Uni Bonn (Germany) Sched10_5

8/13/2019 Uni Bonn (Germany) Sched10_5

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Scheduling

Shop Scheduling

Tim Nieberg


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Shop models: General Introduction

Remark: Consider non preemptive problems with regular objectives

Notation Shop Problems:m machines, n jobs 1, . . . , noperationsO = {(i , j )| j = 1 , . . . , n; i M j M := {1, . . . , m}} withprocessing times p ij M j is the set of machines where job j has to be processed on

PREC species the precedence constraints on the operations


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Notation Shop Problems:m machines, n jobs 1, . . . , noperationsO = {(i , j )| j = 1 , . . . , n; i M j M := {1, . . . , m}} withprocessing times p

ij M j is the set of machines where job j has to be processed onPREC species the precedence constraints on the operationsFlow shop: M j = M andPREC = {(i , j ) (i + 1 , j )|i = 1 , . . . , m 1; j = 1 , . . . , n}


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Notation Shop Problems:

m machines, n jobs 1, . . . , noperationsO = {(i , j )| j = 1 , . . . , n; i M j M := {1, . . . , m}} withprocessing times p ij

M j

is the set of machines where job j has to be processed onPREC species the precedence constraints on the operationsFlow shop: M j = M andPREC = {(i , j ) (i + 1 , j )|i = 1 , . . . , m 1; j = 1 , . . . , n}

Open shop: M j

= M and PREC = Job shop: PREC contain a chain (i 1 , j ) . . . , (i |M j | , j ) foreach j


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Disjunctive Formulation of the constraintsC ij denotes completion time of operation (i , j )PREC have to be respected:


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C ij p ij C kl for all (k , l ) (i , j ) PREC

no two operations of the same job are processed at the sametime:


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C ij p ij C kj or C kj p kj C ij for all i , k M j ; i = k

no two operations are processed jointly on the same machine:


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Disjunctive Formulation of the constraintsPREC have to be respected:



C ij p ij C kj or C kj p kj C ij for all i , k M j ; i = k

no two operations are processed jointly on the same machine:

C ij p ij C il or C il p il C ij for all (i , j ), (i , l ) O ; j = l

C ij p ij 0the or constraints are called disjunctive constraintssome of the disjunctive constraints are overruled by the

PREC constraints


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Disjunctive Formulation - makes pan objective

min C max s . t .

C max C ij (i , j ) O C ij p ij C kl (k , l ) (i , j ) PREC C ij p ij C kj or C kj p kj C ij i , k M j ; i = k C ij p ij C il or C il p il C ij (i , j ), (i , l ) O ; j = l C ij p ij 0 (i , j ) O


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Disjunctive Formulation - sum objective

min w j L j s . t .L j C ij d j (i , j ) O C ij p ij C kl (k , l ) (i , j ) PREC

C ij p ij C kj or C kj p kj C ij i ,k M

j

; i = k C ij p ij C il or C il p il C ij (i , j ), (i , l ) O ; j = l C ij p ij 0 (i , j ) O

Remark:

also other constraints, like e.g. release dates, can beincorporatedthe disjunctive constraints make the problem hard (lead to anILP formulation)


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Disjunctive Graph Formulationgraph representation used to represent instances and solutionsof shop problemscan be applied for regular objectives only

h d l l d


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Disjunctive Graph G = ( V , C , D )

V set of vertices representing the operations O a vertex is labeled by the corresponding processing time;Additionally, a source node 0 and a sink node belong to V ;their weights are 0C set of conjunctive arcs reecting the precedence constraints:for each (k , l ) (i , j ) PREC a directed arc belongs to C additionally 0 O and O are added to C D set of disjunctive arcs representing conicting operations:between each pair of operations belonging to the same job orto be processed on the same machine, for which no orderfollows from PREC , an undirected arc belongs to D

Sh d l G l I d i


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Disjunctive Graph - Example Job Shop

Data: 3 jobs, 3 machines;

M1 M2 M3

32 (1, 2) (3, 2)

(3, 1) (2, 1) (1, 1)

(2, 3) (1, 3) (3, 3)

p 31 = 4 , p 21 = 2 , p 11 = 1

p 12 = 3 , p 32 = 3p 23 = 2 , p 13 = 4 , p 33 = 1

Jobs: 1

Sh d l G l I t d ti


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Disjunctive Graph - Example Job Shop

3

2 (1, 2) (3, 2)

(3, 1) (2, 1) (1, 1)

