Consider two different implementations P1 and P2, of the sameinstruction set. Th
ID: 3609660 • Letter: C
Question
Consider two different implementations P1 and P2, of the sameinstruction set. There are five classes of instructions (A,B,C,D,E)in the instruction set.P1's clock rate is 4 Ghz. P2 has a clock rate of 6 GHz. Theaverage number of cycles for ezch instruction class for P1 and P2is as follows:
Class | CPI on P1 | CPI on P2 A | 1 | 2 B | 2 | 2 C | 3 | 2 D | 4 | 4 E | 3 | 4
Assume that peak performansce is defined as the fastest ratethat a computer can execute any inctruction sequence. What are thepeak performances of P1 and P2 expressed in instructions persecond?
Thanks.
P1's clock rate is 4 Ghz. P2 has a clock rate of 6 GHz. Theaverage number of cycles for ezch instruction class for P1 and P2is as follows:
Class | CPI on P1 | CPI on P2 A | 1 | 2 B | 2 | 2 C | 3 | 2 D | 4 | 4 E | 3 | 4
Assume that peak performansce is defined as the fastest ratethat a computer can execute any inctruction sequence. What are thepeak performances of P1 and P2 expressed in instructions persecond?
Thanks.
Explanation / Answer
Dear, The ideal instruction sequencefor P1 is one composed entirely of instructionsFrom class A (which have CPI of 1). So M1's peak performance is(4x109cycles/second)/(1 cycle/instruction) = 4000 MIPS.
Similarly, the ideal sequence for M2 contains only instructionsfrom A, B, and C (which all have a CPI of 2). So M2'speak performance is (6×109cycles/second)/(2cycles/instruction) = 3000 MIPS.
The average CPI of P1 is (1x2 + 2 + 3 + 4 + 3)/6 = 7/3.
The average CPI ofP2 is (2x2 + 2 + 2 + 4 + 4)/6 = 8/3.
P2 is((6x109cycles/second)/(8/3cycles/instruction))/((4x109cycles/second)/(7/3cycles/instruction)) = 21/16 times faster than P1. I hope this will helpfulfor you................... I hope this will helpfulfor you................... I hope this will helpfulfor you...................