1. Operations SchedulingJ
Operations Scheduling
For Operations Management, 9e by Krajewski/Ritzman/Malhotra © 2010 Pearson Education
PowerPoint Slides by Jeff Heyl
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2. Scheduling Service and Manufacturing Processes
The scheduling techniques cut across the various process types found in services and manufacturing
Front-office process with high customer contact, divergent work flows, customization, and a complex scheduling environment
Back-office process has low customer involvement, uses more line work flows, and provides standardized services
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3. Time of completion of last job Starting time of first job
Performance Measures
Flow time is the amount of time a job spends in the service or manufacturing system
Past due (tardiness) is the amount of time by which a job missed its due date
Makespan is the total amount of time required to complete a group of jobs
Makespan =
Time of completion of last job –
Starting time of first job
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4. Scheduled receipts for all items On-hand inventories of all items
Performance Measures
Total inventory is used to measure the effectiveness of schedules for manufacturing processes.
Total Inventory =
Scheduled receipts for all items +
On-hand inventories of all items
Utilization is the percentage of work time that is productively spent by an employee or a machine
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Sequencing Jobs
Operations schedules are short-term plans designed to implement the sales and operations plan
An operation with divergent flows is often called a job shop
Low-to medium-volume production
Utilizes job or batch processes
The front office would be the equivalent for a service provider
Difficult to schedule because of the variability in job routings and the continual introduction of new jobs to be processed
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Sequencing Jobs
An operation with line flow is often called a flow shop
Medium- to high-volume production
Utilizes line or continuous flow processes
The back office would be the equivalent for a service provider
Tasks are easier to schedule because the jobs have a common flow pattern through the system
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Job Shop Sequencing
Raw Materials
Shipping Department
Legend:
Batch of parts
Workstation
Figure J.1 – Diagram of a Manufacturing Job Shop Process
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8. Priority Sequencing Rules
First-come, first-served (FCFS)
Earliest due date (EDD)
Critical ratio (CR)
CR =
(Due date) – (Today’s date)
Total shop time remaining
A ratio less than 1.0 implies that the job is behind schedule
A ratio greater than 1.0 implies the job is ahead of schedule
The job with the lowest CR is scheduled next
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9. Priority Sequencing Rules
Shortest processing time (SPT)
Slack per remaining operations (S/RO)
S/RO =
Due date
Today’s date
Total shop time remaining
– –
Number of operations remaining
The job with the lowest S/RO is scheduled next
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10. Sequencing One Workstation
Single-dimension rules
A job’s priority assignment based only on information waiting for processing at the individual workstation (e.g., FCFS, EDD, and SPT)
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11. Comparing EDD and SPT Rules
EXAMPLE J.1
The Taylor Machine Shop rebores engine blocks. Currently, five engine blocks are waiting for processing. At any time, the company has only one engine expert on duty who can do this type of work. The engine problems have been diagnosed, and the processing times for the jobs have been estimated. Expected completion times have been agreed upon with the shop’s customers. The accompanying table shows the current situation. Because the Taylor Machine Shop is open from 8:00 A.M. until 5:00 P.M. each weekday, plus weekend hours as needed, the customer pickup times are measured in business hours from the current time. Determine the schedule for the engine expert by using (a) the EDD rule and (b) the SPT rule. For each rule, calculate the average flow time, average hours early, and average hours past due. If average past due is most important, which rule should be chosen?
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12. Comparing EDD and SPT Rules
Engine Block
Business Hours Since Order Arrived
Processing Time, Including Setup (hours)
Business Hours Until Due Date (customer pickup time)
Ranger 12 8 10
Explorer 6
Bronco 1 15 20
Econoline 150 3 18
Thunderbird 22
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13. Comparing EDD and SPT Rules
SOLUTION
a. The EDD rule states that the first engine block in the sequence is the one with the closest due date. Consequently, the Ranger engine block is processed first. The Thunderbird engine block, with its due date furthest in the future, is processed last. The sequence is shown in the following table, along with the flow times, the hours early, and the hours past due.
