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Optimal page size=√(2*S*E)

where S=process size and E= Table entry size

so for given instance S=32MB=2^25 Byte

E=4B=2^2

so optimal size=√(2*(2^25)*(2^2)

=2^14

=2^4 KB

=16KB

so 16 should be ans.

 

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page size=p bytes

pte=e bytes     s=process size

overheads=p/2 + (s/p)* e

for minimal overhead   d(overhead)/d(p)=0  

p^2=2se=2*2^25*4=2^28    so p=2^14B=16KB
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Given, Process size =32 MB = 2^25 B

PTE= 4 B = 2^2 B.

For minimizing total overload due to page table and internal fragmentation

Page size >= Page table size.

Let page size be p

We know that, Page table size= no. Of pages * PTE

So, p >= (2^25)/p * 2^2 B

p^2 >= 2^27 B

p >= 2^(27/2) B

So, p min will be 2^14 B = 16 KB

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