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