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; } }

Question

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Answer 1

Question

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=( number of miss/total reads )×100%

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Check out a sample textbook solution

See solution

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using r language

Answer 2

Question

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=( number of miss/total reads )×100%

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

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=( number of miss/total reads )×100%

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

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=( number of miss/total reads )×100%

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

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=( number of miss/total reads )×100%

Answer 6

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;

}

}

Answer 7

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;

}

}

Answer 8

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;

}

}

Answer 9

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;

}

}

Answer 10

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=( number of miss/total reads )×100%

Answer 11

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=( number of miss/total reads )×100%

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 15

Ch. 6.1 - Prob. 6.1PP Ch. 6.1 - Prob. 6.2PP Ch. 6.1 - Prob. 6.3PP Ch. 6.1 - Prob. 6.4PP Ch. 6.1 - Prob. 6.5PP Ch. 6.1 - Prob. 6.6PP Ch. 6.2 - Prob. 6.7PP Ch. 6.2 - Prob. 6.8PP Ch. 6.4 - Prob. 6.9PP Ch. 6.4 - Prob. 6.10PP

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=( number of miss/total reads )×100%

Want to see the full answer?

Chapter 6 Solutions

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=( number of miss/total reads )×100%