 In this video, we provide the solution to question number 13 for practice exam number 2 for math 1030 in which case we are given a project digraph. So we have task ABCDEFGH and that actually is going to be our priority list for this project here. We're just going to do alphabetical order. So using this priority list, we have to construct the schedule of events using two processors on this project. And for convenience, the priority list is also listed down here. So at the start of the project, we're looking at the graph here. G would be ready. So is A, so is B. Okay? So by the priority list, job A is the highest priority. So we're going to give that to processor 1. That'll take four units of time to do A. Then we would get, so that one's going to be in execution now. B, we're going to give to processor 2. That'll take seven days to do that one. B, like so. And so B is now in execution. So then when we get to time four, A will then be done. Now A precedes C, but so is this B, which is not done yet. It also precedes H, which is not done yet because G is not done. So that means that G will be given to A. And so G takes five units of time to do. We're currently at time four. So this is going to move us up to time nine. Fill that in. And then this is job G, like so. So then at time seven, B will then be done. Now since B is done, B precedes C, so does A. So C is going to open up. Likewise, B precedes D and D has no other precedence. So D is now ready as well. By our priority list, C is considered the higher priority. So C will be given to processor 2. It takes three units of time. We're currently at time seven. So this will go up until time 10 as we fill that thing in there. This was task C. And so C is now in execution. Fast forward to time nine. At time nine, processor one will finish task G. Now that G is done, that opens up task H because both G and A are now done. H of course has lowest priority in our list here. So next we would then choose task D. D will then be given to processor one. D takes four units of time. We're at time nine right now. So this will take us up to 13, like so. This was task D, like so. So then we fast forward to time 10 now. At 10, C will be completed. C precedes E. So E is now open. C also precedes F, but D also has precedence over F and D is currently in execution. So F does not open up right now. So then E is the highest priority. So that's going to be given to processor two. It'll take two units of time. We are currently at time 10. So that's going to go until time 12. So this was E, like so. So that makes E in execution. Then if we fast forward to time 12, that is when E will be completed. E is completed. And now that E is completed, well, it doesn't open up anything else. So then the next, the only available task will be H. So we're going to give H to processor two here. It's currently time 12 and H has a processing time of eight. So that's going to bring us up to 20 at the very end. Fill that in. This is task H, like so. So H is now in execution. So then if we come to time 13, D will be completed. D precedes F. C is already done. So F is now open up. That is the last task. We're going to give that task to processor one. We're at time 13. And this will take six units of time. So 13 plus six is 19. So it'll finish this task at time 19. Like so. That was task F. And so then fast forwarding to time 19. At this moment, F will be then done. And there's no other tasks to do. So processor one will have to idle. It's going to have to idle for one time unit. Then at the, then one more time unit later, H will be done. That is the last task. So the project is now complete. So we see that the finishing time for this project will be 20 units of time. And this is the schedule we constructed. Processor one will do job A, then G, then D, then F. And processor two will do B, then C, then E, then H.