How many bits required for page frame number on 32-bit system with 2^20 Bytes physical memory and 4KB page size?
is it 11 bits??
Need an answer please help! Thank you.
If your page size is 4K, then that means the page frame number must specify all but the low 12 bits of the page address (4K == 2^12). So, in a 32-bit system, the page frame number will generally specify the upper 20 bits (32 - 12).
Strictly speaking since you only have 2^20 bytes of total memory, the lowest 8 (20 - 12) of those is sufficient to unambiguously define which page is being referenced, but it's likely that your system is designed to support more than 2^20.