Cache

Question

Answer 1

System is like

CPU ->LEVEL1 CACHE -> LEVEL2 CACHE

So effective memory access time

=Hit of level 1 *(Access time of level 1) + (Miss of level 1)(Access of level 1 + access of level 2)

=.95*(10) + 0.05 *(100+10)=15 sec

But in question they asked for AVERAGE ACCESS TIME FOR WORD

so its = Total time /Number of words.

total time = 100*10 + 4900*110= 540000

number of words are 5000

so time = 540000/5000=108.nano sec

Comments

why 0.5 is taken it should be 0.05?

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oh!! yes i went wrong there

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SHUBHAM SHASTRI

Is this is a correct answer ?? I don't think so

here , they didn't given that number of words  = 5000 words

they say that L2 holds : 5000 words

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but number of words in Level 2 cache are 5000 and we are supposed to access them ..

Definately i agree with you but thing is like..they have mention that only ..so we should take that

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@shubham I think 110 should be multiplied by 5000 and overall division. By 5100

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Should we consider word level transfer or block transfer?

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@himanshu19 i also thought the same that there are 5100 Words ...but as you know that Level 1 cache conaitns those words which are in Level 2  ..

because we always copy words/Blocks from lower level memory to upper level memory ... correct me i am wrong

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i have edited answer please check once

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@shubham I find nothing wrong in your explanation .....answer should be mentioned for such questions ....

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@shivangi can you share the solution ??