Under demand paging, memory access time is 100 ns, servicing a page fault takes 5 ms, and the page-fault rate is 0.0002. What is the effective access time, approximately?
Correct answer B. 1.1 microseconds
Explanation
EAT = (1 - p) x 100 ns + p x 5,000,000 ns = 99.98 + 1000 = 1099.98 ns, about 1.1 microseconds. Even one fault in 5000 accesses makes memory roughly eleven times slower, because a disk access is so slow compared with RAM. The lesson is that the page-fault rate must be kept extremely low for demand paging to perform well.
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