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Consider a two-level memory hierarchy with separate instruction and data caches in level 1, and main memory in level 2. The clock cycle time in 1 ns. The miss penalty is 20 clock cycles for both read and write. 2% of the instructions are not found in I-cache, and 10% of data references not found in D-cache. 25% of the total memory accesses are for data, and cache access time (including hit detection) is 1 clock cycle. The average access time of the memory hierarchy will be …………. nanoseconds

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Average access time = 0.75 (0.98 x 1 + 0.02 x 20) + 0.25 (0.90 x 1 + 0.10 x 20) = 1.76 ns
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