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Non pipelined system takes 130ns to process an instruction . A program of 1000 instructions is executed in non pipelined system. Then same program is processed with processor with 5 segment pipeline with clock cycle of 30 ns/stage.

Determine speed up ratio of pipeline.

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For a non-pipelined system: 

  • Total number of instruction/task $(n)=1000$
  • Total time required to perform a single task in pipelined processor $(T_{np})=130$ ns

For a pipelined system:

  • Total number of stages $(k)=5$
  • Total number of instruction/task $(n)=1000$
  • Total time required to perform a single task in pipelined processor $(T_p)=30$ ns

$\because \text{Speedup ($S_k$)=$\frac{ET_{np}}{ET_p}$}$

$\implies S_k= \frac{(n*T_{np})}{(k+(n-1))T_p}$ 

$\implies S_k=\left[\frac{1000*130}{(5+(1000-1))*30}\right] $ ns

$\implies S_k=\left[\frac{130000}{(5+999)*30}\right]$ ns

$\implies S_k=\left[\frac{130000}{30120}\right]$ ns

$\implies S_k= 4.316$ ns

speedup =$4.316$ ns is the answer.

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