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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?

in CO and Architecture 309 views
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

0
Correct
0

@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??
0

@srestha  ma'am

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

1 Answer

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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

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