Consider a machine with 64 MB physical memory and a 32-bit virtual address space. If the page size is 4 KB, what is the approximate size of the page table?
A |
16 MB |
B |
8 MB |
C |
2 MB |
D |
24 MB |
No. of entries in pge table = 232/ 212 = 220
Frame size = 226 / 212 = 214
PT have to be stored in one frame so entry size must be 2 bytes, hence size of PT = 220 * 2 = 2 MB