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A non-pipeline system takes $50$ns to process a task. The same task can be processed in six-segment pipeline with a clockcycle of $10$ns. Determine approximately the speedup ratio of the pipeline for $500$ tasks.

  1. $6$
  2. $4.95$
  3. $5.7$
  4. $5.5$
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time for non pipeline system  1 task =50 ns  500 tasks =500x50 =25000 ns

time with k(=6) segment pipeline      first task 6x10 ns  subsequent task =10 ns  so 499x10 =4990

              total =60+4990=5050

Speedup ratio = time taken without pipeline / time with pipeline

                      =25000/5050 =4.95

hence option 2) is the right ans
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For a non-pipelined system: 

  • Total number of instruction/task $(n)=500$
  • Total time required to perform a single task in pipelined processor $(T_{np})=50$ ns

For a pipelined system:

  • Total number of stages $(k)=6$
  • Total number of instruction/task $(n)=500$
  • Total time required to perform a single task in pipelined processor $(T_p)=10$ 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{500*50}{(6+(500-1))*10}\right] $ ns

$\implies S_k=\left[\frac{25000}{(6+499)*10}\right]$ ns

$\implies S_k=\left[\frac{25000}{5050}\right]$ ns

$\implies S_k= 4.95$ ns

Option $(B)$ is correct.

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