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Sagot :
Answer:
480,026ns
Explanation:
the time it would take to access word = 20ns
time coipied in cache = 60 ns
20ns +60ns = 80ns
this is the time to acess from main memory
it takes 12ms to move memory word to main memory
1ms is 1000000
so 12ms = 12000000ns
1200000+20+60
= 12000080 nanoseconds
hit ratio = 0.9(cache)
hit ratio = 0.6 (main memory)
miss rate = 0.1
probability of word being in main memory
= 0.1 * 0.6
= 0.06
the hit ratio of disk = 1
prob = 0.1 * 0.4 * 1 = 0.04
.1 = miss rate of cache
.4 = miss rate of main memory
1 = hit ratio of hard disk
In cache with prob 0.9, total ns = 20
in main memory prob = 0.60, total ns = 80
in hard disk prob = 0.04, total ns = 12000080ns
average time = [pr(cache)*access time] + [pr(main memory)*access time] + [pr(hard disk)*access time]
= (0.9*20)+(0.06*80)+(0.04*12000080)
= 18+4.8+480,003.2
= 480,026 ns
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