Question

A processor is designed such that the clock of the processor runs at 1 GHz. The following table gives the instruction frequencies for the benchmark and how many cycles each instruction takes.

Instruction Type	Frequency	Cycles
Load & Stores	25%	10 cycles
Arithmetic Instructions	65%	6 cycles
Branch instructions	10%	4 cycles

(a) Calculate the CPI for the above benchmark.

(b) Suppose the amount of registers are doubled, such that clock cycle time increases by 40%. What is the new clock speed (in GHz)?

(c) Assume only the load & stores instructions are speed up by 5 times and their frequency is increased to 50% (Arithmetic instructions drop to 40%), what is the overall speedup of the processor?

(d) Assume only the load & stores instructions are speed up by 5 times and their frequency is increased to 50% (Arithmetic instructions drop to 40%), what is the overall speedup of the processor using Amdahl's Law this time?

Answer 1

(a)

If we say that there are 100 instructions, then:

25 of them will be loads and stores.

65 of them will be arithmetic instructions.

10 of them will be branch instructions.

Clock cycles per 100 instructions are

(25 * 10) + (65 * 6) + (10 * 4) = 680 cycles/100 instructions

Therefore, there are 6.8 Cycles per instruction.

(b)

Clock time = 1/Cycle Time

Cycle Time = 1/Clock Time

Cycle Time = 1/(10⁹) = 10^-9

The cycle time is then increased by 40%:

(10^-9) * 1.4 = 1.4 * 10^-9

The new clock rate is thus:

1/(1.4 * 10^-9) = 0.71 GHz.

(c)

CPI old = 6.8

CPI new = ((.5 * 10)/5) + (.4 * 6) + (.1 * 4) = 3.8

SpeedUp = 6.8 / 3.8 = 1.7

(d)

1.66 1-5)+ SpeedUp-( F.E 1