cache memory

Question

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 being read-only instructions. What is the average access time for the system (in ns) if it uses the WRITETHROUGH technique?

Comments

0.4*200ns + 0.6*(5ns + 0.2*100ns) = 95ns

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(0.8*5+0.2*(5+100))*0.6+0.4*200=95ns

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$(0.6\times 5)+\left \{ 0.4(0.8\times 20+0.2(20+200)) \right \}=27$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

Answer 1

read:0.8(5)+0.2(5+100)=25 ns

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

total time=0.6*25+0.4*200

=95 ns