7030 Job Shop Scheduling

8/3/2019 7030 Job Shop Scheduling

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NSY 7030 - Manufacturing Systems Design 3/22/20

Industrial and Systems Engineering

Job Shop Scheduling

3/22/2004 Job Shop Scheduling 2

Scheduling Allocation of limited resources to jobs over time

Sequencing

Permutation of the job set (order)

Scheduling

Assigning start/end/preemption times to individual jobs on each

machine


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Pinedo, p. 5


Notation / Definitions General Notation

N number of jobs to be scheduled

M Number of machines

di due date of job i

pij processing time for job ion machinej(includes any sequence independent

setups)

ri ready time for job i

Performance Measures

Ci Completion time for job i

Fi length of time job iis in the shop

Li lateness of job i

Ti tardiness of job i

Common Objectives

min average flow time

min makespan

min average tardiness

min max tardiness

min number of tardy jobs


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Single-machine Scheduling

No sequence-dependentsetups:

Facts:

Cmax is independent ofsequence

Average flowtime is

minimized by an SPT

sequence

Lmax is minimized by an EDD

sequence

Book presents a simple

construction procedure to

minimize the average flowtime

subject to tardiness

constraints

Sequence-dependent setups

Cmax now depends on the

sequence, since the individual

process times depend on a

sequence.

3/22/2004 Sequence-dependent Setup Times 6

Sequence-dependent Setup Times Njobs on one machine

Total processing time is fixed

Processing time is independent of sequence

Total setup time depends on the sequence

Individual setup times depend on the current and next products

Part 1 2 N

1 c12 c1N2 c21 c2N

N cN1 cN2

cij is the setup time required

for productjgiven that product i

was just produced on the machine


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Sequence-dependent Setup Times

Traveling Salesperson Problem (TSP) Formulation Salesperson must visit each city in a territory and return to the

starting point

Tour is a complete cycle that visits each city

Tour length is the sum of the individual arc lengths

Want to find the shortest length tour

1

2 3

4 5

c13c21

c52 c34

c45

Tour length = c13 + c34 + c45 + c52 + c21


Task Sequencing

minimize

subject to:

no subtours

or

c x

x j

x i

x

ij ij

j

N

i

N

ij

i

N

ij

j

N

ij

==

=

=

=

=

=

11

1

1

1

1

0 1

TSP Formulation

xij = 1 if cityj is visited immediately after city i


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TSP Formulation

Subtour Return to a city prior to visiting all other cities

Can lead to unconnected subgraphs

Considerably complicates the formulation

Without the subtour constraint, problem reduces to LAP

LAP provides a lower bound, but it is relatively weak

1

2 3

4 5


Job Shop Scheduling Environment

nparts to be processed on mmachines

sequence of machines for each part is prescribed (operation or job)

each machine can process only one job at a time

each part visits each machine at most once (no recirculation)

p4

M1

M3

M2

M4

p1 p2

p3

p4

p1

p1

p1

p2

p2 p3

p3

p3

p3

p4


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Dispatching

Askinand Standridge, p.115

Work in next queueWINQSelect a job at randomRANDOM

Most work remainingMWKR

Most operations remainingMOPNR

Least work remainingLWKR

Least total workLTWK

Order jobs based on ration of slack-based priority to

processing time

Covert

Slack per remaining operation Select a job with the smallest

ratio of slack to operations remaining to be performed

S/RO

First in system, first servedFISFS

First come, first servedFCFS

Earliest due dateEDDShortest processing timeSPT


Job Shop Scheduling Shifting Bottleneck Procedure (Adams et al., 1988)

Basic approach is to iteratively solve a single machine

scheduling problem and propagate the constraints imposed on

other machines

Sequencing one machine will fix ready times and due dates for

other machines

Questions

Which machine to choose?

How do you sequence that machine?

How many iterations to perform?


