Scheduling 155

© David Lippman Creative Commons BY-SA

Scheduling An event planner has to juggle many workers completing different tasks, some of which must

be completed before others can begin. For example, the banquet tables would need to be

arranged in the room before the catering staff could begin setting out silverware. The event

planner has to carefully schedule things so that everything gets done in a reasonable amount

of time.

The problem of scheduling is fairly universal. Contractors need to schedule workers and

subcontractors to build a house as quickly as possible. A magazine needs to schedule

writers, editors, photographers, typesetters, and others so that an issue can be completed on

time.

Getting Started To begin thinking about scheduling, let us consider an auto shop that is converting a car from

gas to electric. A number of steps are involved. A time estimate for each task is given.

Task 1: Remove engine and gas parts (2 days)

Task 2: Steam clean the inside of the car (0.5 day)

Task 3: Buy an electric motor and speed controller (2 days for travel)

Task 4: Construct the part that connects the motor to the car’s transmission (1 day)

Task 5: Construct battery racks (2 days)

Task 6: Install the motor (0.5 day)

Task 7: Install the speed controller (0.5 day)

Task 8: Install the battery racks (0.5 day)

Task 9: Wire the electricity (1 day)

Some tasks have to be completed before others – we certainly can’t install the new motor

before removing the old engine! There are some tasks, however, that can be worked on

simultaneously by two different people, like constructing the battery racks and installing the

motor.

To help us visualize the ordering of tasks, we will create a digraph.

Digraph

A diagraph is a graphical representation of a set of tasks in which tasks are represented

with dots, called vertices, and arrows between vertices are used to show ordering.

For example, this digraph shows that Task 1, notated T1 for compactness, needs to be

completed before Task 2. The number in parentheses after the task name is the time required

for the task.

T1 (2) T2 (0.5)

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Example 1

A complete digraph for our car conversion would look like this:

The time it takes to complete this job will partially depend upon how many people are

working on the project.

Processors

In scheduling jargon, the workers completing the tasks 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.

Finishing Time

The finishing time is how long it will take to complete all the tasks. The finishing time

will depend upon the number of processors and the specific schedule.

If we had only one processor working on this task, it is easy to determine the finishing time;

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.

Time: 0 1 2 3 4 5 6 7 8 9 10

P1 T1 T3 T4 T5 T2 T6 T7 T8 T9

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.

T1 (2) T2 (0.5)

T3 (2)

T4 (1)

T5 (2)

T6 (0.5)

T7 (0.5)

T8 (0.5)

T9 (1)

Scheduling 157

Optimal Schedule

An optimal schedule is the schedule with the shortest possible finishing time.

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:

Time: 0 1 2 3 4 5 6 7 8 9 10

P1 T1 T3 T6 T9

P2 T5 T4 T2 T8 T7

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.

Idle Time

Idle time is time in the schedule when there are no tasks available for the processor to

work on, so they sit idle.

Is this schedule optimal? Could it have been completed in 5 days? Because every other task

had to be completed before task 9 could start, there would be no way that both processors

could be busy during task 9, so it is not possible to create a shorter schedule.

So how long will it take if we use three processors? About 10/3 = 3.33 days? Again we will

guess-and-check a schedule:

Time: 0 1 2 3 4 5 6 7 8 9 10

P1 T1 T2 T6 T8 T9

P2 T3 T7

P3 T4 T5

With three processors, the job still took 4.5 days. It is a little harder to tell whether this

schedule is optimal. However, it might be helpful to notice that since Task 1, 2, 6, and 9

have to be completed sequentially, there is no way that this job could be completed in less

than 2+0.5+0.5+1 = 4 days, regardless of the number of processors. Four days is, for this

digraph, the absolute minimum time to complete the job, called the critical time.

Critical Time

The critical time is the absolute minimum time to complete the job, regardless of the

number of processors working on the tasks.

Critical time can be determined by looking at the longest sequence of tasks in the

digraph, called the critical path

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Adding an algorithm Up until now, we have been creating schedules by guess-and-check, which works well

enough for small schedules, but would not work well with dozens or hundreds of tasks. To

create a more procedural approach, we might begin by somehow creating a priority list.

Once we have a priority list, we can begin scheduling using that list and the list processing

algorithm.

Priority List

A priority list is a list of tasks given in the order in which we desire them to be

completed.

The List Processing Algorithm turns a priority list into a schedule

List Processing Algorithm

1. On the digraph or priority list, circle all tasks that are ready, meaning that all pre-

requisite tasks have been completed.

2. Assign to each available processor, in order, the first ready task. Mark the task as

in progress, perhaps by putting a single line through the task.

3. Move forward in time until a task is completed. Mark the task as completed,

perhaps by crossing out the task. If any new tasks become ready, mark them as

such.

