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Main-Memory Size=64MB, Logical address-32bits long, Page size-4KB,

What is the total space wasted in maintaining the page table?

  Soln- 220(No. of pages) * 2B(14bits~2B)(frame-address-bits) = 2MB

In this question why we consider the total no of pages in the secondary memory as the number of  entries in the page table? 

The Main Memory can hold only that much pages as much frames it has.

So, shouldn't we consider the no. of frames in main memory as the no. of page table entries?

I don't know why such confusions come in such a crucial time! Please answer and clear my confusion asap!

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Page table size =  Total no of page table entries * single entry size ( no of bits required to represent main memory).

                       = 2^20 * 26 bits
Note:- Each page table contains information about all page entries of a process present in the virtual memory.

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