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
`* 10^(-9)` s

Clock cycle for 5- segment pipeline = 40 ns = 40
`* 10^(-9)` s

Pipeline speed up ratio = 200

Formula used:

`(Speed-up)_(max) = (System to process a task* Speed-up ratio)/((speed-up ratio -1)clock cycle + segment)`

Solution:

Using formula:

`(Speed-up)_(max) = (5* 40* 200)/(((200 - 1)40 + 5)`

`(Speed-up)_(max)` = 5