GATE - 2008 | OS | A processor uses 36 bit physical addresses and 32 bit virtual addresses

Question:

A processor uses 36 bit physical addresses and 32 bit virtual addresses, with a page frame size of 4 Kbytes. Each page table entry is of size 4 bytes. A three level page table is used for virtual to physical address translation, where the virtual address is used as follows

• Bits 30-31 are used to index into the first level page table
• Bits 21-29 are used to index into the second level page table
• Bits 12-20 are used to index into the third level page table, and
• Bits 0-11 are used as offset within the page

The number of bits required for addressing the next level page table (or page frame) in the page table entry of the first, second and third level page tables are respectively

 A 20, 20 and 20 B 24, 24 and 24 C 24, 24 and 20 D 25, 25 and 24

Solution:

Option (D) is Correct.

Virtual address size = 32 bits
From the question we can see the below info:
Physical address size = 36 bits
Physical memory size = 236 bytes
Page frame size = 4K bytes = 212 bytes
No. of bits for offset = 12
No. of bits required to access physical memory frame = 36 – 12 = 24
So in third level of page table, 24 bits are required to access an entry.

In second level page table entry -- 9 bits of virtual address are used to access second level page table entry and size of pages in second level is 4 bytes.

So size of second level page table is (29)*4 = 211 bytes.

It means there are (236)/(211) possible locations to store this page table.

Therefore the second page table requires 25 bits to address it.

the first page table needs 25 bits.