Eng. Haneen El-masry
2013
CPU Scheduling Algorithms
 First Come First Served (FCFS)
 Shortest Job First (SJF)
 Priority
 Round Robin (RR)
 Multi-Level Queue
 Multi-Level Feedback Queue
Example 1
 Consider the following set of processes with the indicated
arrival time. The length of the CPU-burst time is given in
ms.
Process
Arrival Time
CPU-Burst Time
Priority
P1
0
10
7
P2
1
1
2
P3
3
4
4
P4
5
2
3
P5
11
3
5
For each of the following scheduling algorithms, draw the
Gantt chart and compute the avg. waiting time, avg. Response
time, avg. turnaround time.
FCFS
 Gantt Chart
P1
0
P2
P3
10 11
P4
15
P5
17
20
 Calculation
Process
Waiting Time Response Time Turnaround Time
P1
0
0
10-0=10
P2
10-1=9
10-1=9
11-1=10
P3
11-3=8
11-3=8
15-3=12
P4
15-5=10
15-5=10
17-5=12
P5
17-11=6
17-11=6
20-11=9
Avg. W. T.= (0+9+8+10+6)/5= 6.6 ms.
Avg. R. T.= (0+9+8+10+6)/5= 6.6 ms.
Avg. T. T.= (10+10+12+12+9)/5= 10.6 ms.
Non-Preemptive SJF
 Gantt Chart
P1
P2
0
10
P4
11
P5
13
P3
16
20
 Calculation
P
W.T.
R.T.
T.T.
P1
0
0
10
P2
10-1=9
10-1=9
11-1=10
P3
16-3=13
16-3=13
20-3=17
P4
11-5=6
11-5=6
13-5=8
P5
13-11=2
13-11=2
16-11=5
Avg. W. T.= (0+9+13+6+2)/5 = 6 ms.
Avg. R. T. = (0+9+13+6+2)/5 = 6 ms.
Avg. T. T. = (10+10+17+8+5)/5 = 10 ms.
Preemptive SJF
 Gantt Chart
P1 P2 P1
0
1
2
P3
3
P4
7
P1
P5
9
11
P1
14
20
 Calculation
P
W.T.
R.T.
T.T.
P1
(2-1)+(9-3)+(14-11)= 10 0
20-0=20
P2
1-1=0
1-1=0
2-1=1
P3
3-3=0
3-3=0
7-3=4
P4
7-5=2
7-5=2
9-5=4
P5
11-11=0
11-11=0
14-11=3
Avg. W. T.= (10+0+0+2+0)/5 = 2.4 ms.
Avg. R. T. = (0+0+0+2+0)/5 = 0.4 ms.
Avg. T. T. = (20+1+4+4+3)/5 = 6.4 ms.
You can conclude that:
Waiting Time = Turnaround Time - CPU Burst
Non-Preemptive Priority
 Note: Smaller value means a higher priority.
 Gantt Chart
P1
0
 Calculation
P2
10 11
P4
P3
13
P5
17
P
T.T.
P1
10-0=10
10-10=0
0
P2
11-1=10
10-1=9
10-1=9
P3
17-3=14
14-4=10
13-3=10
P4
13-5=8
8-2=6
11-5=6
P5
20-11=9
9-3=6
17-11=6
Avg.
10.2
W.T.
20
6.2
R.T.
6.2
Preemptive Priority
 Gantt Chart
P1 P2 P1
P3
0 1 2 3
 Calculation
P
P4
5
T.T.
P3
7
P1
9
P5
P1
11
14
W.T.
R.T.
P1
20-0=20
20-10=10
0
P2
2-1=1
1-1=0
1-1=0
P3
9-3=6
6-4=2
3-3=0
P4
7-5=2
2-2=0
5-5=0
P5
14-11=3
3 - 3=0
11-11=0
Avg.
6.4
2.4
0
20
RR (q=2 ms)
 Note: If two process need to enter the ready queue at the same time,
use the FCFS to choose between them.
 Gantt Chart
P1
0
P2
2
P1
3
P3
5
P1
7
P4
9
P3
11
P1
13
P5
15
 Calculation
P
T.T.
