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.
Note : This question was asked as Numerical Answer Type.
Explanation: Size of memory = 240
Page size = 16KB = 214
No of pages= size of Memory/ page size = 240 / 214 = 226
Size of page table = 226 * 48/8 bytes = 26*6 MB =384 MB
Thus, A is the correct choice.