Made Easy Operating System

Question

Consider a 32 bit virtual address used for paging with page size of 1024 bytes. Two level paging is implemented with equal number of entries in every page table of the system. If page table entry size is 2 bytes, what is the maximum size of main memory supported by the above system?

a)16 MB

b)32 MB

c)64 MB

d)None

Answer 1

Page table entry must contain frame number.

Answer 2

size of physical Memory Address = No.of Frames * Size of the Frame

Size of the Frame = Size of the Page ----> It is given ===> fixed

To increase No.of Frames ===> Increase No.bits represent it

Given that PTE = 2 Bytes = 16 Bits ===> PTE = Bits to represent Frame + (Some bits for additional information)

Bits  to represent Frame =16 Bits when (Some bits for additional information = 0) ----> doing this for getting more no.of bits for frames

∴ Physical Address Size = 2¹⁶ * 2¹⁰ =2⁶ MB = 64 MB, This is theoretically but practically it's not possible