Consider a physical address with a page frame size of 2kb. how many bits must be used to represent the page-frame offset of the

Question

andrew11 [14]

4 years ago

7

Consider a physical address with a page frame size of 2kb. how many bits must be used to represent the page-frame offset of the

physical address?

Computers and Technology

1 answer:

MArishka [77] · Answer 1 · 2021-12-08T06:06:35+03:00

MArishka [77]4 years ago

7 0

11 bits would be enough for 0 - 2047 decimal, 2048 (2k) possible values. 2048 itself would need a 12 bit, which would allow from 0 - 4095 possible values. Useful tidbit to remember: 1024, 1k is ten bits.