Test series

Question

Consider the program which was made to run on two different machines the machine-2 results 20% speeder compared to machine-1 however the CPI of 2nd machine is 25% more than the CPI of machine 1. If the clock frequency of the machine 1 is 2 MHZ what is the clock frequency of machine 2?

Comments

I got what hs_yadav did.

@Anu won't the relation be T(M1) = 1.2 * T(M2) ?

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@Warlock

T(M2) = 0.80T(M1) is correct, because machine T2 is 20% faster, this means it will take 20% less time.

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i think something is missing please complete the question.

Answer 1

Comments

is this way correct?

Speed1=no of operation/Time1

Speed1= x

Speed2=no of operation/Time2

Speed2=0.2x+x=1.2x