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please explain

please explain

4Comments
Arul @innovwelt
4 Jan 2015 11:14 pm

I think like this.

when there is a cache miss = cache access time +  time to transfer a block from main m/y

= 5 + (50 + 15*10) = 205

when the cache is accessed next time , there will be a hit (5 ms).

so, total = 210 ns

TarGate @tar_gate
5 Jan 2015 12:05 am

205 it should be. 

Re-execution time is not counted in access times.

But 370 does not seem to be the answer from any calculation.

Arul @innovwelt
5 Jan 2015 05:27 am

the question is "what is the access time when there is a cache miss".

provided the assumption that "when there is a cache miss, the cache waits till the block is fetched from Main memory & re-executes for a hit". In this case, cache is accessed twice.

1. before fetching block from main memory ( to check if  block is available in cache)

2. after fetching block from main memory ( for cache hit)

What do you think?

TarGate @tar_gate
5 Jan 2015 03:18 pm

I think its a bit ambiguous. Normally we do not take into account the re-execution. Because then it becomes the case of cache hit :)

Thus making our calculated cache miss time erroneous.