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Job Shop Scheduling

Problem Representation N = {0, 1, 2, , n} is a set of operations

where 0 and nare dummy operations

M is the set of machines

A is the set of pairs of operations (i,j) where operation iprecedes operationj(precedence constraints)

Ek is the set of operations performed on machine k

di is the processing time for operation i

ti is the start time for operation i


Job Shop Scheduling Assume objective is to minimize makespan

Formulation

minimize

subject to:

t

t t d i j

t i

t t d t t d i j k

n

j i i

i

j i i i j j k

, ( , )

,

, ( , ) ,

A

N

E M

0

Disjunctive Graph Representation G = (N, A, E) N - nodes (one for each operation)

A - normal arcs (task precedence)

E -disjunctive arcs(precedence caused by machine capacity constraints)


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Job Shop Scheduling

Disjunctive Graph ExampleJob Sequence

machine (node no.)

1 1 (1) 2 (2)

2 1 (3) 3 (4) 2 (5)

3 1 (6) 2 (7) 4 (8)

4 1 (9) 4 (10) 3 (11)

5 4 (12) 3 (13)

0

4

6

5

7 8

9 10

1312

3

21

14

11

precedence constraints


Disjunctive Graph Representation The set E decomposes in to cliques, Ek, one for each machine k,

where the set of cliques covers E

Sk is a selectionin Ekand contains exactly one member of eachdisjunctive arc pair in Ek

6

9

3

1

seq. = 3, 1, 6, 9

Each acyclic selection Skcorrespondsto a unique sequence of the operationsperformed on machine k

Some arcs in the selection giveredundant information

Sequencing a machine corresponds to

choosing an acyclic selection in Ek


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Disjunctive Graph Representation

A complete selection, S, is the union of selections Sk, one for eachEk

A partial selectionis taken over some subset M0 ofM

Every acyclic complete selectionS, defines a familyof schedules andevery schedule belongs to such a family

DS

= (N, A S) is an ordinary graph obtained by picking a complete selection S

replacing the disjunctive arc set E by an ordinary arc set S

The makespanof a schedule that is optimal forS is equal to thelength of a longest path in D

S

Thus, the objective is to find an acyclic complete selectionS thatminimizes the length of a longest path inD

S


Shifting Bottleneck Procedure Given a partial selection S = (Sk | kM0) and the corresponding

graph DS, machine kis criticalwith respect to the schedule

associated with S ifSkhas some arc on the longest path in DS This is a direct result of the fact that any schedule better than the one

associated with S uses a selection in which at least one arc of every

longest path in DS

is reversed (Balas, 1969)

Our basic strategy is to iteratively select and optimally schedule the

bottleneck machine

However, given our definition of criticality, there can be multiple critical

machines -- as a result, we need to quantifythe criticality


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U Industrial and Systems Engineering 1


Shifting Bottleneck Procedure

Define M0 as the set of machines alreadysequenced by choosing selections Sp, p M0 (M0 = initially)

For any k M\M0, we want to define a new singlemachine problem by

replacing each disjunctive arc set Ep, p M0, by thecorresponding selection Sp

deleting each disjunctive arc set Ep, p M\M0, p k This assumes that the associated machines can process jobs in

parallel (i.e., they have infinite capacity)


Single Machine Subproblem Example

Let k=2 and M0={1} with S1= {3, 1, 6, 9}

0

4

6

5

7 8

9 10

1312

3

21

14

11


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Single Machine Subproblem

Formulation P(k,M0)

minimize

subject to:

t

t t d i j

t i

t t d t t d i j

n

j i i p

p

i

j i i i j j k

, ( , )

,

, ( , )

S A

N

E

M 0

0

To generate this problem, all disjunctive arc pairs are

removed for all machines not scheduled except machine k

The makespanof this relaxed problem is equal to the lengthof a longest path through the associated graph


Single Machine Subproblem This problem can be viewed as a single machine

problem (for machine k) with ready times and due datesand an objective of minimizing maximum lateness

Ready time is the length of a longest path from node 0 to nodej

rj= L(0,j)

Due date is the latest completion time that will not increase the

maximum lateness

dj= L(0, n) - L(j, n) + pj


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Shifting Bottleneck Procedure

The bottleneck machineis defined as the machine withthe maximum solution to (P(k,M0)).

If v(k,M0) is an optimal solution to (P(k,M0)), then mis the

bottleneck among M\M0 ifv(m,M0)=max{v(k,M0) | k M\M0}

Procedure:

Step 1: Identify a bottleneck machine mamong machines kM\M0 and sequence it optimally. Set M0 M0 {m}.

