1. Operations Scheduling
Chapter 8

2. The Hierarchy of Production Decisions
The logical sequence of operations in factory planning corresponds to the sequencing of chapters in a production management text book.
All planning starts with the demand forecast.
Demand forecasts are the basis for the top level (aggregate) planning.
The Master Production Schedule (MPS) is the result of disaggregating aggregate plans down to the individual item level.
Based on the MPS, MRP is used to determine the size and timing of component and subassembly production.
Detailed shop floor schedules are required to meet production plans resulting from the MRP.

3. Hierarchy of Production Decisions

4. Scheduling
Scheduling: Establishes the timing of the use of equipment, facilities and human activities in an organization
Effective scheduling can yield
Cost savings
Increases in productivity
Improved customer satisfaction

5. Scheduling Techniques
Scheduling techniques are designed to disaggregate the master production schedule into time-phased daily or hourly activities.
A detailed production schedule must include when and where each activity must take place in order to meet the master schedule.

6. Scheduling Activities
Scheduling involves the following major activities:
Routing (determining where the work is going to be done).
Short-run capacity planning.
Short-run machine, manpower and production scheduling.
Determining the sequence of activities (determining when the work is to be done).
Dispatching (issuing the order to begin work).
Controlling the progress of orders and monitoring the process to determine that operations are running according to plan.
Revising the schedule based on changes in order status of jobs, material and/or capacity availability and various other reasons.
Expediting (speeding the progress of the work order) late, critical jobs.

7. Elements of Scheduling
Elements of Scheduling Problems:
1. Job arrival patterns (static vs. dynamic). Dynamic arrival pattern means that more jobs will arrive in the system during the time those currently in the system are being processed. In a static system, all jobs that will ever enter the system are known. Most job shops are dynamic.
2. Ratio of workers to machines (machine limited vs. labor limited environment).
3. Priority rules for scheduling.
4. Flow patterns of jobs through the plant.
a. Flow shop: All jobs follow the same pattern of flow through the system. In a flow shop, routing is not typically a problem.
b. Job shop: Each job follows its own specified pattern. Job shop is more difficult to analyze than the flow shop.

8. Goals of Production Scheduling
High Customer Service: on-time delivery
Low Inventory Levels: WIP and FGI
High Utilization: of machines and labor

9. Meeting Due Dates – Measures
Service Level:
Used typically in make to order systems.
Fraction of orders which are filled on before their due dates.
Fill Rate:
Used typically in make to stock systems.
Fraction of demands met from stock.
Lateness:
Used in shop floor control.
Difference between order due date and completion date.
Average lateness has little meaning.
Better measure is lateness variance.
Tardiness:
Is equal to the lateness of a job if it is late and zero, otherwise.
Average tardiness is meaningful but unintuitive.

10. Basic Definitions
Throughput (TH): for a line, throughput is the average quantity of good (non-defective) parts produced per unit time.
Work in Process (WIP): inventory between the start and endpoints of a product routing.
Raw Material Inventory (RMI): material stocked at beginning of routing.
Finished Goods Inventory (FGI): material held in inventory prior to shipping to the customer.
Cycle Time (CT): time between release of the job at the beginning of the routing until it reaches an inventory point at the end of the routing.
Makespan: The total amount of time to process a fixed number of jobs.
Little’s Law: TH = WIP/CT  WIP=TH*CT  (L=λw)
where λ=TH and w=CT

11. Reducing WIP and Cycle Time
Less WIP Equals Shorter Cycle Times (Little’s Law)
Shorter cycle time means:
Less WIP
Better responsiveness
All of which reduce costs and improve sales revenue

12. Classic Scheduling – Assumptions
MRP/ERP:
Benefits – Simple paradigm, hierarchical approach.
Problems –
MRP assumes that lead times are an attribute of the part, independent of the status of the shop.
MRP uses pessimistic lead time estimates.

