Question

I need help with the following Computer Architecture 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 90 MHz and M2 has a clock rate of 80 MHz. The average number of cycles for each instruction class and their frequencies (for a typical program) are as follows:

Instruction Class	Machine M1 – Cycles/Instruction Class	Machine M2 – Cycles/Instruction Class	Frequency
A	2	1	60%
B	3	2	30%
C	1	3	10%

a) Calculate the average CPI for each machine, M1 and M2.

b) Which machine has the better performance? Show your reasoning.

// SIMILAR PROBLEM BELOW:

Two compilers are used for a 1GHz machine. There are three classes of instructions:   Class A, Class B, and Class C, which requires 2, 4 and 5 cycles (respectively). Both compilers are used to produce code.

Compiler 1 generates code with 1 million Class A instructions, 3 million Class B instructions, and 2 million Class C instructions.

Compiler 2 generates code with 5 million Class A instructions, 1 million Class B instructions, and 1 million Class C instructions.

Which sequence will be faster (higher performance)? Explain.

Solution:

CLOCK CYCLES compiler1:

CC1 = (1,000,000 * 2) + (3,000,000 * 4) + (2,000,000 * 5)

CC1 = 24,000,000 clock cycles

CLOCK CYCLES compiler2:

CC2 = (5,000,000 * 2) + (1,000,000 * 4) + (1,000,000 + 5)

CC2 = 19,000,000

Compiler 2 will clearly be faster. Compiler 1 requires 24e6 instructions while compiler 2 only requires 19e6 instructions to execute the same set of instructions.

Execution Time 1 = 24,000,000 / 1 GHz

                  = 24 x 10^-3 seconds

Execution Time 2 = 19,000,000 / 1GHz

                  = 19 x 10^-3 seconds

Compiler 2 utilizes class A, which has the highest CPI of the three classes, to the do bulk of the computational work. The result is an efficiency advantage versus compiler1.

Answer 1

a)

Average CPI for machine M1 =

Average CPI for machine M2 = 1* 60/100 +2*30/100+3*10/10 = 4.2

b)

Average time per instruction for M1= 3.1/(1000000*90) seconds=3.4444e-08 seconds

Average time per instruction for M2= 4.2/(1000000*80) seconds=5.2500e-08 seconds

So the M1 is better