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Operations Scheduling Chapter 8

Operations Scheduling Chapter 8. The Hierarchy of Production Decisions. The logical sequence of operations in factory planning corresponds to the sequencing of chapters in a production management text book. All planning starts with the demand forecast.

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Operations Scheduling Chapter 8

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  1. Operations Scheduling Chapter 8

  2. The Hierarchy of Production Decisions The logical sequence of operations in factory planning corresponds to the sequencing of chapters in a production management text book. • All planning starts with the demand forecast. • Demand forecasts are the basis for the top level (aggregate) planning. • The Master Production Schedule (MPS) is the result of disaggregating aggregate plans down to the individual item level. • Based on the MPS, MRP is used to determine the size and timing of component and subassembly production. • Detailed shop floor schedules are required to meet production plans resulting from the MRP.

  3. Hierarchy of Production Decisions

  4. Detailedproduction schedule A detailed production schedule must include when and where each activity must take place in order to meet the master schedule.

  5. Scheduling ServiceOperationsVsManufacturing Operations Scheduling service systemspresentscertainproblems not generallyencountered in manufacturingsystems. This is primarilydueto: • Theinabilitytostoreservices • Therandomnature of customerrequests Toavoidproblemssuch as longdelays, unsatisfiedcustomers, service systemsrely on appoinmentsystemsandreservationsystems.

  6. Goals of Production Scheduling • High Customer Service: on-time delivery • Low Inventory Levels: WIP and FGI • High Utilization: of machines and labor

  7. MeasurestoEvaluatePerformance of a SchedulingMethod • Service Level: Fraction of orders filled on before their due dates (used in make-to-order systems) • Fill Rate: Fraction of demand that are met from inventory without backorder (used in make-to-stock systems) • Job Flow Time: Time elapsed from the release of a job until it is completed.

  8. Measures to Evaluate Performance of a Scheduling Method • Lateness: Difference between completion time and due date of a job (may be negative). • Tardiness: The positive difference between the completion time and the due date of a job. • Makespan:The total length of the schedule (that is, when all the jobs have finished processing).

  9. Reducing WIP and Flow Time • Shorter flow time means: • Less WIP • Better responsiveness All of which reduce costs and improve sales revenue

  10. Measures to Evaluate Performance of a Scheduling Method • Production Rate • Utilization Keep in mind that high utilization means high return on investment. This is good provided that the equipment is utilized to increase revenue. Otherwise, high utilization only helps to increase inventory, not profits.

  11. Types of ProductionSystems • High Volume Systems (MassProduction) -ContinousProduction(petroleumrefining, sugarrefining) -DiscreteProduction(autos, personalcomputers, televisons) • Intermediate-Volume Systems (Batchproduction) • Low-Volume Systems(Job-Shops)

  12. WorkCenter #1 Work Center #2 Output High-Volume Systems • Flow system: High-volume system with Standardized equipment and activities

  13. High-Volume Systems • High-volume systems are often referred to as flow systems; scheduling in these systems is referred to as flow-shop scheduling. Major aspects of system design include line balancing and flow system design.

  14. High-Volume Systems • Line balancing concerns allocating the required tasks to workstations so that they satisfy technical (sequencing) constraints and are balanced with respect to equal work times among stations. • It results in the maximum utilization of equipment and personnel as well as the highest possible rate of output.

  15. Intermediate-Volume Systems • Outputs are between standardized high-volume systems and made-to-order job shops • The volume of output is not large enough to justify continuous production. • Examples include canned foods, paint and cosmetics.

  16. Intermediate-Volume Systems The three basic issues in these systems are: 1.Run size, 2.Timing, and 3.Sequence of jobs • Economic run size: h’ is defined as h’= h(1- λ/P)

  17. Scheduling Low-Volume Systems • Loading - assignment of jobs to process centers • Sequencing - determining the order in which jobs will be processed • Job-shop scheduling • Scheduling for low-volume systems with many variations in requirements

  18. Ganttcharts • Gantt charts are used as visual aid for loading and scheduling purposes. • The name was derived from Henry Gantt in the early 1900s. • The purpose of Gantt charts is to organize the use of resources in a time framework. • In most cases, a time scale is represented horizontally, and resources to be scheduled are listed vertically.

  19. Gantt Load Chart Figure 15.2 • Gantt chart - used as a visual aid for loading and scheduling

  20. Loading

  21. LOADING (The Assignment Problem) • In many business situations, management needs to assign - personnel to jobs, - jobs to machines, - machines to job locations, or - salespersons to territories. • Consider the situation of assigning n jobs to n machines. • When a job i (=1,2,....,n) is assigned to machine j (=1,2, .....n) that incurs a cost Cij. • The objective is to assign the jobs to machines at the least possible total cost.

  22. The Assignment Problem • This situation is a special case of the Transportation Model and it is known as the assignment problem. • Here, jobs represent “sources” and machines represent “destinations.” • The supply available at each source is 1 unit and demand at each destination is 1 unit.

  23. The Assignment Problem The assignment model can be expressed mathematically as follows: Xij= 0, if the job j is not assigned to machine i 1, if the job j is assigned to machine i

  24. The Assignment Problem

  25. The Assignment Problem Example • Ballston Electronics manufactures small electrical devices. • Products are manufactured on five different assembly lines (1,2,3,4,5). • When manufacturing is finished, products are transported from the assembly lines to one of the five different inspection areas (A,B,C,D,E). • Transporting products from five assembly lines to five inspection areas requires different times (in minutes)

  26. The Assignment Problem Example • Ballston Electronics manufactures small electrical devices. • Products are manufactured on five different assembly lines (1,2,3,4,5). • When manufacturing is finished, products are transported from the assembly lines to one of the five different inspection areas (A,B,C,D,E). • Transporting products from five assembly lines to five inspection areas requires different times (in minutes)

  27. The Assignment Problem Example Under current arrangement, assignment of inspection areas to the assembly lines are 1 to A, 2 to B, 3 to C, 4 to D, and 5 to E. This arrangement requires 10+7+12+17+19 = 65 man minutes.

