Title: Management Science Course 20032004 Steef van de Velde professor of Operations Management
1Management Science Course2003-2004Steef van
de Veldeprofessor of Operations Management
TechnologyRSM
2Lecturers
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
Raf Jans rjans_at_fbk.eur.nl office F2-53 tel.
4082774
3Management Science
- All about better managerial decision making using
quantitative techniques - Science of Better
- aka Operations Research, Decision Science,
Systems Engineering
4Teaching 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
5Teaching Objectives
- To convince you that Management Science can be a
very powerful and practical tool - for doing things better (optimization)
- for doing better things (scenario analysis)
6From last years EMBA essays
- The awareness about the existence and power of
OR at higher management levels is actually quite
poor, which I find amazing in a complex
environment like ours, begging for optimization. - A key issue is the acceptance of OR in our
company. The reason proved to be that they do not
trust the models and outcomes. Certainly, the
fear of being made redundant by a computer (not
unrealistic in some departments like production
scheduling) does not help the acceptance of OR.
7Teaching Objectives
- To convince you that Management Science can be a
very powerful and practical tool - for doing things better (optimization)
- for doing better things (scenario analysis)
- At the end of the course you are (better) able
- to identify opportunities for better decision
making through Management Science - to solve (some) MS problems on your own
- to manage Management Science/Operations Research
projects and consultants
8COURSE 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
9COURSE 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
10REQUIREMENTS
- 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)
11Course 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
12Todays Program
- Introduction to the world of Management Science
- Operations Research
- Introduction to LINEAR PROGRAMMING (LP)
- Introduction to Excel Solver (to solve linear
programming problems)
13A 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.
14A Flavor of Techniques
- MATHEMATICAL PROGRAMMING
- linear programming
- integer linear programming
- quadratic programming
- dynamic programming
- COMBINATORIAL OPTIMIZATION
- QUEUEING THEORY
- DECISION ANALYSIS
- INVENTORY THEORY
- MARKOV THEORY
- NEURAL NETWORKS
- DEA
15Linear 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
16An Introductory Example
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
17What do you want to know?
- How many tons of
- Fuel Additive
- Solvent Base
- to produce in order to
- maximize profit
18(No Transcript)
19Formulas
- LHS Left Hand Side
- RHS Right Hand Side
- Profit (I6) E6E5 F6F5
- LHS material 1 (H10) E10E5 F10F5
- LHS material 2 (H11) F11F5
- LHS material 3 (H12) E12E5 F12F5
20Verbal formulation as an Optimization Problem
- DETERMINE THE NUMBER OF TONS OF FUEL
- ADDITIVE AND SOLVENT BASE TO PRODUCE
- SO AS TO
- MAXIMIZE PROFIT
- SUBJECT TO
- MATERIAL AVAILABILITY CONSTRAINTS
21.. continued ...
- SPECIFY THE DECISION VARIABLES
- DESCRIBE THE CONSTRAINTS
- (in terms of the decision variables)
- DESCRIBE THE OBJECTIVE FUNCTION
- (in terms of the decision variables)
22LP Modeling
- Decision variables
- Objective function
- Constraints
F the number of tons of Fuel Additive to be
produced S the number of tons of Solvent Base
to be produced
Maximize ???????
23LP Modeling
- Decision variables
- Objective function
- Constraints
F the number of tons of Fuel Additive to be
produced S the number of tons of Solvent Base
to be produced
Maximize 40 F 30 S
- Material availability constraints
- Non-negative constraints
24MAX 40 F 30 S
Subject to
(1) material availability constraints
Material 1 Material 2 Material 3
????????
????????
????????
(2) non-negativity constraints
F gt 0 S gt 0
25MAX 40 F 30 S
Subject to
(1) material availability constraints
Material 1 Material 2 Material 3
0.4 F 0.5 S
lt 20
0.2S
lt 5
0.6 F 0.3 S
lt 21
(2) non-negativity constraints
F gt 0 S gt 0
26Linear Programming Model
MAX 40 F 30 S Subject to
0.4 F 0.5 S
lt 20
0.2 S
lt 5
0.6 F 0.3 S
lt 21
F gt 0
S gt 0
27An investment example
- Kathy Allen, an individual investor, has
70,000 to divide among several investments - municipal bonds with 8.5 annual return
- certificates of deposit with a 5 return
- treasury bills with a 6.5 return
- growth stock fund with a 13 annual return
- Kathy wants to know how much to invest in each
alternative to - maximize returns and lessening the risk perceived
by the investor -
- No more than 20 in municipal bonds
- The amount invested in certificates of deposit
should not exceed - the amount invested in the other three
alternatives - At least 30 should be in treasury bills and
certificates of deposit
28Decision variables
- MB in municipal bonds
- CD in certificates of deposit
- TB in treasury bills
- SF in growth stock fund
29Constraints
- MB TB CD SF 70,000
- MB lt 14,000
- CD lt MB TB SF
- TB CD gt 21,000
- MB, CD, TB, SF gt 0
30MAX ?????
