doubt os

Question

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Same thing right? :P They have given 4096 bytes. I said 4KB

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oh :P i didn't see :P

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plzzz explain

Answer 1

Now, we calculate total size.

Total page table entries at 3rd level = 8 pages * (2⁶ entries per page) = 512

Total page table entries at 2nd level = 1 page * (2⁸ entries per page) = 256

Total page tabl entries at 1st level = Total page table entries at 2nd level = 256

$\therefore$ Total entries = 512 + 256 + 256 = 1024 entries

Each page table entry occupies 4B

$\therefore$ Total size of page table for the process = Total page table entries * size of Page table entry = 4KB

Comments

thanks a ton :)