Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

for (i =0; i < 16; i++)

{

for (j = 0; j < 16; j++)

{

//add values of x into grid

total_x += grid[i][j].x;

//add values of y into grid

total_y += grid[i][j].y;

}

}

C)

Explanation of Solution

Miss rate:

• The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0].  Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.
• Hence, each iteration will have one hit and one miss. 
• This means one will have 128 hits and 128 misses.

Hence,

$\begin{array}{l}\text{miss rate = (}\text{number}\text{of}\text{miss/total}\text{reads}\end{array}$

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...