In a computer if the page fault service time is 10ms and average memory access time is 30ns . If one page fault is generated for every \(10^6\) memory accesses. What is the effective access time for the memory?
Effective access time (EAT) = memory access time + (page fault rate × page fault service time)
Page fault rate \(p = \frac{1}{10^6}\). Service time = 10 ms = \(10^7\) ns.
EAT = \(30 + \frac{1}{10^6} \times 10^7 = 30 + 10 = 40\) ns
Answer: (D) 40 ns approximately
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