Engine Block Sequence
Hours Since Order Arrived
Begin Work
Processing Time, (hr)
Finish Time (hr)
Flow Time (hr)
Scheduled Customer Pickup Time
Actual Customer Pickup Time
Hours Early
Hours Past Due
Ranger
Explorer
Econoline 150
Bronco
Thunderbird 12 + 8 = 20 10 2 — 10 8 + 6 = 14 24 12 13 — 2 3 14 + 15 = 17 20 18 1 — 32 33 12 44 22
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14. Comparing EDD and SPT Rules
The flow time for each job is its finish time, plus the time since the job arrived.1 For example, the Explorer engine block’s finish time will be 14 hours from now (8 hours waiting time before the engine expert started to work on it plus 6 hours processing). Adding the 10 hours since the order arrived at this workstation (before the processing of this group of orders began) results in a flow time of 24 hours. You might think of the sum of flow times as the total job hours spent by the engine blocks since their orders arrived at the workstation until they were processed.
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15. Comparing EDD and SPT Rules
The performance measures for the EDD schedule for the five engine blocks are 5
= 28.2 hrs
Average flow time = 5
= 0.6 hrs
Average hours early = 5
= 7.2 hrs
Average hours past due =
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16. Comparing EDD and SPT Rules
b. Under the SPT rule, the sequence starts with the engine block that has the shortest processing time, the Econoline 150, and it ends with the engine block that has the longest processing time, the Bronco. The sequence, along with the flow times, early hours, and past due hours, is contained in the following table:
Engine Block Sequence
Hours Since Order Arrived
Begin Work
Processing Time, (hr)
Finish Time (hr)
Flow Time (hr)
Scheduled Customer Pickup Time
Actual Customer Pickup Time
Hours Early
Hours Past Due
Econoline 150
Explorer
Ranger
Thunderbird
Bronco 3 + = 6 19 18 15 — 10 3 + 6 = 9 19 12 — 12 9 + 8 = 17 29 10 — 7 22 1 15 44 45 20 24
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17. Comparing EDD and SPT Rules
The performance measures are 5
= 25.6 hrs
Average flow time = 5
= 3.6 hrs
Average hours early = 5
= 7.6 hrs
Average hours past due =
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18. Comparing Sequencing Rules
EDD rule
Performs well with respect to the percentage of jobs past due and the variance of hours past due
Popular with firms that are sensitive to achieving due dates
SPT rule
Tends to minimize the mean flow and maximize shop utilization
For single-workstations will always provide the lowest mean finish time
Could increase total inventory
Tends to produce a large variance in past due hours
FCFS rule
Considered fair
It performs poorly with respect to all performance measures
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19. Standard Time, Including Setup (hr)
Application J.1
Given the following information, devise an SPT schedule for the automatic routing machine.
Order
Standard Time, Including Setup (hr)
Due Date (hrs from now)
AZ135 14
DM246 8 20
SX435 10 6
PC088 3 18
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20. Scheduled Customer Pickup Time
Application J.1
Order Sequence
Hours
Since Order Arrived
Begin Work
Finish Time (hr)
Flow
Time
(hr)
Scheduled Customer Pickup Time
Actual Pickup Time
Hours Early
Hours Past Due 1. 2. 3. 4.
Total
Average
PC088
DM246
SX435
AZ135 2 3 5 18 15 4 3 11 15 20 9 1 11 21 22 6 15 5 35 40 14 25 82 94 24 40
20.5
23.5 6 10
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21. Multiple-Dimension Rules
The priority rules CR and S/RO incorporate information about the remaining workstations
S/RO is better than EDD with respect to the percentage of jobs past due but usually worse than SPT and EDD with respect to average job flow times
CR results in longer job flow times than SPT, but CR also results in less variance in the distribution of past due hours
No choice is clearly best; each rule should be tested in the environment for which it is intended
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22. Sequencing with the CR and S/RO Rules
EXAMPLE J.2
The first five columns of the following table contain information about a set of four jobs that just arrived (end of hour 0 or beginning of hour 1) at an engine lathe. They are the only ones now waiting to be processed. Several operations, including the one at the engine lathe, remain to be done on each job. Determine the schedule by using (a) the CR rule and (b) the S/RO rule. Compare these schedules to those generated by FCFS, SPT, and EDD.
