Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

Question

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Answer 1

Question

Chapter 18, Problem 1P

(a)

Program Plan Intro

Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

(a)

Expert Solution

Explanation of Solution

In worse case for every stack operation it requires disk access for implementation sothe number of disk accesses for n stack operations using this simple implementation is asymptotically equals to θ(m) .

The CPU time in worse case for n -stack operations is θ(mn) .

(b)

Program Plan Intro

Describethe number of diskaccesses and the CPU time required for n -PUSH operations in worse-case.

(b)

Expert Solution

Explanation of Solution

The new page starts only when everym^t^hpushes therefore the number of disk operations is n/m. The diskaccess requires time of asymptotically θ(n) .

Hence,the number of disk accesses and CPU time for n - PUSH operations in worse-case is θ(n) .

(c)

Program Plan Intro

Describe the disk accesses and the CPU time required for n -stack operations in worse-case.

(c)

Expert Solution

Explanation of Solution

Stack operations include PUSH and POP. The PUSH and POP operations requires alternate disk accesses that means mdisk accesses for performing stack operation so the number of disk accesses asymptotically is θ(m) .

The CPU time for performing stack operations is equal to θ(mn) ,where m is number of words andn is number of stack operations.

(d)

Program Plan Intro

Describe that the amortized number of disk accesses is O (1 /m ) and the amortized CPU time is O (1) for any stack operation .

(d)

Expert Solution

Explanation of Solution

At the initial function called as Potential Function of stack defined by difference of size of stack and recent pass which is multiple of m. The value of potential function is 0 at initial state.

When any stack operation is performed the value is decreased and increased according to the operation by one.Loading new pages requires the position which is at leastm position away from recent position to cross the page boundary therefore loading and storing of new stack page requires CPU time of O ( m ).

By dropping appropriate Potential function of order O ( m ) and selecting the suitable value for potential value results to managed the stack pages so that the amortized number of disk accesses for any stack operation is O (1 /m ) and the amortized CPU time for any stack operation is O (1).

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Answer 2

Question

Chapter 18, Problem 1P

(a)

Program Plan Intro

Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

(a)

Expert Solution

Explanation of Solution

In worse case for every stack operation it requires disk access for implementation sothe number of disk accesses for n stack operations using this simple implementation is asymptotically equals to θ(m) .

The CPU time in worse case for n -stack operations is θ(mn) .

(b)

Program Plan Intro

Describethe number of diskaccesses and the CPU time required for n -PUSH operations in worse-case.

(b)

Expert Solution

Explanation of Solution

The new page starts only when everym^t^hpushes therefore the number of disk operations is n/m. The diskaccess requires time of asymptotically θ(n) .

Hence,the number of disk accesses and CPU time for n - PUSH operations in worse-case is θ(n) .

(c)

Program Plan Intro

Describe the disk accesses and the CPU time required for n -stack operations in worse-case.

(c)

Expert Solution

Explanation of Solution

Stack operations include PUSH and POP. The PUSH and POP operations requires alternate disk accesses that means mdisk accesses for performing stack operation so the number of disk accesses asymptotically is θ(m) .

The CPU time for performing stack operations is equal to θ(mn) ,where m is number of words andn is number of stack operations.

(d)

Program Plan Intro

Describe that the amortized number of disk accesses is O (1 /m ) and the amortized CPU time is O (1) for any stack operation .

(d)

Expert Solution

Explanation of Solution

At the initial function called as Potential Function of stack defined by difference of size of stack and recent pass which is multiple of m. The value of potential function is 0 at initial state.

When any stack operation is performed the value is decreased and increased according to the operation by one.Loading new pages requires the position which is at leastm position away from recent position to cross the page boundary therefore loading and storing of new stack page requires CPU time of O ( m ).

By dropping appropriate Potential function of order O ( m ) and selecting the suitable value for potential value results to managed the stack pages so that the amortized number of disk accesses for any stack operation is O (1 /m ) and the amortized CPU time for any stack operation is O (1).

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Answer 3

Question

Chapter 18, Problem 1P

(a)

Program Plan Intro

Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

(a)

Expert Solution

Explanation of Solution

In worse case for every stack operation it requires disk access for implementation sothe number of disk accesses for n stack operations using this simple implementation is asymptotically equals to θ(m) .

The CPU time in worse case for n -stack operations is θ(mn) .

(b)

Program Plan Intro

Describethe number of diskaccesses and the CPU time required for n -PUSH operations in worse-case.

(b)

Expert Solution

Explanation of Solution

The new page starts only when everym^t^hpushes therefore the number of disk operations is n/m. The diskaccess requires time of asymptotically θ(n) .

Hence,the number of disk accesses and CPU time for n - PUSH operations in worse-case is θ(n) .

(c)

Program Plan Intro

Describe the disk accesses and the CPU time required for n -stack operations in worse-case.

