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

Want to see the full answer?

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%

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

I need help in construct a matlab code to find the voltage, the currents, and the watts based on that circuit.

Objective Implement Bottom-Up Iterative MergeSort and analyze its efficiency compared to recursive MergeSort. Unlike the recursive approach, which involves multiple function calls and stack overhead, the bottom-up version sorts iteratively by merging small subarrays first, reducing recursion depth and improving performance. Task 1. Implement Bottom-Up Iterative MergeSort о Start with single-element subarrays and iteratively merge them into larger sorted sections. Use a loop-based merging process instead of recursion. ○ Implement an efficient in-place merging strategy if possible. 2. Performance Analysis Compare execution time with recursive MergeSort on random, nearly sorted, and reversed datasets. ○ Measure and plot time complexity vs. input size. O Submission Explain why the iterative version reduces function call overhead and when it performs better. • Code implementation with comments. • A short report (1-2 pages) comparing performance. • Graph of execution time vs. input size for…

Given a shared data set, we allow multiple readers to read at the same time, and only one single writer can access the shared data at the same time. In the lecture slides, a solution is given. However, the problem is that the write cannot write forever, if there are always at least one reader. How to ensure that the writer can eventually write? Propose your solution by using semaphores and implemented in Python from threading import Thread, Semaphore from time import sleep from sys import stdout class Reader(Thread): def__init__(self, name): self.n=name; Thread.__init__(self) defrun(self): globalnr, nw, dr, dw whileTrue: # ⟨await nw == 0 then nr += 1⟩ e.acquire() ifnw>0: #if nw > 0 or dw > 0 : dr+=1; e.release(); r.acquire() nr+=1 ifdr>0: dr-=1; r.release() else: e.release() # read data stdout.write(self.n+' reading\n') sleep(1) # ⟨nr -= 1⟩ e.acquire() nr-=1 ifnr==0anddw>0: dw-=1 ; w.release() else: e.release() class Writer(Thread): def__init__(self, name):…

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%