UGC NET 2021 | What is the size of page

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?
i➥ 12KB
ii➥ 2KB
iii➥ 2MB
iv➥ 32MB
Answer: Option III 


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 = 2page number.


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

Page Table size operating system
Number of pages = 2page number = 220,
Page Table size = Number of pages * Page Table Entry size
                = 220 * 2B
                = 2 MB

So, Option(III)is correct. 

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UGC NET 2021 | What is the size of page

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