence of evidence to contrary (N79T-1-27m) 204 Justifies accruals and deferrals (M78T-1-7m) 45 Goodwill See also Consolidated financial statements See also Intangible assets Computation of capitalization of excess earnings at 10% should not use extraordinary gains (N78PII-1-7m) 66 Consolidated Computation of amount, given purchased all (number of shares) shares of company, price per share, and amount of fair value of assets in excess of book value (N79PII-1-3m) 167 Matched on basis other than cause and effect (M78T-1-20m) 47 Recording goodwill from revaluation is not acceptable (M78A-1-lm) 23 Result of investing for 45% ownership and 30% ownership in two companies, computation of amount (considering fair value of fixed assets of one company exceeds book value), amortization, and accumulated amortiza tion (M79PI-3) 112 Governmental accounting See also Municipalities See also State and local government Budget First of given steps in acquisition of goods and services (N79T-1-47m) 206 Recording credit to fund balance— estimated revenues exceed appropriations (M78T-2-26m) 4 7 Budgetary account Appropriations (M79T-1-49m) 151 Debt Service Fund Interest expense on bonds payable recorded when legally payable (N79T-1-45m) 206 Enterprise fund Encumbrances would not appear (N79T-1-44m) 206 Fixed assets accounted for in a manner similar to “for-profit” organization (M78T-2-28m) 4 7 Expenditure Actual cost rather than an estimate (N79T-1-48m) 206 Funds (three) accounted for in a manner similar to a for-profit entity are Special Assessment, Enterprise, and Intragovernmental Service (M79T-1-41m) 150 General Fund Expenditure Repairs made and bill received (N79T-1-41m) 206 Truck received (N79T-1-49m) 206 Wages earned by employees but not paid (N79T-1-50m) 206 Property taxes recorded directly in general fund (M78T-2-31m) 48 Intragovernmental service fund Data processing center to service all agencies within a governmental unit (M78T-2-29m) 47 Not an expendable fund (N79T-1-43m) 206 Modified accrual method of accounting Property tax is susceptible to accrual but not state sales tax, income tax, or business licenses (M78T-2-27m) 4 7 Property taxes usually recorded before received (N79T-1-42m) 206 No counterpart to the land, building, and equipment fund in a 233 Governmental accounting No counterpart to the land (cont.) voluntary health and welfare organization (M79T-1-47m) I5I Reserve for encumbrances— past year Expenditures for which purchase orders were made in prior year but disbursement will be in current year (M78T-2-30m) 47 Revenue Business licenses are not susceptible to accrual using modified accrual method (M78T-2-27m) 47 Income tax is not susceptible to accrual using modified accrual method (M78T-2-27m) 47 Property tax Recorded directly in the general fund (M78T-2-31m) 48 Susceptible to accrual using modified accrual method (M78T-2-27m) 47 State sales tax is not susceptible to accrual using modified accrual method (M78T-2-27m) 47 Separate funds used because diverse nature of services offered and legal provisions make it necessary to segregate activities by functional nature (M79T-1-50m) 151 Special revenue fund Expenditures account appears in (N79T-1-46m) 206 University accounting Entries to record summary of transactions (N78PII-5) 73 Statement of changes in fund balances preparation (N78PII-5) 73 Governmental auditing Extends beyond opinion and includes audits of efficiency, economy, effectiveness, and compliance (N78A-1-18m) 77 Governmental order to remove patented product from market because of health hazard results in expense (rather than extraordinary) (M79PII-1-14m) 118 H Health and welfare organizations See Voluntary health and welfare organizations I Illegal acts by clients Auditor believes client may have committed illegal acts (as result of audit procedures) and should immediately inquire of management and consult with client’s legal counsel or other specialists, as necessary, to understand nature of acts and possible effects (M79A-1-39m) 129 Auditor unable to determine amounts associated Qualify or disclaim opinion (M79A-1-36m) 128 Examination in accordance with GAAS should not be relied upon to provide assurance that such acts will be detected (N79A-1-25m) 180 Review of internal administrative control may bring to auditor’s attention (M79A-1-29m) 128 Incentive commission expense Computation involving addition of amounts applicable to particular year—accrual accounting, and matching (N78PII-1-2m) 65 Incentive compensation plans See Bonus Income statement Discontinued operations Net loss includes estimated loss of disposal and loss of segment during year of statement (M78PI-1-6m) 3 Income statement (cont.) Overall review Most important of four given procedures was compare actual revenues and expenses with previous year and investigate significant differences (N78A-1-10m) 76 Independence See Generally accepted auditing standards— General standards See Quality control— Independence Independent auditor Appointed by board of directors or elected by stockholders of publicly-traded corporations to emphasize auditor independence from management (M79A-1-49m) 130 Role and responsibilities (M79A-1-28m) 128 Industry segments See Segments Installment sales See also Quantitative methods— Present value— Present value of an annuity Interest income Computation for second year, given cash price, interest rate, and annual payments (N78PI-1-9m) 55 Computation for third year, given cash sale price at beginning of first year when sold, amount of annual payments (which included interest) at end of each of five years, and interest rate (N79PI-1-19m) 158 Insurance Adjusting entry for expensing three-year insurance premium (N78PII-3) 71 Fire Coinsurance requirement met (M78L-1-17m) 35 Partial loss not fully recoverable on policy for full value of property if insured had a policy with another company for a total of insurance exceeding that full value (M79L-1-2m) 135 Recovery-amount computation, given building value, coinsurance clause, face of policy, and loss (N78L-1-7m) 87 Sells insured property, but retains policy, will not be able to collect in event of destruction by fire (N78L-1-13m) 88 Total loss entitles lessee to recover damages to extent of the value of his leasehold interest (M78L-1-20m) 55 Total loss must be paid by insurance company, but it will be subrogated to insured’s rights against disgruntled employee who caused fire (N79L-1-12m) 189 Total loss of 40% insured property with 80% coinsurance clause would result in collecting face of policy (N79L-1-38m) 194 Usual policy does have to meet insurable interest test, provide for subrogation of insurer to insured’s rights upon payment of covered loss, and cover losses caused by negligence of insured’s agent, but does not permit assignment of policy without consent of insurer (N79L-1-45m) 196 Insurable interest Nonexistent in property of general trade debtor with the debt unsecured (M78L-1-18m) 35 When furniture is ordered to design and specification is marked or otherwise designated by the seller as goods to which contract refers (M78L-1-14m) 3 4 Life Assigned to secure loan (M79L-1-3m) 135 Incontestability clause Policy with two-year clause not voided by misrepresentation of physical condition when death occurred by accident three years later (M79L-1-4m) 135 Would make unsuccessful a defense by a life insurance company that the insured made material misstatements of fact relating to his health (N78L-1-9m) 87 234 Insurance Life (cont.) Insurable interest— partners cross-insured each other’s lives, but if they were never partners recovery would be denied (no insurable interest) (M79L-1-4m) 135 Intangible assets See also Leasehold improvements Amortization Forty years used if benefit for an indeterminate but very long period (M79T-1-12m) 148 Straight-line normally recommended (M78T-1-12m) 46 Goodwill See also Consolidated financial statements See also Goodwill Computation of capitalization of excess earnings at 10% should not use extraordinary gains (N78PII-1-7m) 66 Consolidated— computation of amount, given purchased all (number of shares) shares o f company, price per share, and amount of fair value of assets in excess of book value (N78Pn-1-7m) 167 Matched on basis other than cause and effect (M78T-1-20m) 47 Recording goodwill from revaluation is not acceptable (M78A-1-lm) 23 Result of investing for 45% ownership and 30% ownership in two companies, computation of amount (considering fair value o f fixed assets of one company exceeds book value), amortization, and accumulated amortization (M79PI-3) 112 Income statement effects of amortizing patent and franchise, and expensing research and development costs (M79PI-3) 112 Prepare intangibles section of balance sheet involving patent and franchise and accumulated amortization on each (M79PI-3) 112 Trademarks Amortized over maximum period of forty years, using straight- line (M79T-2) 151 Interest income See Installment sales Interim financial information Limited review Consists primarily of making inquiries and performing analytical procedures concerning significant accounting matters (M78A-1-36m) 26 GAAP not followed— advise board of directors of particulars (M78A-1-37m) 2 7 Interim financial reporting Advertising expense Expenditures in August promoting subscriptions to skiing magazine with only two issues (October and November) should all be expensed on the quarterly income statement for the three months ended with December (N79PI-1-11m) 158 How revenue, cost, and expense recognized for such reporting relate to those recognized for year-end reporting (N79T-4) 207 Income tax recognition at such dates (N79T-4) 207 Inventory Estimated gross profit rates are allowable at interim dates but not at year-end (M79T-1-3m) 147 Inventory and cost of goods sold special accounting treatment (N78T-6) 103 Inventory loss from market decline occurring in second quarter should be reflected in second quarter income statement (M79PI-1-7m) 107 Product and period cost recognition discussion (N78T-6) 103 Provision for income taxes computation and reflection in statements (N78T-6) 103 Interim financial reporting (cont.) Recognition of revenue discussion and specifically for seasonal business and for percentage-of-completion (N78T-6) 103 Interim financial statements Limited review Does not provide basis for opinion because not accord with GAAS (N79A-3) 184 Procedures and purpose of each (N79A-3) 184 Internal auditors Objectivity may be judged by a review of the recommendations made in their reports (N78A-1-50m) 81 Internal control See also Electronic data processing See also Errors or irregularities See also Flowcharts Accounting See also Audit procedures— Internal control Authorization of transactions (M79A-1-30m) 128 Cash disbursements Require all checks signed by officer after support examined to protect against improper or inaccurate (M78A-1-60m) 29 Signed checks mailed under supervision of check signer (N78A-1-2m) 75 Compliance tests (N79A-1-21m) 180, (M78A-1-46m) 28 Equipment— weakness would be a policy requiring all purchases of factory equipment to be made by the department in need of the equipment (N78A-1-34m) 79 Flowcharts record the auditor’s understanding of client’s system (N78A-1-42m) 79 Inherent limitations (N78A-1-49m) 80 Narrative memorandum, flowchart, and decision table are all mediums that are normally used by auditor to record information concerning the client’s system (M79A-1-26m) 127 Normally includes procedures to provide reasonable assurance that transactions executed accord with management’s general or specific authorization (N78A-1-58m) 82 Payroll Lack of sufficient segregation to afford effective control is reason for observation of actual distribution of payroll checks (M79A-1-20m) 127 Segregate duties of making salary payments to employees and hiring new employees (M78A-1-41m) 27 Postdated checks received from customers should be restrictively endorsed (N79A-1-10m) 179 Purchases— procedures expected if effective (N78A-4) 83 Reasonable assurance is basic concept, which recognizes cost should not exceed benefits (M79A-1-llm ) 126 Study and evaluation (M78A-1-49m) 28 Weakness discovered during audit, auditor required to communicate the material weakness to senior management and board of directors (N78A-1-39m) 39 Accounts receivable Prevention of concealment of cash shortage from improper write off of an account most likely (of listed procedures) by write-off approval by responsible officer after review of credit department recommendations and supporting evidence (M78A-1-32m) 26 Written off transferred to a separate ledger (N79A-1-28m) 181 Administrative Review not primarily for expressing an opinion, but may bring possible illegal acts to auditor’s attention (M79A-1-29m) 128 Compliance testing Statistical sampling— attribute more useful than variable, unrestricted random, or stratified random sampling (M78A-1-2m) 23 235 Internal control Compliance testing (cont.) To provide reasonable assurance that procedures are being applied as prescribed (N79A-5) 185 Documentation of auditor’s understanding of system in form of questionnaires, narrative memorandums, decision tables, and flowcharts (N79A-5) 185 Effective Based on tests during year, auditor concluded records, procedures, and representations could be relied on, but auditor should test these again at year end if inquiries and observations lead auditor to believe that conditions have changed significantly (M79A-1-16m) 126 Overriding principles include responsibility for performance of each duty must be fixed as result of concept of charge and discharge of responsibility and duty (M79A-1-20m) 127 Electronic data processing On-line, real-time (OLRT) strengthened by making a validity check of an identification number before a user can obtain access to the computer files (N78A-1-llm) 76 Evaluation to determine the extent of substantive tests that must be performed (N79A-1-29m) 181 Flowcharts Example symbols mean a document has been generated by a manual operation (M79A-1-34m) 128 Interpret symbols and partially complete a flowchart on charge sales activities telling the internal control procedure or internal document (M79A-2) 131 Inventory Cycle physical count does not indicate weakness (M78A-1-21m) 25 Expensing items of little value does not indicate weakness (M78A-1-21m) 25 Perpetual (M78A-1-21m) 25 Marketable securities Bank custodial agent for safekeeping is best of several given actions (M78A-1-26m) 24 Organizational independence of departments Not impaired by internal auditors reporting to audit committee of board of directors, by payroll reporting to chief accountant, by cashier reporting to treasurer, but impaired by controller reporting to vice president of production (M79A-1-42m) 129 Payroll Accounting department— appropriate that it be responsible for preparation of periodic governmental reports regarding employees’ earnings and withholding taxes, but not for approval of employee time records, for maintenance of records of employment, discharges, and pay increases, or for temporary retention of unclaimed employee paychecks (N78A-1-25m) 78 Cash— each employee should be asked to sign a receipt (N79A-1-54m) 183 Department supervisor may have responsibility for reviewing and approving time reports for subordinates, hiring subordinates, and initiating requests for salary adjustments for subordinates, but not for distributing payroll checks to employees (N79A-1-59m) 184 Distribution of checks— receptionist preferably responsible rather than bookkeeper, payroll clerk, or cashier (N79A-1-5m) 178 Distribution responsibility for payroll checks should belong to the company paymaster (M79A-1-59m) 131 Less than effective if payroll supervisor distributes payroll checks to employees (N78A-1-22m) 77 Total time spent on jobs should be compared with total time indicated on time-clock punch cards (M79A-1-37m) 129 Preliminary evaluation objectives (N79A-5) 185 Property, plant and equipment (N78A-1-5m) 23 Weakness— all purchases of factory equipment required to be made by department in need of the equipment (N79A-1-45m) 182 Internal control (cont.) Purchasing of raw materials (M79A-1-48m) 130 Reasonable assurance Cost should not exceed benefits expected (N78A-1-4m) 75 Report based on an audit is an indication of changing role of the CPA that calls for extention of auditor’s attest function (M79A-1-22m) 127 Returns of defective merchandise by customers should be delivered to receiving clerk (N79A-1-55m) 183 Stock certificates cancelled Defaced and attached to their corresponding stubs, where no independent stock transfer agents are employed and corporation issues its own stocks and maintains stock records (N79A-1-42m) 182 Study and evaluation Audit program developed after evaluation of internal accounting control (M78A-1-49m) 28 Basis for reliance in determining nature, extent, and timing of tests, and basis for suggestions for improvements (N79A-5) 185 Evaluation of (not compliance with, review of, or study of) existing system focuses directly on purpose of preventing or detecting errors or irregularities (M78A-1-51m) 28 Purpose is to aid in determining nature, timing and extent of tests, to provide basis for constructive service suggestions, and to establish a basis for reliance thereon, but not to provide training and development for staff accountants (M78A-1-22m) 25 Secondary objective is to provide a basis for constructive suggestions concerning improvements (M78A-1-40m) 27 To determine the extent of substantive tests that must be performed (N79A-1-29m) 181 Vouchers Official signing check should compare check with voucher and deface voucher documents to prevent a paid voucher from being presented for payment a second time (M79A-1-21m) 127 Weakness Storekeeper keeps the perpetual records of large number of individual items (M78A-1-21m) 25 Internal Revenue Code sec. 305 (M79L-3) 144 Inventory See also Consignments See also Cost accounting See also Electronic data processing See also Internal control Accounting changes Computation of net income resulting from change in principle from FIFO to LIFO (M78PII-1-7m) 14 Audit procedures Damaged merchandise discovery— observation is best of given procedures (M78A-1-6m) 23 Cash-disbursements-for-month-estimate computation, given estimated purchases, estimated percentage payments for purchases of month within month, and estimated payments during month for each of previous two months’ purchases (N79PI-2-21m) 160 Cash discounts always taken not included in cost or accounts payable (N78PI-4) 61 Consignments-in not included in inventory or accounts payable (N78PI-4) 61 Consignments-out included in inventory (N78PI-4) 61 Cost computation would include cost of merchandise out on consignment, goods purchased and shipped f.o.b. shipping point and cost of goods out on approval, but exclude any markup and goods held on consignment (M78PII-1-17m) 16 Dollar-value LIFO method computation (M79PII-3) 122 Estimated gross profit rates Allowable at interim dates but not at year- end (M79T-1-3m) 147 236 Inventory (cont.) FIFO Computation for one product (N79PI-4) 164 Finished goods Cost accounting tests primarily to determine that costs properly assigned to finished goods, work-in-process, and cost of goods sold (M79A-1-6m) 125 F.O.B. shipping point goods recorded as sale (or purchase) upon shipment (N78PI-4) 61 Freight included in cost (N78PI-4) 61 Gross margin (profit) method Fire loss estimated computation involving salvage (N79PI-4) 164 Internal control Weakness if storekeeper keeps the perpetual records of large number of individual items (M78A-1-21m) 25 LIFO computation for one product (N79PI-4) 164 LIFO dollar-value method computation (M79PII-3) 122 Lower of cost or market is an example of conservatism (N79T-1-19m) 204 Overstatement Adjusting entry for errors in opening and closing (N78PII-3) 71 Permanent decline in value Net realizable value is proper valuation (N78T-1-12m) 99 Perpetual Non-manufacturing company—debit verification (M78A-1-12m) 24 Perpetual of only more valuable items with frequent physical verification would best (of four choices) detect theft of such items from inventory consisting of hundreds of different items selling for $1 to hundreds of dollars (M79A-1-47m) 130 Physical is lower—could be result of failure to record sales (M78A-1-3m) 23 Physical Lower than perpetual—could be result of failure to record sales (M78A-1-3m) 23 Prices decreased if FIFO cost of goods sold exceeds LIFO cost of goods sold (M79T-1-llm) 148 Purchase cut-off Best of given items for auditor to review and test to ascertain merchandise properly included (M78A-1-50m) 28 Retail method Computation to approximate LCM would include markups, but not markdowns, at point of computing cost ratio, but markdowns and sales subtracted from merchandise available at retail before applying cost ratio (N78PII-1-12m) 67 Estimate (M79PII-3) 122 Lower of average cost or market estimated ending inventory computation (N79PI-4) 164 Lower of FIFO or market computation (N79PII-1-9m) 169 Sold goods (not shipped) not included in inventory (N78PI-4) 61 Turnover analysis may detect obsolete merchandise (N79A-1-22m) 180 Weighted average for one product (N79PI-4) 164 Work in process Cost accounting tests primarily to determine that costs properly assigned to finished goods, work-in-process, and cost of goods sold (M79A-1-6m) 125 Estimation of loss by flood computation involving gross profit method estimation of cost of finished goods manufactured which were not destroyed (M79PII-3) 122 Investments See also Marketable securities Investment committee of board of directors are best persons to make periodic reviews of investments of additional temporary funds (M79A-1-10m) 126 Land acquired as potential future building site (N79PII-3) 174 Long-term Bonds—interest income (M78PI-1-18m) 5 Cost method—computation of income Investments Long-term (cont.) reported (M78PI-1-11m) 4 Equity method Computation of income reported (M78PI-1-4m) 2 Valuation computation one month after purchase, given 45% interest, amount of share of month’s income, dividend per share amount and record date during that month, and amount of goodwill amortization for that month (M79PI-1-9m) 107 Permanent decline in market value—decline other than temporary must be recognized (N78A-1-53m) 81 Short-term Equity securities—computation of balance of account, given cost of shares, number of shares, 25% investment, ability to significantly influence, per share dividend received, and net income of invested company (N78PI-1-12m) 55 Reduced to market (N78PII-1-4m) 65 Irregularities See Errors or irregularities Joint cost See Cost accounting K Kiting See Auditing procedures—Cash Knowledge Obtain a knowledge of matters relating to nature of entity’s business if accept audit engagement without industry expertise (N79A-1-llm) 179 Land Analysis of account computation, given beginning balance, issuance of stock (quoted market price given) for land and building, and appraised value for the land and the building (N79PII-3) 174 Book value computation, given exchange of given number of shares of stated fair market value for it, and sale of scrap removed from site (N79PII-1-10m) 169 Cost computation (M78PI-5) 10 Cost computation includes price, demolition of old building, legal fees for title investigation and purchase contract, less salvaged materials from building, and problem includes cost of new building (N78PII-1-10m) 66 Land held for resale shown separately (M78PI-5) 10 Tract acquired as potential future building site shown under investments (N79PII-3) 174 Lapping See Accounts receivable—Lapping Lawyer’s response to letter of audit inquiry May be limited to matters considered individually or collectively material if the lawyer and auditor have reached an understanding on limits of materiality for this purpose (M79A-1-33m) 128 Leasehold improvements Analysis of account, given beginning balance and cost of new parking lots, streets, and sidewalks (N79PII-3) 174 Capitalizable amount (M78PI-5) 10 237 J L Leases See also Sale and leaseback Lessee Capital Accounts created or affected by transaction and how costs related to earnings (N78T-4) 103 Amount computation should utilize incremental borrowing rate of lessee in absence of information on lessor’s implicit rate (N78T-4) 103 Computation of interest expense and depreciation, given present value of lease payments, effective interest rate, and useful life of asset (N78PII-1-3m) 65 Disclosures that must be made (N78T-4) 103 Period costs (M78T-1-15m) 46 Present value of minimum lease payments greater than 90% of fair value of asset at inception of lease (N78T-4) 103 Operating Expense computation includes monthly rent and amortization of leasehold improvement over 120 months (M79PIM-llm) 117 Expense is amount of monthly payments for number of months leased during year, but lessor’s income is total payments received less depreciation (M79PI-1-llm ) 107 Expense is payments for fraction of year leased (N79PI-1-15m) 158 Valuation on balance sheet is zero (M78T-1-13m) 46 Lessor Capital Not properly capitalized, amortized, or disclosed (M79T-2) 151 Direct financing Determination of appropriate amount of earnings to be recognized from each lease payment (N78T-4) 103 Disclosures that must be made (N78T-4) 103 Reasons for so classifying given lease (N78T-4) 103 Recording and determination of amounts (N78T-4) 103 Unearned income amortized over period of lease using interest method (N79T-1-12m) 203 Operating Income-before-income-taxes computation, given (for fraction of year leased) rental payments (earned), depreciation, and related costs (N79PI-1-14m) 158 Income is two lessees’ total payments received from for the respective months leased, less depreciation, but lessee’s expense is amount of monthly payments for number of months leased (M79PI-1-llm ) 107 Sale by lessor Computation of amount of profit (difference between cash selling price and cost) and interest income (cash selling price minus first payment made beginning date of lease, multiplied by the interest rate contemplated and by the one-half year it was leased during first fiscal-calendar year) (M79PI-1-14m) 108 Interest income computed on cash sales price minus first (down) payment for fraction of year leased (M78PI-1-16m) 5 Sales or direct financing Bargain purchase provision, collectibility of lease payments assured, no important uncertainties, and no transfer of ownership by end of lease (M79T-1-6m) 147 Legal capital Corporate capital required by statute to be retained to minimally protect creditors (M79T-6) 153 Liability See also Accountant’s legal liability See also Audit procedures Premium claims outstanding Computation of estimated, given number of coupons issued, number necessary to redeem for a premium, premium cost, estimated percentage of coupons which will be redeemed.
Liability Premium claims outstanding (cont.) and number already redeemed (M79PII-1-2m) 116 Unredeemed coupon computation, given total value of coupons issued, amount disbursed as of end of year, percentage expected to be redeemed, and one month lapse between date grocer received coupon and company receives it (N79PII-1-2m) 167 Warranty Computation of estimate for two-year (M78PII-1-lm) 13 Limited review ‘ Interim financial information Consists of primarily making inquiries and perform analytical procedures concerning significant accounting matters (M78A-1-36m) 26 GAAP not followed— advise board of directors of particulars (M78A-1-37m) 27 Interim financial statements Do not provide basis for opinion because not accord with GAAS (N79A-3) 184 Procedures and purpose of each (N79A-3) 184 Linear programming See Quantitative methods Litigation See Audit procedures Local government See State and local government Long-term contracts Completed-contract method Billings on uncompleted contract in excess of related costs computation (N78PI-4) 61 Gross profit to be recognized is none when no contract is completed (N78PI-4) 61 Interim billings not GAAP for recognizing earnings because often no meaningful relationship to work performed (M79T-4) 153 Percentage of completion Billings on uncompleted contract in excess of related costs computation (N78PI-4) 61 Computation of income for each of two years, given costs incurred in each year, estimated cost to complete in first year, and contract price (N78PII-1-13m) 6 7 Gross profit recognized first year would be estimated total gross profit multiplied by percentage of costs incurred to total estimated cost (N79T-1-llm) 203 Gross-profit-to-be-realized computation (N78PI-4) 61 Preferable when estimates of costs to complete and extent of progress toward completion are reasonably dependable (N79T-1-15m) 203, (M79T-4) 153 Percentage-of-completion and completed-contract methods of computation and recognition of earnings (M79T-4) 153 Percentage-of-completion and completed-contract methods of reflecting job costs and interim billings on the balance sheet (M79T-4) 153 Progress billings and construction in progress shown at reporting dates prior to completion net, as a current asset if debit balance and current liability if credit (N78T-1-40m) 101 Long-term investments See also Marketable securities Bonds Computation of valuation at end of year, given purchase date, face, amount paid, including amount of accrued interest, straight-line amortization of premium, and maturity date (M79PI-1-8m) 107 Marketable equity securities Net unrealized gain—footnote or parenthetical disclosure only (M78T-1-11m) 46 Net unrealized loss— valuation allowance in equity section of balance sheet (M78T-1-10m) 46 238 Loss on sale of asset Proceeds were less than book value (M79T-1-13m) 148 Loss on sale of machine computation, given purchase date and cost, salvage, monthly straight-line depreciation and sale price (M79PII-1-15m) 118 Loss on scrapping of machine is book value on that date and should be shown as a normal operating expense (N79PII-3) 174 Losses See also Expropriation Lump-sum purchase Land, warehouse, and office building purchased for given total price allocated on relative current assessed values, rather than vendor’s original costs (M79PII-1-19m) 119 M Machine See Equipment Machinery and equipment See also Equipment Analysis of account, given beginning balance, cost (invoice price, freight and unloading, sales and use taxes, and installation costs) of new items added, cost of machine scrapped, and cost of machine sold (N79PII-3) 174 Make or buy Relevant costs computation, given per unit materials, labor, variable and fixed overhead costs, reduction in fixed overhead, and number of units (N78PI-2-35m) 59 Relevant costs include direct labor, variable overhead, and fixed overhead avoided if the part is bought, but not fixed overhead that will continue even if the part is bought (N79T-1-32m) 205 Relevant costs including computation of amount that would have to be saved by using released facilities manufacture another part in order to decide to buy, given number of units, variable costs per unit, fixed overhead applied per unit eliminated, cost per unit to buy, and total amount of savings desired to make buy decision (N79PI-2-36m) 162 Relevant-costs-total computation, given per unit variable costs and fixed costs, number of units, and percentage change of per unit fixed costs that would continue (M79PI-2-36m) 111 Management advisory services See also Cost accounting— Direct costing See also Quantitative methods— Probability See also Special order Implementation of CPA’s recommendations Not undertaken unless client had made firm decision to proceed based on a complete understanding and consideration of alternatives (M79A-1-35m) 128 Management information system Designed to ensure management possesses information needed to cany out its functions through integrated actions of data- gathering, analysis, and reporting functions (N79A-1-52m) 183 Management representation letter See Representation letter of management Managerial accounting Decision making Operating-income-increase-as-result-of-accepting computation, given brief absorption costing income statement, fixed costs in manufacturing expenses, number of units and price in special order, no additional selling expenses, and per unit sales price and manufacturing Managerial accounting Decision making (cont.) cost (M79PI-2-38m) 112 Operating-income-increase-as-result-of-accepting-special-order computation, given variable manufacturing cost, selling price, and number of units (N79PI-2-35m) 162 Definition and identifying characteristics of variable, fixed and semivariable costs, and the three basic methods to “break down’’ semivariable costs into their component parts (N78T-3) 102 Sell or process further Relevant costs (M78PI-2-29m) 7 Marketable securities See also Internal control See also Investments See also Long-term investments Current asset classification Must have a ready market and be available to management for use in short run operations (M79T-1-15m) 148 Equity Cost method— dividends from earnings since acquisition do not change carrying value (N79PI-1-13m) 158 Current portfolio— valuation allowance (N79T-1-30m) 205 Equity method Amount initially recorded and subsequent adjustments (M79T-3) 153 Change from cost method (when owned 10%) to equity (owning 25%) by retroactively restating all prior periods in which the investment was held as equity used from inception, and current and subsequent reporting discussion (M79T-3) 153 Circumstances in which applied (M79T-3) 153 Equity in net income of 30% owned company computation, given cost, fair value of net assets at acquisition date, excess of cost over share of net assets has indeterminate life, and net income of investee (N79PII-1-20m) 171 Recognition, presentation, and determination of amount of earnings considering earnings reported by investee, dividends, prior period adjustments, extraordinary items, excess of price over net equity (depreciation or amortization), and income tax deferral (M79T-3) 153 Recorded at cost on acquisition date (N79PII-1-19m) 171 Income statement not affected by increase in market valuation of long-term portfolio over cost less allowance to reduce to market shown in previous balance sheet (M79PII-1-3m) 116 Noncurrent reclassified to current—valuation allowance recognized as loss in current period (N78T-1-llm ) 99 Short-term— adjusting entry to write down to market (N78PII-3) 71 Valuation (M78T-1-19m) 46 Unrealized loss charged to income (M78PI-1-14m) 4 Matching See also Advertising expense See also Expense recognition Bad debt expense estimated on basis o f percentage of past actual losses from bad debts to past net credit sales, adjusted for anticipated conditions (N79T-1-13m) 203 Costs with revenues is purpose of capitalization or expensing (M78T-1-5m) 45 Goodwill on basis other than cause and effect (M78T-1-20m) 47 Incentive-commission-expense computation involving addition of amounts applicable to particular year, ignoring item accrued in previous year (N78PII-1-2m) 65 Materiality Auditor’s concept of what is meant (N78A-3) 82 239 Materiality (con t.) Effect on planning and execution of audit program (N78A-3) 82 Relationships and other considerations used in judging (N78A-3) 82 Municipalities Civic Center construction fund Balance sheet preparation (M79PII-5) 123 Entries for transactions and closing entries (M79PII-5) 123 General bonded debt and interest group of accounts Adjusting entries (N79PII-5) 176 General fixed assets group of accounts Adjusting entries (M78PII-4) 2 0 , (N79PII-5) 176 General fund Adjusting and closing entries (N79PII-5) 176 Budget—revenues in excess of appropriations initially will result in an increase in fund balance (M79T-1-47m) 151 Entries— reclassification, adjusting, and closing (also affecting receivable from water utility fund) (M78PII-4) General long-term debt group of accounts (M78PII-4) Special assessment fund (N79PII-5) 176 Water utility fund (N79PII-5) 176 20 20 N Negotiable instruments S ee Commercial paper Net income Computation, given working capital provided from operations, depreciation on fixed assets, and amortization of goodwill (M79PI-1-16m) 108 Schedule to compute corrected net income for two years after various errors and adjustments (N78PII-3) 71 Nonmonetary transactions Common stock exchanged for land at fair market value of stock, with sale of scrap credited to the land (N79PII-1-10m) 169 Exchange not culmination of earning process (N78PI-1-17m) 56 Issuance of stock for land and building with allocation of quoted market of stock to land and building on basis of appraised values (N79PII-3) 174 Notes receivable Net receivable computation involving noninterest-bearing note with six annual payments, implicit interest, and two years (M79PI-4) 113 Profit on sale for noninterest-bearing note with six annual payments, and implicit interest on it for each of two years (M79PI-4) 113 Notes receivable discounted S ee Audit procedures— Contingent liabilities Not-for-profit accounting S ee a lso Municipalities S ee a lso Voluntary health and welfare organizations University accounting Entries to record summary of transactions (N78PII-5) 73 Statement of changes in fund balances preparation (N78PII-5) 73 o Objectives of examination of revenues S ee Audit procedures— Revenues Objectivity Arm’s-length transaction between independent interests (M78T-1-lm) 4 5 Operational audit Concentrates on seeking out aspects of operations in which waste would be reduced by the introduction of controls (M79A-1-7m) 125 Owners’ equity Expected to require substantially less audit time than other areas listed (M78A-1-57m) 29 Objectives Earnings— setting (M78T-2-48m) 4 9 Paid-in capital in excess of par Computation of amount, given par value, two sales at given amounts, which were above par, and a sale of treasury stock above cost (N79PII-1-8m) 168 Partnership Admission of partner Contribution-of-new-partner computation, given balance sheet of old partnership, assets and liabilities fairly valued, no goodwill or bonus to be recorded, and new partner is to be admitted with a one-fifth interest (M79PII-1-9m) 117 Investment— bonus to new partner (N79PII-1-5m) 168 Sale of portion of interests of old partners— computation of gain (M78PII-1-2m) 13 Assignment of a partner’s interest as security for a personal loan is allowed in absence of any such restriction in the partnership agreement, but that does not cause dissolution or make the assignee a partner (M79L-4) 145 Bankrupt Solvent partner liable in full for partnership deficit (N79L-1-15m) 190 Capital account balance computation after contribution of assets and liabilities (M78PII-1-10m) 15 Created Profits divided equally on quarterly basis is most important of four factors in determining whether partnership has been created by sole proprietor regularly joining another in marketing certain items (N78L-1-17m) 8 8 Criteria for determining existence Door-nameplates, directory listings, business cards, stationery, and newspaper announcements do not establish partnership with nonconsenting person (M78L-5) 4 3 Death of partner Left to wife his share of business, and she is entitled to the value of his interest in the partnership (N79L-1-16m) 190 Dissolution Breach of fiduciary duty by partner (by padding expenses and taking kickbacks) gives any partner a right to a formal accounting of partnership affairs and grant of application for dissolution appropriate (M79L-4) 145 Partner’s financial problems will not precipitate dissolution (unless he is bankrupt) (M79L-4) 145 Result from bankruptcy of a partner, from death of a partner, and from wrongful withdrawal of a partner, but not from assignment by partner of his entire partnership interest (N79L-1-17m) 190 Entity recognition for taking of title and ownership of property (N78L-1-16m) 88 Estoppel Has been held that persons are partners (and liable to creditors of the partnership) if they know they are held out to be so and take no action to prevent (M78L-5) 4 3 240 p Partnership (cont.) Incorporation Liabilities including contingent—corp. may expressly assume, may be liable because presumed intention of parties, or may be disregarded and treated as alter ego or sham (M78L-5) 43 Parties dealing with corporation only afterwards not able to assert rights against former partners (M78L-5) 43 Partners will remain jointly and severally liable for the debts of the partnership (M78L-5) 43 Tax implications of transfer of partnership assets to newly-created corp. are generally no recognition of gain or loss— affects basis and holding period of stock (M78L-5) 43 Liability Acts of partner without authorization will bind partnership if he renews an existing supply contract that the other partners had decided to terminate and that they had specifically voted against (N78L-1-23m) 89 Limited Corporation similar in that both can only be created pursuant to statute and each must file a copy of their respective certificates with proper state authorities (N78L-1-18m) 88, (N79L-1-30m) 193 General partners may invest the entire contributions of the limited partners in a single venture without the written consent of the limited partners, if the partnership agreement is silent on the point (N79L-1-18m) 190 Limited partner may assign his partnership interest to whomsoever he wishes at any time (N78L-1-26m) 90 Rights of limited partners (N79L-1-19m) 190 Valid creation of such organization requires a state statute that permits it, limited partnership certificate must be signed and sworn to by participants and filed in proper office in state, and there must be one or more general partners and one or more limited partners, but it is not necessary that limited partners’ capital be paid in cash (N79L-1-14m) 190 Liquidation Cash-distribution-to-each-partner computation, given profit and loss ratio, balance sheet, and sale price of all noncash assets (M79Pn-1-10m) 117 Cash-distribution-to-partners computation (N79PII-1-7m) 168 Liquidation by installments Computation o f cash distribution to each partner (M78PII-1-8m) 14 Name Filing or registration of fictitious or assumed names used in trade or business to disclose the real parties required by virtually all states (M79L-4) 145 Remaining partners could probably continue to use par tnership name where the par tnership had a fixed duration and one of the partners has caused a dissolution wrongfully and in contravention of the partnership agreement (M79L-4) 145 Personal creditors of partner in financial difficulty have no rights to partnership property directly or indirectly but may obtain right to payments that would ordinarily go to that partner (M79L-4) 145 Retirement Capital-balances-of-remaining-partners computation, given beginning capital balances and profit and loss ratio, assets cost and fair value adjustment, cash (in excess of his balance) to be paid to retiring partner for his capital, and no goodwill to be recorded (N79PII-1-6m) 168 Patent Balance computation that was charged to expense as the result of governmental health hazard order, given cost, useful life, purchase date, and that amortization is recorded at end of each year (M79PII-1-14m) 118 Payroll See also Audit procedures Payroll (cont.) Accounting department Appropriate that it be responsible for preparation of periodic governmental reports as to employees’ earnings and withholding taxes, but not for approval of employee time records, for maintenance of records of employment, discharges, and pay increases, or for temporary retention of unclaimed employee paychecks (N78A-1-25m) 78 Pension plans Defined maximum for pension cost includes normal cost, interest on overfunding, 10% of prior service costs, but not provision for vested benefits (N78T-1-16m) 99 Expense Define normal cost, past service cost, prior service cost, and interest, and discuss how each is treated under GAAP (M78T-4) 50 Funded, noncontributory Computation of total pension expense for first year includes normal cost and amortization of past service costs (N78PII-1-14m) 6 7 Entry at end of second year recording normal cost and amortization of past service costs (less interest on funding) as pension expense, increase in deferred charge— funding in excess of costs by given amount, and cash (N78Pn-1-15m) 67 Normal cost is always included in both minimum and maximum limit on annual provision for pension cost (N79T-1-21m) 204 Past service costs Amortized ratably as a charge to operations over ten years, even though funded over fifteen (M79PII-1-18m) 118 Arguments in favor of accruing only to extent funded, and regardless of amount funded (M78T-4) 50 Pension cost assigned to years prior to inception of pension plan (N79T-1-20m) 204 Reduced for interest presumed earned when fully funded prior to accrual (M79T-1-8m) 147 Pay-as-you-go or terminal funding not GAAP (M79T-2) 151 Vested benefits are not contingent on employee’s continuing in the employer’s service (N79T-1-8m) 203 Periodicity Statements prepared at arbitrary points in time (N78T-1-6m) 98 PERT See Quantitative methods Predecessor and successor auditor communications See Audit engagement—First examination Predecessor auditor See also Successor auditor Working paper review Should permit review relative to matters of continuing accounting significance, such as analysis of contingencies (M78A-1-33m) 26 Preferred stock See Stock— Preferred Premium claims outstanding See Liability— Premium claims outstanding Prepaid Subscription income not recognized on prepaid subscription until subscription actually begins (N78PI-1-llm ) 55 Prepaid expense Adjusting entry for expensing three-year insurance premium (N78PII-3) 71 241 Prepaid insurance See Audit procedures Price-level accounting Auditor’s report Special purpose report on supplementary financial statements prepared in accordance with a price-level basis of accounting that has substantial support (M79A-1-24m) 127 Current value accounting Gain includes holding gain and gain on sale (M78PI-1-l3m) 4 Gain-on-sale-of-machine computation, given purchase at beginning of year, sale price at end of year, and replacement cost (current value) at end of year (N79PI-1-l7m) 159 Holding gains are recognized (N79T-1-24m) 204 Income statement— gain-or-loss-on-sale-of-machine computation, given current fair value of machine and sales price (M78PI-1-l5m) 56 May be replacement cost, market value, net present value of expected future cash flows, but not restatement of cost for changes in general price level (N78T-1-22m) 100 Nonmonetary assets— theory in unadjusted and in dollars restated for changes in purchasing power, effect of each on “relationship between resources” and “standard of comparison,” and effect of each on recognition of gain or loss during life and from sale or abandonment (M78T-6) 51 Replacement cost and present value of future cash flows described and pros and cons of various procedures used to arrive at valuation for each (M78T-6) 51 Stockholders’ equity would increase by difference between cost of land and its fair value at end of its first year of ownership, and by difference between the last fair value and the selling price in the next month (second year) (M79PI-1-20m) 109 Depreciation restated for changes in general purchasing power computation, given depreciation amount, and general price- level indices in year of acquisition and in year of statements (M79PI-1-15m) 108 General price-level income statement Computation of general price-level gain or loss, given issuance of stock for cash and land, index at beginning and end of year, and that no other transactions took place (N78PI-1-13m) 55 General price-level restatement is not a violation of historical cost (M79T-1-19m) 148 Historical cost restated for changes in general price level attempts to eliminate the effect of changing value of dollar (N78T-1-3m) 98 Historical dollars restated for changes in purchasing power Nonmonetary assets— theory, effect on “relationship between resources” and “standard of comparison,” and effect on recognition of gain or loss during life and from sale or abandonment (M78T-6) 51 Marketable securities adjusted for price-level changes but not cash or long-term receivables (M78PI-1-12m) 4 Prices decreased Cost of goods sold using FIFO exceeds that using LIFO (M79T-1-llm) 148 Special report (N78A-1-21m) 77 (M78L-1-8m) 33 Price maintenance agreement Verbally illegally entered Principal auditor See Audit procedures Prior period adjustment Change in amount of opening retained earnings balance due to an error in a prior period (N78T-1-8m) 98 Privileged communication See Accountant’s legal liability Property Adverse possession Adjoining land owner fenced off land, mistakenly believing it belonged to him, installed well, storage shed, and garage and continuously farmed and occupied it for 22 years prior to another’s purchase, and the occupier has title (N79L-1-41m) 195 Deed ambiguous regarding type of estate created and interest of each of two sons (M78L-1-27m) 36 Deed that recites a consideration of $1 and other valuable consideration is valid (M78L-1-31m) 37 Lease Prohibiting subletting does not prohibit assignment, but lessor is still obligated to pay the rent if assignee defaults (N78L-1-48m) 93 Leasehold interest assignment is generally allowed in absence of restriction specifically stated in lease (M79L-5) 145 Leasehold interest assignor remains liable in absence of a release in the nature of a surety (M79L-5) 145 Lessor can assert rights against assignee on promise of assignee of lease to assignor of lease to pay the rent over life of the lease (M79L-5) 145 Mortgage Mortgage promptly recorded will have priority over earlier unrecorded mortgage (N79L-1-40m) 195 On the land, buildings, and “all other property that is added to the real property or which is considered as real property as a matter of law” will not include detachable trade fixtures (M78L-1-29m) 37 Personal liability Buyer acquired property subject to mortgage but did not assume the mortgage, abandoned the property and did not pay mortgage, with result seller must satisfy mortgage debt in event that foreclosure yields less than the unpaid balance (N79L-1-39m) 194 Purchaser acquired subject to mortgage on which seller was personally liable, and later abandoned the property, leaving the seller to satisfy the mortgage if foreclosure yields less than unpaid balance (M78L-1-28m) 36 Seller of land expressly warrants that there is no mortgage, buyer purchases unaware of a mortgage, seller disappears, buyer refuses to pay mortgage, and mortgage holder will prevail in foreclosure (N78L-1-43m) 92 Real Bank vault door probable so classified— reasons for conclusion stated (M79L-5) 145 Real versus personal Disputes arise involving categorization between real property mortgagees and personal creditors, landlord versus tenant, tax collectors versus tax payers, takers under a will versus executor, seller versus purchaser of real property, and mortgagor versus mortgagee (M79L-5) 145 Sale Binding contract may be established by purchaser by having a writing signed by owner stating essential terms (M78L-1-26m) 36 Tenants in common Each of the co-tenants owns an undivided interest in the whole (N79L-1-10m) 189 Each owns an undivided interest in the whole, which he or she may dispose of by deed or by will (N78L-1-41m) 92 Property, plant and equipment See Fixed assets Provision for product warranties Accrued when probable (N79T-1-3m) 202 Public warehouses See Audit procedures 242 Purchases See also Audit procedures—Supplies See also Internal control—Accounting Factors in determining time and quantity to order (N78A-4) 45 Purchases discount See Audit procedures Quality control Audit Includes assigning personnel, consultation with appropriate persons, and supervision, but not keeping records of quality control policies and procedures (M78A-1-47m) 28 Includes policies and procedures for professional development which provide reasonable assurance that all entry-level personnel have knowledge required to enable fulfilling responsibilities assigned (N79A-1-60m) 184 Independence Notify professional personnel of names of audit clients having publicly held securities, and confirm periodically with such personnel that prohibited relations do not exist (M79A-1-50m) 130 Timely identification of staffing requirements of specific engagements, so that enough qualified personnel can be made available (N78A-1-7m) 76 Quantitative methods See also Breakeven point See also Capital budgeting See also Ratios Correlation Coefficient of is a measure of extent to which two variables are related linearly (M79T-1-27m) 149 Economic order quantity Computation (M78PI-2-27m) 6, (M79PI-2-30m) 110 Cost-of-carrying-one-unit-in-inventory-for-one-year computation, using economic-order-quantity approach, given set-up costs, number of units manufactured evenly throughout the year, and optimal production run (N79PI-2-31m) 161 Factors in determining quantity to order (N78A-4) 83 Formula, given cost to carry, cost of placing order, and number needed per year (N78PI-2-28m) 58 Information necessary includes annual demand, cost of placing an order, and annual cost of carrying a unit in stock (N79T-1-36m) 205 Set up costs computation, given cost of carrying one item in inventory for one year, number of units produced evenly throughout the year, and optimal production run (N78PI-2-38m) 60 Variable costs of processing a purchase order is relevant factor (M78T-2-32m) 48 Future amount Investment amount at end of ten years, given same amount invested at end of each year, interest-earned rate, and future amount of ordinary annuity for ten periods at rate (M79PI-2-40m) 112 Least squares Used to determine fixed and variable elements of a semivariable cost (N78T-1-45m) 102 Will separate semivariable cost into fixed and variable components with highest degree of precision under all circumstances (M79T-1-26m) 149 Linear programming Choose one of four stated equations that would not represent a constraint (restriction) based on given information (M79PI-2-33m) 111 Constraint formula (M78PI-2-21m) 5, (N78PI-2-31m) 58, (N79T-I-40m) 205 Quantitative methods Linear programming (cont.) Contribution margin (total) maximization formula determination for two products, given the contribution margin for each product (N79PI-2-32m) 161 Graphic—optimal solution always a comer point described by the feasible area (M79T-1-35m) 150 Objective function (maximization of total contribution margin) (N78PI-2-30m) 58 Simplex method—maximization of contribution margin optimum solution reached when values in index row of matrix are all positive or zero (N78T-1-46m) 102 Net present value See also Capital budgeting Computation, given cost of machine, annual cash flow from operations (net of income taxes), time-adjusted rate of return % to be used, present value of ordinary annuity for useful life at that rate, and no salvage (M79PI-2-29m) 110 Computation, given cost of machine, useful life, no residual value, annual cash flow, income tax rate and present value of an annuity (M78PI-2-20m) 5 Investment determination, given four-year project, expected rate of return, cash inflow and present value amounts for each of four years, and a positive net present value amount (N78PI-2-27m) 58 Payback Cash flow from operations for last payback year of one machine computation, given cost of machine, payback period, and cash flow from operations, net of tax, in each year prior to the last (N79PI-2-23m) 160 Period computation given cost of machine, useful life, no residual value, annual cash flow and income tax rate (M78PI-2-19m) 5 PERT Primary objective is to define a critical path (M79T-1-36m) 150 Uses network analysis and critical path methods (N79T-1-38m) 205 Present value Annuity due—used in calculating present value of a capital lease entered into with the initial lease payment due upon the signing of the lease agreement (M79T-1-10m) 148 Bonds payable Amount-received computation for sale of serial bonds, given face of five-year 5% bonds sold to yield 6%, amount and date of principal payments, interest payment dates, present value of an annuity value at 6%, and present value of $1 at 6% table (N78PI-3) 60 Proceeds equal to present value of principal and of interest, each discounted at prevailing market rate of interest (M78T-1-4m) 45 Sale-receipts computation, given face, contract percentage interest, yield rate, interest paid annually, life of bonds, and present value of $1 and of annuity of $1 for life of bonds at yield rate (N79PI-5) 165 Computation of cash flow in third year of three-year project, given investment, rate of return, present value of return of first two years, and present value factor for third year (M78PI-2-28m) 7 Determination that a rate of return is “less than 10%, but more than 0%,” given cost of machine, the present value of an annuity for the life, and annual cash inflow net of tax (N78PI-2-26m) 58 Interest rate implicit in lease computation, given cost of machine, number and amount of payments in annuity due, receive title at end of lease, and present value of ordinary annuity for each of five periods at three interest rates (N78PI-2-39m) 60 Present value of annuity due of $1 for five periods used to compute installment payments (N79T-1-39m) 205 Sales price computation involving noninterest-bearing note and present value of an annuity (M79PI-4) 113 Probability Contribution margin expected if given number of units 243 Q Quantitative methods Probability (cont.) manufactured and all sales orders filled either from those or h'om others purchased (units manufactured and not sold by end of month must be discarded) (N79PII-4) 175 Contribution margin expected if probabilities of strike at supplier of cheaper ingredient’s plant is considered and all sales are filled either from manufactured items or from purchases from competitors (N79PII-4) 175 Sales-probability-for-product computation, given product has been selling for 20 months at three different volumes and the number of months it has sold at each volume level (N79PII-4) 175 Rate of return See Capital budgeting Safety stock computation to maintain lowest cost, given stockout cost, carrying cost of each unit of safety stock, number of orders a year, and the probabilities of running out of stock during another period at various safety stock levels (M79Pn-2-28m) 110 Simple regression analysis Involves use of two variables (N79T-1-37m) 205 Useful in determining fixed and variable elements of semi variable expenses (M78T-2-33m) 48 Statistical sampling Attribute More useful than variable, unrestricted random, or stratified random sampling when testing for compliance (M78A-1-2m) 23 When auditor concerned with number of client’s sales invoices that contain math errors, this is most likely (of four choices) utilized (N78A-1-55m) 81 Distinguishing feature Provides a means for measuring mathematically the degree of uncertainty that results from examining only part of a population (M78A-1-30m) 26 Provides for measuring mathematically the degree of uncertainty that results from examining only a part of the data (N78A-1-52m) 81 Estimation sampling for variables— confidence level desired must be known to estimate the appropriate sample size to meet the auditor’s needs (M79A-1-12m) 126 Internal control Compliance testing— attribute sampling more useful than variable, unrestricted random, or stratified random sampling when testing for compliance (M78A-1-2m) 23 Compliance testing— conclusion about whether procedural deviations are within an acceptable range (M78A-1-56m) 29 Minimum-acceptable-dollar-value-of-population computation, given number in population, sample adequate for auditor’s purposes, arithmetic mean of sample, and desired precision (N79A-1-38m) 182 Principal advantage flows from its unique ability to provide a mathematical measurement of uncertainty (N79A-1-6m) 178 Ratio estimation, and when to use to estimate dollar value of population (N78A-1-14m) 76 Sampling for variables to estimate a numerical measurement of a population (M79A-1-18m) 727 (N79T-I-23m) 204 Quasi-reorganization Retained earnings zero immediately after R Ratios See also Financial statement analysis Accounts receivable turnover Total-net-sales computation, given net accounts receivable at beginning and at end of year, net cash sales, and accounts Ratios Accounts receivable turnover (cont.) receivable turnover (M79PI-1-18m) 108 Analytical review Substantive tests of financial information by study and comparison of relationships among data (N79A-1-50m) 183 Common size statements Enables financial analyst to compare mix of assets, liabilities, capital, revenue, and expenses within a company or over a period or between companies without respect to relative size (M78T-2-40m) 48 Current ratio Computation, given current assets and current liabilities (N78PI-1-10m) 55 Higher current ratio may not be better, because it may indicate inefficient use of various assets and liabilities (M78T-2-42m) 49 Dividend-payout ratio involved in computation of total common dividends (N78PI-1-7m) 55 Earnings per share See Earnings per share Gross profit to sales Unexplained decrease may suggest unrecorded sales (M79A-1-8m) 126 Inventory turnover Computation, given purchases, cost of goods sold, and ending inventory (M79PI-1-6m) 107 May detect existence of obsolete merchandise (N79A-1-22m) 180 Number-of-days’-sales-in-average-inventories computation, given cost of goods sold, inventories at beginning and end of year, and number of days in a business year (N79PI-1-7m) 157 Overall profitability of a manufacturing company Net income to working capital less useful than net income to net worth, net income to total assets, or net income to sales (N78A-1-60m) 82 Pay-back period See Capital budgeting Pr ice-earnings computation (M78PI-1-10m) 4 Quick (acid-test) Computation, given total current liabilities, and current asset amounts for cash, marketable securities, and accounts receivable (N79PI-1-8m) 157 Computations involved (M78PI-1-8m) 3 Rate of return See Capital budgeting Real estate Paid real estate tax bills are better evidence of ownership at balance sheet date than title insurance policy, original deed in client’s safe, or closing statement (N79A-1-14m) 179 Receivables See Accounts receivable See Internal control— Accounts receivable See Notes receivable Relevant costs See Special order Rent revenue collected one month in advance is current liability (N79T-1-14m) 203 Repairs and maintenance See Audit procedures— Fixed assets Replacement cost information Unaudited Footnote indicates auditor performed certain procedures but does not indicate whether auditor expresses an opinion on this information, with the result that the auditor’s report on audited statements should be expanded to disclaim this information (M79A-1-17m) 126 244 Report See Accountant’s report See Auditor’s report See Special reports Reporting standards See Generally accepted auditing standards— Standards of reporting Representation letter See Audit procedures Representation letter of management Date should coincide with date of auditor’s report (N79A-1-34m) 181 Research and development Costs Items included and excluded (M78PI-1-lm) 2 Define “research” and “development” (M78T-5) 50 Expense as incurred (M79PI-3) 113 Expense as incurred design and engineering studies, prototype manufacturing costs, administrative costs related solely, cost of equipment produced solely for development, but capitalize and depreciate (as research and development expense while manufacturing pre-production prototype) equipment, and consider market research and related administrative expenses as period costs (not research and development costs) (M78T-5) 50 Expense as incurred, including materials used and compensation to consultants (M79PII-1-17m) 118 Expense immediately (N78T-1-9m) 98 General laboratory building should be capitalized and amortized over current and future periods (N78T-1-19m) 99 Immediate recognition as expense (N79T-1-9m) 203 Practical and conceptual reasons for FASB conclusion on accounting and reporting practices (M78T-5) 50 Retained earnings See Stockholders’ equity Revenue recognition See also Audit procedures Exchange has taken place and earnings process virtually complete (M78T-1-3m) 45 Prior to sale (M78T-1-8m) 45, (N79T-1-5m) 202 Subscription income not recognized on prepaid subscription until subscription actually begins (N78PI-1-11m) 55 Review of statements of nonpublic entity See Accountant’s report Robinson-Patman Act (M78L-1-9m) 33 Role and responsibilities of independent auditors Oversight of professional practice should remain with the accounting profession is a conclusion of the Commission on Auditors’ Responsibilities (M79A-1-28m) 128 Sale and leaseback Rent expense Computation for one-half year, given sale price, book value, life of lease, and rental charge per month (M78PII-1-14m) 15 Sales Installment See Installment sales Setting earnings objectives must consider sales volume attainable in the present plant, breakeven point, and earnings or losses for given sales-volume levels, but not indices of industrial Sales Setting earnings objectives (con t.) production (M78T-2-48m) 49 Subject to sales tax Computation of amount, given sales tax rate, and amount remitted (in time to keep 2% of amount collected according to law) (N79PII-1-llm ) 169 SASs No.
1 (M78A-1-13m) 24, (M78A-1-15m) (M78A-1-19m) 25, (M78A-1-22m) 25, (M78A-1-31m) 26, (M78A-1-35m) 26, (M78A-1-40m) 27, (M78A-1-46m) 28, (M78A-1-54m) 28, (N78A-1-3m) 75, (N78A-1-5m) 75, (N78A-1-9m) 76, (N78A-1-13m) 76, (N78A-1-15m) 77, (N78A-1-16m) 77, (M79A-1-lm) .125, (M79A-1-3m) 125, (M79A-1-llm ) 126, (M79A-1-39m) 129, (M79A-5) 133, (N79A-1-lm) 178, (N79A-1-9m) 179, (N79A-1-26m) 180, (N79A-1-41m) 182, (N79A-1-56m) 184 24, No. 2 (N78A-1-lm) 75, (M79A-1-17m) 126 No. 3 (M79A-3) 133 No. 4 (M78A-1-45m) 27, (M78A-1-47m) 28, (N78A-1-7m) 76, (M79A-1-50m) 130, (N79A-1-60m) 184 No. 7 (M78A-1-33m) 26, (M78A-1-44m) 27, (M79A-1-44m) 129 No. 9 (N78A-1-50m) 81, (N78A-1-51m) 81 No. 10 (M78A-1-36m) 26, (M78A-1-37m) 27, (N79A-3) 184 No. 12 (N78A-1-59m) 82, (M79A-1-33m) 128, (M79A-4) 133 No. 13 (N78A-1-47m) 80 No. 14 (M78A-1-10m) 24, (M78A-1-43m) 27, (N78A-1-21m) 77, (M79A-1-24m) 127, (M79A-1-43m) 129, (M79A-1-56m) 131, (N79A-1-13m) 179, (N79A-1-18m) 180 No. 15 (N79A-1-36m) 181 No. 16 (M78A-1-4m) 13, (M78A-1-28m) 25, (M78A-1-51m) 28, (M78A-1-59m) 29, (M79A-1-5m) 125, (N79A-1-7m) 178, (N79A-1-37m) 181 No. 17 (M79A-1-29m) 128, (M79A-1-36m) 128, (M79A-1-38m) 129, (N79A-1-25m) 180 No. 18 (M79A-1-17m) 126 No. 19 (N79A-1-12m) 179 No. 22 (M78A-1-45m) 27, (N78A-1-35m) 79, (N79A-1-llm ) 179 Schedule of changes in working capital Preparation (M78PII-3) 19 Scope limitation Unavailability of sufficient competent evidential matter (N78A-1-32m) 78 SEC See Securities and Exchange Commission Secured transactions Attachment relates primarily to rights against debtor, and perfection relates primarily to rights against third parties (N79L-1-23m) 191 Attachment requirements include debtor must have rights in collateral, value must be given by creditor, and either creditor must take possession or debtor must sign security agreement which describes collateral, but not proper filing (N79L-1-24m) 192 Buyer in ordinary course of business of inventory goods (N79L-1-22m) 191 Perfected security interest Can’t protect against all parties without obtaining possession— 245 Secured transactions Perfected security interest (con t.) where goods are inventory of debtor, purchaser takes free of creditor’s interest (N78L-2) 94 Consignor must file a financing statement and give notice to certain creditors where consignment agreement clearly states intended to create a security interest in the consigned merchandise in desire to protect against other creditors of and purchasers from franchised retailers (M79L-1-20m) 138 Creditor must take possession of negotiable stock certificates traded on a local exchange (M79L-1-21m) 138 If not against all parties, file financing statement to protect against subsequent purchasers f r om the debtor (N78L-2) 94 In present and future inventory and proceeds therefrom— security interest in certified check debtor received for inventory item is perfected against debtor’s other creditors (N79L-1-26m) 192 Seller is secured creditor that has right, if not paid upon bankruptcy of debtor, to assert rights against the machine sold (M78L-1-23m) 36 Various type loans— whether such interest exists and against whom (N78L-2) 94 Perfection by attachment Manufacturing company will not have to file financing statement to perfect its security interest on sales made to consumers who purchase for their own personal use (M79L-1-14m) 137 Since discount store takes a purchase money security interest in consumer goods sold, its security interest is perfected upon attachment (M79L-1-15m) 137 Purchase money security interest Credit advanced to buyer must be used to obtain property that serves as collateral (N79L-1-25m) 192 Debtor has right to have creditor dispose of repossessed goods if debtor has paid sixty percent or more of the purchase price of consumer goods in satisfaction of (M78L-1-50m) 40 Secured creditor Must protect itself against debtor, debtor’s creditors, trustee in bankruptcy, and subsequent purchasers for value from debtor (N78L-2) 94 Released sole-owner surety of the business’ debt by accepting in full settlement proceeds of the sale of the security from the trustee in bankruptcy of the business (M79L-1-6m) 136 Security interest Attaches in all of given cases because (1) collateral is in possession of the secured party pursuant to agreement, or debtor has signed a security agreement that contains a description of the collateral, (2) value has been given by the creditor, and (3) the debtor has rights in the collateral (N78L-2) 94 Subrogation of surety Surety paid debt of obligor bankrupt to secured creditor and has the right of a secured creditor to the amount for which the security was sold and the standing of a general creditor for the balance (M79L-1-5m) 136 ucc Applies to equipment lease, sale of accounts, and field warehousing, but not to a second mortgage on borrower’s home (M79L-1-18m) 138 Warehouse receipt marked “negotiable” and pledged with bank for personal loan cannot be obtained by business creditors of the borrower without paying the loan (M78L-1-49m) 40 Securities See Audit procedures See Marketable securities Securities Act of 1933 (M78L-1-llm) 34, (M78L-3) 42, (N78L-1-llm) 88, (N78L-4) 95, (N79L-1-28m) 192, (N79L-1-29m) 192 Securities and Exchange Commission Regulation A (N78L-4) 95 (M78L-3) 42 Securities Exchange Act of 1934 (N78L-1-33m) 91, (N78L-4) 95, (N79L-1-27m) 192, (N79L-1-32m) 193 Securities Exchange Act of 1934, sec. 16(b) (M79L-1-49m) 143 Segments See also Disposal of segment Component (horizontal) of an enterprise providing a product or service or group of such items primarily to unaffiliated customers for profit (M79T-5) 153 Identifiable assets Explanation (M79T-5) 153 Operating profit and loss Explanation (and items not included by SFAS 14) (M79T-5) 153 Operating profit computation, given sales, traceable costs, total sales and total common costs for all segments combined, and allocation of common costs based on ratio of segment’s sales to total sales (N79PI-1-16m) 159 Profitability information reported for each reportable segment consists of operating profit or loss figure consisting of segment revenues less traceable costs and allocated common costs (N79T-1-25m) 204 Revenue Sales to unaffiliated and intersegment sales or transfers of products and services similar to those sold unaffiliated, and interest on the segment’s assets earned outside (M79T-5) 153 Ten or fewer should be separately reported so that the total of revenues from sales to unaffiliated customers for those segments equals or exceeds 75% of the combined revenues from sales to unaffiliated customers for the entire enterprise (M79T-5) 153 Tests to determine whether or not reportable (M79T-5) 153 SFAS See FASB Statements Short-term investments See Marketable securities Small Business Administration (N79L-1-49m) 197 SMAS No. 3 (M79A-1-35m) 128 Social Security Act (N79L-1-49m) 197 Software See Electronic data processing—General purpose computer software package Special order Idle capacity Relevant costs include materials, direct labor, and variable overhead, but not depreciation (M79T-1-21m) 149 Variable costs would be used to determine the lowest price (M79T-1-24m) 149 Sufficient capacity available Operating-income-increase-as-result-of-accepting computation, given brief absorption costing income statement, fixed costs in manufacturing expenses, number of units and price in special order, no additional selling expenses, and per unit sales price and manufacturing cost (M79PI-2-38m) 112 Operating-income-increase-as-result-of-accepting-special-order computation, given variable manufacturing cost, selling price, and number of units (N79PI-2-35m) 162 Special reports See also Auditor’s report 246 Special reports (cont.) GAAP reporting standard does not apply when asked to perform an examination to express opinion on one or more specified elements, accounts, or items of a statement (N79A-1-18m) 180 Includes reports prepared for limited purposes, such as one that relates to only certain aspects of statements, reports of not-for- profit organizations, which accounting differs from profit- making, and reports prepared on a cash basis (M78A-1-10m) 24 Prescribed form Assertion not justified— reword or attach a separate report (M78A-1-43m) 27 Price-level basis financial statements (N78A-1-21m) 77 Term does not include reports on financial statements where the client has limited the auditor’s scope (M78A-1-11m) 24 SSARS No. 1 (N79A-1-39m) 182 Standard Bank Confirmation Inquiry (N79A-1-15m) 179 Standards of field work See Generally accepted auditing standards Standards of reporting See Generally accepted auditing standards State and local government See also Governmental accounting See also Municipalities Actuarial deficiency would appear in a trust fund (N78T-1-28m) 100 General fixed assets Investment in general fixed assets accounts would appear (N78T-I-32m) 101 General fund Accounting for special revenue funds is most similar to (of four choices) (N78T-1-30m) 100 Personal services normally recorded on the accrual basis (N78T-1-33m) 101 Intragovernmental service fund Excess of net billings to departments over costs would appear in the financial statement (N78T-1-29m) 100 L o cal General fund— sales taxes held by the state that will be remitted to the local government should be accrued as revenues (N78T-1-26m) 100 Modified accrual property taxes susceptible to accrual (M79T-1-48m) 151 Non-enterprise fund Equity is fund balance (N78T-1-27m) 100 Special assessment fund Revenues from interest on assessments receivable recorded when legally due (N78T-1-34m) 101 Trust fund Actuarial deficiency would appear in this (N78T-1-28m) 100 Can be either expendable or non expendable (N78T-1-31m) 100 Stated capital Computation and discussion of reasons involving no-par common that sold for $20 and board of directors voted $19 into capital surplus, and par value preferred (N79L-2) 197 Statement of changes in financial position See also Working capital Audit procedures (M78A-5) 30 Cash basis Net increase in inventory adjusts reported net earnings because it reduced cost of goods sold but represents an assumed use of cash (N78T-1-21m) 100 Change from cash to working capital Consistency exception (N79A-1-lm) 178 Statement of changes in financial position (cont.) Computation of sources (proceeds from issuance of common and from preferred, and other sales of stock to officers and employees) and uses (dividends on common and on preferred, and purchases of treasury stock) (M79PI-1-17m) 108 Illustrated statements are incomplete with this one missing (M79T-2) 151 Inventory increase reported as a use of cash (M79T-1-16m) 148 Net income computation, given working capital provided from operations, depreciation on fixed assets, and amortization of goodwill (M79PI-1-16m) 108 Omission requires qualified opinion, with middle paragraph explaining company declined to present (M78A-1-35m) 26 Primary function is summarization of financing and investing aspects of all significant transactions affecting financial position (N78T-1-lm) 98 Required whenever balance sheet and income statement presented (M78A-2) 29 Retirement of long-term debt by issuance of preferred included only because of the all-financial-resources concept (M78T-1-21m) 47 Sale of fixed assets at gain (N78PI-1-16m) 56 Working capital Increase— computation (M78PI-I-3m) 2 Machinery accounts information indicates that there should be an addition to net income for depreciation and a use from purchase (N79PI-1-19m) 159 Preparation (M78PII-3) 19 Working-capital-from-operations computation, given net income, goodwill amortization, bond discount amortization, depreciation, and provision for doubtful accounts on long-term receivables (N78PI-1-18m) 56 Statement of changes in fund balances Preparation (N78PII-5) 73 Statement of Financial Accounting Standards See FASB Statements Statement on Auditing Standards See SAS Statement on Management Advisory Services See SMAS Statements on Responsibilities in Tax Practice, Sec.
161 (N79A-1-20m) 180 Statement on Standards for Accounting and Review Services See SSARS Statistical sampling See Quantitative methods Stock See also Dividends See also Stock dividends See also Stock options See also Stock warrants See also Treasury stock Common Basic rights inherent—preemptive, voting, share in profits and residual assets (M79T-6) 153 Dividends with preferred stock (M78PI-3) 9 Issued shares exceed outstanding shares as a result of purchase of treasury stock (N79T-1-7m) 202 Sale (M78PI-3) 8 Preferred Convertible— additional paid-in capital credited upon issuance and upon conversion computation, given number of preferred shares issued, par, issuance price, all converted, conversion ratio, and par of common (N78PI-1-20m) 57 247 Stock Preferred (cont.) Cumulative and fully participating— dividends (M78PI-3) 9 Definition, explanation, and discussion of various preferences (M79T-6) 153 Noncumulative and nonparticipating— dividends (M78PI-3) 9 Sale (M78PI-3) 8 Stock dividends See also Federal income tax— Corporations See also Stockholders’ equity— Retained earnings Issued on outstanding shares (M78PI-3) 8 Small Computation of amount capitalized, given number of shares issued and outstanding, percent of dividend and market price date of declaration (M78PII-1-4m) 13 Stock dividends from treasury stock and cash See Federal income tax— Corporations Stock options Exercised (M78PI-3) 8 Stock right Privilege extended by corporation to acquire additional (or fractional) shares of its capital stock (M79T-6) 153 Stock split See also Federal income tax— Corporations Effect upon conversion of convertible bonds issued prior (N78PI-3) 60 Stock warrants See also Bonds payable— Detachable stock purchase warrants Detachable issued with debt Computation of amount credited to outstanding account, given amount of debt securities issued, number of warrants issued with given amount, one share entitled per warrant, and fair market value of warrants date of issuance (M78PII-1-5m) 14 Exercised (M78PI-3) 8 Issuance (M78PI-3) 8 Issued with convertible preferred Exercise entry after recording at market value upon issuance (N78PI-1-19m) 56 Physical evidence of stock rights, specifying number of shares, rights, price, life, and other terms of the rights (M79T-6) 153 Stockholders’ equity See also Audit procedures See also Capital in excess of par See also Federal income tax—C orporations— Stock dividends from treasury stock and cash See also Internal control— Stock certificates cancelled See also Legal capital See also Paid-in capital in excess of par See also Price-level accounting— Current value accounting See also Stated capital See also Stock See also Stock dividends See also Stock right See also Stock split See also Stock warrant See also Treasury stock Common stock Adjusting entry to transfer excess of par to separate account (N78PII-3) 71 Contributed capital Total includes amounts paid for common and preferred plus additional shares subscribed at given price (N78PII-1-5m) 66 Stockholders’ equity (cont.) Retained earnings is reduced by total cash dividends declared during year and by market value use in 5% stock dividend declared, the stock dividend results in crediting common stock at the par value of stock issued and additional paid-in capital being credited for the balance of the market value capitalized (M79PI-1-13m) 108 Total computation, given total at end of previous year, cost method used for treasury stock, two purchases and one sale of treasury stock (FIFO used in computations), and net income (M79PII-1-1m) 116 Worksheet to summarize transactions (M78PI-3) 8 Subsequent discovery of facts Contingency resolved subsequent to report qualified with respect to it Auditor should take no action (N79A-1-56m) 184 Subsequent events Affecting realization of assets ordinarily require adjustment of the statements under examination because such events typically represent culmination of conditions existing at balance sheet date (N78A-1-29m) 78 After fieldwork but before issuance of report Report dual-dated and auditor’s responsibility for events subsequent to field work limited to specific event referred to (N79A-1-26m) 180 Requiring disclosure would include sale of bond or capital stock issue, loss of plant or inventories in fire or flood, settlement of litigation when event giving rise to claim took place after balance sheet date, but not major drop in market price of stock of the corporation (M79A-1-lm) 125 Successor and predecessor auditor communications See Audit engagement— First examination Successor auditor Predecessor auditor Statements reported on may require revision (M78A-1-44m) 27 Predecessor auditor working paper review Should permit review relative to matters of continuing accounting significance, such as analysis of contingencies (M78A-1-33m) 26 Supplies See Audit procedures Suretyship Bankruptcy of obligor Surety company pays workers of obligor in full and is entitled to the same priority in the bankruptcy that the workers would have had (M78L-1-46m) 40 Contractual prerequisites Signed writing, separate consideration for surety’s promise, and legal capacity of the surety, but not solvency of the principal debtor (N78L-1-25m) 89 Cosureties Liable in default of entire loan in the amounts of their relative surety relationship to the total of surety relationships (N78L-1-21m) 89 Release by the creditor of one cosurety (each was for full amount of the loan) also released the other to the extent of one half the loan (M79L-1-7m) 136 Shareholders who orally guaranteed repayment of a loan to the corporation are cosureties and, as such, their surety undertaking must be in a signed writing (N78L-1-24m) 89 Default and payment by surety Surety entitled to subrogation to rights of creditor or reimbursement from the obligor (M78L-1-45m) 39 No liability of surety since failure to give notice of first embezzlement (after which a larger second was made with disappearance) of employee is valid defense (N78L-1-20m) 89 248 Suretyship (cont.) 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11171 | http://physics.bu.edu/~duffy/PY106/Electricfield.html | Electric field
Electric field
7-7-99
Electric field
To help visualize how a charge, or a collection of charges, influences the region around it, the concept of an electric field is used. The electric field E is analogous to g, which we called the acceleration due to gravity but which is really the gravitational field. Everything we learned about gravity, and how masses respond to gravitational forces, can help us understand how electric charges respond to electric forces.
The electric field a distance r away from a point charge Q is given by:
Electric field from a point charge : E = k Q / r 2
The electric field from a positive charge points away from the charge; the electric field from a negative charge points toward the charge. Like the electric force, the electric field E is a vector. If the electric field at a particular point is known, the force a charge q experiences when it is placed at that point is given by :
F = qE
If q is positive, the force is in the same direction as the field; if q is negative, the force is in the opposite direction as the field.
Learning from gravity
Right now you are experiencing a uniform gravitational field: it has a magnitude of 9.8 m/s 2 and points straight down. If you threw a mass through the air, you know it would follow a parabolic path because of gravity. You could determine when and where the object would land by doing a projectile motion analysis, separating everything into x and y components. The horizontal acceleration is zero, and the vertical acceleration is g. We know this because a free-body diagram shows only mg, acting vertically, and applying Newton's second law tells us that mg = ma, so a = g.
You can do the same thing with charges in a uniform electric field. If you throw a charge into a uniform electric field (same magnitude and direction everywhere), it would also follow a parabolic path. We're going to neglect gravity; the parabola comes from the constant force experienced by the charge in the electric field. Again, you could determine when and where the charge would land by doing a projectile motion analysis. The acceleration is again zero in one direction and constant in the other. The value of the acceleration can be found by drawing a free-body diagram (one force, F = qE) and applying Newton's second law. This says:
qE = ma, so the acceleration is a = qE / m.
Is it valid to neglect gravity? What matters is the size of qE / m relative to g. As long as qE / m is much larger than g, gravity can be ignored. Gravity is very easy to account for, of course : simply add mg to the free-body diagram and go from there.
The one big difference between gravity and electricity is that m, the mass, is always positive, while q, the charge, can be positive, zero, or negative.
What does an electric field look like?
An electric field can be visualized on paper by drawing lines of force, which give an indication of both the size and the strength of the field. Lines of force are also called field lines. Field lines start on positive charges and end on negative charges, and the direction of the field line at a point tells you what direction the force experienced by a charge will be if the charge is placed at that point. If the charge is positive, it will experience a force in the same direction as the field; if it is negative the force will be opposite to the field.
The fields from isolated, individual charges look like this:
When there is more than one charge in a region, the electric field lines will not be straight lines; they will curve in response to the different charges. In every case, though, the field is highest where the field lines are close together, and decreases as the lines get further apart.
