Virtual Memory

Question

A computer uses 30 bit physical address, 38 bit virtual address and uses 2-level paging. The page table base register stores the base address of the first level, Each page table, occupied exact one page. Each entry of first level store base address of second level table and each entry of second level table store a page table entry which is of size 32 bits. The size of page in KB in computer is ________.

Comments

is answer $16KB$?

Answer 1

actually, my convention is go outer page table is n, inner page table is n-1. and so on... ( process --> PT₁ ---> PT₂ --->....PT_n  but note that given notation is process ---> PT_n ---> PT_n-1 --->...PT₁ )

According to my convention, the formula

Page Table size at level n = $\frac{process\:  size \: * \: PTE^n}{Page\:  size^n}$

Given that Outer page table fit in 1 Page ===> size of outer page table = Page size

$\frac{process\:  size \: * \: PTE^n}{Page\:  size^n}$ = Page size

===> Page size ⁽ⁿ⁺¹⁾ = Process size * PTEⁿ

substitute the values according to your question

Page size ⁽²⁺¹⁾ = 2³⁸ * 4²

Page size³ = 2³⁸⁺⁴ 

Page size³ = 2⁴²

Page size = 2^42/3 =  2¹⁴ =16 KB

Comments

https://gateoverflow.in/379/gate2013-52

sachin sir answer 

Answer 2

A more descriptive approach I read somewhere,