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The speed up of a pipeline processing over an equivalent non-pipeline processing is defined by the ration:

where $n \rightarrow$ no. of tasks

$t_n \rightarrow$ time of completion of each tasks

$k \rightarrow$ no. of segments of pipeline

$t_p \rightarrow $ clock cycle time

$S \rightarrow$ speed up ratio

  1. $S=\frac{n t_n}{(k+n-1)t_p}$
  2. $S=\frac{n t_n}{(k+n+1)t_p}$
  3. $S=\frac{n t_n}{(k-n+1)t_p}$
  4. $S=\frac{(k+n-1)t_p}{n t_n}$
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Answer A:

Without pipeline one task needs tn time.

  • n tasks need ntn time.

With pipeline,

  • First task needs  k cycles to finish. So time will be k tp
  • Other n-1 tasks needs tp time only to finish.
  • Total time = (k+ n -1 ) tp

Speed up = T pipeline / T without pipeline = n t n / (k+n-1) t p

Answer:

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