Question

D. find the instructions per cycle (ipc) for each processor.

Answer 1

Find the IPC (instructions per cycle) for each processor.

CPI = CPU Clock Cycle / Instruction Count

IPC = 1/CPI

case a.

IPC for P1 = (1 / (7s x 3 x 109Hz / 20 x 109 ) )      =  0.952

IPC for P2 = (1 / (10s x 2.5 x 109Hz / 30 x 109 ) ) =  1.20

IPC for P3 = (1 / (9s x 4 x 109Hz / 90 x 109 ) )      =  2.50

 

case b.

IPC for P1 = (1 / (5s x 2 x 109Hz / 20 x 109 ) )    =  2

IPC for P2 = (1 / (8s x 3 x 109Hz / 30 x 109 ) )    =  1.25

IPC for P3 = (1 / (7s x 4 x 109Hz / 25 x 109 ) )    =  0.89

 

Find the clock rate for P2 that reduces its execution to that of P1.

 

Using P1 and P2 from block a.

CPU time = IC x CPI / clock rate

CPI of P2 = CPU time of P2 x clock rate of P2 / IC of P2

CPI of P2 = 10s x 2.5 x 109 / 30 x 109 = 0.833

 

Clock rate for P2 that reduces its execution to that of P1 is thus:

CPU time of P1 = IC of P2 x CPI of P2 / clock rate of P2

Clock rate of P2 = IC of P2 x CPI of P2 / CPU time of P1

Clock rate of P2 = 30 x 109 x 0.833 / 7s

Clock rate of P2 = 3.57GHz

 

Find the number of instructions for P2 that reduces its execution time to that of P3.

 

Using P2 and P3 from block a.

CPI of P2 = CPU time of P2 x clock rate of P2 / IC of P2

CPI of P2 = 10s x 2.5 x 109 / 30 x 109 = 0.833

 

Number of instructions for P2 that reduces its execution time to that of P3 thus:

IC (Number of instructions of P2) = CPU time of P3 x clock rate of P2 / CPI of P2

IC = 9s x 2.5 x 109 / 0.833

IC = 27 x 109