Step 2: Reoptimize the sequence of each critical machine, lM0, in turn, while keeping the other sequences fixed; that is, set

M0=M0 - {k} and solve P(k,M0)

Step 3: ifM0 = M, stop; otherwise, goto Step 1


Shifting Bottleneck Example

Sequence (m/c, time) node

Job 1 2 3 4

1 (1, 10) 1 (2, 8) 2 (3, 4) 3

2 (2, 8) 4 (1, 3) 5 (4, 5) 6 (3, 6) 7

3 (1, 4) 8 (2, 7) 9 (4, 3) 10

0 4 65 7

8 9 10

321

11

10 84

8 3 5 6

4 73

0

0

0


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Iteration 1. M0 =

0 4 65 7

8 9 10

321

11

10 84

8 3 5 6

4 73

0

0

0

Cmax = 22 (parts 1 & 2)



Machine 1: {1, 5, 8}rj = L(0,j)

dj = L(0, n) - L(j, n) +pjOp rj pj dj1 0 10 22-22+10 = 10

5 8 3 22-14+3 = 11

8 0 4 22-14+4 = 12


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0 4 65 7

8 9 10

321

11

10 84

8 3 5 6

4 73

0

0

0 10

3

M0 ={1}

Cmax{1} = Cmax{} +Lmax{1} = 22+5 = 27Path {1, 5, 8, 9, 10, 11}



Machine 2: {2, 4, 9}rj = L(0,j)


4 0 8 27-25+8 = 10

9 17 7 27-10+7 = 24

Seq max lateness

2, 4, 9 max{-5, 16, 1} = 16

2, 9, 4 max{-5, 1, 23} = 23

4, 2, 9 max{-2, -5, 1} = 1

4, 9, 2 max{-2, 0 , 9} = 9

9, 2, 4 max{0, 9 , 30} = 30

9, 4, 2 max{0, 22, 17} = 22

Lmax(2) = 1

Lmax(3) = 1

Lmax(4) = 0

machines 2&3

are bottlenecks


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0 4 65 7

8 9 10

321

11

10 84

8 3 5 6

4 73

0

0

010

3

M0 ={1,2}

Cmax{M0} = Cmax{1} +Lmax{2} = 27+1 = 28

Path {1, 2, 9, 10, 11}

88



Reoptimize Machine 1: {1, 5, 8}rj = L(0,j)


5 8 3 28-14+3 = 17

8 0 4 28-14+4 = 18

Lmax(1) = 0

sequence {1, 5, 8}

no changego to iteration 3


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Iteration 3

Neither machine 3 or machine 4 is a bottleneck

Final solution

M1: {1, 5, 8}

M2: {4, 2, 9}

M3: {7, 3}

M4: {6, 10}

0 4 65 7

8 9 10

321

11

10 84

8 3 5 6

4 73

0

0

010

3

88 6

5


Schedule Generation

*

Job Shop Scheduling Environment

Basic Idea

Create a list of schedulable operations

Based on operation precedence and machine capacity

While the list is not empty

Schedule an operation

Update the list

*This material adapted from Storer et al. 1992 and Leon and Balakrishnan 1995.


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Schedule Generation Example

Operation (m/c, time)

Part 1 2 3

1 (1, 5) (2, 2) (3, 3)

2 (1, 4) (3, 3) (2, 4)

3 (2, 8) (1, 3) (3, 4)

Number of possible schedules (3!)3 = 216

This number grows quickly, e.g., (10!)3 = 4.77x1019




Part 1 2 3

1 (1, 5) (2, 2) (3, 3)

2 (1, 4) (3, 3) (2, 4)

3 (2, 8) (1, 3) (3, 4)

Idea is to create a Gantt chart that represents the schedule

1,1M1

M2

M35 8

2,1 3,2

3,1 1,2

2,2

2,3

1,3 3,39 10 12 13 15 16 19


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Schedule Generation Definitions

PSt is the partial schedule at time t

is defined as the delay parameter = 0 generates an active schedule

= 1 generates a non-delay schedule

Pj is the set of predecessors for operationj

dj is the duration of taskj

mj is the number of predecessors already scheduled (0 mj |Pj|)Cj is the completion time of operationj

