Question

A system uses a two-level paging scheme. If the logical address space is 32 bits, the page size is 4 KB, and the outer page table has 1024 entries, how many bits are used for the outer page number?

• A. 10
• B. 12
• C. 20
• D. 22

Hint:

In multi-level paging, subtract the offset bits first. Then split remaining bits among page tables. Use \(\log_2(\textnumber of entries)\) to determine bit length.

Answer 1

The Correct Option is A

iven:
- Logical address space = 32 bits → total addressable memory = \( 2^{32} \) bytes
- Page size = 4 KB = \( 2^{12} \) bytes → So, 12 bits for offset
That leaves: \[ 32 - 12 = 20 \text{ bits for the page number} \] 
Now, two-level paging splits this 20-bit page number into:
- Outer page table bits (say, \( x \))
- Inner page table bits (remaining bits = \( 20 - x \))
We are told that the outer page table has 1024 entries. So:
\[ \text{Number of entries} = 2^x = 1024 \Rightarrow x = 10 \] 
Hence, the number of bits used for the outer page number is 10.