(2, 3) (1, 3) (3, 3)

p 31 = 4 , p 21 = 2 , p 11 = 1

p 12 = 3 , p 32 = 3

p 23 = 2 , p 13 = 4 , p 33 = 1

Jobs: 1

Graph:

0

Conjunctive arcs3,1 2,1

1,2

1,1

3,2

2,3 1,3 3,3

Disjunctive arcs

Sh d l G l I t d ti


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Disjunctive Graph - Example Open Shop

Data: 3 jobs, 3 machines;

3

2Jobs: 1 (1, 1), (2, 1), (3, 1)

(1, 2), (2, 2), (3, 2)

(1, 3), (2, 3), (3, 3)

p 11 = 4 , p 21 = 2 , p 31 = 1p 12 = 3 , p 22 = 1 , p 32 = 3

p 13 = 2 , p 23 = 4 , p 33 = 1

Shop models: General Introd ction


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Disjunctive Graph - Example Open Shop

3

2

Jobs: 1 (1, 1), (2, 1), (3, 1)

(1, 2), (2, 2), (3, 2)

(1, 3), (2, 3), (3, 3)

p 11 = 4 , p 21 = 2 , p 31 = 1

p 12 = 3 , p 22 = 1 , p 32 = 3

p 13 = 2 , p 23 = 4 , p 33 = 1

Graph:

0

Conjunctive arcs2,1

1,2

3,3

Disjunctive arcs

1,1 3,1

2,2

1,3 2,3

3,2



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Disjunctive Graph - Selection

basic scheduling decision for shop problems (see disj.formulation):dene an ordering for operations connected by a disjunctivearc

turn the undirected disjunctive arc into a directed arcselection S : a set of directed disjunctive arcs(i.e. S D together with a chosen direction for each a S )disjunctive arcs which have been directed are called xed

a selection is a complete selection if each disjunctive arc has been xedthe graph G (S ) = ( V , C S ) is acyclic



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Selection - Remarks

a feasible schedule induces a complete selectiona complete selection leads to sequences in which operationshave to be processed on machinesa complete selection leads to sequences in which operations of a job have to be processedDoes each complete selection leads to a feasible schedule?



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Calculate a Schedule for a Complete Selection S

calculated longest paths from 0 to all other vertices in G (S )Technical description:

length of a path i 1 , i 2 , . . . , i r = sum of the weights of thevertices i 1 , i 2 , . . . , i r calculate length l ij of the longest path from 0 to (i , j ) (usinge.g. Dijkstra)start operation ( i , j ) at time l ij p ij (i.e. C ij = l ij )the length of a longest path from 0 to (such paths are calledcritical paths ) is equal to the makespan of the schedule

resulting schedule is the semiactive schedule which respects allprecedence given by C and S



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Reformulation Shop Problem

nd a complete selection for which the corresponding scheduleminimizes the given (regular) objective function

Flow Shop models


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Flow Shop models

Makespan MinimizationLemma: For problem F || C max an optimal schedule exists with

the job sequence on the rst two machines is the samethe job sequence on the last two machines is the same

(Proof as Exercise)Consequence: For F 2|| C max and F 3|| C max an optimal solutionexists which is a permutation solutionFor Fm || C max , m 4, instances exist where no optimalsolution exists which is a permutation solution(Exercise)

Flow Shop models


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Flow Shop models

Problem F 2|| C max

solution can be described by a sequence problem was solved by Johnson in 1954

Johnsons Algorithm:1 L = set of jobs with p 1 j < p 2 j ;2 R = set of remaining jobs;3 sort L by SPT w.r.t. the processing times on rst machine

(p 1 j )4 sort R by LPT w.r.t. the processing times on second machine

(p 2 j )5 sequence L before R (i.e. = ( L, R ) where L and R are

sorted)

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Flow Shop models

Example solution problem F 2|| C max

n = 5; p = 4 3 3 1 88 3 4 4 7

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Flow Shop models


n = 5; p = 4 3 3 1 88 3 4 4 7L

= {1,3

,4};

R = {2

,5}sorting leads to L = {4, 3, 1}; R = {5, 2}

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ow S op ode s


n = 5; p = 4 3 3 1 88 3 4 4 7L = {1, 3, 4}; R = {2, 5}

sorting leads to L = {4, 3, 1}; R = {5, 2} = (4 , 3, 1, 5, 2)

M 2

3 1 525134

4M 1

5 10 15 20 25

2

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p

Problem F 2|| C max

Lemma 1: If

min{p 1i , p 2 j } < min{p 2i , p 1 j }

then job i is sequenced before job j by Johnsons algorithm.

