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Management Science Course2003Class 1 Linear

Programming

Teachers

Steef van de Velde svelde_at_fbk.eur.nl office

F2-62 tel. 4081719

Moritz Fleischmann mfleischmann_at_fbk.eur.nl office

F1-38 tel. 4082277

Jo van Nunen jnunen_at_fbk.eur.nl office

F1-21 tel. 4082032

Raf Jans rjans_at_fbk.eur.nl office F2-53 tel.

4082774

Teaching Objectives

- To teach you how managerial problems from

marketing, finance, production, logistics etc.

can be modeled with quantitative modeling

techniques - To illustrate how these techniques are used in

practice in Decision Support Systems, ERP systems

etc. - To show you how model formulations are solved

with standard commercial software - To let you interpret model solutions
- To provide you insight into the advantages and

limitations of model-based decision making

COURSE FORMAT

- 10 classes, mostly case-based
- Extensive use of Excel (add-ins) as modeling

tool - Material grouped around 3 major themes
- 1 group assignment per theme
- 3 optional workshops
- ? Opportunity to
- gain hands-on modeling practice
- recap course concepts
- obtain tailored feedback
- work on group assignments

COURSE MATERIAL

- Winston Albright, Practical Management Science,

Duxbury, 2nd edition - including companion CD-ROM containing Palisade

Decision Tools Suite (Excel add-ins) - Cases distributed as hardcopies or via

blackboard - Blackboard
- Handout slides before class (if appropriate)
- Complete slides after each class
- Excel files of class examples
- Additional course material
- Announcements

REQUIREMENTS

- Prepare each lecture

- Practice hands-on modeling in Excel

- Active class participation
- (20 OF FINAL GRADE)

- 3 group assignments (30 OF FINAL GRADE)

- Exams Mid-term
- (20 OF FINAL GRADE)
- Final
- (30 OF FINAL GRADE)

Course Program

- Theme I LINEAR PROGRAMMING
- Classes 1 3, Workshop 1
- Theme II DECISION ANALYSIS
- Classes 2 4, Workshop 2
- Theme III SIMULATION
- Classes 5 6, Workshop 3
- Comprehensive Applications
- Classes 7 - 9
- Behavioral Perspective
- Class 10

Todays Program

- Introduction to the world of Management Science
- Operations Research
- Introduction to LINEAR PROGRAMMING (LP)
- Introduction to modeling

A Flavor of Quantitative Modeling Applications

- strategic positioning of activities
- asset liability
- production planning scheduling
- fleet management (routing, scheduling etc.)
- blending problems (food process industry)
- revenue management
- cutting packing problems
- supply chain optimization
- urban transportation planning
- scheduling of trains, drivers, conductors
- human resource mgmt / personnel planning
- risk analysis
- etc.

Models

- Abstraction
- Strategic
- Tactical
- operational

Models DSS

- Communication tool
- Power tool
- Learning tool
- Information system
- Decision taking
- Decision support

A Flavor of Techniques

- MATHEMATICAL PROGRAMMING
- linear programming
- integer linear programming
- quadratic programming
- dynamic programming
- COMBINATORIAL OPTIMIZATION
- QUEUEING THEORY
- DECISION ANALYSIS
- INVENTORY THEORY

- SIMULATION

- REGRESSION
- ANALYSIS

- CPM PERT

- MARKOV THEORY
- NEURAL NETWORKS
- DEA

Linear Programming is an Important Mathematical

Optimization Tool

- Many business problems can be modeled as
- linear programming problems.
- STATE-OF-THE-ART LP-SOLVERS are able to
- solve LPs of huge dimensions

Dryer Washer Production Plant

Profit

200 100

What would you want to know?

- How many washers to produce - How many dryers

to produce in order to .. .. MAXIMIZE

PROFIT!

Verbal Formulation as an Optimization Problem

Determine the number of washers and dryers to be

produced So as to MAXIMIZE PROFIT

Subject to RESOURCE CONSTRAINTS

continued

- Specify the DECISION VARIABLES
- Describe the CONSTRAINTS
- (in terms of the decision variables)
- Describe the OBJECTIVE FUNCTION
- (in terms of the decision variables)

