8 Job Sequencing & Operations Scheduling CHAPTER Arranged by Dr Sh Salleh bin Sh Ahmad Originated from: Operations Management, Eighth Edition, by William J. Stevenson Copyright © 2005 by The McGraw-Hill Companies, Inc. All rights reserved.

Learning Outcomes After completing the chapter, students should be able to: 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 Use and interpret scheduling Gantt charts

Learning Outcomes (continue) After completing the chapter, students should be able to: 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 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

High-Volume Systems Flow system: High-volume system with Standardized equipment and activities Flow-shop scheduling: Scheduling for high-volume flow system Work Center #1 Work Center #2 Output

Scheduling Manufacturing Operations High-volume Intermediate- volume Low-volume Service operations Build A A Done Build B B Done Build C C Done Build D Ship JAN FEB MAR APR MAY JUN On time!

High-Volume Success Factors Process and product design Preventive maintenance Rapid repair when breakdown occurs Optimal product mixes Minimization of quality problems Reliability and timing of supplies

Intermediate-Volume Systems Outputs are between standardized high-volume systems and made-to-order job shops Run size, timing, and sequence of jobs Economic run size:

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

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

Loading Infinite loading – Jobs are assigned to work centers without regard to the capacity of the work centre Finite loading – Jobs are assigned to work centers taking into account the work centre capacity and the processing times Forward scheduling – Scheduling ahead, from some point in time Backward scheduling – Scheduling by working backwards from the due dates Schedule chart

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.

Sequencing Priority rules: Simple heuristics used to select the order in which jobs will be processed. Job time: Time needed for setup and processing of a job. Everything is #1 Priority

Priority Rules FCFS - first come, first served Table 15.2 FCFS - first come, first served SPT - shortest processing time EDD - earliest due date CR - critical ratio S/O - slack per operation Rush - emergency Top Priority

Simple Sequencing Rules PROCESSING DUE JOB TIME DATE A 2 7 B 8 16 C 4 4 D 10 17 E 5 15 F 12 18

Simple Sequencing Rules: FCFS Sequence: A-B-C-D-E-F Job Processing Flow DUE TARDINESS SEQUENCE TIME TIME DATE Flow – dd (0 if negative) A 2 2 7 0 B 8 10 16 0 C 4 14 4 10 D 10 24 17 7 E 5 29 15 14 F 12 41 18 23 41 120 54 Average flow time = 120/6 = 20 days Utilization = 41/120 = 34% Average Tardiness = 54/6 = 9 days Makespan = 41 days Average number of jobs at the work center = 120/41 = 2.93

Simple Sequencing Rules: SPT Sequence: A-C-E-B-D-F Job Processing Flow DUE TARDINESS SEQUENCE TIME TIME DATE Flow – dd (0 if negative) A 2 2 7 0 C 4 6 4 2 E 5 11 15 0 B 8 19 16 3 D 10 29 17 12 F 12 41 18 23 41 108 40 Average flow time = 108/6 = 18 days Utilization = 41/108 = 38% Average Tardiness = 40/6 = 6.67 days Makespan = 41 days Average number of jobs at the work center = 108/41 = 2.63

Simple Sequencing Rules: EDD Sequence: C-A-E-B-D-F Job Processing Flow DUE TARDINESS SEQUENCE TIME TIME DATE Flow – dd (0 if negative) C 4 4 4 0 A 2 6 7 0 E 5 11 15 0 B 8 19 16 3 D 10 29 17 12 F 12 41 18 23 41 110 38 Average flow time = 110/6 = 18.33 days Utilization = 41/110 = 37% Average Tardiness = 38/6 = 6.33 days Makespan = 41 days Average number of jobs at the work center = 110/41 = 2.68

Sequencing Rules: Critical Ratio Job Processing DUE Critical Ratio SEQUENCE TIME DATE 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 The lowest CR is Job C, therefore it is selected as the first sequenced job

Sequencing Rules: Critical Ratio At day 4 [Job C completed], the CR are Job Processing DUE Critical Ratio SEQUENCE TIME DATE Calculation A 2 7 (7-4)/2 =1.5 B 8 16 (16-4)8 =1.5 C - - - D 10 17 (17-4)/10=1.3 E 5 15 (15-4)/5=2.2 F 12 18 (18-4)/12=1.17 The lowest CR is Job F, therefore it is selected as the 2nd sequenced job

Sequencing Rules: Critical Ratio At day 16 [Job C and F completed], the CR are Job Processing DUE Critical Ratio SEQUENCE TIME DATE Calculation A 2 7 (7-16)/2 =-4.5 B 8 16 (16-16)8 =0.0 C - - - D 10 17 (17-16)/10=0.1 E 5 15 (15-16)/5=-0.2 F - - - The lowest CR is Job A, therefore it is selected as the 3rd sequenced job

