1. Chapter 16: Learning Objectives Instructor Slides You should be able to: 1.Explain what scheduling involves and the importance of good scheduling 2.Describe scheduling needs in high-volume and intermediate-volume systems 3.Describe scheduling needs in job shops 4.Use and interpret Gantt charts, and use the assignment method for loading 5.Give examples of commonly used priority rules 6.Summarize some of the unique problems encountered in service systems, and describe some of the approaches used for scheduling service systems 16-1

2. Scheduling Scheduling: – Establishing the timing of the use of equipment, facilities and human activities in an organization Effective scheduling can yield Cost savings Increases in productivity Other benefits Instructor Slides 16-2

3. Scheduling Context Instructor Slides Scheduling is constrained by multiple system design decisions – System capacity – Product and/or service design – Equipment selection – Worker selection and training – Aggregate planning and master scheduling 16-3

4. Scheduling Hierarchies Instructor Slides 16-4

5. High Volume Systems Instructor Slides Flow System – High-volume system in which all jobs follow the same sequence – Flow system scheduling Scheduling for flow systems The goal is to achieve a smooth rate of flow of goods or customers through the system in order to get high utilization of labor and equipment Workstation 1 Workstation 2 Output 16-5

6. High-Volume: Scheduling Difficulties Instructor Slides Few flow systems are entirely dedicated to a single product or service – Each product change requires Slightly different inputs of parts Slightly different materials Slightly different processing requirements that must be scheduled into the line – Need to avoid excessive inventory buildup – Disruptions may result in less-than-desired output 16-6

7. High-Volume Success Factors Instructor Slides The following factors often dictate the success of high- volume systems: Process and product design Preventive maintenance Rapid repair when breakdowns occur Optimal product mixes Minimization of quality problems Reliability and timing of supplies 16-7

8. Intermediate-Volume Systems Instructor Slides Outputs fall between the standardized type of output of high-volume systems and the make-to-order output of job shops Output rates are insufficient to warrant continuous production Rather, it is more economical to produce intermittently Work centers periodically shift from one product to another 16-8

9. Intermediate-Volume Systems Three basic issues: – Run size of jobs – The timing of jobs – The sequence in which jobs will be produced Instructor Slides 16-9

10. Low-Volume Systems Instructor Slides Job shop scheduling Scheduling for low-volume systems with many variations in requirements Make-to-order products Processing requirements Material requirements Processing time Processing sequence and steps A complex scheduling environment It is impossible to establish firm schedules until actual job orders are received 16-10

11. Intermediate-Volume Systems Instructor Slides Important considerations – Setup cost – Usage is not always as smooth as assumed in the economic lot size model Alternative scheduling approach – Base production on a master schedule developed from customer orders and forecasted demand 16-11

12. Low-Volume Systems: Loading Instructor Slides Loading – the assignment of jobs to processing centers – Gantt chart Used as a visual aid for loading and scheduling purposes Purpose of the Gantt chart is to organize and visually display the actual or intended use of resources in a time framework Managers may use the charts for trial-and-error schedule development to get an idea of what different arrangements would involve 16-12

13. Gantt Charts Load chart – A Gantt chart that shows the loading and idle times for a group of machines or list of departments Instructor Slides 16-13

14. Loading Approaches Infinite loading – Jobs are assigned to workstations without regard to the capacity of the work center Finite loading – Jobs are assigned to work centers taking into account the work center capacity and job processing times Instructor Slides 1 2 3 45 6 over Capacity Infinite loading 1 2 3 4 5 6 Capacity Finite loading 16-14

15. Scheduling Approaches Instructor Slides  Forward scheduling  Scheduling ahead from some point in time.  Used when the question is:  “How long will it take to complete this job?  Backward scheduling  Scheduling backwards from some due date  Used when the question is:  “When is the latest this job can be started and still be completed on time?” 16-15

16. Gantt Charts Schedule chart – A Gantt chart that shows the orders or jobs in progress and whether they are on schedule Instructor Slides 16-16

17. Managing Work Flows Instructor Slides Input/Output (I/O) control – Managing work flow and queues at work centers Without I/O control: – If demand exceeds processing capacity, a work center overload is created – If work arrives more slowly than a work center can handle, work center underutilization results The goal is to strike a balance between input and output rates in order to minimize queues and maximize utilization 16-17

18. I/O Chart Instructor Slides 16-18

19. Assignment Instructor Slides Assignment model – A linear programming model for optimal assignment of tasks and resources Hungarian method – Method of assigning jobs by a one-for-one matching to identify the lowest cost solution 16-19

20. Hungarian Method Instructor Slides 1.Row reduction: subtract the smallest number in each row from every number in the row a.Enter the result in a new table 2.Column reduction: subtract the smallest number in each column from every number in the column a.Enter the result in a new table 3.Test whether an optimum assignment can be made a.Determine the minimum number of lines needed to cross out all zeros b.If the number of lines equals the number of rows, an optimum assignment is possible. Go to step 6 c.Else, go to step 4 16-20

