Compilers can have a profound impact on the performance of an application. Assume that for a program, compiler A results in a dynamic instruction count of 1.0E9 and has an execution time of 1.1 s, while compiler B results in a dynamic instruction count of 1.2E9 and an execution time of 1.5 s. a. Find the average CPI for each program given that the processor has a clock cycle time of 1 ns.

Answer

1.

CPI for Compiler A = 1.1

CPI for Compiler B = 1.25

2. The clock rate of A is 27% slower than clock rate of B

3. C is 1.67 times faster than A and 2.27 times faster than B

Explanation:

1.

Given

Cycle Time = 1ns = 10^-9

CPU Time for A = 1.1s

Instruction count = 10^9

CPU Time for B = 1.5s

Instruction Count = 1.2*10^9

CPU Time = Instructions * Cycle Time * CPI

So, CPI = CPU Time / (Instructions * Cycle Time)

CPI for A = 1.1/(10^9 * 10^-9)

CPI = 1.1

CPI for B = 1.5/(1.2 * 10^9 * 10^-9)

CPI = 1.5/12

CPI = 1.25

2.

Given

Execution Time on A = Execution Time on B

CPU Time for A = 1.1s

Instruction count = 10^9

CPI = 1.1

CPU Time for B = 1.5s

Instruction Count = 1.2*10^9

CPI = 1.25

Execution Time = Instructions * CPI/ Clock Rate

Execution Time A = Execution Time B

Instruction A * CPI A / Clock Rate A = Instruction B * CPI B/ Clock Rate B. Make Clock Rate A the subject of formula

Clock Rate A = (Instruction A * CPI A * Clock Rate B) / (Instruction B * CPI B)

Clock Rate A = (10^9 * 1.1 * Clock Rate B) / (1.2 * 10^9 * 1.25)

Clock Rate A = 0.73 Clock Rate B

So the clock rate of A is 27% slower than clock rate of B

3.

For compiler A and C,

CPU Time = Instructions * Cycle Time * CPI

CPU Time of C = 6 * 10^8 * 1.1 * 10^-9 = 0.66s

Using ratio of CPU time and Compiler performance

i,e.

CPU Time of A * Performance of A = CPU Time of C * Performance of C

Given

CPU Time of A = 1.1

CPU Time of C = 0.66

1.1 * Performance of A = 0.66 * Performance of C

Performance of C: Performance of A = 1.1/0.66

Performance of C : Performance of A = 1.67

So, C is 1.67 times faster than A

Also,

CPU Time of B = 1.5

CPU Time of C = 0.66

1.5 * Performance of B = 0.66 * Performance of C

Performance of C: Performance of B = 1.5/0.66

Performance of C : Performance of B = 2.27

So, C is 2.27 times faster than B

Answer

1.

CPI for Compiler A = 1.1

CPI for Compiler B = 1.25

2. The clock rate of A is 27% slower than clock rate of B

3. C is 1.67 times faster than A and 2.27 times faster than B

Explanation:

1.

Given

Cycle Time = 1ns = 10^-9

CPU Time for A = 1.1s

Instruction count = 10^9

CPU Time for B = 1.5s

Instruction Count = 1.2*10^9

CPU Time = Instructions * Cycle Time * CPI

So, CPI = CPU Time / (Instructions * Cycle Time)

CPI for A = 1.1/(10^9 * 10^-9)

CPI = 1.1

CPI for B = 1.5/(1.2 * 10^9 * 10^-9)

CPI = 1.5/12

CPI = 1.25

2.

Given

Execution Time on A = Execution Time on B

CPU Time for A = 1.1s

Instruction count = 10^9

CPI = 1.1

CPU Time for B = 1.5s

Instruction Count = 1.2*10^9

CPI = 1.25

Execution Time = Instructions * CPI/ Clock Rate

Execution Time A = Execution Time B

Instruction A * CPI A / Clock Rate A = Instruction B * CPI B/ Clock Rate B. Make Clock Rate A the subject of formula

Clock Rate A = (Instruction A * CPI A * Clock Rate B) / (Instruction B * CPI B)

Clock Rate A = (10^9 * 1.1 * Clock Rate B) / (1.2 * 10^9 * 1.25)

Clock Rate A = 0.73 Clock Rate B

So the clock rate of A is 27% slower than clock rate of B

3.

For compiler A and C,

CPU Time = Instructions * Cycle Time * CPI

CPU Time of C = 6 * 10^8 * 1.1 * 10^-9 = 0.66s

Using ratio of CPU time and Compiler performance

i,e.

