Activity

1. Activity Duration Predecessor Activity A 3 - B 5 - C 4 A D 7 B E 10 B F 3 E G 8 C, D Example AOA (Activity On Arrow) In this example, we will carry out a calculation by using an AOA-network and the information given in the table to the left. The project consists of 7 activities. The durations and sequences of the activities are shown in the table.

2. 1 Node nr. Activity Duration Predecessor activity A 3 - B 5 - C 4 A Earliest Latest D 7 B Time point Time point E 10 B F 3 E G 8 C, D In order to start the project, we need a start node (event). For each node, we divide a circle into upper and lower halves. The upper half represents the activity number. We then divide the lower part into halves. We write the earliest start point on the left half, and the latest start point on the right half.

3. 2 5 C (4) G (8) D (7) 1 3 6 E (10) F (3) 4 Activity Duration Predecessor activity A 3 - B 5 - C 4 A D 7 B E 10 B F 3 E G 8 C, D A (3) B (5) Activity B has a duration of 5 units and lies between nodes 1 and 3. Both activities A and B have predecessor activities and begin with start-node 1. By using the information in the table, we can draw a network. The lines in the network represent the activities given in the table. The durations that are given in the table are given in parenthesis in the network as well. Activity A, which is shown between nodes 1 and 2, has a duration of 3 units. These relationships are also illustrated for activities C, D, E, F and G. With this network, we will carry out a calculation.

4. 2 5 C (4) G (8) A (3) D (7) 1 3 6 B (5) E (10) F (3) 4 Activity Duration Predecessor activity A 3 - B 5 - Node nr. C 4 A D 7 B E 10 B F 3 E G 8 C, D Earliest Latest Time point Time point The calculation is carried out in two steps. First, we move forward through the network. In this step we will determine the earliest start points for all nodes. In the second step, we move backward from the final node in the network to the start node. In this step we determine the latest start points for each activity in the network.

5. 2 5 C (4) G (8) A (3) D (7) 1 3 6 B (5) Activity Duration Predecessor activity A 3 - E (10) F (3) B 5 - 4 C 4 A D 7 B E 10 B F 3 E G 8 C, D 3 0+3=3 0+5=5 If event 1 has earliest start point at time 0, event 2 cannot start before the end of activity A. Activity B has a duration of 5, so the earliest start point for event 3 becomes 5. The duration of activity A is 3. Therefore, event 2 has an earliest start point at 3. In this way, we can calculate the rest of the earliest start points in the network. The only way to reach node (event) 3 is to accomplish activity B. The duration of activity E is 10, so the earliest start point for event 4 is 10 plus the earliest start point for event 3, which is 5. This gives us 15 as the earliest start point for event 4. We can reach node (event) 4 by passing through activity E. Thus far, we have not said anything about the starting time of the project. For this exercise,we will say that the project starts at time 0. 5 0 15 5+10=15

6. 2 5 C (4) G (8) A (3) D (7) 1 3 6 B (5) Activity Duration Predecessor activity A 3 - E (10) F (3) B 5 - 4 C 4 A D 7 B E 10 B F 3 E G 8 C, D 3+4=7 12 3 5+7=12 12+8=20 We have two different values for the earliest start point, but both activities C and D must be finished before event 5 is begun. By choosing the larger value, both activities will be finished before event 5 begins. Therefore, the earliest start point for event 5 becomes 12. Again we have two different values, and again we should choose the largest of them. Therefore, the earliest start for activity 6 becomes 20. Now we have determinedall earliest start points for all activities. There are two possible routes for reach event 5: via activities C or D. If we go via activity C, the earliest start point for event 5 becomes 3 plus the duration of activity C, which is 4. This gives us an earliest start poing of 7. By going through activity D, the earliest start point for the event becomes five plus the duration of activity D, which is 7. This gives us an earliest start point of 12 for event 5. The same applies to event 6 where we have two routes to choose from in order to reach it: via activities F or G. By passing through activity F we add 15 plus 3, the duration of F. This gives us an earliest start point of 18. By passing through activity G we add 12 plus 8, the duration of G. The earliest start point for event 6 becomes 20. 5 20 0 15+3=18 15

