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Consider an instruction pipeline with four stages (S1, S2, S3 and S4) each with combinational circuit only. The pipeline registers are required between each stage and at the end of the last stage. Delays for the stages and for the pipeline registers are as given in the figure.
What is the approximate speed up of the pipeline in steady state under ideal conditions when compared to the corresponding non-pipeline implementation?
  • a)
    4.0
  • b)
    2.5
  • c)
    1.1
  • d)
    3.0
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Consider an instruction pipeline with four stages (S1, S2, S3 and S4) ...
Concept:
The segments are separated by registers Ri that holds the intermediate results between the stages.
Data:
Stage delay and corresponding register delay given
S1 = 5 ,
S2 = 6 ,
S3 = 11,
S4 = 8,
And corresponding register delay is 1 for each stage
Number of stage = 4
Time is taken to execute N instructions in non-pipelined implementation will be = (5 + 6 + 11 + 8)N = 30 × N
Clock period for pipelined implementation = max(5, 6, 11, 8) + 1 = 12 ns
Time is taken to execute N instructions in pipelined implementation will be = (4 + N-1)12 ≈ 12×N (N is very large)
Speedup = 30 N/12N = 2.5
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Consider an instruction pipeline with four stages (S1, S2, S3 and S4) each with combinational circuit only. The pipeline registers are required between each stage and at the end of the last stage. Delays for the stages and for the pipeline registers are as given in the figure.What is the approximate speed up of the pipeline in steady state under ideal conditions when compared to the corresponding non-pipeline implementation?a)4.0b)2.5c)1.1d)3.0Correct answer is option 'B'. Can you explain this answer?
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