Tj is the earliest time the machine required for operationjis available

j is the earliest start time for operationjj= max{max{Ck|k Pj}, Tj}

St is the set of schedulable operations


Schedule Generation Algorithm Step 1

t= 0

PSt= NULL (empty partial schedule)

Tj= 0; mj= 0 j St= {j| |Pj| = 0} as the set ofschedulable operations j= 0 j St and j= M j St

Step 2

X* is the cutoff start time for updating the potential operations list. It

ranges from the earliest start time to the earliest completion

( ){ }X dj S

j jt

* min= +

1


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Schedule Generation Algorithm

Step 3 Set st = {j|j St, j< X*}, if 0 < 1

if = 0, X* is the minimum completion time

Set st= {j|j St, j= X*}, if = 1 X* is the minimum start time

Step 4

Selectj* stusing a heuristic H (e.g., SPT or LPT) Addj* to PSt to form PSt+1


Schedule Generation Algorithm Step 5 -- Update data

Cj* = j* + dj* Tj* = Cj* mk= mk+1 kwherej* Pk t= t+ 1

St= {j| mj= |Pj|,j PSt}; if NULL, stop j= max{max{Ck | k Pj}, Tj} j PSt

Step 6

Go to Step 2


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Operation (m/c, time) node

Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 0 - 0 (1)

2 {1} 2 M - 0 (2)

3 {2} 3 M - 0 (3)

4 - 4 0 - 0 (1)

5 {4} 3 M - 0 (3)

6 {5} 4 M - 0 (2)

7 - 8 0 - 0 (2)

8 {7} 3 M - 0(1)9 {8} 4 M - 0 (3)

Assume = 1




Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 0 - 0 (1)

2 {1} 2 M - 0 (2)

3 {2} 3 M - 0 (3)

4 - 4 0 - 0 (1)

5 {4} 3 M - 0 (3)

6 {5} 4 M - 0 (2)

7 - 8 0 - 0 (2)

8 {7} 3 M - 0(1)

9 {8} 4 M - 0 (3)

S0 = {1, 4, 7}

X* = min{0, 0, 0} = 0

s0 = {1, 4, 7}

select operation 4 by SPTPS0 = {4}


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Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 4 - 4 (1)

2 {1} 2 M - 0 (2)

3 {2} 3 M - 0 (3)

4 - 4 - 4 4 (1)

5 - 3 4 - 0 (3)

6 {5} 4 M - 0 (2)

7 - 8 0 - 0 (2)8 {7} 3 M - 4 (1)

9 {8} 4 M - 0 (3)




Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

S1 = {1, 5, 7}

X* = min{4, 4, 0} = 0

s1 = {7}

PS1 = {4,7}

Op pred dj j Cj Tj1 - 5 4 - 4 (1)

2 {1} 2 M - 0 (2)

3 {2} 3 M - 0 (3)

4 - 4 - 4 4 (1)

5 - 3 4 - 0 (3)6 {5} 4 M - 0 (2)

7 - 8 0 - 0 (2)

8 {7} 3 M - 4 (1)

9 {8} 4 M - 0 (3)


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Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 4 - 4 (1)

2 {1} 2 M - 8 (2)

3 {2} 3 M - 0 (3)

4 - 4 - 4 4 (1)

5 - 3 4 - 0 (3)

6 {5} 4 M - 8 (2)

7 - 8 - 8 8 (2)

8 - 3 8 - 4 (1)

9 {8} 4 M - 0 (3)

S2 = {1, 5, 8}

X* = min{4, 4, 8} = 4

s2 = {1, 5}

select operation 5 by SPT

PS2 = {4,7, 5}




Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 4 - 4 (1)

2 {1} 2 M - 8 (2)

3 {2} 3 M - 7 (3)

4 - 4 - 4 4 (1)

5 - 3 - 7 7 (3)

6 - 4 8 - 8 (2)

7 - 8 - 8 8 (2)

8 - 3 8 - 4 (1)

9 {8} 4 M - 7 (3)

S3 = {1, 6, 8}

X* = min{4, 8, 8} = 4

s3 = {1}

PS3 = {4, 7, 5, 1}


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Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 - 9 9 (1)

2 - 2 9 - 8 (2)

3 {2} 3 M - 7 (3)

4 - 4 - 4 4 (1)

5 - 3 - 7 7 (3)

6 - 4 8 - 8 (2)

