Post on 22-Jan-2016
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Activity networks – Example 3
Task Duration (days) Immediate predecessors
A 2 -
B 4 -
C 5 A, B
D 3 B
E 6 C
F 3 C
G 8 D
H 2 D, F
I 3 E, G, H
The table below shows the tasks involved in a project, with their durations and immediate predecessors.
Draw an activity network and use it to find the critical activities and the minimum duration of the project.
Activity networks – Example 3
A
2
B
4
Start by placing the two activities that have no immediate predecessors.
Activity networks – Example 3
A
2
B
4
C
5
D
3
Activity C depends on A and B.
Activity D depends on B only.
Activity networks – Example 3
A
2
B
4
E
6
F
3
C
5
D
3G
8
Activity E depends on C only.
Activity F depends on C only.
Activity G depends on D only.
Activity networks – Example 3
A
2
B
4
E
6
H
2
F
3
C
5
D
3G
8
Activity H depends on both D and F.
Activity networks – Example 3
A
2
B
4
E
6
H
2
F
3
C
5
D
3G
8
I
3
Finally, activity I depends on E, H and G.
Notice that no END activity is needed here, since I is the only final activity.
Activity networks – Example 3
A
2
B
4
E
6
H
2
F
3
C
5
D
3G
8
I
3
The next step is to find the early times. The early time for an activity is the earliest time that the activity can begin. The activity cannot begin until all activities leading into the activity have finished.
Activity networks – Example 3
A
2
B
4
E
6
H
2
F
3
C
5
D
3G
8
I
3
Activities A and B can both begin at time zero.
0
0
Activity networks – Example 3
A
0 2
B
0 4
E
6
H
2
F
3
C
5
D
3G
8
I
3
4
Activity C cannot begin until both activities A and B have finished.
The earliest that A can finish is at time 0 + 2 = 2. The earliest that B can finish is at time 0 + 4 = 4.
The earliest that C can start is at the later of these two times, which is 4.
Activity networks – Example 3
A
0 2
B
0 4
E
6
H
2
F
3
C
4 5
D
3G
8
I
3
4
Activity D cannot begin until activity B has finished.
The earliest that B can finish is at time 0 + 4 = 4, so the earliest that D can start is at time 4.
Activity networks – Example 3
A
0 2
B
0 4
E
6
H
2
F
3
C
4 5
D
4 3G
8
I
3
9
Activities E and F cannot begin until activity C has finished.
9
The earliest that C can finish is at time 4 + 5 = 9, so the earliest that E and F can start is at time 9.
Activity networks – Example 3
A
0 2
B
0 4
E
9 6
H
2
F
9 3
C
4 5
D
4 3G
8
I
3
7
Activity G cannot begin until activity D has finished.
The earliest that D can finish is at time 4 + 3 = 7, so the earliest that G can start is at time 7.
Activity networks – Example 3
A
0 2
B
0 4
E
9 6
H
2
F
9 3
C
4 5
D
4 3G
7 8
I
3
12
Activity H cannot begin until both activities D and F have finished.
The earliest that D can finish is at time 4 + 3 = 7. The earliest that F can finish is at time 9 + 3 = 12.
The earliest that H can start is at the later of these two times, which is 12.
Activity networks – Example 3
A
0 2
B
0 4
E
9 6
H
12 2
F
9 3
C
4 5
D
4 3G
7 8
I
315
Activity I cannot begin until all of activities E, H and G have finished.
The earliest that E can finish is at time 9 + 6 = 15. The earliest that H can finish is at time 12 + 2 = 14. The earliest that G can finish is at time 7 + 8 = 15.
The earliest that activity I can start is at the latest of these three times, which is 15.
Activity networks – Example 3
A
0 2
B
0 4
E
9 6
H
12 2
F
9 3
C
4 5
D
4 3G
7 8
I
15 3 18
The next stage is to find the late times. The late time for an activity is the latest time at which the activity can finish without delaying the whole project.
The late time for activity I is 15 + 3 = 18.
Activity networks – Example 3
A
0 2
B
0 4
E
9 6
H
12 2
F
9 3
C
4 5
D
4 3G
7 8
I
15 3 18
15
Activities E, H and G all lead directly into activity I. So the latest time that each of E, H and G must finish is at time 18 – 3 = 15, otherwise the project will be delayed.
15
15
Activity networks – Example 3
A
0 2
B
0 4
E
9 6 15
H
12 2 15
F
9 3
C
4 5
D
4 3G
7 8 15
I
15 3 1813
Activity F leads directly into activity H only.
The latest time that activity H can begin is at time 15 – 2 = 13. So the latest time that F must finish is at time 13.
Activity networks – Example 3
A
0 2
B
0 4
E
9 6 15
H
12 2 15
F
9 3 13
C
4 5
D
4 3G
7 8 15
I
15 3 18
7
Activity D leads into two activities: H and G.
The latest that H can start is at time 15 – 2 = 13. The latest that G can start is at time 15 – 8 = 7.
The latest time that C can finish is the earlier of these times, which is 7.
Activity networks – Example 3
A
0 2
B
0 4
E
9 6 15
H
12 2 15
F
9 3 13
C
4 5
D
4 3 7G
7 8 15
I
15 3 18
9
Activity C leads into two activities: E and F.
The latest that E can start is at time 15 – 6 = 9. The latest that F can start is at time 13 – 3 = 10.
The latest time that C can finish is the earlier of these times, which is 9.
Activity networks – Example 3
A
0 2
B
0 4
E
9 6 15
H
12 2 15
F
9 3 13
C
4 5 9
D
4 3 7G
7 8 15
I
15 3 18
4
Activity A leads directly into activity C only.
The latest time that activity C can begin is at time 9 – 5 = 4. So the latest time that A must finish is at time 4.
Activity networks – Example 3
A
0 2 4
B
0 4
E
9 6 15
H
12 2 15
F
9 3 13
C
4 5 9
D
4 3 7G
7 8 15
I
15 3 184
Activity B leads into two activities: C and D.
The latest that C can start is at time 9 – 5 = 4. The latest that D can start is at time 7 – 3 = 4.
The latest time that B can finish is the earlier of these times, which is 4.
Activity networks – Example 3
A
0 2 4
B
0 4 4
E
9 6 15
H
12 2 15
F
9 3 13
C
4 5 9
D
4 3 7G
7 8 15
I
15 3 18
The completed network shows that the project can be completed in 18 days.
The critical activities are the ones for which the difference between the late time and the early time is equal to the duration of the activity.
The critical activities are B, C, D, E, G and I, and there are two critical paths: BCEI and BDGI.