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Effective address time for a cache comprising of L1 and L2 cache =9ns

hit ratio of L1 cache = 0.8 

hit ratio of L2 cache = 0.9

memory access time =100ns

Miss penalty of L1 cache = 25ns

 Access times for L1 and L2 caches are x and y ns;

let z=x+y; what’s z??

I am getting 20.33 but the given answer is different!

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Given,

Effective access time=Hit time L1*Hit Ratio L1+Miss ratio L1*Miss Penalty

9ns=0.8*x+0.2*(0.9(x+y)+0.1(x+y+100))

Miss penalty =25

0.9(x+y)+0.1(x+y+100)=25

x+y+10=25

x+y=15=z
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h4kr asked Dec 27, 2022
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In $T_{Read Avg}$, shouldn’t it be $T_{Read Avg}$ = (10*0.9) + 0.1*(10+100), because it must be checking the cache in case of cache miss too, right?