Let Logical Address Space (LAS) = L bytes
LAS (process size) = L bytes = bytes (We take as a power of 2)
Page size = P bytes = bytes (We take as a power of 2)
So,
And the number of bits for the page number in LAS = m - n
Now let consider,
Let Physical Address Space (PAS) = P bytes
Now,
And
So, the number of bits required for frame number in PAS = x - y bits = k bits.
Number of entries in the page table (PT) = number of pages present in the process
Let each entry of the page contain e bit.
So, page table size = number of pages in LAS X Entry size.
Important point:
Sometimes page entry is not given in that case we user simply frame number as page entry because the page contains frame number.
So, we get the number of bits for frame number = x-y
Though in general the page entry not only contain frame numbers it also contains some important other information which we will discuss later.