13. Classic Scheduling – Assumptions (cont.)
Classic Scheduling: (only classic in academia)
Benefits – “Optimal” schedules
Problems – Bad assumptions.
All jobs available at the start of the problem.
Deterministic processing times.
No setups.
No machine breakdowns.
No preemption.
No cancellation.

14. Objectives in Job Shop Scheduling
Meet due dates
Minimize work-in-process (WIP) inventory
Minimize average flow time
Maximize machine/worker utilization
Reduce set-up times for changeovers
Minimize direct production and labor costs
(note: that these objectives can be conflicting)

15. Measures to Evaluate Performance of a Scheduling Method
Service Level: Fraction of orders filled on before their due dates (used in make-to-order systems)
Fill Rate: Fraction of demand that are met from inventory without backorder (used in make-to-stock systems)
Job Flow Time: Time elapsed from the release of a job until it is completed.
Lateness: Difference between completion time and due date of a job (may be negative).
Tardiness: The positive difference between the completion time and the due date of a job.
Makespan: Flow time of the job completed last.

16. Measures to Evaluate Performance of a Scheduling Method
Production Rate
Utilization
Keep in mind that high utilization means high return on investment. This is good provided that the equipment is utilized to increase revenue. Otherwise, high utilization only helps to increase inventory, not profits.

17. Terminology
Flow shop: n jobs processed through m machines in the same sequence
Job shop: the sequencing of jobs through machines may be different, and there may be multiple operations on some machines.
Parallel processing vs. sequential processing: parallel processing means that the machines are identical, any job can be processed on any machine.

18. Common Sequencing Rules
FCFS. First Come First Served. Jobs processed in the order they come to the shop.
SPT. Shortest Processing Time. Jobs with the shortest processing time are scheduled first.
EDD. Earliest Due Date. Jobs are sequenced according to their due dates.
CR. Critical Ratio. Compute the ratio of processing time of the job and remaining time until the due date. Schedule the job with the largest CR value next.

19. Scheduling Service Operations Vs Manufacturing Operations
Scheduling service systems presents certain problems not generally encountered in manufacturing systems. This is primarily due to:
The inability to store services
The random nature of customer requests
To avoid problems such as long delays, unsatisfied customers, service systems rely on appoinment systems and reservation systems.

20. High-Volume Systems
Flow system: High-volume system with Standardized equipment and activities
Flow-shop scheduling: Scheduling for high-volume flow system
Work Center #1
Work Center #2
Output

21. High-Volume Systems
Examples of high-volume products include autos, personal computers, televisons.
In process industries, examples include petroleum refining, sugar refining.
A major issue in design of high-volume (flow) systems is line balancing.

22. Success Factors in High-Volume Systems
Process and product design
Preventive maintenance
Rapid repair when breakdown occurs
Minimization of quality problems
Reliability and timing of supplies

23. Intermediate-Volume Systems
Outputs are between standardized high-volume systems and made-to-order job shops
The volume of output is not large enough to justify continuous production.
Examples include canned foods, paint and cosmetics.

24. Intermediate-Volume Systems
The three basic issues in these systems are:
1.Run size, 2.Timing, and 3.Sequence of jobs
Economic run size:
h’ is defined as h’= h(1- λ/P)

25. Scheduling Low-Volume Systems
Loading - assignment of jobs to process centers
Sequencing - determining the order in which jobs will be processed
Job-shop scheduling
Scheduling for low-volume systems with many variations in requirements

26. Gantt Load Chart
Figure 15.2
Gantt chart - used as a visual aid for loading and scheduling

27. Loading
Infinite loading: Assigning jobs to work centers without considering the capacity of work center.
Finite loading: Takes into acccount the capacity of work center.
Forward scheduling: Scheduling ahead from some point in time.
Backward scheduling. Scheduling backwards from due dates.

