24 views
Consider a system with a $2$ $level$ paging applicable the page table is divided into $8K$ pages each of size $16KB$. The memory is Byte addressable if the physical address space is $128MB$ which is divided into $4KB$ frames. The PTE size of the outer page table is $32bits$ and the page table table entry size of the inner page table is $64bits$. What will be the page table size of the inner and outer page table?

edited | 24 views
0
0

What a question. Not only is the language hard to understand - but is page size really different from frame size ??

2 level paging applicable the page table is divided into 8K pages each of size 16KB

What ? Its a weird piece of information to provide in a question. Which page table ? There are 2 levels.

The memory is Byte addressable if the physical address space is 128MB which is divided into 4KB frame.

What ?? So the memory is not byte addressable if the physical address space is not 128 MB ? The grammar used leaves the question open to all kinds of interpretation.

This is an awful question - don't waste your time. There are tons of better questions out there testing multi-level paging. Best way to forcefully learn bad ideas and stress yourself.

0

Yah

I am also having trouble understanding the language, and let alone solve it

The original source https://gateoverflow.in/93582/paging

+1

Going by the interpretation of the folk who have claimed to solve it in that link, here is my guess:

'Page table' (as a whole) is divided as follows : 8K pages, each with 16K entries (not 16KB as given in the question).

Interpret this from the CPU looking in to the page tables - you have a logical addres, now you will pick one of the 8K pages of the outer table, which will index you into a 2nd level page table. Then that 2nd level page table will have 16k entries.

Going by this, VAS will be

 1st Level PT 2nd Level PT Page Offset 8K = 13 bits 16K = 14 bits 4KB = 12 bits because common sense dictates page size = frame size

Then,

Size of 1st level page table = $2^{13}*32\:bits = 32KB$
Size of 2nd level page table = $2^{14}*64\:bits = 128KB$

I'm sure this matches with one of the given 'options'.

But wait - there's more. Notice how the page size itself is $4KB$. The whole point of multilevel paging is to fit page tables into a single page. Clearly the above sizes are incongruent.

I think some of my brain cells just died.

0
I think it is best to leave such ambiguous questions
if your brain cells died and as I am pretty muck weak in this area, mine will explode I guess.

Anyway, Thanks.. :)