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JOB SHOPSCHEDULING

Job Shop scheduling

Problem with m machines and n jobsEach job visits some or all of the machines

Only once (or multiple times if recirculation is allowed)Each customer order is unique and of small batches

Wafer fabrication in semiconductor industryHospital (patients are jobs or machines?)

Special case of project scheduling with workforceconstraintsVery difficult to solve! (NP-hard)

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Classic scheduling theory

Look at a specific machine environment with a specificobjectiveAnalyze to prove an optimal policy or to show that nosimple optimal policy existsThousands of problems have been studied in detailwith mathematical proofs!

But often, scheduling solutions are found usingdispatching rules, which are heuristics.

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Dispatching rules

Release/due date relatedEarliest release date first (ERD) rule

variance in throughput timesEarliest due date first (EDD) rule

maximum latenessMinimum slack first (MS) rule

maximum latenessApparent tardiness cost first (ATC)

maximum total weighted lateness

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max ,0j jd p t

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Dispatching rules

Processing time relatedLongest Processing Time first (LPT) rule

balance load on parallel machinesmakespan

Shortest Processing Time first (SPT) rulesum of completion timesWIP

Weighted Shortest Processing Time first (WSPT) rule

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Discussion

Very simple to implementOptimal for special casesOnly focus on one objectiveCombine several dispatching rules:

Composite Dispatching Rules

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Single machine models

With n jobs with pj, rj and dj, andTotal weighted completion time should be minimized:

Solution: Weighted Shortest Processing Time (WSPT)first is optimal.

Schedules jobs in decreasing order of wj /pj.SPT rule starts with job with the shortest pj, moves onto job with second shortest pj, and so on.

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j jw C

Single machine models

If rj = 0Each job has its own dj

Objective: minimize lateness Lmax

Earliest Due Date (EDD) results in optimal scheduleOrder operations in increasing order of dj

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Types of dispatching rules

WSPT and EDD are static.Static – basis for ordering operations does not changebased on scheduling decisions

All operations can be sorted onceDynamic – scheduling decisions change the order ofremaining operations

Need to resort operations in queue (potentially) afterevery decision

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Single machine models

If rj = 0, andObjective: minimize lateness Lmax

Minimum Slack (MS) orders operations at time t indescending order of:

max(dj – pj – t, 0)MS does not guarantee optimal schedule!

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Composite rule

Two good heuristics:Weighted Shorted Processing Time (WSPT)

optimal with due dates zeroMinimum Slack (MS)

Optimal when due dates are “spread out”Any real problem is somewhere in betweenCombine the characteristics of these rules into onecomposite dispatching rule.

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Composite dispatching rule

One-machine, rj = 0Objective: minimize weighted tardiness wjTj

Apparent Tardiness Cost (ATC) rule orders operationsin descending order (K is a parameter):

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max( ,0)( ) expj j j

jj

w d p tI t

p K p

WSPT MSMean pj

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Special cases

If K is very large:ATC reduces to WSPT

If K is very small and no overdue jobs:ATC reduces to MS

If K is very small and overdue jobs:ATC reduces to WSPT applied to overdue jobs

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Jobs with different release dates

One-machine problem with different rj

Objective: minimize lateness Lmax

Problem is NP-hardPossible algorithms to solve the problem (from theOptimization and Decision course)

Branch-and-bound (see Appendix B of Pinedo’s book)Dynamic Programming

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Parallel machines

A set of m machines in parallel is available.Objective: minimize makespan Cmax

P2||Cmax is already NP-Hard!

Longest Processing Time (LPT) firstpick operations in descending order of processing time

LPT balances the loads of the machines (why?).LPT does not guarantee optimality.

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Parallel machines

Pm|| Cj; Objective: minimize completion time Cj

SPT assures optimality, even when preemptions areallowed.

Objective: minimize weighted completion time wjCj

WSPT does not assures optimality.

Objective: minimize total weighted tardiness wjTj

This more general problem is even harder. ATC can beapplied, but solutions can be poor.

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Single and parallel machine problems

Dispatching rules for single machine problems:

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Problem Heuristic Complexity1||Lmax Earliest Due Date (OPT) P1|| Cj Shortest Processing Time (OPT) P1|| jCj Weighted Shortest Processing Time (OPT) P1||rj| Lmax Branch and Bound (OPT) NP-Hard1|| Tj Branch and Bound (OPT) NP-HardPn||Cmax Longest Processing Time NP-Hard

FLOW SHOP MODELS

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Flow shop

m machines in series and each job has to be processedon each of the m machinesF2 ||Cmax

Heuristic: Johnson´s rule (SPT(1) – LPT(2))Use SPT for set of jobs (p1j< p2j)Use LPT for set of jobs (p1j> p2j)

Provides optimal solutionF3 ||Cmax

NP- Hard Problem

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JOB SHOP MODELS

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Job Shop scheduling

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precedence constraint

job

operation/task/activity

Dispatching rule for Job Shop

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J0

J1

J2

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Dispatching rule for Job Shop

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J0

J1

J2

Dispatching rule for Job Shop

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J0

J1

J2

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Dispatching rule for Job Shop

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J0

J1

J2

What is the dispatchingrule being used?