28. Loading

29. LOADING (The Assignment Problem)
In many business situations, management needs to assign - personnel to jobs, - jobs to machines, - machines to job locations, or - salespersons to territories.
Consider the situation of assigning n jobs to n machines.
When a job i (=1,2,....,n) is assigned to machine j (=1,2, .....n) that incurs a cost Cij.
The objective is to assign the jobs to machines at the least possible total cost.

30. The Assignment Problem
This situation is a special case of the Transportation Model and it is known as the assignment problem.
Here, jobs represent “sources” and machines represent “destinations.”
The supply available at each source is 1 unit And demand at each destination is 1 unit.

31. The Assignment Problem
The assignment model can be expressed mathematically as follows:
Xij= 0, if the job j is not assigned to machine i
1, if the job j is assigned to machine i

32. The Assignment Problem

33. The Assignment Problem Example
Ballston Electronics manufactures small electrical devices.
Products are manufactured on five different assembly lines (1,2,3,4,5).
When manufacturing is finished, products are transported from the assembly lines to one of the five different inspection areas (A,B,C,D,E).
Transporting products from five assembly lines to five inspection areas requires different times (in minutes)

34. The Assignment Problem Example
Ballston Electronics manufactures small electrical devices.
Products are manufactured on five different assembly lines (1,2,3,4,5).
When manufacturing is finished, products are transported from the assembly lines to one of the five different inspection areas (A,B,C,D,E).
Transporting products from five assembly lines to five inspection areas requires different times (in minutes)

35. The Assignment Problem Example
Under current arrangement, assignment of inspection areas to the assembly lines are 1 to A, 2 to B, 3 to C, 4 to D, and 5 to E.
This arrangement requires = 65 man minutes.

36. The Assignment Problem Example
Management would like to determine whether some other assignment of production lines to inspection areas may result in less cost.
This is a typical assignment problem. n = 5 And each assembly line is assigned to each inspection area.
It would be easy to solve such a problem when n is 5, but when n is large all possible alternative solutions are n!, this becomes a hard problem.

37. The Assignment Problem Example
Assignment problem can be either formulated as a linear programming model, or it can be formulated as a transportation model.
However, An algorithm known as Hungarian Method has proven to be a quick and efficient way to solve such problems.
This technique is programmed into many computer modules such as the one in WINQSB.

38. The Assignment Problem Example
WINQSB solution for this problem is as follows:

39. Hungarian Method Example
Step 1: Select the smallest value in each row.
Subtract this value from each value in that row
Step 2: Do the same for the columns that do not have any zero value.

40. Hungarian Method Example
If not finished, continue with other columns.

41. Hungarian Method Example
Step 3: Assignments are made at zero values.
Therefore, we assign job 1 to machine 1; job 2 to machine 3, and job 3 to machine 2.
Total cost is = 30.
It is not always possible to obtain a feasible assignment as in here.

42. Hungarian Method Example 2

43. Hungarian Method Example 2
A feasible assignment is not possible at this moment.
In such a case, The procedure is to draw a minimum number of lines through some of the rows and columns, Such that all zero values are crossed out.

44. Hungarian Method Example 2
The next step is to select the smallest uncrossed out element. This element is subtracted from every uncrossed out element and added to every element at the intersection of two lines.

45. Hungarian Method Example 2
We can now easily assign to the zero values. Solution is to assign (1 to 1), (2 to 3), (3 to 2) and (4 to 4).
If drawing lines do not provide an easy solution, then we should perform the task of drawing lines one more time.
Actually, we should continue drawing lines until a feasible assignment is possible.

46. Sequencing
Sequencing: Determine the order in which jobs at a work center will be processed.
Workstation: An area where one person works, usually with special equipment, on a specialized job.

47. Sequencing n jobs on a Single Machine
Priority rules:
Simple heuristics such as FCFS, SPT, DD, CR are used to select the order in which jobs will be processed.
CR= (Due Date – Current Time)/ Processing Time
Job time: Time needed for setup and processing of a job.

