1. © 2007 Pearson Education Scheduling Chapter 16

2. © 2007 Pearson Education Scheduling  Scheduling: The allocation of resources over time to accomplish specific tasks.  Demand scheduling: A type of scheduling whereby customers are assigned to a definite time for order fulfillment.  Workforce scheduling: A type of scheduling that determines when employees work.  Operations scheduling: A type of scheduling in which jobs are assigned to workstations or employees are assigned to jobs for specified time periods.

3. © 2007 Pearson Education  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 Scheduling Service and Manufacturing Processes

4. © 2007 Pearson Education Performance Measures  Job flow time : The amount of time a job spends in the service or manufacturing system. Also referred to as throughput time or time spent in the system, including service.  It is the sum of waiting time for servers or machines; the process time, including setup; the time spent moving between operations; and delays resulting from machine breakdown, unavailability of facilitating goods or components, and the like  Job flow time = time of completion – time was available for the first processing operation  Makespan : The total amount of time required to complete a group of jobs. Makespan= Time of completion of last job – Starting time of first job

5. © 2007 Pearson Education Performance Measures  Past due (Tardiness): The amount of time by which a job missed its due date or the percentage of total jobs processed over some period of time that missed their due dates.  Work-in-process (WIP) inventory: Any job that is waiting in line, moving from one operation to the next, being delayed, being processed, or residing in a semi-finished state.  Average WIP inventory = sum of flow times Makespan

6. © 2007 Pearson Education Performance Measures…  Total inventory : The sum of scheduled receipts and on-hand inventories.  Average total inventory = sum of time in system Makespan  Utilization : The percentage of work time that is productively spent by an employee or machine. Total Inventory = Scheduled receipts for all items + On-hand inventories of all items

7. © 2007 Pearson Education Gantt Charts  Gantt chart : Used as a tool to monitor the progress of work and to view the load on workstations.  The chart takes two basic forms: (1) the job or activity progress chart, and (2) the workstation chart.  The Gantt progress chart graphically displays the current status of each job or activity relative to its scheduled completion date.  The Gantt workstation chart shows the load on the workstations and the nonproductive time.

8. © 2007 Pearson Education Gantt Progress Chart Plymouth Ford Pontiac Job4/204/224/234/244/254/264/214/174/184/19 Current date Scheduled activity time Actual progress Start activity Finish activity Nonproductive time Gantt Progress Chart for an Auto Parts Company

9. © 2007 Pearson Education Gantt Workstation Chart Gantt Workstation Chart for Hospital Operating Rooms

10. © 2007 Pearson Education 1) Scheduling Customer Demand  Three methods are commonly used to schedule customer demand: (1)Appointments assign specific times for service to customers. (2)Reservations are used when the customer actually occupies or uses facilities associated with the service. (3)Backlogs: The customer is given a due date for the fulfillment a product order, or Allow a backlog to develop as customers arrive at the system. Customers may never know exactly when their orders will be fulfilled

11. © 2007 Pearson Education 2) Scheduling Employees  Rotating schedule: A schedule that rotates employees through a series of workdays or hours.  Fixed schedule: A schedule that calls for each employee to work the same days and hours each week.  Constraints: The technical constraints imposed on the workforce schedule are the resources provided by the staffing plan and the requirements placed on the operating system.  Other constraints, including legal and behavioral considerations, also can be imposed.

12. © 2007 Pearson Education DayMTWThFSSu Number of employees64891032 Required employees The Amalgamated Parcel Service is open 7 days a week. The schedule of requirements is: The manager needs a workforce schedule that provides two consecutive days off and minimizes the amount of total slack capacity. To break ties in the selection of off days, the scheduler gives preference to Saturday and Sunday if it is one of the tied pairs. If not, she selects one of the tied pairs arbitrarily. Workforce Scheduling Example 16.1

13. © 2007 Pearson Education Required employees DayMTWThFSSu Number of employees6489 10* 32 Employee1 XXXXX Workforce Scheduling Example 16.1 Steps 1 & 2 Step 1. Find all the pairs of consecutive days that exclude the maximum daily requirements. Select the unique pair that has the lowest total requirements for the 2 days. Friday contains the maximum requirements (10), and the pair S–Su has the lowest total requirements. Therefore, Employee 1 is scheduled to work Monday through Friday. Step 2. If a tie occurs, choose one of the tied pairs or ask the employee to make a choice.

