Operations scheduling
- 2. Scheduling • Prescribing when and where each operation necessary to manufacture the product is to be performed • Establishing of times at which to begin and complete each event or operation comprising the procedure
- 3. Strategic Importance of Scheduling Effective and efficient scheduling can be a competitive advantage Effective Scheduling helps in faster movement of goods through a facility which helps in better use of assets and lower costs Additional capacity resulting from faster throughput improves customer service through faster delivery Good schedules result in more dependable deliveries
- 4. Objectives of Scheduling Meet due date Minimize Work In Process inventory Minimize the average flow time through the systems Provide for high machine/worker utilization (Minimize the idle time) Reduce setup cost Minimize production and worker costs
- 5. Scheduling Issues Scheduling deals with the timing of operations The task is the allocation and prioritization of demand Significant issues are The type of scheduling, forward or backward The criteria for priorities
- 6. Scheduling Decisions Organization Scheduling Issues Hospital Operation room use Patient admissions Outpatient treatments College Classrooms and audiovisual equipment Student and instructor schedules Graduate and undergraduate courses Manufacturer Production of goods Purchase of materials Restaurant Delivery of fresh foods Opening of dining areas Cooking Air lines Maintenance of aircraft Departure timetables Catering, ticketing
- 7. Forward Scheduling Used when customers place their demand on as soon as possible basis Forward scheduling starts as soon as the requirements are known Produces a feasible schedule though it may not meet due dates Frequently results in buildup of work-in- process inventory It gets job done in shorter lead times than backward scheduling Due DateNow
- 8. Backward Scheduling Backward scheduling begins with the due date and schedules the final operation first Schedule is produced by working backwards though the processes Involves assigning jobs as late as possible Minimizes inventory Resources may not be available to accomplish the schedule Due DateNow
- 9. Often these approaches are combined to develop a trade-off between a feasible schedule and customer due dates
- 10. Scheduling Criteria Following four criteria are used to evaluate scheduling performance 1.Minimize completion time Average completion time = sum of total flow time / number of jobs 2.Maximize utilization of facilities Utilization = total job work time / sum of total flow time
- 11. 3. Minimize work-in-process (WIP) inventory Average number of jobs in the system = sum of total flow time / total job work time 4. Minimize customer waiting time Average job lateness = Total late days / number of jobs
- 12. Loading Jobs •Assignment of jobs or works to work or processing centers •Assign jobs so that costs, idle time, or completion time are minimized •Work centers may be a single machine, a group of machines or an area where a particular type of work is done •Infinite Loading/vertical loading • It ignores capacity constraints, but helps identify bottlenecks in a proposed schedule • Jobs are assigned to work centers without regard for capacity of the work center
- 13. •Finite loading/horizontal loading • Jobs are assigned to work centers within the available capacity • It considers the capacity of each work center and compares the processing time so that process time does not exceed capacity
- 15. Input Output Control Under this system, facility work flow is managed by tracking work added to a work center and its work completed Identifies overloading and under-loading conditions Prompts managerial action to resolve scheduling problems
- 16. •Backlog: work that has not been completed
- 17. Input-Output Control Example Week Ending 6/6 6/13 6/20 6/27 7/4 7/11 Planned Input 280 280 280 280 280 Actual Input 270 250 280 285 280 Cumulative Deviation –10 –40 –40 –35 Planned Output 320 320 320 320 Actual Output 270 270 270 270 Cumulative Deviation –50 –100 –150 –200 Cumulative Change in Backlog 0 –20 –10 +5 Work Center DNC Milling (in standard hours)
