Size of page table - Example - Operating System

Discussion

• Nirja Shah   -Posted on 21 Nov 15
  - A page entry is used to get the address of physical memory.
  
  - At this stage it can be assumed that single level of Paging is happening.
  
  - Hence, the resulting page table will contain entries for all the pages of the Virtual address space.
  
  Number of entries in page table = (virtual address space size)/(page size)
  
  - By using the above formula we can say that there will be 2∧(32-12) = 2∧20 entries in page table.
  
  - The no. of bits required to address the 64MB Physical memory = 26.
  
  - So there will be 2∧(26-12) = 2∧14 page frames in the physical memory.
  
  - The page table needs to store the address of all these 2∧14 page frames.
  
  - Therefore, each page table entry will contain 14 bits address of the page frame and 1 bit for valid-invalid bit. Since memory is byte addressable.
  
  - So we take that each page table entry is 16 bits i.e. 2 bytes long.
  
  Size of page table = (total number of page table entries) *(size of a page table entry)
  = (2∧20 *2) = 2MB
  

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