Job
Processing Time
at Engine Lathe (hours)
Time Remaining Until Due Date (days)
Number of Operations Remaining
Shop Time Remaining (days) CR
S/RO 1
2.3 15 10
6.1
2.46
0.89 2
10.5
7.8
1.28
1.10 3
6.2 20 12
14.5
1.38
0.46 4
15.6 8 5
10.2
0.78
–0.44
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23. Sequencing with the CR and S/RO Rules
SOLUTION
a. Using CR to schedule the machine, we divide the time remaining until the due date by the shop time remaining to get the priority index for each job. For job 1,
CR =
Time remaining until the due date
Shop time remaining
= = 2.46 15
6.1
By arranging the jobs in sequence with the lowest critical ratio first, we determine that the sequence of jobs to be processed by the engine lathe is 4, 2, 3, and finally 1, assuming that no other jobs arrive in the meantime.
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24. Sequencing with the CR and S/RO Rules
b. Using S/RO, we divide the difference between the time remaining until the due date and the shop time remaining by the number of remaining operations. For job 1,
S/RO =
Time remaining
until the due date
Shop time
remaining –
Number of operations remaining
= = 0.89
15 – 6.1 10
Arranging the jobs by starting with the lowest S/RO yields a 4, 3, 1, 2 sequence of jobs.
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25. Sequencing with the CR and S/RO Rules
Priority Rule Summary
FCFS
SPT
EDD CR
S/RO
Average flow time
17.175
16.100
26.175
27.150
24.025
Average early time
3.425
6.050
Average past due
7.350
8.900
12.925
13.900
10.775
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26. Application J.2
The following four jobs have just arrived at an idle drill process and must be scheduled.
Job
Processing Time at Drill Press (wk)
Time Remaining to Due Date (wks)
Number of Operations Remaining*
Shop Time Remaining* (wks) AA 4 5 3 BB 8 11 6 CC 13 16 10 9 DD 18 12 EE 2 7
* including drill press
Create the sequences for two schedules, one using the Critical Ratio rule and one using the S/RO rule.
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27. Application J.2 Critical Ratio Slack/Remaining Operation Job
Processing Time at Drill Press (wk)
Time Remaining to Due Date (wks)
Number of Operations Remaining*
Shop Time Remaining* (wks) AA 4 5 3 BB 8 11 6 CC 13 16 10 9 DD 18 12 EE 2 7
Critical Ratio
Slack/Remaining Operation
Job
Priority Index
Sequence on Drill Press AA BB CC DD EE
1.25
1.83
1.44
1.50
2.33
First
Fourth
Second
Third
Fifth AA BB CC DD EE
0.33
1.25
0.70
2.00
0.80
First
Fourth
Second
Fifth
Third
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28. Multiple Workstations
Identifying the best priority rule to use at a particular operation in a process is a complex problem because the output from one operation becomes the input to another
Computer simulation models are effective tools to determine which priority rules work best in a given situation
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29. Scheduling a Two-Station Flow Shop
In single-workstation scheduling, the makespan is the same regardless of the priority rule chosen
In the scheduling of two or more workstations in a flow shop, the makespan varies according to the sequence chosen
Determining a production sequence for a group of jobs to minimize the makespan has two advantages
The group of jobs is completed in minimum time
The utilization of the two-station flow shop is maximized
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Johnson’s Rule
Minimizes makespan when scheduling a group of jobs on two workstations
Step 1: Scan the processing time at each workstation and find the shortest processing time among the jobs not yet scheduled. If two or more jobs are tied, choose one job arbitrarily.
Step 2: If the shortest processing time is on workstation 1, schedule the corresponding job as early as possible. If the shortest processing time is on workstation 2, schedule the corresponding job as late as possible.
Step 3: Eliminate the last job scheduled from further consideration. Repeat steps 1 and 2 until all jobs have been scheduled.
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31. Scheduling Jobs on Two Workstations
EXAMPLE J.3
The Morris Machine Company just received an order to refurbish five motors for materials handling equipment that were damaged in a fire. The motors have been delivered and are available for processing. The motors will be repaired at two workstations in the following manner.
Workstation 1: Dismantle the motor and clean the parts.
Workstation 2: Replace the parts as necessary, test the motor, and make adjustments.
