Question 2136 – Nielit Scientific Assistance CS 15-10-2017
December 12, 2023Question 21 – ISRO-2007
December 12, 2023Question 2138 – Nielit Scientific Assistance CS 15-10-2017
The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260,
0320, 0370, What is the number of page faults?
Correct Answer: C
Question 58 Explanation:
Memory page numbers will be calculated by using corresponding to given address as follows
The page number is 1 from the record numbers 1-100,
The page number is 2 from the record numbers 101 -200
The page number is 3 from the record numbers 201 -300
The page number is 4 from the record numbers 301 -400
The page number is 5 from the record numbers 401 -500
The page number is 6 from the record numbers 501 -600
The page number is 7 from the record numbers 601 -700
For a given address sequence 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240,0260, 0320, 0370
The page number numbers sequence is 1,2,5,5,6,6,6,2,3,3,3,4,4
The number of available frame is 1, So we need to use only one frame for executing demand page process.
So total number of page faults are 7.
The page number is 1 from the record numbers 1-100,
The page number is 2 from the record numbers 101 -200
The page number is 3 from the record numbers 201 -300
The page number is 4 from the record numbers 301 -400
The page number is 5 from the record numbers 401 -500
The page number is 6 from the record numbers 501 -600
The page number is 7 from the record numbers 601 -700
For a given address sequence 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240,0260, 0320, 0370
The page number numbers sequence is 1,2,5,5,6,6,6,2,3,3,3,4,4
The number of available frame is 1, So we need to use only one frame for executing demand page process.
So total number of page faults are 7.
15,4
6,4
7,2
4,6