- 18. Input-Output Control Example Work Center DNC Milling (in standard hours) Week Ending 6/6 6/13 6/20 6/27 7/4 7/11 Planned Input 280 280 280 280 280 Actual Input 270 250 280 285 280 Cumulative Deviation –10 –40 –40 –35 Planned Output 320 320 320 320 Actual Output 270 270 270 270 Cumulative Deviation –50 –100 –150 –200 Cumulative Change in Backlog 0 –20 –10 +5 Explanation: 270 input, 270 output implies 0 change Explanation: 250 input, 270 output implies –20 change
- 19. Input-Output Control Example Options available to operations personnel include 1. Correcting performances 2. Increasing capacity 3. Increasing or reducing input to the work center
- 20. Gantt Charts • Visual aids for loading and scheduling Load chart shows the loading and idle times of departments, machines, or facilities Displays relative workloads over time Schedule chart monitors jobs in process All Gantt charts need to be updated frequently to account for changes
- 21. Gantt Load Chart Example Day Monday Tuesday Wednesday Thursday FridayWork Center Metalworks Mechanical Electronics Painting Job 349 Job 349 Job 349 Job 408 Job 408 Job 408 Processing Unscheduled Center not available Job 350 Job 349 Job 295
- 22. Gantt Schedule Chart Example Figure 15.4 Job Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Day 7 Day 8 A B C Now Maintenance Start of an activity End of an activity Scheduled activity time allowed Actual work progress Nonproduction time Point in time when chart is reviewed
- 23. Assignment Method A special class of linear programming models that assign tasks or jobs to resources Objective is to minimize cost or time Only one job (or worker) is assigned to one machine (or project)
- 24. Assignment Method Build a table of costs or time associated with particular assignments Typesetter Job A B C R-34 $11 $14 $ 6 S-66 $ 8 $10 $11 T-50 $ 9 $12 $ 7
- 25. Assignment Method 1. Create zero opportunity costs by repeatedly subtracting the lowest costs from each row and column 2. Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the table. If the number of lines equals either the number of rows or the number of columns, proceed to step 4. Otherwise proceed to step 3.
- 26. Assignment Method 3. Subtract the smallest number not covered by a line from all other uncovered numbers. Add the same number to any number at the intersection of two lines. Return to step 2. 4. Optimal assignments are at zero locations in the table. Select one, draw lines through the row and column involved, and continue to the next assignment.
- 27. Special Cases • when the number of labors/machines does not equal the number of jobs, add a dummy row with zero values. Rest process is same. • Assignment problems are normally minimization problems. If the goal is to maximize profit, convert the matrix in Relative Loss Matrix by subtracting all the elements on the matrix from the maximum value in the matrix. • If any job can not be assigned to a person due to certain reasons, we need to solve that assignment problem by placing dash (-) sign so that no calculations will be made at that place.
- 28. Assignment Example A B C Job R-34 $11 $14 $ 6 S-66 $ 8 $10 $11 T-50 $ 9 $12 $ 7 Typesetter A B C Job R-34 $ 5 $ 8 $ 0 S-66 $ 0 $ 2 $ 3 T-50 $ 2 $ 5 $ 0 Typesetter Step 1a - Rows A B C Job R-34 $ 5 $ 6 $ 0 S-66 $ 0 $ 0 $ 3 T-50 $ 2 $ 3 $ 0 Typesetter Step 1b - Columns
- 29. Assignment Example Step 2 - Lines A B C Job R-34 $ 5 $ 6 $ 0 S-66 $ 0 $ 0 $ 3 T-50 $ 2 $ 3 $ 0 Typesetter Because only two lines are needed to cover all the zeros, the solution is not optimal Step 3 - Subtraction A B C Job R-34 $ 3 $ 4 $ 0 S-66 $ 0 $ 0 $ 5 T-50 $ 0 $ 1 $ 0 Typesetter The smallest uncovered number is 2 so this is subtracted from all other uncovered numbers and added to numbers at the intersection of lines
- 30. Assignment Example Because three lines are needed, the solution is optimal and assignments can be made Step 2 - Lines A B C Job R-34 $ 3 $ 4 $ 0 S-66 $ 0 $ 0 $ 5 T-50 $ 0 $ 1 $ 0 Typesetter Start by assigning R-34 to worker C as this is the only possible assignment for worker C. Step 4 - Assignments A B C Job R-34 $ 3 $ 4 $ 0 S-66 $ 0 $ 0 $ 5 T-50 $ 0 $ 1 $ 0 Typesetter Job T-50 must go to worker A as worker C is already assigned. This leaves S-66 for worker B.
