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ISRO-2013-15

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12 12 votes

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. $9$

7 Answers

Best answer
26 26 votes

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
 

• selected by
11 11 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.
2 2 votes
efficiency = speed_up/ stages

70/100=6/k

k=600/70

k=8.571 ≃9 stages
0 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.

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