A. A non pipeline system takes 50 ns to process a task. The same task can be processed in a 6 segment pipeline with a clock cycle of 10 ns. Determine the speed up ratio of the pipeline for 100 tasks. What is the maximum speed up that can be achieved.
Answers
The speed up ratio of the pipeline for 100 tasks is 4.76
Explanation:
Total Number of tasks "n" = 100
Time taken by non pipeline "Tn" = 50 ns
Time period of 100 tasks = ntn
= 100 x 50 = 5000 ns
Number of segment pipeline "K" = 6
Time period of 1 clock cycle = 10 ns
Total time required = ( k + n - 1)tp
= ( 6 + 100 - 1)10
= 1050 ns
Speed up ratio " S" = 5000/1050
= 4.76
Total Number of tasks "n" = 100
Time taken by non pipeline "Tn" = 50 ns
Time period of 100 tasks = ntn
= 100 x 50 = 5000 ns
Number of segment pipeline "K" = 6
Time period of 1 clock cycle = 10 ns
Total time required = ( k + n - 1)tp
= ( 6 + 100 - 1)10
= 1050 ns
Speed up ratio " S" = 5000/1050
= 4.76