Given scenario: In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

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Answer 1

Question

Chapter 5, Problem 5.1.1E

Program Plan Intro

Given scenario:

In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

Expert Solution & Answer

Explanation of Solution

Number of 32-bit integers that can be stored in a 16-byte cache block:

Formula to find number of 32-bit integers is as follows:

(cache block × number of loop execution)/32 bit integers(16×8)/32128/324

Hence, the number of 32-bit integers can stored in 16-byte cache block is “4”.

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Solve the following questions by WRITING YOUR INSTRUCTION CODE in Assembly language. 1. Modify the Lab 4 Tutorial 1 to allow USERS TO ENTER the following parameter and displaythefinal result. You are also required to store the results in the memory named ‘output’. Findthesum of the last four numbers in the array. For example, assume the value of n is 10, the sumwillbe 10+9+8+7= 34 (in decimal). Your console should display the following:Please enter the value of n: 10The sum of the last four numbers is 34Thank you. Before computing the difference of the numbers in the array, your code must performachecking; if n < 3, you should display a message ‘The value of n is less than 4’, and no operationwill be performed. In this case, your console should display the following: Please enter the value of n: 1 The entered n is less than 3. Thank you Type the…

Answer 2

Question

Chapter 5, Problem 5.1.1E

Program Plan Intro

Given scenario:

In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

Expert Solution & Answer

Explanation of Solution

Number of 32-bit integers that can be stored in a 16-byte cache block:

Formula to find number of 32-bit integers is as follows:

(cache block × number of loop execution)/32 bit integers(16×8)/32128/324

Hence, the number of 32-bit integers can stored in 16-byte cache block is “4”.

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Answer 3

Question

Chapter 5, Problem 5.1.1E

Program Plan Intro

Given scenario:

In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

Expert Solution & Answer

Explanation of Solution

Number of 32-bit integers that can be stored in a 16-byte cache block:

Formula to find number of 32-bit integers is as follows:

(cache block × number of loop execution)/32 bit integers(16×8)/32128/324

Hence, the number of 32-bit integers can stored in 16-byte cache block is “4”.

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Answer 4

Question

Chapter 5, Problem 5.1.1E

Program Plan Intro

Given scenario:

In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

Expert Solution & Answer

Explanation of Solution

Number of 32-bit integers that can be stored in a 16-byte cache block:

Formula to find number of 32-bit integers is as follows:

(cache block × number of loop execution)/32 bit integers(16×8)/32128/324

Hence, the number of 32-bit integers can stored in 16-byte cache block is “4”.

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Answer 5

Chapter 5, Problem 5.1.1E

Program Plan Intro

Given scenario:

In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

Expert Solution & Answer

Explanation of Solution

Number of 32-bit integers that can be stored in a 16-byte cache block:

Formula to find number of 32-bit integers is as follows:

(cache block × number of loop execution)/32 bit integers(16×8)/32128/324

Hence, the number of 32-bit integers can stored in 16-byte cache block is “4”.

Answer 6

Chapter 5, Problem 5.1.1E

Program Plan Intro

Given scenario:

In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

Expert Solution & Answer

Explanation of Solution

Number of 32-bit integers that can be stored in a 16-byte cache block:

Formula to find number of 32-bit integers is as follows:

(cache block × number of loop execution)/32 bit integers(16×8)/32128/324

Hence, the number of 32-bit integers can stored in 16-byte cache block is “4”.

Answer 7

Chapter 5, Problem 5.1.1E

Program Plan Intro

Given scenario:

In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

Answer 8

Expert Solution & Answer

Explanation of Solution

Number of 32-bit integers that can be stored in a 16-byte cache block:

Formula to find number of 32-bit integers is as follows:

(cache block × number of loop execution)/32 bit integers(16×8)/32128/324

Hence, the number of 32-bit integers can stored in 16-byte cache block is “4”.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. 5 - Prob. 5.1.1E Ch. 5 - Prob. 5.1.2E Ch. 5 - Prob. 5.1.3E Ch. 5 - Prob. 5.1.4E Ch. 5 - Prob. 5.1.5E Ch. 5 - Prob. 5.1.6E Ch. 5 - Prob. 5.2.1E Ch. 5 - Prob. 5.2.2E Ch. 5 - Prob. 5.2.3E Ch. 5 - Prob. 5.2.4E

Given scenario: In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

Given scenario: In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.

Explanation of Solution

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Chapter 5 Solutions

Given scenario:

In given C code, the Outer “for” loop executes eight times and inner “for” loop executes 8000 times for matrix addition.