2. After viewing this presentation, you
should be able to:
Explain why production scheduling must be done
by every organization whether it manufactures or
provides services.
Discuss the application of the loading function.
Draw a Gantt chart and explain its information
display.
Describe the role of sequencing and how to apply
sequencing rules for one facility and for more than
one facility. 2
3. After viewing this presentation, you
should be able to (continued)
Classify scheduling problems according to
various criteria that are used in practice.
Explain the purpose of priority sequencing rules.
Describe various priority rules for sequencing.
Apply Johnson’s rule to the 2-machine flow shop
problem.
Analyze dynamic scheduling problems.
3
4. Loading, Sequencing and Scheduling
The production-schedules are developed
by performing the following functions:
Loading
Sequencing
Scheduling
4
5. Loading, Sequencing and Scheduling
(continued)
Loading: Which department is going to do what
work?
Sequencing: What is the order in which the work will
be done?
Scheduling: What are the start and finish times of
each job?
5
6. Loading
Loading, also called shop loading assigns the work to various
facilities like divisions, departments, work centers, load centers,
stations, machines and people.
We will often use the term “machines” in this presentation when we
refer to a facility.
Loading is done for both manufacturing and services.
6
7. Loading vs. Aggregate Planning
Aggregate planning is based on forecasts.
However, the loading function loads the real jobs and
not the forecast.
If the aggregate scheduling job was done well, then
the appropriate kinds and amounts of resources will
be available for loading.
7
8. Loading Objectives
Each facility carries a backlog of work, which is its ‘‘load’’—hardly a case of
perfect just-in-time in which no waiting occurs.
The backlog is generally much larger than the work in process, which can be
seen on the shop floor.
The backlog translates into an inventory investment which is idle and
receiving no value-adding attention.
A major objective of loading is to spread the load so that waiting is
minimized, flow is smooth and rapid, and congestion is avoided.
8
9. Sequencing
Sequencing models and methods follow the discussion of loading models and
methods.
Sequencing establishes the order for doing the jobs at each facility.
Sequencing reflects job priorities according to the way that jobs are arranged in the
queues.
Say that Jobs x, y, and z have been assigned to workstation 1 (through loading
function).
Jobs x, y, and z are in a queue (waiting line). Sequencing rules determine which job
should be first in line, which second, etc.
9
10. Sequencing (continued)
A good sequence provides less waiting time, decreased delivery delays, and
better due date performance.
There are costs associated with waiting and delays.
There are many other costs associated with the various orderings of jobs, for
example, set up cost and in-process inventory costs.
The objective function can be to minimize system’s costs, or to minimize total
system’s time, or (if margin data are available) to maximize total system’s profit.
We discuss several objective functions later in the presentation.
10
11. Sequencing (continued)
Total savings from regularly sequencing the right way,
the first time, can accumulate to substantial sums.
Re-sequencing can be significantly more costly. When
there are many jobs and facilities, sequencing rules have
considerable economic importance.
Sequencing also involves shop floor control, which
consists of communicating the status of orders and the
productivity of workstations.
11
12. Scheduling
A production schedule is the time table that
specifies the times at which the jobs in a
production department will be processed on
various machines.
The schedule gives the starting and ending times
of each job on the machines on which the job has
to be processed.
.
12
13. Scheduling Example
Suppose there are three jobs in a production department that are
to be processed on four categories (types) of machines. We
designate the jobs as A, B, and C; and the machine types are
designated as M1, M2, M3, and M4.
The three jobs consist of 4, 3, and 4 operations respectively; and
there are four machines - one machine of each type. We designate
them as M1, M2, M3, and M4 based on their categories.
The operations for job A are designated as A1, A2, A3, and A4. The
operations of job B are designated as B1, B2, and B3. Similarly the
four operations of job C are designated as C1, C2, C3, and C4.
13
14. Scheduling Example (continued)
Each job is characterized by its routing that specifies the
information about the number of operations to be performed, the
sequence of these operations, and the machines required for
processing these operations.
The times required for processing these operations are also
required for developing a production schedule.
14
15. Scheduling Example - Data
The table on right hand side (RHS) gives the data
for this example.
The table gives the machine required for each
operation of each job. For example, the first
operation of job A, A1, is processed on machine
M1; second operation, A2, is processed on machine
M3 and so on.
