Pipeline and Vector Processing Q.16

16Consider 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 ......

Option : D
Explanation : 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+(2 25/100)] = 3/2 = 1.5
Speed up = Non– pipeline time/Pipeline time = 6/1.5

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