An example
Two charges are placed on the x axis. The first, with a charge of +Q, is at the origin. The second, with a charge of -2Q, is at x = 1.00 m. Where on the x axis is the electric field equal to zero?
This question involves an important concept that we haven't discussed yet: the field from a collection of charges is simply the vector sum of the fields from the individual charges. To find the places where the field is zero, simply add the field from the first charge to that of the second charge and see where they cancel each other out.
In any problem like this it's helpful to come up with a rough estimate of where the point, or points, where the field is zero is/are. There is no such point between the two charges, because between them the field from the +Q charge points to the right and so does the field from the -2Q charge. To the right of the -2Q charge, the field from the +Q charge points right and the one from the -2Q charge points left. The field from the -2Q charge is always larger, though, because the charge is bigger and closer, so the fields can't cancel. To the left of the +Q charge, though, the fields can cancel. Let's say the point where they cancel is a distance x to the left of the +Q charge.
Cross-multiplying and expanding the bracket gives:
Solving for x using the quadratic equation gives: x = 2.41 m or x = -0.414 m
The answer to go with is x = 2.41 m. This corresponds to 2.41 m to the left of the +Q charge. The other point is between the charges. It corresponds to the point where the fields from the two charges have the same magnitude, but they both point in the same direction there so they don't cancel out.
The field around a charged conductor
A conductor is in electrostatic equilibrium when the charge distribution (the way the charge is distributed over the conductor) is fixed. Basically, when you charge a conductor the charge spreads itself out. At equilibrium, the charge and electric field follow these guidelines:
the excess charge lies only at the surface of the conductor
the electric field is zero within the solid part of the conductor
the electric field at the surface of the conductor is perpendicular to the surface
charge accumulates, and the field is strongest, on pointy parts of the conductor
Let's see if we can explain these things. Consider a negatively-charged conductor; in other words, a conductor with an excess of electrons. The excess electrons repel each other, so they want to get as far away from each other as possible. To do this they move to the surface of the conductor. They also distribute themselves so the electric field inside the conductor is zero. If the field wasn't zero, any electrons that are free to move would. There are plenty of free electrons inside the conductor (they're the ones that are canceling out the positive charge from all the protons) and they don't move, so the field must be zero.
A similar argument explains why the field at the surface of the conductor is perpendicular to the surface. If it wasn't, there would be a component of the field along the surface. A charge experiencing that field would move along the surface in response to that field, which is inconsistent with the conductor being in equilibrium.
Why does charge pile up at the pointy ends of a conductor? Consider two conductors, one in the shape of a circle and one in the shape of a line. Charges are distributed uniformly along both conductors. With the circular shape, each charge has no net force on it, because there is the same amount of charge on either side of it and it is uniformly distributed. The circular conductor is in equilibrium, as far as its charge distribution is concerned.
With the line, on the other hand, a uniform distribution does not correspond to equilbrium. If you look at the second charge from the left on the line, for example, there is just one charge to its left and several on the right. This charge would experience a force to the left, pushing it down towards the end. For charge distributed along a line, the equilibrium distribution would look more like this:
The charge accumulates at the pointy ends because that balances the forces on each charge.
Electric flux
A clever way to calculate the electric field from a charged conductor is to use Gauss' Law, which is explained in Appendix D in the textbook. Gauss' Law can be tricky to apply, though, so we won't get into that. What we will do is to look at some implications of Gauss' Law. It's also a good time to introduce the concept of flux. This is important for deriving electric fields with Gauss' Law, which you will NOT be responsible for; where it'll really help us out is when we get to magnetism, when we do magnetic flux.
Electric flux is a measure of the number of electric field lines passing through an area. To calculate the flux through a particular surface, multiply the surface area by the component of the electric field perpendicular to the surface. If the electric field is parallel to the surface, no field lines pass through the surface and the flux will be zero. The maximum flux occurs when the field is perpendicular to the surface.
Permittivity
Even though we won't use this for anything, we should at least write down Gauss' law:
Gauss' Law - the sum of the electric flux through a surface is equal to the charge enclosed by a surface divided by a constant , the permittivity of free space.
What is the permittivity of free space? It's a constant related to the constant k that appears in Coulomb's law. The relationship between the two is this:
Implications of Gauss' Law
Gauss' Law is a powerful method of calculating electric fields. If you have a solid conducting sphere (e.g., a metal ball) that has a net charge Q on it, you know all the excess charge lies on the outside of the sphere. Gauss' law tells us that the electric field inside the sphere is zero, and the electric field outside the sphere is the same as the field from a point charge with a net charge of Q. That's a pretty neat result.
The result for the sphere applies whether it's solid or hollow. Let's look at the hollow sphere, and make it more interesting by adding a point charge at the center.
What does the electric field look like around this charge inside the hollow sphere? How is the negative charge distributed on the hollow sphere? To find the answers, keep these things in mind:
The electric field must be zero inside the solid part of the sphere
Outside the solid part of the sphere, you can find the net electric field by adding, as vectors, the electric field from the point charge alone and from the sphere alone
We know that the electric field from the point charge is given by kq / r 2. Because the charge is positive, the field points away from the charge.
If we took the point charge out of the sphere, the field from the negative charge on the sphere would be zero inside the sphere, and given by kQ / r 2 outside the sphere.
The net electric field with the point charge and the charged sphere, then, is the sum of the fields from the point charge alone and from the sphere alone (except inside the solid part of the sphere, where the field must be zero). This is shown in the picture:
How is the charge distributed on the sphere? The electrons must distribute themselves so the field is zero in the solid part. This means there must be -5 microcoulombs of charge on the inner surface, to stop all the field lines from the +5 microcoulomb point charge. There must then be +2 microcoulombs of charge on the outer surface of the sphere, to give a net charge of -5+2 = -3 microcoulombs.
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11172 | https://cseweb.ucsd.edu/classes/wi16/cse140-a/ds/Mid1Review.pdf | Midterm1 Review Jan 24 Armita Outline ● Boolean Algebra ○ Axioms ■ closure, Identity elements, complements, commutativity, distributivity ○ theorems ■ Associativity, Duality, De Morgan, ● Consensus theorem ● Shannon Expansion ● Canonical SOP & POS, minterms, maxterms ● Implicant, Prime Implicant, Essential Prime Implicant ● Implicates, Prime Implicates, Essential Prime Implicates ● Minimized two level SOP/POS using Kmaps Boolean Algebra Boolean Algebra is a multiple valued logic {a0, a1, a2, ...an}, defined over two operations {op1, op2}, that satisfies the following properties: ● It is closed with respect to each of the operations.
● Each operation has an identity element: ○ there exist x such that: A op1 x = A , for each A ∈ {a0, a1, a2, ...an} ■ we call x the identity element of op1 ○ there exist y such that: A op2 y = A , for each A ∈ {a0, a1, a2, ...an} ■ we call y the identity element of op2 for example: 1 AND X = X → “1” is the identity element of AND operation.
(in switching algebra) 0 OR X = X → “0” is the identity element of OR operation.
x ∈ {0,1} complements ● Each Element (M) has a unique complement element (N), defined as follows: ○ M op1 N = Identity element of op2 ○ M op2 N = Identity element of op1 ■ for each M from {a0, a1, a2, ...an} we can find a unique N from {a0, a1, a2, ...an} that satisfies the above equalities and we call them complements of each other.
■ We cannot have Boolean algebra that has odd number of values!
for example: (in switching algebra) 0 OR 1 = 1 = Identity element of AND operation 0 AND 1 = 0 = Identity element of OR operation duality 0 and 1 are complement of each other!
Axioms ● Commutativity: for each X,Y∈ {a0, a1, a2, ...an} : X op1 Y = Y op1 X X op2 Y = X op2 Y ● Distributivity: for each X,Y, Z ∈ {a0, a1, a2, ...an} : X op1 ( Y op2 Z) = (X op1 Y) op2 (X op1 Z) X op2 ( Y op1 Z) = (X op2 Y) op1 (X op2 Z) for each X,Y ∈ {0,1} : X AND Y = Y AND X X OR Y = X OR Y for each X,Y, Z ∈ {0,1} : X AND ( Y OR Z) = (X AND Y) OR (X AND Z) X OR ( Y AND Z) = (X OR Y) AND (X OR Z) switching Algebra switching Algebra duality duality Knowing these Axioms, we can prove useful theorems for Boolean Algebra, ex: De morgan law, Consensus theorems ...
Consensus Theorem consensus theorem : AB + A’C + BC = AB + A’C [BC is called the consensus term of AB and A’C] In a general form: AN + A’ M + MN = AN + A’M where M and N are boolean expressions, for example: abcd + a’ce + bcde = abcd + a’ce [(bcd)(ce) is the consensus term, so its redundant.] abc + a’ (c + d) + bc + bcd = abc + a’(c+d) [ (bc)(c+d) is the consensus term (which is equal to bc +bcd), so its redundant. ] Consensus Theorem In the dual form : consensus theorem: (A+B) (A’ + C) (B + C ) = (A+B) (A’+C) [ (B+C) is called the consensus term of (A+B) and (A’+C)] In a general form: (A + N)(A’ + M)(M+N) = (A+N)(A’+M) where M and N are boolean expressions, for example: (a + b + c + d) (a’ + c + e ) ( b + c + d + e) = (a + b + c + d) (a’ + c + e) [(b+c+d)+(c+e) is the consensus term of (a + b + c + d) and (a’ + c + e) so its redundant.] Shannon Expansion f (x, y, z) = x.f(1, y, z) + x’.f(0, y, z) in “ f ” replace x with 1 and x’ with 0 in “ f ” replace x with 0 and x’ with 1 = y.f(x, 1, z) + y’.f(x, 0, z) = z.f(x, y, 1) + z’.f(x, y, 0) example: simplify the given function: f(x, y, z) = xyz + x’y + z’y On which variable should be apply the Shannon expansion theorem?
only for switching algebra!!
Shannon Expansion example: simplify the given function: f(x, y, z) = xyz + x’y + z’y Lets choose x ! = x. f(1, y, z) + x’.f(0,y,z) = x . (yz + 0 + z’y) + x’ . ( 0 + y + z’y) = x . (y (z+z’)) + x’ . (y (1+z’)) = x. y + x’. y = y (x+x’) = y Shannon Expansion example: simplify the given function: f(x, y, z) = xyz + x’y + z’y what if we choose z ?! = z. f(x, y, 1) + z’.f(x,y,0) = z . (xy + x’y + 0) + z’ . ( 0 + x’y + y) = z . (y (x+x’)) + z’ . (y (x’+1)) = z.y + z’. y = y (z+z’) = y Shannon Expansion f (x, y, z) = (x + f(0, y, z)) . (x’ + f(1, y, z)) in “ f ” replace x with 0 and x’ with 1 in “ f ” replace x with 1 and x’ with 0 = (y + f(x, 0, z)) . (y’ + f(x, 1, z)) = (z + f(x, y, 0)) . (z’ + f(x, y, 1)) example: simplify the given function: f(x, y, z) = (x + y + z’) ( x’ + y’ + z ) ( z + y) Shannon Expansion example: simplify the given function: f(x, y, z) = (x + y + z’) ( x’ + y’ + z ) ( z + y) = (x + f(0, y, z)) . (x’ + f(1, y, z)) = (x + (y+z’) (1) (z+y) ) . (x’ + (1) (y’+z) (z+y) ) = (x + y + z’z ) . (x’ + z + y’y) = (x + y + 0 ) . (x’ + z + 0 ) = (x+y) (x’+z) Definitions: literal → a variable in the direct or complemented form: x , x’ , y , y’ product term → AND of some literals: (xy’z) , (xy) can differ in their size sum term → OR of some literals: (x + y’ + z) , (y+z) , can differ in their size Minterm minterm → a product term that has all the variables in either the direct form or the complement form. For a n-variable system we have 2^n minterms and all have the same size (one AND gate with n input): {a, b,c,d} → abcd, abcd’, abc’d, …, a’b’c’d’ Each minterm is equal to value “ one ” only for a unique input combination and is “ zero ” otherwise: “ abcd ” will be one if and only if (a,b,c,d) = (1,1,1,1) “ abcd’ ” will be one if and only if (a,b,c,d) = (1,1,1,0) “ abc’d ” will be one if and only if (a,b,c,d) = (1,1,0,1) …..
“ a’b’c’d’ ” will be one if and only if (a,b,c,d) = (0,0,0,0) Maxterm maxterm → a sum term that has all the variables in either the direct form or the complement form. For a n-variable system we have 2^n maxterms and all have the same size (one OR gate with n input): {a,b,c, d} → (a+b+c+d) , (a+b+c+d’) , (a+b+c’+d ) , …, (a’+b’+c’+d’) Each maxterm is equal to value “ zero ” only for a unique input combination and is “ one ” otherwise: “ a+b+c+d ” will be zero if and only if (a,b,c,d) = (0,0,0,0) “ a+b+c+d’ ” will be zero if and only if (a,b,c,d) = (0,0,0,1) “ a+b+c’+d ” will be zero if and only if (a,b,c,d) = (0,0,1,0) ….
“ a’+b’+c’+d’ ” will be zero if and only if (a,b,c,d) = (1,1,1,1) Canonical form Based on the definition of minterms and maxterms, we can represent any switching function with two canonical form: canonical sum of products: Sum of its minterms (i.e ORing all places where the function is one) canonical product of sums: Product of its maxterms (i.e ANDing all places where the function is zero) canonical form Example: Show the canonical forms for a switching function that receives 3 binary inputs {a,b,c} and the output {F} is one if and only if the number of ones in the input are even: ● draw the truth table representing F(a,b,c) ● find the minterms → write canonical SOP form ● find the maxterms → write the canonical POS form 3 inputs {a,b,c} → the truth table has 2^3 = 8 rows: a b c F 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 a’b’c’ a+b+c’ a+b’+c a’bc a’+b+c ab’c abc’ a’+b’+c’ canonical SOP F (a,b,c) = a’b’c’ + a’bc + ab’c + abc’ = ∑m (0 , 3, 5, 6 ) = m0 + m3 + m5 + m6 canonical POS F (a,b,c) = (a+b+c’) (a+b’+c) (a’+b+c) (a’+b’+c’) = ∏M (1, 2, 4, 7) = M1 . M2 . M4 . M7 Logic Minimization Representing a function in the canonical form is not efficient. Hence, we are looking for ways to merge two or more minterms (maxterms) together in order to get a smaller product (sum) term which is more efficient.
Adjacency: two minterms (maxterms) are adjacent, if they differ only in one variable: abcd and ab’cd are adjacent. (a+b+c) and (a+b+c’) are adjacent.
We can merge each Adjacent pair into one term !
abcd + ab’cd = acd (b+b’) = acd (a+b+c)(a+b+c’) = (a+b) + (cc’) = a+b K-maps!
Searching for adjacent pairs are the basis of logic minimization. Karnaugh maps present a visual way to efficiently look for this pairs!
All neighbour cells in a Kmaps are adjacent. Dont forget the edges!! ab\cd 00 01 11 10 00 0 1 3 2 01 4 5 7 6 11 12 13 15 14 10 8 9 11 10 c d b a We can use Kmaps for 5 or 6 variables function.
In a 5 variable kmap, each cell has 5 neighbours and in a 6 variable kmap each cell has 6 neighbours.
Definitions: on-set : set of all the “ones” in the function off-set : set of all the “zeros” in the function Implicant: A product term that has non-empty intersection with on-set and does not intersect with the off-set. It can also contain “dont cares” ● Being a product term means that it should include 1 minterm, or 2 minterms, or 4, …(powers of two)... . It CANNOT have 3 minterms! Definitions: Prime Implicant: Is an Implicant which is not a proper subset of any other Implicant. i.e you cannot find any other implicant that completely covers a prime implicant. For logic minimization we use Prime Implicants.
Essential Prime Implicant: A Prime Implicant that has at least one element from the on-set that is cannot be covered by any other Prime Implicant. By this definition, these Implicants are essential and will show up in all minimal SOPs. Minimal SOP ● Draw the kmap representing the given function ● Find all the Prime Impilcants using the on-set.
● Find all the Essential Prime Implicants.
● Write all the Essential Prime Implicants in the final cover, ● For the ones in the on-set which are not covered yet, search through the remaining non-essential prime implicants that will cover the remaining ones in the map. By trial and error at this step you can find all the minimal covers.
Example ab\cd 00 01 11 10 00 0 1 1 3 2 1 01 4 1 5 1 7 1 6 1 11 12 13 1 15 x 14 10 8 9 1 11 10 b a d c Prime Implicants: ?
Essential Prime Implicants: ?
Minimal SOP: ?
The empty cells are “0”.
Example ab\cd 00 01 11 10 00 0 1 1 3 2 1 01 4 1 5 1 7 1 6 1 11 12 13 1 15 x 14 10 8 9 1 11 10 b a d c Is ∑m (0,4) a Prime Implicant?
NO!! it can be covered by ∑m (0,4,2,6) How about ∑m (5,9,13)?
NO!! Its not a product term, we cannot have product terms with 3 minterms!
The empty cells are “0”.
Example ab\cd 00 01 11 10 00 0 1 1 3 2 1 01 4 1 5 1 7 1 6 1 11 12 13 1 15 x 14 10 8 9 1 11 10 b a d c Prime Implicants: ∑ m (0,2,4,6) = a’d’ ∑ m (4,5,6,7) = a’b ∑ m (5,7,13,15) = bd ∑ m (9,13) = ac’d Essential Prime Implicants: ∑ m (0,2,4,6) = a’d’ ∑ m (9,13) = ac’d Minimal SOP: ?
The empty cells are “0”.
Example ab\cd 00 01 11 10 00 0 1 1 3 2 1 01 4 1 5 1 7 1 6 1 11 12 13 1 15 x 14 10 8 9 1 11 10 b a d c Prime Implicants: ∑ m (0,2,4,6) = a’d’ ∑ m (4,5,6,7) = a’b ∑ m (5,7,13,15) = bd ∑ m (9,13) = ac’d Essential Prime Implicants: ∑ m (0,2,4,6) = a’d’ ∑ m (9,13) = ac’d Minimal SOP: (#1) F (a,b,c,d) = a’d’ + ac’d + a’b The empty cells are “0”.
Example ab\cd 00 01 11 10 00 0 1 1 3 2 1 01 4 1 5 1 7 1 6 1 11 12 13 1 15 x 14 10 8 9 1 11 10 b a d c Prime Implicants: ∑ m (0,2,4,6) = a’d’ ∑ m (4,5,6,7) = a’b ∑ m (5,7,13,15) = bd ∑ m (9,13) = ac’d Essential Prime Implicants: ∑ m (0,2,4,6) = a’d’ ∑ m (9,13) = ac’d Minimal SOP: (#2) F (a,b,c,d) = a’d’ + ac’d + bd The empty cells are “0”.
Example ab\cd 00 01 11 10 00 0 1 1 3 2 1 01 4 1 5 1 7 1 6 1 11 12 13 1 15 x 14 10 8 9 1 11 10 b a d c Prime Implicants: ∑ m (0,2,4,6) = a’d’ ∑ m (4,5,6,7) = a’b ∑ m (5,7,13,15) = bd ∑ m (9,13) = ac’d Essential Prime Implicants: ∑ m (0,2,4,6) = a’d’ ∑ m (9,13) = ac’d Minimal SOP: F (a,b,c,d) = a’d’ + ac’d + a’b OR F (a,b,c,d) = a’d’ + ac’d + bd The empty cells are “0”.
Definitions: Another approach is covering the “zeros” and finiding the minimal POS form: Implicate: A sum term that has non-empty intersection with off-set and does not intersect with the on-set. It can also contain “dont cares” ● Being a sum term means that it should include 1 maxterm, or 2 maxterms, or 4, or ...(powers of two)... It CANNOT have 3 maxterms! Definitions: Prime Implicate: Is an Implicate which is not subset of any other Implicate. i.e you cannot find any other Implicate that completely cover a Prime Implicate. For logic POS minimization we use Prime Implicates.
Essential Prime Implicate: A Prime Implicate that has at least one element from the off-set that is cannot be covered by any other Prime Implicate. By this definition, these Implicates are essential and will show up in all minimal POSs. Minimal POS ● Draw the kmap representing the given function ● Find all the Prime Impilcate using the off-set.
● Find all the Essential Prime Implicates.
● Write all the Essential Prime Implicates in the final cover, ● For the zeros in the off-set which are not covered yet, search through the remaining non-essential Prime Implicates that will cover the remaining zero in the map. By trial and error at this step you can find all the minimal covers.
Example ab\cd 00 01 11 10 00 0 0 1 3 0 2 0 01 4 0 5 7 0 6 0 11 12 13 0 15 x 14 10 8 9 11 0 10 b a d c Prime Implicates: ∏M(0,2,4,6) = a+d ∏M (3,5,7,11) = c’+d’ ∏M (2,3,4,6) = a+c’ ∏M(13,15) = a’+b’+d’ Essential Prime Implicates: ∏M(0,2,4,6) = a+d ∏M (3,5,7,11) = c’+d’ ∏M(13,15) = a’+b’+d’ Minimal POS: F (a,b,c,d) = (a+d)(c’+d’)(a’+b’+d’) The empty cells are “1”.
Boolean Algebra VS. K-maps F (a,b,c) = c’b + a’b + a’b’c + a’bc c\ab 00 01 11 10 0 0 1 1 0 1 1 1 0 0 Boolean Algebra VS. K-maps F (a,b,c) = c’b + a’b + a’b’c + a’bc = c’b + a’b + a’b’c + a’bc = c’b + a’b + a’c (b’+b) = c’b + a’b + a’c c\ab 00 01 11 10 0 0 1 1 0 1 1 1 0 0 c\ab 00 01 11 10 0 0 1 1 0 1 1 1 0 0 not prime!
not prime!
adjacent Boolean Algebra VS. K-maps F (a,b,c) = c’b + a’b + a’b’c + a’bc = c’b + a’b + a’b’c + a’bc = c’b + a’b + a’c (b’+b) = c’b + a’b +a’c (consensus ) = c’b + a’c c\ab 00 01 11 10 0 0 1 1 0 1 1 1 0 0 c\ab 00 01 11 10 0 0 1 1 0 1 1 1 0 0 essential essential QUESTIONS? GOOD LUCK! |
11173 | https://en.wikipedia.org/wiki/Snub_cube | Snub cube - Wikipedia
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1 ConstructionToggle Construction subsection
1.1 Cartesian coordinates
2 Properties
3 Graph
4 References
5 External links
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From Wikipedia, the free encyclopedia
Archimedean solid with 38 faces
| Snub cube |
| |
| Type | Archimedean solid |
| Faces | 38 |
| Edges | 60 |
| Vertices | 24 |
| Symmetry group | Rotational octahedral symmetryO{\displaystyle \mathrm {O} } |
| Dihedral angle (degrees) | triangle-to-triangle: 153.23° triangle-to-square: 142.98° |
| Dual polyhedron | Pentagonal icositetrahedron |
| Properties | convex, chiral |
| Vertex figure |
| |
| Net |
| |
In geometry, the snub cube, or snub cuboctahedron, is an Archimedean solid with 38 faces: 6 squares and 32 equilateral triangles. It has 60 edges and 24 vertices. Kepler first named it in Latin as cubus simus in 1619 in his Harmonices Mundi.H. S. M. Coxeter, noting it could be derived equally from the octahedron as the cube, called it snub cuboctahedron, with a vertical extended Schläfli symbols{4 3}{\displaystyle s\scriptstyle {\begin{Bmatrix}4\3\end{Bmatrix}}}, and representing an alternation of a truncated cuboctahedron, which has Schläfli symbol t{4 3}{\displaystyle t\scriptstyle {\begin{Bmatrix}4\3\end{Bmatrix}}}.
Construction
[edit]
The snub cube can be generated by taking the six faces of the cube, pulling them outward so they no longer touch, then giving them each a small rotation on their centers (all clockwise or all counter-clockwise) until the spaces between can be filled with equilateral triangles.
Process of snub cube's construction by rhombicuboctahedron
The snub cube may also be constructed from a rhombicuboctahedron. It started by twisting its square face (in blue), allowing its triangles (in red) to be automatically twisted in opposite directions, forming other square faces (in white) to be skewed quadrilaterals that can be filled in two equilateral triangles.
The snub cube can also be derived from the truncated cuboctahedron by the process of alternation. 24 vertices of the truncated cuboctahedron form a polyhedron topologically equivalent to the snub cube; the other 24 form its mirror-image. The resulting polyhedron is vertex-transitive but not uniform.
Uniform alternation of a truncated cuboctahedron
Cartesian coordinates
[edit]
Cartesian coordinates for the vertices of a snub cube are all the even permutations of (±1,±1 t,±t),{\displaystyle \left(\pm 1,\pm {\frac {1}{t}},\pm t\right),} with an even number of plus signs, along with all the odd permutations with an odd number of plus signs, where t≈1.83929{\displaystyle t\approx 1.83929} is the tribonacci constant. Taking the even permutations with an odd number of plus signs, and the odd permutations with an even number of plus signs, gives a different snub cube, the mirror image. Taking them together yields the compound of two snub cubes.
This snub cube has edges of length α=2+4 t−2 t 2{\displaystyle \alpha ={\sqrt {2+4t-2t^{2}}}}, a number which satisfies the equation α 6−4 α 4+16 α 2−32=0,{\displaystyle \alpha ^{6}-4\alpha ^{4}+16\alpha ^{2}-32=0,} and can be written as α=4 3−16 3 β+2 β 3≈1.609 72 β=26+6 33 3.{\displaystyle {\begin{aligned}\alpha &={\sqrt {{\frac {4}{3}}-{\frac {16}{3\beta }}+{\frac {2\beta }{3}}}}\approx 1.609\,72\\beta &={\sqrt[{3}]{26+6{\sqrt {33}}}}.\end{aligned}}} To get a snub cube with unit edge length, divide all the coordinates above by the value α given above.
Properties
[edit]
3D model of a snub cube
For a snub cube with edge length a{\displaystyle a}, its surface area and volume are:A=(6+8 3)a 2≈19.856 a 2 V=8 t+6 3 2(t 2−3)a 3≈7.889 a 3.{\displaystyle {\begin{aligned}A&=\left(6+8{\sqrt {3}}\right)a^{2}&\approx 19.856a^{2}\V&={\frac {8t+6}{3{\sqrt {2(t^{2}-3)}}}}a^{3}&\approx 7.889a^{3}.\end{aligned}}}
The snub cube is an Archimedean solid, meaning it is a highly symmetric and semi-regular polyhedron, and two or more different regular polygonal faces meet in a vertex. It is chiral, meaning there are two distinct forms whenever being mirrored. Therefore, the snub cube has the rotational octahedral symmetryO{\displaystyle \mathrm {O} }. The polygonal faces that meet for every vertex are four equilateral triangles and one square, and the vertex figure of a snub cube is 3 4⋅4{\displaystyle 3^{4}\cdot 4}. The dual polyhedron of a snub cube is pentagonal icositetrahedron, a Catalan solid.
Graph
[edit]
The graph of a snub cube
The skeleton of a snub cube can be represented as a graph with 24 vertices and 60 edges, an Archimedean graph.
References
[edit]
^Conway, John H.; Burgiel, Heidi; Goodman-Struss, Chaim (2008). The Symmetries of Things. CRC Press. p.287. ISBN978-1-4398-6489-0.
^Holme, A. (2010). Geometry: Our Cultural Heritage. Springer. p.99. doi:10.1007/978-3-642-14441-7. ISBN978-3-642-14441-7.
^Conway, Burgiel & Goodman-Struss (2008), p.287–288.
^Collins, Julian (2019). Numbers in Minutes. Hachette. p.36–37. ISBN978-1-78747-730-8.
^Berman, Martin (1971). "Regular-faced convex polyhedra". Journal of the Franklin Institute. 291 (5): 329–352. doi:10.1016/0016-0032(71)90071-8. MR0290245.
^Diudea, M. V. (2018). Multi-shell Polyhedral Clusters. Carbon Materials: Chemistry and Physics. Vol.10. Springer. p.39. doi:10.1007/978-3-319-64123-2. ISBN978-3-319-64123-2.
^Koca, M.; Koca, N. O. (2013). "Coxeter groups, quaternions, symmetries of polyhedra and 4D polytopes". Mathematical Physics: Proceedings of the 13th Regional Conference, Antalya, Turkey, 27–31 October 2010. World Scientific. p.49.
^Cromwell, Peter R. (1997). Polyhedra. Cambridge University Press. p.386. ISBN978-0-521-55432-9.
^Williams, Robert (1979). The Geometrical Foundation of Natural Structure: A Source Book of Design. Dover Publications, Inc. p.85. ISBN978-0-486-23729-9.
^Read, R. C.; Wilson, R. J. (1998), An Atlas of Graphs, Oxford University Press, p.269
Jayatilake, Udaya (March 2005). "Calculations on face and vertex regular polyhedra". Mathematical Gazette. 89 (514): 76–81. doi:10.1017/S0025557200176818. S2CID125675814.
Williams, Robert (1979). The Geometrical Foundation of Natural Structure: A Source Book of Design. Dover Publications, Inc. ISBN0-486-23729-X. (Section 3-9)
External links
[edit]
Weisstein, Eric W., "Snub cube" ("Archimedean solid") at MathWorld.
Weisstein, Eric W."Snub cubical graph". MathWorld.
Klitzing, Richard. "3D convex uniform polyhedra s3s4s - snic".
The Uniform Polyhedra
Virtual Reality Polyhedra The Encyclopedia of Polyhedra
Editable printable net of a Snub Cube with interactive 3D view
| v t e Archimedean solids |
| Truncated tetrahedron (Truncate) Truncated tetrahedron (Zip) Truncated cube (Truncate) Truncated octahedron (Zip) Truncated dodecahedron (Truncate) Truncated icosahedron (Zip) Tetratetrahedron (Ambo) Cuboctahedron (Ambo) Icosidodecahedron (Ambo) Rhombitetratetrahedron (Expand) Truncated tetratetrahedron (Bevel) Rhombicuboctahedron (Expand) Truncated cuboctahedron (Bevel) Rhombicosidodecahedron (Expand) Truncated icosidodecahedron (Bevel) Snub tetrahedron (Snub) Snub cube (Snub) Snub dodecahedron (Snub) | Catalan duals | | Triakis tetrahedron (Needle) Triakis tetrahedron (Kis) Triakis octahedron (Needle) Tetrakis hexahedron (Kis) Triakis icosahedron (Needle) Pentakis dodecahedron (Kis) Rhombic hexahedron (Join) Rhombic dodecahedron (Join) Rhombic triacontahedron (Join) Deltoidal dodecahedron (Ortho) Disdyakis hexahedron (Meta) Deltoidal icositetrahedron (Ortho) Disdyakis dodecahedron (Meta) Deltoidal hexecontahedron (Ortho) Disdyakis triacontahedron (Meta) Pentagonal dodecahedron (Gyro) Pentagonal icositetrahedron (Gyro) Pentagonal hexecontahedron (Gyro) | |
| v t e Convex polyhedra |
| Platonic solids(regular) | tetrahedron cube octahedron dodecahedron icosahedron |
| Archimedean solids (semiregular or uniform) | truncated tetrahedron cuboctahedron truncated cube truncated octahedron rhombicuboctahedron truncated cuboctahedron snub cube icosidodecahedron truncated dodecahedron truncated icosahedron rhombicosidodecahedron truncated icosidodecahedron snub dodecahedron |
| Catalan solids (duals of Archimedean) | triakis tetrahedron rhombic dodecahedron triakis octahedron tetrakis hexahedron deltoidal icositetrahedron disdyakis dodecahedron pentagonal icositetrahedron rhombic triacontahedron triakis icosahedron pentakis dodecahedron deltoidal hexecontahedron disdyakis triacontahedron pentagonal hexecontahedron |
| Dihedral regular | dihedron hosohedron |
| Dihedral uniform | prisms antiprisms duals: bipyramids trapezohedra |
| Dihedral others | pyramids truncated trapezohedra gyroelongated bipyramid cupola bicupola frustum bifrustum rotunda birotunda prismatoid scutoid |
| Degenerate polyhedra are in italics. |
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11174 | https://domath.surju.ee/apex-angles/ | Apex angles of an isosceles triangle – DoMath Dictionary
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Apex angles of an isosceles triangle
Apex angle (vertex angle) of an isosceles triangle
An apex angle is the angle between the lines that define the apex.
Apex angle of an isosceles triangle.
In an isosceles triangle, the apex is the vertex where the two sides of equal length meet, opposite the unequal third side
The angles at the base are called the base angles, and the angle opposite the base is called the vertex angle.
An isosceles triangleis a triangle that has two sides of equal length.
The two equal sides are called the legs and the third side is called the base of the triangle.
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11175 | https://www.kimballphysics.com/learning_center/topic-materials-and-magnetic-properties/ | Table of Contents
| | |
--- |
| Introduction | Extraneous Magnetic Fields |
| Electron Optics | Sources of Unintended Magnetic_Fields |
| Electric Field Optics | Eddy Currents |
| Magnetic Field Optics | Materials Used Vacuum Hardware |
| Beam Focus | Machining and Manufacturing Overview |
| Beam Deflection | Welds and Casting |
| Electron Spin | Materials Used Kimball Physics Systems |
| Magnetic Permeability | Stainless Steel |
| Relative Magnetic Permeability Units | Titanium |
| | Table- Magnetic Permeability Various Materials |
| Categories of Magnetic Permeability | FAQ |
| Ferromagnetic Materials | 1) Does titanium have any changes with its magnetic permeability if it is processed mechanically, such as by cold working or bending? |
| Paramagnetic Materials | 2) Does welding of stainless steel affect its magnetic permeability ? |
| Diamagnetic Materials | 3) Are there effects on material properties of 316L SS (specifically magnetic permeability) with CNC Machining vs. Welding? |
| Other Factors | 4) What are advantages of Titanium for ultra (UHV) and extreme (XHV) high vacuum chambers? |
| | 5) Do thread forming fabrication techniques in 316L SS cause working hardening of the metal and possible changes in its crystal structure from austenitic to a crystal structure that is ferromagnetic? |
| | 6) What effects on permeability would welding two pieces of 316L SS cause. Is its effect significantly larger the effects of work hardening? |
Introduction
Electron beams are used in a variety of applications, such as in particle accelerators, electron microscopes, and electron beam lithography. In these applications, precise control over the path and spatial distribution of the electron beam is essential for accurate and reliable performance.
Various materials are in proximity to the electron beam in vacuum systems. These materials can influence the electron beam in both intended and unintended ways.
In this document, we will briefly review 1) simplified electron beam control with electron optics, 2) materials used in vacuum systems and their magnetic properties, and 3) various unintended magnetic effects and how they can be mitigated.
Later in the document we will be discussing the material and magnetic properties of titanium and will also briefly discuss the properties of hydrogen diffusion with these materials in ultra-high (UHV) and extremely high vacuum (XHV) systems. At the end of the document, we also provide some comments to frequently asked questions.
Please note that the content in this document is only intended to provide a starting point for the reader to further research this topic.
Electron Optics
The electron optics used in an electron gun system utilize both electric fields (deflection plates) and magnetic fields (coils). The optical elements are typically carefully designed to precisely alter the trajectory of the beam as well as its cross-sectional form and focus.
Electric Field Optics
Electric fields generated across the opposing deflection plates of the electron gun system optics exert a force on the negative charges in the electron beam, causing an acceleration perpendicular to their original path, with the electron deflecting away from the negatively charged plate. The magnitude of this acceleration depends on 1) the strength of the electric field (the voltage across the plates, separation distance of the plates), 2) the mass and velocity of the particle, 3) plate geometry, and 4) the dielectric of the material present between the plates (in this case, vacuum).
By manipulating the electric field across the deflection plates, the trajectory of the electron beam can be altered to direct it at specific targets or to raster it in various pre-determined patterns.
Magnetic Field Optics
The magnetic fields generated by the coils in electron optics have a significant impact on both the trajectory and form of the electron beam.
Beam Focus
One key function of these magnetic fields is to focus the electron beam. Magnetic lenses, formed by cylindrical coils, are used to achieve this. The current passing through these coils generates a magnetic field that exerts a force on the electrons, known as the Lorentz force. The direction of this force is perpendicular to both the velocity of the electrons and the magnetic field direction. The orientation of the magnetic field and the resulting Lorentz force forces electrons in the beam to converge to a small focal spot. Careful design of magnetic fields inside the lens provides precise control of beam convergence and focused spot size.
Beam Deflection
Magnetic fields can also be used to deflect the path of the electron beam. Deflection coils generate magnetic fields that are perpendicular to the direction of electron motion. The fields are linear, and not cylindrically symmetric. As the electrons move through these magnetic fields, the Lorentz force causes them to experience a perpendicular acceleration, resulting in a change in their trajectory. By varying the strength and direction of the magnetic field, the electron beam can be precisely steered and scanned to specific positions.
Electron Spin
Additionally, magnetic fields find applications in manipulating the spin of electrons, as seen in electron spin resonance spectroscopy. By applying a magnetic field along a specific axis, the alignment of electron spins can be controlled, influencing their behavior and interactions with other particles or materials.
In summary, the magnetic fields generated by the coils in electron gun system optics serve to focus, deflect, and control the path and spin of electrons in the electron beam. This control is essential for various scientific and technological applications where precise manipulation of the beam is required.
Magnetic Permeability
Magnetic permeability is a physical property of materials that describes their ability to respond to a magnetic field. It is a measure of how much a material can be magnetized in the presence of an external magnetic field. Specifically, magnetic permeability is defined as the ratio of magnetic flux density to the magnetic field strength in a material. In electromagnetism, permeability is the measure of magnetization produced in a material in response to an applied magnetic field. Permeability is typically represented by the Greek letter µ. It is ratio of the magnetic induction B to the magnetizing field H in the material.
In SI units, permeability is measured in henries per meter (H/m), or equivalently in newtons per ampere squared (N/A2). The permeability constant μ0, also known as the magnetic constant or the permeability of free space, is the proportionality between magnetic induction and magnetizing force when forming a magnetic field in a classical vacuum.
Materials with high magnetic permeability are easily magnetized by an external magnetic field, while materials with low magnetic permeability are more difficult to magnetize. The magnetic permeability of a material also affects its ability to conduct magnetic flux, which is the flow of magnetic field lines through a material.
Relative Magnetic Permeability Units
Relative Permeability is a dimensionless quantity that compares the permeability of a material (µ) to the permeability of free space (µo). It is calculated by dividing the permeability of the material by the permeability of free space:
Relative Permeability = µ / µo
Since it is a ratio of two permeability values, relative permeability has no units. It is a convenient measure to compare magnetic permeability of materials without the units.
If we know the permeability of a material in H/m, we can calculate its relative permeability by dividing it by the permeability of free space (4π × 10-7 H/m). If we know the relative permeability of a material, we can calculate its permeability in H/m by multiplying it by the permeability of free space (4π × 10-7 H/m).
The unit of magnetic permeability is henries per meter (H/m) in International System of Units (SI), which is equivalent to 1 kgm/sec2A2.
Categories of Magnetic Permeability
In physics, several different types of material magnetism have been described. Ferromagnetism (along with the similar effect ferrimagnetism) is the strongest form of magnetism and is responsible for the common phenomenon of magnetism diversly seen in everyday life- from refrigerator magnets to the electromagnetic effects seen in electric motors and transformers.
Other substances can respond weakly to magnetic fields with three other basic forms of magnetism- paramagnetism, diamagnetism, and antiferromagnetism These forces are typically very weak and can only be detected by sensitive instruments in a laboratory. However, these effects can be a factor in sensitive electron beam systems.
Ferromagnetic Materials
These materials have a very high magnetic susceptibility and are strongly attracted to a magnetic field (e.g., magnet- permanent or electro). They can also retain their magnetization even after the external magnetic field is removed, a phenomenon referred to as remanence. They typically have a relative magnetic permeability value very much greater than 1.0 with values often of several thousand (see table below).
All common carbon steels (including mild steel), low alloy steels and tool steels are ferromagnetic. Some other metals such as nickel and cobalt are also ferromagnetic.
All stainless steels with the exception of the austenitic grades (such as 316L SS) are also magnetic – this includes: 1) all ferritic grades (e.g. 430, AtlasCR12, 444, F20S), 2) all duplex grades (e.g. 2205, 2304, 2101, 2507), 3) all martensitic grades (e.g. 431, 416, 420, 440C) and 4) all precipitation hardening grades (e.g. 630/17-4PH). Even though the duplex grades are mixtures of austenite and ferrite they are still strongly attracted to a magnet. (Ref: Atlas Steels- Tech Note No. 11).
In summary, carbon steels, low alloy steels, and tool steel, including Stainless Steel with ferritic and martensitic grades are ferromagnetic and are considered “magnetic” materials. Stainless Steel 316L, an alloy that contains iron but with an austenitic crystal structure (grade) is not magnetic, unless there have been changes to its crystal structure during a manufacturing operation (see below).
Paramagnetic Materials
These materials have a minimally positive magnetic susceptibility, which means that they are weakly attracted to a magnetic field. Examples of paramagnetic materials include aluminum, magnesium, platinum, 316L stainless steel (austenitic grade) and titanium. They typically have very low relative magnetic permeability values, usually just a fraction above 1.0 (see table below)
In summary, most non-ferrous metals such as aluminum, titanium, and their alloys are non-magnetic. Austenitic stainless steels, both the common 300-series (Cr-Ni) and the lower nickel 200-series (Cr-Mn-Ni) are non-magnetic. It is common for wrought austenitic stainless steels to contain a very small amount of ferrite, but this is not sufficient to significantly affect magnetic performance except in very critical applications (Ref Atlas Steels- Tech Note No. 11). Inconel is also paramagnetic, even though it has a large proportion of nickel (which is magnetic) as part of its alloy.
Diamagnetic Materials
These materials have a negative magnetic susceptibility (permeability less than 1.0), which means that they are weakly repelled by a magnetic field. Examples of diamagnetic materials include water, most organic compounds, copper, silver, gold, carbon, mercury, and lead.
Other Factors
The magnetic permeability of a material also depends on factors such as temperature and the strength of the external magnetic field. In addition, some materials can exhibit different magnetic properties depending on their microstructure, such as the presence of impurities or defects. In some metals, at elevated temperature (their Curie Temperature) they can change from ferromagnetic to non-magnetic. For common carbon steels this happens at about 768°C.
Overall, the magnetic permeability of a material is an important factor in determining its behavior in the presence of a magnetic field and is a key consideration in many technological applications.
Extraneous Magnetic Fields
If there are extraneous unintended magnetic fields along the path of the electron beam, they can deflect or distort the beam from its intended path and form or profile, which can result in inaccurate or unreliable results.
Sources of Unintended Magnetic Fields
Magnetic fields can unintentionally be introduced from structures used to create the vacuum environment for the electron beam, such as the metal vacuum chamber structures, flanges, and support materials and fittings within the vacuum space. As previously noted, these poorly defined and uncharacterized fields can deflect the path or distort the form of the electron beam. This is especially a concern when the electron beam application requires precision and accuracy such as in particle accelerators, electron microscopes, and electron beam lithography. In these applications, precise control over the path and form of the electron beam is essential for accurate and reliable performance.
Unintended sources of magnetic field in the electron beam environment include 1) eddy currents, 2) materials with high magnetic permeabilities such as ferromagnetic materials, and 3) unintended changes to magnetic properties caused by machining and manufacturing operations.
Eddy Currents
Metal vacuum chamber structures can create magnetic fields due to a phenomenon known as eddy currents. When a changing magnetic field passes through a conductive material, such as metal, it can induce electric currents to flow in the material. These currents, in turn, create their own magnetic fields, which can interact with the electron beam.
To avoid these problems, it is important to design vacuum chambers and other equipment used in electron beam applications with magnetic shielding and other measures to minimize the impact of eddy currents and other sources of magnetic fields. By minimizing magnetic field interference, researchers and engineers can ensure that their electron beam applications are accurate, reliable, and effective.
Materials Used in Vacuum Systems
Typically, Stainless Steel 316L and Titanium (and it alloys), materials commonly used in vacuum hardware, have relative magnetic permeability values of only a fraction above 1.0. In certain instances, however, parts of a Stainless Steel structure can have values much greater than 1.0. This is often related to various manufacturing operations such as bending and welding that can alter its normal crystal structure.
As noted in the table below, 316L Stainless Steel, titanium and its alloy, and aluminum all have relatively very low permeability values, which makes them good choices for vacuum chambers used in electron beam systems.
In contrast, high permeability materials such as iron and carbon steel should be avoided in these applications to minimize interference with the electron beam trajectory, profile, and focus.
Ceramic and glass materials have a permeability value of 1.0, which means they are non-magnetic and suitable for use in vacuum chamber hardware where magnetic interference must be minimized.
Machining / Manufacturing Operations Overview
CNC (computer numerical control) operations controlling milling machines and lathes using optimized cutting tool pressures with lubrication are typically used at Kimball Physics with no measurable effects on 316L Stainless Steel and Titanium magnetic permeability properties.
Several other methods are available to create the final form of a metallic structure. The terms used to describe these various processes can be confusing. The magnetic permeability in some materials can be influenced by how the material is manufactured- cold working, work hardening, tempering, bending, casting, and welding to name a few. We will briefly cover the more common processes below.
“Wrought” is a term that simply means “worked”. In ancient times, wrought iron was produced by hammering a heated metal repeatedly.
Metalworking is the process of shaping and forming metals to create various metal components.
Recrystallization is a process in metals by which deformed grains are replaced by a new set of non-deformed grains that nucleate until the original deformed grains have been entirely consumed. This is important concept when taking about cold and hot work shaping processes.
Cold working involves various forming and shaping processes that are applied below the metal’s recrystallization temperature, usually at the ambient temperature. This process increases ultimate tensile strength, yield point, and fatigue strength, but decreases resistance to corrosion.
In the precision machining industry, cold working processes can include thread rolling, thread forming, swaging, crimping, staking, planishing and metal spinning. Steel bars, commonly used as a starting material, are typically machined by cold drawn (cold worked). Unlike hot working, cold working causes the crystal grains and inclusions to distort following the flow of the metal, which may cause work hardening and anisotropic material properties. Work hardening makes the metal harder, stiffer, and stronger, but less plastic, and may cause microscopic cracks in the piece.
Hot working or forming involves various forming and shaping processes that are applied above the metal’s recrystallization temperature.
Examples of these forming techniques include hot rolling, forging, welding, etc. These processes can alter the grains to a different crystal structure- and in some cases also change the magnetic permeability.
Welds and Castings
Castings in austenitic stainless steels have slightly different compositions compared to their wrought counterparts. The cast version of 316L for instance is grade CF-3M. Most “austenitic” cast alloys are very deliberately made so that they have a few percent of ferrite – this helps prevent hot cracking during casting.
A weld can be viewed as a small, long casting, and for the same reason as detailed above, austenitic welds have about 4 – 8% ferrite. In the case of both welds and castings, the small amount of ferrite results in a small amount of magnetic response, but it can be readily detected with a good hand-held magnet. With a suitable “ferrite meter” this magnetic response can in fact be used to measure the amount of ferrite in a weld.
Materials Used in Kimball Physics Vacuum Systems
Stainless Steel
316L Stainless Steel is predominately composed of iron, with 16-18% chromium, 10-14% nickel, 2-3% molybdenum, and small percentages of carbon, manganese, and silicon. The magnetic permeability of stainless steels can vary, depending on its composition and microstructure. Austenitic stainless steels (316L SS), generally have a low magnetic permeability, typically around 1.0 (to learn more about the units to measure magnetic permeability, and the values of various materials, please see the content above). However, the magnetic permeability of stainless steel can increase if it undergoes certain types of mechanical and heat processing as noted above.
Welding and casting can cause regions of ferrite formation within the austenitic material, which can result in a small magnetic response which is measurable and can have an impact on certain applications, such as in magnetic sensors or electromagnetic shielding.
Only a limited number of Kimball Physics Multi-CF hardware parts require welding. The method of welding we use limits the introduction unintended magnetic effects.
Titanium
Titanium and its alloys are known for their high strength-to-weight ratio, with high strength and low density, low thermal expansion, and relatively high melting point. It is a non-magnetic (paramagnetic) metal and has a magnetic permeability of approximately 1.0. This means that they are not affected by magnetic fields and does not exhibit any magnetic properties. It readily forms a tenacious oxide surface on the bulk metal, providing excellent resistance to corrosion and significantly reduces diffusion of hydrogen from the bulk metal.
Table of Relative and Magnetic Permeability of Selected Materials
| | | | |
--- --- |
| Material | Relative Magnetic Permeability (µ/µ0) | Magnetic Permeability (µ) (H/m) | Comments |
| 316L Stainless Steel (Austenitic) | 1.003 – 1.05 | 1.260×10-6 – 8.8 x 10-6 | Paramagnetic |
| Titanium (Pure) | 1.00001 | | Paramagnetic |
| Titanium Alloy (Ti-6Al-4V) | 1.00001 | | Paramagnetic |
| Aluminum (Pure) | 1.000022 | 1.256665×10−6 | Paramagnetic |
| Inconel 600 (6-10Fe, 14.7-17Cr, 0.15C, rem Ni) | 1.01 | | Paramagnetic even with high nickel content of 50-55% |
| Iron (Pure) | 200000 | 2.5×10−1 | Ferromagnetic |
| Carbon Steel | 100 | 1.26×10−4 | Ferromagnetic |
| Stainless Steel, Ferritic (annealed) | 1000 – 1800 | 1.26×10−3 – 2.26×10−3 | Ferromagnetic |
| Stainless Steel, Martensitic (annealed) | 750-950 | 9.42×10−4 – 1.19×10−3 | Ferromagnetic |
| Stainless Steel, Martensitic (hardened) | 40 – 95 | 5.0×10−5 – 1.2×10−4 | Ferromagnetic |
| Nickel | 100-600 | 1.26 x 10-4 – 7.54 x 10-4 | Ferromagnetic |
| Alumina(Ceramic) | 1.00001 | | Paramagnetic |
| Glass | 1.0 | | Paramagnetic |
| Air | 1.00000037 | 1.25663753×10−6 | |
| Vacuum | 1.0 | 1.25663706212 × 10−6= µo | |
| Copper | 0.999994 | 1.256629×10−6 | Diamagnetic |
| Water | 0.999992 | 1.256627×10−6 | Diamagnetic |
| Notes: | | | |
In summary, materials and processes used by Kimball Physics for vacuum hardware, including austenitic stainless steels (316L SS) and titanium and its alloys, have very low magnetic permeability with either minimal or no magnetic field artifacts introduced into our electron beam systems. Please see the FAQ section below for more information about Titanium and hydrogen diffusion in ultra-high (UHV) and extreme-high vacuum applications.
Frequently Asked Questions (FAQ)
FAQ
1) Does titanium have any changes with its magnetic permeability if it is processed mechanically, such as by cold working or bending?
No, titanium is a non-magnetic metal and is not affected by magnetic fields, regardless of its mechanical processing. The magnetic permeability of titanium is very close to that of a vacuum, which is approximately 1.0, meaning that it is not magnetic and does not exhibit any magnetic properties.
Titanium’s non-magnetic properties make it a useful material in applications where magnetic interference is a concern, such as in electronic devices or medical implants that require magnetic resonance imaging (MRI) scans. In fact, titanium is one of the preferred materials for implants in MRI-compatible medical devices because it does not produce any artifacts or distortions in the images.
In summary, titanium is a non-magnetic metal and does not experience any changes to its magnetic permeability, even when it is processed mechanically.
2) Does welding of stainless steel affect its magnetic permeability ?
Yes, welding of stainless steel can affect its magnetic permeability, depending on the welding process and the specific composition of the stainless steel being used.
Welding involves heating the material to a high temperature, which can cause changes in the microstructure of the metal, including changes in its magnetic properties. In particular, the heat from welding can cause the formation of ferrite, which is a magnetic phase of iron that can be present in some types of stainless steel.
Stainless steels are typically classified into different grades based on their microstructure and chemical composition. The two most common
types of stainless steel are austenitic and ferritic. Austenitic stainless steels are generally non-magnetic, while ferritic stainless steels can be magnetic.
If an austenitic stainless steel is welded, it can become more magnetic if the welding process causes the formation of ferrite in the weld zone. This is known as weld decay, and it can result in a material that is more susceptible to corrosion and other types of degradation.
However, it is possible to minimize the magnetic effects of welding on stainless steel by using the appropriate welding technique and selecting a grade of stainless steel that is less susceptible to weld decay. For example, some grades of stainless steel are designed to be more resistant to the formation of ferrite during welding, which can help to minimize any changes in magnetic permeability.
In summary, welding can affect the magnetic permeability of stainless steel, particularly if the welding process causes the formation of ferrite in the weld zone. However, the degree of change in magnetic properties will depend on the specific type of stainless steel being used and the welding process employed.
3) Are there effects on material properties of 316L SS (specifically magnetic permeability) with CNC Machining vs. Welding?
When manufacturing vacuum chambers, there are two common methods: 1) welded components, joining together multiple individual components to create the vacuum chamber, and 2) CNC machining which involves manufacturing the vacuum chamber from a single block or monolith of material with no welding operations.
From a magnetic permeability and structural precision standpoint, CNC machining is superior to welding. Welding and cold working can change the crystal structure of the metal, which can result in increased magnetic permeability. This increase in magnetic permeability can introduce artifacts into the beam trajectory, profile, and focus.
CNC machining, on the other hand, ensures that the vacuum chamber has a uniform crystal structure, resulting in minimal magnetic permeability. CNC machining also ensures that the vacuum chamber has precise structural dimensions, orientations, and tolerances, resulting in a vacuum chamber that is structurally sound and meets the required specifications.
In conclusion, 316L stainless steel is an excellent choice of material for vacuum chambers used in electron beam systems due to its high corrosion resistance, high-temperature resistance, and low magnetic permeability. When manufacturing vacuum chambers, CNC machining is a superior method compared to welding, as it ensures that the vacuum chamber has a uniform crystal structure, resulting in minimal magnetic permeability and precise structural dimensions and tolerances. By using 316L stainless steel and CNC machining, we can manufacture vacuum chambers that meet the required specifications and ensure that electron beam systems operate with maximum efficiency.
4) What are advantages of Titanium for ultra (UHV) and extreme (XHV) high vacuum chambers?
Titanium is a material that is often used for ultra-high (UHV) and extreme-high (XHV) vacuum chambers. It is a popular choice because of its low outgassing rate, low magnetic permeability, excellent corrosion resistance, and high strength-to-weight ratio. Titanium also has a low thermal expansion coefficient, making it a good choice for applications that require stability over a wide temperature range. In addition, titanium has a lower magnetic permeability than stainless steel, which makes it ideal for electron beam systems that may be sensitive to extraneous magnetic fields.
While we are discussing titanium and XHV systems, one issue that can arise with vacuum chambers at extreme vacuum is the diffusion of hydrogen into the vacuum space from out of the bulk metal. Hydrogen can diffuse through the metal of the chamber and contaminate the vacuum space. This can be particularly problematic in ultra and extreme high vacuum systems, where even small amounts of contamination can cause issues.
Titanium has better hydrogen diffusion properties than 316L stainless steel. This is because titanium forms a protective oxide layer on its surface that prevents hydrogen from diffusing through the metal surface. In contrast, stainless steel is more prone to hydrogen diffusion, particularly if it has been cold worked or welded.
The advantages of titanium and TiN coatings are nicely described in the following experiment (see reference: Vacuum characteristics of Titanium). In the measurement of the outgassing rate of small samples, it was found that the outgassing rate of titanium decreased 1.5 times faster than that of stainless steel. Furthermore, a TiN coating was found to reduce the total amount outgassed from 1.78×10−2 to 1.17×10−2 Pa m3in the measurement of titanium and from 1.02×10−1 to 9.22×10−3Pa m3 in the measurement of stainless steel. In a mass analysis of the gases from the samples, the ion current of hydrogen from chemically polished titanium, chemically polished and TiN‐coated titanium, and TiN‐coated stainless steel decreased after attaining a maximum, while that from buffed and electropolished stainless steel increased continuously during the baking at 200 °C.
Considering these results, a vacuum chamber made of titanium coated with TiN and a vacuum chamber made of electropolished stainless steel were prepared and pumped down into XHV region. The pressure of the titanium chamber reached 6×10−11 Pa after a 312 h baking at 250 °C, while the stainless steel chamber required a 434 h baking at 250 °C to attain the ultimate pressure of 7×10−11 Pa. The pressure of the titanium chamber decreased more rapidly than that of the stainless steel chamber during the baking at 250 °C.
In conclusion, titanium is an excellent choice for ultra and extreme high vacuum chambers due to its low outgassing rate, low magnetic permeability, excellent corrosion resistance, and high strength-to-weight ratio. It also has better hydrogen diffusion properties than 316L stainless steel. When designing vacuum chambers, it is important to consider the properties of the materials used and how they will impact the performance of the electron beam system. By selecting the appropriate material and manufacturing method, we can ensure that vacuum chambers perform optimally and maintain the required vacuum levels.
5) Do thread forming fabrication techniques in 316L SS cause working hardening of the metal and possible changes in its crystal structure from austenitic to a crystal structure that is ferromagnetic?
Thread forming fabrication techniques can cause working hardening of 316L stainless steel, but it is unlikely to change its crystal structure from austenitic to a crystal structure that is ferromagnetic.
316L stainless steel is an austenitic stainless steel, which means it has a face-centered cubic (FCC) crystal structure at room temperature. This crystal structure is non-magnetic, and the material is generally considered to be non-magnetic in its annealed condition.
However, when stainless steel is subjected to mechanical deformation, such as during thread forming, the metal can become work hardened. Work hardening occurs when the metal is deformed by cold working, which causes the dislocation density to increase and the material to become harder and stronger. This work hardening effect can increase the tensile strength of the material, but it can also reduce the ductility and toughness of the metal.
While it is possible for working hardening to cause a minimal increase in magnetic permeability, it is unlikely to change the crystal structure of the 316L stainless steel from austenitic to ferromagnetic. For this to occur, the metal would need to be heated above its critical temperature (around 1450°F) and then rapidly cooled, which is known as quenching. This process would cause the crystal structure to transform from austenitic to ferritic, which is ferromagnetic. However, thread forming processes typically do not involve heating the metal to such high temperatures, and so the crystal structure of the 316L stainless steel is unlikely to change from austenitic to ferromagnetic.
6)What effects on permeability would welding two pieces of 316L SS cause. Is its effect significantly larger the effects of work hardening?
Welding two pieces of 316L SS can cause an increase in magnetic permeability compared to the base metal due to the formation of ferrite in the weld zone. This effect can be significant and is generally larger than the effects of work hardening.
316L SS is an austenitic stainless steel, which has a low magnetic permeability in its annealed condition. However, during welding, the material undergoes thermal cycling, which can cause the formation of ferrite in the weld zone. Ferrite is a different crystal structure than the austenitic structure of the base metal, and it has a higher magnetic permeability than austenitic stainless steel. Therefore, the presence of ferrite in the weld zone can cause an increase in magnetic permeability.
The amount of ferrite in the weld zone depends on several factors, such as the composition of the base metal and the welding process parameters, including welding temperature and cooling rate. The ferrite content can be controlled by adjusting the welding parameters and by using filler metals with a specific ferrite content.
The increase in magnetic permeability due to welding can be significantly larger than the effects of work hardening, especially in the weld zone. Work hardening can cause a small increase in magnetic permeability, but it is generally much smaller than the increase due to the presence of ferrite. However, the effects of work hardening can be significant in areas adjacent to the weld zone where the material has been deformed during welding.
In summary, welding two pieces of 316L SS can cause an increase in magnetic permeability due to the formation of ferrite in the weld zone. This effect can be significant and is generally larger than the effects of work hardening. The amount of ferrite in the weld zone depends on several factors, and it can be controlled by adjusting the welding parameters and by using filler metals with a specific ferrite content.
| | |
--- |
| 1) | Permeability (electromagnetism)- Wikipedia |
| | |
| 2) | Atlas Tech Note No. 11 Magnetic Response of Stainless Steels- Excellent discussion of magnetic response of various Stainless Steels |
| | |
| 3) | Everything About CNC Machining Stainless Steel |
| | |
| 4) | Stainless Steels for Machining |
| | |
| 5) | Ferromagnetism- Wikipedia. |
| | |
| 6) | Overview of E-Beam Optics with illustrations |
| | |
| 7) | HyperPhysics- Ferromagnetism |
| | |
| 8) | Vacuum Characteristics of Titanium, Journal of Vacuum Science, Volume 13, Issue 3, May 1995 |
| | |
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人工智能第六章——约束满足问题（CSP）
最新推荐文章于 2025-08-12 17:17:40 发布
宇内虹游于 2018-07-05 19:52:58 发布
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本文介绍了约束满足问题(CSP)的基本概念，包括其定义、优势及形式化表示方法。探讨了约束传播技术，如节点相容、弧相容及路径相容等，并讨论了如何通过变量和值的选择优化搜索效率。
摘要
本文会讲清楚：
1）什么是CSP（约束满足问题）
2）约束传播与局部相容性
3）CSP形式化为一个搜索问题（回溯法）
4）如何提高搜索效率（变量/值的顺序，提前检查失败等）
一、CSP
使用要素化来描述状态：一组变量，每个变量有自己的值。当每个变量都有自己的赋值同时满足所有关于变量的约束时，问题就得到了解决。这类问题就叫做约束满足问题（CSP），全称Constraint Satisfaction Problem。
CSP利用了状态结构的优势，使用的是通用策略而不是问题专业启发式来求解复杂问题。
主要思想：通过识别违反约束的变量/值的组合迅速消除大规模的搜索空间。
1.2 定义CSP
CSP包含三个成分X,D,C：
X：变量集合 （variables）
D：值域集合，每个变量有自己的值域 （domain）
C：描述变量取值的约束集合 （constraint）
C的形式一般是：
1.3 CSP的优点
图1.3.1 CSP的优点
对比局部搜索（全分配，每次必须考虑整个状态），但是CSP，是部分分配，每次只需要考虑部分赋值。（一旦不是解，立即丢弃）
1.4 CSP的形式化
用约束语言来表示CSP的约束条件。
一元约束：只限制单个变量的取值
二元约束：与两个变量有关。
变量个数任意的约束称为全局约束。
图1.4.1 约束优化问题（COP）
二、约束传播
核心思想：局部相容性。
2.1 节点相容
单个变量（对应一个节点）值域中的所有取值满足它的一元约束，就是节点相容的。
2.2 弧相容
如果CSP中某变量值域中所有取值满足该变量所有二元约束，则此变量弧相容。
比如：对变量X1，X2，如果D1中每个数值在D2中都存在一些数值满足弧（X1，X2）的二元约束，那么X1相对X2是弧相容的。
如果每个变量相对其他变量都是弧相容的，则称该网络是弧相容的。
图2.2.1 弧相容算法AC-3
2.3 路径相容
弧相容可能缩小变量的值域，有时甚至还能找到解（每各变量值域大小都为1时），或者有时发现CSP无解（一些变量的值域大小=0）。
但是弧相容也会失败（比如澳大利亚地图着色问题，如果只有两种颜色，弧相容意义不大），所以才要用更强的相容概念。
路径相容：观察变量得到隐式约束，并以此来加强二元约束。
图2.3.1 路径相容的定义
2.4 k-相容
如果对于任何k-1个变量的相容赋值，第k个变量总能被赋予一个和前k-1个变量相容的值，那么这个CSP就是k相容的。
三、全局约束
1）Alldiff约束：表示所有相关变量必须取 不同的 值。
2）atmost约束（另一个重要的高阶约束）：也叫资源约束。
如果对于每个变量X和它的取值上下界，每个变量Y都存在某个取值满足X和Y之间的约束，则称该CSP是边界相容（此边界传播广泛应用于实际CSP）。
四、CSP的回溯搜索
很多CSP只用 推理 是无法求解的，还需要通过搜索来求解。
部分赋值的回溯搜索 算法：
可以用标准的深度优先搜索，状态可能是部分赋值，行动是将 var=value加入到赋值中。
回溯搜索用于深度优先搜索中，每次为一个变量选一个赋值，没有合法的值的时候就回溯。
图4.1 CSP的简单回溯算法
五、有效解决CSP
5.1 变量和取值顺序
变量：
1）选择“合法”取值最少的变量——称为最少剩余值（MRV）启发式。（做一个强有力的引导，方便提早遇到失败，从而剪枝）
2）度启发式：通过选择与其他未赋值变量约束最多的变量来试图降低未来的分支因子。（用来打破僵局，如选择第一个着色区域）
值：
最少约束至：优先选择的赋值是给邻居变量留下更多的选择（为了找到一个解，所以没必要排序，二十要最少约束）
5.2 搜索与推理交错进行
前向检验：只要变量X被赋值，就对它进行弧相容检查，对每个通过约束与X相关的未赋值变量Y，从Y值域中删去与X不相容的值。
5.3 智能回溯：向后看
主要概念：前向检验；冲突集；回跳
六、CSP的局部搜索
局部搜索算法对求解许多CSP都是很 有效的。它们使用完整状态的形式化：初始状态是给每个变量都赋一个值，搜索过程是一次改变一个变量的取值。
七、文末诗词
满恨游丝兼落絮。红杏开时，一霎（sha，四声）清明雨。浓睡觉来莺乱语。惊残好梦无寻处。
——晏殊《蝶恋花·六曲阑干偎碧树》
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人智导（四）：约束满足问题 约束满足问题 ( Constraint Satisfaction Problem ) 一种结构化的、简单而标准模式的问题表示 状态被描述为一系列变量XiX_iXi（对应的值域为DiD_iDi） 目标测试：一个 约束 集，描述这些变量或子集允许的取值 CSP 问题的定义 约束满足问题 的定义： 变量集合：{X1,X2,…,Xn}{X_1,X_2,\dots ,X_n}{X1,X2,…,Xn}；约束 集合：{C1,C2,…,Cn}{C_1,C_2,\dots ,C_n}{C1
人工智能 讲义：约束满足问题 CSP 解析
"该资源是一份关于搜索 约束满足问题（CSP）的PPT，主要讲解了 CSP 在 人工智能 中的应用。内容包括 CSP 的定义、建模、求解算法以及实际应用，如地图着色问题和3-SAT问题。" 在 人工智能 领域，搜索 约束满足问题...
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摘要
一、CSP
1.2 定义CSP
1.3 CSP的优点
1.4 CSP的形式化
二、约束传播
2.1 节点相容
2.2 弧相容
2.3 路径相容
2.4 k-相容
三、全局约束
四、CSP的回溯搜索
五、有效解决CSP
5.1 变量和取值顺序
5.2 搜索与推理交错进行
5.3 智能回溯：向后看
六、CSP的局部搜索
七、文末诗词
展开全部
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目录
摘要
一、CSP
1.2 定义CSP
1.3 CSP的优点
1.4 CSP的形式化
二、约束传播
2.1 节点相容
2.2 弧相容
2.3 路径相容
2.4 k-相容
三、全局约束
四、CSP的回溯搜索
五、有效解决CSP
5.1 变量和取值顺序
5.2 搜索与推理交错进行
5.3 智能回溯：向后看
六、CSP的局部搜索
七、文末诗词
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11177 | https://www.ahajournals.org/doi/10.1161/01.cir.0000146906.42375.d3 | Cardiovascular Effects of the 3 Phosphodiesterase-5 Inhibitors Approved for the Treatment of Erectile Dysfunction | Circulation
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Reference #1
Review Article
Originally Published 9 November 2004
Free Access
Cardiovascular Effects of the 3 Phosphodiesterase-5 Inhibitors Approved for the Treatment of Erectile Dysfunction
Robert A.Kloner, MD, PhDAuthor Info & Affiliations
Circulation
Volume 110, Number 19
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Contents
Effect of PDE5 Inhibitors on Blood Pressure and Heart Rate
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Within the last year, 2 new phosphodiesterase-5 (PDE5) inhibitors have been approved by the US Food and Drug Administration (FDA) for the treatment of erectile dysfunction (ED). Currently, sildenafil (Viagra), vardenafil (Levitra), and tadalafil (Cialis) are on the market. These agents have been shown to be effective in a broad population of men with ED, including patients with vascular disease, coronary artery disease, hypertension, and diabetes.1–5 Because the enzyme that they inhibit, PDE-5, is found in smooth muscle cells of the systemic arteries and veins throughout the body, these agents have mild vasodilator effects and thus, have the potential to impact the cardiovascular system.6 This fact is especially important for the patient with ED, because risk factors for ED include many of the same risk factors that are associated with coronary artery disease: lipid abnormalities, hypertension, smoking, diabetes, and lack of physical exercise.7,8 Because erection is a vascular event, endothelial dysfunction may inhibit it.9 Endothelial dysfunction, an early component of atherosclerosis, is rarely confined to the arteries supplying blood to the penis but more likely occurs throughout the vascular bed. Kaiser et al10 studied 30 men with ED and observed that brachial artery flow–mediated vasodilation and nitroglycerin-mediated vasodilation were reduced in these patients compared with men without ED. Thus, ED may be an early marker of vascular disease.10 Patients with frank coronary artery disease, known to be associated with endothelial dysfunction, and frank atherosclerosis often have ED, as we recently observed in 1 study, in which three fourths of the men with chronic stable angina also reported some degree of ED.11 Hence, the vasodilator effect of these PDE5 agents should be taken into consideration for the cardiac patient, both as a possible concern in some cases or a possible beneficial effect in others. The purpose of the present review is to describe the cardiovascular effects of the 3 available PDE5 inhibitors, the issue of nitrate interaction, differences and similarities in labeling regarding concomitant use of nitrates and α-blockers, their effect on the QT interval, their safety in regard to cardiac events, and the concept that these agents may eventually play a role in therapies for various cardiac conditions, including pulmonary hypertension and congestive heart failure.
Effect of PDE5 Inhibitors on Blood Pressure and Heart Rate
Sildenafil
Sildenafil was approved as an effective agent for treating ED in 1998, and its cardiovascular effects have been studied in great depth. It has a half-life of ≈4 hours (Table). In 1 study of 8 volunteers,12 80 mg of intravenous sildenafil reduced systolic blood pressure from 131±12 mm Hg with placebo to 122±13 mm Hg (P<0.001) with sildenafil. Diastolic blood pressure fell from 71±12 to 64±11, respectively, but heart rate did not change significantly (64±10 to 66±14 beats/min; mean±SD). Jackson et al12 gave 8 healthy male subjects 100, 150, and 200 mg of oral sildenafil. The mean maximum reduction in supine systolic/diastolic blood pressure was 10/7 mm Hg at 3 hours after dosing, with no difference in the 200-mg versus the 100-mg dose. In most subjects, blood pressure returned to baseline within 6 hours. The authors did not observe a significant orthostatic drop in blood pressure or increase in heart rate in the standing position. Sildenafil, as expected, reduced vascular resistance.
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Aspects of 3 PDE5 Inhibitors
| | Sildenafil | Vardenafil | Tadalafil |
---
| T max indicates time to maximum plasma concentration; C max, maximum plasma concentration. |
| Starting dose | 50 mg taken ≈1 hour before sexual activity. May be taken 4 hours to 0.5 hour before sexual activity. Dose may be increased to 100 mg or decreased to 25 mg | 10 mg taken ≈60 minutes before sexual activity. May be increased to 20 mg or decreased to 5 mg | 10 mg taken before anticipated sexual activity. May be increased to 20 mg or decreased to 5 mg |
| Effect on other PDEs | 4000 times more selective for PDE5 than PDE3; >80-fold more selective than for PDE1; >700-fold more selective for PDE5 than for PDE2–4 and 7–11. Only ≈10-fold as potent for PDE5 compared with PDE6 | >1000-fold more selective for PDE5 than PDE2–4 and 7–10; >300-fold relative to PDE11; >130-fold relative to PDE1; >15-fold relative to PDE6 | >10 000-fold more potent for PDE5 than for PDE1–4 and 7; >9000-fold more potent for PDE5 than PDE8–10; >700-fold more potent for PDE5 than PDE6; 14-fold more potent for PDE5 than PDE11 A1 |
| T max | Median, 60 minutes; high-fat meal reduces C max by 29% and delays T max | Median, 60 minutes; high-fat meal reduces C max by 18% to 50% | Median, 2 hours; not influenced by food |
| Mean terminal half-life | 4 hours | 4–5 hours. | 17.5 hours |
| Dosage adjustment | Lower starting doses in elderly, hepatic impairment, severe renal impairment, potent cytochrome P450 3A4 inhibitors; consider lower doses (25 mg) with concomitant α-blockers | Lower doses in patients with moderate hepatic impairment, strong CYP3A4 inhibitors, erythromycin, elderly | Lower doses in patients with severe renal insufficiency, mild to moderate hepatic impairment, potent CYP3A4 inhibitors |
| Contraindication | Organic nitrates (regular or intermittent use) | Nitrates and NO donors (regular or intermittent use), α-blockers | Organic nitrates (regular or intermittent use), α-blockers other than 0.4 mg tamsulosin |
| Adverse events reported by ≥2% of patients | Headache, flushing, dyspepsia, nasal congestion, urinary tract infection, abnormal vision, diarrhea, dizziness | Headache, flushing, rhinitis, dyspepsia, accidental injury, sinusitis, flu syndrome, dizziness, increased creatine kinase, nausea | Headache, dyspepsia, back pain, myalgia, nasal congestion, flushing, pain in limb |
| Other cardiac precautions and warnings | Doses of 50 or 100 mg sildenafil should not be taken within 4 hours of α-blocker administration | Patients with congenital QT prolongation and those taking class IA or Class III antiarrhythmic agents should avoid vardenafil | Warning regarding left ventricular outflow obstruction, in that patients may be sensitive to vasodilators in this situation |
| | Caution in patients with left ventricular outflow obstruction or severely impaired autonomic control of blood pressure | Caution in patients with left ventricular outflow obstruction and those patients for whom physician would be concerned about giving any vasodilator | Avoid excess alcohol Precaution in patients in whom vasodilators may be a concern, including patients with severely impaired autonomic control of blood pressure |
Expand Table
In patients with stable ischemic heart disease 40 mg of intravenous sildenafil reduced systolic arterial pressure from 150±12 to 141±16 mm Hg (mean±SD) and diastolic blood pressure from 74±8 to 66±10 mm Hg but had no effect on heart rate. Cardiac output was 5.6±0.9 L/min at baseline and 5.2±1.1 L/min after intravenous sildenafil. Pulmonary artery pressure fell from 16.7±4.0 to 12.1±3.9 mm Hg. Subsequent studies verified that sildenafil results in small reductions in systolic and diastolic blood pressure, with little or no effect on heart rate.13–15 In addition, sildenafil was shown in numerous studies of pulmonary hypertension to reduce pulmonary vascular resistance and pulmonary artery pressure (PDE5 is rich within the pulmonary vasculature).16–18
Vardenafil
Vardenafil was the second PDE5 inhibitor to come to market in the United States, and it was approved by the FDA in 2003. It has many similarities to sildenafil; in fact, its molecular structure is very similar (Figure), and they share a half-life of ≈4 hours. In healthy volunteers, vardenafil caused a transient reduction in supine blood pressure with a mean maximum decrease of 7 mm Hg systolic and 8 mm Hg diastolic. Maximum reduction occurred between 1 and 4 hours after dosing.19 A 20-mg dose of vardenafil resulted in a mean maximum increase in heart rate of 4 beats/min.
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Molecular structures of sildenafil, vardenafil, and tadalafil. Note similarity in structure between sildenafil and vardenafil. Reprinted with permission from Rosen RC, Kostis JB. Overview of phosphodiesterase 5 inhibition in erectile dysfunction. Am J Cardiol. 2003;92:(suppl):9M–18M.
