Consider a 6-stage instruction pipeline, where all stages are perfectly balanced.Assume that there is no cycle-time overhead of pipelining. When an application is executing on this 6-stage pipeline, the speedup achieved with respect to non-pipelined execution if 25% of the instructions incur 2 pipeline stall cycles is
(A) 4
(B) 8
(C) 6
(D) 7


Answer: (A)

Explanation:

It was a numerical digit type question so answer must be 4.

As for 6 stages, non-pipelining takes 6 cycles.

There were 2 stall cycles for pipelining for 25% of the instructions

So pipe line time = (1+(25/100)*2) = 1.5

Speed up = Non pipeline time/Pipeline time = 6/1.5 = 4


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  • Last Updated : 19 Nov, 2018

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