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

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