Tadalafil
Tadalafil was the third PDE5 inhibitor to come to market in the United States and was approved by the FDA at the end of 2003. Its chemical structure differs substantially from that of sildenafil and vardenafil, and it has a unique, long half-life of 17.5 hours. In healthy subjects, a single oral dose of 10 or 20 mg of tadalafil did not drop systolic blood pressure significantly compared with placebo. The decrease in systolic blood pressure with 20 mg tadalafil versus placebo was −0.2 mm Hg (95% confidence interval [CI], −7.1 to 6.6) in the standing position.20,21 The mean decrease in diastolic blood pressure compared with placebo was −4.6 mm Hg (95% CI, −7.9 to −1.2) in the standing position. Diastolic blood pressure was generally lower for 12 hours after administration of the drug, but the decrease in pressure remained small. Changes were even less marked in the supine position. Standing and supine heart rates were not significantly different from those with placebo.20
In patients with coronary artery disease, who had higher systolic blood pressures at baseline than did healthy volunteers, 10 mg tadalafil reduced standing systolic blood pressure from 134 to 127 mm Hg (−7 mm Hg) and diastolic blood pressure from 78 to 74 mm Hg (−4 mm Hg) at time to maximum plasma concentration (T max; 2 hours) after dosing.21
Hence, all 3 of the PDE5 inhibitors are mild vasodilators and may cause small drops in arterial pressure. However, the degree of drop in blood pressure is usually small. Although this is rarely a concern in healthy individuals, in cardiovascular patients with low blood pressure or hypotension at baseline, this could be a concern, as described in recent guidelines.22 This vasodilating property would also be a concern for patients with left ventricular outflow obstruction or aortic stenosis.22,23 This property also becomes an issue regarding the concomitant use of nitrates and α-blockers, as described next.
Nitrate Interaction
PDE5 inhibitors work to improve ED by preventing the breakdown of cyclic GMP, the substance that promotes relaxation of smooth muscle cells in the arteries, arterioles, and sinusoids of the corpus cavernosum of the penis.24 Nitric oxide (NO) donors, such as nitroglycerin (short or long acting), isosorbide dinitrate, isosorbide mononitrate, and others stimulate guanylate cyclase, which increases the production of cyclic GMP. When NO donors are given at the same time as PDE5 inhibitors, there is both an increase in the production of cyclic GMP (due to the NO donor) and an inhibition of its breakdown (due to the PDE5 inhibitor) that can lead to a build-up of cyclic GMP with pronounced vasodilation and, in some patients, frank hypotension. Organic nitrates are contraindicated with all 3 PDE5 inhibitors on the market.19–21,25,26 Details of interaction studies are presented in the online supplement to this article.
Duration of PDE5 Inhibitor–Nitrate Interaction
If a patient with coronary artery disease takes a PDE5 inhibitor and then develops angina pectoris, when is it safe to administer nitroglycerin?
Sildenafil
A consensus statement from the American College of Cardiology/American Heart Association suggested that for sildenafil, 6 half-lives, or 24 hours (4 hours × 6), pass between sildenafil intake and the administration of a nitrate.22 This time frame was chosen to allow full washout of drug from the tissues. In a previous analysis,20 at 24 hours after sildenafil, there was no interaction with sublingual nitroglycerin. In a recent preliminary study of healthy individuals, the sildenafil-nitroglycerin interaction appeared to be gone as early as 4 hours.27
Vardenafil
Although a detailed time-course study of the interaction between vardenafil and nitrates is not available, 1 unpublished study19 did suggest a lack of interaction at 24 hours, consistent with a 4-hour half-life.
Tadalafil
A nitrate-tadalafil interaction study was recently reported.28 In a population of patients who were healthy or had controlled hypertension or diabetes, 20 mg tadalafil or placebo was given daily for 7 days. On the seventh day, sublingual nitroglycerin was administered at 2, 4, 8, 24, 48, 72, or 96 hours. There was evidence that tadalafil augmented the nitrate-induced decrease in blood pressure within the first 24 hours of its administration. At and beyond 48 hours, the interaction was not detectable, as assessed by mean maximal falls in blood pressure or outlier criteria. Thus, the manufacturer suggests that if a patient has taken tadalafil and develops chest pain and it is deemed medically necessary that nitrates be given, at least 48 hours should elapse after the last dose of tadalafil before nitrates are given. Even then, nitrates should be given only under close medical supervision and appropriate hemodynamic monitoring.
What should a patient do if he/she develops chest pain after having taken a PDE5 inhibitor? The patient should call for emergency assistance for hospital transport29 and apprise the emergency medical technicians and emergency physicians regarding use of the PDE5 inhibitor. Organic nitrates are initially contraindicated. The American College of Cardiology/American Heart Association Consensus document22 reminds physicians that other antianginal agents can be considered in such situations. β-Blockers, calcium channel blockers, morphine, and other anti-ischemic agents, such as oxygen, aspirin, and statins, may be safely given in the setting of a PDE5 inhibitor. Furthermore, should a patient develop a myocardial infarction while taking a PDE5 inhibitor, usual care—thrombolytics, antiplatelet agents, and percutaneous coronary intervention—is indicated. Only organic nitrates should be avoided. If a patient has taken a PDE5 inhibitor, inadvertently receives a nitrate, and becomes hypotensive, the American College of Cardiology/American Heart Association recommends common-sense measures: placing the patient in the Trendelenburg position, administering intravenous fluids and α-agonists, and finally using intra-aortic balloon counterpulsation. At the present time, there are no specific antidotes to PDE5 inhibition.
Interaction With Common Antihypertensive Agents Other Than α-Blockers
Several studies investigated the interactions of sildenafil, vardenafil, and tadalafil on blood pressure in patients already taking usual antihypertensive medicines, such as calcium channel blockers, diuretics, β-blockers, angiotensin-converting-enzyme inhibitors, and angiotensin receptor blockers. In general, these studies suggested that when a PDE5 inhibitor was administered to patients already taking antihypertensive medications, small additive drops in blood pressure occurred that were roughly equivalent to or less than the drops in pressure that occurred when the PDE5 inhibitors were taken without concomitant antihypertensive agents.
Sildenafil
Webb et al25 studied men who were already receiving amlodipine for the treatment of hypertension for at least 1 month. These authors showed that 100 mg sildenafil given 2 hours after a morning dose of 5 or 10 mg amlodipine resulted in a mean maximal change from baseline blood pressure of −8 mm Hg supine systolic and −7 mm Hg supine diastolic compared with men who received placebo plus amlodipine. The differences in standing systolic and diastolic blood pressures were −10 and −8 mm Hg, respectively.
Even smaller additional reductions in blood pressure were reported when the PDE5 inhibitor sildenafil was administered to patients already taking β-blockers, diuretics, or angiotensin-converting-enzyme inhibitors.30 There was no increase in adverse events in patients treated with sildenafil who received antihypertensive medicines, even multiple antihypertensive medicines.30–32 No increase in syncope or postural hypotension was reported when sildenafil was given to patients already taking these common antihypertensive drugs. Furthermore, the most common side effects of sildenafil (headache, flushing, dyspepsia, visual disturbance, rhinitis) were not worsened in patients on concomitant antihypertensive medicines, even if the patients were taking 2 or more antihypertensive medicines.31 The efficacy of sildenafil to improve ED remained high (≈72%) in hypertensive patients taking antihypertensive agents.31,32
Vardenafil
Vardenafil was associated with a small or no fall in blood pressure when administered to patients taking their usual antihypertensive medicines.33,34 Vardenafil 20 mg produced an additional fall in mean systolic/diastolic blood pressure of 6/5 mm Hg versus placebo when it was concomitantly administered with slow-release nifedipine (30 or 60 mg) in patients whose blood pressure had been controlled with this calcium channel blocker.35
Tadalafil
Kloner et al36 reported a series of studies in which 10 or 20 mg tadalafil was administered to patients with hypertension taking antihypertensive medications or to healthy individuals who received an antihypertensive medicine. In healthy individuals receiving 5 mg amlodipine daily for at least 14 days, 20 mg tadalafil had no significant additional effect with a mean change in systolic/diastolic blood pressure of −1/1 mm Hg compared with placebo. This ambulatory blood pressure monitoring study showed no interaction between 5 mg amlodipine and 20 mg tadalafil. Tadalafil 10 mg given to hypertensive patients who had been receiving the thiazide diuretic bendrofluazide was associated with an additional reduction in supine systolic/diastolic blood pressure of 6/4 mm Hg and in standing systolic/diastolic pressure of the same degree. Tadalafil 10 mg administered to patients taking metoprolol (25 to 200 mg daily) resulted in an additional mean reduction in supine systolic/diastolic blood pressure of 5/3 mm Hg and in standing systolic/diastolic blood pressure of 7/4 mm Hg. Tadalafil 10 mg administered to patients taking enalapril (10 to 20 mg per day) was associated with a mean reduction in supine systolic/diastolic blood pressure of 4/1 mm Hg and in standing systolic/diastolic blood pressure of 3/1 mm Hg. In an ambulatory BP monitoring study, 20 mg tadalafil given to patients already receiving various angiotensin receptor blockers was associated with an additional mean reduction in ambulatory systolic/diastolic blood pressure of 8/4 mm Hg. The incidence of adverse events in tadalafil-treated patients in general was similar to that in patients receiving or not receiving concomitant antihypertensive medicines, similar to observations with the other PDE5 inhibitors.
In general, sildenafil, vardenafil, and tadalafil appear to be well tolerated and safe in patients receiving most concomitant antihypertensive agents. There appears to be no or only small additive drops in blood pressure that occur when the PDE5 inhibitor is administered in addition to the antihypertensive medicines. The 1 exception to this appears to be α-blockers, as described next.
Interaction of PDE5 Inhibitors With α-Blockers
Sildenafil
There were recent changes to the labeling of sildenafil regarding concomitant use in patients receiving α-blockers.37 One study assessed concomitant administration of the nonselective α 1-blocker doxazosin with sildenafil in patients with benign prostatic hypertrophy who had been taking α-blockers for at least 8 weeks. Doxazosin is given for both hypertension and benign prostatic hypertrophy. Two of 20 patients receiving 4 mg doxazosin and 50 mg sildenafil developed symptomatic hypotension. Interestingly, no (0/20) patients receiving 4 mg doxazosin and 100 mg sildenafil had symptomatic hypotension. Also, the 25-mg dose of sildenafil, when coadministered with 4 mg doxazosin, was not associated with hypotension. The label precaution (not contraindication) now advises that 50 or 100 mg sildenafil should not be taken within a 4-hour window of α-blocker administration. A 25-mg dose of sildenafil may be taken at any time in relation to an α-blocker.
Vardenafil
Vardenafil is contraindicated in patients taking α-blockers.19 When 10 or 20 mg vardenafil was administered to healthy volunteers either simultaneously or 6 hours after a 10-mg dose of the α 1-blocker terazosin (used for both hypertension and benign prostatic hypertrophy), significant hypotension was observed. When 10 mg vardenafil was given at the same time as 10 mg terazosin, 6 of 8 volunteers developed a standing systolic blood pressure <85 mm Hg. In a study with 20 mg vardenafil, simultaneous administration of 10 mg terazosin resulted in 2 of 9 subjects dropping their standing systolic blood pressure to <85 mm Hg. When dosing was separated by 6 hours, 7 of 28 subjects dropped their standing systolic blood pressure to below 85 mm Hg. Results with tamsulosin (a selective α 1a-blocker, which is more selective for prostatic tissue and mainly used to treat benign prostatic hypertrophy) also showed an interaction with vardenafil but to a lesser extent. Two of 16 subjects who received 10 mg vardenafil plus 0.4 mg tamsulosin developed a standing systolic blood pressure below 85 mm Hg. Thus, when vardenafil was released in the United States, it carried an absolute contraindication in patients taking α-blockers.19 However, a recent analysis suggests that this interaction is much less marked when vardenafil is administered to patients with benign prostatic hypertrophy who have been on long-term α-blocker therapy.38
Tadalafil
Tadalafil is contraindicated in patients taking α-blockers except for 0.4 mg tamsulosin (α 1a-blocker). In 1 study,39 20 mg tadalafil augmented the hypotensive effect of 8 mg doxazosin with a mean maximal decrease in standing systolic blood pressure that was greater than placebo (mean difference of 9.8 mm Hg). The number of subjects with a standing systolic blood pressure of <85 mm Hg was greater after doxazosin plus tadalafil (28%) versus doxazosin plus placebo (6%). In a separate analysis,39 0.4 mg tamsulosin was given with 10 or 20 mg tadalafil. In subjects taking this α-blocker, tadalafil produced mean maximal reductions in standing systolic blood pressure that were similar to those seen with placebo (mean difference of 1.7 mm Hg with 10 mg tadalafil and of 2.3 mm Hg with 20 mg tadalafil). Furthermore, none of the subjects receiving tamsulosin plus tadalafil dropped their standing systolic blood pressure to <85 mm Hg.
QT Interval
Vardenafil was associated with small increases in QTc in 1 study19,40 and carries the label warning that it should be avoided in patients with congenital QT prolongation and patients taking class IA (eg, quinidine, procainamide) or class III (eg, amiodarone, sotalol) antiarrhythmic medications. Tadalafil20,41 and sildenafil37 do not carry this statement. To date, there have been no known cases of torsade de pointes in patients receiving PDE5 inhibitors. In our experimental animal studies of acute coronary artery occlusion, administration of sildenafil was not associated with an increase in ventricular arrhythmia.42 Additional details on QT interval may be found in the online supplement to this article.
Do the PDE5 Inhibitors Exacerbate Ischemia?
Studies with sildenafil, vardenafil, and tadalafil showed that when these agents were given to patients with coronary artery disease during exercise stress testing, there was no exacerbation of ischemia compared with placebo.43–46 A very important aspect of these studies was that the level of exercise was similar to or exceeded that achieved during sexual intercourse. In a few of these studies, sildenafil43 and vardenafil45 actually appeared to have a beneficial effect on exercise-induced ischemia. A study performed in the cardiac catheterization laboratory by Hermann et al14 observed that when oral sildenafil was administered to patients with severe coronary artery disease, it had no adverse effects on coronary artery diameter or coronary artery flow velocity and actually improved coronary artery vasodilator reserve induced by adenosine. Halcox et al47 demonstrated a slight vasodilating effect of sildenafil on epicardial coronary arteries. The effects of vardenafil and tadalafil on coronary artery hemodynamics are unknown at this time.
Do PDE5 Inhibitors Cause Myocardial Infarction or Death?
After the initial release of sildenafil, there were anecdotal reports of myocardial infarction and death. However, sexual activity itself may be associated with a small but definite increase in risk of myocardial infarction.48 Patients with ED are more likely to have risk factors for coronary artery disease (such as age ≥45, smoking, lipid abnormalities, diabetes, hypertension, and lack of physical activity) or have frank heart disease.6–8,49 Analysis of double-blind, randomized, controlled studies and open-label studies did not show an increase in the rates of myocardial infarction or death in patients receiving sildenafil versus placebo or when compared with age-matched expected events.2,50–55 Similar results have been reported with tadalafil20,21 and vardenafil.56 Postmarketing surveys in the United Kingdom54 and in the United States, including 1 by the FDA,52 did not show a rate of myocardial infarction, ischemic heart disease, or mortality in men who took sildenafil to be greater than expected than in an age-matched population. In fact, death rates tended to be less in the population taking sildenafil (perhaps suggesting that healthier men were receiving the drug). Postmarketing data regarding cardiovascular event rates in men taking vardenafil or tadalafil have yet to be reported and will be important.
Other Cardiovascular Effects
PDE5 inhibitors have potential therapeutic use for a number of cardiovascular disorders, including pulmonary hypertension, congestive heart failure, hypertension, and endothelial dysfunction. Details are described in the online supplement.
Summary
There are currently 3 PDE5 inhibitors available for the treatment of ED. All are effective in patients with organically based ED, including ED due to vascular disease. Sildenafil and vardenafil have relatively short half-lives of ≈4 hours, whereas the half-life of tadalafil is 17.5 hours. These agents are all mild vasodilators and have minimal effects on blood pressure. Sildenafil, vardenafil, and tadalafil are all contraindicated with nitrate use. Should a patient develop chest pain while taking sildenafil or vardenafil, nitrate may be administered under supervision at 24 hours (and possibly earlier); should a patient take tadalafil and develop chest pain, nitrate may not be given until at least 48 hours after the tadalafil dose and then again, only under close monitoring. α-Blockers are a contraindication to the use of vardenafil. Except for 0.4 mg tamsulosin, α-blockers are a contraindication to the use of tadalafil. Sildenafil >25 mg should not be given within 4 hours of an α-blocker. Doses of sildenafil of 25 mg may be administered at any time in relation to an α-blocker. Vardenafil should not be administered to patients taking class IA or III antiarrhythmic drugs or to patients with congenital QT prolongation. The PDE5 inhibitors do not adversely affect total exercise time or time to ischemia during exercise stress testing. The PDE5 inhibitors are safe to administer to patients taking antihypertensive medicines (except for the α-blockers, as described earlier). The PDE5 inhibitors are effective for the treatment of ED in patients with cardiac disease, including patients with chronic coronary artery disease57 and hypertension. Data from controlled clinical trials do not suggest an increase in myocardial infarction or death rates in patients taking PDE5 inhibitors. These agents appear safe and effective in most patient populations but have not been studied extensively in patients with baseline severe or unstable cardiac conditions.
Acknowledgments
Disclosure
Dr Kloner is a compensated speaker for and consultant to Pfizer, Bayer-GSK, and Lilly ICOS, and receives research support from Lilly ICOS.
Footnote
An online-only Data Supplement is available at
Supplemental Material
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Published In
Circulation
Volume 110 • Number 19 • 9 November 2004
Pages: 3149 - 3155
PubMed: 15533876
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© 2004.
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Published online: 9 November 2004
Published in print: 9 November 2004
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Subjects
Congenital Heart Disease
Endothelium/Vascular Type/Nitric Oxide
Epidemiology
Pharmacology
Pulmonary Hypertension
Notes
Received May 17, 2004; revision received June 22, 2004; accepted June 22, 2004.
Authors
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Robert A.Kloner, MD, PhD
From the Heart Institute, Good Samaritan Hospital, and the Division of Cardiovascular Medicine, Keck School of Medicine, University of Southern California, Los Angeles.
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Notes
Correspondence to Robert A. Kloner, MD, PhD, Heart Institute, Good Samaritan Hospital, 1225 Wilshire Blvd, Los Angeles, CA 90017. E-mail rkloner@goodsam.org
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11178 | https://www.geeksforgeeks.org/maths/sum-of-angles-in-a-polygon/ | Sum of Angles in a Polygon
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23 Jul, 2025
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Polygon is defined as a two-dimensional geometric figure that has a finite number of line segments connected to form a closed shape. The line segments of a polygon are called edges or sides, and the point of intersection of two edges is called a vertex. The angle of a polygon is referred to as the space formed at the intersection point (vertex) of two adjacent sides.
A polygon is of two types: a regular polygon and an irregular polygon. A regular polygon is a polygon whose all sides and all interior angles are measured the same, whereas an irregular polygon is a polygon whose all sides and all interior angles do not measure the same. And we also have different types of polygons like triangles, quadrilaterals, pentagons, hexagons, etc, based on the number of sides of a polygon. Every polygon has interior angles and exterior angles, where an interior angle is the one that lies inside the polygon and the exterior angle is the one that lies outside the polygon.
Table of Content
What is Polygons
Angles in Polygons
Interior Angles
Exterior Angles
Sum of Interior Angles of a Polygon
Interior Angle Formulae
Using the Number of Sides
Using Exterior Angle
Using Sum of Interior Angles
Interior Angles of Regular Polygons
Sum of Interior Angle of Polygon Theorem
Solved Examples on Interior Angles Formula
What is Polygons
A polygon is of two types: a regular polygon and an irregular polygon.
A regular polygon- Regular polygon is a polygon whose all sides and all interior angles are measured the same,
An irregular polygon - Irregular polygon is a polygon whose all sides and all interior angles do not measure the same.
We also have different types of polygons like triangles, quadrilaterals, pentagons, hexagons, etc., based on the number of sides of a polygon. Every polygon has interior angles and exterior angles, where an interior angle is the one that lies inside the polygon and the exterior angle is the one that lies outside the polygon.
Angles in Polygons
Angles in polygons are the angles formed by the intersection of two adjacent sides of the polygon. There are two main types of angles associated with polygons: interior angles and exterior angles.
Interior Angles
The angle of a polygon is referred to as the space formed at the intersection point (vertex) of two adjacent sides. Now, the interior angle of a polygon is the one that lies inside the polygon. The number of angles in a polygon having "n" sides is "n". For example, a triangle has three sides, so it has three interior angles.
We know that a polygon is of two types of the polygon: a regular polygon and an irregular polygon. The measurement of all interior angles is the same, whereas in an irregular polygon the measurement of each angle may differ. In the figure given below, ABCD is a square whose interior angles are ∠1, ∠2, ∠3, and ∠4.
Read More about Interior Angles of a Polygon
Exterior Angles
The angle that lies at the outside of a polygon, which is formed by one side of the polygon and the extension of the other side, is referred to as the exterior angle of a polygon. The sum of an adjacent interior angle and exterior angle is equal to 180°. And the sum of all the exterior angles of a polygon is always equal to 360°.
Exterior angle of a regular polygon = 360° ÷ number of sides = 360°/n
Sum of Interior Angles of a Polygon
Every polygon has interior angles and exterior angles, where an interior angle is the one that lies inside the polygon and the exterior angle is the one that lies outside the polygon. The number of angles and their measurements differ from one polygon to another based on the number of sides of a polygon.
For example, a triangle has three interior angles, and its sum is equal to 180°, whereas a square has four interior angles, and its sum is equal to 360°. But the sum of the interior angles of a polygon remains the same whether it is a regular or an irregular polygon.
The formula to determine the sum of the interior angles of a polygon is given as follows:
S = (n - 2) × 180°
S is the sum of interior angles, and
n is the number of sides or number of angles of polygons
Interior Angle Formulae
The interior angle of a polygon is the one that lies inside the polygon. The interior angle of a polygon can be found in the following ways:
Using the Number of Sides
The measure of each interior angle of a regular polygon having n sides is given as follows:
I = [(n – 2) × 180° / n]
Where,
I is the value of the interior angle, and
n is the number of sides or number of angles of polygons.
Using Exterior Angle
We know that the sum of an adjacent interior angle and exterior angle is equal to 180°. So, when the exterior angle of a polygon is given, the polygon formula to determine the interior angle of a polygon is given as follows:
I = 180° – E
Where,
I is the value of the interior angle, and
E is the corresponding exterior angle for that interior angle.
Using Sum of Interior Angles
We know that all the interior angles of a regular polygon are equal. So, the measure of each interior angle of a regular polygon is equal to the sum of the interior angles of the polygon divided by the number of sides.
I = S / n
Where,
S is the sum of interior angles, and
n is the number of sides or number of angles of polygons
Interior Angles of Regular Polygons
Interior angles of first 10 regular polygons are listed in the following table:
| Name of the polygon | Number of sides (n) | Sum of interior angles S = (n – 2) × 180° | Measurement of each interior angle of a regular polygon [(n – 2) × 180°/n] |
--- --- |
| Triangle | 3 | 180° | 180/3 = 60° |
| Quadrilateral | 4 | 360° | 360/4 = 90° |
| Pentagon | 5 | 540° | 540/5 = 108° |
| Hexagon | 6 | 720° | 720/6 = 120° |
| Heptagon | 7 | 900° | 900/7 = 128.57° |
| Octagon | 8 | 1080° | 1080/8 = 135° |
| Nonagon | 9 | 1260° | 1260/9 = 140° |
| Decagon | 10 | 1440° | 1440/10 = 144° |
Sum of Interior Angle of Polygon Theorem
Statement: The interior angle theorem states the sum of the interior angles of a polygon with n vertices is S = (n – 2) × 180°.
Derivation:
To prove the interior angle theorem, we need the statement that the sum of the interior angles of a triangle is 180°.
Now, let us consider a polygon having "n" sides. A polygon having "n" sides forms "n" triangles. For example, look at the figure given below, where a polygon having 8 sides has formed 8 triangles.
As we know that the sum of the interior angles of a triangle is 180°, the sum of all the interior angles of "n" triangles will be n × 180°.
Sum of all the interior angles of "n" triangles = n × 180°
So, we can conclude that,
Sum of interior angles of the polygon + Sum of the angles at the point O = n × 180° ——— (1)
We know that the sum of the angles at the point O = 360° ——— (2)
By substituting equation (2) in equation (1), we get
Sum of interior angles of the polygon + 360°= n × 180°
So, the sum of interior angles of the polygon = n × 180° – 360°
= (n – 2) × 180°
Thus, the sum of interior angles of the polygon = (n – 2) × 180°
The measure of each interior angle of a regular polygon = ((n – 2) × 180°/n) (Since the measure of each angle is the same for a regular polygon)
Hence proved.
Similar Reads
| Sum of internal angles of a Polygon |
| Program to find the Interior and Exterior Angle of a Regular Polygon |
| Understanding Quadrilaterals |
| Angle between 3 given vertices in a n-sided regular polygon |
| Area of Triangle |
| Area of Square |
| Area of Rectangle |
Solved Examples on Interior Angles Formula
Example 1: Determine the sum of interior angles of a regular heptagon.
Solution:
We know that a heptagon has seven sides, i.e., n = 7.
From the sum of the interior angles of a given polygon,
We have, S = (n − 2) × 180°
= (7 − 2) × 180°
= 5 × 180° = 900°
Hence, the sum of interior angles of a regular heptagon is 900°.
Example 2: Find the interior angle at vertex C in the figure given below.
Solution:
The given polygon is a pentagon. We know that a pentagon has seven sides, i.e., n = 5.
From the sum of the interior angles of a given polygon,
We have, S = (n − 2) × 180°
= (5 − 2) × 180° = 540°
Now, we have the sum of all the interior angles in the given polygon is 540°.
∠A + ∠B + ∠C + ∠D + ∠E = 540°
x + 111° + (x - 20°) + 105° + (x - 10°) = 540°
3x + 186° = 540°
3x = 354°
x = 118°
Now, let the angle at ∠C = x - 20°
= 118° - 20°
∠C = 98°
Thus, the interior angle at vertex C is ∠C = 98°.
Example 3: What is the sum of the interior angles of a polygon having 12 sides?
Solution:
Given data:
The number of sides of the given polygon (n) = 12
From the sum of the interior angles of a given polygon,
We have, S = (n − 2) × 180°
= (12 − 2) × 180°
= 10 × 180° = 1800°
The sum of interior angles of a polygon having 12 sides is equal to 1800°.
Example 4: What is the measure of the interior angles of a regular polygon having 8 sides?
Solution:
Given data:
The number of sides of the given polygon (n) = 8
Since the given polygon is regular, all the angles are equal.
We have,
The measure of each interior angle of a regular polygon = (n – 2) × 180°/n
= [(8 − 2) × 180°]/8
= 1,080°/8 = 135°
Hence, the measure of each interior angle of a regular polygon having 8 sides is 135°.
Example 5: Find the interior angle at vertex F in the figure given below.
Solution:
The given polygon is a pentagon. We know that a pentagon has seven sides, i.e., n = 5.
From the sum of the interior angles of a given polygon,
We have, S = (n − 2) × 180°
= (6 − 2) × 180° = 720°
Now, we have the sum of all the interior angles in the given polygon is 720°.
∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 720°
132° + y° + (y - 15)° + 110° + (y - 20)° + 99° = 720°
3y + 306° = 720°
3y = 414°
y = 138°
Now, let the angle at ∠F = y - 20°
= 138° - 20°
∠F = 118°
Thus, the interior angle at vertex F is ∠F = 118°.
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2024 AMC 12B Problems/Problem 20
Contents
1 Problem 20
2 Solution 1
3 Solution 2 (Geometry)
4 Solution 3 (Trigonometry)
5 Solution 4 (Apollonius)
6 Solution 5 (Median length formula)
7 Solution 6 (AM-GM Inequality)
8 Video Solution by SpreadTheMathLove
9 See also
Problem 20
Suppose , , and are points in the plane with and , and let be the length of the line segment from to the midpoint of . Define a function by letting be the area of . Then the domain of is an open interval , and the maximum value of occurs at . What is ?
Solution 1
Let the midpoint of be , and let the length . We know there are limits to the value of , and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length to and , and doesn't contain any information about the median. Therefore we're going to have to write the side in terms of and then use the triangle inequality to find bounds on .
We use Stewart's theorem to relate to the median : . In this case , , , , , .
Therefore we get the equation
.
Notice that since is a pythagorean triple, this means .
By triangle inequality, and
Let's tackle the first inequality:
Here we use the property that .
Therefore in this case, .
For the second inequality,
Therefore we have , so the domain of is .
The area of this triangle is . The maximum value of the area occurs when the triangle is right, i.e. . Then the area is . The length of the median of a right triangle is half the length of it's hypotenuse, which squared is . Thus the length of is .
Our final answer is
~KingRavi
Solution 2 (Geometry)
Let midpoint of as , extends to and ,
triangle has sides , based on triangle inequality, so
so which is achieved when , then
~luckuso
Solution 3 (Trigonometry)
Let A = (0, 0) , B =(b, 0) , C= ()
When :
When : The domain of is the open interval:
The rest follows Solution 2
Solution 4 (Apollonius)
Here's a faster way to solve this problem using Apollonius's Theorem (which is a special case of Stewart's Theorem for medians). In this case, . So,
We know that, by the Triangle Inequality, . Applying these to Apollonius, we have that the minimum value of is and the maximum value is (both cannot be reached, however).
The rest of the solution follows Solution 1.
~xHypotenuse
Solution 5 (Median length formula)
Let the midpoint of be . Then, by the Median Length Formula: . If we let and , then we get the relationship that: . By the Triangle Inequality , so . This means that the domain of x is .
The rest follows Solution 1.
~mathwizard123123
Solution 6 (AM-GM Inequality)
By letting BC equal , we can use Heron's formula to calculate the area. Notice the semi-perimeter is just which is just . Next, by Heron's formula, the area of ABC is: which simplifies to the . We now know that the domain of is just the domain of . This domain is very easy to calculate. We see that 1 and . Because is always positive, we see that is in the open interval . Now, we find the maximum of . By the AM-GM inequality, we have: ≥ . Simplifying and letting = , we get that ≤ = . We know by AM-GM that = if and only if 1 = . Solving, = . Therefore, we have found the domain of is the open interval and the maximum of is which occurs at = (Apply Stewart's to triangle ABC when knowing that BC = .) Adding these up, we get = or .
~ilikemath247365
Video Solution by SpreadTheMathLove
See also
2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19Followed by
Problem 21
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
All AMC 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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11180 | https://dr282zn36sxxg.cloudfront.net/datastreams/f-d%3A162f89dad808e4fdce45a409c7df77aa08e3b290c47b1b046b27802b%2BATTACHMENT%2BATTACHMENT.1 | CK-12 Physics FlexBook® 2.0 Answer Key Chapter 2: Motion in One-Dimension 2.1 Position and Displacement Review Questions 1.
Explain the difference between distance and displacement in your own words. 2.
Suppose that John lives on a square block that is 180 yards per side, and in the evenings, he walks with his dog around the block for a little exercise. 1.
If John walks once around the block, what distance does he travel? 2.
If John walks once around the block, what is his final displacement? 3.
Joanna’s house is 8000 feet due west of her school. If her house is assigned the position of zero and her school is assigned the position of +8000, what would Joanna’s position be if she walked 100 feet west of her house? Answers 1.
Distance is the total units traveled by an object as it changes position, while displacement is the net change in position. 2. 1.
John will have traveled a distance of 720 yards. 2.
John’s displacement is zero because he ends up exactly where he started – his house. 3.
Joanna will be at -100. Explore More Questions 1.
What is position? 2.
Can two objects be the same distance from a single point but be in different positions? Why or why not? 3.
What is the difference between distance and displacement? 4.
Does distance have direction? Does displacement have direction? Answers 1.
Position is an object’s location in relation to a reference point. 2.
Yes. As in the movie theater example, an object can be one unit positive from the reference point and a different object can be one unit negative from the reference point. Both objects are one unit away (the same distance), but they occupy different locations in space because they are in different directions. 3.
Distance is the total amount an object has traveled, while displacement is the shortest distance between the object’s starting and finishing point. 4.
Distance does not have direction. Objects can be one unit away from a point in every direction. However, displacement does have a direction. Displacement is defined as a specific distance in a specific direction. 2.2 Average Velocity Review Questions 1.
On a one-day vacation, Jane traveled 340 miles in 8.0 hours. What was her average speed? 2.
An object on a number line moved from x = 12 m to x = 124 m and moved back to x = 98 m. The time interval for all the motion was 10. s. What was the average velocity of the object? 3.
An object on a number line moved from x = 15 cm to x = 165 cm and then moved back to x = 25 cm all in a time of 100 seconds. a.
What was the average velocity of the object? b.
What was the average speed of the object? Answers 1.
340miles / 8 hours = 42.5 mph 2.
x1=12 m, x2=98 m: Δ x=86 m; Δt =10 s. Δ x/Δt = 86m/10s = 8.6 m/s 3. a. Δ x/Δt = 10cm/100s = 0.1 m/s b. Δ x/Δt = (165-15cm) + (165-25) / 100s = 290 cm / 100s = 2.9 cm/s Explore More Questions 1.
What is the main difference between average speed and average velocity? Answers 1.
Average speed is calculated by dividing the total distance travelled by the time interval. Average velocity involves total displacement, instead of distance. 2.3 Instantaneous Velocity Review Questions Draw a velocity versus time graph for an object whose constant velocity is 15 m/s and whose position starts at x= 0 when t = 0. Graph the motion for the first 5.0 seconds. Use the graph below to answer the following questions: 1. For the motion graphed in the position versus time graph shown above, what is the average velocity in the time interval 1 to 3 seconds? 2.
For the motion graphed in the position versus time graph shown above, what is the average velocity in the time interval 3 to 4 seconds? 3.
For the motion graphed in the position versus time graph shown above, what is the average velocity in the time interval 5 to 6 seconds? Answers 1.
Δx/Δt = 40m / 2s = 20 m/s. The average velocity from 1 to 3 seconds is 20 m/s. 2.
Δx/Δt = 0m / 1s = 0 m/s. The average velocity from 3 to 4 seconds is 0 m/s. 3.
The position at t=5s must be estimated. 47m is a good approximation, but answers will vary. Assuming 47m gives: Δx/Δt = (20-47)m / 1s = -27 m/s. The average velocity from 5 to 6 seconds is -27 m/s. Explore More Questions Use this resource to answer the questions that follow: . 1.
In the graph on the video, what is graphed on the vertical axis? 2.
What is graphed on the horizontal axis? 3.
What does the slope of this graph represent? Answers 1.
Velocity is graphed on the vertical axis. 2.
Time is graphed on the horizontal axis. 3.
The slope of the graph represents the acceleration. 2.4 Average Acceleration Review Questions 1.
The velocity of a car increases from 2.0 m/s to 16.0 m/s in a time period of 3.5 s. What was the average acceleration? 2.
If an automobile slows from 26 m/s to 18 m/s in a period of 4.0 s, what was the average acceleration? 3.
If a runner increases his velocity from 0 m/s to 20 m/s in 2.0 s, what was his average acceleration? 4.
If a runner decreases his velocity from 20 m/s to 10 m/s in 2.0 s, what was his average acceleration? Answers 1.
Δv/Δt = 14m/s / 3.5 s = 4 m/s 2 2.
Δv/Δt = -8m/s / 4.0 s = -2 m/s 2 3.
Δv/Δt = 20m/s / 2.0 s = 10 m/s 2 4.
Δv/Δt = -10m/s / 2.0 s = -5 m/s 2 2.5 Uniform Acceleration Review Questions 1.
If an object has zero acceleration, does that mean it has zero velocity? Give an example. 2.
If an object has zero velocity, does that mean it has zero acceleration? Give an example. 3.
If the acceleration of a motorboat is 4.0 m/s 2, and the motorboat starts from rest, what is its velocity after 6.0 s? 4.
The friction of the water on a boat produces an acceleration of -10.0 m/s 2. If the boat is traveling at 30.0 m/s and the motor is shut off, how long it take the boat to slow down to 5.0 m/s? Answers 1.
No. An object with zero acceleration is not changing its velocity. However, it can have any velocity. For example, a car travelling at 60 mph on a straight section of the freeway has a consistent velocity of 60 mph, but it has zero acceleration. 2.
No. An object can have a velocity of zero but still have an acceleration. For example, take a ball that has been thrown into the air. At the moment the ball has reached its highest location, the ball has zero velocity. Nonetheless, its direction is changing, which means it has a changing (non-zero) acceleration. 3. V f = V i + at = 0 m /s + (4.0 m/s 2)( 6.0 s) = 24 m/s 4. V f = V i + at; 5.0 m/s = 30.0 m/s + (-10.0 m/s 2)(t); t = 2.5 s 2.6 Displacement During Constant Acceleration Review Questions 1.
An airplane accelerates with a constant rate of 3.0 m/s 2 starting at a velocity of 21 m/s. If the distance traveled during this acceleration was 535 m, what is the final velocity? 2.
An car is brought to rest in a distance of 484 m using a constant acceleration of -8.0 m/s 2. What was the velocity of the car when the acceleration first began? 3.
An airplane starts from rest and accelerates at a constant 3.00 m/s 2 for 20.0 s. What is its displacement in this time? 4.
A driver brings a car to a full stop in 2.0 s. a.
If the car was initially traveling at 22 m/s, what was the acceleration? b.
How far did the car travel during braking? Answers 1.
Using the formula v 2 = v 0 2 + 2ax, the final velocity is 60 m/s 2.
Using the formula v 2 = v 0 2 + 2ax, the initial velocity was 88 m/s 3.
Using the formula x = ½ at 2, the displacement is 600 m. 4. a.
Since , a = 11 m/s 2 b.
Using the formula x = ½ at 2, the displacement is 22 m Explore More Question Use this resource to answer the questions that follow: 1.
What does the area bounded by a velocity versus time graph represent? Answer 1.
The area bounded by a velocity versus time graph represents the object’s displacement. 2.7 Acceleration Due to Gravity Review Questions 1.
A baseball is thrown vertically into the air with a speed of 24.7 m/s. a.
How high does it go? b.
How long does the round trip up and down require? 2.
A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top? 3.
A kangaroo jumps to a vertical height of 2.8 m. How long will it be in the air before returning to earth? Answers 1. a.
a = 9.8 m/s2 ; v i=24.7 m/s; v f=0 m/s; v f 2 = v i 2 + 2ad; d= (v f 2 - v i 2) / 2a; d = (610 - 0 m 2/s 2) / 2( 9.8 m/s2) = 31 m b.
d = 31m; v i=24.7 m/s; v f=0 m/s; d= 1/2(v f - v i) t; t = 2d / (v f + v i); t = 62 m / 24.7 m/s = 2.5 s 2. a = -9.8 m/s2 ; v f=0 m/s; d=2.4m; v f 2 = v i 2 + 2ad; v i 2= v f 2 - 2ad; v i 2 = 0 m 2/s 2 - 2( -9.8 m/s2 )(2.4m) = m 2/s 2 = 6.9 m/s √47 3.
d = 2.8m; 9.8 m/s2 ; d = ½ at 2; t= = = 0.76s; 0.76s x 2 = 1.5 s √a 2d √9.8m/s2 2 (2.8m) Explore More Use this resource to answer the questions that follow: . Questions 1.
What is the gravitational acceleration given in the video? Why does it differ from that given in this text? 2.
Why does the ball travel further in later time intervals than in the earlier ones? Answers 1.
The gravitational acceleration given in the video is 9.81 m/s. This is different because of the altitude of the classroom. 2.
The ball travels further later because it has a faster velocity. The change in velocity (acceleration) is consistent, but at higher velocities, this change makes the ball get faster and faster. 2.8 Position vs. Time Graphs Review Questions 1.
Describe how to make a position-time graph. 2.
What is the slope of a line graph? What does the slope of a position-time graph represent? 3.
Can a line on a position-time graph have a negative slope, that is, can it slope downward from left to right? Why or why not? 4.
In Graph 1 in the Figure above, what is the object’s average velocity? Answers 1.
To make a position-time graph, you plot position relative to the starting point on the y-axis against the corresponding time on the x-axis. 2.
The slope of a line graph is its steepness. The slope of a position-time graph represents velocity. 3.
Yes, a position-time graph can have a negative slope. This would represent the motion of an object that is getting closer to the starting position. 4.
The object’s velocity is 10 m/s. 2.9 Velocity vs. Time Graphs Review Questions 1.
Describe a velocity-time graph. What does the slope of the graph line represents? 2.
In the Figure above, the sprinter reaches a velocity of 2 m/s in just 1 second. At a constant rate of acceleration, how long does it take for her to double this velocity? What is her acceleration during this time period? 3.
Create a velocity-time graph by plotting the data in the Table 1.1 below. Answers 1.
A velocity-time graph shows how velocity changes over time. It plots velocity on the y-axis and time on the x-axis. Each point on the graph represents the velocity at a given time. The slope of the graph represents acceleration. 2.
The runner doubles her velocity to 4 m/s by 2 seconds into the race, so it takes her just 1 second more to reach this velocity. Her acceleration during this period is 2 m/s2 . 3.
See sketch of graph below: |
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Walker HK, Hall WD, Hurst JW, editors. Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition. Boston: Butterworths; 1990.
Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition.
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Walker HK, Hall WD, Hurst JW, editors.
Boston: Butterworths; 1990.
Contents
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Chapter 31Cholesterol, Triglycerides, and Associated Lipoproteins
Rafael A. Cox and Mario R. García-Palmieri.
Definition
Cholesterol, triglycerides, and high-density lipoproteins are important constituents of the lipid fraction of the human body. Cholesterol is an unsaturated alcohol of the steroid family of compounds; it is essential for the normal function of all animal cells and is a fundamental element of their cell membranes. It is also a precursor of various critical substances such as adrenal and gonadal steroid hormones and bile acids. Triglycerides are fatty acid esters of glycerol and represent the main lipid component of dietary fat and fat depots of animals.
Cholesterol and triglycerides, being nonpolar lipid substances (insoluble in water), need to be transported in the plasma associated with various lipoprotein particles. Plasma lipoproteins are separated by hydrated density; electrophretic mobility; size; and their relative content of cholesterol, triglycerides, and protein into five major classes: chylomicrons, very-low-density lipoproteins (VLDL), intermediate-density lipoproteins (IDL), low-density lipoproteins (LDL), and high-density lipoproteins (HDL).
Since the levels of plasma lipids have a bell-shaped distribution in the general population, the definition of either a high or a low value of these substances has remained an arbitrary statistical decision. High values have been traditionally considered as those in the 90th and 95th percentiles; low values were considered to be those below the 5th percentile. The NIH Concensus Conference has recently revised the values concerning cholesterol, however, in view of clear evidence of an increased risk of coronary atherosclerosis in persons falling in the 75th to 90th percentiles. According to this last statement, cholesterol levels below 200 mg/dl are classified as "desirable blood cholesterol," those 200 to 239 mg/dl as "borderline-high blood cholesterol," and those 240 mg/dl and above as "high blood cholesterol."
Technique
Analysis of plasma or serum total cholesterol, triglycerides, and lipoproteins is usually performed on blood specimens obtained by venipuncture in tubes containing EDTA. Because of a larger content of water, plasma levels have been found to be around 3% lower than matched serum levels. Cholesterol levels are fairly constant, but triglyceride levels fluctuate considerably from day to day and are highest 1 to 4 hours after meals.
Collection of blood for triglyceride testing should be done after a 12-hour fasting period, when chylomicrons have ordinarily been cleared from the circulation. Measurements must be ideally done while the patients are on their usual diet and taking no medications that could alter blood lipid levels. Sampling should not be performed during periods of stress or within 6 weeks after a major illness, such as an acute myocardial infarction, as plasma cholesterol may be reduced and triglyceride levels increased in these instances.
The standard chemical determination for plasma cholesterol is the Abell-Kendall (A-K) technique, which involves the Liebermann-Burchardt reaction after hydrolysis and extraction of cholesterol. Samples for plasma cholesterol and triglyceride determinations are commonly analyzed by completely automated methods at clinical laboratories. As the results obtained at commercial laboratories differ systematically from those obtained by research methods, it is advisable to inquire how the method used by the particular laboratory compares with reference values, and to employ the same laboratory facility for serial measurements. The best reference values available today for North American populations are those obtained in the Lipid Research Clinics Program, which surveyed 12,595 Americans aged 20 to 79 in the United States and Canada. Reference values for plasma total cholesterol were obtained by utilizing autoanalyzer systems to which either the ferric chloride–sulfuric acid method or the Liebermann-Burchardt test was adapted. A fluorometric analysis was utilized to determine triglyceride reference values. Plasma high-density lipoprotein cholesterol (HDL-C) determination in this program was measured by the same procedures used for plasma total cholesterol, after precipitation of apoprotein B-containing lipoproteins in whole plasma (LDL and VLDL) by heparin–manganese chloride.
A useful additional test is examination of a refrigerated sample of fasting plasma at 4°C; chylomicrons, if present, form a creamy layer at the top of the test tube. This test could help to distinguish type 1 hyperlipoproteinemia (creamy layer on top with clear infranatant) from type 5 hyperlipoproteinemia (creamy layer on top with turbid infranatant).
Lipoproteins can be separated by ultracentrifugation, precipitation, and electrophoresis. Lipoprotein fractions can be quantified by preparative or density-adjusted ultracentrifugation. This method is usually employed in elaborate studies and not as a routine clinical tool. The quantitative reference method employs precipitation and ultracentrifugation sequentially to determine each lipoprotein component.
Usually, measurement of plasma lipid levels is sufficient for evaluation of patterns of lipoprotein elevations. Occasionally paper electrophoresis of the plasma is helpful. The correspondence of electrophoretic fractions with lipoproteins is shown in Table 31.1.
Table 31.1
Lipoproteins Present on Electrophoresis.
Quantitative determination of apoproteins is performed in sophisticated laboratories utilizing radioimmunoassays (for apoprotein AI and apoprotein B) and isoelectric focusing (for apoproteins C and E).
Other specialized tests are: (1) the intravenous heparin test for determination of postheparin lipolytic activity (this test measures the preheparin and postheparin liberation of lipoprotein lipase and is principally used to confirm congenital lipoprotein lipase deficiency), and (2) evaluation of LDL receptor status on cultured skin fibroblasts or isolated lymphocytes.
Basic Science
Plasma lipoprotein particles contain variable proportions of four major elements: cholesterol, triglycerides, phospholipids and specific proteins called apoproteins (Table 31.2). An alphabetical nomenclature (A, B, C, D, E.) is generally used to designate the apoproteins. The varying composition of these elements determines the density, size, and electrophoretic mobility of each particle. These factors in turn have been used for the clinical and biochemical classification of lipoprotein disorders. Schematically, lipoproteins have been described as globular or spherical units in which a nonpolar core lipid (consisting mainly of cholesterol esters and triglycerides) is surrounded by a layer containing phospholipids, apoproteins, and small amounts of unesterified cholesterol. Apoproteins, in addition to serving as carrier proteins, have other important functions such as being co-factors for enzymes involved in lipoprotein metabolism, acting as specific ligands for binding of the particles to cellular receptor sites, and intervening in the exchange of lipid constituents between lipoprotein particles.
Table 31.2
Characteristics and Percentage Content of the Various Lipoprotein Particles Relative to Total Weight.
The fact that all the cholesterol required by the body can be produced by biosynthesis points to the essential nature of this substance. As an estimated loss of 1.0 to 1.5 g of cholesterol occurs daily through desquamation and fecal loss, this amount must be replaced. Usually this replacement is obtained from dietary sources, but another portion is synthesized in multiple cells of the body. Triglycerides are also obtained from the diet as well as synthesized by the liver.
The origin of circulating lipoproteins is less understood than is their uptake, transport, and degradation. The lipid transport system in plasma has been described as involving two pathways: an exogenous route for the transport of cholesterol and triglycerides absorbed from dietary fat in the intestine, and an endogenous system through which cholesterol and triglycerides reach the plasma from the liver and other nonintestinal tissues (Figure 31.1).
Figure 31.1
Exogenous and endogenous fat-transport pathways are diagrammed. Dietary cholesterol is absorbed through the wall of the intestine and is packaged, along with triglyceride (glycerol ester-linked to three fatty acid chains), in chylomicrons. In the capillaries (more...)
Exogenous Pathway
The exogenous pathway starts with the intestinal absorption of triglycerides and cholesterol from dietary sources. Its end result is the transfer of triglycerides to adipose and muscle tissue and of cholesterol to the liver. After absorption, triglycerides and cholesterol are re-esterified in the intestinal mucosal cells and then coupled with various apoproteins, phospholipids, and unesterified cholesterol into lipoprotein particles called chylomicrons. The chylomicrons in turn are secreted into intestinal lymph, enter the bloodstream through the thoracic: duct, and bind to the wall of capillaries in adipose and skeletal muscle tissue. At these binding sites the chylomicrons interact with the enzyme lipoprotein lipase, which causes hydrolysis of the triglyceride core and liberation of free fatty acids. These fatty acids then pass through the capillary endothelial cells and reach the adipocytes and skeletal muscle cells for storage or oxidation, respectively.
After removal of the triglyceride core, remnant chylomicron particles are formed. These are high in cholesterol esters and characterized by the presence of apoproteins B, CIII, and E. These remnants are cleared from the circulation by binding of their E apoprotein to a receptor present only on the surface of hepatic cells. Subsequently, the bound remnants are taken to the inside of hepatic cells by endocytosis and then catabolized by lysosomes. This process liberates cholesterol, which is then either converted into bile acids, excreted in bile, or incorporated into lipoproteins originated in the liver (VLDL).
Under normal physiologic conditions, chylomicrons are present in plasma for 1 to 5 hours after a meal and may give it a milky appearance. They are usually cleared from the circulation after a 12-hour fast.
Endogenous Pathway
The liver constantly synthesizes triglycerides by utilizing as substrates free fatty acids and carbohydrates; these endogenous triglycerides are secreted into the circulation in the core of very-low-density lipoprotein particles (VLDL). The synthesis and secretion of VLDL at cellular level occur in a process similar to that of chylomicrons, except that a different B apoprotein (B-100 instead of B-48) together with apoproteins C and E intervene in their secretion. Subsequent interaction of the VLDL particles with lipoprotein lipase in tissue capillaries leads to hydrolysis of the core triglycerides and production of smaller remnant VLDL particles rich in cholesterol esters (intermediate-density lipoproteins, IDL) and liberation of free fatty acids. Around half of these remnant particles are removed from the circulation in 2 to 6 hours as they bind tightly to hepatic cells. The rest undergo modifications with detachment of the remaining triglycerides and its substitution by cholesterol esters and removal of all the apoproteins except apoprotein B. This process results in transformation of the remnant VLDL particles into low-density lipoprotein particles (LDL) rich in cholesterol. In fact, these last particles contain around three-fourths of the total cholesterol in human plasma, although they constitute only some 7% of the total cholesterol pool. Their predominant function is to supply cholesterol to cells with LDL receptors, like those in the adrenal glands, skeletal muscle, lymphocytes, gonads, and kidneys. The quantity of cholesterol freed from LDL is said to control cholesterol metabolism in the cell through the following mechanisms: (1) increased LDL cholesterol in the cell decreases synthesis of the enzyme 3-hydroxy-3 methylglutaryl coenzyme A (HMG-CoA) reductase, which modulates the intracellular synthesis of cholesterol; (2) increased LDL cholesterol may enhance the storage of cholesterol within the cell by activation of another enzyme; and (3) increased cholesterol within the cell diminishes the synthesis of LDL receptors through a negative feedback process.
Besides the above described route for LDL degradation in extrahepatic sites, a so-called scavenger cell pathway has been described. This consists of cells in the reticuloendothelial system which, by phagocytosis, dispose of the excess concentrations of this lipoprotein in plasma.
Transport of High-Density Lipoprotein Cholesterol
High-density lipoproteins are a heterogeneous group of macromolecules with different physical properties and chemical components; two subclasses of HDL have been identified (HDL2 and HDL3) within which several subspecies have also been demonstrated. The predomination function of HDL seems to be the reverse transport of cholesterol from different tissues into the liver, where it is eventually removed. Subclass HDL2 has been reported to have a better correlation with coronary artery disease protection than total HDL cholesterol.
The serum concentration of HDL and its components derives from various complex intravascular and cellular metabolic events. These events include secretion of precursor HDL particles from the liver and small intestine, interaction of these particles with lipids and proteins released during the catabolism of triglyceride-rich lipoproteins, and production of cholesteryl esters (the core substance in HDL) from the action of lecithin–cholesterol acyltransferase (LCAT), an enzyme that originates in the liver. This enzyme acts on unesterified cholesterol released into plasma from cellular turnover. The cholesterol esters formed in this reaction are in turn transferred to VLDL and subsequently appear in LDL. The end result is a system that allows the transfer of cholesterol through LDL to peripheral cells and its return to the liver through HDL, and that prevents excessive accumulation of cholesterol in the body.
Clinical Significance
Cholesterol and triglycerides, like many other essential components of the body, attract clinical attention when present in abnormal concentrations. Increased or decreased levels usually occur because of abnormalities in the synthesis, degradation, and transport of their associated lipoprotein particles. When hyperlipidemia or hypolipidemia are defined in terms of the class or classes of increased or decreased plasma lipoproteins, the names hyperlipoproteinemia or hypolipoproteinemia are preferentially employed.
Hyperlipoproteinemia is the lipid disturbance of major relevance clinically because of its association with an increased risk of atherosclerotic cardiovascular disease. Multiple epidemiologic studies have demonstrated that increased levels of plasma total cholesterol and low-density lipoproteins are strongly and directly related to a greater incidence of coronary heart disease. Elevated plasma triglycerides and very-low-density lipoproteins are directly associated with the risk of atherosclerotic heart disease, although not as independent risk factors. In contrast, high levels of high-density lipoprotein cholesterol have been found to be a protective factor for the development of that disease, so that decreased levels constitute a risk factor.
Clinical manifestations of hyperlipoproteinemia include a greater incidence of ischemic vascular disease, acute pancreatitis, and visible accumulations of lipid deposits (xanthomas and xanthelasmas). The localization of these lesions is of great help in many instances to categorize the lipoprotein dysfunction present.
Increased concentration of plasma lipids is etiologically related mainly to genetic disorders, dietary factors (such as ingestion of excessive calories, saturated fatty acids and cholesterol), or ingestion of drugs, or it may occur as a secondary phenomenon in a large variety of diseases. In any of these instances the elevation of the different plasma lipoproteins usually occurs in a number of combinations that have led to their classification into six different patterns or phenotypes (Table 31.3). Genetic or acquired disorders may be related to one or more of these lipoprotein patterns, so the identification of a particular pattern gives no specific information regarding the cause of the hyperlipidemic disorder in question. A practical approach is to classify the different hyperlipidemic: states into those that mainly cause hypercholesterolemia (Table 31.4) or those that predominantly cause hypertriglyceridemia (Table 31.5). In each of these categories primary (genetic) as well as secondary disorders are included. The genetic disorders in turn could result from a single gene disturbance either of dominant or recessive inheritance, or a polygenic derangement in which multiple genes interact with environmental factors. The main characteristics of primary hyperlipoproteinemias are included in Table 31.6.
Table 31.3
Lipoprotein Patterns Resulting from Elevation of Different Plasma Lipid Fractions.
Table 31.4
Disorders Mainly Causing Hypercholesterolemia.
Table 31.5
Disorders Mainly Causing Hypertriglyceridemia.
Table 31.6
Characteristics of Primary Hyperlipoproteinemias.
For the diagnosis of the different lipoprotein disorders, one can usually rely on simple plasma values of total cholesterol and triglycerides. These measurements should be repeated at least once, so as to obtain a more reliable estimate of the patient's level. Obtained lipid values should be interpreted in relation to age and sex according to reference values established by the Lipid Research Clinics Program (Tables 31.7, 31.8, and 31.9). As stated previously, the diagnosis of hyperlipidemia is based on arbitrary reference values above the 95th percentile for the distribution of cholesterol, triglyceride or LDL cholesterol for a given population. Low reference values (5th percentile for the distribution of cholesterol, LDL and HDL) identify persons with low lipid levels.
Table 31.7
Percentiles for Plasma Total Cholesterol (mg/dl) in the Normal White North American Population.
Table 31.8
Percentiles for Plasma Triglycerides (mg/dl) in the Normal White North American Population.
Table 31.9
Percentiles for Plasma Total Low-Density Lipoprotein Cholesterol (mg/dl) in the Normal White North American Population.
Plasma cholesterol and triglyceride values provide information concerning the lipoprotein particles that are increased. Plasma cholesterol values represent total cholesterol, which includes both unesterified cholesterol and cholesteryl esters. Isolated elevation of plasma cholesterol usually indicates that LDL is increased. Isolated elevation of plasma triglyceride points to an elevation in chylomicrons, VLDL, and/or remnants.
The bedside inspection of a refrigerated fasting plasma sample, the measurement of HDL cholesterol (advisable if total plasma cholesterol is persistently elevated) and the evaluation of associated clinical findings provide useful information. In all cases, a detailed history and physical examination is imperative. Evaluation and exclusion of secondary disorders, if necessary, could require additional laboratory tests as indicated in each particular case. Screening of first-degree relatives is advisable, especially if no secondary disorder is evident, as it may reveal a genetic problem. Only in selected instances will special tests or lipoprotein electrophoresis be required for exact genotypic diagnosis.
Hypercholesterolemia
Three primary disorders causing hypercholesterolemia have been identified. They are: (1) polygenic hypercholesterolemia, (2) familial hypercholesterolemia, and (3) a variant of familial combined hyperlipidemia.
Polygenic hypercholesterolemia is the term utilized to describe the most common primary disorder causing an increase in plasma cholesterol. It includes a group of related disorders in which multiple genes apparently interact to cause an elevation in LDL above the 95th percentile in the general population. Increased rate of formation of LDL, defective clearance of LDL, or both could be responsible for this elevation. Recent evidence suggests that this may result from a reduced activity of hepatic LDL receptors. Clinical distinction between this disorder and familial hypercholesterolemia and familial combined hyperlipidemia is mainly obtained by family studies and examination for tendon xanthomas. Hyperlipidemia occurs in only 10% of the first-degree relatives in the polygenic disorder, while it occurs in around 50% of the relatives in the other two conditions. Tendon xanthomas are not found in both the polygenic disorder and familial combined hyperlipidemia but occur in around 75% of adult heterozygotes with familial hypercholesterolemia. Epidemiologic studies indicate that subjects with this disorder are at particularly high risk for coronary artery disease.
Familial hypercholesterolemia is a common autosomal dominant disorder that affects approximately 1:500 persons in the general population. Its principal defect lies in the gene for the LDL receptor on the surface of cells so that clearance of LDL from plasma is delayed. Homozygotes are rare and usually attain a six- to eightfold increase in total plasma cholesterol due to an elevation in LDL; heterozygotes may have a two- to threefold elevation and can be diagnosed at birth with analysis of umbilical cord blood. The most important clinical characteristic of this disorder is the presence of premature and accelerated coronary artery disease. Heterozygotes manifest myocardial infarctions around the third decade, while homozygotes usually die of myocardial infarction complications before age 20. The second most important finding is the presence of xanthomas over tendons. Although tendon xanthomas are virtually diagnostic of the condition, their absence does not exclude it. Other associated findings include xanthelasmas and arcus corneae.
Familial hypercholesterolemia can be diagnosed in sophisticated laboratories by direct measurement of the number of LDL receptors on freshly isolated lymphocytes or on cultured skin fibroblasts. It has been diagnosed in utero by the absence of LDL receptors on cultured amniotic fluid cells.
Familial combined hyperlipidemia is another common disorder that has an autosomal dominant inheritance. It can present clinically as hypercholesterolemia (type 2a), hypertriglyceridemia (type 4) or both (type 2b) and has also been called multiple-type hyperlipoproteinemia. It is characterized clinically by the absence of hyperlipoproteinemia during childhood, and its development occurs around puberty in association with variable and mild elevation in plasma lipid levels. There is no specific clinical or laboratory test to determine if an individual has this disorder, and family screening is needed in order to make the diagnosis. Nevertheless, it should be suspected in any individual with a mild hyperlipoproteinemia whose lipoprotein type changes with time. This condition is also characterized by an increased incidence of premature atherosclerosis and myocardial infarction. The basic abnormality apparently is overproduction of lipoproteins that contain apolipoprotein B (apo B).
Various common disorders are associated with hypercholesterolemia. Among these are hypothyroidism, nephrotic syndrome, dysproteinemias, obstructive liver disease, and use of thiazide diuretics.
Hypertriglyceridemia
The primary disorders predominantly causing hypertriglyceridemia are: (1) familial hypertriglyceridemia, (2) familial combined hyperlipidemia, (3) congenital deficiency of lipoprotein lipase, (4) deficiency of apoprotein CII and (5) familial dysbetalipoproteinemias.
Familial hypertriglyceridemia is a common autosomal dominant disorder characterized by increased plasma concentration of VLDL (type 4 lipoprotein pattern). Moderate elevations of triglycerides usually occur during early adulthood, and a triad of obesity, hyperglycemia, and hyperinsulinemia can be seen in affected individuals. In individuals with moderate elevation in plasma triglycerides associated with a normal cholesterol level, the possibility of familial hypertriglyceridemia should be suspected.
Although patients with this disorder usually have moderate elevation of plasma triglycerides, whenever conditions known to elevate this lipid fraction, such as diabetes mellitus or alcohol or oral contraceptive consumption, concurrently occur, they may develop a profound hypertriglyceridemia and a mixed hyperlipidemia (type 5 lipoprotein pattern). An increased incidence of atherosclerosis has been reported. Hypertension and hyperuricemia have also been frequently observed.
Familial combined hyperlipidemia has already been discussed under Hypercholesterolemia.
Congenital lipoprotein lipase deficiency is a rare autosomal recessive disorder secondary to absence or severe diminution in the activity of lipoprotein lipase. Affected individuals are homozygous for a mutation that prevents normal expression of lipoprotein lipase activity. The parents, although clinically normal, are obligate heterozygotes. This enzymatic disorder is reflected in a massive accumulation of chylomicrons in the plasma without elevation of VLDL (type 1 lipoprotein pattern). Triglycerides may reach levels of 2000 to 10,000 mg/dl. This disorder usually appears in childhood with recurrent bouts of abdominal pain secondary to pancreatitis. Another characteristic feature of the disorder is the occurrence of eruptive xanthomas, predominantly in the buttocks and other pressure-sensitive areas of the body, attributed to the deposition of chylomicron triglycerides in histiocytes. The elevation in chylomicrons may also cause a lipemic serum, a picture of lipemia retinalis, and hepatosplenomegaly. The diagnosis is suspected by the presence of a lipemic plasma in a young patient who has fasted for 12 hours. Overnight incubation of a test tube of such plasma in a refrigeration at 4°C demonstrates a visible white layer at the top of the tube. Failure of lipoprotein lipase levels to increase after an infusion of heparin confirms the diagnosis. The occurrence of accelerated atherosclerosis is not reported in this condition.
Apoprotein CII deficiency is a rare autosomal recessive disorder caused by absence of apoprotein CII, a required co-factor for the activity of lipoprotein lipase. The ensuing functional deficiency in this enzyme leads to a clinical picture similar to that described above for congenital lipoprotein lipase deficiency. However, in contrast to what occurs in the latter disorder, affected individuals are diagnosed at later age and rarely present eruptive xanthomas. The usual presentation is also with recurrent abdominal pains secondary to acute pancreatitis. At times the diagnosis is made by chance discovery of a milky serum.
Due to the inherent defect in this condition, in which lipoprotein lipase is not activated, both chylomicrons and VLDL are elevated in the blood causing a type 1 or type 5 lipoprotein pattern.
Familial dysbetalipoproteinemia, also called familial type 3 hyperlipoproteinemia, is a condition inherited through a single gene mechanism whose clinical presentation requires the presence of other genetic or environmental factors. Elevation of both plasma cholesterol and triglycerides occurs because of accumulation of remnant VLDL particles in the plasma. The metabolic defect in most patients occurs in apolipoprotein E. This has three common alleles, designated E2, E3, and E4. Patients with this disorder have only apolipoprotein E2 in VLDL, which is less effective in facilitating clearance of remnants than E3 or E4. The condition occurs only in individuals who are homozygous for E2, that is, those with an E2/E2 genotype. Clinical evidence of hyperlipoproteinemia usually appears after the second decade. The characteristic clinical findings are xanthoma striata palmaris and tuberous and tuberoeruptive xanthomas over the elbows and knees. The disorder is associated with severe atherosclerosis of the coronary arteries, abdominal aorta, and peripheral arteries.
The diagnosis is facilitated by encountering a broad beta band on lipoprotein electrophoresis. Confirmation can be obtained in specialized laboratories either by measuring the chemical composition of the VLDL fraction after ultracentrifugation of the plasma or by determining for the E2 allele after isoelectric focusing of remnant proteins.
Table 31.10Percentiles for Plasma Total High-Density Lipoprotein Cholesterol (mg/dl) in the Normal White North American Population
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| Men | Women |
--- |
| Age (yr) | 5 | 95 | Age (yr) | 5 | 95 |
| 5–14 | 35 | 75 | 5–19 | 35 | 70 |
| 15–19 | 30 | 65 | 20–24 | 35 | 80 |
| 20–24 | 30 | 65 | 25–34 | 35 | 80 |
| 25–29 | 30 | 65 | 35–39 | 35 | 80 |
| 30–34 | 30 | 65 | 40–44 | 35 | 90 |
| 35–39 | 30 | 60 | 45–49 | 35 | 85 |
| 40–44 | 25 | 65 | 50–54 | 35 | 90 |
| 45–69 | 30 | 70 | 55 + | 35 | 95 |
| 70 + | 30 | 75 |
: Source: Abridged from Lipid Research Clinic Data, 1983.
A variety of disorders (Table 31.6) may cause a secondary hypertriglyceridemic state by several mechanisms. The most common of these disorders are non-insulin-dependent diabetes mellitus, excessive alcohol consumption, and intake of oral contraceptives.
Hyperalphalipoproteinemia
Another clinical condition associated with elevation in plasma lipoproteins is hyperalphalipoproteinemia, characterized by elevated plasma levels of high-density lipoproteins. The elevation in HDL leads to slight increase in total plasm cholesterol values. Other plasma lipid components (LDL, VLDL, and triglycerides) are normal.
The majority of cases of hyperalphalipoproteinemia are genetic with either a dominant or polygenic inheritance. Secondary elevations of HDL have been related to various factors such as weight reduction, regular exercise, moderate alcohol intake, estrogen administration, exposure to chlorinated hydrocarbon pesticides, and biliary cirrhosis. Patients with this condition usually do not present any distinguishing clinical features. As previously mentioned, hyperalphalipoproteinemia is associated with a decreased risk of coronary atherosclerosis and with increased longevity.
Hypolipoproteinemias
Hypoalphalipoproteinemia probably is the most clinically significant hypolipoproteinemia in view of the fact that considerable evidence suggests that low levels of plasma HDL cholesterol are related to an increased incidence of coronary heart disease in high-risk populations.
According to the NIH Consensus Conference, the finding of HDL cholesterol values below 35 mg/dl constitutes an independent risk factor for coronary artery disease. Several factors have been identified as causing a decrease in HDL cholesterol. These include ill-defined genetic factors, obesity, cigarette smoking, physical inactivity, hypertriglyceridemia, oral contraceptives, beta-adrenergic blocking drugs, thiazide diuretics, and cholesterol-reducing diets.
Other hypolipoproteinemias comprise two rare disorders characterized by a decrease in the concentration of lipids in plasma and an autosomal recessive inheritance.
Abetalipoproteinernia usually appears early in childhood, and because of a defective production of apoprotein B, there is absence of chylomicrons, VLDL, and LDL in the plasma. The plasma cholesterol level is usually less than 75 mg/dl and that of triglycerides less than 15 mg/dl. The main clinical features are malabsorption of fats, peripheral neuropathy, ataxia, retinitis pigmentosa, and acanthocytosis.
Tangier disease is a condition that also manifests in childhood and is characterized by the absence of HDL from the plasma. This defect leads to the production of abnormal chylomicron remnants, which are stored as cholesterol esters in cells of the phagocytic system. Levels of plasma cholesterol are usually less than 100 mg/dl and that of triglycerides range from 100 to 250 mg/dl. The main clinical features are enlarged orange tonsils, corneal opacities, and infiltration of the bone marrow and the intestinal mucosa. Patients with this illness are at increased risk for premature atherosclerosis.
Dyslipoproteinemia, is the term utilized for conditions in which structurally abnormal lipoproteins circulate in plasma. Such a defect is seen in lecithin cholesterol acyltransferase (LCAT) deficiency. This is a rare disorder in which decreased activity of this enzyme leads to a large accumulation of unesterified cholesterol in plasma and body tissues. Laboratory findings include a variable level of total plasma cholesterol with decreased esterified cholesterol, an increase in unesterified cholesterol and increased VLDL. The structure of all the lipoproteins is abnormal. The condition usually presents in young adulthood with corneal opacities, renal insufficiency, hemolytic anemia, and premature atherosclerosis.
References
Ad Hoc Committee of AHA to Design a Dietary Treatment of Hyperlipoproteinemia. Recommendations for Treatment of Hyperlipidemia in Adults. A Joint Statement of the Nutrition Committee and the Council on Arteriosclerosis. Circulation. 1984;72:1067A–90A. [PubMed: 6368042]
Brewer HB, Zech LA, Greg , RE, Schwartz D, Schaefer EJ. Type III hyperlipoproteinemia: diagnosis, molecular defects, pathology, and treatment. Ann Intern Med. 1983;98:623–40. [PubMed: 6846977]
Brown MS, Goldstein JS. How LDL receptors influence cholesterol and atherosclerosis. Sci Am. 1984;251:57–66. [PubMed: 6390676]
Castelli WP, Cooper GR, Doyle JT, Garcia-Palmieri MR, Gordon T, Homes C, Hudley SB, Kagan A, Kuckmak M, McGee D, Vicic WJ. Distribution of triglyceride and total LDL and HDL cholesterol in several populations: a cooperative lipoprotein phenotyping study. J Chron Dis. 1977;30:147–69. [PubMed: 191465]
Council on Scientific Affairs, Division of Scientific Analysis and Technology, American Medical Association. Dietary and pharmacology therapy for the lipid risk factors. JAMA. 1983;250:1873–79. [PubMed: 6620484]
Goldstein JL, Kita T, Brown MS. Defective lipoprotein receptors and atherosclerosis. N Engl J Med. 1983;309:288–96. [PubMed: 6306464]
Gordon T, Castelli WP, Hjortland MC, Kannel WB, Dawber TR. High density lipoprotein as a protective factor against coronary heart disease. Am J Med. 1977;62:707–14. [PubMed: 193398]
Havel RJ. Approach to the patient with hyperlipidemia. Med Clin N Amer. 1982;66:319–33. [PubMed: 7070147]
Hulley SB, Lo B. Choice and use of blood lipid tests. Arch Intern Med. 1984;143:667–73. [PubMed: 6340621]
Lipid Research Clinics Program. Lipid and lipoprotein analysis. Vol. 1, in Manual of laboratory operations. NIH publication 75–628. Bethesda, Md.: U.S. Department of Health and Human Services, 1974 (revised October 1982).
Naito H. Reliability of lipid and lipoprotein testing 1. Am J Cardiol. 1985;56:9 J.
National Institutes of Health Consensus Conference. Lowering blood cholesterol to prevent heart disease. JAMA. 1985;253:2080. [PubMed: 3974099]
Report of the National Cholesterol Education Program Expert Panel on Detection, Evaluation, and Treatment of High Blood Cholesterol in Adults. The Expert Panel. Arch Intern Med. 1988;148:36–69. [PubMed: 3422148]
Rifkind BM, Segal P. Lipid Research Clinics Program reference values for hyperlipidemia and hypolipidemia. JAMA. 1983;250:1869–72. [PubMed: 6578354]
Stanbury JB, et al, eds. The metabolic basis of inherited disease, 5th ed. New York: McGraw-Hill, 1983, Chaps. 29, 31, 34.
Copyright © 1990, Butterworth Publishers, a division of Reed Publishing.
Bookshelf ID: NBK351PMID: 21250192
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Cox RA, García-Palmieri MR. Cholesterol, Triglycerides, and Associated Lipoproteins. In: Walker HK, Hall WD, Hurst JW, editors. Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition. Boston: Butterworths; 1990. Chapter 31.
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11182 | https://www.youtube.com/watch?v=d1U4Z7Dcs14 | How to determine the Discriminant of a Quadratic Equation
Algebra-1 with Mr. Peters
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In today's video we will review the process of identifying the number of real solutions to any quadratic equation. Whether you're a math enthusiast, a student grappling with algebra, or a teacher looking for effective teaching methods, this video is your go-to resource.
📚 What You'll Learn:
1️⃣The Quadratic Formula:We'll start with the basics, reviewing how the quadratic formula can be used to find solutions to any quadratic equation. Whether the solutions are real or complex, this method never fails to provide an answer.
2️⃣The Discriminant Formula:Next, we dive into the discriminant formula, a powerful tool that allows you to determine the number of real solutions without actually solving the equation. It's a quick, efficient method that every math aficionado should know.
3️⃣Beyond Factoring:While factoring is a commonly taught method for solving quadratics, it's not always the most effective, especially for complex equations. We'll explore why factoring might not be your best option and when to use alternative methods.
4️⃣Graphing Quadratics:Finally, we connect the dots between solving equations and graphing quadratics. Understanding how to find x-intercepts on a graph is crucial, and we'll show you how this relates to determining the number of real solutions.
Why It Matters:
Grasping these concepts is not just about solving equations; it's about enhancing your mathematical intuition and problem-solving skills. Whether you're preparing for an exam, teaching a class, or simply love math, knowing how to determine the number of real solutions to a quadratic equation is an invaluable skill.