7 - 8 - 8 8 (2)

8 - 3 9 - 9 (1)

9 {8} 4 M - 7 (3)

S4 = {2, 6, 8}

X* = min{9, 8, 9} = 8

s4 = {6}

PS4 = {4, 7, 5, 1, 6}




Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 - 9 9 (1)

2 - 2 12 - 8 (2)

3 {2} 3 M - 7 (3)

4 - 4 - 4 4 (1)

5 - 3 - 7 7 (3)

6 - 4 - 12 12 (2)

7 - 8 - 8 8 (2)

8 - 3 9 - 9 (1)

9 {8} 4 M - 7 (3)

S5 = {2, 8}

X* = min{12, 9} = 9

s5 = {8}

PS5 = {4, 7, 5, 1, 6, 8}


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Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 - 9 9 (1)

2 - 2 12 - 0 (2)

3 {2} 3 M - 0 (3)

4 - 4 - 4 4 (1)

5 - 3 - 7 7 (3)

6 - 4 - 12 12 (2)

7 - 8 - 8 8 (2)

8 - 3 - 12 12 (1)

9 - 4 12 - 0 (3)

S6 = {2, 9}

X* = min{12, 12} = 12

s6 = {2, 9}

select operation 2 by SPT

PS6 = {4, 7, 5, 1, 6, 8, 2}




Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 - 9 9 (1)

2 - 2 - 14 14 (2)

3 - 3 14 - 0 (3)

4 - 4 - 4 4 (1)

5 - 3 - 7 7 (3)

6 - 4 - 12 12 (2)

7 - 8 - 8 8 (2)

8 - 3 - 12 12 (1)

9 - 4 12 - 0 (3)

S7 = {3, 9}

X* = min{14, 12} = 12

s7 = {9}

select operation 2 by SPTPS7 = {4, 7, 5, 1, 6, 8, 2, 9}


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Part 1 2 3

1 (1, 5) 1 (2, 2) 2 (3, 3) 3

2 (1, 4) 4 (3, 3) 5 (2, 4) 6

3 (2, 8) 7 (1, 3) 8 (3, 4) 9

Op pred dj j Cj Tj1 - 5 - 9 9 (1)

2 - 2 - 14 14 (2)

3 - 3 16 19 0 (3)

4 - 4 - 4 4 (1)

5 - 3 - 7 7 (3)

6 - 4 - 12 12 (2)

7 - 8 - 8 8 (2)

8 - 3 - 12 12 (1)

9 - 4 - 16 16 (3)

PS8 = {4, 7, 5, 1, 6, 8, 2, 9, 3}




Part 1 2 3

1 (1, 5) (2, 2) (3, 3)

2 (1, 4) (3, 3) (2, 4)

3 (2, 8) (1, 3) (3, 4)

Final Solution

1,1M1

M2

M38

2,1 3,2

3,1 1,2

2,2

2,3

1,33,39 12 14 16 194 7


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Problem Space-based Search with

Schedule Generation


Reference Leon, V. J. and R. Balakrishnan, Strength and

Adaptability of Problem-Space Based Neighborhoods for

Resource-Constrained Scheduling, OR Spektrum, Issue2/3, Vol. 17, 1995.

Resource-Constrained Scheduling (RCS) Similar to thejob-shop scheduling problem except that each activity

may require multiple resources.


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RCS Formulation

Notation:

Ci - Completion time

for task i

di Duration of task i

bk Number of units

of resource k

Dt Set of activities in

process at time t

Zis a regular measureof performance

A regular measureis

non-decreasing incompletion time of the

activities

{ }

1 2minimize ( , , , )

subject to:

max ,

, ,

j

k

n

i i jj P

k ik

i D

Z C C C

C d C i

b B k t

max{ | 1,2, , }

1 max{0, }

i

i i

i

Z C i n

Z C d n

= =

=

Consider makespan and mean tardiness


RCS Solution with Local Search

If we assume that each activity requires exactly one resource and

that there is one unit of each resource available, we have our job

shop problem.

Local Search

Given an initial feasible solution and the corresponding value of the

objective function, local search consists of constructing similar

solutions, evaluating them, moving to more promising regionsof the

neighborhood, and repeating until no further improvement is found (or

until the procedure runs out of available time).

