General Scheduling Model.ppt - University of Iowauser.engineering.uiowa.edu/~coneng/lectures/General_Scheduling_Model.pdf · General Scheduling Model and Algorithm Andrew Kusiak Intelligent

General Scheduling Modeland

Algorithmg

Andrew KusiakIntelligent Systems Laboratory

2139 Seamans Center The University of Iowa

I Cit I 52242 1527

The University of Iowa Intelligent Systems Laboratory

Iowa City, Iowa 52242 - 1527Tel: 319 - 335 5934 Fax: 319 - 335 5669

[email protected]://www.icaen.uiowa.edu/~ankusiak

MPS

MRP

Balancing

MANUFACTURING SCHEDULING

n Jobs (Operations)Schedulingm Machines

+ other resources

Scheduling ScheduleSchedulingData


SchedulingAlgorithm

DataNo of operationsProcessing timePrecedence constraintsDue time

2 4 1 3 6M1 5

Example

6 11 18 22 25 292 4 1 3 61 5

6 14 18 22 30

2 4 5M2

1 63

25118

2M3 Time16

34 5


18 30248 35

What type of schedule is it -jobshop or flowshop?

Scheduling Model: m Machines and n Operations

n = number of partsm = number of machinesm number of machinesRi = set of pairs of operations [k, l] for part Pi, where operation k precedes

operation l, i = 1,..., nQi = set of pairs of operations [k, l] for part Pi, where k and l can be

performed in any order, i = 1,..., nIi = set of operations without precedence constraints, i = 1,...., n.Np = set of operations to be performed on machine p, p = 1,..., mni = number of operations in part Pi, i = 1,..., nil i i f i l f i l 1 i 1


til = processing time of operation l of part Pi, l = 1,..., n , i = 1,..., nM = an arbitrary large positive number

xik = completion time of operation k of part Pi, k = 1,..., n , i = 1,..., nx[i] = completion time of the final operation of part Pi,

i = 1,..., n

1 if operation k precedes operation lykl = {

0 otherwise

for all [k, l] ∈ Qi, i = 1,..., n.

1 if operation k precedes operation lzkl = {


zkl = {0, otherwisefor all [k, l] ∈Np, p = 1,..., m

Basic Notation

P t iPart i:Operations k, l,...Precedence constraints

k l

xik xilCompletion time


Part j

The ModelnMin x[i] Min the total completion time

i =1subject to:

∑

xil - xik til for all [k, l] ∈ Ri, for all i

Operations k and l of part i are processed according to the precedence required

≥

k lP t i


k lPart i

xik xilk l

til

TimeSchedule

xil - xik + M(1- ykl ) tilfor all [k, l] ∈ Qifor all i

xik - xil + Mykl tik≥

≥

Two operations of part i can not be processed at the same time

Part i k l l k

TWO POSSIBILITIES (Same part)

OR


xik xilk l

till k

xil xikl k

tikTime

OR

xjl - xik + M(1- zkl ) tjl for all [k, l] ∈ Np for all p

xik - xjl + Mzkl tik i j

≥

≥ ≠

where Np = Set of operations to be performed on machine p A machine can not process more than one part at the same time

TWO POSSIBILITIES (Different parts)

Part i Part j Part j Part i


Part i

xik xjlk l

til

xjl xikl k

tikTime

ORPart j Part j Part i

≥tik for [i, k ] ∈ Ii, i =1,..., n

xik {0 for all other i, k

ktik

xik

Part iTime For operations without

precedence constraints


ykl = 0, 1 for all [k, l] ∈ Qi, for all i

zkl = 0, 1 for all [k, l] ∈Np, for all p

!!!!

Go over this model at least twice!


