rolling pem m c02 scheduling

PEM M C02: Operations ManagementDr. ANAND PRAKASHPost Graduate Programme in Project Engineering and Management

Operations scheduling job order schedulingSESSION 11

Introduction-What is Operations Scheduling ?

Objectives of operations schedulingMeet due date;Minimize WIP inventory;Minimize the average flow time through the systems;Provide for high machine/worker (time) utilization (minimize idle time); Reduce setup cost;Minimize production and worker costs

Functions of Scheduling and ControlAllocating orders, equipments, and personnel to work centers or other specified locationShort term capacity planningDetermining the sequence of ordersJob prioritiesInitializing performance of the scheduled work, commonly termed the dispatching of jobsShop-floor control, involvingReviewing the status and controlling the progress of orders as they are being worked onExpediting the late and critical ordersRevising the schedules in light of changes in order status.

Elements of the Shop Floor Scheduling ProblemsThe classic approaches to shop floor scheduling focuses on the following six elements:Job arrival patterns: static or dynamicStatic: jobs arrive in batch;Dynamic: jobs arrive over time interval according to some statistical distribution.

Numbers and variety of machines in the shop floorIf there is only one machine or if a group of machines can be treated as one machine, the scheduling problem is much more simplified.As number of variety of machines increase, the more complex the scheduling problems is likely to become.Ratio of workers to machinesMachine limited system: more workers than machine or equal number workers and machines;Labor-limited system: more machines than worker.Flow pattern of jobs: flow shop or job shopFlow shop: all jobs follow the same paths from one machine to the next;Job shop: no similar pattern of movement of jobs from one machine to the next.

Job sequencingSequencing or priority sequencing: the process of determining which job is started first on some machines or work center by priority rule;Priority rule: the rule used for obtaining a job sequencing;Priority rule evaluation criteria To meet corresponding objectives of scheduling;Common standard measures:Meeting due date of customers or downstream operations;Minimizing flow time (the time a job spends in the shop flow);Minimizing WIP;Minimizing idle time of machines and workers (Maximizing utilization).

A job shop is organized by machines which are grouped according to their functions.A Job Shop

Not all jobs are assumed to require exactly the same number of operations, and some jobs may require multiple operations on a single machine (Reentrant system, Job B twice in work center 3).Each job may have a different required sequencing of operations.No all-purpose solution algorithms for solving general job shop problems;Operations scheduling of shop floor usually means job shop scheduling; Job AJob BA Job Shop

Parallel processing versus sequential processingSequencing Processing: the m machines are distinguishable, and different operations are performed by different machines.Parallel processing: The machines are identical, and any job can be processed on any machine.M1, M2, M3, and M4 are different;Job A has 2 operations which should be processed on different Machines: M1and M2;Job B has 3 operations which should be processed on different Machines: M3, M2 and M4;M1, M2, M3, and M4 are identical;Jobs A and B can be processed on any one of the 4 machinesJob Shop Scheduling Terminology

2Flow timeThe flow time of job i is the time that elapses from the initiation of that job on the first machine to the completion of job i.The mean flow time, which is a common measure of system performance, is the arithmetic average of the flow times for all n jobs

Job 1Job 2Job 1Job 3Job 2Job 3MachinesM1M2TimeF1: FT of Job 1F2: FT of Job 2F3: FT of Job 3Mean Flow Time=(F1+F2+F3)/3Job Shop Scheduling Terminology

3. Make-spanThe make-span is the time required to complete a group of jobs (all n jobs).Minimizing the make-span is a common objective in multiple-machine sequencing problems.Make-span of the 3 jobsJob Shop Scheduling Terminology

4. Tardiness and latenessTardiness is the positive difference between the completion time and the due date of a job.Lateness refers to the difference between the job completion time and its due date and differs from tardiness in that lateness can be either positive or negative.If lateness is positive, it is tardiness; when it is negative, it is earlinessWhen the completion of Job is earlier than due date, the tardiness is 0Lateness>0---TardinessLateness

Criteria and Objective Functions for Job Shop SchedulingMean Flow TimeAverage amount of time that a job spends in the system.Minimization is appropriate for rapid turnaround for fast deliveries.Mean TardinessUseful when there is a penalty per unit of time if job completion is delayed beyond a specified due date.Maximum TardinessUseful when the penalty per day for tardiness increases with the amount of tardiness.Number of tardy jobsSimply counts jobs that are not completed by due dates.

