in CO and Architecture
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In an instruction pipeline of 10 ns clock, the memory instruction takes 2 stalls while branch related instruction takes 3 stalls. If the frequency of memory related and branch instructions is 20% and 30% respectively. What is the average time taken by an instruction?

a) 11.3 ns   b) 13 ns

c) 23 ns d) 13.3 ns
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23 nsec is correct answer
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2 Answers

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Best answer
ans would be like this:

only memory=20%

only Branch=30%

so 1+ 20%*2 + 30% * 3

1+0.4 + 0.9 =2.3 clocksp=2.3 * 10ns =23ns

PS:here no relation has been given between memory related and branch related instructions

so i have not consider it.
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1 vote
1 vote
use formula :
 Average time taken=(1 + No of Stalls per Instruction)*Cycle time

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