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Transcript:
what's good math family in today's video I'm going to show you guys two different methods and how to determine the amount of real solutions we have from a quadratic equation as we start this video out we're we're really specifically focusing on discriminant formula which is that b^2 minus 4 a c that's a quick way for us to determine right and our answer if it's greater than zero it means we have two real solutions if it's equal to zero we have one real solution and if it's less than zero no real solutions so what does that mean when we look at problem number one right see we could Factor this as x - 3 squared right and once we once we solve for it we know that our answer will be X is equal to three right and even though we have two factors because they're the same we know we only have one real solution so let's break that down right so let's say we followed our discriminant formula so we have -6 SAR - 4 1 POS 9 so now we have 36 - 36 which is equal to zero right so what does that mean if it's equal to zero we have one real solution now let's say if we went over to the quadratic formula right so now we're talking about let's switch colors right X is equal to the opposite of B plus or minus the square < TK of b^ 2 minus 4 a c all over 2 a once we simplify right to find our Solutions you're going to notice that we only have one answer it it follows this discriminant all right so let's go through and solve so now when I go through this is what I'm going to have X is equal to posi 6 plus or minus 6^ 2 - 4 1 9 all over 2 1 right so as we simplify we have 6 plus or minus 0 all over 2 so when we split this answer right X is = to 6 + 0 / 2 and X is equal to 6 - 0/ 2 no matter what we do we're going to get one solution of three now that we understood just that basic one let's go to problem number two right because when we use a discriminant this is what we're probably going to use it for and we have to understand what the answer essentially means all right so let's go back to uh problem number two and we're going to stay in Black so I follow my discriminant right first off typically with these problem s try to factor it first if you cannot Factor it then go into your discriminant or into the quadratic equation okay guys just a tip for you all right so now we're going to go ahead do the same thing so I have here b^2 which is 2^ 2 minus 4 a which is 4 C which is five so I'm going to get four minus 80 which is going to give me -76 so what does this 76 represent so when we do the discriminant and our answer is less than zero that means we have no real solutions so what exactly does this mean so this means that if you're trying to factor you can't Factor this unless you're using a quadratic formula and two that means there's no x intercept so if we were to graph this quadratic some way it would not touch or cross the x-axis and just to prove that let's do the quadratic formula and see what we get all right so going back to the quadratic formula right we have X is equal to the opposite of B so that's -2 plus or minus right 2^ 2 - 4 4 5 all over 2 a which is just8 believe that's right yep so as we solve this now we're going to get the same thing X is = -2 plus or minus the square T and let's let's erase some more here guys so I got some more space the square root of -76 all over eight so what we need to understand guys is even after we split this we this this is that that negative under the radical sign means that we are now working with imaginary numbers and typically you don't see that in Algebra 1 we typically see it in Algebra 2 so when we're talking about the discriminant and determining whether we have real solutions or how many this is exactly what we mean and understand real solutions is important for can we graph it is it going to touch the x axis really hope this video was helpful on you guys on how to determine if there are real solutions using a discriminate thank you guys so much for joining today on alga one with Mr Peters smash the like button subscribe to the channel and leave comments for future videos you guys would like to see on our Channel or if you have questions on today's video |
11183 | https://en.wikipedia.org/wiki/Logarithm_of_a_matrix | Jump to content
Logarithm of a matrix
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From Wikipedia, the free encyclopedia
Mathematical operation on invertible matrices
In mathematics, a logarithm of a matrix is another matrix such that the matrix exponential of the latter matrix equals the original matrix. It is thus a generalization of the scalar logarithm and in some sense an inverse function of the matrix exponential. Not all matrices have a logarithm and those matrices that do have a logarithm may have more than one logarithm. The study of logarithms of matrices leads to Lie theory since when a matrix has a logarithm then it is in an element of a Lie group and the logarithm is the corresponding element of the vector space of the Lie algebra.
Definition
[edit]
The exponential of a matrix A is defined by
: .
Given a matrix B, another matrix A is said to be a matrix logarithm of B if eA = B.
Because the exponential function is not bijective for complex numbers (e.g. ), numbers can have multiple complex logarithms, and as a consequence of this, some matrices may have more than one logarithm, as explained below. If the matrix logarithm of exists and is unique, then it is written as in which case
Power series expression
[edit]
If B is sufficiently close to the identity matrix, then a logarithm of B may be computed by means of the power series
: ,
which can be rewritten as
: .
Specifically, if , then the preceding series converges and .
Example: Logarithm of rotations in the plane
[edit]
The rotations in the plane give a simple example. A rotation of angle α around the origin is represented by the 2×2-matrix
For any integer n, the matrix
is a logarithm of A.
| Proof |
| ⇔ where ... qed. |
Thus, the matrix A has infinitely many logarithms. This corresponds to the fact that the rotation angle is only determined up to multiples of 2π.
In the language of Lie theory, the rotation matrices A are elements of the Lie group SO(2). The corresponding logarithms B are elements of the Lie algebra so(2), which consists of all skew-symmetric matrices. The matrix
is a generator of the Lie algebra so(2).
Existence
[edit]
The question of whether a matrix has a logarithm has the easiest answer when considered in the complex setting. A complex matrix has a logarithm if and only if it is invertible. The logarithm is not unique, but if a matrix has no negative real eigenvalues, then there is a unique logarithm that has eigenvalues all lying in the strip . This logarithm is known as the principal logarithm.
The answer is more involved in the real setting. A real matrix has a real logarithm if and only if it is invertible and each Jordan block belonging to a negative eigenvalue occurs an even number of times. If an invertible real matrix does not satisfy the condition with the Jordan blocks, then it has only non-real logarithms. This can already be seen in the scalar case: no branch of the logarithm can be real at -1. The existence of real matrix logarithms of real 2×2 matrices is considered in a later section.
Properties
[edit]
If A and B are both positive-definite matrices, then
Suppose that A and B commute, meaning that AB = BA. Then
if and only if , where is an eigenvalue of and is the corresponding eigenvalue of . In particular, when A and B commute and are both positive-definite. Setting B = A−1 in this equation yields
Similarly, for non-commuting and , one can show that
More generally, a series expansion of in powers of can be obtained using the integral definition of the logarithm
applied to both and in the limit .
Further example: Logarithm of rotations in 3D space
[edit]
A rotation R ∈ SO(3) in 3 is given by a 3×3 orthogonal matrix.
The logarithm of such a rotation matrix R can be readily computed from the antisymmetric part of Rodrigues' rotation formula, explicitly in Axis angle. It yields the logarithm of minimal Frobenius norm, but fails when R has eigenvalues equal to −1 where this is not unique.
Further note that, given rotation matrices A and B,
is the geodesic distance on the 3D manifold of rotation matrices.
Calculating the logarithm of a diagonalizable matrix
[edit]
A method for finding log A for a diagonalizable matrix A is the following:
: Find the matrix V of eigenvectors of A (each column of V is an eigenvector of A).
: Find the inverse V−1 of V.
: Let
: Then A′ will be a diagonal matrix whose diagonal elements are eigenvalues of A.
: Replace each diagonal element of A′ by its (natural) logarithm in order to obtain .
: Then
That the logarithm of A might be a complex matrix even if A is real then follows from the fact that a matrix with real and positive entries might nevertheless have negative or even complex eigenvalues (this is true for example for rotation matrices). The non-uniqueness of the logarithm of a matrix follows from the non-uniqueness of the logarithm of a complex number.
Logarithm of a non-diagonalizable matrix
[edit]
The algorithm illustrated above does not work for non-diagonalizable matrices, such as
For such matrices one needs to find its Jordan decomposition and, rather than computing the logarithm of diagonal entries as above, one would calculate the logarithm of the Jordan blocks.
The latter is accomplished by noticing that one can write a Jordan block as
where K is a matrix with zeros on and under the main diagonal. (The number λ is nonzero by the assumption that the matrix whose logarithm one attempts to take is invertible.)
Then, by the Mercator series
one gets
This series has a finite number of terms (Km is zero if m is equal to or greater than the dimension of K), and so its sum is well-defined.
Example. Using this approach, one finds
which can be verified by plugging the right-hand side into the matrix exponential:
A functional analysis perspective
[edit]
A square matrix represents a linear operator on the Euclidean space Rn where n is the dimension of the matrix. Since such a space is finite-dimensional, this operator is actually bounded.
Using the tools of holomorphic functional calculus, given a holomorphic function f defined on an open set in the complex plane and a bounded linear operator T, one can calculate f(T) as long as f is defined on the spectrum of T.
The function f(z) = log z can be defined on any simply connected open set in the complex plane not containing the origin, and it is holomorphic on such a domain. This implies that one can define ln T as long as the spectrum of T does not contain the origin and there is a path going from the origin to infinity not crossing the spectrum of T (e.g., if the spectrum of T is a circle with the origin inside of it, it is impossible to define ln T).
The spectrum of a linear operator on Rn is the set of eigenvalues of its matrix, and so is a finite set. As long as the origin is not in the spectrum (the matrix is invertible), the path condition from the previous paragraph is satisfied, and ln T is well-defined. The non-uniqueness of the matrix logarithm follows from the fact that one can choose more than one branch of the logarithm which is defined on the set of eigenvalues of a matrix.
A Lie group theory perspective
[edit]
In the theory of Lie groups, there is an exponential map from a Lie algebra to the corresponding Lie group G
For matrix Lie groups, the elements of and G are square matrices and the exponential map is given by the matrix exponential. The inverse map is multivalued and coincides with the matrix logarithm discussed here. The logarithm maps from the Lie group G into the Lie algebra .
Note that the exponential map is a local diffeomorphism between a neighborhood U of the zero matrix and a neighborhood V of the identity matrix .
Thus the (matrix) logarithm is well-defined as a map,
An important corollary of Jacobi's formula then is
Constraints in the 2 × 2 case
[edit]
If a 2 × 2 real matrix has a negative determinant, it has no real logarithm. Note first that any 2 × 2 real matrix can be considered one of the three types of the complex number z = x + y ε, where ε2 ∈ { −1, 0, +1 }. This z is a point on a complex subplane of the ring of matrices.
The case where the determinant is negative only arises in a plane with ε2 =+1, that is a split-complex number plane. Only one quarter of this plane is the image of the exponential map, so the logarithm is only defined on that quarter (quadrant). The other three quadrants are images of this one under the Klein four-group generated by ε and −1.
For example, let a = log 2 ; then cosh a = 5/4 and sinh a = 3/4.
For matrices, this means that
: .
So this last matrix has logarithm
: .
These matrices, however, do not have a logarithm:
: .
They represent the three other conjugates by the four-group of the matrix above that does have a logarithm.
A non-singular 2 × 2 matrix does not necessarily have a logarithm, but it is conjugate by the four-group to a matrix that does have a logarithm.
It also follows, that, e.g., a square root of this matrix A is obtainable directly from exponentiating (logA)/2,
For a richer example, start with a Pythagorean triple (p,q,r)
and let a = log(p + r) − log q. Then
: .
: .
has the logarithm matrix
: ,
where a = log(p + r) − log q.
See also
[edit]
Matrix function
Square root of a matrix
Matrix exponential
Baker–Campbell–Hausdorff formula
Derivative of the exponential map
Notes
[edit]
^ Hall 2015 Theorem 2.8
^ Higham (2008), Theorem 1.27
^ Higham (2008), Theorem 1.31
^ Culver (1966)
^ APRAHAMIAN, MARY; HIGHAM, NICHOLAS J. (2014). "The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential". SIAM Journal on Matrix Analysis and Applications. 35 (1): 97. doi:10.1137/130920137.
^ Unpublished memo by S Adler (IAS)
^ Hall 2015 Theorem 3.42
^ Abstract Algebra/2x2 real matrices at Wikibooks
References
[edit]
Gantmacher, Felix R. (1959), The Theory of Matrices, vol. 1, New York: Chelsea, pp. 239–241.
Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations An Elementary Introduction, Graduate Texts in Mathematics, vol. 222 (2nd ed.), Springer, ISBN 978-3319134666
Culver, Walter J. (1966), "On the existence and uniqueness of the real logarithm of a matrix", Proceedings of the American Mathematical Society, 17 (5): 1146–1151, doi:10.1090/S0002-9939-1966-0202740-6, ISSN 0002-9939.
Higham, Nicholas (2008), Functions of Matrices. Theory and Computation, SIAM, ISBN 978-0-89871-646-7.
Engø, Kenth (June 2001), "On the BCH-formula in so(3)", BIT Numerical Mathematics, 41 (3): 629–632, doi:10.1023/A:1021979515229, ISSN 0006-3835, S2CID 126053191
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11184 | https://sogc.org/en/en/content/events/HUB-Pages/Endometriosis.aspx | Endometriosis
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Endometriosis
Endometriosis is an inflammatory disease characterized by the growth of endometrial-like tissue outside of the uterus that causes chronic pelvic pain, painful periods, painful sexual intercourse, bowel and bladder symptoms, and infertility. Endometriosis is most found on the ovaries, fallopian tubes, uterosacral ligaments, and in the gastrointestinal tract, but has been found on every organ in the body except the spleen.
Endometriosis impacts approximately 10% of women and an unknown number of transgender, non-binary, and gender-diverse individuals. It accounts for 50% of pelvic pain in Canadian women and an unmeasured amount in transgender, non-binary, and gender-diverse individuals. Endometriosis impacts patients’ quality of life, productivity, relationships, and has long-term implications on patients’ overall health and wellbeing. Endometriosis also carries a large economic burden to the health care system as well as indirect costs resulting from disease symptomology and presentation. Yet, a recent survey found that only 7% of Canadian women received a clinical diagnosis of endometriosis.
Due to the complex nature and presentation of the disease; lack of provider knowledge about endometriosis, diagnostic criterion, and disease management; stigma; and other systemic considerations, the current diagnostic delay is at least five years in Canada.
The Society of Obstetricians and Gynaecologists of Canada (SOGC) has published clinical guidance related to endometriosis diagnosis and management:
SOGC Guideline: Surgical Management of Endometriosis
SOGC Guideline No. 445: Management of Chronic Pelvic Pain
SOGC Guideline No. 449: Diagnosis and Impact of Endometriosis – A Canadian Guideline
The SOGC’s website, yourperiod.ca, has information about endometriosis for patients and the public.
INDEX
Pathogenisis
Symptoms
Diagnosis
Clinical Guidelines
Management
Gender-Inclusive Language & 2SLGBTQIA+ Care
Addressing Racial and Socioeconomic Disparities
Trauma-Informed Considerations
Resources for Health Care Providers
Resources for Patients
Resources for African, Caribbean and Black Patients
Resources for Gender-Diverse Patients
Resources for Indigenous Patients
Pathogenesis
Endometriosis causes, presentations, and mechanisms continue to be explored, although none of the hypotheses fully explain the pathogenesis. Current theories include:
Retrograde menstruation
Genetic disposition
Lymphatic or circulatory spread
Metaplasia
Immune dysfunction
Mullerian abnormalities
Environmental factors
Symptoms
Endometriosis affects women, girls, transgender men, and gender-diverse and non-binary people of all races, ethnicities, and socio-economic groups. Symptoms and disease presentation often begins to appear after menarche and may progress with age.
Disease presentation is highly individual and may depend on the location and severity of endometriosis. However, there is not always a connection between the extent of the disease and the symptoms. The most common symptom of endometriosis is pain, which is inclusive of:
Chronic pelvic pain (pain that persists for more than three months)
Dysmenorrhea
Dyspareunia
Dyschezia
Dysuria
Patients may also present with post-coital bleeding, bloating, diarrhea, constipation, obstructive symptoms, chest or thoracic pain, sciatica, general fatigue, and infertility. Concomitant conditions include uterine fibroids, polycystic ovarian syndrome, irritable bowel disease, migraines, fibromyalgia, anxiety, and depression, among others.
Persistent symptoms can impact patients’ quality of life, including their mental and physical health. Endometriosis can affect patients across the life course, impacting their social life, school or professional aspirations, health, and finances. Feelings of uncertainty, anxiety, and depression may be higher among people who feel that they have been dismissed by friends, family, or medical professionals, and among those who have experienced systemic oppressions, delays to diagnosis, and medical trauma. Said feelings may also appear in realms where culturally safe, trauma-informed, and/or gender-affirming care is not available.
Diagnosis
Early diagnosis is extremely important to help slow and mitigate worsening disease progression, identify treatment options, limit unnecessary interventions, and address patient concerns. However, research has continuously shown that patients do not feel as though their symptoms are believed. The diagnostic delay in Canada is at least five years. Diagnostic delay is associated with self-diagnosis and management, misdiagnosis, and non-discriminatory tests and treatment. It can adversely influence the patient–provider relationship, as patients may not feel as though they are being validated or receiving adequate care.
Barriers to diagnosis include a lack of education about menstrual health in schools, menstrual stigma and the normalization of menstrual pain, attitudes and a lack of training for health care professionals, delayed referrals to specialist services, overlap of symptoms with other gynaecologic and gastrointestinal conditions, lack of a non-invasive method of diagnosis, and concerns about the value of a diagnosis. Patients from underserved communities (BIPOC and 2SLGBTQIA+ Peoples) may experience additional barriers, such as differences in disease presentation and reduced access to primary and specialized care. Youth may also face unique barriers, both due to atypical lesions presenting upon direct visualization and due to a lack of support and being met with skepticism.
Collecting patient history, symptomology, and conducting a physical exam are important to the diagnostic process. A trauma-informed approach should be adopted, bearing in mind that some patients may feel discomfort or unsafe in medical settings with a pelvic or other physical examination due to previous trauma. A physical exam may support:
diagnosing endometriosis by looking for signs of the disease, including deep, ovarian, and extrapelvic endometriosis;
evaluating the degree and localization of pain;
and determining the process for further investigation.
It is important to bear in mind that there is not always a correlation between symptomology and disease severity. If the clinician is not confident in their assessment, it is crucial to refer patients to an appropriate clinician.
Clinical Guidelines
SOGC Guideline No. 449: Diagnosis and Impact of Endometriosis – A Canadian Guideline provides recommendations for health care providers on the following topics.
The clinical diagnosis, whereby endometriosis is suspected through patient history, physical exam, and transvaginal, transrectal, and limited abdominal ultrasound (depending on patient comfort and preference) without advanced imaging or surgical confirmation.
The imaging diagnosis, whereby advanced abdominal and pelvic ultrasound or MRI is used to determine if ovarian endometriomas and deep endometriosis are present, as interpreted by an imaging specialist. An MRI may also be used to look for disease implantation on the abdominal wall, thoracic or diaphragmatic endometriosis, nerve endometriosis, or genito-urinary tract endometriosis.
The surgical diagnosis, whereby direct visualization at the time of laparoscopy and laparotomy is completed and diagnosis is confirmed if a histologic specimen is evaluated by pathology.
While surgical diagnosis was long considered the gold standard, it should not be used as the primary investigation tool as it leads to delays in treatment and can exacerbate acute and chronic pain.
It is critical that the diagnostic approach be individualized to each patient based on their clinical presentation, priorities, and specific needs that consider their gender, sexual orientation, education level, ability, race, culture, or ethnicity. Patients have the right to make informed decisions about their care and should be provided with accurate, up-to-date information about the diagnostic approaches including the advantages and disadvantages of each diagnostic approach. Patient decisions regarding the diagnostic pathway should be discussed and respected.
Management
Source: Williams Gyne fourth edition: Hoffman, B. L., Schorge, J. O., Halvorson, L. M., Hamid, C. A., Corton, M. M., & Schaffer, J. I. (2020).
Williams gynecology (4th ed.). McGraw-Hill Education.
Disease management should consider patients’ age, gender identity, disease progression and symptomology, family-planning goals, values, and beliefs.
There are several first- and second-line hormonal therapies available to suppress ovulation and menstruation. They can be paired with surgical management.
First-line therapies include:
Combined hormonal contraceptives
Progestins
Second-line therapies include:
GnRH agonists
GnRH antagonists
Aromatase inhibitors
Diagnosis and management of endometriosis, published in CMAJ, includes a summary of hormonal therapies for endometriosis including their dosage, adverse effects and special considerations, and relative costs.
Surgical management, using ablative or excision techniques, may be discussed. While surgery may increase fertility rates and reduce symptomology, not all patients will be responsive to surgical management, and are both surgical risks and risk of pain recurrence over time. It is important to clearly articulate the risks of surgical management with patients.
Additional management options may include non-steroidal anti-inflammatory agents (NSAIDs) as a first-line therapy for pain. Pain modifiers, opioids, and cannabis may also be recommended to patients for pain and other symptom management. Adopting a multidisciplinary approach may be beneficial to supporting patients and supporting biopsychosocial pain management. Evidence-based approaches include physiotherapy, counselling, and mindfulness techniques. There is emerging evidence that suggests acupuncture may be beneficial for chronic pelvic pain, and that diet and exercise can help reduce inflammation associated with endometriosis, though more research is needed on these topics.
It is important that patients are provided with comprehensive management options that consider their unique circumstances. For instance, transmasculine individuals may choose to have gender-affirming surgery as part of simultaneous surgical transition and endometriosis management.
As with the diagnostic process, patients have the right to make informed decisions about their care. Patients should be provided with information about the management options including the advantages and disadvantages of management pathways. Patients may choose management options based on perceived benefits and harms, affordability, beliefs, values, and availability. Patient decisions should be respected.
Gender-Inclusive Language & 2SLGBTQIA+ Care
Endometriosis, traditionally viewed as a condition affecting cisgender women, also significantly impacts transgender men. Testosterone therapy, commonly used during gender transition, can influence the symptoms and presentation of endometriosis in transgender men. Despite testosterone's role in inducing amenorrhea, studies have shown that endometrial activity may persist, leading to active lesions. This underscores the need for healthcare providers to recognize that endometriosis can occur in transgender men, even those undergoing hormone therapy. Limited research and awareness contribute to diagnostic challenges, often resulting in delays and inadequate care.
Addressing Racial and Socioeconomic Disparities
Racial and socioeconomic inequities significantly impact the diagnosis and management of endometriosis and uterine fibroids. Black women are disproportionately affected by uterine fibroids, experiencing higher prevalence, earlier onset, and more severe symptoms compared to Caucasian women.
Despite similar prevalence rates of endometriosis across racial groups, Black and Hispanic women often face longer diagnostic delays , sometimes taking twice as long to receive a diagnosis compared to Caucasian women. These delays can lead to prolonged pain and reduced quality of life. To address these disparities, it is crucial to collect race-based data to identify and mitigate gaps in care. Additionally, dedicated research and resources focusing on the intersection of fibroids and endometriosis in Black patient populations are essential to develop culturally sensitive and effective healthcare interventions.
Trauma-Informed Considerations
Trauma-informed practice considers the enduring effects of trauma and focuses on relational approaches and changes to organizational culture to provide support. Patients present with unique histories that directly and indirectly affect their health and how they may engage with health care services. Working in a trauma-informed manner is not based on patients disclosing trauma but is instead focused on creating safety, promoting choice and collaboration, and building on existing patient knowledge and strength.
Trauma can influence patients’ pain and disease presentation, their relationships, health care utilization, their direct care, and endometriosis management. Embedded power relations in patient–provider care may create an unequal power structure, where patients do not feel comfortable sharing personal information or personal experiences out of fear for re-traumatization or discrimination.
A trauma-informed approach to care may:
Consider patients’ comfortability in medical settings
Consider patients’ comfortability with physical examinations
Consider geographic barriers including access to clinics, hospitals, and costs associated with travel
Asking patients permission before starting or continuing with the examination
Detailing the steps of the physical exam
Pausing or stopping the examination if patients appear hesitant or frozen during your discussion or physical exam
Providing patients with comprehensive information about treatment and diagnostic pathways
Integrating a multidisciplinary approach to care
Providing educational materials and resources to help patients self-advocate
Continuing unbiased care regardless of patient decision-making and/or treatment adherence
Providing a continuity in care
Working from a life-course perspective that recognizes endometriosis as a chronic disease
Working in a gender-affirming and a culturally safe way.
Offering a social worker, interpreter, or cultural liaison upon request
Adopting a trauma- and violence-informed approachvalidates patient perspectives and experiences, something that patients may not have previously experienced.
Endometriosis Resources for Health Care Providers
Websites:
The Society of Obstetricians and Gynaecologists of Canada
YourPeriod
PregnancyInfo
Sex & U
World Endometriosis Society
The Endometriosis Network of Canada
Centre for Pelvic Pain and Endometriosis
Endometriosis UK
Rainbow Health Ontario
University of New South Wales NECST Registry
New global classification system for chronic pelvic pain | FIGO
Clinical Guidelines and Statements:
SOGC Guideline: Surgical Management of Endometriosis
SOGC Guideline: Canadian Contraception Consensus (Part 3 of 4): Chapter 7 – Intrauterine Contraception
SOGC Guideline No. 286: Surgical Safety Checklist in Obstetrics and Gynaecology
SOGC Guideline No. 322: Endometrial Ablation in the Management of Abnormal Uterine Bleeding
SOGC Guideline No. 445: Management of Chronic Pelvic Pain
SOGC Guideline No. 449: Diagnosis and Impact of Endometriosis – A Canadian Guideline
SOGC Statement on Intrauterine Devices, Counselling and Pain Management
Research:
JOGC Virtual Special Issue on Endometriosis
Online Courses:
Endometriosis: Impact and Diagnosis
Intrauterine Contraception (IUC) Insertion Preceptorship
Trauma- and Violence-Informed Care: Putting Principles into Practice
Multimedia:
Focus on: Minimally invasive surgery
Management of endometriosis - Comparing cases: When is surgery and when is medical appropriate
Virtual workshop to enhance assessment of pelvic floor myalgia in patients with chronic pelvic pain
Endometrial thickness and polyps
The care of underserved populations: Who loses out in a pandemic
Respectful care and informed consent
Teaching TIC Down Under
"Just try to relax”: Practical approaches to integrating trauma-informed care into your clinical practice
Taking care of ourselves when providing trauma informed care
La santé de la femme autochtone; un pas de plus vers la réconciliation (video in French)
Cultural safety and trauma informed care with indigenous lenses
Care for the Gender Diverse Community: An Educational Video
The SOGC Women’s Health Podcast:
The Impact of Endometriosis and Fibroids on Black Communities
Endometriosis: A Patient’s Journey to Symptom Validation
Imaging for the Diagnosis and Treatment of Advanced Endometriosis
Tools:
Generalized Anxiety Tool (GAD-7)
Patient Heath Questionnaire (PHQ-9)
Endometriosis Impact Questionnaire (EIQ)
Pain Logbook
Endometriosis Resources for Patients
Websites:
The Endometriosis Network Canada
The Endo Educational Organization of Canada
Endo Black, Incorporated
Minocare
Ottawa Minimally Invasive Gynecology
Tame the Beast
World Health Organization's Endometriosis Fact Sheet
Mme L’Ovary
The Endometriosis Resource Portal for People of Color
EndoFound’s LGBTQIA Resource Hub for Endo Patients
Multimedia:
Endometriosis
Understanding pain in less than 5 minutes, and what to do about it!
Care for the Gender Diverse Community: An Educational Video
2024 Endometriosis Summit
WHEMS Canada Pelvic Health Series
The SOGC Women’s Health Podcast:
The Impact of Endometriosis and Fibroids on Black Communities
Endometriosis: A Patient’s Journey to Symptom Validation
Previous Live Events
Endometriosis summit March 5 2025
Information and Tools:
Pain and Symptom Diary
Pelvic Pain Symptom Tracker
IUD Insertion: A Guide to Easing Discomfort
My Endo Report
Endometriosis Symptom Checker
Pain Logbook
Decision Aids
SOGC What is Endometriosis Infographic
Endometriosis for patients infographic - 1
Fact Sheet
Endometriosis
Endometriosis Resources for African, Caribbean, and Black Patients:
The Endometriosis Resource Portal for People of Color
Endo Black, Incorporated
The Impact of Endometriosis and Fibroids on Black Communities
Endometriosis Resources for Indigenous Patients:
Indigenous Self-Care and Medicine Wheel Teachings
Endometriosis Resources for Gender-Diverse Patients:
EndoFound’s LGBTQIA Resource Hub for Endo Patients
Rainbow Health Ontario
Endometriosis: A Patient’s Journey to Symptom Validation
New SOGC Resources
The SOGC, in collaboration with the Centre of Excellence for Women’s Health (CEWH), has created a series of educational resources to help you better understand and manage endometriosis.
These materials are evidence-based, and designed with your needs in mind. Whether you're newly diagnosed, exploring treatment options, planning for a family, or looking for support, these resources can help guide you.
What is Endometriosis?
What is endometriosis?
Common Symptoms of Endometriosis
Endometriosis at different ages and stages
How is endometriosis diagnosed?
References and Additional Resources
Teens & Endometriosis
Menopause & Endometriosis
Fertility and Endometriosis
Fertility and Endometriosis
Endometriosis and Parenting
Pregnancy Safe Endometriosis Management
Questions to ask your doctor about fertility and endometriosis
References and Additional Resources
Endometriosis Pain and Symptom Management
Endometriosis Pain and Symptom Management
What other symptom management options have worked for you?
Hormone Therapies for Endometriosis
Surgical Management of Endometriosis
When might surgery be recommended for endometriosis?
Complementary Therapies for Endometriosis
Additional Treatment and Management Options for Endometriosis
Questions to consider when deciding on a treatment plan
Non-Hormonal Medical Management for Endometriosis
References and Additional Resources
Mental Health and Emotional Wellbeing
Mental Health & Emotional Wellbeing
Self-Care and Endometriosis
Finding Support for Endometriosis
Why is an endometriosis diagnosis important?
Supporting a Loved One with Endometriosis
Endometriosis and Intimacy
References and Additional Resources
Getting ready for your appointment
Preparing for a doctor’s visit if you think you have endometriosis
Questions to ask your doctor about endometriosis
Inclusive and Respectful Care
Culturally Sensitive Endometriosis Care
Gender Affirming Endometriosis Care
Professional Resources
Resource Pages
Abortion
Medication Abortion
Endometriosis
Menopause
Perinatal Mental Health
Pregnancy Loss
Prevention of Maternal Mortality in Canada
Trauma and Violence Informed Care
Virtual Care for Substance Use
Printable Public Resource Cards
Guidelines & JOGC
Member Practice Resources
Position Statements
RSV Resource Page
SOGC Women's Health Podcast
White Papers
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11185 | https://journals.lww.com/stdjournal/fulltext/2024/10000/prevalence_of_genital_herpes_and_antiviral.9.aspx | Published Time: Wed, 03 Sep 2025 15:13:47 GMT
Sexually Transmitted Diseases
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October 2024 - Volume 51 - Issue 10
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Outline
Abstract
MATERIALS AND METHODS
Study Sample and Follow-Up
Study Outcomes: Genital Herpes, Recurrent Genital Herpes, and Genital Herpes Treatment
Statistical Analysis
RESULTS
DISCUSSION
CONCLUSIONS
REFERENCES
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Original Studies
Prevalence of Genital Herpes and Antiviral Treatment
Jain, Purva PhD, MPH∗; Embry, Alan PhD†; Arakaki, Brent BS∗; Estevez, Irisdaly MPH∗; Marcum, Zachary A. PharmD, PhD∗; Viscidi, Emma PhD, MHS†
Author Information
From the ∗Aetion, Inc., New York City, NY
†Moderna, Inc., Cambridge, MA
Acknowledgments: Editorial assistance was provided by Alejandra De Angulo, PhD, of MEDiSTRAVA in accordance with Good Publication Practice (GPP 2022) guidelines, funded by Moderna, Inc., and under the direction of the authors.
Conflict of Interest and Source of Funding: Drs. Viscidi and Embry are both employees of Moderna, Inc., and hold stock/stock options in this company. All other authors are employees of Aetion, Inc., which receives funding from Moderna. This study was funded by Moderna, Inc.
Correspondence: Emma Viscidi, PhD, MHS, 325 Binney St, Cambridge, MA 02141. E-mail: emma.viscidi@modernatx.com.
Received for publication December 19, 2023, and accepted March 17, 2024.
Supplemental digital content is available for this article. Direct URL citations appear in the printed text, and links to the digital files are provided in the HTML text of this article on the journal’s Web site (
This is an open-access article distributed under the terms of the Creative Commons Attribution-Non Commercial-No Derivatives License 4.0 (CCBY-NC-ND), where it is permissible to download and share the work provided it is properly cited. The work cannot be changed in any way or used commercially without permission from the journal.
Sexually Transmitted Diseases 51(10):p 686-693, October 2024. | DOI: 10.1097/OLQ.0000000000001990
Open
SDC
Abstract In Brief
Background
Genital herpes is a common sexually transmitted infection caused by the herpes simplex virus. Contemporary US population–based epidemiologic data on genital herpes are limited. This study aimed to provide nationally representative estimates of genital herpes prevalence and treatment using a large US health insurance claims database.
Methods
This observational cohort study used administrative claims data from HealthVerity. Crude and age- and sex-standardized prevalence rates of genital herpes and recurrent genital herpes were calculated for the years 2019 to 2021. The distribution of patients with prevalent genital herpes who received episodic or suppressive antiviral therapy was also estimated.
Results
From 2019 to 2021, the standardized prevalence of genital herpes and recurrent genital herpes ranged from 236 to 280 cases per 100,000 person-years and 81 to 98 cases per 100,000 person-years, respectively. The prevalence of genital herpes was highest among those aged 25 to 29 years (prevalence range, 497–582 years), female patients (prevalence range, 348–404 years), and those with a history of HIV infection (prevalence range, 1608–2080 years). The prevalence of recurrent genital herpes was also highest in these groups. From 2019 to 2021, two-thirds of patients (65%–68%) with prevalent genital herpes received antiviral medications; the majority received episodic therapy (80%) rather than suppressive therapy (20%).
Conclusions
The burden of genital herpes and recurrent genital herpes in the United States is substantial, with the highest rates observed in young adults, women, and immunocompromised individuals. About two-thirds receive antiviral treatment each year.
This US claims study found a significant burden of diagnosed genital herpes and recurrent genital herpes, with the highest prevalence observed in young adults, women, and immunocompromised individuals.
Genital herpes is a common sexually transmitted infection caused by the herpes simplex virus (HSV). Genital herpes is frequently caused by HSV type 2, although an increasing number of primary episodes are due to HSV type 1.1 In 2018, there were an estimated 18.6 million prevalent cases and 572,000 incident cases of genital herpes annually among adults aged 18 to 49 years in the United States.2 Herpes simplex virus infection is lifelong, and thus, periodic reactivations may occur, leading to recurrent outbreaks.3 Genital herpes recurrences, which are typically less severe than primary infections, are common and can have a significant impact on quality of life and mental health.4 Although antiviral medications can treat symptoms and reduce rates of recurrence, there is currently no cure for genital herpes.5
A majority of genital HSV infections are asymptomatic or unrecognized, and an estimated one-third of infected individuals have at least 1 episode of genital ulcer disease.6 Symptoms of primary infection vary widely in severity and can include painful genital lesions, itching, dysuria, urinary retention, vaginal/urethral discharge, fever, headache, and other systemic symptoms. Herpes simplex virus transmission can occur whether patients with genital herpes are symptomatic or asymptomatic.7 All individuals with a primary infection of genital herpes are recommended to receive antiviral therapy. Individuals with recurrent genital herpes are prescribed either episodic antiviral therapy, taken only when an outbreak occurs to ameliorate or shorten the duration of lesions and viral shedding, or daily suppressive antiviral therapy to reduce the frequency of or prevent recurrences and reduce the risk of transmitting HSV to a sexual partner.8
Although previous characterization of the epidemiology of genital herpes has been based mainly on HSV-2 seroprevalence data, there are limited contemporary data on symptomatic patients who are diagnosed and seek medical care for genital herpes and recurrent genital herpes.2,9 Specifically, little evidence exists on the disease burden of diagnosed and medically attended genital herpes in the United States from large, population-based data sources, such as administrative claims data. Moreover, little is known about the number of recurrent outbreaks or the proportion of individuals with genital herpes being treated with episodic or suppressive antiviral therapy.10,11 Understanding the burden of genital herpes and treatment patterns in the United States, especially among high-risk populations such as immunocompromised patients, is important to understand potential unmet medical needs. The objective of this study was to provide nationally representative estimates of the prevalence as well as the treatment landscape, of diagnosed genital herpes in the United States using claims data.
MATERIALS AND METHODS
This observational retrospective cohort study used secondary deidentified medical and pharmacy claims data from HealthVerity from December 1, 2017, to December 31, 2021. The HealthVerity data used in this study were limited to 2 closed claims databases (Private Source 17 and Private Source 20) within the HealthVerity data ecosystem. These sources include individuals in the United States insured under commercial, Medicare Advantage, or Medicaid plans. Private Source 20 is a closed claims source that includes both medical and pharmacy claims. Private Source 17 is an adjudicated pharmacy claims source from a pharmacy benefit manager that includes an associated pharmacy enrollment file. Hospitalizations are included in the data at the summary level. Race and ethnicity data are not available. This study was deemed exempt by the WIRB-Copernicus Group institutional review board. All research data were deemed compliant with the Health Insurance Portability and Accountability Act.
Study Sample and Follow-Up
This study included data from December 1, 2017, to December 31, 2021. The prevalence of genital herpes and recurrent genital herpes was estimated each year from 2019 to 2021. The study design for measuring prevalence of genital herpes in 2019 is depicted in Figure 1. The same design was used to assess prevalence of genital herpes and recurrent genital herpes in 2020 and 2021.
Figure 1:
Genital herpes prevalence study diagram. This study diagram illustrates the method to assess prevalent genital herpes within the 2019 calendar year. The same study design was used to assess cases of prevalent genital herpes in 2020 and 2021, as well as cases of prevalent recurrent genital herpes in 2019, 2020, and 2021. Genital herpes and recurrent genital herpes were assessed separately. Note: This study design figure depicts a single calendar year (2019). The same design was used for subsequent calendar years (2020 and 2021).
The baseline period was from the start of all available data to the start of the calendar year of interest (2019, 2020, or 2021). This period was used to capture the prevalence of clinical characteristics in participants before the start of the study period. To ensure observability throughout a given calendar year and mitigate under ascertainment of the outcome, continuous enrollment of 180 days (no gaps allowed) was required before calendar year of interest as well as 365 days (no gaps allowed) of continuous enrollment during follow-up within the calendar year of interest. Patients were excluded if they had missing or conflicting sex data or missing age data on the index date. Immunocompromised individuals were identified using a claims-based algorithm that aligns with the Centers for Disease Control and Prevention definition of immunocompromised individuals.12 Immunocompromised status was classified as ≥1 of the following criteria (relative to index): blood or stem cell transplantation in prior 2 years; history of organ transplant and immunosuppressive therapy in prior 60 days; active cancer treatment on cohort entry, with an active cancer diagnosis in the prior year; any history of primary immunodeficiency; or any history of HIV infection.13 The codes for each immunocompromised criterion are listed in Supplemental Table S1,
Prevalence was assessed in 2019, 2020, and 2021 separately to examine any potential temporal trends due to the COVID-19 pandemic. Outcomes were assessed over the course of the calendar year of interest from January 1 to December 31. Patients were followed from the index date to a genital herpes diagnosis or recurrent genital herpes diagnosis or until the end of the calendar year of interest, whichever came first.
Study Outcomes: Genital Herpes, Recurrent Genital Herpes, and Genital Herpes Treatment
Genital herpes and recurrent genital herpes were measured using International Classification of Diseases , 10th Edition (ICD-10) codes (Supplemental Table S1, Genital herpes was defined as having at least 1 of the corresponding ICD-10 diagnosis codes (any position, inpatient, or outpatient setting). Recurrent genital herpes was defined as having 2 or more claims for genital herpes (any position, inpatient, or outpatient setting) at least 28 days apart or 1 genital herpes diagnosis claim (any position, inpatient, or outpatient setting) followed by 2 or more nonoverlapping prescription claims for antiviral treatment of at least 7-day supply (Supplemental Fig. S1, If an individual met the recurrent genital herpes case definition more than once within a calendar year, they contributed to the prevalence estimate only once. The start and end date of an event had to occur within the same calendar year. To assess prevalence, there was no washout for a prior genital herpes diagnosis (i.e., people with a previous diagnosis code were not excluded) for the definitions of both genital herpes and recurrent genital herpes.
Antiviral treatments included acyclovir (oral), valacyclovir, or famciclovir and were identified via National Drug Codes (Supplemental Table S1, Based on Centers for Disease Control and Prevention guidelines for the treatment of genital herpes,8 a single antiviral treatment episode was defined as at least 7 days of medication supply per episode, allowing for a maximum of 30 days between prescription claims. Episodic therapy was defined as having a record of fewer than 6 months of total antiviral treatment, whereas suppressive therapy was defined as having at least 6 months of total antiviral treatment. Treatment was required to start within the calendar year of interest and allowed to carry over to the following calendar year. Because the antiviral treatments in this study are prescribed to treat conditions aside from genital herpes, a genital herpes diagnosis was required during the calendar year of interest (Supplemental Figs. S2, and S3,
Statistical Analysis
Baseline demographic and patient characteristics, including age, sex, health care resource utilization, and clinical conditions, were assessed for individuals with and without genital herpes and with and without recurrent genital herpes in 2019, 2020, and 2021. Descriptive comparisons of these characteristics were made for individuals with and without genital herpes and with and without recurrent genital herpes. Summary statistics for continuous variables include mean and SD and counts and proportions for categorical variables.
Crude prevalence of genital herpes and recurrent genital herpes was reported per 100,000 person-years by dividing the number of individuals identified as meeting the case definition of genital herpes or recurrent genital herpes by the total person-time observed within a given calendar year among individuals who met the cohort observability and entry criteria. Analyses were stratified by age, sex, and immunocompromised status. The overall estimates were standardized by age and sex using the 2019 US Census distributions. The count and proportion of individuals who received episodic and suppressive therapy for genital herpes were reported for each calendar year.
RESULTS
A total of 23,441,464 (2019), 34,353,848 (2020), and 31,525,253 (2021) patients were included in the analytic cohort. In 2019, the mean age and SD for the entire cohort was 39.2 (23.2) years, and 51.5% of individuals were female. The mean (SD) age among those with genital herpes was 42.2 (15.8) years, and 75.2% were female; similarly, the mean (SD) age among those with recurrent genital herpes was 42.7 (15.3) years, and 73.7% were female (Table 1). In descriptive comparisons between those with genital herpes or recurrent genital herpes, those without genital herpes or recurrent genital herpes were slightly older and more likely to be female from the Northeast; have comorbidities such as cardiovascular disease, chronic lung disease, hypertension, and obesity; have a history of tobacco use/smoking and autoimmune conditions; and be immunocompromised (Table 1). The distributions of baseline characteristics in the 2020 and 2021 cohorts were similar to those in the 2019 cohort (Supplemental Tables S2, and S3, Individuals with genital herpes and recurrent genital herpes had a higher frequency of inpatient and outpatient visits than persons without genital herpes or recurrent genital herpes (Supplemental Tables S2, and S3,
TABLE 1 - Baseline Characteristics of the Study Population, Stratified by Diagnosed Genital Herpes and Recurrent Genital Herpes (2019)
Scroll left or right to view entire table.
| Characteristic | No Genital Herpes (n = 23,378,645) | Genital Herpes (n = 62,819) | No Recurrent Genital Herpes(n = 23,420,018) | Recurrent Genital Herpes (n = 21,446) |
:---: :---:
| Age, mean (SD), y | 39.22 (23.22) | 42.19 (15.77) | 39.23 (23.21) | 42.65 (15.31) |
| Age range, y, no. (%) |
| <5 | 1,006,453 (4.3) | 55 (0.1) | 1,006,506 (4.3) | N/A |
| 5–9 | 2,017,742 (8.6) | 84 (0.1) | 2,017,819 (8.6) | N/A |
| 10–14 | 1,742,049 (7.5) | 188 (0.3) | 1,742,205 (7.4) | 32 (0.1) |
| 15–19 | 1,511,036 (6.5) | 2745 (4.4) | 1,513,004 (6.5) | 777 (3.6) |
| 20–24 | 1,343,152 (5.7) | 6453 (10.3) | 1,347,519 (5.8) | 2086 (9.7) |
| 25–29 | 1,226,777 (5.2) | 6809 (10.8) | 1,231,261 (5.3) | 2325 (10.8) |
| 30–34 | 1,350,062 (5.8) | 6986 (11.1) | 1,354,722 (5.8) | 2326 (10.8) |
| 35–39 | 1,446,193 (6.2) | 6558 (10.4) | 1,450,514 (6.2) | 2237 (10.4) |
| 40–44 | 1,405,055 (6.0) | 5577 (8.9) | 1,408,617 (6.0) | 2015 (9.4) |
| 45–49 | 1,576,192 (6.7) | 5833 (9.3) | 1,579,861 (6.7) | 2164 (10.1) |
| 50–54 | 1,628,542 (7.0) | 5538 (8.8) | 1,632,109 (7.0) | 1971 (9.2) |
| 55–59 | 1,766,930 (7.6) | 5629 (9.0) | 1,770,496 (7.6) | 2063 (9.6) |
| 60–64 | 1,757,261 (7.5) | 4951 (7.9) | 1,760,485 (7.5) | 1727 (8.1) |
| 65–69 | 1,314,267 (5.6) | 2897 (4.6) | 1,316,253 (5.6) | 911 (4.2) |
| 70–74 | 940,773 (4.0) | 1451 (2.3) | 941,748 (4.0) | 476 (2.2) |
| 75–79 | 656,340 (2.8) | 678 (1.1) | 656,815 (2.8) | 203 (0.9) |
| 80–84 | 205,687 (0.9) | 159 (0.3) | 205,792 (0.9) | 54 (0.3) |
| ≥85 | 484,134 (2.1) | 228 (0.4) | 484,292 (2.1) | 70 (0.3) |
| Sex, no. (%) |
| Male | 11,346,598 (48.5) | 15,576 (24.8) | 11,356,543 (48.5) | 5631 (26.3) |
| Female | 12,032,047 (51.5) | 47,243 (75.2) | 12,063,475 (51.5) | 15,815 (73.7) |
| US region, no. (%) | | | | |
| Northeast | 3,455,590 (14.8) | 10,273 (16.4) | 3,462,242 (14.8) | 3621 (16.9) |
| Midwest | 5,389,985 (23.1) | 13,029 (20.7) | 5,398,530 (23.1) | 4484 (20.9) |
| South | 8,726,169 (37.3) | 23,839 (37.9) | 8,741,414 (37.3) | 8594 (40.1) |
| West | 5,705,516 (24.4) | 15,449 (24.6) | 5,716,288 (24.4) | 4677 (21.8) |
| US territories | 73,236 (0.3) | 188 (0.3) | 73,369 (0.3) | 55 (0.3) |
| Missing | 28,149 (0.1) | 41 (0.1) | 28,175 (0.1) | 15 (0.1) |
| Payer type, no. (%) | | | | |
| Commercial | 12,502,037 (53.5) | 34,754 (55.3) | 12,525,686 (53.5) | 11,105 (51.8) |
| Medicaid | 9,417,291 (40.3) | 26,407 (42.0) | 9,433,991 (40.3) | 9707 (45.3) |
| Medicare | 2,592,375 (11.1) | 4869 (7.8) | 2,595,447 (11.1) | 1797 (8.4) |
| Other/Unknown | 126,116 (0.5) | 502 (0.8) | 126,486 (0.5) | 132 (0.6) |
| Quan-Charlson Comorbidity Index score, no. (%) |
| 0 | 19,427,143 (83.1) | 47,986 (76.4) | 19,459,453 (83.1) | 15,676 (73.1) |
| 1 | 1,987,335 (8.5) | 7792 (12.4) | 1,992,254 (8.5) | 2873 (13.4) |
| 2 | 1,964,167 (8.4) | 7041 (11.2) | 1,968,311 (8.4) | 2897 (13.5) |
| Clinical characteristics, no. (%) |
| Asthma | 929,964 (4.0) | 4681 (7.5) | 1,349,877 (5.8) | 2479 (11.6) |
| Cardiovascular disease | 4,609,874 (19.7) | 15,366 (24.5) | 5,509,903 (23.5) | 7004 (32.7) |
| Chronic lung disease | 1,171,465 (5.0) | 5493 (8.7) | 1,705,060 (7.3) | 2974 (13.9) |
| Hypertension | 4,021,165 (17.2) | 12,848 (20.5) | 4,800,136 (20.5) | 5818 (27.1) |
| Obesity (BMI ≥30 kg/m 2) | 1,033,519 (4.4) | 5086 (8.1) | 1,639,358 (7.0) | 3030 (14.1) |
| History of tobacco use/smoking | 1,164,669 (5.0) | 6680 (10.6) | 1,668,028 (7.1) | 3591 (16.7) |
| Immunocompromised, no. %† |
| Overall | 230,675 (1.0) | 2307 (3.7) | 231,891 (1.0) | 1091 (5.1) |
| HIV infection | 81,580 (0.3) | 1706 (2.7) | 82,425 (0.4) | 861 (4.0) |
| Organ transplant or immunosuppressive therapy | 27,695 (0.1) | 159 (0.3) | 27,788 (0.1) | 66 (0.3) |
| Blood transplant/stem cell transplant | 22,633 (0.1) | 152 (0.2) | 22,709 (0.1) | 76 (0.4) |
| Primary immunodeficiency | 19,640 (0.1) | 69 (0.1) | 19,688 (0.1) | 21 (0.1) |
| Active malignancy | 192,339 (0.8) | 679 (1.1) | 192,802 (0.8) | 216 (1.0) |
| HCRU during baseline, n (%)‡ |
| ICU/CCU | 272,161 (1.2) | 922 (1.5) | 272,694 (1.2) | 389 (1.8) |
| Inpatient visits | 2,277,392 (9.7) | 9548 (15.2) | 2,283,518 (9.8) | 3422 (16.0) |
| Outpatient visits | 19,127,193 (81.8) | 59,576 (94.8) | 19,166,223 (81.8) | 20,546 (95.8) |
| Hospitalizations | 2,009,475 (8.6) | 8893 (14.2) | 2,015,182 (8.6) | 3186 (14.9) |
Genital herpes and recurrent genital herpes prevalence was assessed from January 1, 2019, to December 31, 2019. Clinical characteristics were limited to those who were present among at least 10% of individuals in at least one group. Cells with ≤10 individuals were suppressed and marked as N/A to ensure patients remain unidentifiable. Payer type is not mutually exclusive and may sum to greater than 100%.
The Quan-Charlson Comorbidity Index score includes the following clinical conditions: myocardial infarction, congestive heart failure, peripheral vascular disease, cerebrovascular disease, dementia, chronic pulmonary disease, rheumatologic disease, peptic ulcer disease, mild liver disease, diabetes, diabetes with chronic complications, hemiplegia or paraplegia, renal disease, any malignancy (including leukemia or lymphoma), moderate or severe liver disease, metastatic solid tumor, and AIDS (2005). Comorbidity weights were taken from the original Charlson comorbidity CI Index.
†The ICD-10, procedure, HCPCS, CPT codes, and the immunosuppressive therapies used to define the components of immunocompromised status are detailed in the Appendix,
‡Inpatient visit is captured using the following: Inpatient Indicator = Yes. Hospitalization is captured using the following: Inpatient Indicator = Yes and Place of Service = Inpatient Hospital and any of the following bill types have been recorded; 113, 114, 118, 119, 11C, 11G, 11H, 11I, 11 J, 11P, 11Q, 11Z, 110, 111, 112, 115, 117, 11A, 11B, 11D, 11E, 11F, 11K, 11M, 11N, 11O, 11X, and 11Y. A patient who is recorded as having a hospitalization will also be recorded as having an inpatient visit.
BMI indicates body mass index; COPD, chronic obstructive pulmonary disease; CCU, critical care unit; CPT, Current Procedural Terminology; HCPCS, Healthcare Common Procedure Coding System; HCRU, health care resource utilization; HIV, human immunodeficiency virus; ICD-10, International Classification of Diseases, 10th Edition; ICU, intensive care unit.
A total of 62,819 (0.3% of the analytic cohort) genital herpes cases were observed in 2019, 80,573 (0.2%) in 2020, and 85,835 (0.3%) in 2021. The crude and standardized prevalence per 100,000 person-years, respectively, were 269 and 280 in 2019, 235 and 236 in 2020, and 273 and 276 in 2021. The highest prevalence was observed among individuals aged 25 to 39 years (Fig. 2; Supplemental Table S4, Supplemental Figures S4, and S5, The crude genital herpes prevalence per 100,000 person-years for male patients was 138 in 2019, 109 in 2020, and 124 in 2021. The crude genital herpes prevalences per 100,000 person-years for female patients were 393 in 2019, 348 in 2020, and 405 in 2021 (Fig. 2; Supplemental Table S4, Supplemental Figs. S4, and S5, The crude prevalences of genital herpes per 100,000 person-years for immunocompromised patients were 999 in 2019, 714 in 2020, and 733 in 2021. Among immunocompromised patients, the highest prevalence was observed in those with a history of HIV infection; 2080 in 2019, 1608 in 2020, and 1617 in 2021 (Supplemental Table S4,
Figure 2:
Crude prevalence (overall, male patients, and female patients) of genital herpes and recurrent genital herpes, 2019. Note: The scale of the x axis for the presentation of genital herpes and recurrent genital herpes prevalence in this figure differ for ease of visualization. These forest plots display the 2019 genital herpes and recurrent genital herpes prevalence. The top 2 panels display the overall and age-stratified prevalence for all individuals, the middle 2 panels display the overall and age-stratified prevalence for male patients, and the bottom 2 panels display the overall and age-stratified prevalence for female patients. Note: The standardized prevalences of genital herpes and recurrent genital herpes in 2019 were 280 and 95, respectively.
In 2019, there were 21,446 (<0.1% of the analytic cohort) recurrent genital herpes cases; 27,713 (<0.1%) were observed in 2020 and 30,452 (<0.1%) in 2021. The crude and standardized prevalence per 100,000 person-years, respectively, were 92 and 95 in 2019, 81 and 81 in 2020, and 97 and 98 in 2021. The highest rates were observed among individuals aged 20 to 39 years (Fig. 2; Supplemental Table S5, Supplemental Figs. S4, and S5, The crude recurrent genital herpes prevalences per 100,000 person-years for male patients were 50 in 2019, 40 in 2020, and 47 in 2021. The crude recurrent genital herpes prevalences per 100,000 person-years for female patients were 131 in 2019, 117 in 2020, and 141 in 2021 (Fig. 2; Supplemental Table S5, Supplemental Figs. S4, and S5, The crude prevalences per 100,000 person-years for immunocompromised patients were 471 in 2019, 345 in 2020, and 352 in 2021. Among immunocompromised patients, the highest prevalence was observed among those with a history of HIV infection; 1043 in 2019, 832 in 2020, and 847 in 2021 (Supplemental Table S5,
From 2019 to 2021, the proportion of individuals with genital herpes who received episodic antiviral therapy ranged from 54.9% to 56.9% and suppressive therapy ranged from 13.3% to 14.3%. From 32.3% to 35.0% of individuals with genital herpes did not receive antiviral therapy. Among individuals with genital herpes, suppressive (vs. episodic) therapy was more common among male patients and older individuals, and episodic (vs. suppressive) therapy was more common among female and immunocompromised patients (Fig. 3). The findings across age, sex, and immunocompromised subgroups were consistent from 2019 to 2021 (Tables 2, 3).
Figure 3:
Frequency of episodic and suppressive antiviral treatment among patients with prevalent genital herpes, 2019, 2020, and 2021. This bar graph displays the proportion of individuals with prevalent genital herpes who are untreated, and among individuals with evidence of treatment, the proportion who received episodic or suppressive therapy. A patient could have had distinct treatment patterns that classify for both suppressive and episodic therapy within a calendar year.
TABLE 2 - Frequency of Episodic and Suppressive Therapy Among Patients With Prevalent Genital Herpes, 2019
Scroll left or right to view entire table.
| 2019 |
| | Prevalent Genital Herpes | Untreated | Episodic Therapy | Suppressive Therapy |
| Overall, no. % | 62,819 | 21,995 (35.0) | 34,494 (54.9) | 8379 (13.3) |
| Age, y, no. (%) | | | | |
| <5 | 55 | 41 (74.5) | 14 (25.5) | N/A |
| 5–9 | 84 | 68 (81.0) | 15 (17.9) | N/A |
| 10–14 | 188 | 100 (53.2) | 86 (45.7) | N/A |
| 15–19 | 2745 | 805 (29.3) | 1850 (67.4) | 118 (4.3) |
| 20–24 | 6453 | 2068 (32.0) | 4087 (63.3) | 424 (6.6) |
| 25–29 | 6809 | 2235 (32.8) | 4191 (61.6) | 545 (8.0) |
| 30–34 | 6986 | 2306 (33.0) | 4127 (59.1) | 762 (10.9) |
| 35–39 | 6558 | 2255 (34.4) | 3652 (55.7) | 885 (13.5) |
| 40–44 | 5577 | 1878 (33.7) | 3026 (54.3) | 884 (15.9) |
| 45–49 | 5833 | 1903 (32.6) | 3110 (53.3) | 1066 (18.3) |
| 50–54 | 5538 | 1922 (34.7) | 2912 (52.6) | 945 (17.1) |
| 55–59 | 5629 | 1977 (35.1) | 2834 (50.3) | 1060 (18.8) |
| 60–64 | 4951 | 1858 (37.5) | 2391 (48.3) | 894 (18.1) |
| 65–69 | 2897 | 1365 (47.1) | 1191 (41.1) | 433 (14.9) |
| 70–74 | 1451 | 659 (45.4) | 616 (42.5) | 212 (14.6) |
| 75–79 | 678 | 344 (50.7) | 263 (38.8) | 89 (13.1) |
| 80–84 | 159 | 81 (50.9) | 59 (37.1) | 24 (15.1) |
| ≥85 | 228 | 130 (57.0) | 70 (30.7) | 32 (14.0) |
| Sex | | | | |
| Male | 15,576 | 5635 (36.2) | 8006 (51.4) | 2533 (16.3) |
| Female | 47,243 | 16,360 (34.6) | 26,488 (56.1) | 5846 (12.4) |
| Immunocompromised | | | | |
| Overall | 2307 | 830 (36.0) | 1097 (47.6) | 511 (22.1) |
| HIV infection | 1706 | 564 (33.1) | 849 (49.8) | 399 (23.4) |
| Organ transplant or immunosuppressive therapy | 159 | 61 (38.4) | 74 (46.5) | 32 (20.1) |
| Blood transplant/stem cell transplant | 152 | 48 (31.6) | 81 (53.3) | 35 (23.0) |
| Primary immunodeficiency | 69 | 31 (44.9) | 30 (43.5) | N/A |
| Active malignancy | 679 | 355 (52.3) | 239 (35.2) | 102 (15.0) |
“Untreated” is defined as individuals with no evidence of treatment or those with treatment episodes less than 7 days in length after a diagnosis of genital herpes. “Suppressive therapy” is defined as at least 6 months of total antiviral treatment with at least 7 days treated per treatment episode while allowing for 30 days between prescription claims. “Episodic therapy” is defined as fewer than 6 months of total antiviral treatment with at least 7 days treated per treatment episode while allowing for 30 days between prescription claims. Individuals can contribute both episodic and suppressive therapy episodes within the same calendar year; thus, summing Untreated + Episodic Therapy + Suppressive Therapy can sum to >100%. Cells with ≤10 individuals were suppressed and marked as N/A to ensure patients remain unidentifiable.
The ICD-10, procedure, HCPCS, CPT codes, and the immunosuppressive therapies used to define the components of immunocompromised status are detailed in the Appendix,
CPT indicates Current Procedural Terminology; HCPCS, Healthcare Common Procedure Coding System; ICD-10, International Classification of Diseases, 10th Edition.
TABLE 3 - Crude Overall Prevalence of Genital Herpes and Recurrent Genital Herpes, 2019 to 2021
Scroll left or right to view entire table.
| Year | 2019 | 2020 | 2021 |
:---: :---: |
| | N | N Person-Years | N Patients With an Event | Crude Prevalence Per 100,000 Person-Years | N | N Person-Years | N Patients With an Event | Crude Prevalence Per 100,000 Person-Years | N | N Person-Years | N Patients With an Event | Crude Prevalence Per 100,000 Person-Years |
| Genital herpes | 23,441,464 | 23,342,406 | 62,819 | 269 | 34,353,848 | 34,309,063 | 80,573 | 235 | 31,525,253 | 31,392,078 | 85,835 | 273 |
| Recurrent genital herpes | 23,441,464 | 23,364,235 | 21,446 | 92 | 34,353,848 | 34,336,767 | 27,713 | 81 | 31,525,253 | 31,420,289 | 30,452 | 97 |
DISCUSSION
This observational cohort study using administrative claims data from privately and publicly insured individuals estimated the prevalence of genital herpes and recurrent genital herpes from 2019 to 2021 and described the use of antiviral treatment among medically attended patients, providing some of the first claims-based evidence on the disease burden and treatment landscape of genital herpes in the United States. The standardized prevalence of genital herpes from 2019 to 2021 ranged from 236 to 280 per 100,000 person-years, whereas the standardized prevalence for recurrent genital herpes ranged from 81 to 98 per 100,000 person-years. The prevalence declined from 2019 to 2020, followed by an increase from 2020 to 2021, possibly due to a decline in health care resource utilization during the COVID-19 pandemic.14 The prevalences of genital herpes and recurrent genital herpes were highest among those in early adulthood, female patients, and those with a history of HIV infection. Approximately two-thirds of individuals with genital herpes were treated. Among those who received treatment for genital herpes, most individuals had episodic antiviral therapy rather than suppressive therapy. Episodic therapy was more common than suppressive therapy overall and among all age, sex, and immunocompromised subgroups. Moreover, episodic therapy was most frequently observed among those who were younger, female, or immunocompromised. The results of this study provide unique insights into the disease burden and treatment landscape of genital herpes across a contemporary period, which includes the COVID-19 pandemic.
To the best of our knowledge, this is the first study to estimate the prevalence of genital herpes and recurrent genital herpes in the United States using a claims database. Although not directly comparable, a few studies have estimated the number of incident or prevalent cases of genital herpes in the United States based on HSV-2 seroprevalence data. Kreisel et al.10 estimated the incidence and prevalence of genital herpes using 2015–2018 National Health and Nutrition Examination Survey (NHANES) data. In 2018, among individuals of all ages (18–49 years in the 2015–2018 NHANES cycle), the estimated median number of prevalent genital herpes cases in the United States was 18.6 million overall, 6.3 million among men, and 12.2 million among women. In 2018, the estimated median number of incident genital herpes cases was 572,000 overall, 300,000 among men, and 260,000 among women. Using NHANES data from 2015 to 2018, Spicknall et al.2 estimated a similar number of prevalent and incident genital herpes cases, and women accounted for two-thirds of the prevalent infections. Using US Census total population data from 2019 (n = 328,239,523) and the standardized prevalence estimated from our study, we estimate 919,071 prevalent cases of genital herpes and 311,828 prevalent cases of recurrent genital herpes in the United States in 2019.15 These estimates may be lower than prior estimates because of only capturing medically attended genital herpes cases defined using medical claims diagnoses. Spicknall et al. used a modeling approach based on HSV-2 seroprevalence and estimated approximately 18,547,000 prevalent cases of genital herpes in 2018 in the United States among individuals aged 18 to 49 years. This study estimated 40,961 prevalent cases of diagnosed genital herpes in 2019 in the United States among individuals aged 15 to 49 years, which approximates to <1% of genital herpes cases in the United States being diagnosed and receiving medical care. Consistent with previous findings, women accounted for about three-fourths of the prevalent genital herpes cases in our study. Of note, NHANES only estimates genital herpes cases that are caused by HSV-2 and does not estimate genital herpes cases caused by HSV-1, which is likely an underestimation of the true burden of genital herpes in the United States. Furthermore, NHANES reports patients with HSV-2 infection, not all of whom have symptomatic genital herpes. In contrast, our study provides prevalence estimates of diagnosed genital herpes, which is informative for an understanding of the burden of disease and health care resource utilization as well as identifying populations who may benefit from a therapeutic HSV-2 vaccine.
Although not curative, acyclovir, valacyclovir, and famciclovir are the recommended antiviral treatments for genital herpes and are effective in treating genital herpes symptoms when used episodically and in reducing recurrences and the risk of transmission to uninfected sexual partners when taken as daily suppressive therapy.3,8,16 There is limited evidence describing the proportions of persons with genital herpes who receive episodic or suppressive therapy. Similar to a 2004 web-based survey conducted by Patrick et al.,17 which included responses from 2075 patients across 75 countries, results of our study suggest that approximately two-thirds of patients had evidence of either episodic or suppressive therapy and one-third did not receive antiviral treatment. In addition, in our study, more individuals with genital herpes received episodic therapy (approximately 80% across all 3 years) than suppressive therapy (approximately 20% across all 4 years). Patients who were immunocompromised were also more likely to have episodic therapy than suppressive therapy. Individuals may receive episodic versus suppressive therapy for a variety of reasons, such as clinician recommendation, patient preference, frequency of outbreaks, and time from outbreak to start of treatment.8 Further research to determine the reasons for differences in treatment between patients with genital herpes may be warranted.
This study has several strengths. This was a large study that used contemporary data from commercial, Medicare Advantage, and Medicaid plans. Because of the large sample size, this study was able to estimate prevalence and treatment utilization among subgroups of interest defined by age, sex, and immunocompromised status. This data source also provided several years of data that allowed for the examination of prevalence and episodic and suppressive therapy by calendar year. However, one limitation of the study is that it included only diagnosed cases of genital herpes and may underestimate the true burden of genital herpes in the United States. Asymptomatic cases and cases of genital herpes in patients who do not seek medical care were not captured in this study. In addition, although less common than HSV-2, HSV-1 is also a cause of genital herpes, and we were unable to determine whether cases of genital herpes were due to HSV-1 or HSV-2 and, consequently, the receipt of episodic versus suppressive therapy by HSV type. In addition, we were unable to assess outcomes stratified by race or ethnicity because the dataset used in this study did not provide this information to protect patient privacy. This study included patients with continuous enrollment in a health insurance plan and therefore is not representative of the uninsured population. Furthermore, laboratory data and electronic health records were unavailable for analysis, preventing the measurement of laboratory-confirmed genital herpes. As with all claims-based measures of medication utilization, prescription fills were measured rather than medication ingestion, so it is not known how well patients adhered to their antiviral medication. Furthermore, the medications assessed in this study are not used exclusively for genital herpes treatment and may have been used for the treatment of other conditions, such as herpes zoster. To increase the likelihood that patients were being treated for genital herpes, medication utilization was only examined in patients with 1 or more genital herpes diagnosis codes within a given calendar year. Finally, the definitions of episodic and suppressive therapy in this study were defined using a data-driven approach and expert opinion along with treatment guidelines for genital herpes. The selection of different operational definitions for antiviral treatment frequency, length of treatment regimen, and total duration would result in different results.
CONCLUSIONS
In conclusion, genital herpes and recurrent genital herpes were most common among female patients in early adulthood and immunocompromised individuals, with approximately two-thirds of those with genital herpes being treated with antiviral medications in a given year. These results suggest a sizable disease burden of genital herpes in the United States and point to a need for more effective preventive and therapeutic interventions, including vaccines.
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Copyright © 2024 The Author(s). Published by Wolters Kluwer Health, Inc. on behalf of the American Sexually Transmitted Diseases Association.
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Prevalence of Genital Herpes and Antiviral Treatment
Sexually Transmitted Diseases51(10):686-693, October 2024.
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11186 | https://math.libretexts.org/Bookshelves/Differential_Equations/Elementary_Differential_Equations_with_Boundary_Value_Problems_(Trench)/07%3A_Series_Solutions_of_Linear_Second_Order_Equations/7.02%3A_Review_of_Power_Series | Skip to main content
7.2: Review of Power Series
Last updated
: Jun 23, 2024
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7.1: Prelude to Series Solutions of Linear Second Order Equations
7.2E: Review of Power Series (Exercises)
Page ID
: 9439
William F. Trench
Trinity University
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Many applications give rise to differential equations with solutions that can’t be expressed in terms of elementary functions such as polynomials, rational functions, exponential and logarithmic functions, and trigonometric functions. The solutions of some of the most important of these equations can be expressed in terms of power series. We’ll study such equations in this chapter. In this section we review relevant properties of power series. We’ll omit proofs, which can be found in any standard calculus text.
Definition 7.2.1
An infinite series of the form
∞∑n=0an(x−x0)n,
∑n=0∞an(x−x0)n,(7.2.1)
where x0x0 and a0a0, a1,a1, …, an,an, …are constants, is called a power series in x−x0.x−x0. We say that the power series Equation 7.2.17.2.1 converges for a given xx if the limit
limN→∞N∑n=0an(x−x0)n
limN→∞∑n=0Nan(x−x0)n
exists;; otherwise, we say that the power series diverges for the given x.x.
A power series in x−x0x−x0 must converge if x=x0x=x0, since the positive powers of x−x0x−x0 are all zero in this case. This may be the only value of xx for which the power series converges. However, the next theorem shows that if the power series converges for some x≠x0x≠x0 then the set of all values of xx for which it converges forms an interval.
Theorem 7.2.2
For any power series
∞∑n=0an(x−x0)n,
∑n=0∞an(x−x0)n,
exactly one of the these statements is true:
The power series converges only for x=x0.x=x0.
The power series converges for all values of x.x.
There’s a positive number RR such that the power series converges if |x−x0|R|x−x0|>R.
In case (iii) we say that RR is the radius of convergence of the power series. For convenience, we include the other two cases in this definition by defining R=0R=0 in case (i) and R=∞R=∞ in case (ii). We define the open interval of convergence of ∑∞n=0an(x−x0)n∑∞n=0an(x−x0)n to be
(x0−R,x0+R)if0<R<∞,or(−∞,∞)ifR=∞.
(x0−R,x0+R)if0<R<∞,or(−∞,∞)ifR=∞.
If RR is finite, no general statement can be made concerning convergence at the endpoints x=x0±Rx=x0±R of the open interval of convergence; the series may converge at one or both points, or diverge at both.
Recall from calculus that a series of constants ∑∞n=0αn∑∞n=0αn is said to converge absolutely if the series of absolute values ∑∞n=0|αn|∑∞n=0|αn| converges. It can be shown that a power series ∑∞n=0an(x−x0)n∑∞n=0an(x−x0)n with a positive radius of convergence RR converges absolutely in its open interval of convergence; that is, the series
∞∑n=0|an||x−x0|n
∑n=0∞|an||x−x0|n
of absolute values converges if |x−x0|<R|x−x0|<R. However, if R<∞R<∞, the series may fail to converge absolutely at an endpoint x0±Rx0±R, even if it converges here.
The next theorem provides a useful method for determining the radius of convergence of a power series. It’s derived in calculus by applying the ratio test to the corresponding series of absolute values. For related theorems see Exercises 7.2.2 and 7.2.4.
Theorem 7.2.3
Suppose there’s an integer NN such that an≠0an≠0 if n≥Nn≥N and
limn→∞|an+1an|=L,
limn→∞∣∣∣an+1an∣∣∣=L,
where 0≤L≤∞.0≤L≤∞. Then the radius of convergence of ∑∞n=0an(x−x0)n∑∞n=0an(x−x0)n is R=1/L,R=1/L, which should be interpreted to mean that R=0R=0 if L=∞,L=∞, or R=∞R=∞ if L=0L=0.
Example 7.2.1
Find the radius of convergence of the series:
∞∑n=0n!xn∑n=0∞n!xn
∞∑n=10(−1)nxnn!∑n=10∞(−1)nxnn!
∞∑n=02nn2(x−1)n.∑n=0∞2nn2(x−1)n.
Solution a
Here an=n!an=n!, so
limn→∞|an+1an|=limn→∞(n+1)!n!=limn→∞(n+1)=∞.
limn→∞∣∣∣an+1an∣∣∣=limn→∞(n+1)!n!=limn→∞(n+1)=∞.
Hence, R=0R=0.
Solution b
Here an=(1)n/n!an=(1)n/n! for n≥N=10n≥N=10, so
limn→∞|an+1an|=limn→∞n!(n+1)!=limn→∞1n+1=0.
limn→∞∣∣∣an+1an∣∣∣=limn→∞n!(n+1)!=limn→∞1n+1=0.
Hence, R=∞R=∞.
Solution c
Here an=2nn2an=2nn2, so
limn→∞|an+1an|=limn→∞2n+1(n+1)22nn2=2limn→∞(1+1n)2=2.
limn→∞∣∣∣an+1an∣∣∣=limn→∞2n+1(n+1)22nn2=2limn→∞(1+1n)2=2.
Hence, R=1/2R=1/2.
Taylor Series
If a function ff has derivatives of all orders at a point x=x0x=x0, then the Taylor series of f about x0 is defined by
∞∑n=0f(n)(x0)n!(x−x0)n.
In the special case where x0=0, this series is also called the Maclaurin series of f.
Taylor series for most of the common elementary functions converge to the functions on their open intervals of convergence. For example, you are probably familiar with the following Maclaurin series:
ex=∞∑n=0xnn!,−∞<x<∞sinx=∞∑n=0(−1)nx2n+1(2n+1)!,−∞<x<∞cosx=∞∑n=0(−1)nx2n(2n)!−∞<x<∞11−x=∞∑n=0xn−1<x<1
Differentiation of Power Series
A power series with a positive radius of convergence defines a function
f(x)=∞∑n=0an(x−x0)n
on its open interval of convergence. We say that the series represents f on the open interval of convergence. A function f represented by a power series may be a familiar elementary function as in Equations 7.2.2 - 7.2.5; however, it often happens that f isn’t a familiar function, so the series actually defines f.
The next theorem shows that a function represented by a power series has derivatives of all orders on the open interval of convergence of the power series, and provides power series representations of the derivatives.
Theorem 7.2.4 : A Power Series
A power series
f(x)=∞∑n=0an(x−x0)n
with positive radius of convergence R has derivatives of all orders in its open interval of convergence, and successive derivatives can be obtained by repeatedly differentiating term by term; that is,
f′(x)=∞∑n=1nan(x−x0)n−1,f″(x)=∞∑n=2n(n−1)an(x−x0)n−2,⋮f(k)(x)=∞∑n=kn(n−1)⋯(n−k+1)an(x−x0)n−k.
Moreover, all of these series have the same radius of convergence R.
Example 7.2.2
Let f(x)=sinx. From Equation 7.2.3,
f(x)=∞∑n=0(−1)nx2n+1(2n+1)!.
From Equation 7.2.6,
f′(x)=∞∑n=0(−1)nddx[x2n+1(2n+1)!]=∞∑n=0(−1)nx2n(2n)!,
which is the series Equation 7.2.4 for cosx.
Uniqueness of Power Series
The next theorem shows that if f is defined by a power series in x−x0 with a positive radius of convergence, then the power series is the Taylor series of f about x0.
Theorem 7.2.5
If the power series
f(x)=∞∑n=0an(x−x0)n
has a positive radius of convergence, then
an=f(n)(x0)n!;
that is, ∑∞n=0an(x−x0)n is the Taylor series of f about x0.
This result can be obtained by setting x=x0 in Equation 7.2.9, which yields
f(k)(x0)=k(k−1)⋯1⋅ak=k!ak.
This implies that
ak=f(k)(x0)k!.
Except for notation, this is the same as Equation 7.2.10.
The next theorem lists two important properties of power series that follow from Theorem 7.2.4
.
Theorem 7.2.6
If ∞∑n=0an(x−x0)n=∞∑n=0bn(x−x0)nfor all x in an open interval that contains x0, then an=bn for n=0, 1, 2, ….
If ∞∑n=0an(x−x0)n=0for all x in an open interval that contains x0, then an=0 for n=0, 1, 2, ….
To obtain (a) we observe that the two series represent the same function f on the open interval; hence, Theorem 7.2.4
implies that
an=bn=f(n)(x0)n!,n=0,1,2,….
(b) can be obtained from (a) by taking bn=0 for n=0, 1, 2, ….
Taylor Polynomials
If f has N derivatives at a point x0, we say that
TN(x)=N∑n=0f(n)(x0)n!(x−x0)n
is the N-th Taylor polynomial of f about x0. This definition and Theorem 7.2.4
imply that if
f(x)=∞∑n=0an(x−x0)n,
where the power series has a positive radius of convergence, then the Taylor polynomials of f about x0 are given by
TN(x)=N∑n=0an(x−x0)n.
In numerical applications, we use the Taylor polynomials to approximate f on subintervals of the open interval of convergence of the power series. For example, Equation 7.2.2 implies that the Taylor polynomial TN of f(x)=ex is
TN(x)=N∑n=0xnn!.
The solid curve in Figure 7.2.1
is the graph of y=ex on the interval [0,5]. The dotted curves in Figure 7.2.1
are the graphs of the Taylor polynomials T1, …, T6 of y=ex about x0=0. From this figure, we conclude that the accuracy of the approximation of y=ex by its Taylor polynomial TN improves as N increases.
