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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.

Sagot :

Answer:

The answer is "1.25"

Explanation:

[tex]{CPU \ time}= {instructions \times CPI \times cycle\ time}[/tex]

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

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

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

For compiler B, we have

[tex]CPI_{B}= \frac{CPU \ times_{B}} {instructions_{B} \times cycle \ time}[/tex]

         [tex]= \frac{1.5\ s}{ 1.2 \times 10^{9} \times 10^{-9} \ s}\\\\= \frac{1.5}{ 1.2 }\\\\= 1.25[/tex]

abiolataiwo2015

The average CPI for each program given that the processor has a clock cycle time of 1 ns is : 1.1, 1.25.

Average CPI for each program

CPI (Complier A)=CPU time/Instruction×Cycle time

Where:

Cycle time=1ns=10^-9s

Hence:

CPI (Complier A)=1.1s/10^9×10^-9s

CPI (Complier A)=1.1

CPI (Complier B)=CPU time/Instruction×Cycle time

CPI (Complier B)=1.5s/1.2×10^9×10^-9s

CPI (Complier B)=1.25

Inconclusion the average CPI for each program given that the processor has a clock cycle time of 1 ns is : 1.1, 1.25.

Learn more about average CPI here:https://brainly.com/question/24723238

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