1. ActivityDurationPredecessor Activity A3- B5- C4A D7B E10B F3E G8C, D In this example we will carry out a calculation by using AOA-network, on the basis of information given in the table. The Project consists of 7 activities. Durations and sequences are shown in the table. Example AOA (Activity On Arrow)

2. 1 Node nr. Latest Time point Earliest Time point ActivityDurationPredecessor activity A3- B5- C4A D7B E10B F3E G8C, D For being able to start the project, we should have a start node(event). For each node, we have divided the circle into upper and lover halves. The upper half represents the activity number. We have divided the lower part into halves as well. In the left half, we will write the earliest start and in the right half, the latest start that the activity can happen.

3. 1 2 3 A (3) B (5) 4 5 6 C (4) D (7) E (10)F (3) G (8) ActivityDurationPredecessor activity A3- B5- C4A D7B E10B F3E G8C, D By using information in the table, we can draw a network. Lines In the network represent activities given in the table. The durations which stand in the table are given in the parenthesis in the network as well. Activity A which is shown between node 1 and 2, has a duration of 3 units. Activity B has a duration of five units and lies between node 1 and 3. Both activities of A and B have predecessor activities and begin from start-node 1. Accordingly it is illustrated for activities C,D,E,F and G. In this network, we will carry out a calculation.

4. ActivityDurationPredecessor activity A3- B5- C4A D7B E10B F3E G8C, D Latest Time point Node nr. Earliest Time point The calculation is carried out in two steps. First we go forwards in the network. In this step we will determine earliest start for all nodes. In the second step, we go backwards from final node in the network to the start node. In this step we determine the latest time for each activity in the network. 1 2 3 A (3) B (5) 4 5 6 C (4) D (7) E (10)F (3) G (8)

5. 1 2 3 4 5 6 A (3) B (5) C (4) D (7) E (10)F (3) G (8) 0 3 5 15 5+10=15 0+3=3 0+5=5 ActivityDurationPredecessor activity A3- B5- C4A D7B E10B F3E G8C, D In our project, we have not said anything on startpoint of the project but we will calculate relatively and say that the project starts at time 0. If event 1 happens earliest at time 0, event 2 can not happen before end of activity A. The duration of activity A is 3. Therefore event 2 can happen earliest at 3. We will go on with forward calculation in the network. The only way for reaching to node 3 is to accomplish activity B. Activity B has a duration of 5 so the earliest start for event 3 becomes 5. We can reach Node(event) 4, by passing through activity E. The Duration of activity E is 10 so the earliest start for the event 4 is 10 pluss earliest start for event 3 which is 5. this gives 15 as the earliest start for event 4.

6. 1 2 3 4 5 6 A (3) B (5) C (4) D (7) E (10)F (3) G (8) 0 3 5 5+7=12 3+4=7 12 15 15+3=18 12+8=20 20 ActivityDurationPredecessor activity A3- B5- C4A D7B E10B F3E G8C, D For event 5, there are two possibilities for reaching to: via activity C or D. If we go via activity C the earliest start for event 5 becomes 3 pluss the duration of activity C equal to 4 that gives 7. But there was another way for reaching event 5. By going through activity D, the earliest event time becomes five pluss the duration of activity D which is 7. So we will have the earliest time of event 5 equal to 12. We have two different values, but both activities should be finished before event 5 occurrence. It results in the need to choose one of two and we choose the larger value.so the earliest start for event 5 becomes 12. The same applies to event 6 where we have two alternative ways for reaching to it: via activity F or Activity G. By passing through activity F we gain 15 pluss 3,the duration of F, that becomes 18. By going through, the activity G we obtain 12 pluss 8 which is the duration of G, so the earliest start for event 6 becomes 20. Now we have two different values and should choose the largest of them. Therefore the earliest start for activity 6 becomes 6. Now we have determined all the earliest starts for activities.

7. 1 2 3 4 5 6 A (3) B (5) C (4) D (7) E (10)F (3) G (8) 0 3 5 12 15 20 20-8=12 12 ActivityDurationPredecessor activity A3- B5- C4A D7B E10B F3E G8C, D we will calculate beckwards in the network. In the case that a finish date had been determined for the project, we would insert this value in the right half of event 6.but no date is given so we assume that project should be finished as soon as possible. Therefore we say that the latest time for event 6 is equal to its earliest time and we set 20 as the latest time for that. We will now carry out a similar process used for caculating ealiest time points but this time we will go bavkwards whole the way by subtracting For reaching to event 5, we can just go through activity G with a duration of 8. 20-3=17 17 For event 4, we go from event 6 through activity F with a duration of 3. Therefore the latest time for event 5 becomes 20 minus 8 equal to 12. If event 5 happens later than 12, event 6 will not have 20 as the latest start. The latest time for event 4 becomes 20 minus 3 that gives 17.

