# Scheduling. Getting Started. Scheduling 79

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2 80 Example: A complete digraph for our car conversion would look like this: T 1 (2) T 2 (0.5) (2) (1) T 6 (0.5) T (0.5) T 9 (1) T 5 (2) T 8 (0.5) The time it takes to complete this job will partially depend upon how many people are working on the project. In scheduling jargon, the workers are called processors. While in this example the processors are humans, in some situations the processors are computers, robots, or other machines. For simplicity, we are going to make the very big assumptions that every processor can do every task, that they all would take the same time to complete it, and that only one processor can work on a task at a time. If we had only one processor working on this task, it is easy to determine the finishing time - how long it will take to complete all the tasks; just add up the individual times. We assume one person can t work on two tasks at the same time, ignore things like drying times during which someone could work on another task. Scheduling with one processor, a possible schedule would look like this, with a finishing time of 10 days. T 1 T 5 T 2 T 6 T T 8 T 9 In this schedule, all the ordering requirements are met. This is certainly not the only possible schedule for one processor, but no other schedule could complete the job in less time. Because of this, this is an optimal schedule with optimal finishing time there is nothing better. For two processors, things become more interesting. For small digraphs like this, we probably could fiddle around and guess-and-check a pretty good schedule. Here would be a possibility: T 1 T 6 T 9 P 2 T 5 T 2 T 8 T With two processors, the finishing time was reduced to 5.5 days. What was processor 2 doing during the last day? Nothing, because there were no tasks for the processor to do. This is called idle time.

5 Scheduling 83 Time 4: Both processors complete their tasks. The completion of T 5 allows T 8 to become ready. We assign T to and T 8 to P 2. Priority list: T 1,,, T 5, T 6, T, T 8, T 2, T 9 T 1 T 2 T 6 T P 2 T 5 T 8 Time 4.5: Both processors complete their tasks. T 9 becomes ready, and is assigned to. There is no ready task for P 2 to work on, so P 2 idles. Priority list: T 1,,, T 5, T 6, T, T 8, T 2, T 9 T 1 T 2 T 6 T T 9 P 2 T 5 T 8 With the last task completed, we have a schedule. It is worth noting that the list processing algorithm itself does not influence the resulting schedule the schedule is completely determined by the priority list followed. The list processing, while do-able by hand, could just as easily be computed by a computer. The interesting part of scheduling, then, is how to create the best priority list possible. Choosing a priority list We will explore two algorithms for selecting a priority list. Decreasing time algorithm The decreasing time algorithm takes the approach of trying to get the very long tasks out of the way as soon as possible by putting them first on the priority list. Decreasing Time Algorithm Create the priority list by listing the tasks in order from longest completion time to shortest completion time.

9 Scheduling 8 I m sure you can imagine that searching for the critical path every time you remove a task from the digraph would get really tiring, especially for a large digraph. In practice, the critical path algorithm is implementing by first working from the end backwards. This is called the backflow algorithm. Backflow Algorithm 1) Introduce an end vertex, and assign it a time of 0, shown in [brackets] 2) Move backwards to every vertex that has an arrow to the end and assign it a critical time 3) From each of those vertices, move backwards and assign those vertices critical times. Notice that the critical time for the earlier vertex will be that task s time plus the critical time for the later vertex. Example: T 1 (5) T 2 (4) [10] In this case, if T 2 has already been determined to have a critical time of 10, then T 1 will have a critical time of 5+10 = 15 T 1 (5) [15] T 2 (4) [10] If you have already assigned a critical time to a vertex, replace it only if the new time is larger. Example: In the digraph below, T 1 should be labeled with a critical time of 16, since it is the longer of 5+10 and T 1 (5) T 2 (4) [10] T 2 (3) [11] 4) Repeat until all vertices are labeled with their critical times One you have completed the backflow algorithm, you can easily create the critical path priority list by using the critical times you just found. Critical Path Algorithm (version 2) 1. Apply the backflow algorithm to the digraph 2. Create the priority list by listing the tasks in order from longest critical time to shortest critical time This version of the Critical Path Algorithm will usually be the easier to implement.