Shifting the Summation Index
In Definition 7.2.1
of a power series in x−x0, the n-th term is a constant multiple of (x−x0)n. This isn’t true in Equation 7.2.6, Equation 7.2.7, and Equation 7.2.9, where the general terms are constant multiples of (x−x0)n−1, (x−x0)n−2, and (x−x0)n−k, respectively. However, these series can all be rewritten so that their n-th terms are constant multiples of (x−x0)n. For example, letting n=k+1 in the series in Equation 7.2.6 yields
f′(x)=∞∑k=0(k+1)ak+1(x−x0)k,
where we start the new summation index k from zero so that the first term in Equation 7.2.11 (obtained by setting k=0) is the same as the first term in Equation 7.2.6 (obtained by setting n=1). However, the sum of a series is independent of the symbol used to denote the summation index, just as the value of a definite integral is independent of the symbol used to denote the variable of integration. Therefore we can replace k by n in Equation 7.2.11 to obtain
f′(x)=∞∑n=0(n+1)an+1(x−x0)n,
where the general term is a constant multiple of (x−x0)n.
It isn’t really necessary to introduce the intermediate summation index k. We can obtain Equation 7.2.12 directly from Equation 7.2.6 by replacing n by n+1 in the general term of Equation 7.2.6 and subtracting 1 from the lower limit of Equation 7.2.6. More generally, we use the following procedure for shifting indices.
Shifting the Summation Index in a Power Series
For any integer k, the power series
∞∑n=n0bn(x−x0)n−k
can be rewritten as
∞∑n=n0−kbn+k(x−x0)n
that is, replacing n by n+k in the general term and subtracting k from the lower limit of summation leaves the series unchanged.
Example 7.2.3
Rewrite the following power series from Equation 7.2.7 and Equation 7.2.9 so that the general term in each is a constant multiple of (x−x0)n:
(a)∞∑n=2n(n−1)an(x−x0)n−2(b)∞∑n=kn(n−1)⋯(n−k+1)an(x−x0)n−k.
Solution a
Replacing n by n+2 in the general term and subtracting 2 from the lower limit of summation yields
∞∑n=2n(n−1)an(x−x0)n−2=∞∑n=0(n+2)(n+1)an+2(x−x0)n.
Solution b
Replacing n by n+k in the general term and subtracting k from the lower limit of summation yields
∞∑n=kn(n−1)⋯(n−k+1)an(x−x0)n−k=∞∑n=0(n+k)(n+k−1)⋯(n+1)an+k(x−x0)n.
Example 7.2.4
Given that
f(x)=∞∑n=0anxn,
write the function xf″ as a power series in which the general term is a constant multiple of xn.
Solution
From Theorem 7.2.4
with x0=0,
f″(x)=∞∑n=2n(n−1)anxn−2.
Therefore
xf″(x)=∞∑n=2n(n−1)anxn−1.
Replacing n by n+1 in the general term and subtracting 1 from the lower limit of summation yields
xf″(x)=∞∑n=1(n+1)nan+1xn.
We can also write this as
xf″(x)=∞∑n=0(n+1)nan+1xn,
since the first term in this last series is zero. (We’ll see later that sometimes it is useful to include zero terms at the beginning of a series.)
Linear Combinations of Power Series
If a power series is multiplied by a constant, then the constant can be placed inside the summation; that is,
c∞∑n=0an(x−x0)n=∞∑n=0can(x−x0)n.
Two power series
f(x)=∞∑n=0an(x−x0)n and g(x)=∞∑n=0bn(x−x0)n
with positive radii of convergence can be added term by term at points common to their open intervals of convergence; thus, if the first series converges for |x−x0|<R1 and the second converges for |x−x0|<R2, then
f(x)+g(x)=∞∑n=0(an+bn)(x−x0)n
for |x−x0|<R, where R is the smaller of R1 and R2. More generally, linear combinations of power series can be formed term by term; for example,
c1f(x)+c2f(x)=∞∑n=0(c1an+c2bn)(x−x0)n.
Example 7.2.5
Find the Maclaurin series for coshx as a linear combination of the Maclaurin series for ex and e−x.
Solution
By definition,
coshx=12ex+12e−x.
Since
ex=∞∑n=0xnn! and e−x=∞∑n=0(−1)nxnn!,
it follows that
coshx=∞∑n=012[1+(−1)n]xnn!.
Since
12[1+(−1)n]={1 if n=2m, an even integer,0 if n=2m+1, an odd integer,
we can rewrite Equation 7.2.13 more simply as
coshx=∞∑m=0x2m(2m)!.
This result is valid on (−∞,∞), since this is the open interval of convergence of the Maclaurin series for ex and e−x.
Example 7.2.6
Suppose
y=∞∑n=0anxn
on an open interval I that contains the origin.
Express (2−x)y″+2yas a power series in x on I.
Use the result of (a) to find necessary and sufficient conditions on the coefficients {an} for y to be a solution of the homogeneous equation
(2−x)y″+2y=0
on I.
Solution a
From Equation 7.2.7 with x0=0,
y″=∞∑n=2n(n−1)anxn−2.
Therefore
(2−x)y″+2y=2y″−xy′+2y=∑∞n=22n(n−1)anxn−2−∑∞n=2n(n−1)anxn−1+∑∞n=02anxn.
To combine the three series we shift indices in the first two to make their general terms constant multiples of xn; thus,
∞∑n=22n(n−1)anxn−2=∞∑n=02(n+2)(n+1)an+2xn
∞∑n=2n(n−1)anxn−1=∞∑n=1(n+1)nan+1xn=∞∑n=0(n+1)nan+1xn,
where we added a zero term in the last series so that when we substitute from Equation 7.2.16 and Equation 7.2.17 into Equation 7.2.15 all three series will start with n=0; thus,
(2−x)y″+2y=∞∑n=0[2(n+2)(n+1)an+2−(n+1)nan+1+2an]xn.
Solution b
From Equation 7.2.18 we see that y satisfies Equation 7.2.14 on I if
2(n+2)(n+1)an+2−(n+1)nan+1+2an=0,n=0,1,2,….
Conversely, Theorem 7.2.5
b implies that if y=∑∞n=0anxn satisfies Equation 7.2.14 on I, then Equation 7.2.19 holds.
Example 7.2.7
Suppose
y=∞∑n=0an(x−1)n
on an open interval I that contains x0=1. Express the function
(1+x)y″+2(x−1)2y′+3y
as a power series in x−1 on I.
Solution
Since we want a power series in x−1, we rewrite the coefficient of y″ in Equation 7.2.20 as 1+x=2+(x−1), so Equation 7.2.20 becomes
2y″+(x−1)y″+2(x−1)2y′+3y.
From Equation 7.2.6 and Equation 7.2.7 with x0=1,
y′=∞∑n=1nan(x−1)n−1 and y″=∞∑n=2n(n−1)an(x−1)n−2.
Therefore
2y″=∞∑n=22n(n−1)an(x−1)n−2,(x−1)y″=∞∑n=2n(n−1)an(x−1)n−1,2(x−1)2y′=∞∑n=12nan(x−1)n+1,3y=∞∑n=03an(x−1)n.
Before adding these four series we shift indices in the first three so that their general terms become constant multiples of (x−1)n. This yields
2y″=∞∑n=02(n+2)(n+1)an+2(x−1)n,(x−1)y″=∞∑n=0(n+1)nan+1(x−1)n,2(x−1)2y′=∞∑n=12(n−1)an−1(x−1)n,3y=∞∑n=03an(x−1)n,
where we added initial zero terms to the series in Equation 7.2.22 and Equation 7.2.23. Adding Equations 7.2.21–7.2.24 yields
(1+x)y″+2(x−1)2y′+3y=2y″+(x−1)y″+2(x−1)2y′+3y=∞∑n=0bn(x−1)n,
b0=4a2+3a0,bn=2(n+2)(n+1)an+2+(n+1)nan+1+2(n−1)an−1+3an,n≥1.
The formula Equation 7.2.25 for b0 can’t be obtained by setting n=0 in Equation 7.2.26, since the summation in Equation 7.2.23 begins with n=1, while those in Equation 7.2.21, Equation 7.2.22, and Equation 7.2.24 begin with n=0.
7.1: Prelude to Series Solutions of Linear Second Order Equations
7.2E: Review of Power Series (Exercises) |
11187 | https://www.movable-type.co.uk/scripts/latlong.html | Calculate distance and bearing between two Latitude/Longitude points using haversine formula in JavaScript
Movable Type Scripts
Calculate distance, bearing and more between Latitude/Longitude points
This page presents a variety of calculations for latitude/longitude points, with the formulas and code fragments for implementing them.
All these formulas are for calculations on the basis of a spherical earth (ignoring ellipsoidal effects) – which is accurate enough for most purposes… [In fact, the earth is very slightly ellipsoidal; using a spherical model gives errors typically up to 0.3%1 – see notes for further details].
Great-circle distance between two points
Enter the co-ordinates into the text boxes to try out the calculations. A variety of formats are accepted, principally:
deg-min-sec suffixed with N/S/E/W (e.g. 40°44′55″N, 73 59 11W), or
signed decimal degrees without compass direction, where negative indicates west/south (e.g. 40.7486, -73.9864):
Point 1:,
Point 2:,
Distance:968.9 km (to 4 SF)
Initial bearing:009° 07′ 11″
Final bearing:011° 16′ 31″
Midpoint:54° 21′ 44″ N, 004° 31′ 50″ W
And you can see it on a map
... hide map
Distance
This uses the ‘haversine’ formula to calculate the great-circle distance between two points – that is, the shortest distance over the earth’s surface – giving an ‘as-the-crow-flies’ distance between the points (ignoring any hills they fly over, of course!).
Haversine
formula:a = sin²(Δφ/2) + cos φ 1 ⋅ cos φ 2 ⋅ sin²(Δλ/2)
c = 2 ⋅ atan2( √a, √(1−a) )
d = R ⋅ c
where:φ is latitude, λ is longitude, R _is earth’s radius (mean radius = 6,371km);
note that angles need to be in radians to pass to trig functions!_
JavaScript:const R = 6371e3; // metresconst φ1 = lat1 Math.PI/180; // φ, λ in radiansconst φ2 = lat2 Math.PI/180;const Δφ = (lat2-lat1) Math.PI/180;const Δλ = (lon2-lon1) Math.PI/180;const a = Math.sin(Δφ/2) Math.sin(Δφ/2) + Math.cos(φ1) Math.cos(φ2) Math.sin(Δλ/2) Math.sin(Δλ/2);const c = 2 Math.atan2(Math.sqrt(a), Math.sqrt(1-a));const d = R c; // in metres
Note in these scripts, I generally use lat/lon for latitude/longitude in degrees, and φ/λ for latitude/longitude in radians – having found that mixing degrees & radians is often the easiest route to head-scratching bugs...
Historical aside: The height of technology for navigator’s calculations used to be log tables. As there is no (real) log of a negative number, the ‘versine’ enabled them to keep trig functions in positive numbers. Also, the sin²(θ/2) form of the haversine avoided addition (which entailed an anti-log lookup, the addition, and a log lookup). Printed tables for the haversine/inverse-haversine (and its logarithm, to aid multiplications) saved navigators from squaring sines, computing square roots, etc – arduous and error-prone activities.
The haversine formula1 ‘remains particularly well-conditioned for numerical computation even at small distances’ – unlike calculations based on the spherical law of cosines. The ‘(re)versed sine’ is 1−cosθ, and the ‘half-versed-sine’ is (1−cosθ)/2 or sin²(θ/2) as used above. Once widely used by navigators, it was described by Roger Sinnott in Sky & Telescope magazine in 1984 (“Virtues of the Haversine”): Sinnott explained that the angular separation between Mizar and Alcor in Ursa Major – 0°11′49.69″ – could be accurately calculated in Basic on a TRS-80 using the haversine.
For the curious, c is the angular distance in radians, and a is the square of half the chord length between the points.
If atan2 is not available, c could be calculated from 2 ⋅ asin( min(1, √a) ) (including protection against rounding errors).
Using Chrome on an aging Core i5 PC, a distance calculation takes around 2 – 5 microseconds (hence around 200,000 – 500,000 per second). Little to no benefit is obtained by factoring out common terms; probably the JIT compiler optimises them out.
Spherical Law of Cosines
In fact, JavaScript (and most modern computers & languages) use ‘IEEE 754’ 64-bit floating-point numbers, which provide 15 significant figures of precision. By my estimate, with this precision, the simple spherical law of cosines formula (cos c = cos a cos b + sin a sin b cos C) gives well-conditioned results down to distances as small as a few metres on the earth’s surface. (Note that the geodetic form of the law of cosines is rearranged from the canonical one so that the latitude can be used directly, rather than the colatitude).
This makes the simpler law of cosines a reasonable 1-line alternative to the haversine formula for many geodesy purposes (if not for astronomy). The choice may be driven by programming language, processor, coding context, available trig functions (in different languages), etc – and, for very small distances an equirectangular approximation may be more suitable.
Law of cosines:d = acos( sin φ 1 ⋅ sin φ 2 + cos φ 1 ⋅ cos φ 2 ⋅ cos Δλ ) ⋅ R
JavaScript:const φ1 = lat1 Math.PI/180, φ2 = lat2 Math.PI/180, Δλ = (lon2-lon1) Math.PI/180, R = 6371e3;const d = Math.acos( Math.sin(φ1)Math.sin(φ2) + Math.cos(φ1)Math.cos(φ2) Math.cos(Δλ) ) R;
Excel:=ACOS( SIN(lat1)SIN(lat2) + COS(lat1)COS(lat2)COS(lon2-lon1) ) 6371000
(or with lat/lon in degrees):=ACOS( SIN(lat1PI()/180)SIN(lat2PI()/180) + COS(lat1PI()/180)COS(lat2PI()/180)COS(lon2PI()/180-lon1PI()/180) ) 6371000
While simpler, the law of cosines is slightly slower than the haversine, in my tests.
Equirectangular approximation
If performance is an issue and accuracy less important, for small distances Pythagoras’ theorem can be used on an equirectangular projection:
Formula x = Δλ ⋅ cos φ m
y = Δφ
d = R ⋅ √x² + y²
JavaScript:const x = (λ2-λ1) Math.cos((φ1+φ2)/2);const y = (φ2-φ1);const d = Math.sqrt(xx + yy) R;
This uses just one trig and one sqrt function – as against half-a-dozen trig functions for cos law, and 7 trigs + 2 sqrts for haversine. Accuracy is somewhat complex: along meridians there are no errors, otherwise they depend on distance, bearing, and latitude, but are small enough for many purposes (and often trivial compared with the spherical approximation itself).
Alternatively, the polar coordinate flat-earth formula can be used: using the co-latitudes θ 1 = π/2−φ 1 and θ 2 = π/2−φ 2, then d = R ⋅ √θ 1 ² + θ 2 ² − 2 ⋅ θ 1 ⋅ θ 2 ⋅ cos Δλ. I’ve not compared accuracy.
Baghdad to Osaka –
not a constant bearing!
Bearing
In general, your current heading will vary as you follow a great circle path (orthodrome); the final heading will differ from the initial heading by varying degrees according to distance and latitude (if you were to go from say 35°N,45°E (≈Baghdad) to 35°N,135°E (≈Osaka), you would start on a heading of 60° and end up on a heading of 120°!).
This formula is for the initial bearing (sometimes referred to as forward azimuth) which if followed in a straight line along a great-circle arc will take you from the start point to the end point:1
Formula:θ = atan2( sin Δλ ⋅ cos φ 2 , cos φ 1 ⋅ sin φ 2 − sin φ 1 ⋅ cos φ 2 ⋅ cos Δλ )
where:φ 1 , λ 1 is the start point, φ 2 , λ 2 the end point ( Δλ is the difference in longitude)
JavaScript:
(all angles
in radians)const y = Math.sin(λ2-λ1) Math.cos(φ2);const x = Math.cos(φ1)Math.sin(φ2) - Math.sin(φ1)Math.cos(φ2)Math.cos(λ2-λ1);const θ = Math.atan2(y, x);const brng = (θ180/Math.PI + 360) % 360; // in degrees
Excel:
(all angles
in radians)=ATAN2(COS(lat1)SIN(lat2)-SIN(lat1)COS(lat2)COS(lon2-lon1),
SIN(lon2-lon1)COS(lat2))
note that Excel reverses the arguments to ATAN2 – see notes below
Since atan2 returns values in the range -π ... +π (that is, -180° ... +180°), to normalise the result to a compass bearing (in the range 0° ... 360°, with −ve values transformed into the range 180° ... 360°), convert to degrees and then use (θ+360)%360, where % is (floating point) modulo.
For final bearing, simply take the initial bearing from the end point to the start point and reverse it (using θ = (θ+180) % 360).
Midpoint
This is the half-way point along a great circle path between the two points.1
Formula:B x = cos φ 2 ⋅ cos Δλ
B y = cos φ 2 ⋅ sin Δλ
φ m = atan2( sin φ 1 + sin φ 2, √(cos φ 1 + B x)² + B y ² )
λ m = λ 1 + atan2(B y, cos(φ 1)+B x)
JavaScript:
(all angles
in radians)const Bx = Math.cos(φ2) Math.cos(λ2-λ1);const By = Math.cos(φ2) Math.sin(λ2-λ1);const φ3 = Math.atan2(Math.sin(φ1) + Math.sin(φ2), Math.sqrt( (Math.cos(φ1)+Bx)(Math.cos(φ1)+Bx) + ByBy ) );const λ3 = λ1 + Math.atan2(By, Math.cos(φ1) + Bx);
The longitude can be normalised to −180…+180 using (lon+540)%360-180
Just as the initial bearing may vary from the final bearing, the midpoint may not be located half-way between latitudes/longitudes; the midpoint between 35°N,45°E and 35°N,135°E is around 45°N,90°E.
Intermediate point
An intermediate point at any fraction along the great circle path between two points can also be calculated.1
Formula:a = sin((1−f)⋅δ) / sin δ
b = sin(f⋅δ) / sin δ
x = a ⋅ cos φ 1 ⋅ cos λ 1 + b ⋅ cos φ 2 ⋅ cos λ 2
y = a ⋅ cos φ 1 ⋅ sin λ 1 + b ⋅ cos φ 2 ⋅ sin λ 2
z = a ⋅ sin φ 1 + b ⋅ sin φ 2
φ i = atan2(z, √x² + y²)
λ i = atan2(y, x)
where:f is fraction along great circle route (f=0 is point 1, f=1 is point 2), δ is the angular distance d/R between the two points.
Destination point given distance and bearing from start point
Given a start point, initial bearing, and distance, this will calculate the destination point and final bearing travelling along a (shortest distance) great circle arc.
Destination point along great-circle given distance and bearing from start point
Start point:,
Bearing:
Distance:km
Destination point:53° 11′ 18″ N, 000° 08′ 00″ E
Final bearing:097° 30′ 52″
view map
hide map
Formula:φ 2 = asin( sin φ 1 ⋅ cos δ + cos φ 1 ⋅ sin δ ⋅ cos θ )
λ 2 = λ 1 + atan2( sin θ ⋅ sin δ ⋅ cos φ 1, cos δ − sin φ 1 ⋅ sin φ 2 )
where:φ is latitude, λ is longitude, θ is the bearing (clockwise from north), δ is the angular distance d/R ; d being the distance travelled, R the earth’s radius
JavaScript:
(all angles
in radians)const φ2 = Math.asin( Math.sin(φ1)Math.cos(d/R) + Math.cos(φ1)Math.sin(d/R)Math.cos(brng) );const λ2 = λ1 + Math.atan2(Math.sin(brng)Math.sin(d/R)Math.cos(φ1), Math.cos(d/R)-Math.sin(φ1)Math.sin(φ2));
The longitude can be normalised to −180…+180 using (lon+540)%360-180
Excel:
(all angles
in radians)lat2: =ASIN(SIN(lat1)COS(d/R) + COS(lat1)SIN(d/R)COS(brng))
lon2: =lon1 + ATAN2(COS(d/R)-SIN(lat1)SIN(lat2), SIN(brng)SIN(d/R)COS(lat1))
Remember that Excel reverses the arguments to ATAN2 – see notes below
For final bearing, simply take the initial bearing from the end point to the start point and reverse it with (brng+180)%360.
Intersection of two paths given start points and bearings
This is a rather more complex calculation than most others on this page, but I've been asked for it a number of times. This comes from Ed William’s aviation formulary. See below for the JavaScript.
Intersection of two great-circle paths
Point 1:, Brng 1:
Point 2:, Brng 2:
Intersection point:50° 54′ 27″ N, 004° 30′ 31″ E
Formula:δ 12 = 2⋅asin( √(sin²(Δφ/2) + cos φ 1 ⋅ cos φ 2 ⋅ sin²(Δλ/2)) )angular dist. p1–p2
θ a = acos( ( sin φ 2 − sin φ 1 ⋅ cos δ 12 ) / ( sin δ 12 ⋅ cos φ 1 ) )
θ b = acos( ( sin φ 1 − sin φ 2 ⋅ cos δ 12 ) / ( sin δ 12 ⋅ cos φ 2 ) )initial / final bearings
between points 1 & 2
if sin(λ 2−λ 1) > 0
θ 12 = θ a
θ 21 = 2π − θ b
else
θ 12 = 2π − θ a
θ 21 = θ b
α 1 = θ 13 − θ 12
α 2 = θ 21 − θ 23 angle p2–p1–p3
angle p1–p2–p3
α 3 = acos( −cos α 1 ⋅ cos α 2 + sin α 1 ⋅ sin α 2 ⋅ cos δ 12 )angle p1–p2–p3
δ 13 = atan2( sin δ 12 ⋅ sin α 1 ⋅ sin α 2 , cos α 2 + cos α 1 ⋅ cos α 3 )angular dist. p1–p3
φ 3 = asin( sin φ 1 ⋅ cos δ 13 + cos φ 1 ⋅ sin δ 13 ⋅ cos θ 13 )p3 lat
Δλ 13 = atan2( sin θ 13 ⋅ sin δ 13 ⋅ cos φ 1 , cos δ 13 − sin φ 1 ⋅ sin φ 3 )long p1–p3
λ 3 = λ 1 + Δλ 13 p3 long
where _φ 1, λ 1, θ 13 : 1st start point & (initial) bearing from 1st point towards intersection point
φ 2, λ 2, θ 23 : 2nd start point & (initial) bearing from 2nd point towards intersection point
φ 3, λ 3 : intersection point_
% = (floating point) modulo
note –if sin α 1 = 0 and sin α 2 = 0: infinite solutions
if sin α 1 ⋅ sin α 2< 0: ambiguous solution
this formulation is not always well-conditioned for meridional or equatorial lines
This is a lot simpler using vectors rather than spherical trigonometry: see latlong-vectors.html.
Cross-track distance
Here’s a new one: I’ve sometimes been asked about distance of a point from a great-circle path (sometimes called cross track error).
Formula:d xt = asin( sin(δ 13) ⋅ sin(θ 13−θ 12) ) ⋅ R
where _δ 13 is (angular) distance from start point to third point
θ 13 is (initial) bearing from start point to third point
θ 12 is (initial) bearing from start point to end point
R is the earth’s radius_
JavaScript:const δ13 = d13 / R;const dXt = Math.asin(Math.sin(δ13)Math.sin(θ13-θ12)) R;
Here, the great-circle path is identified by a start point and an end point – depending on what initial data you’re working from, you can use the formulas above to obtain the relevant distance and bearings. The sign of d xt tells you which side of the path the third point is on.
The along-track distance, from the start point to the closest point on the path to the third point, is
Formula:d at = acos( cos(δ 13) / cos(δ xt) ) ⋅ R
where _δ 13 is (angular) distance from start point to third point
δ xt is (angular) cross-track distance
R is the earth’s radius_
JavaScript:const δ13 = d13 / R;const dAt = Math.acos(Math.cos(δ13)/Math.cos(dXt/R)) R;
Closest point to the poles
And: ‘Clairaut’s formula’ will give you the maximum latitude of a great circle path, given a bearing θ and latitude φ on the great circle:
Formula:φ max = acos( | sin θ ⋅ cos φ | )
JavaScript:const φMax = Math.acos(Math.abs(Math.sin(θ)Math.cos(φ)));
Rhumb lines
A ‘rhumb line’ (or loxodrome) is a path of constant bearing, which crosses all meridians at the same angle.
Sailors used to (and sometimes still) navigate along rhumb lines since it is easier to follow a constant compass bearing than to be continually adjusting the bearing, as is needed to follow a great circle. Rhumb lines are straight lines on a Mercator Projection map (also helpful for navigation).
Rhumb lines are generally longer than great-circle (orthodrome) routes. For instance, London to New York is 4% longer along a rhumb line than along a great circle – important for aviation fuel, but not particularly to sailing vessels. New York to Beijing – close to the most extreme example possible (though not sailable!) – is 30% longer along a rhumb line.
Rhumb-line distance between two points
Point 1:,
Point 2:,
Distance:5198 km
Bearing:260° 07′ 38″
Midpoint:46° 21′ 32″ N, 038° 49′ 00″ W
view map
hide map
Destination point along rhumb line given distance and bearing from start point
Start point:,
Bearing:
Distance:km
Destination point:50° 57′ 48″ N, 001° 51′ 09″ E
view map
hide map
Key to calculations of rhumb lines is the inverse Gudermannian function¹, which gives the ‘isometric latitude’ (ψ), equivalent to the height on a Mercator projection map of a given geodetic latitude – this can be expessed in various ways:
ψ = ln( tan(π/4 + φ/2) )
ψ = ln( tanφ + secφ ) [i.e. ln( tanφ + 1/cosφ )]
ψ = asinh( tanφ ) [≡ atanh( sinφ )]
The isometric latitude of course tends to infinity at the poles (in keeping with the Mercator projection).
For obsessives, this is a simplification of the full ellipsoidal version:
ψ = ln( tan(π/4+φ/2) ⋅ [ (1−e⋅sinφ) / (1+e⋅sinφ)]e/2)
ψ = asinh(tanφ) − e⋅atanh(e⋅sinφ)
where e is the eccentricity
The formulas to derive Mercator projection easting and northing coordinates from spherical latitude and longitude are then:
E = R ⋅ λ
N = R ⋅ ln( tan(π/4 + φ/2) )
The following formulas are from Ed Williams’ aviation formulary¹.
Distance
Since a rhumb line is a straight line on a Mercator projection, the distance between two points along a rhumb line is the length of that line (by Pythagoras); but the distortion of the projection needs to be compensated for.
On a constant latitude course (travelling east-west), this compensation is simply cosφ; in the general case, it is Δφ/Δψ where Δψ is the ‘isometric’ (projected) latitude difference.
Formula:Δψ = ln( tan(π/4 + φ 2/2) / tan(π/4 + φ 1/2) )
q = Δφ / Δψ (or cosφ for E-W line)
d = √(Δφ² + q²⋅Δλ²) ⋅ R(Pythagoras)
where:φ is geodetic latitude, ψ is isometric latitude, λ is longitude, Δλ is taking shortest route (<180°), R is the earth’s radius, ln is natural log
JavaScript:
(all angles
in radians)const Δψ = Math.log(Math.tan(Math.PI/4+φ2/2) / Math.tan(Math.PI/4+φ1/2));const q = Number.isFinite(Δφ/Δψ) ? Δφ/Δψ : Math.cos(φ1); // E-W course becomes ill-conditioned with 0/0// if dLon over 180° take shorter rhumb line across the anti-meridian:if (Math.abs(Δλ) > Math.PI) Δλ = Δλ>0 ? -(2Math.PI-Δλ) : (2Math.PI+Δλ);const dist = Math.sqrt(ΔφΔφ + qqΔλΔλ) R;
Bearing
A rhumb line is a straight line on a Mercator projection, with an angle on the projection equal to the compass bearing.
Formula:Δψ = ln( tan(π/4 + φ 2/2) / tan(π/4 + φ 1/2) )
θ = atan2(Δλ, Δψ)
where:φ is geodetic latitude, ψ is isometric latitude, λ is longitude, Δλ is taking shortest route (<180°), R is the earth’s radius, ln is natural log
JavaScript:
(all angles
in radians)const Δψ = Math.log(Math.tan(Math.PI/4+φ2/2)/Math.tan(Math.PI/4+φ1/2));// if dLon over 180° take shorter rhumb line across the anti-meridian:if (Math.abs(Δλ) > Math.PI) Δλ = Δλ>0 ? -(2Math.PI-Δλ) : (2Math.PI+Δλ);const brng = Math.atan2(Δλ, Δψ) 180/Math.PI;
Destination
Given a start point and a distance d along constant bearing θ, this will calculate the destination point. If you maintain a constant bearing along a rhumb line, you will gradually spiral in towards one of the poles.
Formula:φ 2 = φ 1 + δ ⋅ cos θ
Δψ = ln( tan(π/4 + φ 2/2) / tan(π/4 + φ 1/2) )
q = Δφ / Δψ (or cos φ for E-W line)
Δλ = δ ⋅ sin θ / q
λ 2 = λ 1 + Δλ
where:δ is angular distance, φ is geodetic latitude, ψ is isometric latitude, λ is longitude, Δλ is taking shortest route (<180°), ln is natural log, R is the earth’s radius
JavaScript:
(all angles
in radians)const δ = d/R;const Δφ = δ Math.cos(θ);const φ2 = φ1 + Δφ;const Δψ = Math.log(Math.tan(φ2/2+Math.PI/4)/Math.tan(φ1/2+Math.PI/4));const q = Math.abs(Δψ) > 10e-12 ? Δφ / Δψ : Math.cos(φ1); // E-W course becomes ill-conditioned with 0/0const Δλ = δMath.sin(θ)/q;const λ2 = λ1 + Δλ;// check for some daft bugger going past the pole, normalise latitude if soif (Math.abs(φ2) > Math.PI/2) φ2 = φ2>0 ? Math.PI-φ2 : -Math.PI-φ2;
The longitude can be normalised to −180…+180 using (lon+540)%360-180
Mid-point
This formula for calculating the ‘loxodromic midpoint’, the point half-way along a rhumb line between two points, is due to Robert Hill and Clive Tooth1 (thx Axel!).
Formula:φ m = (φ 1+φ 2) / 2
f 1 = tan(π/4 + φ 1/2)
f 2 = tan(π/4 + φ 2/2)
f m = tan(π/4+φ m/2)
λ m = [ (λ 2−λ 1) ⋅ ln(f m) + λ 1 ⋅ ln(f 2) − λ 2 ⋅ ln(f 1) ] / ln(f 2/f 1)
where:φ is latitude, λ is longitude, ln is natural log
JavaScript:
(all angles
in radians)if (Math.abs(λ2-λ1) > Math.PI) λ1 += 2Math.PI; // crossing anti-meridianconst φ3 = (φ1+φ2)/2;const f1 = Math.tan(Math.PI/4 + φ1/2);const f2 = Math.tan(Math.PI/4 + φ2/2);const f3 = Math.tan(Math.PI/4 + φ3/2);const λ3 = ( (λ2-λ1)Math.log(f3) + λ1Math.log(f2) - λ2Math.log(f1) ) / Math.log(f2/f1);if (!isFinite(λ3)) λ3 = (λ1+λ2)/2; // parallel of latitude
The longitude can be normalised to −180…+180 using (lon+540)%360-180
Using the scripts in web pages
Using these scripts in web pages would be something like the following:
html
Using the scripts in web pages import LatLon from ' document.addEventListener('DOMContentLoaded', function() { document.querySelector('#calc-dist').onclick = function() { calculateDistance(); } }); function calculateDistance() { const p1 = LatLon.parse(document.querySelector('#point1').value); const p2 = LatLon.parse(document.querySelector('#point2').value); const dist = parseFloat(p1.distanceTo(p2).toPrecision(4)); document.querySelector('#result-distance').textContent = dist + ' metres'; } Point 1: Point 2: Calculate distance
Convert between degrees-minutes-seconds & decimal degrees
Latitude Longitude
d 1° ≈ 111 km (110.57 eq’l — 111.70 polar)
dm 1′ ≈ 1.85 km (= 1 nm)0.01° ≈ 1.11 km
dms 1″ ≈ 30.9 m 0.0001° ≈ 11.1 m
Display calculation results as: degrees deg/min deg/min/sec
Notes:
Accuracy: since the earth is not quite a sphere, there are small errors in using spherical geometry; the earth is actually roughly ellipsoidal (or more precisely, oblate spheroidal) with a radius varying between about 6,378km (equatorial) and 6,357km (polar), and local radius of curvature varying from 6,336km (equatorial meridian) to 6,399km (polar). 6,371 km is the generally accepted value for the earth’s mean radius. This means that errors from assuming spherical geometry might be up to 0.55% crossing the equator, though generally below 0.3%, depending on latitude and direction of travel (whuber explores this in excellent detail on stackexchange). An accuracy of better than 3m in 1km is mostly good enough for me, but if you want greater accuracy, you could use the Vincenty formula for calculating geodesic distances on ellipsoids, which gives results accurate to within 1mm. (Out of sheer perversity – I’ve never needed such accuracy – I looked up this formula and discovered the JavaScript implementation was simpler than I expected).
Trig functions take arguments in radians, so latitude, longitude, and bearings in degrees (either decimal or degrees/minutes/seconds) need to be converted to radians, rad = deg⋅π/180. When converting radians back to degrees (deg = rad⋅180/π), West is negative if using signed decimal degrees. For bearings, values in the range -π to +π [-180° to +180°] need to be converted to 0 to +2π [0°–360°]; this can be done by (brng+360)%360 where % is the (floating point) modulo operator (note that different languages implement the modulo operation in different ways).
All bearings are with respect to true north, 0°=N, 90°=E, etc; if you are working from a compass, magnetic north varies from true north in a complex way around the earth, and the difference has to be compensated for by variances indicated on local maps.
The atan2() function widely used here takes two arguments, atan2(y, x), and computes the arc tangent of the ratio y/x. It is more flexible than atan(y/x), since it handles x=0, and it also returns values in all 4 quadrants -π to +π (the atan function returns values in the range -π/2 to +π/2).
If you implement any formula involving atan2 in a spreadsheet (Microsoft Excel, LibreOffice Calc, Google Sheets, Apple Numbers), you will need to reverse the arguments, as Excel etc have them the opposite way around from JavaScript – conventional order is atan2(y, x), but Excel uses atan2(x, y). To use atan2 in a (VBA) macro, you can use WorksheetFunction.Atan2().
If you are using Google Maps, several of these functions are now provided in the Google Maps API V3 ‘spherical’ library (computeDistanceBetween(), computeHeading(), computeOffset(), interpolate(), etc; note they use a default Earth radius of 6,378,137 meters).
If you use UK Ordnance Survey Grid References, I have implemented a script for converting between Lat/Long & OS Grid References.
If you use UTM coordinates or MGRS grid references, I have implemented scripts for converting between Lat/Long, UTM, & MGRS.
I learned a lot from the US Census Bureau GIS FAQ which is no longer available, so I’ve made a copy.
Thanks to Ed Williams’ Aviation Formulary for many of the formulas.
For miles, divide km by 1.609344
For nautical miles, divide km by 1.852
See below for the JavaScript source code, also available on GitHub. Full documentation is available, as well as a test suite.
Note I use Greek letters in variables representing maths symbols conventionally presented as Greek letters: I value the great benefit in legibility over the minor inconvenience in typing (if you encounter any problems, ensure your <head> includes <meta charset="utf-8">), and use UTF-8 encoding when saving files).
With its untyped C-style syntax, JavaScript reads remarkably close to pseudo-code: exposing the algorithms with a minimum of syntactic distractions. These functions should be simple to translate into other languages if required, though can also be used as-is in browsers and Node.js.
For convenience & clarity, I have extended the base JavaScript Number object with toRadians() and toDegrees() methods: I don’t see great likelihood of conflicts, as these are ubiquitous operations.
I also have a page illustrating the use of the spherical law of cosines for selecting points from a database within a specified bounding circle – the example is based on MySQL+PDO, but should be extensible to other DBMS platforms.
Several people have asked about example Excel spreadsheets, so I have implemented the distance & bearing and the destination point formulas as spreadsheets, in a form which breaks down the all stages involved to illustrate the operation.
February 2019: I have refactored the library to use ES modules, as well as extending it in scope; see the GitHub README and CHANGELOG for details.
Performance: as noted above, the haversine distance calculation takes around 2 – 5 microseconds (hence around 200,000 – 500,000 per second). I have yet to complete timing tests on other calculations.
Other languages: I cannot support translations into other languages, but if you have ported the code to another language, I am happy to provide links here.
Brian Lambert has made an Objective-C version.
Jean Brouwers has made a Python version.
Vahan Aghajanyan has made a C++ version.
Bahadır Arslan has made a Kotlin version and a Dart version.
I offer these scripts for free use and adaptation to balance my debt to the open-source info-verse. You are welcome to re-use these scripts [under an MIT licence, without any warranty express or implied] provided solely that you retain my copyright notice and a link to this page.
If you would like to show your appreciation and support continued development of these scripts, I would most gratefully accept donations.
If you need any advice or development work done, I am available for consultancy.
If you have any queries or find any problems, contact me at ku.oc.epyt-elbavom@oeg-stpircs.
© 2002-2022 Chris Veness
/ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - // Latitude/longitude spherical geodesy tools (c) Chris Veness 2002-2021 // MIT Licence // www.movable-type.co.uk/scripts/latlong.html // www.movable-type.co.uk/scripts/geodesy-library.html#latlon-spherical // - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - /import Dms from './dms.js';const π = Math.PI;/ Library of geodesy functions for operations on a spherical earth model. Includes distances, bearings, destinations, etc, for both great circle paths and rhumb lines, and other related functions. All calculations are done using simple spherical trigonometric formulae. @module latlon-spherical /// note greek letters (e.g. φ, λ, θ) are used for angles in radians to distinguish from angles in// degrees (e.g. lat, lon, brng)/ LatLonSpherical - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - // Latitude/longitude points on a spherical model earth, and methods for calculating distances, bearings, destinations, etc on (orthodromic) great-circle paths and (loxodromic) rhumb lines. /class LatLonSpherical { / Creates a latitude/longitude point on the earth’s surface, using a spherical model earth. @param {number} lat - Latitude (in degrees). @param {number} lon - Longitude (in degrees). @throws {TypeError} Invalid lat/lon. @example import LatLon from '/js/geodesy/latlon-spherical.js'; const p = new LatLon(52.205, 0.119); / constructor(lat, lon) { if (isNaN(lat)) throw new TypeError(invalid lat ‘${lat}’); if (isNaN(lon)) throw new TypeError(invalid lon ‘${lon}’); this._lat = Dms.wrap90(Number(lat)); this._lon = Dms.wrap180(Number(lon)); } / Latitude in degrees north from equator (including aliases lat, latitude): can be set as numeric or hexagesimal (deg-min-sec); returned as numeric. / get lat() { return this._lat; } get latitude() { return this._lat; } set lat(lat) { this._lat = isNaN(lat) ? Dms.wrap90(Dms.parse(lat)) : Dms.wrap90(Number(lat)); if (isNaN(this._lat)) throw new TypeError(invalid lat ‘${lat}’); } set latitude(lat) { this._lat = isNaN(lat) ? Dms.wrap90(Dms.parse(lat)) : Dms.wrap90(Number(lat)); if (isNaN(this._lat)) throw new TypeError(invalid latitude ‘${lat}’); } / Longitude in degrees east from international reference meridian (including aliases lon, lng, longitude): can be set as numeric or hexagesimal (deg-min-sec); returned as numeric. / get lon() { return this._lon; } get lng() { return this._lon; } get longitude() { return this._lon; } set lon(lon) { this._lon = isNaN(lon) ? Dms.wrap180(Dms.parse(lon)) : Dms.wrap180(Number(lon)); if (isNaN(this._lon)) throw new TypeError(invalid lon ‘${lon}’); } set lng(lon) { this._lon = isNaN(lon) ? Dms.wrap180(Dms.parse(lon)) : Dms.wrap180(Number(lon)); if (isNaN(this._lon)) throw new TypeError(invalid lng ‘${lon}’); } set longitude(lon) { this._lon = isNaN(lon) ? Dms.wrap180(Dms.parse(lon)) : Dms.wrap180(Number(lon)); if (isNaN(this._lon)) throw new TypeError(invalid longitude ‘${lon}’); } / Conversion factors; 1000 LatLon.metresToKm gives 1. / static get metresToKm() { return 1/1000; } / Conversion factors; 1000 LatLon.metresToMiles gives 0.621371192237334. / static get metresToMiles() { return 1/1609.344; } / Conversion factors; 1000 LatLon.metresToMiles gives 0.5399568034557236. / static get metresToNauticalMiles() { return 1/1852; } / Parses a latitude/longitude point from a variety of formats. Latitude & longitude (in degrees) can be supplied as two separate parameters, as a single comma-separated lat/lon string, or as a single object with { lat, lon } or GeoJSON properties. The latitude/longitude values may be numeric or strings; they may be signed decimal or deg-min-sec (hexagesimal) suffixed by compass direction (NSEW); a variety of separators are accepted. Examples -3.62, '3 37 12W', '3°37′12″W'. Thousands/decimal separators must be comma/dot; use Dms.fromLocale to convert locale-specific thousands/decimal separators. @param {number|string|Object} lat|latlon - Latitude (in degrees) or comma-separated lat/lon or lat/lon object. @param {number|string} [lon] - Longitude (in degrees). @returns {LatLon} Latitude/longitude point. @throws {TypeError} Invalid point. @example const p1 = LatLon.parse(52.205, 0.119); // numeric pair (≡ new LatLon) const p2 = LatLon.parse('52.205', '0.119'); // numeric string pair (≡ new LatLon) const p3 = LatLon.parse('52.205, 0.119'); // single string numerics const p4 = LatLon.parse('52°12′18.0″N', '000°07′08.4″E'); // DMS pair const p5 = LatLon.parse('52°12′18.0″N, 000°07′08.4″E'); // single string DMS const p6 = LatLon.parse({ lat: 52.205, lon: 0.119 }); // { lat, lon } object numeric const p7 = LatLon.parse({ lat: '52°12′18.0″N', lng: '000°07′08.4″E' }); // { lat, lng } object DMS const p8 = LatLon.parse({ type: 'Point', coordinates: [ 0.119, 52.205] }); // GeoJSON / static parse(...args) { if (args.length == 0) throw new TypeError('invalid (empty) point'); if (args===null || args===null) throw new TypeError('invalid (null) point'); let lat=undefined, lon=undefined; if (args.length == 2) { // regular (lat, lon) arguments [ lat, lon ] = args; lat = Dms.wrap90(Dms.parse(lat)); lon = Dms.wrap180(Dms.parse(lon)); if (isNaN(lat) || isNaN(lon)) throw new TypeError(invalid point ‘${args.toString()}’); } if (args.length == 1 && typeof args == 'string') { // single comma-separated lat,lon string [ lat, lon ] = args.split(','); lat = Dms.wrap90(Dms.parse(lat)); lon = Dms.wrap180(Dms.parse(lon)); if (isNaN(lat) || isNaN(lon)) throw new TypeError(invalid point ‘${args}’); } if (args.length == 1 && typeof args == 'object') { // single { lat, lon } object const ll = args; if (ll.type == 'Point' && Array.isArray(ll.coordinates)) { // GeoJSON [ lon, lat ] = ll.coordinates; } else { // regular { lat, lon } object if (ll.latitude != undefined) lat = ll.latitude; if (ll.lat != undefined) lat = ll.lat; if (ll.longitude != undefined) lon = ll.longitude; if (ll.lng != undefined) lon = ll.lng; if (ll.lon != undefined) lon = ll.lon; lat = Dms.wrap90(Dms.parse(lat)); lon = Dms.wrap180(Dms.parse(lon)); } if (isNaN(lat) || isNaN(lon)) throw new TypeError(invalid point ‘${JSON.stringify(args)}’); } if (isNaN(lat) || isNaN(lon)) throw new TypeError(invalid point ‘${args.toString()}’); return new LatLonSpherical(lat, lon); } / Returns the distance along the surface of the earth from ‘this’ point to destination point. Uses haversine formula: a = sin²(Δφ/2) + cosφ1·cosφ2 · sin²(Δλ/2); d = 2 · atan2(√a, √(a-1)). @param {LatLon} point - Latitude/longitude of destination point. @param {number} [radius=6371e3] - Radius of earth (defaults to mean radius in metres). @returns {number} Distance between this point and destination point, in same units as radius. @throws {TypeError} Invalid radius. @example const p1 = new LatLon(52.205, 0.119); const p2 = new LatLon(48.857, 2.351); const d = p1.distanceTo(p2); // 404.3×10³ m const m = p1.distanceTo(p2, 3959); // 251.2 miles / distanceTo(point, radius=6371e3) { if (!(point instanceof LatLonSpherical)) point = LatLonSpherical.parse(point); // allow literal forms if (isNaN(radius)) throw new TypeError(invalid radius ‘${radius}’); // a = sin²(Δφ/2) + cos(φ1)⋅cos(φ2)⋅sin²(Δλ/2) // δ = 2·atan2(√(a), √(1−a)) // see mathforum.org/library/drmath/view/51879.html for derivation const R = radius; const φ1 = this.lat.toRadians(), λ1 = this.lon.toRadians(); const φ2 = point.lat.toRadians(), λ2 = point.lon.toRadians(); const Δφ = φ2 - φ1; const Δλ = λ2 - λ1; const a = Math.sin(Δφ/2)Math.sin(Δφ/2) + Math.cos(φ1)Math.cos(φ2) Math.sin(Δλ/2)Math.sin(Δλ/2); const c = 2 Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); const d = R c; return d; } / Returns the initial bearing from ‘this’ point to destination point. @param {LatLon} point - Latitude/longitude of destination point. @returns {number} Initial bearing in degrees from north (0°..360°). @example const p1 = new LatLon(52.205, 0.119); const p2 = new LatLon(48.857, 2.351); const b1 = p1.initialBearingTo(p2); // 156.2° / initialBearingTo(point) { if (!(point instanceof LatLonSpherical)) point = LatLonSpherical.parse(point); // allow literal forms if (this.equals(point)) return NaN; // coincident points // tanθ = sinΔλ⋅cosφ2 / cosφ1⋅sinφ2 − sinφ1⋅cosφ2⋅cosΔλ // see mathforum.org/library/drmath/view/55417.html for derivation const φ1 = this.lat.toRadians(); const φ2 = point.lat.toRadians(); const Δλ = (point.lon - this.lon).toRadians(); const x = Math.cos(φ1) Math.sin(φ2) - Math.sin(φ1) Math.cos(φ2) Math.cos(Δλ); const y = Math.sin(Δλ) Math.cos(φ2); const θ = Math.atan2(y, x); const bearing = θ.toDegrees(); return Dms.wrap360(bearing); } / Returns final bearing arriving at destination point from ‘this’ point; the final bearing will differ from the initial bearing by varying degrees according to distance and latitude. @param {LatLon} point - Latitude/longitude of destination point. @returns {number} Final bearing in degrees from north (0°..360°). @example const p1 = new LatLon(52.205, 0.119); const p2 = new LatLon(48.857, 2.351); const b2 = p1.finalBearingTo(p2); // 157.9° / finalBearingTo(point) { if (!(point instanceof LatLonSpherical)) point = LatLonSpherical.parse(point); // allow literal forms // get initial bearing from destination point to this point & reverse it by adding 180° const bearing = point.initialBearingTo(this) + 180; return Dms.wrap360(bearing); } / Returns the midpoint between ‘this’ point and destination point. @param {LatLon} point - Latitude/longitude of destination point. @returns {LatLon} Midpoint between this point and destination point. @example const p1 = new LatLon(52.205, 0.119); const p2 = new LatLon(48.857, 2.351); const pMid = p1.midpointTo(p2); // 50.5363°N, 001.2746°E / midpointTo(point) { if (!(point instanceof LatLonSpherical)) point = LatLonSpherical.parse(point); // allow literal forms // φm = atan2( sinφ1 + sinφ2, √( (cosφ1 + cosφ2⋅cosΔλ)² + cos²φ2⋅sin²Δλ ) ) // λm = λ1 + atan2(cosφ2⋅sinΔλ, cosφ1 + cosφ2⋅cosΔλ) // midpoint is sum of vectors to two points: mathforum.org/library/drmath/view/51822.html const φ1 = this.lat.toRadians(); const λ1 = this.lon.toRadians(); const φ2 = point.lat.toRadians(); const Δλ = (point.lon - this.lon).toRadians(); // get cartesian coordinates for the two points const A = { x: Math.cos(φ1), y: 0, z: Math.sin(φ1) }; // place point A on prime meridian y=0 const B = { x: Math.cos(φ2)Math.cos(Δλ), y: Math.cos(φ2)Math.sin(Δλ), z: Math.sin(φ2) }; // vector to midpoint is sum of vectors to two points (no need to normalise) const C = { x: A.x + B.x, y: A.y + B.y, z: A.z + B.z }; const φm = Math.atan2(C.z, Math.sqrt(C.xC.x + C.yC.y)); const λm = λ1 + Math.atan2(C.y, C.x); const lat = φm.toDegrees(); const lon = λm.toDegrees(); return new LatLonSpherical(lat, lon); } / Returns the point at given fraction between ‘this’ point and given point. @param {LatLon} point - Latitude/longitude of destination point. @param {number} fraction - Fraction between the two points (0 = this point, 1 = specified point). @returns {LatLon} Intermediate point between this point and destination point. @example const p1 = new LatLon(52.205, 0.119); const p2 = new LatLon(48.857, 2.351); const pInt = p1.intermediatePointTo(p2, 0.25); // 51.3721°N, 000.7073°E / intermediatePointTo(point, fraction) { if (!(point instanceof LatLonSpherical)) point = LatLonSpherical.parse(point); // allow literal forms if (this.equals(point)) return new LatLonSpherical(this.lat, this.lon); // coincident points const φ1 = this.lat.toRadians(), λ1 = this.lon.toRadians(); const φ2 = point.lat.toRadians(), λ2 = point.lon.toRadians(); // distance between points const Δφ = φ2 - φ1; const Δλ = λ2 - λ1; const a = Math.sin(Δφ/2) Math.sin(Δφ/2) + Math.cos(φ1) Math.cos(φ2) Math.sin(Δλ/2) Math.sin(Δλ/2); const δ = 2 Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); const A = Math.sin((1-fraction)δ) / Math.sin(δ); const B = Math.sin(fractionδ) / Math.sin(δ); const x = A Math.cos(φ1) Math.cos(λ1) + B Math.cos(φ2) Math.cos(λ2); const y = A Math.cos(φ1) Math.sin(λ1) + B Math.cos(φ2) Math.sin(λ2); const z = A Math.sin(φ1) + B Math.sin(φ2); const φ3 = Math.atan2(z, Math.sqrt(xx + yy)); const λ3 = Math.atan2(y, x); const lat = φ3.toDegrees(); const lon = λ3.toDegrees(); return new LatLonSpherical(lat, lon); } / Returns the destination point from ‘this’ point having travelled the given distance on the given initial bearing (bearing normally varies around path followed). @param {number} distance - Distance travelled, in same units as earth radius (default: metres). @param {number} bearing - Initial bearing in degrees from north. @param {number} [radius=6371e3] - (Mean) radius of earth (defaults to radius in metres). @returns {LatLon} Destination point. @example const p1 = new LatLon(51.47788, -0.00147); const p2 = p1.destinationPoint(7794, 300.7); // 51.5136°N, 000.0983°W / destinationPoint(distance, bearing, radius=6371e3) { // sinφ2 = sinφ1⋅cosδ + cosφ1⋅sinδ⋅cosθ // tanΔλ = sinθ⋅sinδ⋅cosφ1 / cosδ−sinφ1⋅sinφ2 // see mathforum.org/library/drmath/view/52049.html for derivation const δ = distance / radius; // angular distance in radians const θ = Number(bearing).toRadians(); const φ1 = this.lat.toRadians(), λ1 = this.lon.toRadians(); const sinφ2 = Math.sin(φ1) Math.cos(δ) + Math.cos(φ1) Math.sin(δ) Math.cos(θ); const φ2 = Math.asin(sinφ2); const y = Math.sin(θ) Math.sin(δ) Math.cos(φ1); const x = Math.cos(δ) - Math.sin(φ1) sinφ2; const λ2 = λ1 + Math.atan2(y, x); const lat = φ2.toDegrees(); const lon = λ2.toDegrees(); return new LatLonSpherical(lat, lon); } / Returns the point of intersection of two paths defined by point and bearing. @param {LatLon} p1 - First point. @param {number} brng1 - Initial bearing from first point. @param {LatLon} p2 - Second point. @param {number} brng2 - Initial bearing from second point. @returns {LatLon|null} Destination point (null if no unique intersection defined). @example const p1 = new LatLon(51.8853, 0.2545), brng1 = 108.547; const p2 = new LatLon(49.0034, 2.5735), brng2 = 32.435; const pInt = LatLon.intersection(p1, brng1, p2, brng2); // 50.9078°N, 004.5084°E / static intersection(p1, brng1, p2, brng2) { if (!(p1 instanceof LatLonSpherical)) p1 = LatLonSpherical.parse(p1); // allow literal forms if (!(p2 instanceof LatLonSpherical)) p2 = LatLonSpherical.parse(p2); // allow literal forms if (isNaN(brng1)) throw new TypeError(invalid brng1 ‘${brng1}’); if (isNaN(brng2)) throw new TypeError(invalid brng2 ‘${brng2}’); // see www.edwilliams.org/avform.htm#Intersection const φ1 = p1.lat.toRadians(), λ1 = p1.lon.toRadians(); const φ2 = p2.lat.toRadians(), λ2 = p2.lon.toRadians(); const θ13 = Number(brng1).toRadians(), θ23 = Number(brng2).toRadians(); const Δφ = φ2 - φ1, Δλ = λ2 - λ1; // angular distance p1-p2 const δ12 = 2 Math.asin(Math.sqrt(Math.sin(Δφ/2) Math.sin(Δφ/2) + Math.cos(φ1) Math.cos(φ2) Math.sin(Δλ/2) Math.sin(Δλ/2))); if (Math.abs(δ12) < Number.EPSILON) return new LatLonSpherical(p1.lat, p1.lon); // coincident points // initial/final bearings between points const cosθa = (Math.sin(φ2) - Math.sin(φ1)Math.cos(δ12)) / (Math.sin(δ12)Math.cos(φ1)); const cosθb = (Math.sin(φ1) - Math.sin(φ2)Math.cos(δ12)) / (Math.sin(δ12)Math.cos(φ2)); const θa = Math.acos(Math.min(Math.max(cosθa, -1), 1)); // protect against rounding errors const θb = Math.acos(Math.min(Math.max(cosθb, -1), 1)); // protect against rounding errors const θ12 = Math.sin(λ2-λ1)>0 ? θa : 2π-θa; const θ21 = Math.sin(λ2-λ1)>0 ? 2π-θb : θb; const α1 = θ13 - θ12; // angle 2-1-3 const α2 = θ21 - θ23; // angle 1-2-3 if (Math.sin(α1) == 0 && Math.sin(α2) == 0) return null; // infinite intersections if (Math.sin(α1) Math.sin(α2) < 0) return null; // ambiguous intersection (antipodal/360°) const cosα3 = -Math.cos(α1)Math.cos(α2) + Math.sin(α1)Math.sin(α2)Math.cos(δ12); const δ13 = Math.atan2(Math.sin(δ12)Math.sin(α1)Math.sin(α2), Math.cos(α2) + Math.cos(α1)cosα3); const φ3 = Math.asin(Math.min(Math.max(Math.sin(φ1)Math.cos(δ13) + Math.cos(φ1)Math.sin(δ13)Math.cos(θ13), -1), 1)); const Δλ13 = Math.atan2(Math.sin(θ13)Math.sin(δ13)Math.cos(φ1), Math.cos(δ13) - Math.sin(φ1)Math.sin(φ3)); const λ3 = λ1 + Δλ13; const lat = φ3.toDegrees(); const lon = λ3.toDegrees(); return new LatLonSpherical(lat, lon); } / Returns (signed) distance from ‘this’ point to great circle defined by start-point and end-point. @param {LatLon} pathStart - Start point of great circle path. @param {LatLon} pathEnd - End point of great circle path. @param {number} [radius=6371e3] - (Mean) radius of earth (defaults to radius in metres). @returns {number} Distance to great circle (-ve if to left, +ve if to right of path). @example const pCurrent = new LatLon(53.2611, -0.7972); const p1 = new LatLon(53.3206, -1.7297); const p2 = new LatLon(53.1887, 0.1334); const d = pCurrent.crossTrackDistanceTo(p1, p2); // -307.5 m / crossTrackDistanceTo(pathStart, pathEnd, radius=6371e3) { if (!(pathStart instanceof LatLonSpherical)) pathStart = LatLonSpherical.parse(pathStart); // allow literal forms if (!(pathEnd instanceof LatLonSpherical)) pathEnd = LatLonSpherical.parse(pathEnd); // allow literal forms const R = radius; if (this.equals(pathStart)) return 0; const δ13 = pathStart.distanceTo(this, R) / R; const θ13 = pathStart.initialBearingTo(this).toRadians(); const θ12 = pathStart.initialBearingTo(pathEnd).toRadians(); const δxt = Math.asin(Math.sin(δ13) Math.sin(θ13 - θ12)); return δxt R; } / Returns how far ‘this’ point is along a path from from start-point, heading towards end-point. That is, if a perpendicular is drawn from ‘this’ point to the (great circle) path, the along-track distance is the distance from the start point to where the perpendicular crosses the path. @param {LatLon} pathStart - Start point of great circle path. @param {LatLon} pathEnd - End point of great circle path. @param {number} [radius=6371e3] - (Mean) radius of earth (defaults to radius in metres). @returns {number} Distance along great circle to point nearest ‘this’ point. @example const pCurrent = new LatLon(53.2611, -0.7972); const p1 = new LatLon(53.3206, -1.7297); const p2 = new LatLon(53.1887, 0.1334); const d = pCurrent.alongTrackDistanceTo(p1, p2); // 62.331 km / alongTrackDistanceTo(pathStart, pathEnd, radius=6371e3) { if (!(pathStart instanceof LatLonSpherical)) pathStart = LatLonSpherical.parse(pathStart); // allow literal forms if (!(pathEnd instanceof LatLonSpherical)) pathEnd = LatLonSpherical.parse(pathEnd); // allow literal forms const R = radius; if (this.equals(pathStart)) return 0; const δ13 = pathStart.distanceTo(this, R) / R; const θ13 = pathStart.initialBearingTo(this).toRadians(); const θ12 = pathStart.initialBearingTo(pathEnd).toRadians(); const δxt = Math.asin(Math.sin(δ13) Math.sin(θ13-θ12)); const δat = Math.acos(Math.cos(δ13) / Math.abs(Math.cos(δxt))); return δatMath.sign(Math.cos(θ12-θ13)) R; } / Returns maximum latitude reached when travelling on a great circle on given bearing from ‘this’ point (‘Clairaut’s formula’). Negate the result for the minimum latitude (in the southern hemisphere). The maximum latitude is independent of longitude; it will be the same for all points on a given latitude. @param {number} bearing - Initial bearing. @returns {number} Maximum latitude reached. / maxLatitude(bearing) { const θ = Number(bearing).toRadians(); const φ = this.lat.toRadians(); const φMax = Math.acos(Math.abs(Math.sin(θ) Math.cos(φ))); return φMax.toDegrees(); } / Returns the pair of meridians at which a great circle defined by two points crosses the given latitude. If the great circle doesn't reach the given latitude, null is returned. @param {LatLon} point1 - First point defining great circle. @param {LatLon} point2 - Second point defining great circle. @param {number} latitude - Latitude crossings are to be determined for. @returns {Object|null} Object containing { lon1, lon2 } or null if given latitude not reached. / static crossingParallels(point1, point2, latitude) { if (point1.equals(point2)) return null; // coincident points const φ = Number(latitude).toRadians(); const φ1 = point1.lat.toRadians(); const λ1 = point1.lon.toRadians(); const φ2 = point2.lat.toRadians(); const λ2 = point2.lon.toRadians(); const Δλ = λ2 - λ1; const x = Math.sin(φ1) Math.cos(φ2) Math.cos(φ) Math.sin(Δλ); const y = Math.sin(φ1) Math.cos(φ2) Math.cos(φ) Math.cos(Δλ) - Math.cos(φ1) Math.sin(φ2) Math.cos(φ); const z = Math.cos(φ1) Math.cos(φ2) Math.sin(φ) Math.sin(Δλ); if (z z > x x + y y) return null; // great circle doesn't reach latitude const λm = Math.atan2(-y, x); // longitude at max latitude const Δλi = Math.acos(z / Math.sqrt(xx + yy)); // Δλ from λm to intersection points const λi1 = λ1 + λm - Δλi; const λi2 = λ1 + λm + Δλi; const lon1 = λi1.toDegrees(); const lon2 = λi2.toDegrees(); return { lon1: Dms.wrap180(lon1), lon2: Dms.wrap180(lon2), }; } / Rhumb - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - / / Returns the distance travelling from ‘this’ point to destination point along a rhumb line. @param {LatLon} point - Latitude/longitude of destination point. @param {number} [radius=6371e3] - (Mean) radius of earth (defaults to radius in metres). @returns {number} Distance in km between this point and destination point (same units as radius). @example const p1 = new LatLon(51.127, 1.338); const p2 = new LatLon(50.964, 1.853); const d = p1.distanceTo(p2); // 40.31 km / rhumbDistanceTo(point, radius=6371e3) { if (!(point instanceof LatLonSpherical)) point = LatLonSpherical.parse(point); // allow literal forms // see www.edwilliams.org/avform.htm#Rhumb const R = radius; const φ1 = this.lat.toRadians(); const φ2 = point.lat.toRadians(); const Δφ = φ2 - φ1; let Δλ = Math.abs(point.lon - this.lon).toRadians(); // if dLon over 180° take shorter rhumb line across the anti-meridian: if (Math.abs(Δλ) > π) Δλ = Δλ > 0 ? -(2 π - Δλ) : (2 π + Δλ); // on Mercator projection, longitude distances shrink by latitude; q is the 'stretch factor' // q becomes ill-conditioned along E-W line (0/0); use empirical tolerance to avoid it (note ε is too small) const Δψ = Math.log(Math.tan(φ2 / 2 + π / 4) / Math.tan(φ1 / 2 + π / 4)); const q = Math.abs(Δψ) > 10e-12 ? Δφ / Δψ : Math.cos(φ1); // distance is pythagoras on 'stretched' Mercator projection, √(Δφ² + q²·Δλ²) const δ = Math.sqrt(ΔφΔφ + qq ΔλΔλ); // angular distance in radians const d = δ R; return d; } / Returns the bearing from ‘this’ point to destination point along a rhumb line. @param {LatLon} point - Latitude/longitude of destination point. @returns {number} Bearing in degrees from north. @example const p1 = new LatLon(51.127, 1.338); const p2 = new LatLon(50.964, 1.853); const d = p1.rhumbBearingTo(p2); // 116.7° / rhumbBearingTo(point) { if (!(point instanceof LatLonSpherical)) point = LatLonSpherical.parse(point); // allow literal forms if (this.equals(point)) return NaN; // coincident points const φ1 = this.lat.toRadians(); const φ2 = point.lat.toRadians(); let Δλ = (point.lon - this.lon).toRadians(); // if dLon over 180° take shorter rhumb line across the anti-meridian: if (Math.abs(Δλ) > π) Δλ = Δλ > 0 ? -(2 π - Δλ) : (2 π + Δλ); const Δψ = Math.log(Math.tan(φ2 / 2 + π / 4) / Math.tan(φ1 / 2 + π / 4)); const θ = Math.atan2(Δλ, Δψ); const bearing = θ.toDegrees(); return Dms.wrap360(bearing); } / Returns the destination point having travelled along a rhumb line from ‘this’ point the given distance on the given bearing. @param {number} distance - Distance travelled, in same units as earth radius (default: metres). @param {number} bearing - Bearing in degrees from north. @param {number} [radius=6371e3] - (Mean) radius of earth (defaults to radius in metres). @returns {LatLon} Destination point. @example const p1 = new LatLon(51.127, 1.338); const p2 = p1.rhumbDestinationPoint(40300, 116.7); // 50.9642°N, 001.8530°E / rhumbDestinationPoint(distance, bearing, radius=6371e3) { const φ1 = this.lat.toRadians(), λ1 = this.lon.toRadians(); const θ = Number(bearing).toRadians(); const δ = distance / radius; // angular distance in radians const Δφ = δ Math.cos(θ); let φ2 = φ1 + Δφ; // check for some daft bugger going past the pole, normalise latitude if so if (Math.abs(φ2) > π / 2) φ2 = φ2 > 0 ? π - φ2 : -π - φ2; const Δψ = Math.log(Math.tan(φ2 / 2 + π / 4) / Math.tan(φ1 / 2 + π / 4)); const q = Math.abs(Δψ) > 10e-12 ? Δφ / Δψ : Math.cos(φ1); // E-W course becomes ill-conditioned with 0/0 const Δλ = δ Math.sin(θ) / q; const λ2 = λ1 + Δλ; const lat = φ2.toDegrees(); const lon = λ2.toDegrees(); return new LatLonSpherical(lat, lon); } / Returns the loxodromic midpoint (along a rhumb line) between ‘this’ point and second point. @param {LatLon} point - Latitude/longitude of second point. @returns {LatLon} Midpoint between this point and second point. @example const p1 = new LatLon(51.127, 1.338); const p2 = new LatLon(50.964, 1.853); const pMid = p1.rhumbMidpointTo(p2); // 51.0455°N, 001.5957°E / rhumbMidpointTo(point) { if (!(point instanceof LatLonSpherical)) point = LatLonSpherical.parse(point); // allow literal forms // see mathforum.org/kb/message.jspa?messageID=148837 const φ1 = this.lat.toRadians(); let λ1 = this.lon.toRadians(); const φ2 = point.lat.toRadians(), λ2 = point.lon.toRadians(); if (Math.abs(λ2 - λ1) > π) λ1 += 2 π; // crossing anti-meridian const φ3 = (φ1 + φ2) / 2; const f1 = Math.tan(π / 4 + φ1 / 2); const f2 = Math.tan(π / 4 + φ2 / 2); const f3 = Math.tan(π / 4 + φ3 / 2); let λ3 = ((λ2 - λ1) Math.log(f3) + λ1 Math.log(f2) - λ2 Math.log(f1)) / Math.log(f2 / f1); if (!isFinite(λ3)) λ3 = (λ1 + λ2) / 2; // parallel of latitude const lat = φ3.toDegrees(); const lon = λ3.toDegrees(); return new LatLonSpherical(lat, lon); } / Area - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - / / Calculates the area of a spherical polygon where the sides of the polygon are great circle arcs joining the vertices. @param {LatLon[]} polygon - Array of points defining vertices of the polygon. @param {number} [radius=6371e3] - (Mean) radius of earth (defaults to radius in metres). @returns {number} The area of the polygon in the same units as radius. @example const polygon = [new LatLon(0,0), new LatLon(1,0), new LatLon(0,1)]; const area = LatLon.areaOf(polygon); // 6.18e9 m² / static areaOf(polygon, radius=6371e3) { // uses method due to Karney: osgeo-org.1560.x6.nabble.com/Area-of-a-spherical-polygon-td3841625.html; // for each edge of the polygon, tan(E/2) = tan(Δλ/2)·(tan(φ₁/2)+tan(φ₂/2)) / (1+tan(φ₁/2)·tan(φ₂/2)) // where E is the spherical excess of the trapezium obtained by extending the edge to the equator // (Karney's method is probably more efficient than the more widely known L’Huilier’s Theorem) const R = radius; // close polygon so that last point equals first point const closed = polygon.equals(polygon[polygon.length-1]); if (!closed) polygon.push(polygon); const nVertices = polygon.length - 1; let S = 0; // spherical excess in steradians for (let v=0; v<nVertices; v++) { const φ1 = polygon[v].lat.toRadians(); const φ2 = polygon[v+1].lat.toRadians(); const Δλ = (polygon[v+1].lon - polygon[v].lon).toRadians(); const E = 2 Math.atan2(Math.tan(Δλ/2) (Math.tan(φ1/2)+Math.tan(φ2/2)), 1 + Math.tan(φ1/2)Math.tan(φ2/2)); S += E; } if (isPoleEnclosedBy(polygon)) S = Math.abs(S) - 2π; const A = Math.abs(S RR); // area in units of R if (!closed) polygon.pop(); // restore polygon to pristine condition return A; // returns whether polygon encloses pole: sum of course deltas around pole is 0° rather than // normal ±360°: blog.element84.com/determining-if-a-spherical-polygon-contains-a-pole.html function isPoleEnclosedBy(p) { // TODO: any better test than this? let ΣΔ = 0; let prevBrng = p.initialBearingTo(p); for (let v=0; v<p.length-1; v++) { const initBrng = p[v].initialBearingTo(p[v+1]); const finalBrng = p[v].finalBearingTo(p[v+1]); ΣΔ += (initBrng - prevBrng + 540) % 360 - 180; ΣΔ += (finalBrng - initBrng + 540) % 360 - 180; prevBrng = finalBrng; } const initBrng = p.initialBearingTo(p); ΣΔ += (initBrng - prevBrng + 540) % 360 - 180; // TODO: fix (intermittant) edge crossing pole - eg (85,90), (85,0), (85,-90) const enclosed = Math.abs(ΣΔ) < 90; // 0°-ish return enclosed; } } / - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - / / Checks if another point is equal to ‘this’ point. @param {LatLon} point - Point to be compared against this point. @returns {bool} True if points have identical latitude and longitude values. @example const p1 = new LatLon(52.205, 0.119); const p2 = new LatLon(52.205, 0.119); const equal = p1.equals(p2); // true / equals(point) { if (!(point instanceof LatLonSpherical)) point = LatLonSpherical.parse(point); // allow literal forms if (Math.abs(this.lat - point.lat) > Number.EPSILON) return false; if (Math.abs(this.lon - point.lon) > Number.EPSILON) return false; return true; } / Converts ‘this’ point to a GeoJSON object. @returns {Object} this point as a GeoJSON ‘Point’ object. / toGeoJSON() { return { type: 'Point', coordinates: [ this.lon, this.lat ] }; } / Returns a string representation of ‘this’ point, formatted as degrees, degrees+minutes, or degrees+minutes+seconds. @param {string} [format=d] - Format point as 'd', 'dm', 'dms', or 'n' for signed numeric. @param {number} [dp=4|2|0] - Number of decimal places to use: default 4 for d, 2 for dm, 0 for dms. @returns {string} Comma-separated formatted latitude/longitude. @throws {RangeError} Invalid format. @example const greenwich = new LatLon(51.47788, -0.00147); const d = greenwich.toString(); // 51.4779°N, 000.0015°W const dms = greenwich.toString('dms', 2); // 51°28′40.37″N, 000°00′05.29″W const [lat, lon] = greenwich.toString('n').split(','); // 51.4779, -0.0015 / toString(format='d', dp=undefined) { // note: explicitly set dp to undefined for passing through to toLat/toLon if (![ 'd', 'dm', 'dms', 'n' ].includes(format)) throw new RangeError(invalid format ‘${format}’); if (format == 'n') { // signed numeric degrees if (dp == undefined) dp = 4; return ${this.lat.toFixed(dp)},${this.lon.toFixed(dp)}; } const lat = Dms.toLat(this.lat, format, dp); const lon = Dms.toLon(this.lon, format, dp); return ${lat}, ${lon}; }}/ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - /export { LatLonSpherical as default, Dms };
/ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - // Geodesy representation conversion functions (c) Chris Veness 2002-2020 // MIT Licence // www.movable-type.co.uk/scripts/latlong.html // www.movable-type.co.uk/scripts/js/geodesy/geodesy-library.html#dms // - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - // eslint no-irregular-whitespace: [2, { skipComments: true }] // Latitude/longitude points may be represented as decimal degrees, or subdivided into sexagesimal minutes and seconds. This module provides methods for parsing and representing degrees / minutes / seconds. @module dms // Degree-minutes-seconds (& cardinal directions) separator character / let dmsSeparator = '\u202f'; // U+202F = 'narrow no-break space'/ Functions for parsing and representing degrees / minutes / seconds. /class Dms { // note Unicode Degree = U+00B0. Prime = U+2032, Double prime = U+2033 / Separator character to be used to separate degrees, minutes, seconds, and cardinal directions. Default separator is U+202F ‘narrow no-break space’. To change this (e.g. to empty string or full space), set Dms.separator prior to invoking formatting. @example import LatLon, { Dms } from '/js/geodesy/latlon-spherical.js'; const p = new LatLon(51.2, 0.33).toString('dms'); // 51° 12′ 00″ N, 000° 19′ 48″ E Dms.separator = ''; // no separator const pʹ = new LatLon(51.2, 0.33).toString('dms'); // 51°12′00″N, 000°19′48″E / static get separator() { return dmsSeparator; } static set separator(char) { dmsSeparator = char; } / Parses string representing degrees/minutes/seconds into numeric degrees. This is very flexible on formats, allowing signed decimal degrees, or deg-min-sec optionally suffixed by compass direction (NSEW); a variety of separators are accepted. Examples -3.62, '3 37 12W', '3°37′12″W'. Thousands/decimal separators must be comma/dot; use Dms.fromLocale to convert locale-specific thousands/decimal separators. @param {string|number} dms - Degrees or deg/min/sec in variety of formats. @returns {number} Degrees as decimal number. @example const lat = Dms.parse('51° 28′ 40.37″ N'); const lon = Dms.parse('000° 00′ 05.29″ W'); const p1 = new LatLon(lat, lon); // 51.4779°N, 000.0015°W / static parse(dms) { // check for signed decimal degrees without NSEW, if so return it directly if (!isNaN(parseFloat(dms)) && isFinite(dms)) return Number(dms); // strip off any sign or compass dir'n & split out separate d/m/s const dmsParts = String(dms).trim().replace(/^-/, '').replace(/[NSEW]$/i, '').split(/[^0-9.,]+/); if (dmsParts[dmsParts.length-1]=='') dmsParts.splice(dmsParts.length-1); // from trailing symbol if (dmsParts == '') return NaN; // and convert to decimal degrees... let deg = null; switch (dmsParts.length) { case 3: // interpret 3-part result as d/m/s deg = dmsParts/1 + dmsParts/60 + dmsParts/3600; break; case 2: // interpret 2-part result as d/m deg = dmsParts/1 + dmsParts/60; break; case 1: // just d (possibly decimal) or non-separated dddmmss deg = dmsParts; // check for fixed-width unseparated format eg 0033709W //if (/[NS]/i.test(dmsParts)) deg = '0' + deg; // - normalise N/S to 3-digit degrees //if (/[0-9]{7}/.test(deg)) deg = deg.slice(0,3)/1 + deg.slice(3,5)/60 + deg.slice(5)/3600; break; default: return NaN; } if (/^-|[WS]$/i.test(dms.trim())) deg = -deg; // take '-', west and south as -ve return Number(deg); } / Converts decimal degrees to deg/min/sec format - degree, prime, double-prime symbols are added, but sign is discarded, though no compass direction is added. - degrees are zero-padded to 3 digits; for degrees latitude, use .slice(1) to remove leading zero. @private @param {number} deg - Degrees to be formatted as specified. @param {string} [format=d] - Return value as 'd', 'dm', 'dms' for deg, deg+min, deg+min+sec. @param {number} [dp=4|2|0] - Number of decimal places to use – default 4 for d, 2 for dm, 0 for dms. @returns {string} Degrees formatted as deg/min/secs according to specified format. / static toDms(deg, format='d', dp=undefined) { if (isNaN(deg)) return null; // give up here if we can't make a number from deg if (typeof deg == 'string' && deg.trim() == '') return null; if (typeof deg == 'boolean') return null; if (deg == Infinity) return null; if (deg == null) return null; // default values if (dp === undefined) { switch (format) { case 'd': case 'deg': dp = 4; break; case 'dm': case 'deg+min': dp = 2; break; case 'dms': case 'deg+min+sec': dp = 0; break; default: format = 'd'; dp = 4; break; // be forgiving on invalid format } } deg = Math.abs(deg); // (unsigned result ready for appending compass dir'n) let dms = null, d = null, m = null, s = null; switch (format) { default: // invalid format spec! case 'd': case 'deg': d = deg.toFixed(dp); // round/right-pad degrees if (d<100) d = '0' + d; // left-pad with leading zeros (note may include decimals) if (d<10) d = '0' + d; dms = d + '°'; break; case 'dm': case 'deg+min': d = Math.floor(deg); // get component deg m = ((deg60) % 60).toFixed(dp); // get component min & round/right-pad if (m == 60) { m = (0).toFixed(dp); d++; } // check for rounding up d = ('000'+d).slice(-3); // left-pad with leading zeros if (m<10) m = '0' + m; // left-pad with leading zeros (note may include decimals) dms = d + '°'+Dms.separator + m + '′'; break; case 'dms': case 'deg+min+sec': d = Math.floor(deg); // get component deg m = Math.floor((deg3600)/60) % 60; // get component min s = (deg3600 % 60).toFixed(dp); // get component sec & round/right-pad if (s == 60) { s = (0).toFixed(dp); m++; } // check for rounding up if (m == 60) { m = 0; d++; } // check for rounding up d = ('000'+d).slice(-3); // left-pad with leading zeros m = ('00'+m).slice(-2); // left-pad with leading zeros if (s<10) s = '0' + s; // left-pad with leading zeros (note may include decimals) dms = d + '°'+Dms.separator + m + '′'+Dms.separator + s + '″'; break; } return dms; } / Converts numeric degrees to deg/min/sec latitude (2-digit degrees, suffixed with N/S). @param {number} deg - Degrees to be formatted as specified. @param {string} [format=d] - Return value as 'd', 'dm', 'dms' for deg, deg+min, deg+min+sec. @param {number} [dp=4|2|0] - Number of decimal places to use – default 4 for d, 2 for dm, 0 for dms. @returns {string} Degrees formatted as deg/min/secs according to specified format. @example const lat = Dms.toLat(-3.62, 'dms'); // 3°37′12″S / static toLat(deg, format, dp) { const lat = Dms.toDms(Dms.wrap90(deg), format, dp); return lat===null ? '–' : lat.slice(1) + Dms.separator + (deg<0 ? 'S' : 'N'); // knock off initial '0' for lat! } / Convert numeric degrees to deg/min/sec longitude (3-digit degrees, suffixed with E/W). @param {number} deg - Degrees to be formatted as specified. @param {string} [format=d] - Return value as 'd', 'dm', 'dms' for deg, deg+min, deg+min+sec. @param {number} [dp=4|2|0] - Number of decimal places to use – default 4 for d, 2 for dm, 0 for dms. @returns {string} Degrees formatted as deg/min/secs according to specified format. @example const lon = Dms.toLon(-3.62, 'dms'); // 3°37′12″W / static toLon(deg, format, dp) { const lon = Dms.toDms(Dms.wrap180(deg), format, dp); return lon===null ? '–' : lon + Dms.separator + (deg<0 ? 'W' : 'E'); } / Converts numeric degrees to deg/min/sec as a bearing (0°..360°). @param {number} deg - Degrees to be formatted as specified. @param {string} [format=d] - Return value as 'd', 'dm', 'dms' for deg, deg+min, deg+min+sec. @param {number} [dp=4|2|0] - Number of decimal places to use – default 4 for d, 2 for dm, 0 for dms. @returns {string} Degrees formatted as deg/min/secs according to specified format. @example const lon = Dms.toBrng(-3.62, 'dms'); // 356°22′48″ / static toBrng(deg, format, dp) { const brng = Dms.toDms(Dms.wrap360(deg), format, dp); return brng===null ? '–' : brng.replace('360', '0'); // just in case rounding took us up to 360°! } / Converts DMS string from locale thousands/decimal separators to JavaScript comma/dot separators for subsequent parsing. Both thousands and decimal separators must be followed by a numeric character, to facilitate parsing of single lat/long string (in which whitespace must be left after the comma separator). @param {string} str - Degrees/minutes/seconds formatted with locale separators. @returns {string} Degrees/minutes/seconds formatted with standard Javascript separators. @example const lat = Dms.fromLocale('51°28′40,12″N'); // '51°28′40.12″N' in France const p = new LatLon(Dms.fromLocale('51°28′40,37″N, 000°00′05,29″W'); // '51.4779°N, 000.0015°W' in France / static fromLocale(str) { const locale = (123456.789).toLocaleString(); const separator = { thousands: locale.slice(3, 4), decimal: locale.slice(7, 8) }; return str.replace(separator.thousands, '⁜').replace(separator.decimal, '.').replace('⁜', ','); } / Converts DMS string from JavaScript comma/dot thousands/decimal separators to locale separators. Can also be used to format standard numbers such as distances. @param {string} str - Degrees/minutes/seconds formatted with standard Javascript separators. @returns {string} Degrees/minutes/seconds formatted with locale separators. @example const Dms.toLocale('123,456.789'); // '123.456,789' in France const Dms.toLocale('51°28′40.12″N, 000°00′05.31″W'); // '51°28′40,12″N, 000°00′05,31″W' in France / static toLocale(str) { const locale = (123456.789).toLocaleString(); const separator = { thousands: locale.slice(3, 4), decimal: locale.slice(7, 8) }; return str.replace(/,([0-9])/, '⁜$1').replace('.', separator.decimal).replace('⁜', separator.thousands); } / Returns compass point (to given precision) for supplied bearing. @param {number} bearing - Bearing in degrees from north. @param {number} [precision=3] - Precision (1:cardinal / 2:intercardinal / 3:secondary-intercardinal). @returns {string} Compass point for supplied bearing. @example const point = Dms.compassPoint(24); // point = 'NNE' const point = Dms.compassPoint(24, 1); // point = 'N' / static compassPoint(bearing, precision=3) { if (![ 1, 2, 3 ].includes(Number(precision))) throw new RangeError(invalid precision ‘${precision}’); // note precision could be extended to 4 for quarter-winds (eg NbNW), but I think they are little used bearing = Dms.wrap360(bearing); // normalise to range 0..360° const cardinals = [ 'N', 'NNE', 'NE', 'ENE', 'E', 'ESE', 'SE', 'SSE', 'S', 'SSW', 'SW', 'WSW', 'W', 'WNW', 'NW', 'NNW' ]; const n = 4 2(precision-1); // no of compass points at req’d precision (1=>4, 2=>8, 3=>16) const cardinal = cardinals[Math.round(bearingn/360)%n 16/n]; return cardinal; } / Constrain degrees to range -90..+90 (for latitude); e.g. -91 => -89, 91 => 89. @private @param {number} degrees @returns degrees within range -90..+90. / static wrap90(degrees) { if (-90<=degrees && degrees<=90) return degrees; // avoid rounding due to arithmetic ops if within range // latitude wrapping requires a triangle wave function; a general triangle wave is // f(x) = 4a/p ⋅ | (x-p/4)%p - p/2 | - a // where a = amplitude, p = period, % = modulo; however, JavaScript '%' is a remainder operator // not a modulo operator - for modulo, replace 'x%n' with '((x%n)+n)%n' const x = degrees, a = 90, p = 360; return 4a/p Math.abs((((x-p/4)%p)+p)%p - p/2) - a; } / Constrain degrees to range -180..+180 (for longitude); e.g. -181 => 179, 181 => -179. @private @param {number} degrees @returns degrees within range -180..+180. / static wrap180(degrees) { if (-180<=degrees && degrees<=180) return degrees; // avoid rounding due to arithmetic ops if within range // longitude wrapping requires a sawtooth wave function; a general sawtooth wave is // f(x) = (2ax/p - p/2) % p - a // where a = amplitude, p = period, % = modulo; however, JavaScript '%' is a remainder operator // not a modulo operator - for modulo, replace 'x%n' with '((x%n)+n)%n' const x = degrees, a = 180, p = 360; return (((2ax/p - p/2)%p)+p)%p - a; } / Constrain degrees to range 0..360 (for bearings); e.g. -1 => 359, 361 => 1. @private @param {number} degrees @returns degrees within range 0..360. / static wrap360(degrees) { if (0<=degrees && degrees<360) return degrees; // avoid rounding due to arithmetic ops if within range // bearing wrapping requires a sawtooth wave function with a vertical offset equal to the // amplitude and a corresponding phase shift; this changes the general sawtooth wave function from // f(x) = (2ax/p - p/2) % p - a // to // f(x) = (2ax/p) % p // where a = amplitude, p = period, % = modulo; however, JavaScript '%' is a remainder operator // not a modulo operator - for modulo, replace 'x%n' with '((x%n)+n)%n' const x = degrees, a = 180, p = 360; return (((2ax/p)%p)+p)%p; }}// Extend Number object with methods to convert between degrees & radiansNumber.prototype.toRadians = function() { return this Math.PI / 180; };Number.prototype.toDegrees = function() { return this 180 / Math.PI; };/ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - /export default Dms; |
11188 | https://array.aami.org/content/news/standard-ppe-updated-additional-equipment-types-safety-guidance | Standard for PPE Updated for Additional Equipment Types and Safety Guidance | AAMI News
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News Article 18 January 2023
Standard for PPE Updated for Additional Equipment Types and Safety Guidance
Authors:
Brian Stallard
Topics:
Patient Safety
Regulations
Standards
Sterilization
The Association for the Advancement of Medical Instrumentation (AAMI) has released a revised American National Standard containing new and crucial information for the manufacturers and users of personal protective equipment (PPE) in the healthcare space. The new edition of the standard, ANSI/AAMI PB70:2022,Liquid barrier performance and classification of protective apparel and drapes intended for use in health care facilities, accounts for a decade of new knowledge and the evolution of technologies keep our frontline healthcare workers safe.
