Explain how the address below is come up with, and explain formulas for the particular numbers?

 

Even two-level paging scheme not sufficient If page size is 4 KB (22) Then page table has 252 entries If two level scheme, inner page tables could be 210 4-byte entries Address would look like outer page P 42 inner page P2 10 offset d 12 Outer page table has 242 entries or 244 bytes One solution is to add a 2nd outer page table But in the following example the 2nd outer page table is still 234 bytes in size And possibly 4 memory access to get to one physical memory location