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Operations Management

Operations Management. Chapter 15 – Short-Term Scheduling. PowerPoint presentation to accompany Heizer/Render Operations Management, 8e . © 2006 Prentice Hall, Inc. Strategic Importance of Short-Term Scheduling. Effective and efficient scheduling can be a competitive advantage

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Operations Management

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  1. Operations Management Chapter 15 – Short-Term Scheduling PowerPoint presentation to accompany Heizer/Render Operations Management, 8e © 2006 Prentice Hall, Inc.

  2. Strategic Importance of Short-Term Scheduling • Effective and efficient scheduling can be a competitive advantage • Faster movement of goods through a facility means better use of assets and lower costs • Additional capacity resulting from faster throughput improves customer service through faster delivery • Good schedules result in more reliable deliveries

  3. Scheduling Decisions Table 15.1

  4. Activity in Sequencing • Sequence the following cars into as many work days as needed. Garage can work on two cars simultaneously • Assume first come first serve sequencing; 8 hour workday. Customers arrive in the following orderDAY 1 • Car 3: Maintenance ; time needed 6 hours • Car 4: Maintenance ; time needed 10 hours • Car 1: Repair ; time needed 2 hours • Car 2: Repair ; time needed 2.5 hours DAY 2 • Car 5: Maintenance ; time needed 3.5 hours • Car 6: Repair ; time needed 3.5 hrs • Car 7: Maintenance ; time needed 4 hours

  5. Solution: Sequencing Repair Track 2 Car 4: Repair – 8 hours Repair Track 1 Car 3: Repair - 6 hours Car 1: Maintenance 2 hours Day 1 Repair Track 1 Car 2: Maintenance 2.5 hoursCar 6:Maintenance 3.5 hours Repair Track 2 Car 4: Repair – 2 hours Car 5: Repair – 3.5 hours Car 7:Maintenance 2.5 hours Day 2 Repair Track 2 Car 7:Maintenance 1.5 hours Repair Track 1 Day 3

  6. Activity in Sequencing_2 • Schedule the following cars into 2 work days. Garage can work on two cars simultaneously • Method: Garage controlled scheduling (First assigned first serve; or capacity-based scheduling). 8 hours per day work time. • Car 3: Maintenance ; time needed 6 hours • Car 4: Maintenance ; time needed 10 hours • Car 1: Repair ; time needed 2 hours • Car 2: Repair ; time needed 2.5 hours • Car 5: Maintenance ; time needed 3.5 hours • Car 6: Repair ; time needed 3.5 hrs • Car 7: Maintenance ; time needed 4 hours

  7. Solution: Sequencing _2 • Schedule for days 1 and 2. Notice one track for longduration work and the other for fast jobs! Long turnaround jobs Fast turnaround jobs Repair Track 1 Car 1: Repair - 2 hours Car 5: Maintenance 3.5 hours Car 2: Maintenance 2.5 hours Repair Track 2 Car 4: Repair – 8 hours Day 1 Repair Track 2 Car 4: Repair – 2 hours Car 3: Repair – 6 hours Repair Track 1 Car 6: Maintenance 3.5 hoursCar 7:Maintenance 4.0 hours Day 2

  8. Definitions • Scheduling is the assignment of due dates to specific work or jobs. • Loading is the assignment of jobs to work centers. • Sequencing: Determining the order in which jobs should be done at each work center so that due dates are met. • Input-Output control: Any technique that enables managers to manage workflows at each work center by comparing work added to work completed.

