# cache memory

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Consider a two level memory hierarchy, L1  (cache) has an accessing time of 5 ns and main memory has an accessing time of 100 ns. Writing or updating contents takes 20 ns and 200 ns for L1  and main memory respectively. Assume L1 gives misses 20% of the time with 60% of the instructions are read only instructions. What is the average access time for system (in ns) if it uses WRITETHROUGH technique?

2
0.4*200ns + 0.6*(5ns + 0.2*100ns) = 95ns
1
(0.8*5+0.2*(5+100))*0.6+0.4*200=95ns
0
$(0.6\times 5)+\left \{ 0.4(0.8\times 20+0.2(20+200)) \right \}=27$ns
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@Shivangi Parashar 2

i am getting 103.4 can anyone confirm this answer.... as everyone seem to get different answer

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Correct
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@srestha what is correct ans ma'am

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what should be?? Can u give the correct logic?

" 60% of the instructions are read only instructions. "

right??
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@srestha  ma'am

avg time ={(0.8*5+0.2*(100+5)}*0.60 + {(0.8*20+0.2*(200+20)}*0.40

write:0.8* max(20,200)+0.2(200)=200 ns

total time=0.6*25+0.4*200

=95 ns

## Related questions

1
374 views
consider two level cache hierarchies with L1 and L2 cache.Programs refer memory 1000 times out of which 40 misses are in L1 cache and 10 misses are in L2 cache.If the miss penalty of L2 is 200 clock cycles,hit time of L1 is 1 clock cycle,and hit time of L2 is 15 clock cycles,the average memory access time is__________clock cycles.