Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600MHz. Computer B has a CPI of 2.5 and can be run at a clock rate of 750 Mhz. We have a particular program we wish to run. When compiled for computer A, this program has exactly 100,000 instructions. How many instructions would the program need to have when compiled for Computer B, in order for the two computers to have exactly the same execution time for this program?

Question

Quizplus · Accepted Answer

The Answer is (CPUTime)A = (Instruction count)A * (CPI)A * (Clock cycle Time)A
= (100,000)*(1.3)/(600*106) ns
(CPUTime)B = (Instruction count)B * (CPI)B * (Clock cycle Time)B
= (I) *(2.5)/(750*106) ns

(CPUTime)A = (Instruction count)A * (CPI)A * (Clock cycle Time)A = (100,000)*(1.3)/(600*106) ns (CPUTime)B = (Instruction count)B * (CPI)B * (Clock cycle Time)B = (I) *(2.5)/(750*106) ns Since (CPUTime)A = (CPUTime)B, we have to solve for (I)B and get 65000

Since (CPUTime)A = (CPUTime)B, we have to solve for (I)B and get 65000

Computer a Has an Overall CPI of 1