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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See solution

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4.1.1 Complete solution and answer only no need explanation Given three data points (1,6), (3,28), and (10, 231), What is the value of y at x = 2 if the function y = 2x2 + 3x + 1 passes through the three data points.

Problem 10

Problem 1. Two DNA strands of length 100 (number of bases in the sequence) are given in DNA1.txt and DNA2.txt file. Find the longest common subsequence (LCS) between the two sequences using dynamic programming bottom-up (tabulation) approach. Submit your code to solve the problem. How much time (in seconds or milliseconds) is required by your computer to run the algorithm? Solve the same LCS problem using dynamic programming top-down (memorization) approach. Submit your code to solve the problem. How much time (in seconds or milliseconds) is required by your computer to run the algorithm. Which solution is faster and why? Discuss your solutions.

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%