UGC NET CSE | December 2007 | Part 2 | Question: 38

Question

Average process size = $s$ bytes. Each page entry requires $e$ bytes. The optimum page size is given by :

• A. $\sqrt{(se)}$
• B. $\sqrt{(2se)}$
• C. $s$
• D. $e$

Answer 1

option b is the answer.

Reference:

http://dysphoria.net/OperatingSystems1/4_page_size.html

Answer 2

The optimum page size is optimised with bare minimum overheads. Its size in terms of page table entry and average process size is given by =square root of 2*average process sisze *page enry size. hence the option (B) is correct.