GATE IT 2004 | Question: 66

Question

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?

• A. $2$
• B. $10$
• C. $12$
• D. $14$

Comments

no. of bits for frames = 18 , so leftover => 32-18 = 14 

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in this question the size of each entry is already given 32 bits and it is asking about other than the storage of frame how many bits required of additional storage.

so for frame number=2^30/2^12=2^18

size of frame allocation frames is 18

So additional storage = 32-18= 14 bits
 

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References : OS book Tanenbaum 

Structure of a Page Table Entry

Answer 1

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

Comments

2 level paging is used here right?

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Why do you think so rohith1001 that 2 level paging is implemented here?

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@richa116 I'm also having the same doubt as you. Could you please tell any insights found in this gap?