To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

Question

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Answer 1

Question

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If i=2b−1 then it shows an overflow error. Otherwise, it increases the counter by 1 with probability 1/(ni+1−ni) and leaves the counter with probability 1−1/(ni+1−ni) .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from ni to ni+1 with a probability of 1ni+1−ni and leaving the value unchanged.

The expected increase is calculated as,

ni+1−nini+1−ni=1

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: ni=100i for all i≥0 and variance of the counter depends on the ni .

Explanation:

Consider that Vn and Xj are the variant and pairwise independent events. The variance Vn is calculated as,

Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]

Since ni=100i and ni+1−ni=100(i+1)−100i=100 . Therefore, the value represented by the counter with probability 99/100 due to the jth INCREMENT is 0 and with the probability of 1/100 .

The value represented increases by 100. Therefore, by the equation (C.27)

Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99

Now, adding the variances of the Xj gives Var[Vn]=99n .

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Students have asked these similar questions

Explian this C program #include <stdio.h> unsigned int rotateRight(unsigned int num, unsigned int bits) { unsignedint bit_count =sizeof(unsignedint) *8; bits = bits % bit_count; // Handle cases where bits >= bit_count return (num >> bits) | (num << (bit_count - bits)); } int main() { unsignedint num, bits; printf("Enter a number: "); scanf("%u", &num); printf("Enter the number of bits to shift: "); scanf("%u", &bits); printf("After rotation: %u\n", rotateRight(num, bits)); return0; }

Explian thiS C program #include<stdio.h> int countSetBits(int n) { int count = 0; while (n) { count += n & 1; n >>= 1; } return count;} int main() { int num; printf("Enter a number: "); scanf("%d", &num); printf("Output: %d units\n", countSetBits(num)); return 0;}

Please provide the Mathematica code

Answer 2

Question

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If i=2b−1 then it shows an overflow error. Otherwise, it increases the counter by 1 with probability 1/(ni+1−ni) and leaves the counter with probability 1−1/(ni+1−ni) .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from ni to ni+1 with a probability of 1ni+1−ni and leaving the value unchanged.

The expected increase is calculated as,

ni+1−nini+1−ni=1

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: ni=100i for all i≥0 and variance of the counter depends on the ni .

Explanation:

Consider that Vn and Xj are the variant and pairwise independent events. The variance Vn is calculated as,

Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]

Since ni=100i and ni+1−ni=100(i+1)−100i=100 . Therefore, the value represented by the counter with probability 99/100 due to the jth INCREMENT is 0 and with the probability of 1/100 .

The value represented increases by 100. Therefore, by the equation (C.27)

Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99

Now, adding the variances of the Xj gives Var[Vn]=99n .

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Answer 3

Question

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If i=2b−1 then it shows an overflow error. Otherwise, it increases the counter by 1 with probability 1/(ni+1−ni) and leaves the counter with probability 1−1/(ni+1−ni) .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from ni to ni+1 with a probability of 1ni+1−ni and leaving the value unchanged.

The expected increase is calculated as,

ni+1−nini+1−ni=1

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: ni=100i for all i≥0 and variance of the counter depends on the ni .

Explanation:

Consider that Vn and Xj are the variant and pairwise independent events. The variance Vn is calculated as,

Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]

Since ni=100i and ni+1−ni=100(i+1)−100i=100 . Therefore, the value represented by the counter with probability 99/100 due to the jth INCREMENT is 0 and with the probability of 1/100 .

The value represented increases by 100. Therefore, by the equation (C.27)

Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99

Now, adding the variances of the Xj gives Var[Vn]=99n .

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Answer 4

Question

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If i=2b−1 then it shows an overflow error. Otherwise, it increases the counter by 1 with probability 1/(ni+1−ni) and leaves the counter with probability 1−1/(ni+1−ni) .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from ni to ni+1 with a probability of 1ni+1−ni and leaving the value unchanged.

The expected increase is calculated as,

ni+1−nini+1−ni=1

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: ni=100i for all i≥0 and variance of the counter depends on the ni .

