Q39: Bits for page table entry (PTE)

Posted on by bytesoftheday

Question 39: A machine has 48-bit virtual address and 32-bit physical addresses. Pages are 8 KB. How many bits are needed for the page table entry?

Options:

  1. 20 bits
  2. 24 bits
  3. 34 bits
  4. 48 bits

 

Solution:

For an 8 KB page, 14 bits would be used for offset within the page (2^14 = 8K). Hence, out of 48, remaining 48-14 = 34 bits would be needed for the page table entry. Therefore, the correct answer is option 3.