The operations of all jobs have to follow their
processing sequences. For example operation A3
of job A can not be processed before operation A2.
The processing time for each operation is also
given in this table.
Job
Operation
Number
Machine
Number
Processing Time
(Days)
A A1 M1 5
A2 M3 3
A3 M4 7
A4 M2 4
B B1 M2 2
B2 M3 6
B3 M4 8
C C1 M1 4
C2 M2 6
C3 M3 8
C4 M4 2
15
16. Scheduling Example – Objective
Function
The objective is to schedule these jobs so
as to minimize the time to complete all
jobs. This time is called make-span or the
schedule time. We will use the term make-
span in this presentation.
16
17. Scheduling Example Solution – Gantt Chart
One of the schedules for this example is presented below in the form of a
Gantt Chart.
The Gantt chart, for each machine, shows the start and finish times of all
operations scheduled on that machine.
17
18. Scheduling Example - Alternative schedules
Several alternative schedules can be generated for this example.
The schedules differ in the order in which the jobs are processed on the four
machines. Three of these schedules are:
o The first schedule orders jobs as: A first, then B and then C (A-B-C).
o The second schedule orders jobs as: B first, then A, and then C (B-A-C).
o The third schedule orders jobs as: C first, then A, and then B (C-A-B).
The Gantt charts for these schedules are shown in next slide.
The values of make-span for these three schedules are 25, 27 and 30 days
respectively. Schedule A-B-C is the best of these three schedules.
18
20. Scheduling Example - Alternative Schedules
(continued)
Is sequence A-B-C the global optimal? Can we find a better
sequence than this? The scheduling techniques attempt to
answer these questions.
It should be mentioned that there are different effectiveness
measures of a schedule in different situations. Minimizing
make-span is only one of them. We will study other
effectiveness measures also.
20
21. Scheduling Example - Assumptions
Once a job is started on a machine, its processing can not
be interrupted, that is, preemption is not allowed.
The machines are continuously available and will not
break down during the planning horizon. This
assumption is rather unrealistic but we make this
assumption to avoid complexity in discussing scheduling
concepts.
A machine is not kept idle if a job is available to be
processed.
Also, each machine can process only one job at a time. 21
22. Classification of Scheduling Problems
The scheduling problems can be classified based on the
following criteria:
• Sequence of machines
• Number of machines
• Processing times
• Job arrival time
• Objective functions
22
23. Sequence of Machines
The sequencing problems, based on the sequence
of machines, are classified as:
Flow Shops
Job Shops
23
24. Flow Shop
In a flow-shop , processing of all jobs require machines in the
same order.
The following table gives an example of a flow-shop in which
three jobs, A, B, and C are processed on four machines, M1, M2,
M3, and M4.
The sequences of machines to process these jobs are same (M1-
M3-M4-M2).
24
Example of a Flow Shop
Job
Operation
# 1
Operation
# 2
Operation
# 3
Operation
# 4
Machine for
Operation # 1
Machine for
Operation # 2
Machine for
Operation # 3
Machine for
Operation # 4
A A1 A2 A3 A4 M1 M3 M4 M2
B B1 B2 B3 B4 M1 M3 M4 M2
C C1 C2 C3 C4 M1 M3 M4 M2
25. Job Shop
In a job shop the sequence of machines will be mixed,
that is, the jobs may require machines in different
sequences.
25
Example of a Job Shop
Job
Operation
# 1
Operation
# 2
Operation
# 3
Operation
# 4
Machine for
Operation # 1
Machine for
Operation # 2
Machine for
Operation # 3
Machine for
Operation # 4
A A1 A2 A3 A4 M1 M3 M4 M2
B B1 B2 B3 M2 M3 M4
C C1 C2 C3 C4 M1 M2 M3 M4
26. Number of Machines
Based on the number of machines, the
scheduling problems are classified as:
Single machine problems
Two-machine problems
Multiple (3 or more) machine problems
26
27. Processing Times
Deterministic: If processing times of all jobs are
known and constant the scheduling problem is
called a deterministic problem.
Probabilistic: The scheduling problem is called
probabilistic (or stochastic) if the processing
times are not fixed; i.e., the processing times
must be represented by a probability
distribution.
