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Official PYQ · 2015

GATE CS 2015 Set 1 Q1 — Memory Management and Virtual Memory

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Consider a system with byte-addressable memory, 32-bit32\text{-bit} logical addresses, 4  kilobyte4\;\text{kilobyte} page size and page table entries of 4  bytes4\;\text{bytes} each. The size of the page table in the system in megabytes\text{megabytes} is___________.
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Answer: 4

To find the size of the page table, we first determine the total number of pages in the logical address space.

1. Calculate the number of pages:

  • Logical address size = 32 bits32\text{ bits}
  • Logical Address Space (LAS) = 232 bytes2^{32}\text{ bytes}
  • Page size = 4 KB=212 bytes4\text{ KB} = 2^{12}\text{ bytes}
  • Number of pages = Logical Address SpacePage Size=232212=220\frac{\text{Logical Address Space}}{\text{Page Size}} = \frac{2^{32}}{2^{12}} = 2^{20}

2. Calculate the page table size:

  • Page Table Entry (PTE) size = 4 bytes4\text{ bytes}
  • Page Table Size = Number of pages×PTE size\text{Number of pages} \times \text{PTE size}
  • Page Table Size = 220×4 bytes=4 MB2^{20} \times 4\text{ bytes} = 4\text{ MB}

Therefore, the size of the page table in megabytes is 4.

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