Learning Objectives 16 Scheduling McGraw-Hill/Irwin Explain what scheduling involves and the importance of good scheduling. Discuss scheduling needs in high-volume and intermediate-volume systems. Discuss scheduling needs in job shops. Copyright © 2007 by The McGraw-Hill Companies, Inc. All rights reserved. 16-2 Scheduling Learning Objectives Use and interpret Gantt charts, and use the assignment method for loading. Discuss and give examples of commonly used priority rules. Describe some of the unique problems encountered in service systems, and describe some of the approaches used for scheduling service systems. 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 16-3 16-4 1 High-Volume Success Factors High-Volume Systems Process and product design Flow system: High-volume system with Standardized equipment and activities Preventive maintenance Flow-shop scheduling: Scheduling for high- Rapid repair when breakdown occurs volume flow system Optimal product mixes Work Center #1 Work Center #2 Minimization of quality problems Output Reliability and timing of supplies 16-5 Intermediate-Volume Systems Scheduling Low-Volume Systems Outputs are between standardized highvolume systems and made-to-order job shops Loading - assignment of jobs to process centers Sequencing - determining the order in which jobs will be processed Job-shop scheduling Run size, timing, and sequence of jobs Economic run size: Q0 16-7 Scheduling for low-volume systems with many variations in requirements 2DS p H p u 16-8 16-9 2 Loading Gantt Load Chart Figure 16.2 Infinite loading – jobs are assigned to work centers without regard for the capacity of the work center. Gantt chart - used as a visual aid for loading and scheduling Work Mon. Tues. Wed. Thurs. Fri. Center 1 Job 3 Job 4 2 Job 3 Job 7 3 Job 1 Job 6 Job 7 4 Job 10 Finite loading – jobs are assigned to work centers taking into account the work center capacity and job processing times 16-10 16-11 Assignment Method of linear programming Loading (cont’d) Forward scheduling – scheduling ahead from some point in time. Backward scheduling – scheduling by working backwards in time from the due date(s). Schedule chart – a form of Gantt chart that shows the orders or jobs in progress and whether they are on schedule. The assignment model is a special-purpose linear programming model that is useful in situations that call for assigning tasks or other work requirements to resources. Typical examples include assigning jobs to machines or workers, territories to salespeople, and repair jobs to repair crews. The idea is to obtain an optimum matching of tasks and resources. 16-12 16-13 3 Example 1 Example 1 Determine the optimum assignment of jobs to machines for the following data : a. Subtract the smallest number in each row from every number in the row, and enter the results in a new table. The result of this row reduction is b. Subtract the smallest number in each column from every number in the column, and enter the results in a new table. The result of this column reduction is c. Determine the minimum number of lines needed to cross out all zeros. (Try to cross out as many zeros as possible when drawing lines.) 16-14 Example 1 d. Since only three lines are needed to cross out all zeros and the table has four rows, this is not the optimum. Note that the smallest uncovered value is 1. e. Subtract the smallest uncovered value from every uncovered number that hasn’t been crossed out, and add it to numbers that are at the intersections of covering lines. The results are as follows: f. Determine the minimum number of lines needed to cross out all zeros (four). Since this equals the number of rows, you can make the optimum assignment. g. 16-15 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. Make assignments: Start with rows and columns with only one zero. Match jobs with machines that have a zero cost. 16-16 16-17 4 Sequencing Priority Rules Table 16.2 FCFS - first come, first served Priority rules: Simple heuristics used to select the order in which jobs will be processed. SPT - shortest processing time EDD - earliest due date Everything is #1 Priority CR - critical ratio Job time: Time needed for setup and processing of a job. S/O - slack per operation Rush - emergency Top Priority 16-18 16-19 Assumptions of Priority Rules Performance Measures Job flow time. The setup of jobs is known Setup time is independent of