  9. Positioning Scheduling Figure 15.1

  10. Defining Scheduling • Scheduling deals with the assignment of activities (demand) to resources (supply) (or vice-versa) and timing of activities. E.g.supply could be production capacity of a firm) • Types of scheduling situations • Type I: Supply options (M) are fewer than demand options (N) • Type II: Supply options (M) are equal to demand options (N) • Type III: Supply options (M) exceed the number of demand options (N)

  11. Objectives of Scheduling • Goals of scheduling • Type I: Supply options (M) are fewer than demand options (N) • Assign scarce supply to demand to minimize cost or maximize benefits • Type II: Supply options (M) are equal to demand options (N) • Assign supply to demand to minimize cost or maximize benefits for total process • Type III: Supply options (M) exceed the number of demand options (N) • Scheduling is done for limited capacity and excess capacity is outsourced.

  12. Methods of Scheduling • Forward scheduling concept • Scheduling begins as soon as customer requests and requirements are known • Scheduling begins from the estimated start date of the project and works forward to determine the start and finish dates for each of the activities that make up the order. • Backward scheduling concept • Scheduling begins from the expected delivery date and works backwards to determine the finish and start dates for the activities that make up the order.[Usually this method is available for projects that have long completion times, large number of units or parts, and have the completion of project on or before the delivery deadline as a key objective] .

  13. Loading_Activity_A • We own a hotel which has a large ballroom. We have to schedule activities for two Saturdays in May. The closing time of the ballroom each Saturday is 10 pm. Which activities would you schedule? We charge per hour for the time a client spend using room. No charges for cleaning and preparation times. • Event A: 9 am - 1 pm, Cleanup needed after event 2 hrs. • Event B: 4 pm – 7 pm, preparation needed before event 0.5 hours. Cleanup needed after event 1 hrs. • Event C: 5 pm - 10 pm, preparation needed before event 1 hours. Cleanup after event 2 hrs. • Event D: 9 am -12 pm, preparation needed before event 1 hours. Cleanup after event 1 hrs. • Event E: 11 am – 8 pm, preparation needed before event 2 hours. Cleanup after event 2 hrs.

  14. Activity_A Scheduling Criteria and Options • Option1 • Event A: 9 am - 1 pm AND Event B: 4 – 7 pm • Event D: 9 am -12 pm AND Event C: 5 - 10 pm. • Option 2 • Event A: 9 am - 1 pm AND Event C 5 - 10 pm. • Event D: 9 am -12 pm AND Event B: 4 – 7 pm • Option 3 • Event A: 9 am - 1 pm AND Event B OR Event C • Event E: 11 am – 8 pm Scheduling Criteria: Why did we schedule the way we did? We are tried to maximize the utilization of the ballroom (maximize utilization)! Other criteria; Min. Cost, Min. waiting time or Work in progress (WIP)

  15. Comparing Options_Activity_A • Option1 • Event A: 9 am - 1 pm AND Event B: 4 – 7 pm • Event D: 9 am -12 pm AND Event C: 5 - 10 pm. • Option 2 • Event A: 9 am - 1 pm AND Event C 5 - 10 pm. • Event D: 9 am -12 pm AND Event B: 4 – 7 pm • Option 3 • Event A: 9 am - 1 pm AND Event B OR Event C • Event E: 11 am – 8 pm OPTION 3: Less switching costs, maximize utilization, minimize waiting times, maximize profits

  16. Activity_Scheduling • If you could change one thing in the operations scheduling of this case, what would you change? • We own a hotel which has a large ballroom. We have to schedule activities for two Saturdays in May. The closing time of the ballroom each Saturday is 10 pm. Which activities would you schedule? We charge per hour for time spent in room. No charges for cleaning and preparation times. • Event A: 9 am - 1 pm, Cleanup needed after event 2 hrs. • Event B: 4 pm – 7 pm, preparation needed before event 0.5 hours. Cleanup needed after event 1 hrs. • Event C: 5 pm - 10 pm, preparation needed before event 1 hours. Cleanup after event 2 hrs. • Event D: 9 am -12 pm, preparation needed before event 1 hours. Cleanup after event 1 hrs. • Event E: 11 am – 8 pm, preparation needed before event 2 hours. Cleanup after event 2 hrs.

