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andriy [413]
3 years ago
5

Compilers can have a profound impact on the performance of an application. Assume that for a program, compiler A results in a dy

namic 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.
Computers and Technology
1 answer:
Naddik [55]3 years ago
3 0

Answer:

The answer is "1.25"

Explanation:

{CPU \ time}= {instructions \times  CPI \times  cycle\  time}

\therefore\\\CPI= \frac{CPU \ time}{instructions \times cycle \ time} \\\\cycle \ time = 1 \\\\ns = 10^{-9} s \\

Also for this context, it executes the time =  CPU time. So, the compiler A, we has

CPI_{A}= \frac{CPU \ time_{A}} {instructions_{A} \times cycle \ time}= \frac{1.1 s}{10^{9} \times 10^{-9} s}= 1.1

For compiler B, we have

CPI_{B}= \frac{CPU \ times_{B}} {instructions_{B} \times cycle \ time}

         = \frac{1.5\ s}{ 1.2 \times 10^{9} \times 10^{-9} \ s}\\\\= \frac{1.5}{ 1.2 }\\\\= 1.25

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<em />

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// In this exercise, you will be given a variable, it will be called: value
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Answer:

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public class Solution {

   public static void main(String args[]) {      

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Explanation:

The first line import the Scanner which is use to enable to accept user input.

The class is defined as Solution in the next line. The main method is declared in the next line which signify the beginning of the program.

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