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. which implementation (m1 or m2) is faster?
c. find the clock cycles required for both processors.

a) the average CPI for machine M1 = 1.6

the average CPI for machine M2 = 2.5

b) M1 implementation is faster.

c) the clock cycles required for both processors.52.6*10^6.

Explanation:

(a)

The average CPI for M1 = 0.6 x 1 + 0.3 x 2 + 0.1 x 4

= 1.6

The average CPI for M2 = 0.6 x 2 + 0.3 x 3 + 0.1 x 4

= 2.5

(b)

The average MIPS rate is calculated as: Clock Rate/ averageCPI x 10^6

Given 80MHz = 80 * 10^6

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

= 50

Given 100MHz = 100 * 10^6

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

= 40

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.