ANSI/AAMI PB70 specifies requirements for the performance of surgical gowns, isolation gowns, other gowns and protective apparel, surgical drapes, and drape accessories designed to protect health care workers during surgery and other health care procedures. That performance is based on the ability of these protective barriers to prevent exposure to blood, body fluids, and other potentially infectious materials. It also establishes an industry-wide classification system and associated labeling requirements for these various gowns and drapes.
“It is that variety of protective apparel that inspired many of the changes you will find in this updated guidance document,” explained Jill Holdsworth, manager of Infection Prevention for Emory University Hospital Midtown. Holdsworth currently co-chairs the AAMI Protective Barriers (PB) Committee, which is responsible for the standard and developed the revised new edition. The PB Committee consists of experts hailing from manufacturing, academia, regulatory bodies, and the frontlines of healthcare.
“Barrier claims under the previous edition were limited to surgical gowns, surgical drapes, and isolation gowns,” added former committee co-chair Shane Woody, who served as the senior manager for the Global PPE Portfolio Strategy at Aspen Surgical during his tenure leading the committee developing the revision.
“As there are many other protective apparel products used in healthcare settings which provide barrier protection, it was necessary to add other categories of products for which PB70 barrier level claims could be made,” the Committee wrote. “This will allow healthcare professionals to have a better understanding of the barrier protection properties of the protective apparel items available.”
Additional protective apparel categories added to PB70 include:
surgical gown–E - A specialized gown which provides extended protection of critical zones—defined as the areas where direct contact with blood, bodily fluids, and other potentially infectious material is most likely to happen.
Non-surgical protective gowns including
procedure gown
decontamination gown
These non-surgical gown categories also include “protective full coverage gowns, protective gowns with non-protective backs, and protective gowns with open backs,” the authors explained. “Protective hoods and togas were also added along with a general category of other protective apparel items. (e.g., aprons, footwear covers, sleeves, etc.).”
Labeling requirements were also added for each non-surgical gown, stating that the gown packaging should inform users of the area of protection:
full coverage gown
non-protective back gown
open back gown
The standard does not cover other types of PPE such as medical gloves, surgical masks, or goggles, nor is there an industry consensus for best practices concerning the gown-to-glove interface.
And while PB70 is intended to be used primarily by device manufacturers in qualifying, classifying, and labeling the barrier performance of their products, understanding the protective properties each label promises will also prove invaluable for healthcare professionals, so they may make informed decisions when selecting appropriate PPE.
A Post-Pandemic Project
With the protective barrier standard updated to reflect modern scientific and industry consensus, the PB committee is now turning its attention to gathering valuable information and lessons learned from the COVID-19 pandemic—a time when PPE shortages across the world left frontline healthcare workers unprotected.
The PB Committee’s pandemic preparedness task group is seeking additional members, data, and materials relevant to PPE practices during a time of emergency. Prospective participants should contact standards@aami.org for more information.
“We have a learned a great deal from the COVID-19 pandemic’s successes and challenges, including the need for timely and consistence guidance in PPE and emergency preparedness,” said Holdsworth. “This group will be turning our focus to meeting this need for our members, healthcare providers, customers and the public.”
For more information and to purchase PB70, visitwww.aami.org/PB70.
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11189 | https://www.youtube.com/watch?v=MIvgQRHObCQ | Moment of Inertia: Wheel and Spokes
Physics Ninja
232000 subscribers
145 likes
Description
6018 views
Posted: 18 Nov 2023
Physics Ninja looks at calculating the moment of inertia of a wheel with 3 spokes. The moment of inertia is calculated about three different axes. The first case we consider the moment of inertia through the center of mass of the system. In the second and third case we use the parallel axis theorem to calculate the moment of inertia through axes located on the rim of the wheel.
3 comments
Transcript:
hi everybody physics ninja here today's problem we're going to look at calculating the moment of inertia of this wheel with has uh which has three spokes uh the goal is going to be to calculate it with respect to three different axes the axis a goes um right through the center of mass of the object uh axis B and C are along the rim so how would you calculate the moment of inertia if I was rotating this wheel about those three different axes now we're going to assume that the mass of each spoke uh there's three of them is M and it's also equal to the mass of the wheel so how would you kind of use our formulas for moment of inertia to calculate the total moment of inertia of this more complex object all right like with all my videos if you like it give it a thumbs up consider subscribing to physics ninja and if you have the means consider giving a super thanks a dollar or two goes a long way into helping me all right let's get started all right so first going to consider the the case of the moment of inertia uh through the axis a that is through the center you can see it right here okay in the center of those three Spokes and in the center of the hoop the next thing we need is we're going to look up the values of the moment of inertia of different objects and for that you go in your textbook and you should find a table that might look similar to this one so typically there are a lot of different objects they could be diss Rings rods spheres for example or just point masses Like Satellites and what you do is these moment of inertias are typically through the center of each object now sometimes like this case over here they also represent the moment of inertia of a rod or a plank through the end you can see the axis here is through the end of the object so the value is a little bit different but most of the other cases are always through the center so now we consider our our wheel here so what do we have we have our wheel is made up of what one hoop and and for the first case we're rotating it through the center so the moment of inertia you can represent the hoop by a ring so the moment of inertia is simply the mass of the object which is M right in my case the mass of the hoop is equal to the mass of each spoke and the radius is simply R so you do Mr R squ is for the hoop now what do we have we have three spokes well what is the spokes moment of inertia now think about one of these spokes right here um there are two values for the moment of inertia of a Spoke uh there's this guy right here that is if I was rotating a rod through the center but I'm not I am rotating each one of these spokes through the end of the spoke so actually for the spokes I really should use this second result which is 1/3 mass of the spoke and L is the total length of the spoke for us the length of the spoke is simply equal to the radius of the wheel so that simplifies our result so for each spoke guess what what for each spoke I get 1/3 M and the length is R so you have to do R 2 and now I also have three of those so I have to multiply each one um uh the result of one spoke by three to get the total moment of inertia of the spokes all right now what do you do well all you have to do now is simply add them up so our total moment of inertia and I'm going to write the little a here through that axis through the center of mass is the moment of inertia of the hoop plus the moment of inertia of the spokes so this is pretty straightforward so you get Mr squared here plus now I add this you multiply the three and the three can cancel over here um then you're left with another term which is Mr 2 okay uh so in this case here my final moment of inertia through the center of mass is simply going to be 2 Mr 2 okay uh this is how you can calculate the moment of inertia of a more complicated object by building it up and using our formulas from a table all right for our second part we're interested in what is the moment of inertia through axis B uh axis B is right here and again this axis runs in and out of the page right here now the one thing that is important about axis B is that it is parallel to axis a that is really really important now if you wanted to calculate the moment of inertia of this entire object with respect to axis B um a trick here and one of the keys to understanding this problem is to use the parallel axis theorem okay the parallel axis theorem says this the moment of inertia through an axis that is parallel so I'm just going to write here parallel uh to the axis that runs through the center of mass okay so in this case axis B is parallel to axis a the way you calculate it is is you simply take the moment of inertia of your complicated object through the center of mass so for us guess what this term here is simply the moment of inertia through axis a that's what we calculated in part one but we have to add a correction term and the correction term looks like this it's the total mass of the object multiplied by it's sometimes written as D2 okay and D here is the distance between both of those axes so in this case it's the distance between what it's the distance between A and B that's it so for us our value D is simply equal to the radius of the wheel that is the distance between axis a and axis B so this is becomes a pretty straightforward problem if you use the parallel axis theorem so I would say that the moment of inertia through axis B in this case is equal to the moment of inertia through axis a plus plus now let's think about this correction term the correction term is the total mass What's the total mass of the object for us it is 4 M that's the total mass of the object and again I said my value D was simply equal to R so here you have r squared all right put everything together now so what do we have for moment of inertia of a that's written up top we have 2 Mr R 2 and then plus 4 Mr 2 so at the end I get that my moment of inertia through axis B is equal to 6 m 2 one thing we notice is that the moment of inertia through any axis that is parallel to the center of mass is going to result that the moment of inertia that we get is bigger than IIA all right that's a great problem and a great application of this parallel axis theorem here all right in the last case we're going to consider the moment of inertia through axis C axis C is located right here on the rim of the wheel now one thing we can use again have a look this axis runs in and out of the page the axis is also parallel to axis C so guess what we're doing we're using the parallel axis theorem and one thing you should note is that well if you just blindly just apply this formula this is what you would get you'd get the moment of inertia through the center of mass is I a plus the total mass which is M4 and then the distance from axis C to axis a is this guy right here that is also the radius right so you do radius squared here at the end guess what you get you get 6 m r squar that is the moment of inertia through axis C all right pretty straightforward you notice some things here that our moment of inertia through axis B is equal to axis C it would be equal to any axis that is anywhere on this rim okay because they are the same distance away from the center of mass of the object and we also notice that both of those moment of inertias are bigger than the moment of inertia through the center of mass okay so this is the relationship between all of those three values all right folks that's it for me hopefully you appreciate this video and you learn how to use the parallel axis theorem to find the moment of inertia through a complicated object we'll see you next time |
11190 | https://sbartoloma.files.wordpress.com/2013/05/worksheet-52-volo-nolo-malo.pdf | Latin 8 Name: ______ May 22, 2013 Section: ____ Worksheet 52 Worksheet 52: Volō, Nōlō, Mālō A. Volō, Nōlō, and Mālō • The irregular verb volō, velle, volui means to be willing, to wish, to want. It is often seen with a complementary infinitive. There are two important compound of volō: • nōlō, nolle, noluī means to be unwilling, to not wish, to not want. It is made up of nōn + volō. • mālō, malle, maluī means to prefer, to be more willing. It is made up of māgis (more) + volō. • These verbs are irregular in the present tense. Their imperfect (-ēbam, -ēbās, -ēbat) and future (-am, -ēs, -et) tenses follow the patterns of the 3rd conjugation. • Of these three verbs, only nolō has present imperatives: nōlī (singular) and nōlīte (plural). These forms become negative commands when followed by a present infinitive. Present Indicative volō Singular Plural nōlō Singular Plural mālō Singular Plural 1st volō volumus 1st nōlō nōlumus 1st mālō mālumus 2nd vīs vultis 2nd nōn vīs nōn vultis 2nd māvis māvultis 3rd vult volunt 3rd nōn vult nōlunt 3rd māvult mālunt Present Subjunctive volō Singular Plural nōlō Singular Plural mālō Singular Plural 1st velim velīmus 1st nōlim nōlīmus 1st mālim mālīmus 2nd velīs velītis 2nd nōlīs nōlītis 2nd mālīs mālītis 3rd velit velint 3rd nōlit nōlint 3rd mālit mālint B. Give the Latin form of volō, nolō, malō and its complementary infinitive for the following sentences: 1. She wants to be a famous author. __________ 2. The Romans preferred to worship many gods. _______ 3. You all will not want to see this. _________ 4. We had wanted to read every book in the library. _______ 5. The soldiers don’t want to die. __________ 6. Are you willing to sail with me? _________ 7. We would rather follow you. ___________ 8. Did you want to say something? _________ 9. Our children will prefer to use computers. ____________ 10. The lion is so fierce that Hercules wants to kill it. _______ 11. There were so many people that we preferred to go home. ______ 12. Since we had been unwilling to help, the plan failed. _________ 13. Although we prefer to eat, still we are not willing to steal food. ______ 14. Do not attack us, soldier! ___________ 15. Don’t leave, everyone! _____________ |
11191 | https://study.com/skill/learn/how-to-write-a-quadratic-equation-given-the-roots-the-leading-coefficient-explanation.html | How to Write a Quadratic Equation Given the Roots & the Leading Coefficient | Algebra | Study.com
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How to Write a Quadratic Equation Given the Roots & the Leading Coefficient
Algebra 2 Skills Practice
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00:04 Example 1
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Lisa Stowe, Amy McKenney
Instructors Lisa Stowe
Lisa has tutored grade school, high school and post-secondary mathematics for over 30 years. She earned an Honors Bachelor of Mathematics from the University of Waterloo in Waterloo, Ontario, Canada.
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Amy McKenney
Amy has taught high school mathematics for over 14 years. She has a master's degree in education from Plymouth State University and her undergraduate degree in mathematics. She is certified to teach grades 7-12 mathematics.
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Example SolutionsPractice Questions
Write a Quadratic Equation Given the Roots and a Leading Coefficient
Step 1: Write the roots as factors.
Step 2: Input the factors from step 1, and the leading coefficient, into the factored form of the equation. (If you are interested in the factored form you are finished at this step!)
Step 3: Rewrite the equation from step 2 into standard form by using the distributive property and simplification.
Write a Quadratic Equation Given the Roots and a Leading Coefficient Equations
Factored form: The factored form of a quadratic equation looks like:
f(x)=a(x−p)(x−q)
Where a is the leading coefficient, and (x−p),(x−q) are factors.
We use the factors to solve for the roots as follows:
x−p=0 x−q=0 x−p+p=0+p x−q+q=0+q x=p x=q
So that roots of the equation are p and q.
Tip: This is important to note because we can always work backward from the roots to get the factors!
Standard form: The standard form of a quadratic equation looks like:
f(x)=a x 2+b x+c
Where a is the leading coefficient.
Let's try two example problems to practice writing a quadratic equation given the roots and a leading coefficient. For the first question, we will write the equation in factored form. For the second question, we will write the equation in standard form.
Example 1: Write a Quadratic Equation Given the Roots and a Leading Coefficient: Factored Form
Find the factored form of the equation of a quadratic with roots of 6 and -12 and a leading coefficient of -7.
Step 1: Write the roots as factors.
For this first step, we need to take the roots we've been given and rewrite them as factors. Normally when given the factored form of the equation we can pull the roots from the equation by setting the factors equal to zero and solving (as we did in the equations section above!). But we now have the roots and want the factors, so we will need to perform those same steps but in reverse!
We have two roots as follows:
x=6 x=−12
Let's rewrite each of these as factors, or said another way so that they are an expression equal to zero. We do this by putting x and the root, on the same side of the equation.
x=6 x=−12 x−6=6−6 x+12=−12+12 x−6=0 x+12=0
So, we now have two factors for our equation (x−6),(x+12).
Step 2: Input the factors from step 1, and the leading coefficient, into the factored form of the equation.
For the second step, we will take the factors and the leading coefficient and put it into the factored form of the equation. Keeping in mind that the factored form looks like:
f(x)=a(x−b)(x−c)
Taking the factors from step 1, and the leading coefficient of a=−7, we then have:
f(x)=−7(x−6)(x+12)
Therefore, the factored form of the equation of a quadratic with roots of 6 and -12 and a leading coefficient of -7 is f(x)=−7(x−6)(x+12).
Note: We don't need step 3 here because we want to keep the equation in the factored form!
Example 2: Write a Quadratic Equation Given the Roots and a Leading Coefficient: Standard Form
Find the standard form of the equation of a quadratic with roots of 3 and 11, and a leading coefficient of 4.
Step 1: Write the roots as factors.
We determine the factors of the equation by using the roots as we did above.
x=3 x=11 x−3=3−3 x−11=11−11 x−3=0 x−11=0
This gives us our two factors of (x−3),(x−11).
Step 2: Input the factors from step 1, and the leading coefficient, into the factored form of the equation.
Inputting these two factors along with the leading coefficient of 4 we have:
f(x)=4(x−3)(x−11)
Step 3: Rewrite the equation from step 2 into standard form by using the distributive property and simplification.
Since the equation above is in a factored form, we need to perform the extra steps of distribution and simplification to get the equation into standard form as follows:
f(x)=4(x−3)(x−11)=4(x 2−11 x−3 x+33)=4(x 2−14 x+33)=4(x 2)+4(−14 x)+4(33)=4 x 2−56 x+132
Therefore, the standard form of the equation of a quadratic with roots of 3 and 11 and a leading coefficient of 4 is f(x)=4 x 2−56 x+132.
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Table of Contents
Write a Quadratic Equation Given the Roots and a Leading Coefficient
Write a Quadratic Equation Given the Roots and a Leading Coefficient Equations
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Writing a Quadratic Equation Given the Roots & the Leading Coefficient
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11192 | https://arxiv.org/abs/1809.02430 | Published Time: Mon, 10 Sep 2018 00:09:51 GMT
[1809.02430] Arithmetic Progressions with Restricted Digits
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Mathematics > Number Theory
arXiv:1809.02430 (math)
[Submitted on 7 Sep 2018]
Title:Arithmetic Progressions with Restricted Digits
Authors:Aled Walker, Alexander Walker
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Abstract:For an integer b \geqslant 2and a set S\subset {0,\cdots,b-1}, we define the Kempner set \mathcal{K}(S,b)to be the set of all non-negative integers whose base-bdigital expansions contain only digits from S. These well-studied sparse sets provide a rich setting for additive number theory, and in this paper we study various questions relating to the appearance of arithmetic progressions in these sets. In particular, for all bwe determine exactly the maximal length of an arithmetic progression that omits a base-bdigit.
Comments:11 pages, submitted to American Mathematical Monthly
Subjects:Number Theory (math.NT); Combinatorics (math.CO)
Cite as:arXiv:1809.02430 [math.NT]
(or arXiv:1809.02430v1 [math.NT] for this version)
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From: Aled Walker [view email]
[v1] Fri, 7 Sep 2018 12:14:17 UTC (17 KB)
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11193 | https://www.enchantedlearning.com/subjects/astronomy/planets/earth/Mass.shtml | Login | Sign Up | Print Page
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The Calculation
How is the mass of the Earth determined? Newton, Galileo, Henry Cavendish, and Eratosthenes contributed to this amazing calculation.
This calculation is done using Newton’s Law of Gravity, which formulates the attractive force (gravitational force) that two masses exert on each other:
F=GmM/r2
In Newton’s equation, F is the gravitational force, G is a constant of proportionality, M and m are the two masses exerting the forces, and r is the distance between the two objects.
G was calculated by Henry Cavendish in 1798, and was determined to be 6.67 x 10-11 m3/(kg sec2).
Also needed is Newton’s second law of motion, F=ma, where F is the force applied to an object, m is the mass of the object, and a is its acceleration due to the force.
Galileo determined that the acceleration due to the force of gravity of Earth was a constant equal to 9.8 m/sec2 near the surface of the Earth.
Lastly, you need to know the radius of the Earth; this was first calculated by the Greek Eratosthenes thousands of years ago (by comparing shadows in wells during the summer solstice about 230 B.C.).
Calculating the mass of the Earth
F = GmM/r2 = ma, where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, r is the radius of the Earth, and m is the mass of another object (near the surface of the Earth).
GM/r2= a (The m’s canceled out.) Now solve for M, the mass of the Earth.
M = ar2/G, where a = 9.8m/sec2, r = 6.4 x 106m, and G = 6.67 x 10-11m3/(kg sec2).
M = 9.8 x (6.4 x 106)2/(6.67 x 10-11) = 6.0 x 1024 kg
The People Behind the Calculation
CAVENDISH, HENRY
: Henry Cavendish (1731-1810) was an English chemist and physicist. Cavendish discovered that hydrogen gas was a substance different from ordinary air (whose components he analyzed), described the composition of water (hydrogen and oxygen) and other important works. Cavendish was the first person to determine Newton’s gravitational constant gravitational constant and accurately measured of the Earth’s mass and density.
ERATOSTHENES
: Eratosthenes (276-194 BC) was a Greek scholar who was the first person to determine the circumference of the Earth. He compared the midsummer’s noon shadow in deep wells in Syene (now Aswan on the Nile in Egypt) and Alexandria. He properly assumed that the Sun’s rays are virtually parallel (since the Sun is so far away ). Knowing the distance between the two locations, he calculated the circumference of the Earth to be 250,000 stadia. Exactly how long a stasia is is unknown, so his accuracy is uncertain. He also accurately measured the tilt of the Earth’s axis and the distance to the sun and moon, and devised a method for determining the prime numbers up to a given number (the Sieve of Eratosthenes). Eratosthenes made numerous contributions to the sciences and arts in many fields, including geography, mathematics, astronomy, chronography (calendars), music, and literature. Eratosthenes was a brilliant all-around scholar; although not the top expert on any topic, he was well-versed in all subjects, and therefore nicknamed “Beta” (which is the second letter of the Greek alphabet).
GALILEI, GALILEO
: Galileo Galilei (1564-1642) was an Italian mathematician, astronomer, and physicist. He was the first person to use a telescope to observe the skies (in 1609). Galileo discovered four moons of Jupiter, observed the phases of Venus, studied sunspots, and discovered many other important phenomena. After publishing the many discoveries he made using his telescope, including the motion of the Earth around the Sun (the Copernican System), Galileo was accused of heresy by the Inquisition (in 1633).
NEWTON, ISAAC
: Isaac Newton (1642-1727) was an English mathematician and physicist who invented calculus (simultaneously, but independently of Leibniz), formulated the laws of gravitation and mechanics, investigated the nature of light (he discovered that sunlight is made of light of different colors).
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11194 | https://www.scribd.com/document/629249226/ch-071-Acute-Kidney-Injury-in-Children | CH 071 Acute Kidney Injury in Children | PDF | Medicine | Clinical Medicine
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CH 071 Acute Kidney Injury in Children
The document discusses acute kidney injury in children. It defines AKI and its stages according to the KDIGO classification system. It describes the causes of AKI which can be prerenal, intr…
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Indian Academy of Pediatrics (IAP)
Under the Auspices of the IAP Action Plan 2022
Remesh Kumar R
IAP President 2022
Vineet Saxena
IAP HSG 2022–2023
Piyush Gupta
IAP President 2021
Upendra Kinjawadekar
IAP President-Elect 2022
STANDARD T R E A T M E N T
GUIDELINES 2022
Acute Kidney Injury in Children
Lead Author
Mukta Manthan
Co-Authors
Bipin Jose, Mihir Sarkar
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© Indian Academy of Pediatric s
IAP Standard Tr eatment Guidelines Committee
Chairperson
Remesh Kumar R
IAP Coordinator
Vineet Saxena
National Coordinators
SS Kamath, Vinod H Ratageri
Member Secretaries
Krishna Mohan R, Vishnu Mohan PT
Members
Santanu Deb, Surender Singh Bisht, Prashant Kariya, Narmada Ashok, Pawan Kalyan
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Acu te Kidney Injury in Children
1
71
I n t r o d u c t i o n
Acute kidney injury (AKI) previously known as acute renal failure (ARF), is an important emergency where prompt and appropriate management is life-saving. AKI usually occurs in patients with previously normal renal function but may occasionally be superimposed on preexisting renal disease (acute-on-chronic renal failure). The incidence of AKI in pediatric intensive care unit (PICU) is around 30–40% with mortal ity rates of 40–50%.
N o m e n c l a t u r e a n d C l a s s i f i c a t i o n
;
Acute kidney injury is defined as an increase in serum creatinine by ≥0.3 mg/dL within 48 hours; or an increase in serum creatinine to ≥1.5 times baseline, which is known or presumed to have occurr ed within last 7 days; or a urine volume <0.5 mL/kg/h for 6 hours.
;
Acute kidney injury is further classified into three stages based on rise of serum creatinine or change in urine volume
(Table 1)
. It is recommended to use Kidney Disease: Improving Global Outcomes (KDIGO) classification for an early diagnosis of AKI so that measures are taken to prevent the progre ssion of the condition.
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Acute Kidney Injury in Children
4
N o m e n c l a t u r e a n d C l a s s i f i c a t i o n
TABLE 1:
Classification of acute kidney injury (KDIGO classific ation).
AKI severity Serum creatinine criteria Urine output criteria
Stage I
1.5–1.9 times baseline Or
≥
0.3 mg/dL increase<0.5 mL/kg/h for 6–12 hours
Stage II
Increase
≥
2–2.9 t i m e s b a s e l i n e<0.5 m L/k g/h f o r ≥1 2 h o u r s
Stage III
3.0 times baseline Or Increase in serum creatinine to
≥
4.0 mg/dL Or Initiation of renal replacement therapy Or In patients <18 years, decrease in eGFR to <35 mL/min per 1.73 m
2
<0.3 mL/kg/h for 24 hours or anuria for 12 hours(AKI: acute kidney injury; eGFR: estimated glomerular filtration rate; KDIGO: Kidney Disease: Improving Global Outcomes)
C a u s e s
The etiology of AKI has conventionally been classified as prerenal, intrinsic renal, or postrenal
(Box 1)
. Prerenal AKI occurs due to inadequate systemic and/or renal circulation, due to either systemic hypovolemia or renal hypoperfusion. Both pre- and postrenal categories can, if prolonged, lead to intrinsic renal failure.
BOX 1:
Causes of AKI.
Prerenal
;
Hypovolemia (dehydrat ion, blood loss, and diabetic ketoacidosis)
;
Third space losses (sept icemia and nephrotic syndr ome)
;
Congestive heart failure
;
Perinatal asphyxia
;
Drugs (ACE inhibitors, NSAIDs, and diuretics)
Intrinsic
Acute tubular necrosis
;
Prolong ed prerenal insult (see above)
;
Medications:
Aminoglycoside, radiocontrast, and NSAIDs
;
Exogenous toxins:
Diethylene glycol and methanol
;
Intravascular hemolysis and hemoglobinuria
;
Snake bite
;
Tumo r lysis syndrome
Contd...
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11195 | https://encompass.eku.edu/cgi/viewcontent.cgi?article=1247&context=etd | Published Time: Tue, 13 Oct 2020 21:22:29 GMT
Eastern Kentucky University
Encompass
Online Theses and Dissertations Student Scholarship January 2015
General Factoring Algorithms for Polynomials over Finite Fields
Wade Combs
Eastern Kentucky University
Follow this and additional works at:
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Combs, Wade, "General Factoring Algorithms for Polynomials over Finite Fields" (2015). Online Theses and Dissertations . 249. GENERAL FACTORING ALGORITHMS FOR POLYNOMIALS OVER FINITE FIELDS By Wade Combs Bachelor of Science, Mathematics Teaching Eastern Kentucky University Richmond, Kentucky 2013 Submitted to the Faculty of the Graduate School of Eastern Kentucky University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE May, 2015 Copyright c©Wade Combs, 2015 All rights reserved ii DEDICATION I dedicate this thesis to my parents, Harold Combs and Pamela Combs, whose love has made the many great things in my life possible. iii ACKNOWLEDGEMENTS I would like to begin by thanking my thesis advisor, Dr. Patrick Costello, for giving me the support and confidence necessary to take on the challenge of writing a paper of this magnitude. Before guiding me through the thesis writing process, Dr. Costello was very influential as my undergraduate academic advisor. He sparked my interest in mathematics and continues to ignite it every day. I would also like to thank the other members of my thesis committee, Dr. Bangteng Xu and Dr. Rachel Bishop-Ross, for their instructive and insightful comments regarding my thesis. Any academic work is made possible in part by those who take the time to carefully read it and value it. I am extremely thankful to have had a committee that read my work with great care. I want to thank Dr. Jeffrey Neugebauer and Dr. Kirk Jones for being excellent instructors of advanced mathematics. Both of these professors have contributed a great deal to my growth as a student and as a researcher. Finally, I want to thank my graduate advisor, Dr. Lisa Kay, for giving me thorough and detailed guidance throughout my entire time in graduate school at Eastern Kentucky University. Dr. Kay definitely made each of the many tasks I have had to complete in graduate school as smooth and feasible as possible. iv ABSTRACT In this paper, we generate algorithms for factoring polynomials with coefficients in finite fields. In particular, we develop one deterministic algorithm due to El-wyn Berlekamp and one probabilistic algorithm due to David Cantor and Hans Zassenhaus. While some authors present versions of the algorithms that can only factor polynomials of a certain form, the algorithms we give are able to factor
any polynomial over any finite field. Hence, the algorithms we give are the most general algorithms available for this factorization problem. After formulating the algorithms, we look at various ways they can be applied to more specialized in-quiries. For example, we use the algorithms to develop two tests for irreducibility and a process for finding the roots of a polynomial over a finite field. We conclude our work by considering how the Berlekamp and Cantor-Zassenhaus methods can be combined to develop a more efficient factoring process. vTable of Contents
Introduction 11 Preliminaries 3
1.1 Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Polynomials over Fields . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Field Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2 Berlekamp’s Method 28
2.1 Square-Free Factorization . . . . . . . . . . . . . . . . . . . . . . . 28 2.2 The General Factoring Algorithm . . . . . . . . . . . . . . . . . . . 36
3 The Cantor-Zassenhaus Method 51
3.1 Distinct Degree Factorization . . . . . . . . . . . . . . . . . . . . . 51 3.2 Equal Degree Factorization . . . . . . . . . . . . . . . . . . . . . . . 57 3.3 Applications of the Cantor-Zassenhaus Method . . . . . . . . . . . 67
Bibliography 77
vi Introduction
In 1967, Elwyn Berlekamp developed the first efficient method for finding factorizations of polynomials with coefficients in finite fields. His method is de-terministic and primarily relies on solving systems of linear equations using row reduction of matrices. The concept behind his factoring strategy is unbelievably clever, yet very accessible to students of mathematics at all levels. Over the years, mathematicians have formulated various algorithms based on Berlekamp’s factor-ing scheme that have the ability to completely factor polynomials over finite fields. Actually, a few researchers have been so motivated by the findings of Berlekamp that they have formulated their own separate strategies for factorization. Citing Berlekamp as a major influence, in 1981, David Cantor and Hans Zassenhaus developed a new probabilistic method for factoring. Their method is deeply rooted in the theory of fields but is ultimately easy to apply in specific problems. In their original paper, Cantor and Zassenhaus only demonstrated how to find nontrivial factorizations of polynomials using their method. Hence, they did not give a full algorithm for finding a polynomial’s complete factorization. However, as Cantor and Zassenhaus probably suspected, many mathematicians and computer scientists have since used their findings to formulate various algo-rithms and comprehensive factoring strategies. In this paper, our primary objective will be to thoroughly develop two fac-toring algorithms for polynomials over finite fields. The first will be deterministic and rely on the method of Berlekamp, while the second will be probabilistic and rely on the method of Cantor-Zassenahaus. The algorithms we present will be distinguishable by the fact that they represent the most general factoring algo-rithms available. What we mean by “general” is that our algorithms will have the ability to factor any polynomial over any finite field. Some authors present more specialized algorithms and exclude the generality that we will seek here. For example, Childs presents a factoring technique based on Berlekamp’s method that only handles polynomials over fields that have prime order. This technique excludes polynomials whose coefficients come from finite fields that have order pv,where p is a prime number and v is a positive integer greater than 1. Further, Shoup gives an algorithm based on the findings of Berlekamp that can only factor polynomials which are square-free. Hence, this algorithm cannot directly handle a polynomial that has repeated factors in its factorization. 1Of course, Childs and Shoup are both aware that general versions of their algorithms can be formulated. They present such specialized algorithms because factoring generally runs much better when the input polynomial is square-free and/or has coefficients that come from fields of prime order. While our algorithms will be able to handle such specialized cases, generality will be our primary desire. With that said, over the course of the paper, we will offer various tips in regard to the best ways to factor polynomials in practice. In order to generate factoring algorithms, we will require many preliminary results relating to finite fields and polynomials over fields. In fact, our first chapter will act as a stand-alone introduction to these concepts. Then, in Chapters 2 and 3, we will use the results from Chapter 1 to develop the algorithms. 2Chapter 1 Preliminaries
Our preliminaries will build the theory that is necessary for developing factor-ing algorithms in later chapters. We will begin by looking at some basic properties of finite fields. Then we will consider properties of polynomials whose coefficients come from fields. Finally, we will end the chapter by using field extensions to further delve into the structure of finite fields. Throughout all of our work, it is assumed that the reader has a good knowl-edge of the standard terms and theorems given in a first-semester course over group theory. However, with that said, we will explicitly state all results relating to finite fields and polynomials over fields that are used in the paper. For more information on any result given in this chapter, refer to Childs and Dummit.
1.1 Finite Fields
We begin by formulating the definition of a field in terms of the definition of a ring.
Definition 1.1: A ring R is a set equipped with the binary operations + and ·
(called addition and multiplication) that satisfies the following axioms: (1) R is an abelian group under addition. (2) Multiplication is associative: ( a · b) · c = a · (b · c) for all a, b, c ∈ R.(3) Multiplication distributes over addition: for all a, b, c ∈ R,
a · (b + c) = a · b + a · c and (a + b) · c = a · c + b · c.Note that the additive identity of a ring R will always be denoted by 0, and the additive inverse of an element a ∈ R will be denoted by −a. Now, we give the definition of a field. 3Definition 1.2: A field F is a ring that satisfies the following axioms: (1) F has a nonzero multiplicative identity, i.e., there is an element 1 F ∈ F with 1F 6 = 0 and 1 F · a = a · 1F = a for all a ∈ F .(2) Every nonzero element a ∈ F has a multiplicative inverse, i.e., there exists an element c ∈ F with a · c = c · a = 1 F . (We will typically denote the element c
by a−1.) (3) Multiplication is commutative: a · b = b · a for all a, b ∈ F .We will use F × to denote the set of all nonzero elements of F . The elements of F × will often be referred to as units . With respect to the first field axiom, we will generally write 1 in short for 1 F , but in cases where it may be unclear whether 1 represents an integer or a field element, we will use the notation 1 F . Also, in regard to the third field axiom, we will typically write ab instead of a · b.The most commonly studied infinite fields are the complex numbers C, the real numbers R, and the rational numbers Q. However, we will focus our attention on fields which have finitely many elements. For every prime number p, the inte-gers modulo p, usually denoted by Z/p Z, is a field; these are the most commonly used finite fields. In order to describe the general structure of a finite field, we require a few more definitions.
Definition 1.3: Let R be a ring. A nonzero element a ∈ R is called a zero divisor
if there exists a nonzero element b ∈ R such that either ab = 0 or ba = 0. It is not difficult to see that there are no zero divisors in a field. Let a be a nonzero element of the field F . Suppose there exists a member b ∈ F with
ab = ba = 0. Then b = 1 F b = ( a−1a)b = a−1(ab ) = a−10 = 0. Hence, a is not a zero divisor. Next, we define the characteristic of a ring.
Definition 1.4: Let R be a ring with multiplicative identity 1 R. The character-istic of R, denoted char( R), is defined to be the smallest positive integer m such that m · 1R = 0 if such an m exists, and 0 otherwise. Infinite fields, such as R and C, have characteristic 0. As we will see in our first major theorem regarding finite fields, the characteristic of any finite field 4is a prime number. Before giving this theorem, however, we recall the following standard result from group theory:
If G is a finite abelian group of order n and p is a prime dividing n, then G
contains an element of order p.
Now for the theorem:
Theorem 1.5: Let F be a finite field. Then char( F ) = p for some prime number
p. Moreover, the order of F is pv for some positive integer v.
Proof : Since F is finite, the characteristic of F must be a positive integer. Assume, by way of contradiction, that char( F ) is not a prime number. By the definition of a field, 0 6 = 1 F = 1 · 1F , and so char( F ) 6 = 1. Then char( F ) is composite. Say char( F ) = st , where s and t are positive integers with 0 < s <
char( F ) and 0 < t < char( F ). By the definition of characteristic, s · 1F 6 = 0 and
t · 1F 6 = 0. Since F has no zero divisors, it follows that ( s · 1F )( t · 1F ) 6 = 0. But, this implies 0 = ( st ) · 1F = ( s · 1F )( t · 1F ) 6 = 0 ,
a contradiction. Thus, char( F ) = p for some prime p.Next, let n be the order of F . Considering F as a group under addition, let ord( g) denote the order of an element g ∈ F . Since n ≥ 2, there must exist at least one prime number dividing n. Suppose m is a prime dividing n. Then since F is a finite abelian group under addition, by the mentioned result from group theory, there is an element a ∈ F with ord( a) = m. Applying the division algorithm for integers, we can find integers q and r with p = mq + r and 0 ≤ r < m . Notice that p · a = p · (1 F · a) = ( p · 1F ) · a = 0 · a = 0. Now, 0 = p · a
= ( mq + r) · a
= ( mq ) · a + r · a
= ( qm ) · a + r · a
= q(m · a) + r · a
= q · 0 + r · a
= 0 + r · a
= r · a.
5Since r < ord( a), it follows that r = 0. Thus, p = mq , and m divides p. But, since p is prime and m 6 = 1, it must be that m = p. Hence, p is the only prime divisor of n, which means n = pv for some positive integer v.
Theorem 1.5 shows us that the order of any finite field is some power of a prime number. In Section 1.3, we will establish the following related (and remark-able!) fact:
For every prime number p and positive integer v, up to isomorphism, there exists a unique finite field of order pv.
Notice that for any element a in a ring R and any integer n, since R is closed under addition, we have that na ∈ R. Using this observation, we now present a theorem which will aid us in establishing some important results in later chapters.
Theorem 1.6: Let R be a commutative ring of prime characteristic p. Then for any elements a, b ∈ R,(a + b)p = ap + bp.
Proof : Since R is commutative, we can apply the Binomial Theorem to write (a + b)p =
p
∑
k=0
(pk
)
ap−kbk,
where (pk
) = p!
k!( p−k)!
for each k. Because (pk
) is an integer, k!( p − k)! divides p!. Notice that for k ∈ { 1, · · · , p − 1}, the prime p is a factor of neither k! nor ( p − k)!, and so p and k!( p − k)! are relatively prime integers. Hence, for k ∈ { 1, · · · , p − 1},
k!( p − k)! divides ( p − 1)!, and we can write p! = pj k, where jk = (p−1)!
k!( p−k)!
∈ Z.Since R has characteristic p, it now follows that (a + b)p =
(p
0
)
ap +
p−1
∑
k=1
(pj k)ap−kbk +
(pp
)
bp
= ap +
p−1
∑
k=1
0 + bp
= ap + bp.
6For any integer v, it is important to see that applying Theorem 1.6 repeat-edly gives that ( a + b)pv
= apv
bpv
in R. Theorem 1.6 can always be applied to finite fields, but we will also require this theorem when dealing with commutative rings that are not fields and have prime characteristic. To conclude our introduction to finite fields, let us recall one of the most famous theorems from number theory. Fermat’s Little Theorem (FLT) says that if p is a prime number and a is a nonzero element of Z/p Z, then ap−1 = 1 in Z/p Z.We generalize this result in Theorem 1.7.
Theorem 1.7 (Generalized FLT): Let F be a field of order q. Then aq−1 = 1 for all a ∈ F ×.
Proof : Observe that F × is a multiplicative group of order q − 1. Then it is a con-sequence of Lagrange’s Theorem from group theory that aq−1 = 1 for all a ∈ F ×.
Multiplying both sides of the equation in Theorem 1.7 by a gives that aq = a
for all a ∈ F ×. In fact, since 0 q = 0, we have that aq = a for all a ∈ F . We will directly cite the Generalized FLT whenever this property of the elements of F is applied. In the upcoming section, we begin looking at polynomials over fields.
1.2 Polynomials over Fields
Let F be a field and x an indeterminate. We will use F [x] to denote the set of all finite sums anxn + an−1xn−1 + · · · + a1x + a0, called polynomials , where n
is a nonnegative integer and each ai ∈ F . If an 6 = 0, then the polynomial is of
degree n. The polynomial is called monic if an = 1. Notice that F ⊂ F [x]. The elements of F are called constant polynomials with respect to their membership in F [x]. We define addition in F [x] to be componentwise:
n
∑
i=1
aixi +
n
∑
i=0
bixi =
n
∑
i=1
(ai + bi)xi,
where some of the ai and bi terms may be 0, so that addition of polynomials of different degrees is defined. We define multiplication in F [x] by first defining (ax i)( bx j ) = abx i+j and then distributing multiplication over addition to get
( n∑
i=0
aixi
)
·
( m∑
i=0
bixi
)
=
n+m
∑
k=0
( k∑
i=0
aibk−i
)
xk.
7For an arbitrary polynomial f (x) ∈ F [x], let deg( f (x)) denote the degree of
f (x). Here are a few straightforward properties of F [x]:
• F [x] is a commutative ring having multiplicative identity 1 F and no zero divisors.
• The characteristic of F [x] is the same as the characteristic of F .
• For nonzero polynomials f (x), g (x) ∈ F [x], deg( f (x)g(x)) = deg( f (x)) + deg( g(x)).
• The elements of F [x] with multiplicative inverses are precisely the elements of F ×.We now provide some standard terminology and traditional results for ele-ments in F [x].
Definition 1.8: Let f (x), g (x) ∈ F [x]. If f (x) = g(x)h(x) for some h(x) ∈ F [x], then g(x) is said to divide f (x), and we write g(x)|f (x). The polynomial g(x) is called a factor or divisor of f (x). Recall from elementary algebra that dividing a polynomial in Q[x] by an-other (nonzero) polynomial in Q[x] yields a quotient and remainder. This still holds true over any field.
Theorem 1.9 (Division Algorithm): Let f (x), g (x) ∈ F [x] with g(x) 6 = 0. Then there exist unique polynomials q(x), r (x) ∈ F [x] such that
f (x) = g(x)q(x) + r(x) with r(x) = 0 or deg( r(x)) < deg( g(x)) . (∗)
Proof : Let g(x) 6 = 0 be fixed. We will prove that for any f (x), there exist polynomials q(x) and r(x) satisfying ( ∗). Let deg( f (x)) = m and deg( g(x)) = n.Then n is fixed, and we must show the necessary polynomials exist for all integers
m.If m < n , then the choices of q(x) = 0 and r(x) = f (x) satisfy the desired conditions. For m ≥ n, we proceed by strong induction on m. First, write
f (x) = amxm +am−1xm−1 +· · · +a1x+a0 and g(x) = bnxn +bn−1xn−1 +· · · +b1x+b0.Suppose that m = n. Set q(x) = am · b−1
m
and r(x) = f (x) − q(x)g(x). Observe that q(x) is well-defined since bm 6 = 0 and that the coefficient of the xm
term vanishes in r(x), which means deg( r(x)) = 0 or deg( r(x)) < deg( g(x)). Now 8clearly we have f (x) = g(x)q(x) + r(x). This takes care of the base case. For the inductive step, assume polynomials satisfying ( ∗) exist for all m
with n ≤ m < k , where k is a positive integer. Consider the case of m = k. Set
f0(x) = f (x) − amb−1
n
xm−ng(x). Notice that the xm term vanishes in f0(x) so that deg( f0(x)) < k . Then, by the inductive hypothesis, there exist polynomials q0(x)and r(x) such that
f0(x) = g(x)q0(x) + r(x) with r(x) = 0 or deg( r(x)) < n.
Now, letting q(x) = q0(x) + amb−1
n
xm−n, we get
f (x) = g(x)q(x) + r(x) with r(x) = 0 or deg( r(x)) < n.
So the case of m = k holds. Therefore, by strong induction, there exist polynomials q(x) and r(x) satisfying ( ∗) for all m ≥ n.For uniqueness, suppose the pairs q(x), r (x) and q1(x), r 1(x) both satisfy ( ∗). Then g(x)q(x) + r(x) = g(x)q1(x) + r1(x), and we get the equation
g(x)( q(x) − q1(x)) = r(x) − r1(x).
If r(x) = r1(x) = 0, then q(x) = q1(x) since g(x) is nonzero. So, assume that either r(x) 6 = 0 or r1(x) 6 = 0. This implies that both r(x) and r1(x) have degree
< n . Then r(x) − r1(x) = g(x)( q(x) − q1(x)) clearly has degree < n = deg( g(x)). Since the degree of the product of two nonzero polynomials is the sum of their degrees, it must be that q(x) − q1(x) = 0. Thus, q(x) = q1(x) and r(x) = r1(x).
Let f (x) and g(x) be polynomials over F with g(x) non-constant. We write the congruence f (x) ≡ h(x) (mod g(x)) for polynomials h(x) ∈ F [x] such that
g(x)|f (x) − h(x). Notice that the Division Algorithm guarantees that there is a unique polynomial r(x) over F which satisfies f (x) ≡ r(x) (mod g(x)) and deg( r(x)) < deg( g(x)). The polynomial r(x) is called the least residue of f (x)mod g(x). Sometimes we will simply write f (x)(mod g(x)) to denote the least residue. Since a division algorithm can be established for F [x], it follows that F [x]has many of the same properties as the integers. This starts to become apparent when investigating the greatest common divisor of two polynomials. 9Definition 1.10: The greatest common divisor of polynomials f (x), g (x) ∈ F [x]with g(x) 6 = 0 is the unique monic polynomial d(x) ∈ F [x] satisfying: (i) d(x)|f (x) and d(x)|g(x), and (ii) if h(x)|f (x) and h(x)|g(x) for some h(x) ∈ F [x], then h(x)|d(x). The greatest common divisor of f (x) and g(x) 6 = 0 will be denoted by gcd( f (x), g (x)). Informally speaking, gcd( f (x), g (x)) is the monic polynomial of largest degree which divides both f (x) and g(x). In the case gcd( f (x), g (x)) = 1, we say f (x) and g(x) are relatively prime .Considering the unique polynomials q(x), r (x) ∈ F [x] such that f (x) =
g(x)q(x) + r(x) with r(x) = 0 or deg( r(x)) < deg( g(x)) guaranteed by the Divi-sion Algorithm, it is easy to see that
• if r(x) = 0, then gcd( f (x), g (x)) = α·g(x), where α ∈ F is the multiplicative inverse of the leading coefficient of g(x).
• if r(x) 6 = 0, then gcd( f (x), g (x)) = gcd( g(x), r (x)). This suggests the following iterative algorithm for finding the gcd of f (x) and
g(x), which mirrors the Euclidean Algorithm for integers.
Euclidean Algorithm for Polynomials over F :
(1) Let f0(x) = f (x) and g0(x) = g(x). (2) Find the unique q(x), r (x) ∈ F [x] such that f0(x) = g0(x)q(x) + r(x) with
r(x) = 0 or deg( r(x)) < deg( g0(x)). (3) If r(x) = 0, then stop: gcd( f (x), g (x)) = α · g0(x), where α ∈ F is the multiplicative inverse of the leading coefficient of g0(x). (4) If r(x) 6 = 0, then replace f0(x) by g0(x) and g0(x) by r(x), and go back to (2). This process does indeed terminate in a finite number of steps, since the degree of
g0(x) decreases each time we cycle through (4). When the algorithm terminates, notice that we need to multiply the current value for g0(x) by the multiplicative inverse in F of its leading coefficient in order to meet the requirement that the gcd be monic. We apply the Euclidean Algorithm in the upcoming example. 10 Example 1.11: Consider the polynomials f (x) = x8 +3 x7 +x6 +x5 +4 x3 +3 x2 +3 and g(x) = x5 + x4 + 3 x3 + 4 x + 2 in ( Z/5Z)[ x]. Using polynomial long division, we find
f (x) = g(x)( x3 + 2 x2 + x + 4) + (4 x4 + 2 x3 + 2 x)
g(x) = (4 x4 + 2 x3 + 2 x)(4 x + 2) + (4 x3 + 2 x2 + 2) 4x4 + 2 x3 + 2 x = (4 x3 + 2 x2 + 2) x + 0 .
Thus, gcd( f (x), g (x)) = 4 · (4 x3 + 2 x2 + 2) = x3 + 3 x2 + 3. Since a Euclidean Algorithm can be established for F [x], we can formulate a result that parallels B´ ezout’s identity for the integers:
Theorem 1.12: Let f (x), g (x) ∈ F [x] with g(x) 6 = 0. Then there exist polyno-mials a(x), b (x) ∈ F [x] such that gcd( f (x), g (x)) = a(x)f (x) + b(x)g(x).
Theorem 1.12 can be proven in exactly the same manner as B´ ezout’s iden-tity by using the Euclidean Algorithm and backwards substitution. Utilizing this theorem, we now present a proposition that will be very valuable throughout the rest of our work.
Proposition 1.13: Let f be a polynomial in F [x]. If g and h are relatively prime polynomials in F [x], then gcd( f, gh ) = gcd( f, g ) · gcd( f, h ).
Proof : Let
d0 = gcd( f, gh ),d1 = gcd( f, g ),d2 = gcd( f, h ).
By Theorem 1.12, d1 = a1f + b1g and d2 = a2f + b2h for some a1, b 1, a 2, b 2 ∈ F [x]. Multiplying d1 and d2 results in the equality
d1d2 = ( a1a2f + a1b2h + b1a2g)f + ( b1b2)gh.
11 Note that d0|f and d0|gh . So d0 divides both terms in the sum on the right hand side of the above equation, and it follows that d0|d1d2.Also, since g and h are relatively prime, gcd( g, h ) = 1. Hence, applying Theorem 1.12 again, 1 = a3g + b3h for some a3, b 3 ∈ F [x]. Multiplying both sides of this equation by d0 gives
d0 = a3(d0g) + b3(d0h).
By the definition of greatest common divisor, d1|d0 and d2|d0. Additionally, d1|g
and d2|h. Thus, d1d2|d0g and d1d2|d0h, which implies d1d2|d0. Now since d0 and
d1d2 are both monic polynomials, it must be that d0 = d1d2.
Theorem 1.12 can also be used to get this significant result:
Theorem 1.14: Let f (x), g (x), h (x) be polynomials in F [x]. If f (x)|g(x)h(x) and gcd( f (x), g (x)) = 1, then f (x)|h(x). The strategy of proof for Theorem 1.14 is nearly identical to the strategy used in the second part of the proof of Proposition 1.13. Recall that any positive integer can be factored uniquely into a product of prime numbers. We will see that unique factorization also holds in F [x]. First, consider the definition:
Definition 1.15: Suppose p(x) is a non-constant polynomial in F [x]. Then p(x)is called irreducible if whenever p(x) = a(x)b(x) with a(x), b (x) ∈ F [x], either
a(x) ∈ F × or b(x) ∈ F ×. Otherwise, p(x) is said to be reducible .Essentially, a non-constant polynomial in F [x] is irreducible if it cannot be written as the product of two positive degree polynomials. For instance, the poly-nomial x + 1 ∈ (Z/3Z)[ x] is irreducible, since x + 1 = g(x)h(x) surely implies that either g(x) or h(x) is a unit. On the other hand, x2 + 2 ∈ (Z/3Z)[ x] is reducible, since x2 + 2 = ( x + 1)( x + 2) over Z/3Z.Irreducible elements in F [x] carry many of the same properties as prime numbers. Consider, for example, the upcoming theorem, which extends Euclid’s Lemma for integers to F [x].
Theorem 1.16: Suppose p(x) is an irreducible element of F [x] and p(x)|g(x)h(x)for some g(x), h (x) ∈ F [x]. Then either p(x)|g(x) or p(x)|h(x). 12 Proof : Suppose p(x) - g(x). Then, since p(x) is irreducible, gcd( p(x), g (x)) = 1. By hypothesis, p(x)|g(x)h(x). Thus, by Theorem 1.14, p(x)|h(x).
Naturally, we get the following corollary.
Corollary 1.17: Suppose p(x) is irreducible and p(x)|g1(x)g2(x) · · · gn(x) over F .Then p(x)|gi(x) for some i = 1 , 2, · · · , n .Now we are ready to establish unique factorization for F [x].
Theorem 1.18: Every non-constant polynomial in F [x] can be factored into a product of irreducible polynomials. The factorization is unique up to rearrange-ment of the irreducibles and multiplication by elements of F ×.
Proof : Let S be the set of all non-constant polynomials in F [x] which cannot be factored into a product of irreducibles. Assume, by way of contradiction, that S 6 = ∅. Let D = {deg( s(x)) : s(x) ∈ S}. Since D is a non-empty set of positive integers, it follows from the well-ordering principle that D has a least element, say n. Let p(x) be an element of S with deg( p(x)) = n. Since p(x)cannot be written as a product of irreducibles, p(x) is clearly not irreducible itself. Hence, p(x) = g(x)h(x) for some h(x), g (x) ∈ F [x] with 1 ≤ deg( g(x)) < n
and 1 ≤ deg( h(x)) < n . Then g(x), h(x) /∈ S, and so both g(x) and h(x)can be written as a product of irreducibles. Say g(x) = g1(x)g2(x) · · · gr(x) and
h(x) = h1(x)h2(x) · · · ht(x), where the gi(x) and hi(x) are irreducibles. Then
p(x) = g1(x)g2(x) · · · gr(x)h1(x)h2(x) · · · hs(x).
is a product of irreducibles, a contradiction. Therefore, S = ∅.For uniqueness, suppose
a1(x)a2(x) · · · an(x) = b1(x)b2(x) · · · bm(x), () where the ai(x) and bi(x) are irreducibles. Now, since a1(x)|b1(x)b2(x) · · · bm(x), by Corollary 1.17, a1(x)|bi(x) for some 1 ≤ i ≤ m. If necessary, we can reindex the
bi(x)’s to get a1(x)|b1(x). Since b1(x) is irreducible, it follows that b1(x) = β1 ·a1(x)for some β1 ∈ F ×. Then dividing both sides of the equation ( ∗∗ ) by a1 gives
a2(x)a3(x) · · · an(x) = β1b2(x)b3(x) · · · bm(x).
13 Since a2(x)|b2(x)b3(x) · · · bm(x), a2(x)|bi(x) for some 2 ≤ i ≤ m. Reindexing the
bi(x)’s again, if necessary, we may write a2(x)|b2(x). Then a2(x) = β2 · b2(x) for some β2 ∈ F ×. Continuing this process, we get that ai(x) = βi · bi(x), βi ∈ F ×,for each i = 1 , 2, · · · n. In particular, this shows that n ≤ m.For a contradiction, suppose that n < m and let d = m − n. Now, dividing both sides of equation ( ∗∗ ) by a1(x)a2(x) · · · an(x), we get 1 = ( β1 · · · βn) · bn+1 (x) · · · bn+d(x).
But, the left hand side of this equation has degree 0, while the right hand side has positive degree. This is a contradiction. Thus, n = m.
Let f (x) be a non-constant polynomial in F [x]. Note that for each irreducible factor g(x) of f (x), g(x) = β · h(x) for some monic irreducible polynomial h(x)and some β ∈ F ×. In view of Theorem 1.18, this suggests that f (x) can be uniquely factored into the product of a nonzero constant and monic irreducibles. Collecting repeated monic irreducibles in this factorization, it follows that f (x)can be written uniquely in the form
f (x) = α · f1(x)k1 f2(x)k2 · · · fm(x)km , (#) where α ∈ F ×, the fi(x) are pairwise distinct monic irreducibles, and the ki are positive integers satisfying ki ≤ kj for i ≤ j. We will call the form (#) the com-plete factorization of f (x). In particular, if f (x) is monic, then α = 1 in (#). In Chapters 2 and 3, our primary objective will be to develop algorithms which can find the complete factorization of an arbitrary polynomial over a finite field. Before inputting a polynomial into any algorithm, however, we can often inspect the polynomial’s roots to gain information about its factorization.
Proposition 1.19: Let f (x) ∈ F [x]. Then f (x) has a factor of degree 1 if and only if f (x) has a root in F , i.e., there exists an α ∈ F such that f (α) = 0.
Proof : Suppose f (x) has a factor of degree 1. Since F is a field, we may assume the factor is monic and hence is of the form x − α with α ∈ F . Then f (x) = (x − α)q(x) for some q(x) ∈ F [x], and f (α) = 0 · q(α) = 0. For the converse, suppose f (α) = 0 for some α ∈ F . By the Division Algorithm,
f (x) = ( x − α)q(x) + r,
14 for some q(x) ∈ F [x] and constant r. Then 0 = f (α) = r, and so ( x − α) is a degree 1 factor of f (x).
We now use Proposition 1.19 to establish a very useful irreducibility test for small degree polynomials.
Proposition 1.20: A polynomial of degree 2 or 3 in F [x] is reducible if and only if it has a root in F .
Proof : A polynomial of degree 2 or 3 is reducible if and only if it has a factor of degree 1 if and only if it has a root in F .
This proposition is applied in the upcoming example.
Example 1.21: Consider the polynomial f (x) = x3 + 2 x + 1 ∈ (Z/3Z)[ x]. Notice that
f (0) = 1
f (1) = 1
f (2) = 1 .
So f (x) has no root in Z/3Z and hence is irreducible by Proposition 1.20. We will end our overview of the basic properties of F [x] with a powerful result known as the Chinese Remainder Theorem. Before giving the theorem, however, we need this fairly intuitive proposition.
Proposition 1.22: Let f (x), a (x), b (x) ∈ F [x] with a(x) and b(x) relatively prime. Then a(x)b(x)|f (x) if and only if a(x)|f (x) and b(x)|f (x).
Proof : The first direction of the statement is trivial. For the other direction, suppose a(x)|f (x) and b(x)|f (x). Then f (x) = a(x)h(x) for some h(x) ∈ F [x], and b(x)|a(x)h(x). Since gcd( a(x), b (x)) = 1, it now follows from Theorem 1.14 that b(x)|h(x). So, a(x)b(x)|a(x)h(x). That is, a(x)b(x)|f (x).
Now, we are ready for the theorem. 15 Theorem 1.23 (Chinese Remainder Theorem): Let g1(x), g 2(x), · · · , g n(x)be arbitrary polynomials, and m1(x), m 2(x), · · · , m n(x) be non-constant pairwise relatively prime polynomials in F [x]. Set m(x) = m1(x)m2(x) · · · mn(x) and let
d = deg( m(x)). Then there exists a unique polynomial r(x) ∈ F [x] with
r(x) ≡ g1(x) (mod m1(x))
r(x) ≡ g2(x) (mod m2(x)) ...
r(x) ≡ gn(x) (mod mn(x)) ,
and deg( r(x)) < d .
Proof : Since mi(x) is relatively prime to mj (x) for all j 6 = i, mi(x) is relatively prime to the product
pi(x) = ∏
j6=i
mj (x).
Then gcd( mi(x), p i(x)) = 1, and so, by Theorem 1.12, there exist ai(x), b i(x) in
F [x] with 1 = ai(x)mi(x) + bi(x)pi(x).
Observe that bi(x)pi(x) satisfies the congruences
bi(x)pi(x) ≡ 1 (mod mi(x)) ,bi(x)pi(x) ≡ 0 (mod mj (x)) for every j 6 = i.
Set
f (x) = g1(x)b1(x)p1(x) + g2(x)b2(x)p2(x) + · · · + gn(x)bn(x)pn(x).
Then clearly f (x) ≡ gi(x) (mod mi(x)) for each i = 1 , 2, · · · , n .Since m1(x), m 2(x), · · · , m n(x) are pairwise relatively prime, it follows from Proposition 1.22 that f (x) ≡ h(x) (mod mi(x)) for each i = 1 , 2, · · · , n if and only if f (x) ≡ h(x) (mod m(x)). By the Division Algorithm, there exists a unique polynomial r(x) with f (x) ≡ r(x) (mod m(x)) and deg( r(x)) < d . Then r(x) is the unique polynomial of degree less than d satisfying r(x) ≡ gi(x) (mod mi(x)) for each i = 1 , 2, · · · , n .
16 The Chinese Remainder Theorem will be applied multiple times in Chapter 2. In particular, the theorem will be very useful in the case that the gi(x)’s in the statement of theorem are constant polynomials.
1.3 Field Extensions
Here we will delve further into the structure of finite fields by looking at field extensions. Many of the results in this section can only be proven using high-powered facts from ring theory. For our purposes, there is little consequence of stating or investigating such facts. Hence, in order to naturally progress our work, we will omit quite a few proofs over the course of the section. Most omitted proofs can be found in Dummit, pages 509-545. Let F be a field and f (x) a polynomial in F [x] of degree n > 0. Consider the set (f (x)) = {f (x)g(x) : g(x) ∈ F [x]}.
In ring theory, ( f (x)) is called the principal ideal of F [x] generated by f (x). Now, we equip the (additive) quotient group
F [x]/(f (x)) = {h(x) + ( f (x)) : h(x) ∈ F [x]}
with the binary operations [a(x) + ( f (x))] + [ b(x) + ( f (x))] = [ a(x) + b(x)] + ( f (x)) and [a(x) + ( f (x))] · [b(x) + ( f (x))] = a(x)b(x) + ( f (x)) .
It is well-known that these operations are well-defined in F [x]/(f (x)), and that, under these operations, F [x]/(f (x)) is a ring (generally called a quotient ring ). Recall from the properties of quotient groups that h(x) + ( f (x)) = g(x) + ( f (x)) iff h(x) − g(x) ∈ (f (x)) iff h(x) ≡ g(x) (mod f (x)). By the Division Algo-rithm, for any h(x) ∈ F [x], there is a unique r(x) with h(x) ≡ r(x) (mod f (x)) and deg( r(x)) < n . Hence, r(x) is the unique polynomial in F [x] satisfying
h(x) + ( f (x)) = r(x) + ( f (x)) and deg( r(x)) < n . This shows that every ele-ment of F [x]/(f (x)) is represented by a polynomial in F [x] of degree < n . So, when referring to the elements of F [x]/(f (x)), we will usually refer to their rep-resentatives of degree < n in F [x]. Furthermore, we will do computations in 17 F [x]/(f (x)) by computing congruences mod f (x). Thus, when we compute the congruence h(x) ≡ g(x) (mod f (x)), it should be interpreted as the equality
h(x) + ( f (x)) = g(x) + ( f (x)). We now present the relationship between quotients rings and irreducible polynomials.
Proposition 1.24: Let p(x) be an irreducible polynomial over F . Then the quo-tient ring F [x]/(p(x)) is a field. We observe that the result of Proposition 1.24 does not hold for reducible polynomials. To see this, suppose that f (x) is reducible over F . Then f (x) =
a(x)b(x) for some non-constant polynomials a(x) and b(x) of degree < deg( f (x)). Now, a(x) and b(x) correspond to nonzero elements of F [x]/(f (x)) and satisfy
a(x)b(x) ≡ 0 (mod f (x)). This shows that F [x]/(f (x)) has zero divisors and hence cannot be a field. Note that if F and L are fields with F ⊆ L, then we say that F is a subfield
of L. Oftentimes, we only use the term subfield when we find a subset of a given field which is itself a field. On the other hand, we use the term extension field
when we find a superset of a given field which is itself a field. This is formalized in the upcoming defintion.
Definition 1.25: If L is a field containing the subfield F , then L is said to be an
extension field (or simply an extension ) of F , denoted L : F .Before proceeding any further, we make a few notes about vector spaces. When referencing a vector space V over the field F , we may use some of the fol-lowing terms: basis , dimension , linearly independent , scalar , span , subspace , and
vector . All of these terms, including vector space , have the same meaning as they do in a first year course in linear algebra. The only difference is that our vectors come from the arbitrary additive abelian group V rather than coming from Rn
exclusively, and our scalars come from the arbitrary field F rather than coming from R exclusively. If the reader is unfamiliar with vector spaces and any related terms, a wealth of information about these concepts is available online and in standard algebra texts. We observe that if L : F is some extension of fields, then the multiplication defined in L makes L into a vector space over F .18 Definition 1.26: The degree of a field extension L : F , denoted [ L : F ], is the dimension of L as a vector space over F . The extension is called finite if [ L : F ]is finite and is called infinite otherwise. In the next theorem, we give a very important property of extension degrees:
Theorem 1.27: Let F ⊆ K ⊆ L be fields. Then [ L : F ] = [ L : K][ K : F ]. Let F = Z/p Z, with p a prime number, and suppose that L is an extension of F with [ L : F ] = v ∈ Z+. Then since L is a v-dimensional vector space over F ,there is a subset {l1, l 2, · · · , l v} of L that forms a basis for L over F . Hence, each element l ∈ L can be written uniquely in the form l = a1l1 +a2l2 +· · · +avlv, where
a1, a 2, · · · , a v ∈ F . Notice that when forming an arbitrary element of L, there are
p choices for each of the ai’s, 1 ≤ i ≤ v. Thus, L has exactly pv elements. This demonstrates a strategy for finding new finite fields: Starting with Z/p Z, we can find a finite field of order pv if we can find an extension of Z/p Z that has degree
v.Naturally, we now ask: is it possible to find extensions of Z/p Z with spec-ified (finite) degrees? Remarkably, the answer to this question does turn out to be “yes”. In fact, we often use irreducible polynomials to construct extensions of a particular degree. To demonstrate how this is done, we start out with a result that says any irreducible polynomial over a field F has a root in some extension of F .
Theorem 1.28: Let F be a field and p(x) ∈ F [x] an irreducible polynomial. Then there is an extension L : F and an element θ ∈ L such that p(θ) = 0.
Proof Idea : Let I = ( p(x)) and L = F [x]/I . By Proposition 1.24, L is a field. Define the map φ : F → L by φ(a) = a + I. It can be seen that the image of this map is isomorphic to the field F , i.e., φ(F ) ∼= F . Hence, L contains an isomorphic copy of F , and so we can think of L as an extension of F . Let θ = x + I, and suppose p(x) = a0 + a1x + · · · + anxn ∈ F [x]. Then
p(θ) = ( a0 + I) + a1(x + I) + · · · + an(x + I)n
= ( a0 + a1x + · · · + anxn) + I
= p(x) + I = 0 + I = 0 ∈ L.
19 Now, given an irreducible p(x) over F , we know the field L = F/ (p(x)) is an extension of F which contains a root of p(x). The next theorem gives the degree of the extension L : F as well as a way to explicitly represent L.
Theorem 1.29: Let p(x) be an irreducible polynomial of degree n over the field F , and let L = F [x]/(p(x)). Let θ = x(mod p(x)) ∈ L. Then the set
{1, θ, θ 2, · · · , θ n−1} is a basis for L as a vector space over F . Hence, the degree of the extension L : F is n, and
L = {a0 + a1θ + a2θ2 + · · · + an−1θn−1 : a0, a 1, a 2, · · · , a n−1 ∈ F }. (##) Notice that Theorem 1.29 looks at the specific extension F [x]/(p(x)) of F
and the specific root x(mod p(x)) of p(x) in this extension. It turns out that we do not have to be so specific; we can get an extension of F identical to (##) by simply defining an arbitrary root of p(x) in some arbitrary extension of F (note that Theorem 1.28 guarantees the existence of such a root). Before formally pre-senting this result, however, we give the following definition.
