How can we determine Intel Pentium page table size?

Question

Let's suppose we have a computer with Intel Pentium CPU. Its OS implements virtual memory with paging. If a process has access to 16 MBytes of memory in total, how can I calculate the minimum (best case) and the maximum (worst case) size of page tables in kBytes?

My calculations so far: Minimum - Best case: Size of PageTable (bytes) = Number of Entries in PageTable * Size of Entry => Size of PageTable (bytes) = 2^20 * 2^10 = 2^30 bytes

Maximum - Worst case: Size of PageTable (bytes) = Number of Entries in PageTable * Size of Entry => Size of PageTable (bytes) = 2^20 * 2^14 = 2^34 bytes

Are those calculations correct and if yes how can I proceed to find the total size of page tables?

Thanks!

Answer 1

This will solve your problem

Getting Intel Pentium page Table Size