In a virtual memory system, page size is 4KB. If a process has a virtual address space of 1MB, how many page table entries are required?
Q1. In a virtual memory system, page size is 4KB. If a process has a virtual address space of 1MB, how many page table entries are required?
Answer: 256
Explanation: Number of page table entries = virtual address space / page size = 1MB / 4KB = 1024 / 4 = 256