processing sequence Setup time is deterministic There will be no interruptions in processing such as: This is the length of time a job is at a particular workstation or work center. It includes not only actual processing time but also any time waiting to be processed, transportation time between operations, and any waiting time related to equipment breakdowns, unavailable parts, quality problems, and so on. Job lateness. (tardiness) This is the length of time the job completion date is expected to exceed the date the job was due or promised to a customer. Makespan. Machine breakdowns Accidents Worker illness Makespan is the total time needed to complete a group of jobs. Average number of jobs. Jobs that are in a shop are considered to be work-in-process inventory. Average number of jobs = Total flow time ÷ Makespan 16-20 16-21 5 Example 2-a) FCFS Example 2 Processing times (including setup times) 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 the work center, for each of these rules: a. FCFS Job Processing Due Dates b. SPT Time (days) (days) c. EDD A 2 7 d. CR B Assume jobs arrived in the order shown. 8 The FCFS sequence is simply A-B-C-D-E-F. The measures of effectiveness are as follows (see table): Job (1) Processing Time 4 4 D 10 17 E 5 15 F 12 18 (2) - (3) Days Tardy [0 if negative] 0 A 2 2 7 8 10 16 0 C 4 14 4 10 D 10 24 17 7 E 5 29 15 14 F 12 41 18 23 41 (3) Due Dates B 16 C (2) Flow Time 54 120 Average flow time: 120÷6 = 20 days. Average tardiness: 54÷6 = 9 days. The makespan is 41 days. Average number of jobs at the work center: 120÷41 = 2.93. 16-22 16-23 Example 2-b) SPT The FCFS sequence is simply A-B-C-D-E-F. The measures of effectiveness are as follows (see table): The FCFS sequence is simply A-B-C-D-E-F. The measures of effectiveness are as follows (see table): Job (1) Processing Time (2) Flow Time (3) Due Dates (2) - (3) Days Tardy [0 if negative] Job (1) Processing Time (2) Flow Time (3) Due Dates (2) - (3) Days Tardy [0 if negative] A 2 2 7 0 C 4 4 4 0 C 4 6 4 2 A 2 6 7 0 E 5 11 15 0 E 5 11 15 0 B 8 19 16 3 B 8 19 16 3 D 10 29 17 12 D 10 29 17 12 F 12 41 18 23 F 12 41 18 23 41 Example 2-c) EDD 108 40 41 Average flow time: 108÷6 = 18 days. Average tardiness: 40÷6 = 6.67 days. The makespan is 41 days. Average number of jobs at the work center: 108÷41 = 2.63. 16-24 110 38 Average flow time: 110÷6 = 18.33 days. Average tardiness: 38÷6 = 6.33 days. The makespan is 41 days. Average number of jobs at the work center: 110÷41 = 2.68. 16-25 6 Example 2-d) Critical Ratio Using the critical ratio we find Job Example 2-d) Critical Ratio Processing Time Due Dates Critical Ratio Calculation A 2 7 (7-0)/2 = 3.5 B 8 16 (16-0)/8 = 2.0 C 4 4 (4-0)/4 = 1.0 D 10 17 (17-0)/10 = 1.7 E 5 15 (15-0)/5 = 3.0 F 12 18 (18-0)/12 =1.5 C At day 16 [C and F completed], the critical ratios are Job Processing Time Due Dates Critical Ratio Calculation A 2 7 (7-16)/2 = -4.5 B 8 16 (16-16)/8 = 0 D 10 17 (17-16)/10 = 0.1 E 5 15 (15-16)/5 = -0.2 Job Processing Time Due Dates Critical Ratio Calculation B 8 16 (16-18)/8 = -0.25 Job Processing Time Due Dates Critical Ratio Calculation D 10 17 (17-18)/10 = -0.1 A 2 7 (7-4)/2 = 1.5 E 5 15 (15-18)/5 = -0.6 B 8 16 (16-4)/8 = 1.5 D 10 17 (17-4)/10 = 1.3 E 5 15 (15-4)/5 = 2.2 12 18 (18-4)/12 =1.17 F A At day 18 [C,F and A completed], the critical ratios are At day 4 [C completed], the critical ratio are F C C F A E At day 23 [C ,F,A, and E completed], the critical ratios are C F 16-26 Job Processing Time Due Dates Critical Ratio Calculation B 8 16 (16-23)/8 = -0.875 D 10 17 (17-23)/10 = -0.6 Example 2-d) Critical ratio C F A E B D 16-27 Example 2 Table 16.4 The FCFS sequence is simply A-B-C-D-E-F. The measures of effectiveness are as follows (see table): Job (1) Processing Time (3) Due Dates (2) - (3) Days Tardy [0 if negative] C 4 4 4 0 F 12 16 18 0 A 2 18 7 11 E 5 23 15 8 B 8 31 16 15 D 10 41 17 24 41 (2) Flow Time 133 Rule Average Flow Time (days) Average Tardiness (days) Average Number of Jobs at the Work Center FCFS 20.00 9.00 2.93 SPT 18.00 6.67 2.63 EDD 18.33 6.33 2.68 CR 22.17 9.67 3.24 58 Average flow time: 133÷6 = 22.17 days. Average tardiness: 58÷6 = 9.67 days. Average number of jobs at the work center: 133÷41 = 3.24. 