Question

Consider a computer system with 40-bit virtual addressing and page size of sixteen kilobytes. If the computer system has a one- level page table per process and each page table entry requires 48 bits, then the size of the per-process page table is _____ megabytes.

• A. 384

Answer 1

The correct option is A 384
Size of memory = 2⁴⁰
Page size = 16KB = 2¹⁴
Page table size = Number of entries in page table $\times$ Page table entry size
= $\left(\frac{2^{40}}{2^{14}}\right)\times 48bits$
= $2^{26}\times 6bytes$
= 64 M $\times$ 6 B
= 384 MB