Analysing a network Example 12
A(3)1
D(5)
C(6) B(4) E(1)
4 G(7) 5
3
F(6)
This is the network found in Example 1. We want to find the
minimum duration of this project.The first step is to find the
earliest event times. This is the earliest time that an event
(denoted by the numbered nodes) can occur. The event cannot occur
until all activities leading into the event node have finished.
Analysing a network Example 12
A(3)1 0 B(4)
D(5)
C(6) E(1)
4 G(7) 5
3
F(6)
Event 1 occurs at time zero.
Analysing a network Example 13 2 D(5)
A(3)1 0 B(4)
C(6) E(1)
4 G(7) 5
3
F(6)
Event 2 cannot occur until A is finished. The earliest time for
this is 3.
Analysing a network Example 13 2 D(5)
A(3)1 0 B(4)
C(6) E(1)
4 G(7) 5
34 Event 3 cannot occur until B is finished. The earliest time
for this is 4.
F(6)
Analysing a network Example 13 2 D(5)
A(3)1 0 B(4)
8C(6) E(1) 5 4 G(7)
34
F(6)
Event 4 cannot occur until C, D and E are all finished. The
earliest C can finish is 6. The earliest D can finish is 3 + 5 = 8.
The earliest E can finish is 4 + 1 = 5.
So the earliest time for event 4 is 8.
Analysing a network Example 13 2 D(5)
A(3)1 0 B(4)
8C(6) E(1) 5 4 G(7)
34
F(6)
15
Event 5 cannot occur until F and G are both finished. The
earliest F can finish is 4 + 6 = 10. The earliest G can finish is 8
+ 7 = 15.
So the earliest time for event 5 is 15.
Analysing a network Example 13 2 D(5)
A(3)1 0 B(4)
8C(6) E(1) 5 4 G(7)
34
F(6)
15
The next step is to find the late event times, working backwards
through the network. A latest event time is the latest time that an
event can occur without delaying the project. It is found by
finding the latest time that each activity leading out of the event
can begin the latest event time is the earliest of these.
Analysing a network Example 13 2 D(5)
A(3)1 0 B(4)
8C(6) E(1) 5 4 G(7)
34 Event 5 must occur by time 15, or the project will not finish
in the minimum possible time.
F(6)
15
15
Analysing a network Example 13 2 D(5)
A(3)1 0 B(4)
8C(6) E(1) 4
8
G(7) 5
34
F(6)
15
15
The only activity leading from event 4 is G, which must start by
time 8 if the project is not to be delayed. So event 4 must occur
by time 8.
Analysing a network Example 13 2 D(5)
A(3)1 0 B(4)
8C(6) E(1) 4
8
G(7) 5
34 7
F(6)
15
15
The activities leading from event 3 are E (which must start by
time 7) and F (which must start by time 9). So event 3 must occur
by time 7.
Analysing a network Example 13 3 2 D(5)
A(3)1 0 B(4)
8C(6) E(1) 4
8
G(7) 5
34 7
F(6)
15
15
The only activity leading from event 2 is D, which must start by
time 3. So event 2 must occur by time 3.
Analysing a network Example 13 3 2 D(5)
A(3)1 0 0 B(4)
8C(6) E(1) 4
8
G(7) 5
34 7
F(6)
15
15
Finally, event 1 must occur by time zero.
Analysing a network Example 13 3 2 D(5)
A(3)1 0 0 B(4)
8C(6) E(1) 4
8
G(7) 5
34 7
F(6)
15
15
The completed network shows that the project can be completed in
15 hours.
The critical activities are the activities (i, j) for which the
LET for j the EET for i is equal to the activity duration. The
critical activities are A, D and G.
For analysis of the float in this example, see the Notes and
Examples.
Analysing a network Example 22
A(2)1 3 D(6) 5 G(4) 6
B(3)
C(5)
E(8) F(2) 4
This is the network found in Example 2. We want to find the
minimum duration of this project. The first step is to find the
earliest event times.
Analysing a network Example 22 2
A(2)1 0 C(5) E(8) F(2) 4 3 D(6) 5 G(4) 6
B(3)
Event 1 occurs at time zero. The earliest that event 2 can occur
is after A has finished, at time 2.