16-28 16-29 7 Example 3 Example 3- Solution Use the S/O rule to schedule the following jobs. Note that processing time includes the time remaining for the current and subsequent operations. In addition, you will need to know the number of operations remaining, including the current one. Job Processing Time (days) Due Dates (days) Remaining number of operations A 4 14 3 B 16 32 6 C 8 8 5 D 20 34 2 E 10 30 4 F 18 30 2 Determine the difference between the due date and the processing time for each operation. Divide the difference by the number of remaining operations, and rank them from low to high. This yields the sequence of jobs: Job (1) (2) Remaining Due Date Processing Time (3) (2)-(1) Slack (4) Remaining Number of Operations (5) (3)/(4) Ratio (6) Rank A 4 14 10 3 3.33 3 B 16 32 16 6 2.67 2 C 8 8 0 5 0 1 D 20 34 14 2 7.00 6 E 10 30 20 4 5.00 4 F 18 30 12 2 6.00 5 16-30 16-31 Johnson’s Rule Conditions Two Work Center Sequencing Johnson’s Rule: technique for minimizing Job time must be known and constant completion time for a group of jobs to be processed on two machines or at two work centers. Job times must be independent of sequence Minimizes total idle time Jobs must follow same two-step sequence Several conditions must be satisfied Job priorities cannot be used All units must be completed at the first work center before moving to second 16-32 16-33 8 Johnson’s Rule Optimum Sequence Scheduling Difficulties Variability in 1.List the jobs and their times at each work center 2.Select the job with the shortest time 3.Eliminate the job from further consideration 4.Repeat steps 2 and 3 until all jobs have been scheduled 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 16-34 16-35 Example 4 Example 4- Solution 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: 1st Processing Time (Hours) Job Work Center 1 Work Center 2 A 5 5 B 4 3 C 8 9 D 2 a. Select the job with the shortest processing time. It is job D, with a time of two hours. b. Since the time is at the first center, schedule job D first. Eliminate job D from further consideration. 2nd 3rd 4th 5th 6th D c. 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 7 E 6 8 1st F 12 15 D 16-36 2nd 3rd 4th 5th 6th B t Job C1 C2 A 5 5 C 8 9 D 2 7 E 6 8 F 12 15 Job C1 C2 t A 5 5 B 4 3 C 8 9 E 6 8 F 12 15 16-37 9 Example 4- Solution Example 4- Solution d. The remaining jobs and their times are 1st 2nd 3rd 4th D 5th 6th A B e. 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, 1st 2nd D E 3rd 4th 5th 6th A B f. 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 t Job C1 C2 A 5 5 C 8 9 E 6 8 F 12 15 Note that there is a tie for the shortest remaining time: job A has the same time at each work center. It makes no difference, then, whether we place it toward the beginning or the end of the sequence. Suppose it is placed arbitrarily toward the end. We now have t 1st D 2nd E 3rd C 4th F 5th A 6th B Job C1 C 8 9 F 12 C2 15 g. One way to determine the throughput time and idle times at the work centers is to construct a chart: t Job C1 C2 C 8 9 E 6 8 F 12 15 16-38 16-39 Theory of Constraints Minimizing Scheduling Difficulties Set realistic due dates The Theory of Constraints Goal is to maximize flow through the entire system Emphasizes balancing flow Improve performance of bottleneck: Focus on bottleneck operations Consider lot splitting of large jobs 16-40 Determine what is constraining the operation Exploit the constraint Subordinate everything to the constraint Determine how to overcome the constraint Repeat the process for the next constraint 16-41 10 Scheduling Services Considerations Scheduling Services Appointment systems Scheduling services different from manufacturing Inability to store or inventory services Random nature of customer requests for service Point of customer contact Controls customer arrivals for service Reservation systems Estimates demand for service Scheduling the workforce Back-office operations Front-office operations Manages capacity for service Scheduling multiple resources Coordinates use of more than one resource 16-43 16-44 Yield Management Cyclical Scheduling Yield Management – the application of pricing strategies to allocate capacity among various categories of demand. The goal is to maximize the revenue generated by the fixed capacity Fixed capacity Hotel, motel rooms Airline seats Hospitals, police/fire departments, restaurants, supermarkets Rotating schedules Set a scheduling horizon Identify the work pattern Develop a basic employee schedule Assign employees to the schedule Unsold rooms or seats cannot be carried over 16-45 16-46 11