2 votes 2 votes 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? CO and Architecture co-and-architecture + – rahuldb asked Nov 17, 2016 retagged Nov 13, 2017 by Arjun rahuldb 15.2k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 4 votes 4 votes Speedup = time required without pipeline/ with pipeline =100*50ns / (6+99)10ns = 4.76 Maximum speedup = k = number of stages = 6 here papesh answered Nov 17, 2016 selected Nov 17, 2016 by rahuldb papesh comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes 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 topper98 answered Mar 23, 2020 topper98 comment Share Follow See all 0 reply Please log in or register to add a comment.