FCFS (first come-first served)Jobs are processed in the sequence in which they entered the shop;The simplest and nature way of sequencing as in queuing of a bankSPT (shortest processing time)Jobs are sequenced in increasing order of their processing time;The job with shortest processing time is first, the one with the next shortest processing time is second, and so on;EDD (earliest due date)Jobs are sequenced in increasing order of their due dates;The job with earliest due date is first, the one with the next earliest due date is second, and so on;Scheduling Procedures / Rules

CR (Critical ratio)Critical ratio is the remaining time until due date divided by processing time;Scheduling the job with the smallest CR next;CRi=Remaining time of Job i/Processing time of Job i =(Due date of Job i-current time)/Processing time of Job i CR provides the balance between SPT and EDD, such that the task with shorter remaining time and longer processing time takes higher priority; CR will become smaller as the current time approaches due date, and more priority will given to one with longer processing time; For a job, if the numerator of its CR is negative ( the job has been already later), it is naturally scheduled next; If more than one jobs are later, higher priority is given to one that has shorter processing time (SPT).Scheduling Procedures / Rules

A machine center in a job shop for a local fabrication company has five unprocessed jobs remaining at a particular point in time. The jobs are labeled 1, 2, 3, 4, and 5 in the order that they entered the shop. The respective processing times and due dates are given in the table below.Sequence the 5 jobs by above 4 rules and compare results based on mean flow time, average tardiness, and number of tardy jobsScheduling Procedures / RulesExample 1

Job numberProcessing TimeDue Date12345112931126145313332

1 11 61 02 40 45 03 71 31 404 72 33 395 74 32 42Totals 268 121Mean Flow time=268/5=53.6Average tardiness=121/5=24.2No. of tardy jobs=3.Scheduling RulesFCFS


JobCompletion TimeDue DateTardiness

4 1 1 33 05 2 3 32 01 11 14 61 02 29 43 45 03 31 74 31 43Totals 135 43Mean Flow time=135/5=27.0Average tardiness=43/5=8.6No. of tardy jobs=1.Scheduling RulesSPT


JobProcessing TimeCompletion TimeDue DateTardiness

3 31 31 31 05 2 33 32 14 1 34 33 12 29 63 45 181 11 74 61 13Totals 235 33Mean Flow time=235/5=47.0Average tardiness=33/5=6.6No. of tardy jobs=4.Scheduling RulesEDD


JobProcessing TimeCompletion TimeDue DateTardiness

Current time should be reset after scheduling one jobScheduling RulesCR

Current time: t=0Job numberProcessing TimeDue DateCritical Ratio1234511293112614531333261/11(5.545)45/29(1.552)31/31(1.000)33/1 (33.00)32/2 (16.00)

Current time: t=31Job numberProcessing TimeDue Date-Current TimeCritical Ratio1245112912 30 14 2 130/11(2.727)14/29(0.483) 2/1 (2.000)1/2 (0.500)

Mean Flow time=289/5=57.8Average tardiness=87/5=17.4No. of tardy jobs=4.Both Jobs 4 and 5 are later, however Job 4 has shorter processing time and thus is scheduled first; Finally, job 1 is scheduled last.Scheduling RulesCR

Current time=60Job numberProcessing TimeDue Date-Current TimeCritical Ratio14511121-27-281/11(0.0909)-27/1

DiscussionsSPT results in smallest mean flow time; EDD yields the minimum average;Always true? Yes!Scheduling RulesSUMMARY

RuleMean Flow TimeAverage TardinessNumber of Tardy JobsFCFSSPTEDDCR53.627.047.057.824.28.66.617.43144

