In a virtual memory system, size of virtual address is 32-bit, size of…
2004
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
Attempted by 119 students.
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Correct answer: D
Key insight: compute the page offset, then determine how many bits the physical frame number occupies, and subtract that from the page table entry size.
Page offset bits = log2(page size) = log2(4096) = 12 bits.
Physical frame number bits = physical address bits - offset bits = 30 - 12 = 18 bits.
Each page table entry is 32 bits. The bits required to store the frame number are 18, so the remaining bits available for protection and other information = 32 - 18 = 14 bits.
Answer: 14 bits.