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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 ________.

### 1 comment

is answer $16KB$?

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

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 = 242/3 =  214 =16 KB

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A more descriptive approach I read somewhere,