GATE | GATE-IT-2004 | Question 66

In a virtual memory system, size of virtual address is 32-bit, size of physical address is 30-bit, page size is 4 Kbyte and size of each page table entry is 32-bit. The main memory is byte addressable. Which one of the following is the maximum number of bits that can be used for storing protection and other information in each page table entry?
(A) 2
(B) 10
(C) 12
(D) 14

Answer: (D)
Explanation:
Virtual memory = 2³² bytes
Physical memory = 2³⁰ bytes

Page size = Frame size = 4 * 10³ bytes = 2² * 2¹⁰ bytes = 2¹² bytes

Number of frames = Physical memory / Frame size = 2³⁰/2¹² = 2¹⁸

Therefore, Numbers of bits for frame = 18 bits

Page Table Entry Size = Number of bits for frame + Other information
Other information = 32 – 18 = 14 bits


Thus, option (D) is correct.


Please comment below if you find anything wrong in the above post.

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