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+19 votes

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@Digvijay @Arjun

then answer in the following question should be 7?

https://gateoverflow.in/75627/pipeline-efficiency

+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.

(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 vote

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