In a virtual memory system, if the page size is 8KB and the virtual address space is 2MB, how many page table entries are required?
Q1. In a virtual memory system, if the page size is 8KB and the virtual address space is 2MB, how many page table entries are required?
Answer: 256
Explanation: Number of page table entries = virtual address space / page size = 2MB / 8KB = 2048 / 8 = 256