This document contains 7 cases of different CPU scheduling algorithms, including FCFS, SJF (preemptive and non-preemptive), priority scheduling (preemptive and non-preemptive), and round robin. For each case, it provides the process details, draws the Gantt chart, and calculates the average waiting time. The key information provided includes the scheduling algorithm used, processes with burst times and arrival times, the Gantt chart showing process scheduling order and times, and the average waiting time calculation.

1. 1st Case : FCFS (First Come First Served)
CPU Scheduling
Process Burst Time
P1 3
P2 9
P3 5
P4 7
Draw Gantt Chart and calculate the average waiting time using the given table ??
Suppose that the processes arrive at time 0, in the order: P1 , P3 , P2 , P4
P1 P3 P2 P4
0 3 8 17 24
Waiting time :
P1 = 0
P2 = 8
P3 = 3
P4 = 17
Average waiting time = (0 + 8 + 3 + 17) / 4 = 7

2. 2nd Case : FCFS (First Come First Served)
Process
Burst
Time
Arrival
Time
P1 20 0
P2 12 3
P3 4 2
P4 9 5
Draw Gantt Chart and calculate the average waiting time using the given table ??
0
Waiting time : start time – arrival time
P1 = 0 – 0 = 0
P2 = 24 – 3 = 21
P3 = 20 – 2 = 18
P4 = 36 – 5 = 31
20 24
P1 P3 P2 P4
36 45
Average waiting time = (0 + 21 + 18 + 31) / 4 = 70 / 4

3. 3rd Case : SJF (short job first) non-Preemptive
Process
Burst
Time
Arrival
Time
P2 12 0
P3 8 3
P4 4 5
P1 10 10
P5 6 12
Draw Gantt Chart and calculate the average waiting time using the given table ??
0
Waiting time : start time – arrival time
P1 = 30 – 10 = 20
P2 = 0 – 0 = 0
P3 = 22 – 3 = 19
P4 = 12 – 5 = 7
12 16
P2 P4 P5 P3
22 40
Average waiting time = (20 + 0 + 19 + 7 + 4) / 5 = 50 / 5 = 10
P1
30
P5 = 16 – 12 = 4

4. 4th Case : SJF (short job first) Preemptive
Process
Burst
Time
Arrival
Time
P2 12 0
P3 8 3
P4 4 5
P1 10 10
P5 6 12
Draw Gantt Chart and calculate the average waiting time using the given table ??
0
Waiting time : start time – arrival time
P1 = 30 – 10 = 20
P2 = (0 – 0) + (21 - 3) = 18
P3 = (3 – 3) + (9 - 5) = 4
P4 = (5 – 5) = 0
3 5
P2
9 21
Average waiting time = (20 + 18 + 4 + 0 + 3) / 5 = 45 / 5 = 9
15
P5 = 15 – 12 = 3
30 40
P3 P4 P3 P5 P2 P1
9
6

5. 5th Case : Priority Scheduling non-Preemptive
Process
Burst
Time
Priority
Arrival
Time
P1 10 3 All
Processes
Arrived at
The
Same
Time
P2 1 1
P3 2 4
P4 1 5
P5 5 2
Draw Gantt Chart and calculate the average waiting time using the given table ??
0
Waiting time :
start time – arrival time
P1 = 6
P2 = 0
P3 = 16
P4 = 18
1 6
P2
16
Average waiting time = (6 + 0 + 16 + 18 + 1) / 5 = 41 / 5 = 8.2
18
P5 = 1
19
P5 P1 P3 P4

6. 6th Case : Priority Scheduling Preemptive
Process
Burst
Time
Priority
Arrival
Time
P1 10 3 0.0
P2 1 1 1.0
P3 2 4 2.0
P4 1 5 3.0
P5 5 2 4.0
Draw Gantt Chart and calculate the average waiting time using the given table ??
Waiting time :
start time – arrival time
P1 = (0 - 0)+(2 - 1)+(9 - 4) = 6
P2 = 1 – 1 = 0
P3 = 16 – 2 = 14
P4 = 18 – 3 = 15
Average waiting time = (6 + 0 + 14 + 15 + 0) / 5 = 35 / 5 = 7
P5 = 4 – 4 = 0
0 1 2
P1
4 16
9 18 19
P2 P1 P5 P1 P3 P4
9 7

7. 7th Case : Round Robin (RR)
Process Burst Time
P1 12
P2 8
P3 4
P4 10
P5 5
Draw Gantt Chart and Calculate The Average Waiting Time , where
P1 P2 P5 P4
0 5 10 14 19
Waiting time :
P1 = 0 + (24 - 5) + (37 - 29) = 27
P2 = 5 + (29 - 10) = 24
P3 = 10
P4 = 14 + (32 - 19) = 27
Average waiting time = (27 + 24 + 10 + 27 + 19) / 5 = 107 / 5 = 21.4
P3 P4 P1 P2 P1
24 29 32 37 39
7 2
3
5
Quantum = 5 ms
P5 = 19

8. Thank You For Watching
Good Luck
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