Question

Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600 MHz. 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

Answer 1

Answer:

Check the explanation

Explanation:

CPI means Clock cycle per Instruction

given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M

Execution time is given by following Formula.

Execution Time(CPU time) = CPI*Instruction Count * clock time = $\frac{CPI*Instruction Count}{ClockRate}$

a)

for system A CPU time is 1.3 * 100, 000 600 106

= 216.67 micro sec.

b)

for system B CPU time is $=\frac{2.5*100,000}{750*10^6}$

= 333.33 micro sec

c) Since the system B is slower than system A, So the system A executes the given program in less time

Hence take CPU execution time of system B as CPU time of System A.

therefore

216.67 micro = =$\frac{2.5*Instruction}{750*10^6}$

Instructions = 216.67*750/2.5

= 65001

hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A