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?
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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