Given a logical address of 16 bits with 4 KB page size, how many bits are used…
2025
Given a logical address of 16 bits with 4 KB page size, how many bits are used for the page offset?
- A.
6 bits
- B.
10 bits
- C.
12 bits
- D.
8 bits
Attempted by 34 students.
Show answer & explanation
Correct answer: C
Page size = 4 KB = 4096 bytes
4096 = 2¹²
Therefore, number of bits needed for page offset = 12 bits
In paging:
Page offset bits = log2(page size in bytes)
So:
log2(4096) = 12
Final Answer: 12 bits