Consider a 5-segment pipeline with a clock cycle time 20 ns in each sub…
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?
Answer: C. 4.81 — Solution: (iii) is correct Explanation: cycle time (in pipeline) = register delay +max stage delay =0+20 =20ns In the given question no register delay are…
- A.
5
- B.
4.03
- C.
4.81
- D.
4.17
Attempted by 4 students.
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Correct answer: C
Solution: (iii) is correct
Explanation: cycle time (in pipeline) = register delay +max stage delay =0+20 =20ns
In the given question no register delay are given so consider it 0.
According to given data, speedup is
= (Time taken by non-pipeline) / (Time taken by pipeline)
= (5*100*20) / {(5+100- 1)*20}
= 4.81
Option (iii) is correct.
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