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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?

Respuesta :

Debel

Answer:

1.25

Explanation:

#instructions = [tex]5*10^{5}[/tex]  

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*[tex]10^{5}[/tex] /x) / (3.2*5*[tex]10^{5}[/tex] /2x)  = 4/3.2 = 1.25

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