Perspective of supply chain optimization

Tokyo University of Marine Science and Technology

Mikio Kubo

Agenda

• Supply Chain and Analytic IT• Mathematical Programming Solver Gurobi

with Python language• Constrained Programming Solver SCOP• Scheduling Solver OptSeq

What’s the Supply Chain?

IT(Information Technology)＋ Logistics ＝ Supply Chain

配送計画

スケジューリングロットサイズ決定

安全在庫配置在庫方策

ロジスティクス・ネットワーク設計

需要予測収益管理

Logistics Network Design

Safety Stock Allocation

Vehicle Routing/Scheduling

ForecastingRevenue Management

Lot-SizingScheduling

Logistic System, Transactional IT, Analytic IT

実 シ ス テ ム

IT処 理 的

IT解 析 的

Logistic System=Truck, Ship, Plant, Product, Machine, …

Transactional ITPOS, ERP, MRP, DRP…Automatic Information Flow

Analytic ITModel ＋ Algorithm=Decision Support System

brain

muscle

nerve

Levels of Decision Making

Strategic LevelStrategic Level

Tactical LevelTactical Level

Operational LevelOperational Level

A year to several years; long-term decision making

A week to several months; mid-term decision making

Real time to several days; short-term decision making

Transactional IT

Analytic IT

Models in Analytic IT

Logistics Network Design

Inventory

Safety stock allocationInventory policy

optimization

Production

Lot-sizingScheduling

TransportationDelivery

Vehicle Routing

Multi-period Logistics Network Design

Strategic

Tactical

Operational

Plant DCSupplier Retailer

How to Solve Real SC Optimization Problems Quickly

• Mixed Integer Programming (MIP) SolverGurobi =>Logistics Network Design, Lot-Sizing

• Constraint Programming (CP) Solver SCOP =>Lot-Sizing, Staff Scheduling

• Scheduling Solver OptSeq =>Scheduling• Vehicle Routing Solver • Inventory Policy Optimization Solver

... using Python Language

Why Python?

• We can do anything by importing some modules

• Optimization import gurobipy (MIP)import SCOP (CP)

• Draw graphsimport networkX

• Also fly! import antigravity ?

http://xkcd.com/353/

What’s Gurobi?MIP solver Developed by: Zonghao Gu, Edward Rothberg ， Robert Bixby

Free academic license

Introduction to Gurobi (1)

• Create a model objectmodel = Model("Wine Blending")

• Add variable objectsx1 = model.addVar(name="x1")x2 = model.addVar(name="x2")x3 = model.addVar(name="x3")

• Model update (needed before adding constraints; lazy update!)model.update()

Introduction to Gurobi (2)

Introduction to Gurobi (3)• Set the objective

model.setObjective(15*x1 + 18*x2 + 30*x3, GRB.MAXIMIZE)

• Add constraints model.addConstr(2*x1 + x2 + x3 <= 60)model.addConstr(x1 + 2*x2 + x3 <= 60)model.addConstr(x3 <= 30)

• Optimizemodel.optimize()

Mr. Python is too lazy. His room is always mess.He asked to the God of Python.“How can I tide up my toys?”

The God replied from the heaven.“Use the list. The charm is “Abracadabra [ ].”Mr. Python said “L=[ ].”What a miracle! Some boxes were fallen from the heaven.The God said. “Tide up the toys using these boxes.”

OK. Everything is packed into the list.To pick up “Teddy”, just say “L[3].”

To sort the toys in alphabetical order, just say “L.sort().”

Note that L is a list object and sort() is a function definedin the object called “method.”

Modeling with Lists

• Add variable objects into list x=[] for i in range(3): var=model.addVar(name=“x[%s]”%i) x.append(var)

• Add constraint “x1 + x2 + x3 <= 2”model.addConstr( sum(x) <= 2 ) or model.addConstr( quicksum(x) <= 2 )

Mr. Python is impatient, too. He complained to the God. “I’d like to pick my toys immediately.” The God replied from the heaven again. “Use the dictionary. The charm is “Abracadabra {}.”

Using lists and dictionaries, Mr. Python could manage his toysefficiently and lived happily ever after.