Moores procedure, Johnsons procedure, ECC schedulingSESSION 12

Assuming that n jobs are to be processed through one machine. For each job i, define the following quantities: ti=Processing time for job i, constant for job i; di=Due date for job i, constant for job i; Wi=Waiting time for job i, the amount of time that the job must wait before its processing can begin. When all the jobs are processed continuously, Wi is the sum of the processing times for all of the preceding jobs; Fi=Flow time for job i, the waiting time plus the processing time: Fi= Wi+ ti; Li=Lateness of job i , Li= Fi- di, either positive or negative; Ti=Tardiness of job i, the positive part of Li, Ti=max[Li,0] ; Ei=Earliness of job i, the negative part of Li, Ei =max[- Li,0]Sequencing Theory for A Single Machines

Minimizing the number of Tardy Jobs: An algorithm from Moore(1968) that minimizes the number of tardy jobs for the single machine problem.

Step1. Sequence the jobs according to the earliest due date to obtain the initial solution. That is d[1] d[2],, d[n];Step2. Find the first tardy job in the current sequence, say job [i]. If none exists go to step 4.Step3. Consider jobs [1], [2], , [i]. Reject the job with the largest processing time. Return to step2. (Why ?)Reason: It has the largest effect on the tardiness of the Job[i].Step4. Form an optimal sequence by taking the current sequence and appending to it the rejected jobs. (Can be appended in any order?)Yes, because we only consider the number of tardiness jobs rather than tardiness.Sequencing Theory for A Single Machines

Example 2Solution Sequencing Theory for A Single Machines

Job123456Due date1569232030Processing time10348106

Job231546Due date6915202330Processing time34101086Completion time3717273541

Example 2 :Solution (Cont.)The optimal sequence: 2, 3, 4, 6, 5, 1 or 2, 3, 4, 6, 1, 5. In each case the number of tardy jobs is exactly 2.Sequencing Theory for A Single Machines



Example 3Sequencing Theory for A Single Machines

Job123456Processing time234321Due date3697117

Example 3Step1: find the job scheduled last(sixth) =2+3+4+3+2+1=15Step2: find the job scheduled fifth =15-2=13Sequencing Theory for A Single Machines


356Tardiness15-9=615-11=415-7=8


36Tardiness13-9=413-7=6

Example 3Step3: find the job scheduled fourth =13-4=9Step4: find the job scheduled third =9-1=8 Because job3 is no longer on the list, Job 2 now because a candidate.

Because job6 has been scheduled, Job 4 now because a candidate along with Job 2.

Sequencing Theory for A Single Machines


26Tardiness9-6=39-7=2


24Tardiness8-6=28-7=1

Example 3Step5: find the job scheduled secondThe optimal sequence: 1-2-4-6-3-5Sequencing Theory for A Single Machines


JobProcessing timeFlow timeDue dateTardiness124635233142258913153677911001244

Assume that n jobs are to be processed through m machines. The number of possible schedules is astonishing, even for moderate values of both n and m. For each machine, there is n! different ordering of the jobs; if the jobs may be processed on the machines in any order, there are totally (n!)m possible schedules. (n=5, m=5, 25 billion possible schedules) Even with the availability of inexpensive computing today, enumerating all feasible schedules for even moderate-sized problems is impossible or, at best, impractical.Sequencing Theory for Multiple Machines

Gantt chartSuppose that two jobs, I and J, are to be scheduled on two machines, 1 and 2, the processing times are Assume that both jobs must be processed first on machine 1 and then on machine 2. There are four possible schedules.Sequencing Theory for Multiple Machines

Machine 1Machine 2Job I41Job J14

Sequencing Theory for Multiple Machines

ScheduleTotal flow timeMean flow timeMean idle time19(5+9)/2=7(4+4)/2=4265.51310854109.55

Scheduling n Jobs on Two MachinesThe optimal solution for scheduling n jobs on two machines is always a permutation schedule. A very efficient algorithm for solving the two-machine problem was discovered by Johnson(1954). Denote the machines by A and BThe jobs must be processed first on machine A and then on machine B.DefineAi=Processing time of job i on machine ABi=Processing time of job i on machine B Rule: Job i precedes job i+1 if min(Ai, Bi-1)

Example 4Optimal sequence : 2 4 3 5 1Sequencing Theory for Multiple Machines

JobMachine AMachine B1522163974385104

rolling pem m c02 scheduling

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