Ace Test Series | Operating System | Paging

Question

A machine supports a 48 bit virtual address and 32 bit physical address. Pages are 8KB. If the page table entry contains besides other information, one V/I bit, 1R, 1M and 2 bits for page protection. What is the page table size in bytes?

1. 86 MB
2. 128 MB
3. 96 MB
4. 512 MB

Answer 1

There is mistake in question.According to the  question page table size should be 3*2^35B which is not in option.

But if we assume VA is of 38 bit then page table size will be 96MB .

So I think there is typo in question.

Comments

96MB is correct

VA= 48 bit

PAGE SIZE= 2^13 so no. of pages = 2^35

physical addrss=32 bits so no. of frames= 2^19

now table entry sie will be = 19+1+1+1+2=24bits =3byte

page table sie= no. of pages *entry size

so it will be 2^35*3= 96 MB