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4 votes
4 votes

For the implementation of a paging scheme, suppose the average process size be $x$ bytes, the page size be $y$ bytes, and each page entry requires $z$ bytes. The optimum page size that minimizes the total overhead due to the page table and the internal fragmentation loss is given by

  1. $\frac{x}{2}$
  2. $\frac{xz}{2}$
  3. $\sqrt{2xz}$
  4. $\frac{\sqrt{xz}}{2}$
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1 Answer

5 votes
5 votes

Average amount of internal fragmentation per segment is y/2.

Average no of  pages per process segment is x/y.

Each page requires 'z' bytes of page table.

so each process segment requires page table size of xz/y bytes.

Total overhead per segment, therefore, due to internal fragmentation and page table entries, is

    xz/y + y/2

To minimise the overhead, differentiate with respect to page size, p, and equate to 0:
d(xz/y + y/2)/dy =    -xz/y2 + ½ = 0
    => y = √(2xz)

Answer (C)

So for example, if the average segment size were 256K, and the page table entry size were 8 bytes, the optimum page size, to minimise overhead due to page table entries and internal fragmentation, would be sqrt(2 × 256K × 8) = 2048 = 2K.


1 comment

Can you please post the resource you used to solve this question?

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