So let's take another example here. Let's draw also an activity on node network diagram and perform the forward and the backward pass calculations for the following activities. We try to make it a little bit more challenging this time, so I would suggest for you to study the table, try to do it yourself, and then let's solve it together. So let's look at the table here, we have activity A, B, C all the way to I. These are all the predecessors of the activities, and these are the durations we have on each of activities. So it says, we have activity, A here, which will be the beginning of our project. And activity B, C, and D, they are the successors of activity A. So we have activity B, Activity C, activity D. And then, we do have activity E, a successor of B. So let's have activity E here. And then, we do have activity F, the successor for B and C. We said B has one predecessor, which is B. And activity F has two predecessors, which is B and C. And then, we do have activity G, the predecessors C and D. And then, we have two last activities which will be H and I. H predecessor is E and F. E and F, and we have activity I, which the predecessors G and H. So I, Once H finishes and G finishes, then I can start. So that will make activity I the last activity we have for our project. So this is the activity on node diagram for our project. So let's do and start performing the forward and the backward pass calculations for this project. Let's remind you again that we have the key of here's, Activity and thus the duration. And then, we do have here the early start, early finish, late start, and late finish. Let's put the durations of all the activities. We have 2 here. We have 5 for B. C, we have 9. 3 for D. We have 6 for E. We have 4 for F. We have 2 for G. We have 4 for H. And 2 for I. So for the forward pass calculations, we said, we start with all was, 0 plus 2 equal 2. The early start of first activity 0 plus 2 early finish of activity A is 2. The early start for activity B, C, and D will be the early finish of the predecessor of A. And then we have, then the early finish of all these activities is the early start plus the duration of each of the activities. So we have 7. We have 11, we have 5. So let's move forward. We have activity E, only one predecessor, so that sentel, early start of seven which is equal to the early finish of their predecessor. Then we have 7 plus 6, 13 is the early finish of activity E. For both activity F and G, they do have two predecessors. So for F is the maximum number of both early finishes, which would be 11. And for activity G, the maximum is also 11 and 5, we'll take 11. Again, the reasoning is before we start any of these activities, we want to make sure both activities in the scales B and C will finish before we start F. And both activities will finish from C and D before we start G. So in this case, we have the early finish for F is 15. The early finish for G is 13. So we have two activities left, activity H and I. We have early start will be the maximum number of the early finishes of E and F, which will be 15. And I can't jump to activity I yet because activity I depends on both activities H and G, and I still need to do the math here. So 15 plus 4 is 19, and the maximum between 19 and 13 for activity I is then 19 plus 2 is 21. So with that is the forward pass calculations we have for this activity on node diagram. So let's perform the backward pass calculations. The first step as maybe now you know is to take the minimum duration of the project. The early finish date of the last node or the last activity we have in the project which is I in this case. We have 21 minus 2 to find the late start of activity I, 19. So the late finish for H and G will be also equal to the late start of activity I 19, 19, and then we have 19 minus 4, 15. 19 minus 2, 17. The late start of H, 15. The late start of G, 17. And then, let's keep going from that backward pass calculations. The late start 15 will go as a late finish for E and F also as 15. And then, we have 15 minus 6, 9. And then, we have 15 minus 4, 11 for activity F. So the late start for E is 9, the late start of F, 11. And that will leave us for only these activities here. Activity B is the minimum late start of activity E and F, as we explained before. So we have 9, and then, C also we do have two activities, 11 and 17. The minimum will be 11. So the late finish of activity C is 11. At activity D only we have 1 activity here. So the late finish will be the same as the late start, 17. 17 minus 3, 14, which will be the late start of activity D. And then, we have 11 minus 9, 2, which will be the late start of activity C. The late start of activity B, 9 minus 5, 4, and that will leave us with the last activity. Or the first activity we have of the entire project, the late finish date for that activity will be the minimum between the late start date of activity B, which is 4, activity C, which is 2, and activity D, which is 14. So we'll take number 2, 2 minus 2 equals 0, which will be the late start of activity A. So that will be the forward and the pass forward of the backward pass calculations, We have for this simple projects.