  28. The Assignment Problem Example • Management would like to determine whether some other assignment of production lines to inspection areas may result in less cost. • This is a typical assignment problem. n = 5 And each assembly line is assigned to each inspection area. • It would be easy to solve such a problem when n is 5, but when n is large all possible alternative solutions are n!, this becomes a hard problem.

  29. The Assignment Problem Example • Assignment problem can be either formulated as a linear programming model, or it can be formulated as a transportation model. • However, An algorithm known as Hungarian Method has proven to be a quick and efficient way to solve such problems. • This technique is programmed into many computer modules such as the one in WINQSB.

  30. The Assignment Problem Example WINQSB solution for this problem is as follows:

  31. 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.

  32. Common Sequencing Rules • FCFS. First Come First Served. Jobs processed in the order they come to the shop. • SPT. Shortest Processing Time. Jobs with the shortest processing time are scheduled first. • EDD. Earliest Due Date. Jobs are sequenced according to their due dates. • CR. Critical Ratio. Compute the ratio of processing time of the job and remaining time until the due date. Schedule the job with the largest CR value next.

  33. Sequencing n jobs on a Single Machine Priority rules: Simple heuristics such as FCFS, SPT, DD, CR are usedto select the order in which jobs will be processed. CR= (Due Date – Current Time)/ Processing Time

  34. Example: Sequencing Rules Use the FCFS, SPT, and Critical Ratio rules to sequence the five jobs below. Evaluate the rules on the bases of average flow time, average number of jobs in the system, and average job lateness. (Due Date) JobProcessing TimeTime to Promised Completion A 6 hours 10 hours B 12 16 C 9 8 D 14 14 E 8 7

  35. Example: Sequencing Rules • FCFS Rule A > B > C > D > E Processing DueFlow JobTime DateTimeLateness A 6 10 6 0 B 12 16 18 2 C 9 8 27 19 D 14 14 41 27 E 8 7 49 42 49 141 90

  36. Example: Sequencing Rules • FCFS Rule Performance • Average flow time: 141/5 = 28.2 hours • Average number of jobs in the system: 141/49 = 2.88 jobs • Average job lateness: 90/5 = 18.0 hours

  37. Example: Sequencing Rules • SPT Rule A > E > C > B > D Processing Due Flow JobTime Date TimeLateness A 6 10 6 0 E8 7 14 7 C 9 8 23 15 B 12 16 35 19 D14 14 4935 49 127 76

  38. Example: Sequencing Rules • SPT Rule Performance • Average flow time: 127/5 = 25.4 hours • Average number of jobs in the system: 127/49 = 2.59 jobs • Average job lateness: 76/5 = 15.2 hours

  39. Example: Sequencing Rules • Critical Ratio Rule E > C > D > B > A Processing Promised Flow JobTimeCompletionTimeLateness E (.875) 8 7 8 1 C (.889) 9 8 17 9 D (1.00) 14 14 31 17 B (1.33) 12 16 43 27 A (1.67) 6 10 4939 49 148 93

  40. Example: Sequencing Rules • Critical Ratio Rule Performance • Average flow time: 148/5 = 29.6 hours • Average number of jobs in the system: 148/49 = 3.02 jobs • Average job lateness: 93/5 = 18.6 hours

  41. Example: Sequencing Rules • Comparison of Rule Performance Average Average Average Flow Number of Jobs Job Rule Time in System Lateness FCFS 28.2 2.88 18.0 SPT 25.4 2.59 15.2 CR 29.6 3.02 18.6 SPT rule was superior for all 3 performance criteria.

  42. Sequencing n jobs on two machines • Johnson’s Rule: technique for minimizing completion time for a group of n jobs to be processed on two machines or at two work centers. • Minimizes total idle time • Johnson’s Rule requires satisfying the following conditions:

  43. Johnson’s Rule Conditions • Job time must be known and constant • Job times must be independent of sequence • Jobs must follow same two-step sequence • Job priorities cannot be used • All units must be completed at the first work center before moving to second

  44. Johnson’s Rule Optimum Sequence • List the jobs and their times at each work center • Find the smallest processing time. If it belongs to the first operation of a job schedule that job next, otherwise schedule that job last. • Eliminate the job from further consideration • Repeat steps 2 and 3 until all jobs have been scheduled

  45. Johnson’s Algorithm Example • Data: • Iteration 1: min time is 4 (job 1 on M1); place this job first and remove from lists:

  46. Johnson’s Algorithm Example (cont.) • Iteration 2: min time is 5 (job 3 on M2); place this job last and remove from lists: • Iteration 3: only job left is job 2; place in remaining position (middle). • Final Sequence: 1-2-3 • Makespan: 28

  47. Gantt Chart for Johnson’s Algorithm Example Short task on M2 to “clear out” quickly. Short task on M1 to “load up” quickly.

  48. Example 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:

  49. Select the job with the shortest processing time. It is job D, with a time of two hours. Since the time is at the first center, schedule job D first. Eliminate job D from further consideration. 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 The remaining jobs and their times are

  50. http://www.baskent.edu.tr/~kilter 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, 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

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