سوال ۹۸ کنکور ارشد مهندسی کامپیوتر ۱۴۰۲

The effective (memory) access time under demand paging is a weighted average of a normal access (no fault) and a page-fault service:

$\text{EMAT} = (1-P)\,t_{mem} + P\,t_{fault}$

with $t_{mem} = 100$ ns and $t_{fault} = 25\text{ ms} = 25{,}000{,}000$ ns.

Since the fault service is enormously larger than a memory access, simplify by approximating $1-P \approx 1$ (the $-P\,t_{mem}$ term is negligible):

$\text{EMAT} \approx t_{mem} + P\,t_{fault} = 100 + P\,(25{,}000{,}000)$

Impose the requirement $\text{EMAT} < 110$ ns:

$100 + P\,(25{,}000{,}000) < 110$ $P\,(25{,}000{,}000) < 10$ $P < \frac{10}{25{,}000{,}000} = 4 \times 10^{-7} = 0.0000004$

Keeping the exact form $(1-P)(100) + P(25{,}000{,}000) < 110$ gives $P < \dfrac{10}{24{,}999{,}900} \approx 4.00002 \times 10^{-7}$, the same bound to four significant figures.

So the page-fault rate must satisfy $P < 0.0000004$, which is option 2.

Pitfall: Watch the units — convert the $25$ ms service time to nanoseconds ($2.5 \times 10^{7}$ ns) before mixing it with the $100$ ns memory time. Forgetting this (or mis-counting the leading zeros in the answer choices) flips you to a wrong option like $3 \times 10^{-7}$.