Aspire Faculty ID #18917 · Topic: UGC NET Computer Science Nov 2020 (Paper II) · Just now
UGC NET Computer Science Nov 2020 (Paper II)

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

Solution

Non-pipeline time for $500$ tasks:

$500\times50=25000\ ns$

For pipeline:

Number of stages $=6$

Clock cycle $=10ns$

Pipeline time:

$(k+n-1)\times clock$

$=(6+500-1)\times10$

$=505\times10=5050\ ns$

Speedup:

$\frac{25000}{5050}=4.95$

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