Abstract Formulation

DECISION VARIABLES

D the number of DRYERS to be produced W

the number of WASHERS to be produced

CONSTRAINTS

Metal department Electronic department Assembl

y Washers Assembly Dryers

5D 4W lt 4,000

9D 10W lt 9,000

10D lt 6,000

10W lt 8,000

PROFIT

200D 100W

Linear Programming Format

MAXIMIZE 200D 100W

Subject to

5D 4W lt 4,000 9D 10W lt

9,000 10D lt 6,000 10W

lt 8,000

D gt 0 W gt 0

Modeling Assumptions

- Proportionality
- Additivity
- Divisibility

So, Modeling as an LP Involves

Determining the appropriateness of LP

- A SYMBOLIC LANGUAGE The Decision Variables
- - Describing the constraints
- - Describing the objective function

- EXPERIENCE
- Background reading (SEE TEXTBOOK, BLACKBOARD)
- Exercises and assignments

JOHN BEASLEYS MBA COURSE ON INTERNET

How to Solve LP Problems

- Graphically, with two decision variables
- Simplex (algebraic) method
- STATE-OF-THE-ART SOFTWARE like CPLEX
- (e.g. http//www.cplex.com) solves tens of

thousands - of variables and constraints
- EXCEL Solver for problems of moderate size

What next?

- The Graphical Solution Procedure
- Sensitivity Analysis
- Solving LP using EXCEL
- ..

Linear Programming The Graphical Method

Problem description

Product Fuel Additive Solvent Base

Material 1 Material 2 Material 3

Profit

0.4 0.0 0.6

40 30

0.5 0.2 0.3

Amount Available

20 5 21

Example 0.4 ton of Material 1 is used

in each ton of Fuel Additive

Formulation of the Problem of Maximizing Profit

as a Linear Programming Problem

Decision variables

F the number of tons of Fuel Additive to be

produced S the number of tons of Solvent Base

to be produced

Objective function

Maximize 40 F 30 S

Constraints

(1) material availability (2) non-negativity

Formulation of the Problem of Maximizing Profit

as a Linear Programming Problem

Maximize 40 F 30 S

Subject to

(1) material availability constraints

Material 1 Material 2 Material 3

lt 20

0.4 F 0.5 S

0.2 S

lt 5

0.6 F 0.3 S

lt 21

(2) non-negativity constraints

F gt 0 S gt 0

Non-negativity constraints

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

A solution point with F 10 and S 40

INFEASIBLE

Tons of Solvent Base

30

20

A solution point with F 20 and S 15

FEASIBLE

10

0

10

20

30

40

50

Tons of Fuel Additive

Material 1 constraint

Material 1 constraint line 0.4 F 0.5 S 20

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

FEASIBLE REGION FOR THE MATERIAL 1 CONSTRAINT

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

Material 2 constraint line 0.2 S 5

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

FEASIBLE REGION FOR THE MATERIAL 2 CONSTRAINT

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

MATERIAL 3 CONSTRAINT LINE 0.6 F 0.3 S 21

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

FEASIBLE REGION FOR THE MATERIAL 3 CONSTRAINT

LINE

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

MATERIAL 3

Tons of Solvent Base

30

MATERIAL 2

20

MATERIAL 1

FEASIBLE REGION

10

0

10

20

30

40

50

Tons of Fuel Additive

240 PROFIT LINE

40

(40F 30S 240)

F 0, S 8 Profit?

Tons of Solvent Base

240

30

20

F 6, S 0 Profit

240

10

0

10

20

30

40

50

Tons of Fuel Additive

1200

40

Tons of Solvent Base

720

30

240

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

(40F 30S 1600)

Tons of Solvent Base

OPTIMAL SOLUTION!

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

40

EXTREME POINT (INTERSECTION OF TWO OR

MORE CONSTRAINTS)

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

HOW TO FIND THE OPTIMAL SOLUTION (VALUE)?

40

Tons of Solvent Base

30

The intersection of the Material 1 and Material

3 constraint lines

20

10

0

10

20

30

40

50

Tons of Fuel Additive

Calculating the OptimalSolution Value

The values of the decision variables must satisfy

the following equations simultaneously 0.4 F

0.5 S 20 0.6 F 0.3 S 21

gt S 40 - 0.8 F (1)

gt S 70 - 2.0 F (2)

Substituting (1) into (2) gives 40 - 0.8 F

70 - 2.0 F

gt F 25 gt S 20

OPTIMAL SOLUTION VALUE 1600

Summary of Optimal Solution

Materials Tons Required Tons Available Slack Mate

rial 1 20 20 0 Material 2 4 5

1 Material 3 21 21 0

Sensitivity Analysis

WHY SENSITIVITY ANALYSIS ?