Explanation:

Consider that Vn and Xj are the variant and pairwise independent events. The variance Vn is calculated as,

Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]

Since ni=100i and ni+1−ni=100(i+1)−100i=100 . Therefore, the value represented by the counter with probability 99/100 due to the jth INCREMENT is 0 and with the probability of 1/100 .

The value represented increases by 100. Therefore, by the equation (C.27)

Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99

Now, adding the variances of the Xj gives Var[Vn]=99n .

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Answer 5

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If i=2b−1 then it shows an overflow error. Otherwise, it increases the counter by 1 with probability 1/(ni+1−ni) and leaves the counter with probability 1−1/(ni+1−ni) .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from ni to ni+1 with a probability of 1ni+1−ni and leaving the value unchanged.

The expected increase is calculated as,

ni+1−nini+1−ni=1

Hence, the expected increment represented by the counter is 1.

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: ni=100i for all i≥0 and variance of the counter depends on the ni .

Explanation:

Consider that Vn and Xj are the variant and pairwise independent events. The variance Vn is calculated as,

Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]

Since ni=100i and ni+1−ni=100(i+1)−100i=100 . Therefore, the value represented by the counter with probability 99/100 due to the jth INCREMENT is 0 and with the probability of 1/100 .

The value represented increases by 100. Therefore, by the equation (C.27)

Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99

Now, adding the variances of the Xj gives Var[Vn]=99n .

Answer 6

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If i=2b−1 then it shows an overflow error. Otherwise, it increases the counter by 1 with probability 1/(ni+1−ni) and leaves the counter with probability 1−1/(ni+1−ni) .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from ni to ni+1 with a probability of 1ni+1−ni and leaving the value unchanged.

The expected increase is calculated as,

ni+1−nini+1−ni=1

Hence, the expected increment represented by the counter is 1.

Answer 7

Chapter 5, Problem 1P

(a)

Program Plan Intro

To show that the expected value represented by the counter after n INCREMENT operations that have been performed is exactly n.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given Information: The INCREMENT operation works on a counter containing the value i in a probabilistic manner. If i=2b−1 then it shows an overflow error. Otherwise, it increases the counter by 1 with probability 1/(ni+1−ni) and leaves the counter with probability 1−1/(ni+1−ni) .

Explanation:

The counter after n INCREMENT operations is performed exactly n times for the expected value.

Consider that the initial value of the counter is i and increasing the number representation from ni to ni+1 with a probability of 1ni+1−ni and leaving the value unchanged.

The expected increase is calculated as,

ni+1−nini+1−ni=1

Hence, the expected increment represented by the counter is 1.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: ni=100i for all i≥0 and variance of the counter depends on the ni .

Explanation:

Consider that Vn and Xj are the variant and pairwise independent events. The variance Vn is calculated as,

Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]

Since ni=100i and ni+1−ni=100(i+1)−100i=100 . Therefore, the value represented by the counter with probability 99/100 due to the jth INCREMENT is 0 and with the probability of 1/100 .

The value represented increases by 100. Therefore, by the equation (C.27)

Var[Xj]=E[Xj2]−E2[Xj]=(( 0 2 ⋅ 99 100 )+( 100 2 ⋅ 1 100 ))−12=100−1=99

Now, adding the variances of the Xj gives Var[Vn]=99n .

Answer 13

(b)

Program Plan Intro

To calculate the variance in the value represented by the register after n INCREMENT operations have been performed.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given Information: Consider a simple case: ni=100i for all i≥0 and variance of the counter depends on the ni .

Explanation:

Consider that Vn and Xj are the variant and pairwise independent events. The variance Vn is calculated as,

Var[Vn]=Var[X1]+Var[X2]+⋯Var[Xn]

Since ni=100i and ni+1−ni=100(i+1)−100i=100 . Therefore, the value represented by the counter with probability 99/100 due to the jth INCREMENT is 0 and with the probability of 1/100 .