Consider a page size of 16384 bits and page table entry size is 2 byte. What is the number of levels of paging required to map a 32 bit virtual space if a page table can be fit into a single page?
(A) 1 levels
(B) 2 levels
(C) 3 levels
(D) 4 levels


Answer: (C)

Explanation: Virtual address = 2³²
Page size = 16384 bits = 2048 byte
Page table entry size = 2 Byte.
Number of page table entries = virtual address space / page size
2³² / 2¹¹ = 2²¹
1^st level page table = Number of page table entries * Page table entry size
= 2²¹ * 2B = 2²² = 4 MB
Since 1^st level page table is greater then page size, thats why we need further level of paging.
2^nd level page table = (First level address space /page size) * Page table entry size
= 2²² / 2¹¹ * 2B = 4KB
Since 2^nd level page table is greater then page size, thats why we need further level of paging.
3^rd level page table = (Second level address space /page size) * Page table entry size
2¹² / 2¹¹ * 2B = 4B.
Since 3^rd level page table is less than page size, that\’s why we don\’t need a further level of paging.
So, option (C) is correct.

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