- WITH LINEAR PROGRAMMING, YOU GET
- TWO TYPES OF SENSITIVITY INFORMATION
- WHAT HAPPENS IF ONE OF THE
- OBJECTIVE COEFFICIENTS CHANGES
- WHAT HAPPENS IF ONE OF THE
- RIGHT HAND SIDE VALUES CHANGES

OBJECTIVE FUNCTION LINE

40

HOW LONG WILL THECURRENT EXTREMEPOINT REMAIN

OPTIMAL IF THE OBJECTIVECOEFFICIENTS AREGOING

TO CHANGE?

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

MATERIAL 3 CONSTRAINT LINE

40

Tons of Solvent Base

30

MATERIAL 1 CONSTRAINT LINE

20

10

0

10

20

30

40

50

Tons of Fuel Additive

EXTREME POINT WILL BE OPTIMAL AS LONG AS SLOPE

OF MATERIAL 3 CONSTRAINT LINE lt SLOPE OBJ.

FUNCT. LINE lt SLOPE OF MATERIAL 1 CONSTRAINT

LINE

The equation for Material 1 constraint line in

its slope intercept form 0.5 S - 0.4F 20

S - 0.8F 40

Intercept of line on S axis

Slope of line

The equation for Material 3 constraint line in

its slope intercept form S -2F 70

CURRENT SOLUTION REMAINS OPTIMAL AS LONG AS -2

lt SLOPE OF THE OBJECTIVE FUNCT. LINE lt -0.8

The objective function line is a F b S

iso-profit

Hence, the slope intercept form of the objective

function line is S - a/b F iso-profit/b

The current solution will be optimal as long

as -2 lt -a/b lt -0.8

Computing the RANGE OF OPTIMALITY of the Fuel

Additive Coefficient

-2 lt -a/30 lt -0.8 gt 24 lt a lt 60

What is the RANGE OF OPTIMALITY of the Solvent

Base Coefficient (the b coefficient)?

The current solution will be optimal as long

as -2 lt -a/b lt -0.8

Hence, with a 40

-2 lt -40/b lt -0.8 so 20 lt b lt 50

How will a change in the right hand side

valueaffect the solution?That is, in this

example,what happens if you would have more or

less of any material?

OBJECTIVE FUNCTION LINE

40

HOW WILL A CHANGE IN THE RIGHT-HAND SIDE

VALUE FOR A CONSTRAINT AFFECT THE

FEASIBLEREGION?

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

FOR INSTANCE, WHAT HAPPENS IF AN ADDITIONAL 3

TONS OF MATERIAL 3 BECOMES AVAILABLE?

Tons of Solvent Base

30

20

10

Current Material 3 line

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

NEW Material 3 line

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

NEW Material 3 line

Tons of Solvent Base

30

20

10

ADDITIONAL FEASIBLE REGION

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

NEW Material 3 line

Tons of Solvent Base

NO LONGER ANEXTREME POINT . AND THUS NOLONGER

OPTIMAL

30

20

10

ADDITIONAL FEASIBLE REGION

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

NEW Material 3 line

Tons of Solvent Base

30

20

10

ADDITIONAL FEASIBLE REGION

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

NEW Material 3 line

Tons of Solvent Base

30

20

10

ADDITIONAL FEASIBLE REGION

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

NEW Material 3 line

Tons of Solvent Base

30

20

10

ADDITIONAL FEASIBLE REGION

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

NEW Material 3 line

Tons of Solvent Base

30

NEW OPTIMAL SOLUTION

20

10

ADDITIONAL FEASIBLE REGION

0

10

20

30

40

50

Tons of Fuel Additive

SHADOW PRICE

The new optimal solution is F 100/3 S

40/3

The new objective value is 1733.33

Since the value of the optimal solution to the

orginal problem is 1600, increasing the RHS of

the material 3 constraint by 3 tons provides an

increase in profit of 1733.33 - 1600 133.33

Thus, the increased profit occurs at a rate

of 133.33/3 tons 44.44

SHADOW (DUAL) PRICE OF THE MATERIAL 3 CONSTRAINT

IS 44.44

OBJECTIVE FUNCTION LINE

40

WHAT IS THESHADOW PRICEFOR AN ADDITIONAL6 TONS

OF MATERIAL 3?

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OBJECTIVE FUNCTION LINE

40

WHAT IS THESHADOW PRICEFOR AN ADDITIONALTON OF

MATERIAL 2 ?

Tons of Solvent Base

30

20

10

0

10

20

30

40

50

Tons of Fuel Additive

OUTLOOK

- Class 2 (Oct.29/30), Decision Analysis 1?

Prepare Freemark Abbey Winery Case (see

handout blackboard) - Class 3 (Nov.5), Linear Programming 2? Prepare

Red Brand Canners Case (see handout)

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