Awesome q2a theme
0 votes
20 views
Q. consider a system with 2k pages ,512 frames ,pa=22 bits. calculate the page table size in bytes?
in Operating System by (49 points) | 20 views
0
I know here page table entry size is not giving.but if not giving then we consider with the help of frame numbers in bytes= one page table entry size ?

1 Answer

+1 vote
Given,

Number of pages $= 2k = 2*2^{10} = 2^{11}$

Number of frames $= 512 = 2^9$

Number of bits to represent physical address $= 22$

Page table size = Number of pages * Page table entry size

If no additional information is given then,

Page table entry size = Number of bits to represent frames.

Number of bits to represent frames $=9$

Page table size $= 2^{11} * 9$ bits $= \cfrac{2^{11} * 9}{8}$ bytes
by (35 points)
edited by
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
Welcome to GATE CSE Doubts, where you can ask questions and receive answers from other members of the community.
9,270 questions
3,204 answers
14,751 comments
96,304 users