0 votes 0 votes A non-pipeline system takes $25$ ns to process a task. The same task can be processed in a six-segment pipeline in a clock cycle of $10$ ns. The speed-up rotation of the pipeline for $10$ tasks will be _______. CO and Architecture tbb-coa-1 co-and-architecture pipelining speedup numerical-answers + – Bikram asked Nov 25, 2016 retagged Sep 14, 2020 by ajaysoni1924 Bikram 315 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 1 votes 1 votes speed up = ntn/ (k+n-1)tp where n--no. of inst tn-time taken in non pipelined system k-- no.of stages in pipeline tp -- time of one stage(1 clock) On substitution .. ans is 1.67 David answered Dec 1, 2016 selected Aug 26, 2019 by Bikram David comment Share Follow See 1 comment See all 1 1 comment reply Vijay Thakur commented Dec 9, 2016 reply Follow Share Yes answer is 1.67, and i updated it from 16.66(old) to 1.67. thanks!! 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes For 10 instructions, non-pipelined architecture = $250ns$ For 10 instructions, pipelined architecture = $60+9(10)=150ns$ Speedup = $\frac{250}{150}=1.66$ JashanArora answered Dec 19, 2019 JashanArora comment Share Follow See all 0 reply Please log in or register to add a comment.