edited by
794 views
1 votes
1 votes
suppose we have a memory system with 32 bit virtual addresses and 4KB pages.If the 20 level paging is used and the memory required to store all the page tables if page table entry size is 4bytes is 2^a(2^b-1)bytes then a*b is
edited by

1 Answer

0 votes
0 votes

Given page table entry size is 4 bytes = 22 B

the memory required to store all the page tables is 2^a(2^b-1) bytes .

32 bit virtual addresses.

page size = 4KB = 22 * 210 B = > offset bits = 12

Number of entry in inner page table = page size / page table entry size 

= 212 / 22

= 210

outer page table will require 32-12(page offset)-10(inner page table)= 10 bits

so for outer page tables there will be 210 entries, for inner page tables there will be 210*210 entries

For 20 level paging , total entires including all the page tables= 21+22+23+24.......220 

=2*(220-1) entries,

now each entry is of 4 bytes so total size = 2*(220-1)*4

memory required to store all the page tables = (220-1) * 23

=> 2a * (2b-1) = (220   - 1 ) *  23

a= 3 and b = 20

hence a*b = 3 * 20 = 60 

edited by
Answer:

Related questions

1 votes
1 votes
2 answers
3
ajayraho asked Oct 29, 2022
639 views
Is the page table entry size same for each entry in page tables of all levels in multilevel paging?