Question

Consider a logical address space of 512 pages with a 4-KB page size, mapped onto a physical memory of 256 frames. How many bits are required in the logical address? How many bits are required in the physical address?

Answer 1

A)
Given page size= 4KB=2^12 bytes
offset bits=12 bits
given logical address=512 pages
512=2^9
bits for page no = 9 bits
therefore no of bits required = bits for page no+offset bits
=9+12
=21 bits.
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B)
Given page size= 4KB
so total size is = 256*4 KB(frames * page size)
=(2^8) * (2^2)*(2^10)
=2^20
total bits in physical address= 20 bits.
  

bits required in logical address is =21 bits.
bits required in Physical address is=20 bits.

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