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Question 1 (Multicore Performance) Part A Machine A Machine B 1 core 2 core 4 GHz clock 2 GHz clock Assume a computation take

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Answer #1

Processor A execution time=1000/4*(10^9)

=250*(10^-9)

=0.25*(10^-6)

=0.25μ

Processor B execution time=2000/2*(10^9)

=1000*(10^-9)

=1*(10^-6)

=1μ

so from execution times of processor A and processor B i,we can specify that processor A is faster than processor B because execution time is Minimum.

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