GATE CS 2024 Set 1 — Question 62

NAT+2 / -0EasyPaging & Address TranslationMemory ManagementOperating System

Operating System → Memory Management → Paging & Address Translation

Last updated 20 August 2026

Question

Consider a memory management system that uses a page size of

2\text{ KB}

. Assume that both the physical and virtual addresses start from

0

. Assume that the pages

0, 1, 2,

and

3

are stored in the page frames

1, 3, 2,

and

0,

respectively. The physical address (in decimal format) corresponding to the virtual address

2500

(in decimal format) is _________

6596 to 6596

1.Determine Page Size in bytes:

Page size =

2\text{ KB} = 2 \times 1024\text{ bytes} = 2048\text{ bytes}

2.Find the Page Number ( $PN$ ) and Offset ( $d$ ) for the given Virtual Address ( $VA$ ):

VA = 2500

PN = \lfloor VA / \text{Page Size} \rfloor = \lfloor 2500 / 2048 \rfloor = 1

d = VA \pmod{\text{Page Size}} = 2500 - 2048 = 452

3.Map the Page Number to the Frame Number ( $FN$ ):

According to the given mapping:

For

PN = 1

, the corresponding

FN = 3

4.Calculate the Physical Address ( $PA$ ):

PA = (FN \times \text{Page Size}) + d

PA = (3 \times 2048) + 452

PA = 6144 + 452 = 6596

Therefore, the physical address is

6596