14. © 2007 Pearson Education Required employees Step 3. Subtract the requirements satisfied by the Employee 1 from the net requirements for each day the employee is to work and repeat step one. Again the pair S–Su has the lowest total requirements. Therefore, Employee 2 is scheduled to work Monday through Friday. DayMTWThFSSu Number of employees648910*32 Employee1XXXXX Requirements53789*32 Employee2XXXXX Workforce Scheduling Example 16.1 Step 3

15. © 2007 Pearson Education Required employees DayMTWThFSSu Number of employees648910*32 Employee1XXXXX Requirement53789*32 Employee2XXXXX Requirement42678*32 Employee3XXXXX Requirement31567*32 Workforce Scheduling Example 16.1 Step 4 Step 4. Repeat steps 1 through 3 until all the requirements have been satisfied. After Employees 1, 2, and 3 have reduced the requirements, the pair with the lowest requirements changes, and Employee 4 will be scheduled for Wednesday through Sunday.

16. © 2007 Pearson Education DayMTWThFSSu Number of employees648910*32 Employee1XXXXX Requirement53789*32 Employee2XXXXX Requirement42678*32 Employee3XXXXX Requirement31567*32 Employee4XXXXX Requirement31456*21 Employee5XXXXX Required employees Workforce Scheduling Example 16.1 Step 4 continued

17. © 2007 Pearson Education DayMTWThFSSu Requirement20345*21 Employee6XXXXX Requirement20234*10 Employee7XXXXX Requirement10123*10 Employee8XXXXX Requirement00012*10 Employee9XXXXX Requirement00001*00 Employee10XXXXX Required employees Workforce Scheduling Example 16.1 Step 4 continued

18. © 2007 Pearson Education DayMTWThFSSu Employee1XXXXX offoff Employee2XXXXX offoff Employee3XXXXX offoff Employee4 offoff XXXXX Employee5XXXXX offoff Employee6 offoff XXXXX Employee7XXXXX offoff Employee8XXXXX offoff Employee9 off XXXXX off Employee10XXXXX offoff Final Schedule Workforce Scheduling Example 16.1

19. © 2007 Pearson Education MTWThFSSu Employee1XXXXX offoff Employee2XXXXX offoff Employee3XXXXX offoff Employee4 offoff XXXXX Employee5XXXXX offoff Employee6 offoff XXXXX Employee7XXXXX offoff Employee8XXXXX offoff Employee9 off XXXXX off Employee10XXXXX offoff Workforce Scheduling Example 16.1 Final Schedule Capacity, C781010103250 Requirements, R6489103242 Slack, C – R14210008 Total Final Schedule

20. © 2007 Pearson Education 3) Operations Scheduling  Operations schedules are short-term plans designed to implement the master production schedule.  Operations scheduling focuses on how best to use existing capacity.  Often, several jobs must be processed at one or more workstations. Typically, a variety of tasks can be performed at each workstation.  Job shop: A firm that specializes in low- to medium-volume production and utilizes job or batch processes.  Flow shop: A firm that specializes in medium- to high-volume production and utilizes line or continuous processes.

21. © 2007 Pearson Education Shipping Department Manufacturing Process Raw Materials Legend: Batch of parts Workstation

22. © 2007 Pearson Education Job Shop Dispatching  Dispatching: A method of generating schedules in job shops whereby the decision about which job to process next is made using simple priority rules whenever the workstation becomes available for further processing.  Priority sequencing rules: The rules that specify the job processing sequence when several jobs are waiting in line at a workstation. (Due date – Today’s date)/Total shop time remaining  1) Critical ratio (CR): A ratio that is calculated by dividing the time remaining until a job’s due date by the total shop time remaining for the job. CR = (Due date – Today’s date)/Total shop time remaining  Total Shop Time = Setup, processing, move, and expected waiting times of all remaining operations, including the operation being scheduled.

23. © 2007 Pearson Education Priority sequencing rules  Critical ratio (CR):  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  2) Earliest due date (EDD): A priority sequencing rule that specifies that the job with the earliest due date is the next job to be processed.

24. © 2007 Pearson Education  3) First-come, first-served (FCFS): A priority sequencing rule that specifies that the job arriving at the workstation first has the highest priority.  4) Shortest processing time (SPT): A priority sequencing rule that specifies that the job requiring the shortest processing time is the next job to be processed. Job Shop Dispatching

25. © 2007 Pearson Education  5) Slack per remaining operations (S/RO): A priority sequencing rule that determines priority by dividing the slack by the number of operations that remain, including the one being scheduled. The job with the lowest S/RO is scheduled next Job Shop Dispatching S/RO = ((Due date – Today’s date) – Total shop time remaining) Number of operations remaining

26. © 2007 Pearson Education  Single-dimension rules: A set of rules such as FCFS, EDD, and SPT, that bases the priority of a job on a single aspect of the job, such as arrival time at the workstation, the due date, or the processing time.  Priority rules, such as CR and S/RO, incorporate information about the remaining workstations at which the job must be processed. We call these rules multiple-dimension rules.  Multiple-dimension rules: A set of rules that apply to more than one aspect of a job. Scheduling Jobs for One Workstation