Analysing a network Example 22 2
A(2)1 0 C(5) 3 3 E(8) F(2) 4 D(6) 5 G(4) 6
B(3)
Event 3 cannot occur until both A and B have finished, so the
earliest time at which event 3 can occur is 3.
Analysing a network Example 22 2
A(2)1 0 C(5) 3 3 E(8) F(2) 4 D(6) 5 G(4) 6
B(3)
Event 4 cannot occur until C has finished, so the earliest time
at which event 5 can occur is 5.
5
Analysing a network Example 22 2
A(2) 131 0 C(5) B(3) 3 E(8) F(2) 4 3 D(6) 5 G(4) 6
The earliest that D can finish is at time 9, and the earliest
that E can finish is at time 13, so the earliest that event 5 can
occur is time 13.
5
Analysing a network Example 22 2
A(2) 131 0 C(5) B(3) 3 E(8) F(2) 4 3 D(6) 5 G(4) 6 17
The earliest that F can finish is at time 7, and the earliest
that G can finish is at time 17, so the earliest that event 5 can
occur is time 17.
5
Analysing a network Example 22 2
A(2) 131 0 C(5) B(3) 3 E(8) F(2) 4 3 D(6) 5 G(4) 6 17
The next step is to find the latest event times, starting from
the sink node and working backwards.
5
Analysing a network Example 22 2
A(2) 131 0 C(5) B(3) 3 E(8) F(2) 4 3 D(6) 5 G(4) 6 17 17
Event 6 must not occur later than time 17, or the project will
be delayed .
5
Analysing a network Example 22 2
A(2) 131 0 C(5) B(3) 3 E(8) F(2) 4 3 D(6)
135 G(4) 6 17 17
The latest G can start is at time 13, so the latest time for
event 5 is 13.
5
Analysing a network Example 22 2
A(2) 131 0 C(5) B(3) 3 E(8) F(2) 4 3 D(6)
135 G(4) 6 17 17
The latest that E can start is at time 5, and the latest that F
can start is at time 15, so the latest possible time for event 4 is
5.
5
5
Analysing a network Example 22 2
A(2) 131 0 C(5) B(3) 3 3 7 E(8) F(2) 4 D(6)
135 G(4) 6 17 17
The latest that D can start is at time 7, so the latest possible
time for event 3 is 7.
5
5
Analysing a network Example 22 2 7
A(2) 131 0 C(5) B(3) 3 3 7 E(8) F(2) 4 D(6)
135 G(4) 6 17 17
The latest that the dummy activity (with zero duration) can
start is at time 7, so the latest possible time for event 2 is
7.
5
5
Analysing a network Example 22 2 7
A(2) 131 0 0 C(5) B(3) 3 3 7 E(8) F(2) 4 D(6)
135 G(4) 6 17 17
The latest that event 1 can start is at time 0.
5
5
Analysing a network Example 22 2 7
A(2) 131 0 0 C(5) B(3) 3 3 7 E(8) F(2) 4 D(6)
135 G(4) 6 17 17
The critical activities are 5 activities for which the float is
zero: i.e. the latest event time for activity j the earliest event
time for activity i is equal to the activity duration.
5
Analysing a network Example 22 2 7
A(2) 131 0 0 C(5) B(3) 3 3 7 E(8) F(2) 4 D(6)
135 G(4) 6 17 17
The critical activities are C, E and G.The project can be
completed in 17 days.
5
5
For analysis of the float in this example, see Example 2 in the
Notes and Examples.
Analysing a network Example 3C(5)3 4
A(2)1
E(6)F(3) H(2)
6B(4) G(8) 2 5
7
D(3) This is the network found in Example 3. We want to find the
minimum duration of this project.The first step is to find the
early event times.
Analysing a network Example 3C(5)3 4
A(2)1
E(6)F(3) H(2)
60 B(4) G(8) 2 5
7
D(3) Event 1 occurs at time zero.
Analysing a network Example 3C(5)3 4
A(2)1
E(6)F(3) H(2)
60 B(4) G(8) 2 5
7
D(3)4
Event 2 cannot occur until B has finished, so the earliest time
for event 2 is 4.
Analysing a network Example 34
C(5)3 4
A(2)1
E(6)F(3) H(2)
60 B(4) G(8) 2 5
7
D(3)4
Event 3 cannot occur until both A and B have finished, so the
earliest time for event 3 is 4.