7. 2 5 C (4) 12 3 G (8) A (3) D (7) 1 3 6 B (5) Activity Duration Predecessor 5 20 0 activity A 3 - E (10) F (3) B 5 - 4 C 4 A D 7 B 15 E 10 B F 3 E G 8 C, D We will now carry out a process similar to the one we used to caculate the earliest start points, but this time we will go bavkwards through the network by subtracting In order to reach event 5 we can go through activity G, which has a duration of 8. In order to find the latest start points for each event, we will now calculate backwards through the network. If a finish date had already been determined for the project, we would insert this value into the blank space on the right part of event 6. Since no date has been given, we assume that the project should be finished as soon as possible. Therefore, we say that the latest start pointfor event 6 is equal to its earliest start point, which is20. Therefore, the latest start point for event 4 becomes: 20 - 3 = 17 Therefore, the latest start point for event 5 becomes: 20 - 8 = 12 If event 5 starts later than 12, the latest start for event 6 will have to change. For event 4, we go from event 6 through activity F, which has a duration of 3. 12 20-8=12 20 17 20-3=17

8. 2 5 C (4) 12 3 G (8) A (3) D (7) 1 3 6 B (5) Activity Duration Predecessor 5 20 0 activity A 3 - E (10) F (3) B 5 - 4 C 4 A D 7 B 15 E 10 B F 3 E G 8 C, D 12-4=8 12-7=5 8-3=5 12 8 From event 3, we subtract 5 (the duration of activity B) from 5, which gives us a latest start point of 0 for event 1. The backward calculation for event 3 is slightly similar to the forward calculations for event 5 and 6, where we had to choose between two values for the earliest start point. The latest start point for event 3 will therefore be one of the two values reached through either event 4 or 5. Initially, we will go from event 4 and subtract 10 (the duration of activity E) from 17. This gives us a latest start point of 7 for event 3. Now we must look at the alternative calculation. By coming from event 5, we get 12 minus 7 (the duration of activity D), which gives us a latest start point of 5. We now have two values, and we should choose the lower value as the latest start point. Therefore, the latest start point for event 3 is 5. We can reach event 2 by going through activity C. The latest start point for event 2 is 12 minus 4 (the duration of activity C), which equals 8. For event 1, we again have two alternative values to choose from. From event 2, we subtract 3 (the duration of activity A) from 8, which gives us a latest start point of 5. However, we also need to look at the value from event 3. Again, we have two values and must choose the smaller one. Therefore, the latest start point for event 1 is 0. 5 20 0 5-5=0 17 17-10=7

9. Activity Duration ES EF LS LF FL A 3 0 3 5 8 5 B 5 0 5 0 5 0 C 4 3 7 8 12 5 D 7 5 12 5 12 0 E 10 5 15 7 17 2 F 3 15 18 17 20 2 G 8 12 20 12 20 0 Activity F has event 4 as its starting node. Therefore, the earliest start point for activity F is 15. Activity C has event 2 as its starting event. Therefore, the earliest start point for activity C is 3. Both activities D and E have node 3 as their starting node, where the earliest start point is 5. By looking at the network, we see that activities A and B have event 1 as their starting event, where the earliest start point is 0. In other words, the earliest start point for activities A and B is also 0. We can extract the earliest start points for all activities by looking at the earliest start points of the start-nodes for each activity. Now we will calculate the earliest start points for all of the activities. This value is labeled ES (earliest start). First, we list all of the activities in the leftmost column. In the next column, the respective durationsfor the avtivities are shown. For example, activity A has a duration of 3. Similarly, activity G has event 5 as its starting event, which gives it an earliest start point of 12. We have now carried out calculations on all of the project nodes. Next, we will perform calculations for each of the activities, which we will do in the table.

10. Activity Duration ES EF LS LF FL A 3 0 3 5 8 5 B 5 0 5 0 5 0 C 4 3 7 8 12 5 D 7 5 12 5 12 0 E 10 5 15 7 17 2 F 3 15 18 17 20 2 G 8 12 20 12 20 0 For activity B, the EF is equal to 5 plus 0, which gives us 5. For activity C, the EF is 4 plus 3, which gives us 7. The calculations for all of the remaining activities are carried out similarly. For activity A, we add 0 and 3, which gives us an EF of 3. To find this value for each activity, we take the earliest start point of its starting event and add it to the duration of that activity. In other words, since the earliest start points of the activities are equal to the earliest start points of their starting events, we can find the EF by aggregating the ES and the duration of the activities. Now we will calculate the earliest time points at which the activities can end. These time points will be put into the column labeled EF (earliest finish). EFi=ESi+ti

