Madeeasy test series

Question

In a virtual memory system, size of virtual address is 40 bit, size of physical address is 32 bit, page size is 8 kb and size of each page table entry is 40 bit. Assume the main memory to be Byte addressable. The maximum number of bits that can be used for storing protection and other information in each page table entry are ___

Comments

21 ?

Answer 1

Page table Entry contains additional bits also along with frame bits.(PTE bits = Frame bits  + Other bits )

No of frames = $\frac{PAS}{PS} =\frac{2^{32}}{2^{13}} = 2^{19}$

So,19 bits are required for frames , 

No.of bits for PTE = 40.

So,No of bits for additional information = 40-19 = 21.

Comments

Here page size is 8kb and small b ('b') represents bit not byte.
So i think answer would be 18.

Correct me if I am Wrong

Answer 2

Given page size (and memory is byte addressable) 8KB=8k*1B

therefore 2^13(8K) no of locations in each page (it is offset=13 bit)

Virtual address 40bit(x+13)

x=27 (No of page table entries in page table)

physical address 32(y+13)

y=19 no of bits required to address a frame

each page table entry contains 40 bit(bits required to address a frame+extra information)

-> 40bit=(19bit+extra)

extra =21 bit for other information in page table

Comments

Here page size is 8kb and small b ('b') represents bit not byte.
So i think answer would be 18.

Correct me if I am Wrong