Lemma 2: If job j is scheduled immediately after job i and

min{p 1 j , p 2i } < min{p 2 j , p 1i }

then swapping job i and j does not increase C max .Theorem: Johnsons algorithm solves problem F 2|| C max optimal in O (n log(n)) time.

(Proofs on the board)

Flow Shop models


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p

Problem F 3|| C max

F 3|| C max is NP-hard in the strong senseReduction using 3-PARTITIONProof on the board

Open Shop models


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p p

Algorithm Problem O 2|| C max 1 I = set of jobs with p 1 j p 2 j ; J = set of remaining jobs;2 IF p 1r = max{max j I p 1 j , max j J p 2 j } then

order on M 1: (I \ { r }, J , r ); order on M 2: (r , I \ { r }, J )r rst on M

2, than on M

1; all other jobs vice versa

M 2

M 1r

rI \ { r } JJI \ { r }

Open Shop models


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p p

Algorithm Problem O 2|| C max 1 I = set of jobs with p 1 j p 2 j ; J = set of remaining jobs;2 IF p 1r = max{max j I p 1 j , max j J p 2 j } then

order on M 1: (I \ { r }, J , r ); order on M 2: (r , I \ { r }, J )r rst on M 2, than on M 1; all other jobs vice versa

3 ELSE IF p 2r = max{max j I p 1 j , max j J p 2 j } thenorder on M 1: (r , J \ { r }, I ); order on M 2: (J \ { r }, I , r )r rst on M 1, than on M 2; all other jobs vice versa

M 2

M 1 J \ { r }J \ { r }

Ir

rI

Open Shop models


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Remarks Algorithm Problem O 2|| C max

complexity: O (n)algorithm solves problem O 2|| C max optimallyProof builds on fact that C max is either

n j =1 p 1 j orn j =1 p 2 j or

p 1r + p 2r

Open Shop models


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Remarks Algorithm Problem O 2|| C max

complexity: O (n)algorithm solves problem O 2|| C max optimallyProof builds on fact that C max is either

n j

=1 p 1 j or

n j =1 p 2 j or

p 1r + p 2r

Problem O 3|| C max Problem O 3|| C max is NP-hardProof as Exercise (Reduction using PARTITION)

Open Shop models


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Problem O |pmtn |C max

dene MLi := n j =1 p ij (load of machine i )dene JL j := mi =1 p ij (load of job j )

LB := max {maxmi =1 MLi , max

n j =1 JL j } is a lower bound onC max

Open Shop models


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Problem O |pmtn |C max

dene MLi := n j =1 p ij (load of machine i )dene JL j := mi =1 p ij (load of job j )LB := max {maxmi =1 MLi , max

n j =1 JL j } is a lower bound on

C max Theorem: For problem O |pmtn |C max a schedule withC max = LB exists.Proof of the theorem is constructive and leads to a polynomial

algorithm for problem O |pmtn |C max

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Notations for Algorithm O |pmtn |C max job j (machine i ) is called tight if JL j = LB (MLi = LB ) job j (machine i ) has slack if JL j < LB (MLi < LB )a set D of operataions is called a decrementing set if it contain

for each tight job and machine exactly one operation and foreach job and machine with slack at most one operationTheorem: A decrementing set always exists and can becalculated in polynomial time(Proof based on maximal cardinality matchings; see e.g. P.Brucker: Scheduling Algorithms)

Open Shop models


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Algorithm O |pmtn |C max REPEAT

1 Calculate a decrementing set D ;2 Calculate maximum value with

min( i , j )D p ij LB MLi if machine i has slack and no operation in D

LB JL j if job j has slack and no operation in D ;3 schedule the operations in D for time units in parallel;4 Update values p , LB , JL, and ML

UNTIL all operations have been completely scheduled.

Open Shop models


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Correctness Algorithm O |pmtn |C max

after an iteration we have: LB new = LB old in each iteration a time slice of time units is scheduledthe algorithm terminates after at most nm + n + m iterationssince in each iteration either

an operation gets completely scheduled orone additional machine or job gets tight

Open Shop models


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Example Algorithm O |pmtn |C max p ML

2 4 3 2 11p 3 1 2 3 9

2 3 3 2 10JL 7 8 8 7 LB = 11

Open Shop models


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Example Algorithm O |pmtn |C max = 3 p ML