  17. Activity_Scheduling • If you could change one thing in the operations scheduling of this case, what would you change? • Demand Options: • Charge for cleaning time • Set minimum reservation time • Capacity Options: • Build new ballroom • Extend working hours per day

  18. Scheduling Criteria • Types of scheduling/sequencing criteria • Goal-based approaches • Minimize cost, waiting times • Minimize work-in-process • Maximize profits • Priorities-based approaches • First-come first serve or Last-in-first-out • Longest processing time • Earliest due date • Shortest processing time

  19. Job Loading Methods • Types of scheduling methods • Arbitrary approaches • Useful when there are no constraints of resources (Supply exceeds demand) • Rule-based approaches • Useful when there are constraints of resources • Priorities-based approaches • Useful when there are constraints of resources and there are priorities among suppliers or customers

  20. Assignment Method (Type II Scheduling) • 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)

  21. Assignment Method Create zero opportunity costs by repeatedly subtracting the lowest costs from each row and column 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. 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. 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.

  22. Least numbersper row Step 1a - Rows Step 1b - Columns Typesetter Typesetter Typesetter A B C Job R-34 $11 $14 $ 6 S-66 $ 8 $10 $11 T-50 $ 9 $12 $ 7 A B C Job R-34 $ 5 $ 8 $ 0 S-66 $ 0$ 2 $ 3 T-50 $ 2 $ 5 $ 0 A B C Job R-34 $ 5 $ 6 $ 0 S-66 $ 0 $ 0 $ 3 T-50 $ 2 $ 3 $ 0 Least numbersper column Assignment Example

  23. Step 2 - Lines Step 3 - Subtraction Typesetter Typesetter A B C Job R-34 $ 5 $ 6 $ 0 S-66 $ 0 $ 0 $ 3 T-50 $ 2 $ 3 $ 0 A B C Job R-34 $ 3 $ 4 $ 0 S-66 $ 0 $ 0 $ 5 T-50 $ 0 $ 1 $ 0 Assignment Example The smallest uncovered number is 2 so this is subtracted from all other uncovered numbers and added to numbers at the intersection of lines Because only two lines are needed to cover all the zeros, the solution is not optimal (it is fewer than the number of jobs to assign)

  24. Step 2 - Lines Step 4 - Assignments Typesetter Typesetter A B C Job R-34 $ 3 $ 4 $ 0 S-66 $ 0 $ 0 $ 5 T-50 $ 0 $ 1 $ 0 A B C Job R-34$ 3 $ 4 $ 0 S-66$ 0 $ 0 $ 5 T-50 $ 0 $ 1 $ 0 Assignment Example Start by assigning S-66 for worker B. Job T-50 must go to worker A. This leaves R-34 to worker C as this is the least cost assignment for worker C. Because three lines are needed to cover all the numbers, the solution is optimal and job assignments can now be made

  25. From the original cost table Minimum cost = $6 + $10 + $9 = $25 Typesetter Typesetter A B C Job R-34$ 3 $ 4 $ 0 S-66$ 0 $ 0 $ 5 T-50 $ 0 $ 1 $ 0 A B C Job R-34 $11 $14 $ 6 S-66 $ 8 $10 $11 T-50 $ 9 $12 $ 7 Assignment Example Step 4 - Assignments Costs Table

  26. Typesetter Typesetter Typesetter A B C Job R-34 $11 $14 $ 6 S-66 $ 8 $10 $11 T-50 $ 9 $12 $ 7 A B C Job R-34 $ 4 $ 1 $ 9 S-66 $ 7 $ 5 $ 4 T-50 $ 5 $ 2 $ 7 A B C Job R-34 $15-$11 $15-$14 $15-$6 S-66 $15-$8 $15-$10 $15-$11 T-50 $14-$9 $14-$12 $14-$7 Opportunity Loss: Example 2 (Deriving Opportunity Loss Table) Assignment Costs Table Assume that the fixed sale price for each job is as follows : R-34 = $ 15 /unit; S-66 = $ 15 /unit; T-50 = $ 14 /unit; Opportunity Loss Table (Sales – Costs) Table

