edited by
1,161 views
1 votes
1 votes
In a virtual memory system, size of the virtual address is 32-bit, size of the physical address is 30-bit, page size is 4 KByte and size of each page table entry is 32-bit. The main memory is word addressable and word size is 4Bytes. Find the maximum number of bits that can be used for storing other information ?
edited by

1 Answer

4 votes
4 votes
Here,

Assuming Word size = 1B (as the size is not mentioned in the question)

VAS = $2^{32}$

PAS = $2^{30}$

Page size = Frame size = 4K words = $2^{12}$

#frames = $\frac{PAS}{PS} = \frac{2^{30}}{2^{12}} = 2^{18}$

We require 18 bits to recognize every frame.

Total PTE size = #bits to recognize frames + Extra bits

Extra bits = 32 – 18 = 14 bits

Related questions

0 votes
0 votes
0 answers
1
SSR17 asked Feb 8
185 views
we have 8 pages (each side 32B) to store in physical memory of 2^32 bits how many bits are required to identify each page , according to me 3 bits are required but that i...
2 votes
2 votes
2 answers
2
0 votes
0 votes
0 answers
3
naveen81 asked Feb 1, 2017
1,071 views
2 votes
2 votes
0 answers
4