So, let's go through two examples in the module here highlighting first to draw the activity on node and to build and perform the calculations on the forward and the backward pass. So if we can have here a simple project and we want to draw an activity on node, and perform the CPM forward and CPM backward calculations. Activity A, as we can see here, it doesn't have any predecessors, which highlighted, it's the first activity on the projects. And here the duration of each of the activities, and so on from B, C, all the way to G. So let's go one by one. So the first activity we said, we have activity A here, and then activity B is a predecessor of A, and then we have activity C. The predecessor A, and then we have activity E or activity D, predecessors of B. So in this case, that's the relationship between the activities. And then we have activity E, the predecessor of B and C. So let's say, if we have E here and activity B and activity C, and then we have activity F, which are predecessor of C. So let's have F here, so C is the predecessor of F. And the last activity, G, the predecessors of D, E and F. We have 1 here. So here's activity or not. Let's put the durations of, I think four here. 7 days, 5 for C, E also 5 days. D around 8 days and then we have 2 days here and the 1 day here. So that been said, let's go through what we went through previous example from the forward and the past ward, or the backward I'm sorry, past calculations. We assumed that we have an early start of 0 + 4, which will give 4. And then before we move forward, let's remind you about the key to have it, as this is the activity, and this is the duration. And we have here the early start, the early finish, the late start, and the late finish. So in this case, the early start 0 plus the duration of 4, the early finish 4 and then the early start for the successors will be the same as 4 here, so we have 4, we have 4. 4 + 7, we have 11 which will be the early finish. 4 + 5, 9, the early finish for C. Then we have for D, the early start is 11. 11 + 8, you have 19, the early finish. Let's go for F, 9, we took it from the early finish of activity C, its only predecessor. And 9 + 2, 11. And then we have for activity E is the maximum between 11 and 9, because these are the two predecessors for E whenever which one finishes, the last one to finish will be the early start for here, 11 + 5, 16. In this case, what's left in the forward pass calculations is the maximum number between all the predecessors of activity G, which will be 19, then 19 + 1, you have 20. So in this case we do have here the forward pass calculations, and these are the earliest dates, start and finish, of all the activities of the projects. And the minimum duration to execute the project is going to be in this case, 20. So let's go from the backward past calculations as we mentioned, this is the only number we can get from the four past calculations and the only connection we have with the four past calculations, so we have 20. And you go the same calculations but backward 20- 1, 19. And then you going to take the late start of the activity to go to the late finish of all the predecessors. So we have 19 here, 19 here, 19 here, 19- 8, 11. 19- 5, 14, and then we have 17. So what's left for us is the last three activities here, from B. We can have between 14 and 11. So you take that minimum of 11, and 14 and 17, you take the minimum of 14. And then you have 4 and you have 9. So the minimum between both for activity A to find their late finish is 4 and then 0. So that would be the forward and the backward past calculations for our example in the project here.