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A non-pipeline system takes $25$ ns to process a task. The same task can be processed in a six-segment pipeline in a clock cycle of $10$ ns.

The speed-up rotation of the pipeline for $10$ tasks will be _______.
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speed up = ntn/ (k+n-1)tp

where n--no. of inst

tn-time taken in non pipelined system

k-- no.of stages in pipeline

tp -- time of one stage(1 clock)

On substitution .. ans is 1.67

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Yes answer is 1.67, and i updated it from 16.66(old) to 1.67. thanks!!
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For 10 instructions, non-pipelined architecture = $250ns$

For 10 instructions, pipelined architecture = $60+9(10)=150ns$

Speedup = $\frac{250}{150}=1.66$
Answer:

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