displacement offset

Question

A paging system uses 32-bit address, each of which specifies 1 byte of memory. The system has a main memory unit of 1024 MB, and a page size of 32 KB. How many bits does the system use to maintain displacements (offset)?

Comments

I know that..

how do you come know that displacement offset  = no.of pages.

any reference, please ?

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It should be 15.It cannot  be 17.page size 32kb = 2^15.

Address size 32 bits.This is how the virtual address will be decoded.

|17 bits --page nos for a process | 15 bits -- offset or displacement|

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I think displacement offset is nothing but number of bits required to represent number of pages..

because displacement we usually calculate in page accessing and there is some limit like zero to 199 etc

so to represent that 200 pages what will be number of bits require that is displacement offset..

In given question number of pages require (32-15) 17 bit so 17 is answer..