Question

Assume that we have an application with a total of 500,000 instructions where 20% of them are the load/store instructions with an average CPI of 6 cycles, and the rest instructions are ALU instructions with average CPI of 1 cycle. If we double the clock rate without optimizing the memory latency, the average CPI for load/store instruction will also be doubled to 12 cycles. What is the speedup after this change?

Answer 1

1.25

Step-by-step explanation:

#instructions =
`5*10^(5)`

Average CPI (old) = 0.2*6 + 0.8*1 = 1.2+0.8 = 2.0

Average CPI (new) = 0.2*12 + 0.8*1 = 2.4 + 0.8 = 3.2

Assuming Clock Rate = x

Speedup = Execution Time (old) / Execution time (new) = (2.0*5*
`10^(5)` /x) / (3.2*5*
`10^(5)` /2x) = 4/3.2 = 1.25