2 4 3 2 11p 3 1 2 3 9

2 3 3 2 10JL 7 8 8 7 LB = 11

Open Shop models


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2 4 3 2 11p 3 1 2 3 9

2 3 3 2 10JL 7 8 8 7 LB = 11

M 1M 2

M 3

31

2

3

Open Shop models


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2 4 3 2 11p 3 1 2 3 9

2 3 3 2 10

JL 7 8 8 7 LB = 11

M 1

M 2

M 3

312

3

p ML2 4 0 2 8

p 0 1 2 3 62 0 3 2 7

JL 4 5 5 7 LB = 8

Open Shop models


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2 4 3 2 11p 3 1 2 3 9

2 3 3 2 10

JL 7 8 8 7 LB = 11

M 1

M 2

M 3

312

3

= 1 p ML2 4 0 2 8

p 0 1 2 3 62 0 3 2 7

JL 4 5 5 7 LB = 8

M 1

M 2

M 3

312

3 42

Open Shop models


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2 4 0 2 8p 0 1 2 3 6

2 0 3 2 7

JL 4 5 5 7 LB = 8

M 1

M 2

M 3

312

3 42

= 3 p ML2 3 0 2 7

p 0 1 2 3 6

2 0 3 2 7JL 4 4 5 7 LB = 7

M 1

M 2

M 3

312

3 4 7243

2

Open Shop models


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2 3 0 2 7p 0 1 2 3 6

2 0 3 2 7

JL 4 4 5 7 LB = 7

M 1

M 2

M 3

312

3 4 7243

2

= 2 p ML2 0 0 2 4

p 0 1 2 0 3

2 0 0 2 4JL 4 1 2 4 LB = 4

M 1

M 2

M 3

312

3 4 7243

9431

2

Open Shop models


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2 0 0 2 4p 0 1 2 0 3

2 0 0 2 4JL

4 1 2 4 LB

= 4

M 1

M 2

M 3

312

3 4 7243

9431

2

= 1 p ML2 0 0 0 2

p 0 1 0 0 1

0 0 0 2 2JL 2 1 0 2 LB = 2

M 1

M 2

M 3

312

3 4 7243

9431

101

4

2

Open Shop models


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2 0 0 0 2p 0 1 0 0 1

0 0 0 2 2JL

2 1 0 2 LB

= 2

M 1

M 2

M 3

312

3 4 7243

9431

101

4

2

= 1 p ML1 0 0 0 1

p 0 1 0 0 1

0 0 0 1 1JL 1 1 0 1 LB = 1

M 1

M 2

M 3

312

3 4 7243

9431

101

4 42

1112

Open Shop models


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Final Schedule Example Algorithm O |pmtn |C max p ML

2 4 3 2 11p 3 1 2 3 9

2 3 3 2 10JL 7 8 8 7 LB = 11

M 1

M 2

M 3

312

3 4 7243

9431

101

4 42

1112

6 iterationsC max = 11 = LB sequence of time slices may be changed arbitrary

Job Shop models


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Problem J 2|| C max

I 1: set of jobs only processed on M 1I 2: set of jobs only processed on M 2

I 12: set of jobs processed rst on M 1 and than on M 2I 21: set of jobs processed rst on M 2 and than on M 1 12: optimal ow shop sequence for jobs from I 12 21: optimal ow shop sequence for jobs from I 21

Job Shop models


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Algorithm Problem J 2|| C max 1 on M 1 rst schedule the jobs from I 12 in order 12, than the

jobs from I 1, and last the jobs from I 21 in order 212 on M 2 rst schedule the jobs from I 21 in order 21, than the

jobs from I 2, and last the jobs from I 12 in order 12

M 1

M 2I 12 I 1 I 21

I 2 I 12I 21

Job Shop models


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Algorithm Problem J 2|| C max 1 on M 1 rst schedule the jobs from I 12 in order 12, than the

jobs from I 1, and last the jobs from I 21 in order 212 on M 2 rst schedule the jobs from I 21 in order 21, than the

jobs from I 2, and last the jobs from I 12 in order 12

M 1

M 2I 12 I 1 I 21

I 2 I 12I 21

Theorem: The above algorithm solves problem J 2|| C max

optimallyin O (n log(n)) time.Proof: almost straightforward!