Dispatching rule for Job Shop

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J0

J1

J2

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Dispatching rule for Job Shop

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J0

J1

J2

Dispatching rule for Job Shop

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J0

J1

J2

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Dispatching rule for Job Shop

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J0

J1

J2

Dispatching rule for Job Shop

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J0

J1

J2

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Dispatching rule: SPT

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makespan

j = 2

j = 2

j = 2

J0

J1

J2

Job Shop models

Jm - m machines and each job has its ownpredetermined route to follow

Processing of job j in machine i is operation (i, j) withduration pij, and (i, j) N nodes.

J2 ||Cmax can be reduced to 2 problems F2 ||Cmax

Heuristic: SPT(1) - LPT(2) (provides optimal solution)

Jm ||Cmax can be represented in a disjunctive graphG(N,A,B).

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Job Shop models

Direct graph G = (N, A, B) with a set of N operations.Arcs A - conjunctive arcs represent the precedencerelationships between processing operations of a job.Arcs B - disjunctive arcs connect two operations whichbelong to two different jobs, that are to be processedon the same machine.Source U and sink V:

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k,lU V

operation (k,l)source sink

Disjunctive graph

Example of a Job Shop problem:4 machines3 jobs

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U

1,3

1,1

1,2

2,3

2,1

4,2

4,3

3,1

3,2 V2,2

p43

0

0

0

p23

p42

p43

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Solution methods

Branch-and-BoundOptimal solutionWorks until 20 jobs, 20 machines

Search heuristics (Tabu search, Simulated Annealing,Genetic Algorithms, etc.)

Do not guarantee the optimal solutionShifting Bottleneck

Easy and successful

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Shifting Bottleneck heuristic

Minimize makespan in a job shopLet M denote the set of machinesLet M0 M be machines for which disjunctive arcshave been selectedBasic idea:

Select a machine in M – M0 to be included in M0

Sequence the operations on this machine

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Shifting Bottleneck algorithm

Step 1: Set the initial conditionsSet M0 = . Graph G is the graph with all theconjunctive arcs and no disjunctive arcs.Set Cmax(M0) equal to the longest path in graph G.

Step 2: Analysis of the machines still to be scheduledFor each machine i in M – M0: formulate a singlemachine problem with all operations subject to releasedates and due dates. Release date is the longest path inG from the source to the node. Due date is the longestpath in G from the node to the sink and subtracting pij.Minimize Lmax in each machine.

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Shifting Bottleneck algorithm

Step 3: Bottleneck selectionThe machine with the highest cost is designated thebottleneck.Insert all the corresponding disjunctive arcs in graph G.Insert machine which is the bottleneck in M0.

Step 4: Resequencing all machines scheduled earlierFind the sequence that minimized the cost and insertthe corresponding disjunctive arcs in graph G.

Step 5: Stopping conditionIf all machines are scheduled (M0 = M) then STOP, elsego to Step 2.

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Example 5.4.2 (p. 89)

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Jobs Machine sequence Processing Times1 1,2,3 P11=10, P21=8, P31=42 2,1,4,3 P22=8, P12=3, P42=5,P32=63 1,2,4 P13=4, P23=7, P43=3

U

1,3

1,1

1,2

2,3

2,1

4,2

4,3

3,1

3,2 V2,2

p43

0

0

0

p23

p42

p43

Example

Another form of presenting processing times:

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J1

J2

J3

10 8 4

8 3 5 6

4 7 3

M1

M2

M4

M3

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Initial conditions

M0 = Ø(set of machines with determined sequence)Set graph G with conjunctive arcs:

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S

1,3

1,1

1,2

2,3

2,1

4,2

4,3

3,1

3,2 T2,2

10 8

0 4

0

0

8 3 5 6

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Step 1

Find release date and due date of each operationRemove all sequence constraints among activities inM – M0, use CPM to find critical path and minimizestart time, maximize ending time for each activity

Since M0 is initially empty, only conjunctive arcs appearin the graph.