27. © 2007 Pearson Education Five engine blocks are waiting for processing. The processing times have been estimated. Expected completion times have been agreed. The table shows the situation as of Monday morning. Customer pickup times are measured in business hours from Monday morning. Determine the schedule by using the EDD rule and then the SPT rule. Calculate the average hours early, hours past due, WIP inventory, and total inventory for each method. If low job flow times and WIP inventories are critical, which rule should be chosen? Example 16.2 Single-Dimension Rule Sequencing

28. © 2007 Pearson Education 8 + 14 + 17 + 32 + 44 44 Average hours early = 0.6 hour JobScheduledActual EngineProcessingFlowCustomerCustomerHours BlockBeginTimeTimePickupPickupHoursPast SequenceWork(hr)(hr)TimeTimeEarlyDue Ranger 810 Explorer 612 Econoline 150 318 Bronco 1520 Thunderbird 1222 0+=8102 17+=323212 8+=14142 14+=17181 32+=444422 Example 16.2 Single-Dimension Rule – EDD Average job flow time = 23 hours Average hours past due = 7.2 hours Average WIP = 2.61 blocks Average total inventory = 2.68 engine blocks Average total inventory = 10 + 14 + 18 + 32 + 44 44

29. © 2007 Pearson Education 3 + 9 + 17 + 29 + 44 44 Average hours early = 3.6 hour JobScheduledActual EngineProcessingFlowCustomerCustomerHours BlockBeginTimeTimePickupPickupHoursPast SequenceWork(hr)(hr)TimeTimeEarlyDue Ranger 810 Explorer 612 Econoline 150 318 Bronco 1520 Thunderbird 1222 0+= 31815 17+= 2929 7 8+= 9123 14+= 17177 29+= 4444 24 Example 16.2 Single-Dimension Rule – SPT Average job flow time = 20.4 hours Average hours past due = 7.6 hours Average WIP = 2.32 blocks Average total inventory = 2.73 engine blocks Average total inventory = 18 + 12 + 17 + 20 + 44 44 Econoline 150 Explorer Ranger Thunderbird Bronco 0 3 9 17 29 3 6 8 12 15 18 12 10 22 20

30. © 2007 Pearson Education Comparing the EDD and SPT Rules EDDSPT Average job flow time23.0020.40 Average hours early0.603.60 Average hours past due7.207.60 Average WIP2.612.32 Average total inventory2.682.73 Using the previous example, a comparison of the EDD and SPT sequencing is shown below. The SPT schedule has a lower average job flow time and lower WIP inventory. The EDD schedule has better customer service, (average hours past due) and lower maximum hours past due. EDD also has a lower total inventory because fewer hours were spent waiting for customers to pick up their engine blocks after they had been completed.

31. © 2007 Pearson Education Example 16.3 Multiple-Dimension Rule – CR 12.315106.12.46 210.51027.81.28 36.2201214.51.38 415.68510.2.78 OperationTime Time atRemainingNumber of Engineto Due DateOperationsShop Time JobLathe (hr)(hrs)RemainingRemainingCRS/RO CR = Time remaining to due date Shop time remaining

32. © 2007 Pearson Education 12.315106.12.460.89 210.51027.81.281.10 36.2201214.51.380.46 415.68510.2.78– 0.44 OperationTime Time atRemainingNumber of Engineto Due DateOperationsShop Time JobLathe (hr)(hrs)RemainingRemainingCRS/RO S/RO = Time remaining to due date – Shop time remaining Number of operations remaining Example 16.3 Multiple-Dimension Rule – S/RO

33. © 2007 Pearson Education 12.315106.12.460.89 210.51027.81.281.10 36.2201214.51.380.46 415.68510.2.78– 0.44 OperationTime Time atRemainingNumber of Engineto Due DateOperationsShop Time JobLathe (hr)(hrs)RemainingRemainingCRS/RO CR Sequence= 4 – 2 – 3 – 1 S/RO Sequence= 4 – 3 – 1 – 2 CR Sequence= Comparing the CR and S/RO Rules

34. © 2007 Pearson Education Priority Rule Summary FCFS=1 – 2 – 3 – 4 SPT=1 – 3 – 2 – 4 EDD=4 – 2 – 1 – 3 CR=4 – 2 – 3 – 1 S/RO=4 – 3 – 1 – 2 Avg Flow Time 17.17516.10026.17527.15024.025 Avg Early Time 3.4256.050000 Avg Past Due 7.3508.90012.92513.90010.775 Avg WIP 1.9861.8613.0263.1292.777 Avg Total Inv 2.3822.5613.0263.1292.777 ShortestSlack per ProcessingEarliestCriticalRemaining FCFSTimeDue DateRatioOperation The S/RO rule is better than the EDD rule and the CR rule but it is much worse than the SPT rule and the FCFS rule for this example. S/RO has the advantage of allowing schedule changes when due dates change. These results cannot be generalized to other situations because only four jobs are being processed.

