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1. Consider three processors (P1, P2, P3), each with the same instruction set. P

ID: 3883780 • Letter: 1

Question

1. Consider three processors (P1, P2, P3), each with the same instruction set. P1 has a 2.0 GHz clock rate and a CPI of 1.5. P2 has a 2.5 GHz clock rate and a CPI of 2.0. P3 has a 4.0 GHz clock rate and a CPI of 2.5.

a) Compute the instructions per second for each processor.

b) For each processor executing a different program in 10 seconds, compute the number of cycles and the number of instructions that were performed.

c) For each processor executing the programs in part b, the CPI is increased by 20%. We now wish to reduce the execution time by 30%. What should the new clock rate be for each processor?

Explanation / Answer

ANSWER

A )

Given,

P1 has 2.0 GHz clock rate and CPI of 1.5

P2 has 2.5 GHz clock rate and CPI of 2.0

P1 has 4.0 GHz clock rate and CPI of 2.5

  performance of P1, instructions per second = 2.0 * (109/1.5) = 1.33* 109

  performance of P2, instructions per second = 2.5 * (109/2.0) = 1.25* 109

  performance of P3, instructions per second = 4.0 * (109/2.5) = 1.6 * 109

B )

   Given

each proccer executing a different program in 10 secs,

cycles;

number of cycles for P1 = 10* 2.0* 109 = 20* 109

number of cycles for P2 =  10* 2.5* 109 =25* 109

number of cycles for P3 =  10* 4.0* 109 =40* 109

instructions;

number of instructions for P1 = 20* (109/1.5) = 13.33* 109

number of instructions for P2 =  25* (109/2.0)  = 12.5* 109

number of instructions for P3 =  40* (109/2.5)  = 16* 109

C )

  Reducing execution time by 30% means that new execution time is 7 seconds.

  CPI (new) = CPI(old) * 1.2 .. So,

CPI (P1) = 1.5 * 1.2 =1.8,

CPI (P2) = 2.0 * 1.2 = 2.4,

CPI (P3) = 2.5 * 1.2 = 3

f = No.Instructions * CPI time , where

number of instructions for P1 =  20* (109/1.5) = 13.33* 109

number of instructions for P2 =  25* (109/2.0)  = 12.5* 109

number of instructions for P3 =  40* (109/2.5)  = 16* 109

then ,

  f(P1) = 13.33 * 109 * (1.8/7) = 3.427 GHz

f(P2) = 12.5 * 109 * (2.4/7 )=4.285 GHz

f(P3) = 16* 109 * (3/7) =6.857 GHz