27
28. Job Arrival Times
Based on this criterion, scheduling problems are classified as
static and dynamic problems.
Static: In the case of static problems the number of jobs is
fixed and will not change until the current set of jobs has
been processed.
Dynamic: In the case of dynamic problems, new jobs enter
the system and become part of the current set of unprocessed
jobs. The arrival rate of jobs is given in the case of dynamic
problems.
28
29. Objective Functions
Scheduling researchers have studied a large variety of objective
functions. In this presentation, we will focus on the following
objectives.
Minimize make-span
Minimize average flow time (or job completion time)
Average number of jobs in the system
Minimize average tardiness
Minimize maximum tardiness
Minimize number of tardy jobs
29
30. Objective Functions (continued)
Minimizing make span is relevant for two or more machines.
In this presentation we will discuss the scheduling rule for static and deterministic
flow shop problems consisting of two machines where the objective is to minimize
make-span.
The other five objectives can be used for any number of machines, both deterministic
and probabilistic processing times, and for static as well as dynamic problems.
However, we will study these objective functions for a single machine, deterministic
and static problems.
The scheduling rule for job shops and for more than three machines are complex and
beyond the scope of this presentation.
30
31. Example: 2-Machines Flow Shop
Consider a problem with five jobs (A, B, C, D, and E); and
two machines designated as M1 and M2.
All five jobs consist of two operations each. The first
operation of each job is processed on machine M1; and the
second operation is processed on machine M2.
The next slide gives the machines required for each job; and
the processing times for each operation of each job.
31
32. Data for a 2-Machine Flow Shop
Job
Operation
# 1
Operation
# 2
Machine for
Operation # 1
Machine for
Operation # 2
Time for
Operation # 1
(Days)
Time for
Operation #
2 (Days)
A A1 A2 M1 M2 8 3
B B1 B2 M1 M2 5 7
C C1 C2 M1 M2 6 9
D D1 D2 M1 M2 7 1
E E1 E2 M1 M2 4 6
Data for a 5-Job 2-Machine Flow Shop Problem
32
33. Scheduling Objective
The scheduling objective is to find an optimal sequence that gives the
order in which the five jobs will be processed on the two machines.
Once we know a sequence, the time to complete all jobs can be determined.
This time is called the schedule time, or the make-span. The optimal sequence
is the one that minimizes make-span.
For example, A-B-C-D-E is a sequence order that tells us that A is the first
job to be processed; B is the second job and so on. E is the last job to be
processed. Is it optimal?
Another sequence could be B-C-A-E-D. Is it optimal?
What is the optimal sequence?
33
34. Number of Sequences
For this 5-job problem there are 120 (5!) different sequences.
For a six-job problem, the number of sequences will be 720 (6!).
In general, for a “n” job problem there are n! (n-factorial)
sequences.
Our goal is to find the best sequence that minimizes make-span.
Let us find the make-span for one of these sequences, say A-B-
C-D-E. We will draw a Gantt chart to find make-span.
34
35. Gantt Chart
Sequence A-B-C-D-E
The Gantt chart for the sequence A-B-C-D-E is given below. We are assuming that
the sequence is the same on both machines.
The value of make-span (time to complete all jobs) is 36 days.
Our objective is to identify the sequence that minimizes the value of make-span.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
M1 A1 A1 A1 A1 A1 A1 A1 A1 B1 B1 B1 B1 B1 C1 C1 C1 C1 C1 C1 D1 D1 D1 D1 D1 D1 D1 E1 E1 E1 E1
M2 A2 A2 A2 B2 B2 B2 B2 B2 B2 B2 C2 C2 C2 C2 C2 C2 C2 C2 C2 D2 E2 E2 E2 E2 E2 E2
Gantt Chart for Sequence A-B-C-D-E
Time (Days)
35
36. Identifying the best sequence
There may be multiple optimal sequences.
We will study Johnson’s rule that identifies one of these
optimal sequence.
There are five sequence positions 1 through 5.
Johnson’s rule assigns each job to one of these positions
in an optimal manner.
Position 1 Position 2 Position 3 Position 4 Position 5
36
37. Johnson’s Rule
to Minimize Make-span
We use the following four step process to
find the optimal sequence.
Step 1: Find the minimum processing
time considering times on both machines.
Step 2: Identify the corresponding job and
the corresponding machine for the
minimum time identified at Step 1.
