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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
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In a virtual memory system, size of virtual address is 32-bit, size of...
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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In a virtual memory system, size of virtual address is 32-bit, size of...
Virtual Memory System with 32-bit Virtual Address and 30-bit Physical Address

Virtual address size = 32-bit
Physical address size = 30-bit
Page size = 4 Kbyte
Size of each page table entry = 32-bit

To determine the maximum number of bits that can be used for storing protection and other information in each page table entry, we need to calculate the number of bits required to represent the page offset and the page number.

Page Offset Calculation

Since the page size is 4 Kbyte, the page offset requires 12 bits (2^12 = 4 Kbyte). Therefore, the lower 12 bits of the virtual address are used to represent the page offset.

Page Number Calculation

To determine the number of bits required to represent the page number, we need to calculate the maximum number of pages in the virtual address space and physical address space.

Maximum number of pages in virtual address space = (2^32) / (2^12) = 2^20
Maximum number of pages in physical address space = (2^30) / (2^12) = 2^18

Since the virtual address space has a larger number of pages, we need 20 bits to represent the page number in the virtual address space. However, since the physical address space has a smaller number of pages, we only need 18 bits to represent the page number in the physical address space.

Therefore, the remaining 32 - 12 - 20 = 0 bits are available for storing protection and other information in each page table entry.

Answer: d) 14
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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)2b)10c)12d)14Correct answer is option 'D'. Can you explain this answer?
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