Example Structure of Three PartsPart P

1, 1 1, 2 1, 3

1

Part P

2, 1 2, 2

2

Part P

3 1

3


3, 1

3, 3

3, 2

Scheduling Data

______________________________________________P

______________________________________________Partnumber 1 2 3

______________________________________________Operationnumber 1 2 3 1 2 1 2 3

______________________________________________Machinenumber 1 1 2 1 2 2 1 2


______________________________________________Processingtime 3 5 6 8 4 9 2 7______________________________________________

LINDO Input FileMIN X13+X21+X22+X33SUBJECT TOX12-X11 >= 5X13-X12 >= 6X33 X31 > 7

Operations of part i are processed according to the precedencerequired

Part P

1, 1 1, 2 1, 3

1

Part P 2X33-X31 >= 7X33-X32 >= 7

X21-X22+999Y1 >= 8X22-X21- 999Y1 >= -995X31-X32+999Y2 >= 9

xil - xik til

Two operations of the same part i can not be processed atthe same time

≥2, 1 2, 2

Part P

3, 1

3, 3

3

3, 2


X32-X31- 999Y2 >= -997the same time

xik - xil + Mykl tikxil - xik + M(1- ykl) til

≥≥

X22-X21 + 999(1-Y1) >= 4

Derived from

______________________________________________Partnumber 1 2 3


______________________________________________Machine

b 1 1 2 1 2 2 1 2

X11-X21+999Z1 >= 3X21-X11- 999Z1 >=-991X11-X32+999Z2 >= 3

A machine can not process more than one part at the same time

number 1 1 2 1 2 2 1 2____________________________ _________________

Processingtime 3 5 6 8 4 9 2 7______________________________________________


X32-X11- 999Z2 >= -997X12-X21+999Z3 >= 5X21-X12- 999Z3 >= -991X12-X32+999Z4 >= 5X32-X12- 999Z4 >= -997

time

xik - xjl+ Mzkl tikxjl - xik + M(1- zkl) tjl

≥≥

X21 - X32 + 999Z5 >= 8X32 - X21 - 999Z5 >= -997

X13-X22+999Z6 >= 6X22-X13 -999Z6 >= -995X13-X31+999Z7 >= 6X31-X13-999Z7 >= -990

Partnumber 1 2 3


______________________________________________Machine

b 1 1 2 1 2 2 1 2

A machine can not process more than one part at the same

X13-X33+999Z8 >= 6X33-X13-999Z8 >= -992X22-X31+999Z9 >= 4

number 1 1 2 1 2 2 1 2____________________________ _________________

Processingtime 3 5 6 8 4 9 2 7______________________________________________


timexjl - xik + M(1- zkl) tjlxik - xjl+ Mzkl tik

≥≥

X22 X31+999Z9 > 4X31-X22-999Z9 >= -990X22-X33+999Z10 >= 4X33-X22-999Z10 >= -992

X11 >= 3X21 >= 8X22 >= 4

Operation finish time t ll th itX22 >= 4

X31 >= 9X32 >= 2END

not smaller than its processing time


INTEGER Y1INTEGER Y2INTEGER Z1INTEGER Z2INTEGER Z2INTEGER Z3INTEGER Z4INTEGER Z5INTEGER Z6INTEGER Z7INTEGER Z8INTEGER Z9


INTEGER Z10

The Optimal Schedule

x32 = 2 , x21 = 12, x11 = 15, x12 = 20, x22 = 4, x31 = 13, x33 = 20, and x13 = 26, , ,

2 12 15 20

M4

3,2 2,1 1,1 1,21


4 13 20

M2 Time

262,2 3,1 3,3 1,3

0

m = number of resource types

rs = number of resources of type s, s = 1,..., m

Scheduling Model with Multiple Resources

d[i] = due date of part number i

Nq = set of operations using resource q , q = 1,..., rs , s = 1,..., mn

min x[i]i=1

bj

∑


subject to:xil - xik til for all [k, l] ∈ Ri, for all i

xil - xik + M(1- ykl ) til for all [k, l] ∈ Qi

xik - xil + Mykl tik for all i

≥

≥

≥

xjl - xik + M(1- zkl ) tjl for all [k, l] ∈ Nq,for all q,

≥

xik - xjl + Mzkl tjl i j

x[i] d[i] for all i

tik for [i, k] ∈ Ii, i = 1,..., nxik {

0 for all other i, k

≥

≥

≤

≠


ykl = 0, 1 for all [k, l] ∈ Qi, for all Izkl = 0, 1 for all [k, l] ∈ Nq, for all q

