Question

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

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

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

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