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Question

Comments

16??

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yes

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just to let you know there will be some fragmentation for 16KB as pagesize :)

Answer 1

We try to find the optimum page size due to two factors.

• Overhead due to page table entries.
• Overhead due to internal fragmentation.

Now if  we go for,

• Small page size:
  • Unused space due to internal fragmentation can be reduced. 
  • But this leads to process wanting more pages resulting in a larger page table.
• Larger page size:
  • Efficient for IO
  • Smaller page table.
  • But increases the space wasted due to internal fragmentation.

To find an optimum page size we need to minimize the overhead due to page table entries and internal fragmentation.

• If the process segment size is, S 
• Page table entry size is, E bytes
• And page size is, P
  • The average amount of internal fragmentation = P/2
  • Average number of pages per process segment = S/P
    • Each page required E bytes of the page table.
    • So page table size this process segment requires = ES/P
• Total overhead due to internal fragmentation and page table size = ES/P + P/2
  • Now we need to minimize this.Differentiate this w.r.t page size P and equate to 0
    • => -SE/P² + 1/2 = 0
    • => P = sqrt( 2*S*E )

 

Applying this here,

S = 32MB => 2²⁵

E = 4 B 

Now, P = sqrt ( 2* 2²⁵ * 2²) = 2¹⁴ = 16KB

Comments

amazing explanation!!

Answer 2

Optimal Page size = sqrt(2*2^25*2^2), should be the answer.