Sequencing Rules: Critical Ratio At day 18 [Job C, F and A completed], the CR are Job Processing DUE Critical Ratio SEQUENCE TIME DATE Calculation A - - - B 8 16 (16-18)8 =-0.25 C - - - D 10 17 (17-18)/10=-0.1 E 5 15 (15-18)/5=-0.6 F - - - The lowest CR is Job E, therefore it is selected as the 4th sequenced job

Sequencing Rules: Critical Ratio At day 23 [Job C, F, A and E completed], the CR are Job Processing DUE Critical Ratio SEQUENCE TIME DATE Calculation A - - - B 8 16 (16-23)8 =-0.88 C - - - D 10 17 (17-23)/10=-0.6 E - - - F - - - The lowest CR is Job B, therefore it is selected as the 4th sequenced job. The last sequence is set to Job D. Therefore the final sequence is C-F-A-E-B-D

Sequence: C-F-A-E-B-D Sequencing Rules: CR Sequence: C-F-A-E-B-D Job Processing Flow DUE TARDINESS SEQUENCE TIME TIME DATE Flow – dd (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 41 133 58 Average flow time = 133/6 = 22.17 days Utilization = 41/133 = 31% Average Tardiness = 58/6 = 9.67 days Makespan = 41 days Average number of jobs at the work center = 133/41 = 3.24

Comparison of the four sequence rules 3.24 9.67 22.17 CR 2.68 6.33 18.33 EDD 2.63 6.67 18.00 SPT 2.93 9.00 20.00 FCFS Average Number of Jobs at the Work Center Average Tardiness (days) Average Flow Time (days) Rule

Sequencing Jobs Through Multiple Serial Process Example In a manufacturing process, there are three type of components that must go through the machines of A, B and C respectively. Recommend the best components sequencing that will minimize the idle and makespan. Components Processing Time (min) Machines C1 C2 C3 A 2 4 B 8 3 C 6

Components Processing Time (min) Machines C1 C2 C3 A 2 4 B 8 3 C 6 With 3 jobs (components) to be schedule, we can arrange the sequence with: C1-C2-C3 or C1-C3-C2 or C2-C1-C3 or ….. Total possible sequences is equal to n!, n = number of jobs. n! = 3! = 3 x 2 x 1 = 6

Components Processing Time (min) Machines C1 C2 C3 A 2 4 B 8 3 C 6 Scheduling sequence of C1-C2-C3: makespan = 22 min C1 C2 C3 2 4 8 10 12 Idle 15 14 20 22 A B C

Components Processing Time (min) Machines C1 C2 C3 A 2 4 B 8 3 C 6 Ask the student to complete the task of solving for scheduling sequence of C2-C1-C3

Components Processing Time (min) Machines C1 C2 C3 A 2 4 B 8 3 C 6 Scheduling sequence of C2-C1-C3: makespan = 18 min C2 C1 C3 2 4 8 12 Idle 15 10 16 18 A B C

Two Work Center Sequencing Johnson’s Rule: technique for minimizing completion time for a group of jobs to be processed on two machines or at two work centers. Minimizes total idle time Several conditions must be satisfied

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

Sequencing Jobs Through Two Serial Process Johnson’s Rule List time required to process each job at each machine. Set up a one-dimensional matrix to represent desired sequence with # of slots equal to # of jobs. Select smallest processing time at either machine. If that time is on machine 1, put the job as near to beginning of sequence as possible. If smallest time occurs on machine 2, put the job as near to the end of the sequence as possible. Remove job from list. Repeat steps 2-4 until all slots in matrix are filled and all jobs are sequenced.

Johnson’s Rule A B C D E JOB PROCESS 1 PROCESS 2 A 6 8 B 11 6 C 7 3

Johnson’s Rule (cont.) E A D B C JOB PROCESS 1 PROCESS 2 (sanding) 5 11 20 31 38 Process 2 (painting) 15 23 30 37 41 Idle time Completion time = 41 Idle time = 5+1+1+3=10

Scheduling Difficulties Variability in 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

Minimizing Scheduling Difficulties Set realistic due dates Focus on bottleneck operations Consider lot splitting of large jobs

Scheduling Service Operations Appointment systems Controls customer arrivals for service Reservation systems Estimates demand for service Scheduling the workforce Manages capacity for service Scheduling multiple resources Coordinates use of more than one resource

Cyclical Scheduling 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

Service Operation Problems Cannot store or inventory services Customer service requests are random Scheduling service involves Customers Workforce Equipment