A nonpipelined system takes 100 ns to process a task. The same task can be processed in a five-stage pipeline with a clock cycle of 20 ns. Determine the speedup ratio of the pipeline for 100 tasks. What is the theoretical speedup that could be achieved with the pipeline system over a nonpipelined system

Time Taken Pipe line Processor = 2080 ns

Speedup = 4.807

Explanation:

given data

time non pipe lined system tn = 100ns

time taken pipe lined system tp = 20 ns

no of segment k = 5

no of task n = 100

solution

we get here first time taken by Non-Pipeline to process 100 task is

time taken = 100 × 100 = 10000 ns

so here Time Taken by Pipe lined Processor will be

Time Taken Pipe line Processor = (k + n - 1 ) × t p   ..............................1

put here value and we will get

Time Taken Pipe line Processor = (5+ 100 - 1 ) × 20

Time Taken Pipe line Processor = 2080 ns

so here speedup will be  

Speedup =

Speedup = 4.807

efficiency of reversible power cycles operating between a hot re

explanation:

answer:

a

explanation: