Question

A non-pipeline system 50ns to process a task. The same task can be processed in a 6 segment pipeline with a clock cycle of 10ns. Determine the speed of ratio of pipeline system for 100 tasks . What is the maximum speed up that can be achieved?

Answer 1

Speedup = time required without pipeline/ with pipeline

=100*50ns / (6+99)10ns

= 4.76

Maximum speedup = k = number of stages = 6 here

Answer 2

Speed up ratio (S)​ :
It is defined as the speedup of a pipeline processing with respect to the equivalent non-pipeline processing.
S =nt n/(k+n−1)t p
Number of tasks n = 100
For Non-pipeline:
Time taken by non-pipeline to process a task t n = 50ns
Total time taken by non-pipeline to process 100 task = n t n
= 100 × 50
= 5000ns
For Pipeline:
Number of segment pipeline k = 6
Time period of 1 clock cycle t p = 10ns
Total time required to complete n tasks in k segment pipeline with tp clock cycle time:
= ( k + n − 1 )t p
= ( 6 + 100 − 1 )10
= 1050ns
Speed up Ratio:
When total time taken by the pipeline to process 100 tasks is divided by the total time required to complete n tasks in k segment pipeline with t p clock cycle time then speed up ratio is obtained.
S =5000/1050
= 4 .76