Modeling with Dictionaries

• Dictionary that maps keys (“Dry”, “Medium”, “Sweet”) to variable objects x={} x[“Dry”]= model.addVar(name=“Dry”) x[“Medium”]= model.addVar(name=“Medium”) x[“Sweet”]= model.addVar(name=“Sweet”)

• Add constraintmodel.addConstr( 2*x[“Dry”]+ x[“Medium”] +x[“Sweet”] <=30 )

Wine Blending with Dictionaries (1)

Blends, Profit = multidictmultidict({"Dry":15, "Medium":18, "Sweet":30})

=> Blends=["Dry", "Medium“, "Sweet“] List of Keys Profit[“Dry”]=15, Profit[“Medium”]=18, ...

Grapes, Inventory = multidict({"Alfrocheiro":60, "Baga":60, "Castelao":30})

Use = { ("Alfrocheiro","Dry"):2, ("Alfrocheiro","Medium"):1, ("Alfrocheiro","Sweet"):1, ("Baga","Dry"):1, .... }

Wine Blending with Dictionaries (2)

x = {}for j in Blends: x[j] = model.addVar(vtype="C", name="x[%s]"%j)model.update()

model.setObjective(quicksum(Profit[j]*x[j] for j in Blends), GRB.MAXIMIZE)

for i in Grapes: model.addConstr(quicksum(Use[i,j]*x[j] for j in Blends)

<= Inventory[i], name="use[%s]"%i)

model.optimize()

k-median problem• A facility location problem with min-sum

object• Number of customers n=200, number of

facilities selected from customer sites k=20• Euclidian distance ， coordinates are

random

Formulation

weak formulation

Python Code （ 1 ）from gurobipy import *

model = Model("k-median")

x, y = {}, {} # empty dictionaries

Key“Hanako”,

(1,2)

Value“127cm”

Variable ObjectMapping

Dictionary Data Structure

I=range(n)J=range(n)for j in J:

y[j] = model.addVar(vtype="B", name="y[%s]"%j)

for i in I:

x[i,j] = model.addVar( vtype="B",name="x[%s,%s]"%(i,j))

model.update()

Python Code （２）

model.setObjective(quicksum(c[i,j]*x[i,j] for i in I for j in J))

Add variable objects“B” means binary variable or GRB.BINARY

Set the objective

Python Code （３） for i in I:

model.addConstr(quicksum(x[i,j] for j in J) = = 1, "Assign[%s]"%i)

for j in J:

model.addConstr(x[i,j] <= y[j], "Strong[%s,%s]"%(i,j))

model.addConstr(quicksum(y[j] for j in J) = = k, "k_median")

Optimize a model with 401 Rows, 40200 Columns and 80400 NonZeros

…

Explored 1445 nodes (63581 simplex iterations) in 67.08 seconds

Thread count was 2 (of 2 available processors)

Optimal solution found (tolerance 1.00e-04)

Best objective 1.0180195861e+01, best bound 1.0179189780e+01, gap 0.0099%

Opt.value= 10.1801958607

Weak formulation (result)n=200,k=20

Upper and lower bounds (Weak Formulation)

0

2

4

6

8

10

12

14

16

18

0 10 20 30 40 50 60 70

CPU

Obj. F

unc.

Value

Optimize a model with 40401 Rows, 40200 Columns and 160400 NonZeros

…

Explored 0 nodes (1697 simplex iterations) in 3.33 seconds(No branching ！）

Thread count was 2 (of 2 available processors)

Optimal solution found (tolerance 1.00e-04)

Best objective 1.0180195861e+01, best bound 1.0180195861e+01, gap 0.0%

Opt.value= 10.1801958607

Strong formulation (result)

k-center problem

• A facility location problem with min-max object

• 100 customers ， 10 facilities

k-center (n=30,k=3) k-median (n=30,k=3)

Formulation

Upper and lower bounds

(n=100,k=10)

0

0.2

0.4

0.6

0.8

1

1.2

0 50 100 150 200 250 300 350 400

CPU Time

Obj. F

un. V

alue

k-Covering Problem

=1 if customer is not covered

# of uncovered customers

parameter that is =1 if distance is less than or equal to θ

k-Covering+Binary Search

Upper and Lower Bounds UB ， LB while UB – LB >ε:

θ= (UB+LB)/2 if opt. val. of k-covering is 0 then

UB = θ else LB = θ

Computational Experiments

Traveling salesman problem• Find a minimum cost (distance) Hamiltonian

circuit • World record 85,900 nodes (symmetric

instance) -> We try to solve asymmetric ones.

