Let X=page table size overhead + internal fragmentation overhead
We need to minimize X.
If we take very large P i.e. large page size then chances of internal fragmentation will be high.
If we take very small P then no. of pages required for a process will be high which means no. of page table entry will increase which in turn means the page table size will increase.
We need to manage both.
If P is the page size then log(P) is the page offset bits. Since logical address bits is "d" so no. of bits for pages= (d-logP).
So no. of pages= 2(d-logP) = 2d/2logP = 2d/P
Size of page table= (2d/P)*4 B
internal fragmentation in page table:
Min internal frag=0 i.e. when the page table is full
Max internal frag=P-4 i.e. when only one entry(i.e. of size 4B) is present in page table (it can be approximated to P)
So average internal frag=0+P/2=P/2.
X= (2d/P)*4 + P/2
To minimize X(i.e. overall overhead) we need to find dX/dP
dX/dP= (-4*2d)/P2 + 1/2
dX/dP=0 then P= (2(d+3))1/2
Since d2X/dP2 > 0 at P =(2(d+3))1/2 so we know that X is minimum of that value of P.