  27. Typesetter Typesetter Typesetter A B C Job R-34 $ 4 $ 1 $ 9 S-66 $ 7 $ 5 $ 4 T-50 $ 5 $ 2 $ 7 A B C Job R-34 -$ 5 -$ 8 $ 0 S-66 $ 0-$ 2 -$ 3 T-50 -$ 2 -$ 5 $ 0 A B C Job R-34 -$ 5 -$ 6 $ 0 S-66 $ 0 $ 0 -$ 3 T-50 -$ 2 -$ 3 $ 0 Assignment: Example 2 (Deriving Opportunity Loss Table) The table has profit margins that are earned for each unit made. To find the optimal assignment, use the method but subtract the highest score of each row not the least one. 1 2 Opportunity Loss Table 3 Take highest number from each columnand subtract from all the numbersin the column. Note -2 is the highest number in column B!

  28. Typesetter A B C Job R-34 -$ 5 -$ 6 $ 0 S-66 $ 0 $ 0 -$ 3 T-50 -$ 2 -$ 3 $ 0 Assignment: Example 2 (Deriving Opportunity Loss Table) Draw lines across the zeros. As only two lines cross all the zeros,solution is not yet optimal. Opportunity Loss Table 4 Take highest number from uncrossed cells and subtract it from all other uncrossed numbers in each column. Add the number to number on the intersection Intersection to get table 5. This is not an optimal solution – 2 lines through all zeros

  29. Typesetter Typesetter Typesetter A B C Job R-34 $ 4 $ 1 $ 9 S-66 $ 7 $ 5 $ 4 T-50 $ 5 $ 2 $ 7 A B C Job R-34 -$ 3 -$ 4 $ 0 S-66 $ 0 $ 0 -$ 5 T-50 $ 0 -$ 1 $ 0 A B C Job R-34 -$ 5 -$ 6 $ 0 S-66 $ 0 $ 0 -$ 3 T-50 -$ 2 -$ 3 $ 0 Opportunity Loss: Example 2 Largest uncrossed number Table 5 now has three lines going through all the zeros. An optimal assignment can now be Made for our problem! 4 Assign C to R-34; assign A to T-50; assign B to S-66; The profit margin of the assignment is taken from first table: = $5 + $ 5 + $ 9 = $ 19 Gross Margin - Opportunity Loss Table 1 5

  30. Day Work Center Monday Tuesday Wednesday Thursday Friday Metalworks Job 349 Job 350 Mechanical Job 349 Job 408 Electronics Job 408 Job 349 Painting Job 295 Job 408 Job 349 Processing Unscheduled Center not available Gantt Load Chart Method (Type III Scheduling) Figure 15.3

  31. Plan 1: Gantt Staffing Chart (Type III Scheduling) Tue Wed Thu Sat Sun Mon Fri Bill Off Off Mary Off Off Sue Off Off Schedule Will Off Off Bob Off Off Mon Off Off Josh Off Off 1. Required Capacity 5 6 5 9 9 5 8 2. Max available staff 7 7 7 7 7 7 7 3. Max off duty limits 2 1 2 -2 -2 2 -1 4. Scheduled off-duty 3 2 3 1 0 3 2 5. Extra staff needed 1 1 1 3 2 1 3 Scheduled off duty minus Max. off duty limit ( row 4. Minus row 3.) What would you advise the manager to do?