37
38. Johnson’s Rule (continued)
Step 3: Scheduling Rule
(a) If the machine identified in Step 2 is machine
M1 then the job identified in Step 2 will be scheduled
in the first available schedule position.
(b) If the machine identified in Step 2 is machine
M2 then the job identified in Step 2 will be scheduled
in the last available schedule position.
Step 4: Remove the job from consideration whose position
has been fixed in Step 3; and go to Step 1.
Continue this process until all jobs have been scheduled.
38
39. Johnson’s Rule (continued)
Johnson’s rule makes the following assumptions:
The same optimal sequence is used on both
machines.
Preemption is not allowed, that is, once a job is
started it is not interrupted.
39
40. Iteration 1
Step 1: The minimum time is 1.
Step 2: The job is D and the machine is
M2.
Step 3: Since the machine identified at
Step 2 is machine M2, job D will be
assigned to the last available
sequence position which is position 5;
and the resulting partial sequence is
given below.
Step 4: Delete job D from
consideration.
Position 1 Position 2 Position 3 Position 4 D
Job
Operation
#
1
Operation
#
2
Machine
for
Operation
#
1
Machine
for
Operation
#
2
Time
for
Operation
#
1
(Days)
Time
for
Operation
#
2
(Days)
A A1 A2 M1 M2 8 3
B B1 B2 M1 M2 5 7
C C1 C2 M1 M2 6 9
D D1 D2 M1 M2 7 1
E E1 E2 M1 M2 4 6
40
41. Iteration 2
Step 1: The next minimum time
is 3.
Step 2: The job is A and the
machine is M2.
Step 3: The job A will be
assigned to the last available
schedule position, which is
position 4. After assigning job A
to position 4, the partial
sequence is given below.
Step 4: Delete job A from
consideration.
Job
Operation
#
1
Operation
#
2
Machine
for
Operation
#
1
Machine
for
Operation
#
2
Time
for
Operation
#
1
(Days)
Time
for
Operation
#
2
(Days)
A A1 A2 M1 M2 8 3
B B1 B2 M1 M2 5 7
C C1 C2 M1 M2 6 9
D D1 D2 M1 M2 7 1 Scheduled
E E1 E2 M1 M2 4 6
Position 1 Position 2 Position 3 A D
41
42. Iteration 3
Step 1: The minimum time is 4.
Step 2: The job is E and the
machine is M1.
Step 3: The job E will be
assigned to the first available
schedule position, which is
position 1. The partial sequence
after assigning job E to position
1 is given below.
Step 4: Delete job E from
consideration
Job
Operation
#
1
Operation
#
2
Machine
for
Operation
#
1
Machine
for
Operation
#
2
Time
for
Operation
#
1
(Days)
Time
for
Operation
#
2
(Days)
A A1 A2 M1 M2 8 3 Scheduled
B B1 B2 M1 M2 5 7
C C1 C2 M1 M2 6 9
D D1 D2 M1 M2 7 1 Scheduled
E E1 E2 M1 M2 4 6
E Position 2 Position 3 A D
42
43. Iteration 4
Step 1: The minimum time is 5.
Step 2: The job is B and the
machine is M1.
Step 3: The job B will be
assigned to the first available
schedule position, which is
position 2. The partial sequence
after assigning job B to position
2 is given below.
Step 4: Delete job B from
consideration
Job
Operation
#
1
Operation
#
2
Machine
for
Operation
#
1
Machine
for
Operation
#
2
Time
for
Operation
#
1
(Days)
Time
for
Operation
#
2
(Days)
A A1 A2 M1 M2 8 3 Scheduled
B B1 B2 M1 M2 5 7
C C1 C2 M1 M2 6 9
D D1 D2 M1 M2 7 1 Scheduled
E E1 E2 M1 M2 4 6 Scheduled
E B Position 3 A D
43
44. Iteration 5
The only unscheduled job at
this stage is C and; it will be
assigned to the remaining
unassigned position 3.
The final sequence is given
below.
The value of make-span for this
sequence will be determined by
drawing the Gantt chart.
