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 60MHz and M2 has a clock rate of 80MHz. The average number of cycles of each instruction class and their frequencies (for a typical program) are as follows: Instruction Class M1-cycles M2-cycles/ Frequency
/instruction class instruction class
A 1 2 50%
B 2 3 20%
C 3 4 30%
a. Calculate the average CPI for each machine, M1 and M2.
b. Calculate the CPU execution time of M1 and M2.

Explanation:

A.)

we have two machines M1 and M2

cpi stands for clocks per instruction.

to get cpi for machine 1:

= we multiply frequencies with their corresponding M1 cycles and add everything up

50/100 x 1 = 0.5

20/100 x 2 = 0.4

30/100 x 3 = 0.9

CPI for M1 = 0.5 + 0.4 + 0.9 = 1.8

We find CPI for machine 2

we use the same formula we used for 1 above

50/100 x 2 = 1

20/100 x 3 = 0.6

30/100 x 4 = 1.2

CPI for m2 =  1 + 0.6 + 1.2 = 2.8

B.)

CPU execution time for m1 and m2

this is calculated by using the formula;

I * CPI/clock cycle time

execution time for A:

= I * 1.8/60X10⁶

= I x 30 nsec

execution time b:

I x 2.8/80x10⁶

= I x 35 nsec

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1. a

2. a

3. d

explanation: