UGC NET CSE | December 2014 | Part 2 | Question: 39

Question

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

• A. $\frac{x}{2}$
• B. $\frac{xz}{2}$
• C. $\sqrt{2xz}$
• D. $\frac{\sqrt{xz}}{2}$

Answer 1

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.

Comments

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