# GATE | GATE CS 2008 | Question 67

• Difficulty Level : Hard
• Last Updated : 14 Feb, 2018

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

Explanation: Virtual address size = 32 bits

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

Physical address size = 36 bits

Physical memory size = 2^36 bytes

Page frame size = 4K bytes = 2^12 bytes

No. of bits for offset (or number of bits required for accessing location within a page frame) = 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.

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 (2^9)*4 = 2^11 bytes. It means there are (2^36)/(2^11) possible locations to store this page table. Therefore the second page table requires 25 bits to address it.

Similarly, the first page table needs 25 bits to address it.

My Personal Notes arrow_drop_up