Job
Operation
#
1
Operation
#
2
Machine
for
Operation
#
1
Machine
for
Operation
#
2
Time
for
Operation
#
1
(Days)
Time
for
Operation
#
2
(Days)
A A1 A2 M1 M2 8 3 Scheduled
B B1 B2 M1 M2 5 7 Scheduled
C C1 C2 M1 M2 6 9
D D1 D2 M1 M2 7 1 Scheduled
E E1 E2 M1 M2 4 6 Scheduled
E B C A D
44
45. Finding Make-span
The sequence E-B-C-A-D identified by Johnson’s
rule guarantees the minimum value of make-
span.
However, Johnson’s rule does not give the value of
make-span. It only identifies the best sequence.
The value of make-span is obtained either by
drawing the Gantt chart or a computerized
algorithm can be developed.
The Gantt chart for this optimal sequence is given
in the next slide.
45
46. Sequence E-B-C-A-D
The Gantt chart for the sequence E-B-C-A-
D is given below. The value of make-span
is 31 days.
The optimal (minimum) value of make-
span for this problem is therefore, 31
days.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
M1 E1 E1 E1 E1 B1 B1 B1 B1 B1 C1 C1 C1 C1 C1 C1 A1 A1 A1 A1 A1 A1 A1 A1 D1 D1 D1 D1 D1 D1 D1
M2 E2 E2 E2 E2 E2 E2 B2 B2 B2 B2 B2 B2 B2 C2 C2 C2 C2 C2 C2 C2 C2 C2 A2 A2 A2 D2
Gantt Chart for Sequence E-B-C-A-D
Time (Days)
46
47. Multiple Sequences
It may be noted that multiple optimal
sequences are possible for a given problem.
It means that several sequences can have
the same minimum value of make-span.
For example, for the problem studied in
previous slides, the sequence E-C-B-A-D
also gives a make-span of 31 days. TRY IT.
However, Johnson’s rule identifies only one
of these sequences.
47
48. Breaking Ties
It might happen at Step 1, that there are more than one
minimum times. In such a situation, which job should be
picked up for assignment.
We will discuss three different cases of these ties.
Case 1: Minimum time is on both machines but for
different jobs.
Case2: Minimum time is on the same machine but for
different jobs.
Case 3: Minimum time is on both machines but for the
same job.
48
49. Ties: Case 1 - Data
Consider the problem given below. The minimum time is 1 but occurs at
two places – Job A at M1 and Job D at M2.
Job
Operation
# 1
Operation
# 2
Machine for
Operation #
1
Machine for
Operation #
2
Time for
Operation #
1 (Days)
Time for
Operatio
n # 2
(Days)
A A1 A2 M1 M2 1 3
B B1 B2 M1 M2 5 7
C C1 C2 M1 M2 6 9
D D1 D2 M1 M2 7 1
E E1 E2 M1 M2 4 6
49
50. Ties: Case 1 - Solution
The ties are broken at random.
A may be selected before D or D may be selected before A.
In either case, the resulting partial sequence after both A and D are
scheduled, is given below.
The scheduling algorithm continues until all jobs are scheduled. The
final sequence is also shown below.
50
.
A Position 2 Position 3 Position 4 D
A E B C D
Final sequence
Partial scequence after scheduling A and D
51. Ties: Case 2 - Data
Consider the problem given below. The minimum time is 1 but occurs at two places
– Job B at M2 and Job D at M2.
Job
Operation
# 1
Operation
# 2
Machine for
Operation #
1
Machine for
Operation #
2
Time for
Operation #
1 (Days)
Time for
Operatio
n # 2
(Days)
A A1 A2 M1 M2 8 3
B B1 B2 M1 M2 5 1
C C1 C2 M1 M2 6 9
D D1 D2 M1 M2 7 1
E E1 E2 M1 M2 4 6
51
52. Ties: Case 2 – Solution
The ties are broken at random.
B may be selected before D or D may be selected before B.
In this case, two different partial sequences will result based on which
job is selected first. These partial sequences are shown below.
The scheduling algorithm continues until all jobs are scheduled.
The two final sequences are also shown below.
52
Partial Sequence if B is selected first and then D
Position 1 Position 2 Position 3 D B
Final sequence if B is selected first and then D
E C A D B
Partial Sequence D is selected first and then B
Position 1 Position 2 Position 3 B D
Final sequence if D is selected first and then B
E C A B D
53. Ties: Case 3 - Data
Consider the problem given below. The minimum time is 2 but
occurs at two places – Job C at M1 and also Job C at M2.