Job Shop models


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Problem J || C max

as a generalization of F || C max , this problem is NP-hardit is one of the most treated scheduling problems in literaturewe presented

a branch and bound approacha heuristic approach called the Shing Bottleneck Heuristic

for problem J || C max which both depend on the disjunctivegraph formulation

Job Shop models


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Base of Branch and BoundThe set of all active schedules contains an optimal scheduleSolution method: Generate all active schedules and take thebest

Improvement: Use the generation scheme in a branch andbound settingConsequence: We need a generation scheme to produce allactive schedules for a job shop

Approach: extend partial schedules

Job Shop models


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Generation of all active schedulesNotations: (assuming that already a partial schedule S isgiven)

: set of all operations which predecessors have already been

scheduled in S r ij :earliest possible starting time of operation (i , j ) w.r.t. S : subset of

Remark: r ij can be calculated via longest path calculations inthe disjunctive graph belonging to S

Job Shop models


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Generation of all active schedules (cont.)1

(Initial Conditions) := {rst operations of each job}; r ij := 0 for all (i , j ) ;2 (Machine selection)

Compute for current partial schedulet () := min (i , j ){r ij + p ij }; i := machine on whichminimum is achieved;

3 (Branching) := {(i , j )|r i j < t ()}FOR ALL (i , j ) DO

1 extend partial schedule by scheduling (i , j ) next on machinei ;

2 delete (i , j ) from ;3 add job-successor of (i , j ) to ;4 Return to Step 2

Job Shop models


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Generation of all active schedules - example

3

2 (1, 2) (3, 2)

(3, 1) (2, 1) (1, 1)

(2, 3) (1, 3) (3, 3)

p 31 = 4 , p 21 = 2 , p 11 = 1

p 12 = 3 , p 32 = 3

p 23 = 2 , p 13 = 4 , p 33 = 1

Jobs: 1

Partial Schedule:

M 3

M 2

M 1

4 6

Job Shop modelsGeneration of all active schedules example


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Generation of all active schedules - example

32 (1, 2) (3, 2)

(3, 1) (2, 1) (1, 1)

(2, 3) (1, 3) (3, 3)

p 31 = 4 , p 21 = 2 , p 11 = 1

p 12 = 3 , p 32 = 3p 23 = 2 , p 13 = 4 , p 33 = 1

Jobs: 1

Partial Schedule:

M 3

M 2

M 1

4 6

= {(1, 1), (3, 2), (1, 3)};r 11 = 6 , r 32 = 4 , r 13 = 3;t () = min {6 + 1 , 4 + 3 , 3 + 4} = 7;i = M 1; = {(1, 1), (1, 3)}

Job Shop modelsGeneration of all active schedules - example (cont )


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Generation of all active schedules example (cont.)Partial Schedule:

M 3

M 2

M 1

4 6= {(1, 1), (3, 2), (1, 3)};r 11 = 6 , r 32 = 4 , r 13 = 3;t () = min {6 + 1 , 4 + 3 , 3 + 4} = 7;i = M 1; = {(1, 1), (1, 3)}Extended partial schedules:

M 1

M 2

M 36 6

M 1

M 2

M 3

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Remarks on the generation:the given algorithm is the base of the branchingnodes of the branching tree correspond to partial schedulesStep 3 branches from the node corresponding to the current

partial schedulethe number of branches is given by the cardinality of a branch corresponds to the choice of an operation (i , j ) tobe schedules next on machine i

a branch xes new disjunctions

Job Shop models


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Disjunctions xed by a branching

Node v

Node v with = {(i , j ) , (i , l )}

Node v

selection (i , j )Add disjunctions (i , j ) (i k ) for all unscheduled operations (i , k )

Consequence: Each node in the branch and bound tree ischaracterized by a set S of xed disjunctions

Job Shop models


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Lower bounds for nodes of the branch and bound tree

Consider node V with xed disjunctions S :Simple lower bound:

calculate critical path in G (S ) Lower bound LB (V )

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Lower bounds for nodes of the branch and bound treeConsider node V with xed disjunctions S :Simple lower bound:

calculate critical path in G (S )

Lower bound LB (V )Better lower bound:

consider machine i allow parallel processing on all machines = i solve problem on machine i

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1-machine problem resulting for better LB1 calculate earliest starting times r ij of all operations (i , j ) on

machine i (longest paths from source in G (S ))2 calculate minimum amount q ij of time between end of (i , j )

and end of schedule (longest path to sink in G

(S

))3 solve single machine problem on machine i :respect release datesno preemptionminimize maximum value of C ij + q ij

Result: head-body-tail problem (see Lecture 3)

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Better lower bound

solve 1-machine problem for all machines

this results in values f 1 , . . . , f mLB new (V ) = max mi =1 f i

Job Shop models


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Better lower bound

solve 1-machine problem for all machinesthis results in values f 1 , . . . , f mLB new (V ) = max mi =1 f i

Remarks:1-machine problem is NP-hardcomputational experiments have shown that it pays of to solvethese m NP-hard problems per node of the search tree20 20 job-shop instances are already hard to solve by branchand bound

Job Shop models

Better lower bound - example


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Better lower bound - examplePartial Schedule:

M 1

M 2

M 33 6

Corresponding graph G (S ):3,1 2,1 1,1

1,2 3,20 *

2,3 1,3 3,3

Conjunctive arcs xed disj.

