سوال ۱۰۳ کنکور ارشد مهندسی کامپیوتر ۱۴۰۰

We need each page table (at every level) to occupy exactly one page, then count how many levels are required to map the entire virtual page number.

Step 1 — Offset (bits within a page). $\text{page size} = 16\text{ kB} = 2^{14}\text{ bytes} \;\Rightarrow\; \text{offset} = 14\text{ bits}.$

Step 2 — Virtual page number (VPN) width. $\text{VPN} = 47 - 14 = 33\text{ bits}.$

Step 3 — Entries that fit in one page. Each entry is $8 = 2^{3}$ bytes, so one page holds $\frac{2^{14}}{2^{3}} = 2^{11}\text{ entries}.$ A single-page table therefore resolves $11$ bits of the VPN per level.

Step 4 — Number of levels. $\left\lceil \frac{33}{11} \right\rceil = \frac{33}{11} = 3.$

The $33$-bit VPN splits cleanly into three $11$-bit index fields, so a 3-level page table is needed. This matches option 2.

Exam trick: "each page table fits in one page" means each level indexes exactly $\log_2(\text{page size}/\text{PTE size})$ bits; the number of levels is $\lceil \text{VPN bits} / \text{bits per level}\rceil$.