Job
Operation
# 1
Operation
# 2
Machine for
Operation #
1
Machine for
Operation #
2
Time for
Operation #
1 (Days)
Time for
Operatio
n # 2
(Days)
A A1 A2 M1 M2 8 3
B B1 B2 M1 M2 5 7
C C1 C2 M1 M2 2 2
D D1 D2 M1 M2 7 9
E E1 E2 M1 M2 4 6
53
54. Ties: Case 3 - Solution
The ties are broken at random.
Job C at M1 may be selected before Job C at M2 or vice versa.
In this case, two different partial sequences will result based on which
combination is selected first. These partial sequences are shown below.
The scheduling algorithm continues until all jobs are scheduled.
The two final sequences are also shown below.
Note: C can be the first job or the last job in the optimal sequence
54
Partial Sequence if C on M1 is selected first
C Position 2 Position 3 Position 4 Position 5
Final sequence if C on M1 is selected first
C E B D A
Partial Sequence if C on M2 is selected first
Position 1 Position 2 Position 3 Position 4 C
Final sequence if C on M2 is selected first
E B D A C
55. Comments About Ties
In general all three cases of ties may exist in a given problem.
The examples that we considered show ties in the first iteration.
However, the ties may occur during any iteration.
The general rule is to break the ties at random. However, we
will break the ties in the alpha order, that is A before B etc. For
case 3, the ties will be broken by the rule machine M1 before
machine M2.
All resulting sequences, irrespective of the tie breaking rule,
will give the same minimum value of the make-span.
55
56. Example with Multiple Ties
Consider the problem given below.
Johnson’s rule will give four different sequences for this problem because of various ties.
This problem is included on students’ website.
TRY IT.
Job
Operation
# 1
Operation
# 2
Machine for
Operation #
1
Machine for
Operation #
2
Time for
Operation #
1 (Days)
Time for
Operatio
n # 2
(Days)
A A1 A2 M1 M2 2 3
B B1 B2 M1 M2 5 7
C C1 C2 M1 M2 2 2
D D1 D2 M1 M2 8 9
E E1 E2 M1 M2 4 2
56
57. 57
Single Machine Scheduling
There is one machine on which several jobs have
to be processed.
The order in which these jobs will be processed
needs to be specified. This schedule will not be
changed until all jobs have been processed. This is
the “static” version of the problem.
In the “dynamic” version, the schedule can be
altered. The dynamic version is studied later in
this presentation.
57
58. 58
Scheduling Rules
There are several rules that can be used to
find the order of processing. We will study the
following three rules.
First Come First Served (FCFS)
Shortest Processing Time (SPT). This is also
called as Shortest Operation Time.
Earliest (shortest) Due Date (EDD)
58
59. 59
Objective Functions
There are several objective functions that can be minimized
in a single machine problem. We will study the following
objective functions.
Minimize average completion (flow) time.
Minimize average number of jobs in the system.
Minimize average tardiness.
Minimize maximum tardiness
Minimize number of tardy jobs.
Note: A job is tardy if it is not completed by its due date.
59
60. 60
Example
Consider the example given on the
RHS.
There are five jobs A, B, C, D, and E.
A is the first job that arrived in the
production department. B,C, D, and
E followed A in this order.
The processing times and due
dates of all jobs are also given.
The order in which these jobs
have to be processed needs to be
specified.
Days
Job Time
Due
Date
A 17 45
B 12 35
C 22 27
D 18 54
E 26 47
60
61. First Come First Served (FCFS)
The table on RHS gives answers by
using the FCFS rule. The order of
processing is A, B, C, D, and E.
A is the first job to be processed
and will be completed at time 17.
Its due date is 45. So job A is not
late (tardy); tardiness is zero.
Job B starts after job A, and is
completed at time 29 (17+12). This
is also not tardy.
In this way the completion time
and tardiness of all jobs are
completed.
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First Come First Served
Job Time Due Date
Completion
Time
Tardiness
A 17 45 17 0
B 12 35 29 0
C 22 27 51 24
D 18 54 69 15
E 26 47 95 48
Tardiness = Completion Time – Due Date
Make it zero if you get a negative value.
