# Optimizing Page Size

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Given that the logical address is "d" bits.Page table entry size is 4  bytes,what must be the optimal page size P by minimizing  the page table size overhead and internal fragmentation in paging ?

a) (4d)1/2                   b) (4*d)2

C) (8*2(2d))1/2        d) (2(d+3))1/2

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it will be D...

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.

Option D

F(o)=(V.A/P)*e +P/2

P=page size

e=page table entry

Differentiating with respect to P

F'(o)=-K*e/$p^{2}$+1/2

F'(o)=0

P=$(2*k*e)^{1/2}$         (1)

Where K=virtual address=$(2)^{d}$

Putting this value in eq (1)

Option (D) will satisfied.

## Related questions

1
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I studied from book william stalling ,it was written there if we increase the size of page then pagefault first increases and then when pagesize become size of process then pagefault decreases. Can someone explain with an example why this happens?
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