Miller-Tucker-Zemlin formulation

Upper and lower bounds（ 80 nodes ， Euclid TSP ）

0

5

10

15

20

25

30

35

40

45

0 50 100 150 200 250 300 350 400

CPU

Obj. F

unc.

Value

Non-lifted MTZ constraints-> Out of memoryafter running 1 day

ResultOptimize a model with 6480 Rows, 6400 Columns and 37762 NonZeros

…

Cutting planes:

Gomory: 62

Implied bound: 470

MIR: 299

Zero half: 34

Explored 125799 nodes (2799697 simplex iterations) in 359.01 seconds

Optimal solution found (tolerance 1.00e-04)

Best objective 7.4532855108e+00, best bound 7.4525704995e+00, gap 0.0096%

Opt.value= 7.45328551084

Graph coloring problem

• An example that has symmetric structure of solutions

• Number of nodes n=40 ， maximum number of colors Kmax=10

• Random graph G(n,p=0.5)

Formulation

Weak formulation

n=40 ， Kmax=10

0

2

4

6

8

10

12

0 200 400 600 800 1000 1200 1400

CPU Time

Obj. F

unc.

Value

Optimize a model with 3820 Rows, 410 Columns and 11740 NonZeros

Explored 17149 nodes (3425130 simplex iterations) in 1321.63 seconds

Improvement of the formulation

Avoid symmetric variables

SOS: Special Ordered Set ） Type 1

model.addSOS(1,list of var.s)

Avoid symmetric variables

0

2

4

6

8

10

12

0 50 100 150 200 250 300 350 400 450

CPU

Obj. F

unc.

Value

Optimize a model with 3829 Rows, 410 Columns and 11758 NonZeros

Explored 4399 nodes (1013290 simplex iterations) in 384.53 seconds

MIPFocus=2 （ priority=proving the optimality) 67secMIPFocus=3 (priority=lower bound) 70 sec.

+SOS constraints

Optimize a model with 3829 Rows, 410 Columns and 11758 NonZerosExplored 109 nodes (58792 simplex iterations) in 22.02 seconds

MIPFocus=2 65 sec. ， MIPFocus=3 126 sec.

0

2

4

6

8

10

12

0 5 10 15 20 25

CPU

Obj. F

unc.

Val.

Fixed-K approach

If the end vertices of an edge have the same color, the edge is called bad, and the corresponding variable z is equal to 1

Number of “bad” edges

Fixed-K Approach +Binary Search

UB, LB := Upper and lower bounds of Kwhile UB – LB >1:

K= [ (UB+LB)/2 ] [ ] : Gauss notation if obj. fun. of Fixd-K model = 0 then UB = K

else LB = K

Fixed-K Approach +Binary Search

Improved Formulation

SCOP 　Solver for COnstraint Programming

Log Opt Co., Ltd.

Agenda

• Introduction • SCOP Interfaces• Demonstration• Benchmark• Case Study

What’s Constraint ProgrammingConstraint Programming is a paradigm to solve

combinatorial problems efficiently

• Constraint Satisfaction Problem (CSP) – seeks an assignment of values to variables

such that all the given constraints are satisfied

Constraint Satisfaction ProblemConstraint satisfaction problem consists of:

• variable: variable chooses a value from its domain.

• domain: a finite set of values for each variable.

• constraint: a logical relation among several variables.

Weighted Constraint Satisfaction Problem (WCSP)

• WCSP is an enhanced framework of CSP • Not just seeks an assignment that satisfies all of the

constraints but to minimize the weighted sum of infeasibility of constraints (called penalties): – Hard Constraint

weight is infinity, i.e., constraint must be satisfied– Soft Constraint

weight is a positive integer, i.e., constraint may be violated by paying penalty

• The goal is to minimize the total penalties =>optimization model.

