retagged by
4,250 views
0 votes
0 votes
Two level memory contains cache and main memory. Cache access time is 20ns and main memory access time is 120ns/word.The size of block is 4 words. Main memory is referred  20% of times.What is the average access time:

a) 120ns. b) 20ns c) 116ns d) 150ns
retagged by

2 Answers

Best answer
2 votes
2 votes

Here memory hierarchy is followed , so for each main memory reference block need to be copied back to cache

So 1 word memory access time =120 ns

so 4 word memory access time=Tm=4*120=480 ns

Hc=cache hit ratio= 80%=0.8

(1-Hc)=memory hit ratio=cacge miss ratio=20%=0.2

Tc=20ns

Tm=480ns

Average Access time=Hc*T+ (1-Hc)*(Tm+Tc)=(0.8*20)+0.2*(480+20)=116ns

–2 votes
–2 votes
I think its b. 20ns

Related questions

1 votes
1 votes
1 answer
3