(c)

Expert Solution

Explanation of Solution

Stack operations include PUSH and POP. The PUSH and POP operations requires alternate disk accesses that means mdisk accesses for performing stack operation so the number of disk accesses asymptotically is θ(m) .

The CPU time for performing stack operations is equal to θ(mn) ,where m is number of words andn is number of stack operations.

(d)

Program Plan Intro

Describe that the amortized number of disk accesses is O (1 /m ) and the amortized CPU time is O (1) for any stack operation .

(d)

Expert Solution

Explanation of Solution

At the initial function called as Potential Function of stack defined by difference of size of stack and recent pass which is multiple of m. The value of potential function is 0 at initial state.

When any stack operation is performed the value is decreased and increased according to the operation by one.Loading new pages requires the position which is at leastm position away from recent position to cross the page boundary therefore loading and storing of new stack page requires CPU time of O ( m ).

By dropping appropriate Potential function of order O ( m ) and selecting the suitable value for potential value results to managed the stack pages so that the amortized number of disk accesses for any stack operation is O (1 /m ) and the amortized CPU time for any stack operation is O (1).

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Answer 4

Question

Chapter 18, Problem 1P

(a)

Program Plan Intro

Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

(a)

Expert Solution

Explanation of Solution

In worse case for every stack operation it requires disk access for implementation sothe number of disk accesses for n stack operations using this simple implementation is asymptotically equals to θ(m) .

The CPU time in worse case for n -stack operations is θ(mn) .

(b)

Program Plan Intro

Describethe number of diskaccesses and the CPU time required for n -PUSH operations in worse-case.

(b)

Expert Solution

Explanation of Solution

The new page starts only when everym^t^hpushes therefore the number of disk operations is n/m. The diskaccess requires time of asymptotically θ(n) .

Hence,the number of disk accesses and CPU time for n - PUSH operations in worse-case is θ(n) .

(c)

Program Plan Intro

Describe the disk accesses and the CPU time required for n -stack operations in worse-case.

(c)

Expert Solution

Explanation of Solution

Stack operations include PUSH and POP. The PUSH and POP operations requires alternate disk accesses that means mdisk accesses for performing stack operation so the number of disk accesses asymptotically is θ(m) .

The CPU time for performing stack operations is equal to θ(mn) ,where m is number of words andn is number of stack operations.

(d)

Program Plan Intro

Describe that the amortized number of disk accesses is O (1 /m ) and the amortized CPU time is O (1) for any stack operation .

(d)

Expert Solution

Explanation of Solution

At the initial function called as Potential Function of stack defined by difference of size of stack and recent pass which is multiple of m. The value of potential function is 0 at initial state.

When any stack operation is performed the value is decreased and increased according to the operation by one.Loading new pages requires the position which is at leastm position away from recent position to cross the page boundary therefore loading and storing of new stack page requires CPU time of O ( m ).

By dropping appropriate Potential function of order O ( m ) and selecting the suitable value for potential value results to managed the stack pages so that the amortized number of disk accesses for any stack operation is O (1 /m ) and the amortized CPU time for any stack operation is O (1).

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Answer 5

Chapter 18, Problem 1P

(a)

Program Plan Intro

Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

(a)

Expert Solution

Explanation of Solution

In worse case for every stack operation it requires disk access for implementation sothe number of disk accesses for n stack operations using this simple implementation is asymptotically equals to θ(m) .

The CPU time in worse case for n -stack operations is θ(mn) .

(b)

Program Plan Intro

Describethe number of diskaccesses and the CPU time required for n -PUSH operations in worse-case.

(b)

Expert Solution

Explanation of Solution

The new page starts only when everym^t^hpushes therefore the number of disk operations is n/m. The diskaccess requires time of asymptotically θ(n) .

Hence,the number of disk accesses and CPU time for n - PUSH operations in worse-case is θ(n) .

(c)

Program Plan Intro

Describe the disk accesses and the CPU time required for n -stack operations in worse-case.

(c)

Expert Solution

Explanation of Solution

Stack operations include PUSH and POP. The PUSH and POP operations requires alternate disk accesses that means mdisk accesses for performing stack operation so the number of disk accesses asymptotically is θ(m) .

The CPU time for performing stack operations is equal to θ(mn) ,where m is number of words andn is number of stack operations.

(d)

Program Plan Intro

Describe that the amortized number of disk accesses is O (1 /m ) and the amortized CPU time is O (1) for any stack operation .

(d)

Expert Solution

Explanation of Solution

At the initial function called as Potential Function of stack defined by difference of size of stack and recent pass which is multiple of m. The value of potential function is 0 at initial state.

When any stack operation is performed the value is decreased and increased according to the operation by one.Loading new pages requires the position which is at leastm position away from recent position to cross the page boundary therefore loading and storing of new stack page requires CPU time of O ( m ).

By dropping appropriate Potential function of order O ( m ) and selecting the suitable value for potential value results to managed the stack pages so that the amortized number of disk accesses for any stack operation is O (1 /m ) and the amortized CPU time for any stack operation is O (1).

