Given Information : The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Question

Want to see the full answer?

Answer 1

Question

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Consider the data path below for a single cycle 32-bits MIPS processor Assume that we are executing the following instruction lw St1, 48($10) What is the value of the element pointed by arrow number 1 by in hexadecimal? Note that the PC and the content of registers $10 and $t1 are found in bottom left of the figure below Data Memory Select one a. Ox00001724 Ob. 0x00002DE4 Oc 0x000016F4 Od 0x00001720 Oe. 0x00002DEO Clear my choice Rea " Address Content 0x000016EC 0x0000ABOD 0x000016F0 0x00A01245 0x000016F4 0x00001A42 0x000171C 0x0008124F 0x0001720 0x00021345 0x0001724 0x000067AB ALU Ox0001734 0x0000AB35 0x0001738 0x0000FA72 0x0000ABOC ALS 0x000174C Register File and PC $to = 0x000016FO $t1 = 0x000016F4 PC = 0x0000148A0 (Before executing LW)

Consider the data path below for a single cycle 32-bits MIPS processor Assume that we are executing the following instruction SLL S10, $3,4 What is the value of the element pointed by arrow number 1 by in hexadecimal or binary? Note that the PC and the content of registers are found in bottom left of the figure below Address Select one Oa. Ox00000000 Ob. Ob 10011' Oc 0x0000AF00 ® d. 06/01000 Oe. None of the options Of Ox000000BA LEY Content 0x000016EC Dx0000ABOD Data Memory 0x000016F0 0x00A01245 0x000016F4 0x00001A42 Ox000171C 0x0008124 0x0001720 0x00021345 0x0001724 0x000067AB 0x0001734 0x0001738 0x0000A835 0x0000FA72 0x000174C 0x0000AC $63 0x000000BA PC = 0x000FAC04 Register File and PC (Before executing SLL)

Consider the data path below for a single cycle 32-bits MIPS processor Assume that we are executing the following instruction lw $10, 48(S1) What is the value of the element pointed by arrow number 1 by in hexadecimal? Note that the PC and the content of registers $10 and $t1 are found in bottom left of the figure below Select one a. None of the options Ob 0x00014C64 Oc 0x00014CC4 Od 0x00014BA4 Oe. Ox14C1C Of 0x00014C60 Og 0x00014CCO D Data Memory Address Content 0x000016EC 0x0000ABOD 0x000016F0 0x00A01245 0x000016F4 0x00001A42 0x000171C 0x0008124F 0x0001720 0x00021345 0x0001724 0x000067AB 0x0001734 0x0000AB35 0x0001738 0x0000FA72 0x000174C 0x0000ABOC ALU Register File and PC Sto Ox000016F0 St1 = 0x000016F4 PC = 0x00014BA0 (Before executing LW).

Answer 2

Question

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Answer 6

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

Answer 7

Chapter 6.5, Problem 6.20PP

A)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

Answer 8

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

Answer 9

B)

Program Plan Intro

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

Answer 10

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

Answer 11

C)

Explanation of Solution

Miss rate:

The cache can only hold half of the elements in the array, so that means that a read to grid[8][0] will evict the block that was loaded when we read grid[0][0]. Since this block also contained grid[0][1], the first read of grid[0][1] will be a miss.

Hence, each iteration will have one hit and one miss.

This means one will have 128 hits and 128 misses.

Hence,

miss rate = ( number of miss/total reads

Answer 12

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Answer 13

D)

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 17

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

Given Information : The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Solution Summary: The author assumes that sizeof(int)=4 Grid begins at memory address 0. The cache is empty and only memory accesses are to the array grid entries.

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

Explanation of Solution

Explanation of Solution

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...

Want to see the full answer?

Chapter 6 Solutions

Given Information : The definition for the code is mentioned below: //Traverse through the grid for (i =0; i &lt; 16; i++) { for (j = 0; j &lt; 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Solution Summary: The author assumes that sizeof(int)=4 Grid begins at memory address 0. The cache is empty and only memory accesses are to the array grid entries.

Given Information: The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Given Information: The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Explanation of Solution

Explanation of Solution

New Miss Rate: If the cache were twice as big them also no change will be made to the mi...

Want to see the full answer?

Chapter 6 Solutions

Given Information : The definition for the code is mentioned below: //Traverse through the grid for (i =0; i < 16; i++) { for (j = 0; j < 16; j++) { //add values of x into grid total_x += grid[i][j].x; //add values of y into grid total_y += grid[i][j].y; } }

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

Given Information:

The definition for the code is mentioned below:

//Traverse through the grid

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

{

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

{

//add values of x into grid

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

//add values of y into grid

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

}

}

New Miss Rate:

If the cache were twice as big them also no change will be made to the mi...