In OS

Question

In OS 
Consider the following page reference string:
7,2,3,1,2,5,3,4,6,7,7,1,0,5,4,6,2,3,0,1.
Assuming demand paging with three frames,how many page faults would occur for the following replacement algorithms?
A,LRU replacement
B.FIFO replacement
C.Optimal replacement

GPT-4o mini · Answer

To analyze the page reference string using three frames (assuming demand paging) and count the page faults for the three different page replacement algorithms (LRU, FIFO, and Optimal), we can step through the reference string and maintain the set of frames while tracking page faults.

Page reference string: 7, 2, 3, 1, 2, 5, 3, 4, 6, 7, 7, 1, 0, 5, 4, 6, 2, 3, 0, 1

A. LRU Replacement (Least Recently Used)

7 → Page fault (Frames: [7])
2 → Page fault (Frames: [7, 2])
3 → Page fault (Frames: [7, 2, 3])
1 → Page fault (Replace 7, Frames: [2, 3, 1])
2 → No page fault (Frames: [2, 3, 1])
5 → Page fault (Replace 3, Frames: [2, 1, 5])
3 → Page fault (Replace 2, Frames: [1, 5, 3])
4 → Page fault (Replace 1, Frames: [5, 3, 4])
6 → Page fault (Replace 5, Frames: [3, 4, 6])
7 → Page fault (Replace 4, Frames: [3, 6, 7])
7 → No page fault (Frames: [3, 6, 7])
1 → Page fault (Replace 3, Frames: [6, 7, 1])
0 → Page fault (Replace 6, Frames: [7, 1, 0])
5 → Page fault (Replace 7, Frames: [1, 0, 5])
4 → Page fault (Replace 1, Frames: [0, 5, 4])
6 → Page fault (Replace 0, Frames: [5, 4, 6])
2 → Page fault (Replace 5, Frames: [4, 6, 2])
3 → Page fault (Replace 4, Frames: [6, 2, 3])
0 → Page fault (Replace 6, Frames: [2, 3, 0])
1 → Page fault (Replace 2, Frames: [3, 0, 1])

Total page faults using LRU: 15

B. FIFO Replacement (First In, First Out)

7 → Page fault (Frames: [7])
2 → Page fault (Frames: [7, 2])
3 → Page fault (Frames: [7, 2, 3])
1 → Page fault (Replace 7, Frames: [2, 3, 1])
2 → No page fault (Frames: [2, 3, 1])
5 → Page fault (Replace 2, Frames: [3, 1, 5])
3 → No page fault (Frames: [3, 1, 5])
4 → Page fault (Replace 3, Frames: [1, 5, 4])
6 → Page fault (Replace 1, Frames: [5, 4, 6])
7 → Page fault (Replace 5, Frames: [4, 6, 7])
7 → No page fault (Frames: [4, 6, 7])
1 → Page fault (Replace 4, Frames: [6, 7, 1])
0 → Page fault (Replace 6, Frames: [7, 1, 0])
5 → Page fault (Replace 7, Frames: [1, 0, 5])
4 → Page fault (Replace 1, Frames: [0, 5, 4])
6 → Page fault (Replace 0, Frames: [5, 4, 6])
2 → Page fault (Replace 5, Frames: [4, 6, 2])
3 → Page fault (Replace 4, Frames: [6, 2, 3])
0 → Page fault (Replace 6, Frames: [2, 3, 0])
1 → Page fault (Replace 2, Frames: [3, 0, 1])

Total page faults using FIFO: 17

C. Optimal Replacement

7 → Page fault (Frames: [7])
2 → Page fault (Frames: [7, 2])
3 → Page fault (Frames: [7, 2, 3])
1 → Page fault (Replace 7, Frames: [1, 2, 3]) (7 is not needed in future)
2 → No page fault (Frames: [1, 2, 3])
5 → Page fault (Replace 3, Frames: [1, 2, 5]) (3 is not needed in future)
3 → Page fault (Replace 1, Frames: [3, 2, 5]) (1 is not needed till the end)
4 → Page fault (Replace 2, Frames: [3, 4, 5]) (2 is not needed till the end)
6 → Page fault (Replace 5, Frames: [3, 4, 6]) (5 is not needed till the end)
7 → Page fault (Replace 3, Frames: [7, 4, 6]) (3 is not needed till the end)
7 → No page fault (Frames: [7, 4, 6])
1 → Page fault (Replace 4, Frames: [7, 1, 6]) (4 is not needed again)
0 → Page fault (Replace 6, Frames: [7, 1, 0]) (6 is not needed again)
5 → Page fault (Replace 7, Frames: [5, 1, 0]) (7 is not needed again)
4 → Page fault (Replace 1, Frames: [5, 4, 0])
6 → Page fault (Replace 0, Frames: [5, 4, 6])
2 → Page fault (Replace 5, Frames: [2, 4, 6])
3 → Page fault (Replace 4, Frames: [2, 3, 6])
0 → Page fault (Replace 6, Frames: [2, 3, 0])
1 → Page fault (Replace 2, Frames: [1, 3, 0])

Total page faults using Optimal: 12

Summary

LRU: 15 page faults
FIFO: 17 page faults
Optimal: 12 page faults

These calculations provide the total number of page faults for each page replacement algorithm based on the given reference string.