UGC-NET COMPUTER SCIENCE OPERATING SYSTEM QUESTIONS WITH EXPLANATION

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q :  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) x/2

(B) xz/2

(C) √(2xz)

(D) √(xz)/2

(UGC-NET Computer Science December 2014 )

Answer: (C)

Explanation:

Important consideration is size of pages to use. OS often has choice in this matter---many MMUs allow various different page sizes.

• Small size: less wasted space due to internal fragmentation; fewer unused pages in memory.
• Large page size: more efficient disk I/O, smaller page tables

If average process segment size is x; page table entry size is z bytes; page size is y,

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

Average number 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 minimize the overhead, differentiate with respect to page size, y, and equate to 0:

-xz/y² + 1/2 = 0
=> y = √(2xz)

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

Note: that this calculation ignores the need to keep page sizes large in order to speed up paging operations; it only considers memory overheads.


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