CPU Time of A * Performance of A = CPU Time of C * Performance of C

Given

CPU Time of A = 1.1

CPU Time of C = 0.66

1.1 * Performance of A = 0.66 * Performance of C

Performance of C: Performance of A = 1.1/0.66

Performance of C : Performance of A = 1.67

So, C is 1.67 times faster than A

Also,

CPU Time of B = 1.5

CPU Time of C = 0.66

1.5 * Performance of B = 0.66 * Performance of C

Performance of C: Performance of B = 1.5/0.66

Performance of C : Performance of B = 2.27

So, C is 2.27 times faster than B

1.1

1.25

Clock rate A = 0.733 clock rate B

Explanation:

Given the following :

Clock cycle time = 1ns = 10^-9 s

COMPILER A:

Instruction count = 1.0E9

Execution time = 1.1 s

COMPILER B:

Instruction count = 1.2E9

Execution time = 1.5 s

Using the relation :

CPI = [Execution time / (instruction count × cycle time)]

CPI for compiler A :

[1.1 / (10^9 × 10^-9)] = 1.1 / 10^0 = 1.1 / 1 = 1.1

CPI for compiler B:

[1.5 / (1.2 ×10^9 * 10^-9)]

(1.5 / 1.2 * 1) = 1.5 / 1.2 = 1.25

b. Assume the compiled programs run on two different processors. If the execution times on the two processors are the same, how much faster

Since, execution time is the same on both processors

Execution time = [(Instruction count × CPI) / clock rate] ;

[(Instruction count × CPI) / clock rate] A = [(Instruction count × CPI) / clock rate] B

Make clock rate of A the subject of the formula :

(Instruction count A * CPI A * clock rate B) / (instruction count B * CPI B)

= (10^9 * 1.1 * clock rate B / 1.2 × 10^9 * 1.25)

1.1 × 10^9 clock rate B / 1.5 × 10^9

Clock rate A = 0.733 clock rate B

Incomplete question as we have not told what to find.So I have assumed to find average CPI for each program.So complete question is here

Compilers can have a profound impact on the performance of an application. Assume that for a program, compiler A results in a dynamic instruction count of 1.0E9 and has an execution time of 1.1 s, while compiler B results in a dynamic instruction count of 1.2E9 and an execution time of 1.5 s.

Find the average CPI for each program given that the processor has a clock cycle time of 1 ns.

Step-by-step explanation:

As we know CPU time given as

For Compiler A

For Compiler B

Incomplete question as we have not told what to find.So I have assumed to find average CPI for each program.So complete question is here

Compilers can have a profound impact on the performance of an application. Assume that for a program, compiler A results in a dynamic instruction count of 1.0E9 and has an execution time of 1.1 s, while compiler B results in a dynamic instruction count of 1.2E9 and an execution time of 1.5 s.

Find the average CPI for each program given that the processor has a clock cycle time of 1 ns.

Step-by-step explanation:

As we know CPU time given as

For Compiler A

For Compiler B

The solution that would help the client pool their data together is a data integration solution. This type of solution allows for the collection and consolidation of data from various sources into a single, unified database. This can be achieved through a variety of methods, including Extract, Transform, Load (ETL) processes, data warehousing, and master data management. By implementing a data integration solution, the client will be able to access and analyze their data more efficiently, leading to better insights and informed decision-making.

The correct answer to the following question will be "Auto-encoder".

Explanation:

It is indeed a form of artificial neural net that utilizes in an unmonitored way to practice successful information coding.

boolean rsvp;

int selection;

String option1;

String option2;

(a) if(rsvp){

System.out.println("Attending");

} else {

System.out.println("Not attending");

}

(b) if (selection == 1){

System.out.println("Beef");

} else if (selection == 2){

System.out.println("Chicken");

} else if (selection == 3) {

System.out.println("Pasta");

} else {

System.out.println("Fish");

}

(c) switch(selection){

case 1:

if(rsvp){

option1 = "Thanks for attending. You will be served beef.";

} else{

option1 = "Sorry you can't make it";

}

break;

case 2:

if(rsvp){

option1 = "Thanks for attending. You will be served chicken.";

} else{

option1 = "Sorry you can't make it";

}

break;

case 3:

if(rsvp){

option1 = "Thanks for attending. You will be served pasta.";

} else{

option1 = "Sorry you can't make it";

}

break;

default:

if(rsvp){

option1 = "Thanks for attending. You will be served fish.";

} else{

option1 = "Sorry you can't make it";

}

break;

}

(d) if (option1 == option2){

System.out.println("True");

} else {

System.out.println("False");

}

Explanation:

This code snippet is written in Java.

a) Uses if statement to print "attending" if rsvp is true and "not attending" if rsvp is false.

b) Uses if-else statement to print the variety of food based on user selection. If selection is 1, it print beef. If selection is 2, it print chicken. If selection is 3, it prints pasta else it print fish for any other value of selection.

c) Uses switch statement to determine the value of selection and then uses inner if statement to assign "Thanks for attending. You will be served beef." if rsvp is true to option1. The value of beef changes depending on the value of selection. If rsvp is false, then "Sorry you can't make it." is assigned to option1.

d) Uses if statement to print true if option1 and option2 are equal else it print false.