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A computer system has a cache with access time 10 ns, a hit ratio of 80% and average memory access time is 20 ns. Then what is the access time for physical memory?
1. 50 ns
2. 40 ns
3. 30 ns
4. 20 ns

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Correct Answer - Option 1 : 50 ns

Concept:

TM = h × TC + (1 – h) × TP

TM is the average memory access time

TC is the cache access time

TP is the access time for physical memory

h is the hit ratio

Calculation:

Given: TC = 10 ns, TM = 20 ns, h = 80% = 0.8

TM = h × TC + (1 – h) × TP

20 = 0.8 × 10 + (1 – 0.8) × TP

20 = 8 + 0.2 TP

TP = 60 ns

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