8. 1 2 3 4 5 6 A (3) B (5) C (4) D (7) E (10)F (3) G (8) 0 3 5 12 15 20 17 12 12-7=5 17-10=7 5 12-4=8 8 5-5=0 8-3=5 0 ActivityDurationPredecessor activity A3- B5- C4A D7B E10B F3E G8C, D Event 3 is slightly similar to event 5 and 6 when we calculate forwardly. The latest time for event 3 should be chosen between two alternatives because there are two acctivities branching from event 3. Initially we go from event 4 and gain 17 minus 10 which is the duration of activity E. This gives the latest time of 7 for event 3. But we should look at the other alternative as well. We gain 12 minus the duration of activity D equal to 7 that gives 5. Then we have values of 5 and 7 and should choose the lower value. So the earliest time for event 3 is 5. We can reach to event 2 just by going through activity C. By this way we gain 12 minus the duration of C which is 4. This results in a latest time equal to 8 for event 2. For event 1 we have a case with two alternatives as well. From event 2 we have 8 and the duration of activity A is 3, so we gain 5. but we need to look at other alternative as well. By starting from event 3 with 5 and subtracting the duration of activity B(5), we gain the latest time for event 1 that is equal to 0. Again we have two values, 0 and 5 and we should choose the smaller value.

9. Now we have carried out calculations of all of project nodes. we will now look at calculation of activities we will do this in the table. First we list all activities in a column in extreme left. In its adjacent column, the respective durations are shown. For example activity A has a duration of 3. Now we will calculate the earliest start of all the activities. This value is demonstrated by ES(earliest start) We can extract the earliest start of all activities by looking at earliest start of start-node of each activity. From network, we see that activity A and B have event 1 as their start event, where the earliest time is 0.In other words, the start time for activities A and B is 0. Both activity D and E have node 3 as their start node where the earliest start is 5. Activity C has event 2 as its start- event. Therefore the earliest start for activity C becomes 3. Activity F has event 4 as its start node. Therefore the earliest start for activity F becomes 15. Similarly, Activity G has the event 5 as start-event and it gives 12 as the earliest start for activity G.

10. Now we will calculate the earliest time points at which activities can end. These time points are supplied in the column under EF(Earliest Finish) For finding this value for each activity, we take the earliest time of its start event and add to the duration of that activity. In other words since the earliest start of activities is equal to earliest time of its start event, we can find EF by aggregating ES and the duration of activity. For activity A we gain 0 plus 3 that gives an EF equal to 3. Calculations of all remaining activities are carried out similarly. Activity C has the event 2 as its start event therefore the earliest start for activity C becomes 3. For activity B, EF is equal to 5 pluss 0 that gives 5 EF i =ES i +t i

11. Now we have accomplished the calculation of earliest start and earliest finish for all activities. From now on we will look at latest start and finish times. First we look at latest finish(LF). We can find this value directly from the network diagram as well. The latest time that an activity can finish is same as the latest time of the end- event of that activity. Activity A has event 2 as its final event and the latest time that the event 2 can happen is 8. Therefore the latest finish of A becomes 8. For activity E, event 4 is the final event and the latest finish becomes 17. Event 5 is the final event for both activities C and D and this gives latest finish of 12 for both activities. For activity B, event 3 with latest start of 5 is the final event. So the latest finish for activity B becomes 5. Activities F and G have event 6 as their end-event so the latest finish for both activities is equal to 20.

12. Now it remains to calculate the latest start of each activity. In the table it is demonstrated by LS(latest start) The latest start is the latest finish minus the duration of each activity. For activity A it becomes 8 minus 3 equal to 5. 8 is the latest finish and 3 is the duration of activity. The latest start for activity C becomes 8 or 12 minus 4. Activity B has a latest finish equal to 5 and a duration of 5. This gives a latest start of 0. For remaining activities, same calculations are carried out as illustrated in the table. LS i =LF i -t i

13. Now we have calculated sufficiently in order to know when earliest and latest times of each event occurs. We also know when each activity can start and finish at earliest and latest.we have found out that the project becomes accomplished in 20 days. Which activities influence in this end date? Which activities have importance for determining the end date. Fl i =LF i -Es i -t i =LF i -(Es i +t i ) =LF i -Ef i Fl i =(LF i -t i )-ES i =LS i -ES i We can determine this by looking at the float of different activities.In order to calculate float of each activity, we take the latest time of its end-node and subtract from that the earliest time of its start-node. Now we have the total period in which the activity will be carried out. From this value, we subtract duration of same activity. The remaining value is called float. The float is the surplus time and represents the amount that activities have freedom for planning. In our case, the FL(Float) is equal to LF minus ES minus the duration of activity 1.

14. For activity A LF is equal to 8 minus EF equal to 3 that gives a float of 5. Activity C has a float of 5, when LF is 12 and EF is 7. For activity B, the float is 5 minus 5=0.Similarly it is carried out for the remaining activities. Fl i =LF i -Es i -t i =LF i -(Es i +t i ) =LF i -Ef i Fl i =(LF i -t i )-ES i =LS i -ES i Now we have two equivalent formulas for calculation of the float. In this example We use LF minus EF.

15. 1 2 3 4 5 6 A (3) B (5) C (4) D (7) E (10)F (3) G (8) 0 3 5 12 15 20 17 12 5 8 0 Activities B,D, and G which all have float of zero are critical activities. We can draw a continous chain of critical activities from first event to the last event. This chain includes activities B, D and G and we call it critical path. Pay attention that the float that we are speaking about belongs to whole chain in the network. For instance activities E and F lie in a chain between event 3 and 6. They both have a float of 2 days; but we can not simultanously use 2 days float for each of them.In other words these two activities altogether have a float of two days. 2 days for E, 2 days for F or 1 day for each of them. The same applies to the chain in which A and C lie. In this case these two, altogether share a float of five days. Some of activities have float of zero. It means that there is no freedom in planning of such activities, in other words they can not be extended. We call this type of activities, critical activities. In contrast, activities with positive float have possibility to be extended without affecting the finish date of network.