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GATE | GATE CS 2021 | Set 1 | Question 63
  • Last Updated : 22 Mar, 2021
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A five-stage pipeline has stage delays of 150,120,150,160 and 140 nanoseconds. The registers that are used between the pipeline stages have a delay of 5 nanoseconds each.
The total time to execute 100 independent instructions on this pipeline, assuming there are no pipeline stalls, is _______ nanoseconds.
(A) 17160
(B) 16640
(C) 17640
(D) 15000


Answer: (A)

Explanation: Given : k=5, n=100

Total time=(k+n-1)*tp ,
where k=number of pipeline stages,
n=number of instructions,
tp=pipeline cycle time.

tp=max(stage delays) + register delay
tp=max(150,120,150,160,140)+5ns
tp=160+5=165ns

Total time =(5+100-1)*165=104*165=17160 ns.

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