Question

A nonpipelined system takes 200ns to process a task. The same task can be processed in a 5-segment pipeline with a clock cycle of 40ns. Determine the speedup ratio of the pipeline for 200 tasks. What is the maximum speedup that could be achieved with the pipeline unit over the nonpipelined unit?

Answer 1

Given:

Time taken by non pipelined system = 200 ns = 200$\times 10^{-9}$ s

Clock cycle for 5- segment pipeline = 40 ns = 40$\times 10^{-9}$ s

Pipeline speed up ratio = 200

Formula used:

$(Speed-up)_{max} = \frac{System to process a task\times Speed-up ratio}{(speed-up ratio -1)clock cycle + segment}$

Solution:

Using formula:

$(Speed-up)_{max} = \frac{5\times 40\times 200}{((200 - 1)40 + 5}$

$(Speed-up)_{max}$ = 5