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

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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

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Check out a sample textbook solution

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Problem 1.8 The following code segment, consisting of six instructions, needs to be executed 64 times for the evaluation of vector arithmetic expression: D(I) = A(I) + B(I) xC(I) for 0 ≤ I ≤ 63. Load R1, B(I) /R1 - Memory (a + I)/ Load R2, C(I) Multiply R1, R2 Load R3, A(I) Add R3, R1 Store D(I), R3 t /R2 Memory (8 + 1)/ /R1 - (R1) × (R2)/ /R3 - Memory (7 + I)/ - /R3 (R3) + (R1)/ /Memory (0 + I) ← (R3)/ where R1, R2, and R3 are CPU registers, (R1) is the content of R1, a, ß,7, and are the starting memory addresses of arrays B(1), C(I), A(I), and D(I), respectively. Assume four clock cycles for each Load or Store, two cycles for the Add, and eight cycles for the Multiply on either a uniprocessor or a single PE in an SIMD machine. (a) Calculate the total ber of CPU cycles needed to execute the above code seg- ment repeatedly 64 times on an SISD uniprocessor computer sequentially, ignoring all other time delays. (b) Consider the use of an SIMD computer with 64 PEs to execute the above…

Problem 1.8 The following code segment, consisting of six instructions, needs to be executed 64 times for the evaluation of vector arithmetic expression: D(I) = A(I) + B(I) xC(I) for 0 ≤ I≤ 63. Load R1, B(I) /R1 - Memory (a + I)/ Load R2, C(I) Multiply R1, R2 Load R3, A(I) Add R3, R1 Store D(I), R3 t /R2 Memory (8 + 1)/ /R1 - (R1) × (R2)/ /R3 - Memory (7 + I)/ - /R3 (R3) + (R1)/ /Memory (0 + I) ← (R3)/ where R1, R2, and R3 are CPU registers, (R1) is the content of R1, a, ß,7, and are the starting memory addresses of arrays B(1), C(I), A(I), and D(I), respectively. Assume four clock cycles for each Load or Store, two cycles for the Add, and eight cycles for the Multiply on either a uniprocessor or a single PE in an SIMD machine. (a) Calculate the total ber of CPU cycles needed to execute the above code seg- ment repeatedly 64 times on an SISD uniprocessor computer sequentially, ignoring all other time delays. (b) Consider the use of an SIMD computer with 64 PEs to execute the above…

(d) Given memory holes (i.e., unused memory blocks) of 100K, 500K, 200K, 300K and 600K (in address order) as shown below, how would each of the first-fit, next-fit, best- fit algorithms allocate memory requests of 120K, 320K, 280K, 90K and 210K (in this order). The shaded areas are used/allocated regions that are not available. 100k 500k 200k 300k 600k Figure 2: Current status of main memory

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

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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...