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Graph G (S ) with processing times:

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

3

3

2 4 1

0 *

LB (V )= l (0, (1, 2), (1, 3), (3, 3) ,)=8

Job Shop models

Graph G(S ) with processing times:


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Graph G (S ) with processing times:

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

3

3

2 4 1

0 *

LB (V )= l (0, (1, 2), (1, 3), (3, 3) ,)=8

Data for jobs on Machine 1:green blue redr 12 = 0 r 13 = 3 r 11 = 6q 12 = 5 q 13 = 1 q 11 = 0

Opt. solution: Opt = 8, LB new (V ) = 8M 1

3 7 8

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Change p 11 from 1 to 2!

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 2

3

3

2 4 1

0 *

LB (V ) = l (0, (1, 2), (1, 3), (3, 3), ) = l (0, (3, 1), (2, 1), (1, 1),) = 8

Job Shop models

Change p11 from 1 to 2!


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Change p 11 from 1 to 2!

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 2

3

3

2 4 1

0 *

LB (V ) = l (0, (1, 2), (1, 3), (3, 3), ) = l (0, (3, 1), (2, 1), (1, 1),) = 8

Data for jobs on Machine 1:green blue redr 12 = 0 r 13 = 3 r 11 = 6q 12 = 5 q 13 = 1 q 11 = 0

Opt. solution: OPT = 9, LB new (V ) = 9

3 7 9

M 1

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The Shifting Bottleneck Heuristic

successful heuristic to solve makespan minimization for jobshopiterative heuristicdetermines in each iteration the schedule for one additional

machineuses reoptimization to change already scheduled machinescan be adapted to more general job shop problems

other objective functionsworkcenters instead of machinesset-up times on machines...

Shifting Bottleneck Heuristic for Job Shop


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Basic Idea

Notation: M set of all machinesGiven: xed schedules for a subset M 0 M of machines (i.e.a selection of disjunctive arcs for cliques corresponding tothese machines)

Actions in one iteration:select a machine k which has not been xed (i.e. a machinefrom M \ M 0)determine a schedule (selection) for machine k on the base of the xed schedules for the machines in M 0

reschedule the machines from M 0

based on the other xedschedules



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Selection of a machine

Idea: Chose unscheduled machine which causes the mostproblems (bottleneck machine)Realization:

Calculate for each operation on an unscheduled machine theearliest possible starting time and the minimal delay betweenthe end of the operation and the end of the complete schedulebased on the xed schedules on the machines in M 0 and the job orderscalculate for each unscheduled machine a schedule respectingthese earliest release times and delayschose a machine with maximal completion time and x theschedule on this machine



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Technical realization

Dene graph G = ( N , A ):N same node set as for the disjunctive graph

A contains all conjunctive arcs and the disjunctive arcscorresponding to the selections on the machines in M 0

C max (M 0) is the length of a critical path in G



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Technical realization

Dene graph G = ( N , A ):N same node set as for the disjunctive graphA contains all conjunctive arcs and the disjunctive arcscorresponding to the selections on the machines in M 0

C max (M 0) is the length of a critical path in G Comments:

with respect to G operations on machines from M \ M 0 maybe processed in parallel

C max (M 0) is the makespan of a corresponding schedule



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Technical realization (cont.)for an operation (i , j ); i M \ M 0 let

r ij be the length of the longest path from 0 to ( i , j ) (withoutp ij ) in G q ij be the length of the longest path from ( i , j ) to (withoutp ij ) in G

Comments:r ij is the release time of (i , j ) w.r.t. G q ij is the tail (minimal time till end) of (i , j ) w.r.t. G



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Technical realization (cont.)

