NAT
Q.
A system uses demand paging with a page size of 4 KB. The memory access time is 100 ns, and the page fault service time (including swap in and restart) is 10 ms. If the effective access time (EAT) must not exceed 200 ns, what is the maximum acceptable page fault rate (in terms of number of faults per memory access, rounded to the nearest integer power of 10)?
Correct Answer: 0.00001
Effective Access Time (EAT) = (1 - p) × memory access time + p × page fault service time. Substituting values: EAT = (1 - p) × 100 ns + p × 10,000,000 ns. For EAT ≤ 200 ns, solving gives p ≤ 1 × 10^-5. Thus, the maximum acceptable page fault rate is 10^-5.