EBK COMPUTER SYSTEMS
EBK COMPUTER SYSTEMS
3rd Edition
ISBN: 8220101459107
Author: O'HALLARON
Publisher: YUZU
Question
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Chapter 6.5, Problem 6.18PP

A.

Program Plan Intro

Given Information:

The heart of the recent hit game SimAquarium is a tight loop that calculates the average position of 256 algae. One needs to determine the cache performance on a machine that has a 1024 byte direct mapped data cache with 16 byte blocks (B=16).

The definition for the game is mentioned below:

//Declare structure algae_position

struct algae_position{

int x;

int y;

};

//declare grid

struct algae_position grid[16][16];

//variable declaration

int total_x=0, total_y=0;

int i,j;

//traverse through grid to add x

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

{

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

{

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

}

}

//traverse through grid to add y

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

{

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

{

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

}

}

B.

Program Plan Intro

Given Information:

The heart of the recent hit game SimAquarium is a tight loop that calculates the average position of 256 algae. One needs to determine the cache performance on a machine that has a 1024 byte direct mapped data cache with 16 byte blocks (B=16).

The definition for the game is mentioned below:

//Declare structure algae_position

struct algae_position{

int x;

int y;

};

//declare grid

struct algae_position grid[16][16];

//variable declaration

int total_x=0, total_y=0;

int i,j;

//traverse through grid to add x

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

{

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

{

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

}

}

//traverse through grid to add y

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

{

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

{

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

}

}

C.

Explanation of Solution

Miss rate:

It is observed that for all 256 reads, 128 of them are misses. Similarly, for the second loops another 128 reads will be missed.

Hence, total number of misses= 128+128=256

Hence,

miss rate=(numberofmiss/totalreads)×100%

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