paging

Question

Comments

20 bit??

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Yes. please explain

Answer 1

Let P be page size

PAGE table size=no of page *PTE size={2^32/p}*3B

p=2^32*3B/24*2^20=512 B=P

 

system has 2K support so 2*2^10*2^9=2^20=2 MB=20 bit

Comments

please explain this part in 2nd line : 2^32/p*3B

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@ Parshu gate

simple(  Logical address space/page size)*page table entry size

got it

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Given VA= 32bit, e= 3B, PTS= 24 MB, No of frames = $2^{11}$

PA= f+d

f is 11 bit, we have to calculate page offset d now

PTS= No of pages * e

No of pages= 22MB/3B = 8M = $2^{23}$

Page Size= VAS/No of pages = $2^{32}/2^{23}$= $2^{9}$, d =9 bit

PA = f+d = 11+9 = 20 bit

Answer 2

We have 32 bit virtual address,

to know the number of bits required for identifying page number we need to calculate the number of entries in the page table

Page table entries = Size of Page table / size per entry = 2MB/3bytes = 2²³ entries

so from that we get the 23 bits for Page number and remaining for offset = 32 - 23 = 9

offset bits are same for physical and virtual address hence,

we have 2k frames and to identify 2k (2¹¹) frames we need 11 bits

physcial address = 11 bits + 9(offset) = 20 bits