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32. Scheduling Jobs on Two Workstations
The customer’s shop will be inoperable until all the motors have been repaired, so the plant manager is interested in developing a schedule that minimizes the makespan and has authorized around-the-clock operations until the motors have been repaired. The estimated time to repair each motor is shown in the following table:
Time (hr)
Motor
Workstation 1
Workstation 2 M1 12 22 M2 4 5 M3 6 3 M4 15 16 M5 10 8
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33. Scheduling Jobs on Two Workstations
SOLUTION
The logic for the optimal sequence is shown in the following table:
Establishing a Job Sequence
Iteration
Job Sequence
Comments 1 M3
The shortest processing time is 3 hours for M3 at workstation 2. Therefore, M3 is scheduled as late as possible. 2 M2
Eliminate M3 from the table of estimated times. The next shortest processing time is 4 hours for M2 at workstation 1. M2 is therefore scheduled first. 3 M5
Eliminate M2 from the table. The next shortest processing time is 8 hours for M5 at workstation 2. Therefore, M5 is scheduled as late as possible. 4 M1
Eliminate M5 from the table. The next shortest processing time is 12 hours for M1 at workstation 1. M1 is scheduled as early as possible. 5 M4
The last motor to be scheduled is M4. It is placed in the last remaining position, in the middle of the schedule.
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34. Scheduling Jobs on Two Workstations5 10 15 20 25 30
Hour 35 40 45 50 55 60 65 1 2 M2
(4) M1
(12) M4
(15) M5
(10) M3
(5)
Idle—available
for further work
Idle M2
(5)
Idle M1
(22) M4
(16) M5
(8)
(3) M3
Figure J.2 – Gantt Chart for the Morris Machine Company Repair Schedule
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Application J.3
Use the following data to schedule two workstations arranged as a flow shop
Time (hr)
Job
Workstation 1
Workstation 2 A 4 3 B 10 20 C 2 15 D 8 7 E 14 13
Sequence: C B E D A
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Application J.3
Workstation 1
Workstation 2
Start
Finish C B E D A 2 2 17 2 12 17 37 12 26 37 50 26 34 50 57 34 38 57 60
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37. Labor-Limited Environment
The resource constraint is the amount of labor available, not the number of machines or workstations
The scheduler must also assign workers to their next workstations
Some possible labor assignment rules
Assign personnel to the workstation with the job that has been in the system longest
Assign personnel to the workstation with the most jobs waiting for processing
Assign personnel to the workstation with the largest standard work content
Assign personnel to the workstation with the job that has the earliest due date
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38. Solved Problem 1
The Neptune’s Den Machine Shop specializes in overhauling outboard marine engines. Some engines require replacement of broken parts, whereas others need a complete overhaul. Currently, five engines with varying problems are awaiting service. The best estimates for the labor times involved and the promise dates (in number of days from today) are shown in the following table. Customers usually do not pick up their engines early.
Engine
Time Since Order Arrived (days)
Processing Time, Including Setup (days)
Promise Date (days from now)
50-hp Evinrude 4 5 8
7-hp Johnson 6 15
100-hp Mercury 10 12
50-hp Honda 1 20
75-hp Nautique 3
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Solved Problem 1
a. Develop separate schedules by using the SPT and EDD rules
b. Compare the two schedules on the basis of average flow time, percentage of past due jobs, and maximum past due days for any engine
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40. Days Since Order Arrived
Solved Problem 1
SOLUTION
a. Using the SPT rule, we obtain the following schedule:
Repair Sequence
Days Since Order Arrived
Processing Time
Finish Time
Flow Time
Promise Date
Actual Pickup Date
Days Early
Days Past Due
50-hp Honda 1 2 20 19 —
75-hp Nautique 15 3 4 10 6
7-hp Johnson 8 14 7
50-hp Evinrude 5 13 17
100-hp Mercury 23 31 12 11
Total 83
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41. Days Since Order Arrived
Solved Problem 1
Using the EDD rule we obtain this schedule:
Repair Sequence
Days Since Order Arrived
Processing Time
Finish Time
Flow Time
Promise Date
Actual Pickup Date
Days Early
Days Past Due
50-hp Evinrude 4 5 9 8 3 —
75-hp Nautique 15 23 10 2
100-hp Mercury 18 26 12 6
7-hp Johnson 22 28 7
50-hp Honda 1 24 20
Total
110
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Solved Problem 1
b. Performance measures are as follows:
Average flow time is 16.6 (or 83/5) days for SPT and 22.0 (or 110/5) days for EDD. The percentage of past due jobs is 40 percent (2/5) for SPT and 60 percent (3/5) for EDD. For this set of jobs, the EDD schedule minimizes the maximum days past due but has a greater flow time and causes more jobs to be past due.