48. Example: Sequencing Rules
Use the FCFS, SPT, and Critical Ratio rules to sequence the five jobs below. Evaluate the rules on the bases of average flow time, average number of jobs in the system, and average job lateness.
(Due Date)
Job Processing Time Time to Promised Completion
A hours 10 hours B C D E

49. Example: Sequencing Rules
FCFS Rule A > B > C > D > E
Processing Due Flow
Job Time Date Time Lateness A B C D E

50. Example: Sequencing Rules
FCFS Rule Performance
Average flow time:
141/5 = 28.2 hours
Average number of jobs in the system:
141/49 = 2.88 jobs
Average job lateness:
90/5 = 18.0 hours

51. Example: Sequencing Rules
SPT Rule A > E > C > B > D
Processing Due Flow
Job Time Date Time Lateness A E C B D

52. Example: Sequencing Rules
SPT Rule Performance
Average flow time:
127/5 = 25.4 hours
Average number of jobs in the system:
127/49 = 2.59 jobs
Average job lateness:
76/5 = 15.2 hours

53. Example: Sequencing Rules
Critical Ratio Rule E > C > D > B > A
Processing Promised Flow
Job Time Completion Time Lateness
E (.875)
C (.889)
D (1.00)
B (1.33)
A (1.67)

54. Example: Sequencing Rules
Critical Ratio Rule Performance
Average flow time:
148/5 = 29.6 hours
Average number of jobs in the system:
148/49 = 3.02 jobs
Average job lateness:
93/5 = 18.6 hours

55. Example: Sequencing Rules
Comparison of Rule Performance
Average Average Average
Flow Number of Jobs Job
Rule Time in System Lateness
FCFS
SPT CR
SPT rule was superior for all 3 performance criteria.

56. Sequencing n jobs on two machines
Johnson’s Rule: technique for minimizing completion time for a group of n jobs to be processed on two machines or at two work centers.
Minimizes total idle time
Johnson’s Rule requires satisfying the following conditions:

57. Johnson’s Rule Conditions
Job time must be known and constant
Job times must be independent of sequence
Jobs must follow same two-step sequence
Job priorities cannot be used
All units must be completed at the first work center before moving to second

58. Johnson’s Rule Optimum Sequence
List the jobs and their times at each work center
Find the smallest processing time. If it belongs to the first operation of a job schedule that job next, otherwise schedule that job last.
Eliminate the job from further consideration
Repeat steps 2 and 3 until all jobs have been scheduled

59. Johnson’s Algorithm Example
Data:
Iteration 1: min time is 4 (job 1 on M1); place this job first and remove from lists:

60. Johnson’s Algorithm Example (cont.)
Iteration 2: min time is 5 (job 3 on M2); place this job last and remove from lists:
Iteration 3: only job left is job 2; place in remaining position (middle).
Final Sequence: 1-2-3
Makespan: 28

61. Gantt Chart for Johnson’s Algorithm Example
Short task on M2 to
“clear out” quickly.
Short task on M1 to
“load up” quickly.

62. Example
A group of six jobs is to be processed through a two-machine flow shop. The first operation involves cleaning and the second involves painting. Determine a sequence that will minimize the total completion time for this group of jobs. Processing times are as follows:

63. The remaining jobs and their times are
Select the job with the shortest processing time. It is job D, with a time of two hours.
Since the time is at the first center, schedule job D first. Eliminate job D from further consideration.
Job B has the next shortest time. Since it is at the second work center, schedule it last and eliminate job B from further consideration. We now have
The remaining jobs and their times are

64. The shortest remaining time is six hours for job E at work center 1
The shortest remaining time is six hours for job E at work center 1. Thus, schedule that job toward the beginning of the sequence (after job D). Thus,
Job C has the shortest time of the remaining two jobs. Since it is for the first work center, place it third in the sequence. Finally, assign the remaining job (F) to the fourth position and the result is

