Multilevel Pagetable

Question

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

Answer 1

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 !!

Comments

Mam I understood , WE ALWAYS DO PAGING OF PAGE TABLE FOR FURTHER LEVEL OF HIERARCHY ,Suppose had been the page size less then 4KB we would have again divided it further like

2^12/Page Size = 2^x

So now for P1 = x , P2 = x ,P3 = 10 ,Offset = 12

Mam am i right plz Clear me Up?????

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see in memory everything resides within a page . so for processing  a process also control will go and encounter  a page of memory , (where the page table is stored page table contains details of mapping ex. which page of process is in which fame ). That's why we keep doing paging on process until  the page table fits within one page.  

and yes when size of page table is less than or equal to page size for ex. here it is  4 KB , then their is no need of further paging . bcz it already fits within a page .