2,588 views
4 votes
4 votes
consider a system using 3-level page-table where it uses first 2 bit to index into first level page-table, 9 bit to index into second level page table , 9 bit to index into third level page-table.  processor uses 36 bit physical address and 32 bit virtual address . each page-table entry is of size 4 bytes. calculate total overhead for the page-table ?

2 Answers

Best answer
4 votes
4 votes

Given three level page table,

 2    ||     9     ||     9   ||    12   ||

PTE size =  4B

Page Table Size = No.of entries in page in page table * entry size

PTS of 1st level page table = $2^{2}* 4B = 16B$

PTS of 2nd level page table =  $2^{9}* 4B$ and we have  $2^{2}$ 2nd level Page tables.

So PTS of 2nd level page table =  $2^{9}* 4B * 2^{2}  =  8KB$

PTS of 3rd level page table  =  $2^{9}* 4B $ and we have  $2^{2}* 2^{9}$ 3rd level page tables.

So PTS of 2nd level page table =  $2^{9}* 4B * 2^{2} * 2^{9} = 2^{20} * 4B = 4MB$

So Total overhead = 16B + 8KB + 4MB $\approx$ 4MB.  

selected by

Related questions

0 votes
0 votes
1 answer
1
Markzuck asked Dec 22, 2018
1,363 views
for memory overhead in Multi level paging, for innermost table only 1 page size shall be counted na? and NOT the complete page table size?please explain the concept, than...
1 votes
1 votes
2 answers
2
!KARAN asked May 18, 2019
588 views
What is the mode(User/Kernel) of the Operating System when a Page Table or TLB are being accessed?
0 votes
0 votes
0 answers
4
srestha asked Jul 26, 2018
672 views
Consider a system with byte-addressable memory, 32 bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the s...