Cache Organization

Question

Can any one help me out with this question :

This was asked in MadeEasy CBT held on 23^rd jan 

Comments

140ns?

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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

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Sorry for low quality image....

The answer given was 200 ns

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

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$Tm=40+0.4*2*50+0.4*0.5*4*150=200 ns$

Answer 1

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

Comments

Block size = word per block

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Still not able to interpret that table.

The way we are taking care of destination size, here 

https://gateoverflow.in/43329/gate2010-49?show=44176#a44176

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

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This solution is being provided is very correct...

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