Answer 6

Chapter 18, Problem 1P

(a)

Program Plan Intro

Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

(a)

Expert Solution

Explanation of Solution

In worse case for every stack operation it requires disk access for implementation sothe number of disk accesses for n stack operations using this simple implementation is asymptotically equals to θ(m) .

The CPU time in worse case for n -stack operations is θ(mn) .

Answer 7

Chapter 18, Problem 1P

(a)

Program Plan Intro

Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

Answer 8

(a)

Expert Solution

Explanation of Solution

In worse case for every stack operation it requires disk access for implementation sothe number of disk accesses for n stack operations using this simple implementation is asymptotically equals to θ(m) .

The CPU time in worse case for n -stack operations is θ(mn) .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Program Plan Intro

Describethe number of diskaccesses and the CPU time required for n -PUSH operations in worse-case.

(b)

Expert Solution

Explanation of Solution

The new page starts only when everym^t^hpushes therefore the number of disk operations is n/m. The diskaccess requires time of asymptotically θ(n) .

Hence,the number of disk accesses and CPU time for n - PUSH operations in worse-case is θ(n) .

Answer 13

(b)

Program Plan Intro

Describethe number of diskaccesses and the CPU time required for n -PUSH operations in worse-case.

Answer 14

(b)

Expert Solution

Explanation of Solution

The new page starts only when everym^t^hpushes therefore the number of disk operations is n/m. The diskaccess requires time of asymptotically θ(n) .

Hence,the number of disk accesses and CPU time for n - PUSH operations in worse-case is θ(n) .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Program Plan Intro

Describe the disk accesses and the CPU time required for n -stack operations in worse-case.

(c)

Expert Solution

Explanation of Solution

Stack operations include PUSH and POP. The PUSH and POP operations requires alternate disk accesses that means mdisk accesses for performing stack operation so the number of disk accesses asymptotically is θ(m) .

The CPU time for performing stack operations is equal to θ(mn) ,where m is number of words andn is number of stack operations.

Answer 19

(c)

Program Plan Intro

Describe the disk accesses and the CPU time required for n -stack operations in worse-case.

Answer 20

(c)

Expert Solution

Explanation of Solution

Stack operations include PUSH and POP. The PUSH and POP operations requires alternate disk accesses that means mdisk accesses for performing stack operation so the number of disk accesses asymptotically is θ(m) .

The CPU time for performing stack operations is equal to θ(mn) ,where m is number of words andn is number of stack operations.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Program Plan Intro

Describe that the amortized number of disk accesses is O (1 /m ) and the amortized CPU time is O (1) for any stack operation .

(d)

Expert Solution

Explanation of Solution

At the initial function called as Potential Function of stack defined by difference of size of stack and recent pass which is multiple of m. The value of potential function is 0 at initial state.

When any stack operation is performed the value is decreased and increased according to the operation by one.Loading new pages requires the position which is at leastm position away from recent position to cross the page boundary therefore loading and storing of new stack page requires CPU time of O ( m ).

By dropping appropriate Potential function of order O ( m ) and selecting the suitable value for potential value results to managed the stack pages so that the amortized number of disk accesses for any stack operation is O (1 /m ) and the amortized CPU time for any stack operation is O (1).

Answer 25

(d)

Program Plan Intro

Describe that the amortized number of disk accesses is O (1 /m ) and the amortized CPU time is O (1) for any stack operation .

Answer 26

(d)

Expert Solution

Explanation of Solution

At the initial function called as Potential Function of stack defined by difference of size of stack and recent pass which is multiple of m. The value of potential function is 0 at initial state.

When any stack operation is performed the value is decreased and increased according to the operation by one.Loading new pages requires the position which is at leastm position away from recent position to cross the page boundary therefore loading and storing of new stack page requires CPU time of O ( m ).

By dropping appropriate Potential function of order O ( m ) and selecting the suitable value for potential value results to managed the stack pages so that the amortized number of disk accesses for any stack operation is O (1 /m ) and the amortized CPU time for any stack operation is O (1).

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Ch. 18.1 - Prob. 1E Ch. 18.1 - Prob. 2E Ch. 18.1 - Prob. 3E Ch. 18.1 - Prob. 4E Ch. 18.1 - Prob. 5E Ch. 18.2 - Prob. 1E Ch. 18.2 - Prob. 2E Ch. 18.2 - Prob. 3E Ch. 18.2 - Prob. 4E Ch. 18.2 - Prob. 5E

Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

Describe thenumber of disk accesses and CPU time for n stack operations in worse-case by using the simple implementation.

Explanation of Solution

Describethe number of diskaccesses and the CPU time required for n -PUSH operations in worse-case.

Explanation of Solution

Describe the disk accesses and the CPU time required for n -stack operations in worse-case.

Explanation of Solution

Describe that the amortized number of disk accesses is O (1 /m ) and the amortized CPU time is O (1) for any stack operation .

Explanation of Solution

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Chapter 18 Solutions