A non-pipelined system takes 30 ns to process a task. The same task can be processed in a four-segment pipeline with a clock cycle of 10 ns. Determine the speed up of the pipeline for 100 tasks.

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Question

This question was previously asked in

UGC NET Computer Science (Paper 2) Dec 2019 Paper

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Answer 1

Concept:

\({\rm{Speed\;up}} = \frac{{{\rm{n}}\; \times \;{{\rm{t}}_{\rm{n}}}}}{{\left( {{\rm{n}}\; +\; {\rm{k}}\; - \;1} \right){{\rm{t}}_{\rm{p}}}}}\)

Given:

For a non-pipelines system,

time to process a task, t_n = 30 ns

For a pipelined system,

number of segments, k = 4

clock cycle of each segment, t_p = 10 ns

For n = 100 tasks,

Calculation:

\({\rm{Speed\;up}} = \frac{{100\; \times \;30}}{{\left( {100\; + \;4\; - \;1} \right)10}}\)

∴ speed up = 2.91