21. Hungarian Method (contd.) Instructor Slides 4.If the number of lines is less than the number of rows, modify the table: a.Subtract the smallest number from every uncovered number in the table b.Add the smallest uncovered number to the numbers at intersections of cross-out lines c.Numbers crossed out but not at intersections of cross-out lines carry over unchanged to the next table 5.Repeat steps 3 and 4 until an optimal table is obtained 6.Make the assignments a.Begin with rows or columns with only one zero b.Match items that have zeros, using only one match for each row and each column c.Eliminate both the row and the column after the match 16-21

22. Example: Hungry Owner A contractor pays his subcontractors a fixed fee plus mileage for work performed. On a given day the contractor is faced with three electrical jobs associated with various projects. Given below are the distances between the subcontractors and the projects. Project A B C Westside 50 36 16 Subcontractors Federated 28 30 18 Goliath 35 32 20 Universal 25 25 14 How should the contractors be assigned to minimize total distance (and total cost)?

23. Example: Hungry Owner LP Formulation – Decision Variables Defined x ij = 1 if subcontractor i is assigned to project j = 0 otherwise where: i = 1 (Westside), 2 (Federated), 3 (Goliath), and 4 (Universal) j = 1 (A), 2 (B), and 3 (C)

24. Example: Hungry Owner LP Formulation – Objective Function Minimize total distance: Min 50x 11 + 36x 12 + 16x 13 + 28x 21 + 30x 22 + 18x 23 + 35x 31 + 32x 32 + 20x 33 + 25x 41 + 25x 42 + 14x 43

25. Example: Hungry Owner LP Formulation – Constraints x 11 + x 12 + x 13 < 1 (no more than one x 21 + x 22 + x 23 < 1 project assigned x 31 + x 32 + x 33 < 1 to any one x 41 + x 42 + x 43 < 1 subcontractor) x 11 + x 21 + x 31 + x 41 = 1 (each project must x 12 + x 22 + x 32 + x 42 = 1 be assigned to just x 13 + x 23 + x 33 + x 43 = 1 one subcontractor) all x ij > 0 (non-negativity)

26. Example: Hungry Owner Optimal Assignment Subcontractor Project Distance Westside C 16 Federated A 28 Universal B 25 Goliath (unassigned) Total Distance = 69 miles

27. Example: Hungry Owner Initial Tableau Setup Since the Hungarian algorithm requires that there be the same number of rows as columns, add a Dummy column so that the first tableau is: A B C Dummy Westside 50 36 16 0 Federated 28 30 18 0 Goliath 35 32 20 0 Universal 25 25 14 0

28. Example: Hungry Owner Step 1: Subtract minimum number in each row from all numbers in that row. Since each row has a zero, we would simply generate the same matrix above. Step 2: Subtract the minimum number in each column from all numbers in the column. For A it is 25, for B it is 25, for C it is 14, for Dummy it is 0. This yields: A B C Dummy Westside 25 11 2 0 Federated 3 5 4 0 Goliath 10 7 6 0 Universal 0 0 0 0

29. Example: Hungry Owner Step 3: Draw the minimum number of lines to cover all zeroes. Although one can "eyeball" this minimum, use the following algorithm. If a "remaining" row has only one zero, draw a line through the column. If a remaining column has only one zero in it, draw a line through the row. A B C Dummy Westside 25 11 2 0 Federated 3 5 4 0 Goliath 10 7 6 0 Universal 0 0 0 0 Step 4: The minimum uncovered number is 2 (circled).

30. Example: Hungry Owner Step 5: Subtract 2 from uncovered numbers; add 2 to all numbers covered by two lines. This gives: A B C Dummy Westside23 9 0 0 Federated 1 3 2 0 Goliath 8 5 4 0 Universal 0 0 0 2

31. Example: Hungry Owner Step 3: Draw the minimum number of lines to cover all zeroes. A B C Dummy Westside 23 9 0 0 Federated 1 3 2 0 Goliath 8 5 4 0 Universal 0 0 0 2 Step 4: The minimum uncovered number is 1 (circled).

