Linxu Consider a physical address with a page frame size of

Linxu..

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?

A. 8

B. 10

C. 12

D. 14

Answer:

Page size = 2KB = 2 * 2^10 = 2^11 = 11 Bits must be used to represent the page offset of the physical address.

We know that page offset = log[ page size] = log [ 2^11] = 11 *log2 = 11.