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in case of hierarchical memory organization when there is a miss in cache , we need to bring the entire block from main memory to cache so in the formula-

AMAT= H1*T1+(1-H1(T1+T2))

T1- cache access time/word

T2= memory access time/word

T2 should be the entire block transfer time right? Why in some cases we just take word access time of main memory.

also please tell me what should be T2 in case of simultaneous organization?

retagged | 48 views
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None of the reference books have anything about simultaneous organisation of cache memory. I think you can ignore it - at least that's what I did. I haven't heard that term anywhere outside of GO.
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For hierarchical, there's only one formula:

$\text{AMAT = Hit Time + Miss Rate*Miss Penalty}$.

This can be expanded to any number of levels.

If there are two levels of cache,

AMAT = HT + MR (of L1)*Miss Penalty.

Now, Miss Penalty of L1 can be given as: HT(of L2) + MR(of L2)*MP(of L2).

Substituting this in the original equation, we get:

AMAT = HT + MR(of L1)*(HT(of L2) + MR(of L2)*MP(of L2)) and so on.
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Write allocate and no write allocate are write mudd policies and not read. In read when we do hierarchical access we always bring in the desired block from cache to memory and then access the cache. Please check it
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Oh right, I didn't see that, you're correct. It's for write policies. Yes, you're correct. Whenever a read is a miss, it is brought to the cache and read. I'll edit my comment so that other people don't get confused.
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For your other doubt, maybe this will help.

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Thanks. Can u please share if u know about memory access time for write back policy?