610 views
3 votes
3 votes

can some one provide me procedure to solve this que?

1 Answer

Best answer
5 votes
5 votes

Suppose page size is 2B

Level 1:  2B =( 32B / 2B ) * 4B 

              2  =  232 - X + 2

Level 2 :  2x B  = [ ( 232 - X + 2 ) / 2B  ] * 4B

              2  =  232 - 2X + 2 * 2

 So x =12 .  PageSize = 212B  

------------------------------------------------------------

Let the page size = 2x B

$\therefore$ Total pages in virtual memory

= VA space / page size

= $\frac{4 * 2^{30}}{2^{x}}$

= $2^{32-x}$

$\therefore$ Total entries in the innermost page table level

= $2^{32-x}$

Now, if 1 level of paging is used:

Total entries in the innermost page table level * PTE size = page size

$\therefore$ $2^{32-x}$ * 4B = $2^{x}$

$\therefore$ x=17

Now, if 2 level of paging is used:

Total pages in the innermost level of paging               

= Total entries in the innermost page table level * PTE size / page size

= $2^{34-2x}$

Thus, size of outermost page table

= Total  pages in the innermost level of paging * PTE size

= $2^{36-2x}$

$\therefore$ size of outermost page table = page size

$\therefore$ x=12

This, minimum page size = 212B

selected by

Related questions

0 votes
0 votes
1 answer
2
Markzuck asked Dec 22, 2018
1,329 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...
0 votes
0 votes
1 answer
3