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Official PYQ · 2025Set 1 · Q4Memory Management and Virtual MemoryMCQ1 mark

GATE CS 2025 Set 1 Q4 — Memory Management and Virtual Memory

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Consider a demand paging memory management system with 3232-bit logical address, 2020-bit physical address, and page size of 20482048 bytes. Assuming that the memory is byte addressable, what is the maximum number of entries in the page table?

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Answer: A

To find the maximum number of entries in the page table, we need to determine the total number of pages in the logical address space.

We are given:

  • Logical Address Size = 3232 bits
  • Page Size = 2048 bytes=211 bytes2048 \text{ bytes} = 2^{11} \text{ bytes}

Since the memory is byte-addressable, the logical address is divided into two parts:

Logical Address=Page Number (p)+Page Offset (d)\text{Logical Address} = \text{Page Number (} p \text{)} + \text{Page Offset (} d \text{)}

The page offset dd is determined by the page size:

d=log⁡2(Page Size)=log⁡2(211)=11 bitsd = \log_2(\text{Page Size}) = \log_2(2^{11}) = 11 \text{ bits}

The number of bits required for the page number pp is:

p=Logical Address Size−d=32−11=21 bitsp = \text{Logical Address Size} - d = 32 - 11 = 21 \text{ bits}

The maximum number of entries in the page table is equal to the total number of pages in the logical address space:

Maximum Page Table Entries=2p=221\text{Maximum Page Table Entries} = 2^p = 2^{21}

Therefore, the correct option is A.

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