GATE CS 2025 Set 1 Q.14: Memory Management and Virtual Memory

Consider a demand paging memory management system with 32-bit logical address, 20-bit physical address, and page size of 2048 bytes. Assuming that the memory is byte addressable, what is the maximum number of entries in the page table?

$2^{21}$

$2^{20}$

$2^{22}$

$2^{24}$

Official answer and solution

Official answer: A
From: Official GATE 2025 answer key

Solution

Divide the logical address space size by the page size to find the number of virtual pages.

Logical address length $= 32$ bits $\implies$ Logical address space size $= 2^{32}$ bytes.
Page size $= 2048\text{ bytes} = 2^{11}$ bytes.
Maximum number of entries in a conventional page table equals the total number of virtual pages in the logical address space:

\text{Number of page table entries} = \frac{2^{32}}{2^{11}} = 2^{21}

Final answer

Option (A)

Common trap

Using the 20-bit physical address or assuming a 4 KB ( $2^{12}$ ) page size instead of 2048 bytes ( $2^{11}$ ).

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 1 (CS1)
Question: 14 of 65
Section: Computer Science and Information Technology

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