GATE CS 2025 Set 2 Q.58: Memory Management and Virtual Memory

A computer system supports a logical address space of $2^{32}$ bytes. It uses two-level hierarchical paging with a page size of 4096 bytes. A logical address is divided into a $b$ -bit index to the outer page table, an offset within the page of the inner page table, and an offset within the desired page. Each entry of the inner page table uses eight bytes. All the pages in the system have the same size.

The value of $b$ is ___________ . (Answer in integer)

Your answer

Official answer and solution

Official answer: 11
From: Official GATE 2025 answer key

Solution

Subtract the page offset bits ( $\log_2 4096 = 12$ ) and the inner page table index bits ( $\log_2(4096/8) = 9$ ) from the 32-bit logical address.

Total logical address bits: $\log_2(2^{32}) = 32\text{ bits}$ .
Page offset bits: Since page size $= 4096 = 2^{12}\text{ bytes}$ , the page offset requires $12\text{ bits}$ .
Inner page table offset bits: Each inner page table page has size $4096\text{ bytes}$ and each entry uses $8\text{ bytes}$ , so one page of the inner page table holds $\frac{4096}{8} = 512 = 2^9$ entries, requiring $9\text{ bits}$ .
Outer page table index bits ( $b$ ): $b = 32 - 12 - 9 = 11$

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 2 (CS2)
Question: 58 of 65
Section: Computer Science and Information Technology

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