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.