Ace Test Series: Operating System - Virtual Memory And Paging

Question

 

Comments

36... Entries in 1st level page table = the virtual address has that much pages

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how 36???? can explain

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@altamash, i added the answer

Answer 1

Size of Physical address Space = no.of frames * Size of Page

2³⁰ = 4K * P ===> P = $ \frac{2^{30}}{2^{12}} $ ===> P = $ {2^{18}} $

 

Size of the Logical Address Space = no.of pages in it * Size of Pages

                                             = 256 K * $ {2^{18}} $ = $ {2^{18}} * {2^{18}} $ = $ {2^{36}} $

therefore size of logical Address = 36 bits.

 

Note that

 no.of pages in a Process = no.of entries in 1st level page table

Answer 2

page index

page offset

Virtual address bits can be divided into two parts page index and page offset bits.

Page index bits are used to index into the page table.

There are total 256k entries in the page table.

So the number of page index bits=18bits(as 256k=2^18)

Size of the page frame=2^30/2^12=2^18

We know the size of the page is equal to size  of page frame.

So the size of the page = 2^18

So the number of bits for page offset=18 bits

Total number of bits in the virtual address=18+18=36bits