[Solution] UGC NET 2021 | What is the size of page SamagraCS

UGC NET 2021 | What is the size of page

Comprehension:
Read the following and answer the questions:
Consider a machine with 16 GB main memory and 32-bits virtual address space, with page size as 4KB. Frame size and page size is same for the given machine.
What is the size of page table for handling the given virtual address space, given that each page table entry is of size 2 bytes?

NTA UGC NET June 2021

Operating System

Memory Management

i➥ 12KB

ii➥ 2KB

iii➥ 2MB

iv➥ 32MB

Answer: Option III 

Solution:

Given,
Virtual Address Space = 32 bits,
Page size = 4KB = 2¹²,
Page table entry size = 2B.

Formula used,
Page offset = ⌈log₂⁡(Page Size)⌉ ,
Page Table size = Number of pages * Page Table Entry,
Number of pages = 2^{page number}.

Calculation,

Virtual Address Space:

Page offset = ⌈log₂⁡(Page Size)⌉ = ⌈log₂⁡ 2¹²⌉ = 12 bits,
Page number = Virtual address space-page offset = 32-12 = 20 bits,


Number of pages = 2^{page number} = 2²⁰,
Page Table size = Number of pages * Page Table Entry size
                = 2²⁰ * 2B
                = 2 MB
                            

So, Option(III)is correct.