Question

In a virtual memory system, size of virtual address is 32-bit, size of physical address is 30-bit, page size is 4 Kbyte and size of each table entry is 32-bit. The main memory is byte addressable. Which one of the followiong is the maximum number of bits that can be used for storing protection and other information in each page table entry?

Answer 1

In page table, number of entries equal to number of pages in virtual memory and page table entry is equal to number of frames and other information i.e.

So, number of frames $=\frac{Physicalmemory}{Pagesize}$
$=\frac{2^{30}}{2^{12}}=2^{18}$
So, number of bits for frames $=lo{g}_2\left(2^{18}\right)=18$
We know that,
Page table entry = Number of bits for frames + Other information
32 = 18 + x
x = 32 - 18
x = 14

So, 14 bits can be used for storing protection and other information.

Note : Each page table entry must contain frame number.