For the implementation of a paging scheme, suppose the average process size be x bytes, the page size be y bytes, and each entry requires z bytes. The…
For the implementation of a paging scheme, suppose the average process size be x bytes, the page size be y bytes, and each 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 :
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Correct answer: (C) √(2xz)
Explanation
Average internal fragmentation per process is y/2. A process has x/y pages and each page-table entry needs z bytes, so the page table takes xz/y bytes. Total overhead = xz/y + y/2. Setting d/dy (xz/y + y/2) = –xz/y² + 1/2 = 0 gives y = √(2xz).