For each machine from M \ M 0 solve the nonpreemptiveone-machine head-body-tail problem 1|r j , d j < 0|Lmax Result: values f (i ) for all i M \ M 0

Action:Chose machine k as the machine with the largest f (i ) valueschedule machine k according to the optimal schedule of theone-machine problemadd k to M 0 and the corresponding disjunctive arcs to G

C max (M 0

k ) f (k )


T h i l li i E l


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Technical realization - Example

Given: M 0

= {M 3} and on M 3 sequence(3, 2) (3, 1) (3, 3)Graph G :

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

3 3

2 4 1

0 *

Conjunctive arcs Selection M

3

C max (M 0) = 13



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Technical realization - Example (cont.)Machine M 1:

(i , j ) (1, 1) (1, 2) (1, 3)r ij 12 0 2q ij 0 10 1p ij 1 3 4



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Technical realization - Example (cont.)Machine M 1:

(i , j ) (1, 1) (1, 2) (1, 3)r ij 12 0 2q ij 0 10 1p ij 1 3 4

5 10 13M 1

10

1

1,2 1,3 1,1

f (M 1) = 13


Technical realization - Example (cont.)


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Technical realization Example (cont.)

Machine M 1:(i , j ) (1, 1) (1, 2) (1, 3)r ij 12 0 2q ij 0 10 1

p ij 1 3 4

5 10 13M 1

101

1,2 1,3 1,1

f (M 1) = 13

Machine M 2:

(i , j ) (2, 1) (2, 3)r ij 10 0

q ij 1 5p ij 2 2


T h i l li i E l ( )


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Machine M 1:

(i , j ) (1, 1) (1, 2) (1, 3)r ij 12 0 2q ij 0 10 1p ij 1 3 4

5 10 13M 1

101

1,2 1,3 1,1

f (M 1) = 13Machine M 2:

(i , j ) (2, 1) (2, 3)r ij 10 0q ij 1 5p ij 2 2

M 25 10 13

5 1

2,12,3

f (M 2) = 13


T h i l li i E l ( )


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Choose machine M 1 as the machine to x the schedule:add (1 , 2) (1 , 3) (1 , 1) to G M 0 = {M 1 , M 3}

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

3 3

2 4 1

0 *

Conjunctive arcs Selection M 1 , M 3

C max (M 0) = 13



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Reschedule Machines

try to reduce the makespan of the schedule for the machinesin M 0

Realization:consider the machines from M 0 one by oneremove the schedule of the chosen machine and calculate anew schedule based on the earliest starting times and delaysresulting from the other machines of M 0 and the job orders



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Technical realization reschedulingFor a chosen machine l M 0 \ { k } do:

remove the arcs corresponding to the selection on machine l from G

call new graph G calculate values r ij , q ij in graph G reschedule machine l according to the optimal schedule of thesingle machine head-body-tail problem



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Technical realization rescheduling - Example

M 0 \ { k } = {M 3}, thus l = M 3removing arcs corresponding to M 3 leads to graph G :

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

3 3

2 4 1

0 *

Conjunctive arcs Selection M 1

C max (G ) = 8;

Shifting Bottleneck Heuristic for Job ShopTechnical realization rescheduling - Example

M 0 \ { k } = {M 3}, thus l = M 3


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removing arcs corresponding to M 3 leads to graph G :

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

33

2 4 1

0 *


C max (G ) = 8;(i , j ) (3, 1) (3, 2) (3, 3)r ij 0 3 7q ij 3 0 0p ij 4 3 1

Shifting Bottleneck Heuristic for Job ShopTechnical realization rescheduling - Example

M 0 \ { k} = {M3} thus l = M3


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M \ { k } = {M 3}, thus l = M 3

removing arcs corresponding to M 3 leads to graph G :3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

33

2 4 1

0 *


C max (G ) = 8;(i , j ) (3, 1) (3, 2) (3, 3)r ij 0 3 7q ij 3 0 0p ij 4 3 1

33

M 3

5 8

3,1 3,2 3,3

f (M 3) = 8


Technical realization rescheduling Example (cont )


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Technical realization rescheduling - Example (cont.)

add (3 , 1) (3, 2) (3, 3) to G New graph:

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

3

3

2 4 1

0 *


C new max (M 0) = 8


Heuristic: summary


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Heuristic: summary1

Initialization:1 M 0 := ;2 G := graph with all conjunctive arcs;3 C max (M 0) := length longest path in G :

2 Analyze unscheduled machines:

FOR ALL i M \ M 0 DO

FOR ALL operation (i , j ) DO1 r ij := length longest path from 0 to ( i , j ) in G ;

2 q ij := length longest path from ( i , j ) to in G ;solve single machine head body tail problem f (i )



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Heuristic: summary (cont.)3 Bottleneck selection:

1 determine k such that f (k ) = max i M \ M 0 f (i );2 schedule machine k according to the optimal solution in Step

2;3 add corresponding disjunctive arcs to G ;4 M 0 := M 0 {k };


Heuristic: summary (cont.)


