Question

Consider two different implementations, M1 and M2, of the same instruction set. There are three classes of instructions (A, B, and C) in the instruction set. M1 has a clock rate of 80 MHz and M2 has a clock rate of 100 MHz. The average number of cycles for each instruction class and their frequencies (for a typical program) are as follows:

(a) Calculate the average CPI for each machine, M1, and M2.
(b) Calculate the average MIPS ratings for each machine, M1 and M2.
(c) Which machine has a smaller MIPS rating? Which individual instruction class CPI do you need to change, and by how much, to have this machine have the same or better performance as the machine with the higher MIPS rating (you can only change the CPI for one of the instruction classes on the slower machine)?

Instruction Class DC Machine M1 Cycles/Instruction Class Machine M2 Cycles/Instruction Class Frequency 60% 30% 10%

Answer 1

cpi means clocks per instruction

average cpi can be obtained by multiplying the cpi with frequency with their respective machines

1. Avg CPI for M1 is defined as below:

average clocks per Instruction for M1 = (60/100)* 1 + (30/100)*2 + (10/100)*4

= 1.6

Avg CPI for M2 is defined as below:

average clocks per Instruction for M2 = (60/100)*2 + (30/100)*3 + (10/100)*4

= 2.5

b). The average MIPS rate is calculated as: Clock Rate/ averageCPI x

The average MIPS ratings for M1 = 80 / 1.6 x 10^6

                                                       = 50 x 10^6

The average MIPS ratings for M2 = 100 / 2.5 x 10^6

                                                       = 40 x 10^6

c)

   Machine M2 has a smaller MIPS rating

Changing instruction set A from 2 to 1

The CPI will be increased to 1.9 (1*.6+3*.3+4*.1)

and hence MIPS Rating will now be (100/1.9)*10^6 =52.6*10^6.