  32. Plan 2: Gantt Staffing Chart (Type III Scheduling) Tue Wed Thu Sat Sun Mon Fri Bill Off Off Mary Off Off Sue Off Off Schedule Will Off Off Bob Off Off Mon Off Off Josh Off Off 1. Required Capacity 5 6 5 9 9 5 8 2. Max available staff 7 7 7 7 7 7 7 3. Max off duty staff 2 1 2 -2 -2 2 -1 4. Scheduled off-duty 2 1 2 3 3 1 2 5. Extra staff needed 0 0 0 5 5 -1 3 This solution shifts all temp staff requirement to weekends 4. Minus 3 What could be the benefit/problem with this plan?

  33. Start of an activity End of an activity Scheduled activity time allowed Maintenance Actual work progress Nonproduction time Point in time when chart is reviewed Now Gantt Schedule Chart Example Figure 15.4

  34. Sequencing • 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

  35. Sequencing Example Apply the four popular sequencing rules to these five jobs

  36. Sequencing: FCFS Example FCFS: Sequence A-B-C-D-E (assume that all jobs arrived on same day in the sequence given).

  37. Total flow time Number of jobs Average completion time = = 77/5 = 15.4 days Utilization = = 28/77 = 36.4% Total job work time Total flow time Total late days Number of jobs Average job lateness = = 11/5 = 2.2 days Average number of jobs in the system = = 77/28 = 2.75jobs/month Total flow time Total job work time Sequencing Example FCFS: Sequence A-B-C-D-E

  38. Sequencing Example SPT (Shortest processing time): Sequence B-D-A-C-E The sequence changes with the priority rule

  39. Total flow time Number of jobs Average completion time = = 65/5 = 13 days Utilization = = 28/65 = 43.1% Total job work time Total flow time Total late days Number of jobs Average job lateness = = 9/5 = 1.8 days Average number of jobs in the system = = 65/28 = 2.32jobs/months Total flow time Total job work time Sequencing Example SPT: Sequence B-D-A-C-E

  40. Sequencing Example EDD (Earliest due date) : Sequence B-A-D-C-E

  41. Total flow time Number of jobs Average completion time = = 68/5 = 13.6 days Utilization = = 28/68 = 41.2% Total job work time Total flow time Total late days Number of jobs Average job lateness = = 6/5 = 1.2 days Average number of jobs in the system = = 68/28 = 2.43 jobs/ month Total flow time Total job work time Sequencing Example EDD: Sequence B-A-D-C-E

  42. Sequencing Example LPT (Longest processing time): Sequence E-C-A-D-B

  43. Total flow time Number of jobs Average completion time = = 103/5 = 20.6 days Utilization = = 28/103 = 27.2% Total job work time Total flow time Total late days Number of jobs Average job lateness = = 48/5 = 9.6 days Average number of jobs in the system = = 103/28 = 3.68 jobs Total flow time Total job work time Sequencing Example LPT: Sequence E-C-A-D-B

  44. Summary Sequencing Examples Summary of Rules

  45. 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 on any criteria (or does poorly on most criteria) but it is perceived as fair by customers • EDD minimizes lateness

  46. Improving Performance of System • Changing setting of due dates • Changing process serial to parallel form A B C E D A B E C D

  47. Example from Service Industry There are two doctors in a specialist clinic, one is a dermatologist the other is a neurologist. The patients call-in the order shown. Create a schedule for the clinic assuming that each specialist works 8-12 pm and 1-4pm daily. Assign slots to patients using First come First servepriority rule. Assume that the appointments slots are one hour each.

  48. Resolution from Service Industry There are two doctors in a specialist clinic, one is a dermatologist the other is a neurologist. The patients call-in the order shown. Create a schedule for the clinic assuming that each specialist works 8-12 pm and 1-4pm daily. Assign slots to patients using First come First servepriority rule. Assume that the appointments slots are one hour each.

  49. Resolution from Service Industry There are two doctors in a specialist clinic, one is a dermatologist the other is a neurologist. The patients call-in the order shown. Create a schedule for the clinic assuming that each specialist works 8-12 pm and 1-4pm daily. Assign slots to patients using First come First servepriority rule combined with SPT and LPT time slots

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