Analysing a network Example 34
C(5)3 4 F(3)
9
A(2)1
E(6)
H(2)
60 B(4) G(8) 2 5
7
D(3)4
Event 4 cannot occur until C has finished, so the earliest time
for event 4 is 4 + 5 = 9.
Analysing a network Example 34
C(5)3 4 F(3)
9
A(2)1
E(6)
H(2)
60 B(4) G(8) 2 5
7
D(3)4 7
Event 5 cannot occur until D has finished, so the earliest time
for event 5 is 4 + 3 = 7.
Analysing a network Example 34
C(5)3 4 F(3)
9
A(2)1 0 B(4)
E(6)
H(2) 12
6
7
G(8) 2 5
D(3)4 7
Event 6 cannot occur until both D and F have finished. The
earliest time that D can finish is 7, and the earliest time that F
can finish is 9 + 3 = 12, so the earliest time for event 6 is
12.
Analysing a network Example 34
C(5)3 4 F(3)
9
A(2)1 0 B(4)
E(6)
H(2) 12
6
715 G(8)
2
5
D(3)4 7
Event 7 cannot occur until E, G and H have all finished. The
earliest time that E can finish is 9 + 6 = 15, the earliest time
that G can finish is 7 + 8 = 15, and the earliest time that H can
finish is 12 + 2 = 14, so the earliest time for event 7 is 15.
Analysing a network Example 34
C(5)3 4 F(3)
9
A(2)1 0 B(4)
E(6)
H(2) 12
6
715 G(8)
2
5
D(3)4 7
The next step is to find the late event times, working backwards
through the network.
Analysing a network Example 34
C(5)3 4 F(3)
9
A(2)1 0 B(4)
E(6)
H(2) 12
6
715 15 G(8)
2
5
D(3)4 7
The latest time that event 7 can occur without delaying the
project is 15.
Analysing a network Example 34
C(5)3 4 F(3)
9
A(2)1 0 B(4)
E(6)
H(2) 12 13
6
715 15 G(8)
2
5
D(3)4 7
The latest time that activity H can start is 13, so the latest
time for event 6 is 13.
Analysing a network Example 34
C(5)3 4 F(3)
9
A(2)1 0 B(4)
E(6)
H(2) 12 13
6
715 15 G(8)
2
5
D(3)4 7 7
Activity G and a dummy activity both lead out of event 5. The
latest time that activity G can start is 7, and the latest time
that the dummy activity can start is 13, so the latest time for
event 5 is 7.
Analysing a network Example 34
C(5)3 4 F(3)
9
9
A(2)1 0 B(4)
E(6)
H(2) 12 13
6
715 15 G(8)
2
5
D(3)4 7 7
Activities E and F both lead out of event 4. The latest time
that activity E can start is 9, and the latest time that activity F
can start is 9, so the latest time for event 4 is 9.
Analysing a network Example 34 4
C(5)3 4 F(3)
9
9
A(2)1 0 B(4)
E(6)
H(2) 12 13
6
715 15 G(8)
2
5
D(3)4 7 7
The latest time that activity C can start is 4, so the latest
time for event 3 is 4.
Analysing a network Example 34 4
C(5)3 4 F(3)
9
9
A(2)1 0 B(4)
E(6)
H(2) 12 13
6
715 15 G(8)
2
5
D(3)4 4 7 7
Activity D and a dummy activity both lead out of event 2. The
latest time that activity D can start is 6, and the latest time
that the dummy activity can start is 4, so the latest time for
event 2 is 4.
Analysing a network Example 34 4
C(5)3 4 F(3)
9
9
A(2)1 0 0 B(4)
E(6)
H(2) 12 13
6
715 15 G(8)
2
5
D(3)4 4 7 7
Finally, the latest time for event 1 is zero.
Analysing a network Example 34 4
C(5)3 4 F(3)
9
9
A(2)1 0 0 B(4)
E(6)
H(2) 12 13
6
715 15 G(8)
2
5
D(3)4 4 7 7 The float in this example is analysed in the Notes
and Examples.
The minimum duration of the project is 15 days. The critical
activities are B, C, D, E and G.Notice that there are two critical
paths: BCE and BDG.