10 88 Example: Applying this to our digraph from earlier, we add an end vertex and give it a critical time of 0. T 1 (6) T 5 (5) T 9 (2) T 2 (3) () T 6 (10) T 8 (3) T 10 () End [0] (4) T (4) We then move back to, T 9, and T 10, labeling them with their critical times T 1 (6) T 5 (5) T 9 (2) [2] T 2 (3) T 6 (10) T 8 (3) T 10 () [] () End [0] (4) [4] T (4) From each vertex marked with a critical time, we go back. T, for example, will get labeled with a critical time 11 the task time of 4 plus the critical time for T 10. For T5, there are two paths to the end. We use the longer, labeling T 5 with critical time 5+ = 12. (This value will end up getting replaced later with the even longer path through T 8 ). T 1 (6) T 5 (5) [12] T 9 (2) [2] T 2 (3) T 6 (10) T 8 (3) [10] T 10 () [] () End [0] (4) [4] T (4) [14]

11 Scheduling 89 Continue the process until all vertices are labeled. T 1 (6) [21] T 5 (5) [15] T 9 (2) [2] T 2 (3) [28] T 6 (10) [25] T 8 (3) [10] T 10 () [] () [21] End [0] (4) [4] T (4) [14] We can now quickly create the critical path priority list by listing the tasks in decreasing order of critical time: Priority list: T 2, T 6, T 1,, T 5, T, T 8, T 10,, T 9 Applying this priority list using the list processing algorithm, we get the schedule: T 2 T 6 T 10 T 5 T T In this particular case, we were able to achieve the minimum possible completion time with this schedule, suggesting that this schedule is optimal. This is certainly not always the case. Example: This example is designed to show that the critical path algorithm doesn t always work wonderfully. T9 T 1 (1) T 6 (2) T 2 (1) T (2) T 9 (6) (1) T 8 (2) (6) T 5 (6)

12 90 To create a critical path priority list, we could first apply the backflow algorithm: T 1 (1) [9] T 6 (2) [8] T 2 (1) [1] T (2) [8] T 9 (6) [6] (1) [1] T 8 (2) [8] (6) [6] T 5 (6) [6] This yields the critical-path priority list: T 1, T 6, T, T 8,, T 5, T 9, T 2,. Applying the list processing algorithm to this priority list leads to the schedule: T 1 1 T 6 3 T 5 T 9 13 T 8 T2 P 2 P 3 T 5 6 By observation, we can see that a much better schedule exists: 9 P 2 T 1 1 T 6 3 T 2 T 9 T T 5 P 3 T 8 In most cases the critical path algorithm will lead to a very good schedule. There are cases, like this, where it will not. Unfortunately, there is no known algorithm to always produce the optimal schedule.

13 Scheduling 91 Exercises Skills 1. Create a digraph for the following set of tasks: Task Time required Tasks that must be completed first A 3 B 4 C D 6 A, B E 5 B F 5 D, E G 4 E 2. Create a digraph for the following set of tasks: Task Time required Tasks that must be completed first A 3 B 4 C D 6 A E 5 A F 5 B G 4 D, E Use this digraph for the next 6 problems. T 1 (8) T 5 (9) T 9 (2) T 2 (6) T 6 (3) T 8 (5) T 10 () () (4) T (2) 3. Using the priority list,, T 9, T 10, T 8, T 5, T 6, T 1, T, T 2 schedule the project with two processors. 4. Using the priority list T 2,, T 6, T 8, T 10, T 1,, T 5, T, T 9 schedule the project with two processors. 5. Using the priority list,, T 9, T 10, T 8, T 5, T 6, T 1, T, T 2 schedule the project with three processors.

14 92 6. Using the priority list T 2,, T 6, T 8, T 10, T 1,, T 5, T, T 9 schedule the project with three processors.. Use the decreasing time algorithm to create a priority list for the digraph above, and schedule with two processors. 8. Use the decreasing time algorithm to create a priority list for the digraph above, and schedule with three processors. 9. Use the decreasing time algorithm to create a priority list for the problem from #1, and schedule with two processors. 10. Use the decreasing time algorithm to create a priority list for the problem from #2, and schedule with two processors. 11. With the digraph from above question 3: a. Apply the backflow algorithm to find the critical time for each task b. Find the critical path for the project and the minimum completion time c. Use the critical path algorithm to create a priority list and schedule on two processors. 12. With the digraph from above question 3, use the critical path algorithm to schedule on three processors. 13. Use the critical path algorithm to schedule the problem from #1 on two processors. 14. Use the critical path algorithm to schedule the problem from #2 on two processors. Concepts 15. If an additional order requirement is added to a digraph, can the optimal finishing time ever become longer? Can the optimal finishing time ever become shorter? 16. Will an optimal schedule always have no idle time? 1. Consider the digraph below. a. How many priority lists could be created for these tasks? b. How many unique schedules are created by those priority lists? T 1 (6) T 2 (5) () (3) T 5 (4)

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