- 31. Assignment Example From the original cost table Minimum cost = $6 + $10 + $9 = $25 Step 4 - Assignments A B C Job R-34 $ 3 $ 4 $ 0 S-66 $ 0 $ 0 $ 5 T-50 $ 0 $ 1 $ 0 Typesetter A B C Job R-34 $11 $14 $ 6 S-66 $ 8 $10 $11 T-50 $ 9 $12 $ 7 Typesetter
- 32. Sequencing Jobs Specifies the order in which jobs should be performed at work centers Priority rules are used to dispatch or sequence jobs FCFS: First come, first served SPT: Shortest processing time EDD: Earliest due date LPT: Longest processing time
- 33. Key Terminologies • Processing time(job work time) • Time required to process a job • Setup time + actual processing time • Due date • Date at which a job is to be completed • Flow time • Waiting time + processing time • Job lateness • Time at which job is actually completed – scheduled finishing time • Flow time(days) – Job due date(days)
- 34. Evaluate the Priority Techniques in following Criteria 1.Minimize completion time Average completion time = sum of total flow time / number of jobs 2.Maximize utilization of facilities Utilization = total job work time / sum of total flow time
- 35. 3. Minimize work-in-process (WIP) inventory Average number of jobs in the system = sum of total flow time / total job work time 4. Minimize customer waiting time Average job lateness = Total late days / number of jobs
- 36. Sequencing Example Job Job Work (Processing) Time (Days) Job Due Date (Days) A 6 8 B 2 6 C 8 18 D 3 15 E 9 23 Apply the four popular sequencing rules to these five jobs
- 37. Sequencing Example Job Sequence Job Work (Processing) Time Flow Time Job Due Date Job Lateness A 6 6 8 0 B 2 8 6 2 C 8 16 18 0 D 3 19 15 4 E 9 28 23 5 28 77 11 FCFS: Sequence A-B-C-D-E
- 38. Sequencing Example Job Sequence Job Work (Processing) Time Flow Time Job Due Date Job Lateness A 6 6 8 0 B 2 8 6 2 C 8 16 18 0 D 3 19 15 4 E 9 28 23 5 28 77 11 FCFS: Sequence A-B-C-D-E Average completion time = = 77/5 = 15.4 days Sum of total flow time Number of jobs Utilization = = 28/77 = 36.4% Total job work time Sum of total flow time Average number of jobs in the system = = 77/28 = 2.75 jobs Sum of total flow time Total job work time Average job lateness = = 11/5 = 2.2 days Total late days Number of jobs
- 39. Sequencing Example Job Sequence Job Work (Processing) Time Flow Time Job Due Date Job Lateness B 2 2 6 0 D 3 5 15 0 A 6 11 8 3 C 8 19 18 1 E 9 28 23 5 28 65 9 SPT: Sequence B-D-A-C-E
- 40. Sequencing Example Job Sequence Job Work (Processing) Time Flow Time Job Due Date Job Lateness B 2 2 6 0 D 3 5 15 0 A 6 11 8 3 C 8 19 18 1 E 9 28 23 5 28 65 9 SPT: Sequence B-D-A-C-E Average completion time = = 65/5 = 13 days Sum of total flow time Number of jobs Utilization = = 28/65 = 43.1% Total job work time Sum of total flow time Average number of jobs in the system = = 65/28 = 2.32 jobs Sum of total flow time Total job work time Average job lateness = = 9/5 = 1.8 days Total late days Number of jobs
- 41. Sequencing Example Job Sequence Job Work (Processing) Time Flow Time Job Due Date Job Lateness B 2 2 6 0 A 6 8 8 0 D 3 11 15 0 C 8 19 18 1 E 9 28 23 5 28 68 6 EDD: Sequence B-A-D-C-E
- 42. Sequencing Example Job Sequence Job Work (Processing) Time Flow Time Job Due Date Job Lateness B 2 2 6 0 A 6 8 8 0 D 3 11 15 0 C 8 19 18 1 E 9 28 23 5 28 68 6 EDD: Sequence B-A-D-C-E Average completion time = = 68/5 = 13.6 days Sum of total flow time Number of jobs Utilization = = 28/68 = 41.2% Total job work time Sum of total flow time Average number of jobs in the system = = 68/28 = 2.43 jobs Sum of total flow time Total job work time Average job lateness = = 6/5 = 1.2 days Total late days Number of jobs
