1. Consider the following set of processes, whit the length of the CPU- burst time given in milliseconds:
Process
Burst time (ms)
Priority
p1
p2
p3
p4
p5
10 (ms)
1 (ms)
2 (ms)
1 (ms)
5 (ms)
3
1
3
4
2
The processes are assumed to have arrived in the order p1, p2, p3, p4, p5, all at time 0.
a-Draw four Gantt charts illustrating the execution of these processes using FCFS, SJF a non-preemptive
priority (a smaller priority number implies a higher priority), and RR (quantum =1 ) scheduling.
b-What is the turnaround time of each process for each of the scheduling algorithms in part a?
c-What is the waiting time of each process for each of the scheduling algorithms in part a?
ANSWER
1. FCFS
P1
p2
0
10
p3
11
p4
p5
13
14
19
2. SJF
p2
0
p4
1
p3
2
p5
p1
4
9
19
3. Priority
p2
0
p5
1
p1
6
p3
16
p4
18
19
4. RR
p
1
0
p
2
1
p
3
2
p
4
3
p
5
4
p
1
5
p
3
6
1. Turnaround time for FCFS
p1
p2
p3
p4
p5
=
=
=
=
=
10
11
13
14
19
ms
ms
ms
ms
ms
2. Turnaround time for SJF
p1
p2
p3
p4
p5
=
=
=
=
=
19
1
4
2
9
ms
ms
ms
ms
ms
3. Turnaround time for priority
p1
p2
p3
p4
p5
=
=
=
=
=
16 ms
1 ms
18 ms
19 ms
6 ms
4. Turnaround time for RR
p1
p2
p3
p4
p5
=
=
=
=
=
19
2
7
4
14
ms
ms
ms
ms
ms
p
5
7
p
1
8
p
5
9
p
1
10
p
5
11
p
1
12
p
5
13
p
1
14
p
1
15
p
1
16
p
1
17
p
1
18
19
1. Waiting time for FCFS
p1
p2
p3
p4
p5
=
=
=
=
=
0 ms
10 ms
11 ms
13 ms
14 ms
Average waiting time (FCFS ) = ( 0 + 10 + 11 + 13 + 14 ) / 5 = 9.6 ms
2. Waiting time for SJF
p1
p2
p3
p4
p5
=
=
=
=
=
9
0
2
1
4
ms
ms
ms
ms
ms
Average waiting time (SJF) = (9 + 0 + 2 + 1 + 4) / 5 = 3.2 ms
3. Waiting time for priority
p1
p2
p3
p4
p5
=
=
=
=
=
6
0
16
18
1
ms
ms
ms
ms
ms
Average waiting time ( priority ) = ( 6 + 0 + 16 + 18 + 1 ) / 5 = 8.2 ms
4. Waiting time for RR
p1
p2
p3
p4
p5
=
=
=
=
=
14 - 5 = 9
1
6-1 = 5
3
13 - 4 = 9
ms
ms
ms
ms
ms
Average waiting time ( RR ) = ( 9 + 1 + 5 + 3 + 9 ) / 5 = 5.4 ms
Average waiting time;
SJF
=
( 9 +0+ 2+1+ 4 )/5
=
3.2
RR
Priority
=
=
5.4
8.2
FCFS
=
( 9 +1+ 5+3+ 9 )/5 =
( 6 + 0 + 16 + 18 + 1 ) =
/5
( 0 + 10 + 11 + 13 + 14 =
)/5
is the
min
ms
ms
9.6
ms