Consider a 5-segment pipeline with a clock cycle time 20 ns in each sub operation. Find out the approximate speed-up ratio between pipelined and non-pipelined system to execute 100 instructions. (If an average, every five cycles, a bubble due to data hazard has to be introduced in the pipeline.).
(A) 5
(B) 4.03
(C) 4.81
(D) 4.17
Answer: (B)
Explanation: According to given data, speedup is
= (Time taken by non-pipeline) / (Time taken by pipeline)
= (5*100*20) / {(100+20+5 - 1)*20}
= 4.03
Option (B) is correct.
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Last Updated :
10 May, 2020
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