UGC NET Computer Science December 2019 | Question 59

Question 59

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A non-pipelined system takes 30ns to process a task. The same task can be processed in a four-segment pipeline with a clock evcle of 10ns. Determine the speed up of the pipeline for 100 tasks.

1. 1. 5
2. 2. 4
3. 3. 3.91
4. 4. 2.91

Explanation

Speed up ratio (S):
It is defined as the speedup of a pipeline processing with respect to the equivalent non-pipeline processing.

Formula for calculating Speed up ratio is : nt_n/(n+k-1)t_p
Number of tasks, n = 100

For Nonpipeline:
Time taken by nonpipeline to process a task, t_n = 30 ns

Total time taken by nonpipeline to process 100 task = nt_n
= 100 * 30
= 3000 ns

For Pipeline:
Number of segment pipeline k = 4
Time period of 1 clock cycle, t_p = 10 ns

Total time requires to complete n tasks in k segment pipeline with t_p clock cycle time:
= (n + k - 1)t_p
= (100 + 4 -1)10
= 1030

Speed up Ratio:
When total time taken by the pipeline to process 100 tasks is divided by the total time requires to complete n tasks in k segment pipeline with t_p clock cycle time then speed up ratio is obtained.
= 3000/1030
= 2.9161

So, option 4 is correct answer

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