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 page table entry is 32-bit. The main memory is byte addressable. Which one of the following is the maximum number of bits that can be used for storing protection and other information in each page table entry?

Answer 1

Answer: 14 Bits

Explanation:

Each page Table Entry contains bits for representing frame number to which a page maps ,along with that it contains other bits for storing information like dirty bit, reference bit ,flag bits etc.

Number of frames =  $\frac{Physical Memory Size}{Page Size}$

                              = $\frac{2^{30} }{2^{12} }$ = $2^{18}$

So any frame number can be represented by at most 18 bits

But, we know each page table entry is of 32 bits

This implies, maximum number of bits that can be used for storing protection and other information in each page table entry is :

(32-18) = 14 bits.