pipelining - branch delay slot

Question

Comments

cpi 1.2 for machine 1

cpi 1.14 for machine 2 ??

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2nd seems correct,

1st one , tell ur approach ?

 for first one , I think we need help.. :)

Answer 1

for first machine -

there will be 1 delay slot for branch instruction weather a branch will be taken or not but because 70% of time it will get filled by useful instruction so only 30% of time we have to pay branch penalty. there for cpi will be

CPI = 1 + 0.2x0.3 = 1.06 

for second machine- 

cpi = 1 + 0.2x0.35x2 = 1.14

Answer 2

I'm not sure of the explanation however here is a try

For machine 70% of the delay slots are utilized so only 30% actually causes stalls. So taking stalls caused in machine 1 as

1(stage in which address is resolved -1 )*0.30≐.30 stalls/instruction.

Now  avg stall created is

0.30*.20*.35≐0.021

therefore cpi≐1.021

For machine 2

avg stall created ≐ 2*.20*.35≐0.14

therefor cpi≐1.14

Hope it helps..:)

Comments

Ya , definitely it helps :)

Second one seems to be correct.

For First one, we need an expert answer .

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In first case there is no prediction. and only delays. So, no need to do multiplication by 0.35.

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So, it shud be 1.06 for machine 1.

Thanks :)

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@Arjun sir, I didn't get it. why not multiplied by .35 in machine 1. It also uses the same set of instruction as machine 2 in which there are 20% branch instruction and 35% of branches are taken and rest are normal. what does branches are taken imply here..??