System Software and Operating System
System Software and Operating System
Important Points
- Burst time is total time required for execution, which is given in problem
- Arrival time is moment at which job is submitted into a queue , which is given in problem
- start time is time at which procees starts its execution
- Finish time of every process is calulated by addition of start time and burst time of that process
- wait time of every process is calculated by subtraction of start time and arrival time of that process
- Turn around time of every process is calculated by subtraction of finish time and arrival time
- Avearge wait time is calculated by average of wait time of all processes.'
- Average turn around time is calculated by average of turn around time of all processes.
- Belady's Anomaly - If there are three free frames and we are finding the total page faults , this page fault should be reduced if we increase number of free frames then number of page faults should be decreased. When this does not happen , then exception is called belady's anomaly.
Consider the following set of processes with the length of CPU burst time and arrival time given in milli seconds
|
Process |
Burst Time |
Arrival Time |
|
P1 |
5 |
1.5 |
|
P2 |
1 |
0 |
|
P3 |
2 |
2 |
|
P4 |
4 |
3 |
Illustrate the execution of these process using FCFS CPU Scheduling algorithems. Also calculate wait timeand turn around timeof each process and calculate average waiting time and average turn around time for above situation. Also draw a gantt chart.
Solution :
|
Process (1) |
Burst Time (2) |
Arrival Time (3) |
Start Time (4) |
Wait Time (5) (4)-(3)
|
Finish Time (6) (4)+(2) |
Turn Around Time (7) (6)-(3) |
|
P1 |
5 |
1.5 |
1.5 |
0 |
6.5 |
5 |
|
P2 |
1 |
0 |
0 |
0 |
1 |
1 |
|
P3 |
2 |
2 |
6.5 |
5.5 |
8.5 |
6.5 |
|
P4 |
4 |
3 |
8.5 |
5.5 |
12.5 |
9.5 |
Now calculate average wait time
Average Wait time =10/4=2.5
average turn around time=22/4=5.5
Gantt Chart
|
P2 |
|
P1 |
P3 |
P4 |
0 1 1.5 6.5 8.5 12.5
In the shaded area CPU is idle and not doing any thing because no process is in ready queue
Consider the following set of processes with the length of CPU burst time and arrival time given in milli seconds
|
Process |
Burst Time |
Arrival Time |
|
P1 |
2 |
1 |
|
P2 |
3 |
2 |
|
P3 |
5 |
0 |
Illustrate the execution of these process using pre-emptive SJF CPU Scheduling algorithems. Also calculate wait time and turn around timeof each process and calculate average waiting time and average turn around time for above situation
Solution
|
Process (1) |
Burst Time (2) |
Arrival Time (3) |
Start Time (4) |
Wait Time (5) (4)-(3)
|
Finish Time (6) (4)+(2) |
Turn Around Time (7) (6)-(3) |
|
P1 |
2 |
1 |
1 |
0 |
3 |
2 |
|
P2 |
3 |
2 |
3 |
1 |
6 |
4 |
|
P3 |
5 |
0 |
0 |
0+(6-1)=5 |
10 |
10 |
Total wait time =6
Total turn around time =20
Average wait time=6/3=2
Average turn around time=20/3=6.66
Consider the following set of processes with the length of CPU burst time and arrival time given in milli seconds
|
Process |
Burst Time |
Arrival Time |
Priority |
|
P1 |
3 |
1 |
3(highest) |
|
P2 |
2 |
2 |
2 |
|
P3 |
5 |
0 |
1 |
Illustrate the execution of these process using non-premptive priority CPU Scheduling algorithems. Also calculate wait time and turn around time of each process and calculate average waiting time and average turn around time for above situation.
Solution :
|
Process (1) |
Burst Time (2) |
Arrival Time (3) |
Priority |
Start Time (4) |
Wait Time (5) (4)-(3)
|
Finish Time (6) (4)+(2) |
Turn Around Time (7) (6)-(3) |
|
P1 |
3 |
1 |
3 |
5 |
4 |
8 |
7 |
|
P2 |
2 |
2 |
2 |
8 |
6 |
10 |
8 |
|
P3 |
5 |
0 |
1 |
0 |
0 |
5 |
5 |
Total wait time= 10
Average Wait time =10/3=3.33
Total Turn around time = 20
average turn around time=20/3=6.66
Assume there are three free frames , find out belady's anomaly for 4 free frames, If reference string is 1,2,3,4,1,2,5,1,2,3,4,5. Use LRU
Solution
|
RS |
1 |
2 |
3 |
4 |
1 |
2 |
5 |
1 |
2 |
3 |
4 |
5 |
|
F1 |
1 |
1 |
1 |
4 |
4 |
4 |
5 |
5 |
5 |
3 |
3 |
3 |
|
F2 |
|
2 |
2 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
4 |
4 |
|
F3 |
|
|
3 |
3 |
3 |
2 |
2 |
2 |
2 |
2 |
2 |
5 |
|
PF |
1 |
2 |
3 |
4 |
1 |
2 |
5 |
--- |
---- |
3 |
4 |
5 |
Total Page Fault = 10
Now page fault for 4 frames
|
RS |
1 |
2 |
3 |
4 |
1 |
2 |
5 |
1 |
2 |
3 |
4 |
5 |
|
F1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
5 |
|
F2 |
|
2 |
2 |
2 |
2 |
2 |
2 |
2 |
2 |
2 |
2 |
2 |
|
F3 |
|
|
3 |
3 |
3 |
3 |
5 |
5 |
5 |
5 |
4 |
4 |
|
F4 |
|
|
|
4 |
4 |
4 |
4 |
4 |
4 |
3 |
3 |
3 |
|
PF |
1 |
2 |
3 |
4 |
--- |
--- |
5 |
--- |
---- |
3 |
4 |
5 |
Total page fault =8
Hence we can say there is no belady's anomaly.
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