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# CO-Grand Test -Q7

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+1 vote
A computer has a cache, main memory, and a disk. If a referenced word is in the cache, 10 ns are required to access it. If it is in the main memory but not in the cache, 50 ns are required to load it into the cache, and then the reference is started again. If the word is not in the main memory, 10 ms are needed to load it from the disk to main memory, followed by 50 ns to copy it to the cache, and then the reference is started again. The cache hit ratio is 0.8 and the main memory hit ratio is 0.7. What is the average time needed to access a referenced word (in nsec)?
reshown Jun 18, 2020

This Question has explicitely defined what Hierarchical access does.

Cache access time = 10ns

The average time needed to access a referenced word :

$10ns + 0.2[ 50ns + (0.30)(10ms) ]$

NOTE that for every word we need 10ns for cache. For 0.2 times, we need some extra time.

That extra time is like this : 50ns is fixed because of Memory access time And 0.30 times, we need to access disk.

Answer is asked in nsec so, convert msec into nsec by multiplying with 10^6.