Given Information : The given code is: // define structure square point_color { // variable declaration int c; int m; int y; int k; }; // declare structure array struct point_color square[16][16]; int i,j; // traverse through the array for(i=0;i<16;i++) { //traverse through elements for(j=0;j<16;j++) { //square all elements of the 2-D array square[i][j].c=0; square[i][j].m=0; square[i][j].y=1; square[i][j].k=0; } } Write hit: If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty. If the state of information is in valid state then it executes a write-through operation. It then updates the memory and block and changes its blocked state to reserved state. Write miss : A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Question

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Answer 1

Question

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

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Students have asked these similar questions

3M decides to make Post-its by printing yellow squares on whitepieces of paper. As part of the printing process, they need to set theCMYK (cyan, magenta, yellow, black) value for every point in thesquare. 3M hires you to determine the efficiency of the followingalgorithms on a machine with a 2,048-byte direct-mapped data cachewith 32-byte blocks. You are given the following definitions:1 struct point_color {2 int c;3 int m;4 int y;5 int k;6 };78 struct point_color square[16][16];9 int i, j;Assume the following:sizeof(int) = 4.square begins at memory address 0.The cache is initially empty.The only memory accesses are to the entries of the array square.Variables i and j are stored in registers.Determine the cache performance of the following code:1 for (i = 0; i < 16; i++){2 for (j = 0; j < 16; j++) {3 square[i][j].c = 0;4 square[i][j].m = 0;5 square[i][j].y = 1;6 square[i][j].k = 0;7 }8 }A. What is the total number of writes?B. What is the total number of writes that miss in the…

We have a fully associative cache memory with 216entries). Each entry is hosting 32 data. How manybits are necessary for the TAG for a 28 bits address?5 A12 B23 C None of previous answers D , also is there a difference if it was fully associative mapping cache memory or no? solve this question in great detail explaining every single bit of info as i'm a complete beginner and i need to understand how to solve questions of this type. also provide small examples of how the solution would differ if the question was slightly altered.

Answer 2

Question

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Answer 6

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 7

Chapter 6, Problem 6.38HW

A)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 8

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 9

B)

Program Plan Intro

Given Information:

The given code is:

// define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Write hit:

If the information in the cache is reserved or in dirty state then the cache line is updated in its place without updating memory set from its state to dirty.

If the state of information is in valid state then it executes a write-through operation.

It then updates the memory and block and changes its blocked state to reserved state.

Write miss:

A partial cache line write is handed as a read miss followed by a write hit. All the other caches are left in the invalid state and the reserved state is occupied by the current state.

Answer 10

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Answer 11

C)

Program Plan Intro

Given Information:

The given code is:

//define structure

square point_color

{

// variable declaration

int c;

int m;

int y;

int k;

};

// declare structure array

struct point_color square[16][16];

int i,j;

// traverse through the array

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

{

//traverse through elements

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

{

//square all elements of the 2-D array

square[i][j].c=0;

square[i][j].m=0;

square[i][j].y=1;

square[i][j].k=0;

}

}

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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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

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Chapter 6 Solutions