UGC NET CSE | December 2019 | Part 2 | Question: 9

Question

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

• A. $3$
• B. $4$
• C. $3.91$
• D. $2.91$

Answer 1

Time taken by non pipelined processor to  complete 100 tasks=100* time taken to complete each task

=100*30

=3000

Time taken by pipelined processor to  complete 100 tasks=(100+number of stages-1)*clock cycle time

=(100+4-1)*10

=103*10

=1030

speed up=(Time taken by non pipelined processor to  complete 100 tasks)/(Time taken by pipelined processor to  complete 100 tasks)

=3000/1030

=2.91

Answer 2

For a non-pipelined system: 

• Total number of instruction/task $(n)=100$
• Total time required to perform a single task in pipelined processor $(T_{np})=30$ ns

For a pipelined system:

• Total number of stages $(k)=4$
• Total number of instruction/task $(n)=100$
• Total time required to perform a single task in pipelined processor $(T_p)=10$ ns

$\because \text{Speedup ($S_k$)=$\frac{ET_{np}}{ET_p}$}$

$\implies S_k= \frac{(n*T_{np})}{(k+(n-1))T_p}$ 

$\implies S_k=\left[\frac{30*100}{(4+(100-1))*10}\right] $ ns

$\implies S_k=\left[\frac{3000}{(4+99)*10}\right]$ ns

$\implies S_k=\left[\frac{3000}{1030}\right]$ ns

$\implies S_k= 2.91$ ns

Option $(D)$ is correct.