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.