Paging- Memory Management

Question

A system support 32 bit virtual address with a page size of power of 2. Page table entries 32 bit wide. The page size in KB of the system should be ____ in order to make sure that the page table fits exactly in one frame of memory.

Comments

128KB?

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please explain

Answer 1

Let page size(also block size) be 2^x.

Number of frames = Virtual Address Space / Page Size

Assuming single level paging,byte addressable and the given condition,

Page Table Entry * Number of Frames <= Page Size

or, PTE * Virtual Address Space / Page Size  <= Page Size

or, (Page Size)^2 >= PTE * Virtual Address Space

or, 2^(2x) >= 2^2 * 2^32

or, 2x >= 34

or, x >= 17  Therefore x should be atleast 17.

Minimum required Page Size = 2^x= 2^17= 128 KB (Answer)

Any page size >= 128 KB is possible.

Answer 2

Assume page size=2^x

No of entries in page table=2^(32-x)

Now it is given that page table should  fit in one frame

So,    2^(32-x)*4=2^x

x=17

128KB