4. Repeat until all tasks have been scheduled.

Example 2

Using our digraph from above, schedule it using the priority list below:

T1, T3, T4, T5, T6, T7, T8, T2, T9

Time 0: Mark ready tasks

Priority list: T1, T3, T4, T5, T6, T7, T8, T2, T9

We assign the first task, T1 to the first processor, P1, and the second ready task, T3, to the

second processor. Making those assignments, we mark those tasks as in progress:

Priority list: T1, T3, T4, T5, T6, T7, T8, T2, T9

Schedule up to here:

We jump to the time when the next task completes, which is at time 2.

Time: 0 1 2 3 4 5 6 7 8 9 10

P1 T1

P2 T3

Scheduling 159

Time 2: Both processors complete their tasks. We mark those tasks as complete. With Task

1 complete, Task 2 becomes ready:

Priority list: T1, T3, T4, T5, T6, T7, T8, T2, T9

We assign the next ready task on the list, T4 to P1, and T5 to P2.

Priority list: T1, T3, T4, T5, T6, T7, T8, T2, T9

Time 3: Processor 1 has completed T4. Completing T4 does not make any other tasks ready

(note that all the rest require that T2 be completed first).

Priority list: T1, T3, T4, T5, T6, T7, T8, T2, T9

Since the next three tasks are not yet ready, we assign the next ready task, T2 to P1

Priority list: T1, T3, T4, T5, T6, T7, T8, T2, T9

Time 3.5: Processor 1 has completed T2. Completing T2 causes T6 and T7 to become ready.

We assign T6 to P1

Priority list: T1, T3, T4, T5, T6, T7, T8, T2, T9

Time 4: Both processors complete their tasks. The completion of T5 allows T8 to become

ready. We assign T7 to P1and T8 to P2.

Priority list: T1, T3, T4, T5, T6, T7, T8, T2, T9

Time: 0 1 2 3 4 5 6 7 8 9 10

P1 T1 T4

P2 T3 T5

Time: 0 1 2 3 4 5 6 7 8 9 10

P1 T1 T4 T2

P2 T3 T5

Time: 0 1 2 3 4 5 6 7 8 9 10

P1 T1 T4 T2 T6

P2 T3 T5

Time: 0 1 2 3 4 5 6 7 8 9 10

P1 T1 T4 T2 T6 T7

P2 T3 T5 T8

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Time 4.5: Both processors complete their tasks. T9 becomes ready, and is assigned to P1.

There is no ready task for P2 to work on, so P2 idles.

Priority list: T1, T3, T4, T5, T6, T7, T8, T2, T9

With the last task completed, we have a completed schedule, with finishing time 5.5 days.

Try it Now 1

Using the digraph below, create a schedule using the priority list:

T1, T2, T3, T4, T5, T6, T7, T8, T9

It is important to note 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 executed by a computer. The

interesting part of scheduling, then, is how to choose a priority list that will create the best

possible schedule.

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.

Time: 0 1 2 3 4 5 6 7 8 9 10

P1 T1 T4 T2 T6 T7 T9

P2 T3 T5 T8

T1 (10) T2 (7)

T4 (9)

T7 (13)

T5 (5)

T8 (4)

T6 (4)

T9 (6)

T3 (4)

Scheduling 161

Example 3

Consider the scheduling problem represented by the digraph below. Create a priority list

using the decreasing time list algorithm, then use it to schedule for two processors using the

list processing algorithm.

To use the decreasing time list algorithm,

we create our priority list by listing the

tasks in order from longest task time to

shortest task time. If there is a tie, we will

list the task with smaller task number first

(not for any good reason, but just for

consistency).

For this digraph, the decreasing time

algorithm would create a priority list of:

T6 (10), T3 (7), T10 (7), T1 (6), T5 (5), T4 (4), T7 (4), T2 (3), T8 (3), T9 (2)

Once we have the priority list, we can create the schedule using the list processing algorithm.

With two processors, we’d get:

Time 0: We identify ready tasks, and assign T3 to P1 and T1 to P2

Priority list: T6, T3, T10, T1, T5, T4, T7, T2, T8, T9

Time 6: P2 completes T1. No new tasks become ready, so T4 is assigned to P2.

Priority list: T6, T3, T10, T1, T5, T4, T7, T2, T8, T9

Time 7: P1 completes T3. No new tasks become ready, so T2 is assigned to P1.

Priority list: T6, T3, T10, T1, T5, T4, T7, T2, T8, T9

T1 (6) T5 (5)

T2 (3)

T3 (7)

T4 (4)

T6 (10)

T7 (4)

T10 (7) T8 (3)

T9 (2)

T3 P1

P2 T1

7

6

T3 P1

P2 T1

7

6

T4

10

T3 P1

P2 T1

7

6

T4

10

T2

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Time 10: Both processors complete their tasks. T6 becomes ready, and is assigned to P1.