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43. Shop Time Remaining (days)
Solved Problem 2
The following data were reported by the shop floor control system for order processing at the edge grinder. The current date is day 150. The number of remaining operations and the total work remaining include the operation at the edge grinder. All orders are available for processing, and none have been started yet. Assume the jobs were available for processing at the same time.
Current Order
Processing Time (hr)
Due Date (day)
Remaining Operations
Shop Time Remaining (days)
A101 10
162 9
B272 7
158 6
C106 15
152 1
D707 4
170 8 18
E555
154 5
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Solved Problem 2
a. Specify the priorities for each job if the shop floor control system uses slack per remaining operations (S/RO) or critical ratio (CR).
b. For each priority rule, calculate the average flow time per job at the edge grinder.
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Solved Problem 2
SOLUTION
a. We specify the priorities for each job using the two sequencing rules. The sequence for S/RO is shown in the brackets.
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Solved Problem 2
The sequence of production for CR is shown in the brackets.
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Solved Problem 2
b. We are sequencing a set of jobs at a single machine, so each job’s finish time equals the finish time of the job just prior to it in sequence plus its own processing time. Further, all jobs were available for processing at the same time, so each job’s finish time equals its flow time. Consequently, the average flow times at this single machine are
In this example, the average flow time per job is lower for the CR rule, which is not always the case. For example, the critical ratios for B272 and A101 are tied at If we arbitrarily assigned A101 before B272, the average flow time would increase to ( )/5 = 23.0 hours.
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Solved Problem 3
The Rocky Mountain Arsenal, formerly a chemical warfare manufacturing site, is said to be one of the most polluted locations in the United States. Cleanup of chemical waste storage basins will involve two operations.
Operation 1: Drain and dredge basin.
Operation 2: Incinerate materials.
Management estimates that each operation will require the following amounts of time (in days):
Storage Basin A B C D E F G H I J
Dredge 3 4 6 1 2 8
Incinerate
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Solved Problem 3
Management’s objective is to minimize the makespan of the cleanup operations. All storage basins are available for processing right now. First, find a schedule that minimizes the makespan. Then calculate the average flow time of a storage basin through the two operations. What is the total elapsed time for cleaning all 10 basins? Display the schedule in a Gantt machine chart.
SOLUTION
We can use Johnson’s rule to find the schedule that minimizes the total makespan. Four jobs are tied for the shortest process time: A, D, E, and H. E and H are tied for first place, while A and D are tied for last place. We arbitrarily choose to start with basin E, the first on the list for the drain and dredge operation. The 10 steps used to arrive at a sequence are as follows:
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Solved Problem 3
1. Select basin E first (tied with basin H); put it at the front. E —
2. Select basin H next; put it toward the front. E H —
3. Select basin A next (tied with basin D); put it at the end. E H — A
4. Put basin D toward the end. E H — D A
5. Put basin G toward the front. E H G — D A
6. Put basin C toward the end. E H G — C D A
7. Put basin I toward the end. E H G — I C D A
8. Put basin F toward the front. E H G F — I C D A
9. Put basin B toward the front. E H G F B — I C D A
10. Put basin J in the remaining space. E H G F B J I C D A
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Solved Problem 3
Several optimal solutions are available to this problem because of the ties at the start of the scheduling procedure. However, all have the same makespan. The schedule would be as follows:
Operation 1
Operation 2
Basin
Start
Finish E 1 3 H 2 4 G 8 F 7 14 B 11 18 J 15 26 I 23 28 C 32 D 35 A 36
Total
200
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Solved Problem 3
The makespan is 36 days. The average flow time is the sum of incineration finish times divided by 10, or 200/10 = 20 days. The Gantt machine chart for this schedule is given in Figure J.3.
Storage Basin
Dredge
Incinerate E H G F B J I C D A E H G F B J I C D A
Figure J.3
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