UGC NET 2021 | What is the size of page
| Comprehension: Read the following and answer the questions: Consider a machine with 16 GB main memory and 32-bits virtual address space, with page size as 4KB. Frame size and page size is same for the given machine. What is the size of page table for handling the given virtual address space, given that each page table entry is of size 2 bytes? |
| i➥ 12KB |
| ii➥ 2KB |
| iii➥ 2MB |
| iv➥ 32MB |
Answer: Option III Solution: Given, Virtual Address Space = 32 bits, Page size = 4KB = 212, Page table entry size = 2B. Formula used, Page offset = ⌈log2(Page Size)⌉ , Page Table size = Number of pages * Page Table Entry, Number of pages = 2page number. Calculation, Virtual Address Space:Page offset = ⌈log2(Page Size)⌉ = ⌈log2 212⌉ = 12 bits, Page number = Virtual address space-page offset = 32-12 = 20 bits,
Number of pages = 2page number = 220, Page Table size = Number of pages * Page Table Entry size = 220 * 2B = 2 MB So, Option(III)is correct.
Watch Video For Better Explanation
Read More
- Operating System
- Computer Network
- Data Structure
- Theory of Computation
- Algorithms
- Digital Logic Design
- Computer System Architecture
- Discrete Math
- Computer Graphics
- Compiler Design
- Software Engineering
- Artificial Intelligence
- Database Management System
- Data mining
- Programming Languages
- Microprocessor
- Linear Programming Problems
| More related Questions on Memory Management |
Read more
| More related Questions on Operating System |
Read more
| More Questions on June 2021 |