3,290 views
1 votes
1 votes
Ques. On a System with 32 bit address and 4KB pages how many levels are required in multilevel page table(assuming page table entry is of 4 bytes of storage)?

(A). 2

(B). 3

(C). 1

(D) None

 

plz explain the solution

1 Answer

Best answer
4 votes
4 votes

logical address=32 bit

page size= 4KB =2^12 byte   , 

  now we can write logical address = !!20!!12!!  ,  12 bits are for representing pages and 20 bits are for no of pages in 1st level page table.

so,

 page table 1 = 2^20 * PTE  = 2^20 * 4 byte = 4 MB > 1 page size , so  we will do paging again on this page table. that it can fit in 1 page.

now, # of pages of page table = 2^22 / 2^12 = 2^10 pages 

page table 2 = # of pages of page table1 * PTE 

                   = 2^10  * 2^2 = 2^12 = 4KB  which is equal to page size  

so level of  page table =2

 now logical address = !! 10 !! 10 !! 12 !!

selected by

Related questions

1 votes
1 votes
1 answer
2
jenny101 asked Jul 25, 2016
839 views
I am not able to understand the concept of multilevel paging.plz can someone tell what is stored in 1st level and 2nd level page table and how do we access it??
3 votes
3 votes
0 answers
4
sushmita asked Jan 5, 2017
1,037 views
A computer system has a 36-bit virtual address space with a page size of 8K, and 4 bytes per page table entry.What is the maximum size of addressable physical memory in t...