سوال ۹۷ کنکور ارشد مهندسی کامپیوتر ۱۴۰۲

A single-level page table has one entry per virtual page, so its size is

$\text{page-table size} = (\text{number of pages}) \times (\text{entry size}).$

Step 1 — Number of pages. The virtual address space is $2^{32}$ bytes and each page is

$16\ \text{KiB} = 2^{4} \cdot 2^{10} = 2^{14}\ \text{bytes},$

so the page offset uses $14$ bits and the page number uses $32 - 14 = 18$ bits. Hence

$\text{number of pages} = \frac{2^{32}}{2^{14}} = 2^{18}.$

Step 2 — Multiply by entry size. Each entry is $4 = 2^{2}$ bytes, so

$\text{page-table size} = 2^{18} \times 2^{2} = 2^{20}\ \text{bytes} = 1\ \text{MiB}.$

Thus the page table occupies exactly $1$ megabyte, which is option 1.

Exam trick: Work entirely in powers of two. The number of page-table entries equals $2^{(\text{address bits} - \text{offset bits})}$; multiply by the entry size and convert $2^{20}$ bytes $= 1$ MiB.