Management Science Course 2003 Class 1: Linear Programming

1
Management Science Course2003Class 1 Linear
Programming
2
Teachers
Steef van de Velde svelde_at_fbk.eur.nl office
F2-62 tel. [phone]
Moritz Fleischmann mfleischmann_at_fbk.eur.nl office
F1-38 tel. [phone]
Jo van Nunen jnunen_at_fbk.eur.nl office
F1-21 tel. [phone]
Raf Jans rjans_at_fbk.eur.nl office F2-53 tel.
4082774
3
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

4
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

5
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

6
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)

7
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

8
Todays Program

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

9
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.

10
Models

• Abstraction
• Strategic
• Tactical
• operational

11
Models DSS

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

12
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

13
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

14
Dryer Washer Production Plant





















Profit
200 100
15
What would you want to know?
- How many washers to produce - How many dryers
to produce in order to .. .. MAXIMIZE
PROFIT!
16
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
17
continued

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

18

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
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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
20
Modeling Assumptions

• Proportionality
• Additivity
• Divisibility

21
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
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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

23
What next?

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

24
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
25
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
26
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
27
Non-negativity constraints
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
28
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
29
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
30
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
31
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
32
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
33
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
34
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
35
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
36
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
37
1200
40
Tons of Solvent Base
720
30
240
20
10
0
10
20
30
40
50
Tons of Fuel Additive
38
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
39
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
40
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
41
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
42
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
43
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
44
40
(40F 30S 1600)
Tons of Solvent Base
OPTIMAL SOLUTION!
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
45
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
46
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
47
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
48
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
49
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

50
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
51
OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
52
OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
53
OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
54
OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
55
OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
56
OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
57
OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
58
OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
59
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
60
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
61
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
62
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
63
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?
64
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
65
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
66
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
67
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
68
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
69
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
70
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
71
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
72
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
73
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
74
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
75
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
76
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)