Total address bits available to represent virtual address space = 44 bits
Page size = 16KB
Single page table entry size = 2B
The important thing we have to note here is that the page table must fit into a single page.
- Page table size = 16KB
- Number of entries in the page table
- Page table size/size of the single page table entry
- => 214/22 = 212
- To address these entries we need 12 bits
- Total Number of bits available after page offset
- So the total number of levels required =>
So the total levels of paging required = 3.