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Consider a processor has virtual address 52 bits, the physical address is 32 bits and the page size is 16KB. Each individual table at any level of the page table occupies a single page of physical memory. Using multi level paging, how many entries are there in each individual page table page. Consider page table entry requires additional 14 bits for the special purpose?

  1.    < 2, 12, 12, 12 >
  2.   <10,14,14,14>
  3.    <4,14,14,14>
  4.    <10,12,14,14>

1 Answer

3 votes
3 votes

virtual address = 52 bits

page size =  16kB = 14 bits

no. of pages = 238

hence sum of all bits (in ans) = 38

also for frames , 

no. of frames = 32-14=218

page table entry = 18+ (additional) 14 bits = 32 bits = 22B

in each page = 214/22=212

i.e. at max there can be 212 pages in any page table .

hence, ans is option A)

< 2, 12, 12, 12 >

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