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A computer has a cache, main memory and a hard disk used for virtual memory. If referenced word is in cache, 20 ns are required to access it. If it is in main memory but not in cache 60 ns are needed to load it into cache and then reference is started again. If word is not in main memory, 12 ms are required to fetch the word from disk followed by 60 ns to copy into cache, the reference is started again. The cache hit ratio is 0.9 and main memory hit ratio is 0.6. The average time in nano seconds required to access a referenced word on this system is ___________.
 
Solution: 480026

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tavg = .9 * 20 + .1( .6 * (60 +20) + .4 ( 60+20+12000000) ) = 480026 ns.
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In Cache Pr(Hit)=0.9 Access Time=20ns
In MM not in Cache 0.1*0.6 60+20=80ns
Not MM but in Hard disk 0.1*0.4 12ms+20ns+80ns=12000080ns

So Average Time to access it=0.9*20+0.06*80+0.04*12000080=480026ns

Answer:

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