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A computer has a single cache (off-chip) with a 3 ns hit time and a 95% hit rate. Main memory has a 50 ns access time. If we add an on-chip cache with a 0.6 ns hit time and a 98% hit rate, the computer's effective access time:
1. 2.8 ns
2. 5.5 ns
3. 0.7 ns
4. None of these

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Correct Answer - Option 3 : 0.7 ns

Data:

Cache

Cache = T1 = 0.6 ns

Hit ratio = H1 = p = 0.98

Miss ratio = 1 – p = 1 – 0.98 = 0.02

Cache = T2 = 3 ns

Hit ratio = H2 = q = 0.95

Miss ratio = 1 – q = 1 – 0.95 = 0.05

Main memory

Main memory = T3 = 50 ns

Hit ratio = 100% = 1 (assume since nothing is given)

Formula:

Tavg = T1 + (1 – p)T2 + (1 – p)(1 – q)T3

Calculation:

Tavg = 0.06 + (0.02) × 3 + ((0.02) × (0.05) × 50

Tavg = 0.662 ≈ 0.7 ns

Important Point:

Serial addition of chip is taken

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