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In a virtual memory system, size of the virtual address is $32$-bit, size of the 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?

  1. $2$
  2. $10$
  3. $12$
  4. $14$
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Answer is (D).

Page table entry must contain bits for representing frames and other bits for storing information like dirty bit, reference bit etc.

No. of frames (no. of possible pages) $= \frac{\text{Physical memory size}}{\text{Page size}} = \frac{2^{30}}{2^{12}}= 2^{18}$

$18+x=32$     (PT entry size=$32$ bit)

$x =14$ bits

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