BARC2017

Question

L1 and L2 cache access time 2ns and 5ns respt.

Hit percentage 3% and 5% respt.

If Average access time =2.4ns then memory access time=?

Comments

Something is wrong with the question. If L2 access time is 5ns then how can average access time be 2.4ns despite such poor hit rate. Still if we were to solve it i'm getting memory access time as 2.27ns.

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I dont able to recall the the actual data but question was like this only.

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It is a very common question in COA. just try to recollect the correct data

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yes may be some wrong data

I am getting -ve ans

equation will be

$0.03\times 2+0.97\times 0.05(2+5)+0.97\times 0.05(2+5+m)=2.4$

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yes it's possible . if we add the access time of previous memory hierarchy then memory time comes out to be -ve and if you dont, you still get a value very close to access time of L1 which is impossible.

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some books have the convention of adding L1 time to L2 and some books don't add. It is still a grey area in COA

Answer 1

solution...

Comments

H1T1+(1-H1)H2T2+(1-H1)(1-H2)H3T3=T

.03*2   +  .97*.05*5  +   .97*.95*T3  =   2.4

T3 = 2.276 ns