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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]
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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