35. © 2007 Pearson Education Scheduling Jobs for Multiple Workstations  Priority sequencing rules can be used to schedule more than one operation. Each operation is treated independently.  Identifying the best priority rule to use at a particular operation in a process is a complex problem because the output from one process becomes the input for another.  Computer simulation models are effective tools to determine which priority rules work best in a given situation.  When a workstation becomes idle, the priority rule is applied to the jobs waiting for that operation, and the job with the highest priority is selected.  When that operation is finished, the job is moved to the next operation in its routing, where it waits until it again has the highest priority.

36. © 2007 Pearson Education Johnson’s Rule  Johnson’s rule : A procedure that minimizes makespan when scheduling a group of jobs on two workstations.  Step 1. 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.

37. © 2007 Pearson Education Example 16.5  The Morris machine company just received an order to refurbish five motors for materials handling equipment that were damaged in a fire. 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.  The customer 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.

38. © 2007 Pearson Education Eliminate M3 from consideration. The next shortest time is M2 at Workstation 1, so schedule M2 first. Eliminate M5 from consideration. The next shortest time is M1 at workstation #1, so schedule M1 next. Eliminate M1 and the only job remaining to be scheduled is M4. Johnson’s Rule at the Morris Machine Co. Example 16.5 Johnson’s Rule at the Morris Machine Co. Time (hr) MotorWorkstation 1Workstation 2 M11222 M245 M353 M41516 M5108 Sequence = M1M2M3M4M5 Shortest time is 3 hours at workstation 2, so schedule job M3 last. Eliminate M2 from consideration. The next shortest time is M5 at workstation #2, so schedule M5 next to last.

39. © 2007 Pearson Education Workstation M2 (4) M1 (12) M4 (15) M5 (10) M3 (5) Idle—available for further work 051015202530 Day 35404550556065 Idle2 M2 (5) M1 (22) M4 (16) M5 (8) M3 (3) Idle 1 Gantt Chart for the Morris Machine Company Repair Schedule The schedule minimizes the idle time of workstation 2 and gives the fastest repair time for all five motors. No other sequence will produce a lower makespan. Johnson’s Rule at the Morris Machine Co. Example 16.5 Johnson’s Rule at the Morris Machine Co.

40. © 2007 Pearson Education Labor-limited Environments  The limiting resource thus far has been the number of machines or workstations available. A more typical constraint is the amount of labor available.  Labor-limited environment : An environment in which the resource constraint is the amount of labor available, not the number of machines or workstations. 1.Assign personnel to the workstation with the job that has been in the system longest. 2.Assign personnel to the workstation with the most jobs waiting for processing. 3.Assign personnel to the workstation with the largest standard work content. 4.Assign personnel to the workstation with the job that has the earliest due date.

41. © 2007 Pearson Education  The Food Bin grocery store operates 24 hours per day, 7 days per week. At the end of the month, they calculated the average number of checkout registers that should be open during the first shift each day. Results showed peak needs on Saturdays and Sundays. 1.Develop a schedule that covers all requirements while giving two consecutive days off to each clerk. How many clerks are needed? 2.Plans can be made to use the clerks for other duties if slack or idle time resulting from this schedule can be determined. How much idle time will result from this schedule, and on what days? Solved Problem 1

42. © 2007 Pearson Education Solved Problem 1

43. © 2007 Pearson Education Solved Problem 1

44. © 2007 Pearson Education  The Neptune’s Den Machine Shop specializes in overhauling outboard marine engines. 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. Solved Problem 2 Develop separate schedules using SPT and then EDD rules. Compare them using average job flow time, % of past due jobs, and maximum past due days. Calculate average WIP inventory (in engines) and average total inventory.

45. © 2007 Pearson Education Solved Problem 2 SPT EDD

46. © 2007 Pearson Education SPTEDD Average job flow time9.8015.20 % of past due jobs40%60% Maximum past due days117 Average WIP inventory2.133.30 Average total inventory3.52 Solved Problem 2

47. © 2007 Pearson Education Cleanup of chemical waste storage basins involves 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): Find a schedule that minimizes the makespan. Calculate the average job flow time of a storage basin through the two operations. What is the total elapsed time for cleaning all 10 basins? Solved Problem 4

48. © 2007 Pearson Education DredgeIncinerate A31 B44 C32 D61 E12 F36 G24 H11 I82 J48 Solved Problem 4 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 EHGFBJICDA

49. © 2007 Pearson Education DredgeEHGFBJICDA IncinerateEHGFBJICDA Storage basin EHGFBJICDA Solved Problem 4 The Gantt machine chart for this schedule