Definition 1.30: Let L : F be an extension of fields and let α ∈ L. Denote by
F (α) the smallest subfield of L which contains both F and the element α. We call
F (α) the field generated by α over F .Now, we give the mentioned result.
Theorem 1.31: Let F be a field and p(x) an irreducible polynomial over F . Let
α be a root of p(x) in some extension L of F . Then
F (α) ∼= F [x]/(p(x)) .
We apply Theorem 1.31 in the upcoming example.
Example 1.32: Let p(x) = x4 + 2 x + 2 ∈ (Z/3Z)[ x]. It can be shown that p(x)has no linear factors or quadratic factors over Z/3Z. So, p(x) must be irreducible. Now, we let α be an arbitrary root of p(x) in some extension of Z/3Z. Then, it 20 follows from Theorem 1.31 that (Z/3Z)( α) = {a + bα + cα 2 + dα 3 : a, b, c ∈ Z/3Z}
is a field with 3 4 = 81 elements. Furthermore, we can do computations in (Z/3Z)( α) by using the fact that α is a root of p(x). Hence, we use the fact that α4 = −2α − 2 = α + 1 over Z/3Z. For instance, we multiply the elements
α + 1 and α3 + 2 as follows: (α + 1)( α3 + 2) = α4 + α3 + 2 α + 2 = ( α + 1) + α3 + 2 α + 2 = α3.
For a positive integer v, we have established that if there exists an irreducible polynomial over Z/p Z of degree v, then there exists a corresponding finite field of order pv, namely ( Z/p Z)[ x]/(p(x)). However, we have not established that for every positive integer v, an irreducible polynomial over Z/p Z of degree v actually exists. So, we have not yet shown there exists a finite field of order pv. To get this result (as was promised in Section 1.1), we need to look at a specific class of extension fields called splitting fields . Preceding our look at splitting fields, we will give a few more results related to roots of polynomials.
Definition 1.33: Let L : F be an extension of fields. The element α ∈ L is said to be algebraic over F if α is a root of some nonzero polynomial f (x) ∈ F [x], i.e.,
f (α) = 0. The next proposition gives the relationship between algebraic elements and irreducible polynomials.
Proposition 1.34: Let L : F be an extension of fields, and let α ∈ L be algebraic over F . Then there is a unique monic irreducible polynomial mα(x) ∈ F [x] which has α as a root. Moreover, a polynomial f (x) ∈ F [x] has α as root if and only if
mα(x) divides f (x) in F [x].
Proof : Let g(x) ∈ F [x] be a polynomial of minimal positive degree having α
as root. Multiplying g(x) by a unit in F , we may assume that g(x) is monic. Now, for a contradiction, suppose that g(x) is reducible. Then g(x) = a(x)b(x)for some polynomials a(x), b (x) ∈ F [x], both of degree smaller than the degree of 21 g(x). Then 0 = g(α) = a(α)b(α) in L. Since L has no zero divisors, it follows that a(α) = 0 or b(α) = 0, which contradicts the minimality of the degree of g(x). Thus, g(x) is irreducible. Next, suppose that f (x) ∈ F [x] has α as a root. By the Division Algorithm, there are polynomials q(x), r (x) ∈ F [x] such that
f (x) = g(x)q(x) + r(x) with r(x) = 0 or deg( r(x)) < deg( g(x)) .
Then 0 = f (α) = g(α)q(α) + r(α) = 0 + r(α) = r(α) in L. Now, if r(x) 6 = 0, then
r(x) is a non-constant polynomial of degree less than g(x) having α as root, which contradicts the minimality of the degree of g(x). Hence, r(x) = 0, and it follows that g(x) divides f (x) in F [x]. We have established that g(x) divides any polynomial over F which has α as root. In particular, g(x) would divide any other monic irreducible polynomial over
F having α has a root. So, it immediately follows that mα(x) = g(x) is unique.
Definition 1.35 The polynomial mα(x) in Proposition 1.34 is called the minimal polynomial for α over F .We now turn our attention to splitting fields.
Definition 1.36: Let F be a field and f (x) a polynomial of degree n > 0 in F [x]. An extension K of F is called a splitting field for f (x) (over F ) if (i) f (x) factors completely into linear factors (or splits completely ) in K[x], i.e.,
f (x) = c(x − a1)( x − a2) · · · (x − an) for some c, a 1, a 2, · · · , a n ∈ K, and (ii) f (x) does not factor completely into linear factors over any proper subfield of K containing F .Here is the major theorem regarding splitting fields.
Theorem 1.37: Let F be a field and f (x) ∈ F [x]. Then f (x) has a splitting field over F which is unique up to isomorphism. Often, when we are looking at a polynomial f (x) over its splitting field K,we are interested in whether f (x) has any duplicate linear factors over K. Intu-itively, we say that α ∈ K is a multiple root of f (x) if ( x − α)2|f (x) in K[x]. 22 In order to test f (x) for multiple roots, and hence test for duplicate linear factors, we typically inspect the value of its derivative .
Definition 1.38: The derivative of the polynomial
f (x) = anxn + an−1xn−1 + · · · + a1x + a0 ∈ K[x]is defined to be the polynomial
f ′(x) = na nxn−1 + ( n − 1) an−1xn−2 + · · · + 2 a2x + a1 ∈ K[x].
Notice that the definition of the derivative of a polynomial over an arbitrary field is purely algebraic. Hence, the analytic notion of a limit plays no part in this definition of a derivative. The reason for this is that limits, which are con-tinuous operations, cannot be taken in certain fields. With that said, the same differentiation formulas given in a first-year single variable Calculus course still hold true. For example, we get the following formulas for differentiating a sum and a product:
• (f + g)′(x) = f ′(x) + g′(x)
• (f g )′(x) = f ′(x)g(x) + f (x)g′(x). We now show how derivatives can be used to test for multiple roots.
Proposition 1.39: Let f (x) be a polynomial over the field K. If f ′(x) has no root in K, then f (x) has no multiple root in K.
Proof: Suppose that f (x) has a multiple root at α ∈ K. Then f (x) = ( x−α)2h(x)for some h(x) ∈ K[x]. Now, by the formula for differentiating a product,
f ′(x) = 2( x − α)h(x) + ( x − α)2h′(x)= ( x − α)(2 h(x) + ( x − α)h′(x)) .
Thus, α is a root of f ′(x).
In the next example, we use splitting fields to demonstrate the existence and uniqueness of a particular finite field. 23 Example 1.40: Consider x4 − x ∈ (Z/2Z)[ x]. Notice that
x4 − x = x(x − 1)( x2 + x + 1) .
Since x2 + x + 1 is irreducible over Z/2Z, it follows that Z/2Z is not the splitting field for x4 − x. Now, let α be an arbitrary root of x2 + x + 1 in some extension of Z/2Z, and consider the field (Z/2Z)( α) = {a + bα : a, b ∈ Z/2Z}.
Notice that
x4 − x = x(x − 1)( x − α)( x − (1 + α)) .
Clearly x4 − x does not split completely over any proper subfield of ( Z/2Z)( α). Hence, ( Z/2Z)( α) is the splitting field for x4 − x over Z/2Z.Moreover, we note that ( Z/2Z)( α) is a finite field containing 4 elements. Suppose that F is another finite field of order 4. Then it follows from Theorem 1.5 that
F has characteristic 2 and hence must contain an isomorphic copy of Z/2Z as a subfield. Furthermore, by the Generalized FLT, each element a ∈ F is a root of the polynomial x4 − x in F[x]. Correspondingly, x4 − x has four distinct linear factors in F[x]. Since x4 − x is also of degree 4, it must be that F is a splitting field for
x4 −x over Z/2Z. By Theorem 1.37, splitting fields are unique up to isomorphism, and so we have that ( Z/2Z)( α) ∼= F. This shows that, up to isomorphism, there exists a unique finite field of order 4. With the strategy of Example 1.40 in mind, we prove there exists a unique finite field of any prime power order.
Theorem 1.41: For every prime number p and positive integer v, up to isomor-phism, there exists a unique finite field of order pv.
Proof: Let q = pv, and consider the polynomial xq − x ∈ (Z/p Z)[ x]. Let K be the splitting field for xq − x over Z/p Z. Since K has Z/p Z as a subfield, K must have characteristic p. So, the derivative of xq − x over K is qx q−1 − 1 = −1. Since the derivative of xq − x has no roots, by Proposition 1.38, xq − x has no multiple roots in K. Because K is also the splitting field for xq − x, it follows that xq − x has exactly q distinct roots in K. Let A be the set containing these q distinct roots, i.e., A = {a ∈ K : aq = a}. We note the following properties of A:24 • For any a, b ∈ A, it follows from Theorem 1.6 that ( a − b)q = aq − bq = a − b,and so a − b ∈ A. This shows that A is closed under addition and additive inverses;
• For any a, b ∈ A with b 6 = 0, since multiplication is commutative in the field
K, ( ab −1)q = aq(b−1)q = aq(bq)−1 = ab −1, and so ab −1 ∈ A. This establishes that A is closed under multiplication and multiplicative inverses. Since A inherits all of the other properties of a field from K, it follows that A is a subfield of K. Clearly A has characteristic p, and so contains Z/p Z. Furthermore,
xq − x splits completely in A, since A contains all of its roots. Hence, the splitting field of xq − x over Z/p Z is a subfield of A; that is, K is a subfield of A. Thus,
K = A, and K is a finite field of order q.For uniqueness, suppose that F is a finite field with q elements. Then, by Theorem 1.5, F has characteristic p and hence contains Z/p Z. Now, it follows from the Generalized FLT that aq − a = 0 for all elements a ∈ F. So each of the
q elements of F is a root of xq − x. Correspondingly, xq − x has q distinct linear factors in F[x]. Since xq − x also has degree q, it must be that F is a splitting field for xq − x over Z/p Z. Since splitting fields are unique up to isomorphism, we have that K ∼= F. Thus, up to isomorphism, K is the unique finite field of order q.
Observe that the proof of Theorem 1.41 shows us that the splitting field for the polynomial xq − x over Z/p Z is precisely the unique field of order q = pv .Throughout the remainder of our work, we will denote the unique finite field of order q = pv by Fq. Specifically, in the case where v = 1, we will use the notation
Fp in place of Z/p Z.The final results of this section give a few additional properties of Fq.
Proposition 1.42: The multiplicative group of nonzero elements of Fq, denoted (Fq)×, is cyclic.
Proof: Since ( F2)× contains only one element, it is clearly cyclic. So, we may assume that q ≥ 3. Set r = q − 1, the order of ( Fq)×, and let r = p1w1 p2w2 · · · pkwk
be its factorization into powers of distinct primes. We will denote the order of an element a in the multiplicative group ( Fq)× by ord( a). Recall from group theory that if m is a positive integer with am = 1, then ord( a)|m. In particular, since
ar = 1, we have that ord( a)|r.Now, for each 1 ≤ i ≤ k, the polynomial xr/p i − 1 has at most r/p i roots in
Fq. Then since r/p i < r , there exists an element ai ∈ (Fq)× which is not a root of xr/p i − 1. Set bi = air/p wii . Then bipiwi
= air = 1, and it follows that ord( bi)25 divides piwi . But,
bipiwi−1
= air/p i 6 = 1 ,
which shows that ord( bi) cannot be a proper divisor of piwi . So it must that ord( bi) = piwi .Let b = b1b2 · · · bk ∈ (Fq)×. Assume, by way of contradiction, that ord( b) 6 = r.Then ord( b) is a proper divisor of r and hence must divide r/p i for some 1 ≤ i ≤ k.Without loss of generality, assume that ord( b) divides r/p 1. Then 1 = br/p 1 = b1r/p 1 b2r/p 1 · · · bkr/p 1 .
Notice that for 2 ≤ i ≤ k, ord( bi) divides r/p 1, which implies bir/p 1 = 1. This forces b1r/p 1 = 1, and it follows that ord( b1) divides r/p 1. But, this contradicts that fact that ord( b1) = p1w1 . Thus, ord( b) = r = q − 1, and b is a generator for (Fq)×. This shows that ( Fq)× is cyclic.
We can use Proposition 1.42 to establish an important result related to ir-reducible polynomials in Fq[x].
Proposition 1.43: For every positive integer n, there exists an irreducible poly-nomial of degree n over Fq.
Proof: Let α be a generator of the cyclic group ( Fqn )×. Then clearly Fqn = Fq(α). Note that since α is a root of the polynomial xqn
− x ∈ Fq[x], α is algebraic over
Fq. Now, consider the minimal polynomial, mα(x), of α over Fq. By definition,
mα(x) is irreducible. Suppose that deg( mα(x)) = d. Then Fq[x]/(mα(x)) ∼= Fqd .But, by Theorem 1.31,
Fq[x]/(mα(x)) ∼= Fq(α) = Fqn .
Thus, qd = qn, and d = n.
In particular, for any prime number p and positive integer v, Proposition 1.43 says that there exists an irreducible polynomial, say f (x), over Fp of degree
v. Then Fp[x]/(f (x)) ∼= Fq, where q = pv. Now, if f (x) is known, then we can use Theorem 1.29 to do computations in Fp[x]/(f (x)), which is equivalent to doing computations in the finite field Fq. The hard part is actually finding the irreducible polynomial f (x). In Chapter 3, we will use results relating to our factoring algo-rithms to develop a process for finding irreducible polynomials of any given degree 26 over any finite field. In particular, the significance of being able to find irreducible polynomials of any degree over Fp is that we can do computations in any finite field of our choosing. Now, in the following chapters, we use the theory that we have built to generate factoring algorithms for polynomials over Fq.27 Chapter 2 Berlekamp’s Method
In this chapter, we will develop a deterministic algorithm for factoring poly-nomials over the finite field Fq, where q = pv with p a prime number and v a posi-tive integer. The general factoring method we will present is due to Berlekamp and will provide us with a way to completely factor any polynomial over Fq. Before generating the main algorithm, however, we offer an optional pre-processing stage called Square-Free Factorization (SFF). As mentioned in our introduction, factor-ing algorithms generally run much better when the input polynomial is square-free. This is so much the case that some authors actually formulate condensed algo-rithms that are restricted to square-free inputs. While no algorithm in this paper has such a restriction, we often recommend that SFF be employed as the initial factoring step. We note that all algorithms we present will only accept input polynomials which are monic. We do not lose any generality with this restriction. For example, if f (x) = ∑ni=0 aixi is a polynomial over Fq of degree n with an 6 = 1, we start by factoring the monic polynomial a−1
n
f (x). Then multiplying the factorization of
a−1
n
f (x) by an will give the factorization of f (x).
2.1 Square-Free Factorization
Loosely speaking, a polynomial f (x) ∈ Fq[x] is square-free if it has no re-peated non-constant factors. We give the formal definition:
Definition 2.1: A polynomial f (x) ∈ Fq[x] is square-free if g(x)2 - f (x) for each non-constant polynomial g(x) ∈ Fq[x]. The conventional way to determine whether f (x) is square-free is to inspect the value of its derivative, f ′(x). The major relationship between f (x) and f ′(x)is given in the upcoming theorem. 28 Theorem 2.2: If f (x) is a polynomial over Fq with gcd( f (x), f ′(x)) = 1, then
f (x) is square-free.
Proof : Suppose f (x) is not square-free. Then there is a non-constant polynomial
g(x) ∈ Fq[x] such that g(x)2 divides f (x). So, f (x) = g(x)2h(x) for some h(x) ∈
Fq[x], and applying the formula for differentiating a product, we have
f ′(x) = 2 g(x)g′(x) · h(x) + g(x)2 · h′(x).
Clearly g(x) is a common divisor of f (x) and f ′(x). Thus, gcd( f (x), f ′(x)) 6 = 1.
If f (x) is a polynomial over Fq that is not square-free, then, by Theorem 2.2, gcd( f (x), f ′(x)) 6 = 1. But, is it necessarily the case that gcd( f (x), f ′(x)) is a nontrivial factor of f (x)? The answer here is ”no”; it may very well be the case that gcd( f (x), f ′(x)) = f (x). Consider, for example, the polynomial f (x) = x14 + 3 x7 + 2 over F7. Then
f ′(x) = 14 x13 + 21 x6 = 0, and hence gcd( f (x), f ′(x)) = f (x). Notice here that each exponent on x in f (x) is a multiple of 7, which is the characteristic of F7.This observation is generalized in the following Theorem.
Theorem 2.3: Let f (x) ∈ Fq[x] be a polynomial of degree n > 0 such that gcd( f (x), f ′(x)) = f (x). Then there exists a polynomial g(x) ∈ Fq[x] with
f (x) = g(xp). Furthermore, if g(x) = ∑
j
bj xj , then f (x) =
(∑
j
bpv−1
j
xj
)p
so that f (x) 1
p
= ∑
j
bpv−1
j
xj .
Proof : Since deg( f ′(x)) < deg( f (x)) and gcd( f (x), f ′(x)) = f (x), it must be that
f ′(x) = 0. Suppose f (x) has summation representation f (x) = ∑ni=0 aixi, where some of the ai values may be zero. Then f ′(x) = ∑ni=1 ia ixi−1, and hence ia i = 0 in Fq for all i. Since Fq has characteristic p, it now follows that each i ∈ { 1, · · · , n }
with ai 6 = 0 is some multiple p. Because f (x) has degree n, observe, in particular, that n is some multiple of p. Let g(x) = ∑n/p j=0 bj xj , where bj = apj for each j.Then clearly f (x) = g(xp). Moreover, applying Theorem 1.6 and the Generalized FLT,
n/p
∑
j=0
bpv−1
j
xj
p
=
n/p
∑
j=0
bpv
j
xjp
=
n/p
∑
j=0
bqj (xp)j
29 =
n/p
∑
j=0
bj (xp)j
= g(xp)= f (x).
We will now turn our attention to finding the square-free factorization of a non-constant monic polynomial f (x) ∈ Fq[x]. To begin, note that since we can combine irreducibles that are raised to the same power in the complete factor-ization of f (x), there exist unique pairs ( gi(x), s i), 1 ≤ i ≤ r, with the following properties:
• f (x) = r∏
i=1
gi(x)si
• each gi(x) is a square-free, non-constant polynomial over Fq
• the gi(x) are pairwise relatively prime
• si < s j for i < j .Observe that each gi(x) is the product of all distinct irreducible factors h(x) of
f (x) with h(x)si |f (x) and h(x)si+1 - f (x). The goal of SFF is to identify the pairs ( gi(x), s i). To accomplish this goal, we need to consider two scenarios.
Scenario 1: In this scenario, we suppose gcd( f (x), f ′(x)) 6 = f (x). First, notice that
f ′(x) =
r
∑
i=1
[( sigi(x)si−1g′
i
(x)) · ∏
1≤j≤rj6=i
gj (x)sj ].
Now, for each i with si 6 ≡ 0 (mod p), gcd( gi(x)si , f ′(x)) = gi(x)si−1. On the other hand, for each i with si ≡ 0 (mod p), gcd( gi(x)si , f ′(x)) = gi(x)si . Since the gi(x)si
are pairwise relatively prime, we can apply Proposition 1.13 inductively to get gcd( f (x), f ′(x)) =
r
∏
i=1
gcd( gi(x)si , f ′(x)) = ∏
1≤i≤rsi≡0( p)
gi(x)si · ∏
1≤i≤rsi6≡0( p)
gi(x)si−1 .
30 Next, set
d(x) = f (x)gcd( f (x), f ′(x)) = ∏
1≤i≤rsi6≡0( p)
gi(x).
Observe that since gcd( f (x), f ′(x)) 6 = f (x), we have that d(x) 6 = 1. We can now begin a process for identifying the pairs ( gi(x), s i) for i with si 6 ≡ 0 (mod p). Let
f1(x) = gcd( f (x), f ′(x))
h1(x) = gcd( f1(x), d (x))
m1(x) = d(x)/h 1(x).
If s1 = 1, then clearly m1(x) = g1(x), and we get the pair ( g1(x), 1). Otherwise
m1(x) = 1. Regardless, we go on to compute
f2(x) = f1(x)/h 1(x)
h2(x) = gcd( f2(x), h 1(x))
m2(x) = h1(x)/h 2(x).
If 2 ∈ { si : 1 ≤ i ≤ r, s i 6 ≡ 0(mod p)}, then m2(x) = gi(x) with i = 1 or i = 2 (depending on whether s1 = 1 or s1 6 = 1), and we get the pair ( gi(x), 2). Otherwise
m2(x) = 1. Regardless, we go on to compute
f3(x) = f2(x)/h 2(x)
h3(x) = gcd( f3(x), h 2(x))
m3(x) = h2(x)/h 3(x),
and m3(x) either gives the pair ( gi(x), 3) for some i ∈ { 1, 2, 3} or m3(x) = 1. Continuing on, in general, for the kth step we get
fk(x) = fk−1(x)/h k−1(x)
hk(x) = gcd( fk(x), h k−1(x))
mk(x) = hk−1(x)/h k(x),
where 1 < k ≤ sr. If k ∈ { si : 1 ≤ i ≤ r, s i 6 ≡ 0(mod p)}, then mk(x) = gi(x) for some i ∈ { 1, · · · , k } , and we get the pair ( gi(x), k ). Otherwise mk(x) = 1. The process terminates when we reach a k value for which hk(x) = 1. By the end of the process, we will have necessarily collected the pairs ( gi(x), s i) for all i values with si 6 ≡ 0 (mod p). 31 Suppose k0 is the largest number in {si : 1 ≤ i ≤ r, s i 6 ≡ 0 (mod p)}. Then in step k0, we get hk0 (x) = 1, and the process terminates. Our remaining task is to find the pairs ( gi(x), s i) for i values with si ≡ 0 (mod p). To accomplish this, we need to find the SFF of the polynomial
fk0 (x) = gcd( f (x), f ′(x))
∏
1≤i≤rsi6≡0( p)
gi(x)si−1 .
If {i : si ≡ 0 (mod p)} = ∅, then fk0 (x) = 1, and there is nothing left to do. However, if {i : si ≡ 0 (mod p)} 6 = ∅, then we must handle the polynomial
fk0 (x) = ∏
1≤i≤rsi≡0( p)
gi(x)si ,
whose derivative is 0 over Fq. How we go about the SFF of such a polynomial is described in the upcoming scenario.
Scenario 2: Here we suppose that gcd( f (x), f ′(x)) = f (x), which occurs precisely when f ′(x) = 0. By Theorem 2.3, f (x) is a pth power. So, we can compute
z1(x) = ( f (x)) 1
p
, the pth root of f (x). However, it is possible that z1(x) has a derivative of 0 and hence is another pth power. In such a situation, we would need to compute z2(x) = z1(x) 1
p
. So, we continue computing pth roots until we get a polynomial whose derivative is nonzero. Say it takes w p th root computations to get a polynomial with a nonzero derivative. Then we need to consider two subcases for zw(x): (1) If gcd( zw(x), z ′
w
(x)) = 1, then zw(x) is square-free, and zw(x)pw
will give us the SFF of f (x). (2) If gcd( zw(x), z ′
w
(x)) 6 = 1, then we enter zw(x) into the process described in scenario 1 to begin finding the SFF of zw(x). Once the SFF of zw(x) is found, we raise zw(x) to the pw power in order to obtain the SFF of f (x). Next we offer an iterative algorithm, developed by Shoup, that condenses the general strategy we have developed for SFF. Note that the algorithm takes as input a non-constant monic polynomial f ∈ Fq[x]. 32 SFF Algorithm over Fq:
k ← 1
repeat
j ← 1, d ← f / gcd( f, f ′)
repeat
f ← f /d , h ← gcd( f, d ), m ← d/h
if m 6 = 1, then output ( m, jk )
end if
d ← h, j ← j + 1
until d = 1
if f 6 = 1, then
f ← f 1
p
, s ← ps
end if until f = 1 The SFF algorithm is equivalent to the process we developed earlier for SFF and hence outputs the desired pairs ( gi(x), s i). In the upcoming example, we find the SFF of a specific polynomial over F5.
Example 2.4: Consider f (x) = x13 + 3 x10 + 3 x8 + 2 x6 + x5 + 2 x3 + 2 x + 3 ∈ F5[x]. First, we compute
f ′(x) = 3 x12 + 4 x7 + 2 x5 + x2 + 2 .
Using the Euclidean Algorithm, we find gcd( f (x), f ′(x)) = x8 + 4 x7 + x5 + x3 + 4 x2 + 1 .
Since gcd( f (x), f ′(x)) 6 = f (x), we enter the process described in Scenario 1. Set
d(x) = f (x)gcd( f (x), f ′(x)) = x5 + x4 + x3 + 3 x2 + 2 x + 3 .
For the first step, we have
f1(x) = gcd( f (x), f ′(x))
h1(x) = gcd( f1(x), d (x)) = x3 + 4 x2 + 1
m1(x) = d(x)/h 1(x) = x2 + 2 x + 3 .
33 Thus, we get the pair ( x2 + 2 x + 3 , 1). Now, for the next step, we have
f2(x) = f1(x)/h 1(x) = x5 + 1
h2(x) = gcd( f2(x), h 1(x)) = 1
m2(x) = h1(x)/h 2(x) = x3 + 4 x2 + 1 .
This gives the pair ( x3 + 4 x2 + 1 , 2). Since h2(x) = 1, the process terminates, and we are left with the task of finding the SFF of f2(x) = x5 + 1. For this, we enter Scenario 2. Applying Theorem 2.3, we see that f2(x) = ( x + 1) 5, and so (f2(x)) 15 = x + 1. Now since x + 1 is square-free, f2(x) = ( x + 1) 5 is the SFF of
f2(x). With respect to f (x), this gives us the pair ( x + 1 , 5). Thus,
f (x) = ( x2 + 2 x + 3)( x3 + 4 x2 + 1) 2(x + 1) 5
is the SFF of f (x). Note that this is not the complete factorization of f (x) into irreducibles. In order to find the complete factorization, we need to write the square-free factors
x2 +2 x+3, x3 +4 x2 +1, and x+1 as products of irreducibles. Clearly x+1 is itself irreducible. Since x2 + 2 x + 3 has no roots in F5, it is also irreducible. However,
x3 + 4 x2 + 1 has a root at 2, and it turns out that x3 + 4 x2 + 1 = ( x + 3)( x2 + x + 2) is the complete factorization of this polynomial. Hence,
f (x) = ( x2 + 2 x + 3)( x + 3) 2(x2 + x + 2) 2(x + 1) 5
is the complete factorization of f (x) over F5.In Example 2.4, SFF alone nearly yielded a complete factorization of the given polynomial. This will not always be the case. When a polynomial has many distinct irreducible factors raised to the same power in its complete factorization, we will surely require the algorithm in the upcoming section to separate these irreducibles. With that said, however, for non-square-free high degree polynomials with few distinct irreducible factors, like the one in Example 2.4, SFF is a very powerful factoring tool. To conclude this section, we look at polynomials which are known to be powers of a single irreducible. While our SFF process can find the complete factorization of such polynomials, it requires more work than is actually needed; we will provide a simpler factoring strategy. Suppose it is known that f (x) is the power of a single irreducible. Then f (x) = g(x)s for some irreducible g(x) and some positive integer s. In seeking to identify g(x) and s, we consider two cases: 34 (i) Suppose f ′(x) 6 = 0. Then g(x) = f (x)gcd( f (x), f ′(x)) , and s = deg( f (x)) deg( g(x)) .(ii) Suppose f ′(x) = 0. Then f (x) is a pth power, and s = kp w for some positive integers k and w with k 6 ≡ 0 (mod p). Now, we take pth roots until we obtain a polynomial, say h(x), which has a nonzero derivative. The number of pth roots taken gives the value of w. Since f (x) = h(x)pw
, it follows that
h(x) = g(x)k. So we use the strategy described in case (i) above on h(x) to determine g(x) and k.In Example 2.5, we utilize this method.
Example 2.5: Given that f (x) = x42 +2 x35 +2 x28 +3 x21 +2 x14 +2 x7+1 ∈ F7[x] can be expressed as the power of a single irreducible, we find the complete factorization of f (x). It is easily seen that f ′(x) = 0. So we calculate the 7 th root
h(x) = f (x)17 = x6 + 2 x5 + 2 x4 + 3 x3 + 2 x2 + 2 x + 1 .
Now
h′(x) = 6 x5 + 3 x4 + x3 + 2 x2 + 4 x + 2 6 = 1 .
Then we compute gcd( h(x), h ′(x)) = x4 + 6 x3 + 4 x2 + 6 x + 1 and
d(x) = h(x)gcd( h(x), h ′(x)) = x2 + 3 x + 1 .
Notice that (deg( h(x))) /(deg( d(x))) = 3. So h(x) = ( x2 + 3 x + 1) 3, and we get
f (x) = ( x2 + 3 x + 1) 3·7 = ( x2 + 3 x + 1) 21 , which is the desired factorization. The reader may be wondering: Under what circumstances might we know a polynomial is the power of a single irreducible before finding its factorization? Well, in the upcoming section, our General Factoring Algorithm will output poly-nomials of just this form. We will require the method of Example 2.5 in conjunc-tion with our General Factoring Algorithm to have a complete factoring process over Fq.35 2.2 The General Factoring Algorithm
We will now use the method of Berlekamp to develop our first large-scale factoring algorithm. This method of factoring is deterministic, and the algorithm presented here will have the ability to decompose any polynomial over Fq into pair-wise relatively prime factors so that each factor can be expressed as the power of a single irreducible. After applying the algorithm to a polynomial, we will require the method of factoring powers of irreducibles that was presented in the previous section to find the complete factorization of the polynomial. This will be the only technique in our factoring process that is independent of the algorithm itself. Berlekamp’s method relies on polynomial long division and solving systems of equations using matrices. In particular, the efficiency of the algorithm rests on the efficiency of gcd computations using the Euclidean Algorithm and the effi-ciency of finding the reduced row echelon form of matrices with entries in Fq.Throughout this section, let f (x) ∈ Fq[x] be a non-constant monic polyno-mial of degree n with complete factorization f (x) = f1(x)k1 f2(x)k2 · · · fm(x)km ,where k1, k 2, ..., k m are positive integers and f1(x), f 2(x), ..., f m(x) are distinct monic irreducibles. Our goal is to generate an algorithm for determining the factors fi(x)ki . To accomplish this, we need to investigate polynomials g(x) ∈ Fq[x] of degree < n
with the property that f (x) divides g(x)q − g(x). The following proposition begins to lay the framework for how such polynomials can be used as factoring tools.
Proposition 2.6: Let g(x) = b0 + b1x + b2x2 + · · · + bdxd ∈ Fq[x] be a polynomial of degree d < n . Let R(h(x)) denote the unique remainder of h(x) after division by f (x). The following are equivalent. (a) f (x) divides the product ∏
s∈Fq
(g(x) − s). (b) R(g(xq)) = g(x). (c) For each i = 1 , 2, ..., m , there is a unique si ∈ Fq with g(x) ≡ si (mod fi(x)ki ).
Proof : Applying Theorem 1.6 and the Generalized FLT, notice
g(x)q = ( b0 + b1x + b2x2 + · · · + bdxd)q
= bq
0
bq
1
xq + bq
2
x2q + · · · + bqdxdq
= b0 + b1xq + b2x2q + · · · + bdxdq
= g(xq).
36 Now, by the definition of R(g(xq)) = R(g(x)q), we have that
g(x)q = f (x)q(x) + R(g(xq)) for some q(x) ∈ Fq[x]. Notice that f (x) divides
g(x)q − g(x) = f (x)q(x) − (R(g(xq)) − g(x)) if and only if f (x) divides R(g(xq)) − g(x). By the Division Algorithm,
R(g(xq)) < n = deg( f (x)). Since we also have deg( g(x)) < n , it follows that deg[ R(g(xq)) − g(x)] < n.
Thus, f (x) divides R(g(xq)) − g(x) iff R(g(xq)) − g(x) = 0. Recall that the polynomial uq − u has a root at each s ∈ Fq. So, uq − u
factors as
uq − u = ∏
s∈Fq
(u − s).
Setting u = g(x) gives
g(x)q − g(x) = ∏
s∈Fq
(g(x) − s).
Thus, f (x) divides ∏
s∈Fq
(g(x) − s) iff R(g(x)q) = g(x). This shows that (a) and (b) are equivalent. Suppose that f (x) divides ∏
s∈Fq
(g(x) − s). Then for each i = 1 , 2, · · · , m ,both fi(x)ki and fi(x) divide ∏
s∈Fq
(g(x) − s). Since fi(x) is irreducible, it follows from Corollary 1.17 that fi(x)|g(x) − si for some si ∈ Fq. Now, if fi(x)ki does not divide g(x) − si, then it must be that fi(x) also divides g(x) − si0 for some si0 ∈ Fq
with si0 6 = si. However, this cannot be the case, since g(x) − si and g(x) − si0
are relatively prime. So, fi(x)ki |g(x) − si, and hence g(x) ≡ si (mod fi(x)ki ). Furthermore, this si is unique due to the observation that g(x) − si and g(x) − si0
are relatively prime for si 6 = si0 .Next, suppose that for each i = 1 , 2, · · · , m , there is an si ∈ Fq such that g(x) ≡ si (mod fi(x)ki ). Then clearly for each i = 1 , 2, · · · , m , fi(x)ki di-vides ∏
s∈Fq
(g(x) − s). Since f1(x)k1 , f 2(x)k2 , · · · , f m(x)km are relatively prime, it now follows from Proposition 1.22 that f1(x)k1 f2(x)k2 · · · fm(x)km = f (x) divides
∏
s∈Fq
(g(x) − s). Thus, (a) and (c) are equivalent.
37 In the upcoming theorem, Proposition 2.6 will be used to show that the set
V = {g(x) ∈ Fq[x] : deg( g(x)) < n and g(x)q ≡ g(x) (mod f (x)) } is a vector space over Fq whose dimension is equivalent to the number of distinct irreducible factors in the complete factorization of f (x).
Theorem 2.7: The set V is a vector space over Fq of dimension m.
Proof : To show V is a vector space over Fq, it needs only to be shown that V
is a subspace of Fq[x]. Hence, we must show that V is non-empty, closed under addition, and satisfies tV ⊆ V for each t ∈ Fq. Notice that 0 q ≡ 0 (mod f (x)). So 0 ∈ V , and V 6 = ∅. Let g(x), h (x) ∈ V and t ∈ Fq. Then g(x)q ≡ g(x) (mod f (x)) and h(x)q ≡ h(x) (mod f (x)). Now, applying Theorem 1.6 and the Generalized FLT, we get [( g + h)( x)] q = [ g(x) + h(x)] q
= g(x)q + h(x)q
≡ g(x) + h(x) (mod f (x))
≡ (g + h)( x) (mod f (x)) and [( tg )( x)] p = [ tg (x)] q
= tq · g(x)q
= tg (x)q
≡ tg (x) (mod f (x))
≡ (tg )( x) (mod f (x)) .
Thus, ( g + h)( x), (tg )( x) ∈ V . This shows that V is a subspace of Fq[x]. Let S = {(s1, s 2, · · · , s m) : si ∈ Fq}. Now, construct a one-to-one correspondence between V and S as follows. Let g(x) ∈ V . Then f (x) divides g(x)q − g(x) = ∏
s∈Fq
(g(x) − s). By Proposition 2.6, for each i = 1 , 2, · · · , m , there exists a unique si ∈ Fq with
g(x) ≡ si (mod fi(x)ki ). To g(x) correspond the unique m-tuple ( s1, s 2, · · · , s m). Next, establish the inverse map. Let ( s1, s 2, · · · , s m) ∈ S. Since f1(x)k1 , f 2(x)k2 , · · · , f m(x)km are pairwise relatively prime, by the Chinese Remainder Theorem (Theorem 1.23), there is a unique polynomial g(x) ∈ Fq[x] of degree < n with g(x) ≡ si (mod fi(x)ki ) for 38 each i = 1 , 2, · · · , m . Now, by Proposition 2.6, f (x) divides ∏
s∈Fq
(g(x) − s) =
g(x)q − g(x). To ( s1, s 2, · · · , s m) ∈ S correspond the unique element g(x) ∈ V .Thus, there is a one-to-one correspondence between V and S. Since S is a vector space over Fq of dimension m, S has qm elements. Due to the one-to-one correspondence between V and S, it follows that V also has qm elements. Suppose now that V has dimension w over Fq. Then V must have qw ele-ments. So qw = qm, and hence w = m.
In the following corollary to Theorem 2.7, we establish that f (x) is the power of a single irreducible if and only if V = Fq.
Corollary 2.8: f (x) is the power of a single irreducible polynomial over Fq[x] iff the vector space V over Fq has dimension 1 iff V = Fq.
Proof : By Theorem 2.7, m = 1 in the factorization of f (x) if and only if the dimension of V is 1 over Fq.Now, for the second part of the statement, suppose that V has dimension 1 over Fq. Since aq ≡ a (mod f (x)) for each a ∈ Fq, any basis for V must contain a unit in Fq. But, the dimension of V over Fq is 1. So any basis for V contains only a unit in Fq, and hence V = Fq.For the converse, it is trivial to see that V = Fq only if V has dimension 1 over Fq.
The vector space V gives us information about the number of irreducible factors of f (x). Now, the next theorem gives us a method by which we can actu-ally use a non-constant element of V to obtain a nontrivial factorization of f (x)(in the case that f (x) is divisible by two or more irreducibles).
Theorem 2.9: Let g(x) ∈ Fq[x] be a polynomial with 1 ≤ deg( g(x)) < n such that f (x) divides g(x)q − g(x). Then f (x) = ∏
s∈Fq
gcd( f (x), g (x) − s) is a non-trivial factorization of f (x) in Fq[x].
Proof : Since f (x) divides g(x)q −g(x), notice that gcd( f (x), g (x)q −g(x)) = f (x). Now, because g(x) − s and g(x) − j are relatively prime for s 6 = j, it follows from Proposition 1.13 that
f (x) = gcd( f (x), g (x)q − g(x)) = gcd( f (x), ∏
s∈Fq
(g(x) − s)) = ∏
s∈Fq
gcd( f (x), g (x) − s). (∗)39 Since 1 ≤ deg( g(x) − s) < n , we clearly have that gcd( f (x), g (x) − s) 6 = f (x)for each s ∈ Fq. Hence, the factorization ( ∗) only involves polynomials of degree
< n = deg( f (x)), and so must be a nontrivial factorization of f (x) .
Notice that we can utilize Theorem 2.9 to factor f (x) only if we can find a non-constant polynomial in V . To demonstrate a strategy for doing this, let
g(x) = b0 + b1x + b2x2 + · · · + bn−1xn−1 ∈ Fq[x]. Recall from Proposition 2.6 that
f (x) divides g(x)q − g(x) = ∏
s∈Fq
(g(x) − s) iff 0 = R(g(xq)) − g(x),
where R(g(xq)) is the unique remainder of g(xq) after division by f (x). Now, we explicitly find R(g(xq)). Note that g(xq) = b0 + b1xq + b2x2q + · · · + bn−1x(n−1) q.Dividing xjq by f (x) for each j = 1 , 2, · · · , n − 1, it follows from the Division Algorithm that there are polynomials qj (x), r j (x) ∈ Fq[x] with
xjq = f (x)qj (x) + rj (x) and deg( rj (x)) < n.
Thus,
g(xq) = b0 + b1[f (x)q1(x) + r1(x)] + · · · + bn−1[f (x)qn−1(x) + rn−1(x)] = [ b1q1(x) + · · · + bn−1qn−1(x)] f (x) + [ b0 + b1r1(x) + · · · + bn−1rn−1(x)] .
Since deg[( b0 + b1r1(x) + · · · bdrd(x))] < n , it also is a consequence of the Division Algorithm that R(g(xq)) = b0 + b1r1(x) + · · · bn−1rn−1(x). Thus, f (x) divides
g(x)q − g(x) iff 0 = [ b0 + b1r1(x) + · · · + bn−1rn−1(x)] − [b0 + b1x + b2x2 + · · · + bn−1xn−1]. (∗∗ )Collecting the coefficients of 1 , x, x 2, · · · , x n−1 in this equation and setting them equal to 0 produces a homogenous system of n equations in the n unknowns
b0, b 1, · · · , b n−1. Finding values for the coefficients that satisfy the system will produce a polynomial g(x) with degree < n such that f (x) divides g(x)q − g(x). This process will be applied in the following example to factor a polynomial over F3.
Example 2.10: Let f (x) = x5 + 2 x2 + 2 x + 2 ∈ F3[x]. We desire to find a non-constant polynomial of the form g(x) = b0 + b1x + b2x2 + b3x3 + b4x4 such that f (x)divides g(x)3 − g(x). To accomplish this, we first find the remainder polynomials 40 rj (x) by computing xj·3(mod f (x)) for j = 1 , 2, 3, 4:
x1·3 = x3 ≡ x3 (mod f (x)) ,x2·3 = x6
= f (x) · x + ( x3 + x2 + x)
≡ x3 + x2 + x (mod f (x)) ,x3·3 = x9
= x6 · x3
≡ (x3 + x2 + x)x3 (mod f (x))
≡ x6 + x5 + x4 (mod f (x))
≡ (x3 + x2 + x) + ( x2 + x + 1) + x4 (mod f (x))
≡ x4 + x3 + 2 x2 + 2 x + 1 (mod f (x)) ,x4·3 = x12
= x6x6
= ( x3 + x2 + x)( x3 + x2 + x) (mod f (x)) = x6 + 2 x5 + 2 x3 + x2 (mod f (x))
≡ (x3 + x2 + x) + (2 x2 + 2 x + 2) + 2 x3 + x2 (mod f (x))
≡ x2 + 2 (mod f (x)) .
Thus, r1(x) = x3, r 2(x) = x3 + x2 + x, r 3(x) = x4 + x3 + 2 x2 + 2 x + 1 , and
r4(x) = x2 + 2. Now, referring to ( ∗∗ ), f (x) divides g(x)3 − g(x) iff 0 = b0 + b1x3 + b2(x3 + x2 + x) + b3(x4 + x3 + 2 x2 + 2 x + 1) + b4(x2 + 2)
− [b0 + b1x + b2x2 + b3x3 + b4x4].
Next we collect the coefficients of 1 , x, x 2, x 3, and x4 to get a homogenous system of equations:
b0 − b0 = 0 (2 b1 + b2 + 2 b3)x = 0 ⇒ 2b1 + b2 + 2 b3 = 0 (2 b3 + b4)x2 = 0 ⇒ 2b3 + b4 = 0 (b1 + b2)x3 = 0 ⇒ b1 + b2 = 0 (b3 + 2 b4)x4 = 0 ⇒ b3 + 2 b4 = 0 .
41 These reduce to b1 = 2 b2 = b3 = b4, b0 free. Select b1 = b3 = b4 = 1, b2 = 2, and
b0 = 0 to form the polynomial g(x) = x4 + x3 + 2 x2 + x. Then f (x) necessarily divides g(x)3 − g(x). Applying the Euclidean Algorithm, it can be found that gcd( f (x), g (x)) = 1 gcd( f (x), g (x) − 1) = x2 + 1 gcd( f (x), g (x) − 2) = x3 + 2 x + 2 .
And applying Theorem 2.9, we get the nontrivial factorization
f (x) = ( x2 + 1)( x3 + 2 x + 2) .
In fact, since x2 + 1 and x3 + 2 x + 2 have no roots in F3, we can conclude that this is the complete factorization of f (x) into irreducibles. Example 2.10 suggests the following corollary to Theorem 2.9.
Corollary 2.11: Let h(x) ∈ Fq[x] be a reducible polynomial of degree 5 with no factors of degree 1. If g(x) is a polynomial of degree ≥ 1 and < 5 such that h(x)divides g(x)q − g(x), then ∏
s∈Fq
gcd( h(x), g (x) − s) is the complete factorization of h(x).
Proof : Since h(x) is a reducible polynomial over Fq of degree 5 and has no factors of degree 1, h(x) must be the product of an irreducible quadratic polynomial and an irreducible cubic polynomial. By Theorem 2.8, ∏
s∈Fq
gcd( h(x), g (x) − s) is a nontrivial factorization of h(x), which implies that at least two terms of this prod-uct must be non-constant. But, h(x) is the product of precisely two irreducibles. Thus, h(x) = ∏
s∈Fq
gcd( h(x), g (x) − s) is the complete factorization of h(x).
Corollary 2.11 is a special case where Theorem 2.9 gives the complete factor-ization of a polynomial. However, for the general polynomial f (x) = f1(x)k1 f2(x)k2
· · · fm(x)km , the nontrivial factorization f (x) = ∏
s∈Fq
gcd( f (x), g (x)−s) is usually not the complete factorization of f (x). In particular, ∏
s∈Fq
gcd( f (x), g (x) − s) is not the complete factorization of f (x) when there is an s0 ∈ Fq such that g(x) − s0
is divisible by fi(x)ki fj (x)kj for some i 6 = j. This situation occurs regularly. For example, consider the product h(x) = x(x + 1) 2(x2 + x + 1) ∈ F2[x]. If
g(x) is a non-constant polynomial of degree < 5 such that h(x)|g(x)2 − g(x), then Theorem 2.9 gives that x(x + 1) 2(x2 + x + 1) = gcd( h(x), g (x)) · gcd( h(x), g (x) − 1). 42 Now, one of the two gcd’s on the right hand side of this equation must be divisible by two of the three factors x, (x + 1) 2, and x2 + x + 1. So, in order to develop a process by which we can separate the factors
f1(x)k1 , f 2(x)k2 , · · · , f m(x)km , we must account for the strong likelihood that the factorization ∏
s∈Fq
gcd( f (x), g (x) − s) has terms which are divisible by multiple distinct irreducibles. With this in mind, we present the following formulation of Berlekamp’s algorithm.
General Factoring Algorithm:
Assume m > 1. Let {g0(x), g 1(x), · · · , g m−1(x)} be a basis for the vector space V
with g0(x) = 1. (We are allowed to let g0(x) = 1, since aq ≡ a (mod f (x)) for each a ∈ Fq.) Note that g1(x), · · · , g m−1(x) are non-constant. Now, complete the following steps:
Step 1) Compute gcd( f (x), g 1(x) − s) for each s ∈ Fq[x]. Let A1 be the set containing each of these factors which has degree ≥ 1. If |A1| = m, then stop and output A1. Otherwise, continue to Step 2.
Step 2) Compute gcd( h(x), g 2(x) − s) for each h(x) ∈ A1 and s ∈ Fq. Let A2 be the set containing each of these factors which has degree ≥ 1. If |A2| = m, then stop and output A2. Otherwise, continue to Step 3. ...
Step m-2) Compute gcd( h(x), g m−2(x) − s) for each h(x) ∈ Am−3 and s ∈ Fq.Let Am−2 be the set containing each of these factors which has degree ≥ 1. If
|Am−2| = m, then stop and output Am−2. Otherwise, continue to Step m-1.
Step m-1) Compute gcd( h(x), g m−1(x) − s) for each h(x) ∈ Am−2 and s ∈ Fq.Let Am−1 be the set containing each of these factors which has degree ≥ 1. Output
Am−1.
Theorem 2.12: Suppose the General Factoring Algorithm stops on the jth step. Then the output Aj contains precisely the elements f1(x)k1 , f 2(x)k2 , · · · , f m(x)km .
Proof : Suppose j < m − 1. Then, by construction, Aj contains m relatively prime polynomials of degree ≥ 1. Applying Proposition 1.13 and Theorem 2.9, it follows that f (x) must be the product of these m polynomials. Since the factor-ization of f (x) into powers of distinct irreducibles is unique, it must be that the
m polynomials in Aj are precisely f1(x)k1 , f 2(x)k2 , · · · , f m(x)km .Now, suppose j = m − 1. Since the elements of Aj are relatively prime and have a product which equals f (x), |Aj | ≤ m. Assume, by way of contra-diction, that |Aj | < m . Then there exists an h(x) ∈ Aj which is divisible by at least two of the powers of irreducibles that are factors of f (x). Without loss of 43 generality, say both f1(x)k1 and f2(x)k2 divide h(x). Then f1(x)k1 and f2(x)k2
must both divide exactly one gcd in each step of the algorithm. Thus, for each
i = 0 , 1, 2, · · · , m − 1, there exists an si ∈ Fq such that gi(x) ≡ si (mod f1(x)k1 )and gi(x) ≡ si (mod f2(x)k2 ). Now, by the Chinese Remainder Theorem (Theorem 1.23), there exists a g(x) ∈ V (not necessarily unique) with g(x) ≡ 0 (mod f1(x)k1 )and g(x) ≡ 1 (mod f2(x)k2 ). Since {g0(x), g 1(x), · · · , g m−1(x)} forms a basis for
V , there exist c0, c 1, · · · , c m−1 ∈ Fq with g(x) =
m−1
∑
i=0
cigi(x). Let s =
m−1
∑
i=0
cisi ∈ Fq.Then
s =
m−1
∑
i=0
cisi
≡
m−1
∑
i=0
cigi(x) (mod f1(x)k1 )
≡ g(x) (mod f1(x)k1 )
≡ 0 (mod f1(x)k1 ).
This implies s = 0. But,
s =
m−1
∑
i=0
cisi
≡
m
∑
i=1
cigi(x) (mod f2(x)k2 )
≡ g(x) (mod f2(x)k2 )
≡ 1 (mod f2(x)k2 ),
which implies s = 1, a contradiction. Thus, |Aj | = m, and Aj contains precisely the elements f1(x)k1 , f 2(x)k2 , · · · , f m(x)km .
The final thing we need to do before applying the General Factoring Algo-rithm is describe a process for determining a basis for V . Recall that determining such a basis is necessary to begin the algorithm. Consider again the equation
R(g(xq)) − g(x) = [ b0 + b1r1(x) + · · · + bn−1rn−1(x)]
− [b0 + b1x + b2x2 + · · · + bn−1xn−1]for a polynomial g(x) = b0 + b1x + b2x2 + · · · + bn−1xn−1 ∈ Fq[x]. Set r0(x) = 1. 44 For each j = 0 , 1, 2, · · · , n − 1, let rj (x) = r0,j + r1,j x + · · · + rn−1,j xn−1. Form the matrices
B =
b0
b1
...
bn−1
, Q =
r0,0 r0,1 · · · r0,n −1
r1,0 r1,1 · · · r1,n −1
... ... . . . ...
rn−1,0 rn−1,1 · · · rn−1,n −1
, and
I =
1 0 · · · 00 1 · · · 0... ... . . . ...0 0 · · · 1
.Then g(x) ∈ V iff R(g(xq)) − g(x) = 0 iff ( Q − I)B = 0. Let
V ′ = {B : B is an n × 1 column matrix with entries in Fq and ( Q − I)B = 0 }.
Clearly V ′ is a vector space over Fq, called the null space of the matrix Q − I,with dimension equivalent to that of V , which was shown in Theorem 2.7 to be
m. Since Q − I is an n × n matrix over the field Fq, we get the following standard relationship between the rank of Q − I and the dimension of V ′ over Fq:
n = rank( Q − I) + dim( V ′)= rank( Q − I) + m,
which implies
m = n − rank( Q − I).
So, the number of distinct irreducible factors of f (x) can be found by computing the difference between the degree of f (x) and the number of linearly independent rows in the matrix Q − I.Let A be the reduced row echelon form of Q − I. The rank of Q − I
can be identified by counting the number of non-zero rows in A. Furthermore, a basis for V ′ is most efficiently found by finding a basis for the vector space
{B : B is an n × 1 column matrix with entries in Fq and AB = 0 } over Fq. We note that finding such a basis comes down to finding the free variables in a ho-mogenous system of equations. 45 Finally, if
Bi =
b0,i
b1,i
...
bn−1,i
is an element in a basis for V ′, then gi(x) = b0,i + b1,i x + · · · + bn−1,i xn−1 is in the corresponding basis for V .Notice that only one step of the General Factoring Algorithm would have been required to find the complete factorization of the polynomial in Example 2.10. The next example gives a situation in which we require more than one step of the algorithm.
Example 2.13: Let f (x) = x9 + 2 x8 + x7 + x4 + 2 x3 + 2 x2 + 2 x + 1 ∈ F3[x]. Using polynomial long division to reduce modulo f (x), it is found that
x1·3 = x3
x2·3 = x6
x3·3 = x9 ≡ x8 + 2 x7 + 2 x4 + x3 + x2 + x + 2 (mod f (x))
x4·3 = x12 ≡ 2x8 + 2 x5 + 2 x3 + x2 + 2 x + 1 (mod f (x))
x5·3 = x15 ≡ x5 + 2 x3 + x (mod f (x))
x6·3 = x18 ≡ x8 + 2 x6 + x4 (mod f (x))
x7·3 = x21 ≡ x8 + 2 x7 + 2 x6 + x3 + 2 x2 + x + 1 (mod f (x))
x8·3 = x24 ≡ x8 + 2 x7 + x4 + x + 2 (mod f (x)) .
So form the remainder polynomials:
r0(x) = 1
r1(x) = x3
r2(x) = x6
r3(x) = 2 + x + x2 + x3 + 2 x4 + 2 x7 + x8
r4(x) = 1 + 2 x + x2 + 2 x3 + 2 x5 + 2 x8
r5(x) = x + 2 x3 + x5
r6(x) = x4 + 2 x6 + x8
r7(x) = 1 + x + 2 x2 + x3 + 2 x6 + 2 x7 + x8
r8(x) = 2 + x + x4 + 2 x7 + x8.
46 Now the coefficient of xj in ri(x) represents the j + 1 , i + 1 entry in the following matrix Q:
Q =
1 0 0 2 1 0 0 1 20 0 0 1 2 1 0 1 10 0 0 1 1 0 0 2 00 1 0 1 2 2 0 1 00 0 0 2 0 0 1 0 10 0 0 0 2 1 0 0 00 0 1 0 0 0 2 2 00 0 0 2 0 0 0 2 20 0 0 1 2 0 1 1 1
. Form Q−I =
0 0 0 2 1 0 0 1 20 2 0 1 2 1 0 1 10 0 2 1 1 0 0 2 00 1 0 0 2 2 0 1 00 0 0 2 2 0 1 0 10 0 0 0 2 0 0 0 00 0 1 0 0 0 1 2 00 0 0 2 0 0 0 1 20 0 0 1 2 0 1 1 0
.
Elementary row operations over F3 give that the reduced row echelon form of the matrix Q − I is
A =
0 1 0 0 0 2 0 1 00 0 1 0 0 0 0 0 10 0 0 1 0 0 0 2 10 0 0 0 1 0 0 0 00 0 0 0 0 0 1 2 20 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
.
Notice that A has rank 5. Thus, the dimension of both V and V ′ is 9 − 5 = 4. This tells us that there are exactly four powers of distinct irreducibles in the factorization of f (x). Now, set
000000000
= A
b0
b1
b2
b3
b4
b5
b6
b7
b8
=
b1 + 2 b5 + b7
b2 + b8
b3 + 2 b7 + b8
b4
b6 + 2 b7 + 2 b8
0000
.
47 This implies
b1 = b5 + 2 b7
b2 = 2 b8
b3 = b7 + 2 b8
b4 = 0
b6 = b7 + b8.
Thus,
b0
b1
b2
b3
b4
b5
b6
b7
b8
=
b0
b5 + 2 b7
2b8
b7 + 2 b8
0
b5
b7 + b8
b7
b8
= b0
100000000
b5
010001000
b7
020100110
b8
002200101
.
It follows that
100000000
,
010001000
,
020100110
, and
002200101
form a basis for V ′. Correspondingly, we have that g0(x) = 1, g1(x) = x + x5,
g2(x) = 2 x + x3 + x6 + x7, and g3(x) = 2 x2 + 2 x3 + x6 + x8 form a basis for V .48 Next, we use the Euclidean Algorithm to obtain the necessary gcd’s for the first step of the General Factoring Algorithm. These gcd’s are: gcd ( f (x), g 1(x)) = x2 + x + 2 ,
gcd ( f (x), g 1(x) − 1) = x5 + x + 2 ,
gcd ( f (x), g 1(x) − 2) = x2 + x + 1 .
Only three of the four needed factors were found in the Step 1, so we proceed to Step 2. The necessary gcd’s for this step are: gcd ( x2 + x + 2 , g 2(x)) = 1 ,
gcd ( x2 + x + 2 , g 2(x) − 1) = 1 ,
gcd ( x2 + x + 2 , g 2(x) − 2) = x2 + x + 2; gcd ( x5 + x + 2 , g 2(x)) = x5 + x + 2 ,
gcd ( x5 + x + 2 , g 2(x) − 1) = 1 ,
gcd ( x5 + x + 2 , g 2(x) − 2) = 1; gcd ( x2 + x + 1 , g 2(x)) = 1 ,
gcd ( x2 + x + 1 , g 2(x) − 1) = 1 ,
gcd ( x2 + x + 1 , g 2(x) − 2) = x2 + x + 1 .
Step 2 only produced three factors as well, so we proceed to compute the necessary gcd’s for the third step (which is the final possible step). gcd ( x2 + x + 2 , g 3(x)) = x2 + x + 2 ,
gcd ( x2 + x + 2 , g 3(x) − 1) = 1 ,
gcd ( x2 + x + 2 , g 3(x) − 2) = 1; gcd ( x5 + x + 2 , g 3(x)) = 1 ,
gcd ( x5 + x + 2 , g 3(x) − 1) = x3 + x2 + 2 ,
gcd ( x5 + x + 2 , g 3(x) − 2) = x2 + 2 x + 1; gcd ( x2 + x + 1 , g 3(x)) = x2 + x + 1 ,
gcd ( x2 + x + 1 , g 3(x) − 1) = 1 ,
gcd ( x2 + x + 1 , g 3(x) − 2) = 1 .
49 Now, by Theorem 2.12, x2 + x + 2 , x 3 + x2 + 2 , x 2 + 2 x + 1 , and x2 + x + 1 are pairwise relatively prime and multiply to give f (x). Further, each of these four factors can be written as the power of a single irreducible. Using the method given in Section 2.1 for factoring such polynomials, we find: x2 + x + 2 and x3 + x2 + 2 are both irreducible, x2 + 2 x + 1 = ( x + 1) 2, and x2 + x + 1 = ( x + 2) 2. Thus,
f (x) = ( x2 + x + 2)( x3 + x2 + 2)( x + 1) 2(x + 2) 2
is the complete factorization of f (x). After this lengthy example, we must pose the question: Is the General Fac-toring Algorithm alone our best factoring tool thus far? The answer is clearly no! Practically speaking, it is strongly recommended that any polynomial undergo SFF in the first step of the factoring process. Then, if necessary, the separate square-free parts of the polynomial can be reduced using the algorithm of this section. There are many reasons for this recommendation. First, notice that implementing SFF allows us to avoid inserting polynomials into the General Fac-toring Algorithm which are divisible by high powers of irreducibles. Also, the algorithm works very nicely with square-free polynomials - it has the ability to directly separate all of the irreducibles of such polynomials. Consider how implementing SFF speeds up the factoring process for the polynomial in Example 2.13. SFF gives
f (x) = x9 + 2 x8 + x7 + x4 + 2 x3 + 2 x2 + 2 x + 1 = ( x5 + 2 x4 + x2 + 2 x + 1)( x2 + 2) 2
over F3. Using Corollary 2.11, we find x5+2 x4+x2+2 x+1 = ( x2+x+2)( x3+x2+2). Searching for roots of x2 + 2 in F3, we also see that x2 + 2 = ( x + 1)( x + 2). Thus,
f (x) = ( x2 + x + 2)( x3 + x2 + 2)( x + 1) 2(x + 2) 2.
The process of factorization described here is a much quicker process than the application of the General Factoring Algorithm in Example 2.13. Even though applying SFF is the recommended first step in factoring, the formulation of Berelekamp’s algorithm in this section is of great theoretical inter-est. It has the ability to separate any polynomial over Fq into powers of distinct irreducibles. This is stronger than formulations of the algorithm which only accept square-free polynomials. 50 Chapter 3 The Cantor-Zassenhaus Method
The next factoring scheme we will develop is due to Cantor and Zassenhaus and has two stages: 1. Distinct Degree Factorization (DDF): The input polynomial is decom-posed into factors so that each factor can be expressed as the product of distinct irreducibles that all have the same degree (and this degree is found). 2. Equal Degree Factorization (EDF): Each of the “equal degree” factors produced in the DDF stage is completely factored. The algorithm we give for DDF is deterministic, while the algorithm we give for EDF is probabilistic. Combined, the two algorithms will provide us with another method by which we can completely factor an arbitrary polynomial over Fq, where
q = pv with p a prime number and v a positive integer.
3.1 Distinct Degree Factorization
To begin this section, let f (x) be a monic polynomial over Fq of degree ≥ 1. Formally, in order to find a distinct degree factorization of f (x), we need a list of pairs (g1(x), n 1), (g2(x), n 2), · · · , (gk(x), n k)such that f (x) = g1(x)g2(x) · · · gk(x) and each gi(x) is the product of deg[ gi(x)]
ni
distinct irreducibles that all have degree ni. Notice, in particular, that the degrees
ni are not necessarily pairwise distinct. In fact, if f (x) is not square-free, then it will be the case that ni = nj for some i 6 = j.To develop an algorithm for DDF, we start with the following proposition. 51 Proposition 3.1: Let r be a positive integer and h(x) a monic irreducible poly-nomial over Fq of degree n. Then h(x) divides xqr
− x in Fq[x] if and only if n|r.
Proof : Suppose h(x) divides xqr
− x in Fq[x]. Note that Fqr is the splitting field for xqr
− x, and each of the qr distinct elements of Fqr is a root of xqr
− x. So, we have that xqr
− x = ∏
a∈Fqr
(x − a). Now, since h(x) divides ∏
a∈Fqr
(x − a) in
Fqr [x] and is of degree ≥ 1, there must exist an a′ ∈ Fqr such that a′ is a root of
h(x). Consider Fq(a′), the smallest subfield of Fqr containing Fq and the element
a′. Since h(x) is irreducible over Fq, it follows from Theorems 1.29 and 1.31 that
Fq(a′) ∼= Fq[x]/(h(x)) ∼= Fqn . This shows that Fq(a′) is a field extension of Fq, and [Fqr (a′) : Fq] = [ Fqn : Fq] = n. Thus, by Theorem 1.27,
r = [ Fqr : Fq]= [ Fqr : Fq(a′)] · [Fq(a′) : Fq]= [ Fqr : Fq(a′)] · n,
and n|r.Now, suppose that n|r. Let θ = x (mod h(x)) ∈ Fq[x]/(h(x)) ∼= Fqn . Note that θ is a root of xqn
− x. So, it follows from Proposition 1.34 that xqn
− x is divisible by the minimal polynomial for θ over Fq, which is h(x). Since n|r, r = nj
for some j ∈ Z, and hence
qr − 1 = qnj −1 = ( qn − 1)( qn(j−1) + qn(j−2) + · · · + 1) .
So qn − 1|qr − 1 in Z, and an identical argument shows that xqn−1 − 1|xqr −1 − 1in Fq[x]. Thus, xqn
− x = x(xqn−1 − 1) divides xqr
− x = x(xqr −1 − 1). Since
h(x)|xqn
− x, it now follows that h(x)|xqr
− x in Fq[x].
From this proposition follows a very important theorem, which we will use to generate our DDF algorithm.
Theorem 3.2: For n ≥ 1, let hn denote the product of all distinct monic irre-ducibles in Fq[x] of degree n. Then for all positive integers r,
xqr
− x = ∏
n|r
hn,
where the product is over all positive divisors of r.52 Proof : By Proposition 3.1, every monic irreducible polynomial over Fq of degree
n with n|r divides xqr
−x. So ∏
n|r
hn|xqr
−x. Furthermore, each monic irreducible polynomial in Fq[x] that divides xqr
− x must have degree n such that n|r. So,
xqr
− x is the product of monic irreducibles whose degree divides r, and all such polynomials appear at least once in this product. Now, we compute the derivative
ddx (xqr
− x) = qrxqr −1 − 1= 0 − 1 ∈ Fq[x]= −1.
So gcd( xqr
− x, ddx (xqr
− 1)) = 1, which shows that no irreducible polynomial ap-pears more than once in the complete factorization of xqr
− x. This means the complete factorization of xqr
− x consists of one copy of each of the disinct monic irreducibles over Fq whose degree divides r. Hence, xqr
− x = ∏
n|r
hn.
Utilizing Theorem 3.2, we can now develop a process for finding a distinct degree factorization of f (x). First, note that xq − x is the product of all distinct linear monic polynomials over Fq. So we can compute g(x) = gcd( xq − x, f (x)) = gcd( xq − x(mod f (x)) , f (x)) to get the product of all distinct linear factors of
f (x). We then remove the factor g(x) from f (x) to obtain a new f (x). But, if the original polynomial was not square-free, f (x) may still have some linear factors, and so we have to repeat the step with xq −x until we get a gcd of 1 and hence f (x)has no more linear factors. Once all linear factors have been removed, we com-pute gcd( xq2
− x, f (x)), which is the product of all distinct quadratic irreducible factors of f (x). Note that even though xq2
− x is the product of all distinct linear and quadratic monic irreducibles over Fq, since we removed all linear factors from
f (x) beforehand, we know gcd( xq2
− x, f (x)) will only give us the product of the distinct quadratic irreducibles that divide f (x). Like before, it may be necessary to repeat the step with xq2
− x multiple times to remove all quadratic factors from
f (x). In general, the strategy of our DDF algorithm will be this: Starting with k = 1, once all factors of degree less than k have been removed from f (x), we compute gcd( xqk
− x, f (x)) to get the product of all distinct degree
k monic irreducible factors of f (x). Following this, we may need to compute mul-tiple gcd’s with xqk
− x in order to remove all factors of degree k from f (x). This process will be completed until no more factors can be removed from f (x), which will occur when our redefined value for f (x) is 1. We now give the algorithm. 53 DDF Algorithm over Fq:
The input is a monic polynomial f ∈ Fq[x].
r ← 1
while f 6 = 1 do
h ← xqr
− x (mod f )
g ← gcd( h, f )
while g 6 = 1 do
output ( g, r )
f ← f /g g ← gcd( h, f )
end while
r ← r + 1
end while
The polynomials g in the outputs ( g, r ) in the DDF algorithm clearly multi-ply to give the input polynomial f . Further, each pair ( g, r ) represents a polyno-mial g which is the product of deg( g)/r distinct irreducibles of degree r.Notice that the the degrees of the polynomials xqr
− x blow up very quickly. So, repeatedly computing xqr
− x(mod f ) = xqr
(mod f ) − x(mod f ) is one of the more difficult aspects of applying the algorithm. In view of this, we present a fairly intuitive binary exponentiation algorithm for computing xM (mod f ).
Binary Exponentiation Algorithm for Computing xM (mod f ):
Let M = bn · 2n + bn−1 · 2n−1 + · · · + b1 · 2 + b0 be the binary representation for M
(where bj = 0 or 1).
g ← xj ← n − 1
while j ≥ 0 do
g ← g2 (mod f )
if bj = 1, then
g ← g · x (mod f)
end if
j ← j − 1
end while
output g
We are now ready to apply the DDF algorithm. 54 Example 3.3: We find the DDF of the polynomial
f (x) = x11 + 2 x6 + 2 x4 + 2 x3 + x2 + 1 over F3.First, we use the Euclidean Algorithm to compute that gcd( x31
− x, f (x)) = x2 + 2 .
This gives the pair ( x2 + 2 , 1). Now, compute
f (x)/(x2 + 2) = x9 + x7 + x5 + 2 x4 + x3 + x2 + 2 .
We must test this new polynomial for linear factors: gcd( x3 − x, x 9 + x7 + x5 + 2 x4 + x3 + x2 + 2) = x + 2 .
This gives the pair ( x + 2 , 1), and we get that (x9 + x7 + x5 + 2 x4 + x3 + x2 + 2) /(x + 2) = x8 + x7 + 2 x6 + 2 x5 + 2 x3 + x + 1 .
Testing again for linear factors: gcd( x3 − x, x 8 + x7 + 2 x6 + 2 x5 + 2 x3 + x + 1) = 1 .
This shows that we have removed all of the linear factors from f (x). Polynomial long division easily gives that
x32
− x = x9 − x
≡ 2x7 + 2 x5 + x4 + 2 x3 + 2 x2 + 2 x + 1 (mod x8 + x7 + 2 x6 + 2 x5 + 2 x3 + x + 1) ,
and we use this congruence to compute the product of the distinct quadratic factors of f (x): gcd(2 x7 + 2 x5 + x4 + 2 x3 + 2 x2 + 2 x + 1 , x 8 + x7 + 2 x6 + 2 x5 + 2 x3 + x + 1) = x2 + 1 .
This gives the pair ( x2 + 1 , 2). Now, compute (x8 + x7 + 2 x6 + 2 x5 + 2 x3 + x + 1) /(x2 + 1) = x6 + x5 + x4 + x3 + 2 x2 + x + 1 .
55 Before testing again for quadratic factors, notice that
x32
− x = x9 − x
≡ 2x7 + 2 x5 + x4 + 2 x3 + 2 x2 + 2 x + 1 (mod x6 + x5 + x4 + x3 + 2 x2 + x + 1)
≡ 2x5 + x4 + x2 + 2 x (mod x6 + x5 + x4 + x3 + 2 x2 + x + 1) .
Now, gcd(2 x5 + x4 + x2 + 2 x, x 6 + x5 + x4 + x3 + 2 x2 + x + 1) = 1 .
Hence, all of the quadratic factors have been removed from f (x). Next, we let
p(x) = x6 + x5 + x4 + x3 + 2 x2 + x + 1 and use binary exponentiation to compute x33
(mod p(x)). Notice that 3 3 = 27 = 1 · 24 + 1 · 23 + 0 · 22 + 1 · 21 + 1 · 20.Step 1: Since 2 3 has a coefficient of 1, compute g = x2, and then g = x2 · x = x3.Step 2: Since 2 2 has a coefficient of 0, compute
g = ( x3)2 = x6 ≡ 2x5 + 2 x4 + 2 x3 + x2 + 2 x + 2 (mod p(x)). Step 3: Since 2 1 has a coefficient of 1, compute
g = (2 x5 + 2 x4 + 2 x3 + x2 + 2 x + 2) 2 ≡ x5 + x4 + x3 + x2 + 2 x + 1 (mod p(x)), and then g = ( x5 + x4 + x3 + x2 + 2 x + 1) · x ≡ 2 (mod p(x)). Step 4: Since 2 0 has a coefficient of 1, compute g = 2 2 = 1 ∈ F3, and then
g = 1 · x = x.Thus, x33
≡ x (mod p(x)), and so x33
− x ≡ 0 (mod p(x)). Clearly, gcd(0 , p (x)) = p(x), which gives the pair ( p(x), 3). Now, p(x)/p (x) = 1, and the algorithm terminates. In summary, f (x) = ( x2 + 2)( x + 2)( x2 + 1)( x6 + x5 + x4 + x3 + 2 x2 + x + 1), and these four factors are products of distinct irreducibles of degrees 1 , 1, 2, and 3, respectively. In particular, the middle two factors are themselves irreducible. 56 Observe that in Example 3.3, the input polynomial was not square-free since it had x + 2 as a repeated linear factor. This repeated linear factor forced us to compute an extra gcd when applying the DDF algorithm, giving us a total of three gcd computations before all linear factors were removed from the input polynomial. For the general polynomial f (x), suppose g(x)k is a factor of f (x), where
g(x) is an irreducible polynomial over Fq of degree r, and k is the largest power on any irreducible factor of f (x) with degree r. In order to remove all irreducible factors of degree r from f (x) using the algorithm, we would have to compute exactly k + 1 gcd’s with xqr
− x. The first k of these gcd computations would actually remove all the factors of degree r, and the final gcd computation (of 1) would verify that there are no factors of degree r left. It is quite obvious that as
k grows large, this process becomes extremely time consuming. The tediousness here is most likely why Cantor and Zassenhaus formulate a DDF method which only accepts square-free polynomials. As a general rule, we also suggest that a polynomial undergo SFF before being separated with our algorithm. When it is known up front that the input polynomial is square free, it is never necessary to compute gcd’s in the second while loop of the algorithm. Hence, in such a case, we need only compute one gcd in each new degree iteration of the DDF process. In the next section, we delve into the second stage of the Cantor-Zassenhaus factoring scheme, Equal Degree Factorization.
3.2 Equal Degree Factorization
Recall again that the an output ( g, r ) in the DDF algorithm represents a polynomial g which can be written as the product of distinct monic irreducibles that all have the same (known) degree, namely r. So, if we can develop an algo-rithm that separates the irreducible factors of such an “equal degree” polynomial, we will have generated a complete factoring process over Fq.Throughout this stage, let f (x) ∈ Fq[x] be a monic polynomial of degree n
that is the product of s > 1 distinct monic irreducibles fi(x), 1 ≤ i ≤ s, with deg( fi(x)) = d for each i. Notice, in particular, that n = sd . Working under the assumption that d is known, our goal for the section is to use the fact that each irreducible factor of f (x) has degree d to develop a method for factoring f (x) . Note that a factor p(x) of f (x) with 0 < deg( p(x)) < n will be referred to as a proper factor of f (x). In seeking a strategy for finding proper factors of f (x), we start with the following intuitive proposition (which is not dependent on the fact that the irreducible factors of f (x) all have the same degree). 57 Proposition 3.4: Let g(x) be a non-constant polynomial over Fq with deg( g(x)) <n. If g(x) ≡ 0 (mod fi0 (x)) for some i0 ∈ { 1, · · · , s }, then gcd( f (x), g (x)) is a proper factor of f (x).
Proof : Since g(x) is non-constant and deg( g(x)) < n , g(x) 6 ≡ 0 (mod f (x)). Hence, gcd( f (x), g (x)) 6 = f (x). Also, since g(x) ≡ 0 (mod fi0 (x)), fi0 (x) is a com-mon factor of g(x) and f (x). So, gcd( f (x), g (x)) is some multiple of fi0 (x), which means gcd( f (x), g (x)) 6 = 1. Thus, gcd( f (x), g (x)) is a proper factor of f (x).
If we can find a polynomial g(x) that satisfies the conditions of Proposition 3.4, we will obtain a nontrivial factorization of f (x). We first restrict ourselves to the case that p > 2, so that q = pv is odd. With this restriction in place, set
c = qd−12 . Notice that the definition of c brings into play the fact that all irre-ducible factors of f (x) have degree d. Now the next theorem will provide us with a way to efficiently generate polynomials that satisfy the hypotheses of Proposition 3.4.
Theorem 3.5: Let h(x) ∈ Fq[x] be a polynomial with gcd( f (x), h (x)) = 1. Let
g(x) = h(x)c − 1(mod f (x)). If h(x)c 6 ≡ ± 1 (mod f (x)), then gcd( f (x), g (x)) is a proper factor of f (x).
Proof : Since gcd( f (x), h (x)) = 1, h(x) 6 ≡ 0 (mod fi(x)) for each i ∈ { 1, · · · , s }.Now, for each i, Fq[x]/(fi(x)) is a field with qd elements, and so by the Generalized FLT, (h(x)c)2 = h(x)qd−1 ≡ 1 (mod fi(x)) ,
which gives (h(x)c − 1)( h(x)c + 1) ≡ 0 (mod fi(x)) .
Because Fq[x]/(fi(x)) has no zero divisors, it follows that h(x)c ≡ 1 (mod fi(x)) or h(x)c ≡ − 1 (mod fi(x)). Observe that since h(x)c 6 ≡ ± 1 (mod f (x)), h(x)c is clearly non-constant over
Fq. In turn, g(x) is also non-constant. Furthermore, since h(x)c 6 ≡ − 1 (mod f (x)), there exists an i0 ∈ { 1, · · · , s } with h(x)c 6 ≡ − 1 (mod fi0 (x)). Then it must be that h(x)c ≡ 1 (mod fi0 (x)), which means g(x) ≡ 0 (mod fi0 (x)). Hence, by Proposition 3.4, gcd( f (x), g (x)) is a proper factor of f (x).
