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pav-90 [236]
3 years ago
13

The utilization of a subset of the performance equation as a performance metric is a pitfall. To illustrate this, assume the fol

lowing two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requi res the execution of 1.0E9 instructions. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2.
Computers and Technology
1 answer:
Troyanec [42]3 years ago
5 0

Answer:

Given Data:

Clock rate of P1 = 4 GHz

Clock rate of P2 = 3 GHz

Average CPI of P1 = 0.9

Number of Instructions = 5.0E9 =  5 × 10^9

Clock rate of P2 = 3 GHz

Average CPI of P2 = 0.75

Number of Instructions = 1.0E9 = 10^9

To find: If the computer with largest clock rate has the largest performance?

Explanation:

Solution:

As given in the question, clock rate of P1 = 4 GHz which is greater than clock rate of P2 = 3 GHz

According to the performance equation:

CPU Time = instruction count * average cycles per instruction/ clock rate

CPU Time = I * CPI / clock rate

Where instruction count refers to the number of instructions.

Performance of P1:

CPU Time (P1) = 5 * 10^9 * 0.9 / (4 * 10^9 )

                        = 5000000000 * 0.9 / 4000000000

                        = 4500000000 / 4000000000

                        =  1.125s

Performance of P2:

CPU Time (P2) = 10^9 * 0.75/ (3 * 10^9 )

                        = 750000000 / 3000000000

                        = 0.25s

So the Performance of P2 is larger than that of P1,

                          performance (P2) > performance (P1)

                                         0.25 is better than 1.125

But clock rate of P1 was larger than P2

                             clock rate of P1 > clock rate of P2

                                                  4 GHz > 3 GHz

So this is a misconception about P1 and P2.

It is not true that computer with the largest clock rate as having the largest performance.

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In Java.Use a single for loop to output odd numbers, even numbers, and an arithmetic function of an odd and even number. Use the
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Answer:

<em>This program is written using java programming language</em>

<em>Difficult lines are explained using comments (See Attachment for Source file)</em>

<em>Program starts here</em>

import java.util.*;

import java.lang.Math;

public class oddeven{

public static void main(String [] args)

{

 double even,odd;//Declare variables even and odd as double

 //The next iteration prints odd numbers

 for(int i = 1;i<=173;i+=2)

 {

  odd = i;

  System.out.format("%.4f",odd);//Round to 4 decimal places and print

System.out.print("\t");

 }

 System.out.print('\n');//Start printing on a new line

 //The next iteration prints even numbers

 for(int i = 0;i<=173;i+=2)

 {

  even=i;

  System.out.format("%.4f",even);//Round to 4 decimal places and print

System.out.print("\t");

 }

 System.out.print('\n');//Start printing on a new line

 double  ssqrt;//Declare ssqrt to calculate the square root of sum of even and odd numbers

 for(int i = 0;i<=173;i+=2)

 {

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System.out.print("\t");

 }

}

}

Explanation:

Libraries are imported into the program

import java.util.*;

import java.lang.Math;

The following line declares variables even and odd as double

double even,odd;

The following iteration is used to print odd numbers  with in the range of 0 to 173

for(int i = 1;i<=173;i+=2)  {

odd = i;

System.out.format("%.4f",odd); This particular line rounds up each odd numbers to 4 decimal places before printing them

System.out.print("\t"); This line prints a tab

}

The following code is used to start printing on a new line

System.out.print('\n');

The following iteration is used to print even numbers  with in the range of 0 to 173

for(int i = 0;i<=173;i+=2)

{

even=i;

System.out.format("%.4f",even); This particular line rounds up each even numbers to 4 decimal places before printing them

System.out.print("\t"); This line prints a tab

}

The following code is used to start printing on a new line

System.out.print('\n');

The next statement declares ssqrt as double to calculate the square root of the sum of even and odd numbers

double  ssqrt;

for(int i = 0;i<=173;i+=2)

{

ssqrt = Math.sqrt(2*i+1);This line calculates the square root of sum of even and odd numbers

System.out.format("%.4f",ssqrt); This particular line rounds up each even numbers to 4 decimal places before printing them

System.out.print("\t"); This line prints a tab

}

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