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6 votes
6 votes

What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid and invalid bit) and a page size of 512 bytes?

  1. $2^{11}$
  2. $2^{15}$
  3. $2^{19}$
  4. $2^{20}$
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5 Answers

Best answer
15 votes
15 votes

without valid/invalid bit

page table entry size is 11 bits  to address..page frames...11 bits means 2^11 frames available...

no. of frames = physical address space/frame size

frame size is same as page size

 physical address space = no. of frames* frame size

       = 211* 512 byte = 220 B

valid/invalid bit included...10 bits are remaining...for addressing ...page frames..1bit to identify..page is valid or invalid...

 physical address space = 210 * 512 byte = 219 byte....

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5 votes
5 votes
Out of 11 bits take one bit for valid and invalid bit.

so 10 bits left

Hence, 2^PA = 2^10 * 512

PA = 19 bits
4 votes
4 votes
Page table entry=frame bits+others=11 bit

                             =10 frame bits+1 (V/I bit)

So f=10 bit and

 Page size=512 B ; so d=9 bit

So PA=f+d=10+9=19 bit

So, PAS=2^19

Hence,Option C is the Correct Ans
Answer:

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