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2 Answers

3 votes
3 votes
First Level = 244/214 *2 =231>214

Second Level = 231/214 *2= 218>214

Third Level = 218/214 *2= 25 < 214

So, Level 3 should be the answer.
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Total address bits available to represent virtual address space = 44 bits

Page size = 16KB

Single page table entry size = 2B

The important thing we have to note here is that the page table must fit into a single page.

  • Page table size = 16KB
  • Number of entries in the page table 
    • Page table size/size of the single page table entry
      • => 214/22 = 212
      • To address these entries we need 12 bits
  • Total Number of bits available after page offset
    • 44 - 14 = 30 bits
  • So the total number of levels required =>
    • 30/12 = 3 (round) 

So the total levels of paging required = 3.

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SHOULDN'T THE ANSWER BE 792??