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Best answer
21 votes
21 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

 

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7 votes
7 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.
1 votes
1 votes
efficiency = speed_up/ stages

70/100=6/k

k=600/70

k=8.571 ≃9 stages
1 votes
1 votes

Since we know that 

Speedup = m x efficiency 

6 = m x 0.7

m=6/0.7

m=8.58

m=9(approx)

Answer:

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