- MB TB CD SF 70,000
- MB lt 14,000
- CD lt MB TB SF
- TB CD gt 21,000
- MB, CD, TB, SF gt 0
31MAX 8.5MB 5CD 6.5TB 13SF
- MB TB CD SF 70,000
- MB lt 14,000
- CD lt MB TB SF
- TB CD gt 21,000
- MB, CD, TB, SF gt 0
32Solving the Fuel Additive and Solvent Base
problem using Excel Solver
33Modeling Assumptions
- Proportionality
- Additivity
- Divisibility
34So, 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 THE INTERNET
35SO, MODELING INVOLVES ...
Model of Reality
OUTPUTDATA
INPUTDATA
- What if your model oversimplifies reality?
- What if your input data are unreliable?
36SO, MODELING INVOLVES ...
Model of Reality
CRAP OUT!
CRAP IN
OUTPUTDATA
INPUTDATA
37SO, MODELING INVOLVES ...
MODEL OFREALITY
OUTPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
INPUTDATA
38EXACTLY BECAUSE GETTING THE INPUT DATA IS SO
DIFFICULT ...
MODEL OFREALITY
OUTPUTDATA
INPUTDATA
INPUTDATA
CONFUSION ABOUT WHAT IS INPUT AND WHAT IS OUTPUT
39EXAMPLE A LOCATION-ALLOCATION MODEL FOR A
GLOBAL-PLAYER IN INDUSTRIAL COATINGS
- thousands of products
- a dozen of plants
- hundreds of customers
40So, modeling also involves making a choice
between
- PRECISION and RELEVANCE
- RELEVANCE and COMPLEXITY
- PRECISION and ROBUSTNESS
41How 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
42What next?
- The Graphical Solution Procedure
- Sensitivity Analysis
- Solving LP using EXCEL
- ..
43Formulation 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
44Non-negativity constraints
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
4540
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
46Material 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
47FEASIBLE REGION FOR THE MATERIAL 1 CONSTRAINT
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
48Material 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
49FEASIBLE REGION FOR THE MATERIAL 2 CONSTRAINT
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
50MATERIAL 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
51FEASIBLE 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
5240
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
53240 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
541200
40
Tons of Solvent Base
720
30
240
20
10
0
10
20
30
40
50
Tons of Fuel Additive
5540
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
5640
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
5740
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
5840
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
5940
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
6040
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
6140
(40F 30S 1600)
Tons of Solvent Base
OPTIMAL SOLUTION!
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
6240
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
63HOW 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
64Calculating 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
65Summary 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
66Sensitivity 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
67OBJECTIVE 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
68OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
69OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
70OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
71OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
72OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
73OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
74OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
75OBJECTIVE FUNCTION LINE
40
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
76MATERIAL 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
77EXTREME 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
78The 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
79What 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
80How 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?
81OBJECTIVE 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
82OBJECTIVE 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
83OBJECTIVE FUNCTION LINE
40
NEW Material 3 line
Tons of Solvent Base
30
20
10
0
10
20
30
40
50
Tons of Fuel Additive
84OBJECTIVE 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
85OBJECTIVE 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
86OBJECTIVE 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
87OBJECTIVE 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
88OBJECTIVE 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
89OBJECTIVE 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
90SHADOW 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
91OBJECTIVE 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
92OBJECTIVE 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
93Solving the Fuel Additive and Solvent Base
problem using Excel Solver
94How to best try and master linear programming
- Study the slides well
- Master the use of Excel Solver
- Use Beasleys internet course to practice,
practice, practice modeling, in particular if you
are a poet . - Browse the textbook chapters for good examples
95OUTLOOK
- 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)