ExampleStructure of Three Parts

Part P

1, 1 1, 2

1

Part P

2, 1 2, 2

2

Part P 3


3, 1

3, 3

3, 2

Scheduling Data___________________________________________

Partnumber 1 2 3u be 3

___________________________________________Operationnumber 1 2 1 2 1 2 3

___________________________________________Machinenumber 1 2 2 1 1 2 1

___________________________________________Processingtime 2 3 1 2 3 2 1


___________________________________________Toolnumber 1 1 2 2 2 1 2

___________________________________________

The Optimal Schedule with Limited Number of Tools

(1) (2) (2) (2)

7

M

M2 Time

1, 1 2, 2 3, 3

2, 1 1, 2 3, 2

0

1

10

3, 1

1 2 5 9

(2) (1) (1)


3C = 10

i, j(k)

Operation j of uses tools k

The Optimal Schedule with Unlimited Number of Tools

(1) (2) (2) (2)(1)M

M2 Time

1, 1 2, 2 3, 3

3, 1 1, 20

1

8

3, 2

3 6

(1) (2) (2) (2)

(2)

(1)

(1)

2, 1


C' = 83

Two copies of tool 1 and 2 are needed.

General Scheduling Heuristic

Part P with6 operations

3 5

21 6

An operation is schedulable at time t if the following three conditions are satisfied:• No other operation that belongs to the same part is being processed at the time t

Definition4


processed at the time t.• All operations preceding the operation considered have been completed before the time t.

• All resources (for example, machines, tools, fixtures) required for performing the operation are available at the time t.

Two Scheduling Scenarios

• No single operation is• No single operation is schedulable (a deadlock)

• Many operations are h d l bl


schedulable

How to select the most promising operations?

PRIORITY RULES


DefinitionsPart P with 6 operations

3 5

Operations 3, 4, 5, and 6 are successive operationsof operation 2

421 6


Operations 3 and 4 are immediate successive operations of operation 2

General HeuristicStep 1. Initialize:• Current time• Set of schedulable operations• Set of completed operations

Step 2. From the set of schedulable operations, select an p poperation using the following priority rules:• P1: with the largest number of successive operations• P2: belonging to a part with the minimum number

of schedulable operations• P3: with the largest number of immediate successive

operations


• P4: belonging to a part with the largest number of unprocessed operations

• P5: with the shortest processing time• P6: belonging to a part with the shortest slack time.

Step 3. Schedule the operation selected in step 2.Update:

• The resource status

• P7: arbitrary choice

• The set of schedulable operations

If the set of schedulable operations is empty, go to step 4; otherwise, go to step 2.

Step 4. Calculate the completion time of each operation scheduled but not completed at the current time.


Set the current time equal to the completion time of the operation with the least remaining processing time. Add this operation (or operations in case of a tie) to the set of completed operations.

Update:• The resource status• The set of schedulable operations

If there is no unprocessed operations, stop; otherwise, go to step 5.

Step 5. If the set of schedulable operations is empty, go to step 4; h i 2


otherwise go to step 2.

Heuristic Structure

Operation with no

Operation belongs to a part

Manufacturingresources

Determine Schedulable Set

with noprecedences being

not-processed

resources available

Use Priority Rules: P1 - P7

Schedule an operation


UpdateCurrent time, machine status, part status

All operations are scheduledyes

Endno

Priority Rules: P1 - P7Schedulable Set

P1P2

>1 operations>1 operations

>1 tiP2P3

P4

P5P6

>1 operations>1 operations

>1 operations

>1 operations

>1 operations


P7

One Operation Selected Arbitrary select 1 operation

Priority Rule P6: Min slack time

Part P

1 2

1

Part P

3

5

2

4

Part P

6 7 8

3

M

M

3 6 82 Time

1 47

93 6 2 8

0

1 5

12P3 P3

tP2 sP2dP2


tP3dP3

sP3

Slack = Due time - Current time

ExamplePart P

1 2

1

Part P 2

Three Parts3

5

4


Part P

6 7 8

3

___________________________________________Partnumber 1 2 3

Scheduling Data

Part P 1

__________________________________________Operationnumber 1 2 3 4 5 6 7 8

__________________________________________Machinenumber 1 2 2 1 1 2 1 2

__________________________________________Machiningtime 4 2 3 2 2 3 2 1

1 2

Part P

3

5

2

4


__________________________________________Part P

6 7 8

3

Iterations of the Heuristic

Step 1. Initialize:• Current time t = 0• The set of schedulable operations S1 = {1 3 4 6 7}

