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Can any one help me out with this question :

This was asked in MadeEasy CBT held on 23rd jan 

in CO and Architecture by Active (1.3k points) | 178 views
basically it's given that Access time per word  = 50 ns

and 3rd col it given that --->  block size in blocks  ---- > means here they want to say 1 block contains  2 words

and we know that if there is a miss in L1 then it retrieve the whole 1 block from L2 and transfer in into L1

So access time of 1 word = 50ns

it retrieve 2 word from L2  = 1 block = 50*2   = 100

and rest calculation part is simple wrt me
Sorry for low quality image....

The answer given was 200 ns

(here, L2 Cache access time per word = 80ns)

1 Answer

+1 vote
Best answer
$=0.6*40 + 0.4*0.5(40+50*2) +0.4*0.5(40+ 50*2 + 150*4)$
$=24 + 0.2(140) + .2(740)$
$=24 + 28 + 148$
$= 200$ ns

Note that, on a miss in L1 and L2, blocks will be transferred from one level to next level.

PS: As pic quality is low so i assumed that respective hit ratios are 0.6, 0.5 and 1.
by Active (4.8k points)
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Why are we not considering the size where the block is copied?

Copy 4 blocks into L2 cache where size of cache is 2 blocks and then copy into cache whose size is 1 block. ?
i didn't get your doubt?
What is block size in blocks.. the 2nd column.
Block size = word per block

Still not able to interpret that table.

The way we are taking care of destination size, here

why it's not done here. How are we copying 4 blocks to a cache with size 2 blocks.

This solution is being provided is very correct...

I think by default the DATA BUS capacity is 1 Block...

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