What’s SCOP

• Based on Metaheuristics by Ibaraki (Kyotyo-University), Nonobe (Hosey-University)

• Trial version （１５ variables available ） http://www.logopt.com/scop.htm

• 1. Simple modeling language (text file) • 2. Python interface• 3. Library （ C++, .Visual Basic, C# ）

Solver to solve a large combinatorial problems effectively

SCOP is a solver for solving WCSP

SCOP Model

• variable 　 • domain• constraint:

–weight = positive integer （ soft constraint ）　　　　　 inf （ hard constraint ）

– type = linear ， alldiff, quadratic

Assignment Problem• Assign 5 workers A,B,C,D,E to 3 jobs 0,1,2• Each job (0,1,2) needs at least (1,2,2) workers• Worker A and worker C have to assign to different

jobs• Assignment cost ＝ 0 1 2

A 15 20 30

B 7 15 12

C 25 10 13

D 15 18 3

E 5 12 17

The object is to minimize sum of the

assignment cost

variable A in {0, 1, 2}variable B in {0, 1, 2}variable C in {0, 1, 2}variable D in {0, 1, 2}variable E in {0, 1, 2}

Variable and domain• Variables are workers A,B,C,D,E• Domain for each variables is set of jobs {0,1,2}

variable var-name in { domain }

domain = value1, value2, value3, ...

constraint0: weight=inf type=linear　　　　　 　 1(A,0) 1(B,0) 1(C,0) 1(D,0) 1(E,0) >=1constraint1: weight=inf type=linear 　　　　　　 　 1(A,1) 1(B,1) 1(C,1) 1(D,1) 1(E,1) >=2constraint2: weight=inf type=linear 　　　 　 1(A,2) 1(B,2) 1(C,2) 1(D,2) 1(E,2) >=2

Linear constraint （ 1 ）• Each job (0,1,2) needs at least (1,2,2) workers

0 1 2

A 15 20 30

B 7 15 12

C 25 10 13

D 15 18 3

E 5 12 17

Con-name ： weight= integer or inf 　 type= linear

coeff1(var-name1,value1) coeff2(var-name2,value2).... <= (>=, =) right hand side

hard constraint

obj: weight=1 type=linear 　　 15 (A, 0) 20(A, 1) 30(A, 2) 7 (B, 0) 15(B, 1) 12(B, 2) 　　 25 (C, 0) 10(C, 1) 13(C, 2) 15 (D, 0) 18(D, 1) 3(D, 2) 　　 5 (E, 0) 12(E, 1) 17(E, 2) 　 <=0

Linear constraint （ 2 ）Wright objective function as linear constraint（ object value = sum of the penalty cost ） 0 1 2

A 15 20 30

B 7 15 12

C 25 10 13

D 15 18 3

E 5 12 17

hard constraintSum of the assignment cost <=0

quad: weight=100 type=quadratic 1 (A,0) (C,0) 1 (A,1) (C,1) 1 (A,2) (C,2) <=0

Quadratic constraint

Con-name ： weight= integer or inf 　 type= quadraticcoeff1(var-name1,value1) (var-name2,value2) coeff2(var-name3,value3) (var-name4,value4) ....

Worker A and worker C have to assign to different jobs

larger than the weitht in objective function

AllDiff: weight= 100 type=alldiff A C ;

Alldifferent constraint

Con-name ： weight= integer or inf 　 type= alldiffvar-name1 var-name2 var-name3 .... ;

Worker A and worker C have to assign to different jobs

How to use SCOP solver

• Start with Command Poompt (or Colsole)

• scop < file name

• scop - （ option ） <file name

• scop -help (display all of the SCOP options)

SCOP options-noadjust deactivate weight adjustment mechanism-display # set log display level-iteration # set iteration limit-interval # set log display interval-noimprovement # set iteration limit for no improvement-seed # set random seed-target # set target-time #.# set CPU time limit in second

Result (solved by SCOP)scop <ex3-scop.dat# reading data ... done: 0.00(s)

penalty = 1/52 (hard/soft), time = 0.00(s), iteration = 0# improving the initial solution greedilypenalty = 0/157 (hard/soft), time = 0.00(s), iteration =

0# start tabu searchpenalty = 0/60 (hard/soft), time = 0.01(s), iteration = 2penalty = 0/52 (hard/soft), time = 0.01(s), iteration = 8# penalty = 0/52 (hard/soft)# cpu time = 0.01/0.03(s)# iteration = 8/100[best solution]A: 0B: 2C: 1D: 2E: 1

penalty: 0/52 (hard/soft)[Violated constraints]obj: 52

Obj : 52 =15+12+10+3+12

0 1 2A 15 20 30

B 7 15 12C 25 10 13

D 15 18 3E 5 12 17

Summary 1. variable & domain

variable var-name in { value1, value2, ... }

2. target （ sum of the penalty ） target = target-value

3. constraintscon-name: weight = interger / inftype = linear / quadratic / alldiff

cnstraint

Steps 1 and 2 have to declare before Step 3 ！

Graph Coloring

Have to assign to different color class!

