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Assume that for a certain processor, a read request takes 50 nanoseconds on a cache miss and 5 nanoseconds on a cache hit. Suppose while running a program, it was observed that 80% of the processor's read requests result in a cache hit. The average read access time in nanoseconds is __________.

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Data

For Cache Hit,

Hit rate = H1 = 0.8

T1 = 5 ns

For Cache Miss,

Miss rate = (1 – H1) = 0.2

T2 = 50 ns

Formula:

Average read access time = H1 × T1 + (1 – H1) × T2

Calculation:

Average read access time = 0.8 × 5 + 0.2 × 50

                                              = 4 + 10 = 14 ns

The average read access time in nanoseconds is 14.

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