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Unit 2 · Computer System Architecture Paper 2 CS · December 2019

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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Correct answer: D. 2.91

Explanation

100 × 30 ns = 3000 ns non-pipelined versus (4 + 99) × 10 = 1030 ns pipelined. Speed-up ≈ 2.91.

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