Question

Consider a logical address space of 256 pages with a 4-KB page size, mapped onto a physical memory of 64 frames.

a. How many bits are required in the logical address?

b. How many bits are required in the physical address?

Answer 1

a):-

Logical address space =number of pages *page size

=256*4*2¹⁰B

=2⁸*2²*2¹⁰B=2²⁰B

=1 MB

to represent 1MB, logical address has to be 20 bits.

b) :-

Size of physical memory =number of frames *size of frame.

=64*4*2¹⁰B

=2¹⁸B

=256 KB

To rereprese 256KB, physical address has to be 18 bits.

Note that frame size = page size and problem is done by assuming system is byte addressable.

"if you have any doubts, please use comment section. "

Answer 2

Answer:

4KB page size, 2^12 bits.

Page Size + Number of (Pages or Frames)
a) 256 Pages = 2^8 pages, 12 + 8 = 20 bits
b) 64 Frames = 2^6 Frames, 12 + 6 = 18 bits