11. Activity Duration ES EF LS LF FL A 3 0 3 5 8 5 B 5 0 5 0 5 0 C 4 3 7 8 12 5 D 7 5 12 5 12 0 E 10 5 15 7 17 2 F 3 15 18 17 20 2 G 8 12 20 12 20 0 For activity B, event 3 is the final event and has a latest start point of 5. Therefore, the latest finish point for activity B is 5. Event 5 is the final event for both activities C and D. This gives us a latest finish point of 12 for both activities. For activity E, event 4 is the final event. Therefore, the latest finish point for activity E becomes 17. Activity A has event 2 as its final event. The latest time that event 2 can start is 8, therefore the latest finish of A is 8. We can find the latest finish points directly from the network diagram as well. The latest time that an activity can finish is the same as the latest start point of the end-event of that activity. Activities F and G have event 6 as their end-event, so the latest finish point for both activities is 20. Now we have completed the calculations for the earliest start and finish points for all of the activities. From now on, we will look at the latest start and finish points. First, we will look at the latest finish points (LF).

12. Activity Duration ES EF LS LF FL A 3 0 3 5 8 5 B 5 0 5 0 5 0 C 4 3 7 8 12 5 D 7 5 12 5 12 0 E 10 5 15 7 17 2 F 3 15 18 17 20 2 G 8 12 20 12 20 0 For the remaining activities, the same calculations are carried out. The results are illustrated in the table. Activity B has a latest finish point of 5 and a duration of 5. This gives us a latest start point of 0. The latest start point for activity C is: 12 - 4 = 8 For activity A, the latest start becomes: 8 - 3 = 5 Where 8 is the latest finish point and 3 is the duration of the activity. The latest start point is the latest finish point minus the duration of each activity. Now the latest start point for each activity must be calculated. In the table this is labeled LS (latest start). LSi=LFi-ti

13. Activity Duration ES EF LS LF FL A 3 0 3 5 8 5 B 5 0 5 0 5 0 C 4 3 7 8 12 5 D 7 5 12 5 12 0 E 10 5 15 7 17 2 F 3 15 18 17 20 2 G 8 12 20 12 20 0 We can determine this by looking at the floats of the different activities. In order to calculate the float of each activity, we take the latest time of its end-node and subtract the earliest time of its start-node. This gives us the total time period in which the activity will be carried out. From this value, we subtract the duration of the same activity. The remaining value is called the float. The float is surplus time and represents the amount of freedom available for planning each activity. Now we must find out which activities are important in determining the end date. In our case, the FL (float) is equal to LF minus ES minus the duration of the activity. Now we have calculated all the necessary figures for the earliest and latest start and finish points for each event. We now know that the project will be accomplished in 20 days. Fli =LFi-Esi -ti =LFi-(Esi +ti) =LFi-Efi Fli =(LFi-ti)-ESi =LSi-ESi

14. Activity Duration ES EF LS LF FL A 3 0 3 5 8 5 B 5 0 5 0 5 0 C 4 3 7 8 12 5 D 7 5 12 5 12 0 E 10 5 15 7 17 2 F 3 15 18 17 20 2 G 8 12 20 12 20 0 The calculations are carried out similarly for the remaining activities. For activity B, the float is equal to: 5- 5 = 0 For activity C, the float is equal to: 12 - 7 = 5 Now we have two equivalent formulas for calculating the floats. In this example, we will use the LF minus EF formula. For activity A, the float is equal to: 8 - 3 = 5 Fli =LFi-Esi -ti =LFi-(Esi +ti) =LFi-Efi Fli =(LFi-ti)-ESi =LSi-ESi

15. 2 5 C (4) 12 3 G (8) A (3) D (7) 1 3 6 B (5) 5 20 Activity Duration ES EF LS LF FL 0 A 3 0 3 5 8 5 E (10) F (3) B 5 0 5 0 5 0 4 C 4 3 7 8 12 5 15 D 7 5 12 5 12 0 E 10 5 15 7 17 2 F 3 15 18 17 20 2 G 8 12 20 12 20 0 12 8 Some of the activities have a float of zero. This means that there is no freedom in planning these activities and they can not be extended. We call these types of activities critical activities. In contrast, the activities with positive floats can be extended without affecting the finish date of the network. Pay attention that the float that we speak about is incorporated into the continuous chain in the network. For example, activities E and F lie in a chain between events 3 and 6. They both have a float of 2 days, but we cannot use 2 days float for each of them. In other words, these two activities altogether have a float of two days. 2 days can be used for E or F, or 1 day for each of them. The same applies to the chain in which A and C lie. In this case, A and C share a float of five days. Activities B, D,and G, which all have floats of zero, are critical activities. We can draw a line linking the critical activities from the first event to the last event. This chain includes activities B, D and G and we call it the critical path. 5 20 0 17