・・・

Graph Coloring

Graph Coloring Programs

SCOP Gurobi

Test results (Graph Coloring)• Fixed the number of color

*(1) 　 16 constraint violation after 828 seconds 　

# of nodes # of edges # of colors SCOP (seconds)

Gurobi (seconds)

30 0.5 7 0.01 0.0750 0.5 9 0.16 29100 0.5 15 0.66 - *(1)

500 0.5 54 50.9 -800 0.5 80 74.45 -

Quadratic Assignment Problem

The number of times who meets each other in one week

3

1 2

- -

Distances between two houses

2km5km

1km

Quadratic Assignment Problem

3

1 2

- -

2km5km

1km

2×2＋ 5×1＋ 1×3＝ 12km

- -

Test result (QAP)nodes Gurobi (seconds) SCOP

(seconds)Gaps % (Gurobi/SCOP)

5 0.03 0.01 08 12.84 0.23 010 1589 1.76 020 140 138.2 4.925 300 298.2 3.530 540 534.4 1.5

Application of SCOP• Staff scheduling problem• Time tabling problem• Graph partition problem• Maximum stable set problem• Graph coloring problem 　• Quadric assignment problem 　• Travelling sails man problem• Multiple knapsack problem• Process Scheduling in Chemical Plants

Scheduling optimization systemOptSeq

http://www.logopt.com/OptSeq/OptSeq.htm

Log Opt Co., Ltd.

What is the scheduling?

• Allocation of activities (jobs, tasks) over time – Resource constraints. For example, machines,

workers, raw material, etc. may be scare resources. 　

– Precedence (or temporal) relation between activities. For example., some activities cannot start unless other activities finish.

Take off the airplane a.s.a.p. !

You’re a consultant of an airline company. Your job is to find a schedule that takes off the airplane as soon as possible. We need several activities before taking off the airplane. Firstly let the passengers get off and unload their baggage; secondly clean up the cabin; finally let new passengers get on and load their baggage. What is the best (shortest) schedule for taking off the airplane?

PERT• PERT: Program Evaluation and Review Technique that was used to

get a schedule during World War II

22分25分

13分 15分 27分

1作 業

2作 業

3作 業 4作 業

5作 業

ダ ミ ー 作 業

Temporal relation

Activity1 ： Get off passengers (13 min.)Activity2 ： Unload baggage (25 min.)Activity3 ： Clean up the cabin (15 min.)Activity4 ： Get on passengers (27 min.)Activity5 ： Load baggage (22 min.)

Completion time (make-span) is minimized

Dummy activity(take-off)

Act 1

Act 2

Act 3 Act 4

Act 5

Act sink

Modeling with OptSeq （１）Activity

activity act1 mode duration 13

activity Activity-Name duedate Integer+ mode duration Integer+

Description ofActivity

E.g., to add an activity named act1 with duration (processing time) 13 :

Activity must have at least one mode (the way to execute the activity)

Duedate is optional

Modeling with OptSeq （ 2 ）Temporal Constraint

temporal act1 act3

temporal Predecessor Successor Description ofTemporal Constraint

E.g., to define that activity act3 must start after finishing activity act1:

Modeling with OptSeq （ 3 ）Objective

activity sink duedate 0

Objective Minimize the latest completion time (makespan)

sink: the dummy activity that must be executed after all activitiesWe minimize the delay time of sink:

activity Activity-Name duedate Integer+Description ofActivity

Optimization and Result

optseq -time 3 < Input-file-nameTime Limit is 3 sec.

--- best solution --- source ---: 0 0 dummy source starts and finishes at time 0 sink ---: 55 55 dummy sink starts and finished at time 55 activity[1] ---: 0 0--13 13activity[2] ---: 0 0--25 25activity[3] ---: 13 13--28 28activity[4] ---: 28 28--55 55activity[5] ---: 25 25--47 47

objective value = 55 latest completion time (makespan) is 55cpu time = 0.00/3.00(s) computational time is 0 sec.iteration = 1/62605 number of iteration of tabu search is 1

Result

PERT with Resource Constraint

resource Resource-Name interval Time1 Time2 capacity Integer+ ...