No other tasks are ready, so P2 idles.

Priority list: T6, T3, T10, T1, T5, T4, T7, T2, T8, T9

Time 20: With T6 complete, T5 and T7 become ready, and are assigned to P1 and P2

respectively.

Priority list: T6, T3, T10, T1, T5, T4, T7, T2, T8, T9

Time 24: P2 completes T7. No new items become ready, so P2 idles.

Time 25: P1 completes T5. T8 and T9 become ready, and are assigned.

Priority list: T6, T3, T10, T1, T5, T4, T7, T2, T8, T9

Time 27: T9 is completed. No items ready, so P2 idles.

Time 28: T8 is completed. T10 becomes ready, and is assigned to P1.

Priority list: T6, T3, T10, T1, T5, T4, T7, T2, T8, T9

This is our completed schedule, with a finishing time of 35.

T3 P1

P2 T1

7

6

T4

10

T2 T6

20

T3 P1

P2 T1

7

6

T4

10

T2 T6

20

T5

T7

25

24

T3 P1

P2 T1

7

6

T4

10

T2 T6

20

T5

T7

25

24

T8

T9

28

27

T3 P1

P2 T1

7

6

T4

10

T2 T6

20

T5

T7

25

24

T8

T9

28

27

T10

35

Scheduling 163

Using the decreasing time algorithm, the priority list led to a schedule with a finishing time

of 35. Is this good? It certainly looks like there was a lot of idle time in this schedule. To

get some idea how good or bad this schedule is, we could compute the critical time, the

minimum time to complete the job. To find this, we look for the sequence of tasks with the

highest total completion time. For this digraph that sequence would appear to be: T2, T6, T5,

T8, T10, with total sequence time of 28. From this we can conclude that our schedule isn’t

horrible, but there is a possibility that a better schedule exists.

Try it Now 2

Determine the priority list for the digraph from Try it Now 1 using the decreasing time

algorithm.

Critical path algorithm

A sequence of tasks in the digraph is called a path. In the previous example, we saw that the

critical path dictates the minimum completion time for a schedule. Perhaps, then, it would

make sense to consider the critical path when creating our schedule. For example, in the last

schedule, the processors began working on tasks 1 and 3 because they were longer tasks, but

starting on task 2 earlier would have allowed work to begin on the long task 6 earlier.

The critical path algorithm allows you to create a priority list based on idea of critical paths.

Critical Path Algorithm (version 1)

1. Find the critical path.

2. The first task in the critical path gets added to the priority list.

3. Remove that task from the digraph

4. Repeat, finding the new critical path with the revised digraph.

Example 4

The original digraph from Example 3 has critical path T2, T6, T5, T8, T10, so T2 gets added

first to the priority list. Removing T2 from the digraph, it now looks like:

The critical path (longest path) in the remaining digraph is now T6, T5, T8, T10, so T6 is added

to the priority list and removed.

T1 (6) T5 (5)

T3 (7)

T4 (4)

T6 (10)

T7 (4)

T10 (7) T8 (3)

T9 (2)

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Now there are two paths with the same longest length: T1, T5, T8, T10 and T3, T7, T8, T10. We

can add T1 to the priority list (or T3 – we usually add the one with smaller item number) and

remove it, and continue the process.

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: Consider this segment of digraph.

In this case, if T2 has already been determined to have a critical time of 10, then T1

will have a critical time of 5+10 = 15

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, T1 should be labeled with a critical time of 16,

since it is the longer of 5+10 and 5+11.

4. Repeat until all vertices are labeled with their critical times

T1 (6) T5 (5)

T3 (7)

T4 (4) T7 (4)

T10 (7) T8 (3)

T9 (2)

T1 (5) T2 (4) [10]

T1 (5) [15] T2 (4) [10]

T1 (5) T2 (4) [10]

T3 (5) [11]

Scheduling 165

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.

Example 5

Applying this to our digraph from the earlier example, we start applying the backflow

algorithm.

We add an end vertex and give it a critical time of 0.

We then move back to T4, T9, and T10, labeling them with their critical times

From each vertex marked with a critical time, we go back. T7, for example, will get labeled

with a critical time 11 – the task time of 4 plus the critical time of 7 for T10. For T5, there are

two paths to the end. We use the longer, labeling T5 with critical time 5+7 = 12.

T1 (6) T5 (5)

T2 (3)

T3 (7)

T4 (4)

T6 (10)

T7 (4)

T10 (7) T8 (3)

T9 (2)

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