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Step 1: Find Cmax(M0)

For each machine in M-M0 generate an instance of1|rj|Lmax

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J1

J2

J3

10 8 4

8 3 5 6

4 7 3

Cmax(M0)0 22

[16 22]

J1

J2

J3

10 8 4

8 3 5 6

4 7 3

[0 10] [10 18] [18 22]

[0 8] [8 11] [11 16]

[0 12] [4 19] [11 22]

[rj dj]

Step 1: Find release and due dates

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J1

J2

J3

10 8 4

8 3 5 6

4 7 3

Cmax(M0)0 22

[16 22]

J1

J2

J3

10 8 4

8 3 5 6

4 7 3

[0 10] [10 18] [18 22]

[0 8] [8 11] [11 16]

[0 12] [4 19] [11 22]

[rj dj]

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Step 2: Find optimal 1-machine schedules

minimize Lmax for each subproblem

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Lj = Cj – djLmax = max(Lj)

[16 22]

M1

M2

M3

10

8

4

8

3

5

6

4

7

3

[0 10]

[10 18]

[18 22]

[0 8]

[8 11]

[11 16]

[0 12]

[4 19]

[11 22]M4

J2 J3

10 3 4 Lmax(1) = 5

0 10 13

88 7J2 J3 J1 Lmax(2) = 40 8 15

46 Lmax(3) = 416 22

5 311 16

Lmax(4) = 0

J1

Step 3: Add machine to M0

Pick machine with highest Lmax

Use sequence found in Step 2Lmax(1) = 5Machine 1 must be added to M0

Pinedo’s book has a small mistake (Lmax(2) = 5 in the book)

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Graph after iteration 1

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S

1,3

1,1

1,2

2,3

2,1

4,2

4,3

3,1

3,2 T2,2

10 8

0 4

0

0

8 3 5 6

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Cmax({1}) = Cmax( ) + Lmax(1) = 22 + 5 = 27

Iteration 2, Step 1

M0 = {1}Cmax(M0) = 27, find release and due dates

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[18 27]M3 4 6

[18 27]

5 3[13 21] [24 27]

M4

M2 8 8 7[10 23] [0 10] [17 24]

J2J1 J3J1

J2

J3

10 8 4

8 3 5 6

4 7 3

Cmax(M0)0 27

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Iteration 2, Step 1

M0 = {1}Cmax(M0) = 27, find release and due dates

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[18 27]M3 4 6

[18 27]

5 3[13 21] [24 27]

M4

M2 8 8 7[10 23] [0 10] [17 24]

J2J1 J3J1

J2

J3

10 8 4

8 3 5 6

4 7 3

Cmax(M0)0 27

Iteration 2, Step 2

Using release and due dates, minimize Lmax

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46 Lmax(3) = 118 24

5 313 24

Lmax(4) = 0

[18 27]M3 4 6

[18 27]

5 3[13 21] [24 27]

M4

M2 8 8 7[10 23] [0 10] [17 24]

J2J1 J3 88 7J2 J3J1 Lmax(2) = 10 8 1810

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Iteration 2, Step 3

Assign machine to M0

Pick machine with highest Lmax

Use sequenc found in Step 2Lmax(2) = Lmax(3) = 1Arbitrarily choose to add machine 2

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Graph after iteration 2

M0 = {1,2}

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1,3

1,1

1,2

2,3

2,1

4,2

4,3

3,1

3,2 T2,2

10 8

0 4

0

0

8 3 5 6

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Cmax({1,2}) = Cmax({1}) + Lmax(2) = 27 + 1 = 28

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Iteration 3, Step 1

M0 = {1, 2}Cmax(M0) = 28, find release and due dates

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[18 28]M3 4 6

[18 28]

5 3[13 22] [25 28]

M4

J1

J2

J3

10 8 4

8 3 5 6

4 7 3

Cmax(M0+)0 28

Iteration 3, Step 1

M0 = {1, 2}Cmax(M0) = 28, find release and due dates

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[18 28]M3 4 6

[18 28]

5 3[13 22] [25 28]

M4

J1

J2

J3

10 8 4

8 3 5 6

4 7 3

Cmax(M0+)0 28

28

Iteration 3, Step 2

Using release and due dates, min. Lmax

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46 Lmax(3) = 018 24

5 313 24

Lmax(4) = 0

[18 28]M3 4 6

[18 28]

5 3[13 22] [25 27]

M4

Iteration 3, Step 3

Lmax(3) = Lmax(4) = 0Thus, a final schedule was found in Step 2

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8J2

6

5

M1

M2

M4

M3

10 3 4

8J1

4

3

7J3

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Iteration 3, Step 4

If M0 = M, STOPCmax({1,2,3,4}) = Cmax({1})+Lmax(2)+Lmax(3)+Lmax(4)=27+1+0+0 = 28

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S

1,3

1,1

1,2

2,3

2,1

4,2

4,3

3,1

3,2 T2,2

10 8

0 4

0

0

8 3 5 6

4 73

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3

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Discussion

Very effectiveRelatively fastGood solutionsMore general Job Shop problems can be solved as well

‘Just a heuristic’No guarantee of optimum

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LEKIN

On disk with bookGeneric job shop scheduling systemUser friendly windows environmentC++ object oriented designCan add own routinesLook at slides from University of Nottingham comingwith book! or:

http://www.cs.nott.ac.uk/~sxp/

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LEKIN

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