62. 62
FCFS: Calculation of Objective
Functions
Average Completion Time: Add completion
times of all jobs and divide by the number of
jobs (261/5). It is 52.5.
Average Number of Jobs in the System: This is
obtained by dividing the total of completion
times of all jobs by the completion time of
the last job (261/95). It is 2.75.
Average Tardiness : This is obtained by
adding the tardiness of all jobs and dividing
it by the number of jobs (87/5). It is 17.4.
Maximum Tardiness: This is the maximum of
all tardiness values, It is 48.
Number of Tardy Jobs: Count the number of
jobs that are tardy. Three jobs (C, D, and E)
are tardy for this problem. It is 3.
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63. 63
Shortest Processing Time (SPT)
The jobs are processed in the
increasing order of their
processing times.
The job with the minimum
processing time (B) is processed
first. B is followed by A, D, C,
and E.
The calculations of the
objective functions follow the
same procedure as for the FCFS
rule.
Job Time Due Date
Completion
Time
Tardiness
B 12 35 12 0
A 17 45 29 0
D 18 54 47 0
C 22 27 69 42
E 26 47 95 48
Total 252 90
Average Completion Time 50.4
Average Number of Jobs in System 2.65
Average Tardiness 18
Maximum Tardiness 48
Number of Tardy Jobs 2
Shortest Processing Time
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64. 64
Earliest Due Date (EDD)
The jobs are processed in the
increasing order of their due
dates.
The job with the minimum due
date (C) is processed first; and
is followed by B, A, E, and D.
The calculations of the
objective functions follow the
same procedure as for the FCFS
rule
Job Time Due Date
Completion
Time
Tardiness
C 22 27 22 0
B 12 35 34 0
A 17 45 51 6
E 26 47 77 30
D 18 54 95 41
Total 279 77
Average Completion Time 55.8
Average Number of Jobs in System 2.94
Average Tardiness 15.4
Maximum Tardiness 41
Number of Tardy Jobs 3
Earliest Due Date
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65. 65
Dynamic Scheduling Problems
A scheduling problem is classified as a dynamic problem if the
number of jobs is not fixed.
The examples include new production orders, customers in a bank,
shoppers in a store, and cars at a gas station etc.
The new jobs (production orders, customers, cars etc.) keep on
coming into the system; and the schedule needs to integrate the new
arrivals when an updated schedule is prepared.
A new schedule is prepared every time a job is completed.
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66. 66
Example – Dynamic Scheduling
Problem
Consider a single machine problem for
which the data are given on RHS.
There are five jobs A, B, C, D, and E that are
waiting to be processed.
Suppose Shortest Processing Time (SPT)
rule is being used.
SPT sequence is given on RHS.
B is the first job to be processed.
When B is being processed, suppose two
new jobs F and G arrive.
F arrives on the 5th
day and G arrives on
the 10th
day. Also assume that the
processing time of job F is 8 days and that
of G is 20 days.
Job Time (Days)
A 17
B 12
C 22
D 18
E 26
Data for Dynamic Problem
Job Time (Days)
B 12
A 17
D 18
C 22
E 26
SPT Sequence
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67. 67
Example – Dynamic Scheduling Problem
(continued)
After job B has been processed, there are six jobs
(A, C, D, E, F, and G) that are waiting to be
processed.
Since the scheduling rule is SPT, the job with the
minimum processing time from among the jobs
that are waiting to be processed will be
scheduled next. Table 3 gives the schedule at this
time.
The next job is F. This will be completed at time
20 (12+8) where 12 is the completion time of job
B. If more jobs arrive in the system while F is
being processed, they will be integrated with the
current jobs and a new schedule will be
developed.
These are called dynamic problems since the
schedule is continuously updated. Dynamic
situations are faced in multiple machines
problems also.
Job Time (Days)
F 8
A 17
D 18
G 20
C 22
E 26
SPT Sequence - New
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68. 68
Objective Functions – Dynamic Scheduling
Problems
Objective functions for dynamic problems are defined in the same
way as for single machine static problems.
Minimize average completion (flow) time.
Minimize average number of jobs in the system.
Minimize number of late (tardy) jobs.
Minimize average lateness (tardiness).
The values of completion time and tardiness of each job are recorded;
and the values of the objective functions can be calculated at any
time based on the number of jobs completed at that time.
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