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?
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.
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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
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