Hi im reading the Silberschatz book on OS concepts.

I tried to do an exercise on memory management:

physical address = 32 bit;

physical address space is 4 times smaller than the logical address space;

if a page is 2^22 byte,

what is the number of the entries in the page table of a process?

What is the number of frames?

 

To solve this exercise i do this:

physical address = number of frames + offset

number of frames = physical address - offset = 32 - 22 = 10

now i know the number of frames that is 2^10.

A frame has the same size of a page so the physical address space is

2^22 * 2^10 = 2^32 byte.

The logical address space is 4 times bigger than the physical and

2^32 * 4 = 2^32 * 2^2 = 2^34 byte

The number of entries of a page table of a process is 2^34/2^22 = 2 ^12 entries.

The book doesn't do examples on how to solve this kind of exercises, it's only explain a lot of theory.

I don't know if my solution is correct, what do you think about?

looks okay to me...