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A pipeline $P$ operating at $400$ MHz has a speedup factor of $6$ and operating at $70$% efficiency. How many stages are there in the pipeline?

  1. $5$
  2. $6$
  3. $8$
  4. $9$
in CO and Architecture by Boss (30.8k points)
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6 Answers

+19 votes
Best answer

Efficiency of K stage Pipeline = SpeedUp Factor (Sk) / Number of Stages (K)
0.70 = 6/K
K = 6/0.70 = 8.57  ≌ 9

Number of Stages = 9

 

by Veteran (60.9k points)
selected by
0
why took 9 and not 8?
+10

Number of stages < 8.57 then efficiency will be >70%.

0
Yup got it...
0

@Digvijay @Arjun
then answer in the following question should be 7?
https://gateoverflow.in/75627/pipeline-efficiency

 

+1
@Digvijay, Why we cannot have more than 70% efficiency if we can have less?
0
Digvijay, can we  say "operating at 70% efficiency" when E<70% ? No I guess ..but for E>70% we can say efficiency is 70%( means efficiency is at least 70%).. So, 8 should be the answer here. Right ?
+1
How did you derive the formula relating efficiency, speedup and stages?
0
What is advantage of giving pipeline frequency 400MHz ??? No 1 is considering it , (will ans always be same in any case of of operating frequency , time clock...
+3 votes
we know speedup = (time taken to complete n instructions in non pipeline/ time taken to complete n instructions in pipeline)

                                (nt/ (k+ n - 1)p)    where t = time taken to process one segment in non pipipeline and p = time taken to process one segment in pipeline

                          =   ( nt / np) when n approaches to very large number of instructions

                           = ( t / p )  = ( kp / p) = k  when k = number of segments in pipeline and we assume ideal case that each segment in nonpipeline takes p cycles

 

theoretically, maximum speedup is achieved when it is operating with 100 % efficiency then speedup = number of segments in pipeline = k

so, if it is operating at 70% efficiency, then speedup = 0.7 k which is equal to 6 as given in the question.

therefore, 0.7k = 6

k = 8.something which is equal to 9 segments .

because if we take k = 8, then it is being operated at less than 70% efficiency.

so, we take ceil value. therefore, answer is equal to 9 stages.
by (345 points)
0
well explained!
+1 vote
efficiency = speed_up/ stages

70/100=6/k

k=600/70

k=8.571 ≃9 stages
by Boss (20.1k points)
0
0
u have to take ceil value ryt because with 8 stages the process will not be executed
+1 vote

Since we know that 

Speedup = m x efficiency 

6 = m x 0.7

m=6/0.7

m=8.58

m=9(approx)

by Boss (10.2k points)
0 votes
S=Efficieny(n)*M(Stages)

M=S/n

M=ceil(60/7)=~9
by Boss (10.2k points)
0 votes

Speedup = m * efficiency.

6 = 0.7m

m = 8.57

 

If we take 8 stages, efficiency becomes 75%. If we take 9 stages, efficiency becomes 66%.

Less-than-mentioned efficiency sounds more practical than more-than-mentioned; so 9

 

Option D.

by Loyal (6.4k points)
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