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Consider a 4 stage pipeline that consists of IF,ID,EX and WB stages.The time taken by these stages are 50ns,60ns,110ns and 80ns respectively.The pipeline registers are required after every pipeline stage and each of these pipeline registers consumes 10ns delay.What is the speedup of the pipeline under ideal conditions compare to the corresponding non-pipelined implementation?

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3 votes
$Time_{pipe}$ = max(stage delay) + Register delay

= 110 + 10 = 120 ns

$Time_{Non-pipe}$ = sum of all stage delay (Note that register delay is not considered in Non-Pipeline implementation)

= 60 + 50 + 110 + 80 = 300 ns

Speedup = $\frac{Time_{Non-pipe}}{Time_{pipe}}$

= $\frac{300}{120}$

= 2.5
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  • Register delay is used in pipeline . 

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