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

frame Size = Page Size = 4Kbytes = 2'12 bytes

Physical Address Size = 2'30 bytes

number of frames = Physical Address/Frame Size

                          = 2'30/2'12=2'18 frames

numbers of bits required for frame = 18 bits

Page Table Entry Size =  number of bits for frame  + other information

other info = 32 - 18 = 14 bits

So Ans is (d)

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

@raushan sah

It is already given in the question ('size of each page table entry is 32-bits').It is 32 bits.

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2 level paging is used here right?

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