5 weeks
DESCRIPTION
B. 10 WEEKS. A. D. E. 5 WEEKS. 8 WEEKS. 6 WEEKS. C. 12 WEEKS. ES. EF. ACTIVITY. FL. (t). LS. LF. We have a network with five activities: A, B, C, D and E. The activities have different durations that are shown in the figure. For example, activity A has a duration of 5 weeks. - PowerPoint PPT PresentationTRANSCRIPT
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A
5 WEEKS
B
10 WEEKS
C
12 WEEKS
D
8 WEEKS
E
6 WEEKS
ACTIVITY
(t)
ES EF
FL
LS LF
We have a network with five activities: A, B, C, D and E. The activities have different durations that are shown in the figure. For example, activity A has a duration of 5 weeks.
The figure in the lower left corner shows the different values that are calculated in the network.First, we calculate early positions for all of the activities. These show the earliest start points (ES) and earliest finish points (EF).
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0 5
5 15
5 17
17 25 25 31
A
5 WEEKS
B
10 WEEKS
C
12 WEEKS
D
8 WEEKS
E
6 WEEKS
ACTIVITY
(t)
ES EF
FL
LS LF
First, we take the starting point in activity A and assume that it starts in the timepoint 0.The earliest start point (ES) for A is 0, and we can directly find the earliest finish (EF) by adding to it the duration of A:
0 + 5 = 5
By looking at the network, we see that the earliest time point for activity A gives the earliest time point that both activities B and C can start on.
Activity B has a duration of 10 weeks and its’ earliest finish is in week 15.
5 + 10 = 15
Activity C has a duration of 12 weeks and its’ earliest finish is in week 17.
5 + 12 = 17
The earliest time point that activity D can start on is the maximum of the earliest finish values for B and C. Therefore, activity D starts on week 17.
The duration of activity D is 8 weeks, so its earliest finish is 25.
17 + 8 = 25
Activity E therefore has an earliest start point in week 25.The earliest finish for activity E is week 31 because the duration of the activity is 6 weeks.
25 + 6 = 31
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0 5
7 17
5 17
17 25 25 31
A
5 WEEKS
0 5
B
10 WEEKS
5 15
C
12 WEEKS
5 17
D
8 WEEKS
17 25
E
6 WEEKS
25 31
ACTIVITY
(t)
ES EF
FL
LS LF
Now we can calculate backwards in the network to find the latest position for all activities. This means that we will calculate the latest start points and latest finish points.
The finish date of the project is not given, so we assume that the latest finish point is equal to the earliest finish point. Thus, we set the latest finish equal to 31.
Now we can subtract to find the latest start for all activities.
For activity E, which has a duration of 6 weeks, the latest start is week 25.
31 - 6 = 25
Additionally, by looking at the network structure we can see that activity D at has a latest finish point in week 25.Activity D has a duration of 8 weeks, which gives us a latest start point of 17.
25 - 8 = 17
Since 17 is the latest time point at which activity D can start, both activities B and C should be finished by this time. If either activity B or C are finished later, the entire project does not finish in week 31.
Activity B has a duration of 10 weeks and the latest start point is 7.
17 - 10 = 7
Activity C has a duration of 12 weeks and has a latest start point of 5.
17 - 12 = 5
Now we have two values for activity A that can give us the latest finish. We should choose the lowest value, which is value 5.The latest start for activity A is 0 and the duration is 5.
5 - 5 = 0
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For activity A, the float becomes 0.
5 - 5 = 0
0
2
0
0 0A
5 WEEKS
0 5
0 5
B
10 WEEKS
5 15
7 17
C
12 WEEKS
5 17
5 17
D
8 WEEKS
17 25
17 25
E
6 WEEKS
25 31
25 31
ACTIVITY
(t)
ES EF
FL
LS LF
Now we will calculate the float of the different activities in the network. We find the floats by looking at the differences between the latest finish points (LF) and the earliest finish points (EF).
For activity E, the float becomes 0.
31 - 31 = 0
For activity D, the float becomes 0.
25 - 25 = 0
For activity C, the float becomes 0.
17 - 17 = 0
For activity B, the float becomes 2.
17 - 15 = 2
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0
2
0
0 0A
5 WEEKS
0 5
0 5
B
10 WEEKS
5 15
7 17
C
12 WEEKS
5 17
5 17
D
8 WEEKS
17 25
17 25
E
6 WEEKS
25 31
25 31
ACTIVITY
(t)
ES EF
FL
LS LF
This means that all activities in the network are critical, except activity B. Therefore, we find a chain of critical activities A, C , D and E, which form a critical path in the network.
At the project start at time 0, the project will have a total duration of 31 weeks.