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4 Resequencing of machines:

FOR ALL i M 0 \ { k } DO

1 delete disjunctive arcs corresponding to machine k from G ;2 FOR ALL operation (i , j ) DO

1 r ij := length longest path from 0 to ( i , j ) in G ;2 q ij := length longest path from ( i , j ) to in G ;

3 solve single machine head body tail problem f (i )4

insert corresponding disjunctive arcs to G

;5 Stopping condition

IF M 0 = M THEN Stop ELSE go to Step 2;


SBH - Example (cont.)


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p ( )M 0 = {M 1 , M 3}; thus M 2 is bottleneckgraph G :

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

33

2 4 1

0 *


C max (G ) = 8;

Shifting Bottleneck Heuristic for Job ShopSBH - Example (cont.)

M 0 = {M 1 , M 3}; thus M 2 is bottleneck


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graph G :

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

33

2 4 1

0 *


C max (G ) = 8;(i , j ) (2, 1) (2, 3)r ij 4 0q ij 1 5p ij 2 2

Shifting Bottleneck Heuristic for Job ShopSBH - Example (cont.)

M 0 = {M 1 , M 3}; thus M 2 is bottleneck


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graph G :

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

33

2 4 1

0 *


C max (G ) = 8;

(i , j

) (2,1) (2

,3)r ij 4 0

q ij 1 5p ij 2 2

2,3 2,1

15

M 2

5 8

f (M 2) = 7


SBH Example (cont )


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Situation after Step 3: M 0 = {M 1 , M 2 , M 3}, C max (M 0) = 8Graph G :

3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

33

2 4 1

0 *



Situation after Step 3: M0 = {M M M } C (M 0) = 8


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Situation after Step 3: M 0 = {M 1 , M 2 , M 3}, C max (M 0) = 8

Graph G :3,1 2,1 1,1

1,2 3,2

2,3 1,3 3,3

4 2 1

33

2 4 1

0 *

Corresponding Schedule:

5 1080

M 1

M 2

M 3


An Important Subproblem


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An Important Subproblem

within the SBH the one-machine head-body-tail problemoccurs frequently:this problem was also used within branch and bound tocalculate lower bounds

the problem is NP-hard (see Lecture 3)there are efficient branch and bound methods for smallerinstances (see also Lecture 3)the actual one-machine problem is a bit more complicated

than stated in Lecture 3 (see following example)

Shifting Bottleneck Heuristic for Job ShopExample Delayed Precedences

Jobs: (1, 1) (2, 1)

(2 2) (1 2)


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(2, 2) (1, 2)

(3, 3)(3, 4)


Example Delayed Precedences (cont.)


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after 2 iterations SBH we get:M 0 = {M 3 , M 1}; (3, 4) (3, 3) and (1 , 2) (1, 1)Resulting graph G : (C max (M 0) = 8)

1,1 2,1

2,2 1,2

3,3

3,4

1 1

1 1

4

4

0 *


Example Delayed Precedences (cont )


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3. iteration: only M 2 unscheduled(i , j ) p ij r ij q ij (2, 1) 1 3 0(2, 2) 1 0 3

Possible schedules for M 2:

50 8M 2

50 8M 2




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Both schedules are feasible and might be added to currentsolutionBut: second schedule leads to

1,1 2,1

2,2 1,2

3,3

3,4

1 1

1 1

4

4

0 *

which contains a cycle


Delayed Precedences


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The example shows:not all solutions of the one-machine problem t to the givenselections for machines from M 0the given selections for machines from M 0 may induceprecedences for machines from M \ M 0

Example:scheduling operation (1 , 2) before (1, 1) on M 1 induces adelayed precedence constraint between (2 , 2) and (2 , 1) of length 3 operation (2 , 1) has to start at least 3 time units after (2 , 2)

this time is needed to process operations (2,

2),

(1,

2), and(1 , 1)


Rescheduling Machines


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Rescheduling Machines

after adding a new machine to M 0, it may be worth to putmore effort in rescheduling the machines:

do the rescheduling in some specic order (e.g. based on theirhead-body-tail values)repeat the rescheduling process until no improvement is foundafter rescheduling one machine, make a choice which machineto reschedule next (allowing that certain machines arerescheduled more often)...

practical test have shown that these extra effort often pays off

Uni Bonn (Germany) Sched10_5

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