Ardavan Asef-Vaziri:	Thank you very much for attending this session. Today I'm going to talk about constructing a project network and computing the critical path. Critical path is the path in a network which takes time more than any other path and the duration of the project is defined by the critical path. Let's first construct a project network. Activity B has its precedence as activity A. Activity C also has its precedence as activity A and also activity D. Activity E has its precedence activity B, precedence of F is C, precedence of G is also C. Now we want to construct a network represent this explanation in a visualization form. We can start activity A. It has no activity as precedence. When activity A completes, then we can start activity B, we can start activity C, and we can start activity D. When we complete activity B, we can start activity E. When we complete activity C, we can start F and G. To start H we need to complete E and F, as soon as they are completed, we can start H. To start I, we need to complete D and G. When H and I are completed, then we can start J. Start of A is a start of the project, end of J is end of the project. There are one path here, one path here, one path here, and one path here. The project has four paths. We need to compute each and find it out which one is the longest. Let's solve a second problem. Here, we have immediate precedence, and here we have activity. As soon as activity A is completed, we can start activity B. When B is completed, we can start C and D. When D is completed, we can start E, F, and G. When C, E, F, and G are completed, we can start H. When H is completed, we can start I and J. When I is completed, we can start K and when J is completed, we can start L. When K and L are completed, we can start M. When we finish M, the project is over. To how many paths this network has. Let's see if my formula works. My formula is here, we have two. At any of the following nodes, if you have more than one arc going out, that arc is added to the number of patterns. When I come here, I have three arcs going out. Three is going out. One is already counted, so 3-1, two more are added to that. Now here, four paths, not four arcs, four paths. Here there are four paths coming into this node. The number of arcs is the same as the number of paths, but in some instances, they are different. I have 2+2. Now, four arcs are coming here. If one arc was going out, if I didn't have these two, then the number of paths were four. But now I have two arcs. One of them has already been accounted for, so four paths are added. Therefore, this project has eight paths. Let's see if my formula works. One, 2, 3, 4, 5, 6, 7, and 8. It worked. In some complicated networks, you shouldn't count the number of arcs which come here but number of paths. But here, number of paths is equal to number of arcs. We have eight paths here. The critical path is the longest one. Here, we only have provided you with technical precedents. We just gave you immediate precedence of each activity. It was the same here, too. For each activity, we gave you immediate precedence of that activity. We have not talked about times yet. Let's look at the next project to construct its network. Here again, we have immediate precedences, and also we have durations, which allows us to compute the lengths of each path. Here are all activities. Activity A1 and A2 do not have any precedents, then we can start them. A3 has its precedence as A1, A4 also has its precedence at A1, A5 has its precedence as A2, A6 has its precedence as A4 and A5. When A3 and A6 are completed, the project ends. When A1 or A2 is started, the project starts. A3 and A6, both end project ends. A1 or A2 starts, project starts. Here we have the duration of each activity. How many paths do we have? Two. Here we have two going out. One of them has been already accounted for, so one is added, and we do not have anything else like this or three paths. Blue path. Ten units of time, 10 days, 10 weeks, whatever, 10 hours. Red path, 4+4+3, 11 hours or 11 days or 11 weeks. Green, 3+2+3, eight. This is the critical path because its length is more than any other path. Critical path is the longest path. It is the shortest time to complete the project. The longest path is the shortest time to complete the project. Now, let's go and compute the critical path in a formal way. We can start project at Day 0. Let's assume these numbers are days. Activity A1 can start in Day 0 and activity A2 can start at Day 0. This is the earliest start. A1 can start at zero and its duration is four. Earliest finish is four. A2 can start at zero and duration three, so earliest finish three. This one can start at three. This one can start at four. This one can start at four. 4+6, earliest finish 10, 4+4, earliest finish, eight, 3+2, earliest finish five. Now, this is finished at eight. This is finished at five. But when can this start because it depends on both of these two activities? It cannot start until Day 8. 8+3, 11. This one is finished in Day 11, this one is finished in Day 10. The project ends in 11 days. This is the critical paths. I have four and three. Three here, four here. Click. Now we have this data here, 4+6, will go here, 4+4, will go here, 3+2, will go here. Click. Now, this one has two precedence. One would be here at Day 8. The other one would be here in Day 5. They both should be there to be able to start this one. Therefore, it is 8+3 will be 11. Click. You may practice it yourself if you did not fully follow. But I think the concept is quite straightforward and project takes 11 days. The only key idea was this. If more than one arc comes into a node, if more than one precedence an activity has, then the start of that activity is the maximum of the finish of its all precedence so if this ends at Day 10, if it ends at Day 13 and 20, then we will have max of 10. Then, for example, if the activity takes five days, then the earliest time to finish this activity is Day 45. Those rows which go up can start going up in day 35 and not earlier. Now let's do backward computation. Now, we assume that the earliest finish is the same as latest finish. This activity, the latest time that it can be finished is Day 11. Therefore, this activity, the latest time that it can be finished is Day 11, and this activity, the latest time that it can be finished is Day 11. But this activity by itself takes three days. Therefore, the latest time that this activity can start is Day 8. For this activity, the latest time that it can start is 11-6, which is five. The latest time that this one can start is eight. Therefore, the latest time that this one can finish and the latest time that this one can finish is eight. 8-4 is four, 8-2 is six. This is six, or the latest time for this one to finish is six, and the latest time to start is 6-3, which is three. The latest time for this activity to finish, this one, the latest to start is five, this one, the latest to start is four. Therefore, for this one, the latest finish is four because it should satisfy both of these. This one cannot start later than four, this one cannot start later than five, both together cannot start later than four. If the latest start for this activity is four and it takes four, therefore, the latest is a zero. Therefore, as soon as the project start, we need to immediately start A1, but we can delay A2 for a little bit. Let's see it again. Click 11. 11-6 is five, 11-3 is eight. Eight and five. Eight, 8, 4, 8-2, 6. 6-3, 3. The latest finish of this activity from this perspective is four from this perspective is five, but it should satisfy both of them. Therefore, the latest finish for this point is the minimum of these two, which is four. 4-4 is zero. Here, when the activity is connected to several activities after it, the latest finish of this activity is equal to the minimum of the latest start of the following activities which are directly connected to this activity. Therefore, here, I have 35, minimum of them. Then, for example, if this activity takes five days, then the latest start would be 35-5, which would be 30. This 30 will be applied to all activities which are directly associated with this activity. This is the final structure of our project, the graphical illustration of the project. Activities which have their earliest start, equal to latest start or earliest finish, equal to latest finish. These all have slack of zero. That means if we delay any of these activities for one day, the whole project will be enlonged by one day. We also have some other activities which the gap between earliest finished and latest finished, or earliest start and latest start is greater than zero. That means this activity A3 has one day as flexibility. If instead of Day 5, this activity has a slack of three. This one also has a slack of three. That means any of these two activities, but not both of them can be delayed by 1, 2, or 3, or a combination of the two activities could be delayed by one or two or three and still they will not increase the length of the project. Instead of Day 0, we may start it at Day 1. Instead of Day 3, we may start this in Day 5. Nothing will happen to the length of the entire project. These two activities together, they have three days slack and this one has one day slack, and this is our critical paths. In my next talk, I will talk about construction of a little bit more complicated project and also compute forward path and backward path for a little bit more complicated project. I will also discuss the concept of total flow or total slack and free slack or free flow. Thank you very much for attending this topic.