NAT
Q.
Consider a system with 32-bit virtual addresses and a page size of 8 KB. The system uses a two-level paging scheme where the first-level page table has 1024 entries. Each page table entry is 4 bytes. What is the size (in KB) of the second-level page table?
Correct Answer: 2
Virtual address space = 2^32 bytes. Page size = 8 KB = 2^13 bytes. Thus, total pages = 2^32 / 2^13 = 2^19. First-level page table has 1024 = 2^10 entries. Each entry points to a second-level table. Total pages covered per second-level table = 2^19 / 2^10 = 2^9 = 512 pages. Each second-level table must have 512 entries. Each entry = 4 bytes, so size = 512 × 4 = 2048 bytes = 2 KB.