Description ofResource

Period

Time

“Interval 1 3” means period 1, 2

Modeling Resources with OptSeq

resource worker interval 0 inf capacity 1Resource “worker”can be used from 0 to infinity （ inf ）with capacity 1

activity Activity-Name Resource-Name interval Time1 Time2 requirement Integer+ 　　　 ...

Activity requiresresource

activity act1 mode duration 13 worker interval 0 13 requirement 1

Activity “act1”requires resource“worker” by 1 unit

Optimization and Result

source ---: 0 0sink ---: 102 102activity[1] ---: 47 47--60 60activity[2] ---: 0 0--25 25activity[3] ---: 60 60--75 75activity[4] ---: 75 75--102 102activity[5] ---: 25 25--47 47

objective value = 102cpu time = 0.00/3.00(s)iteration = 0/64983

Result

Optimal solution （ Gantt’s chart ）

13分

1作 業15分

3作 業27分

4作 業

25分

2作 業22分

5作 業

時 間０ 55

Resource constrained schedule (unit resource upper bound)

15分 3作 業

27分 4作 業

13分 1作 業

25分 2作 業

22分 5作 業

時 間０ 102

Example 2Minimize the Pit-in time!

You’re a staff of F1 race. There are three pit staffs who have to do some tasks for your F1 car. Find the best schedule that minimizes the pit-in time.

Temporal (Precedence) relation and the processing time

gas

4

4

4

4

2

3

2

2

2

11作業１

作業２

作業３

作業４

作業５

作業６

作業７

作業８

作業９

作業１０

秒

秒

秒

秒

秒

秒

秒

秒

秒

秒

3 staffs(resource constraint)

3 sec.

Task

Task

Task

TaskTask

Task Task

Task

Task

Task

11 sec.

2 sec.

2 sec.

2 sec.

2 sec.

4 sec.

4 sec.

4 sec.

4 sec.

Modeling with OptSeq -using a resource with capacity 3-

resource worker interval 0 inf capacity 3

activity prepare mode duration 3 worker interval 0 3 requirement 1．．．

temporal prepare oiltemporal jackup tire1．．．

activity sink duedate 0

Define a resource“worker” with capacity 3

Temporal constraints

Minimize themakespan

Optimization by OptSeqand Gantt’s chart

source ---: 0 0sink ---: 14 14prepare ---: 0 0--3 3water ---: 0 0--2 2front ---: 0 0--2 2jackup ---: 2 2--4 4tire1 ---: 8 8--12 12tire2 ---: 4 4--8 8tire3 ---: 8 8--12 12tire4 ---: 4 4--8 8oil ---: 3 3--14 14jackdown ---: 12 12--14 14

objective value = 14cpu time = 0.00/3.00(s)iteration = 0/37644

Another Model - using multiple modes -

1. Convenience Store Mode ： Duration (Processing Time) 20 min. requires Money Resource 110 yen and human resource 1

2. Vending Machine Mode ： Duration 20 min. requires Money 120 yen and human resource 13. Supermarket Mode: Duration 20 min.

requires Money 120 yen, human resource 1 and car resource 1

Mode ： The way for processing an activity

Example ： Activity “Buy a juice”

Mode (1)

mode Mode-Name duration Processing-Time Resource-Name interval Time1 Time2 requirement Integer+　　　 ...

Description of Mode

activity Activity-Name Mode-Name1 Mode-Name2 ....