- 43. Sequencing Example Job Sequence Job Work (Processing) Time Flow Time Job Due Date Job Lateness E 9 9 23 0 C 8 17 18 0 A 6 23 8 15 D 3 26 15 11 B 2 28 6 22 28 103 48 LPT: Sequence E-C-A-D-B
- 44. Sequencing Example Job Sequence Job Work (Processing) Time Flow Time Job Due Date Job Lateness E 9 9 23 0 C 8 17 18 0 A 6 23 8 15 D 3 26 15 11 B 2 28 6 22 28 103 48 LPT: Sequence E-C-A-D-B Average completion time = = 103/5 = 20.6 days Sum of total flow time Number of jobs Utilization = = 28/103 = 27.2% Total job work time Sum of total flow time Average number of jobs in the system = = 103/28 = 3.68 jobs Sum of total flow time Total job work time Average job lateness = = 48/5 = 9.6 days Total late days Number of jobs
- 45. Sequencing Example Rule Average Completion Time (Days) Utilization (%) Average Number of Jobs in System Average Lateness (Days) FCFS 15.4 36.4 2.75 2.2 SPT 13.0 43.1 2.32 1.8 EDD 13.6 41.2 2.43 1.2 LPT 20.6 27.2 3.68 9.6 Summary of Rules
- 46. Comparison of Sequencing Rules No one sequencing rule excels on all criteria SPT does well on minimizing flow time and number of jobs in the system But SPT moves long jobs to the end which may result in dissatisfied customers FCFS does not do especially well (or poorly) on any criteria but is perceived as fair by customers EDD minimizes lateness
- 47. Critical Ratio (CR) An index number found by dividing the time remaining until the due date by the work time remaining on the job Jobs with low critical ratios are scheduled ahead of jobs with higher critical ratios Performs well on average job lateness criteria CR = = Due date - Today’s date Work (lead) time remaining Time remaining Workdays remaining
- 48. • If CR = 1, then the job is exactly on schedule • If CR < 1, then the job is behind the schedule • If CR > 1, then the job is ahead of schedule
- 49. Critical Ratio Example Job Due Date Workdays Remaining Critical Ratio Priority Order A 30 4 (30 - 25)/4 = 1.25 3 B 28 5 (28 - 25)/5 = .60 1 C 27 2 (27 - 25)/2 = 1.00 2 Currently Day 25 With CR < 1, Job B is late. Job C is just on schedule and Job A has some slack time.
- 50. Critical Ratio Technique 1. Helps determine the status of specific jobs 2. Establishes relative priorities among jobs on a common basis 3. Relates both stock and make-to-order jobs on a common basis 4. Adjusts priorities automatically for changes in both demand and job progress 5. Dynamically tracks job progress
- 51. Sequencing N Jobs on Two Machines: Johnson’s Rule Works with two or more jobs that pass through the same two machines or work centers Minimizes total production time and idle time
- 52. Johnson’s Rule 1. List all jobs and times for each work center 2. Choose the job with the shortest activity time. If that time is in the first work center, schedule the job first. If it is in the second work center, schedule the job last. 3. Once a job is scheduled, it is eliminated from the list 4. Repeat steps 2 and 3 working toward the center of the sequence
- 53. Johnson’s Rule Example Job Work Center 1 (Drill Press) Work Center 2 (Lathe) A 5 2 B 3 6 C 8 4 D 10 7 E 7 12
- 54. B E D C A Johnson’s Rule Example Job Work Center 1 (Drill Press) Work Center 2 (Lathe) A 5 2 B 3 6 C 8 4 D 10 7 E 7 12
- 55. Johnson’s Rule Example Job Work Center 1 (Drill Press) Work Center 2 (Lathe) A 5 2 B 3 6 C 8 4 D 10 7 E 7 12 Time 0 3 10 20 28 33 B ACDE WC 1 WC 2 B ACDE
