NIELIT 2016 DEC Scientist B (CS) - Section B: 23

Question

A nonpipeline system taken $50ns$ to process a task. The same task can be processed in a six-segment pipeline with a clock cycle of $10ns.$ Determinant the speedup ration of the pipeline for $100$ tasks. What is the maximum speedup that can be achieved?

• A. $4.90,5$
• B. $4.76,5$
• C. $3.90,5$
• D. $4.30,5$

Comments

option B is the answer.

---

Ans : B

---

@haralk10, maximum speed up is equal to no of stages only when each stage takes equal time, or in other words, in case of uniform delay pipeline

 

Answer 1

Speed Up Ratio(S) = n*Tn/(k+n-1)Tp

where n=no. of tasks per segment

Tn=Total time required to performed a single task in non pipeline

Tp=Total time required to performed a single task in pipeline

k=No. of segment in pipeline

so n=100

Tn=50 ns

Tp=10 ns

k=6

so

S=(100*50) / (100+6-1)*10

S= (5000)/1050

S= 4.76

So Speed up ratio is 4.76

In this case maximum speed up =  Tn/Tp = 50 ns/10 ns=5

 

So the correct answer is option (B) 4.76, 5

Answer 2

for non pipelined system ET=(100*50)=5000ns

speed up=5000/((6+100-1)*10)=4.76

maximum speed up=(50/10)=5

So,option b is correct.

Answer 3

The speed up ratio = non -piped line time implementation for a task/pipedline time implementation for the same task

In this case speed up = 50 ns/10 ns=5

further for 100 task----in  this case of non-piped line processor takes 50*100 ns= 5000 ns

for the piped line processor with 6 stage ,task= 100, time /task=10 ns,the total time is taken (100+6-1)*10=1050 ns.

hence speed up for 100 task = 5000 ns/1050 ns=4.76

since theritically maximum speed up is equal to nos of stages which is 6 in this which is never achieveable, hence max speed up in this case is 5 as above. hence the option (B) is correct.