P1
19-0=19
19-10=9
0
P2
3-1=2
2-1=1
2-1=1
P3
13-3=10
10-4=6
5-3=2
P4
11-5=6
6-2=4
9-5=4
P5
20-11=9
9-3=6
15-11=4
Avg.
9.2
W.T.
5.2
R.T.
2.2
P1
17
P5
19 20
RR (q=2 ms)
 Note: If two process need to enter the ready queue at the same time,
the new process will be served first.
 Gantt Chart
P1
0
P2
2
P1
3
P3
5
P4
7
P1
9
P3
11
P5
13
P1
15
 Calculation
P
T.T.
W.T.
P1
20-0=20
20-10=10
0
P2
3-1=2
2-1=1
2-1=1
P3
13-3=10
10 - 4=6
5 - 3=2
P4
9-5=4
4 -2=2
7 - 5=2
P5
18-11=7
7 - 3=4
13 -11=2
Avg.
8.6
4.6
R.T.
1.4
P5
17 18
P1
20
Multi-level Queue
 Assume both of P1 and P3 are batch processes, P2 is interactive
processes, and both of P4 and P5 are system processes.
 Assume also that the batch processes queue is scheduled using the
preemptive SJF and the other queues use FCFS.
System Process
(FCFS)
Interactive Process
(FCFS)
Batch Process
(SJF)
 Gantt Chart
P1
0
P2
3
P3
4
P4
7
P1
9
P5
11
P3
14 15
 Calculation
P
T.T.
W.T.
P1
20-0=20
20-10=10
0
P2
4-1=3
3-1=2
3-1=2
P3
15-3=12
12 - 4=8
4 - 3=1
P4
9-5=4
4 -2=2
7 - 5=2
P5
14-11=3
3 - 3=0
11-11=0
Avg.
8.4
4.4
R.T.
1
P1
20
Multi-level feedback queue
 Assume both of the first and second levels use the RR
scheduling with quantum size as shown below.
Q=3
Q=2
FCFS
 Gantt Chart
P1 P2 P1
0
1
2
P3
3
P4
5
P3
7
P1
9
P5
11
14
 Calculation
P
T.T.
W.T.
P1
20-0=20
20-10=10
0
P2
2-1=1
1-1=0
1-1=0
P3
9-3=6
6- 4=2
3 - 3=0
P4
7-5=2
2 -2=0
5 - 5=0
P5
14-11=3
3 - 3=0
11-11=0
Avg.
6.4
2.4
p1
R.T.
0
20
CPU Utilization
 CPU Utilization = (CPU time/ Total Time) *100%
Q: Compute the CPU Utilization for part (f). Assume the contextswitching time is 0.5 ms.
Solution
Context-Switching Time = 0.5*10= 5 ms
CPU Utilization = (20/ (20+5) )*100%
= 80 %
Next CPU Burst Length
Q: If the length of the next CPU burst was predicted to be 8ms, the actual
length used in this prediction was 5ms. According to the average
exponential formula, what was the previous predicted value of CPU
burst if α=0.4.
Solution
ιn+1= 8ms.
tn= 5ms.
ιn= ?
α= 0.4
ιn+1= α*tn + (1- α)*ιn
8= 0.4*5+0.6* ιn
ιn= (8-2)/0.6=10 ms.
Starvation

 Starvation: low priority processes may never execute.
Q. Which of the following scheduling algorithms could result in
starvation ?
A. FCFS
B. SJF
C. RR
D. Priority
Exercise 1
 Consider five processes P1, P2, P3, P4 and P5 with the following
Process Burst Times and Priorities:
P
CPU Burst
Priority
P1
15
2
P2
2
0
P3
4
2
P4
2
3
P5
8
1
 The processes are assumed to have arrived in the order P1, P2, P3,
P4, and P5 all at time 0. Smaller value means a higher priority.
 Draw a Gantt diagram representing the allocation of the CPU and
calculate the average Turnaround time and average wait time using
FCFS, SJF, Priority, and Round Robin (q = 3). Show all calculations.
With My Best Wishes 