58. A system has a memory access time of 50 ns and a cache access time of 2 ns. The instruction time is 4 ns (not counting memory access) and the average instruction requires 0.25 memory accesses. If the hit ratio is 0.90, what is the average instruction time?
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The average time is 4 ns (instruction) + 0.25 (memory access rate) × (htc + (1 ? h)tm)
= 4.0 + 0.25 (0.9 × 2 + 0.1 × 50) = 4.0 + 0.25(1.8 + 5.0) = 4.0 + 0.25 × 6.8 = 5.7 ns.
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