Multilevel Paging

Question

Comments

8kB?

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How?@arvin

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see brother its a 3level paging...

and size of outer page table (level 1) = one frame size (=page size)...

let page size =2^x(given that page size should be in powers of 2)

virtual address = 46bits = 2⁴⁶ bytes.

when paging is done 1st (level3) : #pages = 2⁴⁶/ 2x : 

                               page table size=  2⁴⁶/ 2^x * 4B ( 32 bits= pte size)

similarly : level 2 : page table size = ((2⁴⁶/ 2^x * 4) / 2x ) *4

               level 1 = page table size = ( ((2⁴⁶/ 2^x * 4) / 2^x ) *4 ) / 2^x ) *4

now this should be fitting in one frame = page size =2^x

 = >   (2⁴⁶ * 2⁶)/2^3x= 2^x

  => 2⁵² = 2^4x

therefore x = 13bits = 8KB.. answer.

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Thank bro

Answer 1

a simple formula :- 

let there are 'n' levels in paging.

last level page table size = $\LARGE\frac{(\;process\;size\;) \;*\; (\; PTE\;)^{n}}{ (\; Page\;Size\;)^{n}}$

given that last level page table exactly fit in one Page.

 

==>  $\LARGE\frac{(\;process\;size\;) \;*\; (\; PTE\;)^{n}}{ (\; Page\;Size\;)^{n}}$ = Page Size

process size = 2⁴⁶ B

PTE = 32 bits = 4 Bytes = 2² B

Page Size = 2^P B

n = 3

substitute you get P = 13 ===> Page Size = 2¹³ B = 8 KB