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https://gateoverflow.in/872/gate2002-19

IN THE SOLUTION GIVEN BY Akash Kanase  I am not getting the reason behind  using  this formula

No. of bits in first level page table to address a second level page table is log2 of

Physical memory size /(#Entries in a Second level page table × PTE size) ...why we are using PA

AND HOW THE MAXIMUM NUMBER OF PAGETABLES IN 2ND LEVEL CAN BE 220 ....??

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Question can be solved without that formula too!

PTE = 4B = 32bits

A PTE contains frame number + protection & other info, so if we first find Frame number, which is

Frame Number = Physical Address - Page offset

= 32 - 12 = 20.

Now, PTE = Frame Number + protection & other info

32   =           20               + protection & other info

protection & other info = 12 bits.
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log2 of Physical memory size /(#Entries in a Second level page table × PTE size)

in-direct formula for finding no.of bits required for identifying a frame

How?

(#Entries in a Second level page table × PTE size) = page size

Physical memory size/ page size = no.of frames

log2 ( no.of frames ) =  no.of bits required for identifying a frame

HOW THE MAXIMUM NUMBER OF PAGETABLES IN 2ND LEVEL CAN BE 220

i didn't get this

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OK THANKS FOR FORMULA  I GOT ...........BUT AT MADEEASY TEACHER TOLD THAT THE NUMBER OF BITS REMAINS SAME WHATEVER BE THE PAGING LEVEL SO GOT CONFUSED NOW CLEAR

AND YOU DIDNT GET HOW THE MAXIMUM NUMBER OF PAGETABLE IN 2ND LEVEL CAN BE 220 OR U DIDNT GET WHAT I AM ASKING???

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U DIDNT GET WHAT I AM ASKING???

yes

MAXIMUM NUMBER OF PAGETABLE IN 2ND LEVEL

check this term

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JUST LOOK THE DIAGRAM IN THE ANSWER OF THE QUESTION ....iT IS WRITTEN THAT UPTO MAXIMUM 220

2ND LEVEL PAGETABLES ARE POSSIBLE....

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now i understood your question but i didn't understand that point in the answer.

i hope it would be

UPTO MAXIMUM 220 2ND LEVEL PAGE TABLES ENTRIES ARE POSSIBLE due to only 220 frames in the memory

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YA SAME I WAS THINKING