Operating System: Paging Question

Question

I have a question that I am trying to answer that gives the following situation:

16K Pages 32-bit Virtual Addresses 512MB hard disk, sector-addressable with 16K sectors 8 processes currently running

I am asked:

i) How many process page tables are required?

I think this is a trick question? Surely the answer is just 1.

ii) If a process address register PAR can be up to 32 bits, what is the maxmimum amount of physical memory that can be supported on this machine?

iii) How wide in bits should each entry in a process table be if 64MB physical memory is installed?

Please could anyone give me help/hint with the last two parts as I'm really stuck on them? Thanks!

Answer 1

In case you look on here before the exam later today, it is because it doesn't mean Process address register, it means Page address register!

Try looking at http://cseweb.ucsd.edu/classes/fa03/cse120/Lec08.pdf for some more information including help about segmentation and paging combined

Also, the book in the IC library called Operating Systems concepts with code 005.43SIL says that each process has it's own process page table and can even be segmented itself!

i) I said 8

ii) Well, 32 bits of virtual memory addressing with 14 bits of offset in the page table (2^14 = 16K page length) means there are 18 bits left for the page number. In 32 bits of PAR, this means 14 bits for the page location. If you multiple the amount of page locations by the page size, you get 2^14 * 2^14 = 2^18 which is 256MB of RAM

iii) I got 30 bits. 64MB is 2^26 divided by the page size is 2^26/2^14 = 2^12 which means 12 bits for the page location. From (ii) I calculated that 18 bits are left in the virtual memory address for the page number meaning that it should be 30 bits wide. I also put a comment that since it should be byte-aliged maybe the extra 2 bits can be used so that we know whether it has been written to and whether it is currently being stored on the disk.

Hope this helps!