Two critical issues for local search

Where to search generation of the neighborhoods

How to search finding similarsolutions in the neighborhood.


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Local Search

Characteristics of neighborhoods Strength

A neighborhood is strongif solutions in it are close to optimal

Adaptability

A neighborhood is adaptableif the procedure used to generate it

can easily be modified to accommodate changes in the

performance measure.

Common local search procedures

Simulated annealing

Genetic algorithms

Tabu search

Problem space-based search focus of Leon and Balakrishnan


Local Search Characteristics of local search procedures:

Neighbors are generated by perturbing a feasible solution

A neighborhood of certain size has a certain strength,

depending on the quality of the solutions

One either moves to a new neighborhood as soon as a better

solution is found (first improvement) or after the complete

neighborhood has been evaluated (steepest descent).

The search of neighboring solutions either follows a

lexicographical ordering induced by indexing, or a random

ordering.


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Perturbing the Solution

Many local search procedures involve finding a feasible

solution (like the above Gantt chart) and perturbingthesolution to try to improve the objective function value.

Problem-Space Based search perturbs the problemrather than the solution.

1,1M1

M2

M35 8

2,1 3,2

3,1 1,2

2,2

2,3

1,3 3,39 10 12 13 15 16 19


Problem-Space Based Search Define (h, p) as a heuristic-problem pair where:

his a fast problem-specific heuristic (like schedule generation),and

pis a specific problem instance.

(h, p) is an encodingof a specific solution

(h, p) = (schedule generation with SPT, specific problem)

(h, p) solution (Gantt chart with the associated makespan)

We want to perturb the original problem data to define a

neighborhood of solutions (by applying hto a

neighborhood of problems).


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Problem-Space Based Search

Let p0 be a vector of the original processing times. (h, p0) gives the initial solution

Define a neighborof p0:

p= p0 + , where is a pertubation vector, and

p= {pj= pj0+j,j= 1, 2, , n}

Define:

Np= {p+ | |||| < } as a problem space neighborhood

Ns

= {h(q) | qNp

} as a neighborhood of solutions


Problem-Space Based Search Job shop scheduling:

Let hbe schedule generation with SPT

Define the pertubation operation as:

where is a constant0 0 0UNIFORM( , )i i i ip p p p = +


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Search Methods

Simple Neighborhood Search1. Sol = h(p0)

2. pI = p0

3. for (i = 0; i < NBRS; i++)

4. p = p0 +

5. if (h(p) < Sol)

6. Sol = h(p)

7. pI = p

8. endif

9. endfor


Search Methods Steepest Descent Search

1. for (i = 0; i < NHOODS; i++)

2. Sol = h(p0)

3. pI = p0

4. for (j = 0; j < NBRS; j++)

5. p = p0 +

6. if (h(p) < Sol)

7. Sol = h(p)

8. pI = p

9. endif

10. endfor

11. p0 = pI

12. endfor


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Search Methods

First Improvement Search1. Sol = h(p0)

2. pI = p0

3. for (i = 0; i < Iterations; i++)

4. p = pI +

5. if (h(p) < Sol)

6. Sol = h(p)

7. pI = p

8. endif

9. endfor


Problem-Space Based Search Wanted to investigate the use ofsimpleintuitive

heuristics to generate neighborhoods

Using the preceding search techniques, Leon and

Balakrishnan found the following keys to getting good

solutions:

1. Having a good starting point, and

2. Investigating a large number of neighbors of the starting

point.

In order to achieve these keys, hmust encode problem-

specific data and it must be fast

Also developed and tested a genetic algorithm


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Computational Testing

Objectives of the computational study: Evaluate the quality of the proposed problem-space based

neighborhoods

Evaluate the improvements achievable with a genetic algorithm

Evaluate the adaptability of the proposed neighborhoods to

different objective functions

Tested using 110 problems from the literature. Optimal

solutions were known for all test problems.


Conclusions Leon and Balakrishnan reported the following

conclusions to their study.

1. The quality of the search neighborhoods was found to be good

and the solutions obtained were close to optimal (strongneighborhoods)

2. Minimal improvements in the solution quality were achieved

using the genetic algorithm

3. The adaptability of the search procedures were established by

experimenting with makespanand mean tardinessas themeasures of performance

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