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?