65. Sequencing Jobs When Setup Times Are Sequence-Dependent

66. Scheduling Difficulties
Randomness in job arrival times
Variability in
Setup times
Processing times
Interruptions
Changes in the set of jobs
No method for identifying optimal schedule
Scheduling is not an exact science
Ongoing task for a manager

67. Classic Dispatching Results
Optimal Schedules: Impossible to find for most real problems.
Dispatching: sorts jobs as they arrive at a machine.
Dispatching rules:
FIFO – simplest, seems “fair”.
SPT – Actually works quite well with tight due dates.
EDD – Works well when jobs are mostly the same size.
Many (100?) others.
Problems with Dispatching:
Cannot be optimal (can be bad).
Tends to be myopic.

68. The Difficulty of Scheduling Problems
Dilemma:
Too hard for optimal solutions.
Need something anyway.
Classifying “Hardness”:
Class P: has a polynomial solution.
Class NP: has no polynomial solution.
Example: Sequencing problems grow as n!.
Compare en/10000 and 10000n10.
At n = 40, en/10000 = 2.4  1013, 10000n10 = 1.0  1020
At n = 80, en/10000 = 5.5  1030, 10000n10 = 1.1  1023
3! = 6, 4! = 24, 5! = 120, 6! = 720, … 10! =3,628,800, while
13! = 6,227,020,800
25!= 15,511,210,043,330,985,984,000,000
en/10000
10000n10

69. The Difficulty of Scheduling Problems
NP stands for non polynomial, meaning that the time required to solve such problems is an exponential function of the number of jobs rather than a polynomial function.
The problems for which total enumeration is hopeless are known in mathematics as NP hard.

70. Computation Times
Current situation: computer can examine 1,000,000 sequences per second and we wish to build a scheduling system that has response time of no longer than one minute. How many jobs can we sequence optimally?

71. Effect of Faster Computers
Future Situation: New computer is 1,000 times faster, i.e. it can do 1 billion comparisons per second. How many jobs can we sequence optimally now?

72. Implications for Real Problems
Computation: NP algorithms are slow to use.
No Technology Fix: Faster computers don’t help on NP algorithm.
Scheduling is Hard: Real scheduling problems tend to be NP Hard.
Scheduling is Big: Real scheduling problems also tend to be quite large; impossible to solve optimally.

73. Implications for Real Problems (cont.)
Robustness? NP hard problems have many solutions, and presumably many “good” ones.
Our task is to find one of these.
Role of Heuristics: Polynomial algorithms can be used to obtain “good” solutions. Example heuristics include:
Simulated Annealing
Tabu Search
Genetic Algorithms

74. The Bad News
Violation of Assumptions: Most “real-world” scheduling problems violate the assumptions made in the classic literature:
There are always more than two machines.
Process times are not deterministic.
All jobs are not ready at the beginning of the problem.
Process time are sequence dependent.
Problem Difficulty: Most “real-world” production scheduling problems are NP-hard.
We cannot hope to find optimal solutions of realistic sized scheduling problems.
Polynomial approaches, like dispatching, may not work well.

75. The Good News Due Dates: We can set the due dates.
Job Splitting: We can get smaller jobs by splitting larger ones.
Single machine SPT results imply small jobs “clear out” more quickly than larger jobs.
Mechanics of Johnson’s algorithm implies we should start with a small job and end with a small job.
Small jobs make for small “move” batches and can be combined to form larger “process” batches.

76. The Good News (cont.)
Feasible Schedules: We do not need to find an optimal schedule, only a good feasible one.
Focus on Bottleneck: We can often concentrate on scheduling the bottleneck process, which simplifies problem closer to single machine case.
Capacity: Capacity can be adjusted dynamically (overtime, floating workers, use of vendors, etc.) to adapt facility (somewhat) to schedule.

77. Minimizing Scheduling Difficulties
Set realistic due dates
Focus on bottleneck operations
Consider lot splitting of large jobs