- 56. Johnson’s Rule Example Job Work Center 1 (Drill Press) Work Center 2 (Lathe) A 5 2 B 3 6 C 8 4 D 10 7 E 7 12 Time 0 3 10 20 28 33 Time 0 1 3 5 7 9 10 11 12 13 17 19 21 22 2325 27 29 31 33 35 B ACDE B ACDE WC 1 WC 2 B E D C A B ACDE
- 57. Scheduling Services Service systems differ from manufacturing Manufacturing Services Schedules machines and materials Schedule staff Inventories used to smooth demand Seldom maintain inventories Machine-intensive and demand may be smooth Labor-intensive and demand may be variable Scheduling may be bound by union contracts Legal issues may constrain flexible scheduling Few social or behavioral issues Social and behavioral issues may be quite important
- 58. Techniques for Scheduling Services Demand Management • Appointments • Reservations • Posted schedules • Delayed services or backlogs Scheduling employees • Staffing for peak demand • Floating employees • Employees on call • Temporary employees • Seasonal employees • Part-time employees
- 59. Demand Management Appointment or reservation systems FCFS sequencing rules Discounts or other promotional schemes When demand management is not feasible, managing capacity through staffing flexibility may be used
- 60. Scheduling Service Employees With Cyclical Scheduling Objective is to meet staffing requirements with the minimum number of workers Schedules need to be smooth and keep personnel happy Many techniques exist from simple algorithms to complex linear programming solutions
- 61. Cyclical Scheduling Example 1. Determine the staffing requirements 2. Identify two consecutive days with the lowest total requirements and assign these as days off 3. Make a new set of requirements subtracting the days worked by the first employee 4. Apply step 2 to the new row 5. Repeat steps 3 and 4 until all requirements have been met
- 62. Cyclical Scheduling Example M T W T F S S Employee 1 5 5 6 5 4 3 3 Capacity (Employees) Excess Capacity
- 63. Cyclical Scheduling Example M T W T F S S Employee 1 5 5 6 5 4 3 3 Employee 2 4 4 5 4 3 3 3 Capacity (Employees) Excess Capacity
- 64. Cyclical Scheduling Example M T W T F S S Employee 1 5 5 6 5 4 3 3 Employee 2 4 4 5 4 3 3 3 Employee 3 3 3 4 3 2 3 3 Capacity (Employees) Excess Capacity
- 65. Cyclical Scheduling Example M T W T F S S Employee 1 5 5 6 5 4 3 3 Employee 2 4 4 5 4 3 3 3 Employee 3 3 3 4 3 2 3 3 Employee 4 2 2 3 2 2 3 2 Capacity (Employees) Excess Capacity
- 66. Cyclical Scheduling Example M T W T F S S Employee 1 5 5 6 5 4 3 3 Employee 2 4 4 5 4 3 3 3 Employee 3 3 3 4 3 2 3 3 Employee 4 2 2 3 2 2 3 2 Employee 5 1 1 2 2 2 2 1 Capacity (Employees) Excess Capacity
- 67. Cyclical Scheduling Example M T W T F S S Employee 1 5 5 6 5 4 3 3 Employee 2 4 4 5 4 3 3 3 Employee 3 3 3 4 3 2 3 3 Employee 4 2 2 3 2 2 3 2 Employee 5 1 1 2 2 2 2 1 Employee 6 1 1 1 1 1 1 0 Capacity (Employees) Excess Capacity
- 68. Cyclical Scheduling Example M T W T F S S Employee 1 5 5 6 5 4 3 3 Employee 2 4 4 5 4 3 3 3 Employee 3 3 3 4 3 2 3 3 Employee 4 2 2 3 2 2 3 2 Employee 5 1 1 2 2 2 2 1 Employee 6 1 1 1 1 1 1 0 Employee 7 1 Capacity (Employees) Excess Capacity
- 69. Cyclical Scheduling Example M T W T F S S Employee 1 5 5 6 5 4 3 3 Employee 2 4 4 5 4 3 3 3 Employee 3 3 3 4 3 2 3 3 Employee 4 2 2 3 2 2 3 2 Employee 5 1 1 2 2 2 2 1 Employee 6 1 1 1 1 1 1 0 Employee 7 1 Capacity (Employees) 5 5 6 5 4 3 3 Excess Capacity 0 0 0 0 0 1 0
- 70. Methods of Minimizing Non-sampling Errors • Checking the sampling process • Preparation of questionnaire • Pilot survey • Fix procedures • Use of competent manpower • Provide information • Provide training • Use of experts • Checking data processing and analysis