paging

Question

Consider a system using multi level paging environment. Address bits are segregated as follows: Here first 8 bits represent the required bits for first level, next 8 bits represent the required bits for second level, next 6 bits represent the required bits for third level and last 10 bits represents the offset. 

What is the size of memory required for storing the all page tables for a process that has 512 K of memory starting at address 0? Assume each page table entry is 4 bytes.

1.    2048
2.    3192
3.    1024
4.    4096

Answer 1

Ans is D

Number of Pages in process is 2^19/2^10=  2^9 pages are there

 

so the 3rd level have to contain 2^9 enteries

and each table in 3rd level contain 2^6

so we need 8 tables in 3rd level to point 2^9 enteries

 

on second level we only need 8 entry but we have to use 1 whole page table which contain 256 entry

on first level we only need 1 entry but we have to use 1 whole page table which contain 256 entry

each entry is of size 4bytes

total size needed by page table is= size taken by first level+ size taken by second level + size taken by third level

                                                = 256*4 + 256*4 + (64*4)*8=4096