7/24/2019 CSE530_1

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Q1.

Sol.

Let IC of Computer B = x

IC of Computer A =1.5*x

Average CPI of Computer A= 1.8

Average CPI of Computer B =4

Clock frequency of Computer B= 3GHz

To be as fast as Computer B,

CPU time of Computer A = CPU time of Computer B

CPU Time = IC * CPI * Clock time

1.5*x*1.8*(1/freq)=x*4*(1/3GHz)

Frequency of Computer A= 2.025GHz

Q2.

Instructions

25% loads

10% stores

35% adds (adds, compares, and, sub, etc.)

5% multiplies

1% divides

20% conditional branches (assume perfect prediction)

4% unconditional branches (jumps, calls, returns, etc.)

Latencies

loads: 8 cycles

stores: 6 cycles

adds: 4 cycles

multiplies: 10 cycles

divides: 40 cycles

cond branches: 4 cycles

uncond branches: 2 cycles

I would pick up LoadInstruction to make twice as fast as it the first major contributor to the CPU time . By making

load cycle to 0.25, processor is 50.42% faster.