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GATE | GATE-CS-2014-(Set-1) | Question 65

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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 : 30 Sep, 2021
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