Question

Consider a virtual memory system with the following properties: 36 bit virtual byte address, 8 KB
pages size, and 32 bit physical byte address. Please explain how you determined your answer.
a. What is the size of main memory for this system if all addressable frames are used?
b. What is the total size of the page table for each process on this processor, assuming that the
valid, protection, dirty, and use bits take a total of 4 bits and that all the virtual pages are in
use? (Assume that disk addresses are not stored in the page table).

Answer 1

a. What is the size of main memory for this system if all addressable frames are used?

Main memory size = 2^32 = 4GB

b. What is the total size of the page table for each process on this processor, assuming that the

valid, protection, dirty, and use bits take a total of 4 bits and that all the virtual pages are in

use? (Assume that disk addresses are not stored in the page table).

8KB = 2^3 x 2^10 = 2^13
Total number of bits in virtual page number = 36 - 13 = 23

total size of the page table = 2^23 x 1/2 byte => 2^22 = 4MB

Thanks, PLEASE COMMENT if there is any concern.