Determine page table size for virtual memory

Question

Consider a virtual memory system with a 38-bit virtual byte address, 1KB pages and 512 MB of physical memory. What is the total size of the page table for each process on this m

Answer 1

1KB pages = 2^10, 512MB = 2^29 => Offset = 29 - 10 = 19 bit.

virtual includes two part: page frame + offset => page frame + dirty bit = 38 - 19 = 29 bit. 29 bit includes 4 bit dirty (above) => 25 bit for real page frame, each page frame has 10 bit long.

So, page table size: 2^25 * 10 = 320M.

Hope this correct.