Important Points
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.