32. Example: Hungry Owner Step 5: Subtract 1 from uncovered numbers. Add 1 to numbers covered by two lines. This gives: A B C Dummy Westside 23 9 0 1 Federated 0 2 1 0 Goliath 7 4 3 0 Universal 0 0 0 3

33. Example: Hungry Owner Step 4: The minimum number of lines to cover all 0's is four. Thus, there is a minimum-cost assignment of 0's with this tableau. The optimal assignment is: Subcontractor Project Distance Westside C 16 Federated A 28 Goliath (unassigned) Universal B 25 Total Distance = 69 miles

34. Sequencing Instructor Slides Sequencing – Determine the order in which jobs at a work center will be processed Priority rules – Simple heuristics used to select the order in which jobs will be processed – The rules generally assume that job setup cost and time are independent of processing sequence Job time – Time needed for setup and processing of a job 16-34

35. Priority Rules FCFS - first come, first served SPT- shortest processing time EDD - earliest due date CR - critical ratio S/O - slack per operation Rush - emergency Instructor Slides 16-35

36. Priority Rules: Assumptions Instructor Slides The set of jobs is known; no new orders arrive after processing begins and no jobs are canceled Setup time is independent of processing time Setup time is deterministic Processing times are deterministic There will be no interruptions in processing such as machine breakdowns or accidents 16-36

37. Priority Rules: Local v. Global Local priority rules: – Focus on information pertaining to a single workstation when establishing a job sequence Global priority rules: – Incorporate information from multiple workstations when establishing a job sequence Instructor Slides 16-37

38. Sequence: Performance Metrics Instructor Slides Common performance metrics: Job flow time This is the amount of time it takes from when a job arrives until it is complete It includes not only processing time but also any time waiting to be processed Job lateness This is the amount of time the job completion time is expected to exceed the date the job was due or promised to a customer Makespan The total time needed to complete a group of jobs from the beginning of the first job to the completion of the last job Average number of jobs Jobs that are in a shop are considered to be WIP inventory 16-38

39. Example: one work center sequencing The processing times (including setup time) and due dates for six jobs waiting to be processed at a work center are given in the following table. Determine the sequence of jobs, the average flow time, average tardiness, and average number of jobs at work center for each of these rules: -FCFS -SPT -EDD -CR Instructor Slides39 JobProcessing time (days) Due date (days from present time A27 B816 C44 D1017 E515 F1218

40. Example: one work center sequencing The FCFS sequence is simply A-B-C-D-E-F. the measures of effectiveness are as follows: 1.Average flow time = 120/6 20 days 2.Average tardiness = 54/6 = 9 days 3.The Makespan 41 days 4.Average number of jobs at the work center = 120/41 = 2.93 Instructor Slides40 Job sequence Processing time Flow time Due date Days tardy A2270 B810160 C414410 D 24177 E5291514 F12411823 Sum4112054

41. Example: one work center sequencing Using the SPT rule, the job sequence is A-C-E-B-D-F. the measures of effectiveness are: 1.Average flow time = 108/6 = 18 days 2.Average tardiness = 40/6 = 6.67 days 3.Average number of jobs at work center =108/41 = 2.63 Instructor Slides41 Job sequence Processing time Flow timeDue date Days tardy A2270 C4642 E511150 B819163 D10291712 F 411823 Sum4110840

42. Example: one work center sequencing Using EDD, the jobs sequence is : C-A-E-B-D-F. the measures of effectiveness are as follows: 1. average flow time = 110/6 =18.33 days 2. Average tardiness = 38/6 = 6.33 days 3. Average number of jobs at the work center = 110/41 = 2.68 days Instructor Slides42 Jobs sequence Processing time Flow time Due date Days tardy C4440 A2670 E511150 B819163 D10291712 F 411823 sum4111038

43. Two Work Center Sequencing Johnson’s Rule – Technique for minimizing makespan for a group of jobs to be processed on two machines or at two work centers. Minimizes total idle time Several conditions must be satisfied Instructor Slides 16-43

44. Johnson’s Rule Conditions Instructor Slides Job time must be known and constant for each job at the work center Job times must be independent of sequence Jobs must follow same two-step sequence All jobs must be completed at the first work center before moving to second work center 16-44

45. Johnson’s Rule: Optimum Sequence Instructor Slides 1.List the jobs and their times at each work center 2.Select the job with the shortest time a.If the shortest time is at the first work center, schedule that job first b.If the shortest time is at the second work center, schedule the job last. c.Break ties arbitrarily 3.Eliminate the job from further consideration 4.Repeat steps 2 and 3, working toward the center of the sequence, until all jobs have been scheduled 16-45

46. Example: two work centers sequencing The following table shows the processing time (in minutes) required for 5 jobs on two work centers: Solution Instructor Slides46 JobTime (work center 1) Time (work center 2) A25 B79 C31 D48 E94 SequenceADBEC

47. Scheduling Difficulties Instructor Slides Variability in – Setup times – Processing times – Interruptions – Changes in the set of jobs Except for small job sets, there is no method for identifying an optimal schedule Scheduling is not an exact science It is an ongoing task for a manager 16-47

48. Minimizing Scheduling Difficulties Set realistic due dates Focus on bottleneck operations – First, try to increase the capacity of the operations – If that is not possible Schedule bottleneck operations first Then, schedule non-bottleneck operations around the bottleneck operations Consider lot splitting of large jobs – Often works best when there are large differences in job times Instructor Slides 16-48