58 In hopes of utilizing Theorem 3.5 to factor f (x), we randomly select a poly-nomial h(x) from the qn − q non-constant polynomials in Fq[x] of degree < n . We now will calculate the probability that h(x) cannot be used with the tools that we have to find a proper factor of f (x). Notice that if gcd( f (x), h (x)) 6 = 1, then we automatically get a proper factor of f (x). Furthermore, if gcd( f (x), h (x)) = 1 with h(x)c 6 ≡ ± 1 (mod f (x)), then we can use Theorem 3.5 to get a proper fac-tor. It follows that h(x) cannot be used to get a proper factor of f (x) only if
h(x)c ≡ ± 1 (mod f (x)). Now, by the results of Cantor and Zassenhaus, for each
i ∈ { 1, · · · , s }, there are c polynomials pi(x), of degree < d , such that pi(x)c ≡
1 (mod fi(x)), and c such that pi(x)c ≡ − 1 (mod fi(x)). This means there are 2 cs
polynomials p(x) of degree < n in Fq[x] satisfying p(x)c ≡ ± 1 (mod f (x)), q − 1of which are constant. Since h(x) was chosen to be non-constant, it follows that the probability that h(x)c ≡ ± 1 (mod f (x)) is 2cs − (q − 1)
qn − q =2
(qd−12
)s
− (q − 1)
qn − q
= 12s−1 · (qd − 1) s − 2s−1(q − 1)
qn − q< 12s−1 · qds − 2( q − 1)
qn − q
= 12s−1 · qn − 2( q − 1)
qn − q< 12s−1
≤ 12.
Correspondingly, the probability that we can use h(x) to obtain a proper factor of f (x) is > 1 − 12s−1 . Notice that as the number of factors, s, of f (x) grows large, the probability that h(x) can be used to get a non-trivial factorization of
f (x) approaches 1. Suppose now that we can use h(x) along with Theorem 3.5 to get the proper factor gcd( f (x), h (x)c − 1) of f (x). Then we observe that gcd( f (x), h (x)c − 1) and f (x)/ gcd( f (x), h (x)c − 1) are polynomials with irreducible factors that all have equal degree, namely d. So we can apply Theorem 3.5 again in conjunction with randomly selected polynomials to find proper factors of gcd( f (x), h (x)c − 1) and f (x)/ gcd( f (x), h (x)c − 1), and hence come one step closer to finding all the irreducible factors of f (x). Continuing this process iteratively suggests a full al-gorithm for EDF. In view of the fact that proceeding at random gives us a high probability of further separating the input polynomial f (x) at each stage, the up-coming EDF algorithm is a highly efficient factoring tool. 59 EDF Algorithm over Fq with q odd:
A ← { f }
while |A| < s do
for each p ∈ A with deg( p) > d do
choose h ∈ Fq[x] with 0 < deg( h) < deg( p) at random
g ← gcd( p, h )
if g = 1, then
g ← hc − 1 (mod p)
end if if gcd( p, g ) 6 = 1 and gcd( p, g ) 6 = pA ← (A − { p}) ∪ { gcd( p, g ), p/ gcd( p, g )}
end if end while
output A Notice that the algorithm runs until |A| = s. Hence, the algorithm runs until all s of the equal degree irreducible factors of f have been obtained. We apply EDF in the upcoming example.
Example 3.6: Given that f (x) = x6 +4 x3 +3 x2 +2 x+1 ∈ F5[x] can be written as the product of distinct irreducible polynomials of degree 2, we find the complete factorization of f (x). In this scenario, the number of irreducible factors of f (x) is s = 6 /2 = 3. Since each of these factors has degree d = 2, set c = 52−12 = 12. Note that there is a > 34 probability that a randomly chosen non-constant polynomial over F5 with degree < 6 will yield a non-trivial factorization of f (x). We randomly choose
x3 + 2 ∈ F5[x]. The Euclidean Algorithm can be applied to find that gcd( f (x), x 3 + 2) = 1 .
Next, using binary exponentiation, we compute (x3 + 2) 12 − 1 ≡ 3x5 + 4 x3 + 3 x2 + 4 (mod f (x)) .
And we find that gcd( f (x), 3x5 + 4 x3 + 3 x2 + 4) = x4 + 4 x3 + 4 x2 + 2 x + 3 .
60 Now,
f (x)/(x4 + 4 x3 + 4 x2 + 2 x + 3) = x2 + x + 2 ,
and so
f (x) = ( x4 + 4 x3 + 4 x2 + 2 x + 3)( x2 + x + 2) .
Continuing with the algorithm, we factor the polynomial p(x) = x4 + 4 x3 + 4 x2 +2x + 3. Note that there is a > 12 chance that a randomly chosen non-constant polynomial over F5 with degree < 4 will yield a non-trivial factorization of p(x). We randomly select x + 2 ∈ F5[x] and find that gcd( p(x), x + 2) = 1 .
Then we compute the following congruence: (x + 2) 12 − 1 ≡ 3 (mod p(x)) .
(How is the congruence to 3 in the last line consistent with our previous results? Well, we know that for a polynomial h(x) that is nonzero mod p(x), h(x)12 is congruent to a constant mod p(x) if and only if h(x)12 ≡ ± 1 (mod p(x)) if and only if h(x) − 1 ≡ 0 (mod p(x)) or h(x) − 1 ≡ − 2 (mod p(x)). Now −2 = 3 ∈ F5.) So it follows that gcd( p(x), 3) = 1 ,
and we must randomly choose another polynomial. We choose x3 + 2 x2 + 4 ∈ F5,and compute gcd( p(x), x 3 + 2 x2 + 4) = 1 .
Now, (x3 + 2 x2 + 4) 12 − 1 ≡ x3 + 2 x2 + 3 x + 1 (mod p(x)) and gcd( p(x), x 3 + 2 x2 + 3 x + 1) = x2 + 3 .
61 Dividing we get
p(x)/(x2 + 3) = x2 + 4 x + 1 .
Thus,
f (x) = ( x2 + 3)( x2 + 4 x + 1)( x2 + x + 2) is the complete factorization of f (x) over F5.Next we describe Cantor and Zassenhaus’ original strategy for factoring the polynomial f (x) defined at the beginning of this section in the case that p = 2, and hence q = pv is even. First we treat the subcase that q ≡ 1 (mod 3). We will demonstrate a strategy for finding a proper factor of f (x). Recall from Proposition 1.42 that (Fq)× is a cyclic group under multiplication with q − 1 elements. Now, since 3 divides q −1, it is a consequence of the fact that ( Fq)× is cyclic that there exists an element ρ ∈ (Fq)× of (multiplicative) order 3. Note that to proceed, the element
ρ must be known. Observe also that for each i ∈ { 1, · · · , s }, ρ is an element of order 3 in the cyclic group ( Fq[x]/(fi(x))) ×. So, {1, ρ, ρ 2} is the unique subgroup of (Fq[x]/(fi(x))) × of order 3 and hence contains all elements p(x)(mod fi(x)) ∈ Fq[x]satisfying p(x)3 ≡ 1 (mod fi(x)). Set c = qd−13 and suppose h(x) is a non-constant polynomial over Fq satisfying gcd( f (x), h (x)) = 1 and h(x)c 6 ∈ { 1, ρ, ρ 2} (mod f (x)). Since gcd( f (x), h (x)) = 1,
h(x) 6 ≡ 0 (mod fi(x)) for each i, and so (h(x)c)3 = h(x)qd−1 ≡ 1 (mod fi(x)) .
Thus, h(x)c ∈ { 1, ρ, ρ 2} (mod fi(x)) for each i.Let g1(x) = h(x)c − 1(mod f (x)) and g2(x) = h(x)c − ρ(mod f (x)). Since
h(x)c 6 ∈ { 1, ρ, ρ 2} (mod f (x)), clearly h(x)c is non-constant over Fq. Then g1 and
g2 are also non-constant. Moreover, since h(x)c 6 ≡ ρ2 (mod f (x)), there exists an
i0 ∈ { 1, · · · , s } with either h(x)c ≡ 1 (mod fi0 (x)) or h(x)c ≡ ρ (mod fi0 (x)). Correspondingly, either g1(x) ≡ 0 (mod fi0 (x)) or g2(x) ≡ 0 (mod fi0 (x)). Hence, by Proposition 3.4, either gcd( f (x), g 1(x)) or gcd( f (x), g 2(x)) is a proper factor of
f (x). We now give an upper bound for the probability that a polynomial h(x)randomly chosen from the qn − q non-constant polynomials of degree < n in Fq[x]cannot be used in conjunction with the strategy given above to yield a proper factor of f (x). To accomplish this, we need only calculate the probability that 62 h(x)c ∈ { 1, ρ, ρ 2} (mod f (x)). By the results of Cantor and Zassenhaus, for each j = 0 , 1, 2, there exist c polynomials pi(x), of degree < d , which satisfy
pi(x)c ≡ pj (mod fi(x)), 1 ≤ i ≤ s. This means there are 3 cs polynomials p(x)over Fq of degree < n satisfying p(x)c ∈ { 1, ρ, ρ 2} (mod f (x)), q − 1 of which are constant. Since h(x) was chosen to be non-constant, there is a 3cs − (q − 1)
qn − q = 3( qd−13 )s − (q − 1)
qn − q< 13s−1
≤ 13probability that h(x) ∈ { 1, ρ, ρ 2} (mod f (x)). Hence, there is > 1 − 13s−1 chance that the randomly selected polynomial h(x) can be used to find a proper factor of
f (x). Consider the other subcase that q ≡ 2 (mod 3). Since 3 does not divide q − 1in this case, there exists no element of order 3 in ( Fq)×. Notice that if there were an element θ ∈ Fq with θ2 + θ + 1 = 0, then θ3 − 1 = ( θ − 1)( θ2 + θ + 1) = 0 would contradict the fact that ( Fq)× has no element of order 3. Thus, the polynomial
x2 + x + 1 has no root in Fq and is therefore irreducible over Fq. Let ρ be a root of
x2 + x + 1 in some extension of Fq. We now factor f (x) in the quadratic extension field
Fq(ρ) = {a + bρ : a, b ∈ Fq} ∼= Fq2 .
It is possible that some of the equal degree factors of f (x) which are irreducible over Fq are not irreducible over Fq(ρ). So, we should perform DDF as the first step in factoring f (x) over Fq(ρ). Then, since q2 ≡ 22 (mod 3) ≡ 1 (mod 3), we can use the EDF process described earlier to find all of the irreducible factors of
f (x) over Fq(ρ) ∼= Fq2 , where ρ is our known element of order 3. After obtaining the irreducible factors of f (x) in Fq(ρ)[ x], we can combine factors with coefficients lying outside of Fq to obtain a non-trivial factorization of f (x) over Fq.We observe that Cantor and Zassenhaus’ original strategy for factoring f (x)when p = 2 is very difficult to apply in practice, especially in the case where
q ≡ 2 (mod 3). So, we will develop an alternate factoring approach for the case of p = 2, which draws from the work of Shoup. We start with the following proposition. 63 Proposition 3.7: Let q = 2 v and a ∈ Fq. Then
v−1
∑
j=0
a2j
= 0 or
v−1
∑
j=0
a2j
= 1 .
Proof: By Theorem 1.6 and the Generalized FLT,
v−1
∑
j=0
a2j
= a +
v−1
∑
j=1
a2j
= a2v
+
v−1
∑
j=1
a2j
=
v
∑
j=1
a2j
=
v
∑
j=1
a2·2j−1
=
( v∑
j=1
a2j−1
)2
=
(v−1∑
j=0
a2j
)2
.
Hence,
v−1
∑
j=0
a2j
(v−1∑
j=0
a2j
− 1
)
= 0 ,
and so it must be that
v−1
∑
j=0
a2j
= 0 or
v−1
∑
j=0
a2j
= 1 .
Now, we get a theorem which will give us a new strategy for finding a proper factor of f (x). 64 Theorem 3.8: Let q = 2 v and h(x) a polynomial over Fq. Set
g(x) =
vd −1
∑
j=0
h(x)2j
(mod f (x)) .
If vd −1∑
j=0
h(x)2j
/∈ { 0, 1} (mod f (x)), then gcd( f (x), g (x)) is a proper factor of f (x).
Proof: For each i ∈ { 1, 2, · · · , s }, note that Fq[x]/(fi(x)) is an isomorphic copy of the finite field containing qd = 2 vd elements. Thus, by Proposition 3.7,
vd −1
∑
j=0
h(x)2j
≡ 0 (mod fi(x)) or
vd −1
∑
j=0
h(x)2j
≡ 1 (mod fi(x)) .
Since ∑vd −1
j=0
h(x)2j
/∈ { 0, 1}(mod f (x)), clearly ∑vd −1
j=0
h(x)2j
is non-constant over
Fq. Hence, g(x) is non-constant. Now, since ∑vd −1
j=0
h(x)2j
6 ≡ 1 (mod f (x)), there must exist an i0 ∈ { 1, · · · , s } with ∑vd −1
j=0
h(x)2j
≡ 0 (mod fi0 (x)). So,
g(x) ≡ 0 (mod fi0 (x)), and it follows from Proposition 3.4 that gcd( f (x), g (x)) is a proper factor of f (x).
For a polynomial h(x) randomly chosen from the qn − q non-constant poly-nomials in Fq[x] of degree < n , q = 2 v, Theorem 3.8 cannot be used to get a proper factor of f (x) only if ∑vd −1
j=0
h(x)2j
∈ { 0, 1} (mod f (x)), which oc-curs precisely when gcd( f (x), ∑vd −1
j=0
h(x)2j
) ∈ { 1, f (x)}. Using the results of Shoup, we have that for each i ∈ { 1, · · · , s }, there are qd/2 polynomials pi(x), of degree < d , such that ∑vd −1
j=0
pi(x)2j
≡ 0 (mod fi(x)), and qd/2 such that
∑vd −1
j=0
pi(x)2j
≡ 1 (mod fi(x)). This means there are 2( qd/2) s polynomials p(x)of degree < n in Fq[x] with ∑vd −1
j=0
pi(x)2j
∈ { 0, 1} (mod f (x)), q of which are constant. Since h(x) was chosen to be non-constant, it follows that there is a 2
(qd
2
)s
− qqn − q < 12s−1
chance that ∑vd −1
j=0
h(x)2j
∈ { 0, 1} (mod f (x)). So, there is > 1 − 12s−1 chance that gcd( f (x), g (x)) is a proper factor of f (x), where g(x) = ∑vd −1
j=0
h(x)2j
(mod f (x)). Using the strategy of finding a proper factor of f (x) given in Theorem 3.8 with randomly chosen polynomials, we now present an EDF algorithm for the case of p = 2. Apart from the definition of the polynomial g, this algorithm is identical to the EDF algorithm for the case that q is odd. 65 EDF Algorithm over Fq with q = 2 v:
A ← { f }
while |A| < s do
for each p ∈ A with deg( p) > d do
choose h ∈ Fq[x] with 0 < deg( h) < deg( p) at random
g ← ∑vd −1
j=0
h2j
(mod p)
if gcd( p, g ) 6 = 1 and gcd( p, g ) 6 = pA ← (A − { p}) ∪ { gcd( p, g ), p/ gcd( p, g )}
end if end while
output A We apply this algorithm over F2 in the upcoming example.
Example 3.9: Given that f (x) = x8 + x7 + x5 + x4 + x3 + x + 1 ∈ F2[x] can be written as the product of distinct irreducibles of degree d = 4, we find the complete factorization of f (x). Note that the number of irreducible factors of f (x) is s = 8 /4 = 2. So, there is a > 12 probability that a randomly chosen non-constant polynomial over
F2 of degree < 8 will yield a proper factor of f (x) using the EDF algorithm. We randomly choose x3 + x2 + 1 ∈ F2[x]. Now, we compute
1·4−1
∑
j=0
(x3 + x2 + 1) 2j
=
3
∑
j=0
(x3 + x2 + 1) 2j
≡ x7 + x5 + x3 + x2 (mod f (x)) .
Observe that since ∑3
j=0
(x3 + x2 + 1) 2j
/∈ { 0, 1} (mod f (x)), we are guaranteed to get a proper factor of f (x) by applying Theorem 3.8. So, we compute gcd( f (x), x 7 + x5 + x3 + x2) = x4 + x3 + 1 ,
and then
f (x)/(x4 + x3 + 1) = x4 + x + 1 .
Thus,
f (x) = ( x4 + x3 + 1)( x4 + x + 1) is the complete factorization of f (x) over F2.66 This concludes our discussion of equal degree factorization. In the final section of Chapter 3, we will look at a few useful applications of the Cantor-Zassenhaus factoring method.
3.3 Applications of the Cantor-Zassenhaus Method
Our work so far in Chapter 3 has provided us with quite a few tools for gaining information about the factorization of a polynomial over Fq. When used together, we have established that these tools result in a complete factoring pro-cess. However, we can also use our tools to answer more specialized questions about polynomials. Specifically, in this section we will develop a root finding pro-cess over Fq, two tests for irreducibility, and a method by which we can generate irreducible polynomials of any given degree. We will then conclude the section by giving an interesting application of the Cantor-Zassenahus method to Berlekamp’s method of factoring. Let f (x) be a non-constant monic polynomial over Fq. Recall from Section 3.1 that xq − x is the product of all distinct monic polynomials of degree 1 in
Fq[x]. Then gcd( f (x), x q − x) gives us the product of all distinct linear factors of
f (x). Notice that we can use our EDF algorithm to factor gcd( f (x), x q − x) and, in turn, separate all these linear factors. This suggests the following process for finding all of the roots of f (x).
Steps for Root Finding:
(1) Let h(x) = xq −x(mod f (x)), and find gcd( f (x), h (x)). If gcd( f (x), h (x)) = 1, conclude that f (x) has no roots. Otherwise, proceed to (2). (2) Use the EDF algorithm to factor gcd( f (x), h (x)). (3) Find the roots associated with the linear factors identified in (2). This will give all the roots of f (x) in Fq.
Example 3.10: Let f (x) = x12 +3 x11 +4 x10 +5 x8+x6+3 x5+6 x4+6 x3+10 ∈ F13 [x]. We desire to find all of the roots of f (x) in F13 . Let
h(x) = x13 − x(mod f (x)) = 5 x11 + 12 x10 + 8 x9 + 2 x8 + 12 x7 + 3 x5 + 12 x4 + 5 x3 + 2 x + 4 .
67 Now, we compute gcd( f (x), h (x)) = x4 + 3 x3 + 4 x2 + 5 .
Finally, we use the EDF algorithm to find that
x4 + 3 x3 + 4 x2 + 5 = ( x + 2)( x + 7)( x + 8)( x + 12) = ( x − 11)( x − 6)( x − 5)( x − 1) .
Hence, 1 , 5, 6, and 11 are all the roots of f (x) in F13 .Next we turn our attention to generating tests for irreducibility. For a posi-tive integer r, recall that xqr
−x is the product of all the distinct monic irreducibles in Fq[x] of degree d, where d runs through all of the positive divisors of r. We will use polynomials of the form xqr
−x to develop a couple of methods for determining whether the arbitrary polynomial f (x) is irreducible. Note that if f (x) is reducible, then it is not hard to see that f (x) must have an irreducible factor of degree ≤ deg( f (x)) /2. With this observation in mind, we give the following proposition.
Proposition 3.11 (General Irreducibility Test): Let f (x) ∈ Fq[x] be a monic polynomial of degree n > 1. Then f (x) is irreducible over Fq if and only if gcd( f (x), x qr
− x) = 1 for all integers r with n
4
< r ≤ n
2
.
Proof : First, suppose that f (x) is irreducible over Fq, and let r be a positive integer with n
4
< r ≤ n
2
. Note that the degree of any monic irreducible factor of
xqr
− x must divide r. Then since n - r, the irreducible f (x) is not a factor of
xqr
− x. So it must be that gcd( f (x), x qr
− x) = 1. We now prove the other direction of the statement by proving its contraposi-tive. Suppose that f (x) is reducible. We seek to show there exists an integer r with
n
4
< r ≤ n
2
such that gcd( f (x), x qr
−x) 6 = 1. Since f (x) is reducible, f (x) must have an irreducible factor, say g(x), of degree k ≤ n
2
. Clearly g(x)| gcd( f (x), x qk
− x), and so gcd( f (x), x qk
− x) 6 = 1. Hence, if the positive integer k satisfies n
4
< k ≤ n
2
,then we are done. So, assume that k ≤ n
4
. Let j = [ n
4k
], where [ n
4k
] is the largest integer ≤ n
4k
, and let s = ( j + 1) k. Then
s > n
4k · k = n
4 ,
68 and
s ≤
( n
4k + 1
)
k
= n
4 + k
≤ n
4 + n
4= n
2 .
Now, since k divides the integer s and g(x) is an irreducible of degree k, it follows that g(x) is a factor of xqs
− x. Hence, g(x)| gcd( f (x), x qs
− x), and gcd( f (x), x qs
− x) 6 = 1.
We now use Proposition 3.11 to develop an algorithm for irreducibility test-ing.
General Irreducibility Test Algorithm:
The input is a monic polynomial f over Fq of degree n > 1.
for r = [n
4
] + 1 , [n
4
] + 2 , · · · , [n
2
] do
h ← xqr
− x (mod f (x))
g ← gcd( f, h )
if g 6 = 1 then output “reducible” and STOP end if end for
output “irreducible” The basic idea of this algorithm is that starting with r = [n
4
]+1, we compute
xqr
(mod f (x)) using binary exponentiation and then take the corresponding gcd. If we reach the end of the for loop, then by Proposition 3.11, we know that f (x)is irreducible. This algorithm is applied in Example 3.12.
Example 3.12: Consider the polynomial f (x) = x7 +2 x6 +x3 +x2 +x+2 over F3.We seek to discover whether or not f (x) is irreducible. To start the irreducibility test, notice that the only integers r satisfying 74 < r ≤ 72 are r = 2 and r = 3. First, we use binary exponentiation and the Euclidean Algorithm to find that
x32
− x ≡ x6 + 2 x5 + x4 + 2 x2 + 2 x + 1 (mod f (x)) , and gcd( f (x), x 6 + 2 x5 + x4 + 2 x2 + 2 x + 1) = 1 .
69 Next, we compute
x33
− x ≡ x6 + 2 x4 + 2 x2 + x + 2 (mod f (x)) , and gcd( f (x), x 6 + 2 x4 + 2 x2 + x + 2) = 1 .
Thus, by the General Irreducibility Test, f (x) is irreducible over F3.When n is large, we observe that the General Irreducibility Test requires a great deal of gcd computations before concluding that a polynomial is irreducible. We will now formulate an alternate irreducibility test, due to Rabin, that does not require nearly as many gcd computations for large degree inputs.
Proposition 3.13 (Rabin’s Irreducibility Test): Let n > 1 be an integer and w1, w 2, · · · , w k be all the distinct prime divisors n. Denote ni = n/w i for 1 ≤ i ≤ k. A monic polynomial f (x) ∈ Fq[x] of degree n is irreducible in Fq[x] if and only if gcd( f (x), x qni − x) = 1 for each 1 ≤ i ≤ k, and f (x) divides xqn
− x.
Proof : First, suppose f (x) is irreducible over Fq. For each ni, the degree of each irreducible factor of xqni −x divides ni. Since n - ni, clearly f (x) not an irreducible factor of xqni − x. Hence, gcd( f (x), x qni − x) = 1 for each i. Furthermore, since each irreducible in Fq[x] of degree n divides xqn
− x, it follows that f (x) divides
xqn
− x.Next we prove the contrapositive of the other direction of the statement. Suppose that f (x) is reducible over Fq and f (x) divides xqn
− x. Since f (x)is reducible, f (x) has an irreducible factor in Fq[x], say g(x), of degree d < n .Now, since g(x) divides xqn
− x it follows that d|n. Suppose n = wα1
1
wα2
2
· · · wαk
k
is the prime factorization of the integer n, where α1, α 2, · · · , α k are positive in-tegers. Then since d|n, we can write d = wβ1
1
wβ2
2
· · · wβk
k
for some nonnegative integers β1, β 2, · · · , β k. Furthermore, since d < n , it must be that βi < α i for some 1 ≤ i ≤ k. Then clearly d|ni. Hence, g(x) is a factor of xqni − x, and it follows that gcd( f (x), x qni − x) 6 = 1.
Using Proposition 3.13, on the next page we present an alternate algorithm for irreducibility testing. 70 Rabin’s Irreducibility Test Algorithm:
Let w1, w 2, · · · , w k be all the distinct prime divisors of an integer n > 1 ordered so that w1 > w 2 > · · · > w k. The input is a monic polynomial f ∈ Fq[x] of degree
n.
for j = 1 , 2, · · · , k do
nj ← n/w j
end for for i = 1 , 2, · · · , k do
h ← xqni − x (mod f )
g ← gcd( f, h )
if g 6 = 1 then output “reducible” and STOP end if end for
g ← xqn
− x (mod f )
if g 6 = 0 then output “reducible” and STOP end if
output “irreducible” Note that the prime divisors w1, w 2, · · · , w k of n are ordered so that w1 >w2 > · · · > w k to give n1 < n 2 < · · · < n k. In the case that the input f is reducible, this ensures that we do not reduce xqni − x mod f for any unnecessarily large values of ni.Now, let’s look at an example.
Example 3.14: Let f (x) = x10 + x9 + x7 + x5 + x4 + x2 + 1 ∈ F2[x]. To apply Rabin’s Irreducibility Test, first notice that the only prime divisors of 10 are 5 and 2. So, we let n1 = 10 /5 = 2 and n2 = 10 /2 = 5 in the algorithm. First, we compute
x22
− x ≡ x4 − x (mod f (x)) , and gcd( f (x), x 4 − x) = 1 .
Next we compute
x25
− x ≡ x7 + x6 + x4 (mod f (x)) , and gcd( f (x), x 7 + x6 + x4) = 1 .
71 So, all of the necessary gcd’s are 1. Now, using binary exponentiation at great length, it can be found that x210
≡ x (mod f (x)). Hence,
x210
− x ≡ 0 (mod f (x)) ,
which means that f (x) divides x210
− x. Thus, by Rabin’s Irreducibility Test, f (x)is irreducible over F2.A major disadvantage of Rabin’s Test is that we must always compute
xqn
− x (mod f (x)), where n is the degree of f (x). This computation can be quite tedious, even in cases where n is not relatively large. Recall that when using the General Irreducibility Test, the highest value for r for which xqr
−x (mod f (x)) must be computed is r = n
2
. So, in cases where n is not very large, the General Irreducibility Test is preferable from a computational standpoint. For example, the computations required by the General Irreducibility Test to show the degree 10 polynomial of Example 3.14 is irreducible are much less tiresome than the com-putations required by Rabin’s Test. Now that we have established irreducibility tests, we can potentially generate irreducible polynomials of a given degree by using trail and error. For example, if we want an irreducible of degree 7, we can begin randomly selecting degree 7 poly-nomials in Fq[x] and hope that we eventually find one that is deemed irreducible by an irreducibility test. With a little luck, this process may work sometimes, but, in general, it is not even close to being an efficient method for generating irreducibles. We seek to develop a better method. We begin by setting
h1(x) = xq − x.
Then h1(x) is the product of all monic degree 1 polynomials over Fq. We can ac-tually separate all the monic linear polynomials by applying EDF to h1(x). Notice that since xq2
− x is the product of all monic degree 1 and degree 2 irreducibles, dividing xq2
− x by h1(x) gives the product of only the degree 2 irreducibles. Now we let
h2(x) = xq2
− xh1(x) .
Applying EDF to h2(x) will separate all degree 2 irreducibles. Further, since
xq3
− x is the product of all monic degree 1 and degree 3 irreducibles over Fq,dividing xq3
− x by h1(x) gives the product of just the degree 3 irreducibles. So, 72 we let
h3(x) = xq3
− xh1(x) .
We then can apply EDF to h3(x) to generate all the distinct degree 3 irreducibles. Next, since xq4
−x is the product of all degree 1, degree 2, and degree 4 irreducibles, dividing xq4
− x by h1(x)h2(x) gives the product of all degree 4 irreducibles. Then we let
h4(x) = xq4
− xh1(x)h2(x),
and we can apply EDF to h4(x) to separate all the distinct degree 4 irreducibles. Continuing this process recursively, for a positive integer k, we have
hk(x) = xqk
− x
∏
d|kd<k
hd(x),
where hk(x) is the product of all disinct monic irreducibles of degree k over Fq.These irreducibles can be separated by applying EDF to hk(x). We have now developed a process for generating all irreducibles in Fq[x] of any given degree. This process is applied in Example 3.15.
Example 3.15: We will find all monic irreducible polynomials of degree ≤ 5 in
F2[x]. Applying EDF in each step of our process, we find
h1(x) = x2 − x
= x(x − 1) ,h2(x) = x4 − xh1(x)= x2 + x + 1 ,h3(x) = x8 − xh1(x)= ( x3 + x + 1)( x3 + x2 + 1) ,
73 h4(x) = x16 − xh1(x)h2(x)= ( x4 + x + 1)( x4 + x3 + 1)( x4 + x3 + x2 + x + 1) ,h5(x) = x32 − xh1(x)= ( x5 + x2 + 1)( x5 + x3 + 1)( x5 + x3 + x2 + x + 1)( x5 + x4 + x2 + x + 1)
· (x5 + x4 + x3 + x + 1)( x5 + x4 + x3 + x2 + 1) .
The factors in these five products give all monic irreducible polynomials of degree
≤ 5 over F2.Recall that being able to find irreducibles allows us to explicitly construct finite fields. For example, if we can find an irreducible g(x) of degree v over Fp,then we can let α be an arbitrary root of g(x) and construct the finite field
Fp(θ) = {a0 + a1α + a2α2 + · · · + av−1αv−1 : a0, a 1, a 2, · · · , a v−1 ∈ Fp}∼= Fpv
= Fq.
We can then use the fact that α is a root of g(x) to perform operations in Fp(θ). Computationally speaking, when v > 1, the field Fq is useless to us unless we can find irreducibles that yield concrete representations of it. Hence, the fact that we have developed a way to generate irreducibles is of great importance - it basically means we can do computations in any finite field of our choosing. We will conclude the chapter by applying the strategies of Cantor and Zassen-haus to Berlekamp’s technique for factoring. Specifically, for the case that q is odd, we will generate a probabilistic method for finding a proper factor of an arbitrary polynomial over Fq. Note that the work we do here will closely resemble our work in Section 3.2 over EDF. Suppose q is odd and let f (x) ∈ Fq[x] be a non-constant monic polynomial of degree n with complete factorization f (x) = f1(x)k1 f2(x)k2 · · · fm(x)km , where
m ≥ 2. Recall from Chapter 2 that we sought to factor f (x) by finding polynomials in the vector space
V = {g(x) ∈ Fq[x] : deg( g(x)) < n and g(x)q ≡ g(x) (mod f (x)) }
over Fq. It follows from Theorem 2.7 that V has qm elements, where m is the 74 number of distinct irreducibles which divide f (x). We now choose a random polynomial g(x) from the qm − 1 nonzero polynomials contained in V (remember that selecting a random element from V amounts to randomly selecting coefficients that solve a homogenous system of equations). We seek to use g(x) in some way to get a proper factor of f (x). Notice that if gcd( f (x), g (x)) 6 = 1, then gcd( f (x), g (x)) gives us a proper factor of f (x). So, we assume that gcd( f (x), g (x)) = 1. By Proposition 2.6, for i = 1 , 2, · · · , m , we have that g(x) ≡ gi (mod fi(x)ki )for some gi ∈ Fq. Set c = q−12 . Since gcd( f (x), g (x)) = 1, it follows that g(x) 6 ≡
0 (mod fi(x)ki ) for each i, which implies gi 6 = 0. Thus, by the Generalized FLT, (g(x)c)2 = g(x)q−1
≡ gq−1
i
(mod fi(x)ki )
≡ 1 (mod fi(x)ki ),
and so g(x)c ≡ 1 (mod fi(x)ki ) or g(x)c ≡ − 1 (mod fi(x)ki ) for each i.For the time being, we additionally assume that g(x)c 6 ≡ ± 1 (mod f (x)) (note that this automatically guarantees g(x) is non-constant in Fq[x]). In partic-ular, since g(x)c 6 ≡ − 1 (mod f (x)), we have that g(x)c 6 ≡ − 1 (mod fi0 (x)ki0 ) for some 1 ≤ i0 ≤ m. Then g(x)c ≡ 1 (mod fi0 (x)ki0 ), and so fi0 (x)ki0 is a common factor of f (x) and g(x)c − 1. Since we also have that g(x)c − 1 6 ≡ 0 (mod f (x)), it follows that gcd( f (x), g (x)c − 1) is a proper factor of f (x). Removing all of our assumptions, we now calculate the probability that for a random, nonzero element g(x) ∈ V , neither gcd( f (x), g (x)) nor gcd( f (x), g (x)c−1) is a proper factor of f (x). Considering our previous results, we need only calculate the probability that g(x)c ≡ ± 1 (mod f (x)). Recall from the proof of Theorem 2.7 that there is a one-to-one correspondence between V and the set
S = {(s1, s 2, · · · , s m) : si ∈ Fq}.
The nature of this correspondence is that s(x) ∈ V iff there exists a unique m-tuple ( s1, s 2, · · · , s m) ∈ S with s(x) ≡ si (mod fi(x)ki ) for each i and deg( s(x)) <n. Now, as noted by Cantor and Zassenhaus, it can be shown that there are cm m-tuples ( s1, s 2, · · · , s m) such that sci = 1 for each i, and cm such that
sci = −1 for each i. Correspondingly, there are 2 cm polynomials s(x) ∈ V with
s(x) ≡ ± 1 (mod f (x)). Thus, the probability that neither gcd( f (x), g (x)) nor gcd( f (x), g (x)c − 1) is a proper factor of f (x) is 75 2cm
qm − 1 = 12m−1 · (q − 1) m
qm − 1
< 12m−1 .
So, there is a > 1 − 12m−1 chance that either gcd( f (x), g (x)) or gcd( f (x), g (x)c − 1) is a proper factor of f (x). As m, the number of distinct irreducible factors of f (x), grows large, the probability of getting a proper factor of f (x) using this method approaches 1. Recall that for a non-constant element g(x) ∈ V , Theorem 2.9 guarantees that at least one element of the set
{gcd( f (x), g (x) − s) : s ∈ Fq}
is a proper factor of f (x). Notice that if q is large, we may have to compute gcd( f (x), g (x) − s) for many values of s before finding a proper factor. How-ever, with our new results, we know that there is a high probability that either gcd( f (x), g (x)) or gcd( f (x), g (x)c − 1) will be a proper factor of f (x). So, our new probabilistic factorization technique using the elements of V only requires two gcd computations, no matter the size of q. Since there are not as many gcd computations required by this probabilistic technique, we recommend using it to get a nontrivial factorization of f (x). By meshing together the Cantor-Zassenhaus and Berlekamp methods for factoring, we got a considerable result. Hopefully, even more progess in factor-ing polynomials over Fq can be made by looking at these methods together. For example, it would be an interesting endeavor to attempt to generate a complete factoring algorithm that utilizes the probabilistic method we just formulated for finding a proper factor of the arbitrary polynomial f (x) ∈ Fq[x]. Furthermore, it might be productive to explore how DDF can be used in conjunction with Berlekamp’s method to formulate a better deterministic algorithm for factoring. As always, the search for new ideas goes on. 76 Bibliography
Berlekamp, E.R. (1967). “Factoring Polynomials Over Finite Fields.” Bell System Technical Journal , 46(8), 1853-1859. Cantor, D. & Zassenhaus, H. (1981). “A New Algorithm for Factoring Poly-nomials Over Finite Fields.” Mathematics of Computation , 36(154), 587-592. Childs, L. (1995). “A Concrete Inroduction to Higher Algebra.” 2nd ed. New York: Springer. Dummit, D. & Foote, R. (2004). “Abstract Algebra.” 3rd ed. New Jersey: Wiley. Rabin, M. (1980). “Probabilistic Algorithms in Finite Fields.” SIAM Journal on Computing , 9(2), 273-280. Shoup, V. (2005). “A Computational Introduction to Number Theory and Algebra.” Cambridge: Cambridge University Press. 77 |
11196 | https://www.youtube.com/watch?v=IyYHOhBCqFE | 3-4-5 Triangle Method For Finding Square
Training Hands Academy
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Posted: 21 Mar 2021
Today you will learn the 3-4-5 triangle method for an effective way to find square. The 3-4-5 triangle method works great for small projects, and it works great for large projects such as flooring, home construction, and landscaping/gardening.
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0:00 Intro
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My name is Josh Fedorka, and I’m the founder of Training Hands Academy™. I have been a carpenter and woodworker for over 25 years. I have also held certifications in home energy auditing and have built several LEED certified homes in New England.
God has gifted me with many “hands on” skills and it is my calling and purpose to share those skills by teaching others. Whether it is to seek a career in carpentry, become a general contractor or improve one’s DIY skills and knowledge, the motivation behind THA is to help others successfully learn how to work with their hands.
diy #345trianglemethod #constructionlayout
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Transcript:
Intro Today i'm answering a subscriber's question by
showing how to use the 3 4 5 triangle method for finding square this method is generally used
when a square such as a framing square is just too small to get an accurate measurement of squareness
for example this method can be used when framing walls decks porches it can be used for laying out
flooring as well as large woodworking projects it has many uses and it's fantastic for establishing
a 90 degree angle when you don't have one so what What is the 3-4-5 method? exactly is the three four five method well it's
actually just a ratio between whole numbers and the smallest ratio that creates a right triangle
happens to be three four and five this triple number or ratio has a name and it's called
pythagorean triples what might be the coolest thing about these triples is that it's just a
ratio and that means that as long as you keep the ratio there's an infinite amount of possibilities
for example if i multiply each side by 2 i get 6 8 and 10 which is the same ratio and will produce
a right triangle or i could multiply each side by 6 and get 18 24 and 30. the possibilities
are endless so why is it so helpful to be able to change these numbers well it's because you
can scale these triples up for large projects or you can scale them down for smaller projects
for example if i'm building a small project i could use three four and five knowing that those
measurements will actually take up most of the project's footprint or if i was building a large
project where three four and five would only cover a small portion of the project i could use 18 24
and 30. what's also nice about these numbers is you can make them any unit of measurement you want
for example 3 4 and 5 could be inches feet meters or any unit of measurement that works for
you let's take a look now at two different ways to use this method the first way to use
it is for squaring projects like for frames Method #1 - Finding square built-ins cabinets really any woodworking
or carpentry related project for example here today i have a large wooden frame that
needs check for square because of its size i'm going to use three feet four feet and five feet
because it covers most of the size of the project and again that's a good thing because that means
the accuracy will be greater the first measure will be taken from this corner on the outer edge
over to the left three feet and let's make a mark the next measurement will be taken from that same
corner now up the long side of the frame up to four feet and we'll make another mark the last
measurement is taken between those two points which if our frame is square that measurement
will be exactly five feet but in this case as you can see it's short by about a quarter
inch or six millimeters therefore in order to square this frame up the frame would have
to move at this end to the right until we've reached our measurement of five feet and as
you can imagine if projects are large enough you will need a partner to hold the other end of
the tape again this method can be used for large projects as well like home construction floor
layout walls decks porches just to name a few i do have to mention at this point that there
is another way to check this frame for square and that's to use diagonals this is where you
measure from one set of corners diagonally and then from the opposite set of corners diagonally
if these numbers match then you know the project is square we've actually already done a video
detailing exactly how to do this and i'll leave a link in the notes section below so you can watch
it after this one now that this frame is square as a whole we can also be certain then that this
side here is coming off of this side squarely meaning that this is an exact 90 degree angle and
no matter if we extend this line all the way out this piece will remain 90 degrees to this line
regardless of what side we measure it from now that we cover the first method let's talk about
the second method which is laying out 90 degree Method #2 - Laying out 90 degree lines lines to each other some of the most common places
for this to be used would be in floor layout like tile all aspects of home construction from rough
to finish and of course some woodworking projects to demonstrate this method we're going to be using
a chalk line so let's pretend that the line on the bench represents the threshold of an opening say
into another room or space now in order to get a square line that runs through that other room
we'll need to use the three four five method and to do this is pretty simple because it's very
similar to the first method we start by making a reference point on this line and then measure over
from that point three feet and make another mark with that take the chalk line now and extend it
out which would be into that hypothetical room or in our case up the bench from there measure
from the starting point up the string four feet make sure that the string is pulled taut and make
a mark and i've always found it very helpful just to use a sharpie now all i have to do is swing the
string either to the left or to the right until your diagonal measurement reads five feet and
then snap your line with that now you know that these two lines are 90 degrees to each other and
you can use that new line to help with whatever layout you need for your next project well that's
it for today's video i hope it's been helpful of Outro course if you have any questions or need specific
help and like to share some photos and videos reach out to me on instagram thank you so much
for watching i'll see you in the next video |
11197 | https://www.sparknotes.com/chemistry/electrochemistry/thermo/section1/ | Thermodynamics: Energy, Concentration, and Potential | SparkNotes
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Thermodynamics ==============
Chemistry ---------
Title Home
Study Guide
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Introduction and Summary
Energy, Concentration, and Potential
Problems and Solutions
Terms and Formulae
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Topics
Thermodynamics Energy, Concentration, and Potential
Energy, Concentration, and Potential
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Topics Thermodynamics Energy, Concentration, and Potential
Page 1Page 2Page 3
PreviousNext
Electrical Work and Cell Potential
So far we have cited the possibility of doing useful electrical work as the major reason for constructing galvanic cells. Now, we will define exactly what "useful electrical work" means and derive a relationship between work, free energy (G), and cell potential.
From physics, we know that: Potential (E) = - Work (w) / Charge (q) therefore, w = -qE
Let us now define a quantity called a faraday (F) and let F equal the charge in coulombs per mole of electrons (96,485 C). Then q = nF and w= -nFE.
From thermodynamics we know that ΔG = ΔU -TΔS + Δ(PV) and U = heat + w. Therefore, at constant T and P: ΔG = w. Therefore: ΔG = -nFE and at standard state: ΔG o = -nFE o
Because the sign of ΔG predicts the direction of spontaneous reaction and G and E are directly related by the above equation, we can also use E to predict the direction of spontaneous reaction. As you know, if ΔG is greater than zero, the reaction is non-spontaneous but spontaneous in the reverse direction but if ΔG is less than zero, the reaction is spontaneous as written. ΔG and E have opposite signs from the above equation, therefore, spontaneous reactions will have positive cell potentials.
Adding Reduction Potentials
As alluded to in Galvanic Cells, Heading , E's are intrinsic properties of reactions and therefore do not need to be multiplied by any factors when computing the overall cell potential. However, when adding reduction potentials to calculate the potential of a new reduction reaction there are additional mathematical complications. Those complications arise because potentials are not thermodynamic quantities. According to Hess's Law, only state functions (G, H, S) and not path functions (w, q, E) of a series of reactions may be summed to generate a new value for the overall reaction. Because G is a state function and we have a relationship between G and E (ΔG = -nFE) we can add the ΔG's of the reactions to compute the G of the overall reaction which can then be converted into a value of E. For example, that you can compute the overall cell potential by adding the reduction and oxidation potentials of the half-reactions.
Previous sectionNext page Energy, Concentration, and Potential page 2
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11198 | https://www.mayocliniclabs.com/test-catalog/overview/113145 | QUAD1 - Overview: Quad Screen (Second Trimester) Maternal, Serum
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Useful For
Suggests clinical disorders or settings where the test may be helpful
Prenatal screening for open neural tube defect (alpha-fetoprotein only), trisomy 21 (alpha-fetoprotein, human chorionic gonadotropin, estriol, and inhibin A) and trisomy 18 (alpha-fetoprotein, human chorionic gonadotropin, and estriol)
Method Name
A short description of the method used to perform the test
Immunoenzymatic Assay
NY State Available
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Yes
Reporting Name
Lists a shorter or abbreviated version of the Published Name for a test
QUAD SCRN (2nd Tri) Maternal, S
Aliases
Lists additional common names for a test, as an aid in searching
AFP 4 Marker Screen
AFP Maternal Screening
AFP Neural Tube Defects
E3 (Estriol)
Maternal Screening, AFP Four Marker
QUAD
Triple Test
UE3 (Unconjugated Estriol)
Specimen Type
Describes the specimen type validated for testing
Serum
Necessary Information
To provide the best result interpretation, either answer the order entry questions or provide the required information using the Second Trimester Maternal Screening Alpha-Fetoprotein / Quad Screen Patient Information (T595).
ORDER QUESTIONS AND ANSWERS
| Question ID | Description | Answers |
---
| DRPHN | Physician Phone Number | |
| ESTDD | Patient's Estimated Due Date (EDD) | |
| MTHOD | Method Used to Determine EDD | Ultrasound LMP |
| MTWT | Patient Weight | |
| LBKGS | Units (lbs or kg) | lbs kg |
| IDD | Insulin dependent diabetes | None Diabetic |
| RACE1 | Patients race | Black non-Black |
| MULTF | Number of Fetuses | 1 2 3 or more |
| CHOR_ | Number of Chorions | Monochorionic Dichorionic Unknown Not applicable |
| IVFP | IVF pregnancy | No Yes |
| EGGDR | IVF Egg Donor Date of Birth | |
| EGGFR | IVF Egg or Embryo Freeze Date | |
| PRHIS | Prev Down (T21) / Trisomy Pregnancy | No Yes Unknown If MAFP - Not applicable |
| PRNTD | Prev Pregnancy w/ Neural Tube Defect | No Yes Unknown |
| PTNTD | Patient or father of baby has a NTD | No Yes Unknown |
| SMKNG | Current Cigarette smoking status | Non-smoker Smoker |
| INTL | Initial or repeat testing | Initial Repeat |
Specimen Required
Defines the optimal specimen required to perform the test and the preferred volume to complete testing
Collection Container/Tube:
Preferred:Serum gel
Acceptable:Red top
Submission Container/Tube: Plastic vial
Specimen Volume: 1 mL
Collection Instructions:
1. Do not collect specimen after amniocentesis as this could affect results.
Within 2 hours of collection, centrifuge and aliquot serum into a plastic vial.
Additional Information:
For an assessment that includes neural tube defect results, gestational age must be between 15 weeks, 0 days and 22 weeks, 6 days.
Assessments for trisomy 21 (Down syndrome) and trisomy 18 (Edwards syndrome) only are available between 14 weeks, 0 days and 22 weeks, 6 days.
Initial or repeat testing is determined in the laboratory at the time of report and will be reported accordingly. To be considered a repeat test for the patient, the testing must be within the same pregnancy and trimester, with interpretable results for the same tests, and both tests are performed at Mayo Clinic.
Maternal Serum Screening patient education brochure (T522) is available upon request.
Special Instructions
Library of PDFs including pertinent information and forms related to the test
Second Trimester Maternal Screening Alpha-Fetoprotein / Quad Screen Patient Information
Forms
If not ordering electronically, Second Trimester Maternal Screening Alpha-Fetoprotein / Quad Screen Patient Information(T595) is required.
Specimen Minimum Volume
Defines the amount of sample necessary to provide a clinically relevant result as determined by the testing laboratory. The minimum volume is sufficient for one attempt at testing.
0.75 mL
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Gross hemolysis Reject
Gross lipemia OK
Gross icterus OK
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| Specimen Type | Temperature | Time | Special Container |
--- --- |
| Serum | Refrigerated (preferred) | 7 days | |
| | Ambient | 7 days | |
| | Frozen | 90 days | |
Useful For
Suggests clinical disorders or settings where the test may be helpful
Prenatal screening for open neural tube defect (alpha-fetoprotein only), trisomy 21 (alpha-fetoprotein, human chorionic gonadotropin, estriol, and inhibin A) and trisomy 18 (alpha-fetoprotein, human chorionic gonadotropin, and estriol)
Clinical Information
Discusses physiology, pathophysiology, and general clinical aspects, as they relate to a laboratory test
Maternal serum screening is used to identify pregnancies that may have an increased risk for certain birth defects, including neural tube defects (NTD), trisomy 21 (Down syndrome), and trisomy 18 (Edwards syndrome). The screen is performed by measuring analytes in maternal serum that are produced by the fetus and the placenta. The analyte values along with maternal demographic information such as age, weight, gestational age, diabetic status, and race are combined in a mathematical model to derive a risk estimate. A specific cutoff for each condition is used to classify the risk estimate as either screen-positive or screen-negative. A screen-positive result indicates that the value obtained exceeds the established cutoff. A positive screen does not provide a diagnosis but rather indicates that further evaluation should be considered.
Analytes:
Alpha-Fetoprotein:
Alpha-fetoprotein (AFP) is a fetal protein that is initially produced in the fetal yolk sac and liver. A small amount is produced by the gastrointestinal tract. By the end of the first trimester, nearly all AFP is produced by the fetal liver. The concentration of AFP peaks in fetal serum between 10 to 13 weeks. Fetal AFP diffuses across the placental barrier into the maternal circulation. A small amount also is transported from the amniotic cavity.
The AFP concentration in maternal serum rises throughout pregnancy, from a non-pregnancy level of 0.2 ng/mL to about 250 ng/mL at 32 weeks gestation. If the fetus has an open NTD, AFP is thought to leak directly into the amniotic fluid causing unexpectedly high concentrations of AFP. Subsequently, the AFP reaches the maternal circulation, thus producing elevated serum levels. Other fetal abnormalities such as omphalocele, gastroschisis, congenital kidney disease, esophageal atresia, and other fetal distress situations (eg, threatened abortion and fetal demise) also may result in maternal serum AFP elevations. Increased maternal serum AFP concentrations also may be seen in multiple pregnancies and in unaffected singleton pregnancies in which the gestational age has been underestimated.
Lower maternal serum AFP concentrations have been associated with an increased risk for genetic conditions such as trisomy 21 and trisomy 18.
Estriol:
Estriol (E3), the principal circulatory estrogen hormone in the blood during pregnancy, is synthesized by the intact feto-placental unit. E3r exists in maternal blood as a mixture of the unconjugated form and a number of conjugates. The half-life of unconjugated estriol (uE3) in the maternal blood system is 20 to 30 minutes because the maternal liver quickly conjugates E3 to make it more water soluble for urinary excretion. E3 levels increase during the course of pregnancy. Decreased uE3 has been shown to be a marker for trisomy 21 and trisomy 18. Low levels of E3 also have been associated with pregnancy loss, Smith-Lemli-Opitz, and X-linked ichthyosis (placental sulfatase deficiency).
Decreased second trimester uE3 has been shown to be a marker for trisomy-21 and trisomy-18 syndromes. uE3 is a part of multiple marker prenatal biochemical screening, together with alpha-fetoprotein, human chorionic gonadotropin, and inhibin-A measurements. Low levels of uE3 also have been associated with pregnancy loss, Smith-Lemli-Opitz syndrome (defect in cholesterol biosynthesis), X-linked ichthyosis and contiguous gene syndrome (placental sulfatase deficiency disorders), aromatase deficiency, and primary or secondary fetal adrenal insufficiency.
Human Chorionic Gonadotropin:
Human chorionic gonadotropin (hCG) is a glycoprotein consisting of 2 noncovalently bound subunits. The alpha subunit is identical to that of luteinizing hormone (LH), follicle-stimulating hormone (FSH), and thyrotropin (TSH, previously thyroid-stimulating hormone), while the beta subunit has significant homology to the beta subunit of LH and limited similarity to the FSH and TSH beta subunits. The beta subunit determines the unique physiological, biochemical, and immunological properties of hCG. hCG is synthesized by placental cells, starting very early in pregnancy, and serves to maintain the corpus luteum, and hence, progesterone production, during the first trimester. Thereafter, the concentration of hCG begins to fall as the placenta begins to produce steroid hormones and the role of the corpus luteum in maintaining pregnancy diminishes.
Increased total hCG levels are associated with trisomy 21, while decreased levels may be seen in trisomy 18. Elevations of hCG also can be seen in multiple pregnancies, unaffected singleton pregnancies in which the gestational age has been overestimated, triploidy, fetal loss, and hydrops fetalis.
Inhibin A:
Inhibins are a family of heterodimeric glycoproteins, primarily secreted by ovarian granulosa cells and testicular Sertoli cells, which consist of disulfide-linked alpha and beta subunits. While the alpha subunits are identical in all inhibins, the beta subunits exist in 2 major forms, termed A and B, each of which can occur in different isoforms. Depending on whether an inhibin heterodimer contains a beta A or a beta B chain, they are designated as inhibin A or inhibin B, respectively. Together with the related activins, which are homodimers or heterodimers of beta A and B chains, the inhibins are involved in gonadal-pituitary feedback and in paracrine regulation of germ cell growth and maturation. During pregnancy, inhibins and activins are produced by the feto-placental unit in increasing quantities, mirroring fetal growth. Their physiological role during pregnancy is uncertain. They are secreted into the coelomic and amniotic fluid, but only inhibin A is found in appreciable quantities in the maternal circulation during the first and second trimesters.
Maternal inhibin A levels are correlated with maternal hCG levels and are abnormal in the same conditions that are associated with abnormal hCG levels (eg, inhibin A levels are typically higher in trisomy 21 pregnancies). However, despite their similar behavior, measuring maternal serum inhibin A concentrations in addition to maternal serum hCG concentrations further improves the sensitivity and specificity of maternal multiple marker screening for trisomy 21.
Reference Values
Describes reference intervals and additional information for interpretation of test results. May include intervals based on age and sex when appropriate. Intervals are Mayo-derived, unless otherwise designated. If an interpretive report is provided, the reference value field will state this.
Neural Tube Defect Risk Estimate:
An alpha-fetoprotein (AFP) multiple of the median (MoM) <2.5 is reported as screen negative.
AFP MoM > or =2.5 (singleton and twin pregnancies) are reported as screen positive.
Down Syndrome Risk Estimate:
Calculated screen risks <1/270 are reported as screen negative, risks > or =1/270 are reported as screen positive.
Trisomy 18 Risk Estimate:
Calculated screen risks <1/100 are reported as screen negative, risks > or =1/100 are reported as screen positive.
An interpretive report will be provided.
Interpretation
Provides information to assist in interpretation of the test results
Neural Tube Defects
A screen-negative result indicates that the calculated alpha-fetoprotein (AFP) multiple of the median (MoM) falls below the established cutoff of 2.50 MoM. A negative screen does not guarantee the absence of neural tube defects (NTD).
A screen-positive result indicates that the calculated AFP MoM is 2.50 or greater and may indicate an increased risk for open NTD. The actual risk depends on the level of AFP and the individual's pretest risk of having a child with NTD based on family history, geographical location, maternal conditions such as diabetes and epilepsy, and use of folate prior to conception. A screen-positive result does not infer a definitive diagnosis of NTD but indicates that further evaluation should be considered. Approximately 80% of pregnancies affected with NTD have elevated AFP, MoM values greater than 2.5.
Trisomy 21 (Down syndrome) and Trisomy 18 (Edwards syndrome):
A screen-negative result indicates that the calculated screen risk is below the established cutoff of 1/270 for trisomy 21 and 1/100 for trisomy 18. A negative screen does not guarantee the absence of trisomy 21 or trisomy 18.
When a trisomy 21 second-trimester risk cutoff of 1/270 is used for follow-up, the combination of maternal age, AFP, estriol, human chorionic gonadotropin, and inhibin A has an overall detection rate of approximately 77% to 81% with a false-positive rate of 6% to 7%. In practice, both the detection rate and false-positive rate increase with age. The detection rate ranges from 66% (early teens) to 99% (late 40s), with false-positive rates of between 3% and 62%, respectively.
The detection rate for trisomy 18 is 60% to 80% using a second trimester cutoff of 1/100.
Follow-up
Upon receiving maternal serum screening results, all information used in the risk calculation should be reviewed for accuracy (maternal date of birth, gestational dating, etc). If any information is incorrect, the laboratory should be contacted for a recalculation of the estimated risks.
Screen-negative results typically do not warrant further evaluation.
Ultrasound is recommended to confirm dates for NTD or trisomy 21 screen-positive results. Many pregnancies affected with trisomy 18 are small for gestational age. Recalculations that lower the gestational age may decrease the detection rate for trisomy 18. If ultrasound yields new dates that differ by at least 7 days, a recalculation should be considered. If dates are confirmed, high-resolution ultrasound and amniocentesis (including amniotic fluid AFP and acetylcholinesterase measurements for NTD) are typically offered.
Cautions
Discusses conditions that may cause diagnostic confusion, including improper specimen collection and handling, inappropriate test selection, and interfering substances
Variables Affecting Marker Levels:
Race, weight, smoking, multiple fetus pregnancy, insulin-dependent diabetes (IDD), and in vitro fertilization (IVF) may affect marker concentrations. Black mothers tend to have higher alpha-fetoprotein (AFP) levels but lower risk of neural tube defects and are assigned to a separate AFP median set. All multiples of the median (MoM) are adjusted for maternal weight (to account for dilution effects in heavier mothers). The AFP, unconjugated estriol (uE3), and inhibin MoM are adjusted upward in IDD to account for lower values in diabetic pregnancies. Human chorionic gonadotropin (hCG) levels are higher and uE3 levels are lower in pregnancies conceived by IVF, MoM are adjusted accordingly to account for the alterations. Smoking results in higher second trimester maternal serum AFP and inhibin A levels and lower uE3 and hCG levels. MoM are adjusted accordingly to account for analyte differences in smokers.
The estimated risk calculations and screen results are dependent on accurate information for gestation, maternal age, race, IDD, and weight. Inaccurate information can lead to significant alterations in the estimated risk. In particular, erroneous assessment of gestational age can result in false-positive or false-negative screen results. Because of its increased accuracy, the determination of gestational age by ultrasound is recommended, when possible, rather than by last menstrual period.
A screen-negative result does not guarantee the absence of fetal defects. A screen-positive result does not provide a diagnosis but indicates that further diagnostic testing should be considered (an unaffected fetus may have screen-positive result for unknown reasons).
Valid measurements of AFP in maternal serum cannot be made after amniocentesis.
Triplet and higher multiple pregnancies cannot be interpreted.
Each center offering maternal serum screening to patients should establish a standard screening protocol that provides pre- and post-screening education and appropriate follow-up for screen-positive results.
In a small percentage of samples, there is potential for alkaline phosphatase associated positive interference in the Beckman Access uE3 assay. This potential interference does not appear to be related to the amount of alkaline phosphatase in the patient sample. A falsely elevated uE3 test result can lead to inaccurately underestimating the relative risk of chromosomal abnormalities, such as trisomy 21 and 18.
In rare cases, some individuals can develop antibodies to mouse or other animal antibodies (often referred to as human anti-mouse antibodies [HAMA] or heterophile antibodies), which may cause interference in some immunoassays. Caution should be used in interpretation of results, and the laboratory should be alerted if the result does not correlate with the clinical presentation.
Clinical Reference
Recommendations for in-depth reading of a clinical nature
Wald NJ, Cuckle HS, Densem JW, Stone RB. Maternal serum unconjugated oestriol and human chorionic gonadotrophin levels in pregnancies with insulin-dependent diabetes: implications for screening for Down's syndrome. Br J Obstet Gynaecol. 1992;99(1):51-53
American College of Obstetricians and Gynecologists. Practice Bulletin No. 163: Screening for Fetal Aneuploidy. Obstet Gynecol. 2016;127(5):e123-137
Malone FD, Canick JA, Ball RH, et al. First-trimester or second-trimester screening, or both, for Down's syndrome. N Engl J Med. 2005;353(19):2001-2011
Wald NJ, Rodeck C, Hackshaw AK, et al. SURUSS in perspective. Semin Perinatol. 2005;29(4):225-235
Rudnicka AR, Wald NJ, Huttly W, Hackshaw AK. Influence of maternal smoking on the birth prevalence of Down syndrome and on second trimester screening performance. Prenat Diagn. 2002;22(10):893-897
Zhang J, Lambert-Messerlian G, Palomaki GE, Canick JA. Impact of smoking on maternal serum markers and prenatal screening in the first and second trimesters. Prenat Diagn. 2011;31(6):583-588
Yarbrough ML, Stout M, Gronowski AM. Pregnancy and its disorders. In: Rifai N, Horvath AR, Wittwer CT, eds. Tietz Textbook of Clinical Chemistry and Molecular Diagnostics. 6th ed. Elsevier; 2018:1655-1696
Method Description
Describes how the test is performed and provides a method-specific reference
This 4-marker screen includes alpha-fetoprotein (AFP), estriol (uE3), human chorionic gonadotropin (total beta-hCG: ThCG), and inhibin A. Analyte values are compared to median values at a given gestational age and multiple of the median (MoM) results obtained. The MoM results are used in a multivariate algorithm that includes the mother's age to derive risk factors for trisomy 21 (Down syndrome) and trisomy 18 (Edwards syndrome). The screen for neural tube defects (NTD) uses the AFP MoM only. An interpretive report will be provided. The Beckman Access AFP, ThCG, uE3, and inhibin A assays are automated immunoenzymatic assays with paramagnetic separation and chemiluminescent detection.(Package inserts: Access AFP. Beckman Coulter; 2024; Access Total bhCG. Beckman Coulter; 2024; Access Unconjugated Estriol. Beckman Coulter; 2021; Access Inhibin A. Beckman Coulter; 2024)
PDF Report
Indicates whether the report includes an additional document with charts, images or other enriched information
No
Day(s) Performed
Outlines the days the test is performed. This field reflects the day that the sample must be in the testing laboratory to begin the testing process and includes any specimen preparation and processing time before the test is performed. Some tests are listed as continuously performed, which means that assays are performed multiple times during the day.
Monday through Friday
Report Available
The interval of time (receipt of sample at Mayo Clinic Laboratories to results available) taking into account standard setup days and weekends. The first day is the time that it typically takes for a result to be available. The last day is the time it might take, accounting for any necessary repeated testing.
4 to 6 days
Specimen Retention Time
Outlines the length of time after testing that a specimen is kept in the laboratory before it is discarded
3 months
Performing Laboratory Location
Indicates the location of the laboratory that performs the test
Mayo Clinic Laboratories - Rochester Superior Drive
CLIA Number: 24D1040592
Fees :
Several factors determine the fee charged to perform a test. Contact your U.S. or International Regional Manager for information about establishing a fee schedule or to learn more about resources to optimize test selection.
Authorized users can sign in to Test Prices for detailed fee information.
Clients without access to Test Prices can contact Customer Service 24 hours a day, seven days a week.
Prospective clients should contact their account representative. For assistance, contact Customer Service.
Test Classification
Provides information regarding the medical device classification for laboratory test kits and reagents. Tests may be classified as cleared or approved by the US Food and Drug Administration (FDA) and used per manufacturer instructions, or as products that do not undergo full FDA review and approval, and are then labeled as an Analyte Specific Reagent (ASR) product.
This test was developed and its performance characteristics determined by Mayo Clinic in a manner consistent with CLIA requirements. It has not been cleared or approved by the US Food and Drug Administration.
CPT Code Information
Provides guidance in determining the appropriate Current Procedural Terminology (CPT) code(s) information for each test or profile. The listed CPT codes reflect Mayo Clinic Laboratories interpretation of CPT coding requirements. It is the responsibility of each laboratory to determine correct CPT codes to use for billing.
CPT codes are provided by the performing laboratory.
81511
LOINC® Information
Provides guidance in determining the Logical Observation Identifiers Names and Codes (LOINC) values for the order and results codes of this test. LOINC values are provided by the performing laboratory.
| Test Id | Test Order Name | Order LOINC Value |
---
| QUAD1 | QUAD SCRN (2nd Tri) Maternal, S | 48800-7 |
| Result Id | Test Result Name | Result LOINC Value Applies only to results expressed in units of measure originally reported by the performing laboratory. These values do not apply to results that are converted to other units of measure. |
---
| IDD | Insulin dependent diabetes | 44877-9 |
| IVFP | IVF pregnancy | 47224-1 |
| MULTF | Number of Fetuses | 55281-0 |
| 10334 | Down syndrome screen risk estimate | 43995-0 |
| 10335 | Down syndrome maternal age risk | 49090-4 |
| 10337 | Trisomy 18 screen risk estimate | 43994-3 |
| 10356 | INTERPRETATION | 49092-0 |
| 10248 | Additional comments | 48767-8 |
| 10357 | RECOMMENDED FOLLOW UP | 80615-8 |
| 10358 | GENERAL TEST INFORMATION | 62364-5 |
| 7058 | Recalculated Maternal Serum Screen | 32399-8 |
| 3009 | Specimen collection date | 33882-2 |
| 7823 | Maternal date of birth | 21112-8 |
| 7834 | Calculated age at EDD | 43993-5 |
| 26717 | Maternal Weight | 29463-7 |
| 26718 | Maternal Weight | 29463-7 |
| 10353 | hCG, TOTAL | 83086-9 |
| 10054 | EDD by U/S scan | 11781-2 |
| 7753 | EDD by LMP | 11779-6 |
| 7203 | GA on collection by U/S scan | 11888-5 |
| 7204 | GA on collection by dates | 11885-1 |
| 7830 | GA used in risk estimate | 21299-3 |
| 10351 | AFP | 83073-7 |
| 10352 | uE3 | 2250-9 |
| 10354 | INHIBIN | 2478-6 |
| 113146 | Results Summary | 32399-8 |
| 113147 | Neural tube defect risk estimate | 48803-1 |
| 113148 | AFP MoM | 23811-3 |
| 113149 | uE3 MoM | 21264-7 |
| 113150 | hCG, TOTAL MoM | 23841-0 |
| 113151 | INHIBIN MoM | 36904-1 |
| RACE1 | Patient race | 21484-1 |
| SMKNG | Current cigarette smoking status | 64234-8 |
| CHOR_ | Number of Chorions | 92568-5 |
| PRHIS | Prev Down (T21) / Trisomy Pregnancy | 53826-4 |
| PRNTD | Prev Pregnancy w/ Neural Tube Defect | 53827-2 |
| PTNTD | Patient or father of baby has a NTD | 53827-2 |
| INTL | Initial or repeat testing | 77202-0 |
| DRPHN | Physician Phone Number | 68340-9 |
| 601921 | AFP MoM (14,0-14,6) | 23811-3 |
Test Setup Resources
Setup Files
Test setup information contains test file definition details to support order and result interfacing between Mayo Clinic Laboratories and your Laboratory Information System.
Excel| Pdf
Sample Reports
Normal and Abnormal sample reports are provided as references for report appearance.
Normal Reports | Abnormal Reports
SI Sample Reports
International System (SI) of Unit reports are provided for a limited number of tests. These reports are intended for international account use and are only available through MayoLINK accounts that have been defined to receive them.
SI Normal Reports | SI Abnormal Reports
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IN THIS TOPIC
Proton Pump Inhibitors
Histamine-2 (H2) Blockers
Antacids
Other Medications for Stomach Acid
OTHER TOPICS IN THIS CHAPTER
Introduction to Gastritis and Peptic Ulcer Disease
Gastritis
Helicobacter pylori Infection
Peptic Ulcer Disease
Medications for the Treatment of Stomach Acid
Medications for the Treatment of Stomach Acid
ByNimish Vakil, MD, University of Wisconsin School of Medicine and Public Health
Reviewed/Revised Modified Jan 2025
v47645951
VIEW PROFESSIONAL VERSION
Proton Pump Inhibitors|
Histamine-2 (H2) Blockers|
Antacids|
Other Medications for Stomach Acid|
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Medications Used to Treat...
The Stomach
Stomach acid plays a role in a number of disorders of the stomach, including peptic ulcer, gastritis, and gastroesophageal reflux disease (GERD). Although the amount of acid present in the stomach is usually normal in people with these disorders, reducing the amount of acid in the stomach is important in treating the damage to the stomach and intestines as well as in relieving symptoms.
The Stomach
video
Table
Medications Used to Treat Stomach Acid
Table
Medications Used to Treat Stomach Acid
Medications Used to Treat Stomach Acid| Type of Medication | Some Side Effects | Comments |
---
| Antacids |
| 1. Aluminum hydroxide Aluminum hydroxide 2. Calcium carbonate Calcium carbonate 3. Magnesium hydroxide Magnesium hydroxide 4. Sodium bicarbonate Sodium bicarbonate | Aluminum hydroxide: Nausea, headache, weakness, loss of appetite, and constipation Aluminum hydroxide: Nausea, headache, weakness, loss of appetite, and constipation Magnesium hydroxide: Diarrhea Magnesium hydroxide: Diarrhea | These medications are used mainly to relieve symptoms, not as a cure. |
| Histamine-2 (H2) blockers |
| 1. Cimetidine Cimetidine 2. Famotidine Famotidine 3. Nizatidine Nizatidine | Rash, fever, diarrhea, muscle pains, and confusion Cimetidine: May cause breast enlargement and erectile dysfunction in men and may interfere with elimination of certain medications | The once-daily dose is taken in the evening or at bedtime. Doses taken in the morning are less effective. |
| Proton pump inhibitors |
| 1. Esomeprazole Esomeprazole 2. Lansoprazole Lansoprazole 3. Omeprazole Omeprazole 4. Pantoprazole Pantoprazole 5. Rabeprazole Rabeprazole | Diarrhea, constipation, and headache | These medications are usually well tolerated and are most effective means of reducing stomach acid. |
| Other medications for stomach acid |
| 1. Misoprostol Misoprostol 2. Sucralfate Sucralfate | Misoprostol: Abdominal cramping, spontaneous abortion, diarrhea Sucralfate: May reduce absorption of other medications, constipation | Misoprostol is rarely used for treatment of stomach disorders due to the side effects. Sucralfate does not affect acid production but coats stomach ulcers to protect them from acid. |
Proton Pump Inhibitors
The proton pump is the name for the chemical process by which the stomach secretes acid. Proton pump inhibitors are the most potent of the medications that reduce acid production. Proton pump inhibitors promote healing of ulcers in a greater percentage of people in a shorter period of time than do histamine-2 (H2) blockers and thus are typically preferred to H2 blockers for treating ulcers. They are used for severe forms of gastritis (such as when bleeding is present) and for severe GERD. Proton pump inhibitors are also very useful in treating conditions that cause excessive stomach acid secretion, such as Zollinger-Ellison syndrome.
Proton pump inhibitors can be given by mouth or by vein (IV). These drugs are usually very well tolerated but may cause diarrhea, constipation, and headache. Long-term use of proton pump inhibitors may cause reduced absorption of vitamin B12, iron, magnesium, and calcium.
Histamine-2 (H2) Blockers
Histamine is a substance naturally produced in the body that has several roles. Histamine is one of the main substances responsible for allergic reactions, which is why antihistamines (histamine blockers) are given to people who are having an allergic reaction to something. Histamine also helps signal the body to produce stomach acid, which is why a certain type of antihistamine, called a histamine-2 blocker, is used to decrease stomach acid. Thus, H2 blockers are used for many of the same disorders as proton pump inhibitors.
H2 blockers are taken once or twice a day and can be given by mouth or by vein (IV). H2 blockers usually do not cause serious side effects. However, all H2 blockers may cause diarrhea, rash, fever, muscle pains, and confusion. The H2 blocker cimetidine may cause H2 blockers are taken once or twice a day and can be given by mouth or by vein (IV). H2 blockers usually do not cause serious side effects. However, all H2 blockers may cause diarrhea, rash, fever, muscle pains, and confusion. The H2 blocker cimetidine may causebreast enlargement and erectile dysfunction in men. In addition, cimetidine and, to a lesser extent, the other H2 blockers may interfere with the body's elimination of certain medications, such as theophylline for asthma, warfarin for excessive blood clotting, and phenytoin for seizures.and, to a lesser extent, the other H2 blockers may interfere with the body's elimination of certain medications, such as theophylline for asthma, warfarin for excessive blood clotting, and phenytoin for seizures.
Antacids
Antacids are chemical substances that neutralize stomach acid that has already been secreted and thereby raise the pH level in the stomach (make it less acidic). Antacids may be used alone for mild symptoms caused by stomach acid. But by themselves, antacids are not adequate treatment for serious acid-related disorders such as ulcers and severe gastritis. In those disorders, antacids are typically taken in addition to proton pump inhibitors or H2 blockers to help relieve symptoms in the early stage of treatment. Their effectiveness varies with the amount of antacid taken and with the amount of acid a person produces. Almost all antacids can be purchased without a doctor's prescription and are available in tablet, soft chew, or liquid form. However, antacids can interfere with the absorption of many different medications, so a pharmacist or doctor should be consulted about possible drug-drug interactions before antacids are taken.
Sodium bicarbonate Sodium bicarbonate (baking soda) and calcium carbonate,calcium carbonate, the strongest antacids, may be taken occasionally for fast, short-term relief. However, because they are absorbed by the bloodstream, continual use of these antacids may make the blood too alkaline (alkalosis), resulting in nausea, headache, and weakness. Therefore, these antacids generally should not be used in large amounts for more than a few days. These antacids also contain a lot of salt and should not be used by people who need to follow a low-sodium diet or who have heart failure or high blood pressure.
Aluminum hydroxide Aluminum hydroxide is a relatively safe, commonly used antacid. However, aluminum may bind with phosphate in the digestive tract, causing weakness, nausea, and a loss of appetite. The risk of these side effects is greater in people who have an alcohol use disorder, who are undernourished, and who have kidney disease, including those receiving dialysis. Aluminum hydroxide may also cause constipation.
Magnesium hydroxide Magnesium hydroxide is a more effective antacid than aluminum hydroxide. This antacid acts fast and neutralizes acids effectively. However, magnesium is also a laxative. Bowel movements usually remain regular if only a few tablespoons a day are taken. More than 4 doses a day may cause diarrhea. Because small amounts of magnesium are absorbed into the bloodstream, people with kidney damage should take magnesium hydroxide only in small doses. To limit diarrhea, many antacids contain both magnesium hydroxide and aluminum hydroxide.
Anyone who has heart disease, high blood pressure, or a kidney disorder should consult a doctor before selecting an antacid.
Other Medications for Stomach Acid
Sucralfate Sucralfate may work by forming a protective coating in the base of an ulcer to promote healing. It works well on peptic ulcers and is a reasonable alternative to antacids. Sucralfate is taken 2 to 4 times a day and is not absorbed into the bloodstream, so it causes few side effects. It may, however, cause constipation, and in some cases it reduces the effectiveness of other medications.
Misoprostol Misoprostol may be used to reduce the likelihood of developing stomach and duodenal ulcers caused by nonsteroidal anti-inflammatory drugs (NSAIDs). Misoprostol may work by reducing production of stomach acid and by making the stomach lining more resistant to acid. Older adults, people taking corticosteroids, and people who have a history of ulcers or of complications resulting from ulcers are at higher risk of developing an ulcer when they take NSAIDs. These people can take misoprostol with food and their NSAID. However, misoprostol causes diarrhea and other digestive problems in 30% of people who take it. In addition, this medication can cause spontaneous abortions in pregnant women. Alternatives to misoprostol are available for people taking aspirin, NSAIDs, or corticosteroids. These alternatives, such as proton pump inhibitors, are just as effective for reducing the likelihood of developing an ulcer and cause fewer side effects.
Potassium-competitive acid inhibitors (PCABs) such as vonoprazan cause rapid and reversible inhibition of proton pumps and therefore acid secretion. such as vonoprazan cause rapid and reversible inhibition of proton pumps and therefore acid secretion.
Drugs Mentioned In This Article
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