• P1: with the largest number of successive operations• P2: belonging to a part with the minimum number


operations• P4: belonging to a part with the largest number of

unprocessed operations• P5: with the shortest processing time• P6: belonging to a part with the corresponding

shortest slack time.

The set of schedulable operations S1 {1, 3, 4, 6, 7}• The set of completed operations F = Ø

Step 2. Using rule P1, operations 1, 3, 4, 7 have been selected. Priority rule P2 selects operation 1.

Step 3. Operation 1 is scheduled on machine 1.

Part P

1 2

1

Part P

3

2


5

4

Part P

6 7 8

3

M1 1

Time0

Step 4. Operations status update

Scheduled operation

S hi




unprocessed operations• P5: with the shortest processing time• P6: belonging to a part with the corresponding


1 2number

1 2 3Part number

1 2 3

number1 3Operation

Machine number

Machining

1 3 2 4 5 6 7 8

1 2 2 1 1 2 1 2

4 2 3 2 2 3 2 1

Same machineas operation 1

1

3

2

4

87

5

6


gtime

4 2 3 2 2 3 2 1

The set of schedulable operations is updated to S1 = {3, 6}.Go to step 2.

Step 2. Rule P1 selects operation 3 from S1 = {3, 6}.


M1 1




M2 3

number 1 2 3Part number 1 2 3

number 1 2 3Operation numberMachine

1 3 2 4 5 6 7 8

1 2 2 1 1 2 1 2

1

3

2

5

g g p gunprocessed operations•

• P5: with the shortest processing time• P6: belonging to a part with the corresponding



numberMachining time

1 2 2 1 1 2 1 2

4 2 3 2 2 3 2 1

The set of schedulable operations is updated to S1= Ø. Go to step 4.

4

876

Step 4. The completion time of operations 1 and 3 is 4 and 3, respectively, and current time is set to 3. Operation 3 is added to the set of completed operations F, F = {3}. Machine 2 becomes available and its status has been updated. The set of schedulable operations is updated to S1= {6}. Go to step 5.

Partnumber 1 2 3Partnumber 1 2 3

1

3

2

Step 5. Since the set of schedulable operations S1={6} Ø, go to Step 2.

Step 2. Operation 6 is selected.Step 3. Operation 6 is scheduled on machine 2.

≠

M1

M2

1

3 Time t=3

Partnumber 1 2 3OperationnumberMachinenumberMachiningtime

1 32 4 5 6 7 8

1 2 2 1 1 2 1 2

4 2 3 2 2 3 2 1

3

4

87

5

6


M1

M2

1

3 6

Time t=4

number1 2 3Part

number1 2 3

number1 2 3Operation

numberMachine numberMachining

1 3 2 4 5 6 7 8

1 2 2 1 1 2 1 2

1

3

2

45M1 1

Time t=4

Machining time 4 2 3 2 2 3 2 1

The set of schedulable operations S1={4} Ø.

Step 2. Operation 4 is selected.


≠

876M2 3 6


p p

M1

M2

1

3 6

4

Time t=6

The Final Schedule Part number 1 2 3Part number 1 2 3Part number 1 2 3Operation numberMachine numberMachining

1 3 2 4 5 6 7 8

1 2 2 1 1 2 1 2

4 2 3 2 2 3 2 1g

time 4 2 3 2 2 3 2 1

M 1 741 5


The makespan = 103 6 8

M2 Time

93 6 2 8

0 10

Documents

General Scheduling Model.ppt - University of Iowauser.engineering.uiowa.edu/~coneng/lectures/General_Scheduling_Model.pdf · General Scheduling Model and Algorithm Andrew Kusiak Intelligent