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5 WEEKS

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.

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5 WEEKS

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  1. 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. First, we calculate early positions for all of the activities. These show the earliest start points (ES) and earliest finish points (EF). The figure in the lower left corner shows the different values that are calculated in the network. Http://www.prosjektledelse.ntnu.no

  2. B 10 WEEKS A D E 5 WEEKS 8 WEEKS 6 WEEKS C 12 WEEKS ES EF ACTIVITY FL (t) LS LF 5 15 0 5 17 25 25 31 5 17 Activity E therefore has an earliest start point in week 25. 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 The earliest finish for activity E is week 31 because the duration of the activity is 6 weeks. 25 + 6 = 31 Activity C has a duration of 12 weeks and its’ earliest finish is in week 17. 5 + 12 = 17 The duration of activity D is 8 weeks, so its earliest finish is 25. 17 + 8 = 25 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. 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 Http://www.prosjektledelse.ntnu.no

  3. 5 15 B 10 WEEKS 0 5 17 25 25 31 A D E 5 WEEKS 8 WEEKS 6 WEEKS 5 17 C 12 WEEKS ES EF ACTIVITY FL (t) LS LF 7 17 0 17 25 5 25 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 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. 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. The latest start for activity A is 0 and the duration is 5. 5 - 5 = 0 Activity D has a duration of 8 weeks, which gives us a latest start point of 17. 25 - 8 = 17 Activity C has a duration of 12 weeks and has a latest start point of 5. 17 - 12 = 5 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 Additionally, by looking at the network structure we can see that activity D at has a latest finish point in week 25. 5 17 Http://www.prosjektledelse.ntnu.no

  4. 5 15 B 10 WEEKS 0 5 17 25 25 31 7 17 A D E 5 WEEKS 8 WEEKS 6 WEEKS 5 17 0 17 25 5 25 31 C 12 WEEKS 5 17 ES EF ACTIVITY FL (t) LS LF 2 0 0 0 0 For activity A, the float becomes 0. 5 - 5 = 0 For activity D, the float becomes 0. 25 - 25 = 0 For activity E, the float becomes 0. 31 - 31 = 0 For activity B, the float becomes 2. 17 - 15 = 2 For activity C, the float becomes 0. 17 - 17 = 0 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). Http://www.prosjektledelse.ntnu.no

  5. 5 15 B 2 10 WEEKS 0 5 17 25 25 31 7 17 A D E 0 0 0 5 WEEKS 8 WEEKS 6 WEEKS 5 17 0 17 25 5 25 31 C 0 12 WEEKS 5 17 ES EF ACTIVITY FL (t) 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. Http://www.prosjektledelse.ntnu.no

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