Physical address space =64MB
No. of bits to represent the physical memory =26 bits
Logical address space =32 bits
page size =4 KB
So , the number of pages =2^32 / 2^12 = 2^20
Concept : Page table only consist of frame number . So , page table size will depend on frame number and we don’t know how many bits do we require to represent the frame number in page table.
We know that , page offset = frame offset and page offset =12 bits
bits req to represent the frame number = 26 -12 =14 bits
Additionally , In page table we have valid and invalid bit , for that we need 2 bits to represent valid (0) or invalid(1)
The bits now required in page table are = 14 + 2 bits =16 bits = 2 bytes
Total size of page table = 2^20 * 2 bytes = 2 MB