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Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some program and computer B has a clock cycle time of 250 ps and an effective CPI of 1.7 for the same program. Which computer is faster and by how much?

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Answer:

Computer A is 1.41 times faster than the Computer B

Step-by-step explanation:

Assume that number of instruction in the program is 1

Clock time of computer A is
CT_(A) =200 ps

Clock time of computer B is
CT_(B) =250 ps

Effective CPI of computer A is
CPI_(A) =1.5

Effective CPI of computer B is
CPI_(B) =1.7

CPU time of A is


CPU_(time)=instructions * CPA_(A) * CT_(A)\\CPU_(time)=1 * 1.5 * 200=300 sec

CPU time of B is


CPU_(time)=instructions * CPA_(B) * CT_(B)\\CPU_(time)=1 * 1.7 * 250=425 sec

Hence Computer A is Faster by
(425)/(300) =1.41

Computer A is 1.41 times faster than the Computer B

User Charlie Skilbeck
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