Add modesto activity

Mode (2)

mode m1 duration 3 worker interval 0 3 requirement 1

mode m2 duration 2 worker interval 0 2 requirement 2

mode m3 duration 1 worker interval 0 1 requirement 3

E.g., an activity has 3 difference modes; m1, m2 and m3 ：

activity prepare m1 m2 m3

Add 3 modes to activity “prepare”

Mode Resource

single worker mode

double worker mode

triple worker mode

Result and Gantt’s chart

--- best solution ---source ---: 0 0sink ---: 13 13prepare m3: 0 0--1 1water ---: 1 1--3 3front ---: 11 11--13 13jackup ---: 1 1--3 3tire1 ---: 7 7--11 11tire2 ---: 3 3--7 7tire3 ---: 7 7--11 11tire4 ---: 3 3--7 7oil ---: 1 1--12 12jackdown ---: 11 11--13 13

objective value = 13cpu time = 0.00/3.00(s)iteration = 7/23318

activity “prepare” was done by mode m3, i.e., execute by 3 worker with duration 1

Resource Constrained Project Scheduling

土 台

１ 階

内 装

屋 根

完 成 ！

時 間 （ 日 ）

資 源 量 （ 人 ）

30

2

1 worker rests onthe 3rd day

Duration : 2 days １ worker on the 1st dat 2 workers on the 2nd day

BasementFirst Floor Ceiling

Interior

Workers

Completion!

Modeling with OptSeq (1)

resource worker interval 0 2 capacity 2 interval 2 3 capacity 1 interval 3 inf capacity 2

Time Dependent Resource Capacity

０　　　　　　　　２　　　３

１

Resource Capacity

Modeling with OptSeq (2)

Time DependentResource Usage ofActivity

activity first mode duration 3 worker interval 0 1 requirement 2 worker interval 1 3 requirement 1

Optimal Solution

時 間 （ 日 ）

資 源 量 （ 人 ）

Resource Constrained Project Scheduling (RCPSP) is a generic model （ job shop and flow shop scheduling problems are special cases.)

Resource Capacity

Critical Path Method (CPM)

Consider “Take off the airplane” again! However, each activity has an emergency modeActivity1 ： Get off passengers (13min.) => Emergency mode (10 min. ) requires 10000 yenActivity2 ： Unload baggage (25min.) => Emergency mode (20 min. ) requires 10000 yenActivity3 ： Clean up the cabin (15min.) => Emergency mode (10 min. ) requires 10000 yenActivity4 ： Get on passengers (27 min.) => Emergency mode (25 min. ) requires 10000 yenActivity5 ： Load baggage (22 min.) => Emergency mode (20 min. ) requires 10000 yen

Renewable Resources andNon-renewable Resources

• Renewable resourcescan be used again after the completion of activities; for example, machines and/or workers are renewable

• Non-renewable resourcesare not renewed after the completion of activities; for example, money and/or ingredients are non-renewable

CPM can be modeled using a non-renewable resource

mode m[1][1] duration 13 mode m[1][2] duration 10

activity activity[1] m[1][1] m[1][2]

Modeling with OptSeq (1)

Activity 1 “Get off passengers” takes:13 minutes (normal mode) m[1][1] or10 minutes (emergency mode) m[1][2]

nonrenewable +1 (activity[1],m[1][2]) +1 (activity[2],m[2][2]) +1 (activity[3],m[3][2]) +1 (activity[4],m[4][2]) +1 (activity[5],m[5][2]) <= 4

Modeling with OptSeq (2)

Declaration ofnon-renewableresource

nonrenewable Amount1 (Activity-Name1,Mode-Name1) Amount2 (Activity-Name2,Mode-Name2) ... <= Upper-Bound (or Capacity)

Budget (Capacity of money resource) is 40000 yen.

Note that Amounts and Capacity may be 0 or negative integer.

Results10分

1作業

13分

1作業

10分

3作業

10分

3作業

25分

4作業

27分

4作業

20分

2作業

25分

2作業

22分

5作業

22分

5作業

時間

時間

０

０

45

13分

1作業15分

3作業27分

4作業

25分

2作業22分

5作業

時間０ 55

52

4万円

1万円

0万円

Budget is40000 yen.

Budget is10000 yen.

Budget is 0 yen.

Emergency mode

Type of precedence relation

– ＝１： Finish ->Start– ＝２： Finish ->Finish– ＝３： Start ->Start– ＝４： Start ->Finish

Predecessor

Successor

Set-up time lower boundSet-up time upper bound

Finish Start

OptSeq Capability

• OptSeq is a solver for Resource Constrained Project Scheduling Problem with:– Time-Dependent Resource Usage and Capacity– Non-renewable Resource Constraints– Generic Temporal Constraintsalso it is possible to:– declare activities are breakable and/or processed in parallel – add “states” to represents the generic states of the system